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PARTIAL DIFFERENTIALEQUATIONS FOR

MATHEMATICAL PHYSICISTS

Bijan Kumar Bagchi

ISBN 978-0-367-22702-9

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Partial Differential Equations for Mathematical Physicists is intended for graduatestudents, researchers of theoretical physics and applied mathematics, and professionalswho want to take a course in partial differential equations. This book offers the essentialsof the subject with the prerequisite being only an elementary knowledge of introductorycalculus, ordinary differential equations, and certain aspects of classical mechanics. Wehave stressed more the methodologies of partial differential equations and how they canbe implemented as tools for extracting their solutions rather than dwelling on thefoundational aspects. After covering some basic material, the book proceeds to focusmostly on the three main types of second order linear equations, namely those belongingto the elliptic, hyperbolic, and parabolic classes. For such equations a detailed treatmentis given of the derivation of Green's functions, and of the role of characteristics andtechniques required in handling the solutions with the expected amount of rigor. In thisregard we have discussed at length the method of separation variables, application ofGreen's function technique and employment of Fourier and Laplace's transforms. Alsocollected in the appendices are some useful results from the Dirac delta function, Fouriertransform, and Laplace transform meant to be used as supplementary materials to thetext. A good number of problems are worked out and an equally large number of exerciseshave been appended at the end of each chapter keeping in mind the needs of thestudents. It is expected that this book will provide a systematic and unitary coverage ofthe basics of partial differential equations.

Key Features

An adequate and substantive exposition of the subject.

Covers a wide range of important topics.

Maintains mathematical rigor throughout.

Organizes materials in a self-contained way with each chapter ending with asummary.

Contains a large number of worked out problems.

PARTIALDIFFERENTIAL

EQUATIONSFOR

MATHEM

ATICALPHYSICISTS

Bijan

Kum

arBagchi

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Partial DifferentialEquations forMathematical

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Bijan Kumar Bagchi

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In my mother’s memory

Contents

Preface ix

Acknowledgments xi

Author xiii

1 Preliminary concepts and background material 1

1.1 Notations and definitions . . . . . . . . . . . . . . . . . . . . 21.2 Generating a PDE . . . . . . . . . . . . . . . . . . . . . . . . 61.3 First order PDE and the concept of characteristics . . . . . . 81.4 Quasi-linear first order equation: Method of

characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Second order PDEs . . . . . . . . . . . . . . . . . . . . . . . 181.6 Higher order PDEs . . . . . . . . . . . . . . . . . . . . . . . 191.7 Cauchy problem for second order linear PDEs . . . . . . . . 211.8 Hamilton-Jacobi equation . . . . . . . . . . . . . . . . . . . . 271.9 Canonical transformation . . . . . . . . . . . . . . . . . . . . 281.10 Concept of generating function . . . . . . . . . . . . . . . . . 311.11 Types of time-dependent canonical transformations . . . . . 33

1.11.1 Type I Canonical transformation . . . . . . . . . . . . 331.11.2 Type II Canonical transformation . . . . . . . . . . . 341.11.3 Type III Canonical transformation . . . . . . . . . . . 351.11.4 Type IV Canonical transformation . . . . . . . . . . . 35

1.12 Derivation of Hamilton-Jacobi equation . . . . . . . . . . . . 381.13 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Basic properties of second order linear PDEs 45

2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.2 Reduction to normal or canonical form . . . . . . . . . . . . 472.3 Boundary and initial value problems . . . . . . . . . . . . . . 602.4 Insights from classical mechanics . . . . . . . . . . . . . . . . 702.5 Adjoint and self-adjoint operators . . . . . . . . . . . . . . . 732.6 Classification of PDE in terms of eigenvalues . . . . . . . . . 752.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

vii

viii Contents

3 PDE: Elliptic form 81

3.1 Solving through separation of variables . . . . . . . . . . . . 833.2 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . 903.3 Maximum-minimum principle for Poisson’s and Laplace’s

equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923.4 Existence and uniqueness of solutions . . . . . . . . . . . . . 933.5 Normally directed distribution of doublets . . . . . . . . . . 943.6 Generating Green’s function for Laplacian operator . . . . . 973.7 Dirichlet problem for circle, sphere and half-space . . . . . . 1003.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4 PDE: Hyperbolic form 109

4.1 D’Alembert’s solution . . . . . . . . . . . . . . . . . . . . . . 1104.2 Solving by Riemann method . . . . . . . . . . . . . . . . . . 1134.3 Method of separation of variables . . . . . . . . . . . . . . . 1174.4 Initial value problems . . . . . . . . . . . . . . . . . . . . . . 1214.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

5 PDE: Parabolic form 137

5.1 Reaction-diffusion and heat equations . . . . . . . . . . . . . 1375.2 Cauchy problem: Uniqueness of solution . . . . . . . . . . . . 1405.3 Maximum-minimum principle . . . . . . . . . . . . . . . . . 1415.4 Method of separation of variables . . . . . . . . . . . . . . . 1435.5 Fundamental solution . . . . . . . . . . . . . . . . . . . . . . 1545.6 Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . 1575.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

6 Solving PDEs by integral transform method 165

6.1 Solving by Fourier transform method . . . . . . . . . . . . . 1656.2 Solving by Laplace transform method . . . . . . . . . . . . . 1726.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

A Dirac delta function 185

B Fourier transform 203

C Laplace transform 213

Bibliography 221

Index 223

Preface

This book aims at providing an introduction to partial differential equations(PDEs) and expects to serve as a textbook for the young graduate students oftheoretical physics and applied mathematics who have had an initial trainingin introductory calculus and ordinary differential equations (ODEs) and anelementary grasp of classical mechanics. The presentation of the book does notalways follow the conventional practice of classifying the PDE and then takingup the treatment of the representative Laplace equation, wave equation andheat conduction equation one by one. On the contrary, certain basic featuresof these equations are presented in the introductory chapter itself with an aimto provide an appropriate formulation of mathematical methods associatedwith them as and when they emerge from the class of PDE they belong to.

Numerous examples have been worked out in the book which aim to illus-trate the key mathematical concepts involved and serve as a means to improvethe problemsolving skills of the students. We have purposefully kept the com-plexities of the mathematical structure to a minimum, often at the expense ofgeneral and abstract formulation, but put greater stress on the application-side of the subject. We believe that the book will provide a systematic andcomprehensive coverage of the basic theory of PDEs.

The book is organized into six chapters and three appendices. Chapter 1contains a general background of notations and preliminaries required thatwould enable one to follow the rest of the book. In particular, it contains adiscussion of the first order PDEs and the different types of equations thatone often encounters in practice. The idea of characteristics is presented andalso second and higher order PDEs are introduced. We also comment brieflyon the Cauchy problem and touch upon the classification of a second orderpartial differential equation in two variables.

Chapter 2 shows how a PDE can be reduced to a normal or canonicalform and the utility in doing it. We give some insights from the classicalmechanics as well. We also discuss the construction of the adjoint and self-adjoint operators.

The elliptic form of a partial differential equation is discussed in Chapter 3.Topics like method of separation of variables for the two-dimensional planepolar coordinates, three-dimensional spherical polar coordinates and cylin-drical polar coordinates, harmonic functions and their properties, maximum-minimum principle, existence, uniqueness and stability of solutions, normallydirected distribution of doublets, Green’s equivalent layer theorem, generation

ix

x Preface

of Green’s function, Dirichlet’s problem of a circle, sphere and half-space areconsidered.

PDEs of hyperbolic type are taken up in Chapter 4 which begins withthe D’Alembert’s solution of the linear second order wave equation. The moregeneral Riemann’s method is introduced next. The role of Riemann func-tion is pointed out to help us solve the PDE in the form of a quadrature.Solutions by the method of separation of variables is illustrated for the three-dimensional wave equation both for the spherical polar and cylindrical coor-dinates. This chapter also consists of detailed discussions of the initial valueproblems related to the three-dimensional and two-dimensional wave equa-tions. We give here the basic derivation of the Poisson/Kirchoff solution forthe three-dimensional homogeneous equation supplemented by a set of inho-mogeneous conditions and then focus on the solution of the inhomogeneouswave equation by means of the superposition principle, which gives us thePoisson formula. Hadamard’s method of descent is employed to solve com-pletely the two-dimensional counterpart.

Parabolic equations in Chapter 5 is our next topic of inquiry. This chap-ter covers the Cauchy problem for the heat equation wherein we discuss theuniqueness criterion of the solution, the method of separation of variables forthe Cartesian coordinates, spherical polar and cylindrical polar coordinates,and derive the fundamental solution and give a formulation of the Green’sfunction.

Chapter 6 is concerned with solving different types of PDEs by the inte-gral transform method focusing on the use of Fourier transform and Laplacetransforms only. Some problems are worked out for their asymptotic nature.

Finally in the three appendices we address respectively some importantissues of the delta function, the Fourier transform and Laplace transform. Iwish to remark that we added a summary at the end of each chapter for thelay reader to have a quick look at the materials that have been covered andalso appended a reasonable collection of homework problems relevant to thechapter. In addition, we solved in each chapter a wide range of problems toclarify the basic ideas involved. It is believed that these will help the readersin furthering the understanding of the subject.

Acknowledgments

I would like to express my gratitude to Prof. Rupamanjari Ghosh, Vice Chan-cellor, Shiv Nadar University for her steadfast support without which it wouldnot have been possible to complete the book. I also appreciate the continualencouragement from Prof. Sankar Dhar, Head, Department of Physics, ShivNadar University during the preparation of this book.

It gives me great pleasure to acknowledge the kind help that I received,time and time again, from Prof. Kalipada Das, Department of Applied Math-ematics, University of Calcutta that saw me through this book. I am verygrateful to him. I also recall the insightful comments of the late Mithil RanjanGupta of the same department that helped my understanding enormously. Ithank Dr. Santosh Singh, Department of Mathematics, Shiv Nadar Univer-sity for many helpful conversations and countless acts of support, Dr. SantoshKumar, Department of Physics, Shiv Nadar University for frequent help, andProf. Somnath Sarkar, Department of Electronics, University of Calcutta forhis interest in the development of this book. I owe gratitude to my studentsMr. Supratim Das and Ms. Debanjana Bose for organizing the first few chap-ters of my lecture notes, and Mr. Yogesh Yadav for carefully reading cer-tain portions and offering valuable suggestions. I would also like to thankMs. Aastha Sharma, Commissioning Editor, CRC Press, Taylor & FrancisGroup, and her editorial team, especially Ms. Shikha Garg for extending sup-port toward finalizing the draft of the book.

Finally, I thank my wife Minakshi, and daughter Basabi for always urgingme to turn the idea of writing a book on partial differential equations into areality.

Bijan Kumar BagchiDepartment of Physics,Shiv Nadar University,

Greater Noida

xi

Author

Bijan Kumar Bagchi received his B.Sc., M.Sc., and Ph.D. from the Uni-versity of Calcutta. He has a variety of research interests and involvementsranging from spectral problems in quantum mechanics to exactly solvablemodels, supersymmetric quantum mechanics, parity-time symmetry and re-lated non-Hermitian phenomenology, nonlinear dynamics, integrable modelsand high energy phenomenology.

Dr. Bagchi has published more than 150 research articles in refereed jour-nals and held a number of international visiting positions. He is the authorof the books entitled Advanced Classical Mechanics and Supersymmetry inQuantum and Classical Mechanics both published by CRC Press, respectively,in 2017 and 2000. He was formerly a professor in applied mathematics at theUniversity of Calcutta, and is currently a professor in the Department ofPhysics at Shiv Nadar University.

xiii

Chapter 1

Preliminary concepts andbackground material

1.1 Notations and definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Generating a PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

(a) Eliminating arbitrary constants from a given relation . . . . . . 6(b) Elimination of an arbitrary function . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 First order PDE and the concept of characteristics . . . . . . . . . . . . . 81.4 Quasi-linear first order equation: Method of characteristics . . . . . 9

(a) Lagrange’s method of seeking a general solution . . . . . . . . . . . . 9(b) Integral lines and integral surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.5 Second order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.6 Higher order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.7 Cauchy problem for second order linear PDEs . . . . . . . . . . . . . . . . . . 211.8 Hamilton-Jacobi equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.9 Canonical transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.10 Concept of generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.11 Types of time-dependent canonical transformations . . . . . . . . . . . . 33

1.11.1 Type I Canonical transformation . . . . . . . . . . . . . . . . . . . . . . . . 331.11.2 Type II Canonical transformation . . . . . . . . . . . . . . . . . . . . . . . 341.11.3 Type III Canonical transformation . . . . . . . . . . . . . . . . . . . . . . 351.11.4 Type IV Canonical transformation . . . . . . . . . . . . . . . . . . . . . . 35

1.12 Derivation of Hamilton-Jacobi equation . . . . . . . . . . . . . . . . . . . . . . . . . 381.13 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

An ordinary differential equation (ODE) is an expression that involves oneor more functions defined with respect to one independent variable alongwith the derivatives of such functions while a partial differential equation(PDE) is concerned with a relationship involving an unknown function ofseveral independent variables and their partial derivatives with respect tothese variables. The utility of an ODE is limited because a physical systemvery often depends on more than one independent variable. On the otherhand, with the presence of a multitude of independent variables (at leasttwo), PDEs address a far wider audience of various branches of mathematicaland theoretical physics.

PDEs find numerous applications for different ranges of physical phenom-ena. Some commonplace examples are the Navier-Stokes equation of fluid

1

2 Partial Differential Equations for Mathematical Physicists

mechanics, Maxwell’s equations of electrodynamics, problems of populationdynamics and dynamical systems. PDEs are also encountered in various otherdisciplines such as plasma dynamics, quantum mechanics, mathematical biol-ogy, computational chemistry, areas dealing with differential geometry, com-plex analysis, symplectic geometry and algebraic geometry apart from otherpursuits seeking analytical or computational results arising in problems ofmathematical methods.

The study of differential equations can be traced back to the basic worksof Issac Newton (1642-1727) and Gottfried Leibniz (1646-1716). While New-ton dealt with what he called fluxional equations, Leibniz made an extensivestudy of the first order equations like a(x, y)∂u∂x + b(x, y)∂u∂y = c(x, y) and

introduced shorthands like δ to denote partial derivatives ∂∂x or ∂

∂y . Wideranging applications of differential equations to classical mechanics, hydro-dynamics, astronomy and related areas were subsequently pioneered by theBernoulli brothers Jakob (1654-1705) and Johann (1667-1748), Johann’s illus-trious son Daniel (1700-1782), Leonhard Euler (1707-1783), Jean D’Alembert(1717-1783), Joseph Lagrange (1736-1813) and Pierre Simon de Laplace (1749-1827). Certain aspects of mathematical rigour were exploited by AugustinCauchy (1789-1859) who proved a number of existence theorems while otherimportant contributions came from Johann Pfaff (1765-1825), Joseph Fourier(1768-1830), Simeon Poisson (1781-1840), Carl Jacobi (1804-1851) and SophusLie (1842-1899).

1.1 Notations and definitions

We shall normally adopt the following notations :

Independent variables : x, y, z, ξ, η, ...

Dependent variables : φ, ψ, u, v, w, ...

Partial derivatives : φx(≡ ∂φ∂x ), φy(≡ ∂φ

∂y , φxx(≡ ∂2φ∂x2 ), φxy(≡ ∂2φ

∂x∂y ), φyy(≡∂2φ∂y2 ), ...

The general form of a partial differential equation involving two independentvariables, x and y, is given by

F (x, y, φx, φy, φxx, φyy, φxy, φxxx, φyyy, ...) = 0, x, y ∈ Ω (1.1)

Preliminary concepts and background material 3

where Ω is some specified domain, F is a functional of the arguments indicatedand φ is an arbitrary function of (x, y). By the order of a PDE we meanthe order of the highest derivative appearing in (1.1). A solution of (1.1)corresponds to the function φ(x, y) obeying (1.1) for all values of x and y.

If we are dealing with n independent variables, x1, x2, ..., xn, the domainΩ will refer to an n− dimensional space containing an (n − 1)-dimensionalhypersurface. A hypersurface of dimension (n − 1) is given by an equationof the form x2

1 + x22 + ... + x2

n = 1 residing in an n-dimensional Euclideanspace. For instance, in an Euclidean space of dimension two, a hypersurface isa plane curve while in an Euclidean space of dimension three, a hypersurfaceis a surface.

Some simple examples of a PDE are

Example 1.1 φx−bφy = 0 : b is a constant: the PDE is first order in x and y.

Example 1.2 φxx+φy = 0 : the PDE is second order in x but first order in y.

Example 1.3 φx+φyy = 0 : the PDE is first order in x but second order in y.

In this book we will be interested mostly in the linear forms of the PDE.Adopting an operator notation a linear PDE can be put in the form

Lφ(x, y) = ρ(x, y) (1.2)

where L denotes a suitable operator, all terms involving φ and its derivativesare grouped in the left side while the term ρ(x, y) in the right side, whichacts as a source term, is assumed to be known. Formally, a linear PDE is theone which is linear in the dependent variable and all of its partial derivatives.Linearity is governed by the following properties of L

(i)L(φ+ ψ) = Lφ+ Lψ

(ii)L(cφ) = cLφ

for arbitrary functions φ, ψ and a constant c.

Example 1.4

The heat conduction equation

φt − kφxx = 0, t > 0 0 < x < l

where k is a constant, is clearly linear.


Example 1.5

The PDE

φφt + φ = sinx

is nonlinear because of the product term φφt.

Taking L to be a linear operator, if ρ(x, y) = 0 in (1.2), the PDE is classifiedas a linear homogeneous equation. The latter then reads simply

Lφ(x, y) = 0 (1.3)

When this is not the case i.e. if ρ(x, y) 6= 0, the PDE is said to be an inho-mogeneous (or non-homogeneous) equation. We therefore observe that everyinhomogeneous equation has a corresponding homogeneous counterpart. InExample (1.5), sinx stands for the inhomogeneous term. However its pres-ence does not alter the linear or nonlinear character of the equation. A generalsolution of the homogeneous equation when superposed with the particularsolution of the inhomogeneous equation produces the general solution of theinhomogeneous equation.

For a first order linear equation, Lφ reads

Lφ(x, y) ≡ a(x, y)φx + b(x, y)φy + c(x, y)φ

where the coefficients a, b, c are functions of x and y while for a second orderlinear equation, Lφ can be projected in its general form

Lφ(x, y) ≡ A(x, y)φxx + 2B(x, y)φxy + C(x, y)φyy

+D(x, y)φx + E(x, y)φy + F (x, y)φ

where the coefficients A, B, C, D, E and F are functions of x and y.

If the PDE is not linear then it is a nonlinear PDE. Note however that ifan equation is linear in the highest derivatives of φ then such an equation iscalled a quasi-linear equation. A couple of typical examples with respect totwo independent variables x and y are the first order equation

(1 + φ2)φx + φy = x2

and the second order equation

φxφxx + φyφyy = φ3

On the other hand, if the coefficients of the highest order derivatives of aquasi-linear equation are functions only of the independent variables then suchan equation is called an almost linear, or half-linear, or simply a semi-linearPDE. An appropriate example of a semi-linear equation is


x2φxx + 4xyφyy + φφx + φ2 = 0

where x and y are the independent variables.To extract a solution of a PDE one needs to have some knowledge of

the associated initial or boundary conditions. If the PDE is equipped with acondition at an initial time, say at t = 0, for a certain portion of the regionunder consideration, it is referred to as an initial condition. A condition thatholds on any other curve for all times in that region is called a boundarycondition. Sometimes we encounter an eigenvalue problem too. An appropriateset of boundary or initial condition is expected to yield a unique and stablesolution.

Let us explain all this by means of some examples from the theory of ODE.Suppose that the temperature T of a given material is given to be T0 at someinitial time t = 0. We can solve such an initial value problem by assigning forthe temperature variation a law of cooling that gives the rate of cooling asdirectly proportional to the temperature difference between the object and itssurroundings. We assume that the temperature of the surroundings is zero.Then the guiding differential equation takes the form

T ′(t) = −kT (t) t > 0

T (0) = T0

where k is a constant of proportionality. The solution turns out to be expo-nentially damping

T (t) = T0e−kt t ≥ 0

Note that the complementary case corresponds to the differential equationT ′(t) = kT (t) which provides the character of the solution as an exponentiallyexpanding one.

Next, consider the simple case of fitting a straight line as given by thestandard equation y(x) = mx + c which connects two points, say x1 and x2

in a certain given plane. This criterion determines both the unknowns m andc enabling us to express the straight line equation as

y(x) = (y2 − y1

x2 − x1)x+ (

y1x2 − y2x1

x2 − x1), x1 < x < x2

where y1 and y2 denote the y values for x = x1 and x2 respectively.This is an example of a boundary value problem corresponding to a given

prescription on the function y(x) in the domain [x1, x2]. Mathematically wecan formulate the whole scheme in the form of a second order ODE namely,

y′′(x) = 0 x1 < x < x2

where the primes correspond to derivatives with respect to the independentvariable x.


Finally, we can think of oscillating functions such as φ(x) = sin kx, wherek is a constant, which emerge as solutions of the differential equation φ′′(x) =−k2φ(x), −∞ < x <∞. As is evident, we have an eigenvalue problem whichcan be solved if suitable initial inputs are provided.

1.2 Generating a PDE

We can generate PDEs in different ways. We discuss here two possibilities.

(a) Eliminating arbitrary constants from a given relation

Consider a two-dimensional quadratic equation

(x− α)2 + (y − β)2 + φ2 = 1

where φ is a function of two independent variables x and y and α, β are twoconstants. Partially differentiating both sides with respect to x gives

(x− α) + φφx = 0

while partially differentiating both sides with respect to y results in

(y − β) + φφy = 0

If we eliminate the constants α and β from the last two equations we obtaina first order PDE

φ2[1 + (φx)2 + (φy)2] = 1

Next, consider a somewhat different type of quadratic equation namely

x2

a2+y2

b2+φ2

c2= 1

where φ is a function of two independent variables x and y but the equationinvolves three constants a, b and c.

Partially differentiating both sides of this equation with respect to x gives

2x

a2+

2φ

c2φx = 0 ⇒ c2

a2= −φ

xφx

If partially differentiated again with respect to x we get

2

a2+

2

c2(φx)2φxx = 0 ⇒ c2

a2= −(φx)2 − φφxx


Eliminating c2

a2 from the last two equations leads to a second order PDE interms of the independent variable x

xφφxx + x(φx)2 − φφx = 0

In a similar way if we carry out partial differentiations with respect tothe variable y twice we get a second order PDE in terms of the independentvariable y

yφφyy + y(φy)2 − φφy = 0

What do we infer from the two examples considered?If the number of constants to be eliminated equals the number of indepen-

dent variables, as was the case in the first example, then we run into a firstorder PDE. However, if the number of constants to be eliminated is greaterthan the number of independent variables, as we saw in the case of the secondexample, then we obtain an equation of second or higher order.

(b) Elimination of an arbitrary function

Let us now deal with two functions F = F (x, y, φ), G = G(x, y, φ) andassume a certain connection χ existing between them

χ(F,G) = 0

The above relation, on taking partial derivatives with respect to x and y,yields respectively

χF (Fx + Fφφx) + χG(Gx +Gφφx) = 0

χF (Fy + Fφφy) + χG(Gy +Gφφy) = 0

It is an easy task to get rid of χF and χG which yields

(Fx + Fφφx)(Gy +Gφφy)− (Fy + Fφφy)(Gx +Gφφx) = 0

The above expression simplifies because the term FφφxGφφy cancels out andwe are left with the form

(FφGy − FyGφ)φx + (FxGφ − FφGy)φy = FyGx − FxGyThis gives the Lagrange’s form of a first order quasi-linear PDE

a(x, y, φ)φx + b(x, y, φ)φy = c(x, y, φ)

where the coefficients a and b do not involve the arbitrary connecting func-tion χ.


1.3 First order PDE and the concept of characteristics

We inquire into solving a linear first order PDE of the form

a(x, y)φx + b(x, y)φy = 0

where we regard φ = φ(x, y) as a surface which is generated by the family ofcurves C1, C2, ....

Let us parametrize a typical curve Ci in terms of a parameter s as givenby the following pair of differential equations

dx

ds= a(x, y)

dy

ds= b(x, y)

Then the derivative of φ with respect to s reads

dφ

ds= φx

dx

ds+ φy

dy

ds= a(x, y)φx + b(x, y)φy = 0;

In other words, along Ci, φ is a function of s only. Its solution is φ = k =constant and so along Ci, φ is a constant. Ci is referred to as a characteristiccurve.

The equations of the characteristics are obtained by solving the subsidiaryequations:

dx

a=dy

b

in which the parameter s does not appear. Since the solution involves onearbitrary constant, the characteristics, at the first order level, generate a one-parameter family of curves.

Example 1.6

Consider the first order PDE for φ(x, y)

φx + xφy = 0

subject to φ(0, y) = 1 + y2 where 0 < y < 2.

The pair of characteristics are readily seen to be given by the equations

dx

ds= 1

dy

ds= x

Elimination of ds gives

x2 − 2y = k, k = constant

implying that the resulting characteristics are parabolas.


On each of the parabolas, φ is constant. The form of φ in the region Rbounded by the characteristics passing through the origin (0, 0) and (0, 2) canbe determined as follows.

Let (x, y) be any point in R. The equation of the characteristic passingthrough the point (x, y) is given by

x2 − 2y = x2 − 2y = k

On the above parabola, φ must be a constant say, λ

φ = λ

This parabola will intersect the line segment between (0, 0) and (0, 2) at thepoint where x = 0 and the corresponding y coordinate y0 given by

02 − 2y0 = x2 − 2y ⇒ y0 =1

2(2y − x2)

Comparing with the given condition φ(0, y) = 1 + y2 we see that

λ = 1 +1

4(2y − x2)2

Since (x, y) is any point in R, the solution of φ in R is

φ = 1 +1

4(2y − x2)2

Here the stringent requirement has been that the line segment connecting(0, 0) and (0, 2) is not a part of any characteristic as shown in Figure 1.1.

1.4 Quasi-linear first order equation: Method ofcharacteristics

(a) Lagrange’s method of seeking a general solution

Let us address Lagrange’s form of the quasi-linear first order PDE

a(x, y, φ)φx + b(x, y, φ)φy = c(x, y, φ) (1.4)

where a, b, c are functions of x, y and φ. Note that the functions a, b, c donot involve the derivatives of φ.


FIGURE 1.1: The parabolas through the points (0, 0) and (0, 2).

In what follows we introduce Lagrange’s method to demonstrate that thegeneral solution to (1.4) is of the type

χ(F,G) = 0

where χ is an arbitrary function of F and G with F (x, y, φ) = k1, k1 is a con-stant and G(x, y, φ) = k2, k2 is a constant, corresponding to two independentsolutions of the subsidiary equations

dx

a(x, y, φ)=

dy

b(x, y, φ)=

dφ

c(x, y, φ)(1.5)

Proof:

Partially differentiating χ(F,G) with respect to x and y gives respectively

χx + χφφx = 0, χy + χφφy = 0 (1.6)

The above equations imply that

φx = −χxχφ, φy = −χy

χφ(we let χφ 6= 0) (1.7)


As a result the Lagrange’s equation is converted to the form

aχx + bχy = cχφ (1.8)

On the other hand, the subsidiary equations associated with the aboveform read

dy

dx=b

a,

dφ

dx=c

a(1.9)

To prove that F (x, y, φ) = k1 and G(x, y, φ) = k2 are solutions of the La-grange’s equation we take the differential of F and G to write down

dF ≡ Fxdx+ Fydy + Fφdφ = 0, dG ≡ Gxdx+Gydy +Gφdφ = 0 (1.10)

Due to the subsidiary equations (1.9) these equations assume the forms

aFx + bFy + cFφ = 0, aGx + bGy + cGφ = 0 (1.11)

showing both F and G to be solutions of Lagrange’s equation. The generalsolution is thus expressible in the functional form

χ(F,G) = 0 (1.12)

with χ being an arbitrary function.

(b) Integral lines and integral surfaces

Let φ(x, y) be a solution of (1.4). We can represent the latter equation inthe form

(a, b, c) · (φx, φy,−1) = 0 (1.13)

Since (φx, φy,−1) is normal to the surface S = [x, y, φ(x, y)], it follows thatthe vector field V = (a, b, c) lies in the tangent plane to S. This is indeed afeature of the solution φ(x, y) we are looking for. The surface for which thevector field V lies in its tangent plane, at each point in the surface, is calledan integral surface. In fact any point on the integral surface has the form(x, y, φ(x, y)).

We now address the Cauchy problem associated with the Lagrange’s equa-tion (1.4). It is concerned with the determination of its unique solution φ(x, y)such that it takes given values on a regular arc1 γ in the xy-plane.

1A plane arc is defined as a collection of points which can be parametrized by thecoordinates x = f(t) and y = g(t) with the parameter t belonging to some closed interval


Let the parametric form of the curve γ be

γ : x = f(t), y = g(t) (1.14)

We can look upon γ as a projection in the xy-plane of the curve Γ- defined ina three-dimensional xyφ-space by

Γ : x = f(t), y = g(t), h = h(t) (1.15)

where

h(t) = φ(f(t), g(t)) (1.16)

The coefficient functions in (1.4) are now all expressible in terms of a singleparameter t and we assume them to be analytic: in other words, differentiablesufficient numbers of times. The functions a, b and c are also assumed to bedifferentiable sufficient numbers of times.

At the point (x0, y0). equation (1.4) takes the form

a0(φx)0 + b0(φy)0 = c0 (1.17)

where the suffix (0) indicates the value at (x0, y0). Next, differentiating (1.16)with respect to t and evaluating the resulting expression at (x0, y0) yields

(φx)0f′(t0) + (φy)0g

′(t0) = h′(t0) (1.18)

where the prime stands for differentiation with respect to t.It is clear that a unique solution of (φx)0 and (φy)0 would emerge from

(1.17) and (1.18) provided the discriminant

∆ ≡ a0g′(t0)− b0f ′(to) 6= 0 (1.19)

The condition (1.19) is crucial for the solvability of the Cauchy problem.We also see that if we eliminate (φx)0 from (1.17) and (1.18) we obtain

[a0g′(t0)− b0f ′(to)](φy)0 = a0h

′(t0)− c0f ′(to) (1.20)

Thus ∆ = 0 is equivalent to

∆ = 0 : a0h′(t0)− c0f ′(to) = 0 (1.21)

as well.

[a, b], f(t) and g(t) being continuous functions of t. A simple arc is one in which no point onit corresponds to two different values of t. If f(t) and g(t) are also continuously differentiable(with the derivatives being one-sided at the end points) then the simple arc is termed aregular arc.


We therefore have for the vanishing of the discriminant the following cri-terion

∆ = 0 :f ′(t0)

a0=g′(t0)

b0=h′(t0)

c0(1.22)

which from (1.15) points to the subsidiary or auxiliary equations

∆ = 0 :dx

a=dy

b=dφ

c(1.23)

These auxiliary equations amount to a set of ODEs.

We can state equivalently that the curve Γ = [x(t), y(t), h(t)] satisfies theODEs namely,

dx

dt= a(x(t), y(t)),

dy

dt= b(x(t), y(t)),

dφ

dt= c(x(t), y(t)) (1.24)

We call this curve an integral curve for the vector field having the compo-nents (a, b, c). The integral curves are known as the characteristic curves for(1.4). The notable feature of a characteristic curve Γ is that it would supportinfinitely many solutions for the Cauchy problem. A unique solution of theCauchy problem will only exist as long as Γ is non-characteristic.

The characteristic curves are obtained by solving (1.24). The projectedcurves on the xy- plane is determined by solving the differential equationdydx = b(x,y)

a(x,y) . Union of the characteristic curves gives the integral surface. For-

mally, it corresponds to assembling the surface φ = φ(x, y) through each pointof the characteristics. The following examples will make the point clear.

Example 1.7

Find the integral lines and integral surface of the one-dimensional trans-port equation

φτ + cφx = 0

where c is a constant. The initial condition is provided by φ(x, 0) = f(x).Here the corresponding equations to (1.24) are

dτ

dt= 1,

dx

dt= c,

dφ

dt= 0


Their solutions are rather simple

x(t) = ct+ λ, τ(t) = t+ µ, z(t) = ν

where λ. µ and ν are constants of integration. From the first two equationswe can get rid of the parameter t to have x− cτ = k, k = λ− cµ, which alongwith z(t) = ν give the integral curves as straight lines in the three-dimensionalxyφ-space. The integral surface is a collection of these lines. Note that alongthese lines the solution φ(x, τ) is constant (which can be checked by takingthe τ - derivative: dφ

dτ = cφx + φτ = 0) and we may set φ(x, τ) = f(x − cτ)which gives the general solution.

Example 1.8

Discuss the general solution of the quasi-linear equation

(y + xφ)φx + (x+ yφ)φy = φ2 − 1

Find the integral surface through the parabola x = t, y = 1, φ = t2.Here the equations corresponding to (1.24) are

dx

y + xφ=

dy

x+ yφ=

dφ

φ2 − 1

By summing and subtracting the first two equalities and equating with thethird we can easily deduce

d(x+ y)

x+ y=

dφ

φ− 1

and

d(x− y)

x− y=

dφ

φ+ 1

Solving the above two differential equations we find for the respectiveequations

φ = 1 + λ(x+ y), φ = −1 + µ(x− y)

where λ and µ are two arbitrary constants of integration.The general solution is therefore given by

F (φ− 1

x+ y,φ+ 1

x− y) = 0

where F is an arbitrary function of its arguments.


To get the integral surface we note that for the given parabola the constantλ and µ turn out to be

λ = t− 1, µ =t2 + 1

t− 1

which on elimination of t implies that λ and µ are subjected to the constraint

λµ = (1 + λ)2 + 1

Hence using the relations for λ and µ we find for the integral surface theequation

(φ− 1)(φ+ 1)

(x+ y)(x− y)= (

φ− 1

x+ y+ 1)2 + 1

That is

(φ+ x+ y − 1)2 + (x+ y)2 =(x+ y)(φ2 − 1)

x− yExample 1.9

Find the integral surface of the following quasi-linear equation

(y − φ)∂φ

∂x+ (φ− x)

∂φ

∂y= x− y

which goes through the curve φ = 0, xy = 1 and through the circle x+y+φ =0, x2 + y2 + φ2 = a2.

Here the equations corresponding to (1.24) are

dx

y − φ=

dy

φ− x=

dφ

x− y

By summing the numerator and denominator each fraction turns out to be

=dx+ dy + dφ

0

On the other hand, if we multiply the numerator and denominator of the firstfraction by x, the second fraction by y and the third fraction by φ and sumthem then each fraction takes the form

=xdx+ ydy + φdφ

0


It at once follows that

dx+ dy + dφ = 0⇒ x+ y + φ = λ

and

xdx+ ydy + φdφ = 0⇒ x2 + y2 + φ2 = µ

where λ and µ are two arbitrary constants of integration.

Therefore the general solution is

F (x+ y + φ, x2 + y2 + φ2) = 0

where F is an arbitrary function of its arguments.

On the given curve φ = 0 and xy = 1, let us parametrize the latter byx = t, y = 1

t . Then the two constraints above read

t+1

t= λ, t2 +

1

t2= µ

and we have the relation

µ = λ2 − 2

This implies

x2 + y2 + φ2 = (x+ y + φ)2 − 2

On simplification the integral surface is

xy + yφ+ xφ = 1

Sometimes solving in the parametric form proves tedious. Consider the fol-lowing example.


Example 1.10

Find the general solution and the characteristic curves of the followingPDE

φφy + 2yφx = 2yφ2

Here the corresponding equations to (1.4) are

dy

dt= φ,

dx

dt= 2y,

dφ

dt= 2yφ2

Here if one tries to eliminate the parameter t from the first and third equationone runs into a nonlinear differential equation y = 2yy2, where the dot refersto a derivative with respect to t. We therefore take an alternative approach.We see that the first two equalities give dy

dx = φ2y while the first and third

equalities lead to dφdy = 2yφ. While the solution of the second differential

equation gives φ = λey2, the first one gives (xφ + 1)e−y2

= µ when the lastsolution is used. Note that λ and µ are arbitrary constants.

The general solution is thus

F ((xφ+ 1)e−y2

, φe−y2

) = 0

implying

φ(x, y) = ey2

f((xφ+ 1)e−y2

)

where f is an arbitrary function and the characteristics are φe−y2

= constantand (xφ+ 1)e−y

2

= constant.We conclude this section by making a few remarks on the Taylor series ex-

pansion of the function φ(x, y). Expanding about a point t = t0 correspondingto which x0 = f(t0) and y0 = g(t0) we can write

φ(x, y) = φ0+[(x− x0)(φx)0 + (y − y0)(φy)0]+

+1

2![(x−x0)2(φxx)0+2(x−x0)(y−y0)(φxy)0 + (y−y0)2(φyy)0]+ ...

(1.25)

If the condition (1.19) holds, in other words, Γ is non-characteristic at thepoint (x0, y0), then all the terms in the right side (1.25) corresponding to thefirst and higher order derivatives of φ at t = t0 can be uniquely determinedfrom the given values on γ.


In fact on partially differentiating with respect to x we have at t = t0

(φxx)0f′(t0) + (φxy)0g

′(t0) = φ′x(t0) (1.26)

where the prime denotes a derivative with respect to t.

Next, differentiating (1.4) partially with respect to x

aφxx + bφxy = Q(x, y, φ, φx, φy) (1.27)

where Q stands for

Q(x, y, φ, φx, φy) = cx + cφφx − (ax + aφφx)ux − (bx + bφφx)φy (1.28)

At t = t0, (1.27) yields

a0(φxx)0 + b0(φxy)0 = Q(x0, y0, φ0, (φx)0, (φy)0) (1.29)

Since Γ is non-characteristic at t = t0 and in its neighbourhood, we canuniquely determine the quantities (φxx)0 and (φxy)0 from (1.26) and (1.29).Note that (φyy)0 is uniquely determined by taking a derivative partially withrespect to y instead of x and derive analogous equations to (1.27) and (1.29)containing the quantity (φyy)0 . Higher derivatives at t = t0 are estimatedby proceeding in a similar manner. Thus a unique solution of (1.4) can beobtained by forming a Taylor expansion for φ(x, y) which assumes given valueson a non-characteristic curve γ.

1.5 Second order PDEs

We give now some typical examples of PDE which we often come acrossin problems of mathematical physics. These are

(1) Poisson’s equation and Laplace’s equation (electrostatics, Newtoniangravity)

∇2φ(r, t) = −(4π)ρ

where ρ is a charge distribution indicating the presence of a source term. Withρ 6= 0, Poisson’s equation is an inhomogeneous PDE.

When ρ = 0, we get the Laplace’s equation

∇2φ(r, t) = 0


Laplace’s equation appears in problems of electrostatics, Newtonian gravityand hydrodynamics.

(2) Wave equation (acoustics, electrodynamics)

∇2φ(r, t)− 1

c2φtt(r, t) = 0

The vibration of a string of constant density under constant tension is guidedby an equation of the above form. In acoustics, c stands for the velocity ofsound while in electrodynamics, c refers to the velocity of light.

(3) Helmholtz’s equation (vibrating string problem)

(∇2 + k2)φ(r, t) = 0

where k is a constant. It appears straightforwardly when the wave equationis subjected to a Fourier transform with respect to the time variable. Separa-tion of variable when applied to the wave equation also results in Helmholtz’sequation to which we will come later.

(4) Heat conduction equation (equalization of energy process)

∇2φ(r, t)− 1

kφt(r, t) = 0

where k is called the temperature conductivity. It has relevance in the equal-ization of energy process.

(5) Schrodinger equation (non-relativistic quantum mechanics)

i~∂ψ(r, t)

∂t= [− ~2

2µ∇2 + V (r, t)]ψ(r, t)

where ~(= h2π ) is the reduced Planck’s constant, ψ is the wave function

that guides the quantum system and V is the potential energy. Schrodingerequation provides a description of the dynamics of a micro-particle in non-relativistic situations. It is also used to solve for the allowed energy levels ofthe particle.

1.6 Higher order PDEs

(1) Korteweg-de Vries equation (shallow water wave problem)

φt = 6φφx − φxxx


It is a nonlinear PDE and tells of an interesting balance between nonlinearityand dispersion to cause appearance of soliton-like solutions.

(2) Fourth order diffusion equation

φt = −φxxxx

Fourth order PDEs appear in many places of physical problems such asin thin film models, surface diffusion on solids, etc. Perhaps the simplest ex-ample of a fourth order diffusion equation is the above form where the minussign makes it a dissipative equation. Such a PDE is generated by the energy

functional H =∫ baφ2xdx on L2((a, b)). Solving a forth or higher order PDE

analytically is not always an easy task because of the difficulty in implement-ing the boundary conditions which are more in number as compared to thesecond order equations. Thus, one often has to take recourse to computationalcalculations.

From a class of solutions that a PDE may enjoy, imposition of boundaryconditions or initial conditions or both picks out the one which is relevant tothe problem at hand. A PDE when subjected to initial conditions constitutesan initial value problem while if boundary conditions are imposed, we havea boundary value problem. If both types of conditions are necessary then wespeak of an initial-boundary value problem. One has to take care that adequatenumbers of such conditions are prescribed otherwise we may not arrive at anysolution at all or run into too many solutions. The solution we generally lookfor needs to have a unique character, depend continuously on the prescribeddata (otherwise the solution will not be stable) and has to be well posed.

For the Poisson or Laplace’s equation, time is not involved and so we havea boundary value problem at disposal. We normally take for the hypersurfacea closed curve or surface (see Figure 1.2). The type of boundary value problemis distinguished by the nature of data that is prescribed. For instance, if φ isprescribed on the boundary of a closed curve or surface it is a Dirichlet’sproblem. On the other hand, if the normal derivative of φ is prescribed onthe boundary of a closed curve or surface, we run into a Neumann problem.Sometimes a problem requires specification of the function φ on certain partof the hypersurface and the normal derivative of φ in the remaining part ofthe hypersurface. In such a case we have a mixed or Robin condition.

There is also an additional possibility when both φ and ∂φ∂n are specified

on the hypersurface. We then have the Cauchy problem. In fact, for the waveequation which is an evolution equation, time is an open boundary and so oneneeds both φ and its derivative at an initial time. To these are to be supple-mented by the values of φ at the boundary which is closed (see Figure 1.3).

For the heat conduction or diffusion equation time is an open boundary too.However, it is a first order equation and so the initial value of φ is sufficient.Of course since the spatial boundary is closed φ needs to be prescribed (seeFigure 1.4).


FIGURE 1.2: A closed curve with a boundary.

In general it is not possible to solve a Dirichlet problem for the wave equa-tion nor a Cauchy problem for the Laplace or Poisson equation. This is a no-gosituation. These will be illustrated with examples in Chapter 2.

1.7 Cauchy problem for second order linear PDEs

Let us adopt the following form of a second order linear PDE

Aφxx + 2Bφxy + Cφyy +Dφx + Eφy + Fφ = 0 (1.30)

where the coefficients A, B, C, D, E, F are functions of the two indepen-dent variables x and y possessing sufficiently many derivatives. Let us use thenotations

p = φx, q = φy, r = φxx, s = φxy, t = φyy (1.31)

This puts the above PDE in the following form

Ar + 2Bs+ Ct = Φ (1.32)

where Φ contains at most first order partial derivative terms.


FIGURE 1.3: Open time boundary for the wave equation.

Suppose we are given a curve Γ along which both φ and its normal deriva-tive ∂φ

∂n are prescribed at some initial time. The Cauchy problem asks thequestion whether a solution of the PDE (1.30) exists that satisfies these ini-tial conditions. Since knowing φ on Γ implies that the partial derivative withrespect to the arc length ∂φ

∂s is known, we can therefore claim2 that along itφ as well as its partial derivatives φx and φy are also known.

The differentials which follow from (1.31) namely

dp = rdx+ sdy, dq = sdx+ tdy (1.33)

also hold on Γ. From (1.32) and (1.33) we thus have three equations which,in principle, should determine r, s and t unless the coefficient determinantgiven by

∆ = Ady2 − 2Bdxdy + Cdy2 (1.34)

vanishes.The vanishing of ∆ however always takes place if the derivative dy

dx obeys

dy

dx=B ±

√J

AC, J = B2 −AC (1.35)

From (1.35) we can determine two directions for which ∆ = 0. These repre-sent two families of curves which are called the characteristic curves for thesecond order system. Thus the solvability of the problem demands that Γ

2See the discussion in Chapter 2.


FIGURE 1.4: Open time boundary for the heat conduction equation.

shall nowhere be a tangent to the characteristic. Should this be so (i.e. Γ isa non-characteristic) we can undertake a Taylor series expansion as we did inthe first order case and evaluate all the terms corresponding to various ordersof derivatives of φ arising in it at some given point in an arbitrarily chosenneighbourhood in the xy-plane and thus uniquely determine the solution.

A second order linear PDE is classified by the sign of the quantity J .When J < 0, the PDE is of elliptic type and the characteristics are conjugatecomplex in character.

Real characteristics forming two distinct families of curves appear whenJ > 0. The PDE is then classified as of hyperbolic type.

Only one real family of a characteristic curve exists when J = 0 in whichcase the PDE is said to be of parabolic type.

In general, A, B and C are functions of x and y. So in the xy-plane therecould be various regions where different types of characteristics occur for agiven PDE.

The method of characteristics is a powerful tool for solving a PDE. Let usillustrate it by considering the following examples.

Example 1.11

Find the characteristics of the following PDE

φxx(x, y) + xyφyy(x, y) = 0

Hence transform the equation in terms of the characteristic variable.


This is a typical PDE in which it is of one type ( hyperbolic) in a certainregion of xy-plane and different type (elliptic) in another region of xy-plane.In fact, since here A = 1, B = 0 and C = xy, the quantity J is

J = −xy

signifying that the PDE is hyperbolic in the second and fourth quadrants ofthe xy-plane but elliptic in the first and third quadrant of the xy-plane.

The characteristic curves follow from the equation

(dy

dx)2 + xy = 0 → dy

dx= ±√−xy

We first focus on the hyperbolic case and take the second quadrant (x < 0,y > 0):

dy

dx=√−x√y

Simple integration furnishes the solution

y12 − 1

3(−x)

32 = constant

We now set

ξ = y12 , η =

1

3(−x)

32

which implies

η + ξ = constant, η − ξ = constant

In terms of ξ and η the quantities φxx and φyy can be evaluated using thechain rule of differentiation:

φxx = −x4

(φηη +φη3η

), φyy =1

4y(φξξ −

φξξ

)

Hence the given PDE becomes in the hyperbolic case

φηη − φξξ +φη3η

+φξξ

= 0

in terms of ξ and η variables. As we will learn in a later chapter this refers tothe normal form of the equation in terms of the characteristic variables (ξ, η).

In a similar way, for the elliptic case we set

ξ = y12 , η =

1

3(x)

32

which implies


η + iξ = constant, η − iξ = constant

Proceeding as in the hyperbolic case we get for the elliptic counterpart thecorresponding normal form

φηη + φξξ +φη3η− phiξ

ξ= 0

in terms of ξ and η variables.

Example 1.12

Solve the two-dimensional Laplace’s equation

φxx(x, y) + φyy(x, y) = 0

within a square of length a subject to the boundary conditions

φ(0, y) = 0, φ(a, y) = 0, φ(x, 0) = 0, φ(x, a) = φ0 sinπx

a

With the conditions given on the boundary of the square, let us note thatthe problem is a well-posed one. There can be various methods to solve it butlet us proceed with the method of characteristics.

Since the equation is of elliptic in nature the characteristics are conjugatecomplex with the derivative dy

dx satisfying

dy

dx= 0

Indeed these provide the characteristic variables

ξ = x+ iy = λ, η = x− iy = µ

where λ and µ are constants. In terms of ξ and η the Laplace’s equationassumes the normal form

∂2φ

∂ξ∂η− 0

Its solution can be expressed in terms of two arbitrary functions χ and ψ i.e.φ(ξ, η) = χ(ξ) + ψ(η). Projecting in terms of the variables x and y we canexpress φ as

φ(x, y) = χ(x+ iy) + ψ(x− iy)


To be justified as a solution of the Laplace’s equation we need to fix χ andψ properly. Keeping in mind the boundary conditions, the following periodicchoice is natural

φ(x, y) = ik(cos[π

a(x+ iy)]− cos[

π

a(x− iy)])

where k is an overall constant. We keep the real part

φ(x, y) = k sin(πx

a) sinh(

πy

a)

All except the boundary conditions φ(x, a) = φ0 sin πxa are fulfilled. We come

to terms with this one by fixing k = φ0

sinhπ . It gives the required solution

φ(x, y) =φ0

sinhπsin(

πx

a) sinh(

πy

a)

Example 1.13

Find the characteristics of the parabolic equation

φt(x, t) = α2φxx(x, t)

where α is a constant. Transform the equation in terms of the characteristicvariable.

Here A = α2, B = 0 = C implying J = 0. Therefore the characteristicequation is

dt

dx= 0

We thus see that the t = c lines represent the characteristics. We can transformthe variables according to

ξ = t, η = x

and notice that the equation is already in normal form.


1.8 Hamilton-Jacobi equation

In this section we focus on the occurrence of a typical linear first order PDEwhich is of immense utility in problems of classical mechanics. It is called theHamilton-Jacobi equation. As is well known the analytical structure of classi-cal dynamics describing the motion of a particle or a system of particles restson two important but almost equivalent principles namely, those of Lagrangianand Hamiltonian mechanics. While the Lagrangian for system of n degrees offreedom is a function of the generalized coordinates qi, i = 1, 2, ..., n and gen-eralized velocities qi, i = 1, 2, ..., n and appears as the difference of kinetic andpotential energies of the particles, the Hamiltonian, which can be derived fromthe Lagrangian by effecting a Legendre transformation, is regarded as a func-tion of the generalized coordinates qi, i = 1, 2, ..., n and generalized momentapi, i = 1, 2, ..., n. It appears, in a conservative system, as the sum of kineticand potential energies. The underlying equations of motion are given by a setof n first order partial differential equations each for the time derivative of thegeneralized coordinates and momenta. The Hamiltonian procedure is carriedout in a phase space which is 2n-dimensional being made up of positions andthe accompanying canonical momenta. The Hamiltonian equations give theclue as to how a system evolves in phase space. Both Lagrangian and Hamil-tonian methods address the laws of mechanics without any preference to theselection of any particular coordinate system.

Use of Hamilton’s equations does not always offer any simplification tothe understanding of the concerned dynamics. But the almost symmetricalappearance of the coordinates and momenta in them facilitates developmentof formal theories such as the canonical transformations and Hamilton-Jacobiequation. We will discuss, in the following, how canonical transformations areconstructed that offer considerable simplifications in writing down the equa-tions of motion. The task is to transform the Hamiltonian to a new formthrough the introduction of a new set of coordinates and momenta such thatwith respect to them the form of canonical equations is preserved. The ad-vantage of employing canonical transformation lies in the fact that it is oftenpossible to adopt new sets of conjugate variables through which the basicequations get more simplified thus generating solutions which may otherwisebe very complicated to determine. The special case when all the coordinatesare cyclic yields a single partial differential equation of first order which iscalled the Hamilton-Jacobi equation.


1.9 Canonical transformation

For simplicity we will be frequently restricting to a system of one degreeof freedom described by a single pair of canonical variables (q, p). The phasespace is then of dimension two. The Hamiltonian equations are given by

q =∂H

∂p, p = −∂H

∂q(1.36)

We ask the question whether a transformation from (q, p) to a new set ofvariables (Q,P ) is feasible in a way that the above form of the Hamiltonianequations is preserved

Q =∂K

∂P, P = −∂K

∂Q(1.37)

We emphasize that our intention in looking for new variables (Q,P ) is toinquire if these could also serve in a new but simplified means of describingthe system but at the same time not disturb the essential physics content. In(1.36), K is a transformed version of H

H(q(Q,P ), p(Q,P ))→ K(Q,P ) ≡ K (1.38)

defined in terms of the variable Q and P . It is obvious that the equations(1.36) may not always hold except for some special situations. Such restrictedtransformations for which these equations hold are called canonical transfor-mations implying that the new variables (Q,P ) too form a canonical set.

Noting that the Poisson bracket of any two functions F and G is definedto be

F,G(q,p) =∂F

∂q

∂G

∂p− ∂F

∂p

∂G

∂q(1.39)

where (q, p) is the set of canonical variables, Hamiltonian equations for Q =Q(q, p), P = P (q, p) read

Q = Q,H(q,p) =∂Q

∂q

∂H

∂p− ∂Q

∂p

∂H

∂q

P = P,H(q,p) =∂P

∂q

∂H

∂p− ∂P

∂p

∂H

∂q(1.40)

Interpreting H(q, p) as K(Q,P ) in terms of the new variables (Q,P ) andemploying the chain rule of partial derivatives gives

∂H

∂q=

∂K

∂Q

∂Q

∂q+∂K

∂P

∂P

∂q

∂H

∂p=

∂K

∂Q

∂Q

∂p+∂K

∂P

∂P

∂p(1.41)


As a result we are led to the following expressions for Q and P

Q =∂Q

∂q

(∂K

∂Q

∂Q

∂p+∂K

∂P

∂P

∂p

)− ∂Q

∂p

(∂K

∂Q

∂Q

∂q+∂K

∂P

∂P

∂q

)=

∂K

∂P

(∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q

)=

∂K

∂PQ,P(q,p)

P = −∂K∂QQ,P(q,p) (1.42)

These equations coincide with (1.37) provided we set Q,P(q,p) equal tounity:

Q,P(q,p) = 1 (1.43)

Hamiltonian forms of canonical equations are then also valid for the new pairof variables (Q,P ).

For a system of n degrees of freedom the condition for the canonical trans-formation is given by the Poisson bracket conditions

Qi, Qj(q,p) = 0, Pi, Pj(q,p) = 0, Qi, Pj(q,p) = δij (1.44)

The invariance of Poisson bracket relations is a fundamental feature of canon-ical transformations. In fact the above properties of the Poisson brackets serveas the necessary and sufficient conditions for a transformation to be canonical.Note that out of 4n variables, (qi, pi) and (Qi, Pi), i = 1, 2, ..., n, only 2n ofthese are independent.

It needs to be pointed out that the Poisson bracket Q,P(q,p) is the sameas the Jacobian determinant :

Q,P(q,p) =∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q

=∂(Q,P )

∂(q, p)(1.45)

Conversely we also have

q, p(Q,P ) =∂(q, p)

∂(Q,P )=

[∂(Q,P )

∂(q, p)

]−1

=[Q,P(q,p)

]−1(1.46)

Consider a region D in the phase space plane (q, p) that is bounded by aclosed curve A. Using (1.45) we can express


∫D

dQdP =

∫D

∂(Q,P )

∂(q, p)dqdp =

∫D

[Q,P ]q,pdqdp =

∫D

dqdp (1.47)

where we have exploited (1.43). In terms of closed line integrals we then haveby Stokes’s theorem ∮

A

pdq =

∮A

PdQ (1.48)

In general, for a system with n generalized coordinates qi together withtheir associated momenta pi, the volume element transforms as

n∏i=1

dQidPi = ∆n∏i=1

dqidpi =n∏i=1

dqidpi (1.49)

where ∆ stands for the determinant of the Jacobian matrix

∆ = det[∂(Q1, Q2, ..., Qn;P1, P2, ..., Pn)

∂(q1, q2, ..., qn; p1, p2, ..., pn)] (1.50)

and we employed Liouville’s theorem which points to the preservation of vol-ume in the phase space. Thus phase space volume remains invariant undercanonical transformations.

Example 1.14

Consider the transformation (q, p)→ (Q,P ) given by

q =

√2P

mωsinQ, p =

√2mωP cosQ

Inverting

Q = tan−1

(mω

q

p

), P =

1

2ω

(p2

m+mω2q2

)⇒ ∂Q

∂q=

mωp

p2 +m2ω2q2,

∂Q

∂p= − mωq

p2 +m2ω2q2

As a result

Q,P(q,p) =∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q

=1

p2 +m2ω2q2

(mωp

∂P

∂p+mωq

∂P

∂q

)


Since ∂P∂p = p

mω ,∂P∂q = mωq it follows that Q,P(q,p) = 1. Hence the trans-

formation is canonical.

Example 1.15: Harmonic oscillator problem

Let us now concentrate on the specific case of the harmonic oscillatordescribed by the Hamiltonian H = 1

2mp2 + 1

2mω2q2. It is clear from the form

of P in the previous example that if employed the Hamiltonian takes a verysimple form H → K = Pω, where K is the transformed Hamiltonian. Theaccompanying Hamilton’s equations for the new variables Q and P followeasily

Q =∂K

∂P= ω

P = −∂K∂Q

= 0

Solving for Q and P we find Q = ωt + t0, P = b where t0 and b areconstants of integration. Switching to the original variables we get for q and p

q =

√2b

mωsin (ωt+ t0)

p =√

2mωb cos (ωt+ t0)

which conform to their standard forms. This example serves to illustrate theadvantage of employing canonical variables judiciously that allow the basicequations to get simplified and so obtaining the solutions becomes an easytask.

1.10 Concept of generating function

That the quantity (pdq − PdQ) is an exact differential can be establishedby noting

pdq − PdQ = pdq − P(∂Q

∂qdq +

∂Q

∂pdp

)=

(p− P ∂Q

∂q

)dq − P ∂Q

∂pdp (1.51)


from which the condition for an exact differential namely,

∂

∂p

(p− P ∂Q

∂q

)=

∂

∂q

(−P ∂Q

∂p

)(1.52)

holds because of Q,P(q,p) = 1.Setting pdq − PdQ = dG1, we call G1 to be the generating function of

the transformation (q, p) → (Q,P ). Such a class of generating functions isreferred to as the Type I generating function. G1 being a function of q and Q[i.e. G1 = G1(q,Q)] gives

dG1 =∂G1

∂qdq +

∂G1

∂QdQ (1.53)

Matching with the left hand side of (1.51)

p =∂G1

∂q, P = −∂G1

∂Q(1.54)

Consider the case of Example 1.14. We first of all check whether pdq−PdQis a perfect differential for this problem. For the generating function G1, theindependent variables are q and Q. Expressing p and P in terms of thesenamely, p = mωq cotQ, P = 1

2mωq2cosec2Q we can express

pdq − PdQ = d

(1

2mωq2 cot θ

)(1.55)

which is indeed an exact differential. Hence the generating function G1 forthis problem is G1(q,Q) = 1

2mωq2 cot θ.

There can be other types of generating functions depending on what com-binations of old and new canonical variables we choose. For instance a Type IIgenerating function G2 is a function of q and P and the counterpart of (1.54)reads

p =∂G2

∂q, Q =

∂G2

∂P(1.56)

For the Type III generating function G3 which is a function of p and Qthe underlying relations are

q = −∂G3

∂p, P = −∂G3

∂Q(1.57)

Finally, a Type IV generating function G4 depends on the pair (p, P ) withthe complementary variables q and Q satisfying

q = −∂G4

∂p, Q = −∂G4

∂P(1.58)


The transformations given in (1.54), (1.56), (1.57) and (1.58) hold for thetime-independent canonical transformations. These are restricted canonicaltransformations. In the time-dependent case, which we discuss in the next sec-tion, the generating functions possess additionally a function of the variable t.

1.11 Types of time-dependent canonical transformations

Consider a transformation (q, p, t)→ (Q,P, t) for a general time-dependentsituation. Since adding a total differential does not change the essential dy-namics of the system, we define a time-dependent canonical transformation(q, p, t)→ (Q,P, t) according to the following criterion:

n∑i=1

pidqi −Hdt =n∑i=1

PidQi −Kdt+ dG1 (1.59)

with G1 = G1(q,Q, t) to be the time-dependent Type I generating function ofthe transformation. The time-independent generating function was consideredearlier from the exact differentiability of the quantity (pdq − PdQ).

We now turn to four kinds of canonical transformations as induced by thecorresponding time-dependent generating functions.

1.11.1 Type I Canonical transformation

A Type I canonical transformation treats the old and new coordinates qiand Qi as independent variables. For G1(qi, Qi, t) we have

dG1 =

n∑i=1

(∂G1

∂qidqi +

∂G1

∂QidQi

)+∂G1

∂tdt (1.60)

Putting this form in (1.59) we find on comparing the differentials

pi =∂G1(qi, Qi, t)

∂qi, Pi = −∂G1(qi, Qi, t)

∂Qi(1.61)

along with

K = H +∂G1

∂t(1.62)

The first of (1.61) gives Qi in terms of qi and pi which when substituted inthe second equation determines Pi. We of course assume that the Jacobian of

the transformations det | ∂2G1

∂qi∂Qj| 6= 0). K is the new Hamiltonian.


It is worthwhile to note that the Type I canonical transformation inducesthe exchange transformation. For instance if we choose G1(q,Q, t) = qiQi thenpi and Pi turn out to be Qi and -qi respectively while K = H reflects thatthe coordinates and momenta are exchanged.

1.11.2 Type II Canonical transformation

For a Type II canonical transformation the independent variables are qiand Pi which stand respectively for the old coordinates and the new momenta.In this case we express

n∑i=1

PidQi = d

(n∑i=1

PiQi

)−

n∑i=1

QidPi (1.63)

which results in

n∑i=1

pidqi +

n∑i=1

QidPi −Hdt = d

(n∑i=1

PiQi

)−Kdt+ dG1 (1.64)

The accompanying generating function G2 can therefore be defined by

G2 = G1 +n∑i=1

PiQi (1.65)

Viewing G2 as a function of qi, Pi and t we write

dG2(qi, Pi, t) =

n∑i=1

(∂G2

∂qidqi +

∂G2

∂PidPi

)+∂G2

∂tdt (1.66)

which implies from (1.59) and (1.65)

pi =∂G2(qi, Pi, t)

∂qi, Qi =

∂G2(qi, Pi, t)

∂Pi(1.67)

along with

K = H +∂G2

∂t(1.68)

The first of (1.67) gives Pi in terms of qi and pi which when substituted inthe second equation provides for Qi. The Jacobian of the transformation is

assumed to be nonvanishing: det | ∂2G2

∂qi∂Pj| 6= 0 . K as given by (1.68) is the

transformed Hamiltonian.


The Type II canonical transformation has in its embedding both the iden-tity as well as the point transformations. In fact, corresponding toG2(q, P, t) =qiPi we see that pi = Pi, Qi = qi and K = H showing for the identity trans-formation while for G2(q, P, t) = φ(qi, t)Pi we have Qi = φ(qi, t) implying thatthe new coordinates are functions of old coordinates.

1.11.3 Type III Canonical transformation

In Type III canonical transformation the underlying generating functionG3 is a function of the independent variables pi, Qi which are respectively theold momenta and new coordinates. G3 is defined by

G3(pi, Qi, t) = G1 −n∑i=1

qipi (1.69)

which leads to

qi = −∂G3

∂pi, Pi = −∂G3

∂Qi, K = H +

∂G3

∂t(1.70)

Here the Jacobian of the transformation is assumed to be nonvanishing:

det | ∂2G3

∂pi∂Qj| 6= 0.

1.11.4 Type IV Canonical transformation

In the Type IV canonical transformation the generating function is a func-tion of old and new momenta and given by

G4(pi, Pi, t) = G1 +

n∑i=1

QiPi −n∑i=1

qipi (1.71)

in which pi and Pi are treated as independent variables. We then have

qi = −∂G4

∂pi, Qi =

∂G4

∂Pi, K = H +

∂G4

∂t(1.72)

Here the Jacobian of the transformation is assumed to be nonvanishing:

det | ∂2G4

∂pi∂Pj| 6= 0.

Example 1.17

Consider the transformation

Q = −p, P = q + λp2


where λ is a constant. By the Poisson bracket test namely [Q,P ](q,p) = 1 weconclude that the transformation is a canonical transformation.

The Type I generating function is determined by showing pdq − PdQ tobe a perfect differential:

pdq − PdQ = (−Q)dq − (q + λQ2)dQ

= d(−qQ− 1

3λQ3)

Thus

G1(q,Q) = −qQ− 1

3λQ3

On the other hand the Type II generating function is obtained by treatingq and P as independent variables. From (1.65) we have

G2(q, p) = G1 + PQ

= −qQ− 1

3λQ3 + (q + λQ2)Q

=2

3λQ3

=2

3λ

(P − qλ

) 32

Check that (∂G2

∂q

)P

= − 1√λ

(P − q) 12 = p(

∂G2

∂P

)q

=1√λ

(P − q) 12 = Q

which are as required.

Example 1.18

The transformations

Q = q cos θ − p

mωsin θ

P = mωq sin θ + p cos θ

are easily seen to be canonical due to Q,P(q,p) = 1. We also find

p = mω

(q cot θ − Q

sin θ

)P = mω

( q

sin θ−Q cot θ

)


Hence the quantity pdq − PdQ can be expressed as

pdq − PdQ = d

[1

2mω(q2 +Q2) cot θ −mωqQcosecθ

]The Type I generating function G1(q,Q) is identified as

G1(q,Q) =1

2mω(q2 +Q2) cot θ −mωqQcosecθ

On other other hand, the Type II generating function can be obtainedfrom G2 = G1 + PQ. We get

G2(q,Q) =1

2mω(q2 −Q2) cot θ

To assign the right variable dependence on G2 namely q and P we note that

Q =q

cos θ− P

mωtan θ

by eliminating p from the given transformations. Substituting for Q, G2 turnsout to be

G2(q, P ) =qP

cos θ− 1

2mω

(q2 +

P 2

m2ω2

)tan θ

We can verify that (∂G2

∂q

)P

=P

cos θ−mωq tan θ = p(

∂G2

∂P

)q

=q

cos θ− P

mωtan θ = Q

Example 1.19

We consider the harmonic oscillator Hamiltonian H which is invariantunder the set of canonical transformations considered in the previous example:

H(q, p) =1

2mp2 +

1

2mω2q2

→ 1

2mP 2 +

1

2mω2Q2 = H(Q,P )

The generating function G2(q, P ) which was found to be

G2(q, P ) =qP

cos θ− 1

2mω

(q2 +

P 2

m2ω2

)tan θ


gives for a partial derivative with respect to time(∂G2

∂t

)q,P

=

[qP sin θ − 1

2mω

(q2 +

p2

m2ω2

)]sec2 θθ

= −(P 2

2mω+

1

2mωQ2

)θ

= −H(Q,P )θ

ω

Therefore the transformed Hamiltonian reads

K(Q,P, t) =

(1− θ

ω

)H(Q,P )

which vanishes for θ = ω i.e. θ = ωt. As a consequence Q = ∂K∂P = 0 and

P = −∂K∂Q = 0. Hence Q and P are constants which we write as

Q = Q0, P = P0

where Q0 and P0 stand for the initial values of q and p respectively.Reverting to the old coordinates we get

q(t) = Q0 cosωt+P0

mωsinωt

p(t) = −mωQ0 sinωt+ P0 cosωt

which give the time evolution for q and p.

1.12 Derivation of Hamilton-Jacobi equation

The idea of deriving Hamilton-Jacobi equation rests on applying a canoni-cal transformation that maps one basis of known canonical variables to a newset of coordinates and momenta such that the latter are cyclic with respectto the transformed Hamiltonian. What does it mean?

Consider the specific case of a Type II generating function G2(qi, Pi, t), i =1, 2, ..., n which facilitates the transformations according to (1.67) where thetime derivatives of Qi and Pi stand as

Qi =∂K

∂Pi, Pi = − ∂K

∂Qi, i = 1, 2, ..., n (1.73)


as defined by (1.37). Now if Qi and Pi are cyclic coordinates in the transformedHamiltonian K, then it follows that these have to be constants in time. Sucha validity is assured if we assume, without loss of generality, K to be zero. Itthen gives from (1.68) the equation

H(q1, q2, ..., qn; p1, p2, ..., pn; t) +∂G2

∂t= 0 (1.74)

A few remarks are in order:From (1.66) if we take the time-derivative of the generating function

G2(qi, pi, t) then in view of the above equation it follows that we can write

dG2

dt=

n∑i=1

∂G2

∂qiqi =

n∑i=1

piqi −H(qi, pi, t) = L (1.75)

where we employed (1.67) and noted that Pi’s are cyclic coordinates in Kand hence treated as constants. Thus integrating between times say, t1 andt2, the quantity S ≡ G2 defined by the action integral S =

∫ t2t1Ldt follows

from (1.75). We therefore arrive at the form

pi =∂S

∂qi: H(q1, q2, ..., qn;

∂S

∂q1,∂S

∂q2, ...,

∂S

∂qn; t) +

∂S

∂t= 0 (1.76)

Equation (1.76) is called the time-dependent Hamilton-Jacobi equation.3 Morecompactly it is also expressed as

H(qi,∂S

∂qi, t) +

∂S

∂t= 0 (1.77)

A function S which is a solution of the above equation is often referred to asthe Hamilton Principal function.

Equation (1.77) is first order differential and involves the n coordinatesqi’s and t. It is not linear in ∂S

∂qibecause the partial derivatives of S appear in

higher degree than the first. Associated with the (n+ 1) variables (n qi’s andt) we expect (n+ 1) constants of motion namely α1, α2, ..., αn, αn+1. Howeverwe notice one curious thing in (1.77) is that the dependent variable S itselfdoes not appear in it: only its partial derivatives do. So one of the constantshas no bearing on the solution i.e. the solution has an additive constant.Disregarding such an irrelevant additive constant we write for the completeintegral of (1.77) the form

S = S(q1, q2, ..., qn;α1, α2, ..., αn; t) (1.78)

3There could be other variants of Hamilton-Jacobi equation resulting Type I, Type IIIand Type IV generating functions.


where it is ensured that none of the constant α’s is additive to the solution. Ac-tually we identify these constants with the constant momenta Pi, i = 1.2..., n,which, according to our choice, were assumed to be cyclic.

The cyclic coordinates Qi’s provide another set of constants β′is, i =1, 2, ..., n and read from the second equation of (1.67)

Qi =∂S

∂αi= βi, i = 1, 2, ..., n (1.79)

We clarify the above issues by considering two examples: one for the free par-ticle and the other that of the harmonic oscillator.

Example 1.20: Free particle case

The Hamilton-Jacobi equation for the free particle problem is obviously

1

2m(∂S

∂q)2 +

∂S

∂t= 0

Its complete integral can be easily ascertained by inspection which reads

S(q, α, t) =√

2mαq − αtwhere α is a non-additive constant.

From (1.79)

β =∂S

∂α= k (say)

i.e. k =

√m

2αq − t ⇒ q =

√2E

m(t+ k) and p =

∂S

∂q=√

2mα.

which conform to their expected forms.It is to be noted that Hamilton-Jacobi equation being a partial differential

equation can admit of multiple solutions. For instance, in the above case, thereis also a legitimate solution given by

S(q, α, t) =m(q − α)2

2t

where α is a non-additive constant. This implies

β =∂S

∂α= −m

t(q − α)

giving the time-dependence: q(t) = α − βm t where α can be identified as the

initial value of q while β is the negative of the momentum: β = −p.


Example 1.22

For the harmonic oscillator problem the Hamilton-Jacobi equation has theform

H =1

2m

(p2 +m2ω2q2

)The relation p = ∂S

∂q gives

1

2m

[(∂S

∂q

)2

+m2ω2q2

]+∂S

∂t= 0

To solve for the above equation we try separation of variables for S:

S(q : α; t) = W (q : α)− αt

where α is a non-additive constant.The function W (q : α) is the time-independent part called the Hamilton’s characteristic function. We get

1

2m

[(∂W

∂q

)2

+m2ω2q2

]= α

and gives the integral

W =√

2mα

∫dq

√1− mω2q2

2α

It implies for β

β =∂S

∂α=

√m

2α

∫dq√

1− mω2q2

2α

− t

which easily integrates to

t+ β =1

ωsin−1 q

√mω2

2α

Hence the solutions for q and p are

q =

√2α

mω2sinω(t+ β)

p =∂S

∂q=∂W

∂q=√

2mα−mω2q2 =√

2mα cosω(t+ β)

which are agreement with their well known forms.


1.13 Summary

In this chapter we began by familiarizing the readers with some of the back-ground formalities of a PDE with a view to fixing the notations and explainingthem by accompanying examples. We showed how a PDE can be generated bysome elementary means namely, elimination of constants or elimination of anarbitrary function. We pursued in some detail the linear first order equation,in particular we dealt with the quasi-linear case. We then gave some examplesof second and higher order PDE and discussed the Cauchy problem at length.The method of characteristics was illustrated by means of examples. Finally,from an application point of view, we discussed the canonical transformationsalong with the Hamilton-Jacobi equation which is a linear first order PDEand of immense importance in classical mechanics.


Exercises

1. Find the general solution of the PDE

xφφx + yφφy + (x2 + y2) = 0

2. Find the general solution of the linear equation

(y − z)∂φ∂x

+ (z − x)∂φ

∂y+ (x− y)

∂φ

∂z= 0

3. Find the general solution of the linear equation

2xφx + 2yφφy = φ2 − x2 − y2

Also determine the integral surface which passes through the circle

x+ y + φ = 0, x2 + y2 + z2 = a2

where a is constant.

4. Solve the following PDE

(φ+ y + z)φx + (φ+ z + x)φy + (φ+ x+ y)φz = x+ y + z

by employing Lagrange’s method.

5. Consider the quasi-linear equation

x(y2 + z)φx − y(x2 + z)φy = (x2y2)φ

Determine the integral surface which passes through the line x+ y = 0, z = 1.

6. Show that integral surface of the PDE

(x− y)y2φx + (y − x)x2φy = (x2 + y2)φ

which contains the straight line x+ y = 0, φ = 1 is given by the form f(x2 +yφ, x2 + y2) = 0, where f is an arbitrary function.


7. Show that the integral surface of the PDE

x(y2 + φ)φx − y(x2 + φ) = (x2 − y2)φ

which passes through the curve xφ = a3, y = 0 is given by the formφ3(x3 + y3)2 = a9(x− y)3, where a is a constant.

8. Show that a local solution of the partial differential equation φt+aφx =φ2 subject to φ(x, 0) = cos(x), where a is a constant can be written as

φ(x, t) =cos(x− at)

1− t cos(x− at)

9. Find the characteristic curves and the general solution of the linearequation

φx + exφy + ezφz = (2x+ ex)eφ

10. Find the characteristics of the following PDE

φxx + yφyy = 0

If the characteristic variables are ξ and η for the hyperbolic case show thatthe transformed equation is

φξη +1

η − ξ(φη − φξ) = 0

On the other hand, if the characteristic variables are σ and β for the ellipticcase show that the transformed equation is

φσσ + φββ −1

βφβ = 0

Explain the significance of the singularity at η = ξ for the hyperbolic case andβ = 0 for the elliptic case.

Chapter 2

Basic properties of second orderlinear PDEs

2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.2 Reduction to normal or canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . 472.3 Boundary and initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

(a) Different types of boundary and initial value problems . . . . . . 60(b) Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Dirichlet problem on a rectangular domain:0 ≤ x ≤ a, 0 ≤ y ≤ b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Neumann problem on a rectangular domain:0 ≤ x ≤ a, 0 ≤ y ≤ b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64Cauchy initial value problem on an interval: 0 ≤ x ≤ l, t > 0 . . . 64(c) Well-posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Hadamard example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2.4 Insights from classical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.5 Adjoint and self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Two-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742.6 Classification of PDE in terms of eigenvalues . . . . . . . . . . . . . . . . . . . 752.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

2.1 Preliminaries

Given a domain Ω in the n-dimensional Euclidean space consisting ofpoints x1, x2, ..., xn, the following functional representation

F (x1, x2, ..., xn;φ;φx1 , φx2 , ..., φxn ;φx1x1 , φx1x2 , ....φx1xn , ..., ) = 0,

x1, x2, ..., xn(≡ x) ∈ Ω (2.1)

furnishes a PDE with respect to an unknown function φ(x). The order of thehighest derivative occurring in (2.1) determines the order of the PDE.

Introducing a multi-index notation an m-th order PDE can be presentedin the form

F (x;φ;Dmφ,Dm−1φ, ...,Dφ) = 0, x ∈ Ω (2.2)

45


in which the differential operator Dm stands for the successive partial deriva-tives

Dm ≡ ∂m11

∂xm11

...∂mnn∂xmnn

≡ ∂|m|

∂xm11 ...∂xmnn

, xm = xm11 ...xmnn

where m1, ...,mn are non-negative integers and |m| = m1 + ... + mn. As atypical example, if m = (4, 1, 2) then |m| = 4 + 1 + 2 = 7, m! = 4!1!2! = 48

and Dm = ∂7

∂4x1∂x2∂2x3= ∂4

x1∂x2

∂2x3

.A PDE expressed as

Lφ = ρ(x, y)

is linear when the operator L involves only the first powers of the partialdifferential operators ∑

|k|≤m

pk(x)Dkφ(x) = ρ(x)

where the coefficient functions pk(x)’s are assumed to be known. As alreadynoted earlier, a inhomogeneous PDE is characterized by the non-vanishing ofthe term ρ(x, y). Otherwise it is a homogeneous equation.

An arrangement of (2.2) in the form

∑|k|=m

pk(x)(Dm−1φ, ...,Dφ, φ, x)Dkφ+ p0(x)(Dm−1φ, ...,Dφ, φ, x) = 0

defines a quasi-linear equation where it is clear that the highest-derivativesappear only linearly, whatever the functional nature of their coefficients.

A semi-linear equation, on the other hand, has the feature

∑|k|=m

pk(x)Dkφ+p0(x)(Dm−1φ, ...,Dφ, φ, x)Dkφ+p0(Dm−1φ, ...,Dφ, φ, x) = 0

which has a quasi-linear structure but the coefficients of the highest-derivativesare restricted to be functions of the multi-index vector x only. Typical exam-ples of a quasi-linear and semi-linear PDE have already been given in chap-ter 1.

Any deviation from a linear character of a PDE would classify the equationas a nonlinear equation. The second order Monge-Ampere PDE

det | ∂2φ

∂xi∂xj|+

∑i,j=1

Bij(x, φ,Dφ)∂2φ

∂xi∂xj+ C(x, φ,Dφ) = 0

Basic properties of second order linear PDEs 47

where the coefficients Bij and C are functions of x, φ and Dφ with Dφ =(D1φ, ...,Dnφ), Di = ∂

∂xi, i = 1, 2, ..., n, is an example1 of a nonlinear PDE.

This equation has much relevance in optical transport problems. However, itssevere nonlinear character makes an analytic determination of its solutionsrather a formidable task2. To seek a solution for it one often has to takerecourse to a numerical treatment.

Expressed in terms of two independent variables x and y, the Monge-Ampere PDE takes the simplified form

φxxφyy − φ2xy = ξφxx + 2ηφxy + ζφyy + χ

where ξ, η, ζ and χ are functions of x, y, φ along with the first derivatives ofφ: φx and φy.

In the rest of this chapter we will be interested mostly in the linear PDEsof two independent variables. As we already know from Chapter 1 the operatorL for a first order linear PDE has the form

Lφ ≡ a(x, y)φx + b(x, y)φy + c(x, y)φ (2.3)

the coefficients a, b, c being assumed to be continuous differentiable functionsof x and y while for the second order linear PDE L has a general representation

Lφ ≡ A(x, y)φxx+2B(x, y)φxy+C(x, y)φyy+D(x, y)φx+E(x, y)φy+F (x, y)φ(2.4)

where A, B, C, D, E and F are assumed to be sufficiently smooth and con-tinuously differentiable functions of x and y in a certain given domain.

2.2 Reduction to normal or canonical form

We focus on the second order linear PDE in two independent variablesx and y where L is defined by (2.4). The second order linear PDEs can beclassified much in the same way as is done when dealing with different types ofconics in two-dimensional coordinate geometry. Recall that the three types of

1C. E. Gutierrez, The Monge–Ampere Equation, Progress in Nonlinear Differential Equa-tions and Their Applications, vol. 44, Birkhauser, Basel, 2001.

2See, for instance, K. Brix, Y. Hafizogullari and A. Platen, Solving the Monge–Ampereequations for the inverse reflector problems, arXiv:1404.7821 /math.NA


conics are the hyperbola, ellipse and parabola according to whether the signof the underlying discriminant3 is positive, negative or zero.

The relevant discriminant for the second order linear PDEs is the quantityJ defined by

J(x, y) = B2(x, y)−A(x, y)C(x, y) (2.5)

A second order linear PDE Lφ = 0 can thus be classified to be of hyperbolictype, elliptic type or parabolic type in an arbitrary chosen neighbourhood inthe xy-plane according to the signs of J . The different types of classificationsgo as follows:

B2 −AC < 0 : elliptic

B2 −A > 0 : hyperbolic

B2 −AC = 0 : parabolic (2.6)

It is evident that with A,B,C depending on the independent variables x and y,the same PDE can reflect different characters for different neighbourhoods inthe xy-plane. We already encountered, in the previous chapter, the three sim-plest representative equations of the elliptic, hyperbolic and parabolic equa-tions as respectively the Laplace’s equation, the wave equation and the heatconduction equation. One might be curious to know about the character ofthe nonlinear Monge-Ampere equation involving two independent variables.We just mention in passing that its type4 depends on the sign of the quan-tity Q ≡ ξζ + η2 + χ. Specifically, if Q > 0 then it is elliptic, if Q < 0 it ishyperbolic and if Q = 0 it is parabolic.

We now take up the issue of the reduction of a PDE to its normal form orcanonical form. To this end, we ask the question that, given a regular curveγ in the xy-plane which is referred to as a boundary curve, if both φ and itsderivative ∂φ

∂n in a direction normal to the curve are prescribed, can we find asolution of L(φ) = 0 near the curve?

To proceed with this aim we note, first of all, that if φ(x, y) is knownalong the boundary curve (or on some portion of it) where both x and y areparametrized by the same parameter s, where we identify s as the arc lengthmeasuring the distance along the curve (see Figure 2.1), then the derivative∂φ∂s in the tangent direction to the boundary curve is known too. From ∂φ

∂n

and ∂φ∂s we can therefore determine both the partial derivatives φx and φy as

explained below.

3S.L.Loney, The Elements of Coordinate Geometry, Arihant Publications (2016).4Encyclopedia of Mathematics, Springer, The European Mathematical Society.


FIGURE 2.1: The boundary curve γ parametrized by s.

From the the theory of vector analysis, a vector having components(dxds ,

dyds ) is tangent to the boundary curve which has a perpendicular vector

having components (−dyds ,dxds ). We can therefore write

∂φ

∂n= (φx, φy).(−dy

ds,dx

ds) = −φx

dy

ds+ φy

dx

ds(2.7)

Also

∂φ

∂s= φx

dx

ds+ φy

dy

ds(2.8)

where the quantities φx and φy are evaluated corresponding to x = x(s) andy = y(s). From the above equations we therefore solve for φx and φy at some

required point (x0, y0) in terms of the derivatives ∂φ∂s and ∂φ

∂n known there.

Turning to the derivatives of φx and φy we use the chain rule of calculusto write down

d

dsφx = φxx

dx

ds+ φxy

dy

ds(2.9)

d

dsφy = φxy

dx

ds+ φyy

dy

ds(2.10)

From the above two equations and the given L(φ) = 0 in which the secondorder partial derivatives also appear, it is evident that after arranging themappropriately we can always solve for the quantities φxx, φxy, φyy uniquely


unless the determinant of the coefficient vanishes:

4 ≡

∣∣∣∣∣∣dxds

dyds 0

0 dxds

dyds

A B C

∣∣∣∣∣∣ = 0

∆ when expanded yields

A dy2 − 2B dx dy +C dx2 = 0→ A(dy

dx)2 − 2B

dy

dx+C = 0, A 6= 0 (2.11)

It yields two families of curves each with its own constant of integration.Indeed there exist two directions given by the corresponding values of dy

dx forwhich two families of curves (real or conjugate complex) exist. Such curvesare called the characteristic curves. Of course on such curves (or lines) thequantity 4 vanishes i.e. 4 = 0 and so as a necessary condition of solvabilityof the second derivatives φxx, φxy, φyy at a particular point the boundarycurve must nowhere be tangent to a characteristic at that point. It shouldbe remarked that if A = 0 but C 6= 0 then we should go for the equation indxdy (as worked out in Example 2.5). Of course if both A = 0 and C = 0 wealready are in a reduced form. The relevance of the characteristic curves tothe propagation of singularities is discussed in Example 2.1 in the context ofthe one-dimensional wave equation.

Thus to know the values of the second partial derivatives at a particularpoint it is necessary to demand that 4 6= 0. To find the third order andhigher partial derivatives we have to carry out repeated operations of furtherdifferentiations. Assuming the Taylor series to be convergent we can projectthe solution of φ(x, y) about the point (x0, y0) in the manner

φ(x, y) = φ0+[(x− x0)(φx)0 + (y − y0)(φy)0]+

+1

2![(x−x0)2(φxx)0 + 2(x−x0)(y−y0)(φxy)0 + (y−y0)2(φyy)0] + ...

(2.12)

It is worthwhile repeating that prescribing data on a characteristic curve (onwhich ∆ = 0) has generally no meaning since it does not lead to a viablesolution. It is only when the curve γ is non-characteristic (∆ 6= 0) that thedata on it uniquely determine the second and higher-derivatives appearing inthe Taylor expansion of φ(x, y) about some suitable point of expansion.

We now examine the following examples.


Example 2.1

Consider the one-dimensional wave equation

φxx −1

c2φtt = 0

Here A = 1, B = 0 and C = − 1c2 . The characteristics therefore obey the

equationsdx

dt= ±c

and thus turn out to be straight lines

x− ct = ξ = constant

x+ ct = η = constant

A sketch of the family of lines is shown. See Figure 2.2.

FIGURE 2.2: The characteristics lines.

The general solution of the wave equation is given by the sum of left andright travelling wave solutions

φ(x, t) = f(x+ ct) + g(x− ct)

If the initial values are assigned as

φ(x, 0) = 0, φt(x, 0) = δ(x)


where δ(x) is Dirac’s delta function, then remembering that the deriva-tive of the Heaviside function is the delta function (see equation (A.91) inAppendix A), the resultant solution which obey these conditions is easilyworked out to be

φ(x, t) =1

2c[H(x+ ct)−H(x− ct)]

where H is the Heaviside function. The above solution of φ serves5 as an inter-esting example of propagating singularity showing propagation to the left andright. In fact, for a hyperbolic equation every singularity (i.e. discontinuitiesof the boundary values) is continued along the characteristics and isolatedsingularities are not admitted. In contrast for an elliptic type of PDE thepropagation of singularities can only be thought as being carried out in animaginary domain.

Example 2.2

Consider the one-dimensional heat conduction equation

φt = α2φxx

The characteristic curves are given by dt2 = 0 or t =constant which areparallel lines as shown the Figure 2.3.

FIGURE 2.3: The characteristics lines.

As already touched upon in the previous chapter we can effectively usethe characteristic variables to convert the PDE in (x, y) variables to a special

5Jeffrey Rauch, Lecture notes on Hyperbolic PDEs and geometrical optics, AmericanMathematical Society (2012).


form called the normal or canonical form. In the present context we solve thecondition 4 = 0 as furnished by (2.11) to obtain the pair of solutions

y± =

∫x

dy

dxdx =

∫x

(B ±

√B2 −ACA

) dx (2.13)

These provide the two equations of characteristics which are functions of xand y when the given forms of A, B and C are substituted in (2.13) andintegration is carried out.

Let us express the characteristic equations according to the prescriptions

ξ(x, y) = constant, η(x, y) = constant (2.14)

Our task would be to transform the independent variables (x, y) in terms ofthe new ones (ξ, η). With a non-vanishing Jacobian of the transformatiomwhich reads

4 ≡∣∣∣∣ ξx ξyηx ηy

∣∣∣∣ 6= 0

a one-to-one relation between the old and new variables is assured.

The first and second order derivatives of φ can be transformed in terms ofthe variables ξ and η by employing the chain rule of partial derivatives. Wefind in this way for the first derivatives

φx = φξξx + φηηx, φy = φξξy + φηηy (2.15)

and subsequently for the second derivatives the set of relations

φxx = φξξξ2x + 2φξηξxηx + φηηη

2x + ... (2.16)

φxy = φξξξxξy + φξη(ξxηy + ξyηx) + φηηηxηy + ... (2.17)

φyy = φξξξ2y + 2φξηξyηy + φηηη

2y + ... (2.18)

where the dots in the right side of each equation represent the lower orderderivative terms.

Hence, in terms of the variables (ξ, η), the second order partial derivativeterms present in (2.4) go over to the form

Aφxx + 2Bφxy + Cφyy → Aφξξ + +2Bφξη + Cφηη

where the new coefficients A, B and C are given by


A = Aξ2x + 2Bξxξy + Cξ2

y , B = Aξxηx +B(ξxηy + ξyηx) + Cξyηy,

C = Aη2x + 2Bηxηy + Cη2

y (2.19)

Notice that since

B2 − AC = (ξxηy − ξyηx)2(B2 −AC) (2.20)

the intrinsic character of the PDE is not changed under the transformation(x, y)→ (ξ, η).

For the characteristic family given by ξ(x, y) = constant we have

ξxdx+ ξydy = 0 → dy

dx= −ξx

ξy(2.21)

which transforms the condition (2.11) to

Aξ2x + 2Bξxξy + Cξ2

y = 0 (2.22)

This implies from (2.19) that the coefficient A of φξξ vanishes.

Similarly for the other characteristic family given by η(x, y) = constantone runs into the form

Aη2x + 2Bηxηy + Cη2

y = 0 (2.23)

which points to the vanishing of the coefficient C of φηη in (2.19).

Combining the above two results one arrives at the following canonicalform of the hyperbolic equation

φξη = f(φ, φξ, φη, ξ, η) (2.24)

where f is an arbitrary function of its arguments. This form corresponds tothe case B 6= 0 and A = 0 = C which ensures that the hyperbolic conditionB2 −AC > 0 is automatically fulfilled.

On the other hand, the condition B2 − AC > 0 is also met through hav-ing B = 0 and A = −C. The manifestation of this case comes through thetransformations

ξ = x+ y, η = x− y (2.25)


which provides a different canonical form of the hyperbolic equation namely,

φxx − φyy = g(φ, φx, φy, x, y) (2.26)

where g is the transformed function of its arguments.

We need not deal with the elliptic case separately for it follows from thehyperbolic case on employing the complex transformations

ξ = x+ iy, η = x− iy (2.27)

Indeed (2.27) results in

φxx + φyy = h(φ, φx, φy, x, y) (2.28)

where h is the transformed function of its arguments. It is clear that in theelliptic case the characteristics are conjugate complex. Example 1.12 workedout in the previous chapter will help to clear up this point.

For the parabolic case, we set η = x but keep ξ free. In such a situationthe left side of the relation (2.22) becomes a perfect square i.e.

(Aξx +Bξy)2 = 0 (2.29)

Here we employed the parabolic condition B2 −AC = 0.

Now with the help of the relations (2.16), (2.17) and (2.18) we find

Aφxx + 2Bφxy + Cφyy = 2φξη(Aξx +Bξy) +Aφηη = Aφηη (2.30)

where we have used (2.29). As a result we get the parabolic normal form

φηη = G(φ, φξ, φη, ξ, η) (2.31)

where G is an arbitrary function of its arguments.

The following examples serve to illustrate the reduction of a PDE to itsnormal form.


Example 2.3

Reduce the following PDE to its normal form

yφxx + (x+ y)φxy + xφyy = 0

Here A = y, B = x+y2 and C = x. As a result

B2 −AC =(x− y)2

4> 0

Hence the equation is hyperbolic except on the line x = y on which it isparabolic. In the following we focus on the hyperbolic case.

The characteristic curves are obtainable from (2.11) which gives

dy

dx=x

yor 1

They therefore correspond to the equations x2 − y2 = constant and x − y =constant. Let us set

ξ = x2 − y2, η = x− y

Transforming to the variables ξ and η we find for the first derivatives φxand φy

φx = 2xφξ + φη, φy = −2yφξ − φη

and for the second derivatives φxx, φxy and φyy

φxx = 4x2φξξ + 4xφξη + φηη − 2φξ

φxy = −4xyφξξ − 2(x+ y)φξη − φηηφyy = 4y2φξξ + 4yφξη + φηη − 2φξ

On substitution of the above results the given equation is converted to

(x− y)2φξη + (x+ y)φξ = 0

which implies that the normal form is

η3φξη + ξφξ = 0


Example 2.4

Solve the PDE

∂

∂y(φx + φ) + 2x2y(φx + φ) = 0

by transforming it to the normal form.The given equation reads

φxy + φy + 2x2y(φx + φ) = 0

Here A = 0, B = 12 and C = 0. Hence B2 −AC = 1

4 > 0 and so the equationis hyperbolic. The equations of the characteristics are given by

dy

dx= 0,

dx

dy= 0

and imply that y = constant and x = constant are the characteristic lines.We therefore put

ξ = x = constant, η = y = constant

Thus the given equation has the same form in terms of ξ and η variables:

∂

∂η(φξ + φ) + 2ξ2η(φξ + φ) = 0

Let us put

ζ = φξ + φ

which transforms the above equation to a rather simple representation

∂ζ

∂η+ 2ξ2ηζ = 0

Integrating partially with respect to η gives

ζ(ξ, η) = f(ξ)e−ξ2η2

where f is an arbitrary function of ξ.

Transforming back the φ variable and integrating partially with respect toξ we obtain


φ(ξ, η) = e−ξ[g(η) +

∫f(ξ)eξ−ξ

2η2dξ]

where g is an arbitrary function of η.

Example 2.5

Solve the following PDE by transforming it to the normal form

eyφxy − φyy + φy = 0

where it is given that at y = 0, φxy = −x2

2 and φy = − sinx.

Here A = 0, B = ey

2 and C = −1. Since A = 0, as noted before, we need

to consider the equation in dxdy which reads

A− 2B(dx

dy) + C(

dx

dy)2 = 0

We thus have

dx

dy(−2B + C

dx

dy) = 0

Plugging in the forms of B and C we find

dx

dy= 0,

dx

dy= −ey

The first equation imples x = constant and the second one on integrationgives x+ ey = constant. We therefore identify the ξ and η variables to be

ξ = ey + x− 1, η = x

where −1 has been inserted without any loss of generality.It is an easy task to work out the first derivative conversions

φx = φξ + φη, φy = eyφξ

and in consequence have the follwoing results for the second derivatives

φxx = φξξ + 2φξη + φηη, φxy = ey(φξξ + φξη), φyy = eyφξ + e2yφξξ


When substituted in the given equation we are led to the normal form

φξη = 0

Integrating easily we find the solution for φ to be

φ(ξ, η) = f(ξ) + g(η)

where f and g are two arbitrary functions of their arguments.In terms of the variables x and y we thus have

φ(x, y) = f(ey + x− 1) + g(x)

Employing the given conditions we can determine f and g. This gives thesolution

φ(x, y) = −x2

2+ cos(ey + x− 1)− cosx

We conclude this section by making a few remarks on a fluid motion wherewe can categorize the flow in the following way. For a steady viscous flow itis elliptic whereas it is parabolic for the unsteady case. For an inviscid flow,the elliptic or hyperbolic nature depends on the Mach number being lessthan or greater than one. However, for the unsteady inviscid flow it is alwayshyperbolic.

As an illustration let us consider a two-dimensional potential equation ofa steady, irrotational, inviscid flow6 in terms of the velocity potential Φ

(1− Φ2x

c2)Φxx − 2

ΦxΦyc2

Φxy + (1−Φ2y

c2)Φyy = 0

where c is the local speed of sound, Φx and Φy are the x and y components of

the flow velocity and the Mach number is M =

√Φ2x+Φ2

y

c . The square of thediscriminant of the equation is easily seen to be

Φ2xΦ2

y

c4− (1− Φ2

x

c2)(1−

Φ2y

c2) =

Φ2x

c2+

Φ2y

c2− 1 = M2 − 1

We therefore find that for a subsonic flow, when M < 1, the equation iselliptic while for a supersonic flow, when M > 1, the equation is hyperbolic.The parabolic character occurs when the sonic condition holds.

6R. Vos and S. Farokhi, Introduction to Transonic Aerodynamics, Fluid Mechanics andits Applications, Springer Science+Business Media, Dordrecht, 2015.


2.3 Boundary and initial value problems

(a) Different types of boundary and initial value problems

An extended form of elliptic PDE that includes as particular cases thePoisson equation, Laplace equation and Helmholtz equation is given by

−∇ · (α∇φ) + βφ = f (2.32)

where α, β and f are functions of the space variable ~r. In fact, (2.32) reducesto the Poisson’s form when α = 1, β = 0 and f = 4πρ while Laplace’s equa-tion follows when α = 1, β = 0 and f = 0. The case when α = 1, f = 0 andβ = −k2 corresponds to the Helmholtz’s form.

In a general way, let us consider a domain Ω specified by its boundary ∂Ωsuch that a solution φ of (2.32) has continuous second order derivatives inΩ and is also continuous in Ω, the closure of Ω, with the following conditionbeing valid on ∂Ω i.e.

∂Ω : aφ+ b∂φ

∂n= v (2.33)

where a, b and v are given continuous functions on ∂Ω and the derivative in(2.33) is along the outward drawn normal to ∂Ω. We read off from (2.33) threedistinct boundary conditions as given by the following possibilities

(i) If a = 1 and b = 0 then we have

∂Ω : φ = v (2.34)

It is referred to as the boundary condition of the first kind accompanying theboundary value problem of the first kind.

(ii) If a = 0 and b = 1 then

∂Ω :∂φ

∂n= v (2.35)

It is referred to as the boundary condition of the second kind accompanyingthe boundary value problem of the second kind.

(ii) If a ≥ 0 and b = 1 which implies


∂Ω : aφ+∂φ

∂n= v (2.36)

It is referred to as the boundary condition of the third kind7 accompanyingthe boundary value problem of the third kind.

Two standard types of boundary value problems that one meets in solvingphysical problems are those which go by the classifications of Dirichlet andthe Neumann types. In the Dirichlet’s case we need to find a solution φ whichhas continuous second order derivatives in Ω and also continuous in Ω. Weexpress this problem by

Ω : ∇2φ = f, ∂Ω : φ = g (2.37)

where g is a continuous function.

In the Neumann’s case we have to find a solution φ which has continuoussecond order derivatives in Ω and also continuous in Ω. We express it as

Ω : ∇2φ = f, ∂Ω :∂φ

∂n= h (2.38)

where h is a continuous function.

As opposed to an elliptic equation which addresses a boundary value prob-lem, the hyperbolic case calls for the specification of initial conditions. Theprototype of a hyperbolic PDE is the wave equation which in three dimensionsreads

∇2φ− 1

c2∂2φ

∂t2= 0 (2.39)

In a standard initial value problem, the dependent function and its time deriva-tive are specified on the boundary ∂Ω of some region Ω at some initial time.For instance, in a three-dimensional region (x, y, z) it is given that

φ(x, y, z, 0) = f(x, y, z), φt(x, y, z, 0) = g(x, y, z), on ∂Ω (2.40)

As we shall see later these conditions determine a unique solution to the waveequation.

7Such a boundary condition is also referred to as Robin’s type or mixed type.


An initial value problem as stated above defines a Cauchy problem for a sec-ond order linear PDE. As should be clear the related hypersurface in sucha case should nowhere be a characteristic surface for the solvability of theCauchy problem.

(b) Applications

We now turn to the two-dimensional Laplace’s equation

φxx + φyy = 0 (2.41)

Its solution is sought in the separable form

φ(x, y) = X(x)Y (y) (2.42)

Substitution in the Laplace equation gives

X ′′

X= −Y

′′

Y= k (2.43)

where k is a separation constant arising due to one side being a function of xwhile the other is a function of y and dashes refer to derivatives with respectto the relevant variable.

We can distinguish three cases of the solutions

(i) k = 0⇒ φ(x, y) = (ax+ b)(cx+ d)

(ii) k = p2 > 0⇒ φ(x, y) = (r epx + f e−px)(g cos(py) + h sin(py))

(iii) k = p2 < 0⇒ φ(x, y) = (j cos(px) + s sin(px))(l epy +m e−py)

where a, b, c, d, r, f, g, h, j, s, l, m are arbitrary constants.

With this little background let us first address the Dirichlet problem on arectangular domain.

Dirichlet problem on a rectangular domain: 0 ≤ x ≤ a, 0 ≤y ≤ b

Let us solve the two-dimensional Laplace’s equation (2.41) by imposingthe boundary conditions:

φ(a, y) = 0 = φ(x, b), φ(0, y) = 0, φ(x, 0) = f(x) (2.44)


In this problem the case (iii) above is relevant and so on applying the condi-tions φ(0, y) = 0 and φ(a, y) = 0 we are led to

φ(0, y) = 0⇒ j = 0φ(a, y) = 0⇒ s 6= 0, sin(pa) = 0⇒ p = nπ

a (n = 1, 2, ...)

Thus φ(x, y) takes the form

φ(x, y) =∞∑n=1

An sinnπx

a(lne

nπya +mne

−nπya ) (2.45)

where An is an overall coefficient and ln, mn, n = 1, 2, ... are constants.

We now exploit the condition φ(x, b) = 0 to have

φ(x, b) = 0 ⇒ mn = −lnenπba

e−nπba

(2.46)

As a result we can express φ(x, y) as

φ(x, y) =∞∑n=1

An sinnπx

asinh[

nπ

a(y − b)] (2.47)

where the coefficient An can be determined from the remaining conditionφ(x, 0) = f(x):

φ(x, 0) = f(x)⇒∞∑n=1

An sinnπx

asinh(−nπb

a) = f(x) (2.48)

On inversion we readily find

An =2

a

1

sinh(−nπba )

∫ a

0

f(x) sin(nπx

a) dx (2.49)

The final form of φ(x, y) satisfying all the given boundary conditions is givenby (2.47) where for An the integral provided by (2.49) holds.

Next we look at the analogous problem for the Neumann problem in arectangular domain.


Neumann problem on a rectangular domain: 0 ≤ x ≤ a, 0 ≤y ≤ b

Here the Laplace’s equation (2.41) is solved subject to the boundary con-ditions

∂φ(0, y)

∂x= 0,

∂φ(a, y)

∂x= 0;

∂φ(x, 0)

∂y= 0,

∂φ(x, b)

∂y= f(x) (2.50)

Here again case (iii) is relevant. Proceeding as in the previous case we findfor φ(x, y) the form

φ(x, y) = B0 +

∞∑n=1

Bn cosnπx

acosh

nπy

a(2.51)

where Bn is given by the integral

Bn =2

nπ

1

sinh(nπba )

∫ a

0

f(x) cosnπx

adx, (n = 1, 2, ...) (2.52)

It may be remarked that in an exterior boundary value problem for anelliptic type of equation, the problem is to find a solution in the region Ω′

exterior to Ω which is unbounded. In such a case, φ has to satisfy a conditionat infinity controlling the asymptotic behaviour of the solution.

Cauchy initial value problem on an interval: 0 ≤ x ≤ l, t > 0

We consider the one-dimensional wave equation

φtt − c2φxx = 0, 0 ≤ x ≤ l, t > 0 (2.53)

subject to the following Cauchy data

φ(x, 0) = g(x), φt(x, 0) = h(x), 0 ≤ x ≤ l (2.54)

We also implement a set of boundary conditions

φ(0, t) = 0, φ(l, t) = 0, t ≥ 0 (2.55)


Let us first of all write the general solution in the form

φ(x, t) =A0(t)

2+∞∑n=1

[An(t) cosnπx

l+Bn(t) sin

nπx

l] (2.56)

where An and Bn are appropriate coefficients. Because of the boundary con-dition φ(0, t) = 0, the quantities An, n = 0, 1, 2, ... have to vanish leaving uswith the representation

φ(x, t) =∞∑n=1

Bn(t) sinnπx

l(2.57)

In it the other boundary condition φ(l, t) = 0 automatically holds.Substituting in the given equation we therefore find that Bn’s obey the

differential equation

Bn(t) + (nπc

l)2Bn(t) = 0 (2.58)

where the overhead dots denote derivatives with respect to t. The generalsolution of Bn turns out to be

Bn(t) = cn sinnπct

l+ dn cos

nπct

l(2.59)

where cn and dn are arbitrary constants to be determined from suitable inputs.Towards this end we exploit the initial conditions:

g(x) = φ(x, 0) =∞∑n=1

Bn(0) sinnπx

l=∞∑n=1

dn sinnπx

l(2.60)

h(x) = φt(x, 0) =

∞∑n=1

Bn(0) sinnπx

l=

∞∑n=1

cnnπc

lsin

nπx

l(2.61)

We now multiply both sides of the above equations by sin mπxl and inte-

grate between 0 and l. Then by orthogonality

∫ l

0

g(x) sinmπx

ldx =

∫ l

0

∞∑n=1

dn sinnπx

lsin

mπx

ldx = dm(

l

2) (2.62)


∫ l

0

h(x) sinmπx

ldx =

∫ l

0

∞∑n=1

cnnπc

lsin

nπx

lsin

mπx

ldx = cm(

mπc

2) (2.63)

where we have used∫ l

0sin nπx

l sin mπxl dx = l

2 if n = m but 0 if n 6= m.

Hence the constants cn and dn are

dn =2

l

∫ l

0

g(x) sinnπx

ldx, cn =

2

nπc

∫ l

0

h(x) sinnπx

ldx (2.64)

With the knowledge of cn and dn, Bn is known from (2.59) which in turn givesφ(x, t) from (2.57).

(c) Well-posedness

A PDE is said to be well-posed if a solution to it exists (i.e. at least one), itis unique (i.e. at most one) and that it depends continuously on the data (i.e.a small change in the initial data produces a correspondingly small changein the solution). For the existence and uniqueness of the solution appropriateinitial and/or boundary conditions need to be imposed.

Take for instance the Laplace’s equation defined over a rectangular re-gion (a, b)

φxx + φyy = 0, 0 < x < a, 0 < y < b

subject to obeying the boundary conditions

φ(x, 0) = x2, φ(x, b) = x2 − b2, φ(0, y) = −y2, φ(a, y) = a2 − y2

It is straightforward to see that a solution exists in the form

φ(x, y) = x2 − y2

which satisfies the given boundary conditions and is unique as well. One cannotafford to do away with any of the boundary conditions as that would lead tomultiple solutions and the problem would be rendered ill-posed.

A few comments on the appropriateness of the boundary conditions are inorder by focusing on the following Hadamard’s example.


Hadamard example

It is evident that the solution of the PDE

φxx(x, t) + φtt(x, t) = 0 −∞ < x <∞, t > 0

subject to the homogeneous conditions

φ(x, 0) = 0 −∞ < x <∞,φt(x, 0) = 0 −∞ < x <∞

is given by

φ(x, t) = 0

In contrast the solution of the same PDE i.e.

ψxx(x, t) + ψtt(x, t) = 0, −∞ < x <∞, t > 0

which is elliptic and being subjected to Cauchy-like initial conditions

ψ(x, 0) = 0 −∞ < x <∞

ψt(x, 0) = ε sin(x

ε) −∞ < x <∞

where ε > 0 is a small quantity, has the solution

φ(x, t) = ε2 sin(x

ε) sinh(

t

ε)

It follows that while

|ψt(0)− φt(0)| = ε→ 0 as ε→ 0

the difference of the time-dependent solutions behaves

|ψ(t)− φ(t)| = ε2| sinh(t

ε)| → ∞ as ε→ 0

i.e. exponentially blows up. We thus see that a small change in the initial dataproduces a massive change in the solution. We conclude that there is a failureto depend continuously on the supplied data and the problem is ill-posed.

We remind the readers that if we had replaced the Laplace equation bythe hyperbolic wave equation

φtt(x, t)− φxx(x, t) = 0 −∞ < x <∞, t > 0

and subjected it to the homogeneous conditions

φ(x, 0) = 0 −∞ < x <∞φt(x, 0) = 0 −∞ < x <∞


a trivial solution would be

φ(x, t) = 0

However, the same form of the PDE i.e.

ψxx(x, t)− ψtt(x, t) = 0 −∞ < x <∞, t > 0

which is now hyperbolic be subjected to Cauchy-like conditions

ψ(x, 0) = 0 −∞ < x <∞

ψt(x, 0) = ε sin(x

ε) −∞ < x <∞

would result in the solution

ψ(x, t) = ε2 sin(x

ε) sin(

t

ε)

Comparing with previous case we find while

|ψt(0)− φt(0)| = ε→ 0 as ε→ 0

the difference of the time-dependent solutions behaves

|ψ(t)− φ(t)| = ε2| sin(t

ε)| ≤ ε2 as ε→ 0

We thus infer that if we made a small change in the initial data it would reflecta corresponding small change in the solution as well. Thus unlike the previouscase the problem here is well-posed.

We conclude this section explaining the perspective of the well knownCauchy-Kowalevski theorem that gives a sufficient condition for the local solv-ability of the Cauchy problem in some neighbourhood of an initial point. Weconsider a Cauchy system which consists of a set of (n+ 1)- variables distin-guished by a set of n spatial coordinates x1, x2, ..., xn and additionally a timeparameter t. In terms of these we look at the following set of PDEs in termsof the functions φ1, φ2, ..., φN written in a compact but self-evident form

∂niφi∂tni

= Φi(x1, x2, ..., xn; t;φ1, φ2, ..., φN ;Dkφj ; ...), 1 ≤ i, j ≤ N (2.65)

where in the left side there are partial derivatives of t of order ni operatingon φ1, φ2, ..., φN each of which is a function of the variables x1, x2, ..., xn andin the right side the Φi’s are arbitrary functions depending on the variablesx1, x2, ..., xn, the time t along with φ1, φ2, ..., φN . Adopting the multi-indexednotation introduced earlier in this chapter, the differential operator Dk de-notes


Dk ≡ ∂|k|

∂tk0∂xk11 ...∂xknn

(2.66)

where

|k| = k0 + k1 + ...+ kn ≤ nj , k0 < nj (2.67)

and that Dk involves involves the partial derivatives of t as well as of thevariables x1, x2, ..., xn as shown by the respective superscripts.

An an example of a Cauchy system we may think of the following PDE8

in two variables x and y

∂2φ

∂t2= t+ x2 − y2 + x

∂φ

∂x+ 2

∂2φ

∂t∂x− ∂2φ

∂x∂y+∂2φ

∂x2+∂2φ

∂y2(2.68)

One can see that in this scheme we have just one equation which means N = 1,n = 2 because the number of variables are three t, x and y, the order of thepartial-time derivative in the left side being two implies ni = 2. The highestderivative in the right side of the space variables is also two while the highestderivative of time there is one.

The following coupled system also explains the notations

∂3φ1

∂t3=∂2φ1

∂x21

− ∂3φ2

∂x21∂x2

+ x1x2∂3φ1

∂x1∂x2+ t2x1 (2.69)

∂2φ2

∂t2=∂2φ1

∂x22

+∂2φ2

∂x1∂t− ∂φ1

∂t− ∂φ2

∂t(2.70)

Here it is easy to identify that N = 2, n = 2 and n1 = 3, n2 = 2.

Coming back to the set of N PDEs at hand, let us suppose that they aresubjected to the initial conditions at t = t0

∂kφi∂tk|t=t0 = fi,k(x1, x2, ..., xn), i = 1, 2, ..., N (2.71)

At k = 0 we have the zeroth order partial derivative ∂0φi∂t0 |t=t0 which is taken

as the function of itself at (t0, x1, x2, ..., xn).Suppose now that the arguments of the functions Φ1,Φ2, ...,Φn are analytic

in an open region containing the point (t0, x01, x

02, ...x

0n), the arguments reading

explicitly

8P. V. O’Neil, Beginning Partial Differential Equations (1969), John Wiley and Sons,Inc., USA.


(x01, x

02, ..., x

0n; t0, ..., [D

k−k0φi,k0 ]k1=x01,...,kn=x0

n) (2.72)

and that around an open region of (x01, x

02, ..., x

0n) the functions fi,k are an-

alytic too, then the Cauchy-Kowalevski theorem states that the initial-valueproblem for the Cauchy system, as defined by (2.66) together with the condi-tions (2.72), admits of exactly one unique solution which is analytic around(x0

1, x02, ..., x

0n; t0).

2.4 Insights from classical mechanics

Let us consider the motion9 of a particle in one-dimension. We discuss twocases of the force acting on the particle.

• Elastic force −kx (k > 0)

The equation of motion can be written as

p = −kx where p = mx

The general solution is

x = a cos(ωt+ φ) p = −mωa sin(ωt+ φ)

where ω =√

km .

In the (x, p) phase plane the representative point traces out an ellipse

of semi-axes (√

2Ek ,√

2mE), E being the energy. The ellipse is a closed

curve. See Figure 2.4. In the theory of an elliptic PDE we already know thatfor a well meaning solution to exist we generally have either a Dirichlet orNeumann condition to be specified on the closed boundary.

• Repulsive force +kx (k > 0)

Here we have two sets of solutions :

x = ±a cosh(ωt) p = ±mωa sinh(ωt)

x = ±a sinh(ωt) p = ±mωa cosh(ωt)

The first case is for the particle with initial velocity towards the center of re-pulsion coming from x = ±∞ and stopping at ±a before rebounding. The energy

E = p2

2m− kx2

2= − ka

2

2is negative.

9M.G. Calkin, Lagrangian and Hamiltonian Mechanics, World Scientific Publishing Co.Pte. Ltd. (1996).


FIGURE 2.4: The ellipse.

The second case corresponds to E = + ka2

2which is positive. For t = 0 we have

x = 0, p = ±mωa, and the particle is with a finite velocity. Note that unlike ellipse,hyperbola is an open curve. See Figure 2.5. It is therefore clear why for the Cauchyproblem, in the hyperbolic case, time appears as an open boundary. A summary ofvarious features of L(φ) = 0 and the conic represented by the equation S(x, y) = 0is presented in Table 2.1 and Table 2.2 respectively.

TABLE 2.1: A summary of the various features of L(φ) = 0.

System Equation Discriminant SignaturePDE L(φ) = 0 B2 −AC (i) −

(function of E : ( ∂2

∂x2 + ∂2

∂y2 )φ = 0

x, y only) (Laplace equation)imaginary characteristics

Closed boundary(Dirichlet or Neumann)

(ii) +

H : ( ∂2

∂x2 − ∂2

∂y2 )φ = 0

(Wave equation)real characteristics

Open boundary(Cauchy)

(iii) 0

P : ∇2φ = α2 ∂φ∂t

(Heat conduction equation)real characteristicOpen boundary

(Dirichlet or Neumann)


FIGURE 2.5: The hyperbola.

TABLE 2.2: A summary of the various features of the conic equationS(x, y) = 0 where S(x, y) ≡ ax2 + 2bxy + cy2 + dx+ ey + f .

System Equation Discriminant SignatureConic S(x, y) = 0 b2 − ac (i) −

(constant) E: x2

a2 + y2

b2 = 1imaginary asymptotes

(ii) +

H: x2

a2 −y2

b2 = 1real asymptotes

(iii) 0P: y2 = 4ax

no asymptote


2.5 Adjoint and self-adjoint operators

We consider in a general way a second order linear operator L defined in termsof n independent variables

L =

n∑i=1

n∑j=1

Aij∂2

∂xi∂xj+

n∑i=1

Bi∂

∂xi+ C (2.73)

Note that the coefficient Aij is symmetric and we assume that Aij , Bi and C possesscontinuous second order derivatives to the independent variables (x1, x2, ..., xn).

Let φ and ψ be two arbitrary functions of the independent variables(x1, x2, ..., xn) which possess continuous derivatives. It then transpires that we canwrite

ψAij∂2φ

∂xi∂xj=

∂

∂xi(ψAij

∂φ

∂xj)− ∂

∂xj[φ

∂

∂xi(ψAij)] + φ

∂2

∂xi∂xj(ψAij) (2.74)

and

ψBi∂φ

∂xi=

∂

∂xi(Biψφ)− φ ∂

∂xi(ψBi) (2.75)

Taking now L between the functions φ and ψ gives

ψLφ = φ[

n∑i=1

n∑j=1

∂2

∂xi∂xj(ψAij)−

n∑i=1

∂

∂xi(ψBi) + Cψ]+

+

n∑i=1

∂

∂xi[

n∑j=1

(ψAij∂φ

∂xj−

n∑j=1

φ∂

∂xj(Aijψ) +Biφψ] (2.76)

where since Aij is symmetric i.e. Aij = Aji we have used

n∑i=1

n∑j=1

[∂

∂xjφ∂

∂xi(Aijψ)] =

n∑i=1

n∑j=1

[∂

∂xiφ∂

∂xj(Aijψ)] (2.77)

Calling

Mψ =

n∑i=1

n∑j=1

∂2

∂xi∂xj(ψAij)−

n∑i=1

∂

∂xi(ψBi) + Cψ (2.78)

we can express ψLφ as

ψLφ = φMψ +

n∑i=1

∂

∂xi[

n∑j=1

Aij(ψ∂φ

∂xj− φ ∂ψ

∂xj) + φψ(Bi −

n∑j=1

∂Aij∂xj

)] (2.79)


We thus have the Lagrange’s identity

ψLφ− φMψ =

n∑i=1

∂Pi∂xi

(2.80)

where Pi stands for

Pi =

n∑j=1

Aij(ψ∂φ

∂xj− φ ∂ψ

∂xj) + φψ(Bi −

n∑j=1

∂Aij∂xj

) (2.81)

The operator M is called the adjoint to L. Note that the right side of (2.80) looks like

a divergence. For instance, in two-dimensions, it looks like ( ∂Px∂x

+∂Py∂y

). However, thequantities Px and Py are functions of φ and ψ rather than the vector components.Still, the divergence form enables us to add and subtract to Px and Py respectivelythe quantities −Fy and Fx, where F is an arbitrary function of (φ, ψ, x, y), withoutchanging the right side of (2.80).

If L is a self-adjoint operator then L = M . In such a case we can express

Mφ =

n∑i=1

n∑j=1

∂2

∂xi∂xj(φAij)−

n∑i=1

∂

∂xi(φBi) + Cφ = Lφ+R (2.82)

where R is

R = −2

n∑j=1

∂φ

∂xi(Bi −

n∑j=1

∂Aij∂xj

)− φn∑j=1

∂

∂xi(Bi −

n∑j=1

∂Aij∂xj

) (2.83)

It enables us to conclude that the necessary and sufficient condition for the operatorL to be self adjoint is that R vanishes which amounts to Bi being in the form

Bi =

n∑j=1

∂Aij∂xj

(2.84)

This means that L is expressible as

L =

n∑i=1

n∑j=1

∂

∂xi(Aij

∂

∂xj) + C (2.85)

Two-dimensional case

In the two-dimensional case corresponding to

Lφ = Aφxx + 2Bφxy + Cφyy +Dφx + Eφy + Fφ (2.86)


the expression for its adjoint is

Mψ = (Aψ)xx + 2(Bψ)xy + (Cψ)yy − (Dψ)x − (Eψ)y + Fψ (2.87)

where we have identified A11 = A, A12 = B = A21, A22 = C, B1 = D, B2 =E, C = F . Also the quantities Px and Py are given by

Px = A(ψφx − φψx) +B(ψφy − φψy) + (D −Ax −By)φψ,

Py = B(ψφx − φψx) + C(ψφy − φψy) + (E −Bx − Cy)φψ (2.88)

Example 2.6

Given the L-operator to be L = ∂2

∂x2− p(x) ∂

∂x+ q(x), where p(x) and q(x) are

twice continuously differentiable functions of x, find its adjoint M .

Comparing with the definition of the L− operator given in the previous example,here A = 1, D = −p, F = q and B = C = E = 0. Hence from the expression of Mwe have

Mψ = ψxx + (pψ)x + qψ

Explicitly the M-operator reads

M =∂2

∂x2+ p(x)

∂

∂x+ px + q(x)

2.6 Classification of PDE in terms of eigenvalues

If we focus on the second order terms of the L-operator (i.e. its principal part)we can identify

A = A11, B = A12 = A21, C = A22 (2.89)

Then the corresponding 2× 2 matrix is(A BB C

)whose eigenvalues λ1, λ2 are determined from the vanishing of its underlying deter-minant ∣∣∣∣ A− λ B

B C − λ

∣∣∣∣ = 0


These are

λ1, λ2 =A+ C ±

√(A− C)2 + 4B2

2(2.90)

which are clearly real because of the symmetric nature of the matrix. Their product is

λ1λ2 = −(B2 −AC) = −J (2.91)

If J > 0 then the roots λ1 and λ2 are non-zero and are of opposite sign. ThePDE is of hyperbolic type.

If J < 0 then the roots λ1 and λ2 are non-zero and are of same sign. The PDEis of elliptic type.

If J = 0 then one of the roots λ1 and λ2 is zero and the PDE is of parabolic type.

In general, for a nth order linear PDE we have to take into consideration itshomogeneous part which corresponds to a real symmetric matrix. If we denote byP its number of positive roots and Z its number of zero roots, the classification ofPDE goes as follows:

If Z = 0 and P = 1 or Z = 0 and P = n− 1, the PDE is of hyperbolic type.If Z > 0 and P = 0 or Z = 0 and P = n, the PDE is of elliptic type.If Z = 0, the PDE is of parabolic type.

Additionally if Z > 0 and 1 < P < n, the PDE is of ultra-hyperbolic type. It isevident that for a PDE to be of ultra-hyperbolic type, it must involve at least fourindependent variables.

Example 2.7

The PDE

φxx + φyy − φzz − φtt − φww = 0

where (x, y, z, t, w) is a set of independent variables in <5, is ultra-hyerbolic. Thiscan be easily checked by examining the character of eigenvalues. The eigenvalueequation reads

(1− λ)(1− λ)(1 + λ)(1 + λ)(1 + λ) = 0

So there are 2 positive roots and no zero root. Since 0 < 2 < 5, the PDE is ultra-hyperbolic.


2.7 Summary

The main purpose of this chapter was to show how the method of characteristicscan be employed to reduce a second order linear PDE in two variables to a normalor canonical form. Indeed such a procedure provides an elegant way to solve theequation. We discussed in some detail the boundary and initial value problems andclarified the issues by working out a few typical problems. In this regard the feature ofa well-posed problem was pointed out. We then provided some insights from classicalmechanics by considering the oscillator problem under the influence of an elastic forceand a repulsive one in the phase plane. In the former case the representative pointmoves in a closed curve while in the latter case an open curve is traced out. The rolesof Dirichlet’s (or Neumann) and Cauchy’s conditions were discussed and analogiessought. We derived general expressions for the adjoint and self-adjoint operators. Wealso showed how a PDE can be classified in terms of the eigenvalues to the principalpart of the second order operator.


Exercises

1. Consider the Tricomi equation given by

yφxx + φyy = 0, y < 0

Bring it to a normal form.

2. Transform the telegraph equation

φtt + aφt + bφ = φxx

where a and b are constants to the normal form

ψξη +a2 − 4b2

16c2ψ = 0

using the transformations ξ(x, t) = x + ct, η(x, t) = x − ct and putting

φ[x(ξ, η), t(ξ, η)] = ψ(ξ, η)e−at2 .


φxy + φy + 2x2y(φx + φ) = 0


e−2xφxx − e−2yφyy − e−2xφx + e−2yφy + 8ey = 0

5. Solve the PDE

φxx − 2 sin(x)φxy − (3 + cos2(x))φyy + φx + (2− sin(x)− cos(x))φy = 0

subject to the conditions

φ(x, cos(x)) = 0, φy(x, cos(x)) = e−x2

cos(x)

6. Solve the PDE

eyφxy − φyy + φy = 0


φxy(x, 0) = 0, φy(x, 0) = − sin(x)


7. Discuss the solution of the PDE

φt + φφx = −x, t ≥ 0

subject to the condition

φ(x, 0) = f(x), −∞ < x <∞

8. Show that the L-operator operating as Lφ(x, t) = φtt(x, t) + φxxxx(x, t) isself-adjoint.

9. If, as in non-relativistic quantum mechanics, the momentum operator has arepresentation p = −i∂x, then show that the Hamiltonian H = x3p+px3 is formallyself-adjoint.

Chapter 3

PDE: Elliptic form

3.1 Solving through separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . 83(a) Two dimensions: plane polar coordinates (r, θ) . . . . . . . . . . . . . . 83(b) Three dimensions: spherical polar coordinates (r, θ, φ) . . . . . . 85(c) Cylindrical polar coordinates (r, θ, z) . . . . . . . . . . . . . . . . . . . . . . . . 88

3.2 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Gauss’ mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

3.3 Maximum-minimum principle for Poisson’s and Laplace’sequations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

3.4 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.5 Normally directed distribution of doublets . . . . . . . . . . . . . . . . . . . . . . 943.6 Generating Green’s function for Laplacian operator . . . . . . . . . . . . 973.7 Dirichlet problem for circle, sphere and half-space . . . . . . . . . . . . . . 100

(a) Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100(b) Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102(c) Half-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

As already noted in Chapter 2, the Poisson, Helmholtz and Laplace’s equations arethe three basic examples of an elliptic type of PDE. In practice they are used tomodel the time-invariant response of physical systems. Without a source term thePoisson equation reduces to the Laplace’s equation whose solution, irrespective ofits dimension, is referred to as a harmonic function.

The Laplace’s equation

∇2φ = 0 in Ω ⊂ <n (3.1)

finds its natural appearence in problems of gravitation, electrostatics, magnetostaticsand fluid dynamics. In two-dimensions, the function φ depends on the variables(x, y). Defining z = x+ iy, the analyticity of a function f(z) at a point z = z0 meansthat it has a derivative at every point within a region encircling the point z = z0.Let f(x, y) = ψ(x, y) + iφ(x, y). Then if the functions φ(x, y) and ψ(x, y) satisfy theCauchy-Riemann equations namely,

∂ψ

∂x=∂φ

∂y,

∂ψ

∂y= −∂φ

∂x(3.2)

81


in a certain region where the partial derivatives are continuous, then f(z) is an-alytic in that region. However, the converse may not be true i.e. if f = ψ + iφsatisfies Cauchy-Riemann equations then f may not be analytic. Take for examplethe function f(x, y) defined by

f(x) =

x3(1+i)−y3(1−i)

x2+y2if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

then f(x, y) satisfies the Cauchy-Riemann equations but f is not an analytic func-tion.

Simple differentiations tell us that both φ and ψ obey Laplace’s equations

∇2φ = φxx + φyy = 0, ∇2ψ = ψxx + ψyy = 0 (3.3)

We call the solutions φ and ψ of the Laplace’s equation satisfying Cauchy-Riemannequations to be harmonic conjugates. Note that there is no dependence on time inφ which means that the Laplace’s equation is concerned with a steady state picture.Because of symmetry reasons it is often advantageous to refer the Laplace’s equa-tion to polar coordinates. For instance, when we are dealing with an isolated pointcharge located at the origin, the potential produced by it at any point in space ap-pears as inversely proportional to the radial distance from the origin whereas if thesame problem is referred to in the Cartesian coordinates, a more complicated form,involving the square root of the sum of the squares of all the Cartesian coordinatesfor the point, emerges.

Poisson’s equation also appears in problems of electrostatics, among other areasin theoretical physics. Let us consider the scalar potential χ at the point x in the form

χ(x) =

∫ρ(x’)

|x-x’|d3x′ (3.4)

for a continuous charge distribution characterized by the function ρ. Operating onboth sides by ∇2 and using the formula ∇2( 1

|x-x’| ) = −4πδ(x-x’), Poisson’s equationis easily seen to hold

∇2χ(x) = −4π

∫ρ(x’)δ(x-x’)d3x′ = −4πρ(x) (3.5)

where ρ accounts for the inhomogeneity of the equation.

Conversely, Poisson’s equation can be exploited to generate the type of chargedistribution that could be relevant for a particular class of scalar potential. For in-stance if we look for the spherically symmetric Yukawa potential φ at the point P (r)

χ(r) =e−αr

r(3.6)

PDE: Elliptic form 83

where r corresponds to the radial distance of P from the origin and α is a constant,then the charge distribution can be found out by substituting (3.6) into (3.5). Tothis end, we convert (3.5) to the polar form by referring to spherical coordinates

1

r2

d

dr(r2 dχ

dr) = −4πρ (3.7)

and employ (3.6) in it. We then obtain for the continuous charge distribution thesolution

ρ =α2

4π

e−αr

r(3.8)

Helmholtz’s equation is given by

(∇2 + k2)φ = 0 (3.9)

where k is the wave number and φ represents the amplitude. Its solution1 dependson the spatial boundary conditions and its applications are typically in the problemsof wave mechanics and membranes. The paraxial approximation of the Helmholtzequation also plays an important role in quantum optics.

One of the objectives in this chapter will be to look at the boundary valueproblems that go with the elliptic equation. These problems are typically classifiedas the Dirichlet and Neumann problems. Consider a metallic plate having a closedcurve C for its boundary which is kept at a constant temperature φ0 i.e. φ = φ0 onC. The temperature profile φ(x, y) obeys the two-dimensional Laplace’s equation

φxx + φyy = 0 (3.10)

Dirichlet problem associated with the two-dimensional Laplace’s equation is the onewhen we are required to find the solution that takes prescribed values on the closedcurve C. On the other hand, if the normal derivative of φ is prescribed on C, theproblem is identified as a Neumann type. Of course in three-dimensions, the closedcurve C is replaced by some closed surface.

We now turn to a discussion of the solvability criterion for the Laplace’s equation.We begin with the method of separation of variables.

3.1 Solving through separation of variables

(a) Two dimensions: plane polar coordinates (r, θ)

In two-dimensional plane polar coordinates (r, θ), ∇2 reads

∇2 =∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2(3.11)

1The separable solution of Helmholtz’s equation is discussed in Chapter 4 in connectionwith the separable spatial solution of the wave equation.


where in terms of Cartesian coordinates the variables r and θ read

r =√x2 + y2, θ = tan−1(

y

x) (3.12)

Note that r ∈ (0,∞) and θ ∈ (0, 2π].

We seek solutions of the two-dimensional Laplace’s equation

∇2φ(r, θ) = 0 (3.13)

in the variable-separated product form namely,

φ(r, θ) = F (r)G(θ) (3.14)

Substituting (3.14) in (3.13) and using (3.11) gives the ODE

1

F(r2 d

2F

dr2+ r

dF

dr) = − 1

G

d2G

dθ2(3.15)

Since the left side of (3.15) is a function of only r and the right side is a functionof only θ, consistency requires that each side must assume a constant value whichwe take as n2. (3.15) thus leads to the pair of decoupled equations

r2 d2F

dr2+ r

dF

dr− n2F = 0 (3.16)

and

d2G

dθ2+ n2G = 0 (3.17)

(3.17) resembles the simple harmonic motion in classical mechanics. The solutionsfor F and G are straightforward to find and are given respectively by

F (r) = Arn +Br−n, n = 1, 2, ... (3.18)

and

G(θ) = C cosnθ +D sinnθ, n = 1, 2, (3.19)

In the case when n = 0, the forms of the solutions are different

F (r) = A0 ln r +B0, n = 0 (3.20)

and

G(θ) = C0, n = 0 (3.21)

Noting that if one added 2π to θ we would get back the same point we imposed

φ(r, θ) = φ(r, θ + 2π) (3.22)


which implied that the integer values of n are relevant. We also omitted the θ-termin (3.21) because θ and θ + 2π represented the same point. Further, in (3.18) and(3.19) A, B, C, D and in (3.20) and (3.21) A0, B0, C0 are constants.

Combining the above solutions we can put the separable form of the generalsolution of a two-dimensional Laplace’s equation in the manner

φ0(r, θ) = A0 ln r +B0, n = 0 (3.23)

where C0 has been absorbed in A0 and B0 and

φn(r, θ) = (Arn +Br−n)(Cn cosnθ +Dn sinnθ), n = 1, 2, .. (3.24)

The complete solution for φ(r, θ) reads

φ(r, θ) = A0 ln r+ Σnrn(An cosnθ+Bn sinnθ) + Σnr

−n(Cn cosnθ+Dn sinnθ) + Λ0

(3.25)where Λ0 is a constant. Observe that (3.25) contains the following fundamentalsolution of the two dimensional Laplace’s equation

φ(r) = − 1

2πln(

1

r) (3.26)

by fixing A0 = − 12π

.If we wish to avoid blowing up of the solution at r = 0 then φ would read from

(3.23) and (3.24)

φ(r, θ) =

B0, n = 0

Σnrn(An cosnθ +Bn sinnθ), n = 1, 2, ...

(b) Three dimensions: spherical polar coordinates (r, θ, φ)

In three-dimensional spherical polar coordinates (r, θ, φ), ∇2 reads

∇2 =1

r2

∂

∂r(r2 ∂

∂r) +

1

r2 sin θ

∂

∂θ(sin θ

∂

∂θ) +

1

r2 sin2 θ

∂2

∂φ2(3.27)

where in terms of Cartesian coordinates the variables r, θ and φ read

r =√x2 + y2 + z2, θ = cos−1(

z√x2 + y2 + z2

), φ = tan−1(y

x) (3.28)

Note that r ∈ (0,∞), θ ∈ [0, π) and φ ∈ [0, 2π).

We seek solutions of the three-dimensional Laplace’s equation

∇2ψ(r, θ, φ) = 0 (3.29)



ψ(r, θ, φ) = R(r)f(θ, φ) (3.30)

Substituting (3.30) in (3.29) and using (3.27) gives

1

R

d

dr(r2 dR

∂r) = − 1

f[

1

sin θ

∂

∂θ(sin θ

∂f

∂θ) +

1

sin2 θ

∂2f

∂φ2] = −p(say) (3.31)

where p is a constant because the left side of the first equality deals with a functionof r only while the second equality depends on a function of θ and φ only and theseare consistent if each is equal to a constant.

The R-equation is given by

1

R

d

dr(r2 dR

∂r) + p = 0 (3.32)

which amounts to the form

d2R

ds2+dR

ds− l(l + 1)R = 0 (3.33)

where we have defined p = −l(l + 1) and set s = ln r.

The second order ODE (3.33) has the general solution

R(r) = Arl +Br−(l+1) (3.34)

where A and B are arbitrary constants.

We now separate f(r, θ) in the product form

f(r, θ) = g(θ)h(φ) (3.35)

Substituting in (3.31) one obtains from the second equation

sin θ

g

d

dθ(sin θ

dg

dθ) + l(l + 1) sin2 θ = − 1

h

d2h

dφ2= ν2(say) (3.36)

where we have applied a similar reasoning as noted below (3.31) that since we havetwo equations, one in θ and another in φ, the two can be compromised if each isequal to a constant which we have set as ν2.

As a result of (3.36), the differential equations for h and g assume respectivelythe forms

d2h

dφ2= −hν2 (3.37)


and

d

dµ[(1− µ2)

dg

dµ] + [l(l + 1)− ν2

1− µ2]g == 0 (3.38)

where we have put µ = cos θ.

The general solution of (3.37) is given by the periodic form

h(φ) = E cos(νφ) + F sin(νφ) (3.39)

where E and F are arbitrary constants.

On the other hand, equation (3.38) can be immediately identified as an associatedLegendre equation whose general solution is well known to be

g(µ) = CP νl (µ) +DQνl (µ) (3.40)

where P νl (µ) and Qνl (µ) are the associated Legendre functions which are expressiblein terms of hypergeometric functions. In (3.40), C and D are two arbitrary constants.

Since we are often interested in a physically meaningful finite and single-valuedsolution2 in the ranges 0 ≤ φ < 2π and 0 ≤ θ < π we restrict ν = m, m =0, 1, 2, .... In other words, ν must be taken as a non-negative integer. Further, becauseof logarithmic singularities at µ = ±1 in Qνl (µ), we set D = 0, For the seriesrepresentation of Pml (µ) to be convergent the restriction on l is that it must be apositive integer or zero. In such a case the infinite series for Pml (µ) terminates andis given by the relation

Pml (µ) = (1− µ2)m2dm

dµmPl(µ), l = 0, 1, 2, ... (3.41)

where Pl(µ) obeys the Rodrigues formula

Pl(µ) =1

2ll!

dl

dµl(µ2 − 1)l (3.42)

The general solution of the Laplace’s equation in spherical polar coordinates thatremains finite and single-valued in a region that includes the origin is

ψ(r,θ,φ)=

∞∑l=0

rll∑

m=0

(Almcos(mφ)+Blm sin(mφ))Pml (µ)=

∞∑l=0

rlYl(θ, φ), l=0, 1, 2, ...

where Yl, the surface spherical harmonics of degree n, and stands for the angularpart

2The physical requirement of single valuedness concerning h(φ) is h(φ + 2π) = h(φ)resulting in ν to be zero or integer.


Yl(θ, φ) =

l∑m=0

(Alm cos(mφ) +Blm sin(mφ))Pml (µ)

The spherical harmonics are orthogonal and normalized∫ 2π

0

∫ π

0

dφdθ sin θY ml (θ, φ)∗Y m′

l′ (θ, φ) = δll′δmm′

where we took the form Y ml (θ, φ) = NeimφPml (cos θ), N is a normalization con-stant to be chosen so as to have the right side of the integral equal unity whenl = l′,m = m′.

On the other hand, the general solution of the Laplace’s equation in sphericalpolar coordinates that remains finite and single-valued in a region that includes thepoint at infinity is

φ(r, φ) =

∞∑l=0

r−l−1Yl(θ, φ), l = 0, 1, 2, ...

The fundamental solution of Laplace’s equation in three-dimensions corre-sponds to

φ(r) = − 1

4π

1

r, r 6= 0 (3.43)

while for the n-dimensional case, the fundamental solution is

φ(r) =1

n(n− 2)B(n)

1

rn−2, r 6= 0, n ≥ 3 (3.44)

where B(n) stands for the volume of a unit ball in <n.

(c) Cylindrical polar coordinates (r, θ, z)

In cylindrical polar coordinates (r, θ, z), ∇2 reads

∇2 =∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2+

∂2

∂z2(3.45)

where in terms of the Cartesian coordinates the variables r, θ and z are given by

r =√x2 + y2, θ = tan−1(

y

x), z = z (3.46)

Note that r ∈ [0,∞), θ ∈ [0, 2π) and z ∈ (∞,∞).

We seek solutions of the three-dimensional Laplace’s equation

∇2φ(r, θ, z) = 0 (3.47)


φ(r, θ) = R(r)f(θ)h(z) (3.48)


Substituting (3.48) in (3.47) and using (3.45) gives the form

1

h

d2h

dz2= − 1

R(d2R

dr2+

1

r

dR

dr)− 1

fr2

d2f

dθ2(3.49)

Since the left side is a function of z only and the right side is a function of r and θconsistency requires that each side is equal to a constant which we set as p2. Thenwe get the following pair of equations

d2h

dz2= p2h (3.50)

andr2

R

d2R

dr2+r

R

dR

dr+ p2r2 = − 1

f

d2f

dθ2(3.51)

The solution of (3.50) is known to be in the form

h(z) = αepz + βe−pz (3.52)

where α and β are arbitrary constants.

In (3.51) the variables are already separated and so we set each side as equal tothe constant ν2. This gives

d2f

dθ2+ ν2f = 0 (3.53)

and

r2 d2R

dr2+ r

dR

dr+ (p2r2 − ν2)R = 0 (3.54)

Equation (3.53) has the simple harmonic form whose general solution reads

f(θ) = γ cos(νθ) + δ sin(νθ) (3.55)

where γ and δ are arbitrary constants. If one requires a unique θ then f(θ+2π) = f(θ)results in ν to be zero or an integer.

The differential equation (3.54) can be easily converted to the Bessel form byapplying a transformation

y = pr (3.56)

which gives

y2 d2R

dy2+ y

dR

dy+ (y2 − ν2)R = 0 (3.57)

Its general solution for all values of ν is

R(r) = λJν(pr) + µYν(pr) (3.58)

where Jν and Yν are respectively Bessel and Neumann functions of order ν.


It is useful to note that had we chosen a negative separation constant −p2 wewould have gotten a modified Bessel equation

y2 d2R

dy2+ y

dR

dy− (y2 + ν2)R = 0 (3.59)

whose solutions are the modified Bessel functions Iν and Kν

R(r) == λ′Iν(pr) + µ′Kν(pr) (3.60)

Both Kν and Yν are divergent at r = 0 and so are to be excluded if we are interestedin a solution around r = 0. On the other hand, because Jν and Iν diverge at r →∞they are to be disregarded for an exterior solution.

3.2 Harmonic functions

Simple algebraic functions like φ = x3−3xy2 and φ = 3x2y−y3 or transcendentalfunctions such as φ = ex cos(y) and φ = ex sin(y) which satisfy Laplace’s equationare examples of harmonic functions in two variables while a unit point charge atorigin described by the function 1

rin a three-dimensional region (r 6= 0) and a line

of unit charge density on the whole of z-axis in terms of the function − ln(r2 − z2),r 6= z, are examples of harmonic functions in three variables.

We shall now prove Gauss’ mean value theorem for a harmonic function.

Gauss’ mean value theorem

Let φ be a harmonic function in a three-dimensional region <3 and P be a pointinside it. We imagine a sphere S of radius r with centre at P which is immersedwholly inside <3. Then φ at P is given by the integral

φ(P ) =1

4πr2

∫ ∫S

φ(Q)ds (3.61)

where ds is the surface element and Q is a point on S. The above result is calledGauss’ mean value theorem.

Proof : The spherical mean of φ over the surface of the sphere S is

φ(r) =1

4πr2

∫ ∫S

φ(Q)ds (3.62)

If the Cartesian coordinates of P are given by (x, y, z) then those of Q are evidently(ξ, η, ζ) where


ξ = x+ lr, η = y +mr, ζ = z + nr (3.63)

with the quantities l,m, n, in terms of spherical polar angles θ and φ, stand for

l = sin θ cosφ, m = sin θ sinφ, n = cos θ (3.64)

Noting that the surface element ds is r2 sin θdθdφ, φ(r) reads

φ(r) =1

4π

∫ 2π

0

∫ π

0

φ(x+ lr, y +mr, z + nr) sin θdθdφ (3.65)

By differentiating with respect to r one finds

dφ(r)

dr=

1

4π

∫ 2π

0

∫ π

0

(l∂φ

∂ξ+m

∂φ

∂η+ n

∂φ

∂ζ) sin θdθdφ (3.66)

which can be recast as

dφ(r)

dr=

1

4πr2

∫ 2π

0

∫ π

0

(l∂φ

∂ξ+m

∂φ

∂η+ n

∂φ

∂ζ)ds (3.67)

The above representation prompts us to make use of the divergence theorem toconvert the surface integral into the following volume integral

dφ(r)

dr=

1

4πr2

∫ ∫ ∫∇2φdv (3.68)

where dv is the volume element.

The harmonic property of φ makes the right side vanish and we have the simpleresult

dφ(r)

dr= 0 (3.69)

In other words, φ(r) is a constant function independent of r. Hence it is possible tointerpret

φ(r) = φ(0)

= limr−>0

φ(r)

= limr−>0

1

4π

∫ 2π

0

∫ π

0

φ(x+ lr, y +mr, z + nr) sin θdθdφ

=1

4π

∫ 2π

0

∫ π

0

limr−>0

φ(x+ lr, y +mr, z + nr) sin θdθdφ

=1

4πφ(x, y, z)

∫ 2π

0

∫ π

0

sin θdθdφ

= φ(x, y, z)

= φ(P ) (3.70)


where use has been made of the double-integral having the value of 4π. From (3.62)we therefore conclude that

φ(P ) =1

4πr2

∫ ∫S

φ(Q)ds (3.71)

By a similar reasoning the result in two-dimensions can be proved.

3.3 Maximum-minimum principle for Poisson’s andLaplace’s equations

Let us consider a two-dimensional Poisson equation

∇2φ = −4πρ(x, y) (3.72)

in a given bounded region Ω having the boundary ∂Ω. We assume φ to be con-tinuous in Ω as well as on ∂Ω. The maximum principle states that if ρ < 0 in Ωand φ is a solution of the Poisson equation then its maximum value is attained on∂Ω. On the other hand, if ρ > 0 in Ω and φ is a solution of the Poisson equationthen its minimum value is attained on ∂Ω. We give a simple proof of these assertions.

Proof: First let ρ < 0. The continuity of φ implies that its maximum oughtto exist anywhere in Ω + ∂Ω. If it occurs at a point (x, y) inside Ω then evidently(φx)(x,y) = 0 = (φy)(x,y) along with (φxx)(x,y) < 0 and (φyy)(x,y) < 0 which by(3.72) contradicts the assertion that ρ < 0. Hence the maximum of φ is attained onthe boundary.

Similarly we can establish that the minimum of φ(x, y) is attained on the bound-ary for the opposite case ρ > 0. Note that the latter amounts to replacing φ by −φin (3.72) . Since ρ > 0 is equivalent to −ρ < 0, the preceding arguments lead to−φ assuming its maximum value on ∂Ω. In other words, the minimum of φ(x, y) isattained on the boundary ∂Ω.

We now turn to Laplace’s equation which corresponds to ρ = 0. Let us supposethat the maximum M of φ occurs on the boundary ∂Ω. Define a function ψ =φ+ ε(x2 + y2), ε > 0 which implies ψ ≥ φ. The function is often referred to as thehelper function.

Operating on both sides of ψ by ∇2 we find ∇2ψ = 4ε > 0 because of ∇2(x2 +y2) = 4. Thus ψ satisfies the Poisson’s equation in Ω and so the maximum-minimumprinciple for ψ holds. We therefore conclude that the maximum of ψ has to occuron the boundary ∂Ω : ψ ≤M + εR2 where x2 + y2 = R2. With ψ ≥ φ it thus followsthat φ ≤ M + εR2 ⇒ φ ≤ M in ∂Ω on letting ε → 0. In other words, if φ ≤ M on∂Ω then φ ≤M in Ω too.

Now if φ is a solution of Laplace’s equation, then − φ is a solution too and sothe minimum principle follows trivially.

As a corollary, we can comment on the stability of the solution. Let φ be asolution of the Poisson’s equation in Ω+∂Ω. Consider φ′ to be another solution that


results on perturbing the Dirichlet boundary condition on ∂Ω by a small amount.Evidently |φ− φ′| is small too on ∂Ω ⇒ |φ− φ′|∂Ω < ε⇒ |φ− φ′| < ε throughout Ωand the stability of the solution is implied. (Note that (φ− φ′) satisfies the Laplaceequation in ∂Ω and so the maximum of |φ − φ′| is attained on ∂Ω where it is lessthan ε.)

3.4 Existence and uniqueness of solutions

Let us focus on the Poisson’s equation ∇2φ = −4π (we have put ρ = 1) andassume the existence of its solution in the region Ω which has the boundary ∂Ω. Ouraim is to establish the following theorem:

Theorem

A unique solution exists within Ω which on its closed boundary ∂Ω satisfies eitherthe condition of Dirichlet (or a mixed one) or Neumann (up to a constant). How-ever, should ∂Ω be open, Dirichlet or Neumann boundary condition is insufficientto produce a unique solution. The problem is overdetermined if we apply Cauchyboundary conditions on the closed surface.

Proof : Given that the Poisson’s equation is defined for the region Ω havingthe boundary ∂Ω, let, if possible, two solutions exist which are specified by φ1 andφ2. Then their difference Φ ≡ φ1 − φ2 satisfies Laplace’s equation ∇2Φ = 0 on Ω,the closure of Ω, along with Φ = 0 on ∂Ω for the Dirichlet boundary conditionand ∂Φ

∂n= 0 on ∂Ω for the Neumann boundary condition. For the mixed boundary

condition on Ω = Ω1 + Ω2 with ∂Ω1 and ∂Ω2 corresponding to two open surfacesthat constitute the whole of ∂Ω i.e. ∂Ω = ∂Ω1 + ∂Ω2 such that Φ = 0 on ∂Ω1 and∂Φ∂n

= 0 on ∂Ω2, it follows that in these cases Φ ∂Φ∂n

= 0 on ∂Ω. Taking cue from thelatter we observe that

0 =

∮∂Ω

Φ∂Φ

∂ndS

=

∮∂Ω

Φ∇Φ.n dS

=

∫Ω

∇.(Φ∇Φ) dV (using Gauss’ theorem)

=

∫Ω

(Φ∇2Φ + |∇Φ|2) dV

=

∫Ω

|∇Φ|2 dV (∵ ∇2Φ = 0) (3.73)

Consistency of both sides requires that |∇Φ|2 = 0 ⇒ ∇Φ = 0 on Ω. This meansthat Φ =constant on Ω. For the Dirichlet case, this constant is zero. The constantis zero too in the mixed case because in the Ω1 portion of Ω, Φ = 0. However, for


the Neumann case any value of the constant is feasible because whatever the valueof the constant, ∂Φ

∂n= 0.

Consider employing a Dirichlet condition on the open surface ∂Ω1

φ|∂Ω1 = f(x, y, z) (3.74)

and suppose for ∂Ω2 the following feature holds

φ|∂Ω2 = g1(x, y, z) (3.75)

The above conditions on ∂Ω1 and ∂Ω2 thus provide a Dirichlet condition on theclosed ∂Ω. The solution pertaining to this Dirichlet condition is evidently unique inΩ as we learnt from the previous proposition. Let us specify this unique solution byφ = φ1(x, y, z).

We now modify the condition on ∂Ω2 to read

φ|∂Ω2 = g2(x, y, z) (3.76)

Applying a similar reasoning on this along with the condition on ∂Ω1 we have an-other Dirichlet condition on the closed surface ∂Ω which is also unique in Ω. Let uscall this solution φ = φ2(x, y, z). Note that φ1 and φ2 are different since they differon ∂Ω2. We therefore observe that both φ1 and φ2 satisfy the same PDE supportedby a Dirichlet condition on an open ∂Ω1. We are thus led to the non-uniqueness ofthe solution. Carrying on the argument we would run into an infinity of solutions.

Regarding the Cauchy boundary conditions on the closed surface Ω, let us specifythem by

φ|∂Ω = f1(x, y, z),∂φ

∂n|∂Ω = f2(x, y, z) (3.77)

If we consider the first condition only then we have a Dirichlet problem for thegiven PDE and from what we discussed above, a unique solution for it exists withinΩ. Let us call this solution φ1(x, y, z). This solution allows us to determine thenormal derivative ∂φ1

∂non Ω. If it coincides with the given function f2(x, y, z) then

we have a solution for the Cauchy boundary conditions but at the cost of renderingone of the two prescribed conditions to be unnecessary. The problem is thereforeoverdetermined. If the coincidence does not happen then we do not have any solution.

3.5 Normally directed distribution of doublets

Consider a circle of radius ε with its centre at the fixed point P (x, y) and havinga boundary C lying entirely inside a two-dimensional region Ω whose boundary is∂Ω. Then in the sub-region Ω′ formed by omitting C, the fundamental solution −1

2πln 1

rof the two-dimensional Laplace’s equation holds


(∂2

∂ξ2+

∂2

∂η2) ln

1

r= 0, (ξ, η) ∈ Ω′ (3.78)

where r2 = (x− ξ)2 + (y− η)2 and (ξ, η) are the coordinates of the variable integra-tion point.

We now employ the following Green’s second identity∫ ∫Ω

(u∇2v − v∇2u)dΩ =

∫∂Ω

(u∂v

∂n− v ∂u

∂n)dl (3.79)

where the functions u and v have continuous second derivatives in Ω. In Ω′, which isa multiply-connected region because of the non-intersecting closed boundaries ∂Ω′

and C, if we identify u = φ and v = ln 1r, then we obtain the formula

−∫ ∫

Ω′(ln

1

r)∇2φdΩ′ =

∫∂Ω′

[φ∂

∂n(ln

1

r)−(ln

1

r)∂φ

∂n]dl′+

∫C

[φ∂

∂n(ln

1

r)−(ln

1

r)∂φ

∂n]dl

(3.80)where we have used (3.78).

We make a couple of observations:

(i) Since φ has continuous derivatives in Ω, it gives the bound | ∂φ∂n| ≤ Λ, where

Λ is a constant in Ω. Accordingly,

∫C

| − (ln1

r)∂φ

∂n|dl ≤ ln ε

∫C

|∂φ∂n|dl = 2πΛε ln ε→ 0 as ε→ 0 (3.81)

(ii) Since on C, ∂∂n

= − ∂∂r, we have on taking the derivative

∫C

φ∂

∂n(ln

1

r)dl =

1

ε

∫ 2π

0

φ(x+ε cos θ, y+ε sin θ)εdθ =

∫ 2π

0

φ(x+ε cos θ, y+ε sin θ)dθ

(3.82)With φ being a continuous function, the integrand above is a continuous function

too. So taking the limit through the sign of integration we can write

limε→0

∫C

φ∂

∂n(ln

1

r)dl =

∫ 2π

0

limε→0

φ(x+ ε cos θ, y + ε sin θ)dθ (3.83)

implying

limε→0

∫C

φ∂

∂n(ln

1

r)dl = 2πφ(x, y) (3.84)

Inserting (3.81) and (3.84) in (3.80) we have the result

φ(x, y) = − 1

2π

∫ ∫Ω

(ln1

r)∇2φdΩ− 1

2π

∫∂Ω

[φ∂

∂n(ln

1

r)− (ln

1

r)∂φ

∂n]dl (3.85)

in the limit ε→ 0.


The interpretation of the right side of (3.85) goes as follows:

The first and third terms represent respectively the potential of surface distribu-tion and line distribution of matter of density − 1

2π∇2φ and 1

2π∂φ∂n

, one is distributedover the surface Ω and the other along ∂Ω. The second term is due to a normallydirected distribution of doublets of strength per unit length − φ

2πalong ∂Ω.

For the case when φ(x, y) is a harmonic function then the first term in the rightside of (3.85) drops out and we have the reduced expression

φ(x, y) = − 1

2π

∫∂Ω

[φ∂

∂n(ln

1

r)− (ln

1

r)∂φ

∂n]dl (3.86)

It implies a superposition of potentials of normally directed distribution of doubletsand a distribution of matter.

For the three-dimensional case the procedure is similar. Here we replace thecircle by a sphere S of radius ε which lies entirely inside a three-dimensional regionΩ having a boundary ∂Ω. The sphere has its centre at the fixed point P (x, y, z) andhas a boundary S. Then in the sub-region Ω′ formed by omitting S, the fundamentalsolution − 1

4π1r

of the three-dimensional Laplace’s equation holds

(∂2

∂ξ2+

∂2

∂η2+

∂2

∂ζ2)1

r= 0, (ξ, η, ζ) ∈ Ω′ (3.87)

where r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 and (ξ, η, ζ) are the coordinates of thevariable integration point.

We now use Green’s second identity (3.79) for the region Ω′ which has non-interacting closed boundaries ∂Ω′ and S:

∫ ∫ ∫Ω′

(u∇2v − v∇2u)dΩ′ =

∫ ∫∂Ω′

(u∂v

∂n− v ∂u

∂n)dS′ +

∫ ∫S

(u∂v

∂n− v ∂u

∂n)dS

(3.88)

where we have identified u = φ and v = 1r. Because of (3.87) we are led to

−∫ ∫ ∫

Ω′(1

r)∇2φdΩ′ =

∫ ∫∂Ω′

[φ∂

∂n(1

r)− (

1

r)∂φ

∂n]dS′

+

∫ ∫S

[φ∂

∂n(ln

1

r)− (ln

1

r)∂φ

∂n]dS (3.89)

Since φ has continuous derivatives in Ω, employing similar arguments as in thetwo-dimensional case, we can see that the last term in the right side of (3.89) goesto zero as ε → 0. Similarly on S, the derivative ∂

∂n= − ∂

∂R. The arguments of φ

being (x+ ε sin θ cosφ, y+ ε sin θ sinφ, z+ ε cos θ) we have on taking the limit ε→ 0

limε→0

∫ ∫S

φ∂

∂n(1

r) = 4πφ(x, y, z) (3.90)


Thus, in the limit ε→ 0, we have from (3.89), the result

φ(x, y, z) = − 1

4π

∫ ∫ ∫Ω

(1

r)∇2φdΩ− 1

4π

∫ ∫∂Ω

[φ∂

∂n(1

r)− (

1

r)∂φ

∂n]dS (3.91)

The terms in the right hand side of (3.91) can be interpreted as follows:

The first and third terms represent the potential of volume distribution andsurface distribution of matter of density − 1

4π∇2φ and 1

4π∂φ∂n

respectively, one is dis-tributed over the volume Ω and the other over ∂Ω. The second term is due to thepotential of a double layer of density − φ

4πdistributed over ∂Ω.

Finally, should φ(x, y) be a harmonic function the first term in the right side of(3.91) drops out and we have the formula

φ(x, y, z) = − 1

4π

∫ ∫∂Ω

[φ∂

∂n(1

r)− (

1

r)∂φ

∂n]dS (3.92)

which is known as the Green’s equivalent layer theorem. It gives the solution of theboundary value problem for Laplace’s equation in three-dimensions when the valuesof φ and ∂φ

∂nare given on the surface ∂Ω.

3.6 Generating Green’s function for Laplacianoperator

In two-dimensions we already have for φ(x, y) the result (3.85). Let us take inaddition a two-dimensional harmonic function ψ(ξ, η) in Ω which is twice continu-ously differentiable in Ω (≡ Ω + ∂Ω). We then obtain from Green’s second identity(3.79) the relation

− 1

2π

∫ ∫Ω

ψ∇2φdΩ = − 1

2π

∫∂Ω

(φ∂ψ

∂n− ψ∂φ

∂n)dl (3.93)

on putting u = φ and v = ψ.

Subtracting it from (3.85) we derive another representation for φ(x, y)

φ(x, y) = − 1

2π

∫ ∫Ω

(ψ − ln r)∇2φdΩ− 1

2π

∫∂Ω

[(φ∂

∂n(ψ − ln r)− (ψ − ln r)

∂φ

∂n)]dl

(3.94)

Notice that in the above equation both φ and ∂φ∂n

are present in the integrand.However, in the Dirichlet problem only φ is prescribed on the boundary. So we


choose the harmonic function ψ in such a way as to ensure that a function G definedby the difference

G(ξ, η;x, y) = ψ(ξ, η;x, y)− ln r (3.95)

vanishes on ∂Ω:

G(~ξ, ~x) = 0 on ∂Ω (3.96)

where ~ξ = (ξ, η) and ~x = (x, y). G is called the Dirichlet Green’s function associ-ated with the two dimensional Laplacian operator for the region Ω. G satisfies theLaplace’s equation for ~ξ 6= ~x. (3.96) gives the boundary condition G = 0 on ∂Ω.

In terms of G, (3.94) reads

φ(P ) = − 1

2π

∫ ∫Ω

G(Q;P )∇2φ(Q)dΩ− 1

2π

∫∂Ω

φ(Q)∂

∂nG(Q;P )dl (3.97)

Like ψ, if φ is also a two-dimensional harmonic function a further reduction for φ(P )is possible

φ(P ) = − 1

2π

∫∂Ω

φ(Q)∂

∂nG(Q;P )dl (3.98)

Thus it is reasonable to claim from the above relation that φ(P ) represented inthe form

φ(P ) = − 1

2π

∫∂Ω

f(x, y)∂

∂nG(Q;P )dl (3.99)

is a solution of the two-dimensional Dirichlet problem for the Lapalce’s equationnamely,

(∂2

∂x2+

∂2

∂y2)φ(x, y) = 0 on Ω (3.100)

where

φ = f(x, y) on ∂Ω (3.101)

To establish that (3.99) is a plausible solution of the Dirichlet problem, it is neces-sary to show that the Green’s function G exists and further that φ(~x) approachesthe limit f(~x) as any point ~x in the two-dimensional region Ω tends to its surfacevalue. The demonstration of this proposition is omitted here.

In the three dimensional case we have in place of (3.97) the result


φ(x, y, z) = − 1

4π

∫ ∫ ∫Ω

(ψ+1

r)∇2φdΩ− 1

4π

∫ ∫∂Ω

[(φ∂

∂n(ψ+

1

r)− (ψ+

1

r)∂φ

∂n)]dl

(3.102)

where ψ is a harmonic function Ω, r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 and (ξ, η, ζ)refers to the variable point Q ∈ Ω. Here 1

rreplaces ln r because 1

ris a fundamental

solution of the Laplace’s equation in three-dimensions.

Choosing φ in such a way that the Green’s function G defined by

G(ξ, η, ζ;x, y, z) = ψ(ξ, η, ζ;x, y, z) +1

r(3.103)

vanishes on ∂Ω similar to (3.96) , we obtain for φ(P )

φ(P ) = − 1

4π

∫ ∫Ω

G(Q;P )∇2φ(Q)dΩ− 1

4π

∫∂Ω

φ(Q)∂

∂nG(Q;P )dl (3.104)

Like ψ, if φ is also a three-dimensional harmonic function we get the reduced formfor φ(P )

φ(P ) = − 1

4π

∫∂Ω

φ(Q)∂

∂nG(Q;P )dl (3.105)

We can therefore reasonably claim that φ(P ) as given by

φ(P ) = − 1

4π

∫∂Ω

f(~ξ)∂

∂nG(~ξ; ~x)dl (3.106)

is a solution of the three-dimensional Dirichlet problem namely,

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2)φ(x, y, z) = 0 on Ω (3.107)

where

φ = f(x, y, z) on ∂Ω (3.108)

As with the two-dimensional case, to establish that (3.106) is a plausible so-lution of the Dirichlet problem, it is necessary to show that the Green’s functionG exists and further that φ(~x) approaches the limit f(~x) as any point ~x in thethree-dimensional region Ω approaches the surface ∂Ω. The demonstration of thisproposition is omitted here.

We now move to some applications of the above results.


3.7 Dirichlet problem for circle, sphere andhalf-space

(a) Circle

Consider a circle C of radius a, centred at O as shown in Figure 3.1. The pointsQ and P are taken, as usual, to be the variable point and fixed point respectively.In polar form we take their coordinates to be Q = (ρ, θ) and P = (r, θ′). We adopt ageometrical approach and to this end we identify P ′ as the inverse point of P withrespect to the circle C . We denote the distances as

FIGURE 3.1: Circle C of radius a.

PQ = R, P ′Q−R′, OP = r (3.109)

We address the Dirichlet problem for the two-dimensional Laplace’s equation

(∂2

∂x2+

∂2

∂y2)φ(x, y) = 0 (3.110)

in the region Ω that is bounded by the circle C. The boundary condition is pro-vided by

φ = f(θ) on r = a (3.111)

where r =√x2 + y2 and θ ∈ (0, 2π].

We take the following choice of ψ to make use of the Green’s function as notedin (3.95)

ψ = ln(rR′

a) (3.112)


It works from the geometrical point of view. The reason is that if Q is any point onthe boundary of the circle, the triangles POQ and P ′OQ become similar and so P ′

as an inverse point of P leads to

OP ·OP ′ = a2 ⇒ OP

OQ=

OQ

OP ′(3.113)

The above equality signals that the angles OQP and OP ′Q are equal. Therefore wecan write down

OQ

OP=R′

R=a

r(3.114)

Thus the ratio rR′

aR= 1 and hence its log vanishes. An implication is that if the point

Q is on the boundary then the Green’s function given by

G(Q,P ) = ln(1

R)− ln(

a

rR′) = ln(

rR′

aR) (3.115)

vanishes. (3.115) is the desired form of the Green’s function for the problem of thecircle

From Figure 3.1 it is clear that the angle ∠QOP = θ− θ′. Further, the followingtrigonometrical equalities hold

R2 = r2 + ρ2 − 2rρ cos(θ − θ′), R′2 =1

r2[a4 + r2ρ2 − 2a2rρ cos(θ − θ′)] (3.116)

So substituting these expressions in (3.115) we obtain the corresponding expressionfor the Green’s function

G(Q,P ) =1

2ln

[a4 + r2ρ2 − 2a2rρ cos(θ − θ′)][a2(r2 + ρ2 − 2rρ cos(θ − θ′))] (3.117)

On C, the derivative ∂G(Q,P )∂n

assumes the form

∂G(Q,P )

∂n|C =

∂G(Q,P )

∂n|ρ=a =

r2 − a2

a[r2 + a2 − 2ra cos(θ − θ′)] (3.118)

We already have shown that if φ is harmonic in a region Ω bounded by a closedcurve then at any point in Ω, φ is given by (3.98). In the present case, φ = f(θ) onthe circle C and so on using (3.118) it turns out to be

φ(r) =a2 − r2

2π

∫ 2π

0

f(a, θ)dθ

r2 + a2 − 2ra cos(θ − θ′) (3.119)

The above formula is called Poisson integral for the disc or the complete circle andsolves the Dirichlet’s problem, summarized in (3.110) and (3.111), in the interior ofthe circle.


It is simple to deduce from the above formula the value of φ at the point r = 0

φ(0) =1

2π

∫ 2π

0

f(θ)dθ (3.120)

We see that the value of φ at the centre of the circle is equal to the average value off on the boundary.

We leave as an exercise the proof of the following equality:

a2 − r2

2π

∫ 2π

0

dθ

r2 + a2 − 2ra cos(θ − θ′) = 1 (3.121)

(b) Sphere

We now proceed to determine the Green’s function for the Dirichlet problem inthe interior of the sphere. Consider a sphere S of radius a, centred at O. The pointsQ and P correspond to the variable point and fixed point respectively. We assigntheir spherical polar coordinates to be Q = (ρ, θ, λ) and P = (r, θ′, λ′) respectively.As in the case of the circle we take recourse to a geometrical analysis of the problem.Let P ′ be the inverse point of P with respect to the sphere S as shown in Figure3.2. We denote the distances as

PQ = R, P ′Q−R′, OP = r (3.122)

We consider the Dirichlet problem for the three-dimensional Laplace’s equation

FIGURE 3.2: Sphere S of radius a.

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2)φ(x, y, z) = 0 (3.123)


in the region Ω bounded by the sphere S. The boundary condition for φ is given by

φ = f(θ, λ) on r = a (3.124)

where r =√x2 + y2 + z2, θ ∈ (0, π] and λ ∈ (0, 2π].

To facilitate the use of Green’s function which for the three-dimensional case isgiven by (3.103) we consider the following choice

ψ = − a

rR′(3.125)

This ensures that ψ is harmonic.

As in the circle’s case we see that if Q is any point on the boundary of the sphere,the triangles POQ and P ′OQ become similar and so P ′ as an inverse point of Pleads to

OP ·OP ′ = a2 ⇒ OP

OQ=

OQ

OP ′(3.126)

indicating that the angles OQP and OP ′Q are equal. Therefore we can write down

OQ

OP=QP ′

QP=R′

R=a

r(3.127)

Thus the ratio rR′

aR= 1 implying that if the point Q is on the boundary then the

Green’s function given by the form

G(Q,P ) =1

R− a

rR′(3.128)

vanishes. (3.128) gives the general form of the Green’s function for the problem ofsphere.

From Figure 3.2 it is clear that the following equalities hold

R2 = r2 + ρ2 − 2rρ cosα, R′2 =1

r2[a4 + r2ρ2 − 2a2rρ cosα] (3.129)

where cosα iscosα = cos θ cos θ′ + sin θ sin θ′ cos(λ− λ′) (3.130)

So substituting these expressions in the expression of the Green’s function (3.128)we obtain

G(Q,P ) = (r2 + ρ2 − 2rρ cosα)−12 − a(a4 + r2ρ2 − 2a2rρ cosα)−

12 (3.131)


On S, the derivative ∂G(Q,P )∂n

is given by

∂G(Q,P )

∂n|S =

∂G(Q,P )

∂n|ρ=a =

r2 − a2

a[r2 + a2 − 2ra cosα]32

(3.132)

We have already shown that if φ is harmonic in a region Ω bounded by a closedsurface then at any point in Ω, φ is given by (3.106). In the present case φ(P ) turnsout to be

φ(P ) =a(a2 − r2)

4π

∫ 2π

0

∫ π

0

f(a, θ, λ) sin θdθdλ

(r2 + a2 − 2ra cosα)32

(3.133)

on using (3.132). The above formula is the Poisson integral for the sphere and solvesthe Dirichlet’s problem summarized in (3.123) and (3.124).

(c) Half-space

Here the Dirichlet’s problem seeks to find the solution of the Laplace’s equation

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2)φ(x, y, z) = 0 (3.134)

in the half-space region z ≥ 0. The boundary condition is given on the plane z = 0namely

φ = f(x, y), on z = 0 (3.135)

The function f is assumed to be twice differentiable with each derivative being con-tinuous.

In Figure 3.3, the points Q(ξ, η, ζ) and P (x, y, z) correspond to the variable pointand fixed point respectively in the region z ≥ 0. Let P ′(x, y,−z) be the image of Pwith respect to the plane z − 0. We denote the distances as

PQ = R, P ′Q = R′ (3.136)

In three-dimensions the Green’s function reads from (3.103)

G(Q;P ) =1

R+ ψ (3.137)

where in the present case r = R and ψ is harmonic in z ≥ 0 i.e.

(∂2

∂ξ2+

∂2

∂η2+

∂2

∂η2)ψ(x, y, z) = 0, z ≥ 0 (3.138)


FIGURE 3.3: The half space in the region z ≥ 0.

Taking ψ = − 1R′ the relevant Green’s function is

G(Q,P ) =1

R− 1

R′(3.139)

This is so because if the point Q is on the boundary z = 0 then from Figure 3.3, itis clear that R = R′ implying G(Q,P ) = 1

R− 1

R′ = 0.

The expressions for R and R′ are

R2 = (ξ − x)2 + (η − y)2 + (ζ − z)2, R′2 = (ξ − x)2 + (η − y)2 + (ζ + z)2 (3.140)

and the normal derivative of G(Q;P ) is given by

∂G(Q,P )

∂n|ζ=0 = −∂G(Q,P )

∂ζ|ζ=0 =

1

R2

∂R

∂ζ|ζ=0 −

1

R′2∂R′

∂ζ|ζ=0 (3.141)

A little algebra yields

∂G(Q,P )

∂n|ζ=0 = − 2z

[(ξ − x)2 + (η − y)2 + z2]32

(3.142)

We already know that if φ is harmonic in a region Ω bounded by a closed surfacethen at any point in Ω, φ is given by (3.106). In the present case this gives for φ(P )

φ(P ) =z

2π

∫ ∞−∞

∫ ∞−∞

f(ξ, η)dξdη

[(ξ − x)2 + (η − y)2 + z2]32

(3.143)

which solves the Dirichlet’s problem posed by (3.134) and (3.135).


3.8 Summary

To summarize, we introduced in this chapter the elliptic form of PDE. We de-scribed the method of solutions through the separation of variables touching uponthe plane polar coordinates, spherical polar coordinates and cylindrical coordinates.We discussed the harmonic function and gave a formal proof of Gauss’ mean valuetheorem. We also furnished the proofs of the maximum-minimum principle for Pois-son’s and Laplace’s equations and commented on the existence and uniqueness ofthe solutions for these equations. Subsequently we addressed the normally directeddistribution of doublets and outlined the procedure of generating Green’s functionfor the Laplacian operator both in two and three dimensions. As applications wederived the solutions of the Dirichlet problem for the circle, the sphere and thehalf-space.


Exercises

1. Solve Laplace’s equation

φxx(x, y) + φyy(x, y) = 0

subject to the boundary conditions

φ(x, 0) = sin(πx), φ(x, 1) = sin(πx)e−π, φ(0, y) = 0, φ(1, y) = 0

and show that the solution is φ(x, y) = sin(πx)e−πy.

2. Show that φ(x, y) = 16x4 − x2y2 + 1

6y4 is a harmonic function.

3. The planes x = 0, x = π and y = 0 form the boundary of a semi-infinitetwo-dimensional space. The potential φ satisfies the Poisson’s equation

φxx(x, y) + φyy(x, y) = −ρ

in the presence of a uniform distribution of charge density ρ4π

. While φ is boundedover the concerned region, the boundary conditions are

φ(0, y) = 0, φ(π, y) = 1, φ(x, 0) = 0

Show that φ(x, y) is given by the form

φ(x, y) =2

π

∞∑n=1

[ρ((−1)n − 1)

n3+

(−)n

n](e−ny − 1) sin(nx)

4. Consider the Beltrami equations, which are a generalization of Cauchy-Riemann equations, as given by the pair of functions φ and ψ satisfying the equations

ζφx − bψx − cψy = 0 ζφy + aψx + bψy = 0

where ζ 6= 0, the matrix

M =

[a bc d

]is positive definite and the coefficients a, b, c and d are given functions of x and y.Show that such a system is elliptic.

5. Show that sin(k|x|)|x| solves the Helmholtz’s equation 4φ+k2φ = 0 in <n in the

deleted neighbourhood of the origin.


6. By analyzing the coefficient matrix of the following PDE

φxx − φxy + φyy = f(x, y)

show that the equation is of elliptic type.

7. Establish that a bounded harmonic function in <n is a constant.

8. Consider the Lapalce’s equation φxx(x, y) + φyy(x, y) = 0 in the region

x2 + y2 < 1 along with the boundary condition φ = xyex2+y4+2 for x2 + y2 = 1.

Show by the mean value theorem that φ(0, 0) = 2.

9. Solve Laplace’s equation

φxx(x, y) + φyy(x, y) = 0, 0 < y <∞

with φ = f(x) on y = 0 and show that the solution can be expressed as

φ(x, y) =y

π

∫ ∞0

f(ξ)dξ

(x− ξ)2 + y2

10. In the semi-annulus region defined by r ∈ (1, 2) and θ ∈ (0, π) show that thetwo-dimensional Lapalce’s equation

1

r

∂

∂r(r∂φ

∂r) +

1

r2

∂2φ

∂θ2= 0

subject to the boundary conditions

φ(r, 0) = 0 = φ(r, π) in 1 < r < 2

and

φ(1, θ) = sin θ, φ(2, θ) = 0 in 0 < θ < π

has the solution φ(r, θ) = ( 43r− r

3) sin θ.

Chapter 4

PDE: Hyperbolic form

4.1 D’Alembert’s solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.2 Solving by Riemann method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1134.3 Method of separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

(a) Three dimensions: spherical polar coordinates (r, θ, φ) . . . . . . 117(b) Cylindrical polar coordinates (r, θ, z) . . . . . . . . . . . . . . . . . . . . . . . . 120

4.4 Initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121(a) Three dimensional wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121(b) Two dimensional wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

The wave equation which was introduced in Chapter 1 is a prototypical example forthe class of hyperbolic PDEs. It arises in many branches of physics. A quick reviewof the derivation of the wave equation goes as follows. Consider a region of surfaceS enclosing a volume V . Newton’s force-acceleration relation gives on integration

∂2

∂t2

∫V

φ(~r, t)dv = −∫S

~F (~r, t) · ~nds

where ~n is the outward drawn unit normal to S and we have set the mass-densityfactor to be unity. Employing Gauss’s divergence theorem∫

V

div ~F (~r, t)dv =

∫S

~F (~r, t) · ~nds

allows us to re-cast the previous equation to the form∫V

div ~F (~r, t)dv = − ∂2

∂t2

∫V

φ(~r, t)dv = −∫V

∂2φ(~r, t)

∂t2dv

which implies

div ~F (~r, t) = −∂2φ(~r, t)

∂t2

Taking F to be proportional to the gradient of φ but acting in the opposite directioni.e. given by ~F (~r, t) = −c2∇φ(~r), where c is a constant we then have the waveequation

φtt − c2∇2φ = 0

where c is the propagation speed.

109


The above equation is obtained in the homogeneous form. A non-homogeneousequation of course is more general where one can have a term on the right side asa function of the spatial variables as in the case of Poisson equation. If we considerthe problem of small oscillations about some stationary or equilibrium point thenin one-dimension the wave equation stands for the displacement from the stationaryposition. In two-dimensions the wave equation is relevant to the vibrations of amembrane. In three dimensions the wave equation appears in electromagnetic theorywhere φ could represent a component of say the electric field E. Wave equations alsoplay important roles in theory of hydromechanics, optical problems, heat transfer etc.

In this chapter we are going to concentrate on the wave equation as the typicalrepresentative equation of the PDE in the hyperbolic form and study some of itsdifferent properties. We begin, first of all, by taking up the derivation of D’Alembert’ssolution by focusing on the one-dimensional wave equation.

4.1 D’Alembert’s solution

The one-dimensional wave equation is categorized as a classical example forstudying travelling-wave phenomena. It is also relevant in accounting for the trans-verse displacements of a vibrating string.

We will be interested in solving the Cauchy problem defined on an infinite domainand towards this pursuit we will consider the following one-dimensional form

φtt − c2φxx = 0, −∞ < x < +∞, t ≥ 0 (4.1)

being subjected to a pair of initial conditions on the position and velocity

φ(x, 0) = f(x), φt(x, 0) = g(x) (4.2)

We assume that the function f(x) is twice differentiable and that g(x) is differen-tiable.

We will determine the solution in some domain of (t, x) plane. Setting

ξ = x− ct, η = x+ ct (4.3)

the PDE (4.1) is readily converted to the mixed form

∂2φ

∂ξ ∂η= 0 (4.4)

Equation (4.4) has the immediate solution

PDE: Hyperbolic form 111

φ = χ(ξ) + ψ(η) = χ(x− ct) + ψ(x+ ct) (4.5)

Let us recall from Chapter 1 that the families of straight lines ξ = x − ct =constant and η = x+ ct = constant are referred to as the characteristics. It is acrossthem that a solution can have jump discontinuities. If φ is given on the segments oftwo of the characteristics and if we complete a rectangle by drawing the parallel linesto ξ = constant and η = constant, then any discontinuity on the characteristics willbe propagated into the interior of the rectangle. The functions χ and ψ representrespectively a right- and left-travelling wave with velocity c.

Employing the given Cauchy conditions we obtain for φ and ψ the expressions

χ(x) =1

2[f(x)−

∫ x

x0

g(s) ds− k] (4.6)

and

ψ(x) =1

2[f(x) +

∫ x

x0

g(s) ds+ k] (4.7)

where x0 is arbitrary and k is a constant of integration. Inserting (4.6) and (4.7) in(4.5) we obtain

φ(x, t) =1

2[f(x− ct) + f(x+ ct) +

∫ x+ct

x−ctg(s) ds] (4.8)

which is called the D’Alembert’s solution.

Some remarks are in order about the energy equation for the infinite string forwhich we got the D’Alembert’s solution. Identifying for the kinetic energy the termφ2t and for the potential energy the term c2φ2

x we can readily write down for theenergy E the integral

E =1

2

∫ +∞

−∞(φ2t + c2φ2

x)dx (4.9)

where we assume φt and φx to decay rapidly and asymptotically. Note that we setthe mass density to be equal to unity. Taking the time derivative

dE

dt=

∫ +∞

−∞(φtφtt + c2φxφxt)dx (4.10)

gives on integrating by parts

dE

dt=

∫ +∞

−∞φt(φtt − c2φxx)dx = 0 (4.11)


where we have dropped the boundary term because of the decaying condition atinfinity and used the wave equation. (4.11) reveals that the energy is conserved.

From a physical point of view it is clear that the given set of Cauchy initialconditions (4.2) are appropriate for the uniqueness of the solution (4.8). Indeed ifwe take two solutions labeled by φ1 and φ2 with the corresponding f = f1 and f2 andg = g1 and g2 for the initial conditions, then the difference Φ(x, t) ≡ φ1−φ2 satisfiesthe one-dimensional wave equation with Φ(x, 0) = 0 and Φt(x, 0) = 0. It shows thatthe energy for Φ at t = 0 is zero and stays so for all time by the conservation of theenergy proved above. Further Φ(x, 0) = 0 and hence Φ ≡ 0 justifying the uniquenessof the solution obtained in (4.8).

That the D’Alembert’s solution is well-posed can be established by consideringtwo solutions such that their difference is small :|f1 − f2| < ε and |g1 − g2| < ε.D’Alembert’s formula (4.8) gives

|φ1 − φ2| <1

2(ε+ ε) +

1

2c

∫ x+ct

x−ctε ds = ε(1 + t) (4.12)

We therefore see that for a finite time span, a small change in the initial conditionsproduces a proportionately small change in the solution signifying that the solutionis well posed.

Consider an arbitrary point P (x, ct) in the xt−plane as shown in Figure 4.1.The characteristics through P are given by the equation x − x = ±c(t − t) whichintersects the x axis at A and B whose abscissae are respectively x− ct and x+ ct.Thus we see that D’Alembert’s solution at P depends on f at A and B and g betweenA and B. The closed segment AB, which is also the base of an isosceles triangle orthe characteristic triangle PAB, is called the domain of dependence of P .

FIGURE 4.1: Domain of dependence.


Example 4.1

Construct the solution of the following wave equation

φtt = c2φxx, x > 0, t > 0

subject to the following initial-boundary conditions

φ(x, 0) = 0 = φt(x, 0), x > 0, φ(0, t) = s(t), t > 0

Demonstrate that the solution can be expressed in the form

φ(x, t) =

0 if 0 < t ≤ x

c

s(t− xc) if t > x

c

(4.13)

4.2 Solving by Riemann method

We take the following normal form of a hyperbolic PDE

L(φ) ≡ φxy +Dφx + Eφy + Fφ = 0 (4.14)

where D, E and F are functions of the variables x and y. The adjoint to L from(2.78) is

M(ψ) ≡ ψxy − (Dψ)x − (Eψ)y + Fψ = 0 (4.15)

The essence of Riemann method consists in exploiting the two-dimensional formof Gauss’ theorem according to which one equates the flux integral of a vector ~Athrough the positively oriented closed boundary curve C with the double integral ofthe divergence of ~A over the full region S namely,

∫CAndl =

∫ ∫Sdiv ~Ads, where n is

an outwardly drawn normal to C. Actually this divergence form of Gauss’ theoremis the counterpart of Green’s theorem for the curl. Indeed if the components of ~A are(P,Q) then the divergence theorem assumes the form

∮CPdy−Qdx =

∫ ∫S

( ∂P∂x

+ ∂Q∂y

).

Since we can write

ψLφ− φMψ =∂P

∂x+∂Q

∂y(4.16)

employing the two-dimensional divergence theorem we connect

∫S

[ψLφ− φMψ]ds =

∫S

(∂P

∂x+∂Q

∂y)ds =

∫C

[P cos(n, x) +Q cos(n, y)]dl (4.17)

where from the given expressions of L and M


FIGURE 4.2: The curve γ in the region S.

P =1

2(ψφy − φψy) +Dφψ, Q =

1

2(ψφx − φψx) + Eφψ (4.18)

We determine ψ from the solution of the homogeneous equation

M(ψ) = 0 (4.19)

subject to the requirement that

ψ = 1 at P (x = ξ, y = η) (4.20)

and that on the characteristics it obeys the following conditions

ψy −Dψ = 0 on x = ξ and ψx − Eψ = 0 on y = η (4.21)

We refer to ψ as the Riemann function or characteristic function. Riemann’s methodis essentially a reduction of the problem to a more easily tractable boundary-valueproblem proposed in (4.18), (4.19) and (4.20).

To proceed with the relation (4.16), we look at the Figure 4.2 in which a regionis formed when the non-charateristic curve γ is intersected by the hands of thecharacteristics AP and PB at the points A and B. We take the coordinates of thepoint P to be (ξ, η), with y = η and x = ξ being the equations of the lines APand PB respectively. We assume φ and ∂φ

∂nto be given on γ. Employing (4.18) and

noting that L(φ) = 0 is the given PDE, we obtain on integrating over the peripheryof the region bounded by the curve γ and the lines PB and PA


∫(γ+PB+PA)

[P cos(n, x) +Q cos(n, y)]dl = 0

where P and Q are given by (4.17).

In the integration over PB, only the P -term survives because cos(n, y) = 0.Writing

1

2

∫ B

P

ψφydy =1

2φψ|BP −

1

2

∫ B

P

φψydy

we obtain ∫ B

P

Pdy =1

2[(φψ)|B − (φψ)|P ]−

∫ B

P

φ(ψy −Dψ)dy (4.22)

For the integral∫ PA

, we notice that cos(n, x) = 0 and cos(n, y) = −1, n beingthe outer normal, we obtain

−∫ P

A

Qdx =1

2[(φψ)|A − (φψ)|P ] +

∫ P

A

φ(ψx − Eψ)dx (4.23)

The two integrals in the right sides of (4.21) and (4.22) vanish because of theconditions (4.20) that the conditions fulfill. Further, making use of (4.19) we arriveat the central result of the Riemann’s method

φP =

∫γ

[P cos(n, x) +Q cos(n, y)]dl +1

2[(φψ)|A + (φψ)B ] (4.24)

giving the value of φ at an arbitrary point P .

To see how Riemann’s method works in practice we consider an example fromhydro-mechanics where it is given that

D = − λ

x+ y= E, F = 0

where λ is a constant.Substituting in the two PDEs written down in (4.20) we immediately obtain two

solutions for ψ

x = ξ : ψ = c1(ξ + y)−λ; y = η : ψ = c2(x+ η)−λ (4.25)

where c1 and c2 are constants.Requiring the condition (4.19) to hold implies c1 = c2 = (ξ + η)λ which means

ψ = (ξ + η

x+ y)λ (4.26)

However such a ψ does not provide a general solution to M(ψ) = 0 which is givenby a more complicated form

ψ = (ξ + η

x+ y)λ 2F1(λ+ 1,−λ, 1, ; z) (4.27)


where 2F1(λ+ 1,−λ, 1, ; z) is the Gauss hypergeometric series and the quantity z isdefined to be

z = − (x− ξ)(y − η)

(x+ y)(ξ + η)(4.28)

Proof: We briefly outline the steps. Taking the form (4.27) for ψ we computefirst the derivatives ψx, ψy and ψxy. Substituting them in the equation for M(ψ) = 0which has the form

M(ψ) ≡ ψxy +λ

x+ y(ψx + ψy)− 2λ

ψ

(x+ y)2= 0 (4.29)

results in

zxzyF′′(z) + zxyF

′(z)− λ(λ+ 1)

(x+ y)2F = 0 (4.30)

where 2F1(λ + 1,−λ, 1, ; z) ≡ F (z). Using (4.28) we can now easily convert (4.30)into the standard differential equation z(1− z)F ′′(z) + (1− 2z)F ′+ (λ(λ+ 1)F = 0.

Example 4.2

Find the Riemann function when L is given in the form

Lφ ≡ φxy + λφ = ρ(x, y)

where λ is a constant and ρ is the inhomogeneous term.

The Riemann function satisfies the adjoint equation

Mψ ≡ ψxy + λψ = 0

along with ψx = 0 on y = η, ψy = 0 on x = ξ and ψ(x = ξ, y = η) = 1.

A plausible form for ψ is

ψ = χ(t), t = (x− ξ)(y − η)

Substituting in the above equation for ψ gives the differential equation

td2χ

dt2+dχ

dt+ λχ = 0

A change of variable z = 2√λt yields the form

d2χ

dz2+

1

r

dχ

dz+ χ = 0


which can be recognized as the Bessel equation of order zero. We see that the solution

χ = J0(z), χ = J0[2√λ(x− ξ)(y − η)]

satisfies χ = 1 at t = 0 i.e. z = 0 and hence all requirements of ψ enlisted in (4.19)-(4.21) are met.

4.3 Method of separation of variables

(a) Three dimensions: spherical polar coordinates (r, θ, φ)

In this section we consider solving the wave equation through the method of sep-aration of variables. The procedure is similar to what was adopted for the Laplaceequation.

Consider the three-dimensional wave equation

∇2φ− 1

c2φtt = 0 (4.31)

where the form of ∇2 in three-dimensional spherical polar coordinates (r, θ, φ) hasbeen noted in (3.26). Further in (3.27) the relations of the variables r, θ and φ aregiven in terms of the Cartesian coordinates including the ranges of r, θ and φ.

Let us assume that the solution of φ exists in the variable-separated product form

φ(r, θ, φ, t) = f(r, θ, φ)T (t) (4.32)

where f is a function of (r, θ, φ) and T is a function of t only.Substituting (4.32) in (4.31) gives for f

∇2f + k2f = 0 (4.33)

which is known as the Helmholtz equation and an ODE for T

d2T

dt2+ c2k2T = 0 (4.34)

where k2 is the separation constant and arises because t has been separated out fromthe remaining variables.

The general solution of (4.34) is

T (t) = A cos(kct) +B sin(kct) (4.35)

where A and B are arbitrary constants.


We now further separate f(r, θ, φ) in the product form

f(r, θ, φ) = R(r)g(θ, φ) (4.36)

where R is a function of r only and g(θ, φ) is a function of the variables θ and φ only.Substituting (4.36) in (4.33) one obtains the following equations

d

dr(r2 dR

dr) + (k2r2 − p)R = 0 (4.37)

and

1

sin θ

∂

∂θ(sin θ

∂g

∂θ) +

1

sin2 θ

∂2g

∂φ2+ pg = 0 (4.38)

where p is a separation constant and arises because the variable r has been separatedout from the remaining ones (θ, φ) as is clear from (4.37) and (4.38).

Finally, we separate g(θ, φ) in the form

g(θ, φ) = s(θ)h(φ) (4.39)

where s is a function of θ only and h(φ) is a function of φ only. Substituting in (4.38)leads to the following pair of equations

sin θ∂

∂θ(sin θ

ds

dθ) + (p sin2 θ − ν2)s = 0 = 0 (4.40)

and

d2h

dφ2+ ν2h = 0 (4.41)

To summarize, all the variables have been separated out and we have one individualequation for each of them namely, (4.34), (4.37), (4.40) and (4.41).

Making a change of variable U =√rR converts (4.37) to

r2 d2U

dr2+ r

dU

dr+ (k2r2 − p− 1

4)U = 0 (4.42)

Substitution of y = kr then gives

y2 d2U

dr2+ y

dU

dy+ (y2 − α2)U = 0, α2 = p+

1

4(4.43)

which is a Bessel differential equation of order α.The general solution to (4.43) can be expressed as a combination of the Bessel

function Jα and Neumann function Yα. As such the radial function R(r) has the form


R(r) =1√r

[]CJα(kr) +DYα(kr)] (4.44)

where C and D are constants.Addressing now the equation (4.41) its general solution is

h(φ) = E cos(νφ) + F sin(νφ) (4.45)

where E and F are constants.Finally for the equation (4.40) if we make the substitutions µ = cos θ and p =

β(β + 1) which implies α = β + 12, it can be represented by

d

dµ[(1− µ2)

dg

dµ] + [β(β + 1)− ν2

1− µ2]g = 0 (4.46)

The above form is readily recognized as an associated Legendre equation whosegeneral solution is

g(µ) = GP νβ (µ) +HQνβ(µ) (4.47)

where G, H are constants and P νβ (µ) and Qνβ(µ) are the associated Legendre func-tions which we know to be expressible in terms of hypergeometric functions.

The general solution of the three-dimensional wave equation is given by theproduct of the functions T (t), R(r), h(φ) and g(µ) whose explicit forms are shownin (4.35), (4.44), (4.45) and (4.47) respectively involving the arbitrary constants A,B, C, D, E, F , G and H. In particular for the Helmholtz part if we look for a finiteand single-valued solution in the ranges 0 ≤ φ < 2π and 0 ≤ θ < π on the surfaceof a sphere having radius a, we have to restrict D = 0 and H = 0 since logarithmicsingularity affects Yβ+ 1

2(kr) at r = 0 and Qνβ(µ) at µ = ±1. Single-valuedness of the

functions sin(νφ) and cos(νφ) implies ν be a positive integer or zero. Convergenceof the series for P νβ (µ) in the range −1 ≤ µ ≤ +1 restricts β to a positive integer orzero in which case the series terminates..

Putting the above arguments together and keeping in mind the transformationY =

√rR the solution of the spherical Helmholtz equation (4.33) which is regular

at the origin takes the form

f = r−12 Jn+ 1

2(kr)Pmn (µ)[Enm cos(mφ+ Fnm sin(mφ)] (4.48)

where β = n = 0, 1, 2, ... and m = 0, 1, 2, ..., n. Further, Pmn (µ) = 0 if m > n. Notethat the portion

Y mn (θ, φ) =

n∑m=0

Pmn (µ)[Enm cos(mφ) + Fnm sin(mφ)] (4.49)

denotes the spherical harmonics of degree n, the θ-dependence arising from thedefinition µ = cos(θ).


(b) Cylindrical polar coordinates (r, θ, z)

We now inquire into the solution of the Helmholtz equation in cylindrical polarcoordinates (r, θ, z), It reads

∂2f

∂r2+

1

r

∂f

∂r+

1

r2

∂2f

∂θ2+∂2f

∂z2+ k2f = 0 (4.50)

where we have used ∇2 from (3.44). In (3.45) the relations of the variables r, θ andz are given in terms of the Cartesian coordinates including the ranges of r, θ and z.

We seek solutions of (4.50) in the variable-separated product form namely,

f(r, θ, z) = h(r, θ)g(z) (4.51)

where f is a function of r, θ while g is a function of z only. It is readily found thath and g obey the equations

∂2h

∂r2+

1

r

∂h

∂r+

1

r2

∂2h

∂θ2+ p2h = 0 (4.52)

and

d2g

dz2+ α2g = 0 (4.53)

where p2 is a separation constant and α2 = k2 − p2.

The general solution to the equation (4.53) is provided by

g(z) = A cos(αz) +B sin(αz) (4.54)

where A and B are constants.

To tackle (4.52) we subject h to a further separation of variables

h(r, θ) = R(r)Θ(θ) (4.55)

where R is a function of r only and Θ is a function of θ only.

Substituting (4.55) in (4.52) we deduce for the equation of R the form

d2R

dr2+

1

r

dR

dr+ (p2 − ν2

r2)R = 0 (4.56)

which is Bessel’s equation of order ν and for the equation of Θ the form

d2Θ

dθ2+ ν2Θ = 0 (4.57)

where ν2 is a separation constant.


The general solution to (4.56) is a combination of Bessel and Neumann func-tions of order ν as already discussed while seeking solution of Laplace’s equation incylindrical coordinates in the previous chapter namely

R(r) = CJν(pr) +DYν(pr) (4.58)

where C and D are constant, the one for (4.57) is of the periodic type

Θ(θ) = E cos(νθ) + F sin(νθ) (4.59)

where E and F are constants. The condition θ + 2π = θ requires ν to be zero or aninteger.

The general solution of the Helmholtz equation is given by the product of thefunctions g(z), R(r) and Θ(θ) whose explicit forms are shown in (4.54), (4.58) and(4.59) respectively involving the arbitrary constants A, B, C, D, E and F .

4.4 Initial value problems

(a) Three dimensional wave equation

We analyze first the following initial value problem for the three-dimensionalinhomogeneous wave equation

φtt − c2∇2φ(x, y, z, t) = ρ(x, y, z, t) (4.60)

where ρ is an inhomogeneous term. We subject (4.60) to the initial conditions

φ(x, y, z, 0) = f(x, y, z), φt(x, y, z, 0) = g(x, y, z) (4.61)

where f is assumed to have continuous derivatives up to second order while onlycontinuous derivatives of g are sufficient.

To tackle the problem we proceed in two steps. In the first step we set up ascheme in which a homogeneous counterpart of the equation (4.60) is solved sub-ject to the inhomogeneous initial conditions (4.61). In the second step we solve theinhomogeneous equation (4.60) on which a set of homogeneous initial conditions isimposed. By the principle of superposition we then arrive at a unique solution of(4.60) being accompanied by the initial conditions (4.61).

Step 1

Let us take a sphere S of radius ct whose centre is at the point (x, y, z). Itsequation is given by


(x− x′)2 + (y − y′)2 + (z − z′)2 = c2t2 (4.62)

where (x′, y′, z′) is any point on the surface. The radius of the sphere being ct, wecan express the coordinates (x′, y′, z′) as

x′ = x+ lct ≡ x+ (sin θ cosφ)ct (4.63)

y′ = y +mct ≡ y + (sin θ sinφ)ct (4.64)

z′ = z + nct ≡ z + (cos θ)ct (4.65)

Consider the function g(x, y, z). Its spherical mean or the average value over Sis by definition

g(x, y, z, t) =1

4πc2t2

∫S

g(x′, y′, z′)ds′ (4.66)

Since ds′ = c2t2 sin θdθφ, g reads

g =1

4π

∫ π

0

∫ 2π

0

g(x+ lct, y +mct, z + nct) sin θdθdφ (4.67)

We now prove the following assertion.

A function ψ defined by

ψ(x, y, z, t) = tg(x, y, z, t) (4.68)

satisfies the homogeneous wave equation

ψtt − c2∇2ψ(x, y, z, t) = o (4.69)

obeying the initial conditions

ψ(x, y, z, 0) = 0, ψt(x, y, z, 0) = g(x, y, z) (4.70)

Proof: Partially differentiating both sides of (4.67) with respect to t gives

∂g

∂t=

c

4π

∫ π

0

∫ 2π

0

[l∂g

∂x′+m

∂g

∂y′+ n

∂g

∂z′] sin θdθdφ (4.71)

This can be re-expressed as

∂g

∂t=

c

4πc2t2

∫ π

0

∫ 2π

0

[l∂g

∂x′+m

∂g

∂y′+ n

∂g

∂z′]ds (4.72)


Applying the divergence theorem gives

∂g

∂t=

c

4πc2t2

∫ ∫ ∫V

(gx′x′ + gy′y′ + gz′z′)dv (4.73)

where the integration is carried over the corresponding volume V of S and dv is thevolume element.

Partially differentiating with respect to t again we can write

∂2g

∂t2+

2

t

∂g

∂t=

c

4πc2t2∂

∂t

∫ ∫ ∫V

(gx′x′ + gy′y′ + gz′z′)dv (4.74)

Inserting the specific form of the volume element namely, dv = r2 sin θdrdφ wherer = ct implies

∂2g

∂t2+

2

t

∂g

∂t=

c

4πc2t2∂

∂t

∫ ct

0

∫ π

0

∫ 2π

0

(gx′x′ + gy′y′ + gz′z′)r2 sin θdrdθdφ (4.75)

The partial derivative of t gets rid of the t-integral yielding the simple result

∂2g

∂t2+

2

t

∂g

∂t=c2

4π

∫ π

0

∫ 2π

0

(gx′x′ + gy′y′ + gz′z′) sin θdθdφ (4.76)

Using now the definition (4.68) of ψ which points to

ψtt − c2∇2ψ = (tg)tt − c2∇2(tg)

gives the form

ψtt − c2∇2ψ = t[(2

t

∂g

∂t+∂2g

∂t2)− c2∇2g]

Employing (4.67) and (4.76), the right side of the above can be seen to vanish provingthat ψ satisfies the homogeneous wave equation

ψtt − c2∇2ψ = 0 (4.77)

We now note the following points:

(i) The continuity of g implies from (4.66) that g is continuous too and hence itfollows from the definition of ψ that the first criterion of (4.70) holds.

(ii) If we differentiate (4.68) with respect to t we obtain

ψt = g + tgt (4.78)


Since the derivative of g is continuous so is gt and is therefore bounded. Hence itfollows from (4.78) that the second criterion of (4.70) holds too.

So far we have dealt with the function g. We now focus on f . Let us define afunction χ by

χ(x, y, z, t) = tf(x, y, z, t) (4.79)

where f is the spherical mean over S

f(x, y, z, t) =1

4πc2t2

∫S

f(x′, y′, z′)ds′ (4.80)

which, as similar to (4.67), is given by

f =1

4π

∫ π

0

∫ 2π

0

f(x+ lct, y +mct, z + nct) sin θdθdφ (4.81)

We now show that the following proposition holds:

The function χ is a solution of the homogeneous equation

χtt − c2∇2χ(x, y, z, t) = 0 (4.82)

obeying the initial conditions

χ(x, y, z, 0) = 0, χt(x, y, z, 0) = f(x, y, z) (4.83)

Proof: To proceed with the proof, we focus on the function χt which has contin-uous derivatives up to second order. Moreover it obeys, on using (4.82)

(χt)tt = (χtt)t = (c2∇2χ)t = c2∇2(χt) (4.84)

If for simplicity we set ξ = χt then from the above equation we see that ξ obeys thehomogeneous equation

ξtt(x, y, z, t)− c2∇2ξ(x, y, z, t) = 0 (4.85)

to be considered along with it the initial condition

ξ(x, y, z, 0) = f(x, y, z) (4.86)

as is clear from the second equation of (4.83).


On the other hand, if we twice differentiate the first equation of (4.83) withrespect to the variables x, y and z we obtain

χxx(x, y, z, 0) = 0, χyy(x, y, z, 0) = 0 χzz(x, y, z, 0) = 0 (4.87)

This has the implication from (4.85) that

ξt(x, y, z, 0) = χtt(x, y, z, 0) = c2∇2 = c2[χxx + χyy + χzz](x,y,z,0) = 0 (4.88)

We therefore find that ξ is a solution of the wave equation (4.85) subject to theinitial conditions

ξ(x, y, z, 0) = f(x, y, z), ξt(x, y, z, 0) = 0 (4.89)

If we now superpose the ψ-equation from (4.77) and ξ-equation from (4.85)and call

Φ = ψ + ξ (4.90)

then Φ obeys the wave equation

Φtt(x, y, z, t)− c2∇2Φ(x, y, z, t) = 0 (4.91)

To find the accompanying initial conditions to (4.91), we simply have to add (4.70)and (4.89).

To summarize, the initial value problem for the homogeneous form of the PDE

Φtt(x, y, z, t)− c2∇2Φ(x, y, z, t) = 0 (4.92)

subject to the inhomogeneous conditions

Φ(x, y, z, 0) = f(x, y, z), Φt(x, y, z, 0) = g(x, y, z) (4.93)

has the following solution for Φ(x, y, z, t)

Φ(x, y, z, t) =1

4π

∂

∂t

∫ π

0

∫ 2π

0

tf(x+ lct, y +mct, z + nct) sin θdθdφ

+1

4π

∫ π

0

∫ 2π

0

tg(x+ lct, y +mct, z + nct) sin θdθdφ (4.94)

where l,m, n are defined in (4.63) - (4.65) and we have used (4.68) and (4.79) alongwith the definition ξ = χt. The above integral representation of Φ is called Poisson’sformula and gives the propagation of an initial disturbance over the entire space.(4.94) is also called the Kirchhoff’s formula and stands for the specific solution ofthe initial value problem formulated in (4.92) and (4.93). Note that (4.92) is thehomogeneous counterpart of the equation (4.60) but tied up here with the set ofinhomogeneous initial conditions (4.93).


Step 2

We now proceed to find the solution of the inhomogeneous equation

Ψtt(x, y, z, t)− c2∇2Ψ(x, y, z, t) = ρ(x, y, z, t) (4.95)

which is subject to the homogeneous conditions

Ψ(x, y, z, 0) = 0, Ψt(x, y, z, 0) = 0 (4.96)

Suppose t is translationally shifted by a parameter τ and we identifyg(x, y, z, t) ≡ ρ(x, y, z, t− τ). Then by (4.68) ψ given by

ψ(x, y, z, t; τ) = (t− τ)ρ(x, y, z, t− τ) (4.97)

and reads explicitly

ψ(x, y, z, t; τ) =t− τ4π

∫ π

0

∫ 2π

0

ρ(x+ lc(t−τ), y+mc(t−τ), z+nc(t−τ); τ) sin θdθdφ

(4.98)

and serves as a solution of the wave equation (4.69). The accompanying condi-tions are

ψ(x, y, z, τ) = 0, ψt(x, y, z, τ) = ρ(x, y, z, τ) (4.99)

where because of the factor in (4.97) the condition t = 0 has been transformed tot = τ .

Concerning Ψ we take its form as the integral

Ψ(x, y, z, t) =

∫ t

0

ψ(x, y, z, t; τ)dτ (4.100)

which implies

Ψ(x, y, z, 0) = 0 (4.101)

Further partially differentiating both sides of (4.100) with respect to t gives

Ψt = ψ(x, y, z, t; t) +

∫ t

0

ψt(x, y, z, t; τ)dτ (4.102)

Because of the vanishing of the first term in the right side of (4.102) due to the firstcondition of (4.99) at t = τ , Ψt corresponds to the integral

Ψt =

∫ t

0

ψt(x, y, z, t; τ)dτ (4.103)


and we conclude that

Ψt(x, y, z, 0) = 0 (4.104)

If we partially differentiate again the expression of Ψt, we obtain from (4.103)

Ψtt = ψt(x, y, z, t; t) +

∫ t

0

ψtt(x, y, z, t; τ)dτ (4.105)

Here the first term in the right side is ρ(x, y, z, t) by the second condition of (4.99)while we can replace ψtt by c2∇2ψ in the second term. This means that Ψ satisfiesthe inhomogeneous equation

Ψtt = ρ(x, y, z, t) + c2∇2

∫ t

0

ψ(x, y, z, t; τ)dτ = c2∇2Ψ (4.106)

where we have used (4.100). Note that the explicit expression for Ψ, by (4.98) and(4.100), is

Ψ(x, y, z, t) =

∫ t

0

dτt− τ4π

∫ π

0

∫ 2π

0

ρ(x+lc(t−τ), y+mc(t−τ), z+nc(t−τ); τ) sin θdθdφ

(4.107)We can set r = c(t − τ) to represent (4.107) in a more meaningful form from a

physical point of view. In fact we find

Ψ(x, y, z, t) =1

4πc2

∫ ct

0

∫ π

0

∫ 2π

0

1

rρ(x+ lr, y +mr, z + nr; t− r

c)dv (4.108)

where dv = r2 sin θdθdφ. Equivalently, (4.108) can be put as

Ψ(x, y, z, t) =1

4πc2

∫V

ρ(x′, y′, z′; t− rc)

rdx′dy′dz′ (4.109)

where r =√

(x− x′)2 + (y − y′)2 + (z − z′)2 and quite clearly represents a retardedpotential.

We therefore find that while Φ as given by (4.94) is a solution1 of the problem(4.92) and (4.93), Ψ as given by (4.108) is a solution of the problem (4.95) and(4.96). By applying the principle of superposition φ(≡ Φ + Ψ) is a solution of theproblem (4.60) and (4.61).

1For a qualitative analysis and a discussion of the uniqueness of the problem seeR.B.Guenther and J.W.Lee, Partial differential equations and mathematical physics andintegral equations, Dover Publications, New York (1988), pp. 407–410.


(b) Two dimensional wave equation

The Cauchy problem for the two-dimensional inhomogeneous wave equation isgiven by the following set of equations

φtt − c2(φxx + φyy) = ρ(x, y, t), −∞ < x, y < +∞ (4.110)

where ρ is the inhomogeneous term, subject to the initial conditions

φ(x, y, 0) = f(x, y), φt(x, y, 0) = g(x, y), −∞ < x, y < +∞ (4.111)

We require that f has continuous derivatives up to second order and g has continu-ous derivatives only.

To tackle this problem we follow Hadamard’s method of descent wherein welook upon the above problem as a reduced version of the three-dimensional Cauchyproblem in the sense that the solution does not depend on the z-variable.

With this aim in mind, let us first focus on the following two-dimensional initialvalue problem defined in terms of ψ(x, y, t) which satisfies the homogeneous equation

ψtt − c2∇2ψ(x, y, t)− = 0 (4.112)

subject to the inhomogeneous initial conditions

ψ(x, y, 0) = f(x, y), ψt(x, y, 0) = g(x, y) (4.113)

Let us recall the Poisson’s or Kirchoff’s formula for the three-dimensional case weencountered in the previous section. We found for the three-dimensional counterpartof (4.112) the solution (4.94). In terms of ψ it reads

ψ(x, y, z, t) =1

4π

∂

∂t

∫ π

0

∫ 2π

0

tα(x+ lct, y +mct, z + nct) sin θdθdφ

+1

4π

∫ π

0

∫ 2π

0

tβ(x+ lct, y +mct, z + nct) sin θdθdφ (4.114)

where we call ψ(x, y, z, 0) = α(x, y, z) and ψt(x, y, z, 0) = β(x, y, z)Note that the above formula could also have been expressed as

ψ(x, y, z, t) =1

4πc2∂

∂t

∫S

α(x′, y′, z′)

tds+

1

4πc2

∫S

β(x′, y′, z′)

tds (4.115)

where the integration is over the surface of the sphere S of radius ct with its centreat (x, y, z) and the point (x′, y′, z′) lying anywhere on the surface of S.

If we suppress the dependence on z for ψ and let the functions α and β beindependent of the variable z′, the above representation reduces to


ξ(x, y, t) =1

4πc2∂

∂t

∫S

α(x′, y′)

tds+

1

4πc2

∫S

β(x′, y′)

tds (4.116)

where we have replaced ψ by ξ to avoid confusion of notation. Writing f and g inplace of α and β respectively it is clear that the following form for ξ(x, y, t)

ξ(x, y, t) =1

4πc2∂

∂t

∫S

f(x′, y′)

tds+

1

4πc2

∫S

g(x′, y′)

tds (4.117)

is a solution of the problem posed in (4.112) and (4.113).

The solution can be expressed in a different form by considering the projectionof S on the z′ = 0 plane. It is evidently S′ which stands for the area of the circle(x− x′)2 + (y − y′)2 = c2t2. Now the normal to S at the point (x′, y′, z′) makes anangle θ with the z′-axis given by

cos θ =z − z′

ct=

1

ct

√c2t2 − (x− x′)2 − (y − y′)2 =

√c2t2 −R2

ct(4.118)

where R2 = (x− x′)2 + (y − y′)2. Therefore the projection ds′ of ds on the z′ = 0-plane is

ds′ = | cos θ|ds =

√c2t2 −R2

ctds (4.119)

Consequently we can write

∫S

f(x′, y′)

tds = 2c

∫ ct

0

∫ 2π

0

f(x+R cosφ, y +R sinφ)√c2t2 −R2

RdφdR (4.120)

and

∫S

g(x′, y′)

tds = 2c

∫ ct

0

∫ 2π

0

g(x+R cosφ, y +R sinφ)√c2t2 −R2

RdφdR (4.121)

where the factor of 2 in the right side appears because of the integration over sin θin the range between 0 and π.

As a result ξ acquires the following form

ξ(x, y, t) =1

2πc

∂

∂t

∫ ct

0

∫ 2π

0

f(x+R cos θ, y +R sin θ)√c2t2 −R2

RdθdR

+1

2πc

∫ ct

0

∫ 2π

0

g(x+R cos θ, y +R sin θ)√c2t2 −R2

RdθdR (4.122)

This above formula is called Poisson-Parseval formula.


Having derived the solution of the two-dimensional homogeneous wave equationwith inhomogeneous initial conditions, we attend to the inhomogeneous form

ηtt − c2(ηxx(x, y, t) + ηyy(x, y, t)) = ρ(x, y, t) (4.123)

where ρ is the inhomogeneous term, subject to the homogeneous initial conditions

η(x, y, 0) = 0, ηt(x, y, 0) = 0 (4.124)

As before taking cue from the three-dimensional case where we found for theinhomogeneous equation

ψtt(x, y, z, t)− c2(ψxx + ψyy + ψzz) = ρ(x, y, z, t) (4.125)

which is subject to the homogeneous conditions

ψ(x, y, z, 0) = 0, ψt(x, y, z, 0) = 0 (4.126)

the solution

ψ(x, y, z, t)=1

4π

∫ t

0

∫ π

0

∫ 2π

0

(t−τ)ρ(x+lc(t−τ), y+mc(t−τ), z+nc(t−τ);τ) sin θdθdφdτ

(4.127)

where l, m, n have the usual meanings as noted before.

If we now assume ψ and ρ to be independent of the variable z the reduced formof (4.127) in terms of η is

η(x, y, t) =1

4π

∫ t

0

∫ π

0

∫ 2π

0

(t−τ)ρ(x+lc(t−τ), y+mc(t−τ); τ) sin θdθdφdτ (4.128)

and serves as a solution of the problem (4.123) and (4.124). In (4.128) we have re-placed ψ by η to avoid confusion of notation.

Notice the symmetry of the integrand in (4.127) under θ → π − θ allows us tore-express it as

η(x, y, t) =1

2π

∫ t

0

∫ π2

0

∫ 2π

0

(t− τ ′)ρ(x+ lc(t− τ ′), y +mc(t− τ ′); τ ′) sin θdθdφdτ ′

(4.129)

where l = sin θ cosφ and m = sin θ cosφ.


Let us now go for the following change of variables

X = x+ c(t− τ ′) sin θ cosφ, Y = y +mc(t− τ ′) sin θ sinφ, τ = τ ′ (4.130)

which means that the region of integration is restricted to the zone

(X − x)2 + (Y − y)2 ≤ c2(t− τ)2, 0 ≤ τ ≤ t (4.131)

and the Jacobian matrix of the transformation

J =∂(X,Y, τ)

∂(θ, φ, τ ′)(4.132)

takes the form

J =

∂X∂θ

∂Y∂θ

∂τ∂θ

∂X∂φ

∂Y∂φ

∂τ∂φ

∂X∂τ ′

∂Y∂τ ′

∂τ∂τ ′

(4.133)

On performing the partial derivatives we obtain for J

J =

c(t− τ ′) cos θ cosφ c(t− τ ′) cos θ sinφ 0−c(t− τ ′) sin θ sinφ c(t− τ ′) sin θ cosφ 0− sin θ cosφ −c sin θ sinφ 1

(4.134)

and as a result the element dXdY dτ becomes

dXdY dτ = c2(t− τ ′)2 sin θ cos θdθdφdτ ′ (4.135)

It implies

(t−τ ′) sin θdθdφdτ ′=1

c2(t− τ ′) cos θdXdY dτ=

1

c√c2(t− τ)2 − (X − x)2 − (Y − y)2

(4.136)

where (4.130) has been used.By the above result the expression (4.128) for η becomes

η(x, y, t) =1

2πc

∫ ∫ ∫V

ρ(X,Y, τ)√c2(t− τ)2 − ε2

dXdY dτ (4.137)

where ε2 = (X − x)2 + (Y − y)2 and by the principle of superposition

φ = ξ + η (4.138)

is a solution of the problem posed in (4.110) and (4.111).


4.5 Summary

The main goal of this chapter was to introduce the wave equation as a typi-cal PDE of hyperbolic type. We started with the D’Alembert solution of the one-dimensional case and inquired into its special properties. Subsequently we discussedthe Riemann method of solving the normal form of a hyperbolic equation and therole played by the Riemann function with the characteristic boundary conditions.The Riemann function was found by prescribing the initial data on any smoothnon-characteristic curve and the solution was obtained in the form of quadratures.Next, we took up the problem of finding the solution of the wave equation by themethod of separation of variables. We considered both the cases of spherical andcylindrical coordinates and wrote down the exact forms of the general nature of thesolutions. We then turned to addressing in some detail the initial value problemsfor the homogeneous and non-homogeneous cases of the three and two-dimensionalwave equation and derived Poisson-Kirchoff and Poisson-Parseval formulas.


Exercises

1. Consider the problem of forced vibrations of a string represented by the in-homogeneous equation

φxx(x, t) =1

c2φtt(x, t) + ρ(x, t)

subject to

φ(0, t) = 0 = φ(l, t)

where x = 0 and x = l are the end points of the string and the function ρ is as-sumed to be given. If ρ(x, t) = ρ(x) sinωt then employing the separation of variablesφ(x, t) = φ(x) sinωt show that the given problem can be converted to the ODE form

φxx(x) +ω2

c2φ(x) = ρ(x), φ(0) = 0 = φ(l)

with the solution φ(x) = λn sin nπxl

, λn 6= 0 are arbitrary constants.

2. Show that the Cauchy problem for the following hyperbolic equation

y2φxx − yφyy +φy2

= 0


φ(x, 0) = f(x), φy(x, 0) = g(x)

is not well-posed.

3. Show that the function φ(x, y, t) = xyt− 16xyt3 solves the initial value prob-

lem of

φxx + φyy − φtt = xyt


φ(x, y, 0) = 0, φt(x, y, 0) = xy

4. The equation of an oscillatory string is solved by φ(x, t) satisfying the equation

φxx − φtt = 0

Show that the following function with the specified arguments

φ(x

x2 − t2 ,t

x2 − t2 )

also solves the above equation.


5. Consider the Cauchy problem wherein the one-dimensional inhomogeneouswave equation

φtt = c2φxx + ρ(x, t), −∞ < x <∞, t > 0

is guided by the initial conditions

φ(x, 0) = 0 = φt(x, 0), −∞ < x <∞

Show that if φ(x, t) is an odd function with respect to the variable x then φ(x, t) isvanishing at x = 0 while if φ(x, t) is an even function with respect to the variable xthen the spatial-derivative φx(x, t) is vanishing at x = 0.

6. Suppose that we are given a second order operator

L ≡ A ∂2

∂x2+ 2B

∂2

∂x∂y+ C

∂2

∂y2

where A, B, C are constant coefficients. By factorizing L in terms of first orderentities ξ1 and ξ2 as given by

ξi = ai∂

∂x+ bi

∂

∂y+ ci, i = 1, 2

show that a general solution to Lφ = 0 can be written as the superposition φ =φ1 + φ2, where φ1 and φ2 are solutions of ξ1φ1 = 0 and ξ2φ2 = 0.

Hence establish that for the PDE

∂2φ

∂x∂y+ 2

∂2φ

∂y2− ∂φ

∂x− 2

∂φ

∂y= 0

its general solution is expressible as

φ(x, y) = f(2x− y) + eyg(x)

for arbitrary functions f and g.

7. Show that for the one-dimensional Klein-Gordon equation

φtt − c2φxx + k2φ = 0

where k2 is a constant, the Riemann function is of the form

χ(x, y; ξ, η) = J0[ik√

(x− ξ)(y − η)]

where J0 is the Bessel function of order zero.


8. Show for the Euler-Poisson equation

φxy +k

x+ 1(φx + φy) = 0

where k is a constant, the Riemann function is of the form

χ(x, y; ξ, η) = (x+ y

ξ + η)kF (1− k, k, 1;

(ξ − x)(η − y)

(ξ + η)(x+ y)

where F is the hypergeometric function.

9. Given the inhomogeneous PDE

φtt = c2φxx + cos(x)

with the initial conditions

φ(x, 0) = sin(x), φt(x, 0) = 1 + x

find the following solution

φ(x, t) = sin(x) cos(ct) + xt+ t+1

c2(cos(x)− cos(x) cos(ct))

10. Consider the inhomogeneous hyperbolic equation in 1 + 1 dimensions

φtt = φxx + sin(ωt) sin(x)

Depending on whether the frequency ω is unity or not show that that the solutionassumes two different forms

φ(x, t) =

sin(ωt)−ω sin(t)

1−ω2 sin(x), 0 < ω 6= 1sin(t)−t cos(t)

2sin(x), ω = 1

(4.139)

Note that the case ω = 1 reveals a bounded nature of the solution whereas for ω = 1the solution blows up asymptotically with respect to t. This case corresponds to ωbeing a resonant frequency.

11. The nonlinear hyperbolic sine-Gordon equation in 1+1 dimensions is given by

φtt − φxx + sinφ = 0


Its solutions are defined on the whole real line and possess the decay propertylim|x|−>∞ φ(x, t) = 0. Obtain by substitution the following solution

φ(x, t) = 4 tan−1[

√1− ω2 cos(ωt)

ω cosh(√

1− ω2x)]

Such a solution is referred to as the breather.2 Notice that it is time-periodic andlocalized in space. The sine-Gordon equation is the sine version of the well knownKlein-Gordon equation.

2The above is a 1-soliton solution. Sine-Gordon equation enjoys multi-soliton solutionstoo. The Lagrangian for the sine-Gordon equation is given by L = 1

2(φ2t − φ2

x) + cosφ− 1.

Chapter 5

PDE: Parabolic form

5.1 Reaction-diffusion and heat equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.2 Cauchy problem: Uniqueness of solution . . . . . . . . . . . . . . . . . . . . . . . . 1405.3 Maximum-minimum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.4 Method of separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

(a) Cartesian coordinates (x, y, z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143(b) Three dimensions: spherical polar coordinates (r, θ, φ) . . . . . . 149(c) Cylindrical polar coordinates (r, θ, z) . . . . . . . . . . . . . . . . . . . . . . . . 151

5.5 Fundamental solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545.6 Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1575.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

5.1 Reaction-diffusion and heat equations

Reaction-diffusion equation

We start with a brief derivation of the reaction-diffusion equation. Let a uniformcircular tube of cross-sectional area A be held in such a way that that the x-axisis chosen to coincide with its axial direction. We focus on the portion of the tubemarked with x = a and x = b (see Figure 5.1). We consider the movement ofsome substance in this tube. Let the substance be governed by the density functionφ(x, t), x and t denoting respectively the location and time. The total amount of thesubstance inside the tube would be

Φ(t) =

∫ b

a

φ(x, t)Adx (5.1)

Typically, the substance can correspond to a population in which case φ(x, t) standsfor the population density. But here we will be interested with a flow possessingproperties like continuity and differentiability up to sufficient orders.

To inquire into the change of Φ with evolution of time, we note that two thingscan contribute to it. One is that there is a continuous creation or destruction of thesubstance due to certain physical reasons that we attribute to the presence of somesome external or internal agency. Specifically, if there is some source function f asdefined by

137


FIGURE 5.1: A tube of cross-sectional area A.

f > 0 : source (5.2)

it would give the rate r of generation of the substance in the tube

r =

∫ b

a

fAdx (5.3)

where, in principle, f = f(x, t, φ).

The other factor influencing the motion of the flow is the flux which denotes theamount of the substance that passes through the area at the point x at time t. Ifξ(x, t) define the flux function, then the rate of flux across the boundary is

q = A[ξ(a, t)− ξ(b, t)] (5.4)

By the principle of conservation we have from (5.1), (5.3) and (5.4) the bal-ance law

dΦ

dt= q + r (5.5)

which implies ∫ b

a

[φt(x, t) + ξx(x, t)− f(x, t, φ)]dx = 0 (5.6)

The continuity of the integrand allows us to conclude

φt(x, t) + ξx(x, t)− f(x, t, φ) = 0, t > 0 (5.7)

PDE: Parabolic form 139

About ξ(x, t), Fick’s law makes an educated approximation that it is proportionalto the gradient of the density as is often found to be true in many physical processes.Thus

ξ(x, t) = −κφx(x, t) (5.8)

where κ is a diffusion coefficient which can be a position-dependent quantity. Itsdimension is the ratio of the square of the length and time. We are therefore led tothe form

φt(x, t)− (κφx)x = f(x, t, φ), t > 0 (5.9)

(5.9) is called the reaction-diffusion equation.

In the absence of a source term we have the reduced version of the diffusionequation

φt(x, t)− (κφx)x = 0, t > 0 (5.10)

An appropriate boundary condition can be imposed if an insulation condition isassumed at the end point x = 0 i.e. there is no flux at x = 0:

φx(0, t) = 0, x = 0 (5.11)

Note that the two or three-dimensional analog of (5.10) is

φt(x, t)−∇× (κ∇φ) = 0, t > 0 (5.12)

where ∇ is a two or three-dimensional vector differential operator.

Heat conduction equation

In the heat equation κ is treated as a constant. The one-dimensional heat con-duction equation has the form

φt(x, t) = α2φxx(x, t), |x| <∞, t > 0 (5.13)

where α2 = κ.In two or three dimensions the heat conduction equation has the form

φt(~r, t) = α2∇2φ(~r, t), t > 0 (5.14)


5.2 Cauchy problem: Uniqueness of solution

In this section we attempt to establish the uniqueness of the solution for theone-dimensional heat equation with respect to the Cauchy problem. It states that ifφ(x, t) is a solution of the heat equation (5.13) obeying

|∫ +∞

−∞φ(x, t)dx| <∞ (5.15)

along with fulfilling the conditions

φ(x, 0) = f(x), x ∈ <, φx(x, t)→ 0 as x→ ±∞ (5.16)

then the heat equation has a unique solution.

Proof: The proof is rather simple and goes as follows. We assume, if possible,two plausible solutions of the heat equation and name them as φ1(x, t) and φ2(x, t).If ψ(x, t) is the difference of the two solutions then it satisfies the boundary valueproblem

ψt(x, t) = α2ψxx(x, t), t > 0 (5.17)

subject to

ψ(x, 0) = 0, x ∈ <, ψx(x, t)→ 0 as x→ ±∞, t > 0 (5.18)

Define a time-dependent quantity Ψ(t) given by the integral

Ψ(t) =

∫ +∞

−∞ψ2(x, t)dx ≥ 0, Ψ(0) = 0 (5.19)

Taking the time-derivative of both sides results in

dΨ

dt= 2

∫ +∞

−∞ψψtdx = 2α2

∫ +∞

−∞ψψxxdx (5.20)

Integrating by parts and noting that ψx(x, t) → 0 as x → ±∞ immediatelygives the inequality

dΨ

dt= −2α2

∫ +∞

−∞ψ2xdx ≤ 0 (5.21)

It follows that Ψ(t) is a non-increasing function of t to be compared with the claimin (5.19) that Ψ(t) ≥ 0 and Ψ(0) = 0. The only way of reconcilement is Ψ(t) = 0which implies that ψ = 0. In other words, φ1(x, t) = φ2(x, t) and we have provedthe uniqueness criterion.


5.3 Maximum-minimum principle

We focus on the one-dimensional inhomogeneous heat equation defined over aspace-time rectangular domain Ω : [0, l]× [0, τ ]

φt(x, t)− α2φxx(x, t) = ρ(x, t), 0 ≤ x ≤ l, 0 ≤ t ≤ τ (5.22)

where ρ is the inhomogeneous term and φ(x, t) is continuous in the region Ω. Weare going to show that if ρ < 0 then the maximum of φ(x, t) is attained either att = 0 or on one of the boundaries x = 0 or x = l. However, the maximum is neverattained in the interior of Ω or at t = τ . On the other hand, if ρ > 0 then theminimum of φ(x, t) is attained either at t = 0 or on one of the boundaries x = 0 orx = l. However, the minimum is never attained in the interior of Ω or at t = τ .

Proof: We prove by contradiction. First we take ρ < 0. Then if the maximum ofφ occurs at an interior point of Ω then at this point φt = 0, φx = 0 and φxx ≤ 0and we run into a contradiction by looking for consistency with the inhomogeneousequation (5.20). However, if the maximum of φ occurs at t = τ , then at this pointwhich is on the boundary of Ω, φ could be increasing and we have φt ≥ 0, φx = 0and φxx ≤ 0. Again we run into a contradiction on looking for consistency with theinhomogeneous equation. Hence we conclude that the maximum must occur at t = 0or on one of the boundaries x = 0 or x = l and not in the interior.

The lines of arguments for the case ρ > 0 are exactly similar and we arrive atthe conclusion that the minimum must occur at t = 0 or on one of the boundariesx = 0 or x = l and not in the interior.

We now turn to the homogeneous portion

φt(x, t)− α2φxx(x, t) = 0, 0 ≤ x ≤ l, 0 ≤ t ≤ τ (5.23)

where φ(x, t) is continuous in the region Ω. We will presently see that the maximumand minimum of φ(x, t) are attained at t = 0 or on one of the points on the boundaryx = 0 or x = l. As with the inhomogeneous case here too the maximum or minimumis never attained in the interior of Ω or at t = τ .

Proof: We first address the maximum case. Let us follow the simplest approachby defining a function χ(x, t)

χ(x, t) = φ(x, t) + εx2, ε > 0 (5.24)

where φ(x, t) solves the homogeneous equation (5.23). Since φ(x, t) is continuous inΩ, χ(x, t) too is continuous in Ω and attains a maximum in Ω, say, at (x.t) in Ω.Further, since φ(x, t) is a solution of the homogeneous equation we observe that

χt(x, t)− α2χxx(x, t) = φt(x, t)− α2φxx(x, t)− 2α2ε = −2α2ε < 0 (5.25)


We thus find that χ(x, t) solves an inhomogeneous equation.

If (x.t) is an interior point of Ω then in it

χt(x.t) ≥ 0, χxx(x.t) ≤ 0, (5.26)

The two inequalities in (5.26) imply χt(x, t) − α2χxx(x, t) ≥ 0 which is to be com-pared with the validity of (5.25). Taken together it follows that

0 ≤ χt(x, t)− α2χxx(x, t) < 0 (5.27)

which is clearly a contradiction. Since χ(x, t) is a solution of an inhomogeneousequation we can use the earlier result to claim that χ(x, t) attains its maximumvalue M on the initial line t = 0 or on the boundary x = 0 or x = l.

From the definition of χ(x, t) we can write

χ(x, t) = φ(x, t) + εx2 ≤M + εl2, 0 ≤ x ≤ l, 0 ≤ t ≤ τ (5.28)

and as a consequence obtain

χ(x, t) ≤M, 0 ≤ x ≤ l, 0 ≤ t ≤ τ (5.29)

since ε is an arbitrary parameter. Thus the maximum principle holds for the homo-geneous heat equation.

The result for the minimum case follows from the criterion that if φ(x, t) is asolution of the homogeneous heat equation then so is −φ(x, t). Since the maximumprinciple holds for φ(x, t), it follows that if m is the minimum of φ(x, t), then −mis the maximum value of −φ(x, t) in Ω and so our conclusion would be that φ(x, t)assumes its minimum on t = 0 or on the boundary x = 0 or x = l. In other wordswe have the result

m ≤ χ(x, t) ≤M, 0 ≤ x ≤ l, 0 ≤ t ≤ τ (5.30)

Example 5.1

Make use of the maximum-minimum principle to estimate the maximum andminimum of the function f(x, t) = 1 − x2 − 2α2t(1 + 3νx) − νx3 where ν > 0 andx ∈ [0, 1] in the context of the heat equation φt = α2φxx, 0 ≤ t ≤ T , T is fixed.

We first note that since fxx = −2 − 6νx and ft = −6να2x − 2α2, f(x, t) is asolution of the heat equation. Also, the given function f(x, t) < 1 unless x = 0, t = 0in which case it is equal to 1. Hence, f(x, t) ≤ 1 and 1 is the maximum value off(x, t) at (x = 0, t = 0) . On the other hand, in the given intervals of x and t,f(x, t) ≥ −2α2T (1 + 3ν) − ν indicating that −6να2T − 2α2T − ν is the minimumvalue of f(x, t) at (x = 1, t = T ).

The above estimates of the maximum and minimum can be verified from themaximum-minimum principle of the heat equation. We know that, given the whole


rectangle [0, 1] × [0, T ], the maximum or minimum is attained either on t = 0 oron one of the other sides x = 0 or x = 1. Specifically, we see that at t = 0,f(x, 0) = 1−x2− νx3 which clearly shows its maximum is 1 at x = 0 and minimumis −ν at x = 1. For the side where x = 0, f(0, t) = 1− 2α2t, maximum or minimumoccurs respectively at 1 when t = 0 and 1−2α2T where t = T while for the side x = 1for which f(1, t) = −6να2t − 2α2t − ν, maximum or minimum occurs respectivelyat −ν when t = 0 and −6να2T − 2α2T − ν at t = T . We therefore conclude thatthe maximum value of 1 for f occurs at (x = 0, t = 0) and the minimum value of−6να2T − 2α2T − ν for f occurs at (x = 1, t = T ) thereby justifying the maximum-minimum principle of the heat equation.

5.4 Method of separation of variables

(a) Cartesian coordinates (x, y, z)

We focus on the three-dimensional Cartesian coordinates (x, y, z) in terms ofwhich the three-dimensional heat conduction equation (5.14) reads

φxx + φyy + φzz = α2φt (5.31)

where φ = φ(x, y, z, t). We seek solutions in the variable-separated product formnamely,

φ(x, y, z, t) = F (x)G(y)H(z)T (t) (5.32)

This gives (5.31) the transformed ODE

1

F

d2F

dx2+

1

G

d2G

dy2+

1

H

d2H

dz2=α2

T

dT

dt(5.33)

Since the left-side of (5.33) is a function of the coordinates x, y and z and theright side is a function of t, consistency requires that each side must be a constantwhich we take as −k2. The t-integration at once separates out yielding the solution

T (t) = λe− k

2

α2 t (5.34)

and the remaining part reads

1

F

d2F

dx2+

1

G

d2G

dy2+

1

H

d2H

dz2= −k2 (5.35)

The left side of the above equation has three second order ordinary derivatives in x, yand z respectively. Taking any one to the right side, say, the z-dependent piece, gives


the right side in a variable-separated form with respect to the left side. Equating itto a constant, say −r2 we get

d2H

dz2+ s2H = 0 (5.36)

where s2 = k2 − r2.

Carrying on the separation process with respect to the remaining two variablesx and y yields two more ODEs namely,

1

F

d2F

dx2= − 1

G

d2G

dy2− r2 = −a2 (5.37)

where −a2 is a separation constant, This leads to the pair of equations

d2F

dx2+ a2F = 0,

d2G

dy2+ b2G = 0 (5.38)

where b2 = r2 − a2.

Since the differential equation (5.36) for the function H and the ones in (5.38)for F and G have solutions given by a combination of sine and cosine functions, thegeneral solution φ(x, y, z, t), which includes the solution (5.34) of the T -equation,emerges as

φ(x, y, z, t) = (c1 cos ax+ c2 sin ax)(c3 cos by + c4 sin by)(c5 cos sz + c6 sin sz)e− k

2

α2 t

(5.39)where k2 = a2 + b2 + s2, c1, c2, c3, c4, c5, c6 are constants and the constant λ hasbeen absorbed in one of the coefficients.

The procedure to handle the one-dimensional and two-dimensional cases is muchsimpler and is illustrated in the two worked-out problems below.

Example 5.2

The temperature φ(x, t) in a rod of length l i.e. 0 ≤ x ≤ l is initially given bythe constant φ0 and the ends of the rod are kept at zero temperature

φ(0, t) = φ(l, t) = 0

Obtain the temperature profile of the rod at time t and at position x.

We assume that φ(x, t) can be expressed in a separation of variable form :

φ(x, t) = X(x)T (t)

Then the one-dimensional heat conduction equation (5.13) gives

X ′′

X=

T ′

α2T= k (constant)


Solving for X reveals the following possibilities

X(x) =

λ exp(√kx) + µ exp(−

√kx) k > 0

λx+ µ k = 0

λ cos(√−kx) + µ sin(

√−kx) k < 0

where λ and µ are constants.

Since we have to meet the boundary conditions X(0) = X(l) = 0, only the sinepart of the third solution is relevant :

X(x) = µ sin(nπx

l)

where√−k = nπ

l.

Employing now the solution of the T-equation which reads T ′(t) = kα2T wededuce for φ(x, t) the form

φ(x, t) =

∞∑n=1

µn sin(nπx

l) exp[−n

2π2α2t

l2]

on using the above value of k = −n2π2

l.

To meet the condition φ(x, 0) = φ0, we find from the above representation

φ0 =

∞∑n=1

µn sin(nπx

l)

Inverting gives

µn =2φ0

l

∫ l

0

sinnπx

ldx =

2φ0

l

l

nπ[cos(

nπx

l)]l0

=2φ0

nπ[(−1)n − 1]

=

− 4φ0nπ

n odd0 n even

Hence the required form for φ(x, t) is

φ(x, t) = −4φ0

π

∑n=odd

1

nsin(

nπx

l) exp[−n

2π2α2t

l2]

→ 0 as t→∞

at all points of the rod.

Example 5.3

Solve the problem of thermal waves in half space (z > 0) wherein the temperaturedistribution obeys the PDE

φt(z, t) = α2φzz(z, t), t > 0


subject to the boundary condition at z = 0

φ(0, t) = φ0 cosωt

where the frequency ω is assumed to be known.

To obtain the solution we apply the process of separation of variables and assumethat it is of oscillatory type

φ(z, t) = Reφ(z)e−iωt

at the applied frequency ω. Note that at z = 0 the given condition is

φ(0, t) = Reφ0e−iωt

with the complex amplitude φ(z) satisfying the ODE

d2φ

dz2= − iω

α2φ

Its solution can be put in the form

φ(z) = Aekz

where A is an overall constant and k2 is constrained by

k2 =ω

α2e−

iπ2 → k± = ±

√ω

α2e−i

π4 = ±(

ω

2α2)12 (1− i)

The solution corresponding to the positive sign blows up as z →∞ and so is tobe discarded. The acceptable form of the solution is then

φ(z) = A exp [−(ω

2α2)12 (1− i)z] = A exp(

iz

d− z

d)

where

d = (2α2

ω)12 = (

τα2

π)12

stands for the penetration depth corresponding to the frequency ω and periodτ = 2π

ω.

Identifying A = φ0 gives the desired solution

φ(z, t) = Reφ(z)e−iωt = φ0e− zd cos(

z

d− ωt)

indicating an exponentially damped nature of the temperature profile.


Example 5.4

In the context of an anisotropic diffusivity solve the heat equation

φt(x, y, t) = α2φxx(x, y, t) + β2φyy(x, y, t), α 6= β, t > 0

on a rectangular region defined by x ∈ (0, a) and y ∈ (0, b). The given boundaryconditions are

φ(0, y, t) = 0, φy(x, 0, t) = 0, φ(a, y, t) = 0, φy(x, b, t) = 0

while the initial condition is

φ(x, y, 0) = s(x, y)

To obtain the solution we apply the process of separation of variables in the followingproduct form

φ(x, y, t) = F (x)G(y)T (t)

This implies that the boundary conditions are converted to

F (0) = 0, F (a) = 0, G′(0) = 0, G′(b) = 0

We now substitute our chosen form for φ(x, y, t) in the given equation. Takingthe separation constant as −k we readily obtain the set of ODEs

α2F′′(x)

F (x)+ β2G

′′(y)

G(y)=T ′′(t)

T (t)= −k

where the primes denote the corresponding derivatives.

Solving for the T -equation we find

T (t) = λe−kt

where λ is an overall constant.The F -equation has the form

F ′′(x) +q

α2F (x) = 0, F (0) = 0, F (l) = 0

where q is another separation constant. Its solution that fits the boundary condi-tions is

Fn(x) = sin(mπx

a), m = 1, 2, ...

Next, the G-equation has the form

G′′(y) +κ

β2G(y) = 0, G′(0) = 0, G′(b) = 0


where κ is given by

κ = k − α2(nπ

a)2, n = 1, 2, ...

The solution which fits the derivative boundary conditions of G at y = 0 and y = b is

Gn(y) = cos(nπy

b), n = 1, 2, ...

Thus for a specific pair of indices m and n, φ(x, y, t) assumes the form

φ(x, y, t) = sin(mπx

a) cos(

nπy

b)e−kmnt, m, n = 1, 2, ...

where kmn is given by

kmn = π2(α2m2

a2+ β2 n

2

b2)

The general solution is the result of superposition over all m and n

φ(x, y, t) =

∞∑m=1

∞∑n=1

ξmn sin(mπx

a) cos(

nπy

b)e−kmnt

where ξmn is a set of unknown coefficients. The latter can be fixed on using theinitial condition at t = 0

s(x, y) =

∞∑m=1

∞∑n=1

ξmn sin(mπx

a) cos(

nπy

b)

To invert the above expression for the determination of ξmn we need to multiply

both sides by the product sin( lπxa

) cos( l′πyb

) and integrate over the whole rectangle

∫ a

0

∫ b

0

dxdys(x, y) sin(lπx

a) cos(

l′πy

b)

=

∞∑m=1

∞∑n=1

ξmn

∫ a

0

dx sin(mπx

a) sin(

lπx

a)

∫ b

0

dy cos(nπy

b) cos(

l′πy

b)

Since the two integrals in the right side are respectively equal to a2δml and b

2δnl′ ,

from the orthogonality conditions of sine and cosine function, the entire right sidehas the value ξll′

ab4

. Hence we arrive at the following result for ξmn

ξmn =4

ab

∫ a

0

∫ b

0

dxdys(x, y) sin(mπx

a) cos(

nπy

b)


(b) Three dimensions: spherical polar coordinates (r, θ, φ)

We will now be concerned with seeking solutions of the three-dimensional heatconduction equation

φt(~r, t) = α2∇2φ(~r, t), t > 0 (5.40)

in spherical polar coordinates. To this end we assume that the following decompo-sition holds

φ(~r, t) = R(r)F (θ)G(φ)T (t) (5.41)

Here ~r ≡ (r, θ, φ) are the spherical polar coordinates and ∇2 has the known repre-sentation

∇2 =1

r2

∂

∂r(r2 ∂

∂r) +

1

r2 sin θ

∂

∂θ(sin θ

∂

∂θ) +

1

r2 sin2 θ

∂2

∂φ2(5.42)

as already furnished in (3.27).Substitution of the ∇2 in the heat equation facilitates conversion to the follow-

ing ODE

1

R

d

dr(r2 dR

∂r) +

1

F

1

r2 sin θ

d

dθ(sin θ

dF

dθ) +

1

r2 sin2 θ

1

G

d2G

dφ2=α2

T

dT

dt(5.43)

Since the left-side of (5.43) is a function of r, θ and φ and the right side is afunction of t, consistency requires that each side must be a constant which we takeas −k2. The t-integration at once separates out and provides the solution

T (t) = λe− k

2

α2 t (5.44)

The remaining part has the form

r2 sin2 θ[1

R

d

dr(r2 dR

∂r) +

1

F

1

r2 sin θ

d

dθ(sin θ

dF

dθ) + k2] = − 1

G

d2G

dφ2= m2 (5.45)

where we have put the constant m2 to signal that the variable φ has been sepa-rated out.

The G-equation reads explicitly

d2G

dφ2+m2G = 0 (5.46)

whose general solution is

G(φ) = a cos(mφ) + b sin(mφ) (5.47)


where a and b are constants. There is a restriction on m to be m = 0.1.2., , , becausefor a physical system m and m+ 2π need to represent the same point.

Turning to r and θ part we see that they can be readily disentangled giving

1

R

d

dr(r2 dR

∂r) + k2r2 =

m2

sin2 θ− 1

F sin θ

d

dθ(sin θ

dF

dθ) = n(n+ 1) (5.48)

where to acquire a standard representation we have put the separation constant asn(n+ 1). As a result the R-equation takes the form

d2R

dr2+

2

r

dR

dr+ [k2 − n(n+ 1)

r2]R = 0 (5.49)

while the F -equation becomes

d2F

dθ2+ cot θ

dF

dθ+ [n(n+ 1)− m2

sin2 θ]F = 0 (5.50)

One can recognize the R-equation to be a spherical Bessel equation and itssolutions are called spherical Bessel functions. We write the solution R(r) as

R(r) = Ajn(kr) +Byn(kr) (5.51)

where A and B are constants and the links of jn(kr) and yn(kr) to Bessel functionJn and Neumann function Yn are respectively given by

jn(kr) =

√π

2rJn+ 1

2(r), yn(kr) =

√π

2rYn+ 1

2(r) (5.52)

Note that in the steady state system when k vanishes, R(r) reduces to the sim-ple form

R(r) = Crn +Dr−n−1 (5.53)

where C and D are constants.

On the other hand, the F equation depicts an associated Legendre equation andits solution can be expressed in terms of associated Legendre functions

F (θ) = EPmn (cos θ) + FQmn (cos θ) (5.54)

where E and F are constants. It may be remarked that we need to restrict n to integervalues to avoid a singularity at θ = 0 and that we have to have n > m because ofthe divergence problem of F at θ = 0. Further, θ = π points to a singularity forQmn (cos θ). Hence for a finite solution for F we are led to the form

F (θ) = EPmn (cos θ) (5.55)


where m is an integer. In such a case the series solution in (5.55) terminates at somedesired order.

Hence a finite and single-valued general solution of φ(r, θ, φ) is

φ(~r, t) =

∞∑n=0

n−1∑m=0

[Anjn(kr) +Bnyn(kr)][am cos(mφ) + bm sin(mφ)]Pmn (cos θ)e− k

2

α2 t

(5.56)

Or equivalently

φ(~r, t) = r−12

∞∑n=0

n−1∑m=0

[AnJn+ 12(kr) +BnJn+ 1

2(kr)][am cos(mφ)

+ bm sin(mφ)]Pmn (cos θ)e− k

2

α2 t (5.57)

where the boundary conditions determine the various unknown coefficients.

(c) Cylindrical polar coordinates (r, θ, z)

Let us proceed to seek solutions of the three-dimensional heat conduction equa-tion in cylindrical polar coordinates (r, θ, z)

φt(r, θ, z, t) = α2∇2φ(r, θ, z, t), t > 0 (5.58)

where from (3.45) ∇2 is given by

∇2 =∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2+

∂2

∂z2(5.59)

and the ranges of the variables are r ∈ [0,∞), θ ∈ [0, 2π) and z ∈ (∞,∞).

Let us assume that a solution exists in the variable-separated form

φ(r, θ, z, t) = R(r)F (θ)G(z)T (t) (5.60)

Substitution in (5.58) and using (5.59) gives the ODE

1

G

d2G

dz2+

1

R(d2R

dr2+

1

r

dR

dr) +

1

Fr2

d2F

dθ2=

1

α2

1

T

dT

dt(5.61)

Since the left side is a function of z, r and θ and the right side is a function of tconsistency requires that each side be equal to a constant which is set here as −k2.Then we get the following solution for T (t)

T (t) = Ae−α2k2t (5.62)


leaving for the remaining portion

1

R(d2R

dr2+

1

r

dR

dr) +

1

Fr2

d2F

dθ2+ k2 = − 1

G

d2G

dz2(5.63)

In the above equation the left side is a function of r and θ and the right side is afunction of z. Consistency requires that each side is equal to a constant which weset as −q2. Then we get the following equation for G(z)

d2G

dz2− q2z = 0 (5.64)

whose general solution is

G(z) = Beqz + Ce−qz (5.65)

where A and B are arbitrary constants. The solution corresponding to the first termin the right side blows up as z →∞ and so we need to set B = 0. This means thatG(z) reduces to

Z(z) = Ce−qz (5.66)

We are thus left with the differential equation

1

R(d2R

dr2+

1

r

dR

dr) + s2 = − 1

Fr2

d2F

dθ2(5.67)

where s2 = k2 + q2.In the above equation the left side is a function of r only and the right side is

a function of θ only. So consistency requires that each side is equal to a constantwhich we set as w2. Then we get the following pair of equations

d2R

dr2+

1

r

dR

dr+ (s2 − w2

r2)R = 0 (5.68)

andd2F

dθ2+ w2F = 0 (5.69)

Let us observe that with the substitution ρ = sr, the R-equation can be converted to

d2R

dρ2+

1

ρ

dR

dρ+ (1− w2

ρ2)R = 0 (5.70)

where R is now a function of the new independent variable ρ. This equation can bereadily recognized as the Bessel equation whose two linearly independent solutionsare Jw(ρ) and Yw(ρ) which are Bessel and Neumann functions respectively. Herew is to be looked upon as the order of the Bessel equation. Also note that Yw(ρ)diverges as ρ→ 0.

Reverting to the r variable, the solution of the R-equation is thus controlled bythe form


R(r) ∝ Jw(sr) (5.71)

because, from physics point of view, we are only interested in a finite solution asr → 0.

Finally, the F-equation can be recognized as the harmonic oscillator equationwhose solution is

F (θ) = D cos(wθ) + S sin(wθ) (5.72)

where D and S are arbitrary constants.

Hence the separable solution of φ(r, θ, z, t) that remains finite as r → 0 has thefollowing form

φ(r, θ, z, t) = ΛJw(sr)[M cos(wθ) +N sin(wθ)]e−(qz+α2k2t) (5.73)

where Λ, M and N are constants of integration.

Example 5.5

A cylinder of radius a is defined between the planes z = 0 and z = l. Its planefaces are maintained at zero temperature while the temperature on the curved sur-face is f(z). Find the steady symmetric temperature distribution within the cylinder.

Since we need to find the steady temperature distribution we can assume ∂φ∂t

= 0.Also because of the requirement of symmetric temperature distribution we can put∂φ∂θ

= 0. Hence the governing equation for the problem would read from (5.63)

1

G

d2G

dz2+

1

R(d2R

dr2+

1

r

dR

dr) = 0

Adopting a variable-separable form for φ

φ(r, z) = R(r)G(z)

we are guided to the following ODE

1

R(d2R

dr2+

1

r

dR

dr) = − 1

G

d2G

dz2= k2

where k2 is a separation constant. This yields the following pair of equations

d2R

dr2+

1

r

dR

dr− k2R = 0,

d2G

dz2+ k2G = 0

Since we are interested in a finite solution, the solution of the R-equation isobviously governed by the Bessel function of order zero of the first kind J0(kr)


while the boundary conditions φ(r, 0) = 0 = φ(r, l) restrict the solution of the G-equation only to the sine part with k = nπ

l, n = 1, 2, ... Hence from the principle of

superposition we can write the solution in the form

φ(r, z) =

∞∑n=1

ΛnJ0(nπr

l) sin(

nπz

l)

where Λn is an overall constant.To estimate Λn we make use of the given temperature function f(z) at the

surface. On putting r = a we have

f(z) =

∞∑n=1

ΛnJ0(nπa

l) sin(

nπz

l)

We now multiply both sides by sin(nπzl

), integrate between the limits 0 and l andmake use of the orthogonality condition for the sine function. This gives for Λn

Λn =2l

∫ l0dzf(z) sin(nπz

l)

J0(nπal

)

and leads to the required solution

φ(r, z) =

∞∑n=1

2

l[

∫ l

0

dzf(z) sin(nπz

l)]J0(nπr

l)

J0(nπal

)sin(

nπz

l)

5.5 Fundamental solution

We address the three-dimensional problem first. Let us express φ(~r, t) in termsof its Fourier transform φ(~p, t) and vice-versa through the pair of relations

φ(~r, t) = (2π)−32

∫ ∞−∞

φ(~p, t)ei~p·~rd3p, φ(~p, t) = (2π)−32

∫ ∞−∞

φ(~r, t)e−i~p·~rd3r

(5.74)

where ~p is the transform variable. Substituting the first equation in (5.14) gives

(2π)−32

∫ ∞−∞

d3p[∂φ(~p, t)

∂t+ α2p2φ(~p, t)]ei~p·~r = 0 (5.75)

Multiplying now the left side of (5.75) by e−i~p′·~r and performing integration over

the entire space gives

(2π)−32

∫ ∞−∞

d3p

∫ ∞∞

[d3rei(~p−~p′)·~r][

∂φ(~p, t)

∂t+ α2p2φ(~p, t)] = 0 (5.76)


Using the Fourier transform of unity yielding the form of the delta function

δ3(~p− ~p′) = (2π)−32

∫ ∞−∞

d3re−i(~p−~p′)·~r (5.77)

(5.76) reduces to the form∫ ∞−∞

d3p[∂φ(~p, t)

∂t+ α2p2φ(~p, t)]δ3~p− ~p′) = 0 (5.78)

(5.78) shows that φ(~p, t) obeys the equation

∂φ(~p, t)

∂t+ α2p2φ(~p, t) = 0 (5.79)

The solution of the PDE (5.79) can be expressed as

φ(~p, t) = f(~p)e−α2p2t (5.80)

where f is an arbitrary function of ~p. When (5.80) is substituted in the first equationof (5.74) we obtain for φ(~r, t)

φ(~r, t) = (2π)−32

∫ ∞−∞

d3pf(~p)ei~p·~r−α2p2t (5.81)

We find an interesting result from here. At t = 0, setting φ(~r, 0) ≡ ξ(~r) gives

ξ(~r) = (2π)−32

∫ ∞−∞

d3pf(~p)ei~p·~r (5.82)

whose inversion points to

f(~p) = (2π)−32

∫ ∞−∞

d3rξ(~r)e−i~p·~r (5.83)

In other words, knowing the initial form ξ(~r) of φ(~r, t), we can determine f(~p) andhence from (5.81), know φ(~r, t) for all times.

Next, let us take the delta-function representation of ξ(~r) to write it as ξ(~r) =δ3(~r − ~a) at some point ~r = ~a. Then by (5.77) f(~p) corresponds to

f(~p) = (2π)−32 e−i~p·~a, ~a ∈ <3 (5.84)

It therefore follows from (5.81)

φ(~r, t) = (2π)−32

∫ ∞−∞

d3pei~p·(~r−~a)−α2p2t (5.85)


To evaluate the integrals corresponding to the three components of ~p, we use thefollowing well-known results corresponding to the x, y and z components∫ ∞

∞dpxe

ipx(x−ax)−α2p2xt =

√π

α2te− (x−ax)2

4α2t , t > 0 (5.86)

and ∫ ∞−∞

dpyeipy(y−ay)−α2p2yt =

√π

α2te− (y−ay)2

4α2t , t > 0,∫ ∞−∞

dpzeipz(z−az)−α2p2zt =

√π

α2te− (z−az)2

4α2t , t > 0 (5.87)

These provide for φ(~r, t) the closed-form solution

φ(~r, t) = (2π)−3(π

α2t)32 e− |~r−~a|

2

4α2t , ~a ∈ <3, t > 0 (5.88)

It is regarded as the fundamental or principal solution of the heat equation.

In one-dimension the solution reduces to

φ(x, t) =1√

4πα2te− (x−a)2

4α2t , a ∈ <, t > 0 (5.89)

We can also carry out a superposition to express φ(~r, t) in a general sense

φ(x, t) =

∫ +∞

−∞g(a)

1√4πα2t

e− (x−a)2

4α2t da, a ∈ <, t > 0 (5.90)

where g(a) is some continuous and bounded function of a. On carrying out differen-tiation under the sign of integration it is clear that φ(x, t) is a solution of the heatequation.

Some remarks are in order concerning the asymptotic behaviour of φ(x, t) with

respect to t. For large values of both x and t, the ratios x2

tand a2

tbehave respectively

like∼ 1 and∼ O(ε), where ε << 1. As a result the square root of the product behaveslike xa

t∼ O(

√ε). It is thus straightforward to deduce from (5.90) that asymptotically

φ(x, t)→ e− x2

4α2t

√4πα2t

∫ +∞

−∞g(a)da, a ∈ < (5.91)

implying

φ(x, t)→ g(0)√2te− x2

4α2t , t > 0 (5.92)

where the appearance of g(0) follows from the definition of the Fourier transform ofg(a) on setting the transform parameter to be zero. Thus asymptotically a damping

nature of the solution is signaled with respect to the diffusion variable η = x2

t.


5.6 Green’s function

The Green’s function is defined in the following manner. In three-dimensions itis given by

G0(|~r − ~a|, t) = (2π)−3(π

α2t)32 e− |~r−~a|

2

4α2t , a ∈ <, t > 0 (5.93)

while in one-dimension it has the form1

G0(x− a, t) =1√

4πα2te− (x−a)2

4α2t , a ∈ <, t > 0 (5.94)

Note that the following properties for G0 hold:

(i) G0(|~r − ~a|, t) satisfies the homogeneous heat equation ∂G0

∂t= α2∇2G0. This

is because for t > 0, G0 coincides with φ(~r, t) according to (5.88) and since φ(~r, t)obeys the homogeneous heat equation, the same holds for G0.

(ii) As t→ 0+, the delta function limit is implied: G0(|~r − ~a|, t)→ δ3(~r − ~a).

(iii) G0(|~r − ~a|, t) obeys the integral∫ +∞

−∞(

1

4πα2t)32 e− |~r−~a|

2

4α2t dxdydz = 1 (5.95)

The proof easily follows by separating out the components and projecting the prod-uct form

e− |~r−~a|

2

4α2t = e− (x−ax)2

4α2t e− (y−ay)2

4α2t e− (z−az)2

4α2t (5.96)

The individual integrals with respect to x, y and z are evaluated by using the result∫ +∞

−∞e−k(ζ−aζ)2dζ =

√π

k, ζ = x, y, z and k =

1

4α2t(5.97)

(iv) The link between φ(~r, t) and φ(~r, 0), where the latter is assumed to beknown, is provided by the integral

φ(~r, t) =

∫ ∫ ∫dx′dy′dz′G0(~r − ~r′, t)φ(~r′, 0), t ≥ 0 (5.98)

1It corresponds to the Gaussian form discussed in Appendix A.


This is seen by operating with ( ∂∂t− α2∇2) on both sides of (5.98) and making

use of the first property that G0 satisfies the homogeneous wave equation. Thisin turn reflects that φ(~r, t) too is a solution of the homogeneous heat equation:( ∂∂t− α2∇2)φ(~r, t) = 0.

(v) As t→ 0, φ(~r, t)→ φ(~r, 0) on using property (ii) and (5.98) while as r →∞,φ(~r, t)→ 0 since G0(~r − ~r′, t)→ 0.

Consider the inhomogeneous case when φ(~r, t) solves the equation

∂φ

∂t− α2∇2φ = ρ(~r, t) (5.99)

where ρ is the inhomogeneous term. In such a case we can show that a particularsolution of (5.99) enjoys an integral representation

φ(~r, t) =

∫ ∫ ∫dx′dy′dz′

∫dt′G0(~r − ~r′, t− t′)ρ(~r′, t′) (5.100)

where G0 obeys the equation

(∂

∂t− α2∇2)G0(~r − ~a, t− t0) = δ3(~r − ~a)δ(t− t0) (5.101)

G0 is called the inhomgeneous Green’s function.

To establish our claim, we define

G0(~r − ~a, t− t0) = G0(~r − ~a, t− t0), t > t0 (5.102)

G0(~r − ~a, t− t0) = 0, t < t0 (5.103)

Then it is clear from the inhomogeneous equation (5.101) that for t > t0 both thesides vanish, the left side due to G0 being a solution of the homogeneous heat equa-tion by property (i) and the right side due to the property of the delta function.Likewise for t < t0, the left side vanishes by the definition of G0 and the right sidedue to the property of delta function.

Thus we have at hand a form of the particular solution of the inhomogeneousheat equation. Before we proceed to write down the general solution let us look atthe behaviour of the particular solution as t′ → 0+. If we integrate (5.101) over t′

around an interval (−ε,+ε), ε being an infinitesimal quantity, we find as ε proceedsto zero

∫ +ε

−εdt′

∂

∂t′G0(~r − ~a, t′)−α2

∫ +ε

−εdt′∇2G0(~r − ~a, t′) =

∫ +ε

−εdt′δ3(~r−~a)δ(t′) = δ3(~r−~a)

(5.104)


where we have suppressed t0. The first term in the left side when integrated outgives G0(~r − ~a,+ε) which, by property (ii) as mentioned earlier, acquires the formδ3(~r−~a) as ε goes to zero while the second term points to −α2ε(∇2G0)|t→0+ whichvanishes as ε goes to zero. Hence the behaviours of both sides are similar and con-sistency is maintained.

We are now in a position to write down the general solution for the inhomoge-neous heat equation as the superposition of the general solution of the homogeneouspart and the particular solution which we alluded to just now. In fact, if the initialcondition on φ(~r, t) is given and ρ(~r, t) is prescribed then from (5.98) and (5.100)the general solution for the inhomogeneous heat equation has the form

φ(~r, t) =

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞dx′dy′dz′G0(~r − ~r′, t)φ(~r′, 0)

+

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞dx′dy′dz′

∫ +∞

0

dt′G0(~r − ~r′, t− t′)ρ(~r′, t′), t > 0

(5.105)

The proof is evident. The main points to be noted are that φ(~r, t) solves the in-homogeneous heat equation (5.99), that with the limit t→ 0, φ(~r, t) goes to φ(~r, 0)and further if both φ(~r, 0) = 0 and ρ(~r, t) = 0 then φ(~r, t) also vanishes for all t.Moreover from the behaviour of G0(~r−~r′, t) and G0(~r−~r′, t) for both the inequalitiest > 0 and t < 0 already mentioned, φ(~r, t) → 0 as r → ∞. All this also speaks forthe required boundary conditions on φ(~r, t).

Since G0(~r − ~r′, t− t′) vanishes for t < t′, we can also project φ(~r, t) given bythe sum of the integrals

φ(~r, t) =

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞dx′dy′dz′G0(~r − ~r′, t)φ(~r′, 0)

+

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞dx′dy′dz′

∫ t

0

dt′G0(~r − ~r′, t− t′)ρ(~r′, t′), t ≥ 0

(5.106)

as the general solution.

5.7 Summary

Starting with a brief generation of the reaction-diffusion equation which yieldsthe form of the heat equation when the diffusion constant is treated as a constant, weexamined the Cauchy problem for the heat equation focusing also on the uniquenessof the solution. We explored the maximum-minimum principle for the heat equation


giving a detailed account of the proof by the approach of contradiction for both theinhomogeneous and the homogeneous forms. Subsequently we looked at the methodof solution of the heat equation by separation of variables, We considered the in-dividual case of the three-dimensional Cartesian coordinates which also guided usto the simpler versions of the one-dimensional and two-dimensional cases. We alsoderived the separable solutions of the three-dimensional polar coordinates and cylin-drical polar coordinates. Seeking the fundamental solution of the heat equation wasour next topic of inquiry when we extracted the form of the standard representationformula. We then solved the problem for the Green’s function that allows us to writethe general solution for the inhomogeneous heat equation as the superposition of thegeneral solution of the homogeneous equation along with a particular solution of theinhomogeneous equation both in their typical integral representations.


Exercises

1. Consider the one-dimensional parabolic equation

φt(x, t) = φxx(x, t), −∞ < x <∞, t > 0

obeying the condition

φ(x, 0) = f(x), −∞ < x <∞

Show that the following integral

φ(x, t) =1

2√πt

∫ +∞

−∞φ(w)e−

(x−w)2

4w dw, t > 0

is a solution, where φ(y) may be assumed to be bounded in −∞ < y <∞.

2. Solve the one-dimensional equation of heat conduction as given by

φt(x, t) = φxx(x, t), 0 < x <∞

where x denotes the distance measured along a semi-infinite rod from a fixed point,φ(x, t) is the temperature at any point of the rod at t > 0. Assume that the equationis subject to the initial condition

φ(x, 0) = q(x)

along with the boundary condition

φ(0, t) = 0

Assume also that φ and all of its x- derivatives vanish asymptotically.Examine the specific case of heat flow in a bar of length l when q(x) is defined by

q(x) =

1 x ∈ [1, 2]

0 otherwise(5.107)

and show that the profile is

φ(x, t) =

∫ ∞0

2

πp(cos p− cos 2p)e−p

2t sin pxdp

3. Solve the following initial-boundary value problem of heat flow in a bar oflength l


φt(x, t)− α2φxx(x, t) = q(x), 0 < x < l, t > 0

whose end points are maintained at zero temperature

φx(0, t) = 0, φx(l, t) = 0, φ(x, 0) = s(x), 0 < x < l

Show that the solution is

φx(x, t) =

∫ l

0

(2

l

∞∑n=1

e−n2π2t

l2 sinnπx

lsin

nπξ

l)s(ξ)dξ

4. Solve the following initial-boundary value problem of heat flow in a bar oflength l

φt(x, t)− α2φxx(x, t) = q(x), 0 < x < l, t > 0

whose end points are maintained at constant temperature

φx(0, t) = γ, φx(l, t) = δ, φ(x, 0) = s(x)

where γ and δ are constants.

5. Solve the one-dimensional equation of heat conduction

φt(x, t) = φxx(x, t), 0 < x < l

whose end-points are maintained at zero-temperature

φ(0, t) = 0, φ(1, t) = 0 for all t

along with the initial conditions

φ(x, 0) = x, 0 ≤ x ≤ l

2and φ(x, 0) = l − x, l

2≤ x ≤ l

6. Suppose that the one-dimensional parabolic equation

φt(x, t) = α2φxx(x, t), t > 0

has a positive solution φ(x, t). Show that the function ψ given by the form −2α2 φxφ

solves Burger’s equation2

2The general form of Burger’s equation is ψt(x, t) +ψψx = Dψxx(x, t). The term in theright side is the diffusion term.


ψt + ψψx = α2ψxx, t > 0

7. Consider the two-dimensional parabolic equation

φt(x, y, t) = φxx(x, y, t) + φyy(x, y, t), t > 1

where it is given that φ(x, y, 1) = 1− (x2 + y2)2. Show that

φ(x, y, t) = 1− (x2 + y2)2 − 16(x2 + y2)(t− 1)− 32(t− 1)2

solves the equation.

8. The two-dimensional diffusion equation is given by

φt = D(φxx + φyy), D > 0

Applying the transformation to polar coordinates x = r cos θ, y = r sin θ, obtainthe form

φt = D(φrr +1

rφr)

Show that the functions e−tJk(r) cos(kφ) and e−tJk(r) sin(kφ), k = 0, 1, 2, ... satisfythe above equation for D = 1. Here Jk is Bessel function of integral order k.

9. Consider a homogeneous solid sphere of radius a possessing an initial temper-ature distribution f(r), r being the distance from the centre of the sphere. Solve forthe temperature distribution T (r) in the sphere from the equation

φt = D(Trr +2

rTr)

where it may be assumed that the surface of the sphere is maintained at zero tem-perature.

10. Determine the temperature T (r, t) for an infinite cylinder 0 ≤ r ≤ a giventhat the initial temperature is T (r, 0) = f(r) and that the surface r = a is main-tained at zero degree.

Chapter 6

Solving PDEs by integraltransform method

6.1 Solving by Fourier transform method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1656.2 Solving by Laplace transform method . . . . . . . . . . . . . . . . . . . . . . . . . . . 1726.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

The integral transform method is an effective tool of reducing the complexity ofPDEs to a clear tractable form that allows many problems of mathematical physicsto be tackled straightforwardly. In this chapter we will be primarily interested in solv-ing problems of PDEs that requires an employment of either the Fourier transformor the Laplace transform technique. The essence of the method of either transform isto reduce a given PDE into an ODE in a somewhat similar spirit as that of the sepa-ration of variables. However, here the main difference lies in the fact that while in thecase of the Fourier transform, the domain of the differential operator is defined overthe full-line (−∞,+∞) and hence the method is applicable when the given functionis defined over an entire real line equipped with appropriate boundary conditions,for the Laplace transform, since the domain of the differential operator is only thehalf-line (0,∞), the method becomes relevant when we are dealing with an initialvalue problem. Such a half-line-full-line contrast of the two transforms is due to thefact that it often happens that Fourier transforms of certain classes of functions maynot exist, such as for example x2, and a control factor is necessary to be appendedto the function to ensure better convergence behaviour on at least on one side ofthe interval. We have discussed and collected some useful analytical results of theFourier and Laplace transforms in Appendix B and Appendix C respectively. In thefollowing we begin first with the Fourier transform method.

6.1 Solving by Fourier transform method

Let us apply the Fourier transform method to deduce solutions of some typicalPDEs that we generally encounter in standard physical situations. The followingillustrative examples will make the strategy clear.

165


Example 6.1

Solve the one-dimensional heat conduction equation

φt(x, t) = α2φxx(x, t), −∞ < x <∞, t > 0 (6.1)

where φ(x, t) is the temperature at any point of the rod and x is the distancemeasured along the rod from a fixed point, subject to the initial condition

φ(x, 0) = φ0(x) (6.2)

To solve by the Fourier transform method we assume that the relevant Fouriertransform exists and the Dirichlet’s conditions hold for its convergence (see Appendix

B). Multiplying both sides of the given heat equation by (2π)−12 e−ipx and integrating

over the full range (−∞,+∞) we have

(2π)−12

∫ +∞

−∞φt(x, t)e

−ipxdx = (2π)−12α2

∫ +∞

−∞φxx(x, t)e−ipxdx

= −(2π)−12α2

∫ +∞

−∞φx(x, t)[(ip)]e−ipx]dx

= (2π)−12α2(ip)2

∫ +∞

−∞φ(x, t)e−ipxdx

= −α2p2φ (6.3)

where φ is the Fourier transform of φ and in the second and third steps we integratedby parts and assumed φ and φx → 0 as x → ±∞. Taking the time-derivative outof the integral, the left side gets converted to an ordinary derivative and we get theform of an ODE

dφ

dt= −α2p2φ (6.4)

Next, we take the Fourier transform of (6.2) with respect to x and obtain

φ(p, 0) = (2π)−12

∫ +∞

−∞φ0(x)e−ipxdx ≡ φ0 (6.5)

As a result we can express the solution of (6.4) as

φ(p, t) = φ0e−α2p2t (6.6)

where we have accounted for (6.5). Notice that the nature of φ(p, t) is of decayingtype. The task now is to invert (6.4) to the coordinate representation of φ.

To proceed from here we make use of the fact that the quantity e−α2p2t can be

expressed as a Fourier transform

Solving PDEs by integral transform method 167

e−α2p2t = F [

1

a√

2te− x2

4tα2 ] (6.7)

This means that the right side of (6.6) is basically a product of two Fourier trans-forms i.e.

φ(p, t) = φ0F [1

a√

2te− x2

4tα2 ] (6.8)

By the convolution theorem of Fourier transform φ(x, t) can be provided anintegral representation

φ(x, t) =1

2α√πt

∫ ∞∞

dξφ0(ξ)e− (x−ξ)2

4tα2 (6.9)

and gives the temperature distribution at any point x in the rod. It can be solvedcompletely if we know the form of φ0.

Example 6.2

Find the solution of the two-dimensional Laplace’s equation in the half-space

φxx(x, y) + φyy(x, y) = 0, x ∈ (−∞,∞), y ≥ 0 (6.10)

which is governed by the boundary condition

φ(x, 0) = f(x), x ∈ (−∞,∞), (6.11)

along with φ(x, y)→ 0 for both |x| and y →∞.

Multiplying both sides of the Laplace’s equation by 1√2πe−ipx and integrating

over (−∞,∞) with respect to the variable x gives the conversion to an ODE

d2φ

dy2− p2φ = 0 (6.12)

where we have assumed φ and φx → 0 as x→ ±∞, φ stands for the Fourier transform

φ(p, y) =1√2π

∫ ∞−∞

φ(x, y)e−ipxdx (6.13)

and p is the transform parameter.

Taking Fourier transform of the boundary condition (6.11) gives

φ(p, 0) = f(p) (6.14)

where f(p) is the Fourier transform of f(x)


f(p) =1√2π

∫ ∞−∞

f(x)e−ipxdx (6.15)

Further

φ(p, y) → 0 as y →∞ (6.16)

Now the general solution of the differential equation (6.12) is of the form

φ(p, y) = Ae−py +Bepy (6.17)

where A and B are are arbitrary constants. Because of (6.16), we have to imposeB = 0 for p > 0 and A = 0 for p < 0. Combining we can write

φ(p, y) = f(p)e−|p|y for all p (6.18)

Setting g(p, y) = e−|p|y, the inversion formula of Fourier transform gives forg(x, y)

g(x, y) =1√2π

∫ ∞−∞

e−|p|yeipxdp (6.19)

Decomposing the integral range into two parts (−∞, 0) and (0,∞), g(x, y) turns outto be

g(x, y) =1√2π

∫ ∞0

e−p(y+ix)dp+1√2π

∫ ∞0

e−p(y−ix)dp =2y√

2π(x2 + y2)(6.20)

where the integrals are straightforward to evaluate.

Since by (6.18)

φ(p, y) = f(p)g(p, y) (6.21)

the convolution theorem of Fourier transform gives (see Appendix B)

φ(x, y) =1√2π

∫ ∞−∞

f(ξ)g(x− ξ, y)dξ (6.22)

which, by using (6.20), points to the integral

φ(x, y) =y

π

∫ ∞−∞

f(ξ)dξ

y2 + (x− ξ)2(6.23)

The exact evaluation of (6.14) depends on the specific form of f(x).


Example 6.3: D’Alembert’s solution

Solve by employing the method of Fourier transform the following Cauchy prob-lem of free vibration of an infinitely stretched string

φtt(x, t) = c2φxx(x, t), −∞ ≤ x <∞, t ≥ 0 (6.24)

subject to the initial conditions

φ(x, 0) = f(x) and φt(x, 0) = g(x) (6.25)

Multiplying both sides of (6.24) by 1√2πe−ipx, integrating with respect to x over

(−∞,∞) and assuming the vanishing limits φ, φx → 0 as x→ ±∞ gives

d2φ

dt2= −c2p2φ (6.26)

where φ(t, p) is the Fourier transform of φ with respect to x

φ(t, p) =1√2π

∫ ∞−∞

φ(x, t)e−ipxdx (6.27)

The general solution of the above equation is given by

φ(t, p) = Aeipct +Be−ipct (6.28)

where A and B are arbitrary constants and need to be determined from the givenconditions (6.25).

Defining the Fourier transforms of f and g to be

f(p) =1√2π

∫ ∞−∞

f(x)e−ipxdx, g(p) =1√2π

∫ ∞−∞

g(x)e−ipxdx (6.29)

along with their corresponding inversions

f(x) =1√2π

∫ ∞−∞

f(p)eipxdp, g(x) =1√2π

∫ ∞−∞

g(p)eipxdp (6.30)

we obtain from the initial conditions

φ = f ,dφ

dt= g at t = 0 (6.31)

From (6.26) and (6.31) the following estimates of A and B then follow

A =1

2(f − i g

pc), B =

1

2(f + i

g

pc) (6.32)

Substituting them in (6.28) we are guided to the form


φ(x, t) =1

2

1√2π

∫ ∞−∞

f(p)[e−ip(x−ct) + e−ip(x+ct)]dp

− i

2

1√2π

∫ ∞−∞

g(p)

p[e−ip(x−ct) − e−ip(x+ct)]dp (6.33)

Since using (6.30) it is possible to express for f(x+ ct) and f(x− ct) the respectiveintegrals

f(x+ ct) =1

2

1√2π

∫ ∞−∞

f(p)[e−ip(x+ct)dp (6.34)

f(x− ct) =1

2

1√2π

∫ ∞−∞

f(p)[e−ip(x−ct)dp (6.35)

and for the integral∫ x+ct

x−ct g(u)du the expression∫ x+ct

x−ctg(u)du = − i√

2π

∫ ∞−∞

g(p)

p[e−ip(x−ct) − e−ip(x+ct)]dp (6.36)

the final form of φ(x, t) is given by

φ(x, t) =1

2[f(x+ ct) + f(x− ct)] +

1

2c

∫ x+ct

x−ctg(u)du (6.37)

which coincides with the D’Alembert’s solution already determined in Chapter 4.

Example 6.4

Solve the linear flow of heat problem in a semi-infinite rod

φt(x, t) = α2φxx(x, t), −∞ < x <∞ (6.38)

where x denotes the distance measured along the rod from a fixed point, φ(x, t) isthe temperature at any point of the rod at t > 0. It is given that the equation issubject to the initial condition

φ(x, 0) = q(x), x > 0 (6.39)

and the following boundary condition at x = 0

φ(0, t) = r(t), t > 0 (6.40)

Further φ− > 0 as x− > ∞ for t > 0. Find the solution when r(t) = φo, where φ0

is a constant.

Here taking Fourier sine transform would be worthwhile

φs(p, t) =

√2

π

∫ ∞0

φ(x, t) sin pxdx (6.41)


Multiplying both sides of (6.24) by√

2π

sin px and integrating from 0 to ∞, the

equation is converted to the form

∂φs(p, t)

∂t= α2[

√2

πpr(t)− p2φs] (6.42)

where we have used the formula (B.24) given in Appendix B

[fxx]s =

√2

πpf(0)− p2fs (6.43)

the suffix s indicating the sine transform and f(x, t) is any given function.If we now apply the initial condition in (6.39), we obtain from (6.41)

φs(p, 0) =

√2

π

∫ ∞0

q(x) sin pxdx (6.44)

Solving (6.42) in the presence of (6.44) yields φs(p, t). Hence φ(x, t), on takingthe inverse sine transform, is given by

φ(x, t) =

∫ ∞0

φs(p, t) sin pxdp (6.45)

where φs(p, t) is known from the following argument. When r(t) = φ0, (6.28) be-comes

∂φs(p, t)

∂t+ α2p2φs(p, t) = α2p

√2

πφ0 (6.46)

which has the solution

φs(p, t) =

√2

π

φ0

p(1− e−α

2p2t) (6.47)

Employing (6.47), φ(x, t), from (6.45), turns out to be

φ(x, t) =

√2

πφ0

∫ ∞0

sin px

p(1− e−α

2p2t)dp (6.48)

which is the integral representation of the solution. We can express φ(x, t) in termsof the error function by observing that the latter is defined by

erf(y) =2√π

∫ y

0

e−u2

du (6.49)

and obeys the following property∫ ∞0

e−u2 sin(2uy)

udu =

π

2erf(y) (6.50)


A little manipulation then shows that φ(x, t) can be expressed in the form

φ(x, t) =

√2

πφ0[

π

2− π

2erf(

x

2α√t)] (6.51)

which can also be recast as

φ(x, t) =

√π

2φ0(1− 2√

π

∫ x2α√t

0

e−u2

du) (6.52)

6.2 Solving by Laplace transform method

We now proceed to find the solution of a class of initial-value problems by em-ploying the Laplace transform method. We focus here on the one-dimensional prob-lems only. In such models the governing PDE invariably contains a function havingthe time parameter t as one of the independent variables with a prescribed initialcondition. By looking for the Laplace transform over the interval (0,∞), we canconvert the given equation along with the initial-boundary conditions into an ODEof a complex-valued function in terms of the Laplace variable s. Once this equationis solved we can make use of the inversion formula outlined in Appendix C to arriveat the desired solution in a certain tractable form. We note in passing that for thehigher-dimensional PDEs one may have to take recourse to the application of theLaplace transform more than once and subsequently go for an adequate number ofinversions. It sometimes could happen that the Laplace transform method may notprove to be suitable for a given PDE in extracting its solution in a closed form. Insuch cases the limiting behaviour of the solution, subject to certain conditions, maybe extracted. A couple of illustrative examples have been considered in this regardas well.

Example 6.5

Solve the heat conduction equation by employing the method of Laplacetransform

φt(x, t) = α2φxx(x, t), x ≥ 0, t > 0 (6.53)

subject to the initial-boundary conditions

φ(x, 0) = 0, x > 0, φ(0, t) = f(t) (6.54)


Multiplying both sides of the diffusion equation by e−st and integrating withrespect to the variable t over (0,∞) we get

s

∫ ∞0

e−stφdt = α2 d2

dx2

∫ ∞0

e−stφdt (6.55)

where the boundary term in the left side has been dropped by employing the vanish-ing initial condition on φ and also noting that the latter does not show any singularbehaviour at t→∞. We thus arrive at the ODE

φxx =s

α2φ (6.56)

where φ is the Lapalce transform of φ

φ(x, s) =

∫ ∞0

e−stφ(x, t)dt (6.57)

with respect to t.The general solution of the above ODE is of the type

φ(x, s) = Ae−√sxα +Be

√sxα (6.58)

in which we have to set B = 0 to avoid blowing up of the solution at x → ∞. Weare thus led to the reduced form

φ(x, s) = Ae−√sxα (6.59)

in terms of an overall constant A.Taking the Laplace transform of the second boundary condition in (6.54) gives

φ(0, s) = F (s) (6.60)

where F (s) is the Laplace transform of f(t). Comparing (6.59) and (6.60) fixes A tobe A = F (s). We thus have for φ the form

φ(x, s) = F (s)e−√sxα (6.61)

We notice that the right side is a product of two Laplace transforms of whichthe second one stands for

e−x√s

α = L[x

2αt√πte− x2

4α2t ] (6.62)

where we have utilized the following standard result (see the table in Appendix C)

e−a√s = L[

a

2t√πte−

a2

4t ] (6.63)

by scaling the parameter a as a→ xα

.


The following integral form for φ(x, t) is thus suggested from (6.61) and (6.62)by employing the convolution integral (C.65) of Laplace theorem (see Appendix C)

φ(x, t) =

∫ t

0

f(τ)x

2α√π(t− τ)

√t− τ

e− x2

4α2(t−τ) dτ (6.64)

It also corresponds to

φ(x, t) =x

2α√π

∫ t

0

f(τ)

(t− τ)√t− τ

e− x2

4α2(t−τ) dτ (6.65)

by taking the variable x out of the integral sign. (6.64) or (6.65) provides the re-quired solution for φ(x, t).

Example 6.6

Solve the heat equation in the closed interval [0, a], a > 0 by employing themethod of Laplace transform

φt(x, t) = α2φxx(x, t), 0 ≤ x ≤ a, t > 0 (6.66)

subject to the initial condition

φ(x, 0) = 0, 0 ≤ x ≤ a (6.67)

and the boundary conditions

φ(0, t) = 0, φ(a, t) = φ0, t > 0 (6.68)

As in the previous problem we multiply both sides of the given PDE by e−st,integrate with respect to the variable t over (0,∞) and drop the boundary term onapplying the initial condition (6.67). Thus we arrive at the ODE

φxx =s

α2φ (6.69)

where φ is the Laplace transform of φ with respect to the variable t

φ(x, s) =

∫ ∞0


The general solution of (6.69) is

φ(x, s) = Ae√sxα +Be−

√sxα (6.71)

where A and B are the arbitrary constants.


Next, taking the Laplace transform of the two boundary conditions in (6.68)we have

φ(0, s) = 0, φ(a, s) =φ0

s(6.72)

By bringing to terms with the above conditions we can solve for A and B toobtain

A =φ0

2s sinh(√saα

), B = − φ0

2s sinh(√saα

)(6.73)

Hence φ(x, s) assumes the form

φ(x, s) =φ0

s

sinh(√sxα

)

sinh(√saα

)(6.74)

Now, through inversion, we can write

φ(x, t) =1

2iπ

∫ r+i∞

r−i∞est

φ0

s

sinh(√sxα

)

sinh(√saα

)(6.75)

The singularities in the integrand of (6.75) are seen to be located at

s = 0,ia√s

α⇒ s = −α

2n2π2

a2, n = 0, 1, 2, ... (6.76)

The results for the residues are as follows

at s = 0 :xφ0

a(6.77)

and

at s = −α2n2π2

a2, n = 1, 2, ... :

2φ0

nπ(−1)ne

−α2n2π2ta2 sin

nπx

a(6.78)

Therefore

φ(x, t) = φ0[x

a+

2

π

∞∑n=1

(−1)n

ne−α

2n2π2ta2 sin

nπx

a] (6.79)

solves the given PDE satisfying the prescribed conditions of the problem.


Example 6.7

Find the asymptotic expansion of φ(t) when φ(t) satisfies the heat equation

φt(x, t) = α2φxx(x, t), x ≥ 0, t > 0 (6.80)


φ(x, 0) = 0, x ≥ 0 (6.81)

and the boundary condition

φx(0, t) = h(φ− φ0), t > 0 (6.82)

where h and φ0 are constants.

The Laplace transform of the given equation with respect to t yields, as usual,the ODE

φxx =s

α2φ (6.83)

in which the given initial condition is used and φ(x, s) denotes the Laplace transformintegral of φ(x, t) with respect to the variable t

φ(x, s) =

∫ ∞0


The general solution of (6.83) is of course similar to (6.71) as provided in terms ofthe two arbitrary constants A and B. Requiring a finite solution of φ at x → ∞restricts B = 0 and we have

φ(x, s) = Ae−√sxα (6.85)

in terms of A only.Taking next the Laplace transform of the boundary condition (6.82) we have at

x = 0

φx(0, s) = h(φ|x=0 −φ0

s) = h(A− φ0

s) (6.86)

Seeking consistency between (6.85) and (6.86) at x = 0 boundary fixes A

A =hφ0

s(h+√

sk

)(6.87)

Thus φ(x, s) acquires the form

φ(x, s) =hφ0

s(h+√

sk

)e−√sxα (6.88)


Note that the branch point s = 0 is the only singularity of φ(x, s) since for h > 0the quantity h+

√sk

cannot vanish. Let us take up the case when x = 0.

From (6.88) we expand φ(s) as follows

φ(s) = φ0(1

s+

1

α2h2+

1

α4h4+ ...)− φ0s

− 12 (

1

αh+

s

α3h3+

s2

α5h5) (6.89)

By the asymptotic inversion formula of Laplace transform given in (C8) of Ap-pendix C we have for large t

φ(t) = φ0 − φ0

sin π2

π[

Γ( 12)

αh√t− 1

α3h3

Γ( 32)

t32

+1

α5h5

Γ( 52)

t52

− ...] (6.90)

where we have set β = 12. A more compact form is

φ(t) = φ0 −φ0√π

[1

αh√t− 1

α3h3t32

+1 · 3

4α5h4t52

− ...] (6.91)

giving the asymptotic nature of φ(t).The case x 6= 0 where the exponential in (6.88) contributes some additional

terms is handled similarly and is left as an exercise.

Example 6.8

Find the asymptotic expansion of φ(t) for when φ(t) satisfies the heat equation

φt(x, t) = α2φxx(x, t), x ≥ 0, t > 0 (6.92)


φ(x, 0) = 0, x ≥ 0 (6.93)

and an oscillatory fluctuation at x = 0

φ(0, t) = a cos(ωt), t > 0 (6.94)

where a is a constant and ω is the frequency of oscillation.

Taking the Laplace transform of both sides of (6.92) yields the form similar to(6.83) in which (6.94) is used. Its bounded solution is given by (6.85).

Taking now the Laplace transform of the boundary condition (6.94) we find

φ(0, s) =as

s2 + ω2(6.95)

where we have employed the Laplace transform of the cosine function from Table C1.


The bounded solution (6.85) when confronted with (6.95) yields the followingresult

φ(x, s) =as

s2 + ω2e−x√

sα (6.96)

Notice that the singularities of φ(x, s) are the simple poles located at s = ±iωand the branch point s = 0. Writing

φ(t) = φ+(t) + φ−(t) + φ0(t) (6.97)

where +,− and 0 denote the contributions from the three singularities, we havespecifically for φ+(t)

φ+(t) = residue of estas

s2 + ω2e−x√

sα at s = iω (6.98)

The result is

φ+(t) = eiωt(a · 2iω

2iω)e−x√ωαei π4

=a

2ei(ωt−x

√ω2α

)e−x√

ω2α (6.99)

Similarly φ−(t) is given by

φ−(t) =a

2e−i(ωt−x

√ω2α

)e−x√

ω2α (6.100)

To estimate φ0(t) the best we can do is to go for an asymptotic evaluation. To-wards this end we note the following expansion of the right side of (6.96) around s = 0

as

s2 + ω2e−x√

sα =

a

ω2(s+

s2x2

2α+ ...)− a

ω2s−

12 (xs2

α+x3s3

6α3+ ...) (6.101)

This shows from the asymptotic inversion formula

φ0(t) ∼ − a

ω2

sin π2

π[x

α

Γ( 52)

t52

− x3

6α3

Γ( 72)

t72

+ ...] (6.102)

Combining (6.99), (6.100) and (6.102) we arrive at the result

φ(t) ∼ ae−x√

ω2α cos(ωt− x

√ω

2α)− a

πω2[x

α

Γ( 52)

t52

− x3

6α3

Γ( 72)

t72

+ ...] (6.103)

which stands for the asymptotic form for φ(t).


6.3 Summary

The general aim in this chapter was to solve some typical problems of PDEs bythe use of integral transform methods as a convenient mathematical tool. We howeverconcentrated only on the Fourier and Laplace transforms although one should notethat other transforms could also be used as a means to generate solutions of a PDE.We made use of their properties of the transforms collected in Appendix B andAppendix C. We also considered a couple of examples for which the asymptoticform of the solution could be derived. Such asymptotic forms are often found usefulin many classes of problems holding practical interest.


Exercises

1. Find the Fourier integral representation of the single pulse function

f(x) =

1 if |x| < 1

0 if |x| > 1

Deduce the Dirichlet integral ∫ ∞0

sin p

pdp =

π

2

2. Find by using Fourier transform the fundamental solution of the PDE

φt = φxx − xφx

as φ(x, t) = Cetex2

2 , where C is a constant.

3. Small amplitude water waves produced by an inertial surface displacementare controlled by a two-dimensional Laplace’s equation

φxx + φzz = 0, −∞ < z < η

where x and z are respectively the horizontal and upward vertical axis, φ(x, z, t) isthe velocity potential and η(x, t) is the elevation of the free surface subject to theinitial conditions

φ(x, z, 0) = 0, η(x, 0) = f(x),

where f(x) is an even function of x, the boundary conditions

ηt − φz = 0, φt + gη = 0 both at z = 0

and the smoothness condition

φ(x, z, t)→ 0 as z → −∞

Use the method of Fourier transform, find an integral expression of η(x, t) inthe form

η(x, t) =1√2π

∫ ∞0

f(p)[cos(px−√gpt) + cos(px+√gpt)]

where p is the transform parameter.


4. Consider a three-dimensional Laplace’s equation

φxx + φyy + φzz = 0, −∞ < x <∞, z ≥ 0

subject to the boundary condition

φ(x, y, 0) = f(x, y)

and the smoothness conditions

φ(x, y, z) → 0 as z → ∞ and also as x, y → ±∞

along withφx, φy → 0 as x, y → ±∞

Find the solution in the form

φ(x, y, z) =z

2π

∫ +∞

−∞

∫ +∞

−∞

f(ξ, η)

[(x− ξ)2 + (y − η)2 + z2]32

dξdη

5. Solve the following generalized form of the telegraph equation by the Fouriertransform method

φtt + (α+ β)φt + αβφ = c2φxx

where α and β are non-zero constants and show that the solution reveals dispersion.Discuss the case when α = β.

6. Use Laplace transform to write down the solution of the PDE

φxt + xφx + 2φ = 0

subject to the initial-boundary conditions

φ(x, 0) = 1, φ(0, t) = e−at, a > 0

in the form

φ(x, t) =1

2iπ

∫ r+i∞

r−i∞estζ(s)ds, s > α

where ζ(s) = s2

s(s+a)(x+s)2.

7. Consider the problem of a semi-infinite string described by the PDE

φtt = c2φxx + φ0, 0 < x <∞, t > 0


where one end of the string is held fixed and φ0 is the constant external force actingon it. Further, the string is subjected to the initial boundary conditions

φ(x, 0) = 0, φt(x, 0) = 0, φ(0, t) = 0

Assume that φx(x, t)→ 0 as x → ∞. Solve by the Laplace transform methodto derive φ(x, t) in the form

φ(x, t) =

φ02

[t2 − (t− x2)2] if t ≥ x

cφ02t2 if t ≤ x

c

8. Solve the wave equation

φtt = c2φxx, 0 < x <∞, t > 0

which is subjected to the initial boundary conditions

φ(x, 0) = 0, φt(x, 0) = 0, φ(0, t) = f(t)

where f(t) is a given function. Assume that φx(x, t) → 0 as x → ∞. Solveby the Laplace transform method to derive φ(x, t) in the form

φ(x, t) =

sin(t− x

c) if x

c< t < x

c+ 2π

0 otherwise

9. Solve the following fourth order PDE

φtt = λ2φxxxx, 0 < x < l, t > 0

which is subjected to the initial conditions

φ(x, 0) = 0, φt(x, 0) = φ0 sin(πx

l)

and boundary conditions

φ(0, t) = 0, φ(l, t) = 0, φxx(0, t) = 0. φxx(l, t) = 0

where λ is a constant.

10. Focus on the homogeneous heat equation

φt(x, t) = φxx(x, t), −∞ < x <∞, t > 0


subjected to the following initial data on time at t = s

φ(x, s; s) = f(x, s), −∞ < x <∞

In terms of the Green’s function G(x, s) = 1√4πt

e−x2

4t the solution of the problem isgiven by

φ(x, t; s) =

∫ ∞−∞

G(x− ξ, t− s)f(ξ, s)dξ, −∞ < x <∞

Establish Duhamel’s principle which states that in an infinite domain the solutionof the corresponding inhomogeneous problem namely,

φt(x, t) = φxx(x, t) + f(t, x), −∞ < x <∞, t > 0

subject to the vanishing initial condition φ(x, 0) = 0 is given by

φ(x, t) =

∫ t

0

φ(x, t; s)ds

i.e. φ(x, t) has the form

φ(x, t) =

∫ t

s=0

∫ ∞ξ=−∞

G(x− ξ, t− s)f(ξ, s)φ(x, t; s)ds

which translates to

φ(x, t) =

∫ t

s=0

∫ ∞ξ=−∞

1√4π(t− s)

e− (x−ξ)2

4(t−s) f(ξ, s)dξds

A problem along similar lines can be set up and solved for the heat equationdefined over a finite interval. Duhamel’s principle also holds for the solvability ofthe inhomogeneous wave equation where the source term is transferred to the initialvelocity for the corresponding homogeneous problem.

Appendix A

Dirac delta function

Dirac delta function δ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185Other results using delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194Test function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

Dirac delta function δ(x)

The delta function was introduced by P.A.M. Dirac in the development of thetheory of quantum mechanics in his 1930 book The Principles of Quantum Mechanicsand hence it goes by his name. His point was that, as he wrote, δ(x) is not a quantitywhich can be generally used in mathematical analysis like an ordinary function, butits use must be confined to certain simple types of expression for which it is obviousthat no inconsistency can arise. However, the history1 of delta function is far older2

and dates back to the work of Gustav Kirchhoff who studied it in 1882 as the limitof the functions

δn(x) =

√n

πe−nx

2

, n = 1, 2, 3, ...

Delta function was also used a decade later in 1892 by the electrical engineer OliverHeaviside in a paper entitled On operations in physical mathematics. Later, in awider setting, the famous Russian mathematician Sergei Lvovich Sobolev introducedthe concept of distribution theory in 1935 and subsequently the French mathemati-cian Laurent Schwartz gave an elaborate exposition of it in the two volumes of thebook entitled Theory of Distributions appearing in 1950 and 1951. In the literaturea distribution is also referred to as a generalized function. The delta function is theanalog of the piecewise function Kronecker delta for the continuous case. The latteris defined for two indices i and j by δij = 1, i = j but = 0, i 6= j. Both for the deltafunction and Kronecker delta the same symbol δ is retained.

The Dirac delta function, denoted by δ(x), is a mathematical object equippedwith the property

1J. Lighthill, Fourier Analysis and Generalized functions, Tributes to Paul Dirac, Ed.J.G.Taylor, Adam Hilger, IOP Publishing, Bristol, 1987.

2Although no in-depth study was made, the presence of delta function was noted byPoisson in 1815, Fourier in 1822 and Cauchy in 1823 and 1827.

185


∫ +∞

−∞f(x)δ(x)dx = f(0) (A.1)

for any continuous function f(x). Furthermore, δ(x) is vanishing everywhere exceptat x = 0

δ(x) = 0 for all x 6= 0 (A.2)

On the other hand, setting f(x) = 1 implies that the integral of the delta functionover (−∞,+∞) is unity ∫ +∞

−∞δ(x)dx = 1 (A.3)

(A.1), (A.2) and (A.3) are the essential characteristics that distinguish a delta func-tion from an ordinary function.

Instead of x = 0 if the argument of the delta function is at the point x = ξ thenthe above properties take the forms

δ(x− ξ) = 0, x 6= ξ (A.4)

∫ b

a

δ(x− ξ)dx = 1, a ≤ ξ ≤ b,∫ b

a

δ(x− ξ)dx = 0, a, b < ξ and a, b > ξ

(A.5)∫ +∞

−∞f(x)δ(x− ξ)dx = f(ξ) (A.6)

where if f(x) = 1 then ∫ +∞

−∞δ(x− ξ)dx = 1 (A.7)

We refer to (A.1) or (A.6) as the reproductive property of the delta function. It isalso called the sifting property or selector property.

If for a sequence of strongly peaked classical functions gk(x) such that as k →∞the following integral character is revealed

limk→∞

∫ +∞

−∞gk(x)f(x)dx = f(0), k = 1, 2, ...

then such a sequence gk(x), k = 1, 2, .... is called a delta sequence and the deltafunction may be viewed as a limit form of it.

Dirac delta function 187

The delta function has a relevance to multi-dimensional problems too. Supposewe are dealing with a set of spherical polar coordinates r, θ, φ. The Jacobian matrixwhich is defined by

J =

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

(A.8)

becomes for x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ

J =

sin θ cosφ r cos θ cosφ −r sin θ sinφsin θ sinφ r cos θ sinφ r sin θ cosφ

cos θ −r sin θ 0

(A.9)

The value of the determinant of the matrix is r2 sin θ. Hence the delta function inspherical polar coordinates reads

(r2 sin θ)−1δ(r − r′)δ(θ − θ′)δ(φ− φ′) (A.10)

In two dimensions since the Jacobian is J = r, the delta function is

(r)−1δ(r − r′)δ(θ − θ′) (A.11)

In cylindrical coordinates x = r cosφ, y = r sinφ, z = z, making appropriatechanges in variables in (A.8), the Jacobian is easily worked out to be simply J = r.Hence the delta function in cylindrical coordinates reads

(r)−1δ(r − r′)δ(φ− φ′)δ(z − z′) (A.12)

(A.10) and (A.12) are to be compared with the Cartesian situation where it isδ(x− x′)δ(y − y′)δ(z − z′).

We now look for sequence of functions, the delta sequence, as mentioned a littleearlier, that satisfies the reproductive property or sifting property. One such sequenceis provided by the Dirichlet’s formula

limk→∞

∫ +∞

−∞f(x)

sin(kx)

πxdx = f(0), k = 1, 2, ... (A.13)

where f(x) is a bounded and differentiable function.The representation of rectangular blocks defined as the limit

limk→∞

∫ +∞

−∞f(x)uk(x)dx = f(0), k = 1, 2, ... (A.14)

of a family of functions uk(x)(x)

uk(x) =

k2

if |x| < 1k

0 otherwise,

is another example.


One can also visualize the limit of the Gaussian

limk→∞

∫ +∞

−∞f(x)ξk(x)dx = f(0), k = 1, 2, ... (A.15)

where ξk(x) is (see Figure A.1)

FIGURE A.1: Gaussian profile.

ξk(x) =

√k

πe−kx

2

, k = 1, 2, ... (A.16)

and the so-called Lorentzian

limk→∞

∫ +∞

−∞f(x)ζk(x)dx = f(0), k = 1, 2, ... (A.17)

where ζk(x) is (see Figure A.2)


FIGURE A.2: Lorentzian profile.

ζk(x) =1

π

k

1 + k2x2, k = 1, 2, ... (A.18)

as legitimate candidates for the delta sequence. We also note some of the otherchoices which include the following functions

1

2εexp(−|x|

ε),

1

2√πεexp(−x

2

4ε),

1

2ε

1

cosh2(xε)

(A.19)

which in the limit ε→ 0 reflect the standard characteristics of a delta function.Note that the Gaussian and Lorentzian can also be respectively translated to

the form

limα→0

1√2πα

e− x2

2α2 (A.20)

These can be also looked upon as the limit of the Cauchy distribution

limε→0

1

π

ε

x2 + ε2(A.21)


The sequence of functions sin(kx)πx

or uk(x) or ξ(x) or ζ(x), k = 1, 2, ..., is a represen-tative of a delta sequence.

Let us attend to sin(kx)πx

first. The following integral values are well known∫ +∞

0

sin(x)

πxdx =

1

2(A.22)

and ∫ 0

−∞

sin(x)

πxdx =

1

2(A.23)

When put together we have ∫ +∞

−∞

sin(x)

πxdx = 1 (A.24)

and a change of variable allows us to write∫ +∞

−∞

sin(kx)

πxdx = 1 (A.25)

We now look at the finite range of integral over (a, b) where b > a > 0. Then theparameter k can be easily transferred to the limits of the integral

1

π

∫ b

a

sin(kx)

xdx =

1

π

∫ bk

ak

sin(t)

tdt, b > a > 0 (A.26)

where the right side → 0 as k →∞. Likewise

1

π

∫ bk

ak

sin(t)

tdt → 0 as k →∞, 0 < a < b (A.27)

Next, due to the above vanishing results on either side of zero, we have for anyε > 0, however small

limk→∞

∫ +∞

−∞f(x)

sin(kx)

πxdx = lim

k→∞

∫ +ε

−εf(x)

sin(kx)

πxdx (A.28)

where the right side can be put as

f(0) limk→∞

∫ +ε

−ε

sin(kx)

πxdx (A.29)

with the help of the mean value theorem of integral calculus.In (A.29) we can, without any loss of generality, extend the range of integration

to (−∞,+∞) resulting in


f(0) limk→∞

∫ +∞

−∞

sin(kx)

πxdx (A.30)

which when (A.25) is used boils down to simply f(0) and we conclude

limk→∞

∫ +∞

−∞f(x)

sin(kx)

πxdx = f(0) (A.31)

We thus recover the Dirichlet’s formula signalling that in the limit k → ∞,sin(kx)πx

is a representation of the Dirac delta function

limk→∞

sin(kx)

πx= δ(x) (A.32)

Equivalently,

limε→0+

sin(xε)

πx= δ(x) (A.33)

Further since1

2π

∫ +k

−keirxdr =

sin(kx)

πx(A.34)

we also have another representation of the delta function

1

2π

∫ +∞

−∞eikxdk = δ(x) (A.35)

We now turn to the behaviour of uk(x). Given its definition we find

limk→∞

∫ +∞

−∞uk(x)f(x)dx =

k

2

∫ + 1k

− 1k

f(x)dx (A.36)

where the integral in the right side, by the mean value theorem of integral calculusis 2

kf(ξ), where |ξ| < 1

k, and as such

limk→∞

∫ +∞

−∞uk(x)f(x)dx = f(ξ) (A.37)

But as k →∞ the right side goes to f(0) and hence we can write

limk→∞

∫ +∞

−∞uk(x)f(x)dx→ f(0) (A.38)

which reflects the property of a delta sequence.

In a similar way if the argument of uk is different from zero, then a change ofvariable gives


limk→∞

∫ +∞

−∞uk(x− x0)f(x)dx = lim

k→∞

∫ +∞

−∞uk(y)f(y + x0)dy → f(x0) (A.39)

where we employed the mean value theorem of integral calculus to express∫ +∞−∞ uk(y)f(y+x0)dy = k

22kf(ξ+x0), where |ξ| < 1

kand exploited the limit k →∞.

In essence this is the reproductive or sifting property of the delta function projectingout the value of f at x = x0.

If the argument of the delta function involves a scaling x→ ηx then we have forη > 0

limk→∞

∫ +∞

−∞uk(ηx)f(x)dx = lim

k→∞

∫ +∞

−∞uk(y)f(

y

η)dy

η→ 1

ηf(0) (A.40)

while for η < 0

limk→∞

∫ +∞

−∞uk(ηx)f(x)dx = − lim

k→∞

∫ +∞

−∞uk(y)f(

y

η)dy

η→ −1

ηf(0) (A.41)

The above two results when united furnish an important result of scaling

δ(ηx) =1

|η|δ(x) (A.42)

Next, we look at the Gaussian ξk(x). We first of all transform the integral∫ +∞

−∞e−x

2

dx =√π (A.43)

to the form ∫ +∞

−∞e−kx

2

dx =

√π

k(A.44)

on replacing x by x√k. This means∫ +∞

−∞ξk(x)dx = 1, k = 1, 2, ... (A.45)

Let us consider at the finite range of integration over (a, b), where b > a > 0.Then ∫ b

a

ξk(x)dx =1√π

∫ b√k

a√k

e−y2

dy → 0 as k →∞ (A.46)

The above limiting result also holds for a < b < 0.


Decomposing (−∞,+∞) to pieces (−∞,−ε), (−ε,+ε) and (+ε,+∞), ε > 0 nomatter how small, and making use of the above results we find that we can write

limk→∞

∫ +∞

−∞ξk(x)f(x)dx = lim

k→∞

∫ +ε

=ε

ξk(x)f(x)dx (A.47)

where the right side by the use of mean value theorem of integral calculus can bewritten as

f(0) limk→∞

∫ +ε

=ε

ξk(x)dx (A.48)

Stretching the integral domain to (−∞,+∞), which as shown earlier has the valueequal to unity, we arrive at the desired reproductive result

limk→∞

∫ +∞

=∞ξk(x)f(x)dx = f(0) (A.49)

Finally we deal with the Lorentzian case. If we set g(x) = f(x) − f(0) whichimplies g(0) = 0, then to establish the reproductive property

limk→∞

∫ +∞

=∞ζk(x)f(x)dx = f(0) (A.50)

it suffices if we can show

limk→∞

∫ +∞

=∞ζk(x)g(x)dx = 0 (A.51)

Proceeding towards this aim, we choose some appropriate number a > 0 and de-compose (−∞,+∞) around (−a,+a) to write

∫ +∞

=∞ζk(x)g(x)dx =

∫ −a=∞

ζk(x)g(x)dx+

∫ +a

=a

ζk(x)g(x)dx+

∫ +∞

+a

ζk(x)g(x)dx

(A.52)

Concerning the integral over (−a,+a), if |g(x)| ≤ µ(a) there, µ(a) being the maxi-mum value, then

|∫ +a

=a

ζk(x)g(x)dx| ≤∫ +a

=a

ζk(x)|g(x)|dx ≤ µ(a)

∫ +a

=a

ζk(x)dx (A.53)

where the definite integral in the right side has the value 2π

tan−1(ak), it follows that

|∫ +a

=a

ζk(x)g(x)dx| ≤ µ(a) (A.54)


Since by definition of g(x), g(x) = 0 we can claim that lima→0 µ(a) = 0 andwrite

|∫ +a

=a

ζk(x)g(x)dx| < ε

2(A.55)

where ε is a quantity no matter how small.For the other two integrals we have

|∫ −a

=∞ζk(x)g(x)dx+

∫ +∞

+a

ζk(x)g(x)dx| ≤ ν[

∫ −a=∞

ζk(x)dx+

∫ +∞

+a

ζk(x)dx] (A.56)

where since f(x) is bounded in (−∞,+∞) and g(0) = 0, g(x) is bounded too in(−∞,+∞) and given by |g(x)| ≤ ν. After the easy evaluation of the integrals, theright side has the value ν(1− 2

πtan−1(ak) which → 0 as k →∞. Thus we can write

|∫ −a

=∞ζk(x)g(x)dx+

∫ +∞

+a

ζk(x)g(x)dx| ≤ ε

2(A.57)

in terms of the small quantity ε.All this means that we have∫ +∞

=∞ζk(x)g(x)dx ≤ ε

2+ε

2= ε (A.58)

This being so the reproductive property holds.

Other results using delta function

The following results often prove to be useful:

(i) xδ(x) = 0

(ii) δ(x) = δ(−x)

(iii)

∫δ(x− ξ)δ(x− η)dx = δ(ξ − η) (A.59)

The proof of the first result is trivial. For the second result since we can write

δ(−x) =1

2π

∫ +∞

−∞e−ikxdk (A.60)

replacing k by −k in the right side and changing the limits of the integral accordinglya representation of δ(x) easily follows and shows that δ(x) = δ(−x) implying thatthe delta function is an even function.


To prove (iii) let us represent the integral on the left side as

∫dxδ(x− ξ)δ(x− η) =

1

(2π)2

∫dx

∫ +∞

−∞eiy(x−ξ)dy

∫ +∞

−∞eit(x−η)dt (A.61)

where the right side can be arranged to the following form

1

2π

∫dy

∫dt(

1

2π

∫ei(y+t)xdx)e−iξy−iηt (A.62)

It is clear that the integral over x is simply δ(y + t). So if we perform the integralover t we will be left with

1

2π

∫e−i(ξ−iη)ydy (A.63)

This integral, on using the property that the delta function is an even function,stands for δ(ξ − η) and we have justified the required assertion.

We often have to face situations when the argument of the delta function rep-resents a function such as δ(f(x)). Setting aside complicated cases let us look ata monotonic function which possesses a simple zero at x = ξ. If f ′(ξ) > 0 we canexpress for a smooth function g(x)∫ +∞

−∞δ(f(x))g(x)dx =

∫ +∞

−∞δ(y)g(x(y))

dy

f ′(x)=

g(ξ)

f ′(ξ)(A.64)

where we have replaced f(x) by the variable y and noted that the vanishing of theargument of δ(y) takes place for the corresponding x = ξ value. Since the right sidecan also be expressed in the form∫ +∞

−∞

δ(x− ξ)g(x)

f ′(ξ)dx (A.65)

it therefore follows that

δ(f(x)) =δ(x− ξ)f ′(ξ)

dx (A.66)

On the other hand, if f ′(ξ) < 0 we encounter a sign change only and so combiningboth the cases we have the result

δ(f(x)) =δ(x− ξ)|f ′(ξ)| dx (A.67)

When f(x) runs into multiple zeros located at (ξ1, ξ2, , ξn) we get the generalform

δ(f(x)) =

n∑k=1

δ(x− ξk)

|f ′(ξk)| dx (A.68)


As applications of the above formula it is easy to show

(i) δ(x2 − ξ2) =1

2ξ[δ(x− ξ) + δ(x+ ξ)]

(ii) δ(tanx) = δ(x), x ∈ (−π2,π

2)

(iii) δ(cosx) = δ(x− π

2), 0 < x < π

(iv)

∫ +2π

−2π

eπxδ(x2 − π2) =1

πcosh(π2) (A.69)

Test function

Consider a set S of real numbers. Its limit point α is such that every neighbour-hood of it, no matter how small, contains at least one point of S. The set S with allits limit points constitutes the closure of S and denoted by S.

The support of a function addresses the smallest closed set where it takes onnonzero values but vanishes identically outside of it. The support of a function f isdenoted by supp(f). A compact support is the one in which the supp(f) is compacti.e. it is a bounded set.

The example of f(x) = sin(x) is an interesting one in the sense that its supportruns over the entire real axis although sin(x) vanishes at all points x = nπ, n =1, 2, .... Another example is (see Figure A.3)

FIGURE A.3: The function f(x).


f(x) =

0, if −∞ < x <≤ −1

1 + x, if − 1 < x < 0

1− x, if 0 ≤ x < 1

0, if 1 ≤ x <∞

It is clear that the function f(x) is different from zero in the open interval(−1, 1), the closure of the latter is [−1, 1] and so [−1, 1] stands for the compactsupport of f(x).

With these introductory remarks we are in a position to define a test function.Real valued functions denoted by φ(x) = φ(x1, x2, ..., xn) whose derivatives of allorders exist (in other words, they are infinitely differentiable at all points in <)and have a compact support in a closed bounded subset of < are called test func-tions. Test functions are smooth, with no sharp features in their graphs. An explicitillustration of a test function is (see Figure A.4)

φ(x) =

e− 1

1−|x|2 if |x| < 1

0 otherwise,

FIGURE A.4: Test function φ(x).

It gives a compactly supported test function φ(x) which is differentiable to anyorder. Note, however, that there exist numerous classes of smooth functions (for


example, a polynomial or an exponential) which do not have a compact support. Thespace is denoted by D and addresses the family of infinitely differentiable compactfunctions.

As another example consider the function g(x) defined by

g(x) =

exp(− 1

x) if x > 0

0 otherwise,

One checks that φ(x) = g(x)g(1 − x) is a test function with the criterion of itsbeing infinitely differentiable and vanishing outside a finite interval holding.

We know that an integrable function is not always differentiable, Sometimes wemay be interested in locally integrable functions. A locally integrable function f(x)defined over a certain domain (a, b) is the one for which, around every point of (a, b),there exists a neighbourhood on which it is integrable. Mathematically it means thatthere exist sub-intervals such as (α, β) of (a, b) such that∫ β

α

|f(x)dx <∞ (A.70)

Some examples of a locally integrable functions are continuous functions, integrablefunctions, a piece-wise continuous function or a constant. However, the delta functionis not locally integrable. This can be understood if we consider a sequence of nestedintervals I1, I2, ..., In, ... i.e. each interval is included in the preceding one whoselength shrinks to zero:

limk→∞

∫Ik

f(x)dx = 0 (A.71)

If δ(x) replaces f(x) we would get a finite value.

Formally, a generalized function or a distribution is any linear functional whichis continuous on D. The set of continuous linear functionals is denoted by D1 andits elements are called generalized functions or distributions.

A locally integrable function f(x) generates a regular distribution through thefollowing linear functional

(f, φ) =

∫ +∞

−∞f(x)φ(x)dx (A.72)

for all test functions ∈ D.The linearity follows straightfowardly. For all test functions φ1 and φ2 ∈ D and

arbitrary constants α1 and α2 we have

(f, αφ1 + βφ2) = (α

∫ +∞

−∞f(x)φ1(x) + β

∫ +∞

−∞f(x)φ2(x))dx) (A.73)


and the right side is basically α(f, φ1) + β(f, φ2). The continuity follows from thebound

|f(x)| ≤ max(x∈suppφ)

|φ(x)|∫ +∞

−∞|f(x)|dx <∞ (A.74)

All ordinary continuous functions are regular generalized functions.The delta function δ(x), however, is a singular distribution. Its support is pro-

vided by just the point x = 0 : (supp)δ(x) = (0). But we often use the symbolicoperation to express

(δ(x), φ) = (δξ=0, φ) = φ(0) (A.75)

which in the integral form reads∫δ(x)φ(x)dx = φ(0) (A.76)

For a shifted argument the symbolic notation means∫δ(x− ξ)φ(x)dx = φ(ξ) (A.77)

We have already addressed the issue of delta-convergent sequence or simplydelta sequence. In the present symbolic context we can think of the delta functionas being obtained in the limit of a sequence of ordinary integrals generated by locallyintegrable functions fk(x) ∫

fk(x)φ(x)dx = φ(0) (A.78)

The Gausssian, Cauchy or Lorentzian are some examples in this regard.Finally, we can define the distributional derivative in the following way. If f and

f ′ are two locally integrable functions then (f ′, φ) stands for

(f ′, φ) =

∫ b

a

f ′φdx = −∫ b

a

fφ′dx = −(f, φ′) (A.79)

where we have integrated by parts and exploited the feature of the test function sothat it vanishes outside a closed, bounded interval. Thus we see that the derivativehas been transferred to the test function implying the property

(f ′, φ) = −(f, φ′), φ ∈ D (A.80)

On integrating by parts n times, the n-th distributional derivative f (n) is given by

(f (n), φ) = (−1)n(f, φ(n)), φ ∈ D (A.81)

For the delta function we have symbolically the following operations for itsderivative


∫δ′(x− ξ)φ(x)dx = −φ′(ξ), φ ∈ D (A.82)

and for its n-th derivative∫δ(n)(x− ξ)φ(x)dx = (−1)nφ(n)(ξ), φ ∈ D (A.83)

To illustrate how a distributional derivative might work in actual practice we takethe example of the Heaviside function H(x). It is locally integrable and defined by

H(x) =

+1 if x > 0

0 ifx < 0

Heaviside function is also called3 the unit step function. The jumped discontinuity ofH(x) is obvious. It is also clear that H(−x) = 1 = H(x). If the jumped discontinuityis located at the point x = ξ say, then the notation H(x−ξ) is used and the relationH(ξ − x) = 1−H(x− ξ) holds.

We can represent H(x) by means of a distribution∫H(x)φ(x)dx =

∫ ∞0

φ(x)dx (A.84)

The distributional derivative of H looks like

−∫H(x)φ(x)dx = −

∫ ∞0

φ′(x)dx (A.85)

where the right side is φ(0)− φ(∞) = φ(0) by the definition of a test function. Butφ(0) can be expressed symbolically∫

δ(x)φ(x)dx (A.86)

So in a distributional sense the derivative4 of H(x) is given by

H ′(x) = δ(x) (A.87)

Consider the following example in which a function f(x) is defined by

f(x) =

x2 if x < a

x3 if x ≥ a

where a is a constant. Expressing f(x) in the manner f(x) = x2 +(x3−x2)H(x−a),it is straightforward to work out the derivative f ′(x) in the distributional sense

3Often the notation θ(x) is used.4Note that the Heaviside function is discontinuous at x = 0 and hence not differentiable

in the ordinary sense. The derivative vanishes everywhere except at the point x = 0 whereit does not exist.


namely, f ′(x) = 2x + (3x2 − 2x)H(x − a) + (a3 − a2)δ(x − a). The latter can beprojected as

f ′(x) = a2(a− 1)δ(x− a) +

2x if x < a

3x2 if x ≥ a

When dealing with differential equations, say a PDE with an inhomogeneousterm, a distribution can act for latter. In the book we have discussed at many placesthe role of the delta function inhomogeneity as given by the form

Lφ = δ (A.88)

The distributional solution resulting from it is the fundamental solution or Green’sfunction.

Green’s function

The use of Green’s function method facilitates solving a non-homogeneous linearPDE. The idea of a Green’s function goes as follows.

Let φ(ξ1, ξ2, ..., ξn) satisfy a non-homogeneous PDE in a certain n-dimensionalregion <n

4φ ≡ (∂2

∂ξ12 +

∂2

∂ξ22 + ...+

∂2

∂ξn2 )φ(ξ1, ξ2, ..., ξn) = ρ(ξ1, ξ2, ..., ξn) (A.89)

where ρ is a source term. We assume that (A.89) is accompanied by the Dirichletboundary condition on the corresponding hypersurface. We refer to ξ ≡ (ξ1, ξ2, ..., ξn)as the coordinates of a variable point Q.

Consider a solution G(ξ,x) of (A.89) when ρ is replaced by the Dirac deltafunction

4G(ξ,x) = δ(ξ − x) (A.90)

where x corresponds to a fixed point P in Ω. Like φ, G also obeys the Dirchletboundary condition on the hypersurface. G(ξ,x) is called the Green’s function ofthe operator 4.


It is easy to check that φ as given by the following integral

φ(ξ) =

∫Ω

G(ξ,x)ρ(x)dx (A.91)

satisfies (A.90), by operating on both sides of (A.91), by 4 and using (A.89).

We state the Dirichlet boundary problem for φ as follows

4φ(ξ) = ρ(ξ), ξ ∈ Ω, φ(ξ) = 0, ξ ∈ ∂Ω (A.92)

Note that the vanishing of φ on the boundary is consistent with the vanishingof G on the same from the given integral representation (A.91) of φ in terms of G.

Appendix B

Fourier transform

Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203Convolution theorem and Parseval relation . . . . . . . . . . . . . . . . . . . . . . . . 210

Fourier transform

A periodic function with period 2l can be represented by a Fourier series in thecomplex form 1

f(x) =

∞∑n=−∞

aneinπxl (B.1)

where the complex coefficients are given by the integral

an =1

2l

∫ +l

−lf(z)e−

inπzl dz (B.2)

z being a dummy variable.Letting (−l,+l)− > (−∞,+∞) we can think of connecting to functions which

are not periodic. Towards this end, we define

pn =nπ

l⇒ 4p ≡ pn+1 − pn =

π

l⇒ 1

2l=4p2π

(B.3)

which puts (B.1) and (B.2) in the forms

f(x) =

∞∑n=−∞

aneipnx (B.4)

along with

an =4p2π

∫ +l

−lf(z)e−ipnzdz (B.5)

1It is well known that a periodic function of period 2π can be represented by the expan-

sion f(x) = a02

+∑∞n=−∞ an cos(nx) + bn sin(nx). Noting that (cos(t), sin(t)) = eit±e−it

2,

it straightforwardly follows, on setting d0 = a02, dn = an−ibn

2, d−n = an+ibn

2, that

f(x) =∑∞n=−∞ dneinx.

203


As a consequence we can express

f(t) =

∞∑n=−∞

[4p2π

∫ +l

−lf(z)e−ipnzdz]eipnt (B.6)

where we have used the symbol t in place of x to avoid confusion. f(t) can be putin the form of the sum

f(t) =1

2π

∞∑n=−∞

G(pn)4p (B.7)

where

G(pn) =

∫ +l

−lf(z)eipn(t−z)dz (B.8)

Of course we can transform the infinite sum to an integral by letting l → ∞which implies by (B.3) 4p→ 0. This gives us from (B.8)

G(p) =

∫ ∞−∞

f(z)eip(t−z)dz (B.9)

where p now represents a continuous variable and the suffix n on p put previously(which indicted discreteness) has been dropped.

The summation in (B.7) is also changed to an indefinite integral and f(t) reads

f(t) =1

2π

∫ ∞−∞

G(p)dp =1

2π

∫ ∞−∞

eiptdp

∫ ∞−∞

f(z)e−ipzdz (B.10)

Denoting f(p) to stand for the integral

f(p) =

∫ ∞−∞


f(t) takes the form

f(t) =1

2π

∫ ∞−∞

f(p)eiptdp (B.12)

We call f(p) to be the Fourier transform of f(t) and (B.12) as the inversion formulaor inversion theorem for Fourier transform. Note that the Fourier transformation isa linear transformation.

From (B.11) and (B.12) it is easy to convince oneself that the Fourier transformof 1 is 2πδ(p) while the Fourier transform of the Heaviside step function is 1

ip+πδ(p).

For a complex function f , by separating into its real and imaginary parts, it canbe easily shown that the Fourier transforms of the complex conjugate (denoted byoverlines) of f(t) and f(−t) are respectively

F [f(t)] = f(−p), F [f(−t)] = f(p) for all p (B.13)

Fourier transform 205

For the existence of a Fourier transform certain conditions need to be satisfiedwhich we state below. These are called Dirichlet’s conditions. First, f(t) has to beabsolutely integrable. In other words,

∫∞−∞ |f(t)|dt converges. Second, f has a finite

number of points of maxima and minima in any finite interval. This means thatfunctions like f(t) = sin( 1

t) are excluded from consideration. Third, f(t) has only

a finite number of points of discontinuities in any finite interval. This means thatfunctions such as

f(t) =

0 if t rational

1 if t irrational

are disregarded from consideration. In this way, Dirichlet’s conditions give a set ofsufficient conditions for the existence of a Fourier transform.

In some cases the factor 12π

is symmetrized between the Fourier transform f(p)and its inverse to read as a pair

f(p) =1√2π

∫ ∞−∞


f(t) =1√2π

∫ ∞−∞

f(p)eiptdp (B.15)

In this book we have adopted such a convention.Let us consider the following examples.

(1) Let f(t) be defined by

f(t) = e−γ|t|, γ > 0

Then its Fourier transform is given by

f(p) =1√2π

∫ ∞−∞

e−γ|t|e−iptdt

The above can be expressed as

f(p) =1√2π

2 limR→∞

∫ R

0

e−γt cos(pt)dt

where because of its odd character sin(pt) term vanishes. Performing the integraland taking the limit R→∞ we find

f(p) =

√2

π

γ

p2 + γ2

(2) Let f(t) be given by the Heaviside function

f(t) = H(a− |t|), a > 0

i.e.

f(t) =

0 if |t| > a

1 if |t| < a


Then the Fourier transform of f(t) is given by

f(p) =1√2π

∫ ∞−∞

H(a− |t|)e−iptdt

which reduces to simply

f(p) =1√2π

∫ a

−ae−iptdt

Evaluating the integral it turns out that

f(p) =

√2

π

sin(pa)

p

Depending on the odd-ness and even-ness of a function it proves useful to definethe sine transform and cosine transform of the function respectively. The Fouriersine transform and its inverse are defined by

fs(p) =

√2

π

∫ ∞0

f(z) sin(pz)dz, 0 ≤ p <∞ (B.16)

fs(t) =

√2

π

∫ ∞0

fs(p) sin(pt)dp, 0 ≤ t <∞ (B.17)

On the other hand, the Fourier cosine transform and its inverse are defined by

fc(p) =

√2

π

∫ ∞0

f(z) cos(pz)dz 0 ≤ p <∞ (B.18)

fc(t) =

√2

π

∫ ∞0

fs(p) cos(pt)dp, 0 ≤ t <∞ (B.19)

where we have assumed Fourier sine and cosine transforms to exist which is ensuredby assuming

∫∞0f(t)dt to be absolutely convergent.

The linearity of Fourier sine and cosine transforms is self-evident. In the followingwe work out the formula for the derivatives. Let us assume that a function f(t) isdifferentiable n times with respect to t such that f(t) and its (n − 1)-derivativeswith respect to t all tend to zero as t→∞. The Fourier sine transform for the firstderivative is given by

[fx]s(p) =

√2

π

∫ ∞0

f ′(z) sin(pz)dz (B.20)

which when integrated by parts yields


[fx]s(p) = −p√

2

π

∫ ∞0

f(z) cos(pz)dz = −pfc(p) (B.21)

by the assumed conditions on f(t).Integrating again we find

[fxx]s(p) =

√2

πpf(0)− p2fs(p) (B.22)

Similarly for the first derivative of the cosine transform the following holds

[fx]c(p) = pfs(p)−√

2

πf(0) (B.23)

and for the second derivative we have the result

[fxx]c(p) = −p2fc(p)−√

2

πf ′(0) (B.24)

More generally, we can derive for multiple derivatives the formulae

(fxx...(2n)−times)c(p) = −n−1∑m=0

(−1)ma2n−2m−1p2m + (−1)np2nfc(p) (B.25)

(fxx...(2n+1)−times)c(p) = −n∑

m=0

(−1)ma2n−2mp2m + (−1)np2n+1fs(p) (B.26)

(fxx...(2n)−times)s(p) = −n∑

m=0

(−1)ma2n−2mp2m−1 + (−1)np2nfs(p) (B.27)

(fxx...(2n+1)−times)s(p) = −n∑

m=1

(−1)ma2n−2m+1p2m−1 + (−1)np2n+1fc(p) (B.28)

where the quantity ar denotes

ar =

√2

πlimt→0+

fr(t) (B.29)

As simple applications of sine and cosine transforms we note that since

fs[f(αt)] =

√2

π

∫ ∞0

f(αt) sin(pt)dt (B.30)

fc[f(αt)] =

√2

π

∫ ∞0

f(αt) cos(pt)dt (B.31)


we have directly

fs(e−bt) =

√2

π

∫ ∞0

e−bt sin(pt)dt =

√2

π

b

b2 + p2(B.32)

fc(e−bt) =

√2

π

∫ ∞0

e−bt cos(pt)dt =

√2

π

p

b2 + p2(B.33)

Similarly when a decaying exponential is appended

fs(e−bt cos(at)) =

1

2

√2

π[

p− ab2 + (p− a)2

+p+ a

p2 + (p+ a)2] (B.34)

fc(e−bt cos(at)) =

b

2[

1

b2 + (a− p)2+

1

b2 + (a+ p)2] (B.35)

In both cases above a, b > 0 and t ≥ 0.

As another example consider the exponential function f(x) = e−xa , x > 0 where

a > 0. That the Fourier transform is a complex function can be seen from a directevaluation

f(p) =

∫ ∞0

e−xa e−ipxdx =

a

1 + ipa(B.36)

This implies that

|f(p)|2 =a2

1 + p2a2(B.37)

which corresponds to a bell-shaped Lorentz curve discussed in Appendix A.We furnish in Table B.1 certain standard results of Fourier transform.


TABLE B.1: Some standard results in Fourier transform.

Function Fourier transform

√2πf(t) =

∫∞−∞ f(p)eiptdp

√2πf(p) =

∫∞−∞ f(z)e−ipzdz

1 2πδ(p)

f ′(t) ipf(p)

f(t− t0) f(p)e−ipt0

cos(λt) π[δ(p− λ) + δ(p+ λ)]

sin(λt) −iπ[δ(p− λ)− δ(p+ λ)]

e−|γ|t|√

2π

γp2+γ2

1t2+a2

1a

√π2 e−a|p|, a > 0

e−at2

, a > 0 1√2ae−

p2

4a

δ(x) 1√2π

H(a− |t|), a > 0√

2π

sin pxp

∑∞n=−∞ δ(t− nT )

√2πT

∑∞m=−∞ δ(p− 2mπ

T )


Convolution theorem and Parseval relation

To derive the convolution theorem for the Fourier transform let us define theconvolution of two functions f(t) and g(t) as the integral∫ +∞

−∞f(τ)g(t− τ)dτ = f ∗ g (B.38)

where f ∗ g (= g ∗ f) speaks of the convolution integral of Fourier transform.

Let the functions f and g have their Fourier transforms as f(p) and g(p) respec-tively. Then

f ∗ g =1√2π

∫ ∞−∞

f(τ)dτ

∫ ∞−∞

g(p)eip(t−τ)dp (B.39)

where g(p) is the Fourier transform of g(t− τ). Writing

f ∗ g =

∫ ∞−∞

g(p)eiptdp(1√2π

∫ ∞−∞

f(τ)e−ipτdτ) (B.40)

it transpires that

f ∗ g =

∫ ∞−∞

f(p)g(p)eiptdp (B.41)

where f(p) is the Fourier transform of f(t). This is one form of convolution theoremshowing that f ∗ g is the inverse Fourier transform of

√2πf(p)g(p).

Let us now put t = 0. From (B.40) and (B.42) we arrive at the form∫ +∞

−∞f(τ)g(−τ)dτ =

∫ ∞−∞

f(p)g(p)dp (B.42)

If we put in place of g(−τ) the function g(τ) and note that F [g(−t)] = g(p) wededuce ∫ +∞

−∞f(τ)g(τ)dτ =

∫ ∞−∞

f(p)g(p)dp (B.43)

If f and g are equal i.e. f(t) = g(t) and f(p) is the common Fourier transform then∫ +∞

−∞f(t)f(t)dt =

∫ ∞−∞

f(p)f(p)dp (B.44)

which implies ∫ +∞

−∞|f(t)|2dt =

∫ ∞−∞|f(p)|2dp (B.45)

(B.43) - (B.45) are called Parseval’s relation.


As an application of one of Parseval’s relations let us consider the function f(x)as defined by

f(x) =

1− |x| in (-1,1)

0 in |x| > 1

Here the Fourier transform of f(x) is given by

f(p) =1√2π

∫ 0

−1

(1 + x)e−ipxdx+1√2π

∫ 1

0

(1− x)e−ipxdx (B.46)

Evaluation of the integrals gives

f(p) =1√2π

[sin( p

2)

p2

]2 (B.47)

Applying now the Parseval’s relation (B.47)

∫ +∞

−∞[sin( p

2)

p2

]4dp = 2π

∫ 1

−1

(1− |x|)2dx = 4π

∫ 1

0

(1− x)2dx =4π

3(B.48)

Appendix C

Laplace transform

Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213Inversion theorem for Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 214Asymptotic form for Laplace’s inversion integral . . . . . . . . . . . . . . . . . . 215

Laplace transform

For any function f(t), which is

(i) piecewise continuous in the closed interval [0, a], a > 0 and(ii) obeys |f(t)| ≤ Keαt, where t > T , K, α and T are constants with K,T > 0,

i.e. f(t) is of exponential order as t→∞then the Laplace transform of f(t) is said to exist which is a function of a complexvariable s and is given by

F (s) =1

2iπ

∫ ∞0

e−stf(t)dt, Re(s) > α (C.1)

The above proposition is only a sufficient condition for the existence of the Laplacetransform but by no means necessary. This basically means that a function mayhave a Laplace transform without satisfying the above two conditions. Logarithmicfunction and unit impulse function are examples in this regard. For the logarith-mic function the result is L(ln(t)) = Γ′(1) − ln(s), where Γ is the Eulerian gammafunction while for the unit impulse function, which is essentially the delta function,the Laplace transform is unity. The integral (C.1) does not converge for functionsthat increase faster than the exponential and these functions do not have a Laplacetransform.

Laplace transform of the derivatives is easily worked out in the following forms.For the first and second derivatives the expressions are

L[f ′(t)] = sL[f(t)]− f(0) (C.2)

and

L[f ′′(t)] = s2L[f(t)]− sf(0)− f ′(0) (C.3)

213


while for the n-th derivative it is given by

L[f (n)(t)] = snL[f(t)]− sn−1f(0)− sn−2f ′(0)− fn−1(0) (C.4)

The convolution theorem for the Laplace transform states that if L[f(t)] = F (s)and L[g(t)] = G(s) then

L[

∫ t

0

f(t)g(t− τ)dτ ] = F (s)G(s) (C.5)

where the integral stands for the convolution integral of Laplace transform.We furnish in Table C.1 certain standard results of Laplace transform.

Inversion theorem for Laplace transform

Here the main aim is to construct f(t) from the form of F (s). The inversion the-orem for Laplace transform which is also referred to as the Mellin’s inverse formulaor Bromwich integral gives sufficient conditions under which F (s) is the Laplacetransform of f(t).

Let F (s) be an analytic function of a complex variable s and of order O(s−k), k >1 for r = Re(s) > α. Further, F (s) is real for r = Re(s) > α. Then an integralformula for the inverse Laplace transform is given by

f(t) =1

2iπ

∫ r+i∞

r−i∞estF (s)ds, Re(s) > α (C.6)

As would be realized in solving problems of inverse formula Cauchy’s residuetheorem greatly facilitates evaluation of the complex integral. It can be shown thatthe function f(t) is independent of r at points of continuity of f(t) whenever r > aand has the following properties

(i) f(t) is continuous for all t,

(ii) f(t) = 0 for t < 0,

(iii) f(t) is of exponential order i.e. O(eαt) at t→∞ and

(iv) F (s) is the Laplace transform of f(t).

Laplace transform 215

Asymptotic form for Laplace’s inversion integral

To evaluate the asymptotic form (i.e. for large t) for the inversion integral

f(t) =1

2iπ

∫ r+i∞

r−i∞estF (s)ds, 0 < β < 1 (C.7)

from the behaviour of F (s) near its singularity with the largest real part, let s = x+iyand the singularity of F (s) be located at the point s = s0 = a + ib where a < r.Subject to the following assumptions that

(i) F(s) is analytic in the region where Re(s) ≥ a− δ, δ > 0 except at s = s0,(ii) F (s)→ 0 uniformly with x for a− δ ≤ x < r as y → ±∞,(iii)

∫ y |F (s)|dy converges for y → ±∞ for a− δ ≤ x < r and(iv) near s = s0, F (s) can be expanded as

F (s) =

∞∑n=−1

an(s− s0)n + (s− s0)−β∞∑n=0

bn(s− s0)n, 0 < β < 1 (C.8)

where the two series converge for 0 < |s− s0| ≤ l, f(t) has, for large t, the followingrepresentation

f(t) ∼ es0t[a−1 +sinβπ

π

∞∑n=0

(1)nbnΓ(n+ 1− β)1

tn+1−β ] (C.9)

Before we outline the proof, it may be mentioned that the above result alsoaddresses the particular case of Watson lemma which states that if F (s) can beexpanded as

F (s) ∼∞∑n=0

bn1

sβ−n, Re(β) > 1 (C.10)

as s− > 0+, then the asymptotic expansion of f(t) around t = 0 is provided by

f(t) ∼∞∑n=0

bnΓ(n+ 1− β)1

tn+1−β Re(β) > 1 (C.11)

Watson’s lemma has applications in problems dealing with the asymptotic natureof integrals.

To get along with the proof of the expansion (C.8), we focus on the contour Cas sketched in Figure C.1.

C is defined by the line segments LM, QR which are portions of the straight linex = a− δ (note s = x+ iy), parallel segments MN, PQ and a small circle cρ, ρ beingits radius fulfilling s− s0 = ρ ≡ ε

t, ε being an infinitesimal quantity.


FIGURE C.1: The contour C.

For the validity of the expansion (C.7) over MN, PQ and cρ, we choose δ < l.Because of the assumptions made in (i) and (ii), we replace (r− i∞, r+ i∞) by thecontour C to write

e−s0tf(t) =1

2iπ

∫C

e(s−s0)tF (s)ds, 0 < β < 1

Taking into account the various components of the contour C as shown in FigureC.1, we split the right side of (C.11) into the following parts I1, I2 and I3

I1 =1

2iπ

∫LM+QR

e(s−s0)tF (s)ds, 0 < β < 1 (C.12)

I2 =1

2iπ

∫MN+PQ

e(s−s0)tF (s)ds, 0 < β < 1 (C.13)


I3 =1

2iπ

∫cρ

e(s−s0)tF (s)ds, 0 < β < 1 (C.14)

For I1 we see that

|I1| ≤1

2π

∫ b− εt

−∞e−δt|F (s)|ds+

1

2π

∫ ∞b+ ε

t

e−δt|F (s)|ds

which because of the condition (iii) reduces to

|I1| ≤ Ke−δt

where K is a constant.

For I3 we see that

I3 =1

2π

∫ 2π

0

eρteiθ

[

∞∑n=−1

anρneinθ + ρ−βe−iβθ

∞∑n=0

bnρneinθ]ρeiθdθ

which as ρ tends to zero gives

limρ→0

I3 =1

2π

∫ 2π

0

a−1dθ = a−1

Finally, for I2, since on MN , s = s0 + xe−iπ and on PQ, s = s0 + xeiπ, wededuce

IMN2 =

1

2iπ

∫ δ

εt

e−xt[

∞∑n=−1

an(−1)nxn + x−βeiπβ∞∑n=0

bn(−1)nxn]dx

and

IPQ2 = − 1

2iπ

∫ δ

εt

e−xt[

∞∑n=−1

an(−1)nxn + x−βe−iπβ∞∑n=0

bn(−1)nxn]dx

On putting xt = z the above two can be combined to represent I2 = IMN2 + IPQ2 as

I2 =sinπβ

π

∫ δt

ε

dze−z

z

∞∑n=0

bn(−1)n(z

t)n+1−β

Collecting the above evaluations of I1, I2 and I3 we find on proceeding to thelimit ε→ 0

e−s0tf(t) ∼ a−1 +sinπβ

π

∫ δt

0

dze−z

z

∞∑n=0

bn(−1)n(z

t)n+1−β


We now proceed to show that as t→∞ the quantity ζ defined by

ζ = [

∫ δt

0

dze−z

z

∞∑n=0

bn(−1)n(z

t)n+1−β −

∞∑n=0

(−1)nbnΓ(n+ 1− β)

tn+1−β ]tn+1−β (C.15)

goes to 0. Towards this end, we use the integral representation of the gamma function

Γ(λ+ 1) =

∫ ∞0

e−ttλdt, λ > −1

to re-write ζ as

ζ= tn+1−β [

∫ δt

0

dze−z

z

∞∑m=0

bm(−1)m(z

t)m+1−β −

n∑m=0

bm(−1)m∫ ∞

0

dze−z

z(z

t)m+1−β ]

Splitting the integral appearing in the second term into portions (0, δt) and (δt,∞)and subtracting the piece of summation over m = 0 to n from the summation in thefirst integral we can write

ζ = tn+1−β [

∫ δt

0

dze−z

z

∞∑m=n+1

bm(−1)m(z

t)m+1−β

−n∑

m=0

bm(−1)m∫ ∞δt

dze−z

z(z

t)m+1−β ]

This can be expressed in the manner

ζ = t−1I −n∑

m=0

I ′m

where I and I ′m stand for

I =

∫ δt

0

dzzn+1−βe−z∞∑

m=n+1

bm(−1)m(z

t)m−n−1

and

I ′m = tn−βe−δtn∑

m=0

bm(−1)m∫ ∞

0

dx(δ +x

t)m−βe−x

where we have put z = δt+ x. It is clear that I ′m → 0 for m = 1, 2, ..., n.About I we find

|I| ≤∫ δt

0


m=n+1

|bm|(z

t)m−n−1

≤∫ δt

0


m=n+1

|bm|δm−n−1


≤ B∫ δt

0

dzzn+1−βe−z

≤ BΓ(n+ 2− β)

In the above steps we have noted that the series∑n bn(s − s0)n is convergent and

its radius of convergence is < δ. Also∫ δt

0≤∫∞

0and B is a constant. Since there is

a factor t−1 in (C.28) it follows that ζ → 0 as t→∞.From (C.23), (C.24) and using the limiting result of ζ we arrive at (C.9).


TABLE C.1: Some standard results of Laplace transform. In the text wehave used the notation f in place of F . Note that F (s) = L(f(t)) and f(t) =L−1F (s).

f(t) F (s), s ∈ C

1 1s

eαt 1s−α

tn n!sn+1 , n = 0, 1, 2, ...

tα Γ(α+1)sα+1 , α ≥ 0

f ′(t) sL(f)− f(0)

f ′′(t) s2L(f)− sf(0)− f ′(0)

f(t− t0) F (s)e−st0

cos(αt) ss2+α2

sin(αt) αs2+α2

cosh(αt) ss2−α2

sinh(αt) αs2−α2

δ(t− α) e−αs

ae−a2

4t

2√πt

e−a√s

√s

ae−a2

4t

2t√πt

e−a√s

Bibliography

1. Bagchi, B., (2017) Advanced Classical Mechanics, CRC/Taylor and Francis,London.2. Balakrishnan, V., (2003) All about the Dirac delta function, Resonance,Volume 8, Issue 8, 48.3. Bitsadze, A.V. and Kalinichenko, D.F., (1983) A Collection of Problems onthe Equations of Mathematical Physics, MIR Publishers, Moscow.4. Boas, M.L., (1966) Mathematical Methods in the Physical Sciences, JohnWiley & Sons, Inc., New York.5. Corinaldesi, E., (1998) Classical Mechanics for Physics Graduate Students,World Scientific, Singapore.6. Costin, O., (2008) Asmptotics and Borel Summability, Chapman andHall/CRC, London.7. Dijk, G. van, (2013) Distribution Theory, Walter de Gruyter GmbH, Berlin.8. James, J.F., (2011) A Student’s Guide to Fourier Transforms, CambridgeUniversity Press, Cambridge.9. Joglekar, S.D., (2006) Mathematical Physics: Advanced Topics, UniversitiesPress, India.10. John, F., (1982) Partial Differential Equations, Applied Mathematics Se-ries, Springer, New York.11. Johnson, S.G., (2017) When functions have no value(s): Delta functionsand distributions, MIT Course 18.303 Notes.12. Kanwal, R.P., (2004) Generalized Functions - Theory and Applications,Birkhauser Boston.13. Kersale, E., (2003) Analytic Solutions of Partial Differential Equations,http://www.maths.leeds.ac.uk/˜kersale/14. Kinnunen, J., (2017) Partial Differential Equations, Lecture notes, AaltoUniversity, Finland.15. Kumaran, V., Lecture Notes on Steady and Unsteady Diffusion. IISc., Ban-galore.16. Lamoureux, M.P., (2006) The Mathematics of Partial Differential Equa-tions and the Wave Equation, Lecture at Seismic Imaging Summer School(unpublished).17. Lass, H.,(1950) Vector and Tensor Analysis, McGraw-Hill, New York.18. Lindenbaum, S.D., (1996) Mathematical Methods in Physics, World Scien-tific.19. Logan, J.D., (2006) Beginning Partial Differential Equations, John Wiley& Sons., Inc., New York.20. Muthukumar, T., (2014) Partial Differential Equations, Indian Instituteof Technology, Kanpur, Lecture notes.

221

http://www.maths.leeds.ac.uk/

222 Bibliography

21. Olver, P., (2013) Introduction to Partial Differential Equations, Springer,Berlin.22. O’Neil, P.V., (1999)Begining Partial Differential Equations, John Wiley &Sons, Inc., New York.23. Petrovsky, I.G., (1991) Lectures on Partial Differential Equations, DoverPublications, Inc., New York.24. Sneddon, I.N., (1957) Elements of Partial Differential Equations, McGraw-Hill, New York.25. Sommerfeld, A., (1949) Partial Differential Equations in Physics, Aca-demic Press, New York.26. Stavroulakis, I.P. and Tersian, S.A., (2004) Partial Differential Equations:An Introduction with Mathematica and MAPLE, World Scientific, Singapore.27. Strauss, W.A., (1992) Partial Differential Equations, An Introduction,John Wiley & Sons., Inc., New York.28. Williams, W.E., (1980) Partial Differential Equations, Oxford UniversityPress, Oxford.29. Yanovsky, I., (2005) Partial Differential Equations: Graduate Level Prob-lems and Solutions (unpublished).30. Zachmann, D. and Duchateau P., Schaum’s Outline of Partial DifferentialEquations, McGraw-Hill, New York.

Index

1-soliton solution, 136

adjoint, 73asymptotic form for Laplace’s inversion

integral, 215

Beltrami equations, 107boundary condition of first kind, 60boundary condition of second kind,

60boundary condition of third kind, 61boundary-initial value problems, 60breather, 136Bromwich integral, 214Burger’s equation, 162

canonical form, 48canonical transformations, 28Cauchy, 199Cauchy problem for second order linear

PDE, 21Cauchy’s residue theorem, 214Cauchy-Kowalevski theorem, 68Cauchy-Riemann equations, 81, 107characteristic curve, 8, 13, 50characteristic function, 114characteristic triangle, 112circle, 100classical mechanics, 70compact support, 196convolution integral of Fourier

transform, 210convolution integral of Laplace

transform, 214convolution theorem for Fourier

transform, 210convolution theorem for Laplace

transform, 214

D’Alembert’s solution, 111, 170delta function, 185

delta sequence, 186, 187diffusion equation, 139Dirichlet integral, 180Dirichlet problem on rectangular

domain, 62Dirichlet’s conditions, 205Dirichlet’s formula, 187Dirichlet’s problem, 20distributions, 198Duhamel principle, 183

elliptic, 76Euler-Poisson equation, 135existence and uniqueness of solutions,

93

Fick’s law, 139Fourier cosine transform, 206Fourier sine transform, 206Fourier transform, 203fourth order diffusion equation, 20fundamental solution of Laplace’s

equation, 88

Gauss’ mean value theorem, 90Gaussian, 188, 199generalized functions, 185, 198generating function, 31Green’s equivalent layer theorem, 97Green’s function, 201

Hadamard’s example, 66, 67Hadamard’s method of descent, 128Half-space, 104Hamilton’s characteristic function,

41Hamilton principal function, 39Hamilton-Jacobi equation, 27, 38, 40harmonic functions, 90heat conduction equation, 19Helmholtz’s equation, 19, 117

223

224 Index

hyperbolic, 76hypersurface, 20

inhomogeneous equation, 4initial value problems, 121integral curve, 13integral lines, 11integral surfaces, 11inversion theorem for Fourier transform,

204inversion theorem for Laplace

transform, 214

Kirchhoff’s formula, 125Korteweg-de Vries equation, 19

Lagrange’s identity, 74Lagrange’s method, 9Laplace’s equation, 19Laplace’s transform, 213Liouville’s theorem, 30locally integrable functions, 198Lorentzian, 188, 199

Neumann problem, 20, 63, 64non-homogeneous, 4normal form, 48normally directed distribution of

doublets, 94

parabolic, 76Parseval’s relation, 211plane arc, 11Poisson bracket, 28Poisson’s equation, 18Poisson’s formula, 125Poisson-Parseval formula, 129principal solution of heat equation, 156propagating singularity, 52

reaction-diffusion equation, 139regular arc, 12regular distribution, 198reproductive property, 186resonant frequency, 135retarded potential, 127Riemann function, 114

Schrodinger equation, 19self-adjoint, 73separation of variables, 83, 117simple arc, 12sine-Gordon equation, 135singular distribution, 199sphere, 102spherical harmonics, 119spherical harmonics of degree n, 87spherical polar coordinates, 85, 117support, 196

telegraph equation, 78, 181test function, 197time-dependent Hamilton-Jacobi

equation., 39Tricomi equation, 78type I canonical transformation, 33type I generating function, 32type II canonical transformation, 34type II generating function, 32type III canonical transformation, 35type III generating function, 32type IV canonical transformation, 35type IV generating function, 32

ultra-hyperbolic, 76

Watson lemma, 215wave equation, 19well-posedness, 66

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