Authors Konstantinos N. Anagnostopoulos
License CC-BY-SA-4.0
Computational Physics
Konstantinos N. Anagnostopoulos
COMPUTATIONAL PHYSICS
A Practical Introduction to Computational Physics
and Scientific Computing
Athens, 2014
KONSTANTINOS N. ANAGNOSTOPOULOS
National Technical University of Athens
National Technical University of Athens
COMPUTATIONAL PHYSICS
A Practical Introduction to Computational Physics and Scientific Computing
AUTHORED BY KONSTANTINOS N. ANAGNOSTOPOULOS
Physics Department, National Technical University of Athens, Zografou Campus, 15780 Zografou, Greece
konstant@mail.ntua.gr, www.physics.ntua.gr/˜konstant/
PUBLISHED BY KONSTANTINOS N. ANAGNOSTOPOULOS
and the
NATIONAL TECHNICAL UNIVERSITY OF ATHENS
Book Website:
www.physics.ntua.gr/˜konstant/ComputationalPhysics
©Konstantinos N. Anagnostopoulos 2014, 2016
First Published 2014
Version¹ 1.1.20161207095000
Cover: Design by K.N. Anagnostopoulos. The front cover picture is a snapshot taken during Monte Carlo sim-
ulations of hexatic membranes. Work done with Mark J. Bowick. Relevant video at youtu.be/Erc7Q6YXfLk
⃝
CC This book and its cover(s) are subject to copyright. They are licensed under the Creative Commons
Attribution-ShareAlike 4.0 International License. To view a copy of this license, visit
creativecommons.org/licenses/by-sa/4.0/
The book is accompanied by software available at the book’s website. All the software, unless the copyright
does not belong to the author, is open source, covered by the GNU public license, see www.gnu.org/licenses/.
This is explicitly mentioned at the end of the respective source files.
ISBN 978-1-312-46441-4 (lulu.com, vol. I)
ISBN 978-1-312-46488-9 (lulu.com, vol. II)
¹The first number is the major version, corresponding to an “edition” of a conventional book. Versions
differing by major numbers have been altered substantially. Chapter numbers and page references are not
guaranteed to match between different versions. The second number is the minor version. Versions differing
by a minor version may have serious errors/typos corrected and/or substantial text modifications. Versions
differing by only the last number may have minor typos corrected, added references etc. When reporting
errors, please mention the version number you are referring to.
Contents
Foreword vii
1 The Computer 1
1.1 The Operating System . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Filesystem . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.2 Commands . . . . . . . . . . . . . . . . . . . . . . . 11
1.1.3 Looking for Help . . . . . . . . . . . . . . . . . . . . 14
1.2 Text Processing Tools – Filters . . . . . . . . . . . . . . . . 16
1.3 Programming with Emacs . . . . . . . . . . . . . . . . . . . 21
1.3.1 Calling Emacs . . . . . . . . . . . . . . . . . . . . . . 21
1.3.2 Interacting with Emacs . . . . . . . . . . . . . . . . 23
1.3.3 Basic Editing . . . . . . . . . . . . . . . . . . . . . . 24
1.3.4 Cut and Paste . . . . . . . . . . . . . . . . . . . . . . 27
1.3.5 Windows . . . . . . . . . . . . . . . . . . . . . . . . 29
1.3.6 Files and Buffers . . . . . . . . . . . . . . . . . . . . 30
1.3.7 Modes . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.3.8 Emacs Help . . . . . . . . . . . . . . . . . . . . . . . 32
1.3.9 Emacs Customization . . . . . . . . . . . . . . . . . 34
1.4 The Fortran Programming Language . . . . . . . . . . . . 35
1.4.1 The Foundation . . . . . . . . . . . . . . . . . . . . 35
1.4.2 Details . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.4.3 Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . 50
1.5 Gnuplot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
1.6 Shell Scripting . . . . . . . . . . . . . . . . . . . . . . . . . 62
2 Kinematics 77
2.1 Motion on the Plane . . . . . . . . . . . . . . . . . . . . . . 77
2.1.1 Plotting Data . . . . . . . . . . . . . . . . . . . . . . 85
2.1.2 More Examples . . . . . . . . . . . . . . . . . . . . . 88
2.2 Motion in Space . . . . . . . . . . . . . . . . . . . . . . . . 99
2.3 Trapped in a Box . . . . . . . . . . . . . . . . . . . . . . . . 107
iii
iv CONTENTS
2.3.1 The One Dimensional Box . . . . . . . . . . . . . . 108
2.3.2 Errors . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.3.3 The Two Dimensional Box . . . . . . . . . . . . . . 117
2.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
3 Logistic Map 145
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
3.2 Fixed Points and 2n Cycles . . . . . . . . . . . . . . . . . . 147
3.3 Bifurcation Diagrams . . . . . . . . . . . . . . . . . . . . . 154
3.4 The Newton-Raphson Method . . . . . . . . . . . . . . . . 157
3.5 Calculation of the Bifurcation Points . . . . . . . . . . . . . 163
3.6 Liapunov Exponents . . . . . . . . . . . . . . . . . . . . . . 167
3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
4 Motion of a Particle 191
4.1 Numerical Integration of Newton’s Equations . . . . . . . . 191
4.2 Prelude: Euler Methods . . . . . . . . . . . . . . . . . . . . 192
4.3 Runge–Kutta Methods . . . . . . . . . . . . . . . . . . . . . 204
4.3.1 A Program for the 4th Order Runge–Kutta . . . . . 208
4.4 Comparison of the Methods . . . . . . . . . . . . . . . . . . 212
4.5 The Forced Damped Oscillator . . . . . . . . . . . . . . . . 215
4.6 The Forced Damped Pendulum . . . . . . . . . . . . . . . . 222
4.7 Appendix: On the Euler–Verlet Method . . . . . . . . . . . 229
4.8 Appendix: 2nd order Runge–Kutta Method . . . . . . . . 233
4.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
5 Planar Motion 241
5.1 Runge–Kutta for Planar Motion . . . . . . . . . . . . . . . 241
5.2 Projectile Motion . . . . . . . . . . . . . . . . . . . . . . . . 246
5.3 Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . 252
5.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
5.4.1 Rutherford Scattering . . . . . . . . . . . . . . . . . 260
5.4.2 More Scattering Potentials . . . . . . . . . . . . . . . 267
5.5 More Particles . . . . . . . . . . . . . . . . . . . . . . . . . . 270
5.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
6 Motion in Space 285
6.1 Adaptive Stepsize Control for Runge–Kutta Methods . . . . 286
6.2 Motion of a Particle in an EM Field . . . . . . . . . . . . . 295
6.3 Relativistic Motion . . . . . . . . . . . . . . . . . . . . . . . 296
CONTENTS v
6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
7 Electrostatics 311
7.1 Electrostatic Field of Point Charges . . . . . . . . . . . . . . 311
7.2 The Program – Appetizer and ... Desert . . . . . . . . . . . 314
7.3 The Program – Main Dish . . . . . . . . . . . . . . . . . . . 323
7.4 The Program - Conclusion . . . . . . . . . . . . . . . . . . . 329
7.5 Electrostatic Field in the Vacuum . . . . . . . . . . . . . . . 334
7.6 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
7.7 Poisson Equation . . . . . . . . . . . . . . . . . . . . . . . . 342
7.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
8 Diffusion Equation 353
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
8.2 Heat Conduction in a Thin Rod . . . . . . . . . . . . . . . . 355
8.3 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . 356
8.4 The Program . . . . . . . . . . . . . . . . . . . . . . . . . . 358
8.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
8.6 Diffusion on the Circle . . . . . . . . . . . . . . . . . . . . . 363
8.7 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
8.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
9 The Anharmonic Oscillator 373
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
9.2 Calculation of the Eigenvalues of Hnm (λ) . . . . . . . . . . 375
9.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
9.4 The Double Well Potential . . . . . . . . . . . . . . . . . . . 390
9.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
10 Time Independent Schrödinger Equation 401
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
10.2 The Infinite Potential Well . . . . . . . . . . . . . . . . . . . 404
10.3 Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . 415
10.4 Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . 425
10.5 The Anharmonic Oscillator - Again... . . . . . . . . . . . . 431
10.6 The Lennard–Jones Potential . . . . . . . . . . . . . . . . . 435
10.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438
11 The Random Walker 443
11.1 (Pseudo)Random Numbers . . . . . . . . . . . . . . . . . . 444
11.2 Using Pseudorandom Number Generators . . . . . . . . . . 456
vi CONTENTS
11.3 Random Walks . . . . . . . . . . . . . . . . . . . . . . . . . 463
11.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
12 Monte Carlo Simulations 475
12.1 Statistical Physics . . . . . . . . . . . . . . . . . . . . . . . . 476
12.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479
12.3 Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
12.4 Correlation Functions . . . . . . . . . . . . . . . . . . . . . 485
12.5 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
12.5.1 Simple Sampling . . . . . . . . . . . . . . . . . . . . 488
12.5.2 Importance Sampling . . . . . . . . . . . . . . . . . 489
12.6 Markov Processes . . . . . . . . . . . . . . . . . . . . . . . . 490
12.7 Detailed Balance Condition . . . . . . . . . . . . . . . . . . 491
12.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
13 Simulation of the d = 2 Ising Model 495
13.1 The Ising Model . . . . . . . . . . . . . . . . . . . . . . . . 495
13.2 Metropolis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
13.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 504
13.3.1 The Program . . . . . . . . . . . . . . . . . . . . . . 509
13.3.2 Towards a Convenient User Interface . . . . . . . . . 515
13.4 Thermalization . . . . . . . . . . . . . . . . . . . . . . . . . 527
13.5 Autocorrelations . . . . . . . . . . . . . . . . . . . . . . . . 529
13.6 Statistical Errors . . . . . . . . . . . . . . . . . . . . . . . . 535
13.6.1 Errors of Independent Measurements . . . . . . . . 537
13.6.2 Jackknife . . . . . . . . . . . . . . . . . . . . . . . . 539
13.6.3 Bootstrap . . . . . . . . . . . . . . . . . . . . . . . . 542
13.7 Appendix: Autocorrelation Function . . . . . . . . . . . . . 543
13.8 Appendix: Error Analysis . . . . . . . . . . . . . . . . . . . 550
13.8.1 The Jackknife Method . . . . . . . . . . . . . . . . . 550
13.8.2 The Bootstrap Method . . . . . . . . . . . . . . . . . 555
13.8.3 Comparing the Methods . . . . . . . . . . . . . . . . 558
13.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
14 Critical Exponents 571
14.1 Critical Slowing Down . . . . . . . . . . . . . . . . . . . . . 573
14.2 Wolff Cluster Algorithm . . . . . . . . . . . . . . . . . . . . 574
14.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 581
14.3.1 The Program . . . . . . . . . . . . . . . . . . . . . . 583
14.4 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . 588
14.5 Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 591
CONTENTS vii
14.6 Autocorrelation Times . . . . . . . . . . . . . . . . . . . . . 599
14.7 Temperature Scaling . . . . . . . . . . . . . . . . . . . . . . 604
14.8 Finite Size Scaling . . . . . . . . . . . . . . . . . . . . . . . 608
14.9 Calculation of βc . . . . . . . . . . . . . . . . . . . . . . . . 611
14.10Studying Scaling with Collapse . . . . . . . . . . . . . . . . 615
14.11Binder Cumulant . . . . . . . . . . . . . . . . . . . . . . . . 625
14.12Appendix: Scaling . . . . . . . . . . . . . . . . . . . . . . . 629
14.12.1Binder Cumulant . . . . . . . . . . . . . . . . . . . . 629
14.12.2Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . 634
14.12.3Finite Size Scaling . . . . . . . . . . . . . . . . . . . 636
14.13Appendix: Critical Exponents . . . . . . . . . . . . . . . . . 639
14.13.1Definitions . . . . . . . . . . . . . . . . . . . . . . . . 639
14.13.2Hyperscaling Relations . . . . . . . . . . . . . . . . . 640
14.14Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 640
Bibliography 644
Index 652
viii CONTENTS
This book has been written assuming that the reader executes
all the commands presented in the text and follows all the
instructions at the same time. If this advice is neglected, then
the book will be of little help and some parts of the text may
seem incomprehensible.
The book’s website is at
http://www.physics.ntua.gr/˜konstant/ComputationalPhysics/
From there, you can can download the accompanying software, which con-
tains, among other things, all the programs presented in the book.
Some conventions: Text using the font shown below refers to com-
mands given using a shell (the “command line”), input and output of
programs, code written in Fortran (or any other programming language),
as well as to names of files and programs:
> echo Hello world
Hello world
When a line starts with the prompt
>
then the text that follows is a command, which can be given from the
command line of a terminal. The second line, Hello World, is the output
of the command.
The contents of a file with Fortran code is listed below:
program add
z = 1.0
y = 2.0
x = z + y
print * , x
end program add
What you need in order to work on your PC:
CONTENTS ix
• An operating system of the GNU/Linux family and its basic tools.
• A Fortran compiler. The gfortran compiler is freely available
for all major operating systems under an open source license at
http://www.gfortran.org.
• An advanced text editor, suitable for editing code in several pro-
gramming languages, like Emacs².
• A good plotting program, suitable for data analysis, like gnuplot³.
• The shell tcsh⁴.
• The programs awk⁵, grep, sort, cat, head, tail, less. Make sure
that they are available in your computer environment.
If you have installed a GNU/Linux distribution on your computer,
all of the above can be installed easily. For example, in a Debian like
distribution (Ubuntu, ...) the commands
> sudo apt−get install tcsh emacs gnuplot−x11 gnuplot−doc
> sudo apt−get install gfortran gawk gawk−doc binutils
> sudo apt−get install manpages−dev coreutils liblapack3
install all the necessary tools.
If you don’t wish to install GNU/Linux on your computer, you can
try the following:
• Boot your computer using a usb/DVD live GNU/Linux, like Ubuntu⁶.
This will not make any permanent changes in your hard drive but
it will start and run slower. On the other hand, you may save all
your computing environment and documents and use it on any
computer you like.
• Install Cygwin⁷ in your Microsoft Windows. It is a very good solu-
tion for Microsoft-addicted users. If you choose the full installation,
then you will find all the tools needed in this book.
²http://www.gnu.org/software/emacs/
³http://www.gnuplot.info
⁴http://www.tcsh.org
⁵http://www.gnu.org/software/gawk
⁶http://www.ubuntu.com
⁷http://www.cygwin.com
x CONTENTS
• Mac OS X is based on Unix. It is possible to install all the software
needed in this book and follow the material as presented. Search
the internet for instructions, e.g. google “gfortran for Mac”, “emacs
for Mac”, “tcsh for Mac”, etc.
Foreword
This book is the culmination of my ten years’ experience in teaching
three introductory, undergraduate level, scientific computing/computational
physics classes at the National Technical University of Athens. It is suit-
able mostly for junior or senior level science courses, but I am currently
teaching its first chapters to sophomores without a problem. A two
semester course can easily cover all the material in the book, including
lab sessions for practicing.
Why another book in computational physics? Well, when I started
teaching those classes there was no bibliography available in Greek, so I
was compelled to write lecture notes for my students. Soon, I realized that
my students, majoring in physics or applied mathematics, were having
a hard time with the technical details of programming and computing,
rather than with the physics concepts. I had to take them slowly by the
hand through the “howto” of computing, something that is reflected in
the philosophy of this book. Hoping that this could be useful to a wider
audience, I decided to translate these notes in English and put them in
an order and structure that would turn them into “a book”.
I also decided to make the book freely available on the web. I was
partly motivated by my anger caused by the increase of academic (e)book
prices to ridiculous levels during times of plummeting publishing costs.
Publishers play a diminishing role in academic publishing. They get an
almost ready-made manuscript in electronic form by the author. They
need to take no serious investment risk on an edition, thanks to print-
on-demand capabilities. They have virtually zero cost ebook publishing.
Moreover, online bookstores have decreased costs quite a lot. Academic
books need no advertisement budget, their success is due to their aca-
demic reputation. I don’t see all of these reflected on reduced book
prices, quite the contrary, I’m afraid.
My main motivation, however, is the freedom that independent pub-
lishing would give me in improving, expanding and changing the book
in the future. It is great to have no length restrictions for the presenta-
xi
xii FOREWORD
tion of the material, as well as not having to report to a publisher. The
reader/instructor that finds the book long, can read/print the portion of
the book that she finds useful for her.
This is not a reference book. It uses some interesting, I hope, physics
problems in order to introduce the student to the fundamentals of solv-
ing a scientific problem numerically. At the same time, it keeps an eye
in the direction of advanced and high performance scientific computing.
The reader should follow the instructions given in each chapter, since
the book teaches by example. Several skills are taught through the solution
of a particular problem. My lectures take place in a (large) computer
lab, where the students are simultaneously doing what I am doing (and
more). The program that I am editing and the commands that I am
executing are shown on a large screen, displaying my computer monitor
and actions live. The book provides no systematic teaching of a program-
ming language or a particular tool. A very basic introduction is given in
the first chapter and then the reader learns whatever is necessary for the
solution of her problem. There is more than one way to do it⁸ and the
problems can be solved by following a basic or a fancy way, depending
on the student’s computational literacy. The book provides the necessary
tools for both. A bibliography is provided at the end of the book, so that
the missing pieces of a puzzle can be sought in the literature.
This is also not a computational physics playground. Of course I
hope that the reader will have fun doing what is in the book, but my
goal is to provide an experience that will set the solid foundation for
her becoming a high performance computing, number crunching, heavy
duty data analysis expert in the future. This is why the programming
language of the core numerical algorithms has been chosen to be Fortran,
a highly optimized, scientifically oriented, programming language. The
computer environment is set in a Unix family operating system, enriched
by all the powerful GNU tools provided by the FSF⁹. These tools are
indispensable in the complicated data manipulation needed in scientific
research, which requires flexibility and imagination. Of course, Fortran
is not the best choice for heavy duty object oriented programming, and is
not optimal for interacting with the operating system. The philosophy¹⁰
⁸A Perl moto!
⁹Free Software Foundation, www.fsf.org.
¹⁰Java and C++ have been popular choices in computational physics courses. But
object oriented programming is usually avoided in the high performance part of a com-
putation. So, one usually uses those languages in a procedural style of programming,
cheating herself that she is actually learning the advantages of object oriented program-
ming.
xiii
is to let Fortran do what is best for, number crunching, and leave data
manipulation and file administration to external, powerful tools. Tools,
like awk, shell scripting, gnuplot, Perl and others, are quite powerful
and complement all the weaknesses of Fortran mentioned before. The
plotting program is chosen to be gnuplot, which provides very powerful
tools to manipulate the data and create massive and complicated plots. It
can also create publication quality plots and contribute to the “fun part”
of the learning experience by creating animations, interactive 3d plots
etc. All the tools used in the book are open source software and they are
accessible to everyone for free. They can be used in a Linux environment,
but they can also be installed and used in Microsoft Windows and Mac
OS X.
The other hard part in teaching computational physics to scientists
and engineers is to explain that the approach of solving a problem nu-
merically is quite different from solving it analytically. Usually, students
of this level are coming with a background in analysis and fundamental
physics. It is hard to put them into the mode of thinking about solving
a problem using only additions, multiplications and some logical opera-
tions. The hardest part is to explain the discretization of a model defined
analytically, which can be done in many ways, depending on the accu-
racy of the approximation. Then, one has to extrapolate the numerical
solution, in order to obtain a good approximation of the analytic one.
This is done step by step in the book, starting with problems in simple
motion and ending with discussing finite size scaling in statistical physics
models in the vicinity of a continuous phase transition.
The book comes together with additional material which can be found
at the web page of the book¹¹. The accompanying software contains all the
computer programs presented in the book, together with useful tools and
programs solving some of the exercises of each chapter. Each chapter has
problems complementing the material covered in the text. The student
needs to solve them in order to obtain hands on experience in scientific
computing. I hope that I have already stressed enough that, in order for
this book to be useful, it is not enough to be read in a café or in a living
room, but one needs to do what it says.
Hoping that this book will be useful to you as a student or as an
instructor, I would like to ask you to take some time to send me feedback
for improving and/or correcting it. I would also appreciate fan mail or,
if you are an expert, a review of the book. If you use the book in a
class, as a main textbook or as supplementary material, I would also be
¹¹www.physics.ntua.gr/˜konstant/ComputationalPhysics/
xiv FOREWORD
thrilled to know about it. Send me email at konstantmail.ntua.gr and
let me know if I can publish, anonymously or not, (part of) what you say
on the web page (otherwise I will only use it privately for my personal
ego-boost). Well, nothing is given for free: As one of my friends says,
some people are payed in dollars and some others in ego-dollars!
Have fun computing scientifically!
Athens, 2014.
Chapter 1
The Computer
The aim of this chapter is to lay the grounds for the development of
the computational skills which are necessary in the following chapters.
It is not an in depth exposition but a practical training by example.
For a more systematic study of the topics discussed, we refer to the
bibliography. Many of the references are freely available in the web.
The are many choices that one has to make when designing a com-
puter project. These depend on the needs for numerical efficiency, on
available programming hours, on the needs for extensibility and upgrad-
ability and so on. In this book we will get the flavor of a project that is
mostly scientifically and number crunching oriented. One has to make
the best of the available computing resources and have powerful tools
available for a productive analysis of the data. Such an environment,
found in most of today’s supercomputers, that offers flexibility, depend-
ability, simplicity, powerful tools for data analysis and effective compilers
is provided by the family of the Unix operating systems. The GNU/Linux
operating system is a Unix variant that is freely available and most of its
utilities are open source software. The voluntary work of millions of
excellent programmers worldwide has built the most stable, fastest and
highest quality software available for scientific computing today. Thanks
to the idea of the open source software pioneered by Richard Stallman¹
this giant collaboration has been made possible.
Another choice that we have to make is the programming language,
and this is going to be Fortran. Fortran has been built mainly for numer-
ical applications and it has been used by many scientists and engineers
because of its efficiency in high performance computing. The language
is simple and compilers are able to optimize, parallelize and vectorize the
¹www.stallman.org
1
2 CHAPTER 1. THE COMPUTER
code very efficiently. There is a lot of scientific and engineering software
available in libraries written in Fortran, which has been used and tested
extensively for many years. This is a crucial factor for scientific software,
so that it can be trusted to be efficient and free of errors. Fortran is
not the best choice for interacting with the operating system or for text
processing. This shortcoming can be easily overcome by the use of ex-
ternal tools and Fortran can be left to do what she has been designed
for: number crunching. Its structure is simple and can be used both
for procedural and object oriented programming, in such a way that, it
will not make the life of an inexperienced programmer difficult, and at
the same time provide high level, abstract and powerful tools for high
performance, modular, object oriented, programming needed in a large
and complicated project.
Fortran, as well as other languages like C, C++ and Java, is a language
that needs to be compiled by a compiler. Other languages, like perl, awk,
shell scripting, Macsyma, Mathematica, Octave, Matlab, . . ., are interpreted
line by line. These languages can be simple in their use, but they can be
prohibitively slow when it comes to a numerically demanding program.
A compiler is a tool that analyzes the whole program and optimizes the
computer instructions executed by the computer. But if programming
time is more valuable, then a simple, interpreted language can lead to
faster results.
Another choice that we make in this book, and we mention it because
it is not the default in most Linux distributions, is the choice of shell.
The shell is a program that “connects” the user to the operating system.
In this book, we will teach how to use a shell² to “send” commands to the
operating system, which is the most effective way to perform complicated
tasks. We will use the shell tcsh, although most of the commands can be
interpreted by most popular shells. Shell scripting is simpler in this shell,
although shells like bash provide more powerful tools, mostly needed
for complicated system administration tasks. That may cause a small
inconvenience to some readers, since tcsh is not preinstalled in Linux
distributions³.
²It is more popular to be called “the command line”, or the “terminal”, or the
“console”, but in fact the user interaction is through a shell.
³See www.tcsh.org. On Debian like systems, like Ubuntu, installation is very simple
through the software center or by the command sudo apt-get install tcsh.
1.1. THE OPERATING SYSTEM 3
1.1 The Operating System
The Unix family of operating systems offer an environment where com-
plicated tasks can be accomplished by combining many different tools,
each of which performs a distinct task. This way, one can use the power
of each tool, so that trivial but complicated parts of a calculation don’t
have to be programmed. This makes the life of a researcher much easier
and much more productive, since research requires from us to try many
things before we understand how to compute what we are looking for.
In the Unix operating system everything is a file, and files are or-
ganized in a unique and unified filesystem. Documents, pictures, music,
movies, executable programs are files. But also directories or devices,
like hard disks, monitors, mice, sound cards etc, are, from the point of
view of the operating system, files. In order for a music file to be played
by your computer, the music data needs to be written to a device file,
connected by the operating system to the sound card. The characters
you type in a terminal are read from a file “the keyboard”, and written
to a file “the monitor” in order to be displayed. Therefore, the first thing
that we need to understand is the structure of the Unix filesystem.
1.1.1 Filesystem
There is at least one path in the filesystem associated with each file. There
are two types of paths, relative paths and absolute paths. These are two
examples:
bin / RungeKutta / rk . exe
/ home / george / bin / RungeKutta / rk . exe
The paths shown above may refer to the same or a different file. This
depends on “where we are”. If “we are” in the directory /home/george,
then both paths refer to the same file. If on the other way “we are” in
a directory /home/john or /home/george/CompPhys, then the paths refer⁴
to two different files. In the last two cases, the paths refer to the files
/ home / john / bin / RungeKutta / rk . exe
/ home / george / CompPhys / bin / RungeKutta / rk . exe
⁴Some times two or more paths refer to the same file, or as we say, a file has two or
more “links” in the same filesystem, but let’s keep it simple for the moment.
4 CHAPTER 1. THE COMPUTER
respectively. How can we tell the difference? An absolute path always
begins with the / character, whereas a relative path does not. When we
say that “we are in a directory”, we refer to a position in the filesystem
called the current directory, or working directory. Every process in the
operating system has a unique current directory associated with it.
Figure 1.1: The Unix filesystem. It looks like a tree, with the root directory / at the
top and branches that connect directories with their parents. Every directory contains
files, among them other directories called its subdirectories. Every directory has a unique
parent directory, noted by .. (double dots). The parent of the root directory is itself.
The filesystem is built on its root and looks like a tree positioned
upside down. The symbol of the root is the character / The root is
a directory. Every directory is a file that contains a list of files, and it
is connected to a unique directory, its parent directory . Its list of files
contains other directories, called its subdirectories, which all have it as
their parent directory. All these files are the contents of the directory.
Therefore, the filesystem is a tree of directories with the root directory
at its top which branch to its subdirectories, which in their turn branch
into other subdirectories and so on. There is practically no limit to how
1.1. THE OPERATING SYSTEM 5
large this tree can become, even for a quite demanding environment⁵.
A path consists of a string of characters, with the characters / sep-
arating its components, and refers to a unique location in the filesystem.
Every component refers to a file. All, but the last one, must be directories
in a hierarchy, from parent directory to subdirectory. The only exception
is a possible / in the beginning, which refers to the root directory. Such
an example can be seen in figure 1.1.
In a Unix filesystem there is complete freedom in the choice of the loca-
tion of the files⁶. Fortunately, there are some universally accepted conven-
tions respected by almost everyone. One expects to find home directories
in the directory /home, configuration files in the directory /etc, appli-
cation executables in directories with names such as /bin, /usr/bin,
/usr/local/bin, software libraries in directories with names such as
/lib, /usr/lib etc.
There are some important conventions in the naming of the paths. A
single dot “.” refers to the current directory and a double dot “..” to the
parent directory. Similarly, a tilde “~” refers to the home directory of the
user. Assume, e.g., that we are the user george running a process with
a current directory /home/george/Music/Rock (see figure 1.1). Then, the
following paths refer to the same file /home/george/Doc/lyrics.doc:
. . / . . / Doc / lyrics . doc
~/ Doc / lyrics . doc
~george / Doc / lyrics . doc
. / . . / . . / Doc / lyrics . doc
Notice that ~ and ~george refer to the home directory of the user george
(ourselves), whereas ~mary refer to the home directory of another user,
mary.
We are now going to introduce the basic commands for filesystem
navigation and manipulation⁷. The command cd (=change directory)
⁵Of course, the capacity of the filesystem is finite, issue the command “df -i .” in
order to see the number of inodes available in your filesystem. Every file corresponds
to one and only one inode of the filesystem. Every path is mapped to a unique inode,
but an inode maybe pointed to by more than one paths.
⁶This gives a great sense of freedom, but historically this was a important factor that
led the Unix operating systems, although superior in quality, not to win a fair share
of the market! The Linux family tries to keep things simple and universal to a large
extent, but one should be aware that because of this freedom files in different version
of Linuxes or Unices can be in different places.
⁷Remember that lines that begin with the > character are commands. All other lines
refer to the output of the commands.
6 CHAPTER 1. THE COMPUTER
changes the current directory, whereas the command pwd (=print working
directory) prints the current directory:
> cd / usr / bin
> pwd
/ usr / bin
> cd / usr / local / lib
> pwd
/ usr / local / lib
> cd
> pwd
/ home / george
> cd −
> pwd
/ usr / local / lib
> cd . . / . . /
> pwd
/ usr
The argument of the command cd is an absolute or a relative path. If
the path is correct and we have the necessary permissions, the command
changes the current directory to this path. If no path is given, then
the current directory changes to the home directory of the user. If the
character - is given instead of a path, then the command changes the
current directory to the previous current directory.
The command mkdir creates new directories, whereas the command
rmdir removes empty directories. Try:
> mkdir new
> mkdir new / 0 1
> mkdir new / 0 1 / 0 2 / 0 3
mkdir : cannot create directory ‘ new / 0 1 / 0 2 / 0 3 ’ : No such file or
directory
> mkdir −p new / 0 1 / 0 2 / 0 3
> rmdir new
rmdir : ‘ new ’ : Directory not empty
> rmdir new / 0 1 / 0 2 / 0 3
> rmdir new / 0 1 / 0 2
> rmdir new / 0 1
> rmdir new
Note that the command mkdir cannot create directories more than one
level down the filesystem, whereas the command mkdir -p can. The
“switch” -p makes the behavior of the command different than the default
one.
1.1. THE OPERATING SYSTEM 7
In order to list the contents of a directory, we use the command ls
(=list):
> ls
BE . eps Byz . eps Programs srBE_xyz . eps srB_xyz . eps
B . eps Bzy . eps srBd_xyz . eps srB_xy . eps
> l s Programs
Backup rk3_Byz . f90 rk3 . f90
plot−commands rk3_Bz . f90 rk3_g . f90
The first command is given without an argument and it lists the con-
tents of the current directory. The second one, lists the contents of the
subdirectory of the current directory Programs. If the argument is a list
of paths pointing to regular files, then the command prints the names of
the paths. Another way of giving the command is
total 252
-rw-r--r-- 1 george users 24284 May 1 12:08 BE . eps
-rw-r--r-- 1 george users 22024 May 1 11:53 B . eps
-rw-r--r-- 1 george users 29935 May 1 13:02 Byz . eps
-rw-r--r-- 1 george users 48708 May 1 12:41 Bzy . eps
drwxr -xr-x 4 george users 4096 May 1 23:38 Programs
-rw-r--r-- 1 george users 41224 May 1 22:56 srBd_xyz . eps
-rw-r--r-- 1 george users 23187 May 1 21:13 srBE_xyz . eps
-rw-r--r-- 1 george users 24610 May 1 20:29 srB_xy . eps
-rw-r--r-- 1 george users 23763 May 1 20:29 srB_xyz . eps
The switch -l makes ls to list the contents of the current directory to-
gether with useful information on the files in 9 columns. The first column
lists the permissions of the files (see below). The second one lists the num-
ber of links of the files⁸. The third one lists the user who is the owner of
each file. The fourth one lists the group that is assigned to the files. The
fifth one lists the size of the file in bytes (=8 bits). The next three ones
list the modification time of the file and the last one the paths of the files.
File permissions⁹ are separated in three classes: owner permissions,
group permissions and other permissions. Each class is given three spe-
cific permissions, r=read, w=write and x=execute. For regular files, read
permission effectively means access to the file for reading/copying, write
permission means permission to modify the contents of the file and ex-
⁸For a directory it means the number of its subdirectories plus 2 (the parent directory
and itself). For a regular file, it shows how many paths in the filesystem point to this
file.
⁹See the “File system permissions” entry in en.wikipedia.org.
8 CHAPTER 1. THE COMPUTER
ecute permission means permission to execute the file as a command¹⁰.
For directories, read permission means that one is able to read the names
of the files in the directory (but not make it as current directory with the
cd command), write permission means to be able to modify its contents
(i.e. create, delete, and rename files) and execute permission grants per-
mission to access/modify the contents of the files (but not list the names
of the files, this is granted by the read permission).
The command ls -l lists permissions in three groups. The owner
(positions 2-4), the group (positions 5-7) and the rest of the world (others
- positions 8-10). For example
-rw-r--r--
-rwxr-----
drwx--x--x
In the first case, the owner has read and write but not execute permissions
and the group+others have only read permissions. In the second case,
the user has read, write and execute permissions, the group has read
permissions and others have no permissions at all. In the last case, the
user has read, write and execute permissions, whereas the group and the
world have only execute permissions. The first character d indicates a
special file, which in this case is a directory. All special files have this
position set to a character, while regular files have it set to -.
File permissions can be modified by using the command chmod:
> chmod u+x file
> chmod og−w file1 file2
> chmod a+r file
Using the first command, the owner (u≡ user) obtains (+) permission
to execute (x) the file named file. Using the second one, the rest of
the world (o≡ others) and the group (g≡group) loose (-) the write (w)
permission to the files named file1 and file2. Using the third one,
everyone (a≡all) obtain read (r) permission on the file named file.
We will close this section by discussing some commands which are
used for administering files in the filesystem. The command cp (copy)
copies the contents of files into other files:
> cp file1 . f90 file2 . f90
¹⁰Of course it is the user’s responsibility to make sure the file with execute permission
is actually a program that is possible to execute. An error results if this is not the case.
1.1. THE OPERATING SYSTEM 9
> cp file1 . f90 file2 . f90 file3 . f90 Programs
If the file file2.f90 does not exist, the first command copies the contents
of file1.f90 to a new file file2.f90. If it already exists, it replaces its
contents by the contents of the file file2.f90. In order for the second
command to be executed, Programs needs to be a directory. Then, the
contents of the files file1.f90, file2.f90, file3.f90 are copied to
indentical files in the directory Programs. Of course, we assume that
the user has the appropriate privileges for the command to be executed
successfully.
The command mv “moves”, or renames, files:
> mv file1 . f90 file2 . f90
> mv file1 . f90 file2 . f90 file3 . f90 Programs
The first command renames the file file1.f90 to file2.f90. The second
one moves files file1.f90, file2.f90, file3.f90 into the directory
Programs.
The command rm (remove) deletes files¹¹. Beware, the command is
unforgiving: after deletion, a file cannot be restored into the filesystem¹².
Therefore, after executing successfully the following commands
> ls
file1 . f90 file2 . f90 file3 . f90 file4 . csh
> rm file1 . f90 file2 . f90 file3 . f90
> ls
file4 . csh
the files file1.f90, file2.f90, file3.f90 do not exist in the filesystem
anymore. A more prudent use of the command demands the flag -i.
Then, before deletion we are asked for confirmation:
> rm −i *
rm : remove regular file ‘ file1 . f90 ’ ? y
rm : remove regular file ‘ file2 . f90 ’ ? y
rm : remove regular file ‘ file3 . f90 ’ ? y
rm : remove regular file ‘ file4 . csh ’ ? n
¹¹Actually it removes “links” from files. A file may have more than one links in the
same partition of a filesystem. A file is deleted when its last link is removed.
¹²This does not mean that its contents have been deleted from the disk. Deletion
means marking for overwriting. Until the data is overwritten it can be recovered by the
use of special tools. Shredding sensitive data can be tricky business...
10 CHAPTER 1. THE COMPUTER
> ls
file4 . csh
When we type y, the file is deleted, when we type n, the file is not deleted.
We cannot remove directories the same way. It is possible to use
the command rmdir in order to remove empty directories. In order to
delete directories together with their contents (including subdirectories
and their contents) use the command¹³ rm -r. For example, assume that
the contents of the directories dir1 and dir1/dir2 are the files:
. / dir1
. / dir1 / file2 . f90
. / dir1 / file1 . f90
. / dir1 / dir2
. / dir1 / dir2 / file3 . f90
Then the results of the following commands are:
> rm dir1
rm : cannot remove ‘ dir1 ’ : Is a directory
> rm dir1 / dir2
rm : cannot remove ‘ dir1 / dir2 ’ : Is a directory
> rmdir dir1
rmdir : dir1 : Directory not empty
> rmdir dir1 / dir2
rmdir : dir1 / dir2 : Directory not empty
> rm −r dir1
The last command removes all files (assuming that we have write per-
missions for all directories and subdirectories). Alternatively, we can
empty the contents of all directories first, and then remove them with the
command rmdir:
> cd dir1 / dir2 ; rm file3 . f90
> cd . . ; rmdir dir2
> rm file1 . f90 file2 . f90
> cd . . ; rmdir dir1
Note that by using a semicolon, we can execute two or more commands
on the same line.
¹³A small mistake, like rm -rf * and your data is ... history!
1.1. THE OPERATING SYSTEM 11
1.1.2 Commands
Commands in a Unix operating system are files with execute permission.
When we write a sentence on the command line, like
> l s −l test . f90 test . dat
the shell reads its and interprets it. The shell is a program that creates a
interface between a user and the operating system. The first word (ls) of
the sentence is interpreted as a command. The rest of the words are the
arguments of the command and the program can use them (or not) at the
discretion of its programmer. There is a special convention for arguments
that begin with a - (e.g. -l, --help, --version, -O3). They are called
options or switches, and they act as virtual switches that make the program
act in a particular way. We have already seen that the program ls gives
a different output with the switch -l.
In order for a command to be executed, the shell looks for a file that
has the same name as the command (here a file named ls). In order
to understand where the shell looks for such a file, we should digress
a little bit and explain the use of shell variables and environment variables.
These have a name, which is a string of permissible characters, and their
values are obtained by preceding their name with the $ character. For
example the variable PATH has value $PATH. The values of the environment
variables can be set with the command¹⁴ setenv and of the shell variables
with the command set:
> s e t e n v MYVAR test−env
> s e t myvar = test−s h e l l
> echo $MYVAR $myvar
test−env test−s h e l l
Two special variables are the variables PATH and path:
>echo $PATH
/ usr / local / bin : / usr / bin : / bin : / usr / X11 / bin
>echo $path
/ usr / local / bin / usr / bin / bin / usr / X11 / bin
The first one is an environment variable and the second one is a shell
variable. Their values are set by the shell, and we don’t need to worry
¹⁴The command setenv is special to the tcsh shell. For example the bash shell uses
the syntax MYVAR=test-env in order to set the value of an environment variable.
12 CHAPTER 1. THE COMPUTER
about them, unless we want to change them. Their value is a string of
characters whose components should be valid paths to directories. In
the first case, the components are separated by a :, while in the second
case, by one or more spaces. In the example shown above, the shell
searches each component of the path or PATH variables (in this order)
until it finds a file ls in their contents. If it succeeds and the file has
execute permissions, then the program in this file is executed. If it fails,
then it prints an error message. Try the commands:
> which l s
/ bin / l s
> l s −l / bin / l s
−rwxr−xr−x 1 root root 93560 Sep 28 2006 / bin / l s
We see that the program that the ls command executes the program in
the file /bin/ls.
The arguments of a command are passed on to the program that the
command executes for possible interpretation. For example:
> l s −l test . f90 test . dat
The argument -l is the switch that results in a long listing of the files.
The arguments test.f90 and test.dat are interpreted by the program
ls as paths that it will look up for file information.
You can use the * (wildcard) character as a shorthand notation for a
group of files. For example, in the command shown below
> l s −l * . f90 * . dat
the shell will expand *.f90 and *.dat to a list of all files whose names
end with .f90 or .dat. Therefore, if the current directory contains the
files test.f90, test1.f90, myprog.f90, test.dat, hello.dat, the ar-
guments that will be passed on to the command ls are
> l s −l myprog . f90 test1 . f90 test . f90 hello . dat test . dat
For each command there are three special files associated with it. The
first one is the standard input (stdin), the second one is the standard output
(stdout) and the third one the standard error (stderr). These are files
where the program can print or read data from. By default, these files
are the terminal that the user uses to execute the command. In this case,
1.1. THE OPERATING SYSTEM 13
when the program reads data from the stdin, then it reads the data
that we type to the terminal using the keyboard. When the program
writes data to the stdout or to the stderr, then the data is written to the
terminal.
The advantage of using these special files in order to read/write data
is that the user can redirect the input/output to these files to any file she
wants. Using the character > at the end of a command redirects the
stdout to the file whose name is written after >. For example:
> ls
file1 . f90 file2 . f90 file3 . f90 file4 . csh
> l s > results
> ls
file1 . f90 file2 . f90 file3 . f90 file4 . csh results
The first of the above commands, prints the contents of the current work-
ing directory to the terminal. The second command redirects data written
to the stdout to the file results. After executing the command, the file
results is created and its contents are the names of the files file1.f90
file2.f90 file3.f90 file4.csh. If the file results does not exist (as in
the above example), the file is created. If it already exists, it is truncated
and its contents replaced by the data written to the stdout of the com-
mand. If we want to append data without erasing the existing contents,
then we should use the string of characters >>. Therefore, if we give the
command
> l s >> results
after executing the previous commands, then the contents of the file
results will be
file1 . f90 file2 . f90 file3 . f90 file4 . csh
file1 . f90 file2 . f90 file3 . f90 file4 . csh results
The redirection of the stdin is accomplished by the use of the char-
acter < while that of the stderr by the use of the string of characters¹⁵
>&. We will see more examples in section 1.2.
It is possible to redirect the stdout of a command to be the stdin
of another command. This is very useful for creating filters. A filter is
¹⁵This syntax is particular to the tcsh shell. For other shells (bash, sh, ...) read
their documentation.
14 CHAPTER 1. THE COMPUTER
a command that creates a flow of data between two or more programs.
This process is called piping. Pipes are creating by using the character |
> cmd1 | cmd2 | cmd3 | ... | cmdN
Using the syntax shown above, the stdout of the command cmd1 is redi-
rected to the stdin of the command cmd2, the stdout of the command
cmd2 is redirected to the stdin of the command cmd3 etc. More examples
will be presented in section 1.2.
1.1.3 Looking for Help
Unix got itself a reputation for not being user friendly. This is far from the
truth. Although there is a steep learning curve, detailed documentation
for almost everything is available online.
The key for a comfortable ride is to learn how to use the help system
available on your computer and on the internet. Most of the commands
are self documented. A simple test, like the one shown below, will help
you with the basic usage of most of the commands:
> cmd --help
> cmd -h
> cmd -help
> cmd -\?
For example, try the command ls --help. For a window application,
start from the menu “Help”. You should not be afraid and/or lazy and
you should proceed with careful searching and reading.
For example, let’s assume that you have heard about a command that
sounds like printf, or something like that. The first level of online help
is the man (=manual) command that searches the “man pages”. Read the
output of the command
> man p r i n t f
The command info usually provides more detailed and user friendly
documentation. It has basic browsing capabilities like the browsers you
use to read pages from the internet. Try the command
> info printf
1.1. THE OPERATING SYSTEM 15
Furthermore, the commands
> man −k p r i n t f
> whatis p r i n t f
will inform you that there are other, possibly related, commands with
names like fprintf, fwprintf, wprintf, sprintf...:
> whatis p r i n t f
printf (1) − format and p r i n t data
printf (1 p ) − write formatted output
printf (3) − formatted output conversion
printf (3 p ) − p r i n t formatted output
p r i n t f [ builtins ] (1) − bash built−in commands , see bash←-
(1)
The second column printed by the whatis command is the “section” of
the man pages. In order to gain access to the information in a particular
section, you have to give it as an argument to the man command:
> man 1 printf
> man 1p p r i n t f
> man 3 printf
> man 3p p r i n t f
> man bash
Section 1 of the man pages contains information of ordinary command
line commands, section 3 contains information on functions in libraries
of the C language. Section 2 contains information on commands used for
system administration. You may browse the directory /usr/share/man,
or read the man page of the man command (use the command man man
for that!).
By using the command
> p r i n t f --help
we obtain plenty of memory refreshing information. The command
> locate printf
shows us many files related to the command printf. The commands
16 CHAPTER 1. THE COMPUTER
> which p r i n t f
> where p r i n t f
give information on the location of the executable(s) of the command
printf.
Another useful feature of the shell is the command or it filename com-
pletion. This means that we can write only the first characters of the
name of a command or filename and then press simultaneously the keys
[Ctrl-d]¹⁶ (i.e. press the key Ctrl and the key of the letter d at the same
time). Then the shell will complete the name of the command up to the
point that is is unique with the given string of characters¹⁷:
> pri [ Ctrl−d ]
printafm printf printenv printnodetest
Try to type an x on the command line and then type [Ctrl-d]. You will
learn all the commands that are available and whose name begins with
an x: xterm, xeyes, xclock, xcalc, ...
Finally, the internet contains a wealth of information. Google your
blues... and you will be rewarded!
1.2 Text Processing Tools – Filters
For doing data analysis, we will need powerful tools for manipulating
data in text files. These are files that consist solely of printable charac-
ters. Some tools that can be used in order to construct complicated and
powerful filters are the programs cat, less, head, tail, grep, sort
and awk.
Suppose that we have data in a file named data¹⁸ which contains
information on the contents of a food warehouse and their prices:
bananas 100 pieces 1.45
apples 325 boxes 1.18
pears 34 kilos 2.46
bread 62 kilos 0.60
¹⁶If you use the bash shell press [Tab] once or twice.
¹⁷Use the same procedure to auto-complete the names of files in the arguments of
commands.
¹⁸The particular file, as well as most of the files in this section, can be found in the
accompanying software of the chapter. It is highly recommended that you try all the
commands in this section by using all the provided files.
1.2. TEXT PROCESSING TOOLS – FILTERS 17
ham 85 kilos 3.56
The command
> c a t data
prints the contents of the file data to the stdout. In general, this com-
mand prints the contents of all files given in its arguments or the stdin
if none is given. Since the stdin and the stdout can be redirected, the
command
> cat < data > data1
takes the contents of the file data from the stdin and prints them to the
stdout, which in this case is the file data1. This command has the same
result as the command:
> cp data data1
The command
> c a t data data1 > data2
prints the contents of the file data and then the contents of the file data1
to the stdout. Since the stdout is redirected to the file data2, data2
contains the data of both files.
By giving the command
> l e s s g f o r t r a n . txt
you can browse the data contained in the file gfortran.txt one page at
a time. Press [space] in order to “turn” a page, [b] to turn back a page.
Press the up and down arrows to move one line backwards/forward.
Press [g] in order to jump to the beginning of the file and press [G] in
order to jump to the end. Press [h] in order to get a help message and
press [q] in order to quit.
The commands
> head −n 1 data
bananas 100 pieces 1 . 4 5
> t a i l −n 2 data
18 CHAPTER 1. THE COMPUTER
bread 62 kilos 0.60
ham 85 kilos 3.56
> t a i l −n 2 data | head −n 1
bread 62 kilos 0.60
print the first line, the last two lines and the second to the last line of
the file data to the stdout respectively. Note that, by piping the stdout
of the command tail to the stdin of the command head, we are able to
construct the filter “print the line before the last one”.
The command sort sorts the contents of a file by comparing each line
of its text with all others. The sorting is alphabetical, unless otherwise
set by using options. For example
> s o r t data
apples 325 boxes 1.18
bananas 100 pieces 1.45
bread 62 kilos 0.60
ham 85 kilos 3.56
pears 34 kilos 2.46
For reverse sorting, try sort -r data. We can also sort by comparing
specific fields of each line. By default, fields are words separated by one
or more spaces. For example, in order to sort w.r.t. the second column
of the file data, we can use the switch -k 2 (=second field). Furthermore,
we can use the switch -n for numerical sorting:
> s o r t −k 2 −n data
pears 34 kilos 2.46
bread 62 kilos 0.60
ham 85 kilos 3.56
bananas 100 pieces 1.45
apples 325 boxes 1.18
If we omit the switch -n, the comparison of the lines is performed based
on character sorting of the second field and the result is
> s o r t −k 2 data
bananas 100 pieces 1.45
apples 325 boxes 1.18
pears 34 kilos 2.46
bread 62 kilos 0.60
ham 85 kilos 3.56
1.2. TEXT PROCESSING TOOLS – FILTERS 19
The last column contains floating point numbers (not integers). In order
to sort by the values of such numbers we should use the switch -g:
> s o r t −k 4 −g data
bread 62 kilos 0.60
apples 325 boxes 1.18
bananas 100 pieces 1.45
pears 34 kilos 2.46
ham 85 kilos 3.56
The command grep processes a text file line by line, searching for a
given string of characters. When this string is found anywhere in a line,
this line is printed to the stdout. The command
> grep kilos data
pears 34 kilos 2.46
bread 62 kilos 0.60
ham 85 kilos 3.56
prints each line containing the string “kilos”. If we want to search for all
line not containing the string “kilos”, then we add the switch -v:
> grep −v kilos data
bananas 100 pieces 1 . 4 5
apples 325 boxes 1 . 1 8
We can use a regular expression for searching a whole family of strings
of characters. These monsters need a full book for discussing them in
detail! But it is not hard to learn how to use some simple forms of
regular expressions. Here are some examples:
> grep ^b data
bananas 100 pieces 1.45
bread 62 kilos 0.60
> grep ’0$ ’ data
bread 62 kilos 0.60
> grep ’ 3 [ 2 4 ] ’ data
apples 325 boxes 1.18
pears 34 kilos 2.46
The first one, prints each line whose first character is a b. The second
one, prints each line that ends with a 0. The third one, prints each line
contaning the strings 32 or 34.
20 CHAPTER 1. THE COMPUTER
By far, the strongest tool in our toolbox is the awk program. By
default, awk analyzes a text file line by line. Each word (or field in the
awk jargon) of these lines is stored in a set of variables with names
$1, $2, .... The variable $0 contains the full line currently processed,
whereas the variable NF counts the number of fields in the current line.
The variable NR counts the number of lines of the file processed so far by
awk.
An awk program can be written in the command line. A set of com-
mands within { ... } is executed for each line of input. The constructs
BEGIN{ ... } and END{ ... } contain commands executed, only once,
before and after the processing of the file respectively. For example, the
command
> awk ’{ p r i n t $1 , ” t o t a l v a l u e= ” , $2 * $4 } ’ data
bananas total value= 145
apples total value= 383.5
pears total value= 83.64
bread total value= 3 7 . 2
ham total value= 302.6
prints the name of the product (1st column = $1) and the total value
stored in the warehouse (2nd column = $2) × (4th column = $4). More
examples are given below:
> awk ’{ value += $2 * $4 }END{ p r i n t ” T o t a l= ” , value } ’ data
Total= 951.94
> awk ’{ av += $4 }END{ p r i n t ” Average P r i c e = ” , av / NR } ’ data
Average Price= 1.85
> awk ’{ p r i n t $2^2 * sin ( $4 ) + exp ( $4 ) } ’ data
The first one calculates the total value of all products: The processing
of each line results in the increment (+=) of the variable value by the
product of the second and fourth fields. In the end (END{ ... }),
the string Total= is printed, together with the final value of the variable
value. This is an easy way for computing the sum of the values calculated
for each line. The second command, calculates and prints an average.
The sum is calculated in each line and stored in the variable av. In the
end, we print the quotient of the sum of all values by the number of
lines that have been processed (NR). The last command shows a (crazy)
mathematical expression based on numerical values found in each line
of the file data: It computes the square of the second field times the sine
of the fourth field plus the exponential of the fourth field.
1.3. PROGRAMMING WITH EMACS 21
There is much more potential in the commands presented above.
Reading the documentation and getting experience by using them will
provide you with very strong tools in order to accomplish complicated
tasks.
1.3 Programming with Emacs
For a programmer that spends many hours programming every day, the
environment and the tools available for editing the commands of a large
and complicated program determine, to a large extent, the quality of
her life! An editor edits the contents of a text file, that consists solely of
printable characters. Such editors, available in most Linux environments,
are the programs gedit, vim, pico, nano, zile... They provide basic
functionality such as adding, removing or changing text within a file as
well as more complicated functions, such as copying, pasting, searching
and replacing text etc. There are many functions that are particularly
useful to a programmer, such as detecting and formatting keywords of
a particular programming language, pretty printing, closing scopes etc,
which can be very useful for comfortable programming and for spotting
errors. A very powerful and “knowledgeable” editor, offering many such
functions for several programming languages, is the GNU Emacs editor¹⁹.
Emacs is open source software, it is available for free and can be used
in most available operating systems. It is programmable²⁰ and the user
can automate most of her everyday repeated tasks and configure it to her
liking. There is a full interaction with the operating system, in fact Emacs
has been built with the ambition of becoming an operating system. For
example, a programmer can edit a Fortran file, compile it, debug it and
run it, everything done with Emacs commands.
1.3.1 Calling Emacs
In the command line type
> emacs &
¹⁹http://www.gnu.org/software/emacs/ (main site),
http://www.emacswiki.org/ (expert tips), http://en.wikipedia.org/wiki/Emacs
(general info)
²⁰Emacs is written in a dialect of the programming language Lisp, called Elisp. There
is no need of an in-depth knowledge of the language in order to program simple
functions, just see how others are doing it...
22 CHAPTER 1. THE COMPUTER
Note the character & at the end of the line. This makes the particular
command to run in the background. Without it, the shell waits until a
command exits in order to return the prompt.
In a desktop environment, Emacs starts in its own window. For a
quick and dirty editing session, or in the case that a windows environ-
ment is not available²¹, we can run Emacs in a terminal mode. Then, we
omit the & at the end of the line and we run the command
> emacs −nw
The switch -nw forces Emacs to run in terminal mode.
Figure 1.2: The Emacs window in a windows environment. The buttons of very
basic functions found on its toolbar are shown and explained.
²¹Quite handy when we edit files in a remote computer.
1.3. PROGRAMMING WITH EMACS 23
Figure 1.3: Emacs in a non-window mode running on the console. In this figure,
we have typed the command save-buffers-kill-emacs in the minibuffer, a command
that exits Emacs after saving edited data from all buffers. The same command can be
given using the keyboard shortcut C-x C-c. We can see the mode line and the name of
the buffer toy.f written on it, the percentage of the buffer (6%) shown in the window,
the line and columns (33,0) where the point lies and the editing mode which is active
on the buffer (Fortran mode (Fortran), Abbreviation mode (Abbrev), Auto Fill mode
(Fill)).
1.3.2 Interacting with Emacs
We can interact with Emacs in various ways. Newbies will prefer buttons
and menus that offer a simple and intuitive interface. For advanced
usage, however, we recommend that you make an effort to learn the
keyboard shortcuts. There are also thousands of functions available to
be used interactively. They are called from a “command line”, called the
minibuffer in the Emacs jargon.
Keyboard shortcuts are usually combinations of keystrokes that con-
sist of the simultaneous pressing of the Ctrl or Alt keys together with
other keys. Our convention is that a key sequence starting with a C-
means that the characters that follow are keys simultaneously pressed
with the Ctrl key. A key sequance starting with a M- means that the
24 CHAPTER 1. THE COMPUTER
characters that follow are keys simultaneously pressed with the Alt key²².
Some commands have shortcuts consisting of two or more composite
keystrokes. For example by C-x C-c we mean that we have to press
simultaneously the Ctrl key together with x and then press simultane-
ously the Ctrl key together with c. This sequence is a shortcut to the
command that exits Emacs. Another example is C-x 2 which means to
press the Ctrl key together with x and then press only the key 2. This
is a shortcut to the command that splits a window horizontally to two
equal parts.
The most useful shortcuts are M-x (press the Alt key siumutaneously
with the x key) and C-g. The first command takes us to the minibuffer
where we can give a command by typing its name. For example, type
M-x and then type save-buffers-kill-emacs in the minibuffer (this will
terminate Emacs). The second one is an “SOS button” that interrupts
anything Emacs does and returns control to the working buffer. This
can be pretty handy when a command hangs or destroys our work and
we need to interrupt it.
The conventions for the mouse events are as follows: With Mouse-1,
Mouse-2 and Mouse-3 we denote a simple click with the left, middle and
right buttons of the mouse respectively. With Drag-Mouse-1 we mean to
press the left button of the mouse and at the same time drag the mouse.
We summarize the possible ways of giving a command in Emacs with
the following examples that have the same effect: Open a file and put its
contents in a buffer for editing.
• By pressing the toolbar button that looks like a white sheet of paper
(see figure 1.2).
• By choosing the File→Visit New File menu entry.
• By typing the keyboard shortcut C-x C-f.
• By typing the name of the command in the minibuffer: M-x find-file
The number of available commands increases from the top to the bottom
of the above list.
1.3.3 Basic Editing
In order to edit a file, Emacs places the contents of a file in a buffer. Such
a buffer is a chunk of computer memory where the contents of the file
²²Actually, M- is the so called Meta key, usually bound to the Alt key. It is also bound
to the Esc and C-[ keys. The latter can be our only choices available in dumb terminals.
1.3. PROGRAMMING WITH EMACS 25
Figure 1.4: The basic menus found in Emacs when run in a desktop environment. We
can see the basic commands and the keyboard shortcut reminders in the parentheses.
E.g. the command File → Visit New File can be given by typing C-x C-f. Note
the commands File → Visit New File (open a file), File→Save (write contents of
a buffer to a file), File→Exit Emacs, File → Split Window (split window in two),
File→New Frame (open a new Emacs desktop window) and of course the well known
commands Cut, Copy, Paste, Undo from the Edit menu. We can choose different
buffers from the menu Buffers, which contain the contents of other files that we have
opened for editing. We recommend trying the Emacs Tutorial and Read Emacs Manual
in the Help menu.
are copied and it is not the file itself. When we make changes to the
contents of a buffer, the file remains intact. For our changes to take effect
and be written to the file, we have to save the buffer. Then, the contents
of the buffer are written back to the file. It is important to understand
the following cycle of events:
• Read a file’s contents to a buffer.
• Edit buffer contents.
• Write (save) buffer’s contents back into the file.
Emacs may have more than one buffers open for editing simultaneously.
By default, the name of the buffer is the same as the name of the file
26 CHAPTER 1. THE COMPUTER
that is edited, although this is not necessary²³. The name of a buffer is
written in the modeline of the window of the buffer, as can be seen in
figure 1.3.
If Emacs crashes or exits before we save our edits, it is possible to
recover (part of) them. There is a command M-x recover-file that will
guide us through the necessary recovery steps, or we can look for a file
that has the same name as the buffer we were editing surrounded by two
#. For example, if we were editing the file file.f90, the automatically
saved changes can be found in the file #file.f90#. Auto saving is done
periodically by Emacs and its frequency can be controlled by the user.
The point where we insert text while editing is called “the point”.
This is right before the blinking cursor²⁴. Each buffer has another posi-
tion marked by “the mark”. A point and the mark define a “region”
in the buffer. This is a part of the text in the buffer where the func-
tions of Emacs can act (e.g. copy, cut, change case, spelling etc.). We
can set the region by setting a point and then press C-SPC²⁵ or give the
command M-x set-mark-command. This defines the current point to be
the mark. Then we can move the cursor to another point which will
define a region together with the mark that we set. Alternatively we can
use Drag-Mouse-1 (hold the left mouse button and drag the mouse) and
mark a region. The mark can be set with Mouse-3, i.e. with a simple
click of the right button of the mouse. Therefore by Mouse-1 at a point
and then Mouse-3 at a different point will set the region between the two
points.
We can open a file in a buffer with the command C-x C-f, and then
by typing its path. If the file already exists, its contents are copied to a
buffer, otherwise a new buffer is created. Then:
• We can browse the buffer’s contents with the Up/Down/Left/Right
arrows. Alternatively, by using the commands C-n, C-p, C-f and
C-b.
• If the buffer is large, we can browse its contents one page at a time
²³The user can change the name of the buffer without affecting the name of the file
it edits. Also, if we open more than one files with the same name, emacs gives each
buffer a unique name. E.g. if we edit more than one files named index.html then the
corresponding buffers are named index.html, index.html<2>, index.html<3>, ... .
²⁴Strictly speaking, the point lies between two characters and not on top of a character.
The cursor lies on the character immediately to the right of the point. A point is assigned
to every window, therefore a buffer can have multiple points, one for each window that
displays its contents.
²⁵Press the Ctrl and spacebar keys simultanesouly.
1.3. PROGRAMMING WITH EMACS 27
by using the Page Up/Page Dn keys. Alternatively, by using the
commands C-v, M-v.
• Enter text at the points simply by typing it.
• Delete characters before the point by using the Backspace key and
after the point by using the Delete key. The command C-d deletes
a forward character.
• Erase all the characters in a line that lie ahead of the point by using
the command C-k.
• Open a new line by using Enter or C-o.
• Go to the first character of a line by using Home and the last one
by using End. Alternatively, by using the commands C-a and C-e,
respectively.
• Go to the first character of the buffer with the key C-Home and the last
one with the key C-End. Alternatively, with M-x beginning-of-buffer
and M-x end-of-buffer.
• Jump to any line we want: Type M-x goto-line and then the line
number.
• Search for text after the point: Press C-s and then the text you
are looking for. This is an incremental search and the point jumps
immediately to the first string that matches the search. The same
search can be repeated by pressing C-s repeatedely.
When we finish editing (or frequently enough so that we don’t loose
our work due to an unfortunate event), we save the changes in the buffer,
either by pressing the save icon on the toolbar, or by pressing the keys
C-s, or by giving the command M-x save-buffer.
1.3.4 Cut and Paste
Use the instructions below for slightly more advanced editing:
• Undo! Some of the changes described below can be catastrophic.
Emacs has a great Undo function that keeps in its memory many
of the changes inflicted by our editing commands. By repeatedely
pressing C-/, we undo the changes we made. Alternatively, we
can use C-x u or the menu entry Edit→Undo. Remember that C-g
interrupts any Emacs process currently running in the buffer.
28 CHAPTER 1. THE COMPUTER
• Cut text by using the mouse: Click with Mouse-1 at the point before
the beginning of the text and then Mouse-3 at the point after the
end. A second Mouse-3 and the region is ... gone (in fact it is
written in the “kill ring” and it is available for pasting)!
• Cut text by using a keyboard shortcut: Set the mark by C-SPC at the
point before the beginning of the text that you want to cut. Then
move the cursor after the last character of the text that you want to
cut and type C-w.
• Copy text by using the mouse: Drag the mouse Drag-Mouse-1 and
mark the region that you want to copy. Alternatively, Mouse-1 at
the point before the beginning of the text and then Mouse-3 at the
point after the end.
• Copy text by using a keyboard shortcut: Set the mark at the begin-
ning of the text with C-SPC and then move the cursor after the last
character of the text. Then type M-w.
• Pasting text with the mouse: We click the middle button²⁶ Mouse-2
at the point that we want to insert the text from the kill ring (the
copied text).
• Pasting text with a keyboard shortcut: We move the point to the
desired insertion point and type C-y.
• Pasting text from previous copying: A fast choice is the menu entry
Edit→Paste from kill manu and then select from the copied texts.
The keyboard shortcut is to first type C-y and then M-y repeatedly,
until the text that we want is yanked.
• Insert the contents of a file: Move the point to the desired place and
type C-x i and the path of the file. Alternatively, give the command
M-x insert-file.
• Insert the contents of a buffer: We can insert the contents of a whole
buffer at a point by giving the command M-x insert-buffer.
• Replace text: We can replace text interactively with the command
M-x query-replace, then type the string we want to replace, and
then the replacement string. Then, we will be asked whether we
want the change to be made and we can answer by typing y (yes),
²⁶If it is a two button mouse, try clicking the left and right buttons simultaneously.
1.3. PROGRAMMING WITH EMACS 29
n (no), q (quit the replacements). A , (comma) makes only one
replacement and quits (useful if we know that this is the last change
that we want to make). If we are confident, we can change all
string in a buffer, no questions asked, by giving the command M-x
replace-string.
• Change case: We can change the case in the words of a region with
the commands M-x upcase-region, M-x capitalize-region and
M-x downcase-region. Try it.
We note that cutting and pasting can be made between different windows
of the same or different buffers.
1.3.5 Windows
Sometimes it is very convenient to edit one or more different buffers in
two or more windows. The term “windows” in Emacs refers to regions
of the same Emacs desktop window. In fact, a desktop window running
an Emacs session is referred to as a frame in the Emacs jargon. Emacs
can split a frame in two or more windows, horizontally or/and vertically.
Study figure 1.5 on page 69 for details. We can also open a new frame
and edit several buffers simultaneously²⁷. We can manipulate windows
and frames as follows:
• Position the point at the center of the window and clear the screen
from garbage: C-l (careful: l not 1).
• Split a window in two, horizontally: C-x 2.
• Split a window in two, vertically: C-x 3.
• Delete all other windows (remain only with the current one): C-x
1.
• Delete the current windows (the others remain): C-x 0.
• Move the cursor to the other window: Mouse-1 or C-x o.
• Change the size of window: Use Drag-Mouse-1 on the line sepa-
rating two windows (the mode line). Use C-^, C-} for making a
change of the horizontal/vertical size of a window respectively.
²⁷Be careful not to start a new Emacs session each time that all you need is a new
frame. A new Emacs process takes time to start, binds computer resources and does
not communicate with a different Emacs process.
30 CHAPTER 1. THE COMPUTER
• Create a new frame: C-x 5 2.
• Delete a frame: C-x 5 0.
• Move the cursor to a different frame: With Mouse-1 or with C-x 5
o.
You can have many windows in a dumb terminal. This is a blessing
when a dekstop environment is not available. Of course, in that case you
cannot have many frames.
1.3.6 Files and Buffers
• Open a file: C-x C-f or M-x find-file.
• Save a buffer: C-x C-s or M-x save buffer. With C-x C-c or
M-x save-buffers-kill-emacs we can also exit Emacs. From the
menu: File→Save. From the toolbar: click on the save icon.
• Save buffer contents to a different file: C-x C-w or M-x write-file.
From the menu: File→Save As. From the toolbar: click on the
“save as” icon.
• Save all buffers: C-x s or M-x save-some-buffers.
• Connect a buffer to a different file: M-x set-visited-filename.
• Kill a buffer: C-x k.
• Change the buffer of the current window: C-x b. Also, use the
menu Buffers, then choose the name of the buffer.
• Show the list of all buffers: C-x C-b. From the menu: Buffers
→ List All Buffers. By typing Enter next to the name of the
buffer, we make it appear in the window. There are several buffer
administration commands. Learn about them by typing C-h m when
the cursor is in the Bufer List window.
• Recover data from an edited buffer: If Emacs crashed, do not de-
spair. Start a new Emacs and type M-x recover-file and follow
the instructions. The command M-x recover-session recovers all
unsaved buffers.
1.3. PROGRAMMING WITH EMACS 31
• Backup files: When you save a buffer, the previous contents of the
file become a backup file. This is a file whose path is the same as
the original’s file with a ˜ appended in the end. For example a
file test.f90 will have as a backup the file test.f90˜. Emacs has
version control, and you can configure it to keep as many versions
of your edits as you want.
• Directory browsing and directory administration commands: C-x
d or M-x dired. You can act on the files of a directory (open,
delete, rename, copy etc) by giving appropriate commands. When
the cursor is in the dired window, type C-h m to read the relevant
documentation.
1.3.7 Modes
Each buffer can be in different modes. Each mode may activate different
commands or editing environment. For example each mode can color
keywords relevant to the mode and/or bind keys to different commands.
There exist major modes, and a buffer can be in only one of them. There
are also minor modes, and a buffer can be in one or more of them. Emacs
activates major and minor modes by default for each file. This usually
depends on the filename but there are also other ways to control this. The
user can change both major and minor modes at will, using appropriate
commands.
Active modes are shown in a parenthesis on the mode line (see figures
1.3 and 1.5.
• M-x f90-mode: This mode is of special interest in this book since we
will edit a lot of Fortran code. We need it activated in buffers that
contain a Fortran program and its most useful characteristics are
automatic code alignment by pressing the key TAB, the coloring of
Fortran commands, variables and other structural constructs (sub-
routines, if statements, do loops, variable declarations, statement
labels etc). Another interesting function is the one that comments
out a whole region of code, as well as the inverse function.
• M-x c-mode: For files containing programs written in the C lan-
guage. Related modes are the c++-mode, java-mode, perl-mode,
awk-mode, python-mode, makefile-mode, octave-mode, gnuplot-mode,
mathematica-mode and others.
• latex-mode: For files containing LATEX text formatting commands.
32 CHAPTER 1. THE COMPUTER
• text-mode: For editing simple text files (.txt).
• fundamental-mode: The basic mode, when one that fits better doesn’t
exist...
Some interesting minor modes are:
• M-x auto-fill-mode: When a line becomes too long, it is wrapped
automatically. A related command to do that for the whole region
is M-x fill-region, and for a paragraph M-x fill-paragraph.
• M-x overwite-mode: Instead of inserting characters at the point,
overwrite the existing ones. By giving the command several times,
we toggle between activating and deactivating the mode.
• M-x read-only mode: When visiting a file with valuable data that
we don’t want to change by mistake, we can activate this mode so
that changes will not be allowed by Emacs. When we open a file
with the command C-x C-r or M-x find-file-read-only this mode
is activated. We can toggle this mode on and off with the command
C-x C-q (M-x toggle-read-only). See the mode line of the buffer
jack.c in figure 1.5 which contains a string %%. By clicking on the
%% we can toggle the read-only mode on and off.
• flyspell-mode: Spell checking as we type.
• font-lock-mode: Colors the structural elements of the buffer which
are defined by the major mode (e.g. the commands of a Fortran
program).
In a desktop environment, we can choose modes from the menu of
the mode line. By clicking with Mouse-3 on the name of a mode we are
offered options for (de)activating minor modes. With a Mouse-1 we can
(de)activate the read-only mode with a click on :%% or :-- respectively.
See figure 1.5.
1.3.8 Emacs Help
Emacs’ documentation is impressive. For newbies, we recommend to
follow the mini course offered by the Emacs tutorial. You can start the
tutorial by typing C-h t or select Help → Emacs Tutorial from the
menu. Enjoy... The Emacs man page (give the man emacs command in
the command line) will give you a summary of the basic options when
calling Emacs from the command line.
1.3. PROGRAMMING WITH EMACS 33
A quite detailed manual can be found in the Emacs info pages²⁸.
Using info needs some training, but using the Emacs interface is quite
intuitive and similar to using a web browser. Type the command C-h r
(or choose Help→Emacs Tutorial from the menu) and you will open the
front page of the emacs manual in a new window. By using the keys SPC
and Backspace we can read the documentation page by page. When you
find a link (similar to web page hyperlinks), you can click on it in order
to open to read the topic it refers to. Using the navigation icons on the
toolbar, you can go to the previous or to the next pages, go up one level
etc. There are commands that can be given by typing single characters.
For example, type d in order to jump to the main info directory. There
you can find all the available manuals in the info system installed on
your computer. Type g (emacs) and go to the top page of the Emacs
manual. Type g (info) and read the info manual.
Emacs is structured in an intuitive and user friendly way. You will
learn a lot from the names of the commands: Almost all names of Emacs
commands consist of whole words, separated by a hyphen “-”, which
almost form a full sentence. These make them quite long sometimes,
but by using auto completion of their names this does not pose a grave
problem.
• auto completion: The names of the commands are auto completed
by typing a TAB one or more times. E.g., type M-x in order to go to
the minibuffer. Type capi[TAB] and the command autocompletes
to capitalize-. By typing [TAB] for a second time, a new window
opens and offers the options for completing to two possible com-
mands: capitalize-region and capitalize-word. Type an extra
r[TAB] and the command auto completes to the only possible choice
capitalize-region. You can see all the commands that start with
an s by typing M-x s[TAB][TAB]. Sure, there are many... Click on
the *Completions* buffer and browse the possibilities. A lot will
become clear just by reading the names of the commands. By typ-
ing M-x [TAB][TAB], all available commands will appear in your
buffer!
• keyboard shortcuts: If you don’t remember what happens when
you type C-s, no problem: Type C-h k and then the ... forgotten key
sequence C-s. Conversely, have you forgotten what is the keyboard
²⁸If you prefer books in the form of PDF visit the page www.gnu.org/software/emacs
and click on Documentation. You will find a 600 page book that has almost everything!
34 CHAPTER 1. THE COMPUTER
shortcut of the command save-buffer? Type C-h w and then the
command.
• functions: Are you looking for a command, e.g. save-something
-I-forgot? Type C-h f and then save-[TAB] in order to browse
over different choices. Use Mouse-2 in order to select the command
you are interested in, or type and complete the rest of its name (you
may use [TAB] again). Read about the function in the *Help* buffer
that opens.
• variables: Do the same after typing C-h v in order to see a vari-
able’s value and documentation.
• command apropos: Have you forgotten the exact name of a com-
mand? No problem... Type C-h a and a keyword. All commands
related to the keyword you typed will appear in a buffer. Use C-h
d for even more information.
• modes: When in a buffer, type C-h m and read information about
the active modes of the buffer.
• info: Type C-h i
• Have you forgotten everything mentioned above? Just type C-h ?
1.3.9 Emacs Customization
You can customize everything in Emacs. From key bindings to program-
ming your own functions in the Elisp language. The most common way
for a user to customize her Emacs sessions, is to put all her customization
commands in the file ∼/.emacs in her home directory. Emacs reads and
executes all these commands just before starting a session. Such a .emacs
file is given below:
; D e f i n e F1 key t o save t h e b u f f e r
( global-set-key [ f1 ] ’ save-buffer )
; D e f i n e Control−c s t o save t h e b u f f e r
( global-set-key ”\C−cs ” ’ save-some-buffers )
; D e f i n e Meta−s ( Alt−s ) t o i n t e r a c t i v e l y s e a r c h forward
( global-set-key ”\M−s” ’ isearch-forward )
; D e f i n e M−x i s t o i n t e r a c t i v e l y s e a r c h forward
( defalias ’ is ’ isearch-forward )
; D e f i n e M−x fm t o s e t fortran−mode f o r t h e b u f f e r
( defun fm ( ) ( interactive ) ( f90-mode ) )
1.4. THE FORTRAN PROGRAMMING LANGUAGE 35
; D e f i n e M−x s i g n t o s i g n my name
( defun sign ( ) ( interactive ) ( insert ”K. N. Anagnostopoulos ” ) )
Everything after a ; is a comment. Functions/commands are enclosed
in parentheses. The first three ones bind the keys F1, C-c s and M-s to
the commands save-buffer, save-some-buffers and isearch-forward
respectively. The next one defines an alias of a command. This means
that, when we give the command M-x is in the minibuffer, then the
command isearch-forward will be executed. The last two commands
are the definitions of the functions (fm) and (sign), which can be called
interactively from the minibuffer.
For more complicated examples google “emacs .emacs file” and you
will see other users’ .emacs files. You may also customize Emacs from the
menu commands Options→Customize Emacs. For learning the Elisp lan-
guage, you can read the manual “Emacs Lisp Reference Manual” found
at the address
www.gnu.org/software/emacs/manual/elisp.html
1.4 The Fortran Programming Language
In this section, we give a very basic introduction to the Fortran program-
ming language. This is not a systematic exposition and you are expected
to learn what is needed in this book by example. So, please, if you have
not already done it, get in front of a computer and do what you read.
You can find many good tutorials and books introducing Fortran in a
more complete way in the bibliography.
1.4.1 The Foundation
The first program that one writes when learning a new programming
language is the “Hello World!” program. This is the program that prints
“Hello World!” on your screen:
program hello
! p r i n t a message t o t h e world :
p r i n t * , ’ H e l l o World ! ’ ! t h i s i s a comment
end program hello
36 CHAPTER 1. THE COMPUTER
Commands, or statements, in Fortran are strings of characters separated by
blanks (“words”) that we are allowed to write from the 1st to the 132nd
column of a file. Each line starts a new command²⁹. We can put more
than one command on each line by separating them with a semicolon (;).
Everything after an exclamation mark (!) is a comment. Proliferation of
comments is necessary for documenting our code. Good documentation
of our code is an integral part of programming. If the code is planned to
be read by others, or by us at a later time, make sure to explain in detail
what each line is supposed to do. You and your collaborators will save
a lot of time in the process of debugging, improving and extending your
code.
The main entry to the program is defined by the command program
name, where name can be any string of alphanumeric characters and an
underscore. When the program runs, it starts executing commands at
this point. The end of the program, as well as of any other program unit
(functions, subroutines, modules), is defined by the line end program
name.
The first (and only) command given in the above program is the print
command. It prints the string of characters “Hello World!” to the stdout.
The “*,” is part of the syntax and it is not printed, of course. Fortran does
not distinguish capital from small letters, so we could have written PRINT,
Print, prINt, ... A string of characters in Fortran is enclosed in single or
double quotes ('Hello World!' or "Hello World!" is equivalent).
In order to execute the commands in a program, it is necessary to com-
pile it. This is a job done by a program called the compiler that translates
the human language programming statements into binary commands that
can be loaded to the computer memory for execution. There are many
Fortran compilers available, and you should learn which compilers are
available for use in your computing environment. Typical names for
Fortran compilers are gfortran, f90, ifort, g95, .... You should
find out which compiler is best suited for your program and spend time
reading its documentation carefully. It is important to learn how to use a
new compiler so that you can finely tune it to optimize the performance
of your program.
We are going to use the open source and freely available compiler
gfortran, which can be installed on most popular operating systems³⁰.
The compilation command is:
²⁹It is possible to break long lines by putting a & at the end of each broken line and
continue the same command in the next one. More on that later.
³⁰http://gcc.gnu.org/fortran/
1.4. THE FORTRAN PROGRAMMING LANGUAGE 37
> g f o r t r a n hello . f90 −o hello
The switch -o defines the name of the executable file, which in our case
is hello. If the compilation is successful, the program runs with the
command:
> . / hello
Hello world !
Now, we will try a simple calculation. Given the radius of a circle we
will compute its length and area. The program can be found in the file
area_01.f90:
program circle_area
PI = 3.141593
R = 4.0
p r i n t * , ’ P e r i m e t e r= ’ , 2 . 0 * PI * R
p r i n t * , ’ Area= ’ , PI * R * * 2
end program circle_area
The first two commands define the values of the variables PI and R. These
variables are of type REAL, which are floating point numbers. Fortran
has implicit rules that can be used to define the type of variables. By
default, variables whose name starts with i, j, k, l, m and n are of
INTEGER type. These are exact whole numbers. All other variables are of
type REAL³¹. We can override these implicit rules by explicitly declaring
the type of a variable or by changing the implicit rules with the use of
the implicit statement. The following two commands have two effects:
Computing the length 2πR and the area πR2 of the circle and printing
the results. The expressions 2.0*PI*R and PI*R**2 are evaluated before
being printed by the print command. The multiplication and raising to
a power operators are * and **, respectively. Note the explicit decimal
points at the constants 2.0 and 4.0. If we write 2 or 4 instead, then
these are going to be constants of the INTEGER type and by using them
³¹Don’t confuse REAL variables with the real numbers. REAL variables take values
that are finite approximations to real numbers and take values that are a subset of the
rational numbers. This approximation becomes better with increasing the amount of
memory allocated to REALs. In most computing environments, REALs are allocated 4 or
8 bytes of memory, in which case they approximate real numbers with, more or less, 7
or 17 significant digits, respectively.
38 CHAPTER 1. THE COMPUTER
the wrong way we may obtain surprising results³². We compile and run
the program with the commands:
> g f o r t r a n area_01 . f90 −o area
> . / area
Perimeter= 25.13274
Area= 50.26548
We will now try a process that repeats itself for many times. We will
calculate the length and area of 10 circles of different radii Ri = 1.28 + i,
i = 1, 2, . . . , 10. We will store the values of the radii in an array R(10) of
the REAL type. The code can be found in the file area_02.f90:
program circle_area
dimension R ( 1 0 )
PI = 3.141593
R ( 1 ) = 2.28
do i =2 ,10
R ( i ) = R ( i−1) + 1 . 0
enddo
do i = 1 , 1 0
perimeter = 2* PI * R ( i )
area = PI * R ( i ) * * 2
p r i n t * , i , ’ ) R= ’ , R ( i ) , ’ p e r i m e t e r= ’ , perimeter
p r i n t * , i , ’ ) R= ’ , R ( i ) , ’ a r e a = ’ , area
enddo
end program circle_area
The command dimension R(10) defines an array of length 10. This way,
the elements of the array are referred by an index that takes value from
1 to 10. For example R(4) is the fourth element of the array.
Between the lines
do i = 2 , 10
...
enddo
we can write commands that are repeatedly executed while the INTEGER
³²Try adding the command print *,2/4, 2.0/4.0 and check the results.
1.4. THE FORTRAN PROGRAMMING LANGUAGE 39
variable i takes values from 2 to 10 with increasing step³³ equal to 1.
The command:
R ( i ) = R ( i−1) + 1 . 0
defines the i-th radius to have a value which is larger by the (i-1)-th
by 1. For the loop to work correctly, we must define the initial value
of R(1), otherwise the final result is undefined³⁴. The second loop uses
the defined R-values in order to do the computation and printing of the
results.
Now, we will write an interactive version of the program. Instead of
hard coding the values of the radii, we will interact with the user asking
her to give her own values. The program will read the 10 values of the
radii from the standard input (stdin). The program can be found in the
file area_03.f90:
program circle_area
i m p l i c i t none
i n t e g e r , parameter :: N=10
real , parameter :: PI =3.141593
real , dimension ( N ) :: R
real :: area , perimeter
integer :: i
do i =1 , N
p r i n t * , ’ Enter r a d i u s o f c i r c l e : ’
read * , R ( i )
p r i n t * , ’ i = ’ , i , ’ R( i )= ’ , R ( i )
enddo
open (UNIT=13 ,FILE= ’AREA.DAT’ )
do i = 1 , N
perimeter = 2* PI * R ( i )
area = PI * R ( i ) * * 2
w r i t e ( 1 3 , * ) i , ’ ) R= ’ , R ( i ) , ’ a r e a= ’ , area ,&
’ p e r i m e t e r= ’ , perimeter
enddo
³³The step can change by adding one more entry to the do line: do i=0,12,4 runs
the loop for i=0,4,8,12, whereas do i=10,6,-2 for i=10,8,6.
³⁴That means that different compilers and/or runs can give different results.
40 CHAPTER 1. THE COMPUTER
close (13)
end program circle_area
The first statement in the above program is implicit none! This state-
ment deactivates the implicit rules of Fortran, and the programmer is
obliged to declare all variables in a program unit. It is highly recom-
mendable that you always use this option... You might spend a little
more time typing the declarations, but this effort cannot be compared to
the pain looking for bugs due to typos in the names of variables³⁵! We
will follow this practice throughout the book.
The declarations of the variables follow this statement. The variables
N and i are declared to be of the INTEGER type, whereas the variables
PI, area, perimeter and R(N) are declared to be of the REAL type.
The variables PI and N are specified to be parameters. Parameters are
given specific values which cannot be changed during the execution of
the program.
The array elements R(i) are read using the command read:
read * , R ( i )
The command read reads from the stdin. The user types the values
at the terminal and then presses [Enter]. We can read more than one
variables with one read command.
In order to print data to a file, we have to connect it to a unit. Each
unit is represented by any number between 0 and 99. Some numbers
are reserved for special units³⁶. The connection of a unit to a file is done
with the open command. When this is done, we can write to the file
with the command³⁷ write(n,*), where n is the unit number. When we
are done writing to a file we should use the command close(n). Then
the unit number is available to be used for a different file. The flow of
commands is like
open (UNIT=13 ,FILE= ’AREA.DAT’ )
...
write (13 ,*) . . . .
...
close (13)
³⁵Can you see the difference between the names pl1 and p11?
³⁶E.g., 5 is the stdin, 6 is the stdout and 0 is the stderr.
³⁷Try to see what happens when you write to a unit what has not been connected to
a file via an open command!
1.4. THE FORTRAN PROGRAMMING LANGUAGE 41
The name of the file is determined by the option FILE='AREA.DAT' of
the open statement. Uppercase or lowercase characters in the filename
make a difference. The option FILE='path' can use any valid path in
the filesystem, provided that we have the necessary permissions.
The line
w r i t e ( 1 3 , * ) i , ’ ) R= ’ , R ( i ) , ’ a r e a= ’ , area ,&
’ p e r i m e t e r= ’ , perimeter
shows us how to continue a line containing a long statement to the next
one. We place a & at the end of the line and then continue writing the
statement to the next. This can happen up to 39 times.
The next step will be to learn how to define and use functions and
subroutines. The program below shows how to define a subroutine
area_of_circle, which computes the length and area of a circle of given
radius. The following program can be found in the file area_04.f90:
program circle_area
i m p l i c i t none
i n t e g e r , parameter :: N=10
real , parameter :: P =3.141593
real , dimension ( N ) :: R
real :: area , perimeter
integer :: i
do i =1 , N
p r i n t * , ’ Enter r a d i u s o f c i r c l e : ’
read * , R ( i )
p r i n t * , ’ i = ’ , i , ’ R( i )= ’ , R ( i )
enddo
open (UNIT=13 ,FILE= ’AREA.DAT’ )
do i = 1 , N
c a l l area_of_circle ( R ( i ) , perimeter , area )
w r i t e ( 1 3 , * ) i , ’ ) R= ’ , R ( i ) , ’ a r e a= ’ , area ,&
’ p e r i m e t e r= ’ , perimeter
enddo
close (13)
end program circle_area
42 CHAPTER 1. THE COMPUTER
s u b r o u t i n e area_of_circle ( R , L , A )
i m p l i c i t none
real :: R,L,A
r e a l , parameter : : PI = 3.141593 , PI2 = 2 . 0 * PI
L= PI2 * R
A= PI * R * R
return
end s u b r o u t i n e area_of_circle
The calculation of the length and the area of the circle is performed by
the call to the subroutine:
c a l l area_of_circle ( R ( i ) , perimeter , area )
The command call calls a subroutine and transfers the control of the
program within the subroutine. The above subroutine has the arguments
(R(i),perimeter,area). The argument R(i) is an input variable. It
provides the necessary data to the subroutine in order to perform its
computation. The arguments perimeter and area are intended for out-
put. Upon return of the subroutine to the main program, they store the
result of the computation. The user of a subroutine must learn how to
use its arguments in order to be able to call it in her program. These
must be documented carefully by the programmer of the subroutine.
The actual program executed by the subroutine is between the lines:
s u b r o u t i n e area_of_circle ( R , L , A )
...
end s u b r o u t i n e area_of_circle
The arguments (R,L,A) must be declared in the subroutine and need not
have the same names as the ones that we use when we call it. A change
of their values within the subroutine will change the values of the cor-
responding variables in the calling program³⁸. Therefore, the statements
L=PI2*R and A=PI*R*R change the values of the variables perimeter and
area to the desired values. The command return returns the control
to the calling program. The parameters PI and PI2 are “private” to the
subroutine. Their names and values are invisible outside the subroutine.
³⁸We say that variables in Fortran are passed to subroutines by reference and not by
value as in C.
1.4. THE FORTRAN PROGRAMMING LANGUAGE 43
Similarly, the variables i, N, ..., defined in the main program, are
invisible within the subroutine.
We summarize all of the above in a program trionymo.f90, which
computes the roots of a second degree polynomial:
! =============================================================
! Program t o compute r o o t s o f a 2nd order polynomial
! Tasks : Input from user , l o g i c a l s t a t e m e n t s ,
! use o f f u n c t i o n s , s t o p
! Accuracy i n f l o a t i n g p o i n t a r i t h m e t i c
! e . g . IF ( x . eq . 0 . 0 )
!
! T e s t s : a , b , c= 1 2 3 D= −8
! a , b , c= 1 −8 16 D= 0 x1= 4
! a , b , c= 1 −1 −2 D= 9 . x1= 2 . x2= −1.
! a , b , c= 2.3 −2.99 −16.422 x1= 3. 4 x2= −2.1
! But : 6 . 8 ( x −4.3) * * 2 = 6.8 x * * 2 −58.48*x +125.732
! a , b , c= 6.8 −58.48 125.73199
! D= 0.000204147349 x1= 4.30105066 x2= 4.29894924
! a , b , c= 6.8 −58.48 1 2 5 . 7 3 2 , D= −0.000210891725 < 0 ! !
! =============================================================
program trionymo
i m p l i c i t none
real : : a , b , c , D
r e a l : : x1 , x2
r e a l : : Discriminant
p r i n t * , ’ Enter a , b , c : ’
read * , a , b , c
! T e s t i f we have a w e l l d e f i n e d polynomial o f 2nd degree :
i f ( a . eq . 0 . 0 ) s t o p ’ trionymo : a=0 ’
! Compute t h e d i s c r i m i n a n t (= d i a k r i n o u s a )
D = Discriminant ( a , b , c )
p r i n t * , ’ D i s c r i m i n a n t : D= ’ , D
! Compute t h e r o o t s i n each c a s e : D>0 , D=0 , D<0 ( no r o o t s )
i f ( D . g t . 0.0 ) then
c a l l roots ( a , b , c , x1 , x2 )
p r i n t * , ’ Roots : x1= ’ , x1 , ’ x2= ’ , x2
e l s e i f ( D . eq . 0 . 0 ) then
c a l l roots ( a , b , c , x1 , x2 )
p r i n t * , ’ Double Root : x1= ’ , x1
else
p r i n t * , ’No r e a l r o o t s ’
44 CHAPTER 1. THE COMPUTER
endif
end program trionymo
! =============================================================
! This i s t h e f u n c t i o n t h a t computes t h e d i s c r i m i n a n t
! A f u n c t i o n r e t u r n s a v a l u e . This v a l u e i s a s s i g n e d with t h e
! statement :
! D i s c r i m i n a n t = <value >
! i . e . we simply a s s i g n anywhere i n t h e program a v a r i a b l e with
! t h e name o f t h e f u n c t i o n .
! =============================================================
r e a l f u n c t i o n Discriminant ( a , b , c )
i m p l i c i t none
real : : a , b , c
Discriminant = b * * 2 − 4.0 * a * c
end f u n c t i o n Discriminant
! =============================================================
! The s u b r o u t i n e t h a t computes t h e r o o t s .
! =============================================================
s u b r o u t i n e roots ( a , b , c , x1 , x2 )
i m p l i c i t none
real : : a , b , c
r e a l : : x1 , x2
r e a l : : D , Discriminant
i f ( a . eq . 0 . 0 ) s t o p ’ r o o t s : a=0 ’
D = Discriminant ( a , b , c )
i f ( D . ge . 0 . 0 ) then
D = sqrt (D)
else
p r i n t * , ’ r o o t s : Sorry , cannot compute r o o t s , D<0= ’ , D
stop
endif
x1 = (−b + D ) / ( 2 . 0 * a )
x2 = (−b − D ) / ( 2 . 0 * a )
end s u b r o u t i n e roots
The program reads the coefficients of the polynomial ax2 + bx + c. After
a check whether a ̸= 0, it computes the discriminant D = b2 − 4ac by
calling the function Discriminant(a,b,c). The only difference between
a function and a subroutine is that the first one returns a value of a
given type. We don’t need to use the command call in order to run
the commands of a function, this is done by computing its value in an
1.4. THE FORTRAN PROGRAMMING LANGUAGE 45
expression. The type of the value returned must be declared both in the
program that uses the function (real :: Discriminant) and at the entry
point of its program unit (real function Discriminant(a,b,c)). The
value returned to the calling program is the one assigned to the variable
that has the same name as the function:
r e a l f u n c t i o n Discriminant ( a , b , c )
...
Discriminant = b * * 2 − 4.0 * a * c
...
end f u n c t i o n Discriminant
Notice the use of the comparison operators .gt. (strictly greater than)
and .eq. (equal to)³⁹:
i f ( D . g t . 0.0 ) then
...
e l s e i f ( D . eq . 0 . 0 ) then
...
else
...
endif
1.4.2 Details
You may skip this paragraph during a first reading of the book. It is
intended mainly to be a reference when reading the later chapters.
There are more types of variables built in Fortran. In the program
listed below, we show how to use CHARACTER variables, floating point
numbers of double precision REAL(8) and complex numbers of single
and double precision, COMPLEX and COMPLEX(8) respectively:
program f90_vars
i m p l i c i t none
c h a r a c t e r ( 1 0 0 ) : : string
real (4) :: x ! s i n g l e p r e c i s i o n , same as r e a l : : x
real (8) : : x8 ! e q u i v a l e n t t o : double p r e c i s i o n x8
! re al (16) : : x16 ! may not be supported by a l l c o m p i l e r s
³⁹Other operators are .lt., .ge. .le. (strictly less, greater or equal, less or equal),
.ne. (not equal) and .or., .and., .not. (logical or, and and negation).
46 CHAPTER 1. THE COMPUTER
! Complex Numbers :
complex ( 4 ) : : z ! s i n g l e p r e c i s i o n , same as complex : : z
complex ( 8 ) : : z8 ! double p r e c i s i o n
!A s t r i n g : a c h a r a c t e r a r r a y :
string = ’ H e l l o World ! ’ ! s t r i n g s m a l l e r s i z e , l e a v e s blanks
! TRIM : t r i m blanks
p r i n t * , ’A s t r i n g : : ’ , string , ’ : : ’ ,TRIM( string ) , ’ : : ’
p r i n t * , ’ j o i n them : : ’ , string // string , ’ : : ’
p r i n t * , ’ j o i n them : : ’ , TRIM( string ) / / TRIM( string ) , ’ : : ’
! R e a l s with i n c r e a s i n g a c c u r a c y : Determine PI = 3 . 1 4 1 5 9 . . .
x = 4.0 * atan ( 1 . 0 )
! Use D f o r double p r e c i s i o n exponent
x8 = 4.0 D0 * atan ( 1 . 0 D0 )
! Use Q f o r q u a d r i p l e p r e c i s i o n exponent
! x16 = 4.0Q0* atan ( 1 . 0 Q0)
p r i n t * , ’ x4= ’ , x , ’ x8= ’ , x8 ! , ’ x16= ’ , x16
p r i n t * , ’ x4 : ’ , range ( x ) , p r e c i s i o n ( x ) , EPSILON( x ) ,&
TINY( x ) ,HUGE( x )
p r i n t * , ’ x8 : ’ , range ( x8 ) , p r e c i s i o n ( x8 ) , EPSILON( x8 ) ,&
TINY( x8 ) ,HUGE( x8 )
! Complex numbers : s i n g l e p r e c i s i o n
z = ( 2 . 0 , 1 . 0 ) * cexp ( ( 3 . 0 , − 1 . 0 ) )
p r i n t * , ’ z= ’ , z , ’ Re ( z )= ’ ,REAL( z ) , ’ Im( z )= ’ ,IMAG( z ) ,&
’ | z | = ’ ,ABS( z ) , ’ z *= ’ , CONJG( z )
! Complex numbers : double p r e c i s i o n
z8 = ( 2 . 0 D0 , 1 . 0 D0 ) * cdexp ( ( 3 . 0 D0 , −1.0 D0 ) )
p r i n t * , ’ z= ’ , z8 , ’ Re ( z )= ’ ,DBLE( z8 ) , ’ Im( z )= ’ ,DIMAG( z8 ) ,&
’ | z | = ’ ,CDABS( z8 ) , ’ z *= ’ ,DCONJG( z8 )
p r i n t * , ’ z4 : ’ , range ( z ) , p r e c i s i o n ( z )
p r i n t * , ’ z8 : ’ , range ( z8 ) , p r e c i s i o n ( z8 )
end program f90_vars
Some interesting points of the program above are:
• The number K in the declaration REAL(K):: x refers to the number
of bytes allocated to the variable x. For K=4 we have single precision
(same as REAL), for K=8 double precision and for K=16 quadruple
precision. The latter is not always available. In the declarations
COMPLEX(K), K refers to the number of bytes allocated to the real
and imaginary parts of the complex number.
• We always use the exponent notation D in double precision con-
stants, even if the exponent 0. Otherwise the constants are of single
precision and we loose the desired accuracy.
1.4. THE FORTRAN PROGRAMMING LANGUAGE 47
• When we want to state the precision of the return value of an in-
trinsic function explicitly, we usually add a d at the beginning of its
name (e.g. exp→dexp, ABS→DABS. When we want to use the com-
plex version of a function, we usually add a c at the beginning of its
name (e.g. exp→cexp, ABS→CABS). Modify the program in order
to achieve higher accuracy in the calculation of π and z = (2+i)e3−i ,
by using double precision variables.
• The maximum number of characters in the CHARACTER variable
string is 100, and this is declared by the statement CHARACTER(100).
• When we print a CHARACTER variable, all its characters are printed,
including trailing blanks. This is very annoying and we can use
the function TRIM in order to remove them.
• The operator // joins two CHARACTER variables or constants. Notice
the effect of the function TRIM in the above program.
Another important point to discuss is how to be able to access the same
variables from different program units. So far, we simply mentioned that
variables have a scope within each function and subroutine. If we wish
to have access to the same location of memory⁴⁰ from different program
units, then we use the COMMON statement which defines a common block.
See the following example:
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program f90_common
i m p l i c i t none
r e a l : : k1 = 1 . 0 , k2 = 1 . 0 , k3 =1.0
common / CONSTANTS / k1 , k2
p r i n t * , ’ main : k1= ’ , k1 , ’ k2= ’ , k2 , ’ k3= ’ , k3
c a l l s1 ! p r i n t s k1 and k2 but not k3
c a l l s2 ! changes t h e v a l u e o f k2 but not k3
p r i n t * , ’ main : k1= ’ , k1 , ’ k2= ’ , k2 , ’ k3= ’ , k3
end program f90_common
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
s u b r o u t i n e s1 ( )
i m p l i c i t none
r e a l k1 , k2 , k3
⁴⁰Common blocks are supposed to be obsolescent in Fortran and programmers are
encouraged to avoid them and use modules instead. Due to their simplicity and pop-
ularity we will show their usage and also use them in this book.
48 CHAPTER 1. THE COMPUTER
common / CONSTANTS / k1 , k2
p r i n t * , ’ s 1 : k1= ’ , k1 , ’ k2= ’ , k2 , ’ k3= ’ , k3
end s u b r o u t i n e s1
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
s u b r o u t i n e s2 ( )
i m p l i c i t none
r e a l k1 , k2 , k3
common / CONSTANTS / k1 , k2
k2 = 2.0
k3 = 2.0
end s u b r o u t i n e s2
The common block has the name CONSTANTS and we can refer to it from
any program unit. Each program unit that uses this common block must
use the same declaration, although the names of variables are allowed to
be different. The common block CONSTANTS points to the same location
in the computer memory, where we expect to find the values of two real
variables. These variables (k1 and k2) are used and have their values
changed in the subroutines s1 and s2. The variable k3, is a different
variable in each program unit. The program prints
main : k1= 1.000000 k2= 1.000000 k3= 1.000000
s1 : k1= 1.000000 k2= 1.000000 k3= −2.8117745E−05
main : k1= 1.000000 k2= 2.000000 k3= 1.000000
One of the weaknesses of Fortran is that it does not have a convenient
control for Input/Output (I/O). For complicated I/O and text manipulation
we will use other programs that can do a better job, like awk, perl,
shell scripting, or programs written in C/C++. It is important to know
some details about I/O commands in Fortran, mainly the specifications
that control the accuracy of printed floating point numbers. So far, I/O
commands, like print, write, read, used a * in order to control the
printing of numbers. But we can replace the * with explicit format
directives as follows:
program f90_format1
i m p l i c i t none
integer :: i
real :: x
r e a l , dimension ( 1 0 ) :: a
real (8) :: x8
1.4. THE FORTRAN PROGRAMMING LANGUAGE 49
i = 123456
x = 2.0 * atan2 ( 1 . 0 , 0 . 0 )
p r i n t ’ (A5 , I6 , F12 . 7 ) ’ , ’ x , i = ’ , i , x
x8 = 2.0 D0 * atan2 ( 1 . 0 D0 , 0 . 0 D0 )
w r i t e ( 6 , ’ ( F18 . 1 6 , E24 . 1 7 , G24 . 1 7 , G24 . 1 7 ) ’ ) x8 , x8 ,&
1 . 0 D15 * x8 , 1 . 0 D18 * x8
w r i t e ( 6 , ’ (3F20 . 1 6 ) ’ ) x8 , x8 / 2 . 0 , c o s ( x8 )
w r i t e ( 6 , ’ (200F12 . 6 ) ’ ) ( a ( i ) , i = 1 , 1 0 )
end program f90_format1
Note the parentheses within the single quotes: (A5,I6,F12.7) is a format
directive for the print statement. The A is for printing a CHARACTER,
the I for printing an INTEGER and the F for printing a floating point
number. The numbers after the letter declare the number of spaces used
for printing each one. Beware! If the printing space is not enough,
Fortran will not print and you will find a series of * in place of the
value of your result! Bummer... In order to estimate the number of
spaces needed for a floating point number, you have to include the space
taken by the decimal point, the sign, the exponent character, the sign of
the exponent and the digits needed for the exponent. Plus a space to
separate the numbers in between... So, be generous and give plenty of
printing space. In the example shown above, A5 denotes a character of 5
spaces, I6 and integer of 6 spaces and F12 a floating point number of 12
spaces. The decimal point in F12.7 means that we want a floating point
with the accuracy of 7 significant digits.
The format directive (F18.16,E24.17,G24.17,G24.17) shows how to
print double precision variables. These provide an accuracy of 16-17
significant digits and there is no need for keeping more digits. The
command E prints a number in scientific form with an exponent. The
command G prints the exponent when it is needed. The numbers before
the letters denote multiplicity. Therefore 3F20.16 instructs the printing of
3 floating point numbers by reserving 20 spaces and using 16 significant
digits for each one of them.
The command write(6,'(200F12.6)')(a(i), i=1,10) shows how to
print a large array using an implicit loop. We used many more spaces
than actually needed (200F12.16) which is OK. If the array gets larger
by increasing the range of i, then we will have enough room for printing
in the same line. The program prints (we have folded the long line in
order to make it visible):
x , i= 123456 3.1415927
3.1415926535897931 0.31415926535897931E+01 3141592653589793.0
50 CHAPTER 1. THE COMPUTER
0.31415926535897933E+19
3.1415926535897931 1.5707963267948966 −1.0000000000000000
0.000000 0.000000 0.000000 . . . .
We can organize the format commands by using the FORMAT statement.
Then, we use labeled statements in order to refer to them. Labels are
numbers put in the beginning of a line which should be unique to a
program unit and are within the range 1-99999. We can transfer the
control of the program to such a line with a goto command or by using
the label in the I/O statements print, write and read as in the example
shown below:
program f90_format2
i m p l i c i t none
integer i
real x , a (10)
r e a l *8 x8
i = 123456
x = 2.0 * atan2 ( 1 . 0 , 0 . 0 )
p r i n t 100 , ’ x , i = ’ , i , x
x8 = 2.0 D0 * atan2 ( 1 . 0 D0 , 0 . 0 D0 )
w r i t e ( 6 , 1 2 3 ) x8 , x8 ,&
1 . 0 D15 * x8 , 1 . 0 D18 * x8
w r i t e ( 6 , 4 4 4 4 ) x8 , x8 / 2 . 0 , c o s ( x8 )
w r i t e (6 ,9999) ( a ( i ) , i = 1 , 1 0 )
100 FORMAT( A5 , I6 , F12 . 7 )
123 FORMAT( F18 . 1 6 , E24 . 1 7 , G24 . 1 7 , G24 . 1 7 )
4444 FORMAT(3 F20 . 1 6 )
9999 FORMAT(200 F12 . 6 )
end program f90_format2
The reader should also study the Fortran intrinsic functions shown
in table 1.2, page 71.
1.4.3 Arrays
You may skip this section during the first reading of this book. It will
be useful to come back here later.
Arrays are related data of the same type which can be accessed
by using one or more indices. For example, after a declaration real,
dimension(10) :: A, the expressions
A (1) , A (2) , . . . , A (10)
1.4. THE FORTRAN PROGRAMMING LANGUAGE 51
refer to its 10 real values. The indices can be integer expressions, for
example
A ( i ) , B ( 2 * i+3) , C ( INT ( x+y ( j ) ) )
where in the last case we used the integer value of the intrinsic function
INT(x), which returns the integer part of x. Note that, arrays and func-
tions enclose indices and arguments between parentheses (...) which
are of the same style, and the compiler must look at their declarations in
order to tell the difference. Examples of array declarations are
r e a l , dimension ( 1 0 ) : : a , b
r e a l , dimension (20) : : c , d
which declare the arrays a, b, c, d, which are of the real kind, with
elements a(1) ... a(10), b(1) ... b(10), c(1) ... c(20) and d(1)
... d(20). An equivalent declaration is
r e a l : : a ( 1 0 ) , b ( 1 0 ) , c (20) , d (20 )
or
i n t e g e r , parameter : : n1 = 1 0 , n2 = 20
real , dimension ( n1 ) : : a , c ( n2 )
real : : b ( n1 ) , d ( n2 )
In the last form, we show how to use constant parameters for declaring
the size of arrays. For the declarations shown above, the lower bound of
all arrays is 1 and the upper bound for a and b is 10 and for c and d is
20. The upper and lower bound of arrays can be explicitly determined.
The declarations
i n t e g e r , parameter : : n1 = 1 0 , n2 = 20
real , dimension ( 0 : n1 ) :: a
real , dimension(−n1 : n2 ) :: c
define the real array a with 11 values a(0) ... a(10) and the array c
with 31 values c(-10) c(-9) ... c(-1) c(0) c(1) ... c(20).
The arrays shown above have dimension 1 and they are like vectors.
We can declare arrays of more than one dimension. This means that we
52 CHAPTER 1. THE COMPUTER
need more than one indices in order to determine an array element⁴¹.
Therefore, the declaration
i n t e g e r , dimension ( 2 , 2 ) :: a
defines an integer array with values a(1,1), a(1,2), a(2,1) and a(2,2).
The following declarations define two three dimensional real arrays a and
b:
i n t e g e r , parameter : : n1 = 1 0 , n2 = 20 , n3 = 2* n1+n2
r e a l , dimension ( n1 , n2 , n3 ) :: a
r e a l , dimension(−n1 : n1 , 0 : n2 , 1 3 : n3 ) : : b
Some important definitions used in the bibliography are:
• array: Variables of the same type to which we refer with one or
more indices. Variables with only one value are called scalar.
• An array’s dimension has an upper bound and a lower bound
which define the allowed range of index values. If the lower bound
is omitted in a declaration, then it takes the value 1.
• The rank of an array is the number of its dimensions, i.e. the
number of indices needed to determine its values.
• The extent of a dimension it the number of its elements. It is equal
to (upper bound)-(lower bound)+1.
• The size of an array is the total number of its elements. For a one
dimensional array, its size is equal to its extent, whereas for a multi
dimensional one, it is equal to the product of the extents of all of
its dimensions.
• The shape of an array is its rank and extents of all its dimensions.
The values of arrays can be set the same way as scalars:
integer : : i
real : : a (4) , b (2 ,2)
b ( 1 , 1 ) = 2.0 ; b ( 1 , 2 ) = 4.0
b ( 2 , 1 ) = 3.4 ; b ( 2 , 2 ) = 7 . 8
⁴¹Fortran allows up to seven indices in an array.
1.4. THE FORTRAN PROGRAMMING LANGUAGE 53
do i =1 ,4
a(i) = 1.0
enddo
Alternatively we can use the name of the array as one object:
a = ( / 1 . 0 , 2 . 0 , 3 . 0 , 4.0 / )
b = 0.0
The first line defines the values of an array by using an array constructor.
The second line defines all elements of the array b to be equal to 0. This
is an example of a very convenient feature of the Fortran language. If all
the arrays in an expression are conformable, then we can use the intrinsic
Fortran operations to act on whole arrays. Two arrays are conformable
if they have the same shape or if one of them is a scalar. Therefore the
program
integer : : i , j
real : : x , y , a (10) , b (10) , c (4 ,4) , d (4 ,4)
do i =1 ,10
a(i) = b(i)
enddo
do j =1 ,4
do i =1 ,4
c ( i , j ) = x * d ( i , j )+y
enddo
enddo
is equivalent to
integer : : i , j
real : : x , y , a (10) , b (10) , c (4 ,4) , d (4 ,4)
a = b
c = x * d+y
Many Fortran intrinsic functions are elemental. This means that their
arguments can be arrays, in which case the function acts on each array
element separately. For example, the commands
integer : : i , j
real : : x , y , a (10) , b (10) , c (4 ,4) , d (4 ,4)
54 CHAPTER 1. THE COMPUTER
c = s i n ( d ) + x * exp ( −2.0* d )
c a l l random_number ( a )
set c(i,j) = sin(d(i,j))+x*exp(-2.0*d(i,j)) for all i and j, and the
elements of a(i) equal to a random number uniformly distributed in
the interval [0, 1). We should stress that in order for two arrays to be
conformable, it is not necessary that they have the same lower and upper
bounds. For example, the command b=c*d in the following program has
the same effect as the do loop:
integer : : i
real : : b ( 0 : 1 9 ) , c ( 1 0 : 2 9 ) , d ( −9:10)
b = c*d
do i =1 ,20
b ( i−1) = c ( i+9) * d ( i−10)
enddo
In the following, we mention some useful functions that act on arrays.
Assume that
real : : a ( −10:10) , b ( −10:10) , c ( 1 0 , 1 0 ) , d ( 1 0 , 1 0 ) , e ( 1 0 , 1 0 )
then
• LBOUND(a) and UBOUND(a) return the lower bound and the upper
bound of the array a. In the above example LBOUND(a) = -10 and
UBOUND(a) = 10.
• c = TRANSPOSE(d) sets c(i,j)=d(j,i).
• e = MATMUL(c,d)
∑ sets the array e equal to the matrix product c, d.
I.e. e(i,j)= 10
k=1 c(i,k)*d(k,j). Be careful, the command e=c*d
sets e(i,j)=c(i,j)*d(i,j).
• SUM(a) computes
∑ the sum of all the elements of a.
I.e. SUM(a) = 10
i=−10 a(i)
• PRODUCT(a) computes
∏10 the product of all the elements of a.
I.e. PRODUCT(a) = i=−10 a(i)
• DOT_PRODUCT(a,b) computes
∑10the inner product of a, b.
I.e. DOT_PRODUCT(a,b) = i=−10 a(i)*b(i)
1.4. THE FORTRAN PROGRAMMING LANGUAGE 55
• MAXVAL(a) and MINVAL(a) return the maximum and minimum val-
ues in the array a respectively.
You can find more functions and documentation in the bibliography
[11, 10]. In the following, we provide some information related to the
Input/Output (I/O) of arrays. Input (“reading”) and output (“writing”)
of array values can be done by reading and writing their elements in any
order we want. In the example below, we read the array a and write the
array b in two different ways:
integer : : i , j
real : : a (4) , b (2 ,2)
do i =1 ,4
read * , a ( i )
enddo
read * , ( a ( i ) , i = 1 , 4 )
do j =1 ,2
do i =1 ,2
print * , b(i , j)
enddo
enddo
p r i n t * , ( ( b ( i , j ) , i =1 ,2) , j =1 ,2 )
Inside the do loops, input and output is done one element per line from/to
standard input/output. The commands (a(i), i=1,4) and ( (b(i,j)
i=1,2), j=1,2) are implied do loops and read/write from/to the same
line. During input, if the number of values for a are exhausted, then the
program tries to read values from the following line(s). Similarly, if the
output of b exhausts the maximum number of characters per line, then
the output continues in the next line⁴². Try it...
We can also preform I/O of arrays without explicit reference to their
elements. In this case, the arrays are read/written in a specified order.
For example, the program
real : : a (4) , b (2 ,2)
read *, a
read *, b
⁴²If we want to force a long output to be written in one line, then we must replace the
* by an explicit format directive, e.g. print '(100I6)',( (c(i,j), i=1,10), j=1,10)
56 CHAPTER 1. THE COMPUTER
print * , a , b
reads the values a(1) a(2) a(3) a(4) from the stdin. Then, it continues
reading b(1,1), b(2,1), b(1,2), b(2,2) from the next line (record).
Notice that the array b is read in a column major way. Printing a and b,
will print a(1) a(2) a(3) a(4) and b(1,1), b(2,1), b(1,2), b(2,2)
in two different records (also in column major mode).
Finally, we summarize some of the Fortran capabilities in array ma-
nipulation. More details can be found in the bibliography. Read the
comments in the program for a partial explanation of each command:
program arrays
i m p l i c i t none
integer : : i , j , n , m
real : : a ( 3 ) , b ( 3 , 3 ) , c ( 3 , 3 ) = −99.0 , d ( 3 , 3 ) = −99.0 , s
i n t e g e r : : semester (1000) , grade (1000)
l o g i c a l : : pass (1000)
! c o n s t r u c t t h e matrix : use t h e RESHAPE f u n c t i o n
! | 1 . 1 −1.2 −1.3|
! | 2 . 1 2.2 −2.3|
! | 3 . 1 3.2 3 . 3 |
b = RESHAPE ( ( / 1 . 1 , 2 . 1 , 3 . 1 , & ! ( n o t i c e rows<−>columns )
−1.2 , 2 . 2 , 3 . 2 , &
−1.3 , −2.3 , 3.3 / ) , ( / 3 , 3 / ) )
! same matrix , now exchange rows and columns : ORDER= ( / 2 , 1 / )
b = RESHAPE ( ( / 1 . 2 , −1.2 , −1.3 , &
2 . 1 , 2 . 2 , −2.3 , &
3 . 1 , 3 . 2 , 3.3 / ) , ( / 3 , 3 / ) , ORDER = ( / 2 , 1 / ) )
a = b ( : , 2 ) ! a a s s i g n e d t h e second column o f b : a ( i )=b ( i , 2 )
a = b ( 1 , : ) ! a a s s i g n e d t h e f i r s t row o f b : a ( i )=b ( 1 , i )
a = 2 . 0 * b ( : , 3 ) + s i n ( b ( 2 , : ) ) ! a ( i )= 2* b ( i , 3 ) + s i n ( b ( 2 , i ) )
a = 1 . 0 + 2 . 0 * exp(−a )+b ( : , 3 ) ! a ( i )= 1+2* exp(−a ( i ) )+b ( i , 3 )
s = SUM( b ) ! r e t u r n s sum o f a l l elements of b
s = SUM( b , MASK =(b . g t . 0 ) ) ! r e t u r n s sum o f p o s i t i v e e l e m e n t s o f b
a = SUM( b , DIM=1) ! each a ( i ) i s t h e sum o f t h e columns o f b
a = SUM( b , DIM=2) ! each a ( i ) i s t h e sum o f t h e rows of b
! r e p e a t a l l t h e above using PRODUCT!
! a l l i n s t r u c t i o n s may be e x e c u t e d i n p a r a l l e l a t any order !
FORALL( i = 1 : 3 ) c ( i , i ) = a ( i ) ! s e t t h e d i a g o n a l o f c
! compute upper bounds o f i n d i c e s i n b :
n=UBOUND( b , DIM=1) ; m=UBOUND( b , DIM=2)
! l o g needs p o s i t i v e argument , add a r e s t r i c t i o n ( ” mask ” )
FORALL( i =1: n , j =1: m , b ( i , j ) . g t . 0 . 0 ) c ( i , j ) = l o g ( b ( i , j ) )
! upper t r i a n g u l a r p a r t o f ma t rix :
! c a r e f u l , j = i +1 :m NOT p e r m i t t e d
FORALL( i =1: n , j =1: m , i . l t . j ) c(i , j) = b(i , j)
! each s t a t e m e n t e x e c u t e d BEFORE t h e n e x t one !
1.5. GNUPLOT 57
FORALL( i =2:n −1 , j =2:n−1)
! a l l r i g h t hand s i d e e v a l u a t e d BEFORE t h e assignment
! i . e . , t h e OLD v a l u e s o f b averaged and then a s s i g n e d t o b
b ( i , j ) =(b ( i +1 , j )+b ( i −1 , j )+b ( i , j +1)+b ( i , j−1) ) / 4 . 0
c ( i , j ) = 1 . 0 / b ( i +1 , j +1) ! t h e NEW v a l u e s o f b a r e a s s i g n e d
END FORALL
! assignment but only f o r el e m e n t s b ( i , j ) which a r e not 0
WHERE ( b . ne . 0 . 0 ) c = 1.0/b
!MATMUL( b , c ) i s e v a l u a t e d , then d i s a s s i g n e d t h e r e s u l t only
! a t p o s i t i o n s where b >0.
WHERE ( b . gt . 0.0) d = MATMUL( b , c )
WHERE ( grade . ge . 5 )
semester = semester + 1 ! student ’ s s e m e s t e r i n c r e a s e s by 1
pass = . true .
ELSEWHERE
pass = . false .
END WHERE
end program arrays
The code shown above can be found in the file f90_arrays.f90 of the
accompanying software.
1.5 Gnuplot
Plotting data is an indispensable tool for their qualitative, but also quanti-
tative, analysis. Gnuplot is a high quality, open source, plotting program
that can be used for generating publication quality plots, as well as for
heavy duty analysis of a large amount of scientific data. Its great ad-
vantage is the possibility to use it from the command line, as well as
from shell scripts and other programs. Gnuplot is programmable and it
is possible to call external programs in order manipulate data and cre-
ate complicated plots. There are many mathematical functions built in
gnuplot and a fit command for non linear fitting of data. There exist
interactive terminals where the user can transform a plot by using the
mouse and keyboard commands.
This section is brief and only the features, necessary for the fol-
lowing chapters, are discussed. For more information visit the offi-
cial page of gnuplot http://gnuplot.info. Try the rich demo gallery
at http://gnuplot.info/screenshots/, where you can find the type of
graph that you want to create and obtain an easy to use recipe for it. The
book [14] is an excellent place to look for many of gnuplot’s secrets⁴³.
⁴³A the time of the writing of this book, there was a very nice site
www.gnuplotting.org which shows how to create many beautiful and complicated
58 CHAPTER 1. THE COMPUTER
You can start a gnuplot session with the gnuplot command:
> gnuplot
G N U P L O T
Version X . XX
....
The gnuplot FAQ is available from www . gnuplot . i n f o / faq /
....
Terminal type s e t to ’ wxt ’
gnuplot >
There is a welcome message and then a prompt gnuplot> is issued wait-
ing for your command. Type a command an press [Enter]. Type quit
in order to quit the program. In the following, when we show a prompt
gnuplot>, it is assumed that the command after the prompt is executed
from within gnuplot.
Plotting a function is extremely easy. Use the command plot and x
as the independent variable of the function⁴⁴. The command
gnuplot > p l o t x
plots the function y = f (x) = x which is a straight line with slope 1. In
order to plot many functions simultaneously, you can write all of them
in one line:
gnuplot > p l o t [ −5:5][ −2:4] x , x * * 2 , s i n ( x ) , b e s j 0 ( x )
The above command plots the functions x, x2 , sin x and J0 (x). Within the
square brackets [:], we set the limits of the x and y axes, respectively. The
bracket [-5:5] sets −5 ≤ x ≤ 5 and the bracket [-2:4] sets −2 ≤ y ≤ 4.
You may leave the job of setting such limits to gnuplot, by omitting some,
or all of them, from the respective positions in the brackets. For example,
typing [1:][:5] changes the lower and upper limits of x and y and leaves
the upper and lower limits unchanged⁴⁵.
plots.
⁴⁴You can change the symbol of the independent variable. For example, the command
set dummy t sets the independent variable to be t.
⁴⁵By default, the x and y ranges are determined automatically. In order to force them
to be automatic, you can insert a * in the brackets at the corresponding position(s). For
example plot [1:*][*:5] sets the upper and lower limits of x and y to be determined
automatically.
1.5. GNUPLOT 59
In order to plot data points (xi , yi ), we can read their values from files.
Assume that a file data has the following numbers recorded in it:
# x y1 y2
0.5 1.0 0.779
1.0 2.0 0.607
1.5 3.0 0.472
2.0 4.0 0.368
2.5 5.0 0.287
3.0 6.0 0.223
The first line is taken by gnuplot as a comment line, since it begins with
a #. In fact, gnuplot ignores everything after a #. In order to plot the
second column as a function of the first, type the command:
gnuplot > p l o t ” data ” using 1 : 2 with points
The name of the file is within double quotes. After the keyword using,
we instruct gnuplot which columns to use as the x and y coordinates,
respectively. The keywords with points instructs gnuplot to add each
pair (xi , yi ) to the plot with points.
The command
gnuplot > p l o t ” data ” using 1 : 3 with lines
plots the third column as a function of the first, and the keywords with
lines instruct gnuplot to connect each pair (xi , yi ) with a straight line
segment.
We can combine several plots together in one plot:
gnuplot > p l o t ” data ” using 1 : 3 with points , exp ( −0.5* x )
gnuplot > r e p l o t ” data ” using 1 : 2
gnuplot > r e p l o t 2* x
The first line plots the 1st and 3rd columns in the file data together with
the function e−x/2 . The second line adds the plot of the 1st and 2nd
columns in the file data and the third line adds the plot of the function
2x.
There are many powerful ways to use the keyword using. Instead of
column numbers, we can put mathematical expressions enclosed inside
brackets, like using (...):(...). Gnuplot evaluates each expression
within the brackets and plots the result. In these expressions, the values
60 CHAPTER 1. THE COMPUTER
of each column in the file data are represented as in the awk language. $i
are variables that expand to the number read from columns i=1,2,3,....
Here are some examples:
gnuplot > p l o t ” data ” using 1 : ( $2 * s i n ( $1 ) * $3 ) with points
gnuplot > r e p l o t 2* x * s i n ( x ) * exp(−x / 2 )
The first line plots the 1st column of the file data together with the
value yi sin(xi )zi , where yi , xi and zi are the numbers in the 2nd, 1st and
3rd columns respectively. The second line adds the plot of the function
2x sin(x)e−x/2 .
gnuplot > p l o t ” data ” using ( l o g ( $1 ) ) : ( l o g ( $2 * * 2 ) )
gnuplot > r e p l o t 2* x+ l o g ( 4 )
The first line plots the logarithm of the 1st column together with the
logarithm of the square of the 2nd column.
We can plot the data written to the standard output of any command.
Assume that there is a program called area that prints the perimeter and
area of a circle to the stdout in the form shown below:
> . / area
R= 3.280000 area= 33.79851
R= 6.280000 area= 123.8994
R= 5.280000 area= 87.58257
R= 4.280000 area= 57.54895
The interesting data is at the second and fourth columns. These can be
plotted directly with the gnuplot command:
gnuplot > p l o t ”< . / a r e a ” using 2:4
All we have to do is to type the full command after the < within the
double quotes. We can create complicated filters using pipes as in the
following example:
gnuplot > p l o t \
”< . / a r e a | s o r t −g −k 2 | awk ’{ p r i n t l o g ( $2 ) , l o g ( $4 ) } ’ ” \
using 1 : 2
The filter produces data to the stdout, by combining the action of the
commands area, sort and awk. The data printed by the last program is
1.5. GNUPLOT 61
in two columns and we plot the results using 1:2.
In order to save plots in files, we have to change the terminal that gnu-
plot outputs the plots. Gnuplot can produce plots in several languages
(e.g. PDF, postscript, SVG, LATEX, jpeg, png, gif, etc), which can be inter-
preted and rendered by external programs. By redirecting the output to
a file, we can save the plot to the hard disk. For example:
gnuplot > p l o t ” data ” using 1 : 3
gnuplot > s e t t e r m i n a l jpeg
gnuplot > s e t output ” data . j p g ”
gnuplot > replot
gnuplot > s e t output
gnuplot > s e t t e r m i n a l wxt
The first line makes the plot as usual. The second one sets the output
to be in the JPEG format and the third one sets the name of the file to
which the plot will be saved. The fourth lines repeats all the previous
plotting commands and the fifth one closes the file data.jpg. The last
line chooses the interactive terminal wxt to be the output of the next
plot. High quality images are usually saved in the PDF, encapsulated
postcript or SVG format. Use set terminal pdf,postscript eps or svg,
respectively.
And now a few words for 3-dimensional (3d) plotting. The next
example uses the command splot in order to make a 3d plot of the
function f (x, y) = e−x −y . After you make the plot, you can use the
2 2
mouse in order to rotate it and view it from a different perspective:
gnuplot > s e t pm3d
gnuplot > s e t hidden3d
gnuplot > s e t s i z e ratio 1
gnuplot > s e t i s o s a m p l e s 50
gnuplot > s p l o t [ −2:2][ −2:2] exp(−x**2−y * * 2 )
If you have data in the form (xi , yi , zi ) and you want to create a plot
of zi = f (xi , yi ), write the data in a file, like in the following example:
−1 −1 2.000
−1 0 1.000
−1 1 2.000
0 −1 1.000
0 0 0.000
0 1 1.000
62 CHAPTER 1. THE COMPUTER
1 −1 2.000
1 0 1.000
1 1 2.000
Note the empty line that follows the change of the value of the first
column. If the name of the file is data3, then you can plot the data with
the commands:
gnuplot > s e t pm3d
gnuplot > s e t hidden3d
gnuplot > s e t s i z e ratio 1
gnuplot > s p l o t ” data3 ” with lines
We close this section with a few words on parametric plots. A para-
metric plot on the plane (2-dimensions) is a curve (x(t), y(t)), where t
is a parameter. A parametric plot in space (3-dimensions) is a surface
(x(u, v) , y(u, v), z(u, v)), where (u, v) are parameters. The following com-
mands plot the circle (sin t, cos t) and the sphere (cos u cos v, cos u sin v,
sin u):
gnuplot > s e t p a r a m e t r i c
gnuplot > p l o t s i n ( t ) , c o s ( t )
gnuplot > s p l o t c o s ( u ) * c o s ( v ) , c o s ( u ) * s i n ( v ) , s i n ( u )
1.6 Shell Scripting
Complicated system administration tasks are not among the strengths of
the Fortran programming language. But in a typical GNU/Linux envi-
ronment, there exist many powerful tools that can be used very effectively
for this purpose. This way, one can use Fortran for the high performance
and scientific computing part of the project and leave the administration
and trivial data analysis tasks to other, external, programs.
One can avoid repeating the same sequence of commands by coding
them in a file. An example can be found in the file script01.csh:
# ! / bin / t c s h −f
g f o r t r a n area_01 . f90 −o area
. / area
g f o r t r a n area_02 . f90 −o area
. / area
1.6. SHELL SCRIPTING 63
g f o r t r a n area_03 . f90 −o area
. / area
g f o r t r a n area_04 . f90 −o area
. / area
This is a very simple shell script. The first line instructs the operating
system that the lines that follow are to be interpreted by the program
/bin/tcsh⁴⁶. This can be any program in the system, which in our case
is the tcsh shell. The following lines are valid commands for the shell,
one in each line. They compile the Fortran programs found in the files
that we created in section 1.4 with gfortran, and then they run the
executable ./area. In order to execute the commands in the file, we
have to make sure that the file has the appropriate execute permissions.
If not, we have to give the command:
> chmod u+x script01 . csh
Then we simply type the path to the file script01.csh
> . / script01 . csh
and the above commands are run the one after the other. Some of the
versions of the programs that we wrote are asking for input from the
stdin, which, normally, you have to type on the terminal. Instead of
interacting directly with the program, we can write the input data to a
file Input, and run the command
. / area < Input
A more convenient solution is to use the, so called, “Here Document”. A
“Here Document” is a section of the script that is treated as if it were a
separate file. As such, it can be used as input to programs by sending its
“contents” to the stdin of the command that runs the program⁴⁷. The
“Here Document” does not appear in the filesystem and we don’t need to
administer it as a regular file. An example of using a “Here Document”
can be found in the file script02.csh:
# ! / bin / t c s h −f
⁴⁶Use #!/bin/bash if you prefer the bash shell.
⁴⁷Their great advantage is that we can use variable and command substitution in
them, therefore sending this information to the program that we want to run.
64 CHAPTER 1. THE COMPUTER
g f o r t r a n area_04 . f90 −o area
. / area <<EOF
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
EOF
The stdin of the command ./area is redirected to the contents between
the lines
. / area <<EOF
...
EOF
The string EOF marks the beginning and the end of the “Here Document”,
and can be any string you like. The last EOF has to be placed exactly in
the beginning of the line.
The power of shell scripting lies in its programming capabilities: Vari-
ables, arrays, loops and conditionals can be used in order to create a
complicated program. Shell variables can be used as discussed in section
1.1.2: The value of a variable name is $name and it can be set with the
command set name = value. An array is defined, for example, by the
command
s e t R = ( 1 . 0 2.0 3.0 4.0 5.0 6.0 7 . 0 8.0 9.0 1 0 . 0 )
and its data can be accessed using the syntax $R[1] ... $R[10].
Lets take a look at the following script:
# ! / bin / t c s h −f
s e t files = ( area_01 . f90 area_02 . f90 area_03 . f90 area_04 . f90 )
set R = ( 1 . 0 2.0 3.0 4.0 5.0 6.0 7 . 0 8.0 9.0 1 0 . 0 )
echo ” H e l l o $USER Today i s ” ‘ date ‘
f o r e a c h file ( $files )
echo ” # −−−−−−−−−−− Working on f i l e $ f i l e ”
1.6. SHELL SCRIPTING 65
g f o r t r a n $file −o area
. / area <<EOF
$R [ 1 ]
$R [ 2 ]
$R [ 3 ]
$R [ 4 ]
$R [ 5 ]
$R [ 6 ]
$R [ 7 ]
$R [ 8 ]
$R [ 9 ]
$R [ 1 0 ]
EOF
echo ” # −−−−−−−−−−− Done ”
i f ( −f AREA . DAT ) c a t AREA . DAT
end
The first two lines of the script define the values of the arrays files (4
values) and R (10 values). The command echo echoes its argument to
the stdin. $USER is the name of the user running the script. `date` is an
example of command substitution: When a command is enclosed between
backquotes and is part of a string, then the command is executed and its
stdout is pasted back to the string. In the example shown above, `date`
is replaced by the current date and time in the format produced by the
date command.
The foreach loop
f o r e a c h file ( $files )
...
end
is executed once for each of the 4 values of the array files. Each time the
value of the variable file is set equal to one of the values area_01.f90,
area_02.f90, area_03.f90, area_04.f90. These values can be used by
the commands in the loop. Therefore, the command gfortran $file -o
area compiles a different file each time that it is executed by the loop.
The last line in the loop
i f ( −f AREA . DAT ) c a t AREA . DAT
is a conditional. It executes the command cat AREA.DAT if the condition
-f AREA.DAT is true. In this case, -f constructs a logical expression which
is true when the file AREA.DAT exists.
We close this section by presenting a more complicated and advanced
66 CHAPTER 1. THE COMPUTER
script. It only serves as a demonstration of the shell scripting capabilities.
For more information, the reader is referred to the bibliography [16, 17,
18,19,20]. Read carefully the commands, as well as the comments which
follow the # mark. Then, write the commands to a file script04.csh⁴⁸,
make it an executable file with the command chmod u+x script04.csh
and give the command
> . / script04 . csh This is my first serious tcsh script
The script will run with the words “This is my first serious tcsh script”
as its arguments. Notice how these arguments are manipulated by the
script. Then, the script asks for the values of the radii of ten or more
circles interactively, so that it will compute their perimeter and area. Type
them on the terminal and then observe the script’s output, so that you
understand the function of each command. You will not regret the time
investment!
# ! / bin / t c s h −f
# Run t h i s s c r i p t as :
# . / s c r i p t 0 4 . csh H e l l o t h i s i s a t c s h s c r i p t
#−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
# ‘command‘ i s command s u b s t i t u t i o n : i t i s r e p l a c e d by s t d o u t o f command
s e t now = ‘ date ‘ ; s e t mypc = ‘ uname −a ‘
# P r i n t i n f o r m a t i o n : v a r i a b l e s a r e expanded w i t h i n double q u o t e s
echo ” I am u s e r $user working on t h e computer $HOST” #HOST i s p r e d e f i n e d
echo ”Today t h e d a t e i s : $now” #now i s d e f i n e d above
echo ”My home d i r e c t o r y i s : $home” #home i s p r e d e f i n e d
echo ”My c u r r e n t d i r e c t o r y i s : $cwd” #cwd changes with cd
echo ”My computer runs : $mypc” #mypc i s d e f i n e d above
echo ”My p r o c e s s i d i s : $$ ” #$$ i s predefined
# Manipulate t h e command l i n e : ( $# argv i s number o f e l e m e n t s i n a r r a y argv )
echo ”The command l i n e has $# argv arguments ”
echo ”The name o f t h e command I am running i s : $0 ”
echo ” Arguments 3rd t o l a s t o f t h e command : $argv [3 −] ” # third to l a s t
echo ”The l a s t argument i s : $argv [ $# argv ] ” # l a s t element
echo ” A l l arguments : $argv ”
# Ask u s e r f o r i n p u t : e n t e r r a d i i o f c i r c l e s
echo −n ” Enter r a d i i o f c i r c l e s : ” # v a r i a b l e $< s t o r e s one l i n e o f i n p u t
s e t Rs = ( $ <) #Rs i s now an a r r a y with a l l words e n t e r e d by u s e r
i f ( $#Rs < 10 ) then #make a t e s t , need a t l e a s t 10 o f them
echo ”Need more than 10 r a d i i . E x i t i n g . . . . ”
exit (1)
endif
echo ”You e n t e r e d $#Rs r a d i i , t h e f i r s t i s $Rs [ 1 ] and t h e l a s t $Rs [ $#Rs ] ”
echo ” Rs= $Rs ”
# Now, compute t h e p e r i m e t e r o f each c i r c l e :
f o r e a c h R ( $Rs )
# −v rad=$R s e t t h e awk v a r i a b l e rad equal t o $R . p i=atan2 (0 , −1) = 3 . 1 4 . . .
s e t l = ‘awk −v rad=$R ’BEGIN{ p r i n t 2* atan2 (0 , −1) * rad } ’ ‘
⁴⁸You will find it also in the accompanying software
1.6. SHELL SCRIPTING 67
echo ” C i r c l e with R= $R has p e r i m e t e r $ l ”
end
# a l i a s d e f i n e s a command t o do what you want : use awk as a c a l c u l a t o r
a l i a s acalc ’awk ”BEGIN{ p r i n t \ ! * } ” ’ # \ ! * s u b s t i t u t e s a r g s o f a c a l c
echo ” Using a c a l c t o compute 2+3=” ‘ acalc 2+3‘
echo ” Using a c a l c t o compute c o s ( 2 * p i )=” ‘ acalc cos ( 2 * atan2 (0 , −1) ) ‘
# Now do t h e same loop over r a d i i as above i n a d i f f e r e n t way
# while ( e x p r e s s i o n ) i s e x e c u t e d as long as ” e x p r e s s i o n ” i s t r u e
while ( $#Rs > 0) # e x e c u t e d as long as $Rs c o n t a i n s r a d i i
s e t R = $Rs [ 1 ] # t a k e f i r s t element o f $Rs
s h i f t Rs #now $Rs has one l e s s element : old $Rs [ 1 ] has vanished
s e t a = ‘ acalc atan2 (0 , −1) * ${R }* ${R } ‘ # =p i *R*R c a l c u l a t e d by a c a l c
# c o n s t r u c t a f i l e n a m e t o save t h e r e s u l t from t h e v a l u e o f R :
s e t file = area$ {R } . dat
echo ” C i r c l e with R= $R has a r e a $a ” > $file # save r e s u l t i n a f i l e
end #end while
# Now look f o r our f i l e s : save t h e i r names i n an a r r a y f i l e s :
s e t files = ( ‘ l s −1 area * . dat ‘ )
i f ( $# f i l e s == 0) echo ” Sorry , no a r e a f i l e s found ”
echo ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−”
echo ” f i l e s : $ f i l e s ”
l s −l $files
echo ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−”
echo ”And t h e r e s u l t s f o r t h e a r e a a r e : ”
f o r e a c h f ( $files )
echo −n ” f i l e ${ f } : ”
c a t $f
end
# now play a l i t t l e b i t with f i l e names :
echo ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−”
s e t f = $files [ 1 ] # t e s t p e r m i s s i o n s on f i r s t f i l e
# −f , −r , −w, −x , −d t e s t e x i s t e n c e o f f i l e , rwxd p e r m i s s i o n s
# t h e ! n e g a t e s t h e e x p r e s s i o n ( t r u e −> f a l s e , f a l s e −> t r u e )
echo ” t e s t i n g p e r m i s s i o n s on f i l e s : ”
i f ( −f $f ) echo ” $ f i l e e x i s t s ”
i f ( −r $f ) echo ” $ f i l e i s r e a d a b l e by me”
i f ( −w $f ) echo ” $ f i l e i s w r i t a b l e by be ”
i f ( ! −w / bin / l s ) echo ” / bin / l s i s NOT w r i t a b l e by me”
i f ( ! −x $f ) echo ” $ f i l e i s NOT an e x e c u t a b l e ”
i f ( −x / bin / l s ) echo ” / bin / l s i s e x e c u t a b l e by me”
i f ( ! −d $f ) echo ” $ f i l e i s NOT a d i r e c t o r y ”
i f ( −d / bin ) echo ” / bin i s a d i r e c t o r y ”
echo ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−”
# t r a n s f o r m t h e name o f a f i l e
s e t f = $cwd / $f # add t h e f u l l path i n $ f
s e t filename = $f : r # removes e x t e n s i o n . dat
s e t extension = $f : e # g e t s e x t e n s i o n . dat
s e t fdir = $f : h # g e t s directory of $f
s e t base = ‘ basename $f ‘ # removes d i r e c t o r y name
echo ” f i l e i s : $f ”
echo ” f i l e n a m e i s : $ f i l e n a m e ”
echo ” e x t e n s i o n i s : $ e x t e n s i o n ”
echo ” d i r e c t o r y i s : $ f d i r ”
echo ” basename i s : $base ”
# now t r a n s f o r m t h e name t o one with d i f f e r e n t e x t e n s i o n :
s e t newfile = ${ filename } . jpg
echo ” j p e g name i s : $ n e w f i l e ”
echo ” j p e g base i s : ” ‘ basename $newfile ‘
i f ( $newfile : e == jpg ) echo ‘ basename $newfile ‘ ” i s a p i c t u r e ”
echo ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−”
# Now save a l l data i n a f i l e using a ” here document ”
# A here document s t a r t s with <<EOF and ends with a l i n e
68 CHAPTER 1. THE COMPUTER
# s t a r t i n g e x a c t l y with EOF (EOF can be any s t r i n g as below )
# In a ” here document ” we can use v a r i a b l e s and command
# substitution :
c a t <<AREAS >> areas . dat
# This f i l e c o n t a i n s t h e a r e a s o f c i r c l e o f gi v e n r a d i i
# Computation done by ${ u s e r } on ${HOST} . Today i s ‘ date ‘
‘ c a t $files ‘
AREAS
# now s e e what we g o t :
i f ( −f areas . dat ) c a t areas . dat
# You can use a ” here document ” as standard i n p u t t o any command :
# use gnuplot t o save a p l o t : gnuplot does t h e j o b and e x i t s . . .
gnuplot <<GNU
s e t terminal jpeg
s e t output ” areas . jpg ”
plot ” a r e a s . dat ” using 4 : 7 title ” a r e a s . dat ” ,\
pi * x * x title ” p i *R^2 ”
s e t output
GNU
# check our r e s u l t s : d i s p l a y t h e j p e g f i l e using eog
i f ( −f areas . jpg ) eog areas . jpg &
1.6. SHELL SCRIPTING 69
Figure 1.5: In this figure, the Emacs window has been split in three windows. The
splitting was done horizontally first (C-x 2), and then vertically (C-x 3). By dragging
the mouse (Drag-Mouse-1) on the horizontal mode lines and vertical lines that separate
the windows, we can change window sizes. Notice the useful information diplayed on
the mode lines. Each window has one point and the cursor is on the active window (in
this case the window of the buffer named ELines.f). A buffer with no active changes
in its contents is marked by a --, an edited buffer is marked by ** and a buffer in read
only mode with (%%). With a mouse click on a %%, we can change them to -- (so that we
can edit) and vice versa. With Mouse-3 on the name of a mode we can activate a choice
of minor modes. With Mouse-1 on the name of a mode we ca have access to commands
relevant to the mode. The numbers (17,31), (16,6) and (10,15) on the mode lines show
the (line,column) of the point location on the respective windows.
70 CHAPTER 1. THE COMPUTER
awk search for and process patterns in a file,
cat display, or join, files
cd change working directory
chmod change the access mode of a file
cp copy files
date display current time and date
df display the amount of available disk space
diff display the differences between two files
du display information on disk usage
echo echo a text string to output
find find files
grep search for a pattern in files
gzip compress files in the gzip (.gz) format (gunzip to uncompress)
head display the first few lines of a file
kill send a signal (like KILL) to a process
locate search for files stored on the system (faster than find)
less display a file one screen at a time
ln create a link to a file
lpr print files
ls list information about files
man search information about command in man pages
mkdir create a directory
mv move and/or rename a file
ps report information on the processes run on the system
pwd print the working directory
rm remove (delete) files
rmdir remove (delete) a directory
sort sort and/or merge files
tail display the last few lines of a file
tar store or retrieve files from an archive file
top dynamic real-time view of processes
wc counts lines, words and characters in a file
whatis list man page entries for a command
where show where a command is located in the path (alternatively: whereis)
which locate an executable program using ”path”
zip create compressed archive in the zip format (.zip)
unzip get/list contents of zip archive
Table 1.1: Basic Unix commands.
1.6. SHELL SCRIPTING 71
Table 1.2: Some intrinsic functions in Fortran.
Function Description
ABS modulus of a complex number, absolute
value of number
ACOS arccosine of a number
ADJUSTL moves non blank characters of a string to
the left
ADJUSTR moves non blank characters of a string to
the right
AIMAG imaginary part of a complex number
AINT truncates fractional part but preserves
data type
ANINT rounds to nearest whole number but pre-
serves data type
ASIN arcsine of a number
ATAN arctangent of a number
ATAN2 arctangent of arg1 divided by arg2 re-
solved into the correct quadrant
CMPLX converts to the COMPLEX data type arg1
+ i arg2
CONJG complex conjugate of a complex number
COS cosine of an angle in radians
COSH hyperbolic cosine
DATE_AND_TIME returns current date and time
DBLE converts to the real(8) data type
DIM if arg1 > arg2, then returns arg1 - arg2;
otherwise 0
DPROD double precision product of two single
precision numbers
EXP exponential
EPSILON Returns a positive number that is negligi-
ble compared to 1.0
HUGE Returns the largest number of the same
kind as the argument
INT converts to the INTEGER data type by
truncation
KIND Returns the KIND value of argument
LEN Returns the length of a string
Continued...
72 CHAPTER 1. THE COMPUTER
Table 1.2: Continued...
Function Description
LEN_TRIM returns the length of a string without trail-
ing blanks
LGE,LGT,LLE,LLT string comparison functions
LOG natural logarithm
LOG10 common logarithm
MAX maximum value of arguments
MAXEXPONENT returns the maximum exponent of the
same kind as the argument
MIN minimum value of arguments
MINEXPONENT returns the minimum exponent of the
same kind as the argument
MOD arg1 modulo arg2
NINT converts to the INTEGER data type by
rounding
RANDOM_NUMBER returns pseudo-random numbers 0 ≤ r <
1
RANDOM_SEED starts random number generator or returns
generator parameters
PRECISION returns the decimal precision of the same
kind as the argument
REAL real part of a complex number
REAL converts to the REAL data type
SIGN if arg2 < 0, then returns -arg1; else +arg1
SIN sine of an angle in radians
SINH hyperbolic sine
SQRT square root
TAN tangent of an angle in radians
TANH hyperbolic tangent
TINY returns the smallest positive number of
the same kind as the argument
TRIM returns string with trailing blanks re-
moved
Array functions
ALL true if all values are true
ALLOCATED array allocation status
ANY true if any values are true
COUNT number of elements in an array
Continued...
1.6. SHELL SCRIPTING 73
Table 1.2: Continued...
Function Description
DOT_PRODUCT dot product of two rank-one arrays
LBOUND lower dimension bounds of an array
MATMUL matrix multiplication
MAXLOC location of a maximum value in an array
MAXVAL maximum value in an array
MERGE merge arrays under mask
MINLOC location of a minimum value in an array
MINVAL minimum value in an array
PACK pack an array into an array of rank one
under a mask
PRODUCT product of array elements
RESHAPE reshape an array
SHAPE shape of an array or scalar
SIZE size of an array
SPREAD replicate an array by adding a dimension
SUM sum of array elements
TRANSPOSE transpose an array of rank two
UBOUND upper dimension bounds of an array
UNPACK unpack an array of rank one into an array
under a mask
74 CHAPTER 1. THE COMPUTER
Table 1.3: Basic Emacs commands.
Leaving Emacs
suspend Emacs (or iconify it under X) C-z
exit Emacs permanently C-x C-c
Files
read a file into Emacs C-x C-f
save a file back to disk C-x C-s
save all files C-x s
insert contents of another file into this buffer C-x i
toggle read-only status of buffer C-x C-q
Getting Help
The help system is simple. Type C-h (or F1) and follow the directions. If you
are a first-time user, type C-h t for a tutorial.
remove help window C-x 1
apropos: show commands matching a string C-h a
describe the function a key runs C-h k
describe a function C-h f
get mode-specific information C-h m
Error Recovery
abort partially typed or executing command C-g
recover files lost by a system crash M-x recover-session
undo an unwanted change C-x u, C-_ or C-/
restore a buffer to its original contents M-x revert-buffer
redraw garbaged screen C-l
Incremental Search
search forward C-s
search backward C-r
regular expression search C-M-s
abort current search C-g
Use C-s or C-r again to repeat the search in either direction. If Emacs is still
searching, C-g cancels only the part not matched.
Motion
entity to move over backward forward
character C-b C-f
word M-b M-f
line C-p C-n
go to line beginning (or end) C-a C-e
go to buffer beginning (or end) M-< M->
scroll to next screen C-v
Continued...
1.6. SHELL SCRIPTING 75
Table 1.3: Continued...
scroll to previous screen M-v
scroll left C-x <
scroll right C-x >
scroll current line to center of screen C-u C-l
Killing and Deleting
entity to kill backward forward
character (delete, not kill) DEL C-d
word M-DEL M-d
line (to end of) M-0 C-k C-k
kill region C-w
copy region to kill ring M-w
yank back last thing killed C-y
replace last yank with previous kill M-y
Marking
set mark here C-@ or C-SPC
exchange point and mark C-x C-x
mark paragraph M-h
mark entire buffer C-x h
Query Replace
interactively replace a text string M-% or M-x query-replace
using regular expressions M-x query-replace-regexp
Buffers
select another buffer C-x b
list all buffers C-x C-b
kill a buffer C-x k
Multiple Windows
When two commands are shown, the second is a similar command for a frame
instead of a window.
delete all other windows C-x 1 C-x 5 1
split window, above and below C-x 2 C-x 5 2
delete this window C-x 0 C-x 5 0
split window, side by side C-x 3
switch cursor to another window C-x o C-x 5 o
grow window taller C-x ^
shrink window narrower C-x {
grow window wider C-x }
Continued...
76 CHAPTER 1. THE COMPUTER
Table 1.3: Continued...
Formatting
indent current line (indent code etc) TAB
insert newline after point C-o
fill paragraph M-q
Case Change
uppercase word M-u
lowercase word M-l
capitalize word M-c
uppercase region C-x C-u
lowercase region C-x C-l
The Minibuffer
The following keys are defined in the minibuffer.
complete as much as possible TAB
complete up to one word SPC
complete and execute RET
abort command C-g
Type C-x ESC ESC to edit and repeat the last command that used the minibuffer.
Type F10 to activate menu bar items on text terminals.
Spelling Check
check spelling of current word M-$
check spelling of all words in region M-x ispell-region
check spelling of entire buffer M-x ispell-buffer
On the fly spell checking M-x flyspell-mode
Info – Getting Help Within Emacs
enter the Info documentation reader C-h i
scroll forward SPC
scroll reverse DEL
next node n
previous node p
move up u
select menu item by name m
return to last node you saw l
return to directory node d
go to top node of Info file t
go to any node by name g
quit Info q
Chapter 2
Kinematics
In this chapter we show how to program simple kinematic equations of
motion of a particle and how to do basic analysis of numerical results.
We use simple methods for plotting and animating trajectories on the
two dimensional plane and three dimensional space. In section 2.3 we
study numerical errors in the calculation of trajectories of freely moving
particles bouncing off hard walls and obstacles. This will be a prelude to
the study of the integration of the dynamical equations of motion that we
will introduce in the following chapters.
2.1 Motion on the Plane
When a particle moves on the plane, its position can be given in Cartesian
coordinates (x(t), y(t)). These, as a function of time, describe the particle’s
trajectory. The position vector is ⃗r(t) = x(t) x̂ + y(y) ŷ, where x̂ and ŷ are
the unit vectors on the x and y axes respectively. The velocity vector is
⃗v (t) = vx (t) x̂ + vy (t) ŷ where
d⃗r(t)
⃗v (t) =
dt
dx(t) dy(t)
vx (t) = vy (t) = , (2.1)
dt dt
The acceleration ⃗a(t) = ax (t) x̂ + ay (t) ŷ is given by
d⃗v (t) d2⃗r(t)
⃗a(t) = =
dt dt2
dvx (t) d2 x(t) dvy (t) d2 y(t)
ax (t) = = ay (t) = = . (2.2)
dt dt2 dt dt2
77
78 CHAPTER 2. KINEMATICS
y vy v
11
00
00
11
ax vx
ay
r
a
^y
x^ x
Figure 2.1: The trajectory of a particle moving in the plane. The figure shows its
position vector ⃗r, velocity ⃗v and acceleration ⃗a and their Cartesian components in the
chosen coordinate system at a point of the trajectory.
In this section we study the kinematics of a particle trajectory, there-
fore we assume that the functions (x(t), y(t)) are known. By taking
their derivatives, we can compute the velocity and the acceleration of
the particle in motion. We will write simple programs that compute the
values of these functions in a time interval [t0 , tf ], where t0 is the initial
and tf is the final time. The continuous functions x(t), y(t), vx (t), vy (t)
are approximated by a discrete sequence of their values at the times
t0 , t0 + δt, t0 + 2δt, t0 + 3δt, . . . such that t0 + nδt ≤ tf .
We will start the design of our program by forming a generic template
to be used in all of the problems of interest. Then we can study each
problem of particle motion by programming only the equations of mo-
tion without worrying about the less important tasks, like input/output,
user interface etc. Figure 2.2 shows a flowchart of the basic steps in the
algorithm. The first part of the program declares variables and defines
the values of the fixed parameters (like π = 3.1459 . . ., g = 9.81, etc). The
program starts by interacting with the user (“user interface”) and asks
for the values of the variables x0 , y0 , t0 , tf , δt . . .. The program prints
these values to the stdout so that the user can check them for correctness
and store them in her data.
The main calculation is performed in a loop executed while t ≤ tf .
The values of the positions and the velocities x(t), y(t), vx (t), vy (t) are
2.1. MOTION ON THE PLANE 79
Declare variables
Define fixed parameters (PI,...)
User Interface:
Get input from user
x0,y0, t0, tf, dt, ...
Print parameters to stdout
Initialize variables and other
parameters of the motion
Open data file
t = t0
END t < tf
NO
YES
Calculate
x, y, vx, vy
Print results in data file
t = t + dt
Figure 2.2: The flowchart of a typical program computing the trajectory of a particle
from its (kinematic) equations of motion.
calculated and printed in a file together with the time t. At this point we
fix the format of the program output, something that is very important
to do it in a consistent and convenient way for easing data analysis. We
choose to print the values t, x, y, vx, vy in five columns in each line of
the output file.
The specific problem that we are going to solve is the computation of
the trajectory of the circular motion of a particle on a circle with center
(x0 , y0 ) and radius R with constant angular velocity ω. The position on
the circle can be defined by the angle θ, as can be seen in figure 2.3. We
define the initial position of the particle at time t0 to be θ(t0 ) = 0.
The equations giving the position of the particle at time t are
x(t) = x0 + R cos (ω(t − t0 ))
y(t) = y0 + R sin (ω(t − t0 )) . (2.3)
Taking the derivative w.r.t. t we obtain the velocity
vx (t) = −ωR sin (ω(t − t0 ))
vy (t) = ωR cos (ω(t − t0 )) , (2.4)
80 CHAPTER 2. KINEMATICS
y
(R cos θ , R sin θ )
θ
(x,y)
(x0, y0)
^
y
x^ x
Figure 2.3: The trajectory of a particle moving on a circle with constant angular
velocity calculated by the program Circle.f90.
and the acceleration
ax (t) = −ω 2 R cos (ω(t − t0 )) = −ω 2 (x(t) − x0 )
ay (t) = −ω 2 R sin (ω(t − t0 )) = −ω 2 (y(t) − y0 ) . (2.5)
We note that the above equations imply that R ⃗ · ⃗v = 0 (R
⃗ ≡ ⃗r − ⃗r0 , ⃗v ⊥ R,
⃗
⃗v tangent to the trajectory) and ⃗a = −ω R (R and ⃗a anti-parallel, ⃗a ⊥ ⃗v ).
2⃗ ⃗
The data structure is quite simple. The constant angular velocity ω
is stored in the REAL variable omega. The center of the circle (x0 , y0 ), the
radius R of the circle and the angle θ are stored in the REAL variables x0,
y0, R, theta. The times at which we calculate the particle’s position
and velocity are defined by the parameters t0 , tf , δt and are stored in the
REAL variables t0, tf, dt. The current position (x(t), y(t)) is calculated
and stored in the REAL variables x, y and the velocity (vx (t), vy (t)) in the
REAL variables vx, vy. The declarations of the variables are put in the
beginning of the program:
r e a l : : x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt
r e a l : : theta , omega
r e a l , parameter : : PI =3.1415927
were we defined the value¹ of π = 3.1415927 by using the parameter
specification.
¹The reader is reminded that REAL variables are stored in 4 bytes and have an
accuracy of about 7 decimal digits.
2.1. MOTION ON THE PLANE 81
The user interface of the program is the interaction of the program
with the user and, in our case, it is the part of the program where the
user enters the parameters omega, x0, y0, R, t0, tf, dt. The program
issues a prompt with the names the variables expected to be read. This
is done using simple print statements. The variables are read from the
stdin by simple read statements and the values entered by the user are
printed to the stdout²:
print * , ’ # Enter omega : ’
read * , omega
print * , ’ # Enter c e n t e r o f c i r c l e ( x0 , y0 ) and r a d i u s R : ’
read * , x0 , y0 , R
print * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
print * , ’ # omega= ’ , omega
print * , ’ # x0= ’ , x0 , ’ y0= ’ , y0 , ’ R= ’ , R
print * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
Next, the program initializes the state of the computation. This includes
checking the validity of the parameters entered by the user, so that the
computation will be possible. For example, the program computes the
expression 2.0*PI/omega, where it is assumed that omega has a non zero
value. We will also demand that R > 0 and ω > 0. An if statement
will make those checks and if the parameters have illegal values, the
stop statement will stop the program execution³. The program opens
the file Circle.dat for writing the calculated values of the position and
the velocity of the particle.
i f (R . l e . 0.0) stop ’ I l l e g a l value of R ’
i f ( omega . l e . 0 . 0 ) s t o p ’ I l l e g a l v a l u e o f omega ’
p r i n t * , ’ # T= ’ , 2 . 0 * PI / omega
open ( u n i t =11 , f i l e = ’ C i r c l e . dat ’ )
If R ≤ 0 or ω ≤ 0 the corresponding stop statements are executed which
end the program execution. The optional error messages are included
after the stop statements which are printed to the stdout. The value of
the period T = 2π/ω is also calculated and printed for reference.
The open statement uses unit 11 for writing to the file Circle.dat.
²This is done so that the used can check for typos and see the actual value read by
the program. By redirecting the stdout of a file on the hard disk, the parameters can
be saved for future reference and used in data analysis.
³Note that there are more assumptions that need to be checked by the program. We
leave this as an exercise for the reader.
82 CHAPTER 2. KINEMATICS
The choice of the unit number is free for the programmer to choose. We
recommend using the units 10 to 99 for input/output to files⁴.
The main calculation is performed within the loop
t = t0
do while ( t . l e . tf )
.........
t = t + dt
enddo
The first statement sets the initial value of the time. The statements be-
tween the do while(condition) and enddo are executed as long as condition
has a .TRUE. value. The statement t=t+dt increments the time and this
is necessary in order not to enter into an infinite loop. Τhe commands
put in place of the dots ......... calculate the position and the velocity
and print them to the file Circle.dat:
theta = omega * ( t−t0 )
x = x0+R * c o s ( theta )
y = y0+R * s i n ( theta )
vx = −omega * R * s i n ( theta )
vy = omega * R * c o s ( theta )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
Notice the use of the intrinsic functions sin and cos that calculate the sine
and cosine of an angle expressed in radians. We use the intermediate
variable theta in order to store the phase θ(t) = ω(t − t0 ). The command
write(11,*) writes the variables t,x,y,vx,vy to the unit 11, which has
been associated to the file Circle.dat with the open statement shown
above.
The program is stored in the file Circle.f90 and can be found in
the accompanied software. The extension .f90 is used by the compiler
in order to denote source code written in free format Fortran language.
Compilation and running can be done using the commands:
> g f o r t r a n Circle . f90 −o cl
> . / cl
The switch -o cl forces the compiler gfortran to write the binary com-
⁴Some numbers can be reserved for special files, like 5 for stdin, 6 for stdout and
0 for stderr. Using numbers larger than 99 can lead to portability problems.
2.1. MOTION ON THE PLANE 83
mands executed by the program to the file⁵ cl. The command ./cl loads
the program instructions to the computer memory for execution. When
the programs starts execution, it first asks for the parameter data and
then performs the calculation. A typical session looks like:
> g f o r t r a n Circle . f90 −o cl
> . / cl
# Enter omega :
1.0
# Enter c e n t e r o f c i r c l e ( x0 , y0 ) and r a d i u s R :
1 . 0 1 . 0 0.5
# Enter t0 , t f , dt :
0.0 20.0 0.01
# omega= 1.
# x0= 1 . y0= 1 . R= 0.5
# t 0= 0. t f = 20. dt= 0.00999999978
# T= 6.28318548
The lines shown above that start with a # character are printed by the
program and lines without # are the values of the parameters entered
interactively by the user. The user types in the parameters and then
presses the Enter key in order for the program to read them. Here we
have ω = 1.0, x0 = y0 = 1.0, R = 0.5, t0 = 0.0, tf = 20.0 and δt = 0.01.
You can execute the above program many times for different values of
the parameter by writing the parameter values in a file using an editor.
For example, in the file Circle.in type the following data:
1.0 omega
1 . 0 1 . 0 0.5 ( x0 , y0 ) , R
0.0 20.0 0.01 t0 tf dt
Each line has the parameters read by the program with a read statement
(a record). The rest of the line is ignored by the program and the user can
write anything she likes as a comment on how to use the parameters. The
program can read the above values of the parameters with the command:
> . / cl < Circle . in > Circle . out
The command ./cl runs the commands found in the executable file ./cl.
The < Circle.in redirects the contents of the file Circle.in to the stan-
dard input (stdin) of the command ./cl. This way the program reads
⁵If omitted, the executable file has the default name a.out.
84 CHAPTER 2. KINEMATICS
in the values of the parameters from the contents of the file Circle.in.
The > Circle.out redirects the standard output (stdout) of the com-
mand ./cl to the file Circle.out. Its contents can be inspected after the
execution of the program with the command cat:
> c a t Circle . out
# Enter omega :
# Enter c e n t e r o f c i r c l e ( x0 , y0 ) and r a d i u s R :
# Enter t0 , t f , dt :
# omega= 1.
# x0= 1 . y0= 1 . R= 0.5
# t 0= 0. t f = 20. dt= 0.00999999978
# T= 6.28318548
We list the full program in Circle.f90 below:
! ============================================================
! F i l e C i r c l e . f90
! C o n s t a n t angular v e l o c i t y c i r c u l a r motion
! S e t ( x0 , y0 ) c e n t e r o f c i r c l e , i t s r a d i u s R and omega .
! At t =t0 , t h e p a r t i c l e i s a t t h e t a =0
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program Circle
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of va r ia b le s
r e a l : : x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt
r e a l : : theta , omega
r e a l , parameter : : PI =3.1415927
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter omega : ’
read * , omega
p r i n t * , ’ # Enter c e n t e r o f c i r c l e ( x0 , y0 ) and r a d i u s R : ’
read * , x0 , y0 , R
p r i n t * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
p r i n t * , ’ # omega= ’ , omega
p r i n t * , ’ # x0= ’ , x0 , ’ y0= ’ , y0 , ’ R= ’ , R
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i f (R . l e . 0.0) stop ’ I l l e g a l value of R ’
i f ( omega . l e . 0 . 0 ) s t o p ’ I l l e g a l v a l u e o f omega ’
p r i n t * , ’ # T= ’ , 2 . 0 * PI / omega
open ( u n i t =11 , f i l e = ’ C i r c l e . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
2.1. MOTION ON THE PLANE 85
! Compute :
t = t0
do while ( t . l e . tf )
theta = omega * ( t−t0 )
x = x0+R * c o s ( theta )
y = y0+R * s i n ( theta )
vx = −omega * R * s i n ( theta )
vy = omega * R * c o s ( theta )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
t = t + dt
enddo
close (11)
end program Circle
2.1.1 Plotting Data
We use gnuplot for plotting the data produced by our programs. The
file Circle.dat has the time t and the components x, y, vx, vy in five
columns. Therefore we can plot the functions x(t) and y(t) by using the
gnuplot commands:
gnuplot > p l o t ” C i r c l e . dat ” using 1 : 2 with lines t i t l e ” x ( t ) ”
gnuplot > r e p l o t ” C i r c l e . dat ” using 1 : 3 with lines t i t l e ” y ( t ) ”
1.5 4
x(t) theta(t)
y(t) pi
1.4 -pi
3
1.3
2
1.2
1
1.1
1 0
0.9
-1
0.8
-2
0.7
-3
0.6
0.5 -4
0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20
Figure 2.4: The plots (x(t), y(t)) (left) and θ(t) (right) from the data in Circle.dat
for ω = 1.0, x0 = y0 = 1.0, R = 0.5, t0 = 0.0, tf = 20.0 and δt = 0.01.
The second line puts the second plot together with the first one. The
results can be seen in figure 2.4.
Let’s see now how we can make the plot of the function θ(t). We can
do that using the raw data from the file Circle.dat within gnuplot, with-
out having to write a new program. Note that θ(t) = tan−1 ((y − y0 )/(x − x0 )).
86 CHAPTER 2. KINEMATICS
The function atan2 is available in gnuplot⁶ as well as in Fortran. Use
the online help system in gnuplot in order to see its usage:
gnuplot > help atan2
The ‘ atan2 ( y , x ) ‘ f u n c t i o n returns the arc tangent ( inverse
tangent ) of the ratio of the r e a l parts of its arguments .
‘ atan2 ‘ returns its argument in radians or degrees , as
selected by ‘ s e t a n g l e s ‘ , in the correct quadrant .
Therefore, the right way to call the function is atan2(y-y0,x-x0). In
our case x0=y0=1 and x, y are in the 2nd and 3rd columns of the file
Circle.dat. We can construct an expression after the using command as
in page 60, where $2 is the value of the second and $3 the value of the
third column:
gnuplot > x0 = 1 ; y0 = 1
gnuplot > p l o t ” C i r c l e . dat ” using 1 : ( atan2 ( $3−y0 , $2−x0 ) ) \
with lines t i t l e ” t h e t a ( t ) ” , pi ,−pi
The second command is broken in two lines by using the character \
so that it fits conveniently in the text⁷. Note how we defined the val-
ues of the variables x0, y0 and how we used them in the expression
atan2($3-x0,$2-y0). We also plot the lines which graph the constant
functions f1 (t) = π and f2 (t) = −π which mark the limit values of θ(t).
The gnuplot variable⁸ pi is predefined and can be used in formed ex-
pressions. The result can be seen in the left plot of figure 2.4.
The velocity components (vx (t), vy (t)) as function of time as well as
the trajectory ⃗r(t) can be plotted with the commands:
gnuplot > p l o t ” C i r c l e . dat ” using 1 : 4 t i t l e ” v_x ( t ) ” \
with lines
gnuplot > r e p l o t ” C i r c l e . dat ” using 1 : 5 t i t l e ” v_y ( t ) ” \
with lines
gnuplot > p l o t ” C i r c l e . dat ” using 2:3 t i t l e ”x−y ”
with lines
We close this section by showing how to do a simple animation of the
particle trajectory using gnuplot. There is a file animate2D.gnu in the
accompanied software which you can copy in the directory where you
⁶The command help functions will show you all the available functions in gnuplot.
⁷This can be done on the gnuplot command line as well.
⁸Use the command show variables in order to see the current/default values of
gnuplot variables.
2.1. MOTION ON THE PLANE 87
t= 20.000000 (x,y)= (1.208431,1.454485)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Figure 2.5: The particle trajectory plotted by the gnuplot program in the file
animate2D.gnu of the accompanied software. The position vector is shown at a given
time t, which is marked on the title of the plot together with the coordinates (x,y).
The data is produced by the program Circle.f90 described in the text.
have the data file Circle.dat. We are not going to explain how it works⁹
but how to use it in order to make your own animations. The final result
is shown in figure 2.5. All that you need to do is to define the data file¹⁰,
the initial time t0, the final time tf and the time step dt. These times
can be different from the ones we used to create the data in Circle.dat.
A full animation session can be launched using the commands:
gnuplot > file = ” C i r c l e . dat ”
gnuplot > s e t xrange [ 0 : 1 . 6 ] ; s e t yrange [ 0 : 1 . 6 ]
gnuplot > t0 = 0 ; tf = 20 ; dt = 0 . 1
gnuplot > load ” animate2D . gnu”
The first line defines the data file that animate2D.gnu reads data from.
The second line sets the range of the plots and the third line defines
the time parameters used in the animation. The final line launches the
animation. If you want to rerun the animation, you can repeat the last
two commands as many times as you want using the same or different
parameters. E.g. if you wish to run the animation at “half the speed”
⁹You are most welcome to study the commands in the script and guess how it works
of course!
¹⁰It can be any file that has (t, x, y) in the 1st, 2nd and 3rd columns respectively.
88 CHAPTER 2. KINEMATICS
you should simply redefine dt=0.05 and set the initial time to t0=0:
gnuplot > t0 = 0 ; dt = 0.05
gnuplot > load ” animate2D . gnu”
2.1.2 More Examples
We are now going to apply the steps described in the previous section
to other examples of motion on the plane. The first problem that we are
going to discuss is that of the small oscillations of a simple pendulum.
Figure 2.6 shows the single oscillating degree of freedom θ(t), which
is the small angle that the pendulum forms with the √ vertical direction.
The motion is periodic with angular frequency ω = g/l and period
y
x
l
θ
T
111
000
111
000
111
000
111
000
111
000
mg
Figure 2.6: The simple pendulum whose motion for θ ≪ 1 is described by the
program SimplePendulum.f90.
T = 2π/ω. The angular velocity is computed from θ̇ ≡ dθ/dt which gives
θ(t) = θ0 cos (ω(t − t0 ))
θ̇(t) = −ωθ0 sin (ω(t − t0 )) (2.6)
We have chosen the initial conditions θ(t0 ) = θ0 and θ̇(t0 ) = 0. In order to
write the equations of motion in the Cartesian coordinate system shown
in figure 2.6 we use the relations
x(t) = l sin (θ(t))
y(t) = −l cos (θ(t))
dx(t)
vx (t) = = lθ̇(t) cos (θ(t))
dt
dy(t)
vy (t) = = lθ̇(t) sin (θ(t)) . (2.7)
dt
2.1. MOTION ON THE PLANE 89
These are similar to the equations (2.3) and (2.4) that we used in the case
of the circular motion of the previous section. Therefore the structure of
the program is quite similar. Its final form, which can be found in the
file SimplePendulum.f90, is:
! ==============================================================
! F i l e SimplePendulum . f90
! S e t pendulum o r i g i n a l p o s i t i o n a t t h e t a 0 with no i n i t i a l speed
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program SimplePendulum
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l : : l , x , y , vx , vy , t , t0 , tf , dt
r e a l : : theta , theta0 , dtheta_dt , omega
r e a l , parameter : : PI =3.1415927 , g =9.81
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter l : ’
read * , l
p r i n t * , ’ # Enter t h e t a 0 : ’
read * , theta0
p r i n t * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
p r i n t * , ’ # l = ’ , l , ’ t h e t a 0= ’ , theta0
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
omega = s q r t ( g / l )
p r i n t * , ’ # omega= ’ , omega , ’ T= ’ , 2 . 0 * PI / omega
open ( u n i t =11 , f i l e = ’ SimplePendulum . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = t0
do while ( t . l e . tf )
theta = theta0 * c o s ( omega * ( t−t0 ) )
dtheta_dt = −omega * theta0 * s i n ( omega * ( t−t0 ) )
x = l * s i n ( theta )
y = −l * c o s ( theta )
vx = l * dtheta_dt * c o s ( theta )
vy = l * dtheta_dt * s i n ( theta )
w r i t e ( 1 1 , 1 0 0 ) t , x , y , vx , vy , theta , dtheta_dt
t = t + dt
enddo
close (11)
100 FORMAT( 7 G15 . 7 )
end program SimplePendulum
90 CHAPTER 2. KINEMATICS
We note that the acceleration of gravity g is hard coded in the program
and that the user can only set the length l of the pendulum. The data
file SimplePendulum.dat produced by the program, contains two extra
columns with the current values of θ(t) and the angular velocity θ̇(t). The
statement write(11,100) writes to the unit 11 according to the format
set by the FORMAT statement, found in the line labeled by the label 100.
This is done so that we can be sure that the data is printed in one line
for each value of t (see the discussion on page 48).
A simple session for the study of the above problem is shown below¹¹:
> g f o r t r a n SimplePendulum . f90 −o sp
> . / sp
# Enter l :
1.0
# Enter t h e t a 0 :
0.314
# Enter t0 , t f , dt :
0 20 0.01
# l= 1. t h e t a 0= 0.31400001
# t 0= 0. tf= 20. dt= 0.00999999978
# omega= 3.132092 T= 2.0060668
> gnuplot
gnuplot > p l o t ” SimplePendulum . dat ” u 1 : 2 w l t ” x ( t ) ”
gnuplot > p l o t ” SimplePendulum . dat ” u 1 : 3 w l t ” y ( t ) ”
gnuplot > p l o t ” SimplePendulum . dat ” u 1 : 4 w l t ” v_x ( t ) ”
gnuplot > r e p l o t ” SimplePendulum . dat ” u 1 : 5 w l t ” v_y ( t ) ”
gnuplot > p l o t ” SimplePendulum . dat ” u 1 : 6 w l t ” t h e t a ( t ) ”
gnuplot > r e p l o t ” SimplePendulum . dat ” u 1 : 7 w l t ” t h e t a ’ ( t ) ”
gnuplot > p l o t [ − 0 . 6 : 0 . 6 ] [ − 1 . 1 : 0 . 1 ] ” SimplePendulum . dat ” \
u 2:3 w l t ”x−y ”
gnuplot > file = ” SimplePendulum . dat ”
gnuplot > t0 =0; tf =20.0; dt =0.1
gnuplot > s e t xrange [ − 0 . 6 : 0 . 6 ] ; s e t yrange [ − 1 . 1 : 0 . 1 ]
gnuplot > load ” animate2D . gnu”
The next example is the study of the trajectory of a particle shot near
the earth’s surface¹² when we consider the effect of air resistance to be
negligible. Then, the equations describing the trajectory of the particle
¹¹Notice that we replaced the command “using 1:2 with lines title” with “u
1:2 w lines t”. These abbreviations can be done with every gnuplot command if an
abbreviation uniquely determines a command.
¹²I.e. ⃗g = const. and the Coriolis force can be ignored.
2.1. MOTION ON THE PLANE 91
and its velocity are given by the parametric equations
x(t) = v0x t
1
y(t) = v0y t − gt2
2
vx (t) = v0x
vy (t) = v0y − gt , (2.8)
where t is the parameter. The initial conditions are x(0) = y(0) = 0,
vx (0) = v0x = v0 cos θ and vy (0) = v0y = v0 sin θ, as shown in figure 2.7.
Figure 2.7: The trajectory of a particle moving under the influence of a constant
gravitational field. The initial conditions are set to x(0) = y(0) = 0, vx (0) = v0x = v0 cos θ
and vy (0) = v0y = v0 sin θ.
The structure of the program is similar to the previous ones. The user
enters the magnitude of the particle’s initial velocity and the shooting
angle θ in degrees. The initial time is taken to be t0 = 0. The program
calculates v0x and v0y and prints them to the stdout. The data is written
to the file Projectile.dat. The full program is listed below and it can
be found in the file Projectile.f90 in the accompanied software:
! ============================================================
! F i l e P r o j e c t i l e . f90
! Shooting a p r o j e c t i l e near t h e e a r t h s u r f a c e .
! No a i r r e s i s t a n c e .
! S t a r t s a t ( 0 , 0 ) , s e t ( v0 , t h e t a ) .
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program Projectile
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l : : x0 , y0 , R , x , y , vx , vy , t , tf , dt
92 CHAPTER 2. KINEMATICS
r e a l : : theta , v0x , v0y , v0
r e a l , parameter : : PI =3.1415927 , g =9.81
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter v0 , t h e t a ( i n d e g r e e s ) : ’
read * , v0 , theta
p r i n t * , ’ # Enter t f , dt : ’
read * , tf , dt
p r i n t * , ’ # v0= ’ , v0 , ’ t h e t a = ’ , theta , ’ o ( d e g r e e s ) ’
p r i n t * , ’ # t 0= ’ , 0 . 0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i f ( v0 . l e . 0 . 0 ) s t o p ’ I l l e g a l v a l u e o f v0<=0 ’
i f ( theta . l e . 0.0 . or . theta . ge . 9 0 . 0 ) &
stop ’ I l l e g a l value of the ta ’
theta = ( PI / 1 8 0 . 0 ) * theta ! c o n v e r t t o r a d i a n s
v0x = v0 * c o s ( theta )
v0y = v0 * s i n ( theta )
p r i n t * , ’ # v0x = ’ , v0x , ’ v0y= ’ , v0y
open ( u n i t =11 , f i l e = ’ P r o j e c t i l e . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = 0.0
do while ( t . l e . tf )
x = v0x * t
y = v0y * t − 0 . 5 * g * t * t
vx = v0x
vy = v0y − g*t
w r i t e ( 1 1 , * ) t , x , y , vx , vy
t = t + dt
enddo
close (11)
end program Projectile
A typical session for the study of this problem is shown below:
> g f o r t r a n Projectile . f90 −o pj
> . / pj
# Enter v0 , t h e t a ( i n d e g r e e s ) :
10 45
# Enter t f , dt :
1.4416 0.001
# v0= 10.0000000 t h e t a = 45.000000 o ( d e g r e e s )
# t 0= 0.0000000 tf= 1.4416000 dt= 1.00000005E−03
# v0x = 7.0710678 v0y= 7.0710678
> gnuplot
gnuplot > p l o t ” P r o j e c t i l e . dat ” using 1 : 2 w l t ” x ( t ) ”
gnuplot > r e p l o t ” P r o j e c t i l e . dat ” using 1 : 3 w l t ” y ( t ) ”
2.1. MOTION ON THE PLANE 93
gnuplot > plot ” P r o j e c t i l e . dat ” using 1 : 4 w l t ” v_x ( t ) ”
gnuplot > r e p l o t ” P r o j e c t i l e . dat ” using 1 : 5 w l t ” v_y ( t ) ”
gnuplot > plot ” P r o j e c t i l e . dat ” using 2:3 w l t ”x−y ”
gnuplot > file = ” P r o j e c t i l e . dat ”
gnuplot > s e t xrange [ 0 : 1 0 . 3 ] ; s e t yrange [ 0 : 1 0 . 3 ]
gnuplot > t0 =0; tf = 1 . 4 4 1 6 ; dt =0.05
gnuplot > load ” animate2D . gnu”
Next, we will study the effect of air resistance of the form F⃗ = −mk⃗v .
The solutions to the equations of motion
Figure 2.8: The forces that act on the particle of figure 2.7 when we assume air
resistance of the form F⃗ = −mk⃗v .
dvx
ax = = −kvx
dt
dvy
ay = = −kvy − g (2.9)
dt
with initial conditions x(0) = y(0) = 0, vx (0) = v0x = v0 cos θ and vy (0) =
v0y = v0 sin θ are¹³
vx (t) = v0x e−kt
( g ) −kt g
vy (t) = v0y + e −
k k
v0x ( −kt
)
x(t) = 1−e
k
1( g)( ) g
y(t) = v0y + 1 − e−kt − t (2.10)
k k k
Programming the above equations is as easy as before, the only dif-
ference being that the user needs to provide the value of the constant k.
¹³The proof of equations (2.10) is left as an exercise for the reader.
94 CHAPTER 2. KINEMATICS
The full program can be found in the file ProjectileAirResistance.f90
and it is listed below:
! ============================================================
! F i l e P r o j e c t i l e A i r R e s i s t a n c e . f90
! Shooting a p r o j e c t i l e near t h e e a r t h s u r f a c e
! with a i r r e s i s t a n c e
! S t a r t s a t ( 0 , 0 ) , s e t k , ( v0 , t h e t a ) .
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program ProjectileAirResistance
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of va r ia b le s
r e a l : : x0 , y0 , R , x , y , vx , vy , t , tf , dt , k
r e a l : : theta , v0x , v0y , v0
r e a l , parameter : : PI =3.1415927 , g =9.81
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter k , v0 , t h e t a ( i n d e g r e e s ) : ’
read * , k , v0 , theta
p r i n t * , ’ # Enter t f , dt : ’
read * , tf , dt
print * , ’# k = ’ , k
p r i n t * , ’ # v0= ’ , v0 , ’ t h e t a = ’ , theta , ’ o ( d e g r e e s ) ’
p r i n t * , ’ # t 0= ’ , 0 . 0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i f ( v0 . l e . 0 . 0 ) s t o p ’ I l l e g a l v a l u e o f v0<=0 ’
if ( k . l e . 0 . 0 ) s t o p ’ I l l e g a l v a l u e o f k <=0 ’
i f ( theta . l e . 0.0 . or . theta . ge . 9 0 . 0 ) &
stop ’ I l l e g a l value of the ta ’
theta = ( PI / 1 8 0 . 0 ) * theta ! c o n v e r t t o r a d i a n s
v0x = v0 * c o s ( theta )
v0y = v0 * s i n ( theta )
p r i n t * , ’ # v0x = ’ , v0x , ’ v0y= ’ , v0y
open ( u n i t =11 , f i l e = ’ P r o j e c t i l e A i r R e s i s t a n c e . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = 0.0
do while ( t . l e . tf )
x = ( v0x / k ) *(1.0 − exp(−k * t ) )
y = ( 1 . 0 / k ) * ( v0y +( g / k ) ) *(1.0 − exp(−k * t ) ) −(g / k ) * t
vx = v0x * exp(−k * t )
vy = ( v0y +( g / k ) ) * exp(−k * t ) −(g / k )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
t = t + dt
enddo
close (11)
2.1. MOTION ON THE PLANE 95
end program ProjectileAirResistance
8
x(t) v_x(t)
v0x/k 0
1.4 y(t) v_y(t)
7
-(g/k)*x+(g/k**2)+v0y/k -g/k
6
1.2
5
1
4
0.8 3
2
0.6
1
0.4
0
-1
0.2
-2
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Figure 2.9: The plots of x(t),y(t) (left) and vx (t),vy (t) (right) from the data produced
by the program ProjectileAirResistance.f90 for k = 5.0, v0 = 10.0, θ = π/4, tf =
0.91 and δt = 0.001. We also plot the asymptotes of these functions as t → ∞.
We also list the commands of a typical session of the study of the
problem:
> g f o r t r a n ProjectileAirResistance . f90 −o pja
> . / pja
# Enter k , v0 , t h e t a ( i n d e g r e e s ) :
5.0 10.0 45
# Enter t f , dt :
0.91 0.001
# k = 5.
# v0= 10. theta= 45. o ( d e g r e e s )
# t 0= 0. t f = 0.910000026 dt= 0.00100000005
# v0x = 7.07106781 v0y= 7.07106781
> gnuplot
gnuplot > v0x = 10* c o s ( pi / 4 ) ; v0y = 10* s i n ( pi / 4 )
gnuplot > g = 9.81 ; k = 5
gnuplot > p l o t [ : ] [ : v0x / k + 0 . 1 ] ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” \
using 1 : 2 with lines t i t l e ” x ( t ) ” , v0x / k
gnuplot > r e p l o t ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” \
using 1 : 3 with lines t i t l e ” y ( t ) ” ,\
−(g / k ) * x +( g / k * * 2 ) +v0y / k
gnuplot > p l o t [ : ] [ − g / k − 0 . 6 : ] ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” \
using 1 : 4 with lines t i t l e ” v_x ( t ) ” , 0
gnuplot > r e p l o t ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” \
using 1 : 5 with lines t i t l e ” v_y ( t ) ” ,−g / k
gnuplot > p l o t ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” \
using 2:3 with lines t i t l e ” With a i r r e s i s t a n c e k=5.0 ”
gnuplot > r e p l o t ” P r o j e c t i l e . dat ” \
96 CHAPTER 2. KINEMATICS
3
With air resistance k=5.0
No air resistance k=0.0
2.5
2
1.5
1
0.5
0
0 2 4 6 8 10 12
Figure 2.10: Trajectories of the particles shot with v0 = 10.0, θ = π/4 in the absence
of air resistance and when the air resistance is present in the form F⃗ = −mk⃗v with
k = 5.0.
using 2:3 with lines t i t l e ”No a i r r e s i s t a n c e k=0.0 ”
gnuplot > file = ” P r o j e c t i l e A i r R e s i s t a n c e . dat ”
gnuplot > s e t xrange [ 0 : 1 . 4 ] ; s e t yrange [ 0 : 1 . 4 ]
gnuplot > t0 =0; tf = 0 . 9 1 ; dt =0.01
gnuplot > load ” animate2D . gnu”
Long commands have been continued to the next line as before. We
defined the gnuplot variables v0x, v0y, g and k to have the values that
we used when running the program. We can use them in order to
construct the asymptotes of the plotted functions of time. The results are
shown in figures 2.9 and 2.10.
The last example of this section will be that of the anisotropic har-
monic oscillator. The force on the particle is
Fx = −mω12 x Fy = −mω22 y (2.11)
where the “spring constants” k1 = mω12 and k2 = mω22 are different in the
directions of the axes x and y. The solutions of the dynamical equations
of motion for x(0) = A, y(0) = 0, vx (0) = 0 and vy (0) = ω2 A are
x(t) = A cos(ω1 t) y(t) = A sin(ω2 t)
vx (t) = −ω1 A sin(ω1 t) vy (t) = ω2 A cos(ω2 t) . (2.12)
If the angular frequencies ω1 and ω2 satisfy certain relations, the trajec-
tories of the particle are closed and self intersect at a given number of
2.1. MOTION ON THE PLANE 97
points. The proof of these relations, as well as their numerical confirma-
tion, is left as an exercise for the reader. The program listed below is in
the file Lissajoux.f90:
! ============================================================
! F i l e L i s s a j o u s . f90
! Lissajous curves ( s p e c i a l case )
! x ( t )= c o s ( o1 t ) , y ( t )= s i n ( o2 t )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program Lissajous
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt
r e a l o1 , o2 , T1 , T2
r e a l , parameter : : PI =3.1415927
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter omega1 and omega2 : ’
read * , o1 , o2
p r i n t * , ’ # Enter t f , dt : ’
read * , tf , dt
p r i n t * , ’ # o1= ’ , o1 , ’ o2= ’ , o2
p r i n t * , ’ # t 0= ’ , 0 . 0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i f ( o1 . l e . 0 . 0 . or . o2 . l e . 0 . 0 ) s t o p ’ omega1 or omega2<=0 ’
T1 = 2 . 0 * PI / o1
T2 = 2 . 0 * PI / o2
p r i n t * , ’ # T1= ’ , T1 , ’ T2= ’ , T2
open ( u n i t =11 , f i l e = ’ L i s s a j o u s . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = 0.0
do while ( t . l e . tf )
x = c o s ( o1 * t )
y = s i n ( o2 * t )
vx = −o1 * s i n ( o1 * t )
vy = o2 * c o s ( o2 * t )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
t = t + dt
enddo
close (11)
end program Lissajous
We have set A = 1 in the program above. The user must enter the two
angular frequencies ω1 and ω2 and the corresponding times. A typical
98 CHAPTER 2. KINEMATICS
session for the study of the problem is shown below:
> g f o r t r a n Lissajous . f90 −o lsj
> . / lsj
# Enter omega1 and omega2 :
3 5
# Enter t f , dt :
10.0 0.01
# o1= 3 . o2= 5.
# t 0= 0. t f = 1 0 . dt= 0.00999999978
# T1= 2.09439516 T2= 1.2566371
>gnuplot
gnuplot > p l o t ” L i s s a j o u s . dat ” using 1 : 2 w l t ”x ( t ) ”
gnuplot > r e p l o t ” L i s s a j o u s . dat ” using 1 : 3 w l t ”y ( t ) ”
gnuplot > p l o t ” L i s s a j o u s . dat ” using 1 : 4 w l t ” v_x ( t ) ”
gnuplot > r e p l o t ” L i s s a j o u s . dat ” using 1 : 5 w l t ” v_y ( t ) ”
gnuplot > p l o t ” L i s s a j o u s . dat ” using 2:3 w l t ”x−y f o r 3:5 ”
gnuplot > file = ” L i s s a j o u s . dat ”
gnuplot > s e t xrange [ − 1 . 1 : 1 . 1 ] ; s e t yrange [ −1.1:1.1]
gnuplot > t0 =0; tf =10; dt =0.1
gnuplot > load ” animate2D . gnu”
The results for ω1 = 3 and ω2 = 5 are shown in figure 2.11.
t= 6.400000 (x,y)= (0.949047,0.509265)
1
0.5
0
-0.5
-1
-1 -0.5 0 0.5 1
Figure 2.11: The trajectory of the anisotropic oscillator with ω1 = 3 and ω2 = 5.
2.2. MOTION IN SPACE 99
2.2 Motion in Space
By slightly generalizing the methods described in the previous section,
we will study the motion of a particle in three dimensional space. All
we have to do is to add an extra equation for the coordinate z(t) and the
component of the velocity vz (t). The structure of the programs will be
exactly the same as before.
y
x
θ l
z
T
r θ
v
ωt Txy 00
11
00
11
00
11
mg
Figure 2.12: The conical pendulum of the program ConicalPendulum.f90.
The first example is the conical pendulum, which can be seen in figure
2.12. The particle moves on the xy plane with constant angular velocity
ω. The equations of motion are derived from the relations
Tz = T cos θ = mg Txy = T sin θ = mω 2 r , (2.13)
where r = l sin θ. Their solution¹⁴ is
x(t) = r cos ωt
y(t) = r sin ωt
z(t) = −l cos θ , (2.14)
¹⁴One has to choose appropriate initial conditions. Exercise: find them!
100 CHAPTER 2. KINEMATICS
where we have to substitute the values
g
cos θ =
ω2l
√
sin θ = 1 − cos2 θ
g sin θ
r = . (2.15)
ω 2 cos θ
For the velocity components we obtain
vx = −rω sin ωt
vy = rω cos ωt
vz = 0 . (2.16)
Therefore we must have
√
g
ω ≥ ωmin = , (2.17)
l
and when ω → ∞, θ → π/2.
In the program that we will write, the user must enter the parameters
l, ω, the final time tf and the time step δt. We take t0 = 0. The convention
that we follow for the output of the results is that they should be written
in a file where the first 7 columns are the values of t, x, y, z, vx , vy and
vz . Each line in this file is long and, in order to prevent Fortran from
breaking it into two separate lines, we have to give an explicit format
specification. See the discussion on page 48. The full program is listed
below:
! ============================================================
! F i l e ConicalPendulum . f90
! S e t pendulum angular v e l o c i t y omega and d i s p l a y motion i n 3D
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program ConicalPendulum
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of va r ia b le s
r e a l : : l , r , x , y , z , vx , vy , vz , t , tf , dt
r e a l : : theta , cos_theta , sin_theta , omega
r e a l , parameter : : PI =3.1415927 , g =9.81
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter l , omega : ’
read * , l , omega
p r i n t * , ’ # Enter t f , dt : ’
2.2. MOTION IN SPACE 101
read * , tf , dt
print * , ’# l= ’ , l , ’ omega= ’ , omega
p r i n t * , ’ # T= ’ , 2 . 0 * PI / omega , ’ omega_min= ’ , s q r t ( g / l )
p r i n t * , ’ # t 0= ’ , 0 . 0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
cos_theta = g / ( omega * omega * l )
i f ( cos_theta . ge . 1 ) s t o p ’ c o s ( t h e t a )>= 1 ’
sin_theta = s q r t (1.0 − cos_theta * cos_theta )
z = −g / ( omega * omega ) ! they remain c o n s t a n t throught
vz= 0.0 ! t h e motion
r = g / ( omega * omega ) * sin_theta / cos_theta
open ( u n i t =11 , f i l e = ’ ConicalPendulum . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = 0.0
do while ( t . l e . tf )
x = r * c o s ( omega * t )
y = r * s i n ( omega * t )
vx = −r * s i n ( omega * t ) * omega
vy = r * c o s ( omega * t ) * omega
w r i t e ( 1 1 , 1 0 0 ) t , x , y , z , vx , vy , vz
t = t + dt
enddo
close (11)
100 FORMAT(20 G15 . 7 )
end program ConicalPendulum
In order to compile and run the program we can use the commands
shown below:
> g f o r t r a n ConicalPendulum . f90 −o cpd
> . / cpd
# Enter l , omega :
1 . 0 6.28
# Enter t f , dt :
10.0 0.01
# l= 1 . omega= 6.28000021
# T= 1.00050724 omega_min= 3.132092
# t 0= 0. t f = 1 0 . dt= 0.00999999978
The results are recorded in the file ConicalPendulum.dat. In order to
plot the functions x(t), y(t), z(t), vx (t), vy (t), vz (t) we give the following
gnuplot commands:
> gnuplot
gnuplot > p l o t ” ConicalPendulum . dat ” u 1 : 2 w l t ” x ( t ) ”
102 CHAPTER 2. KINEMATICS
gnuplot > replot ” ConicalPendulum . dat ” u 1:3 w l t ”y ( t ) ”
gnuplot > replot ” ConicalPendulum . dat ” u 1:4 w l t ”z( t )”
gnuplot > plot ” ConicalPendulum . dat ” u 1:5 w l t ” v_x ( t ) ”
gnuplot > replot ” ConicalPendulum . dat ” u 1:6 w l t ” v_y ( t ) ”
gnuplot > replot ” ConicalPendulum . dat ” u 1:7 w l t ” v_z ( t ) ”
The results are shown in figure 2.13. In order to make a three dimen-
1 8
x(t) v_x(t)
y(t) v_y(t)
0.8 z(t) v_z(t)
6
0.6
4
0.4
2
0.2
0 0
-0.2
-2
-0.4
-4
-0.6
-6
-0.8
-1 -8
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
Figure 2.13: The plots of the functions x(t), y(t), z(t), vx (t), vy (t), vz (t) of the program
ConicalPendulum.f90 for ω = 6.28, l = 1.0.
sional plot of the trajectory, we should use the gnuplot command splot:
gnuplot > s p l o t ” ConicalPendulum . dat ” u 2 : 3 : 4 w l t ” r ( t ) ”
The result is shown in figure 2.14. We can click on the trajectory and
rotate it and view it from a different angle. We can change the plot limits
with the command:
gnuplot > s p l o t [ − 1 . 1 : 1 . 1 ] [ − 1 . 1 : 1 . 1 ] [ − 0 . 3 : 0 . 0 ] \
” ConicalPendulum . dat ” using 2 : 3 : 4 w l t ” r ( t ) ”
We can animate the trajectory of the particle by using the file animate3D.gnu
from the accompanying software. The commands are similar to the ones
we had to give in the two dimensional case for the planar trajectories
when we used the file animate2D.gnu:
gnuplot > file = ” ConicalPendulum . dat ”
gnuplot > s e t xrange [ − 1 . 1 : 1 . 1 ] ; s e t yrange [ − 1 . 1 : 1 . 1 ]
gnuplot > s e t zrange [ − 0 . 3 : 0 ]
gnuplot > t0 =0; tf =10; dt =0.1
gnuplot > load ” animate3D . gnu”
2.2. MOTION IN SPACE 103
Figure 2.14: The plot of the particle trajectory ⃗r(t) of the program
ConicalPendulum.f90 for ω = 6.28, l = 1.0. We can click and drag with the mouse on
the window and rotate the curve and see it from a different angle. At the bottom left of
the window, we see the viewing direction, given by the angles θ = 55.0 degrees (angle
with the z axis) and ϕ = 62 degrees (angle with the x axis).
The result can be seen in figure 2.15. The program animate3D.gnu can
be used on the data file of any program that prints t x y z as the first
words on each of its lines. All we have to do is to change the value of
the file variable in gnuplot.
Next, we will study the trajectory of a charged particle in a homoge-
neous magnetic field B ⃗ = B ẑ. At time t0 , the particle is at ⃗r0 = x0 x̂ and
its velocity is ⃗v0 = v0y ŷ + v0z ẑ, see figure 2.16. The magnetic force on the
particle is F⃗ = q(⃗v × B) ⃗ = qBvy x̂ − qBvx ŷ and the equations of motion
are
dvx qB
ax = = ωvy ω≡
dt m
dvy
ay = = −ωvx
dt
az = 0. (2.18)
By integrating the above equations with the given initial conditions we
obtain
vx (t) = v0y sin ωt
vy (t) = v0y cos ωt
vz (t) = v0z . (2.19)
104 CHAPTER 2. KINEMATICS
t= 10.100000 (x,y,z)= (0.964311,-0.090732,-0.248742)
0
-0.05
-0.1
z-0.15
-0.2
-0.25
-0.3
-1 0.5 1
-0.5 0 0
x 0.5 -0.5 y
1 -1
Figure 2.15: The particle trajectory ⃗r(t) computed by the program
ConicalPendulum.f90 for ω = 6.28, l = 1.0 and plotted by the gnuplot script
animate3D.gnu. The title of the plot shows the current time and the particles coor-
dinates.
Integrating once more, we obtain the position of the particle as a function
of time
( v0y ) v0y
x(t) = x0 + − cos ωt = x0 cos ωt
ω ω
v0y v0y
y(t) = sin ωt = −x0 sin ωt με x0 = −
ω ω
z(t) = v0z t , (2.20)
where we have chosen x0 = −v0y /ω. This choice places the center of the
circle, which is the projection of the trajectory on the xy plane, to be at
the origin of the coordinate system. The trajectory is a helix with radius
R = −x0 and pitch v0z T = 2πv0z /ω.
We are now ready to write a program that calculates the trajectory
given by (2.20). The user enters the parameters v0 and θ, shown in
figure 2.16, as well as the angular frequency ω (Larmor frequency). The
components of the initial velocity are v0y = v0 cos θ and v0z = v0 sin θ.
The initial position is calculated from the equation x0 = −v0y /ω. The
program can be found in the file ChargeInB.f90:
2.2. MOTION IN SPACE 105
z
B
v0z v0
x0 θ
v0y
y
x
Figure 2.16: A particle at time t0 = 0 is at the position ⃗r0 = x0 x̂ with velocity
⃗ = B ẑ.
⃗v0 = v0y ŷ + v0z ẑ in a homogeneous magnetic field B
! ===========================================================
! F i l e ChargeInB . f90
!A charged p a r t i c l e o f mass m and charge q e n t e r s a magnetic
! f i e l d B i n +z d i r e c t i o n . I t e n t e r s with v e l o c i t y
! v0x =0 , v0y=v0 c o s ( t h e t a ) , v0z=v0 s i n ( t h e t a ) , 0<= t h e t a < p i / 2
! a t t h e p o s i t i o n x0=−v0y / omega , omega=q B/m
!
! Enter v0 and t h e t a and s e e t r a j e c t o r y from
! t 0=0 t o t f a t s t e p dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program ChargeInB
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l : : x , y , z , vx , vy , vz , t , tf , dt
r e a l : : x0 , y0 , z0 , v0x , v0y , v0z , v0
r e a l : : theta , omega
r e a l , parameter : : PI =3.1415927
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter omega : ’
read * , omega
p r i n t * , ’ # Enter v0 , t h e t a ( d e g r e e s ) : ’
read * , v0 , theta
p r i n t * , ’ # Enter t f , dt : ’
read * , tf , dt
p r i n t * , ’ # omega= ’ , omega , ’ T= ’ , 2 . 0 * PI / omega
p r i n t * , ’ # v0= ’ , v0 , ’ t h e t a = ’ , theta , ’ o ( d e g r e e s ) ’
p r i n t * , ’ # t 0= ’ ,0.0 , ’ tf= ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
106 CHAPTER 2. KINEMATICS
! Initialize
i f ( theta . l t . 0 . 0 . or . theta . ge . 9 0 . 0 ) s t o p ’ I l l e g a l 0< t h e t a <90 ’
theta = ( PI / 1 8 0 . 0 ) * theta ! c o n v e r t t o r a d i a n s
v0y = v0 * c o s ( theta )
v0z = v0 * s i n ( theta )
p r i n t * , ’ # v0x= ’ , 0 . 0 , ’ v0y= ’ , v0y , ’ v0z= ’ , v0z
x0 = − v0y / omega
p r i n t * , ’ # x0= ’ , x0 , ’ y0= ’ , 0 . 0 , ’ z0= ’ , 0 . 0
p r i n t * , ’ # xy plane : C i r c l e with c e n t e r ( 0 , 0 ) and R= ’ ,ABS( x0 )
p r i n t * , ’ # s t e p o f h e l i x : s=v0z *T= ’ , v0z * 2 . 0 * PI / omega
open ( u n i t =11 , f i l e = ’ ChargeInB . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
t = 0.0
vz = v0z
do while ( t . l e . tf )
x = x0 * c o s ( omega * t )
y = −x0 * s i n ( omega * t )
z = v0z * t
vx = v0y * s i n ( omega * t )
vy = v0y * c o s ( omega * t )
w r i t e ( 1 1 , 1 0 0 ) t , x , y , z , vx , vy , vz
t = t + dt
enddo
close (11)
100 FORMAT(20 G15 . 7 )
end program ChargeInB
A typical session in which we calculate the trajectories shown in figures
2.17 and 2.18 is shown below:
3.5 1
x(t) v_x(t)
y(t) v_y(t)
z(t) 0.8 v_z(t)
3
0.6
2.5
0.4
2
0.2
1.5 0
-0.2
1
-0.4
0.5
-0.6
0
-0.8
-0.5 -1
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
Figure 2.17: The plots of the x(t), y(t), z(t), vx (t), vy (t), vz (t) functions calculated by
the program in ChargeInB.f90 for ω = 6.28, x0 = 1.0, θ = 20 degrees.
2.3. TRAPPED IN A BOX 107
> g f o r t r a n ChargeInB . f90 −o chg
> . / chg
# Enter omega :
6.28
# Enter v0 , t h e t a ( d e g r e e s ) :
1 . 0 20
# Enter t f , dt :
10 0.01
# omega= 6.28000021 T= 1.00050724
# v0= 1 . theta= 20. o ( d e g r e e s )
# t 0= 0. t f = 1 0 . dt= 0.00999999978
# v0x= 0 . v0y= 0.939692616 v0z= 0.342020124
# x0= −0.149632573 y0= 0 . z0= 0.
# xy plane : C i r c l e with c e n t e r ( 0 , 0 ) and R= 0.149632573
# s t e p o f h e l i x : s=v0z *T= 0.342193604
> gnuplot
gnuplot > p l o t ” ChargeInB . dat ” u 1 : 2 w l t i t l e ”x ( t ) ”
gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 3 w l t i t l e ”y ( t ) ”
gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 4 w l t i t l e ”z ( t )”
gnuplot > p l o t ” ChargeInB . dat ” u 1 : 5 w l t i t l e ” v_x ( t ) ”
gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 6 w l t i t l e ” v_y ( t ) ”
gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 7 w l t i t l e ” v_z ( t ) ”
gnuplot > s p l o t ” ChargeInB . dat ” u 2 : 3 : 4 w l t i t l e ” r ( t ) ”
gnuplot > file = ” ChargeInB . dat ”
gnuplot > s e t xrange [ − 0 . 6 5 : 0 . 6 5 ] ; s e t yrange [ −0 . 6 5 : 0 . 6 5 ]
gnuplot > s e t zrange [ 0 : 1 . 3 ]
gnuplot > t0 =0; tf = 3 . 5 ; dt =0.1
gnuplot > load ” animate3D . gnu”
2.3 Trapped in a Box
In this section we will study the motion of a particle that is free, except
when bouncing elastically on a wall or on certain obstacles. This motion
is calculated by approximate algorithms that introduce systematic errors.
These types of errors¹⁵ are also encountered in the study of more compli-
cated dynamics, but the simplicity of the problem will allow us to control
them in a systematic and easy to understand way.
¹⁵In the previous sections, our calculations had a small systematic error due to the
approximate nature of numerical floating point operations which approximate exact real
number calculations. But the algorithms used were not introducing systematic errors
like in the cases discussed in this section.
108 CHAPTER 2. KINEMATICS
t= 3.500000 (x,y,z)= (0.149623,0.001671,1.197069)
1.2
1
z
0.8
0.6
0.4
0.2
0
-0.6 0.6
-0.4 0.4
-0.2 0.2
0 0
x 0.2 -0.2 y
0.4 -0.4
0.6 -0.6
Figure 2.18: The trajectory ⃗r(t) calculated by the program in ChargeInB.f90 for
ω = 6.28, v0 = 1.0, θ = 20 degrees as shown by the program animate3D.gnu. The
current time and the coordinates of the particle are printed on the title of the plot.
2.3.1 The One Dimensional Box
The simplest example of such a motion is that of a particle in a “one
dimensional box”. The particle moves freely on the x axis for 0 < x < L,
as can be seen in figure 2.19. When it reaches the boundaries x = 0 and
x = L it bounces and its velocity instantly reversed. Its potential energy
is {
0 0<x<L
V (x) = , (2.21)
+∞ elsewhere
which has the shape of an infinitely deep well. The force F = −dV (x)/dx =
0 within the box and F = ±∞ at the position of the walls.
Initially we have to know the position of the particle x0 as well as
its velocity v0 (the sign of v0 depends on the direction of the particle’s
motion) at time t0 . As long as the particle moves within the box, its
motion is free and
x(t) = x0 + v0 (t − t0 )
v(t) = v0 . (2.22)
2.3. TRAPPED IN A BOX 109
x 1111
0000
v
1111
0000
1111
0000
1111111111111
0000000000000
1111
0000
1111
0000
11111111111111111111111111111111
00000000000000000000000000000000
L
Figure 2.19: A particle in a one dimensional box with its walls located at x = 0 and
x = L.
For a small enough change in time δt, so that there is no bouncing on
the wall in the time interval (t, t + δt), we have that
x(t + δt) = x(t) + v(t)δt
v(t + δt) = v(t) . (2.23)
Therefore we could use the above relations in our program and when
the particle bounces off a wall we could simple reverse its velocity v(t) →
−v(t). The devil is hiding in the word “when”. Since the time interval
δt is finite in our program, there is no way to know the instant of the
collision with accuracy better than ∼ δt. However, our algorithm will
change the direction of the velocity at time t + δt, when the particle will
have already crossed the wall. This will introduce a systematic error,
which is expected to decrease with decreasing δt. One way to implement
the above idea is by constructing the loop
do while ( t . l e . tf )
write (11 ,*) t , x , v
x = x + v * dt
t = t + dt
i f ( x . l t . 0.0 . or . x . g t . L ) v = −v
enddo
where the last line gives the testing condition for the wall collision and
the subsequent change of the velocity.
The full program that realizes the proposed algorithm is listed below
and can be found in the file box1D_1.f90. The user can set the size of
the box L, the initial conditions x0 and v0 at time t0, the final time tf
and the time step dt:
110 CHAPTER 2. KINEMATICS
! ============================================================
! F i l e box1D_1 . f90
! Motion o f a f r e e p a r t i c l e i n a box 0<x<L
! Use i n t e g r a t i o n with time s t e p dt : x = x + v * dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program box1D
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of va r ia b le s
r e a l : : L , x0 , v0 , t0 , tf , dt , t , x , v
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter L : ’
read * , L
print * , ’# L = ’ , L
i f ( L . l e . 0 . 0 ) s t o p ’L must be p o s i t i v e . ’
p r i n t * , ’ # Enter x0 , v0 : ’
read * , x0 , v0
p r i n t * , ’ # x0= ’ , x0 , ’ v0= ’ , v0
i f ( x0 . l t . 0.0 . or . x0 . g t . L ) s t o p ’ i l l e g a l v a l u e o f x0 . ’
i f ( v0 . eq . 0.0 ) s t o p ’ i l l e g a l v a l u e o f v0 = 0 . ’
p r i n t * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
t = t0
x = x0
v = v0
open ( u n i t =11 , f i l e = ’ box1D_1 . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( t . l e . tf )
write (11 ,*) t , x , v
x = x + v * dt
t = t + dt
i f ( x . l t . 0.0 . or . x . g t . L ) v = −v
enddo
close (11)
end program box1D
The computed data is recorded in the file box1D_1.dat in three columns.
Compiling, running and plotting the trajectory using gnuplot can be done
as follows:
> g f o r t r a n box1D_1 . f90 −o box1
> . / box1
# Enter L :
2.3. TRAPPED IN A BOX 111
10
# L = 10.
# Enter x0 , v0 :
0 1.0
# x0= 0 . v0= 1.
# Enter t0 , t f , dt :
0 100 0.01
# t 0= 0. t f = 100. dt= 0.00999999978
> gnuplot
gnuplot > p l o t ” box1D_1 . dat ” using 1 : 2 w l t i t l e ” x ( t ) ” ,\
0 notitle , 1 0 notitle
gnuplot > p l o t [ : ] [ − 1 . 2 : 1 . 2 ] ” box1D_1 . dat ” \
using 1 : 3 w l t i t l e ” v ( t ) ”
12
x(t)
10
8
x(t)
10.0002
6 10.0001
10.0001
4
10
10
2
9.99995
0
9.9999
9.99985
-2
0 10 20 30 40 50 60 70 80 90 100 90 90.0005 90.001 90.0015
Figure 2.20: The trajectory x(t) of a particle in a box with L = 10, x0 = 0.0, v0 = 1.0,
t0 = 0, δt = 0.01. The plot to the right magnifies a detail when t ≈ 90 which exposes
the systematic errors in determining the exact moment of the collision of the particle
with the wall at tk = 90 and the corresponding maximum value of x(t), xm = L = 10.0.
The trajectory x(t) is shown in figure 2.20. The effects of the system-
atic errors can be easily seen by noting that the expected collisions occur
every T /2 = L/v = 10 units of time. Therefore, on the plot to the right
of figure 2.20, the reversal of the particle’s motion should have occurred
at t = 90, x = L = 10.
The reader should have already realized that the above mentioned
error can be made to vanish by taking arbitrarily small δt. Therefore,
we naively expect that as long as we have the necessary computer power
to take δt as small as possible and the corresponding time intervals as
many as possible, we can achieve any precision that we want. Well,
that is true only up to a point. The problem is that the next position is
determined by the addition operation x+v*dt and the next moment in
112 CHAPTER 2. KINEMATICS
time by t+dt. Floating point numbers of the REAL type have a maximum
accuracy of approximately 7 significant decimal digits. Therefore, if the
operands x and v*dt are real numbers differing by more than 7 orders
of magnitude (v*dt≲ 10−7 x), the effect of the addition x+v*dt=x, which
is null! The reason is that the floating point unit of the processor has
to convert both numbers x and v*dt into a representation having the
same exponent and in doing so, the corresponding significant digits of
the smaller number v*dt are lost. The result is less catastrophic when
v*dt≲ 10−a x with 0 < a < 7, but some degree of accuracy is also lost at
each addition operation. And since we have accumulation of such errors
over many intervals t→t+dt, the error can become significant and destroy
our calculation for large enough times. A similar error accumulates in
the determination of the next instant of time t+dt, but we will discuss
below how to make this contribution to the total error negligible. The
above mentioned errors can become less detrimental by using floating
point numbers of greater accuracy than the REAL type. For example
REAL(8) numbers have approximately 16 significant decimal digits. But
again, the precision is finite and the same type of errors are there only
to be revealed by a more demanding and complicated calculation.
The remedy to such a problem can only be a change in the algorithm.
This is not always possible, but in the case at hand this is easy to do.
For example, consider the equation that gives the position of a particle
in free motion
x(t) = x0 + v0 (t − t0 ) . (2.24)
Let’s use the above relation for the parts of the motion between two
collisions. Then, all we have to do is to reverse the direction of the
motion and reset the initial position and time to be the position and time
of the collision. This can be done by using the loop:
t = t0
do while ( t . l e . tf )
x = x0 + v0 * ( t−t0 )
w r i t e ( 1 1 , * ) t , x , v0
i f ( x . l t . 0.0 . or . x . g t . L ) then
x0 = x
t0 = t
v0 = −v0
endif
t = t + dt
In the above algorithm, the error in the time of the collision is not van-
2.3. TRAPPED IN A BOX 113
ishing but we don’t have the “instability” problem of the dt→ 0 limit¹⁶.
Therefore we can isolate and study the effect of each type of error. The
full program that implements the above algorithm is given below and
can be found in the file box1D_2.f90:
! ============================================================
! F i l e box1D_2 . f90
! Motion o f a f r e e p a r t i c l e i n a box 0<x<L
! Use c o n s t a n t v e l o c i t y e q u a t i o n : x = x0 + v0 * ( t−t 0 )
! Re verse v e l o c i t y and r e d e f i n e x0 , t 0 on boundaries
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program box1D
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l : : L , x0 , v0 , t0 , tf , dt , t , x , v
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter L : ’
read * , L
print * , ’# L = ’ , L
i f ( L . l e . 0 . 0 ) s t o p ’L must be p o s i t i v e . ’
p r i n t * , ’ # Enter x0 , v0 : ’
read * , x0 , v0
p r i n t * , ’ # x0= ’ , x0 , ’ v0= ’ , v0
i f ( x0 . l t . 0.0 . or . x0 . g t . L ) s t o p ’ i l l e g a l v a l u e o f x0 . ’
i f ( v0 . eq . 0.0 ) s t o p ’ i l l e g a l v a l u e o f v0 = 0 . ’
p r i n t * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
t = t0
open ( u n i t =11 , f i l e = ’ box1D_2 . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( t . l e . tf )
x = x0 + v0 * ( t−t0 )
w r i t e ( 1 1 , * ) t , x , v0
i f ( x . l t . 0.0 . or . x . g t . L ) then
x0 = x
t0 = t
v0 = −v0
endif
t = t + dt
¹⁶We still have this problem in the t=t+dt operation. See discussion in the next
section.
114 CHAPTER 2. KINEMATICS
enddo
close (11)
end program box1D
Compiling and running the above program is done as before and the
results are stored in the file box1D_2.dat. The algorithm can be improved
in order to compute the exact solution. We leave that as an exercise for
the reader¹⁷.
2.3.2 Errors
In this section we will study the effect of the systematic errors that we
encountered in the previous section in more detail. We considered two
types of errors: First, the systematic error of determining the instant
of the collision of the particle with the wall. This error is reduced by
taking a smaller time step δt. Then, the systematic error that accumulates
with each addition of two numbers with increasing difference in their
orders of magnitude. This error is increased with decreasing δt. The
competition of the two effects makes the optimal choice of δt the result of
a careful analysis. Such a situation is found in many interesting problems,
therefore it is quite instructive to study it in more detail.
When the exact solution of the problem is not known, the systematic
errors are controlled by studying the behavior of the solution as a function
of δt. If the solutions are converging in a region of values of δt, one gains
confidence that the true solution has been determined up to the accuracy
of the convergence.
In the previous sections, we studied two different algorithms, pro-
grammed in the files box1D_1.f90 and box1D_2.f90. We will refer to
them as “method 1” and “method 2” respectively. We will study the
convergence of the results as δt → 0 by fixing all the parameters except δt
and then study the dependence of the results on δt. We will take L = 10,
v0 = 1.0, x0 = 0.0, t0 = 0.0, tf = 95.0, so that the particle will collide
with the wall every 10 units of time. We will measure the position of
the particle x(t ≈ 95)¹⁸ as a function of δt and study its convergence to a
limit¹⁹ as δt → 0.
The analysis requires a lot of repetitive work: Compiling, setting the
parameter values, running the program and calculating the value of x(t ≈
¹⁷See the file box1D_3.dat.
¹⁸Note the ≈!
¹⁹Of course we know the answer: x(95) = 5.
2.3. TRAPPED IN A BOX 115
95) for many values of δt. We write the values of the parameters read by
the program in a file box1D_anal.in:
10 L
0 1.0 x0 v0
0 95 0.05 t0 tf dt
Then we compile the program
> g f o r t r a n box1D_1 . f90 −o box
and run it with the command:
> c a t box1D_anal . in | . / box
By using the pipe |, we send the contents of box1D_anal.in to the stdin
of the command ./box by using the command cat. The result x(t ≈ 95)
can be found in the last line of the file box1D_1.dat:
> t a i l −n 1 box1D_1 . dat
94.9511948 5.45000267 −1.
The third number in the above line is the value of the velocity. In a
file box1D_anal.dat we write δt and the first two numbers coming out
from the command tail. Then we decrease the value δt → δt/2 in
the file box1D_anal.in and run again. We repeat for 12 more times
until δt reaches the value²⁰ 0.000012. We do the same²¹ using method 2
and we place the results for x(t ≈ 95) in two new columns in the file
box1D_anal.dat. The result is
# −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
# dt t1_95 x1 (95) x2 (9 5 )
# −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
0.050000 94.95119 5.450003 5.550126
0.025000 94.97849 5.275011 5.174837
0.012500 94.99519 5.124993 5.099736
0.006250 94.99850 4.987460 5.063134
0.003125 94.99734 5.021894 5.035365
²⁰Try the command sed 's/0.05/0.025/' box1D_anal.in | ./box by changing
0.025 with the desired value of δt.
²¹See the shell script box1D_anal.csh as a suggestion on how to automate this boring
process.
116 CHAPTER 2. KINEMATICS
0.001563 94.99923 5.034538 5.017764
0.000781 94.99939 4.919035 5.011735
0.000391 94.99979 4.695203 5.005493
0.000195 95.00000 5.434725 5.001935
0.000098 94.99991 5.528124 5.000745
0.000049 94.99998 3.358000 5.000330
0.000024 94.99998 2.724212 5.000232
0.000012 94.99999 9.240705 5.000158
Convergence is studied in figure 2.21. The 1st method maximizes its
accuracy for δt ≈ 0.01, whereas for δt < 0.0001 the error becomes > 10%
and the method becomes useless. The 2nd method has much better
behavior that the 1st one.
We observe that as δt decreases, the final value of t approaches the
expected tf = 95. Why don’t we obtain t = 95, especially when t/δt is an
integer? How many steps does it really take to reach t ≈ 95, when the
expected number of those is ≈ 95/δt? Each time you take a measurement,
issue the command
> wc −l box1D_1 . dat
which measures the number of lines in the file box1D_1.dat and compare
this number with the expected one. The result is interesting:
# −−−−−−−−−−−−−−−−−−−−−−
# dt N N0
# −−−−−−−−−−−−−−−−−−−−−−
0.050000 1900 1900
0.025000 3800 3800
0.012500 7601 7600
0.006250 15203 15200
0.003125 30394 30400
0.001563 60760 60780
0.000781 121751 121638
0.000391 243753 242966
0.000195 485144 487179
0.000098 962662 969387
0.000049 1972589 1938775
0.000024 4067548 3958333
0.000012 7540956 7916666
where the second column has the number of steps computed by the
program and the third one has the expected number of steps. We
observe that the accuracy decreases with decreasing δt and in the end
the difference is about 5%! Notice that the last line should have given
2.3. TRAPPED IN A BOX 117
tf = 0.000012 × 7540956 ≈ 90.5, an error comparable to the period of the
particle’s motion.
We conclude that one important source of accumulation of system-
atic errors is the calculation of time. This type of errors become more
significant with decreasing δt. We can improve the accuracy of the calcu-
lation significantly if we use the multiplication t=t0+i*dt instead of the
addition t=t+dt, where i is a step counter:
! t = t + dt ! Not a c c u r a t e , avoid
t = t0 + i * dt ! B e t t e r accuracy , p r e f e r
The main loop in the program box1D_1.f90 becomes:
t = t0
x = x0
v = v0
i = 0
do while ( t . l e . tf )
write (11 ,*) t , x , v
i = i + 1
x = x + v * dt
t = t0 + i * dt
i f ( x . l t . 0.0 . or . x . g t . L ) v = −v
enddo
The full program can be found in the file box1D_4.f90 of the accompa-
nying software. We call this “method 3”. We perform the same change
in the file box1D_2.f90, which we store in the file box1D_5.f90. We call
this “method 4”. We repeat the same analysis using methods 3 and 4
and we find that the problem of calculating time accurately practically
vanishes. The result of the analysis can be found on the right plot of fig-
ure 2.21. Methods 2 and 4 have no significant difference in their results,
whereas methods 1 and 3 do have a dramatic difference, with method 3
decreasing the error more than tenfold. The problem of the increase of
systematic errors with decreasing δt does not vanish completely due to
the operation x=x+v*dt. This type of error is harder to deal with and one
has to invent more elaborate algorithms in order to reduce it significantly.
This will be discussed further in chapter 4.
2.3.3 The Two Dimensional Box
A particle is confined to move on the plane in the area 0 < x < Lx and
0 < y < Ly . When it reaches the boundaries of this two dimensional
118 CHAPTER 2. KINEMATICS
100 100
10 10
1 1
δx (%)
δx (%)
0.1 0.1
0.01 0.01
0.001 0.001
method 1 method 3
method 2 method 4
0.0001 0.0001
1e-05 0.0001 0.001 0.01 0.1 1e-05 0.0001 0.001 0.01 0.1
δt δt
Figure 2.21: The error δx = 2|xi (95) − x(95)|/|xi (95) + x(95)| × 100 where xi (95) is
the value calculated by method i = 1, 2, 3, 4 and x(95) the exact value according to the
text.
box, it bounces elastically off its walls. The particle is found in an infinite
depth orthogonal potential well. The particle starts moving at time t0
from (x0 , y0 ) and our program will calculate its trajectory until time tf
with time step δt. Such a trajectory can be seen in figure 2.23.
If the particle’s position and velocity are known at time t, then at time
t + δt they will be given by the relations
x(t + δt) = x(t) + vx (t)δt
y(t + δt) = y(t) + vy (t)δt
vx (t + δt) = vx (t)
vy (t + δt) = vy (t) . (2.25)
The collision of the particle off the walls is modeled by reflection of the
normal component of the velocity when the respective coordinate of the
particle crosses the wall. This is a source of the systematic errors that we
discussed in the previous section. The central loop of the program is:
i = i + 1
t = t0 + i * dt
x = x + vx * dt
y = y + vy * dt
i f ( x . l t . 0.0 . or . x . g t . Lx ) vx = −vx
i f ( y . l t . 0.0 . or . y . g t . Ly ) vy = −vy
The full program can be found in the file box2D_1.f90. Notice that we
introduced two counters nx and ny of the particle’s collisions with the
walls:
2.3. TRAPPED IN A BOX 119
! ============================================================
! F i l e box2D_1 . f90
! Motion o f a f r e e p a r t i c l e i n a box 0<x<Lx 0<y<Ly
! Use i n t e g r a t i o n with time s t e p dt : x = x + vx * dt y=y+vy * dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program box2D
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l ( 8 ) : : Lx , Ly , x0 , y0 , v0x , v0y , t0 , tf , dt , t , x , y , vx , vy
i n t e g e r : : i , nx , ny
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter Lx , Ly : ’
read * , Lx , Ly
p r i n t * , ’ # Lx = ’ , Lx , ’ Ly= ’ , Ly
i f ( Lx . l e . 0 . 0 ) s t o p ’Lx must be p o s i t i v e . ’
i f ( Ly . l e . 0 . 0 ) s t o p ’Ly must be p o s i t i v e . ’
p r i n t * , ’ # Enter x0 , y0 , v0x , v0y : ’
read * , x0 , y0 , v0x , v0y
p r i n t * , ’ # x0= ’ , x0 , ’ y0= ’ , y0 , ’ v0x= ’ , v0x , ’ v0y= ’ , v0y
i f ( x0 . l t . 0.0 . or . x0 . g t . Lx ) s t o p ’ i l l e g a l v a l u e x0 ’
i f ( y0 . l t . 0.0 . or . y0 . g t . Ly ) s t o p ’ i l l e g a l v a l u e y0 ’
i f ( v0x **2+ v0y * * 2 . eq . 0.0 ) s t o p ’ i l l e g a l v a l u e v0=0 ’
p r i n t * , ’ # Enter t0 , t f , dt : ’
read * , t0 , tf , dt
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i = 0
nx = 0 ; ny = 0
t = t0
x = x0 ; y = y0
vx = v0x ; vy = v0y
open ( u n i t =11 , f i l e = ’ box2D_1 . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( t . l e . tf )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
i = i + 1
t = t0 + i * dt
x = x + vx * dt
y = y + vy * dt
i f ( x . l t . 0.0 . or . x . g t . Lx ) then
vx = −vx
nx = nx + 1
endif
i f ( y . l t . 0.0 . or . y . g t . Ly ) then
120 CHAPTER 2. KINEMATICS
vy = −vy
ny = ny + 1
endif
enddo
close (11)
print * , ’# Number o f c o l l i s i o n s : ’
print * , ’# nx= ’ , nx , ’ ny= ’ , ny
end program box2D
A typical session for the study of a particle’s trajectory could be:
> g f o r t r a n box2D_1 . f90 −o box
> . / box
# Enter Lx , Ly :
10.0 5.0
# Lx = 1 0 . Ly= 5.
# Enter x0 , y0 , v0x , v0y :
5.0 0.0 1 . 2 7 1.33
# x0= 5 . y0= 0 . v0x= 1 . 2 7 v0y= 1.33
# Enter t0 , t f , dt :
0 50 0.01
# t 0= 0. t f = 50. dt= 0.01
# Number o f c o l l i s i o n s :
# nx= 6 ny= 13
> gnuplot
gnuplot > p l o t ” box2D_1 . dat ” using 1 : 2 w l title ”x ( t ) ”
gnuplot > r e p l o t ” box2D_1 . dat ” using 1 : 3 w l title ”y ( t ) ”
gnuplot > p l o t ” box2D_1 . dat ” using 1 : 4 w l title ” vx ( t ) ”
gnuplot > r e p l o t ” box2D_1 . dat ” using 1 : 5 w l title ” vy ( t ) ”
gnuplot > p l o t ” box2D_1 . dat ” using 2:3 w l title ”x−y ”
Notice the last line of output from the program: The particle bounces off
the vertical walls 6 times (nx=6) and from the horizontal ones 13 (ny=13).
The gnuplot commands construct the diagrams displayed in figures 2.22
and 2.23.
In order to animate the particle’s trajectory, we can copy the file
box2D_animate.gnu of the accompanying software to the current direc-
tory and give the gnuplot commands:
gnuplot > file = ” box2D_1 . dat ”
gnuplot > Lx = 10 ; Ly = 5
gnuplot > t0 = 0 ; tf = 50; dt = 1
gnuplot > load ” box2D_animate . gnu”
gnuplot > t0 = 0 ; dt = 0 . 5 ; load ” box2D_animate . gnu”
The last line repeats the same animation at half speed. You can also
2.4. APPLICATIONS 121
12 1.5
x(t) vx(t)
y(t) vy(t)
10 1
8
0.5
6
0
4
-0.5
2
0 -1
-2 -1.5
0 10 20 30 40 50 0 10 20 30 40 50
Figure 2.22: The results for the trajectory of a particle in a two dimensional box
given by the program box2D_1.f90. The parameters are Lx = 10, Ly = 5, x0 = 5,
y0 = 0, v0x = 1.27, v0y = 1.33, t0 = 0, tf = 50, δt = 0.01.
use the file animate2D.gnu discussed in section 2.1.1. We add new com-
mands in the file box2D_animate.gnu so that the plot limits are calculated
automatically and the box is drawn on the plot. The arrow drawn is not
the position vector with respect to the origin of the coordinate axes, but
the one connecting the initial with the current position of the particle.
The next step should be to test the accuracy of your results. This can
be done by generalizing the discussion of the previous section and is left
as an exercise for the reader.
2.4 Applications
In this section we will study simple examples of motion in a box with
different types of obstacles. We will start with a game of ... mini golf.
The player shoots a (point) “ball” which moves in an orthogonal box of
linear dimensions Lx and Ly and which is open on the x = 0 side. In
the box there is a circular “hole” with center at (xc , yc ) and radius R. If
the “ball” falls in the “hole”, the player wins. If the ball leaves out of the
box through its open side, the player loses. In order to check if the ball
is in the hole when it is at position (x, y), all we have to do is to check
whether (x − xc )2 + (y − yc )2 ≤ R2 .
Initially we place the ball at the position (0, Ly /2) at time t0 = 0. The
player hits the ball which leaves with initial velocity of magnitude v0 at
an angle θ degrees with the x axis. The program is found in the file
MiniGolf.f90 and is listed below:
122 CHAPTER 2. KINEMATICS
t= 48.000000 (x,y)= (5.901700,3.817100)
5
4
3
y
2
1
0
0 2 4 6 8 10
x
Figure 2.23: The trajectory of the particle of figure 2.22 until t = 48. The origin of
the arrow is at the initial position of the particle and its end is at its current position.
The bold lines mark the boundaries of the box.
! ============================================================
! F i l e MiniGolf . f
! Motion o f a f r e e p a r t i c l e i n a box 0<x<Lx 0<y<Ly
! The box i s open a t x=0 and has a h o l e a t ( xc , yc ) o f r a d i u s R
! B a l l i s s h o t a t ( 0 , Ly / 2 ) with speed v0 , a n g l e t h e t a ( d e g r e e s )
! Use i n t e g r a t i o n with time s t e p dt : x = x + vx * dt y=y+vy * dt
! B a l l s t o p s i n h o l e ( s u c c e s s ) or a t x=0 ( f a i l u r e )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program MiniGolf
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of va r ia b le s
real (8) : : Lx , Ly , x0 , y0 , v0x , v0y , t0 , tf , dt , t , x , y , vx , vy
real (8) : : v0 , theta , xc , yc , R , R2
r e a l ( 8 ) , parameter : : PI =3.14159265358979324 D0
integer : : i , nx , ny
character (7) : : result
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter Lx , Ly : ’
read * , Lx , Ly
p r i n t * , ’ # Lx = ’ , Lx , ’ Ly= ’ , Ly
2.4. APPLICATIONS 123
t= 45.300000 (x,y)= (7.854117,2.982556)
5
4
3
y
2
1
0
0 2 4 6 8 10
x
Figure 2.24: The trajectory of the particle calculated by the program MiniGolf.f90
using the parameters chosen in the text. The moment of ... success is shown. At time
t = 45.3 the particle enters the hole’s region which has its center at (8, 2.5) and its
radius is 0.5.
i f ( Lx . l e . 0 . 0 ) s t o p ’Lx must be p o s i t i v e . ’
i f ( Ly . l e . 0 . 0 ) s t o p ’Ly must be p o s i t i v e . ’
p r i n t * , ’ # Enter h o l e p o s i t i o n and r a d i u s : ( xc , yc ) , R : ’
read * , xc , yc , R
p r i n t * , ’ # ( xc , yc )= ( ’ , xc , ’ , ’ , yc , ’ ) R= ’ , R
p r i n t * , ’ # Enter v0 , t h e t a ( d e g r e e s ) : ’
read * , v0 , theta
p r i n t * , ’ # v0= ’ , v0 , ’ t h e t a = ’ , theta , ’ d e g r e e s ’
i f ( v0 . l e . 0.0 D0 ) s t o p ’ i l l e g a l v a l u e o f v0 . ’
i f (ABS( theta ) . ge . 90.0 D0 ) s t o p ’ i l l e g a l v a l u e o f t h e t a . ’
p r i n t * , ’ # Enter dt : ’
read * , dt
p r i n t * , ’ # dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
t0 = 0.0 D0
x0 = 0.00001 D0 ! s m a l l but non−z e r o
y0 = Ly / 2 . 0
R2 = R * R
theta = ( PI / 1 8 0 . 0 D0 ) * theta
v0x = v0 * c o s ( theta )
124 CHAPTER 2. KINEMATICS
v0y = v0 * s i n ( theta )
p r i n t * , ’ # x0= ’ , x0 , ’ y0= ’ , y0 , ’ v0x= ’ , v0x , ’ v0y= ’ , v0y
i = 0
nx = 0 ; ny = 0
t = t0
x = x0 ; y = y0
vx = v0x ; vy = v0y
open ( u n i t =11 , f i l e = ’ MiniGolf . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( .TRUE. ) ! f o r e v e r !
w r i t e ( 1 1 , * ) t , x , y , vx , vy
i = i + 1
t = t0 + i * dt
x = x + vx * dt
y = y + vy * dt
i f ( x . g t . Lx ) then
vx = −vx
nx = nx + 1
endif
i f ( y . l t . 0.0 . or . y . g t . Ly ) then
vy = −vy
ny = ny + 1
endif
i f ( x . l e . 0.0 D0 ) then
result = ’ Failure ’
e x i t ! e x i t do loop
endif
i f ( ( ( x−xc ) * ( x−xc ) +(y−yc ) * ( y−yc ) ) . l e . R2 ) then
r e s u l t = ’ Success ’
e x i t ! e x i t do loop
endif
enddo
close (11)
p r i n t * , ’ # Number o f c o l l i s i o n s : ’
p r i n t * , ’ # R e s u l t= ’ , r e s u l t , ’ nx= ’ , nx , ’ ny= ’ , ny
end program MiniGolf
In order to run it, we can use the commands:
> g f o r t r a n MiniGolf . f90 −o mg
> . / mg
# Enter Lx , Ly :
10 5
# Lx = 1 0 . Ly= 5.
# Enter h o l e p o s i t i o n and r a d i u s : ( xc , yc ) , R :
8 2.5 0.5
# ( xc , yc )= ( 8. , 2.5 ) R= 0.5
2.4. APPLICATIONS 125
# Enter v0 , t h e t a ( d e g r e e s ) :
1 80
# v0= 1 . theta= 80. d e g r e e s
# Enter dt :
0.01
# dt= 0.01
# x0= 1 . E−05 y0= 2.5 v0x= 0.173648178 v0y= 0.984807753
# Number o f c o l l i s i o n s :
# R e s u l t= S u c c e s s nx= 0 ny= 9
You should construct the plots of the position and the velocity of the
particle. You can also use the animation program found in the file
MiniGolf_animate.gnu for fun. Copy it from the accompanying software
to the current directory and give the gnuplot commands:
gnuplot > file = ” MiniGolf . dat ”
gnuplot > Lx = 1 0 ; Ly = 5
gnuplot > xc = 8 ; yc = 2.5 ; R = 0.5
gnuplot > t0 = 0 ; dt = 0 . 1
gnuplot > load ” MiniGolf_animate . gnu”
The results are shown in figure 2.24.
The next example with be three dimensional. We will study the mo-
tion of a particle confined within a cylinder of radius R and height L.
The collisions of the particle with the cylinder are elastic. We take the
axis of the cylinder to be the z axis and the two bases of the cylinder to
be located at z = 0 and z = L. This is shown in figure 2.26.
The collisions of the particle with the bases of the cylinder are easy to
program: we follow the same steps as in the case of the simple box. For
the collision with the cylinder’s side, we consider the projection of the
motion on the x − y plane. The projection of the particle moves within
a circle of radius R and center at the intersection of the z axis with the
plane. This is shown in figure 2.25. At the collision, the r component
of the velocity is reflected vr → −vr , whereas vθ remains the same. The
velocity of the particle before the collision is
⃗v = vx x̂ + vy ŷ
= vr r̂ + vθ θ̂ (2.26)
and after the collision is
⃗v ′ = vx′ x̂ + vy′ ŷ
= −vr r̂ + vθ θ̂ (2.27)
126 CHAPTER 2. KINEMATICS
From the relations
r̂ = cos θx̂ + sin θŷ
θ̂ = − sin θx̂ + cos θŷ , (2.28)
and vr = ⃗v · r̂, vθ = ⃗v · θ̂, we have that
vr = vx cos θ + vy sin θ
vθ = −vx sin θ + vy cos θ . (2.29)
The inverse relations are
vx = vr cos θ − vθ sin θ
vy = vr sin θ + vθ cos θ . (2.30)
With the transformation vr → −vr , the new velocity in Cartesian coordi-
nates will be
vx′ = −vr cos θ − vθ sin θ
vy′ = −vr sin θ + vθ cos θ . (2.31)
The transformation vx → vx′ , vy → vy′ will be performed in the subroutine
reflectVonCircle(vx,vy,x,y,xc,yc,R). Upon entry to the subroutine,
we provide the initial velocity (vx,vy), the collision point (x,y), the
center of the circle (xc,yc) and the radius of the circle²² R. Upon exit
from the subroutine, (vx,vy) have been replaced with the new values²³
(vx′ , vy′ ).
The program can be found in the file Cylinder3D.f90 and is listed
below:
! ============================================================
! F i l e Cylinder3D . f90
! Motion o f a f r e e p a r t i c l e i n a c y l i n d e r with a x i s t h e z−a x i s ,
! r a d i u s R and 0<z<L
! Use i n t e g r a t i o n with time s t e p dt : x = x + vx * dt
! y = y + vy * dt
! z = z + vz * dt
! Use s u b r o u t i n e r e f l e c t V o n C i r c l e f o r c o l l i s i o n s a t r=R
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
program Cylinder3D
²²Of course one expects R2 = (x − xc )2 + (y − yc )2 , but because of systematic errors,
we require R to be given.
²³Note that upon exit, the particle is also placed exactly on the circle.
2.4. APPLICATIONS 127
v
^ y^
θ vθ ^r vr
v’ ^x
(x,y)
R
− vr
θ
(xc,yc)
Figure 2.25: The elastic collision of the particle moving within the circle of radius
R = |R|⃗ and center ⃗rc = xc x̂ + yc ŷ at the point ⃗r = xx̂ + y ŷ. We have that R ⃗ =
(x − xc )x̂ + (y − yc )ŷ. The initial velocity is ⃗v = vr r̂ + vθ θ̂ where r̂ ≡ R/R. After
⃗
reflecting vr → −vr the new velocity of the particle is ⃗v ′ = −vr r̂ + vθ θ̂.
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
real (8) : : x0 , y0 , z0 , v0x , v0y , v0z , t0 , tf , dt , t , x , y , z , vx , vy , vz
real (8) : : L , R , R2 , vxy , rxy , r2xy , xc , yc
integer : : i , nr , nz
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter R , L : ’
read * , R , L
p r i n t * , ’ # R= ’ , R , ’ L= ’ , L
i f ( R . l e . 0 . 0 ) s t o p ’R must be p o s i t i v e . ’
i f ( L . l e . 0 . 0 ) s t o p ’L must be p o s i t i v e . ’
p r i n t * , ’ # Enter x0 , y0 , z0 , v0x , v0y , v0z : ’
read * , x0 , y0 , z0 , v0x , v0y , v0z
rxy = DSQRT ( x0 * x0+y0 * y0 )
p r i n t * , ’ # x0 = ’ , x0 , ’ y0 = ’ , y0 , ’ z0= ’ , z0 , ’ rxy= ’ , rxy
p r i n t * , ’ # v0x= ’ , v0x , ’ v0y= ’ , v0y , ’ v0z= ’ , v0z
i f ( rxy . g t . R ) s t o p ’ i l l e g a l v a l u e o f rxy > R ’
i f ( z0 . l t . 0.0 D0 ) s t o p ’ i l l e g a l v a l u e o f z0 < 0 ’
i f ( z0 . g t . L ) s t o p ’ i l l e g a l v a l u e o f z0 > L ’
i f ( v0x **2+ v0y **2+ v0z * * 2 . eq . 0 . 0 ) s t o p ’ i l l e g a l v a l u e o f v0 = 0 . ’
p r i n t * , ’ # Enter t0 , t f , dt : ’
128 CHAPTER 2. KINEMATICS
read * , t0 , tf , dt
p r i n t * , ’ # t 0= ’ , t0 , ’ t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
i = 0
nr = 0 ; nz = 0
t = t0
x = x0 ; y = y0 ; z = z0
vx = v0x ; vy = v0y ; vz = v0z
R2 = R * R
xc = 0.0 D0 ! c e n t e r o f c i r c l e which i s t h e p r o j e c t i o n o f t h e
yc = 0.0 D0 ! c y l i n d e r on t h e xy plane
open ( u n i t =11 , f i l e = ’ Cylinder3D . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( t . l e . tf )
w r i t e ( 1 1 , 1 0 0 ) t , x , y , z , vx , vy , vz
i = i + 1
t = t0 + i * dt
x = x + vx * dt
y = y + vy * dt
z = z + vz * dt
i f ( z . l t . 0.0 . or . z . g t . L ) then
vz = −vz ! r e f l e c t i o n on c y l i n d e r caps
nz = nz + 1
endif
r2xy = x * x+y * y
i f ( r2xy . g t . R2 ) then
c a l l reflectVonCircle ( vx , vy , x , y , xc , yc , R )
nr = nr + 1
endif
enddo
close (11)
p r i n t * , ’ # Number o f c o l l i s i o n s : ’
p r i n t * , ’ # nr= ’ , nr , ’ nz= ’ , nz
100 FORMAT(100 G28 . 1 6 )
end program Cylinder3D
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! ============================================================
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
s u b r o u t i n e reflectVonCircle ( vx , vy , x , y , xc , yc , R )
i m p l i c i t none
real (8) : : vx , vy , x , y , xc , yc , R
real (8) : : theta , cth , sth , vr , vth
theta = atan2 ( y−yc , x−xc )
cth = c o s ( theta )
sth = s i n ( theta )
2.4. APPLICATIONS 129
vr = vx * cth + vy * sth
vth = −vx * sth + vy * cth
vx = −vr * cth − vth * sth ! r e f l e c t vr −> −vr
vy = −vr * sth + vth * cth
x = xc + R * cth ! put x , y on t h e c i r c l e
y = yc + R * sth
end s u b r o u t i n e reflectVonCircle
Notice that the function atan2 is used for computing the angle theta.
This function, when called with two arguments atan2(y,x), returns the
angle θ = tan−1 (y/x) in radians. The correct quadrant of the circle where
(x, y) lies is chosen. The angle that we want to compute is given by
atan2(y-yc,x-xc). Then we apply equations (2.29) and (2.31) and in
the last two lines we enforce the particle to be at the point (xc +R cos θ, yc +
R sin θ), exactly on the circle.
t= 500.000000 (x,y,z)= (2.227212,0.469828,7.088600)
10
8
z 6
4
2
0
10
5
-10 0
-5 y
0 -5
x 5
-10
10
Figure 2.26: The trajectory of a particle moving inside a cylinder with R = 10, L = 10,
computed by the program Cylinder3D.f90. We have chosen ⃗r0 = 1.0x̂ + 2.2ŷ + 3.1ẑ,
⃗v0 = 0.93x̂ − 0.89ŷ + 0.74ẑ, t0 = 0, tf = 500.0, δt = 0.01.
A typical session is shown below:
130 CHAPTER 2. KINEMATICS
> g f o r t r a n Cylinder3D . f90 −o cl
> . / cl
# Enter R , L :
10.0 10.0
# R= 1 0 . L= 10.
# Enter x0 , y0 , z0 , v0x , v0y , v0z :
1 . 0 2.2 3 . 1 0.93 −0.89 0 . 7 4
# x0 = 1 . y0 = 2.2 z0= 3 . 1 rxy= 2.41660919
# v0x= 0.93 v0y= −0.89 v0z= 0.74
# Enter t0 , t f , dt :
0.0 500.0 0.01
# t 0= 0. t f = 500. dt= 0.01
# Number o f c o l l i s i o n s :
# nr= 33 nz= 37
In order to plot the position and the velocity as a function of time, we
use the following gnuplot commands:
gnuplot > file=” Cylinder3D . dat ”
gnuplot > p l o t file using 1 : 2 with lines title ” x ( t ) ” ,\
file using 1 : 3 with lines title ” y ( t ) ” ,\
file using 1 : 4 with lines title ” z( t )”
gnuplot > p l o t file using 1 : 5 with lines title ” v_x ( t ) ” ,\
file using 1 : 6 with lines title ” v_y ( t ) ” ,\
file using 1 : 7 with lines title ” v_z ( t ) ”
We can√also compute the distance of the particle from the cylinder’s axis
r(t) = x(t)2 + y(t)2 as a function of time using the command:
gnuplot > p l o t file using 1 : ( s q r t ( $2 **2+ $3 * * 2 ) ) w l t ” r ( t ) ”
In order to plot the trajectory, together with the cylinder, we give the
commands:
gnuplot > L = 10 ; R = 10
gnuplot > s e t urange [ 0 : 2 . 0 * pi ]
gnuplot > s e t vrange [ 0 : L ]
gnuplot > s e t parametric
gnuplot > s p l o t file using 2 : 3 : 4 with lines notitle , \
R * c o s ( u ) , R * s i n ( u ) , v notitle
The command set parametric is necessary if one wants to make a para-
metric plot of a surface ⃗r(u, v) = x(u, v) x̂ + y(u, v) ŷ + z(u, v) ẑ. The cylin-
der (without the bases) is given by the parametric equations ⃗r(u, v) =
R cos u x̂ + R sin u ŷ + v ẑ with u ∈ [0, 2π), v ∈ [0, L].
2.4. APPLICATIONS 131
We can also animate the trajectory with the help of the gnuplot script
file Cylinder3D_animate.gnu. Copy the file from the accompanying soft-
ware to the current directory and give the gnuplot commands:
gnuplot > R =10; L =10; t0 =0; tf =500; dt=10
gnuplot > load ” Cylinder3D_animate . gnu”
The result is shown in figure 2.26.
The last example will be that of a simple model of a spacetime worm-
hole. This is a simple spacetime geometry which, in the framework of
the theory of general relativity, describes the connection of two distant
areas in space which are asymptotically flat. This means, that far enough
from the wormhole’s mouths, space is almost flat - free of gravity. Such
a geometry is depicted in figure 2.27. The distance traveled by someone
through the mouths could be much smaller than the distance traveled
outside the wormhole and, at least theoretically, traversable wormholes
could be used for interstellar/intergalactic traveling and/or communica-
tions between otherwise distant areas in the universe. Of course we
should note that such macroscopic and stable wormholes are not known
to be possible to exist in the framework of general relativity. One needs
an exotic type of matter with negative energy density which has never
been observed. Such exotic geometries may realize microscopically as
quantum fluctuations of spacetime and make the small scale structure of
the geometry²⁴ a “spacetime foam”.
We will study a very simple model of the above geometry on the plane
with a particle moving freely in it²⁵. We take the two dimensional plane
and cut two equal disks of radius R with centers at distance d like in
figure 2.28. We identify the points on the two circles such that the point
1 of the left circle is the same as the point 1 on the right circle, the point 2
on the left with the point 2 on the right etc. The two circles are given by
the parametric equations x(θ) = d/2 + R cos θ, y(θ) = R sin θ, −π < θ ≤ π
for the right circle and x(θ) = −d/2 − R cos θ, y(θ) = R sin θ, −π < θ ≤ π
for the left. Points on the two circles with the same θ are identified.
A particle entering the wormhole from the left circle with velocity v is
immediately exiting from the right with velocity v ′ as shown in figure
2.28.
²⁴See K.S. Thorne “Black Holes and Time Wraps: Einstein’s Outrageous Legacy”,
W.W. Norton, New York for a popular review of these concepts.
²⁵This idea can be found as an exercise in the excellent introductory general relativ-
ity textbook J. B. Hartle, “Gravity: An Introduction to Einstein’s General Relativity”,
Addison Wesley 2003, Ch. 7, Ex. 25.
132 CHAPTER 2. KINEMATICS
Figure 2.27: A typical geometry of space near a wormhole. Two asymptotically
flat regions of space are connected through a “neck” which can be arranged to be of
small length compared to the distance of the wormhole mouths when traveled from the
outside space.
Then we will do the following:
1. Write a program that computes the trajectory of a particle moving
in the geometry of figure 2.28. We set the limits of motion to be
−L/2 ≤ x ≤ L/2 and −L/2 ≤ y ≤ L/2. We will use periodic
boundary conditions in order to define what happens when the
particle attempts to move outside these limits. This means that
we identify the x = −L/2 line with the x = +L/2 line as well
as the y = −L/2 line with the y = +L/2 line. The user enters the
parameters R, d and L as well as the initial conditions (x0 , y0 ), (v0 , ϕ)
where ⃗v0 = v0 (cos ϕx̂ + sin ϕŷ). The user will also provide the time
parameters tf and dt for motion in the time interval t ∈ [t0 = 0, tf ]
with step dt.
2. Plot the particle’s trajectory with (x0 , y0 ) = (0, −1), (v0 , ϕ) = (1, 10o )
με tf = 40, dt = 0.05 in the geometry with L = 20, d = 5, R = 1.
3. Find a closed trajectory which does not cross the boundaries |x| =
L/2, |y| = L/2 and determine whether it is stable under small per-
turbations of the initial conditions.
4. Find other closed trajectories that go through the mouths of the
2.4. APPLICATIONS 133
y
v’
4 4 θ
θ
v
3 1 1 3 x
2 2
Figure 2.28: A simple model of the spacetime geometry of figure 2.27. The particle
moves on the whole plane except withing the two disks that have been removed. The
neck of the wormhole is modeled by the two circles x(θ) = ±d/2±R cos θ, y(θ) = R sin θ,
−π < θ ≤ π and has zero length since their points have been identified. There is a
given direction in this identification, so that points with the same θ are the same (you
can imagine how this happens by folding the plane across the y axis and then glue the
two circles together). The entrance of the particle through one mouth and exit through
the other is done as shown for the velocity vector ⃗v → ⃗v ′ .
134 CHAPTER 2. KINEMATICS
wormhole and study their stability under small perturbations of
the initial conditions.
5. Add to the program the option to calculate the distance traveled by
the particle. If the particle starts from (−x0 , 0) and moves in the +x
direction to the (x0 , 0), x0 > R + d/2 position, draw the trajectory
and calculate the distance traveled on paper. Then confirm your
calculation from the numerical result coming from your program.
6. Change the boundary conditions, so that the particle bounces off
elastically at |x| = L/2, |y| = L/2 and replot all the trajectories
mentioned above.
Define the right circle c1 by the parametric equations
d
x(θ) = + R cos θ , y(θ) = R sin θ , −π < θ ≤ π , (2.32)
2
and the left circle c2 by the parametric equations
d
x(θ) = − − R cos θ , y(θ) = R sin θ , −π < θ ≤ π . (2.33)
2
The particle’s position changes at time dt by
ti = idt
xi = xi−1 + vx dt
yi = yi−1 + vy dt
(2.34)
for i = 1, 2, . . . for given (x0 , y0 ), t0 = 0 and as long as ti ≤ tf . If the
point (xi , yi ) is outside the boundaries |x| = L/2, |y| = L/2, we redefine
xi → xi ± L, yi → yi ± L in each case respectively. Points defined by
the same value of θ are identified, i.e. they represent the same points of
space. If the point (xi , yi ) crosses either one of the circles c1 or c2 , then
we take the particle out from the other circle.
Crossing the circle c1 is determined by the relation
( )2
d
xi − + yi2 ≤ R2 . (2.35)
2
The angle θ is calculated from the equation
( )
y
θ = tan−1
i
, (2.36)
xi − d2
2.4. APPLICATIONS 135
y
v’
^ ^
e
e’θ θ
^
er
^
e’r (x’,y’) v (x,y)
θ θ x
Figure 2.29: The particle crossing the wormhole through the right circle c1 with
velocity ⃗v . It emerges from c2 with velocity ⃗v ′ . The unit vectors (êr , êθ ), (ê′r , ê′θ ) are
computed from the parametric equations of the two circles c1 and c2 .
and the point (xi , yi ) is mapped to the point (x′i , yi′ ) where
d
x′i = − − R cos θ , yi′ = yi , (2.37)
2
as can be seen in figure 2.29. For mapping ⃗v → ⃗v ′ , we first calculate the
vectors
} { ′
êr = cos θ x̂ + sin θ ŷ êr = − cos θ x̂ + sin θ ŷ
→ , (2.38)
êθ = − sin θ x̂ + cos θ ŷ ê′θ = sin θ x̂ + cos θ ŷ
so that the velocity
⃗v = vr êr + vθ êθ → ⃗v ′ = −vr ê′r + vθ ê′θ , (2.39)
where the radial components are vr = ⃗v · êr and vθ = ⃗v · êθ . Therefore,
the relations that give the “emerging” velocity ⃗v ′ are:
vr = vx cos θ + vy sin θ
vθ = −vx sin θ + vy cos θ
. (2.40)
vx′ = vr cos θ + vθ sin θ
vy′ = −vr sin θ + vθ cos θ
Similarly we calculate the case of entering from c2 and emerging from
c1 . The condition now is:
( )2
d
xi + + yi2 ≤ R2 . (2.41)
2
136 CHAPTER 2. KINEMATICS
The angle θ is given by
( )
yi
θ = π − tan−1 d
, (2.42)
xi + 2
and the point (xi , yi ) is mapped to the point (x′i , yi′ ) where
d
x′i = + R cos θ , yi′ = yi . (2.43)
2
For mapping ⃗v → ⃗v ′ , we calculate the vectors
} { ′
êr = − cos θ x̂ + sin θ ŷ êr = cos θ x̂ + sin θ ŷ
→ ′ , (2.44)
êθ = sin θ x̂ + cos θ ŷ êθ = − sin θ x̂ + cos θ ŷ
so that the velocity
⃗v = vr êr + vθ êθ → ⃗v ′ = −vr ê′r + vθ ê′θ . (2.45)
The emerging velocity ⃗v ′ is:
vr = −vx cos θ + vy sin θ
vθ = vx sin θ + vy cos θ
′ . (2.46)
vx = −vr cos θ − vθ sin θ
vy′ = −vr sin θ + vθ cos θ
Systematic errors are now coming from crossing the two mouths of the
wormhole. There are no systematic errors from crossing the boundaries
|x| = L/2, |y| = L/2 (why?). Try to think of ways to control those errors
and study them.
The closed trajectories that we are looking for come from the initial
conditions
(x0 , y0 , v0 , ϕ) = (0, 0, 1, 0) (2.47)
and they connect points 1 of figure 2.28. They are unstable, as can be
seen by taking ϕ → ϕ + ϵ.
The closed trajectories that cross the wormhole and “wind” through
space can come from the initial conditions
(x0 , y0 , v0 , ϕ) = (−9, 0, 1, 0)
(x0 , y0 , v0 , ϕ) = (2.5, −3, 1, 90o )
and cross the points 3 → 3 and 2 → 2 → 4 → 4 respectively. They are
also unstable, as can be easily verified by using the program that you will
write. The full program is listed below:
2.4. APPLICATIONS 137
! ============================================================
program WormHole2D
i m p l i c i t none
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Declaration of variables
r e a l ( 8 ) , parameter : : PI =3.14159265358979324 D0
r e a l ( 8 ) : : Lx , Ly , L , R , d
r e a l ( 8 ) : : x0 , y0 , v0 , theta
r e a l ( 8 ) : : t0 , tf , dt
r e a l ( 8 ) : : t , x , y , vx , vy
r e a l ( 8 ) : : xc1 , yc1 , xc2 , yc2 , r1 , r2
integer : : i
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Ask u s e r f o r i n p u t :
p r i n t * , ’ # Enter L , d , R : ’
read * , L , d , R
p r i n t * , ’ # L= ’ , L , ’ d= ’ , d , ’ R= ’ , R
i f ( L . l e . d +2.0 D0 * R ) s t o p ’L <= d+2*R ’
i f ( d . le . 2.0 D0 * R ) s t o p ’d <= 2*R ’
p r i n t * , ’ # Enter ( x0 , y0 ) , v0 , t h e t a ( d e g r e e s ) : ’
read * , x0 , y0 , v0 , theta
p r i n t * , ’ # x0= ’ , x0 , ’ y0 = ’ , y0
p r i n t * , ’ # v0= ’ , v0 , ’ t h e t a = ’ , theta , ’ d e g r e e s ’
i f ( v0 . l e . 0.0 D0 ) s t o p ’ i l l e g a l v a l u e o f v0 . ’
p r i n t * , ’ # Enter t f , dt : ’
read * , tf , dt
p r i n t * , ’ # t f = ’ , tf , ’ dt= ’ , dt
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Initialize
theta = ( PI / 1 8 0 . 0 D0 ) * theta
i = 0
t = 0.0 D0
x = x0 ; y = y0
vx = v0 * c o s ( theta ) ; vy = v0 * s i n ( theta )
p r i n t * , ’ # x0= ’ , x , ’ y0= ’ , y , ’ v0x= ’ , vx , ’ v0y= ’ , vy
! Wormhole ’ s c e n t e r s :
xc1 = 0.5 D0 * d ; yc1 = 0.0 D0
xc2 = −0.5D0 * d ; yc2 = 0.0 D0
! Box l i m i t s c o o r d i n a t e s :
Lx = 0.5 D0 * L ; Ly = 0.5 D0 * L
! Test i f already inside cut region :
r1 = s q r t ( ( x−xc1 ) * * 2 + ( y−yc1 ) * * 2 )
r2 = s q r t ( ( x−xc2 ) * * 2 + ( y−yc2 ) * * 2 )
i f ( r1 . l e . R ) s t o p ’ r 1 <= R ’
i f ( r2 . l e . R ) s t o p ’ r2 <= R ’
! T e s t i f o u t s i d e box l i m i t s :
i f (ABS( x ) . ge . Lx ) s t o p ’ | x | >= Lx ’
i f (ABS( y ) . ge . Ly ) s t o p ’ | y | >= Ly ’
138 CHAPTER 2. KINEMATICS
open ( u n i t =11 , f i l e = ’Wormhole . dat ’ )
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Compute :
do while ( t . l t . tf )
w r i t e ( 1 1 , * ) t , x , y , vx , vy
i = i+1
t = i * dt
x = x + vx * dt ; y = y + vy * dt
! T o r o i d a l boundary c o n d i t i o n s :
i f ( x . g t . Lx ) x = x − L
i f ( x . l t . −Lx ) x = x + L
i f ( y . g t . Ly ) y = y − L
i f ( y . l t . −Ly ) y = y + L
! Test i f i n s i d e the cut disks
r1 = s q r t ( ( x−xc1 ) * * 2 + ( y−yc1 ) * * 2 )
r2 = s q r t ( ( x−xc2 ) * * 2 + ( y−yc2 ) * * 2 )
i f ( r1 . l t . R ) then
! N o t i c e : we pass r 1 as r a d i u s o f c i r c l e , not R
c a l l crossC1 ( x , y , vx , vy , dt , r1 , d )
e l s e i f ( r2 . l t . R ) then
c a l l crossC2 ( x , y , vx , vy , dt , r2 , d )
endif
! s m a l l chance here t h a t s t i l l i n C1 or C2 , but OK s i n c e
! another dt−advance giv e n a t t h e beginning o f do−loop
enddo ! do while ( t . l t . t f )
end program WormHole2D
! =======================================================
s u b r o u t i n e crossC1 ( x , y , vx , vy , dt , R , d )
i m p l i c i t none
r e a l ( 8 ) : : x , y , vx , vy , dt , R , d
r e a l ( 8 ) : : vr , v0 ! v0 −> v t h e t a
r e a l ( 8 ) : : theta , xc , yc
p r i n t * , ’ # I n s i d e C1 : ( x , y , vx , vy , R)= ’ , x , y , vx , vy , R
xc = 0.5 D0 * d ! c e n t e r o f C1
yc = 0.0 D0
theta = atan2 ( y−yc , x−xc )
x = −xc − R * c o s ( theta ) ! new x−value , y i n v a r i a n t
! Velocity transformation :
vr = vx * c o s ( theta )+vy * s i n ( theta )
v0 = −vx * s i n ( theta )+vy * c o s ( theta )
vx = vr * c o s ( theta )+v0 * s i n ( theta )
vy = −vr * s i n ( theta )+v0 * c o s ( theta )
! advance x , y , h o p e f u l l y o u t s i d e C2 :
x = x + vx * dt
y = y + vy * dt
print * , ’# Exit C2 : ( x , y , vx , vy )= ’ , x , y , vx , vy
end s u b r o u t i n e crossC1
! =======================================================
s u b r o u t i n e crossC2 ( x , y , vx , vy , dt , R , d )
2.4. APPLICATIONS 139
implicit none
real (8) , parameter : : PI =3.14159265358979324 D0
real (8) : : x , y , vx , vy , dt , R , d
real (8) : : vr , v0 ! v0 −> v t h e t a
real (8) : : theta , xc , yc
p r i n t * , ’ # I n s i d e C2 : ( x , y , vx , vy , R)= ’ , x , y , vx , vy , R
xc = −0.5D0 * d ! c e n t e r o f C2
yc = 0.0 D0
theta = PI−atan2 ( y−yc , x−xc )
x = −xc + R * c o s ( theta ) ! new x−value , y i n v a r i a n t
! Velocity transformation :
vr = −vx * c o s ( theta )+vy * s i n ( theta )
v0 = vx * s i n ( theta )+vy * c o s ( theta )
vx = −vr * c o s ( theta )−v0 * s i n ( theta )
vy = −vr * s i n ( theta )+v0 * c o s ( theta )
! advance x , y , h o p e f u l l y o u t s i d e C1 :
x = x + vx * dt
y = y + vy * dt
print * , ’# Exit C1 : ( x , y , vx , vy )= ’ , x , y , vx , vy
end s u b r o u t i n e crossC2
It is easy to compile and run the program. See also the files Wormhole.csh
and Wormhole_animate.gnu of the accompanying software and run the
gnuplot commands:
gnuplot > file = ”Wormhole . dat ”
gnuplot > R =1; d =5; L=20;
gnuplot > ! . / Wormhole . csh
gnuplot > t0 =0; dt = 0 . 2 ; load ” Wormhole_animate . gnu”
You are now ready to answer the rest of the questions that we asked in
our list.
140 CHAPTER 2. KINEMATICS
2.5 Problems
2.1 Change the program Circle.f90 so that it prints the number of full
circles traversed by the particle.
2.2 Add all the necessary tests on the parameters entered by the user
in the program Circle.f90, so that the program is certain to run
without problems. Do the same for the rest of the programs given
in the same section.
2.3 A particle moves with constant angular velocity ω on a circle that
has the origin of the coordinate system at its center. At time t0 = 0,
the particle is at (x0 , y0 ). Write the program CircularMotion.f90
that will calculate the particle’s trajectory. The user should enter the
parameters ω, x0 , y0 , t0 , tf , δt. The program should print the results
like the program Circle.f90 does.
2.4 Change the program SimplePendulum.f90 so that the user could
enter a non zero initial velocity.
2.5 Study the k → 0 limit in the projectile motion given by equations
(2.10). Expand e−kt = 1−kt+ 2!1 (kt)2 +. . . and keep the non vanish-
ing terms as k → 0. Then keep the next order leading terms which
have a smaller power of k. Program these relations in a file
ProjectileSmallAirResistance.f90. Consider the initial condi-
tions ⃗v0 = x̂ + ŷ and calculate the range of the trajectory numerically
by using the two programs
ProjectileSmallAirResistance.f90, ProjectileAirResistance.f90.
Determine the range of values of k for which the two results agree
within 5% accuracy.
2.6 Write a program for a projectile which moves through a fluid with
fluid resistance proportional to the square of the velocity. Compare
the range of the trajectory with the one calculated by the program
ProjectileAirResistance.f90 for the parameters shown in figure
2.10.
2.7 Change the program Lissajous.f90 so that the user can enter a
different amplitude and initial phase in each direction. Study the
case where the amplitudes are the same and the phase difference
in the two directions are π/4, π/2, π, −π. Repeat by taking the am-
plitude in the y direction to be twice as much the amplitude in the
x direction.
2.5. PROBLEMS 141
2.8 Change the program ProjectileAirResistance.f90, so that it can
calculate also the k = 0 case.
2.9 Change the program ProjectileAirResistance.f90 so that it can
calculate the trajectory of the particle in three dimensional space.
Plot the position coordinates and the velocity components as a func-
tion of time. Plot the three dimensional trajectory using splot
in gnuplot and animate the trajectory using the gnuplot script
animate3D.gnu.
2.10 Change the program ChargeInB.f90 so that it can calculate the
number of full revolutions that the projected particle’s position on
the x − y plane makes during its motion.
2.11 Change the program box1D_1.f90 so that it prints the number of
the particle’s collisions on the left wall, on the right wall and the
total number of collisions to the stdout.
2.12 Do the same for the program box1D_2.f90. Fill the table on page
115 the number of calculated collisions and comment on the results.
2.13 Run the program box1D_1.f90 and choose L= 10, v0=1. Decrease
the step dt up to the point that the particle stops to move. For
which value of dt this happens? Increase v0=10,100. Until which
value of dt the particle moves now? Why?
2.14 Change the REAL declarations to REAL(8) in the program box1D_1.f90.
Add explicit exponents D0 to all constants (e.g. 0.0→0.0D0). Com-
pare your results to those obtained in section 2.3.2. Repeat problem
2.13. What do you observe?
2.15 Change the program box1D_1.f90 so that you can study non elastic
collisions v ′ = −ev, 0 < e ≤ 1 with the walls.
2.16 Change the program box2D_1.f90 so that you can study inelastic
collisions with the walls, such that vx′ = −evx , vy′ = −evy , 0 < e ≤ 1.
2.17 Use the method of calculating time in the programs box1D_4.f90
and box1D_5.f90 in order to produce the results in figure 2.21.
2.18 Particle falls freely moving in the vertical direction. It starts with
zero velocity at height h. Upon reaching the ground, it bounces
inelastically such that vy′ = −evy with 0 < e ≤ 1 a parameter. Write
the necessary program in order to study numerically the particle’s
motion and study the cases e = 0.1, 0.5, 0.9, 1.0.
142 CHAPTER 2. KINEMATICS
2.19 Generalize the program of the previous problem so that you can
study the case ⃗v0 = v0x x̂. Animate the calculated trajectories.
2.20 Study the motion of a particle moving inside the box of figure 2.30.
Count the number of collisions of the particle with the walls before
it leaves the box.
Lx a
Ly
Figure 2.30: Problem 2.20.
2.21 Study the motion of the point particle on the “billiard table” of
figure 2.31. Count the number of collisions with the walls before
the particle enters into a hole. The program should print from
which hole the particle left the table.
a a
a a
Ly
a a a
a
Lx
Figure 2.31: Problem 2.21.
2.22 Write a program in order to study the motion of a particle in the
box of figure 2.32. At the center of the box there is a disk on
which the particle bounces off elastically (Hint: use the routine
reflectVonCircle of the program Cylinder3D.f90).
2.23 In the box of the previous problem, put four disks on which the
particle bounces of elastically like in figure 2.33.
2.5. PROBLEMS 143
Ly
2R
Lx
Figure 2.32: Problem 2.22.
2R
111
000
111
000
111
000
111
000
111
000
111
000
111 111
000 Ly
000
111
000
111
000
111
000
2R 111
000
111
000
111
000
111
000 2R
2R
111
000
111
000
111
000
111
000
111
000
2a
Lx
Figure 2.33: Problem 2.23.
2.24 Consider the arrangement of figure 2.34. Each time the particle
bounces elastically off a circle, the circle disappears. The game is
over successfully if all the circles vanish. Each time the particle
bounces off on the wall to the left, you lose a point. Try to find
trajectories that minimize the number of lost points.
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
Ly
111
000 111
000
111
000
111 111
000
000
111
000
111
000
111
000
111
000 111
000
111
000
111 111 111
000 000
000
111
000
111
000
111
111
000 000
111
000
111
000 111
000
000 000
111 111
000 111 111
000
Lx
Figure 2.34: Problem 2.24.
144 CHAPTER 2. KINEMATICS
Chapter 3
Logistic Map
Nonlinear differential equations model interesting dynamical systems in
physics, biology and other branches of science. In this chapter we per-
form a numerical study of the discrete logistic map as a “simple math-
ematical model with complex dynamical properties” [21] similar to the
ones encountered in more complicated and interesting dynamical sys-
tems. For certain values of the parameter of the map, one finds chaotic
behavior giving us an opportunity to touch on this very interesting topic
with important consequences in physical phenomena. Chaotic evolu-
tion restricts out ability for useful predictions in an otherwise fully deter-
ministic dynamical system: measurements using slightly different initial
conditions result in a distribution which is indistinguishable from the dis-
tribution coming from sampling a random process. This scientific field is
huge and active and we refer the reader to the bibliography for a more
complete introduction [21, 22, 23, 24, 25, 26, 27, 38].
3.1 Introduction
The most celebrated application of the logistic map comes from the study
of population growth in biology. One considers populations which re-
produce at fixed time intervals and whose generations do not overlap.
The simplest (and most naive) model is the one that makes the rea-
sonable assumption that the rate of population growth dP (t)/dt of a
population P (t) is proportional to the current population:
dP (t)
= kP (t) . (3.1)
dt
The general solution of the above equation is P (t) = P (0)ekt showing
an exponential population growth for k > 0 an decline for k < 0. It
145
146 CHAPTER 3. LOGISTIC MAP
is obvious that this model is reasonable as long as the population is
small enough so that the interaction with its environment (adequate food,
diseases, predators etc) can be neglected. The simplest model that takes
into account some of the factors of the interaction with the environment
(e.g. starvation) is obtained by the introduction of a simple non linear
term in the equation so that
dP (t)
= kP (t)(1 − bP (t)) . (3.2)
dt
The parameter k gives the maximum growth rate of the population and
b controls the ability of the species to maintain a certain population level.
The equation (3.2) can be discretized in time by assuming that each gen-
eration reproduces every δt and that the n-th generation has population
Pn = P (tn ) where tn = t0 + (n − 1)δt. Then P (tn+1 ) ≈ P (tn ) + δtP ′ (tn ) and
equation (3.1) becomes
Pn+1 = rPn , (3.3)
where r = 1 + kδt. The solutions of the above equation are well ap-
proximated by Pn ∼ P0 ektn ∝ e(r−1)n so that we have population growth
when r > 1 and decline when r < 1. Equation (3.2) can be discretized
as follows:
Pn+1 = Pn (r − bPn ) . (3.4)
Defining xn = (b/r)Pn we obtain the logistic map
xn+1 = rxn (1 − xn ) . (3.5)
We define the functions
f (x) = rx(1 − x), F (x, r) = rx(1 − x) (3.6)
(their only difference is that, in the first one, r is considered as a given
parameter), so that
xn+1 = f (xn ) = f (2) (xn−1 ) = . . . = f (n) (x1 ) = f (n+1) (x0 ) , (3.7)
where we use the notation f (1) (x) = f (x), f (2) (x) = f (f (x)), f (3) (x) =
f (f (f (x))), . . . for function composition. In what follows, the derivative
of f will be useful:
∂F (x, r)
f ′ (x) = = r(1 − 2x) . (3.8)
∂x
3.2. FIXED POINTS AND 2N CYCLES 147
Since we interpret xn to be the fraction of the population with respect
to its maximum value, we should have 0 ≤ xn ≤ 1 for each¹ n. The
function f (x) has one global maximum for x = 1/2 which is equal to
f (1/2) = r/4. Therefore, if r > 4, then f (1/2) > 1, which for an appro-
priate choice of x0 will lead to xn+1 = f (xn ) > 1 for some value of n.
Therefore, the interval of values of r which is of interest for our model
is
0 < r ≤ 4. (3.9)
The logistic map (3.5) may be viewed as a finite difference equation
and it is a one step inductive relation. Given an initial value x0 , a sequence
of values {x0 , x1 , . . . , xn , . . . } is produced. This will be referred² to as
the trajectory of x0 . In the following sections we will study the properties
of these trajectories as a function of the parameter r.
The solutions of the logistic map are not known except in special
cases. For r = 2 we have
1( )
xn = 1 − (1 − x0 )2n , (3.10)
2
and for³ r = 4
1 √
xn = sin2 (2n πθ) , θ= sin−1 x0 . (3.11)
π
For r = 2, limn→∞ xn = 1/2 whereas for r = 4 we have periodic trajectories
resulting in rational θ and non periodic resulting in irrational θ. For other
values of r we have to resort to a numerical computation of the trajectories
of the logistic map.
3.2 Fixed Points and 2n Cycles
It is obvious that if the point x∗ is a solution of the equation x = f (x), then
xn = x∗ ⇒ xn+k = x∗ for every k ≥ 0. For the function f (x) = rx(1 − x)
we have two solutions
x∗1 = 0 and x∗2 = 1 − 1/r . (3.12)
¹Note that if xn > 1 then xn+1 < 0, so that if we want xn ≥ 0 for each n, then we
should have xn ≤ 1 for each n.
²In the bibliography, the term “splinter of x0 ” is frequently used.
³E. Schröder, “Über iterierte Funktionen”, Math. Ann. 3 (1870) 296; E. Lorenz,
“The problem of deducing the climate from the governing equations”, Tellus 16 (1964)
1
148 CHAPTER 3. LOGISTIC MAP
We will see that for appropriate values of r, these solutions are attractors
of most of the trajectories. This means that for a range of values for the
initial point 0 ≤ x0 ≤ 1, the sequence {xn } approaches asymptotically one
of these points as n → ∞. Obviously the (measure zero) sets of initial
values {x0 } = {x∗1 } and {x0 } = {x∗2 } result in trajectories attracted by x∗1
and x∗2 respectively. In order to determine which one of the two values
is preferred, we need to study the stability of the fixed points x∗1 and x∗2 .
For this, assume that for some value of n, xn is infinitesimally close to
the fixed point x∗ so that
xn = x∗ + ϵn
xn+1 = x∗ + ϵn+1 . (3.13)
Since
xn+1 = f (xn ) = f (x∗ + ϵn ) ≈ f (x∗ ) + ϵn f ′ (x∗ ) = x∗ + ϵn f ′ (x∗ ) , (3.14)
where we used the Taylor expansion of the analytic function f (x∗ + ϵn )
about x∗ and the relation x∗ = f (x∗ ), we have that ϵn+1 = ϵn f ′ (x∗ ). Then
we obtain
ϵn+1
= |f ′ (x∗ )| . (3.15)
ϵn
Therefore, if |f ′ (x∗ )| < 1 we obtain limn→∞ ϵn = 0 and the fixed point x∗ is
stable: the sequence {xn+k } approaches x∗ asymptotically. If |f ′ (x∗ )| > 1
then the sequence {xn+k } deviates away from x∗ and the fixed point is
unstable. The limiting case |f ′ (x∗ )| = 1 should be studied separately and
it indicates a change in the stability properties of the fixed point. In the
following discussion, these points will be shown to be bifurcation points.
For the function f (x) = rx(1 − x) with f ′ (x) = r(1 − 2x) we have that
f (0) = r and f ′ (1 − 1/r) = 2 − r. Therefore, if r < 1 the point x∗1 = 0
′
is an attractor, whereas the point x∗2 = 1 − 1/r < 0 is irrelevant. When
r > 1, the point x∗1 = 0 results in |f ′ (x∗1 )| = r > 1, therefore x∗1 is unstable.
Any initial value x0 near x∗1 deviates from it. Since for 1 < r < 3 we have
that 0 ≤ |f ′ (x∗2 )| = |2 − r| < 1, the point x∗2 is an attractor. Any initial
value x0 ∈ (0, 1) approaches x∗2 = 1 − 1/r. When r = rc = 1 we have the
(1)
limiting case x∗1 = x∗2 = 0 and we say that at the critical value rc = 1 the
(1)
fixed point x∗1 bifurcates to the two fixed points x∗1 and x∗2 .
As r increases, the fixed points continue to bifurcate. Indeed, when
r = rc = 3 we have that f ′ (x∗2 ) = 2 − r = −1 and for r > rc the point
(2) (2)
x∗2 becomes unstable. Consider the solution of the equation x = f (2) (x).
If 0 < x∗ < 1 is one of its solutions and for some n we have that xn = x∗ ,
3.2. FIXED POINTS AND 2N CYCLES 149
then xn+2 = xn+4 = . . . = xn+2k = . . . = x∗ and xn+1 = xn+3 = . . . =
xn+2k+1 = . . . = f (x∗ ) (therefore f (x∗ ) is also a solution). If 0 < x∗3 <
x∗4 < 1 are two such different solutions with x∗3 = f (x∗4 ), x∗4 = f (x∗3 ), then
the trajectory is periodic with period 2. The points x∗3 , x∗4 are such that
they are real solutions of the equation
f (2) (x) = r2 x(1 − x)(1 − rx(1 − x)) = x , (3.16)
and at the same time they are not the solutions x∗1 = 0 x∗2 = 1 − 1/r of the
equation⁴ x = f (2) (x), the polynomial above can be written in the form
(see [22] for more details)
( ( ))
1
x x− 1− (Ax2 + Bx + C) = 0 . (3.17)
r
By expanding the polynomials (3.16), (3.17) and comparing their coef-
ficients we conclude that A = −r3 , B = r2 (r + 1) and C = −r(r + 1).
The roots of the trinomial in (3.17) are determined by the discriminant
∆ = r2 (r + 1)(r − 3). For the values of r of interest (1 < r ≤ 4), the dis-
(2)
criminant becomes positive when r > rc = 3 and we have two different
solutions
√
x∗α = ((r + 1) ∓ r2 − 2r − 3)/(2r) α = 3, 4 . (3.18)
(2)
When r = rc we have one double root, therefore a unique fixed point.
The study of the stability of the solutions of x = f (2) (x) requires
the same steps that led to the equation (3.15) and we determine if the
absolute value of f (2)′ (x) is greater, less or equal to one. By noting
that⁵ f (2)′ (x3 ) = f (2)′ (x4 ) = f ′ (x3 )f ′ (x4 ) = −r2 + 2r + 4, we see that for
√
r = rc = 3, f (2)′ (x∗3 ) = f (2)′ (x∗4 ) = 1 and for r = rc = 1 + 6 ≈ √
(2) (3)
3.4495,
f (x3 ) =f (x4 ) = −1. For the intermediate values 3 < r < 1 + 6 the
(2)′ (2)′
derivatives |f (2)′ (x∗α )| < 1 for α = 3, 4. Therefore, these points are stable
solutions of x = f (2) (x) and the points x∗1 , x∗2 bifurcate to x∗α , α = 1, 2, 3, 4
(2)
for r = rc = 3. Almost all trajectories with initial points in the interval
[0, 1] are attracted by the periodic trajectory with period 2, the “2-cycle”
{x∗3 , x∗4 }.
⁴Because, if x∗ = f (x∗ ) ⇒ f (2) (x∗ ) = f (f (x∗ )) = f (x∗ ) = x∗ etc, the point x∗ is also
a solution of x∗ = f (n) (x∗ ).
⁵The chain rule dh(g(x))/dx = h′ (g(x))g ′ (x) gives that f (2)′ (x∗3 ) = df (f (x∗3 ))/dx =
f (f (x∗3 ))f ′ (x∗3 ) = f ′ (x∗4 )f ′ (x∗3 ) and similarly for f (2)′ (x∗4 ). We can prove by induction
′
that for the n solutions x∗n+1 , x∗n+2 , . . . , x∗2n that belong to the n-cycle of the equation x =
f (n) (x) we have that f (n)′ (xn+i ) = f ′ (xn+1 ) f ′ (xn+2 ) . . . f ′ (x2n ) for every i = 1, . . . , n.
150 CHAPTER 3. LOGISTIC MAP
Using similar arguments we find that the fixed points x∗α , α = 1, 2, 3, 4
√
bifurcate to the eight fixed points x∗α , α = 1, . . . , 8 when r = rc = 1 + 6.
(3)
These are real solutions of the equation that gives the 4-cycle x = f (4) (x).
For rc < r < rc ≈ 3.5441, the points x∗α , α = 5, . . . , 8 are a stable 4-
(3) (4)
cycle which is an attractor of almost all trajectories of the logistic map⁶.
(4) (5)
Similarly, for rc < r < rc the 16 fixed points of the equation x = f (8) (x)
(5) (6)
give a stable 8-cycle, for rc < r < rc a stable 16-cycle etc⁷. This
is the phenomenon which is called period doubling which continues ad
(n)
infinitum. The points rc are getting closer to each other as n increases
(n)
so that limn→∞ rc = rc ≈ 3.56994567. As we will see, rc marks the onset
of the non-periodic, chaotic behavior of the trajectories of the logistic
map.
0.9
0.14
0.8
0.12
0.7
0.1
0.6
0.08
0.5
0.06 0.4
0.04 r= 0.5 0.3
r= 0.99
0.02 r= 1/0.9 0.2
r= 1/0.88
r= 1/0.86
0 0.1
0 20 40 60 80 100 120 140 160 180 200 0 10 20 30 40 50
Figure 3.1: (Left) Some trajectories of the logistic map with x0 = 0.1 and various
values of r. We can see the first bifurcation for rc = 1 from x∗1 = 0 to x∗2 = 1 − 1/r.
(1)
(2) (3)
(Right) Trajectories of the logistic map for rc < r = 3.5 < rc . The three curves start
from three different initial points. After a transient period, depending on the initial
point, one obtains a periodic trajectory
√ which is a 2-cycle. The horizontal lines are the
expected values x∗3,4 = ((r + 1) ∓ r2 − 2r − 3)/(2r) (see text).
Computing the bifurcation points becomes quickly intractable and we
have to resort to a numerical computation of their values. Initially we will
write a program that computes trajectories of the logistic map for chosen
values of r and x0 . The program can be found in the file logistic.f90
and is listed below:
⁶The points x∗α , α = 1, . . . , 4 are unstable fixed points and 2-cycle.
(n) (n+1)
⁷Generally, for rc < r < rc < rc ≈ 3.56994567 we have 2n fixed points of
n−1
the equation x = f (2 ) (x) and stable 2n−1 -cycles, which are attractors of almost all
trajectories.
3.2. FIXED POINTS AND 2N CYCLES 151
! ===========================================================
! D i s c r e t e L o g i s t i c Map
! ===========================================================
program logistic_map
i m p l i c i t none
i n t e g e r : : NSTEPS , i
r e a l ( 8 ) : : r , x0 , x1
! −−−−− Input :
p r i n t * , ’ # Enter NSTEPS , r , x0 : ’
read * , NSTEPS , r , x0
p r i n t * , ’ # NSTEPS = ’ , NSTEPS
print * , ’# r = ’ ,r
p r i n t * , ’ # x0 = ’ , x0
! −−−−− I n i t i a l i z e :
open ( u n i t =33 , f i l e = ’ l o g . dat ’ )
w r i t e ( 3 3 , * ) 0 , x0
! −−−−− C a l c u l a t e :
do i =1 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
w r i t e ( 3 3 , * ) i , x1
x0 = x1
enddo
c l o s e (33)
end program logistic_map
The program is compiled and run using the commands:
> g f o r t r a n logistic . f90 −o l
> echo ” 100 0.5 0 . 1 ” | . / l
The command echo prints to the stdout the values of the parameters
NSTEPS=100, r=0.5 and x0=0.1. Its stdout is redirected to the stdin
of the command ./l by using a pipe via the symbol |, from which the
program reads their value and uses them in the calculation. The results
can be found in two columns in the file log.dat and can be plotted
using gnuplot. The plots are put in figure 3.1 and we can see the first
(1) (2)
two bifurcations when r goes past the values rc and rc . Similarly, we
(n−1)
can study trajectories which are 2n -cycles when r crosses the values rc .
Another way to depict the 2-cycles is by constructing the cobweb plots:
We start from the point (x0 , 0) and we calculate the point (x0 , x1 ), where
x1 = f (x0 ). This point belongs on the curve y = f (x). The point (x0 , x1 ) is
then projected on the diagonal y = x and we obtain the point (x1 , x1 ). We
repeat n times obtaining the points (xn , xn+1 ) and (xn+1 , xn+1 ) on y = f (x)
152 CHAPTER 3. LOGISTIC MAP
0.7 0.8
0.6 0.7
0.5 0.6
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1 0.1
0 0
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
Figure 3.2: Cobweb plots of the logistic map for r = 2.8 and 3.3. (Left) The left plot
is an example of a fixed point x∗ = f (x∗ ). The green line is y = f (x) and the blue line
is y = f (2) (x). The trajectory ends at the unique non zero intersection of the diagonal
and y = f (x) which is x∗2 = 1 − 1/r. The trajectory intersects the curve y = f (2) (x) at
the same point. y = f (2) (x) does not intersect the diagonal anywhere else. (Right) The
right plot shows an example of a 2-cycle. y = f (2) (x) intersects the diagonal at two
additional points determined by x∗3 and x∗4 . The trajectory ends up on the orthogonal
(x∗3 , x∗3 ), (x∗4 , x∗3 ), (x∗4 , x∗4 ), (x∗3 , x∗4 ).
and y = x respectively. The fixed points x∗ = f (x∗ ) are at the intersections
of these curves and, if they are attractors, the trajectories will converge
on them. If we have a 2n -cycle, we will observe a periodic trajectory
n
going through points which are solutions to the equation x = f (2 ) (x).
This exercise can be done by using the following program, which can be
found in the file logistic1.f90:
! ===========================================================
! D i s c r e t e L o g i s t i c Map
! Map t h e t r a j e c t o r y i n 2d spa c e ( plane )
! ===========================================================
program logistic_map
i m p l i c i t none
i n t e g e r : : NSTEPS , i
r e a l ( 8 ) : : r , x0 , x1
! −−−−− Input :
p r i n t * , ’ # Enter NSTEPS , r , x0 : ’
read * , NSTEPS , r , x0
p r i n t * , ’ # NSTEPS = ’ , NSTEPS
print * , ’# r = ’ ,r
p r i n t * , ’ # x0 = ’ , x0
! −−−−− I n i t i a l i z e :
open ( u n i t =33 , f i l e = ’ t r j . dat ’ )
! −−−−− C a l c u l a t e :
3.2. FIXED POINTS AND 2N CYCLES 153
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4 0.4
0.3 0.3
0.2 0.2
0.1 0.1
0 0
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
Figure 3.3: (Left) A 4-cycle for r = 3.5. The blue curve is y = f (4) (x) which
intersects the diagonal at four points determined by xα , α = 5, 6, 7, 8. The four cycle
passes through these points. (Right) a non periodic orbit for r = 3.7 when the system
exhibits chaotic behavior.
w r i t e ( 3 3 , * ) 0 , x0 , 0
do i =1 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
w r i t e ( 3 3 , * ) 2* i−3,x0 , x1
w r i t e ( 3 3 , * ) 2* i−2,x1 , x1
x0 = x1
enddo
c l o s e (33)
end program logistic_map
Compiling and running this program is done exactly as in the case of the
program in logistic.f90. We can plot the results using gnuplot. The
plot in figure 3.2 can be constructed using the commands:
gnuplot > s e t s i z e square
gnuplot > f ( x ) = r * x *(1.0 − x )
gnuplot > r = 3.3
gnuplot > p l o t ”<echo 50 3.3 0 . 2 | . / l ; c a t t r j . dat ” using 2:3 w l
gnuplot > replot f(x) ,f(f(x) ) ,x
The plot command shown above, runs the program exactly as it is done
on the command line. This is accomplished by using the symbol <,
which reads the plot from the stdout of the command "echo 50 3.3
0.2|./l;cat trj.dat". Only the second command "echo trj.dat"
writes to the stdout, therefore the plot is constructed from the contents of
the file trj.dat. The following line adds the plots of the functions f (x),
f (2) (x) = f (f (x)) and of the diagonal y = x. Figures 3.2 and 3.3 show
examples of attractors which are fixed points, 2-cycles and 4-cycles. An
154 CHAPTER 3. LOGISTIC MAP
example of a non periodic trajectory is also shown, which exhibits chaotic
behavior which can happen when r > rc ≈ 3.56994567.
3.3 Bifurcation Diagrams
The bifurcations of the fixed points of the logistic map discussed in the
previous section can be conveniently shown on the “bifurcation diagram”.
We remind to the reader that the first bifurcations happen at the critical
values of r
rc(1) < rc(2) < rc(3) < . . . < rc(n) < . . . < rc , (3.19)
(1) (2) (3) √ (n)
where rc = 1, rc = 3, rc = 1 + 6 and rc = limn→∞ rc ≈ 3.56994567.
we have 2n fixed points x∗α , α = 1, 2, ..., 2n of x =
(n) (n+1)
For rc < r < rc
f (2 ) (x). By plotting these points x∗α (r) as a function of r we construct the
n
bifurcation diagram. These can be calculated numerically by using the
program bifurcate.f90. In this program, the user selects the values of
r that she needs to study and for each one of them the program records
the point of the 2n−1 -cycles⁸ x∗α (r), α = 2n−1 + 1, 2n−1 + 2, . . . , 2n . This
is easily done by computing the logistic map several times until we are
sure that the trajectories reach the stable state. The parameter NTRANS
in the program determines the number of points that we throw away,
which should contain all the transient behavior. After NTRANS steps, the
program records NSTEPS points, where NSTEPS should be large enough
to cover all the points of the 2n−1 -cycles or depict a dense enough set of
values of the non periodic orbits. The program is listed below:
! ===========================================================
! B i f u r c a t i o n Diagram o f t h e L o g i s t i c Map
! ===========================================================
program bifurcation_diagram
i m p l i c i t none
r e a l ( 8 ) , parameter : : rmin = 2.5 D0
r e a l ( 8 ) , parameter : : rmax = 4.0 D0
i n t e g e r , parameter : : NTRANS = 500 ! Number o f d i s c a r d e d s t e p s
i n t e g e r , parameter : : NSTEPS = 100 ! Number o f recorded s t e p s
i n t e g e r , parameter : : RSTEPS = 2000 ! Number o f v a l u e s o f r
integer :: i
real (8) : : r , dr , x0 , x1
! −−−−− I n i t i a l i z e :
⁸If we want to be more precise, the bifurcation diagram contains also the unstable
points. What we really construct is the orbit diagram which contains only the stable
points.
3.3. BIFURCATION DIAGRAMS 155
open ( u n i t =33 , f i l e = ’ b i f . dat ’ )
dr = ( rmax−rmin ) / RSTEPS ! Increment i n r
! −−−−− C a l c u l a t e :
r = rmin
do while ( r . l e . rmax )
x0 = 0.5 D0
! −−−− T r a n s i e n t s t e p s : s k i p
do i =1 , NTRANS
x1 = r * x0 * ( 1 . 0 D0−x0 )
x0 = x1
enddo
do i =1 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
w r i t e ( 3 3 , * ) r , x1
x0 = x1
enddo
r = r + dr
enddo ! do while
c l o s e (33)
end program bifurcation_diagram
1
0.52
0.9
0.8 0.515
0.7 0.51
0.6 0.505
0.5
0.5
0.4
0.495
0.3
0.49
0.2
0.485
0.1
0 0.48
2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 3.6295 3.63 3.6305 3.631 3.6315 3.632 3.6325 3.633 3.6335 3.634
Figure 3.4: (Left) The bifurcation diagram computed by the program bifurcate.f90
for 2.5 < r < 4. Notice the first bifurcation points followed by intervals of chaotic, non-
periodic orbits interrupted by intermissions of stable periodic trajectories. The chaotic
trajectories take values in subsets of the interval √
(0, 1). For r = 4 they take values within
the whole (0, 1). One can see that for r = 1 + 8 ≈ 3.8284 we obtain a 3-cycle which
subsequently bifurcates to 3 · 2n -cycles. (Right) The diagram on the left is magnified in
a range of r showing the self-similarity of the diagram at all scales.
The program can be compiled and run using the commands:
> g f o r t r a n bifurcate . f90 −o b
> ./ b;
156 CHAPTER 3. LOGISTIC MAP
The left plot of figure 3.4 can be constructed by the gnuplot commands:
gnuplot > p l o t ” b i f . dat ” with dots
We observe the fixed points and the 2n -cycles for r < rc . When r goes
past rc , the trajectories become non-periodic and exhibit chaotic behavior.
Chaotic behavior will be discussed more extensively in the next section.
For the time being, we note that if we measure the distance between the
(n+1) (n)
points ∆r(n) = rc − rc , we find that it decreases constantly with n so
that
∆r(n)
lim = δ ≈ 4.669 201 609 , (3.20)
n→∞ ∆r (n+1)
where δ is the Feigenbaum constant. An additional constant α, defined
by the quotient of the separation of adjacent elements ∆wn of period
doubled attractors from one double to the next ∆wn+1 , is
∆wn
lim = α ≈ 2.502 907 875 . (3.21)
n→∞ ∆wn+1
It is √also interesting to note the appearance of a 3-cycle right after r =
1 + 8 ≈ 3.8284 > rc ! By using the theorem of Sharkovskii, Li and
Yorke⁹ showed that any one dimensional system has 3-cycles, therefore
it will have cycles of any length and chaotic trajectories. The stability of
the 3-cycle can be studied from the solutions of x = f (3) (x) in exactly the
same way that we did in equations (3.16) and (3.17) (see [22] for details).
Figure 3.5 magnifies a branch of the 3-cycle. By magnifying different
regions in the bifurcation plot, as shown in the right plot of figure 3.4, we
find similar shapes to the branching of the 3-cycle. Figure 3.4 shows that
between intervals of chaotic behavior we obtain “windows” of periodic
trajectories. These are infinite but countable. It is also quite interesting
to note that if we magnify a branch withing these windows, we obtain a
diagram that is similar to the whole diagram! We say that the bifurcation
diagram exhibits self similarity. There are more interesting properties of
the bifurcation diagram and we refer the reader to the bibliography for
a more complete exposition.
We close this section by mentioning that the qualitative properties
of the bifurcation diagram are the same for a whole class of functions.
Feigenbaum discovered that if one takes any function that is concave and
⁹T.Y. Li, J.A. Yorke, “Period Three Implies Chaos”, American Mathematical Monthly
82 (1975) 985.
3.4. THE NEWTON-RAPHSON METHOD 157
0.52
0.51
0.5
0.49
0.48
0.47
0.46
3.83 3.835 3.84 3.845 3.85
√
Figure 3.5: Magnification of one of the three branches of the 3-cycle for r > 1 + 8.
To the left, we observe the temporary halt of the chaotic behavior of the trajectory, which
comes back as shown in the plot to the right after an intermission of stable periodic
trajectories.
has a unique global maximum, its bifurcation diagram behaves qualita-
tively the same way as that of the logistic map. Examples of such func-
tions¹⁰ studied in the literature are g(x) = xer(1−x) , u(x) = r sin(πx) and
w(x) = b − x2 . The constants δ and α of equations (3.20) and (3.21)
are the same of all these mappings. The functions that result in chaotic
behavior are studied extensively in the literature and you can find a list
of those in [28].
3.4 The Newton-Raphson Method
In order to determine the bifurcation points, one has to solve the nonlin-
ear, polynomial, algebraic equations x = f (n) (x) and f (n)′ (x) = −1. For
this reason, one has to use an approximate numerical calculation of the
roots, and the simple Newton-Raphson method will prove to be a good
choice.
Newton-Raphson’s method uses an initial guess x0 for the solution of
the equation g(x) = 0 and computes a sequence of points x1 , x2 , . . . , xn ,
xn+1 , . . . that presumably converges to one of the roots of the equation.
The computation stops at a finite n, when we decide that the desired level
of accuracy has been achieved. In order to understand how it works, we
assume that g(x) is an analytic function for all the values of x used in
¹⁰ The function x exp(r(1 − x)) has been used as a model for populations whose large
density is restricted by epidemics. The populations are always positive independently
of the (positive) initial conditions and the value of r.
158 CHAPTER 3. LOGISTIC MAP
the computation. Then, by Taylor expanding around xn we obtain
g(xn+1 ) = g(xn ) + (xn+1 − xn )g ′ (x) + . . . . (3.22)
If we wish to have g(xn+1 ) ≈ 0, we choose
g(xn )
xn+1 = xn − . (3.23)
g ′ (xn )
The equation above gives the Newton-Raphson method for one equation
g(x) = 0 of one variable x. Different choices for x0 will possibly lead to
different roots. When g ′ (x), g ′′ (x) are non zero at the root and g ′′′ (x) is
bounded, the convergence of the method is quadratic with the number
of iterations. This means that there is a neighborhood of the root α such
that the distance ∆xn+1 = xn+1 − α is ∆xn+1 ∝ (∆xn )2 . If the root α
has multiplicity larger than 1, convergence is slower. The proofs of these
statements are simple and can be found in [29].
The Newton-Raphson method is simple to program and, most of the
times, sufficient for the solution of many problems. In the general case
it works well only close enough to a root. We should also keep in mind
that there are simple reasons for the method to fail. For example, when
g ′ (xn ) = 0 for some n, the method stops. For functions that tend to
0 as x → ±∞, it is easy to make a bad choice for x0 that does not
lead to convergence to a root. Sometimes it is a good idea to combine the
Newton-Raphson method with the bisection method. When the derivative
g ′ (x) diverges at the root we might get into trouble. For example, the
equation |x|ν = 0 with 0 < ν < 1/2, does not lead to a convergent
sequence. In some cases, we might enter into non-convergent cycles [8].
For some functions the basin of attraction of a root (the values of x0 that
will converge to the root) can be tiny. See problem 13.
As a test case of our program, consider the equation
√
ϵ tan ϵ = ρ2 − ϵ2 (3.24)
which results from the solution of Schrödinger’s equation for the en-
ergy spectrum of a quantum mechanical particle of mass m in a one
dimensional
√ potential well √
of depth V0 and width L. The parameters
ϵ = mL2 E/(2ℏ) and ρ = mL2 V0 /(2ℏ). Given ρ, we solve for ϵ which
gives the energy E. The function g(x) and its derivative g ′ (x) are
√
g(x) = x tan x − ρ2 − x2
x x
g ′ (x) = √ + + tan x . (3.25)
ρ −x
2 2 cos2 x
3.4. THE NEWTON-RAPHSON METHOD 159
The program of the Newton-Raphson method for solving the equation
g(x) = 0 can be found in the file nr.f90:
! ===========================================================
! Newton Raphson f o r a f u n c t i o n o f one v a r i a b l e
! ===========================================================
program NewtonRaphson
i m p l i c i t none
r e a l ( 8 ) , parameter : : rho = 15.0 D0
r e a l ( 8 ) , parameter : : eps = 1D−6
i n t e g e r , parameter : : NMAX = 1000
r e a l ( 8 ) : : x0 , x1 , e r r , g , gp
integer : : i
p r i n t * , ’ Enter x0 : ’
read * , x0
e r r = 1 . 0 D0
print * , ’ i t e r x error ’
p r i n t * , ’−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−’
p r i n t * , 0 , x0 , e r r
do i =1 , NMAX
! value of function g ( x ) :
g = x0 * tan ( x0 )−s q r t ( rho * rho−x0 * x0 )
! value of the d e r i v a t i v e g ’ ( x ) :
gp = x0 / s q r t ( rho * rho−x0 * x0 )+x0 / ( c o s ( x0 ) * * 2 ) +tan ( x0 )
x1 = x0 − g / gp
e r r = ABS( x1−x0 )
p r i n t * , i , x1 , e r r
i f ( e r r . l t . eps ) e x i t
x0 = x1
enddo
end program NewtonRaphson
In the program listed above, the user is asked to set the initial point x0 .
We fix ρ = rho = 15. It is instructive to make the plot of the left and right
hand sides of (3.24) and make a graphical determination of the roots
from their intersections. Then we can make appropriate choices of the
initial point x0 . Using gnuplot, the plots are made with the commands:
gnuplot > g1 ( x ) = x * tan ( x )
gnuplot > g2 ( x ) = s q r t ( rho * rho−x * x )
gnuplot > p l o t [ 0 : 2 0 ] [ 0 : 2 0 ] g1 ( x ) , g2 ( x )
The compilation and running of the program can be done as follows:
> g f o r t r a n nr . f90 −o n
160 CHAPTER 3. LOGISTIC MAP
20
15
10
5
ε tanε
(ρ2-ε2)1/2
0
0 5 10 15 20
ε
Figure 3.6: Plots of the right and left hand sides of equation (3.24). The intersections
of the curves determine the solutions of the equation and their approximate graphical
estimation can serve as initial points x0 for the Newton-Raphson method.
> echo ” 1 . 4 ” | . / n
Enter x0 :
iter x error
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
0 1.3999999999999999 1.0000000000000000
1 1.5254292024457967 0.12542920244579681
2 1.5009739120496131 2.4455290396183660E−002
3 1.4807207017202200 2.0253210329393090E−002
4 1.4731630533073483 7.5576484128716537E−003
5 1.4724779331237687 6.8512018357957949E−004
6 1.4724731072313519 4.8258924167932093E−006
7 1.4724731069952235 2.3612845012621619E−010
We conclude that one of the roots of the equation is ϵ ≈ 1.472473107.
The reader can compute more of these roots by following these steps by
herself.
The method discussed above can be easily generalized to the case
of two equations. Suppose that we need to solve simultaneously two
algebraic equations g1 (x1 , x2 ) = 0 and g2 (x1 , x2 ) = 0. In order to compute
a sequence (x10 , x20 ), (x11 , x21 ), . . ., (x1n , x2n ), (x1(n+1) , x2(n+1) ), . . . that may
converge to a root of the above system of equations, we Taylor expand
3.4. THE NEWTON-RAPHSON METHOD 161
the two functions around (x1n , x2n )
∂g1 (x1n , x2n )
g1 (x1(n+1) , x2(n+1) ) = g1 (x1n , x2n ) + (x1(n+1) − x1n )
∂x1
∂g1 (x1n , x2n )
+ (x2(n+1) − x2n ) + ...
∂x2
∂g2 (x1n , x2n )
g2 (x1(n+1) , x2(n+1) ) = g2 (x1n , x2n ) + (x1(n+1) − x1n )
∂x1
∂g2 (x1n , x2n )
+ (x2(n+1) − x2n ) + . . . . (3.26)
∂x2
Defining δx1 = (x1(n+1) − x1n ) and δx2 = (x2(n+1) − x2n ) and setting
g1 (x1(n+1) , x2(n+1) ) ≈ 0, g2 (x1(n+1) , x2(n+1) ) ≈ 0, we obtain
∂g1 ∂g1
δx1 + δx2 = −g1
∂x1 ∂x2
∂g2 ∂g2
δx1 + δx2 = −g2 . (3.27)
∂x1 ∂x2
This is a linear 2 × 2 system of equations
A11 δx1 + A12 δx2 = b1
A21 δx1 + A22 δx2 = b2 , (3.28)
where Aij = ∂gi /∂xj and bi = −gi , with i, j = 1, 2. Solving for δxi we
obtain
x1(n+1) = x1n + δx1
x2(n+1) = x2n + δx2 . (3.29)
The iterations stop when δxi become small enough.
As an example, consider the equations with g1 (x) = 2x2 − 3xy + y − 2,
g2 (x) = 3x + xy + y − 1. We have A11 = 4x − 3y, A12 = 1 − 3x, A21 = 3 + y,
A22 = 1 + x. The program can be found in the file nr2.f90:
! ===========================================================
! Newton Raphson o f two f u n c t i o n s o f two v a r i a b l e s
! ===========================================================
program NewtonRaphson2
i m p l i c i t none
r e a l ( 8 ) , parameter : : eps = 1D−6
i n t e g e r , parameter : : NMAX = 1000
r e a l ( 8 ) : : A ( 2 , 2 ) , b ( 2 ) , dx ( 2 )
162 CHAPTER 3. LOGISTIC MAP
real (8) : : x , y , err
integer : : i
p r i n t * , ’ Enter x0 , y0 : ’
read * , x , y
e r r = 1 . 0 D0
print * , ’ i t e r x y error ’
p r i n t * , ’−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−’
print * , 0 ,x , y , err
do i =1 , NMAX
b(1) = −(2.0 D0 * x * x−3.0D0 * x * y+y−2.0D0 ) ! −g1 ( x , y )
b (2) = −(3.0 D0 * x + x * y + y − 1 . 0 D0 ) ! −g2 ( x , y )
! dg1 / dx dg1 / dy
A ( 1 , 1 ) = 4.0 D0 * x−3.0D0 * y ; A ( 1 , 2 ) = 1 . 0 D0 −3.0D0 * x
! dg2 / dx dg2 / dy
A ( 2 , 1 ) = 3.0 D0+y ; A ( 2 , 2 ) = 1 . 0 D0+x
c a l l solve2x2 ( A , B , dx )
x = x + dx ( 1 )
y = y + dx ( 2 )
e r r = 0.5 D0 *SQRT( dx ( 1 ) **2+ dx ( 2 ) * * 2 )
print * , i , x , y , err
i f ( e r r . l t . eps ) e x i t
enddo
end program NewtonRaphson2
! ===========================================================
s u b r o u t i n e solve2x2 ( A , b , dx )
i m p l i c i t none
r e a l ( 8 ) : : A ( 2 , 2 ) , b ( 2 ) , dx ( 2 )
r e a l ( 8 ) : : num1 , num2 , det
num1 = A ( 2 , 2 ) * b ( 1 )−A ( 1 , 2 ) * b ( 2 )
num2 = A ( 1 , 1 ) * b ( 2 )−A ( 2 , 1 ) * b ( 1 )
det = A ( 1 , 1 ) * A ( 2 , 2 )−A ( 1 , 2 ) * A ( 2 , 1 )
i f ( det . eq . 0.0 D0 ) s t o p ’ s o l v e 2 x 2 : d e t=0 ’
dx ( 1 ) = num1 / det
dx ( 2 )= num2 / det
end s u b r o u t i n e solve2x2
In order to guess the region where the real roots of the systems lie, we
make a 3-dimensional plot using gnuplot:
gnuplot > s e t i s o s a m p l e s 20
gnuplot > s e t hidden3d
gnuplot > s p l o t 2* x**2 −3*x * y+y −2 ,3* x+y * x+y −1 ,0
We plot the functions gi (x, y) together with the plane x = 0. The in-
tersection of the three surfaces determine the roots we are looking for.
Compiling and running the program can be done by using the com-
mands:
3.5. CALCULATION OF THE BIFURCATION POINTS 163
> g f o r t r a n nr2 . f90 −o n
> echo 2.2 1 . 5 | . / n
Enter x0 , y0 :
iter x y error
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
0 2.20000000 1.50000000 1.0000
1 0.76427104 0.26899383 0.9456
2 0.73939531 −0.68668275 0.4780
3 0.74744506 −0.71105605 1.2834 E−002
4 0.74735933 −0.71083147 1.2019 E−004
5 0.74735932 −0.71083145 1.2029 E−008
> echo 0 1 | . / n
.................
5 −0.10899022 1.48928857 4.3461 E−012
> echo −5 0 | . / n
6 −6.13836909 −3.77845711 3.2165 E−013
The computation above leads to the roots (0.74735932, −0.71083145),
(−0.10899022, 1.48928857), (−6.13836909, −3.77845711).
The Newton-Raphson method for many variables becomes hard quite
soon: One needs to calculate the functions as well as their derivatives,
which is prohibitively expensive for many problems. It is also hard to
determine the roots, since the method converges satisfactorily only very
close to the roots. We refer the reader to [8] for more information on
how one can deal with these problems.
3.5 Calculation of the Bifurcation Points
In order to determine the bifurcation points for r < rc we will solve
the algebraic equations x = f (k) (x) and f (k)′ (x) = −1. At these points,
k-cycles become unstable and 2k-cycles appear and are stable. This hap-
pens when r = rc , where k = 2n−2 . We will look for solutions (x∗α , rc )
(n) (n)
for α = k + 1, k + 2, . . . , 2k.
We define the functions F (x, r) = f (x) = rx(1 − x) and F (k) (x, r) =
f (k) (x) as in equation (3.6). We will solve the algebraic equations:
g1 (x, r) = x − F (k) (x, r) = 0
∂F (k) (x, r)
g2 (x, r) = + 1 = 0. (3.30)
∂x
According to the discussion of the previous section, in order to calculate
the roots of these equations we have to solve the linear system (3.28),
164 CHAPTER 3. LOGISTIC MAP
where the coefficients are
b1 = −g1 (x, r) = −x + F (k) (x, r)
∂F (k) (x, r)
b2 = −g2 (x, r) = − −1
∂x
∂g1 (x, r) ∂F (k) (x, r)
A11 = =1−
∂x ∂x
(k)
∂g1 (x, r) ∂F (x, r)
A12 = =−
∂r ∂r
∂g2 (x, r) ∂ 2 F (k) (x, r)
A21 = =
∂x ∂x2
2 (k)
∂g2 (x, r) ∂ F (x, r)
A22 = = . (3.31)
∂r ∂x∂r
The derivatives will be calculated approximately using finite differences
∂F (k) (x, r) F (k) (x + ϵ, r) − F (k) (x − ϵ, r)
≈
∂x 2ϵ
(k)
∂F (x, r) F (x, r + ϵ) − F (k) (x, r − ϵ)
(k)
≈ , (3.32)
∂r 2ϵ
and similarly for the second derivatives
∂F (k) (x+ 2ϵ ,r) ∂F (k) (x− 2ϵ ,r)
∂ 2 F (k) (x, r) −
≈ ∂x ∂x
∂x2 2 2ϵ
{ }
1 F (k) (x + ϵ, r) − F (k) (x, r) F (k) (x, r) − F (k) (x − ϵ, r)
= −
ϵ ϵ ϵ
1 { (k) }
= 2
F (x + ϵ, r) − 2F (k) (x, r) + F (k) (x − ϵ, r)
ϵ
∂F (k) (x+ϵx ,r) ∂F (k) (x−ϵx ,r)
∂ 2 F (k) (x, r) −
≈ ∂r ∂r
∂x∂r 2ϵx
{ (k)
1 F (x + ϵx , r + ϵr ) − F (k) (x + ϵx , r − ϵr )
=
2ϵx 2ϵr
}
F (x − ϵx , r + ϵr ) − F (k) (x − ϵx , r − ϵr )
(k)
−
2ϵr
1 { (k)
= F (x + ϵx , r + ϵr ) − F (k) (x + ϵx , r − ϵr )
4ϵx ϵr
}
−F (k) (x − ϵx , r + ϵr ) + F (k) (x − ϵx , r − ϵr ) (3.33)
We are now ready to write the program for the Newton-Raphson method
like in the previous section. The only difference is the approximate cal-
culation of the derivatives using the relations above and the calculation
3.5. CALCULATION OF THE BIFURCATION POINTS 165
of the function F (k) (x, r) by a routine that will compose the function f (x)
k-times. The program can be found in the file bifurcationPoints.f90:
! ===========================================================
! bifurcationPoints . f
! C a l c u l a t e b i f u r c a t i o n p o i n t s o f t h e d i s c r e t e l o g i s t i c map
! a t p e r i o d k by s o l v i n g t h e c o n d i t i o n
! g1 ( x , r ) = x − F( k , x , r ) = 0
! g2 ( x , r ) = dF( k , x , r ) / dx+1 = 0
! determining when t h e Floquet m u l t i p l i e r becomes 1
! F( k , x , r ) i t e r a t e s F( x , r ) = r * x * ( x−1) k t i m e s
! The e q u a t i o n s a r e s o l v e d by using a Newton−Raphson method
! ===========================================================
program bifurcationPoints
i m p l i c i t none
r e a l ( 8 ) , parameter : : tol =1.0 D−10
i n t e g e r : : k , iter
r e a l ( 8 ) : : r0 , x0
r e a l ( 8 ) : : A ( 2 , 2 ) , B ( 2 ) , dX ( 2 )
r e a l ( 8 ) : : error
r e a l ( 8 ) : : F , dFdx , dFdr , d2Fdx2 , d2Fdrdx
! −−−− Input :
p r i n t * , ’ # Enter k , r0 , x0 : ’
read * , k , r0 , x0
p r i n t * , ’ # Pe ri od k= ’ , k
p r i n t * , ’ # r0= ’ , r0 , ’ x0= ’ , x0
! −−−− I n i t i a l i z e
error = 1 . 0 D0 ! i n i t i a l l a r g e value of error > t o l
iter = 0
do while ( error . g t . tol )
! −−−− C a l c u l a t e j a c o b i a n matrix
A ( 1 , 1 ) = 1 . 0 D0−dFdx ( k , x0 , r0 )
A ( 1 , 2 ) = −dFdr ( k , x0 , r0 )
A ( 2 , 1 ) = d2Fdx2 ( k , x0 , r0 )
A ( 2 , 2 ) = d2Fdrdx ( k , x0 , r0 )
B(1) = −x0 + F ( k , x0 , r0 )
B (2) = −dFdx ( k , x0 , r0 ) −1.0D0
! −−−− S o l v e a 2x2 l i n e a r system :
c a l l solve2x2 ( A , B , dX )
x0 = x0 + dX ( 1 )
r0 = r0 + dX ( 2 )
error = 0.5 D0 * s q r t ( dX ( 1 ) **2+ dX ( 2 ) * * 2 )
iter = iter+1
p r i n t * , iter , ’ x0= ’ , x0 , ’ r0= ’ , r0 , ’ e r r = ’ , error
enddo ! do while ( e r r o r . g t . t o l )
end program bifurcationPoints
! ===========================================================
! Function F( k , x , r ) and i t s d e r i v a t i v e s
166 CHAPTER 3. LOGISTIC MAP
real (8) function F(k , x , r)
i m p l i c i t none
r e a l ( 8 ) : : x , r , x0
integer k , i
x0 = x
do i =1 , k
x0 = r * x0 * ( 1 . 0 D0−x0 )
enddo
F = x0
end f u n c t i o n F
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n dFdx ( k , x , r )
i m p l i c i t none
r e a l ( 8 ) : : x , r , eps
real (8) : : F
integer k
eps = 1 . 0 D−6*x
dFdx = ( F ( k , x+eps , r )−F ( k , x−eps , r ) ) / ( 2 . 0 D0 * eps )
end f u n c t i o n dFdx
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n dFdr ( k , x , r )
i m p l i c i t none
r e a l ( 8 ) : : x , r , eps
real (8) : : F
integer k
eps = 1 . 0 D−6*r
dFdr = ( F ( k , x , r+eps )−F ( k , x , r−eps ) ) / ( 2 . 0 D0 * eps )
end f u n c t i o n dFdr
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n d2Fdx2 ( k , x , r )
i m p l i c i t none
r e a l ( 8 ) : : x , r , eps
real (8) : : F
integer k
eps = 1 . 0 D−6*x
d2Fdx2 = ( F ( k , x+eps , r ) −2.0D0 * F ( k , x , r )+F ( k , x−eps , r ) ) / ( eps * eps )
end f u n c t i o n d2Fdx2
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n d2Fdrdx ( k , x , r )
i m p l i c i t none
r e a l ( 8 ) : : x , r , epsx , epsr
real (8) : : F
integer k
3.5. CALCULATION OF THE BIFURCATION POINTS 167
epsx = 1 . 0 D−6*x
epsr = 1 . 0 D−6*r
d2Fdrdx = ( F ( k , x+epsx , r+epsr )−F ( k , x+epsx , r−epsr ) &
−F ( k , x−epsx , r+epsr )+F ( k , x−epsx , r−epsr ) ) &
/ ( 4 . 0 D0 * epsx * epsr )
end f u n c t i o n d2Fdrdx
! ===========================================================
s u b r o u t i n e solve2x2 ( A , b , dx )
i m p l i c i t none
r e a l ( 8 ) : : A ( 2 , 2 ) , b ( 2 ) , dx ( 2 )
r e a l ( 8 ) : : num1 , num2 , det
num1 = A ( 2 , 2 ) * b ( 1 ) − A ( 1 , 2 ) * b ( 2 )
num2 = A ( 1 , 1 ) * b ( 2 ) − A ( 2 , 1 ) * b ( 1 )
det = A ( 1 , 1 ) * A ( 2 , 2 )− A ( 1 , 2 ) * A ( 2 , 1 )
i f ( det . eq . 0.0 D0 ) s t o p ’ s o l v e 2 x 2 : d e t = 0 ’
dx ( 1 ) = num1 / det
dx ( 2 ) = num2 / det
end s u b r o u t i n e solve2x2
Compiling and running the program can be done as follows:
> g f o r t r a n bifurcationPoints . f90 −o b
> echo 2 3.5 0.5 | . / b
# Enter k , r0 , x0 :
# Period k= 2
# r0= 3.5000000000000 x0= 0.50000000000
1 x0= 0.4455758353187 r0= 3.38523275827 err= 6.35088E−002
2 x0= 0.4396562547624 r0= 3.45290970406 err= 3.39676E−002
3 x0= 0.4399593001407 r0= 3.44949859951 err= 1.71226 E−003
4 x0= 0.4399601690333 r0= 3.44948974267 err= 4.44967 E−006
5 x0= 0.4399601689937 r0= 3.44948974281 err= 7.22160 E−011
> echo 2 3.5 0.85 | . / b
.................
4 x0= 0.8499377795512 r0= 3.44948974275 err= 1.85082E−011
> echo 4 3.5 0.5 | . / b
.................
5 x0= 0.5235947861540 r0= 3.54409035953 err= 1.86318 E−011
> echo 4 3.5 0.35 | . / b
.................
5 x0= 0.3632903374118 r0= 3.54409035955 err= 5.91653E−013
The above listing shows the points of the 2-cycle and some of the points
(3)
of the 4-cycle. It is also possible to compare the calculated value rc =
(3) √
3.449490132 with the expected one rc = 1+ 6 ≈ 3.449489742. Improving
the accuracy of the calculation is left as an exercise for the reader who
has to control the systematic errors of the calculations and achieve better
(4)
accuracy in the computation of rc .
168 CHAPTER 3. LOGISTIC MAP
3.6 Liapunov Exponents
We have seen that when r > rc ≈ 3.56994567, the trajectories of the lo-
gistic map become non periodic and exhibit chaotic behavior. Chaotic
behavior mostly means sensitivity of the evolution of a dynamical system
to the choice of initial conditions. More precisely, it means that two dif-
ferent trajectories constructed from infinitesimally close initial conditions,
diverge very fast from each other. This implies that there is a set of initial
conditions that densely cover subintervals of (0, 1) whose trajectories do
not approach arbitrarily close to any cycle of finite length.
Assume that two trajectories have x0 , x̃0 as initial points and ∆x0 =
x0 − x̃0 . When the points xn , x̃n have a distance ∆xn = x̃n − xn that for
small enough n increases exponentially with n (the “time”), i.e.
∆xn ∼ ∆x0 eλn , λ > 0, (3.34)
the system is most likely exhibiting chaotic behavior¹¹. The exponent λ
is called a Liapunov exponent. A useful equation for the calculation of λ
is
1∑
n−1
λ = lim ln |f ′ (xk )| . (3.35)
n→∞ n
k=0
This relation can be easily proved by considering infinitesimal ϵ ≡ |∆x0 |
¹¹Sensitivity to the initial condition alone does not necessarily imply chaos. It is
necessary to have topological mixing and dense periodic orbits. Topological mixing
means that every open set in phase space will evolve to a set that for large enough time
will have non zero intersection with any open set. Dense periodic orbits means that
every point in phase space lies infinitesimally close to a periodic orbit.
3.6. LIAPUNOV EXPONENTS 169
so that λ = lim lim n1 ln |∆xn |/ϵ. Then we obtain
n→∞ ϵ→0
x̃1 = f (x̃0 ) = f (x0 + ϵ) ≈ f (x0 ) + ϵf ′ (x0 )
= x1 + ϵf ′ (x0 ) ⇒
∆x1 x̃1 − x1
= ≈ f ′ (x0 )
ϵ ϵ
x̃2 = f (x̃1 ) = f (x1 + ϵf ′ (x0 )) ≈ f (x1 ) + (ϵf ′ (x0 ))f ′ (x1 )
= x2 + ϵf ′ (x0 )f ′ (x1 ) ⇒
∆x2 x̃2 − x2
= ≈ f ′ (x0 )f ′ (x1 )
ϵ ϵ
x̃3 = f (x̃2 ) = f (x2 + ϵf ′ (x0 )f ′ (x1 )) ≈ f (x2 ) + (ϵf ′ (x0 )f ′ (x1 ))f ′ (x2 )
= x3 + ϵf ′ (x0 )f ′ (x1 )f ′ (x2 ) ⇒
∆x3 x̃3 − x3
= ≈ f ′ (x0 )f ′ (x1 )f ′ (x2 ) . (3.36)
ϵ ϵ
We can show by induction that |∆xn |/ϵ ≈ f ′ (x0 )f ′ (x1 )f ′ (x2 ) . . . f ′ (xn−1 )
and by taking the logarithm and the limits we can prove (3.35).
A first attempt to calculate the Liapunov exponents could be made
by using the definition (3.34). We modify the program logistic.f90 so
that it calculates two trajectories whose initial distance is ϵ = epsilon:
! ===========================================================
! D i s c r e t e L o g i s t i c Map :
! Two t r a j e c t o r i e s with c l o s e i n i t i a l c o n d i t i o n s .
! ===========================================================
program logistic_map
i m p l i c i t none
i n t e g e r : : NSTEPS , i
r e a l ( 8 ) : : r , x0 , x1 , x0t , x1t , e p s i l o n
! −−−−− Input :
p r i n t * , ’ # Enter NSTEPS , r , x0 , e p s i l o n : ’
read * , NSTEPS , r , x0 , e p s i l o n
p r i n t * , ’ # NSTEPS = ’ , NSTEPS
print * , ’# r = ’ ,r
p r i n t * , ’ # x0 = ’ , x0
print * , ’# epsilon = ’ , epsilon
x0t = x0+ e p s i l o n
! −−−−− I n i t i a l i z e :
open ( u n i t =33 , f i l e = ’ l i a . dat ’ )
w r i t e ( 3 3 , * ) 1 , x0 , x0t , ABS( x0t−x0 ) / e p s i l o n
170 CHAPTER 3. LOGISTIC MAP
1e+16
1e+14
1e+12
1e+10
|∆xn|/ε
1e+08
1e+06
10000 ε = 10-15
ε = 10-12
100 ε = 10-10
ε = 10-8
1 ε = 10-6
ε = 10-4
0.01
0 20 40 60 80 100 120 140 160 180 200
n
Figure 3.7: A plot of |∆xn |/ϵ for the logistic map for r = 3.6, x0 = 0.2. Note the
convergence of the curves as ϵ → 0 and the approximate exponential behavior in this
limit. The two lines are fits to the equation (3.34) and give λ = 0.213(4) and λ = 0.217(6)
respectively.
! −−−−− C a l c u l a t e :
do i =2 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
x1t = r * x0t * ( 1 . 0 D0−x0t )
w r i t e ( 3 3 , * ) i , x1 , x1t , ABS( x1t−x1 ) / e p s i l o n
x0 = x1 ; x0t = x1t
enddo
c l o s e (3 3)
end program logistic_map
After running the program, the quantity |∆xn |/ϵ is found at the fourth
column of the file lia.dat. The curves of figure 3.7 can be constructed
by using the commands:
> gfortran liapunov1 . f90 −o l
> gnuplot
gnuplot > s e t l o g s c a l e y
gnuplot > p l o t \
”<echo 200 3.6 0.2 1 e−15 | . / l ; c a t l i a . dat ” u 1 : 4 w l
3.6. LIAPUNOV EXPONENTS 171
The last line plots the stdout of the command "echo 200 3.6 0.2 1e-15
|./l;cat lia.dat", i.e. the contents of the file lia.dat produced after
running our program using the parameters NSTEPS = 200, r = 3.6, x0
= 0.2 and epsilon = 10−15 . The gnuplot command set logscale y,
puts the y axis in a logarithmic scale. Therefore an exponential function
is shown as a straight line and this is what we see in figure 3.7: The
points |∆xn |/ϵ tend to lie on a straight line as ϵ decreases. The slopes of
these lines are equal to the Liapunov exponent λ. Deviations from the
straight line behavior indicates corrections and systematic errors, as we
point out in figure 3.7. A different initial condition results in a slightly
different value of λ, and the true value can be estimated as the average
over several such choices. We estimate the error of our computation
from the standard error of the mean. The reader should perform such a
computation as an exercise.
One can perform a fit of the points |∆xn |/ϵ to the exponential function
in the following way: Since |∆xn |/ϵ ∼ C exp(λn) ⇒ ln(|∆xn |/ϵ) = λn + c,
we can make a fit to a straight line instead. Using gnuplot, the relevant
commands are:
gnuplot > f i t [ 5 : 5 3 ] a * x+b \
”<echo 500 3.6 0.2 1 e−15 | . / l ; c a t l i a . dat ”\
using 1 : ( l o g ( $4 ) ) via a , b
gnuplot > r e p l o t exp ( a * x+b )
The command shown above fits the data to the function a*x+b by taking
the 1st column and the logarithm of the 4th column (using 1:(log($4)))
of the stdout of the command that we used for creating the previous plot.
We choose data for which 5 ≤ n ≤ 53 ([5:53]) and the fitting parameters
are a,b (via a,b). The second line, adds the fitted function to the plot.
Now we are going to use equation (3.35) for calculating λ. This
equation is approximately correct when (a) we have already reached the
steady state and (b) in the large n limit. For this reason we should
study if we obtain a satisfactory convergence when we (a) “throw away”
a number of NTRANS steps, (b) calculate the sum (3.35) for increasing
NSTEPS= n (c) calculate the sum (3.35) for many values of the initial point
x0 . This has to be carefully repeated for all values of r since each factor
will contribute differently to the quality of the convergence: In regions
that manifest chaotic behavior (large λ) convergence will be slower. The
program can be found in the file liapunov2.f90:
! ===========================================================
172 CHAPTER 3. LOGISTIC MAP
0.48
0.46
(1/n)Σk=NN+n-1 ln|f’(xk)|
0.44
0.42
0.4
0.38 r=3.8 x0=0.20
r=3.8 x0=0.35
0.36 r=3.8 x0=0.50
r=3.8 x0=0.75
r=3.8 x0=0.90
0.34
0 10000 20000 30000 40000 50000 60000 70000
n
∑N +n−1
Figure 3.8: Plot of the sum (1/n) ln |f ′ (xk )| as a function of n for the logistic
k=N
map with r = 3.8, N = 2000 for different initial conditions x0 = 0.20, 0.35, 0.50, 0.75, 0.90.
The different curves converge in the limit n → ∞ to λ = 0.4325(10).
! D i s c r e t e L o g i s t i c Map :
! Liapunov exponent from sum_i l n | f ’ ( x _ i ) |
! NTRANS: number o f d i s c a r d e d i t e r a t i o n i n order t o d i s c a r d
! t r a n s i e n t behavior
! NSTEPS : number o f terms i n t h e sum
! ===========================================================
program logistic_map
i m p l i c i t none
i n t e g e r : : NTRANS , NSTEPS , i
r e a l ( 8 ) : : r , x0 , x1 , sum
! −−−−− Input :
p r i n t * , ’ # Enter NTRANS, NSTEPS , r , x0 : ’
read * , NTRANS , NSTEPS , r , x0
p r i n t * , ’ # NTRANS = ’ , NTRANS
p r i n t * , ’ # NSTEPS = ’ , NSTEPS
print * , ’# r = ’ ,r
p r i n t * , ’ # x0 = ’ , x0
do i =1 , NTRANS
x1 = r * x0 * ( 1 . 0 D0−x0 )
x0 = x1
3.6. LIAPUNOV EXPONENTS 173
enddo
sum = l o g (ABS( r * ( 1 . 0 D0 −2.0D0 * x0 ) ) )
! −−−−− I n i t i a l i z e :
open ( u n i t =33 , f i l e = ’ l i a . dat ’ )
w r i t e ( 3 3 , * ) 1 , x0 , sum
! −−−−− C a l c u l a t e :
do i =2 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
sum = sum + l o g (ABS( r * ( 1 . 0 D0 −2.0D0 * x1 ) ) )
w r i t e ( 3 3 , * ) i , x1 , sum / i
x0 = x1
enddo
c l o s e (33)
end program logistic_map
After NTRANS steps, the program calculates NSTEPS times the sum of the
terms ln |f ′ (xk )| = ln |r(1 − 2xk )|. At each step the sum divided by the
number of steps i is printed to the file lia.dat. Figure 3.6 shows the
results for r = 3.8. This is a point where the system exhibits strong
chaotic behavior and convergence is achieved after we compute a large
number of steps. Using NTRANS = 2 000 and NSTEPS ≈ 70 000 the achieved
accuracy is about 0.2% with λ = 0.4325 ± 0.0010 ≡ 0.4325(10). The main
contribution to the error comes from the different paths followed by
each initial point chosen. The plot can be constructed with the gnuplot
commands:
> gfortran liapunov2 . f90 −o l
> gnuplot
gnuplot > p l o t \
”<echo 2000 70000 3.8 0.20 |./ l ; cat lia . dat ” u 1:3 w l ,\
”<echo 2000 70000 3.8 0.35 |./ l ; cat lia . dat ” u 1:3 w l ,\
”<echo 2000 70000 3.8 0.50 |./ l ; cat lia . dat ” u 1:3 w l ,\
”<echo 2000 70000 3.8 0.75 |./ l ; cat lia . dat ” u 1:3 w l ,\
”<echo 2000 70000 3.8 0.90 |./ l ; cat lia . dat ” u 1:3 w l
The plot command runs the program using the parameters NTRANS =
2 000, NSTEPS = 70 000, r = 3.8 and x0 = 0.20, 0.35, 0.50, 0.75, 0.90 and
plots the results from the contents of the file lia.dat.
In order to determine the regions of chaotic behavior we have to study
the dependence of the Liapunov exponent λ on the value of r. Using our
experience coming from the careful computation of λ before, we will run
the program for several values of r using the parameters NTRANS = 2 000,
NSTEPS = 60 000 from the initial point x0 = 0.2. This calculation gives
accuracy of the order of 1%. If we wish to measure λ carefully and
estimate the error of the results, we have to follow the steps described in
174 CHAPTER 3. LOGISTIC MAP
figures 3.7 and 3.8. The program can be found in the file liapunov3.f90
and it is a simple modification of the previous program so that it can
perform the calculation for many values of r.
! ===========================================================
! D i s c r e t e L o g i s t i c Map :
! Liapunov exponent from sum_i l n | f ’ ( x _ i ) |
! C a l c u l a t i o n f o r r i n [ rmin , rmax ] with RSTEPS s t e p s
! RSTEPS : v a l u e s or r s t u d i e d : r=rmin +(rmax−rmin ) / RSTEPS
! NTRANS: number o f d i s c a r d e d i t e r a t i o n i n order t o d i s c a r d
! t r a n s i e n t behavior
! NSTEPS : number o f terms i n t h e sum
! x s t a r t : v a l u e o f i n i t i a l x0 f o r eve ry r
! ===========================================================
program logistic_map
i m p l i c i t none
r e a l ( 8 ) , parameter : : rmin = 2.5 D0
r e a l ( 8 ) , parameter : : rmax = 4.0 D0
r e a l ( 8 ) , parameter : : xstart = 0.2 D0
i n t e g e r , parameter : : RSTEPS = 1000
i n t e g e r , parameter : : NSTEPS = 60000
i n t e g e r , parameter : : NTRANS = 2000
i n t e g e r : : i , ir
r e a l ( 8 ) : : r , x0 , x1 , sum , dr
open ( u n i t =33 , f i l e = ’ l i a . dat ’ )
dr = ( rmax−rmin ) / ( RSTEPS −1)
do ir =0 , RSTEPS−1
r = rmin+ir * dr
x0= xstart
do i =1 , NTRANS
x1 = r * x0 * ( 1 . 0 D0−x0 )
x0 = x1
enddo
sum = l o g (ABS( r * ( 1 . 0 D0 −2.0D0 * x0 ) ) )
! −−−−− C a l c u l a t e :
do i =2 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
sum = sum + l o g (ABS( r * ( 1 . 0 D0 −2.0D0 * x1 ) ) )
x0 = x1
enddo
w r i t e ( 3 3 , * ) r , sum / NSTEPS
enddo ! do i r =0 ,RSTEPS−1
c l o s e (3 3)
end program logistic_map
The program can be compiled and run using the commands:
3.6. LIAPUNOV EXPONENTS 175
> g f o r t r a n liapunov3 . f90 −o l
> ./l &
The character & makes the program ./l to run in the background. This
is recommended for programs that run for a long time, so that the shell
returns the prompt to the user and the program continues to run even
after the shell is terminated.
0.6
0.4
0.2
λ
0
-0.2
-0.4
2.6 2.8 3 3.2 3.4 3.6 3.8 4
r
Figure 3.9: The Liapunov exponent λ of the logistic map calculated via equation
(3.35). Note the chaotic behavior that manifests for the values of r where λ > 0 and
the windows of stable k-cycles where λ < 0. Compare this plot with the bifurcation
diagram of figure 3.4. At the points where λ = 0 we have onset of chaos (or “edge of
chaos”) with manifestation of weak chaos (i.e. |∆xn | ∼ |∆x0 |nω ). At these points we
have transitions from stable k-cycles to strong chaos. We observe the onset of chaos for
the first time when r = rc ≈ 3.5699, at which point λ = 0 (for smaller r the plot seems
to touch the λ = 0 line, but in fact λ takes negative values with |λ| very small).
The data are saved in the file lia.dat and we can make the plot
shown in figure 3.7 using gnuplot:
gnuplot > p l o t ” l i a . dat ” with lines notitle , 0 notitle
Now we can compare figure 3.9 with the bifurcation diagram shown in
figure 3.4. The intervals with λ < 0 correspond to stable k-cycles. The
176 CHAPTER 3. LOGISTIC MAP
intervals where λ > 0 correspond to manifestation of strong chaos. These
intervals are separated by points with λ = 0 where the system exhibits
weak chaos. This means that neighboring trajectories diverge from each
other with a power law |∆xn | ∼ |∆x0 |nω instead of an exponential, where
ω = 1/(1 − q) is a positive exponent that needs to be determined. The
parameter q is the one usually used in the literature. Strong chaos is
obtained in the q → 1 limit. For larger r, switching between chaotic and
stable periodic trajectories is observed each time λ changes sign. The
critical values of r can be computed with relatively high accuracy by
restricting the calculation to a small enough neighborhood of the critical
point. You can do this using the program listed above by setting the
parameters rmin and rmax.
5 5
4 4
3 3
p(x)
p(x)
2 2
1 1
0 0
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
x x
Figure 3.10: The distribution functions p(x) of x of the logistic map for r = 3.59
(left) and 3.8 (right). The chaotic behavior appears to be weaker for r = 3.59, and this
is reflected on the value of the entropy. One sees that there exist intervals of x with
p(x) = 0 which become smaller and vanish as r gets close to 4. This distribution is very
hard to be distinguished from a truly random distribution.
We can also study the chaotic properties of the trajectories of the
logistic map by computing the distribution p(x) of the values of x in
the interval (0, 1). After the transitional period, the distribution p(x)
for the k cycles will have support only at the points of the k cycles,
whereas for the chaotic regimes it will have support on subintervals of
(0, 1). The distribution function p(x) is independent for most of the initial
points of the trajectories. If one obtains a large number of points from
many trajectories of the logistic map, it will be practically impossible to
understand that these are produced by a deterministic rule. For this
reason, chaotic systems can be used for the production of pseudorandom
numbers, as we will see in chapter 11. By measuring the entropy, which is
a measure of disorder in a system, we can quantify the “randomness” of
3.6. LIAPUNOV EXPONENTS 177
the distribution. As we will see in chapter 12, it is given by the equation
∑
S=− pk ln pk , (3.37)
k
where pk is the probability of observing the state k. In our case, we can
make an approximate calculation of S by dividing the interval (0, 1) to
N subintervals of width ∆x. For given r we obtain a large number M
of values xn of the logistic map and we compute the histogram hk of
their distribution in the intervals (xk , xk + ∆x). The probability density
is obtained from the limit of pk = hk /(M ∆x) as M becomes large and ∆x
∑ ∫1
small (large N ). Indeed, N pk ∆x = 1 converges to 0 p(x) dx = 1. We
∑ k=1
will define S = − N k=1 pk ln pk ∆x.
5
4
3
p(x)
2
1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
Figure 3.11: The distribution p(x) of x for the logistic map for r = 4. We observe
strong chaotic behavior, p(x) has support over the whole interval (0, 1) and the entropy
is large. The solid line is the analytic form of the distribution p(x) = π −1 x−1/2 (1−x)−1/2
which is known for r = 4 [30]. This is the beta distribution for a = 1/2, b = 1/2.
The program listed below calculates pk for chosen values of r, and
then the entropy S is calculated using (3.37). It is a simple modification
of the program in liapunov3.f90 where we add the parameter NHIST
counting the number of intervals N for the histograms. The probability
density is calculated in the array p(NHIST). The program can be found
in the file entropy.f90:
178 CHAPTER 3. LOGISTIC MAP
! ===========================================================
! D i s c r e t e L o g i s t i c Map :
! Entropy c a l c u l a t i o n from S=−sum_i p _ i l n p _ i
! C a l c u l a t i o n f o r r i n [ rmin , rmax ] with RSTEPS s t e p s
! RSTEPS : v a l u e s or r s t u d i e d : r=rmin +(rmax−rmin ) / RSTEPS
! NHIST : number o f histogram b i n s f o r c a l c u l a t i o n o f p _ i
! NSTEPS : number o f v a l u e s o f x i n t h e h i s t o g r a m s
! NTRANS: number o f d i s c a r t e d i t e r a t i o n i n order t o d i s c a r d
! t r a n s i e n t behavior
! x s t a r t : v a l u e o f i n i t i a l x0 f o r eve ry r
! ===========================================================
program logistic_map
i m p l i c i t none
r e a l ( 8 ) , parameter : : rmin = 2.5 D0
r e a l ( 8 ) , parameter : : rmax = 4.0 D0
r e a l ( 8 ) , parameter : : xstart = 0.2 D0
i n t e g e r , parameter : : RSTEPS = 1000
i n t e g e r , parameter : : NHIST = 10000
i n t e g e r , parameter : : NTRANS = 2000
i n t e g e r , parameter : : NSTEPS = 5000000
r e a l ( 8 ) , parameter : : xmin =0.0 D0 , xmax =1.0 D0
i n t e g e r : : i , ir , isum , n
r e a l ( 8 ) : : r , x0 , x1 , sum , dr , dx
r e a l ( 8 ) : : p ( NHIST ) , S
open ( u n i t =33 , f i l e = ’ entropy . dat ’ )
p = 0.0 D0
dr = ( rmax−rmin ) / ( RSTEPS −1)
dx = ( xmax−xmin ) / ( NHIST −1)
do ir =0 , RSTEPS−1
r = rmin+ir * dr
x0= xstart
do i =1 , NTRANS
x1 = r * x0 * ( 1 . 0 D0−x0 )
x0 = x1
enddo
! make histogram :
n=INT ( x0 / dx ) +1; p ( n )=p ( n ) +1.0 D0
do i =2 , NSTEPS
x1 = r * x0 * ( 1 . 0 D0−x0 )
n = INT ( x1 / dx ) +1
p ( n )=p ( n ) +1.0 D0
x0 = x1
enddo
! p ( k ) i s now histogram o f x−v a l u e s .
! Normalize so t h a t sum_k p ( k ) * dx=1
! to get probability d i s t r i b u t i o n :
p = p / NSTEPS / dx
! sum a l l non z e r o terms : p ( n ) * l o g ( p ( n ) ) * dx
3.6. LIAPUNOV EXPONENTS 179
S = −SUM( p * l o g ( p ) , MASK=p . g t . 0 . 0 D0 ) * dx
write (33 ,*) r , S
enddo ! do i r =0 ,RSTEPS−1
c l o s e (33)
! pr i n t the l a s t p r o b a b i l i t y d i s t r i b u t i o n :
open ( u n i t =34 , f i l e = ’ e n t r o p y _ h i s t . dat ’ )
do n =1 , NHIST
x0 = xmin +(n−1) * dx + 0.5 D0 * dx
w r i t e ( 3 4 , * ) r , x0 , p ( n )
enddo
c l o s e (34)
end program logistic_map
0
-1
-2
-3
-4
S
-5
-6
-7
-8
-9
3.5 3.55 3.6 3.65 3.7 3.75 3.8 3.85 3.9 3.95 4
r
∑
Figure 3.12: The entropy S = − k pk ln pk ∆x for the logistic map as a function
of r. The vertical line is rc ≈ 3.56994567 which marks the beginning of chaos and the
horizontal is the corresponding entropy. The entropy is low for small values of r, where
we have the stable 2n cycles, and large in the chaotic regimes. S drops suddenly when
we pass√to a (temporary) periodic behavior interval. We clearly observe the 3-cycle for
r = 1 + 8 ≈ 3.8284 and the subsequent bifurcations that we observed in the bifurcation
diagram (figure 3.4) and the Liapunov exponent diagram (figure 3.9).
For the calculation of the distribution functions and the entropy we
have to choose the parameters which control the systematic error. The
parameter NTRANS should be large enough so that the transitional behav-
ior will not contaminate our results. Our measurements must be checked
180 CHAPTER 3. LOGISTIC MAP
for being independent of its value. The same should be done for the ini-
tial point xstart. The parameter NHIST controls the partitioning of the
interval (0, 1) and the width ∆x, so it should be large enough. The pa-
rameter NSTEPS is the number of “measurements” for each value of r and
it should be large enough in order to reduce the “noise” in pk . It is obvi-
ous that NSTEPS should be larger when ∆x becomes smaller. Appropriate
choices lead to the plots shown in figures 3.10 and 3.11 for r = 3.59, 3.58
and 4. We see that stronger chaotic behavior means a wider distribution
of the values of x.
The entropy is shown in figure 3.12. The stable periodic trajectories
lead to small entropy, whereas the chaotic ones lead to large entropy.
There is a sudden increase in the value of the entropy at the beginning
of chaos at r = rc , which increases even further as the chaotic behavior
becomes stronger. During the intermissions of the chaotic behavior there
are sudden drops in the value of the entropy. It is quite instructive to
compare the entropy diagrams with the corresponding bifurcation dia-
grams (see figure 3.4) and the Liapunov exponent diagrams (see figure
3.9). The entropy is increasing until r reaches its maximum value 4, but
this is not done smoothly. By magnifying the corresponding areas in the
plot, we can see an infinite number of sudden drops in the entropy in
intervals of r that become more and more narrow.
3.7. PROBLEMS 181
3.7 Problems
Several of the programs that you need to write for solving the problems of
this chapter can be found in the Problems directory of the accompanying
software of this chapter.
3.1 Confirm that the trajectories of the logistic map for r < 1 are falling
off exponentially for large enough n.
(a) Choose r = 0.5 and plot the trajectories for x0 = 0.1 − 0.9 with
step 0.1 for n = 1, . . . , 1000. Put the y axis in a logarithmic
scale. From the resulting curves discuss whether you obtain
an exponential falloff.
(b) Fit the points xn for n > 20 to the function c e−ax and deter-
mine the fitting parameters a and c. How do these parameters
depend on the initial point x0 ? You can use the following
gnuplot commands for your calculation:
gnuplot > ! gfortran logistic . f90 −o l
gnuplot > a = 0 . 7 ; c = 0 . 4 ;
gnuplot > f i t [ 1 0 : ] c * exp(−a * x ) \
”<echo 1000 0.5 0 . 5 | . / l ; c a t l o g . dat ” via a , c
gnuplot > p l o t c * exp(−a * x ) ,\
”<echo 1000 0.5 0 . 5 | . / l ; c a t l o g . dat ” w l
As you can see, we set NSTEPS = 1000, r = 0.5, x0 = 0.5. By
setting the limits [10:] to the fit command, the fit includes
only the points xn ≥ 10, therefore avoiding the transitional
period and the deviation from the exponential falloff for small
n.
(c) Repeat for r = 0.3 − 0.9 with step 0.1 and for r = 0.99, 0.999.
As you will be approaching r = 1, you will need to discard
more points from near the origin. You might also need to
increase NSTEPS. You should always check graphically whether
the fitted exponential function is a good fit to the points xn for
large n. Construct a table for the values of a as a function of
r.
The solutions of the equation (3.3) is e(r−1)x . How is this related to
the values that you computed in your table?
3.2 Consider the logistic map for r = 2. Choose NSTEPS=100 and cal-
culate the corresponding trajectories for x0=0.2, 0.3, 0.5, 0.7,
182 CHAPTER 3. LOGISTIC MAP
0.9. Plot them on the same graph. Calculate the fixed point x∗2
and compare your result to the known value 1 − 1/r. Repeat for
x0= 10−α for α = −1, −2, −5, −10, −20, −25. What do you conclude
about the point x∗1 = 0?
3.3 Consider the logistic map for r = 2.9, 2.99, 2.999. Calculate the stable
point x∗2 and compare your result to the known value 1 − 1/r. How
large should NSTEPS be chosen each time? You may choose x0=0.3.
3.4 Consider the logistic map for r = 3.2. Take x0=0.3, 0.5, 0.9 and
NSTEPS=300 and plot the resulting trajectories. Calculate the fixed
points x∗3 and x∗4 by using the command tail log.dat. Increase
NSTEPS and repeat so that you make sure that the trajectory has
converged to the 2-cycle. Compare their values to the ones given
by equation (3.18). Make the following plots:
gnuplot > p l o t \
”<echo 300 3.2 0 . 3 | . / l ; awk ’NR%2==0’ l o g . dat ” w l
gnuplot > r e p l o t \
”<echo 300 3.2 0 . 3 | . / l ; awk ’NR%2==1’ l o g . dat ” w l
What do you observe?
3.5 Repeat the previous problem for r = 3.4494. How big should NSTEPS
be chosen so that you obtain x3∗ and x∗4 with an accuracy of 6 sig-
nificant digits?
3.6 Repeat the previous problem for r = 3.5 and 3.55. Choose NSTEPS =
1000, x0 = 0.5. Show that the trajectories approach a 4-cycle and
an 8-cycle respectively. Calculate the fixed points x∗5 -x∗8 and x∗9 -x∗16 .
3.7 Plot the functions f (x), f (2) (x), f (4) (x), x for given r on the same
graph. Use the commands:
gnuplot > s e t samples 1000
gnuplot > f ( x ) = r * x*(1 −x )
gnuplot > r =1; p l o t [ 0 : 1 ] x , f ( x ) , f ( f ( x ) ) , f ( f ( f ( f ( x ) ) ) )
√
The command r=1 sets the value of r. Take r = 2.5, 3, 3.2, 1+ 6, 3.5.
Determine the fixed points and the k-cycles from the intersections
of the plots with the diagonal y = x.
3.7. PROBLEMS 183
3.8 Construct the cobweb plots of figures 3.2 and 3.4 for r = 2.8, 3.3
and 3.5. Repeat by dropping from the plot an increasing number
of initial points, so that in the end only the k-cycles will remain.
Do the same for r = 3.55.
3.9 Construct the bifurcation diagrams shown in figure 3.4.
3.10 Construct the bifurcation diagram of the logistic map for 3.840 < r <
3.851 and for 0.458 < x < 0.523. Compute the first four bifurcation
points with an accuracy of 5 significant digits by magnifying the
appropriate parts of the plots. Take NTRANS=15000.
3.11 Construct the bifurcation diagram of the logistic map for 2.9 < r <
(n)
3.57. Compute graphically the bifurcation points rc for n = 2, 3, 4,
5, 6, 7, 8. Make sure that your results are stable against variations
of the parameters NTRANS, NSTEPS as well as from the choice of
(n)
branching point. From the known values of rc for n = 2, 3, and
from the dependence of your results on the choices of NTRANS,
NSTEPS, estimate the accuracy achieved by this graphical method.
(n) (n−1) (n+1) (n)
Compute the ratios (rc − rc )/(rc − rc ) and compare your
results to equation (3.20).
3.12 Choose the values of ρ in equation (3.24) so that you obtain only
one energy level. Compute the resulting value of the energy. When
do we have three energy levels?
3.13 Consider the polynomial g(x) = x3 − 2x2 − 11x + 12. Find the roots
obtained by the Newton-Raphson method when you choose x0 =
2.35287527, 2.35284172, 2.35283735, 2.352836327, 2.352836323. What
do you conclude concerning the basins of attraction of each root of
the polynomial? Make a plot of the polynomial in a neighborhood
of its roots and try other initial points that will converge to each
one of the roots.
3.14 Use the Newton-Raphson method in order to compute the 4-cycle
x∗5 , . . . , x∗8 of the logistic map. Use appropriate areas of the bifur-
cation diagram so that you can choose the initial points correctly.
Check that your result for rc is the same for all x∗α . Tune the
(4)
parameters chosen in your calculation on order to improve the ac-
curacy of your measurements.
3.15 Repeat the previous problem for the 8-cycle x∗9 , . . . , x∗16 and rc .
(5)
184 CHAPTER 3. LOGISTIC MAP
(n) (n)
n rc n rc
2 3.0000000000 10 3.56994317604
3 3.4494897429 11 3.569945137342
4 3.544090360 12 3.5699455573912
5 3.564407266 13 3.569945647353
6 3.5687594195 14 3.5699456666199
7 3.5696916098 15 3.5699456707464
8 3.56989125938 16 3.56994567163008
9 3.56993401837 17 3.5699456718193
rc = 3.56994567 . . .
Table 3.1: The values of rc(n) for the logistic map calculated for problem 17. rc(∞) ≡ rc
is taken from the bibliography.
3.16 Repeat the previous problem for the 16-cycle x∗17 , . . . , x∗32 and rc .
(6)
(n)
3.17 Calculate the critical points rc for n = 3, . . . , 17 of the logistic map
using the Newton-Raphson method. In order to achieve that, you
should determine the bifurcation points graphically in the bifurca-
tion diagram first and then choose the initial points in the Newton-
Raphson method appropriately. The program in bifurcationPoints.f90
should read the parameters eps, epsx, epsr from the stdin so that
they can be tuned for increasing n. If these parameters are too small
the convergence will be unstable and if they are too large you will
have large systematic errors. Using this method, try to reproduce
table 3.1
3.18 Calculate the ratios ∆r(n) /∆r(n+1) of equation (3.20) using the re-
sults of table 3.1. Calculate Feigenbaum’s constant and comment
on the accuracy achieved by your calculation.
3.19 Estimate Feigenbaum’s constant δ and the critical value rc by as-
suming that for large enough n, rc ≈ rc − Cδ −n . This behavior
(n)
is a result of equation (3.20). Fit the results of table 3.1 to this
function and calculate δ and rc . This hypothesis is confirmed in
figure 3.13 where we can observe the exponential convergence of
(n)
rc to rc . Construct the same plot using the parameters of your
calculation.
Hint: You can use the following gnuplot commands:
3.7. PROBLEMS 185
gnuplot > nmin =2; nmax =17
gnuplot > r ( x )= rc−c * d **( − x )
gnuplot > f i t [ nmin : nmax ] r ( x ) ” r c r i t ” u 1 : 2 via rc , c , d
gnuplot > plot ” r c r i t ” , r(x)
gnuplot > p r i n t rc , d
The file rcrit contains the values of table 3.1. You should vary the
parameters nmin, nmax and repeat until you obtain a stable fit.
1
C δ-n
0.01
0.0001
rc(n) - rc
1e-06
1e-08
1e-10
1e-12
2 4 6 8 10 12 14 16 18
n
Figure 3.13: Test of the relation rc(n) ≈ rc − Cδ −n discussed in problem 17. The
parameters used in the plot are approximately rc = 3.5699457, δ = 4.669196 and C =
12.292.
3.20 Use the Newton-Raphson method to calculate the first three bifur-√
cation points after the appearance of the 3-cycle for r = 1 + 8.
Choose one bifurcation point of the 3-cycle, one of the 6-cycle and
one of the 12-cycle and magnify the bifurcation diagram in their
neighborhood.
3.21 Consider the map describing the evolution of a population
xn+1 = p(xn ) = xn er(1−xn ) . (3.38)
(a) Plot the functions x, p(x), p(2) (x), p(4) (x) for r = 1.8, 2, 2.6, 2.67,
2.689 for 0 < x < 8. For which values of r do you expect to
obtain stable k-cycles?
186 CHAPTER 3. LOGISTIC MAP
(b) For the same values of r plot the trajectories with initial points
x0 = 0.2, 0.5, 0.7. For each r make a separate plot.
(c) Use the Newton-Raphson method in order to determine the
(n)
points rc for n = 3, 4, 5 as well as the first two bifurcation
points of the 3-cycle.
(d) Construct the bifurcation diagram for 1.8 < r < 4. Determine
the point marking the onset of chaos as well as the point where
the 3-cycle starts. Magnify the diagram around a branch that
you will choose.
(e) Estimate Feigenbaum’s constant δ as in problem 17. Is your
result compatible with the expectation of universality for the
value of δ? Is the value of rc the same as that of the logistic
map?
3.22 Consider the sine map:
xn+1 = s(xn ) = r sin(πxn ) . (3.39)
(a) Plot the functions x, s(x), s(2) (x), s(4) (x), s(8) (x) for r = 0.65,
0.75, 0.84, 0.86, 0.88. Which values of r are expected to lead to
stable k-cycles?
(b) For the same values of r, plot the trajectories with initial points
x0 = 0.2, 0.5, 0.7. Make one plot for each r.
(c) Use the Newton-Raphson method in order to determine the
(n)
points rc for n = 3, 4, 5 as well as the first two bifurcation
points of the 3-cycle.
(d) Construct the bifurcation diagram for 0.6 < r < 1. Within
which limits do the values of x lie in? Repeat for 0.6 < r < 2.
What do you observe? Determine the point marking the onset
of chaos as well as the point where the 3-cycle starts. Magnify
the diagram around a branch that you will choose.
3.23 Consider the map:
xn+1 = 1 − rx2n . (3.40)
(a) Construct the bifurcation diagram for 0 < r < 2. Within which
limits do the values of x lie in? Determine the point marking
the onset of chaos as well as the point where the 3-cycle starts.
Magnify the diagram around a branch that you will choose.
3.7. PROBLEMS 187
(b) Use the Newton-Raphson method in order to determine the
(n)
points rc for n = 3, 4, 5 as well as the first two bifurcation
points of the 3-cycle.
3.24 Consider the tent map:
{
rxn 0 ≤ xn ≤ 12
xn+1 = r min{xn , 1 − xn } = . (3.41)
r(1 − xn ) 12 < xn ≤ 1
Construct the bifurcation diagram for 0 < r < 2. Within which lim-
its do the values of x lie in? On the same graph, plot the functions
r/2, r − r2 /2.
Magnify the diagram in the area 1.407 < r < 1.416 and 0.580 <
x < 0.588. At which point do the two disconnected intervals within
which xn take their values merge into one? Magnify the areas 1.0 <
r < 1.1, 0.4998 < x < 0.5004 and 1.00 < r < 1.03, 0.4999998 < x <
0.5000003 and determine the merging points of two disconnected
intervals within which xn take their values.
3.25 Consider the Gauss map (or mouse map):
xn+1 = e−rxn + q .
2
(3.42)
Construct the bifurcation diagram for −1 < q < 1 and r = 4.5, 4.9,
7.5. Make your program to take as the initial point of the new
trajectory to be the last one of the previous trajectory and choose
x0 = 0 for q = −1. Repeat for x0 = 0.7, 0.5, −0.7. What do you
observe? Note that as q is increased, we obtain bifurcations and
“anti-bifurcations”.
3.26 Consider the circle map:
xn+1 = [xn + r − q sin(2πxn )] mod 1 . (3.43)
(Make sure that your program keeps the values of xn so that 0 ≤
xn < 1). Construct the bifurcation diagram for 0 < q < 2 and
r = 1/3.
3.27 Use the program in liapunov.f90 in order to compute the distance
between two trajectories of the logistic map for r = 3.6 that origi-
nally are at a distance ∆x0 = 10−15 . Choose x0 = 0.1, 0.2, 0.3, 0.4,
0.5, 0.6, 0.7, 0.8, 0.9, 0.99, 0.999 and calculate the Liapunov exponent
by fitting to a straight line appropriately. Compute the mean value
and the standard error of the mean.
188 CHAPTER 3. LOGISTIC MAP
3.28 Calculate the Liapunov exponent for r = 3.58, 3.60, 3.65, 3.70, 3.80
for the logistic map. Use both ways mentioned in the text. Choose
at least 5 different initial points and calculate the mean and the
standard error of the mean of your results. Compare the values of
λ that you obtain with each method and comment.
(n)
3.29 Compute the critical value rc numerically as the limit lim rc for
n→∞
the logistic map with an accuracy of nine significant digits. Use the
calculation of the Liapunov exponent λ given by equation (3.35).
3.30 Compute the values of r of the logistic map numerically for which
we (a) enter a stable 3-cycle (b) reenter into the chaotic behavior.
Do the calculation by computing the Liapunov exponent λ and
compare your results with the ones obtained from the bifurcation
diagram.
3.31 Calculate the Liapunov exponent using equation (3.35) for the fol-
lowing maps:
xn+1 = xn er(1−xn ) , 1.8 < r < 4
xn+1 = r sin(πxn ) , 0.6 < r < 1
xn+1 = 1 − rx2n , 0<r<2
xn+1 = e−rxn + q ,
2
r = 7.5, −1 < q < 1
[ ]
1
xn+1 = xn + − q sin(2πxn ) mod 1 , 0 < q < 2 ,(3.44)
3
and construct the diagrams similar to the ones in figure 3.9. Com-
pare your plots with the respective bifurcation diagrams (you may
put both graphs on the same plot). Use two different initial points
x0 = 0, 0.2 for the Gauss map (xn+1 = e−rxn + q) and observe the dif-
2
ferences. For the circle map (xn+1 = [xn + 1/3 − q sin(2πxn )] mod 1)
study carefully the values 0 < q < 0.15.
3.32 Reproduce the plots in figures 3.10, 3.11 and 3.12. Compute the
function p(x) for r = 3.68, 3.80, 3.93 and 3.98. Determine the
points where you have stronger chaos by observing p(x) and the
corresponding values of the entropy. Compute the entropy for
r ∈ (3.95, 4.00) by taking RSTEPS=2000 and estimate the values of r
where we enter to and exit from chaos. Compare your results with
the computation of the Liapunov exponent.
3.7. PROBLEMS 189
3.33 Consider the Hénon map:
xn+1 = yn + 1 − ax2n
yn+1 = bxn (3.45)
(a) Construct the two bifurcation diagrams for xn and yn for b =
0.3, 1.0 < a < 1.5. Check if the values a = 1.01, 1.4 that we will
use below correspond to stable periodic trajectories or chaotic
behavior.
(b) Write a program in a file attractor.f90 which will take NINIT
= NL × NL initial conditions (x0 (i), y0 (i)) i = 1, . . . ,NL on a
NL×NL lattice of the square xm ≤ x0 ≤ xM , ym ≤ y0 ≤ yM .
Each of the points (x0 (i), y0 (i)) will evolve according to equa-
tion (3.45) for n = NSTEPS steps. The program will print the
points (xn (i), yn (i)) to the stdout. Choose xm = ym = 0.6,
xM = yM = 0.8, NL= 200.
(c) Choose a = 1.01, b = 0.3 and plot the points (xn (i), yn (i)) for
n = 0, 1, 2, 3, 10, 20, 30, 40, 60, 1000 on the same diagram.
(d) Choose a = 1.4, b = 0.3 and plot the points (xn (i), yn (i)) for
n = 0, . . . , 7 on the same diagram.
(e) Choose a = 1.4, b = 0.3 and plot the points (xn (i), yn (i)) for
n = 999 on the same diagram. Observe the Hénon strange
attractor and its fractal properties. It is characterized by a
Hausdorff¹² dimension dH = 1.261 ± 0.003. Then magnify the
regions
{(x, y)| −1.290 < x < −1.270, 0.378 < y < 0.384} ,
{(x, y)| 1.150 < x < −1.130, 0.366 < y < 0.372} ,
{(x, y)| 0.108 < x < 0.114, 0.238 < y < 0.241} ,
{(x, y)| 0.300 < x < 0.320, 0.204 < y < 0.213} ,
{(x, y)| 1.076 < x < 1.084, 0.090 < y < 0.096} ,
{(x, y)| 1.216 < x < 1.226, 0.032 < y < 0.034} .
3.34 Consider the Duffing map:
xn+1 = yn
yn+1 = −bxn + ayn − yn3 . (3.46)
¹²D.A. Russel, J.D. Hanson, and E. Ott, “Dimension of strange attractors”, Phys. Rev.
Lett. 45 (1980) 1175. See “Hausdorff dimension” in Wikipedia.
190 CHAPTER 3. LOGISTIC MAP
(a) Construct the two bifurcation diagrams for xn and yn for b =
0.3, 0 < a < √ 2.78. Choose
√ four different initial conditions
(x0 , y0 ) = (±1/ 2, ±1/ 2). What do you observe?
(b) Use the program attractor.f90 from problem 33 in order to
study the attractor of the map for b = 0.3, a = 2.75.
3.35 Consider the Tinkerbell map:
xn+1 = x2n − yn2 + axn + byn
yn+1 = 2xn yn + cxn + dyn . (3.47)
(a) Choose a = 0.9, b = −0.6013, c = 2.0, d = 0.50. Plot a trajectory
on the plane by plotting the points (xn , yn ) for n = 0, . . . , 10 000
with (x0 , y0 ) = (−0.72, −0.64).
(b) Use the program attractor.f90 from problem 33 in order to
study the attractor of the map for the values of the parameters
a, b, c, d given above. Choose xm = −0.68, xM = −0.76, ym =
−0.60, yM = −0.68, n = 10 000.
(c) Repeat the previous question by taking d = 0.27.
Chapter 4
Motion of a Particle
In this chapter we will study the numerical solution of classical equations
of motion of one dimensional mechanical systems, e.g. a point particle
moving on the line, the simple pendulum etc. We will make an introduc-
tion to the numerical integration of ordinary differential equations with
initial conditions and in particular to the Euler and Runge-Kutta meth-
ods. We study in detail the examples of the damped harmonic oscillator
and of the damped pendulum under the influence of an external peri-
odic force. The latter system is nonlinear and exhibits interesting chaotic
behavior.
4.1 Numerical Integration of Newton’s Equa-
tions
Consider the problem of the solution of the dynamical equations of mo-
tion of one particle under the influence of a dynamical field given by
Newton’s law. The equations can be written in the form
d2⃗x
= ⃗a(t, ⃗x, ⃗v ) , (4.1)
dt2
where
F⃗ d⃗x
⃗a(t, ⃗x, ⃗v ) ≡ ⃗v = . (4.2)
m dt
From the numerical analysis point of view, the problems that we will dis-
cuss are initial value problems for ordinary differential equations where
the initial conditions
⃗x(t0 ) = ⃗x0 ⃗v (t0 ) = ⃗v0 , (4.3)
191
192 CHAPTER 4. MOTION OF A PARTICLE
determine a unique solution ⃗x(t). The equations (4.1) are of second order
with respect to time and it is convenient to write them as a system of
twice as many first order equations:
d⃗x d⃗v
= ⃗v = ⃗a(t, ⃗x, ⃗v ) . (4.4)
dt dt
In particular, we will be interested in the study of the motion of a particle
moving on a line (1 dimension), therefore the above equations become
dx dv
=v = a(t, x, v) 1-dimension
dt dt
x(t0 ) = x0 v(t0 ) = v0 . (4.5)
When the particle moves on the plane (2 dimensions) the equations of
motion become
dx dvx
= vx = ax (t, x, vx , y, vy ) 2-dimensions
dt dt
dy dvy
= vy = ay (t, x, vx , y, vy )
dt dt
x(t0 ) = x0 vx (t0 ) = v0x
y(t0 ) = y0 vy (t0 ) = v0y , (4.6)
4.2 Prelude: Euler Methods
As a first attempt to tackle the problem, we will study a simple pendulum
of length l in a homogeneous gravitational field g (figure 4.1). The
equations of motion are given by the differential equations
d2 θ g
2
= − sin θ
dt l
dθ
= ω, (4.7)
dt
which can be rewritten as a first order system of differential equations
dθ
= ω
dt
dω g
= − sin θ , (4.8)
dt l
The equations above need to be written in a discrete form appropriate
for a numerical solution with the aid of a computer. We split the interval
4.2. PRELUDE: EULER METHODS 193
θ
l
00
11
00
11
0
1
00
11
0
1 m
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
g
Figure 4.1: A simple pendulum of length l in a homogeneous gravitational field g.
of time of integration [ti , tf ] to N − 1 equal intervals¹ of width ∆t ≡ h,
where h = (tf − ti )/(N − 1). The derivatives are approximated by the
relations (xn+1 − xn )/∆t ≈ x′n , so that
ωn+1 = ωn + αn ∆t
θn+1 = θn + ωn ∆t . (4.9)
where α = −(g/l) sin θ is the angular acceleration. This is the so-called
Euler method. The error at each step is estimated to be of order (∆t)2 .
This is most easily seen by Taylor expanding around the point tn and
neglecting all terms starting from the second derivative and beyond².
What we are mostly interested in is in the total error of the estimate of the
functions we integrate for at time tf ! We expect that errors accumulate in
an additive way at each integration step, and since the number of steps is
N ∝ 1/∆t the total error should be ∝ (∆t)2 × (1/∆t) = ∆t. This is indeed
what happens, and we say that Euler’s method is a first order method.
Its range of applicability is limited and we only study it for academic
reasons. Euler’s method is asymmetric because it uses information only
from the beginning of the integration interval (t, t + ∆t). It can be put
in a more balanced form by using the velocity at the end of the interval
(t, t + ∆t). This way we obtain the Euler-Cromer method with a slightly
¹We have N discrete time points ti ≡ t1 , . . . , tN −1 , tN ≡ tf
²See appendix 4.7 for retails.
194 CHAPTER 4. MOTION OF A PARTICLE
0.5
50
100
1000
0.4 10000
50000
80000
0.3 100000
θs(t)
0.2
0.1
θ
0
-0.1
-0.2
-0.3
-0.4
0 1 2 3 4 5 6 7
t
Figure 4.2: Convergence of Euler’s method for a simple pendulum with period
T ≈ 1.987(ω 2 = 10.0) for several values of the time step ∆t which is determined by the
number of integration steps Nt= 50−100, 000. The solution is given for θ0 = 0.2, ω0 = 0.0
and we compare it with the known solution for small angles with α(t) ≈ −(g/l) θ.
improved behavior, but which is still of first order
ωn+1 = ωn + αn ∆t
θn+1 = θn + ωn+1 ∆t . (4.10)
An improved algorithm is the Euler–Verlet method which is of second
order and gives total error³ ∼ (∆t)2 . This is given by the equations
θn+1 = 2θn − θn−1 + αn (∆t)2
θn+1 − θn−1
ωn = . (4.11)
2∆t
The price that we have to pay is that we have to use a two step
relation in order to advance the solution to the next step. This implies
that we have to carefully determine the initial conditions of the problem
which are given only at one given time ti . We make one Euler time step
backwards in order to define the value of θ0 . If the initial conditions are
θ1 = θ(ti ), ω1 = ω(ti ), then we define
1
θ0 = θ1 − ω1 ∆t + α1 (∆t)2 . (4.12)
2
³See appendix 4.7 for details.
4.2. PRELUDE: EULER METHODS 195
0.025
50
100
0.02 1000
10000
50000
0.015 80000
100000
θs(t)
0.01
0.005
0
θ
-0.005
-0.01
-0.015
-0.02
-0.025
0 1 2 3 4 5 6 7
t
Figure 4.3: Convergence of the Euler-Cromer method, similarly to figure 4.2. We
observe a faster convergence compared to Euler’s method.
It is important that at this step the error introduced is not larger than
O(∆t2 ), otherwise it will spoil and eventually dominate the O(∆t2 ) total
error of the method introduced by the intermediate steps. At the last
step we also have to take
θN − θN −1
ωN = . (4.13)
∆t
Even though the method has smaller total error than the Euler method,
it becomes unstable for small enough ∆t due to roundoff errors. In
particular, the second equation in (4.11) gives the angular velocity as the
ratio of two small numbers. The problem is that the numerator is the
result of the subtraction of two almost equal numbers. For small enough
∆t, this difference has to be computed from the last digits of the finite
representation of the numbers θn+1 and θn in the computer memory. The
accuracy in the determination of (θn+1 − θn ) decreases until it eventually
becomes exactly zero. For the first equation of (4.11), the term αn ∆t2 is
smaller by a factor ∆t compared to the term αn ∆t in Euler’s method.
At some point, by decreasing ∆t, we obtain αn ∆t2 ≪ 2θn − θn−1 and the
accuracy of the method vanishes due to the finite representation of real
number in the memory of the computer. When the numbers αn ∆t2 and
2θn −θn−1 differ from each other by more that approximately seven orders
of magnitude, adding the first one to the second is equivalent to adding
196 CHAPTER 4. MOTION OF A PARTICLE
0.025
50
100
0.02 1000
10000
50000
0.015 80000
100000
θs(t)
0.01
0.005
0
θ
-0.005
-0.01
-0.015
-0.02
-0.025
0 1 2 3 4 5 6 7
t
Figure 4.4: Convergence of the Euler-Verlet method, similarly to figure 4.2. We
observe a faster convergence than Euler’s method, but the roundoff errors make the
results useless for Nt≳ 50, 000 (note what happens when Nt= 100, 000. Why?).
zero and the contribution of the acceleration vanishes⁴.
Writing programs that implement the methods discussed so far is quite
simple. We will write a program that compares the results from all three
methods Euler, Euler–Cromer and Euler–Verlet. The main program is
mainly a user interface, and the computation is carried out by three
subroutines euler, euler_cromer and euler_verlet. The user must
provide the function accel(x) which gives the angular acceleration as a
function of x. The variable x in our problem corresponds to the angle
theta. For starters we take accel(x)= -10.0 * sin(x), the acceleration
of the simple pendulum.
The data structure is very simple: Three real arrays REAL T(P), X(P)
and V(P) store the times tn , the angles θn and the angular velocities ωn for
n = 1, . . . , Nt. The user determines the time interval for the integration
from ti = 0 to tf = Tfi and the number of discrete times Nt. The latter
should be less than P, the size of the arrays. She also provides the initial
conditions θ0 = Xin and ω0 = Vin. After this, we call the main integration
⁴Numbers of type real have approximately seven significant digits. The accuracy of
the operations described above is determined by the number ϵ, which is the smallest
positive number such that 1 + ϵ > 1. For a variable x of some type, this number is
given by a call to the Fortran intrinsic function epsilon(x). For variables of type real,
ϵ ≈ 1.2 × 10−7 and for variables of type real(8) ϵ ≈ 2.2 × 10−16 .
4.2. PRELUDE: EULER METHODS 197
8
50
100
6 1000
10000
15000
4 18000
20000
100000
2
0
-2
v
-4
-6
-8
-10
-12
0 1 2 3 4 5 6 7
t
Figure 4.5: Convergence of Euler’s method for the simple pendulum like in figure
4.2 for θ0 = 3.0, ω0 = 0.0. The behavior of the angular velocity is shown and we notice
unstable behavior for Nt≲ 1, 000.
routines which take as input the initial conditions, the time interval of
the integration and the number of discrete times Xin,Vin,Tfi,Nt. The
output of the routines is the arrays T,X,V which store the results for the
time, position and velocity respectively. The results are printed to the
files euler.dat, euler_cromer.dat and euler_verlet.dat.
After setting the initial conditions and computing the time step ∆t ≡
h = Tfi/(Nt − 1), the integration in each of the subroutines is performed
in do loops which advance the solution for time ∆t. The results are
stored at each step in the arrays T,X,V. For example, the central part of
the program for Euler’s method is:
T ( 1 ) = 0.0
X ( 1 ) = Xin
V ( 1 ) = Vin
h = Tfi / ( Nt −1)
do i = 2 , Nt
T ( i ) = T ( i−1)+h
X ( i ) = X ( i−1)+V ( i−1) * h
V ( i ) = V ( i−1)+accel ( X ( i−1) ) * h
enddo
Some care has to be taken in the case of the Euler–Verlet method where
one has to initialize the first two steps, as well as take special care for the
198 CHAPTER 4. MOTION OF A PARTICLE
8
50
100
1000
6 10000
15000
18000
20000
4 100000
2
0
v
-2
-4
-6
-8
0 1 2 3 4 5 6 7
t
Figure 4.6: Convergence of Euler-Cromer’s method, like in figure 4.5. We observe a
faster convergence than for Euler’s method.
last step for the velocity:
T(1) = 0.0
X(1) = Xin
V(1) = Vin
X0 = X(1) − V ( 1 ) * h + accel ( X ( 1 ) ) *h*h/2.0
T (2) = h
X (2) = 2.0* X (1) − X0 + accel ( X ( 1 ) ) *h*h
do i = 3 , Nt
..............
enddo
V ( Nt )= ( X ( Nt )−X ( Nt −1) ) / h
The full program can be found in the file euler.f90 and is listed below:
! =========================================================
! Program t o i n t e g r a t e e q u a t i o n s o f motion f o r a c c e l e r a t i o n s
! which a r e f u n c t i o n s o f x with t h e method o f Euler ,
! Euler−Cromer and Euler−V e r l e t .
! The u s e r s e t s i n i t i a l c o n d i t i o n s and t h e s u b r o u t i n e s r e t u r n
!X( t ) and V( t )=dX( t ) / dt i n a r r a y s T ( 1 . . Nt ) ,X ( 1 . . Nt ) ,V ( 1 . . Nt )
! The u s e r p r o v i d e s number o f t i m e s Nt and t h e f i n a l
! time T f i . I n i t i a l time i s assumed t o be t _ i =0 and t h e
! i n t e g r a t i o n s t e p h = T f i / ( Nt −1)
! The u s e r programs a r e a l f u n c t i o n a c c e l ( x ) which g i v e s t h e
4.2. PRELUDE: EULER METHODS 199
8
50
100
1000
6 10000
15000
18000
20000
4 100000
2
0
v
-2
-4
-6
-8
0 1 2 3 4 5 6 7
t
Figure 4.7: Convergence of the Euler-Verlet method, similarly to figure 4.5. We
observe a faster convergence compared to Euler’s method but that the roundoff errors
make the results unstable for Nt≳ 18, 000.
! a c c e l e r a t i o n dV( t ) / dt as f u n c t i o n o f X.
!NOTE: T ( 1 ) = 0 T( Nt ) = T f i
! =========================================================
program diff_eq_euler
i m p l i c i t none
i n t e g e r , parameter : : P=110000 ! The s i z e o f t h e a r r a y s
r e a l , dimension ( P ) : : T , X , V ! time t , x ( t ) , v ( t )=dx / dt
real : : Xin , Vin , Tfi ! i n i t i a l conditions
integer : : Nt , i
! The u s e r p r o v i d e s i n i t i a l c o n d i t i o n s X_0 , V_0 f i n a l time t _ f
! and Nt :
p r i n t * , ’ Enter X_0 , V_0 , t _ f , Nt ( t _ i =0) : ’
read ( 5 , * ) Xin , Vin , Tfi , Nt
! This check i s n e c e s s a r y i n order t o avoid memory
! access violations :
i f ( Nt . ge . P ) then
p r i n t * , ’ Nt must be s t r i c t l y l e s s than P . Nt , P= ’ , Nt , P
stop
endif
! Xin= X( 1 ) , Vin=V( 1 ) , T ( 1 ) =0 and t h e r o u t i n e g i v e s e v o l u t i o n i n
! T ( 2 . . Nt ) , X ( 2 . . Nt ) , V ( 2 . . Nt ) which we p r i n t i n a f i l e
c a l l euler ( Xin , Vin , Tfi , Nt , T , X , V )
open ( u n i t =20 , f i l e =” e u l e r . dat ” )
do i =1 , Nt
! Each l i n e i n data f i l e has time , p o s i t i o n , v e l o c i t y :
200 CHAPTER 4. MOTION OF A PARTICLE
write (20 ,*) T(i) , X(i) , V(i)
enddo
c l o s e (2 0) ! we c l o s e t h e u n i t t o be reused below
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
!We r e p e a t e v e r y t h i n g f o r each method
c a l l euler_cromer ( Xin , Vin , Tfi , Nt , T , X , V )
open ( u n i t =20 , f i l e =” e u l e r _ c r o m e r . dat ” )
do i =1 , Nt
write (20 ,*) T(i) , X(i) , V(i)
enddo
c l o s e (2 0)
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
c a l l euler_verlet ( Xin , Vin , Tfi , Nt , T , X , V )
open ( u n i t =20 , f i l e =” e u l e r _ v e r l e t . dat ” )
do i =1 , Nt
write (20 ,*) T(i) , X(i) , V(i)
enddo
c l o s e (2 0)
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
end program diff_eq_euler
! =========================================================
! Function which r e t u r n s t h e v a l u e o f a c c e l e r a t i o n a t
! p o s i t i o n x used i n t h e i n t e g r a t i o n s u b r o u t i n e s
! e u l e r , e u l e r _ c r o m e r and e u l e r _ v e r l e t
! =========================================================
r e a l f u n c t i o n accel ( x )
i m p l i c i t none
real x
accel = −10.0* s i n ( x )
end f u n c t i o n accel
! =========================================================
! D r i v e r r o u t i n e f o r i n t e g r a t i n g e q u a t i o n s o f motion
! using t h e Euler method
! Input :
! Xin=X( 1 ) , Vin=V( 1 ) −− i n i t i a l c o n d i t i o n a t t =0 ,
! T f i t h e f i n a l time and Nt t h e number o f t i m e s
! Output :
! The a r r a y s T ( 1 . . Nt ) , X ( 1 . . Nt ) , V ( 1 . . Nt ) which
! g i v e s x ( t _ k )=X( k ) , dx / dt ( t _ k )=V( k ) , t _ k=T( k ) k = 1 . . Nt
! where f o r k=1 we have t h e i n i t i a l c o n d i t i o n .
! =========================================================
s u b r o u t i n e euler ( Xin , Vin , Tfi , Nt , T , X , V )
i m p l i c i t none
i n t e g e r : : Nt
r e a l , dimension ( Nt ) : : T , X , V ! time t , x ( t ) , v ( t )=dx / dt
real : : Xin , Vin , Tfi
integer : : i
real : : h , accel ! ** declare the function a c c e l **
! I n i t i a l c o n d i t i o n s s e t here :
4.2. PRELUDE: EULER METHODS 201
T ( 1 ) = 0.0
X ( 1 ) = Xin
V ( 1 ) = Vin
! h i s t h e time s t e p Dt
h = Tfi / ( Nt −1)
do i = 2 , Nt
T ( i ) = T ( i−1)+h ! time advances by Dt=h
X ( i ) = X ( i−1)+V ( i−1) * h ! advancement o f p o s i t i o n
V ( i ) = V ( i−1)+accel ( X ( i−1) ) * h ! and v e l o c i t y .
enddo
end s u b r o u t i n e euler
! =========================================================
! D r i v e r r o u t i n e f o r i n t e g r a t i n g e q u a t i o n s o f motion
! using t h e Euler−Cromer method
! Input :
! Xin=X( 1 ) , Vin=V( 1 ) −− i n i t i a l c o n d i t i o n a t t =0 ,
! T f i t h e f i n a l time and Nt t h e number o f t i m e s
! Output :
! The a r r a y s T ( 1 . . Nt ) , X ( 1 . . Nt ) , V ( 1 . . Nt ) which
! g i v e s x ( t _ i )=X( i ) , dx / dt ( t _ i )=V( i ) , t _ i =T( i ) i = 1 . . Nt
! where f o r i =1 we have t h e i n i t i a l c o n d i t i o n .
! =========================================================
s u b r o u t i n e euler_cromer ( Xin , Vin , Tfi , Nt , T , X , V )
i m p l i c i t none
i n t e g e r : : Nt
r e a l , dimension ( Nt ) : : T , X , V ! time t , x ( t ) , v ( t )=dx / dt
real : : Xin , Vin , Tfi
integer : : i
real : : h , accel
T ( 1 ) = 0.0
X ( 1 ) = Xin
V ( 1 ) = Vin
h = Tfi / ( Nt −1)
do i = 2 , Nt
T ( i ) = T ( i−1)+h
V ( i ) = V ( i−1)+accel ( X ( i−1) ) * h
! here i s t h e d i f f e r e n c e compared t o Euler
X ( i ) = X ( i−1)+V ( i ) * h
enddo
end s u b r o u t i n e euler_cromer
! =========================================================
! D r i v e r r o u t i n e f o r i n t e g r a t i n g e q u a t i o n s o f motion
! using t h e Euler − V e r l e t method
! Input :
! Xin=X( 1 ) , Vin=V( 1 ) −− i n i t i a l c o n d i t i o n a t t =0 ,
! T f i t h e f i n a l time and Nt t h e number o f t i m e s
202 CHAPTER 4. MOTION OF A PARTICLE
! Output :
! The a r r a y s T ( 1 . . Nt ) , X ( 1 . . Nt ) , V ( 1 . . Nt ) which
! g i v e s x ( t _ i )=X( i ) , dx / dt ( t _ i )=V( i ) , t _ i =T( i ) i = 1 . . Nt
! where f o r i =1 we have t h e i n i t i a l c o n d i t i o n .
! =========================================================
s u b r o u t i n e euler_verlet ( Xin , Vin , Tfi , Nt , T , X , V )
i m p l i c i t none
i n t e g e r : : Nt
r e a l , dimension ( Nt ) : : T , X , V ! time t , x ( t ) , v ( t )=dx / dt
real : : Xin , Vin , Tfi
integer : : i
real : : h , h2 , X0 , o2h
real : : accel
! I n i t i a l c o n d i t i o n s s e t here :
T(1) = 0.0
X(1) = Xin
V(1) = Vin
h = Tfi / ( Nt −1) ! time s t e p
h2 = h*h ! time s t e p squared
o2h = 0.5/h ! h/2
!We have t o i n i t i a l i z e one more s t e p : X0 c o r r e s po n d s t o ’X( 0 ) ’
X0 = X(1) − V ( 1 ) * h + accel ( X ( 1 ) ) * h2 / 2 . 0
T (2) = h
X (2) = 2.0* X (1) − X0 + accel ( X ( 1 ) ) * h2
!Now i s t a r t s from 3 :
do i = 3 , Nt
T(i) = T ( i−1)+h
X(i) = 2 . 0 * X ( i−1) − X ( i−2) + accel ( X ( i−1) ) * h2
V ( i−1) = o2h * ( X ( i )−X ( i−2) )
enddo
! N o t i c e t h a t we have one more s t e p f o r t h e v e l o c i t y :
V ( Nt )= ( X ( Nt )−X ( Nt −1) ) / h
end s u b r o u t i n e euler_verlet
Compiling the running the program can be done with the commands:
> g f o r t r a n euler . f90 −o euler
> . / euler
Enter X_0 , V_0 , t_f , Nt ( t_i =0) :
0.2 0.0 6.0 1000
> l s euler * . dat
euler_cromer . dat euler . dat euler_verlet . dat
> head −n 5 euler . dat
0.000000 0.2000000 0.000000
6.0060062E−03 0.2000000 −1.1932093E−02
1.2012012 E−02 0.1999283 −2.3864185E−02
1.8018018 E−02 0.1997850 −3.5792060E−02
2.4024025E−02 0.1995700 −4.7711499E−02
4.2. PRELUDE: EULER METHODS 203
The last command shows the first 5 lines of the file euler.dat. We see
the data for the time, the position and the velocity stored in 3 columns.
We can graph the results using gnuplot:
gnuplot > p l o t ” e u l e r . dat ” using 1 : 2 with lines
gnuplot > p l o t ” e u l e r . dat ” using 1 : 3 with lines
These commands result in plotting the positions and the velocities as a
function of time respectively. We can add the results of all methods to
the last plot with the commands:
gnuplot > r e p l o t ” e u l e r _ c r o m e r . dat ” using 1 : 3 with lines
gnuplot > r e p l o t ” e u l e r _ v e r l e t . dat ” using 1 : 3 with lines
The results can be seen in figures 4.2–4.7. Euler’s method is unsta-
ble unless we take a quite small time step. The Euler–Cromer method
behaves impressively better. The results converge and remain constant
for Nt∼ 100, 000. The Euler–Verlet method converges much faster, but
roundoff errors kick in soon. This is more obvious in figure 4.7 where
the initial angular position is large. For small angles we can compare
with the solution one obtains for the harmonic pendulum (sin(θ) ≈ θ):
g
α(θ) = − θ ≡ −Ω2 θ
l
θ(t) = θ0 cos(Ωt) + (ω0 /Ω) sin(Ωt)
ω(t) = ω0 cos(Ωt) − (θ0 Ω) sin(Ωt) . (4.14)
In figures 4.2–4.4 we observe that the results agree with the above for-
mulas for the values of ∆t where the methods converge. This way we
can check our program for bugs. The plot of the functions above can be
done with the following gnuplot commands⁵:
gnuplot > s e t dummy t
gnuplot > omega2 = 10
gnuplot > X0 = 0.2
gnuplot > V0 = 0.0
gnuplot > omega = s q r t ( omega2 )
gnuplot > x(t) = X0 * c o s ( omega * t ) +( V0 / omega ) * s i n ( omega * t )
gnuplot > v(t) = V0 * c o s ( omega * t ) −(omega * X0 ) * s i n ( omega * t )
gnuplot > plot x(t) , v(t)
⁵The command set dummy t sets the independent variable for functions to be t
instead of x which is the default.
204 CHAPTER 4. MOTION OF A PARTICLE
The results should not be compared only graphically since subtle differ-
ences can remain unnoticed. It is more desirable to plot the differences of
the theoretical values from the numerically computed ones which can be
done using the commands:
gnuplot > p l o t ” e u l e r . dat ” using 1 : ( $2−x ( $1 ) ) with lines
gnuplot > p l o t ” e u l e r . dat ” using 1 : ( $3−v ( $1 ) ) with lines
The command using 1:($2-x($1)) puts the values found in the first
column on the x axis and the value found in the second column minus
the value of the function x(t) for t equal to the value found in the first
column on the y axis. This way, we can make the plots shown in⁶ figures
4.11-4.14.
4.3 Runge–Kutta Methods
Euler’s method is a one step finite difference method of first order. First
order means that the total error introduced by the discretization of the
integration interval [ti , tf ] by N discrete times is of order ∼ O(h), where
h ≡ ∆t = (tf − ti )/N is the time step of the integration. In this section we
will discuss a generalization of this approach where the total error will
be of higher order in h. This is the class of Runge-Kutta methods which
are one step algorithms where the total discretization error is of order
∼ O(hp ). The local error introduced at each step is of order ∼ O(hp+1 )
leading after N = (tf − ti )/∆t steps to a maximum error of order
tf − ti 1
∼ O(hp+1 ) × N = O(hp+1 ) × ∼ O(hp+1 ) × = O(hp ) . (4.15)
∆t h
In such a case we say that we have a Runge-Kutta method of pth order.
The price one has to pay for the increased accuracy is the evaluation of
the derivatives of the functions in more than one points in the interval
(t, t + ∆t).
Let’s consider for simplicity the problem with only one unknown
function x(t) which evolves in time according to the differential equation:
dx
= f (t, x) . (4.16)
dt
Consider the first order method first. The most naive approach would
⁶A small modification is necessary in order to plot the absolute value of the differences.
4.3. RUNGE–KUTTA METHODS 205
x
2
01
11
00
1
00
11
tn t n+1 t n+2
t
h
Figure 4.8: The geometry of the step of the Runge-Kutta method of 1st order given
by equation (4.17).
be to take the derivative to be given by the finite difference
dx xn+1 − xn
≈ = f (tn , xn ) ⇒ xn+1 = xn + hf (tn , xn ) . (4.17)
dt ∆t
By Taylor expanding, we see that the error at each step is O(h2 ), therefore
the error after integrating from ti → tf is O(h). Indeed,
dx
xn+1 = x(tn + h) = xn + h (xn ) + O(h2 ) = xn + hf (tn , xn ) + O(h2 ) . (4.18)
dt
The geometry of the step is shown in figure 4.8. We start from point 1 and
by linearly extrapolating in the direction of the derivative k1 ≡ f (tn , xn )
we determine the point xn+1 .
We can improve the method above by introducing an intermediate
point 2. This process is depicted in figure 4.9. We take the point 2
in the middle of the interval (tn , tn+1 ) by making a linear extrapolation
from xn in the direction of the derivative k1 ≡ f (tn , xn ). Then we use the
slope at point 2 as an estimator of the derivative within this interval, i.e.
k2 ≡ f (tn+1/2 , xn+1/2 ) = f (tn + h/2, xn + (h/2)k1 ). We use k2 to linearly
206 CHAPTER 4. MOTION OF A PARTICLE
x
3
2
k2
01
11
00
1
00k
11 1
tn t n+1/2 t n+1
t
h/2 h/2
Figure 4.9: The geometry of an integration step of the 2nd order Runge-Kutta method
given by equation (4.19).
extrapolate from xn to xn+1 . Summarizing, we have that
k1 = f (tn , xn )
h h
k2 ≡ f (tn + , xn + k1 )
2 2
xn+1 = xn + hk2 . (4.19)
For the procedure described above we have to evaluate f twice at each
step, thereby doubling the computational effort. The error at each step
(4.19) becomes ∼ O(h3 ), however, giving a total error of ∼ O(h2 ) ∼
O(1/N 2 ). So for given computational time, (4.19) is superior to (4.17).
We can further improve the accuracy gain by using the Runge–Kutta
method of 4th order. In this case we have 4 evaluations of the derivative
f per step, but the total error becomes now ∼ O(h4 ) and the method is su-
perior to that of (4.19)⁷. The process followed is explained geometrically
in figure 4.10. We use 3 intermediate points for evolving the solution
from xn to xn+1 . Point 2 is determined by linearly extrapolating from xn
⁷Not always though! Higher order does not necessarily mean higher accuracy, al-
though this is true in the simple cases considered here.
4.3. RUNGE–KUTTA METHODS 207
x
2
k2
01
k4
11
00
1 4
00k
11 1 3
k3
tn t n+1/2 t n+1
t
h/2 h/2
Figure 4.10: The geometry of an integration step of the Runge-Kutta method of 4th
order given by equation (4.20).
to the midpoint of the interval (tn , tn+1 = tn + h) by using the direction
given by the derivative k1 ≡ f (tn , xn ), i.e. x2 = xn + (h/2)k1 . We calculate
the derivative k2 ≡ f (tn + h/2, xn + (h/2)k1 ) at the point 2 and we use it
in order to determine point 3, also located at the midpoint of the interval
(tn , tn+1 ). Then we calculate the derivative k3 ≡ f (tn + h/2, xn + (h/2)k2 )
at the point 3 and we use it to linearly extrapolate to the end of the in-
terval (tn , tn+1 ), thereby obtaining point 4, i.e. x4 = xn + hk3 . Then we
calculate the derivative k4 ≡ f (tn + h, xn + hk3 ) at the point 4, and we
use all four derivative k1 , k2 , k3 and k4 as estimators of the derivative of
the function in the interval (tn , tn+1 ). If each derivative contributes with
a particular weight in this estimate, the discretization error can become
208 CHAPTER 4. MOTION OF A PARTICLE
∼ O(h5 ). Such a choice is
k1 = f (tn , xn )
h h
k2 = f (tn + , xn + k1 )
2 2
h h
k3 = f (tn + , xn + k2 )
2 2
k4 = f (tn + h, xn + h k3 )
h
xn+1 = xn + (k1 + 2k2 + 2k3 + k4 ) . (4.20)
6
We note that the second term of the last equation takes an average of
the four derivatives with weights 1/6, 1/3, 1/3 and 1/6 respectively. A
generic small change in these values will increase the discretization error
to worse than h5 .
We remind to the reader the fact that by decreasing h the discretization
errors decrease, but that roundoff errors will start showing up for small
enough h. Therefore, a careful determination of h that minimizes the
total error should be made by studying the dependence of the results as
a function of h.
4.3.1 A Program for the 4th Order Runge–Kutta
Consider the problem of the motion of a particle in one dimension. For
this, we have to integrate a system of two differential equations (4.5) for
two unknown functions of time x1 (t) ≡ x(t) and x2 (t) ≡ v(t) so that
dx1 dx2
= f1 (t, x1 , x2 ) = f2 (t, x1 , x2 ) (4.21)
dt dt
4.3. RUNGE–KUTTA METHODS 209
In this case, equations (4.20) generalize to:
k11 = f1 (tn , x1,n , x2,n )
k21 = f2 (tn , x1,n , x2,n )
h h h
k12 = f1 (tn + , x1,n + k11 , x2,n + k21 )
2 2 2
h h h
k22 = f2 (tn + , x1,n + k11 , x2,n + k21 )
2 2 2
h h h
k13 = f1 (tn + , x1,n + k12 , x2,n + k22 )
2 2 2
h h h
k23 = f2 (tn + , x1,n + k12 , x2,n + k22 )
2 2 2
k14 = f1 (tn + h, x1,n + h k13 , x2,n + h k23 )
k24 = f2 (tn + h, x1,n + h k13 , x1,n + h k23 )
h
x1,n+1 = x1,n + (k11 + 2k12 + 2k13 + k14 )
6
h
x2,n+1 = x1,n + (k21 + 2k22 + 2k23 + k24 ) . (4.22)
6
Programming this algorithm is quite simple. The main program is an
interface between the user and the driver routine of the integration. The
user enters the initial and final times ti = Ti and tf = Tf and the number
of discrete time points Nt. The initial conditions are x1 (ti ) = X10, x2 (ti ) =
X20. The main data structure consists of three real arrays T(P), X1(P),
X2(P) which store the times ti ≡ t1 , t2 , . . . , tNt ≡ tf and the corresponding
values of the functions x1 (tk ) and x2 (tk ), k = 1, . . . , Nt. The main program
calls the driver routine RK(T,X1,X2,Ti,Tf,X10,X20,Nt) which “drives”
the heart of the program, the subroutine RKSTEP(t,x1,x2,dt) which per-
forms one integration step using equations (4.22). RKSTEP evolves the
functions x1, x2 at time t by one step h = dt. The routine RK stores
the calculated values in the arrays T, X1 and X2 at each step. When RK
returns the control to the main program, all the results are stored in T,
X1 and X2, which are subsequently printed in the file rk.dat. The full
program is listed below and can be found in the file rk.f90:
! ========================================================
! Program t o s o l v e a 2 ODE system using Runge−Kutta Method
! User must supply d e r i v a t i v e s
! dx1 / dt= f 1 ( t , x1 , x2 ) dx2 / dt=f 2 ( t , x1 , x2 )
! as r e a l f u n c t i o n s
! Output i s w r i t t e n i n f i l e rk . dat
! ========================================================
210 CHAPTER 4. MOTION OF A PARTICLE
program rk_solve
i m p l i c i t none
i n t e g e r , parameter : : P=110000
r e a l , dimension ( P ) : : T , X1 , X2
real : : Ti , Tf , X10 , X20
i n t e g e r : : Nt
integer : : i
! Input :
p r i n t * , ’ Runge−Kutta Method f o r 2−ODEs I n t e g r a t i o n ’
p r i n t * , ’ Enter Nt , Ti , TF , X10 , X20 : ’
read * , Nt , Ti , Tf , X10 , X20
p r i n t * , ’ Nt = ’ , Nt
p r i n t * , ’ Time : I n i t i a l Ti = ’ , Ti , ’ F i n a l Tf= ’ , Tf
print * , ’ X1( Ti )= ’ , X10 , ’ X2( Ti )= ’ , X20
i f ( Nt . g t . P ) s t o p ’ Nt>P ’
! The C a l c u l a t i o n :
c a l l RK ( T , X1 , X2 , Ti , Tf , X10 , X20 , Nt )
! Output :
open ( u n i t =11 , f i l e = ’ rk . dat ’ )
do i =1 , Nt
w r i t e ( 1 1 , * ) T ( i ) , X1 ( i ) , X2 ( i )
enddo
close (11)
end program rk_solve
! ========================================================
! The f u n c t i o n s f1 , f 2 ( t , x1 , x2 ) provided by t h e u s e r
! ========================================================
r e a l f u n c t i o n f1 ( t , x1 , x2 )
i m p l i c i t none
r e a l : : t , x1 , x2
f1=x2 ! dx1 / dt= v = x2
end f u n c t i o n f1
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l f u n c t i o n f2 ( t , x1 , x2 )
i m p l i c i t none
r e a l : : t , x1 , x2
f2=−10.0D0 * x1 ! harmonic o s c i l l a t o r
end f u n c t i o n f2
! ========================================================
!RK(T , X1 , X2 , Ti , Tf , X10 , X20 , Nt ) i s t h e d r i v e r
! f o r t h e Runge−Kutta i n t e g r a t i o n r o u t i n e RKSTEP
! Input : I n i t i a l and f i n a l t i m e s Ti , Tf
! I n i t i a l v a l u e s a t t =Ti X10 , X20
! Number o f s t e p s o f i n t e g r a t i o n : Nt−1
! S i z e o f a r r a y s T , X1 , X2
! Output : r e a l a r r a y s T( Nt ) , X1 ( Nt ) ,X2( Nt ) where
! T ( 1 ) = Ti X1 ( 1 ) = X10 X2 ( 1 ) = X20
! X1 ( k ) = X1 ( a t t =T( k ) ) X2( k ) = X2( a t t =T( k ) )
4.3. RUNGE–KUTTA METHODS 211
! T( Nt )=TF
! ========================================================
s u b r o u t i n e RK ( T , X1 , X2 , Ti , Tf , X10 , X20 , Nt )
i m p l i c i t none
i n t e g e r : : Nt
r e a l , dimension ( Nt ) : : T , X1 , X2
real : : Ti , Tf , X10 , X20
real : : dt
real : : TS , X1S , X2S ! v a l u e s o f time and X1 , X2 a t gi v e n s t e p
integer : : i
! I n i t i a l i z e variables :
dt = ( Tf−Ti ) / ( Nt −1)
T (1) = Ti
X1 ( 1 ) = X10
X2 ( 1 ) = X20
TS = Ti
X1S = X10
X2S = X20
! Make RK s t e p s : The arguments o f RKSTEP
! a r e r e p l a c e d with t h e new ones !
do i =2 , Nt
c a l l RKSTEP ( TS , X1S , X2S , dt )
T ( i ) = TS
X1 ( i ) = X1S
X2 ( i ) = X2S
enddo
end s u b r o u t i n e RK
! ========================================================
! S u b r o u t i n e RKSTEP( t , x1 , x2 , dt )
! Runge−Kutta I n t e g r a t i o n r o u t i n e o f ODE
! dx1 / dt= f 1 ( t , x1 , x2 ) dx2 / dt=f 2 ( t , x1 , x2 )
! User must supply d e r i v a t i v e f u n c t i o n s :
! r e a l f u n c t i o n f 1 ( t , x1 , x2 )
! r e a l f u n c t i o n f 2 ( t , x1 , x2 )
! Given i n i t i a l p o i n t ( t , x1 , x2 ) t h e r o u t i n e advances i t
! by time dt .
! Input : I n i t a l time t and f u n c t i o n v a l u e s x1 , x2
! Output : F i n a l time t +dt and f u n c t i o n v a l u e s x1 , x2
! C a r e f u l ! : v a l u e s o f t , x1 , x2 a r e o v e r w r i t t e n . . .
! ========================================================
s u b r o u t i n e RKSTEP ( t , x1 , x2 , dt )
i m p l i c i t none
r e a l : : t , x1 , x2 , dt
r e a l : : f1 , f2
r e a l : : k11 , k12 , k13 , k14 , k21 , k22 , k23 , k24
r e a l : : h , h2 , h6
h =dt ! h =dt , i n t e g r a t i o n s t e p
h2 =0.5 D0 * h ! h2=h / 2
212 CHAPTER 4. MOTION OF A PARTICLE
h6 =h / 6 . 0 ! h6=h / 6
k11=f1 ( t , x1 , x2 )
k21=f2 ( t , x1 , x2 )
k12=f1 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 )
k22=f2 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 )
k13=f1 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 )
k23=f2 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 )
k14=f1 ( t+h , x1+h * k13 , x2+h * k23 )
k24=f2 ( t+h , x1+h * k13 , x2+h * k23 )
t =t+h
x1 =x1+h6 * ( k11 +2.0 D0 * ( k12+k13 )+k14 )
x2 =x2+h6 * ( k21 +2.0 D0 * ( k22+k23 )+k24 )
end s u b r o u t i n e RKSTEP
4.4 Comparison of the Methods
100
50
500
5000
50000
1
0.01
1e-04
δx
1e-06
1e-08
1e-10
1e-12
0.1 1 10
t
Figure 4.11: The discrepancy of the numerical results of the Euler method from the
analytic solution for the simple harmonic oscillator. The parameters chosen are ω 2 = 10,
ti = 0, tf = 6, x(0) = 0.2, v(0) = 0 and the number of steps is N = 50, 500, 5, 000, 50, 000.
Observe that the error becomes approximately ten times smaller each time according to
the expectation of being of order ∼ O(∆t).
In this section we will check our programs for correctness and ac-
curacy w.r.t. discretization and roundoff errors. The simplest test is to
4.4. COMPARISON OF THE METHODS 213
0.1
50
500
5000
0.01 50000
0.001
1e-04
1e-05
δx
1e-06
1e-07
1e-08
1e-09
1e-10
0.1 1 10
t
Figure 4.12: Like in figure 4.11 for the Euler-Cromer method. The error becomes
approximately ten times smaller each time according to the expectation of being of order
∼ O(∆t).
check the results against a known analytic solution of a simple model.
This will be done for the simple harmonic oscillator. Our programs will
need small changes which are summarized below. First, we will need
to use higher accuracy variables and we will change all variables of type
REAL to REAL(8). For this we need to change the corresponding dec-
larations in the beginning of each (sub)program. For each numerical
constant in the program we need to put an explicit exponent with the
letter D instead of an E. For example 0.5 → 0.5D0, 1.2E-3 → 1.2D-3 etc.
Then we need to alter the functions that compute the acceleration of the
particle to give a = −ω 2 x. We will take ω 2 = 10 (T ≈ 1.987). Therefore
the relevant part of the program in euler.f90 becomes
r e a l ( 8 ) f u n c t i o n accel ( x )
i m p l i c i t none
real (8) : : x
accel = −10.0D0 * x
end f u n c t i o n accel
and that of the program in rk.f90 becomes
r e a l ( 8 ) f u n c t i o n f2 ( t , x1 , x2 )
i m p l i c i t none
214 CHAPTER 4. MOTION OF A PARTICLE
1
50
500
5000
50000
0.01
1e-04
1e-06
δx
1e-08
1e-10
1e-12
1e-14
0.1 1 10
t
Figure 4.13: Like in figure 4.11 for the Euler-Verlet method. The error becomes
approximately 100 times smaller each time according to the expectation of being of
order ∼ O(∆t2 ).
r e a l ( 8 ) : : t , x1 , x2
f2=−10.0D0 * x1
end f u n c t i o n f2
The programs are run for a given time interval ti = 0 to tf = 6 with
the initial conditions x0 = 0.2, v0 = 0. The time step ∆t is varied by
varying the number of steps Nt-1. The computed numerical solution is
compared to the well known solution for the simple harmonic oscillator
a(x) = −ω 2 x
xh (t) = x0 cos(ωt) + (v0 /ω) sin(ωt)
vh (t) = v0 cos(ωt) − (x0 ω) sin(ωt) , (4.23)
We study the deviation δx(t) = |x(t) − xh (t)| and δv(t) = |v(t) − vh (t)| as a
function of the time step ∆t. The results are shown in figures 4.11–4.14.
We note that for the Euler method and the Euler–Cromer method, the
errors are of order O(∆t) as expected. However, the latter has smaller
errors compared to the first one. For the Euler–Verlet method, the error
turns out to be of order O(∆t2 ) whereas for the 4th order Runge–Kutta
is of order⁸ O(∆t4 ).
⁸The reader should confirm these claims, initially by looking at the figures 4.11-4.14
and then by reproducing these results. A particular time t can be chosen and the errors
can be plotted against ∆t, ∆t2 and ∆t4 respectively.
4.5. THE FORCED DAMPED OSCILLATOR 215
0.01
50
500
5000
1e-04 50000
1e-06
1e-08
δx
1e-10
1e-12
1e-14
1e-16
1e-18
0.1 1 10
t
Figure 4.14: Like in figure 4.11 for the 4th order Runge–Kutta method. The error
becomes approximately 10−4 times smaller each time according to the expectation of
being of order ∼ O(∆t4 ). The roundoff errors become apparent for 50, 000 steps.
Another way for checking the numerical results is by looking at a
conserved quantity, like the energy, momentum or angular momentum,
and study its deviation from its original value. In our case we study the
mechanical energy
1 1
E = mv 2 + mω 2 x2 (4.24)
2 2
which is computed at each step. The deviation δE = |E − E0 | is shown
in figures 4.15–4.18.
4.5 The Forced Damped Oscillator
In this section we will study a simple harmonic oscillator subject to a
damping force proportional to its velocity and an external periodic driving
force, which for simplicity will be taken to have a sinusoidal dependence
in time,
d2 x dx
2
+γ + ω02 x = a0 sin ωt , (4.25)
dt dt
where F (t) = ma0 sin ωt and ω is the angular frequency of the driving
force.
216 CHAPTER 4. MOTION OF A PARTICLE
1000
50
500
5000
100 50000
10
1
δE
0.1
0.01
0.001
1e-04
1e-05
0.1 1 10
t
Figure 4.15: Like in figure 4.11 for the case of mechanical energy for the Euler
method.
Consider initially the system without the influence of the driving force,
i.e. with a0 = 0. The real solutions of the differential equation⁹ which
are finite for t → +∞ are given by
√ 2 2 √ 2 2
x0 (t) = c1 e−(γ+ γ −4ω0 )t/2 + c2 e−(γ− γ −4ω0 )t/2 , γ 2 − 4ω02 > 0 , (4.26)
x0 (t) = c1 e−γt/2 + c2 e−γt/2 t , γ 2 − 4ω02 = 0 , (4.27)
(√ )
−γt/2
x0 (t) = c1 e cos −γ + 4ω0 t/2
2 2
(√ )
−γt/2
+c2 e sin −γ + 4ω0 t/2 ,
2 2
γ 2 − 4ω02 < 0 .(4.28)
In the last case, the solution oscillates with an amplitude decreasing ex-
ponentially with time.
In the a0 > 0 case, the general solution is obtained from the sum
of a special solution xs (t) and the solution of the homogeneous equation
x0 (t). A special solution can be obtained from the ansatz xs (t) = A sin ωt+
⁹These are easily obtained by substituting the ansatz x(t) = Ae−Ωt and solving for
Ω.
4.5. THE FORCED DAMPED OSCILLATOR 217
0.1
50
500
5000
0.01 50000
0.001
1e-04
1e-05
δE
1e-06
1e-07
1e-08
1e-09
1e-10
0.1 1 10
t
Figure 4.16: Like in figure 4.11 for the case of mechanical energy for the Euler–
Cromer method.
B cos ωt, which when substituted in (4.25) and solved for A and B we
find that
a0 [(ω02 − ω 2 ) cos ωt + γω sin ωt]
xs (t) = , (4.29)
(ω02 − ω 2 )2 + ω 2 γ 2
and
x(t) = x0 (t) + xs (t) . (4.30)
The solution x0 (t) decreases exponentially with time and eventually only
xs (t) remains. The only case where this is not true, is when we have
resonance without damping for ω = ω0 , γ = 0. In that case the solution
is
a0
x(t) = c1 cos ωt + c2 sin ωt + 2 (cos ωt + 2(ωt) sin ωt) . (4.31)
4ω
The first two terms are the same as that of the simple harmonic oscillator.
The last one increases the amplitude linearly with time, which is a result
of the influx of energy from the external force to the oscillator.
Our program will be a simple modification of the program in rk.f90.
The main routines RK(T,X1,X2,T0,TF,X10,X20,Nt) and RKSTEP(t,x1,x2,dt)
remain as they are. We only change the user interface. The basic param-
eters ω0 , ω, γ, a0 are entered interactively by the user from the standard
input stdin. These parameters should be accessible also by the function
f2(t,x1,x2), and one way to be able to do this, is to store them in vari-
ables which are placed in a common block. Such variables are accessible to
218 CHAPTER 4. MOTION OF A PARTICLE
1
50
500
5000
0.01 50000
1e-04
1e-06
δE
1e-08
1e-10
1e-12
1e-14
1e-16
0.1 1 10
t
Figure 4.17: Like in figure 4.11 for the case of mechanical energy for the Euler–Verlet
method.
all subprograms that declare a common block with the same name using
a COMMON declaration. Such a declaration is shown in the following lines
real (8) : : omega_0 , omega , gamma , a_0 , omega_02 , omega2
common / params / omega_0 , omega , gamma , a_0 , omega_02 , omega2
which when written in a (sub)program, the (sub)program gains ac-
cess to the “memory position” params where the values of the vari-
ables are stored. Another point that needs our attention is the function
f2(t,x1,x2) which now takes the velocity v → x2 in its arguments:
r e a l ( 8 ) f u n c t i o n f2 ( t , x1 , x2 )
i m p l i c i t none
real (8) omega_0 , omega , gamma , a_0 , omega_02 , omega2
common / params / omega_0 , omega , gamma , a_0 , omega_02 , omega2
r e a l ( 8 ) t , x1 , x2 , a
a = a_0 * c o s ( omega * t )
f2=−omega_02 * x1−gamma * x2+a
end f u n c t i o n f2
The main program found in the file dlo.f90 is listed below. The subrou-
tines RK, RKSTEP are the same as in rk.f90 and should also be included
in the same file.
4.5. THE FORCED DAMPED OSCILLATOR 219
0.01
50
500
5000
50000
1e-04
1e-06
1e-08
δE
1e-10
1e-12
1e-14
1e-16
0.1 1 10
t
Figure 4.18: Like in figure 4.11 for the case of mechanical energy for the 4th order
Runge–Kutta method. Roundoff errors appear for large enough number of steps.
! ========================================================
! Program t o s o l v e Damped Lin ea r O s c i l l a t o r
! using 4 th order Runge−Kutta Method
! Output i s w r i t t e n i n f i l e dlo . dat
! ========================================================
program dlo_solve
i m p l i c i t none
i n t e g e r , parameter : : P=110000
r e a l ( 8 ) , dimension ( P ) : : T , X1 , X2
r e a l ( 8 ) : : Ti , Tf , X10 , X20
r e a l ( 8 ) : : Energy
real (8) : : omega_0 , omega , gamma , a_0 , omega_02 , omega2
common / params / omega_0 , omega , gamma , a_0 , omega_02 , omega2
i n t e g e r : : Nt , i
! Input :
p r i n t * , ’ Runge−Kutta Method f o r DLO I n t e g r a t i o n ’
p r i n t * , ’ Enter omega_0 , omega , gamma , a_0 : ’
read * , omega_0 , omega , gamma , a_0
omega_02 = omega_0 * omega_0
omega2 = omega * omega
p r i n t * , ’ omega_0= ’ , omega_0 , ’ omega= ’ , omega
p r i n t * , ’gamma= ’ , gamma , ’ a_0= ’ , a_0
p r i n t * , ’ Enter Nt , Ti , TF , X10 , X20 : ’
read * , Nt , Ti , Tf , X10 , X20
p r i n t * , ’ Nt = ’ , Nt
p r i n t * , ’ Time : I n i t i a l Ti = ’ , Ti , ’ F i n a l Tf= ’ , Tf
220 CHAPTER 4. MOTION OF A PARTICLE
print * , ’ X1( Ti )= ’ , X10 , ’ X2( Ti )= ’ , X20
i f ( Nt . g t . P ) s t o p ’ Nt>P ’
! The C a l c u l a t i o n :
c a l l RK ( T , X1 , X2 , Ti , Tf , X10 , X20 , Nt )
! Output :
open ( u n i t =11 , f i l e = ’ dlo . dat ’ )
w r i t e ( 1 1 , * ) ’ # Damped Lin ea r O s c i l l a t o r − dlo ’
w r i t e ( 1 1 , * ) ’ # omega_0= ’ , omega_0 , ’ omega= ’ , omega ,&
’ gamma= ’ , gamma , ’ a_0= ’ , a_0
do i =1 , Nt
Energy = 0.5 D0 * X2 ( i ) * X2 ( i ) +0.5 D0 * omega_02 * X1 ( i ) * X1 ( i )
w r i t e ( 1 1 , * ) T ( i ) , X1 ( i ) , X2 ( i ) , Energy
enddo
close (11)
end program dlo_solve
! ========================================================
! The f u n c t i o n s f1 , f 2 ( t , x1 , x2 ) provided by t h e u s e r
! ========================================================
r e a l ( 8 ) f u n c t i o n f1 ( t , x1 , x2 )
i m p l i c i t none
r e a l ( 8 ) t , x1 , x2
f1=x2 ! dx1 / dt= v = x2
end f u n c t i o n f1
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f2 ( t , x1 , x2 )
i m p l i c i t none
real (8) omega_0 , omega , gamma , a_0 , omega_02 , omega2
common / params / omega_0 , omega , gamma , a_0 , omega_02 , omega2
r e a l ( 8 ) t , x1 , x2 , a
a = a_0 * c o s ( omega * t )
f2=−omega_02 * x1−gamma * x2+a
end f u n c t i o n f2
The results are shown in figures 4.19–4.22. Figure 4.19 shows the
transition from a damped motion for γ > 2ω0 to an oscillating motion
with damping amplitude for γ < 2ω0 . The exponential decrease of the
amplitude is shown in figure 4.21, whereas the dependence of the period
T from the damping coefficient γ is shown in figure 4.22. Motivated by
equation (4.28), written in the form
( )
2π
4ω0 −
2
= γ2 , (4.32)
T
we construct the plot in figure 4.22. The right hand side of the equation
is put on the horizontal axis, whereas the left hand side on the vertical.
Equation (4.32) predicts that both quantities are equal and all measure-
ments should lie on a particular line, the diagonal y = x. The period T
4.5. THE FORCED DAMPED OSCILLATOR 221
5
6.29
3.15
4 1.57
0.32
3
2
1
0
x
-1
-2
-3
-4
-5
0 2 4 6 8 10
t
Figure 4.19: The position as a function of time for the damped oscillator for several
values of γ and ω0 = 3.145.
can be estimated from the time between two consecutive extrema of x(t)
or two consecutive zeros of the velocity v(t) (see figure 4.19).
Finally it is important to study the trajectory of the system in phase
space. This can be seen¹⁰ in figure 4.20. A point in this space is a state of
the system and a trajectory describes the evolution of the system’s states
in time. We see that all such trajectories end up as t → +∞ to the point
(0, 0), independently of the initial conditions. Such a point is an example
of a system’s attractor.
Next, we add the external force and study the response of the system
to it. The system exhibits a transient behavior that depends on the initial
conditions. For large enough times it approaches a steady state that does
not depend on (almost all of) the initial conditions. This can be seen in
figure 4.23. This is easily understood for our system by looking at equa-
tions (4.26)–(4.28). We see that the steady state xs (t) becomes dominant
when the exponentials have damped away. xs (t) can be written in the
form
x(t) = x0 (ω) cos(ωt + δ(ω))
a0 ωγ
x0 (ω) = √ 2 , tan δ(ω) = . (4.33)
(ω0 − ω 2 )2 + γ 2 ω 2 ω2 − ω02
¹⁰To be precise, phase space is the space of positions-momenta, but in our case the
difference is trivial.
222 CHAPTER 4. MOTION OF A PARTICLE
15
6.29
3.15
1.57
0.32
10
5
0
v
-5
-10
-15
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
Figure 4.20: The phase space trajectory for the damped oscillator for several values
of γ and ω0 = 3.145. Note the attractor at (x, v) = (0, 0) where all trajectories are
“attracted to” as t → +∞.
These equations are verified in figure 4.24 where we study the depen-
dence of the amplitude x0 (ω) on the angular frequency of the driving
force. Finally we study the trajectory of the system in phase space. As
we can see in figure 4.20, this time the attractor is an ellipse, which is
a one dimensional curve instead of a zero dimensional point. For large
enough times, all trajectories approach their attractor asymptotically.
4.6 The Forced Damped Pendulum
In this section we will study a non-linear dynamical system which ex-
hibits interesting chaotic behavior. This is a simple model which, despite
its deterministic nature, the prediction of its future behavior becomes in-
tractable after a short period of time. Consider a simple pendulum in a
constant gravitational field whose motion is damped by a force propor-
tional to its velocity and it is under the influence of a vertical, harmonic
external driving force:
d2 θ dθ
2
+ γ + ω02 sin θ = −2A cos ωt sin θ . (4.34)
dt dt
4.6. THE FORCED DAMPED PENDULUM 223
10
1.0
0.8
0.6
0.4
0.2
0.1
1
Amplitude
0.1
0.01
0 5 10 15 20 25 30 35 40
t
Figure 4.21: The amplitude of oscillation for the damped oscillator for several
values of γ and ω0 = 3.145. Note the exponential damping of the amplitude with time.
In the equation above, θ is the angle of the pendulum with the vertical
axis, γ is the damping coefficient, ω02 = g/L is the pendulum’s natural
angular frequency, ω is the angular frequency of the driving force and
2A is the amplitude of the external angular acceleration caused by the
driving force.
In the absence of the driving force, the damping coefficient drives the
system to the point (θ, θ̇) = (0, 0), which is an attractor for the system.
This continues to happen for small enough A, but for A > Ac the behavior
of the system becomes more complicated.
The program that integrates the equations of motion of the system can
be obtained by making trivial changes to the program in the file dlo.f90.
This changes are listed in detail below, but we note that X1 ↔ θ, X2 ↔ θ̇,
a_0 ↔ A. The final program can be found in the file fdp.f90. It is listed
below, with the understanding that the commands in between the dots
are the same as in the programs found in the files dlo.f90, rk.f90.
! ========================================================
! Program t o s o l v e Forced Damped Pendulum
! using 4 th order Runge−Kutta Method
224 CHAPTER 4. MOTION OF A PARTICLE
10
2
4 ω0 - (2 π/T)
2
1
1 10
γ2
Figure 4.22: The period of oscillation of the damped oscillator for several values of
γ and ω0 = 3.145. The axes are chosen so that equation (4.28) (2π/T )2 = 4ω02 − γ 2 can
be easily verified. The points in the plot are our measurements whereas the straight
line is the theoretical prediction, the diagonal y = x
! Output i s w r i t t e n i n f i l e fdp . dat
! ========================================================
program dlo_solve
i m p l i c i t none
i n t e g e r , parameter : : P=1010000
................................
Energy = 0.5 D0 * X2 ( i ) * X2 ( i )+omega_02 * ( 1 . 0 D0−c o s ( X1 ( i ) ) )
................................
end program dlo_solve
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f2 ( t , x1 , x2 )
i m p l i c i t none
real (8) omega_0 , omega , gamma , a_0 , omega_02 , omega2
common / params / omega_0 , omega , gamma , a_0 , omega_02 , omega2
r e a l ( 8 ) t , x1 , x2
f2=−(omega_02 +2.0 D0 * a_0 * c o s ( omega * t ) ) * s i n ( x1 )−gamma * x2
end f u n c t i o n f2
! ========================================================
s u b r o u t i n e RKSTEP ( t , x1 , x2 , dt )
i m p l i c i t none
................................
4.6. THE FORCED DAMPED PENDULUM 225
1
x0=1 v0=0
x0=0 v0=1
0.8
0.6
0.4
0.2
x(t)
0
-0.2
-0.4
-0.6
-0.8
0 10 20 30 40 50 60
t
Figure 4.23: The period of oscillation for the forced damped oscillator for different
initial conditions. We have chosen ω0 = 3.145, ω = 2.0, γ = 0.5 and a0 = 1.0. We
note that after the transient behavior the system oscillates harmonically according to
the relation x(t) = x0 (ω) cos(ωt + δ).
r e a l ( 8 ) , parameter : : pi =3.14159265358979324 D0
r e a l ( 8 ) , parameter : : pi2 =6.28318530717958648 D0
................................
x1 =x1+h6 * ( k11 +2.0 D0 * ( k12+k13 )+k14 )
x2 =x2+h6 * ( k21 +2.0 D0 * ( k22+k23 )+k24 )
i f ( x1 . g t . pi ) x1 = x1 − pi2
i f ( x1 . l t . −pi ) x1 = x1 + pi2
end s u b r o u t i n e RKSTEP
The final lines in the program are added so that the angle is kept
within the interval [−π, π].
In order to study the system’s properties we will set ω0 = 1, ω = 2,
and γ = 0.2 unless we explicitly state otherwise. The natural period
of the pendulum is T0 = 2π/ω0 = 2π ≈ 6.28318530717958648 whereas
that of the driving force is T = 2π/ω = π ≈ 3.14159265358979324. For
A < Ac , with Ac ≈ 0.18, the point (θ, θ̇) = (0, 0) is an attractor, which
means that the pendulum eventually stops at its stable equilibrium point.
For Ac < A < 0.71 the attractor is a closed curve, which means that
the pendulum at its steady state oscillates indefinitely without circling
226 CHAPTER 4. MOTION OF A PARTICLE
0.7
0.6
0.5
0.4
x0(ω)
0.3
0.2
0.1
0
1 1.5 2 2.5 3 3.5 4 4.5 5
ω
Figure 4.24: The oscillation amplitude x0 (ω) as a function of ω for the forced
damped oscillator, where ω0 = 3.145, γ = 0.5 and a0 = 1.0. We observe a resonance
for ω ≈ ω0 . The points of the plot are our measurements and the line is the theoretical
prediction given by equation (4.33).
through its unstable equilibrium point at θ = ±π. The period of motion
is found to be twice that of the driving force. For 0.72 < A < 0.79
the attractor is an open curve, because at its steady state the pendulum
crosses the θ = ±π point. The period of the motion becomes equal to
that of the driving force. For 0.79 < A ≲ 1.033 we have period doubling
for critical values of A, but the trajectory is still periodic. For even larger
values of A the system enters into a chaotic regime where the trajectories
are non periodic. For A ≈ 3.1 we find the system in a periodic steady
state again, whereas for A ≈ 3.8 – 4.448 we have period doubling. For
A ≈ 4.4489 we enter into a chaotic regime again etc. These results can
be seen in figures 4.27–4.29. The reader should construct the bifurcation
diagram of the system by solving problem 20 of this chapter.
We can also use the so called Poincaré diagrams in order to study the
chaotic behavior of a system. These are obtained by placing a point in
phase space when the time is an integer multiple of the period of the
driving force. Then, if for example the period of the motion is equal
to that of the period of the driving force, the Poincaré diagram consists
of only one point. If the period of the motion is an n–multiple of the
4.6. THE FORCED DAMPED PENDULUM 227
2
1.5
1
0.5
0
-0.5
v
-1
-1.5
-2
-2.5
-3
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
x
Figure 4.25: A phase space trajectory of the forced damped oscillator with ω0 =
3.145, ω = 2.0, γ = 0.5 and a0 = 1.0. The harmonic oscillation which is the steady state
of the system is an ellipse, which is an attractor of all the phase space trajectories that
correspond to different initial conditions.
period of the driving force then the Poincaré diagram consists of only
n points. Therefore, in the period doubling regime, the points of the
Poincaré diagram double at each period doubling point. In the chaotic
regime, the Poincaré diagram consists of an infinite number of points
which belong to sets that have interesting fractal structure. One way to
construct the Poincaré diagram numerically, is to process the data of the
output file fdp.dat using awk¹¹:
awk −v o=$omega −v nt=$Nt −v tf=$TF \
’BEGIN{T =6.283185307179/ o ; dt=tf / nt ; } $1%T<dt { p r i n t $2 , $3 } ’\
fdp . dat
where $omega, $Nt, $TF are the values of the angular frequency ω, the
number of points of time and the final time tf . We calculate the period T
and the time step dt in the program. Then we print those lines of the file
¹¹The command can be written in one line without the final \ of the first and second
lines.
228 CHAPTER 4. MOTION OF A PARTICLE
0.4
0.3
0.2
0.1
0
v
-0.1
-0.2
-0.3
-0.4
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2
x
Figure 4.26: The trajectory shown in figure 4.25 for t > 100. The trajectory is
almost on top of an ellipse corresponding to the steady state motion of the system. This
ellipse is an attractor of the system.
where the time is an integer multiple of the period¹². This is accomplished
by the modulo operation $1 % T. The value of the expression $1 % T <
dt is true when the remainder of the division of the first column ($1) of
the file fdp.dat with the period T is smaller than dt. The results in the
chaotic regime are displayed in figure 4.30.
We close this section by discussing another concept that helps us in
the analysis of the dynamical properties of the pendulum. This is the
concept of the basin of attraction which is the set of initial conditions in
phase space that lead the system to a specific attractor. Take for example
the case for A > 0.79 in the regime where the pendulum at its steady
state has a circular trajectory with a positive or negative direction. By
taking a large sample of initial conditions and recording the direction of
the resulting motion after the transient behavior, we obtain figure 4.31.
¹²The accuracy of this condition is limited by dt, which makes the points in the
Poincaré diagram slightly fuzzy.
4.7. APPENDIX: ON THE EULER–VERLET METHOD 229
2 -1.4
1.5
-1.6
1
-1.8
0.5
-2
0
-2.2
-0.5
-2.4
-1
-2.6
-1.5
-2 -2.8
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -4 -3 -2 -1 0 1 2 3 4
3 -0.5
2.8
-1
2.6
2.4 -1.5
2.2
-2
2
1.8 -2.5
1.6
-3
1.4
1.2 -3.5
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 4.27: A phase space trajectory of the forced damped pendulum. The
parameters chosen are ω0 = 1.0, ω = 2.0, γ = 0.2 and A = 0.60, 0.72, 0.85, 1.02. We
observe the phenomenon of period doubling.
4.7 Appendix: On the Euler–Verlet Method
Equations (4.11) can be obtained from the Taylor expansion
(∆t)2 ′′ (∆t)3 ′′′
θ(t + ∆t) = θ(t) + (∆t)θ′ (t) + θ (t) + θ (t) + O((∆t)4 )
2! 3!
2 3
(∆t) (∆t)
θ(t − ∆t) = θ(t) − (∆t)θ′ (t) + θ′′ (t) − θ′′′ (t) + O((∆t)4 ) .
2! 3!
By adding and subtracting the above equations we obtain
θ(t + ∆t) + θ(t − ∆t) = 2θ(t) + (∆t)2 θ′′ (t) + O((∆t)4 )
θ(t + ∆t) − θ(t − ∆t) = 2(∆t)θ′ (t) + O((∆t)3 ) (4.35)
which give equations (4.11)
θ(t + ∆t) = 2θ(t) − θ(t − ∆t) + (∆t)2 α(t) + O((∆t)4 )
θ(t + ∆t) − θ(t − ∆t)
ω(t) = + O((∆t)2 ) (4.36)
2(∆t)
230 CHAPTER 4. MOTION OF A PARTICLE
-0.5 -0.5
-1 -1
-1.5 -1.5
-2 -2
-2.5 -2.5
-3 -3
-3.5 -3.5
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-0.5 5
4
-1
3
2
-1.5
1
-2 0
-1
-2.5
-2
-3
-3
-4
-3.5 -5
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 4.28: A phase space trajectory of the forced damped pendulum. The
parameters chosen are ω0 = 1.0, ω = 2.0, γ = 0.2 and A = 1.031, 1.033, 1.04, 1.4. We
observe the chaotic behavior of the system.
From the first equation and equations (4.9) we obtain:
θ(t + ∆t) = θ(t) + ω(t)(∆t) + O((∆t)2 ) (4.37)
When we perform a numerical integration, we are interested in the
total error accumulated after N − 1 integration steps. In this method,
these errors must be studied carefully:
• The error in the velocity ω(t) does not accumulate because it is given
by the difference of the positions θ(t + ∆t) − θ(t − ∆t).
• The accumulation of the errors for the position is estimated as fol-
lows: Assume that δθ(t) is the total accumulated error from the
integration from time t0 to t. Then according to the expansions
4.7. APPENDIX: ON THE EULER–VERLET METHOD 231
3.2 5
3 4
3
2.8
2
2.6
1
2.4
0
2.2
-1
2
-2
1.8
-3
1.6 -4
1.4 -5
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
6 8
6
4
4
2
2
0 0
-2
-2
-4
-4
-6
-6 -8
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 4.29: A phase space trajectory of the forced damped pendulum. The
parameters chosen are ω0 = 1.0, ω = 2.0, γ = 0.2 and A = 1.568, 3.8, 4.44, 4.5. We
observe the system exiting and reentering regimes of chaotic behavior.
(4.36) the error for the first step is δθ(t0 + ∆t) = O((∆t)4 ). Then¹³
θ(t0 + 2∆t) = 2θ(t0 + ∆t) − θ(t0 ) + ∆t2 α(t0 + ∆t) + O((∆t)4 ) ⇒
δθ(t0 + 2∆t) = 2δθ(t0 + ∆t) − δθ(t0 ) + O((∆t)4 )
= 2O((∆t)4 ) − 0 + O((∆t)4 )
= 3O((∆t)4 ) .
For the next steps we obtain
θ(t0 + 3∆t) = 2θ(t0 + 2∆t) − θ(t0 + ∆t) + ∆t2 α(t0 + 2∆t) + O((∆t)4 ) ⇒
δθ(t0 + 3∆t) = 2δθ(t0 + 2∆t) − δθ(t0 + ∆t) + O((∆t)4 )
= 6O((∆t)4 ) − O((∆t)4 ) + O((∆t)4 )
= 6O((∆t)4 ) ,
¹³Remember that the acceleration α(t) is given, therefore δα(t) = 0.
232 CHAPTER 4. MOTION OF A PARTICLE
5 8
4
6
3
4
2
2
1
0 0
-1
-2
-2
-4
-3
-6
-4
-5 -8
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 4.30: A Poincaré diagram for the forced damped pendulum in its chaotic
regime. The parameters chosen are ω0 = 1.0, ω = 2.0, γ = 0.2 and A = 1.4, 4.5.
8 8
6 6
4 4
2 2
0 0
-2 -2
-4 -4
-6 -6
-8 -8
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 4.31: Basin of attraction for the forced damped pendulum. The parameters
chosen are ω0 = 1.0, ω = 2.0, γ = 0.2 and A = 0.85, 1.4.
θ(t0 + 4∆t) = 2θ(t0 + 3∆t) − θ(t0 + 2∆t) + ∆t2 α(t0 + 3∆t) + O((∆t)4 ) ⇒
δθ(t0 + 4∆t) = 2δθ(t0 + 3∆t) − δθ(t0 + 2∆t) + O((∆t)4 )
= 12O((∆t)4 ) − 3O((∆t)4 ) + O((∆t)4 )
= 10O((∆t)4 ) .
Then, inductively, if δθ(t0 + (n − 1)∆t) = (n−1)n
2
O((∆t)4 ), we obtain
θ(t0 + n∆t) = 2θ(t0 + (n − 1)∆t) − θ(t0 + (n − 2)∆t) + ∆t2 α(t0 + (n − 1)∆t)
+O((∆t)4 ) ⇒
δθ(t0 + n∆t) = 2δθ(t0 + (n − 1)∆t) − δθ(t0 + (n − 2)∆t) + O((∆t)4 )
(n − 1)n (n − 2)(n − 1)
= 2 O((∆t)4 ) − O((∆t)4 ) + O((∆t)4 )
2 2
n(n + 1)
= O((∆t)4 ) .
2
4.8. APPENDIX: 2ND ORDER RUNGE–KUTTA METHOD 233
Finally
n(n + 1) 1
δθ(t0 + n∆t) = O((∆t)4 ) ∼ O((∆t)4 ) ∼ O((∆t)2 ) .
2 ∆t2
(4.38)
Therefore the total error is O((∆t)2 ).
We also mention the Velocity Verlet method or the Leapfrog method.
In this case we use the velocity explicitly:
1
θn+1 = θn + ωn ∆t + αn ∆t2
2
1
ωn+ 1 = ωn + αn ∆t
2 2
1
ωn+1 = ωn+ 1 + αn+1 ∆t . (4.39)
2 2
The last step uses the acceleration αn+1 which should depend only on
the position θn+1 and not on the velocity.
The Verlet methods are popular in molecular dynamics simulations of
many body systems. One of their advantages is that the constraints of
the system of particles are easily encoded in the algorithm.
4.8 Appendix: 2nd order Runge–Kutta Method
In this appendix we will show how the choice of the intermediate point
2 in equation (4.17) reduces the error by a power of h. This choice is
special, since by choosing another point (e.g. t = tn + 0.4h) the result
would have not been the same. Indeed, from the relation
∫ tn+1
dx
= f (t, x) ⇒ xn+1 = xn + f (t, x) dx . (4.40)
dt tn
By Taylor expanding around the point (tn+1/2 , xn+1/2 ) we obtain
df
f (t, x) = f (tn+1/2 , xn+1/2 ) + (t − tn+1/2 ) (tn+1/2 ) + O(h2 ) . (4.41)
dt
234 CHAPTER 4. MOTION OF A PARTICLE
Therefore
∫ tn+1
f (t, x) dx
tn
tn+1
df (t − tn+1/2 )2
= f (tn+1/2 , xn+1/2 )(tn+1 − tn ) + (tn+1/2 )
dt 2 tn
+O(h )(tn+1 − tn )
2
{ }
df (tn+1 − tn+1/2 )2 (tn − tn+1/2 )2
= f (tn+1/2 , xn+1/2 )h + (tn+1/2 ) −
dt 2 2
2
+O(h )h
{ 2 }
df h (−h)2
= f (tn+1/2 , xn+1/2 )h + (tn+1/2 ) − + O(h3 )
dt 2 2
= f (tn+1/2 , xn+1/2 )h + O(h3 ) . (4.42)
Note that for the vanishing of the O(h) term it is necessary to place the
intermediate point at time tn+1/2 .
This is not a unique choice. This can be most easily seen by a different
analysis of the Taylor expansion. Expanding around the point (tn , xn )
we obtain
2
dxn 1 2 d xn
xn+1 = xn + (tn+1 − tn ) + (tn+1 − tn ) 2
+ O(h3 )
dt 2 dt
h2 dfn
= xn + hfn + + O(h3 )
2 ( dt )
h2 ∂fn ∂fn dxn
= xn + hfn + + + O(h3 )
2 ∂t ∂x dt
2
( )
h ∂fn ∂fn
= xn + hfn + + fn + O(h3 ) , (4.43)
2 ∂t ∂x
where we have set fn ≡ f (tn , xn ), dxn
dt
≡ dx
dt
(xn ) etc. We define
k1 = f (tn , xn ) = fn
k2 = f (tn + ah, xn + bhk1 )
xn+1 = xn + h(c1 k1 + c2 k2 ) . (4.44)
and we will determine the conditions so that the terms O(h2 ) of the last
equation in the error are identical with those of equation (4.43). By
4.8. APPENDIX: 2ND ORDER RUNGE–KUTTA METHOD 235
expanding k2 we obtain
k2 = f (tn + ah, xn + bhk1 )
∂f
= f (tn , xn + bhk1 ) + ha (tn , xn + bhk1 ) + O(h2 )
∂t
∂f ∂f
= f (tn , xn ) + hbk1 (tn , xn ) + ha (tn , xn ) + O(h2 )
{ ∂x } ∂t
∂fn ∂fn
= fn + h a + bk1 + O(h2 )
∂t ∂x
{ }
∂fn ∂fn
= fn + h a + bfn + O(h2 ) (4.45)
∂t ∂x
Substituting in (4.44) we obtain
xn+1 = xn + h(c1 k1 + c2 k2 )
{ ( ) }
∂fn ∂fn
= x n + h c 1 fn + c 2 fn + c 2 h a + bfn + O(h )2
∂t ∂x
( )
h2 ∂fn ∂fn
= xn + h(c1 + c2 )fn + (2c2 a) + (2c2 b)fn
2 ∂t ∂x
3
+O(h ) . (4.46)
All we need is to choose
c1 + c2 = 1
2c2 a = 1
2c2 b = 1 . (4.47)
The choice c1 = 0, c2 = 1, a = b = 1/2 leads to equation (4.19). Some
other choices in the bibliography are c2 = 1/2 and c2 = 3/4.
236 CHAPTER 4. MOTION OF A PARTICLE
4.9 Problems
4.1 Prove that the total error in the Euler–Cromer method is of order
∆t.
4.2 Reproduce the results in figures 4.11–4.18
4.3 Improve your programs so that there is no accumulation of roundoff
error in the calculation of time when h is very small for the methods
Euler, Euler-Cromer, Euler-Verlet and Runge-Kutta. Repeat the
analysis of the previous problem.
4.4 Make the appropriate changes in your programs of the Euler, Euler-
Cromer, Euler-Verlet and Runge-Kutta methods so that all floating
variables change from REAL→REAL(8). Repeat the analysis of the
previous problem.
4.5 Compare the results obtained from the Euler, Euler-Cromer, Euler-
Verlet, Runge-Kutta methods for the following systems where the
analytic solution is known:
(a) Particle falling in a constant gravitational field. Consider the
case v(0) = 0, m = 1, g = 10.
(b) Particle falling in a constant gravitational field moving in a fluid
from which exerts a force F = −kv on the particle. Consider
the case v(0) = 0, m = 1, g = 10 k = 0.1, 1.0, 2.0. Calculate the
limiting velocity of the particle numerically and compare the
value obtained to the theoretical expectation.
(c) Repeat for the case of a force of resistance of magnitude |F | =
kv 2 .
4.6 Consider the damped harmonic oscillator
d2 x dx
2
+γ + ω02 x = 0 . (4.48)
dt dt
Take ω0 = 3.145, γ = 0.5 and calculate its mechanical energy as a
function of time. Is it monotonic? Why? (show that d(E/m)/dt =
−γv 2 ). Repeat for γ = 4, 5, 6, 7, 8. When is the system oscillating
and when it’s not? Calculate numerically the critical value of γ
for which the system passes from a non oscillating to an oscillating
regime. Compare your results with the theoretical expectations.
4.9. PROBLEMS 237
4.7 Reproduce the results of figures 4.19–4.22.
4.8 Reproduce the results of figures 4.23–4.26. Calculate the phase δ(ω)
numerically and compare with equation (4.33).
4.9 Consider a simple model for a swing. Take the damped harmonic
oscillator and a driving force which periodically exerts a momen-
tary push with angular frequency ω. Define “momentary” to be an
impulse given by the acceleration a0 by an appropriately small time
interval ∆t. The acceleration is 0 for all other times. Calculate the
amplitude x0 (ω) for ω0 = 3.145 and γ = 0.5.
4.10 Consider a “half sine” driving force on a damped harmonic oscilla-
tor {
a0 cos ωt cos ωt > 0
a(t) =
0 cos ωt ≤ 0
Study the transient behavior of the system for several initial con-
ditions and calculate its steady state motion for ω0 = 3.145 and
γ = 0.5. Calculate the amplitude x0 (ω).
4.11 Consider the driving force on a damped oscillator given by
1 1 2 2
a(t) = + cos ω + cos 2ωt − cos 4ωt
π 2 3π 15π
Study the transient behavior of the system for several initial con-
ditions and calculate its steady state motion for ω0 = 3.145 and
γ = 0.5. Calculate the amplitude x0 (ω). Compare your results with
those of the previous problem and comment about.
4.12 Write a program that simulates N identical, independent harmonic
oscillators. Take N = 20 and choose random initial conditions for
each one of them. Study their trajectories in phase space and check
whether they cross each other. Comment on your results.
4.13 Place the N = 20 harmonic oscillators of the previous problem in
a small square in phase space whose center is at the origin of the
axes. Consider the evolution of the system in time. Does the shape
of the rectangle change in time? Does the area change in time?
Explain...
4.14 Repeat the previous problem when each oscillator is damped with
γ = 0.5. Take ω0 = 3.145.
238 CHAPTER 4. MOTION OF A PARTICLE
4.15 Consider the forced damped oscillator with ω = 2, ω0 = 1.0, γ = 0.2.
Study the transient behavior of the system in the plots of θ(t), θ̇(t)
for A = 0.1, 0.5, 0.79, 0.85, 1.03, 1.4.
4.16 Consider the forced damped pendulum with ω = 2, ω0 = 1.0, γ = 0.2
and study the phase space trajectories for A = 0.1, 0.19, 0.21, 0.25,
0.5, 0.71, 0.79, 0.85, 1.02, 1.031, 1.033, 1.05, 1.08, 1.1, 1.4, 1.8, 3.1,
3.5, 3.8, 4.2, 4.42, 4.44, 4.445, 4.447, 4.4488. Consider both the
transient behavior and the steady state motion.
4.17 Reproduce the results in figures 4.30.
4.18 Reproduce the results in figures 4.31.
4.19 Consider the forced damped oscillator with
ω0 = 1 , ω = 2, γ = 0.2
After the transient behavior, the motion of the system for A = 0.60,
A = 0.75 and A = 0.85 is periodic. Measure the period of the
motion with an accuracy of three significant digits and compare it
with the natural period of the pendulum and with the period of
the driving force. Take as initial conditions the following pairs:
(θ0 , θ̇0 ) = (3.1, 0.0), (2.5, 0.0), (2.0, 0.0), (1.0, 0.0), (0.2, 0.0), (0.0, 1.0),
(0.0, 3.0), (0.0, 6.0). Check if the period is independent of the initial
conditions.
4.20 Consider the forced damped pendulum with
ω0 = 1 , ω = 2, γ = 0.2
Study the motion of the pendulum when the amplitude A takes
values in the interval [0.2, 5.0]. Consider specific discrete values of
A by splitting the interval above in subintervals of width equal to
δA = 0.002. For each value of A, record in a file the value of A, the
angular position and the angular velocity of the pendulum when
tk = kπ with k = ktrans , ktrans + 1, ktrans + 2, . . . , kmax :
A θ(tk ) θ̇(tk )
The choice of ktrans is made so that the transient behavior will be
discarded and study only the steady state of the pendulum. You
may take kmax = 500, ktrans = 400, ti = 0, tf = 500π, and split the
intervals [tk , tk + π] to 50 subintervals. Choose θ0 = 3.1, θ̇0 = 0.
4.9. PROBLEMS 239
(a) Construct the bifurcation diagram by plotting the points (A, θ(tk )).
(b) Repeat by plotting the points (A, θ̇(tk )).
(c) Check whether your results depend on the choice of θ0 , θ̇0 .
Repeat your analysis for θ0 = 0, θ̇0 = 1.
(d) Study the onset of chaos: Take A ∈ [1.0000, 1.0400] with δA =
0.0001 and A ∈ [4.4300, 4.4500] with δA = 0.0001 and compute
with the given accuracy the value Ac where the system enters
into the chaotic behavior regime.
(e) The plot the points (θ(tk ), θ̇(tk )) for A = 1.034, 1.040, 1.080,
1.400, 4.450, 4.600. Put 2000 points for each value of A and
commend on the strength of the chaotic behavior of the pen-
dulum.
240 CHAPTER 4. MOTION OF A PARTICLE
Chapter 5
Planar Motion
In this chapter we will study the motion of a particle moving on the
plane under the influence of a dynamical field. Special emphasis will be
given to the study of the motion in a central field, like in the problem
of planetary motion and scattering. We also study the motion of two
or more interacting particles moving on the plane, which requires the
solution of a larger number of dynamical equations. These problems
can be solved numerically by using Runge–Kutta integration methods,
therefore this chapter extends and applies the numerical methods studied
in the previous chapter.
5.1 Runge–Kutta for Planar Motion
In two dimensions, the initial value problem that we are interested in, is
solving the system of equations (4.6)
dx dvx
= vx = ax (t, x, vx , y, vy )
dt dt
dy dvy
= vy = ay (t, x, vx , y, vy ) . (5.1)
dt dt
The 4th order Runge-Kutta method can be programmed by making
small modifications of the program in the file rk.f90. In order to facil-
itate the study of many different dynamical fields, for each field we put
the code of the respective acceleration in a different file. The code which
is common for all the forces, namely the user interface and the imple-
mentation of the Runge–Kutta method, will be put in the file rk2.f90.
The program that computes the acceleration will be put in a file named
rk_XXX.f90, where XXX is a string of characters that identifies the force.
241
242 CHAPTER 5. PLANAR MOTION
For example, the file rk2_hoc.f90 contains the program computing the
acceleration of the simple harmonic oscillator, the file rk2_g.f90 the ac-
celeration of a constant gravitational field ⃗g = −g ŷ etc.
Different force fields will require the use of one or more coupling
constants which need to be accessible to the code in the main program
and some subroutines. For this reason, we will provide two variables
k1, k2 in a common block:
real (8) : : k1 , k2
common / couplings / k1 , k2
This common block will be accessed by the acceleration functions f3 and
f4, the function energy and the main program where the user will enter
the values of k1 and k2. The initial conditions are stored in the variables
X10 ↔ x0 , X20 ↔ y0 , V10 ↔ vx0 , V20 ↔ vy0 , and the values of the functions
of time will be stored in the arrays X1(P) ↔ x(t), X2(P) ↔ y(t), V1(P)
↔ vx (t), V2(P) ↔ vy (t). The integration is performed by a call to the
subroutine
c a l l RK ( T , X1 , X2 , V1 , V2 , Ti , Tf , X10 , X20 , V10 , V20 , Nt )
The results are written to the file rk2.dat. Each line in this file contains
the time, position, velocity and the total mechanical energy, where the
energy is calculated by the function energy(t,x1,x2,v1,v2):
open ( u n i t =11 , f i l e = ’ rk2 . dat ’ )
do i =1 , Nt
w r i t e ( 1 1 , * ) T ( i ) , X1 ( i ) , X2 ( i ) , V1 ( i ) , V2 ( i ) ,&
energy ( T ( i ) , X1 ( i ) , X2 ( i ) , V1 ( i ) , V2 ( i ) )
enddo
The code for the function energy, which is different for each force field, is
written in the same file with the acceleration. The code for the subroutine
RKSTEP(t,x1,x2,x3,x4,dt) should be extended in order to integrate four
instead of two functions. The full code is listed below:
! ========================================================
! Program t o s o l v e a 4 ODE system using Runge−Kutta Method
! User must supply d e r i v a t i v e s
! dx1 / dt= f 1 ( t , x1 , x2 , x3 , x4 ) dx2 / dt=f 2 ( t , x1 , x2 , x3 , x4 )
! dx3 / dt=f 3 ( t , x1 , x2 , x3 , x4 ) dx4 / dt=f 4 ( t , x1 , x2 , x3 , x4 )
! as r e a l ( 8 ) f u n c t i o n s
5.1. RUNGE–KUTTA FOR PLANAR MOTION 243
! Output i s w r i t t e n i n f i l e rk2 . dat
! ========================================================
program rk2_solve
i m p l i c i t none
i n t e g e r , parameter : : P=1010000
r e a l ( 8 ) , dimension ( P ) : : T , X1 , X2 , V1 , V2
r e a l ( 8 ) : : Ti , Tf , X10 , X20 , V10 , V20
i n t e g e r : : Nt , i
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : energy , E0 , EF , DE
! Input :
p r i n t * , ’ Runge−Kutta Method f o r 4−ODEs I n t e g r a t i o n ’
p r i n t * , ’ Enter c o u p l i n g c o n s t a n t s : ’
read * , k1 , k2
p r i n t * , ’ k1= ’ , k1 , ’ k2= ’ , k2
p r i n t * , ’ Enter Nt , Ti , Tf , X10 , X20 , V10 , V20 : ’
read * , Nt , Ti , TF , X10 , X20 , V10 , V20
p r i n t * , ’ Nt = ’ , Nt
p r i n t * , ’ Time : I n i t i a l Ti = ’ , Ti , ’ F i n a l Tf= ’ , Tf
print * , ’ X1( Ti )= ’ , X10 , ’ X2( Ti )= ’ , X20
print * , ’ V1 ( Ti )= ’ , V10 , ’ V2( Ti )= ’ , V20
! The C a l c u l a t i o n :
c a l l RK ( T , X1 , X2 , V1 , V2 , Ti , Tf , X10 , X20 , V10 , V20 , Nt )
! Output :
open ( u n i t =11 , f i l e = ’ rk2 . dat ’ )
do i =1 , Nt
w r i t e ( 1 1 , * ) T ( i ) , X1 ( i ) , X2 ( i ) , V1 ( i ) , V2 ( i ) ,&
energy ( T ( i ) , X1 ( i ) , X2 ( i ) , V1 ( i ) , V2 ( i ) )
enddo
close (11)
! Rutherford s c a t t e r i n g a n g l e s :
p r i n t * , ’ v−a n g l e : ’ , atan2 ( V2 ( Nt ) , V1 ( Nt ) )
p r i n t * , ’ b−a n g l e : ’ , 2 . 0 D0 * atan ( k1 / ( V10 * V10 * X20 ) )
E0 = energy ( Ti , X10 , X20 , V10 , V20 )
EF = energy ( T ( Nt ) , X1 ( Nt ) , X2 ( Nt ) , V1 ( Nt ) , V2 ( Nt ) )
DE = ABS( 0 . 5 D0 * ( EF−E0 ) / ( EF+E0 ) )
p r i n t * , ’E0 , EF , DE/ E= ’ , E0 , EF , DE
end program rk2_solve
! ========================================================
! The v e l o c i t y f u n c t i o n s f1 , f 2 ( t , x1 , x2 , v1 , v2 )
! ========================================================
r e a l ( 8 ) f u n c t i o n f1 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
f1=v1 ! dx1 / dt= v1
end f u n c t i o n f1
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f2 ( t , x1 , x2 , v1 , v2 )
244 CHAPTER 5. PLANAR MOTION
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
f2=v2 ! dx2 / dt= v2
end f u n c t i o n f2
! ========================================================
!RK(T , X1 , X2 , V1 , V2 , Ti , Tf , X10 , X20 , V10 , V20 , Nt ) i s t h e d r i v e r
! f o r t h e Runge−Kutta i n t e g r a t i o n r o u t i n e RKSTEP
! Input : I n i t i a l and f i n a l t i m e s Ti , Tf
! I n i t i a l v a l u e s a t t =Ti X10 , X20 , V10 , V20
! Number o f s t e p s o f i n t e g r a t i o n : Nt−1
! S i z e o f a r r a y s T , X1 , X2 , V1 , V2
! Output : r e a l a r r a y s T( Nt ) , X1 ( Nt ) ,X2( Nt ) ,
! V1 ( Nt ) ,V2( Nt ) where
! T ( 1 ) = Ti X1 ( 1 ) = X10 X2 ( 1 ) = X20 V1 ( 1 ) = V10 V2 ( 1 ) = V20
! X1 ( k ) = X1 ( a t t =T( k ) ) X2( k ) = X2( a t t =T( k ) )
! V1 ( k ) = V1 ( a t t =T( k ) ) V2( k ) = V2( a t t =T( k ) )
! T( Nt )= Tf
! ========================================================
s u b r o u t i n e RK ( T , X1 , X2 , V1 , V2 , Ti , Tf , X10 , X20 , V10 , V20 , Nt )
i m p l i c i t none
i n t e g e r : : Nt
r e a l ( 8 ) , dimension ( Nt ) : : T , X1 , X2 , V1 , V2
r e a l ( 8 ) : : Ti , Tf
r e a l ( 8 ) : : X10 , X20
r e a l ( 8 ) : : V10 , V20
r e a l ( 8 ) : : dt
r e a l ( 8 ) : : TS , X1S , X2S ! v a l u e s o f time and X1 , X2 a t gi v e n s t e p
real (8) : : V1S , V2S
integer : : i
! I n i t i a l i z e variables :
dt = ( Tf−Ti ) / ( Nt −1)
T ( 1 ) = Ti
X1 ( 1 ) = X10 ; X2 ( 1 ) = X20
V1 ( 1 ) = V10 ; V2 ( 1 ) = V20
TS = Ti
X1S = X10 ; X2S = X20
V1S = V10 ; V2S = V20
! Make RK s t e p s : The arguments o f RKSTEP a r e
! r e p l a c e d with t h e new ones
do i =2 , Nt
c a l l RKSTEP ( TS , X1S , X2S , V1S , V2S , dt )
T ( i ) = TS
X1 ( i ) = X1S ; X2 ( i ) = X2S
V1 ( i ) = V1S ; V2 ( i ) = V2S
enddo
end s u b r o u t i n e RK
! ========================================================
! S u b r o u t i n e RKSTEP( t , x1 , x2 , dt )
! Runge−Kutta I n t e g r a t i o n r o u t i n e o f ODE
5.1. RUNGE–KUTTA FOR PLANAR MOTION 245
! dx1 / dt= f 1 ( t , x1 , x2 , x3 , x4 ) dx2 / dt=f 2 ( t , x1 , x2 , x3 , x4 )
! dx3 / dt=f 3 ( t , x1 , x2 , x3 , x4 ) dx4 / dt=f 4 ( t , x1 , x2 , x3 , x4 )
! User must supply d e r i v a t i v e f u n c t i o n s :
! r e a l f u n c t i o n f 1 ( t , x1 , x2 , x3 , x4 )
! r e a l f u n c t i o n f 2 ( t , x1 , x2 , x3 , x4 )
! r e a l f u n c t i o n f 3 ( t , x1 , x2 , x3 , x4 )
! r e a l f u n c t i o n f 4 ( t , x1 , x2 , x3 , x4 )
! Given i n i t i a l p o i n t ( t , x1 , x2 ) t h e r o u t i n e advances i t
! by time dt .
! Input : I n i t a l time t and f u n c t i o n v a l u e s x1 , x2 , x3 , x4
! Output : F i n a l time t +dt and f u n c t i o n v a l u e s x1 , x2 , x3 , x4
! C a r e f u l ! : v a l u e s o f t , x1 , x2 , x3 , x4 a r e o v e r w r i t t e n . . .
! ========================================================
s u b r o u t i n e RKSTEP ( t , x1 , x2 , x3 , x4 , dt )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , x3 , x4 , dt
r e a l ( 8 ) : : f1 , f2 , f3 , f4
r e a l ( 8 ) : : k11 , k12 , k13 , k14 , k21 , k22 , k23 , k24
r e a l ( 8 ) : : k31 , k32 , k33 , k34 , k41 , k42 , k43 , k44
r e a l ( 8 ) : : h , h2 , h6
h =dt ! h =dt , i n t e g r a t i o n s t e p
h2 =0.5 D0 * h ! h2=h / 2
h6=h / 6 . 0 D0 ! h6=h / 6
k11=f1 ( t , x1 , x2 , x3 , x4 )
k21=f2 ( t , x1 , x2 , x3 , x4 )
k31=f3 ( t , x1 , x2 , x3 , x4 )
k41=f4 ( t , x1 , x2 , x3 , x4 )
k12=f1 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 , x3+h2 * k31 , x4+h2 * k41 )
k22=f2 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 , x3+h2 * k31 , x4+h2 * k41 )
k32=f3 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 , x3+h2 * k31 , x4+h2 * k41 )
k42=f4 ( t+h2 , x1+h2 * k11 , x2+h2 * k21 , x3+h2 * k31 , x4+h2 * k41 )
k13=f1 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 , x3+h2 * k32 , x4+h2 * k42 )
k23=f2 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 , x3+h2 * k32 , x4+h2 * k42 )
k33=f3 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 , x3+h2 * k32 , x4+h2 * k42 )
k43=f4 ( t+h2 , x1+h2 * k12 , x2+h2 * k22 , x3+h2 * k32 , x4+h2 * k42 )
k14=f1 ( t+h , x1+h * k13 , x2+h * k23 , x3+h * k33 , x4+h * k43 )
k24=f2 ( t+h , x1+h * k13 , x2+h * k23 , x3+h * k33 , x4+h * k43 )
k34=f3 ( t+h , x1+h * k13 , x2+h * k23 , x3+h * k33 , x4+h * k43 )
k44=f4 ( t+h , x1+h * k13 , x2+h * k23 , x3+h * k33 , x4+h * k43 )
t =t+h
x1=x1+h6 * ( k11 +2.0 D0 * ( k12+k13 )+k14 )
x2=x2+h6 * ( k21 +2.0 D0 * ( k22+k23 )+k24 )
x3=x3+h6 * ( k31 +2.0 D0 * ( k32+k33 )+k34 )
246 CHAPTER 5. PLANAR MOTION
x4=x4+h6 * ( k41 +2.0 D0 * ( k42+k43 )+k44 )
end s u b r o u t i n e RKSTEP
5.2 Projectile Motion
Consider a particle in the constant gravitational field near the surface of
the earth which moves with constant acceleration ⃗g = −g ŷ so that
x(t) = x0 + v0x t , y(t) = y0 + v0y t − 12 gt2
vx (t) = v0x , vy (t) = v0y − gt (5.2)
ax (t) = 0 , ay (t) = −g
The particle moves on a parabolic trajectory that depends on the initial
conditions
( )
v0y 1 g
(y − y0 ) = (x − x0 ) − 2
(x − x0 )2
v0x 2 v0x
tan2 θ
= tan θ (x − x0 ) − (x − x0 )2 , (5.3)
4hmax
where tan θ = v0y /v0x is the direction of the initial velocity and hmax is
the maximum height of the trajectory.
The acceleration ax (t) = 0 ay (t) = −g (ax ↔ f3 , ay ↔ f4) and the
mechanical energy is coded in the file rk2_g.f90:
! ========================================================
! The a c c e l e r a t i o n f u n c t i o n s f3 , f 4 ( t , x1 , x2 , v1 , v2 ) provided
! by t h e u s e r
! ========================================================
! Free f a l l i n c o n s t a n t g r a v i t a t i o n a l f i l e d with
! g = −k2
r e a l ( 8 ) f u n c t i o n f3 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
f3 =0.0 D0 ! dx3 / dt=dv1 / dt=a1
end f u n c t i o n f3
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f4 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
5.2. PROJECTILE MOTION 247
0.25 0.06
x(t) y(t)
0.05
0.2
0.04
0.15
0.03
0.1
0.02
0.05
0.01
0 0
0 0.05 0.1 0.15 0.2 0.25 0 0.05 0.1 0.15 0.2 0.25
1.01 1
vx(t) vy(t)
1.005 0.5
1 0
0.995 -0.5
0.99 -1
0.985 -1.5
0 0.05 0.1 0.15 0.2 0.25 0 0.05 0.1 0.15 0.2 0.25
Figure 5.1: Plots of x(t), y(t), vx (t), vy (t) for a projectile fired in a constant gravita-
tional field ⃗g = −10.0 ŷ with initial velocity ⃗v0 = x̂ + ŷ.
real (8) : : k1 , k2
common / couplings / k1 , k2
f4=−k1 ! dx4 / dt=dv2 / dt=a2
end f u n c t i o n f4
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
energy = 0.5 D0 * ( v1 * v1+v2 * v2 ) + k1 * x2
end f u n c t i o n energy
In order to calculate a projectile’s trajectory you may use the following
commands:
> g f o r t r a n −O2 rk2 . f90 rk2_g . f90 −o rk2
> . / rk2
Runge−Kutta Method for 4−ODEs Integration
Enter coupling constants :
10.0 0.0
k1= 10.000000 k2= 0.000000
248 CHAPTER 5. PLANAR MOTION
0.06 1.8e-05
E(t)-E(0)
1.6e-05
0.05
1.4e-05
0.04 1.2e-05
0.03 1e-05
y
8e-06
0.02
6e-06
0.01 4e-06
2e-06
0
0 0.05 0.1 0.15 0.2 0.25 0
x 0 0.05 0.1 0.15 0.2 0.25
Figure 5.2: (Left) The parabolic trajectory of a projectile fired in a constant gravi-
tational field ⃗g = −10.0 ŷ with initial velocity ⃗v0 = x̂ + ŷ. (Right) The deviation of the
projectile’s energy from its initial value is due to numerical errors.
Enter Nt , Ti , Tf , X10 , X20 , V10 , V20 :
20000 0.0 0.2 0.0 0.0 1 . 0 1 . 0
Nt= 20000
Time : Initial Ti = 0.000000 Final Tf= 0.200000
X1 ( Ti )= 0.000000 X2 ( Ti )= 0.000000
V1 ( Ti )= 1.000000 V2 ( Ti )= 1.000000
The analysis of the results contained in the file rk2.dat can be done using
gnuplot:
gnuplot > s e t t e r m i n a l x11 1
gnuplot > p l o t ” rk2 . dat ” using 1 : 2 with lines t i t l e ”x ( t ) ”
gnuplot > s e t t e r m i n a l x11 2
gnuplot > p l o t ” rk2 . dat ” using 1 : 3 with lines t i t l e ”y ( t ) ”
gnuplot > s e t t e r m i n a l x11 3
gnuplot > p l o t ” rk2 . dat ” using 1 : 4 with lines t i t l e ” vx ( t ) ”
gnuplot > s e t t e r m i n a l x11 4
gnuplot > p l o t ” rk2 . dat ” using 1 : 5 with lines t i t l e ” vy ( t ) ”
gnuplot > s e t t e r m i n a l x11 5
gnuplot > p l o t ” rk2 . dat ” using 1 : ( $6 −1.0) w lines t ”E( t ) Ε−(0) ”
gnuplot > s e t t e r m i n a l x11 6
gnuplot > s e t s i z e square
gnuplot > set t i t l e ” Trajectory ”
gnuplot > p l o t ” rk2 . dat ” using 2:3 with lines notit
The results can be seen in figures 5.1 and 5.2. We note a small increase
in the mechanical energy which is due to the accumulation of numerical
errors.
We can animate the trajectory by writing a script of gnuplot com-
mands in a file rk2_animate.gpl
5.2. PROJECTILE MOTION 249
icount = icount+skip
p l o t ”< c a t −n rk2 . dat ” \
using 3 : ( $1<= icount ? $4 : 1 / 0 ) with lines notitle
# pause 1
i f ( icount < nlines ) r e r e a d
Before calling the script, the user must set the values of the variables
icount, skip and nlines. Each time gnuplot reads the script, it plots
icount number of lines from rk2.dat. Then the script is read again and
a new plot is made with skip lines more than the previous one, unless
icount < nlines. The plotted “file” "<cat -n rk2.dat" is the standard
output (stdout) of the command cat -n rk2.dat which prints to the
stdout the contents of the file rk2.dat line by line, together with the
line number. Therefore the plot command reads data which are the line
number, the time, the coordinate x, the coordinate y etc. The keyword
using in
using 3 : ( $1<= icount ? $4 : 1 / 0 )
instructs the plot command to use the 3rd column on the horizontal axis
and if the first column is less than icount ($1<= icount) put on the
vertical axis the value of the 4th column if the first column is less than
icount. Otherwise ($1 > icount) it prints an undefined number (1/0)
which makes gnuplot print nothing at all. You may also uncomment the
command pause if you want to make the animation slower. In order to
run the script from gnuplot, issue the commands
gnuplot > icount = 10
gnuplot > skip = 200
gnuplot > nlines = 20000
gnuplot > load ” rk2_animate . gpl ”
The scripts shown above can be found in the accompanying software.
More scripts can be found there that automate many of the boring pro-
cedures. The usage of two of these is explained below. The first one is
in the file rk2_animate.csh:
> . / rk2_animate . csh −h
Usage : rk2_animate . csh −t [ sleep time ] −d [ skip points ] <file >
Default file is rk2 . dat
Other options :
−x : s e t lower value in xrange
250 CHAPTER 5. PLANAR MOTION
−X : s e t lower value in xrange
−y : s e t lower value in yrange
−Y : s e t lower value in yrange
−r : automatic determination of x−y range
> . / rk2_animate . csh −r −d 500 rk2 . dat
The last line is a command that animates a trajectory read from the
file rk2.dat. Each animation frame contains 500 more points than the
previous one. The option -r calculates the plot range automatically. The
option -h prints a short help message.
A more useful script is in the file rk2.csh.
> . / rk2 . csh −h
Usage : rk2 . csh −f <force > k1 k2 x10 x20 v10 v20 STEPS t0 tf
Other Options :
−n Do not animate trajectory
Available forces ( value of <force >) :
1 : ax=−k1 ay= −k2 y Harmonic oscillator
2 : ax= 0 ay= −k1 Free fall
3 : ax= −k2 vx ay= −k2 vy − k1 Free fall + \
air resistance ~ v
4 : ax= −k2 | v | vx ay= −k2 | v | vy − k1 Free fall + \
air resistance ~ v^2
5 : ax= k1 * x1 / r^3 ay= k1 * x2 / r^3 Coulomb Force
....
The option -h prints operating instructions. A menu of forces is available,
and a choice can be made using the option -f. The rest of the command
line consists of the parameters read by the program in rk2.f90, i.e. the
coupling constants k1, k2, the initial conditions x10, x20, v10, v20
and the integration parameters STEPS, t0 and tf. For example, the
commands
> rk2 . csh -f 2 -- 10.0 0.0 0.0 0.0 1 . 0 1 . 0 20000 0.0 0.2
> rk2 . csh -f 1 -- 16.0 1 . 0 0.0 1 . 0 1 . 0 0.0 20000 0.0 6.29
> rk2 . csh -f 5 -- 10.0 0.0 -10 0.2 1 0 . 0.0 20000 0.0 3.00
compute the trajectory of a particle in the constant gravitational field
discussed above, the trajectory of an anisotropic harmonic oscillator (k1
= ax = −ω12 x, k2 = ay = −ω22 y) and the scattering of a particle in a
Coulomb field – try them! I hope that you will have enough curiosity to
look “under the hood” of the scripts and try to modify them or create
new ones. Some advise to the lazy guys: If you need to program your
own force field follow the recipe: Write the code of your acceleration field
5.2. PROJECTILE MOTION 251
in a file named e.g. rk2_myforce.f90 as we did with rk2_g.f90. Edit
the file rk2.csh and modify the line
s e t forcecode = ( hoc g vg v2g cb )
to
s e t forcecode = ( hoc g vg v2g cb myforce )
(the variable $forcecode may have more entries than the ones shown
above). Count the order of the string myforce, which is 6 in our case. In
order to access this force field from the command line, use the option -f
6:
> rk2 . csh −f 6 −− . . . . . . .
Now, we will study the effect of the air resistance on the motion of the
projectile. For small velocities this is a force proportional to the velocity
F⃗r = −mk⃗v , therefore
ax = −kvx
ay = −kvy − g . (5.4)
By taking
v0x ( )
x(t) = x0 + 1 − e−kt
k
1( g)( ) g
y(t) = y0 + v0y + 1 − e−kt − t
k k k
−kt
vx (t) = v0x e
( g ) −kt g
vy (t) = v0y + e − , (5.5)
k k
we obtain the motion of a particle with terminal velocity vy (+∞) = −g/k
(x(+∞) = const., y(+∞) ∼ t).
The acceleration caused by the air resistance is programmed in the
file (k1 ↔ g, k2 ↔ k ) rk2_vg.f90:
! ========================================================
! The a c c e l e r a t i o n f u n c t i o n s f3 , f 4 ( t , x1 , x2 , v1 , v2 ) provided
! by t h e u s e r
! ========================================================
252 CHAPTER 5. PLANAR MOTION
! Free f a l l i n c o n s t a n t g r a v i t a t i o n a l f i l e d with
! ax = −k2 vx ay = −k2 vy − k1
r e a l ( 8 ) f u n c t i o n f3 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
f3=−k2 * v1 ! dx3 / dt=dv1 / dt=a1
end f u n c t i o n f3
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f4 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
f4=−k2 * v2−k1 ! dx4 / dt=dv2 / dt=a2
end f u n c t i o n f4
The results are shown in figure 5.3 where we see the effect of an in-
creasing air resistance on the particle trajectory. The effect of a resistance
force of the form F⃗r = −mkv 2 v̂ is shown in figure 5.4.
1.4
0.0 0.0
0.05 0.2 0.2
1.0 1.2 1.0
5.0 5.0
0.04 10.0 1 10.0
20.0 20.0
30.0 0.8 30.0
0.03
y
y
0.6
0.02
0.4
0.01
0.2
0 0
0 0.05 0.1 0.15 0.2 0 0.2 0.4 0.6 0.8 1
x x
Figure 5.3: The trajectory of a projectile moving in a constant gravitational field
⃗g = −10 ŷ with air resistance causing acceleration ⃗ar = −k⃗v for k = 0, 0.2, 1, 5, 10, 20, 30.
The left plot has ⃗v (0) = x̂ + ŷ and the right plot has ⃗v (0) = 5x̂ + 5ŷ.
5.3 Planetary Motion
Consider the simple planetary model of a “sun” of mass M and a planet
“earth” at distance r from the sun and mass m such that m ≪ M . Ac-
cording to Newton’s law of gravity, the earth’s acceleration is
GM GM
⃗a = ⃗g = − 2
r̂ = − 3 ⃗r , (5.6)
r r
5.3. PLANETARY MOTION 253
1.4
0.0 0.0
0.05 0.2 0.2
1.0 1.2 1.0
5.0 5.0
0.04 10.0 1 10.0
20.0 20.0
30.0 0.8 30.0
0.03
y
y
0.6
0.02
0.4
0.01
0.2
0 0
0 0.05 0.1 0.15 0.2 0 0.2 0.4 0.6 0.8 1
x x
Figure 5.4: The trajectory of a projectile moving in a constant gravitational field ⃗g =
−10 ŷ with air resistance causing acceleration ⃗ar = −kv 2 v̂ for k = 0, 0.2, 1, 5, 10, 20, 30.
The left plot has ⃗v (0) = x̂ + ŷ and the right plot has ⃗v (0) = 5x̂ + 5ŷ.
where G = 6.67 × 10−11 kgrm·sec2 , M = 1.99 × 1030 kgr, m = 5.99 × 1024 kgr.
3
When the hypothesis m ≪ M is not valid, the two body problem is
reduced to that of the one body problem with the mass replaced by the
reduced mass µ
1 1 1
= + .
µ m M
The force of gravity is a central force. This implies conservation of the
angular momentum L ⃗ = ⃗r × p⃗ with respect to the center of the force,
which in turn implies that the motion is confined on one plane. We
choose the z axis so that
⃗ = Lz k̂ = m(xvy − yvx )k̂ .
L (5.7)
The force of gravity is conservative and the mechanical energy
1 GmM
E = mv 2 − (5.8)
2 r
is conserved. If we choose the origin of the coordinate axes to be the
center of the force, the equations of motion (5.6) become
GM
ax = − x
r3
GM
ay = − 3 y, (5.9)
r
where r2 = x2 + y 2 . This is a system of two coupled differential equations
for the functions x(t), y(t). The trajectories are conic sections which are
either an ellipse (bound states - “planet”), a parabola (e.g. escape to
254 CHAPTER 5. PLANAR MOTION
infinity when the particle starts moving with speed equal to the escape
velocity) or a hyperbola (e.g. scattering).
Kepler’s third law of planetary motion states that the orbital period
T of a planet satisfies the equation
4π 2 3
T2 = a , (5.10)
GM
where a is the semi-major axis of the elliptical trajectory. The eccentricity
is a measure of the deviation of the trajectory from being circular
√
b2
e= 1− 2 , (5.11)
a
where b is the semi-minor axis. The eccentricity is 0 for the circle and
tends to 1 as the ellipse becomes more and more elongated. The foci F1
and F2 are located at a distance ea from the center of the ellipse. They
have the property that for every point on the ellipse
P F1 + P F2 = 2a . (5.12)
The acceleration given to the particle by Newton’s force of gravity is
programmed in the file rk2_cb.f90:
! ========================================================
! The a c c e l e r a t i o n f u n c t i o n s f3 , f 4 ( t , x1 , x2 , v1 , v2 ) provided
! by t h e u s e r
! ========================================================
! Motion i n Coulombic p o t e n t i a l :
! ax= k1 * x1 / r ^3 ay= k1 * x2 / r ^3
r e a l ( 8 ) f u n c t i o n f3 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : r2 , r3
r2=x1 * x1+x2 * x2
r3=r2 * s q r t ( r2 )
i f ( r3 . g t . 0 . 0 D0 ) then
f3=k1 * x1 / r3 ! dx3 / dt=dv1 / dt=a1
else
f3 =0.0 D0
endif
end f u n c t i o n f3
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f4 ( t , x1 , x2 , v1 , v2 )
5.3. PLANETARY MOTION 255
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : r2 , r3
r2=x1 * x1+x2 * x2
r3=r2 * s q r t ( r2 )
i f ( r3 . g t . 0 . 0 D0 ) then
f4=k1 * x2 / r3 ! dx4 / dt=dv2 / dt=a2
else
f4 =0.0 D0
endif
end f u n c t i o n f4
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
real (8) : : r
r= s q r t ( x1 * x1+x2 * x2 )
i f ( r . g t . 0.0 D0 ) then
energy = 0.5 D0 * ( v1 * v1+v2 * v2 ) + k1 / r
else
energy = 0.0 D0
endif
end f u n c t i o n energy
We set k1= −GM and take special care to avoid hitting the center of the
force, the singular point at (0, 0). The same code can be used for the
electrostatic Coulomb field with k1= qQ/4πϵ0 m.
At first we study trajectories which are bounded. We set GM = 10,
x(0) = 1.0, y(0) = 0, v0x = 0 and vary v0y . We measure the period T and
the length of the semi axes of the resulting ellipse. The results can be
found in table 5.1. Some of the trajectories are shown in figure 5.5. There
we can see the dependence of the size of the ellipse on the period. Figure
5.6 confirms Kepler’s third law of planetary motion given by equation
(5.10).
In order to confirm Kepler’s third law of planetary motion numeri-
cally, we take the logarithm of both sides of equation (5.10)
( 2)
3 1 4π
ln T = ln a + ln . (5.13)
2 2 GM
Therefore, the points (ln a, ln T ) lie on a straight line. Using a linear least
squares fit we calculate the slope and the intercept which should be equal
to 32 and 1/2 ln (4π 2 /GM ) respectively. This is left as an exercise.
256 CHAPTER 5. PLANAR MOTION
v0x T /2 2a
3.2 1.030 2.049
3.4 1.281 2.370
3.6 1.682 2.841
3.8 2.396 3.597
4.0 3.927 5.000
4.1 5.514 6.270
4.2 8.665 8.475
4.3 16.931 13.245
4.3 28.088 18.561
4.38 42.652 24.522
4.40 61.359 31.250
4.42 99.526 43.141
Table 5.1: The results for the period T and the length of the semi-major axis a of
the trajectory of planetary motion for GM = 10, x(0) = 1.0, y(0) = 0, v0y = 0.
In the case where the initial velocity of the particle becomes larger
than the escape velocity ve , the particle escapes from the influence of the
gravitational field to infinity. The escape velocity corresponds to zero
mechanical energy, which gives
2GM
ve2 = . (5.14)
r
When GM = 10, x(0) = 1.0, y(0) = 0, we obtain ve ≈ 4.4721 . . .. The
numerical calculation of ve is left as an exercise.
5.4 Scattering
In this section we consider scattering of particles from a central potential¹.
We assume particles that follow unbounded trajectories that start from
infinity and move almost free from the influence of the force field towards
its center. When they approach the region of interaction they get deflected
and get off to infinity in a new direction. We say that the particles have
been scattered and that the angle between their original and final direction
is the scattering angle θ. Scattering problems are interesting because we
can infer to the properties of the scattering potential from the distribution
of the scattering angle. This approach is heavily used in today’s particle
¹We refer the reader to [38], chapter 4.
5.4. SCATTERING 257
8
1.68
6 2.40
3.93
4 5.52
16.95
2
0
y
-2
-4
-6
-8
-14 -12 -10 -8 -6 -4 -2 0 2
x
Figure 5.5: Planetary trajectories for GM = 10, x(0) = 1.0, y(0) = 0, v0y = 0 and
v0x = 3.6, 3.8, 4.0, 4.1, 4.3. The numbers are the corresponding half periods.
accelerators for the study of fundamental interactions between elementary
particles.
First we will discuss scattering of small hard spheres of radius r1 by
other hard spheres or radius R2 . The interaction potential² is given by
{
0 r > R2 + r1
V (r) = , (5.15)
∞ r < R2 + r1
where r is the distance between the center of r1 from the center of R2 .
Assume that the particles in the beam do not interact with each other
and that there is only one collision per scattering. Let J be the intensity
of the beam³ and A its cross sectional area. Assume that the target has
n particles per unit area. The cross sectional area of the interaction is
σ = π(r1 + R2 )2 where r1 and R2 are the radii of the scattered particles
and targets respectively (see figure (5.8)): All the spheres of the beam
which lie outside this area are not scattered by the particular target. The
²The so called hard core potential.
³The number of particles crossing a surface perpendicular to the beam per unit time
and unit area.
258 CHAPTER 5. PLANAR MOTION
100000
10000
1000
T2
100
10
1
1 10 100 1000 10000
3
a
Figure 5.6: Kepler’s third law of planetary motion for GM = 10. The points are
the measurements taken from table 5.1. The solid line is the known analytic solution
(5.10).
total interaction cross section is
Σ = nAσ, (5.16)
where nA is the total number of target spheres which lie within the beam.
On the average, the scattering rate is
N = JΣ = JnAσ . (5.17)
The above equation is the definition of the total scattering cross section
σ of the interaction. The differential cross section σ(θ) is defined by the
relation
dN = JnAσ(θ) dΩ , (5.18)
where dN is the number of particles per unit time scattered within the
solid angle dΩ. The total cross section is
∫ ∫ ∫
σtot = σ(θ) dΩ = σ(θ) sin θ dθdϕ = 2π σ(θ) sin θ dθ . (5.19)
Ω
5.4. SCATTERING 259
1
0.8
0.6
0.4
0.2
y
0
-0.2
-0.4
-0.6
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
x
Figure 5.7: The spiral orbit of a particle moving under the influence of a central
force F = −k/r r̂.
⃗ 3
In the last relation we used the cylindrical symmetry of the interaction
with respect to the axis of the collision. Therefore
1 dN
σ(θ) = . (5.20)
nAJ 2π sin θ dθ
This relation can be used in experiments for the measurement of the
differential cross section by measuring the rate of detection of particles
within the space contained in between two cones defined by the angles
θ and θ + dθ. This is the relation that we will use in the numerical
calculation of σ(θ).
Generally, in order to calculate the differential cross section we shoot
a particle at a target as shown in figure 5.9. The scattering angle θ
depends on the impact parameter b. The part of the beam crossing the
ring of radius b(θ), thickness db and area 2πb db is scattered in angles
between θ and θ + dθ. Since there is only one particle at the target we
have that nA = 1. The number of particles per unit time crossing the
ring is J2πb db, therefore
2πb(θ) db = −2πσ(θ) sin θ dθ (5.21)
260 CHAPTER 5. PLANAR MOTION
11
00
11
00
11
00
11
00
θ
11
00
11
00
11
00
11
1111111
0000000 00
11
00
11 00
11
1111111
0000000 00
11
00
00
11 00
11
1111111
0000000 11
00
00
11 00
11
1111111
0000000 11
00
1111111
0000000
1111111
0000000
1111111
0000000
R2 11
00
11
00
11
00
1111111
0000000 11
00
1111111
0000000 11
00
1111111
0000000 11
00
1111111
0000000 11
00
r1 11
00
11
00
11
00
1111111
0000000
11
00
1111111
0000000
11
00
11
00
11
00
11
00
11
00
R+r
2 1 11
00
11
00
11
00
11
00
σ
11
00 11
00 11
00 11
00
00
11 00
11 00
11
00
11 00
11 00
11
00
11 00
11 00
11
Figure 5.8: Scattering of hard spheres. The scattering angle is θ. The cross sectional
area σ is shown to the right.
(the − sign is because as b increases, θ decreases). From the potential we
can calculate b(θ) and from b(θ) we can calculate σ(θ). Conversely, if we
measure σ(θ), we can calculate b(θ).
5.4.1 Rutherford Scattering
The scattering of a charged particle with charge q (“electron”) in a Coulomb
potential of a much heavier charge Q (“nucleus”) is called Rutherford
scattering. In this case, the interaction potential is given by
1 Q
V (r) = , (5.22)
4πϵ0 r
which accelerates the particle with acceleration
qQ r̂ ⃗r
⃗a = ≡ α . (5.23)
4πϵ0 m r2 r3
The energy of the particle is E = 12 mv 2 and the magnitude of its angular
momentum is l = mvb, where v ≡ |⃗v |. The dependence of the impact
parameter on the scattering angle is [38]
α θ
b(θ) = 2
cot . (5.24)
v 2
5.4. SCATTERING 261
vf
vi
b 111
000
111
000
θ
111
000
111
000
111
000
db
Figure 5.9: Beam particles passing through the ring 2πbdb are scattered within the
solid angle dΩ = 2πsinθ dθ.
Using equation (5.21) we obtain
α2 1 θ
σ(θ) = 4
sin−4 . (5.25)
4 v 2
Consider the scattering trajectories. The results for same charges are
shown in figure 5.10. A similar figure is obtained in the case of opposite
charges. In the latter case we have to take special care for small impact
parameters b < 0.2 where the scattering angle is ≈ 1. A large number
of integration steps is needed in order to obtain the desired accuracy. A
useful monitor of the accuracy of the calculation is the measurement of
the energy of the particle which should be conserved. The results are
shown in table 5.2. We will now describe a method for calculating the
cross section by using equation (5.20). Alternatively we could have used
equation (5.21) and perform a numerical calculation of the derivatives.
This is left as an exercise for the reader. Our calculation is more like
an experiment. We place a “detector” that “detects” particles scattered
within angles θ and θ+δθ. For this reason we split the interval [0, π] in Nb
bins so that δθ = π/Nb . We perform “scattering experiments” by varying
b ∈ [bm , bM ] with step δb. Due to the symmetry of the problem we fix ϕ to
be a constant, therefore a given θ corresponds to a cone with an opening
angle θ and an apex at the center of scattering. For given b we measure
the scattering angle θ and record the number of particles per unit time
δN ∝ bδb. The latter is proportional to the area of the ring of radius
b. All we need now is the beam ∑ intensity J which is the total number
of particles per unit time J ∝ i bδb (note than in the ratio δN /J the
262 CHAPTER 5. PLANAR MOTION
40
35
30
25
20
15
10
5
0
-20 -15 -10 -5 0 5 10 15 20
Figure 5.10: Rutherford scattering trajectories. We set k1 ≡ qQ
= 1 (see code in
4πϵ0 m
the file rk2_cb.f90) and b = 0.08, 0.015, 0.020, 0.035, 0.080, 0.120, 0.200, 0.240, 0.320,
0.450, 0.600, 1.500. The initial position of the particle is at x(0) = −50 and its initial
velocity is v = 3 in the x direction. The number of integration steps is 1000, the initial
time is 0 and the final time is 30.
proportionality constant and δb cancel) and the solid angle 2π sin(θ) δθ.
Finally we can easily use equation (5.19) in order to calculate the total
cross section σtot . The program that performs this calculation is in the file
scatter.f90 and it is a simple modification of the program in rk2.f90:
! ========================================================
! Program t h a t computes s c a t t e r i n g c r o s s −s e c t i o n o f a c e n t r a l
! f o r c e on t h e plane . The u s e r should f i r s t check t h a t t h e
! parameters used , l e a d t o a f r e e s t a t e i n t h e end .
! * * X20 i s t h e impact parameter b * *
!A 4 ODE system i s s o l v e d using Runge−Kutta Method
! User must supply d e r i v a t i v e s
! dx1 / dt= f 1 ( t , x1 , x2 , x3 , x4 ) dx2 / dt=f 2 ( t , x1 , x2 , x3 , x4 )
! dx3 / dt=f 3 ( t , x1 , x2 , x3 , x4 ) dx4 / dt=f 4 ( t , x1 , x2 , x3 , x4 )
! as r e a l ( 8 ) f u n c t i o n s
! Output i s w r i t t e n i n f i l e s c a t t e r . dat
! ========================================================
program scatter_cross_section
i m p l i c i t none
5.4. SCATTERING 263
b θn θa ∆E/E Nt
0.008 2.9975 2.9978 2.8 10−9 5000
0.020 2.7846 2.7854 2.7 10−9 5000
0.030 2.6131 2.6142 2.5 10−9 5000
0.043 2.4016 2.4031 2.3 10−9 5000
0.056 2.2061 2.2079 2.0 10−9 5000
0.070 2.0152 2.0172 1.7 10−9 5000
0.089 1.7887 1.7909 1.4 10−9 5000
0.110 1.5786 1.5808 1.0 10−9 5000
0.130 1.4122 1.4144 0.8 10−9 5000
0.160 1.2119 1.2140 0.5 10−9 5000
0.200 1.0123 1.0142 0.3 10−9 5000
0.260 0.8061 0.8077 0.1 10−9 5000
0.360 0.5975 0.5987 2.9 10−11 5000
0.560 0.3909 0.3917 0.3 10−11 5000
1.160 0.1905 0.1910 5.3 10−14 5000
Table 5.2: Scattering angles of Rutherford scattering. We set k1 ≡ qQ
= 1 (see file
4πϵ0 m
rk2_cb.f90) and study the resulting trajectories for the values of b shown in column
1. θn is the numerically calculated scattering angle and θa is the one calculated from
equation (5.24). The ratio ∆E/E shows the change in the particle’s energy due to
numerical errors. The last column is the number of integration steps. The particle’s
initial position is at x(0) = −50 and initial velocity ⃗v = 3x̂.
i n t e g e r , parameter : : P=1010000
r e a l ( 8 ) , dimension ( P ) : : T , X1 , X2 , V1 , V2
r e a l ( 8 ) : : Ti , Tf , X10 , X20 , V10 , V20
r e a l ( 8 ) : : X20F , dX20 ! max impact parameter and s t e p
i n t e g e r : : Nt
integer : : i
real (8) : : k1 , k2
common / couplings / k1 , k2
i n t e g e r , parameter : : Nbins=20
i n t e g e r : : index
r e a l ( 8 ) : : angle , bins ( Nbins ) , Npart
r e a l ( 8 ) , parameter : : PI =3.14159265358979324 D0
r e a l ( 8 ) , parameter : : rad2deg =180.0 D0 / PI
r e a l ( 8 ) , parameter : : dangle =PI / Nbins
r e a l ( 8 ) R , density , dOmega , sigma , sigmatot
! Input :
p r i n t * , ’ Runge−Kutta Method f o r 4−ODEs I n t e g r a t i o n ’
p r i n t * , ’ Enter c o u p l i n g c o n s t a n t s : ’
read * , k1 , k2
p r i n t * , ’ k1= ’ , k1 , ’ k2= ’ , k2
264 CHAPTER 5. PLANAR MOTION
b θn θa ∆E/E STEPS
0.020 2.793 2.785 0.02 1 000 000
0.030 2.620 2.614 8.2 10−3 300 000
0.043 2.405 2.403 7.2 10−4 150 000
0.070 2.019 2.017 3.2 10−7 150 000
0.089 1.793 1.791 8.2 10−7 60 000
0.110 1.583 1.581 1.2 10−6 30 000
0.130 1.417 1.414 9.4 10−7 20 000
0.160 1.216 1.214 6.0 10−5 5 000
0.200 1.016 1.014 4.1 10−6 5 000
0.260 0.8093 0.8077 2.2 10−7 5 000
0.360 0.6000 0.5987 7.6 10−9 5 000
0.560 0.3926 0.3917 1.2 10−10 5 000
1.160 0.1913 0.1910 2.9 10−13 5 000
Table 5.3: Rutherford scattering of opposite charges with = −1. The table is
qQ
4πϵ0 m
similar to table 5.2. We observe the numerical difficulty for small impact parameters.
print *, ’ Enter Nt , Ti , Tf , X10 , X20 , V10 , V20 : ’
read *, Nt , Ti , TF , X10 , X20 , V10 , V20
print *, ’ Enter f i n a l impact parameter X20F and s t e p dX20 : ’
read *, X20F , dX20
print *, ’ Nt = ’ , Nt
print *, ’ Time : I n i t i a l Ti = ’ , Ti , ’ F i n a l Tf= ’ , Tf
print *, ’ X1( Ti )= ’ , X10 , ’ X2( Ti )= ’ , X20
print *, ’ V1 ( Ti )= ’ , V10 , ’ V2( Ti )= ’ , V20
print *, ’ Impact par X20F = ’ , X20F , ’ dX20 = ’ , dX20
open ( u n i t =11 , f i l e = ’ s c a t t e r . dat ’ )
bins = 0.0 d0
! The C a l c u l a t i o n :
Npart = 0.0 D0
X20 = X20 + dX20 / 2 . 0 D0 ! s t a r t s i n middle o f f i r s t i n t e r v a l
do while ( X20 . l t . X20F )
c a l l RK ( T , X1 , X2 , V1 , V2 , Ti , Tf , X10 , X20 , V10 , V20 , Nt )
! Take a b s o l u t e v a l u e due t o symmetry :
angle = DABS ( atan2 ( V2 ( Nt ) , V1 ( Nt ) ) )
! Output : The f i n a l a n g l e . Check i f almost c o n s t a n t
w r i t e ( 1 1 , * ) ’@ ’ , X20 , angle ,&
DABS ( atan2 ( V2 ( Nt−50) , V1 ( Nt−50) ) ) ,&
k1 / V10 * * 2 / tan ( angle / 2 . 0 D0 )
! Update histogram :
index = i n t ( angle / dangle ) +1
! Number o f incoming p a r t i c l e s per u n i t time
5.4. SCATTERING 265
100
10
1
σ(θ)
0.1
0.01
0.001
0 0.5 1 1.5 2 2.5 3
θ
Figure 5.11: Differential cross section of the Rutherford scattering. The solid line
qQ
is the function (5.25) for α = 1, v = 3. We set 4πϵ 0m
= 1. The particle’s initial position
is x(0) = −50 and its initial velocity is ⃗v = 3x̂. We used 5000 integration steps, initial
time equal to 0 and final time equal to 30. The impact parameter varies between 0.02
and 1 with step equal to 0.0002.
! i s proportional to radius of ring
! o f r a d i u s X20 , t h e impact parameter :
! db i s c a n c e l l e d from d e n s i t y
bins ( index ) = bins ( index ) + X20
Npart = Npart + X20 !<−− i . e . from here
X20 = X20 + dX20
enddo
! Print scattering cross section :
R = X20 ! beam r a d i u s
density = Npart / ( PI * R * R ) ! beam f l u x d e n s i t y J
sigmatot = 0.0 D0 ! t o t a l cross section
do i =1 , Nbins
angle = ( i−0.5D0 ) * dangle
dOmega = 2.0 D0 * PI * s i n ( angle ) * dangle ! d ( S o l i d Angle )
sigma = bins ( i ) / ( density * dOmega )
i f ( sigma . g t . 0 . 0 D0 ) w r i t e ( 1 1 , * ) ’ ds= ’ ,&
angle , angle * rad2deg , sigma
sigmatot = sigmatot + sigma * dOmega
enddo
w r i t e ( 1 1 , * ) ’ s i g m a t o t= ’ , sigmatot
266 CHAPTER 5. PLANAR MOTION
10
1
σ(θ)
0.1
0.01
0.001
1 10 100 1000
-4
sin (θ/2)
Figure 5.12: Differential cross section of the Rutherford scattering like in figure 5.11.
The solid line is the function 1/(4 × 34 )x from which we can deduce the functional form
of σ(θ).
close (11)
end program scatter_cross_section
The results are recorded in the file scatter.dat. An example session
that reproduces figures 5.11 and 5.12 is
> g f o r t r a n scatter . f90 rk2_cb . f90 −o scatter
> . / scatter
Runge−Kutta Method for 4−ODEs Integration
Enter coupling constants :
1 . 0 0.0
k1= 1.00000 k2= 0.00000
Enter Nt , Ti , Tf , X10 , X20 , V10 , V20 :
5000 0 30 −50 0.02 3 0
Enter final impact parameter X20F and step dX20 :
1 0.0002
Nt= 5000
Time : Initial T0 = 0.00000 Final TF= 30.00000
X1 ( T0 )= −50.00000 X2 ( T0 )= 2.00000E−002
V1 ( T0 )= 3.00000 V2 ( T0 )= 0.00000
Impact par X20F = 1.00000 dX20 = 2.00000E−004
5.4. SCATTERING 267
The results can be plotted with the gnuplot commands:
gnuplot > s e t l o g
gnuplot > p l o t [ : 1 0 0 0 ] ”<grep ds= s c a t t e r . dat ” \
u ( ( s i n ( $2 / 2 ) ) **( −4) ) : ( $4 ) notit , \
( 1 . / ( 4 . * 3 . * * 4 ) ) * x notit
gnuplot > unset l o g
gnuplot > s e t l o g y
gnuplot > p l o t [ : ] ”<grep ds= s c a t t e r . dat ” u 2:4 notit , \
( 1 . / ( 4 . * 3 . * * 4 ) ) * ( s i n ( x / 2 ) ) **( −4) notit
The results are in a very good agreement with the theoretical ones given
by (5.25). The next step will be to study other central potentials whose
solution is not known analytically.
5.4.2 More Scattering Potentials
Consider scattering from a force field
{ 1
− r
r≤a
F⃗ = f (r) r̂ , f (r) = r2 a3 . (5.26)
0 r>a
This is a very simple classical model of the scattering of a positron e+
by the hydrogen atom. The positron has positive charge +e and the
hydrogen atom consists of a positively charged proton with charge +e
in an electron cloud of opposite charge −e. We set the scales so that
me+ = 1 and e2 /4πϵ0 = 1. We will perform a numerical calculation of
b(θ), σ(θ) and σtot .
The potential energy is given by
dV (r) 1 r2 3
f (r) = − ⇒ V (r) = + 2 − . (5.27)
dr r 2a 2a
where V (r) = 0 for r ≥ a. The program containing the calculation of the
acceleration caused by this force can be found in the file rk_hy.f90:
! ========================================================
! The a c c e l e r a t i o n f u n c t i o n s f3 , f 4 ( t , x1 , x2 , v1 , v2 ) provided
! by t h e u s e r
! ========================================================
! Motion i n hydrogen atom + p o s i t r o n :
! f ( r ) = 1 / r^2−r / k1^3
! ax= f ( r ) * x1 / r ay= f ( r ) * x2 / r
r e a l ( 8 ) f u n c t i o n f3 ( t , x1 , x2 , v1 , v2 )
268 CHAPTER 5. PLANAR MOTION
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : r2 , r , fr
r2=x1 * x1+x2 * x2
r = s q r t ( r2 )
i f ( r . l e . k1 . and . r2 . g t . 0 . 0 D0 ) then
fr = 1 / r2−r / k1 * * 3
else
fr = 0.0 D0
endif
i f ( fr . g t . 0 . 0 D0 . and . r . g t . 0 . 0 D0 ) then
f3=fr * x1 / r ! dx3 / dt=dv1 / dt=a1
else
f3 =0.0 D0
endif
end f u n c t i o n f3
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n f4 ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : r2 , r , fr
r2=x1 * x1+x2 * x2
r = s q r t ( r2 )
i f ( r . l e . k1 . and . r2 . g t . 0 . 0 D0 ) then
fr = 1 / r2−r / k1 * * 3
else
fr = 0.0 D0
endif
i f ( fr . g t . 0 . 0 D0 . and . r . g t . 0 . 0 D0 ) then
f4=fr * x2 / r ! dx3 / dt=dv1 / dt=a1
else
f4 =0.0 D0
endif
end f u n c t i o n f4
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( t , x1 , x2 , v1 , v2 )
i m p l i c i t none
r e a l ( 8 ) : : t , x1 , x2 , v1 , v2
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : r , Vr
r= s q r t ( x1 * x1+x2 * x2 )
i f ( r . l e . k1 . and . r . g t . 0 . 0 D0 ) then
5.4. SCATTERING 269
Vr = 1 / r + 0.5 D0 * r * r / k1 * * 3 − 1 . 5 D0 / k1
else
Vr = 0.0 D0
endif
energy = 0.5 D0 * ( v1 * v1+v2 * v2 ) + Vr
end f u n c t i o n energy
The results are shown in figures 5.13–5.14. We find that σtot = πa2
(see problem 5.10).
3.5
2.0
1.5
3 1.0
0.5
2.5 0.25
0.125
2
θ
1.5
1
0.5
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
b
Figure 5.13: The impact parameter b(θ) for the potential given by equation (5.27)
for different values of the initial velocity v. We set a = 1, x(0) = −5 and made 4000
integration steps from ti = 0 to tf = 40.
Another interesting dynamical field is given by the Yukawa potential.
This is a phenomenological model of nuclear interactions:
e−r/a
V (r) = k . (5.28)
r
This field can also be used as a model of the effective interaction of
electrons in metals (Thomas–Fermi) or as the Debye potential in a classic
plasma. The resulting force is
e−r/a ( r)
F⃗ (r) = f (r) r̂ , f (r) = k 1 + (5.29)
r2 a
270 CHAPTER 5. PLANAR MOTION
100
2.0
1.5
1.0
10 0.5
0.25
0.125
σ(θ)
1
0.1
0.01
0 0.5 1 1.5 2 2.5 3 3.5
θ
Figure 5.14: The function σ(θ) for the potential given by equation (5.27) for different
values of the initial velocity v. We set a = 1, x(0) = −5 and the integration is performed
by making 4000 steps from ti = 0 to tf = 40.
The program of the resulting acceleration can be found in the file rk2_yu.f90.
The results are shown in figures 5.15–5.16.
5.5 More Particles
In this section we will generalize the discussion of the previous para-
graphs in the case of a dynamical system with more degrees of freedom.
The number of dynamical equations that need to be solved depends on
the number of degrees of freedom and we have to write a program that
implements the 4th order Runge–Kutta method for an arbitrary number
of equations NEQ. We will explain how to allocate memory dynamically, in
which case the necessary memory storage space, which depends on NEQ,
is allocated at the time of running the program and not at compilation
time.
Until now, memory has been allocated statically. This means that
arrays have sizes which are known at compile time. For example, in
the program rk2.f90 the integer parameter P had a given value which
determined the size of all arrays using the declarations:
5.5. MORE PARTICLES 271
10
yu v= 4.0
cb v= 4.0
yu v=15.0
1 cb v=15.0
0.1
θ
0.01
0.001
1e-04
0 0.5 1 1.5 2 2.5 3
b
Figure 5.15: The function b(θ) for the Yukawa scattering for several values of the
initial velocity v. We set a = 1, k = 1, x(0) = −50 and the integration is performed with
5000 steps from ti = 0 to tf = 30. The lines marked as cb are equation (5.24) of the
Rutherford scattering.
i n t e g e r , parameter : : P=1010000
r e a l ( 8 ) , dimension ( P ) : : T , X1 , X2 , V1 , V2
Changing P after compilation is impossible and if this becomes necessary
we have to edit the file, change the value of P and recompile. Dynamical
memory allocation allows us to read in Nt and NEQ at execution time and
then ask from the operating system to allocate the necessary memory. All
we have to do is to declare the shape of the arrays (i.e. how many indices
they take) and give them the allocatable attribute. The needed memory
can be asked for at execution time by calling the function ALLOCATE. Here
is an example:
integer Nt , NEQ
real (8) , allocatable :: T (:) ! Rank−1 a r r a y
real (8) , a l l o c a t a b l e : : X ( : , : ) ! Rank−2 a r r a y
real (8) , a l l o c a t a b l e : : X0 ( : ) ! Rank−1 a r r a y
read * , Nt
272 CHAPTER 5. PLANAR MOTION
10
0.25
0.5
1 1.0
1.5
0.1 2.0
0.01
θ
0.001
1e-04
1e-05
1e-06
0 0.5 1 1.5 2 2.5 3
b
Figure 5.16: The function b(θ) for the Yukawa scattering for several values of the
range a of the force. We set v = 4.0, k = 1, x(0) = −50 and the integration is performed
with 5000 steps from ti = 0 to tf = 30.
c a l l finit ( NEQ )
a l l o c a t e ( X0 ( NEQ ) )
a l l o c a t e ( T ( Nt ) )
a l l o c a t e ( X ( Nt , NEQ ) )
...
( compute with X0 , T , X )
...
d e a l l o c a t e ( X0 )
deallocate (X )
deallocate (T )
( X0 , T , X are not usable anymore )
...
....................
s u b r o u t i n e finit ( NEQ )
NEQ = 4
end s u b r o u t i n e finit
In this program the arrays have the allocatable attribute and for each :
they have an extra index. Therefore the arrays T,X0 are rank-1 arrays
and have only one index, whereas the array X is a rank-2 array and has
two indices. The user enters the value of Nt and the subroutine finit
5.5. MORE PARTICLES 273
sets the value of NEQ. The calls to the function ALLOCATE allocate the nec-
essary memory⁴. If memory allocation is successful, then the arrays can
be used in the same way as the statically allocated ones. When allocatable
arrays are not necessary anymore we should make a call to the function
DEALLOCATE which returns the unused memory back to the system. Oth-
erwise our program might suffer from “memory leaks” if e.g. the memory
is repeatedly asked in a loop that calls a function that allocates memory
without deallocating it in the end. Dynamical memory allocation is very
convenient but for high performance computing static allocation might
be preferable so that the compiler performs a more efficient optimization.
The main program will be written in the file rkA.f90, whereas the
force-dependent part of the code will be written in files with names of
the form rkA_XXX.f90. In the latter, the user must program a subrou-
tine f(t,X,dXdt) which takes as input the time t and the values of the
functions X(NEQ) and outputs the values of their derivatives dXdt(NEQ)
at time t. The function finit(NEQ) sets the number of functions in f
and it is called once during the initialization phase of the program.
The program in the file rkA.f90⁵ is listed below:
! ========================================================
! Program t o s o l v e an ODE system using t h e
! 4 th order Runge−Kutta Method
!NEQ: Number o f e q u a t i o n s
! User s u p p l i e s two s u b r o u t i n e s :
! f ( t , x , xdot ) : with r e a l ( 8 ) : : t , x (NEQ) , xdot (NEQ) which
! giv en t h e time t and c u r r e n t v a l u e s o f f u n c t i o n s x (NEQ)
! i t r e t u r n s t h e v a l u e s o f d e r i v a t i v e s : xdot = dx / dt
! The v a l u e s o f two c o u p l i n g c o n s t a n t s k1 , k2 may be used
! i n f which a r e read i n t h e main program and s t o r e d i n
! common / c o u p l i n g s / k1 , k2
! f i n i t (NEQ) : s e t s t h e v a l u e o f NEQ
!
! User I n t e r f a c e :
! k1 , k2 : r e a l ( 8 ) c o u p l i n g c o n s t a n t s
! Nt , Ti , Tf : Nt−1 i n t e g r a t i o n s t e p s , i n i t i a l / f i n a l time
! X0 : r e a l ( 8 ) , dimension (NEQ) : i n i t i a l c o n d i t i o n s
! Output :
⁴We assume that Nt, NEQ are positive and small enough so that the requested
memory is available. It is better to use the call allocate(T(Nt),STAT=IERR). The non
zero value of IERR after the call indicates a successful allocation and the following test
stops the program otherwise: IF(IERR .eq. 0) STOP 'Memory allocation for T
failed'
⁵In the accompanying software you will find the files rkN.f90 and rkN_XXX.f90
which show you how to write the same program using static memory allocation.
274 CHAPTER 5. PLANAR MOTION
! rkA . dat with Nt l i n e s c o n s i s t i n g o f : T( Nt ) ,X( Nt ,NEQ)
! ========================================================
program rk2_solve
i m p l i c i t none
real (8) , a l l o c a t a b l e : : T ( : )
real (8) , a l l o c a t a b l e : : X ( : , : )
r e a l ( 8 ) , a l l o c a t a b l e : : X0 ( : )
r e a l ( 8 ) : : Ti , Tf
i n t e g e r : : Nt , NEQ , i
real (8) : : k1 , k2
common / couplings / k1 , k2
!We need e x p l i c i t i n t e r f a c e , s i n c e energy has
! assumed−shape a r r a y s as arguments .
INTERFACE
r e a l ( 8 ) f u n c t i o n energy ( t_intrf , x_intrf )
i m p l i c i t none
r e a l ( 8 ) : : t_intrf , x_intrf ( : )
end f u n c t i o n energy
END INTERFACE
! Input :
p r i n t * , ’ Runge−Kutta Method f o r ODE I n t e g r a t i o n . ’
! Get t h e number o f e q u a t i o n s :
c a l l finit ( NEQ ) ; a l l o c a t e ( X0 ( NEQ ) )
p r i n t * , ’NEQ= ’ , NEQ
p r i n t * , ’ Enter c o u p l i n g c o n s t a n t s : ’
read * , k1 , k2
p r i n t * , ’ k1= ’ , k1 , ’ k2= ’ , k2
p r i n t * , ’ Enter Nt , Ti , Tf , X0 : ’
read * , Nt , Ti , TF , X0
p r i n t * , ’ Nt = ’ , Nt
p r i n t * , ’ Time : I n i t i a l Ti = ’ , Ti , ’ F i n a l Tf= ’ , Tf
p r i n t ’ (A,2000G28 . 1 6 ) ’ , ’ X0 = ’ , X0
a l l o c a t e ( T ( Nt ) ) ; a l l o c a t e ( X ( Nt , NEQ ) )
! The C a l c u l a t i o n :
c a l l RK ( T , X , Ti , Tf , X0 , Nt , NEQ )
! Output :
open ( u n i t =11 , f i l e = ’ rkA . dat ’ )
do i =1 , Nt
w r i t e ( 1 1 , ’ (2000G28 . 1 6 ) ’ ) T ( i ) , X ( i , : ) ,&
energy ( T ( i ) , X ( i , : ) )
enddo
close (11)
end program rk2_solve
! ========================================================
! D r i v e r o f t h e RKSTEP r o u t i n e
! ========================================================
s u b r o u t i n e RK ( T , X , Ti , Tf , X0 , Nt , NEQ )
i m p l i c i t none
i n t e g e r : : Nt , NEQ
5.5. MORE PARTICLES 275
r e a l ( 8 ) , dimension ( Nt ) :: T
r e a l ( 8 ) , dimension ( Nt , NEQ ) : : X
r e a l ( 8 ) , dimension ( NEQ ) : : X0
r e a l ( 8 ) : : Ti , Tf
r e a l ( 8 ) : : dt
r e a l ( 8 ) : : TS , XS ( NEQ ) ! v a l u e s o f time and X a t gi v e n s t e p
integer : : i
! I n i t i a l i z e variables :
dt = ( Tf−Ti ) / ( Nt −1)
T ( 1 ) = Ti
X ( 1 , : ) = X0
TS = Ti
XS = X0
! Make RK s t e p s : The arguments o f RKSTEP a r e
! r e p l a c e d with t h e new ones
do i =2 , Nt
c a l l RKSTEP ( TS , XS , dt , NEQ )
T ( i ) = TS
X ( i , : ) = XS
enddo
end s u b r o u t i n e RK
! ========================================================
! S u b r o u t i n e RKSTEP( t , X, dt )
! Runge−Kutta I n t e g r a t i o n r o u t i n e o f ODE
! ========================================================
s u b r o u t i n e RKSTEP ( t , x , dt , NEQ )
i m p l i c i t none
i n t e g e r : : NEQ
r e a l ( 8 ) , dimension ( NEQ ) : : x
r e a l ( 8 ) : : t , dt , tt
r e a l ( 8 ) , dimension ( NEQ ) : : k1 , k2 , k3 , k4 , xx
r e a l ( 8 ) : : h , h2 , h6
!We need e x p l i c i t i n t e r f a c e , s i n c e f has assumed−shape
! a r r a y s as arguments .
INTERFACE
s u b r o u t i n e f ( t_intrf , x_intrf , xdot_intrf )
i m p l i c i t none
r e a l ( 8 ) : : t_intrf
r e a l ( 8 ) , dimension ( : ) : : x_intrf , xdot_intrf
end s u b r o u t i n e f
END INTERFACE
h =dt ! h =dt , i n t e g r a t i o n s t e p
h2 =0.5 D0 * h ! h2=h / 2
h6=h / 6 . 0 D0 ! h6=h / 6
c a l l f ( t , x , k1 ) ; xx = x + h2 * k1 ; tt =t+h2
c a l l f ( tt , xx , k2 ) ; xx = x + h2 * k2 ; tt =t+h2
c a l l f ( tt , xx , k3 ) ; xx = x + h * k3 ; tt =t+h
276 CHAPTER 5. PLANAR MOTION
c a l l f ( tt , xx , k4 )
t =t+h
x =x +h6 * ( k1 +2.0 D0 * ( k2+k3 )+k4 )
end s u b r o u t i n e RKSTEP
Note the use of array sections:
w r i t e ( 1 1 , ’ (2000G28 . 1 6 ) ’ ) T ( i ) , X ( i , : )
X ( 1 , : ) = X0
X ( i , : ) = XS
The expression X (1,:) refers to the first row of the array X. The ar-
rays X0 and X (1,:) are conformable and we can assign the entries in X
(1,:) equal to the entries in X0, i.e. X(1,1)=X0(1), X(1,2)=X0(2), ...
, X(1,NEQ)=X0(NEQ) in only one statement X(1,:)= X0. Similarly the
statement write(...) X(i,:) prints the whole i-th row of the array X
whereas the statement X(i,:)= XS assigns X(i,1)=XS(1), X(i,2)=XS(2),
... , X(i,NEQ)=XS(NEQ). Note the vector operations:
xx = x + h2 * k1
x = x + h6 * ( k1 +2.0 D0 * ( k2+k3 )+k4 )
which are equivalent to the following do loops
do i =1 , NEQ
xx ( i ) = x ( i ) + h2 * k1 ( i )
enddo
do i =1 , NEQ
x ( i ) = x ( i ) + h6 * ( k1 ( i ) +2.0 D0 * ( k2 ( i )+k3 ( i ) )+k4 ( i ) )
enddo
A few words in order to explain what is an INTERFACE block. Up to
now we declared only the type of the functions in the calling program.
When the arguments of the function are arrays for which we only know
their shape and not their size (assumed-shape arrays), the compiler needs
more information. We need to declare the arguments, their types and, in
case they are arrays, their shapes as well. Each program that calls these
functions should include an INTERFACE block which provides this infor-
mation. For the functions f and energy, the corresponding INTERFACE
block is
INTERFACE
5.5. MORE PARTICLES 277
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
s u b r o u t i n e f ( t_intrf , x_intrf , xdot_intrf )
i m p l i c i t none
r e a l ( 8 ) : : t_intrf
r e a l ( 8 ) , dimension ( : ) : : x_intrf , xdot_intrf
end s u b r o u t i n e f
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( t_intrf , x_intrf )
i m p l i c i t none
r e a l ( 8 ) : : t_intrf , x_intrf ( : )
end f u n c t i o n energy
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
END INTERFACE
You may create files like e.g. interfaces.inc with groups of INTERFACE
blocks and include them in all subprograms that use them with the state-
ment include "interfaces.inc".
Figure 5.17: Three particles of equal mass interact via their mutual gravitational
attraction. The problem is solved numerically using the program in the files rkA.f90,
rkA_3pcb.f90. The same program can be used in order to study the motion of three
equal charges under the influence of their attractive or repulsive electrostatic force.
Consider three particles of equal mass exerting a force of gravitational
attraction on each other⁶ like the ones shown in figure 5.17. The forces
⁶The same program can be used for three equal charges exerting an electrostatic
force on each other, which can be either attractive or repulsive.
278 CHAPTER 5. PLANAR MOTION
exerting on each other are given by
mk1
F⃗ij = 3 ⃗rij , i, j = 1, 2, 3 , (5.30)
rij
where k1 = −Gm and the equations of motion become (i = 1, 2, 3)
dxi dvix ∑ 3
xi − xj
= vix = k1 3
dt dt j=1,j̸=i
rij
dyi dviy ∑ 3
yi − yj
= viy = k1 3
, (5.31)
dt dt j=1,j̸=i
rij
where rij2 = (xi − xj )2 + (yi − yj )2 . The total energy of the system is
1 2 ∑3
k1
2
E/m = (v1 + v2 ) + . (5.32)
2 i,j=1,j<i
rij
The relations shown above are programmed in the file rkA_3pcb.f90
listed below:
! ===============================
! S e t s number o f e q u a t i o n s
! ===============================
s u b r o u t i n e finit ( NEQ )
NEQ = 12
end s u b r o u t i n e finit
! ===============================
! Three p a r t i c l e s o f t h e same
! mass on t h e plane i n t e r a c t i n g
! v i a Coulombic f o r c e
! ===============================
s u b r o u t i n e f ( t , X , dXdt )
i m p l i c i t none
real (8) : : k1 , k2
common / couplings / k1 , k2
r e a l ( 8 ) : : t , X ( : ) , dXdt ( : )
!−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) : : x11 , x12 , x21 , x22 , x31 , x32
r e a l ( 8 ) : : v11 , v12 , v21 , v22 , v31 , v32
r e a l ( 8 ) : : r12 , r13 , r23
!−−−−−−−−−−−−−−−−−−−−−−−
x11 = X ( 1 ) ; x21 = X ( 5 ) ; x31 = X ( 9 )
x12 = X ( 2 ) ; x22 = X ( 6 ) ; x32 = X ( 1 0 )
5.5. MORE PARTICLES 279
v11 = X ( 3 ) ; v21 = X ( 7 ) ; v31 = X ( 1 1 )
v12 = X ( 4 ) ; v22 = X ( 8 ) ; v32 = X ( 1 2 )
!−−−−−−−−−−−−−−−−−−−−−−−
r12 = ( ( x11−x21 ) * ( x11−x21 ) +( x12−x22 ) * ( x12−x22 ) ) * * ( − 1. 5 D0 )
r13 = ( ( x11−x31 ) * ( x11−x31 ) +( x12−x32 ) * ( x12−x32 ) ) * * ( − 1. 5 D0 )
r23 = ( ( x21−x31 ) * ( x21−x31 ) +( x22−x32 ) * ( x22−x32 ) ) * * ( − 1. 5 D0 )
!−−−−−−−−−−−−−−
dXdt ( 1 ) = v11
dXdt ( 2 ) = v12
dXdt ( 3 ) = k1 * ( x11−x21 ) * r12+k1 * ( x11−x31 ) * r13 ! a11=dv11 / dt
dXdt ( 4 ) = k1 * ( x12−x22 ) * r12+k1 * ( x12−x32 ) * r13 ! a12=dv12 / dt
!−−−−−−−−−−−−−−
dXdt ( 5 ) = v21
dXdt ( 6 ) = v22
dXdt ( 7 ) = k1 * ( x21−x11 ) * r12+k1 * ( x21−x31 ) * r23 ! a21=dv21 / dt
dXdt ( 8 ) = k1 * ( x22−x12 ) * r12+k1 * ( x22−x32 ) * r23 ! a22=dv22 / dt
!−−−−−−−−−−−−−−
dXdt ( 9 ) = v31
dXdt ( 1 0 ) = v32
dXdt ( 1 1 ) = k1 * ( x31−x11 ) * r13+k1 * ( x31−x21 ) * r23 ! a31=dv31 / dt
dXdt ( 1 2 ) = k1 * ( x32−x12 ) * r13+k1 * ( x32−x22 ) * r23 ! a32=dv32 / dt
end s u b r o u t i n e f
! ===============================
r e a l ( 8 ) f u n c t i o n energy ( t , X )
i m p l i c i t none
real (8) : : k1 , k2
common / couplings / k1 , k2
real (8) : : t , X ( : )
!−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) : : x11 , x12 , x21 , x22 , x31 , x32
r e a l ( 8 ) : : v11 , v12 , v21 , v22 , v31 , v32
r e a l ( 8 ) : : r12 , r13 , r23
!−−−−−−−−−−−−−−−−−−−−−−−
x11 = X ( 1 ) ; x21 = X ( 5 ) ; x31 = X ( 9 )
x12 = X ( 2 ) ; x22 = X ( 6 ) ; x32 = X ( 1 0 )
v11 = X ( 3 ) ; v21 = X ( 7 ) ; v31 = X ( 1 1 )
v12 = X ( 4 ) ; v22 = X ( 8 ) ; v32 = X ( 1 2 )
!−−−−−−−−−−−−−−−−−−−−−−−
r12 = ( ( x11−x21 ) * ( x11−x21 ) +( x12−x22 ) * ( x12−x22 ) ) **( −0.5 D0 )
r13 = ( ( x11−x31 ) * ( x11−x31 ) +( x12−x32 ) * ( x12−x32 ) ) **( −0.5 D0 )
r23 = ( ( x21−x31 ) * ( x21−x31 ) +( x22−x32 ) * ( x22−x32 ) ) **( −0.5 D0 )
!−−−−−−−−−−−−−−−−−−−−−−−
energy = 0.5 D0*&
( v11 * v11+v12 * v12+v21 * v21+v22 * v22+v31 * v31+v32 * v32 )
energy = energy + k1 * ( r12+r13+r23 )
end f u n c t i o n energy
In order to run the program and see the results look at the commands
in the shell script in the file rkA_3pcb.csh. In order to run the script use
280 CHAPTER 5. PLANAR MOTION
command
> rkA_3pcb . csh −0.5 4000 1 . 5 −1 0 . 1 1 0 1 −0.1 −1 0 0.05 1 0 −1
which will run the program setting k1 = −0.5, ⃗r1 (0) = −x̂+0.1ŷ, ⃗v1 (0) = x̂,
⃗r2 (0) = x̂ − 0.1ŷ, ⃗v2 (0) = −x̂, ⃗r3 (0) = 0.05x̂ + ŷ, ⃗v3 (0) = −ŷ, Nt= 4000 and
tf = 1.5.
5.6. PROBLEMS 281
5.6 Problems
5.1 Reproduce the results shown in figures 5.3 and 5.4. Compare your
results to the known analytic solution.
5.2 Write a program for the force on a charged particle in a constant
magnetic field B ⃗ = B k̂ and compute its trajectory for ⃗v (0) = v0x x̂ +
v0y ŷ. Set x(0) = 1, y(0) = 0, v0y = 0 and calculate the resulting
radius of the trajectory. Plot the relation between the radius and
v0x . Compare your results to the known analytic solution. (assume
non relativistic motion)
5.3 Consider the anisotropic harmonic oscillator ax = −ω12 x, ay = −ω22 y.
Construct the Lissajous curves by setting x(0) = 0, y(0) = 1, vx (0) =
1, vy (0) = 0, tf = 2π, ω22 = 1, ω12 = 1, 2, 4, 9, 16, . . .. What happens
when ω12 ̸= nω22 ?
5.4 Reproduce the results displayed in table 5.1 and figures 5.5 and 5.6.
Plot ln a vs ln T and calculate the slope of the resulting straight line
by using the linear least squares method. Is it what you expect?
Calculate the intercept and compare your result with the expected
one.
5.5 Calculate the angular momentum with respect to the center of the
force at each integration step of the planetary motion and check
whether it is conserved. Show analytically that conservation of
angular momentum implies that the position vector sweeps areas at
constant rate.
5.6 Calculate the escape velocity of a planet ve for GM = 10.0, y(0) = 0.0,
x0 = x(0) = 1 using the following steps: First show that v02 =
−GM (1/a) + ve2 . Then set vx (0) = 0, vy (0) = v0 . Vary vy (0) = v0 and
measure the resulting semi-major axis a. Determine the intercept
of the resulting straight line in order to calculate ve .
5.7 Repeat the previous problem for x0 = 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5,
4.0. From the ve = f (1/x0 ) plot confirm the relation (5.14).
5.8 Check that for the bound trajectory of a planet with GM = 10.0,
x(0) = 1, y(0) = 0.0, vx (0) = 0 , vy (0) = 4 you obtain that F1 P +
F2 P = 2a for each point P of the trajectory. The point F1 is the
center of the force. After determining the semi-major axis a nu-
merically, the point F2 will be taken symmetric to F1 with respect
to the center of the ellipse.
282 CHAPTER 5. PLANAR MOTION
5.9 Consider the planetary motion studied in the previous problem.
Apply a momentary push in the tangential direction after the planet
has completed 1/4 of its elliptical orbit. How stable is the particle
trajectory (i.e. what is the dependence of the trajectory on the
magnitude and the duration of the push?)? Repeat the problem
when the push is in the vertical direction.
5.10 Consider the scattering potential of the positron-hydrogen system
given by equation (5.26). Plot the functions f (r) and V (r) for
different values of a. Calculate the total cross section σtot numerically
and show that it is equal to πa2 .
5.11 Consider the Morse potential of diatomic molecules:
V (r) = D (exp(−2αr) − 2 exp(−αr)) (5.33)
where D, α > 0. Compute the solutions of the problem numerically
in one dimension and compare them to the known analytic solutions
when E < 0:
{ √ √ }
1 D − D(D − |E|) sin(αt 2|E|/m + C)
x(t) = ln , (5.34)
α |E|
where the integration constant as a function of the initial position
and energy is given by
[ ]
D − |E|eαx0
C = sin−1 √ . (5.35)
D(D − |E|)
We obtain
√ a periodic motion with an energy dependent period =
(π/α) 2m/|E|. For E > 0 we obtain
{√ √ }
1 D(D + E) cosh(αt 2E/m + C) − D
x(t) = ln (5.36)
α |E|
whereas for E = 0
{ }
1 1 Dα2
x(t) = ln + (t + C)2 . (5.37)
α 2 m
In these equations, the integration constant C is given by a different
relation and not by equation (5.35). Compute the motion in phase
space (x, ẋ) and study the transition from open to closed trajectories.
5.6. PROBLEMS 283
5.12 Consider the effective potential term Vef f (r) = l2 /2mr2 (l ≡ |L|)
⃗ in
the previous problem. Plot the function Vtot (r) = V (r) + Vef f (r) for
D = 20, α = 1, m = 1, l = 1, and of course for r > 0. Determine the
equilibrium position and the ionization energy.
Calculate the solutions x(t), y(t), y(x), r(t) on the plane for E > 0,
E = 0, and E < 0 numerically. In the E < 0 case consider the
scattering problem and calculate the functions b(θ), σ(θ) and the
total cross section σtot .
5.13 Consider the potential of the molecular model given by the force
F⃗ (r) = f (r) r̂ where f (r) = 24(2/r13 − 1/r7 ). Calculate the potential
V (r) and plot the function Vtot (r) = V (r) + Vef f (r). Determine the
equilibrium position and the ionization energy.
Consider the problem of scattering and calculate b(θ), σ(θ) and σtot
numerically. How much do your results depend on the minimum
scattering angle?
5.14 Compute the trajectories of a particle under the influence of a force
F⃗ = −k/r3 r̂. Determine appropriate initial conditions that give a
spiral trajectory.
5.15 Compute the total cross section σtot for the Rutherford scattering
both analytically and numerically. What happens to your numerical
results as you vary the integration limits?
5.16 Write a program that computes the trajectory of a particle that
moves on the plane in the static electric field of N static point
charges.
5.17 Solve the three body problem described in the text in the case of
three different electric charges by making the appropriate changes
to the program in the file rkA_3cb.f90.
5.18 Two charged particles of equal mass and charge are moving on the
xy plane in a constant magnetic field B ⃗ = B ẑ. Solve the equations of
motion using a 4th order Runge–Kutta Method. Plot the resulting
trajectories for the initial conditions that you will choose.
5.19 Three particles of equal mass m are connected by identical springs.
The springs’ spring constant is equal to k and their equilibrium
length is equal to l. The particles move without friction on a hori-
zontal plane. Solve the equations of motion of the system numeri-
cally by using a 4th order Runge–Kutta Method. Plot the resulting
284 CHAPTER 5. PLANAR MOTION
trajectories for the initial conditions that you will choose. (Hint:
Look in the files rkA_3hoc.f90, rkA_3hoc.csh.)
Figure 5.18: Two identical particles are attached to thin weightless rods of length
l and they are connected by an ideal weightless spring with spring constant k and
equilibrium length l. The rods are hinged to the ceiling at points whose distance is l.
(Problem 5.20).
5.20 Two identical particles are attached to thin weightless rods of length
l and they are connected by an ideal weightless spring with spring
constant k and equilibrium length l. The rods are hinged to the
ceiling at points whose distance is l (see figure 5.18). Compute
the Lagrangian of the system and the equations of motion for the
degrees of freedom θ1 and θ2 . Solve these equations numerically
by using a 4th order Runge–Kutta method. Plot the positions of
the particles in a Cartesian coordinate system and the resulting tra-
jectory. Study the normal modes for small angles θ1 ≲ 0.1 and
compute the deviation of the solutions from the small oscillation
approximation as the angles become larger. (Hint: Look in the files
rk_cpend.f90, rk_cpend.csh)
5.21 Repeat the previous problem when the hinges of the rods slide
without friction on the x axis.
5.22 Repeat problem 5.20 by adding a third pendulum to the right at
distance l.
Chapter 6
Motion in Space
In this chapter we will study the motion of a particle in space (three
dimensions). We will also discuss the case of the relativistic motion,
which is important if one wants to consider the motion of particles moving
with speeds comparable to the speed of light. This will be an opportunity
to use an adaptive stepsize Runge-Kutta method for the numerical solution
of the equations of motion. We will use the open source code rksuite¹
available at the Netlib² repository. Netlib is an open source, high quality
repository for numerical analysis software. The software it contains is
used by many researchers in their high performance computing programs
and it is a good investment of time to learn how to use it.
The technical skill that you will exercise in this chapter is looking for
solutions to your numerical problems provided by software written by
others. It is important to be able to locate the optimal solution to your
problem, find the relevant functions, read the software’s documentation
carefully and filter out the necessary information in order to call and link
the functions to your program.
¹R.W. Brankin, I. Gladwell, and L.F. Shampine, RKSUITE: a suite of Runge-Kutta
codes for the initial value problem for ODEs, Softreport 92-S1, Department of Mathe-
matics, Southern Methodist University, Dallas, Texas, U.S.A, 1992.
²www.netlib.org
285
286 CHAPTER 6. MOTION IN SPACE
6.1 Adaptive Stepsize Control for Runge–Kutta
Methods
The three dimensional equation of motion of a particle is an initial value
problem given by the equations (4.6)
dx dvx
= vx = ax (t, x, vx , y, vy , z, vz )
dt dt
dy dvz
= vy = ay (t, x, vx , y, vy , z, vz )
dt dt
dz dvz
= vz = az (t, x, vx , y, vy , z, vz ) . (6.1)
dt dt
For its numerical solution we will use an adaptive stepsize Runge–
Kutta algorithm for increased performance and accuracy. Adaptive step-
size is used in cases where one needs to minimize computational effort
for given accuracy goal. The method frequently changes the time step
during the integration process, so that it is set to be large through smooth
intervals and small when there are abrupt changes in the values of the
functions. This is achieved by exercising error control either by monitor-
ing a conserved quantity or by computing the same solution using two
different methods. In our case, two Runge-Kutta methods are used, one
of order p and one of order p + 1, and the difference of the results is used
as an estimate of the truncation error. If the error needs to be reduced,
the step size is reduced and if it is satisfactorily small the step size is
increased. For the details we refer the reader to [31]. Our goal is not to
analyze and understand the details of the algorithm, but to learn how to
find and use appropriate and high quality code written by others. The
link http://www.netlib.org/ode/ reads
lib rksuite
alg Runge−Kutta
for initial value problem for first order ordinary ←-
differential
equations . A suite of codes for solving IVPs in ODEs . A
choice of RK methods , is available . Includes an error
assessment facility and a sophisticated stiffness checker .
Template programs and example results provided .
Supersedes RKF 45 , DDERKF , D02PAF .
ref RKSUITE , Softreport 92−S 1 , Dept of Math , SMU , Dallas , ←-
Texas
by R . W . Brankin ( NAG ) , I . Gladwell and L . F . Shampine ( SMU )
lang Fortran
6.1. ADAPTIVE STEPSIZE CONTROL FOR RUNGE–KUTTA METHODS287
prec double
There, we learn that the package provides code for Runge–Kutta methods,
whose source is open and written in the Fortran language. We also learn
that the code is written for double precision variables, which is suitable
for our problem. Last, but not least, we are also happy to learn that it is
written by highly reputable people! We download the files rksuite.f,
rksuite.doc, details.doc, templates, readme.
In order to link the subroutines provided by the suite to our program
we need to read the documentation carefully. In the general case, docu-
mentation is available on the web (html, pdf, ...), bundled files with names
like README, INSTALL, in whole directories with names like doc/, online
help in man or info pages and finally in good old fashioned printed man-
uals. Good quality software is also well documented inside the source
code files, something that is true for the software at hand.
In order to link the suite’s subroutines to our program we need the
following basic information:
• INPUT DATA: This is the necessary information that the program
needs in order to perform the calculation. In our case, the mini-
mal such information is the initial conditions, the integration time
interval and the number of integration steps. The user should also
provide the functions on the right hand side of (6.1). It might also
be necessary to provide information about the desired accuracy goal,
the scale of the problem, the hardware etc.
• OUTPUT DATA: This is the information on how we obtain the
results of the calculation for further analysis. Information whether
the calculation was successful and error free could also be provided.
• WORKSPACE: This is information on how we provide the necessary
memory space used in the intermediate calculations. Such space
needs to be provided by the user in programming languages where
dynamical memory allocation is not possible, like in Fortran 77,
and the size of workspace depends on the parameters of the calling
program.
It is easy to install the software. All the necessary code is in one file
rksuite.f. The file rksuite.doc³ contains the documentation. There we
read that we need to inform the program about the hardware dependent
³This is a simple text file which you can read with the command less rksuite.doc
or with emacs.
288 CHAPTER 6. MOTION IN SPACE
accuracy of floating point numbers. We need to set the values of three
variables:
...
RKSUITE requires three environmental constants OUTCH , MCHEPS ,
DWARF . When you use RKSUITE , you may need to know their
values . You can obtain them by calling the subroutine ENVIRN
in the suite :
CALL ENVIRN ( OUTCH , MCHPES , DWARF )
returns values
OUTCH − INTEGER
Standard output channel on the machine being used .
MCHEPS − DOUBLE PRECISION
The unit of roundoff , that is , the largest
positive number such that 1 . 0 D0 + MCHEPS = 1 . 0 D 0 .
DWARF − DOUBLE PRECISION
The smallest positive number on the machine being
used .
...
* * * * * * * * * * * * * * * * * * * * * * * * * * Installation Details * * * * * * * * * * * *
All machine−dependent aspects of the suite have been
isolated in the subroutine ENVIRN in the rksuite . for file .
Certain environmental parameters must be specified in this
subroutine . The values in the distribution version are
those appropriate to the IEEE arithmetic standard . They
must be altered , if necessary , to values appropriate to the
computing system you are using before calling the codes of
the suite . If the IEEE arithmetic standard values are not
appropriate for your system , appropriate values can often
be obtained by calling routines named in the Comments of
ENVIRN .
...
The variables OUTCH, MCHEPS, DWARF are defined in the subroutine ENVIRN.
They are given generic default values but the programmer is free to
change them by editing ENVIRN. We should identify the routine in the file
rksuite.f and read the comments in it⁴:
...
SUBROUTINE ENVIRN ( OUTCH , MCHEPS , DWARF )
...
⁴These are lines that begin with a C as this is old fixed format Fortran code.
6.1. ADAPTIVE STEPSIZE CONTROL FOR RUNGE–KUTTA METHODS289
C The f o l l o w i n g s i x s t a t e m e n t s a r e t o be Commented out
C a f t e r v e r i f i c a t i o n t h a t t h e machine and i n s t a l l a t i o n
C dependent q u a n t i t i e s a r e s p e c i f i e d c o r r e c t l y .
...
WRITE ( * , * ) ’ Before using RKSUITE , you must v e r i f y t h a t t h e ’
WRITE ( * , * ) ’ machine− and i n s t a l l a t i o n −dependent q u a n t i t i e s ’
WRITE ( * , * ) ’ s p e c i f i e d i n t h e s u b r o u t i n e ENVIRN a r e c o r r e c t , ’
WRITE ( * , * ) ’ and then Comment t h e s e WRITE s t a t e m e n t s and t h e ’
WRITE ( * , * ) ’ STOP s t a t e m e n t out o f ENVIRN . ’
STOP
...
C The f o l l o w i n g v a l u e s a r e a p p r o p r i a t e t o IEEE
C a r i t h m e t i c with t h e t y p i c a l standard output channel .
C
OUTCH = 6
MCHEPS = 1 . 1 1 D−16
DWARF = 2.23 D−308
All we need to do is to comment out the WRITE and STOP commands since
we will keep the default values of the OUTCH, MCHEPS, DWARF variables:
...
C WRITE ( * ,*) ’ Before using RKSUITE , you must v e r i f y t h a t t h e ’
C WRITE ( * ,*) ’ machine− and i n s t a l l a t i o n −dependent q u a n t i t i e s ’
C WRITE ( * ,*) ’ s p e c i f i e d i n t h e s u b r o u t i n e ENVIRN a r e c o r r e c t , ’
C WRITE ( * ,*) ’ and then Comment t h e s e WRITE s t a t e m e n t s and t h e ’
C WRITE ( * ,*) ’ STOP s t a t e m e n t out o f ENVIRN . ’
C STOP
...
In order to check whether the default values are satisfactory, we can
use the Fortran intrinsic functions EPSILON() and TINY(). In the file
test_envirn.f90, we write a small test program
program testme
i m p l i c i t none
i n t e g e r : : OUTCH
r e a l ( 8 ) : : DWARF , MCHEPS
real (8) : : x
OUTCH = 6 ! This i s p r e t t y much a standard
MCHEPS = e p s i l o n ( x ) / 2 . 0 D0
DWARF = tiny (x)
w r i t e ( 6 , 1 0 1 ) OUTCH , MCHEPS , DWARF
101 format ( I4 , 2 E30 . 1 8 )
end program testme
We compile and run the above program as follows:
> g f o r t r a n test_envirn . f90 −o test_envirn
290 CHAPTER 6. MOTION IN SPACE
> . / test_envirn
6 0.111022302462515654E−15 ←-
0.222507385850720138−307
We conclude that our choices are satisfactory.
Next we need to learn how to use the subroutines in the suite. By
carefully reading rksuite.doc we learn the following: The interface to the
adaptive stepsize Runge–Kutta algorithm is the routine UT (UT = “Usual
Task”). The routine can use a 2nd-3rd (RK23) order Runge-Kutta pair
for error control (METHOD=1), a 4th-5th (RK45) order pair (METHOD=2) or
a 7th-8th (RK78) order pair (METHOD=3). We will set METHOD=2 (RK45).
The routine SETUP must be called before UT for initialization. The user
should provide a function F that calculates the derivatives of the functions
we integrate for, i.e. the right hand side of 6.1.
The fastest way to learn how to use the above routines is “by exam-
ple”. The suite include a templates package which can be unpacked by
executing the commands in the file templates using the sh shell:
> sh templates
tmpl1 . out
tmpl1a . f
...
The file tmpl1a.f contains the solution of the simple harmonic oscillator
and has many explanatory comments in it. We encourage the reader to
study it carefully, run it and test its results.
After we become wise enough, we write the driver for the integration
routine UT, which can be found in the file rk3.f90:
! ========================================================
! Program t o s o l v e a 6 ODE system using Runge−Kutta Method
! Output i s w r i t t e n i n f i l e rk3 . dat
! ========================================================
program rk3_solve
i n c l u d e ’ rk3 . i n c ’
r e a l ( 8 ) : : T0 , TF , X10 , X20 , X30 , V10 , V20 , V30
r e a l ( 8 ) : : t , dt , tstep
i n t e g e r : : STEPS
integer : : i
r e a l ( 8 ) : : energy
! Arrays / v a r i a b l e s needed by r k s u i t e :
r e a l ( 8 ) TOL , THRES ( NEQ ) , WORK ( LENWRK ) , Y ( NEQ ) , YMAX ( NEQ ) ,&
YP ( NEQ ) , YSTART ( NEQ ) , HSTART
l o g i c a l ERRASS , MESSAGE
6.1. ADAPTIVE STEPSIZE CONTROL FOR RUNGE–KUTTA METHODS291
i n t e g e r UFLAG
! . . External Subroutines . .
EXTERNAL F , SETUP , STAT , UT
! Input :
p r i n t * , ’ Runge−Kutta Method f o r 6−ODEs I n t e g r a t i o n ’
p r i n t * , ’ Enter c o u p l i n g c o n s t a n t s k1 , k2 , k3 , k4 : ’
read * , k1 , k2 , k3 , k4
p r i n t * , ’ k1= ’ , k1 , ’ k2= ’ , k2 , ’ k3= ’ , k3 , ’ k4= ’ , k4
p r i n t * , ’ Enter STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30 : ’
read * , STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30
p r i n t * , ’No . S t e p s= ’ , STEPS
p r i n t * , ’ Time : I n i t i a l T0 = ’ , T0 , ’ F i n a l TF= ’ , TF
print * , ’ X1(T0)= ’ , X10 , ’ X2(T0)= ’ , X20 , ’ X3(T0)= ’ , X30
print * , ’ V1 (T0)= ’ , V10 , ’ V2(T0)= ’ , V20 , ’ V3(T0)= ’ , V30
! I n i t i a l Conditions
dt = ( TF−T0 ) / STEPS
YSTART ( 1 ) = X10
YSTART ( 2 ) = X20
YSTART ( 3 ) = X30
YSTART ( 4 ) = V10
YSTART ( 5 ) = V20
YSTART ( 6 ) = V30
!
! S e t e r r o r c o n t r o l parameters .
!
TOL = 5.0 D−6
do i = 1 , NEQ
THRES ( i ) = 1 . 0 D−10
enddo
MESSAGE = .TRUE.
ERRASS = . FALSE .
HSTART = 0.0 D0
! Initialization :
c a l l SETUP ( NEQ , T0 , YSTART , TF , TOL , THRES , METHOD , ’ Usual Task ’ ,&
ERRASS , HSTART , WORK , LENWRK , MESSAGE )
open ( u n i t =11 , f i l e = ’ rk3 . dat ’ )
w r i t e ( 1 1 , 1 0 0 ) T0 , YSTART ( 1 ) , YSTART ( 2 ) , YSTART ( 3 ) , YSTART ( 4 ) ,&
YSTART ( 5 ) , YSTART ( 6 ) , energy ( T0 , YSTART )
! Calculation :
do i =1 , STEPS
t = T0 + i * dt
c a l l UT ( F , t , tstep , Y , YP , YMAX , WORK , UFLAG )
i f ( UFLAG . GT. 2 ) e x i t ! e x i t t h e loop : go a f t e r enddo
w r i t e ( 1 1 , 1 0 0 ) tstep , Y ( 1 ) , Y ( 2 ) , Y ( 3 ) , Y ( 4 ) , Y ( 5 ) , Y ( 6 ) ,&
energy ( tstep , Y )
enddo
close (11)
100 format (8 E25 . 1 5 )
end program rk3_solve
292 CHAPTER 6. MOTION IN SPACE
All common parameters and variables are declared in an include file
rk3.inc. This is necessary in order for them to be accessible by the
function F which calculates the derivatives. The contents of this file are
substituted in each line containing the command include 'rk3.inc'.
! Basic d e f i n i t i o n s of v a r i a b l e s fo r the s u i t e r k s u i t e
i m p l i c i t none
!NEQ i s t h e number o f e q u a t i o n s , 6 i n 3 dimensions
!METHOD=2 i s f o r RK45 .
INTEGER NEQ , LENWRK , METHOD
PARAMETER ( NEQ =6 , LENWRK =32*NEQ , METHOD =2)
REAL *8 k1 , k2 , k3 , k4 ! f o r c e c o u p l i n g s
COMMON / COUPLINGS / k1 , k2 , k3 , k4
The number of differential equations is set equal to NEQ=6. The integra-
tion method is set by the choice METHOD=2. The variable LENWRK sets the
size of the workspace needed by the suite for the intermediate calcula-
tions.
The main program starts with the user interface. The initial state of
the particle is stored in the array YSTART in the positions 1 . . . 6. The
first three positions are the coordinates of the initial position and the
last three the components of the initial velocity. Then we set some vari-
ables that determine the behavior of the integration program (see the
file rksuite.doc for details) and call the subroutine SETUP. The main
integration loop is:
do i =1 , STEPS
t = T0 + i * dt
c a l l UT ( F , t , tstep , Y , YP , YMAX , WORK , UFLAG )
i f ( UFLAG . GT. 2 ) e x i t ! e x i t t h e loop : go a f t e r enddo
w r i t e ( 1 1 , 1 0 0 ) tstep , Y ( 1 ) , Y ( 2 ) , Y ( 3 ) , Y ( 4 ) , Y ( 5 ) , Y ( 6 ) ,&
energy ( tstep , Y )
enddo
The function F is the subroutine that calculates the derivatives and it will
be programmed by us later. The variable t stores the desired moment of
time at which we want to calculate the functions. Because of the adaptive
stepsize, it can be different than the one returned by the subroutine UT.
The actual value of time that the next step lands⁵ on is tstep. The array
Y stores the values of the functions. We choose the data structure to be
such that x= Y(1), y= Y(2), z= Y(3) and vx = Y(4), vy = Y(5), vz = Y(6)
⁵When UGLAG ≤ 2, tstep=t and we will not worry about them being different with
our program.
6.1. ADAPTIVE STEPSIZE CONTROL FOR RUNGE–KUTTA METHODS293
(the same sequence as in the array YSTART). The function energy(t,Y)
returns the value of the mechanical energy of the particle and its code
will be written in the same file as that of F. Finally, the variable UFLAG
indicates the error status of the calculation by UT and if UFLAG> 2 we end
the calculation.
Our test code will be on the study of the motion of a projectile in a
constant gravitational field, subject also to the influence of a dissipative
force F⃗r = −mk⃗v . The program is in the file rk3_g.f90. We choose the
parameters k1 and k2 so that ⃗g = -k1 k̂ and k = k2.
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
s u b r o u t i n e F ( T , Y , YP )
i n c l u d e ’ rk3 . i n c ’
real (8) : : t
r e a l ( 8 ) : : Y ( * ) , YP ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3
x1 = Y ( 1 ) ; v1 = Y ( 4 )
x2 = Y ( 2 ) ; v2 = Y ( 5 )
x3 = Y ( 3 ) ; v3 = Y ( 6 )
! Velocities : d x _ i / dt = v _ i
YP ( 1 ) = v1
YP ( 2 ) = v2
YP ( 3 ) = v3
! A c c e l e r a t i o n : dv_i / dt = a _ i
YP ( 4 ) = −k2 * v1
YP ( 5 ) = −k2 * v2
YP ( 6 ) = −k2 * v3−k1
end s u b r o u t i n e F
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( T , Y )
i n c l u d e ’ rk3 . i n c ’
real (8) : : t , e
real (8) : : Y ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3
x1 = Y ( 1 ) ; v1 = Y ( 4 )
x2 = Y ( 2 ) ; v2 = Y ( 5 )
x3 = Y ( 3 ) ; v3 = Y ( 6 )
! K i n e t i c Energy
e = 0 . 5 * ( v1 * v1+v2 * v2+v3 * v3 )
! P o t e n t i a l Energy
e = e + k1 * x3
energy = e
end f u n c t i o n energy
For convenience we “translated” the values in the array Y(NEQ) into
294 CHAPTER 6. MOTION IN SPACE
user-friendly variable names⁶. If the file rksuite.f is in the directory
rksuite/, then the compilation, running and visualization of the results
can be done with the commands:
> gfortran rk3 . f90 rk3_g . f90 rksuite / rksuite . f −o rk3
> . / rk3
Runge−Kutta Method for 6−ODEs Integration
Enter coupling constants k1 , k2 , k3 , k4 :
10 0 0 0
k1= 10.0000 k2= 0.0000 E+000 k3=
0.0000 E+000 k4= 0.0000 E+000
Enter STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30 :
10000 0 3 0 0 0 1 1 1
No . Steps= 10000
Time : Initial T0 = 0.0000 E+000 Final TF= 3.0000
X1 ( T0 )= 0.0000 E+000 X2 ( T0 )= 0.0000 E+000
X3 ( T0 )= 0.0000 E+000
V1 ( T0 )= 1.0000 V2 ( T0 )= 1.0000
V3 ( T0 )= 1.0000
> gnuplot
gnuplot > p l o t ” rk3 . dat ” using 1 : 2 with lines t i t l e ” x1 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 3 with lines t i t l e ” x2 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 4 with lines t i t l e ” x3 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 5 with lines t i t l e ” v1 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 6 with lines t i t l e ” v2 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 7 with lines t i t l e ” v3 ( t ) ”
gnuplot > p l o t ” rk3 . dat ” using 1 : 8 with lines t i t l e ”E( t ) ”
gnuplot > s e t t i t l e ” t r a j e c t o r y ”
gnuplot > s p l o t ” rk3 . dat ” using 2 : 3 : 4 with lines notitle
All the above commands can be executed together using the shell script in
the file rk3.csh. The script uses the animation script rk3_animate.csh.
The following command executes all the commands shown above:
. / rk3 . csh −f 1 −− 10 0 . 0 0 0 0 0 1 1 1 10000 0 3
⁶Note the declaration of the arrays Y, YP: real(8) :: Y(*),YP(*). These arrays are
“assumed-size” arrays for the functions F, energy, i.e. arrays whose size is unknown
to the procedure. For arrays of more than one dimension, only the last index is allowed
to be *. In general it is recommended that assumed-size arrays be avoided and declare
them as assumed-shape like in the program rkA.f90 of page 273. The declaration in
this case is real(8) :: Y(:),YP(:)
6.2. MOTION OF A PARTICLE IN AN EM FIELD 295
6.2 Motion of a Particle in an EM Field
In this section we study the non-relativistic motion of a charged particle
in an electromagnetic (EM) field. The particle is under the influence of
the Lorentz force:
⃗ + ⃗v × B)
F⃗ = q(E ⃗ . (6.2)
⃗ = Ex x̂ + Ey ŷ + Ez ẑ, B
Consider the constant EM field of the form E ⃗ = B ẑ.
The components of the acceleration of the particle are:
ax = (qEx /m) + (qB/m)vy
ay = (qEy /m) − (qB/m)vx
az = (qEz /m) . (6.3)
This field is programmed in the file rk3_B.f90. We set k1 = qB/m, k2
= qEx /m, k3 = qEy /m and k4 = qEz /m:
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! P a r t i c l e i n c o n s t a n t Magnetic and e l e c t r i c f i e l d
! q B/m = k1 z q E /m = k2 x + k3 y + k4 z
s u b r o u t i n e F ( T , Y , YP )
i n c l u d e ’ rk3 . i n c ’
real (8) : : t
r e a l ( 8 ) : : Y ( * ) , YP ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3
x1 = Y ( 1 ) ; v1 = Y ( 4 )
x2 = Y ( 2 ) ; v2 = Y ( 5 )
x3 = Y ( 3 ) ; v3 = Y ( 6 )
! Velocities : d x _ i / dt = v _ i
YP ( 1 ) = v1
YP ( 2 ) = v2
YP ( 3 ) = v3
! A c c e l e r a t i o n : dv_i / dt = a _ i
YP ( 4 ) = k2 + k1 * v2
YP ( 5 ) = k3 − k1 * v1
YP ( 6 ) = k4
end s u b r o u t i n e F
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
r e a l ( 8 ) f u n c t i o n energy ( T , Y )
i n c l u d e ’ rk3 . i n c ’
real (8) : : t , e
real (8) : : Y ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3
x1 = Y ( 1 ) ; v1 = Y ( 4 )
x2 = Y ( 2 ) ; v2 = Y ( 5 )
x3 = Y ( 3 ) ; v3 = Y ( 6 )
296 CHAPTER 6. MOTION IN SPACE
! K i n e t i c Energy
e = 0 . 5 * ( v1 * v1+v2 * v2+v3 * v3 )
! P o t e n t i a l Energy
e = e − k2 * x1 − k3 * x2 − k4 * x3
energy = e
end f u n c t i o n energy
3
2
1
0
1
1
2 0
x
y
3 -1
Figure 6.1: The trajectory of a charged particle in a constant magnetic field B⃗ = B ẑ,
where qB/m = 1.0, ⃗v (0) = 1.0ŷ + 0.1ẑ, ⃗r(0) = 1.0x̂. The integration of the equations of
motion is performed using the RK45 method from t0 = 0 to tf = 40 with 1000 steps.
We can also study space-dependent fields in the same way. The fields
must satisfy Maxwell’s equations. We can study the confinement of a
⃗ = By ŷ + Bz ẑ
particle in a region of space by a magnetic field by taking B
with qBy /m = −k2 y, qBz /m = k1 + k2 z and qBy /m = k3 z, qBz /m =
k1 + k2 y. Note that ∇
⃗ ·B⃗ = 0. You may also want to calculate the current
density from the equation ∇ ⃗ ×B⃗ = µ0⃗j.
The results are shown in figures 6.1–6.4.
6.3 Relativistic Motion
Consider a particle of non zero rest mass moving with speed comparable
to the speed of light. In this case, it is necessary to study its motion using
6.3. RELATIVISTIC MOTION 297
z
4
3
2
1
0-5 1
-4 2
-3 3
-2 4
-1 5 x
y 0 6
1
2 7
Figure 6.2: The trajectory of a charged particle in a constant magnetic field B⃗ = B ẑ,
where qB/m = 1.0 and a constant electric field E ⃗ = Ex x̂+Ey ŷ με qEx /m = qEy /m = 0.1.
⃗v (0) = 1.0ŷ + 0.1ẑ, ⃗r(0) = 1.0x̂. The integration of the equations of motion is performed
using the RK45 method from t0 = 0 to tf = 40 with 1000 steps. Each axis is on a
different scale.
the equations of motion given by special relativity⁷. In the equations
below we√ set c = 1. The particle’s rest mass is m0 > 0, its mass is
m = m0 / 1 √ − v 2 (where v < 1), its momentum is p⃗ = m⃗v and its energy
is E = m = p2 + m20 . Then the equations of motion in a dynamic field
F⃗ are given by:
d⃗p
= F⃗ . (6.4)
dt
In order to write a system of first order equations, we use the relations
p⃗ p⃗ p⃗
⃗v = = =√ . (6.5)
m E p2 + m20
⁷Of course for lower speeds, the special relativity equations of motion are a better ap-
proximation to the particle’s motion, but the corrections to the non relativistic equations
of motion are negligible.
298 CHAPTER 6. MOTION IN SPACE
z
100
0
-100
3
2
-200 0 1
1 0 y
2
3 -1
x 4
5
6 -2
Figure 6.3: The trajectory of a charged particle in a magnetic field B⃗ = By ŷ + Bz ẑ
with qBy /m = −0.02y, qBz /m = 1 + 0.02z, ⃗v (0) = 1.0ŷ + 0.1ẑ, ⃗r(0) = 1.0x̂. The
integration of the equations of motion is performed using the RK45 method from t0 = 0
to tf = 500 with 10000 steps. Each axis is on a different scale.
Using ⃗v = d⃗r/dt we obtain
dx (px /m0 ) d(px /m0 ) Fx
= √ , =
dt 1 + (p/m0 )2 dt m0
dy (py /m0 ) d(py /m0 ) Fy
= √ , =
dt 1 + (p/m0 )2 dt m0
dz (pz /m0 ) d(pz /m0 ) Fz
= √ , = , (6.6)
dt 1 + (p/m0 )2 dt m0
which is a system of first order differential equations for the functions
(x(t), y(t), z(t), (px /m0 )(t), (py /m0 )(t), (pz /m0 )(t)). Given the initial con-
ditions (x(0), y(0), z(0), (px /m0 )(0), (py /m0 )(0), (pz /m0 )(0)) their solution
is unique and it can be computed numerically using the 4th-5th order
Runge–Kutta method according to the discussion of the previous section.
6.3. RELATIVISTIC MOTION 299
z
2
1
0
-1
-21
y 0
-1 -2 0 1 2 3
-1
x
Figure 6.4: The trajectory of a charged particle in a magnetic field B⃗ = By ŷ + Bz ẑ
with qBy /m = 0.08z, qBz /m = 1.4 + 0.08y, ⃗v (0) = 1.0ŷ + 0.1ẑ, ⃗r(0) = 1.0x̂. The
integration of the equations of motion is performed using the RK45 method from t0 = 0
to tf = 3000 with 40000 steps. Each axis is on a different scale.
By using the relations
vx (px /m0 )
(px /m0 ) = √ vx = √
1 − v2 1 + (p/m0 )2
vy (py /m0 )
(py /m0 ) = √ vy = √
1 − v2 1 + (p/m0 )2
vz (pz /m0 )
(pz /m0 ) = √ vz = √ ,
1 − v2 1 + (p/m0 )2
(6.7)
we can use the initial conditions (x(0), y(0), z(0), vx (0), vy (0), vz (0)) in-
stead. Similarly, from the solutions (x(t), y(t), z(t), (px /m0 )(t), (py /m0 )(t),
(pz /m0 )(t)) we can calculate (x(t), y(t), z(t), vx (t), vy (t), vz (t)). We always
have to check that
v 2 = (vx )2 + (vy )2 + (vz )2 < 1 . (6.8)
Since half of the functions that we integrate for are the momentum instead
of the velocity components, we need to make some modifications to the
300 CHAPTER 6. MOTION IN SPACE
program in the file rk3.f90. The main program can be found in the file
sr.f90:
! ========================================================
! Program t o s o l v e a 6 ODE system using Runge−Kutta Method
! Output i s w r i t t e n i n f i l e s r . dat
! I n t e r f a c e t o be used with r e l a t i v i s t i c p a r t i c l e s .
! ========================================================
program sr_solve
include ’ sr . inc ’
r e a l ( 8 ) : : T0 , TF , X10 , X20 , X30 , V10 , V20 , V30
r e a l ( 8 ) : : P10 , P20 , P30
r e a l ( 8 ) : : P1 , P2 , P3 , V1 , V2 , V3
r e a l ( 8 ) : : t , dt , tstep
i n t e g e r : : STEPS
integer : : i
r e a l ( 8 ) : : energy
! Arrays / v a r i a b l e s needed by r k s u i t e :
r e a l ( 8 ) : : TOL , THRES ( NEQ ) , WORK ( LENWRK ) , Y ( NEQ ) , YMAX ( NEQ ) ,&
YP ( NEQ ) , YSTART ( NEQ ) , HSTART
l o g i c a l : : ERRASS , MESSAGE
i n t e g e r : : UFLAG
! . . External Subroutines . .
EXTERNAL F , SETUP , STAT , UT
! Input :
p r i n t * , ’ Runge−Kutta Method f o r 6−ODEs I n t e g r a t i o n ’
print * , ’ Special R e l a t i v i s t i c Particle : ’
p r i n t * , ’ Enter c o u p l i n g c o n s t a n t s k1 , k2 , k3 , k4 : ’
read * , k1 , k2 , k3 , k4
p r i n t * , ’ k1= ’ , k1 , ’ k2= ’ , k2 , ’ k3= ’ , k3 , ’ k4= ’ , k4
p r i n t * , ’ Enter STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30 : ’
read * , STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30
c a l l momentum ( V10 , V20 , V30 , P10 , P20 , P30 )
p r i n t * , ’No . S t e p s= ’ , STEPS
p r i n t * , ’ Time : I n i t i a l T0 = ’ , T0 , ’ F i n a l TF= ’ , TF
print * , ’ X1(T0)= ’ , X10 , ’ X2(T0)= ’ , X20 , ’ X3(T0)= ’ , X30
print * , ’ V1 (T0)= ’ , V10 , ’ V2(T0)= ’ , V20 , ’ V3(T0)= ’ , V30
print * , ’ P1 (T0)= ’ , P10 , ’ P2 (T0)= ’ , P20 , ’ P3 (T0)= ’ , P30
! I n i t i a l Conditions
dt = ( TF−T0 ) / STEPS
YSTART ( 1 ) = X10
YSTART ( 2 ) = X20
YSTART ( 3 ) = X30
YSTART ( 4 ) = P10
YSTART ( 5 ) = P20
YSTART ( 6 ) = P30
!
6.3. RELATIVISTIC MOTION 301
! S e t e r r o r c o n t r o l parameters .
!
TOL = 5.0 D−6
do i = 1 , NEQ
THRES ( i ) = 1 . 0 D−10
enddo
MESSAGE = .TRUE.
ERRASS = . FALSE .
HSTART = 0.0 D0
! Initialization :
c a l l SETUP ( NEQ , T0 , YSTART , TF , TOL , THRES , METHOD , ’ Usual Task ’ ,&
ERRASS , HSTART , WORK , LENWRK , MESSAGE )
open ( u n i t =11 , f i l e = ’ s r . dat ’ )
c a l l velocity ( YSTART ( 4 ) , YSTART ( 5 ) , YSTART ( 6 ) , V1 , V2 , V3 )
w r i t e ( 1 1 , 1 0 0 ) T0 , YSTART ( 1 ) , YSTART ( 2 ) , YSTART ( 3 ) ,&
V1 , V2 , V3 ,&
energy ( T0 , YSTART ) ,&
YSTART ( 4 ) , YSTART ( 5 ) , YSTART ( 6 )
! Calculation :
do i =1 , STEPS
t = T0 + i * dt
c a l l UT ( F , t , tstep , Y , YP , YMAX , WORK , UFLAG )
i f ( UFLAG . GT. 2 ) e x i t
c a l l velocity ( Y ( 4 ) , Y ( 5 ) , Y ( 6 ) , V1 , V2 , V3 )
w r i t e ( 1 1 , 1 0 0 ) tstep , Y ( 1 ) , Y ( 2 ) , Y ( 3 ) ,&
V1 , V2 , V3 ,&
energy ( tstep , Y ) ,&
Y (4) , Y (5) , Y (6)
enddo
close (11)
100 format ( 1 1 E25 . 1 5 )
end program sr_solve
! ========================================================
! momentum −> v e l o c i t y t r a n s f o r m a t i o n
! ========================================================
s u b r o u t i n e velocity ( p1 , p2 , p3 , v1 , v2 , v3 )
i m p l i c i t none
r e a l ( 8 ) : : v1 , v2 , v3 , p1 , p2 , p3 , v , p , vsq , psq
psq = p1 * p1+p2 * p2+p3 * p3
v1 = p1 / s q r t ( 1 . 0 D0+psq )
v2 = p2 / s q r t ( 1 . 0 D0+psq )
v3 = p3 / s q r t ( 1 . 0 D0+psq )
end s u b r o u t i n e velocity
! ========================================================
! v e l o c i t y −> momentum t r a n s f o r m a t i o n
! ========================================================
s u b r o u t i n e momentum ( v1 , v2 , v3 , p1 , p2 , p3 )
302 CHAPTER 6. MOTION IN SPACE
i m p l i c i t none
r e a l ( 8 ) : : v1 , v2 , v3 , p1 , p2 , p3 , v , p , vsq , psq
vsq = v1 * v1+v2 * v2+v3 * v3
i f ( vsq . ge . 1 . 0 D0 ) s t o p ’ sub momentum : vsq >= 1 ’
p1 = v1 / s q r t ( 1 . 0 D0−vsq )
p2 = v2 / s q r t ( 1 . 0 D0−vsq )
p3 = v3 / s q r t ( 1 . 0 D0−vsq )
end s u b r o u t i n e momentum
The subroutines momentum and velocity compute the transformations
(6.7). In the subroutine momentum we check whether the condition (6.8) is
satisfied. These functions are also used in the subroutine F that computes
the derivatives of the functions.
The test drive of the above program is the well known relativistic
motion of a charged particle in a constant EM field. The acceleration of
the particle is given by equations (6.3). The relativistic kinetic energy of
the particle is
( ) (√ )
1
T = √ − 1 m0 = 1 + (p/m0 )2 − 1 m0 (6.9)
1 − v2
These relations are programmed in the file sr_B.f90. The contents of
the file sr_B.f90 are:
! ========================================================
! P a r t i c l e i n c o n s t a n t Magnetic and e l e c t r i c f i e l d
! q B/m = k1 z q E /m = k2 x + k3 y + k4 z
! ========================================================
s u b r o u t i n e F ( T , Y , YP )
include ’ sr . inc ’
real (8) : : t
r e a l ( 8 ) : : Y ( * ) , YP ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3
x1 = Y ( 1 ) ; p1 = Y ( 4 )
x2 = Y ( 2 ) ; p2 = Y ( 5 )
x3 = Y ( 3 ) ; p3 = Y ( 6 )
c a l l velocity ( p1 , p2 , p3 , v1 , v2 , v3 )
! now we can use a l l x1 , x2 , x3 , p1 , p2 , p3 , v1 , v2 , v3
YP ( 1 ) = v1
YP ( 2 ) = v2
YP ( 3 ) = v3
! Acceleration :
YP ( 4 ) = k2 + k1 * v2
YP ( 5 ) = k3 − k1 * v1
YP ( 6 ) = k4
6.3. RELATIVISTIC MOTION 303
end s u b r o u t i n e F
! ========================================================
! Energy per u n i t r e s t mass
! ========================================================
r e a l ( 8 ) f u n c t i o n energy ( T , Y )
include ’ sr . inc ’
real (8) : : t , e
real (8) : : Y ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 , psq
x1 = Y ( 1 ) ; p1 = Y ( 4 )
x2 = Y ( 2 ) ; p2 = Y ( 5 )
x3 = Y ( 3 ) ; p3 = Y ( 6 )
psq= p1 * p1+p2 * p2+p3 * p3
! K i n e t i c Energy / m_0
e = s q r t ( 1 . 0 D0+psq ) −1.0D0
! P o t e n t i a l Energy / m_0
e = e − k2 * x1 − k3 * x2 − k4 * x3
energy = e
end f u n c t i o n energy
The results are shown in figures 6.5–6.6.
z
2
1.6
1.2
0.8
0.4 0.4
0
0 y
1.2
x 1.6 -0.4
Figure 6.5: The trajectory of a relativistic charged particle in a magnetic field
⃗ = Bz ẑ with qBz /m0 = 10.0, ⃗v (0) = 0.95ŷ + 0.10ẑ, ⃗r(0) = 1.0x̂. The integration is
B
performed by using the RK45 method from t0 = 0 to tf = 20 with 1000 steps. Each
axis is on a different scale.
304 CHAPTER 6. MOTION IN SPACE
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7
Figure 6.6: Projection of the trajectory of a relativistic charged particle in a magnetic
⃗ = Bz ẑ with qBz /m0 = 10.0, on the xy plane. ⃗v (0) = 0.95ŷ + 0.10ẑ, ⃗r(0) = 1.0x̂.
field B
The integration is performed by using the RK45 method from t0 = 0 to tf = 20 with
1000 steps. Each axis is on a different scale.
Now we can study a more interesting problem. Consider a simple
model of the Van Allen radiation belt. Assume that the electrons are
moving within the Earth’s magnetic field which is modeled after a mag-
netic dipole field of the form:
( )3 [ ]
RE
⃗ = B0
B 3(dˆ · r̂) r̂ − dˆ , (6.10)
r
where d⃗ = ddˆ is the magnetic dipole moment of the Earth’s magnetic
field and ⃗r = rr̂. The parameter values are approximately equal to B0 =
3.5 × 10−5 T , r ∼ 2RE , where RE is the radius of the Earth. The typical
energy of the moving√particles is ∼ 1 √MeV which corresponds to velocities
of magnitude v/c = E − m0 /E ≈ 1 − 0.5122 /1 = 0.86. We choose the
2 2
coordinate axes so that dˆ = ẑ and we measure distance in RE units⁸. Then
⁸Since c = 1, the unit of time is the time that the light needs to travel distance equal
to RE in the vacuum.
6.3. RELATIVISTIC MOTION 305
z
100
80
60
40
20 0.4
0 0
1.2 y
x 1.6 -0.4
Figure 6.7: ⃗
The influence of an additional electric field q E/m0 = 1.0ẑ on the
trajectory shown in figure 6.5.
we obtain:
3xz
Bx = B0
r5
3yz
By = B0 5
(r )
3zz 1
Bz = B0 − 3 (6.11)
r5 r
The magnetic dipole field is programmed in the file sr_Bd.f90:
! ========================================================
! P a r t i c l e i n Magnetic d i p o l e f i e l d :
! q B_1 /m = k1 (3 x1 x3 ) / r ^5
! q B_2 /m = k1 (3 x2 x3 ) / r ^5
! q B_3 /m = k1 [ ( 3 x3 x3 ) / r ^5 −1/ r ^3]
! ========================================================
s u b r o u t i n e F ( T , Y , YP )
include ’ sr . inc ’
real (8) : : t
r e a l ( 8 ) : : Y ( * ) , YP ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3
306 CHAPTER 6. MOTION IN SPACE
r e a l ( 8 ) : : B1 , B2 , B3
real (8) : : r
x1 = Y ( 1 ) ; p1 = Y ( 4 )
x2 = Y ( 2 ) ; p2 = Y ( 5 )
x3 = Y ( 3 ) ; p3 = Y ( 6 )
c a l l velocity ( p1 , p2 , p3 , v1 , v2 , v3 )
! now we can use a l l x1 , x2 , x3 , p1 , p2 , p3 , v1 , v2 , v3
YP ( 1 ) = v1
YP ( 2 ) = v2
YP ( 3 ) = v3
! Acceleration :
r = s q r t ( x1 * x1+x2 * x2+x3 * x3 )
i f ( r . g t . 0 . 0 D0 ) then
B1 = k1 * ( 3.0 D0 * x1 * x3 ) / r * * 5
B2 = k1 * ( 3.0 D0 * x2 * x3 ) / r * * 5
B3 = k1 * ( ( 3 . 0 D0 * x3 * x3 ) / r **5 −1/ r * * 3 )
YP ( 4 ) = v2 * B3−v3 * B2
YP ( 5 ) = v3 * B1−v1 * B3
YP ( 6 ) = v1 * B2−v2 * B1
else
YP ( 4 ) = 0.0 D0
YP ( 5 ) = 0.0 D0
YP ( 6 ) = 0.0 D0
endif
end s u b r o u t i n e F
! ========================================================
! Energy per u n i t r e s t mass
! ========================================================
r e a l ( 8 ) f u n c t i o n energy ( T , Y )
include ’ sr . inc ’
real (8) : : t , e
real (8) : : Y ( * )
r e a l ( 8 ) : : x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 , psq
x1 = Y ( 1 ) ; p1 = Y ( 4 )
x2 = Y ( 2 ) ; p2 = Y ( 5 )
x3 = Y ( 3 ) ; p3 = Y ( 6 )
psq= p1 * p1+p2 * p2+p3 * p3
! K i n e t i c Energy / m_0
e = s q r t ( 1 . 0 D0+psq ) −1.0D0
energy = e
end f u n c t i o n energy
The results are shown in figure 6.8. The parameters have been exag-
gerated in order to achieve an aesthetically pleasant result. In reality, the
electrons are moving in very thin spirals and the reader is encouraged to
use more realistic values for the parameters ⃗v0 , B0 , ⃗r0 . The problem of
why the effect is not seen near the equator is left as an exercise.
6.4. PROBLEMS 307
2
1.6
1.2
0.8
0.4
0
0.004
0
-0.004
-0.008 0.02
0.012 0.016
-0.012 0 0.004 0.008
Figure 6.8: The trajectory of a charged particle in a magnetic dipole field given by
equation (6.11). We used B0 = 1000, ⃗r = 0.02x̂ + 2.00ẑ, ⃗v = −0.99999ẑ. The integration
was done from t0 = 0 to tf = 5 in 10000 steps.
6.4 Problems
6.1 Compute the trajectory of a projectile moving in space in a con-
stant gravitational field and under the influence of an air resistance
proportional to the square of its speed.
6.2 Two point charges are moving with non relativistic speeds in a
constant magnetic field B ⃗ = B ẑ. Assume that their interaction is
given by the Coulomb force only. Write a program that computes
their trajectory numerically using the RK45 method.
6.3 Write a program that computes the trajectory of the anisotropic
harmonic oscillator F⃗ = −kx xx̂ −ky y ŷ −kz z ẑ. Compute the three
dimensional Lissajous curves which
√ appear for √ appropriate√values
of the angular frequencies ωx = kx /m, ωy = ky /m, ωz = kz /m.
6.4 Two particles of mass M are at the fixed positions ⃗r1 = aẑ and
⃗r2 = −aẑ. A third particle of mass m interacts with them via a
Newtonian gravitational force and moves at non relativistic speeds.
308 CHAPTER 6. MOTION IN SPACE
Compute the particle’s trajectory and find initial conditions that
result in a planar motion.
6.5 Solve problem 5.19 of page 283 using the RK45 method. Choose
initial conditions so that the system executes only translational mo-
tion. Next, choose initial conditions so that the system executes
small vibrations and its center of mass remains stationary. Find the
normal modes of the system and choose appropriate initial condi-
tions that put the system in each one of them.
6.6 Solve the previous problem by putting the system in a box |x| ≤ L
and |y| ≤ L.
Hint: Look in the file springL.f90.
6.7 Solve the problem 5.20 in page 284 by using the RK45 method.
6.8 Solve the problem 5.21 in page 284 by using the RK45 method.
6.9 The electric field of an electric dipole p⃗ = pẑ is given by:
E⃗ = Eρ ρ̂ + Ez ẑ
1 3p sin θ cos θ
Eρ =
4πϵ0 r3
1 p(3 cos2 θ − 1)
Ez = (6.12)
4πϵ0 r3
√
where ρ = x2 + y 2 = r sin θ, Ex = Eρ cos ϕ, Ey = Eρ sin ϕ and
(r, θ, ϕ) are the polar coordinates of the point where the electric field
is calculated. Calculate the trajectory of a test charge moving in this
field at non relativistic speeds. Calculate the deviation between the
relativistic and the non relativistic trajectories when the initial speed
is 0.01c, 0.1c, 0.5c, 0.9c respectively (ignore radiation effects).
6.10 Consider a linear charge distribution with constant linear charge
density λ. The electric field is given by
⃗ = Eρ ρ̂ = 1 2λ
E ρ̂
4πϵ0 ρ
Calculate the trajectories of two equal negative test charges that
move at non relativistic speeds in this field. Consider only the
electrostatic Coulomb forces and ignore anything else.
6.4. PROBLEMS 309
6.11 Consider a linear charge distribution on four straight lines parallel
to the z axis. The linear charge density is λ and it is constant. The
four straight lines intersect the xy plane at the points (0, 0), (0, a),
(a, 0), (a, a). Calculate the trajectory of a non relativistic charge
in this field. Next, compute the relativistic trajectories (ignore all
radiation effects).
6.12 Three particles of mass m interact via their Newtonian gravitational
force. Compute their (non relativistic) trajectories in space.
310 CHAPTER 6. MOTION IN SPACE
Chapter 7
Electrostatics
In this chapter we will study the electric field generated by a static charge
distribution. First we will compute the electric field lines and the equipo-
tential surfaces of the electric field generated by a static point charge dis-
tribution on the plane. Then we will study the electric field generated by
a continuous charge distribution on the plane. This requires the numer-
ical solution of an elliptic boundary value problem which will be done
using successive over-relaxation (SOR) methods.
7.1 Electrostatic Field of Point Charges
Consider N point charges Qi which are located at fixed positions on the
plane given by their position vectors ⃗ri , i = 1, . . . , N . The electric field is
given by Coulomb’s law
1 ∑ Qi
N
⃗ r) =
E(⃗ ρ̂i (7.1)
4πϵ0 i=1 |⃗r − ⃗ri |2
where ρ̂i = (⃗r − ⃗ri )/|⃗r − ⃗ri | is the unit vector in the direction of ⃗r − ⃗ri . The
components of the field are
1 ∑
N
Qi (x − xi )
Ex (x, y) =
4πϵ0 i=1 ((x − xi )2 + (y − yi )2 )3/2
1 ∑
N
Qi (y − yi )
Ey (x, y) = , (7.2)
4πϵ0 i=1 ((x − xi )2 + (y − yi )2 )3/2
311
312 CHAPTER 7. ELECTROSTATICS
The electrostatic potential at ⃗r is
1 ∑
N
Qi
V (⃗r) = V (x, y) = , (7.3)
4πϵ0 i=1 ((x − xi )2 + (y − yi )2 )1/2
and we have that
⃗ r) = −∇V
E(⃗ ⃗ (⃗r) . (7.4)
The electric field lines are the integral curves of the vector field E, ⃗ i.e.
the curves whose tangent lines at each point are parallel to the electric
field at that point. The magnitude of the electric field is proportional to
the density of the field lines (the number of field lines per perpendicular
∫
area). This means that the electric flux ΦE = S E ⃗ · dA
⃗ through a surface
S is proportional to the number of field lines that cross the surface.
Electric field lines of point charge distributions start from positive charges
(sources), end in negative charges (sinks) or extend to infinity.
The equipotential surfaces are the loci of the points of space where
the electrostatic potential takes fixed values. Τhey are closed surfaces.
Equation (7.4) tells us that a strong electric field at a point is equivalent to
a strong spatial variation of the electric potential at this point, i.e. to dense
equipotential surfaces. The direction of the electric field is perpendicular
to the equipotential surfaces at each point¹, which is the direction of
the strongest spatial variation of V , and it points in the direction of
decreasing V . The planar cross sections of the equipotential surfaces are
closed curves which are called equipotential lines.
The computer cannot solve a problem in the continuum and we have
to consider a finite discretization of a field line. A continuous curve is
approximated by a large but finite number of small line segments. The
basic idea is illustrated in figure 7.1: The small line segment ∆l is taken
in the direction of the electric field and we obtain
Ex Ey
∆x = ∆l , ∆y = ∆l , (7.5)
E E
√
where E ≡ |E|⃗ = Ex2 + Ey2 .
In order to calculate the equipotential lines we use the property that
they are perpendicular to the electric field at each point. Therefore, if
(∆x, ∆y) is in the tangential direction of a field line, then (−∆y, ∆x) is
¹Since for every small displacement d⃗r along an equipotential surface the potential
stays constant (dV = 0), we have that 0 = dV = ∇V ⃗ · d⃗r = −E ⃗ · d⃗r, which implies
E ⊥ d⃗r.
⃗
7.1. ELECTROSTATIC FIELD OF POINT CHARGES 313
y E
Ey
∆l
∆y
∆x Ex
x
Figure 7.1: The electric field is tangent at each point of an electric field line and
perpendicular to an equipotential line. By approximating the continuous curve by the
line segment ∆l, we have that ∆y/∆x = Ey /Ex .
in the perpendicular direction since (∆x, ∆y) · (−∆y, ∆x) = −∆x∆y +
∆y∆x = 0. Therefore the equations that give the equipotential lines are
Ey Ex
∆x = −∆l , ∆y = ∆l . (7.6)
E E
The algorithm that will allow us to perform an approximate calcula-
tion of the electric field lines and the equipotential lines is the following:
Choose an initial point that belongs to the (unique) line that you want to
draw. The electric field can be calculated from the known electric charge
distribution and equation (7.2). By using a small enough step ∆l we
move in the direction (∆x, ∆y) to the new position
x → x + ∆x , y → y + ∆y , (7.7)
where we use equations (7.5) or (7.6). The procedure is repeated until
the drawing is finished. The programmer sets a criterion for that, e.g.
when the field line steps out of the drawing area or approaches a charge
closer than a minimum distance.
314 CHAPTER 7. ELECTROSTATICS
7.2 The Program – Appetizer and ... Desert
The hurried, but slightly experienced reader may skip the details of this
section and go directly to section 7.4. There she can find the final form
of the program and brief usage instructions.
In order to program the algorithm described in the previous section,
we will separate the algorithmic procedures into four different but well
defined tasks:
• Main program: The data structure, which is given by the position of
the charges stored in the arrays X(P), Y(P) and the charges stored
in the array Q(P), is defined. It also contains the user interface
which consists of reading data entered by the user, like the number
of charges N, their positions and magnitude. Then the calculation
of a group of field or equipotential lines is performed by calling the
routines eline or epotline respectively.
• subroutine eline(xin,yin,X,Y,Q,N): Calculates the electric field
line passing through the point xin,yin. On entry, the user inputs
the point xin,yin and the data N, X(N), Y(N), Q(N). On exit, the
subroutine prints to the stdout the coordinates of the approximate
electric field line. The line extends up to a point that is either
too close to one of the point charges or until the line leaves the
drawing area². It calls the subroutines efield for the calculation of
the electric field and mdist for the calculation of the minimum and
maximum distance of a point on the field line from all the point
charges.
• subroutine epotline(xin,yin,X,Y,Q,N): Calculates the equipoten-
tial line passing through the point xin,yin. On entry, the user
inputs the point xin,yin and the data N, X(N), Y(N), Q(N). On
exit, the subroutine prints to the stdout the coordinates of the ap-
proximate equipotential line. The subroutine stops calculating an
equipotential line when it comes back close enough to the original
point³ xin,yin or when it leaves the drawing area. It calls the sub-
routines efield for the calculation of the electric field and mdist for
the calculation of the minimum and maximum distance of a point
on the equipotential line from all the point charges.
²Remember that field lines start at sources, end at sinks or extend to infinity.
³Remember that the equipotential lines are closed.
7.2. THE PROGRAM – APPETIZER AND ... DESERT 315
• subroutine efield(x0,y0,X,Y,Q,N,Ex,Ey): Calculates the electric
field Ex, Ey at position x0, y0. On entry, the user provides the
number of charges N, the position of charges X(N), Y(N), the charges
Q(N) and the position x0, y0. On exit, the routine provides the
values Ex, Ey.
• subroutine mdist(x0,y0,X,Y,N,rmin,rmax): Calculates the maxi-
mum and minimum distance of the point x0, y0 from all charges
located at X(N), Y(N). On entry, the user provides the number of
charges N, the position of charges X(N), Y(N) and the point x0, y0.
On exit, the routine provides the minimum and maximum distances
rmin,rmax.
In the main program, the variables N, X(N), Y(N) and Q(N) must
be set. These can be hard coded by the programmer or entered by the
user interactively. The first choice is coded in the program listed below,
which can be found in the file ELines.f90. This is version 1 of the main
program:
! ****************************************************
program Electric_Fields
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
N = 2
X (1) = 1.0
Y ( 1 ) = 0.0
Q (1) = 1.0
X ( 2 ) = −1.0
Y ( 2 ) = 0.0
Q ( 2 ) = −1.0
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
c a l l eline ( 0 . 0 , 0 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 1 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 1 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 2 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 0 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 1 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 1 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 2 . 0 , X , Y , Q , N )
end program Electric_Fields
316 CHAPTER 7. ELECTROSTATICS
The commands
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
N = 2
X (1) = 1.0
Y ( 1 ) = 0.0
Q (1) = 1.0
X ( 2 ) = −1.0
Y ( 2 ) = 0.0
Q ( 2 ) = −1.0
define two opposite charges Q(1)= -Q(2)= 1.0 located at (1, 0) and (−1, 0)
respectively. The next lines call the subroutine eline in order to per-
form the calculation of 8 field lines passing through the points (0, ±1/2),
(0, ±1), (0, ±3/2), (0, ±2):
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
c a l l eline ( 0 . 0 , 0 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 1 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 1 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , 2 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 0 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 1 . 0 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 1 . 5 , X , Y , Q , N )
c a l l eline ( 0 . 0 , − 2 . 0 , X , Y , Q , N )
These commands print the coordinates of the field lines to the stdout
and the user can analyze them further.
The program for calculating the equipotential lines is quite similar.
The calls to the subroutine eline are substituted by calls to epotline.
For the program to be complete, we must program the subroutines
eline, efield, mdist. This will be done later, and you can find the
full code in the file ELines.f90. For the moment, let’s copy the main
program⁴ listed above into the file Elines.f90 and compile and run it
with the commands:
> g f o r t r a n ELines . f90 −o el
> . / el > el . out
The stdout of the program is redirected to the file el.out. We can plot
the results with gnuplot:
⁴See the file ELines_version0.f90.
7.2. THE PROGRAM – APPETIZER AND ... DESERT 317
gnuplot > plot ” e l . out ” with dots
The result is shown in figure 7.2.
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1.5 -1 -0.5 0 0.5 1 1.5
Figure 7.2: Some electric field lines of the electric field of two opposite charges
calculated by the program ELines.f90 (version 1!).
Let’s modify the program so that the user can enter the charge dis-
tribution, as well as the number and position of the field lines that she
wants to draw, interactively. The part of the code that we need to change
is:
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
p r i n t * , ’ # Enter number o f c h a r g e s : ’
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
The first line asks the user to enter the number of charges in the distri-
bution. It proceeds with reading it from the stdin and prints the result
318 CHAPTER 7. ELECTROSTATICS
to the stdout. The following loop reads the positions and charges and
stores them at the position i of the arrays X(i), Y(i), Q(i). The results
are printed to the stdout so that the user can check the values read by
the program.
The drawing of the field lines is now done by modifying the code so
that:
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
p r i n t * , ’ # How many l i n e s t o draw? ’
read * , draw
do i =1 , draw
p r i n t * , ’ # I n i t i a l p o i n t ( x0 , y0 ) : ’
read * , x0 , y0
c a l l eline ( x0 , y0 , X , Y , Q , N )
enddo
As a test case, we run the program for one charge q = 1.0 located at the
origin and we draw one field line passing through the point (0.1, 0.1).
> gfortran ELines . f90 −o el
> . / el
# Enter number of charges :
1
# N= 1
# Charge : 1
# Position and charge : ( X , Y , Q ) :
0.0 0.0 1 . 0
# ( X , Y )= 0.000000 0.000000 Q= 1.000000
# How many lines to draw ?
1
# Initial point ( x 0 , y 0) :
0.1 0.1
9.2928931E−02 9.2928931E−02
8.5857861E−02 8.5857861E−02
7.8786790 E−02 7.8786790 E−02
....
For charge distributions with a large number of point charges, use an
editor to record the charges, their positions and the points where the field
lines should go through.
2 N : Number of Charges
1 . 0 0.0 1 . 0 ( X , Y , Q ) : Position and charge
−1.0 0.0 −1.0 ( X , Y , Q ) : Position and charge
8 Number of lines to draw
7.2. THE PROGRAM – APPETIZER AND ... DESERT 319
0.0 0.5 x0 ,y 0: Initial point of line
0.0 1.0 x0 ,y 0: Initial point of line
0.0 1.5 x0 ,y 0: Initial point of line
0.0 2.0 x0 ,y 0: Initial point of line
0.0 −0.5 x0 ,y 0: Initial point of line
0.0 −1.0 x0 ,y 0: Initial point of line
0.0 −1.5 x0 ,y 0: Initial point of line
0.0 −2.0 x0 ,y 0: Initial point of line
If the data listed above is written into a file, e.g. Input, then the com-
mand
. / el < Input > el . out
reads the data from the file Input and redirects the data printed to the
stdout to the file el.out. This way you can create a “library” of charge
distributions and the field lines of their respective electric fields. The
complete code (version 2) is listed below:
! ****************************************************
program Electric_Fields
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
integer : : i , j , draw
real : : x0 , y0
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
p r i n t * , ’ # Enter number o f c h a r g e s : ’
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
p r i n t * , ’ # How many l i n e s t o draw? ’
read * , draw
do i =1 , draw
p r i n t * , ’ # I n i t i a l p o i n t ( x0 , y0 ) : ’
read * , x0 , y0
c a l l eline ( x0 , y0 , X , Y , Q , N )
enddo
end program Electric_Fields
320 CHAPTER 7. ELECTROSTATICS
If you did the exercises described above, you should have already
realized that in order to draw a nice representative picture of the electric
field can be time consuming. For field lines one can use simple physical
intuition in order to automate the procedure. For distances close enough
to a point charge the electric field is approximately isotropic. The number
of field lines crossing a small enough curve which contains only the
charge is proportional to the charge (Gauss’s law). Therefore we can
draw a small circle centered around each charge and choose initial points
isotropically distributed on the circle as initial points of the field lines.
The code listed below (version 3) implements the idea for charges that
are equal in magnitude. For charges different in magnitude, the program
is left as an exercise to the reader.
! ****************************************************
program Electric_Fields
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
integer : : i , j , nd
real : : x0 , y0 , theta
r e a l , parameter : : PI= 3.14159265359
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
p r i n t * , ’ # Enter number o f c h a r g e s : ’
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
!We draw 2*nd f i e l d l i n e s around each charge
nd = 6
do i = 1 , N
do j = 1 , ( 2 * nd )
theta = ( PI / nd ) * j
x0 = X ( i ) + 0 . 1 * c o s ( theta )
y0 = Y ( i ) + 0 . 1 * s i n ( theta )
c a l l eline ( x0 , y0 , X , Y , Q , N )
enddo
enddo
7.2. THE PROGRAM – APPETIZER AND ... DESERT 321
end program Electric_Fields
We set the number of field lines around each charge to be equal to
12 (nd=6). The initial points are taken on the circle whose center is
(X(i),Y(i)) and its radius is 0.1. The 2*nd points are determined by
the angle theta=(PI/nd)*j.
We record the data of a charge distribution in a file, e.g. Input. Below,
we list the example of four equal charges qi = ±1 located at the vertices
of a square:
4 N : Number of charges
1 1 −1 ( X , Y , Q ) : Position and charge
−1 1 1 ( X , Y , Q ) : Position and charge
1 −1 1 ( X , Y , Q ) : Position and charge
−1 −1 −1 ( X , Y , Q ) : Position and charge
Then we give the commands:
> gfortran ELines . f90 −o el
> . / el < Input > el . out
> gnuplot
gnuplot > p l o t ” e l . out ” with dots
The results are shown in figures 7.3 and 7.4. The reader should deter-
mine the charge distributions that generate those fields and reproduce
the figures as an exercise.
For the computation of the equipotential lines we can work in a similar
way. We will follow a quick and dirty way which will not produce an
accurate picture of the electric field and choose the initial points evenly
spaced on an square lattice. For a better choice see problem 5. The listed
code is in the file EPotential.f90:
! ****************************************************
program Electric_Potential
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
r e a l , parameter : : PI= 3.14159265359
integer : : i , j , nd
real : : x0 , y0 , rmin , rmax , L
p r i n t * , ’ # Enter number o f c h a r g e s : ’
322 CHAPTER 7. ELECTROSTATICS
4
3
2
1
0
-1
-2
-3
-4
-4 -3 -2 -1 0 1 2 3 4
4 4
3 3
2 2
1 1
0 0
-1 -1
-2 -2
-3 -3
-4 -4
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 7.3: Field lines of a static charge distribution of point charges generated by
the program ELines.f90.
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
!We draw l i n e s p a s s i n g through an e q u a l l y
! spaced l a t t i c e o f N=(2*nd+1) x ( 2 * nd+1) p o i n t s
! i n t h e square −L<= x <= L , −L<= y <= L .
nd = 4
L = 1.0
do i = −nd , nd
do j = −nd , nd
x0 = i * ( L / nd )
y0 = j * ( L / nd )
p r i n t * , ’ # @ ’ , i , j , L / nd , x0 , y0
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
! we avoid g e t t i n g t o o c l o s e t o a charge :
i f ( rmin . g t . L / ( nd * 1 0 ) )&
c a l l epotline ( x0 , y0 , X , Y , Q , N )
7.3. THE PROGRAM – MAIN DISH 323
4 4
3 3
2 2
1 1
0 0
-1 -1
-2 -2
-3 -3
-4 -4
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
Figure 7.4: Field lines of a static charge distribution of point charges generated by
the program ELines.f90.
enddo
enddo
end program Electric_Potential
The first and second part of the code is identical to the previous one. In
the third part we call the subroutine epotline for drawing an equipo-
tential line for each initial point. The initial points are on a square lattice
with (2*nd+1)*(2*nd+1)= 81 points (nd=4). The lattice extends within
the limits set by the square (1, 1), (−1, 1), (−1, −1), (1, −1) (L=1.0). For
each point (x0,y0) we calculate the equipotential line that passes through
it. We check that this point is not too close to one of the charges by calling
the subroutine mdist. The call determines the minimum distance rmin
of the point from all the charges which should be larger than L/(nd*10).
You can run the program with the commands:
> g f o r t r a n EPotential . f90 −o ep
> . / ep < Input > ep . out
> gnuplot
gnuplot > p l o t ” ep . out ” with dots
Some of the results are shown in figure 7.5.
7.3 The Program – Main Dish
In this section we look under the hood and give the details of the inner
parts of the program: The subroutines eline and epotline that calculate
the field and equipotential lines, the subroutine efield that calculates
the electric field at a point and the subroutine mdist that calculates the
324 CHAPTER 7. ELECTROSTATICS
1.5
1
0.5
0
-0.5
-1
-1.5
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
10
3
2
5
1
0 0
-1
-5
-2
-3
-10
-3 -2 -1 0 1 2 3 -10 -5 0 5 10
Figure 7.5: Equipotential lines of the electric field generated by a point charge distri-
bution on the plane calculated by the program in EPotential.f90. Beware: the density
of the lines is not correctly calculated and it is not proportional to the magnitude of the
electric field. See problem 7.5.
minimum and maximum distances of a point from the point charges.
The subroutine eline is called by the command:
call eline ( x0 , y0 , X , Y , Q , N )
The input to the routine is the initial point (x0,y0), the number of
charges N, the positions of the charges (X(N),Y(N)) and the charges Q(N).
The routine needs some parameters in order to draw the field line. These
are “hard coded”, i.e. set to fixed values by the programmer that cannot
be changed by the user that calls the routine in her program. One of
them is the step ∆l of equation (7.5) which sets the discretization step
of the field line. It also sets the minimum distance of approaching to
a charge equal to 2∆l. The size of the drawing area of the curve is set
by the parameter max_dist=20.0. We should also provide a check in
the program that checks whether the electric field is zero, in which case
the result of the calculation in equation (7.5) becomes indeterminate. By
taking ∆l > 0, the motion is in the direction of the electric field, which
ends on a negative charge or outside the drawing area (why?). In order
7.3. THE PROGRAM – MAIN DISH 325
to draw the line in both directions, set ∆l < 0 and repeat the calculation.
The code is listed below:
! ****************************************************
s u b r o u t i n e eline ( xin , yin , X , Y , Q , N )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : xin , yin , x0 , y0
r e a l , parameter : : step =0.01
r e a l , parameter : : max_dist =20.0
integer : : i , direction
real : : rmin , rmax , r , dx , dy , dl
real : : Ex , Ey , E
do direction = −1 ,1 ,2 ! d i r e c t i o n = +/− 1
dl = direction * step
x0 = xin
y0 = yin
dx = 0.0
dy = 0.0
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
do while ( rmin . g t . ( 2 . 0 * step ) . and . rmax . l t . max_dist )
p r i n t * , x0 , y0
! We e v a l u a t e t h e E−f i e l d a t t h e midpoint : This r e d u c e s
! systematic errors
c a l l efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey )
E = s q r t ( Ex * Ex+Ey * Ey )
i f ( E . l e . 1 . 0 e−10 ) e x i t
dx = dl * Ex / E
dy = dl * Ey / E
x0 = x0 + dx
y0 = y0 + dy
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
enddo ! do while ( )
enddo ! do d i r e c t i o n = −1 ,1 ,2
end s u b r o u t i n e eline
In the first part of the code we have the variable declarations. We only
note the declaration
r e a l , dimension ( N ) : : X , Y , Q
which declares the dimension of the arrays to be N instead of their true
dimension P. This is fine, as long as the programmer of the calling pro-
gram has already checked that N ≤ P. The necessary memory for the
326 CHAPTER 7. ELECTROSTATICS
arrays is allocated in the calling program and the declaration does not
provide new storage space. The arrays X,Y, Q are passed to the sub-
routine “by reference”, i.e. the routine learns about their position in the
memory to which it can refer to, and not “by value”. The parameters
∆l = step and max_dist are fixed by the parameter attribute:
r e a l , parameter : : step =0.01
r e a l , parameter : : max_dist =20.0
Their values should be the result of a careful study by the programmer
since they determine the accuracy of the calculation.
The outmost loop
do direction = −1 ,1 ,2
dl = direction * step
...
enddo
sets the direction of motion on the field line (i.e. the sign of ∆l). The
command do direction = -1,1,2 executes the loop twice by setting the
variable direction to take values from −1 to 1 with step equal to 2.
The commands x0 = xin, y0 = yin define the initial point on the
field line. (x0, y0) is the current point on the field line which is printed
to the stdout with the command print. The variables (dx, dy) set the
step (x0, y0) → (x0+dx, y0+dy). The drawing of the field line is done
in the inner loop
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
do while ( rmin . g t . ( 2 . 0 * step ) . and . rmax . l t . max_dist )
...
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
enddo
which is executed provided that the logical expression (rmin .gt. (2.0*step)
.and. rmax .lt. max_dist) is .TRUE. This happens as long as the
current point is at a distance greater than 2.0*step and the maximum
distance from all charges is less than max_dist⁵. The minimum and
maximum distances are calculated by calling the subroutine mdist.
The electric field, needed in equation (7.5), is calculated by a call to
efield(x0+0.5*dx,y0+0.5*dy,X,Y,Q,N,Ex,Ey). The first two arguments
give the point at which we want to calculate the electric field, which is
⁵The choice is not unique of course, you may also try e.g. rmin .lt. max_dist.
7.3. THE PROGRAM – MAIN DISH 327
chosen to be the midpoint (x0+dx/2,y0+dy/2) instead of (x0,y0). This
improves the stability and the accuracy of the algorithm.
Equation (7.5) is coded in the commands
E = s q r t ( Ex * Ex+Ey * Ey )
dx = dl * Ex / E
dy = dl * Ey / E
x0 = x0 + dx
y0 = y0 + dy
We also perform checks for the cases E=0.0 and dx=dy=0.0:
i f ( E . l e . 1 . 0 e−10 ) e x i t
When the magnitude of the electric field becomes too small we stop the
calculation by exiting the loop with the command exit. The reader can
improve the code by adding more checks of singular cases.
The subroutine epotline is programmed in a similar way. The rele-
vant code is listed below:
! ****************************************************
s u b r o u t i n e epotline ( xin , yin , X , Y , Q , N )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : xin , yin , x0 , y0
r e a l , parameter : : step =0.02
r e a l , parameter : : max_dist =20.0
integer :: i
real : : r , dx , dy , dl
real : : Ex , Ey , E
dl = step
x0 = xin
y0 = yin
dx = 0.0
dy = 0.0
r = step ! i n order t o s t a r t loop
do while ( r . g t . ( 0 . 9 * dl ) . and . r . l t . max_dist )
p r i n t * , x0 , y0
! We e v a l u a t e t h e E−f i e l d a t t h e midpoint : This r e d u c e s
! systematic errors
c a l l efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey )
E = s q r t ( Ex * Ex+Ey * Ey )
328 CHAPTER 7. ELECTROSTATICS
i f ( E . l e . 1 . 0 e−10 ) e x i t
dx = dl * Ey / E
dy = −dl * Ex / E
x0 = x0 + dx
y0 = y0 + dy
r = s q r t ( ( x0−xin ) * * 2 + ( y0−yin ) * * 2 )
enddo ! do while ( )
end s u b r o u t i n e epotline
The differences are minor: The equipotential lines are closed curves,
therefore we only need to transverse them in one direction. The criterion
for ending the calculation is to approach the initial point close enough
or leave the drawing area:
do while ( r . g t . ( 0 . 9 * dl ) . and . r . l t . max_dist )
...
enddo
The values of dx, dy are calculated according to equation (7.6):
dx = dl * Ey / E
dy = −dl * Ex / E
The subroutine efield is an application of equations⁶ (7.2):
! ****************************************************
s u b r o u t i n e efield ( x0 , y0 , X , Y , Q , N , Ex , Ey )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : x0 , y0 , dx , dy , Ex , Ey
integer :: i
real : : r3 , xi , yi
Ex = 0.0
Ey = 0.0
do i= 1 ,N
xi = x0−X ( i )
yi = y0−Y ( i )
r3 = ( xi * xi+yi * yi ) * * ( − 1 . 5 )
Ex = Ex + Q ( i ) * xi * r3
Ey = Ey + Q ( i ) * yi * r3
enddo
⁶You may improve the program by checking whether ri = 0.
7.4. THE PROGRAM - CONCLUSION 329
end s u b r o u t i n e efield
Finally, the subroutine mdist calculates the minimum and maximum
distance rmin and rmax of a point (x0,y0) from all the point charges in
the distribution:
! ****************************************************
s u b r o u t i n e mdist ( x0 , y0 , X , Y , N , rmin , rmax )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y
real : : x0 , y0 , rmin , rmax
integer :: i
real :: r
rmax = 0.0
rmin = 1000.0
do i = 1 , N
r = s q r t ( ( x0−X ( i ) ) * * 2 + ( y0−Y ( i ) ) * * 2 )
i f ( r . GT . rmax ) rmax = r
i f ( r . LT . rmin ) rmin = r
enddo
end s u b r o u t i n e mdist
The initial value of rmin depends of the limits of the drawing area (why?).
7.4 The Program - Conclusion
In this section we list the programs discussed in the previous sections and
provide short usage information for compiling, running and analyzing
your results. You can jump into this section without reading the previous
ones and go back to them if you need to clarify some points that you
find hard to understand.
First we list the contents of the file ELines.f90:
! ****************************************************
program Electric_Fields
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
integer : : i , j , nd
330 CHAPTER 7. ELECTROSTATICS
real : : x0 , y0 , theta
r e a l , parameter : : PI= 3.14159265359
!−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−−
p r i n t * , ’ # Enter number o f c h a r g e s : ’
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
!We draw 2*nd f i e l d l i n e s around each charge
nd = 6
do i = 1 , N
do j = 1 , ( 2 * nd )
theta = ( PI / nd ) * j
x0 = X ( i ) + 0 . 1 * c o s ( theta )
y0 = Y ( i ) + 0 . 1 * s i n ( theta )
c a l l eline ( x0 , y0 , X , Y , Q , N )
enddo
enddo
end program Electric_Fields
! ****************************************************
s u b r o u t i n e eline ( xin , yin , X , Y , Q , N )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : xin , yin , x0 , y0
r e a l , parameter : : step =0.01
r e a l , parameter : : max_dist =20.0
integer : : i , direction
real : : rmin , rmax , r , dx , dy , dl
real : : Ex , Ey , E
do direction = −1 ,1 ,2 ! d i r e c t i o n = +/− 1
dl = direction * step
x0 = xin
y0 = yin
dx = 0.0
dy = 0.0
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
do while ( rmin . g t . ( 2 . 0 * step ) . and . rmax . l t . max_dist )
p r i n t * , x0 , y0
! We e v a l u a t e t h e E−f i e l d a t t h e midpoint : This r e d u c e s
! systematic errors
c a l l efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey )
E = s q r t ( Ex * Ex+Ey * Ey )
7.4. THE PROGRAM - CONCLUSION 331
i f ( E . l e . 1 . 0 e−10 ) e x i t
dx = dl * Ex / E
dy = dl * Ey / E
x0 = x0 + dx
y0 = y0 + dy
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
enddo ! do while ( )
enddo ! do d i r e c t i o n = −1 ,1 ,2
end s u b r o u t i n e eline
! ****************************************************
s u b r o u t i n e efield ( x0 , y0 , X , Y , Q , N , Ex , Ey )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : x0 , y0 , dx , dy , Ex , Ey
integer :: i
real : : r3 , xi , yi
Ex = 0.0
Ey = 0.0
do i= 1 , N
xi = x0−X ( i )
yi = y0−Y ( i )
! E x e r c i s e : Improve code so t h a t x i * x i + y i * y i =0 i s taken c a r e o f
r3 = ( xi * xi+yi * yi ) * * ( − 1 . 5 )
Ex = Ex + Q ( i ) * xi * r3
Ey = Ey + Q ( i ) * yi * r3
enddo
end s u b r o u t i n e efield
! ****************************************************
s u b r o u t i n e mdist ( x0 , y0 , X , Y , N , rmin , rmax )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y
real : : x0 , y0 , rmin , rmax
integer :: i
real :: r
rmax = 0.0
rmin = 1000.0
do i = 1 , N
r = s q r t ( ( x0−X ( i ) ) * * 2 + ( y0−Y ( i ) ) * * 2 )
i f ( r . GT . rmax ) rmax = r
i f ( r . LT . rmin ) rmin = r
enddo
end s u b r o u t i n e mdist
332 CHAPTER 7. ELECTROSTATICS
Then we list the contents of the file EPotential.f90:
! ****************************************************
program Electric_Potential
! ****************************************************
i m p l i c i t none
i n t e g e r , parameter : : P=20 ! max number o f c h a r g e s
r e a l , dimension ( P ) : : X , Y , Q
integer :: N
r e a l , parameter : : PI= 3.14159265359
integer : : i , j , nd
real : : x0 , y0 , rmin , rmax , L
p r i n t * , ’ # Enter number o f c h a r g e s : ’
read * , N
p r i n t * , ’ # N= ’ , N
do i =1 , N
p r i n t * , ’ # Charge : ’ , i
p r i n t * , ’ # P o s i t i o n and charge : (X, Y , Q) : ’
read * , X ( i ) , Y ( i ) , Q ( i )
p r i n t * , ’ # (X, Y)= ’ , X ( i ) , Y ( i ) , ’ Q= ’ , Q ( i )
enddo
!−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−−
!We draw l i n e s p a s s i n g through an e q u a l l y
! spaced l a t t i c e o f N=(2*nd+1) x ( 2 * nd+1) p o i n t s
! i n t h e square −L<= x <= L , −L<= y <= L .
nd = 4
L = 1.0
do i = −nd , nd
do j = −nd , nd
x0 = i * ( L / nd )
y0 = j * ( L / nd )
p r i n t * , ’ # @ ’ , i , j , L / nd , x0 , y0
c a l l mdist ( x0 , y0 , X , Y , N , rmin , rmax )
! we avoid g e t t i n g t o o c l o s e t o a charge :
i f ( rmin . g t . L / ( nd * 1 0 ) )&
c a l l epotline ( x0 , y0 , X , Y , Q , N )
enddo
enddo
end program Electric_Potential
! ****************************************************
s u b r o u t i n e epotline ( xin , yin , X , Y , Q , N )
! ****************************************************
i m p l i c i t none
integer :: N
r e a l , dimension ( N ) : : X , Y , Q
real : : xin , yin , x0 , y0
r e a l , parameter : : step =0.02
7.4. THE PROGRAM - CONCLUSION 333
r e a l , parameter :: max_dist =20.0
integer :: i
real :: r , dx , dy , dl
real :: Ex , Ey , E
dl = step
x0 = xin
y0 = yin
dx = 0.0
dy = 0.0
r = step ! i n order t o s t a r t loop
do while ( r . g t . ( 0 . 9 * dl ) . and . r . l t . max_dist )
p r i n t * , x0 , y0
! We e v a l u a t e t h e E−f i e l d a t t h e midpoint : This r e d u c e s
! systematic errors
c a l l efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey )
E = s q r t ( Ex * Ex+Ey * Ey )
i f ( E . l e . 1 . 0 e−10 ) e x i t
dx = dl * Ey / E
dy = −dl * Ex / E
x0 = x0 + dx
y0 = y0 + dy
r = s q r t ( ( x0−xin ) * * 2 + ( y0−yin ) * * 2 )
enddo ! do while ( )
end s u b r o u t i n e epotline
...
where ... are the subroutines efield and mdist which are identical to
the ones in the file ELines.f90.
In order to compile the program use the commands:
> g f o r t r a n ELines . f90 −o el
> g f o r t r a n EPotential . f90 −o ep
Then, edit a file and name it e.g. Input and write the data that define a
charge distribution. For example:
4 N : Number of charges
1 1 −1 ( X , Y , Q ) : Position and charge
−1 1 1 ( X , Y , Q ) : Position and charge
1 −1 1 ( X , Y , Q ) : Position and charge
−1 −1 −1 ( X , Y , Q ) : Position and charge
The results are obtained with the commands:
> . / el < Input > el . dat
334 CHAPTER 7. ELECTROSTATICS
> . / ep < Input > ep . dat
> gnuplot
gnuplot > p l o t ” e l . dat ” with dots
gnuplot > p l o t ” ep . dat ” with dots
Have fun!
7.5 Electrostatic Field in the Vacuum
Consider a time independent electric field in an area of space which is
empty of electric charge. Maxwell’s equations are reduced to Gauss’s law
∂Ex ∂Ey ∂Ez
∇
⃗ · E(x,
⃗ y, z) = + + = 0, (7.8)
∂x ∂y ∂z
together with the equation that defines the electrostatic potential⁷
⃗
E(x, y, z) = −∇V
⃗ (x, y, z) . (7.9)
Equations (7.8) and (7.9) give the Laplace equation for the function
V (x, y, z):
∂2V ∂ 2V ∂ 2V
∇2 V (x, y, z) = + + = 0. (7.10)
∂x2 ∂y 2 ∂z 2
The solution of the equation above is a boundary value problem: We
are looking for the potential V (x, y, z) in a region of space S bounded
by a closed surface ∂S. When the potential is known on ∂S the solution
to (7.10) is unique and the potential and the electric field is determined
everywhere in S.
For simplicity consider the problem confined on a plane, therefore
V = V (x, y). In this case the last term in equation (7.10) vanishes, the
region S is a compact subset of the plane and ∂S is a closed curve.
For the numerical solution of the problem, we approximate S by a
discrete, square lattice. The potential is defined on the N sites of the
lattice. We take S to be bounded by a square with sides of length l. The
distance between the nearest lattice √ points is called the lattice constant
a. Then l = (L − 1)a, where L = N is the number of lattice points
on each side of the square. The continuous solution is approximated by
the solution on the lattice, and the approximation becomes exact in the
N → ∞ and a → 0 limits, so that the length l = (L−1)a remains constant.
The curve ∂S is approximated by the lattice sites that are located on the
⁷Equivalent to the equation ∇
⃗ ×E
⃗ = 0.
7.5. ELECTROSTATIC FIELD IN THE VACUUM 335
perimeter of the square and the loci in the square where the potential
takes constant values. This is a simple model of a set of conducting
surfaces (points where V = const. ̸= 0) in a compact region whose
boundary is grounded (points where V = 0). An example is depicted in
figure 7.6.
111
000
000111
000
000 11001100 11001100 11001100 11001100111
000
000111
000 1100 1100 1100111
000
000111
000
000111
000
111111 111000111
111000
000 1100 1100 1100111
111 111000 11001100
111
000
111 1100
000
111 1100
111
000 0011
111
000 0011
111
000 0011 0011 0011
000
111 1100 1100 1100
000
111 1100 1100 11001100
000
111 1100 1100
111
000 0011 0011 0011
111
000 0011 0011 0011
111
000 0011 0011 0011
000
111 1100
000
111 11001100
000
111
111000
000111 0011 0011 0011 0011000 111111
111000000
000 0011 0011 0011000
111 111000
111000
111 0011
Figure 7.6: A lattice which corresponds to a cross section of two parallel conducting
planes inside a grounded cubic box. The black lattice sites are the points of constant,
fixed potential whereas the white ones are sites in the vacuum.
In order to derive a finite difference equation which approximates
equation (7.10), we Taylor expand around a point (x, y) according to the
equations:
∂V 1 ∂2V
V (x + δx, y) = V (x, y) + δx + 2
(δx)2 + . . .
∂x 2 ∂x
∂V 1 ∂ 2V
V (x − δx, y) = V (x, y) − δx + (δx)2 + . . .
∂x 2 ∂x2
∂V 1 ∂ 2V
V (x, y + δy) = V (x, y) + δy + 2
(δy)2 + . . .
∂y 2 ∂y
∂V 1 ∂ 2V
V (x, y − δy) = V (x, y) − δy + 2
(δy)2 + . . . .
∂y 2 ∂y
By summing both sides of the above equations, taking δx = δy and
336 CHAPTER 7. ELECTROSTATICS
ignoring the terms implied by . . ., we obtain
V (x + δx, y) + V (x − δx, y) + V (x, y + δy) + V (x, y − δy)
∂ 2V ∂ 2V
= 4V (x, y) + (δx)2 ( 2 + ) + ...
∂x ∂y 2
≈ 4V (x, y) , (7.11)
The second term in the second line was eliminated by using equation
(7.10).
We map the coordinates of the lattice points to integers (i, j) such that
xi = (i−1)a and yj = (j −1)a where i, j = 1, . . . , L. By taking δx = δy = a
so that xi ± δx = xi ± a = (i − 1 ± 1)a = xi±1 and yj ± δy = yj ± a =
(j − 1 ± 1)a = yj±1 , equation (7.11) becomes:
1
V (i, j) = (V (i − 1, j) + V (i + 1, j) + V (i, j − 1) + V (i, j + 1)) . (7.12)
4
The equation above states that the potential at the position (i, j) is the
arithmetic mean of the potential of the nearest neighbors. We will de-
scribe an algorithm which belongs to the class of “successive overrelax-
ation methods” (SOR) whose basic steps are:
1. Set the size L of the square lattice.
2. Flag the sites that correspond to “conductors”, i.e. the sites where
the potential remains fixed to the boundary conditions values.
3. Choose an initial trial function for V (x, y) on the vacuum sites. Of
course it is not the solution we are looking for. A good choice will
lead to fast convergence of the algorithm to the true solution. A
bad choice may lead to slow convergence, no convergence or even
convergence to the wrong solution. In our case the problem is easy
and the simple choice V (x, y) = 0 will do.
4. Sweep the lattice and enforce equation (7.12) on each visited vacuum
site. This defines the new value of the potential at this site.
5. Sweep the lattice repeatedly until two successive sweeps result in a
very small change in the function V (x, y).
A careful study of the above algorithm requires to test different criteria of
“very small change” and test that different choices of the initial function
V (x, y) result in the same solution.
7.5. ELECTROSTATIC FIELD IN THE VACUUM 337
We write a program that implements this algorithm in the case of a
system which is the projection of two parallel conducting planes inside
a grounded cubic box on the plane. The lattice is depicted in figure 7.6,
where the black dots correspond to the conductors. All the points of
the box have V = 0 and the two conductors are at constant potential
V1 and V2 respectively. The user enters the values V1 and V2 , the lattice
size L and the required accuracy interactively. The latter is determined
by a small number ϵ. The convergence criterion that we set is that the
maximum difference between the values of the potential between two
successive sweeps should be less than ϵ.
The data structure is very simple. We use a real array V(L,L) in
order to store the values of the potential at each lattice site. A logical
array isConductor(L,L) flags each site as a “conductor site” (= .TRUE.)
or as a “vacuum site” (=.FALSE.).
The main program reads in the data entered by the user and then
calls three subroutines:
1. initialize_lattice(V,isConductor,L,V1,V2):
The routine needs at its input the values of the potential V1 and V2
on the left and right plate respectively and the size of the lattice L.
On exit it provides the initial values of the potential V(L,L) and the
flags isConductor(L,L). The geometry of the setting is hard coded
and the user needs to change this subroutine each time that she
wants to study a different geometry.
2. laplace(V,isConductor,L,epsilon):
This is the heart of the program. On entry we provide the initial-
ized arrays V(L,L) and isConductor(L,L), the lattice size L, and
the desired accuracy epsilon. On exit we obtain the final solution
V(L,L). This subroutine calculates the arithmetic mean of the po-
tential of the nearest neighbors Vav and the value V(i,j)=Vav is
changed immediately⁸. The maximum change in the new value of
the potential Vav from the old one V(i,j) is stored in the variable
error. When error becomes smaller than epsilon we assume that
convergence has been achieved.
3. print_results(V,L):
This subroutine prints the potential V(L,L) to the file data. Each
line contains the integers i, j and the value of the potential V(i,j).
⁸A different choice would have been to store the value Vav in a temporary array
Vnew(i,j). After the sweep, the potential V(i,j)=Vnew(i,j) is changed to the new
values. Which method do you expect to have better convergence properties? Try...
338 CHAPTER 7. ELECTROSTATICS
We note that each time that the index i changes, the subroutine
prints an extra empty line. This is done so that the output can
be read easily by the three dimensional plotting function splot of
gnuplot.
The full program is listed below:
! *************************************************************
!PROGRAM LAPLACE_EM
! Computes t h e e l e c t r o s t a t i c p o t e n t i a l around c o n d u c t o r s .
! The computation i s performed on a square l a t t i c e o f l i n e a r
! dimension L . A r e l a x a t i o n method i s used t o converge t o t h e
! s o l u t i o n o f Laplace e q u a t i o n f o r t h e p o t e n t i a l .
!DATA STRUCTURE:
! r e a l ( 8 ) V(L , L) : Value o f t h e p o t e n t i a l on t h e l a t t i c e s i t e s
! l o g i c a l i s C o n d u c t o r (L , L) : I f .TRUE. s i t e has f i x e d p o t e n t i a l
! I f . FALSE . s i t e i s empty sp a c e
! r e a l e p s i l o n : Determines t h e a c c u r a c y o f t h e s o l u t i o n
! The maximum d i f f e r e n c e o f t h e p o t e n t i a l on each s i t e between
! two c o n s e c u t i v e sweeps should be l e s s than e p s i l o n .
!PROGRAM STRUCTURE
! main program :
! . Data Input
! . c a l l s u b r o u t i n e s f o r i n i t i a l i z a t i o n , computation and
! printing of r e s u l t s
! subroutine i n i t i a l i z e _ l a t t i c e :
! . I n i t i l i z a t i o n o f V(L , L) and i s C o n d u c t o r (L , L)
! subroutine laplace :
! . S o l v e s l a p l a c e e q u a t i o n using a r e l a x a t i o n method
! subroutine p r i n t _ r e s u l t s :
! . P r i n t s r e s u l t s f o r V(L , L) i n a f i l e . Uses format c o m p a t i b l e
! with s p l o t o f gnuplot .
! *************************************************************
program laplace_em
i m p l i c i t none
! P d e f i n e s t h e s i z e o f t h e a r r a y s and i s equal t o L
i n t e g e r , parameter : : P=31
l o g i c a l , dimension ( P , P ) : : isConductor
r e a l ( 8 ) , dimension ( P , P ) : : V
! V1 and V2 a r e t h e v a l u e s o f t h e p o t e n t i a l on t h e i n t e r i o r
! conductors . epsilon i s the accuracy desired f o r the
! convergence o f t h e r e l a x a t i o n method i n s u b r o u t i n e
! laplace ()
real (8) : : V1 , V2 , e p s i l o n
integer :: L
!We ask t h e u s e r t o provide t h e n e c e s s a r y data :
! V1 , V2 and e p s i l o n
7.5. ELECTROSTATIC FIELD IN THE VACUUM 339
L = P
p r i n t * , ’ Enter V1 , V2 : ’
read * , V1 , V2
p r i n t * , ’ Enter e p s i l o n : ’
read * , e p s i l o n
p r i n t * , ’ S t a r t i n g Laplace : ’
p r i n t * , ’ Grid S i z e = ’ , L
p r i n t * , ’ Conductors s e t a t V1= ’ , V1 , ’ V2= ’ , V2
p r i n t * , ’ R e l a x i n g with a c c u r a c y e p s i l o n = ’ , e p s i l o n
! The a r r a y s V and i s C o n d u c t o r a r e i n i t i a l i z e d
c a l l initialize_lattice ( V , isConductor , L , V1 , V2 )
!We e n t e r i n i t i a l i z e d V , i s C o n d u c t o r . On e x i t t h e
! routine gives the s o l u t i o n V
c a l l laplace ( V , isConductor , L , e p s i l o n )
!We p r i n t V i n a f i l e .
c a l l print_results ( V , L )
end program laplace_em
! *************************************************************
! subroutine i n i t i a l i z e _ l a t t i c e
! I n i t i a l i z e s a r r a y s V(L , L) and i s C o n d u c t o r (L , L) .
!V(L , L)= 0.0 and i s C o n d u c t o r (L , L)= . FALSE . by d e f a u l t
! i s C o n d u c t o r ( i , j )= .TRUE. on boundary o f l a t t i c e where V=0
! i s C o n d u c t o r ( i , j )= .TRUE. on s i t e s with i = L/ 3 + 1 , 5<= j <= L−5
! i s C o n d u c t o r ( i , j )= .TRUE. on s i t e s with i =2*L/ 3 + 1 , 5<= j <= L−5
!V( i , j ) = V1 on a l l s i t e s with i = L/ 3 + 1 , 5<= j <= L−5
!V( i , j ) = V2 on a l l s i t e s with i =2*L/ 3 + 1 , 5<= j <= L−5
!V( i , j ) = 0 on boundary ( i =1 ,L and j =1 ,L)
!V( i , j ) = 0 on i n t e r i o r s i t e s with i s C o n d u c t o r ( i , j )= . FALSE .
! INPUT :
! i n t e g e r L : Linear s i z e of l a t t i c e
! r e a l ( 8 ) V1 , V2 : Values o f p o t e n t i a l on i n t e r i o r c o n d u c t o r s
!OUTPUT:
! r e a l ( 8 ) V(L , L) : Array provided by u s e r . Values o f p o t e n t i a l
! l o g i c a l i s C o n d u c t o r (L , L) : I f .TRUE. s i t e has f i x e d p o t e n t i a l
! I f . FALSE . s i t e i s empty s p a ce
! *************************************************************
s u b r o u t i n e initialize_lattice ( V , isConductor , L , V1 , V2 )
i m p l i c i t none
integer :: L
l o g i c a l , dimension ( L , L ) : : isConductor
r e a l ( 8 ) , dimension ( L , L ) : : V
real (8) : : V1 , V2
integer :: i,j
! I n i t i a l i z e t o 0 and . FALSE ( d e f a u l t v a l u e s f o r boundary and
! interior sites ) .
V = 0.0 D0
isConductor = . FALSE .
340 CHAPTER 7. ELECTROSTATICS
!We s e t t h e boundary t o be a conductor : (V=0 by d e f a u l t )
do i =1 , L
isConductor ( 1 , i ) = .TRUE.
isConductor ( i , 1 ) = .TRUE.
isConductor ( L , i ) = .TRUE.
isConductor ( i , L ) = .TRUE.
enddo
!We s e t two c o n d u c t o r s a t giv e n p o t e n t i a l V1 and V2
do i =5 ,L−5
V ( L / 3 + 1 , i ) = V1
isConductor ( L / 3 + 1 , i ) = .TRUE.
V ( 2 * L / 3 + 1 , i ) = V2
isConductor ( 2 * L / 3 + 1 , i ) = .TRUE.
enddo
end s u b r o u t i n e initialize_lattice
! *************************************************************
! subroutine laplace
! Uses a r e l a x a t i o n method t o compute t h e s o l u t i o n o f t h e
! Laplace e q u a t i o n f o r t h e e l e c t r o s t a t i c p o t e n t i a l
! on a 2 d i m e n s i o n a l s q u a r e l a t t i c e o f l i n e a r s i z e L .
! At eve ry sweep o f t h e l a t t i c e we compute t h e a v e r a g e
! Vav o f t h e p o t e n t i a l a t each s i t e ( i , j ) and we immediately
! update V( i , j ) .
! The computation c o n t i n u e s u n t i l Max | Vav−V( i , j ) | < e p s i l o n
! INPUT :
! i n t e g e r L : Li ne ar s i z e o f l a t t i c e
! r e a l ( 8 ) V(L , L) : Value o f t h e p o t e n t i a l a t each s i t e
! l o g i c a l i s C o n d u c t o r (L , L) : I f .TRUE. potential is fixed
! I f . FALSE . p o t e n t i a l i s updated
! r e a l ( 8 ) e p s i l o n : i f Max | Vav−V( i , j ) | < e p s i l o n r e t u r n t o
! callingprogram .
!OUTPUT:
! r e a l ( 8 ) V(L , L) : The computed s o l u t i o n f o r t h e p o t e n t i a l
! *************************************************************
s u b r o u t i n e laplace ( V , isConductor , L , e p s i l o n )
i m p l i c i t none
integer : : L
l o g i c a l , dimension ( L , L ) : : isConductor
r e a l ( 8 ) , dimension ( L , L ) : : V
real (8) : : epsilon
integer : : i , j , icount
real (8) : : Vav , error , dV
icount = 0 ! co un ts number o f sweeps
do while ( . TRUE . ) ! an i n f i n i t e loop :
error = 0.0 D0 ! E x i t when e r r o r < e p s i l o n
do j =2 ,L−1
do i =2 ,L−1
7.6. RESULTS 341
!We change V only f o r non c o n d u c t o r s :
i f ( .NOT. isConductor ( i , j ) ) then
Vav = ( V ( i −1 , j )+V ( i +1 , j )+V ( i , j +1)+V ( i , j−1) ) * 0.25 D0
dV = DABS ( V ( i , j )−Vav )
i f ( error . LT . dV ) error = dV ! maximum e r r o r
V ( i , j ) = Vav ! we immendiately update V( i , j )
endif
enddo
enddo
icount = icount + 1
p r i n t * , icount , ’ e r r = ’ , error
i f ( error . LT . e p s i l o n ) r e t u r n ! r e t u r n t o main program
enddo
end s u b r o u t i n e laplace
! *************************************************************
! subroutine p r i n t _ r e s u l t s
! P r i n t s t h e a r r a y V(L , L) i n f i l e ” data ”
! The format o f t h e output i s a p p r o p r i a t e f o r t h e s p l o t f u n c t i o n
! o f gnuplot : Each time i changes an empty l i n e i s p r i n t e d .
! INPUT :
! i n t e g e r L: s i z e of array V
! r e a l ( 8 ) V(L , L) : a r r a y t o be p r i n t e d
!OUTPUT:
! no output
! *************************************************************
s u b r o u t i n e print_results ( V , L )
i m p l i c i t none
integer :: L
r e a l ( 8 ) , dimension ( L , L ) : : V
integer :: i,j
open ( u n i t =11 , f i l e =” data ” )
do i =1 , L
do j =1 , L
write (11 ,*) i , j , V(i , j)
enddo
w r i t e ( 1 1 , * ) ’ ’ ! empty l i n e f o r gnuplot , s e p a r a t e i s o l i n e s
enddo
end s u b r o u t i n e print_results
7.6 Results
The program in the previous section is written in the file LaplaceEq.f90.
Compiling and running is done with the commands:
342 CHAPTER 7. ELECTROSTATICS
> g f o r t r a n LaplaceEq . f90 −o lf
> . / lf
Enter V1 , V2 :
100 −100
Enter epsilon :
0.01
Starting Laplace :
Grid Size= 31
Conductors s e t at V1= 100. V2= −100.
Relaxing with accuracy epsilon= 0.01
1 err= 33.3333333
2 err= 14.8148148
3 err= 9.87654321
.......................
110 err= 0.0106860904
111 err= 0.0101182476
112 err= 0.00958048937
In the example above, the program performs 112 sweeps until the error
becomes 0.00958048937 < 0.01. The results are stored in the file data.
We can make a three dimensional plot of the function V (i, j) with the
gnuplot commands:
gnuplot > s e t pm3d
gnuplot > s e t hidden3d
gnuplot > s e t s i z e ratio 1
gnuplot > s p l o t ” data ” with lines
The results are shown in figure 7.7
7.7 Poisson Equation
This section contains a short discussion of the case where the space is
filled with a continuous static charge distribution given by the charge
density function ρ(⃗r). In this case the Laplace equation becomes the
Poisson equation:
∂ 2V ∂ 2V ∂2V
∇2 V = + + = −4πρ(x, y, z) (7.13)
∂x2 ∂y 2 ∂z 2
The equation on the lattice becomes
1
V (i, j) = (V (i−1, j)+V (i+1, j)+V (i, j −1)+V (i, j +1)+ ρ̃(i, j)) , (7.14)
4
7.7. POISSON EQUATION 343
"data"
100
100 50
50 0
0
-50
-50
-100
-100
35
30
25
0 20
5 15
10
15 10
20
25 5
30
35 0
Figure 7.7: The solution of the equation (7.10) computed by the program
LaplaceEq.f90 for L= 31, V1=100, V2=-100, epsilon=0.01.
where⁹ ρ̃(i, j) = 4πa2 ρ(i, j).
The program in the file PoissonEq.f90 solves equation (7.14) for a
uniform charge distribution (figure 7.10), where we have set a = 1. The
reader is asked to reproduce this figure together with figures 7.8 and
7.9.
! *************************************************************
! s e t t h e boundary o f a square t o giv e n p o t e n t i a l s
! *************************************************************
program poisson_eq
i m p l i c i t none
i n t e g e r , parameter : : P=51
l o g i c a l , dimension ( P , P ) : : isConductor
r e a l ( 8 ) , dimension ( P , P ) : : V , rho
real (8) : : V1 , V2 , V3 , V4 , Q , e p s i l o n
integer :: L
L = P
p r i n t * , ’ Enter V1 , V2 , V3 , V4 : ’
read * , V1 , V2 , V3 , V4
∫ ∑ ∑ ∑
⁹Since Q = ρdA ≈ i,j ρa2 = (1/4π) i,j ρ̃. Therefore i,j ρ̃ ≈ 4πQ.
344 CHAPTER 7. ELECTROSTATICS
"data"
800 800
700 700
600 600
500 500
400 400
300 300
200 200
100 100
0 0
0 10 20
50 60
30 30 40
40 50 20
600 10
Figure 7.8: The solution of the equation (7.13) by the program in the file Poisson.f90
for L= 51, V= 0 on the boundary and the charge 4πQ = 1000 all concentrated at one
point.
print * , ’ Enter 4* PI *Q: ’
read *, Q
print * , ’ Enter e p s i l o n : ’
read * , epsilon
print * , ’ S t a r t i n g Laplace : ’
print * , ’ Grid S i z e = ’ , L
print * , ’ Boundaries s e t a t V1= ’ , V1 , ’ V2= ’ , V2 , ’ V3= ’ , V3 ,&
’ V4= ’ , V4 , ’ and Q= ’ , Q
p r i n t * , ’ R e l a x i n g with a c c u r a c y e p s i l o n = ’ , e p s i l o n
c a l l initialize_lattice ( V , isConductor , rho , L , V1 , V2 , V3 , V4 , Q )
c a l l laplace ( V , isConductor , rho , L , e p s i l o n )
c a l l print_results ( V , L )
end program laplace_sq
! **********************************************************
subroutine &
initialize_lattice ( V , isConductor , rho , L , V1 , V2 , V3 , V4 , Q )
! **********************************************************
7.7. POISSON EQUATION 345
"data"
350
350
300
300
250
250
200 200
150 150
100 100
50 50
0 0
0 60
10 50
20 40
30 30
40 20
50 10
600
Figure 7.9: The solution of equation (7.13) by the program in the file Poisson.f90
for L= 51, V= 0 on the boundary and the charge 4πQ = 1000 uniformly distributed in
a small square with sides made of 10 lattice sites.
i m p l i c i t none
integer :: L
l o g i c a l , dimension ( L , L ) : : isConductor
r e a l ( 8 ) , dimension ( L , L ) : : V , rho
real (8) : : V1 , V2 , V3 , V4 , Q , Area
integer : : i , j , L1 , L2
! I n i t i a l i z e t o 0 and . FALSE .
V = 0.0 D0
isConductor = . FALSE .
rho = 0.0 D0
!We s e t t h e boundary t o be a conductor :
do i =1 , L
isConductor ( 1 , i ) = .TRUE.
isConductor ( i , 1 ) = .TRUE.
isConductor ( L , i ) = .TRUE.
isConductor ( i , L ) = .TRUE.
V ( 1 , i ) = V1
V ( i , L ) = V2
V ( L , i ) = V3
V ( i , 1 ) = V4
enddo
!We s e t t h e p o i n t s with non−z e r o charge
346 CHAPTER 7. ELECTROSTATICS
"data"
80
80 70
70 60
60 50
50
40
40
30 30
20 20
10 10
0 0
0 60
10 50
20 40
30 30
40 20
50 10
600
Figure 7.10: The solution of equation (7.13) by the program in the file Poisson.f90
for L= 51, V= 0 on the boundary and the charge 4πQ = 1000 uniformly distributed on
all internal lattice sites.
!A uniform d i s t r i b u t i o n a t a c e n t e r square
L1 = ( L / 2 )−5
L2 = ( L / 2 ) +5
i f ( L1 . LT . 1 ) s t o p ’ a r r a y rho out o f bounds . Small L1 ’
i f ( L2 . GT . L ) s t o p ’ a r r a y rho out o f bounds . Large L2 ’
Area = ( L2−L1 +1) * ( L2−L1 +1)
do j=L1 , L2
do i=L1 , L2
rho ( i , j ) = Q / Area ! rho i s \ t i l d e \rho i n n o t e s
enddo ! so Q i s 4* PI *Q
enddo
end s u b r o u t i n e initialize_lattice
! *************************************************************
s u b r o u t i n e laplace ( V , isConductor , rho , L , e p s i l o n )
! *************************************************************
i m p l i c i t none
integer : : L
l o g i c a l , dimension ( L , L ) : : isConductor
r e a l ( 8 ) , dimension ( L , L ) : : V , rho
real (8) : : epsilon
integer : : i , j , icount
7.7. POISSON EQUATION 347
real (8) : : Vav , error , dV
icount = 0
do while ( . TRUE . )
error = 0.0 D0
do j =2 ,L−1
do i =2 ,L−1
!We change t h e v o l t a g e only f o r non c o n d u c t o r s :
i f ( .NOT. isConductor ( i , j ) ) then
Vav = ( V ( i −1 , j )+V ( i +1 , j )+V ( i , j +1)+V ( i , j−1)+rho ( i , j ) )&
* 0. 25 D0
dV = DABS ( V ( i , j )−Vav )
i f ( error . LT . dV ) error = dV ! maximum e r r o r
V ( i , j ) = Vav
endif
enddo
enddo
icount = icount + 1
i f ( error . LT . e p s i l o n ) e x i t
enddo
p r i n t * , icount , ’ e r r = ’ , error
end s u b r o u t i n e laplace
! *************************************************************
s u b r o u t i n e print_results ( V , L )
! *************************************************************
i m p l i c i t none
integer :: L
r e a l ( 8 ) , dimension ( L , L ) : : V
integer :: i,j
open ( u n i t =11 , f i l e =” data ” )
do i =1 , L
do j =1 , L
write (11 ,*) i , j , V(i , j)
enddo
w r i t e ( 1 1 , * ) ’ ’ ! empty l i n e f o r gnuplot , s e p a r a t e i s o l i n e s
enddo
end s u b r o u t i n e print_results
In the bibliography the algorithm described above is called the Gauss–
Seidel method. In this method, the right hand side of equation (7.14)
uses the updated values of the potential in the calculation of V (i, j) and
V (i, j) is immediately updated. In contrast, the Jacobi method uses the
old values of the potential in the right hand side of (7.14) and the new
value computed is stored in order to be used in the next sweep. The
Gauss–Seidel method is superior to the Jacobi method as far as speed of
348 CHAPTER 7. ELECTROSTATICS
convergence is concerned. We can generalize Jacobi’s method by defining
the residual Ri,j of equation (7.14)
Ri,j = V (i + 1, j) + V (i − 1, j) + V (i, j + 1) + V (i, j − 1) − 4V (i, j) + ρ̃(i, j) ,
(7.15)
which vanishes when V (i, j) is a solution of equation (7.14). Then, using
Ri,j , Jacobi’s method can be formulated as
1 (n)
V (n+1) (i, j) = V (n) (i, j) + Ri,j , (7.16)
4
where the quantities with index (n) refer to the values of the potential
during the n-th sweep. The successive overrelaxation (SOR) method is
given by:
ω (n)
V (n+1) (i, j) = V (n) (i, j) + Ri,j . (7.17)
4
When ω < 1 we have “underrelaxation” and we obtain slower conver-
gence than the Jacobi method. When 1 < ω < 2 we have “overrelaxation”
and an appropriate choice of ω can lead to an improvement compared
to the Jacobi method. When ω > 2 SOR diverges. Further study of the
SOR methods is left as an exercise to the reader.
7.8. PROBLEMS 349
7.8 Problems
7.1 Reproduce the figures with the electric field lines and equipotential
lines shown in section 7.2.
7.2 Take the charge distributions that you used in the previous prob-
lems, make all the charges to be positive and remake the figures of
the field lines and the equipotential lines. Then repeat by taking
half of the charges to be twice in magnitude than the others.
7.3 The program ELines.f90 gets stuck when you apply it on a charge
distribution of four equal charges located at the vertices of a square.
How can you correct this pathology?
7.4 Make the necessary changes to the program in the file ELines.f90 so
that the number of field lines starting near a charge q is proportional
to q.
7.5 Improve the program in EPotential.f90 so that the equipotential
lines are drawn with a density proportional to the magnitude of the
electric field.
Hint:
(a) Write a subroutine that calculates the potential V (x, y) at the
point (x, y).
(b) From each point charge draw a line in the radial direction and
calculate the potential on points that are at small distance ∆l
from each other.
(c) Calculate the maximum/minimum value of the potential Vmax /Vmin
and use them in order to choose the values of the potential
on the equipotential lines that you plan to draw. If e.g. you
choose to draw 5 equipotential lines, take δV = (Vmax − Vmin )/4
and Vi = Vmin + iδV i = 0, . . . , 4.
(d) Repeat the second step. When the potential at a point takes
approximately one of the values Vi chosen in the previous step,
draw an equipotential line from that point.
7.6 Compute the electric potential using the program in the file LaplaceEq.f90
for
(a) L= 31, V1=100, V2=100
(b) L= 31, V1=100, V2=0
350 CHAPTER 7. ELECTROSTATICS
and construct the corresponding plot for V (i, j).
7.7 Compute the electric potential using the program in the file LaplaceEq.f90
for
(a) V1=100, V2=100
(b) V1=100, V2=100
(c) V1=100, V2=0
for L=31,61,121,241,501 and construct the corresponding plot for
V (i, j). Vary epsilon=0.1, 0.01, 0.001, 0.0001, 0.00001,
0.000001. What is the dependence of the number of sweeps N
on epsilon? Make the plot of N (epsilon). Put the points and
curves of N (epsilon) for all values of L on the same plot.
7.8 Compute the electrostatic potential of a square conductor when the
potential on each side is V1, V2, V3, V4. Repeat what you did in
the previous problem for
(a) V1=10, V2=5, V3=10, V4= 5
(b) V1=10, V2=0, V3=0, V4= -10
(c) V1=10, V2=0, V3=0, V4= 0
7.9 Compute the electrostatic potential of a system of square conductors
where the one is inside the other as shown in figure 7.11. The side
of each conductor has L1, L2 sites respectively and the value of the
potential is V1,V2 respectively. Take L2= L1/5 and repeat the steps
in the previous problem for V1=10, V2=-10 and L1= 25, 50, 100,
200.
7.10 Perform a numerical computation of the capacitance C = Q/V of
the system of conductors of the previous problem when V1 = V ,
V2 = −V . In order to calculate the charge Q, compute the surface
charge density σ using the equation
En
σ= ,
4π
where En is the perpendicular component of the electric field on
the surface. Use the approximation
δV
En = − ,
δr
7.8. PROBLEMS 351
111
000
000111
000
000 11001100 11001100 11001100 11001100111
000
000111
000 11001100 11001100 11001100111
000000
111000
111 11001100
111111 111000111
111000
000
111 000
111000
111000
111
000
111 1100
000
111 1100
111
000 0011
111
000 0011
000
111 11001100000
111000
111000
111 0011 1100 11001100
000
111 1100
000
111 1100 1100 1100
000
111 1100 1100 1100
111
000 0011 0011 0011
111
000 0011000 111000
111000
111 0011 0011 0011
111
000 0011
000
111 1100
000
111 11001100
000
111
111000
000111 0011 0011 0011 0011000 111111
111000000
000 0011 0011 0011000
111 111000
111000
111 0011
Figure 7.11: The square conductors described in problem 7.9.
where δV is the potential difference between a point on the con-
ductor and its nearest neighbor. By integrating (i.e. summing)
you can estimate the total charge on each conductor. If these are
opposite and their absolute value is Q, then the capacitance can be
calculated from the equation C = Q/V . Perform the calculation
described above for V = 10 and L1=25, 75.
7.11 In the system of the previous problem compute the function Q(V ).
Verify that the capacitance is independent of V . Use L1=25,50, V1=
-V2 =1, 2, 5, 10, 15, 20, 25.
7.12 Reproduce figures 7.8, 7.9 and 7.10. Compare the result of the first
case with the known solution of a point charge in empty space.
7.13 Introduce the lattice spacing a in the corresponding equations in the
program in the file PoissonEq.f90. Set the length of each side to be
l = 1 and print the results in the file data as (xi , yi , V (xi , yi )) instead
of (i, j, V (i, j)). Take L=51,101,151,201,251 and plot V (x, y) in the
square 0 < x < 1, 0 < y < 1. Study the convergence of the solutions
by plotting the section V (x, 1/2) for each L.
7.14 Write a program that implements the SOR algorithm given by equa-
352 CHAPTER 7. ELECTROSTATICS
tion (7.16) for the problem solved in LaplaceEq.f90. Compare the
speed of convergence of SOR with that of the Gauss-Seidel method
for L = 51, ω = 1.0, 0.9, 0.8, 0.6, 0.4, 0.2. What happens when
ω > 1?
7.15 Write a program that implements the SOR algorithm given by equa-
tion (7.16) for the problem solved in PoissonEq.f90. Compare the
speed of convergence of SOR with that of the Gauss-Seidel method
for L = 51, ω = 1.0, 0.9, 0.8, 0.6, 0.4, 0.2. What happens when
ω > 1?
Chapter 8
Diffusion Equation
8.1 Introduction
The diffusion equation is related to the study of random walks. Consider
a particle moving on a line (one dimension) performing a random walk.
The motion is stochastic and the kernel
K(x, x0 ; t) , (8.1)
is interpreted as the probability density to observe the particle at position
x at time t if the particle is at x0 at t = 0. The equation that determines
K(x, x0 ; t) is
∂K(x, x0 ; t) ∂ 2 K(x, x0 ; t)
=D , (8.2)
∂t ∂x2
which is the diffusion equation. The coefficient D depends on the details
of the system that is studied. For example, for the Brownian motion of
a dust particle in a fluid which moves under the influence of random
collisions with the fluid particles, we have that D = kT /γ, where T is
the (absolute) temperature of the fluid, γ is the friction coefficient¹ of the
particle in the fluid and k is the Boltzmann constant.
Usually the initial conditions are chosen so that at t = 0 the particle
is localized at one point x0 , i.e.²
K(x, x0 ; 0) = δ(x − x0 ) . (8.3)
¹For a spherical particle of radius R in a Newtonian liquid with viscosity η we have
that γ = 6πηR.
² δ(x − x0 ) is the Dirac delta “function”. It can be defined from the requirement
∫ +∞
that for every function f (x) we have that −∞ f (x)δ(x − x0 ) dx = f (x0 ). Obviously we
∫ +∞
also have that −∞ δ(x − x0 ) dx = 1. Intuitively one can think of it as a function that is
almost zero everywhere except in an infinitesimal neighborhood of x0 .
353
354 CHAPTER 8. DIFFUSION EQUATION
The interpretation of K(x, x0 ; t) as a probability density implies that
for every t we should have that³
∫ +∞
K(x, x0 ; t) dx = 1 . (8.4)
−∞
It is not obvious that this relation can be imposed for every instant of
time. Even if K(x, x0 ; t) is normalized so that (8.4) holds for t = 0, the
time evolution of K(x, x0 ; t) is governed by equation (8.2) which can spoil
equation (8.4) at later times.
If we impose equation (8.4) at t = 0, then it will hold at all times if
∫
d +∞
K(x, x0 ; t)dx = 0 . (8.5)
dt −∞
d
∫ +∞ ∫ +∞ ∂K(x,x0 ;t)
By taking into account that dt −∞
K(x, x0 ; t)dx = −∞ ∂t
dx and
∂K(x,x0 ;t) ∂ 2 K(x,x 0 ;t)
that ∂t
=D ∂x2
we obtain
∫ +∞ ∫ +∞ ( )
d ∂K(x, x0 ; t) ∂
K(x, x0 ; t)dx = D dx
dt −∞ −∞ ∂x ∂x
∂K(x, x0 ; t) ∂K(x, x0 ; t)
=D −D . (8.6)
∂x x→+∞ ∂x x→−∞
The above equation tells us that for functions for which the right hand
side vanishes, the normalization condition will be valid for all t > 0.
A careful analysis of equation (8.2) gives that the asymptotic behavior
of K(x, x0 ; t) for small times is
|x−x0 |2
e− 4Dt ∑
∞
K(x, x0 ; t) ∼ ai (x, x0 )ti . (8.7)
td/2 i=0
This relation shows that diffusion is isotropic (the same in all directions)
and that the probability of detecting the particle drops exponentially with
the distance squared from the initial position of the particle. This relation
cannot hold for all times, since for large enough times the probability of
detecting the particle will be the same everywhere⁴.
³Alternatively, if K(x, x0 ; t) is interpreted as e.g. the mass density of a drop of ink
∫ +∞
of mass mink inside a transparent liquid, we will have that −∞ K(x, x0 ; t) dx = mink
and K(x, x0 ; 0) = mink δ(x − x0 ).
⁴Remember the analogy of an ink drop diffusing in a transparent liquid. After long
enough time, the ink is homogeneously dissolved in the liquid.
8.2. HEAT CONDUCTION IN A THIN ROD 355
The return probability of the particle to its initial position is
1 ∑
∞
PR (t) = K(x0 , x0 ; t) ∼ ai (x0 , x0 )ti . (8.8)
td/2 i=0
The above relation defines the spectral dimension d of space. d = 1 in our
case.
The expectation value of the distance squared of the particle at time
t is easily calculated⁵
∫ +∞
⟨r ⟩ = ⟨(x − x0 ) ⟩(t) =
2 2
(x − x0 )2 K(x, x0 ; t) dx ∼ 2Dt . (8.9)
−∞
This equation is very important. It tells us that the random walk (Brow-
nian motion) is not a classical motion but it can only be given a stochastic
description: A classical particle moving with constant velocity v so that
x − x0 ∼ vt results in r2 ∼ t2 .
In the following sections we take⁶ D = 1 and define
u(x, t) ≡ K(x − x0 , x0 ; t) . (8.10)
8.2 Heat Conduction in a Thin Rod
Consider a thin rod of length L and let T (x, t) be the temperature dis-
tribution within the rod at time t. The two ends of the rod are kept at
constant temperature T (0, t) = T (L, t) = T0 . If the initial temperature
distribution in the rod is T (x, 0), then the temperature distribution at all
times is determined by the diffusion equation
∂T (x, t) ∂ 2 T (x, t)
=α , (8.11)
∂t ∂x2
where α = k/(cp ρ) is the thermal diffusivity, k is the thermal conductivity,
ρ is the density and cp is the specific heat of the rod.
Define 2
T (xL, Lα t) − T0
u(x, t) = , (8.12)
T0
where x ∈ [0, 1]. The function u(x, t), giving the fraction of the tempera-
ture difference to the temperature at the ends of the rod, is dimensionless
and
u(0, t) = u(1, t) = 0 . (8.13)
∫∞
⁵ 0 dr rn e−r /4Dt = 2n Γ( n+1
2 n+1
2 .
2 )(Dt)
⁶According to equation (8.2) this amounts to taking t → Dt.
356 CHAPTER 8. DIFFUSION EQUATION
These are called Dirichlet boundary conditions⁷.
Equation (8.11) becomes
∂u(x, t) ∂ 2 u(x, t)
= (8.14)
∂t ∂x2
Equation (8.6) becomes
∫
d 1 ∂u ∂u
u(x, t)dx = − (8.15)
dt 0 ∂x x=1 ∂x x=0
The relation above cannot be equal to zero at all times due to the
boundary conditions (8.13). This can be easily understood with an ex-
ample. Suppose that
u(x, 0) = sin(πx) , (8.16)
then it is easy to confirm that the boundary conditions are satisfied and
that the function
u(x, t) = sin(πx)e−π t ,
2
(8.17)
is the solution to the diffusion equation. It is easy to see that
∫ 1
2
u(x, t)dx = e−π t
2
0 π
drops exponentially with time and that
∫
d 1
u(x, t)dx = −2πe−π t ,
2
dt 0
which is in agreement with equations (8.15).
The exponential drop of the magnitude of u(x, t) is in agreement with
the expectation that the rod will have constant temperature at long times,
which will be equal to the temperature at its ends (limt→+∞ u(x, t) = 0).
8.3 Discretization
The numerical solution of equation (8.14) will be computed in the interval
x ∈ [0, 1] for t ∈ [0, tf ]. The problem will be defined on a two dimensional
discrete lattice and the differential equation will be approximated by finite
difference equations.
⁷If the derivative ∂u/∂x was given as a boundary condition instead, then we would
have Neumann boundary conditions.
8.3. DISCRETIZATION 357
The lattice is defined by Nx spatial points xi ∈ [0, 1]
xi = 0 + (i − 1)∆x i = 1, . . . , Nx , (8.18)
where the Nx − 1 intervals have the same width
1−0
∆x = , (8.19)
Nx − 1
and by the Nt time points tj ∈ [0, tf ]
tj = 0 + (j − 1)∆t j = 1, . . . , Nt , (8.20)
where the Nt − 1 time intervals have the same duration
tf − 0
∆t = . (8.21)
Nt − 1
We note that the ends of the intervals correspond to
x1 = 0 , xNx = 1 , t1 = 0 , tNt = tf . (8.22)
The function u(x, t) is approximated by its values on the Nx × Nt lattice
ui,j ≡ u(xi , tj ) . (8.23)
The derivatives are replaced by the finite differences
∂u(x, t) u(xi , tj + ∆t) − u(xi , tj ) 1
≈ ≡ (ui,j+1 − ui,j ) , (8.24)
∂t ∆t ∆t
∂ 2 u(x, t) u(xi + ∆x, tj ) − 2u(xi , tj ) + u(xi − ∆x, tj )
2
≈
∂x (∆x)2
1
≡ (ui+1,j − 2ui,j + ui−1,j ) . (8.25)
(∆x)2
By equating both sides of the above relations according to (8.14), we
obtain the dynamic evolution of ui,j in time
∆t
ui,j+1 = ui,j + (ui+1,j − 2ui,j + ui−1,j ) . (8.26)
(∆x)2
This is a one step iterative relation in time. This is very convenient,
because one does not need to store the values ui,j for all j in the computer
memory.
358 CHAPTER 8. DIFFUSION EQUATION
The second term (the “second derivative”) in (8.26) contains only the
nearest neighbors ui±1,j of the lattice point ui,j at a given time slice tj .
Therefore it can be used for all i = 2, . . . , Nx − 1. The relations (8.26)
are not needed for the points i = 1 and i = Nx since the values u1,j =
uNx ,j = 0 are kept constant.
The parameter
∆t
(8.27)
(∆x)2
determines the time evolution in the algorithm. It is called the Courant
parameter and in order to have a time evolution without instabilities it
is necessary to have
∆t 1
2
< . (8.28)
(∆x) 2
This condition will be checked in our analysis empirically.
"d.dat"
1
0.9
1 0.8
0.9 0.7
0.8 0.6
0.7 0.5
u(x,t) 0.6 0.4
0.5 0.3
0.4 0.2
0.3 0.1
0.2 0
0.1
0
1
0.8
0 0.05 0.6
0.1 0.15 0.4 x
0.2 0.25 0.2
t 0.3 0.35
0.4 0
Figure 8.1: The function u(x, t) for Nx=10, Nt=100, tf= 0.4.
8.4 The Program
The fact that equation (8.26) is a one time step iterative relation, leads to
a substantial simplification of the structure of the program. Because of
this, at each time step, it is sufficient to store the values of the second term
(the “second derivative”) in one array. This array will be used in order
8.4. THE PROGRAM 359
to update the values of ui,j . Therefore we will define only two arrays ui ,
i = 1, . . . , Nx and (∂ 2 u/∂x2 )i , i = 1, . . . , Nx which store the values of ui,j
and ∆t/(∆x)2 (ui+1,j − 2ui,j + ui−1,j ) at time tj respectively. In the program
listed below, the names of these arrays are u(P) and d2udx2(P).
The data is stored in the array positions u(1) ... u(Nx) and
d2udx2(1) ... d2udx2(Nx) and the parameter P is taken large enough
so that Nx is always smaller than P.
The user enters the Nx = Nx, Nt =Nt and tf =tf interactively. The
values of ∆x, ∆t and ∆t/∆x2 = courant are calculated during the ini-
tialization.
On exit, we obtain the results in the file d.dat which contains (tj , xi , ui,j )
in three columns. When a time slice is printed, the program prints an
empty line so that the output is easily read by the three dimensional
plotting function splot of gnuplot.
The program is in the file diffusion.f90 and is listed below:
! =======================================================
! 1−dimensional D i f f u s i o n Equation with simple
! D i r i c h l e t boundary c o n d i t i o n s u ( 0 , t )=u ( 1 , t ) =0
! 0<= x <= 1 and 0<= t <= t f
!
! We s e t i n i t i a l c o n d i t i o n u ( x , t =0) t h a t s a t i s f i e s
! t h e g i ven boundary c o n d i t i o n s .
! Nx i s t h e number o f p o i n t s i n s p a t i a l l a t t i c e :
! x = 0 + ( j −1) * dx , j = 1 , . . . , Nx and dx = (1 −0) / ( Nx−1)
! Nt i s t h e number o f p o i n t s i n temporal l a t t i c e :
! t = 0 + ( j −1) * dt , j = 1 , . . . , Nt and dt = ( t f −0) / ( Nt −1)
!
! u(x , 0 ) = sin ( pi * x ) t e s t e d against a n a l y t i c a l solution
! u ( x , t ) = s i n ( p i * x ) * exp(− p i * p i * t )
!
! =======================================================
program diffusion_1d
i m p l i c i t none
i n t e g e r , parameter : : P =100000 ! Max no o f p o i n t s
r e a l ( 8 ) , parameter : : PI =3.1415926535897932 D0
r e a l ( 8 ) , dimension ( P ) : : u , d2udx2
r e a l ( 8 ) : : t , x , dx , dt , tf , courant
i n t e g e r Nx , Nt , i , j
! −−− Input :
p r i n t * , ’ # Enter : Nx , Nt , t f : ( P= ’ , P , ’ Nx must be < P ) ’
read * , Nx , Nt , tf
i f ( Nx . ge . P ) s t o p ’Nx >= P ’
i f ( Nx . l e . 3) s t o p ’Nx <= 3 ’
i f ( Nt . l e . 2) s t o p ’ Nt <= 2 ’
360 CHAPTER 8. DIFFUSION EQUATION
! −−− I n i t i a l i z e :
dx = 1 . 0 D0 / ( Nx −1)
dt = tf / ( Nt −1)
courant = dt / dx * * 2
p r i n t * , ’ # 1d D i f f u s i o n Equation : 0<=x <=1 , 0<=t <= t f ’
p r i n t * , ’ # dx= ’ , dx , ’ dt= ’ , dt , ’ t f = ’ , tf
p r i n t * , ’ # Nx= ’ , Nx , ’ Nt= ’ , Nt
p r i n t * , ’ # Courant Number= ’ , courant
i f ( courant . g t . 0.5 D0 ) p r i n t * , ’ # WARNING: c o u r a n t > 0.5 ’
open ( u n i t =11 , f i l e = ’d . dat ’ ) ! data f i l e
! −−− I n i t i a l c o n d i t i o n a t t =0 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! u(x , 0 ) = sin ( pi x )
do i= 1 , Nx
x = ( i−1) * dx
u ( i ) = s i n ( PI * x )
enddo
u ( 1 ) = 0.0 d0
u ( Nx ) = 0.0 d0
do i= 1 , Nx
x = ( i−1) * dx
w r i t e ( 1 1 , * ) 0.0 D0 , x , u ( i )
enddo
write (11 ,*) ’ ’
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! −−− C a l c u l a t e time e v o l u t i o n :
do j =2 , Nt
t = ( j−1) * dt
! −−−−− second d e r i v a t i v e :
do i =2 , Nx−1
d2udx2 ( i ) = courant * ( u ( i +1) −2.0D0 * u ( i )+u ( i−1) )
enddo
! −−−−− update :
do i =2 , Nx−1
u ( i ) = u ( i ) + d2udx2 ( i )
enddo
do i =1 , Nx
x = ( i−1) * dx
write (11 ,*) t , x , u(i)
enddo
write (11 ,*) ’ ’
enddo ! do j =2 , Nt
close (11)
end program diffusion_1d
8.5. RESULTS 361
8.5 Results
The compilation and running of the program can be done with the com-
mands:
> g f o r t r a n diffusion . f90 −o d
> echo ” 10 100 0.4 ” | . / d
# Enter : Nx , Nt , t f : ( P= 100000 Nx must be < P )
# 1d D i f f u s i o n Equation : 0<=x <=1 , 0<=t <= t f
# dx= 0 . 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 dt= 4.04040404040404040E−3 t f = 0.4
# Nx= 10 Nt= 100
# Courant Number= 0.32727272727272733
The input to the program ./d is read from the stdin and it is given by
the stdout of the command echo through a pipe, as shown in the second
line in the listing above. The lines that follow are the standard output
stdout of the program.
The three dimensional plot of the function u(x, t) can be made with
the gnuplot commands:
gnuplot > s e t pm3d
gnuplot > s e t hidden3d
gnuplot > s p l o t ”d . dat ” with lines
gnuplot > unset pm3d
In order to make the plot of u(x, t) for a fixed value of t we first note that
an empty line in the file d.dat marks a change in time. The following
awk program counts the empty lines of d.dat and prints only the lines
when the number of empty lines that have been encountered so far is
equal to 3. The counter n=0, 1, ..., Nt-1 determines the value of
tj = tn−1 . We save the results in the file tj which can be plotted with
gnuplot. We repeat as many times as we wish:
> awk ’NF<3{n++}n==3 { p r i n t } ’ d . dat > tj
gnuplot > p l o t ” t j ” using 2:3 with lines
The above task can be completed without creating the intermediate file
tj by using the awk filter within gnuplot. For example, the commands
gnuplot > ! echo ”10 800 2” | . / d
gnuplot > plot ”<awk ’NF<3{n++}n==3 { p r i n t } ’ d . dat ” u 2:3 w l
gnuplot > replot ”<awk ’NF<3{n++}n==6 { p r i n t } ’ d . dat ” u 2:3 w l
gnuplot > replot ”<awk ’NF<3{n++}n==10 { p r i n t } ’ d . dat ” u 2:3 w l
362 CHAPTER 8. DIFFUSION EQUATION
gnuplot > replot ”<awk ’NF<3{n++}n==20 { p r i n t } ’ d . dat ” u 2:3 w l
gnuplot > replot ”<awk ’NF<3{n++}n==30 { p r i n t } ’ d . dat ” u 2:3 w l
gnuplot > replot ”<awk ’NF<3{n++}n==50 { p r i n t } ’ d . dat ” u 2:3 w l
gnuplot > replot ”<awk ’NF<3{n++}n==100{ p r i n t } ’ d . dat ” u 2:3 w l
run the program for Nx=10, Nt=800, tf= 2 and construct the plot in
figure 8.2
✂
✁✡
✁✠
✁✟
✎✕✔ ✁✞
✓✒
✑ ✁✝
✎✏✍
✌☞ ✁✆
☛
✁☎
✁✄
✁✂
✁✄ ✁✆ ✁✞ ✁✠ ✂
✖
Figure 8.2: The function u(x, t) for Nx=10, Nt=800, tf= 2 for different values of
the time tj . We take j = 4, 7, 11, 21, 31, 51, 101 and observe that the function u(x, t)
decreases then j increases.
It is instructive to compare the results with the known solution u(x, t) =
sin(πx)e−π t . We compute the relative error
2
ui,j − u(xi , tj )
,
ui,j
which can be done within gnuplot with the commands:
gnuplot > du ( x , y , z ) = ( z − s i n ( pi * x ) * exp(−pi * pi * y ) ) / z
gnuplot > p l o t ”<awk ’NF<3{n++}n==2 ’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
gnuplot > p l o t ”<awk ’NF<3{n++}n==6 ’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
gnuplot > p l o t ”<awk ’NF<3{n++}n==20 ’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
gnuplot > p l o t ”<awk ’NF<3{n++}n==200’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
gnuplot > p l o t ”<awk ’NF<3{n++}n==600’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
gnuplot > p l o t ”<awk ’NF<3{n++}n==780’ d . dat ” u 2 : ( du ( $2 , $1 , $3 ) )
The results can be seen in figure 8.3.
8.6. DIFFUSION ON THE CIRCLE 363
✁✂
☛✒ ✁ ✂
☛☛
☞
☞✑
✎✏✍
✌☞
☛
✁ ✂
✁ ✂
✁✂ ✁✄ ✁☎ ✁✆ ✁✝ ✁✞ ✁✟ ✁✠ ✁✡
✓
Figure 8.3: The absolute value of the relative error of the numerical computation
for Nx=10, Nt=800, tf= 2 for different times tj . We take j = 3, 7, 21, 201, 601, 781 and
observe that the relative error increases with j.
8.6 Diffusion on the Circle
In order to study the kernel K(x, x0 ; t) for the diffusion, or random walk,
problem, we should impose the normalization condition (8.4) for all
times. In the case of the function u(x, t) defined for x ∈ [0, 1] the re-
lation becomes ∫ 1
u(x, t) dx = 1 . (8.29)
0
In order to maintain this relation at all times, it is necessary that the
right hand side of equation (8.15) is equal to 0. One way to impose
this condition is to study the diffusion problem on the circle. If we
parametrize the circle using the variable x ∈ [0, 1], then the points x = 0
and x = 1 are identified and we obtain
∂u(0, t) ∂u(1, t)
u(0, t) = u(1, t) , = . (8.30)
∂x ∂x
The second relation in the above equations makes the right hand side
∫1
of equation (8.15) to vanish. Therefore if 0 u(x, 0) dx = 1, we obtain
∫1
0
u(x, t) dx = 1, ∀t > 0.
Using the above assumptions, the discretization of the differential
equation is done exactly as in the problem of heat conduction. Instead
364 CHAPTER 8. DIFFUSION EQUATION
of keeping the values u(0, t) = u(1, t) = 0, we apply equation (8.26) also
for the points x1 , xNx . In order to take into account the cyclic topology
we take
∆t
u1,j+1 = u1,j + (u2,j − 2u1,j + uNx ,j ) , (8.31)
(∆x)2
and
∆t
uNx ,j+1 = ui,j + (u1,j − 2uNx ,j + uNx −1,j ) , (8.32)
(∆x)2
since the neighbor to the right of the point xNx is the point x1 and the
neighbor to the left of the point x1 is the point xNx . For the rest of the
points i = 2, . . . , Nx − 1 equation (8.26) is applied normally.
The program that implements the problem described above can be
found in the file diffusionS1.f90. The boundary conditions (8.30) are
enforced in the lines
nnr = i+1
i f ( nnr . g t . Nx ) nnr = 1
nnl = i−1
i f ( nnl . l t . 1 ) nnl = Nx
d2udx2 ( i ) = courant * ( u ( nnr ) −2.0D0 * u ( i )+u ( nnl ) )
The initial conditions at t = 0 are chosen so that the particle is located
at xNx /2 . For each instant of time we perform measurements in order to
verify the equations (8.4) and (8.9) and the fact that limt→+∞ u(x, t) =
const. ∑ x
The variable prob = N i=1 ui,j and we should check that its value is
conserved and is always equal to 1.
∑ x
The variable r2 = N i=1 (xi − xNx /2 ) ui,j is a discrete estimator of the
2
expectation value of the distance squared from the initial position. For
small enough times it should follow the law given by equation (8.9).
These variables are written to the file e.dat together with the values
uNx /2,j , uNx /4,j and u1,j . The latter are measured in order to check if for
large enough times they obtain the same constant value according to the
expectation limt→+∞ u(x, t) = const.
The full code is listed below:
! =======================================================
! 1−dimensional D i f f u s i o n Equation with
! p e r i o d i c boundary c o n d i t i o n s u ( 0 , t )=u ( 1 , t )
! 0<= x <= 1 and 0<= t <= t f
!
! We s e t i n i t i a l c o n d i t i o n u ( x , t =0) t h a t s a t i s f i e s
8.6. DIFFUSION ON THE CIRCLE 365
! t h e g i ven boundary c o n d i t i o n s .
! Nx i s t h e number o f p o i n t s i n s p a t i a l l a t t i c e :
! x = 0 + ( j −1) * dx , j = 1 , . . . , Nx and dx = (1 −0) / ( Nx−1)
! Nt i s t h e number o f p o i n t s i n temporal l a t t i c e :
! t = 0 + ( j −1) * dt , j = 1 , . . . , Nt and dt = ( t f −0) / ( Nt −1)
!
! u ( x , 0 ) = \ d e l t a _ {x , 0 . 5 }
!
! =======================================================
program diffusion_1d
i m p l i c i t none
i n t e g e r , parameter : : P =100000 ! Max no o f p o i n t s
r e a l ( 8 ) , parameter : : PI =3.1415926535897932 D0
r e a l ( 8 ) , dimension ( P ) : : u , d2udx2
r e a l ( 8 ) : : t , x , dx , dt , tf , courant , prob , r2 , x0
i n t e g e r Nx , Nt , i , j , nnl , nnr
! −−− Input :
p r i n t * , ’ # Enter : Nx , Nt , t f : ( P= ’ , P , ’ Nx must be < P ) ’
read * , Nx , Nt , tf
i f ( Nx . ge . P ) s t o p ’Nx >= P ’
i f ( Nx . l e . 3) s t o p ’Nx <= 3 ’
i f ( Nt . l e . 2) s t o p ’ Nt <= 2 ’
! −−− I n i t i a l i z e :
dx = 1 . 0 D0 / ( Nx −1)
dt = tf / ( Nt −1)
courant = dt / dx * * 2
p r i n t * , ’ # 1d D i f f u s i o n Equation on S1 : 0<=x <=1 , 0<=t <= t f ’
p r i n t * , ’ # dx= ’ , dx , ’ dt= ’ , dt , ’ t f = ’ , tf
p r i n t * , ’ # Nx= ’ , Nx , ’ Nt= ’ , Nt
p r i n t * , ’ # Courant Number= ’ , courant
i f ( courant . g t . 0.5 D0 ) p r i n t * , ’ # WARNING: c o u r a n t > 0.5 ’
open ( u n i t =11 , f i l e = ’d . dat ’ ) ! data f i l e
open ( u n i t =12 , f i l e = ’ e . dat ’ ) ! data f i l e
! −−− I n i t i a l c o n d i t i o n a t t =0 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
do i= 1 , Nx
x = ( i−1) * dx
u(i) = 0.0 D0
enddo
u ( Nx / 2 ) = 1 . 0 D0
do i= 1 , Nx
x = ( i−1) * dx
w r i t e ( 1 1 , * ) 0.0 D0 , x , u ( i )
enddo
write (11 ,*) ’ ’
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! −−− C a l c u l a t e time e v o l u t i o n :
do j =2 , Nt
t = ( j−1) * dt
! −−−−− second d e r i v a t i v e :
366 CHAPTER 8. DIFFUSION EQUATION
do i =1 , Nx
nnr = i+1
i f ( nnr . g t . Nx ) nnr = 1
nnl = i−1
i f ( nnl . l t . 1 ) nnl = Nx
d2udx2 ( i ) = courant * ( u ( nnr ) −2.0D0 * u ( i )+u ( nnl ) )
enddo
! −−−−− update :
prob = 0.0 D0
r2 = 0.0 D0
x0 = ( ( Nx / 2 ) −1) * dx ! o r i g i n a l p o s i t i o n
do i =1 , Nx
x = ( i−1) * dx
u ( i ) = u ( i ) + d2udx2 ( i )
prob = prob + u ( i )
r2 = r2 + u ( i ) * ( x−x0 ) * ( x−x0 )
enddo
do i =1 , Nx
x = ( i−1) * dx
write (11 ,*) t , x , u(i)
enddo
write (11 ,*) ’ ’
w r i t e ( 1 2 , * ) ’pu ’ , t , prob , r2 , u ( Nx / 2 ) , u ( Nx / 4 ) , u ( 1 )
enddo ! do j =2 , Nt
close (11)
end program diffusion_1d
8.7 Analysis
For each moment of time, the program writes the following quantities to
the file e.dat:
∑
Nx
Uj = ui,j (8.33)
i=1
which is an estimator of (8.29) and we expect to obtain Uj = 1 for all j,
∑
Nx
⟨r2 ⟩j = ui,j (xi − xNx /2 )2 (8.34)
i=1
which is an estimator of (8.9) for which we expect to obtain
⟨r2 ⟩j ∼ 2tj , (8.35)
for small times as well as the values of uNx /2,j , uNx /4,j , u1,j .
8.7. ANALYSIS 367
The values of tj , Uj , ⟨r2 ⟩j , uNx /2,j , uNx /4,j , u1,j are found in columns 2,
3, 4, 5, 6 and 7 respectively of the file e.dat. The gnuplot commands
gnuplot > ! gfortran diffusionS1 . f90 −o d
gnuplot > ! echo ” 10 100 0.4 ” | . / d
compile and run the program within gnuplot. They set Nx = 10, Nt =
100, tf = 0.4, ∆x ≈ 0.111, ∆t ≈ 4.0404, ∆t/∆x2 ≈ 0.327. The gnuplot
✁✝✂ ✠✡☛☞✌✂
✠✡☛☞✌✄
✠✡✝
✁✝
✁ ✆
✁ ☎
✁ ✄
✁ ✂
✁ ✞ ✁✝ ✁✝✞ ✁✂ ✁✂✞ ✁✟ ✁✟✞ ✁✄
Figure 8.4: The functions uNx /2,j , uNx /4,j , u1,j are given as a function of tj for
Nx = 10, Nt = 100, tf = 0.4. We observe that for large times they are consistent with
uniform diffusion.
commands
gnuplot > p l o t ” e . dat ” u 2:5 w l
gnuplot > r e p l o t ” e . dat ” u 2:6 w l
gnuplot > r e p l o t ” e . dat ” u 2 : 7 w l
construct the plot in figure 8.4. We observe that for large times we obtain
uniform diffusion.
The relation Uj = 1 can be easily confirmed by inspecting the values
recorded in the file e.dat.
The asymptotic relation ⟨r2 ⟩j ∼ 2tj can be confirmed with the com-
mands
368 CHAPTER 8. DIFFUSION EQUATION
✁✝
✁ ✆
✁ ☎
✁ ✄
✁ ✂
✠✡☛☞✌✍✎✏
✂✍
✁ ✞ ✁✝ ✁✝✞ ✁✂ ✁✂✞ ✁✟ ✁✟✞ ✁✄
Figure 8.5: The expectation value ⟨r2 ⟩j as a function of tj for Nx = 10, Nt = 100,
tf = 0.4. For small values of tj we obtain ⟨r2 ⟩j ≈ 2tj . The solid line is the straight line
2t.
gnuplot > p l o t [ : ] [ : 0 . 1 1 ] ” e . dat ” u 2 : 4 , 2 * x
which construct the plot in figure 8.5.
Finally we make a plot of the function u(x, t) with the commands
gnuplot > ! echo ” 10 100 0.16 ” | . / d
gnuplot > s e t pm3d
gnuplot > s p l o t [ 0 : 0 . 1 6 ] [ 0 : 1 ] [ 0 : 1 ] ”d . dat ” w l
gnuplot > s p l o t [ 0 : 0 . 1 6 ] [ 0 : 1 ] [ 0 : . 2 ] ”d . dat ” w l
and the result is shown in figure 8.6.
8.7. ANALYSIS 369
✝
✝ ✁✆
✁✆ ✁☎
✁☎
✁✄
✁✄
✁✂ ✁✂
✝
✁ ✂ ✁ ✄ ✁✆
✁ ☎ ✁ ✆ ✁☎
✁✄
✁✝ ✁✝✂ ✁✂
✁✝✄ ✁✝☎
✆✂✝
✆✂✡✠
✆✂✝ ✆✂✡✟
✆✂✡✞
✆✂✡☛ ✆✂✡✝
✆✂✡
✆✂✡ ✆✂✆✠
✆✂✆✟
✆✂✆☛ ✆✂✆✞
✆✂✆✝
✆ ✆
✡
✆ ✆✂✠
✆✂✆✝ ✆✂✟
✆✂✆✞
✆✂✆✟ ✆✂✞
✆✂✆✠
✆✂✡ ✆✂✝
✆✂✡✝ ✁✂✁✄☎
✆✂✡✞ ✆
✆✂✡✟
Figure 8.6: The function u(x, t) for Nx = 10, Nt = 100, tf = 0.16. The second plot
differs only in the scale of the z axis so that we can easily see the details of the diffusion
away from the point x0 ≡ xNx /2 = x5 .
370 CHAPTER 8. DIFFUSION EQUATION
8.8 Problems
8.1 Reproduce the results in figure 8.3.
8.2 The temperature distribution u(x, t) in a thin rod satisfies equation
(8.14) together with the boundary conditions (8.13) at the ends
x = 0, 1. The initial temperature distribution at t = 0 is given by
the function {
0.5 x ∈ [x1 , x2 ]
u(x, 0) = ,
0.3 x ∈/ [x1 , x2 ]
where x1 = 0.25 and x2 = 0.75.
(a) Calculate the temperature distribution u(x, tf ) for tf = 0.0001,
0.001, 0.01, 0.05. Take Nx = 100 and Nt = 1000. Do the same
for tf = 0.1 by choosing appropriate Nx and keeping Nt = 1000.
Plot the functions u(x, tf ) in the same plot.
(b) Calculate the maximum value of the temperature graphically
for tf = 0.0001, 0.001, 0.01, 0.05, 0.1, 0.15, 0.25. Take Nx = 100
and choose an appropriate value for the corresponding Nt .
(c) Calculate the time at which the temperature of the rod becomes
everywhere less than 0.1.
Hint: Make your program print only the final temperature distri-
bution u(x, tf ).
8.3 The temperature distribution u(x, t) in a thin rod satisfies the equa-
tion
∂u ∂ 2u
=α 2.
∂t ∂x
The temperature at the ends of the rod is u(0, t) = u(1, t) = 0, and
when t = 0
{ [ ( )]
0.5 1 − cos 2πxb
0≤x<b
u(x, 0) = .
0 b≤x≤1
(a) Calculate the temperature distribution u(x, tf ) for α = 0.5, b =
0.09 and for tf = 0.0001, 0.001, 0.01, by taking Nx = 300, Nt =
1000. Do the same for tf = 0.05 by choosing appropriate Nx .
Plot the functions u(x, tf ) in the same plot.
(b) Using the same parameters, calculate the time evolution of the
values of the temperature distribution at the points x1 = 0.05,
x2 = 0.50 and x3 = 0.95 for 0 ≤ t ≤ 0.05. Plot the functions
u(x1,2,3 , t) in the same plot.
8.8. PROBLEMS 371
(c) Calculate the temperature distribution u(x, tf ) for b = 0.09 and
α = 5, 2, 1 for tf = 0.001. Plot the functions u(x, tf ) in the
same plot. Comment on the effect of the parameter α on your
results.
8.4 The temperature distribution u(x, t) in a thin rod of length L satisfies
equation
∂u ∂ 2u 4 ∂u
= D(x) 2 − D(x) ,
∂t ∂x L ∂x
where D(x) = ae−4x/L is the x-dependent thermal diffusivity. The
temperature of the rod at its ends is such that u(0, t) = u(L, t) = 0,
and at time t = 0, the temperature distribution is
u(x, 0) = Ce−(x−L/2)
2 /σ 2
.
(a) Write a program where the user enters the parameters L, a, C,
σ, Nx , Nt and tf interactively. On exit, the program calculates
u(x, tf ) and writes the points (xi , u(xi , tf )) in two columns to a
file d.dat.
(b) Run the program for L = 4, a = 0.2, C = 1, σ = 1/2, Nx = 400,
Nt = 20000 and calculate u(x, tf ) for tf = 0.05, 1.0, 5.0. Plot the
functions u(x, tf ) in the same plot.
(c) Using the same parameters, calculate the time evolution of the
temperature distribution at the points x1 = 1 and x2 = 2 for
0 ≤ t ≤ 5. Plot the functions u(x1,2 , t) in the same plot.
8.5 Reproduce the results shown in figures 8.4 and 8.5.
372 CHAPTER 8. DIFFUSION EQUATION
Chapter 9
The Anharmonic Oscillator
In this chapter we will use matrix methods in order to compute the
quantum mechanical energy spectrum of the anharmonic oscillator. This
problem cannot be solved exactly and one has to resort to perturbative or
other approximation methods. We will approach this problem numeri-
cally by representing the Hamiltonian H as a real symmetric matrix in an
appropriately chosen basis of the Hilbert space H of quantum mechani-
cal states. The energy spectrum is obtained from the eigenvalues of this
matrix and the numerical problem reduces to that of the diagonalization
of a real symmetric matrix. Since the Hamiltonian is represented in H
by an infinite size matrix, we have to restrict ourselves to a finite dimen-
sional subspace HN of dimension N . In this space the Hamiltonian is
represented by an N × N real symmetric matrix. The eigenvalues of this
matrix will be calculated numerically using standard methods and the
energy eigenvalues will be obtained in the N → ∞ limit.
For the calculation of the eigenvalues we will use software that is
found in the well known library Lapack which contains high quality,
freely available, linear algebra software. Part of the goals of this chapter
is to show to the reader how to link her programs with software libraries.
In order to solve the same problem using Mathematica or Matlab see [40]
and [41] respectively.
9.1 Introduction
The Hamiltonian of the harmonic oscillator is given by
p2 1
H0 = + mω 2 x2 . (9.1)
2m 2
373
374 CHAPTER 9. THE ANHARMONIC OSCILLATOR
√ √
Define the position and momentum scales x0 = ℏ/(mω), p0 = ℏmω so
that we can express the above equation using dimensionless terms:
( )2 ( )2
H0 1 p 1 x
= + . (9.2)
ℏω 2 p0 2 x0
If we take the units of energy, distance and momentum to be ℏω, x0 and
p0 , then we obtain
1 1
H0 = p2 + x2 , (9.3)
2 2
where H0 , p and x are now dimensionless. The operator H0 can be
diagonalized with the help of the creation and annihilation operators a
and a† , defined by the relations:
1 i
x = √ (a† + a) p = √ (a† − a) , (9.4)
2 2
or
1 1
a = √ (x + ip) a† = √ (x − ip) , (9.5)
2 2
which obey the commutation relation
[a, a† ] = 1 , (9.6)
which leads to
1
H0 = a† a + . (9.7)
2
The eigenstates |n⟩, n = 0, 1, 2, . . . of H0 span the Hilbert space of states
H and satisfy the relations
√ √
a† |n⟩ = n + 1 |n + 1⟩ a |n⟩ = n |n − 1⟩ a |0⟩ = 0 , (9.8)
therefore
a† a |n⟩ = n |n⟩ , (9.9)
and
1
H0 |n⟩ = En |n⟩ , En = n +
. (9.10)
2
The position representation of the eigenstates |n⟩ is given by the wave-
functions:
1
e−x /2 Hn (x) ,
2
ψn (x) = ⟨x|n⟩ = √ √ (9.11)
2n n! π
where Hn (x) are the Hermite polynomials.
9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 375
From equations (9.4) and (9.8) we obtain
1 √ 1 √
xnm = ⟨n| x |m⟩ = √ m + 1 δn,m+1 + √ m δn,m−1 (9.12)
2 2
1 √
= n + m + 1 δ|n−m|,1 (9.13)
2
i √ i √
pnm = ⟨n| p |m⟩ = √ m + 1 δn,m+1 − √ m δn,m−1 . (9.14)
2 2
From the above equations we can easily calculate the Hamiltonian of
the anharmonic oscillator
H(λ) = H0 + λx4 . (9.15)
The matrix elements of H in this representation are:
Hnm (λ) ≡ ⟨n| H(λ) |m⟩ = ⟨n| H0 |m⟩ + λ⟨n| x4 |m⟩ (9.16)
1
= (n + )δn,m + λ(x4 )nm (9.17)
2
where (x4 )nm can be calculated from equation (9.12):
∑
∞
4
(x )nm = xni1 xi1 i2 xi2 i3 xi3 m . (9.18)
i1 ,i2 ,i3 =0
This relation computes the matrix elements of the matrix x4 from the
matrix product of x with itself.
The problem of the calculation of the energy spectrum has now been
reduced to the problem of calculating the eigenvalues of the matrix Hnm .
9.2 Calculation of the Eigenvalues of Hnm(λ)
We start by choosing the dimension N of the subspace HN of the Hilbert
space of states H. We will restrict ourselves to states within this subspace
and we will use the N dimensional representation matrices of x, H0 and
H(λ) in HN . For example, when N = 4 we obtain
0 √12 0 0
√1 0
2 0 1
√
x= 0 1 0
3 (9.19)
√ 2
3
0 0 2
0
376 CHAPTER 9. THE ANHARMONIC OSCILLATOR
1
0 0 02
0 3 0 0
H0 = 2
0 0 5 0
(9.20)
2
0 0 0 27
1 3λ 3λ
√
+ 0 0
2 4 2 √
0 3
+ 15λ 0 3 3
λ
H(λ) =
2 4 2
3λ 5 (9.21)
√ 0 + 27λ 0
2 √ 2 4
0 3 32 λ 0 7
2
+ 15λ
4
Our goal is to write a program that calculates the eigenvalues En (N, λ)
of the N × N matrix Hnm (λ). Instead of reinventing the wheel, we will
use ready made routines that calculate eigenvalues and eigenvectors of
matrices found in the Lapack library. This library can be found in the
high quality numerical software repository Netlib and more specifically
at http://www.netlib.org/lapack/. Documentation can be found at
http://www.netlib.org/lapack/lug/, but it is also easily accessible on-
line by a Google search or by using the man pages¹.
As inexperienced users we will first look for driver routines that per-
form a diagonalization process. Since our task is to diagonalize a real
symmetric matrix, we pick the subroutine DSYEV (D = double precision,
SY = symmetric, EV = eigenvalues with optional eigenvectors). If the
documentation of the library is installed in our system, we may use the
Linux man pages for accessing it:²
> man dsyev
From this page we learn how to use this subroutine:
SUBROUTINE DSYEV ( JOBZ , UPLO , N , A , LDA , W , WORK , LWORK , INFO )
CHARACTER JOBZ , UPLO
INTEGER INFO , LDA , LWORK , N
DOUBLE PRECISION A ( LDA , * ) , W ( * ) , WORK ( * )
ARGUMENTS
JOBZ ( input ) CHARACTER * 1
¹The library can be easily installed in many Linux distributions. For example in
Ubuntu or other Debian like systems you may use the command apt-get install
liblapack3 liblapack-doc liblapack-dev.
²A Google search “dsyev” will easily take you to the same page.
9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 377
= ’N ’ : Compute eigenvalues only ;
= ’V ’ : Compute eigenvalues and eigenvectors .
UPLO ( input ) CHARACTER * 1
= ’ U ’ : Upper triangle of A is stored ;
= ’ L ’ : Lower triangle of A is stored .
N ( input ) INTEGER
The order of the matrix A . N >= 0 .
A ( input / output ) DOUBLE PRECISION array , dimension ( LDA , N←-
)
On entry , the symmetric matrix A . If UPLO = ’ U ’ , the
leading N−by−N upper triangular part of A contains the
upper triangular part of the matrix A . If UPLO = ’ L ’ ,
the leading N−by−N lower triangular part of A contains
the lower triangular part of the matrix A . On exit , if
JOBZ = ’ V ’ , then if INFO = 0 , A contains
the orthonormal eigenvectors of the matrix A . If
JOBZ = ’ N ’ , then on exit the lower triangle ( if UPLO = ’L←-
’)
or the upper triangle ( if UPLO = ’U ’ ) of A , including the
diagonal , is destroyed .
LDA ( input ) INTEGER
The leading dimension of the array A . LDA >= max ( 1 , N ) .
W ( output ) DOUBLE PRECISION array , dimension ( N )
If INFO = 0 , the eigenvalues in ascending order .
WORK ( workspace / output ) DOUBLE PRECISION array , dimension
( LWORK ) .
On exit , if INFO = 0 , WORK ( 1 ) returns the optimal LWORK .
LWORK ( input ) INTEGER
The length of the array WORK . LWORK >= max ( 1 , 3 * N←-
−1) .
For optimal efficiency , LWORK >= ( NB +2) * N , where NB is
the blocksize for DSYTRD returned by ILAENV .
If LWORK = −1 , then a workspace query is assumed ; the
routine only calculates the optimal size of the WORK
array , returns this value as the first entry of the
WORK array , and no error message related to LWORK is
issued by XERBLA .
INFO ( output ) INTEGER
= 0 : successful exit
< 0 : if INFO = −i , the i−th argument had an illegal ←-
value
378 CHAPTER 9. THE ANHARMONIC OSCILLATOR
> 0 : if INFO = i , the algorithm failed to converge ; ←-
i
off−diagonal elements of an intermediate tridiagonal
form did not converge to zero .
These originally cryptic pages contain all the necessary information and
the reader should familiarize herself with its format. For a quick and
dirty use of the routine, all we need to know is the types and usage of its
arguments. These are classified as “input”, “output” and “working space”
variables (some are in more than one classes). Input is the necessary data
that the routine needs in order to perform the computation. Output is
where the results of the computation are stored. And working space
is the memory provided by the user to the routine in order to store
intermediate results.
From the information above we learn that the matrix to be diagonal-
ized is A which is a rectangular matrix with the number of its rows and
columns ≤ N . The number of rows LDA (LDA= “leading dimension of
A”) can be larger than N is which case DSYEV will diagonalize the upper
left N×N part of the matrix³. In our program we define a large matrix
A(LDA,LDA) and diagonalize a smaller submatrix A(N,N). This way we
can study many values of N using the same matrix. The subroutine can
be used in two ways:
• If JOBZ='N', it calculates only the eigenvalues of the matrix A(N,N)
and stores them in the array W(N), sorted in ascending order. We
have to be careful because, upon return, the routine destroys the
upper (UPLO='U') or lower (UPLO='L') triangular part of A. Since
A is symmetric, only this part is needed by DSYEV. If we need to
reuse the matrix A, we have to make a backup copy before the call
to DSYEV.
• If JOBZ='V', it calculates both the eigenvalues and the eigenvectors
of the matrix A(N,N). The eigenvalues are stored in the array W(N)
as before, whereas the corresponding eigenvectors in the columns
of the matrix A(N,N). In order to use the eigenvectors, we can use
a statement like v = A(1:N,j) where the array v(N) stores the
components of the j-th eigenvector of the matrix corresponding to
the eigenvalue λj . The eigenvectors are normalized to unity, i.e.
∑N
i=1 v(i)*v(i)= 1. The matrix A(N,N) is destroyed after the call
³The number LDA is necessary because the matrix element A(i,j) is found after
i+(LDA-1)*j memory positions from A(1,1).
9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 379
to DSYEV and if we need it we have to make a backup copy before
the call.
The reader should also familiarize herself with the use of the workspace
array WORK. This is memory space given to the routine for all its interme-
diate calculations. Determining the size of this array needs some care.
This is given by LWORK and if performance is an issue the reader should
read the documentation carefully for its optimal determination. We will
make the simple choice LWORK=3*LDA-1. The variable INFO is used as a
flag which informs the user whether the calculation was successful, in
which case its value is set to 0. In our case, if INFO takes a non zero
value, the program will abort the calculation.
Before using the program in a complicated calculation, it is necessary
to test its use in a simple, easily controlled problem. We will familiarize
ourselves with the use of DSYEV by writing the following program:
program test_evs
i m p l i c i t none
i n t e g e r , parameter : : P = 100 ! P= LDA
i n t e g e r , parameter : : LWORK = 3* P−1
r e a l ( 8 ) : : A ( P , P ) , W ( P ) , WORK ( LWORK )
i n t e g e r : : N ! DSYEV d i a g o n a l i z e s A(N, N)
integer : : i , j
i n t e g e r : : LDA , INFO
c h a r a c t e r ( 1 ) : : JOBZ , UPLO
! D e f i n e t h e * * symmetric * * matrix t o be d i a g o n a l i z e d
! The s u b r o u t i n e us e s t h e upper t r i a n g u l a r p a r t (UPLO= ’U’ )
! t h e r e f o r e t h e lower t r i a n g u l a r p a r t needs not t o be d e f i n e d
N=4
A ( 1 , 1 ) = −7.7;
A ( 1 , 2 )= 2 . 1 ; A (2 ,2)= 8 . 3 ;
A ( 1 , 3 ) = −3.7; A ( 2 , 3 ) = −16.; A ( 3 , 3 ) =−12.
A ( 1 , 4 ) = 4 . 4 ; A ( 2 , 4 ) = 4 . 6 ; A ( 3 , 4 ) = −1.04; A ( 4 , 4 ) =−3.7
!We p r i n t t h e matrix A b e f o r e c a l l i n g DSYEV s i n c e i t i s
! destroyed a f t e r the c a l l .
do i =1 , N
do j=i , N
p r i n t * , ’A( ’ , i , ’ , ’ , j , ’ )= ’ , A ( i , j )
enddo
enddo
!We ask f o r e i g e n v a l u e s AND e i g e n v e c t o r s ( JOBZ= ’V’ )
JOBZ= ’V ’ ; UPLO= ’U’
p r i n t * , ’COMPUTING WITH DSYEV: ’
LDA=P ! n o t i c e t h a t LDA−> P>N ! !
c a l l DSYEV ( JOBZ , UPLO , N , A , LDA , W , WORK , LWORK , INFO )
p r i n t * , ’DSYEV: DONE. CHECKING NOW: ’
380 CHAPTER 9. THE ANHARMONIC OSCILLATOR
! I f INFO i s nonzero , then t h e r e i s an e r r o r :
i f ( INFO . ne . 0) then
p r i n t * , ’DSYEV FAILED . INF0= ’ , INFO
stop
endif
! P r i n t r e s u l t s : W( I ) has t h e e i g e n v a l u e s :
p r i n t * , ’DSYEV: DONE . : ’
p r i n t * , ’EIGENVALUES OF MATRIX: ’
do i =1 , N
p r i n t * , ’LAMBDA( ’ , i , ’ )= ’ , W ( i )
enddo
! E i g e n v e c t o r s a r e i n s t o r e d i n t h e columns o f A:
p r i n t * , ’EIGENVECTORS OF MATRIX’
do J =1 , N
p r i n t * , ’EIGENVECTOR ’ , j , ’ FOR EIGENVALUE ’ , W ( j )
do i =1 , N
p r i n t * , ’V_ ’ , j , ’ ( ’ , i , ’ )= ’ , A ( i , j )
enddo
enddo
end program test_evs
The next step is to compile and link the program. In order to link
the program to Lapack we have to instruct the linker ld where to find
the libraries Lapack and BLAS⁴ and link them to our program. A library
contains compiled software in archives of object files. The convention for
their names in a Unix environment is to start with the string “lib” fol-
lowed by the name of the library and a .a or .so extension. For example,
in our case the files we are interested in have the names liblapack.so and
libblas.so which can be searched in the file system by the commands:
> l o c a t e libblas
> l o c a t e liblapack
In order to see the files that they contain we give the commands⁵:
> ar −t / usr / lib / libblas . so
> ar −t / usr / lib / liblapack . so
In the commands shown above you may have to substitute /usr/lib
with the path appropriate for your system. If the program is in the file
test.f90, the compilation/linking command is:
⁴The library BLAS contains the basic linear algebra subroutines used by Lapack. In
some versions of the library, one has to only link to Lapack ignoring the link BLAS but
in some other version, linking to BLAS is necessary.
⁵If the .so files don’t exist in your system, try ar -t /usr/lib/libblas.a etc.
9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 381
> g f o r t r a n test . f90 −o test −L / usr / lib −llapack −lblas
The option -L/usr/lib instructs the linker to look for libraries in the
/usr/lib directory⁶, whereas the options -llapack -lblas instructs the
linker to look for any unresolved symbols (i.e. names of external func-
tions and subroutines not coded in our program) first in the library
liblapack.a and then in the library libblas.a.
The command shown above produces the executable file test which,
when run, produces the result:
EIGENVALUES OF MATRIX :
LAMBDA ( 1 )= −21.4119907
LAMBDA ( 2)= −9.93394359
LAMBDA ( 3)= −2.55765591
LAMBDA ( 4)= 18.8035905
EIGENVECTORS OF MATRIX
EIGENVECTOR 1 FOR EIGENVALUE −21.4119907
V_ 1 ( 1 )= −0.197845668
V_ 1 ( 2)= −0.464798676
V_ 1 ( 3)= −0.854691009
V_ 1 ( 4)= 0.119676904
EIGENVECTOR 2 FOR EIGENVALUE −9.93394359
V_ 2( 1 )= 0.824412399
V_ 2( 2)= −0.132429396
V_ 2( 3)= −0.191076519
V_ 2( 4)= −0.516039161
EIGENVECTOR 3 FOR EIGENVALUE −2.55765591
V_ 3( 1 )= 0.502684215
V_ 3( 2)= −0.247784372
V_ 3( 3)= 0.132853329
V_ 3( 4)= 0.817472616
EIGENVECTOR 4 FOR EIGENVALUE 18.8035905
V_ 4( 1 )= 0.168848655
V_ 4( 2)= 0.839659187
V_ 4( 3)= −0.464050682
V_ 4( 4)= 0.226096318
We are now ready to tackle the problem of computing the energy spec-
trum of the anharmonic oscillator. The main program contains the user
interface where the basic parameters for the calculation are read from the
stdin. The user can specify the dimension DIM ≡ N of HN and the cou-
pling constant λ. Then the program computes the eigenvalues En (N, λ)
⁶This is not necessary in our case, since /usr/lib is in the path that ld searches
anyway. This option is useful for libraries located in non conventional paths.
382 CHAPTER 9. THE ANHARMONIC OSCILLATOR
of the N × N matrix Hnm (λ), which represents the action of the operator
H(λ) in the { |n⟩}n=0,1,...,N −1 representation in HN . The tasks are allocated
to the subroutines calculate_X4, calculate_evs and calculate_H. The
subroutine calculate_X4 calculates the N × N matrix (x4 )nm . First, the
matrix xnm is calculated and then (x4 )nm is obtained by computing its
fourth power. The matrix (x4 )nm can also be calculated analytically and
this is left as an exercise to the reader. The subroutine calculate_H calcu-
lates the matrix Hnm (λ) using the result for (x4 )nm given by calculate_X4.
Finally the eigenvalues are calculated in the subroutine calculate_evs
by a call to DSYEV, which are returned to the main program for printing
to the stdout. The program is listed below and can be found in the file
anharmonic.f90:
! ========================================================
program anharmonic_elevels
! ========================================================
i m p l i c i t none
i n t e g e r , parameter :: P = 1000
i n t e g e r , parameter : : LWORK = 3* P−1
integer : : DIM
r e a l ( 8 ) , dimension ( P , P ) : : H , X , X4 ! H a m i l t i o n i a n+ P o s i t i o n Ops
r e a l ( 8 ) , dimension ( P ) :: E ! energy e i g e n v a l u e s
r e a l ( 8 ) , dimension ( LWORK ) : : WORK
real (8) : : lambda
integer :: i
print * , ’ # Enter H i l b e r t Space dimension : ’
read * ,DIM
print * , ’ # Enter lambda : ’
read * , lambda
print * , ’ # lambda= ’ , lambda
! Print Message :
print * , ’ # ################################################ ’
print * , ’ # Energy spectrum o f anharmonic o s c i l l a t o r ’
print * , ’ # using matrix methods . ’
print * , ’ # H i l b e r t Space Dimension DIM = ’ ,DIM
print * , ’ # lambda c o u p l i n g = ’ , lambda
print * , ’ # ################################################ ’
print * , ’ # Outpout : DIM lambda E_0 E_1 . . . . E_{N−1} ’
print * , ’ # −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−’
! C a l c u l a t e X^4 o p e r a t o r :
c a l l calculate_X4 ( X , X4 , DIM)
! Calculate eigenvalues :
c a l l calculate_evs ( H , X4 , E , WORK , lambda , DIM)
w r i t e ( 6 , 1 0 0 ) ’EV ’ ,DIM, lambda , ( E ( i ) , i =1 ,DIM)
9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 383
100 FORMAT( A3 , I8 ,20000 G25 . 1 5 )
end program anharmonic_elevels
! ========================================================
s u b r o u t i n e calculate_evs ( H , X4 , E , WORK , lambda , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter :: P = 1000
i n t e g e r , parameter : : LWORK = 3* P−1
r e a l ( 8 ) , dimension ( P , P ) : : H , X4
r e a l ( 8 ) , dimension ( P ) :: E
r e a l ( 8 ) , dimension ( LWORK ) : : WORK
integer : : DIM
real (8) : : lambda
character (1) : : JOBZ , UPLO
integer : : LDA , INFO , i , j
c a l l calculate_H ( H , X4 , lambda , DIM)
JOBZ= ’V ’ ; UPLO= ’U’
c a l l DSYEV ( JOBZ , UPLO , DIM, H , P , E , WORK , LWORK , INFO )
p r i n t * , ’ # * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * * * ’
do j =1 ,DIM
w r i t e ( 6 , 1 0 1 ) ’ # EVEC ’ , lambda , ( H ( i , j ) , i =1 ,DIM)
enddo
p r i n t * , ’ # * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * * * ’
101 FORMAT( A7 , F15 .3 ,20000 G14 . 6 )
! I f INFO i s nonzero then we have an e r r o r
i f ( INFO . ne . 0) then
p r i n t * , ’ dsyev f a i l e d . INFO= ’ , INFO
stop
endif
end s u b r o u t i n e calculate_evs
! ========================================================
s u b r o u t i n e calculate_H ( H , X4 , lambda , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P = 1000
r e a l ( 8 ) , dimension ( P , P ) : : H , X4
integer : : DIM
real (8) : : lambda
integer :: i,j
do j =1 ,DIM
do i =1 ,DIM
H ( i , j )=lambda * X4 ( i , j )
enddo
H ( j , j ) = H ( j , j ) + DBLE( j ) − 0.5 D0 ! E_n=n + 1 / 2 , n=j −1=>E_n=j −1/2
enddo
384 CHAPTER 9. THE ANHARMONIC OSCILLATOR
print * , ’# * * * * * * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * * * ’
do j =1 ,DIM
w r i t e ( 6 , 1 0 2 ) ’ # HH ’ , ( H ( i , j ) , i =1 ,DIM)
enddo
print * , ’# * * * * * * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * * * ’
102 FORMAT( A5 ,20000 G20 . 6 )
end s u b r o u t i n e calculate_H
! ========================================================
s u b r o u t i n e calculate_X4 ( X , X4 , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P=1000
r e a l ( 8 ) , dimension ( P , P ) : : X , X4 , X2
integer : : DIM
integer :: i,j,m,n
r e a l ( 8 ) , parameter : : isqrt2 =1.0 D0 / s q r t ( 2 . 0 D0 )
! Compute t h e p o s i t i o n o p e r a t o r :
X = 0.0 D0
! Compute t h e nonzero e l e m e n t s
do i =1 ,DIM
n=i−1 ! i n d i c e s 0 , . . . , DIM−1
! The d e l t a _ {n ,m+1} term , i . e . m=n−1
m=n−1 ! t h e energy l e v e l n −> i =n+1 , m−> j =m+1
j=m+1
i f ( j . ge . 1 ) X ( i , j )=isqrt2 * s q r t (DBLE( m +1) )
! The d e l t a _ {n ,m−1} term , i . e . m=n+1
m=n+1
j=m+1
i f ( j . l e .DIM) X ( i , j )=isqrt2 * s q r t (DBLE( m ) )
enddo
! Compute t h e Hamiltonian o p e r a t o r :
! S t a r t with t h e X^4 o p e r a t o r :
X2 = MATMUL( X , X ) ! f i r s t X2 , then X4 :
X4 = MATMUL( X2 , X2 )
end s u b r o u t i n e calculate_X4
9.3 Results
Compiling and running the program can be done with the commands:
> g f o r t r a n −O2 anharmonic . f90 −o an −llapack −lblas
> . / an
# Enter H i l b e r t Space dimension :
4
# Enter lambda :
9.3. RESULTS 385
n=0
0.8
λ=0.9
λ=0.2
0.75
0.7
En
0.65
0.6
0.55
0.5
0 0.1 0.2 0.3 0.4 0.5
1/N
Figure 9.1: The ground state energy E0 (λ) for λ = 0.2, 0.9 is calculated in the large N
limit of the eigenvalues E0 (N, λ). Convergence is satisfactory for relatively small values
of N and it is slightly faster for λ = 0.2 than it is for λ = 0.9.
0.0
.....
# ********************** H *******************
# HH 0.50 0.00 0.00 0.00
# HH 0.00 1.50 0.00 0.00
# HH 0.00 0.00 2.50 0.00
# HH 0.00 0.00 0.00 3.50
# ********************** H *******************
# * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * *
# EVEC 0.000 1.00 0.00 0.00 0.00
# EVEC 0.000 0.00 1.00 0.00 0.00
# EVEC 0.000 0.00 0.00 1.00 0.00
# EVEC 0.000 0.00 0.00 0.00 1.00
# * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * *
EV 4 0.000 0.50 1.50 2.50 3.50
In the above program we used N = 4 and λ = 0. The λ = 0 choice leads
us to the simple harmonic oscillator and we obtain the expected solutions:
Hnm = (n + 1/2)δn,m , E
∑n 3= (n + 1/2) and the eigenstates (eigenvectors of
Hnm ) |n⟩λ=0 = |n⟩ = m=0 δn,m |m⟩. Similar results can be obtained for
larger N .
For non zero values of λ, the finite N calculation contains systematic
errors from neglecting all the matrix elements Hnm (λ) for n ≥ N or
m ≥ N . Our program calculates the eigenvalues En (N, λ) of the finite
matrix Hnm (λ), m, n = 0, . . . , N − 1 and one expects that
En (λ) = lim En (N, λ) , (9.22)
N →∞
386 CHAPTER 9. THE ANHARMONIC OSCILLATOR
n=9
140
λ=0.9
120 λ=0.2
100
80
En
60
40
20
0
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
1/N
Figure 9.2: The 9th excited state E9 (λ) for λ = 0.2, 0.9 is given by the large N limit
of the eigenvalues E9 (N, λ).
where
H(λ) |n⟩λ = En (λ) |n⟩λ , (9.23)
is the true n-th level eigenvalue of the Hamiltonian H(λ). In practice
the limit 9.22 for given λ and n is calculated by computing En (N, λ)
numerically for increasing values of N . If convergence to a desired level
of accuracy is achieved for the accessible values of N , then the approached
limit is taken as an approximation to En (λ). This process is shown
graphically in figures 9.1-9.3 for λ = 0.2, 0.9. Convergence is satisfactory
for quite small N for n = 0, 9 but larger values of N are needed for n = 20.
Increasing the value of n for fixed λ makes the use of larger values of N
necessary. Similarly for a given energy level n, increasing λ also makes
the use of larger values of N necessary. A session that computes this
limit for the ground level energy E0 (λ = 0.9) is shown below⁷:
> tcsh
> g f o r t r a n −O2 anharmonic . f90 −llapack −lblas −o an
> f o r e a c h N (4 8 12 16 24 32)
f o r e a c h ? ( echo $N ; echo 0 . 9 ) | . / an >> data
f o r e a c h ? end
> grep ^ EV data | awk ’{ p r i n t $2 , $4 } ’
4 0.711467845686790
8 0.786328966767866
12 0.785237674919165
⁷The foreach loop construct is special to the tcsh shell. This is why an explicit tcsh
command is shown. For other shells use their corresponding syntax.
9.3. RESULTS 387
n = 20
500
λ=0.9
450 λ=0.2
400
350
300
En
250
200
150
100
50
0
0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
1/N
Figure 9.3: The 20th excited state E20 (λ) for λ = 0.2, 0.9 is given by the large N
limit of the eigenvalues E20 (N, λ). Convergence has not been achieved for the displayed
values of N ≤ 80.
16 0.784964461939594
24 0.785032515135677
32 0.785031492177730
> gnuplot
gnuplot > plot ”<grep ^EV data | awk ’{ p r i n t 1 / $2 , $4 } ’ ”
Further automation of this process can be found in the shell script file
anharmonic.csh in the accompanying software. We note the large N
convergence of E0 (N, 0.9) and that we can take E0 (0.9) ≈ 0.78503. For
higher accuracy, a computation using larger N will be necessary.
We can also compute the expectation values ⟨A⟩n (λ) of an operator
A = A(p, q) when the anharmonic oscillator is in a state |n⟩λ :
⟨A⟩n (λ) = λ ⟨n| A |n⟩λ . (9.24)
In practice, the expectation value will be computed from the limit
⟨A⟩n (λ) = lim ⟨A⟩n (N, λ) ≡ lim N,λ ⟨n| A |n⟩N,λ , (9.25)
N →∞ N →∞
where |n⟩N,λ are the eigenvectors of the finite N × N matrix Hnm (λ) com-
puted numerically by DSYEV. These are determined by their components
cm (N, λ), where
∑
N −1
|n⟩N,λ = cm (N, λ) |m⟩ , (9.26)
m=0
which are stored in the columns of the array H after the call to DSYEV:
cm (N, λ) = H(m + 1, n + 1) . (9.27)
388 CHAPTER 9. THE ANHARMONIC OSCILLATOR
Substituting equation (9.26) to (9.24) we obtain
∑
N −1
⟨A⟩n (λ) = c∗m (N, λ)cm′ (N, λ)Amm′ , (9.28)
m,m′ =0
and we can use (9.27) for the computation of the sum.
As an application, consider the expectation values of the operators
x2 , x4 and p2 . Taking √ ⟨x⟩n = ⟨p⟩n = 0, √
√into account that we obtain the
uncertainties ∆xn ≡ ⟨x ⟩n − ⟨x⟩n = ⟨x ⟩n and ∆pn = ⟨p2 ⟩n . Their
2 2 2
product should satisfy Heisenberg’s uncertainty relation ∆xn · ∆pn ≳ 1/2.
The results are shown in table 9.1 and in figures 9.4-9.5. The calculation
is left as an exercise to the reader.
10
∆X ∆P
<X2>1/2
<P2>1/2
8
6
4
2
0
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1/N
Figure 9.4: The expectation values ⟨x2 ⟩1/2 2 1/2
n (λ), ⟨p ⟩n (λ) and the product of un-
certainties ∆xn · ∆pn for n = 9 and λ = 0.5 calculated from the large N limits of
1/2 1/2
⟨x2 ⟩n (N, λ), ⟨p2 ⟩n (N, λ).
The physics of the anharmonic oscillator can be better understood by
studying the large λ limit. As shown in figure 9.5, the term λx4 dominates
in this limit and the expectation value ⟨x2 ⟩n (λ) decreases. This means
that states that confine the oscillator to a smaller range of x are favored.
This, using the uncertainty principle, implies that the typical momentum
of the oscillator also increases in magnitude. This is confirmed in figure
9.5 where we observe the expectation value ⟨p2 ⟩n (λ) to increase with λ.
In order to understand quantitatively these competing effects we will use
a scaling argument due to Symanzik. We redefine x → λ−1/6 x, p → λ1/6 p
in the Hamiltonian H(λ) = p2 /2 + x2 /2 + λx4 and for large enough λ we
9.3. RESULTS 389
14
12
10
8 ∆X ∆P
<P2>1/2
<X2>1/2
6
4
2
0
0 20 40 60 80 100
λ
Figure 9.5: The expectation values ⟨x2 ⟩1/2 2 1/2
n (λ), ⟨p ⟩n (λ) and the product of uncer-
tainties ∆xn · ∆pn for n = 9.
obtain⁸ the asymptotic behavior
H(λ) ∼ λ1/3 h(1) , λ → ∞, (9.29)
where h(λ) = p2 /2+λx4 is the Hamiltonian of the anharmonic “oscillator”
with ω = 0. Since the operator h(1) is independent of λ, the energy
spectrum will have the asymptotic behavior
En (λ) ∼ Cn λ1/3 , λ → ∞. (9.30)
In reference [42] it is shown that for λ > 100 we have that
( )
E0 (λ) = λ1/3 0.667 986 259 18 + 0.143 67λ−2/3 − 0.0088λ−4/3 + . . . ,
(9.31)
6
with an accuracy better than one part in 10 . For large values of n, the
authors obtain the asymptotic behavior
( )4/3
1
En (λ) ∼ Cλ 1/3
n+ , λ → ∞,n → ∞, (9.32)
2
where C = 34/3 π 2 /Γ(1/4)8/3 ≈ 1.376 507 40. This relation is tested in figure
9.6 where we observe good agreement with our calculations.
⁸For x → λ−1/6 x, H → λ1/3 (p2 /2 + λ−2/3 x2 /2 + x4 ), therefore in the limit λ → ∞ the
second term vanishes and we obtain equation (9.29).
390 CHAPTER 9. THE ANHARMONIC OSCILLATOR
λ = 0.5 λ = 2.0
n ⟨x ⟩ 2
⟨p ⟩
2
∆x · ∆p ⟨x ⟩
2
⟨p2 ⟩ ∆x · ∆p
0 0.305814 0.826297 0.502686 0.21223 1.19801 0.504236
1 0.801251 2.83212 1.5064 0.540792 4.21023 1.50893
2 1.15544 5.38489 2.49438 0.761156 8.15146 2.49089
3 1.46752 8.28203 3.48627 0.958233 12.6504 3.48166
4 1.75094 11.4547 4.47845 1.13698 17.596 4.47285
5 2.01407 14.8603 5.47079 1.30291 22.9179 5.46443
6 2.2617 18.4697 6.4632 1.45905 28.5683 6.45619
7 2.49696 22.2616 7.45562 1.60735 34.5124 7.44805
8 2.72198 26.2196 8.44804 1.74919 40.7234 8.43998
9 2.93836 30.3306 9.44045 1.88558 47.1801 9.43194
Table 9.1: The expectation values ⟨x2 ⟩, ⟨p2 ⟩, ∆x · ∆p for the√
anharmonic oscillator for
√ |n⟩, n = 0, . . . , 9. We observe a decrease of ∆x = ⟨x ⟩ and an increase of
the states 2
∆p = ⟨p ⟩ as λ is increased. The product ∆x · ∆p seems to remain very close to the
2
values (n + 1/2) of the harmonic oscillator for both values of λ.
9.4 The Double Well Potential
We can also use matrix methods in order to calculate the energy spectrum
of a particle in a double well potential given by the Hamiltonian:
p2 x2 x4
H= − +λ . (9.33)
2 2 4
The equilibrium points of the classical motion are located at the minima:
1 1
x0 = ± √ , Vmin = − . (9.34)
λ 4λ
When the well is very deep, then for the lowest energy levels the potential
can be well approximated by that of a harmonic oscillator with angular
frequency ω 2 = V ′′ (x0 ), therefore
1
Emin ≈ Vmin + ω . (9.35)
2
In this case the tunneling effect is very weak and the energy levels are
arranged in almost degenerate pairs. The corresponding eigenstates are
symmetric and antisymmetric linear combinations of states localized near
the left and right minima of the potential. For example, for the two lowest
Table 9.2: Numerical calculation of the energy levels of the anharmonic oscillator given in reference [42].
λ E0 E1 E2 E3 E4
0.002 0.501 489 66 1.507 419 39 2.519 202 12 3.536 744 13 4.559 955 56
0.006 0.504 409 71 1.521 805 65 2.555 972 30 3.606 186 33 4.671 800 37
0.01 0.507 256 20 1.535 648 28 2.590 845 80 3.671 094 94 4.774 913 12
0.05 0.532 642 75 1.653 436 01 2.873 979 63 4.176 338 91 5.549 297 81
0.1 0.559 146 33 1.769 502 64 3.138 624 31 4.628 882 81 6.220 300 90
0.3 0.637 991 78 2.094 641 99 3.844 782 65 5.796 573 63 7.911 752 73
0.5 0.696 175 82 2.324 406 35 4.327 524 98 6.578 401 95 9.028 778 72
0.7 0.743 903 50 2.509 228 10 4.710 328 10 7.193 265 28 9.902 610 70
1 0.803 770 65 2.737 892 27 5.179 291 69 7.942 403 99 10.963 5831
2 0.951 568 47 3.292 867 82 6.303 880 57 9.727 323 19 13.481 2759
50 2.499 708 77 8.915 096 36 17.436 9921 27.192 6458 37.938 5022
200 3.930 931 34 14.059 2268 27.551 4347 43.005 2709 60.033 9933
1000 3.694 220 85 23.972 2061 47.017 3387 73.419 1140 102.516 157
8000 13.366 9076 47.890 7687 93.960 6046 146.745 512 204.922 711
20000 18.137 2291 64.986 6757 127.508 839 199.145 124 278.100 238
λ E5 E6 E7 E8
0.002 5.588 750 05 6.623 044 60 7.662 759 33 8.707 817 30
9.4. THE DOUBLE WELL POTENTIAL
0.006 5.752 230 87 6.846 948 47 7.955 470 29 9.077 353 66
0.01 5.901 026 67 7.048 326 88 8.215 837 81 9.402 692 31
0.05 6.984 963 10 8.477 397 34 10.021 9318 11.614 7761
0.1 7.899 767 23 9.657 839 99 11.487 3156 13.378 9698
0.3 10.166 4889 12.544 2587 15.032 7713 17.622 4482
0.5 11.648 7207 14.417 6692 17.320 4242 20.345 1931
0.7 12.803 9297 15.873 6836 19.094 5183 22.452 9996
1 14.203 1394 17.634 0492 21.236 4362 24.994 9457
2 17.514 1324 21.790 9564 26.286 1250 30.979 8830
50 49.516 4187 61.820 3488 74.772 8290 88.314 3280
391
200 78.385 6232 97.891 3315 118.427 830 139.900 400
1000 133.876 891 167.212 258 202.311 200 239.011 580
8000 267.628 498 334.284 478 404.468 350 477.855 700
20000 363.201 843 453.664 875 548.916 140 648.515 330
392 CHAPTER 9. THE ANHARMONIC OSCILLATOR
1.41
n=1
n=2
n=5
1.405 n=9
n=20
C
1.4
En λ-1/3 (n+1/2)-4/3
1.395
1.39
1.385
1.38
1.375
0 500 1000 1500 2000
λ
Figure 9.6: Test of the asymptotic relation (9.32). The vertical axis is
En λ−1/3 (n + 1/2)−4/3 where for large enough n and λ should approach the value
C = 34/3 π 2 /Γ(1/4)8/3 ≈ 1.376 507 40 (horizontal line).
energy levels we expect that
∆
E0,1 ≈ Emin ± , (9.36)
2
where ∆ ≪ |Emin | and
|+⟩ + |−⟩ |+⟩ − |−⟩
|0⟩λ ≈ √ , |1⟩λ ≈ √ , (9.37)
2 2
where the states |+⟩ and |−⟩ are localized to the left and right well of
the potential respectively (see also figure 10.4 of chapter 10).
We will use equations (9.12) in order to calculate the Hamiltonian
(9.33). We need to make very small modifications to the code in the file
anharmonic.f90. We will only add a routine that calculates the matrices
pnm . The resulting program can be found in the file doublewell.f90:
! ========================================================
program doublewell_elevels
! ========================================================
! H : Hamiltonian o p e r a t o r H0+( lambda / 4 ) *X^4
9.4. THE DOUBLE WELL POTENTIAL 393
1.5 λ=0.2
λ=0.1
1
0.5
0
V(x)
-0.5
-1
-1.5
-2
-2.5
-4 -2 0 2 4
x
Figure 9.7: The potential energy V (x) for the double well potential for λ = 0.1, 0.2.
! H0 : Hamiltonian H0=1/2 P^2 −1/2 X^2
! X, X2 , X4 : P o s i t i o n o p e r a t o r and i t s powers
! iP : i P operator
! P2 : P^2 = −( i P ) ( i P ) o p e r a t o r
! E : Energy e i g e n v a l u e s
! WORK : Workspace f o r l a p a c k r o u t i n e DSYEV
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P=1000
i n t e g e r , parameter : : LWORK =3*P−1
r e a l ( 8 ) , dimension ( P , P ) : : H , H0 , X , X4 , X2 , iP , P2
r e a l ( 8 ) , dimension ( P ) :: E
r e a l ( 8 ) , dimension ( LWORK ) : : WORK
real (8) : : lambda , lambda0 , lambdaf , dlambda
integer : : DIM0 , DIMF , dDIM , DIM
integer :: i
! Minimum and maximum v a l u e s o f H i l b e r t sp a ce dimensions :
p r i n t * , ’ Enter H i l b e r t Space dimensions (DIM0 , DIMF, DDIM) : ’
read * , DIM0 , DIMF , DDIM
! Minimum and maximum v a l u e s o f lambda ( s t e p dlambda ) :
p r i n t * , ’ Enter lambda0 , lambdaf , dlambda : ’
read * , lambda0 , lambdaf , dlambda
p r i n t * , ’ lambda0= ’ , lambda0
! P r i n t Message :
p r i n t * , ’ # ################################################ ’
p r i n t * , ’ # Energy l e v e l s o f double w e l l p o t e n t i a l ’
p r i n t * , ’ # using matrix methods . ’
p r i n t * , ’ # H i l b e r t Space Dimensions = ’ , DIM0 , ’ − ’ , DIMF ,&
’ s t e p= ’ , dDIM
394 CHAPTER 9. THE ANHARMONIC OSCILLATOR
p r i n t * , ’ # lambda c o u p l i n g = ’ , lambda0 , ’ − ’ , lambdaf ,&
’ s t e p= ’ , dlambda
p r i n t * , ’ # ################################################ ’
p r i n t * , ’ # Outpout : DIM lambda E_0 E_1 . . . . E_{N−1} ’
p r i n t * , ’ # −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−’
do DIM=DIM0 , DIMF , dDIM
c a l l calculate_operators ( X , X2 , X4 , iP , P2 , H0 , DIM)
lambda = lambda0
do while ( lambda . l e . lambdaf )
c a l l calculate_evs ( H , H0 , X4 , E , WORK , lambda , DIM)
w r i t e ( 6 , 1 0 0 ) ’EV ’ ,DIM, lambda , ( E ( i ) , i =1 ,DIM)
lambda = lambda+dlambda
enddo
enddo
100 FORMAT( A3 , I5 ,1000 G25 . 1 5 )
end program doublewell_elevels
! ========================================================
s u b r o u t i n e calculate_evs ( H , H0 , X4 , E , WORK , lambda , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P=1000
i n t e g e r , parameter : : LWORK =3*P−1
r e a l ( 8 ) , dimension ( P , P ) : : H , H0 , X4
r e a l ( 8 ) , dimension ( P ) :: E
r e a l ( 8 ) , dimension ( LWORK ) : : WORK
integer : : DIM
real (8) : : lambda
character (1) : : JOBZ , UPLO
integer : : LDA , INFO , i , j
c a l l calculate_H ( H , H0 , X4 , lambda , DIM)
JOBZ= ’V ’ ; UPLO= ’U’
c a l l DSYEV ( JOBZ , UPLO , DIM, H , P , E , WORK , LWORK , INFO )
p r i n t * , ’ # * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * * * ’
do j =1 ,DIM
w r i t e ( 6 , 1 0 1 ) ’ # EVEC ’ ,DIM, lambda , ( H ( i , j ) , i =1 ,DIM)
enddo
p r i n t * , ’ # * * * * * * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * * * ’
101 FORMAT( A7 , I5 , F8 . 4 , 1 0 0 0 G14 . 6 )
i f ( INFO . ne . 0) then
p r i n t * , ’ dsyev f a i l e d . INFO= ’ , INFO
stop
endif
end s u b r o u t i n e calculate_evs
9.4. THE DOUBLE WELL POTENTIAL 395
! ========================================================
s u b r o u t i n e calculate_H ( H , H0 , X4 , lambda , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P=1000
r e a l ( 8 ) , dimension ( P , P ) : : H , H0 , X4
integer : : DIM
real (8) : : lambda
integer :: i,j
do j =1 ,DIM
do i =1 ,DIM
H ( i , j )=H0 ( i , j ) +0.25 D0 * lambda * X4 ( i , j )
enddo
enddo
print * , ’# * * * * * * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * * * ’
do j =1 ,DIM
w r i t e ( 6 , 1 0 2 ) ’ # HH ’ , ( H ( i , j ) , i =1 ,DIM)
enddo
print * , ’# * * * * * * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * * * ’
102 FORMAT( A5 ,1000 G14 . 6 )
end s u b r o u t i n e calculate_H
! ========================================================
s u b r o u t i n e calculate_operators ( X , X2 , X4 , iP , P2 , H0 , DIM)
! ========================================================
i m p l i c i t none
i n t e g e r , parameter : : P=1000
r e a l ( 8 ) , dimension ( P , P ) : : X , X4 , X2 , iP , P2 , H0
integer : : DIM
integer :: i,j,m,n
r e a l ( 8 ) , parameter : : isqrt2 =1.0 D0 / s q r t ( 2 . 0 D0 )
X =0.0 D0 ; X2 =0.0 D0 ; X4 =0.0 D0
iP =0.0 D0 ; P2 =0.0 D0
do i =1 ,DIM
n=i−1 ! i n d i c e s 0 , . . . , DIM−1
! The d e l t a _ {n ,m+1} term , i . e . m=n−1
m=n−1 ! energy l e v e l : n −> i =n+1 , m−> j =m+1
j=m+1
i f ( j . ge . 1 ) X ( i , j ) = isqrt2 * s q r t (DBLE( m +1) )
i f ( j . ge . 1 ) iP ( i , j ) = −isqrt2 * s q r t (DBLE( m +1) )
! The d e l t a _ {n ,m−1} term , i . e . m=n+1
m=n+1
j=m+1
i f ( j . l e .DIM) X ( i , j ) = isqrt2 * s q r t (DBLE( m ) )
i f ( j . l e .DIM) iP ( i , j ) = isqrt2 * s q r t (DBLE( m ) )
396 CHAPTER 9. THE ANHARMONIC OSCILLATOR
enddo ! do i =1 ,DIM
X2 = MATMUL( X , X )
P2 = −MATMUL( iP , iP )
X4 = MATMUL( X2 , X2 )
! The H a m i l t i o n i a n :
H0 = 0.5 D0 * ( P2−X2 )
end s u b r o u t i n e calculate_operators
Where is the particle’s favorite place when it is in the states |+⟩ and |−⟩?
The answer to this question is obtained from the study of the expectation
value of the position operator ⟨x⟩ in each one of them. We know that
when the particle is in one of the energy eigenstates, then we have that
⟨x⟩n (λ) = λ ⟨n| x |n⟩λ = 0 (9.38)
because the potential V (x) = V (−x) is even. Therefore
⟨x⟩± (λ) = ⟨±| x |±⟩
1
= √ (λ ⟨0| x |0⟩λ ± λ ⟨1| x |0⟩λ ± λ ⟨0| x |1⟩λ + λ ⟨1| x |0⟩λ )
2
√
= ± 2⟨1| x |0⟩λ , (9.39)
where in the last line we used the relation (9.38) λ ⟨0| x |0⟩λ = λ ⟨1| x |1⟩λ =
0 and that the amplitudes λ ⟨1| x |0⟩λ = λ ⟨0| x |1⟩λ . Also⁹ we have that
∑∞ (0)
λ ⟨1| x |0⟩λ > 0. Therefore, if we have that |0⟩λ = m=0 cm |m⟩ and |1⟩λ =
∑∞ (1)
m=0 cm |m⟩, we obtain
√ ∑∞
(1)
⟨x⟩± (λ) = ± 2 c(0)
m cm′ Xmm′ . (9.40)
m,m′ =0
Given that for finite N , the subroutine DSYEV returns approximations to
(n) (n)
the coefficients cm in the columns of the matrix H(DIM,DIM) so that cm ≈
H(m+1,n+1), you√ may compare the value of ⟨x⟩± (λ) with the classical
values x0 = ±1/ λ as λ is increased.
⁹You may convince yourselves by looking at the wave functions in figures 10.4 of
chapter 10 and by computing the relevant integrals.
9.4. THE DOUBLE WELL POTENTIAL 397
1
0.01
0.0001
∆n
1e-06
1e-08
1e-10 n=0
n=6
n=30
0.01 0.1 1 10 100
λ
Figure 9.8: Calculation of the difference of the energy levels ∆n = En+1 − En for
n = 0, 6, 30 for the double well potential from the program doublewell.f90. The
difference vanishes√as the well becomes deeper with decreasing λ. The states |±⟩ =
( |n + 1⟩λ ± |n⟩λ )/ 2 are more and more localized to the right or left well respectively.
398 CHAPTER 9. THE ANHARMONIC OSCILLATOR
9.5 Problems
9.1 Calculate the matrix H(λ) for N = 2, 3 analytically. Calculate its
eigenvalues for N = 2. Compare your results with the numerical
values that you obtain from your program.
9.2 Add the necessary code to the program in the file test.f90 so
that it checks that the eigenvectors satisfy their defining relations
A vi = λi vi and that they form an orthonormal basis vi · vj = δij .
9.3 Calculate E5 (λ) and E9 (λ) for λ = 0.8, 1.2 with an accuracy better
than 0.01%.
9.4 For how large n can you calculate En (λ) for λ = 1 with an accuracy
better than 2% when N = 64?
9.5 Calculate E3 (λ) and E12 (λ) for 0 ≤ λ ≤ 4 with step δλ = 0.2 by
achieving accuracy better than 0.01%. How large should N be taken
in each case?
9.6 Calculate the expression that gives the matrix elements of the oper-
ator x4 in the |n⟩ representation analytically. Modify the program
in anharmonic.f90 in order to incorporate your calculation. Verify
that the results are the same and test if it has an effect in the to-
tal computation time with and without calculating the eigenvalues
and eigenvectors of the Hamiltonian. Compute in each case the de-
pendence of the cpu time on N by computing the exponent (cpu
time)∼ N a for N = 40 − 1000.
9.7 Modify the code in the file anharmonic.f90 so that the arrays H, X,
X4, E, WORK are ALLOCATABLE and their dimension is determined
by the variable DIM read by the program interactively.
(Hint: Look at the file anharmonicSPEED.f90.)
9.8 Make an attempt to reproduce the results of Hioe and Montroll [42]
given in table 9.2 for n = 3 and n = 5. What is the largest value of
λ that you can study given your computational resources?
9.9 Make an attempt to reproduce the results of Hioe and Montroll [42]
given by equation (9.31). Calculate the ground state energy E0 for
200 < λ < 20000 and then fit your results to a function of the form
λ1/3 (a + bλ−2/3 + cλ−4/3 ). What is the accuracy in the calculation
of the coefficients a, b and c and how good is the agreement with
equation (9.31)?
9.5. PROBLEMS 399
9.10 Modify the code in the file anharmonic.f90 so that it calculates the
expectation values ⟨x2 ⟩n (N, λ), ⟨p2 ⟩n (N, λ) and the corresponding
products ∆x · ∆p.
(Hint: See the file anharmonicOBS.f90.)
9.11 Reproduce the results shown in figure 9.4. Repeat your calculation
for λ = 2.0, 10.0, 100.0. Repeat your calculations for n = 20.
9.12 Reproduce the results shown in figure 9.5. Repeat your calculations
for n = 20.
9.13 Reproduce the results shown in figure 9.6. Repeat your calculation
for n = 3, 7, 12, 18, 24.
9.14 Write a program that calculates the energy levels of the anharmonic
oscillator
1 1
H(λ, µ) = p2 + x2 + λx4 + µx6 . (9.41)
2 2
Calculate En (λ) for n = 0, 3, 8, 20, λ = 0.2 and µ = 0.2, 0.5, 1.0, 2.0, 10.0.
9.15 Modify the program of the previous problem so that it calculates the
expectation values ⟨x2 ⟩n (N, λ), ⟨p2 ⟩n (N, λ) and the products ∆x · ∆p.
Calculate the expectation values ⟨x2 ⟩n (λ), ⟨p2 ⟩n (λ) and ∆x · ∆p for
n = 0, 3, 8, 20, λ = 0.2 and µ = 0.2, 0.5, 1.0, 2.0, 10.0.
9.16 Use the program doublewell.f90 in order to calculate the energy
level pairs En , En+1 for n = 0, 4, 20 and λ = 0.2, 0.1, 0.05, 0.02. Cal-
culate the difference ∆n = En+1 − En and comment on your results.
9.17 Define the energy values
( )
1 1
ϵn = − + n + .
4λ 2
Compare the results for En , En+1 of the previous problem with ϵn −
∆n /2 and ϵn + ∆n /2 respectively. Explain your results.
9.18 Modify the program doublewell.f90, so that it calculates the ex-
pectation values ⟨x⟩± (λ) given by equation
√ (9.40). Compare ⟨x⟩± (λ)
with the classical values x0 = ±1/ λ for λ = 0.2, 0.1, 0.05, 0.02, 0.01.
√
9.19 Repeat the previous problem when the states |±⟩ = (1/ 2)( |n⟩λ ±
|n + 1⟩λ ) for n = 6 and n = 30.
400 CHAPTER 9. THE ANHARMONIC OSCILLATOR
9.20 For the simple harmonic oscillator, the energy levels are equidis-
tant, i.e. ∆n = En+1 − En = 1, (∆n+2 − ∆n )/∆n = 0. Calculate these
quantities for the anharmonic oscillator and the double well poten-
tial for λ = 1, 10, 100, 1000 and n = 0, 8, 20. What do you conclude
from your results?
Chapter 10
Time Independent Schrödinger
Equation
In this chapter, we will study the time independent Schrödinger equation
for a non relativistic particle of mass m, without spin, moving in one
dimension, in a static potential V (x). We will only study bound states.
The solutions in this case yield the discrete energy spectrum {En } as well
as the corresponding eigenstates of the Hamiltonian {ψn (x)} in position
representation.
From a numerical analysis point of view, the problem consists of
solving for the eigensystem of a differential equation with boundary con-
ditions. Part of the solution is the energy eigenvalue which also needs to
be determined.
As an exercise, we will use two different methods, one that can be
applied to a particle in an infinite well with V (x) = V (−x), and one that
can be applied to more general cases. The first method is introduced
only for educational purposes and the reader may skip section 10.2 to
go directly to section 10.3.
10.1 Introduction
The wave functions ψ(x), which are the position representation of the
energy eigenstates, satisfy the Schrödinger equation
ℏ2 ∂ 2 ψ(x)
− + V (x)ψ(x) = Eψ(x) , (10.1)
2m ∂x2
401
402 CHAPTER 10. SCHRÖDINGER EQUATION
with the normalization condition
∫ +∞
⟨ψ|ψ⟩ = ψ ∗ (x)ψ(x) dx = 1 . (10.2)
−∞
The Hamiltonian operator is given in position representation by
ℏ2 ∂ 2
Ĥ = − + V (x̂) , (10.3)
2m ∂x2
and it is Hermitian, i.e. Ĥ † = Ĥ. Equation (10.1) is an eigenvalue
problem
Ĥψ(x) = Eψ(x) , (10.4)
which, for bound states, has as solutions a discrete set of real functions
ψn∗ (x) = ψn (x) such that Ĥψn (x) = En ψn (x). The numbers E0 ≤ E1 ≤
E2 ≤ . . . are real and they are the (bound) energy spectrum of the particle
in the potential¹ V (x). The minimum energy E0 is called the ground
state energy and the corresponding ground state is given by a non trivial
function ψ0 (x). According to the Heisenberg uncertainty principle, in this
state the uncertainties in momentum ∆p > 0 and position ∆x > 0 so that
∆p · ∆x ≥ ℏ/2.
The eigenstates ψn (x) form an orthonormal basis
∫ +∞
⟨ψn |ψm ⟩ = ψn∗ (x)ψm (x) dx = δn,m . (10.5)
∞
so that any (square integrable) wave function ϕ(x) which represents the
state |ϕ⟩ is given by the linear combination
∑
∞
ϕ(x) = cn ψn (x) . (10.6)
n=0
∫ +∞
The amplitudes cn = ⟨ψn |ϕ⟩ = −∞ ψn∗ (x)ϕ(x) dx are complex numbers
that give the probability pn = |cn |2 to measure energy En in the state |ϕ⟩.
For any state |ϕ⟩ the function
pϕ (x) = |ϕ(x)|2 = ϕ∗ (x)ϕ(x) (10.7)
¹The fact that the energy spectrum of the particle is bounded from below depends
on the form of the potential. We assume that V (x) is such that E0 is finite. Also, in one
dimension, the energy spectrum of a particle for reasonable potentials is non degenerate
(see, however, S. Kar, R. Parwani, arXiv:0706.1135.)
10.1. INTRODUCTION 403
is the probability density of finding the particle at position x, i.e. the
probability of detecting the particle in the interval [x1 , x2 ] is given by
∫ x2 ∫ x2
Pϕ (x1 < x < x2 ) = pϕ (x) dx = ϕ∗ (x)ϕ(x) dx . (10.8)
x1 x1
The normalization condition (10.2) reflects the conservation of probabil-
ity (independent of time, respected by the time dependent Schrödinger
equation) and the completeness (in this case the certainty that the particle
will be observed somewhere on the x axis).
The classical observables A(x, p) of this quantum mechanical system
are functions of the position and the momentum and their quantum
mechanical versions are given by operators Â(x̂, p̂). Their expectation
values when the system in a state |ϕ⟩ are given by
∫ +∞
⟨Â⟩ϕ = ⟨ϕ| Â |ϕ⟩ = ϕ∗ (x)Â(x̂, p̂)ϕ(x) dx . (10.9)
−∞
From a numerical point of view, the eigenvalue problem (10.1) re-
quires the solution of an ordinary second order differential equation.
There are certain differences in this problem compared to the ones stud-
ied in previous sections:
• Instead of an initial value problem (i.e. the values of the function
and its derivative are given at one point), we have a boundary
value problem (values of the function or its derivative given at two
different points).
• The eigenvalue (energy) is unknown and should be determined as
part of the solution.
As an introduction to such classes of problems, we will present some
simple methods which are special to one dimension.
For the numerical solution of the above equation we renormalize x, the
function ψ(x) and the parameters so that we deal only with dimensionless
quantities. Equation (10.1) is rewritten as:
d2 2m
ψ(x) + 2 (E − V (x))ψ(x) = 0 . (10.10)
dx 2 ℏ
Then we choose a length scale L which is defined by the parameters of
the problem² and we redefine x̃ = x/L. We define ψ̃(x̃) = ψ(x) ψ̃ ′ (x̃) =
²There are m, ℏ and the coupling constants in the function V (x). The range of the
potential will determine L in some problems and it is given explicitly in potential wells.
In potentials of real physical systems, however, this is also determined by the coupling
constants.
404 CHAPTER 10. SCHRÖDINGER EQUATION
dψ(x)/dx̃ = L dψ(x)/dx and we obtain
′′ 2mL2
ψ̃ (x̃) + (E − V (x̃L))ψ̃(x̃) = 0 . (10.11)
ℏ2
We define v(x̃) = 2mL2 V (x)/ℏ2 = 2mL2 V (x̃L)/ℏ2 , ϵ = 2mL2 E/ℏ2 and
change notation to x̃ → x, ψ̃ → ψ. We obtain
ψ ′′ (x) = −(ϵ − v(x))ψ(x) . (10.12)
The solutions of equation (10.1) can be obtained from those of equation
(10.12) by using the following “dictionary”³:
x ℏ2 ℏ2
x→ , E= ϵ , V (x) = v(x/L) . (10.13)
L 2mL2 2mL2
The dimensionless momentum is defined as p̃ = −i∂/∂ x̃ = −iL∂/∂x and
we obtain
L
p̃ = p . (10.14)
ℏ
The commutation relation [x, p] = iℏ becomes [x̃, p̃] = i. The kinetic
p2
energy T = is given by
2m
ℏ2 2 ℏ2 ∂ 2
T =
p̃ = − , (10.15)
2mL2 2mL2 ∂ x̃2
and the Hamiltonian H = T + V
( )
ℏ2 ( 2 ) ℏ2 ∂2
H= p̃ + v(x̃) = − 2 + v(x̃) . (10.16)
2mL2 2mL2 ∂ x̃
In what follows, we will omit the tilde above the symbols and write x
instead of x̃.
10.2 The Infinite Potential Well
The simplest model for studying the qualitative features of bound states
is the infinite potential well of width L where a particle is confined within
the interval [−L/2, L/2]:
{
0 |x| < 1
v(x) = (10.17)
+∞ |x| ≥ 1
³If we normalize the solutions ψ̃(x̃) of equation (10.12) according to the relation
∫ +∞ ∗ √
−∞
ψ̃ (x̃)ψ̃(x̃)dx̃ = 1, we should also take ψ(x) = (1/ L)ψ̃(x/L) in order to be prop-
∫ +∞
erly normalized −∞ ψ ∗ (x)ψ(x)dx = 1.
10.2. THE INFINITE POTENTIAL WELL 405
v v v
v0 v0
−1 +1 −1 +1 −1 −a +a +1 x
Figure 10.1: The potentials given by equations (10.17), (10.26) and (10.27).
The length scale chosen here is L/2 and the dimensionless variable x
corresponds to x/(L/2) when x is measured in length units.
The solution of (10.12) can be easily computed. Due to the symmetry
v(−x) = v(x) , (10.18)
of the potential, the solutions have well defined parity. This property will
be crucial to the method used below. The method discussed in the next
section can also be used on non even potentials.
The solutions are divided into two categories, one with even parity
(+) (+)
ψn (x) ≡ ψn (−x) = ψn (x) for n = 1, 3, 5, 7, . . . and one with odd parity
(−) (−)
ψn (x) ≡ −ψn (−x) = ψn (x) for n = 2, 4, 6, 8, . . ..
{
(+)
ψn (x) = cos ( nπ
2
x) |x| < 1 n = 1, 3, 5, 7, . . .
ψn (x) = (−) (10.19)
ψn (x) = sin ( 2 x) |x| < 1 n = 2, 4, 6, 8, . . .
nπ
where ( nπ )2
ϵn = , (10.20)
2
∫1
and the normalization has been chosen so that⁴ −1
(ψn (x))2 dx = 1.
⁴According to the dictionary mentioned in the previous section, for a potential
well where x ∈ [−L/2, L/2] the dimensionless position variable has been chosen to be
ℏ2 ℏ2 π 2 2 (+) √
x/(L/2) ∈ [−1, 1]. Then En = 2m(L/2)2 ϵn = 2mL2 n and ψn (x) = 2/L cos (nπx/L),
(−) √
ψn (x) = 2/L sin (nπx/L). Note that ϵn = p2n according to equations (10.13) and
(10.14).
406 CHAPTER 10. SCHRÖDINGER EQUATION
The solutions can be found by using the parity of the wave functions.
We note that for the positive parity solutions
ψn(+) (0) = A ψn(+) ′ (0) = 0 , (10.21)
whereas for the negative parity solutions
ψn(−) (0) = 0 ψn(−) ′ (0) = A . (10.22)
The constant A depends on the normalization of the wave function.
Therefore we can set A = 1 originally and then renormalize the wave
function so that equation (10.2) is satisfied. If the energy is known, the
relations (10.21) and (10.22) can be taken as initial conditions in relation
(10.12). By using a Runge–Kutta algorithm we can evolve the solution
towards x = ±1. The problem is that the energy ϵ is unknown. If the en-
ergy is not allowed by the quantum theory we will find that the boundary
conditions
ψn(±) (±1) = 0 (10.23)
are violated. As we approach the correct value of the energy, we obtain
(±)
ψn (±1) → 0.
1 1
i= 0 i= 0
i= 1 i= 5
i= 2 i= 10
i= 3 i= 20
0.8 i= 5 0.01 i= 29
i= 12
0.6 0.0001
|ψi(x)-ψn=0(x)|
ψi(x)
0.4 1e-06
0.2 1e-08
0 1e-10
-0.2 1e-12
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
x x
Figure 10.2: Convergence of the solution ψi (x) of (10.12) with the potential (10.17)
as a function of the number of iterations i in the program well.f90. Initially energy
= 2.0 and parity = 1. After 29 iterations the solution converges to the ground state
ψ1 (x) = cos (πx/2) with energy ϵ = (π/2)2 and with relative accuracy ∼ 10−9 . The
bottom plot shows the error as a function of the number of iterations in a logarithmic
scale. For i ≡iter = 1,2,3,5,10,12,20 we obtain energy = 2.4, 2.6, 2.4, 2.4625,
2.46875, 2.4673828125.
Therefore we follow the steps described below:
• We choose an initial value for the energy ϵ that is lower than the one
we are looking for. We can use estimates from known solutions of
10.2. THE INFINITE POTENTIAL WELL 407
similar looking potential wells or simply start from a value slightly
higher than the absolute minimum of the potential.
• We choose the parity of the solution and we set initial conditions
according to equations (10.21) and (10.22).
• We evolve the solutions using a 4th order Runge-Kutta method
from ⁵ x = 0 to x = +1.
• If equation (10.23) is not satisfied, we increase the energy by δϵ and
we repeat.
(±)
• We repeat until ψn (1) changes sign. Then we lower the energy by
δϵ = −δϵ/2.
(±)
• The process is ended when |ψn (1)| < δ for appropriately chosen
small δ.
For the evolution of the solution from x = 0 to x = 1 we use the 4th
order Runge-Kutta method programmed in the file rk.f90 of chapter 4.
We copy the subroutine RKSTEP in a local file rk.f90. The integration of
(10.12) can by done by using the function ϕ(x) ≡ ψ ′ (x)
ψ ′ (x) = ϕ(x)
ϕ′ (x) = (v(x) − ϵ)ψ(x) , (10.24)
with the initial conditions
ψ(0) = 1 , ϕ(0) ≡ ψ ′ (0) = 0 even parity
ψ(0) = 0 , ϕ(0) ≡ ψ ′ (0) = 1 odd parity . (10.25)
We use the notation ψ(x) → psi, ϕ(x) → psip. The functions f1 and f2
correspond to the right hand side of (10.24). They are the derivatives of
ψ(x) and ϕ(x) respectively and f1=psip, f2=(V-energy)*psi. The code
of f1 and f2 is put in a different file so that we can easily reuse the code
for many different potentials v(x). The file wellInfSq.f90 contains the
necessary program for the potential of equation (10.17):
! ===========================================================
! f i l e : wellInfSq . f
!
! Functions used i n RKSTEP r o u t i n e . Here :
⁵The function in [−1, 0) is determined by the parity of the solution.
408 CHAPTER 10. SCHRÖDINGER EQUATION
! f1 = psip ( x ) = psi ( x ) ’
! f 2 = p s i p ( x ) ’= p s i ( x ) ’ ’
!
! A l l one has t o s e t i s V , t h e p o t e n t i a l
! ===========================================================
!−−−−−−−− t r i v i a l f u n c t i o n : d e r i v a t i v e o f p s i
r e a l ( 8 ) f u n c t i o n f1 ( x , psi , psip )
r e a l ( 8 ) : : x , psi , psip
f1=psip
end f u n c t i o n f1
! ===========================================================
!−−−−−−−− t h e second d e r i v a t i v e o f wavefunction :
! p s i p ( x ) ’ = p s i ( x ) ’ ’ = −(E−V) p s i ( x )
r e a l ( 8 ) f u n c t i o n f2 ( x , psi , psip )
i m p l i c i t none
r e a l ( 8 ) : : x , psi , psip , energy , V
common / params / energy
!−−−−−−− p o t e n t i a l , s e t here :
V = 0.0 D0
!−−−−−−− S c h r o e d i n g e r eq : RHS
f2 = ( V−energy ) * psi
end f u n c t i o n f2
! ===========================================================
We stress that the energy ϵ = energy is put in a common block so that it
can be accessed by the main program.
The main program is in the file well.f90. The user enters the pa-
rameters (energy, parity, Nx) and the loop
do while ( iter . l t . 10000)
............
i f ( DABS ( psinew ) . l e . e p s i l o n ) EXIT
i f ( psinew * psiold . l t . 0.0 D0 ) de = −0.5D0 * de
energy = energy + de
............
enddo ! do while
exits when ψ(1) =psinew has an absolute value which is less than epsilon,
i.e. when the condition (10.23) is satisfied to the desired accuracy. The
value of the energy increases up to the point where the sign of the wave
function at x = 1 changes (psinew*psiold< 0). Then the value of the
energy is overestimated and we change the sign of the step de and re-
duce its magnitude by a half. The algorithm described on page 406 is
implemented inside the loop. After exiting the loop, the energy has been
determined with the desired accuracy and the rest of the program stores
the solution in the array psifinal(STEPS). The results are written to the
10.2. THE INFINITE POTENTIAL WELL 409
file psi.dat. Note how the variable parity is used so that both cases
parity= ±1 can be studied. The full program is listed below:
! ===========================================================
! f i l e : well . f
!
! Computation o f energy e i g e n v a l u e s and e i g e n f u n c t i o n s
! o f a p a r t i c l e i n an i n f i n i t e w e l l with V(−x )=V( x )
!
! Input : energy : i n i t i a l guess f o r energy
! p a r i t y : d e s i r e d p a r i t y o f s o l u t i o n (+/ − 1 )
! Nx−1 : Number o f RK4 s t e p s from x=0 t o x=1
! Output : energy : energy e i g e n v a l u e
! p s i . dat : f i n a l p s i ( x )
! a l l . dat : a l l p s i ( x ) f o r t r i a l e n e r g i e s
! ===========================================================
program even_potential_well
i m p l i c i t none
i n t e g e r , parameter : : P=10000
r e a l ( 8 ) : : energy , dx , x , e p s i l o n , de
common / params / energy
i n t e g e r : : parity , Nx , iter , i
r e a l ( 8 ) : : psi , psip , psinew , psiold
r e a l ( 8 ) : : psifinal(−P : P ) , xstep(−P : P )
!−−−−−− Input :
p r i n t * , ’ Enter energy , p a r i t y , Nx : ’
read * , energy , parity , Nx
i f ( Nx . g t . P ) s t o p ’Nx > P ’
i f ( parity . g t . 0) then
parity = 1
else
parity = −1
endif
p r i n t * , ’ # ####################################### ’
p r i n t * , ’ # E s t a r t = ’ , energy , ’ p a r i t y = ’ , parity
dx = 1 . 0 D0 / ( Nx −1)
e p s i l o n = 1 . 0 D−6
p r i n t * , ’ # Nx= ’ , Nx , ’ dx = ’ , dx , ’ eps= ’ , e p s i l o n
p r i n t * , ’ # ####################################### ’
!−−−−− C a l c u l a t e :
open ( u n i t =11 , f i l e = ’ a l l . dat ’ )
iter = 0
psiold = 0.0 D0 ! c a l c u l a t e d v a l u e s o f p s i a t x=1
psinew = 1 . 0 D0
de = 0 . 1 D0 * DABS ( energy ) ! o r i g i n a l change i n energy
do while ( iter . l t . 10000)
!−−−−−−−−−− I n i t i a l c o n d i t i o n s a t x=0
410 CHAPTER 10. SCHRÖDINGER EQUATION
x = 0.0 D0
i f ( parity . eq . 1 ) then
psi = 1 . 0 D0
psip = 0.0 D0
else
psi = 0.0 D0
psip = 1 . 0 D0
endif
w r i t e ( 1 1 , * ) iter , energy , x , psi , psip
! −−−−−−−−− Use Runge−Kutta t o forward t o x=1
do i =2 , Nx
x = ( i−2) * dx
c a l l RKSTEP ( x , psi , psip , dx )
w r i t e ( 1 1 , * ) iter , energy , x , psi , psip
enddo ! do i =2 ,Nx
psinew = psi
p r i n t * , iter , energy , de , psinew
! −−−−−−−−− Stop i f v a l u e o f p s i c l o s e t o 0
i f ( DABS ( psinew ) . l e . e p s i l o n ) EXIT
! −−−−−−−−− Change d i r e c t i o n o f energy s e a r c h :
i f ( psinew * psiold . l t . 0.0 D0 ) de = −0.5D0 * de
energy = energy + de
psiold = psinew
iter = iter + 1
enddo ! do while
close (11)
!We found t h e s o l u t i o n : c a l c u l a t e i t once aga in and s t o r e i t
i f ( parity . eq . 1 ) then
psi = 1 . 0 D0
psip = 0.0 D0
node = 0 ! count number o f nodes o f f u n c t i o n
else
psi = 0.0 D0
psip = 1 . 0 D0
node = 1
endif
x = 0.0 D0
xstep (0) = x
psifinal ( 0 ) = psi ! a r r a y t h a t s t o r e s p s i ( x )
psiold = 0.0 D0
!−−−−−−− Use Runge−Kutta t o move t o x=1
do i =2 , Nx
x = ( i−2) * dx
c a l l RKSTEP ( x , psi , psip , dx )
xstep ( i−1) = x
psifinal ( i−1) = psi
! −−−−−− Use p a r i t y t o compute p s i (−x )
xstep (1−i ) = −x
psifinal(1−i ) = parity * psi
10.2. THE INFINITE POTENTIAL WELL 411
!−−−−−−− P r i n t f i n a l s o l u t i o n :
open ( u n i t =11 , f i l e = ’ p s i . dat ’ )
p r i n t * , ’ F i n a l r e s u l t : E= ’ , energy , ’ n= ’ , node ,&
’ p a r i t y = ’ , parity
w r i t e ( 1 1 , * ) ’ # E= ’ , energy , ’ n= ’ , node ,&
’ p a r i t y = ’ , parity
do i=−(Nx −1) , ( Nx −1)
w r i t e ( 1 1 , * ) xstep ( i ) , psifinal ( i )
enddo
close (11)
end program even_potential_well
The compilation and running of the program can be done with the com-
mands
> gfortran well . f90 wellInfSq . f90 rk . f90 −o well
> . / well
Enter energy , parity , Nx :
2.0 1 400
# #######################################
# Estart= 2.0000000000000000 parity= 1
# Nx= 400 dx = 2.50626566416E−003 eps= 9.9999999999E−007
# #######################################
0 2.0000000000000 0.200000000000 0.15594369476721
1 2.2000000000000 0.200000000000 8.74448016806986E−2
............................................
28 2.4674072265624 1.220703125000E−5 −1.95005436858826E−6
29 2.4674011230468 −6.103515625000E−6 −7.24621589476086E−9
Final result : E= 2.4674011230468746 parity= 1
The energy is determined to be ϵ =2.467401123 which can be compared
to the exact value ϵ = (π/2)2 ≈ 2.467401100. The fractional error is
∼ 10−8 . The convergence can be studied graphically in figure 10.2.
The calculation of the excited states is done by changing the parity and
by choosing the initial energy slightly higher than the one determined in
the previous step⁶. The results are in table 10.1. The agreement with the
exact result ϵn = (nπ/2)2 is excellent.
We close this section with two more examples. First, we study a
potential well with triangular shape at its bottom
{
v0 |x| |x| < 1
v(x) = (10.26)
+∞ |x| > 1
⁶Careful: if the energy levels are too close, we should keep the initial energy constant
and change the sign of parity.
412 CHAPTER 10. SCHRÖDINGER EQUATION
n (nπ/2)2 Square Triangular Double Well
1 2.467401100 2.467401123 5.248626709 15.294378662
2 9.869604401 9.869604492 14.760107422 15.350024414
3 22.2066099 22.2066040 27.0690216 59.1908203
4 39.47841 39.47839 44.51092 59.96887
5 61.6850275 61.6850242 66.6384315 111.3247375
6 88.82643 88.82661 93.84588 126.37628
7 120.902653 120.902664 125.878830 150.745215
8 157.91367 157.91382 162.92569 194.07578
9 199.859489 199.859490 204.845026 235.017471
10 246.74011 246.74060 251.74813 275.67383
11 298.555533 298.555554 303.545814 331.428306
12 355.3057 355.3064 360.3107 388.7444
Table 10.1: Energy eigenvalues for the square, triangular and double well potentials
(equations (10.17), (10.26) with v0 = 10 and equation (10.27) with v0 = 100, a = 0.3).
The agreement of the results for the square potential with the exact ones is excellent.
For the other potentials, we note that as we move further from the bottom of the well
we obtain energy levels very close to those of the square well: The particle does not feel
the influence of the details at the bottom of the well. For the double well potential we
obtain E1 ≈ E2 and E3 ≈ E4 according to the analysis on page 413.
10.2. THE INFINITE POTENTIAL WELL 413
and then a double well potential with
v0 |x| < a
v(x) = 0 a < |x| < 1 (10.27)
+∞ 1 < |x|
where the parameters v0 , a are positive numbers. A qualitative plot of
these functions is shown in figure 10.1.
For the triangular potential we take v0 = 10, whereas for the double
well potential v0 = 100 and a = 0.3. The code in wellInfSq.f90 is appro-
priately modified and saved in the files wellInfTr.f90 and wellInfDbl.f90
respectively. All we have to do is to change the line computing the value
of the potential in the function f2. For example the file wellInfTr.f90
contains the code
!−−−−−−− p o t e n t i a l , s e t here :
V = 10.0 D0 * DABS ( x )
whereas the file wellInfDbl.f90 contains the code
!−−−−−−− p o t e n t i a l , s e t here :
i f ( DABS ( x ) . l e . 0.3 D0 ) then
V = 100.0 D0
else
V = 0.0 D0
endif
The analysis is performed in exactly the same way and the results are
shown in table 10.1. Note that, for large enough n, the energy levels of
all the potentials that we studied above tend to have identical values.
This happens because, when the particle has energy much larger than v0 ,
the details of the potential at the bottom do not influence its dynamical
properties very much. For the triangular potential, the energy levels have
higher values than the corresponding ones of the square potential. This
happens because, on the average, the potential energy is higher and the
potential tends to confine the particle to a smaller region (∆x is decreased,
therefore ∆p is increased). This can be seen in figure 10.3 where the wave
functions of the particle in each of the two potentials are compared.
Similar observations can be made for the double well potential. More-
over, we note the approximately degenerate energy levels, something
which is expected for potentials of this form. This can be understood
in terms of the localized states given by the wave functions ψ+ (x) =
414 CHAPTER 10. SCHRÖDINGER EQUATION
√ √
(1/ 2)(ψ1 (x) + ψ2 (x)) and ψ− (x) = (1/ 2)(ψ1 (x) − ψ2 (x)). The first one
represents a state where the particle is localized in the left well and the
second one in the right. This is shown in figure 10.4. As v0 → +∞ the
two wells decouple and the wave functions ψ± (x) become equal to the en-
ergy eigenstate wave functions of two particles in separate infinite square
wells of width 1 − a with energy eigenvalues ϵ+,1 = ϵ−,1 = (π/(1 − a))2 .
The difference of ϵ1 and ϵ2 from these two values is due to the finite v0
(see problem 4).
1.2 1.5
Square Square
Triangle Triangle
1
1
0.8
0.5
0.6
ψ1(x)
ψ2(x)
0
0.4
-0.5
0.2
-1
0
-0.2 -1.5
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
1 1.5
Square Square
Triangle Triangle
1
0.5
0.5
0
ψ3(x)
ψ4(x)
0
-0.5
-0.5
-1
-1
-1.5 -1.5
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
1.5 1.5
Square Square
Triangle Triangle
1 1
0.5 0.5
ψ12(x)
ψ8(x)
0 0
-0.5 -0.5
-1 -1
-1.5 -1.5
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
Figure 10.3: The wave functions of the energy eigenstates of the infinite square and
triangular well potentials for n = 1, 2, 3, 4, 8, 12 given by equations (10.17) and (10.26)
with v0 = 10. We observe the influence of the shape of the potential on the wave
functions with small n, while for n ≥ 8 the influence becomes weaker.
We will now discuss the limitations of this method. First, the method
can be used only on potential wells that are even, i.e. v(x) = v(−x). We
used this assumption in equations (10.21) and (10.22) giving the initial
conditions for states of well defined parity. When the potential is even,
the energy eigenstates have definite parity. The other problem can be
10.3. BOUND STATES 415
2 2
ψ+(x) ψ-(x)
ψ1(x) ψ1(x)
ψ2(x) ψ2(x)
1.5 1.5
1 1
0.5 0.5
0 0
-0.5 -0.5
-1 -1
-1.5 -1.5
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
2 2
ψ+(x) ψ-(x)
ψ3(x) ψ3(x)
ψ4(x) ψ4(x)
1.5 1.5
1 1
0.5 0.5
0 0
-0.5 -0.5
-1 -1
-1.5 -1.5
-2 -2
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
1.5 1.5
ψ+(x) ψ-(x)
ψ5(x) ψ5(x)
ψ6(x) ψ6(x)
1 1
0.5 0.5
0 0
-0.5 -0.5
-1 -1
-1.5 -1.5
-1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1
x x
√
Figure 10.4: The functions ψ± (x) = (1/ 2)(ψn (x) ± ψn+1 (x)) for n = 1, 3, 5 for the
double well potential (equation (10.27) with v0 = 100, a = 0.3) are plotted using bold red
lines. We observe that the more degenerate the states, the stronger the localization of the
particle to the left or right well. The other plots are those of the energy eigenfunctions
for n = 1, 2, 3, 4, 5, 6.
understood by solving problem 4: When v(0) ≫ ϵ, the wave function is
almost zero around x = 0 and the integration from x = 0 to x = 1 will be
dominated by numerical errors. The same is true when the particle has
to go through high potential barriers.
This method can also we used on potential wells that are not infinite.
In that case we can add infinite walls at points that are far enough so
that the wave function is practically zero there. Then the influence of
this artificial wall will be negligible (see problem 3).
10.3 Bound States
A serious problem with the method discussed in the previous section is
that it is numerically unstable. You should have already realized that if
you tried to solve problem 3. In that problem, when the walls are moved
416 CHAPTER 10. SCHRÖDINGER EQUATION
Figure 10.5: Integration of Schrödinger’s equation by the use of the algorithm of
section 10.3. The wave functions and their derivatives are given small trial values
at xmin and xmax which are in the classically forbidden regions of x. The point xm
is calculated from the equation v(xm ) = ϵ. The wave functions are evolved to xm
according to (10.24) and we obtain the solutions ψ (+) (x) and ψ (−) (x). We renormalize
ψ (−) (x) so that ψ (+) (xm ) = ψ (−) (xm ) and we vary the energy until the derivatives
ψ (+)′ (xm ) ≈ ψ (−)′ (xm ).
further than |x| = 3, the convergence of the algorithm becomes harder.
You can understand this by realizing that in the integration process the
solution is evolved from the classically allowed into the classically forbid-
den region so that an oscillating solution changes into an exponentially
damped one. But as |x| → +∞ there are two solutions, one that is phys-
ically acceptable ψ(x) ∼ e−k|x| and one that is diverging ψ(x) ∼ e+k|x|
which is not acceptable due to (10.2). Therefore, in order to achieve con-
vergence to the physically acceptable solution, the energy has to be finely
tuned, especially when we integrate towards large |x|. For this reason it
is preferable to integrate from the exponentially damped region towards
the oscillating region. The idea is to start integrating from these regions
and try to match the solutions and their derivatives at appropriately cho-
sen matching points. The matching is achieved at a point xm by trying
to determine the value of the energy that sets the ratio
ψ (+)′ (xm )/ψ (+) (xm ) − ψ (−)′ (xm )/ψ (−) (xm )
f (ϵ) = (10.28)
ψ (+)′ (xm )/ψ (+) (xm ) + ψ (−)′ (xm )/ψ (−) (xm )
10.3. BOUND STATES 417
equal to zero, within the attainable numerical accuracy. It is desirable to
choose a point xm within the classical region (ϵ > v(x)) and usually we
pick a turning point ϵ = v(x). By renormalizing ψ (±) (x) we can always set
ψ (+) (xm ) = ψ (−) (xm ), therefore f (ϵ) ≪ 1 means that ψ (+)′ (xm ) ≈ ψ (−)′ (xm ).
The denominator of (10.28) sets the scale of the desired accuracy⁷ The
idea is depicted in figure 10.5. The algorithm is the following:
• Choose the integration interval [xmin,xmax].
• Choose the initial conditions ψ (−) (xmin), ψ (−)′ (xmin), ψ (+) (xmax),
ψ (+)′ (xmax). This choice depends on the potential v(x). Usually
we take xmin and xmax deep enough in the classically forbidden
region and choose the values ψ (−) (xmin), ψ (+) (xmax) to be zero or
exponentially small (e.g. ∼ e−k|x| , k 2 = v(x)−ϵ). The corresponding
values of the derivatives ψ (−)′ (xmin), ψ (+)′ (xmax) are also taken to be
small. The arbitrary normalization of ψ(x) allows these initial val-
ues to be chosen in a crude way. The relative sign of the derivatives
at large |x| (determined e.g. by the parity of the wave function for
even potentials) is also taken care by the renormalization of ψ (−) (x)
when applying the matching condition. For an infinite well, the
points xmin,xmax are the ones where the potential becomes infinite
and ψ (−) (xmin) = ψ (+) (xmax) = 0.
• Choose the initial value of the energy ϵ and of the energy variation
step δϵ.
• Calculate xm from the initial value of the energy and the solution of
v(x) = ϵ. Choose the solution that is at the left most side⁸.
• Evolve the equations (10.24) from xmin to xm and obtain the solu-
tions ψ (−) (x),ψ (−)′ (x).
• Evolve the equations (10.24) from xmax to xm and obtain the solu-
tions ψ (+) (x),ψ (+)′ (x).
( )
• Renormalize ψ (−) (x) → ψ (−) (x) ψ (+) (xm)/ψ (−) (xm) , so that ψ (+) (xm) =
ψ (−) (xm).
• Compute the ratio f (ϵ) of equation (10.28).
⁷If we are unlucky enough to pick a point where ψ ′ (xm ) = 0, this criterion will fail.
⁸Note that this point changes when we vary ϵ
418 CHAPTER 10. SCHRÖDINGER EQUATION
• If |f (ϵ)| < δ for appropriately chosen δ > 0, the calculation ends.
The result for the energy eigenvalue and eigenfunction is considered
to be determined with adequate accuracy and we may proceed with
the analysis of the results.
• If f (ϵ) changes sign it means that we have crossed the energy eigen-
value. Reverse the direction of search by taking δϵ → −δϵ/2.
• Change the energy ϵ → ϵ+δϵ and repeat by going back to the fourth
step.
When we exit the above loop, the current wave function is a good ap-
proximation to the eigenfunction ψn (x) corresponding to the eigenvalue
ϵn . We normalize the wave function according to equation (10.2) and we
calculate the expectation values according to (10.9). It is also interest-
ing to determine the number of nodes⁹ n0 of the wave function which is
related to n by n = n0 + 1.
Our program needs to implement the Runge–Kutta algorithm. We use
the routine RKSTEP (see page 209) which performs a 4th order Runge–
Kutta step. Its code is copied to the file rk.f90.
The potential v(x) is coded in the function V(x). The boundary condi-
tions are programmed in the subroutine boundary(xmin, xmax, psixmin,
psipxmin, psixmax, psipxmax) which returns the values of psixmin =
ψ (−) (xmin), psipxmin = ψ (−)′ (xmin), psixmax = ψ (+) (xmax), psipxmax =
ψ (−)′ (xmax) to the calling program. These routines are put in a separate
file for each potential that we want to study. The name of the file is
related to the form of the potential, e.g. we choose schInfSq.f90 for the
infinite potential well of (10.17). The same file contains the code for the
functions f1, f2:
! ===========================================================
! f i l e : schInfSq . f
!
! Functions used i n RKSTEP r o u t i n e . Here :
! f1 = psip ( x ) = psi ( x ) ’
! f 2 = p s i p ( x ) ’= p s i ( x ) ’ ’
!
! One has t o s e t :
! 1 . V( x ) , t h e p o t e n t i a l
! 2 . The boundary c o n d i t i o n s f o r p s i , p s i p a t x=xmin and x=xmax
!
⁹The number of points x for which ψ(x) = 0 and xmin < x < xmax. The relation
n = n0 + 1 sets ϵ1 to be the ground state for which n0 = 0.
10.3. BOUND STATES 419
! ===========================================================
!−−−−− p o t e n t i a l :
real (8) function V(x)
i m p l i c i t none
real (8) : : x
V = 0.0 D0
end f u n c t i o n V
!−−−−− boundary c o n d i t i o n s :
subroutine &
boundary ( xmin , xmax , psixmin , psipxmin , psixmax , psipxmax )
i m p l i c i t none
r e a l ( 8 ) : : xmin , xmax , psixmin , psipxmin , psixmax , psipxmax , V
! f o r i n f i n i t e square w e l l we s e t p s i =0 a t boundary
! and p s i p =+/−1
psixmin = 0.0 D0
psipxmin = 1 . 0 D0
psixmax = 0.0 D0
psipxmax = −1.0D0
!−−−−− I n i t i a l v a l u e s a t xmin and xmax
end s u b r o u t i n e boundary
! ===========================================================
! ===========================================================
!−−−−− t r i v i a l f u n c t i o n : d e r i v a t i v e o f p s i
r e a l ( 8 ) f u n c t i o n f1 ( x , psi , psip )
r e a l ( 8 ) : : x , psi , psip
f1=psip
end f u n c t i o n f1
! ===========================================================
!−−−−− t h e second d e r i v a t i v e o f wavefunction :
! p s i p ( x ) ’ = p s i ( x ) ’ ’ = −(E−V) p s i ( x )
r e a l ( 8 ) f u n c t i o n f2 ( x , psi , psip )
i m p l i c i t none
r e a l ( 8 ) : : x , psi , psip , energy , V
common / params / energy
!−−−−− S c h r o e d i n g e r eq : RHS
f2 = ( V ( x )−energy ) * psi
end f u n c t i o n f2
! ===========================================================
We note that if the potential becomes infinite for x < xmin and/or x >xmax,
then this will be determined by the boundary conditions at xmin and/or
xmax.
The main program is in the file sch.f90. The code is listed below
and it includes the function integrate(psi, dx, Nx) used for the nor-
malization of the wave function. It performs a numerical integration of
the square of a function whose values psi(i) i=1,...,Nx are given at an
odd number of Nx equally spaced points by a distance dx using Simpson’s
420 CHAPTER 10. SCHRÖDINGER EQUATION
rule.
! ===========================================================
!
! F i l e : sch . f90
!
! I n t e g r a t e 1d S c h r o d i n g e r e q u a t i o n from xmin t o xmax .
! Determine energy e i g e n v a l u e and e i g e n f u n c t i o n by matching
! e v o l v i n g s o l u t i o n s from xmin and from xmax a t a p o i n t xm.
! Matching done by e q u a t i n g v a l u e s o f f u n c t i o n s and t h e i r
! d e r i v a t i v e s a t xm. The p o i n t xm chosen a t t h e l e f t most
! t u r n i n g p o i n t o f t h e p o t e n t i a l a t any gi v e n v a l u e o f t h e
! energy . The p o t e n t i a l and boundary c o n d i t i o n s chosen i n
! different file .
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Input : energy : T r i a l v a l u e o f energy
! de : energy s t e p , i f matching f a i l s de −> e+de , i f
! l o g d e r i v a t i v e changes s i g n de −> −de / 2
! xmin , xmax , Nx
! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Output : F i n a l v a l u e o f energy , number o f nodes o f
! wavefunction i n s t d o u t
! F i n a l e i g e n f u n c t i o n i n f i l e p s i . dat
! A l l t r i a l f u n c t i o n s and e n e r g i e s i n f i l e a l l . dat
! ===========================================================
program schroedinger_equation_1D
i m p l i c i t none
i n t e g e r , parameter : : P=20001
i n t e g e r : : Nx , NxL , NxR
r e a l ( 8 ) : : psi ( P ) , psip ( P )
r e a l ( 8 ) : : dx
r e a l ( 8 ) : : xmin , xmax , xm ! l e f t / r i g h t / matching p o i n t s
r e a l ( 8 ) : : psixmin , psipxmin , psixmax , psipxmax
r e a l ( 8 ) : : psileft , psiright , psistep , psinorm
r e a l ( 8 ) : : psipleft , psipright , psipstep
r e a l ( 8 ) : : energy , de , e p s i l o n , integrate
common / params / energy
r e a l ( 8 ) : : matchlogd , matchold , psiold , norm , x
i n t e g e r : : iter , i , imatch , nodes
real (8) : : V
!−−−−−−−−−− Input :
p r i n t * , ’ # Enter energy , de , xmin , xmax , Nx ’
read * , energy , de , xmin , xmax , Nx
!−−− need even i n t e r v a l s f o r n o r m a l i z a t i o n i n t e g r a t i o n
i f ( mod( Nx , 2 ) . eq . 0 ) Nx=Nx+1
i f ( Nx . g t . P ) s t o p ’ F a t a l Error : Nx>P ’
i f ( xmin . ge . xmax ) s t o p ’ Error : xmin >= xmax ’
dx = ( xmax − xmin ) / ( Nx −1)
10.3. BOUND STATES 421
e p s i l o n = 1 . 0 D−6
c a l l boundary ( xmin , xmax , psixmin , psipxmin , psixmax , psipxmax )
p r i n t * , ’ # ####################################### ’
p r i n t * , ’ # E s t a r t = ’ , energy , ’ de= ’ , de
p r i n t * , ’ # Nx= ’ , Nx , ’ eps= ’ , e p s i l o n
p r i n t * , ’ # xmin= ’ , xmin , ’ xmax= ’ , xmax , ’ dx= ’ , dx
p r i n t * , ’ # p s i ( xmin )= ’ , psixmin , ’ p s i p ( xmin )= ’ , psipxmin
p r i n t * , ’ # p s i ( xmax )= ’ , psixmax , ’ p s i p ( xmax )= ’ , psipxmax
p r i n t * , ’ # ####################################### ’
!−−−−− C a l c u l a t e :
open ( u n i t =11 , f i l e = ’ a l l . dat ’ )
matchold = 0.0 d0
do iter =1 ,10000
!−−−−− Determine matching p o i n t a t t u r n i n g p o i n t from t h e l e f t :
imatch = −1
do i =1 , Nx
x = xmin + ( i−1) * dx
i f ( imatch . l t . 0 . and . ( energy−V ( x ) ) . g t . 0.0 D0 ) imatch = i
enddo
i f ( imatch . l e . 100 . or . imatch . ge . Nx −100) imatch = Nx / 5
xm = xmin + ( imatch −1) * dx
NxL = imatch
NxR = Nx−imatch+1
!−−−−− Evolve wavefunction from t h e l e f t :
psi ( 1 ) = psixmin
psip ( 1 ) = psipxmin
psistep = psixmin
psipstep = psipxmin
do i =2 , NxL
x = xmin + ( i−2) * dx ! t h i s i s x b e f o r e t h e s t e p
c a l l RKSTEP ( x , psistep , psipstep , dx )
psi ( i ) = psistep
psip ( i ) = psipstep
enddo
! use t h i s t o normalize e i g e n f u n c t i o n t o match a t xm
psinorm = psistep
psipleft = psipstep
!−−−−− Evolve wavefunction from t h e r i g h t :
psi ( Nx ) = psixmax
psip ( Nx ) = psipxmax
psistep = psixmax
psipstep = psipxmax
do i =2 , NxR
x = xmax − ( i−2) * dx
c a l l RKSTEP ( x , psistep , psipstep ,−dx )
psi ( Nx−i +1) = psistep
psip ( Nx−i +1) = psipstep
enddo
psinorm = psistep / psinorm
422 CHAPTER 10. SCHRÖDINGER EQUATION
psipright = psipstep
!−−−−− Renormalize p s i l so t h a t p s i l (xm)= p s i r (xm)
do i =1 , NxL−1
psi ( i ) = psinorm * psi ( i )
psip ( i ) = psinorm * psip ( i )
enddo
psipleft = psinorm * psipleft
!−−−−− p r i n t c u r r e n t s o l u t i o n :
do i =1 , Nx
x = xmin + ( i−1) * dx
w r i t e ( 1 1 , * ) iter , energy , x , psi ( i ) , psip ( i )
enddo
!−−−−− matching using d e r i v a t i v e s :
! C a r e f u l : t h i s can f a i l i f p s i ’ (xm) = 0 ! ! ( use a l s o | de | < 1 e−6
! criterion )
matchlogd = &
( psipright−psipleft ) / ( DABS ( psipright )+DABS ( psipleft ) )
p r i n t * , ’ # i t e r , energy , de , xm, logd : ’ ,&
iter , energy , de , xm , matchlogd
!−−−−− E x i t c o n d i t i o n :
i f ( DABS ( matchlogd ) . l e . e p s i l o n . or . DABS ( de / energy ) . l t . 1 . 0 D−12)&
EXIT
i f ( matchlogd * matchold . l t . 0.0 D0 ) de = −0.5D0 * de
energy = energy + de
matchold = matchlogd
enddo ! do i t e r =1 ,10000
close (11)
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
!−−−−− S o l u t i o n has been found and now i t i s s t o r e d :
norm = integrate ( psi , dx , Nx )
norm = 1 . 0 D0 / s q r t ( norm )
do i =1 , Nx
psi ( i ) = norm * psi ( i )
enddo
!−−−−− Cound number o f z e r o e s , add one and g e t energy l e v e l :
nodes = 1
psiold = psi ( 1 )
do i =2 , Nx−1
! should be 0 w i t h i n e p s i l o n
i f ( DABS ( psi ( i ) ) . g t . e p s i l o n ) then
i f ( psiold * psi ( i ) . l t . 0 . 0 D0 ) nodes = nodes+1
psiold = psi ( i )
endif
enddo ! i =2 ,Nx−1
!−−−−−−− P r i n t f i n a l s o l u t i o n :
open ( u n i t =11 , f i l e = ’ p s i . dat ’ )
p r i n t * , ’ F i n a l r e s u l t : E= ’ , energy , ’ n= ’ , nodes ,&
’ norm = ’ , norm
i f ( DABS ( matchlogd ) . g t . e p s i l o n ) p r i n t *&
10.3. BOUND STATES 423
, ’ F i n a l r e s u l t : SOS : logd > e p s i l o n . logd= ’ , matchlogd
w r i t e ( 1 1 , * ) ’ # E= ’ , energy , ’ n= ’ , nodes ,&
’ norm = ’ , norm
do i =1 , Nx
x = xmin + ( i−1) * dx
w r i t e ( 1 1 , * ) x , psi ( i )
enddo
close (11)
end program schroedinger_equation_1D
! ===========================================================
! Simpson ’ s r u l e t o i n t e g r a t e p s i ( x ) * p s i ( x ) f o r proper
! n o r m a l i z a t i o n . For n i n t e r v a l s o f width dx ( n even )
! Simpson ’ s r u l e i s :
! i n t ( f ( x ) dx ) =
! ( dx / 3 ) * ( f ( x_0 ) +4 f ( x_1 ) +2 f ( x_2 ) + . . . + 4 f ( x_ {n−1})+ f ( x_n ) )
!
! Input : D i s c r e t e v a l u e s o f f u n c t i o n p s i (Nx)
! I n t e g r a t i o n s t e p dx
! Returns : I n t e g r a l ( p s i ( x ) p s i ( x ) dx )
! ===========================================================
r e a l ( 8 ) f u n c t i o n integrate ( psi , dx , Nx )
i m p l i c i t none
i n t e g e r : : Nx
!−−−−−−−−−−−−− Note : we need P due t o geometry o f a r r a y
r e a l ( 8 ) : : psi ( Nx ) , dx
!−−−−−−−−−−−−−
real (8) : : int
integer : : i
!−−−−− z e r o t h order p o i n t :
i = 1
i n t = psi ( i ) * psi ( i )
!−−−−− odd order p o i n t s ( i =k+1 i s even ) :
do i =2 , Nx −1 ,2
i n t = i n t + 4.0 D0 * psi ( i ) * psi ( i )
enddo
!−−−−− even order p o i n t s :
do i =3 , Nx −2,2
i n t = i n t + 2.0 D0 * psi ( i ) * psi ( i )
enddo
!−−−−− l a s t p o i n t :
i = Nx
i n t = i n t + psi ( i ) * psi ( i )
!−−−−− measure n o r m a l i z a t i o n :
i n t = i n t * dx / 3 . 0 D0
!−−−−− f i n a l r e s u l t :
integrate = i n t
end f u n c t i o n integrate
! ===========================================================
424 CHAPTER 10. SCHRÖDINGER EQUATION
The reproduction of the results of the previous section for the infinite
potential well is left as an exercise. The compilation and running of the
program can be done with the commands:
> g f o r t r a n sch . f90 schInfSq . f90 rk . f90 −o s
> ./s
# Enter energy , de , xmin , xmax , Nx
1 0.5 −1 1 2000
# #######################################
# Estart= 1.000 de= 0.5
# Nx= 2001 eps= 1 . 0E−006
# xmin= −1.000 xmax= 1.000 dx= 1.000E−003
# p s i ( xmin )= 0.000 p s i p ( xmin )= 1.000
# p s i ( xmax )= 0.000 p s i p ( xmax )= −1.000
# #######################################
# i t e r , energy , de , xm, logd : 1 1.0000 0.500 −0.601 −0.9748
# i t e r , energy , de , xm, logd : 2 1.5000 0.500 −0.601 −0.6412
.....
# i t e r , energy , de , xm, logd : 30 2.4674 −3.815E−6 −0.601 −1.0E−6
# i t e r , energy , de , xm, logd : 31 2.4674 1 . 9 0 7E−6 −0.601 2 . 7E−7
Final result : E= 2.467401504516602 n= 1 norm = 1.5707965025
We set xmin= -1, xmax = 1, Nx= 2000 and ϵ = 1, δϵ = 0.5. The energy
of the ground state is found to be ϵ1 = 2.4674015045166016. The wave
function is stored in the file psi.dat and can be plotted with the gnuplot
command
gnuplot > p l o t ” p s i . dat ” using 1 : 2 with lines
The functions computed during the iterations of the algorithm are stored
in the file all.dat. The first column is the iteration number (here we
have iter = 0, ... 31) and we can easily filter each one of them with
the commands
gnuplot > plot ”<awk ’ $1 ==1’ all . dat ” using 3:4 w l t ” i t e r =1”
gnuplot > replot ”<awk ’ $1 ==2’ all . dat ” using 3:4 w l t ” i t e r =2”
gnuplot > replot ”<awk ’ $1 ==3’ all . dat ” using 3:4 w l t ” i t e r =3”
gnuplot > replot ”<awk ’ $1 ==4’ all . dat ” using 3:4 w l t ” i t e r =4”
.....
which reproduce figure 10.6.
10.4. MEASUREMENTS 425
1
iter=1
0.9 iter=2
iter=3
0.8 iter=4
0.7
0.6
ψ(x)
0.5
0.4
0.3
0.2
0.1
0
-1 -0.5 0 0.5 1
x
Figure 10.6: The convergence of the solutions to the solution of Schrödinger’s equa-
tion for the ground state of the infinite potential well according to the discussion on
page 424.
10.4 Measurements
The action of an operator Â(x̂, p̂) on a state |ψ⟩ can be easily calculated
in the position representation by its action on the corresponding wave
function ψ(x). The action of the operators
∂
x̂ψ(x) = xψ(x) p̂ψ(x) = −i ψ(x) (10.29)
∂x
yield¹⁰
∂
Â(x̂, p̂)ψ(x) = A(x, −i )ψ(x) . (10.30)
∂x
Using equation (10.9) we can calculate the expectation value ⟨A⟩ of the
operator A when the system is at the state |ψ⟩. Interesting examples
are the observables “position” x, “position squared” x2 , “momentum”
p, “momentum squared” p2 , “kinetic energy” T , “potential energy” V ,
“energy” or “Hamiltonian” H = T + V whose expectation values are
¹⁰We do not consider ordering problems of operators formed by products of non
commuting operators, e.g. xp2 .
426 CHAPTER 10. SCHRÖDINGER EQUATION
given by the relations
∫ +∞
⟨x⟩ = ψ ∗ (x) x ψ(x) dx
−∞
∫ +∞
⟨x2 ⟩ = ψ ∗ (x) x2 ψ(x) dx
−∞
∫ +∞ ( )
∗ ∂
⟨p⟩ = ψ (x) −i ψ(x) dx
−∞ ∂x
∫ +∞ ( )
∗ ∂2
⟨p2 ⟩ = ψ (x) − 2 ψ(x) dx
−∞ ∂x
∫ +∞ ( )
ℏ 2
∗ ∂2
⟨T ⟩ = ψ (x) − 2 ψ(x) dx
2mL2 −∞ ∂x
∫ +∞
ℏ 2
⟨V ⟩ = ψ ∗ (x) v(x) ψ(x) dx
2mL2 −∞
∫ +∞ ( )
ℏ2 ∗ ∂2
⟨H⟩ = ψ (x) − 2 + v(x) ψ(x) dx . (10.31)
2mL2 −∞ ∂x
We remind the reader that we used the dimensionless x, p as well as
equations (10.15) and (10.16). Especially interesting are the “uncertain-
ties” ∆x2 = ⟨x2 ⟩ − ⟨x⟩2 , ∆p2 = ⟨p2 ⟩ − ⟨p⟩2 that satisfy the inequality
(“Heisenberg’s uncertainty relation”)
1
∆x · ∆p ≥ . (10.32)
2
In the previous section we described how to calculate numerically the
eigenfunctions of the Hamiltonian. If Ĥψ(x) = Eψ(x), we obtain that
⟨H⟩ = (1/2mL2 )ϵ. Other operators need a numerical approximation for
the calculation of their expectation values. If the values of the wave
function are given at N equally spaced points x1 , x2 , . . . , xN , then we
obtain
∂ψ(xi ) ψ(xi+1 ) − ψ(xi−1 )
≈ (10.33)
∂x 2h
where h = xi+1 − xi and
∂ 2 ψ(xi ) ψ(xi+1 ) − 2ψ(xi ) + ψ(xi−1 )
2
≈ . (10.34)
∂x h2
Both equations entail an error of the order of O(h2 ). Special care should
be taken at the endpoints of the interval [x1 , xN ]. As a first approach we
10.4. MEASUREMENTS 427
will use the naive approximations¹¹
∂ψ(x1 ) ψ(x2 ) − ψ(x1 )
≈
∂x h
∂ψ(xN ) ψ(xN ) − ψ(xN −1 )
≈ (10.35)
∂x h
and
∂ 2 ψ(x1 ) ψ(x3 ) − 2ψ(x2 ) + ψ(x1 )
2
≈
∂x h2
2
∂ ψ(xN ) ψ(xN ) − 2ψ(xN −1 ) + ψ(xN −2 )
≈ . (10.36)
∂x2 h2
The relevant program that calculates ⟨x⟩, ⟨x2 ⟩, ⟨p⟩, ⟨p2 ⟩, ∆x, ∆p can be
found in the file observables.f90 and is listed below:
! ===========================================================
!
! F i l e o b s e r v a b l e s . f90
! Compile : g f o r t r a n o b s e r v a b l e s . f90 −o o
! Usage : . / o < p s i . dat >
!
! Read i n a f i l e with a wavefunction i n t h e format o f p s i . dat :
! # E= <energy > . . . .
! x1 p s i ( x1 )
! x2 p s i ( x2 )
! ............
!
! Outputs e x p e c t a t i o n v a l u e s :
! n o r m a l i z a t i o n Energy <x> <p> <x^2> <p^2> Dx Dp DxDp
! where Dx = s q r t ( < x^2>−<x >^2) Dp = s q r t ( <p^2>−<p>^2)
! DxDp = Dx * Dp
!
! ===========================================================
program observables_expectation
i m p l i c i t none
i n t e g e r , parameter : : P=50000
i n t e g e r Nx , i
r e a l ( 8 ) : : xstep ( P ) , psi ( P ) , obs ( P )
r e a l ( 8 ) : : xav , pav , x2av , p2av , Dx , Dp , DxDp , energy , h , norm
r e a l ( 8 ) : : integrate
¹¹See the files observables.f90, Derivatives.nb of the accompanying software.
There you can find formulas that have errors of O(h2 ). In the examples discussed
below, the influence of the O(h) error on the results is approximately at the fourth
significant digit.
428 CHAPTER 10. SCHRÖDINGER EQUATION
c h a r a c t e r (20) : : psifile , scratch
! t h e f i r s t argument o f t h e command l i n e must be t h e path
! t o t h e f i l e with t h e wavefunction . (GNU f o r t r a n e x t e n s i o n . . . )
i f ( iargc ( ) . ne . 1 ) s t o p ’ Usage : o < f i l e n a m e > ’
c a l l getarg ( 1 , psifile )
! I f t h e f i l e does not e x i s t , we go t o l a b e l 100 ( s t o p ) :
open ( u n i t =11 , f i l e =psifile , s t a t u s = ’OLD’ , e r r =100)
p r i n t * , ” # r e a d i n g wavefunction from f i l e : ” , psifile
! we read t h e f i r s t comment l i n e from t h e f i l e :
read ( 1 1 , * ) scratch , scratch , energy
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Input data : p s i ( x )
Nx = 1
do while ( . TRUE . )
i f ( Nx . ge . P ) s t o p ’Too many p o i n t s ’
read ( 1 1 , * , end =101) xstep ( Nx ) , psi ( Nx )
Nx = Nx+1
enddo ! do while ( . TRUE . )
101 c o n t i n u e
Nx = Nx − 1
i f (mod( Nx , 2 ) . eq . 0) Nx = Nx − 1
h = ( xstep ( Nx )−xstep ( 1 ) ) / ( Nx −1)
!−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
! Calculate :
!−−−−−−−−−− norm :
do i =1 , Nx
obs ( i ) = psi ( i ) * psi ( i )
enddo
norm = integrate ( obs , h , Nx )
!−−−−−−−−−− <x> :
do i =1 , Nx
obs ( i ) = xstep ( i ) * psi ( i ) * psi ( i )
enddo
xav = integrate ( obs , h , Nx ) / norm
!−−−−−−−−−− <p > / i :
obs ( 1 ) = psi ( 1 ) * ( psi ( 2 )−psi ( 1 ) ) / h
do i =2 , Nx−1
obs ( i ) = psi ( i ) * ( psi ( i +1)−psi ( i−1) ) / ( 2 . 0 D0 * h )
enddo
obs ( Nx ) = psi ( Nx ) * ( psi ( Nx )−psi ( Nx −1) ) / h
pav = −integrate ( obs , h , Nx ) / norm
!−−−−−−−−− <x^2>
do i =1 , Nx
obs ( i ) = xstep ( i ) * xstep ( i ) * psi ( i ) * psi ( i )
enddo
x2av = integrate ( obs , h , Nx ) / norm
!−−−−−−−− <p^2>
obs ( 1 ) = psi ( 1 ) * ( psi ( 3 ) −2.0D0 * psi ( 2 ) +psi ( 1 ) ) / ( h * h )
10.4. MEASUREMENTS 429
do i =2 , Nx−1
obs ( i ) = psi ( i ) * ( psi ( i +1) −2.0D0 * psi ( i )+psi ( i−1) ) / ( h * h )
enddo
obs ( Nx ) = psi ( Nx ) *&
( psi ( Nx ) −2.0D0 * psi ( Nx −1)+psi ( Nx−2) ) / ( h * h )
p2av = −integrate ( obs , h , Nx ) / norm
!−−−−−−−− Dx
Dx = s q r t ( x2av − xav * xav )
!−−−−−−−− Dp
Dp = s q r t ( p2av − pav * pav )
!−−−−−−−− Dx . Dp
DxDp = Dx * Dp
! print results :
p r i n t * , ’ # norm E <x> <p > / i <x^2> <p^2> Dx Dp DxDp ’
p r i n t ’ (10G25 . 1 7 ) ’ , norm , energy , xav , pav , x2av , p2av , Dx , Dp , DxDp
s t o p ! normal e x e c u t i o n ends here . Error messages f o l l o w
100 s t o p ’ Cannot open f i l e ’
end program observables_expectation
! ===========================================================
!
! Simpson ’ s r u l e t o i n t e g r a t e p s i ( x ) .
! For n i n t e r v a l s o f width dx ( n even )
! Simpson ’ s r u l e i s :
! i n t ( f ( x ) dx ) =
! ( dx / 3 ) * ( f ( x_0 ) +4 f ( x_1 ) +2 f ( x_2 ) + . . . + 4 f ( x_ {n−1})+ f ( x_n ) )
!
! Input : D i s c r e t e v a l u e s o f f u n c t i o n p s i (Nx)
! I n t e g r a t i o n s t e p dx
! Returns : I n t e g r a l ( p s i ( x ) p s i ( x ) dx )
! ===========================================================
r e a l ( 8 ) f u n c t i o n integrate ( psi , dx , Nx )
i m p l i c i t none
i n t e g e r : : Nx
r e a l ( 8 ) : : psi ( Nx ) , dx
real (8) : : int
integer i
!−−−−− z e r o t h order p o i n t :
i = 1
i n t = psi ( i )
!−−−−− odd order p o i n t s ( i =k+1 i s even ) :
do i =2 , Nx −1 ,2
i n t = i n t + 4.0 D0 * psi ( i )
enddo
!−−−−− even order p o i n t s :
do i =3 , Nx −2,2
i n t = i n t + 2.0 D0 * psi ( i )
enddo
!−−−−− l a s t p o i n t :
i = Nx
430 CHAPTER 10. SCHRÖDINGER EQUATION
i n t = i n t + psi ( i )
!−−−−− measure n o r m a l i z a t i o n :
i n t = i n t * dx / 3 . 0 D0
!−−−−− f i n a l r e s u l t :
integrate = i n t
end f u n c t i o n integrate
! ===========================================================
The program needs to read in the wave function at the points x1 , . . . , xNx
in the format produced by the program in sch.f90. The first line should
have the energy written at the 3rd column, whereas from the 2nd line
and on there should be two columns with the (xi , ψ(xi )) pairs. It is not
necessary to have the wave function properly normalized, the program
will take care of it. If this data is stored in a file psi.dat, then the
program can be used by running the commands
> g f o r t r a n observables . f90 −o obs
> . / obs psi . dat
The program prints the normalization constant of ψ(x), the value of the
energy¹², ⟨x⟩, ⟨x2 ⟩, ⟨p⟩/i, ⟨p2 ⟩, ∆x, ∆p and ∆x · ∆p to the stdout.
Some details about the program: In order to read in the data from the
file psi.dat we use the functions iargc(), getarg(n,string). The for-
mer returns the number of arguments of the command line and the latter
stores the n-th argument to the CHARACTER variable string. Therefore,
the statements
c h a r a c t e r (20) : : psifile , scratch
i f ( iargc ( ) . ne . 1 ) s t o p ’ Usage : o <filename > ’
c a l l getarg ( 1 , psifile )
stop the program if the command line does not have exactly one argument
and store the first argument to the variable file.
The command
open ( u n i t =11 , f i l e =psifile , s t a t u s = ’OLD’ , e r r =100)
100 s t o p ’ Cannot open f i l e ’
opens a file which should already exist (status='OLD'), otherwise an
error message is issued. The option err=100 transfers the control of the
program to the statement labeled '100'. In the example shown above,
¹²The one read from the file. It is not calculated from the data.
10.5. THE ANHARMONIC OSCILLATOR - AGAIN... 431
the program stops and prints an error message 'Cannot open filename'
to the stdout.
The commands
Nx = 1
do while ( . TRUE . )
read ( 1 1 , * , end =101) xstep ( Nx ) , psi ( Nx )
Nx = Nx+1
enddo ! do while ( . TRUE . )
101 c o n t i n u e
read the opened file line by line. The option end=101 at the statement
read(11,*,end=101) transfers the control of the program to the labeled
statement with label 101 (i.e. outside the do loop) when we reach the
end of file.
The rest of the commands are applications of equations (10.33), (10.34),
(10.35) and (10.36) to the formulas (10.31) and the reader is asked to
study them carefully. The program uses the function integrate in order
to perform the necessary integrals.
10.5 The Anharmonic Oscillator - Again...
In the previous chapter 9 we studied the quantum mechanical harmonic
and anharmonic oscillator in the representation of the energy eigenstates
of the harmonic oscillator |n⟩. In this section we will revisit the problem
by using the position representation. We will calculate the eigenfunctions
ψn,λ (x) that diagonalize the Hamiltonian (9.15),
√ which are the solutions
of the Schrödinger equation. By setting L = ℏ/mω in equation (10.13),
equation (10.12) becomes
ψ ′′ (x) = −(ϵ − v(x))ψ(x) , (10.37)
where v(x) = x2 + 2λx4 . For λ = 0 we obtain the harmonic oscillator with
( )
1 −x2 /2 1
ψn (x) = √ √ e Hn (x) , ϵn = 2 n + , (10.38)
2n n! π 2
where Hn (x) are the Hermite polynomials.
We start with the simple harmonic oscillator where the exact solution
is known. The potential and the initial conditions are programmed in
the file schHOC.f90. The changes that we need to make concern the
functions V(x), boundary(xmin, xmax, psixmin, psipxmin, psixmax,
psipxmax):
432 CHAPTER 10. SCHRÖDINGER EQUATION
! ===========================================================
! f i l e : schHOC . f
! ..............
!−−−−− p o t e n t i a l :
real (8) function V(x)
i m p l i c i t none
real (8) : : x
V = x*x
end f u n c t i o n V
!−−−−− boundary c o n d i t i o n s :
subroutine &
boundary ( xmin , xmax , psixmin , psipxmin , psixmax , psipxmax )
i m p l i c i t none
r e a l ( 8 ) : : xmin , xmax , psixmin , psipxmin , psixmax , psipxmax , V
psixmin = exp ( −0.5 D0 * xmin * xmin )
psipxmin = −xmin * psixmin
psixmax = exp ( −0.5 D0 * xmax * xmax )
psipxmax = −xmax * psixmax
end s u b r o u t i n e boundary
! ===========================================================
.................
The code omitted at the dots is identical to the one discussed in the
previous section. The initial conditions are inspired by the asymptotic
behavior of the solutions to Schrödinger’s¹³ equation ψ0 (x) ∼ e−x /2 ,
2
ψn′ (x) ∼ −xψn (x). You are encouraged to test the influence of other
choices on the results. The results are depicted in figure 10.7 where, be-
sides the qualitative agreement, their difference from the known values
(10.38) is also shown. This difference turns out to be of the order of
10−11 –10−7 . The values of the energy ϵn for n ≤ 14 are in agreement with
(10.38) with relative accuracy better than 10−9 .
Then we calculate the expectation values ⟨x⟩, ⟨x2 ⟩, ⟨p⟩, ⟨p2 ⟩, ∆x and
∆p. These are easily calculated
√ using equations (9.4) √ and (9.8). We see
that ⟨x⟩ = ⟨n| (a† + a)/ 2 |n⟩ = 0, ⟨p⟩ = ⟨n| i(a† − a)/ 2 |n⟩ = 0, whereas
( )
1 † † 1
⟨x ⟩ = ⟨p ⟩ = ⟨n| (a a + aa ) |n⟩ = n +
2 2
. (10.39)
2 2
The program observables.f90 calculates ⟨x⟩ = 0 with accuracy ∼ 10−6
and ⟨p⟩ = 0 with accuracy ∼ 10−11 . The expectation values ⟨x2 ⟩, ⟨p2 ⟩ are
shown in table 10.2.
¹³In fact ψn (x) ∼ xn e−x /2 which we neglect. This does not influence the results for
2
the values of n studied here. Examine if this is necessary for larger values of n.
10.5. THE ANHARMONIC OSCILLATOR - AGAIN... 433
0.8 3e-11
ε0=1
0.7 2.5e-11
2e-11
0.6
1.5e-11
0.5 1e-11
∆ψ0(x)
ψ0(x)
0.4 5e-12
0.3 0
-5e-12
0.2
-1e-11
0.1 -1.5e-11
0 -2e-11
-6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
x x
0.6 1.2e-07
ε0=19
1e-07
0.4
8e-08
6e-08
0.2
4e-08
∆ψ9(x)
ψ9(x)
0 2e-08
0
-0.2
-2e-08
-4e-08
-0.4
-6e-08
-0.6 -8e-08
-8 -6 -4 -2 0 2 4 6 8 -8 -6 -4 -2 0 2 4 6 8
x x
Figure 10.7: The eigenfunctions ψ0 (x), ψ9 (x) calculated by the program in sch.f90,
schHOC.f90. The plot to the right shows the difference of the results from the known
values (10.38).
Next, the calculation is repeated for the anharmonic oscillator for λ =
0.5, 2.0. We copy the file schHOC.f90 to schUOC.f90 and change the
potential in the function V(x):
! ===========================================================
! f i l e : schUOC . f
! ...................
!−−−−− p o t e n t i a l :
real (8) function V(x)
i m p l i c i t none
r e a l ( 8 ) : : x , lambda
lambda = 2.0 D0
V = x * x +2.0 D0 * lambda * x * x * x * x
end f u n c t i o n V
....................
The wave functions are plotted in figure 10.8. We see that by increasing
λ the particle becomes more confined in space as expected. In table 10.3
we list the values of the energy ϵn for n = 0, ..., 9. By increasing λ, ϵn (λ) is
increased. Table 10.4 lists the expectation values ⟨x2 ⟩, ⟨p2 ⟩ and ∆x·∆p for
434 CHAPTER 10. SCHRÖDINGER EQUATION
n ⟨x2 ⟩ ⟨p2 ⟩ ∆x · ∆p
0 0.500000000 0.4999977 0.4999989
1 1.500000284 1.4999883 1.4999943
2 2.499999747 2.4999711 2.4999854
3 3.499999676 3.4999441 3.4999719
4 4.499999607 4.4999082 4.4999539
5 5.499999520 5.4998633 5.4999314
6 6.499999060 6.4998098 6.4999044
7 7.499999642 7.4995484 7.4997740
8 8.499999715 8.4994203 8.4997100
9 9.499999837 9.4992762 9.4996380
10 10.500000012 10.4991160 10.4995580
11 11.499999542 11.4994042 11.4997019
12 12.499999610 12.4992961 12.4996479
13 13.499999705 13.4991791 13.4995894
14 14.499999835 14.4990529 14.4995264
Table 10.2: The expectation values ⟨x2 ⟩, ⟨p2 ⟩ and the product ∆x · ∆p for the simple
harmonic oscillator for the states |n⟩, n = 0, . . . , 14.
n ϵn ϵn,λ=0.5 ϵn,λ=2.0
0 1.0000 1.3924 1.9031
1 3.0000 4.6488 6.5857
2 5.0000 8.6550 12.6078
3 7.0000 13.1568 19.4546
4 9.0000 18.0576 26.9626
5 11.0000 23.2974 35.0283
6 13.0000 28.8353 43.5819
7 15.0000 34.6408 52.5723
8 17.0000 40.6904 61.9598
9 19.0000 46.9650 71.7129
Table 10.3: The values of the energy ϵn for the harmonic and anharmonic oscillator
for λ = 0.5, 2.0. The values of the corresponding energy levels are increased with
increasing λ.
10.6. THE LENNARD–JONES POTENTIAL 435
1 1
λ=0.0 λ=0.0
0.9 λ=0.5 0.8 λ=0.5
λ=2.0 λ=2.0
0.8 0.6
0.7 0.4
0.6 0.2
ψ0(x)
ψ1(x)
0.5 0
0.4 -0.2
0.3 -0.4
0.2 -0.6
0.1 -0.8
0 -1
-6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
x x
1 0.8
λ=0.0 λ=0.0
0.8 λ=0.5 0.6 λ=0.5
λ=2.0 λ=2.0
0.6 0.4
0.4
0.2
0.2
ψ2(x)
ψ3(x)
0
0
-0.2
-0.2
-0.4 -0.4
-0.6 -0.6
-0.8 -0.8
-6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
x x
0.8 0.8
λ=0.0 λ=0.0
0.6 λ=0.5 0.6 λ=0.5
λ=2.0 λ=2.0
0.4 0.4
0.2 0.2
ψ4(x)
ψ5(x)
0 0
-0.2 -0.2
-0.4 -0.4
-0.6 -0.6
-0.8 -0.8
-6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
x x
Figure 10.8: The wave functions of the anharmonic oscillator ψn,λ (x) for n =
0, 1, 2, 3, 4, 5 and λ = 0.5, 2.0 compared to the respective ones of the simple harmonic
oscillator. Increasing λ yields stronger confinement of the particle in space.
the anharmonic
√ oscillator for the states√|n⟩, n = 0, . . . , 9. By increasing
λ, ∆x = ⟨x ⟩ is decreased and ∆p = ⟨p2 ⟩ is increased. The product
2
of the uncertainties ∆x · ∆p seems to be quite close to the corresponding
values for the harmonic oscillator. The results should be compared with
the ones obtained in table 9.1 of chapter 9.
10.6 The Lennard–Jones Potential
The Lennard–Jones potential is a simple phenomenological model of the
interaction between two neutral atoms in a diatomic molecule. This is
given by
{( ) }
σ 12 ( σ )6
V (x) = 4V0 − . (10.40)
x x
436 CHAPTER 10. SCHRÖDINGER EQUATION
λ = 0.5 λ = 2.0
n ⟨x ⟩ 2
⟨p ⟩
2
∆x · ∆p ⟨x ⟩
2
⟨p2 ⟩ ∆x · ∆p
0 0.3058 0.8263 0.5027 0.2122 1.1980 0.5042
1 0.8013 2.8321 1.5064 0.5408 4.2102 1.5089
2 1.1554 5.3848 2.4944 0.7612 8.1513 2.4909
3 1.4675 8.2819 3.4862 0.9582 12.6501 3.4816
4 1.7509 11.4545 4.4784 1.1370 17.5955 4.4728
5 2.0141 14.8599 5.4707 1.3029 22.9169 5.4643
6 2.2617 18.4691 6.4631 1.4590 28.5668 6.4560
7 2.4970 22.2607 7.4555 1.6074 34.5103 7.4478
8 2.7220 26.2184 8.4478 1.7492 40.7206 8.4397
9 2.9384 30.3289 9.4402 1.8856 47.1762 9.4316
Table 10.4: The expectation values ⟨x2 ⟩, ⟨p2 ⟩ and the product ∆x · ∆p for √
the anhar-
√ states |n⟩, n = 0, . . . , 9. Note the decrease of ∆x = ⟨x ⟩ and
monic oscillator for the 2
the increase of ∆p = ⟨p ⟩ with increasing λ. The uncertainty product ∆x · ∆p seems
2
to take values close to the corresponding ones of the harmonic oscillator for both values
of λ. Compare the results in this table with the ones in table 9.1.
The repulsive term describes the Pauli interaction due to the overlapping
of the electron orbitals, whereas the attractive term describes the Van der
Waals force. The first one dominates at short distances and the latter at
long distances. We choose L = σ in (10.13) and define v0 = 2mσ 2 V0 /ℏ2 .
Equation (10.40) becomes
{( ) ( )6 }
12
1 1
v(x) = 4v0 − , (10.41)
x x
whereas the eigenvalues ϵn are related to the energy values En by
( )
En
ϵn = 4v0 . (10.42)
V0
The plot of the potential is shown in figure 10.5 for v0 = 250. The
minimum is located at xm = 21/6 ≈ 1.12246 and its value is −v0 . The
code for this potential is in the file schLJ.f90. The necessary changes to
the code discussed in the previous sections are listed below:
! ===========================================================
! f i l e : schLJ . f90 ( Lennard−J o n e s )
! ..................
10.6. THE LENNARD–JONES POTENTIAL 437
n ϵn ⟨x⟩ ⟨p⟩ ⟨x2 ⟩ ⟨p2 ⟩ ∆x ∆p ∆x · ∆p
0 -173.637 1.186 1.0e-10 1.415 34.193 0.091 5.847 0.534
1 -70.069 1.364 6.0e-11 1.893 56.832 0.178 7.539 1.338
2 -18.191 1.699 -4.5e-08 2.971 39.480 0.291 6.283 1.826
3 -1.317 2.679 -2.6e-08 7.586 9.985 0.638 3.160 2.016
Table 10.5: The results for the Lennard-Jones potential with v0 = 250. We find 4
bound states.
!−−−−− p o t e n t i a l :
real (8) function V(x)
i m p l i c i t none
r e a l ( 8 ) : : x , V0
V0 = 250.0 D0
V = 4.0 D0 * V0 * ( 1 . 0 D0 / x **12 −1.0 D0 / x * * 6 )
end f u n c t i o n V
!−−−−− boundary c o n d i t i o n s :
subroutine &
boundary ( xmin , xmax , psixmin , psipxmin , psixmax , psipxmax )
i m p l i c i t none
r e a l ( 8 ) : : xmin , xmax , psixmin , psipxmin , psixmax , psipxmax , V
r e a l ( 8 ) : : energy
common / params / energy
!−−−−− I n i t i a l v a l u e s a t xmin and xmax
psixmin = exp(−xmin * s q r t ( DABS ( energy−V ( xmin ) ) ) )
psipxmin = s q r t ( DABS ( energy−V ( xmin ) ) ) * psixmin
psixmax = exp(−xmax * s q r t ( DABS ( energy−V ( xmax ) ) ) )
psipxmax = −s q r t ( DABS ( energy−V ( xmax ) ) ) * psixmax
end s u b r o u t i n e boundary
............................
For the integration we choose v0 = 250 and xmin = 0.7, 4 <xmax
< 10. The results are plotted in figure 10.9. There are four bound states.
The first two ones are quite confined within the potential well whereas
the last ones begin to “spill” out of it. Table 10.5 lists the results. We
observe that ⟨p⟩ = 0 within the attained accuracy as expected for real,
bound states¹⁴.
¹⁴For ψ(+∞) = ψ(0) = 0 and ψ ∗ (x) = ψ(x) we have that i⟨p⟩/ℏ =
∫ +∞ ∫ +∞
0
ψ(x)(d/dx)ψ(x) dx = − 0 (d/dx)ψ(x)ψ(x) dx = 0.
438 CHAPTER 10. SCHRÖDINGER EQUATION
2.5
v(x)/v0
2 ε1/v0
ε2/v0
1.5 ε3/v0
ε4/v0
ψ1(x)
1 ψ2(x)
ψ3(x)
0.5 ψ4(x)
0
-0.5
-1
-1.5
1 2 3 4 5 6 7
x
Figure 10.9: The four bound states for the Lennard-Jones potential with v0 = 250.
The bold red line is the potential v(x)/v0 . We plot the energy levels ϵn /v0 and the
corresponding wave functions.
10.7 Problems
10.1 Add the necessary code to the program in the file well.f90 so
that the final wave function printed in the file psi.dat is properly
∫1
normalized. The integral −1 ψ(x)ψ(x) dx can be computed using
the Simpson rule
∫ b
f (x) dx = (h/3) (f (x0 ) + 4f (x1 ) + 2f (x2 ) + . . .
a
+2f (xn−2 ) + 4f (xn−1 ) + f (xn ) .)
The interval [a, b] is discretized by n points x0 = a, x1 , x2 , . . . , xn = b
where n is even. Each interval [xi , xi+1 ] has width h.
10.2 Add the necessary code to the program in the file well.f90 in order
to calculate the number of nodes (zeroes) of the wave function. Us-
ing this result, the program should print the level n of the calculated
wave function ψn (x).
10.7. PROBLEMS 439
1 mat 0.8 mat
ψ1 sch(x) ψ6 sch(x)
0.8 ψ1 (x) 0.6 ψ6 (x)
0.6
0.4
0.4
0.2 0.2
0 0
-0.2 -0.2
-0.4
-0.4
-0.6
-0.8 -0.6
-1 -0.8
-6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
x x
0.8 mat 1.2 mat
ψ9 sch(x) ψ14 (x)
0.6 ψ9 (x) 1
0.4 0.8
0.2 0.6
0 0.4
-0.2 0.2
-0.4 0
-0.6 -0.2
-0.8 -0.4
-6 -4 -2 0 2 4 6 -8 -6 -4 -2 0 2 4 6 8
x x
Figure 10.10: Comparison of the results of the calculation of the wave functions
ψn,λ (x) of the anharmonic oscillator for λ = 2.0 using the methods described in problem
12. The wave functions ψ sch (x) are the wave functions ψn,λ (x) calculated using the
methods described in this chapter. The wave functions ψ mat (x) are the wave functions
ψn,λ (x) calculated using the methods described in chapter 9 for Hilbert space dimension
N = 40. Note the difference at large x. This is because the amplitudes ψn,λ (x) = ⟨x|n⟩λ
for large x receive contributions from states |m⟩ with large m (why?).
10.3 Calculate the wave functions of the energy eigenstates for the po-
tential (10.27) with v0 < 0. This is the problem of the (finite)
potential well. Solve the problem for v0 = −100 and a = 0.3.
How many bound states do you find? Next study the influence
of the wall on the solutions. Introduce a parameter b so that
v(x ≥ b) = +∞ and study the dependence of the solutions on
b. Take b = 0.35, 0.4, 0.5, 0.6, 0.8, 1.0, 1.5, 2.0, 2.5, 3.0 and compute the
difference of the first two energy eigenvalues. Estimate the accuracy
of the method. Next lower the value of |v0 | until there is no bound
state. What is the relation between a and v0 when this happens?
Compare with the analytic result which you know from your quan-
tum mechanics course.
Hint: For the largest values of b, take Nx > 1000. When convergence
is not achieved decrease epsilon.
440 CHAPTER 10. SCHRÖDINGER EQUATION
10.4 Set v0 = 1000, 5000 to the double well potential. Observe √
the (almost)
degenerate states and plot the wave functions ψ±,n = (1/ 2)(ψn (x)±
ψn+1 (x)), where n is odd. Compare the results with the correspond-
ing energy levels and eigenfunctions of the infinite square well.
Increase v0 to the point where you cannot solve the problem nu-
merically.
Hint: For large v0 the numerical effort is increased. For |x| < a
the wave function is almost zero and it is hard to obtain the non
trivial wave function for a < |x| < 1. As the accuracy deteriorates,
you should increase epsilon in the program so that convergence is
achieved relatively fast.
10.5 Repeat problems 3 and 4 using the program sch.f90. Compare the
results.
10.6 Study the bound states in the potentials
0 a < |x|
v(x) = −V0 b < |x| < a
−V1 |x| < b
for a = 1, b = 0.2, V0 = 100, V1 = 0, 50 and
V1 x < 0
v(x) = −V0 0 < x < a
0 a<x
for a = 1, V0 = 100, V1 = +∞, 10, 100 and
V1 a < |x|
−V0 b < |x| < a
v(x) =
0 c < |x| < b
−V0 |x| < c
for a = 1, b = 0.7, c = 0.6, 0.3, V0 = 100, V1 = +∞, 10, 0. In each case
calculate ⟨x⟩, ⟨x2 ⟩, ⟨p⟩, ⟨p2 ⟩, ∆x, ∆p, ∆x · ∆p.
10.7 Write a program that calculates the probability that a particle is
found in an interval [x1 , x2 ] given the wave function calculated by
the program in the file sch.f90. Apply your program on the results
of the previous problem and calculate the intervals [−x1 , x1 ] where
the probability to find the particle inside them is equal to 1/3.
10.8 Fill the tables 10.3 and 10.4 with the results for λ = 0.2, 0.7, 1.0,
1.3, 1.6, 2.5, 3.0 and plot each expectation value as a function of λ.
10.7. PROBLEMS 441
10.9 A particle is under the influence of a potential
{ }
ℏ2 2 1 1
V (x) = α λ(λ − 1) − .
2m 2 cosh2 (αx)
The energy spectrum is given by
{ }
ℏ2 2 λ(λ − 1)
En = α − (λ − 1 − n)2
2m 2
for the values of n = 0, 1, 2, . . . for which En > Vmin . Calculate the
energy levels ϵn of the bound states numerically by setting L =
1/α in equation (10.13) and λ = 4. Plot the potential v(x) and
the corresponding eigenfunctions. Calculate the expectation values
of the position and momentum, the uncertainties in position and
momentum and their product. Repeat for λ = 2, 6, 8, 10.
10.10 Write a program that reads in a wavefunction and calculates the
expectation value of the Hamiltonian
∫ +∞ ( )
ℏ2 ∂ 2
⟨Ĥ⟩ = ψ(x) − + V (x) ψ(x) dx ,
−∞ 2m ∂x2
by assuming that ψ(x) is real. Calculate ψn (x) for the harmonic
oscillator for n = 1, . . . , 10 and show (numerically) that ⟨Ĥ⟩n = En .
10.11 Consider a particle in the Morse potential
{( )2 }
V (x) = De 1 − e−a(r−re ) − 1 .
Calculate the energy spectrum of the bound states. Choose L = 1/a,
x = ar, xe = are , λ2 = 2mDe /a2 ℏ2 and obtain
( )
v(x) = λ2 e−2(x−xe ) − 2e−(x−xe ) .
Compare your results with the known analytic solutions
( )2
1
ϵn = λ − n −
2
ψn (z) = Nn z λ−n−1/2 e−z/2 Ln2λ−2n−1 (z)
√
where z = 2λe−(x−xe ) , Nn = n! (2λ − 2n − 1)/(Γ(n + 1)Γ(2λ − n)),
and Lαn (z) is a Laguerre polynomial given by Lαn (z) = (z −α ez /n!)(dn /dz n )(z n+α e−z )
= (Γ(α + 2)/(Γ(n + 2)Γ(α − n + 2))1 F1 (−n, α + 1, z). You can take
λ = 4, xe = 1 and calculate ⟨x⟩, ⟨x2 ⟩, ⟨p⟩, ⟨p2 ⟩, ∆x, ∆p, ∆x · ∆p.
442 CHAPTER 10. SCHRÖDINGER EQUATION
10.12 Calculate the wave functions of the eigenstates of the Hamiltonian
for the anharmonic oscillator for λ = 2.0 and n = 0, . . . , 15. Calculate
the wavefunctions using the program anharmonic.f90 of chapter 9
for N = 15, 40, 100 and compare the two results.
Hint: Write a program that calculates the energy eigenfunctions of
the simple harmonic oscillator
1
e−x /2 Hn (x)
2
ψn (x) = √ √
2n n! π
where the Hermite polynomials satisfy the relations
Hn+1 (x) = 2xHn (x) − 2nHn−1 (x), H0 (x) = 1, H1 (x) = 2x .
The program anharmonic.f90 calculates the eigenstates of the an-
harmonic oscillator
∑
N −1
|n⟩λ = H(m + 1, n + 1) |m⟩
m=0
by storing the coefficients of the linear expansion in the elements
of the array H(N,N). The same relation holds for the corresponding
wave functions ψn,λ (x), ψn (x). From ψn (x) and H(i,j) calculate
ψn,λ (x) for −8 < x < 8 and determine the accuracy achieved by the
calculation for each N . For which values of x do you obtain large
discrepancies between your results? Remember that for large x, the
states of high energy contribute more than for small x. Figure 10.10
can help you understanding this statement.
Chapter 11
The Random Walker
In this chapter we will study the typical path followed by a ... drunk
when he decides to start walking from a given position. Because of
his drunkenness, his steps are in random directions and uncorrelated.
These are the basic properties of the models that we are going to study.
These models are related to specific physical problems like the Brownian
motion, the diffusion, the motion of impurities in a lattice, the large
distance properties of macromolecules etc. In the physics of elementary
particles random walks describe the propagation of free scalar particles
and they most clearly arise in the Feynman path integral formulation
of the euclidean quantum field theory. Random walks are precursors
to the theory of random surfaces which is related to the theory of two
dimensional “soft matter” membranes, two dimensional quantum gravity
and string theory [44].
The geometry of a typical path of a simple random walk is not classical
and this can be seen from two of its non classical properties. First, the
average distance traveled by the random walker is proportional to the
square root of the time traveled, i.e. the classical relation r = vt does
not apply. Second, the geometry of the path of the random walker has
fractal dimension which is larger than one¹. Similar structures arise in
the study of quantum field theories and random surfaces, where the
non classical properties of a typical configuration can be described by
appropriate generalizations of these concepts. For further study we refer
to [7, 43, 44, 45].
In order to simulate a stochastic system on the computer, it is neces-
sary to use random number generators. In most of the cases, these are
deterministic algorithms that generate a sequence of pseudorandom num-
¹More precisely, the Hausdorff dimension of the simple random walk is dH = 2.
443
444 CHAPTER 11. THE RANDOM WALKER
bers distributed according to a desired distribution. The heart of these
algorithms generate numbers distributed uniformly from which we can
generate any other complex distribution. In this chapter we will study
simple random number generators and learn how to use high quality,
research grade, portable, random number generators.
11.1 (Pseudo)Random Numbers
The production of pseudorandom² numbers is at the heart of a Monte
Carlo simulation. The algorithm used in their production is deterministic:
The generator is put in an initial state and the sequence of pseudorandom
numbers is produced during its “time evolution”. The next number in the
sequence is determined from the current state of the generator and it is
in this sense that the generator is deterministic. Same initial conditions
result in exactly the same sequence of pseudorandom numbers. But
the “time evolution” is chaotic and “neighboring” initial states result in
very different, uncorrelated, sequences. The chaotic properties of the
generators is the key to the pseudorandomness of the numbers in the
sequence: the numbers in the sequence decorrelate exponentially fast
with “time”. But this is also the weak point of the pseudorandom number
generators. Bad generators introduce subtle correlations which produce
systematic errors. Truly random numbers (useful in cryptography) can
be generated by using special devices based on e.g. radioactive decay
or atmospheric noise³. Almost random numbers are produced by the
special files /dev/random and /dev/urandom available on unix systems,
which read bits from an entropy pool made up from several external
sources (computer temperature, device noise etc).
Pseudorandom number generators, however, are the source of ran-
dom numbers of choice when efficiency is important. The most popular
generators are the modulo generators (D.H. Lehmer, 1951) because of their
simplicity. Their state is determined by only one integer xi−1 from which
the next one xi is generated by the relation
xi = a xi−1 + c (mod m) (11.1)
²We can’t define what a random process is, only what it isn’t. Outcomes which
lack discernable patterns are assumed to be random. If there is no way to predict
an event, we say it is random...Thus, there is no definition of what randomness is,
only definitions of what it isn’t. See Chris Wetzel, “Can you behave randomly?”,
http://faculty.rhodes.edu/wetzel/random/level23intro.html.
³There are online services which provide such sequences like www.random.org,
www.fourmilab.ch/hotbits/ and others.
11.1. (PSEUDO)RANDOM NUMBERS 445
for appropriately chosen values of a, c and m. In the bibliography, there
is a lot of discussion on the good and bad choices of a, c and m, which
depend on the programming language and whether we are on a 32–bit
or 64–bit systems. For details see the chapter on random numbers in [8].
The value of the integer m determines the maximum period of the
sequence. It is obvious that if the sequence encounters the same num-
ber after k steps, then the exact same sequence will be produced and k
will be the period of the sequence. Since there are at most m different
numbers, the period is at most equal to m. For a bad choice of a, c and
m the period will be much smaller. But m cannot be arbitrarily large
since there is a maximum number of bits that computers use for the
storage of integers. For 4-byte (32 bit) unsigned integers the maximum
number is 232 − 1, whereas for signed integers 231 − 1. One can prove⁴
that a good choice of a, c and m results in a sequence which is a permu-
tation {π1 , π2 , . . . , πm } of the numbers 1, 2, . . . , m. This is good enough
for simple applications that require fast random number generation but
for serious calculations one has to carefully balance efficiency with qual-
ity. Good quality random generators are more complicated algorithms
and their states are determined by more than one integer. If you need
the source code for such generators you may look in the bibliography,
like in e.g. [4], [5], [8], [47]. If portability is an issue, we recommend
the RANLUX random number generator [47] or the Marsaglia, Zaman
and Tsang generator. The Fortran code for RANLUX can also be found in
the accompanying software, whereas the MZT generator can be found in
Berg’s book/site [5].
In order to understand the use of random number generators, but
also in order to get a feeling of the problems that may arise, we list
the code of the two functions naiveran() and drandom(). The first one
is obviously problematic and we will use it in order to study certain
type of correlations that may exist in the generated sequences of random
numbers. The second one is much better and can be used in non–trivial
applications, like in the random walk generation or in the Ising model
simulations studied in the following chapters.
The function naiveran() is a simple application of equation (11.1)
with a = 1277, c = 0 and m = 217 :
! =============================================
! F i l e : n a i v e r a n . f90
! Program t o demonstrate t h e usage o f a modulo
⁴See Knuth [46].
446 CHAPTER 11. THE RANDOM WALKER
! g e n e r a t o r with a bad c h o i c e o f c o n s t a n t s
! r e s u l t i n g i n s t r o n g p a i r c o r r e l a t i o n s between
! g e n e r a t e d numbers
! =============================================
r e a l ( 8 ) f u n c t i o n naiveran ( )
i m p l i c i t none
integer : : iran =13337
common / naiveranpar / iran
i n t e g e r , parameter : : m = 131072 ! equal t o 2 * * 1 7
i n t e g e r , parameter : : a = 1277
iran = a * iran
iran = MOD( iran , m )
naiveran = iran /DBLE( m )
end f u n c t i o n naiveran
The function drandom() is also an application of the same equation, but
now we set a = 75 , c = 0 and m = 231 − 1. This is the choice of
Lewis, Goodman and Miller (1969) and provides a generator that passes
many tests and, more importantly, it has been used countless of times
successfully. One technical problem is that, when we multiply xi−1 by a,
we may obtain a number which is outside the range of 4-byte integers
and this will result in an “integer overflow”. In order to have a fast
and portable code, it is desirable to stay within the range of the 231 − 1
positive, 32-bit (4 byte), signed integers. Schrage has proposed to use
the relation
[ ]
a (xi−1 mod q) − r xi−1 if it is ≥ 0
(axi−1 ) mod m = [ q ]
a (xi−1 mod q) − r xi−1 + m if it is < 0
q
(11.2)
where m = aq + r, q = [m/a] and r = m mod a. One can show that
if r < q and if 0 < xi−1 < m − 1, then 0 ≤ a(xi−1 mod q) ≤ m − 1,
0 ≤ r[xi−1 /q] ≤ m−1 and that (11.2) is valid. The period of the generator
is 231 −2 ≈ 2×109 . The proof of the above statements is left as an exercise
to the reader.
! ====================================================
! F i l e : drandom . f90
! Implementation o f t h e Schrage a l g o r i t h m f o r a
! p o r t a b l e modulo g e n e r a t o r f o r 32 b i t s i g n e d i n t e g e r s
! ( from numerical r e c i p e s )
!
11.1. (PSEUDO)RANDOM NUMBERS 447
! r e t u r n s uniformly d i s t r i b u t e d pseudorandom numbers
! 0.0 < x < 1 . 0 (0 and 1 excluded )
! ====================================================
r e a l ( 8 ) f u n c t i o n drandom ( )
i m p l i c i t none
i n t e g e r , parameter : : a = 16807 ! a = 7**5
i n t e g e r , parameter : : m = 2147483647 ! m = a * q+r = 2**31 −1
i n t e g e r , parameter : : q = 127773 ! q = [m/ a ]
i n t e g e r , parameter : : r = 2836 ! r = MOD(m, a )
r e a l ( 8 ) , parameter : : f = ( 1 . 0 D0 / m )
integer :: p
integer : : seed
real (8) : : dr
common / randoms / seed
101 c o n t i n u e
p = seed / q ! = [ seed / q ]
seed = a * ( seed− q * p ) − r * p ! = a *MOD( seed , q )−r * [ seed / q ]
i f ( seed . l t . 0) seed = seed + m
dr = f * seed
i f ( dr . l e . 0.0 D0 . or . dr . ge . 1 . 0 D0 ) goto 101
drandom = dr
end f u n c t i o n drandom
The line that checks the result produced by the generator is necessary
in order to check for the number 0 which appears once in the sequence.
This adds a 10 − 20% overhead, depending on the compiler. If you don’t
care about that, you may remove the line. Note that the number seed
is put in a common block so it can be accessed by other parts of the
program.
Now we will write a program in order to test the problem of correla-
tions in the sequence of numbers produced by naiveran(). The program
will produce pairs of integers (i, j), where 0 ≤ i, j < 10000, which are sub-
sequently mapped on the plane. This is done by taking the integer part
of the numbers L u with L = 10000 and 0 ≤ u < 1 is the random number
produced by the generator:
! ==========================================================
! Program t h a t produces N random p o i n t s ( i , j ) with
! 0<= i , j < 10000. Simple q u a l i t a t i v e t e s t o f s e r i a l
! c o r r e l a t i o n s o f random number g e n e r a t o r s on t h e plane .
!
! compile :
! g f o r t r a n c o r r e l a t i o n s 2 r a n . f90 n a i v e r a n . f90 drandom . f90
! ==========================================================
program correlations2
448 CHAPTER 11. THE RANDOM WALKER
i m p l i c i t none
i n t e g e r , parameter : : L = 10000
integer :: i,N
character (10) : : arg
real (8) : : naiveran , drandom
integer : : seed
common / randoms / seed
! Read t h e number o f p o i n t s from f i r s t command argument
i f ( IARGC ( ) .EQ. 1 ) then
c a l l GETARG ( 1 , arg ) ; read ( arg , * ) N ! c o n v e r t s t r i n g −> i n t e g e r
e l s e ! d e f a u l t value , i f no N g iven by u s e r :
N=1000
endif
seed = 348325
do i =1 , N
p r i n t * , INT ( L * naiveran ( ) ) , INT ( L * naiveran ( ) )
! p r i n t * , INT (L * drandom ( ) ) , INT (L * drandom ( ) )
enddo
end program correlations2
The program can be found in the file correlations2ran.f90. In order
to test naiveran() we compile with the command
> g f o r t r a n correlations2ran . f90 naiveran . f90 −o naiveran
whereas in order to test drandom() we uncomment the print lines as
follows
! p r i n t * , INT (L * n a i v e r a n ( ) ) , INT (L * n a i v e r a n ( ) )
p r i n t * , INT ( L * drandom ( ) ) , INT ( L * drandom ( ) )
and recompile:
> g f o r t r a n correlations2ran . f90 drandom . f90 −o drandom
These commands result in two executable files naiveran and drandom.
In order to see the results we run the commands
> . / naiveran 100000 > naiveran . out
> . / drandom 100000 > drandom . out
> gnuplot
gnuplot > p l o t ” n a i v e r a n . out ” using 1 : 2 with dots
gnuplot > p l o t ”drandom . out ” using 1 : 2 with dots
11.1. (PSEUDO)RANDOM NUMBERS 449
which produce 105 points used in the plots in figures 11.1 and 11.2. In
the plot of figure 11.1, we see the pair correlations between the num-
bers produced by naiveran(). Figure 11.2 shows the points produced
by drandom(), and we can see that the correlations shown in figure 11.1
have vanished. The plot in figure 11.2 is qualitative, and a detailed,
quantitative, study of drandom() shows that the pairs (ui , ui+1 ) that it
produces, do not pass the χ2 test when we have more than 107 points,
which is much less than the period of the generator. In order to avoid
such problems, there are many solutions that have been proposed and
the simplest among them “shuffle” the results so that the low order se-
rial correlations vanish. Such generators will be discussed in the next
section. The uniform distribution of the random numbers produced
10000
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
Figure 11.1: Pairs of pseudorandom numbers produced by the function naiveran().
The correlations among pairs of such numbers show in the distribution of such pairs
on a clearly seen lattice.
can be examined graphically by constructing a histogram of the relative
frequency of their appearance. In order to construct the histograms we
use the script histogram which is written in the awk language⁵ as shown
⁵See the accompanying software in the Tools directory. Give the command
histogram -- -h which prints short usage instructions. I hope you remember how to
make the file histogram executable and put it in your path...
450 CHAPTER 11. THE RANDOM WALKER
10000
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
Figure 11.2: Pairs of pseudorandom numbers produced by the function drandom().
These points have a random distribution on the plane compared to those generated by
naiveran().
below:
> histogram −v f =0.01 drandom . out > drandom . hst
> gnuplot
gnuplot > p l o t ”drandom . h s t ” using 1 : 3 with histeps
gnuplot > p l o t [ : ] [ 0 : ] ”drandom . h s t ” using 1 : 3 with histeps
The command histogram -v f=0.01 constructs a histogram of the data
so that the bin width is 1/0.01 = 100. The reciprocal of the number
following the option -v f=0.01 defines the bin width. The histogram is
saved in the file drandom.out.
The results are shown in figures 11.3 and 11.4. Next, we study the
variance of the measurements, shown in figure 11.3. The variance is
decreased with the size of the sample of the collected random numbers.
This is seen in the histogram of figure 11.5. For a quantitative study of
the dependence of the variance on the size n of the sample, we calculate
11.1. (PSEUDO)RANDOM NUMBERS 451
0.012
0.01
0.008
0.006
0.004
0.002
0
0 2000 4000 6000 8000 10000
Figure 11.3: The relative frequency distribution of the pseudorandom numbers gen-
erated by drandom(). The distribution is uniform within (0, 1) and we see the deviations
from the average value.
the standard deviation
v
u
u ( )2
1 1 ∑ 2 1∑
n n
u ,
σ=t x − xi (11.3)
n − 1 n i=1 i n i=1
where {xi } is the sequence of random numbers. Figure 11.6 plots this
relation. By fitting
1
ln σ ∼ ln(n) , (11.4)
2
to a straight line, we see that
1
σ∼√ . (11.5)
n
If we need to generate random numbers which are distributed accord-
ing to the probability density f (x) we can use a sequence of uniformly
distributed random numbers in the interval (0, 1) as follows: Consider
the cumulative distribution function
∫ x
0 ≤ u ≡ F (x) = f (x′ ) dx′ ≤ 1 , (11.6)
−∞
452 CHAPTER 11. THE RANDOM WALKER
0.0108
0.0106
0.0104
0.0102
0.01
0.0098
0.0096
0.0094
0.0092
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
Figure 11.4: Same as in figure 11.3, but with the scale enlarged, so that the dispersion
of the histogram values is clearly seen.
which is equal to the area under the curve f (x) in the interval (−∞, x] and
it is equal to the probability P (x′ < x). If u is uniformly distributed in the
interval (0, 1) then we have that P (u′ < u) = u. Therefore x = F −1 (u) is
such that P (x′ < x) = u = F (x) and follows the f (x) distribution. There-
fore, if ui form a sequence of uniformly distributed random numbers,
then the numbers
xi = F −1 (ui ) (11.7)
form a sequence of random numbers distributed according to f (x).
Consider for example the Cauchy distribution
1 c
f (x) = c > 0. (11.8)
π c2 + x2
Then ∫ x ( )
1 1′ −1 x ′
F (x) = f (x ) dx = + tan . (11.9)
−∞ 2 π c
According to the previous discussion, the random number generator is
given by the equation
xi = c tan (πui − π/2) (11.10)
11.1. (PSEUDO)RANDOM NUMBERS 453
0.018
100000
10000
0.016 1000
0.014
0.012
0.01
0.008
0.006
0.004
0.002
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
Figure 11.5: The relative frequency distribution of the pseudorandom numbers gen-
erated by drandom() as a function of the sample size n for n = 1000, 10000, 100000.
or equivalently (for a more efficient generation)
xi = c tan (2πui ) . (11.11)
The generator of Gaussian random numbers is found in many appli-
cations. The Gaussian distribution is given by the probability density
1
e−x /(2σ )
2 2
g(x) = √ (11.12)
2π σ
The cumulative distribution function is
∫ x ( )
′ ′ 1 1 x
G(x) = g(x ) dx = + erf √ (11.13)
−∞ 2 2 2σ
∫x
where erf(x) = −∞ exp{−(x′ )2 } dx′ is the error function. The error func-
tion, as well as its inverse, can be calculated numerically, but this would
result in a slow computation. A trick to make a more efficient calculation
is to consider the probability density ρ(x, y) of two independent Gauss