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Applied Mathematical Methods in Theoretical Physics - Masujima M.

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Published book (Wiley-VCH, 2005) by Michio Masujima, grown out of MIT course notes on integral equations and calculus of variations. Covers function spaces, Green's functions, Volterra and Fredholm equations, Hilbert-Schmidt theory, singular Cauchy-type equations, Wiener-Hopf methods, and nonlinear integral equations. The last chapters treat variational calculus and applications such as Feynman's action principle, Schwinger-Dyson equations and Weyl's gauge principle. A downloaded reference book, not Phil's own work.

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Michio Masujima Applied Mathematical Methods in Theoretical Physics WILEY -VCH Verlag GmbH & Co. KGaATitelei_Masujima 23.12.2004 9:18 Uhr Seite 3 Cover Picture K. SchmidtAll books published by Wiley-VCH are carefully produced. Nevertheless, authors, editors, and publisherdo not warrant the information contained in thesebooks, including this book, to be free of errors.Readers are advised to keep in mind that statements,data, illustrations, procedural details or other itemsmay inadvertently be inaccurate. Library of Congress Card No.: applied for British Library Cataloging-in-Publication Data:A catalogue record for this book is available from theBritish Library Bibliographic information published by Die Deutsche BibliothekDie Deutsche Bibliothek lists this publication in theDeutsche Nationalbibliografie; detailed bibliographicdata is available in theInternet at <http://dnb.ddb.de>. © 2005 WILEY -VCH Verlag GmbH & Co. KGaA, Weinheim All rights reserved (including those of translation into other languages). No part of this book may be repro-duced in any form – nor transmitted or translated intomachine language without written permission fromthe publishers. Registered names, trademarks, etc.used in this book, even when not specifically markedas such, are not to be considered unprotected by law. Printed in the Federal Republic of Germany Printed on acid-free paper Printing Strauss GmbH, Mörlenbach Bookbinding Litges & Dopf Buchbinderei GmbH, Heppenheim ISBN-13: 978- 3-527-40534-3 ISBN-10: 3-527-40534-8Titelei_Masujima 23.12.2004 9:18 Uhr Seite 4 Contents Preface IX Introduction 1 1 Function Spaces, Linear Operators and Green’s Functions 5 1.1 Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Orthonormal System of Functions . . . . . . . . . . . . . . . . . . . . . . . 71 . 3 L i n e a r O p e r a t o r s ................................ 8 1 . 4 E i g e n v a l u e s a n d E i g e n f u n c t i o n s ........................ 1 1 1.5 The Fredholm Alternative . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 . 6 S e l f - a d j o i n t O p e r a t o r s............................. 1 51.7 Green’s Functions for Differential Equations . . . . . . . . . . . . . . . . . 16 1 . 8 R e v i e w o f C o m p l e x A n a l y s i s ......................... 2 1 1 . 9 R e v i e w o f F o u r i e r T r a n s f o r m ......................... 2 8 2 Integral Equations and Green’s Functions 33 2.1 Introduction to Integral Equations . . . . . . . . . . . . . . . . . . . . . . . 33 2.2 Relationship of Integral Equations with Differential Equations and Green’s F u n c t i o n s .................................... 3 9 2.3 Sturm–Liouville System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.4 Green’s Function for Time-Dependent Scattering Problem . . . . . . . . . . 48 2.5 Lippmann–Schwinger Equation . . . . . . . . . . . . . . . . . . . . . . . . 52 2 . 6 P r o b l e m s f o r C h a p t e r 2 ............................ 5 7 3 Integral Equations of Volterra Type 63 3.1 Iterative Solution to V olterra Integral Equation of the Second Kind . . . . . 63 3 . 2 S o l v a b l e c a s e s o f V o l t e r r a I n t e g r a l E q u a t i o n ................. 6 6 3 . 3 P r o b l e m s f o r C h a p t e r 3 ............................ 7 1 4 Integral Equations of the Fredholm Type 75 4.1 Iterative Solution to the Fredholm Integral Equation of the Second Kind . . 75 4 . 2 R e s o l v e n t K e r n e l ................................ 7 84 . 3 P i n c h e r l e – G o u r s a t K e r n e l........................... 8 14.4 Fredholm Theory for a Bounded Kernel . . . . . . . . . . . . . . . . . . . . 86 4 . 5 S o l v a b l e E x a m p l e ............................... 9 3 VI Contents 4.6 Fredholm Integral Equation with a Translation Kernel . . . . . . . . . . . . 95 4.7 System of Fredholm Integral Equations of the Second Kind . . . . . . . . . 1004 . 8 P r o b l e m s f o r C h a p t e r 4 ............................ 1 0 1 5 Hilbert–Schmidt Theory of Symmetric Kernel 109 5 . 1 R e a l a n d S y m m e t r i c M a t r i x.......................... 1 0 95 . 2 R e a l a n d S y m m e t r i c K e r n e l.......................... 1 1 15.3 Bounds on the Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . 1225 . 4 R a y l e i g h Q u o t i e n t............................... 1 2 65.5 Completeness of Sturm–Liouville Eigenfunctions . . . . . . . . . . . . . . 1295 . 6 G e n e r a l i z a t i o n o f H i l b e r t – S c h m i d t T h e o r y .................. 1 3 15.7 Generalization of Sturm–Liouville System . . . . . . . . . . . . . . . . . . 1385 . 8 P r o b l e m s f o r C h a p t e r 5 ............................ 1 4 4 6 Singular Integral Equations of Cauchy Type 149 6 . 1 H i l b e r t P r o b l e m ................................ 1 4 96 . 2 C a u c h y I n t e g r a l E q u a t i o n o f t h e F i r s t K i n d.................. 1 5 36.3 Cauchy Integral Equation of the Second Kind . . . . . . . . . . . . . . . . 1576 . 4 C a r l e m a n I n t e g r a l E q u a t i o n .......................... 1 6 16 . 5 D i s p e r s i o n R e l a t i o n s .............................. 1 6 66 . 6 P r o b l e m s f o r C h a p t e r 6 ............................ 1 7 3 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation 177 7.1 The Wiener–Hopf Method for Partial Differential Equations . . . . . . . . . 1777.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind . . . . . 191 7.3 General Decomposition Problem . . . . . . . . . . . . . . . . . . . . . . . 207 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind . . . . 216 7.5 Toeplitz Matrix and Wiener–Hopf Sum Equation . . . . . . . . . . . . . . . 227 7.6 Wiener–Hopf Integral Equation of the First Kind and Dual Integral Equations 2357 . 7 P r o b l e m s f o r C h a p t e r 7 ............................ 2 3 9 8 Nonlinear Integral Equations 249 8.1 Nonlinear Integral Equation of V olterra type . . . . . . . . . . . . . . . . . 249 8.2 Nonlinear Integral Equation of Fredholm Type . . . . . . . . . . . . . . . . 253 8.3 Nonlinear Integral Equation of Hammerstein type . . . . . . . . . . . . . . 257 8 . 4 P r o b l e m s f o r C h a p t e r 8 ............................ 2 5 9 9 Calculus of Variations: Fundamentals 263 9.1 Historical Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2639 . 2 E x a m p l e s .................................... 2 6 79 . 3 E u l e r E q u a t i o n................................. 2 6 79 . 4 G e n e r a l i z a t i o n o f t h e B a s i c P r o b l e m s..................... 2 7 29 . 5 M o r e E x a m p l e s ................................ 2 7 69.6 Differential Equations, Integral Equations, and Extremization of Integrals . . 2789 . 7 T h e S e c o n d V a r i a t i o n ............................. 2 8 3 Contents VII 9 . 8 W e i e r s t r a s s – E r d m a n n C o r n e r R e l a t i o n .................... 2 9 7 9 . 9 P r o b l e m s f o r C h a p t e r 9 ............................ 3 0 0 10 Calculus of Variations: Applications 303 10.1 Feynman’s Action Principle in Quantum Mechanics . . . . . . . . . . . . . 30310.2 Feynman’s V ariational Principle in Quantum Statistical Mechanics . . . . . 308 10.3 Schwinger–Dyson Equation in Quantum Field Theory . . . . . . . . . . . . 312 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics . . . . . . . 3291 0 . 5 W e y l ’ s G a u g e P r i n c i p l e ............................ 3 3 91 0 . 6 P r o b l e m s f o r C h a p t e r 1 0 ............................ 3 5 6 Bibliography 365 Index 373 Preface This book on integral equations and the calculus of variations is intended for use by senior undergraduate students and ﬁrst-year graduate students in science and engineering. Basic fa- miliarity with theories of linear algebra, calculus, differential equations, and complex analysis on the mathematics side, and classical mechanics, classical electrodynamics, quantum mecha-nics including the second quantization, and quantum statistical mechanics on the physics side,is assumed. Another prerequisite for this book on the mathematics side is a sound understand- ing of local and global analysis. This book grew out of the course notes for the last of the three-semester sequence of Methods of Applied Mathematics I (Local Analysis) ,II (Global Analysis) and III (Integral Equations and Calculus of V ariations) taught in the Department of Mathematics at MIT. About two-thirds of the course is devoted to integral equations and the remaining one-third to the calculus of variations. Professor Hung Cheng taught the course on integral equations and thecalculus of variations every other year from the mid 1960s through the mid 1980s at MIT. Since then, younger faculty have been teaching the course in turn. The course notes evolved in the intervening years. This book is the culmination of these joint efforts. There will be the obvious question: Why yet another book on integral equations and the calculus of variations? There are already many excellent books on the theory of integralequations. No existing book, however, discusses the singular integral equations in detail; inparticular, Wiener–Hopf integral equations and Wiener–Hopf sum equations with the notionof the Wiener–Hopf index. In this book, the notion of the Wiener–Hopf index is discussed in detail. This book is organized as follows. In Chapter 1 we discuss the notion of function space, the linear operator, the Fredholm alternative and Green’s functions, to prepare the reader forthe further development of the material. In Chapter 2 we discuss a few examples of integral equations and Green’s functions. In Chapter 3 we discuss integral equations of the V olterra type. In Chapter 4 we discuss integral equations of the Fredholm type. In Chapter 5 we discuss the Hilbert–Schmidt theories of the symmetric kernel. In Chapter 6 we discuss singular inte- gral equations of the Cauchy type. In Chapter 7, we discuss the Wiener–Hopf method for the mixed boundary-value problem in classical electrodynamics, Wiener–Hopf integral equations,and Wiener–Hopf sum equations; the latter two topics being discussed in terms of the notionof the index. In Chapter 8 we discuss nonlinear integral equations of the V olterra, Fredholm and Hammerstein type. In Chapter 9 we discuss the calculus of variations, in particular, the second variations, the Legendre test and the Jacobi test, and the relationship between integralequations and applications of the calculus of variations. In Chapter 10 we discuss Feyn-man’s action principle in quantum mechanics and Feynman’s variational principle, a system X Preface of the Schwinger–Dyson equations in quantum ﬁeld theory and quantum statistical mechanics, Weyl’s gauge principle and Kibble’s gauge principle. A substantial portion of Chapter 10 is taken from my monograph, “ Path Integral Quanti- zation and Stochastic Quantization ”, V ol. 165, Springer Tracts in Modern Physics, Springer, Heidelberg, published in the year 2000. A reasonable understanding of Chapter 10 requires the reader to have a basic understand- ing of classical mechanics, classical ﬁeld theory, classical electrodynamics, quantum mecha- nics including the second quantization, and quantum statistical mechanics. For this reason, Chapter 10 can be read as a side reference on theoretical physics, independently of Chapters 1through 9. The examples are mostly taken from classical mechanics, classical ﬁeld theory, classical electrodynamics, quantum mechanics, quantum statistical mechanics and quantum ﬁeld the-ory. Most of them are worked out in detail to illustrate the methods of the solutions. Thoseexamples which are not worked out in detail are either intended to illustrate the general meth- ods of the solutions or it is left to the reader to complete the solutions. At the end of each chapter, with the exception of Chapter 1, problem sets are given for sound understanding of the content of the main text. The reader is recommended to solve allthe problems at the end of each chapter. Many of the problems were created by Professor Hung Cheng over the past three decades. The problems due to him are designated by the note ‘(Due to H. C.)’. Some of the problems are those encountered by Professor Hung Cheng inthe course of his own research activities. Most of the problems can be solved by the direct application of the method illustrated in the main text. Difﬁcult problems are accompanied by the citation of the original references. The problems for Chapter 10 are mostly taken from classical mechanics, classical electrodynamics,quantum mechanics, quantum statistical mechanics and quantum ﬁeld theory. A bibliography is provided at the end of the book for an in-depth study of the background materials in physics, beside the standard references on the theory of integral equations and the calculus of variations. The instructor can cover Chapters 1 through 9 in one semester or two quarters with a choice of the topic of his or her own taste from Chapter 10. I would like to express many heart-felt thanks to Professor Hung Cheng at MIT, who appointed me as his teaching assistant for the course when I was a graduate student in the Department of Mathematics at MIT, for his permission to publish this book under my single authorship and also for his criticism and constant encouragement without which this book would not have materialized. I would like to thank Professor Francis E. Low and Professor Kerson Huang at MIT, who taught me many of the topics within theoretical physics. I would like to thank Professor Roberto D. Peccei at Stanford University, now at UCLA, who taught me quantum ﬁeld theory and dispersion theory. I would like to thank Professor Richard M. Dudley at MIT, who taught me real analysis and theories of probability and stochastic processes. I would like to thank Professor Herman Chernoff, then at MIT, now at Harvard University, who taught me many topics in mathematicalstatistics starting from multivariate normal analysis, for his supervision of my Ph. D. thesis at MIT. Preface XI I would like to thank Dr. Ali Nadim for supplying his version of the course notes and Dr. Dionisios Margetis at MIT for supplying examples and problems of integral equations from his courses at Harvard University and MIT. The problems due to him are designated bythe note ‘(Due to D. M.)’. I would like to thank Dr. George Fikioris at the National TechnicalUniversity of Athens for supplying the references on the Yagi–Uda semi-inﬁnite arrays. I would like to thank my parents, Mikio and Hanako Masujima, who made my undergrad- uate study at MIT possible by their ﬁnancial support. I also very much appreciate their moralsupport during my graduate student days at MIT. I would like to thank my wife, Mari, and myson, Masachika, for their strong moral support, patience and encouragement during the period of the writing of this book, when the ‘going got tough’. Lastly, I would like to thank Dr. Alexander Grossmann and Dr. Ron Schulz of Wiley-VCH GmbH & Co. KGaA for their administrative and legal assistance in resolving the copyrightproblem with Springer. Michio Masujima Tokyo, Japan, June, 2004 Introduction Many problems within theoretical physics are frequently formulated in terms of ordinary dif- ferential equations or partial differential equations. We can often convert them into integral equations with boundary conditions or with initial conditions built in. We can formally de-velop the perturbation series by iterations. A good example is the Born series for the potentialscattering problem in quantum mechanics. In some cases, the resulting equations are nonlinear integro-differential equations. A good example is the Schwinger–Dyson equation in quantum ﬁeld theory and quantum statistical mechanics. It is the nonlinear integro-differential equation,and is exact and closed. It provides the starting point of Feynman–Dyson type perturbationtheory in conﬁguration space and in momentum space. In some singular cases, the resulting equations are Wiener–Hopf integral equations. These originate from research on the radiative equilibrium on the surface of a star. In the two-dimensional Ising model and the analysis ofthe Y agi–Uda semi-inﬁnite arrays of antennas, among others, we have the Wiener–Hopf sumequation. The theory of integral equations is best illustrated by the notion of functionals deﬁned on some function space. If the functionals involved are quadratic in the function, the integralequations are said to be linear integral equations, and if they are higher than quadratic in the function, the integral equations are said to be nonlinear integral equations. Depending on the form of the functionals, the resulting integral equations are said to be of the ﬁrst kind, of thesecond kind, or of the third kind. If the kernels of the integral equations are square-integrable,the integral equations are said to be nonsingular, and if the kernels of the integral equations are not square-integrable, the integral equations are then said to be singular. Furthermore, depend- ing on whether or not the endpoints of the kernel are ﬁxed constants, the integral equations aresaid to be of the Fredholm type, V olterra type, Cauchy type, or Wiener–Hopf types, etc. Bythe discussion of the variational derivative of the quadratic functional, we can also establish the relationship between the theory of integral equations and the calculus of variations. The integro-differential equations can best be formulated in this manner. Analogies of the theoryof integral equations with the system of linear algebraic equations are also useful. The integral equation of Cauchy type has an interesting application to classical electro- dynamics, namely, dispersion relations. Dispersion relations were derived by Kramers in 1927 and Kronig in 1926, for X-ray dispersion and optical dispersion, respectively. Kramers-Kronig dispersion relations are of very general validity which only depends on the assumption of the causality. The requirement of the causality alone determines the region of analyticity of dielectric constants. In the mid 1950s, these dispersion relations were also derived fromquantum ﬁeld theory and applied to strong interaction physics. The application of the covari-ant perturbation theory to strong interaction physics was impossible due to the large coupling Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 2 Introduction constant. From the mid 1950s to the 1960s, the dispersion-theoretic approach to strong in- teraction physics was the only realistic approach that provided many sum rules. To cite a few, we have the Goldberger–Treiman relation, the Goldberger–Miyazawa–Oehme formulaand the Adler–Weisberger sum rule. In the dispersion-theoretic approach to strong interac-tion physics, experimentally observed data were directly used in the sum rules. The situa- tion changed dramatically in the early 1970s when quantum chromodynamics, the relativistic quantum ﬁeld theory of strong interaction physics, was invented by the use of asymptotically-free non-Abelian gauge ﬁeld theory. The region of analyticity of the scattering amplitude in the upper-half k-plane in quantum ﬁeld theory, when expressed in terms of the Fourier transform, is immediate since quantum ﬁeld theory has microscopic causality. But, the region of analyticity of the scattering ampli-tude in the upper-half k-plane in quantum mechanics, when expressed in terms of the Fourier transform, is not immediate since quantum mechanics does not have microscopic causality. We shall invoke the generalized triangular inequality to derive the region of analyticity of the scattering amplitude in the upper-half k-plane in quantum mechanics. This region of analytic- ity of the scattering amplitudes in the upper-half k-plane in quantum mechanics and quantum ﬁeld theory strongly depends on the fact that the scattering amplitudes are expressed in terms of the Fourier transform. When another expansion basis is chosen, such as the Fourier–Bessel series, the region of analyticity drastically changes its domain. In the standard application of the calculus of variations to the variety of problems in theo- retical physics, we simply write the Euler equation and are rarely concerned with the second variations; the Legendre test and the Jacobi test. Examination of the second variations and the application of the Legendre test and the Jacobi test becomes necessary in some cases of theapplication of the calculus of variations theoretical physics problems. In order to bring thedevelopment of theoretical physics and the calculus of variations much closer, some historical comments are in order here. Euler formulated Newtonian mechanics by the variational principle; the Euler equation. Lagrange began the whole ﬁeld of the calculus of variations. He also introduced the notionof generalized coordinates into classical mechanics and completely reduced the problem to that of differential equations, which are presently known as Lagrange equations of motion, with the Lagrangian appropriately written in terms of kinetic energy and potential energy.He successfully converted classical mechanics into analytical mechanics using the variational principle. Legendre constructed the transformation methods for thermodynamics which are presently known as the Legendre transformations. Hamilton succeeded in transforming the Lagrange equations of motion, which are of the second order, into a set of ﬁrst-order differen-tial equations with twice as many variables. He did this by introducing the canonical momentawhich are conjugate to the generalized coordinates. His equations are known as Hamilton’s canonical equations of motion. He successfully formulated classical mechanics in terms of the principle of least action. The variational principles formulated by Euler and Lagrange apply only to the conservative system. Hamilton recognized that the principle of least actionin classical mechanics and Fermat’s principle of shortest time in geometrical optics are strik- ingly analogous, permitting the interpretation of optical phenomena in mechanical terms and vice versa. Jacobi quickly realized the importance of the work of Hamilton. He noted thatHamilton was using just one particular set of the variables to describe the mechanical systemand formulated the canonical transformation theory using the Legendre transformation. He Introduction 3 duly arrived at what is presently known as the Hamilton–Jacobi equation. He formulated his version of the principle of least action for the time-independent case. Path integral quantization procedure, invented by Feynman in 1942 in the Lagrangian for- malism, is usually justiﬁed by the Hamiltonian formalism. We deduce the canonical formal-ism of quantum mechanics from the path integral formalism. As a byproduct of the discussion of the Schwinger–Dyson equation, we deduce the path integral formalism of quantum ﬁeld theory from the canonical formalism of quantum ﬁeld theory. Weyl’s gauge principle also attracts considerable attention due to the fact that all forces in nature; the electromagnetic force, the weak force and the strong force, can be uniﬁed with Weyl’s gauge principle by the appropriate choice of the grand unifying Lie groups as the gauge group. Inclusion of the gravitational force requires the use of superstring theory. Basic to these are the integral equations and the calculus of variations. 1 Function Spaces, Linear Operators and Green’s Functions 1.1 Function Spaces Consider the set of all complex valued functions of the real variable x, denoted by f(x),g(x),... and deﬁned on the interval (a,b). We shall restrict ourselves to those func- tions which are square-integrable . Deﬁne the inner product of any two of the latter func- tions by (f,g)≡/integraldisplayb af∗(x)g(x)dx, (1.1.1) in which f∗(x)is the complex conjugate of f(x). The following properties of the inner product follow from the deﬁnition (1.1.1). (f,g)∗=( g,f), (f,g+h)=( f,g)+(f,h), (f,αg)= α(f,g), (αf,g)= α∗(f,g),(1.1.2) withαa complex scalar. While the inner product of any two functions is in general a complex number, the inner product of a function with itself is a real number and is non-negative. This prompts us to deﬁne the norm of a function by /bardblf/bardbl≡/radicalbig (f,f)=/bracketleftBigg/integraldisplayb af∗(x)f(x)dx/bracketrightBigg1 2 , (1.1.3) provided that fissquare-integrable , i.e.,/bardblf/bardbl<∞. Equation (1.1.3) constitutes a proper deﬁnition for a norm since it satisﬁes the following conditions, (i) scalar multiplication /bardblαf/bardbl=|α|·/bardblf/bardbl, for all complex α, (ii) positivity /bardblf/bardbl>0, for all f/negationslash=0, /bardblf/bardbl=0, if and only if f=0, (iii) triangular inequality /bardblf+g/bardbl≤/bardblf/bardbl+/bardblg/bardbl.(1.1.4) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 6 1 Function Spaces, Linear Operators and Green’s Functions A very important inequality satisﬁed by the inner product (1.1.1) is the so-called Schwarz inequality which says |(f,g)|≤/bardblf/bardbl·/bardblg/bardbl. (1.1.5) To prove the latter, start with the trivial inequality /bardbl(f+αg)/bardbl2≥0, which holds for any f(x)andg(x)and for any complex number α. With a little algebra, the left-hand side of this inequality may be expanded to yield (f,f)+α∗(g,f)+α(f,g)+αα∗(g,g)≥0. (1.1.6) The latter inequality is true for any α, and is thus true for the value of αwhich minimizes the left-hand side. This value can be found by writing αasa+iband minimizing the left-hand side of Eq. (1.1.6) with respect to the real variables aandb. A quicker way would be to treat αandα∗as independent variables and requiring ∂/∂α and∂/∂α∗of the left hand side of Eq. (1.1.6) to vanish. This immediately yields α=−(g,f)/(g,g)as the value of αat which the minimum occurs. Evaluating the left-hand side of Eq. (1.1.6) at this minimum then yields /bardblf/bardbl2≥|(f,g)|2 /bardblg/bardbl2, (1.1.7) which proves the Schwarz inequality (1.1.5). Once the Schwarz inequality has been established, it is relatively easy to prove the trian- gular inequality (1.1.4)(iii). To do this, we simply begin from the deﬁnition /bardblf+g/bardbl2=(f+g,f+g)=(f,f)+(f,g)+(g,f)+(g,g). (1.1.8) Now the right-hand side of Eq. (1.1.8) is a sum of complex numbers. Applying the usual triangular inequality for complex numbers to the right-hand side of Eq. (1.1.8) yields |Right-hand side of Eq. (1.1.8) |≤/bardblf/bardbl2+|(f,g)|+|(g,f)|+/bardblg/bardbl2 =(/bardblf/bardbl+/bardblg/bardbl)2.(1.1.9) Combining Eqs. (1.1.8) and (1.1.9) ﬁnally proves the triangular inequality (1.1.4)(iii). We remark ﬁnally that the set of functions f(x),g(x),...is an example of a linear vector space , equipped with an inner product and a norm based on that inner product. A similar set of properties, including the Schwarz and triangular inequalities, can be established for other linear vector spaces. For instance, consider the set of all complex column vectors /vectoru,/vectorv,/vectorw,... of ﬁnite dimension n. If we deﬁne the inner product (/vectoru,/vectorv)≡(/vectoru∗)T/vectorv=n/summationdisplay k=1u∗ kvk, (1.1.10) and the related norm /bardbl/vectoru/bardbl≡/radicalbig (/vectoru,/vectoru), (1.1.11) 1.2 Orthonormal System of Functions 7 then the corresponding Schwarz and triangular inequalities can be proven in an identical man- ner yielding |(/vectoru,/vectorv)|≤/bardbl/vectoru/bardbl/bardbl/vectorv/bardbl, (1.1.12) and /bardbl/vectoru+/vectorv/bardbl≤/bardbl/vectoru/bardbl+/bardbl/vectorv/bardbl. (1.1.13) 1.2 Orthonormal System of Functions Two functions f(x)andg(x)are said to be orthogonal if their inner product vanishes, i.e., (f,g)=/integraldisplayb af∗(x)g(x)dx=0. (1.2.1) Af u n c t i o ni ss a i dt ob e normalized if its norm equals to unity, i.e., /bardblf/bardbl=/radicalbig (f,f)=1. (1.2.2) Consider now a set of normalized functions {φ1(x),φ2(x),φ3(x),...}which are mutually orthogonal. Such a set is called an orthonormal set of functions , satisfying the orthonormality condition (φi,φj)=δij=/braceleftBigg 1,ifi=j, 0,otherwise ,(1.2.3) where δijis the Kronecker delta symbol itself deﬁned by Eq. (1.2.3). An orthonormal set of functions {φn(x)}is said to form a basis for a function space ,o rt o becomplete , if any function f(x)in that space can be expanded in a series of the form f(x)=∞/summationdisplay n=1anφn(x). (1.2.4) (This is not the exact deﬁnition of a complete set but it will do for our purposes.) To ﬁnd the coefﬁcients of the expansion in Eq. (1.2.4), we take the inner product of both sides withφ m(x)from the left to obtain (φm,f)=∞/summationdisplay n=1(φm,anφn) =∞/summationdisplay n=1an(φm,φn) =∞/summationdisplay n=1anδmn=am.(1.2.5) 8 1 Function Spaces, Linear Operators and Green’s Functions In other words, for any n, an=(φn,f)=/integraldisplayb aφ∗ n(x)f(x)dx. (1.2.6) An example of an orthonormal system of functions on the interval (−l,l)is the inﬁnite set φn(x)=1 √ 2lexp/bracketleftbigginπx l/bracketrightbigg ,n=0,±1,±2,... (1.2.7) with which the expansion of a square-integrable function f(x)on(−l,l)takes the form f(x)=∞/summationdisplay n=−∞cnexp/bracketleftbigginπx l/bracketrightbigg , (1.2.8a) with cn=1 2l/integraldisplay+l −lf(x)e x p/bracketleftbigg −inπx l/bracketrightbigg , (1.2.8b) which is the familiar complex form of the F ourier series off(x). Finally the Dirac delta function δ(x−x/prime), deﬁned with xandx/primein(a,b), can be expanded in terms of a complete set of orthonormal functions φn(x)in the form δ(x−x/prime)=/summationdisplay nanφn(x) with an=/integraldisplayb aφ∗ n(x)δ(x−x/prime)dx=φ∗ n(x/prime). That is, δ(x−x/prime)=/summationdisplay nφ∗ n(x/prime)φn(x). (1.2.9) The expression (1.2.9) is sometimes taken as the statement which implies the completeness of an orthonormal system of functions . 1.3 Linear Operators An operator can be thought of as a mapping or a transformation which acts on a member of the function space ( i.e., a function) to produce another member of that space (i.e., another function). The operator, typically denoted by a symbol such as L,i ss a i dt ob e linear if it satisﬁes L(αf+βg)=αLf+βLg, (1.3.1) where αandβare complex numbers, and fandgare members of that function space. 1.3 Linear Operators 9 Some trivial examples of linear operators Lare (i) multiplication by a constant scalar, i.e., Lφ=aφ, (ii) taking the third derivative of a function, i.e., Lφ=d3 dx3φ orL=d3 dx3, which is a differential operator, or, (iii) multiplying a function by the kernel, K(x, x/prime), and integrating over (a,b)with respect tox/prime, i.e., Lφ(x)=/integraldisplayb aK(x, x/prime)φ(x/prime)dx/prime, which is an integral operator. An important concept in the theory of the linear operator is that of the adjoint of the operator which is deﬁned as follows. Given the operator L, together with an inner product deﬁned on a vector space, the adjoint Ladjof the operator Lis that operator for which (ψ,Lφ)=(Ladjψ,φ), (1.3.2) is an identity for any two members φandψof the vector space. Actually, as we shall see later, in the case of the differential operators, we frequently need to worry to some extent about theboundary conditions associated with the original and the adjoint problems. Indeed, there often arise additional terms on the right-hand side of Eq. (1.3.2) which involve the boundary points, and a prudent choice of the adjoint boundary conditions will need to be made in order to avoidunnecessary difﬁculties. These issues will be raised in connection with Green’s functions fordifferential equations. As our ﬁrst example of the adjoint operator, consider the liner vector space of n- dimensional complex column vectors /vectoru,/vectorv,... with their associated inner product (1.1.10). In this space, n×nsquare matrices A,B,...with complex entries are linear operators when multiplied by the n-dimensional complex column vectors according to the usual rules of ma- trix multiplication. Consider now the problem of ﬁnding the adjoint A adjof the matrix A. According to the deﬁnition (1.3.2) of the adjoint operator, we search for the matrix Aadjsatis- fying (/vectoru, A/vectorv)=(Aadj/vectoru,/vectorv). (1.3.3) Now, from the deﬁnition of the inner product (1.1.10), we must have /vectoru∗T(Aadj)∗T/vectorv=/vectoru∗TA/vectorv, i.e., (Aadj)∗T=A orAadj=A∗T. (1.3.4) 10 1 Function Spaces, Linear Operators and Green’s Functions That is, the adjoint Aadjof a matrix Ais equal to the complex conjugate of its transpose, which is also known as its Hermitian transpose , Aadj=A∗T≡AH. (1.3.5) As a second example, consider the problem of ﬁnding the adjoint of the linear integral operator L=/integraldisplayb adx/primeK(x, x/prime), (1.3.6) on our function space. By deﬁnition, the adjoint LadjofLis the operator which satisﬁes Eq. (1.3.2). Upon expressing the left-hand side of Eq. (1.3.2) explicitly with the operator L given by Eq. (1.3.6), we ﬁnd (ψ,Lφ)=/integraldisplayb adx ψ∗(x)Lφ(x)=/integraldisplayb adx/prime/bracketleftBigg/integraldisplayb adxK(x, x/prime)ψ∗(x)/bracketrightBigg φ(x/prime). (1.3.7) Requiring Eq. (1.3.7) to be equal to (Ladjψ,φ)=/integraldisplayb adx(Ladjψ(x))∗φ(x) necessitates deﬁning Ladjψ(x)=/integraldisplayb adξK∗(ξ,x)ψ(ξ). Hence the adjoint of integral operator (1.3.6) is found to be Ladj=/integraldisplayb adx/primeK∗(x/prime,x). (1.3.8) Note that, aside from the complex conjugation of the kernel K(x, x/prime), the integration in Eq. (1.3.6) is carried out with respect to the second argument of K(x, x/prime)while that in Eq. (1.3.8) is carried out with respect to the ﬁrst argument of K∗(x/prime,x). Also, be careful to note which of the variables throughout the above is the dummy variable of integration. Before we end this section, let us deﬁne what is meant by a self-adjoint operator. An oper- atorLis said to be self-adjoint (or Hermitian ) if it is equal to its own adjoint Ladj. Hermitian operators have very nice properties which will be discussed in Section 1.6. Not the least ofthese is that their eigenvalues are real. (Eigenvalue problems are discussed in the next section.) Examples of self-adjoint operators are Hermitian matrices, i.e., matrices which satisﬁes A=A H, and linear integral operators of the type (1.3.6) whose kernel satisfy K(x, x/prime)=K∗(x/prime,x), each on their respective linear spaces and with their respective inner products. 1.4 Eigenvalues and Eigenfunctions 11 1.4 Eigenvalues and Eigenfunctions Given a linear operator Lon a linear vector space, we can set up the following eigenvalue problem Lφn=λnφn(n=1,2,3,...). (1.4.1) Obviously the trivial solution φ(x)=0 always satisﬁes this equation, but it also turns out that for some particular values of λ(called the eigenvalues and denoted by λn), nontrivial solu- tions to Eq. (1.4.1) also exist. Note that for the case of the differential operators on boundeddomains, we must also specify an appropriate homogeneous boundary condition (such thatφ=0 satisﬁes those boundary conditions) for the eigenfunctions φ n(x).W eh a v ea f ﬁ x e dt h e subscript nto the eigenvalues and eigenfunctions under the assumption that the eigenvalues are discrete and that they can be counted (i.e., with n=1 ,2,3,...). This is not always the case. The conditions which guarantee the existence of a discrete (and complete) set of eigenfunctions are beyond the scope of this introductory chapter and will not be discussed. So, for the moment, let us tacitly assume that the eigenvalues λnof Eq. (1.4.1) are discrete and that their eigenfunctions φnform a basis (i.e., a complete set) for their space. Similarly the adjoint Ladjof the operator Lwould posses a set of eigenvalues and eigen- functions satisfying Ladjψm=µmψm(m=1,2,3,...). (1.4.2) It can be shown that the eigenvalues µmof the adjoint problem are equal to complex conju- gates of the eigenvalues λnof the original problem. (We will prove this only for matrices but it remains true for general operators.) That is, if λnis an eigenvalue of L,λ∗ nis an eigenvalue ofLadj. This prompts us to rewrite Eq. (1.4.2) as Ladjψm=λ∗ mψm,(m=1,2,3,...). (1.4.3) It is then a trivial matter to show that the eigenfunctions of the adjoint and original operators are all orthogonal, except those corresponding to the same index ( n=m). To do this, take the inner product of Eq. (1.4.1) with ψmfrom the left, and the inner product of Eq. (1.4.3) withφnfrom the right, to ﬁnd (ψm,L φn)=(ψm,λnφn)=λn(ψm,φn) (1.4.4) and (Ladjψm,φn)=(λ∗ mψm,φn)=λm(ψm,φn). (1.4.5) Subtract the latter two equations and note that their left-hand sides are equal because of the deﬁnition of the adjoint, to get 0=(λn−λm)(ψm,φn). (1.4.6) This implies (ψm,φn)=0 ifλn/negationslash=λm, (1.4.7) 12 1 Function Spaces, Linear Operators and Green’s Functions which proves the desired result. Also, since each φnandψmis determined to within a multi- plicative constant (e.g., if φnsatisﬁes Eq. (1.4.1) so does αφn), the normalization for the latter can be chosen such that (ψm,φn)=δmn=/braceleftBigg 1,forn=m, 0,otherwise .(1.4.8) Now, if the set of eigenfunctions φn(n=1 ,2,...)forms a complete set, any arbitrary function f(x)in the space may be expanded as f(x)=/summationdisplay nanφn(x), (1.4.9) and to ﬁnd the coefﬁcients an, we simply take the inner product of both sides with ψkto get (ψk,f)=/summationdisplay n(ψk,anφn)=/summationdisplay nan(ψk,φn) =/summationdisplay nanδkn=ak, i.e., an=(ψn,f),(n=1,2,3,...). (1.4.10) Note the difference between Eqs. (1.4.9) and (1.4.10) and the corresponding formulas (1.2.4) and (1.2.6) for an orthonormal system of functions. In the present case, neither {φn} nor{ψn}form an orthonormal system, but they are orthogonal to one another. Proof that the eigenvalues of the adjoint matrix are complex conjugates of the eigenvalues of the original matrix. Above, we claimed without justiﬁcation that the eigenvalues of the adjoint of an operator are complex conjugates of those of the original operator. Here we show this for the matrix case. The eigenvalues of a matrix Aare given by det(A−λI)=0. (1.4.11) The latter is the characteristic equation whose nsolutions for λare the desired eigenvalues. On the other hand, the eigenvalues of Aadjare determined by setting det(Aadj−µI)=0. (1.4.12) Since the determinant of a matrix is equal to that of its transpose, we easily conclude that the eigenvalues of Aadjare the complex conjugates of λn. 1.5 The Fredholm Alternative The Fredholm Alternative , which may be also called the Fredholm solvability condition ,i s concerned with the existence of the solution y(x)of the inhomogeneous problem Ly(x)=f(x), (1.5.1) 1.5 The Fredholm Alternative 13 where Lis a given linear operator and f(x)a known forcing term. As usual, if Lis a differ- ential operator, additional boundary or initial conditions must also be speciﬁed. The Fredholm Alternative states that the unknown function y(x)can be determined uniquely if the corresponding homogeneous problem LφH(x)=0 (1.5.2) with homogeneous boundary conditions, has no nontrivial solutions. On the other hand, if the homogeneous problem (1.5.2) does possess a nontrivial solution, then the inhomogeneous problem (1.5.1) has either no solution or inﬁnitely many solutions. What determines the latter is the homogeneous solution ψHto the adjoint problem LadjψH=0. (1.5.3) Taking the inner product of Eq. (1.5.1) with ψHfrom the left, (ψH,L y)=(ψH,f). Then, by the deﬁnition of the adjoint operator (excluding the case wherein Lis a differential operator, to be discussed in Section 1.7.), we have (LadjψH,y)=(ψH,f). The left-hand side of the equation above is zero by the deﬁnition of ψH, Eq. (1.5.3). Thus the criteria for the solvability of the inhomogeneous problem Eq. (1.5.1) is given by (ψH,f)=0. If these criteria are satisﬁed, there will be an inﬁnity of solutions to Eq. (1.5.1) ,o t h e r w i s e Eq. (1.5.1) will have no solution. To understand the above claims, let us suppose that LandLadjpossess complete sets of eigenfunctions satisfying Lφn=λnφn(n=0,1,2,...), (1.5.4a) Ladjψn=λ∗ nψn(n=0,1,2,...), (1.5.4b) with (ψm,φn)=δmn. (1.5.5) The existence of a nontrivial homogeneous solution φH(x)to Eq. (1.5.2), as well as ψH(x) to Eq. (1.5.3), is the same as having one of the eigenvalues λnin Eqs. (1.5.4a), (1.5.4b) to be zero. If this is the case, i.e., if zero is an eigenvalue of Eq. (1.5.4a) and hence Eq. (1.5.4b),we shall choose the subscript n=0 to signify that eigenvalue ( λ 0=0 ), and in that case 14 1 Function Spaces, Linear Operators and Green’s Functions φ0andψ0are the same as φHandψH. The two circumstances in the Fredholm Alternative correspond to cases where zero is an eigenvalue of Eqs. (1.5.4a), (1.5.4b) and where it is not. Let us proceed formally with the problem of solving the inhomogeneous problem Eq. (1.5.1). Since the set of eigenfunctions φnof Eq. (1.5.4a) is assumed to be complete, both the known function f(x)and the unknown function y(x)in Eq. (1.5.1) can presumably be expanded in terms of φn(x): f(x)=∞/summationdisplay n=0αnφn(x), (1.5.6) y(x)=∞/summationdisplay n=0βnφn(x), (1.5.7) where the αnare known (since f(x)is known), i.e., according to Eq. (1.4.10) αn=(ψn,f), (1.5.8) while the βnare unknown. Thus, if all the βncan be determined, then the solution y(x)to Eq. (1.5.1) is regarded as having been found. T ot r yt od e t e r m i n et h e βn, substitute both Eqs. (1.5.6) and (1.5.7) into Eq. (1.5.1) to ﬁnd ∞/summationdisplay n=0λnβnφn=∞/summationdisplay k=0αkφk, (1.5.9) where different summation indices have been used on the two sides to remind the reader that the latter are dummy indices of summation. Next, take the inner product of both sides with ψm(with an index which must be different from the two above) to get ∞/summationdisplay n=0λnβn(ψm,φn)=∞/summationdisplay k=0αk(ψm,φk), or ∞/summationdisplay n=0λnβnδmn=∞/summationdisplay k=0αkδmk, i.e., λmβm=αm. (1.5.10) Thus, for any m=0 ,1,2,..., we can solve Eq. (1.5.10) for the unknowns βmto get βn=αn/λn(n=0,1,2,...), (1.5.11) provided that λnis not equal to zero. Obviously the only possible difﬁculty occurs if one of the eigenvalues (which we take to be λ0) is equal to zero. In that case, equation (1.5.10) with m=0 reads λ0β0=α0(λ0=0 ). (1.5.12) 1.6 Self-adjoint Operators 15 Now if α0/negationslash=0 , then we cannot solve for β0and thus the problem Ly=fhas no solution. On the other hand if α0=0 , i.e., if (ψ0,f)=(ψH,f)=0, (1.5.13) implying that fis orthogonal to the homogeneous solution to the adjoint problem, then Eq. (1.5.12) is satisﬁed by any choice of β0. All the other βn(n=1 ,2,...) are uniquely determined but there are inﬁnitely many solutions y(x)to Eq. (1.5.1) corresponding to the in- ﬁnitely many values possible for β0. The reader must make certain that he or she understands the equivalence of the above with the original statement of the Fredholm Alternative. 1.6 Self-adjoint Operators Operators which are self-adjoint or Hermitian form a very useful class of operators. Theypossess a number of special properties, some of which are described in this section. The ﬁrst important property of self-adjoint operators is that their eigenvalues are real .T o prove this, begin with Lφ n=λnφn, Lφm=λmφm,(1.6.1) and take the inner product of both sides of the former with φmfrom the left, and the latter withφnfrom the right, to obtain (φm,L φn)= λn(φm,φn), (Lφm,φn)= λ∗ m(φm,φn).(1.6.2) For a self-adjoint operator L=Ladj, the two left-hand sides of Eq. (1.6.2) are equal and hence, upon subtraction of the latter from the former, we ﬁnd 0=(λn−λ∗ m)(φm,φn). (1.6.3) Now, if m=n, the inner product (φn,φn)=/bardblφn/bardbl2is nonzero and Eq. (1.6.3) implies λn=λ∗ n, (1.6.4) proving that all the eigenvalues are real. Thus Eq. (1.6.3) can be rewritten as 0=(λn−λm)(φm,φn), (1.6.5) indicating that if λn/negationslash=λm, then the eigenfunctions φmandφnare orthogonal. Thus, upon normalizing each φn, we verify a second important property of self-adjoint operators that (upon normalization) the eigenfunctions of a self-adjoint operator form an orthonormal set . The Fredholm Alternative can also be restated for a self-adjoint operator Lin the following form: The inhomogeneous problem Ly=f(withLself-adjoint) is solvable for y,i ffis orthogonal to all eigenfunctions φ0ofLwith eigenvalue zero (if indeed any exist). If zero is not an eigenvalue of L, the solution is unique. Otherwise, there is no solution if (φ0,f)/negationslash=0 , and an inﬁnite number of solutions if (φ0,f)=0 . 16 1 Function Spaces, Linear Operators and Green’s Functions Diagonalization of Self-adjoint Operators: Any linear operator can be expanded in some sense in terms of any orthonormal basis set. To elaborate on this, suppose that the orthonormal system {ei(x)}i, with(ei,ej)=δijforms a complete set. Any function f(x)can be expanded as f(x)=∞/summationdisplay j=1αjej(x),α j=(ej,f). (1.6.6) Thus the function f(x)can be thought of as an inﬁnite dimensional vector with components αj. Now consider the action of an arbitrary linear operator Lon the function f(x). Obviously Lf(x)=∞/summationdisplay j=1αjLej(x). (1.6.7) ButLacting on ej(x)is itself a function of xwhich can be expanded in the orthonormal basis {ei(x)}i. Thus we write Lej(x)=∞/summationdisplay i=1lijei(x), (1.6.8) wherein the coefﬁcients lijof the expansion are found to be lij=(ei,L ej). Substitution of Eq. (1.6.8) into Eq. (1.6.7) then shows Lf(x)=∞/summationdisplay i=1/parenleftbigg∞/summationdisplay j=1lijαj/parenrightbigg ei(x). (1.6.9) We discover that just as we can think of f(x)as the inﬁnite dimensional vector with compo- nentsαj, we can consider Lto be equivalent to an inﬁnite dimensional matrix with compo- nentslij, and we can regard Eq. (1.6.9) as a regular multiplication of the matrix L(components lij) with the vector f(components αj). However, this equivalence of the operator Lwith the matrix whose components are lij, i.e.,L⇔lij, depends on the choice of the orthonormal set. For a self-adjoint operator L=Ladj, the most natural choice of the basis set is the set of eigenfunctions of L. Denoting these by {φi(x)}i, the components of the equivalent matrix for Ltake the form lij=(φi,L φj)=(φi,λjφj)=λj(φi,φj)=λjδij. (1.6.10) 1.7 Green’s Functions for Differential Equations In this section, we describe the conceptual basis of the theory of Green’s functions . We do this by ﬁrst outlining the abstract themes involved and then by presenting a simple example. More complicated examples will appear in later chapters. Prior to discussing Green’s functions, recall some of the elementary properties of the so- called Dirac delta function δ(x−x/prime). In particular, remember that if x/primeis inside the domain 1.7 Green’s Functions for Differential Equations 17 of integration (a,b), for any well-behaved function f(x),w eh a v e /integraldisplayb aδ(x−x/prime)f(x)dx=f(x/prime), (1.7.1) which can be written as (δ(x−x/prime),f(x)) =f(x/prime), (1.7.2) with the inner product taken with respect to x. Also remember that δ(x−x/prime)is equal to zero for any x/negationslash=x/prime. Suppose now that we wish to solve a differential equation Lu(x)=f(x), (1.7.3) on the domain x∈(a,b)and subject to given boundary conditions, with La differential operator. Consider what happens when a function g(x, x/prime)(which is as yet unknown but will end up being the Green’s function) is multiplied on both sides of Eq. (1.7.3) followed by integration of both sides with respect to xfromatob. That is, consider taking the inner product of both sides of Eq. (1.7.3) with g(x, x/prime)with respect to x. (We suppose everything is real in this section so that no complex conjugation is necessary.) This yields (g(x, x/prime),L u(x)) = (g(x, x/prime),f(x)). (1.7.4) Now by deﬁnition of the adjoint LadjofL, the left-hand side of Eq. (1.7.4) can be written as (g(x, x/prime),L u(x)) = (Ladjg(x, x/prime),u(x)) + boundary terms , (1.7.5) in which, for the ﬁrst time, we explicitly recognize the terms involving the boundary points which arise when Lis a differential operator. The boundary terms on the right-hand side of Eq. (1.7.5) emerge when we integrate by parts. It is difﬁcult to be more speciﬁc than this when we work in the abstract, but our example should clarify what we mean shortly. If Eq. (1.7.5) is substituted back into Eq. (1.7.4), it provides (Ladjg(x, x/prime),u(x)) = ( g(x, x/prime),f(x)) + boundary terms . (1.7.6) So far we have not discussed what function g(x, x/prime)to choose. Suppose we choose that g(x, x/prime)which satisﬁes Ladjg(x, x/prime)=δ(x−x/prime), (1.7.7) subject to appropriately selected boundary conditions which eliminate all the unknown terms within the boundary terms. This function g(x, x/prime)is known as Green’s function. Substituting Eq. (1.7.7) into Eq. (1.7.6) and using property (1.7.2) then yields u(x/prime)=(g(x, x/prime),f(x)) + known boundary terms , (1.7.8) 18 1 Function Spaces, Linear Operators and Green’s Functions x=0 x=1fx() ux() Fig. 1.1: Displacement u(x)of a taut string under the distributed load f(x)withx∈(0,1). which is the solution to the differential equation since everything on the right-hand side is known once g(x, x/prime)has been found. More accurately, if we change x/primetoxin the above and use a different dummy variable ξof integration in the inner product, we have u(x)=/integraldisplayb ag(ξ,x)f(ξ)dξ+known boundary terms . (1.7.9) In summary, to solve the linear inhomogeneous differential equation Lu(x)=f(x) using Green’s function, we ﬁrst solve the equation Ladjg(x, x/prime)=δ(x−x/prime) for Green’s function g(x, x/prime), subject to the appropriately selected boundary conditions, and immediately obtain the solution to our differential equation given by Eq. (1.7.9). The above will we hope become more clear in the context of the following simple example. ❑ Example 1.1. Consider the problem of ﬁnding the displacement u(x)of a taut string under the distributed load f(x)as in Figure 1.1. Solution. The governing ordinary differential equation for the vertical displacement u(x)has the form d2u dx2=f(x)forx∈(0,1) (1.7.10) subject to boundary conditions u(0) = 0 andu(1) = 0 . (1.7.11) To proceed formally, multiply both sides of Eq. (1.7.10) by g(x, x/prime)and integrate from 0to1 with respect to xto ﬁnd /integraldisplay1 0g(x, x/prime)d2u dx2dx=/integraldisplay1 0g(x, x/prime)f(x)dx. 1.7 Green’s Functions for Differential Equations 19 Integrate the left-hand side by parts twice to obtain /integraldisplay1 0d2 dx2g(x, x/prime)u(x)dx +/bracketleftbigg g(1,x/prime)du dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=1−g(0,x/prime)du dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0−u(1)dg(1,x/prime) dx+u(0)dg(0,x/prime) dx/bracketrightbigg =/integraldisplay1 0g(x, x/prime)f(x)dx. (1.7.12) The terms contained within the square brackets on the left-hand side of (1.7.12) are the bound- ary terms. In consequence of the boundary conditions (1.7.11), the last two terms therein vanish. Hence a prudent choice of boundary conditions for g(x, x/prime)w o u l db et os e t g(0,x/prime)=0 andg(1,x/prime)=0. (1.7.13) With that choice, all the boundary terms vanish (this does not necessarily happen for other problems). Now suppose that g(x, x/prime)satisﬁes d2g(x, x/prime) dx2=δ(x−x/prime), (1.7.14) subject to the boundary conditions (1.7.13). Use of Eqs. (1.7.14) and (1.7.13) in Eq. (1.7.12) yields u(x/prime)=/integraldisplay1 0g(x, x/prime)f(x)dx, (1.7.15) as our solution, once g(x, x/prime)has been obtained. Note that, if the original differential operator d2/dx2is denoted by L, its adjoint Ladjis also d2/dx2as found by twice integrating by parts. Hence the latter operator is indeed self-adjoint. The last step involves the actual solution of (1.7.14) subject to (1.7.13). The variable x/prime plays the role of a parameter throughout. With x/primesomewhere between 0and1, Eq. (1.7.14) can actually be solved separately in each domain 0<x<x/primeandx/prime<x< 1. For each of these, we have d2g(x, x/prime) dx2=0 for0<x<x/prime, (1.7.16a) d2g(x, x/prime) dx2=0 forx/prime<x< 1. (1.7.16b) The general solution in each subdomain is easily written down as g(x, x/prime)=Ax+B for0<x<x/prime, (1.7.17a) g(x, x/prime)=Cx+D forx/prime<x< 1, (1.7.17b) 20 1 Function Spaces, Linear Operators and Green’s Functions involving the four unknown constants A,B,CandD. Two relations for the constants are found using the two boundary conditions (1.7.13). In particular, we have g(0,x/prime)=0→B=0, (1.7.18a) g(1,x/prime)=0→C+D=0. (1.7.18b) To provide two more relations which are needed to permit all four of the constants to be determined, we return to the governing equation (1.7.14). Integrate both sides of the latterwith respect to xfromx /prime−εtox/prime+εand take the limit as ε→0to ﬁnd lim ε→0/integraldisplayx/prime+ε x/prime−εd2g(x, x/prime) dx2dx= lim ε→0/integraldisplayx/prime+ε x/prime−εδ(x−x/prime)dx, from which, we obtain dg(x, x/prime) dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime+−dg(x, x/prime) dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime−=1. (1.7.19) Thus the ﬁrst derivative of g(x, x/prime)undergoes a jump discontinuity as xpasses through x/prime. But we can expect g(x, x/prime)itself to be continuous across x/prime, i.e., g(x, x/prime)/vextendsingle/vextendsingle/vextendsingle x=x/prime+=g(x, x/prime)/vextendsingle/vextendsingle/vextendsingle x=x/prime−. (1.7.20) In the above, x/prime+andx/prime−denote points inﬁnitesimally to the right and the left of x/prime, respec- tively. Using solutions (1.7.17a) and (1.7.17b) for g(x, x/prime)in each subdomain, we ﬁnd that Eqs. (1.7.19) and (1.7.20), respectively, imply C−A=1, (1.7.21a) Cx/prime+D=Ax/prime+B. (1.7.21b) Equations (1.7.18a), (1.7.18b) and (1.7.21a), (1.7.21b) can be used to solve for the four con- stants A,B,CandDto yield A=x/prime−1,B =0,C =x/prime,D =−x/prime, from whence our solution (1.7.17) takes the form g(x, x/prime)=/braceleftBigg (x/prime−1)x forx<x/prime, x/prime(x−1) forx>x/prime,(1.7.22a) =x<(x>−1) for  x<=(x+x/prime) 2−|x−x/prime| 2, x>=(x+x/prime) 2+|x−x/prime| 2.(1.7.22b) Physically the Green’s function (1.7.22) represents the displacement of the string subject to a concentrated load δ(x−x/prime)atx=x/primeas in Figure 1.2. For this reason, it is also called the inﬂuence function . 1.8 Review of Complex Analysis 21 x=1 x=0G xx(, ) ′ xx=′ δ()xx−′ Fig. 1.2: Displacement u(x)of a taut string under the concentrated load δ(x−x/prime)atx=x/prime. Having found the inﬂuence function above for a concentrated load, the solution with any given distributed load f(x)is given by Eq. (1.7.15) as u(x)=/integraldisplay1 0g(ξ,x)f(ξ)dξ =/integraldisplayx 0(x−1)ξf(ξ)dξ+/integraldisplay1 xx(ξ−1)f(ξ)dξ =(x−1)/integraldisplayx 0ξf(ξ)dξ+x/integraldisplay1 x(ξ−1)f(ξ)dξ.(1.7.23) Although this example has been rather elementary, we hope it has provided the reader with a basic understanding of what the Green’s function is. More complex, and hence more interesting examples, are encountered in later chapters. 1.8 Review of Complex Analysis Let us review some important results from complex analysis. Cauchy Integral Formula. Letf(z)be analytic on and inside the closed, positively oriented contour C. Then we have f(z)=1 2πi/contintegraldisplay Cf(ζ) ζ−zdζ. (1.8.1) Differentiate this formula with respect to zto obtain d dzf(z)=1 2πi/contintegraldisplay Cf(ζ) (ζ−z)2dζ, and/parenleftbiggd dz/parenrightbiggn f(z)=n! 2πi/contintegraldisplay Cf(ζ) (ζ−z)n+1dζ. (1.8.2) Liouville’s theorem. The only entire functions which are bounded (at inﬁnity) are constants. 22 1 Function Spaces, Linear Operators and Green’s Functions Proof. Suppose that f(z)is entire. Then it can be represented by the Taylor series, f(z)=f(0) + f(1)(0)z+1 2!f(2)(0)z2+···. Now consider f(n)(0). By the Cauchy Integral Formula, we have f(n)(0) =n! 2πi/contintegraldisplay Cf(ζ) ζn+1dζ. Sincef(ζ)is bounded, we have |f(ζ)|≤M. Consider Cto be a circle of radius R, centered at the origin. Then we have /vextendsingle/vextendsingle/vextendsinglef(n)(0)/vextendsingle/vextendsingle/vextendsingle≤n! 2π·2πRM Rn+1=n!·M Rn→0 asR→∞. Thus f(n)(0) = 0 forn=1,2,3,.... Hence f(z)= constant , completing the proof. More generally, (i) Suppose that f(z)is entire and we know |f(z)|≤|z|aasR→∞ , with0<a< 1.W e still ﬁnd f(z)= constant. (ii) Suppose that f(z)is entire and we know |f(z)|≤|z|aasR→∞ , withn−1≤a<n . Thenf(z)is at most a polynomial of degree n−1. Discontinuity theorem. Suppose that f(z)has a branch cut on the real axis from atob.I t has no other singularities and it vanishes at inﬁnity. If we know the difference between the value of f(z)above and below the cut, D(x)≡f(x+iε)−f(x−iε),(a≤x≤b), (1.8.3) withεpositive inﬁnitesimal, then f(z)=1 2πi/integraldisplayb aD(x) (x−z)dx. (1.8.4) 1.8 Review of Complex Analysis 23 Z o1Zx= Z exi=2π Fig. 1.3: The contours of the integration for f(z).CRis the circle of radius Rcen- tered at the origin. Proof. By the Cauchy Integral Formula, we know f(z)=1 2πi/contintegraldisplay Γf(ζ) ζ−zdζ, where Γconsists of the following pieces (see Figure 1.3 ), Γ=Γ 1+Γ2+Γ3+Γ4+CR. The contribution from CRvanishes since |f(z)|→0asR→∞ . Contributions from Γ3 andΓ4cancel each other. Hence we have f(z)=1 2πi/parenleftbigg/integraldisplay Γ1+/integraldisplay Γ2/parenrightbiggf(ζ) ζ−zdζ. OnΓ1,w eh a v e ζ=x+iε with x:a→b, f(ζ)=f(x+iε), /integraldisplay Γ1f(ζ) ζ−zdζ=/integraldisplayb af(x+iε) x−z+iεdx→/integraldisplayb af(x+iε) x−zdx asε→0+. OnΓ2,w eh a v e ζ=x−iε with x:b→a, f (ζ)=f(x−iε), /integraldisplay Γ2f(ζ) ζ−zdζ=/integraldisplaya bf(x−iε) x−z−iεdx→−/integraldisplayb af(x−iε) x−zdx asε→0+. Thus we obtain f(z)=1 2πi/integraldisplayb af(x+iε)−f(x−iε) x−zdx=1 2πi/integraldisplayb aD(x) x−zdx, completing the proof. 24 1 Function Spaces, Linear Operators and Green’s Functions If, in addition, f(z)is known to have other singularities elsewhere, or may possibly be nonzero as |z|→∞ , then it is of the form f(z)=1 2πi/integraldisplayb aD(x) x−zdx+g(z), (1.8.5) withg(z)free of cut on [a,b]. This is a very important result. Memorizing it will give a better understanding of the subsequent sections. Behavior near the endpoints. Consider the case when zis in the vicinity of the endpoint a. The behavior of f(z)asz→ais related to the form of D(x)asx→a. Suppose that D(x) is ﬁnite at x=a,s a yD(a). Then we have f(z)=1 2πi/integraldisplayb aD(a)+D(x)−D(a) x−zdx =D(a) 2πiln/parenleftbiggb−z a−z/parenrightbigg +1 2πi/integraldisplayb aD(x)−D(a) x−zdx.(1.8.6) The second integral above converges as z→aas long as D(x)satisﬁes a Hölder condition (which is implicitly assumed) requiring |D(x)−D(a)|<A|x−a|µ,A , µ > 0. (1.8.7) Thus the endpoint behavior of f(z)asz→ais of the form f(z)=O(ln(a−z)) asz→a, (1.8.8) if D(x)ﬁnite as x→a. (1.8.9) Another possibility is for D(x)to be of the form D(x)→1/(x−a)αwith α<1 asx→a, (1.8.10) since, even with such a singularity in D(x), the integral deﬁning f(z)is well-deﬁned. We claim that in that case, f(z)also behaves as f(z)=O(1/(z−a)α) asz→a, with α<1, (1.8.11) that is, f(z)is less singular than a simple pole. Proof of the claim. Using the Cauchy Integral Formula, we have 1/(z−a)α=1 2πi/integraldisplay Γdζ (ζ−a)α(ζ−z), where Γconsists of the following paths (see Figure 1.4) Γ=Γ 1+Γ2+CR. 1.8 Review of Complex Analysis 25 CR aZ Γ1 Γ2 Fig. 1.4: The contour Γof the integration for 1/(z−a)α. The contribution from CRvanishes as R→∞ . OnΓ1,w es e t ζ−a=r, and(ζ−a)α=rα, 1 2πi/integraldisplay Γ1dζ (ζ−a)α(ζ−z)=1 2πi/integraldisplay+∞ 0dr rα(r+a−z). OnΓ2,w es e t ζ−a=re2πi,and(ζ−a)α=rαe2πiα, 1 2πi/integraldisplay Γ2dζ (ζ−a)α(ζ−z)=e−2πiα 2πi/integraldisplay0 +∞dr rα(r+a−z). Thus we obtain 1/(z−a)α=1−e−2πiα 2πi/integraldisplay+∞ adx (x−a)α(x−z), which may be written as 1/(z−a)α=1−e−2πiα 2πi/bracketleftBigg/integraldisplayb adx (x−a)α(x−z)+/integraldisplay+∞ bdx (x−a)α(x−z)/bracketrightBigg . The second integral above is convergent for zclose to a. Obviously then, we have 1 2πi/integraldisplayb adx (x−a)α(x−z)=O/parenleftbigg1 (z−a)α/parenrightbigg asz→a. A similar analysis can be done as z→b, completing the proof. 26 1 Function Spaces, Linear Operators and Green’s Functions Summary of behavior near the endpoints f(z)=1 2πi/integraldisplayb aD(x)dx x−z,   ifD(x→a)=D(a), then f(z)=O(ln(a−z)), ifD(x→a)=1 (x−a)α,(0<α< 1),then f(z)=O/parenleftBig 1 (z−a)α/parenrightBig ,(1.8.12a)   ifD(x→b)=D(b), then f(z)=O(ln(b−z)), ifD(x→b)=1 (x−b)β,(0<β< 1),then f(z)=O/parenleftBig 1 (z−b)β/parenrightBig .(1.8.12b) Principal Value Integrals. We deﬁne the principal value integral by P/integraldisplayb af(x) x−ydx≡lim ε→0+/bracketleftBigg/integraldisplayy−ε af(x) x−ydx+/integraldisplayb y+εf(x) x−ydx/bracketrightBigg . (1.8.13) Graphically expressed, the principal value integral contour is as in Figure 1.5. As such, to evaluate a principal value integral by doing complex integration, we usually make use ofeither of the two contours as in Figure 1.6. ay−ε y y+ε b Fig. 1.5: The principal value integral contour. a y−ε y+ε b Fig. 1.6: Two contours for the principal value integral (1.8.13). 1.8 Review of Complex Analysis 27 Now, the contour integrals on the right of Figure 1.6 are usually possible and hence the principal value integral can be evaluated. Also, the contributions from the lower semicircle C−and the upper semicircle C+take the forms, /integraldisplay C−f(z) z−ydz=iπf(y),/integraldisplay C+f(z) z−ydz=−iπf(y), asε→0+, as long as f(z)is not singular at y. Mathematically expressed, the principal value integral is given by either of the following two formulae, known as the Plemelj formula , 1 2πiP/integraldisplayb af(x) x−ydx= lim ε→0+1 2πi/integraldisplayb af(x) x−y−iεdx−1 2f(y), (1.8.14a) 1 2πiP/integraldisplayb af(x) x−ydx= lim ε→0+1 2πi/integraldisplayb af(x) x−y+iεdx+1 2f(y). (1.8.14b) These are customarily written as lim ε→0+1 x−y∓iε=P/parenleftbigg1 x−y/parenrightbigg ±iπδ(x−y), (1.8.15a) or may equivalently be written as P/parenleftbigg1 x−y/parenrightbigg = lim ε→0+1 x−y∓iε∓iπδ(x−y). (1.8.15b) Then we interchange the order of the limit ε→0+and the integration over x. The principal value integrand seems to diverge at x=y, but it is actually ﬁnite at x=yas long as f(x)is not singular at x=y. This comes about as follows; 1 x−y∓iε=(x−y)±iε (x−y)2+ε2=(x−y) (x−y)2+ε2±iπ·1 πε (x−y)2+ε2 =(x−y) (x−y)2+ε2±iπδε(x−y),(1.8.16) where δε(x−y)is deﬁned by δε(x−y)≡1 πε (x−y)2+ε2, (1.8.17) with the following properties, δε(x/negationslash=y)→0+asε→0+, δε(x=y)=1 π1 ε→+∞ asε→0+, and /integraldisplay+∞ −∞δε(x−y)dx=1. The ﬁrst term on the right-hand side of Eq. (1.8.16) vanishes at x=ybefore we take the limit ε→0+, while the second term δε(x−y)approaches the Dirac delta function, δ(x−y),a s ε→0+. This is the content of Eq. (1.8.15a). 28 1 Function Spaces, Linear Operators and Green’s Functions 1.9 Review of Fourier Transform The Fourier transform of a function f(x),w h e r e −∞<x< ∞,i sd e ﬁ n e da s ˜f(k)=/integraldisplay∞ −∞dxexp[−ikx]f(x). (1.9.1) There are two distinct theories of the Fourier transforms. (I) Fourier transform of square-integrable functions It is assumed that /integraldisplay∞ −∞dx|f(x)|2<∞. (1.9.2) The inverse Fourier transform is given by f(x)=/integraldisplay∞ −∞dk 2πexp[ikx]˜f(k). (1.9.3) We note that, in this case, ˜f(k)is deﬁned for real k. Accordingly, the inversion path in Eq. (1.9.3) coincides with the entire real axis. It should be borne in mind that Eq. (1.9.1) ismeaningful in the sense of the convergence in the mean, namely, Eq. (1.9.1) means that thereexists ˜f(k)for all real ksuch that lim R→∞/integraldisplay∞ −∞dk/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle˜f(k)−/integraldisplay R −Rdxexp[−ikx]f(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =0. (1.9.4) Symbolically we write ˜f(k)= l.i.m. R→∞/integraldisplayR −Rdxexp[−ikx]f(x). (1.9.5) Similarly in Eq. (1.9.3), we mean that, given ˜f(k), there exists an f(x)such that lim R→∞/integraldisplay∞ −∞dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglef(f)−/integraldisplayR −Rdk 2πexp[ikx]˜f(k)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =0. (1.9.6) We can then prove that /integraldisplay∞ −∞dk/vextendsingle/vextendsingle/vextendsingle˜f(k)/vextendsingle/vextendsingle/vextendsingle2 =2π/integraldisplay∞ −∞dx|f(x)|2, (1.9.7) which is Parseval’s identity for the square-integrable functions. We see that the pair (f(x),˜f(k))deﬁned in this way, consists of two functions with very similar properties. We shall ﬁnd that this situation may change drastically if the condition (1.9.2) is relaxed. 1.9 Review of F ourier Transform 29 (II) Fourier transform of integrable functions We relax the condition on the function f(x)as /integraldisplay∞ −∞dx|f(x)|<∞. (1.9.8) Then we can still deﬁne ˜f(k)for real k. Indeed, from Eq. (1.9.1), we obtain /vextendsingle/vextendsingle/vextendsingle˜f(k: real)/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay∞ −∞dxexp[−ikx]f(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤/integraldisplay∞ −∞dx|exp[−ikx]f(x)|=/integraldisplay∞ −∞dx|f(x)|<∞.(1.9.9) We can further show that the function deﬁned by ˜f+(k)=/integraldisplay0 −∞dxexp[−ikx]f(x) (1.9.10) is analytic in the upper half-plane of the complex kplane, and ˜f+(k)→0 as|k|→∞ with Imk>0. (1.9.11) Similarly, we can show that the function deﬁned by ˜f−(k)=/integraldisplay∞ 0dxexp[−ikx]f(x) (1.9.12) is analytic in the lower half-plane of the complex kplane, and ˜f−(k)→0 as|k|→∞ with Imk<0. (1.9.13) Clearly we have ˜f(k)=˜f+(k)+˜f−(k),k : real. (1.9.14) We can show that ˜f(k)→0 ask→± ∞ ,k : real. (1.9.15) This is a property in common with the Fourier transform of the square-integrable functions. ❑ Example 1.2. Find the Fourier transform of the following function, f(x)=sin(ax) x,a > 0,−∞<x< ∞. (1.9.16) 30 1 Function Spaces, Linear Operators and Green’s Functions Solution. The Fourier transform ˜f(k)is given by ˜f(k)=/integraldisplay∞ −∞dxexp[ikx]sin(ax) x=/integraldisplay∞ −∞dxexp[ikx]exp[iax]−exp[−iax] 2ix =/integraldisplay∞ −∞dxexp[i(k+a)x]−exp[i(k−a)x] 2ix=I(k+a)−I(k−a), where we deﬁne the integral I(b)by I(b)≡/integraldisplay∞ −∞dxexp[ibx] 2ix=/integraldisplay Γdxexp[ibx] 2ix. The contour Γextends from x=−∞ tox=∞with the inﬁnitesimal indent below the realx-axis at the pole x=0 . Noting x=R ex+iImxfor the complex x,w eh a v e I(b)=  2πi·Res/bracketleftBig exp[ibx] 2ix/bracketrightBig x=0=π, b > 0, 0,b < 0. Thus we have ˜f(k)=I(k+a)−I(k−a) =/integraldisplay∞ −∞dxexp[ikx]sin(ax) x=/braceleftBigg π for|k|<a , 0 for|k|>a ,(1.9.17) while at k=±a,w eh a v e ˜f(k=±a)=π 2, which is equal to 1 2[˜f(k=±a+)+˜f(k=±a−)]. ❑ Example 1.3. Find the Fourier transform of the following function, f(x)=sin(ax) x(x2+b2),a , b > 0,−∞<x< ∞. (1.9.18) Solution. The Fourier transform ˜f(k)is given by ˜f(k)=/integraldisplay Γdzexp[i(k+a)z] 2iz(z2+b2)−/integraldisplay Γdzexp[i(k−a)z] 2iz(z2+b2)=I(k+a)−I(k−a),(1.9.19) where we deﬁne the integral I(c)by I(c)≡/integraldisplay∞ −∞dzexp[icz] 2iz(z2+b2)=/integraldisplay Γdzexp[icz] 2iz(z2+b2), (1.9.20) 1.9 Review of F ourier Transform 31 where the contour Γis the same as in Example 1.2. The integrand has simple poles at z=0 andz=±ib. Noting z=R ez+iImz,w eh a v e I(c)=  2πi·Res/bracketleftBig exp[icz] 2iz(z2+b2)/bracketrightBig z=0+2πi·Res/bracketleftBig exp[icz] 2iz(z2+b2)/bracketrightBig z=ib,c > 0, −2πi·Res/bracketleftBig exp[icz] 2iz(z2+b2)/bracketrightBig z=−ib,c < 0, or I(c)=  (π/2b2)(2−exp[−bc]),c > 0, (π/2b2)e x p [bc],c < 0. Thus we have ˜f(k)=I(k+a)−I(k−a) =  (π/b 2) sinh( ab)e x p [bk],k < −a, (π/b2){1−exp[−ab]c o s h( bk)},|k|<a , (π/b2) sinh( ab)e x p [−bk],k > a .(1.9.21) We note that ˜f(k)isstep-discontinuous atk=±ain Example 1.2. We also note that ˜f(k)and˜f/prime(k)are continuous for real k, while ˜f/prime/prime(k)isstep-discontinuous atk=±ain Example 1.3. We note the rate with which f(x)→0 as|x|→+∞ affects the degree of smoothness of ˜f(k). For the square-integrable functions, we usually have f(x)=O/parenleftbigg1 x/parenrightbigg as|x|→+∞⇒ ˜f(k)step-discontinuous , f(x)=O/parenleftbigg1 x2/parenrightbigg as|x|→+∞⇒/braceleftBigg˜f(k)continuous , ˜f/prime(k)step-discontinuous , f(x)=O/parenleftbigg1 x3/parenrightbigg as|x|→+∞⇒/braceleftBigg˜f(k),˜f/prime(k)continuous , ˜f/prime/prime(k)step-discontinuous , and so on. 32 1 Function Spaces, Linear Operators and Green’s Functions *** Having learned in the above the abstract notions relating to linear space, inner product, the operator and its adjoint, eigenvalue and eigenfunction, Green’s function, and having reviewed the Fourier transform and complex analysis, we are now ready to embark on our study ofintegral equations. We encourage the reader to make an effort to connect the concrete example that will follow with the abstract idea of linear function space and the linear operator. This will not be possible in all circumstances. The abstract idea of function space is also useful in the discussion of the calculus of varia- tions where a piecewise continuous, but nowhere differentiable, function and a discontinuousfunction appear as the solution of the problem. We present the applications of the calculus of variations to theoretical physics speciﬁcally, classical mechanics, canonical transformation theory, the Hamilton–Jacobi equation, classicalelectrodynamics, quantum mechanics, quantum ﬁeld theory and quantum statistical mecha-nics. The mathematically oriented reader is referred to the monographs by R. Kress, and I. M. Gelfand and S. V . Fomin for details of the theories of integral equations and the cal-culus of variations. 2 Integral Equations and Green’s Functions 2.1 Introduction to Integral Equations An integral equation is the equation in which the function to be determined appears in an integral. There exist several types of integral equations: Fredholm Integral Equation of the second kind: φ(x)=F(x)+λ/integraldisplayb aK(x, y)φ(y)dy (a≤x≤b), Fredholm Integral Equation of the ﬁrst kind: F(x)=/integraldisplayb aK(x, y)φ(y)dy (a≤x≤b), V olterra Integral Equation of the second kind: φ(x)=F(x)+λ/integraldisplayx 0K(x, y)φ(y)dy with K(x, y)=0 fory>x , V olterra Integral Equation of the ﬁrst kind: F(x)=/integraldisplayx 0K(x, y)φ(y)dy with K(x, y)=0 fory>x . In the above, K(x, y)is the kernel of the integral equation and φ(x)is the unknown function. If F(x)=0 , the equations are said to be homogeneous ,a n di f F(x)/negationslash=0 ,t h e ya r e said to be inhomogeneous . Now, begin with some simple examples of Fredholm Integral Equations. ❑Example 2.1. Inhomogeneous Fredholm Integral Equation of the second kind. φ(x)=x+λ/integraldisplay1 −1xyφ(y)dy, −1≤x≤1. (2.1.1) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 34 2 Integral Equations and Green’s Functions Solution. Since/integraldisplay1 −1yφ(y)dyis some constant, deﬁne A=/integraldisplay1 −1yφ(y)dy. (2.1.2) Then Eq. (2.1.1) takes the form φ(x)=x(1 +λA). (2.1.3) Substituting Eq. (2.1.3) into the right-hand side of Eq. (2.1.2), we obtain A=/integraldisplay1 −1(1 +λA)y2dy=2 3(1 +λA). Solving for A, we obtain /parenleftbigg 1−2 3λ/parenrightbigg A=2 3. Ifλ=3 2,n os u c h Aexists. Otherwise Ais uniquely determined to be A=2 3/parenleftbigg 1−2 3λ/parenrightbigg−1 . (2.1.4) Thus, if λ=3 2, no solution exists. Otherwise, a unique solution exists and is given by φ(x)=x/parenleftbigg 1−2 3λ/parenrightbigg−1 . (2.1.5) We shall now consider the homogeneous counter part of the inhomogeneous Fredholm integral equation of the second kind, considered in Example 2.1. ❑Example 2.2. Homogeneous Fredholm Integral Equation of the second kind. φ(x)=λ/integraldisplay1 −1xyφ(y)dy, −1≤x≤1. (2.1.6) Solution. As in Example 2.1, deﬁne A=/integraldisplay1 −1yφ(y)dy. (2.1.7) Then φ(x)=λAx. (2.1.8) Substituting Eq. (2.1.8) into Eq. (2.1.7), we obtain A=/integraldisplay1 −1λAy2dy=2 3λA. (2.1.9) 2.1 Introduction to Integral Equations 35 The solution exists only when λ=3 2. Thus the nontrivial homogeneous solution exists only forλ=3 2, whence φ(x)is given by φ(x)=αx withαarbitrary. If λ/negationslash=3 2, no nontrivial homogeneous solution exists. We observe the following correspondence in Examples 2.1 and 2.2: Inhomogeneous case .Homogeneous case . λ/negationslash=3/2 Unique solution. Trivial solution. λ=3/2 No solution. Inﬁnitely many solutions.(2.1.10) We further note the analogy of an integral equation to a system of inhomogeneous linear algebraic equations (matrix equations) : (K−µI)/vectorU=/vectorF (2.1.11) where Kis ann×nmatrix, Iis then×nidentity matrix, /vectorUand/vectorFaren-dimensional vectors, andµis a number. Equation (2.1.11) has the unique solution, /vectorU=(K−µI)−1/vectorF, (2.1.12) provided that (K−µI)−1exists, or equivalently that det(K−µI)/negationslash=0. (2.1.13) The homogeneous equation corresponding to Eq. (2.1.11) is (K−µI)/vectorU=0 orK/vectorU=µ/vectorU (2.1.14) which is the eigenvalue equation for the matrix K. A solution to the homogeneous equa- tion (2.1.14) exists for certain values of µ=µn, which are called the eigenvalues. If µis equal to an eigenvalue µn,(K−µI)−1fails to exist and Eq. (2.1.11) has generally no ﬁnite solution. ❑Example 2.3. Change the inhomogeneous term xof Example 2.1 to 1. φ(x)=1+ λ/integraldisplay1 −1xyφ(y)dy,−1≤x≤1. (2.1.15) Solution. As before, deﬁne A=/integraldisplay1 −1yφ(y)dy. (2.1.16) Then φ(x)=1+ λAx. (2.1.17) Substituting Eq. (2.1.17) into Eq. (2.1.16), we obtain A=/integraltext1 −1y(1 +λAy)dy=2 3λA . Thus, forλ/negationslash=3 2, the unique solution exists with A=0 ,a n dφ(x)=1 , while for λ=3 2, inﬁnitely many solutions exist with Aarbitrary and φ(x)=1+3 2Ax . 36 2 Integral Equations and Green’s Functions The above three examples illustrate the Fredholm Alternative : •F o rλ=3 2, the homogeneous problem has a solution, given by φH(x)=αx for any α. •F o rλ/negationslash=3 2, the inhomogeneous problem has a unique solution, given by φ(x)=  x /parenleftbig 1−2 3λ/parenrightbig when F(x)=x, 1 when F(x)=1. •F o r λ=3 2, the inhomogeneous problem has no solution when F(x)=x, while it has inﬁnitely many solutions when F(x)=1 . In the former case, (φH,F)=/integraltext1 −1αx·xd x/negationslash= 0, while in the latter case, (φH,F)=/integraltext1 −1αx·1dx=0 . It is, of course, not surprising that Eq. (2.1.15) has inﬁnitely many solutions when λ=3/2. Generally, if φ0is a solution of an inhomogeneous equation, and φ1is a solution of the corresponding homogeneous equation, then φ0+aφ1is also a solution of the inhomo- geneous equation, where ais any constant. Thus, if λis equal to an eigenvalue, an inhomo- geneous equation has inﬁnitely many solutions as long as it has one solution. The nontrivial question is: Under what condition can we expect the latter to happen? In the present example, the relevant condition is/integraltext1 −1yd y=0 , which means that the inhomogeneous term (which is 1) multiplied by yand integrated from −1to1, is zero. There is a counterpart of this condition for matrix equations. It is well known that, under certain circumstances, the inhomogeneousmatrix equation (2.1.11) has solutions even if µis equal to an eigenvalue. Speciﬁcally this happens if the inhomogeneous term /vectorFis a linear superposition of the vectors each of which forms a column of (K−µI). There is another way to phrase this. Consider all vectors /vectorV satisfying (K T−µI)/vectorV=0, (2.1.18) where KTis the transpose of K. The equation above says that /vectorVis an eigenvector of KT with the eigenvalue µ. It also says that /vectorVis perpendicular to all row vectors of (KT−µI).I f /vectorFis a linear superposition of the column vectors of (K−µI)(which are the row vectors of (KT−µI)), then /vectorFis perpendicular to /vectorV. Therefore, the inhomogeneous equation (2.1.11) has solutions when µis an eigenvalue, if and only if /vectorFis perpendicular to all eigenvectors of KTwith eigenvalue µ. Similarly an inhomogeneous integral equation has solutions even when λis equal to an eigenvalue, as long as the inhomogeneous term is perpendicular to all of the eigenfunctions of the transposed kernel (the kernel with x↔y) of that particular eigenvalue. As we have seen in Chapter 1, just as a matrix, a kernel and its transpose have the same eigenvalues. Consequently the homogeneous integral equation with the transposed kernel has no solution if λis not equal to an eigenvalue of the kernel. Therefore, if λis not an eigen- value, any inhomogeneous term is trivially perpendicular to all solutions of the homogeneousintegral equation with the transposed kernel, since all of them are trivial. Together with the 2.1 Introduction to Integral Equations 37 result in the preceding paragraph, we have arrived at the necessary and sufﬁcient condition for an inhomogeneous integral equation to have a solution: the inhomogeneous term must be perpendicular to all solutions of the homogeneous integral equation with the transposedkernel. There exists another kind of integral equation in which the unknown appears only in the integrals. Consider one more example of a Fredholm Integral Equation. ❑Example 2.4. Fredholm Integral Equation of the ﬁrst kind. Case(A) 1=/integraldisplay 1 0xyφ(y)dy, 0≤x≤1. (2.1.19) Case(B) x=/integraldisplay1 0xyφ(y)dy, 0≤x≤1. (2.1.20) Solution. In both cases, divide both sides of the equations by xto obtain Case(A) 1 x=/integraldisplay1 0yφ(y)dy. (2.1.21) Case(B) 1=/integraldisplay1 0yφ(y)dy. (2.1.22) In the case of Eq. (2.1.21), no solution is possible, while in the case of Eq. (2.1.22), inﬁnitely manyφ(x)are possible. Essentially any function ψ(x)which satisﬁes /integraldisplay1 0yψ(y)dy/negationslash=0 or∞, can be made a solution to Eq. (2.1.20). Indeed, φ(x)=ψ(x) /integraldisplay1 0yψ(y)dy(2.1.23) will do. Therefore, for the kind of integral equations considered in Example 2.4, no solution exists for some inhomogeneous terms, while inﬁnitely many solutions exist for some other inhomo-geneous terms. 38 2 Integral Equations and Green’s Functions Next, we shall consider an example of a V olterra Integral Equation of the second kind with the transformation of an integral equation into an ordinary differential equation. ❑Example 2.5. V olterra Integral Equation of the second kind. φ(x)=ax+λx/integraldisplayx 0φ(x/prime)dx/prime. (2.1.24) Solution. Divide both sides of Eq. (2.1.24) by xto obtain φ(x) x=a+λ/integraldisplayx 0φ(x/prime)dx/prime. (2.1.25) Differentiate both sides of Eq. (2.1.25) with respect to xto obtain d dx/parenleftbiggφ(x) x/parenrightbigg =λφ(x). (2.1.26) By setting u(x)=φ(x) x, the following differential equation results, du(x) u(x)=λx dx. (2.1.27) By integrating both sides, lnu(x)=1 2λx2+constant. Hence the solution is given by u(x)=Ae1 2λx2, orφ(x)=Axe1 2λx2. (2.1.28) To determine the integration constant Ain Eq. (2.1.28), note that as x→0, based on the integral equation (2.1.24), φ(x)above behaves as φ(x)→ax+O(x3) (2.1.29) while our solution (2.1.28) behaves as φ(x)→Ax+O(x3). (2.1.30) Hence, from Eqs. (2.1.29) and (2.1.30), we identify A=a. Thus the ﬁnal form of the solution is φ(x)=axe1 2λx2, (2.1.31) which is the unique solution for all λ. 2.2 Relationship of Integral Equations with Differential Equations and Green’s Functions 39 We observe three points: 1. The integral equation (2.1.24) has a unique solution for all values of λ. It follows that the corresponding homogeneous integral equation, obtained from Eq. (2.1.24) by setting a=0 , does not have a nontrivial solution. Indeed, this can be directly veriﬁed by setting a=0 in Eq. (2.1.31). This means that the kernel for Eq. (2.1.24) has no eigenvalues. This is true for all square-integrable kernels of the V olterra type. 2. While the solution to the differential equation (2.1.26) or (2.1.27) contains an arbitrary constant, the solution to the corresponding integral equation (2.1.24) does not. More pre-cisely, Eq. (2.1.24) is equivalent to Eq. (2.1.26) or Eq. (2.1.27) plus an initial condition. 3. The transformation of the V olterra Integral Equation of the second kind to an ordinary differential equation is possible whenever the kernel of the V olterra integral equation is a sum of the factored terms. In the above example, we solved the integral equation by transforming it into a differential equation. This is not often possible. On the other hand, it is, in general, easy to transform a differential equation into an integral equation. However, to avoid any misunderstanding, let me state that we never solve a differential equation by such a transformation. Indeed,an integral equation is much more difﬁcult to solve than a differential equation in a closed form. Only very rarely can this be done. Therefore, whenever it is possible to transform an integral equation into a differential equation, it is a good idea to do so. On the other hand,there are advantages in transforming a differential equation into an integral equation. This transformation may facilitate the discussion of the existence and uniqueness of the solution, the spectrum of the eigenvalue and the analyticity of the solution. It also enables us to obtainthe perturbative solution of the equation. 2.2 Relationship of Integral Equations with Differential Equations and Green’s Functions To provide the reader with a sense of bearing, we shall discuss the transformation of a dif- ferential equation to an integral equation. This transformation is accomplished by the use ofGreen’s functions . As an example, consider the one-dimensional Schrödinger equation with potential U(x): /parenleftbiggd 2 dx2+k2/parenrightbigg φ(x)=U(x)φ(x). (2.2.1) It is assumed that U(x)vanishes rapidly as |x|→∞ . Although Eq. (2.2.1) is most usually thought of as an initial value problem, let us suppose that we are given φ(0) = a, andφ/prime(0) = b, (2.2.2) and we are interested in the solution for x>0. 40 2 Integral Equations and Green’s Functions Green’s function: First treat the right-hand side of Eq. (2.2.1) as an inhomogeneous term f(x). Namely, consider the following inhomogeneous problem: Lφ(x)=f(x) with L=d2 dx2+k2, (2.2.3) and the boundary conditions speciﬁed by Eq. (2.2.2). Multiply both sides of Eq. (2.2.3) by g(x, x/prime)and integrate with respect to xfrom0to∞. Then /integraldisplay∞ 0g(x, x/prime)Lφ(x)=/integraldisplay∞ 0g(x, x/prime)f(x)dx. (2.2.4) Integrate by parts twice on the left-hand side of Eq. (2.2.4) to obtain /integraldisplay∞ 0(Lg(x, x/prime))φ(x)dx+g(x, x/prime)φ/prime(x)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=∞ x=0−dg(x, x/prime) dxφ(x)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=∞ x=0 =/integraldisplay∞ 0g(x, x/prime)f(x)dx. (2.2.5) In the boundary terms, φ/prime(0) andφ(0) are known. To eliminate unknown terms, we require g(∞,x/prime)=0 anddg dx(∞,x/prime)=0. (2.2.6) Also, choose g(x, x/prime)to satisfy Lg(x, x/prime)=δ(x−x/prime). (2.2.7) Then we ﬁnd from Eq. (2.2.5): φ(x/prime)=bg(0,x/prime)−adg dx(0,x/prime)+/integraldisplay∞ 0g(x, x/prime)f(x)dx. (2.2.8) Solution for g(x, x/prime):The governing equation and the boundary conditions are given by /parenleftbiggd2 dx2+k2/parenrightbigg g(x, x/prime)=δ(x−x/prime) onx∈(0,∞) with x/prime∈(0,∞), (2.2.9) Boundary condition 1, g(∞,x/prime)=0, (2.2.10) Boundary condition 2, dg dx(∞,x/prime)=0. (2.2.11) Forx<x/prime, g(x, x/prime)=Asinkx+Bcoskx. (2.2.12) 2.2 Relationship of Integral Equations with Differential Equations and Green’s Functions 41 Forx>x/prime, g(x, x/prime)=Csinkx+Dcoskx. (2.2.13) Applying the boundary conditions, (2.2.10) and (2.2.11) above, results in C=D=0 . Thus g(x, x/prime)=0 forx>x/prime. (2.2.14) Now, integrate the differential equation (2.2.9) across x/primeto obtain dg dx(x/prime+ε, x/prime)−dg dx(x/prime−ε, x/prime)=1, (2.2.15) g(x/prime+ε, x/prime)=g(x/prime−ε, x/prime). (2.2.16) Letting ε→0, we obtain the equations for AandB. /braceleftBigg Asinkx/prime+Bcoskx/prime=0, −Acoskx/prime+Bsinkx/prime=1 k. ThusAandBare determined to be A=−coskx/prime k,B =sinkx/prime k, (2.2.17) and the Green’s function is found to be g(x, x/prime)=  sink(x/prime−x) kforx<x/prime, 0 forx>x/prime.(2.2.18) Equation (2.2.8) becomes φ(x/prime)=bsinkx/prime k+acoskx/prime+/integraldisplayx/prime 0sink(x/prime−x) kf(x)dx. Changing xtoξandx/primetox, and recalling that f(x)=U(x)φ(x),w eﬁ n d φ(x)=acoskx+bsinkx k+/integraldisplayx 0sink(x−ξ) kU(ξ)φ(ξ)dξ, (2.2.19) which is a V olterra Integral Equation of the second kind. Next consider the very important scattering problem for the Schrödinger equation: /parenleftbiggd2 dx2+k2/parenrightbigg φ(x)=U(x)φ(x) on−∞<x< ∞, (2.2.20) where the potential U(x)→0as|x|→∞ . We might expect that /braceleftBigg φ(x)→Aeikx+Be−ikxasx→− ∞ , φ(x)→Ceikx+De−ikxasx→+∞. 42 2 Integral Equations and Green’s Functions Now (with an e−iωtimplicitly multiplying φ(x)), the term eikxrepresents a wave going to the right while e−ikxis a wave going to the left. In the scattering problem, we suppose that there is an incident wave with amplitude 1(i.e.,A=1 ), a reﬂected wave with amplitude R,( i . e . , B=R) and a transmitted wave with amplitude T(i.e.,C=T). Both RandTare unknown still. Also as x→+∞, there is no left-going wave (i.e., D=0 ). Thus the problem is to solve /parenleftbiggd2 dx2+k2/parenrightbigg φ(x)=U(x)φ(x), (2.2.21) with boundary conditions, /braceleftBigg φ(x→− ∞ )= eikx+Re−ikx, φ(x→+∞)= Teikx.(2.2.22) Green’s function: Multiply both sides of Eq. (2.2.21) by g(x, x/prime), integrate with respect to x from−∞ to+∞, and integrate by parts twice. The result is /integraldisplay+∞ −∞φ(x)/parenleftbiggd2 dx2+k2/parenrightbigg g(x, x/prime)dx+g(∞,x/prime)dφ dx(∞)−g(−∞,x/prime)dφ dx(−∞) −dg dx(∞,x/prime)φ(∞)+dg dx(−∞,x/prime)φ(−∞)=/integraldisplay+∞ −∞g(x, x/prime)U(x)φ(x)dx. (2.2.23) We require that Green’s function satisﬁes /parenleftbiggd2 dx2+k2/parenrightbigg g(x, x/prime)=δ(x−x/prime). (2.2.24) Then Eq. (2.2.23) becomes φ(x/prime)+g(∞,x/prime)Tikeikx−g(−∞,x/prime)/bracketleftbig ikeikx−Rike−ikx/bracketrightbig −dg dx(∞,x/prime)Teikx+dg dx(−∞,x/prime)/bracketleftbig eikx+Re−ikx/bracketrightbig =/integraldisplay+∞ −∞g(x, x/prime)U(x)φ(x)dx. (2.2.25) We require that terms involving the unknowns TandRvanish in Eq. (2.2.25), i.e., /braceleftBiggdg dx(∞,x/prime)= ikg(∞,x/prime), dg dx(−∞,x/prime)= −ikg(−∞,x/prime).(2.2.26) These conditions, (2.2.26), are the appropriate boundary conditions for g(x, x/prime). Hence we obtain φ(x/prime)=/bracketleftbigg ikg(−∞,x/prime)−dg dx(−∞,x/prime)/bracketrightbigg eikx+/integraldisplay+∞ −∞g(x, x/prime)U(x)φ(x)dx. (2.2.27) 2.2 Relationship of Integral Equations with Differential Equations and Green’s Functions 43 Solution for g(x, x/prime):The governing equation for g(x, x/prime)is /parenleftbiggd2 dx2+k2/parenrightbigg g(x, x/prime)=δ(x−x/prime), and the boundary conditions are Eq. (2.2.26). The solution to this problem is found to be g(x, x/prime)=/braceleftBigg A/primeeikxforx>x/prime, B/primee−ikxforx<x/prime. Atx=x/prime, there exists a discontinuity in the ﬁrst derivativedg dxofgwith respect to x. dg dx(x/prime +,x/prime)−dg dx(x/prime −,x/prime)=1,g(x/prime +,x/prime)=g(x/prime −,x/prime). From these two conditions, A/primeandB/primeare determined to be A/prime=e−ikx/prime/2ik and B/prime=eikx/prime/2ik. Thus the Green’s function g(x, x/prime)for this problem is given by g(x, x/prime)=1 2ikeik|x−x/prime|. Now, the ﬁrst term on the right-hand side of Eq. (2.2.27) assumes the following form: ikg(−∞,x/prime)−dg dx(−∞,x/prime)=2ikB/primee−ikx=eik(x/prime−x). Hence Eq. (2.2.27) becomes φ(x/prime)=eikx/prime+/integraldisplay+∞ −∞eik|x−x/prime| 2ikU(x)φ(x)dx. (2.2.28) Changing xtoξandx/primetoxin Eq. (2.2.28), we have φ(x)=eikx+/integraldisplay+∞ −∞eik|ξ−x| 2ikU(ξ)φ(ξ)dξ. (2.2.29) This is the Fredholm Integral Equation of the second kind. Reﬂection: Asx→− ∞ ,|ξ−x|=ξ−xso that φ(x)→eikx+e−ikx/integraldisplay+∞ −∞eikξ 2ikU(ξ)φ(ξ)dξ. From this, the reﬂection coefﬁcient Ris found. R=/integraldisplay+∞ −∞eikξ 2ikU(ξ)φ(ξ)dξ. 44 2 Integral Equations and Green’s Functions Transmission: Asx→+∞,|ξ−x|=x−ξso that φ(x)→eikx/bracketleftbigg 1+/integraldisplay+∞ −∞e−ikξ 2ikU(ξ)φ(ξ)dξ/bracketrightbigg . From this, the transmission coefﬁcient Tis found. T=1+/integraldisplay+∞ −∞e−ikξ 2ikU(ξ)φ(ξ)dξ. These RandTare still unknowns since φ(ξ)is not known, but for |U(ξ)|/lessmuch1(weak po- tential), we can approximate φ(x)byeikx. Then the approximate equations for RandTare given by R/similarequal/integraldisplay+∞ −∞e2ikξ 2ikU(ξ)dξ, andT/similarequal1+/integraldisplay+∞ −∞1 2ikU(ξ)dξ. Also, by approximating φ(ξ)byeikξin the integrand of Eq. (2.2.29) on the right-hand side, we have, as the ﬁrst approximation, φ(x)/similarequaleikx+/integraldisplay+∞ −∞eik|ξ−x| 2ikU(ξ)eikξdξ. By continuing the iteration, we can generate the Born series forφ(x). We shall discuss the Born approximation thoroughly in Section 2.5. 2.3 Sturm–Liouville System Consider the linear differential operator L=1 r(x)/bracketleftbiggd dx/parenleftbigg p(x)d dx/parenrightbigg −q(x)/bracketrightbigg (2.3.1) where r(x),p(x)>0 on0<x< 1, (2.3.2) together with the inner product deﬁned with r(x)as the weight, (f,g)=/integraldisplay1 0f(x)g(x)·r(x)dx. (2.3.3) Examine the inner product (g,Lf)by integral by parts twice, to obtain, (g,Lf)=/integraldisplay1 0dx·r(x)·g(x)1 r(x)/bracketleftbiggd dx/parenleftbigg p(x)df(x) dx/parenrightbigg −q(x)f(x)/bracketrightbigg =p(1)/bracketleftbig f/prime(1)g(1)−f(1)g/prime(1)/bracketrightbig −p(0)/bracketleftbig f/prime(0)g(0)−f(0)g/prime(0)/bracketrightbig +(Lg, f).(2.3.4) 2.3 Sturm–Liouville System 45 Suppose that the boundary conditions on f(x)are f(0) = 0 andf(1) = 0 , (2.3.5) and the adjoint boundary conditions on g(x)are g(0) = 0 andg(1) = 0 . (2.3.6) (Many other boundary conditions of the type αf(0) + βf/prime(0) = 0 (2.3.7) also work.) Then the boundary terms in Eq. (2.3.4) disappear and we have (g,Lf)=(Lg, f), (2.3.8) i.e.,Lisself-adjoint with the given weighted inner product. Now examine the eigenvalue problem. The governing equation and boundary conditions are given by Lφ(x)=λφ(x), with φ(0) = 0 , andφ(1) = 0 , i.e., d dx/bracketleftbigg p(x)d dxφ(x)/bracketrightbigg −q(x)φ(x)=λr(x)φ(x), (2.3.9) with the boundary conditions φ(0) = 0 , andφ(1) = 0 . (2.3.10) Suppose that λ=0 is not an eigenvalue (i.e., the homogeneous problem has no nontrivial solutions) so that the Green’s function exists. (Otherwise we have to deﬁne the modiﬁed Green’s function.) Suppose that the second-order ordinary differential equation d dx/bracketleftbigg p(x)d dxy(x)/bracketrightbigg −q(x)y(x)=0 (2.3.11) has two independent solutions y1(x)andy2(x)such that y1(0) = 0 andy2(1) = 0 . (2.3.12) In order for λ=0 not to be an eigenvalue, we must make sure that the only C1andC2for which C1y1(0) + C2y2(0) = 0 andC1y1(1) + C2y2(1) = 0 are not nontrivial. This requires y1(1)/negationslash=0 andy2(0)/negationslash=0. (2.3.13) Now, to ﬁnd the Green’s function, multiply the eigenvalue equation (2.3.9) by G(x, x/prime) and integrate from 0to1. Using the boundary conditions G(0,x/prime)=0 andG(1,x/prime)=0, (2.3.14) 46 2 Integral Equations and Green’s Functions we obtain, after integrating by parts twice, /integraldisplay1 0φ(x)/bracketleftbiggd dx/parenleftbigg p(x)dG(x, x/prime) dx/parenrightbigg −q(x)G(x, x/prime)/bracketrightbigg dx =λ/integraldisplay1 0G(x, x/prime)r(x)φ(x)dx. (2.3.15) Requiring that the Green’s function G(x, x/prime)should satisfy d dx/parenleftbigg p(x)dG(x, x/prime) dx/parenrightbigg −q(x)G(x, x/prime)=δ(x−x/prime) (2.3.16) with the boundary conditions (2.3.14), we arrive at the following equation φ(x/prime)=λ/integraldisplay1 0G(x, x/prime)r(x)φ(x)dx. (2.3.17) This is an homogeneous Fredholm integral equation of the second kind, once G(x, x/prime)is known. Solution for G(x, x/prime):Recalling Eqs. (2.3.12), (2.3.13) and (2.3.14), we have G(x, x/prime)=/braceleftBigg Ay1(x)+By 2(x) forx<x/prime, Cy 1(x)+Dy 2(x) forx>x/prime. From the boundary conditions (2.3.14) of G(x, x/prime), and (2.3.12) and (2.3.13) of y1(x)and y2(x), we immediately have B=0, andC=0. Thus we have G(x, x/prime)=/braceleftBigg Ay1(x) forx<x/prime, Dy 2(x) forx>x/prime. In order to determine AandD, integrate Eq. (2.3.16) across x/primewith respect to x,a n dm a k e use of the continuity of G(x, x/prime)atx=x/primewhich results in p(x/prime)/bracketleftbiggdG dx(x/prime +,x/prime)−dG dx(x/prime −,x/prime)/bracketrightbigg =1, G(x/prime +,x/prime)=G(x/prime −,x/prime), or, Ay1(x/prime)=Dy 2(x/prime), Dy/prime 2(x/prime)−Ay/prime 1(x/prime)=1/p(x/prime). 2.3 Sturm–Liouville System 47 Noting that W/parenleftbig y1(x),y2(x)/parenrightbig ≡y1(x)y/prime 2(x)−y2(x)y/prime 1(x) (2.3.18) is the Wronskian of the differential equation (2.3.11), we obtain AandDas   A=y 2(x/prime) p(x/prime)W/parenleftbig y1(x/prime),y2(x/prime)/parenrightbig, D=y1(x/prime) p(x/prime)W/parenleftbig y1(x/prime),y2(x/prime)/parenrightbig. Now, it can easily be proved that p(x)W/parenleftbig y1(x),y2(x)/parenrightbig =constant , (2.3.19) for the differential equation (2.3.11). Denoting this constant by p(x)W/parenleftbig y1(x),y2(x)/parenrightbig =C/prime, we simplify AandDas   A=y2(x/prime) C/prime, D=y1(x/prime) C/prime. Thus the Green’s function G(x, x/prime)for the Sturm–Liouville system is given by G(x, x/prime)=  y1(x)y2(x/prime) C/primeforx<x/prime, y1(x/prime)y2(x) C/primeforx>x/prime,(2.3.20) =y1(x<)y2(x>) C/primefor  x<=(x+x/prime) 2−|x−x/prime| 2, x>=(x+x/prime) 2+|x−x/prime| 2.(2.3.21) Thus the Sturm–Liouville eigenvalue problem is equivalent to the homogeneous Fredholm integral equation of the second kind, φ(x)=λ/integraldisplay1 0G(ξ,x)r(ξ)φ(ξ)dξ. (2.3.22) We remark that the Sturm–Liouville eigenvalue problem turns out to have a complete set of eigenfunctions in the space L2(0,1)as long as p(x)andr(x)are analytic and positive on(0,1). 48 2 Integral Equations and Green’s Functions The kernel of Eq. (2.3.22) is K(ξ,x)=r(ξ)G(ξ,x). This kernel can be symmetrized by deﬁning ψ(x)=/radicalbig r(x)φ(x). Then the integral equation (2.3.22) becomes ψ(x)=λ/integraldisplay1 0/radicalbig r(ξ)G(ξ,x)/radicalbig r(x)ψ(ξ)dξ. (2.3.23) Now, the kernel of Eq. (2.3.23), /radicalbig r(ξ)G(ξ,x)/radicalbig r(x) is symmetric since G(ξ,x)is symmetric. Symmetry of the Green’s function, called reciprocity , is true in general for any self-adjoint operator . The proof of this fact is as follows: Consider LxG(x, x/prime)=δ(x−x/prime), (2.3.24) LxG(x, x/prime/prime)=δ(x−x/prime/prime). (2.3.25) Take the inner product of Eq. (2.3.24) with G(x, x/prime/prime)from the left and Eq. (2.3.25) with G(x, x/prime)from the right. (G(x, x/prime/prime),LxG(x, x/prime)) = ( G(x, x/prime/prime),δ(x−x/prime)), (LxG(x, x/prime/prime),G(x, x/prime)) = ( δ(x−x/prime/prime),G(x, x/prime)). SinceLxis assumed to be self-adjoint, subtracting the two equations above results in G∗(x/prime,x/prime/prime)=G(x/prime/prime,x/prime). (2.3.26) IfGis real, we have G(x/prime,x/prime/prime)=G(x/prime/prime,x/prime), i.e.,G(x/prime,x/prime/prime)is symmetric. 2.4 Green’s Function for Time-Dependent Scattering Problem The time-dependent Schrödinger equation takes the following form after setting /planckover2pi=1 and 2m=1 , /parenleftbigg i∂ ∂t+∂2 ∂x2/parenrightbigg ψ(x, t)=V(x, t)ψ(x, t). (2.4.1) 2.4 Green’s Function for Time-Dependent Scattering Problem 49 Assume /braceleftBigg lim|t|→∞V(x, t)=0, limt→−∞exp[iω0t]ψ(x, t)=e x p [ ik0x],(2.4.2) from which, we ﬁnd ω0=k2 0. (2.4.3) Deﬁne the Green’s function G(x, t;x/prime,t/prime)by requiring ψ(x, t)=e x p/bracketleftbig i(k0x−k2 0t)/bracketrightbig +/integraldisplay+∞ −∞dt/prime/integraldisplay+∞ −∞dx/primeG(x, t;x/prime,t/prime)V(x/prime,t/prime)ψ(x/prime,t/prime).(2.4.4) In order to satisfy partial differential equation (2.4.1), we require /parenleftbigg i∂ ∂t+∂2 ∂x2/parenrightbigg G(x, t;x/prime,t/prime)=δ(t−t/prime)δ(x−x/prime). (2.4.5) We also require that G(x, t;x/prime,t/prime)=0 fort<t/prime. (2.4.6) Note that the initial condition at t=−∞ is satisﬁed as well as causality . Note also that the set of equations could be obtained by the methods we used in the previous two examples. To solve the above equations, Eqs. (2.4.5) and (2.4.6), we Fourier transform in time and space, i.e., we write   ˜G(k,ω;x/prime,t/prime)=/integraldisplay+∞ −∞dx/integraldisplay+∞ −∞dt e−ikxe−iωtG(x, t;x/prime,t/prime), G(x, t;x/prime,t/prime)=/integraldisplay+∞ −∞dk 2π/integraldisplay+∞ −∞dω 2πe+ikxe+iωt˜G(k,ω;x/prime,t/prime).(2.4.7) Taking the Fourier transform of the original equation (2.4.5), we ﬁnd G(x, t;x/prime,t/prime)=/integraldisplay+∞ −∞dk 2π/integraldisplay+∞ −∞dω 2π/parenleftbigg−1 ω+k2/parenrightbigg eik(x−x/prime)eiω(t−t/prime). (2.4.8) Where do we use the condition that G(x, t;x/prime,t/prime)=0 fort<t/prime? Consider the ω integration in the complex ωplane as in Figure 2.1, /integraldisplay+∞ −∞dω1 ω+k2eiω(t−t/prime). (2.4.9) We ﬁnd that there is a singularity right on the path of integration at ω=−k2. We either have to go above or below it. Upon writing ωasω=ω1+iω2, we have the following bound, /vextendsingle/vextendsingle/vextendsingleeiω(t−t/prime)/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingleeiω1(t−t/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglee−ω2(t−t/prime)/vextendsingle/vextendsingle/vextendsingle=e−ω2(t−t/prime). (2.4.10) 50 2 Integral Equations and Green’s Functions ω2 ω1−k2ω Fig. 2.1: The location of the singularity of the integrand of Eq. (2.4.9) in the complex ωplane. ω2 ω1ω −k2 Fig. 2.2: The contour of the complex ωintegration of Eq. (2.4.9). Fort<t/prime, we close the contour in the lower half-plane to carry out the contour integral. Since we want Gto be zero in this case, we want no singularities inside the contour in that case. This prompts us to take the contour to be as in Figure 2.2. For t>t/prime, when we close the contour in the upper half-plane, we get the contribution from the pole at ω=−k2. The result of the calculation is given by /integraldisplay+∞ −∞dω1 ω+k2eiω(t−t/prime)=/braceleftBigg 2πi·eik(x−x/prime)−ik2(t−t/prime),t > t/prime, 0,t < t/prime.(2.4.11) We remark that the idea of the deformation of the contour satisfying causality is often expressed by taking the singularity to be at −k2+iε(ε>0) as in Figure 2.3, whence we replace the denominator ω+k2withω+k2−iε, G(x, t;x/prime,t/prime)=/integraldisplay+∞ −∞dk 2π/integraldisplay+∞ −∞dω 2π/parenleftbigg−1 ω+k2−iε/parenrightbigg eik(x−x/prime)+iω(t−t/prime). (2.4.12) 2.4 Green’s Function for Time-Dependent Scattering Problem 51 −+k i2εω2 ω1ω Fig. 2.3: The singularity of the integrand of Eq. (2.4.9) at ω=−k2gets shifted to ω=−k2+iε (ε> 0) in the upper half-plane of the complex ωplane. This shifts the singularity above the real axis and is equivalent, as ε→0+, to our previous solution. After the ωintegration in the complex ωplane is performed, the kintegration can be done by completing the square in the exponent of Eq. (2.4.12), but the resulting Gaussianintegration is a bit more complicated than the diffusion equation. The result is given by G(x, t;x /prime,t/prime)=  /radicalBig i 4π(t−t/prime)ei(x−x/prime)2/4(t−t/prime)fort>t/prime, 0 fort<t/prime,(2.4.13) where /planckover2pi1=1 and2m=1 . In the case of the diffusion equation, /parenleftbigg −1 D∂ ∂t+∂2 ∂x2/parenrightbigg ψ(x, t)=0, (2.4.14) Equation (2.4.13) reduces to the Green’s function for the diffusion equation, G(x, t;x/prime,t/prime)=  /radicalBig 1 4πκ(t−t/prime)e−(x−x/prime)2/4D(t−t/prime)fort>t/prime, 0 fort<t/prime,(2.4.15) which satisﬁes the following equation, /parenleftbigg −1 D∂ ∂t+∂2 ∂x2/parenrightbigg G(x, t;x/prime,t/prime)=δ(t−t/prime)δ(x−x/prime), (2.4.16) G(x, t;x/prime,t/prime)=0 fort<t/prime, where the diffusion constant Dis given by D=K Cρ=(thermal conductivity ) (speciﬁc heat )×(density ). 52 2 Integral Equations and Green’s Functions These two expressions, Eqs. (2.4.13) and (2.4.15), are related by the analytic continuation, t→−it. The diffusion constant Dplays the role of the inverse of the Planck constant /planckover2pi1. We shall devote the next section for the more formal discussion of the scattering problem, namely the orthonormality of the outgoing and incoming wave and the unitarity of the S matrix. 2.5 Lippmann–Schwinger Equation In a nonrelativistic scattering problem of quantum mechanics, we have the Macroscopic causality of Stueckelberg. When we regard the potential V(t,r)as a function oft, we have no scattered wave, ψscatt(t,r)=0 ,f o rt<T ,i fV(t,r)=0 fort<T . We employ the Adiabatic switching hypothesis. W e can take the limit T→− ∞ after the computation of the scattered wave, ψscatt(t,r). We derive the Lippmann–Schwinger equation, and prove the orthonormality of the outgo- ing wave and the incoming wave and the unitarity of the Smatrix. Lippmann–Schwinger Equation We shall begin with the time-dependent Schrödinger equa- tion with the time-dependent potential V(t,r), i/planckover2pi1∂ ∂tψ(t,r)=[H0+V(t,r)]ψ(t,r). In order to use the macroscopic causality, we assume V(t,r)=/braceleftBigg V(r) fort≥T, 0 fort<T . Fort<T , the particle obeys the free equation, i/planckover2pi1∂ ∂tψ(t,r)=H0ψ(t,r). (2.5.1) We write the solution of Eq. (2.5.1) as ψinc(t,r). The wave function for the general time tis written as ψ(t,r)=ψinc(t,r)+ψscatt(t,r), where we have /parenleftbigg i/planckover2pi1∂ ∂t−H0/parenrightbigg ψscatt(t,r)=V(t,r)ψ(t,r). (2.5.2) We introduce the retarded Green’s function for Eq. (2.5.2) as /braceleftBigg (i/planckover2pi1(∂/∂t)−H0)Kret(t,r;t/prime,r/prime)= δ(t−t/prime)δ3(r−r/prime), Kret(t,r;t/prime,r/prime)= 0 fort<t/prime.(2.5.3) 2.5 Lippmann–Schwinger Equation 53 The formal solution to Eq. (2.5.2) is given by ψscatt(t,r)=/integraldisplay∞ −∞/integraldisplay Kret(t,r;t/prime,r/prime)V(t/prime,r/prime)ψ(t/prime,r/prime)dt/primedr/prime. (2.5.4) We note that the integrand of Eq. (2.5.4) survives only for t≥t/prime≥T. We now take the limit T→− ∞ , thus losing the t-dependence of V(t,r), ψscatt(t,r)=/integraldisplay∞ −∞/integraldisplay Kret(t,r;t/prime,r/prime)V(r/prime)ψ(t/prime,r/prime)dt/primedr/prime. When H0has no explicit space–time dependence, we have from the translation invariance that Kret(t,r;t/prime,r/prime)=Kret(t−t/prime;r−r/prime). Adding ψinc(t,r)toψscatt(t,r), Eq. (2.5.4), we have ψ(t,r)=ψinc(t,r)+ψscatt(t,r) =ψinc(t,r)+/integraldisplay∞ −∞/integraldisplay Kret(t−t/prime;r−r/prime)V(r/prime)ψ(t/prime,r/prime)dt/primedr/prime.(2.5.5) This equation is the integral equation determining the total wave function, given the incident wave. We rewrite this equation in a time-independent form. For this purpose, we set ψinc(t,r)=e x p/bracketleftbigg−iEt /planckover2pi1/bracketrightbigg ψinc(r), ψ(t,r)=e x p/bracketleftbigg−iEt /planckover2pi1/bracketrightbigg ψ(r). Then, from Eq. (2.5.5), we obtain ψ(r)=ψinc(r)+/integraldisplay G(r−r/prime;E)V(r/prime)ψ(r/prime)dr/prime. (2.5.6) HereG(r−r/prime;E)is given by G(r−r/prime;E)=/integraldisplay∞ −∞dt/primeexp/bracketleftbiggiE(t−t/prime) /planckover2pi1/bracketrightbigg Kret(t−t/prime;r−r/prime). (2.5.7) Setting Kret(t−t/prime;r−r/prime)=/integraldisplaydEd3p (2π)4exp/bracketleftbigg{ip(r−r/prime)−iE(t−t/prime)} /planckover2pi1/bracketrightbigg K(E,p), δ(t−t/prime)δ3(r−r/prime)=/integraldisplaydEd3p (2π)4exp/bracketleftbigg{ip(r−r/prime)−iE(t−t/prime)} /planckover2pi1/bracketrightbigg , substituting into Eq. (2.5.3), and writing H0=p2/2m, we obtain /parenleftbigg E−p2 2m/parenrightbigg K(E,p)=1. 54 2 Integral Equations and Green’s Functions The solution consistent with the retarded boundary condition is K(E,p)=1 E−(p2/2m)+iε, with εpositive inﬁnitesimal . Namely Kret(t−t/prime;r−r/prime)=1 (2π)4/integraldisplay dEd3pexp/bracketleftBig {ip(r−r/prime)−iE(t−t/prime)} /planckover2pi1/bracketrightBig E−(p2/2m)+iε. Substituting this into Eq. (2.5.7) and setting E=(/planckover2pi1k)2/2m, we obtain G(r−r/prime;E)=1 (2π)3/integraldisplay d3pexp[ip(r−r/prime)//planckover2pi1] E−(p2/2m)+iε=−m 2πexp[ik|r−r/prime|] |r−r/prime|. In Eq. (2.5.6), since the Fourier transform of G(r−r/prime;E)is written as 1 E−H0+iε, the equation (2.5.6) can be written formally as Ψ=Φ+1 E−H0+iεVΨ,E > 0, (2.5.8) where we wrote Ψ=ψ(r),Φ=ψinc(r), and the incident wave Φsatisﬁes the free particle equation, (E−H0)Φ = 0 . Operating (E−H0)on Eq. (2.5.8) from the left, we obtain the Schrödinger equation, (E−H0)Ψ = VΨ. (2.5.9) We note that equation (2.5.9) is the differential equation whereas equation (2.5.8) is the inte- gral equation , which embodies the boundary condition. For the bound state problem ( E<0), since the operator (E−H0)is negative deﬁnite and has a unique inverse, we have Ψ=1 E−H0VΨ,E < 0. (2.5.10) We call Eqs. (2.5.8) and (2.5.10) the Lippmann–Schwinger equation (the L-S equation in short). The +iεin the denominator of Eq. (2.5.8) makes the scattered wave the outgoing spherical wave. The presence of the +iεin Eq. (2.5.8) enforces the outgoing wave condition . It is mathematically convenient also to introduce the −iεinto Eq. (2.5.8), which makes the scattered wave the incoming spherical wave and thus enforces the incoming wave condition . 2.5 Lippmann–Schwinger Equation 55 We construct two kinds of wave functions: Ψ(+) a=Φ a+1 Ea−H0+iεVΨ(+) a, outgoing wave condition , (2.5.11. +) Ψ(−) a=Φ a+1 Ea−H0−iεVΨ(−) a, incoming wave condition . (2.5.11. −) The formal solution to the L-S equation was obtained by G. Chew and M. Goldberger. By iteration of Eq. (2.5.11. +), we have Ψ(+) a=Φ a+1 Ea−H0+iε/parenleftbigg 1+V1 Ea−H0+iε+···/parenrightbigg VΦa =Φ a+1 Ea−H0+iε/parenleftbigg 1−V1 Ea−H0+iε/parenrightbigg−1 VΦa=Φ a+1 Ea−H+iεVΦa. Here we have used the operator identity A−1B−1=(BA)−1andH=H0+Vrepresents the total Hamiltonian. We write the formal solution for Ψ(+) a andΨ(−) a together as: Ψ(+) a=Φ a+1 Ea−H+iεVΦa, (2.5.12. +) Ψ(−) a=Φ a+1 Ea−H−iεVΦa. (2.5.12. −) Orthonormality of Ψ(+) a:We will prove the orthonormality only for Ψ(+) a: (Ψ(+) b,Ψ(+) a)=( Φ b,Ψ(+) a)+/parenleftbigg1 Eb−H+iεVΦb,Ψ(+) a/parenrightbigg =( Φ b,Ψ(+) a)+/parenleftbigg Φb,V1 Eb−H−iεΨ(+) a/parenrightbigg =( Φ b,Ψ(+) a)+1 Eb−Ea−iε/parenleftBig Φb,VΨ(+) a/parenrightBig =( Φ b,Φa)+/parenleftbigg Φb,1 Ea−H0+iεVΨ(+) a/parenrightbigg +1 Eb−Ea−iε/parenleftBig Φb,VΨ(+) a/parenrightBig =δba+/parenleftbigg1 Ea−Eb+iε+1 Eb−Ea−iε/parenrightbigg/parenleftBig Φb,VΨ(+) a/parenrightBig =δba.(2.5.13) ThusΨ(+) a forms a complete and orthonormal basis. The same proof applies for Ψ(−) a also. Frequently, the orthonormality of Ψ(±) a is assumed on the outset. We have proved the or- thonormality of Ψ(±) a using the L-S equation and the formal solution due to G. Chew and M. Goldberger. 56 2 Integral Equations and Green’s Functions In passing, we state that, in relativistic quantum ﬁeld theory in the L.S.Z. formalism, the outgoing wave Ψ(+) a is called the in-state and is written as Ψ(in) a, and the incoming wave Ψ(−) a is called the out-state and is written as Ψ(out) a . There exists some confusion on this matter. Unitarity of the Smatrix: We deﬁne the Smatrix by Sba=( Ψ(−) b,Ψ(+) a)=( Ψ(out) b,Ψ(in) a). (2.5.14) This deﬁnition states that the Smatrix transforms one complete set to another complete set. By ﬁrst making use of the formal solution of G. Chew and M. Goldberger ﬁrst and then using theL−Sequation as before, we obtain Sba=δba+/parenleftbigg1 Ea−Eb+iε+1 Eb−Ea+iε/parenrightbigg/parenleftBig Φb,VΨ(+) a/parenrightBig =δba−2πiδ(Eb−Ea)/parenleftBig Φb,VΨ(+) a/parenrightBig .(2.5.15) We deﬁne the Tmatrix by Tba=/parenleftBig Φb,VΨ(+) a/parenrightBig . (2.5.16) Then we have Sba=δba−2πiδ(Eb−Ea)Tba. (2.5.17) If theSmatrix is unitary, it satisﬁes ˆS†ˆS=ˆSˆS†=1. (2.5.18) These unitarity conditions are equivalent to the following conditions in terms of the Tmatrix: T† ba−Tba=/braceleftBigg 2πi/summationtext nT† bnδ(Eb−En)Tna, 2πi/summationtext nTbnδ(Eb−En)T† na,with Eb=Ea. (2.5.19, 20) In order to prove the unitarity of the Smatrix, Eq. (2.5.18), it sufﬁces to prove Eqs. (2.5.19) and (2.5.20), which are expressed in terms of the Tmatrix. We ﬁrst note T† ba=T∗ ab=( Φ a,VΨ(+) b)∗=(VΨ(+) b,Φa)=( Ψ(+) b,VΦa). Then T† ba−Tba=( Ψ(+) b,VΦa)−(Φb,VΨ(+) a). Inserting the formal solution of G. Chew and M. Goldberger to Ψ(+) a andΨ(+) babove, we have T† ba−Tba=( Φ b,VΦa)+/parenleftbigg1 Eb−H+iεVΦb,VΦa/parenrightbigg −(Φb,VΦa)−/parenleftbigg Φb,V1 Ea−H+iεVΦa/parenrightbigg =/parenleftbigg VΦb,/parenleftbigg1 Eb−H−iε−1 Eb−H+iε/parenrightbigg VΦa/parenrightbigg =(VΦb,2πiδ(Eb−H)VΦa),(2.5.21) 2.6 Problems for Chapter 2 57 where, in the one line above the last line of Eq. (2.5.21), we used the fact that Eb=Ea. Inserting the complete orthonormal basis, Ψ(−), between the product of the operators in Eq. (2.5.21), we have T† ba−Tba=/summationdisplay n(VΦb,Ψ(−) n)2πiδ(Eb−En)(Ψ(−) n,VΦa) =2πi/summationdisplay nT† bnδ(Eb−En)Tna. This is Eq. (2.5.19). If we insert the complete orthonormal basis, Ψ(+), between the product of the operators in Eq. (2.5.21), we obtain T† ba−Tba=/summationdisplay n(VΦb,Ψ(+) n)2πiδ(Eb−En)(Ψ(+) n,VΦa) =2πi/summationdisplay nTbnδ(Eb−En)T† na. This is Eq. (2.5.20). Thus the Smatrix deﬁned by Eq. (2.5.14) is unitary. The unitarity of the Smatrix is equivalent to the fact that the outgoing wave set {Ψ(+) a}and the incoming wave set{Ψ(−) b}form the complete orthonormal basis, respectively. In fact the Smatrix has to be unitary since the Smatrix transforms one complete or- thonormal set to another complete orthonormal set . 2.6 Problems for Chapter 2 2.1. (Due to H. C.) Solve φ(x)=1+ λ/integraldisplay1 0(xy+x2y2)φ(y)dy. a) Show that this is equivalent to a 2×2matrix equation. b) Find the eigenvalues and the corresponding eigenvectors of the kernel. c) Find the solution of the inhomogeneous equation if λ/negationslash=eigenvalues. d) Solve the corresponding Fredholm integral equation of the ﬁrst kind. 2.2. (Due to H. C.) Solve φ(x)=1+ λ/integraldisplayx 0xyφ(y)dy. Discuss the solution of the homogeneous equation. 2.3. (Due to H. C.) Consider the integral equation, φ(x)=f(x)+λ/integraldisplay+∞ −∞e−(x2+y2)φ(y)dy,−∞<x< ∞. 58 2 Integral Equations and Green’s Functions a) Solve this equation for f(x)=0. For what values of λ, does it have non-trivial solutions? b) Solve this equation for f(x)=xm, with m=0,1,2,.... Does this inhomogeneous equation have any solutions when λis equal to an eigen- value of the kernel? Hint: Y ou may express your results in terms of the Gamma function, Γ(z)=/integraldisplay∞ 0tz−1e−tdt,Rez>0. 2.4. (Due to H. C.) Solve the following integral equation, u(θ)=1+ λ/integraldisplay2π 0sin(φ−θ)u(φ)dφ, 0≤θ<2π, where u(θ)is periodic with period 2π. Does the kernel of this equation have any real eigenvalues? Hint: Notice sin(φ−θ)=s i n φcosθ−cosφsinθ. 2.5. (Due to D. M.) Consider the integral equation, φ(x)=1+ λ/integraldisplay1 0xn−yn x−yφ(y)dy,0≤x≤1. a) Solve this equation for n=2 . For what values of λ, does the equation have no solutions? b) Discuss how you would solve this integral equation for arbitrary positive integer n. 2.6. (Due to D. M.) Solve the integral equation, φ(x)=1+/integraldisplay1 0(1 +x+y+xy)νφ(y)dy,0≤x≤1,ν:real. 2.6 Problems for Chapter 2 59 Hint: Notice that the kernel, (1 +x+y+xy)ν, can be factorized. 2.7. In the Fredholm integral equation of the second kind, if the kernel is given by K(x, y)=N/summationdisplay n=1gn(x)hn(y), show that the integral equation is equivalent to an N×Nmatrix equation. 2.8. (Due to H. C.) Consider the motion of an harmonic oscillator with a time-dependent spring constant, d2 dt2x+ω2x=−A(t)x, where ωis a constant and A(t)is a complicated function of t. Transform this differential equation together with the boundary conditions, x(Ti)=xi andx(Tf)=xf, to an integral equation. Hint: Construct a Green’s function G(t, t/prime)satisfying G(Ti,t/prime)=G(Tf,t/prime)=0. 2.9. Generalize the discussion of Section 2.4 to the case of three spatial dimensions and transform the Schrödinger equation with the initial condition, lim t→−∞eiωtψ(/vectorx, t)=eikz, to an integral equation. Hint: Construct a Green’s function G(t, t/prime)satisfying /parenleftbigg i∂ ∂t+/vector∇2/parenrightbigg G(/vectorx, t;/vectorx/prime,t/prime)=δ(t−t/prime)δ3(/vectorx−/vectorx/prime),G(/vectorx, t;/vectorx/prime,t/prime)=0 fort<t/prime. 2.10. (Due to H. C.) Consider the equation /bracketleftbigg −∂2 ∂t2+∂2 ∂x2−m2/bracketrightbigg φ(x, t)=U(x, t)φ(x, t). If the initial and ﬁnal conditions are φ(x,−T)=f(x) andφ(x, T)=g(x), transform the equation to an integral equation. 60 2 Integral Equations and Green’s Functions Hint: Consider the Green’s function /bracketleftbigg −∂2 ∂t2+∂2 ∂x2−m2/bracketrightbigg G(x, t;x/prime,t/prime)=δ(x−x/prime)δ(t−t/prime). 2.11. (Due to D. M.) The time-independent Schrödinger equation with the periodic potential, V(x)=−(a2+k2cos2x), reads as d2 dx2ψ(x)+(a2+k2cos2x)ψ(x)=0. Show directly that even periodic solutions of this equation, which are even Mathieu functions, satisfy the homogeneous integral equation, ψ(x)=λ/integraldisplayπ −πexp[kcosxcosy]ψ(y)dy. Hint: Show that φ(x)deﬁned by φ(x)≡/integraldisplayπ −πexp[kcosxcosy]ψ(y)dy is even and periodic, and satisﬁes the above time-independent Schrödinger equation. Thus, ψ(x)is a constant multiple of φ(x), ψ(x)=λφ(x). 2.12. (Due to H. C.) Consider the differential equation, d2 dt2φ(t)=λe−tφ(t),0≤t<∞,λ=constant , together with the initial conditions, φ(0) = 0 andφ/prime(0) = 1 . a) Find the partial differential equation for the Green’s function G(t, t/prime). Determine the form of G(t, t/prime)whent/negationslash=t/prime. b) Transform the differential equation for φ(t), together with the initial conditions, to an integral equation. Determine the conditions on G(t, t/prime). c) Determine G(t, t/prime). d) Substitute your answer for G(t, t/prime)into the integral equation and verify explicitly that the integral equation is equivalent to the differential equation together with the initial conditions. e) Does the initial value problem have a solution for all λ? If so, is the solution unique? 2.6 Problems for Chapter 2 61 2.13. In the V olterra integral equation of the second kind, if the kernel is given by K(x, y)=N/summationdisplay n=1gn(x)hn(y), show that the integral equation can be reduced to an ordinary differential equation of Nthorder. 2.14. Consider the partial differential equation of the form, /parenleftbigg∂2 ∂t2−∂2 ∂x2/parenrightbigg φ(x, t)=p(x, t)−λ∂2 ∂x2φ(x, t)·∂ ∂xφ(x, t), where −∞<x< ∞,t≥0, andλis a constant, with the initial conditions speciﬁed by φ(x,0) =a(x), and∂ ∂tφ(x,0) =b(x). This partial differential equation describes the displacement of a vibrating string under the distributed load p(x, t). a) Find the Green’s function for this partial differential equation. b) Express this initial value problem in terms of an integral equation using the Green’s function found in a). Explain how you would ﬁnd an approximate solution if λwere small. Hint: By applying the Fourier transform in x, ﬁnd a function φ0(x, t)which satisﬁes the wave equation, /parenleftbigg∂2 ∂t2−∂2 ∂x2/parenrightbigg φ0(x, t)=0, and the given initial conditions. 3 Integral Equations of V olterra Type 3.1 Iterative Solution to V olterra Integral Equation of the Second Kind Consider the inhomogeneous V olterra integral equation of the second kind , φ(x)=f(x)+λ/integraldisplayx 0K(x, y)φ(y)dy, 0≤x, y≤h, (3.1.1) withf(x)andK(x, y)square-integrable , /bardblf/bardbl2=/integraldisplayh 0|f(x)|2dx <∞, (3.1.2) /bardblK/bardbl2=/integraldisplayh 0dx/integraldisplayx 0dy|K(x, y)|2<∞. (3.1.3) Also, deﬁne A(x)=/integraldisplayx 0|K(x, y)|2dy. (3.1.4) Note that the upper limit of yintegration is x. Note also that the V olterra integral equation is a special case of the Fredholm integral equation with the kernel K(x, y)=0 forx<y<h . (3.1.5) We will prove in the following that, for Eq. (3.1.1), (1) A solution exists for all values of λ, (2) The solution is unique for all values of λ, (3) The iterative solution is convergent for all values of λ. We start our discussion with the construction of an iterative solution . Consider a series solu- tion of the usual form, φ(x)=φ0(x)+λφ1(x)+λ2φ2(x)+···+λnφn(x)+···. (3.1.6) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 64 3 Integral Equations of V olterra Type Substituting the series solution (3.1.6) into Eq. (3.1.1), we have ∞/summationdisplay k=0λkφk(x)=f(x)+λ/integraldisplayx 0K(x, y)∞/summationdisplay j=0λjφj(y)dy. Collecting like powers of λ,w eh a v e φ0(x)=f(x), φn(x)=/integraldisplayx 0K(x, y)φn−1(y)dy, n =1,2,3,···. (3.1.7) We now examine convergence of the series solution (3.1.6). Applying the Schwarz inequality to Eq. (3.1.7), we have φ1(x)=/integraldisplayx 0K(x, y)f(y)dy ⇒|φ1(x)|2≤A(x)/integraldisplayx 0|f(y)|2dy≤A(x)/bardblf/bardbl2, φ2(x)=/integraldisplayx 0K(x, y)φ1(y)dy ⇒|φ2(x)|2≤A(x)/integraldisplayx 0|φ1(y)|2dy≤A(x)/integraldisplayx 0dx1A(x1)/bardblf/bardbl2, and φ3(x)=/integraldisplayx 0K(x, y)φ2(y)dy ⇒|φ3(x)|2≤A(x)/integraldisplayx 0dx2|φ2(x2)|2≤1 2A(x)/bracketleftbigg/integraldisplayx 0dx1A(x1)/bracketrightbigg2 /bardblf/bardbl2. In general, we have |φn(x)|2≤1 (n−1)!A(x)/bracketleftbigg/integraldisplayx 0dy A(y)/bracketrightbiggn−1 /bardblf/bardbl2,n=1,2,3,···. (3.1.8) Deﬁne B(x)≡/integraldisplayx 0dy A(y). (3.1.9) Thus, from Eqs. (3.1.8) and (3.1.9), we obtain the following bound on φn(x), |φn(x)|≤1 /radicalbig (n−1)!/radicalbig A(x)[B(x)](n−1)/2/bardblf/bardbl,n=1,2,3,···. (3.1.10) 3.1 Iterative Solution to V olterra Integral Equation of the Second Kind 65 We now examine the convergence of the series solution (3.1.6). |φ(x)−f(x)|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationdisplay n=1λnφn(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤∞/summationdisplay n=1|λ|n 1 /radicalbig (n−1)!/radicalbig A(x)[B(x)](n−1)/2/bardblf/bardbl =/radicalbig A(x)/bardblf/bardbl∞/summationdisplay n=1|λ|n[B(x)](n−1)/2 1 /radicalbig (n−1)!. (3.1.11) Letting an=|λ|n[B(x)](n−1)/2 1 /radicalbig (n−1)!, and applying the ratio test on the right-hand side of Eq. (3.1.11), we have an+1 an=/braceleftBig |λ|n+1[B(x)]n/2/radicalbig (n−1)!/bracerightBig /braceleftBig |λ|n[B(x)](n−1)/2√ n!/bracerightBig =|λ|/radicalbig B(x) √ n, whence we have, lim n→∞an+1 an=0 for all λ. (3.1.12) Thus the series solution (3.1.6) converges for all λ, provided that /bardblf/bardbl,A(x)andB(x)exist and are ﬁnite. We have proved statements (1) and (3) under the condition that the kernel K(x, y)and the inhomogeneous term f(x)are square-integrable , Eqs. (3.1.2) and (3.1.3). To show that Eq. (3.1.1) has a unique solution which is square-integrable (/bardblφ/bardbl<∞),w e shall prove that Rn(x)→0 asn→∞, (3.1.13) where λn+1Rn+1(x)≡φ(x)−k=n/summationdisplay k=0λkφk(x), (3.1.14) and Rn(x)=/integraldisplayx 0K(x, y)Rn−1(y)dy, R 0(x)=φ(x). Repeating the same procedure as above, we can establish [Rn(x)]2≤/bardblφ/bardbl2A(x)[B(x)]n−1 (n−1)!. (3.1.15) 66 3 Integral Equations of V olterra Type ThusRn+1(x)/Rn(x)vanishes as n→∞ . Returning to Eq. (3.1.14), we see that the iterative solution φ(x), Eq. (3.1.6), is unique . The uniqueness of the solution of the inhomogeneous V olterra integral equation implies that the square-integrable solution of the homogeneous V olterra integral equation ψH(x)=λ/integraldisplayx 0K(x, y)ψH(y)dy (3.1.16) is trivial , ψH(x)≡0. Otherwise, φ(x)+cψ H(x)would also be a solution of the inhomogeneous V olterra integral equation (3.1.1), in contradiction to our conclusion that the solution of Eq. (3.1.1) is unique. The result above can be used to prove the existence and uniqueness of the solution of a differential equation. As noted in Chapter 2, a differential equation can always be transformedinto an integral equation. Consider the following second-order ordinary differential equation: /bracketleftbigg a 0(x)d2 dx2+a1(x)d dx+a2(x)/bracketrightbigg u(x)=f(x). (3.1.17) We shall transform this equation together with some initial conditions into a V olterra integral equation. We ﬁrst perform a transformation to φ(x)by setting u(x)=e x p/bracketleftbigg −1 2/integraldisplayxa1(y) a0(y)dy/bracketrightbigg φ(x), to reduce Eq. (3.1.17) into the form /bracketleftbiggd2 dx2+q(x)/bracketrightbigg φ(x)=F(x). Moving the term q(x)φ(x)to the right-hand side, we can immediately convert this differential equation into a V olterra integral equation of the second kind. Hence the existence and theuniqueness of the solution for the differential equation (3.1.17) can be deduced. When the coefﬁcient functions a i(x)(i=0,1,2) of Eq. (3.1.17) depend on some param- eter, say β, and are analytic in β, what can we say about the analyticity of the solution as a function of β? All the terms in the iteration series (3.1.6) are analytic in βby construction. The iteration series is absolutely convergent so that the solution to Eq. (3.1.17) is analytic in β. 3.2 Solvable cases of V olterra Integral Equation We list a few solvable cases of the V olterra integral equation. Case (1): The kernel is equal to a sum of nfactorized terms . We demonstrated the reduction of such an integral equations into an nthorder ordinary differential equation in Problem 6 in Chapter 2. 3.2 Solvable cases of V olterra Integral Equation 67 Case (2): The kernel is translational . K(x, y)=K(x−y). (3.2.1) Consider φ(x)=f(x)+λ/integraldisplayx 0K(x−y)φ(y)dy on0≤x<∞. (3.2.2) We note that the second term on the right-hand side of Eq. (3.2.2) is a convolution integral . We shall use the Laplace transform , L{F(x)}≡¯F(s)≡/integraldisplay∞ 0dxF(x)e−sx. (3.2.3) Taking the Laplace transform of the integral equation (3.2.2), we obtain ¯φ(s)=¯f(s)+λ¯K(s)¯φ(s). (3.2.4) Solving Eq. (3.2.4) for ¯φ(s), we obtain ¯φ(s)=¯f(s) 1−λ¯K(s). (3.2.5) Applying the inverse Laplace transform to Eq. (3.2.5), we obtain φ(x)=/integraldisplayγ+i∞ γ−i∞ds 2πiesx¯f(s) 1−λ¯K(s), (3.2.6) where the inversion path ( γ±i∞) in the complex splane lies to the right of all singularities of the integrand as indicated in Figure 3.1. Before we solve an example, we ﬁrst recall the Laplace transform and some of its proper- ties. Deﬁnition: Suppose F(t)is deﬁned on [0,∞)withF(t)=0 fort<0. Then the Laplace transform of F(t)is deﬁned by ¯F(s)≡L{F(t)}≡/integraldisplay∞ 0F(t)e−stdt. (3.2.7) The inversion is given by F(t)=1 2πi/integraldisplayγ+i∞ γ−i∞¯F(s)estds=L−1/braceleftbig¯F(s)/bracerightbig . (3.2.8) Properties: L/braceleftbiggd dtF(t)/bracerightbigg =s¯F(s)−¯F(0). (3.2.9) 68 3 Integral Equations of V olterra Type γ+∞i γ−∞is s1s2 Fig. 3.1: The inversion path of the Laplace transform ¯φ(s)in the complex splane from s=γ−i∞tos=γ+i∞, which lies to the right of all singularities of the integrand. The Laplace transform of convolution H(t)=/integraldisplayt 0G(t−t/prime)F(t/prime)dt/prime=/integraldisplayt 0G(t/prime)F(t−t/prime)dt/prime(3.2.10) is given by ¯H(s)=¯G(s)¯F(s), (3.2.11) which is already used in deriving Eq. (3.2.4). The Laplace transforms of 1andtnare respectively given by L{1}=1 s, (3.2.12) and L{tn}=Γ(n+1 ) sn+1. (3.2.13) The Gamma Function :Γ(z)is deﬁned by Γ(z)=/integraldisplay∞ 0tz−1e−tdt. (3.2.14) 3.2 Solvable cases of V olterra Integral Equation 69 Properties of the Gamma Function: Γ( 1 )=1 , (3.2.15a) Γ/parenleftbig1 2/parenrightbig =√ π, (3.2.15b) Γ(z+1 )= zΓ(z), (3.2.15c) Γ(n+1 )= n!, (3.2.15d) Γ(z)Γ( 1−z)=π/sin (πz), (3.2.15e) Γ(z)is singular at z=0,−1,−2,.... (3.2.15f) Derivation of the Abel integral equation: The descent time of a frictionless ball on the side of a hill is known as a function of its initial height x. Let us ﬁnd the shape of the hill. Starting with initial velocity zero, the speed of the ball at height yis obtained by solving 1 2mv2=mg(x−y), from which v=/radicalbig 2g(x−y). Let the shape of the hill to be given by ξ=f(y). Then the arc length is given by ds=/radicalbig (dy)2+(dξ)2=/radicalbig 1+(f/prime(y))2|dy|. The descent time to height y=0 is given by T(x)=/integraldisplay dt=/integraldisplaydt dsds=/integraldisplayds ds/dt=/integraldisplayds v=/integraldisplayy=0 y=x/radicalbig 1+(f/prime(y))2 /radicalbig 2g(x−y)|dy|. Sinceyis decreasing, dy is negative so that |dy|=−dy. Thus the descent time is given by T(x)=/integraldisplayx 0φ(y) √ x−ydy (3.2.16) with φ(y)=1 √ 2g/radicalbig 1+(f/prime(y))2. (3.2.17) So, given the descent time T(x)as a function of the initial height x, we solve the Abel integral equation (3.2.16) for φ(x), and then solve (3.2.17) for f/prime(y)which gives the shape of the curve ξ=f(y). We solve an example of the V olterra integral equation with the translational kernel derived above, Eq. (3.2.16). ❑ Example 3.1. Abel Integral Equation . /integraldisplayx 0φ(x/prime) √ x−x/primedx/prime=f(x), with f(0) = 0 . (3.2.18) 70 3 Integral Equations of V olterra Type Solution. Take the Laplace transform of x−1 2. L/braceleftBig x−1 2/bracerightBig =Γ/parenleftbig1 2/parenrightbig s1 2=/radicalbigg π s. Then the Laplace transform of Eq. (3.2.18) is /radicalbigg π s¯φ(s)=¯f(s). Solving for ¯φ(s), we obtain ¯φ(s)=/radicalbigg s π¯f(s). Unfortunately√ sis not the Laplace transform of anything since, for any function g(t),w e must have ¯g(s)→0ass→∞ . (Recall the deﬁnition (3.2.7) of the Laplace transform.) Thus we rewrite φ(x)=1 2πi/integraldisplayγ+i∞ γ−i∞esx(¯f(s)√ s/√ π)ds =1 2πid dx/integraldisplayγ+i∞ γ−i∞1 π/radicalbigg π s¯f(s)esxds =1 πd dxL−1/braceleftbigg/radicalbigg π s¯f(s)/bracerightbigg =1 πd dx/integraldisplayx 0f(x/prime) √ x−x/primedx/prime,(3.2.19) where the convolution theorem, Eqs. (3.2.10) and (3.2.11), has been applied. As an extension of Case(2), we can solve a system of V olterra integral equations of the second kind with the translational kernels, φi(x)=fi(x)+n/summationdisplay j=1/integraldisplayx 0Kij(x−y)φj(y)dy, i =1,···,n, (3.2.20) where Kij(x)andfi(x)are known functions with Laplace transforms ¯Kij(s)and¯f(s).T a k - ing the Laplace transform of (3.2.20), we obtain ¯φi(s)=¯fi(s)+n/summationdisplay j=1¯Kij(s)¯φj(s),i =1,···,n. (3.2.21) Equation (3.2.21) is a system of linear algebraic equations for ¯φi(s). We can solve (3.2.21) for¯φi(s)easily and apply the inverse Laplace transform to ¯φi(s)to obtain φi(x). 3.3 Problems for Chapter 3 71 3.3 Problems for Chapter 3 3.1. Consider the V olterra integral equation of the ﬁrst kind, f(x)=/integraldisplayx 0K(x, y)φ(y)dy, 0≤x≤h. Show that, by differentiating the above equation with respect to x, we can transform this integral equation to a V olterra integral equation of the second kind as long as K(x, x)/negationslash=0. 3.2. Transform the radial Schrödinger equation /bracketleftbiggd2 dr2−l(l+1 ) r2+k2−V(r)/bracketrightbigg ψ(r)=0, with ψ(r)∼rl+1asr→0, to a V olterra integral equation of the second kind. Hint: There are two ways to deﬁne the homogeneous equation. (i) /bracketleftbiggd2 dr2−l(l+1 ) r2+k2/bracketrightbigg ψH(r)=0 ⇒ψH(r)=/braceleftBigg krjl(kr) krhl(kr), where jl(kr)is thelth-order spherical Bessel function and hl(kr)is thelth-order spherical Hankel function. (ii) /bracketleftbiggd2 dr2−l(l+1 ) r2/bracketrightbigg ψH(r)=0 ⇒ψH(r)=rl+1,andr−l. There exist two equivalent V olterra integral equations of the second kind for this problem. 3.3. Solve the generalized Abel equation, /integraldisplayx 0φ(y) (x−y)αdy=f(x),0<α< 1. 3.4. Solve /integraldisplayx 0φ(y)l n (x−y)dy=f(x), with f(0) = 0 . 3.5. Solve φ(x)=1+/integraldisplay∞ xeα(x−y)φ(y)dy, α > 0. Hint: Reduce the integral equation to the ordinary differential equation. 72 3 Integral Equations of V olterra Type 3.6. Solve φ(x)=1+ λ/integraldisplayx 0e−(x−y)φ(y)dy. 3.7. (Due to H. C.) Solve φ(x)=1+/integraldisplayx 11 x+yφ(y)dy, x ≥1. Find the asymptotic behavior of φ(x)asx→∞ . 3.8. (Due to H. C.) Solve the integro-differential equation, ∂ ∂tφ(x, t)=−ixφ(x, t)+λ/integraldisplay+∞ −∞g(y)φ(y,t)dy, with φ(x,0) =f(x), where f(x)andg(x)are given. Find the asymptotic form of φ(x, t)ast→∞ . 3.9. Solve /integraldisplayx 01 √ x−yφ(y)dy+/integraldisplay1 02xyφ(y)dy=1. 3.10. Solve λ/integraldisplayx 01 (x−y)1/3φ(y)dy+λ/integraldisplay1 0φ(y)dy=1. 3.11. Solve λ/integraldisplay1 0K(x, y)φ(y)dy=1, where K(x, y)=/braceleftBigg (x−y)−1/4+xy for0≤y≤x≤1, xy for0≤x<y ≤1. 3.12. (Due to H. C.) Solve φ(x)=λ/integraldisplayx 0J0(xy)φ(y)dy, where J0(x)is the zeroth-order Bessel function of the ﬁrst kind. 3.13. Solve x2φµ(x)−/integraldisplayx 0K(µ)(x, y)φµ(y)dy=/braceleftBigg 0 forµ≥1, −x2forµ=0, where K(µ)(x, y)=−x−(x2−µ2)(x−y). 3.3 Problems for Chapter 3 73 Hint: Setting φµ(x)=∞/summationdisplay n=0a(µ) nxn, ﬁnd the recursive relation of a(µ) n and solve for a(µ) n. 3.14. Solve a system of the integral equations, φ1(x)=1−2/integraldisplayx 0exp[2( x−y)]φ1(y)dy+/integraldisplayx 0φ2(y)dy, φ2(x)=4x−/integraldisplayx 0φ1(y)dy+4/integraldisplayx 0(x−y)φ2(y)dy. 3.15. Solve a system of the integral equations, φ1(x)+φ2(x)−/integraldisplayx 0(x−y)φ1(y)dy=ax, φ1(x)−φ2(x)−/integraldisplayx 0(x−y)2φ2(y)dy=bx2. 3.16. (Due to D. M.) Consider the V olterra integral equation of the second kind, φ(x)=f(x)+λ/integraldisplayx 0exp[x2−y2]φ(y)dy, x > 0. a) Sum up the iteration series exactly and ﬁnd the general solution to this equation. V erify that the solution is analytic in λ. b) Solve this integral equation by converting it into a differential equation. Hint: Multiply both sides by exp[−x2]and differentiate. 3.17. (Due to H. C.) The convolution of f1(x),f2(x),··· ,fn(x)is deﬁned as C(x)≡/integraldisplay∞ 0dxn···/integraldisplay∞ 0dx1n/productdisplay i=1fi(xi)δ/parenleftBigg x−n/summationdisplay i=1xi/parenrightBigg . a) V erify that for n=2 , this is the convolution deﬁned in the text, and that ˜C(s)=n/productdisplay i=1˜fi(s). 74 3 Integral Equations of V olterra Type b) Iff1(x)=f(x),a n df2(x)=f3(x)=···=fn(x)=1 , show that C(x)is thenth integral of f(x). Show also that ˜C(s)=˜f(s)s1−n, and hence C(x)=/integraldisplayx 0(x−y)n−1 (n−1)!f(y)dy. c) With the result in b), can you deﬁne the “one-third integral” of f(x)? 4 Integral Equations of the Fredholm Type 4.1 Iterative Solution to the Fredholm Integral Equation of the Second Kind Consider the inhomogeneous Fredholm Integral Equation of the second kind , φ(x)=f(x)+λ/integraldisplayh 0dx/primeK(x, x/prime)φ(x/prime),0≤x≤h, (4.1.1) and assume that f(x)andK(x, x/prime)are both square-integrable , /bardblf/bardbl2<∞, (4.1.2) and /bardblK/bardbl2≡/integraldisplayh 0dx/integraldisplayh 0dx/prime|K(x, x/prime)|2<∞. (4.1.3) Suppose that we look for an iterative solution inλ, φ(x)=φ0(x)+λφ1(x)+λ2φ2(x)+···+λnφn(x)+···, (4.1.4) which, when substituted into Eq. (4.1.1), yields φ0(x)=f(x), φ1(x)=/integraldisplayh 0dy1K(x, y 1)φ0(y1)=/integraldisplayh 0dy1K(x, y 1)f(y1), φ2(x)=/integraldisplayh 0dy2K(x, y 2)φ1(y2)=/integraldisplayh 0dy2/integraldisplayh 0dy1K(x, y 2)K(y2,y1)f(y1). In general, we have φn(x)=/integraldisplayh 0dynK(x, yn)φn−1(yn)=/integraldisplayh 0dyn/integraldisplayh 0dyn−1···/integraldisplayh 0dy1 ×K(x, yn)K(yn,yn−1)···K(y2,y1)f(y1).(4.1.5) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 76 4 Integral Equations of the Fredholm Type Bounds: First, in order to establish the bound on |φn(x)|,d e ﬁ n e A(x)≡/integraldisplayh 0dy|K(x, y)|2. (4.1.6) Then the square of the norm of the kernel K(x, y)is given by /bardblK/bardbl2=/integraldisplayh 0dx A(x). (4.1.7) Examine the iteration series (4.1.4) and apply the Schwarz inequality. Each term in the itera- tion series (4.1.4) is bounded as follows: φ1(x)=/integraldisplayh 0dy1K(x, y 1)f(y1) ⇒|φ1(x)|2≤A(x)/bardblf/bardbl2, φ2(x)=/integraldisplayh 0dy2K(x, y 2)φ1(y2) ⇒|φ2(x)|2≤A(x)/bardblφ1/bardbl2≤A(x)/bardblf/bardbl2/bardblK/bardbl2, and φ3(x)=/integraldisplayh 0dy3K(x, y 3)φ2(y3) ⇒|φ3(x)|2≤A(x)/bardblφ2/bardbl2≤A(x)/bardblf/bardbl2/bardblK/bardbl4. In general, we have |φn(x)|2≤A(x)/bardblf/bardbl2/bardblK/bardbl2(n−1). Thus the bound on |φn(x)|is established as |φn(x)|≤/radicalbig A(x)/bardblf/bardbl/bardblK/bardbln−1,n =1,2,3,···. (4.1.8) Now examine the whole series (4.1.4). φ(x)−f(x)=λφ1(x)+λ2φ2(x)+λ3φ3(x)+···. Taking the absolute value of both sides, applying the triangular inequality on the right-hand side, and using the bound on |φn(x)|established as in Eq. (4.1.8), we have |φ(x)−f(x)|≤|λ||φ1(x)|+|λ|2|φ2(x)|+··· =∞/summationdisplay n=1|λ|n|φn(x)| ≤∞/summationdisplay n=1|λ|n/radicalbig A(x)·/bardblf/bardbl·/bardblK/bardbln−1 =|λ|/radicalbig A(x)·/bardblf/bardbl·∞/summationdisplay n=0|λ|n·/bardblK/bardbln.(4.1.9) 4.1 Iterative Solution to the Fredholm Integral Equation of the Second Kind 77 Now, the series on the right-hand side of the inequality (4.1.9) ∞/summationdisplay n=0|λ|n·/bardblK/bardbln converges as long as |λ|<1 /bardblK/bardbl, (4.1.10) converging to 1 1−|λ|·/bardblK/bardbl. Therefore, in that case, (i.e., for Eq. (4.1.10)), the assumed series (4.1.4) is a convergent series, giving us a solution φ(x)to the integral equation (4.1.1), which is analytic inside the disk (4.1.10). Symbolically the integral equation (4.1.1) can be written as if it is an algebraic equation, φ=f+λKφ⇒(1−λK)φ=f ⇒φ=f 1−λK=( 1+ λK+λ2K2+···)f, which converges only for |λK|<1. Uniqueness: Inside the disk, Eq. (4.1.10), we can establish the uniqueness of the solution by showing that the corresponding homogeneous problem has no nontrivial solutions. Considerthe homogeneous problem, φ H(x)=λ/integraldisplayh 0K(x, y)φH(y). (4.1.11) Applying the Schwarz inequality to Eq. (4.1.11), |φH(x)|2≤|λ|2A(x)/bardblφH/bardbl2. Integrating both sides with respect to xfrom0toh, /bardblφH/bardbl2≤|λ|2·/bardblφH/bardbl2/integraldisplayh 0A(x)dx=|λ|2·/bardblK/bardbl2·/bardblφH/bardbl2, i.e., /bardblφH/bardbl2·(1−|λ|2·/bardblK/bardbl2)≤0. (4.1.12) Since|λ|·/bardblK/bardbl<1, the inequality (4.1.12) can only be satisﬁed if and only if /bardblφH/bardbl=0 or φH≡0. (4.1.13) Thus the homogeneous problem has only a trivial solution, so the inhomogeneous problem has a unique solution. 78 4 Integral Equations of the Fredholm Type 4.2 Resolvent Kernel Returning to the series solution (4.1.4), we ﬁnd that after deﬁning the iterated kernels as follows, K1(x, y)=K(x, y), (4.2.1) K2(x, y)=/integraldisplayh 0dy2K(x, y 2)K(y2,y), (4.2.2) K3(x, y)=/integraldisplayh 0dy3/integraldisplayh 0dy2K(x, y 3)K(y3,y2)K(y2,y), (4.2.3) and generally Kn(x, y)=/integraldisplayh 0dyn/integraldisplayh 0dyn−1···/integraldisplayh 0dy2K(x, yn)K(yn,yn−1)···K(y2,y),(4.2.4) we may write each term in the series (4.1.4) as φ1(x)=/integraldisplayh 0dy K 1(x, y)f(y), (4.2.5) φ2(x)=/integraldisplayh 0dy K 2(x, y)f(y), (4.2.6) φ3(x)=/integraldisplayh 0dy K 3(x, y)f(y), (4.2.7) and generally φn(x)=/integraldisplayh 0dy K n(x, y)f(y). (4.2.8) Therefore, we have φ(x)=f(x)+∞/summationdisplay n=1λn/integraldisplayh 0dy K n(x, y)f(y). (4.2.9) Now, deﬁne the resolvent kernel H(x, y;λ)to be −H(x, y;λ)≡K1(x, y)+λK 2(x, y)+λ2K3(x, y)+··· =∞/summationdisplay n=1λn−1Kn(x, y).(4.2.10) Then the solution (4.2.9) can be expressed compactly as, φ(x)=f(x)−λ/integraldisplayh 0dy H(x, y;λ)f(y), (4.2.11) 4.2 Resolvent Kernel 79 with the resolvent H(x, y;λ)deﬁned by Eq. (4.2.10). We have in effect shown that H(x, y;λ) exists and is analytic for |λ|<1 /bardblK/bardbl. (4.2.12) ❑ Example 4.1. Solve the Fredholm Integral Equation of the second kind, φ(x)=f(x)+λ/integraldisplay1 0ex−yφ(y)dy. (4.2.13) Solution. We have, for the iterated kernels, K1(x, y)=K(x, y)=ex−y, K2(x, y)=/integraldisplay1 0dξ K(x, ξ)K(ξ,y)=ex−y, K3(x, y)=/integraldisplay1 0dξ K 1(x, ξ)K2(ξ,y)=ex−y, and hence Kn(x, y)=ex−yfor all n. (4.2.14) Then we have as the resolvent kernel of this problem, −H(x, y;λ)=∞/summationdisplay n=1λn−1ex−y=ex−y(1 +λ+λ2+λ3+···). (4.2.15) For |λ|<1, (4.2.16) we have H(x, y;λ)=−ex−y 1−λ. (4.2.17) Thus the solution to this problem is given by φ(x)=f(x)+λ 1−λ/integraldisplay1 0dy ex−yf(y). (4.2.18) We note that, in this case, not only do we know the radius of convergence for the series solution (or for the resolvent kernel) as in Eq. (4.2.16), but we also know the nature of the singularity (a simple pole at λ=1 ). In fact, our solution (4.2.18) is valid for all values ofλ, even those which have |λ|>1, with the exception of λ=1 . The question we now wish to address is whether, in general, we can say more about the nature of the singularity inH(x, y;λ), than simply knowing the radius of the disk in which we have a convergence. 80 4 Integral Equations of the Fredholm Type Properties of the resolvent: We now derive some properties of the resolvent H(x, y;λ) which will be useful to us later. Consider the original integral operator written as ˜K=/integraldisplayh 0dy K(x, y), (4.2.19) and the operator corresponding to the resolvent as ˜H=/integraldisplayh 0dy H(x, y;λ). (4.2.20) Note the integrals are with respect to the second argument in Eqs. (4.2.19) and (4.2.20). Now, the operators ˜K2,o r˜H2,o r˜K˜H,o r˜H˜Kare deﬁned in the usual way: ˜K2φ=˜K(˜Kφ)=˜K/parenleftBigg/integraldisplayh 0dy1K(x, y 1)φ(y1)/parenrightBigg =/integraldisplayh 0dy2K(x, y 2)/integraldisplayh 0dy1K(y2,y1)φ(y1), i.e., ˜K2=/integraldisplayh 0dy2/integraldisplayh 0dy1K(x, y 2)K(y2,y1). (4.2.21) We wish to show that the two operators ˜Kand˜Hcommute, i.e., ˜K˜H=˜H˜K. (4.2.22) The original integral equation can be written as φ=f+λ˜Kφ (4.2.23) while the solution obtained by the resolvent takes the form φ=f−λ˜Hf. (4.2.24) Deﬁning ˜Ito be the identity operator ˜I=/integraldisplayh 0dy δ(x−y), (4.2.25) Eqs. (4.2.23) and (4.2.24) can be written as f=(˜I−λ˜K)φ, (4.2.26) φ=(˜I−λ˜H)f. (4.2.27) Then, combining Eqs. (4.2.26) and (4.2.27), we obtain f=(˜I−λ˜K)(˜I−λ˜H)f andφ=(˜I−λ˜H)(˜I−λ˜K)φ. 4.3 Pincherle–Goursat Kernel 81 In other words, we have (˜I−λ˜K)(˜I−λ˜H)=˜I and(˜I−λ˜H)(˜I−λ˜K)=˜I. Thus we obtain ˜K+˜H=λ˜K˜H and ˜K+˜H=λ˜H˜K. Hence we have established the identity ˜K˜H=˜H˜K, (4.2.28) i.e.,˜Kand˜Hcommute. This can be written explicitly as /integraldisplayh 0dy2/integraldisplayh 0dy1K(x, y 2)H(y2,y1)=/integraldisplayh 0dy2/integraldisplayh 0dy1H(x, y 2)K(y2,y1). Let both of these operators act on the function δ(y1−y). Then we ﬁnd /integraldisplayh 0dξ K(x, ξ)H(ξ,y)=/integraldisplayh 0dξ H(x, ξ)K(ξ,y). (4.2.29) Similarly the operator equation ˜K+˜H=λ˜K˜H (4.2.30) may be written as H(x, y;λ)=−K(x, y)+λ/integraldisplayh 0K(x, ξ)H(ξ,y;λ)dξ. (4.2.31) We will ﬁnd this to be a useful relation later. 4.3 Pincherle–Goursat Kernel Let us now examine the problem of determining a more explicit formula for the resolvent which points out more clearly the nature of the singularities of H(x, y;λ)in the complex λ plane. We do this for two different cases. First, we look at the case of a kernel which is given by a ﬁnite sum of separable terms (the so-called Pincherle–Goursat kernel ). Secondly we examine the case of a general kernel which we decompose into a sum of a Pincherle–Goursatkernel and a remainder, which can be made as small as possible. Pincherle–Goursat kernel: Suppose that we are given the kernel which is a ﬁnite sum of separable terms, K(x, y)= N/summationdisplay n=1gn(x)hn(y), (4.3.1) 82 4 Integral Equations of the Fredholm Type i.e., we are given the following integral equation, φ(x)=f(x)+λ/integraldisplayh 0N/summationdisplay n=1gn(x)hn(y)φ(y)dy. (4.3.2) Deﬁne βnto be βn≡/integraldisplayh 0hn(y)φ(y)dy. (4.3.3) Then the integral equation (4.3.2) takes the form, φ(x)=f(x)+λN/summationdisplay k=1gk(x)βk. (4.3.4) Substituting Eq. (4.3.4) into the expression (4.3.3) for βn,w eh a v e βn=/integraldisplayh 0dy h n(y)f(y)+/integraldisplayh 0dy h n(y)·λN/summationdisplay k=1gk(y)βk. (4.3.5) Hence, upon letting Ank=/integraldisplayh 0dy h n(y)gk(y), (4.3.6) and αn=/integraldisplayh 0dy h n(y)f(y), (4.3.7) Equation (4.3.5) takes the form βn=αn+λN/summationdisplay k=1Ankβk, or N/summationdisplay k=1(δnk−λAnk)βk=αn, (4.3.8) which is equivalent to the N×Nmatrix equation (I−λA)/vectorβ=/vectorα, (4.3.9) where the αs are known and the βs are unknown. If the determinant of the matrix (I−λA)is denoted by ˜D(λ) ˜D(λ)=d e t ( I−λA), 4.3 Pincherle–Goursat Kernel 83 the inverse of (I−λA)can be written as (I−λA)−1=1 ˜D(λ)·D, where Dis a matrix whose ij-th element is the cofactor of the ji-th element of I−λA . (We recall that the cofactor of aijis given by (−1)i+jdetMij,w h e r e detMijis the minor determinant obtained by deleting the row and column to which aijbelongs.) Therefore, βn=1 ˜D(λ)·N/summationdisplay k=1Dnkαk. (4.3.10) From Eqs. (4.3.4) and (4.3.10), we obtain the solution φ(x)as φ(x)=f(x)+λ ˜D(λ)N/summationdisplay n=1N/summationdisplay k=1gn(x)Dnkαk. (4.3.11) Writing out αkexplicitly, we have φ(x)=f(x)+λ ˜D(λ)N/summationdisplay n=1N/summationdisplay k=1gn(x)Dnk/integraldisplayh 0dy h k(y)f(y). (4.3.12) Comparing Eq. (4.3.12) with the deﬁnition of the resolvent H(x, y;λ) φ(x)=f(y)−λ/integraldisplayh 0dy H(x, y;λ)f(y), (4.3.13) we obtain the resolvent for the case of the Pincherle–Goursat kernel as −H(x, y;λ)=1 ˜D(λ)N/summationdisplay n=1N/summationdisplay k=1gn(x)Dnkhk(y). (4.3.14) Note that this is a ratio of two polynomials in λ. We note that the cofactors of the matrix (I−λA)are polynomials in λand hence have no singularities in λ. Thus the numerator of H(x, y;λ)has no singularities. Then the only singularities of H(x, y;λ)occur at the zeros of the denominator ˜D(λ)=d e t ( I−λA), which is a polynomial of degree Ninλ. Therefore, H(x, y;λ)in this case has, at most, N singularities which are poles in the complex λplane. At the poles of H(x, y;λ)inλ,t h e homogeneous problem has nontrivial solutions. 84 4 Integral Equations of the Fredholm Type General kernel: By approximating a general kernel as a sum of a Pincherle–Goursat kernel (plus a small remainder term), we can now prove that in any ﬁnite region of the complex λ plane, there can be, at most, a ﬁnit number of singularities. Consider the integral equation φ(x)=f(x)+λ/integraldisplayh 0K(x, y)φ(y)dy, (4.3.15) with a general square-integrable kernel K(x, y). Suppose we are interested in examining the singularities of H(x, y;λ)in the region |λ|<1/εin the complex λplane (with εpossibly quite small). For this purpose, we can always ﬁnd an approximation to the kernel K(x, y)in the form (with Nsufﬁciently large) K(x, y)=N/summationdisplay n=1gn(x)hn(y)+R(x, y) (4.3.16) with /bardblR/bardbl<ε . (4.3.17) The integral equation (4.3.15) then becomes φ(x)=f(x)+λ/integraldisplayh 0N/summationdisplay n=1gn(x)hn(y)φ(y)dy+λ/integraldisplayh 0R(x, y)φ(y)dy. Deﬁne F(x)=f(x)+λ/integraldisplayh 0N/summationdisplay n=1gn(x)hn(y)φ(y)dy. (4.3.18) Then φ(x)=F(x)+λ/integraldisplayh 0R(x, y)φ(y)dy. LetHR(x, y;λ)be the resolvent kernel corresponding to R(x, y), −HR(x, y;λ)=R(x, y)+λR 2(x, y)+λ2R3(x, y)+···, whence we have φ(x)=F(x)−λ/integraldisplayh 0HR(x, y;λ)F(y)dy. (4.3.19) Substituting the given expression (4.3.18) for F(x)into Eq. (4.3.19), we have φ(x)=f(x)+λ/integraldisplayh 0N/summationdisplay n=1gn(x)hn(y)φ(y)dy −λ/integraldisplayh 0HR(x, y;λ)/bracketleftBigg f(y)+λ/integraldisplayh 0N/summationdisplay n=1gn(y)hn(z)φ(z)dz/bracketrightBigg dy. 4.3 Pincherle–Goursat Kernel 85 Deﬁne ˜F(x)≡f(x)−λ/integraldisplayh 0HR(x, y;λ)f(y)dy. Then φ(x)=˜F(x)+λ/integraldisplayh 0dyN/summationdisplay n=1gn(x)hn(y)φ(y) −λ2/integraldisplayh 0dy/integraldisplayh 0dz H R(x, y;λ)/parenleftBiggN/summationdisplay n=1gn(y)hn(z)/parenrightBigg φ(z).(4.3.20) In the above expression (4.3.20), interchange yandzin the last term on the right-hand side, φ(x) =˜F(x)+λ/integraldisplayh 0dy/bracketleftBiggN/summationdisplay n=1gn(x)hn(y)−λ/integraldisplayh 0dz H R(x, z;λ)N/summationdisplay n=1gn(z)hn(y)/bracketrightBigg φ(y). Then we have φ(x)=˜F(x)+λ/integraldisplayh 0dy/parenleftBiggN/summationdisplay n=1Gn(x;λ)hn(y)/parenrightBigg φ(y) where Gn(x;λ)=gn(x)−λ/integraldisplayh 0dz H R(x, z;λ)gn(z). We have thus reduced the integral equation with the general kernel to one with a Pincherle– Goursat type kernel . The only difference is that the entries in the new kernel N/summationdisplay n=1Gn(x;λ)hn(y) also depend on λthrough the dependence of Gn(x;λ)onλ. We know, however, that for |λ|·/bardblR/bardbl<1, the resolvent HR(x, y;λ)is analytic in λ. Hence Gn(x;λ)is also analytic in λ. Therefore, the singularities in the complex λplane are still found by setting det(I−λA)=0, where the kn-th element of Ais given by Akn=/integraldisplayh 0dy h k(y)Gn(y;λ)=Akn(λ). 86 4 Integral Equations of the Fredholm Type Since the entries Akndepend on λanalytically, the function det(I−λA)is an analytic func- tion of λ(but not necessarily a polynomial of degree N) and hence it has ﬁnitely many zeros in the region |λ|·/bardblR/bardbl<1,o r |λ|<1 ε<1 /bardblR/bardbl. This concludes the proof that in any disk |λ|<1/ε, there are a ﬁnit number of singularities ofλfor the integral equation (4.3.15). 4.4 Fredholm Theory for a Bounded Kernel We now consider the case of a general kernel as approached by Fredholm. We shall show that the resolvent kernel can be written as a ratio of the entire functions of λ, whence the singularities in λoccur when the function in the denominator is zero. Consider φ(x)=f(x)+λ/integraldisplayh 0K(x, y)φ(y)dy,0≤x≤h. (4.4.1) Discretize the above equation by letting ε=h N,x i=iε, y j=jε, i, j=0,1,2,...,N. Also let φi=φ(xi),f i=f(xi),K ij=K(xi,yj). The discrete version of the integral equation (4.4.1) takes the form φi=fi+λN/summationdisplay j=1Kijφjε, i.e., N/summationdisplay j=1(δij−λεK ij)φj=fi. (4.4.2) Deﬁne ˜D(λ)to be ˜D(λ)=d e t ( I−λεK). 4.4 Fredholm Theory for a Bounded Kernel 87 Writing out ˜D(λ)explicitly, we have ˜D(λ)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−λεK 11,−λεK 12,···− λεK 1N −λεK 21,1−λεK 22,···− λεK 2N ·· · ·· ··· ·−λεK N1,−λεK N2,··· 1−λεK NN/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. This determinant can be expanded as ˜D(λ)=˜D(0) + λ˜D /prime(0) +λ2 2!˜D/prime/prime(0) +···+λN N!˜D(N)(0). Using the fact that d dλ|/vectora1,/vectora2,···,/vectoraN|=/vextendsingle/vextendsingle/vextendsingle/vextendsingled dλ/vectora1,/vectora2,···,/vectoraN/vextendsingle/vextendsingle/vextendsingle/vextendsingle +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vectora1,d dλ/vectora2,···,/vectoraN/vextendsingle/vextendsingle/vextendsingle/vextendsingle+···+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vectora 1,/vectora2,···,d dλ/vectoraN/vextendsingle/vextendsingle/vextendsingle/vextendsingle, we ﬁnally obtain (after considerable algebra), ˜D(λ)=1−λεN/summationdisplay i=1Kii+λ2ε2 2!N/summationdisplay i=1N/summationdisplay j=1/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK ii,K ij Kji,K jj/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle −λ3ε3 3!N/summationdisplay i=1N/summationdisplay j=1N/summationdisplay k=1/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK ii,K ij,K ik Kji,K jj,K jk Kki,K kj,K kk/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+···. In the limit as n→∞ , each sum, when multiplied by ε, is approximates a corresponding integral, i.e., N/summationdisplay i=1εKii→/integraldisplayh 0K(x, x)dx, N/summationdisplay i=1N/summationdisplay j=1ε2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK ii,K ij Kji,K jj/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle→/integraldisplay h 0dx/integraldisplayh 0dy/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK(x, x),K(x, y) K(y,x),K(y,y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. 88 4 Integral Equations of the Fredholm Type Deﬁne K/parenleftBigg x1,x 2, ..., x n y1,y 2, ..., y n/parenrightBigg ≡/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK(x 1,y1),K(x1,y2),··· K(x1,yn) K(x2,y1),·· · · K(x2,yn) ·· · ·· · ·· ·K(x n,y1),· K(xn,yn)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Then, in the limit as N→∞ , we ﬁnd (on renaming ˜DasD) D(λ)=1+∞/summationdisplay n=1(−1)nλn n!Dn with Dn=/integraldisplayh 0dx1/integraldisplayh 0dx2···/integraldisplayh 0dxnK/parenleftBigg x1,x 2, ..., x n x1,x 2, ..., x n/parenrightBigg . We expect singularities in the resolvent H(x, y;λ)to occur only when the determinant D(λ) vanishes. Thus we hope to show that H(x, y;λ)can be expressed as the ratio, H(x, y;λ)≡D(x, y;λ) D(λ). (4.4.3) So we need to obtain the numerator D(x, y;λ)and show that it is entire. We also show that the power series given above for D(λ)has an inﬁnite radius of convergence and thus represents an analytic function. To this end, we make use of the fact that K(x, y)isbounded (by assumption) and also invoke the Hadamard inequality which says |det[/vectorv1,/vectorv2,···,/vectorvn]|≤/bardbl/vectorv1/bardbl/bardbl/vectorv2/bardbl···/bardbl/vectorvn/bardbl. This has the interpretation that the volume of the parallelepiped whose edges are /vectorv1through /vectorvnis less than the product of the lengths of those edges. Suppose that |K(x, y)|is bounded byAonx,y∈[0,h]. Then K/parenleftBigg x1, ..., x n x1, ..., x n/parenrightBigg =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK(x 1,x1), ..., K (x1,xn) ·· K(xn,x1), ..., K (xn,xn)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle is bounded by /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK/parenleftBigg x1, ..., x n x1, ..., x n/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤(√ nA)n, 4.4 Fredholm Theory for a Bounded Kernel 89 since the norm of each column is less than√ nA . This implies |Dn|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplayh 0dx1···/integraldisplayh 0dxnK/parenleftBigg x1,···,x n x1,···,x n/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤hnnn/2An. Thus /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationdisplay n=1(−1)nλn n!Dn/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤∞/summationdisplay n=1|λ|nhnnn/2An n!. (4.4.4) Letting an=|λ|nhnnn/2An n!, and applying the ratio test to the right-hand side of the inequality (4.4.4), we have lim n→∞an+1 an= lim n→∞(|λ|n+1hn+1(n+1 )(n+1)/2An+1·n!) (|λ|nhnnn/2An·(n+1 ) ! ) = lim n→∞/bracketleftBigg |λ|hA/parenleftbigg 1+1 n/parenrightbiggn/21 √ n+1/bracketrightBigg =0. Hence the series converges for all λ. We conclude that D(λ)is an entire function of λ. The last step we need to take is to ﬁnd the numerator D(x, y;λ)of the resolvent and show that it, too, is an entire function of λ. For this purpose, we recall that the resolvent itself, H(x, y;λ), satisﬁes the integral equation, H(x, y;λ)=−K(x, y)+λ/integraldisplayh 0K(x, z)H(z,y;λ)dz. (4.4.5) Therefore, upon multiplying the integral equation (4.4.5) by D(λ)and using the deﬁnition (4.4.3) of D(x, y;λ),w eh a v e D(x, y;λ)=−K(x, y)D(λ)+λ/integraldisplayh 0K(x, z)D(z,y;λ)dz. (4.4.6) SinceD(λ)has the expansion D(λ)=∞/summationdisplay n=0(−λ)n n!Dn with D0=1, (4.4.7) we seek an expansion for D(x, y;λ)of the form D(x, y;λ)=∞/summationdisplay n=0(−λ)n n!Cn(x, y). (4.4.8) 90 4 Integral Equations of the Fredholm Type Substituting Eqs. (4.4.7) and (4.4.8) into the integral equation (4.4.6) for D(x, y;λ),w eﬁ n d ∞/summationdisplay n=0(−λ)n n!Cn(x, y)=−∞/summationdisplay n=0(−λ)n n!DnK(x, y) −∞/summationdisplay n=0/integraldisplayh 0(−λ)n+1 n!K(x, z)Cn(z,y)dz. Collecting like powers of λ,w eg e t C0(x, y)=−K(x, y) forn=0, Cn(x, y)=−DnK(x, y)−n/integraldisplayh 0K(x, z)Cn−1(z,y)dz forn=1,2,.... Let us calculate the ﬁrst few of these: C0(x, y)=−K(x, y). From this, we have C1(x, y)=−K(x, y)D1+/integraldisplayh 0K(x, z)K(z,y)dz =/integraldisplayh 0dx1(K(x, x 1)K(x1,y)−K(x1,x1)K(x, y)) =−/integraldisplayh 0dx1K/parenleftBigg x, x 1 y, x 1/parenrightBigg , from which, we obtain C2(x, y)=−/integraldisplayh 0dx1/integraldisplayh 0dx2/bracketleftbigg K(x, y)K/parenleftBigg x1,x 2 x1,x 2/parenrightBigg −K(x, x 1)K/parenleftBigg x1,x 2 y, x 2/parenrightBigg +K(x, x 2)K/parenleftBigg x1,x 2 y, x 1/parenrightBigg/bracketrightbigg =−/integraldisplayh 0dx1/integraldisplayh 0dx2K/parenleftBigg x, x 1,x 2 y, x 1,x 2/parenrightBigg . In general, we have Cn(x, y)=−/integraldisplayh 0dx1/integraldisplayh 0dx2···/integraldisplayh 0dxnK/parenleftBigg x, x 1,x 2,···,x n y, x 1,x 2,···,x n/parenrightBigg . Therefore, we have the numerator D(x, y;λ)ofH(x, y;λ), D(x, y;λ)=∞/summationdisplay n=0(−λ)n n!Cn(x, y), (4.4.9) 4.4 Fredholm Theory for a Bounded Kernel 91 with Cn(x, y)=−/integraldisplayh 0dx1···/integraldisplayh 0dxnK/parenleftBigg x, x 1,x 2, ..., x n y, x 1,x 2, ..., x n/parenrightBigg , n=1,2,..., (4.4.10) and C0(x, y)=−K(x, y). (4.4.11) We prove that the power series for D(x, y;λ)converges for all λ. First, by the Hadamard inequality, we have the following bounds, /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK/parenleftBigg x, x 1, ..., x n y, x 1, ..., x n/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤(√ n+1A)n+1, sinceKabove is a (n+1 )×(n+1 ) determinant with each entry less than A, i.e., |K(x, y)|<A . Then the bound on Cn(x, y)is given by |Cn(x, y)|≤hn(√ n+1A)n+1. Thus we have the bound on D(x, y;λ)as |D(x, y;λ)|≤∞/summationdisplay n=0|λ|n n!hn(√ n+1A)n+1. (4.4.12) Letting an=|λ|n n!hn(√ n+1A)n+1, we apply the ratio test to the right-hand side of inequality (4.4.12). lim n→∞an an−1= lim n→∞/parenleftbig |λ|nhn(n+1 )(n+1)/2An+1(n−1)!/parenrightbig /parenleftBig |λ|n−1hn−1nn/2Ann!/parenrightBig = lim n→∞|λ|hA√ n+1 n/parenleftbiggn+1 n/parenrightbiggn/2 = lim n→∞|λ|hA√ n+1 n/parenleftbigg 1+1 n/parenrightbiggn/2 =0. 92 4 Integral Equations of the Fredholm Type Hence the power series expansion for D(x, y;λ)converges for all λ,a n dD(x, y;λ)is an entire function of λ. Finally, we can prove that whenever H(x, y;λ)exists (i.e., for all λsuch that D(λ)/negationslash=0 ), the solution to the integral equation (4.4.1) is unique . This is best done using the operator notation introduced in Section 4.2. Consider the homogeneous problem φH=λ˜Kφ H. (4.4.13) ˜Hacts on both sides to give ˜Hφ H=λ˜H˜Kφ H. Use the identity λ˜H˜K=˜K+˜H to get ˜Hφ H=˜Kφ H+˜Hφ H. H e n c ew eg e t ˜Kφ H=0, which implies φH=λ˜Kφ H=0. (4.4.14) Thus the homogeneous problem has no nontrivial solutions and the inhomogeneous problem has a unique solution. Summary of the Fredholm theory for a bounded kernel The integral equation φ(x)=f(x)+λ/integraldisplayh 0K(x, y)φ(y)dy has the solution φ(x)=f(x)−λ/integraldisplayh 0H(x, y;λ)f(y)dy with the resolvent kernel given by H(x, y;λ)=D(x, y;λ) D(λ) 4.5 Solvable Example 93 in which D(λ)=∞/summationdisplay n=0(−λ)n n!Dn Dn=/integraldisplayh 0dx1···/integraldisplayh 0dxnK/parenleftBigg x1, ..., x n x1, ..., x n/parenrightBigg ;D0=1 and D(x, y;λ)=∞/summationdisplay n=0(−λ)n n!Cn(x, y), Cn(x, y)=−/integraldisplayh 0dx1···/integraldisplayh 0dxnK/parenleftBigg x, x 1,..., x n y, x 1, ..., x n/parenrightBigg ; C0(x, y)=−K(x, y) where K/parenleftBigg z1,z2, ..., z n w1,w2,..., w n/parenrightBigg ≡/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleK(z 1,w1),K (z1,w2), ..., K (z1,wn) K(z2,w1),K(z2,w2), ..., K (z2,wn) ·· · K(z n,w1),K(zn,w2), ..., K (zn,wn)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. 4.5 Solvable Example Consider the following homogeneous integral equation. ❑ Example 4.2. Solve φ(x)=λ/integraldisplayx 0dy e−(x−y)φ(y)+λ/integraldisplay∞ xdy φ(y),0≤x<∞. (4.5.1) Solution. This is a Fredholm integral equation of the second kind with the kernel K(x, y)=/braceleftBigg e−(x−y)for0≤y<x< ∞, 1 for0≤x≤y<∞.(4.5.2) Note that this kernel (4.5.2) is not square-integrable . Differentiating both sides of Eq. (4.5.1) once, we ﬁnd after a little algebra exφ/prime(x)=−λ/integraldisplayx 0dy eyφ(y). (4.5.3) 94 4 Integral Equations of the Fredholm Type Differentiate both sides of Eq. (4.5.3) once more to obtain the second-order ordinary differen- tial equation of the form φ/prime/prime(x)+φ/prime(x)+λφ(x)=0. (4.5.4) We try a solution of the form φ(x)=Ceαx. (4.5.5) Substituting Eq. (4.5.5) into Eq. (4.5.4), we obtain α2+α+λ=0, or, α=−1±√ 1−4λ 2. In general, we obtain φ(x)=C1eα1x+C2eα2x, (4.5.6) with α1=−1+√ 1−4λ 2,α 2=−1−√ 1−4λ 2. (4.5.7) Now, the expression for φ/prime(x)given above, Eq. (4.5.3), indicates that φ/prime(0) = 0 . (4.5.8) This requires α1C1+α2C2=0 orC2=−α1 α2C1. Hence the solution is φ(x)=C/bracketleftbiggeα1x α1−eα2x α2/bracketrightbigg . However, in order for the integral equation to make sense, we must require the integral /integraldisplay∞ xdy φ(y) to converge. This requires Reα<0, which in turn requires λ>0. (4.5.9) 4.6 Fredholm Integral Equation with a Translation Kernel 95 Thus, for λ≤0, we have no solution, and for λ>0,w eh a v e φ(x)=C/bracketleftbiggeα1x α1−eα2x α2/bracketrightbigg , (4.5.10) withα1andα2g i v e nb yE q .( 4 . 5 . 7 ) . Note that, in this case, we have a continuous spectrum of eigenvalues ( λ>0) for which the homogeneous problem has a solution. The reason the eigenvalue is not discrete is that K(x, y)is not square-integrable. 4.6 Fredholm Integral Equation with a Translation Kernel Suppose x∈(−∞,+∞)and the kernel is translation invariant , i.e., K(x, y)=K(x−y). (4.6.1) Then the inhomogeneous Fredholm integral equation of the second kind is given by φ(x)=f(x)+λ/integraldisplay+∞ −∞K(x−y)φ(y)dy. (4.6.2) Take the F ourier transform of both sides of Eq. (4.6.2) to ﬁnd ˆφ(k)=ˆf(k)+λˆK(k)ˆφ(k). Solve for ˆφ(k)to ﬁnd ˆφ(k)=ˆf(k) 1−λˆK(k). (4.6.3) Solution φ(x)is provided by inverting the Fourier transform ˆφ(k)obtained above. It seems very simple, but there are some subtleties involved in the inversion of ˆφ(k). We present some general discussion of the inversion of the F ourier transform . Suppose that the function F(x)has the asymptotic forms F(x)∼/braceleftBigg eaxasx→+∞(a>0), ebxasx→− ∞ (b>a> 0).(4.6.4) Namely F(x)grows exponentially as x→+∞, and decays exponentially as x→− ∞ . Then the Fourier transform ˆF(k) ˆF(k)=/integraldisplay+∞ −∞e−ikxF(x)dx (4.6.5) exists as long as −b<Imk<−a, (4.6.6) 96 4 Integral Equations of the Fredholm Type since the integrand has magnitude /vextendsingle/vextendsinglee−ikxF(x)/vextendsingle/vextendsingle∼/braceleftBigg e(k2+a)xasx→+∞, e(k2+b)xasx→− ∞ , where we set k=k1+ik2, with k1andk2 real. With −b<k 2<−a, the magnitude of the integrand vanishes exponentially at both ends. The inverse Fourier transformation then becomes F(x)=1 2π/integraldisplay+∞−iγ −∞− iγeikxˆF(k)dk with a<γ0), then the inversion contour is on γ=aand the Fourier transform exists only for k2=−a. Similarly, if the function F(x)decays exponentially as x→+∞and grows as x→− ∞ , we are able to continue deﬁning the Fourier transform and its inverse by proceeding in the upper half-plane. As an example, a function like F(x)=e−α|x|(4.6.8) which decays exponentially as x→± ∞ has a Fourier transform which exists and is analytic in −α<k 2<α . (4.6.9) With these qualiﬁcations, we should be able to invert ˆφ(k)to obtain the solution to the inhomogeneous problem (4.6.2). Now follows the homogeneous problem, φH(x)=λ/integraldisplay+∞ −∞K(x−y)φH(y)dy. (4.6.10) By Fourier transforming Eq. (4.6.10), we obtain (1−λˆK(k))ˆφH(k)=0. (4.6.11) If1−λˆK(k)has no zeros for all k,t h e nw eh a v e ˆφH(k)=0⇒φH(x)=0, (4.6.12) 4.6 Fredholm Integral Equation with a Translation Kernel 97 i.e., no nontrivial solution exists for the homogeneous problem. If, on the other hand, 1−λˆK(k)has a zero of order natk=α,ˆφH(k)can be allowed to be of the form ˆφH(k)=C1δ(k−α)+C2d dkδ(k−α)+···+Cn/parenleftbiggd dk/parenrightbiggn−1 δ(k−α). On inversion, we ﬁnd φH(x)=C1eiαx+C2xeiαx+···+Cnxn−1eiαx=eiαxn/summationdisplay j=1Cjxj−1. (4.6.13) For the homogeneous problem, we choose the inversion contour of the Fourier transform based on the asymptotic behavior of the kernel, a point to be discussed in the following example. ❑ Example 4.3. Consider the homogeneous integral equation, φH(x)=λ/integraldisplay+∞ −∞e−|x−y|φH(y)dy. (4.6.14) Solution. Since the kernel vanishes exponentially as e−yasy→∞ and as e+yasy→− ∞ , we need not require φH(y)to vanish as y→± ∞ , rather, more generally we may permit φH(y)→/braceleftBigg e(1−ε)yasy→∞, e(−1+ε)yasy→− ∞ , and the integral equation still makes sense. So in the Fourier transform, we may allow −1<k 2<1, (4.6.15) and still have a valid solution. The Fourier transform of e−α|x|is given by /integraldisplay+∞ −∞e−ikxe−α|x|dx=2α k2+α2. (4.6.16) Taking the Fourier transform of the homogeneous equation with α=1 ,w eﬁ n d ˆφH(k)=2λ k2+1ˆφH(k), from which, we obtain k2+1−2λ k2+1ˆφH(k)=0. So there exists no nontrivial solution unless k=±i√ 1−2λ. By the inversion formula, φH(x)is a superposition of e+ikxterms with amplitude ˆφH(k).B u t ˆφH(k)is zero for all but k=±i√ 1−2λ. Hence we may conclude tentatively that φH(x)=C1e−√ 1−2λx+C2e+√ 1−2λx. However, we can at most allow φH(x)to grow as fast as exasx→∞ and as e−xasx→− ∞ , as we discussed above. Thus further analysis is in order. 98 4 Integral Equations of the Fredholm Type Case (1) 1−2λ<0,o r λ>1 2. φH(x)is oscillatory and is given by φH(x)=C1e−i√ 2λ−1x+C2e+i√ 2λ−1x. (4.6.17) Case (2) 0<1−2λ<1,o r 0<λ<1 2. φH(x)g r o w sl e s sf a s tt h a n e|x|as|x|→∞ . φH(x)=C1e−√ 1−2λx+C2e+√ 1−2λx. (4.6.18) Case (3) 1−2λ≥1,o r λ≤0. No acceptable solution for φH(x)exists, since e±√ 1−2λxgrows faster than e|x|as |x|→∞ . Case (4) λ=1 2. φH(x)=C1+C2x. (4.6.19) Now consider the corresponding inhomogeneous problem. ❑ Example 4.4. Consider the inhomogeneous integral equation, φ(x)=ae−α|x|+λ/integraldisplay+∞ −∞e−|x−y|φ(y)dy. (4.6.20) Solution. On taking the Fourier Transform of Eq. (4.6.20), we obtain ˆφ(k)=2aα k2+α2+2λ k2+1ˆφ(k). Solving for ˆφ(k), we obtain ˆφ(k)=2aα(k2+1 ) (k2+1−2λ)(k2+α2). To invert the latter transform we note that, depending on whether λis larger or smaller than 1/2, the poles k=±√ 2λ−1could lie on the real or imaginary axis of the complex kplane. What we can do is to choose any contour for the inversion within the strip −min(α,1)<k 2<min(α,1) (4.6.21) to get a particular solution to our equation and we may then add any multiple of the homo- geneous solution when the latter exists. The reason for choosing the strip (4.6.21) insteadof −1<k 2<1 in this case is that, in order for the Fourier transform of the inhomogeneous term e−α|x|to exist, we must also restrict our attention to −α<k 2<α . Consider the ﬁrst three cases given in Example 4.3. 4.6 Fredholm Integral Equation with a Translation Kernel 99 Cases (2) and (3) λ<1 2. In these cases, we have 1−2λ>0. Hence ˆφ(k)has simple poles at k=±iα and k=±i√ 1−2λ. To ﬁnd a particular solution, use the real kaxis as the integration contour for the inverse Fourier transformation. Then φP(x)is given by φP(x)=2aα 2π/integraldisplay+∞ −∞dk eikx (k2+1 ) (k2+1−2λ)(k2+α2). Forx>0, we close the contour in the upper half-plane to obtain φP(x)=2aα 2π·2πi/bracketleftBig Res(iα)+ Res/parenleftBig i√ 1−2λ/parenrightBig/bracketrightBig =a 1−2λ−α2/bracketleftbigg/parenleftbig 1−α2/parenrightbig e−αx−2λα √ 1−2λe−√ 1−2λx/bracketrightbigg . Forx<0, we close the contour in the lower half-plane to get an identical result with x replaced by −x. Thus our particular solution φP(x)is given by φP(x)=a 1−2λ−α2/bracketleftbigg/parenleftbig 1−α2/parenrightbig e−α|x|−2λα √ 1−2λe−√ 1−2λ|x|/bracketrightbigg . (4.6.22) For Case (3), this is the unique solution because there exists no acceptable homogeneous solution, while for Case (2) we must also add the homogeneous part given by φH(x)=C1e−√ 1−2λx+C2e+√ 1−2λx. Case (1) λ>1 2. In this case, we have 1−2λ<0. Hence ˆφ(k)has simple poles at k=±iα and k=±√ 2λ−1. To do the inversion for the particular solution, we can take any of the con- tours (1), (2), (3) or (4) as displayed in Figures 4.1 through 4.4, or Principal V alue contourswhich are equivalent to half the sum of the ﬁrst two contours or half the sum of the latter two contours. Any of these differs by a multiple of the homogeneous solution. Consider a particular choice (4) for the inversion. For x>0, we close the contour in the upper half-plane. Then our particular solution φ P(x)is given by φP(x)=2aα 2π·2πi/bracketleftBig Res(−√ 2λ−1) + Res(+iα)/bracketrightBig =a 1−2λ−α2/bracketleftbigg (1−α2)e−αx+2λαi √ 2λ−1e−i√ 2λ−1x/bracketrightbigg . Forx<0, we close the contour in the lower half-plane to get an identical result with x replaced by −x. So, in general, we can write our particular solution with the inversion contour (4) as, φP(x)=a 1−2λ−α2/bracketleftbigg/parenleftbig 1−α2/parenrightbig e−α|x|+2λαi √ 2λ−1e−i√ 2λ−1|x|/bracketrightbigg , (4.6.23) to which must be added the homogeneous part for Case (1) which reads, φH(x)=C1e−i√ 2λ−1x+C2e+i√ 2λ−1x. 100 4 Integral Equations of the Fredholm Type k k1k2 −− 21λ 21λ−iα −iαk k1k2 iα −iα−− 21λ 21λ− k2 k1 −− 21λ21λ−iα −iαk kk2 k1−− 21λ 21λ−iα −iα Fig. 4.1: The inversion contour (1)–(4) for Case (1) –Case (4) . 4.7 System of Fredholm Integral Equations of the Second Kind We solve the system of Fredholm integral equations of the second kind, φi(x)−λ/integraldisplayb an/summationdisplay j=1Kij(x, y)φj(y)dy=fi(x),i =1,2,···,n, (4.7.1) where the kernels Kij(x, y)are square-integrable . We ﬁrst extend the basic interval from [a,b]to[a,a+n(b−a)], and set x+(i−1)(b−a)=X<a +i(b−a), y+(j−1)(b−a)=Y< a +j(b−a),(4.7.2) φ(X)=φi(x),K (X,Y)=Kij(x, y),f (X)=fi(x). (4.7.3) We then obtain the Fredholm integral equation of the second kind, φ(X)−λ/integraldisplaya+n(b−a) aK(X,Y)φ(Y)dY=f(X), (4.7.4) 4.8 Problems for Chapter 4 101 where the kernel K(X,Y)is discontinuous in general but is square-integrable on account of the square-integrability of Kij(x, y). The solution φ(X)to Eq. (4.7.4) provides the solutions φi(x)to Eq. (4.7.1) with Eqs. (4.7.2) and (4.7.3). 4.8 Problems for Chapter 4 4.1. Calculate D(λ)for a)K(x, y)=/braceleftBigg xy, y ≤x, 0, otherwise . b)K(x, y)= xy, 0≤x, y≤1. c)K(x, y)=/braceleftBigg g(x)h(y),y≤x, 0, otherwise . d)K(x, y)= g(x)h(y),0≤x, y≤1. Find zero of D(λ)for each case. 4.2. (Due to H. C.) Solve φ(x)=λ/braceleftbigg/integraldisplayx 0dyφ(y) (y+1 )2/parenleftBigy x/parenrightBiga +/integraldisplay+∞ xdyφ(y) (y+1 )2/bracerightbigg ,a > 0. Find all eigenvalues and eigenfunctions. 4.3. (Due to H. C.) Solve the Fredholm integral equation of the second kind, given that K(x−y)=e−|x−y|,f (x)=/braceleftBigg x, x > 0, 0,x < 0. 4.4. (Due to H. C.) Solve the Fredholm integral equation of the second kind, given that K(x−y)=e−|x−y|,f (x)=x for−∞<x< +∞. 4.5. (Due to H. C.) Solve φ(x)+λ/integraldisplay+1 −1K(x, y)φ(y)dy=1,−1≤x≤1,K(x, y)=/radicalbigg 1−y2 1−x2. Find all eigenvalues of K(x, y). Calculate also D(λ)andD(x, y;λ). 4.6. (Due to H. C.) Solve the Fredholm integral equation of the second kind, φ(x)=e−x 2+λ/integraldisplay+∞ −∞1 cosh(x−y)φ(y)dy. 102 4 Integral Equations of the Fredholm Type Hint: /integraldisplay+∞ −∞eikx coshxdx=π cosh(πk/2). 4.7. (Due to H. C. and D. M.) Consider the integral equation, φ(x)=λ/integraldisplay+∞ −∞dy √ 2πeixyφ(y),−∞<x< ∞. a) Show that there are only four eigenvalues of the kernel (1/√ 2π)e x p [ixy].W h a t a r e these? b) Show by an explicit calculation that the functions, φn(x)=e x p/bracketleftbigg −x2 2/bracketrightbigg Hn(x), where Hn(x)≡(−1)nexp[x2]dn dxnexp[−x2], are Hermite polynomials, are eigenfunctions with the corresponding eigenvalues, (i)n,(n=0,1,2,...). Why should one expect φn(x)to be Fourier transforms of themselves? Hint: Think of the Schrödinger equation for the harmonic oscillator. c) Using the result in b) and the fact that {φn(x)}nform a complete set, in some sense, show that any square-integrable solution is of the form, φ(x)=f(x)+C˜f(x), where f(x)is an arbitrary even or odd, square-integrable function with Fourier trans- form ˜f(k),a n dCis a suitable constant. Evaluate Cand relate its values to the eigenvalues found in a). d) From c), construct a solution by taking f(x)=e x p [ −ax2/2],a>0. 4.8. (Due to H. C.) Find an eigenvalue and the corresponding eigenfunction for K(x, y)=e x p/bracketleftbig −(ax2+2bxy+cy2)/bracketrightbig ,−∞<x ,y< ∞,a+c>0. 4.9. Consider the homogeneous integral equation, φ(x)=λ/integraldisplay∞ −∞K(x, y)φ(y)dy, −∞<x< ∞, where K(x, y)=1 √ 1−t2exp/bracketleftbiggx2+y2 2/bracketrightbigg exp/bracketleftbigg −x2+y2−2xyt 1−t2/bracketrightbigg ,t ﬁxed,0<t< 1. 4.8 Problems for Chapter 4 103 (a) Show directly that φ0(x)=e x p [ −x2/2]is an eigenfunction of K(x, y)corre- sponding to eigenvalue λ=λ0=1/√ π. (b) Let φn(x)=e x p/bracketleftbigg −x2 2/bracketrightbigg Hn(x). Assume that φn=λnKφn. Show that φn+1=λn+1Kφn+1 withλn=tλn+1. This means that the original integral equation has eigenvalues λn=t−n/√ π, with the corresponding eigenfunctions φn(x). 4.10. (Due to H. C.) Find the eigenvalues and eigenfunctions of the integral equation, φ(x)=λ/integraldisplay∞ 0exp[−xy]φ(y)dy. Hint: Consider the Mellin transform, Φ(p)=/integraldisplay∞ 0xip−1 2φ(x)dx with φ(x)=1 2π/integraldisplay∞ −∞x−ip−1 2Φ(p)dp. 4.11. (Due to H. C.) Solve ψ(x)=ebx+λ/integraldisplayx 0ψ(y)dy+2λ/integraldisplay1 xψ(y)dy. 4.12. Solve φ(x)=λ/integraldisplay+1 −1K(x, y)φ(y)dy−1 2/integraldisplay+1 −1φ(y)dy with φ(±1) = ﬁnite, where K(x, y)=1 2ln/parenleftbigg1+x< 1−x>/parenrightbigg , x<=1 2(x+y)−1 2|x−y| andx>=1 2(x+y)+1 2|x−y|. 4.13. Solve φ(x)=λ/integraldisplay+∞ −∞K(x, y)φ(y)dy with φ(±∞)= ﬁnite, where K(x, y)=/radicalbigg α π/braceleftbigg exp/bracketleftBigα 2(x2+y2)/bracketrightBig/integraldisplayx< −∞exp[−ατ2]dτ·/integraldisplay+∞ x>exp[−ατ2]dτ/bracerightbigg , x<=1 2(x+y)−1 2|x−y| andx>=1 2(x+y)+1 2|x−y|. 104 4 Integral Equations of the Fredholm Type 4.14. Solve φ(x)=λ/integraldisplay∞ 0K(x, y)φ(y)dy with |φ(x)|<∞ for 0≤x<∞, where K(x, y)=exp[−β|x−y|] 2βxy. 4.15. (Due to D. M.) Show that the non-trivial solutions of the homogeneous integral equa- tion, φ(x)=λ/integraldisplayπ −π/bracketleftbigg1 4π(x−y)2−1 2|x−y|/bracketrightbigg φ(y)dy, arecos(mx)andsin(mx),w h e r e λ=m2andmis any integer. Hint for Problems 4.12 through 4.15 : The kernels change their forms continuously asxpasses through y. Differentiate the given integral equations with respect to xand reduce them to the ordinary differential equations. 4.16. (Due to D. M.) Solve the inhomogeneous integral equation, φ(x)=f(x)+λ/integraldisplay∞ 0cos(2xy)φ(y)dy, x ≥0, where λ2/negationslash=4/π. Hint: Multiply both sides of the integral equation by cos(2xξ)and integrate over x. Use the identity, cos(2xy)c o s ( 2 xξ)=1 2/braceleftbig cos[2x(y+ξ)] + cos[2 x(y−ξ)]/bracerightbig , and observe that /integraldisplay∞ 0cos(αx)dx=1 2/integraldisplay∞ 0(exp[iαx]+e x p [ −iαx])dx =1 2/integraldisplay∞ −∞exp[iαx]dx =1 2·2πδ(α)=πδ(α). 4.8 Problems for Chapter 4 105 4.17. (Due to D. M.) In the theoretical search for “ supergain antennas ”, maximizing the di- rectivity in the far ﬁeld of axially invariant currents j(φ)that ﬂow along the surface of inﬁnitely long, circular cylinders of radius a, leads to the following Fredholm integral equation for the current density j(φ), j(φ)=e x p [ ikasinφ]−α/integraldisplay2π 0dφ/prime 2πJ0/parenleftbigg 2kasinφ−φ/prime 2/parenrightbigg j(φ/prime),0≤φ<2π, where φis the polar angle of the circular cross-section, kis a positive wave number, α is a parameter (Lagrange multiplier) which expresses a constraint on the current magni-tude,α≥0,a n dJ 0(x)is the zeroth-order Bessel function of the ﬁrst kind. a) Determine the eigenvalues of the homogeneous equation. b) Solve the given inhomogeneous equation in terms of the Fourier series, j(φ)=∞/summationdisplay n=−∞fnexp[inφ]. Hint: Use the formulas, exp[ikasinφ]=∞/summationdisplay n=−∞Jn(ka)e x p [inφ], and J0/parenleftbigg 2kasinφ−φ/prime 2/parenrightbigg =∞/summationdisplay m=−∞Jm(ka)2exp[im(φ−φ/prime)], where Jn(x)is thenth-order Bessel function of the ﬁrst kind. Substitution of the Fourier series for j(φ), j(φ)=∞/summationdisplay n=−∞fnexp[inφ], yields the decoupled equation for fn, fn=Jn(ka) 1+αJn(ka)2,−∞<n< ∞. 4.18. (Due to D. M.) Problem 4.17 corresponds to the circular loop in two-dimensions. For the circular disk in two dimensions, we have the following Fredholm integral equationfor the current density j(φ), j(r, φ)=e x p [ ikrsinφ]−2α a2/integraldisplay2π 0dφ/prime 2π/integraldisplaya 0r/primedr/prime ×J0/parenleftbigg k/radicalBig r2+r/prime2−2rr/primecos(φ−φ/prime)/parenrightbigg j(r/prime,φ/prime), 106 4 Integral Equations of the Fredholm Type with 0≤r≤a, 0≤φ<2π. Solve the given inhomogeneous equation in terms of the Fourier series, j(r, φ)=n=∞/summationdisplay n=−∞fn(r)e x p [inφ]. Hint: By using the addition formula, J0/parenleftbigg k/radicalBig r2+r/prime2−2rr/primecos(φ−φ/prime)/parenrightbigg =∞/summationdisplay m=−∞Jm(kr)Jm(kr/prime)e x p [im(φ−φ/prime)], it is found that fn(r)satisfy the following integral equation, fn(r)=/bracketleftbigg 1−2α a2/integraldisplaya 0r/primedr/primefn(r/prime)Jn(kr/prime)/bracketrightbigg Jn(kr),−∞<n< ∞. Substitution of fn(r)=λnJn(kr) yields λn=/bracketleftbigg 1+2α a2/integraldisplaya 0r/primedr/primeJn(kr/prime)2/bracketrightbigg−1 =/bracketleftbig 1+α[Jn(ka)2−Jn+1(ka)Jn−1(ka)]/bracketrightbig−1. 4.19. (Due to D. M.) Problem 4.17 corresponds to the circular loop in two dimensions. For the circular loop in three dimensions, we have the following Fredholm integral equationfor the current density j(φ), j(φ)=e x p [ ikasinφ]−α/integraldisplay 2π 0dφ/prime 2πK(φ−φ/prime)j(φ/prime),0≤φ<2π, with K(φ)=sinw w+cosw w2−sinw w3 =1 4/integraldisplay1 −1(1 +ξ2)e x p [iwξ]dξ, and w=w(φ)=2kasinφ 2. Solve the given inhomogeneous equation in terms of Fourier series, j(φ)=n=∞/summationdisplay n=−∞fnexp[inφ]. 4.8 Problems for Chapter 4 107 Hint: Following the step employed in Problem 4.17, substitute the Fourier series into the integral equation. The decoupled equation for fn, fn=Jn(ka) 1+αUn(ka),−∞<n< ∞, results, where Un(ka)=/integraldisplayπ −πdφ 2πK(φ)e x p [−inφ] =1 8π/integraldisplay1 −1dξ(1 +ξ2)/integraldisplayπ −πdφexp[iw(φ)ξ] cos(nφ) =1 2/integraldisplay1 0dξ(1 +ξ2)J2n(2kaξ). The integral for Un(ka)can be further simpliﬁed by the use of Lommel’s function Sµ,ν(x), and Weber’s function Eν(x). Reference for Problems 4.17, 4.18 and 4.19: We cite the following article for the Fredholm integral equations of the second kind in the theoretical search for “ supergain antennas ”. Margetis, D., Fikioris, G., Myers, J.M., and Wu, T.T.: Phys. Rev. E58 ., 2531, (1998). We cite the following article for the Fredholm integral equations of the second kind for the two-dimensional, highly directive currents on large circular loops. Margetis, D. and Fikioris, G.: Jour. Math. Phys. 41., 6130, (2000). We can derive the above-stated Fredholm integral equations of the second kind for the localized, monochromatic, and highly directive classical current distributions in two andthree dimensions by maximizing the directivity Din the far ﬁeld while constraining C=N/T ,w h e r e Nis the integral of the square of the magnitude of the current density andTis proportional to the total radiated power. This derivation is the application of the calculus of variations. We derive the homogeneous Fredholm integral equations of the second kind and the inhomogeneous Fredholm integral equations of the second kind in their general forms in Section 9.6 of Chapter 9. 5 Hilbert–Schmidt Theory of Symmetric Kernel 5.1 Real and Symmetric Matrix We would now like to examine the case of a symmetric kernel (self-adjoint integral operator ) which is also square-integrable . Recalling from our earlier discussions in Chapter 1 that self- adjoint operators can be diagonalized , our principal aim is to accomplish the same goal for the case of symmetric kernels. For this purpose, let us ﬁrst examine the corresponding problem for an n×nreal and symmetric matrix A. Suppose Ahas eigenvalues λkand normalized eigenvectors /vectorvk, i.e., A/vectorvk=λk/vectorvk,k=1,2,···,n, (5.1.1a) /vectorvT k/vectorvm=δkm. (5.1.1b) We may thus write A[/vectorv1,/vectorv2,...,/vectorvn]=[λ1/vectorv1,λ2/vectorv2,...,λ n/vectorvn] =[/vectorv1,...,/vectorvn] λ 10 ·0 0λ2 · ·· · ··· · 0 0 ·0λn .(5.1.2) Deﬁne the matrix Sby S=[/vectorv1,/vectorv2,···,/vectorvn] (5.1.3a) and consider ST ST= /vectorvT 1 ... /vectorvT n . (5.1.3b) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 110 5 Hilbert–Schmidt Theory of Symmetric Kernel Then we have STS= /vectorvT 1 ... /vectorvT n [/vectorv1,···,/vectorvn]= 10 0 01 ...0 00 1 =I, (5.1.4a) since we have Eq. (5.1.1b). Hence we have S T=S−1. (5.1.5) Deﬁne D= λ100 0λ2 ...0 00 λn . (5.1.6) From Eq. (5.1.2), we have AS=SD. (5.1.7a) H e n c ew eh a v e A=SDS−1=SDST. (5.1.7b) The above relation can also be written as A=n/summationdisplay k=1λk/vectorvT k/vectorvk, (5.1.8) which represents the diagonalization of a symmetric matrix. Eq. (5.1.7b) is really convenient for calculation of functions of A, e.g., A2=/parenleftbig SDS−1/parenrightbig/parenleftbig SDS−1/parenrightbig =SD2S−1=S λ2 100 0... ...0 00 λ2 n S−1. eA=I+A+1 2A2+1 3!A3+···=S eλ100 0· ·0 00 eλn S−1. 5.2 Real and Symmetric Kernel 111 f(A)=S f(λ1)0 0 0 · ·0 00 f(λn) S−1. (5.1.9) Finally we have detA=d e t SDS−1=d e t SdetDdetS−1=d e t D=n/productdisplay k=1λk, (5.1.10a) tr(A)= tr/parenleftbig SDS−1/parenrightbig =tr/parenleftbig DS−1S/parenrightbig =tr(D)=n/summationdisplay k=1λk. (5.1.10b) 5.2 Real and Symmetric Kernel Symmetric kernels have the property that when transposed they remain the same as the original kernel. We denote the transposed kernel KTby KT(x, y)=K(y,x), (5.2.1) and note that when the kernel Kis symmetric, we have KT(x, y)=K(y,x)=K(x, y). (5.2.2) The eigenvalues of K(x, y)and eigenvalues of KT(x, y)are the same. This is because an eigenvalue λnis a zero of D(λ). From the deﬁnition of D(λ), we ﬁnd that, since a determinant remains the same as we exchange its rows and columns, D(λ)forK(x, y)=D(λ)forKT(x, y). (5.2.3) Thus the spectrum of K(x, y)coincides with that of KT(x, y). We will need to make use of the orthogonality property held by the eigenfunctions belong- ing to each eigenvalue. To show this property, we start with the eigenvalue equations, φn(x)=λn/integraldisplayh 0K(x, y)φn(y)dy, (5.2.4) ψn(x)=λn/integraldisplayh 0KT(x, y)ψn(y)dy, (5.2.5) and from the deﬁnition of KT(x, y), ψn(x)=λn/integraldisplayh 0K(y,x)ψn(y)dy. (5.2.6) 112 5 Hilbert–Schmidt Theory of Symmetric Kernel Multiplying Eq. (5.2.4) by ψm(x)and integrating over x,w eg e t /integraldisplayh 0ψm(x)φn(x)dx=λn/integraldisplayh 0dx ψ m(x)/integraldisplayh 0K(x, y)φn(y)dy =λn λm/integraldisplayh 0ψm(y)φn(y)dy, so then /parenleftbigg 1−λn λm/parenrightbigg/integraldisplayh 0ψm(x)φn(x)dx=0. (5.2.7) If λn/negationslash=λm, then we have /integraldisplayh 0ψm(x)φn(x)dx=0 forλn/negationslash=λm. (5.2.8) In the case of ﬁnite matrices, we know that the eigenvalues of a symmetric matrix are real and that the matrix is diagonalizable. Also, the eigenvectors are orthogonal to each other. We shall show that the same statements hold true in the case of square-integrable symmetric kernels . IfKis symmetric, then ψn(x)=φn(x). Then, by Eq. (5.2.8), the eigenfunctions of a symmetric kernel are orthogonal to each other, /integraldisplayh 0φm(x)φn(x)dx=0 forλn/negationslash=λm. (5.2.9) Furthermore, the eigenvalues of a symmetric kernel must be real. This is seen by suppos- ing that the eigenvalue λnis complex. Then we have λn/negationslash=λ∗ n. The complex conjugate of Eq. (5.2.4) is given by φ∗ n(x)=λ∗ n/integraldisplayh 0K(x, y)φ∗ n(y)dy, (5.2.10) implying that λ∗ nandφ∗ n(x)are an eigenvalue and eigenfunction of the kernel K(x, y). But, Eq. (5.2.9) with λn/negationslash=λ∗ n 5.2 Real and Symmetric Kernel 113 then requires that /integraldisplayh 0φn(x)φ∗ n(x)dx=/integraldisplayh 0|φn(x)|2dx=0, (5.2.11) implying then that φn(x)≡0, which is a contradiction. Thus the eigenvalue must be real, λn=λ∗ n, to avoid this contradiction. Therefore the eigenfunctions of a symmetric kernel are orthogonal to each other and the eigenvalues are real . We now, rather boldly, expand the symmetric kernel K(x, y)in terms of φn(x), K(x, y)=/summationdisplay nanφn(x). (5.2.12) We then normalize the eigenfunctions such that /integraldisplayh 0φn(x)φm(x)dx=δnm, (5.2.13) (even if there is more than one eigenfunction belonging to a certain eigenvalue, we can choose linear combinations of these eigenfunctions to satisfy Eq. (5.2.13)). From the orthogonality(5.2.13), we ﬁnd a n=/integraldisplayh 0dx φ n(x)K(x, y)=1 λnφn(y), (5.2.14) and thus obtain Hilbert–Schmidt Theorem: K(x, y)=/summationdisplay nφn(y)φn(x) λn. (5.2.15) There is a problem though. We do not know if the eigenfunctions {φn(x)}nare complete. In fact, we are often sure that the set {φn(x)}nis not complete. An example is the kernel in the form of a ﬁnite sum of factorized terms. However, the content of the Hilbert–Schmidt Theorem (which will be proved shortly) is to claim that Eq. (5.2.15) for K(x, y)is valid whether or not {φn(x)}nis complete. The only conditions are that K(x, y)besymmetric and square-integrable . 114 5 Hilbert–Schmidt Theory of Symmetric Kernel We calculate the iterated kernel, K2(x, y)=/integraldisplayh 0K(x, z)K(z,y)dz =/integraldisplayh 0/summationdisplay nφn(x)φn(z) λn/summationdisplay mφm(z)φm(y) λmdz =/summationdisplay n/summationdisplay m1 λnλmφn(x)δnmφm(y) =/summationdisplay nφn(x)φn(y) λ2n,(5.2.16) and in general, we obtain Kj(x, y)=/summationdisplay nφn(x)φn(y) λj n,j =2,3,···. (5.2.17) Now the deﬁnition for the resolvent kernel H(x, y;λ)=−K(x, y)−λK 2(x, y)−···− λjKj+1(x, y)−··· becomes H(x, y;λ)=−/summationdisplay nφn(x)φn(y) λn/bracketleftbigg 1+λ λn+λ2 λ2n+···+λj λj n+···/bracketrightbigg =−/summationdisplay nφn(x)φn(y) λn1 1−λ λn, i.e., H(x, y;λ)=/summationdisplay nφn(x)φn(y) λ−λn. (5.2.18) This elegant expression explicitly shows the analytic properties of H(x, y;λ)in the complex λplane. We can use this resolvent to solve the inhomogeneous Fredholm Integral Equation of the second kind with a symmetric and square-integrable kernel . φ(x)=f(x)+λ/integraldisplayh 0K(x, y)φ(y)dy =f(x)−λ/integraldisplayh 0H(x, y;λ)f(y)dy =f(x)−λ/summationdisplay nφn(x) λ−λn/integraldisplayh 0φn(y)f(y)dy.(5.2.19) 5.2 Real and Symmetric Kernel 115 Denoting fn≡/integraldisplayh 0φn(y)f(y)dy, (5.2.20) we have the solution to the inhomogeneous equation (5.2.19), φ(x)=f(x)−λ/summationdisplay nfnφn(x) λ−λn. (5.2.21) Atλ=λn, the solution does not exist unless fn=0 ,a su s u a l . As an another application of the eigenfunction expansion (5.2.15), we consider the Fred- holm Integral Equation of the ﬁrst kind with a symmetric and square-integrable kernel , f(x)=/integraldisplayh 0K(x, y)φ(y)dy. (5.2.22) Denoting φn≡/integraldisplayh 0φn(y)φ(y)dy, (5.2.23) we have f(x)=/summationdisplay nφn(x) λnφn. (5.2.24) Immediately we encounter the problem. Equation (5.2.24) states that f(x)is a linear com- bination of φn(x). In many cases, the set {φn(x)}nis not complete, and thus f(x)is not necessarily representable by a linear superposition of {φn(x)}nand Eq. (5.2.22) has no solu- tion. Iff(x)is representable by a linear superposition of {φn(x)}n, it is easy to obtain φn. From Eqs. (5.2.20) and (5.2.24), fn=/integraldisplayh 0φn(x)f(x)dx=φn λn, (5.2.25) and so φn=fnλn. (5.2.26) A solution to Eq. (5.2.22) is then given by φ(x)=/summationdisplay nφnφn(x)=/summationdisplay nλnfnφn(x). (5.2.27) If the set {φn(x)}nis not complete, the solution (5.2.27) is not unique. We can add to it any linear combination of {ψi(x)}ithat is orthogonal to {φn(x)}n, φ(x)=/summationdisplay nλnfnφn(x)+/summationdisplay iCiψi(x), (5.2.28) 116 5 Hilbert–Schmidt Theory of Symmetric Kernel where /integraldisplayh 0ψi(x)φn(x)dx=0 for all iandn. (5.2.29) If the set {φn(x)}nis complete, the solution (5.2.27) is the unique solution. It may, however, still diverge since we have λnin the numerator, unless fnvanishes sufﬁciently rapidly as n→∞ to ensure the convergence of the series (5.2.27). We will now prove the Hilbert–Schmidt expansion, (5.2.15), in order to exhibit its efﬁ- ciency, but, to avoid getting too mathematical, we will not be completely rigorous. We will outline a plan of the proof. First, note the following lemma. Lemma. F or a non-zero normed symmetric kernel, ∞>/bardblK/bardbl>0 and K(x, y)=KT(x, y), (5.2.30) there exists at least one eigenvalue λ1and one eigenfunction φ1(x)(which we normalize to unity). To prove Eq. (5.2.15), once this lemma has been established, we can construct a new kernel ¯K(x, y)by ¯K(x, y)≡K(x, y)−φ1(x)φ1(y) λ1. (5.2.31) Nowφ1(x)cannot be an eigenfunction of ¯K(x, y)because we have /integraldisplayh 0¯K(x, y)φ1(y)dy=/integraldisplayh 0/bracketleftbigg K(x, y)−φ1(x)φ1(y) λ1/bracketrightbigg φ1(y)dy =1 λ1φ1(x)−1 λ1φ1(x)=0,(5.2.32) which leaves us two possibilities, (A)/vextenddouble/vextenddouble¯K/vextenddouble/vextenddouble≡0. We have an equality K(x, y)=φ1(x)φ1(y) λ1, (5.2.33) except over a set of points xwhose measure is zero. A proof for this case is shown. (B)/vextenddouble/vextenddouble¯K/vextenddouble/vextenddouble/negationslash=0 . By the Lemma, there exists at least one eigenvalue λ2and one eigenfunction φ2(x)of a kernel ¯K(x, y). λ2/integraldisplayh 0¯K(x, y)φ2(y)dy=φ2(x), (5.2.34) 5.2 Real and Symmetric Kernel 117 i.e., λ2/integraldisplayh 0/bracketleftbigg K(x, y)−φ1(x)φ1(y) λ1/bracketrightbigg φ2(y)dy=φ2(x). (5.2.35) We then show that φ2(x)andλ2are an eigenfunction and eigenvalue of the original kernel K(x, y)orthogonal to φ1(x). To demonstrate the orthogonality of φ2(x)toφ1(x), multiply Eq. (5.2.35) by φ1(x)and integrate over x. /integraldisplayh 0φ1(x)φ2(x)dx=λ2/integraldisplayh 0φ1(x)dx/integraldisplayh 0/bracketleftbigg K(x, y)−φ1(x)φ1(y) λ1/bracketrightbigg φ2(y)dy =λ2/integraldisplayh 0/bracketleftbigg1 λ1φ1(y)−1 λ1φ1(y)/bracketrightbigg φ2(y)dy=0, i.e., /integraldisplayh 0φ1(x)φ2(x)dx=0. (5.2.36) From Eq. (5.2.35), we then have λ2/integraldisplayh 0K(x, y)φ2(y)dy=φ2(x). (5.2.37) Once we ﬁnd φ2(x), we construct a new kernel ˜K(x, y)by ˜K(x, y)≡¯K(x, y)−φ2(x)φ2(y) λ2=K(x, y)−2/summationdisplay n=1φn(x)φn(y) λn. (5.2.38) We then repeat the argument for ˜K(x, y). Ultimately either we ﬁnd after Nsteps, K(x, y)=N/summationdisplay n=1φn(x)φn(y) λn, (5.2.39) or we ﬁnd the inﬁnite series, K(x, y)≈∞/summationdisplay n=1φn(x)φn(y) λn, (5.2.40) and can show that the remainder R(x, y), which is deﬁned by R(x, y)≡K(x, y)−∞/summationdisplay n=1φn(x)φn(y) λn, (5.2.41) 118 5 Hilbert–Schmidt Theory of Symmetric Kernel cannot have any eigenfunction .I fψ(x)is the eigenfunction of R(x, y), λ0/integraldisplayh 0R(x, y)ψ(y)dy=ψ(x), (5.2.42) we know that (1)ψ(x)is distinct from all {φn(x)}n, ψ(x)/negationslash=φn(x) forn=1,2,···, (5.2.43) and that (2)ψ(x)is orthogonal to all {φn(x)}n, /integraldisplayh 0ψ(x)φn(x)dx=0 forn=1,2,···. (5.2.44) Then, substituting the deﬁnition (5.2.41) of R(x, y)into Eq. (5.2.42) and noting the orthogo- nality (5.2.44), we ﬁnd λ0/integraldisplayh 0K(x, y)ψ(y)dy=ψ(x), (5.2.45) which is a contradiction of Eq. (5.2.43). Thus we must have /bardblR/bardbl2=/integraldisplayh 0/integraldisplayh 0R2(x, y)dxdy=0, (5.2.46) which is the meaning of the ≈in Eq. (5.2.40). Therefore, the formula (5.2.15) holds in the sense of the mean square convergence . Proof of Lemma : S ow eo n l yh a v et op r o v et h e Lemma stated with the condition (5.2.30) and the proof of the Hilbert–Schmidt Theorem will be complete. To do so, it is necessary to work with the iterated kernel K2(x, y), which is also symmetric. K2(x, y)=/integraldisplayh 0K(x, z)K(z,y)dz=/integraldisplayh 0K(x, z)K(y,z)dz. (5.2.47) This is because the trace of K2(x, y)is always positive. /integraldisplayh 0K2(x, x)dx=/integraldisplayh 0dx/integraldisplayh 0dz K2(x, z)=/bardblK/bardbl2>0. (5.2.48) First, we will prove that if K2(x, y)has an eigenvalue, then K(x, y)has at least one eigenvalue equalling one of the square-roots of the former. 5.2 Real and Symmetric Kernel 119 Recall the deﬁnition of the resolvent kernel of K(x, y), H(x, y;λ)=−K(x, y)−λK 2(x, y)−···− λjKj+1(x, y)−··· , (5.2.49) H(x, y;−λ)=−K(x, y)+λK 2(x, y)−···− (−λ)jKj+1(x, y)−··· . (5.2.50) Taking the difference of Eqs. (5.2.49) and (5.2.50), we ﬁnd 1 2[H(x, y;λ)−H(x, y;−λ)] =−λ/bracketleftbig K2(x, y)+λ2K4(x, y)+λ4K6(x, y)+···/bracketrightbig =λH 2(x, y;λ2),(5.2.51) which is the resolvent for K2(x, y). The equality (5.2.51), which is valid for sufﬁciently small λwhere the series expansion in λis deﬁned, holds for all λby analytic continuation. Ifcis an eigenvalue of K2(x, y),H2(x, y;λ2)has a pole at λ2=c. From Eq. (5.2.51), either H(x, y;λ)orH(x, y;−λ)must have a pole at λ=±√ c. This means that at least one of±√ cis an eigenvalue of K(x, y). Now we prove that K2(x, y)has at least one eigenvalue. We have /integraldisplayh 0D2(x, x;s)dx/D 2(s)=/integraldisplayh 0H2(x, x;s)dx =−(A2+sA4+s2A6+···)(5.2.52) where Am=/integraldisplayh 0Km(x, x)dx, m =2,3,···. (5.2.53) IfK2(x, y)has no eigenvalues, then D2(s)has no zeros, and the series (5.2.52) must be convergent for all values of s. To this end, consider Am+n=/integraldisplayh 0dx K m+n(x, x)=/integraldisplayh 0dx/integraldisplayh 0dz K m(x, z)Kn(z,x) =/integraldisplayh 0dx/integraldisplayh 0dz K m(x, z)Kn(x, z).(5.2.54) Applying the Schwarz inequality, A2 m+n≤/bracketleftBigg/integraldisplayh 0dx/integraldisplayh 0dz K2 m(x, z)/bracketrightBigg/bracketleftBigg/integraldisplayh 0dx/integraldisplayh 0dz K2 n(x, z)/bracketrightBigg =/bracketleftBigg/integraldisplayh 0dx K 2m(x, x)/bracketrightBigg/bracketleftBigg/integraldisplayh 0dx K 2n(x, x)/bracketrightBigg =A2mA2n, 120 5 Hilbert–Schmidt Theory of Symmetric Kernel i.e., we have A2 m+n≤A2mA2n. (5.2.55) Setting /braceleftBigg m→n−1, n→n+1, in the inequality (5.2.55), we have A2 2n≤A2n−2A2n+2. (5.2.56) Recalling that A2m>0, (5.2.57) which is precisely the reason we consider K2(x, y),w eh a v e A2n A2n−2≤A2n+2 A2n. (5.2.58) Successively, we have A2n+2 A2n≥A2n A2n−2≥A2n−2 A2n−4≥···≥A4 A2≡R1, (5.2.59) so that A4=R1A2, A6≥R2 1A2, A8≥R3 1A2, and generally A2n≥Rn−1 1A2. (5.2.60) Thus we have A2+sA4+s2A6+s3A8+···≥ A2(1 +sR1+s2R2 1+s3R3 1+···). (5.2.61) The right-hand side of the inequality (5.2.61) diverges for those ssuch that s≥1 R1=A2 A4. (5.2.62) Thus /integraldisplayh 0H2(x, x;s)dx 5.2 Real and Symmetric Kernel 121 is divergent for those ssatisfying the inequality (5.2.62). Then K2(x, y)has the eigenvalue s satisfying s≤A2 A4, andK(x, y)has the eigenvalue λ1satisfying |λ1|≤/radicalbigg A2 A4. (5.2.63) This completes the proof of the Lemma and completes the proof of the Hilbert–Schmidt Theorem. The Hilbert–Schmidt expansion, (5.2.15), can be helpful in many problems where a sym- metric and square-integrable kernel is involved. ❑Example 5.1. Solve the integro-differential equation, ∂ ∂tφ(x, t)=/integraldisplayh 0K(x, y)φ(y,t)dy, (5.2.64a) with the initial condition, φ(x,0) =f(x), (5.2.64b) where K(x, y)is symmetric and square-integrable . Solution. The Hilbert–Schmidt expansion, (5.2.15), can be applied giving ∂ ∂tφ(x, t)=/summationdisplay nφn(x) λn/integraldisplayh 0φn(y)φ(y,t)dy. (5.2.65) Deﬁning An(t)by An(t)≡/integraldisplayh 0φn(y)φ(y,t)dy, (5.2.66) and changing the dummy index of summation from ntomin Eq. (5.2.65), we then have ∂ ∂tφ(x, t)=/summationdisplay mφm(x) λmAm(t). (5.2.67) Taking the time derivative of Eq. (5.2.66) yields d dtAn(t)=/integraldisplayh 0dyφn(y)∂ ∂tφ(y,t). (5.2.68) 122 5 Hilbert–Schmidt Theory of Symmetric Kernel Substituting Eq. (5.2.67) into Eq. (5.2.68) and, observing that the orthogonality of {φm(x)}m means that only the m=nterm is left, we get d dtAn(t)=An(t) λn. (5.2.69) The solution to Eq. (5.2.69) is then An(t)=An(0) exp/bracketleftbiggt λn/bracketrightbigg , (5.2.70) with An(0) =/integraldisplayh 0dx φ n(x)f(x). (5.2.71) We can now integrate Eq. (5.2.67) from 0tot, with An(t)from Eq. (5.2.70), /integraldisplayt 0dt∂ ∂tφ(x, t)=/integraldisplayt 0dt/summationdisplay nφn(x) λnAn(0) exp/bracketleftbiggt λn/bracketrightbigg . (5.2.72) The left-hand side of Eq. (5.2.72) is now exact, and we obtain φ(x, t)−φ(x,0) =/summationdisplay nφn(x) λnAn(0)/parenleftBig exp/bracketleftBig t λn/bracketrightBig −1/parenrightBig /parenleftBig 1 λn/parenrightBig . (5.2.73) From the initial condition (5.2.64b) and Eq. (5.2.73), we get ﬁnally φ(x, t)=f(x)+/summationdisplay nAn(0)/parenleftbigg exp/bracketleftbiggt λn/bracketrightbigg −1/parenrightbigg φn(x). (5.2.74) Ast→∞ , the asymptotic form of φ(x, t)is given either by φ(x, t)=f(x)−/summationdisplay nAn(0)φn(x) if all λn<0, (5.2.75) or by φ(x, t)=Ai(0)φi(x)e x p/bracketleftbiggt λi/bracketrightbigg if0<λi<all other λn. (5.2.76) 5.3 Bounds on the Eigenvalues In the process of proving our Lemma in the previous section, we managed to obtain the upper bound on the lowest eigenvalue, |λ1|≤/radicalbigg A2 A4. 5.3 Bounds on the Eigenvalues 123 A better upper bound can be obtained as follows. If we call R2=A6 A4, we note that R2≥R1, or 1 R2≤1 R1. Furthermore, we ﬁnd A2+sA4+s2A6+s3A8+s4A10+··· =A2+sA4/bracketleftbigg 1+s/parenleftbiggA6 A4/parenrightbigg +s2/parenleftbiggA8 A4/parenrightbigg +···/bracketrightbigg ≥A2+sA4[1 +sR2+s2R2 2+···] which diverges if sR2>1. Hence we have a singularity for s≤1 R2≤1 R1. We therefore have an improved upper bound on λ1, |λ1|≤/radicalbigg A4 A6≤/radicalbigg A2 A4. So, we have the successively better upper bounds on |λ1|, /radicalbigg A2 A4,/radicalbigg A4 A6,/radicalbigg A6 A8,···, for the lowest eigenvalue, each better than the previous one, i.e., |λ1|≤···≤/radicalbigg A6 A8≤/radicalbigg A4 A6≤/radicalbigg A2 A4. (5.3.1a) Recall also that with a symmetric kernel, we have A2m=/bardblKm/bardbl2. (5.3.2) 124 5 Hilbert–Schmidt Theory of Symmetric Kernel The upper bounds (5.3.1a), in terms of the norm of the iterated kernels, become |λ1|≤···≤/bardblK3/bardbl /bardblK4/bardbl≤/bardblK2/bardbl /bardblK3/bardbl≤/bardblK/bardbl /bardblK2/bardbl. (5.3.1b) Now consider the question of ﬁnding the lower bounds for the lowest eigenvalue λ1. Con- sider the expansion of the symmetric kernel, K(x, y)≈/summationdisplay nφn(x)φn(y) λn. (5.3.3) This expression (5.3.3) is an equation in the mean , and hence there is no guarantee that it is true at any point as an exact equation. In particular, on the line y=xwhich has zero measure in the square 0≤x, y≤h, it need not be true. The following equality K(x, x)=/summationdisplay nφn(x)φn(x) λn therefore need not be true. Hence /integraldisplayh 0K(x, x)dx=/summationdisplay n1 λn(5.3.4) need not be true. The right-hand side of Eq. (5.3.4) may not converge. However, for K2(x, y)=/integraldisplayh 0K(x, z)K(z,y)dz=/summationdisplay nφn(x)φn(y) λ2n, since we know K(x, y)to be square-integrable , A2=/integraldisplayh 0K2(x, x)dx=/summationdisplay n1 λ2n(5.3.5) must converge, and, in general, for m≥2,w eh a v e Am=/summationdisplay n1 λmn,m =2,3,···, (5.3.6) and we know that the right-hand side of Eq. (5.3.6) converges. Consider now the expansion for A2, namely A2=1 λ2 1+1 λ2 2+1 λ2 3+···=1 λ2 1/bracketleftBigg 1+/parenleftbiggλ1 λ2/parenrightbigg2 +/parenleftbiggλ1 λ3/parenrightbigg2 +···/bracketrightBigg ≥1 λ2 1, i.e., λ2 1≥1 A2. (5.3.7) 5.3 Bounds on the Eigenvalues 125 Hence we have a lower bound for the eigenvalue λ1, |λ1|≥1 √ A2, or|λ1|≥1 /bardblK/bardbl. (5.3.8) This is consistent with our early discussion of the series solution to the Fredholm integral equation of the second kind for which we concluded that when |λ|<1 /bardblK/bardbl, (5.3.9) there are no singularities in λ, so that the ﬁrst eigenvalue λ=λ1must satisfy the inequality (5.3.8). We can obtain better lower bounds for the eigenvalue λ1. Consider A4, A4=1 λ4 1+1 λ4 2+1 λ4 3+···=1 λ4 1/bracketleftBigg 1+/parenleftbiggλ1 λ2/parenrightbigg4 +/parenleftbiggλ1 λ3/parenrightbigg4 +···/bracketrightBigg ≥1 λ4 1, i.e., |λ1|≥1 (A4)1/4. (5.3.10) This is an improvement over the previously established lower bound since we know from Eqs. (5.3.1a) and (5.3.8) that 1 A1/2 2≤|λ1|≤/parenleftbiggA2 A4/parenrightbigg1/2 , so that A4≤A2 2, i.e., 1 (A4)1/4≥1 (A2)1/2. Thus1/(A4)1/4is a better lower bound than 1/(A2)1/2. Proceeding in the same way with A6,A8,··· , we get increasingly better lower bounds, 1 (A2)1/2≤1 (A4)1/4≤1 (A6)1/6≤···≤| λ1|. (5.3.11) Putting both the upper bounds (5.3.1a) and lower bounds (5.3.11) together, we have for the smallest eigenvalue λ1, 126 5 Hilbert–Schmidt Theory of Symmetric Kernel 1 (A2)1/2≤1 (A4)1/4≤1 (A6)1/6≤··· ···≤| λ1|≤···≤/parenleftbiggA6 A8/parenrightbigg1/2 ≤/parenleftbiggA4 A6/parenrightbigg1/2 ≤/parenleftbiggA2 A4/parenrightbigg1/2 .(5.3.12) Strength permitting, we calculate A6,A8,···, to obtain successively better upper bounds and lower bounds from Eq. (5.3.12). 5.4 Rayleigh Quotient Another useful technique for ﬁnding the upper bounds for eigenvalues of self-adjoint operators is based on the Rayleigh quotient . Consider the self-adjoint integral operator, ˜K=/integraldisplayh 0dy K(x, y) with K(x, y)=K(y,x), (5.4.1) with eigenvalues λnand eigenfunctions φn(x),   ˜Kφn=1 λnφn, (φn,φm)=δnm,(5.4.2) where the eigenvalues are ordered such that |λ1|≤|λ2|≤|λ3|≤··· . Consider any given function g(x)such that /bardblg/bardbl/negationslash=0. (5.4.3) Consider the series /summationdisplay nbnφn(x) with bn=(φn,g). (5.4.4) This series expansion is the projection of g(x)onto the space spanned by the set {φn(x)}n, which may not be complete. We can easily verify the Bessel inequality , which says /summationdisplay nb2 n≤/bardblg/bardbl2. (5.4.5) Proof of the Bessel inequality : /vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoubleg−/summationdisplay nbnφn/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble2 ≥0, 5.4 Rayleigh Quotient 127 which implies (g,g)−/summationdisplay mbm(g,φm)−/summationdisplay nbn(φn,g)+/summationdisplay n/summationdisplay mbnbm(φn,φm)≥0. Thus we have (g,g)−/summationdisplay mb2 m≥0, which states /summationdisplay nb2 n≤/bardblg/bardbl2, completing the proof of the Bessel inequality (5.4.5). Now, consider the quadratic form (g,˜Kg), (g,˜Kg)=/integraldisplayh 0dx g(x)˜Kg(x)=/integraldisplayh 0dx/integraldisplayh 0dy g(x)K(x, y)g(y). (5.4.6) Substituting the expansion K(x, y)≈/summationdisplay nφn(x)φn(y) λn into Eq. (5.4.6), we obtain (g,˜Kg)=/integraldisplayh 0dx/integraldisplayh 0dy g(x)/summationdisplay n/parenleftbiggφn(x)φn(y) λn/parenrightbigg g(y)=/summationdisplay nb2 n λn. (5.4.7) Then, taking the absolute value of the above quadratic form (5.4.7), we obtain /vextendsingle/vextendsingle/vextendsingle(g,˜Kg)/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/summationdisplay nb2 n λn/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤/braceleftbiggb2 1 |λ1|+b2 2 |λ2|+b2 3 |λ3|+···/bracerightbigg =1 |λ1|/braceleftbigg b2 1+/vextendsingle/vextendsingle/vextendsingle/vextendsingleλ1 λ2/vextendsingle/vextendsingle/vextendsingle/vextendsingleb2 2+/vextendsingle/vextendsingle/vextendsingle/vextendsingleλ1 λ3/vextendsingle/vextendsingle/vextendsingle/vextendsingleb2 3+···/bracerightbigg ≤1 |λ1|/braceleftbig b2 1+b2 2+b2 3+···/bracerightbig ≤1 |λ1|/bardblg/bardbl2.(5.4.8) Hence, from Eq. (5.4.8), the Rayleigh quotient Q,d e ﬁ n e db y Q≡(g,˜Kg) (g,g), (5.4.9) 128 5 Hilbert–Schmidt Theory of Symmetric Kernel is bounded above by |Q|=/vextendsingle/vextendsingle/vextendsingle(g,˜Kg)/vextendsingle/vextendsingle/vextendsingle /bardblg/bardbl2≤1 |λ1|, i.e., the absolute value of the lowest eigenvalue λ1is bounded above by |λ1|≤1 |Q|. (5.4.10) To ﬁnd a good upper bound on |λ1|, choose a trial function g(x)with adjustable parameters and obtain the minimum of1 |Q|. Namely, we have |λ1|≤min/parenleftbigg1 |Q|/parenrightbigg (5.4.11) withQg i v e nb yE q .( 5 . 4 . 9 ) . ❑Example 5.2. Find an upper bound on the leading eigenvalue of the symmetric kernel K(x, y)=/braceleftBigg (1−x)y, 0≤y<x ≤1, (1−y)x, 0≤x<y ≤1, using the Rayleigh quotient. Solution. Consider the trial function g(x)=ax which is probably not very good. We have (g,g)=/integraldisplay1 0dx a2x2=a2 3, and (g,˜Kg)=/integraldisplay1 0dx/integraldisplay1 0dy g(x)K(x, y)g(y) =/integraldisplay1 0dx ax/bracketleftbigg/integraldisplayx 0dy(1−x)yay+/integraldisplay1 xdy(1−y)xay/bracketrightbigg =a2/integraldisplay1 0dx x2(1−x2)/6=a2 30. So, the Rayleigh quotient Qis given by Q=(g,˜Kg) (g,g)=1 10, and we get min/parenleftbigg1 |Q|/parenrightbigg =1 0. 5.5 Completeness of Sturm–Liouville Eigenfunctions 129 Thus, from Eq. (5.4.11), we obtain |λ1|≤10, which is a reasonable upper bound. The exact value of λ1turns out to be λ1=π2≈9.8696, so it is not too bad, considering that the eigenfunction for λ1turns out to be Asin(πx), not well approximated by ax. 5.5 Completeness of Sturm–Liouville Eigenfunctions Consider the Sturm–Liouville eigenvalue problem , d dx/bracketleftbigg p(x)d dxφ(x)/bracketrightbigg −q(x)φ(x)=λr(x)φ(x) on[0,h], (5.5.1) with φ(0) = φ(h)=0, and p(x)>0,r(x)>0, forx∈[0,h]. We proved earlier that, using the Green’s function G(x, y)deﬁned by d dx/bracketleftbigg p(x)d dxG(x, y)/bracketrightbigg −q(x)G(x, y)=δ(x−y), (5.5.2) with G(0,y)=0,G(h,y)=0, Eq. (5.5.1) is equivalent to the integral equation φ(x)=λ/integraldisplayh 0G(y,x)r(y)φ(y)dy. (5.5.3) Further, we have shown that, since the Sturm–Liouville operator is self-adjoint and symmetric, we have a symmetric Green’s function, G(x, y)=G(y,x). (5.5.4) Now, deﬁne ψ(x)=/radicalbig r(x)φ(x), (5.5.5a) 130 5 Hilbert–Schmidt Theory of Symmetric Kernel and K(x, y)=/radicalbig r(x)G(x, y)/radicalbig r(y). (5.5.5b) Thenψ(x)satisﬁes ψ(x)=λ/integraldisplayh 0K(x, y)ψ(y)dy, (5.5.6) which has a symmetric kernel . Applying the usual Hilbert–Schmidt theorem, we know that K(x, y)deﬁned above is decomposable in the form, K(x, y)≈/summationdisplay nψn(x)ψn(y) λn=/summationdisplay n/radicalbig r(x)r(y)φn(x)φn(y) λn, (5.5.7) withλnreal and discrete ,a n dt h es e t {ψn(x)}northonormal , i.e., /integraldisplayh 0ψn(x)ψm(x)dx=/integraldisplayh 0r(x)φn(x)φm(x)dx=δnm. (5.5.8) Note the appearance of the weight function r(x)in the middle equation of Eq. (5.5.8). To prove the completeness , we will establish that any function f(x)can be expanded in a series of {ψn(x)}nor{φn(x)}n. Let us do this for the differentiable case (which is stronger than square-integrable ), i.e., assume f(x)is differentiable. Therefore, given any f(x),w e can deﬁne g(x)by g(x)≡Lf(x), (5.5.9) where L=d dx/bracketleftbigg p(x)d dx/bracketrightbigg −q(x). (5.5.10) Take the inner product of both sides of Eq. (5.5.9) with G(x, y)to get (G(x, y),g(x)) =f(y), (5.5.11a) i.e., f(x)=/integraldisplayh 0G(x, y)g(y)dy=/integraldisplayh 0/parenleftBigg K(x, y) /radicalbig r(x)r(y)/parenrightBigg g(y)dy. (5.5.11b) Substituting the expression (5.5.7) for K(x, y)into Eq. (5.5.11b), we obtain f(x)=/integraldisplayh 0dy/summationdisplay n/parenleftbiggφn(x)φn(y) λn/parenrightbigg g(y)=/summationdisplay n/parenleftbiggφn(x) λn/parenrightbigg (φn,g) =/summationdisplay n/parenleftbiggβn λn/parenrightbigg φn(x),(5.5.12) 5.6 Generalization of Hilbert–Schmidt Theory 131 where we set βn≡(φn,g)=/integraldisplayh 0φn(y)g(y)dy. (5.5.13) Since the above expansion of f(x)in terms of φn(x)is true for any f(x), this demonstrates that the set {φn(x)}niscomplete . Actually, in addition, we must require that f(x)satisﬁes the homogeneous boundary con- ditions in order to avoid boundary terms. Also, we must make sure that the kernel for the Sturm–Liouville eigenvalue problem is square-integrable . Since the set {φn(x)}nis com- plete, we conclude that there must be an inﬁnite number of eigenvalues to the Sturm–Liouvillesystem. Also, it is possible to prove the asymptotic results, λ n=O(n2) asn→∞. 5.6 Generalization of Hilbert–Schmidt Theory In this section, we consider the generalization of Hilbert–Schmidt theory in ﬁve directions. Direction 1: So far in our discussion of Hilbert–Schmidt theory, we assumed that K(x, y)is real. It is straightforward to extend to the case when K(x, y)iscomplex . We deﬁne the norm /bardblK/bardblof the kernel K(x, y)by /bardblK/bardbl2=/integraldisplayh 0dx/integraldisplayh 0dy|K(x, y)|2. (5.6.1) The iteration series solution to the Fredholm integral equation of the second kind again con- verges for |λ|<1 /bardblK/bardbl. (5.6.2) Also, the Fredholm theory still remains valid. If K(x, y)is, in addition, self-adjoint , i.e., K(x, y)=K∗(y,x), (5.6.3) then the Hilbert–Schmidt expansion holds in the form, K(x, y)≈/summationdisplay nφn(x)φ∗ n(y) λn, (5.6.4) where /integraldisplayh 0φ∗ n(x)φm(x)dx=δnm, (5.6.5) and λn=real, and nis an integer . (5.6.6) 132 5 Hilbert–Schmidt Theory of Symmetric Kernel Direction 2: Next we note that, in all the discussion so far, the variable xis restricted to a ﬁnite basic interval, x∈[0,h]. (5.6.7) As the second generalization, we extend the basic interval [0,h]to[0,∞).W e w a n t t o s o l v e the following integral equation, φ(x)=f(x)+λ/integraldisplay+∞ 0K(x, y)φ(y)dy, (5.6.8) with /integraldisplay+∞ 0dx/integraldisplay+∞ 0dy K2(x, y)<∞,/integraldisplay+∞ 0dx f2(x)<∞. (5.6.9) By a change of the independent variable x, it is always possible to transform the interval [0,∞)ofxinto[0,h]oft, i.e., x∈[0,∞)⇒t∈[0,h]. (5.6.10) For example, the following transformation will do, x=g(t)=t (h−t). (5.6.11) Then, writing ˜φ(t)=φ(g(t)), etc., we have ˜φ(t)=˜f(t)+λ/integraldisplayh 0˜K(t, t/prime)˜φ(t/prime)g/prime(t/prime)dt/prime. By multiplying/radicalbig g/prime(t)on both sides of the equation above, we have /radicalbig g/prime(t)˜φ(t)=/radicalbig g/prime(t)˜f(t)+λ/integraldisplayh 0/radicalbig g/prime(t)˜K(t, t/prime)/radicalbig g/prime(t/prime)/radicalbig g/prime(t/prime)˜φ(t/prime)dt/prime. Deﬁning ψ(t)by ψ(t)=/radicalbig g/prime(t)˜φ(t), we obtain ψ(t)=/radicalbig g/prime(t)˜f(t)+λ/integraldisplayh 0[/radicalbig g/prime(t)˜K(t, t/prime)/radicalbig g/prime(t/prime)]ψ(t/prime)dt/prime. (5.6.12) 5.6 Generalization of Hilbert–Schmidt Theory 133 If the original kernel K(x, y)is symmetric, then the transformed kernel is also symmetric. Furthermore, the transformed kernel/radicalbig g/prime(t)˜K(t, t/prime)/radicalbig g/prime(t/prime)and the transformed inhomoge- neous term/radicalbig g/prime(t)˜f(t)are square-integrable ifK(x, y)andf(x)are square-integrable ,s i n c e /integraldisplayh 0dt/integraldisplayh 0dt/primeg/prime(t)˜K2(t, t/prime)g/prime(t/prime)=/integraldisplay+∞ 0dx/integraldisplay+∞ 0dy K2(x, y)<∞, (5.6.13a) and /integraldisplayh 0dtg/prime(t)˜f2(t)=/integraldisplay+∞ 0dx f2(x)<∞. (5.6.13b) Thus, under appropriate conditions, the Fredholm theory and the Hilbert–Schmidt theory both apply to Eq. (5.6.8). Similarly we can extend these theories to the case of inﬁnite range. Direction 3: As the third generalization, we consider the case where we have multi- dimensional independent variables . φ(/vectorx)=f(/vectorx)+λ/integraldisplay+∞ 0K(/vectorx,/vectory)φ(/vectory)d/vectory. (5.6.14) As long as the kernel K(/vectorx,/vectory)issquare-integrable , i.e., /integraldisplay+∞ 0/integraldisplay+∞ 0K2(/vectorx,/vectory)d/vectorxd/vectory<∞, (5.6.15) all the arguments for establishing the Fredholm theory and the Hilbert–Schmidt theory apply. Direction 4: As the fourth generalization of the theorem, we will relax the condition on the square-integrability of the kernel .W h e nak e r n e l K(x, y)is not square-integrable , the integral equation is said to be singular . Some singular integral equations can be transformed into one with a square-integrable kernel. One method which may work is to try to symmetrize them as much as we can. For example, a kernel of the form H(x, y)withH(x, y)bounded can be made square-integrable by symmetrizing it into H(x, y)/(xy)1 4. Another way is to iterate the kernel . Suppose the kernel is of the form, K(x, y)=H(x, y) |x−y|α,1 2≤α<1, (5.6.16) where H(x, y)is bounded. We have the integral equation of the form, φ(x)=f(x)+λ/integraldisplayh 0K(x, y)φ(y)dy. (5.6.17) Replacing φ(y)in the integrand with the right-hand side of Eq. (5.6.17) itself, we obtain φ(x)=/bracketleftBigg f(x)+λ/integraldisplayh 0K(x, y)f(y)dy/bracketrightBigg +λ2/integraldisplayh 0K2(x, y)φ(y)dy. (5.6.18) 134 5 Hilbert–Schmidt Theory of Symmetric Kernel The kernel in Eq. (5.6.18) is K2(x, y),w h i c hm a yb e square-integrable since /integraldisplayh 01 |x−z|α1 |z−y|αdz=O/parenleftBigg 1 |x−y|2α−1/parenrightBigg . (5.6.19) Indeed, for those αsuch that 1 2≤α<3 4, K2(x, y)issquare-integrable .I fαis such that 3 4≤α<1, thenK3(x, y),K4(x, y),··· ,e t c . m a yb e square-integrable . In general, when αlies in the interval 1−1 2(n−1)≤α<1−1 2n, (5.6.20) Kn(x, y)will be square-integrable . Thus, for those αsuch that 1 2≤α<1, (5.6.21) we can transform the kernel into a square-integrable kernel by the appropriate number of iterations. However, when α≥1, we have no hope whatsoever of transforming it into a square-integrable kernel in this way. For a kernel which cannot be made square-integrable , which properties remain valid? Does the Fredholm theory hold? Is the spectrum of the eigenvalues discrete? Does the Hilbert– Schmidt expansion hold? The following example gives us some insight into these questions. ❑Example 5.3. Suppose we want to solve the homogeneous equation φ(x)=λ/integraldisplay+∞ 0e−|x−y|φ(y)dy. (5.6.22) The kernel in the above equation is symmetric, but not square-integrable ; even so, this equation can be solved in the closed form. Solution. Writing out Eq. (5.6.22) explicitly, we have φ(x)=λ/integraldisplayx 0e−(x−y)φ(y)dy+λ/integraldisplay+∞ xe−(y−x)φ(y)dy. Multiplying both sides by e−x,w eh a v e e−xφ(x)=λe−2x/integraldisplayx 0eyφ(y)dy+λ/integraldisplay+∞ xe−yφ(y)dy. (5.6.23) 5.6 Generalization of Hilbert–Schmidt Theory 135 Differentiating the above equation with respect to x, we obtain e−x(−φ(x)+φ/prime(x)) =−2λe−2x/integraldisplayx 0eyφ(y)dy. Multiplying both sides by e2x,w eh a v e ex(−φ(x)+φ/prime(x)) =−2λ/integraldisplayx 0eyφ(y)dy. (5.6.24) Differentiating the above equation with respect to xand cancelling the factor ex, we obtain φ/prime/prime(x)+( 2 λ−1)φ(x)=0. The solution to the above equation is given by (i)1−2λ>0, φ(x)=C1e√ 1−2λx+C2e−√ 1−2λx, and, (ii)1−2λ<0, φ(x)=C/prime 1ei√ 2λ−1x+C/prime 2e−i√ 2λ−1x. These solutions satisfy the equation (5.6.22) only if φ/prime(0) = φ(0), which follows from the once-differentiated equation (5.6.24). Thus we have √ 1−2λ(C1−C2)=C1+C2. In order to satisfy Eq. (5.6.23), we must require that the integral /integraldisplay+∞ xe−yφ(y)dy (5.6.25) converges. If1 2>λ> 0,φ(x)grows exponentially but the integral in Eq. (5.6.25) converges. Ifλ>1 2,φ(x)oscillates and the integral in Eq. (5.6.25) converges. If λ<0,h o w e v e r ,t h e integral in Eq. (5.6.25) diverges and no solution exists. In summary, a solution exists for λ>0, but no solution exists for λ<0. We note that: 1. the spectrum is not discrete; 2. the eigenfunctions for λ>1 2alone constitute a complete set (very much like a Fourier sine or cosine expansion). Thus not all of the eigenfunctions are necessary to represent anL2function. 136 5 Hilbert–Schmidt Theory of Symmetric Kernel Direction 5: As the last generalization of the theorem, we shall retain the square-integrability of the kernel , but consider the case of the non-symmetric kernel . Since this generalization is not always possible, we shall illustrate the point by presenting one example. ❑Example 5.4. We consider the following Fredholm integral equation of the second kind, φ(x)=f(x)+λ/integraldisplay1 0K(x, y)φ(y)dy, (5.6.26) where the kernel is non-symmetric, K(x, y)=/braceleftBigg 2,0≤y<x ≤1, 1,0≤x<y ≤1,(5.6.27) but is square-integrable . Solution. We ﬁrst consider the homogeneous equation . φ(x)=λ/integraldisplayx 02φ(y)dy+λ/integraldisplay1 xφ(y)dy=λ/integraldisplay1 0φ(y)dy+λ/integraldisplayx 0φ(y)dy. (5.6.28) Differentiating Eq. (5.6.28) with respect to x, we obtain φ/prime(x)=λφ(x). (5.6.29) From this, we obtain φ(x)=Cexp[λx],0≤x≤1,C/negationslash=0. (5.6.30) From Eq. (5.6.28), we get the boundary conditions,   φ(0) = λ/integraldisplay1 0φ(y)dy, φ(1) = 2 λ/integraldisplay1 0φ(y)dy,⇒φ(1) = 2 φ(0). (5.6.31) Hence, we require that Cexp[λ]=2C⇒exp[λ] = 2 = exp[ln(2) + i2nπ],n integer . Thus we should have the eigenvalues, λ=λn= ln(2) + i2nπ, n =0,±1,±2,···. (5.6.32) The corresponding eigenfunctions are φn(x)=Cnexp[λnx],C nreal. Finally, φn(x)=Cnexp[{ln(2) + i2nπ}x],n integer . (5.6.33) 5.6 Generalization of Hilbert–Schmidt Theory 137 Clearly, the kernel is non-symmetric, K(x, y)/negationslash=K(y,x). The transposed kernel KT(x, y)is given by KT(x, y)=K(y,x)=/braceleftBigg 2,0≤x<y ≤1, 1,0≤y<x ≤1.(5.6.34) We next consider the homogeneous equation for the transposed kernel KT(x, y). ψ(x)=λ/integraldisplayx 0ψ(y)dy+2λ/integraldisplay1 xψ(y)dy=2λ/integraldisplay1 0ψ(y)dy−λ/integraldisplayx 0ψ(y)dy. (5.6.35) Differentiating Eq. (5.6.35) with respect to x, we obtain ψ/prime(x)=−λψ(x). (5.6.36) From this, we obtain ψ(x)=Fexp[−λx],0≤x≤1,F/negationslash=0. (5.6.37) From Eq. (5.6.35), we get the boundary conditions,   ψ(0) = 2 λ/integraldisplay1 0ψ(y)dy, ψ(1) = λ/integraldisplay1 0ψ(y)dy,⇒ψ(1) =1 2ψ(0). (5.6.38) Hence, we require that Fexp[−λ]=1 2F⇒exp[λ] = 2 = exp[ln(2) + i2nπ],n integer . Thus we should have the same eigenvalues, λ=λn= ln(2) + i2nπ, n =0,±1,±2,···. (5.6.39) The corresponding eigenfunctions are ψn(x)=Fnexp[−λnx],F nreal. Finally, ψn(x)=Fnexp[−{ln(2) + i2nπ}x],n integer . (5.6.40) These ψn(x)are the solution to the transposed problem. For n/negationslash=m,w eh a v e /integraldisplay1 0φn(x)ψm(x)dx=CnFm/integraldisplay1 0exp[i2π(n−m)x]dx=0,n/negationslash=m. 138 5 Hilbert–Schmidt Theory of Symmetric Kernel The spectral representation of the kernel K(x, y)is given by K(x, y)=∞/summationdisplay n=−∞φn(x)ψn(y) λn=∞/summationdisplay n=−∞exp/bracketleftbig {ln(2) + i2nπ}(x−y)/bracketrightbig ln(2) + i2nπ, (5.6.41) Cn=Fn=1, which we obtain from the following orthogonality, /integraldisplay1 0φn(x)ψm(x)dx=δnm. (5.6.42) In establishing Eq. (5.6.41), we ﬁrst write R(x, y)≡K(x, y)−∞/summationdisplay n=1φn(x)ψn(y) λn, (5.6.43) and demonstrate the fact that the remainder R(x, y)cannot have any eigenfunctions, by ex- hausting all of the possible eigenfunctions. By explicit solution, we already know that the kernel has at least one eigenvalue. Crucial to this generalization is the fact that the original integral equation and the trans- posed integral equation have the same eigenvalues and that the eigenfunctions of the trans- posed kernel are orthogonal to the eigenfunctions of the original kernel. This last generaliza- tion is not always possible for the general nonsymmetric kernel. 5.7 Generalization of Sturm–Liouville System In Section 5.5, we have shown that, if p(x)>0andr(x)>0, the eigenvalue equation d dx/bracketleftbigg p(x)d dxφ(x)/bracketrightbigg −q(x)φ(x)=λr(x)φ(x) where x∈[0,h], (5.7.1) with appropriate boundary conditions has the eigenfunctions which form a complete set {φn(x)}nbelonging to the discrete eigenvalues λn. In this section, we shall relax the condi- tions on p(x)andr(x). In particular, we shall consider the case in which p(x)has simple or double zeros at the end points, which therefore may be regular singular points of the differen-tial equation (5.7.1). LetL xbe a second-order differential operator, Lx≡a0(x)d2 dx2+a1(x)d dx+a2(x), where x∈[0,h], (5.7.2) which is, in general, non self-adjoint . As a matter of fact, we can always transform a second- order differential operator Lxinto a self-adjoint form by multiplying p(x)/a0(x)onLx, with p(x)≡exp/bracketleftbigg/integraldisplayxa1(y) a0(y)dy/bracketrightbigg . 5.7 Generalization of Sturm–Liouville System 139 However, it is instructive to see what happens when Lxis non self-adjoint. So, we shall not transform the differential operator Lx, (5.7.2), into a self-adjoint form. Let us assume that certain boundary conditions are given at x=0 andx=h. Consider the Green’s functions G(x, y)andG(x, y;λ)deﬁned by LxG(x, y)=δ(x−y), (Lx−λ)G(x, y;λ)=δ(x−y), G(x, y;λ=0 )= G(x, y).(5.7.3) We would like to ﬁnd a representation of G(x, y;λ)in a form similar to H(x, y;λ)given by Eq. (5.2.9). Symbolically, we write G(x, y;λ)as G(x, y;λ)=(Lx−λ)−1. (5.7.4) Since the deﬁning equation of G(x, y;λ)depends on λanalytically, we expect G(x, y;λ)to be an analytic function of λby the P oincaré theorem . There are two possible exceptions: (1) At a regular singular point, the indicial equation yields an exponent which, considered as afunction λ, may have branch cuts; (2) For some value of λ, it may be impossible to match the discontinuity at x=y. To elaborate on the second point (2), let φ 1andφ2be the solution of (Lx−λ)φi(x;λ)=0 ( i=1,2). (5.7.5) We can construct G(x, y;λ)to be G(x, y;λ)∝/braceleftBigg φ1(x;λ)φ2(y;λ),0≤x≤y, φ2(x;λ)φ1(y;λ),y < x ≤h,(5.7.6) where φ1(x;λ)satisﬁes the boundary condition at x=0 ,a n dφ2(x;λ)satisﬁes the boundary condition at x=h. The constant of proportionality of Eq. (5.7.6) is given by C W(φ1(y;λ),φ2(y;λ)). (5.7.7) When the Wronskian vanishes as a function of λ,G(x, y;λ)develops a singularity. It may be a pole or a branch point in λ. However, the vanishing of the Wronskian W(φ1(y;λ),φ2(y;λ)) implies that φ2(x;λ)is proportional to φ1(x;λ)for such λ; namely, we have an eigenfunction ofLx. Thus the singularities of G(x, y;λ)as a function of λare associated with the eigen- functions of Lx. Hence we shall treat G(x, y;λ)as an analytic function of λ, except at poles located at λ=λi(i=1 ,··· ,n,··· ) and at a branch point located at λ=λB, from which a branch cut is extended to −∞ . Assuming that G(x, y;λ)behaves as G(x, y;λ)=O/parenleftbigg1 λ/parenrightbigg as|λ|→∞ , (5.7.8a) we obtain lim R→∞1 2πi/contintegraldisplay CRG(x, y;λ/prime) λ/prime−λdλ/prime=0, (5.7.8b) 140 5 Hilbert–Schmidt Theory of Symmetric Kernel where CRis the circle of radius R, centered at the origin of the complex λplane. Invoking the Cauchy Residue Theorem, we have lim R→∞1 2πi/contintegraldisplay CRG(x, y;λ/prime) λ/prime−λdλ/prime=G(x, y;λ)+/summationdisplay nRn(x, y) λn−λ −1 2πi/integraldisplayλB −∞G(x, y;λ/prime+iε)−G(x, y;λ/prime−iε) λ/prime−λdλ/prime,(5.7.9) where Rn(x, y)= ResG(x, y;λ/prime)|λ/prime=λn. (5.7.10) Combining Eqs. (5.7.8b) and (5.7.9), we obtain G(x, y;λ)=−/summationdisplay nRn(x, y) λn−λ +1 2πi/integraldisplayλB −∞G(x, y;λ/prime+iε)−G(x, y;λ/prime−iε) λ/prime−λdλ/prime.(5.7.11) Let us concentrate on the ﬁrst term in Eq. (5.7.11). Multiplying the second equation in Eq. (5.7.3) by (λ−λn)and letting λ→λn, lim λ→λn(λ−λn)(Lx−λ)G(x, y;λ)= l i m λ→λn(λ−λn)δ(x−y), from which we obtain (Lx−λn)Rn(x, y)=0. (5.7.12) Thus we obtain Rn(x, y)∝ψn(y)φn(x), (5.7.13) where φn(x)is the eigenfunction of Lxbelonging to the eigenvalue λn(assuming that the eigenvalue is not degenerate), (Lx−λn)φn(x)=0. (5.7.14) We claim that ψn(x)is the eigenfunction of LT x, belonging to the same eigenvalue λn, (LT x−λn)ψn(x)=0, (5.7.15) where LT xis deﬁned by LT x≡d2 dx2a0(x)−d dxa1(x)+a2(x). (5.7.16) Suppose we want to solve the following equation, (LT x−λ)h(x)=f(x). (5.7.17) 5.7 Generalization of Sturm–Liouville System 141 We construct the following expression, /integraldisplayh 0dx G(x, y;λ)(LT x−λ)h(x)=/integraldisplayh 0dx G(x, y;λ)f(x), (5.7.18) and perform the integral by parts on the left-hand side of Eq. (5.7.18). We obtain, /integraldisplayh 0dx[(Lx−λ)G(x, y;λ)]h(x)=/integraldisplayh 0dx G(x, y;λ)f(x). (5.7.19) The expression in the square bracket on the left-hand side of Eq. (5.7.19) is δ(x−y),a n dw e have (after exchanging the variables xandy) h(x)=/integraldisplayh 0dy G(y,x;λ)f(y). (5.7.20) Operating (LT x−λ)on both sides of Eq. (5.7.20), recalling Eq. (5.7.17), we have f(x)=(LT x−λ)h(x)=/integraldisplayh 0dy(LT x−λ)G(y,x;λ)f(y). (5.7.21) This is true if and only if (LT x−λ)G(y,x;λ)=δ(x−y). (5.7.22) Multiplying (λ−λn)on both sides of Eq. (5.7.22), and letting λ→λn, lim λ→λn(λ−λn)(LT x−λ)G(y,x;λ) = lim λ→λn(λ−λn)δ(x−y), (5.7.23) from which, we obtain (LT x−λn)Rn(y,x)=0. (5.7.24) Since we know from Eq. (5.7.13), Rn(y,x)∝ψn(x)φn(y), (5.7.13b) equation (5.7.24) indeed demonstrates that ψn(x)is the eigenfunction of LT x, belonging to the eigenvalue λnas claimed in Eq. (5.7.15). Thus Rn(x, y)is a product of the eigenfunctions ofLxandLT x, i.e., a product of φn(x)andψn(y). Incidentally, this also proves that the eigenvalues of LT xare the same as the eigenvalues of Lx. Let us now analyze the second term in Eq. (5.7.11), 1 2πi/integraldisplayλB −∞G(x, y;λ/prime+iε)−G(x, y;λ/prime−iε) λ/prime−λdλ/prime. In the limit ε→0, we have, from Eq. (5.7.3), (Lx−λ)G(x, y;λ+iε)=δ(x−y), (5.7.25a) (Lx−λ)G(x, y;λ−iε)=δ(x−y). (5.7.25b) 142 5 Hilbert–Schmidt Theory of Symmetric Kernel Taking the difference of the two expressions above, we have (Lx−λ)[G(x, y;λ+iε)−G(x, y;λ−iε)] = 0 . (5.7.26) Thus we conclude G(x, y;λ+iε)−G(x, y;λ−iε)∝ψλ(y)φλ(x). (5.7.27) Hence we ﬁnally obtain, by choosing proper normalization for ψnandφn, G(x, y;λ)=+/summationdisplay nψn(y)φn(x) (λn−λ)+/integraldisplayλB −∞dλ/primeψλ/prime(y)φλ/prime(x) (λ/prime−λ). (5.7.28) The ﬁrst term in Eq. (5.7.28) represents a discrete contribution toG(x, y;λ)from the poles atλ=λn, while the second term represents a continuum contribution from the branch cut starting at λ=λBand extending to −∞ along the negative real axis. Equation (5.7.28) is the generalization of the formula, Eq. (5.2.9), for the resolvent kernel H(x, y;λ). Equa- tion (5.7.28) is consistent with the assumption (5.7.8a). Setting λ=0 in Eq. (5.7.28), we obtain G(x, y)=/summationdisplay nψn(y)φn(x) λn+/integraldisplayλB −∞dλ/primeψλ/prime(y)φλ/prime(x) λ/prime, (5.7.29) which is the generalization of the formula, Eq. (5.2.7), for the kernel K(x, y). We now anticipate that the completeness of the eigenfunctions will hold with minor mod- iﬁcation to take care of the fact that Lxis non self-adjoint. In order to see this, we operate (Lx−λ)onG(x, y;λ)in Eq. (5.7.28). (Lx−λ)G(x, y;λ)=(Lx−λ)/summationdisplay nψn(y)φn(x) (λn−λ) +(Lx−λ)/integraldisplayλB −∞dλ/primeψλ/prime(y)φλ/prime(x) (λ/prime−λ), from which, we obtain δ(x−y)=/summationdisplay nψn(y)φn(x)+/integraldisplayλB −∞dλ/primeψλ/prime(y)φλ/prime(x). (5.7.30) This is a statement of the completeness of the eigenfunctions ; discrete eigenfunctions {φn(x),ψn(y)}and continuum eigenfunctions {φλ/prime(x),ψλ/prime(y)}together form a complete set. We further anticipate that the orthogonality of the eigenfunctions will survive with minor modiﬁcation. We consider ﬁrst the following integral, /integraldisplayh 0ψn(x)Lxφm(x)dx=λm/integraldisplayh 0ψn(x)φm(x)dx=/integraldisplayh 0(LT xψn(x))φm(x)dx =λn/integraldisplayh 0ψn(x)φm(x)dx, 5.7 Generalization of Sturm–Liouville System 143 from which, we obtain (λn−λm)/integraldisplayh 0ψn(x)φm(x)dx=0. (5.7.31) H e n c ew eh a v e /integraldisplayh 0ψn(x)φm(x)dx=0 when λn/negationslash=λm. (5.7.32) Thus the eigenfunctions {φm(x),ψn(x)}belonging to the distinct eigenvalues are orthogonal to each other. Secondly we multiply ψn(x)on the completeness relation (5.7.30) and integrate overx.S i n c e λn/negationslash=λ/prime, we have by Eq. (5.7.32), /integraldisplayh 0dx ψ n(x)δ(x−y)=/summationdisplay mψm(y)/integraldisplayh 0dx ψ n(x)φm(x), i.e., ψn(y)=/summationdisplay mψm(y)/integraldisplayh 0ψn(x)φm(x)dx. Then we must have /integraldisplayh 0ψn(x)φm(x)dx=δmn. (5.7.33) Thirdly, we multiply the completeness relation (5.7.30) by ψλ/prime/prime(x)and integrate over x.S i n c e λ/prime/prime/negationslash=λn, we have by Eq. (5.7.32), ψλ/prime/prime(y)=/integraldisplayλB −∞dλ/primeψλ/prime(y)/integraldisplayh 0dx ψ λ/prime/prime(x)φλ/prime(x). Then we must have /integraldisplayh 0ψλ/prime/prime(x)φλ/prime(x)dx=δ(λ/prime−λ/prime/prime). (5.7.34) Thus the discrete eigenfunctions {φm(x),ψn(x)}and the continuum eigenfunctions {φλ/prime(x),ψλ/prime/prime(x)}are normalized respectively as Eqs. (5.7.33) and (5.7.34). The statement of the completeness of the discrete and continuum eigenfunctions (5.7.30) is often assumed at the outset in the standard treatment of Nonrelativistic Quantum Mechanics in the Schrödinger Picture. The discrete eigenfunctions are identiﬁed with the bound state wave functions, whereas the continuum eigenfunctions are identiﬁed with the scattering state wavefunctions. For some potential problems, there exist no scattering states. Simple harmonicoscillator potential is such an example. 144 5 Hilbert–Schmidt Theory of Symmetric Kernel 5.8 Problems for Chapter 5 5.1. (Due to H. C.) Find an upper bound and a lower bound for the ﬁrst eigenvalue of K(x, y)=/braceleftBigg (1−x)y,0≤y≤x≤1, (1−y)x,0≤x≤y≤1. 5.2. (Due to H. C.) Consider the Bessel equation (xu/prime)/prime+λxu=0, with the boundary conditions u/prime(0) = u(1) = 0 . Transform this differential equation into an integral equation and ﬁnd approximately the lowest eigenvalue. 5.3. Obtain an upper limit for the lowest eigenvalue of ∇2u+λru=0, where, in three dimensions, 0<r<a , andu=0 onr=a. 5.4. (Due to H. C.) Consider the Gaussian kernel K(x, y)given by K(x, y)=e−x2−y2,−∞<x ,y< +∞. a) Find the eigenvalues and the eigenfunctions of this kernel. b) V erify the Hilbert–Schmidt expansion of this kernel. c) By calculating A2andA4, obtain the exact lowest eigenvalue. d) Solve the integro-differential equation ∂ ∂tφ(x, t)=/integraldisplay+∞ −∞K(x, y)φ(y,t)dy, with φ(x,0) =f(x). 5.5. Show that the boundary condition (5.5.2) of the Sturm–Liouville system can be replaced by α1φ(0) + α2φ/prime(0) = 0 andβ1φ(h)+β2φ/prime(h)=0 where α1,α2,β1andβ2are some constants and the corresponding boundary condition (5.5.5) on G(x, y)is replaced accordingly. 5.8 Problems for Chapter 5 145 5.6. V erify the Hilbert–Schmidt expansion for the case of Direction 1 in Section 5.6, when the kernel K(x, y)isself-adjoint and square-integrable . 5.7. (Due to H. C.) Reproduce all the results of Section 5.7 with the Green’s function G(x, y;λ)deﬁned by (Lx−λr(x))G(x, y;λ)=δ(x−y), by the weight function, r(x)>0 onx∈[0,h]. 5.8. (Due to H. C.) Consider the eigenvalue problem of the fourth-order ordinary differential equation of the form, /parenleftbiggd4 dx4+1/parenrightbigg φ(x)=−λxφ(x),0<x< 1, with the boundary conditions, φ(0) = φ/prime(0) = 0 , φ(1) = φ/prime(1) = 0 . Do the eigenfunctions form a complete set? Hint: Check whether or not the differential operator Lxdeﬁned by Lx≡−/parenleftbiggd4 dx4+1/parenrightbigg is self-adjoint under the speciﬁed boundary conditions. 5.9. (Due to D. M.) Show that, if ˜λis an eigenvalue of the symmetric kernel K(x, y),t h e inhomogeneous Fredholm integral equation of the second kind, φ(x)=f(x)+˜λ/integraldisplayb aK(x, y)φ(y)dy, a ≤x≤b, has no solution, unless the inhomogeneous term f(x)is orthogonal to all of the eigen- functions φ(x)corresponding to the eigenvalue ˜λ. Hint: Y ou may suppose that {λn}is the sequence of the eigenvalues of the symmetric kernel K(x, y)(ordered by increasing magnitude), with the corresponding eigenfunc- tions{φn(x)}. Y ou may assume that the eigenvalue ˜λhas multiplicity 1. The extension to the higher multiplicity k,k≥2, is immediate. 146 5 Hilbert–Schmidt Theory of Symmetric Kernel 5.10. (Due to D. M.) Consider the integral equation, φ(x)=f(x)+λ/integraldisplay1 0sin2[π(x−y)]φ(y)dy, 0≤x≤1. a) Solve the homogeneous equation by setting f(x)=0. Determine all the eigenfunctions and the eigenvalues. What is the spectral representa- tion of the kernel? Hint: Express the kernel, sin2[π(x−y)], which is translationally invariant, periodic and symmetric, in terms of the powers of exp[π(x−y)]. b) Find the resolvent kernel of this equation. c) Is there a solution to the given inhomogeneous integral equation when f(x)=e x p [ imπx],m integer , andλ=2 ? 5.11. (Due to D. M.) Consider the kernel of the Fredholm integral equation of the second kind, which is given by K(x, y)=/braceleftBigg 3,0≤y<x ≤1, 2,0≤x<y ≤1. a) Find the eigenfunctions φn(x)and the corresponding eigenvalues λnof the kernel. b) IsK(x, y)symmetric? Determine the transposed kernel KT(x, y), and ﬁnd its eigenfunctions ψn(x)and the corresponding eigenvalues λn. c) Show by an explicit calculation that any φn(x)is orthogonal to any ψm(x)ifm/negationslash=n. d) Derive the spectral representation of K(x, y)in terms of φn(x)andψn(x). 5.12. (Due to D. M.) Consider the Fredholm integral equation of the second kind, φ(x)=f(x)+λ/integraldisplay1 0/bracketleftbigg1 2(x+y)−1 2|x−y|/bracketrightbigg φ(y)dy,0≤x≤1. a) Find all non-trivial solutions φn(x)and corresponding eigenvalues λnforf(x)≡0. 5.8 Problems for Chapter 5 147 Hint: Obtain a differential equation for φ(x)with the suitable conditions for φ(x) andφ(x)/prime. b) For the original inhomogeneous equation ( f(x)/negationslash=0 ), will the iteration series con- verge? c) Evaluate the series/summationtext nλ−2 n by using an appropriate integral. 5.13. If |h|<1, ﬁnd the non-trivial solutions of the homogeneous integral equation, φ(x)=λ 2π/integraldisplayπ −π1−h2 1−2hcos(x−y)+h2φ(y)dy. Evaluate the corresponding values of the parameter λ. Hint: The kernel of this equation is translationally invariant. Write cos(x−y)as a sum of two exponentials, express the kernel in terms of the complex variable ζ=e x p [ i(y−x)], use the partial fractions, and then expand each fraction in powers ofζ. 5.14. If |h|<1, ﬁnd the solution of the integral equation, f(x)=λ 2π/integraldisplayπ −π1−h2 1−2hcos(x−y)+h2φ(y)dy, where f(x)is the periodic and square-integrable known function. 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation 7.1 The Wiener–Hopf Method for Partial Differential Equations In Sections 6.3, 6.4 and 6.5, we reduced the singular integral equations of Cauchy type and their variants to the inhomogeneous Hilbert problem by the introduction of the function Φ(z) appropriately deﬁned. We found that the boundary values Φ±(x)across the cut [a,b]satisfy a linear equation, Φ+(x)=R(x)Φ−(x)+f(x),a ≤x≤b, which we are able to solve using the argument in Sections 6.1 and 6.2. Suppose now we are given one linear equation involving two unknown functions, φ−(k) andψ+(k),i nt h ec o m p l e x kplane, φ−(k)=ψ+(k)+F(k), (7.1.1) where φ−(k)is analytic in the lower half-plane ( Imk<τ −)a n dψ+(k)is analytic in the upper half-plane ( Imk≥τ+). Can we solve Eq. (7.1.1) for φ−(k)andψ+(k)? As long as φ−(k)andψ+(k)have a common region of analyticity as in Figure 7.1, namely τ+≤τ−, (7.1.2) we can solve for φ−(k)andψ+(k). In the most stringent case, the common region of an- alyticity can be an arc below which (excluding the arc) φ−(k)is analytic and above which (including the arc) ψ+(k)is analytic. We proceed to split F(k)into a sum of two functions, one analytic in the upper half-plane and the other analytic in the lower half-plane, F(k)=F+(k)+F−(k). (7.1.3) This sum-splitting can be carried out either by inspection or by the general method utiliz- ing the Cauchy integral formula, to be discussed in Section 7.3. Once the sum-splitting is accomplished, we write Eq. (7.1.1) in the following form, φ−(k)−F−(k)=ψ+(k)+F+(k)≡G(k). (7.1.4) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 178 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation k τ+τ− k1k2 Fig. 7.1: The common region of analyticity for φ−(k)andψ+(k)in the complex kplane. ψ+(k)andF+(k)are analytic in the upper half-plane, Imk≥τ+.φ−(k)andF−(k)are analytic in the lower half-plane, Imk<τ −. T h ec o m m o nr e g i o no fa n a l y t i c i t yf o r φ−(k)and ψ+(k)is inside the strip, τ+≤Imk<τ −in the complex kplane. We immediately note that G(k)isentire ink. If the asymptotic behavior of F±(k)as|k|→∞ is such that F±(k)→0 as|k|→∞ , (7.1.5) and on physical grounds, φ−(k),ψ+(k)→0 as|k|→∞ , (7.1.6) then the entire function G(k)must vanish by Liouville’s theorem. Thus we obtain φ−(k)=F−(k), (7.1.7a) ψ+(k)=−F+(k). (7.1.7b) We call this method the Wiener–Hopf method . In the following two examples we apply this method to the mixed boundary value problem of a partial differential equation. 7.1 The Wiener–Hopf Method for Partial Differential Equations 179 ❑Example 7.1. Find the solution to the Laplace Equation in the half-plane . /parenleftbigg∂2 ∂x2+∂2 ∂y2/parenrightbigg φ(x, y)=0,y ≥0,−∞<x< ∞, (7.1.8) subject to the boundary condition on y=0 , as displayed in Figure 7.2, φ(x,0) =e−x,x > 0, (7.1.9a) φy(x,0) =cebx,b > 0,x < 0. (7.1.9b) We further assume φ(x, y)→0 asx2+y2→∞. (7.1.9c) ∂φ ∂y(x,0) =cebxφ(x,0) =e−xxy Fig. 7.2: Boundary conditions of φ(x, y)for the Laplace Equation (7.1.8) in the half-plane, y≥0and−∞<x< ∞. Solution. Since−∞<x< ∞, we may take the F ourier transform with respect to x, i.e., ˆφ(k,y)≡/integraldisplay+∞ −∞dx e−ikxφ(x, y), (7.1.10a) φ(x, y)=1 2π/integraldisplay+∞ −∞dk eikxˆφ(k,y). (7.1.10b) So, if we can obtain ˆφ(k,y), we will have the solution. Taking the Fourier transform of the partial differential equation (7.1.8) with respect to x,w eh a v e /parenleftbigg −k2+∂2 ∂y2/parenrightbigg ˆφ(k,y)=0, 180 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation i.e., ∂2 ∂y2ˆφ(k,y)=k2ˆφ(k,y), (7.1.11) from which we obtain ˆφ(k,y)=C1(k)eky+C2(k)e−ky. The boundary condition at inﬁnity, ˆφ(k,y)→0 asy→+∞, implies /braceleftBigg C1(k)=0 fork>0, C2(k)=0 fork<0. We can write more generally ˆφ(k,y)=A(k)e−|k|y. (7.1.12) To obtain A(k), we need to apply the boundary conditions at y=0 , but this is not trivial. Take the Fourier transform of φ(x,0)to ﬁnd ˆφ(k,0)≡/integraldisplay+∞ −∞dx e−ikxφ(x,0) =/integraldisplay0 −∞dx e−ikxφ(x,0) +/integraldisplay+∞ 0dx e−ikxe−x,(7.1.13) where the ﬁrst term is unknown for x<0, while the second term is equal to 1/(1 +ik). Recall that /vextendsingle/vextendsinglee−ikx/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsinglee−i(k1+ik2)x/vextendsingle/vextendsingle/vextendsingle=ek2xwith k=k1+ik2, which vanishes in the upper half-plane ( k2>0)a sx→− ∞ . So, deﬁne φ+(k)≡/integraldisplay0 −∞dx e−ikxφ(x,0), (7.1.14) which is a +function. (Assuming that φ(x,0)∼O(ebx)asx→− ∞ ,φ+(k)is analytic for k2>−b. So, we have from Eqs. (7.1.12) and (7.1.13), ˆφ(k,0) =φ+(k)+1 1+ikorA(k)=φ+(k)+1 1+ik. (7.1.15) 7.1 The Wiener–Hopf Method for Partial Differential Equations 181 Also, take the Fourier transform of ∂φ(x,0)/∂y to ﬁnd ∂ˆφ ∂y(k,0)≡/integraldisplay+∞ −∞dx e−ikx∂φ(x,0) ∂y =/integraldisplay0 −∞dx e−ikxcebx+/integraldisplay+∞ 0dx e−ikx∂φ(x,0) ∂y,(7.1.16) where the ﬁrst term is equal to c/(b−ik), and the second term is unknown for x>0. Then deﬁne ψ−(k)≡/integraldisplay+∞ 0dx e−ikx∂φ(x,0) ∂y, (7.1.17) which is a −function. Assuming that ∂φ(x,0)/∂y∼O(e−x)asx→∞ ,ψ−(k)is analytic fork2<1. Thus we obtain −|k|A(k)=c b−ik+ψ−(k), or we have A(k)=−c |k|(b−ik)−ψ−(k) |k|. (7.1.18) Equating the two expressions (7.1.15) and (7.1.18) for A(k), (assuming that they are both valid in some common region, say k2=0 ), we get φ+(k)+1 1+ik=−c |k|(b−ik)−ψ−(k) |k|. (7.1.19) Now, the function |k|is not analytic and so we cannot proceed with the Wiener–Hopf method unless we express |k|in a suitable form. One such form, often suitable for application, is |k|= lim ε→0+(k2+ε2)1/2= lim ε→0+(k+iε)1/2(k−iε)1/2, (7.1.20) where, in the last expression, (k+iε)1/2is a+function and (k−iε)1/2is a−function, as displayed in Figure 7.3. We can verify that on the real axis, /radicalbig k2+ε2=|k| asε→0+. (7.1.21) We thus have φ+(k)−i k−i=−ci (k+ib)(k+iε)1/2(k−iε)1/2−ψ−(k) (k+iε)1/2(k−iε)1/2.(7.1.22) Since (k+iε)1/2is a+function (i.e., is analytic in the upper half-plane), we multiply the whole equation (7.1.22) by (k+iε)1/2to get (k+iε)1/2φ+(k)−i(k+iε)1/2 k−i=−ci (k+ib)(k−iε)1/2−ψ−(k) (k−iε)1/2, (7.1.23) 182 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation k iε −iε Fig. 7.3: Factorization of |k|into a +function and a −function. The points, k=±iε,a r e the branch points of√ k2+ε2, from which the branch cuts are extended to ±i∞ along the imaginary kaxis in the complex kplane. where the ﬁrst term on the left-hand side is a +function and the second term on the right-hand side is a −function. We shall decompose the remaining terms into a sum of the +function and the−function. Consider the second term of the left-hand side of Eq. (7.1.23). The numerator is a+function but the denominator is a −function. We rewrite −i(k+iε)1/2 k−i=−i(k+iε)1/2+i(i+iε)1/2 k−i−i(i+iε)1/2 k−i, where, in the ﬁrst term on the right-hand side of the above, we removed the singularity at k=iby making the numerator vanish at k=iso that it is a +function. The second term on the right-hand side has a pole at k=iso that it is a −function. Similarly, the ﬁrst term on the right-hand side of Eq. (7.1.23) is rewritten in the following form, −ci (k+ib)(k−iε)1/2=−ci k+ib/bracketleftbigg1 (k−iε)1/2−1 (−ib−iε)1/2/bracketrightbigg −ci (k+ib)(−ib−iε)1/2, where the ﬁrst term on the right-hand side of the above is no longer singular at k=−ib,b u t has a branch point at k=iεso that it is a −function. The second term on the right-hand side has a pole at k=−ibso that it is a +function. 7.1 The Wiener–Hopf Method for Partial Differential Equations 183 Collating all this, we have the following equation, (k+iε)1/2φ+(k)+−i(k+iε)1/2+i(i+iε)1/2 k−i+ci (k+ib)(−ib−iε)1/2 =−ψ−(k) (k−iε)1/2−ci (k+ib)/bracketleftbigg1 (k−iε)1/2−1 (−ib−iε)1/2/bracketrightbigg +i(i+iε)1/2 k−i,(7.1.24) where each term on the left-hand side is a +function while each term on the right-hand side is a−function. Applying the Wiener–Hopf method, and noting that the left-hand side and the right-hand side both go to 0as|k|→∞ ,w eh a v e φ+(k)=i k−i/bracketleftbigg 1−(i+iε)1/2 (k+iε)1/2/bracketrightbigg −ci (k+ib)(−ib−iε)1/2(k+iε)1/2. We can simplify somewhat, (i+iε)1/2=( 1+ ε)1/2eiπ/4=eiπ/4asε→0+. Similarly (−ib−iε)1/2=√ b+εe−iπ/4=√ be−iπ/4asε→0+. H e n c ew eh a v e φ+(k)=i k−i/bracketleftbigg 1−eiπ/4 (k+iε)1/2/bracketrightbigg −(c/√ b)ieiπ/4 (k+ib)(k+iε)1/2. (7.1.25) So ﬁnally, we obtain A(k)from Eqs. (7.1.15) and (7.1.25) as, A(k)=−ieiπ/4 (k+iε)1/2/bracketleftbiggc √ b(k+ib)+1 k−i/bracketrightbigg , (7.1.26) where we note (k+iε)1/2=/braceleftBigg√ k fork>0,/radicalbig |k|eiπ/2=i/radicalbig |k|fork<0. Therefore, ˆφ(k,y)=A(k)e−|k|y can be inverted with respect to kto obtain φ(x, y). 184 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation We now discuss another example. ❑Example 7.2. Sommerfeld diffraction problem . Solve the wave equation in two space dimensions: ∂2 ∂t2u(x, y, t)=c2∇2u(x, y, t) (7.1.27) with the boundary condition, ∂ ∂yu(x, y, t)=0 aty=0 forx<0. (7.1.28) The incident, reﬂected, and diffracted waves are shown in Figure 7.4. oxy Fig. 7.4: Incident wave, reﬂected wave and diffracted wave in the Sommerfeld diffraction prob- lem . Solution. We look for the solution of the form, u(x, y, t)=φ(x, y)e−iωt. (7.1.29) Setting p≡ω/c, (7.1.30) the wave equation assumes the following form, ∇2φ+p2φ=0. (7.1.31) Letting the forcing increase exponentially in time (so that it is absent as t→− ∞ ), we must have Imω>0, which requires p=p1+iε, ε →0+. (7.1.32) 7.1 The Wiener–Hopf Method for Partial Differential Equations 185 So,φ(x, y)satisﬁes /braceleftBigg ∇2φ+p2φ=0, ∂φ ∂y=0 aty=0 forx<0.(7.1.33) Consider the incident waves, uinc=ei(/vectorp·/vectorx−ωt)=φince−iωt, with |/vectorp|=p, so that uincalso satisﬁes the wave equation. The uincis a plane wave moving in the direction of/vectorp.W et a k e /vectorp=−pcosθ·/vectorex−psinθ·/vectorey, and assume 0<θ<π / 2. Thenφinc(x, y)is given by φinc(x, y)=e−ip(xcosθ+ysinθ). (7.1.34) We seek a disturbance solution ψ(x, y)≡φ(x, y)−φinc(x, y). (7.1.35) Thus the governing equation and the boundary condition for ψ(x, y)become ∇2ψ+p2ψ=0, (7.1.36) subject to ∂ψ(x,0) ∂y=−∂φ inc(x,0) ∂y=ipsinθ·e−ip(xcosθ)forx<0 aty=0.(7.1.37) Asymptotic behavior: We replace pbyp1+iε, p=p1+iε. (7.1.38) In the reﬂection region ,w eh a v e ψ(x, y)∼e−ip(xcosθ−ysinθ)fory>0,x→− ∞ . We note the change of sign in front of y. Near y=0+,w eh a v e ψ(x,0+)∼e−ipxcosθasx→− ∞ , or, /vextendsingle/vextendsingleψ(x,0+)/vextendsingle/vextendsingle∼eεxcosθasx→− ∞ . 186 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation In the shadow region (y<0,x→− ∞ ), φ(−∞,y)=0⇒ψ=−φinc, so that ψ(x,0−)∼−e−ipxcosθ, or, /vextendsingle/vextendsingleψ(x,0−)/vextendsingle/vextendsingle∼eεxcosθasx→− ∞ . In summary, |ψ(x, y)|∼eεxcosθasx→− ∞ ,y=0±. (7.1.39) Although ψ(x, y)might be discontinuous across the obstacle, its normal derivative is contin- uous as given by the boundary condition. On the other side, as x→+∞, both ψ(x,0)and ∂ψ(x,0)/∂x are continuous, but the asymptotic behavior at inﬁnity is obtained by approxi- mating the effect of the leading edge of the obstacle as a delta function, ∇2ψ+p2ψ=−4πδ(x)δ(y). (7.1.40) From Eq. (7.1.40), we obtain ψ(x, y)=πiH(1) 0(pr),r =/radicalbig x2+y2, (7.1.41) where H(1) 0(pr)is the zeroth-order Hankel function of the ﬁrst kind. It behaves as H(1) 0(pr)∼1 √ rexp(ipr) asr→∞, which implies that |ψ(x,0)|∼1 √ xe−εxasx→∞ near y=0. (7.1.42) Now, we try to solve the equation for ψ(x, y)using the Fourier transforms, ˆψ(k,y)≡/integraldisplay+∞ −∞dx e−ikxψ(x, y), (7.1.43a) ψ(x, y)=1 2π/integraldisplay+∞ −∞dk eikxˆψ(k,y). (7.1.43b) We take the Fourier transform of the partial differential equation, ∇2ψ+p2ψ=0, 7.1 The Wiener–Hopf Method for Partial Differential Equations 187 kk2 k1iε −iεp i1+ε −−p i1εp 1 −p 1 r1r2 φ1φ2 Fig. 7.5: Branch cuts of (k2−p2)1/2in the complex kplane. The points, k=±(p1+iε),a r e the branch points ofp k2−p2, from which the branch cuts are extended to ±(p1+i∞). resulting in the form, ∂2 ∂y2ˆψ(k,y)=(k2−p2)ˆψ(k,y). (7.1.44) Consider the following branch of (k2−p2)1/2= lim ε→0+[(k−p1−iε)(k+p1+iε)]1/2, (7.1.45) where the branch cuts are drawn in Figure 7.5. Then we have [k2−(p1+iε)2]1/2=√ r1r2ei(φ1+φ2)/2, so that Re[k2−(p1+iε)2]1/2=√ r1r2cos/parenleftbiggφ1+φ2 2/parenrightbigg . On the real axis above, −π<φ 1+φ2<0, so that 0<cos/parenleftbiggφ1+φ2 2/parenrightbigg <1. 188 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Thus the branch chosen has Re[k2−(p1+iε)2]1/2>0 (7.1.46) on the whole real axis. We can then write the solution to ∂2 ∂y2ˆψ(k,y)=[k2−(p1+iε)2]ˆψ(k,y) (7.1.47) as ˆψ(k,y)=/braceleftBigg A(k)e x p (−/radicalbig k2−(p1+iε)2y),y > 0, B(k)e x p ( +/radicalbig k2−(p1+iε)2y),y < 0. But since ψ(x, y)is not continuous across y=0 forx<0, the amplitudes A(k)andB(k) need not be identical. However, we know ∂ψ(x, y)/∂y is continuous across y=0 for all x. We must therefore have ∂ ∂yˆψ(k,0+)=∂ ∂yˆψ(k,0−), from which we obtain B(k)=−A(k). Thus we have ˆψ(k,y)=/braceleftBigg A(k)e x p (−/radicalbig k2−(p+iε)2y),y > 0, −A(k)e x p ( +/radicalbig k2−(p+iε)2y),y < 0.(7.1.48) If we can determine A(k), we will have arrived at the solution. Let us recall everything we know. We know that ψ(x, y)is continuous for x>0when y=0 , but is discontinuous for x<0. So, consider the function, ψ(x,0+)−ψ(x,0−)=/braceleftBigg 0 forx>0, unknown for x<0.(7.1.49) But from one earlier discussion, we know the asymptotic form of the latter unknown function to be like eεxcosθasx→− ∞ . Take the Fourier transform of the above discontinuity (7.1.49) to obtain ˆψ(k,0+)−ˆψ(k,0−)=/integraldisplay0 −∞dx e−ikx(ψ(x,0+)−ψ(x,0−)), the right-hand side of which is a +function, analytic for k2>−εcosθ.W ed e ﬁ n e U+(k)≡ˆψ(k,0+)−ˆψ(k,0−), analytic for k2>−εcosθ. (7.1.50) 7.1 The Wiener–Hopf Method for Partial Differential Equations 189 Now consider the derivative ∂ψ(x, y)/∂y . This function is continuous at y=0 for all x (−∞<x< ∞). Furthermore, we know what it is for x<0. Namely, ∂ ∂yψ(x,0) =/braceleftBigg ipsinθ·e−ipxcosθforx<0, unknown for x>0.(7.1.51) But we know the asymptotic form of the latter unknown function to be like e−εxasx→∞ . Taking the Fourier transform of Eq. (7.1.51), we obtain ∂ ∂yˆψ(k,0) =/integraldisplay0 −∞dx e−ikx·ipsinθ·−ipxcosθ+/integraldisplay+∞ 0dx e−ikx∂ ∂yψ(x,0). Thus we have ∂ ∂yˆψ(k,0) =−psinθ (k+pcosθ)+L−(k), (7.1.52) where in the ﬁrst term on the right-hand side of the above, p=p1+iε, and the second term represented as L−(k)is deﬁned by L−(k)≡/integraldisplay+∞ 0dx e−ikx∂ ∂yψ(x,0). L−(k)is a−function, analytic for k2<ε . We shall now use Eqs. (7.1.50) and (7.1.52), where U+(k)andL−(k)are deﬁned, together with the Wiener–Hopf method to solve for A(k). We know ˆψ(k,y)=/braceleftBigg A(k)e x p (−/radicalbig k2−p2y) fory>0, −A(k)e x p (/radicalbig k2−p2y) fory<0. Inserting these equations into Eq. (7.1.50), we have 2A(k)=U+(k). Inserting these equations into Eq. (7.1.52), we have −/radicalbig k2−p2A(k)=−psinθ (k+pcosθ)+L−(k). Eliminating A(k)from the last two equations, we obtain the Wiener–Hopf problem, −/radicalbig k2−p2U+(k)/2=−psinθ (k+pcosθ)+L−(k), (7.1.53) /braceleftBigg L−(k) analytic for k2<ε , U+(k) analytic for k2>−εcosθ.(7.1.54) 190 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation We divide through both sides of Eq. (7.1.53) by√ k−p, obtaining −√ k+pU+(k) 2=L−(k) √ k−p−psinθ (k+pcosθ)√ k−p. The term involving U+(k)is a+function, while the term involving L−(k)is a−function. We decompose the last term on the right-hand side. −psinθ (k+pcosθ)√ k−p=−psinθ k+pcosθ/bracketleftbigg1 √ k−p−1 √ −pcosθ−p/bracketrightbigg −psinθ (k+pcosθ)√ −pcosθ−p, where the ﬁrst term on the right-hand side is a −function and the second term a +function. Collecting the +functions and the −functions respectively, we obtain by the Wiener–Hopf method, U+(k) 2=psinθ √ k+p(k+pcosθ)√ −pcosθ−p, (7.1.55) and hence we ﬁnally obtain A(k)=psinθ (k+pcosθ)√ k+p√ −pcosθ−p. (7.1.56) Singularities of A(k)in the complex kplane are drawn in Figure 7.6. k k1k2 −p−pcosθ Fig. 7.6: Singularities of A(k)in the complex kplane. A(k)has a simple pole at k=−pcosθ and a branch point at k=−p, from which the branch cut is extended to −p−i∞. 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 191 The ﬁnal solution for the disturbance function ψ(x, y)in the limit as ε→0+,i sg i v e nb y ψ(x, y)=sgn(y) 2π/integraldisplay Cdk A(k)e x p (−/radicalbig k2−p2|y|+ikx), (7.1.57) where Cis the contour speciﬁed in Figure 7.7. k2k k1 Cp −p −pcosθ Fig. 7.7: The contour of the complex kintegration in Eq. (7.1.57) for ψ(x, y). The integrand has a simple pole at k=−pcosθand the branch points at k=±p. The branch cuts are extended fromk=±pto±∞ along the real kaxis. We note that the choice of p=p1+iε, ε > 0 is based on the requirement that uinc(/vectorx, t)→0 ast→− ∞ but it grows exponentially in time, i.e., uinc=ei(/vectorp·/vectorx−ωt)Imω>0, which is known as ‘turning on the perturbation adiabatically’ . 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind The Wiener–Hopf integral equations are characterized by translation kernels ,K(x, y)= K(x−y), and the integral is on the semi-inﬁnite range ,0<x ,y< ∞. We list the Wiener– Hopf integral equations of several types. 192 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Wiener–Hopf integral equation of the ﬁrst kind: F(x)=/integraldisplay+∞ 0K(x−y)φ(y)dy, 0≤x<∞. Homogeneous Wiener–Hopf integral equation of the second kind: φ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy, 0≤x<∞. Inhomogeneous Wiener–Hopf integral equation of the second kind: φ(x)=f(x)+λ/integraldisplay+∞ 0K(x−y)φ(y)dy, 0≤x<∞. Let us begin with the homogeneous Wiener–Hopf integral equation of the second kind . φ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy, 0≤x<∞. (7.2.1) Here, the translation kernel K(x−y)is deﬁned for its argument both positive and negative. Suppose that K(x)→/braceleftBigg e−bxasx→+∞, eaxasx→− ∞ ,a,b > 0, (7.2.2) so that the Fourier transform of K(x), deﬁned by ˆK(k)=/integraldisplay+∞ −∞dx e−ikxK(x), (7.2.3) is analytic for −a<Imk0, 0 for x<0.(7.2.5) But then, although φ(x)is only known for positive x,s i n c e K(x−y)is deﬁned even for negative x, we can certainly deﬁne ψ(x)for negative x, ψ(x)≡λ/integraldisplay+∞ 0K(x−y)φ(y)dy forx<0. (7.2.6) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 193 k −ab Fig. 7.8: Region of analyticity of ˆK(k)in the complex kplane. ˆK(k)is deﬁned and analytic inside the strip, −a<Imk<b . Take the F ourier transforms of Eqs. (7.2.1) and (7.2.6). Adding up the results, and using the convolution property ,w eh a v e /integraldisplay+∞ 0dx e−ikxφ(x)+/integraldisplay0 −∞dx e−ikxψ(x)=λˆK(k)ˆφ−(k), i.e., we have ˆφ−(k)+ˆψ+(k)=λˆK(k)ˆφ−(k), or, we have /bracketleftbig 1−λˆK(k)/bracketrightbigˆφ−(k)=−ˆψ+(k), (7.2.7) where we have deﬁned ˆφ−(k)≡/integraldisplay+∞ 0dx e−ikxφ(x), (7.2.8) ˆψ+(k)≡/integraldisplay0 −∞dx e−ikxψ(x). (7.2.9) Since we have /vextendsingle/vextendsinglee−ikx/vextendsingle/vextendsingle=ek2x,k=k1+ik2, we know that ˆφ−(k)is analytic in the lower half-plane and ˆψ+(k)is analytic in the upper half-plane. Thus, once again, we have one equation involving two unknown functions, ˆφ−(k) 194 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation andˆψ+(k), one analytic in the lower half-plane and the other analytic in the upper half-plane. The precise regions of analyticity for ˆφ−(k)andˆψ+(k)are each determined by the asymptotic behavior of the kernel K(x)asx→− ∞ . In the original equation, Eq. (7.2.1), at the upper limit of the integral, we have y→∞ so thatx−y→− ∞ asy→∞ . By Eq. (7.2.2), we have K(x−y)∼ea(x−y)∼e−ayasy→∞. To ensure that the integral in Eq. (7.2.1) converges, we conclude that φ(x)can grow as fast as φ(x)∼e(a−ε)xwith ε>0 asx→∞. By deﬁnition of ˆφ−(k), the region of the analyticity of ˆφ−(k)is determined by the requirement /vextendsingle/vextendsinglee−ikxφ(x)/vextendsingle/vextendsingle∼e(k2+a−ε)x→0 asx→∞. Thusˆφ−(k)is analytic in the lower half-plane, Imk=k2<−a+ε, ε > 0, which includes Imk≤−a. As for the behavior of ψ(x)asx→− ∞ , we observe that x−y→− ∞ asx→− ∞ ,a n d K(x−y)∼ea(x−y)asx→− ∞ . By deﬁnition of ψ(x),w eh a v e ψ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy∼λeax/integraldisplay+∞ 0e−ayφ(y)dy asx→− ∞ , where the integral above is convergent due to the asymptotic behavior of φ(x)asx→∞ . The region of analyticity of ˆψ+(k)is determined by the requirement, /vextendsingle/vextendsinglee−ikxψ(x)/vextendsingle/vextendsingle∼e(k2+a)x→0 asx→− ∞ . Thusˆψ+(k)is analytic in the upper half-plane, Imk=k2>−a. To summarize, we know /braceleftBigg φ(x)→e(a−ε)xasx→∞, ψ(x)→eaxasx→− ∞ .(7.2.10) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 195 kk2 k1kb2= ka2=−+ε ka2=− Fig. 7.9: Region of analyticity of ˆφ−(k),ˆψ+(k)andˆK(k).ˆφ−(k)is analytic in the lower half-plane, Imk<−a+ε.ˆψ+(k)is analytic in the upper half-plane, Imk>−a.ˆK(k)is analytic inside the strip, −a<Imk−a. (7.2.11b) V arious regions of the analyticity are drawn in Figure 7.9. Recalling Eq. (7.2.7), we write 1−λˆK(k)as the ratio of the −function and the +function, 1−λˆK(k)=Y−(k) Y+(k). (7.2.12) From Eqs. (7.2.7) and (7.2.12), we have Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)≡F(k), (7.2.13) where F(k)is an entire function. The asymptotic behavior of F(k)as|k|→∞ will determine F(k)completely. We know that ˆφ−(k)→0 as|k|→∞ , ˆψ+(k)→0 as|k|→∞ , ˆK(k)→0 as|k|→∞ .(7.2.14) 196 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation By Eq. (7.2.12), we know then Y−(k) Y+(k)→1, as|k|→∞ . (7.2.15) We only need to know the asymptotic behavior of Y−(k)orY+(k)as|k|→∞ in order to determine the entire function F(k). Once F(k)is determined we have, from Eq. (7.2.13), ˆφ−(k)=F(k) Y−(k)(7.2.16) and we have ﬁnished. We note that it is convenient to choose a function Y−(k)which is not only analytic in the lower half-plane, but also has no zeros in the lower half-plane, so that F(k)/Y−(k)is itself a −function for all entire F(k). Otherwise, we need to choose F(k)so as to have zeros exactly at zeros of Y−(k)to cancel the possible poles in F(k)/Y−(k)and to yield the −function ˆφ−(k). The factorization of 1−λˆK(k)is essential in solving the Wiener–Hopf integral equation of the second kind. As noted earlier, it can be done either by inspection or by the general method based on the Cauchy integral formula. As a general rule, we assign Any pole in the lower half-plane (k=pl)t o Y+(k), Any zero in the lower half-plane (k=zl)t o Y+(k), Any pole in the upper half-plane (k=pu)t o Y−(k), Any zero in the upper half-plane (k=zu)t o Y−(k).(7.2.17) We ﬁrst solve the following simple example where the factorization is carried out by in- spection and illustrate the rationale of this general rule for the assignment. ❑Example 7.3. Solve φ(x)=λ/integraldisplay+∞ 0e−|x−y|φ(y)dy, x ≥0. (7.2.18) Solution. Deﬁne ψ(x)=λ/integraldisplay+∞ 0e−|x−y|φ(y)dy, x < 0. (7.2.19) Also, deﬁne φ(x)=/braceleftBigg φ(x) forx≥0, 0 forx<0. Take the Fourier transform of Eqs. (7.2.18) and (7.2.19) and add the results together to obtain ˆφ−(k)+ˆψ+(k)=λ·2 k2+1·ˆφ−(k). (7.2.20) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 197 Now, K(x)=e−|x|→/braceleftBigg e−xasx→∞, exasx→− ∞ .(7.2.21) Therefore, φ(y)can be allowed to grow as fast as e(1−ε)yasy→∞ . Thus ˆφ−(k)is analytic fork2<−1+ε. We also ﬁnd that ψ(x)→exasx→− ∞ . Thus ˆψ+(k)is analytic for k2>−1. To solve Eq. (7.2.20), we ﬁrst write k2+1−2λ k2+1ˆφ−(k)=−ˆψ+(k), (7.2.22) and then decompose k2+1−2λ k2+1(7.2.23) into a ratio of a −f u n c t i o nt oa +function. The common region of analyticity of ˆφ−(k), ˆψ+(k)andˆK(k)of this example are drawn in Figure 7.10. kk2 k1k21= k2 1=−+ ε k2 1=− Fig. 7.10: Common region of analyticity of ˆφ−(k),ˆψ+(k)andˆK(k)of Example 7.3. ˆφ−(k) is analytic in the lower half-plane, Imk<−1+ε.ˆψ+(k)is analytic in the upper half-plane, Imk>−1.ˆK(k)is analytic inside the strip, −1<Imk<1. The designations, the lower half-plane, and the upper half-plane, must be made relative to a line with −1<Imk=k2<−1+ε. (7.2.24) 198 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Referring to Eq. (7.2.23), k2+1=( k+i)(k−i) so that k=iis a pole of Eq. (7.2.23) in the upper half-plane and k=−iis a pole of Eq. (7.2.23) in the lower half-plane. Now look at the numerator of Eq. (7.2.23), k2+1−2λ. Case 1. λ<0.⇒1−2λ>1. k2+1−2λ=(k+i√ 1−2λ)(k−i√ 1−2λ) (7.2.25) The ﬁrst factor corresponds to a zero in the lower half-plane, while the second factor corre- sponds to a zero in the upper half-plane. Case 2. 0<λ< 1/2.⇒0<1−2λ<1. k2+1−2λ=(k+i√ 1−2λ)(k−i√ 1−2λ) (7.2.26) Both factors correspond to a zero in the upper half-plane. Case 3. λ>1/2.⇒1−2λ<0. k2+1−2λ=(k+√ 2λ−1)(k−√ 2λ−1) (7.2.27) Both factors correspond to a zero in the upper half-plane. Now, in general, when we write 1−λˆK(k)=Y−(k) Y+(k), since we will end up with Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)≡G(k) which is entire, we wish to have ˆφ−(k)=G(k) Y−(k) analytic in the lower half-plane. So, Y−(k)must not have any zeros in the lower half-plane. Hence we assign any zeros or poles in the lower half-plane to Y+(k)so that Y−(k)has neither poles nor zeros in the lower half-plane. Now we have 1−λˆK(k)=(k2+1−2λ) (k2+1 ) =(k+i√ 1−2λ)(k−i√ 1−2λ) (k+i)(k−i). 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 199 Case 1. λ<0. k+i√ 1−2λ⇒ zero in the lower half-plane ⇒Y+(k) k−i√ 1−2λ⇒ zero in the upper half-plane ⇒Y−(k) k+i ⇒ pole in the lower half-plane ⇒Y+(k) k−i ⇒ pole in the upper half-plane ⇒Y−(k) Thus we obtain   Y −(k)=(k−i√ 1−2λ) (k−i), Y+(k)=(k+i) (k+i√ 1−2λ).(7.2.28) Hence, in the following equation, Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)=G(k), we know Y−(k)→1,ˆφ−(k)→0, ask→∞, so that G(k)→0 ask→∞. By Liouville’s theorem, we conclude G(k)=0, from which it follows that ˆφ−(k)=0,ˆψ+(k)=0. (7.2.29) So, there exists no nontrivial solution, i.e., φ(x)=0, forλ<0. Case 2. 0<λ< 1/2. With a similar analysis as in Case 1, we obtain   Y−(k)=(k+i√ 1−2λ)(k−i√ 1−2λ) (k−i), Y+(k)=( k+i).(7.2.30) Noting that Y−(k)→k ask→∞, 200 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation and Y−(k)ˆφ−(k)=G(k), we ﬁnd that G(k)g r o w sl e s sf a s tt h a n kask→∞ . By Liouville’s theorem, we ﬁnd G(k)=A, constant , and thus conclude that ˆφ−(k)=A(k−i) (k+i√ 1−2λ)(k−i√ 1−2λ). (7.2.31) Case 3. λ>1/2. With a similar analysis as in Case 1, we obtain   Y−(k)=(k+√ 2λ−1)(k−√ 2λ−1) (k−i)→k ask→∞, Y+(k)=(k+i) →k ask→∞.(7.2.32) Again, we ﬁnd that G(k)=A, constant , and thus conclude that ˆφ−(k)=A(k−i) (k+√ 2λ−1)(k−√ 2λ−1). (7.2.33) To summarize, we ﬁnd the following. Forλ≤0⇒φ(x)=0. (7.2.34) Forλ>0⇒φ(x)=1 2π/integraldisplay Cdk eikxA(k−i)/(k2+1−2λ), (7.2.35) where the inversion contour Cis indicated in Figure 7.11. Forx>0, we close the contour in the upper half-plane and get the contribution from both poles in the upper half-plane in either Case 2 orCase 3 . The result of inversions are the following. Case 2. 0<λ< 1/2. φ(x)=C/parenleftbigg cosh√ 1−2λx+sinh√ 1−2λx √ 1−2λ/parenrightbigg ,x > 0. (7.2.36) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 201 kk2 k1−− 21λ 21λ−i12−λ − − i12 λ −i Fig. 7.11: The inversion contour for φ(x)of Eq. (7.2.35). Simple poles are located at k= ±i√ 1−2λfor Case 2 and at k=±√ 2λ−1for Case 3. Case 3. λ>1/2. φ(x)=C/parenleftbigg cos√ 2λ−1x+sin√ 2λ−1x √ 2λ−1/parenrightbigg ,x > 0. (7.2.37) We shall now consider another example where the factorization also is carried out by inspection after some juggling of the gamma functions. ❑Example 7.4. Solve φ(x)=λ/integraldisplay+∞ 01 cosh[1 2(x−y)]φ(y)dy, x ≥0. (7.2.38) Solution. We begin with the Fourier transform of the kernel K(x). K(x)=1 cosh/parenleftbig1 2x/parenrightbig→2e−1 2|x|, as|x|→∞ . Then ˆK(k)is analytic inside the strip, −1 2<Imk=k2<1 2. (7.2.39) We calculate ˆK(k)as follows. ˆK(k)=/integraldisplay+∞ −∞dx e−ikx 1 cosh/parenleftbig1 2x/parenrightbig=2/integraldisplay+∞ −∞dxe−ikxe1 2x (ex+1 ). 202 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Setting ex=t, x =l nt, dx =dt t, we have ˆK(k)=2/integraldisplay+∞ 0dtt−ik−1 2 (t+1 ). A further change of variable ρ=1 (t+1 ),t=(1−ρ) ρ,d t =−dρ ρ2 results in ˆK(k)=2/integraldisplay1 0dρ ρik−1 2(1−ρ)−ik−1 2. Recalling the deﬁnition of the Beta function B(n,m), B(n,m)=/integraldisplay1 0dρ ρn−1(1−ρ)m−1=Γ(n)Γ(m) Γ(n+m), we have ˆK(k)=2Γ/parenleftbig ik+1 2/parenrightbig Γ/parenleftbig −ik+1 2/parenrightbig Γ/parenleftbig ik+1 2−ik+1 2/parenrightbig =2 Γ/parenleftbigg ik+1 2/parenrightbigg Γ/parenleftbigg −ik+1 2/parenrightbigg . Recalling the property of the gamma function, Γ(z)Γ(1−z)=π sinπz, (7.2.40) we thus obtain the Fourier transform of K(x)as ˆK(k)=2π coshπk. (7.2.41) Deﬁning ψ(x)by ψ(x)=λ/integraldisplay+∞ 01 cosh/bracketleftbig1 2(x−y)/bracketrightbigφ(y)dy, x < 0, (7.2.42) we obtain the following equation as usual, (1−λˆK(k))ˆφ−(k)=−ˆψ+(k), (7.2.43) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 203 where 1−λˆK(k)=1−2πλ coshπk=Y−(k) Y+(k), (7.2.44) and the regions of the analyticity of ˆφ−(k)andˆψ+(k)are such that (i)ˆφ−(k)is analytic in the lower half-plane ( Imk≤−1/2), (ii)ˆψ+(k)is analytic in the upper half-plane ( Imk>−1/2).(7.2.45) Rewriting Eq. (7.2.43) in terms of Y±(k),w eh a v e Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)≡G(k), (7.2.46) where G(k)is entire in k. Factorizing 1−λˆK(k): Y−(k) Y+(k)=1−2πλ coshπk=coshπk−2πλ coshπk. (7.2.47) Case 1. 0<2πλ≤1. Setting cosπα≡2πλ, 0≤α<1 2, (7.2.48) we have Y−(k) Y+(k)=cos(iπk)−cosπα sinπ/parenleftbig ik+1 2/parenrightbig =2s i n/bracketleftbigπ 2(α+ik)/bracketrightbig sin/bracketleftbigπ 2(α−ik)/bracketrightbig sinπ/parenleftbig ik+1 2/parenrightbig .(7.2.49) Replacing all sine functions in Eq. (7.2.49) with the appropriate product of the gamma func- tions by the use of the formula sinπz=π Γ(z)Γ(1−z), (7.2.50) we obtain Y−(k) Y+(k) =2πΓ/parenleftbig1 2+ik/parenrightbig Γ/parenleftbig1 2−ik/parenrightbig Γ/parenleftbigα+ik 2/parenrightbig Γ/parenleftbigα−ik 2/parenrightbig Γ/parenleftbig 1−α+ik 2/parenrightbig Γ/parenleftbig 1−α−ik 2/parenrightbig. (7.2.51) We note that /braceleftBigg Γ(z)has a simple pole at z=0,−1,−2,···, Γ(1−z)has a simple pole at z=1,2,3,···.(7.2.52) 204 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation (1)Γ/parenleftbig1 2+ik/parenrightbig has simple poles at k=i1 2,i3 2,i5 2,i7 2,···, all of which are assigned to Y−(k). (2)Γ/parenleftbig1 2−ik/parenrightbig has simple poles at k=−i1 2,−i3 2,−i5 2,−i7 2,···, all of which are assigned toY+(k). (3)Γ/parenleftbigα+ik 2/parenrightbig has simple poles at k=iα, i(2 +α),i(4 +α),···, all of which are assigned to Y−(k). (4)Γ/parenleftbigα−ik 2/parenrightbig has simple poles at k=−iα,−i(2 +α),−i(4 +α),···.S i n c e 0<α< 1/2, the ﬁrst pole at k=−iαis assigned to Y−(k), while the remaining poles are assigned to Y+(k). Using the property of the gamma function, Γ(z)=Γ(z+1) z, we rewrite Γ/parenleftbiggα−ik 2/parenrightbigg =2 α−ikΓ/parenleftbigg 1+α−ik 2/parenrightbigg , where (α−ik)/2is assigned to Y−(k)whileΓ/parenleftbig 1+α−ik 2/parenrightbig is assigned to Y+(k). (5)Γ(1−α+ik 2)has simple poles at k=−i(2−α),−i(4−α),−i(6−α),···, all of which are assigned to Y+(k). (6)Γ(1−α−ik 2)has simple poles at k=i(2−α),i(4−α),i(6−α),···, all of which are assigned to Y−(k). Then we obtain Y±(k)as follows. Y−(k)=−2πΓ/parenleftbig1 2+ik/parenrightbig Γ/parenleftbigα+ik 2/parenrightbig Γ/parenleftbig −α−ik 2/parenrightbig, (7.2.53) Y+(k)=Γ/parenleftbig 1+α−ik 2/parenrightbig Γ/parenleftbig 1−α+ik 2/parenrightbig Γ/parenleftbig1 2−ik/parenrightbig . (7.2.54) Now follows the determination of G(k), which is determined by the asymptotic behavior of Y±(k)ask→∞ . Making use of the Duplication formula and the Stirling formula , Γ(2z)=22z−1Γ(z)Γ/parenleftbig z+1 2/parenrightbig √ π, (7.2.55) lim |z|→∞Γ(z+β) Γ(z)∼zβ, (7.2.56) we ﬁnd the asymptotic behavior of Y−(k)to be given by Y−(k)∼−i/radicalbigg π 22ik·k. (7.2.57) Deﬁning Z±(k)≡2−ik·Y±(k), (7.2.58) 7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 205 we ﬁnd Z±(k)∼−i/radicalbigg π 2k, (7.2.59) since Z−(k) Z+(k)=Y−(k) Y+(k)=1−λˆK(k)→1 ask→∞. Then Eq. (7.2.46) becomes Z−(k)ˆφ−(k)=−Z+(k)ˆψ+(k)=2−ikG(k)≡g(k), (7.2.60) where g(k)is now entire. Since ˆφ−(k),ˆψ+(k)→0 ask→∞, and Eq. (7.2.59) for Z±(k),g(k)cannot grow as fast as k. By Liouville’s theorem, we then have g(k)=C/prime, constant . Thus we obtain ˆφ−(k)=C/prime Z−(k)=C/prime/prime2ikΓ/parenleftbigα+ik 2/parenrightbig Γ/parenleftbig −α−ik 2/parenrightbig Γ/parenleftbig1 2+ik/parenrightbig . (7.2.61) We now invert ˆφ−(k)to obtain φ(x), φ(x)=C/prime/prime/integraldisplay+∞ −∞dk 2πeikx2ikΓ/parenleftbigα+ik 2/parenrightbig Γ(−α−ik 2) Γ/parenleftbig1 2+ik/parenrightbig,x ≥0, (7.2.62) φ(x)=0,x < 0. Forx>0, we close the contour in the upper half-plane, picking up the pole contributions from Γ/parenleftbigα+ik 2/parenrightbig andΓ/parenleftbig −α−ik 2/parenrightbig . Poles of Γ/parenleftbigα+ik 2/parenrightbig are located at k=i(2n+α),n=0,1,2,···. Poles of Γ/parenleftbig −α−ik 2/parenrightbig are located at k=i(2n−α),n=0,1,2,···.S i n c e Res.Γ(z)|z=−n=(−1)n1 n!, (7.2.63) we have φ(x)=C/prime/prime∞/summationdisplay n=0/braceleftBigg e−(2n+α)x2−(2n+α)(−1)n n!Γ(−n−α) Γ/parenleftbig1 2−2n−α/parenrightbig+(α→−α)/bracerightBigg .(7.2.64) Since Γ(z)=π sinπz1 Γ(1−z), 206 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation we have Γ(−n−α)=(−1)n+1π sinαπ1 Γ(n+1+ α), and with the use of the Duplication formula, Γ/parenleftbigg1 2−2n−α/parenrightbigg =π cosαπ·√ 2π2−(2n+α)·1 Γ/parenleftbig1 4+α 2+n/parenrightbig Γ/parenleftbig3 4+α 2+n/parenrightbig, we have Γ(−n−α) Γ/parenleftbig1 2−2n−α/parenrightbig=(−1)n+1 √ 2π·2(2n+α)·cosαπ sinαπ·Γ/parenleftbig1 4+α 2+n/parenrightbig Γ/parenleftbig3 4+α 2+n/parenrightbig Γ(1 + α+n). Our solution φ(x)is given by φ(x)=C/prime/prime/prime/parenleftBigcosαπ sinαπ/parenrightBig ·∞/summationdisplay n=0/braceleftBigg e−αx/parenleftbig e−2x/parenrightbign n!Γ(1 4+α 2+n)Γ/parenleftbig3 4+α 2+n/parenrightbig Γ( 1+ α+n)−(α→−α)/bracerightBigg .(7.2.65) We recall that the hypergeometric function F(a, b, c;z)is given by F(a, b, c;z)=∞/summationdisplay n=0Γ(α+n) Γ(a)·Γ(b+n) Γ(b)·Γ(c) Γ(c+n)·zn n!. (7.2.66) Setting a=1 4+α 2,b =3 4+α 2,c =1+ α, we have φ(x)=C/prime/prime/prime/parenleftBigcosαπ sinαπ/parenrightBig/bracketleftbigg e−αxΓ/parenleftbig1 4+α 2/parenrightbig Γ/parenleftbig3 4+α 2/parenrightbig Γ( 1+ α) ·F/parenleftbigg1 4+α 2,3 4+α 2,1+α;e−2x/parenrightbigg −(α→−α)/bracketrightbigg .(7.2.67) Recall that the Legendre function of the second kind Qα−1 2(z)is given by Qα−1 2(z)=√ π 2α+1 2·Γ/parenleftbig1 2+α/parenrightbig Γ(1 + α)·z−(1 2+α)·F/parenleftbigg1 4+α 2,3 4+α,1+α;z−2/parenrightbigg =1 2·Γ/parenleftbig1 4+α 2/parenrightbig Γ(3 4+α 2) Γ(1 + α)·z−(1 2+α)·F/parenleftbigg1 4+α 2,3 4+α 2,1+α;z−2/parenrightbigg ,(7.2.68) 7.3 General Decomposition Problem 207 so that our solution given above can be simpliﬁed as φ(x)=C/prime/prime/prime/parenleftBigcosαπ sinαπ/parenrightBig ex 2/braceleftBig Qα−1 2(ex)−Q−α−1 2(ex)/bracerightBig . (7.2.69) Recall that the Legendre function of the ﬁrst kind Pβ(z)is given by Pβ(z)=1 π/parenleftbiggsinβπ cosβπ/parenrightbigg {Qβ(z)−Q−β−1(z)}, (7.2.70) so that Pα−1 2(z)=−1 π/parenleftBigcosαπ sinαπ/parenrightBig/braceleftBig Qα−1 2(z)−Q−α−1 2(z)/bracerightBig . (7.2.71) So, the ﬁnal expression for φ(x)is given by φ(x)=C·/parenleftBig expx 2/parenrightBig ·Pα−1 2(ex),x≥0,0≤α<1/2,2πλ=c o s απ. (7.2.72) A similar analysis can be carried out for the cases, 2πλ > 1,a n dλ<0. We only list the ﬁnal answers for all cases. Summary of Example 7.4 Case 1. 0<2πλ≤1,2πλ=c o s απ,0≤α<1/2. φ(x)=C1·/parenleftBig expx 2/parenrightBig ·Pα−1 2(ex),x ≥0. Case 2. 2πλ > 1,2πλ= cosh απ,α>0. φ(x)=C2·/parenleftBig expx 2/parenrightBig ·Piα−1 2(ex),x ≥0. Case 3. 2πλ≤0. φ(x)=0,x ≥0. 7.3 General Decomposition Problem In the original Wiener–Hopf problem we examined, in Section 7.1, φ−(k)=ψ+(k)+F(k), (7.3.1) we need to make the decomposition, F(k)=F+(k)+F−(k). (7.3.2) 208 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation In the problems we just examined in Section 7.1, i.e., the homogeneous Wiener–Hopf integral equation of the second kind, we need to make the decomposition, 1−λˆK(k)=Y−(k) Y+(k). (7.3.3) Here we discuss how this can be done in general, rather than by inspection. Consider the ﬁrst problem ( sum-splitting ) ﬁrst. (Figure 7.12.) φ−(k)=ψ+(k)+F(k). Assume that φ−(k),ψ+(k),F(k)→0 ask→∞. (7.3.4) kk2 k1 τ+τ− Fig. 7.12: Sum-splitting of F(k).φ−(k)is analytic in the lower half-plane, Imk<τ −. ψ+(k)is analytic in the upper half-plane, Imk>τ +.F(k)is analytic inside the strip, τ+<Imk<τ −. Examine the decomposition of F(k).S i n c e F(k)is analytic inside the strip, τ+<Imk=k2<τ−, (7.3.5) by the Cauchy integral formula, we have F(k)=1 2πi/integraldisplay CF(ζ) ζ−kdζ (7.3.6) where the complex integration contour Cconsists of the following path as in Figure 7.13, C=C1+C2+C↑+C↓. (7.3.7) 7.3 General Decomposition Problem 209 kk2 k1k2= −τ k2=+τC1C2 C ↑C ↓ Fig. 7.13: Sum-splitting contour Cof Eq. (7.3.6) for F(k)inside the strip, τ+<Imk<τ −. The contributions from C↑andC↓vanish as these contours tend to inﬁnity, since |F(ζ)|is bounded (actually →0), /vextendsingle/vextendsingle/vextendsingle/vextendsingle1 (ζ−k)/vextendsingle/vextendsingle/vextendsingle/vextendsingle→0 asζ→∞. Thus we have F(k)=1 2πi/integraldisplay C1F(ζ) ζ−kdζ+1 2πi/integraldisplay C2F(ζ) ζ−kdζ, (7.3.8) where the contribution from C1is a+function, analytic for Imk=k2>τ+, while the contribution from C2is a−function, analytic for Imk=k2<τ−, i.e., F+(k)=1 2πi/integraldisplay+∞+iτ+ −∞+iτ+F(ζ) ζ−kdζ, (7.3.9a) F−(k)=1 2πi/integraldisplay+∞+iτ− −∞+iτ−F(ζ) ζ−kdζ. (7.3.9b) Consider now the factorization of 1−λˆK(k)into a ratio of the −function to the +func- tion . The function 1−λˆK(k)is analytic inside the strip, −a<Imk=k2<b , (7.3.10) and the inversion contour is somewhere inside the strip, −a<Imk=k2<−a+ε. (7.3.11) 210 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation The analytic function 1−λˆK(k)may have some zeros inside the strip, −a<Imk=k2<b . Choose a rectangular contour, as indicated in Figure 7.14, below all zeros of 1−λˆK(k) inside the strip, −a<Imk=k2<b . kk2 k1 C1C2 C ↑C ↓ kA2=kb2= k a2=− Fig. 7.14: Rectangular contour for the factorization of 1−λˆK(k). This contour is chosen below all the zeros of 1−λˆK(k)inside the strip, −a<Imk<b . Note if 1−λˆK(k)has a zero on k2=−a, it is all right, since it just remains in the lower half-plane. The inversion contour k2=Awill be chosen inside this rectangle. Now, 1−λˆK(k)is analytic inside the rectangle C1+C↑+C2+C↓ and has no zeros inside this rectangle. Also since ˆK(k)→0 ask→∞, we know 1−λˆK(k)→1 ask→∞. In order to express 1−λˆK(k)as the ratio Y−(k)/Y+(k), we take the logarithm of (7.3.3) to ﬁnd ln/bracketleftbig 1−λˆK(k)/bracketrightbig =l n/bracketleftbiggY−(k) Y+(k)/bracketrightbigg =l nY−(k)−lnY+(k). (7.3.12) 7.3 General Decomposition Problem 211 Now,ln[1−λˆK(k)]is itself analytic in the rectangle (because it has no branch points since 1−λˆK(k)has no zeros there), so we can apply the Cauchy integral formula ln/bracketleftbig 1−λˆK(k)/bracketrightbig =1 2πi/integraldisplay Cln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ, withkinside the rectangle and Cconsisting of C1+C2+C↑+C↓. Thus we write ln/bracketleftbig 1−λˆK(k)/bracketrightbig =l nY−(k)−lnY+(k) =1 2πi/integraldisplay C1ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ−1 2πi/integraldisplay −C2ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ +1 2πi/integraldisplay C↑+C↓ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ.(7.3.13) In Eq. (7.3.13), it is tempting to drop the contributions from C↑andC↓altogether. It is, however, not always possible to do so. Because of the multivaluedness of the logarithm, we may have, in the limit |ζ|→∞ , ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig →lne2πin=2πin, (n=0,±1,±2,···) (7.3.14) and we have no guarantee that the contributions from C↑andC↓cancel each other. In other words, 1−λˆK(ζ)may develop a phase angle as ζranges from −∞+iA to+∞+iA. Deﬁnition of Wiener–Hopf index . Let us deﬁne the indexνof1−λˆK(ζ)by ν≡1 2πiln/bracketleftBig 1−λˆK(ζ)/bracketrightBig/vextendsingle/vextendsingle/vextendsingleζ=+∞+iA ζ=−∞+iA. (7.3.15) Graphically we do the following: Plot z=[ 1−λˆK(ζ)]asζranges from −∞+iAto+∞+iA in the complex zplane, and count the number of counter-clockwise revolutions zmakes about the origin. The index νis equal to the number of these revolutions. We shall now examine the properties of the index ν; in particular, a relationship between the index νand the zeros and the poles of 1−λˆK(k)in the complex kplane . Suppose 1−λˆK(k)has a zero in the upper half-plane, say, 1−λˆK(k)=k−zu,Imzu>−a. Then the contribution from this zuto the index νis ν(zu)=1 2πi[0−(−iπ)] =1 2. 212 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Similar analysis yields the following results: zero in the upper half-plane ⇒ν(zu)=+1 2, pole in the upper half-plane ⇒ν(pu)=−1 2, zero in the lower half-plane ⇒ν(zl)=−1 2, pole in the lower half-plane ⇒ν(pl)=+1 2.(7.3.16) In many cases, the translation kernel K(x−y)is of the form K(x−y)=K(|x−y|). (7.3.17) Then ˆK(k)is even in k, ˆK(k)=ˆK(−k), (7.3.18) so that 1−λˆK(k)(which is even) has an equal number of zeros (poles) in the upper half-plane and in the lower half-plane, number of zu=number of zl,number of pu=number of pl. Thus the index of 1−λˆK(k)on the real line is equal to zero ( ν≡0)f o rˆK(k)even . Suppose we now lift the path above the real line ( Imk=0 ) into the upper half-plane. As the path C(Imk=A) passes by a zero of 1−λˆK(k)inImk>0, the index νof1−λˆK(k) with respect to the path C(Imk=A) decreases by 1. This is because the point k=z0is the zero in the upper half-plane with respect to the path C<(Imk=A−) while it is the zero in the lower half-plane with respect to the path C>(Imk=A+), and hence /triangleν=ν(zl)−ν(zu)=−1 2−1 2=−1. (7.3.19a) Likewise, for a pole of 1−λˆK(k)inImk>0,w eﬁ n d ∆ν=ν(pl)−ν(pu)=+1 2+1 2=+ 1. (7.3.19b) Consider ﬁrst the case when the index νis equal to zero , ν=0. (7.3.20) Then we choose a branch so that ln[1−λˆK(ζ)]vanishes on C↑andC↓. We then have lnY−(k)−lnY+(k) =1 2πi/integraldisplay C1ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ−1 2πi/integraldisplay −C2ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ. (7.3.21) 7.3 General Decomposition Problem 213 In the ﬁrst integral on the second line of Eq. (7.3.21), we may let kbe anywhere above C1 where C1is arbitrarily close to Imk=k2=−afrom above. Then we conclude that the integral 1 2πi/integraldisplay C1ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ, Imk=k2>−a, is analytic in the upper half-plane, and hence is identiﬁed to be a +function, lnY+(k)=−1 2πi/integraldisplay C1ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ, Imk=k2>−a. (7.3.22) It also vanishes as |k|→∞ in the upper half-plane ( Imk>−a). In the second integral on the second line of Eq. (7.3.21), we may let kbe anywhere below −C2where−C2is arbitrarily close to Imk=k2=−afrom above. Then we conclude that the integral 1 2πi/integraldisplay −C2ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ, Imk=k2≤−a, is analytic in the lower half-plane, and hence is identiﬁed to be a −function, lnY−(k)=−1 2πi/integraldisplay −C2ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ, Imk=k2≤−a. (7.3.23) It also vanishes as |k|→∞ in the lower half-plane ( Imk≤−a). Thus Y+(k)=e x p/bracketleftBigg −1 2πi/integraldisplay C1ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ/bracketrightBigg , (7.3.24) Y−(k)=e x p/bracketleftBigg −1 2πi/integraldisplay −C2ln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−kdζ/bracketrightBigg . (7.3.25) We also note that Y±(k)→1 as|k|→∞ in/braceleftBigg Imk>−a, Imk≤−a.(7.3.26) Then the entire function G(k)in the following equation, Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)=G(k), must vanish identically, by Liouville’s theorem. Hence ˆφ−(k)=0 orφ(x)=0 when ν=0. (7.3.27) 214 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Consider next the case when index νispositive , ν>0. (7.3.28) Instead of dealing with C↑andC↓of the integral (7.3.13), we construct the object whose index is equal to zero , ν/productdisplay i=1/parenleftbiggk−zl(i) k−pu(i)/parenrightbigg/bracketleftbig 1−λˆK(k)/bracketrightbig =Z−(k) Z+(k), (7.3.29) where zl(i)is a point in the lower half-plane ( Imk≤A) which contributes −ν 2in its totality (i=1 ,···,ν) to the index and pu(i)is a point in the upper half-plane ( Imk>A )w h i c h contributes −ν 2in its totality ( i=1 ,···,ν) to the index. Then the expression (7.3.29) has the index equal to zero with respect to Imk=A, −ν 2(fromzl(i)/primes)−ν 2(frompu(i)/primes)+ν(from1−λˆK(k)) = 0 . (7.3.30) By factoring of Eq. (7.3.29), using Eqs. (7.3.24) and (7.3.25), we obtain Z−(k)=e x p −1 2πi/integraldisplay −C2ln/bracketleftBigg (1−λˆK(ζ))ν/productdisplay i=1/parenleftBig ζ−zl(i) ζ−pu(i)/parenrightBig/bracketrightBigg ζ−kdζ , (7.3.31) Z+(k)=e x p −1 2πi/integraldisplay C1ln/bracketleftBigg (1−λˆK(ζ))ν/productdisplay i=1/parenleftBig ζ−zl(i) ζ−pu(i)/parenrightBig/bracketrightBigg ζ−kdζ , (7.3.32) with the properties, (1)Z ±(k)→1 as|k|→∞ , (2)Z−(k)(Z+(k)) is analytic in the lower (upper) half-plane, (3)Z−(k)(Z+(k)) has no zero in the lower (upper) half-plane. We write Eq. (7.3.29) as 1−λˆK(k)=Y−(k) Y+(k)=Z−(k) Z+(k)·/producttextν i=1(k−pu(i)) /producttextν i=1(k−zl(i)). (7.3.33) By the formula stated in Eq. (7.2.17), we obtain Y−(k)=Z−(k)·ν/productdisplay i=1(k−pu(i)), (7.3.34) Y+(k)=Z+(k)·ν/productdisplay i=1(k−zl(i)). (7.3.35) 7.3 General Decomposition Problem 215 We observe that Y±(k)→kνas|k|→∞ . (7.3.36) Thus the entire function G(k)in the following equation, Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)=G(k), cannot grow as fast as kνask→∞ . By Liouville’s theorem, we have G(k)=ν−1/summationdisplay j=0Cjkj,0≤j≤ν−1, (7.3.37) where the Cjare arbitrary νconstants. Then we obtain ˆφ−(k)=G(k) Y−(k)=ν−1/summationdisplay j=0Cjkj Y−(k),Imk≤A. (7.3.38) Inverting this expression along Imk=A, we obtain φ(x)=1 2πi/integraldisplay+∞+iA −∞+iAdk eikxˆφ−(k)=ν−1/summationdisplay j=0Cjφ(j)(x), when ν>0, (7.3.39) where φ(j)(x)=1 2πi/integraldisplay+∞+iA −∞+iAdkeikxkj Y−(k),j =0,···,ν−1. (7.3.40) We have νindependent homogeneous solutions ,φ(j)(x),j=0 ,···,ν−1, which are related to each other by differentiation, /parenleftbigg −id dx/parenrightbigg φ(j)(x)=φ(j+1)(x),0≤j≤ν−2. Thus it is sufﬁcient to compute φ(0)(x), φ(j)(x)=/parenleftbigg −id dx/parenrightbiggj φ(0)(x),j =0,1,···,ν−1. (7.3.41) Note that the differentiation under the integral, Eq. (7.3.40), is justiﬁed by kj+1 Y−(k)→kj+1 kν→0 ask→∞,j≤ν−2, so that the integral converges. Consider thirdly the case when the index νisnegative , ν<0. (7.3.42) 216 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation As before, we construct the object whose index is equal to zero . |ν|/productdisplay i=1(k−zu(i)) (k−pl(i))·/bracketleftbig 1−λˆK(k)/bracketrightbig =Z−(k) Z+(k), (7.3.43) which does indeed have an index of zero as shown below. +|ν| 2(fromzu(i)/primes)+|ν| 2(frompl(i)/primes)+ν(from1−λˆK(k)) = 0 . (7.3.44) We apply the factorization to the left-hand side of Eq. (7.3.43). Then we write 1−λˆK(k)=Y−(k) Y+(k)=Z−(k) Z+(k)·/producttext|ν| i=1(k−pl(i)) /producttext|ν| i=1(k−zu(i)). (7.3.45) By the formula stated in Eq. (7.2.17), we obtain Y−(k)=Z−(k) /producttext|ν| i=1(k−zu(i)), (7.3.46) Y+(k)=Z+(k) /producttext|ν| i=1(k−pl(i)). (7.3.47) Then we have Z±(k)→1 andY±(k)→1 k|ν|, ask→∞. (7.3.48) Thus the entire function G(k)in the following equation, Y−(k)ˆφ−(k)=−Y+(k)ˆψ+(k)=G(k), must vanish identically by Liouville’s theorem. Hence we obtain φ(x)=0, when ν<0. (7.3.49) 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind Let us consider the inhomogeneous Wiener–Hopf integral equation of the second kind , φ(x)=f(x)+λ/integraldisplay+∞ 0K(x−y)φ(y)dy, x ≥0, (7.4.1) where we assume, as in Section 7.3, that the asymptotic behavior of the kernel K(x)is given by K(x)∼/braceleftBigg O(eax) asx→− ∞ , O(e−bx) asx→+∞,a,b > 0, (7.4.2) 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 217 and the asymptotic behavior of the inhomogeneous term f(x)is given by f(x)→O(ecx) asx→+∞. (7.4.3) We deﬁne ψ(x)forx<0as before ψ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy, x < 0. (7.4.4) We take the Fourier transform of φ(x)andψ(x)forx≥0andx<0and add the results together, ˆφ−(k)+ˆψ+(k)=ˆf−(k)+λˆK(k)ˆφ−(k), (7.4.5) where ˆK(k)is analytic inside the strip, −a<Imk=k2<b . (7.4.6) Trouble may arise when the inhomogeneous term f(x)grows too fast as x→∞ so that there may not exist a common region of analyticity for Eq. (7.4.5) to hold. The Fourier transform ˆf−(k)is deﬁned by ˆf−(k)=/integraldisplay+∞ 0dx e−ikxf(x), (7.4.7) where /vextendsingle/vextendsinglee−ikxf(x)/vextendsingle/vextendsingle∼e(k2+c)xasx→∞ with k=k1+ik2. That is, ˆf−(k)is analytic in the lower half-plane, Imk=k2<−c. (7.4.8) We require that aandcsatisfy a>c . (7.4.9) In other words, f(x)grows at most as fast as f(x)∼e(a−ε)x,ε > 0, asx→∞. We try to solve Eq. (7.4.5) in the narrower strip, −a<Imk=k2<min(−c, b). (7.4.10) Writing Eq. (7.4.5) as (1−λˆK(k))ˆφ−(k)=ˆf−(k)−ˆψ+(k), (7.4.11) 218 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation we are content to obtain one particular solution of Eq. (7.4.1). We factorize 1−λˆK(k)as before, 1−λˆK(k)=Y−(k) Y+(k). (7.4.12) Thus we have from Eqs. (7.4.11) and (7.4.12), Y−(k)ˆφ−(k)=Y+(k)ˆf−(k)−Y+(k)ˆψ+(k), (7.4.13) where Y−(k)ˆφ−(k)is analytic in the lower half-plane and Y+(k)ˆψ+(k)is analytic in the upper half-plane. We split Y+(k)ˆf−(k)into a sum of two functions, one analytic in the upper half-plane and the other analytic in the lower half-plane, Y+(k)ˆf−(k)=(Y+(k)ˆf−(k))++(Y+(k)ˆf−(k))−. In order to do this, we must construct Y+(k)such that Y+(k)ˆf−(k)→0 ask→∞, or, Y+(k)→constant as k→∞. (7.4.14) Suppose ˆF(k)is analytic inside the strip, −a<Imk=k2<min(−c, b). (7.4.15) By choosing the contour Cinside the strip as in Figure 7.15, we can then apply the Cauchy integral formula. ˆF(k)=1 2πi/integraldisplay CˆF(ζ) ζ−kdζ=1 2πi/integraldisplay C1ˆF(ζ) ζ−kdζ−1 2πi/integraldisplay −C2ˆF(ζ) ζ−kdζ. (7.4.16) By the same argument as in the previous section, we identify ˆF−(k)=−1 2πi/integraldisplay −C2ˆF(ζ) ζ−kdζ, (7.4.17) ˆF+(k)=1 2πi/integraldisplay C1ˆF(ζ) ζ−kdζ. (7.4.18) Thus, under the assumption that the Y+(k)satisfy the above-stipulated condition (7.4.14), we obtain (Y+(k)ˆf−(k))−=−1 2πi/integraldisplay −C2/bracketleftbiggY+(ζ)ˆf−(ζ) (ζ−k)/bracketrightbigg dζ, (7.4.19) (Y+(k)ˆf−(k))+=1 2πi/integraldisplay C1/bracketleftbiggY+(ζ)ˆf−(ζ) (ζ−k)/bracketrightbigg dζ. (7.4.20) 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 219 ka2=−k b 2= k c2=−kk2 k1 C1C2 C↑ C↓ Fig. 7.15: Region of analyticity and the integration contour CforˆF(k)inside the strip, −a<Imk<min(−c, b). Then we write Y−(k)ˆφ−(k)−(Y+(k)ˆf−(k))−=(Y+(k)ˆf−(k))+−Y+(k)ˆψ+(k)≡G(k),(7.4.21) where G(k)is entire in k. If we are looking for the most general homogeneous solutions, we set ˆf−(k)≡0and determine the most general form of the entire function G(k).N o w we are just looking for one particular solution to the inhomogeneous equation ,s ot h a tw es e t G(k)=0 . Then we have ˆφ−(k)=(Y+(k)ˆf−(k))− Y−(k), (7.4.22) ˆψ+(k)=(Y+(k)ˆf−(k))+ Y+(k). (7.4.23) Choices of Y±(k)for an inhomogeneous solution are different from those for an homogeneous solution. In view of Eqs. (7.4.22) and (7.4.23), we require that (1)1/Y−(k)is analytic in the lower half-plane, Imk<c/prime,−a<c/primec/prime/prime,−a<c/prime/prime<b . (7.4.25) (2) ˆφ−(k)→0,ˆψ+(k)→0 ask→∞. (7.4.26) 220 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation According to this requirement, Y−(k)for an inhomogeneous solution can have a pole in the lower half-plane. This is all right because then 1/Y−(k)has a zero in the lower half-plane. Once requirements (1) and (2) are satisﬁed, ˆφ−(k)andˆψ+(k), given by Eq. (7.4.22) and (7.4.23), are analytic in the respective half-plane. Then we construct the following expressionfrom Eqs. (7.4.22) and (7.4.23). [1−λˆK(k)]ˆφ −(k)=ˆf−(k)−ˆψ+(k). (7.4.27) Inverting for x≥0, we obtain φ(x)−λ/integraldisplay+∞ 0K(x−y)φ(y)dy=f(x),x≥0. Thusˆφ−(k)andˆψ+(k)derived in Eqs. (7.4.22) and (7.4.23) under the requirements (1) and (2) do provide a particular solution to Eq. (7.4.1). Case 1. Index ν=0 . When the index νof1−λˆK(k)with respect to the line Imk=Ais equal to zero ,n o nontrivial homogeneous solution exists. From Eqs. (7.3.22) and (7.3.23), we have 1−λˆK(k)=Y−(k) Y+(k), (7.4.28) where Y−(k)=e x p/parenleftBigg −1 2πi/integraldisplay −C2dζln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−k/parenrightBigg ,Imk≤A, (7.4.29) Y+(k)=e x p/parenleftBigg −1 2πi/integraldisplay C1dζln/bracketleftbig 1−λˆK(ζ)/bracketrightbig ζ−k/parenrightBigg ,Imk>A , (7.4.30) and Y±(k)→1 as|k|→∞ . (7.4.31) Since Y+(k)(Y−(k)) has no zeros in the upper half-plane (the lower half-plane), 1/Y+(k) (1/Y−(k)) is analytic in the upper half-plane (the lower half-plane). Then we have Y+(k)ˆf−(k)→0 ask→∞, so that Y+(k)ˆf−(k)can be split up into the +part and the −part as in Eqs. (7.4.19) and (7.4.20). Using Y±(k)given for the ν≡0case, Eqs. (7.4.29) and (7.4.30), we construct a resolvent kernel H(x, y).S i n c e 1/Y−(k)is analytic in the lower plane and approaches 1ask→∞ , we deﬁne y−(x)by 1 Y−(k)−1≡/integraldisplay+∞ 0dx e−ikxy−(x), (7.4.32) 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 221 where the left-hand side is analytic in the lower half-plane and vanishes as k→∞ . Inverting Eq. (7.4.32) for y−(x),w eh a v e y−(x)=/integraldisplay+∞+iA −∞+iAdk 2πeikx/bracketleftbigg1 Y−(k)−1/bracketrightbigg forx≥0, (7.4.33) y−(x)=0 forx<0. (7.4.34) Similarly we deﬁne y+(x)by Y+(k)−1≡/integraldisplay0 −∞dx e−ikxy+(x), (7.4.35) where the left-hand side is analytic in the upper half-plane and vanishes as k→∞ . Inverting Eq. (7.4.35) for y+(x),w eh a v e y+(x)=/integraldisplay+∞+iA −∞+iAdk 2πeikx[Y+(k)−1] forx<0, (7.4.36) y+(x)=0 forx≥0. (7.4.37) We deﬁne ˆy±(k)by 1 Y−(k)≡1+/integraldisplay+∞ 0dx e−ikxy−(x)≡1+ˆy−(k), (7.4.38) Y+(k)≡1+/integraldisplay0 −∞dx e−ikxy+(x)≡1+ˆy+(k). (7.4.39) Thenˆφ−(k)given by Eq. (7.4.23) becomes ˆφ−(k)=1 Y−(k)(Y+(k)ˆf−(k))−=( 1+ˆ y−(k))(ˆf−(k)+ˆy+(k)ˆf−(k))− =ˆf−(k)+ˆy−(k)ˆf−(k)+(ˆy+(k)ˆf−(k))−+ˆy−(k)(ˆy+(k)ˆf−(k))−.(7.4.40) Inverting Eq. (7.4.40) for x>0, φ(x)=f(x)+/integraldisplay+∞ 0y−(x−y)f(y)dy+/integraldisplay+∞ 0y+(x−y)f(y)dy +/integraldisplay+∞ 0y−(x−z)dz/integraldisplay+∞ 0y+(z−y)f(y)dy =f(x)+/integraldisplay+∞ 0H(x, y)f(y)dy, x ≥0,(7.4.41) where H(x, y)≡y−(x−y)+y+(x−y)+/integraldisplay+∞ 0y−(x−z)y+(z−y)dz. (7.4.42) 222 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation It is noted that the existence of the resolvent kernel H(x, y)given above is solely due to the analyticity of 1/Y−(k)in the lower half-plane and that of Y+(k)in the upper half-plane. Thus, when the index ν=0 ,w eh a v ea unique solution, solely consisting of a single particular solution to Eq. (7.4.11). Case 2. Index ν>0. When the index νispositive ,w eh a v e νindependent homogeneous solutions given by Eq. (7.3.40). We observed in Section 7.3 that Y±(k)→kνask→∞, (7.4.43) where Y±(k)are given by Eqs. (7.3.34) and (7.3.35). On the other hand, in solving for a particular solution, we want Y±(k)to be such that: (1)1/Y−(k)(Y+(k)) is analytic in the lower half-plane (the upper half-plane), Y±(k)→1 as|k|→∞ . (7.4.44) We construct W±(k)as W±(k)=Y±(k) /producttextν j=1(k−pl(j)),Impl(j)≤A,1≤j≤ν, (7.4.45) where the locations of the pl(j)are quite arbitrary as long as Impl(j)≤A. We notice that the W±(k)satisfy requirements ( 1) and ( 2): (2) 1 W−(k)=/producttextν j=1(k−pl(j)) Y−(k) is analytic in the lower half-plane, while W+(k)=Y+(k) /producttextν j=1(k−pl(j)) is analytic in the upper half-plane; (3) W±(k)→1 as|k|→∞ . (7.4.46) Thus we use W±(k), Eq. (7.4.45), instead of Y±(k), in the construction of the resolvent H(x, y). 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 223 Case 3. Index ν<0. When index νisnegative , we have no nontrivial homogeneous solution. From Eqs. (7.3.46) and (7.3.47), we have Y±(k)→1 k|ν|as|k|→∞ . (7.4.47) Then we have 1 Y−(k)→k|ν|as|k|→∞ , (7.4.48) while Y+(k)ˆf−(k)→0 as|k|→∞ . (7.4.49) By Liouville’s theorem, ˆφ−(k)can grow, at most, as fast as k|ν|−1as|k|→∞ , ˆφ−(k)=(Y+(k)ˆf−(k))− Y−(k)∼k|ν|−1as|k|→∞ . (7.4.50) In general, we have ˆφ−(k)/notarrowright0 as|k|→∞ , so that a particular solution to the inhomogeneous problem may not exist. There are some exceptions to this. We analyze (Y+(k)ˆf−(k))−more carefully. We know (Y+(k)ˆf−(k))−=−1 2πi/integraldisplay −C2Y+(ζ)ˆf−(ζ) ζ−kdζ. (7.4.51) Expanding 1/(ζ−k)in power series of ζ/k , 1 (ζ−k)=−/parenleftbigg1 k/parenrightbigg/parenleftbigg 1+ζ k+ζ2 k2+···+ζ|ν|−1 k|ν|−1+···/parenrightbigg ,/vextendsingle/vextendsingle/vextendsingle/vextendsingleζ k/vextendsingle/vextendsingle/vextendsingle/vextendsingle<1, we write (Y+(k)ˆf−(k))−=1 2πi/integraldisplay −C21 k∞/summationdisplay j=0/parenleftbiggζ k/parenrightbiggj Y+(ζ)ˆf−(ζ)dζ =1 k∞/summationdisplay j=01 kj1 2πi/integraldisplay −C2ζjY+(ζ)ˆf−(ζ)dζ.(7.4.52) 224 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation In view of Eqs. (7.4.22) and (7.4.52), we realize that ˆφ−(k)=(Y+(k)ˆf−(k))− Y−(k)→0 as|k|→∞ , (7.4.53) if and only if 1 2πi/integraldisplay −C2ζjY+(ζ)ˆf−(ζ)dζ=0,j=0,···,|ν|−1. (7.4.54) If this condition is satisﬁed, we get from the j=|ν|term onwards, ˆφ−(k)→C k1+|ν|as|k|→∞ , (7.4.55) so that ˆφ−(k)can be inverted for φ(x), which is the unique solution to the inhomogeneous problem. To understand this solvability condition (7.4.54), we ﬁrst recall the Parseval identity , /integraldisplay+∞ −∞dkˆh(k)ˆg(−k)=2π/integraldisplay+∞ −∞h(y)g(y)dy, (7.4.56) where ˆh(k)andˆg(k)are the Fourier transforms of h(y)andg(y), respectively. Then we consider the homogeneous adjoint problem . Recall that for a real kernel, Kadj(x, y)=K(y,x). Thus, corresponding to the original homogeneous problem, φ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy, there exists the homogeneous adjoint problem, φadj(x)=λ/integraldisplay+∞ 0K(y−x)φadj(y)dy, (7.4.57) whose translation kernel is related to the original one by Kadj(ξ)=K(−ξ). (7.4.58) Now, when we take the Fourier transform of the homogeneous adjoint problem, we ﬁnd /bracketleftBig 1−λˆK(−k)/bracketrightBig ˆφadj −(k)=−ˆψadj +(k), (7.4.59) where the only difference from the original equation is the sign of kinside ˆK(−k).H o w e v e r , since1−λˆK(−k)is just the reﬂection of 1−λˆK(k)through the origin, a zero of 1−λˆK(k)in the upper half-plane corresponds to a zero of 1−λˆK(−k)in the lower half-plane, etc. Thus, when the original 1−λˆK(k)has a negative index ν<0with respect to a line, Imk=k2=A, 7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 225 the homogeneous adjoint problem 1−λˆK(−k)has a positive index |ν|relative to the line, Imk=k2=−A. So, in that case, although the original problem may have no solutions, the homogeneous adjoint problem has |ν|independent solutions. Now 1−λˆK(k)was found to have the decomposition, 1−λˆK(k)=Y−(k) Y+(k), with Y±(k)→1 k|ν|ask→∞, we conclude that 1−λˆK(−k)=Y−(−k) Y+(−k)=Yadj −(k) Yadj +(k), (7.4.60) from which, we recognize that   Yadj −(k)=1 Y+(−k)is analytic in the lower half-plane, →k|ν|, Yadj +(k)=1 Y−(−k)is analytic in the upper half-plane, →k|ν|. Thus the homogeneous adjoint problem reads Yadj −(k)ˆφadj −(k)=−Yadj +(k)ψadj +(k)≡G(k), (7.4.61) with G(k)=C0+C1k+···+C|ν|−1k|ν|−1. (7.4.62) We then know that the |ν|independent solutions to the homogeneous adjoint problem are of the form, ˆφadj −(k)=/braceleftBigg 1 Yadj −(k),k Yadj −(k),···,k|ν|−1 Yadj −(k)/bracerightBigg , which is equivalent to ˆφadj −(k)={Y+(−k),kY+(−k),···,k|ν|−1Y+(−k)}. (7.4.63) Therefore, to within a constant factor, which is irrelevant, we can write the solvability condi- tion (7.4.54) as /integraldisplay −C2ζjY+(ζ)ˆf−(ζ)dζ=0,j =0,1,···,|ν|−1, (7.4.64) 226 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation which is equivalent to /integraldisplay −C2ˆφadj −(j)(−ζ)ˆf−(ζ)dζ=0,j =0,1,···,|ν|−1, (7.4.65) where ˆφadj −(j)(ζ)≡ζjY+(−ζ),j =0,1,···,|ν|−1. (7.4.66) By the Parseval identity (7.4.56), the solvability condition (7.4.65) can now be written as /integraldisplay+∞ 0φadj (j)(x)f(x)dx=0,j =0,1,···,|ν|−1, (7.4.67) where φadj (j)(x)=1 2π/integraldisplay+∞−iA −∞− iAeiζxˆφadj −(j)(ζ)dζ, j =0,1,···,|ν|−1,x≥0.(7.4.68) Namely, if and only if the inhomogeneous term f(x)is orthogonal to all of the homogeneous solutions φadj(x)of the homogeneous adjoint problem, the inhomogeneous equation (7.4.1) has a unique solution, when the index νis negative . Summary of Wiener–Hopf integral equation φ(x)=λ/integraldisplay∞ 0K(x−y)φ(y)dy, x ≥0, φadj(x)=λ/integraldisplay∞ 0K(y−x)φadj(y)dy, x ≥0, φ(x)=f(x)+λ/integraldisplay∞ 0K(x−y)φ(y)dy, x ≥0. 1. Index ν=0 : The homogeneous problem and its homogeneous adjoint problem have no solutions. The inhomogeneous problem has a unique solution. 2. Index ν>0: The homogeneous problem has νindependent solutions and its homogeneous adjoint problem has no solutions. The inhomogeneous problem has nonunique solutions (but there are no solvability con- ditions). 3. Index ν<0: The homogeneous problem has no solutions, and its homogeneous adjoint problem has |ν|independent solutions. The inhomogeneous problem has a unique solution, if and only if the inhomogeneous term is orthogonal to all |ν|independent solutions to the homogeneous adjoint problem. 7.5 T oeplitz Matrix and Wiener–Hopf Sum Equation 227 7.5 Toeplitz Matrix and Wiener–Hopf Sum Equation In this section, we consider the application of the Wiener–Hopf method to the inﬁnite system of the inhomogeneous linear algebraic equation , M/vectorX=/vectorf, (7.5.1) or, /summationdisplay mMnmXm=fn, (7.5.2) where the coefﬁcient matrix Misreal and has the Toeplitz structure , Mnm=Mn−m. (7.5.3) We solve Eq. (7.5.1) for two cases. Case A. Inﬁnite Toeplitz matrix . LetM be an inﬁnite matrix. Then the system of the inﬁnite inhomogeneous linear alge- braic equation (7.5.1) becomes ∞/summationdisplay m=−∞Mn−mXm=fn,−∞<n< ∞. (7.5.4) We look for the solution {Xm}+∞ m=−∞ assuming the uniform convergence of{Xm}+∞ m=−∞ and{Mm}+∞ m=−∞ , ∞/summationdisplay m=−∞|Xm|<∞, and∞/summationdisplay m=−∞|Mm|<∞. (7.5.5) Multiplying ξnon Eq. (7.5.4), and summing over n,w eh a v e /summationdisplay n,mξnMn−mXm=/summationdisplay nξnfn. (7.5.6) Assuming uniform convergence, Eq. (7.5.5), the left-hand side of Eq. (7.5.6) can be expressed as /summationdisplay n,mξnMn−mXm=/summationdisplay nξn−mMn−m/summationdisplay mξmXm=M(ξ)X(ξ), with the interchange of the order of the summations, where X(ξ)andM(ξ)are deﬁned by X(ξ)≡∞/summationdisplay n=−∞Xnξn, (7.5.7) M(ξ)≡∞/summationdisplay n=−∞Mnξn. (7.5.8) 228 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation We also deﬁne f(ξ)≡∞/summationdisplay n=−∞fnξn. (7.5.9) Thus Eq. (7.5.6) takes the following form, M(ξ)X(ξ)=f(ξ). (7.5.10) We assume the following bounds on Mnandfn, |Mn|=/braceleftBigg O(a−|n|) asn→− ∞ , O(b−n) asn→+∞,a,b > 0, (7.5.11a) |fn|=/braceleftBigg O(c−|n|) asn→− ∞ , O(d−n) asn→+∞,c, d > 0. (7.5.11b) ThenM(ξ)is analytic in the annulus in the complex ξplane, 1 a<|ξ|<b , (7.5.12a) provided that 1<a b ,a n df(ξ)is analytic in the annulus in the complex ξplane, 1 c<|ξ|<d , (7.5.12b) provided that 1<c d . Hence we obtain X(ξ)=f(ξ) M(ξ)=∞/summationdisplay n=−∞Xnξnprovided M(ξ)/negationslash=0. (7.5.13) X(ξ)is analytic in the annulus max/parenleftbigg1 a,1 c/parenrightbigg <|ξ|<min(b, d). (7.5.12c) By the F ourier series inversion formula on the unit circle , we obtain Xn=1 2π/integraldisplay2π 0dθexp[−inθ]X(exp[iθ]) =1 2πi/contintegraldisplay |ξ|=1dξ ξ−n−1X(ξ), (7.5.14) with M(ξ)/negationslash=0 for|ξ|=1, (7.5.15) which solves Eq. (7.5.4). 7.5 T oeplitz Matrix and Wiener–Hopf Sum Equation 229 Next, consider the eigenvalue problem of the following form, ∞/summationdisplay m=−∞Mn−mXm=µXn. (7.5.16) We try Xm=ξm. (7.5.17) Then we obtain M(ξ)=µ. (7.5.18) The roots of Eq. (7.5.18) provide the solutions to Eq. (7.5.16). For µ=0 , we obtain Xm=(ξ0)m, (7.5.19) where ξ0is a zero of M(ξ). Case B. Semi-inﬁnite Toeplitz matrix . Consider now the system of the semi-inﬁnite inhomogeneous linear algebraic equations , ∞/summationdisplay m=0Mn−mXm=fn,n≥0, (7.5.20) which is the inhomogeneous Wiener–Hopf sum equation . We let M(ξ)≡∞/summationdisplay n=−∞Mnξn,0≤argξ≤2π, (7.5.21) be such that M(exp[iθ])/negationslash=0, for 0≤θ≤2π. We assume that ∞/summationdisplay n=0|fn|<∞, (7.5.22) and |Xm|<(1 +ε)−m,m≥0,ε > 0. (7.5.23) We deﬁne yn≡∞/summationdisplay m=0Mn−mXm forn≤−1, andyn≡0 forn≥0, (7.5.24a) 230 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation fn≡0 forn≤−1, (7.5.24b) ¯X(ξ)≡∞/summationdisplay n=0Xnξn, (7.5.25) ¯f(ξ)≡∞/summationdisplay n=0fnξn, (7.5.26) Y(ξ)≡−1/summationdisplay n=−∞ynξn. (7.5.27) We look for the solution which satisﬁes ∞/summationdisplay n=0|Xn|<∞. (7.5.28) Then Eq. (7.5.20) is rewritten as ∞/summationdisplay m=−∞Mn−mXm=fn+yn for−∞<n< ∞. (7.5.20b) We note that ∞/summationdisplay n=−∞|yn|=∞/summationdisplay n=−∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationdisplay m=0Mn−mXm/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle<∞/summationdisplay n=−∞|Mn|∞/summationdisplay m=0|Xm|<∞, where changing the order of the summation is justiﬁed since the ﬁnal expression is ﬁnite. Multiplying exp[inθ]on both sides of Eq. (7.5.20b) and summing over n, we obtain M(ξ)¯X(ξ)=¯f(ξ)+Y(ξ). (7.5.29) Homogeneous problem :W es e t fn=0 or¯f(ξ)=0. The problem we will solve ﬁrst is M(ξ)¯X(ξ)=Y(ξ), (7.5.30) where ¯X(ξ) analytic for |ξ|<1+ε, Y(ξ) analytic for |ξ|>1.(7.5.31) We deﬁne the indexνofM(ξ)in the counter-clockwise direction by ν≡1 2πiln[M(exp[iθ])]/vextendsingle/vextendsingle/vextendsingleθ=2π θ=0. (7.5.32) 7.5 T oeplitz Matrix and Wiener–Hopf Sum Equation 231 Suppose that M(ξ)has been factorized into the following form, M(ξ)=Nin(ξ)/N out(ξ), (7.5.33a) where /braceleftBigg Nin(ξ) analytic for |ξ|<1+ε, and continuous for |ξ|≤1, Nout(ξ) analytic for |ξ|>1, and continuous for |ξ|≥1.(7.5.33b) Then Eq. (7.5.30) is rewritten as Nin(ξ)¯X(ξ)=Nout(ξ)Y(ξ)≡G(ξ), (7.5.34) where G(ξ)is entire in the complex ξplane. The form of G(ξ)is now examined. Case 1. theindex ν=0 . By the now familiar formula, Eq. (7.3.25), we have /contintegraldisplayln[M(ξ/prime)] ξ/prime−ξdξ/prime 2πi=/parenleftbigg/contintegraldisplay C1−/contintegraldisplay C2/parenrightbigg/bracketleftbiggln[M(ξ/prime)] ξ/prime−ξdξ/prime 2πi/bracketrightbigg , (7.5.35) where the integration contours, C1andC2, are displayed in Figure 7.16. ′ξ′ξ2 ′ξ1C1 C2 Fig. 7.16: Integration contour of ln[M(ξ/prime)]when the index ν=0 . Thus we have Nin(ξ)=e x p/bracketleftbigg/contintegraldisplay C1ln[M(ξ/prime)] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →1 as|ξ|→∞ , (7.5.36a) Nout(ξ)=e x p/bracketleftbigg/contintegraldisplay C2ln[M(ξ/prime)] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →1 as|ξ|→∞ . (7.5.36b) We ﬁnd, by Liouville’s theorem, G(ξ)=0, (7.5.37) and hence we have no nontrivial homogeneous solution whenν=0 . 232 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Case 2. the indexν>0(positive integer ). We construct the object with the index zero ,M(ξ)/ξν, and obtain M(ξ)=ξνexp/parenleftbigg/contintegraldisplay C1−/contintegraldisplay C2/parenrightbigg/bracketleftbiggln[M(ξ/prime)/(ξ/prime)ν] ξ/prime−ξdξ/prime 2πi/bracketrightbigg , (7.5.38) from which, we obtain Nin(ξ)=ξνexp/bracketleftbigg/contintegraldisplay C1ln[M(ξ/prime)/(ξ/prime)ν] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →ξνas|ξ|→∞ , (7.5.39a) Nout(ξ)=e x p/bracketleftbigg/contintegraldisplay C2ln[M(ξ/prime)/(ξ/prime)ν] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →1 as|ξ|→∞ . (7.5.39b) By Liouville’s theorem, G(ξ)cannot grow as fast as ξν. Hence we have G(ξ)=ν−1/summationdisplay m=0Gmξm, (7.5.40) where th Gmareνarbitrary constants. Thus ¯X(ξ)is given by ¯X(ξ)=ν−1/summationdisplay m=0Gmξm Nin(ξ), (7.5.41a) from which, we obtain νindependent homogeneous solutions Xnby the F ourier series inver- sion formula on the unit circle , Xn=1 2π/integraldisplay2π 0dθexp[−inθ]X(exp[iθ]) =1 2πi/contintegraldisplay |ξ|=1dξ ξ−n−1X(ξ). (7.5.41b) Case 3. the indexν<0(negative integer ). We construct the object with the index zero ,M(ξ)ξ|ν|, and obtain M(ξ)=1 ξ|ν|exp/parenleftbigg/contintegraldisplay C1−/contintegraldisplay C2/parenrightbigg/bracketleftbiggln[M(ξ/prime)(ξ/prime)|ν|] ξ/prime−ξdξ/prime 2πi/bracketrightbigg , (7.5.42) from which, we obtain Nin(ξ)=1 ξ|ν|exp/bracketleftbigg/contintegraldisplay C1ln[M(ξ/prime)(ξ/prime)|ν|] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →1 ξ|ν|as|ξ|→∞ , (7.5.43a) Nout(ξ)=e x p/bracketleftbigg/contintegraldisplay C2ln[M(ξ/prime)(ξ/prime)|ν|] ξ/prime−ξdξ/prime 2πi/bracketrightbigg →1 as|ξ|→∞ . (7.5.43b) By Liouville’s theorem, we have G(ξ)=0, (7.5.44) and hence we have no nontrivial solution . 7.5 T oeplitz Matrix and Wiener–Hopf Sum Equation 233 Inhomogeneous problem: We restate the inhomogeneous problem below, M(ξ)¯X(ξ)=¯f(ξ)+Y(ξ), (7.5.45) where we assume that ¯f(ξ)is analytic for |ξ|<1+ε. Factoring M(ξ)as before, M(ξ)=Nin(ξ)/N out(ξ), (7.5.46) and multiplying Nout(ξ)on both sides of Eq. (7.5.45), we have ¯X(ξ)Nin(ξ)=¯f(ξ)Nout(ξ)+Y(ξ)Nout(ξ), (7.5.47a) or, splitting ¯f(ξ)Nout(ξ)into a sum of the infunction and the out function, ¯X(ξ)Nin(ξ)−[¯f(ξ)Nout(ξ)]in=[¯f(ξ)Nout(ξ)]out+Y(ξ)Nout(ξ)≡F(ξ),(7.5.47b) where F(ξ)is entire in the complex ξplane. Since we wish to obtain one particular solution to Eq. (7.5.45), in Eq. (7.5.47b) we set F(ξ)=0, (7.5.48) resulting in the particular solution, ¯Xpart(ξ)=[¯f(ξ)Nout(ξ)]in/N in(ξ), (7.5.49a) Y(ξ)=−[¯f(ξ)Nout(ξ)]out/N out(ξ). (7.5.49b) The fact that Eqs. (7.5.49a) and (7.5.49b) satisfy Eq. (7.5.47a) can be easily demonstrated. From Eq. (7.5.49a), the particular solution, Xn,part, can be obtained by the F ourier series inversion formula on the unit circle . We note that in writing Eq. (7.5.47b), the following property of Nout(ξ)is essential, Nout(ξ)→1 as|ξ|→∞ . (7.5.50) Case 1. The indexν=0 . Since the homogeneous problem has no nontrivial solution, the unique particular solution , Xn,part, is obtained for the inhomogeneous problem. Case 2. The indexν>0(positive integer ). In this case, since the homogeneous problem has νindependent solutions, the solution to the inhomogeneous problem is not unique . Case 3. The indexν<0(negative integer ). In this case, consider the homogeneous adjoint problem , ∞/summationdisplay m=0Mm−nXadj m=0. (7.5.51) 234 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation ItsMfunction, Madj(ξ), is deﬁned by Madj(ξ)≡+∞/summationdisplay n=−∞M−nξn=+∞/summationdisplay n=−∞Mnξ−n=M/parenleftbigg1 ξ/parenrightbigg . (7.5.52) The index νadjofMadj(ξ)is deﬁned by νadj≡1 2πiln[Madj(exp[iθ])]/vextendsingle/vextendsingleθ=2π θ=0=1 2πiln[M(exp[−iθ])]|θ=2π θ=0 =1 2πiln[{M(exp[iθ])}∗]|θ=2π θ=0=−1 2πiln[M(exp[iθ])]|θ=2π θ=0=−ν.(7.5.53) The factorization of Madj(ξ)is carried out as in the case of M(ξ), with the result, Madj(ξ)=Nadj in(ξ)/Nadj out(ξ)=M(1/ξ)=Nin(1/ξ)/N out(1/ξ). (7.5.54) From this, we recognize that /braceleftBigg Nadj in(ξ)=N−1 out(1/ξ) analytic in |ξ|<1,and continuous for |ξ|≤1, Nadj out(ξ)=N−1 in(1/ξ) analytic in |ξ|>1,and continuous for |ξ|≥1.(7.5.55) Then, in this case, the homogeneous adjoint problem has |ν|independent solutions, Xadj(j) m j=1,···,|ν|,m≥0. (7.5.56) Using an argument similar to the derivation of the solvability condition for the inhomoge- neous Wiener–Hopf integral equation of the second kind, discussed in Section 7.4, notingEq. (7.5.50), we obtain the solvability condition for the inhomogeneous Wiener–Hopf sum equation as follows: ∞/summationdisplay m=0fmXadj(j) m=0,j=1,···,|ν|. (7.5.57) Thus, if and only if the solvability condition (7.5.57) is satisﬁed, i.e., the inhomogeneous termfmis orthogonal to all the |ν|independent solutions Xadj(j) m to the homogeneous ad- joint problem (7.5.51), then the inhomogeneous Wiener–Hopf sum equation has the unique solution ,Xn,part. From this analysis of the inhomogeneous Wiener–Hopf sum equation , we ﬁnd that the problem at hand is the discrete analogue of the inhomogeneous Wiener–Hopf integral equation of the second kind, not of the ﬁrst kind , despite its formal appearance. For an interesting application of the Wiener–Hopf sum equation to the phase transition of the two-dimensional Ising model, the reader is referred to the article by T.T. Wu, cited in the bibliography. For another interesting application of the Wiener–Hopf sum equation to the Y agi– Uda semi-inﬁnite arrays, the reader is referred to the articles by W. Wasylkiwskyj andA.L. V anKoughnett, cited in the bibliography. 7.6 Wiener–Hopf Integral Equation of the First Kind and Dual Integral Equations 235 The Cauchy integral formula used in this section should actually be Pollard’s theorem which is the generalization of the Cauchy integral formula. We avoided the mathematical technicalities in the presentation of the Wiener–Hopf sum equation. As for the mathematical details related to the Wiener–Hopf sum equation, Liouville’s the- orem, the Wiener–Lévy theorem, and Pollard’s theorem, we refer the reader to Chapter IX of the book by B. McCoy and T.T. Wu, cited in the bibliography. Summary of the Wiener–Hopf sum equation ∞/summationdisplay m=0Mn−mXm=fn,n ≥0, ∞/summationdisplay m=0Mm−nXadj m=0,n ≥0. 1) Index ν=0 . The homogeneous problem has no nontrivial solution. The homogeneous adjoint problem has no nontrivial solution. The inhomogeneous problem has a unique solution. 2) Index ν>0. The homogeneous problem has νindependent nontrivial solutions. The homogeneous adjoint problem has no nontrivial solution.The inhomogeneous problem has non-unique solutions. 3) Index ν<0. The homogeneous problem has no nontrivial solution. The homogeneous adjoint problem has |ν|independent nontrivial solutions. The inhomogeneous problem has a unique solution, if and only if the inhomogeneous term is orthogonal to all |ν|independent solutions to the homogeneous adjoint problem. 7.6 Wiener–Hopf Integral Equation of the First Kind and Dual Integral Equations In this section, we re-examine the mixed boundary value problem considered in Section 7.1 with some generality and show its equivalence to the Wiener–Hopf integral equation of the ﬁrst kind a n dt ot h e dual integral equations . This demonstration of equivalence by no means constitutes a solution to the original problem; rather it provides a hint for solving the Wiener–Hopf integral equation of the ﬁrst kind, and the dual integral equations, by the methods we developed in Sections 7.1 and 7.3. 236 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation ❑Example 7.5. Solve the mixed boundary value problem of two-dimensional Laplace equation in the half-plane: /parenleftbigg∂2 ∂x2+∂2 ∂y2/parenrightbigg φ(x, y)=0,y ≥0, (7.6.1) with the boundary conditions speciﬁed on the xaxis, φ(x,0) =f(x),x ≥0, (7.6.2a) φy(x,0) =g(x),x < 0, (7.6.2b) φ(x, y)→0 asx2+y2→∞. (7.6.2c) Solution. We write φ(x, y) = lim ε→0+/integraldisplay+∞ −∞dk 2πeikx−√ k2+ε2yˆφ(k),y ≥0. (7.6.3) Setting y=0 in Eq. (7.6.3), /integraldisplay+∞ −∞dk 2πeikxˆφ(k)=/braceleftBigg f(x),x ≥0, φ(x,0),x < 0.(7.6.4) Thus we have ˆφ(k)=/integraldisplay+∞ −∞dx e−ikxφ(x,0) =ˆφ+(k)+ˆf−(k), (7.6.5) where ˆφ+(k)=/integraldisplay0 −∞dx e−ikxφ(x,0), (7.6.6) ˆf−(k)=/integraldisplay+∞ 0dx e−ikxf(x). (7.6.7) We know that ˆφ+(k)(ˆf−(k)) is analytic in the upper half-plane (the lower half-plane). Dif- ferentiating Eq. (7.6.3) with respect to y, and setting y=0 ,w eh a v e /integraldisplay+∞ −∞dk 2πeikx(−/radicalbig k2+ε2)ˆφ(k)=/braceleftBigg φy(x,0),x≥0, g(x),x < 0.(7.6.8) Then, by inversion, we obtain (−/radicalbig k2+ε2)ˆφ(k)=ˆg+(k)+ˆψ−(k), (7.6.9) 7.6 Wiener–Hopf Integral Equation of the First Kind and Dual Integral Equations 237 where ˆg+(k)=/integraldisplay0 −∞dx e−ikxg(x), (7.6.10) ˆψ−(k)=/integraldisplay+∞ 0dx e−ikxφy(x,0). (7.6.11) As before, we know that ˆg+(k)(ˆψ−(k)) is analytic in the upper half-plane (the lower half- plane). Eliminating ˆφ(k)from Eqs. (7.6.5) and (7.6.9), we obtain ˆφ+(k)+ˆf−(k)=/parenleftbigg −1 √ k2+ε2/parenrightbigg ˆg+(k)+/parenleftbigg −1 √ k2+ε2/parenrightbigg ˆψ−(k). (7.6.12) Inverting Eq. (7.6.12) for x>0, we obtain /integraldisplay+∞ −∞dk 2πeikx/parenleftbigg ˆφ+(k)+ˆf−(k)+1 √ k2+ε2ˆg+(k)/parenrightbigg =/integraldisplay+∞ −∞dk 2πeikx/parenleftbigg −1 √ k2+ε2/parenrightbigg ˆψ−(k),x > 0,(7.6.13) where /integraldisplay+∞ −∞dk 2πeikxˆφ+(k)=0, forx>0, (7.6.14) because ˆφ+(k)is analytic in the upper half-plane and the contour of the integration is closed in the upper half-plane for x>0. The remaining terms on the left-hand side of Eq. (7.6.13) are identiﬁed as /integraldisplay+∞ −∞dk 2πeikxˆf−(k)=f(x),x > 0, (7.6.15) /integraldisplay+∞ −∞dk 2πeikx 1 √ k2+ε2ˆg+(k)≡G(x),x > 0. (7.6.16) The right-hand side of Eq. (7.6.13) is identiﬁed as /integraldisplay+∞ −∞dk 2πeikx/parenleftbigg −1 √ k2+ε2/parenrightbigg ˆψ−(k)=/integraldisplay+∞ 0ψ(y)K(x−y)dy, x > 0,(7.6.17) where ψ(x)andK(x)are deﬁned by ψ(x)≡φy(x,0),x > 0, (7.6.18) K(x)≡/integraldisplay+∞ −∞dk 2πeikx/parenleftbigg −1 √ k2+ε2/parenrightbigg . (7.6.19) 238 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation Thus we obtain the integral equation for ψ(x)from Eq. (7.6.13), /integraldisplay+∞ 0K(x−y)ψ(y)dy=f(x)+G(x),x > 0. (7.6.20) This is the integral equation with a translational kernel of semi-inﬁnite range and is called the Wiener–Hopf integral equation of the ﬁrst kind . As noted earlier, this reduction of the mixed boundary value problem, Eqs. (7.6.1) through (7.6.2c), to the Wiener–Hopf integral equationof the ﬁrst kind, by no means constitutes a solution to the original mixed boundary value problem. In order to solve the Wiener–Hopf integral equation of the ﬁrst kind /integraldisplay +∞ 0K(x−y)ψ(y)dy=F(x),x≥0, (7.6.21) we work backwards. Equation (7.6.21) is reduced to the form of Eq. (7.6.12). Deﬁning the left-hand side of Eq. (7.6.21) for x<0by /integraldisplay+∞ 0K(x−y)ψ(y)dy=H(x),x < 0, (7.6.22) we consider the Fourier transforms of Eqs. (7.6.21) and (7.6.22), /integraldisplay+∞ 0dx e−ikx/integraldisplay+∞ 0dyK(x−y)ψ(y)=/integraldisplay+∞ 0dx e−ikxF(x)≡ˆF−(k),(7.6.23a) /integraldisplay0 −∞dx e−ikx/integraldisplay+∞ 0dyK(x−y)ψ(y)=/integraldisplay0 −∞dx e−ikxH(x)≡ˆH+(k), (7.6.23b) where ˆF−(k)(ˆH+(k)) is analytic in the lower half-plane (the upper half-plane). Adding Eqs. (7.6.23a) and (7.6.23b) together, we obtain /integraldisplay+∞ 0dye−ikyψ(y)/integraldisplay+∞ −∞dx e−ik(x−y)K(x−y)=ˆF−(k)+ˆH+(k). H e n c ew eh a v e ˆψ−(k)ˆK(k)=ˆF−(k)+ˆH+(k), (7.6.24) where ˆψ−(k)andˆK(k), respectively, are deﬁned by ˆψ−(k)≡/integraldisplay+∞ 0e−ikxψ(x)dx, (7.6.25) ˆK(k)≡/integraldisplay+∞ −∞e−ikxK(x)dx. (7.6.26) 7.7 Problems for Chapter 7 239 From Eq. (7.6.24), we have ˆψ−(k)=/parenleftBigg 1 ˆK(k)/parenrightBigg (ˆF−(k)+ˆH+(k)). (7.6.27) Carrying out the sum-splitting on the right-hand side of Eq. (7.6.27) either by inspection or by the general method discussed in Section 7.3, we can obtain ˆψ−(k)as in Section 7.1. Returning to Example 7.5, we note that the mixed boundary value problem we examined belongs to the general class of the equation, ˆφ+(k)+ˆf−(k)=ˆK(k)(ˆg+(k)+ˆψ−(k)). (7.6.28) If we directly invert for ˆψ−(k)forx>0in Eq. (7.6.28), we obtain the Wiener–Hopf integral equation of the ﬁrst kind (7.6.20). Instead, we may write Eq. (7.6.28) as a pair of equations, Φ(k)=ˆg+(k)+ˆψ−(k), (7.6.29a) ˆK(k)Φ(k)=ˆφ+(k)+ˆf−(k). (7.6.29b) Inverting Eqs. (7.6.29a) and (7.6.29b) for x<0andx≥0respectively, we ﬁnd a pair of integral equations for Φ(k)of the following form, /integraldisplay+∞ −∞dk 2πeikxΦ(k)=g(x),x < 0, (7.6.30a) /integraldisplay+∞ −∞dk 2πeikxˆK(k)Φ(k)=f(x),x ≥0. (7.6.30b) Such a pair of integral equations, one holding in some range of the independent variable and the other holding in the complementary range, are called the dual integral equations . This pair is equivalent to the mixed boundary value problem, Eqs. (7.6.1) through (7.6.2c). A solutionto the dual integral equations is again provided by the methods we developed in Sections 7.1 and 7.3. 7.7 Problems for Chapter 7 7.1. (Due to H. C.) Solve the Sommerfeld diffraction problem in two dimensions, with the boundary condition, φx(x,0) = 0 forx<0. 7.2. (Due to H. C.) Solve the half-line problem, /parenleftbigg∂2 ∂x2+∂2 ∂y2−p2/parenrightbigg φ(x, y)=0 with φ(x,0) =exforx≤0, and φ(x, y)→0 asx2+y2→∞. It is assumed that φ(x, y)andφy(x, y)are continuous except on the half-line y=0 withx≤0. 240 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation 7.3. (Due to D. M.) Solve the boundary value problem, /parenleftbigg∂2 ∂x2+∂2 ∂y2−p2/parenrightbigg φ(x, y)=0 with φy(x,0) =eiαxforx≥0, and φ(x, y)→0 as/radicalbig x2+y2→∞, by using the Wiener–Hopf method. In this problem, the point (x, y)lies in the region stated in the previous problem. Note that the Sommerfeld radiation condition is nowreplaced by the usual condition of zero limit. Compare your answer with the previous one. 7.4. (Due to H. C.) Solve ∇ 2φ(x, y)=0, with a cut on the positive xaxis, subject to the boundary conditions, φ(x,0) =e−axforx≥0, φ(x, y)→0 asx2+y2→∞. 7.5. (Due to H. C.) Solve ∇2φ(x, y)=0,0<y< 1, subject to the boundary conditions, φ(x,0) = 0 forx≥0, φ(x,1) =e−xforx≥0, and φy(x,1) = 0 forx<0. 7.6. (Due to H. C.) Solve ∇2φ(x, y)=0,0<y< 1, subject to the boundary conditions, φy(x,0) = 0 for−∞<x< ∞,φ y(x,1) = 0 forx<0, and φ(x,1) =e−xforx≥0. 7.7 Problems for Chapter 7 241 7.7. (Due to H. C.) Solve /parenleftbigg∂2 ∂x2+2∂ ∂x+∂2 ∂y2/parenrightbigg φ(x, y)=φ(x, y),y > 0, with the boundary conditions, φ(x,0) =e−xforx>0,φ y(x,0) = 0 forx<0, and φ(x, y)→0 asx2+y2→∞. 7.8. (Due to H. C.) Solve φ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy, x ≥0, with K(x)≡/integraldisplay+∞ −∞e−ikx 1 √ k2+1dk 2πandλ>1. Find also the resolvent H(x, y)of this kernel. 7.9. (Due to H. C.) Solve φ(x)=λ/integraldisplay+∞ 0e−(x−y)2φ(y)dy, 0≤x<∞. 7.10. Solve φ(x)=λ 2/integraldisplay+∞ 0E1(|x−y|)φ(y)dy, x ≥0,0<λ≤1, with E1(x)≡/integraldisplay+∞ x/parenleftbigge−ζ ζ/parenrightbigg dζ. 7.11. (Due to H. C.) Consider the eigenvalue equation, φ(x)=λ/integraldisplay+∞ 0K(x−y)φ(y)dy where K(x)=x2e−x2. a) What is the behavior of φ(x)so that the integral above is convergent? b) What is the behavior of ψ(x)asx→− ∞ (where ψ(x)is the integral above for x<0)? What is the region of analyticity for ˆψ(k)? 242 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation c) Find ˆK(k). What is the region of analyticity for ˆK(k)? d) It is required that φ(x)does not blow up faster than a polynomial of xasx→∞ . Find the spectrum of λand the number of independent eigenfunctions for each eigen- valueλ. 7.12. Solve φ(x)=e−|x|+λ/integraldisplay+∞ 0e−|x−y|φ(y)dy, x ≥0. Hint: /integraldisplay∞ −∞dx eikxe−|x|=2 k2.+1. 7.13. (Due to H. C.) Solve φ(x)=c o s hx 2+λ/integraldisplay+∞ 0e−|x−y|φ(y)dy, x ≥0. Hint: /integraldisplay+∞ −∞eikx coshxdx=π cosh/parenleftbigπk 2/parenrightbig. 7.14. (Due to H. C.) Solve φ(x)=1+ λ/integraldisplay1 01 x+x/primeφ(x/prime)dx/prime,0≤x≤1. Hint: Perform the change of variables from xandx/primetotandt/prime, x=e x p ( −t) andx/prime=e x p ( −t/prime) with t, t/prime∈[0,+∞). 7.15. Solve φ(x)=λ/integraldisplay+∞ 01 α2+(x−y)2φ(y)dy+f(x),x≥0,α > 0. 7.16. Solve Tn+1(z)+Tn−1(z)=2zTn(z),n≥1,−1≤z≤1, with T0(z)=1, andT1(z)=z. Hint: Factorize the M(ξ)function by inspection. 7.7 Problems for Chapter 7 243 7.17. Solve Un+1(z)+Un−1(z)=2zUn(z),n≥1,−1≤z≤1, with U0(z)=0, andU1(z)=/radicalbig 1−z2. 7.18. Solve ∞/summationdisplay k=0exp[iρ|j−k|]ξk−λξj=qj,j=0,1,2,··, with Imρ>0, and|q|<1. 7.19. (Due to H. C.) Solve the inhomogeneous Wiener–Hopf sum equation which originates from the two-dimensional Ising model, ∞/summationdisplay m=0Mn−mXm=fn,n≥0, with M(ξ)=∞/summationdisplay n=−∞Mnξn≡/radicalBigg (1−α1ξ)(1−α2ξ−1) (1−α1ξ−1)(1−α2ξ), andfn=δn0. Consider the following ﬁve cases, a)α1<1<α2, b)α1<α2<1, c)α1<α2=1 , d)1<α1<α2, e)α1=α2. Hint: Factorize the M(ξ)function by inspection for the above ﬁve cases and determine the functions, Nin(ξ)andNout(ξ). 7.20. Solve the Wiener–Hopf integral equation of the ﬁrst kind, /integraldisplay+∞ 01 2πK0(α|x−y|)φ(y)dy=1,x ≥0, where the kernel is given by K0(x)≡/integraldisplay+∞ 0coskx √ k2+1dk=1 2/integraldisplay+∞ −∞eikx √ k2+1dk. 244 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation 7.21. (Due to D. M.) Solve the Wiener–Hopf integral equation of the ﬁrst kind, /integraldisplay∞ 0K(x−y)φ(y)dy=1,x ≥0, where the kernel is given by K(x)=|x|exp[−|x|]. 7.22. Solve the Wiener–Hopf integral equation of the ﬁrst kind, /integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z ≥0, K(z)≡1 2[H(1) 0(k|z|)+H(1) 0(k/radicalbig d2+z2)], where H(1) 0(k|z|)is the zeroth-order Hankel function of the ﬁrst kind. 7.23. Solve the Wiener–Hopf integral equation of the ﬁrst kind, /integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z ≥0, K(z)≡1 2[H(1) 0(k|z|)−H(1) 0(k/radicalbig d2+z2)], where H(1) 0(k|z|)is the zeroth-order Hankel function of the ﬁrst kind. 7.24. Solve the integro-differential equation, /parenleftbigg∂2 ∂z2+k2/parenrightbigg/integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z≥0, K(z)≡1 2/bracketleftBig H(1) 0(k|z|)+H(1) 0(k/radicalbig d2+z2)/bracketrightBig , where H(1) 0(k|z|)is the zeroth-order Hankel function of the ﬁrst kind. 7.25. Solve the integro-differential equation, /parenleftbigg∂2 ∂z2+k2/parenrightbigg/integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z≥0, K(z)≡1 2/bracketleftBig H(1) 0(k|z|)−H(1) 0(k/radicalbig d2+z2)/bracketrightBig , where H(1) 0(k|z|)is the zeroth-order Hankel function of the ﬁrst kind. 7.7 Problems for Chapter 7 245 Hint: for Problems 7.23 through 7.26: The zeroth-order Hankel function of the ﬁrst kind H(1) 0(kD)is given by H(1) 0(kD)=1 πi/integraldisplay∞ −∞exp/bracketleftbig ik/radicalbig D2+ξ2/bracketrightbig /radicalbig D2+ξ2dξ, D > 0, and hence its Fourier transform is given by 1 2/integraldisplay∞ −∞H(1) 0(k/radicalbig D2+z2)e x p [iωz]dz=exp[iv(ω)D] v(ω),v(ω)=/radicalbig k2−ω2, Imv(ω)>0. The problems are thus reduced to factorizing the following functions, ψ(ω)≡1+e x p [ iv(ω)d]=ψ+(ω)ψ−(ω), ϕ(ω)≡1−exp[iv(ω)d]=ϕ+(ω)ϕ−(ω), where ψ+(ω)(ϕ+(ω)) is analytic and has no zeroes in the upper half-plane, Imω≥0, andψ−(ω)(ϕ−(ω)) is analytic and has no zeroes in the lower half-plane, Imω≤0. For the integro-differential equations, the differential operator ∂2 ∂z2+k2 can be brought inside the integral symbol and we obtain the extra factor, v2(ω)=k2−ω2, for the Fourier transforms, multiplying onto the functions to be factorized. The func- tions to be factorized are given by ˜K(ω)Prob. 7.23 =ψ(ω) v(ω),˜K(ω)Prob. 7.24 =ϕ(ω) v(ω), v(ω)˜K(ω)Prob. 7.25 =v(ω)ψ(ω),v(ω)˜K(ω)Prob. 7.26 =v(ω)ϕ(ω). 7.26. Solve the Wiener–Hopf integral equation of the ﬁrst kind, /integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z ≥0, K(z)≡a 2/integraldisplay∞ −∞J1(v(ω)a)H(1) 1(v(ω)a)e x p [iωz]dω, with v(ω)=/radicalbig k2−ω2, where J1(va)is the1storder Bessel function of the ﬁrst kind and H(1) 1(va)is the ﬁrst- order Hankel function of the ﬁrst kind. 246 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation 7.27. Solve the integro-differential equation, /parenleftbigg∂2 ∂z2+k2/parenrightbigg/integraldisplay+∞ 0K(z−ς)φ(ς)dς=0,z ≥0, K(z)≡a 2/integraldisplay∞ −∞J0(v(ω)a)H(1) 0(v(ω)a)e x p [iωz]dω, with v(ω)=/radicalbig k2−ω2, where J0(va)is the 0thorder Bessel function of the ﬁrst kind and H(1) 0(va)is the zeroth-order Hankel function of the ﬁrst kind. Hint: for Problems 7.27 and 7.28: The functions to be factorized are ˜K(ω)Prob. 7.27 =πaJ1(va)H(1) 1(va),v2˜K(ω)Prob. 7.28 =πav2J0(va)H(1) 0(va). The factorization procedures are identical to those in the previous problems. For the details of the factorizations for Problems 7.23 through 7.28, we refer the reader to the following monograph. Weinstein, L.A.: “ The theory of diffraction and the factorization method ”, Golem Press, (1969). Chapters 1 and 2. 7.28. Solve the dual integral equations of the following form, /integraldisplay∞ 0yf(y)Jn(yx)dy=xnfor0≤x<1, /integraldisplay∞ 0f(y)Jn(yx)dy=0 for1≤x<∞, where nis the non-negative integer and Jn(yx)is thenth-order Bessel function of the ﬁrst kind. Hint: Jackson, J.D.:“ Classical Electrodynamics ”, 3rdedition, John Wiley & Sons, New Y ork, (1999). Section 3.13. 7.29. Solve the dual integral equations of the following form, /integraldisplay∞ 0f(y)Jn(yx)dy=xnfor0≤x<1, /integraldisplay∞ 0yf(y)Jn(yx)dy=0 for1≤x<∞, where nis the non-negative integer and Jn(yx)is thenth-order Bessel function of the ﬁrst kind. Hint: Jackson, J.D.:“ Classical Electrodynamics ”, 3rdedition, John Wiley & Sons, New Y ork, (1999). Section 5.13. 7.7 Problems for Chapter 7 247 7.30. Solve the dual integral equations of the following form, /integraldisplay∞ 0yαf(y)Jµ(yx)dy=g(x) for0≤x<1, /integraldisplay∞ 0f(y)Jµ(yx)dy=0 for1≤x<∞. Hint: Kondo, J.: “ Integral Equations ”, Kodansha Ltd., Tokyo, (1991). p412. 8 Nonlinear Integral Equations 8.1 Nonlinear Integral Equation of V olterra type In Chapter 3, the integral equations of V olterra type, all of which are linear , are examined. We applied the Laplace transform technique for a translation kernel. As an application of theLaplace transform technique , we can solve a nonlinear V olterra integral equation of convolu- tion type : φ(x)=f(x)+λ/integraldisplay x 0φ(y)φ(x−y)dy. (8.1.1) Taking the Laplace transform of Eq. (8.1.1), we obtain ¯φ(s)=¯f(s)+λ[¯φ(s)]2. (8.1.2) H e n c ew eh a v e ¯φ(s)=1±(1−4λ¯f(s))1/2 2λ. We assume that ¯f(s)→0asRes→∞ , and require that ¯φ(s)→0asRes→∞ . Then only one of the two solutions survives. Speciﬁcally it is ¯φ(s)=1−(1−4λ¯f(s))1/2 2λ, (8.1.3a) and φ(x)=L−1/parenleftbigg1−(1−4λ¯f(s))1/2 2λ/parenrightbigg =/integraldisplayγ+i∞ γ−i∞ds 2πiesx1−(1−4λ¯f(s))1/2 2λ,(8.1.3b) where the inversion path is to the right of all singularities of the integrand. We examine two speciﬁc cases. ❑ Example 8.1. f(x)=0 . Solution. In this case, Eq. (8.1.3b) gives φ(x)=0. Thus there is no nontrivial solution. Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 250 8 Nonlinear Integral Equations C o 4λs s1s2 Fig. 8.1: Branch cut of the integrand of Eq. (8.1.5a) from s=0 tos=4λ. ❑ Example 8.2. f(x)=1 . Solution. In this case, ¯f(s)=1 s, (8.1.4) and Eq. (8.1.3b) gives φ(x)=1 2λ/integraldisplayγ+i∞ γ−i∞ds 2πiesx/bracketleftBigg 1−/radicalbigg s−4λ s/bracketrightBigg . (8.1.5a) The integrand has a branch cut from s=0 tos=4λas in Figure 8.1. Letλ>0, then the branch cut is as illustrated in Figure 8.2. By deforming the contour, we get φ(x)=1 2λ/contintegraldisplay Cds 2πiesx/parenleftBigg 1−/radicalbigg s−4λ s/parenrightBigg =−1 2λ/contintegraldisplay Cds 2πiesx/radicalbigg s−4λ s, (8.1.5b) where Cis the contour wrapped around the branch cut, as shown in Figure 8.2. By evaluating the values of the integrand on the two sides of the branch cut, we get φ(x)=1 2πλ/integraldisplay4λ 0ds esx/radicalbigg 4λ−s s=2 π/integraldisplay1 0dt e4λtx/radicalbigg 1−t t, (8.1.6) where we have made the change of variable, s=4λt,0≤t≤1. 8.1 Nonlinear Integral Equation of V olterra type 251 4λ os s1s2 Fig. 8.2: The contour of integration Cwrapping around the branch cut of Figure 8.1 for λ> 0. The integral in Eq. (8.1.6) can be explicitly evaluated. φ(x)=2 π∞/summationdisplay n=0(4λx)n n!/integraldisplay1 0dt tn/radicalbigg 1−t t=2 π∞/summationdisplay n=0(4λx)n n!Γ/parenleftbig n+1 2/parenrightbig Γ/parenleftbig3 2/parenrightbig Γ(n+2 ), (8.1.7) w h e r ew eh a v em a d eu s eo ft h ef o r m u l a , /integraldisplay1 0dt tn−1(1−t)m−1=Γ(n)Γ(m) Γ(n++m). Now, the conﬂuent hypergeometric function is given by F(a;c;z)=1+a cz 1!+a ca+1 c+1z2 2!+···=∞/summationdisplay n=0Γ(c) Γ(a)Γ(a+n) Γ(c+n)zn n!. (8.1.8) From Eqs. (8.1.7) and (8.1.8), we ﬁnd that φ(x)=F/parenleftbigg1 2;2 ;4λx/parenrightbigg (8.1.9) satisﬁes the nonlinear integral equation, φ(x)=1+ λ/integraldisplayx 0φ(x−y)φ(y)dy. (8.1.10) Although the above is proved only for λ>0, we may verify that it is also true for λ<0 by repeating the same argument. Alternatively, we may prove this in the following way. Let 252 8 Nonlinear Integral Equations us substitute Eq. (8.1.9) into Eq. (8.1.10). Since F/parenleftbig1 2;2 ;4λx/parenrightbig is an entire function of λ, each side of the resulting equation is also an entire function of λ. Since this equation is satisﬁed forλ>0,i tm u s tb es a t i s ﬁ e df o ra l l λby analytic continuation. Thus the integral equation (8.1.10) has the unique solution given by Eq. (8.1.9), for all values of λ. In closing this section, we classify the nonlinear integral equations of V olterra type in the following manner: (1) The kernel part is nonlinear, φ(x)−/integraldisplayx aH(x, y, φ (y))dy=f(x). (VN.1) (2) The particular part is nonlinear, G(φ(x))−/integraldisplayx aK(x, y)φ(y)dy=f(x). (VN.2) (3) Both parts are nonlinear, G(φ(x))−/integraldisplayx aH(x, y, φ (y))dy=f(x). (VN.3) (4) The nonlinear V olterra integral equation of the ﬁrst kind, /integraldisplayx aH(x, y, φ (y))dy=f(x). (VN.4) (5) The homogeneous nonlinear V olterra integral equation of the ﬁrst kind, where the kernel part is nonlinear, φ(x)=/integraldisplayx aH(x, y, φ (y))dy. (VN.5) (6) The homogeneous nonlinear V olterra integral equation of the ﬁrst kind where the particu- lar part is nonlinear, G(φ(x)) =/integraldisplayx aK(x, y)φ(y)dy. (VN.6) (7) Homogeneous nonlinear V olterra integral equation of the ﬁrst kind where the both parts are nonlinear, G(φ(x)) =/integraldisplayx aH(x, y, φ (y))dy. (VN.7) 8.2 Nonlinear Integral Equation of Fredholm Type 253 8.2 Nonlinear Integral Equation of Fredholm Type In this section, we give a brief discussion of the nonlinear integral equation of Fredholm type . Let us recall that the linear algebraic equation φ=f+λKφ (8.2.1) has the solution φ=( 1−λK)−1f. (8.2.2) In particular, the solution of Eq. (8.2.1) exists and is unique as long as the corresponding homogeneous equation φ=λKφ (8.2.3) has no nontrivial solutions. Nonlinear equations behave quite differently. Consider, for example, the nonlinear alge- braic equation obtained from Eq. (8.2.1) by replacing Kwithφ, φ=f+λφ2. (8.2.4a) The solutions of Eq. (8.2.4a) are, φ=1±√ 1−4λf 2λ. (8.2.4b) We ﬁrst observe that the solution is not unique. Indeed, if we require the solutions to be real, then Eq. (8.2.4a) has two solutions if 1−4λf > 0, (8.2.5) a n dn os o l u t i o ni f 1−4λf < 0. (8.2.6) Thus the number of solutions changes from 2 to 0 as the value of λpasses 1/4f. The point λ=1/4fis called a bifurcation point of Eq. (8.2.4a). Note that at the bifurcation point, Eq. (8.2.4a) has only one solution. We also observe from Eq. (8.2.4b) that another special point for Eq. (8.2.4a) is λ=0 .A t this point, one of the two solutions is inﬁnite. Since the number of solutions remains to be twoas the value of λpasses λ=0 , the point λ=0 is not a bifurcation point. We shall call it a singular point . Consider now the equation φ=λφ 2, (8.2.7) obtained from Eq. (8.2.4a) by setting f=0 . This equation always has the nontrivial solution φ=1 λ(8.2.8) 254 8 Nonlinear Integral Equations provided that λ/negationslash=0. (8.2.9) There is no connection between the existence of the solutions for Eq. (8.2.4a) and the absence of the solutions for Eq. (8.2.4a) with f=0 , quite unlike the case of linear algebraic equations. Nonlinear integral equations share these properties. This is evident in the following exam- ples. ❑ Example 8.3. Solve φ(x)=1+ λ/integraldisplay1 0φ2(y)dy. (8.2.10) Solution. The right-hand side of Eq. (8.2.10) is independent of x. Thus φ(x)is constant. Let φ(x)=a. Then Eq. (8.2.10) becomes a=1+ λa2. (8.2.11) Equation (8.2.11) is just Eq. (8.2.4.a ) with f=1 . Thus φ(x)=1±√ 1−4λ 2λ. (8.2.12) There are two real solutions for λ<1/4, and no real solutions for λ>1/4. Thus λ=1/4is a bifurcation point. ❑ Example 8.4. Solve φ(x)=1+ λ/integraldisplay1 0φ3(y)dy. (8.2.13) Solution. The right-hand side of Eq. (8.2.13) is independent of x. Thus φ(x)is constant. Letting φ(x)=a, we get a=1+ λa3. (8.2.14a) Equation (8.2.14a) is cubic, and hence has three solutions. Not all of these solutions are real. Let us rewrite Eq. (8.2.14a) as a−1 λ=a3, (8.2.14b) 8.2 Nonlinear Integral Equation of Fredholm Type 255 and plot (a−1)/λas well as a3in the ﬁgure. The points of intersection between these two curves are the solutions of Eq. (8.2.14a). For λnegative, there is obviously only one real root, while for λpositive and very small, there are three real roots (two positive roots and one negative root). Thus λ=0 is a bifurcation point. For λlarge and positive, there is again only one real root. The change of numbers of roots can be shown to occur at λ=4/27,w h i c hi s another bifurcation point. We may generalize the above considerations to φ(x)=c+λ/integraldisplay1 0K(φ(y))dy. (8.2.15a) The right-hand side of Eq. (8.2.15a) is independent of x. Thus φ(x)is constant. Letting φ(x)=a, we get (a−c) λ=K(a). (8.2.15b) The roots of the equation above can be graphically obtained by plotting K(a)and(a−c)/λ. Obviously, with a proper choice of K(a), the number of solutions as well as the number of bifurcation points may take any value. For example, if K(φ)=φsinπφ, (8.2.16) there are inﬁnitely many solutions as long as |λ|<1. As another example, for K(φ)=sinφ (φ2+1 ), (8.2.17) there are inﬁnitely many bifurcation points. In summary, we have found the following conclusions for nonlinear integral equations. (1) There may be more than one solution. (2) There may be one or more bifurcation points.(3) There is no signiﬁcant relationship between the integral equation with f/negationslash=0 and the one obtained from it by setting f=0 . The above considerations for simple examples may be extended to more general cases. Consider the integral equation φ(x)=f(x)+/integraldisplay 1 0K(x, y, φ (x),φ(y))dy. (8.2.18) IfKis separable, i.e., K(x, y, φ (x),φ(y)) =g(x, φ(x))h(y,φ(y)), (8.2.19) 256 8 Nonlinear Integral Equations then the integral equation (8.2.18) is solved by φ(x)=f(x)+ag(x, φ(x)), (8.2.20) with a=/integraldisplay1 0h(x, φ(x))dx. (8.2.21) We may solve Eq. (8.2.20) for φ(x)and express φ(x)as a function of xanda. There may be more than one solution. Substituting any one of these solutions into Eq. (8.2.21), we may obtain an equation for a. Thus the nonlinear integral equation (8.2.18) is equivalent to one or more nonlinear algebraic equations for a. Similarly, if Kis a sum of the separable terms, K(x, y, φ (x),φ(y)) =N/summationdisplay n=1gn(x, φ(x))hn(y,φ(y)), (8.2.22) then the integral equation is equivalent to one or more systems of Ncoupled nonlinear alge- braic equations. In closing this section, we classify the nonlinear integral equations of Fredholm type in the following manner: (1) The kernel part is nonlinear, φ(x)−/integraldisplayb aH(x, y, φ (y))dy=f(x). (FN.1) (2) The particular part is nonlinear, G(φ(x))−/integraldisplayb aK(x, y)φ(y)dy=f(x). (FN.2) (3) Both parts are nonlinear, G(φ(x))−/integraldisplayb aH(x, y, φ (y))dy=f(x). (FN.3) (4) The nonlinear Fredholm integral equation of the ﬁrst kind, /integraldisplayb aH(x, y, φ (y))dy=f(x). (FN.4) (5) Homogeneous nonlinear Fredholm integral equation of the ﬁrst kind where the kernel part is nonlinear, φ(x)=/integraldisplayb aH(x, y, φ (y))dy. (FN.5) 8.3 Nonlinear Integral Equation of Hammerstein type 257 (6) The homogeneous nonlinear Fredholm integral equation of the ﬁrst kind where the partic- ular part is nonlinear, G(φ(x)) =/integraldisplayb aK(x, y)φ(y)dy. (FN.6) (7) The homogeneous nonlinear Fredholm integral equation of the ﬁrst kind where both parts are nonlinear, G(φ(x)) =/integraldisplayb aH(x, y, φ (y))dy. (FN.7) 8.3 Nonlinear Integral Equation of Hammerstein type The inhomogeneous term f(x)in Eq. (8.2.18) is not particularly meaningful. This is because we may deﬁne ψ(x)≡φ(x)−f(x), (8.3.1) then Eq. (8.2.18) is of the form ψ(x)=/integraldisplay1 0K(x, y, ψ (x)+f(x),ψ(y)+f(y))dy. (8.3.2) The nonlinear integral equation of Hammerstein type is a special case of Eq. (8.3.2), ψ(x)=/integraldisplay1 0K(x, y)f(y,ψ(y))dy. (8.3.3) For the remainder of this section, we discuss this latter equation, (8.3.3). We shall show that iffuniformly satisﬁes a Lipschitz condition of the form |f(y,u 1)−f(y,u 2)|<C(y)|u1−u2|, (8.3.4) and if /integraldisplay1 0A(y)C2(y)dy=M2<1, (8.3.5) then the solution of Eq. (8.3.3) is unique and can be obtained by iteration. The function A(x) in Eq. (8.3.5) is given by A(x)=/integraldisplay1 0K2(x, y)dy. (8.3.6) We begin by setting ψ0(x)=0 (8.3.7) 258 8 Nonlinear Integral Equations and ψn(x)=/integraldisplay1 0K(x, y)f(y,ψn−1(y))dy. (8.3.8) If f(y,0) = 0 , (8.3.9) then Eq. (8.3.3) is solved by ψ(x)=0. (8.3.10) If f(y,0)/negationslash=0, (8.3.11) we have ψ2 1(x)≤A(x)/integraldisplay1 0f2(y,0)dy=A(x)/bardblf/bardbl2, (8.3.12) where /bardblf/bardbl2≡/integraldisplay1 0f2(y,0)dy. (8.3.13) Also, as a consequence of the Lipschitz condition (8.3.4), |ψn(x)−ψn−1(x)|</integraldisplay1 0|K(x, y)|C(y)|ψn−1(y)−ψn−2(y)|dy. Thus [ψn(x)−ψn−1(x)]2≤A(x)/integraldisplay1 0C2(y)[ψn−1(y)−ψn−2(y)]2dy. (8.3.14) From Eqs. (8.3.4) and (8.3.5), we get [ψ2(x)−ψ1(x)]2≤A(x)/bardblf/bardbl2M2. By induction, [ψn(x)−ψn−1(x)]2≤A(x)/bardblf/bardbl2(M2)n−1. (8.3.15) Thus the series ψ1(x)+[ψ2(x)−ψ1(x)] + [ψ3(x)−ψ2(x)] +··· is convergent, due to Eq. (8.3.5). The proof of uniqueness will be left to the reader. *** The most typical nonlinear integral equations in quantum ﬁeld theory and quantum sta- tistical mechanics are Schwinger–Dyson equations, which are actually the coupled nonlinear integro-differential equations. Schwinger–Dyson equations in quantum ﬁeld theory and quan-tum statistical mechanics can be solved iteratively. These topics are discussed in Sections 10.3 and 10.4 of Chapter 10. 8.4 Problems for Chapter 8 259 8.4 Problems for Chapter 8 8.1. (Due to H. C.) Prove the uniqueness of the solution to the nonlinear integral equation of Hammerstein type, ψ(x)=/integraldisplay1 0K(x, y)f(y,ψ(y))dy. 8.2. (Due to H. C.) Consider d2 dt2x(t)+x(t)=1 π2x2(t),t > 0, with the initial conditions, x(0) = 0 ,anddx dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=1. a) Transform this nonlinear ordinary differential equation into an integral equation. b) Obtain an approximate solution, accurate to a few percent. 8.3. (Due to H. C.) Consider /parenleftbigg∂2 ∂x2+∂2 ∂y2/parenrightbigg φ(x, y)=1 π2φ2(x, y),x2+y2<1, with φ(x, y)=1 onx2+y2=1. a) Construct a Green’s function satisfying /parenleftbigg∂2 ∂x2+∂2 ∂y2/parenrightbigg G(x, y;x/prime,y/prime)=δ(x−x/prime)δ(y−y/prime), with the boundary condition, G(x, y;x/prime,y/prime)=0 onx2+y2=1. Prove that G(x, y;x/prime,y/prime)=G(x/prime,y/prime;x, y). Hint : To construct the Green’s function G(x, y;x/prime,y/prime)which vanishes on the unit circle, use the method of images. b) Transform the nonlinear partial differential equation above to an integral equation, and obtain an approximate solution accurate to a few percent. 260 8 Nonlinear Integral Equations 8.4. (Due to H. C.) a) Discuss a phase transition of ρ(θ)which is given by the nonlinear integral equation, ρ(θ)=Z−1exp/bracketleftbigg β/integraldisplay2π 0cos(θ−φ)ρ(φ)dφ/bracketrightbigg ,β=J kBT,0≤θ≤2π, where Zis the normalization constant such that ρ(θ)is normalized to unity, /integraldisplay2π 0ρ(θ)dθ=1. b) Determine the nature of the phase transition. Hint : Y ou may need the following special functions, I0(z)=∞/summationdisplay m=01 (m!)2/parenleftBigz 2/parenrightBig2m , I1(z)=∞/summationdisplay m=01 m!(m+1 ) !/parenleftBigz 2/parenrightBig2m+1 , and generally In(z)=1 π/integraldisplayπ 0ezcosθ(cosnθ)dθ. For|z|→0,w eh a v e I0(z)→1, I1(z)→z 2. 8.5. Solve the nonlinear integral equation of Fredholm type, φ(x)−/integraldisplay0.1 0xy/bracketleftbig 1+φ2(y)/bracketrightbig dy=x+1. 8.6. Solve the nonlinear integral equation of Fredholm type, φ(x)−60/integraldisplay1 0xyφ2(y)dy=1+2 0 x−x2. 8.7. Solve the nonlinear integral equation of Fredholm type, φ(x)−λ/integraldisplay1 0xy/bracketleftbig 1+φ2(y)/bracketrightbig dy=x+1. 8.4 Problems for Chapter 8 261 8.8. Solve the nonlinear integral equation of Fredholm type, φ2(x)−/integraldisplay1 0xyφ(y)dy=4+1 0 x+9x2. 8.9. Solve the nonlinear integral equation of Fredholm type, φ2(x)−λ/integraldisplay1 0xyφ(y)dy=1+ x2. 8.10. Solve the nonlinear integral equation of Fredholm type, φ2(x)+/integraldisplay2 0φ3(y)dy=1−2x+x2. 8.11. Solve the nonlinear integral equation of Fredholm type, φ2(x)=5 2/integraldisplay1 0xyφ(y)dy. Hint for Problems 8.5 through 8.11: The integrals are, at most, linear in x. 8.12. Solve the nonlinear integral equation of V olterra type, φ2(x)−/integraldisplayx 0(x−y)φ(y)dy=1+3 x+1 2x2−1 2x3. 8.13. Solve the nonlinear integral equation of V olterra type, φ2(x)+/integraldisplayx 0sin(x−y)φ(y)dy=e x p [ x]. 8.14. Solve the nonlinear integral equation of V olterra type, φ(x)−/integraldisplayx 0(x−y)2φ2(y)dy=x. 8.15. Solve the nonlinear integral equation of V olterra type, φ2(x)−/integraldisplayx 0(x−y)3φ3(y)dy=1+ x2. 8.16. Solve the nonlinear integral equation of V olterra type, 2φ(x)−/integraldisplayx 0φ(x−y)φ(y)dy=s i nx. Hint for Problems 8.12 through 8.16: Take the Laplace transform of the given nonlinear integral equations of V olterra type. 262 8 Nonlinear Integral Equations 8.17. Solve the nonlinear integral equation, φ(x)−λ/integraldisplay1 0φ2(y)dy=1. In particular, identify the bifurcation points of this equation. What are the non-trivial solutions of the corresponding homogeneous equations? 9 Calculus of Variations: Fundamentals 9.1 Historical Background The calculus of variations was ﬁrst found in the late 17th.century soon after calculus was in- vented. The main people involved were Newton, the two Bernoulli brothers, Euler, Lagrange,Legendre and Jacobi. Isaac Newton (1642–1727) formulated the fundamental laws of motion. The fundamental quantities of motion were established as momentum and force. Newton’s laws of motion state: 1. In the inertial frame, every body remains at rest or in uniform motion unless acted on by af o r c e /vectorF. The condition /vectorF=/vector0implies a constant velocity /vectorvand a constant momentum /vectorp=m/vectorv. 2. In the inertial frame, the application of force /vectorFalters the momentum /vectorpby an amount speciﬁed by /vectorF=d dt/vectorp. (9.1.1) 3. To each action of a force, there is an equal and opposite action of another force. Thus if /vectorF21is the force exerted on particle 1by particle 2, then /vectorF21=−/vectorF12, (9.1.2) and these forces act along the line separating the particles. Contrary to the common belief that Newton discovered the gravitational force by observing that the apple dropped from the tree at Trinity College, he actually deduced Newton’s lawsof motion from the careful analysis of Kepler’s laws. He also invented the calculus, named methodus ﬂuxionum in 1666, about 10 years ahead of Leibniz. In 1687, Newton published his “Philosophiae naturalis principia mathematica ”, often called “ Principia ”. It consists of three parts: Newton’s laws of motion, the laws of the gravitational force, and the laws of motion ofthe planets. The Bernoulli brothers, Jacques (1654–1705) and Jean (1667–1748), came from a family of mathematicians in Switzerland. They solved the problem of Brachistochrone . They estab- lished the principle of virtual work as a general principle of statics with which all problems of equilibrium could be solved. Remarkably, they also compared the motion of a particle in a given ﬁeld of force with that of light in an optically heterogeneous medium and tried to Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 264 9 Calculus of V ariations: Fundamentals provide a mechanical theory of the refractive index. The Bernoulli brothers were the forerun- ners of the theory of Hamilton which has shown that the principle of least action in classical mechanics and Fermat’s principle of shortest time in geometrical optics, are strikingly anal-ogous to each other. They used the notation, g, for the gravitational acceleration for the ﬁrst time. Leonhard Euler (1707–1783) grew up under the inﬂuence of the Bernoulli family in Switzerland. He made an extensive contribution to the development of calculus after Leib- niz, and initiated the calculus of variations. He started the systematic study of isoperimetricproblems. He also contributed in an essential way to the variational treatment of classicalmechanics, providing the Euler equation ∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg =0, (9.1.3) for the extremization problem δI=δ/integraldisplayx2 x1f(x, y, y/prime)dx=0, (9.1.4) with δy(x1)=δy(x2)=0. (9.1.5) Joseph Louis Lagrange (1736–1813) provided the solution to the isoperimetric problems by the method presently known as the method of Lagrange multipliers, quite independently of Euler. He started the whole ﬁeld of the calculus of variations. He also introduced the notion of generalized coordinates, {qr(t)}f r=1, into classical mechanics and completely reduced the mechanical problem to that of the differential equations now known as the Lagrange equationsof motion, d dt/parenleftbigg∂L(qs,˙qs,t) ∂˙qr/parenrightbigg −∂L(qs,˙qs,t) ∂qr=0,r=1,···,f, with ˙qr≡d dtqr,(9.1.6) with the Lagrangian L(qr(t),˙qr(t),t)appropriately chosen in terms of the kinetic energy and the potential energy. He successfully converted classical mechanics into analytical mechanics using the variational principle. He also carried out research on the Fermat problem, the general treatment of the theory of ordinary differential equations, and the theory of elliptic functions. Adrien Marie Legendre (1752–1833) announced his research on the form of the planet in 1784. In his article, the Legendre polynomials were used for the ﬁrst time. He provided theLegendre test in the maximization–minimization problem of the calculus of variations, among his numerous and diverse contributions to mathematics. As one of his major accomplishments, his classiﬁcation of elliptic integrals into three types stated in “ Exercices de calcul intégral ”, published in 1811, should be mentioned. In 1794, he published “ Éléments de géométrie, avec 9.1 Historical Background 265 notes ”. He further developed the method of transformations for thermodynamics which are currently known as the Legendre transformations and are used even today in quantum ﬁeld theory. William Rowan Hamilton (1805–1865) started his research on optics around 1823 and introduced the notion of the characteristic function. His results formed the basis of the later development of the concept of the eikonal in optics. He also succeeded in transforming the Lagrange equations of motion (of the second order) into a set of differential equations ofthe ﬁrst order with twice as many variables, by the introduction of the momenta {p r(t)}f r=1 canonically conjugate to the generalized coordinates {qr(t)}f r=1by pr(t)=∂L(qs,˙qs,t) ∂˙qr,r =1,...,f. (9.1.7) His equations are known as Hamilton’s canonical equations of motion: d dtqr(t)=∂H(qs(t),ps(t),t) ∂pr(t), d dtpr(t)=−∂H(qs(t),ps(t),t) ∂qr(t),r =1,...,f.(9.1.8) He formulated classical mechanics in terms of the principle of least action. The variational principles formulated by Euler and Lagrange apply only to conservative systems. He alsorecognized that the principle of least action in classical mechanics and Fermat’s principle of shortest time in geometrical optics are strikingly analogous, permitting the interpretation of the optical phenomena in terms of mechanical terms and vice versa. He was one step short of discovering wave mechanics by analogy with wave optics as early as 1834, although he did not have any experimentally compelling reason to take such a step. On the other hand, by 1924, L. de Broglie and E. Schrödinger had sufﬁcient experimentally compelling reasons to take this step. Carl Gustav Jacob Jacobi (1804–1851), in 1824, quickly recognized the importance of the work of Hamilton. He realized that Hamilton was using just one particular choice of a set of the variables {q r(t)}f r=1 and{pr(t)}f r=1 to describe the mechanical system and carried out the research on the canonical transformation theory with the Legendre transformation. Heduly arrived at what is now known as the Hamilton–Jacobi equation. His research on thecanonical transformation theory is summarized in “ V orlesungen über Dynamik ”, published in 1866. He formulated his version of the principle of least action for the time-independent case and provided the Jacobi test in the maximization–minimization problem of the calculus of variations. In 1827, he introduced the elliptic functions as the inverse functions of the ellipticintegrals. From our discussions, we may be led to the conclusion that the calculus of variations is the completed subject of the 19 th.century. We note, however, that from the 1940s to 1950s, there was a resurgence of the action principle for the systemization of quantum ﬁeld theory. 266 9 Calculus of V ariations: Fundamentals Feynman’s action principle and Schwinger’s action principle are the main subject matter. In contemporary particle physics, if we begin with the Lagrangian density of the system under consideration, the extremization of the action functional is still employed as the starting pointof the discussion (See Chapter 10). Furthermore, the Legendre transformation is used in thecomputation of the effective potential in quantum ﬁeld theory. We now deﬁne the problem of the calculus of variations. Suppose that we have an un- known function y(x)of the independent variable xwhich satisﬁes some condition C.W e construct the functional I[y]which involves the unknown function y(x)and its derivatives. We now want to determine the unknown function y(x)which extremizes the functional I[y] under the inﬁnitesimal variation δy(x)ofy(x)subject to the condition C. A simple example is the extremization of the following functional: I[y]=/integraldisplayx2 x1L(x, y, y/prime)dx, C :y(x1)=y(x2)=0. (9.1.9) The problem is reduced to solving the Euler equation, which we discuss later. This problem and its solution constitute the problem of the calculus of variations. Many basic principles of physics can be cast in the form of the calculus of variations. Most of the problems in classical mechanics and classical ﬁeld theory are of this form, with certaingeneralizations, and the following replacements: for classical mechanics, we replace   x with t, y with q(t), y/primewith ˙q(t)≡dq(t) dt,(9.1.10) and for classical ﬁeld theory, we replace   x with (t,/vectorr), y with ψ(t,/vectorr), y/primewith/parenleftbigg∂ψ(t,/vectorr) ∂t,/vector∇ψ(t,/vectorr)/parenrightbigg .(9.1.11) On the other hand, when we want to solve some differential equation, subject to the condition C, we may be able to reduce the problem of solving the original differential equation to that of the extremization of the functional I[y]subject to the condition C, provided that the Euler equation of the extremization problem coincides with the original differential equation wewant to solve. With this reduction, we can obtain an approximate solution of the originaldifferential equation. The problem of Brachistochrone , to be deﬁned later, the isoperimetric problem, to be deﬁned later, and the problem of ﬁnding the shape of the soap membrane with the minimumsurface area, are the classic problems of the calculus of variations. 9.2 Examples 267 9.2 Examples We list examples of the problems to be solved. ❑Example 9.1. The shortest distance between two points is a straight line: Minimize I=/integraldisplayx2 x1/radicalbig 1+(y/prime)2dx. (9.2.1) ❑Example 9.2. The largest area enclosed by an arc of ﬁxed length is a circle: Minimize I=/integraldisplay yd x , (9.2.2) subject to the condition that /integraldisplay/radicalbig 1+(y/prime)2dx ﬁxed. (9.2.3) ❑Example 9.3. Catenary .The surface formed by two circular wires dipped in a soap solution: Minimize /integraldisplayx2 x1y/radicalbig 1+(y/prime)2dx. (9.2.4) ❑Example 9.4. Brachistochrone. Determine a path down which a particle falls under gravity in the shortest time: Minimize /integraldisplay/radicalBigg 1+(y/prime)2 ydx. (9.2.5) ❑Example 9.5. Hamilton’s Action Principle in Classical Mechanics: Minimize the ac- tion integral Ideﬁned by I≡/integraldisplayt2 t1L(q,˙q)dt with q(t1)andq(t2) ﬁxed, (9.2.6) where L(q(t),˙q(t))is the Lagrangian of the mechanical system. The last example is responsible for beginning the whole ﬁeld of the calculus of variations. 9.3 Euler Equation We shall derive the fundamental equation for the calculus of variations, the Euler equation. We shall extremize I=/integraldisplayx2 x1f(x, y, y/prime)dx, (9.3.1) 268 9 Calculus of V ariations: Fundamentals with the endpoints y(x1),y(x2) ﬁxed. (9.3.2) We consider a small variation in y(x)of the following form, y(x)→y(x)+εν(x), (9.3.3a) y/prime(x)→y/prime(x)+εν/prime(x), (9.3.3b) with ν(x1)=ν(x2)=0,ε=positive inﬁnitesimal . (9.3.3c) Then the variation in Iis given by δI=/integraldisplayx2 x1/parenleftbigg∂f ∂yεν(x)+∂f ∂y/primeεν/prime(x)/parenrightbigg dx =∂f ∂y/primeεν(x)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=x2 x=x1+/integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg/parenrightbigg εν(x)dx =/integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg/parenrightbigg εν(x)dx=0,(9.3.4) forν(x)arbitrary other than the condition (9.3.3c). We set J(x)=∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg . (9.3.5) We suppose that J(x)is positive in the interval [x1,x2], J(x)>0 forx∈[x1,x2]. (9.3.6) Then, by choosing ν(x)>0 forx∈[x1,x2], (9.3.7) we can make δIpositive, δI >0. (9.3.8) We now suppose that J(x)is negative in the interval [x1,x2], J(x)<0 forx∈[x1,x2]. (9.3.9) Then, by choosing ν(x)<0 forx∈[x1,x2], (9.3.10) 9.3 Euler Equation 269 we can make δIpositive, δI >0. (9.3.11) We lastly suppose that J(x)alternates its sign in the interval [x1,x2]. Then by choosing ν(x)≷0 wherever J(x)≷0, (9.3.12) we can make δIpositive, δI >0. (9.3.13) Thus, in order to have Eq. (9.3.4) for ν(x)arbitrary, together with the condition (9.3.3c), we must have J(x)identically equal to zero, ∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg =0, (9.3.14) which is known as the Euler equation . We now illustrate the Euler equation by solving some of the examples listed above. ❑ Example 9.1. The shortest distance between two points. In this case, fis given by f=/radicalbig 1+(y/prime)2. (9.3.15a) The Euler equation simply gives y/prime /radicalbig 1+(y/prime)2=constant ,⇒y/prime=c. (9.3.15b) In general, if f=f(y/prime), independent of xandy, then the Euler equation always gives y/prime=constant. (9.3.16) ❑ Example 9.4. The brachistochrone problem. In this case, fis given by f=/radicalBigg 1+(y/prime)2 y. (9.3.17) The Euler equation gives −1 2/radicalBigg 1+(y/prime)2 y3−d dxy/prime /radicalbig (1 + ( y/prime)2)y=0. This appears somewhat difﬁcult to solve. However, there is a simple way which is applicable to many cases. 270 9 Calculus of V ariations: Fundamentals Suppose f=f(y,y/prime), independent of x, (9.3.18a) then d dx=y/prime∂ ∂y+y/prime/prime∂ ∂y/prime+∂ ∂x where the last term is absent when acting on f=f(y,y/prime). Thus d dxf=y/prime∂f ∂y+y/prime/prime∂f ∂y/prime. Making use of the Euler equation on the ﬁrst term of the right-hand side, we have d dxf=y/primed dx/parenleftbigg∂f ∂y/prime/parenrightbigg +/parenleftbiggd dxy/prime/parenrightbigg∂f ∂y/prime=d dx/parenleftbigg y/prime∂f ∂y/prime/parenrightbigg , i.e., d dx/parenleftbigg f−y/prime∂f ∂y/prime/parenrightbigg =0. Hence we obtain f−y/prime∂f ∂y/prime=constant . (9.3.18b) Returning to the Brachistochrone problem ,w eh a v e /radicalBigg 1+(y/prime)2 y−y/prime y/prime /radicalbig (1 + ( y/prime)2)y=constant , or, y(1 + ( y/prime)2)=2R. Solving for y/prime, we obtain dy dx=y/prime=/radicalBigg 2R−y y. (9.3.19) H e n c ew eh a v e /integraldisplay dy/radicalbigg y 2R−y=x. We set y=2Rsin2/parenleftbiggθ 2/parenrightbigg =R(1−cosθ). (9.3.20a) 9.3 Euler Equation 271 Then we easily get x=2R/integraldisplay sin2/parenleftbiggθ 2/parenrightbigg dθ=R(θ−sinθ). (9.3.20b) Equations (9.3.20a) and (9.3.20b) are the parametric equations for a cycloid , the curve traced by a point on the rim of a wheel rolling on the xaxis. The shape of a cycloid is displayed in Figure 9.1. x −yπ 2π0 -0.5 -1.0 -1.5 -2.0 Fig. 9.1: The curve traced by a point on the rim of a wheel rolling on the xaxis. We state several facts for Example 9.4: 1. The solution of the fastest fall is not a straight line. It is a cycloid with inﬁnite initial slope. 2. There exists a unique solution. In our parametric representation of a cycloid, the range of θis implicitly assumed to be 0≤θ≤2π. Setting θ=0 , we ﬁnd that the starting point is chosen to be at the origin, (x1,y1)=( 0 ,0). (9.3.21) The question is that, given the endpoint (x2,y2), can we uniquely determine a radius of the wheel R? We put y2=R(1−cosθ0),a n dx2=R(θ0−sinθ0),o r , 1−cosθ0 θ0−sinθ0=y2 x2,2R=2y2 1−cosθ0=y2 sin2(θ0/2), which has a unique solution in the range 0<θ0<π . 3. The shortest time of descent is T=/integraldisplayx2 0dx/radicalBigg 1+(y/prime)2 y=2√ 2R/integraldisplayθ0/2 0dθ=√ y2θ0 sin(θ0/2). (9.3.22) 272 9 Calculus of V ariations: Fundamentals ❑ Example 9.5. Hamilton’s Action Principle in Classical Mechanics. Consider the inﬁnitesimal variation δq(t)ofq(t), vanishing at t=t1andt=t2, δq(t1)=δq(t2)=0. (9.3.23) Then Hamilton’s Action Principle demands that δI=δ/integraldisplayt2 t1L(q(t),˙q(t),t)dt=/integraldisplayt2 t1/parenleftbigg δq(t)∂L ∂q(t)+δ˙q(t)∂L ∂˙q(t)/parenrightbigg dt =/integraldisplayt2 t1dt/parenleftbigg δq(t)∂L ∂q(t)+/parenleftbiggd dtδq(t)/parenrightbigg∂L ∂˙q(t)/parenrightbigg =/bracketleftbigg δq(t)∂L ∂˙q(t)/bracketrightbiggt=t2 t=t1+/integraldisplayt2 t1dtδq(t)/parenleftbigg∂L ∂q(t)−d dt/parenleftbigg∂L ∂˙q(t)/parenrightbigg/parenrightbigg =/integraldisplayt2 t1dtδq(t)/parenleftbigg∂L ∂q(t)−d dt/parenleftbigg∂L ∂˙q(t)/parenrightbigg/parenrightbigg =0,(9.3.24) where δq(t)is arbitrary other than the condition (9.3.23). From this, we obtain the Lagrange equation of motion , d dt/parenleftbigg∂L ∂˙q(t)/parenrightbigg −∂L ∂q(t)=0, (9.3.25) which is nothing but the Euler equation (9.3.14), with the identiﬁcation, t⇒x, q(t)⇒y(x),L(q(t),˙q(t),t)⇒f(x, y, y/prime). When the Lagrangian L(q(t),˙q(t),t)does not depend on texplicitly, the following quantity, ˙q(t)∂L ∂˙q(t)−L(q(t),˙q(t))≡E, (9.3.26) is a constant of motion and is called the energy integral , which is simply Eq. (9.3.18b). Solving the energy integral for ˙q(t), we can obtain the differential equation forq(t). 9.4 Generalization of the Basic Problems ❑Example 9.6. Free endpoint: y2arbitrary. An example is to consider, in the Brachistochrone problem , the dependence of the shortest time of fall as a function of y2. The question is: What is the height of fall y2which, for a givenx2, minimizes this time of fall? We may, of course, start by taking the expression for the shortest time of fall: T=√ y2θ0 sin(θ0/2)=√ 2x2θ0 √ θ0−sinθ0, 9.4 Generalization of the Basic Problems 273 which, for a given x2, has a minimum at θ0=π,w h e r e T=√ 2πx2. We shall, however, give a treatment for the general problem of free endpoint. To extremize I=/integraldisplayx2 x1f(x, y, y/prime)dx, (9.4.1) with y(x1)=y1, andy2arbitrary, (9.4.2) we require δI=∂f ∂y/primeεν(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x2+ε/integraldisplayx2 x1/bracketleftbigg∂f ∂y−d dx/parenleftbigg∂f ∂y/prime/parenrightbigg/bracketrightbigg ν(x)dx=0. (9.4.3) By choosing ν(x2)=0 , we get the Euler equation. Next we choose ν(x2)/negationslash=0 and obtain in addition, ∂f ∂y/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x2=0. (9.4.4) Note that y2is determined by these equations. For the Brachistochrone problem of arbitrary y2,w eg e t ∂f ∂y/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x2=y/prime /radicalbig (1 + ( y/prime)2)y=0⇒y/prime=0. Thusθ0=π,a n dy2=x2(2 π), as obtained previously. ❑Example 9.7. Endpoint on the curve y=g(x): An example is to ﬁnd the shortest time of descent to a curve. Note that, in this problem, neither x2nory2are given. They are to be determined and related by y2=g(x2). Suppose that y=y(x)is the solution. This means that if we make a variation y(x)→y(x)+εν(x), (9.4.5) which intersects the curve at (x2+∆x2,y2+∆y2), then y(x2+∆x2)+εν(x2+∆x2)= g(x2+∆x2),o r , εν(x2)=(g/prime(x2)−y/prime(x2))∆x2, (9.4.6) and that the variation δIvanishes: δI=/integraldisplayx2+∆x2 x1f(x, y+εν, y/prime+εν/prime)dx−/integraldisplayx2 x1f(x, y, y/prime)dx /similarequal/bracketleftbigg f(x, y, y/prime)∆x2+∂f ∂y/primeεν(x)/bracketrightbigg x=x2+/integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx∂f ∂y/prime/parenrightbigg εν(x)dx=0. 274 9 Calculus of V ariations: Fundamentals Thus, in addition to the Euler equation, we have /bracketleftbigg f(x, y, y/prime)∆x2+∂f ∂y/primeεν(x)/bracketrightbigg x=x2=0, or, using Eq. (9.4.6), we obtain /bracketleftbigg f(x, y, y/prime)+∂f ∂y/prime(g/prime−y/prime)/bracketrightbigg x=x2=0. (9.4.7) Applying the above equation to the Brachistochrone problem ,w eg e t y/primeg/prime=−1. This means that the path of fastest descent intersects the curve y=g(x)at a right angle. ❑Example 9.8. The isoperimetric problem :F i n d y(x)which extremizes I=/integraldisplayx2 x1f(x, y, y/prime)dx (9.4.8) while keeping J=/integraldisplayx2 x1F(x, y, y/prime)dx ﬁxed. (9.4.9) An example is the classic problem of ﬁnding the maximum area enclosed by a curve of ﬁxed length. Lety(x)be the solution. This means that if we make a variation of ywhich does not change the value of J, the variation of Imust vanish. Since Jcannot change, this variation is not of the form εν(x), with ν(x)arbitrary. Instead, we must put y(x)→y(x)+ε1ν1(x)+ε2ν2(x), (9.4.10) where ε1andε2are so chosen that δJ=/integraldisplayx2 x1/parenleftbigg∂F ∂y−d dx∂F ∂y/prime/parenrightbigg (ε1ν1(x)+ε2ν2(x))dx=0. (9.4.11) For these kinds of variations, y(x)extremizes I: δI=/integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx∂f ∂y/prime/parenrightbigg (ε1ν1(x)+ε2ν2(x))dx=0. (9.4.12) Eliminating ε2,w eg e t ε1/integraldisplayx2 x1/bracketleftbigg∂ ∂y(f+λF)−d dx∂ ∂y/prime(f+λF)/bracketrightbigg ν1(x)dx=0, (9.4.13a) 9.4 Generalization of the Basic Problems 275 where λ=/bracketleftbigg/integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx∂f ∂y/prime/parenrightbigg ν2(x)dx/bracketrightbigg /bracketleftbigg/integraldisplayx2 x1/parenleftbigg∂F ∂y−d dx∂F ∂y/prime/parenrightbigg ν2(x)dx/bracketrightbigg. (9.4.13b) Thus(f+λF)satisﬁes the Euler equation. The λis determined by solving the Euler equa- tion, substituting yinto the integral for J, and requiring that Jtakes the prescribed value. ❑Example 9.9. The integral involves more than one function: Extremize I=/integraldisplayx2 x1f(x, y, y/prime,z,z/prime)dx, (9.4.14) withyandztaking prescribed values at the endpoints. By varying yandzsuccessively, we get ∂f ∂y−d dx∂f ∂y/prime=0, (9.4.15a) and ∂f ∂z−d dx∂f ∂z/prime=0. (9.4.15b) ❑Example 9.10. The integral involves y/prime/prime: I=/integraldisplayx2 x1f(x, y, y/prime,y/prime/prime)dx (9.4.16) withyandy/primetaking prescribed values at the endpoints. The Euler equation is ∂f ∂y−d dx∂f ∂y/prime+d2 dx2∂f ∂y/prime/prime=0. (9.4.17) ❑Example 9.11. The integral is the multi-dimensional: I=/integraldisplay dxdt f (x, t, y, y x,yt), (9.4.18) withytaking the prescribed values at the boundary. The Euler equation is ∂f ∂y−d dx∂f ∂yx−d dt∂f ∂yt=0. (9.4.19) 276 9 Calculus of V ariations: Fundamentals 9.5 More Examples ❑Example 9.12. Catenary: (a) The shape of a chain hanging on two pegs : The gravitational potential of a chain of uniform density is proportional to I=/integraldisplayx2 x1y/radicalbig 1+(y/prime)2dx. (9.5.1a) The equilibrium position of the chain minimizes I, subject to the condition that the length of the chain is ﬁxed, J=/integraldisplayx2 x1/radicalbig 1+(y/prime)2dx ﬁxed. (9.5.1b) Thus we extremize I+λJ and obtain (y+λ) /radicalbig 1+(y/prime)2=α constant, or, /integraldisplaydy /radicalbig (y+λ)2/α2−1=/integraldisplay dx. (9.5.2) We put y+λ α= cosh θ, (9.5.3) then x−β=αθ, (9.5.4) and the shape of chain is given by y=αcosh/parenleftbiggx−β α/parenrightbigg −λ. (9.5.5) The constants, αβ andλ, are determined by the two boundary conditions and the require- ment that Jis equal to the length of the chain. Let us consider the case y(−L)=y(L)=0. (9.5.6) Then the boundary conditions give β=0,λ=αcoshL α. (9.5.7) 9.5 More Examples 277 The condition that Jis constant gives α LsinhL α=l L, (9.5.8) where 2lis the length of the chain. It is easily shown that, for l≥L, a unique solution is obtained. (b) A soap ﬁlm formed by two circular wires: The surface of the soap ﬁlm takes a minimum area as a result of surface tension. Thus we minimize I=/integraldisplayx2 x1y/radicalbig 1+(y/prime)2dx, (9.5.9) obtaining as in (a), y=αcosh/parenleftbiggx−β α/parenrightbigg , (9.5.10) where αandβare constants of integration, to be determined from the boundary conditions atx=±L. Let us consider the special case in which the two circular wires are of equal radius R. Then y(−L)=y(L)=R. (9.5.11) We easily ﬁnd that β=0 ,a n dt h a t αis determined by the equation R L=α LcoshL α, (9.5.12) which has zero, one, or two solutions depending on the ratio R/L . In order to decide if any of these solutions actually minimizes I, we must study the second variation ,w h i c h we shall discuss in Section 9.7. ❑ Example 9.5. Hamilton’s Action Principle in Classical Mechanics. Let the Lagrangian L(q(t),˙q(t))be deﬁned by L(q(t),˙q(t))≡T(q(t),˙q(t))−V(q(t),˙q(t)), (9.5.13) where TandVare the kinetic energy and the potential energy of the mechanical system, respectively. In general, TandVcan depend on both q(t)and˙q(t). When TandVare given respectively by T=1 2m˙q(t)2,V =V(q(t)), (9.5.14) the Lagrange equation of motion (9.3.25) provides us with Newton’s equation of motion , m¨q(t)=−d dq(t)V(q(t)) with ¨q(t)=d2 dt2q(t). (9.5.15) 278 9 Calculus of V ariations: Fundamentals In other words, the extremization of the action integral Igiven by I=/integraldisplayt2 t1/bracketleftbigg1 2m˙q(t)2−V(q(t))/bracketrightbigg dt withδq(t1)=δq(t2)=0 , leads us to Newton’s equation of motion (9.5.15). With TandV given by Eq. (9.5.14), the energy integral Egiven by Eq. (9.3.26) assumes the following form, E=1 2m˙q(t)2+V(q(t)), (9.5.16) which represents the total mechanical energy of the system, very appropriate for the terminol- ogy, “the energy integral ”. ❑Example 9.13. Fermat’s Principle in Geometrical Optics. The path of a light ray between two given points in a medium is the one which minimizes the time of travel. Thus the path is determined from minimizing T=1 c/integraldisplayx2 x1dx/radicalBigg 1+/parenleftbiggdy dx/parenrightbigg2 +/parenleftbiggdz dx/parenrightbigg2 n(x, y, z), (9.5.17) where n(x, y, z)is the index of refraction. If nis independent of x,w eg e t n(y,z) /radicalbigg 1+/parenleftBig dy dx/parenrightBig2 +/parenleftbigdz dx/parenrightbig2=constant . (9.5.18) From Eq. (9.5.18), we easily derive the law of reﬂection and the law of refraction (Snell’s law ). 9.6 Differential Equations, Integral Equations, and Extremization of Integrals We now consider the inverse problem: If we are to solve a differential or an integral equation, can we formulate the problem in terms of one which extremizes an integral? This will havepractical advantages when we try to obtain approximate solutions for differential equations and approximate eigenvalues. ❑Example 9.14. Solve d dx/bracketleftbigg p(x)d dxy(x)/bracketrightbigg −q(x)y(x)=0,x 1<x<x 2, (9.6.1) with y(x1),y(x2) speciﬁed . (9.6.2) This problem is equivalent to extremizing the integral I=1 2/integraldisplayx2 x1/bracketleftbig p(x)(y/prime(x))2+q(x)(y(x))2/bracketrightbig dx. (9.6.3) 9.6 Differential Equations, Integral Equations, and Extremization of Integrals 279 ❑Example 9.15. Solve the Sturm–Liouville eigenvalue problem d dx/bracketleftbigg p(x)d dxy(x)/bracketrightbigg −q(x)y(x)=λr(x)y(x),x 1<x<x 2, (9.6.4) with y(x1)=y(x2)=0. (9.6.5) This problem is equivalent to extremizing the integral I=1 2/integraldisplayx2 x1/bracketleftbig p(x)(y/prime(x))2+q(x)(y(x))2/bracketrightbig dx, (9.6.6) while keeping J=1 2/integraldisplayx2 x1r(x)(y(x))2dx ﬁxed. (9.6.7) In practice, we ﬁnd the approximation to the lowest eigenvalue and the corresponding eigenfunction of the Sturm–Liouville eigenvalue problem by minimizing I/J . Note that an eigenfunction, good to ﬁrst order, yields an eigenvalue which is good to second order. ❑Example 9.16. Solve d2 dx2y(x)=−λy(x),0<x< 1, (9.6.8) with y(0) = y(1) = 0 . (9.6.9) Solution. We choose the trial function to be the one in Figure 9.2. Then I=2·h2/ε, J=h2(1−4 3ε),a n dI/J=2//bracketleftbig ε(1−4 3ε)/bracketrightbig , which has a minimum value of32 3atε=3 8. This is compared with the exact value, λ=π2. Note that λ=π2is a lower bound for I/J .I f we choose the trial function to be y(x)=x(1−x), we get I/J=1 0 . This is accurate to almost one percent. In order to obtain an accurate estimate of the eigenvalue, it is important to choose a trial function which satisﬁes the boundary condition and looks qualitatively like the expected so- lution. For instance, if we are calculating the lowest eigenvalue, it would be unwise to use atrial function which has a zero inside the interval. Let us now calculate the next eigenvalue. This is done by choosing the trial function y(x) which is orthogonal to the exact lowest eigenfunction u 0(x), i.e., /integraldisplay1 0y(x)u0(x)r(x)dx=0, (9.6.10) 280 9 Calculus of V ariations: Fundamentals εε 12−εh Fig. 9.2: The shape of the trial function for Example 9.16. and ﬁnd the minimum value of I/J with respect to some parameters in the trial function. Let us choose the trial function to be y(x)=x(1−x)(1−ax), and then the requirement that it is orthogonal to x(1−x), instead of u0(x)which is unknown, givesa=2 . Note that this trial function has one zero inside the interval [0,1]. This looks qualitatively like the expected solution. For this trial function, we have I J=4 2, and this compares well with the exact value, 4π2. ❑Example 9.17. Solve the Laplace equation ∇2φ=0, (9.6.11) withφgiven at the boundary. This problem is equivalent to extremizing I=/integraldisplay (/vector∇φ)2dV. (9.6.12) 9.6 Differential Equations, Integral Equations, and Extremization of Integrals 281 ❑Example 9.18. Solve the wave equation ∇2φ=k2φ, (9.6.13a) with φ=0 (9.6.13b) at the boundary. This problem is equivalent to extremizing /integraldisplay (/vector∇φ)2dV /integraldisplay φ2dV. (9.6.14) ❑Example 9.19. Estimate the lowest frequency of a circular drum of radius R. Solution. k2≤/integraldisplay (/vector∇φ)2dV /integraldisplay φ2dV. (9.6.15) Try a rotationally symmetric trial function, φ(r)=1−r R,0≤r≤R. (9.6.16) Then we get k2≤/integraldisplayR 0(φr(r))22πrdr /integraldisplayR 0(φ(r))22πrdr=6 R2. (9.6.17) This is compared with the exact value, k2=5.7832 R2. (9.6.18) Note that the numerator on the right-hand side of Eq. (9.6.18) is the square of the smallest zero in magnitude of the zeroth-order Bessel function of the ﬁrst kind, J0(kR). The homogeneous Fredholm integral equations of the second kind for the localized, monochromatic, and highly directive classical current distributions in two and three dimen-sions can be derived by maximizing the directivity Din the far ﬁeld while constraining C=N/T ,w h e r e Nis the integral of the square of the magnitude of the current density andTis proportional to the total radiated power. The homogeneous Fredholm integral equa- tions of the second kind and the inhomogeneous Fredholm integral equations of the secondkind are now derived from the calculus of variations in general terms. 282 9 Calculus of V ariations: Fundamentals ❑Example 9.20. Solve the homogeneous Fredholm integral equation of the second kind , φ(x)=λ/integraldisplayh 0K(x, x/prime)φ(x/prime)dx/prime, (9.6.19) where the square-integrable kernel is K(x, x/prime), and the projection is unity on ψ(x), /integraldisplayh 0ψ(x)φ(x)dx=1. (9.6.20) This problem is equivalent to extremizing the integral I=/integraldisplayh 0/integraldisplayh 0ψ(x)K(x, x/prime)φ(x/prime)dxdx/prime, (9.6.21) with respect to ψ(x), while keeping J=/integraldisplayh 0ψ(x)φ(x)dx=1 ﬁxed. (9.6.22) The extremization of Eq. (9.6.21) with respect to φ(x), while keeping Jﬁxed, results in the homogeneous adjoint integral equation forψ(x), ψ(x)=λ/integraldisplayh 0ψ(x/prime)K(x/prime,x)dx/prime. (9.6.23) With the real and symmetric kernel, K(x, x/prime)=K(x/prime,x), the homogeneous integral equa- tions for φ(x)andψ(x), Eqs. (9.6.19) and (9.6.23), become identical and Eq. (9.6.20) provides the normalization of φ(x)andψ(x)to unity, respectively. ❑Example 9.21. Solve the inhomogeneous Fredholm integral equation of the second kind , φ(x)−λ/integraldisplayh 0K(x, x/prime)φ(x/prime)dx/prime=f(x), (9.6.24) with the square-integrable kernel K(x, x/prime),0</bardblK/bardbl2<∞. This problem is equivalent to extremizing the integral I=/integraldisplayh 0/bracketleftBigg/braceleftBigg 1 2φ(x)−λ/integraldisplayh 0K(x, x/prime)φ(x/prime)dx/prime/bracerightBigg φ(x)+F(x)d dxφ(x)/bracketrightBigg dx, (9.6.25) where F(x)is deﬁned by F(x)=/integraldisplayx f(x/prime)dx/prime. (9.6.26) 9.7 The Second V ariation 283 9.7 The Second Variation The Euler equation isnecessary to extremize the integral, but it is not sufﬁcient .I n o r d e r t o ﬁnd out whether the solution of the Euler equation actually extremizes the integral, we must study the second variation . This is similar to the case of ﬁnding an extremum of a function. To conﬁrm that the point at which the ﬁrst derivative of a function vanishes is an extremum of the function, we must study the second derivatives. Consider the extremization of I=/integraldisplayx2 x1f(x, y, y/prime)dx, (9.7.1) with y(x1), andy(x2) speciﬁed . (9.7.2) Suppose y(x)is such a solution. Let us consider a weak variation , /braceleftBigg y(x)→y(x)+εν(x), y/prime(x)→y/prime(x)+εν/prime(x),(9.7.3) as opposed to a strong variation in which y/prime(x)is also varied, independent of εν/prime(x). Then I→I+εI1+1 2ε2I2+···, (9.7.4) where I1=/integraldisplayx2 x1/bracketleftbigg∂f ∂yν(x)+∂f ∂y/primeν/prime(x)/bracketrightbigg dx=/integraldisplayx2 x1/bracketleftbigg∂f ∂y−d dx∂f ∂y/prime/bracketrightbigg ν(x)dx, (9.7.5) and I2=/integraldisplayx2 x1/bracketleftbigg∂2f ∂y2ν2(x)+2∂2f ∂y∂y/primeν(x)ν/prime(x)+∂2f ∂y/prime2ν/prime2(x)/bracketrightbigg dx. (9.7.6a) Ify=y(x)indeed minimizes or maximizes I, then I2must be positive or negative for all variations ν(x)vanishing at the endpoints, when the solution of the Euler equation, y=y(x), is substituted into the integrand of I2. The integrand of I2is a quadratic form of ν(x)andν/prime(x). Therefore, this integral is always positive if /parenleftbigg∂2f ∂y∂y/prime/parenrightbigg2 −/parenleftbigg∂2f ∂y2/parenrightbigg/parenleftbigg∂2f ∂y/prime2/parenrightbigg <0, (9.7.7a) and ∂2f ∂y/prime2>0. (9.7.7b) 284 9 Calculus of V ariations: Fundamentals Thus, if the above conditions hold throughout x1<x<x 2, I2is always positive and y(x)minimizes I. Similar considerations hold, of course, for max- imizations. The above conditions are, however, too crude, i.e., are stronger than necessary.This is because ν(x)andν /prime(x)are not independent. Let us ﬁrst state the necessary and sufﬁcient conditions for the solution to the Euler equa- t i o nt og i v et h e weak minimum : Let P(x)≡fyy,Q(x)≡fyy/prime,R(x)≡fy/primey/prime, (9.7.8) where P(x),Q(x)andR(x)are evaluated at the point which extremizes the integral Ideﬁned by I≡/integraldisplayx2 x1f(x, y, y/prime)dx. We express I2in terms of P(x),Q(x)andR(x)as I2=/integraldisplayx2 x1[P(x)ν2(x)+2Q(x)ν(x)ν/prime(x)+R(x)ν/prime2(x)]dx. (9.7.6b) Then we have the following conditions for the weak minimum, Necessary condition Sufﬁcient condition Legendre test R(x)≥0.R (x)>0. Jacobi test ξ≥x2.ξ > x 2,or no such ξexists.(9.7.9) where ξis the conjugate point to be deﬁned later. Both conditions must be satisﬁed for sufﬁ- ciency and both are necessary separately. Before we go on to prove the above assertion, it is perhaps helpful to have an intuitive understanding of why these two tests are relevant. Let us consider the case in which R(x)is positive at x=a, and negative at x=b,w h e r e a andbare both between x1andx2. Let us ﬁrst choose ν(x)to be the function as in Figure 9.3. Note that ν(x)is of the order of ε, while ν/prime(x)is of the order√ ε. If we choose εto be sufﬁciently small, I2is positive. Next, we consider the same variation located at x=b.B y the same consideration, I2is negative for this variation. Thus y(x)does not extremize I. This shows that the Legendre test is a necessary condition. The relevance of the Jacobi test is best illustrated by the problem of ﬁnding the shortest path between two points on the surface of a sphere. The solution is obtained by going along the great circle on which these two points lie. There are, however, two paths connecting these two points on the great circle. One of them is truly the shortest path, while the other is neither the shortest nor the longest. Take the circle on the surface of the sphere which passes one of the two points. The other point at which this circle and the great circle intersect lies on one arcof the great circle which is neither the shortest nor the longest arc. This point is the conjugatepointξof this problem. 9.7 The Second V ariation 285 x1 a−ε a a+ε x2ε Fig. 9.3: The shape of ν(x)for the Legendre test. We discuss the Legendre test ﬁrst. We add to I2, (9.7.6b), the following term which is zero, /integraldisplayx2 x1d dx(ν2(x)ω(x))dx=/integraldisplayx2 x1(ν2(x)ω/prime(x)+2ν(x)ν/prime(x)ω(x))dx. (9.7.10) Then we have I2to be I2=/integraldisplayx2 x1/bracketleftBig/parenleftbig P(x)+ω/prime(x)/parenrightbig ν2(x)+2/parenleftbig Q(x)+ω(x)/parenrightbig ν(x)ν/prime(x) +R(x)ν/prime2(x)/bracketrightBig dx. (9.7.11) We require the integrand to be a complete square , i.e., (Q(x)+ω(x))2=(P(x)+ω/prime(x))R(x). (9.7.12) Hence, if we can ﬁnd ω(x)satisfying Eq. (9.7.12), we will have I2=/integraldisplayx2 x1R(x)/bracketleftbigg ν/prime(x)+Q(x)+ω(x) R(x)ν(x)/bracketrightbigg2 dx. (9.7.13) Now, it is not possible to have R(x)<0in any region between x1andx2for the minimum. IfR(x)<0in some region, we can solve the differential equation ω/prime(x)=−P(x)+(Q(x)+ω(x))2 R(x)(9.7.14) 286 9 Calculus of V ariations: Fundamentals forω(x)in this region. Restricting the variation ν(x)such that /braceleftBigg ν(x)≡0,outside the region where R(x)<0, ν(x)/negationslash=0,inside the region where R(x)<0, we can have I2<0. Thus it is necessary to have R(x)≥0for the entire region for the minimum. If P(x)R(x)−Q2(x)>0,P (x)>0, (9.7.15a) then we have no need to go further to ﬁnd ω(x). In many cases, however, we have P(x)R(x)−Q2(x)≤0, (9.7.15b) and so we have to examine further. The differential equation (9.7.14) for ω(x), can be rewrit- ten as (Q(x)+ω(x))/prime=−P(x)+Q/prime(x)+(Q(x)+ω(x))2 R(x), (9.7.16) which is the Riccatti differential equation .M a k i n gt h e Riccatti substitution , Q(x)+ω(x)=−R(x)u/prime(x) u(x), (9.7.17) we obtain the second-order linear ordinary differential equation for u(x), d dx/bracketleftbigg R(x)d dxu(x)/bracketrightbigg +(Q/prime(x)−P(x))u(x)=0. (9.7.18) Expressing everything in Eq. (9.7.13) in terms of u(x),I2becomes I2=/integraldisplayx2 x1/bracketleftBigg R(x)/parenleftbig ν/prime(x)u(x)−u/prime(x)ν(x)/parenrightbig2 u2(x)/bracketrightBigg dx. (9.7.19) If we can ﬁnd any u(x)such that u(x)/negationslash=0,x 1≤x≤x2, (9.7.20) we will have I2>0 forR(x)>0, (9.7.21) which is the sufﬁcient condition for the minimum. This completes the derivation of the Le- gendre test. We further clarify the Legendre test after the discussion of the conjugate point ξ of the Jacobi test. 9.7 The Second V ariation 287 The ordinary differential equation (9.7.18) for u(x)is related to the Euler equation by the inﬁnitesimal variation of the initial condition . In the Euler equation, fy(x, y, y/prime)−d dxfy/prime(x, y, y/prime)=0 (9.7.22) we make the following inﬁnitesimal variation, y(x)→y(x)+εu(x),ε=positive inﬁnitesimal . (9.7.23) Writing out this variation explicitly, fy(x, y+εu, y/prime+εu/prime)−d dxfy/prime(x, y+εu, y/prime+εu/prime)=0, we have, using the Euler equation, fyyu+fyy/primeu/prime−d dx(fy/primeyu+fy/primey/primeu/prime)=0, or, Pu+Qu/prime−d dx(Qu+Ru/prime)=0, (9.7.24) i.e., d dx/bracketleftbigg R(x)d dxu(x)/bracketrightbigg +/parenleftbig Q/prime(x)−P(x)/parenrightbig u(x)=0, which is simply the ordinary differential equation (9.7.18). The solution to the differential equation (9.7.18) corresponds to the inﬁnitesimal change in the initial condition. We further note that the differential equation (9.7.18) is of the self-adjoint form . Thus its Wronskian W(u1(x),u2(x))is given by W(x)≡u1(x)u/prime 2(x)−u/prime 1(x)u2(x)=C R(x), (9.7.25) where u1(x)andu2(x)are the linearly independent solutions of Eq. (9.7.18) and Cin Eq. (9.7.25) is some constant. We now discuss the conjugate point ξand the Jacobi test . We claim that the sufﬁcient condition for the minimum is that R(x)>0on(x1,x2)and that the conjugate point ξlies outside (x1,x2), i.e., both the Legendre test and the Jacobi test are satisﬁed. Suppose that R(x)>0 on(x1,x2), (9.7.26) which implies that the Wronskian has the same sign on (x1,x2). Suppose that u1(x)vanishes only atx=ξ1andx=ξ2, u1(ξ1)=u1(ξ2)=0,x 1<ξ1<ξ2<x2. (9.7.27) 288 9 Calculus of V ariations: Fundamentals We claim that u2(x)must vanish at least once between ξ1andξ2. The Wronskian Wevaluated atx=ξ1andx=ξ2is given respectively by W(ξ1)=−u/prime 1(ξ1)u2(ξ1), (9.7.28a) and W(ξ2)=−u/prime 1(ξ2)u2(ξ2). (9.7.28b) By the continuity of u1(x)on(x1,x2),u/prime 1(ξ1)has the opposite sign to u/prime 1(ξ2), i.e., u/prime 1(ξ1)u/prime 1(ξ2)<0. (9.7.29) But, we have W(ξ1)W(ξ2)=u/prime 1(ξ1)u/prime 1(ξ2)u2(ξ1)u2(ξ2)>0, (9.7.30) from which we conclude that u2(ξ1)has the opposite sign to u2(ξ2), i.e., u2(ξ1)u2(ξ2)<0. (9.7.31) Hence u2(x)must vanish at least once between ξ1andξ2. Since the u(x)provide the inﬁnitesimal variation of the initial condition, we choose u1(x) andu2(x)to be u1(x)≡∂ ∂αy(x,α,β ),u 2(x)≡∂ ∂βy(x,α,β ), (9.7.32) where y(x,α,β )is the solution of the Euler equation, fy−d dxfy/prime=0, (9.7.33a) with the initial conditions, y(x1)=α, y (x2)=β, (9.7.33b) andui(x)(i=1,2) satisfy the differential equation (9.7.18). We now claim that the sufﬁcient condition for the weak minimum is that R(x)>0, andξ>x 2. (9.7.34) We construct U(x)by U(x)≡u1(x)u2(x1)−u1(x1)u2(x), (9.7.35) which vanishes at x=x1. We deﬁne the conjugate point ξas the solution of the equation, U(ξ)=0,ξ /negationslash=x1. (9.7.36) 9.7 The Second V ariation 289 The function U(x)represents another inﬁnitesimal change in the solution to the Euler equation which would also pass through x=x1.S i n c e U(x1)=U(ξ)=0, (9.7.37) there exists another solution u(x)such that u(x)=0 forx∈(x1,ξ). (9.7.38) We choose x3such that x2<x3<ξ (9.7.39) and another solution u(x)to be u(x)=u1(x)u2(x3)−u1(x3)u2(x),u (x3)=0. (9.7.40) We claim that u(x)/negationslash=0 on(x1,x2). (9.7.41) Suppose that u(x)=0 in this interval. Then any other solution of the differential equation (9.7.18) must vanish between these points. But, U(x)does not vanish. This completes the derivation of the Jacobi test. We further clarify the Legendre test and the Jacobi test. We now assume that ξ<x 2, andR(x)>0 forx∈(x1,x2), (9.7.42) and show that I2<0 for some ν(x) such that ν(x1)=ν(x2)=0. (9.7.43) We choose x3such that ξ<x 3<x2. We construct the solution U(x)to Eq. (9.7.18) such that U(x1)=U(ξ)=0. Also, we construct the solution ±v(x), independent of U(x), such that v(x3)=0, i.e., we choose x3such that U(x3)/negationslash=0. We choose the sign of v(x)such that the Wronskian W(U(x),v(x))is given by W(U(x),v(x)) =U(x)v/prime(x)−v(x)U/prime(x)=C R(x),R(x)>0,C > 0. 290 9 Calculus of V ariations: Fundamentals The function U(x)−v(x)solves the differential equation (9.7.18). It must vanish at least once between x1andξ. We call this point x=a,s ot h a t U(a)=v(a), and x1<a<ξ<x 3<x2. We deﬁne ν(x)to be ν(x)≡  U(x),forx∈(x 1,a), v(x), forx∈(a,x3), 0, forx∈(x3,x2). InI2, we rewrite the term involving Q(x)as 2Q(x)ν(x)ν/prime(x)=Q(x)d(ν2(x)), and we perform integration by parts in I2to obtain I2=/bracketleftbig Q(x)ν2(x)+R(x)ν/prime(x)ν(x)/bracketrightbig/vextendsingle/vextendsingle/vextendsinglex2 x1 −/integraldisplayx2 x1ν(x)/bracketleftbig R(x)ν/prime/prime(x)+R/prime(x)ν/prime(x)+(Q/prime(x)−P(x))ν(x)/bracketrightbig dx. The integral in the second term on the right-hand side is broken up into three separate integrals, each of which vanishes identically since ν(x)satisﬁes the differential equation (9.7.18) in all three regions. The Q(x)term of the integrated part of I2also vanishes, i.e., Q(x)ν2(x)/vextendsingle/vextendsinglea x1+Q(x)ν2(x)/vextendsingle/vextendsinglex3 a+Q(x)ν2(x)/vextendsingle/vextendsinglex2 x3=0. We consider now the R(x)term of the integrated part of I2, I2=R(x)ν/prime(x)ν(x)|a−ε−R(x)ν/prime(x)ν(x)|a+ε =R(a)[U/prime(x)v(x)−v/prime(x)U(x)]x=a, where the continuity of U(x)andv(x)atx=ais used. Thus we have I2=R(a)/bracketleftbig U/prime(a)v(a)−v/prime(a)U(a)/bracketrightbig =−R(a)W(U(a),v(a)) =−R(a)/bracketleftbiggC R(a)/bracketrightbigg =−C<0, i.e., I2=−C<0. This ends the clariﬁcation of both the Legendre and the Jacobi tests. 9.7 The Second V ariation 291 ❑Example 9.22. Catenary . Discuss a solution of soap ﬁlm sustained between two circular wires. Solution. The surface area is given by I=/integraldisplay+L −L2πy/radicalbig 1+y/prime2dx. Thusf(x, y, y/prime)is given by f(x, y, y/prime)=y/radicalbig 1+y/prime2, and is independent of x. Then we have f−y/primefy/prime=α, where αis an arbitrary integration constant. After a little algebra, we have dy /radicalBig y2 α2−1=±dx. We perform a change of variable as follows, y=αcoshθ, dy =αsinhθ·dθ. H e n c ew eh a v e αd θ=±dx, or, θ=±/parenleftbiggx−β α/parenrightbigg , i.e., y=αcosh/parenleftbiggx−β α/parenrightbigg , where αandβare arbitrary constants of integration, to be determined from the boundary conditions at x=±L. We have, as the boundary conditions, R1=y(−L)=αcosh/parenleftbiggL+β α/parenrightbigg ,R 2=y(+L)=αcosh/parenleftbiggL−β α/parenrightbigg . For the sake of simplicity, we assume R1=R2≡R. 292 9 Calculus of V ariations: Fundamentals Then we have β=0,R L=α LcoshL α, and y=αcoshx α. Setting v=L α, the boundary conditions at x=±Lread as R L=coshv v. Deﬁning the function G(v)by G(v)≡coshv v, (9.7.44) G(v)is plotted in Figure 9.4. If the geometry of the problem is such that R L>1.5089, then there exist two candidates for the solution atv=v<andv=v>,w h e r e 0<v<<1.1997<v>, and if R L<1.5089, then there exists no solution . If the geometry is such that R L=1.5089, then there exists one candidate for the solution at v<=v>=v=L α=1.1997. We apply two tests for the minimum . 9.7 The Second V ariation 293 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.51234567891011Gv() v<v >v Fig. 9.4: Plot of G(v). Legendre test: fy/primey/prime=y (1 +y/prime2)3/2=R>0, thus passing the Legendre test . Jacobi test: Since y(x,α,β )=αcosh/parenleftbiggx−β α/parenrightbigg , we have u1(x)=∂y ∂α= cosh/parenleftbiggx−β α/parenrightbigg −/parenleftbiggx−β α/parenrightbigg sinh/parenleftbiggx−β α/parenrightbigg , and u2(x)=−∂y ∂β=s i n h/parenleftbiggx−β α/parenrightbigg , 294 9 Calculus of V ariations: Fundamentals where the minus sign for u2(x)does not matter. We construct a solution U(x)which vanishes atx=x1, U(x)=( c o s h˜ v−˜vsinh ˜v)s i n h˜ v1−sinh ˜v(cosh ˜v1−˜v1sinh ˜v1), where ˜v≡x−β α,˜v1≡x1−β α. Then we have U(x) sinh ˜vsinh ˜v1=( c o t h˜ v−˜v)−(coth ˜v1−˜v1). Deﬁning the function F(˜v)by F(˜v)≡coth ˜v−˜v, (9.7.45) F(˜v)is plotted in Figure 9.5. ~vFv(~) ~v1 ~v 1~v2 ~v2~vξ ~vξ Fig. 9.5: Plot of F(˜v). We have U(x) sinh ˜vsinh ˜v1=F(˜v)−F(˜v1). Note that F(˜v)is an odd function of ˜v, F(−˜v)=−F(˜v). 9.7 The Second V ariation 295 We set ˜v1≡−L α,˜v2≡+L α,˜vξ≡ξ α, where β=0, is used. The equation, U(ξ)=0,ξ /negationslash=x1, which determines the conjugate point ξ, is equivalent to the following equation, F(˜vξ)=F(˜v1),˜vξ/negationslash=˜v1. If F(˜v1)>0, (9.7.46) then, from Figure 9.5, we have ˜v1<˜vξ<˜v2, thus failing the Jacobi test . If, on the other hand, F(˜v1)<0, (9.7.47) then, from Figure 9.5, we have ˜v1<˜v2<˜vξ, thus passing the Jacobi test . The dividing line F(˜v1)=0, corresponds to ˜v1<˜v2=˜vξ, and thus we have one solution at ˜v2=−˜v1=L α=1.1997. Having derived the particular statements of the Jacobi test as applied to this example, Eq. (9.7.46) and (9.7.47), we now test which of the two candidates for the solution, v=v< andv=v>, is actually the minimizing solution. Two functions, G(v)andF(v),d e ﬁ n e db y Eq. (9.7.44) and (9.7.45), are related to each other through d dvG(v)=−/parenleftbiggsinhv v2/parenrightbigg F(v). (9.7.48) 296 9 Calculus of V ariations: Fundamentals Atv=v<, we know from Figure 9.4 that d dvG(v<)<0, which implies F(−v<)=−F(v<)<0, so that one candidate ,v=v<,passes the Jacobi test and is the solution , whereas at v=v>, we know from Figure 9.4 that d dvG(v>)>0, which implies F(−v>)=−F(v>)>0, so that the other candidate, v=v>, fails the Jacobi test and is not the solution . When two candidates, v=v<andv=v>, coalesce to a single point , v=1.1997, where the ﬁrst derivative of G(v)vanishes, i.e., d dvG(v)=0, v<=v>=1.1997 is the solution . We now consider a strong variation and the condition for the strong minimum .I n t h e strong variation, since the varied derivatives behave very differently from the original deriva- tives, we cannot expand the integral Iin a Taylor series. Instead, we consider the W eierstrass Efunction deﬁned by E(x, y0,y/prime 0,p)≡f(x, y0,p)−[f(x, y0,y/prime 0)+(p−y/prime 0)fy/prime(x, y0,y/prime 0)]. (9.7.49) Necessary and sufﬁcient conditions for the strong minimum are given by Necessary condition Sufﬁcient condition 1)fy/primey/prime(x, y0,y/prime 0)≥0,w h e r e y0is the solution of the Euler equation.1)fy/primey/prime(x, y, p)>0, for every (x, y) close to (x, y0), and every ﬁnite p. 2)ξ≥x2.2 ) ξ>x 2. (9.7.50) 3)E(x, y0,y/prime 0,p)≥0, for all ﬁnite p,a n dx∈[x1,x2] 9.8 W eierstrass–Erdmann Corner Relation 297 We note that if p∼y/prime 0, then we have E∼fy/primey/prime(x, y0,y/prime 0)(p−y/prime 0)2 2!, just like the mean value theorem of the ordinary function f(x), which is given by f(x)−f(x0)−f/prime(x0)=(x−x0)2 2!f/prime/prime(x0+λ(x−x0)),0<λ< 1. 9.8 Weierstrass–Erdmann Corner Relation In this section, we consider the variational problem with the solutions which are the piecewise continuous functions with corners , i.e., the function itself is continuous, but the derivative is not. We maximize the integral I=/integraldisplayx2 x1f(x, y, y/prime)dx. (9.8.1) We put a point of the corner at x=a. We write the solution for x≤a,a sy(x)and, for x>a , asY(x). Then we have from the continuity of the solution, y(a)=Y(a). (9.8.2) Now, consider the variation of y(x)andY(x)of the following forms, y(x)→y(x)+εν(x),Y(x)→Y(x)+εV(x), (9.8.3a) and y/prime(x)→y/prime(x)+εν/prime(x),Y/prime(x)→Y/prime(x)+εV/prime(x). (9.8.3b) Under these variations, we require Ito be stationary, I=/integraldisplaya x1f(x, y, y/prime)dx+/integraldisplayx2 af(x, Y, Y/prime)dx. (9.8.4) First, we consider the variation problem with the point of the discontinuity of the derivative atx=aﬁxed. We have ν(a)=V(a). (9.8.5) Performing the above variations, we have δI=/integraldisplaya x1[fyεν+fy/primeεν/prime]dx+/integraldisplayx2 a[fYεV+fY/primeεV/prime]dx =fy/primeεν/vextendsingle/vextendsingle/vextendsinglex=a x=x1+/integraldisplaya x1εν/bracketleftbigg fy−d dxfy/prime/bracketrightbigg dx +fY/primeεV/vextendsingle/vextendsingle/vextendsinglex=x2 x=a+/integraldisplayx2 aεV/bracketleftbigg fY−d dxfY/prime/bracketrightbigg dx=0. 298 9 Calculus of V ariations: Fundamentals If the variations vanish at both ends ( x=x1andx=x2), i.e., ν(x1)=V(x2)=0, (9.8.6) we have the following equations, fy−d dxfy/prime=0,x ∈[x1,x2] (9.8.7) and fy/prime|x=a−=fY/prime|x=a+, namely fy/primeis continuous at x=a. (9.8.8) Next, we consider the variation problem with the point of the discontinuity of the derivative atx=avaried, i.e., a→a+∆a. (9.8.9) The point of the discontinuity becomes shifted, and yet the solutions are continuous at x= a+∆a, i.e., y(a)+y/prime(a)∆a+εν(a)=Y(a)+Y/prime(a)∆a+εV(a), or, [y/prime(a)−Y/prime(a)]∆a=ε[V(a)−ν(a)], (9.8.10) which is the condition on ∆a,V(a)andν(a). The integral Ibecomes changed into I→/integraldisplaya+∆a x1f(x, y+εν, y/prime+εν/prime)dx+/integraldisplayx2 a+∆af(x, Y+εV, Y/prime+εV/prime)dx =/integraldisplaya x1f(x, y+εν, y/prime+εν/prime)dx+f(x, y, y/prime)/vextendsingle/vextendsingle/vextendsingle x=a−∆a +/integraldisplayx2 af(x, Y+εV, Y/prime+εV/prime)dx−f(x, Y, Y/prime)/vextendsingle/vextendsingle/vextendsingle x=a+∆a. The integral parts, after the integration by parts, vanish due to the Euler equation in the re- spective region, and what remain are the integrated parts, i.e., δI=ενfy/prime|x=a−−εV f Y/prime|x=a++(f|x=a−−f|x=a+)∆a =ε(ν−V)fy/prime|x=a+(f|x=a−−f|x=a+)∆a=0,(9.8.11) where the ﬁrst term of the second line above follows from the continuity of fy/prime.F r o m t h e continuity condition (9.8.10) and the expression (9.8.11), by eliminating ∆a, we obtain −[y/prime(a)−Y/prime(a)]fy/prime+(f|x=a−−f|x=a+)=0, 9.8 W eierstrass–Erdmann Corner Relation 299 or, f−y/prime(a)fy/prime|x=a−=f−Y/prime(a)fy/prime|x=a+, i.e., f−y/primefy/prime continuous at x=a. (9.8.12) For the solution of the variation problem with the discontinuous derivative, we have fy−d dxfy/prime=0,Euler equation , fy/prime continuous at x=a, f−y/primefy/prime continuous at x=a,(9.8.13) which are called the W eierstrass–Erdmann corner relation . ❑Example 9.23. Extremize I=/integraldisplay1 0(y/prime+1 )2y/prime2dx with the endpoints ﬁxed as below, y(0) = 2 ,y(1) = 0 . Are there solutions with discontinuous derivatives? Find the minimum value of I. Solution. We have f(x, y, y/prime)=(y/prime+1 )2y/prime2, which is independent of x. Thus f−y/primefy/prime=constant, where fy/primeis calculated to be fy/prime=2 (y/prime+1 )y/prime(2y/prime+1 ). H e n c ew eh a v e f−y/primefy/prime=−(3y/prime+1 ) (y/prime+1 )y/prime2. 1) If we want to have a continuous solution alone, we have y/prime=a(constant), from which, we conclude that y=ax+b. From the endpoint conditions, the solution is y=2 (−x+1 ),y/prime=−2. Thus the integral Iis evaluated to be Icont.=4 . 300 9 Calculus of V ariations: Fundamentals 2) Suppose that y/primeisdiscontinuous atx=a. Setting p=y/prime <,p/prime=y/prime >, we have, from the corner relation at x=a, /braceleftBigg p(p+ 1)(2 p+1 ) = p/prime(p/prime+ 1)(2 p/prime+1 ), (3p+1 ) (p+1 )p2=( 3 p/prime+1 ) (p/prime+1 )p/prime2. Setting u=p+p/prime,v=p2+pp/prime+p/prime2, we have /braceleftBigg 3u+2v+1=0 , 3u(2v−u2)+u+4v=0,⇒/braceleftBigg p=0, p/prime=−1,or/braceleftBigg p=−1, p/prime=0. Thus the discontinuous solution gives y/prime+1=0 ,o ry/prime=0 . Then we have Idisc.=0< Icont.=4 . In the above example, the solution y(x)itself became discontinuous. Depending on the boundary conditions, the discontinuous solution may not be present. 9.9 Problems for Chapter 9 9.1 (Due to H. C.) Find an approximate value for the lowest eigenvalue E0for /bracketleftbigg∂2 ∂x2+∂2 ∂y2−x2y2/bracketrightbigg ψ(x, y)=−Eψ(x, y),−∞<x ,y< ∞, where ψ(x, y)is normalized to unity, /integraldisplay+∞ −∞/integraldisplay+∞ −∞|ψ(x, y)|2dxdy=1. 9.2 (Due to H. C.) A light ray in the x-yplane is incident on the lower half-plane ( y≤0)o f the medium with an index of refraction n(x, y)given by n(x, y)=n0(1 +ay),y ≤0, where n0andaare positive constants. a) Find the path of a ray passing through (0,0)and(l,0). b) Find the apparent depth of the object located at (0,0)when viewed from (l,0). 9.9 Problems for Chapter 9 301 9.3 (Due to H. C.) Estimate the lowest frequency of a circular drum of radius Rwith a rotationally symmetric trial function, φ(r)=1−/parenleftBigr R/parenrightBign ,0≤r≤R. Find the nwhich gives the best estimate. 9.4 Minimize the integral I≡/integraldisplay2 1x2(y/prime)2dx, with the endpoints ﬁxed as below, y(1) = 0 ,y(2) = 1 . Apply the Legendre test and the Jacobi test to determine if your solution is a minimum. Is it a strong minimum or a weak minimum? Are there solutions with discontinuous derivatives? 9.5 Extremize /integraldisplay1 0(1 +y2)2 (y/prime)2dx, with the endpoints ﬁxed as below, y(0) = 0 ,y(1) = 1 . Is your solution a weak minimum? Is it a strong minimum? Are there solutions with discontinuous derivatives? 9.6 Extremize /integraldisplay1 0(y/prime2−y/prime4)dx, with the endpoints ﬁxed as below, y(0) = y(1) = 0 . Is your solution a weak minimum? Is it a strong minimum? Are there solutions with discontinuous derivatives? 9.7 Extremize /integraldisplay2 0(xy/prime+y/prime2)dx, with the endpoints ﬁxed as below, y(0) = 1 ,y(2) = 0 . Is your solution a weak minimum? Is it a strong minimum? Are there solutions with discontinuous derivatives? 302 9 Calculus of V ariations: Fundamentals 9.8 Extremize /integraldisplay2 1x3 y/prime2dx, with the endpoints ﬁxed as below, y(1) = 1 ,y(2) = 4 . Is your solution a weak minimum? Is it a strong minimum? Are there solutions with discontinuous derivatives? 9.9 (Due to H. C.) An airplane with speed v0ﬂies in a wind of speed ax. What is the trajectory of the airplane if it is to enclose the greatest area in a given amount of time? Hint: Y ou may assume that the velocity of the airplane is dx dt/vectorex+dy dt/vectorey=(vx+ax)/vectorex+vy/vectorey, with v2 x+v2 y=v2 0. 9.10 (Due to H. C.) Find the shortest distance between two points on a cylinder. Apply the Jacobi test and the Legendre test to verify if your solution is a minimum. 10 Calculus of Variations: Applications 10.1 Feynman’s Action Principle in Quantum Mechanics In this section, we discuss Feynman’s action principle in quantum mechanics. The deriva- tion of Feynman’s action principle in quantum mechanics can be found in the monograph byM. Masujima, but we shall be content here with the deduction of the canonical formalism ofquantum mechanics from Feynman’s action principle for the nonsingular Lagrangian . The operator ˆq(t)at all time tforms the complete set of the commuting operators, i.e., the quantum mechanical state vector |ψ(t)>can be expressed in terms of the linear superposition of the complete set of the eigenket |q,t > of the commuting operator ˆq(t). Feynman’s action principle asserts that the transformation function <q /prime/prime,t/prime/prime|q/prime,t/prime>is given by <q/prime/prime,t/prime/prime|q/prime,t/prime>=1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]exp/bracketleftBigg i /planckover2pi1/integraldisplayt/prime/prime t/primedt L(q(t),˙q(t))/bracketrightBigg ,(10.1.1) Ω=Ω ( t/prime/prime,t/prime)=the “space–time region” sandwiched between t/primeandt/prime/prime. We state here three assumptions. (A-1) In the “space–time region”, Ω1+Ω2,w h e r e Ω1andΩ2are neighboring each other, we have the principle of superposition in the following form, /integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)] =/integraldisplay t=t/prime/prime/primedq/prime/prime/prime/integraldisplayq2(t/prime/prime)=q/prime/prime q2(t/prime/prime/prime)=q/prime/prime/primeD[q2(t)]/integraldisplayq1(t/prime/prime/prime)=q/prime/prime/prime q1(t/prime)=q/primeD[q1(t)].(10.1.2) (A-2) Functional integration by parts is allowed. The requisite damping factor which kills the contribution from the functional inﬁnities will be supplied by an “ iε” piece which originates from the wave function of the vacuum at t=∓∞. (A-3) We have the following resolution of identity , /integraldisplay |q/prime,t/prime>d q/prime<q/prime,t/prime|=1. (10.1.3) Applied Mathematics inTheor etical Physics. Mic hio Masujima Copyright ©2005 Wiley-VC HVerlag GmbH &Co.KGaA, Weinheim ISBN: 3-527-40534-8 304 10 Calculus of V ariations: Applications From the consistency of (A-1), (A-2) and (A-3), the normalization constant N(Ω)must satisfy the following equation, N(Ω1+Ω2)=N(Ω1)N(Ω2). (10.1.4) Equation (10.1.4) also originates from the additivity of the action functional, I[q;t/prime/prime,t/prime]=I[q;t/prime/prime,t/prime/prime/prime]+I[q;t/prime/prime/prime,t/prime], where the action functional is given by, I[q;t/prime/prime,t/prime]=/integraldisplayt/prime/prime t/primedtL(q(t),˙q(t)). In Feynman’s action principle, the operator ˆq(t)is deﬁned by its matrix elements, <q/prime/prime,t/prime/prime|ˆq(t)|q/prime,t/prime>=1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]q(t)e x p/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg .(10.1.5) Iftofˆq(t)is on the t/prime/prime-surface, we have <q/prime/prime,t/prime/prime|ˆq(t/prime/prime)|q/prime,t/prime>=q/prime/prime<q/prime/prime,t/prime/prime|q/prime,t/prime>, (10.1.6) and from the assumed completeness of the eigenket |q/prime,t/prime>,w eh a v e <q/prime/prime,t/prime/prime|ˆq(t/prime/prime)=q/prime/prime<q/prime/prime,t/prime/prime|. (10.1.7) Equation (10.1.7) is the deﬁning equation of the eigenbra <q/prime/prime,t/prime/prime|ofˆq(t). Consider the variation of the action functional, δI[q;t/prime/prime,t/prime]=/integraldisplayt/prime/prime t/primedt/bracketleftbigg∂L(q(t),˙q(t)) ∂q(t)−d dt/parenleftbigg∂L(q(t),˙q(t)) ∂˙q(t)/parenrightbigg/bracketrightbigg δq(t) +/integraldisplayt/prime/prime t/primedtd dt/bracketleftbigg∂L(q(t),˙q(t)) ∂˙q(t)δq(t)/bracketrightbigg .(10.1.8) We ﬁrst employ the variation which vanishes at the endpoints, δq(t/prime)=δq(t/prime/prime)=0 . Then the second term of Eq. (10.1.8) vanishes and we obtain the Euler derivative, δI[q;t/prime/prime,t/prime] δq(t)=∂L(q(t),˙q(t)) ∂q(t)−d dt/parenleftbigg∂L(q(t),˙q(t)) ∂˙q(t)/parenrightbigg . (10.1.9) We now evaluate the matrix elements of the operator, δI[ˆq;t/prime/prime,t/prime]/δˆq(t), in accordance with Feynman’s action principle. /angbracketleftbigg q/prime/prime,t/prime/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingleδI[ˆq;t/prime/prime,t/prime] δˆq(t)/vextendsingle/vextendsingle/vextendsingle/vextendsingleq/prime,t/prime/angbracketrightbigg =1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]δI[q;t/prime/prime,t/prime] δq(t)exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg =1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]/planckover2pi1 iδ δq(t)exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg =0.(10.1.10) 10.1 Feynman’s Action Principle in Quantum Mechanics 305 From the assumed completeness of the eigenket |q/prime,t/prime>, we shall obtain the Lagrange equa- tion of motion at the operator level ,δI[ˆq]/δˆq(t)=0 , i.e., we obtain, d dt/parenleftBigg ∂L(ˆq(t),˙ˆq(t)) ∂˙ˆq(t)/parenrightBigg −∂L(ˆq(t),˙ˆq(t)) ∂ˆq(t)=0. (10.1.11) As for the time-ordered product , we can obtain the following identity by mathematical induction, starting from n=2, by the repeated use of Eqs. (10.1.2) and (10.1.3). 1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]q(t1)·...·q(tn)e x p/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg =<q/prime/prime,t/prime/prime|T[ˆq(t1)·...·ˆq(tn)]|q/prime,t/prime>.(10.1.12) We deﬁne the momentum operator ˆp(t)as the displacement operator, <q/prime/prime,t/prime/prime|ˆp(t/prime/prime)|q/prime,t/prime>=/planckover2pi1 i∂ ∂q/prime/prime<q/prime/prime,t/prime/prime|q/prime,t/prime>. (10.1.13) In order to express the right-hand side of Eq. (10.1.13) in the form in which we can use Feynman’s action principle, we consider the following variation. (1) Inside Ω, we take the inﬁnitesimal variation of q(t),q(t)→q(t)+δq(t), δq(t/prime)=0,δ q(t)=ξ(t),δ q(t/prime/prime)=ξ/prime/prime. (10.1.14) (2) Inside Ω, we assume that the physical system evolves with time tin accordance with the Lagrange equation of motion . The response of the action functional I[q;t/prime/prime,t/prime]to the variation, (10.1.14), is given by δI[q;t/prime/prime,t/prime]=/integraldisplayt/prime/prime t/primedtd dt/bracketleftbigg∂L(q(t),˙q(t) ∂˙q(t)δq(t)/bracketrightbigg =∂L(q(t/prime/prime),˙q(t/prime/prime)) ∂˙q(t/prime/prime)ξ/prime/prime. (10.1.15) Thus we obtain δI[q;t/prime/prime,t/prime] δq(t/prime/prime)=∂L(q(t/prime/prime),˙q(t/prime/prime)) ∂˙q(t/prime/prime). (10.1.16) 306 10 Calculus of V ariations: Applications Using Eq. (10.1.16), the right-hand side of Eq. (10.1.13) can be expressed as, /planckover2pi1 i∂ ∂q/prime/prime<q/prime/prime,t/prime/prime|q/prime,t/prime>=/planckover2pi1 ilim ξ/prime/prime→0<q/prime/prime+ξ/prime/prime,t/prime/prime|q/prime,t/prime>−<q/prime/prime,t/prime/prime|q/prime,t/prime> ξ/prime/prime =/planckover2pi1 ilim ξ/prime/prime→01 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg ×1 ξ/prime/prime/braceleftbigg exp/bracketleftbiggi /planckover2pi1/parenleftBig I[q+ξ/prime/prime,t/prime/prime,t/prime]−I[q,t/prime/prime,t/prime]/parenrightBig/bracketrightbigg −1/bracerightbigg = lim ξ/prime/prime→01 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg ×1 ξ/prime/prime/bracketleftBig I[q+ξ/prime/prime;t/prime/prime,t/prime]−I[q;t/prime/prime,t/prime]+O/parenleftbig (ξ/prime/prime)2/parenrightbig/bracketrightBig =1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg ×δI[q;t/prime/prime,t/prime] δq(t/prime/prime) =1 N(Ω)/integraldisplayq(t/prime/prime)=q/prime/prime q(t/prime)=q/primeD[q(t)]exp/bracketleftbiggi /planckover2pi1I[q;t/prime/prime,t/prime]/bracketrightbigg ×∂L(q(t/prime/prime),˙q(t/prime/prime)) ∂˙q(t/prime/prime) =<q/prime/prime,t/prime/prime|∂L(ˆq(t/prime/prime),˙ˆq(t/prime/prime)) ∂· ˆq(t/prime/prime)|q/prime,t/prime>.(10.1.17) From Eqs. (10.1.13) and (10.1.17), and the assumed completeness of the eigenket |q/prime,t/prime>, we obtain the operator identity , ˆp(t)=∂L(ˆq(t),˙ˆq(t)) ∂˙ˆq(t). (10.1.18) Equation (10.1.18) is the deﬁnition of the momentum ˆp(t)canonically conjugate to ˆq(t). We now consider the equal time canonical (anti-)commutator ofˆp(t)andˆq(t). For the Bose system ,w eh a v e <q/prime/prime,t/prime/prime|ˆqB(t/prime/prime)|q/prime,t/prime>=q/prime/prime B<q/prime/prime,t/prime/prime|q/prime,t/prime>. (10.1.19) 10.1 Feynman’s Action Principle in Quantum Mechanics 307 From Eqs. (10.1.13) and (10.1.19), we have, <q/prime/prime,t/prime/prime|ˆpB(t/prime/prime)ˆqB(t/prime/prime)|q/prime,t/prime>=/planckover2pi1 i∂ ∂q/prime/prime B(q/prime/prime B<q/prime/prime,t/prime/prime|q/prime,t/prime>) =/planckover2pi1 i<q/prime/prime,t/prime/prime|q/prime,t/prime> +q/prime/prime B<q/prime/prime,t/prime/prime|ˆpB(t/prime/prime)|q/prime,t/prime> =/planckover2pi1 i<q/prime/prime,t/prime/prime|q/prime,t/prime> +<q/prime/prime,t/prime/prime|ˆqB(t/prime/prime)ˆpB(t/prime/prime)|q/prime,t/prime>.(10.1.20) Thus, from the assumed completeness of the eigenket |q/prime,t/prime>, we obtain, [ˆpB(t),ˆqB(t)] =/planckover2pi1 i, (10.1.21) where the commutator, [A,B],i sd e ﬁ n e db y [A,B]≡AB−BA. (10.1.22) For the Fermi system , we have a minus sign in front of the second terms of the third line and the ﬁfth line of Eq. (10.1.20) which originates from the anti-commuting Fermion number, so that we obtain, {ˆpF(t),ˆqF(t)}=/planckover2pi1 i, (10.1.23) where the anti-commutator, {A,B},i sd e ﬁ n e db y {A,B}≡AB+BA. (10.1.24) We deﬁne the Hamiltonian as the Legendre transform of the Lagrangian, H(ˆq(t),ˆp(t)) = ˆp(t)˙ˆq(t)−L(ˆq(t),˙ˆq(t)), (10.1.25) where Eq. (10.1.18) is solved for ˙ˆq(t)as a function of ˆq(t)andˆp(t), and this ˙ˆq(t)is substituted into the right-hand side of Eq. (10.1.25). Proof that the Heisenberg equation of motion follows from Eqs. (10.1.11), (10.1.18), (10.1.21), (10.1.23) and (10.1.25) is left as an exercise for the reader. Another proof that the transformation function <q/prime/prime,t/prime/prime|q/prime,t/prime>satisﬁes the Schrödinger equation with respect toq/prime/primeandt/prime/primewith a Lagrangian of the form, L(q(t),˙q(t)) =1 2m˙q(t)2−V(q(t)), (10.1.26) is discussed in the monograph by M. Masujima, cited at the beginning of this section. 308 10 Calculus of V ariations: Applications We have thus deduced the canonical formalism of quantum mechanics from Feyn- man’s action principle for the nonsingular Lagrangian. The extension of the present discussion to the case of quantum ﬁeld theory with the nonsingular Lagrangian densityL/parenleftbig φ(x),∂ µφ(x)/parenrightbig is straightforward, resulting in the normal dependence of the Hamiltonian density H(ˆφ(x),/vector∇ˆφ(x),ˆπ(x))and the momentum ˆπ(x)canonically conjugate to ˆφ(x). In view of the discussions in this section and Section 10.3 which follows, we eventually show the equivalence of path integral quantization and canonical quantization at least for thenonsingular Lagrangian (density); canonical quantization ⇔path integral quantization. It is in fact astonishingly difﬁcult to establish this equivalence for the singular Lagrangian (density). 10.2 Feynman’s Variational Principle in Quantum Statistical Mechanics In this section, we shall brieﬂy consider Feynman’s variational principle in quantum statistical mechanics which is based on the analytic continuation in time from a real time to an imaginary time of Feynman’s action principle in quantum mechanics. We consider the canonical ensemble with the Hamiltonian ˆH({/vectorqj,/vectorpj}N j=1)at ﬁnite tem- perature. The density matrix ˆρC(β)of this system satisﬁes the Bloch equation, −/planckover2pi1∂ ∂τˆρC(τ)=ˆH({/vectorqj,/vectorpj}N j=1)ˆρC(τ),0≤τ≤β, (10.2.1) with its formal solution given by ˆρC(τ)=e x p/bracketleftBigg −τˆH/parenleftbig {/vectorqj,/vectorpj}N j=1/parenrightbig /planckover2pi1/bracketrightBigg ˆρC(0). (10.2.2) We compare the Bloch equation and the density matrix, Eqs. (10.2.1) and (10.2.2), with the Schrödinger equation for the state vector |ψ,t > , i/planckover2pi1d dt|ψ,t > =ˆH({/vectorqj,/vectorpj}N j=1)|ψ,t > , (10.2.3) and its formal solution given by |ψ,t > =e x p/bracketleftBigg −itˆH({/vectorqj,/vectorpj}N j=1) /planckover2pi1/bracketrightBigg |ψ,0>. (10.2.4) We ﬁnd that by the analytic continuation, t=−iτ, 0≤τ≤β≡/planckover2pi1 kBT, (10.2.5) where kB=Boltzmann constant, T=absolute temperature, the (real time) Schrödinger equation and its formal solution, Eqs. (10.2.3) and (10.2.4), are an- alytically continued into the Bloch equation and the density matrix, Eqs. (10.2.1) and (10.2.2), 10.2 Feynman’s V ariational Principle in Quantum Statistical Mechanics 309 respectively. Under the analytic continuation, Eq. (10.2.5), we divide the interval [0,β]into thenequal subintervals, and use the resolution of the identity in both the q-representation and thep-representation. In this way, we obtain the following list of correspondence. Here, we assume a Hamiltonian ˆH({/vectorqj,/vectorpj}N j=1)of the following form, ˆH({/vectorqj,/vectorpj}N j=1)=N/summationdisplay j=11 2m/vectorp2 j+/summationdisplay j>kV(/vectorqj,/vectorqk). (10.2.6) Table 10.1: List of Correspondence Quantum Mechanics Quantum Statistical Mechanics Schrödinger equation Bloch equation i/planckover2pi1∂ ∂t|ψ,t/angbracketright=H({/vectorqj,/vectorpj}N j=1)|ψ,t/angbracketright −/planckover2pi1∂ ∂τˆρC(τ)=H({/vectorqj,/vectorpj}N j=1)ˆρC(τ) Schrödinger state vector Density matrix |ψ,t/angbracketright =e x p/bracketleftbig −itH({/vectorqj,/vectorpj}N j=1)//planckover2pi1/bracketrightbig |ψ,0/angbracketright ˆρC(τ) =e x p/bracketleftbig −τH({/vectorqj,/vectorpj}N j=1)//planckover2pi1/bracketrightbig ˆρC(0) Minkowskian Lagrangian Euclidean Lagrangian LM({qj(t),˙qj(t)}N j=1) =N/summationdisplay j=11 2m˙q2 j(t)−N/summationdisplay j/angbracketrightkV(/vectorqj,/vectorqk) LE({qj(τ),˙qj(τ)}N j=1) =−N/summationdisplay j=11 2m˙q2 j(τ)−N/summationdisplay j/angbracketrightkV(/vectorqj,/vectorqk) Minkowskian action functional Euclidean action functional iIM[{/vectorqj}N j=1;/vectorqf,/vectorqi] =i/integraldisplaytf tidtL M({qj(t),˙qj(t)}N j=1) IE[{/vectorqj}N j=1;/vectorqf,/vectorqi] =/integraldisplayβ 0dτL E({qj(τ),˙qj(τ)}N j=1) Transformation function Transformation function /angbracketleft/vectorqf,tf|/vectorqi,ti/angbracketright=/integraldisplay/vectorq(tf)=/vectorqf /vectorq(ti)=/vectorqiD[/vectorq]× ×exp/bracketleftbig iIM[{/vectorqj}N j=1;/vectorqf,/vectorqi]//planckover2pi1/bracketrightbig Zf,i=/integraldisplay/vectorq(β)=/vectorqf /vectorq(0)=/vectorqiD[/vectorq]× ×exp/bracketleftbig IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]//planckover2pi1/bracketrightbig Vacuum to vacuum transition amplitude Partition function /angbracketleft0,out|0,in/angbracketright =/integraldisplay D[/vectorq]e xp/bracketleftbig iIM[{/vectorqj}N j=1]//planckover2pi1/bracketrightbig ZC(β)=TrˆρC(β) =“/integraldisplay d/vectorqfd/vectorqiδ(/vectorqf−/vectorqi)Zf,i” Vacuum expectation value Thermal expectation value /angbracketleftO(ˆq)/angbracketright=/integraltext D[/vectorq]O(/vectorq)e x p/bracketleftbig iIM[{/vectorqj}N j=1]//planckover2pi1/bracketrightbig /integraltext D[/vectorq]e xp{iIM[{/vectorqj}N j=1]//planckover2pi1} /angbracketleftO(ˆq)/angbracketright=“T rˆρC(β)O(ˆq) TrˆρC(β)” 310 10 Calculus of V ariations: Applications In the list, we have entries enclosed in double quotes, whose precise expressions are given, respectively, by Partition function: ZC(β)=TrˆρC(β)=“/integraldisplay d3/vectorqfd3/vectorqiδ3(/vectorqf−/vectorqi)Zf,i” =1 N!/summationdisplay PδP/integraldisplay d3/vectorqfd3/vectorqiδ3(/vectorqf−/vectorqPi)Zf,Pi =1 N!/summationdisplay PδP/integraldisplay d3/vectorqfd3/vectorqPiδ3(/vectorqf−/vectorqPi) ×/integraldisplay/vectorq(β)=/vectorqf /vectorq(0)=/vectorqPiD[/vectorq]e xp/bracketleftBigg IE[{/vectorqj}N j=1;/vectorqf,/vectorqPi] /planckover2pi1/bracketrightBigg ,(10.2.7) and Thermal expectation value: /angbracketleftˆO(/vectorq)/angbracketright=TrˆρC(β)ˆO(ˆq) TrˆρC(β) =“/integraltext d3/vectorqfd3/vectorqiδ3(/vectorqf−/vectorqi)Zf,i/angbracketlefti|O(/vectorq)|f/angbracketright” “/integraltext d3/vectorqfd3/vectorqiδ3(/vectorqf−/vectorqi)Zf,i”(10.2.8) =1 ZC(β)1 N!/summationdisplay PδP/integraldisplay d3/vectorqfd3/vectorqPiδ3(/vectorqf−/vectorqPi)Zf,Pi/angbracketleft/vectorqPi|ˆO(/vectorq)|/vectorqf/angbracketright. Here,/vectorqiand/vectorqfrepresent the initial position {/vectorqj(0)}N j=1and the ﬁnal position {/vectorqj(β)}N j=1 ofNidentical particles, Prepresents the permutation of {1,...,N },Pirepresents the per- mutation of the initial position {/vectorq(0)}N j=1andδPrepresents the signature of the permutation P, respectively. In this manner, we obtain the path integral representation of the partition function, ZC(β), and the thermal expectation value, /angbracketleftˆO(/vectorq)/angbracketright. This functional IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]of Eq. (10.2.7) can be obtained from IM[{/vectorqj}N j=1;/vectorqf,/vectorqi]by replacing twith−iτ.S i n c e ˆρC(β)is a solution of Eq. (10.2.1), the asymptotic form of ZC(β)for a large τinterval from τitoτfis ZC(β)∼exp/bracketleftbigg −E0(τf−τi) /planckover2pi1/bracketrightbigg asτf−τi→∞. Therefore we must estimate ZC(β)for large τf−τi. We choose any real I1which approximates IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]and write ZC(β)as /integraldisplay D[/vectorq(ζ)]exp/bracketleftBigg IE[{/vectorqj}N j=1;/vectorqf,/vectorqi] /planckover2pi1/bracketrightBigg =/integraldisplay D[/vectorq(ζ)]exp/bracketleftBigg (IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]−I1) /planckover2pi1/bracketrightBigg exp/bracketleftbiggI1 /planckover2pi1/bracketrightbigg .(10.2.9) 10.2 Feynman’s V ariational Principle in Quantum Statistical Mechanics 311 The expression (10.2.9) can be regarded as the average of exp [(IE−I1)//planckover2pi1]with respect to the positive weight exp [I1//planckover2pi1]. This observation motivates the variational principle based on Jensen’s inequality. Since the exponential function is convex, for any real quantities f,t h e average of exp[f]exceeds the exponential of the average <f> , <exp[f]>≥exp[<f> ]. (10.2.10) Hence, if in Eq. (10.2.9) we replace IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]−I1by its average =/integraldisplay D[/vectorq(ζ)](IE[{/vectorqj}N j=1;/vectorqf,/vectorqi]−I1)e x p/bracketleftbiggI1 /planckover2pi1/bracketrightbigg /integraldisplay D[/vectorq(ζ)]exp/bracketleftbiggI1 /planckover2pi1/bracketrightbigg ,(10.2.11) we will underestimate the value of Eq. (10.2.9). If Eis computed from /integraldisplay D[/vectorq(ζ)]exp/bracketleftBigg /planckover2pi1/bracketrightBigg exp/bracketleftbiggI1 /planckover2pi1/bracketrightbigg ∼exp/bracketleftbigg −E(τf−τi) /planckover2pi1/bracketrightbigg ,(10.2.12) we know that Eexceeds the true E0, E≥E0. (10.2.13) If there are any free parameters in I1, we choose as the best values those which minimize E. Sincedeﬁned in Eq. (10.2.11) is proportional to τf−τi,w e write =s(τf−τi). (10.2.14) The factor exp [//planckover2pi1]in Eq. (10.2.12) is constant and can be taken outside the integral. We suppose the lowest energy E1for the action functional I1is known, /integraldisplay D[/vectorq(ζ)]exp/bracketleftbiggI1 /planckover2pi1/bracketrightbigg ∼exp/bracketleftbigg −E1(τf−τi) /planckover2pi1/bracketrightbigg asτf−τi→∞. (10.2.15) Then we have E=E1−s from Eq. (10.2.12), with sgiven by Eqs. (10.2.11) and (10.2.14). 312 10 Calculus of V ariations: Applications If we choose the following trial action functional , I1=−1 2/integraldisplay/parenleftbiggd/vectorq dτ/parenrightbigg2 dτ, (10.2.16) we have what corresponds to the plane wave Born approximation in standard perturbation theory. Another choice is I1=−1 2/integraldisplay/parenleftbiggd/vectorq dτ/parenrightbigg2 dτ+/integraldisplay Vtrial(/vectorq(τ))dτ, (10.2.17) where Vtrial(/vectorq(τ))is a trial potential to be chosen. This corresponds to the distorted wave Born approximation in standard perturbation theory. If we choose a Coulomb potential as the trial potential, Vtrial(R)=Z R, (10.2.18) we vary the parameter Z. If we choose a harmonic potential as the trial potential, Vtrial(R)=1 2kR2, (10.2.19) we vary the parameter k. One problem with the trial potential used in Eq. (10.2.17) is that the particle with the coordinate /vectorq(τ)is bound to a speciﬁc origin. A better choice would be the inter-particle potential of the form Vtrial(/vectorq(τ)−/vectorq(σ))where an electron at /vectorq(τ)is bound to another electron at/vectorq(σ). 10.3 Schwinger–Dyson Equation in Quantum Field Theory In quantum ﬁeld theory, we also have the notion of Green’s functions, which is quite distinct from the ordinary Green’s functions in mathematical physics in one important aspect: the gov- erning equation of motion of the connected part of the two-point “full” Green’s function inquantum ﬁeld theory is the closed system of the coupled nonlinear integro-differential equa- tions . We illustrate this point in some detail for the Y ukawa coupling of the fermion ﬁeld and the boson ﬁeld. We shall establish the relativistic notation. We employ the natural unit system in which /planckover2pi1=c=1. (10.3.1) We deﬁne the Minkowski space–time metric tensor η µνby ηµν≡diag(1;−1,−1,−1)≡ηµν,µ , ν =0,1,2,3. (10.3.2) 10.3 Schwinger–Dyson Equation in Quantum Field Theory 313 We deﬁne the contravariant components and the covariant components of the space–time co- ordinates xby xµ≡(x0,x1,x2,x3), (10.3.3a) xµ≡ηµνxν=(x0,−x1,−x2,−x3). (10.3.3b) We deﬁne the differential operators ∂µand∂µby ∂µ≡∂ ∂xµ=/parenleftbigg∂ ∂x0,∂ ∂x1,∂ ∂x2,∂ ∂x3/parenrightbigg ,∂µ≡∂ ∂xµ=ηµν∂ν. (10.3.4) We deﬁne the four-scalar product by x·y≡xµyµ=ηµνxµyν=x0·y0−/vectorx·/vectory. (10.3.5) We adopt the convention that the Greek indices µ,ν, . . . run over 0, 1, 2 and 3, the Lattin indices i,j, . . . run over 1, 2 and 3, and the repeated indices are summed over. We consider the quantum ﬁeld theory described by the total Lagrangian density Ltotof the form, Ltot=1 4[/hatwide¯ψα(x),Dαβ(x)ˆψβ(x)] +1 4[DT βα(−x)/hatwide¯ψα(x),ˆψβ(x)] +1 2ˆφ(x)K(x)ˆφ(x)+Lint(ˆφ(x),ˆψ(x),/hatwide¯ψ(x)),(10.3.6) where we have Dαβ(x)=(iγµ∂µ−m+iε)αβ,DT βα(−x)=(−iγT µ∂µ−m+iε)βα,(10.3.7a) K(x)=−∂2−κ2+iε, (10.3.7b) {γµ,γν}=2ηµν,(γµ)†=γ0γµγ0,/hatwide¯ψα(x)=(ˆψ†(x)γ0)α, (10.3.8) Itot[ˆφ,ˆψ,/hatwide¯ψ]=/integraldisplay d4xLtot((10.3.6) ),I int[ˆφ,ˆψ,/hatwide¯ψ]=/integraldisplay d4xLint((10.3.6) ).(10.3.9) We have Euler–Lagrange equations of motion for the ﬁeld operators, ˆψα(x):δˆItot δ/hatwide¯ψα(x)=0,orDαβ(x)ˆψβ(x)+δˆIint δ/hatwide¯ψα(x)=0, (10.3.10a) /hatwide¯ψβ(x):δˆItot δˆψβ(x)=0,or−DT βα(−x)/hatwide¯ψα(x)+δˆIint δˆψβ(x)=0, (10.3.10b) ˆφ(x):δItot δˆφ(x)=0,orK(x)ˆφ(x)+δˆIint δˆφ(x)=0. (10.3.10c) 314 10 Calculus of V ariations: Applications We have the equal time canonical (anti-)commutators, δ(x0−y0){ˆψβ(x),/hatwide¯ψα(x)}=γ0 βαδ4(x−y), (10.3.11a) δ(x0−y0){ˆψβ(x),ˆψα(y)}=δ(x0−y0){/hatwide¯ψβ(x),/hatwide¯ψα(x)}=0, (10.3.11b) δ(x0−y0)[ˆφ(x),∂y 0ˆφ(y)] =iδ4(x−y), (10.3.11c) δ(x0−y0)[ˆφ(x),ˆφ(y)] =δ(x0−y0)[∂x 0ˆφ(x),∂y 0ˆφ(y)] = 0 , (10.3.11d) and the rest of the equal time mixed canonical commutators are equal to 0. We deﬁne the generating functional of the “full” Green’s functions by Z[J,¯η,η]≡<0,out|T(exp/bracketleftBig i{(Jˆφ)+(¯ηˆψ)+(/hatwide¯ψη)}/bracketrightBig )|0,in> ≡∞/summationdisplay l,m,n=0il+m+n l!m!n!<0,out|T((¯ηˆψ)l(Jˆφ)m(/hatwide¯ψη)n)|0,in> ≡∞/summationdisplay l,m,n=0il+m+n l!m!n!J(y1)·...·J(ym)¯ηαl(xl)·...·¯ηα1(x1) ×<0,out|T{ˆψα1(x1)·...·ˆψαl(xl)ˆφ(y1)··ˆφ(ym) ×/hatwide¯ψβ1(z1)·...·/hatwide¯ψβn(zn)}|0,in>ηβn(zn)·...·ηβ1(z1) ≡∞/summationdisplay n=0in n!<0,out|T(¯ηˆψ+Jˆφ+/hatwide¯ψη)n|0,in>,(10.3.12) where the repeated continuous space–time indices x1through zna r et ob ei n t e g r a t e do v e r ,a n d we introduced the abbreviations in Eq. (10.3.12), Jˆφ≡/integraldisplay d4yJ(y)ˆφ(y),¯ηˆψ≡/integraldisplay d4x¯η(x)ˆψ(x),/hatwide¯ψη≡/integraldisplay d4z/hatwide¯ψ(z)η(z). The time-ordered product is deﬁned by T{ˆΨ(x1)·...·ˆΨ(xn)} ≡/summationdisplay all possible permutations PδPθ(x0 P1−x0 P2)·...·θ(x0 P(n−1)−x0 Pn)ˆΨ(xP1)·...·ˆΨ(xPn), with δP=  1Peven and odd for a Boson , 1 Peven for a Fermion , −1 Podd for a Fermion . 10.3 Schwinger–Dyson Equation in Quantum Field Theory 315 We observe that δ δ¯ηβ(x)(¯ηˆψ)l=lˆψβ(x)(¯ηˆψ)l−1, (10.3.13a) δ δηα(x)(/hatwide¯ψη)n=−n/hatwide¯ψα(x)(/hatwide¯ψη)n−1, (10.3.13b) δ δJ(x)(Jˆφ)m=mˆφ(x)(Jˆφ)m−1. (10.3.13c) We also observe from the deﬁnition of Z[J,¯η,η], 1 iδ δ¯ηβ(x)Z[J,¯η,η]=/angbracketleftbigg 0,out/vextendsingle/vextendsingle/vextendsingle/vextendsingleT/parenleftbigg ˆψβ(x)e x p/bracketleftbigg i/integraldisplay d4z/braceleftbig J(z)ˆφ(z) +¯ηα(z)ˆψα(z)+/hatwide¯ψβ(z)ηβ(z)/bracerightbig/bracketrightbigg/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle0,in/angbracketrightbigg ,(10.3.14a) iδ δηα(x)Z[J,¯η,η]=/angbracketleftbigg 0,out/vextendsingle/vextendsingle/vextendsingle/vextendsingleT/parenleftbigg /hatwide¯ψα(x)e x p/bracketleftbigg i/integraldisplay d4z/braceleftbig J(z)ˆφ(z) +¯ηα(z)ˆψα(z)+/hatwide¯ψβ(z)ηβ(z)/bracerightbig/bracketrightbigg/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle0,in/angbracketrightbigg ,(10.3.14b) 1 iδ δJ(x)Z[J,¯η,η]=/angbracketleftbigg 0,out/vextendsingle/vextendsingle/vextendsingle/vextendsingleT/parenleftbigg ˆφ(x)e x p/bracketleftbigg i/integraldisplay d4z/braceleftbig J(z)ˆφ(z) +¯ηα(z)ˆψα(z)+/hatwide¯ψβ(z)ηβ(z)/bracerightbig/bracketrightbigg/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle0,in/angbracketrightbigg .(10.3.14c) From the deﬁnition of the time-ordered product and the equal time canonical (anti-)commuta- tors, Eqs. (10.3.11a through d), we have at the operator level, Fermion: Dαβ(x)T/parenleftBig ˆψβ(x)e x p/bracketleftBig i{Jˆφ+¯ηαˆψα+/hatwide¯ψβηβ}/bracketrightBig/parenrightBig =T/parenleftBig Dαβ(x)ˆψβ(x)e x p/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig −ηα(x)T/parenleftBig exp/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig ,(10.3.15) Anti-Fermion: −DT βα(−x)T/parenleftBig/hatwide¯ψα(x)e x p/bracketleftBig i{Jˆφ+¯ηαˆψα+/hatwide¯ψβηβ}/bracketrightBig/parenrightBig =T/parenleftBig −DT βα(−x)/hatwide¯ψα(x)e x p/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig +¯ηβ(x)T/parenleftBig exp/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig ,(10.3.16) 316 10 Calculus of V ariations: Applications Boson: K(x)T/parenleftBig ˆφ(x)e x p/bracketleftBig i{Jˆφ+¯ηαˆψα+/hatwide¯ψβηβ}/bracketrightBig/parenrightBig =T/parenleftBig K(x)ˆφ(x)e x p/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig −J(x)T/parenleftBig exp/bracketleftBig i{Jˆφ+¯ηˆψ+/hatwide¯ψη}/bracketrightBig/parenrightBig .(10.3.17) Applying Euler–Lagrange equations of motion, Eqs. (10.3.10a through c), to the ﬁrst terms on the right-hand sides of Eqs. (10.3.15), (10.3.16) and (10.3.17), and taking the vacuumexpectation values, we obtain the equations of motion of the generating functional Z[J,¯η,η] of the “full” Green’s functions as   D αβ(x)1 iδ δ¯ηβ(x)+δIint[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig iδ δηα(x)/parenrightBig +ηα(x)  Z[J,¯η,η]=0, (10.3.18a)   −DT βα(−x)iδ δηα(x)+δIint[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig 1 iδ δ¯ηβ(x)/parenrightBig−¯ηβ(x)  Z[J,¯η,η]=0,(10.3.18b)   K(x)1 iδ δJ(x)+δIint[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig 1 iδ δJ(x)/parenrightBig +J(x)  Z[J,¯η,η]=0. (10.3.18c) Equivalently, from Eqs. (10.3.10a), (10.3.10b) and (10.3.10c), we can write Eqs. (10.3.18a), (10.3.18b) and (10.3.18c) as   δItot[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig iδ δηα(x)/parenrightBig +ηα(x)  Z[J,¯η,η]=0, (10.3.19a)   δItot[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig 1 iδ δ¯ηβ(x)/parenrightBig−¯ηβ(x)  Z[J,¯η,η]=0, (10.3.19b)   δItot[1 iδ δJ,1 iδ δ¯η,iδ δη] δ/parenleftBig 1 iδ δJ(x)/parenrightBig +J(x)  Z[J,¯η,η]=0. (10.3.19c) We note that the coefﬁcients of the external hook terms, ηα(x),¯ηβ(x)andJ(x),i n Eqs. (10.3.19a through c) are ±1, which is a reﬂection of the fact that we are deal- ing with canonical quantum ﬁeld theory and originates from the equal time canonical (anti-)commutators. With this preparation, we shall discuss the Schwinger theory of Green’s function with the interaction Lagrangian density Lint(ˆφ(x),ˆψ(x),/hatwide¯ψ(x))of the Yukawa coupling in mind, Lint(ˆφ(x),ˆψ(x),/hatwide¯ψ(x)) =−G0/hatwide¯ψα(x)γαβ(x)ˆψβ(x)ˆφ(x), (10.3.20) 10.3 Schwinger–Dyson Equation in Quantum Field Theory 317 with G0=  g0 f e,γ(x)=  γ5 γ5τi γµ,ˆψ(x)=  ˆψα(x) ˆψN,α(x) ˆψα(x),φ(x)=  ˆφ(x) ˆφi(x) ˆAµ(x).(10.3.21) We deﬁne the vacuum expectation values, <F>J,¯η,ηand<F>J, of the operator function F(ˆφ(x),ˆψ(x),/hatwide¯ψ(x))in the presence of the external hook terms {J,¯η,η}by <F>J,¯η,η≡1 Z[J,¯η,η]F/parenleftbigg1 iδ δJ(x),1 iδ δ¯η(x),iδ δη(x)/parenrightbigg Z[J,¯η,η], (10.3.22a) <F>J=<F>J,¯η,η/vextendsingle/vextendsingle ¯η=η=0. (10.3.22b) We deﬁne the connected parts of the two-point “full” Green’s functions in the presence of the external hook J(x)by Fermion: S/primeJ F,αβ(x1,x2)≡1 iδ δ¯ηα(x1)iδ δηβ(x2)1 ilnZ[J,¯η,η]/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 =1 i/parenleftbigg1 iδ δ¯ηα(x1)/parenrightbigg </hatwide¯ψβ(x2)>J,¯η,η/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 =1 i/braceleftbigg <ˆψα(x1)/hatwide¯ψβ(x2)>J,¯η,η/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 −<ˆψα(x1)>J,¯η,η</hatwide¯ψβ(x2)>J,¯η,η/vextendsingle/vextendsingle/vextendsingle ¯η=η=0/bracerightbigg =1 i<ˆψα(x1)/hatwide¯ψβ(x2)>J ≡1 i<0,out|T(ˆψα(x1)/hatwide¯ψβ(x2))/vextendsingle/vextendsingle0,in>J C,(10.3.23) <ˆψα(x1)>J,¯η,η/vextendsingle/vextendsingle/vextendsingle ¯η=η=0=</hatwide¯ψβ(x2)>J,¯η,η/vextendsingle/vextendsingle/vextendsingle ¯η=η=0=0, (10.3.24) and Boson: D/primeJ F(x1,x2)≡1 iδ δJ(x1)1 iδ δJ(x2)1 ilnZ[J,¯η,η]/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 =1 i/parenleftbigg1 iδ δJ(x1)<ˆφ(x2)>J/parenrightbigg =1 i/braceleftBig <ˆφ(x1)ˆφ(x2)>J−<ˆφ(x1)>J<ˆφ(x2)>J/bracerightBig ≡1 i<0,out|T(ˆφ(x1)ˆφ(x2))/vextendsingle/vextendsingle0,in>J C,(10.3.25) <ˆφ(x)>J/vextendsingle/vextendsingle/vextendsingle J=0=0. (10.3.26) 318 10 Calculus of V ariations: Applications We have the equations of motion of Z[J,¯η,η], Eqs. (10.3.18a through c), when the inter- action Lagrangian density Lint(ˆφ(x),ˆψ(x),/hatwide¯ψ(x))is given by Eq. (10.3.20) as /braceleftbigg Dαβ(x)/parenleftbigg1 iδ δ¯ηβ(x)/parenrightbigg −G0γαβ(x)/parenleftbigg1 iδ δ¯ηβ(x)/parenrightbigg/parenleftbigg1 iδ δJ(x)/parenrightbigg/bracerightbigg Z[J,¯η,η] =−ηα(x)Z[J,¯η,η],(10.3.27a) /braceleftbigg −DT βα(−x)/parenleftbigg iδ δηα(x)/parenrightbigg +G0/parenleftbigg iδ δηα(x)/parenrightbigg γαβ(x)/parenleftbigg1 iδ δJ(x)/parenrightbigg/bracerightbigg Z[J,¯η,η] =+ ¯ηβ(x)Z[J,¯η,η],(10.3.27b) /braceleftbigg K(x)/parenleftbigg1 iδ δJ(x)/parenrightbigg −G0/parenleftbigg iδ δηα(x)/parenrightbigg γαβ(x)/parenleftbigg1 iδ δ¯ηβ(x)/parenrightbigg/bracerightbigg Z[J,¯η,η] =−J(x)Z[J,¯η,η].(10.3.27c) Dividing Eqs. (10.3.27a through c) by Z[J,¯η,η], and referring to Eqs. (10.3.22a) and (10.3.22b), we obtain the equations of motion for <ˆψβ(x)>J,¯η,η,</hatwide¯ψα(x)>J,¯η,ηand<ˆφ(x)>J,¯η,η, as Dαβ(x)<ˆψβ(x)>J,¯η,η−G0γαβ(x)<ˆψβ(x)ˆφ(x)>J,¯η,η=−ηα(x), (10.3.28) −DT βα(−x)</hatwide¯ψα(x)>+G0γαβ(x)</hatwide¯ψα(x)ˆφ(x)>J,¯η,η=+ ¯ηβ(x), (10.3.29) K(x)<ˆφ(x)>J,¯η,η−G0γαβ(x)</hatwide¯ψα(x)ˆψβ(x)>J,¯η,η=−J(x). (10.3.30) We take the following functional derivatives, iδ δηε(y)Eq. (10.3.28)/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0: Dαβ(x)</hatwide¯ψε(y)ˆψβ(x)>J −G0γαβ(x)</hatwide¯ψε(y)ˆψβ(x)ˆφ(x)>J=−iδαεδ4(x−y), 1 iδ δ¯ηε(y)Eq. (10.3.29)/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0: −DT βα(−x)<ˆψε(y)/hatwide¯ψα(x)>J +G0γαβ(x)<ˆψε(y)/hatwide¯ψα(x)ˆφ(x)>J=−iδβεδ4(x−y), 10.3 Schwinger–Dyson Equation in Quantum Field Theory 319 1 iδ δJ(y)Eq. (10.3.30)/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0: K(x){<ˆφ(y)ˆφ(x)>J−<φ(y)>J<φ(x)>J} −G0γαβ(x)1 iδ δJ(y)</hatwide¯ψα(x)ˆψβ(x)>J=iδ4(x−y). These equations are a part of the inﬁnite system of coupled equations. We observe the follow- ing identities, </hatwide¯ψε(y)ˆψβ(x)>J=−iS/primeJ F,βε(x, y), <ˆψε(y)/hatwide¯ψα(x)>J=iS/primeJ F,εα(y,x), </hatwide¯ψε(y)ˆψβ(x)ˆφ(x)>J=−i/parenleftbigg <ˆφ(x)>J+1 iδ δJ(x)/parenrightbigg S/primeJ F,βε(x, y), <ˆψε(y)/hatwide¯ψα(x)ˆφ(x)>J=i/parenleftbigg <ˆφ(x)>J+1 iδ δJ(x)/parenrightbigg S/primeJ F,εα(y,x). With these identities, we obtain the equations of motion of the connected parts of the two-point “full” Green’s functions in the presence of the external hook J(x), /braceleftbigg Dαβ(x)−G0γαβ(x)/parenleftbigg <ˆφ(x)>J+1 iδ δJ(x)/parenrightbigg/bracerightbigg S/primeJ F,βε(x, y) =δαεδ4(x−y),(10.3.31a) /braceleftbigg DT βα(−x)−G0γαβ(x)/parenleftbigg <ˆφ(x)>J+1 iδ δJ(x)/parenrightbigg/bracerightbigg S/primeJ F,εα(y,x) =δβεδ4(x−y),(10.3.32a) K(x)D/primeJ F(x, y)+G0γαβ(x)1 iδ δJ(y)S/primeJ F,βα(x, x±)=δ4(x−y). (10.3.33a) Since the transpose of Eq. (10.3.32a) is Eq. (10.3.31a), we have to consider only Eqs. (10.3.31a) and (10.3.33a). We may get the impression that we have the equations of motion of the two-point “full” Green’s functions, S/primeJ F,αβ(x, y)andD/primeJ F(x, y),i nc l o s e df o r m , at ﬁrst sight. Because of the presence of the functional derivatives δ/iδJ(x)andδ/iδJ(y), however, Eqs. (10.3.31a), (10.3.32a) and (10.3.33a) involve the three-point “full” Green’s functions and are merely a part of the inﬁnite system of the coupled nonlinear equations of motion of the “full” Green’s functions. From this point onward, we use the variables, “1”, “2”, “3”, . . . to represent the continuous space–time indices, x,y,z, ..., t h e s p i n o r i n d i c e s , α,β,γ, ..., a s w e l l a s o t h e r i n t e r n a l indices, i,j,k,.... 320 10 Calculus of V ariations: Applications Using the “free” Green’s functions, SF 0(1−2)andDF 0(1−2), deﬁned by D(1)SF 0(1−2) = 1 , (10.3.34a) K(1)DF 0(1−2) = 1 , (10.3.34b) we rewrite the functional differential equations satisﬁed by the “full” Green’s functions, S/primeJ F(1,2)andD/primeJ F(1,2), Eqs. (10.3.31a) and (10.3.33a), into the integral equations, S/primeJ F(1,2) =SF 0(1−2)+ SF 0(1−3)(G0γ(3))/parenleftbigg <ˆφ(3)>J+1 iδ δJ(3)/parenrightbigg S/primeJ F(3,2),(10.3.31b) D/primeJ F(1,2) =DF 0(1−2) +DF 0(1−3)/parenleftbigg −G0trγ(3)1 iδ δJ(2)S/primeJ F(3,3±)/parenrightbigg .(10.3.33b) We compare Eqs. (10.3.31b) and (10.3.33b) with the deﬁning integral equations of the proper self-energy parts ,Σ∗andΠ∗, due to Dyson, in the presence of the external hook J(x), S/primeJ F(1,2) =SF 0(1−2) +SF 0(1−3)(G0γ(3)<φ(3)>J)S/primeJ F(3,2) +SF 0(1−3)Σ∗(3,4)S/primeJ F(4,2),(10.3.35) D/primeJ F(1,2) =DF 0(1−2) +DF 0(1−3)Π∗(3,4)D/primeJ F(4,2), (10.3.36) obtaining G0γ(1)1 iδ δJ(1)S/primeJ F(1,2) =Σ∗(1,3)S/primeJ F(3,2)≡Σ∗(1)S/primeJ F(1,2), (10.3.37) −G0trγ(1)1 iδ δJ(2)S/primeJ F(1,1±)=Π∗(1,3)D/primeJ F(3,2)≡Π∗(1)D/primeJ F(1,2).(10.3.38) Thus we can write the functional differential equations, Eqs. (10.3.31a) and (10.3.33a), com- pactly as {D(1)−G0γ(1)<ˆφ(1)>J−Σ∗(1)}S/primeJ F(1,2) =δ(1−2), (10.3.39) {K(1)−Π∗(1)}D/primeJ F(1,2) =δ(1−2). (10.3.40) Deﬁning the “Nucleon” differential operator and “Meson” differential operator by DN(1,2)≡{D(1)−G0γ(1)<ˆφ(1)>J}δ(1−2)−Σ∗(1,2), (10.3.41) and DM(1,2)≡K(1)δ(1−2)−Π∗(1,2), (10.3.42) we can write the differential equations, Eqs. (10.3.39) and (10.3.40), as 10.3 Schwinger–Dyson Equation in Quantum Field Theory 321 DN(1,3)S/primeJ F(3,2) =δ(1−2),orDN(1,2) = ( S/primeJ F(1,2))−1, (10.3.43) and DM(1,3)D/primeJ F(3,2) =δ(1−2),orDM(1,2) = ( D/primeJ F(1,2))−1. (10.3.44) Next, we take the functional derivative of Eq. (10.3.39), 1 iδ δJ(3)Eq. (10.3.39): {D(1)−G0γ(1)<ˆφ(1)>J−Σ∗(1)}1 iδ δJ(3)S/primeJ F(1,2) =/braceleftbigg iG0γ(1)D/primeJ F(1,3) +1 iδ δJ(3)Σ∗(1)/bracerightbigg S/primeJ F(1,2).(10.3.45) Solving Eq. (10.3.45) for δS/primeJ F(1,2)/iδJ(3)and using Eqs. (10.3.39), (10.3.41) and (10.3.43), we obtain 1 iδ δJ(3)S/primeJ F(1,2) =S/primeJ F(1,4)/braceleftbigg iG0γ(4)D/primeJ F(4,3) +1 iδ δJ(3)Σ∗(4)/bracerightbigg S/primeJ F(4,2) =iG0S/primeJ F(1,4){γ(4)δ(4−5)δ(4−6) +1 G0δ δ<ˆφ(6)>JΣ∗(4,5)}S/primeJ F(5,2)D/primeJ F(6,3).(10.3.46) Comparing Eq. (10.3.46) with the deﬁnition of the vertex operator Γ(4,5; 6) of Dyson, 1 iδ δJ(3)S/primeJ F(1,2)≡iG0S/primeJ F(1,4)Γ(4,5; 6)S/primeJ F(5,2)D/primeJ F(6,3), (10.3.47) we obtain Γ(1,2; 3) = γ(1)δ(1−2)δ(1−3) +1 G0δ δ<ˆφ(3)>JΣ∗(1,2), (10.3.48) while we can write the left-hand side of Eq. (10.3.47) as 1 iδ δJ(3)S/primeJ F(1,2) =iD/primeJ F(6,3)δ δ<ˆφ(6)>JS/primeJ F(1,2). (10.3.49) From this, we have −1 G0δ δ<ˆφ(6)>JS/primeJ F(1,2) =−S/primeJ F(1,4)Γ(4,5; 6)S/primeJ F(5,2), 322 10 Calculus of V ariations: Applications and we obtain the compact representation of Γ(1,2; 3), Γ(1,2; 3) = −1 G0δ δ<ˆφ(3)>J(S/primeJ F(1,2))−1=−1 G0δ δ<ˆφ(3)>JDN(1,2) =(10.3.48) .(10.3.50) Lastly, from Eqs. (10.3.38) and (10.3.39), which deﬁne Σ∗(1,2)andΠ∗(1,2)indirectly, and the deﬁning equation of Γ(1,2; 3), Eq. (10.3.47), we have Σ∗(1,3)S/primeJ F(3,2) =−iG2 0γ(1)S/primeJ F(1,4)Γ(4,5; 6)S/primeJ F(5,2)D/primeJ F(6,1), (10.3.51) Π∗(1,3)D/primeJ F(3,2) =iG2 0trγ(1)S/primeJ F(1,4)Γ(4,5; 6)S/primeJ F(5,1)D/primeJ F(6,2). (10.3.52) Namely, we obtain Σ∗(1,2) =−iG2 0γ(1)S/primeJ F(1,3)Γ(3,2; 4)D/primeJ F(4,1), (10.3.53) Π∗(1,2) =iG2 0trγ(1)S/primeJ F(1,3)Γ(3,4; 2)S/primeJ F(4,1). (10.3.54) Equation (10.3.30) can be expressed after setting η=¯η=0as K(1)<ˆφ(1)>J+iG0tr(γ(1)S/primeJ F(1,1)) =−J(1). (10.3.55) The system of equations, (10.3.41), (10.3.42), (10.3.43), (10.3.44), (10.3.48), (10.3.53), (10.3.54) and (10.3.55), is called the Schwinger–Dyson equation. This system of nonlinear coupled integro-differential equations is exact and closed. Starting from the zeroth-order termofΓ(1,2; 3), we can develop the covariant perturbation theory by iteration. In the ﬁrst-order approximation, after setting J=0, we have the following expressions, Σ ∗(1−2)∼=−iG2 0γ(1)SF 0(1−2)γ(2)DF 0(2−1), (10.3.56) Π∗(1−2)∼=iG2 0tr{γ(1)SF 0(1−2)γ(2)SF 0(2−1)}, (10.3.57) and Γ(1,2; 3)∼=γ(1)δ(1−2)δ(1−3) −iG2 0γ(1)SF 0(1−3)γ(3)SF 0(3−2)γ(2)DF 0(2−1).(10.3.58) We point out that the covariant perturbation theory based on the Schwinger–Dyson equation is somewhat different in spirit from the standard covariant perturbation theory due to Feynman and Dyson. The former is capable of dealing with the bound state problem in general as will be shown shortly. Its power is demonstrated in the positronium problem. 10.3 Schwinger–Dyson Equation in Quantum Field Theory 323 Summary of Schwinger–Dyson equation DN(1,3)S/primeJ F(3,2) =δ(1−2),D M(1,3)D/primeJ F(3,2) =δ(1−2), DN(1,2)≡{D(1)−G0γ(1)<ˆφ(1)>J}δ(1−2)−Σ∗(1,2), DM(1,2)≡K(1)δ(1−2)−Π∗(1,2), K(1)<ˆφ(1)>J+iG0tr(γ(1)S/primeJ F(1,1)) =−J(1), Σ∗(1,2)≡−iG2 0γ(1)S/primeJ F(1,3)Γ(3,2; 4)D/primeJ F(4,1), Π∗(1,2)≡iG2 0tr{γ(1)S/primeJ F(1,3)Γ(3,4; 2)S/primeJ F(4,1)}, Γ(1,2; 3)≡−1 G0δ δ<ˆφ(3)>J(S/primeJ F(1,2))−1 =γ(1)δ(1−2)δ(1−3) +1 G0δ δ<ˆφ(3)>JΣ∗(1,2). ∑*(1,2) = (1)γ 12 34 (1,3)'J FS(4,1)''J FD Γ(3,2;4) Fig. 10.1: Graphical representation of the proper self-energy part, Σ∗(1,2). Π*(1,2) =(1)γ 12 34 Γ(3,4;2) (1,3)'J FS(4,1)'J FS Fig. 10.2: Graphical representation of the proper self-energy part, Π∗(1,2). 324 10 Calculus of V ariations: Applications 3 )31()21()1( (1,2;3) =1 2• − − Γ δ δγ JG )3(1 0 φδδ⋅ +Γ(4,2;5) 1 2 45 (5,1)''J FD (1,4)'J FS{ } (1)γ Fig. 10.3: Graphical representation of the vertex operator, Γ(1,2; 3) . (1−2) ≅∑* (1)γ (2)γ 12(1−2)F oS(2−1)'F oD Fig. 10.4: The ﬁrst-order approximation for the proper self-energy part, Σ∗(1,2). Π*(1-2) ≅(1)γ (2)γ 12 (1−2)F oS(2−1)F oS Fig. 10.5: The ﬁrst-order approximation of the proper self-energy part, Π∗(1,2). 10.3 Schwinger–Dyson Equation in Quantum Field Theory 325 2⋅ −F o oD(2−1) iG1 23(1)γ (2)γ(3)γ(1−3)F oS (3−2)F oS3 )31()21()1( (1,2;3) ≅1 2• − − Γ δ δγ Fig. 10.6: The ﬁrst-order approximation of the vertex operator, Γ(1,2; 3) . *** We consider the two-body (four-point) nucleon “full” Green’s function S/primeJ F(1,2; 3,4)with the Yukawa coupling, Eqs. (10.3.20) and (10.3.21), in mind, deﬁned by S/primeJ F(1,2; 3,4)≡1 iδ δ¯η(1)1 iδ δ¯η(2)iδ δη(4)iδ δη(3)1 ilnZ[J,¯η,η]/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 =/parenleftbigg1 i/parenrightbigg2/braceleftBig <ˆψ(1)ˆψ(2)/hatwide¯ψ(3)/hatwide¯ψ(4)>J −<ˆψ(1)ˆψ(2)>J</hatwide¯ψ(3)/hatwide¯ψ(4)>J +<ˆψ(1)/hatwide¯ψ(3)>J<ˆψ(2)/hatwide¯ψ(4)>J −<ˆψ(1)/hatwide¯ψ(4)>J<ˆψ(2)/hatwide¯ψ(3)>J/bracerightBig ≡/parenleftbigg1 i/parenrightbigg2 <0,out|T(ˆψ(1)ˆψ(2)/hatwide¯ψ(3)/hatwide¯ψ(4))/vextendsingle/vextendsingle0,in>J C.(10.3.59) Operating the “Nucleon differential operators” ,DN(1,5)andDN(2,6),o nS/primeJ F(5,6; 3,4)and using the Schwinger–Dyson equation derived above, we obtain {DN(1,5)DN(2,6)−I(1,2; 5,6)}S/primeJ F(5,6; 3,4) =δ(1−3)δ(2−4)−δ(1−4)δ(2−3),(10.3.60) where the operator I(1,2; 3,4)is called the proper interaction kernel and satisﬁes the follow- ing integral equations, 326 10 Calculus of V ariations: Applications I(1,2; 5,6)S/primeJ F(5,6; 3,4) =g2 0tr(M)[γ(1)Γ(2) D/primeJ F(5,6)]S/primeJ F(5,6; 3,4) +g2 0tr(M)[γ(1)S/primeJ F(1,5)1 iδ δJ(5)]I(5,2; 6,7)S/primeJ F(6,7; 3,4)(10.3.61) =g2 0tr(M)[γ(2)Γ(1) D/primeJ F(5,6)]S/primeJ F(5,6; 3,4) +g2 0tr(M)[γ(2)S/primeJ F(2,5)1 iδ δJ(5)]I(1,5; 6,7)S/primeJ F(6,7; 3,4).(10.3.62) Here tr(M)indicates that the trace should be taken only over the meson coordinate. Equation (10.3.60) can be cast into the integral equation after a little algebra as S/primeJ F(1,2; 3,4) =S/primeJ F(1,3)S/primeJ F(2,4)−S/primeJ F(1,4)S/primeJ F(2,3) +S/primeJ F(1,5)S/primeJ F(2,6)I(5,6; 7,8)S/primeJ F(7,8; 3,4).(10.3.63) The system of equations, (10.3.60) (or (10.3.63)) and (10.3.61) (or (10.3.62)), is called the Bethe–Salpeter equation. If we set t1=t2=t>t 3=t4=t/primein Eq. (10.3.59), S/primeJ F(1,2; 3,4) represents the transition probability amplitude whereby the nucleons originally located at /vectorx3 and/vectorx4at time t/primeare to be found at /vectorx1and/vectorx2at the later time t. In the integral equations for I(1,2; 3,4), Eqs. (10.3.61) and (10.3.62), the ﬁrst terms on the right-hand sides represent the scattering state and the second terms represent the bound state. The bound state problem isformulated by dropping the ﬁrst terms on the right-hand sides of Eqs. (10.3.61) and (10.3.62).The proper interaction kernel I(1,2; 3,4)assumes the following form in the ﬁrst-order ap- proximation, I(1,2; 3,4)∼=g 2 0γ(1)γ(2)DF 0(1−2)(δ(1−3)δ(2−4)−δ(1−4)δ(2−3)).(10.3.64) 1 23 4(1,2;3,4 )'J FS =1 23 4(1,3)'J FS (2,4)'J FS+−1 23 4(1,4)'J FS (2,3)'J FS +1 23 45 67 8(1,5)'J FS (2,6)'J FSI(5,6;7,8)'J FS(7,8;3,4) Fig. 10.7: Graphical representation of the Bethe–Salpeter equation. 10.3 Schwinger–Dyson Equation in Quantum Field Theory 327 I(1,2;3,4) ≅ (1−2)'F oD1 12 23 34 4(1)γ (2)γ (2)γ(1)γδ (1−3) δ (2−4) δ (2−3) δ (1−4)+−(1−2)'F oD Fig. 10.8: The ﬁrst-order approximation of the proper interaction kernel, I(1,2; 3 ,4). As a byproduct of the derivation of Schwinger–Dyson equations, we can derive the func- tional integral representation of the generating functional Z[J,¯η,η]from Eqs. (10.3.19a), (10.3.19b) and (10.3.19c) in the following way. We deﬁne the functional Fourier transform ˜Z[φ, ψ,¯ψ]of the generating functional Z[J,¯η,η]by Z[J,¯η,η]≡/integraldisplay D[φ]D[ψ]D[¯ψ]˜Z[φ, ψ,¯ψ]e xp/bracketleftbig i(Jφ+¯ηψ+¯ψη)/bracketrightbig . (10.3.65) By functional integral by parts, we obtain the identity,  ηα(x) −¯ηβ(x) J(x) Z[J,¯η,η] =/integraldisplay D[φ]D[ψ]D[¯ψ]˜Z[φ, ψ,¯ψ] 1 iδ δ¯ψα(x) 1 iδ δψβ(x) 1 iδ δφ(x) ×exp/bracketleftbig i(Jφ+¯ηψ+¯ψη)/bracketrightbig (10.3.66) =/integraldisplay D[φ]D[ψ]D[¯ψ]   −1 iδ δ¯ψα(x) −1 iδ δψβ(x) −1 iδ δφ(x) ˜Z[φ, ψ,¯ψ]  exp/bracketleftbig i(Jφ+¯ηψ+¯ψη)/bracketrightbig . 328 10 Calculus of V ariations: Applications With the identity derived above, we have the equations of motion of the functional Fourier transform ˜Z[φ, ψ,¯ψ]of the generating functional Z[J,¯η,η]from Eqs. (10.3.19a through c) as /braceleftbiggδItot[φ, ψ,¯ψ] δ¯ψα(x)−1 iδ δ¯ψα(x)/bracerightbigg ˜Z[φ, ψ,¯ψ]=0, (10.3.67a) /braceleftbiggδItot[φ, ψ,¯ψ] δψβ(x)−1 iδ δψβ(x)/bracerightbigg ˜Z[φ, ψ,¯ψ]=0, (10.3.67b) /braceleftbiggδItot[φ, ψ,¯ψ] δφ(x)−1 iδ δφ(x)/bracerightbigg ˜Z[φ, ψ,¯ψ]=0. (10.3.67c) We divide Eqs. (10.3.67a through c) by ˜Z[φ, ψ,¯ψ], and obtain δ δ¯ψα(x)ln˜Z[φ, ψ,¯ψ]=iδ δ¯ψα(x)Itot[φ, ψ,¯ψ], (10.3.68a) δ δψβ(x)ln˜Z[φ, ψ,¯ψ]=iδ δψβ(x)Itot[φ, ψ,¯ψ], (10.3.68b) δ δφ(x)ln˜Z[φ, ψ,¯ψ]=iδ δφ(x)Itot[φ, ψ,¯ψ]. (10.3.68c) We can immediately integrate Eqs. (10.3.68a through c) with the result, ˜Z[φ, ψ,¯ψ]=1 CVexp/bracketleftbig iItot[φ, ψ,¯ψ]/bracketrightbig =1 CVexp/bracketleftbigg i/integraldisplay d4zLtot(φ(z),ψ(z),¯ψ(z))/bracketrightbigg ,(10.3.69) where CVis the integration constant. From Eq. (10.3.69), we shall obtain the generating functional Z[J,¯η,η]in the functional integral representation, Z[J,¯η,η]=1 CV/integraldisplay D[φ]D[ψ]D[¯ψ]e xp/bracketleftbigg i/integraldisplay d4z/braceleftBig Ltot(φ(z),ψ(z),¯ψ(z)) +J(z)φ(z)+¯η(z)ψ(z)+¯ψ(z)η(z)/bracerightBig/bracketrightbigg .(10.3.70) We employ the following normalization, Z[J=¯η=η=0 ]= <0,out|0,in> with CV=1. (10.3.71) In this manner, as a byproduct of the derivation of the Schwinger–Dyson equations, we have succeeded in deriving the functional integral representation of the generating functionalZ[J,¯η,η]of Green’s functions from the canonical formalism of quantum theory. In Sec- tion 10.1, we have derived the canonical formalism of quantum theory from the functional 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics 329 (path) integral formalism of quantum theory, adopting the transformation function in the func- tional (path) integral representation as Feynman’s action principle, therefore establishing the equivalence of canonical formalism and functional (path) integral formalism at least for thenonsingular Lagrangian (density) system. It is astonishingly hard to establish this equivalencefor the singular Lagrangian (density). 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics We consider the grand canonical ensemble of the Fermion (mass m) and the Boson (mass κ) with Euclidean Lagrangian density in contact with the particle source µ, L/prime E/parenleftbig ψEα(τ,/vectorx),∂µψEα(τ,/vectorx),φ(τ,/vectorx),∂µφ(τ,/vectorx)/parenrightbig =¯ψEα(τ,/vectorx)/braceleftbigg iγk∂k+iγ4/parenleftbigg∂ ∂τ−µ/parenrightbigg −m/bracerightbigg α,βψEβ(τ,/vectorx) +1 2φ(τ,/vectorx)/parenleftbigg∂2 ∂x2ν−κ2/parenrightbigg φ(τ,/vectorx)+1 2gTr/braceleftbig γ[¯ψE(τ,/vectorx),ψE(τ,/vectorx)]/bracerightbig φ(τ,/vectorx).(10.4.1) The density matrix ˆρGC(β)of the grand canonical ensemble in the Schrödinger Picture is given by ˆρGC(β)=e x p/bracketleftBig −β(ˆH−µˆN)/bracketrightBig ,β =1 kBT, (10.4.2) where the total Hamiltonian ˆHis split into two parts, ˆH0=free Hamiltonian for Fermion (mass m) and Boson (mass κ), ˆH1=−/integraldisplay d3/vectorxˆ(/vectorx)ˆφ(/vectorx), (10.4.3) with the “current” given by ˆ(/vectorx)=1 2gTr{γ[/hatwide¯ψE(/vectorx),ˆψE(/vectorx)]}, (10.4.4) and ˆN=1 2/integraldisplay d3/vectorxTr{−γ4[/hatwide¯ψE(/vectorx),ˆψE(/vectorx)]}. (10.4.5) By the standard method of quantum ﬁeld theory, we use the Interaction Picture with ˆNin- cluded in the free part, and obtain ˆρGC(β)=ˆρ0(β)ˆS(β), (10.4.6a) 330 10 Calculus of V ariations: Applications ˆρ0(β)=e x p/bracketleftBig −β(ˆH0−µˆN)/bracketrightBig , (10.4.7) ˆS(β)=Tτ/braceleftBigg exp/bracketleftBigg −/integraldisplayβ 0dτ/integraldisplay d3/vectorxˆH1(τ,/vectorx)/bracketrightBigg/bracerightBigg , (10.4.8a) and ˆH1(τ,/vectorx)=−ˆ(I)(τ,/vectorx)ˆφ(I)(τ,/vectorx). (10.4.9) We know that the Interaction Picture operator ˆf(I)(τ,/vectorx)is related to the Schrödinger Picture operator ˆf(/vectorx)by ˆf(I)(τ,/vectorx)=ˆρ−1 0(τ)·ˆf(/vectorx)·ˆρ0(τ). (10.4.10) We introduce the external hook {J(τ,/vectorx),¯ηα(τ,/vectorx),ηβ(τ,/vectorx)}in the Interaction Picture, and obtain ˆHint 1(τ,/vectorx)=−{[ˆ(I)(τ,/vectorx)+J(τ,/vectorx)]ˆφ(I)(τ,/vectorx) +¯ηα(τ,/vectorx)ˆψ(I) Eα(τ,/vectorx)+/hatwide¯ψEβ(τ,/vectorx)ηβ(τ,/vectorx))}.(10.4.11) We replace Eqs. (10.4.6a), (10.4.7) and (10.4.8a) with ˆρGC(β;[J,¯η,η]) = ˆρ0(β)ˆS(β;[J,¯η,η]), (10.4.6b) ˆρ0(β)=e x p/bracketleftBig −β(ˆH0−µˆN)/bracketrightBig , (10.4.7) ˆS(β;[J,¯η,η]) = Tτ{exp/bracketleftBigg −/integraldisplayβ 0dτ/integraldisplay d3/vectorxˆHint 1(τ,/vectorx)/bracketrightBigg }. (10.4.8b) Here we have (a)0≤τ≤β. δ δJ(τ,/vectorx)ˆρGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =ˆρ0(β)Tτ{ˆφ(I)(τ,/vectorx)e x p/bracketleftBigg −/integraldisplayβ 0dτ/integraldisplay d3/vectorx/bracketrightBigg ˆHint 1(τ,/vectorx)]}/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =ˆρ0(β)Tτ{exp/bracketleftBigg −/integraldisplayβ τdτ/integraldisplay d3/vectorxˆHint 1/bracketrightBigg } ×ˆφ(I)(τ,/vectorx)Tτ{exp[−/integraldisplayτ 0dτ/integraldisplay d3/vectorxˆHint 1]}/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =ˆρ0(β)ˆS(β;[J,¯η,η])ˆS(−τ;[J,¯η,η])ˆφ(I)(τ,/vectorx)ˆS(τ;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =ˆρGC(β;[J,¯η,η]){ˆρ0(τ)ˆS(τ;[J,¯η,η])}−1ˆφ(/vectorx){ˆρ0(τ)ˆS(τ;[J,¯η,η])}/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =ˆρGC(β)ˆφ(τ,/vectorx). 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics 331 Thus we obtain δ δJ(τ,/vectorx)ˆρGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0=ˆρGC(β)ˆφ(τ,/vectorx). (10.4.12) Likewise we obtain δ δ¯ηα(τ,/vectorx)ˆρGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0=ˆρGC(β)ˆψEα(τ,/vectorx), (10.4.13) δ δηβ(τ,/vectorx)ˆρGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0=−ˆρGC(β)/hatwide¯ψEβ(τ,/vectorx), (10.4.14) and δ2 δ¯ηα(τ,/vectorx)δηβ(τ/prime,/vectorx)ˆρGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0 =−ˆρGC(β)Tτ{ˆψEα(τ,/vectorx)/hatwide¯ψEβ(τ,/vectorx)}.(10.4.15) The Heisenberg Picture operator ˆf(τ,/vectorx)is related to the Schrödinger Picture operator ˆf(/vectorx) by ˆf(τ,/vectorx)=ˆρ−1 GC(τ)·ˆf(/vectorx)·ˆρGC(τ). (10.4.16) (b)τ/∈[0,β]. As for τ/∈[0,β], the functional derivative of ˆρGC(β;[J,¯η,η])with respect to {J,¯η,η} vanishes. In order to derive the equation of motion for the partition function, we use the “equation of motion” of ˆψEα(τ,/vectorx)andˆφ(τ,/vectorx), /braceleftbigg iγk∂k+iγ4/parenleftbigg∂ ∂τ−µ/parenrightbigg −m+gγˆφ(τ,/vectorx)/bracerightbigg β,αˆψEα(τ,/vectorx)=0, (10.4.17) /hatwide¯ψEβ(τ,/vectorx)/braceleftbigg iγk∂k+iγ4/parenleftbigg∂ ∂τ−µ/parenrightbigg −m+gγˆφ(τ,/vectorx)/bracerightbiggT β,α=0, (10.4.18) /parenleftbigg∂2 ∂x2ν−κ2/parenrightbigg ˆφ(τ,/vectorx)+gTr/braceleftBig γ/hatwide¯ψE(τ,/vectorx)ˆψE(τ,/vectorx)/bracerightBig =0, (10.4.19) and the equal “time” canonical (anti-)commuters, δ(τ−τ/prime)/braceleftBig ˆψEα(τ,/vectorx),/hatwide¯ψEβ(τ,/vectorx)/bracerightBig =δαβδ(τ−τ/prime)δ(/vectorx−/vectorx/prime), (10.4.20a) δ(τ−τ/prime)/bracketleftbigg ˆφ(τ,/vectorx),∂ ∂τ/primeˆφ(τ/prime,/vectorx/prime)/bracketrightbigg =δ(τ−τ/prime)δ3(/vectorx−/vectory), (10.4.20b) with all the rest of equal “time” (anti-)commutators equal to 0. We obtain the equations of motion of the partition function of the grand canonical ensemble ZGC(β;[J,¯η,η]) = TrˆρGC(β;[J,¯η,η]) (10.4.21) in the presence of the external hook {J,¯η,η}from Eqs. (10.4.12), (10.4.13) and (10.4.14) as 332 10 Calculus of V ariations: Applications /braceleftbigg iγk∂k+iγ4/parenleftbigg∂ ∂τ−µ/parenrightbigg −m+gγ1 iδ δJ(τ,/vectorx)/bracerightbigg β,α1 iδ δ¯ηα(τ,¯x)ZGC(β;[J,¯η,η]) =−ηβ(τ,/vectorx)ZGC(β;[J,¯η,η]), (10.4.22) /braceleftbigg iγk∂k+iγ4/parenleftbigg∂ ∂τ+µ/parenrightbigg +m−gγ1 iδ δJ(τ,/vectorx)/bracerightbiggT β,αiδ δηβ(τ,/vectorx)ZGC(β;[J,¯η,η]) =¯ηα(τ,/vectorx)ZGC(β;[J,¯η,η]), (10.4.23) /braceleftbigg/parenleftbigg∂2 ∂x2ν−κ2/parenrightbigg1 iδ δJ(τ,/vectorx)−gγβαδ2 δ¯ηα(τ,/vectorx)δηβ(τ,/vectorx)/bracerightbigg ZGC(β;[J,¯η,η]) =J(τ,/vectorx)ZGC(β;[J,¯η,η]) (10.4.24) We can solve the functional differential equations, (10.4.22), (10.4.23) and (10.4.24), by the method of Section 10.3. As in Section 10.3, we deﬁne the functional Fourier transform ˜ZGC(β;[φ, ψ E,¯ψE])ofZGC(β;[J,¯η,η])by ZGC(β;[J,¯η,η])≡/integraldisplay D[¯ψE]D[ψE]D[φ]˜ZGC(β;[φ, ψ E,¯ψE]) ×exp/bracketleftBigg i/integraldisplayβ 0dτ/integraldisplay d3/vectorx/braceleftbig J(τ,/vectorx)φ(τ,/vectorx)+¯ηα(τ,/vectorx)ψEα(τ,/vectorx)+¯ψEβ(τ,/vectorx)ηβ(τ,/vectorx)/bracerightbig/bracketrightBigg . We obtain the equations of motion satisﬁed by the functional Fourier transform ˜ZGC(β;[φ, ψ E,¯ψE])from Eqs. (10.4.22), (10.4.23) and (10.4.24), after the functional in- tegral by parts on the right-hand sides involving ¯ηα,ηβandJas δ δψEα(τ,/vectorx)ln˜ZGC(β;[φ, ψ E,¯ψE]) =δ δψEα(τ,/vectorx)/integraldisplayβ 0dτ/integraldisplay d3/vectorxL/prime E((10.4.1) ), δ δ¯ψEβ(τ,/vectorx)ln˜ZGC(β;[φ, ψ E,¯ψE]) =δ δ¯ψEβ(τ,/vectorx)/integraldisplayβ 0dτ/integraldisplay d3/vectorxL/prime E((10.4.1) ), δ δφ(τ,/vectorx)ln˜ZGC(β;[φ, ψ E,¯ψE]) =δ δφ(τ,/vectorx)/integraldisplayβ 0dτ/integraldisplay d3/vectorxL/prime E((10.4.1) ), which we can immediately integrate to obtain ˜ZGC(β;[φ, ψ E,¯ψE]) =Cexp/bracketleftBigg/integraldisplayβ 0dτ/integraldisplay d3/vectorxL/prime E((10.4.1) )/bracketrightBigg . (10.4.25a) 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics 333 Thus we have the path integral representation of ZGC(β;[J,¯η,η])as ZGC(β;[J,¯η,η]) =C/integraldisplay D[¯ψE]D[ψE]D[φ]e xp/bracketleftbigg/integraldisplayβ 0dτ/integraldisplay d3/vectorx/braceleftbigg L/prime E((10.4.1) ) +iJ(τ,/vectorx)φ(τ,/vectorx)+i¯ηα(τ,/vectorx)ψEα(τ,/vectorx)+i¯ψEβ(τ,/vectorx)ηβ(τ,/vectorx)/bracerightbigg/bracketrightbigg =Z0exp/bracketleftBigg −gγβα/integraldisplayβ 0dτ/integraldisplay d3/vectorxiδ δηβ(τ,/vectorx)1 iδ δ¯ηα(τ,/vectorx)1 iδ δJ(τ,/vectorx)/bracketrightBigg ×exp/bracketleftbigg/integraldisplayβ 0dτ/integraldisplay d3/vectorx/integraldisplayβ 0dτ/prime/integraldisplay d3/vectorx/prime/braceleftbigg −1 2J(τ,/vectorx)D0(τ−τ/prime,/vectorx−/vectorx/prime)J(τ/prime,/vectorx/prime) +¯ηα(τ,/vectorx)S0 αβ(τ−τ/prime,/vectorx−/vectorx/prime)ηβ(τ/prime,/vectorx/prime)/bracerightbigg/bracketrightbigg . (10.4.25b) The normalization constant Z0is so chosen that Z0=ZGC(β;J=¯η=η=0,g=0 ) (10.4.26) =/productdisplay |/vectorp|,|/vectork|{1+e x p [ −β(ε/vectorp−µ)]}{1+e x p[ −β(ε/vectorp+µ)]}/braceleftbig 1−exp/bracketleftbig −βω/vectork/bracketrightbig/bracerightbig−1, with ε/vectorp=(/vectorp2+m2)1 2,ω /vectork=(/vectork2+κ2)1 2. (10.4.27) D0(τ−τ/prime,/vectorx−/vectorx/prime)andS0 αβ(τ−τ/prime,/vectorx−/vectorx/prime)are the “free” temperature Green’s functions of the Bose ﬁeld and the Fermi ﬁeld, respectively, and are given by D0(τ−τ/prime,/vectorx−/vectorx/prime)=/integraldisplayd3/vectork (2π)32ω/vectork/braceleftbigg (f/vectork+1 )e x p/bracketleftBig i/vectork(/vectorx−/vectorx/prime)−ω/vectork(τ−τ/prime)/bracketrightBig +f/vectorkexp/bracketleftBig −i/vectork(/vectorx−/vectorx/prime)+ω/vectork(τ−τ/prime)/bracketrightBig/bracerightbigg , S0 αβ(τ−τ/prime,/vectorx−/vectorx/prime)=(iγν∂ν+m)α,β/integraldisplayd3/vectork (2π)32ε/vectork ×  /braceleftbigg (N + /vectork−1) exp/bracketleftBig i/vectork(/vectorx−/vectorx/prime)−(ε/vectork−µ)(τ−τ/prime)/bracketrightBig +N− /vectorkexp/bracketleftBig −i/vectork(/vectorx−/vectorx/prime)+(ε/vectork+µ)(τ−τ/prime)/bracketrightBig/bracerightbigg , forτ>τ/prime,/braceleftbigg N+ /vectorkexp/bracketleftBig −i/vectork(/vectorx−/vectorx/prime)−(ε/vectork−µ)(τ−τ/prime)/bracketrightBig +(N− /vectork−1) exp/bracketleftBig i/vectork(/vectorx−/vectorx/prime)+(ε/vectork+µ)(τ−τ/prime)/bracketrightBig/bracerightbigg , forτ<τ/prime, 334 10 Calculus of V ariations: Applications ∂4≡∂ ∂τ−µ, f /vectork=1 exp/bracketleftbig βω/vectork/bracketrightbig −1,N± /vectork=1 exp[β(ε/vectork∓µ)] + 1. Thef/vectorkis the density of the state of the Bose particles at energy ω/vectork, and the N± /vectorkis the density of the state of the (anti-)Fermi particles at energy ε/vectork. We have two ways of expressing ZGC(β;[J,¯η,η]), Eq. (10.4.25b): ZGC(β;[J,¯η,η]) =Z0exp/bracketleftbigg −1 2/integraldisplayβ 0dτ/integraldisplay d3/vectorx/integraldisplayβ 0dτ/prime/integraldisplay d3/vectorx/prime/braceleftbigg J(τ,/vectorx)−gγβαiδ δηβ(τ,/vectorx)1 iδ δ¯ηα(τ,/vectorx)/bracerightbigg ×D0(τ−τ/prime,/vectorx−/vectorx/prime)/braceleftbigg J(τ/prime,/vectorx/prime)−gγβαiδ δηβ(τ/prime,/vectorx/prime)1 iδ δ¯ηα(τ/prime,/vectorx/prime)/bracerightbigg/bracketrightbigg ×exp/bracketleftBigg/integraldisplayβ 0dτ/integraldisplay d3/vectorx/integraldisplayβ 0dτ/prime/integraldisplay d3/vectorx/prime¯ηα(τ,/vectorx)S0 αβ(τ−τ/prime,/vectorx−/vectorx/prime)ηβ(τ/prime,/vectorx/prime)/bracketrightBigg (10.4.28) =Z0/braceleftbigg Det/parenleftbigg 1+gS0(τ,/vectorx)γ1 iδ δJ(τ,/vectorx)/parenrightbigg/bracerightbigg−1 exp/bracketleftbigg/integraldisplayβ 0dτ/integraldisplay d3/vectorx/integraldisplayβ 0dτ/prime/integraldisplay d3/vectorx/prime ×¯ηα(τ,/vectorx)/parenleftbigg 1+gS0(τ,/vectorx)γ1 iδ δJ(τ,/vectorx)/parenrightbigg−1 αεS0 εβ(τ−τ/prime,/vectorx−/vectorx/prime)ηβ(τ/prime,/vectorx/prime)/bracketrightbigg ×exp/bracketleftBigg −1 2/integraldisplayβ 0dτ/integraldisplay d3/vectorx/integraldisplayβ 0dτ/prime/integraldisplay d3/vectorx/primeJ(τ,/vectorx)D0(τ−τ/prime,/vectorx−/vectorx/prime)J(τ/prime,/vectorx/prime)/bracketrightBigg .(10.4.29) The thermal expectation value of the τ-ordered function fτ-ordered(ˆψ,/hatwide¯ψ,ˆφ) in the grand canonical ensemble is given by <fτ-ordered(ˆψ,/hatwide¯ψ,ˆφ)>≡Tr/braceleftBig ˆρGC(β)fτ-ordered(ˆψ,/hatwide¯ψ,ˆφ)/bracerightBig TrˆρGC(β) ≡1 ZGC(β;[J,¯η,η])f/parenleftbigg1 iδ δ¯η,iδ δη,1 iδ δJ/parenrightbigg ZGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0. (10.4.30) 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics 335 According to this formula, the one-body “full” temperature Green’s functions of the Bose ﬁeld and the Fermi ﬁeld, D(τ−τ/prime,/vectorx−/vectorx/prime)andSαβ(τ−τ/prime,/vectorx−/vectorx/prime), are given, respectively, by D(τ−τ/prime,/vectorx−/vectorx/prime)=Tr/braceleftBig ˆρGC(β)Tτ/parenleftbigˆφ(τ,/vectorx)ˆφ(τ/prime,/vectorx/prime)/parenrightbig/bracerightBig TrˆρGC(β) =−1 ZGC(β;[J,¯η,η])1 iδ δJ(τ,/vectorx)1 iδ δJ(τ/prime,/vectorx/prime)ZGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0,(10.4.31) Sαβ(τ−τ/prime,/vectorx−/vectorx/prime)=Tr/braceleftBig ˆρGC(β)ˆψα(τ,/vectorx)/hatwide¯ψβ(τ/prime,/vectorx/prime)/bracerightBig TrˆρGC(β) =−1 ZGC(β;[J,¯η,η])1 iδ δ¯ηα(τ,/vectorx)iδ δηβ(τ/prime,/vectorx/prime)ZGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=¯η=η=0.(10.4.32) From the cyclicity of Tr and the (anti-)commutativity of ˆφ(τ,/vectorx)(ˆψα(τ,/vectorx)) under the T τ- ordering symbol, we have D(τ−τ/prime<0,/vectorx−/vectorx/prime)=+ D(τ−τ/prime+β,/vectorx−/vectorx/prime), (10.4.33) and Sαβ(τ−τ/prime<0,/vectorx−/vectorx/prime)=−Sαβ(τ−τ/prime+β,/vectorx−/vectorx/prime), (10.4.34) where 0≤τ,τ/prime≤β, i.e., the Boson (Fermion) “full” temperature Green’s function is (anti-)periodic with period β. From this, we have the Fourier decompositions as ˆφ(τ,/vectorx)=1 β/summationdisplay n/integraldisplayd3/vectork (2π)32ω/vectork/braceleftbigg exp/bracketleftBig i(/vectork/vectorx−ωnτ)/bracketrightBig a(ωn,/vectork) +e x p/bracketleftBig −i(/vectork/vectorx−ωnτ)/bracketrightBig a†(ωn,/vectork)/bracerightbigg , ωn=2nπ β,n = integer ,(10.4.35) ˆψα(τ,/vectorx)=1 β/summationdisplay n/integraldisplayd3/vectork (2π)32ε/vectork{exp/bracketleftBig i(/vectork/vectorx−ωnτ)/bracketrightBig unα(/vectork)b(ωn,/vectork) +e x p/bracketleftBig −i(/vectork/vectorx−ωnτ)/bracketrightBig ¯vnα(/vectork)d†(ωn,/vectork)}, ωn=(2n+1 )π β,n = integer ,(10.4.36) 336 10 Calculus of V ariations: Applications where [a(ωn,/vectork),a†(ωn/prime,/vectork/prime)] = 2 ω/vectork(2π)3δ3(/vectork−/vectork/prime)δn,n/prime, (10.4.37) [a(ωn,/vectork),a(ωn/prime,/vectork/prime)] = [a†(ωn,/vectork),a†(ωn/prime,/vectork/prime)] = 0 , (10.4.38) /braceleftBig b(ωn,/vectork),b†(ωn/prime,/vectork/prime)/bracerightBig =/braceleftBig d(ωn,/vectork),d†(ωn/prime,/vectork/prime)/bracerightBig =2ε/vectork(2π)3δ3(/vectork−/vectork/prime)δn,n/prime,(10.4.39) the rest of the anti-commutators =0. (10.4.40) We shall now address ourselves to the problem of ﬁnding the equation of motion of the one-body Boson and Fermion Green’s functions. We deﬁne the one-body Boson and Fermion“full” temperature Green’s functions, D J(x, y)andSJ α,β(x, y), by: for Boson ﬁeld Green’s function DJ(x, y)≡−<Tτ(ˆφ(x)ˆφ(y))>J/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 ≡−δ2 δJ(x)δJ(y)lnZGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 ≡−δ δJ(x)<ˆφ(y)>J/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0,(10.4.41) and for Fermion ﬁeld Green’s function SJ α,β(x, y)≡+<Tτ(ˆψα(x)/hatwide¯ψβ(y))>J/vextendsingle/vextendsingle/vextendsingle ¯η=η=0 ≡−1 ZGC(β;[J,¯η,η])δ δ¯ηα(x)δ δηβ(y)ZGC(β;[J,¯η,η])/vextendsingle/vextendsingle/vextendsingle/vextendsingle ¯η=η=0.(10.4.42) From Eqs. (10.4.22), (10.4.23) and (10.4.24), we obtain a summary of the Schwinger– Dyson equation satisﬁed by DJ(x, y)andSJ α,β(x, y): Summary of Schwinger–Dyson equation in conﬁguration space /parenleftBig iγν∂ν−m+gγ < ˆφ(x)>J/parenrightBig αεSJ εβ(x, y)−/integraldisplay d4zΣ∗ αε(x, z)SJ εβ(z,y) =δαβδ4(x−y), /parenleftbigg∂2 ∂x2ν−κ2/parenrightbigg <ˆφ(x)>J =1 2gγβα/braceleftbig SJ αβ(τ,/vectorx;τ−ε,/vectorx)+SJ αβ(τ,/vectorx;τ+ε,/vectorx)/bracerightbig/vextendsingle/vextendsingle/vextendsingle/vextendsingle ε→0+, /parenleftbigg∂2 ∂x2ν−κ2/parenrightbigg DJ(x, y)−/integraldisplay d4zΠ∗(x, z)DJ(z,y)=δ4(x−y), 10.4 Schwinger–Dyson Equation in Quantum Statistical Mechanics 337 Σ∗ αβ(x, y)=g2/integraldisplay d4ud4vγαδSJ δν(x, u)Γνβ(u, y;v)DJ(v,x), Π∗(x, y)=g2/integraldisplay d4ud4vγαβSJ βδ(x, u)Γδν(u, v;y)SJ να(v,x), Γαβ(x, y;z)=γαβ(z)δ4(x−y)δ4(x−z)+1 gδΣ∗ αβ(x, y) δ<ˆφ(z)>J. This system of nonlinear coupled integro-differential equations is exact and closed. Starting from the zeroth-order term of Γαβ(x, y;z), we can develop Feynman–Dyson type graphical perturbation theory for quantum statistical mechanics in the conﬁguration space, by iteration. We here employed the following abbreviation, x≡(τx,/vectorx),/integraldisplay d4x≡/integraldisplayβ 0dτx/integraldisplay d3/vectorx. 2g ∑(x, y) = α β(x)γα δ* x, δ u,ν y, βv (v,x)'JD (x,u)J δ νSΓ(u,y;v ) Fig. 10.9: Graphical representation of the proper self-energy part, Σ∗ αβ(x, y ). *(x,y) = Γ(u,v;y ) (x,u)J βδS(v,x)JS 2g (x)γαβ x, βΠ u, δyv,ν να Fig. 10.10: Graphical representation of the proper self-energy part, Π∗(x, y ). We note that SJ αβ(x, y)andDJ(x, y)are determined by Eqs. (10.4.3a through f) only for τx−τy∈[−β,β], and we assume that they are deﬁned by the periodic boundary condition with the period 2β for other τx−τy/∈[−β,β]. 338 10 Calculus of V ariations: Applications Next we set J≡0, and hence we have <ˆφ(x)>J≡0≡0, restoring the translational invariance of the system. We Fourier transform Sαβ(x)andD(x), Sαβ(x)=1 β/summationdisplay p4/integraldisplayd3/vectorp (2π)3Sαβ(p4,/vectorp)e x p[ i(/vectorp/vectorx−p4τx)], (10.4.43a) D(x)=1 β/summationdisplay p4/integraldisplayd3/vectorp (2π)3D(p4,/vectorp)e x p[ i(/vectorp/vectorx−p4τx)], (10.4.43b) Γα,β(x, y;z)=Γα,β(x−y,x−z)=1 β2/summationdisplay p4,k4/integraldisplayd3/vectorpd3/vectork (2π)6Γα,β(p,k) ×exp/bracketleftBig i{/vectorp(/vectorx−/vectory)−p4(τx−τy)} −i/braceleftBig /vectork(/vectorx−/vectorz)−k4(τx−τz)/bracerightBig/bracketrightBig ,(10.4.43c) p4=  (2n+1 )π β, Fermion ,n=integer, 2nπ β, Boson ,n=integer.(10.4.43d) We have the Schwinger–Dyson equation in momentum space: Summary of Schwinger–Dyson equation in momentum space /braceleftbig −/vectorγ/vectorp+γ4(p4−iµ)−(m+Σ∗(p))/bracerightbig αεSεβ(p)=δαβ/summationdisplay nδ/parenleftbigg p4−(2n+1 )π β/parenrightbigg , /braceleftbig −k2 ν−(κ2+Π∗(k))/bracerightbig D(k)=/summationdisplay nδ/parenleftbigg k4−2nπ β/parenrightbigg , Σ∗ α,β(p)=g21 β/summationdisplay k4/integraldisplayd3/vectork (2π)3γαδSδε(p+k)Γεβ(p+k,k)D(k), Π∗(k)=g21 β/summationdisplay p4/integraldisplayd3/vectorp (2π)3γµνSνλ(p+k)Γλρ(p+k,k)Sρµ(p), Γαβ(p,k)=/summationdisplay n,,mγαβδ/parenleftbigg p4−(2n+1 )π β/parenrightbigg δ/parenleftbigg k4−(2m+1 )π β/parenrightbigg +Λαβ(p,k), 10.5 W eyl’s Gauge Principle 339 whereΛαβ(p,k)represents the sum of the vertex diagram, except for the ﬁrst term. This system of nonlinear coupled integral equations is exact and closed. Starting from the zeroth- order term of Γαβ(p,k), we can develop a Feynman–Dyson type graphical perturbation theory for quantum statistical mechanics in the momentum space, by iteration. From the Schwinger–Dyson equation, we can derive the Bethe–Goldstone diagram rule of the many-body problems at ﬁnite temperature in quantum statistical mechanics, nuclear physics and condensed matter physics. For details of this diagram rule, we refer the reader toA.L. Fetter and J.D. Walecka. 10.5 Weyl’s Gauge Principle Inelectrodynamics , we have a property known as the gauge invariance .I n t h e theory of the gravitational ﬁeld , we have a property known as the scale invariance . Before the birth of quantum mechanics, H. Weyl attempted to construct the uniﬁed theory of classical electro- dynamics and the gravitational ﬁeld. But he failed to accomplish his goal. After the birth ofquantum mechanics, he realized that the gauge invariance of electrodynamics is not related to the scale invariance of the gravitational ﬁeld, but is related to the invariance of the mat- ter ﬁeld φ(x)under the local phase transformation. The matter ﬁeld in interaction with the electromagnetic ﬁeld has a property known as the charge conservation law or the current con- servation law . In this section, we discuss Weyl’s gauge principle for the U(1)gauge ﬁeld and the non-Abelian gauge ﬁeld, and Kibble’s gauge principle for the gravitational ﬁeld. Weyl’s gauge principle: Electrodynamics is described by the total Lagrangian density with the use of the four-vector potential A µ(x)by Ltot=Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig +Lint/parenleftbig φ(x),Aµ(x)/parenrightbig +Lgauge/parenleftbig Aµ(x),∂νAµ(x)/parenrightbig .(10.5.1) This system is invariant under the local U(1)transformations, Aµ(x)→A/prime µ(x)≡Aµ(x)−∂µε(x), (10.5.2) φ(x)→φ/prime(x)≡exp/bracketleftbig iqε(x)/bracketrightbig φ(x). (10.5.3) The interaction Lagrangian density Lint/parenleftbig φ(x),Aµ(x)/parenrightbig is generated by the substitution, ∂µφ(x)→Dµφ(x)≡/parenleftbig ∂µ+iqAµ(x)/parenrightbig φ(x), (10.5.4) in the original matter ﬁeld Lagrangian density Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig . The derivative Dµφ(x) is called the covariant derivative ofφ(x)and transforms exactly as φ(x), Dµφ(x)→/parenleftbig Dµφ(x)/parenrightbig/prime=e x p/bracketleftbig iqε(x)/bracketrightbig Dµφ(x), (10.5.5) under the local U(1)transformations, Eqs. (10.5.2) and (10.5.3). The physical meaning of this local U(1)invariance lies in its weaker version, the global U(1)invariance, namely, ε(x)=ε, space–time independent constant. 340 10 Calculus of V ariations: Applications The global U(1)invariance of the matter ﬁeld Lagrangian density, Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig , under the global U(1)transformation of φ(x), φ(x)→φ/prime/prime(x)=e x p[ iqε]φ(x),ε=constant , (10.5.6) in its inﬁnitesimal version, δφ(x)=iqεφ(x),ε=inﬁnitesimal constant , (10.5.7) results in ∂Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig ∂φ(x)δφ(x)+∂Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig ∂/parenleftbig ∂µφ(x)/parenrightbig δ/parenleftbig ∂µφ(x)/parenrightbig =0. (10.5.8) With the use of the Euler–Lagrange equation of motion for φ(x), ∂Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig ∂φ(x)−∂µ∂Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig ∂/parenleftbig ∂µφ(x)/parenrightbig =0, we obtain the current conservation law, ∂µJµ matter(x)=0, (10.5.9) εJµ matter(x)=∂Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig ∂/parenleftbig ∂µφ(x)/parenrightbig δφ(x). (10.5.10) This in its integrated form becomes the charge conservation law, d dtQmatter(t)=0, (10.5.11) Qmatter(t)=/integraldisplay d3/vectorxJ0 matter(t,/vectorx). (10.5.12) Weyl’s gauge principle considers the analysis backwards. The extension of the “current conserving” global U(1)invariance, Eqs. (10.5.7) and (10.5.8), of the matter ﬁeld Lagrangian density Lmatter/parenleftbig φ(x),∂µφ(x)/parenrightbig to the local U(1)invariance necessitates: 1) the introduction of the U(1)gauge ﬁeld Aµ(x), and the replacement of the derivative ∂µφ(x)in the matter ﬁeld Lagrangian density with the covariant derivative Dµφ(x), ∂µφ(x)→Dµφ(x)≡/parenleftbig ∂µ+iqAµ(x)/parenrightbig φ(x), (10.5.13) and 2) the requirement that the covariant derivative Dµφ(x)transforms exactly like the matter ﬁeldφ(x)under the local U(1)phase transformation of φ(x), Dµφ(x)→/parenleftbig Dµφ(x)/parenrightbig/prime=e x p/bracketleftbig iqε(x)/bracketrightbig Dµφ(x), (10.5.14) under φ(x)→φ/prime(x)≡exp/bracketleftbig iqε(x)/bracketrightbig φ(x). (10.5.15) 10.5 W eyl’s Gauge Principle 341 From requirement 2), we obtain the transformation law of the U(1)gauge ﬁeld Aµ(x) immediately, Aµ(x)→A/prime µ(x)≡Aµ(x)−∂µε(x). (10.5.16) From requirement 2), the local U(1)invariance of Lmatter/parenleftbig φ(x),Dµφ(x)/parenrightbig is also self-evident. In order to give dynamical content to the U(1)gauge ﬁeld, we introduce the ﬁeld strength tensor Fµν(x)to the gauge ﬁeld Lagrangian density Lgauge/parenleftbig Aµ(x),∂νAµ(x)/parenrightbig by the trick, [Dµ,Dν]φ(x)=iq/parenleftbig ∂µAν(x)−∂νAµ(x)/parenrightbig φ(x)≡iqFµν(x)φ(x). (10.5.17) From the transformation law of the U(1)gauge ﬁeld Aµ(x), Eq. (10.5.2), we observe that the ﬁeld strength tensor Fµν(x)is the locally invariant quantity, Fµν(x)→F/prime µν(x)=∂µA/prime ν(x)−∂νA/prime µ(x) =∂µAν(x)−∂νAµ(x)=Fµν(x).(10.5.18) As the gauge ﬁeld Lagrangian density Lgauge/parenleftbig Aµ(x),∂νAµ(x)/parenrightbig , we choose Lgauge/parenleftbig Aµ(x),∂νAµ(x)/parenrightbig =−1 4Fµν(x)Fµν(x). (10.5.19) In this manner, we obtain the total Lagrangian density of the matter-gauge system which is locally U(1)invariant as Ltot=Lmatter/parenleftbig φ(x),Dµφ(x)/parenrightbig +Lgauge/parenleftbig Fµν(x)/parenrightbig . (10.5.20) We obtain the interaction Lagrangian density Lint/parenleftbig φ(x),Aµ(x)/parenrightbig as Lint/parenleftbig φ(x),Aµ(x)/parenrightbig =Lmatter/parenleftbig φ(x),Dµφ(x)/parenrightbig −L matter/parenleftbig φ(x),∂µφ(x)/parenrightbig , (10.5.21) which is the universal coupling generated by Weyl’s gauge principle. As a result of the local extension of the global U(1)invariance, we have derived the electrodynamics from the current conservation law, Eq. (10.5.9), or the charge conservation law, Eq. (10.5.11). We shall now consider the extension of the present discussion to the non-Abelian gauge ﬁeld. We let the semi-simple Lie group Gbe the gauge group. We let the representation of G in the Hilbert space be U(g), and its matrix representation on the ﬁeld operator ˆψn(x)in the internal space be D(g), U(g)ˆψn(x)U−1(g)=Dn,m(g)ˆψm(x),g∈G. (10.5.22) 342 10 Calculus of V ariations: Applications For the element gε∈Gcontinuously connected to the identity of Gby the parameter {εα}N α=1,w eh a v e U(gε)=e x p[ iεαTα]=1+ iεαTα+···,T α:generator of Lie group G, (10.5.23) D(gε)=e x p[ iεαtα]=1+ iεαtα+···,t α:realization of Tαonˆψn(x),(10.5.24) [Tα,Tβ]=iCαβγTγ, (10.5.25) [tα,tβ]=iCαβγtγ. (10.5.26) We shall assume that the action functional Imatter[ψn]of the matter ﬁeld Lagrangian density Lmatter(ψn(x),∂µψn(x)),g i v e nb y Imatter[ψn]≡/integraldisplay d4xLmatter/parenleftbig ψn(x),∂µψn(x)/parenrightbig , (10.5.27) is invariant under the global Gtransformation, δψn(x)=iεα(tα)n,mψm(x),ε α=inﬁnitesimal constant . (10.5.28) Namely, we have ∂Lmatter/parenleftbig ψn(x),∂µψn(x)/parenrightbig ∂ψn(x)δψn(x) +∂Lmatter/parenleftbig ψn(x),∂µψn(x)/parenrightbig ∂/parenleftbig ∂µψn(x)/parenrightbig δ/parenleftbig ∂µψn(x)/parenrightbig =0.(10.5.29) By the use of the Euler–Lagrange equation of motion, ∂Lmatter/parenleftbig ψ(x),∂µψ(x)/parenrightbig ∂ψn(x)−∂µ/parenleftBigg ∂Lmatter/parenleftbig ψ(x),∂µψ(x)/parenrightbig ∂(∂µψn(x))/parenrightBigg =0, (10.5.30) we have the current conservation law and the charge conservation law, ∂µJµ α,matter(x)=0,α =1,...,N, (10.5.31a) where the conserved matter current Jµ α,matter(x)is given by εαJµ α,matter(x)=∂Lmatter/parenleftbig ψ(x),∂µψ(x)/parenrightbig ∂/parenleftbig ∂µψn(x)/parenrightbig δψn(x), (10.5.31b) and d dtQmatter α(t)=0,α =1,...,N, (10.5.32a) where the conserved matter charge Qmatter α(t)is given by Qmatter α(t)=/integraldisplay d3/vectorxJ0 α,matter(t,/vectorx),α =1,...,N. (10.5.32b) 10.5 W eyl’s Gauge Principle 343 Invoking Weyl’s gauge principle, we extend the global Ginvariance of the matter system to the local Ginvariance of the matter-gauge system under the local Gphase transformation, δψn(x)=iεα(x)(tα)n,mψm(x). (10.5.33) Weyl’s gauge principle requires the following: 1) the introduction of the non-Abelian gauge ﬁeld Aαµ(x)and the replacement of the de- rivative ∂µψn(x)in the matter ﬁeld Lagrangian density with the covariant derivative/parenleftbig Dµψ(x)/parenrightbig n, ∂µψn(x)→/parenleftbig Dµψ(x)/parenrightbig n≡/parenleftbig ∂µδn,m+i(tγ)n,mAγµ(x)/parenrightbig ψm(x), (10.5.34) and 2) the requirement that the covariant derivative/parenleftbig Dµψ(x)/parenrightbig ntransforms exactly as the matter ﬁeldψn(x)under the local Gphase transformation of ψn(x), Eq. (10.5.33), δ/parenleftbig Dµψ(x)/parenrightbig n=iεα(x)(tα)n,m/parenleftbig Dµψ(x)/parenrightbig m, (10.5.35) where tγis the realization of the generator Tγupon the multiplet ψn(x). From Eqs. (10.5.33) and (10.5.35), the inﬁnitesimal transformation law of the non-Abelian gauge ﬁeld Aαµ(x)follows, δAαµ(x)=−∂µεα(x)+iεβ(x)(tadj β)αγAγµ(x) (10.5.36a) =−∂µεα(x)+εβ(x)CβαγAγµ(x). (10.5.36b) Then the local Ginvariance of the gauged matter ﬁeld Lagrangian density Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig becomes self-evident as long as the ungauged matter ﬁeld Lagrangian density Lmatter/parenleftbig ψ(x),∂µψ(x)/parenrightbig is globally Ginvariant. In order to provide dynamical content to the non-Abelian gauge ﬁeld Aαµ(x), we intro- duce the ﬁeld strength tensor Fγµν(x)by the following trick, [Dµ,Dν]ψ(x)≡i(tγ)Fγµν(x)ψ(x), (10.5.37) Fγµν(x)=∂µAγν(x)−∂νAγµ(x)−CαβγAαµ(x)Aβν(x). (10.5.38) We can easily show that the ﬁeld strength tensor Fγµν(x)undergoes local Grotation under localGtransformations, Eqs. (10.5.33) and (10.5.36a), under the adjoint representation, δFγµν(x)=iεα(x)(tadj α)γβFβµν(x) (10.5.39a) =εα(x)CαγβFβµν(x). (10.5.39b) 344 10 Calculus of V ariations: Applications As the Lagrangian density of the non-Abelian gauge ﬁeld Aαµ(x), we choose Lgauge/parenleftbig Aγµ(x),∂νAγµ(x)/parenrightbig ≡−1 4Fγµν(x)Fµν γ(x). (10.5.40) The total Lagrangian density Ltotalof the matter-gauge system is given by Ltotal=Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig +Lgauge/parenleftbig Fγµν(x)/parenrightbig . (10.5.41) The interaction Lagrangian density Lintconsists of two parts due to the nonlinearity of the ﬁeld strength tensor Fγµν(x)with respect to Aγµ(x), Lint=Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig −L matter/parenleftbig ψ(x),∂µψ(x)/parenrightbig +Lgauge/parenleftbig Fγµν(x)/parenrightbig −Lquad gauge/parenleftbig Fγµν(x)/parenrightbig ,(10.5.42) which provides the universal coupling just as the U(1)gauge ﬁeld theory. The conserved current Jµ α,total(x)and the conserved charge/braceleftbig Qtotal α(t)/bracerightbigN α=1after the extension to the local G invariance also consist of two parts, Jµ α,total(x)≡Jµgauged α,matter(x)+Jµ α,gauge(x)≡δItotal[ψ,A αµ] δAαµ(x), (10.5.43a) Itotal[ψ,A αµ]=/integraldisplay d4x/braceleftBig Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig +Lgauge/parenleftbig Fγµν(x)/parenrightbig/bracerightBig , (10.5.43b) Qtotal α(t)=Qmatter α(t)+Qgauge α(t)=/integraldisplay d3/vectorx/braceleftBig J0gauged α,matter(t,/vectorx)+J0 α,gauge(t,/vectorx)/bracerightBig .(10.5.44) We note that the gauged matter current Jµgauged α,matter(x)of Eq. (10.5.43a) is not identical to the ungauged matter current Jµ α,matter(x)of Eq. (10.5.31b): εαJµ α,matter(x)of (10.5.31b) =∂Lmatter/parenleftbig ψ(x),∂µψ(x)/parenrightbig ∂/parenleftbig ∂µψn(x)/parenrightbig δψn(x), whereas after the local Gextension, εαJµgauged α,matter(x)of (10.5.43a) =∂Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig ∂/parenleftbig Dµψn(x)/parenrightbig nδψn(x) =∂Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig ∂/parenleftbig Dµψ(x)/parenrightbig niεα(tα)n,mψm(x) =εα∂Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig ∂(Dνψ(x))n∂/parenleftbig Dνψ(x)/parenrightbig n ∂Aαµ(x) =εα∂Lmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig ∂Aαµ(x) =εαδ δAαµ(x)Igauged matter[ψ,D µψ].(10.5.45) 10.5 W eyl’s Gauge Principle 345 Here we note that Igauged matter[ψ,D µψ]is not identical to the ungauged matter action functional Imatter[ψn]given by Eq. (10.5.27), but is the gauged matter action functional deﬁned by Igauged matter[ψ,D µψ]≡/integraldisplay d4xLmatter/parenleftbig ψ(x),Dµψ(x)/parenrightbig . (10.5.46) We emphasize here that the conserved Noether current after the extension of the global G invariance to the local Ginvariance, is not the gauged matter current Jµgauged α,matter(x)but the total current Jµ α,total(x), Eq. (10.5.43a). At the same time, we note that the strict conservation law of the total current Jµ α,total(x)is enforced at the expense of loss of covariance as we shall see. The origin of this problem is the self-interaction of the non-Abelian gauge ﬁeld Aαµ(x)and the nonlinearity of the Euler–Lagrange equation of motion for the non-Abelian gauge ﬁeld Aαµ(x). We shall make a table of the global U(1)transformation law and the global Gtransforma- tion law. Global U(1)transformation law. Global Gtransformation law. δψn(x)=iεqnψn(x),charged. δψn(x)=iεα(tα)n,mψm(x),charged. δAµ(x)=0, neutral. δAαµ(x)=iεβ(tadj β)αγAγµ(x),charged.(10.5.47) In the global transformation law of internal symmetry, Eq. (10.5.47), the matter ﬁelds ψn(x)which have the group charge undergo global ( U(1)orG) rotation. As for the gauge ﬁelds, Aµ(x)andAαµ(x), the Abelian gauge ﬁeld Aµ(x)remains unchanged under global U(1)transformation while the non-Abelian gauge ﬁeld Aαµ(x)undergoes global Grotation under global Gtransformation. Hence the Abelian gauge ﬁeld Aµ(x)isU(1)-neutral while the non-Abelian gauge ﬁeld Aαµ(x)isG-charged. The ﬁeld strength tensors, Fµν(x)and Fαµν(x),b e h a v ea s Aµ(x)andAαµ(x), under global U(1)andGtransformations. The ﬁeld strength tensor Fµν(x)isU(1)-neutral, while the ﬁeld strength tensor Fαµν(x)isG-charged, which originates from their linearity and nonlinearity in Aµ(x)andAαµ(x), respectively. Global U(1)transformation law. Global Gtransformation law. δFµν(x)=0, neutral. δFαµν(x)=iεβ(tadj β)αγFγµν(x),charged. (10.5.48) When we write the Euler–Lagrange equation of motion for each case, the linearity and the nonlinearity with respect to the gauge ﬁelds become clear. Abelian U(1)gauge ﬁeld. non-Abelian Ggauge ﬁeld. ∂νFνµ(x)=jµ matter(x),linear. Dadj νFνµ α(x)=jµ α,matter(x),nonlinear.(10.5.49) From the anti-symmetry of the ﬁeld strength tensor with respect to the Lorentz indices, µ andν, we have the following current conservation as an identity. Abelian U(1)gauge ﬁeld. non-Abelian Ggauge ﬁeld. ∂µjµ matter(x)=0.Dadj µjµ α,matter(x)=0.(10.5.50) 346 10 Calculus of V ariations: Applications Here we have Dadj µ=∂µ+itadj γAγµ(x). (10.5.51) As the result of the extension to the local ( U(1)orG) invariance, in the case of the Abelian U(1)gauge ﬁeld, due to the neutrality of Aµ(x), the matter current jµ matter(x)alone which originates from the global U(1)invariance is conserved, while in the case of the non-Abelian Ggauge ﬁeld, due to the G-charge of Aαµ(x), the gauged matter current jµ α,matter(x)alone which originates from the local Ginvariance is not conserved, but the sum with the gauge current jµ α,gauge(x)which originates from the self-interaction of the non-Abelian gauge ﬁeld Aαµ(x)is conserved at the expense of the loss of covariance. A similar situation exists for the charge conservation law. Abelian U(1)gauge ﬁeld. non-Abelian Ggauge ﬁeld. d dtQmatter(t)=0.d dtQtot α(t)=0. Qmatter(t)=/integraltext d3/vectorxj0 matter(t,/vectorx).Qtot α(t)=/integraltext d3/vectorxj0 α,tot(t,/vectorx).(10.5.52) Before plunging into the gravitational ﬁeld, we discuss the ﬁnite gauge transformation property of the non-Abelian gauge ﬁeld Aαµ(x). Under the ﬁnite local Gphase transformation ofψn(x), ψn(x)→ψ/prime n(x)=( e x p[ iεα(x)tα])n,mψm(x), (10.5.53) we demand that the covariant derivative Dµψ(x)deﬁned by Eq. (10.5.34) transforms exactly asψ(x), Dµψ(x)→/parenleftbig Dµψ(x)/parenrightbig/prime=(∂µ+itγA/prime γµ(x))ψ/prime(x) =e x p[ iεα(x)tα]Dµψ(x).(10.5.54) From Eq. (10.5.54), we obtain the following equation, exp[−iεα(x)tα](∂µ+itγA/prime γµ(x)) exp[ iεα(x)tα]ψ(x)=(∂µ+itγAγµ(x))ψ(x). Cancelling the ∂µψ(x)term from both sides of the equation above, we obtain exp[−iεα(x)tα](∂µexp[iεα(x)tα]) + exp [ −iεα(x)tα](itγA/prime γµ(x)) exp[ iεα(x)tα] =itγAγµ(x). Solving the above equation for tγA/prime γµ(x), we ﬁnally obtain the ﬁnite gauge transformation law of A/prime γµ(x), tγA/prime γµ(x)=e x p[ iεα(x)tα]/braceleftbig tγAγµ(x) +e x p[ −iεβ(x)tβ](i∂µexp[iεβ(x)tβ])/bracerightbig exp[−iεα(x)tα].(10.5.55) 10.5 W eyl’s Gauge Principle 347 At ﬁrst sight, we get the impression that the ﬁnite gauge transformation law of A/prime γµ(x), (10.5.55), may depend on the speciﬁc realization {tγ}N γ=1of the generator {Tγ}N γ=1upon the multiplet ψ(x). Actually A/prime γµ(x)transforms under the adjoint representation/braceleftbig tadj γ/bracerightbigN γ=1. The inﬁnitesimal version of the ﬁnite gauge transformation, (10.5.55), does reduce to the in- ﬁnitesimal gauge transformation, (10.5.36a) and (10.5.36b), under the adjoint representation. The ﬁrst step of an extension of Weyl’s gauge principle to the non-Abelian gauge group G was carried out by C.N. Yang and R.L. Mills for the SU(2)isospin gauge group and the said gauge ﬁeld is commonly called the Yang–Mills gauge ﬁeld. Furthermore, we can generalize Weyl’s gauge principle to Utiyama’s gauge principle and Kibble’s gauge principle to obtain the Lagrangian density for the gravitational ﬁeld. We note that Weyl’s gauge principle, Utiyama’s gauge principle and Kibble’s gauge principle belong to the category of the invariant variational principle. R. Utiyama derived the theory of the gravitational ﬁeld from his version of the gauge prin- ciple, based on the requirement of the invariance of the action functional I[φ]under the local six-parameter Lorentz transformation . T.W.B. Kibble derived the theory of the gravitational ﬁeld from his version of the gauge principle, based on the requirement of the invariance of the action functional I[φ]under the local ten-parameter Poincaré transformation , extending the treatment of Utiyama. Gravitational ﬁeld: We shall now discuss Kibble’s gauge principle for the gravitational ﬁeld. We let φrepresent the set of generic matter ﬁeld variables φa(x), which we regard as the elements of a column vector φ(x), and deﬁne the matter action functional Imatter[φ]in terms of the matter Lagrangian density Lmatter(φ, ∂µφ)as Imatter[φ]=/integraldisplay d4xLmatter(φ, ∂µφ). (10.5.56) We ﬁrst discuss the inﬁnitesimal transformation of both the coordinates xµand the matter ﬁeld variables φ(x), xµ→x/primeµ=xµ+δxµ,φ(x)→φ/prime(x/prime)=φ(x)+δφ(x), (10.5.57) where the invariance group Gis not speciﬁed. It is convenient to allow the possibility that the matter Lagrangian density Lmatter explicitly depends on the coordinates xµ. Then, under the inﬁnitesimal transformation, (10.5.57), we have δLmatter≡∂Lmatter ∂φδφ+∂Lmatter ∂(∂µφ)δ(∂µφ)+∂Lmatter ∂xµ/vextendsingle/vextendsingle/vextendsingle/vextendsingle φﬁxedδxµ. It is also useful to consider the variation of φ(x)at a ﬁxed value of xµ, δ0φ≡φ/prime(x)−φ(x)=δφ−δxµ∂µφ. (10.5.58) It is obvious that δ0commutes with ∂µ,s ow eh a v e δ(∂µφ)=∂µ(δφ)−(∂µδxν)∂νφ. (10.5.59) 348 10 Calculus of V ariations: Applications The matter action functional, (10.5.56), over a space–time region Ωis transformed under the transformations, (10.5.57), into I/prime matter[Ω]≡/integraldisplay ΩL/prime matter(x/prime)d e t (∂νx/primeµ)d4x. Thus the matter action functional Imatter[Ω]over an arbitrary region Ωis invariant if δLmatter+(∂µδxµ)Lmatter≡δ0Lmatter+∂µ(δxµLmatter)≡0. (10.5.60) We now consider the speciﬁc case of the Poincaré transformation, δxµ=iεµ νxν+εµ,δ φ =1 2iεµνSµνφ, (10.5.61) where {εµ}and{εµν}withεµν=−ενµ, are the 10 inﬁnitesimal constant parameters of the Poincaré group, and SµνwithSµν=−Sνµ, are the mixing matrices of the components of a column vector φ(x)satisfying [Sµν,Sρσ]=i(ηνρSµσ+ηµσSνρ−ηνσSµρ−ηµρSνσ). {Sµν}will be identiﬁed as the spin matrices of the matter ﬁeld φlater. From Eq. (10.5.59), we have δ(∂µφ)=1 2iερσSρσ∂µφ−iερ µ∂ρφ. (10.5.62) Since we have ∂µ(δxµ)=εµ µ=0, the condition, (10.5.60), for the invariance of the matter action functional Imatter[φ]under the inﬁnitesimal Poincaré transformations, (10.5.61), reduces to δLmatter≡0, and results in the 10 identities, ∂Lmatter ∂xρ≡∂ρLmatter−∂Lmatter ∂φ∂ρφ−∂Lmatter ∂(∂µφ)∂ρ∂µφ≡0, (10.5.63) ∂Lmatter ∂φiSρσφ+∂Lmatter ∂(∂µφ)(iSρσ∂µφ+ηµρ∂σφ−ηµσ∂ρφ)≡0. (10.5.64) The conditions (10.5.63) express the translational invariance of the system and are equivalent to the requirement that Lmatter is explicitly independent of xµ. We use the Euler–Lagrange equations of motion in Eqs. (10.5.63) and (10.5.64), obtaining the ten conservation laws,which we write as ∂ µTµ ρ=0,∂ µ(Sµ ρσ−xρTµ σ+xσTµ ρ)=0, (10.5.65) Tµ ρ≡∂Lmatter ∂(∂µφ)∂ρφ−δµ ρLmatter,Sµ ρσ≡−i∂Lmatter ∂(∂µφ)Sρσφ. (10.5.66) 10.5 W eyl’s Gauge Principle 349 The ten conservation laws, Eqs. (10.5.63) and (10.5.64), are the conservation laws of energy, momentum and angular momentum. Thus {Sµν}are the spin matrices of the matter ﬁeld φ(x). We shall also examine the transformations in terms of the variation δ0φ, which in this case is δ0φ=−iερ∂ρφ+1 2iερσ/parenleftbigg Sρσ+xρ1 i∂σ−xσ1 i∂ρ/parenrightbigg φ. (10.5.67) On comparing with Weyl’s gauge principle, the role of the realizations {tα}of the generators {Tα}upon the multiplet φis played by the differential operators, 1 i∂ρ,andSρσ+xρ1 i∂σ−xσ1 i∂ρ. Then, by the deﬁnition of the currents, we expect the currents corresponding to ερandερσto be given, respectively, by Jµ ρ≡∂Lmatter ∂(∂µφ)∂ρφ, andJµ ρσ≡Sµ ρσ−xρ1 iJµ σ+xσ1 iJµ ρ. (10.5.68) In terms of δ0, however, the invariance condition (10.5.60) is not simply δ0Lmatter≡0,a n d the additional term δxρ∂ρLmatter results in the appearance of the term ∂ρLmatter in the identities (10.5.63) and thus for the term δµ ρLmatter inTµ ρ. We shall now consider the local ten-parameter Poincaré transformation in which the ten arbitrary inﬁnitesimal constants, {εµ}and{εµν}, in Eq. (10.5.61) become the 10 arbitrary inﬁnitesimal functions, {εµ(x)}and{εµν(x)}. It is convenient to regard εµν(x)andξµ(x)≡iεµ ν(x)xν+εµ(x), as the 10 independent inﬁnitesimal functions. Such choice avoids the explicit appearance of xµ. Furthermore, we can always choose εµ(x)such that ξµ(x)=0 andεµν(x)/negationslash=0, so that the coordinate and ﬁeld transformations are completely separated. Based on this fact, we use Latin indices for εij(x)and Greek indices for ξµandxµ.T h e Latin indices, i, j, k,··, also assume the values 0, 1, 2 and 3. Then the transformations under consideration are δxµ=ξµ(x),andδφ(x)=1 2iεij(x)Sijφ(x), (10.5.69) or δ0φ(x)=−ξµ(x)∂µφ(x)+1 2iεij(x)Sijφ(x). (10.5.70) This notation emphasizes the similarity of the εij(x)transformations to the linear transforma- tions of Weyl/primes gauge principle. Actually, in Utiyama’s gauge principle, the εij(x)transfor- mations alone are considered in the local six-parameter Lorentz transformation. The ξµ(x) transformations correspond to the general coordinate transformation. 350 10 Calculus of V ariations: Applications According to the convention we have just employed, the differential operator ∂µmust have a Greek index. In the matter Lagrangian density Lmatter, we then have the two kinds of indices, and we shall regard Lmatter as a given function of φ(x)and˜∂kφ(x), satisfying the identities, (10.5.63) and (10.5.64). The original matter Lagrangian density Lmatter is obtained by setting ˜∂kφ(x)=δµ k∂µφ(x). The matter Lagrangian density Lmatter is not invariant under the local ten-parameter transfor- mations, (10.5.69) or (10.5.70), but we will later obtain an invariant expression by replacing ˜∂kφ(x)with a suitable covariant derivative Dkφ(x)in the matter Lagrangian density Lmatter. The transformation of ∂µφ(x)is given by δ∂µφ=1 2iεijSij∂µφ+1 2i(∂µεij)Sijφ−(∂µξν)(∂νφ), (10.5.71) and the original matter Lagrangian density Lmatter transforms according to δLmatter≡−(∂µξρ)Jµ ρ−1 2i(∂µεij)Sµ ij. We note that it is Jµ ρinstead of Tµ ρwhich appears here. The reason for this is that we have not included the extra term (∂µδxµ)Lmatter in Eq. (10.5.60). The left-hand side of Eq. (10.5.60) actually has the value δLmatter+(∂µδxµ)Lmatter≡−(∂µξρ)Tµ ρ−1 2i(∂µεij)Sµ ij. We shall now look for the modiﬁed matter Lagrangian density L/prime matter which makes the matter action functional Imatter[φ]invariant under (10.5.69) or (10.5.70). The extra term just mentioned is of a different kind in that it involves Lmatter and not ∂Lmatter/∂(˜∂kφ). In partic- ular, the extra term includes the contributions from terms in Lmatter which do not contain the derivatives. Thus it is clear that we cannot remove the extra term by replacing the derivative ˜∂µwith a suitable covariant derivative Dµ. For this reason, we shall consider the problem in two stages. First we eliminate the noninvariance arising from the fact that ∂µφ(x)is not a covariant quantity, and second, we obtain an expression L/prime matter satisfying δL/prime matter≡0. (10.5.72) Because the invariance condition (10.5.60) for the matter action functional Imatter requires the matter Lagrangian density L/prime matter to be an invariant scalar density rather than an invariant scalar, we shall make a further modiﬁcation, replacing L/prime matter withL/prime/prime matter, which satisﬁes δL/prime/prime matter+(∂µξµ)L/prime/prime matter≡0. (10.5.73) The ﬁrst part of this program can be accomplished by replacing ˜∂kφinLmatter with a covariant derivative Dkφwhich transforms according to δ(Dkφ)=1 2iεijSij(Dkφ)−iεi k(Diφ). (10.5.74) 10.5 W eyl’s Gauge Principle 351 The condition (10.5.72) follows from the identities, (10.5.63) and (10.5.64). To do this, it is necessary to introduce 40 new ﬁeld variables towards the end. We ﬁrst consider the εij transformations, and eliminate the ∂µεijterm in (10.5.71) by setting D|µφ≡∂µφ+1 2Aij µSijφ, (10.5.75) where Aij µwith Aij µ=−Aji µ are 24 new ﬁeld variables. We can then impose the condition δ(D|µφ)=1 2iεijSij(D|µφ)−(∂µξν)(D|νφ), (10.5.76) which determines the transformation properties of Aij µuniquely. They are δAij µ=−∂µεij+εi kAkjµ+εj kAik µ−(∂µξν)Aij ν. (10.5.77) The position of the last term in Eq. (10.5.71) is rather different. The term involving ∂µεij is inhomogeneous in the sense that it contains φrather than ∂µφ, but this is not true of the last term. Correspondingly, the transformation law for D|µφ, (10.5.76), is already homogeneous. This means that to force the covariant derivative Dkφto transform according to Eq. (10.5.74), we must add to D|µφnot a term in φ, but rather a term in D|µφitself. In other words, we merely multiply by a new ﬁeld, Dkφ≡eµ kD|µφ. (10.5.78) Here, the eµ kare 16 new ﬁeld variables with the transformation properties determined by Eq. (10.5.74) to be δeµ k=(∂νξµ)eν k−iεi keµ i. (10.5.79) We note that the ﬁelds eµ kandAij µare independent and unrelated at this stage, although they will be related by the Euler–Lagrange equations of motion. We ﬁnd the invariant matter Lagrangian density L/prime matter deﬁned by L/prime matter≡L matter(φ, D kφ), which is an invariant scalar. We can obtain the invariant matter Lagrangian density L/prime/prime matter which is an invariant scalar density by multiplying L/prime matter by a suitable function of the new ﬁeld variables, L/prime/prime matter≡E L/prime matter=ELmatter(φ, D kφ). 352 10 Calculus of V ariations: Applications The invariance condition (10.5.73) for L/prime/prime matter is satisﬁed if a factor Eitself is an invariant scalar density, δE+(∂µξµ)E≡0. The only function of the new ﬁeld variable eµ kwhich obeys this transformation law and does not involve the derivatives is E=[ d e t ( eµ k)]−1, (10.5.80) where the arbitrary constant factor has been chosen such that E=1 wheneµ kis set equal to δµ k. The ﬁnal form of the modiﬁed matter Lagrangian density L/prime/prime matter w h i c hi sa ni n v a r i a n t scalar density is given by L/prime/prime matter(φ, ∂µφ, eµ k,Aij µ)≡E L matter(φ, D kφ). (10.5.81) As in the case of Weyl’s gauge principle, we can deﬁne the modiﬁed current densities in terms of Lmatter(φ, D kφ)by Tk µ≡∂L/prime/prime matter ∂eµ k≡Eei µ/bracketleftbigg∂Lmatter ∂(Dkφ)Diφ−δk iLmatter/bracketrightbigg , (10.5.82) Sµ ij≡−2∂L/prime/prime matter ∂Aij µ≡iEeµ k∂Lmatter ∂(Dkφ)Sijφ, (10.5.83) where ei µis the inverse of eµ i, satisfying ei µeνi=δν µ,eiµeµ j=δi j. (10.5.84) In order to express the conservation laws of these currents in a simple form, we extend the deﬁnition of D|µφ. Originally, it was deﬁned for φ(x), and is to be deﬁned for any other quantity which is invariant under the ξµtransformations and transforms linearly under the εijtransformations. We extend D|µto any quantity which transforms linearly under the εij transformations by ignoring the ξµtransformations altogether. Thus we have D|νeµ i≡∂νeµ i−Ak iνeµ k, (10.5.85) according to the εijtransformation law of eµ i. We call this the εcovariant derivative. We calculate the commutator of the εcovariant derivatives as, [D|µ,D|ν]φ=1 2iRij µνSijφ, (10.5.86) where Ri jµνis deﬁned by the following equation, Ri jµν≡∂νAi jµ−∂µAi jν−Ai kµAk jν+Ai kνAk jµ. (10.5.87) 10.5 W eyl’s Gauge Principle 353 This quantity is covariant under the εijtransformations. Ri jµνis closely analogous to the ﬁeld strength tensor Fαµνof the non-Abelian gauge ﬁeld. Rij µνis antisymmetric in both pairs of indices. In terms of the εcovariant derivative, the ten conservation laws of the currents, (10.5.82) and (10.5.83), are expressed as D|µ(Tk νeµ k)+Tk µ(D|νeµ k)=Sµ ijRij µν, (10.5.88) D|µSµ ij=Tiµeµ j−Tjµeµi. (10.5.89) Now we examine our ultimate goal, the Lagrangian density LGof the “free” self- interacting gravitational ﬁeld. We examine the commutator of DkandDlacting on φ(x). After some algebra, we obtain [Dk,Dl]φ=1 2iRij klSijφ−Ci klDiφ, (10.5.90) where Rij kl≡eµ keν lRij µν,Ci kl≡(eµ keν l−eµ leν k)D|νei µ. (10.5.91) We note that the right-hand side of Eq. (10.5.90) is not simply proportional to φbut also involves Diφ. The Lagrangian density LGfor the “free” self-interacting gravitational ﬁeld must be an invariant scalar density. If we set LG=EL0, then L0must be an invariant scalar and a function only of the covariant quantities Rij klandCi kl. All the indices of these expressions are of the same kind, unlike the case of the non-Abelian gauge ﬁeld, so that we can take the contractions of the upper indices with the lower indices. The requirement that L0is an invariant scalar in two separate spaces is reduced to the requirement that it is an invariant scalar in one space. We have a linear invariant scalar whichhas no analogue in the case of the non-Abelian gauge ﬁeld, namely, R≡R ij ij. There exist a few quadratic invariants, but we choose the lowest order invariant. Thus we are led to the Lagrangian density LGfor the “free” self-interacting gravitational ﬁeld, LG=1 2κ2ER, (10.5.92) which is linear in the derivatives. In Eq. (10.5.92), κis Newton’s gravitational constant. So far, we have given neither any geometrical interpretation of the local ten-parameter Poincaré transformation, (10.5.69), nor any interpretation of the 40 new ﬁelds, eµ k(x)and Aij µ(x). We shall now establish the connection of the present theory with the standard metric theory of the gravitational ﬁeld. Under the ξµtransformation which is a general coordinate transformation, eµ k(x)trans- forms as a contravariant vector, while ek µ(x)andAij µ(x)transform as covariant vectors. Then the quantity gµν(x)≡ek µ(x)ekν(x) (10.5.93) 354 10 Calculus of V ariations: Applications is a symmetric covariant tensor, and therefore may be interpreted as the metric tensor of a Riemannian space. It remains invariant under the εijtransformations. We shall abandon the convention that all the indices are to be lowered or raised by the ﬂat-space metric ηµν,a n dw e usegµν(x)instead as the metric tensor. We can easily show that E=/radicalbig −g(x)with g(x)≡det/parenleftbig gµν(x)/parenrightbig . (10.5.94) From Eq. (10.5.93), we realize that eµ k(x)andek µ(x)are the contravariant and covariant components of a tetrad system in Riemannian space. The εijtransformations are the tetrad rotations. The Greek indices are the world tensor indices and the Latin indices are the local tensor indices of this system. The original generic matter ﬁeld φ(x)may be decomposed into local tensors and local spinors. From the local tensors, we can form the corresponding worldtensors by multiplying by e µ k(x)orek µ(x). For example, from a local vector vi(x), we can form the world vector as vµ(x)=eµ i(x)vi(x),andvµ(x)=ei µ(x)vi(x). (10.5.95) We note that vµ(x)=gµν(x)vν(x), so that Eq. (10.5.95) is consistent with the deﬁnition of the metric gµν(x), Eq. (10.5.93). The ﬁeld Ai jµ(x)is regarded as a local afﬁne connection with respect to the tetrad system since it speciﬁes the covariant derivatives of local tensors or local spinors. For a local vector, we have   D|νvi=∂νvi+Ai jνvj, D|νvj=∂νvj−Ai jνvi.(10.5.96) We notice that the relationship between D|µφandDkφ, (10.5.78), could be written simply as Dµφ=D|µφ, (10.5.97) according to the convention (10.5.95). We shall, however, make a distinction between Dµand D|µfor a later purpose. We deﬁne the covariant derivative of a world tensor in terms of the covariant derivative of the associated local tensor. Thus, we have /braceleftBigg Dνvλ≡eλ iD|νvi=∂νvλ+Γλ µνvµ, Dνvµ≡ei µD|νvi=∂νvµ−Γλ µνvλ,(10.5.98) Γλ µν≡eλ iD|νei µ≡−ei µD|νeλ i. (10.5.99) 10.5 W eyl’s Gauge Principle 355 We note that this deﬁnition of Γλ µνis equivalent to the requirement that the covariant derivative of the tetrad components vanish, /braceleftBigg Dνeλ i≡0, Dνei µ≡0.(10.5.100) For a generic quantity α, transforming according to δα=1 2iεijSijα+(∂µξλ)Σµ λα, (10.5.101) the covariant derivative of αis deﬁned by Dνα≡∂να+1 2iAij νSijα+Γλ µνΣµ λα. (10.5.102) Theεcovariant derivative of α, deﬁned by Eq. (10.5.85), is obtained by simply dropping the last term in (10.5.102). We calculate the commutator of the covariant derivative of αwith the result, [Dµ,Dν]α=1 2iRij µνSijα+Rρ σµνΣσ ρα−Cλ µνDλα, where Rρ σµνandCλ µνare deﬁned in terms of Ri jµνandCi klin the usual way. These quantities are the world tensors and can be expressed in terms of Γλ µνin the form, Rρ σµν=∂νΓρ σµ−∂µΓρ σν−Γρ λµΓλ σν+Γρ λνΓλ σµ,Cλ µν=Γλ µν−Γλ νµ.(10.5.103) We see that Rρ σµνis the Riemann tensor formed from the afﬁne connection Γλ µν.F r o m Eq. (10.5.100), we have Dρgµν(x)≡0. It is consistent to interpret Γλ µνas an afﬁne connection in a Riemannian space. The def- inition of Γλ µν, Eq. (10.5.99), does not guarantee that it is symmetric so that it is not the Christoffel symbol in general. In the absence of the matter ﬁeld, however, Γλ µνis symmetric so that it is the Christoffel symbol. The curvature scalar has the usual form, R≡Rµ µ,w h e r e Rµν≡Rλ µλν. The Lagrangian density LGfor the “free” self-interacting gravitational ﬁeld, Eq. (10.5.92), is the usual one, LG/parenleftbig gµν(x),Γλ µν(x)/parenrightbig =1 2κ2√ −ggµν(∂νΓλ µλ−∂λΓλ µν+Γρ µλΓλ νρ−Γλ µνΓρ λρ).(10.5.104) 356 10 Calculus of V ariations: Applications Unity of All Forces: Electro-weak uniﬁcation of Glashow–Weinberg–Salam is based on the gauge group SU(2)weak isospin ×U(1)weak hypercharge . It suffers from the problem of the nonrenormalizability due to the triangular anomaly in the lepton sector. In the early 1970s, it was discovered that non-Abelian gauge ﬁeld theory is asymptotically free at short distance, i.e., it behaves as a free ﬁeld at short distances. Thus the relativistic quantum ﬁeld theory of the strong interaction based on the gauge group SU(3)color is invented and is called quantum chromodynamics. The standard model with the gauge group SU(3)color×SU(2)weak isospin ×U(1)weak hypercharge which describes the weak interaction, the electromagnetic interaction and the strong inter- action, is free from the triangular anomaly. It suffers, however, from a serious defect; theexistence of the classical instanton solution to the ﬁeld equation in the Euclidean metric for the SU(2)gauge ﬁeld theory. In the SU(2)gauge ﬁeld theory, we have the Belavin– Polyakov–Schwartz–Tyupkin instanton solution which is a classical solution to the ﬁeld equa-tion in the Euclidean metric. A proper account for the instanton solution requires the ad-dition of the strong CP-violating term to the QCD Lagrangian density in the path integral formalism. The Peccei–Quinn axion and the invisible axion scenario resolve this strong CP- violation problem. In the grand uniﬁed theories, we assume that the subgroup of the grandunifying gauge group is the gauge group SU(3) color×SU(2)weak isospin ×U(1)weak hypercharge . We now attempt to unify the weak interaction, the electromagnetic interaction and the strong interaction by starting from the much larger gauge group Gwhich is reduced to SU(3)color×SU(2)weak isospin ×U(1)weak hypercharge and further down to SU(3)color×U(1)E.M. as a result of the requisite sequences of the spontaneous symmetry breaking, G⊃SU(3)color×SU(2)weak isospin ×U(1)weak hypercharge ⊃SU(3)color×U(1)E.M.. By now, we are almost certain that the true underlining theory of particle interactions, includ- ing gravitational interaction, is superstring theory. Actually, phenomenological predictionshave followed from superstring theory. 10.6 Problems for Chapter 10 10.1. (Due to H. C.) Find the solution or solutions q(t)which extremize I≡/integraldisplayT 0/bracketleftbiggm 2˙q2−1 6q6/bracketrightbigg dt, subject to q(0) = q(T)=0. 10.6 Problems for Chapter 10 357 10.2. (Due to H. C.) Find the solution q(t)which extremizes I≡/integraldisplayT 0/bracketleftbiggm 2˙q2−λ 3q3/bracketrightbigg dt, subject to q(0) = q(T)=0. 10.3. The action for a particle in a gravitational ﬁeld is given by I≡−m/integraldisplay/radicalbigg gµνdxµ dtdxν dtdt, where gµνis the metric tensor. Show that the motion of this particle is governed by d2xρ ds2=−Γρ µνdxµ dsdxν ds, with Γρ µν≡1 2gρσ(∂µgσν+∂νgσµ−∂σgµν), which is called the Christoffel symbol . 10.4. (Due to H. C.) The space–time structure in the presence of a black hole is given by (ds)2=/parenleftbigg 1−1 r/parenrightbigg (dt)2−(dr)2 /parenleftbig 1−1 r/parenrightbig−r2[sin2θ(dφ)2+(dθ)2], and the motion of a particle is such that/integraltext dsis minimized. Let the particle move in thex-yplane and hence θ=π 2. Then the equations of motion of this particle subject to the gravitational pull of the black hole are obtained by extremizing s≡/integraldisplaytf ti/radicaltp/radicalvertex/radicalvertex/radicalbt /parenleftbigg 1−1 r/parenrightbigg −/parenleftbigdr dt/parenrightbig2 /parenleftbig 1−1 r/parenrightbig−r2/parenleftbiggdφ dt/parenrightbigg2 dt, with initial and ﬁnal coordinates ﬁxed. a) Derive the equation of motion obtained by varying φ. Integrate this equation once to obtain an equation with one integration constant. b) Derive the equation of motion obtained by varying r. Find a way to obtain an equation involvingdr dtanddφ dtand a second integration constant. c) Let the particle be at r=2,φ=0, withdr dt=dφ dt=0 at the initial time t=0. Determine the motion of this particle as best you can. How long does it take for this particle to reach r=1? 358 10 Calculus of V ariations: Applications 10.5. (Due to H. C.) The invariant distance dsin the neigborhood of a black hole is given by (ds)2=/parenleftbigg 1−2M r/parenrightbigg (dt)2−(dr)2 /parenleftbig 1−2M r/parenrightbig, where r,θ,φ(we set θ=φ=constant) are the spherical polar coordinates and Mis the mass of the black hole. The motion of a particle extremizes the invariant distance. a) Write down the integral which should be extremized. From the expression of this integral, ﬁnd a ﬁrst-order equation satisﬁed by r(t). b) If(r−2M)is small and positive, solve the ﬁrst-order equation. How much time does it take for the particle to fall to the critical distance r=2M? 10.6. (Due to H. C.) Find the kink solution by extremizing I≡/integraldisplay+∞ −∞dt/parenleftbigg1 2φ2 t−m2 2φ2+λ 4φ4+1 4m4 λ/parenrightbigg . 10.7. Extremize I≡/integraldisplayx0 0dx/integraldisplay dΩ˜f(x, θ)/bracketleftbigg cosθ∂f(x, θ) ∂x+f(x, θ) −κ 4π/integraldisplay w(/vectorn−/vectorn0)f(x, θ0)dΩ0/bracketrightbigg , treating fand˜fas independent. Here κis a constant, the unit vectors, /vectornand/vectorn0, are pointing in the direction speciﬁed by spherical angles, (θ,ϕ)and(θ0,ϕ0),a n d dΩ0is the differential solid angle at /vectorn0. Obtain the steady-state transport equation for anisotropic scattering from the very heavy scatterers, cosθ∂f(x, θ) ∂x=−f(x, θ)+κ 4π/integraldisplay w(/vectorn−/vectorn0)f(x, θ0)dΩ0, −cosθ∂˜f(x, θ) ∂x=−˜f(x, θ)+κ 4π/integraldisplay w(/vectorn0−/vectorn)˜f(x, θ0)dΩ0. Interpret the result for ˜f(x, θ). 10.8. Extremize I≡/integraldisplay dt d3/vectorxL/parenleftbigg ψ,∂ ∂tψ,/vector∇ψ,ϕ,∂ ∂tϕ,/vector∇ϕ/parenrightbigg , where L=−/vector∇ϕ/vector∇ψ−a2 2/parenleftbigg ϕ∂ ∂tψ−ψ∂ ∂tϕ/parenrightbigg , 10.6 Problems for Chapter 10 359 treating ψandϕas independent. Obtain the diffusion equation, /vector∇2ψ(t,/vectorx)=a2∂ψ(t,/vectorx) ∂t, /vector∇2ϕ(t,/vectorx)=−a2∂ϕ(t,/vectorx) ∂t. Interpret the result for ϕ. 10.9. Extremize I≡/integraldisplay dt d3/vectorx/braceleftbigg −1 2m/parenleftbigg/parenleftbigg/planckover2pi1 i/vector∇−e c/vectorA/parenrightbigg ψ/parenrightbigg∗/parenleftbigg/parenleftbigg/planckover2pi1 i/vector∇−e c/vectorA/parenrightbigg ψ/parenrightbigg +1 2/bracketleftbigg ψ∗/parenleftbigg i/planckover2pi1∂ ∂t−eφ/parenrightbigg ψ+/parenleftbigg/parenleftbigg i/planckover2pi1∂ ∂t−eφ/parenrightbigg ψ/parenrightbigg∗ ψ/bracketrightbigg −ψ∗Vψ/bracerightbigg , /vector∇/vectorA+1 c∂φ ∂t=0, treating ψandψ∗as independent. Obtain the Schrödinger equation, /parenleftbigg i/planckover2pi1∂ ∂t−eφ/parenrightbigg ψ=1 2m/parenleftbigg/planckover2pi1 i/vector∇−e c/vectorA/parenrightbigg2 ψ+Vψ , −/parenleftbigg i/planckover2pi1∂ ∂t+eφ/parenrightbigg ψ∗=1 2m/parenleftbigg/planckover2pi1 i/vector∇+e c/vectorA/parenrightbigg2 ψ∗+Vψ∗. Demonstrate that the Schrödinger equation is invariant under the gauge transformation, /vectorA→/vectorA/prime= /vectorA+/vector∇Λ, φ→φ/prime=φ−/parenleftbigg1 c/parenrightbigg/parenleftbiggΛ ∂t/parenrightbigg , ψ→ψ/prime=e x p/bracketleftbigg/parenleftbiggie /planckover2pi1c/parenrightbigg Λ/bracketrightbigg ψ,where/parenleftbigg /vector∇2−1 c2∂2 ∂t2/parenrightbigg Λ=0 . 10.10. Extremize I≡/integraldisplay dt d3/vectorx/braceleftBigg −/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg1 i/vector∇−e/vectorA/parenrightbigg ψ/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg i∂ ∂t−eφ/parenrightbigg ψ/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 −m2|ψ|2/bracerightBigg , /vector∇/vectorA+∂φ ∂t=0, treating ψandψ∗as independent. Obtain the Klein–Gordon equation, /parenleftbigg i∂ ∂t−eφ/parenrightbigg2 ψ−/parenleftbigg1 i/vector∇−e/vectorA/parenrightbigg2 ψ=m2ψ, /parenleftbigg i∂ ∂t+eφ/parenrightbigg2 ψ∗−/parenleftbigg1 i/vector∇+e/vectorA/parenrightbigg2 ψ∗=m2ψ∗. 360 10 Calculus of V ariations: Applications 10.11. Extremize I≡/integraldisplay d4xLtot, where Ltotis given by Ltot=1 4[¯ψα(x),Dαβ(x)ψβ(x)] +1 4[DT βα(−x)¯ψα(x),ψβ(x)] +1 2φ(x)K(x)φ(x)+Lint(φ(x),ψ(x),¯ψ(x)), withDαβ(x),DT βα(−x)andK(x)given by Dαβ(x)=(iγµ∂µ−m+iε)αβ, DT βα(−x)=(−iγT µ∂µ−m+iε)βα, K(x)=−∂2−κ2+iε, andLintis given by the Yukawa coupling speciﬁed by Lint(φ(x),ψ(x),¯ψ(x)) =−G0¯ψα(x)γαβ(x)ψβ(x)φ(x). Theγµ/primes are the Dirac γmatrices with the property speciﬁed by {γµ,γν}=2ηµν, (γµ)†=γ0γµγ0. Theψ(x)is the four-component Dirac spinor and the ¯ψ(x)is the Dirac adjoint of ψ(x) deﬁned by ¯ψ(x)≡ψ†(x)γ0. Obtain the Euler–Lagrange equations of motion for the ψﬁeld, the ¯ψﬁeld and the φ ﬁeld. 10.12. Extremize the action functional for the electromagnetic ﬁeld Aµ, I=/integraldisplay d4x/parenleftbigg −1 4FµνFµν+B∂µAµ+1 2αB2/parenrightbigg , Fµν≡∂µAν−∂νAµ. Obtain the Euler–Lagrange equations of motion for the Aµﬁeld and the Bﬁeld. Can you perform the q-number gauge transformation after canonical quantization? 10.13. Extremize the action functional for the neutral massive vector ﬁeld Uµ, I=/integraldisplay d4x/parenleftbigg −1 4FµνFµν+1 2m2 0UµUµ/parenrightbigg , Fµν≡∂µUν−∂νUµ. Obtain the Euler–Lagrange equation of motion for the Uµﬁeld. Examine the massless limitm0→0after canonical quantization. 10.6 Problems for Chapter 10 361 10.14. Extremize the action functional for the neutral massive vector ﬁeld Aµ, I=/integraldisplay d4x/parenleftbigg −1 4FµνFµν+1 2m2 0AµAµ+B∂µAµ+1 2αB2/parenrightbigg , Fµν≡∂µAν−∂νAµ. Obtain the Euler–Lagrange equations of motion for the Aµﬁeld and the Bﬁeld. Ex- amine the massless limit m0→0after canonical quantization. Hint for Problems 10.12, 10.13 and 10.14: Lautrup, B.: Mat. Fys. Medd. Dan. Vid. Selsk. 35.No.11. 29. (1967). Nakanishi, N.: Prog. Theor. Phys. Suppl. 51. 1. (1972). Yokoyama, K.: Prog. Theor. Phys. 51. 1956. (1974), 52. 1669. (1974). 10.15. Derive the Schwinger–Dyson equation for the interacting scalar ﬁelds ˆφi(x)(i=1,2) whose Lagrangian density is given by L/parenleftbigˆφ1(x),ˆφ2(x),∂µˆφ1(x),∂µˆφ2(x)/parenrightbig =2/summationdisplay i=1/braceleftbigg1 2∂µˆφi(x)∂µˆφi(x)−1 2m2 iˆφ2 i(x)/bracerightbigg −gˆφ2 1(x)ˆφ2(x). Hint: Introduce the proper self-energy parts Π∗ i(x, y)(i=1,2) and the vertex opera- torΛ(x, y, z), and follow the discussion in Section 10.3. 10.16. Derive the Schwinger–Dyson equation for the self-interacting scalar ﬁeld ˆφ(x)whose Lagrangian density is given by L(ˆφ(x),∂µˆφ(x)) =1 2∂µˆφ(x)∂µˆφ(x)−1 2m2ˆφ2(x)−λ4 4!ˆφ4(x). Hint: Introduce the proper self-energy part Π∗(x, y)and the vertex operator Λ4(x, y, z, w ), and follow the discussion in Section 10.3. 10.17. Derive the Schwinger–Dyson equation for the self-interacting scalar ﬁeld ˆφ(x)whose Lagrangian density is given by L(ˆφ(x),∂µˆφ(x)) =1 2∂µˆφ(x)∂µˆφ(x)−1 2m2ˆφ2(x)−λ3 3!ˆφ3(x)−λ4 4!ˆφ4(x). Hint: Introduce the proper self-energy part Π∗(x, y)and the vertex operators Λ3(x, y, z)andΛ4(x, y, z, w ), and follow the discussion in Section 10.3. 362 10 Calculus of V ariations: Applications 10.18. Consider the bound state problem for a system of two distinguishable spinless bosons of equal mass m, exchanging a spinless and massless boson whose Lagrangian density is given by L=2/summationdisplay i=1/braceleftbigg1 2∂µˆφi(x)∂µˆφi(x)−1 2m2ˆφ2 i(x)/bracerightbigg +1 2∂µˆφ(x)∂µˆφ(x)−gˆφ† 1(x)ˆφ1(x)ˆφ(x)−gˆφ† 2(x)ˆφ2(x)ˆφ(x). a) Show that the Bethe–Salpeter equation for the bound state of the two bosons ˆφ1(x1) andˆφ2(x2)is given by S/prime F(x1,x2;B)=/integraldisplay d4x3d4x4∆F(x1−x3)∆F(x2−x4) ×(−g2)DF(x3−x4)S/prime F(x3,x4;B), where ∆F(x)andDF(x)are given by ∆F(x)=/integraldisplayd4k (2π)4exp[ikx] k2−m2+iε,andDF(x)=/integraldisplayd4k (2π)4exp[ikx] k2+iε. b) Transform the coordinates x1andx2to the center-of-mass coordinate Xand the relative coordinate xby X=1 2(x1+x2),andx=x1−x2, and correspondingly to the center-of-mass momentum Pand the relative momen- tump, P=p1+p2,andp=1 2(p1−p2). Deﬁne the Fourier transform Ψ(p)ofS/prime F(x1,x2;B)by S/prime F(x1,x2;B)=e x p[ −iPX]/integraldisplay d4pexp[−ipx]Ψ(p). Show that the above Bethe–Salpeter equation in momentum space assumes the following form, /bracketleftBigg/parenleftbiggP 2+p/parenrightbigg2 −m2/bracketrightBigg/bracketleftBigg/parenleftbiggP 2−p/parenrightbigg2 −m2/bracketrightBigg Ψ(p)=ig2/integraldisplayd4q (2π)4Ψ(q) (p−q)2+iε. c) Assuming that Ψ(p)can be expressed as Ψ(p)=−/integraldisplay1 −1g(z)dz [p2+zpP−m2+(P2/4) +iε]3, 10.6 Problems for Chapter 10 363 substitute this expression into the Bethe–Salpeter equation in momentum space. Carrying out the qintegration using the formula, /integraldisplay d4q1 (p−q)2+iε·1 [q2+zqP−m2+(P2/4) +iε]3 =iπ2 2[−m2+(P2/4)−z2(P2/4)]·1 [p2+zpP−m2+(P2/4) +iε], and comparing the result with the original expression for Ψ(p), obtain the integral equation for g(z)as g(z)=/integraldisplay1 0ςd ς/integraldisplay1 −1dy/integraldisplay1 −1dxλg(x) 2(1−η2+η2x2)δ/parenleftbig z−{ςy+( 1−ς)x}/parenrightbig , where the dimensionless coupling constant λis given by λ=/parenleftBigg 4πm/parenrightBig2 , and the squared mass of the bound state is given by M2=P2=4m2η2,0<η< 1. d) Carrying out the ςintegration, obtain the integral equation for g(z)as g(z)=λ/integraldisplay1 zdx1+z 1+xg(x) 2(1−η2+η2x2)+λ/integraldisplayz −1dx1−z 1−xg(x) 2(1−η2+η2x2). e) Observe that g(z)satisﬁes the boundary conditions, g(±1) = 0 . Differentiate the integral equation for g(z)obtained in d) twice, and reduce it to a second-order ordinary differential equation for g(z)of the form, d2 dz2g(z)=−λ 1−z2g(z) 1−η2+η2z2. This is the eigenvalue problem. f) Solve the above eigenvalue problem for g(z)in the limit, 1/greatermuch1−η>0,a n ds h o w that the lowest approximate eigenvalue is given by λ≈2 π/radicalbig 1−η2. 364 10 Calculus of V ariations: Applications Hint for Problem 10.18: The Wick–Cutkosky model is discussed in the following articles. Wick, G.C.: Phys. Rev. 96., 1124, (1954). Cutkosky, R.E.: Phys. Rev. 96., 1135, (1954). 10.19. Consider the bound state problem of zero total momentum /vectorP=0 for a system of identical two fermions of mass m, exchanging a spinless and massless boson whose Lagrangian density is given by L=/hatwide¯ψ(x)(iγµ∂µ−m+iε)ˆψ(x)+1 2∂µˆφ(x)∂µˆφ(x)−g/hatwide¯ψ(x)ˆψ(x)ˆφ(x). Deﬁne the bound state wave function of the two fermions by [U/vectorP(x)]αβ=<0|T[ˆψα(x 2)ˆψβ(−x 2)]|B>, [u/vectorP(p)]αβ=/integraldisplay d4xexp[ipx][U/vectorP(x)]αβ, [u/vectorP=0(p)]αβ=δαβχ(p) p2−m2+iε. Show that the Bethe–Salpeter equation for the bound state to the ﬁrst-order approxi- mation is given by χ(p)=ig2/integraldisplayd4q (2π)4/bracketleftbigg1 (p−q)2+iε−1 (p+q)2+iε/bracketrightbiggχ(q) q2−m2+iε. Solve this eigenvalue problem by dropping the antisymmetrizing term in the kernel of the above. The antisymmetrizing term originates from the spin-statistics relation forthe fermions. Hint for Problem 10.19: This problem is discussed in the following article. Goldstein, J.: Phys. Rev. 91., 1516, (1953). Bibliography Local Analysis and Global Analysis We cite the following book for the local analysis and global analysis of ordinary differ- ential equations. [1] Bender, Carl M., and Orszag, Steven A.: “ Advanced Mathematical Methods F or Scien- tists And Engineers: Asymptotic Methods and Perturbation Theory ”, Springer-V erlag, New Y ork, (1999). Integral EquationsWe cite the following book for the theory of Green’s functions and boundary value prob- lems. [2] Stakgold, I.: “ Green’s Functions and Boundary V alue Problems ”, John Wiley & Sons, New Y ork, (1979). We cite the following books for general discussions of the theory of integral equations. [3] Tricomi, F.G.: “ Integral Equations ”, Dover, New Y ork, (1985). [4] Pipkin, A.C.: “ A Course on Integral Equations ”, Springer-V erlag, New Y ork, (1991). [5] Bach, M.: “ Analysis, Numerics and Applications of Differential and Integral Equations ”, Addison Wesley, Reading, Massachusetts, (1996). [6] Wazwaz, A.M.: “ A First Course in Integral Equations ”, World Scientiﬁc, Singapore, (1997). [7] Polianin, A.D.: “ Handbook of Integral Equations ”, CRC Press, Florida, (1998). [8] Jerri, A.J.: “ Introduction to Integral Equations with Applications ”, 2 ndedition, John Wiley & Sons, New Y ork, (1999). We cite the following books for applications of integral equations to the scattering prob- lem in nonrelativistic quantum mechanics, namely, the Lippmann–Schwinger equation. [9] Goldberger, M.L., and Watson, K.M.: “ Collision Theory ”, John Wiley & Sons, New Y ork, (1964). Chapter 5. [10] Sakurai, J.J.; “ Modern Quantum Mechanics ”, Addison-Wesley, 1994, Massachusetts. Chapter 7. [11] Nishijima, K.: “ Relativistic Quantum Mechanics ”, Baifuukan, 1973, Tokyo. Section 4– 11 of Chapter 4. (In Japanese.) Applied Mathematics in Theoretical Physics. Michio Masujima Copyright © 2005 Wiley-VCH V erlag GmbH & Co. KGaA, WeinheimISBN: 3-527-40534-8 366 Bibliography We cite the following book for applications of integral equations to the theory of elastic- ity. [12] Mikhlin, S.G. et al.: “ The Integral Equations of the Theory of Elasticity ”, Teubner, Stuttgart, (1995). We cite the following book for the application of integral equations to microwave engi- neering. [13] Collin, R.E.: “ Field Theory of Guided Waves ”, Oxford Univ. Press, (1996). We cite the following article for the application of integral equations to chemical engi- neering. [14] Bazant, M.Z., and Trout, B.L.: Physica, A300 , 139, (2001). We cite the following book for physical details of the dispersion relations in classical electrodynamics. [15] Jackson, J.D.: “ Classical Electrodynamics ”, 3rdedition, John Wiley & Sons, New Y ork, (1999). Section 7.10. p. 333. We cite the following books for applications of Cauchy-type integral equations to disper- sion relations in the potential scattering problem in nonrelativistic quantum mechanics. [16] Goldberger, M.L., and Watson, K.M.: “ Collision Theory ”, John Wiley & Sons, New Y ork, (1964). Chapter 10 and Appendix G.2. [17] De Alfaro, V ., and Regge, T.: “ Potential Scattering ”, North-Holland, Amsterdam, (1965).We note that Appendix G.2 of the book cited above, [9], discusses Cauchy-type integralequations in the scattering problem in nonrelativistic quantum mechanics in terms of the inhomogeneous Hilbert problems with the complete solution. We cite the following article for the integro-differential equation arising from Bose- Einstein condensation in an external potential at zero temperature. [18] Wu, T.T.: Phys. Rev. A58 ., 1465, (1998). We cite the following book for applications of the Wiener–Hopf method in partial differ- ential equations. [19] Noble, B.: “ Methods Based on the Wiener–Hopf Technique for the Solution of Partial Differential Equations ”, Pergamon Press, New Y ork, (1959). We note that the Wiener–Hopf integral equations originated from research on theradiative equilibrium on the surface of the star. We cite the following articles for the discussion of Wiener–Hopf integral equations and Wiener–Hopf sum equations. [20] Wiener, N., and Hopf, E.: S. B. Preuss. Akad. Wiss. 696, (1931). [21] Hopf, E.: “ Mathematical Problems of Radiative Equilibrium ”, Cambridge, New Y ork, (1934). [22] Krein, M.G.: “ Integral equation on a half-line with the kernel depending upon the dif- ference of the arguments ”, Amer. Math. Soc. Transl. (2), 22, 163, (1962). Bibliography 367 [23] Gohberg, I.C., and Krein, M.G.: “ Systems of integral equations on the half-line with kernels depending on the difference of the arguments ”, Amer. Math. Soc. Transl. (2), 14, 217, (1960). We cite the following article for discussion of the iterative solution for a single Wiener– Hopf integral equation and for a system of coupled Wiener–Hopf integral equations. [24] Wu, T.T. and Wu, T.T.: Quarterly Journal of Applied Mathematics, XX , 341, (1963). We cite the following book for the application of Wiener–Hopf methods to radiation from rectangular waveguides and circular waveguides. [25] Weinstein, L.A.: “ The theory of diffraction and the factorization method ”, Golem Press, (1969). pp. 66-88, and pp. 120-156. We cite the following article and books for application of the Wiener–Hopf method to elastodynamics of the crack motion. [26] Freund, L.B.: J. Mech. Phys. Solids, 20, 129, 141, (1972). [27] Freund, L.B.: “ Dynamic Fracture Mechanics ”, Cambridge Univ. Press, New Y ork, (1990). [28] Broberg, K.B.: “ Cracks and Fracture ”, Academic Press, New Y ork, (1999). We cite the following article and the following book for application of the Wiener–Hopf sum equation to the phase transition of the two-dimensional Ising model. [29] Wu, T.T.: Phys. Rev. 149 ., 380, (1966). [30] McCoy, B., and Wu, T.T.: “ The Two-Dimensional Ising Model ”, Harvard Univ. Press, Cambridge, Massachusetts, (1971). Chapter IX. We note that Chapter IX of the book cited above describes practical methods to solve the Wiener–Hopf sum equation with full mathematical details, including discussions of Pollard’s theorem which is the generalization of Cauchy’s theorem, and the two special cases of the Wiener–Lévy theorem. We cite the following articles for application of the Wiener–Hopf sum equation to Y agi– Uda semi-inﬁnite arrays. [31] Wasylkiwskyj, W.: IEEE Transactions Antennas Propagat., AP-21 , 277, (1973). [32] Wasylkiwskyj, W., and V anKoughnett, A.L.: IEEE Transactions Antennas Propagat., AP-24 , 633, (1974). [33] V anKoughnett, A.L.: Canadian Journal of Physics, 48, 659, (1970). We cite the following book for the historical development of the theory of integral equa- tions, the formal theory of integral equations and a variety of applications of the theory of integral equations to scientiﬁc and engineering problems. [34] Kondo, J.: “ Integral Equations ”, Kodansha Ltd., Tokyo, (1991). We cite the following book for the pure-mathematically oriented reader. [35] Kress, R.: “ Linear Integral Equations ”, 2ndedition, Springer-V erlag, Heidelberg, (1999). 368 Bibliography Calculus of Variations . We cite the following books for an introduction to the calculus of variations. [36] Courant, R., and Hilbert, D.: “ Methods of Mathematical Physics ”, (V ols. 1 and 2.), John Wiley & Sons, New Y ork, (1966). V ol.1, Chapter 4. Reprinted in Wiley Classic Edition, (1989). [37] Akhiezer, N.I.: “ The Calculus of V ariations ”, Blaisdell, Waltham, Massachusetts, (1962). [38] Gelfand, I.M., and Fomin, S.V .: “ Calculus of V ariations ”, Prentice-Hall, Englewood Cliffs, New Jersey, (1963). [39] Mathews, J., and Walker, R.L.: “ Mathematical Methods of Physics ”, Benjamin, Reading, Massachusetts, (1970). Chapter 12. We cite the following book for the variational principle in classical mechanics, the canon- ical transformation theory, the Hamilton–Jacobi equation and the semi-classical approx-imation to nonrelativistic quantum mechanics. [40] Fetter, A.L., and Walecka, J.D.: “ Theoretical Mechanics of Particles and Continua ”, McGraw-Hill, New Y ork, (1980). Chapter 6, Sections 34 and 35. We cite the following books for applications of the variational principle to nonrelativistic quantum mechanics and quantum statistical mechanics. [41] Feynman, R.P ., and Hibbs, A.R.: “ Quantum Mechanics and Path Integrals ”, McGraw- Hill, New Y ork, (1965). Chapters 10 and 11. [42] Feynman, R.P .: “ Statistical Mechanics ”, Benjamin, Reading, Massachusetts, (1972). Chapter 8. [43] Landau, L.D., and Lifshitz, E.M.: “ Quantum Mechanics ”, 3 rdedition, Pergamon Press, New Y ork, (1977). Chapter III, Section 20. [44] Huang, K.: “ Statistical Mechanics ”, 2ndedition, John Wiley & Sons, New Y ork, (1983). Section 10.4. The variational principle employed by R.P . Feynman and the variational principle employed by K. Huang are both based on Jensen’s inequality for the convex function. We cite the following book as a general reference for the theory of the gravitational ﬁeld. [45] Weinberg, S.: “ Gravitation and Cosmology. Principles and Applications of The General Theory of Relativity ”, John Wiley & Sons, 1972, New Y ork. We cite the following book for the genesis of Weyl’s gauge principle and the earlier attempt to unify the electromagnetic force and the gravitational force before the birth of quantum mechanics in the context of classical ﬁeld theory. [46] Weyl, H.: “ Space–Time–Matter ”, Dover Publications, Inc., 1950, New Y ork. Although H. Weyl failed to accomplish his goal of the uniﬁcation of the electromagnetic force and the gravitational force in the context of classical ﬁeld theory, his enthusiasm forthe uniﬁcation of all forces in nature survived, even after the birth of quantum mechanics. Bibliography 369 We cite the following articles and book for Weyl’s gauge principle, after the birth of quantum mechanics, for the Abelian electromagnetic gauge group. [47] Weyl, H.; Proc. Nat. Acad. Sci. 15, (1929), 323. [48] Weyl, H.; Z. Physik. 56, (1929), 330. [49] Weyl, H.; “ Theory of Groups and Quantum Mechanics ”, Leipzig, 1928, Zurich; reprinted by Dover, 1950. Chapter 2, section 12, and Chapter 4, section 5. We cite the following article as the ﬁrst attempt to unify the strong and the weak forces in nuclear physics in the context of quantum ﬁeld theory, without invoking Weyl’s gaugeprinciple. [50] Y ukawa, H.: Proc. Phys. Math. Soc. (Japan), 17, 48, (1935). We cite the following articles for the uniﬁcation of the electromagnetic force and the weak force by invoking Weyl’s gauge principle and the Higgs–Kibble mechanism andthe proposal of the standard model which uniﬁes the weak force, the electromagnetic force, and the strong force with quarks, leptons, the Higgs scalar ﬁeld, the Abelian gauge ﬁeld and the non-Abelian gauge ﬁeld, in the context of quantum ﬁeld theory. [51] Weinberg, S.: Phys. Rev. Lett. 19, 1264, (1967); Phys. Rev. Lett. 27, 1688, (1971); Phys. Rev. D5, 1962, (1971); Phys. Rev. D7, 1068, (1973); Phys. Rev. D7, 2887, (1973); Phys. Rev. D8, 4482, (1973); Phys. Rev. Lett. 31, 494, (1973); Phys. Rev. D9, 3357, (1974). We cite the following book for discussion of the standard model which uniﬁes the weak force, the electromagnetic force and the strong force with quarks, leptons, the Higgs scalar ﬁeld, the Abelian gauge ﬁeld and the non-Abelian gauge ﬁeld by invoking Weyl’s gauge principle and the Higgs–Kibble mechanism in the context of quantum ﬁeld theory. [52] Huang, K.: “ Quarks, Leptons, and Gauge Field ”, 2 ndedition, World Scientiﬁc, Singa- pore, (1992). We cite the following article for the O(3) model. [53] Georgi, H. and Glashow, S.L.: Phys. Rev. Lett. 28, 1494, (1972). We cite the following articles for the instanton, the strong CP violation, the Peccei–Quinn axion hypothesis and the invisible axion scenario. [54] Belavin, A.A., Polyakov, A.M., Schwartz, A.S., and Tyupkin, Y .S.; Phys. Letters. 59B , (1975), 85. [55] Peccei, R.D., and Quinn, H.R.; Phys. Rev. Letters. 38, (1977), 1440; Phys. Rev. D16 , (1977), 1791. [56] Weinberg, S.; Phys. Rev. Letters. 40, (1978), 223. [57] Wilczek, F.; Phys. Rev. Letters. 40, (1978), 279. [58] Dine, M., Fishcler, W., and Srednicki, M.; Phys. Letters. 104B , (1981), 199. 370 Bibliography We cite the following article for the see-saw mechanism. [59] Y anagida, T.; Prog. Theor. Phys. 64, (1980), 1103. We cite the following book for the grand uniﬁcation of the electromagnetic, weak and strong interactions. [60] Ross, G.G.; “ Grand Uniﬁed Theories ”, Perseus Books Publishing, 1984, Massachusetts. Chapters 5 through 11. We cite the following article for the SU (5) grand uniﬁed model. [61] Georgi, H. and Glashow, S.L.: Phys. Rev. Lett. 32, 438, (1974). We cite the following article for discussion of the gauge principle in the differential formalism originally due to H. Weyl and the integral formalism originally due to T.T.Wu and C.N. Y ang. [62] Y ang, C.N.: Ann. N.Y . Acad. Sci. 294 , 86, (1977). We cite the following book for the connection between Feynman’s action principle in nonrelativistic quantum mechanics, and the calculus of variations; in particular, the sec-ond variation, the Legendre test and the Jacobi test. [63] Schulman, L.S.; “ Techniques and Application of Path Integration ”, John Wiley & Sons, New Y ork, (1981). We cite the following book for the use of the calculus of variations in the path integral quantization of classical mechanics and classical ﬁeld theory, Weyl’s gauge principle for the Abelian gauge group and the semi-simple non-Abelian gauge group, the Schwinger–Dyson equation in quantum ﬁeld theory and quantum statistical mechanics, and stochas-tic quantization of classical mechanics and classical ﬁeld theory. [64] Masujima, M.: “ Path Integral Quantization and Stochastic Quantization ”, Springer Tracts in Modern Physics, V ol.165, Springer-V erlag, Heidelberg, (2000). Chapter 1,Section 1.1 and 1.2; Chapter 2, Sections 2.3, 2.4 and 2.5; Chapter 3, Sections 3.1, 3.2 and 3.3; Chapter 4, Sections 4.1 and 4.3; Chapter 5, Section 5.2. We cite the following book for discussion of the Schwinger–Dyson equation, and the Bethe–Salpeter equation from the viewpoint of the canonical formalism and the pathintegral formalism of quantum ﬁeld theory. [65] Huang, K.: “ Quantum Field Theory. From Operators to Path Integrals ”, John Wiley & Sons, New Y ork, (1998). Chapter 10, Sections 10.7 and 10.8; Chapters 13 and 16. [66] Huang, K.: “ Quarks, Leptons, and Gauge Field ”, 2 ndedition, World Scientiﬁc, Singa- pore, (1992). Chapters IX and X. [67] Huang, K.: “ Statistical Mechanics ”, 2ndedition, John Wiley & Sons, New Y ork, (1983). Chapter 18. The Wick–Cutkosky model is the only exactly solvable model for the Bethe–Salpeter equation known to this day. We cite the following articles for this model. [68] Wick, G.C.: Phys. Rev. 96., 1124, (1954). Bibliography 371 [69] Cutkosky, R.E.: Phys. Rev. 96., 1135, (1954). We cite the following books for canonical quantization, path integral quantization, the Smatrix approach to the Feynman rule for any spin J, the proof of the non-Abelian gauge ﬁeld theory based on BRST invariance and Zinn–Justin equation, the electro-weak uniﬁcation, the standard model, the grand uniﬁcation of weak, electromagneticand strong interactions, and the grand uniﬁcation with the graded Lie gauge group. [70] Weinberg, S.: “ Quantum Theory of Fields I ”, Cambridge Univ. Press, New Y ork, (1995). [71] Weinberg, S.: “ Quantum Theory of Fields II ”, Cambridge Univ. Press, New Y ork, (1996). [72] Weinberg, S.: “ Quantum Theory of Fields III ”, Cambridge Univ. Press, New Y ork, (2000). Inclusion of the gravitational force in a uniﬁcation scheme beside the weak force, the electromagnetic force and the strong force, requires the use of superstring theory. Index action functional 266, 304, 305, 311, 342, 345, 347, 348, 350 action integral 267 action principle 265 Feynman 266, 303–305, 308, 329, 370Hamilton 267, 272, 277Schwinger 266 adjoint boundary condition 9, 45 integral equation 282 matrix 9, 10, 12operator 9–13, 17problem 9, 11, 13, 15, 224–226, 233– 235 representation 343, 347 Akhiezer, N.I. 368axion 356, 369 Bach, M. 365 Bazant, M.Z. 366 Belavin, A.A. 369Bender, Carl M. 365Bernoulli, Jacques and Jean 263Bessel inequality 126 Bethe–Salpeter equation 326, 370 bifurcation point 253–255 Born approximation 312 boundary terms 17Brachistochrone 267, 270, 272–274 Broberg, K.B. 367 Carleman integral equation homogeneous 161, 166 inhomogeneous 161, 162, 166 Catenary 267, 276, 291 Cauchy integral equation of the ﬁrst kindhomogeneous 155 inhomogeneous 153 integral equation of the second kind generalization 161inhomogeneous 157, 166 integral formula general decomposition problem 208, 211 generalization 235review of complex analysis 21–24 Wiener–Hopf integral equation 196, 218 Wiener–Hopf method 177 Wiener–Hopf sum equation 235 kernel Carleman integral equation 161 residue theorem Sturm–Liouville system 140 singular integral equation 149 causality 49, 50, 168 charge conserved matter 342conserved Noether 344group 345 Collin, R.E. 366 complete 7 complete square 285completeness of an orthonormal system of functions 8 concentrated load 20conjugate point 284, 286–288, 295conservation law charge 339–342, 346 current 339–342, 345 energy, momentum, and angular momen- tum 348, 349, 352, 353 Coulomb potential 312 Applied Mathematics inTheor etical Physics. Mic hio Masujima Copyright ©2005 Wiley-VC HVerlag GmbH &Co.KGaA, Weinheim ISBN: 3-527-40534-8 374 Index Courant, R. 368 covariant derivative gravitational 350–355 non-Abelian 343, 346U(1) 339, 340 current conserved Noether 344, 345 gauged matter 344–346ungauged matter 344 Cutkosky, R.E. 364, 371 De Alfaro, V . 366 differential equation 54 differential operator 320, 325 Dine, M. 369Dirac delta function 8dispersion relations 166–168, 172 distributed load 18 disturbance solution 185dual integral equations 235, 239 eigenfunction 11–16, 32 completeness 113, 129, 138, 142 continuum 142, 143discrete 142, 143 eigenfunctions 11 eigenvalue problem 11, 45, 47, 129 inﬁnite Toeplitz matrix 229 eigenvalues 11electrodynamics 166, 172, 339, 341 energy integral 272 equal time canonical commutator 306, 314– 316, 331 Euler derivative 304Euler equation 264, 267, 269, 270, 272–275, 283, 284, 287–289, 298 Euler, Leonhard 263, 264Euler–Lagrange equation of motion 272, 313, 316, 340, 342, 345, 348, 351 factorization Wiener–Hopf integral equation 196, 201, 203, 209, 214, 216 Wiener–Hopf sum equation 234 Fetter, A.L. 339, 368 Feynman, R.P . 368ﬁeld strength tensor non-Abelian gauge ﬁeld 343–345, 353 U(1) gauge ﬁeld 341, 345Fikioris, G. 107 Fishcler, W . 369 Fomin, S.V . 32, 368 Fourier series 8 Fredholm alternative 12–15integral equation of the ﬁrst kind 33, 37, 115 integral equation of the second kind 33, 79, 107, 125, 131, 281, 282 exactly solvable examples 93 homogeneous 34, 46, 47, 93inhomogeneous 33, 34, 43, 63, 75, 95, 100, 114, 136 system of 100 nonlinear integral equation 253theory for a bounded kernel 86 Fredholm Alternative 12, 36 Fredholm solvability condition 12Freund, L.B. 367function space 7, 8, 10, 32 Gamma Function 68 gauge ﬁeld gravitational 339, 347, 353, 355, 368non-Abelian 168, 339, 341, 343–347, 353, 356, 369 U(1) 339–341, 344–346, 369 gauge group 341, 347, 356, 369–371gauge principle H. Weyl 339–341, 343, 347, 349, 352, 368–370 R. Utiyama 347, 349T.W .B. Kibble 339, 347 Wu-Y ang formalism 370 Gelfand, I.M. 32, 368 generating functional 314, 316, 327, 328 Georgi, H. 369, 370Glashow, S.L. 369, 370Gohberg, I.C. 367Goldberger, M.L. 172, 365, 366 Goldstein, J. 364 Green’s function 9, 16–18, 20, 21, 32, 39–43, 45–49, 51, 52, 129, 139, 145, 169,312, 314, 316, 317, 319, 320, 325, 328, 333, 335, 336 Hadamard inequality 88 Hamilton, William Rowan 265 Index 375 Hamilton–Jacobi equation 32, 265, 368 Hammerstein nonlinear integral equation 257 harmonic potential 143, 312 Hermitian 10Hermitian transpose 10Hibbs, A.R. 368 Higgs scalar ﬁeld 369 Higgs–Kibble mechanism 369Hilbert problem homogeneous 150, 156, 162 inhomogeneous 150, 154, 158, 162, 177 Hilbert, D. 368Hilbert–Schmidt expansion 116, 121, 131, 134, 136 theorem 113, 118, 121, 130 theory generalization 133 Hölder condition 24homogeneous 33 Hopf, E. 366 Huang, K. 368–370 in-state 56 incoming wave condition 54 index Wiener–Hopf integral equation 211, 212, 214–216, 220, 222–226 Wiener–Hopf sum equation 230–235 inﬁnitesimal variation of the initial condition 287 inﬂuence function 20inhomogeneous 33inner product 5–7, 9–15, 17, 18, 32, 44, 45, 48, 130 instanton 356, 369integral equation 54invariance 338, 339, 347 gauge 339 global G 343, 345 global U(1) 339–341, 346 local G 343–346 local U(1) 339–341, 346 scale 339 isoperimetric problem 264, 274 iterative solution 75 Jackson, J.D. 246, 366 Jacobi test 284, 286, 287Jacobi, Carl Gustav Jacob 263, 265 Jerri, A.J. 365 kernel 33 bounded 86, 92general 84inﬁnite translational 67, 95 iterated 78, 79 Pincherle–Goursat 81resolvent 78, 84, 92, 114, 142 Wiener–Hopf integral equation 220, 222 semi-inﬁnite translational 191, 212, 224, 238 square-integrable 39, 65, 76, 114, 115symmetric 48, 109, 111, 282 transposed 36, 37, 111, 137, 138 Kondo, J. 174, 247, 367Krein, M.G. 366, 367Kress, R. 32, 367Kronecker delta symbol 7 Lagrange equation of motion 305 Lagrange, Joseph Louis 263, 264 Lagrangian 264, 267, 272, 277, 303, 307, 308, 329 Lagrangian density 266, 308, 313, 329, 339 Abelian U(1) gauge ﬁeld 341 gravitational ﬁeld 347, 353, 355 interaction 316, 318, 339, 341, 344 matter ﬁeld 339, 340, 342, 343, 347, 350–352 matter-gauge system 341, 344non-Abelian Ggauge ﬁeld 344 QCD 356 Landau, L.D. 368Laplace transform 67Lautrup, B. 361Legendre test 284, 285 Legendre, Adrien Marie 263, 264 Lifshitz, E.M. 368 linear operator 8, 9, 11, 13, 16, 32linear vector space 6Liouville’s theorem 21 Carleman integral equation 164 Cauchy integral equation 154, 159Wiener–Hopf method 178, 199, 200, 205, 213, 215, 216, 223, 231, 232, 235 376 Index Margetis, D. 107 Masujima, M. 303, 307, 370 Mathews, J. 368 McCoy, B. 235, 367 mean value theorem 297 Mikhlin, S.G. 366 minimum 6, 128, 273, 277, 279, 280, 284– 288, 292, 296, 299 momentum operator 305 Myers, J.M. 107 Nakanishi, N. 361 Newton, Isaac 263 Nishijima, K. 365 Noble, B. 366norm 5–7, 76, 89, 124, 131 norm of a function 5 normalized 7 operator 8 Orszag, Steven A. 365 orthogonal functions 7orthonormal set of functions 7 out-state 56 outgoing wave condition 54 Peccei, R.D. 369 Pincherle–Goursat type kernel 85 Pipkin, A.C. 365 Plemelj formula 27 Polianin, A.D. 365 Polyakov, A.M. 369positivity 5 principle of superposition 303 proper self-energy parts 320 quantum mechanics 143, 168 Quinn, H.R. 369 Rayleigh quotient 126–128 reciprocity 48 reﬂection coefﬁcient 43 Regge, T. 366 remainder 117 resolution of identity 303resolvent Fredholm integral equation 78–80, 83, 86 Hilbert–Schmidt theory 114, 142Wiener–Hopf integral equation 220, 222 resolvent kernel 220retarded boundary condition 54Riccatti differential equation 286 Riccatti substitution 286 Ross, G.G. 370 Sakurai, J.J. 365 scalar multiplication 5 scattering problem 41Schrödinger equation 169, 307, 308 boundary value problem 39time-dependent scattering problem 48time-independent scattering problem 41 Schulman, L.S. 370 Schwartz, A.S. 369Schwarz inequality 6, 7, 64, 76, 77 Schwinger–Dyson equation 312, 322, 325, 329, 336, 338, 339, 370 second variation 277, 283, 370self-adjoint 10, 15, 16, 19, 45, 48, 109, 126, 129, 131, 138, 139, 142 self-adjoint operator 15 singular point 253 solvability condition Wiener–Hopf integral equation of the second kind 226 Wiener–Hopf sum equation 234 square-integrability 101, 133, 171 square-integrable 5, 8, 28, 29, 31, 39, 63, 65, 66, 75, 84, 93, 95, 100, 101, 109, 112–115, 121, 124, 130, 131, 133,134, 136, 171, 282 Srednicki, M. 369 Stakgold, I. 365standard model 356, 369, 371 step-discontinuous 31 strong minimum 296strong variation 283 Sturm–Liouville eigenfunction 129eigenvalue problem 47, 129, 131, 279operator 129system 44, 47, 131, 138 summary behavior near the endpoints 26Example 7.4 207Fredholm theory for a bounded kernel 92 Index 377 Schwinger–Dyson equation 323, 336 Wiener–Hopf integral equation 226 Wiener–Hopf sum equation 235 supergain antennas 105, 107 symmetric kernel 109 time-ordered product 305, 314, 315 Toeplitz matrix 227 inﬁnite 227 semi-inﬁnite 229 transformation gauge 343, 346, 347 global G 342, 345 global U(1) 340, 345 local G 343, 346 local phase 339 local U(1) 339–341 transformation function 303 transmission coefﬁcient 44 trial action functional 312 trial potential 312 triangular inequality 5–7, 76Tricomi, F.G. 365 Trout, B.L. 366 Tyupkin, Y .S. 369 uniﬁcation 356, 368–371 vacuum expectation value 316, 317 V anKoughnett, A.L. 234, 367 variational principle 264, 368 Euler–Lagrange 265 Feynman quantum mechanics 368 quantum statistical mechanics 308, 311, 368 invariant 347 vertex operator 321 V olterra coupled integral equations of the second kind solvable case 70 integral equation of the ﬁrst kind 33 solvable case 69 integral equation of the second kind 39 homogeneous 66 inhomogeneous 33, 38, 41, 63, 66 solvable case 66 nonlinear integral equation of convolu- tion type 249Walecka, J.D. 339, 368 Walker, R.L. 368 Wasylkiwskyj, W . 234, 367 Watson, K.M. 172, 365, 366wave function bound state 143scattering state 143 vacuum 303 Wazwaz, A.M. 365weak minimum 284weak variation 283 Weierstrass Efunction 296 Weierstrass–Erdmann corner relation 297, 299, 300 Weinberg, S. 368, 369, 371Weinstein, L.A. 246, 367 Weyl, H. 339, 368–370 Wick, G.C. 364, 370Wick–Cutkosky model 364, 370 Wiener, N. 366 Wiener–Hopf integral equation 177, 191, 226, 366, 367 integral equation of the ﬁrst kind 192, 235, 238, 239 integral equation of the second kind 191, 196 homogeneous 192, 208inhomogeneous 192, 216, 234 method 177, 366, 367 factorization 227 sum-splitting 178, 181, 183, 189, 190 problem sum-splitting 189, 207 sum equation 227, 234, 235, 366, 367 inhomogeneous 229, 234 Wilczek, F. 369 Wronskian 47, 139, 287–289Wu, Tai Te 367Wu, Tai Tsun 107, 234, 235, 366, 367, 370 Y agi–Uda semi-inﬁnite arrays 234, 367 Y anagida, T. 370Y ang, C.N. 370 Y okoyama, K. 361 Y ukawa coupling 312, 316, 325Y ukawa, H. 369 Zinn–Justin equation 371