**Authors**
Konstantinos N. Anagnostopoulos

**License**
CC-BY-SA-4.0

COMPUTATIONAL PHYSICS A Practical Introduction to Computational Physics and Scientific Computing (using C++) Athens, 2016 KONSTANTINOS N. ANAGNOSTOPOULOS National Technical University of Athens National Technical University of Athens COMPUTATIONAL PHYSICS A Practical Introduction to Computational Physics and Scientiﬁc Computing (C++ version) 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 Second Edition 2016 Version¹ 2.0.20161206201400 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 978-1-365-58322-3 (lulu.com, vol. I) ISBN 978-1-365-58338-4 (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 to the Second Edition vii Foreword to the First Edition ix 1 The Computer 1 1.1 The Operating System . . . . . . . . . . . . . . . . . . . . . 2 1.1.1 Filesystem . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.2 Commands . . . . . . . . . . . . . . . . . . . . . . . 10 1.1.3 Looking for Help . . . . . . . . . . . . . . . . . . . . 14 1.2 Text Processing Tools – Filters . . . . . . . . . . . . . . . . 16 1.3 Programming with Emacs . . . . . . . . . . . . . . . . . . . 20 1.3.1 Calling Emacs . . . . . . . . . . . . . . . . . . . . . . 21 1.3.2 Interacting with Emacs . . . . . . . . . . . . . . . . 22 1.3.3 Basic Editing . . . . . . . . . . . . . . . . . . . . . . 25 1.3.4 Cut and Paste . . . . . . . . . . . . . . . . . . . . . . 27 1.3.5 Windows . . . . . . . . . . . . . . . . . . . . . . . . 28 1.3.6 Files and Buﬀers . . . . . . . . . . . . . . . . . . . . 29 1.3.7 Modes . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1.3.8 Emacs Help . . . . . . . . . . . . . . . . . . . . . . . 32 1.3.9 Emacs Customization . . . . . . . . . . . . . . . . . 34 1.4 The C++ Programming Language . . . . . . . . . . . . . . 35 1.4.1 The Foundation . . . . . . . . . . . . . . . . . . . . 35 1.5 Gnuplot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 1.6 Shell Scripting . . . . . . . . . . . . . . . . . . . . . . . . . 52 2 Kinematics 63 2.1 Motion on the Plane . . . . . . . . . . . . . . . . . . . . . . 63 2.1.1 Plotting Data . . . . . . . . . . . . . . . . . . . . . . 72 2.1.2 More Examples . . . . . . . . . . . . . . . . . . . . . 75 2.2 Motion in Space . . . . . . . . . . . . . . . . . . . . . . . . 87 2.3 Trapped in a Box . . . . . . . . . . . . . . . . . . . . . . . . 96 iii iv CONTENTS 2.3.1 The One Dimensional Box . . . . . . . . . . . . . . 97 2.3.2 Errors . . . . . . . . . . . . . . . . . . . . . . . . . . 105 2.3.3 The Two Dimensional Box . . . . . . . . . . . . . . 109 2.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 3 Logistic Map 137 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 3.2 Fixed Points and 2n Cycles . . . . . . . . . . . . . . . . . . 139 3.3 Bifurcation Diagrams . . . . . . . . . . . . . . . . . . . . . 146 3.4 The Newton-Raphson Method . . . . . . . . . . . . . . . . 150 3.5 Calculation of the Bifurcation Points . . . . . . . . . . . . . 156 3.6 Liapunov Exponents . . . . . . . . . . . . . . . . . . . . . . 160 3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 4 Motion of a Particle 185 4.1 Numerical Integration of Newton’s Equations . . . . . . . . 185 4.2 Prelude: Euler Methods . . . . . . . . . . . . . . . . . . . . 186 4.3 Runge–Kutta Methods . . . . . . . . . . . . . . . . . . . . . 198 4.3.1 A Program for the 4th Order Runge–Kutta . . . . . 202 4.4 Comparison of the Methods . . . . . . . . . . . . . . . . . . 206 4.5 The Forced Damped Oscillator . . . . . . . . . . . . . . . . 209 4.6 The Forced Damped Pendulum . . . . . . . . . . . . . . . . 217 4.7 Appendix: On the Euler–Verlet Method . . . . . . . . . . . 223 4.8 Appendix: 2nd order Runge–Kutta Method . . . . . . . . 227 4.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 5 Planar Motion 235 5.1 Runge–Kutta for Planar Motion . . . . . . . . . . . . . . . 235 5.2 Projectile Motion . . . . . . . . . . . . . . . . . . . . . . . . 240 5.3 Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . 248 5.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 5.4.1 Rutherford Scattering . . . . . . . . . . . . . . . . . 255 5.4.2 More Scattering Potentials . . . . . . . . . . . . . . . 263 5.5 More Particles . . . . . . . . . . . . . . . . . . . . . . . . . . 265 5.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 6 Motion in Space 281 6.1 Adaptive Stepsize Control for RK Methods . . . . . . . . . 282 6.1.1 The rksuite Suite of RK Codes . . . . . . . . . . . . 282 6.1.2 Interfacing C++ Programs with Fortran . . . . . . . 286 CONTENTS v 6.1.3 The rksuite Driver . . . . . . . . . . . . . . . . . . 292 6.2 Motion of a Particle in an EM Field . . . . . . . . . . . . . 297 6.3 Relativistic Motion . . . . . . . . . . . . . . . . . . . . . . . 298 6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 7 Electrostatics 313 7.1 Electrostatic Field of Point Charges . . . . . . . . . . . . . . 313 7.2 The Program – Appetizer and ... Desert . . . . . . . . . . . 316 7.3 The Program – Main Dish . . . . . . . . . . . . . . . . . . . 325 7.4 The Program - Conclusion . . . . . . . . . . . . . . . . . . . 331 7.5 Electrostatic Field in the Vacuum . . . . . . . . . . . . . . . 336 7.6 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 7.7 Poisson Equation . . . . . . . . . . . . . . . . . . . . . . . . 344 7.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 8 Diﬀusion Equation 355 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 8.2 Heat Conduction in a Thin Rod . . . . . . . . . . . . . . . . 357 8.3 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . 358 8.4 The Program . . . . . . . . . . . . . . . . . . . . . . . . . . 360 8.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 8.6 Diﬀusion on the Circle . . . . . . . . . . . . . . . . . . . . . 365 8.7 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 8.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 9 The Anharmonic Oscillator 375 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 9.2 Calculation of the Eigenvalues of Hnm (λ) . . . . . . . . . . 377 9.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 9.4 The Double Well Potential . . . . . . . . . . . . . . . . . . . 392 9.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 10 Time Independent Schrödinger Equation 405 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 10.2 The Inﬁnite Potential Well . . . . . . . . . . . . . . . . . . . 408 10.3 Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . 420 10.4 Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . 430 10.5 The Anharmonic Oscillator - Again... . . . . . . . . . . . . 436 10.6 The Lennard–Jones Potential . . . . . . . . . . . . . . . . . 441 10.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 vi CONTENTS 11 The Random Walker 449 11.1 (Pseudo)Random Numbers . . . . . . . . . . . . . . . . . . 450 11.2 Using Pseudorandom Number Generators . . . . . . . . . . 461 11.3 The MIXMAX Random Number Generator . . . . . . . . . 466 11.4 Random Walks . . . . . . . . . . . . . . . . . . . . . . . . . 470 11.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 12 Monte Carlo Simulations 483 12.1 Statistical Physics . . . . . . . . . . . . . . . . . . . . . . . . 484 12.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 12.3 Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 12.4 Correlation Functions . . . . . . . . . . . . . . . . . . . . . 493 12.5 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 12.5.1 Simple Sampling . . . . . . . . . . . . . . . . . . . . 496 12.5.2 Importance Sampling . . . . . . . . . . . . . . . . . 497 12.6 Markov Processes . . . . . . . . . . . . . . . . . . . . . . . . 498 12.7 Detailed Balance Condition . . . . . . . . . . . . . . . . . . 499 12.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 13 Simulation of the d = 2 Ising Model 503 13.1 The Ising Model . . . . . . . . . . . . . . . . . . . . . . . . 503 13.2 Metropolis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 13.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 512 13.3.1 The Program . . . . . . . . . . . . . . . . . . . . . . 517 13.3.2 Towards a Convenient User Interface . . . . . . . . . 523 13.4 Thermalization . . . . . . . . . . . . . . . . . . . . . . . . . 534 13.5 Autocorrelations . . . . . . . . . . . . . . . . . . . . . . . . 537 13.6 Statistical Errors . . . . . . . . . . . . . . . . . . . . . . . . 543 13.6.1 Errors of Independent Measurements . . . . . . . . 544 13.6.2 Jackknife . . . . . . . . . . . . . . . . . . . . . . . . 547 13.6.3 Bootstrap . . . . . . . . . . . . . . . . . . . . . . . . 549 13.7 Appendix: Autocorrelation Function . . . . . . . . . . . . . 550 13.8 Appendix: Error Analysis . . . . . . . . . . . . . . . . . . . 557 13.8.1 The Jackknife Method . . . . . . . . . . . . . . . . . 557 13.8.2 The Bootstrap Method . . . . . . . . . . . . . . . . . 562 13.8.3 Comparing the Methods . . . . . . . . . . . . . . . . 565 13.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 14 Critical Exponents 577 14.1 Critical Slowing Down . . . . . . . . . . . . . . . . . . . . . 579 14.2 Wolﬀ Cluster Algorithm . . . . . . . . . . . . . . . . . . . . 580 CONTENTS vii 14.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 587 14.3.1 The Program . . . . . . . . . . . . . . . . . . . . . . 589 14.4 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . 594 14.5 Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 597 14.6 Autocorrelation Times . . . . . . . . . . . . . . . . . . . . . 605 14.7 Temperature Scaling . . . . . . . . . . . . . . . . . . . . . . 608 14.8 Finite Size Scaling . . . . . . . . . . . . . . . . . . . . . . . 614 14.9 Calculation of βc . . . . . . . . . . . . . . . . . . . . . . . . 616 14.10Studying Scaling with Collapse . . . . . . . . . . . . . . . . 621 14.11Binder Cumulant . . . . . . . . . . . . . . . . . . . . . . . . 631 14.12Appendix: Scaling . . . . . . . . . . . . . . . . . . . . . . . 634 14.12.1Binder Cumulant . . . . . . . . . . . . . . . . . . . . 634 14.12.2Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . 640 14.12.3Finite Size Scaling . . . . . . . . . . . . . . . . . . . 642 14.13Appendix: Critical Exponents . . . . . . . . . . . . . . . . . 645 14.13.1Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . 645 14.13.2Hyperscaling Relations . . . . . . . . . . . . . . . . . 646 14.14Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646 Bibliography 650 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 C++ code is listed below: i n t main ( ) { double x = 0 . 0 ; f o r ( i n t i =0;i <10; i++){ x += i ; } } 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 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 to the Second Edition This book has been out “in the wild” for more than two years. Since then, its pdf version has been downloaded 2-5000 times/month from the main server and has a few thousand hits from sites that oﬀer science e- books for free. I have also received positive feedback from students and colleagues from all over the world and that gave me the encouragement to devote some time to create a C++ version of the book. As far as scientiﬁc programming is concerned, the material has not changed apart from some typo and error corrections⁸. I have to make it clear that by using this book you will not learn much on the advanced features of C++. Scientiﬁc computing is usually simple at its core and, since it must be made eﬃcient and accurate, it needs to go down to the lowest levels of programming. This also partly the reason of why I chose to use Fortran for the core programming in the ﬁrst edition of the book: It is a language designed for numerical programming and high performance computing in mind. It is simple and a scientist or engineer can go directly into programming her code. C++ is not designed for scientiﬁc applications⁹ in mind and this reﬂects on some trivial omissions in its standard. Still, many scientiﬁc groups are now using C++ for programming and the C++ compilers have improved quite a lot. There is still an advantage in performance using a Fortran compiler on a supercomputer, but this is not going to last for much longer. Still, for a scientist, the programming language is a tool to solve her scientiﬁc problems. One should not bind herself to a speciﬁc language. The treasures of today are the garbage of tomorrow, and the time scale for this happening is small in today’s computing environments. What has really lasting value is the ability to solve problems using a computer and this is what needs to be emphasized. Consistent with this idea is that, in the course of reading this book, you will also learn how to make ⁸Check the errata section at the book’s homepage. ⁹Object oriented languages’ aim is to improve modularity, maintenability and ﬂexi- bility of programs. xi xii FOREWORD TO THE SECOND EDITION your C++ code interact with code written in Fortran, like in the case of the popular library Lapack. This will improve your “multilingual skills” and ﬂexibility with interacting with legacy code. The good news for us scientists is that numerical code usually needs simple data structures and programming is similar in any language. It was simple for me to “translate” my book from Fortran to C++. Un- fortunately I will not touch on all this great stuﬀ in true object oriented programming but you may be happy to know that you will most likely not need it¹⁰. So, I hope that you will enjoy using my book and I remind you that I love fan mail and I appreciate comments/corrections/suggestions sent to me. Now, if you want to learn about the structure and educational procedure in this book, read the foreword to the ﬁrst edition, otherwise skip to the real fun of solving scientiﬁc problems numerically. Athens, 2016. ¹⁰A lot of C++ code out there is realizing procedural and not true object oriented programming. Foreword to the First Edition 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- xiii xiv FOREWORD TO THE FIRST EDITION 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 it¹¹ 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 FSF¹². 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 philosophy¹³ ¹¹A Perl moto! ¹²Free Software Foundation, www.fsf.org. ¹³Java 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. xv 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 book¹⁴. 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 ¹⁴www.physics.ntua.gr/˜konstant/ComputationalPhysics/ xvi FOREWORD TO THE FIRST EDITION 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 οn 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. In this edition of the book we will be programming in C++. C++ is a language with very high level of abstraction designed for projects where modular programming and the use of complicated data structures is of very high priority. A large and complicated project should be divided into ¹www.stallman.org 1 2 CHAPTER 1. THE COMPUTER independent programming tasks (modules), where each task contains everything that it needs and does not interfere with the functionality of other modules. Although it has not been designed for high performance numerical applications, it is becoming more and more popular in the recent years. C++, as well as other languages like C, Java and Fortran, is a language that needs to be compiled by a compiler. Other languages, like python, 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 demand- ing 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. 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. 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 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 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. 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 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 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. 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 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 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. 1.1. THE OPERATING SYSTEM 5 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) 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 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 > 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. 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 . cpp rk3 . cpp plot−commands rk3_Bz . cpp rk3_g . cpp 1.1. THE OPERATING SYSTEM 7 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- 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 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. 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. 8 CHAPTER 1. THE COMPUTER (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 . cpp file2 . cpp > cp file1 . cpp file2 . cpp file3 . cpp Programs If the ﬁle file2.cpp does not exist, the ﬁrst command copies the contents of file1.cpp to a new ﬁle file2.cpp. If it already exists, it replaces its contents by the contents of the ﬁle file2.cpp. In order for the second command to be executed, Programs needs to be a directory. Then, the contents of the ﬁles file1.cpp, file2.cpp, file3.cpp 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: 1.1. THE OPERATING SYSTEM 9 > mv file1 . cpp file2 . cpp > mv file1 . cpp file2 . cpp file3 . cpp Programs The ﬁrst command renames the ﬁle file1.cpp to file2.cpp. The second one moves ﬁles file1.cpp, file2.cpp, file3.cpp 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 . cpp file2 . cpp file3 . cpp file4 . csh > rm file1 . cpp file2 . cpp file3 . cpp > ls file4 . csh the ﬁles file1.cpp, file2.cpp, file3.cpp 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 . cpp ’ ? y rm : remove regular file ‘ file2 . cpp ’ ? y rm : remove regular file ‘ file3 . cpp ’ ? y rm : remove regular file ‘ file4 . csh ’ ? n > 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 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... ¹³A small mistake, like rm -rf * and your data is ... history! 10 CHAPTER 1. THE COMPUTER . / dir1 / file2 . cpp . / dir1 / file1 . cpp . / dir1 / dir2 . / dir1 / dir2 / file3 . cpp 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 . cpp > cd . . ; rmdir dir2 > rm file1 . cpp file2 . cpp > cd . . ; rmdir dir1 Note that by using a semicolon, we can execute two or more commands on the same line. 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 . cpp 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 1.1. THE OPERATING SYSTEM 11 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 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 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 −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 . cpp test . dat The argument -l is the switch that results in a long listing of the ﬁles. The arguments test.cpp 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 * . cpp * . dat the shell will expand *.cpp and *.dat to a list of all ﬁles whose names end with .cpp or .dat. Therefore, if the current directory contains the ﬁles test.cpp, test1.cpp, myprog.cpp, test.dat, hello.dat, the ar- guments that will be passed on to the command ls are > l s −l myprog . cpp test1 . cpp test . cpp 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, 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 . cpp file2 . cpp file3 . cpp file4 . csh > l s > results 1.1. THE OPERATING SYSTEM 13 > ls file1 . cpp file2 . cpp file3 . cpp 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.cpp file2.cpp file3.cpp 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 . cpp file2 . cpp file3 . cpp file4 . csh file1 . cpp file2 . cpp file3 . cpp 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 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. ¹⁵This syntax is particular to the tcsh shell. For other shells (bash, sh, ...) read their documentation. 14 CHAPTER 1. THE COMPUTER 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 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 1.1. THE OPERATING SYSTEM 15 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 > 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 ¹⁶If you use the bash shell press [Tab] once or twice. 16 CHAPTER 1. THE COMPUTER 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 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 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 > 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 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 18 CHAPTER 1. THE COMPUTER > 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 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 1.2. TEXT PROCESSING TOOLS – FILTERS 19 > 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. 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 20 CHAPTER 1. THE COMPUTER > 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. 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 1.3. PROGRAMMING WITH EMACS 21 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 C++ ﬁle, compile it, debug it and run it, everything done with Emacs commands. 1.3.1 Calling Emacs In the command line type > emacs & 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. 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... ²¹Quite handy when we edit ﬁles in a remote computer. 22 CHAPTER 1. THE COMPUTER Figure 1.2: The Emacs window in a windows environment. The buttons of very basic functions found on its toolbar are shown and explained. 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 characters that follow are keys simultaneously pressed with the Alt key²². 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 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)). 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 24 CHAPTER 1. THE COMPUTER can be pretty handy when a command hangs or destroys our work and we need to interrupt it. 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. 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. 1.3. PROGRAMMING WITH EMACS 25 • 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 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 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.cpp, the automatically saved changes can be found in the ﬁle #file.cpp#. 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- 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. 26 CHAPTER 1. THE COMPUTER 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 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. ²⁵Press the Ctrl and spacebar keys simultanesouly. 1.3. PROGRAMMING WITH EMACS 27 • 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. • 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. 28 CHAPTER 1. THE COMPUTER • 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), 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 ²⁶If it is a two button mouse, try clicking the left and right buttons simultaneously. 1.3. PROGRAMMING WITH EMACS 29 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 58 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. • 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. 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 • 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. • 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.cpp will have as a backup the ﬁle test.cpp˜. 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 1.3. PROGRAMMING WITH EMACS 31 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 c++-mode: This mode is of special interest in this book since we will edit a lot of C++ code. We need it activated in buﬀers that contain a C++ program and its most useful characteristics are automatic code alignment by pressing the key TAB, the coloring of C++ statements, variables and other structural constructs (classes, if statements, for loops, variable declarations, comments 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 java-mode, perl-mode, awk-mode, python-mode, makefile-mode, octave-mode, gnuplot-mode and others. • latex-mode: For ﬁles containing LATEX text formatting commands. • 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 32 CHAPTER 1. THE COMPUTER 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 C++ pro- gram). 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. 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 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! 1.3. PROGRAMMING WITH EMACS 33 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 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. 34 CHAPTER 1. THE COMPUTER • 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 cm t o s e t c++−mode f o r t h e b u f f e r ( defun cm ( ) ( interactive ) ( c++−mode ) ) ; 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 C++ PROGRAMMING LANGUAGE 35 1.4 The C++ Programming Language In this section, we give a very basic introduction to the C++ programming 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 C++ 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: # i n c l u d e < iostream > using namespace std ; i n t main ( ) { / / This i s a comment . cout << ” H e l l o World ! \ n” ; } Commands, or statements, in C++ are strings of characters separated by blanks (“words”) and end with a semicolon (;). We can put more than one command on each line by separating them with a semicolon. Everything after two slashes (//) is a comment. Proliferation of com- ments is necessary for documenting our code. Good documentation of our code is an integral part of programming. If we plan to have our code read by others (or by us) at a later time, we have to make sure to explain in detail what each line is supposed to do. You and your col- laborators will save a lot of time in the process of debugging, improving and extending your code. The ﬁrst line of the code shown above is a preprocessor directive. These lines start with a # and are interpreted by a separate program, the pre- processor. The #include directive, inserts the contents of a ﬁle replacing the line where the directive is. This acts like an editor! Actually, the code that will be compiled is not the one shown above, but the result of adding the contents of a ﬁle whose name is iostream²⁹. iostream is an example ²⁹The path to the ﬁle is determined by the compiler. If you are curious to see the 36 CHAPTER 1. THE COMPUTER of a header ﬁle that has many deﬁnitions of functions and symbols used by the program. The particular header has the necessary deﬁnitions in order to perform standard input and standard output operations. The execution of a C++ program starts by calling a function whose name is main(). Therefore, the line int main(){ shows how to actually deﬁne a function in C++. Its name is the word before the parentheses () and the keyword int speciﬁes that the function returns a value of integer type³⁰. Within the parentheses placed after the name of the function, we put the arguments that we pass to the function. In our case the parentheses contain nothing, showing how to deﬁne a function without arguments. The curly brackets { ... } deﬁne the scope or the body of the function and contain the statements to be executed when the function is called. The line cout << ” H e l l o World ! \ n” ; is the only line that contains an executable statement that actually does something. Notice that it ends with a semicolon. This statement performs an output operation printing a string to the standard output. The sentence Hello World!\n is a constant string and contains a sequence of printable characters enclosed in double quotes. The last character \n is a newline character, that prints a new line to the stdout. cout identiﬁes the standard character output device, which gives ac- cess to the stdout. The characters << indicate that we write to cout the expression to the right. In order to make cout accessible to our program, we need both the inclusion of the header ﬁle iostream and the statement using namespace std³¹. Statements in C++ end with a semicolon. Splitting them over a num- ber of lines is only a matter of making the code legible. Therefore, the following parts of the code have equivalent eﬀect as the one written above: ﬁle, search for it with the command locate iostream. In order to see the result of adding the contents of the ﬁle (and, actually several other ﬁles added by preprocessor directives in iostream), call the preprocessor with the command cpp hello.cpp ³⁰The value returned by main() is useless within our program, but it is used by the operating system in order to inform us about the successful (or not) termination of the program. Of course, functions returning values is a very useful feature in general! ³¹Omitting using namespace std does not make cout inaccessible. One can use its “full name” std::cout instead. Remove the statement and try it. cout is part of the C++ Standard Library. All elements of the library belong to the std namespace and their names can be preﬁxed by std:: Using the using namespace std statement, the preﬁx may be omitted. 1.4. THE C++ PROGRAMMING LANGUAGE 37 i n t main ( ) { cout << ” H e l l o World ! \ n” ; } i n t main ( ) { cout <<” H e l l o World ! \ n” ; } Finally notice that, for C++, uppercase and lowercase characters are diﬀerent. Therefore main(), Main() and MAIN() are names of diﬀerent functions. 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 C++ compilers available, and you should learn which compilers are available for use in your computing environment. Typical names for C++ compilers are g++, c++, icc, .... 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 com- piler 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 g++, which can be installed on most popular operating systems³². The compilation command is: > g++ hello . cpp −o hello The extension .cpp to the ﬁle name hello.cpp is important and instructs the compiler that the ﬁle contains source code in C++. Use your editor and edit a ﬁle with the name hello.cpp with the program shown above before executing the above command. 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: ³²g++ is a front end to the GNU collection of compilers gcc. By installing gcc, you obtain a collection of compilers for several languages, like C, C++, Fortran, Java and others. See http://gcc.gnu.org/ 38 CHAPTER 1. THE COMPUTER > . / hello Hello world ! The ./ is not a special symbol for running programs. The dot is the current working directory and ./hello is the full path to the ﬁle hello. 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.cpp: # i n c l u d e < iostream > using namespace std ; i n t main ( ) { double PI = 3.1415926535897932; double R = 4 . 0 ; cout << ” P e r i m e t e r= ” << 2 . 0 * PI * R << ”\n” ; cout << ” Area= ” << PI * R * R << ”\n” ; } The ﬁrst two statements in main() declare the values of the variables PI and R. These variables are of type double, which are ﬂoating point numbers³³. The following two commands have two eﬀects: Computing the length 2πR and the area πR2 of the circle and printing the results. The ex- pressions 2.0*PI*R and PI*R*R are evaluated before being printed to the stdout. 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 int type and by using them the wrong way we may obtain surprising results³⁴. We compile and run the program with the commands: > g++ area_01 . cpp −o area > . / area Perimeter= 25.1327 Area= 50.2655 ³³Don’t confuse double variables with the real numbers. double variables take values that are ﬁnite approximations of real numbers and take values that are a subset of the rational numbers. This approximation becomes better by increasing the amount of memory allocated to store them. In most computing environments, doubles are allocated 8 bytes of memory, in which case they approximate real numbers with, more or less, 17 signiﬁcant digits. ³⁴Try adding the command cout << 2/4 << 2.0/4.0; and check the results. 1.4. THE C++ PROGRAMMING LANGUAGE 39 Now we will 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 double type. The code can be found in the ﬁle area_02.cpp: # i n c l u d e < iostream > using namespace std ; i n t main ( ) { double PI = 3.1415926535897932; double R [ 1 0 ] ; double area , perimeter ; int i; R [0] = 2.18; f o r ( i =1; i <10; i++){ R [ i ] = R [ i−1] + 1 . 0 ; } f o r ( i =0;i <10; i++){ perimeter = 2 . 0 * PI * R [ i ] ; area = PI * R [ i ] * R [ i ] ; cout << ( i +1) << ” ) R= ” << R [ i ] << ” p e r i m e t e r= ” << perimeter << ’\n ’ ; cout << ( i +1) << ” ) R= ” << R [ i ] << ” a r e a = ” << area << ’\n ’ ; } } The declaration double R[10] deﬁnes an array of length 10. This way, the elements of the array are referred to by an index that takes values from 0 to 9. For example, R[0] is the ﬁrst, R[3] is the fourth and R[9] is the last element of the array. Between the lines f o r ( i =1; i <10; i++){ ... } we can write commands that are repeatedly executed while the int vari- able i takes values from 1 to 9 with increasing step equal to 1. The way it works is the following: In the round brackets after the keyword for, there exist three statements separated by semicolons. The ﬁrst, i=1, is the statement executed once before the loop starts. The second, i<10, 40 CHAPTER 1. THE COMPUTER is a statement that is evaluated each time before the loop repeats itself. If it is true, then the statements in the loop are executed. If it is false, the control of the program is transferred after the end of the loop. The last statement, i++, is evaluated each time after the last statement in the loop has been executed. The operator ++ is the increment operator, and its eﬀect is equivalent to the statement: i = i + 1; The value of i is increased by one. The command: R [ i ] = R [ i−1] + 1 . 0 ; deﬁnes the i-th radius from the value R[i-1]. For the loop to work correctly, we must deﬁne the initial value of R[0], otherwise³⁵ R[0]=0.0. The second loop uses the deﬁned R-values in order to do the computation and print 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). We will also see how to write the results directly to a ﬁle instead of the standard output (stdout). The program can be found in the ﬁle area_03.cpp: # i n c l u d e < iostream > # i n c l u d e < fstream > using namespace std ; i n t main ( ) { const int N = 10; c o n s t double PI = 3.1415926535897932; double R [ N ] ; double area , perimeter ; int i; f o r ( i =0;i<N ; i++){ cout << ” Enter r a d i u s o f c i r c l e : ” ; cin >> R [ i ] ; cout << ” i = ” << ( i +1) << ” R( i )= ” << R [ i ] << ’\n ’ ; } ofstream myfile ( ”AREA.DAT” ) ; ³⁵Arrays in C++ are zero-initialized. 1.4. THE C++ PROGRAMMING LANGUAGE 41 f o r ( i =0;i<N ; i++){ perimeter = 2 . 0 * PI * R [ i ] ; area = PI * R [ i ] * R [ i ] ; myfile << ( i +1) << ” ) R= ” << R[i] << ” p e r i m e t e r= ” << perimeter << ’\n ’ ; myfile << ( i +1) << ” ) R= ” << R[i] << ” a r e a = ” << area << ’\n ’ ; } myfile . close ( ) ; } In the above program, the size of the array R is deﬁned by a const int. A const declares a variable to be a parameter whose value does not change during the execution of the program and, if it is of int type, it can be used to declare the size of an array. The array elements R[i] are read using the command: cin >> R [ i ] ; cin is the standard input stream, the same way that cout is the standard output stream³⁶. We read input using the >> operator, which indicates that input is written to the variable on the right. In order to interact with ordinary ﬁles, we need to include the header # i n c l u d e < fstream > In this header, the C++ class ofstream is deﬁned and it can be used in order to write to ﬁles (output stream). An object in this class, like myfile, is deﬁned (“instantiated”) by the statement: ofstream myfile ( ”AREA.DAT” ) ; This object’s constructor is called by placing the parentheses ("AREA.DAT"), and then the output stream myfile is directed to the ﬁle AREA.DAT. Then we can write output to the ﬁle the same way we have already done with cout: myfile << ( i +1) << ” ) R= ” << R [ i ] << ” p e r i m e t e r= ” << perimeter << ’\n ’ ; ³⁶And cerr is the standard error stream. 42 CHAPTER 1. THE COMPUTER When we are done writing to the ﬁle, we can close the stream with the statement: myfile . close ( ) ; Reading from ﬁles is done in a similar way by using the class ifstream instead of ofstream. The next step will be to learn how to deﬁne and use functions. The program below shows how to deﬁne a function 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.cpp: # i n c l u d e < iostream > # i n c l u d e < fstream > using namespace std ; c o n s t double PI = 3.1415926535897932; void area_of_circle ( c o n s t double& R , double& L , double& A ) ; i n t main ( ) { const int N = 10; double R [ N ] ; double area , perimeter ; int i; f o r ( i =0;i<N ; i++){ cout << ” Enter r a d i u s o f c i r c l e : ” ; cin >> R [ i ] ; cout << ” i = ” << ( i +1) << ” R( i )= ” << R [ i ] << ’\n ’ ; } ofstream myfile ( ”AREA.DAT” ) ; f o r ( i =0;i<N ; i++){ area_of_circle ( R [ i ] , perimeter , area ) ; myfile << ( i +1) << ” ) R= ” << R [ i ] << ” p e r i m e t e r= ” << perimeter << ’\n ’ ; myfile << ( i +1) << ” ) R= ” << R [ i ] << ” a r e a = ” << area << ’\n ’ ; } myfile . close ( ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void area_of_circle ( c o n s t double& R , double& L , double& A ) { 1.4. THE C++ PROGRAMMING LANGUAGE 43 L = 2 . 0 * PI * R ; A = PI * R * R ; } The calculation of the length and the area of the circle is performed by the function area_of_circle ( R [ i ] , perimeter , area ) ; Calling a function, transfers the control of the program to the statements within the body of the function. The above function has arguments (R[i], perimeter, area). The argument R[i] is intended to be only an input variable whose value is not going to change during the calculation. The arguments perimeter and area are intended for output. Upon return of the function to the main program, they store the result of the compu- tation. The user of a function 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 function. In order to use a function, we need to declare it the same way we do with variables or, as we say, to provide its prototype. The prototype of a function can be declared without providing the function’s deﬁnition. We may provide just enough details that determine the types of its arguments and the value returned. In our program this is done on the line: void area_of_circle ( c o n s t double& R , double& L , double& A ) ; This is the same syntax used later in the deﬁnition of the function, but replacing the body of the function with a semicolon. The argument list does not need to include the argument names, only their types. We could have also used the following line in order to declare the function’s prototype: void area_of_circle ( c o n s t double& , double& , double& ) ; We could also have used diﬀerent names for the arguments, if we wished so. Including the names is a matter of style that improves legibility of the code. The argument R is intended to be left unchanged during the function execution. This is why we used the keyword const in its declaration. The arguments L and R, however, will return a diﬀerent value to the calling program. This is why const is not used for them. The actual program executed by the function is between the lines: 44 CHAPTER 1. THE COMPUTER void area_of_circle ( c o n s t double& R , double& L , double& A ) { L = 2 . 0 * PI * R ; A = PI * R * R ; } The type of the value returned by a function is declared by the keyword before its name. In our case, this is void which declares that the function does not return a value. The arguments (R,L,A) must be declared in the function and need not have the same names as the ones that we use when we call it. All arguments are declared to be of type double. The character & indicates that they are passed to the function by reference. This makes possible to change their values from within the function. If & is omitted, then the arguments will be passed by value and a statement like L = 2.0*PI*R will not change the value of the variable passed by the calling program. This happens because, in this case, only the value of the variable L of the calling program is copied to a local variable which is used only within the function. This is important to understand and you are encouraged to run the program with and without the & and check the diﬀerence in the computed results. The names of variables in a function are only valid within the scope of the function, i.e. between the curly brackets that contain the body of the function. Therefore the variable const int N is valid only within the scope of main(). You may use any name you like, even if it is already used outside the scope of the function. The names of arguments need not be the same as the ones used in the calling program. Only their types have to match. Variables in the global scope are accessible by all functions in the same ﬁle³⁷. An example of such a variable is PI, which is accessible by main(), as well as by area_of_circle(). We summarize all of the above in a program trionymo.cpp, 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 u s e r , 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 , e x i t // ³⁷If the code is spread over multiple ﬁles, then all ﬁles must use the keyword external in order to make the variable accessible to the functions that they contain. In one of the ﬁles, the variable must be deﬁned without the word external. More on that later... 1.4. THE C++ PROGRAMMING LANGUAGE 45 / / 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 / / ========================================================= # i n c l u d e < iostream > # include <cstdlib > # i n c l u d e <cmath> using namespace std ; double Discriminant ( double a , double b , double c ) ; void roots ( double a , double b , double c , double& x1 , double& x2 ) ; i n t main ( ) { double a , b , c , D ; double x1 , x2 ; cout << ” Enter a , b , c : ” ; cin >> a >> b >> c ; cout << a << ” ” << b << ” ” << c << ” ” << ’\n ’ ; / / 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 == 0.0 ) { cerr << ” Trionymo : a=0\n” ; exit ( 1 ) ; } / / Compute t h e d i s c r i m i n a n t D = Discriminant ( a , b , c ) ; cout << ” D i s c r i m i n a n t : D= ” << D << ’\n ’ ; / / 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 ) if (D > 0.0) { roots ( a , b , c , x1 , x2 ) ; cout << ” Roots : x1= ” << x1 << ” x2= ”<< x2 << ’\n ’ ; } e l s e i f ( D == 0 . 0 ) { roots ( a , b , c , x1 , x2 ) ; cout << ” Double Root : x1= ” << x1 << ’\n ’ ; } else { cout << ”No r e a l r o o t s \n” ; exit ( 1 ) ; } } / / ========================================================= / / 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 46 CHAPTER 1. THE COMPUTER / / 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 r e t u r n e d using / / the return statement / / ========================================================= double Discriminant ( double a , double b , double c ) { r e t u r n b * b − 4.0 * a * c ; } / / ========================================================= / / The f u n c t i o n t h a t computes t h e r o o t s . / / a , b , c a r e passed by v a l u e : Their v a l u e s cannot change // within the function / / x1 , x2 a r e passed by r e f e r e n c e : Their v a l u e s DO change // within the function / / ========================================================= void roots ( double a , double b , double c , double& x1 , double& x2 ) { double D ; D = Discriminant ( a , b , c ) ; i f ( D >= 0 . 0 ) { D = sqrt ( D ) ; } else { cerr << ” r o o t s : Sorry , cannot compute r o o t s , D<0=” << D << ’\n ’ ; } x1 = (−b + D ) / ( 2 . 0 * a ) ; x2 = (−b − D ) / ( 2 . 0 * a ) ; } 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 Discriminant(a,b,c). The type of the value returned must be declared at the function’s prototype double Discriminant ( double a , double b , double c ) ; and at the function’s deﬁnition double Discriminant ( double a , double b , double c ) { r e t u r n b * b − 4.0 * a * c ; } The value returned to the calling program is the value of the expression given as an argument to the return statement. return has also the eﬀect of transferring the control of the program back to the calling statement. 1.5. GNUPLOT 47 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 [16] is an excellent place to look for many of gnuplot’s secrets³⁸. 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 / .... 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 function³⁹. The command ³⁸A 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 plots. ³⁹You can change the symbol of the independent variable. For example, the command set dummy t sets the independent variable to be t. 48 CHAPTER 1. THE COMPUTER 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 unchanged⁴⁰. 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 ⁴⁰By 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 49 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 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 50 CHAPTER 1. THE COMPUTER 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 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 qt 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 qt 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 1.5. GNUPLOT 51 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 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 ) 52 CHAPTER 1. THE COMPUTER 1.6 Shell Scripting A typical GNU/Linux environment oﬀers very powerful tools for compli- cated system administration tasks. They are much more simple to use than to incorporate them into your program. This way, the programmer can concentrate on 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++ area_01 . cpp −o area . / area g++ area_02 . cpp −o area . / area g++ area_03 . cpp −o area . / area g++ area_04 . cpp −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/tcsh⁴¹. 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 C++ programs found in the ﬁles that we created in section 1.4 with g++, 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 ⁴¹Use #!/bin/bash if you prefer the bash shell. 1.6. SHELL SCRIPTING 53 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 program⁴². 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 g++ area_04 . cpp −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 ⁴²Their 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. 54 CHAPTER 1. THE COMPUTER 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 . cpp area_02 . cpp area_03 . cpp area_04 . cpp ) 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 ” g++ $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 1.6. SHELL SCRIPTING 55 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.cpp, area_02.cpp, area_03.cpp, area_04.cpp. These values can be used by the commands in the loop. Therefore, the command g++ $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 script. It only serves as a demonstration of the shell scripting capabilities. For more information, the reader is referred to the bibliography [18, 19, 20,21,22]. Read carefully the commands, as well as the comments which follow the # mark. Then, write the commands to a ﬁle script04.csh⁴³, 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 $us er working on t h e computer $HOST” #HOST i s p r e d e f i n e d ⁴³You will ﬁnd it also in the accompanying software 56 CHAPTER 1. THE COMPUTER 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 } ’ ‘ 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 ” 1.6. SHELL SCRIPTING 57 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 # 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 g ive 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 & 58 CHAPTER 1. THE COMPUTER 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. 1.6. SHELL SCRIPTING 59 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. 60 CHAPTER 1. THE COMPUTER Table 1.2: 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 61 Table 1.2: 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... 62 CHAPTER 1. THE COMPUTER Table 1.2: 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 63 64 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 65 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) 66 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.cpp. 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 2 R ⃗ and ⃗a anti-parallel, ⃗a ⊥ ⃗v ). ⃗ (R The data structure is quite simple. The constant angular velocity ω is stored in the double variable omega. The center of the circle (x0 , y0 ), the radius R of the circle and the angle θ are stored in the double 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 double variables t0, tf, dt. The current position (x(t), y(t)) is calculated and stored in the double variables x, y and the velocity (vx (t), vy (t)) in the double variables vx, vy. The declarations of the variables are put in the beginning of the program: double x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt ; double theta , omega ; 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. The variables are read from the stdin by reading from the stream cin and 2.1. MOTION ON THE PLANE 67 the values entered by the user are printed to the stdout using the stream cout¹: cout << ” # Enter omega : \ n” ; cin >> omega ; getline ( cin , buf ) ; cout << ” # 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: \ n” ; cin >> x0 >> y0 >> R ; getline ( cin , buf ) ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout <<” # omega= ” << omega << endl ; cout <<” # x0= ” << x0 << ” y0= ” << y0 << ” R= ” << R << endl ; cout <<” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; There are a couple of things to explain. Notice that after reading each variable from the standard input stream cin, we call the function getline. By calling getline(cin,buf), a whole line is read from the input stream cin into the string buf². Then the statement cin >> x0 >> y0 >> R ; getline ( cin , buf ) ; has the eﬀect of reading three doubles from the stdin and put the rest of the line in the string buf. Since we never use buf, this is a mechanism to discard the rest of the line of input. The reason for doing so will become clear later. Objects of type string in C++ store character sequences. In order to use them you have to include the header # include <string > and, e.g., declare them like string buf , buf1 , buf2 ; Then you can store data in the obvious way, like buf="Hello World!", manipulate string data using operators like buf=buf1 (assign buf1 to ¹This 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. ²In fact it is possible to call getline(cin,buf,char) and read a line until the char- acter char is encountered. 68 CHAPTER 2. KINEMATICS buf), buf=buf1+buf2 (concatenate buf1 and buf2 and store the result in buf), buf1==buf2 (compare strings) etc. Finally, endl is used to end all the cout statements. This has the eﬀect of adding a newline to the output stream and ﬂush the output³. Next, the program initializes the state of the computation. This in- cludes checking the validity of the parameters entered by the user, so that the computation will be possible. For example, the program com- putes 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 exit statement⁴ will stop the program execution and print an informative message to the standard error stream cerr⁵. The program opens the ﬁle Circle.dat for writing the calculated values of the position and the velocity of the particle. i f (R <=0.0) { cerr <<” I l l e g a l v a l u e o f R \n” ; exit ( 1 ) ; } i f ( omega <=0.0) { cerr <<” I l l e g a l v a l u e o f omega\n” ; exit ( 1 ) ; } cout << ” # T= ” << 2 . 0 * PI / omega << endl ; ofstream myfile ( ” C i r c l e . dat ” ) ; myfile . precision ( 1 7 ) ; The line myfile.precision(17) sets the precision of the ﬂoating point numbers (like double) printed to myfile to 17 signiﬁcant digits accuracy. The default is 6 which is a pity, because doubles have up to 17 signiﬁcant digits accuracy. If R ≤ 0 or ω ≤ 0 the corresponding exit statements are executed which end the program execution. The optional error messages are in- cluded after the stop statements which are printed to the stderr. The value of the period T = 2π/ω is also calculated and printed for reference. The main calculation is performed within the loop ³When buﬀered output is used, it is not written out immediately but stored in a temporary memory location (a buﬀer). When the buﬀer ﬁlls, it is automatically ﬂushed to the output stream. If we want to force ﬂushing before the buﬀer is full, then we have to ﬂush the buﬀer. There are several methods to ﬂush an output stream os (like os.flush()). ⁴The exit(1) statement returns 1 as exit code for the program. This is the int that main() returns. exit(0) is conventionally used for a normal exit and a non zero value is used when an error occurs. In order to use exit() you must include the header cstdlib. ⁵Note that there are more assumptions that need to be checked by the program. We leave this as an exercise for the reader. 2.1. MOTION ON THE PLANE 69 t = t0 ; while ( t <= tf ) { ......... t = t + dt ; } The ﬁrst statement sets the initial value of the time. The statements between within the scope of the while(condition) 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 statements put in place of the dots ......... calculate the position and the velocity and print them to the ﬁle Circle.dat: # i n c l u d e <cmath> ................ theta = omega * ( t−t0 ) ; x = x0+R * cos ( theta ) ; y = y0+R * sin ( theta ) ; vx = −omega * R * sin ( theta ) ; vy = omega * R * cos ( theta ) ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; Notice the use of the functions sin and cos that calculate the sine and cosine of an angle expressed in radians. The header cmath is necessary to be included. The program is stored in the ﬁle Circle.cpp and can be found in the accompanied software. The extension .cpp is used to inform the compiler that the ﬁle contains source code written in the C++ language. Compilation and running can be done using the commands: > g++ Circle . cpp −o cl > . / cl The switch -o cl forces the compiler g++ to write the binary commands executed by the program to the ﬁle⁶ 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: ⁶If omitted, the executable ﬁle has the default name a.out. 70 CHAPTER 2. KINEMATICS > g++ Circle . cpp −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= 0.01 # T= 6.28319 The lines shown above that start with a # character are printed by the program and the 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 used 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 parameters 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 that we want to pass to the program with each call to cout. The rest of the line consists of comments that explain to the user what each number is there for. We want to discard these characters during input and this is the reason for using getline to com- plete reading the rest of the line. 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 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: 2.1. MOTION ON THE PLANE 71 > 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= 0.01 # T= 6.28319 We list the full program in Circle.cpp below: / / ============================================================ / / F i l e C i r c l e . cpp / / 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 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt ; double theta , omega ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter omega : \ n” ; cin >> omega ; getline ( cin , buf ) ; cout << ” # 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: \ n” ; cin >> x0 >> y0 >> R ; getline ( cin , buf ) ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout <<” # omega= ” << omega << endl ; cout <<” # x0= ” << x0 << ” y0= ” << y0 << ” R= ” << R << endl ; cout <<” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i f (R <=0.0) { cerr <<” I l l e g a l v a l u e o f R \n” ; exit ( 1 ) ; } 72 CHAPTER 2. KINEMATICS i f ( omega <=0.0) { cerr <<” I l l e g a l v a l u e o f omega\n” ; exit ( 1 ) ; } cout << ” # T= ” << 2 . 0 * PI / omega << endl ; ofstream myfile ( ” C i r c l e . dat ” ) ; / / S e t p r e c i s i o n f o r numeric output t o m y f i l e t o 17 d i g i t s myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = t0 ; while ( t <= tf ) { theta = omega * ( t−t0 ) ; x = x0+R * cos ( theta ) ; y = y0+R * sin ( theta ) ; vx = −omega * R * sin ( theta ) ; vy = omega * R * cos ( theta ) ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; t = t + dt ; } } / / main ( ) 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 ) ” 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 )). The function atan2 is available in gnuplot⁷ as well as in C++. 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 . ⁷The command help functions will show you all the available functions in gnuplot. 2.1. MOTION ON THE PLANE 73 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. ‘ 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 49, 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 text⁸. 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 variable⁹ 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 ⁸This can be done on the gnuplot command line as well. ⁹Use the command show variables in order to see the current/default values of gnuplot variables. 74 CHAPTER 2. KINEMATICS 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 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.cpp described in the text. 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 have the data ﬁle Circle.dat. We are not going to explain how it works¹⁰ 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 ﬁle¹¹, 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 ¹⁰You are most welcome to study the commands in the script and guess how it works of course! ¹¹It can be any ﬁle that has (t, x, y) in the 1st, 2nd and 3rd columns respectively. 2.1. MOTION ON THE PLANE 75 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” 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.cpp. 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) 76 CHAPTER 2. KINEMATICS 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 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.cpp, is: / / ============================================================ / / F i l e SimplePendulum . cpp / / 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 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 # d e f i n e g 9.81 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s double l , x , y , vx , vy , t , t0 , tf , dt ; double theta , theta0 , dtheta_dt , omega ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter l : \ n” ; cin >> l ; getline ( cin , buf ) ; cout << ” # Enter t h e t a 0 : \ n” ; cin >> theta0 ; getline ( cin , buf ) ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout <<” # l = ” << l << ” t h e t a 0= ” << theta0 << endl ; 2.1. MOTION ON THE PLANE 77 cout <<” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize omega = sqrt ( g / l ) ; cout << ” # omega= ” << omega << ” T= ” << 2 . 0 * PI / omega << endl ; ofstream myfile ( ” SimplePendulum . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = t0 ; while ( t <= tf ) { theta = theta0 * cos ( omega * ( t−t0 ) ) ; dtheta_dt = −omega * theta0 * sin ( omega * ( t−t0 ) ) ; x = l * sin ( theta ) ; y = −l * cos ( theta ) ; vx = l * dtheta_dt * cos ( theta ) ; vy = l * dtheta_dt * sin ( theta ) ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << ” ” << theta << dtheta_dt << endl ; t = t + dt ; } } / / main ( ) 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). A simple session for the study of the above problem is shown below¹²: > g++ SimplePendulum . cpp −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= 0.314 # t 0= 0 t f = 20 dt= 0.01 ¹²Notice 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. 78 CHAPTER 2. KINEMATICS # omega= 3.13209 T= 2.00607 > 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 ” theta ( t ) ” gnuplot > r e p l o t ” SimplePendulum . dat ” u 1:7 w l t ” theta ’( 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 surface¹³ when we consider the eﬀect of air resistance to be negligible. Then, the equations describing the trajectory of the particle 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. ¹³I.e. ⃗g = const. and the Coriolis force can be ignored. 2.1. MOTION ON THE PLANE 79 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.cpp in the accompanied software: / / ============================================================ / / F i l e P r o j e c t i l e . cpp / / Shooting a p r o g 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 ) . / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 # d e f i n e g 9.81 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double x0 , y0 , R , x , y , vx , vy , t , tf , dt ; double theta , v0x , v0y , v0 ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter v0 , t h e t a ( i n d e g r e e s ) : \ n” ; cin >> v0 >> theta ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout <<” # v0= ” << v0 << ” t h e t a = ”<< theta << ” o ( d e g r e e s ) ” << endl ; cout <<” # t 0= ” << 0.0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i f ( v0 <= 0 . 0 ) { cerr <<” I l l e g a l v a l u e o f v0 <= 0\n” ; exit ( 1 ) ; } i f ( theta<= 0 . 0 ) { cerr <<” I l l e g a l v a l u e o f t h e t a <= 0\n” ; exit ( 1 ) ; } i f ( theta >=90.0) { cerr <<” I l l e g a l v a l u e o f t h e t a >=90\n” ; exit ( 1 ) ; } 80 CHAPTER 2. KINEMATICS 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 * cos ( theta ) ; v0y = v0 * sin ( theta ) ; cout << ” # v0x= ” << v0x << ” v0y= ” << v0y << endl ; ofstream myfile ( ” P r o j e c t i l e . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = 0.0; while ( t <= tf ) { x = v0x * t ; y = v0y * t − 0 . 5 * g * t * t ; vx = v0x ; vy = v0y − g*t ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; t = t + dt ; } } / / main ( ) A typical session for the study of this problem is shown below: > g++ Projectile . cpp −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 t h e t a = 45o ( d e g r e e s ) # t 0= 0 t f = 1.4416 dt= 0.001 # v0x= 7 . 0 7 1 0 7 v0y= 7 . 0 7 1 0 7 > 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 ) ” gnuplot > p l o t ” 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 > p l o t ” 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 2.1. MOTION ON THE PLANE 81 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 are¹⁴ vx (t) = v0x e−kt ( g ) −kt g vy (t) = v0y + e − k k v0x ( ) x(t) = 1 − e−kt 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. The full program can be found in the ﬁle ProjectileAirResistance.cpp 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 . cpp / / Shooting a p r o g 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 k , ( v0 , t h e t a ) . / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > ¹⁴The proof of equations (2.10) is left as an exercise for the reader. 82 CHAPTER 2. KINEMATICS # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 # d e f i n e g 9.81 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s double x0 , y0 , R , x , y , vx , vy , t , tf , dt , k ; double theta , v0x , v0y , v0 ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter k , v0 , t h e t a ( i n d e g r e e s ) : \ n” ; cin >> k >> v0 >> theta ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout <<” # k = ” << k << endl ; cout <<” # v0= ” << v0 << ” t h e t a = ”<< theta << ” o ( d e g r e e s ) ” << endl ; cout <<” # t 0= ” << 0.0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i f ( v0 <= 0 . 0 ) { cerr <<” I l l e g a l v a l u e o f v0 <= 0\n” ; exit ( 1 ) ; } i f (k <= 0 . 0 ) { cerr <<” I l l e g a l v a l u e o f k <= 0\n” ; exit ( 1 ) ; } i f ( theta<= 0 . 0 ) { cerr <<” I l l e g a l v a l u e o f t h e t a <= 0\n” ; exit ( 1 ) ; } i f ( theta >=90.0) { cerr <<” I l l e g a l v a l u e o f t h e t a >=90\n” ; exit ( 1 ) ; } 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 * cos ( theta ) ; v0y = v0 * sin ( theta ) ; cout << ” # v0x= ” << v0x << ” v0y= ” << v0y << endl ; ofstream myfile ( ” P r o j e c t i l e A i r R e s i s t a n c e . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = 0.0; while ( t <= 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 ) ; 2.1. MOTION ON THE PLANE 83 vy = ( v0y +( g / k ) ) * exp(−k * t ) −(g / k ) ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; t = t + dt ; } } / / main ( ) 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.cpp 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++ ProjectileAirResistance . cpp −o 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 t h e t a = 45o ( d e g r e e s ) # t 0= 0 t f = 0.91 dt= 0.001 # v0x= 7 . 0 7 1 0 7 v0y= 7 . 0 7 1 0 7 > 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 84 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. 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 ” \ 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 2.1. MOTION ON THE PLANE 85 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 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.cpp: / / ============================================================ / / F i l e L i s s a j o u s . cpp / / Lissajous curves ( s p e c i a l case ) / / x ( t )= c o s ( o1 t ) , y ( t )= s i n ( o2 t ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double x0 , y0 , R , x , y , vx , vy , t , t0 , tf , dt ; double o1 , o2 , T1 , T2 ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter omega1 and omega2 : \ n” ; cin >> o1 >> o2 ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout <<” # o1= ” << o1 << ” o2= ” << o2 << endl ; cout <<” # t 0= ” << 0.0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i f ( o1 <=0.0) { cerr <<” I l l e g a l v a l u e o f o1\n” ; exit ( 1 ) ; } i f ( o2 <=0.0) { cerr <<” I l l e g a l v a l u e o f o2\n” ; exit ( 1 ) ; } T1 = 2 . 0 * PI / o1 ; T2 = 2 . 0 * PI / o2 ; cout << ” # T1= ” << T1 << ” T2= ” << T2 << endl ; 86 CHAPTER 2. KINEMATICS ofstream myfile ( ” L i s s a j o u s . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = t0 ; while ( t <= tf ) { x = cos ( o1 * t ) ; y = sin ( o2 * t ) ; vx = −o1 * sin ( o1 * t ) ; vy = o2 * cos ( o2 * t ) ; myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; t = t + dt ; } } / / main ( ) 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 session for the study of the problem is shown below: > g++ Lissajous . cpp −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 = 10 dt= 0.01 # T1= 2.0944 T2= 1.25664 >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. 2.2. MOTION IN SPACE 87 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 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. 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 solution¹⁵ is x(t) = r cos ωt y(t) = r sin ωt z(t) = −l cos θ , (2.14) ¹⁵One has to choose appropriate initial conditions. Exercise: ﬁnd them! 88 CHAPTER 2. KINEMATICS 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.cpp. 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 . The full program is listed below: 2.2. MOTION IN SPACE 89 / / ============================================================ / / F i l e ConicalPendulum . cpp / / 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 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 # d e f i n e g 9.81 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double l , r , x , y , z , vx , vy , vz , t , tf , dt ; double theta , cos_theta , sin_theta , omega ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter l , omega : \ n” ; cin >> l >> omega ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout << ” # l = ” << l << ” omega= ” << omega << endl ; cout << ” # T= ” << 2 . 0 * PI / omega << ” omega_min= ” << sqrt ( g / l ) << endl ; cout <<” # t 0= ” << 0.0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize cos_theta = g / ( omega * omega * l ) ; i f ( cos_theta >= 1 . 0 ) { cerr << ” c o s ( t h e t a )>= 1\n” ; exit ( 1 ) ; } sin_theta = sqrt (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 ; ofstream myfile ( ” ConicalPendulum . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = 0.0; while ( t <= tf ) { 90 CHAPTER 2. KINEMATICS x = r * cos ( omega * t ) ; y = r * sin ( omega * t ) ; vx = −r * sin ( omega * t ) * omega ; vy = r * cos ( omega * t ) * omega ; myfile << t << ” ” << x << ” ” << y << ” ” << z << ” ” << vx << ” ” << vy << ” ” << vz << ” ” << endl ; t = t + dt ; } } / / main ( ) In order to compile and run the program we can use the commands shown below: > g++ ConicalPendulum . cpp −o cpd > . / cpd # Enter l , omega : 1 . 0 6.28 # Enter t f , dt : 10.0 0.01 # l = 1 omega= 6.28 # T= 1.00051 omega_min= 3.13209 # t 0= 0 t f = 10 dt= 0.01 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 ) ” gnuplot > r e p l o t ” ConicalPendulum . dat ” u 1:3 w l t ”y ( t ) ” gnuplot > r e p l o t ” ConicalPendulum . dat ” u 1:4 w l t ”z( t )” gnuplot > p l o t ” ConicalPendulum . dat ” u 1:5 w l t ” v_x ( t ) ” gnuplot > r e p l o t ” ConicalPendulum . dat ” u 1:6 w l t ” v_y ( t ) ” gnuplot > r e p l o t ” 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- 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: 2.2. MOTION IN SPACE 91 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.cpp for ω = 6.28, l = 1.0. 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” 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 92 CHAPTER 2. KINEMATICS Figure 2.14: The plot of the particle trajectory ⃗r(t) of the program ConicalPendulum.cpp 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). 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) 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) 2.2. MOTION IN SPACE 93 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.cpp 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. 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.cpp: / / ============================================================ / / F i l e ChargeInB . cpp / / 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 94 CHAPTER 2. KINEMATICS 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 / / t 0=0 t o t f a t s t e p dt / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s double x , y , z , vx , vy , vz , t , tf , dt ; double x0 , y0 , z0 , v0x , v0y , v0z , v0 ; double theta , omega ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter omega : \ n” ; cin >> omega ; getline ( cin , buf ) ; cout << ” # Enter v0 , t h e t a ( d e g r e e s ) : \ n” ; cin >> v0 >> theta ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout << ” # omega= ” << omega << ” T= ” << 2 . 0 * PI / omega << endl ; cout << ” # v0= ” << v0 << ” t h e t a = ” << theta 2.2. MOTION IN SPACE 95 << ” o ( d e g r e e s ) ”<< endl ; cout <<” # t 0= ” << 0.0 << ” t f = ” << tf << ” dt= ” << dt << endl ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i f ( theta <0.0 | | theta >=90.0) exit ( 1 ) ; 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 * cos ( theta ) ; v0z = v0 * sin ( theta ) ; cout << ” # v0x= ” << 0.0 << ” v0y= ” << v0y << ” v0z= ” << v0z << endl ; x0 = − v0y / omega ; cout << ” # x0= ” << x0 << ” y0= ” << y0 << ” z0= ” << z0 << endl ; cout << ” # xy plane : C i r c l e with c e n t e r ( 0 , 0 ) and R= ” << abs ( x0 ) << endl ; cout << ” # s t e p o f h e l i x : s=v0z *T= ” << v0z * 2 . 0 * PI / omega << endl ; ofstream myfile ( ” ChargeInB . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : t = 0.0; vz = v0z ; while ( t <= tf ) { x = x0 * cos ( omega * t ) ; y = −x0 * sin ( omega * t ) ; z = v0z * t ; vx = v0y * sin ( omega * t ) ; vy = v0y * cos ( omega * t ) ; myfile << t << ” ” << x << ” ” << y << ” ” << z << ” ” << vx << ” ” << vy << ” ” << vz << ” ” << endl ; t = t + dt ; } } / / main ( ) A typical session in which we calculate the trajectories shown in ﬁgures 2.17 and 2.18 is shown below: > g++ ChargeInB . cpp −o chg > . / chg # Enter omega : 6.28 # Enter v0 , t h e t a ( d e g r e e s ) : 96 CHAPTER 2. KINEMATICS 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.cpp for ω = 6.28, x0 = 1.0, θ = 20 degrees. 1 . 0 20 # Enter t f , dt : 10 0.01 # omega= 6.28 T= 1.00051 # v0= 1 t h e t a = 20o ( d e g r e e s ) # t 0= 0 t f = 10 dt= 0.01 # v0x= 0 v0y= 0.939693 v0z= 0.34202 # x0= −0.149633 y0= 0 z0= 3.11248 e −317 # xy plane : C i r c l e with c e n t e r ( 0 , 0 ) and R= 0.149633 # s t e p o f h e l i x : s=v0z *T= 0.342194 > gnuplot gnuplot > p l o t ” ChargeInB . dat ” u 1 : 2 w l title ”x ( t ) ” gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 3 w l title ”y ( t ) ” gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 4 w l title ”z( t )” gnuplot > p l o t ” ChargeInB . dat ” u 1 : 5 w l title ” v_x ( t ) ” gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 6 w l title ” v_y ( t ) ” gnuplot > r e p l o t ” ChargeInB . dat ” u 1 : 7 w l title ” v_z ( t ) ” gnuplot > s p l o t ” ChargeInB . dat ” u 2 : 3 : 4 w l title ”r( t )” gnuplot > file = ” ChargeInB . dat ” gnuplot > s e t xrange [ − 0 . 6 5 : 0 . 6 5 ] ; s e t yrange [ −0.65:0.65] 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. 2.3. TRAPPED IN A BOX 97 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.cpp 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. These types of errors¹⁶ 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. 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 ¹⁶In 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. 98 CHAPTER 2. KINEMATICS 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. 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. 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) 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 while ( t <= tf ) { x += v * dt ; t += dt ; 2.3. TRAPPED IN A BOX 99 i f ( x < 0.0 | | x > L ) v = −v ; } 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.cpp. 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: / / ============================================================ / / F i l e box1D_1 . cpp / / 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 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e <iomanip > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s f l o a t L , x0 , v0 , t0 , tf , dt , t , x , v ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter L: \ n” ; cin >> L ; getline ( cin , buf ) ; cout << ” # L = ” << L << endl ; cout << ” # Enter x0 , v0 : \ n” ; cin >> x0 >> v0 ; getline ( cin , buf ) ; cout << ” # x0= ” << x0 << ” v0= ” << v0 << endl ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout << ” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; i f ( L <= 0.0 f ) { cerr << ”L <=0\n” ; exit ( 1 ) ; } i f ( x0< 0.0 f ) { cerr << ”x0<=0\n” ; exit ( 1 ) ; } i f ( x0> L ) { cerr << ”x0> L\n” ; exit ( 1 ) ; } i f ( v0== 0.0 f ) { cerr << ” v0 =0\n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize t = t0 ; 100 CHAPTER 2. KINEMATICS x = x0 ; v = v0 ; ofstream myfile ( ” box1D_1 . dat ” ) ; myfile . precision ( 9 ) ; / / f l o a t p r e c i s i o n ( and a b i t more . . . ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t <= tf ) { myfile << setw ( 1 7 ) << t << ” ” / / s e t width o f f i e l d << setw ( 1 7 ) << x << ” ” / / t o 17 c h a r a c t e r s << setw ( 1 7 ) << v << ’\n ’ ; / / using setw ( 1 7 ) x += v * dt ; t += dt ; i f ( x < 0.0 f | | x > L ) v = −v ; } myfile . close ( ) ; } / / main ( ) In this section we will study the eﬀects of roundoﬀ errors in numeri- cal computations. Computers store numbers in memory, which is ﬁnite. Therefore, real numbers are represented in some approximation that de- pends on the amount of memory that is used for their storage. This approximation corresponds to what is termed as ﬂoating point numbers. C++ is supposed to provide at least three basic types of ﬂoating point numbers, float, double and long double. In most implementations¹⁷, float uses 4 bytes of memory and double 8. In this case, float has an accuracy to, approximately, 7 signiﬁcant digits and double 17. See Chapter 1 of [8] and [14] for details. Moreover, float represent num- bers with magnitude in the, approximate, range (10−38 , 1038 ) while double in (10−308 , 10308 ). Note that variables of the integer type (int, long, ...) are exact representations of integers, whereas ﬂoating point numbers are approximations to reals. In the program shown above, we used numbers of the float type instead of double in order to exaggerate roundoﬀ errors. This way we can study the dependence of this type of errors on the accuracy of the ﬂoating point numbers used in a program¹⁸. In order to do that, we declared the ﬂoating point variables as float: ¹⁷Notice that the C++ standard states that the value representation of ﬂoating-point types is implementation-deﬁned. The C standard requires that the type double provides at least as much precision as ﬂoat, and the type long double provides at least as much precision as double. The gcc 5.4 version that we are using in this book represents float using 4 bytes and double with 8, but you should check whether this is true with the compiler that you are using. ¹⁸The use of float can be the preferred choice of a programmer for some applications. First, in order to save memory, because float occupies half the memory of a double. Second, it is not always true that increasing the accuracy of ﬂoating point numbers 2.3. TRAPPED IN A BOX 101 f l o a t L , x0 , v0 , t0 , tf , dt , t , x , v ; We also used numerical constants of type float. This is indicated by the letter f at the end of their names: 2.0 is a constant of type double (the C++ default), whereas 2.0f is a constant of type float. Determining the accuracy of ﬂoating point constants is a thorny issue that can be the cause on introducing subtle bugs in a program and the programmer should be very careful about doing it carefully. Finally we changed the form of the output. Since a float represents a real number with at most 7 signiﬁcant digits, there is no point of printing more. That is why we used the statements myfile . precision ( 9 ) ; myfile . setw ( 1 7 ) ; For purposes of studying the numerical accuracy of our results, we used 9 digits of output, which is, of course, slightly redundant. setw(17) prints the numbers of the next output of the stream myfile using at least 17 character spaces. This improves the legibility of the results when inspecting the output ﬁles. The use of setw requires the header iomanip. 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++ box1D_1 . cpp −o box1 > . / box1 # Enter L : 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= 0.01 > 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 ) ” will increase the accuracy of a computation, although in most of the cases it will. The wisdom of the ﬁeld is to always to use as much accuracy as you need and no more! 102 CHAPTER 2. KINEMATICS 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 time by t+dt. Floating point numbers of the float 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 2.3. TRAPPED IN A BOX 103 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 float type. For example double numbers have approximately 17 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: while ( t <= tf ) { x = x0 + v0 * ( t−t0 ) ; i f ( x < 0.0 f | | x > L ) { x0= x ; t0= t ; v0= −v0 ; } t += dt ; } In the above algorithm, the error in the time of the collision is not vanishing but we don’t have the “instability” problem of the dt→ 0 limit¹⁹. 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.cpp: / / ============================================================ / / F i l e box1D_2 . cpp / / 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 ) / / Rev erse v e l o c i t y and r e d e f i n e x0 , t 0 on boundaries / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > ¹⁹We still have this problem in the t=t+dt operation. See discussion in the next section. 104 CHAPTER 2. KINEMATICS # i n c l u d e <iomanip > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s f l o a t L , x0 , v0 , t0 , tf , dt , t , x , v ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter L: \ n” ; cin >> L ; getline ( cin , buf ) ; cout << ” # L = ” << L << endl ; cout << ” # Enter x0 , v0 : \ n” ; cin >> x0 >> v0 ; getline ( cin , buf ) ; cout << ” # x0= ” << x0 << ” v0= ” << v0 << endl ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout << ” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; i f ( L <= 0.0 f ) { cerr << ”L <=0\n” ; exit ( 1 ) ; } i f ( x0< 0.0 f ) { cerr << ”x0<=0\n” ; exit ( 1 ) ; } i f ( x0> L ) { cerr << ”x0> L\n” ; exit ( 1 ) ; } i f ( v0== 0.0 f ) { cerr << ” v0 =0\n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize t = t0 ; ofstream myfile ( ” box1D_2 . dat ” ) ; myfile . precision ( 9 ) ; / / f l o a t p r e c i s i o n ( and a b i t more . . . ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t <= tf ) { x = x0 + v0 * ( t−t0 ) ; myfile << setw ( 1 7 ) << t << ” ” << setw ( 1 7 ) << x << ” ” << setw ( 1 7 ) << v0 << ’\n ’ ; i f ( x < 0.0 f | | x > L ) { x0= x ; t0= t ; v0= −v0 ; } t += dt ; } myfile . close ( ) ; } / / main ( ) 2.3. TRAPPED IN A BOX 105 Compiling and running the above program is done as before and the results are stored in the ﬁle box1D_2.dat. 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.cpp and box1D_2.cpp. 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)²⁰ as a function of δt and study its convergence to a limit²¹ 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 ≈ 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 ²⁰Note the ≈! ²¹Of course we know the answer: x(95) = 5. 106 CHAPTER 2. KINEMATICS > g++ box1D_1 . cpp −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 value²² 0.000012. We do the same²³ 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 (9 5 ) 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 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% ²²Try the command sed 's/0.05/0.025/' box1D_anal.in | ./box by changing 0.025 with the desired value of δt. ²³See the shell script box1D_anal.csh as a suggestion on how to automate this boring process. 2.3. TRAPPED IN A BOX 107 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 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 108 CHAPTER 2. KINEMATICS 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.cpp becomes: t = t0 ; x = x0 ; v = v0 ; i = 0; while ( t <= ( tf +1.0 e−5f ) ) { i += 1; x += v * dt ; t = t0 + i * dt ; i f ( x < 0.0 f | | x > L ) v = −v ; } The full program can be found in the ﬁle box1D_4.cpp of the accom- panying software. We call this “method 3”. We perform the same change in the ﬁle box1D_2.cpp, which we store in the ﬁle box1D_5.cpp. 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. 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. 2.3. TRAPPED IN A BOX 109 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 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 ++; t = t0 + i * dt ; x += vx * dt ; y += vy * dt ; i f ( x < 0.0 | | x > Lx ) { vx = −vx ; nx ++; } i f ( y < 0.0 | | y > Ly ) { vy = −vy ; ny ++; } The full program can be found in the ﬁle box2D_1.cpp. Notice that we introduced two counters nx and ny of the particle’s collisions with the walls: / / ============================================================ / / F i l e box2D_1 . cpp / / 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 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 110 CHAPTER 2. KINEMATICS # i n c l u d e < iostream > # i n c l u d e <iomanip > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s double Lx , Ly , x0 , y0 , v0x , v0y , t0 , tf , dt , t , x , y , vx , vy ; int i , nx , ny ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter Lx , Ly : \ n” ; cin >> Lx >> Ly ; getline ( cin , buf ) ; cout << ” # Lx = ”<< Lx << ” Ly= ” << Ly << endl ; cout << ” # Enter x0 , y0 , v0x , v0y : \ n” ; cin >> x0 >> y0 >> v0x >> v0y ; getline ( cin , buf ) ; cout << ” # x0= ” << x0 << ” y0= ” << y0 << ” v0x= ” << v0x << ” v0y= ” << v0y << endl ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout << ” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; i f ( Lx<= 0 . 0 ) { cerr << ”Lx<=0 \n” ; exit ( 1 ) ; } i f ( Ly<= 0 . 0 ) { cerr << ”Ly<=0 \n” ; exit ( 1 ) ; } i f ( x0< 0 . 0 ) { cerr << ”x0<=0 \n” ; exit ( 1 ) ; } i f ( x0> Lx ) { cerr << ”x0> Lx\n” ; exit ( 1 ) ; } i f ( y0< 0 . 0 ) { cerr << ”x0<=0 \n” ; exit ( 1 ) ; } i f ( y0> Ly ) { cerr << ”y0> Ly\n” ; exit ( 1 ) ; } i f ( v0x * v0x+v0y * v0y == 0.0 ) { cerr << ” v0 =0\n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize i = 0 ; nx = 0 ; ny = 0 ; t = t0 ; x = x0 ; y = y0 ; vx = v0x ; vy = v0y ; ofstream myfile ( ” box2D_1 . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t <= tf ) { myfile << setw (2 8) << t << ” ” << setw (2 8) << x << ” ” << setw (2 8) << y << ” ” 2.3. TRAPPED IN A BOX 111 << setw (28 ) << vx << ” ” << setw (28 ) << vy << ’\n ’ ; i += 1; t = t0 + i * dt ; x += vx * dt ; y += vy * dt ; i f ( x < 0.0 | | x > Lx ) { vx = −vx ; nx ++; } i f ( y < 0.0 | | y > Ly ) { vy = −vy ; ny ++; } } myfile . close ( ) ; cout << ” # Number o f c o l l i s i o n s : \ n” ; cout << ” # nx= ” << nx << ” ny= ” << ny << endl ; } / / main ( ) A typical session for the study of a particle’s trajectory could be: > g++ box2D_1 . cpp −o box > . / box # Enter Lx , Ly : 10.0 5.0 # Lx = 10 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 112 CHAPTER 2. KINEMATICS 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.cpp. The parameters are Lx = 10, Ly = 5, x0 = 5, y0 = 0, v0x = 1.27, v0y = 1.33, t0 = 0, tf = 50, δt = 0.01. 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 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 it 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 2.4. APPLICATIONS 113 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. 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.cpp and is listed below: / / ============================================================ / / F i l e MiniGolf . cpp / / 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 ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e <iomanip > # i n c l u d e < fstream > 114 CHAPTER 2. KINEMATICS 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.cpp 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. # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.14159265358979324 i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of va ri a bl e s double Lx , Ly , x0 , y0 , v0x , v0y , t0 , tf , dt , t , x , y , vx , vy ; double v0 , theta , xc , yc , R , R2 ; int i , nx , ny ; string result ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter Lx , Ly : \ n” ; cin >> Lx >> Ly ; getline ( cin , buf ) ; cout << ” # Lx = ” << Lx << ” Ly= ” << Ly << endl ; cout << ” # Enter h o l e p o s i t i o n and r a d i u s : ( xc , yc ) , R: \ n” ; 2.4. APPLICATIONS 115 cin >> xc >> yc >> R ; getline ( cin , buf ) ; cout << ” # ( xc , yc )= ( ” << xc << ” , ” << yc << ” ) ” << ” R= ” << R << endl ; cout << ” # Enter v0 , t h e t a ( d e g r e e s ) : \ n” ; cin >> v0 >> theta ; getline ( cin , buf ) ; cout << ” # v0= ” << v0 << ” t h e t a = ” << theta << ” d e g r e e s ” << endl ; cout << ” # Enter dt : \ n” ; cin >> dt ; getline ( cin , buf ) ; i f ( Lx<= 0 . 0 ) { cerr << ”Lx<=0 \n” ; exit ( 1 ) ; } i f ( Ly<= 0 . 0 ) { cerr << ”Ly<=0 \n” ; exit ( 1 ) ; } i f ( v0<= 0 . 0 ) { cerr << ”v0<=0 \n” ; exit ( 1 ) ; } i f ( abs ( theta ) > 9 0 . 0 ) { cerr << ” t h e t a > 90\n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize t0 = 0.0; x0 = 0.00001; / / s m a l l but non−z e r o y0 = Ly / 2 . 0 ; R2 = R*R ; theta = ( PI / 1 8 0 . 0 ) * theta ; v0x = v0 * cos ( theta ) ; v0y = v0 * sin ( theta ) ; cout << ” # x0= ” << x0 << ” y0= ” << y0 << ” v0x= ” << v0x << ” v0y= ” << v0y << endl ; i = 0 ; nx = 0 ; ny = 0 ; t = t0 ; x = x0 ; y = y0 ; vx = v0x ; vy = v0y ; ofstream myfile ( ” MiniGolf . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t r u e ) { myfile << setw (28 ) << t << ” ” << setw (28 ) << x << ” ” << setw (28 ) << y << ” ” << setw (28 ) << vx << ” ” << setw (28 ) << vy << ’\n ’ ; i ++; t = t0 + i * dt ; x += vx * dt ; y += vy * dt ; i f ( x > Lx ) { vx = −vx ; nx ++;} i f ( y < 0 . 0 ) {vy = −vy ; ny ++;} i f ( y > Ly ) { vy = −vy ; ny ++;} i f ( x <=0.0) { result=” F a i l u r e ” ; break ; } / / e x i t loop i f ( ( ( x−xc ) * ( x−xc ) +(y−yc ) * ( y−yc ) ) <= R2 ) 116 CHAPTER 2. KINEMATICS { result=” S u c c e s s ” ; break ; } / / e x i t loop } myfile . close ( ) ; cout << ” # Number o f c o l l i s i o n s : \ n” ; cout << ” # R e s u l t= ” << result << ” nx= ” << nx << ” ny= ” << ny << endl ; } / / main ( ) In order to run it, we can use the commands: > g++ MiniGolf . cpp −o mg > . / mg # Enter Lx , Ly : 10 5 # Lx = 10 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 # Enter v0 , t h e t a ( d e g r e e s ) : 1 80 # v0= 1 t h e t a = 80 d e g r e e s # Enter dt : 0.01 # x0= 1 e−05 y0= 2.5 v0x= 0.173648 v0y= 0.984808 # 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 2.4. APPLICATIONS 117 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) 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 function reflectVonCircle(vx,vy,x,y,xc,yc,R). Upon entry to the function, we provide the initial velocity (vx,vy), the collision point (x,y), the center of the circle (xc,yc) and the radius of the circle²⁴ R. Upon exit from the function, (vx,vy) have been replaced with the new values²⁵ (vx′ , vy′ ). ²⁴Of course one expects R2 = (x − xc )2 + (y − yc )2 , but because of systematic errors, we require R to be given. ²⁵Notice that upon exit, the particle is also placed exactly on the circle. 118 CHAPTER 2. KINEMATICS 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. The program can be found in the ﬁle Cylinder3D.cpp and is listed below: / / ============================================================ / / F i l e Cylinder3D . cpp / / 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 f u n c t i o n r e f l e c t V o n C i r c l e f o r c o l i s i o n s a t r=R / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e <iomanip > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; void reflectVonCircle ( double& vx , double& vy , double& x , double& y , 2.4. APPLICATIONS 119 const double& xc , const double& yc , const double& R ) ; i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double x0 , y0 , z0 , v0x , v0y , v0z , t0 , tf , dt , t , x , y , z , vx , vy , vz ; double L , R , R2 , vxy , rxy , r2xy , xc , yc ; int i , nr , nz ; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter R , L: \ n” ; cin >> R >> L ; getline ( cin , buf ) ; cout << ” # R= ” << R << ” L= ” << L << endl ; cout << ” # Enter x0 , y0 , z0 , v0x , v0y , v0z : \ n” ; cin >> x0>>y0>>z0>>v0x>>v0y>>v0z ; getline ( cin , buf ) ; rxy = sqrt ( x0 * x0+y0 * y0 ) ; cout << ” # x0 = ” << x0 << ” y0 = ” << y0 << ” z0 = ” << z0 << ” rxy= ” << rxy << endl ; cout << ” # v0x= ” << v0x << ” v0y= ” << v0y << ” v0z= ” << v0z << endl ; cout << ” # Enter t0 , t f , dt : \ n” ; cin >> t0 >> tf >> dt ; getline ( cin , buf ) ; cout << ” # t 0= ” << t0 << ” t f = ” << tf << ” dt= ” << dt << endl ; i f (R <= 0 . 0 ) { cerr << ”R<=0 \n” ; exit ( 1 ) ; } i f (L <= 0 . 0 ) { cerr << ”L<=0 \n” ; exit ( 1 ) ; } i f ( z0 < 0 . 0 ) { cerr << ” z0<0 \n” ; exit ( 1 ) ; } i f ( z0 > L ) { cerr << ” z0>L \n” ; exit ( 1 ) ; } i f ( rxy > R ) { cerr << ” rxy >R \n” ; exit ( 1 ) ; } i f ( v0x * v0x+v0y * v0y+v0z * v0z == 0 . 0 ) { cerr << ” v0=0 \n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // 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 ; / / 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 yc = 0 . 0 ; / / o f t h e c y l i n d e r on t h e xy plane ofstream myfile ( ” Cylinder3D . dat ” ) ; myfile . precision ( 1 7 ) ; 120 CHAPTER 2. KINEMATICS / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t <= tf ) { myfile << setw (2 8) << t << ” ” << setw (2 8) << x << ” ” << setw (2 8) << y << ” ” << setw (2 8) << z << ” ” << setw (2 8) << vx << ” ” << setw (2 8) << vy << ” ” << setw (2 8) << vz << ’\n ’ ; i ++; t = t0 + i * dt ; x += vx * dt ; y += vy * dt ; z += vz * dt ; i f ( z <= 0.0 | | z > L ) { vz = −vz ; nz ++;} r2xy = x * x+y * y ; i f ( r2xy > R2 ) { reflectVonCircle ( vx , vy , x , y , xc , yc , R ) ; nr ++; } } myfile . close ( ) ; cout << ” # Number o f c o l l i s i o n s : \ n” ; cout << ” # nr= ” << nr << ” nz= ” << nz << endl ; } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / ============================================================ / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void reflectVonCircle ( double& vx , double& vy , double& x , double& y , const double& xc , const double& yc , const double& R ) { double theta , cth , sth , vr , vth ; theta = atan2 ( y−yc , x−xc ) ; cth = cos ( theta ) ; sth = sin ( theta ) ; 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 ; } / / reflectVonCircle () 2.4. APPLICATIONS 121 Note 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.cpp. 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: > g++ Cylinder3D . cpp −o cl > . / cl # Enter R , L : 10.0 10.0 # R= 10 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= 2.41661 # v0x= 0.93 v0y= −0.89 v0z= 0 . 7 4 # Enter t0 , t f , dt : 0.0 500.0 0.01 122 CHAPTER 2. KINEMATICS # 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 > file=” Cylinder3D . dat ” 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]. 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 > file=” Cylinder3D . dat ” gnuplot > R =10; L =10; t0 =0; tf =500; dt=10 gnuplot > load ” Cylinder3D_animate . gnu” The result is shown in ﬁgure 2.26. 2.4. APPLICATIONS 123 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. 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 geometry²⁶ a “spacetime foam”. We will study a very simple model of the above geometry on the plane ²⁶See K.S. Thorne “Black Holes and Time Wraps: Einstein’s Outrageous Legacy”, W.W. Norton, New York for a popular review of these concepts. 124 CHAPTER 2. KINEMATICS 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 ′ . with a particle moving freely in it²⁷. 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. Then we will do the following: ²⁷This 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. 2.4. APPLICATIONS 125 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 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) 2 126 CHAPTER 2. KINEMATICS 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 . 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 127 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 The angle θ is given by ( ) −1 yi θ = π − tan 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) 128 CHAPTER 2. KINEMATICS 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: / / ============================================================ / / F i l e Wormhole . cpp / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; # d e f i n e PI 3.1415926535897932 void crossC1 ( double& x, double& y , double& vx , double& vy , c o n s t double& dt , c o n s t double& R , c o n s t double& d) ; void crossC2 ( double& x, double& y , double& vx , double& vy , 2.4. APPLICATIONS 129 c o n s t double& dt , c o n s t double& R, c o n s t double& d ) ; i n t main ( ) { / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Declaration of v ar i ab l e s double Lx , Ly , L , R , d ; double x0 , y0 , v0 , theta ; double t0 , tf , dt ; double t , x , y , vx , vy ; double xc1 , yc1 , xc2 , yc2 , r1 , r2 ; int i; string buf ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Ask u s e r f o r i n p u t : cout << ” # Enter L , d , R: \ n” ; cin >> L >> d >> R ; getline ( cin , buf ) ; cout << ” # Enter ( x0 , y0 ) , v0 , t h e t a ( d e g r e e s ) : \ n” ; cin >> x0 >> y0 >> v0 >> theta ; getline ( cin , buf ) ; cout << ” # Enter t f , dt : \ n” ; cin >> tf >> dt ; getline ( cin , buf ) ; cout << ” # L= ” << L << ” d= ” << d << ” R= ” << R << endl ; cout << ” # x0= ” << x0 << ” y0= ” << y0 << endl ; cout << ” # v0= ” << v0 << ” t h e t a = ” << theta << ” d e g r e e s ” << endl ; cout << ” # t f = ” << tf << ” dt= ” << dt << endl ; i f ( L <= d +2.0* R ) { cerr <<”L <= d+2*R \n” ; exit ( 1 ) ; } i f ( d <= 2 . 0 * R ) { cerr <<”d <= 2*R \n” ; exit ( 1 ) ; } i f ( v0<= 0.0 ) { cerr <<”v0<= 0 \n” ; exit ( 1 ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− // Initialize theta = ( PI / 1 8 0 . 0 ) * theta ; i = 0; t = 0.0; x = x0 ; y = y0 ; vx = v0 * cos ( theta ) ; vy = v0 * sin ( theta ) ; cout << ” # x0= ” << x0 << ” y0= ” << y0 << ” v0x= ” << vx << ” v0y= ” << vy << endl ; / / Wormhole ’ s c e n t e r s : xc1 = 0 . 5 * d ; yc1 = 0.0; xc2 = −0.5*d ; yc2 = 0.0; / / Box l i m i t s c o o r d i n a t e s : Lx = 0 . 5 * L ; Ly = 0.5* L ; / / Test i f already inside cut region : r1 = sqrt ( ( x−xc1 ) * ( x−xc1 ) +(y−yc1 ) * ( y−yc1 ) ) ; r2 = sqrt ( ( x−xc2 ) * ( x−xc2 ) +(y−yc2 ) * ( y−yc2 ) ) ; i f ( r1<= R ) { cerr <<” r 1 <= R \n” ; exit ( 1 ) ; } i f ( r1<= R ) { cerr <<” r2 <= R \n” ; exit ( 1 ) ; } 130 CHAPTER 2. KINEMATICS / / T e s t i f o u t s i d e box l i m i t s : i f ( abs ( x ) >= Lx ) { cerr <<” | x | >= Lx \n” ; exit ( 1 ) ; } i f ( abs ( y ) >= Ly ) { cerr <<” | y | >= Ly \n” ; exit ( 1 ) ; } ofstream myfile ( ”Wormhole . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Compute : while ( t < tf ) { myfile << t << ” ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; i++; t = i * dt ; x += vx * dt ; 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 > Lx ) x = x − L ; i f ( x < −Lx ) x = x + L ; i f ( y > Ly ) y = y − L ; i f ( y < −Ly ) y = y + L ; r1 = sqrt ( ( x−xc1 ) * ( x−xc1 ) +(y−yc1 ) * ( y−yc1 ) ) ; r2 = sqrt ( ( x−xc2 ) * ( x−xc2 ) +(y−yc2 ) * ( y−yc2 ) ) ; / / 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 if ( r1 < R ) crossC1 ( x , y , vx , vy , dt , r1 , d ) ; e l s e i f ( r2 < R ) crossC2 ( x , y , vx , vy , dt , r2 , d ) ; / / 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 f o r −loop } / / while ( t <= t f ) } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void crossC1 ( double& x , double& y , double& vx , double& vy , c o n s t double& dt , c o n s t double& R , c o n s t double& d ) { double vr , v0 , theta , xc , yc ; cout << ” # I n s i d e C1 : ( x , y , vx , vy , R)= ” << x << ” ” << y << ” ” << vx << ” ” << vy << ” ” <<R << endl ; xc = 0.5* d ; / / c e n t e r o f C1 yc = 0.0; theta = atan2 ( y−yc , x−xc ) ; x = −xc − R * cos ( theta ) ; / / new x−value , y i n v a r i a n t / / Velocity transformation : vr = vx * cos ( theta )+vy * sin ( theta ) ; v0 = −vx * sin ( theta )+vy * cos ( theta ) ; vx = vr * cos ( theta )+v0 * sin ( theta ) ; vy = −vr * sin ( theta )+v0 * cos ( theta ) ; 2.4. APPLICATIONS 131 / / 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 ; cout << ” # E x i t C2 : ( x , y , vx , vy )= ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; } / / void c r o s s C 1 ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void crossC2 ( double& x , double& y , double& vx , double& vy , c o n s t double& dt , c o n s t double& R , c o n s t double& d ) { double vr , v0 , theta , xc , yc ; cout << ” # I n s i d e C2 : ( x , y , vx , vy , R)= ” << x << ” ” << y << ” ” << vx << ” ” << vy << ” ” <<R << endl ; xc = −0.5*d ; / / c e n t e r o f C2 yc = 0.0; theta = PI−atan2 ( y−yc , x−xc ) ; x = −xc + R * cos ( theta ) ; / / new x−value , y i n v a r i a n t / / Velocity transformation : vr = −vx * cos ( theta )+vy * sin ( theta ) ; v0 = vx * sin ( theta )+vy * cos ( theta ) ; vx = −vr * cos ( theta )−v0 * sin ( theta ) ; vy = −vr * sin ( theta )+v0 * cos ( 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 ; cout << ” # E x i t C1 : ( x , y , vx , vy )= ” << x << ” ” << y << ” ” << vx << ” ” << vy << endl ; } / / void c r o s s C 2 ( ) 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. 132 CHAPTER 2. KINEMATICS 2.5 Problems 2.1 Change the program Circle.cpp 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.cpp, 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.cpp 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.cpp does. 2.4 Change the program SimplePendulum.cpp 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.cpp. Consider the initial condi- tions ⃗v0 = x̂ + ŷ and calculate the range of the trajectory numerically by using the two programs ProjectileSmallAirResistance.cpp, ProjectileAirResistance.cpp. 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.cpp for the parameters shown in ﬁgure 2.10. 2.7 Change the program Lissajous.cpp 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 133 2.8 Change the program ProjectileAirResistance.cpp, so that it can calculate also the k = 0 case. 2.9 Change the program ProjectileAirResistance.cpp 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.cpp 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.cpp 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.cpp. Fill the table on page 106 the number of calculated collisions and comment on the results. 2.13 Run the program box1D_1.cpp 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 float declarations to double in the program box1D_1.cpp. Make sure that all the constants that you use become double pre- cision (e.g. 1.0f changes to 1.0). Compare your results to those obtained in section 2.3.2. Repeat problem 2.13. What do you ob- serve? 2.15 Change the program box1D_1.cpp so that you can study non elastic collisions v ′ = −ev, 0 < e ≤ 1 with the walls. 2.16 Change the program box2D_1.cpp 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.cpp and box1D_5.cpp 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 134 CHAPTER 2. KINEMATICS 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. 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.cpp). 2.5. PROBLEMS 135 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. 2.23 In the box of the previous problem, put four disks on which the particle bounces of elastically like in ﬁgure 2.33. 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 111 000 111 000 111 000 111 111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 111 000 111 000 000 111 000 111 111 000 111 000 111 000 000 111 000 111 111 000 111 000 000 111 000 111 000 Lx Figure 2.34: Problem 2.24. 136 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” [23] 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 [23, 24, 25, 26, 27, 28, 29, 40]. 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 137 138 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 139 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) 1 140 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 141 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 [24] 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. 142 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.cpp 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 143 # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { int NSTEPS , i ; double r , x0 , x1 ; string buf ; / / −−−−− Input : cout << ” # Enter NSTEPS , r , x0 : \ n” ; cin >> NSTEPS >> r >> x0 ; getline ( cin , buf ) ; cout << ” # NSTEPS = ” << NSTEPS << endl ; cout << ” # r = ” << r << endl ; cout << ” # x0 = ” << x0 << endl ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” l o g . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : myfile << 0 << x0 ; f o r ( i =1; i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; myfile << i << ” ” << x1 << ”\n” ; x0 = x1 ; } myfile . close ( ) ; } / / main ( ) The program is compiled and run using the commands: > g++ logistic . cpp −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: 144 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 ). 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) 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.cpp: / / =========================================================== / / D i s c r e t e L o g i s t i c Map : Cobweb diagram / / Map t h e t r a j e c t o r y i n 2d s pace ( plane ) / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { 3.2. FIXED POINTS AND 2N CYCLES 145 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. int NSTEPS , i ; double r , x0 , x1 ; string buf ; / / −−−−− Input : cout << ” # Enter NSTEPS , r , x0 : \ n” ; cin >> NSTEPS >> r >> x0 ; getline ( cin , buf ) ; cout << ” # NSTEPS = ” << NSTEPS << endl ; cout << ” # r = ” << r << endl ; cout << ” # x0 = ” << x0 << endl ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” t r j . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : myfile << 0 << ” ” << x0 << ” ” << 0 << ’\n ’ ; f o r ( i =1; i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; myfile << 2* i−3 << ” ” << x0 << ” ” << x1 << ’\n ’ ; myfile << 2* i−2 << ” ” << x1 << ” ” << x1 << ’\n ’ ; x0 = x1 ; } myfile . close ( ) ; } / / main ( ) Compiling and running this program is done exactly as in the case of the program in logistic.cpp. 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 ) 146 CHAPTER 3. LOGISTIC MAP 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 > r e p l o t 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 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 (x). By plotting these points x∗α (r) as a function of r we construct the (2n ) bifurcation diagram. These can be calculated numerically by using the program bifurcate.cpp. 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: 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 147 / / =========================================================== / / 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 / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { c o n s t double rmin = 2.5; c o n s t double rmax = 4.0; c o n s t double NTRANS = 500; / / Number o f d i s c a r t e d s t e p s c o n s t double NSTEPS = 100; / / Number o f recorded s t e p s c o n s t double RSTEPS = 2000; / / Number o f v a l u e s o f r int i; double r , dr , x0 , x1 ; / / −−−−−− I n i t i a l i z e : dr = ( rmax−rmin ) / RSTEPS ; / / Increment i n r ofstream myfile ( ” b i f . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−−− C a l c u l a t e : r = rmin ; while ( r <= rmax ) { x0 = 0.5; / / −−−− T r a n s i e n t s t e p s : s k i p f o r ( i =1; i<=NTRANS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; x0 = x1 ; } f o r ( i =1; i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; myfile << r << ” ” << x1 << ’\n ’ ; x0 = x1 ; } r += dr ; } / / while ( r <= rmax ) myfile . close ( ) ; } / / main ( ) The program can be compiled and run using the commands: > g++ bifurcate . cpp −o b > ./ b; The left plot of ﬁgure 3.4 can be constructed by the gnuplot commands: 148 CHAPTER 3. LOGISTIC MAP 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.cpp 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. 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 ⁹T.Y. Li, J.A. Yorke, “Period Three Implies Chaos”, American Mathematical Monthly 82 (1975) 985. 3.3. BIFURCATION DIAGRAMS 149 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 [24] 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 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. 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 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 [30]. 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. 150 CHAPTER 3. LOGISTIC MAP 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 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 [31]. 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. 3.4. THE NEWTON-RAPHSON METHOD 151 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) ρ2 − x2 cos2 x The program of the Newton-Raphson method for solving the equation g(x) = 0 can be found in the ﬁle nr.cpp: / / =========================================================== / / Newton Raphson o f f u n c t i o n o f one v a r i a b l e / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { c o n s t double rho = 1 5 . 0 ; c o n s t double eps = 1 . 0 e−6; const int NMAX = 1000; double x0 , x1 , err , g , gp ; int i; string buf ; / / −−−−− Input : cout << ” # Enter x0 : \ n” ; cin >> x0 ; getline ( cin , buf ) ; err = 1 . 0 ; cout << ” i t e r x error \n” ; cout << ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n” ; cout << 0 << ” ” << x0 << ” ” << err << ’\n ’ ; cout . precision ( 1 7 ) ; f o r ( i =1; i<=NMAX ; i++){ / / value of function g ( x ) : g = x0 * tan ( x0 )−sqrt ( rho * rho−x0 * x0 ) ; 152 CHAPTER 3. LOGISTIC MAP / / value of the d e r i v a t i v e g ’( x ) : gp = x0 / sqrt ( rho * rho−x0 * x0 )+x0 / ( cos ( x0 ) * cos ( x0 ) )+tan ( x0 ) ; x1 = x0 − g / gp ; err = abs ( x1−x0 ) ; cout << i << ” ” << x1 << ” ” << err << ’\n ’ ; i f ( err < eps ) break ; x0 = x1 ; } } / / main ( ) 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 ) 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. The compilation and running of the program can be done as follows: > g++ nr . cpp −o n > echo ” 1 . 4 ” | . / n # Enter x0 : iter x error −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 0 1.4 1 3.4. THE NEWTON-RAPHSON METHOD 153 1 1.5254292024457967 0.12542920244579681 2 1.5009739120496131 0.02445529039618366 3 1.48072070172022 0.02025321032939309 4 1.4731630533073483 0.0075576484128716537 5 1.4724779331237687 0.00068512018357957949 6 1.4724731072313519 4.8258924167932093e−06 7 1.4724731069952235 2.3612845012621619e−10 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 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) 154 CHAPTER 3. LOGISTIC MAP 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.cpp: / / =========================================================== / / 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 / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; void solve2x2 ( double A [ 2 ] [ 2 ] , double b [ 2 ] , double dx [ 2 ] ) ; i n t main ( ) { c o n s t double eps = 1 . 0 e−6; const int NMAX = 1000; double A [ 2 ] [ 2 ] , b [ 2 ] , dx [ 2 ] ; double x , y , err ; int i; string buf ; / / −−−−− Input : cout << ” # Enter x0 , y0 : \ n” ; cin >> x >> y ; getline ( cin , buf ) ; err = 1 . 0 ; cout << ” i t e r x y error \n” ; cout << ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n” ; cout << 0 << ” ” << x << ” ” << y << ” ” << err << ’\n ’ ; cout . precision ( 1 7 ) ; f o r ( i =1; i<=NMAX ; i++){ b [ 0 ] = −(2.0* x * x −3.0* x * y + y − 2 . 0 ) ; / / −g1 ( x , y ) b [ 1 ] = −(3.0* x + x * y + y − 1 . 0 ) ; / / −g2 ( x , y ) / / dg1 / dx dg1 / dy A [ 0 ] [ 0 ] = 4 . 0 * x −3.0* y ; A [ 0 ] [ 1 ] = 1.0 −3.0* x ; / / dg2 / dx dg2 / dy A [ 1 ] [ 0 ] = 3.0 + y ; A [ 1 ] [ 1 ] = 1.0+ x; solve2x2 ( A , b , dx ) ; x += dx [ 0 ] ; y += dx [ 1 ] ; err = 0 . 5 * sqrt ( dx [ 0 ] * dx [ 0 ] + dx [ 1 ] * dx [ 1 ] ) ; cout << i << ” ” << x << ” ” << y << ” ” << err << endl ; i f ( err < eps ) break ; } 3.4. THE NEWTON-RAPHSON METHOD 155 } / / main ( ) void solve2x2 ( double A [ 2 ] [ 2 ] , double b [ 2 ] , double dx [ 2 ] ) { double num0 , num1 , det ; num0 = A [ 1 ] [ 1 ] * b[0] − A[0][1] * b [1]; num1 = A[0][0] * b[1] − A[1][0] * b[0]; det = A[0][0] * A [1][1] − A [0][1] * A [1][0]; i f ( det == 0 . 0 ) { cerr << ” s o l v e 2 x 2 : d e t =0\n” ; exit ( 1 ) ; } dx [ 0 ] = num0 / det ; dx [ 1 ] = num1 / det ; } / / 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: > g++ nr2 . cpp −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−2 4 0.74735933 −0.71083147 1.2019 e−4 5 0.74735932 −0.71083145 1.2029 e−8 > echo 0 1 | . / n ................. 5 −0.10899022 1.48928857 4.3461 e−12 > echo −5 0 | . / n 6 −6.13836909 −3.77845711 3.2165 e−13 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, 156 CHAPTER 3. LOGISTIC MAP 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) = (k) f (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), 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 ∂g1 (x, r) ∂F (k) (x, r) A12 = =− ∂r ∂r ∂g2 (x, r) ∂ 2 F (k) (x, r) A21 = = ∂x ∂x2 ∂g2 (x, r) ∂ 2 F (k) (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 3.5. CALCULATION OF THE BIFURCATION POINTS 157 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 { } = F (k) (x + ϵ, r) − 2F (k) (x, r) + F (k) (x − ϵ, r) ϵ2 ∂F (k) (x+ϵx ,r) (k) ∂ 2 F (k) (x, r) − ∂F (x−ϵ x ,r) ≈ ∂r ∂r ∂x∂r 2ϵ { (k) x 1 F (x + ϵx , r + ϵr ) − F (k) (x + ϵx , r − ϵr ) = 2ϵx 2ϵr } F (k) (x − ϵx , r + ϵr ) − F (k) (x − ϵx , r − ϵr ) − 2ϵr 1 { = F (k) (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 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.cpp: / / =========================================================== // b i f u r c a t i o n P o i n t s . cpp / / 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 bacomes 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 / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; double F ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) ; double dFdx ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) ; 158 CHAPTER 3. LOGISTIC MAP double dFdr ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) ; double d2Fdx2 ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) ; double d2Fdrdx ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) ; void solve2x2 ( double A [ 2 ] [ 2 ] , double b [ 2 ] , double dx [ 2 ] ) ; i n t main ( ) { c o n s t double tol = 1 . 0 e −10; double r0 , x0 ; double A [ 2 ] [ 2 ] , B [ 2 ] , dX [ 2 ] ; double error ; i n t k , iter ; string buf ; / / −−−−−− Input : cout << ” # Enter k , r0 , x0 : \ n” ; cin >> k >> r0 >> x0 ; getline ( cin , buf ) ; cout << ” # Period k= ” << k << endl ; cout << ” # r0= ” << r0 << ” x0= ” << x0 << endl ; / / −−−−−− I n i t i a l i z e error = 1 . 0 ; / / i n i t i a l l a r g e v a l u e o f e r r o r > t o l iter = 0; cout . precision ( 1 7 ) ; while ( error > tol ) { / / −−−− C a l c u l a t e j a c o b i a n matri x A [ 0 ] [ 0 ] = 1 . 0 −dFdx ( k , x0 , r0 ) ; A [ 0 ] [ 1 ] = −dFdr ( k , x0 , r0 ) ; A [ 1 ] [ 0 ] = d2Fdx2 ( k , x0 , r0 ) ; A [ 1 ] [ 1 ] = d2Fdrdx ( k , x0 , r0 ) ; B[0] = −x0 + F ( k , x0 , r0 ) ; B[1] = −dFdx ( k , x0 , r0 ) −1.0; / / −−−− S o l v e a 2x2 l i n e a r system : solve2x2 ( A , B , dX ) ; x0 = x0 + dX [ 0 ] ; r0 = r0 + dX [ 1 ] ; error = 0 . 5 * sqrt ( dX [ 0 ] * dX [ 0 ] + dX [ 1 ] * dX [ 1 ] ) ; iter ++; cout <<iter << ” x0= ” << x0 << ” r0= ” << r0 << ” e r r = ” << error << ’\n ’ ; } / / while ( e r r o r > t o l ) } / / main ( ) / / =========================================================== / / Function F( k , x , r ) and i t s d e r i v a t i v e s double F ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) { double x0 ; int i; x0 = x ; f o r ( i =1; i<=k ; i++) x0 = r * x0 *(1.0 − x0 ) ; 3.5. CALCULATION OF THE BIFURCATION POINTS 159 r e t u r n x0 ; } / / −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double dFdx ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) { double eps ; eps = 1 . 0 e−6*x ; r e t u r n ( F ( k , x+eps , r )−F ( k , x−eps , r ) ) / ( 2 . 0 * eps ) ; } / / −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double dFdr ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) { double eps ; eps = 1 . 0 e−6*r ; r e t u r n ( F ( k , x , r+eps )−F ( k , x , r−eps ) ) / ( 2 . 0 * eps ) ; } / / −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double d2Fdx2 ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) { double eps ; eps = 1 . 0 e−6*x ; r e t u r n ( F ( k , x+eps , r ) −2.0*F ( k , x , r )+F ( k , x−eps , r ) ) / ( eps * eps ) ; } / / −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double d2Fdrdx ( c o n s t i n t& k , c o n s t double& x , c o n s t double& r ) { double epsx , epsr ; epsx = 1 . 0 e−6*x ; epsr = 1 . 0 e−6*r ; r e t u r n ( 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 * epsx * epsr ) ; } / / =========================================================== void solve2x2 ( double A [ 2 ] [ 2 ] , double b [ 2 ] , double dx [ 2 ] ) { double num0 , num1 , det ; num0 = A [ 1 ] [ 1 ] * b[0] − A[0][1] * b [1]; num1 = A[0][0] * b[1] − A[1][0] * b[0]; det = A[0][0] * A [1][1] − A [0][1] * A [1][0]; i f ( det == 0 . 0 ) { cerr << ” s o l v e 2 x 2 : d e t =0\n” ; exit ( 1 ) ; } dx [ 0 ] = num0 / det ; dx [ 1 ] = num1 / det ; } / / solve2x2 ( ) Compiling and running the program can be done as follows: > g++ bifurcationPoints . cpp −o b > echo 2 3.5 0.5 | . / b # Enter k , r0 , x0 : # Per i od k= 2 160 CHAPTER 3. LOGISTIC MAP # r0= 3.5000000000000 x0= 0.50000000000 1 x0= 0.4455758353187 r0= 3.38523275827 err= 6.35088e−2 2 x0= 0.4396562547624 r0= 3.45290970406 err= 3.39676e−2 3 x0= 0.4399593001407 r0= 3.44949859951 err= 1.71226 e−3 4 x0= 0.4399601690333 r0= 3.44948974267 err= 4.44967 e−6 5 x0= 0.4399601689937 r0= 3.44948974281 err= 7.22160 e−11 > echo 2 3.5 0.85 | . / b ................. 4 x0= 0.8499377795512 r0= 3.44948974275 err= 1.85082e−11 > echo 4 3.5 0.5 | . / b ................. 5 x0= 0.5235947861540 r0= 3.54409035953 err= 1.86318 e−11 > echo 4 3.5 0.35 | . / b ................. 5 x0= 0.3632903374118 r0= 3.54409035955 err= 5.91653e−13 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 . 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 ¹¹Sensitivity to the initial condition alone does not necessarily imply chaos. It is necessary to have topological mixing and dense periodic orbits. Topological mixing 3.6. LIAPUNOV EXPONENTS 161 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 | 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.cpp 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 . / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; 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. 162 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. i n t main ( ) { int NSTEPS , i ; double r , x0 , x1 , x0t , x1t , epsilon ; string buf ; / / −−−−− Input : cout << ” # Enter NSTEPS , r , x0 , e p s i l o n : \ n” ; cin >> NSTEPS >> r >> x0 >> epsilon ; getline ( cin , buf ) ; cout << ” # NSTEPS = ” << NSTEPS << endl ; cout << ” # r = ” << r << endl ; cout << ” # x0 = ” << x0 << endl ; cout << ” # e p s i l o n = ” << epsilon << endl ; x0t = x0+epsilon ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” l i a . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : myfile << 1 << ” ” << x0 << ” ” << x0t << ” ” << abs ( x0t−x0 ) / epsilon << ”\n” ; f o r ( i =2;i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; 3.6. LIAPUNOV EXPONENTS 163 x1t = r * x0t * (1.0 − x0t ) ; myfile << i << ” ” << x1 << ” ” << x1t << ” ” << abs ( x1t−x1 ) / epsilon << ”\n” ; x0 = x1 ; x0t = x1t ; } myfile . close ( ) ; } / / main ( ) 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: > g++ liapunov1 . cpp −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 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 164 CHAPTER 3. LOGISTIC MAP 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. 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). 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.cpp: / / =========================================================== / / 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 ) | 3.6. LIAPUNOV EXPONENTS 165 / / NTRANS: number o f d i s c a r t e d i t e r a t i o n s i n order t o d i s c a r t // t r a n s i e n t behaviour / / NSTEPS : number o f terms i n t h e sum / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { int NTRANS , NSTEPS , i ; double r , x0 , x1 , sum ; string buf ; / / −−−−− Input : cout << ” # Enter NTRANS, NSTEPS , r, x0 : \ n” ; cin >> NTRANS >> NSTEPS >> r >> x0 ; getline ( cin , buf ) ; cout << ” # NTRANS = ” << NTRANS << endl ; cout << ” # NSTEPS = ” << NSTEPS << endl ; cout << ” # r = ” << r << endl ; cout << ” # x0 = ” << x0 << endl ; f o r ( i =1; i<=NTRANS ; i++){ x1 = r * x0 * ( 1 . 0 − x0 ) ; x0 = x1 ; } sum = log ( abs ( r * ( 1 . 0 − 2 . 0 * x0 ) ) ) ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” l i a . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : myfile << 1 << ” ” << x0 << ” ” << sum << ”\n” ; f o r ( i =2;i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; sum += log ( abs ( r * ( 1 . 0 − 2 . 0 * x1 ) ) ) ; myfile << i << ” ” << x1 << ” ” << sum / i << ”\n” ; x0 = x1 ; } myfile . close ( ) ; } / / main ( ) 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 166 CHAPTER 3. LOGISTIC MAP 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: > g++ liapunov2 . cpp −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 0.7 5 |./ 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 ﬁgures 3.7 and 3.8. The program can be found in the ﬁle liapunov3.cpp 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 t e d i t e r a t i o n s i n order t o d i s c a r t // t r a n s i e n t behaviour / / 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 ev er y r / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; 3.6. LIAPUNOV EXPONENTS 167 i n t main ( ) { c o n s t double rmin = 2.5; c o n s t double rmax = 4.0; c o n s t double xstart = 0 . 2 ; const int RSTEPS = 1000; const int NSTEPS = 60000; const int NTRANS = 2000; int i , ir ; double r , x0 , x1 , sum , dr ; string buf ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” l i a . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : dr = ( rmax−rmin ) / ( RSTEPS −1) ; f o r ( ir = 0 ; ir < RSTEPS ; ir++){ r = rmin+ir * dr ; x0 = xstart ; f o r ( i = 1 ; i <= NTRANS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; x0 = x1 ; } sum = log ( abs ( r * ( 1 . 0 − 2 . 0 * x0 ) ) ) ; / / Calculate : f o r ( i = 2 ; i <= NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; sum+= log ( abs ( r * ( 1 . 0 − 2 . 0 * x1 ) ) ) ; x0 = x1 ; } myfile << r << ” ” << sum / NSTEPS << ’\n ’ ; } / / f o r ( i r =0; i r <RSTEPS ; i r ++) myfile . close ( ) ; } / / main ( ) The program can be compiled and run using the commands: > g++ liapunov3 . cpp −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. The data are saved in the ﬁle lia.dat and we can make the plot shown in ﬁgure 3.7 using gnuplot: 168 CHAPTER 3. LOGISTIC MAP 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). 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 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 3.6. LIAPUNOV EXPONENTS 169 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 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 170 CHAPTER 3. LOGISTIC MAP ∑ ∫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 [32]. 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.cpp 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.cpp: / / =========================================================== / / 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 t e d i t e r a t i o n s i n order t o d i s c a r t // t r a n s i e n t behaviour / / 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 ev er y r / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > 3.6. LIAPUNOV EXPONENTS 171 # include <string > # i n c l u d e <cmath> using namespace std ; i n t main ( ) { c o n s t double rmin = 2.5; c o n s t double rmax = 4.0; c o n s t double xstart = 0 . 2 ; const int RSTEPS = 1000; const int NHIST = 10000; const int NTRANS = 2000; const int NSTEPS = 5000000; c o n s t double xmin =0.0 , xmax = 1 . 0 ; int i , ir , isum , n ; double r , x0 , x1 , sum , dr , dx ; double p [ NHIST ] , S ; string buf ; / / −−−−− I n i t i a l i z e : ofstream myfile ( ” entropy . dat ” ) ; myfile . precision ( 1 7 ) ; / / −−−−− C a l c u l a t e : f o r ( i =0;i<NHIST ; i++) p [ i ] = 0 . 0 ; dr = ( rmax−rmin ) / ( RSTEPS −1) ; dx = ( xmax−xmin ) / ( NHIST −1) ; f o r ( ir =0; ir<RSTEPS ; ir++){ r = rmin+ir * dr ; x0= xstart ; f o r ( i =1; i<=NTRANS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; x0 = x1 ; } / / make histogram : n= i n t ( x0 / dx ) ; p [ n ] + = 1 . 0 ; f o r ( i =2;i<=NSTEPS ; i++){ x1 = r * x0 * (1.0 − x0 ) ; n = i n t ( x1 / dx ) ; p [ n ] += 1 . 0 ; x0 = x1 ; } / / 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 : f o r ( i =0; i < NHIST ; i++) p [ i ] /= ( NSTEPS * dx ) ; / / sum a l l non z e r o terms : p [ n ] * l o g ( p [ n ] ) * dx S = 0.0; f o r ( i =0; i < NHIST ; i++) i f (p [ i ] > 0.0) S −= p [ i ] * log ( p [ i ] ) * dx ; 172 CHAPTER 3. LOGISTIC MAP myfile << r << ” ” << S << ’\n ’ ; } / / f o r ( i r =0; i r <RSTEPS ; i r ++) myfile . close ( ) ; myfile . open ( ” e n t r o p y _ h i s t . dat ” ) ; myfile . precision ( 1 7 ) ; f o r ( n =0;n<NHIST ; n++){ x0 = xmin + n * dx + 0 . 5 * dx ; myfile << r << ” ” << x0 << ” ” << p [ n ] << ’\n ’ ; } myfile . close ( ) ; } / / main ( ) 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 3.6. LIAPUNOV EXPONENTS 173 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. 174 CHAPTER 3. LOGISTIC MAP 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 > ! g++ logistic . cpp −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, 3.7. PROBLEMS 175 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 x∗3 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. 176 CHAPTER 3. LOGISTIC MAP 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) 3.7. PROBLEMS 177 (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.cpp 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: 178 CHAPTER 3. LOGISTIC MAP 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? 3.7. PROBLEMS 179 (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. 180 CHAPTER 3. LOGISTIC MAP (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 ≤ 21 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.cpp 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. 3.7. PROBLEMS 181 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. 182 CHAPTER 3. LOGISTIC MAP 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.cpp 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. 3.7. PROBLEMS 183 (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.cpp 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.cpp 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. 184 CHAPTER 3. LOGISTIC MAP 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) 185 186 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 187 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. 188 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 189 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 sixteen orders of magnitude, adding the ﬁrst one to the second is equivalent to 190 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?). adding 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 pro- gram is mainly a user interface, and the computation is carried out by three functions 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 double arrays 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 functions which take as input the initial conditions, the time interval of ⁴Numbers of type double have approximately sixteen 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 variables of type float, ϵ ≈ 1.2 × 10−7 and for variables of type double ϵ ≈ 2.2 × 10−16 . 4.2. PRELUDE: EULER METHODS 191 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. 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 functions is performed in for loops which advance the solution by 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 [0] = 0.0; X [ 0 ] = Xin ; V [ 0 ] = Vin ; h = Tfi / ( Nt −1) ; f o r ( i =1; i<Nt ; i++){ T [ i ] = T [ i−1]+h ; V [ i ] = V [ i−1]+accel ( X [ i −1]) * h ; X [ i ] = X [ i−1]+V [ i ] * h ; } 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 last step for the velocity: 192 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. T[0] = 0.0; X[0] = Xin ; V[0] = Vin ; X0 = X [ 0 ] − V [ 0 ] * h + accel ( X [ 0 ] ) * h2 / 2 . 0 ; T[1] = h; X[1] = 2 . 0 * X [ 0 ] − X0 + accel ( X [ 0 ] ) * h2 ; f o r ( i =2;i<Nt ; i++){ .............. } V [ Nt −1] = ( X [ Nt −1] − X [ Nt −2]) / h ; The full program can be found in the ﬁle euler.cpp 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 f u n c t i o n s r e t u r n / / X[ t ] and V[ t ]=dX[ t ] / dt i n a r r a y s / / T [ 0 . . Nt −1] ,X [ 0 . . Nt −1] ,V [ 0 . . Nt −1] / / 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 / / 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. 4.2. PRELUDE: EULER METHODS 193 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. In this ﬁgure, float variables have been used instead of double in order to enhance the eﬀect. / / NOTE: T[ 0 ] = 0 T[ Nt −1] = T f i / / ========================================================= # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 110000; double T [ P ] , X [ P ] , V [ P ] ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void euler ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) ; void euler_cromer ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) ; void euler_verlet ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) ; double accel ( c o n s t double& x ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { double Xin , Vin , Tfi ; int Nt , i ; string buf ; / / 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_0 194 CHAPTER 4. MOTION OF A PARTICLE / / f i n a l time t _ f and Nt : cout << ” Enter X_0 , V_0 , t _ f , Nt ( t _ i =0) : \ n” ; cin >> Xin >> Vin >> Tfi >> Nt ; getline ( cin , buf ) ; / / This check i s n e c e s s a r y i n order t o avoid / / memory a c c e s s v i o l a t i o n s : i f ( Nt>=P ) { cerr << ” Error : Nt>=P\n” ; exit ( 1 ) ; } / / Xin= X[ 0 ] , Vin=V[ 0 ] , T[0]=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 [ 1 . . Nt −1] , X [ 1 . . Nt −1] , V [ 1 . . Nt −1] / / which we p r i n t i n a f i l e euler ( Xin , Vin , Tfi , Nt ) ; ofstream myfile ( ” e u l e r . dat ” ) ; myfile . precision ( 1 7 ) ; f o r ( i =0;i<Nt ; i++) / / Each l i n e i n f i l e has time , p o s i t i o n , v e l o c i t y : myfile << T [ i ] << ” ” << X [ i ] << ” ” << V [ i ] << endl ; myfile . close ( ) ; / / we c l o s e t h e stream 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 euler_cromer ( Xin , Vin , Tfi , Nt ) ; myfile . open ( ” e u l e r _ c r o m e r . dat ” ) ; f o r ( i =0;i<Nt ; i++) myfile << T [ i ] << ” ” << X [ i ] << ” ” << V [ i ] << endl ; myfile . close ( ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− euler_verlet ( Xin , Vin , Tfi , Nt ) ; myfile . open ( ” e u l e r _ v e r l e t . dat ” ) ; f o r ( i =0;i<Nt ; i++) myfile << T [ i ] << ” ” << X [ i ] << ” ” << V [ i ] << endl ; myfile . close ( ) ; } / / main ( ) / / ========================================================= / / 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 f u n c t i o n 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 / / ========================================================= double accel ( c o n s t double& x ) { r e t u r n −10.0 * sin ( x ) ; } / / ========================================================= / / 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[ 0 ] , Vin=V[ 0 ] −− 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 [ 0 . . Nt −1] , X [ 0 . . Nt −1] , V [ 0 . . Nt −1] which / / g i v e s x ( t _ k )=X[ k −1] , dx / dt ( t _ k )=V[ k −1] , t _ k=T[ k−1] 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 . / / ========================================================= 4.2. PRELUDE: EULER METHODS 195 void euler ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) { int i; double h ; / / I n i t i a l c o n d i t i o n s s e t here : T [0] = 0.0; X [ 0 ] = Xin ; V [ 0 ] = Vin ; / / h i s t h e time s t e p Dt h = Tfi / ( Nt −1) ; f o r ( i =1; i<Nt ; i++){ T [ i ] = T [ i−1]+h ; / / time advances by Dt=h X [ i ] = X [ i−1]+V [ i −1]* h ; / / advancement and s t o r a g e o f V [ i ] = V [ i−1]+accel ( X [ i −1]) * h ; / / p o s i t i o n and v e l o c i t y } } / / 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[ 0 ] , Vin=V[ 0 ] −− 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 [ 0 . . Nt −1] , X [ 0 . . Nt −1] , V [ 0 . . Nt −1] which / / g i v e s x ( t _ k )=X[ k −1] , dx / dt ( t _ k )=V[ k −1] , t _ k=T[ k−1] 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 . / / ========================================================= void euler_cromer ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) { int i; double h ; / / I n i t i a l c o n d i t i o n s s e t here : T [0] = 0.0; X [ 0 ] = Xin ; V [ 0 ] = Vin ; / / h i s t h e time s t e p Dt h = Tfi / ( Nt −1) ; f o r ( i =1; i<Nt ; i++){ T [ i ] = T [ i−1]+h ; V [ i ] = V [ i−1]+accel ( X [ i −1]) * h ; X [ i ] = X [ i−1]+V [ i ] * h ; / / note d i f f e r e n c e from Euler } } / / 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[ 0 ] , Vin=V[ 0 ] −− i n i t i a l c o n d i t i o n a t t =0 , 196 CHAPTER 4. MOTION OF A PARTICLE / / 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 [ 0 . . Nt −1] , X [ 0 . . Nt −1] , V [ 0 . . Nt −1] which / / g i v e s x ( t _ k )=X[ k −1] , dx / dt ( t _ k )=V[ k −1] , t _ k=T[ k−1] 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 . / / ========================================================= void euler_verlet ( c o n s t double& Xin , c o n s t double& Vin , c o n s t double& Tfi , c o n s t i n t & Nt ) { int i; double h , h2 , X0 , o2h ; / / I n i t i a l c o n d i t i o n s s e t here : T[0] = 0.0; X[0] = Xin ; V[0] = 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 rre s ponds t o ’X[ −1] ’ X0 = X [ 0 ] − V [ 0 ] * h + accel ( X [ 0 ] ) * h2 / 2 . 0 ; T[1] = h; X[1] = 2 . 0 * X [ 0 ] − X0 + accel ( X [ 0 ] ) * h2 ; / / Now i s t a r t s from 2 : f o r ( i =2;i<Nt ; i++){ 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]) ; } / / we have one more s t e p f o r t h e v e l o c i t y : V [ Nt −1] = ( X [ Nt −1] − X [ Nt −2]) / h ; } / / euler_verlet () Compiling the running the program can be done with the commands: > g++ euler . cpp −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 0.20000 0 0.00600 0.20000 −0.01193 0.01201 0.19992 −0.02386 0.01801 0.19978 −0.03579 0.02402 0.19957 −0.04771 The last command shows the ﬁrst 5 lines of the ﬁle euler.dat. We see 4.2. PRELUDE: EULER METHODS 197 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 large⁵. 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 commands⁶: 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) ⁵In this ﬁgure, roundoﬀ errors are enhanced by using float variables instead of double. ⁶The command set dummy t sets the independent variable for functions to be t instead of x which is the default. 198 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 in⁷ 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 ⁷A small modiﬁcation is necessary in order to plot the absolute value of the diﬀerences. 4.3. RUNGE–KUTTA METHODS 199 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 200 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)⁸. 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 ⁸Not 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 201 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 202 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 203 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 global double ar- rays 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(Ti,Tf,X10,X20,Nt) which “drives” the heart of the program, the function RKSTEP(t,x1,x2,dt) which performs one integration step using equations (4.22). RKSTEP evolves the functions x1, x2 at time t by one step h = dt. The func- tion 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.cpp: / / ======================================================== / / 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 204 CHAPTER 4. MOTION OF A PARTICLE / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 110000; double T [ P ] , X1 [ P ] , X2 [ P ] ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) ; double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) ; void RK ( c o n s t double& Ti , c o n s t double& Tf , c o n s t double& X10 , c o n s t double& X20 , c o n s t i n t & Nt ) ; void RKSTEP ( double& t , double& x1 , double& x2 , c o n s t double& dt ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { double Ti , Tf , X10 , X20 ; int Nt ; int i; string buf ; / / Input : cout << ”Runge−Kutta Method f o r 2−ODEs I n t e g r a t i o n \n” ; cout << ” Enter Nt , Ti , TF , X10 , X20 : ” << endl ; cin >> Nt >> Ti >> Tf >> X10 >> X20 ; getline ( cin , buf ) ; cout << ” Nt = ” << Nt << endl ; cout << ”Time : I n i t i a l Ti = ” << Ti << ” F i n a l Tf = ” << Tf << endl ; cout << ” X1( Ti )= ” << X10 << ” X2( Ti )= ” << X20 << endl ; i f ( Nt >= P ) { cerr << ” Error ! Nt >= P\n” ; exit ( 1 ) ; } / / Calculate : RK ( Ti , Tf , X10 , X20 , Nt ) ; / / Output : ofstream myfile ( ” rk . dat ” ) ; myfile . precision ( 1 7 ) ; f o r ( i =0;i<Nt ; i++) myfile << T [ i ] << ” ” << X1 [ i ] << ” ” << X2 [ i ] << ’\n ’ ; myfile . close ( ) ; } / / main ( ) 4.3. RUNGE–KUTTA METHODS 205 / / ======================================================== / / 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 / / ======================================================== double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { r e t u r n x2 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { r e t u r n −10.0* x1 ; / / harmonic o s c i l l a t o r } / / ======================================================== / / RK( 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 / / Output : v a l u e s i n a r r a y s T[ Nt ] , X1 [ Nt ] , X2[ Nt ] where / / T[ 0 ] = Ti X1 [ 0 ] = X10 X2[ 0 ] = X20 // X1 [ k−1] = X1( a t t =T( k ) ) // X2[ k−1] = X2( a t t =T( k ) ) / / T[ Nt −1] = Tf / / ======================================================== void RK ( c o n s t double& Ti , c o n s t double& Tf , c o n s t double& X10 , c o n s t double& X20 , c o n s t i n t & Nt ) { double dt ; double TS , X1S , X2S ; / / time and X1 , X2 a t g i v e n s t e p int i; / / I n i t i a l i z e variables : dt = ( Tf−Ti ) / ( Nt −1) ; T [0] = Ti ; X1 [ 0 ] = X10 ; X2 [ 0 ] = 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 ! f o r ( i =1; i<Nt ; i++){ RKSTEP ( TS , X1S , X2S , dt ) ; T [ i ] = TS ; X1 [ i ] = X1S ; X2 [ i ] = X2S ; } } / / RK( ) / / ======================================================== / / Function RKSTEP( t , x1 , x2 , dt ) 206 CHAPTER 4. MOTION OF A PARTICLE / / 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 . . . / / ======================================================== void RKSTEP ( double& t , double& x1 , double& x2 , c o n s t double& dt ) { double k11 , k12 , k13 , k14 , k21 , k22 , k23 , k24 ; double 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* h ; / / h2=h / 2 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 * ( k12+k13 )+k14 ) ; x2 =x2+h6 * ( k21 + 2 . 0 * ( k22+k23 )+k24 ) ; } / / RKSTEP ( ) 4.4 Comparison of the Methods In this section we will check our programs for correctness and accuracy w.r.t. discretization and roundoﬀ errors. The simplest test is to check the results against a known analytic solution of a simple model. This will be done for the simple harmonic oscillator. We will change 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.cpp becomes 4.4. COMPARISON OF THE METHODS 207 100 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). double accel ( c o n s t double& x ) { r e t u r n −10.0 * x ; } and that of the program in rk.cpp becomes double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { r e t u r n −10.0* x1 ; } 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 208 CHAPTER 4. MOTION OF A PARTICLE 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). 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 order⁹ O(∆t4 ). 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. ⁹The 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 209 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 ). 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. Consider initially the system without the inﬂuence of the driving force, i.e. with a0 = 0. The real solutions of the diﬀerential equation¹⁰ which are ﬁnite for t → +∞ are given by √ √ 2 2 x0 (t) = c1 e−(γ+ γ 2 −4ω02 )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) ¹⁰These are easily obtained by substituting the ansatz x(t) = Ae−Ωt and solving for Ω. 210 CHAPTER 4. MOTION OF A PARTICLE 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. (√ ) −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+ 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 4.5. THE FORCED DAMPED OSCILLATOR 211 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. 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.cpp. The main routines RK(T0,TF,X10,X20,Nt) and RKSTEP(t,x1,x2,dt) re- main 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 they are declared within the global scope. Another point that needs our attention is the function f2(t,x1,x2) which now takes the velocity v → x2 in its arguments: double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { double a ; a = a_0 * cos ( omega * t ) ; r e t u r n −omega_02 * x1−gam * x2+a ; } The main program, found in the ﬁle dlo.cpp, is listed below. The func- tions RK, RKSTEP are the same as in rk.cpp and should also be included in the same ﬁle. 212 CHAPTER 4. MOTION OF A PARTICLE 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. / / ======================================================== / / 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 / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 110000; double T [ P ] , X1 [ P ] , X2 [ P ] ; double omega_0 , omega , gam , a_0 , omega_02 , omega2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) ; double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) ; void RK ( c o n s t double& Ti , c o n s t double& Tf , c o n s t double& X10 , c o n s t double& X20 , c o n s t i n t & Nt ) ; void RKSTEP ( double& t , double& x1 , double& x2 , c o n s t double& dt ) ; 4.5. THE FORCED DAMPED OSCILLATOR 213 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. / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { double Ti , Tf , X10 , X20 ; double Energy ; int Nt ; int i; string buf ; / / Input : cout << ”Runge−Kutta Method f o r DLO I n t e g r a t i o n \n” ; cout << ” Enter omega_0 , omega , gamma , a_0 : \ n” ; cin >> omega_0 >> omega >> gam>> a_0 ; getline ( cin , buf ) ; omega_02 = omega_0 * omega_0 ; omega2 = omega * omega ; cout << ”omega_0= ” << omega_0 << ” omega= ” << omega << endl ; cout << ”gamma= ” << gamma << ” a_0= ” << a_0 << endl ; cout << ” Enter Nt , Ti , TF , X10 , X20 : ” << endl ; cin >> Nt >> Ti >> Tf >> X10 >> X20 ; getline ( cin , buf ) ; cout << ” Nt = ” << Nt << endl ; cout << ”Time : I n i t i a l Ti = ” << Ti << ” F i n a l Tf = ” << Tf << endl ; cout << ” X1 ( Ti )= ” << X10 << ” X2( Ti )= ” << X20 << endl ; i f ( Nt >= P ) { cerr << ” Error ! Nt >= P\n” ; exit ( 1 ) ; } / / Calculate : 214 CHAPTER 4. MOTION OF A PARTICLE 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. RK ( Ti , Tf , X10 , X20 , Nt ) ; / / Output : ofstream myfile ( ” dlo . dat ” ) ; myfile . precision ( 1 7 ) ; myfile << ” # Damped Lin ea r O s c i l l a t o r − dlo \n” ; myfile << ” # omega_0= ” << omega_0 << ” omega= ” << omega << ” gamma= ” << gam << ” a_0= ” << a_0 << ←- endl ; f o r ( i =0;i<Nt ; i++){ Energy = 0 . 5 * X2 [ i ] * X2 [ i ] + 0 . 5 * omega_02 * X1 [ i ] * X1 [ i ] ; myfile << T [ i ] << ” ” << X1 [ i ] << ” ” << X2 [ i ] << ” ” << Energy << ’\n ’ ; } myfile . close ( ) ; } / / main ( ) / / ======================================================== / / 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 / / ======================================================== double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { r e t u r n x2 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double 4.5. THE FORCED DAMPED OSCILLATOR 215 f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { double a ; a = a_0 * cos ( omega * t ) ; r e t u r n −omega_02 * x1−gam * x2+a ; } 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. 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 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 216 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 → +∞. space. This can be seen¹¹ 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 ¹¹To be precise, phase space is the space of positions-momenta, but in our case the diﬀerence is trivial. 4.6. THE FORCED DAMPED PENDULUM 217 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. 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 218 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 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.cpp. 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.cpp. 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.cpp, rk.cpp. / / ======================================================== 4.6. THE FORCED DAMPED PENDULUM 219 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). / / Program t o s o l v e Forced Damped Pendulum / / using 4 th order Runge−Kutta Method / / Output i s w r i t t e n i n f i l e fdp . dat / / ======================================================== ................................ c o n s t i n t P = 1010000; ................................ Energy = 0 . 5 * X2 [ i ] * X2 [ i ]+ omega_02 *(1.0 − cos ( X1 [ i ] ) ) ; ................................ double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 ) { r e t u r n −(omega_02 +2.0* a_0 * cos ( omega * t ) ) * sin ( x1 )−gam * x2 ; } ................................ void RKSTEP ( double& t , double& x1 , double& x2 , c o n s t double& dt ) { ................................ c o n s t double pi = 3.14159265358979324; c o n s t double pi2= 6.28318530717958648; x1 =x1+h6 * ( k11 + 2 . 0 * ( k12+k13 )+k14 ) ; 220 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). x2 =x2+h6 * ( k21 + 2 . 0 * ( k22+k23 )+k24 ) ; i f ( x1 > pi ) x1 −= pi2 ; i f ( x1 < −pi ) x1 += pi2 ; } / / 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 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 4.6. THE FORCED DAMPED PENDULUM 221 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. 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 19 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 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 222 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. 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 awk¹²: 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 where the time is an integer multiple of the period¹³. 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 ¹²The command can be written in one line without the ﬁnal \ of the ﬁrst and second lines. ¹³The 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 223 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. 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. 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! 224 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. 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) 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). 4.7. APPENDIX: ON THE EULER–VERLET METHOD 225 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. • 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 (4.36) the error for the ﬁrst step is δθ(t0 + ∆t) = O((∆t)4 ). Then¹⁴ 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 ) , ¹⁴Remember that the acceleration α(t) is given, therefore δα(t) = 0. 226 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 227 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 228 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 229 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. 230 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 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 . 4.5 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. 4.6 Reproduce the results of ﬁgures 4.19–4.22. 4.7 Reproduce the results of ﬁgures 4.23–4.26. Calculate the phase δ(ω) numerically and compare with equation (4.33). 4.9. PROBLEMS 231 4.8 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. 4.9 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). 4.10 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. 4.11 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. 4.12 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... 4.13 Repeat the previous problem when each oscillator is damped with γ = 0.5. Take ω0 = 3.145. 4.14 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. 232 CHAPTER 4. MOTION OF A PARTICLE 4.15 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. 4.16 Reproduce the results in ﬁgures 4.30. 4.17 Reproduce the results in ﬁgures 4.31. 4.18 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. 4.19 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. (a) Construct the bifurcation diagram by plotting the points (A, θ(tk )). (b) Repeat by plotting the points (A, θ̇(tk )). 4.9. PROBLEMS 233 (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. 234 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.cpp. 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.cpp. The program that computes the acceleration will be put in a ﬁle named rk_XXX.cpp, where XXX is a string of characters that identiﬁes the force. 235 236 CHAPTER 5. PLANAR MOTION For example, the ﬁle rk2_hoc.cpp contains the program computing the acceleration of the simple harmonic oscillator, the ﬁle rk2_g.cpp the ac- celeration of a constant gravitational ﬁeld ⃗g = −g ŷ etc. Diﬀerent force ﬁelds will require the use of one or more coupling con- stants 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 the global scope which will be accessed by the acceleration functions f3 and f4, the function energy and the main program where the user will enter their. 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 function RK(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 func- tion energy(t,x1,x2,v1,v2). The code for the function energy, which is diﬀerent for each force ﬁeld, is written in the same ﬁle with the ac- celeration. The code for the function 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 double 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 rk2 . dat / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 1010000; double T [ P ] , X1 [ P ] , X2 [ P ] , V1 [ P ] , V2 [ P ] ; double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double 5.1. RUNGE–KUTTA FOR PLANAR MOTION 237 f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; void RK ( c o n s t double& Ti , c o n s t double& Tf , c o n s t double& X10 , c o n s t double& X20 , c o n s t double& V10 , c o n s t double& V20 , const int & Nt ) ; void RKSTEP ( double& t , double& x1 , double& x2 , double& x3 , double& x4 , const double& dt ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double Ti , Tf , X10 , X20 , V10 , V20 ; int Nt , i ; double E0 , EF , DE ; / / Input : cout << ”Runge−Kutta Method f o r 4−ODEs I n t e g r a t i o n \n” ; cout << ” Enter c o u p l i n g c o n s t a n t s : \ n” ; cin >> k1 >> k2 ; getline ( cin , buf ) ; cout << ” k1= ” << k1 << ” k2= ” << k2 << endl ; cout << ” Enter Nt , Ti , Tf , X10 , X20 , V10 , V20: \ n” ; cin >> Nt >> Ti >> Tf>> X10 >> X20 >> V10 >> V20 ; getline ( cin , buf ) ; cout << ” Nt = ” << Nt << endl ; cout << ”Time : I n i t i a l Ti = ” << Ti << ” F i n a l Tf= ” << Tf << endl ; cout << ” X1 ( Ti )= ” << X10 << ” X2( Ti )=” << X20 << endl ; cout << ” V1 ( Ti )= ” << V10 << ” V2( Ti )=” << V20 << endl ; / / Calculate : RK ( Ti , Tf , X10 , X20 , V10 , V20 , Nt ) ; ofstream myfile ( ” rk2 . dat ” ) ; myfile . precision ( 1 7 ) ; f o r ( i =0;i<Nt ; i++) myfile << T [ i ] << ” ” 238 CHAPTER 5. PLANAR MOTION << X1 [ i ] << ” ” << X2 [ i ] << ” ” << V1 [ i ] << ” ” << V2 [ i ] << ” ” << energy ( T [ i ] , X1 [ i ] , X2 [ i ] , V1 [ i ] , V2 [ i ] ) << endl ; myfile . close ( ) ; / / Rutherford s c a t t e r i n g a n g l e s : cout . precision ( 1 7 ) ; cout <<”v−a n g l e : ”<< atan2 ( V2 [ Nt −1] , V1 [ Nt −1]) << endl ; cout <<”b−a n g l e : ”<< 2 . 0 * atan ( k1 / ( V10 * V10 * X20 ) ) << endl ; E0=energy ( Ti , X10 , X20 , V10 , V20 ); EF=energy ( T [ Nt −1] , X1 [ Nt −1] , X2 [ Nt −1] , V1 [ Nt −1] , V2 [ Nt −1]) ; DE = abs ( 0 . 5 * ( EF−E0 ) / ( EF+E0 ) ) ; cout << ”E0 , EF , DE/ E= ” << E0 << ” ” << EF << ” ” << DE << endl ; } / / main ( ) / / ======================================================== / / 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 ) / / ======================================================== double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n v1 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n v2 ; } / / ======================================================== / / RK( 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[ 0 ] = Ti X1 [ 0 ] = X10 X2[ 0 ] = X20 V1 [ 0 ] = V10 V2 [ 0 ] = 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 −1]= Tf / / ======================================================== void RK ( c o n s t double& Ti , c o n s t double& Tf , c o n s t double& X10 , c o n s t double& X20 , c o n s t double& V10 , c o n s t double& V20 , const int & Nt ) { 5.1. RUNGE–KUTTA FOR PLANAR MOTION 239 double dt ; double 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 double V1S , V2S ; int i; // Initialize : dt = ( Tf−Ti ) / ( Nt −1) ; T [ 0 ] = Ti ; X1 [ 0 ] = X10 ; X2 [ 0 ] = X20 ; V1 [ 0 ] = V10 ; V2 [ 0 ] = 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 f o r ( i =1; i<Nt ; i++){ RKSTEP ( TS , X1S , X2S , V1S , V2S , dt ) ; T [ i ] = TS ; X1 [ i ] = X1S ; X2 [ i ] = X2S ; V1 [ i ] = V1S ; V2 [ i ] = V2S ; } } / / 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 , 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 . . . / / ======================================================== void RKSTEP ( double& t , double& x1 , double& x2 , double& x3 , double& x4 , const double& dt ) { double k11 , k12 , k13 , k14 , k21 , k22 , k23 , k24 ; double k31 , k32 , k33 , k34 , k41 , k42 , k43 , k44 ; double 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* h ; / / h2 = h / 2 240 CHAPTER 5. PLANAR MOTION h6=h / 6 . 0 ; / / 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 * ( k12+k13 )+k14 ) ; x2=x2+h6 * ( k21 + 2 . 0 * ( k22+k23 )+k24 ) ; x3=x3+h6 * ( k31 + 2 . 0 * ( k32+k33 )+k34 ) ; x4=x4+h6 * ( k41 + 2 . 0 * ( k42+k43 )+k44 ) ; } / / 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 5.2. PROJECTILE MOTION 241 where tan θ = v0y /v0x is the direction of the initial velocity and hmax is the maximum height of the trajectory. 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. The acceleration ax (t) = 0 ay (t) = −g (ax ↔ f3 , ay ↔ f4) and the mechanical energy is coded in the ﬁle rk2_g.cpp: / / ======================================================== / / 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 / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / 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 e l d with / / g = −k2 double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , 242 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. c o n s t double& v1 , c o n s t double& v2 ) { return 0.0; / / dx3 / dt=dv1 / dt=a1 } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n −k1 ; / / dx4 / dt=dv2 / dt=a2 } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n 0 . 5 * ( v1 * v1+v2 * v2 ) + k1 * x2 ; } In order to calculate a projectile’s trajectory you may use the following commands: > g++ −O2 rk2 . cpp rk2_g . cpp −o rk2 > . / rk2 Runge−Kutta Method for 4−ODEs Integration Enter coupling constants : 10.0 0.0 k1= 10 k2= 0 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 Final Tf= 0.2 X1 ( Ti )= 0 X2 ( Ti ) =0 5.2. PROJECTILE MOTION 243 V1 ( Ti )= 1 V2 ( Ti ) =1 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 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 ) 244 CHAPTER 5. PLANAR MOTION 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 −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 5.2. PROJECTILE MOTION 245 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.cpp, 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 in a ﬁle named e.g. rk2_myforce.cpp as we did with rk2_g.cpp. 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 246 CHAPTER 5. PLANAR MOTION 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.cpp: / / ======================================================== / / 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 l e d with / / ax = −k2 vx ay = −k2 vy − k1 # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n −k2 * v1 ; / / dx3 / dt=dv1 / dt=a1 } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n −k2 * v2−k1 ; / / dx4 / dt=dv2 / dt=a2 } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 5.2. PROJECTILE MOTION 247 double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { r e t u r n 0 . 5 * ( v1 * v1+v2 * v2 ) + k1 * x2 ; } 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. 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. 248 CHAPTER 5. PLANAR MOTION 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 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 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 5.3. PLANETARY MOTION 249 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.cpp: / / ======================================================== / / 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 / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Motion i n Coulombic p o t e n t i a l : / / ax= k1 * x1 / r ^3 ay= k1 * x2 / r ^3 double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r2 , r3 ; r2=x1 * x1+x2 * x2 ; r3=r2 * sqrt ( r2 ) ; i f ( r3 >0.0) r e t u r n k1 * x1 / r3 ; / / dx3 / dt=dv1 / dt=a1 else return 0.0; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r2 , r3 ; r2=x1 * x1+x2 * x2 ; 250 CHAPTER 5. PLANAR MOTION r3=r2 * sqrt ( r2 ) ; i f ( r3 >0.0 ) r e t u r n k1 * x2 / r3 ; / / dx4 / dt=dv4 / dt=a4 else return 0.0; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r ; r=sqrt ( x1 * x1+x2 * x2 ) ; i f ( r > 0.0) r e t u r n 0 . 5 * ( v1 * v1+v2 * v2 ) + k1 / r ; else return 0.0; } 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 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. 5.3. PLANETARY MOTION 251 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). 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. 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. 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. 252 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). 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 accelerators for the study of fundamental interactions between elementary particles. First we will discuss scattering of small hard spheres of radius r1 by ¹We refer the reader to [40], chapter 4. 5.4. SCATTERING 253 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 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 total interaction cross section is Σ = nAσ, (5.16) ²The so called hard core potential. ³The number of particles crossing a surface perpendicular to the beam per unit time and unit area. 254 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. 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 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). 5.4. SCATTERING 255 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. 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) (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 = 2 ≡α 3. (5.23) 4πϵ0 m r r 256 CHAPTER 5. PLANAR MOTION 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 [40] α θ b(θ) = cot . (5.24) v2 2 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 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 4πϵ0 m= 1 (see code in the ﬁle rk2_cb.cpp) 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. 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 5.4. SCATTERING 257 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 4πϵ0 m= 1 (see ﬁle rk2_cb.cpp) 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. 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 proportionality constant and δb cancel) and the solid angle 2π sin(θ) δθ. 258 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. 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.cpp and it is a simple modiﬁcation of the program in rk2.cpp: / / ======================================================== / / 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 / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 1010000; double T [ P ] , X1 [ P ] , X2 [ P ] , V1 [ P ] , V2 [ P ] ; 5.4. SCATTERING 259 100 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. double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f1 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double f2 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) ; void RK ( c o n s t double& Ti , c o n s t double& Tf , 260 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). c o n s t double& X10 , c o n s t double& X20 , c o n s t double& V10 , c o n s t double& V20 , const int & Nt ) ; void RKSTEP ( double& t , double& x1 , double& x2 , double& x3 , double& x4 , const double& dt ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double Ti , Tf , X10 , X20 , V10 , V20 ; double X20F , dX20 ; / / max impact parameter and s t e p int Nt , i ; c o n s t i n t Nbins =20; i n t index ; double angle , bins [ Nbins ] , Npart ; c o n s t double PI =3.14159265358979324; c o n s t double rad2deg =180.0/ PI ; c o n s t double dangle =PI / Nbins ; double R , density , dOmega , sigma , sigmatot ; 5.4. SCATTERING 261 / / Input : cout << ”Runge−Kutta Method f o r 4−ODEs I n t e g r a t i o n \n” ; cout << ” Enter c o u p l i n g c o n s t a n t s : \ n” ; cin >> k1 >> k2 ; getline ( cin , buf ) ; cout << ” k1= ” << k1 << ” k2= ” << k2 << endl ; cout << ” Enter Nt , Ti , Tf , X10 , X20 , V10 , V20: \ n” ; cin >> Nt >> Ti >> Tf>> X10 >> X20 >> V10 >> V20 ; getline ( cin , buf ) ; cout << ” Enter f i n a l impact parameter X20F and s t e p dX20: \ n” ; cin >> X20F >> dX20 ; cout << ” Nt = ” << Nt << endl ; cout << ”Time : I n i t i a l Ti = ” << Ti << ” F i n a l Tf= ” << Tf << endl ; cout << ” X1 ( Ti )= ” << X10 << ” X2( Ti )=” << X20 << endl ; cout << ” V1 ( Ti )= ” << V10 << ” V2( Ti )=” << V20 << endl ; cout << ” Impact par X20F =” << X20F << ” dX20 =” << dX20 << endl ; ofstream myfile ( ” s c a t t e r . dat ” ) ; myfile . precision ( 1 7 ) ; f o r ( i =0;i<Nbins ; i++) bins [ i ] = 0 . 0 ; / / Calculate : Npart = 0 . 0 ; X20 = X20 + dX20 / 2 . 0 ; / / s t a r t s i n middle o f f i r s t i n t e r v a l while ( X20 < X20F ) { RK ( 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 = abs ( atan2 ( V2 [ Nt −1] , V1 [ Nt −1]) ) ; / / Output : The f i n a l a n g l e . Check i f almost c o n s t a n t myfile << ”@ ” << X20 << ” ” << angle << ” ” << abs ( atan2 ( V2 [ Nt −51] , V1 [ Nt −51]) ) << ” ” << k1 / ( V10 * V10 ) / tan ( angle / 2 . 0 ) << endl ; / / Update histogram : index = i n t ( angle / dangle ) ; / / Number o f incoming p a r t i c l e s per u n i t time / / i s proportional to radius of ring / / o f r a d i u s X20 , t h e impact parameter : bins [ index ] += X20 ; / / db i s c a n c e l l e d from d e n s i t y Npart += X20 ; / / <−− i . e . from here X20 += dX20 ; } / / while ( X20 < X20F ) / / 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 ; / / t o t a l cross section f o r ( i =0;i<Nbins ; i++){ angle = ( i +0.5) * dangle ; dOmega = 2 . 0 * PI * sin ( angle ) * dangle ; / / d ( S o l i d Angle ) 262 CHAPTER 5. PLANAR MOTION sigma = bins [ i ] / ( density * dOmega ) ; i f ( sigma >0.0) myfile << ” ds= ” << angle << ” ” << angle * rad2deg << ” ” << sigma << endl ; sigmatot += sigma * dOmega ; } / / f o r ( i =0; i <Nbins ; i ++) myfile << ” s i g m a t o t= ” << sigmatot << endl ; myfile . close ( ) ; } / / main ( ) The results are recorded in the ﬁle scatter.dat. An example session that reproduces ﬁgures 5.11 and 5.12 is > g++ scatter . cpp rk2_cb . cpp −o scatter > . / scatter Runge−Kutta Method for 4−ODEs Integration Enter coupling constants : 1 . 0 0.0 k1= 1 k2= 0 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 Ti = 0 Final Tf= 30 X1 ( Ti )= −50 X2 ( Ti ) =0.02 V1 ( Ti )= 3 V2 ( Ti ) =0 Impact par X20F =1 dX20 =0.0002 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. SCATTERING 263 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.cpp: / / ======================================================== / / 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 / / ======================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / 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 double f3 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r2 , r , fr ; r2=x1 * x1+x2 * x2 ; r =sqrt ( r2 ) ; i f ( r <= k1 && r2 > 0 . 0 ) fr = 1 . 0 / r2−r / ( k1 * k1 * k1 ) ; else fr = 0 . 0 ; 264 CHAPTER 5. PLANAR MOTION i f ( fr > 0.0 && r > 0 . 0 ) r e t u r n fr * x1 / r ; / / dx3 / dt=dv1 / dt=a1 else return 0.0; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double f4 ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r2 , r , fr ; r2=x1 * x1+x2 * x2 ; r =sqrt ( r2 ) ; i f ( r <= k1 && r2 > 0 . 0 ) fr = 1 . 0 / r2−r / ( k1 * k1 * k1 ) ; else fr = 0 . 0 ; i f ( fr > 0.0 && r > 0 . 0 ) r e t u r n fr * x2 / r ; / / dx4 / dt=dv2 / dt=a2 else return 0.0; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , c o n s t double& x1 , c o n s t double& x2 , c o n s t double& v1 , c o n s t double& v2 ) { double r , Vr ; r=sqrt ( x1 * x1+x2 * x2 ) ; i f ( r <= k1 && r > 0 . 0 ) Vr = 1 / r + 0 . 5 * r * r / ( k1 * k1 * k1 ) − 1 . 5 / k1 ; else Vr = 0 . 0 ; r e t u r n 0 . 5 * ( v1 * v1+v2 * v2 ) + Vr ; } The results are shown in ﬁgures 5.13–5.14. We ﬁnd that σtot = πa2 (see problem 5.10). 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 5.5. MORE PARTICLES 265 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. plasma. The resulting force is e−r/a ( r) F⃗ (r) = f (r) r̂ , f (r) = k 1 + (5.29) r2 a The program of the resulting acceleration can be found in the ﬁle rk2_yu.cpp. 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. 266 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. 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.cpp the integer parameter P had a given value which determined the size of all arrays using the declarations: c o n s t i n t P = 1010000; double T [ P ] , X1 [ P ] , X2 [ P ] , V1 [ P ] , V2 [ P ] ; 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. The needed memory can be asked for at execution time by using the new operator. Here is an example: double * T ; / / D e c l a r e 1−Dim a r r a y s as p o i n t e r s double * * X ; / / D e c l a r e 2−Dim a r r a y s as p o i n t e r s t o p o i n t e r s i n t NEQ , Nt ; / / V a r i a b l e s o f a r r a y s i z e s / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 5.5. MORE PARTICLES 267 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. finit ( NEQ ) ; / / f u n c t i o n t h a t s e t s NEQ a t run time cin >> Nt ; / / read Nt a t run time / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / a l l o c a t e s 1−Dim a r r a y o f Nt doubles T = new double [ Nt ] ; / / a l l o c a t e s 1−Dim a r r a y o f Nt p o i n t e r s t o doubles X = new double * [ Nt ] ; / / f o r each i , a l l o c a t e an a r r a y o f NEQ doubles a t X[ i ] : f o r ( i n t i =0;i<Nt ; i++) X [ i ] = new double [ NEQ ] ; ..... ( compute with T [ Nt ] , X [ Nt ] [ NEQ ] ) ..... / / r e t u r n a l l o c a t e d memory back t o t h e system : d e l e t e [ ] T ; / / d e a l l o c a t e an a r r a y / / f o r each i , d e a l l o c a t e t h e a r r a y o f doubles X[ i ] [NEQ] f o r ( i n t i =0;i<NEQ ; i++) d e l e t e [ ] X [ i ] ; / / and then d e a l l o c a t e t h e a r r a y o f p o i n t e r s t o doubles X[ Nt ] : delete [] X ; ..................... void finit ( i n t& NEQ ) { / / f u n c t i o n t h a t s e t s v a l u e o f NEQ NEQ = 4 ; 268 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. } In this program, we should remember the fact that in C++, the name T of an array T[Nt] is a pointer to T[0], which is denoted by &T[0]. This is the address of the ﬁrst element of the array in the memory. Therefore, if the array is double T[Nt], then T is a pointer to a double: double *T. Then T+i is a pointer to the address of the (i+1)-th element of the array T[i] and T+i = &T [ i ] as well as T [ i ] = * ( T+i ) We can use pointers with the same notation as we do with arrays: If we declare a pointer to a double *T1, and assign T1=T, then T1[0], T1[1], ... are the same as T[0], T[1], .... For example, T1[0] = 2.0 assigns also T[0] and vice-versa. Conversely, we can declare a pointer to 5.5. MORE PARTICLES 269 a double *T, as we do in our program, and make it point to a region in the memory where we have reserved space for Nt doubles. This is what the operator new does. It asks for the memory for Nt doubles and returns a pointer to it. Then we assign this pointer to T: double * T ; T = new double [ Nt ] ; Then we can use T[0], T[1], ..., T[Nt-1] as we do with ordinary arrays. Two dimensional arrays are slightly trickier: For a two dimensional array, double X[Nt][NEQ], X is a pointer to the value X[0][0], which is &X[0][0]. Then X[i] is a pointer to the one dimensional array X[i][NEQ], therefore X is a pointer to a pointer of a double! X [ i ] [ jeq ] is : double X[i] is : double * X is : double * * Conversely, we can declare a double **X, and use the operator new to return a pointer to an array of Nt pointers to doubles, and then for each element of the array, use new to return a pointer to NEQ doubles: double * * X X = new double * [ Nt ] ; X[0] = new double [ NEQ ] ; X[1] = new double [ NEQ ] ; ... X [ Nt −1] = new double [ NEQ ] ; Then we can use the notation T[i][jeq], i= 0, 1, ..., Nt-1 and jeq= 0, 1, ..., NEQ-1, as we do with statically deﬁned arrays. The memory that we ask to be allocated dynamically is a ﬁnite re- source that can easily be exhausted (heard of memory leaks?). Therefore, we should be careful to return unused memory to the system, so that it can be recycled. This should happen especially within functions that we call many times, which allocate large memory dynamically. The operator delete can be used to deallocate memory that has been allocated with the operator new. For one dimensional arrays, this is particularly simple: double * T ; T = new double [ Nt ] ; 270 CHAPTER 5. PLANAR MOTION .... ( use T [ i ] ) ... delete [] T; ... ( cannot use T after d e l e t e ) For “two dimensional arrays” that have been allocated as we described above, ﬁrst we have to delete the arrays pointed by X[i] for i=0, ... , Nt-1, and then the arrays of pointers pointed by X: X = new double * [ Nt ] ; f o r ( i n t i =0;i<Nt ; i++) X [ i ] = new double [ NEQ ] ; . . . . use X [ i ] [ jeq ] . . . . f o r ( i n t i =0;i<NEQ ; i++) d e l e t e [ ] X [ i ] ; / / d e l e t e a l l a r r a y s X[ i ] delete [] X ; / / d e l e t e array of pointers The main program will be written in the ﬁle rkA.cpp, whereas the force-dependent part of the code will be written in ﬁles with names of the form rkA_XXX.cpp. In the latter, the user must program a function f(t,X,dXdt) which takes as input the time t and the values of the func- tions 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.cpp 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 f u n c t i o n s : / / f ( t , x , xdot ) : with double t , x [NEQ] , xdot [NEQ] which / / gi ve n 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 / / 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 : / / double k1 , k2 : c o u p l i n g c o n s t a n t s i n g l o b a l scope / / 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 / / double X0[NEQ] : i n i t i a l c o n d i t i o n s / / Output : / / 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] / / ======================================================== # i n c l u d e < iostream > 5.5. MORE PARTICLES 271 # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double * T ; / / T[ Nt ] s t o r e s the values of times double * * X ; / / X[ Nt ] [NEQ] s t o r e s t h e v a l u e s o f f u n c t i o n s double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void RK ( c o n s t double& Ti , c o n s t double& Tf , double * X0 , const int & Nt , c o n s t i n t & NEQ ) ; void RKSTEP ( double& t , double * x , c o n s t double& dt , c o n s t i n t & NEQ ) ; / / The f o l l o w i n g f u n c t i o n s a r e d e f i n e d i n rkA_XX . cpp : void finit ( int & NEQ ) ; / / S e t s number o f e q u a t i o n s / / D e r i v a t i v e and energy f u n c t i o n s : void f ( c o n s t double& t , double * X , double * dXdt ) ; double energy ( c o n s t double& t , double * X ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; int NEQ , Nt ; double * X0 ; double Ti , Tf ; / / Get number o f e q u a t i o n s and a l l o c a t e memory f o r X0 : finit ( NEQ ) ; X0 = new double [ NEQ ] ; / / Input : cout << ”Runge−Kutta Method f o r ODE I n t e g r a t i o n . \ n” ; cout << ”NEQ= ” << NEQ << endl ; cout << ” Enter c o u p l i n g c o n s t a n t s : \ n” ; cin >> k1 >> k2 ; getline ( cin , buf ) ; cout << ” k1= ” << k1 << ” k2= ” << k2 << endl ; cout << ” Enter Nt , Ti , Tf , X0: \ n” ; cin >> Nt>>Ti>>Tf ; f o r ( i n t i =0;i<NEQ ; i++) cin >> X0 [ i ] ; getline ( cin , buf ) ; cout << ” Nt = ” << Nt << endl ; cout << ”Time : I n i t i a l Ti =” << Ti << ” ” << ” F i n a l Tf=” << Tf << endl ; cout << ” X0 =” ; f o r ( i n t i =0;i<NEQ ; i++) cout << X0 [ i ] << ” ” ; cout << endl ; / / A l l o c a t e memory f o r data a r r a y s : T = new double [ Nt ] ; X = new double * [ Nt ] ; f o r ( i n t i =0;i<Nt ; i++) X [ i ] = new double [ NEQ ] ; / / The C a l c u l a t i o n : RK ( Ti , Tf , X0 , Nt , NEQ ) ; 272 CHAPTER 5. PLANAR MOTION / / Output : ofstream myfile ( ”rkA . dat ” ) ; myfile . precision ( 1 6 ) ; f o r ( i n t i =0;i<Nt ; i++){ myfile << T [ i ] << ” ” ; f o r ( i n t jeq =0; jeq<NEQ ; jeq++) myfile << X [ i ] [ jeq ] << ” ” ; myfile << energy ( T [ i ] , X [ i ] ) << ’\n ’ ; } myfile . close ( ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / C l e a n i n g up dynamic memory : d e l e t e [ ] each a r r a y / / c r e a t e d with t h e new o p e r a t o r ( Not n e c e s s a r y i n t h i s / / program , i t i s done a t t h e end o f t h e program anyway ) d e l e t e [ ] X0 ; delete [] T ; f o r ( i n t i =0;i<NEQ ; i++) d e l e t e [ ] X [ i ] ; delete [] X ; } / / main ( ) / / ======================================================== / / D r i v e r o f t h e RKSTEP r o u t i n e / / ======================================================== void RK ( c o n s t double& Ti , c o n s t double& Tf , double * X0 , const in t & Nt , c o n s t i n t & NEQ ) { double dt ; double TS ; double * XS ; XS = new double [ NEQ ] ; / / I n i t i a l i z e variables : dt = ( Tf−Ti ) / ( Nt −1) ; T [ 0 ] = Ti ; f o r ( i n t ieq =0; ieq<NEQ ; ieq++) X [ 0 ] [ ieq ]= X0 [ ieq ] ; TS = Ti ; f o r ( i n t ieq =0; ieq<NEQ ; ieq++) XS [ ieq ]= X0 [ ieq ] ; / / 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 f o r ( i n t i =1; i<Nt ; i++){ RKSTEP ( TS , XS , dt , NEQ ) ; T [ i ] = TS ; f o r ( i n t ieq =0; ieq<NEQ ; ieq++) X [ i ] [ ieq ] = XS [ ieq ] ; } / / Clean up memory : d e l e t e [ ] XS ; } / / RK( ) / / ======================================================== / / Function 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 / / ======================================================== 5.5. MORE PARTICLES 273 void RKSTEP ( double& t , double * x , c o n s t double& dt , const int & NEQ ) { double tt ; double * k1 , * k2 , * k3 , * k4 , * xx ; double h , h2 , h6 ; k1 = new double [ NEQ ] ; k2 = new double [ NEQ ] ; k3 = new double [ NEQ ] ; k4 = new double [ NEQ ] ; xx = new double [ NEQ ] ; h =dt ; / / h =dt , i n t e g r a t i o n s t e p h2 =0.5* h ; / / h2=h / 2 h6=h / 6 . 0 ; / / h6=h / 6 / / 1 st step : f ( t , x , k1 ) ; / / 2nd s t e p : f o r ( i n t ieq =0; ieq<NEQ ; ieq++) xx [ ieq ] = x [ ieq ] + h2 * k1 [ ieq ] ; tt =t+h2 ; f ( tt , xx , k2 ) ; / / 3rd s t e p : f o r ( i n t ieq =0; ieq<NEQ ; ieq++) xx [ ieq ] = x [ ieq ] + h2 * k2 [ ieq ] ; tt =t+h2 ; f ( tt , xx , k3 ) ; / / 4 th s t e p : f o r ( i n t ieq =0; ieq<NEQ ; ieq++) xx [ ieq ] = x [ ieq ] + h * k3 [ ieq ] ; tt=t+h ; f ( tt , xx , k4 ) ; / / Update : t += h ; f o r ( i n t ieq =0; ieq<NEQ ; ieq++) x [ ieq ] += h6 * ( k1 [ ieq ] + 2 . 0 * ( k2 [ ieq ]+ k3 [ ieq ] ) +k4 [ ieq ] ) ; / / Clean up memory : d e l e t e [ ] k1 ; d e l e t e [ ] k2 ; d e l e t e [ ] k3 ; d e l e t e [ ] k4 ; d e l e t e [ ] xx ; } / / RKSTEP ( ) Consider three particles of equal mass exerting a force of gravitational attraction on each other⁴ like the ones shown in ﬁgure 5.17. The forces ⁴The same program can be used for three equal charges exerting an electrostatic force on each other, which can be either attractive or repulsive. 274 CHAPTER 5. PLANAR MOTION 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.cpp, rkA_3pcb.cpp. 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. 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 , (5.31) dt dt j=1,j̸=i rij3 where rij2 = (xi − xj )2 + (yi − yj )2 . The total energy of the system is 1 ∑3 k1 E/m = (v12 + v22 ) + . (5.32) 2 i,j=1,j<i rij The relations shown above are programmed in the ﬁle rkA_3pcb.cpp listed below: 5.5. MORE PARTICLES 275 # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e x t e r n double k1 , k2 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / S e t s number o f e q u a t i o n s : void finit ( i n t& NEQ ) { NEQ = 1 2 ; } / / =============================== / / 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 / / =============================== void f ( c o n s t double& t , double * X , double * dXdt ) { double x11 , x12 , x21 , x22 , x31 , x32 ; double v11 , v12 , v21 , v22 , v31 , v32 ; double r12 , r13 , r23 ; / /−−−−−−−−−−−−−−−− x11 = X [ 0 ] ; x21 = X [ 4 ] ; x31 = X [ 8 ] ; x12 = X [ 1 ] ; x22 = X [ 5 ] ; x32 = X [ 9 ] ; v11 = X [ 2 ] ; v21 = X [ 6 ] ; v31 = X [ 1 0 ] ; v12 = X [ 3 ] ; v22 = X [ 7 ] ; v32 = X [ 1 1 ] ; / /−−−−−−−−−−−−−−−− r12 = pow ( ( x11−x21 ) * ( x11−x21 ) +( x12−x22 ) * ( x12−x22 ) , − 3 . 0 / 2 . 0 ) ; r13 = pow ( ( x11−x31 ) * ( x11−x31 ) +( x12−x32 ) * ( x12−x32 ) , − 3 . 0 / 2 . 0 ) ; r23 = pow ( ( x21−x31 ) * ( x21−x31 ) +( x22−x32 ) * ( x22−x32 ) , − 3 . 0 / 2 . 0 ) ; / /−−−−−−−−−−−−−−−− dXdt [ 0 ] = v11 ; dXdt [ 1 ] = v12 ; dXdt [ 2 ] = k1 * ( x11−x21 ) * r12+k1 * ( x11−x31 ) * r13 ; / / a11=dv11 / dt dXdt [ 3 ] = k1 * ( x12−x22 ) * r12+k1 * ( x12−x32 ) * r13 ; / / a12=dv12 / dt / /−−−−−−−−−−−−−−−− dXdt [ 4 ] = v21 ; dXdt [ 5 ] = v22 ; dXdt [ 6 ] = k1 * ( x21−x11 ) * r12+k1 * ( x21−x31 ) * r23 ; / / a21=dv21 / dt dXdt [ 7 ] = k1 * ( x22−x12 ) * r12+k1 * ( x22−x32 ) * r23 ; / / a22=dv22 / dt / /−−−−−−−−−−−−−−−− dXdt [ 8 ] = v31 ; dXdt [ 9 ] = v32 ; dXdt [ 1 0 ] = k1 * ( x31−x11 ) * r13+k1 * ( x31−x21 ) * r23 ; / / a31=dv31 / dt dXdt [ 1 1 ] = k1 * ( x32−x12 ) * r13+k1 * ( x32−x22 ) * r23 ; / / a32=dv32 / dt } / / =============================== double energy ( c o n s t double& t , double * X ) { 276 CHAPTER 5. PLANAR MOTION double x11 , x12 , x21 , x22 , x31 , x32 ; double v11 , v12 , v21 , v22 , v31 , v32 ; double r12 , r13 , r23 ; double e ; / /−−−−−−−−−−−−−−−− x11 = X [ 0 ] ; x21 = X [ 4 ] ; x31 = X [ 8 ] ; x12 = X [ 1 ] ; x22 = X [ 5 ] ; x32 = X [ 9 ] ; v11 = X [ 2 ] ; v21 = X [ 6 ] ; v31 = X [ 1 0 ] ; v12 = X [ 3 ] ; v22 = X [ 7 ] ; v32 = X [ 1 1 ] ; / /−−−−−−−−−−−−−−−− r12 = pow ( ( x11−x21 ) * ( x11−x21 ) +( x12−x22 ) * ( x12−x22 ) , −0.5) ; r13 = pow ( ( x11−x31 ) * ( x11−x31 ) +( x12−x32 ) * ( x12−x32 ) , −0.5) ; r23 = pow ( ( x21−x31 ) * ( x21−x31 ) +( x22−x32 ) * ( x22−x32 ) , −0.5) ; / /−−−−−−−−−−−−−−−− e = 0 . 5 * ( v11 * v11+v12 * v12+v21 * v21+v22 * v22+v31 * v31+v32 * v32 ) ; e += k1 * ( r12+r13+r23 ) ; return e ; } 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 the 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 277 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. 278 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 279 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.cpp. 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 280 CHAPTER 5. PLANAR MOTION trajectories for the initial conditions that you will choose. (Hint: Look in the ﬁles rkA_3hoc.cpp, 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.cpp, 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. Most of it is code written in Fortran and, in order to use it, you should learn how to link a program written in C++ with functions written in a diﬀerent programming language. The main technical skill that you will develop 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 281 282 CHAPTER 6. MOTION IN SPACE 6.1 Adaptive Stepsize Control for RK 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 [33]. 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. 6.1.1 The rksuite Suite of RK Codes 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 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 283 by R . W . Brankin ( NAG ) , I . Gladwell and L . F . Shampine ( SMU ) lang Fortran 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 download³ the ﬁles rksuite.f, rksuite.doc, details.doc, templates, readme. In order to link the subroutines provided by the suite to our pro- gram we need to read the documentation carefully. In the general case, documentation is available on the web (html, pdf, ...), bundled ﬁles with names like README and INSTALL, in whole directories with names like doc/, online help in man and/or info pages and, ﬁnally, in good old fash- ioned printed manuals. 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. ³For the convenience of the reader, these ﬁles can be found bundled in the accom- panied software in a subdirectory rksuite. 284 CHAPTER 6. MOTION IN SPACE It is easy to install the software. All the necessary code is in one ﬁle rksuite.f. The ﬁle rksuite.doc⁴ contains the documentation. There we read that we need to inform the program about the hardware dependent 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 . f file . ... 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 it⁵: ... SUBROUTINE ENVIRN ( OUTCH , MCHEPS , DWARF ) ... 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 ’ ⁴This is a simple text ﬁle which you can read with the command less rksuite.doc or with emacs. ⁵These are lines that begin with a C, as this is old ﬁxed format Fortran code. 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 285 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 C++ template numeric_limits, which is part of the C++ Standard Library. In the ﬁle numericLimits.cpp, we write a small test program⁶: # i n c l u d e < iostream > # include <limits > using namespace std ; i n t main ( ) { double MCHEPS , DWARF ; MCHEPS = numeric_limits <double > : : epsilon ( ) ; DWARF = numeric_limits <double > : : min () ; cout << ”MCHEPS = ” << MCHEPS / 2 . 0 << endl ; cout << ”DWARF = ” << DWARF << endl ; } We compile and run the above program as follows: > g++ numericLimits . cpp −o n > ./n MCHEPS = 1.11022 e−16 ⁶The ﬁle in the accompanying software, shows you how to compute numeric limits for several types of variables. 286 CHAPTER 6. MOTION IN SPACE DWARF = 2.22507 e−308 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. The code is in Fortran, but it is not so hard to read. You may compile it and run it with the commands: > cd rksuite / templates > g f o r t r a n tmpl1a . f . . / rksuite . f −o example1 > . / example1 We encourage the reader to study it carefully, run it and test its results. 6.1.2 Interfacing C++ Programs with Fortran Next, we have to learn how to link the Fortran code in rksuite.f to a C++ program. There is a lot of relevant information that you can ﬁnd with a simple search in the web, we will concentrate on what is relevant to the program that we need to write. The ﬁrst thing that we need to learn is how to call a function written in Fortran from a C++ program. A simple “Hello World” program can teach us how to do it. Suppose that a Fortran function/subroutine HELLO() is coded in a ﬁle helloF.f90: 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 287 SUBROUTINE HELLO ( ) PRINT * , ’ H e l l o World ! ’ END SUBROUTINE HELLO Then, we write in the ﬁle hello.cpp: # i n c l u d e < iostream > using namespace std ; e x t e r n ”C” void hello_ ( ) ; i n t main ( ) { hello_ ( ) ; } The ﬁrst thing that we notice is that we call the function by lowering all letters in its name: HELLO → hello. In Fortran, lowercase and uppercase letters are equivalent, and the compiler creates names with lowercase letters only. Next, we ﬁnd that we need to append an underscore _ to the function’s name⁷: hello → hello_. The Fortran function needs to be declared in the “"C" language linkage”: e x t e r n ”C” void hello_ ( ) ; This is something one has to do both for functions written in Fortran, as well as for functions written in plain old C⁸. In order to compile and run the code we have to run the commands: > g f o r t r a n −c helloF . f90 > g++ hello . cpp helloF . o −o hello −lgfortran > . / hello Hello World ! Compilation is done in two steps: We ﬁrst need to compile the Fortran program using the Fortran compiler⁹ gfortran. The ﬂag -c forces the ⁷Read your compiler’s manual, there could be options that you can use at compile time so that you can avoid it. But the advice is to stick with this convention, so that your code will be portable and ... long lived! ⁸Language linkage encapsulates the set of requirements necessary in order to link with a function written in another programming language, and it should be done for all function types, names and variable names. There could be linkage to other program- ming languages, but only the C and C++ linkage is guaranteed to be available. ⁹Available for free download from gcc.gnu.org/fortran/. On debian based Linux 288 CHAPTER 6. MOTION IN SPACE compiler to perform compilation but not linking. It produces an ob- ject ﬁle, whose extension is .o, helloF.o. These are ﬁles which contain compiled code in non-readable text form, but they are not autonomous executable programs. The functions that they contain, can be linked to other compiled programs that call them. In the second step, g++ is called to compile the C++ source code in hello.cpp and link it to the functions in helloF.o. The ﬂag -lgfortran is necessary in order to link to the standard Fortran functions. If you use a diﬀerent compiler, you should read its manual in order to ﬁnd the correct linking options. Another subtle point that needs to be considered is that, in Fortran, variables are passed to functions by reference and not by value. Therefore, we have to pass variables as pointers or as reference to variables. For example, the following Fortran function¹⁰ REAL ( 8 ) FUNCTION SQUAREMYDOUBLE ( X ) REAL( 8 ) x X = 2.0* X SQUAREMYDOUBLE = X * X END FUNCTION SQUAREMYDOUBLE must be declared and called as: e x t e r n ”C” double squaremydouble_ ( double& x ) ; ... double x , x2 ; x2 = squaremydouble_ ( x ) ; For example, by modifying the hello.cpp program as # i n c l u d e < iostream > using namespace std ; e x t e r n ”C” void hello_ ( ) ; e x t e r n ”C” double squaremydouble_ ( double& x ) ; i n t main ( ) { double x , x2 ; hello_ ( ) ; x = 2.0; x2 = squaremydouble_ ( x ) ; systems, install with sudo apt-get install gfortran. ¹⁰A Fortran function returns the value stored in a variable with the same name as the name of the function, here SQUAREMYDOUBLE. 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 289 cout << ” x = ” << 2.0 << endl ; cout << ”2 x = ” << x << endl ; cout << ” (2 x ) * ( 2 x ) = ” << x2 << endl ; } we obtain the output: > g f o r t r a n −c helloF . f90 ; > g++ hello . cpp helloF . o −o hello −lgfortran > . / hello Hello World ! x = 2 2 x = 4 (2 x ) * ( 2 x ) = 16 Notice that the value of x is modiﬁed in the calling program. The ﬁnal issue that we will consider, it how to pass arrays to Fortran functions. One dimensional arrays are quite easy to handle. In order to pass an array double v[N] to a Fortran function, we only need to declare it and pass it as one of its arguments. If the Fortran program is r e a l ( 8 ) f u n c t i o n make_array1 ( v , N ) integer N real (8) v(N) . . . compute v ( 1 ) . . . v ( N ) . . . end f u n c t i o n make_array1 then the corresponding C++ program should be: c o n s t i n t N =4; e x t e r n ”C” double make_array1_ ( double v [ N ] , c o n s t i n t& N ) ; i n t main ( ) { double v [ N ] , x ; ... x = make_array1_ ( v , N ) ; . . . use v [ 0 ] . . . v [ N−1] . . . } The only point we need to stress is that the array real(8) v(N) is indexed from 1 to N in the Fortran program, whereas the array double v[N] is indexed from 0 to N-1 in the C++ program. The correspondence of the values stored in memory is v(1) → v[0], ... , v(i) → v[i-1], ... v(N) → v[N-1]. Two dimensional arrays need more attention. In C++, arrays are in row-major mode, whereas in Fortran in column-major mode. The contents 290 CHAPTER 6. MOTION IN SPACE of a two dimensional array are stored linearly in memory. In C++, the elements of an array double A[N][M] are stored in the sequence A[0][0], A[0][1], A[0][2], ... , A[0][M-1], A[1][0], A[1][1], ... , A[1][M-1], ... A[N-1][M-1]. In Fortran, the elements of an array real(8) A(N,M) are stored in the sequence A(1,1), A(2,1), A(3,1), ... , A(1,M), A(2,1), A(2,2), ... , A(2,M), ... , A(N,M). Therefore, when we pass an array from the C++ program to a Fortran function, we have to keep in mind that the Fortran function will use it with its indices transposed. For example, if the Fortran code deﬁnes A(i , j) = i + j/10.0 which results into A(i,j) = i.j in decimal notation¹¹ (e.g. A(2,3)= 2.3), then the value of A[i][j] will be (j+1).(i+1) (e.g. A[2][3] = 4.3, A[2][1] = 2.3)! All of the above are summarized in the Fortran program in the ﬁle CandFortranF.f90: r e a l ( 8 ) f u n c t i o n make_array1 ( v , N ) i m p l i c i t none integer N , i real (8) v(N) do i =1 , N v(i) = i end do make_array1 = −11.0 ! a r e t u r n v a l u e end f u n c t i o n make_array1 !−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− r e a l ( 8 ) f u n c t i o n make_array2 ( A , N , M ) i m p l i c i t none integer N , M , i , j r e a l ( 8 ) A ( M , N ) ! C a r e f u l : N and M a r e i n t e r c h a n g e d ! ! do i =1 , M do j =1 , N !A( i , j ) = i . j , e . g . A( 2 , 3 ) =2.3 A(i , j) = i + j/10.0 end do end do ¹¹Of course, we assume i,j<10. 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 291 make_array2 = −22.0 ! a r e t u r n v a l u e end f u n c t i o n make_array2 and the C++ program in the ﬁle CandFortranC.cpp: # include < iostream > # include < fstream > # include <cstdlib > # include <string > # include <cmath> using namespace std ; c o n s t i n t N =4 , M =3; e x t e r n ”C” { double make_array1_ ( double v [ N ] , c o n s t i n t& N ) ; double make_array2_ ( double A [ N ] [ M ] , c o n s t i n t& N , c o n s t i n t& M ) ; } i n t main ( ) { double A [ N ] [ M ] , v [ N ] ; double x ; / / Make a 1D a r r a y using a f o r t r a n f u n c t i o n : x = make_array1_ ( v , N ) ; cout << ” 1D a r r a y : Return v a l u e x= ” << x << endl ; f o r ( i n t i =0;i<N ; i++) cout << v [ i ] << ” ” ; cout << ”\n−−−−−−−−−−−−−−−−−−−−−−−−−\n” ; / / Make an 2D a r r a y using a f o r t r a n f u n c t i o n : x = make_array2_ ( A , N , M ) ; cout << ”2D a r r a y : Return v a l u e x= ” << x << endl ; f o r ( i n t i =0;i<N ; i++){ f o r ( i n t j =0;j<M ; j++) / /A i s . . . transposed ! / / A[ i ] [ j ] = ( j +1) . ( i +1) , e . g . A[ 1 ] [ 2 ] = 3 . 1 cout << A [ i ] [ j ] << ” ” ; cout << ’\n ’ ; } } Note that the array A[N][M] is deﬁned as A(M,N) in the Fortran function, and the roles of N and M are interchanged. You can run the code and see the output with the commands: > g f o r t r a n −c CandFortranF . f90 > g++ CandFortranC . cpp CandFortranF . o −o CandFortran −lgfortran 292 CHAPTER 6. MOTION IN SPACE > . / CandFortran 1D array : Return value x= −11 1 2 3 4 −−−−−−−−−−−−−−−−−−−−−−−−− 2D array : Return value x= −22 1.1 2.1 3.1 1 . 2 2.2 3.2 1 . 3 2.3 3.3 1 . 4 2.4 3.4 Note that the values of the array A(M,N) are transposed when printed as rows in the C++ program. 6.1.3 The rksuite Driver After we become wise enough, we can write the driver for the integration routine UT (called by ut_ from our C++ program), which can be found in the ﬁle rk3.cpp: / / ======================================================== / / 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 / / ======================================================== / / Compile with t h e commands : / / g f o r t r a n −c r k s u i t e / r k s u i t e . f ; / / g++ rk3 . cpp rk3_g . cpp r k s u i t e . o −o rk3 −l g f o r t r a n # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> # i n c l u d e ” rk3 . h” using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double k1 , k2 , k3 , k4 ; double energy ( c o n s t double& t , double * Y ) ; void f ( double& t , double * Y , double * YP ) ; e x t e r n ”C” { void setup_ ( c o n s t i n t& NEQ , double& TSTART , double * YSTART , double& TEND , double& TOL , double * THRES , c o n s t i n t& METHOD , c o n s t char& TASK , bool & ERRASS , double& HSTART , double * WORK , c o n s t i n t& LENWRK , bool& MESSAGE ) ; void ut_ ( void f ( double& t , double * Y , double * YP ) , double& TWANT , double& TGOT , double * YGOT , double * YPGOT , double * YMAX , double * WORK , 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 293 i n t& UFLAG ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double T0 , TF , X10 , X20 , X30 , V10 , V20 , V30 ; double t , dt , tstep ; int STEPS , i ; / / rksuite variables : double TOL , THRES [ NEQ ] , WORK [ LENWRK ] , HSTART ; double Y [ NEQ ] , YMAX [ NEQ ] , YP [ NEQ ] , YSTART [ NEQ ] ; bool ERRASS , MESSAGE ; int UFLAG ; c o n s t char TASK = ’U’ ; / / Input : cout << ”Runge−Kutta Method f o r 6−ODEs I n t e g r a t i o n \n” ; cout << ” Enter c o u p l i n g c o n s t a n t s k1 , k2 , k3 , k4 : \ n” ; cin >> k1 >> k2 >> k3 >> k4 ; getline ( cin , buf ) ; cout << ” Enter STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30: \ n” ; cin >> STEPS >> T0 >> TF >> X10 >> X20 >> X30 >> V10 >> V20 >> V30 ; getline ( cin , buf ) ; cout << ”No . S t e p s= ” << STEPS << endl ; cout << ”Time : I n i t i a l T0 =” << T0 << ” F i n a l TF=” << TF << endl ; cout << ” X1 (T0)=” << X10 << ” X2(T0)=” << X20 << ” X3(T0)=” << X30 << endl ; cout << ” V1 (T0)=” << V10 << ” V2(T0)=” << V20 << ” V3(T0)=” << V30 << endl ; / / I n i t i a l Conditions : dt = ( TF−T0 ) / STEPS ; YSTART [ 0 ] = X10 ; YSTART [ 1 ] = X20 ; YSTART [ 2 ] = X30 ; YSTART [ 3 ] = V10 ; YSTART [ 4 ] = V20 ; YSTART [ 5 ] = V30 ; / / S e t c o n t r o l parameters : TOL = 5.0 e−6; f o r ( i = 0 ; i < NEQ ; i++) THRES [ i ] = 1 . 0 e −10; MESSAGE = t r u e ; ERRASS = f a l s e ; HSTART = 0 . 0 ; // Initialization : setup_ ( NEQ , T0 , YSTART , TF , TOL , THRES , METHOD , TASK , ERRASS , HSTART , WORK , LENWRK , MESSAGE ) ; 294 CHAPTER 6. MOTION IN SPACE ofstream myfile ( ” rk3 . dat ” ) ; myfile . precision ( 1 6 ) ; myfile << T0<< ” ” << YSTART [ 0 ] << ” ” << YSTART [ 1 ] << ” ” << YSTART [ 2 ] << ” ” << YSTART [ 3 ] << ” ” << YSTART [ 4 ] << ” ” << YSTART [ 5 ] << ” ” << energy ( T0 , YSTART ) << ’\n ’ ; / / The c a l c u l a t i o n : f o r ( i =1; i<=STEPS ; i++){ t = T0 + i * dt ; ut_ ( f , t , tstep , Y , YP , YMAX , WORK , UFLAG ) ; i f ( UFLAG > 2) break ; / / e r r o r : break t h e loop and e x i t myfile << tstep << ” ” << Y [ 0 ] << ” ” << Y [ 1 ] << ” ” << Y [ 2 ] << ” ” << Y [ 3 ] << ” ” << Y [ 4 ] << ” ” << Y [ 5 ] << ” ” << energy ( T0 , Y ) << ’\n ’ ; } myfile . close ( ) ; } / / main ( ) All common parameters and variables are declared in an include ﬁle rk3.h. This is done so that they are accessible by the function f which calculates the derivatives: c o n s t i n t NEQ = 6; c o n s t i n t LENWRK = 32* NEQ ; c o n s t i n t METHOD = 2 ; e x t e r n double 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 declaration of functions needs some care. The functions energy() and f() are deﬁned in the C++ program and are declared in the global scope (the function f() will also be passed on to the Fortran function UT). The functions setup_() and ut_() are deﬁned in the Fortran program in the ﬁle rksuite.f as SETUP() and UT(). Therefore, they are declared within the extern "C" language linkage, deﬁned using lowercase letters and an underscore is appended to their names. All arguments are passed by reference. Scalar doubles are passed as references to double&, the double precision arrays are declared to be pointers double *, the Fortran logical variables are declared to be references to bool&, and the Fortran 6.1. ADAPTIVE STEPSIZE CONTROL FOR RK METHODS 295 character* variables¹² as simple references to char&. The main program starts with the user interface. The initial state of the particle is stored in the array YSTART in the positions 0 . . . 5. 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: f o r ( i =1; i<=STEPS ; i++){ t = T0 + i * dt ; ut_ ( f , t , tstep , Y , YP , YMAX , WORK , UFLAG ) ; i f ( UFLAG > 2) break ; / / e r r o r : break t h e loop and e x i t myfile << . . . << energy ( T0 , Y ) << ’\n ’ ; } The function f 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 Fortran subroutine UT. The actual value of time that the next step lands¹³ on is tstep. The array Y stores the values of the functions. We choose the data structure to be such that x= Y[0], y= Y[1], z= Y[2] and vx = Y[3], vy = Y[4], vz = Y([5] (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.cpp. We choose the parameters k1 and k2 so that ⃗g = -k1 k̂ and k = k2. # include < iostream > # include < fstream > # include <cstdlib > # include <string > # include <cmath> # include ” rk3 . h” ¹²The Fortran program uses only their ﬁrst character, so we don’t need to use strings. ¹³When UGLAG ≤ 2, tstep=t and we will not worry about them being diﬀerent with our program. 296 CHAPTER 6. MOTION IN SPACE using namespace std ; void f ( double& t , double * Y , double * YP ) { double x1 , x2 , x3 , v1 , v2 , v3 ; x1 = Y [ 0 ] ; v1 = Y [ 3 ] ; x2 = Y [ 1 ] ; v2 = Y [ 4 ] ; x3 = Y [ 2 ] ; v3 = Y [ 5 ] ; // Velocities : d x _ i / dt = v _ i YP [ 0 ] = v1 ; YP [ 1 ] = v2 ; YP [ 2 ] = v3 ; / / A c c e l e r a t i o n : dv_i / dt = a _ i YP [ 3 ] = −k2 * v1 ; YP [ 4 ] = −k2 * v2 ; YP [ 5 ] = −k2 * v3−k1 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , double * Y ) { double e ; double x1 , x2 , x3 , v1 , v2 , v3 ; x1 = Y [ 0 ] ; v1 = Y [ 3 ] ; x2 = Y [ 1 ] ; v2 = Y [ 4 ] ; x3 = Y [ 2 ] ; v3 = Y [ 5 ] ; / / Kinetic Energy : e = 0 . 5 * ( v1 * v1+v2 * v2+v3 * v3 ) ; / / P o t e n t i a l Energy : e += k1 * x3 ; return e ; } For convenience we “translated” the values in the array Y[NEQ] into user- friendly variable names. 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 −c rksuite / rksuite . f > g++ rk3 . cpp rk3_g . cpp rksuite . o −o rk3 −lgfortran > . / rk3 Runge−Kutta Method for 6−ODEs Integration Enter coupling constants k1 , k2 , k3 , k4 : 10 0 0 0 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 Final TF=3 X1 ( T0 ) =0 X2 ( T0 ) =0 X3 ( T0 ) =0 V1 ( T0 ) =1 V2 ( T0 ) =1 V3 ( T0 ) =1 6.2. MOTION OF A PARTICLE IN AN EM FIELD 297 > 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 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.cpp. 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 / / ======================================================== # i n c l u d e ” s r . h” void f ( double& t , double * Y , double * YP ) { double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; 298 CHAPTER 6. MOTION IN SPACE x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; velocity ( p1 , p2 , p3 , v1 , v2 , v3 ) ; / / now we can use a l l x1 , x2 , x3 , p1 , p2 , p3 , v1 , v2 , v3 YP [ 0 ] = v1 ; YP [ 1 ] = v2 ; YP [ 2 ] = v3 ; / / Acceleration : YP [ 3 ] = k2 + k1 * v2 ; YP [ 4 ] = k3 − k1 * v1 ; YP [ 5 ] = k4 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , double * Y ) { double e ; double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 , psq ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; psq= p1 * p1+p2 * p2+p3 * p3 ; / / Kinetic Energy : e = sqrt ( 1 . 0 + psq ) −1.0; / / P o t e n t i a l Energy / m_0 e += − k2 * x1 − k3 * x2 − k4 * x3 ; return e ; } 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 the equations of motion given by special relativity¹⁴. In the equations ¹⁴Of 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. 6.3. RELATIVISTIC MOTION 299 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. 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 p + m20 2 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 300 CHAPTER 6. MOTION IN SPACE 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. 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. 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), 6.3. RELATIVISTIC MOTION 301 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. (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 program in the ﬁle rk3.cpp. The main program can be found in the ﬁle sr.cpp: / / ======================================================== / / 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 / / ======================================================== / / Compile with t h e commands : / / g f o r t r a n −c r k s u i t e / r k s u i t e . f / / g++ s r . cpp sr_B . cpp r k s u i t e . o −o rk3 −l g f o r t r a n / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− # i n c l u d e ” s r . h” double k1 , k2 , k3 , k4 ; 302 CHAPTER 6. MOTION IN SPACE 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. e x t e r n ”C” { void setup_ ( c o n s t i n t& NEQ , double& TSTART , double * YSTART , double& TEND , double& TOL , double * THRES , c o n s t i n t& METHOD , c o n s t char& TASK , bool & ERRASS , double& HSTART , double * WORK , c o n s t i n t& LENWRK , bool& MESSAGE ) ; void ut_ ( void f ( double& t , double * Y , double * YP ) , double& TWANT , double& TGOT , double * YGOT , double * YPGOT , double * YMAX , double * WORK , i n t& UFLAG ) ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double T0 , TF , X10 , X20 , X30 , V10 , V20 , V30 ; double P10 , P20 , P30 ; double P1 , P2 , P3 , V1 , V2 , V3 ; double t , dt , tstep ; int STEPS , i ; / / rksuite variables : 6.3. RELATIVISTIC MOTION 303 double TOL , THRES [ NEQ ] , WORK [ LENWRK ] , HSTART ; double Y [ NEQ ] , YMAX [ NEQ ] , YP [ NEQ ] , YSTART [ NEQ ] ; bool ERRASS , MESSAGE ; int UFLAG ; c o n s t char TASK = ’U’ ; / / Input : cout << ”Runge−Kutta Method f o r 6−ODEs I n t e g r a t i o n \n” ; cout << ” S p e c i a l R e l a t i v i s t i c P a r t i c l e : \ n” ; cout << ” Enter c o u p l i n g c o n s t a n t s k1 , k2 , k3 , k4 : \ n” ; cin >> k1 >> k2 >> k3 >> k4 ; getline ( cin , buf ) ; cout << ” Enter STEPS , T0 , TF , X10 , X20 , X30 , V10 , V20 , V30: \ n” ; cin >> STEPS >> T0 >> TF >> X10 >> X20 >> X30 >> V10 >> V20 >> V30 ; getline ( cin , buf ) ; momentum ( V10 , V20 , V30 , P10 , P20 , P30 ) ; cout << ”No . S t e p s= ” << STEPS << endl ; cout << ”Time : I n i t i a l T0 =” << T0 << ” F i n a l TF=” << TF << endl ; cout << ” X1 (T0)=” << X10 << ” X2(T0)=” << X20 << ” X3(T0)=” << X30 << endl ; cout << ” V1 (T0)=” << V10 << ” V2(T0)=” << V20 << ” V3(T0)=” << V30 << endl ; cout << ” P1 (T0)=” << P10 << ” P2 (T0)=” << P20 << ” P3 (T0)=” << P30 << endl ; / / I n i t i a l Conditions : dt = ( TF−T0 ) / STEPS ; YSTART [ 0 ] = X10 ; YSTART [ 1 ] = X20 ; YSTART [ 2 ] = X30 ; YSTART [ 3 ] = P10 ; YSTART [ 4 ] = P20 ; YSTART [ 5 ] = P30 ; / / S e t c o n t r o l parameters : TOL = 5.0 e−6; f o r ( i = 0 ; i < NEQ ; i++) THRES [ i ] = 1 . 0 e −10; MESSAGE = t r u e ; ERRASS = f a l s e ; HSTART = 0 . 0 ; // Initialization : setup_ ( NEQ , T0 , YSTART , TF , TOL , THRES , METHOD , TASK , ERRASS , HSTART , WORK , LENWRK , MESSAGE ) ; ofstream myfile ( ” s r . dat ” ) ; myfile . precision ( 1 6 ) ; myfile << T0 << ” ” << YSTART [ 0 ] << ” ” << YSTART [ 1 ] << ” ” 304 CHAPTER 6. MOTION IN SPACE << YSTART [ 2 ] << ” ” << V1 << ” ” << V2 << ” ” << V3 << ” ” << energy ( T0 , YSTART ) << ” ” << YSTART [ 3 ] << ” ” << YSTART [ 4 ] << ” ” << YSTART [ 5 ] << ’\n ’ ; / / The c a l c u l a t i o n : f o r ( i =1; i<=STEPS ; i++){ t = T0 + i * dt ; ut_ ( f , t , tstep , Y , YP , YMAX , WORK , UFLAG ) ; i f ( UFLAG > 2) break ; / / e r r o r : break t h e loop and e x i t velocity ( Y [ 3 ] , Y [ 4 ] , Y [ 5 ] , V1 , V2 , V3 ) ; myfile << tstep << ” ” << Y [ 0 ] << ” ” << Y [ 1 ] << ” ” << Y [ 2 ] << ” ” << V1 << ” ” << V2 << ” ” << V3 << ” ” << energy ( T0 , Y ) << ” ” << Y [ 3 ] << ” ” << Y [ 4 ] << ” ” << Y [ 5 ] << ” ” << ’\n ’ ; } myfile . close ( ) ; } / / main ( ) / / ======================================================== / / momentum −> v e l o c i t y t r a n s f o r m a t i o n / / ======================================================== void velocity ( c o n s t double& p1 , c o n s t double& p2 , c o n s t double& p3 , double& v1 , double& v2 , double& v3 ) { double psq ; psq = p1 * p1+p2 * p2+p3 * p3 ; v1 = p1 / sqrt ( 1 . 0 + psq ) ; v2 = p2 / sqrt ( 1 . 0 + psq ) ; v3 = p3 / sqrt ( 1 . 0 + psq ) ; } / / ======================================================== / / v e l o c i t y −> momentum t r a n s f o r m a t i o n / / ======================================================== void momentum ( c o n s t double& v1 , c o n s t double& v2 , c o n s t double& v3 , double& p1 , double& p2 , double& p3 ) { double vsq ; vsq = v1 * v1+v2 * v2+v3 * v3 ; i f ( vsq >= 1 . 0 ) { cerr << ”momentum : vsq >=1\n” ; exit ( 1 ) ; } p1 = v1 / sqrt (1.0 − vsq ) ; p2 = v2 / sqrt (1.0 − vsq ) ; 6.3. RELATIVISTIC MOTION 305 p3 = v3 / sqrt (1.0 − vsq ) ; } The functions momentum and velocity compute the transformations (6.7). In the function momentum we check whether the condition (6.8) is satis- ﬁed. These functions are also used in the function F that computes the derivatives of the functions. Common declarations are now in an include ﬁle sr.h: # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t NEQ = 6; c o n s t i n t LENWRK = 32* NEQ ; c o n s t i n t METHOD = 2 ; e x t e r n double k1 , k2 , k3 , k4 ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , double * Y ) ; void f ( double& t , double * Y , double * YP ) ; void velocity ( c o n s t double& p1 , c o n s t double& p2 , c o n s t double& p3 , double& v1 , double& v2 , double& v3 ) ; void momentum ( c o n s t double& v1 , c o n s t double& v2 , c o n s t double& v3 , double& p1 , double& p2 , double& p3 ) ; 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.cpp. The contents of the ﬁle sr_B.cpp are: 306 CHAPTER 6. MOTION IN SPACE / / ======================================================== // 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 / / ======================================================== # i n c l u d e ” s r . h” void f ( double& t , double * Y , double * YP ) { double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; velocity ( p1 , p2 , p3 , v1 , v2 , v3 ) ; / / now we can use a l l x1 , x2 , x3 , p1 , p2 , p3 , v1 , v2 , v3 YP [ 0 ] = v1 ; YP [ 1 ] = v2 ; YP [ 2 ] = v3 ; / / Acceleration : YP [ 3 ] = k2 + k1 * v2 ; YP [ 4 ] = k3 − k1 * v1 ; YP [ 5 ] = k4 ; } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , double * Y ) { double e ; double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 , psq ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; psq= p1 * p1+p2 * p2+p3 * p3 ; / / Kinetic Energy : e = sqrt ( 1 . 0 + psq ) −1.0; / / P o t e n t i a l Energy / m_0 e += − k2 * x1 − k3 * x2 − k4 * x3 ; return e ; } The results are shown in ﬁgures 6.5–6.6. 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 ˆ ˆ ⃗ = B0 B 3(d · r̂) r̂ − d , (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 = 6.3. RELATIVISTIC MOTION 307 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 ⃗ B = Bz ẑ with qBz /m0 = 10.0, ⃗v (0) = 0.95ŷ + 0.10ẑ, ⃗r(0) = 1.0x̂. 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. 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 2 2 the coordinate axes so that dˆ = ẑ and we measure distance in RE units¹⁵. Then 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.cpp: / / ======================================================== ¹⁵Since c = 1, the unit of time is the time that the light needs to travel distance equal to RE in the vacuum. 308 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. // 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] / / ======================================================== # i n c l u d e ” s r . h” void f ( double& t , double * Y , double * YP ) { double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 ; double B1 , B2 , B3 ; double r , r5 , r3 ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; velocity ( p1 , p2 , p3 , v1 , v2 , v3 ) ; / / now we can use a l l x1 , x2 , x3 , p1 , p2 , p3 , v1 , v2 , v3 YP [ 0 ] = v1 ; YP [ 1 ] = v2 ; YP [ 2 ] = v3 ; / / Acceleration : r = sqrt ( x1 * x1+x2 * x2+x3 * x3 ) ; r3 = r*r*r ; 6.3. RELATIVISTIC MOTION 309 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. r5 = r * r * r3 ; if ( r > 0.0) { B1 = k1 * ( 3 . 0 * x1 * x3 ) / r5 ; B2 = k1 * ( 3 . 0 * x2 * x3 ) / r5 ; B3 = k1 * ( ( 3 . 0 * x3 * x3 ) / r5 −1/ r3 ) ; YP [ 3 ] = v2 * B3−v3 * B2 ; YP [ 4 ] = v3 * B1−v1 * B3 ; YP [ 5 ] = v1 * B2−v2 * B1 ; } else { YP [ 3 ] = 0.0; YP [ 4 ] = 0.0; YP [ 5 ] = 0.0; } } / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double energy ( c o n s t double& t , double * Y ) { double e ; double x1 , x2 , x3 , v1 , v2 , v3 , p1 , p2 , p3 , psq ; x1 = Y [ 0 ] ; p1 = Y [ 3 ] ; x2 = Y [ 1 ] ; p2 = Y [ 4 ] ; x3 = Y [ 2 ] ; p3 = Y [ 5 ] ; psq= p1 * p1+p2 * p2+p3 * p3 ; / / Kinetic Energy : 310 CHAPTER 6. MOTION IN SPACE e = sqrt ( 1 . 0 + psq ) −1.0; return e ; } 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. 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 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 ⃗ = B ẑ. Assume that their interaction is constant magnetic ﬁeld B 6.4. PROBLEMS 311 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. Compute the particle’s trajectory and ﬁnd initial conditions that result in a planar motion. 6.5 Solve problem 5.19 of page 279 using the RK45 method. Choose ini- tial conditions so that the system executes only translational motion. Next, choose initial conditions so that the system executes small vi- brations and its center of mass remains stationary. Find the normal modes of the system and choose appropriate initial conditions 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. 6.7 Solve the problem 5.20 in page 280 by using the RK45 method. 6.8 Solve the problem 5.21 in page 280 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). 312 CHAPTER 6. MOTION IN SPACE 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. 6.11 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. 6.13 There is a C++ “translation” of rksuite. Download it from netlib .org/ ode/ rksuite and teach yourself how to use it. The docu- mentation is not as explicit as for the Fortran version, part of it is in the source code ﬁle rksuite.cpp. You can teach yourself how to use it by reading the example ﬁle RksuiteTest.cpp and the methods of the class RKSUITE in the ﬁle rksuite.h. Write a program to study the motion of the non relativistic electron in a constant magnetic ﬁeld. Then repeat for the relativistic electron. 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 )3/2 313 314 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 315 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. 316 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. • void function 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 function prints to the stdout the coordinates of the ap- proximate 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 functions 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. • void function epotline(xin,yin,X,Y,Q,N): Calculates the equipo- tential 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 function prints to the stdout the coordinates of the approximate equipotential line. The function 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 functions efield for the calculation of the electric ﬁeld and mdist for the calculation of the minimum and maximum distance of a point on the equipo- tential 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 317 • void function efield(x0,y0,X,Y,Q,N,Ex,Ey): Calculates the elec- tric 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. • void function mdist(x0,y0,X,Y,N,rmin,rmax): Calculates the max- imum 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.cpp. We list a simple version of the main() function below: i n t main ( ) { string buf ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− N = 2; X[0] = 1.0; Y [0] = 0.0; Q[0] = 1.0; X [ 1 ] = −1.0; Y [1] = 0.0; Q [ 1 ] = −1.0; / /−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−− eline ( 0 . 0 , 0 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , 1 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , 1 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , 2 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , − 0 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , − 1 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , − 1 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , − 2 . 0 , X , Y , Q , N ) ; } / / main ( ) The statements 318 CHAPTER 7. ELECTROSTATICS N = 2; X[0] = 1.0; Y[0] = 0.0; Q[0] = 1.0; X[1] = −1.0; Y[1] = 0.0; Q[1] = −1.0; deﬁne two opposite charges Q[0]= -Q[1]= 1.0 located at (1, 0) and (−1, 0) respectively. The next lines call the function eline in order to perform the calculation of 8 ﬁeld lines passing through the points (0, ±1/2), (0, ±1), (0, ±3/2), (0, ±2): eline ( 0 . 0 , 0 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , 1 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , 1 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , 2 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , − 0 . 5 , X , Y , Q , N ) ; eline ( 0 . 0 , − 1 . 0 , X , Y , Q , N ) ; eline ( 0 . 0 , − 1 . 5 , X , Y , Q , N ) ; 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 function eline are substituted by calls to epotline. For the program to be complete, we must program the functions eline, efield, mdist. This will be done later, and you can ﬁnd the full code in the ﬁle ELines.cpp. For the moment, let’s copy the main program⁴ listed above into the ﬁle Elines.cpp and compile and run it with the commands: > g++ ELines . cpp −o el > . / el > el . out The stdout of the program is redirected to the ﬁle el.out. We can plot the results with gnuplot: gnuplot > plot ” e l . out ” with dots The result is shown in ﬁgure 7.2. ⁴See the ﬁle ELines_version0.cpp. 7.2. THE PROGRAM – APPETIZER AND ... DESERT 319 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.cpp (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 −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } 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 320 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 −−−−−−−−−−−−− cout << ” # How many l i n e s t o draw?\n” ; cin >> draw ; getline ( cin , buf ) ; f o r ( i =1; i<=draw ; i++){ cout << ” # I n i t i a l p o i n t ( x0 , y0 ) : \ n” ; cin >> x0 >> y0 ; getline ( cin , buf ) ; eline ( x0 , y0 , X , Y , Q , N ) ; } 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). > g++ ELines . cpp −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 0 Q= 1 # How many lines to draw ? 1 # Initial point ( x 0 , y 0) : 0.1 0.1 0.10000000000000001 0.10000000000000001 0.092928932188134528 0.092928932188134528 0.08585786437626905 0.08585786437626905 .... 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 321 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 main() function (version 2) is listed below: i n t main ( ) { string buf ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , draw ; double x0 , y0 ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } / /−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−− cout << ” # How many l i n e s t o draw?\n” ; cin >> draw ; getline ( cin , buf ) ; f o r ( i =1; i<=draw ; i++){ cout << ” # I n i t i a l p o i n t ( x0 , y0 ) : \ n” ; cin >> x0 >> y0 ; getline ( cin , buf ) ; eline ( x0 , y0 , X , Y , Q , N ) ; } 322 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. i n t main ( ) { string buf ; c o n s t double PI = 2 . 0 * atan2 ( 1 . 0 , 0 . 0 ) ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , nd ; double x0 , y0 , theta ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } / /−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−− / / We draw 2*nd f i e l d l i n e s around each charge nd = 6 ; f o r ( i =0;i<N ; i++) f o r ( j =1; j <=(2* nd ) ; j++){ theta = ( PI / nd ) * j ; x0 = X [ i ] + 0 . 1 * cos ( theta ) ; y0 = Y [ i ] + 0 . 1 * sin ( theta ) ; eline ( x0 , y0 , X , Y , Q , N ) ; } 7.2. THE PROGRAM – APPETIZER AND ... DESERT 323 } 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. We list the example of four equal charges qi = ±1 below, 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: > g++ ELines . cpp −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 function main() from the ﬁle EPotential.cpp is listed below: i n t main ( ) { string buf ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , nd ; double x0 , y0 , rmin , rmax , L ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ 324 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.cpp. cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q[i] << endl ; } / /−−−−−−−−−−−−− 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 = 6 ; L = 1 . 0 ; f o r ( i=−nd ; i<nd ; i++) f o r ( j=−nd ; j<=nd ; j++){ x0 = i * ( L / nd ) ; y0 = j * ( L / nd ) ; cout << ” # @ ” << i << ” ” << j<< ” ” << L / nd << ” ” << x0<< ” ” << y0 << endl ; 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 > L / ( nd * 1 0 ) ) epotline ( x0 , y0 , X , Y , Q , N ) ; 7.3. THE PROGRAM – MAIN DISH 325 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.cpp. } } The ﬁrst and second part of the code is identical to the previous one. In the third part we call the function 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 call- ing the function 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++ EPotential . cpp −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 functions eline and epotline that calculate the ﬁeld and equipotential lines, the function efield that calculates the electric ﬁeld at a point and the function mdist that calculates the mini- mum and maximum distances of a point from the point charges. 326 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.cpp. 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. The function eline is called by the statement: 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 to draw the line in both directions, set ∆l < 0 and repeat the calculation. 7.3. THE PROGRAM – MAIN DISH 327 The code is listed below: void eline ( double xin , double yin , double * X , double * Y , double * Q , const int N) { c o n s t double step =0.01; c o n s t double max_dist =20.0; int i , direction ; double x0 , y0 ; double rmin , rmax , r , dx , dy , dl ; double Ex , Ey , E ; cout . precision ( 1 7 ) ; f o r ( direction=−1;direction <=1; direction +=2){ dl = direction * step ; x0 = xin ; y0 = yin ; dx = 0 . 0 ; dy = 0 . 0 ; mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; while ( rmin > ( 2 . 0 * step ) && rmax < max_dist ) { cout << x0 << ” ” << y0 << ’\n ’ ; / / 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 s y s t e m a t i c e r r o r s efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey ) ; E = sqrt ( Ex * Ex+Ey * Ey ) ; i f ( E <= 1 . 0 e−10 ) break ; dx = dl * Ex / E ; dy = dl * Ey / E ; x0 = x0 + dx ; y0 = y0 + dy ; mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; } / / while ( rmin > ( 2 . 0 * s t e p ) && rmax < max_dist ) } / / f o r ( d i r e c t i o n =−1; d i r e c t i o n <=1; d i r e c t i o n +=2) } In the ﬁrst part of the code we have the variable declarations. We note that the values of the parameters step and max_dist are “hard coded” into our program and the user cannot change them: c o n s t double step =0.01; c o n s t double 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 328 CHAPTER 7. ELECTROSTATICS f o r ( direction=−1;direction <=1; direction +=2){ dl = direction * step ; ... } sets the direction of motion on the ﬁeld line (i.e. the sign of ∆l). The loop is executed twice with direction taking the two values −1 and 1. 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. 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 mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; while ( rmin > ( 2 . 0 * step ) && rmax < max_dist ) { cout << x0 << ” ” << y0 << ’\n ’ ; ... mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; } which is executed provided that the logical expression (rmin > (2.0*step) && rmax < 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 function 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 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 = sqrt ( 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 <= 1 . 0 e−10 ) break ; ⁵The choice is not unique of course, you may also try e.g. rmin < max_dist. 7.3. THE PROGRAM – MAIN DISH 329 When the magnitude of the electric ﬁeld becomes too small we stop the calculation by exiting the loop with the command break. The reader can improve the code by adding more checks of singular cases. The function epotline is programmed in a similar way. The main() function is listed below: i n t main ( ) { string buf ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , nd ; double x0 , y0 , rmin , rmax , L ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } / /−−−−−−−−−−−−− 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 = 6 ; L = 1 . 0 ; f o r ( i=−nd ; i<nd ; i++) f o r ( j=−nd ; j<=nd ; j++){ x0 = i * ( L / nd ) ; y0 = j * ( L / nd ) ; cout << ” # @ ” << i << ” ” << j<< ” ” << L / nd << ” ” << x0<< ” ” << y0 << endl ; 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 > L / ( nd * 1 0 ) ) epotline ( x0 , y0 , X , Y , Q , N ) ; } } 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 330 CHAPTER 7. ELECTROSTATICS or leave the drawing area: while ( r > ( 0 . 9 * dl ) && r < max_dist ) { ... } The values of dx, dy are calculated according to equation (7.6): dx = dl * Ey / E ; dy = −dl * Ex / E ; The function efield is an application of equations⁶ (7.2): void efield ( double x0 , double y0 , double * X , double * Y , double * Q , const int N , double& Ex , double& Ey ) { int i; double r3 , xi , yi ; Ex = 0 . 0 ; Ey = 0 . 0 ; f o r ( i =0;i<N ; i++){ xi = x0−X [ i ] ; yi = y0−Y [ i ] ; r3 = pow ( xi * xi+yi * yi , − 1 . 5 ) ; Ex = Ex + Q [ i ] * xi * r3 ; Ey = Ey + Q [ i ] * yi * r3 ; } } Finally, the function mdist calculates the minimum and maximum distance rmin and rmax of a point (x0,y0) from all the point charges in the distribution: void mdist ( double x0 , double y0 , double * X , double * Y, const in t N , double& rmin , double& rmax ) { int i; double r ; rmax = 0 . 0 ; rmin = 1 0 0 0 . 0 ; f o r ( i =0;i<N ; i++){ ⁶You may improve the program by checking whether ri = 0. 7.4. THE PROGRAM - CONCLUSION 331 r = sqrt ( ( x0−X [ i ] ) * ( x0−X [ i ] ) + ( y0−Y [ i ] ) * ( y0−Y [ i ] ) ) ; i f ( r > rmax ) rmax = r ; i f ( r < rmin ) rmin = r ; } } 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.cpp: # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void eline ( double xin , double yin , double * X , double * Y , double * Q , c o n s t i n t N ) ; void efield ( double x0 , double y0 , double * X , double * Y , double * Q , c o n s t i n t N , double& Ex , double& Ey ) ; void mdist ( double x0 , double y0 , double * X , double * Y, const int N , double& rm , double& rM ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; c o n s t double PI = 2 . 0 * atan2 ( 1 . 0 , 0 . 0 ) ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , nd ; double x0 , y0 , theta ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− 332 CHAPTER 7. ELECTROSTATICS cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } / /−−−−−−−−−−−−− DRAWING LINES −−−−−−−−−−−−− / / We draw 2*nd f i e l d l i n e s around each charge nd = 6 ; f o r ( i =0;i<N ; i++) f o r ( j =1; j <=(2* nd ) ; j++){ theta = ( PI / nd ) * j ; x0 = X [ i ] + 0 . 1 * cos ( theta ) ; y0 = Y [ i ] + 0 . 1 * sin ( theta ) ; eline ( x0 , y0 , X , Y , Q , N ) ; } } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void eline ( double xin , double yin , double * X , double * Y , double * Q , c o n s t i n t N ) { c o n s t double step =0.01; c o n s t double max_dist =20.0; int i , direction ; double x0 , y0 ; double rmin , rmax , r , dx , dy , dl ; double Ex , Ey , E ; cout . precision ( 1 7 ) ; f o r ( direction=−1;direction <=1; direction +=2){ dl = direction * step ; x0 = xin ; y0 = yin ; dx = 0 . 0 ; dy = 0 . 0 ; mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; while ( rmin > ( 2 . 0 * step ) && rmax < max_dist ) { cout << x0 << ” ” << y0 << ’\n ’ ; / / 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 s y s t e m a t i c e r r o r s efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey ) ; E = sqrt ( Ex * Ex+Ey * Ey ) ; i f ( E <= 1 . 0 e−10 ) break ; 7.4. THE PROGRAM - CONCLUSION 333 dx = dl * Ex / E ; dy = dl * Ey / E ; x0 = x0 + dx ; y0 = y0 + dy ; mdist ( x0 , y0 , X , Y , N , rmin , rmax ) ; } / / while ( rmin > ( 2 . 0 * s t e p ) && rmax < max_dist ) } / / f o r ( d i r e c t i o n =−1; d i r e c t i o n <=1; d i r e c t i o n +=2) } / / eline () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void efield ( double x0 , double y0 , double * X , double * Y , double * Q , c o n s t i n t N , double& Ex , double& Ey ) { int i; double r3 , xi , yi ; Ex = 0 . 0 ; Ey = 0 . 0 ; f o r ( i =0;i<N ; i++){ xi = x0−X [ i ] ; yi = y0−Y [ i ] ; r3 = pow ( xi * xi+yi * yi , − 1 . 5 ) ; Ex = Ex + Q [ i ] * xi * r3 ; Ey = Ey + Q [ i ] * yi * r3 ; } } / / efield () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void mdist ( double x0 , double y0 , double * X , double * Y, const int N , double& rmin , double& rmax ) { int i; double r ; rmax = 0 . 0 ; rmin = 1 0 0 0 . 0 ; f o r ( i =0;i<N ; i++){ r = sqrt ( ( x0−X [ i ] ) * ( x0−X [ i ] ) + ( y0−Y [ i ] ) * ( y0−Y [ i ] ) ) ; i f ( r > rmax ) rmax = r ; i f ( r < rmin ) rmin = r ; } } / / mdist ( ) Then we list the contents of the ﬁle EPotential.cpp: # include < iostream > # include < fstream > # include <cstdlib > # include <string > # include <cmath> 334 CHAPTER 7. ELECTROSTATICS using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void epotline ( double xin , double yin , double * X , double * Y , double * Q , c o n s t i n t N ) ; void efield ( double x0 , double y0 , double * X , double * Y , double * Q , c o n s t i n t N , double& Ex , double& Ey ) ; void mdist ( double x0 , double y0 , double * X , double * Y, const int N , double& rm , double& rM ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; const int P = 20; / / max number o f c h a r g e s double X[P] , Y[P] , Q[P ]; int N; int i , j , nd ; double x0 , y0 , rmin , rmax , L ; / /−−−−−−−−−−−−− SET CHARGE DISTRIBUTION −−−− cout << ” # Enter number o f c h a r g e s : ” << endl ; cin >> N ; getline ( cin , buf ) ; cout << ” # N= ” << N << endl ; f o r ( i =0;i<N ; i++){ cout << ” # Charge : ” << i+1 << endl ; cout << ” # P o s i t i o n and charge : (X, Y , Q) : ” << endl ; cin >> X [ i ] >> Y [ i ] >> Q [ i ] ; getline ( cin , buf ) ; cout << ” # (X, Y)= ” << X [ i ] << ” ” << Y [ i ] << ” Q= ” << Q [ i ] << endl ; } / /−−−−−−−−−−−−− 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 = 6 ; L = 1 . 0 ; f o r ( i=−nd ; i<nd ; i++) f o r ( j=−nd ; j<=nd ; j++){ x0 = i * ( L / nd ) ; y0 = j * ( L / nd ) ; cout << ” # @ ” << i << ” ” << j<< ” ” << L / nd << ” ” << x0<< ” ” << y0 << endl ; 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 > L / ( nd * 1 0 ) ) epotline ( x0 , y0 , X , Y , Q , N ) ; 7.4. THE PROGRAM - CONCLUSION 335 } } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void epotline ( double xin , double yin , double * X , double * Y , double * Q , c o n s t i n t N ) { c o n s t double step =0.02; c o n s t double max_dist =20.0; int i; double x0 , y0 ; double r , dx , dy , dl ; double Ex , Ey , E ; cout . precision ( 1 7 ) ; dl = step ; x0 = xin ; y0 = yin ; dx = 0 . 0 ; dy = 0 . 0 ; r = step ; while ( r > ( 0 . 9 * dl ) && r < max_dist ) { cout << x0 << ” ” << y0 << ’\n ’ ; / / 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 s y s t e m a t i c e r r o r s efield ( x0 +0.5* dx , y0 +0.5* dy , X , Y , Q , N , Ex , Ey ) ; E = sqrt ( Ex * Ex+Ey * Ey ) ; i f ( E <= 1 . 0 e−10 ) break ; dx = dl * Ey / E ; dy = −dl * Ex / E ; x0 = x0 + dx ; y0 = y0 + dy ; r = sqrt ( ( x0−xin ) * ( x0−xin ) +( y0−yin ) * ( y0−yin ) ) ; } / / while ( ) } / / epotline () ... where ... are the functions efield and mdist which are identical to the ones in the ﬁle ELines.cpp. In order to compile the program use the commands: > g++ ELines . cpp −o el > g++ EPotential . cpp −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 336 CHAPTER 7. ELECTROSTATICS 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 > . / 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 ⁷Equivalent to the equation ∇ ⃗ ×E ⃗ = 0. 7.5. ELECTROSTATIC FIELD IN THE VACUUM 337 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 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 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 11001100 11001100 11001100 000 111 000 111 1100 1100 1100 000 111 1100 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 338 CHAPTER 7. ELECTROSTATICS 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 + (δy)2 + . . . ∂y 2 ∂y 2 ∂V 1 ∂2V V (x, y − δy) = V (x, y) − δy + (δy)2 + . . . . ∂y 2 ∂y 2 By summing both sides of the above equations, taking δx = δy and 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. 7.5. ELECTROSTATIC FIELD IN THE VACUUM 339 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. 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 an array double V[L][L] in order to store the values of the potential at each lattice site. A logical array bool isConductor[L][L] ﬂags each site as a “conductor site” (= true) or as a “vacuum site” (=false). Both arrays are put in the global scope and are accessible by all functions. The main program reads in the data entered by the user and then calls three functions: 1. initialize_lattice(V1,V2): The routine needs at its input the values of the potential V1 and V2 on the left and right plate respectively. 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 function each time that she wants to study a diﬀerent geometry. 2. laplace(epsilon): This is the heart of the program. On entry we provide the desired accuracy epsilon. On exit we obtain the ﬁnal solution V[L][L]. This function calculates the arithmetic mean of the potential of the nearest neighbors Vav and the value V[i][j]=Vav is changed im- mediately⁸. The maximum change in the new value of the potential ⁸A diﬀerent choice would have been to store the value Vav in a temporary array 340 CHAPTER 7. ELECTROSTATICS 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(): This function 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]. We note that each time that the index i changes, the function 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: / / double 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 / / bool i s C o n d u c t o r [ L ] [ L ] : I f t r u e s i t e has f i x e d p o t e n t i a l // I f f a l s e s i t e i s empty s p a ce / / double 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 f u n c t i o n 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 // function 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 ] // function 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 // function print_results : / / . 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 . // ************************************************************ # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> 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... 7.5. ELECTROSTATIC FIELD IN THE VACUUM 341 using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− const i n t L = 31; bool isConductor [ L ] [ L ] ; double V [L][L ]; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void initialize_lattice ( c o n s t double& V1 , c o n s t double& V2 ) ; void laplace ( c o n s t double& epsilon ) ; void print_results () ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double V1 , V2 , epsilon ; cout << ” Enter V1 , V2 : ” << endl ; cin >> V1 >> V2 ; getline ( cin , buf ) ; cout << ” Enter e p s i l o n : ” << endl ; cin >> epsilon ; getline ( cin , buf ) ; cout << ” S t a r t i n g Laplace : ” << endl ; cout << ” Grid S i z e = ” << L << endl ; cout << ” Conductors s e t a t V1= ” << V1 << ” V2= ” << V2 << endl ; cout << ” R e l a x i n g with a c c u r a c y e p s i l o n = ” << epsilon << endl ; / / 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 initialize_lattice ( V1 , V2 ) ; / / On entry , V , i s C o n d u c t o r i s i n i t i a l i z e d . / / On e x i t t h e r o u t i n e g i v e s t h e s o l u t i o n V laplace ( epsilon ) ; / / We p r i n t V i n a f i l e . print_results ( ) ; } / / main ( ) // ************************************************************ / / function 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]= f a l s e by d e f a u l t / / i s C o n d u c t o r [ i ] [ j ]= t r u e 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 ]= t r u e 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 ]= t r u e 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 ]= f a l s e / / INPUT : / / i n t e g e r L : Linear s i z e of l a t t i c e / / double 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: / / double V[ L ] [ L ] : Array provided by u s e r . Values o f p o t e n t i a l / / bool isConductor [L ] [ L ] : I f true s i t e has f i x e d p o t e n t i a l 342 CHAPTER 7. ELECTROSTATICS // I f f a l s e s i t e i s empty sp a c e // ************************************************************ void initialize_lattice ( c o n s t double& V1 , c o n s t double& V2 ) { / / I n i t i a l i z e t o 0 and f a l s e ( d e f a u l t v a l u e s f o r / / boundary and i n t e r i o r s i t e s ) . f o r ( i n t i =0;i<L ; i++) f o r ( i n t j =0;j<L ; j++){ V [i ][ j ] = 0.0; isConductor [ i ] [ j ] = f a l s e ; } / / We s e t t h e boundary t o be a conductor : (V=0 by d e f a u l t ) f o r ( i n t i =0;i<L ; i++){ isConductor [0 ] [ i ] = t r u e ; isConductor [ i ] [ 0 ] = t r u e ; isConductor [ L −1][ i ] = t r u e ; isConductor [ i ] [ L−1] = t r u e ; } / / We s e t two c o n d u c t o r s a t g iven p o t e n t i a l V1 and V2 f o r ( i n t i =4; i<L−5;i++){ V [ L / 3 ] [ i ] = V1 ; isConductor [ L / 3 ] [ i ] = t r u e ; V [ 2 * L / 3 ] [ i ] = V2 ; isConductor [ 2 * L / 3 ] [ i ] = t r u e ; } } / / i n i t i a l i z e _ l a t t i c e () // ************************************************************ / / function 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 dimensional 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 : Lin e ar s i z e o f l a t t i c e / / double 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 / / bool isConductor [L ] [ L ] : I f true potential is fixed // If false p o t e n t i a l i s updated / / double 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: / / double 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 // ************************************************************ void laplace ( c o n s t double& epsilon ) { int icount ; double Vav , error , dV ; 7.6. RESULTS 343 icount = 0 ; while ( icount < 10000) { icount ++; error = 0 . 0 ; f o r ( i n t i = 1 ; i<L −1;i++){ f o r ( i n t j = 1 ; j<L −1;j++){ / / We change V only f o r non c o n d u c t o r s : i f ( ! isConductor [ i ] [ j ] ) { Vav = 0 . 2 5 * ( V [ i −1][ j ]+ V [ i + 1 ] [ j ]+ V [ i ] [ j−1]+V [ i ] [ j + 1 ] ) ; dV = abs ( V [ i ] [ j]−Vav ) ; i f ( error < dV ) error = dV ; / / maximum e r r o r V [ i ] [ j ] = Vav ; } / / i f ( ! isConductor [ i ] [ j ] ) } / / f o r ( i n t j = 1 ; j <L−1; j ++) } / / f o r ( i n t i = 1 ; i <L−1; i ++) cout << icount << ” e r r = ” << error << endl ; i f ( error < epsilon ) r e t u r n ; } / / while ( i c o u n t < 10000) cerr << ” Warning : l a p l a c e did not converge . \ n” ; } / / laplace () // ************************************************************ / / function print_results / / 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 printed . // ************************************************************ void print_results ( ) { ofstream myfile ( ” data ” ) ; myfile . precision ( 1 6 ) ; f o r ( i n t i = 0 ; i < L ; i++){ f o r ( i n t j = 0 ; j < L ; j++){ myfile << i+1 << ” ” << j+1 << ” ” << V [ i ] [ j ] << endl ; } / / p r i n t 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 : myfile << ” ” << endl ; } myfile . close ( ) ; } / / print_results () 7.6 Results The program in the previous section is written in the ﬁle LaplaceEq.cpp. Compiling and running is done with the commands: 344 CHAPTER 7. ELECTROSTATICS > g++ LaplaceEq . cpp −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= 33.3333 2 err= 14.8148 3 err= 9.87654 .............. 110 err= 0.0106861 111 err= 0.0101182 112 err= 0.00958049 In the example above, the program performs 112 sweeps until the error becomes 0.00958 < 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 345 "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.cpp for L= 31, V1=100, V2=-100, epsilon=0.01. where⁹ ρ̃(i, j) = 4πa2 ρ(i, j). The program in the ﬁle PoissonEq.cpp 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. # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− const i n t L = 51; bool isConductor [ L ] [ L ] ; double V [L][L ]; double rho [L][L ]; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void initialize_lattice ( c o n s t double& V1 , c o n s t double& V2 , c o n s t double& V3 , c o n s t double& V4 , ∫ ∑ ∑ ∑ ⁹Since Q = ρdA ≈ i,j ρa2 = (1/4π) i,j ρ̃. Therefore i,j ρ̃ ≈ 4πQ. 346 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.cpp for L= 51, V= 0 on the boundary and the charge 4πQ = 1000 all concentrated at one point. c o n s t double& Q ) ; void laplace ( c o n s t double& epsilon ) ; void print_results () ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; double V1 , V2 , V3 , V4 , Q , epsilon ; cout << ” Enter V1 , V2 , V3 , V4 : ” << endl ; cin >> V1 >> V2 >> V3 >> V4 ; getline ( cin , buf ) ; cout << ” Enter 4* PI *Q: ” ; cin >> Q; getline ( cin , buf ) ; cout << ” Enter e p s i l o n : ” << endl ; cin >> epsilon ; getline ( cin , buf ) ; cout << ” S t a r t i n g Laplace : ” << endl ; cout << ” Grid S i z e = ” << L << endl ; cout << ” Conductors s e t a t V1= ” << V1 << ” V2= ” << V2 << ” V3= ” << V3 << ” V4= ” << V4 << ” Q = ” << Q << endl ; cout << ” R e l a x i n g with a c c u r a c y e p s i l o n = ” 7.7. POISSON EQUATION 347 "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.cpp 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. << epsilon << endl ; initialize_lattice ( V1 , V2 , V3 , V4 , Q ) ; laplace ( epsilon ) ; print_results ( ) ; } / / main ( ) // ************************************************************ void initialize_lattice ( c o n s t double& V1 , c o n s t double& V2 , c o n s t double& V3 , c o n s t double& V4 , c o n s t double& Q ) { int L1 , L2 ; double Area ; / / 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 i n t e r i o r s i t e s ) . f o r ( i n t i =0;i<L ; i++) f o r ( i n t j =0;j<L ; j++){ V [i ][ j ] = 0.0; isConductor [ i ] [ j ] = f a l s e ; rho [i ][ j ] = 0.0; 348 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.cpp for L= 51, V= 0 on the boundary and the charge 4πQ = 1000 uniformly distributed on all internal lattice sites. } / / We s e t t h e boundary t o be a conductor : (V=0 by d e f a u l t ) f o r ( i n t i =0;i<L ; i++){ isConductor [0 ] [ i ] = t r u e ; isConductor [ i ] [ 0 ] = t r u e ; isConductor [ L −1][ i ] = t r u e ; isConductor [ i ] [ L−1] = t r u e ; V [0 ] [ i ] = V1 ; V [ i ] [ L−1] = V2 ; V [ L −1][ i ] = V3 ; V [ i ] [ 0 ] = V4 ; } / / We s e t t h e p o i n t s with non−z e r o charge / / 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< 0) { cerr <<” Array out o f bounds . L1< 0\n” ; exit ( 1 ) ; } i f ( L2>=L ) { cerr <<” Array out o f bounds . L2>=0\n” ; exit ( 1 ) ; } Area = ( L2−L1 +1) * ( L2−L1 +1) ; f o r ( i n t i=L1 ; i<=L2 ; i++) f o r ( i n t j=L1 ; j<=L2 ; j++) 7.7. POISSON EQUATION 349 rho [ i ] [ j ] = Q / Area ; } / / i n i t i a l i z e _ l a t t i c e () // ************************************************************ void laplace ( c o n s t double& epsilon ) { int icount ; double Vav , error , dV ; icount = 0 ; while ( icount < 10000) { icount ++; error = 0 . 0 ; f o r ( i n t i = 1 ; i<L −1;i++){ f o r ( i n t j = 1 ; j<L −1;j++){ / / We change V only f o r non c o n d u c t o r s : i f ( ! isConductor [ i ] [ j ] ) { Vav = 0 . 2 5 * ( V [ i −1][ j ]+ V [ i + 1 ] [ j ]+ V [ i ] [ j−1]+V [ i ] [ j +1] +rho [ i ] [ j ] ) ; dV = abs ( V [ i ] [ j]−Vav ) ; i f ( error < dV ) error = dV ; / / maximum e r r o r V [ i ] [ j ] = Vav ; } / / i f ( ! isConductor [ i ] [ j ] ) } / / f o r ( i n t j = 1 ; j <L−1; j ++) } / / f o r ( i n t i = 1 ; i <L−1; i ++) cout << icount << ” e r r = ” << error << endl ; i f ( error < epsilon ) r e t u r n ; } / / while ( i c o u n t < 10000) cerr << ” Warning : l a p l a c e did not converge . \ n” ; } / / laplace () // ************************************************************ void print_results ( ) { ofstream myfile ( ” data ” ) ; myfile . precision ( 1 6 ) ; f o r ( i n t i = 0 ; i < L ; i++){ f o r ( i n t j = 0 ; j < L ; j++){ myfile << i+1 << ” ” << j+1 << ” ” << V [ i ] [ j ] << endl ; } / / p r i n t 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 : myfile << ” ” << endl ; } myfile . close ( ) ; } / / 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 350 CHAPTER 7. ELECTROSTATICS 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 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 351 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.cpp 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.cpp so that the number of ﬁeld lines starting near a charge q is proportional to q. 7.5 Improve the program in EPotential.cpp 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. cpp for (a) L= 31, V1=100, V2=100 (b) L= 31, V1=100, V2=0 352 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. cpp 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 353 111 000 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.cpp. 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- 354 CHAPTER 7. ELECTROSTATICS tion (7.16) for the problem solved in LaplaceEq.cpp. 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.cpp. 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 . 355 356 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 357 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. 358 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 359 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. 360 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 361 to update the values of ui,j . Therefore we will deﬁne only two arrays in order to store the values ui,j and ∆t/(∆x)2 (ui+1,j − 2ui,j + ui−1,j ) at time tj . In the program listed below, the names of these arrays are u[P] and d2udx2[P]. Some care must be exercised because of the array indexing in C++. The data is stored in the array positions u[0] ... u[Nx-1] and d2udx2[0] ... d2udx2[Nx-1] 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.cpp 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 gi ve n 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 + i * dx , i = 0 , . . . , Nx−1 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 * dt , j = 0 , . . . , Nt−1 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 ) // / / ======================================================= # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { const int P = 100000; c o n s t double PI = 2 . 0 * atan2 ( 1 . 0 , 0 . 0 ) ; double u [ P ] , d2udx2 [ P ] ; double t , x , dx , dt , tf , courant ; 362 CHAPTER 8. DIFFUSION EQUATION int Nx , Nt , i , j ; string buf ; / / Input : cout << ” # Enter : Nx , Nt , t f : ( P= ” << P << ” Nx must be < P ) ” << endl ; cin >> Nx >> Nt >> tf ; getline ( cin , buf ) ; i f ( Nx >= P ) { cerr << ”Nx >= P\n” ; exit ( 1 ) ; } i f ( Nx <= 3) { cerr << ”Nx <= 3\n” ; exit ( 1 ) ; } i f ( Nt <= 2) { cerr << ”Nx <= 2\n” ; exit ( 1 ) ; } // Initialize : dx = 1 . 0 / ( Nx −1) ; dt = tf / ( Nt −1) ; courant= dt / ( dx * dx ) ; cout << ” # 1d D i f f u s i o n Equation : 0<=x <=1 , 0<=t <= t f \n” ; cout << ” # dx= ” << dx << ” dx= ” << dt << ” t f = ” << tf << endl ; cout << ” # Nx= ” << Nx << ” Nt= ” << Nt << endl ; cout << ” # Courant Number= ” << courant << endl ; i f ( courant > 0 . 5 ) cout << ” # WARNING: c o u r a n t > 0.5\n” ; ofstream myfile ( ”d . dat ” ) ; myfile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / 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 ) f o r ( i =0;i<Nx ; i++){ x = i * dx ; u[i] = sin ( PI * x ) ; } u [0 ] = 0.0; u [ Nx −1] = 0 . 0 ; f o r ( i =0;i<Nx ; i++){ x = i * dx ; myfile << 0.0 << ” ” << x << ” ” << u [ i ] << ’\n ’ ; } myfile << ” \n” ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / C a l c u l a t e time e v o l u t i o n : f o r ( j =1; j<Nt ; j++){ t = j * dt ; / / Second d e r i v a t i v e : f o r ( i =1; i<Nx −1;i++) d2udx2 [ i ] = courant * ( u [ i +1] −2.0* u [ i ]+ u [ i −1]) ; / / Update : f o r ( i =1; i<Nx −1;i++) u[i] += d2udx2 [ i ] ; f o r ( i =0;i<Nx ; i++){ x = i * dx ; myfile << t << ” ” << x << ” ” << u [ i ] << ’\n ’ ; } 8.5. RESULTS 363 myfile << ” \n” ; } / / f o r ( j =1; j <Nt ; j ++) myfile . close ( ) ; } / / main ( ) 8.5 Results The compilation and running of the program can be done with the com- mands: > g++ diffusion . cpp −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 dx= 0.0040404 t f = 0.4 # Nx= 10 Nt= 100 # Courant Number= 0.327273 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 364 CHAPTER 8. DIFFUSION EQUATION 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 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: 8.6. DIFFUSION ON THE CIRCLE 365 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 ) ) ✁✂ ☛✒ ✁ ✂ ☛☛ ☞ ☞✑ ✎✏✍ ✌☞ ☛ ✁ ✂ ✁ ✂ ✁✂ ✁✄ ✁☎ ✁✆ ✁✝ ✁✞ ✁✟ ✁✠ ✁✡ ✓ 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. The results can be seen in ﬁgure 8.3. 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 366 CHAPTER 8. DIFFUSION EQUATION 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 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.cpp. At a given time tj , the boundary conditions (8.30) are enforced in the lines f o r ( i =0;i<Nx ; i++){ nnr = i +1; i f ( nnr > Nx −1) nnr = 0 ; nnl = i −1; i f ( nnl < 0) nnl = Nx −1; d2udx2 [ i ] = courant * ( u [ nnr ] −2.0* 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). 8.6. DIFFUSION ON THE CIRCLE 367 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 / / t h e gi ve n 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 + i * dx , i = 0 , . . . , Nx−1 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 * dt , j = 0 , . . . , Nt−1 and dt = ( t f −0) / ( Nt −1) // / / u ( x , 0 ) = \ d e l t a _ {x , 0 . 5 } // / / ======================================================= # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { const int P = 100000; c o n s t double PI = 2 . 0 * atan2 ( 1 . 0 , 0 . 0 ) ; double u [ P ] , d2udx2 [ P ] ; double t , x , dx , dt , tf , courant ; double prob , r2 , x0 ; int Nx , Nt , i , j , nnl , nnr ; string buf ; / / Input : cout << ” # Enter : Nx , Nt , t f : ( P= ” << P << ” Nx must be < P ) ” << endl ; cin >> Nx >> Nt >> tf ; getline ( cin , buf ) ; i f ( Nx >= P ) { cerr << ”Nx >= P\n” ; exit ( 1 ) ; } i f ( Nx <= 3) { cerr << ”Nx <= 3\n” ; exit ( 1 ) ; } i f ( Nt <= 2) { cerr << ”Nx <= 2\n” ; exit ( 1 ) ; } // Initialize : dx = 1 . 0 / ( Nx −1) ; dt = tf / ( Nt −1) ; courant= dt / ( dx * dx ) ; 368 CHAPTER 8. DIFFUSION EQUATION cout << ” # 1d D i f f u s i o n Equation : 0<=x <=1 , 0<=t <= t f \n” ; cout << ” # dx= ” << dx << ” dx= ” << dt << ” t f = ” << tf << endl ; cout << ” # Nx= ” << Nx << ” Nt= ” << Nt << endl ; cout << ” # Courant Number= ” << courant << endl ; i f ( courant > 0 . 5 ) cout << ” # WARNING: c o u r a n t > 0.5\n” ; ofstream myfile ( ”d . dat ” ) ; myfile . precision ( 1 7 ) ; ofstream efile ( ” e . dat ” ) ; efile . precision ( 1 7 ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / I n i t i a l c o n d i t i o n a t t =0 f o r ( i =0;i<Nx ; i++){ x = i * dx ; u[i] = 0.0; } u [ Nx /2 −1] = 1 . 0 ; f o r ( i =0;i<Nx ; i++){ x = i * dx ; myfile << 0.0 << ” ” << x << ” ” << u [ i ] << ’\n ’ ; } myfile << ” \n” ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / C a l c u l a t e time e v o l u t i o n : f o r ( j =1; j<Nt ; j++){ t = j * dt ; / / Second d e r i v a t i v e : f o r ( i =0;i<Nx ; i++){ nnr = i +1; i f ( nnr > Nx −1) nnr = 0 ; nnl = i −1; i f ( nnl < 0) nnl = Nx −1; d2udx2 [ i ] = courant * ( u [ nnr ] −2.0* u [ i ]+ u [ nnl ] ) ; } / / Update : prob = 0 . 0 ; r2 = 0.0; x0 = ( ( Nx / 2 ) −1) * dx ; / / o r i g i n a l p o s i t i o n f o r ( i =0;i<Nx ; i++){ x = i * dx ; u [ i ] += d2udx2 [ i ] ; prob += u [ i ] ; r2 += u [ i ] * ( x−x0 ) * ( x−x0 ) ; } f o r ( i =0;i<Nx ; i++){ x = i * dx ; myfile << t << ” ” << x << ” ” << u [ i ] << ’\n ’ ; } myfile << ” \n” ; 8.7. ANALYSIS 369 efile << ”pu ” << t << ” ” << prob << ” ” << r2 << ” ” << u [ Nx /2 −1] << ” ” << u [ Nx /4 −1] << ” ” << u [ 0 ] << ’\n ’ ; } / / f o r ( j =1; j <Nt ; j ++) myfile . close ( ) ; efile . close ( ) ; } / / main ( ) 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 ⟨r ⟩j = 2 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 . 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 > ! g++ diffusionS1 . cpp −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 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 370 CHAPTER 8. DIFFUSION EQUATION ✁✝✂ ✠✡☛☞✌✂ ✠✡☛☞✌✄ ✠✡✝ ✁✝ ✁ ✆ ✁ ☎ ✁ ✄ ✁ ✂ ✁ ✞ ✁✝ ✁✝✞ ✁✂ ✁✂✞ ✁✟ ✁✟✞ ✁✄ 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. 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 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 371 ✁✝ ✁ ✆ ✁ ☎ ✁ ✄ ✁ ✂ ✠✡☛☞✌✍✎✏ ✂✍ ✁ ✞ ✁✝ ✁✝✞ ✁✂ ✁✂✞ ✁✟ ✁✟✞ ✁✄ 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. 372 CHAPTER 8. DIFFUSION EQUATION ✝ ✝ ✁✆ ✁✆ ✁☎ ✁☎ ✁✄ ✁✄ ✁✂ ✁✂ ✝ ✁ ✂ ✁ ✄ ✁✆ ✁ ☎ ✁ ✆ ✁☎ ✁✄ ✁✝ ✁✝✂ ✁✂ ✁✝✄ ✁✝☎ ✆✂✝ ✆✂✡✠ ✆✂✝ ✆✂✡✟ ✆✂✡✞ ✆✂✡☛ ✆✂✡✝ ✆✂✡ ✆✂✡ ✆✂✆✠ ✆✂✆✟ ✆✂✆☛ ✆✂✆✞ ✆✂✆✝ ✆ ✆ ✡ ✆ ✆✂✠ ✆✂✆✝ ✆✂✟ ✆✂✆✞ ✆✂✆✟ ✆✂✞ ✆✂✆✠ ✆✂✡ ✆✂✝ ✆✂✡✝ ✁✂✁✄☎ ✆✂✡✞ ✆ ✆✂✡✟ 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 . 8.8. PROBLEMS 373 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. 374 CHAPTER 8. DIFFUSION EQUATION (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. 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 [42] and [43] respectively. 9.1 Introduction The Hamiltonian of the harmonic oscillator is given by p2 1 H0 = + mω 2 x2 . (9.1) 2m 2 375 376 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) 377 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 0 378 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. The programs have been written in the Fortran programming language, therefore the reader should review the discussion in Section 6.1.2. 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 subroutine² 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:³ > 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 ( * ) 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 function of type void in Fortran, is called a subroutine. ³A Google search “dsyev” will easily take you to the same page. 9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 379 ARGUMENTS JOBZ ( input ) CHARACTER * 1 = ’ 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 380 CHAPTER 9. THE ANHARMONIC OSCILLATOR = 0 : successful exit < 0 : if INFO = −i , the i−th argument had an illegal ←- value > 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 matrix⁴. 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 lower (UPLO='U') or upper (UPLO='L') triangular part of A. Be careful: the documentation says the opposite, but I hope that you remember from the discussion in Section 6.1.2 that Fortran arrays are trans- posed from the point of view of C++! 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]. The eigenvectors are stored in the rows of ⁴The 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) 381 A[N][N], i.e. the n-th eigenvector corresponding to the eigenvalue λn =W[n-1] is v=A[n-1]. The eigenvectors are normalized to unity, ∑N −1 i.e. m=0 v[m]*v[m]= 1. The matrix A[N][N] is destroyed after the call 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: # include < iostream > # include < fstream > # include <cstdlib > # include <string > # include <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− const int P = 100; / / P=LDA c o n s t i n t LWORK = 3* P −1; double A [ P ] [ P ] , W [ P ] , WORK [ LWORK ] ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e x t e r n ”C” void dsyev_ ( c o n s t char& JOBZ , c o n s t char& UPLO , const in t & N, double A [ P ] [ P ] , c o n s t i n t & LDA , double W [P] , double WORK [ P ] , c o n s t i n t & LWORK , i n t & INFO ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { int N; int i ,j; int LDA , INFO ; char JOBZ , UPLO ; 382 CHAPTER 9. THE ANHARMONIC OSCILLATOR string buf ; / / D e f i n e t h e * * symmetric * * ma tr ix 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 es t h e upper t r i a n g u l a r p a r t / / (UPLO= ’U ’ ) i n t h e *FORTRAN* column−major mode , / / t h e r e f o r e i n C++ , we need t o d e f i n e i t s * lower * / / triangular part ! N = 4 ; / / an N x N mat r ix A [0][0]= −7.7; A [1][0]= 2.1; A [1][1]= 8.3; A [2][0]= −3.7; A [2][1]= −16.; A [2][2]= −12.; A [3][0]= 4 . 4 ; A [3][1]= 4.6; A [3][2]= −1.04; A [3][3]= −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 . f o r ( i =0;i<N ; i++) f o r ( j =0;j<=i ; j++) cout << ”A( ” << i+1 << ” , ” << j+1 << ” )=” << A [ i ] [ j ] << endl ; / / 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’ ; cout << ”COMPUTING WITH DSYEV: ” << endl ; / / LDA: Leading dimension o f A = number o f rows o f / / f u l l array LDA = P ; dsyev_ ( JOBZ , UPLO , N , A , LDA , W , WORK , LWORK , INFO ) ; cout << ”DSYEV: DONE. CHECKING NOW: ” << endl ; i f ( INFO != 0) { cerr << ”DSYEV f a i l e d . INFO= ” << INFO << endl ; exit ( 1 ) ; } / / 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 : cout << ”DSYEV: DONE . : ” << endl ; cout << ”EIGENVALUES OF MATRIX: ” << endl ; cout . precision ( 1 7 ) ; f o r ( i =0;i<N ; i++) cout << ”LAMBDA( ”<< i+1 << ” )=” << W [ i ] << endl ; / / 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 rows o f A: cout << ”EIGENVECTORS OF MATRIX: ” << endl ; f o r ( i =0;i<N ; i++){ cout << ”EIGENVECTOR ” << i+1 << ” FOR EIGENVALUE ” << W [ i ] << endl ; f o r ( j =0;j<N ; j++) cout << ”V_” << i+1 << ” ( ” << j+1 << ” )= ” << A [ i ] [ j ] << endl ; } } / / main ( ) 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 9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 383 libraries Lapack and BLAS⁵ 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 commands⁶: > 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.cpp, the compilation/linking command is: > g++ test . cpp −o test −L / usr / lib −llapack −lblas The option -L/usr/lib instructs the linker to look for libraries in the /usr/lib directory⁷, 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 ) =−21.411990695806409 LAMBDA ( 2 ) =−9.9339436575643028 LAMBDA ( 3 ) =−2.5576560809720039 LAMBDA ( 4 )= 18.803590434342716 EIGENVECTORS OF MATRIX : EIGENVECTOR 1 FOR EIGENVALUE −21.411990695806409 ⁵The 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. ⁶If the .so ﬁles don’t exist in your system, try ar -t /usr/lib/libblas.a etc. ⁷This 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. 384 CHAPTER 9. THE ANHARMONIC OSCILLATOR V _ 1 ( 1 ) = −0.19784566233322534 V _ 1 ( 2 )= −0.4647986784623277 V _ 1 ( 3 )= −0.85469100929950759 V _ 1 ( 4 )= 0.1196769026094445 EIGENVECTOR 2 FOR EIGENVALUE −9.9339436575643028 V _ 2 ( 1 )= 0.82441241087467854 V _ 2 ( 2 )= −0.13242939824916203 V _ 2 ( 3 )= −0.19107651157591365 V _ 2 ( 4 )= −0.51603914386327754 EIGENVECTOR 3 FOR EIGENVALUE −2.5576560809720039 V _ 3 ( 1 )= 0.50268419698022238 V _ 3 ( 2 )= −0.24778437244453691 V _ 3 ( 3 )= 0.13285333709059657 V _ 3 ( 4 )= 0.81747262565942114 EIGENVECTOR 4 FOR EIGENVALUE 18.803590434342716 V _ 4 ( 1 )= 0.16884865648881003 V _ 4 ( 2 )= 0.83965918547426488 V _ 4 ( 3 )= −0.46405068276047351 V _ 4 ( 4 )= 0.22609632301334964 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) 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.cpp: # include < iostream > # include < fstream > # include <cstdlib > # include <string > # include <cmath> using namespace std ; 9.2. CALCULATION OF THE EIGENVALUES OF HN M (λ) 385 / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− const int P = 1000; / / P=LDA c o n s t i n t LWORK = 3* P −1; int DIM ; double H [ P ] [ P ] , X [ P ] [ P ] , X4 [ P ] [ P ] ; double E [ P ] , WORK [ LWORK ] ; double lambda ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e x t e r n ”C” void dsyev_ ( c o n s t char& JOBZ , c o n s t char& UPLO , const in t & N, double H [ P ] [ P ] , c o n s t i n t & LDA , double E [P] , double WORK [ P ] , c o n s t i n t & LWORK , i n t & INFO ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_X4 ( ) ; void calculate_evs ( ) ; void calculate_H ( ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; cout << ” # Enter H i l b e r t Space dimension : \ n” ; cin >> DIM ; getline ( cin , buf ) ; cout << ” # Enter lambda : \ n” ; cin >> lambda ; getline ( cin , buf ) ; cout << ” # lambda= ” << lambda << endl ; cout << ” # ########################################\n” ; cout << ” # Energy spectrum o f anharmonic o s c i l l a t o r \n” ; cout << ” # using matrix methods . \ n” ; cout << ” # H i l b e r t Space Dimension DIM = ”<<DIM<< endl ; cout << ” # lambda c o u p l i n g = ” << lambda << endl ; cout << ” # ########################################\n” ; cout << ” # Output : DIM lambda E_0 E_1 . . . . E_{N−1} \n” ; cout << ” # −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n” ; cout . precision ( 1 5 ) ; / / C a l c u l a t e X^4 o p e r a t o r : calculate_X4 ( ) ; / / Calculate eigenvalues : calculate_evs ( ) ; cout . precision ( 1 7 ) ; cout << ”EV ” << DIM << ” ” << lambda << ” ” ; f o r ( i n t n =0;n<DIM ; n++) cout << E [ n ] << ” ” ; cout << endl ; } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 386 CHAPTER 9. THE ANHARMONIC OSCILLATOR void calculate_evs ( ) { i n t INFO ; c o n s t char JOBZ= ’V ’ , UPLO= ’U’ ; calculate_H ( ) ; dsyev_ ( JOBZ , UPLO , DIM , H , P , E , WORK , LWORK , INFO ) ; i f ( INFO != 0) { cerr << ” dsyev f a i l e d . INFO= ” << INFO << endl ; exit ( 1 ) ; } cout << ” # * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * \ n” ; f o r ( i n t n =0;n<DIM ; n++){ cout << ” # EVEC ” << lambda << ” ” ; f o r ( i n t m =0;m<DIM ; m++) cout << H [ n ] [ m ] << ” ” ; cout << ’\n ’ ; } } / / calculate_evs () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_H ( ) { double X2 [ P ] [ P ] ; f o r ( i n t n =0;n<DIM ; n++){ f o r ( i n t m =0;m<DIM ; m++) H [ n ] [ m ] = lambda * X4 [ n ] [ m ] ; H [ n ] [ n]+= n + 0 . 5 ; } cout << ” # * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * \ n” ; f o r ( i n t n =0;n<DIM ; n++){ cout << ” # HH ” ; f o r ( i n t m =0;m<DIM ; m++) cout << H [ n ] [ m ] << ” ” ; cout << ’\n ’ ; } cout << ” # * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * \ n” ; } / / calculate_H () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_X4 ( ) { double X2 [ P ] [ P ] ; c o n s t double isqrt2 = 1 . 0 / sqrt ( 2 . 0 ) ; f o r ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++) X [ n ] [ m ]=0.0; f o r ( i n t n =0;n<DIM ; n++){ i n t m=n −1; i f ( m>=0) X [ n ] [ m ] = isqrt2 * sqrt ( double ( m +1) ) ; 9.3. RESULTS 387 m =n +1; i f ( m<DIM ) X [ n ] [ m ] = isqrt2 * sqrt ( double ( m )); } / / X2 = X . X for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ X2 [ n ] [ m ] = 0 . 0 ; f o r ( i n t k =0;k<DIM ; k++) X2 [ n ] [ m ] += X [ n ] [ k ] * X [ k ] [ m ] ; } / / X4 = X2 . X2 for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ X4 [ n ] [ m ] = 0 . 0 ; f o r ( i n t k =0;k<DIM ; k++) X4 [ n ] [ m ] += X2 [ n ] [ k ] * X2 [ k ] [ m ] ; } } / / calculate_X4 () 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. 9.3 Results Compiling and running the program can be done with the commands: > g++ −O2 anharmonic . cpp −o an −llapack −lblas > . / an 388 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). # Enter H i l b e r t Space dimension : 4 # Enter lambda : 0.0 ..... # ***************** H ***************** # HH 0.5 0 0 0 # HH 0 1 . 5 0 0 # HH 0 0 2.5 0 # HH 0 0 0 3.5 # ***************** H ***************** # * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * # EVEC 0 1 0 0 0 # EVEC 0 0 1 0 0 # EVEC 0 0 0 1 0 # EVEC 0 0 0 0 1 EV 4 0 0.5 1 . 5 2.5 3.5 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 9.3. RESULTS 389 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. matrix Hnm (λ), m, n = 0, . . . , N − 1 and one expects that En (λ) = lim En (N, λ) , (9.22) N →∞ 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 below⁸: > tcsh > g++ −O2 anharmonic . cpp −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 ⁸The 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. 390 CHAPTER 9. THE ANHARMONIC OSCILLATOR > grep ^ EV data | awk ’{ p r i n t $2 , $4 } ’ 4 0.711467845686790 8 0.786328966767866 12 0.785237674919165 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[n][m] . (9.27) 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. 9.3. RESULTS 391 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 results are shown in table 9.1 and in ﬁgures 9.4-9.5. The calculation is left as an exercise to the reader. 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 obtain⁹ 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) ⁹For 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). 392 CHAPTER 9. THE ANHARMONIC OSCILLATOR 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. In reference [44] 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) with an accuracy better than one part in 106 . 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. 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 9.4. THE DOUBLE WELL POTENTIAL 393 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. 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 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 Table 9.2: Numerical calculation of the energy levels of the anharmonic oscillator given in reference [44]. 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 CHAPTER 9. THE ANHARMONIC OSCILLATOR 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 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 394 200 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 9.4. THE DOUBLE WELL POTENTIAL 395 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). anharmonic.cpp. We will only add a routine that calculates the matrices pnm . The resulting program can be found in the ﬁle doublewell.cpp: / / ======================================================== // H : Hamiltonian o p e r a t o r H0+( lambda / 4 ) *X^4 / / 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 n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− const int P = 1000; / / P=LDA 396 CHAPTER 9. THE ANHARMONIC OSCILLATOR 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. c o n s t i n t LWORK = 3* P −1; int DIM ; double H [ P ] [ P ] , H0 [ P ] [ P ] ; double X [ P ] [ P ] , X2 [ P ] [ P ] , X4 [ P ] [ P ] ; double iP [ P ] [ P ] , P2 [ P ] [ P ] ; double E [ P ] , WORK [ LWORK ] ; double lambda ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e x t e r n ”C” void dsyev_ ( c o n s t char& JOBZ , c o n s t char& UPLO , const int & N, double H [ P ] [ P ] , c o n s t i n t & LDA , double E [P] , double WORK [ P ] , c o n s t i n t & LWORK , i n t & INFO ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_ops ( ) ; void calculate_evs ( ) ; void calculate_H ( ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { string buf ; cout << ” # Enter H i l b e r t Space dimension : \ n” ; cin >> DIM ; getline ( cin , buf ) ; cout << ” # Enter lambda : \ n” ; cin >> lambda ; getline ( cin , buf ) ; cout << ” # lambda= ” << lambda << endl ; cout << ” # ########################################\n” ; cout << ” # Energy l e v e l s o f double w e l l p o t e n t i a l \n” ; 9.4. THE DOUBLE WELL POTENTIAL 397 cout << ”# using matrix methods . \ n” ; cout << ”# H i l b e r t Space Dimension DIM = ”<<DIM<< endl ; cout << ”# lambda c o u p l i n g = ” << lambda << endl ; cout << ”# ########################################\n” ; cout << ”# Output : DIM lambda E_0 E_1 . . . . E_{N−1} \n” ; cout << ”# −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n” ; cout . precision ( 1 5 ) ; / / Calculate operators : calculate_ops ( ) ; / / Calculate eigenvalues : calculate_evs ( ) ; cout . precision ( 1 7 ) ; cout << ”EV ” << DIM << ” ” << lambda << ” ” ; f o r ( i n t n =0;n<DIM ; n++) cout << E [ n ] << ” ” ; cout << endl ; } / / main ( ) / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_evs ( ) { i n t INFO ; c o n s t char JOBZ= ’V ’ , UPLO= ’U’ ; calculate_H ( ) ; dsyev_ ( JOBZ , UPLO , DIM , H , P , E , WORK , LWORK , INFO ) ; i f ( INFO != 0) { cerr << ” dsyev f a i l e d . INFO= ” << INFO << endl ; exit ( 1 ) ; } cout << ” # * * * * * * * * * * * * * * * * * EVEC * * * * * * * * * * * * * * * * * \ n” ; f o r ( i n t n =0;n<DIM ; n++){ cout << ” # EVEC ” << lambda << ” ” ; f o r ( i n t m =0;m<DIM ; m++) cout << H [ n ] [ m ] << ” ” ; cout << ’\n ’ ; } } / / calculate_evs () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_H ( ) { double X2 [ P ] [ P ] ; f o r ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++) H [ n ] [ m ] = H0 [ n ] [ m ] + 0.2 5 * lambda * X4 [ n ] [ m ] ; cout << ” # * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * \ n” ; f o r ( i n t n =0;n<DIM ; n++){ cout << ” # HH ” ; f o r ( i n t m =0;m<DIM ; m++) cout << H [ n ] [ m ] << ” ” ; 398 CHAPTER 9. THE ANHARMONIC OSCILLATOR cout << ’\n ’ ; } cout << ” # * * * * * * * * * * * * * * * * * H * * * * * * * * * * * * * * * * * \ n” ; } / / calculate_H () / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void calculate_ops ( ) { double X2 [ P ] [ P ] ; c o n s t double isqrt2 = 1 . 0 / sqrt ( 2 . 0 ) ; f o r ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ X [ n ] [ m ] = 0 . 0 ; iP [ n ] [ m ] = 0 . 0 ; } f o r ( i n t n =0;n<DIM ; n++){ i n t m=n −1; i f ( m>=0 ) X [ n ] [ m ] = isqrt2 * sqrt ( double ( m +1) ) ; i f ( m>=0 ) iP [ n ] [ m ] = −isqrt2 * sqrt ( double ( m +1) ) ; m =n +1; i f ( m<DIM ) X [ n ] [ m ] = isqrt2 * sqrt ( double ( m ) ) ; i f ( m<DIM ) iP [ n ] [ m ] = isqrt2 * sqrt ( double ( m ) ) ; } / / X2 = X . X for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ X2 [ n ] [ m ] = 0 . 0 ; f o r ( i n t k =0;k<DIM ; k++) X2 [ n ] [ m ] += X [ n ] [ k ] * X [ k ] [ m ] ; } / / X4 = X2 . X2 for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ X4 [ n ] [ m ] = 0 . 0 ; f o r ( i n t k =0;k<DIM ; k++) X4 [ n ] [ m ] += X2 [ n ] [ k ] * X2 [ k ] [ m ] ; } / / P2 =−i P . i P for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++){ P2 [ n ] [ m ] = 0 . 0 ; f o r ( i n t k =0;k<DIM ; k++) P2 [ n ] [ m ] −= iP [ n ] [ k ] * iP [ k ] [ m ] ; } / / Hamiltonian : for ( i n t n =0;n<DIM ; n++) f o r ( i n t m =0;m<DIM ; m++) H0 [ n ] [ m ] = 0 . 5 * ( P2 [ n ] [ m]−X2 [ n ] [ m ] ) ; } / / calculate_ops () 9.4. THE DOUBLE WELL POTENTIAL 399 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 . Also¹⁰ 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) the coeﬃcients cm in the columns of the matrix H[DIM][DIM] so that (n) cm ≈ H[n][m], you √ may compare the value of ⟨x⟩± (λ) with the classical values x0 = ±1/ λ as λ is increased. ¹⁰You may convince yourselves by looking at the wave functions in ﬁgures 10.4 of chapter 10 and by computing the relevant integrals. 400 CHAPTER 9. THE ANHARMONIC OSCILLATOR 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.cpp. 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. 9.5. PROBLEMS 401 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.cpp 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.cpp 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.cpp so that the arrays H, X, X4, E, WORK are dynamically allocated and their dimension is determined by the variable DIM read by the program interactively. 9.8 Make an attempt to reproduce the results of Hioe and Montroll [44] 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 [44] 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)? 402 CHAPTER 9. THE ANHARMONIC OSCILLATOR 9.10 Modify the code in the ﬁle anharmonic.cpp 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.cpp.) 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.cpp 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.cpp, 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. 9.5. PROBLEMS 403 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? 404 CHAPTER 9. THE ANHARMONIC OSCILLATOR 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 405 406 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 407 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. 408 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 409 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). 410 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.cpp. 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 411 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.cpp of chapter 4. We copy the function RKSTEP in a local ﬁle rk.cpp. 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.cpp contains the necessary program for the potential of equation (10.17): / / =========================================================== / / f i l e : w e l l I n f S q . cpp // / / Functions used i n RKSTEP f u n c t i o n . Here : ⁵The function in [−1, 0) is determined by the parity of the solution. 412 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 / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double energy ; / /−−−−−−−− 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 double f1 ( c o n s t double& x , c o n s t double& psi , c o n s t double& psip ) { r e t u r n psip ; } / / =========================================================== / /−−−−−−−− 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 ) double f2 ( c o n s t double& x , c o n s t double& psi , c o n s t double& psip ) { double V ; / /−−−−−−− p o t e n t i a l , s e t here : V = 0.0; / /−−−−−−− S c h r o e d i n g e r eq : RHS r e t u r n ( V−energy ) * psi ; } We stress that the energy ϵ = energy is put in the global scope so that it can be accessed by the main program. The main program is in the ﬁle well.cpp. The user enters the pa- rameters (energy, parity, Nx) and the loop while ( iter < 10000) { .... i f ( abs ( psinew ) <= epsilon ) break ; i f ( psinew * psiold < 0.0 ) de = −0.5* de ; energy += de ; .... } 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- 10.2. THE INFINITE POTENTIAL WELL 413 duce its magnitude by a half. The algorithm described on page 410 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 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 : w e l l . cpp // / / 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 psi [ x ] for t r i a l energies / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 10000; double energy ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− void RKSTEP ( double& t , double& x1 , double& x2 , const double& dt ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { double dx , x , epsilon , de ; double psi , psip , psinew , psiold ; double array_psifinal [ 2 * P + 1 ] , array_xstep [ 2 * P + 1 ] ; double * psifinal = array_psifinal+P ; double * xstep = array_xstep +P ; / / p s i f i n a l p o i n t s t o ( a r r a y _ p s i f i n a l +P ) and one can / / use p s i f i n a l [−P ] . . . p s i f i n a l [ P ] . S i m i l a r l y f o r x s t e p int parity , Nx , iter , i , node ; string buf ; / /−−−−−− Input : cout << ” Enter energy , p a r i t y , Nx: \ n” ; 414 CHAPTER 10. SCHRÖDINGER EQUATION cin >> energy >> parity >> Nx ; getline ( cin , buf ) ; i f ( Nx > P ) { cerr << ”Nx>P\n” ; exit ( 1 ) ; } i f ( parity > 0) parity= 1 ; else parity=−1; cout << ” # #######################################\n” ; cout << ” # E s t a r t = ” << energy << ” p a r i t y = ” << parity << endl ; dx = 1 . 0 / ( Nx −1) ; epsilon = 1 . 0 e−6; cout << ” # Nx= ” << Nx << ” dx = ” << dx << ” eps= ” << epsilon << endl ; cout << ” # #######################################\n” ; / /−−−−−− C a l c u l a t e : ofstream myfile ( ” a l l . dat ” ) ; myfile . precision ( 1 7 ) ; cout . precision ( 1 7 ) ; iter = 0; psiold = 0.0; psinew = 1.0; de = 0 . 1 * abs ( energy ) ; while ( iter < 10000) { / / I n i t i a l c o n d i t i o n s a t x=0 x = 0.0; i f ( parity == 1 ) { psi = 1 . 0 ; psip = 0 . 0 ; } else { psi = 0 . 0 ; psip = 1 . 0 ; } myfile << iter << ” ” << energy << ” ” << x << ” ” << psi << ” ” << psip << endl ; / / Use Runge−Kutta t o forward t o x=1 f o r ( i =2;i<=Nx ; i++){ x = ( i−2) * dx ; RKSTEP ( x , psi , psip , dx ) ; myfile << iter << ” ” << energy << ” ” << x << ” ” << psi << ” ” << psip << endl ; } psinew = psi ; cout << iter << ” ” << energy << ” ” << de << ” ” << psinew << endl ; / / Stop i f v a l u e o f p s i c l o s e t o 0 i f ( abs ( psinew ) <= epsilon ) break ; / / Change d i r e c t i o n o f energy s e a r c h : i f ( psinew * psiold < 0.0 ) de = −0.5* de ; energy += de ; 10.2. THE INFINITE POTENTIAL WELL 415 psiold = psinew ; iter ++; } / / while ( i t e r < 10000) myfile . close ( ) ; / / 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 again and s t o r e i t i f ( parity == 1 ) { psi = 1.0; psip = 0.0; node = 0 ; / / count number o f nodes o f f u n c t i o n } else { psi = 0.0; psip = 1.0; node = 1; } x = 0.0; 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; / / Use Runge−Kutta t o move t o x=1 f o r ( i=2 ; i<=Nx ; i++ ) { x = ( i−2) * dx ; 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 ] = psi * parity ; / / Determine z e r o e s o f p s i ( x ) : / / p s i should not be z e r o w i t h i n e p s i l o n : i f ( abs ( psi ) > 2 . 0 * epsilon ) { i f ( psi * psiold < 0 . 0 ) node += 2 ; psiold = psi ; } } / / f o r ( i =2; i <=Nx ; i ++) node ++; / / print final solution : myfile . open ( ” p s i . dat ” ) ; cout << ” F i n a l r e s u l t : E= ” << energy << ” n= ” << node << ” p a r i t y = ” << parity << endl ; myfile << ” # E= ” << energy << ” n= ” << node << ” p a r i t y = ” << parity << endl ; f o r ( i=−(Nx −1) ; i<=(Nx −1) ; i++) myfile << xstep [i] << ” ” << psifinal [ i ] << endl ; myfile . close ( ) ; } / / main ( ) 416 CHAPTER 10. SCHRÖDINGER EQUATION The compilation and running of the program can be done with the com- mands > g++ well . cpp wellInfSq . cpp rk . cpp −o well > . / well Enter energy , parity , Nx : 2.0 1 400 # ####################################### # Estart= 2 parity= 1 # Nx= 400 dx = 0.00250627 eps= 1e−06 # ####################################### 0 2 0.200000000 0.155943694 1 2.2000000000 0.200000000 0.087444801 ..... 28 2.4674072265 1.220703125e−05 −1.950054368e−06 29 2.4674011230 −6.103515625e−06 −7.246215909e−09 Final result : E= 2.4674011230468746 n= 1 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 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.cpp is appro- priately modiﬁed and saved in the ﬁles wellInfTr.cpp and wellInfDbl.cpp ⁶Careful: if the energy levels are too close, we should keep the initial energy constant and change the sign of parity. 10.2. THE INFINITE POTENTIAL WELL 417 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 418. 418 CHAPTER 10. SCHRÖDINGER EQUATION 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.cpp contains the code / /−−−−−−− p o t e n t i a l , s e t here : V = 10.0 * abs ( x ) ; whereas the ﬁle wellInfDbl.cpp contains the code / /−−−−−−− p o t e n t i a l , s e t here : i f ( abs ( x ) <= 0 . 3 ) V = 100.0; else V = 0.0; 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) = (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). 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, 10.3. BOUND STATES 419 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. the energy eigenstates have deﬁnite parity. The other problem can be 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). 420 CHAPTER 10. SCHRÖDINGER EQUATION 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. 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 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 10.3. BOUND STATES 421 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 ). 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 ) 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]. ⁷If we are unlucky enough to pick a point where ψ ′ (xm ) = 0, this criterion will fail. 422 CHAPTER 10. SCHRÖDINGER EQUATION • 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). • 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. ⁸Note that this point changes when we vary ϵ 10.3. BOUND STATES 423 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 function RKSTEP (see page 203) which performs a 4th order Runge– Kutta step. Its code is copied to the ﬁle rk.cpp. The potential v(x) is coded in the function V(x). The boundary con- ditions are programmed in the function 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 functions 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.cpp for the inﬁnite potential well of (10.17). The same ﬁle contains the code for the functions f1, f2: / / =========================================================== / / f i l e : s c h I n f S q . cpp / / =========================================================== / /−−−−− p o t e n t i a l : # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; e x t e r n double energy ; / /−−−−−−−− double V ( c o n s t double& x ) { return 0.0; } / / V( ) / /−−−−−−−− boundary c o n d i t i o n s : void boundary ( c o n s t double& xmin , c o n s t double& xmax , double& psixmin , double& psipxmin , double& psixmax , double& psipxmax ) { / / 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 ; 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. 424 CHAPTER 10. SCHRÖDINGER EQUATION psipxmin = 1 . 0 ; psixmax = 0 . 0 ; psipxmax = −1.0; / / I n i t i a l v a l u e s a t xmin and xmax } / /−−−−−−−− 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 double f1 ( c o n s t double& x , c o n s t double& psi , c o n s t double& psip ) { r e t u r n psip ; } / /−−−−−−−− 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 ) double f2 ( c o n s t double& x , c o n s t double& psi , c o n s t double& psip ) { / /−−−−−−− S c h r o e d i n g e r eq : RHS r e t u r n ( V ( x )−energy ) * psi ; } 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.cpp. 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=0,...,Nx-1 are given at an odd number of Nx equally spaced points by a distance dx using Simpson’s rule. / / =========================================================== // / / F i l e : sch . cpp // / / 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. / / Mathing 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 / / −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 10.3. BOUND STATES 425 / / 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 / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− c o n s t i n t P = 20001; double energy ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− double V ( c o n s t double& x ); double integrate ( double * psi , c o n s t double& dx , const int & Nx ); void boundary ( c o n s t double& xmin , c o n s t double& xmax , double& psixmin , double& psipxmin , double& psixmax , double& psipxmax ) ; void RKSTEP ( double& t , double& x1 , double& x2 , c o n s t double& dt ); / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( ) { i n t Nx , NxL , NxR ; double psi [ P ] , psip [ P ] ; double dx ; double xmin , xmax , xm ; / / l e f t / r i g h t / matching p o i n t s double psixmin , psipxmin , psixmax , psipxmax ; double psileft , psiright , psistep , psinorm ; double psipleft , psipright , psipstep ; double de , epsilon ; double matchlogd , matchold , psiold , norm , x ; int iter , i , imatch , nodes ; string buf ; / / Input : cout << ” # Enter energy , de , xmin , xmax , Nx: \ n” ; cin >> energy >>de>>xmin >>xmax >>Nx ; getline ( cin , buf ) ; / / 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 ( Nx %2 == 0) Nx ++; i f ( Nx >P ) { cerr << ” Error : Nx > P \n” ; exit ( 1 ) ; } i f ( xmin>=xmax ) { cerr << ” Error : xmin>=xmax\n” ; exit ( 1 ) ; } dx = ( xmax−xmin ) / ( Nx −1) ; epsilon = 1 . 0 e−6; boundary ( xmin , xmax , psixmin , psipxmin , psixmax , psipxmax ) ; 426 CHAPTER 10. SCHRÖDINGER EQUATION cout <<” # #######################################\n” ; cout <<” # E s t a r t = ”<< energy <<” de= ”<< de << ”\n” ; cout <<” # Nx= ” << Nx <<” eps= ” << epsilon << ”\n” ; cout <<” # xmin= ”<< xmin << ”xmax= ”<< xmax <<” dx= ” << dx << ”\n” ; cout <<” # p s i ( xmin )= ”<< psixmin <<” p s i p ( xmin )= ”<< psipxmin << ”\n” ; cout <<” # p s i ( xmax )= ”<< psixmax <<” p s i p ( xmax )= ”<< psipxmax << ”\n” ; cout <<” # #######################################\n” ; / / Calculate : ofstream myfile ( ” a l l . dat ” ) ; myfile . precision ( 1 7 ) ; cout . precision ( 1 7 ) ; matchold = 0.0; f o r ( iter =1; iter <=10000; iter ++){ / / 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; f o r ( i =0;i<Nx ; i++){ x = xmin + i * dx ; i f ( imatch < 0 && ( energy−V ( x ) ) > 0 . 0 ) imatch = i ; } i f ( imatch < 100 | | imatch >= Nx −100) imatch = Nx /5 −1; xm = xmin + imatch * dx ; NxL = imatch +1; NxR = Nx−imatch ; / / Evolve wavefunction from t h e l e f t : psi [ 0 ] = psixmin ; psip [ 0 ] = psipxmin ; psistep = psixmin ; psipstep = psipxmin ; f o r ( i =1; i<NxL ; i++){ x = xmin +(i−1) * dx ; / / t h i s i s x b e f o r e t h e s t e p RKSTEP ( x , psistep , psipstep , dx ) ; psi [ i ] = psistep ; psip [ i ] = psipstep ; } / / 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 −1]= psixmax ; psip [ Nx −1]= psipxmax ; psistep = psixmax ; psipstep = psipxmax ; f o r ( i =1; i<NxR ; i++){ x = xmax −(i−1) * dx ; RKSTEP ( x , psistep , psipstep ,−dx ) ; psi [ Nx−i−1] = psistep ; 10.3. BOUND STATES 427 psip [ Nx−i−1] = psipstep ; } psinorm = psistep / psinorm ; psipright = psipstep ; / / Renormalize p s i l so t h a t p s i l (xm)= p s i r (xm) f o r ( i =0;i<NxL −1;i++){ psi [ i ] *= psinorm ; psip [ i ] *= psinorm ; } psipleft *= psinorm ; / / print current solution : f o r ( i =0;i<Nx ; i++){ x = xmin + i * dx ; myfile << iter <<” ”<<energy <<” ” << x <<” ”<<psi [ i]<< ” ” <<psip [ i ] <<”\n” ; } / / 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 c r i t e r i o n ) matchlogd = ( psipright−psipleft ) / ( abs ( psipright )+abs ( psipleft ) ) ; cout << ” # i t e r , energy , de , xm, logd : ” << iter << ” ” << energy << ” ” << de << ” ” << xm << ” ” << matchlogd << ”\n” ; / / break c o n d i t i o n : i f ( abs ( matchlogd )<=epsilon | | abs ( de / energy ) < 1 . 0 e−12 ) break ; i f ( matchlogd * matchold < 0.0 ) de = −0.5* de ; energy += de ; matchold = matchlogd ; } / / f o r ( i t e r =1; i t e r <=10000; i t e r ++) myfile . close ( ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / 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 / sqrt ( norm ) ; / / f o r ( i =0; i <Nx ; i ++) p s i [ i ] *= norm ; / / Count 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 [ 0 ] ; f o r ( i =1; i<Nx −1;i++) i f ( abs ( psi [ i ] ) > epsilon ) { i f ( psiold * psi [ i ] < 0.0 ) nodes ++; psiold = psi [ i ] ; } 428 CHAPTER 10. SCHRÖDINGER EQUATION / / Print final solution : myfile . open ( ” p s i . dat ” ) ; cout << ” F i n a l r e s u l t : E= ” << energy << ” n= ” << nodes << ” norm= ” << norm << endl ; i f ( abs ( matchlogd ) > epsilon ) cout << ” F i n a l r e s u l t : SOS : logd > e p s i l o n . logd= ” << matchlogd << endl ; myfile << ” # E= ” << energy << ” n= ” << nodes << ” norm = ” << norm << endl ; f o r ( i =0;i<Nx ; i++){ x = xmin + i * dx ; myfile << x << ” ” << psi [ i ] << ”\n” ; } myfile . close ( ) ; } / / main ( ) / / ======================================================== / / 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 ] , Nx i s odd // 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 ) / / ======================================================== double integrate ( double * psi , c o n s t double& dx , c o n s t i n t& Nx ) { double Integral ; int i; / / z e r o t h order p o i n t : i = 0; Integral = psi [ i ] * psi [ i ] ; / / odd order p o i n t s : f o r ( i =1; i<=Nx −2;i+=2) Integral += 4 . 0 * psi [ i ] * psi [ i ] ; / / even order p o i n t s : f o r ( i =2;i<=Nx −3;i+=2) Integral += 2 . 0 * psi [ i ] * psi [ i ] ; / / l a s t point : i = Nx −1; Integral += psi [ i ] * psi [ i ] ; / / measure n o r m a l i z a t i o n : Integral *= dx / 3 . 0 ; r e t u r n Integral ; } / / integrate () The reproduction of the results of the previous section for the inﬁnite 10.3. BOUND STATES 429 potential well is left as an exercise. The compilation and running of the program can be done with the commands: > g++ sch . cpp schInfSq . cpp rk . cpp −o s > ./s # Enter energy , de , xmin , xmax , Nx : 1 0.5 −1 1 2000 # ####################################### # Estart= 1 de= 0.5 # Nx= 2001 eps= 1 e−06 # xmin= −1 xmax= 1 dx= 0.001 # p s i ( xmin )= 0 p s i p ( xmin )= 1 # p s i ( xmax )= 0 p s i p ( xmax )= −1 # ####################################### # 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. 430 CHAPTER 10. SCHRÖDINGER EQUATION 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 429. 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 . 10.4. MEASUREMENTS 431 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 432 CHAPTER 10. SCHRÖDINGER EQUATION 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.cpp and is listed below: / / =========================================================== // / / F i l e o b s e r v a b l e s . cpp / / Compile : g++ o b s e r v a b l e s . cpp −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 // / / =========================================================== # i n c l u d e < iostream > # i n c l u d e < fstream > # include <cstdlib > # include <string > # i n c l u d e <cmath> using namespace std ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− ¹¹See the ﬁles observables.cpp, 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. 10.4. MEASUREMENTS 433 double integrate ( double * psi , c o n s t double& dx , const int & Nx ); / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− i n t main ( i n t argc , char * * argv ) { c o n s t i n t P = 50000; int Nx , i ; double xstep [ P ] , psi [ P ] , obs [ P ] ; double xav , pav , x2av , p2av , Dx , Dp , DxDp , energy , h , norm ; string buf ; char * psifile ; i f ( argc != 2) { cerr << ” Usage : ” << argv [ 0 ] << ” < fi l e n a m e >\n” ; exit ( 1 ) ; } psifile = argv [ 1 ] ; ifstream ifile ( psifile ) ; i f ( ! ifile ) { cerr << ” Error r e a d i n g from f i l e ” << psifile << endl ; exit ( 1 ) ; } cout << ” # r e a d i n g wavefunction from f i l e : ” << psifile << endl ; ifile >> buf >> buf >> energy ; getline ( ifile , buf ) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Input data : p s i [ x ] Nx = 0 ; while ( ifile >> xstep [ Nx ] >> psi [ Nx ] ) { Nx ++; i f ( Nx == P ) { cerr << ”Too many p o i n t s \n” ; exit ( 1 ) ; } } i f ( Nx % 2 == 0) Nx−−; h = ( xstep [ Nx−1]−xstep [ 0 ] ) / ( Nx −1) ; / /−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− / / Calculate : / /−−−−−−−−−− norm : f o r ( i =0;i<Nx ; i++) obs [ i ] = psi [ i ] * psi [ i ] ; norm= integrate ( obs , h , Nx ) ; / /−−−−−−−−−− <x> : f o r ( i =0;i<Nx ; i++) obs [ i ] = psi [ i ] * psi [ i ] * xstep [ i ] ; xav = integrate ( obs , h , Nx ) / norm ; / /−−−−−−−−−− <p > / i : obs [ 0 ] = psi [ 0 ] * ( psi [1] − psi [ 0 ] ) / h ; f o r ( i =1; i<Nx −1;i++) obs [ i ] = psi [ i ] * ( psi [ i+1]−psi [ i −1]) / ( 2 . 0 * h ) ; obs [ Nx −1]=psi [ Nx − 1 ] * ( psi [ Nx−1]−psi [ Nx −2]) / h ; pav = −integrate ( obs , h , Nx ) / norm ; / /−−−−−−−−−− <x ^2 >: 434 CHAPTER 10. SCHRÖDINGER EQUATION f o r ( i =0;i<Nx ; i++) obs [ i ] = psi [ i ] * psi [ i ] * xstep [ i ] * xstep [ i ] ; x2av= integrate ( obs , h , Nx ) / norm ; / /−−−−−−−−−− <p^2 >: obs [ 0 ] = psi [ 0 ] * ( psi [2] −2.0* psi [ 1 ] + psi [ 0 ] ) / ( h * h ) ; f o r ( i =1; i<Nx −1;i++) obs [ i ] = psi [ i ] * ( psi [ i +1] −2.0* psi [ i ]+ psi [ i −1]) / ( h * h ) ; obs [ Nx −1] = psi [ Nx −1] * ( psi [ Nx −1] −2.0* psi [ Nx−2]+psi [ Nx −3]) / ( h * h ) ; p2av= −integrate ( obs , h , Nx ) / norm ; / /−−−−−−−−−− Dx Dx = sqrt ( x2av − xav * xav ) ; / /−−−−−−−−−− Dp Dp = sqrt ( p2av − pav * pav ) ; / /−−−−−−−−−− Dx . Dp DxDp = Dx * Dp ; / / Print results : cout . precision ( 1 7 ) ; cout << ” # norm E <x> <p > / i <x^2> <p^2> Dx Dp DxDp\n” ; cout << norm << ” ” << energy << ” ” << xav << ” ” << pav << ” ” << x2av << ” ” << p2av << ” ” << Dx << ” ” << Dp << ” ” << DxDp << endl ; } / / main ( ) / / ======================================================== / / 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 ] , Nx i s odd // 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 ) dx ) / / ======================================================== double integrate ( double * psi , c o n s t double& dx , c o n s t i n t& Nx ) { double Integral ; int i; / / z e r o t h order p o i n t : i = 0; Integral = psi [ i ] ; / / odd order p o i n t s : f o r ( i =1; i<=Nx −2;i+=2) Integral += 4 . 0 * psi [ i ] ; 10.4. MEASUREMENTS 435 / / even order p o i n t s : f o r ( i =2;i<=Nx −3;i+=2) Integral += 2 . 0 * psi [ i ] ; / / l a s t point : i = Nx −1; Integral += psi [ i ] ; / / measure n o r m a l i z a t i o n : Integral *= dx / 3 . 0 ; r e t u r n Integral ; } / / integrate () The program needs to read in the wave function at the points x0 , . . . , xNx−1 in the format produced by the program in sch.cpp. 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++ observables . cpp −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 variables argc and argv. These contain the information on the number of arguments and the arguments of the command line. If the command line comprises of n words, then argc=n. These words are stored in an array of C-style strings argv[0], argv[1], ... , argv[argc-1]. The ﬁrst argument argv[0] is the name of the program, therefore the lines i f ( argc != 2) { cerr << ” Usage : ” << argv [ 0 ] << ” < fi l e n a m e >\n” ; exit ( 1 ) ; } check if there are two arguments on the command line, including the path to the executable ﬁle. If not, it prints an error message containing the name of the program and exits: ¹²The one read from the ﬁle. It is not calculated from the data. 436 CHAPTER 10. SCHRÖDINGER EQUATION > ./o Usage : . / o <filename > The variables argc and argv must be declared as arguments to the main() function: i n t main ( i n t argc , char * * argv ) { ..... } The variable argv is an array of C-style strings¹³, i.e. and array of an array of characters and can be declared as a pointer to a pointer to char. The statements ifstream ifile ( ” p s i . dat ” ) ; i f ( ! ifile ) { exit ( 1 ) ; } attempt to open a ﬁle psi.dat for input and, if this fails, the program is terminated. The statements Nx = 0 ; while ( ifile >> xstep [ Nx ] >> psi [ Nx ] ) { Nx ++;} read in a pair of doubles and store them in the arrays xstep and psi. The loop terminates when it reaches the end of ﬁle or when it fails to read input that can be converted to two doubles. In the end, Nx stores the number of pairs read into the arrays. 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 ¹³A C-style string, not to be confused with variables declared as string, is an array of “null terminated” characters. This means that the sequence of characters ends with the character '\0'. Functions that treat such objects, detect the end of the string using this convention. 10.5. THE ANHARMONIC OSCILLATOR - AGAIN... 437 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.cpp. The changes that we need to make concern the functions V(x), boundary(xmin, xmax, psixmin, psipxmin, psixmax, psipxmax): / / =========================================================== / / f i l e : schHOC . cpp ................ double V ( c o n s t double& x ) { return x*x ; } / /−−−−−−−− boundary c o n d i t i o n s : void boundary ( c o n s t double& xmin , c o n s t double& xmax , double& psixmin , double& psipxmin , double& psixmax , double& psipxmax ) { psixmin = exp ( −0.5* xmin * xmin ) ; psipxmin = −xmin * psixmin ; psixmax = exp ( −0.5* xmax * xmax ) ; psipxmax = −xmax * psixmax ; } ............... The code omitted at the dots is identical to the one discussed in the previous section. The initial conditions are inspired by the asymp- totic behavior of the solutions to Schrödinger’s¹⁴ equation ψ0 (x) ∼ e−x /2 , 2 ¹⁴In 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. 438 CHAPTER 10. SCHRÖDINGER EQUATION 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- 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.cpp, schHOC.cpp. The plot to the right shows the diﬀerence of the results from the known values (10.38). 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.cpp 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. Next, the calculation is repeated for the anharmonic oscillator for λ = 0.5, 2.0. We copy the ﬁle schHOC.cpp to schUOC.cpp and change the potential in the function V(x): 10.5. THE ANHARMONIC OSCILLATOR - AGAIN... 439 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 stat