integral transform techniques for green's function

Lecture Notes in Applied and Computational Mechanics 76

Kazumi Watanabe

Integral Transform Techniques for Green’s Function Second Edition

Lecture Notes in Applied and ComputationalMechanics

Volume 76

Series editors

Friedrich Pfeiffer, Technische Universität München, Garching, Germanye-mail: [email protected]

Peter Wriggers, Universität Hannover, Hannover, Germanye-mail: [email protected]

About this Series

This series aims to report new developments in applied and computationalmechanics—quickly, informally and at a high level. This includes the fields of fluid,solid and structural mechanics, dynamics and control, and related disciplines. Theapplied methods can be of analytical, numerical and computational nature.

More information about this series at http://www.springer.com/series/4623

http://www.springer.com/series/4623

Kazumi Watanabe

Integral TransformTechniques for Green’sFunction

Second Edition

123

Kazumi WatanabeYamagata UniversityYonezawaJapan

ISSN 1613-7736 ISSN 1860-0816 (electronic)Lecture Notes in Applied and Computational MechanicsISBN 978-3-319-17454-9 ISBN 978-3-319-17455-6 (eBook)DOI 10.1007/978-3-319-17455-6

Library of Congress Control Number: 2013940095

Springer Cham Heidelberg New York Dordrecht London© Springer International Publishing Switzerland 2014, 2015This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar ordissimilar methodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exemptfrom the relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material containedherein or for any errors or omissions that may have been made.

Printed on acid-free paper

Springer International Publishing AG Switzerland is part of Springer Science+Business Media(www.springer.com)

Dedicated to my teachers,Dr. Akira Atsumi (Late Professor,Tohoku University)

and

Dr. Kyujiro Kino (Late Professor,Osaka Institute of Technology)

Preface to the Second Edition

After publishing the original version, the author noticed that more detailed math-ematical techniques should be included so that the young reader could learn thetraditional analytical techniques without any mathematical skip. That is, the dis-cussion on square root functions. In many dynamic/wave problems, we frequentlyencounter the square root function which is the typical multi-valued function andhave to introduce branch cuts in the complex plane for the inversion integral.A simple elementary technique for the introduction of the branch cut and thediscussion on the argument of the square root function along the cut are included inChap. 1 as Sect. 1.3. This branch cut is employed throughout the book and appliedto the inversion integrals in Sect. 2.5 and other sections. Due to the introductionof the unified branch cut, Sect. 2.5 for the time-harmonic Green’s function is whollyrewritten.

In the revising process the author also noticed five exact closed-form solutions:three are Green’s functions for torsion problems, the fourth one is for the reflectionproblem and the last is for the scattering problem. Green’s functions for the torsionproblem are inserted in Chaps. 3 and 7. SH-wave reflection at a moving boundary inSect. 7.4 is slightly rewritten in order to include the closed-form solution. Section 7.5is newly inserted and shows the exact closed-form solution for a wave scatteringproblem in an inhomogeneous elastic solid. Further, employing the branch cutdescribed in Chap. 1, an excellent application technique of the complex integral isexplained in the last Sect. 7.6. It is the transformation of a semi-infinite integral to afinite one that is suitable for numerical computations. Needless to say, many errorsand mistakes in the original version have also been corrected.

The author hopes the young reader can learn one of the traditional analyticaltechniques, especially the application of the complex integral for the integraltransform. Thus, the present revised version is more instructive than the originalone, and every question and inquiry via email “[email protected]” iswelcome.

Hikoshima Island, Japan, January 2015 Kazumi Watanabe

vii

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_3

http://dx.doi.org/10.1007/978-3-319-17455-6_7

http://dx.doi.org/10.1007/978-3-319-17455-6_7

http://dx.doi.org/10.1007/978-3-319-17455-6_7

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_7

Preface

When I was a senior student, I found a book on the desk of my advisor professorand asked him how to get it. His answer was negative, saying its content was toohard, even for a senior student. Some weeks later, I found it again in a book store,the biggest one in Osaka. This was my first encounter with “Fourier Transforms”written by the late Prof. I. N. Sneddon. Since then, I have learned the power ofintegral transform, i.e. the principle of superposition.

All phenomena, regardless of their fields of event, can be described by differ-ential equations. The solution of the differential equation contains the crucialinformation to understand the essential feature of the phenomena. Unfortunately,we cannot solve every differential equation, and almost all phenomena are governedby nonlinear differential equations, of which most are not tractable. The differentialequations that can be solved analytically are limited to a very small number. Buttheir solutions give us the essence of the event. The typical partial differentialequations that can be solved exactly are the Laplace, the diffusion and the waveequations. These three partial differential equations, which are linearized for sim-plicity, govern many basic phenomena in physical, chemical and social events. Inaddition to single differential equations, some coupled linear partial differentialequations, which govern somewhat complicated phenomena, are also solvable andtheir solutions give much information about, for example, the deformation of solidmedia, propagation of seismic and acoustic waves, and fluid flows.

In a case where phenomena are described by linear differential equations, thesolutions can be expressed by superposition of basic/fundamental solutions. Theintegral transform technique is a typical superposition technique. The integraltransform technique does not require any previous knowledge for solving differ-ential equations. It simply transforms partial or ordinary differential equations toreduced ordinary differential equations or to simple algebraic equations. However, asubstantial difficulty is present regarding the inversion process. Many inversionintegrals are tabulated in various formula books, but typically, this is not enough. Ifa suitable integration formula cannot be found, the complex integral must beconsidered and Cauchy’s integral theorem is applied to the inversion integral. Thus,

ix

integral transform techniques are intrinsically connected with the theory of complexintegrals.

The present book intends to show how to apply integral transforms to partialdifferential equations and how to invert the transformed solution into the actualspace-time domain. Not only the use of integration formula tabulated in books, butalso the application of Cauchy’s integral theorem for the inversion integrals aredescribed concisely and in detail. A particular solution for a differential equationwith a nonhomogeneous term of a point source is called the “Green’s function.”The Green’s functions for coupled differential equations are called “Green’s dyadic.”The Green’s function and Green’s dyadic are the basic and fundamental solutionof the differential equation and give the principal features of the event. Furthermore,these Green’s functions and dyadics have many applications for numerical com-putation techniques such as the Boundary Element Method. However, the Green’sfunction and Green’s dyadic have been scattered in many branches of appliedmechanics and thus, their solution methods are not unified. This book intends topresent and illustrate a unified solution method, namely the method of integraltransform for the Green’s function and Green’s dyadic. Thus, the fundamentalGreen’s function for the Laplace and wave equations and the Green’s dyadic forelasticity equations are gathered in this single book so that the reader can have accessto a proper Green’s function and understand the mathematical process for itsderivation.

Chapter 1 describes roughly the definition of the integral transforms and thedistributions to be used throughout the book. Chapter 2 shows how to apply anintegral transform for solving a single partial differential equation such as theLaplace equation and the wave equation. The basic technique of the integraltransform method is demonstrated. Especially, in the case of the time-harmonicresponse for the wave equation, the integration path for the inversion integral isdiscussed in detail. At the end of the chapter, the obtained Green’s functions arelisted in a table so that the reader can easily find the difference of the functionalform among the Green’s functions. An evaluation technique for a singular inversionintegral which arises in a 2D static problem of Laplace equation is also developed.

The Green’s dyadic for 2D and 3D elastodynamic problems are discussed inChap. 3. Three basic responses, impulsive, time-harmonic and static responses, areobtained by the integral transform method. The time-harmonic response is derivedby the convolution integral of the impulsive response without solving the differ-ential equations for the time-harmonic source.

Chapter 4 presents the governing equations for acoustic waves in a viscous fluid.Introducing a small parameter, the nonlinear field equations are linearized andreduced to a single partial differential equation for velocity potential or pressuredeviation. The Green’s function which gives the acoustic field in a uniform flow isderived by the method of integral transform. A conversion technique for theinversion integral is demonstrated. That is, to transform an inversion integral alongthe complex line to that along the real axis in the complex plane. It enabled us toapply the tabulated integration formula.

x Preface

Chapter 5 presents Green’s functions for beams and plates. The dynamicresponse produced by a point load on the surface of a beam and a plate is discussed.The impulsive and time-harmonic responses are derived by the integral transformmethod. In addition to the tabulated integration formulas, the inversion integrals areevaluated by application of complex integral theory.

Chapter 6 presents a powerful inversion technique for transient problems ofelastodynamics, namely the Cagniard-de Hoop method. Transient response of anelastic half space to a point impulsive load is discussed by the integral transformmethod. Applying Cauchy’s complex integral theorem, the Fourier inversionintegral is converted to an integral of the Laplace transform and then its Laplaceinversion is carried out by inspection without using any integration formula. TheGreen’s function for an SH-wave and Green’s dyadics for P, SV and SH-waves areobtained.

The last Chap. 7 presents three special Green’s functions/dyadics. The 2D staticGreen’s dyadic for an orthotropic elastic solid and that for an inhomogeneous solidare derived. In the last section, a moving boundary problems is discussed. Twodifferent Laplace transforms are applied for a single problem, and a conversionformula between two Laplace transforms is developed with use of Cauchy’s the-orem. This conversion enables us to apply the integral transform technique to amoving boundary problem.

The integral transform technique has been used for many years. The inversionprocess inevitably requires a working knowledge of the theory of complex func-tions. The author finds the challenge of a complex integral amusing, especially thechallenge of choosing the right contour for the inversion integral. He hopes thatyoung researchers will join the fun and carry on with the inversion techniques. Inthis respect it must be mentioned that he feels a lack of mathematical skill in therecent research activities, since some researchers tend to use numerical techniqueswithout considering the possibility of an analytical solution. The more mathemat-ical techniques expand the horizon of the differential equations wider and one canextract more firm knowledge from the nature which is described by the differentialequations. The author hopes that the present book gives one more technique to theyounger researchers.

Finally, the author wishes to express his sincere thanks to Dr. Mikael A.Langthjem, Associate Professor of Yamagata University, for his advice and nicecomments.

Yonezawa, Japan, January 2013 Kazumi Watanabe

Preface xi

Contents

1 Definition of Integral Transforms and Distributions . . . . . . . . . . . 11.1 Integral Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Distributions and Their Integration Formulas . . . . . . . . . . . . . 61.3 Branch Cut and Argument of Square Root Functions . . . . . . . 11

1.3.1 Square Root Function 1: gðzÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffi

z� z0p

. . . . . . . . . . 111.3.2 Square Root Function 2: gðzÞ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

z2 � z20p

. . . . . . . . . 141.4 Comments on Inversion Techniques and Integration

Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2 Green’s Functions for Laplace and Wave Equations . . . . . . . . . . 332.1 1D Impulsive Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 1D Time-Harmonic Source. . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 2D Static Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.4 2D Impulsive Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.5 2D Time-Harmonic Source. . . . . . . . . . . . . . . . . . . . . . . . . . 512.6 3D Static Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.7 3D Impulsive Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.8 3D Time-Harmonic Source. . . . . . . . . . . . . . . . . . . . . . . . . . 73Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3 Green’s Dyadic for an Isotropic Elastic Solid . . . . . . . . . . . . . . . . 773.1 2D Impulsive Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 793.2 2D Time-Harmonic Source. . . . . . . . . . . . . . . . . . . . . . . . . . 873.3 2D Static Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 893.4 3D Impulsive Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.5 3D Time-Harmonic Source. . . . . . . . . . . . . . . . . . . . . . . . . . 1073.6 3D Static Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

xiii

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1#Sec1













http://dx.doi.org/10.1007/978-3-319-17455-6_1#Bib1

http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2



















http://dx.doi.org/10.1007/978-3-319-17455-6_3

http://dx.doi.org/10.1007/978-3-319-17455-6_3













3.7 Torsional Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1093.7.1 Ring Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.7.2 Point Torque Source . . . . . . . . . . . . . . . . . . . . . . . . 113

Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4 Acoustic Wave in a Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . 1214.1 Compressive Viscous Fluid . . . . . . . . . . . . . . . . . . . . . . . . . 1214.2 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.3 Viscous Acoustic Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1264.4 Wave Radiation in a Uniform Flow. . . . . . . . . . . . . . . . . . . . 1294.5 Time-Harmonic Wave in a Uniform Flow . . . . . . . . . . . . . . . 135References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5 Green’s Functions for Beam and Plate. . . . . . . . . . . . . . . . . . . . . 1395.1 An Impulsive Load on a Beam. . . . . . . . . . . . . . . . . . . . . . . 1395.2 A Moving Time-Harmonic Load on a Beam . . . . . . . . . . . . . 1425.3 An Impulsive Load on a Plate . . . . . . . . . . . . . . . . . . . . . . . 1455.4 A Time-Harmonic Load on a Plate . . . . . . . . . . . . . . . . . . . . 148Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

6 Cagniard-de Hoop Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.1 2D Anti-plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . 1546.2 2D In-plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1626.3 3D Dynamic Lamb’s Problem . . . . . . . . . . . . . . . . . . . . . . . 178References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

7 Miscellaneous Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . 2057.1 2D Static Green’s Dyadic for an Orthotropic Elastic Solid . . . . 2057.2 2D Static Green’s Dyadic for an Inhomogeneous

Elastic Solid. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2137.2.1 2D Kelvin’s Solution for Homogeneous Media. . . . . . 221

7.3 Green’s Function for Torsional Waves in a MonoclinicMaterial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

7.4 Reflection of a Transient SH-Wave at a Moving Boundary . . . 2277.5 Wave Scattering by a Rigid Inclusion in an Inhomogeneous

Elastic Solid. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2427.6 An Excellent Application of Cauchy Complex Integral . . . . . . 253References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

xiv Contents









http://dx.doi.org/10.1007/978-3-319-17455-6_4

http://dx.doi.org/10.1007/978-3-319-17455-6_4












http://dx.doi.org/10.1007/978-3-319-17455-6_5

http://dx.doi.org/10.1007/978-3-319-17455-6_5











http://dx.doi.org/10.1007/978-3-319-17455-6_6

http://dx.doi.org/10.1007/978-3-319-17455-6_6








http://dx.doi.org/10.1007/978-3-319-17455-6_7

http://dx.doi.org/10.1007/978-3-319-17455-6_7



















Chapter 1Definition of Integral Transformsand Distributions

This first chapter describes a brief definition of integral transforms, such as Laplaceand Fourier transforms, a rough definition of delta and step functions which arefrequently used as the source function, and a concise introduction of the branch cutfor a multi-valued square root function. The multiple integral transforms and theirnotations are also explained. The introduction of the branch cut and the discussionon the argument of the root function along the branch cut will be helpful forapplying the complex integral to the inverse transform. The last short comment listssome important formula books which are crucial for the inverse transform, i.e. theevaluation of the inversion integral.

1.1 Integral Transforms

For a well-defined function f ðxÞ; x 2 ða; bÞ, when the integral with the kernelfunction Kðn; xÞ,

FðnÞ ¼Zb

a

Kðn; xÞf ðxÞdx ð1:1:1Þ

has its inverse integral with another kernel function K�ðn; xÞ,

f ðxÞ ¼ZL

K�ðn; xÞFðnÞdn ð1:1:2Þ

we call this integration pair an “integral transform.” The function f ðxÞ is an originalfunction and the function FðnÞ is the “image or transformed function” in thetransformed domain. If the reciprocal f ðxÞ , FðnÞ holds, we call FðnÞ the “integral

© Springer International Publishing Switzerland 2015K. Watanabe, Integral Transform Techniques for Green’s Function,Lecture Notes in Applied and Computational Mechanics 76,DOI 10.1007/978-3-319-17455-6_1

1

transform” of f ðxÞ and the two-variable-functions Kðn; xÞ and K�ðn; xÞ the kernelsof the integral transform.

We have already learned many integral transforms which are classified andnamed depending on the kernel function and the integration range. A well-knownintegral transform is the Laplace transform. The (one-sided) Laplace transform is, inthe present book, defined for the time-variable function f ðtÞ; t 2 ½0;1Þ as

f �ðsÞ ¼Z10

f ðtÞ expð�stÞdt ð1:1:3Þ

where “s” is the transform parameter and the transform kernel is expð�stÞ. Theinverse transform is also defined by the integral along the complex line,

f ðtÞ ¼ 12pi

Zcþi1

c�i1f �ðsÞ expðstÞds ð1:1:4Þ

where the integration path from c� i1 to cþ i1 is called the “Bromwich line.”The real constant “c” must be larger than the real part of any singular point of thetransformed function f �ðsÞ. Thus this line is placed at far right from all singularpoints in the complex s-plane.

In these definitions, the variable s is called the transform parameter and the twokernels for the transform and the inverse transform are exponential functions:

Kðs; tÞ ¼ expð�stÞ; K�ðs; tÞ ¼ 12pi

expðstÞ ð1:1:5Þ

Further, the integration ranges are also different from each other. The transformintegral is carried out along the semi-infinite real line ½0;1Þ for the time and theinverse transform is carried out along an infinite line ðc� i1; cþ i1Þ in thecomplex s-plane.

Since any notation for the transform parameter is available, one should be awareof the notation of the parameter since some authors use “p” instead of “s.”When theLaplace inversion is carried out by using some inversion formulas in a referencebook, not performing the inversion integral in the complex plane, the symbolic formof the Laplace inversion

f ðtÞ ¼ L�1 f �ðsÞ½ � ð1:1:6Þ

is used for the sake of simplicity. The present book also employs frequently thissimple expression for the Laplace inversion.

So far, many integral transform pairs have been found and defined. We chooseand use one suitable integral transform depending on the geometry (integrationrange) and the simplicity of its application. The followings are typical integraltransforms which are much used in applications.

2 1 Definition of Integral Transforms and Distributions

(1) Finite Fourier transform (Fourier series): f ðxÞ; x 2 ½�p; p�When an original function is defined within a finite region, the Fourier finite

transform, i.e. Fourier series is used and its transform pair can be defined as follows:

(1.1) Complex Fourier series

fn ¼Zp

�p

f ðxÞ expð�inxÞdx; f ðxÞ ¼ 12p

X1n¼�1

fn expð�inxÞ ð1:1:7Þ

where the index “n” is an integer.

(1.2) Fourier cosine series

a0 ¼ 12

Zp

�p

f ðxÞdx; an ¼Zp

�p

f ðxÞ cosðnxÞdx; f ðxÞ ¼ 1p

X1n¼0

an cosðnxÞ ð1:1:8Þ

(1.3) Fourier sine series

bn ¼Zp

�p

f ðxÞ sinðnxÞdx; f ðxÞ ¼ 1p

X1n¼1

bn sinðnxÞ ð1:1:9Þ

(2) Complex Fourier transform: f ðxÞ; x 2 ð�1;1ÞWhen the function is defined in an infinite region, the complex Fourier transform

pair is defined by

�f ðnÞ ¼Z1�1

f ðxÞ expð�inxÞdx; f ðxÞ ¼ 12p

Z1�1

�f ðnÞ expð�inxÞdn ð1:1:10Þ

(2.1) Fourier transform with a non-uniform parameter.

The Fourier transform with a non-uniform parameter is defined as

�f ðnÞ ¼Zb

a

f ðxÞ expfþixhðnÞgdx ð1:1:11aÞ

1.1 Integral Transforms 3

f ðxÞ ¼ 12p

Zþ1

�1

�f ðnÞ expf�ixhðnÞgh0ðnÞdn ð1:1:11bÞ

where hðnÞ is a monotonically increasing/decreasing function, and the lower andupper integration limits, a and b, are arbitrary constants.

Let us verify the pair of Eq. (1.1.11), briefly. Since hðnÞ is the monotonicallyincreasing or decreasing function, we can define its inverse function uniquely as

g ¼ hðnÞ , n ¼ h�1ðgÞ ð1:1:12Þ

Using the above definition, Eq. (1.1.11a) is rewritten in the form of the regularFourier transform,

�f ðh�1ðgÞÞ ¼Zb

a

f ðxÞ expðþixgÞdx ð1:1:13Þ

Applying the inversion formula of Fourier transform in Eq. (1.1.10) to the aboveequation, the Fourier inversion of Eq. (1.1.13) is given by

f ðxÞ ¼ 12p

Zþ1

�1

�f ðh�1ðgÞÞ expð�ixgÞdg; a\ x\ b ð1:1:14Þ

As we have defined the inverse function in Eq. (1.1.12), the integration variable g isreturned to the original variable n,

n ¼ h�1ðgÞ ) g ¼ hðnÞ; dg ¼ h0ðnÞdn ð1:1:15Þ

Then, Eq. (1.1.14) is rewritten as

f ðxÞ ¼ 12p

Zþ1

�1

�f ðnÞ expf�ixhðnÞgh0ðnÞdn ð1:1:16Þ

This is the second of the transform pair Eq. (1.1.11b). Therefore, the pair of theFourier transform with the non-uniform parameter hðnÞ is verified.(3) Fourier cosine/sine transform: f ðxÞ; x 2 ½0;1Þ

When an original function is even or odd in an infinite region, or the function isdefined in a semi-infinite region, we employ Fourier cosine or sine transform. Theyare defined as follows:


(3.1) Fourier cosine transform

�f ðCÞðnÞ ¼Z10

f ðxÞ cosðnxÞdx; f ðxÞ ¼ 2p

Z10

�f ðCÞðnÞ cosðnxÞdn ð1:1:17Þ

(3.2) Fourier sine transform

�f ðSÞðnÞ ¼Z10

f ðxÞ sinðnxÞdx; f ðxÞ ¼ 2p

Z10

�f ðSÞðnÞ sinðnxÞdn ð1:1:18Þ

(4) Hankel transform: f ðxÞ; x 2 ½0;1ÞAnother semi-infinite integral transform is Hankel transform (Sneddon 1951,

p. 48) defined by

~fnðnÞ ¼Z10

xf ðxÞJnðnxÞdx; f ðxÞ ¼Z10

n~fnðnÞJnðnxÞdn ð1:1:19Þ

where JnðzÞ is the n-th order Bessel function of the first kind.In order to guarantee the application of the integral transform, each integral must

converge. For example, the Fourier transform requires a convergence condition atinfinity,

f ðxÞ )x!�1

Oðjxj�mÞ; m[ 1 ð1:1:20Þ

for the original function. However, if we employ the Hankel transform, it isallowable for the original function to be finite since the Bessel function decays withthe order of the inverse square root at the infinity.

In some applications, where we encounter multi-variable functions, or where amultiple integral transform is employed, a different mark for each transform isdefined such as �f ;~f ; f �, and the multiple transform is denoted by piling up the

transform marks, like ~�f �. In the present book, we employ the Laplace transformwith respect to the time variable t as defined in Eqs. (1.1.3) and (1.1.4). For thespace variables ðx; y; zÞ, we apply three Fourier transforms with the transformparameters ðn; g; fÞ respectively. They are defined as

�f ðnÞ ¼Z1�1

f ðxÞ expðþinxÞdx; f ðxÞ ¼ 12p

Z1�1

�f ðnÞ expð�inxÞdn ð1:1:21aÞ

1.1 Integral Transforms 5

~f ðgÞ ¼Z1�1

f ðyÞ expðþigyÞdy; f ðyÞ ¼ 12p

Z1�1

~f ðgÞ expð�igyÞdg ð1:1:21bÞ

f ðfÞ ¼Z1�1

f ðzÞ expðþifzÞdz; f ðzÞ ¼ 12p

Z1�1

f ðfÞ expð�ifzÞdf ð1:1:21cÞ

Please remember that the pair of the space variable and transform parameter is fixedthroughout the present book, such as the pair, ðx; nÞ; ðy; gÞ and ðz; fÞ.

1.2 Distributions and Their Integration Formulas

For modeling engineering phenomena, many mathematical functions are used.Elementary and some special functions are used for continuous phenomena. But fordiscontinuous phenomena, distributions such as delta and step functions are fre-quently used. This subsection explains briefly the definition of three distributions:Dirac’s delta function, Heaviside’s unit step function and Heisenberg’s deltafunction.

(1) Heaviside’s unit step function: Hðx� aÞHeaviside’s unit step function is defined by the graphical form in Fig. 1.1. This

function takes the value 0 when its argument is negative and +1 when the argumentis positive,

Hðx� aÞ ¼ þ1; x[ a0 ; x\ a

�ð1:2:1Þ

Due to this definition, the zero argument at x ¼ a is a discontinuous point. So, it takestwo limiting values from the positive and negative sides of the discontinuous point,

Hð0þÞ ¼ þ1; Hð0�Þ ¼ 0 ð1:2:2Þ

Then, we have to understand that the step function is not defined at x ¼ a.

1+

0

( )H x a−

x a=

Fig. 1.1 Heaviside’s unitstep function


(2) Dirac’s delta function: dðx� aÞDirac’s delta function is defined as the limit e ! 0 of a rectangular pulse with

width e and height 1=e as shown in Fig. 1.2.The center of the rectangular distribution is fixed at x ¼ a in the limiting processand we understand the point as an application point of the delta function. It isdenoted by dðx� aÞ. In the limit, the width of the delta function vanishes and thefunctional value becomes infinite. But, its internal invisible area is one due to thedefinition ð1=eÞ � e ¼ 1. This nature makes the evaluation of the integral verysimple, and we have the formula where the integrand includes the delta function as

Zb

a

f ðxÞdðx� cÞdx ¼ f ðcÞ ; a\ c\ b0 ; c\ a or b\ c

�ð1:2:3Þ

Using this simple integration formula, we can obtain an integral representationfor the delta function. Applying Fourier transform defined by Eq. (1.1.10) to thedelta function, f ðxÞ ¼ dðx� cÞ, we use the integration formula (1.2.3). The trans-form integral is evaluated as

�f ðnÞ ¼Z1�1

dðx� cÞ expðþinxÞdx ¼ expðþincÞ ð1:2:4Þ

The inverse Fourier transform is also applied to the above �f ðnÞ and its integrationrange is reduced to the semi-infinite,

f ðxÞ ¼ dðx� cÞ ¼ 12p

Z1�1

expf�inðx� cÞgdn ¼ 1p

Z10

cosfnðx� cÞgdn ð1:2:5Þ

ε 1/ε

x=a

0 0

x=a

( )x aδ −

0ε →

↑

0 0

(a) (b)

Fig. 1.2 Schematic definition of Dirac’s delta function a rectangular pulse b Dirac’s deltafunction

1.2 Distributions and Their Integration Formulas 7

Then, we have the integral representation for Dirac’s delta function, i.e.

dðx� cÞ ¼ 1p

Z10

cosfnðx� cÞgdn ð1:2:6Þ

Finally, we would like to add one useful formula between delta and step func-tions. That is

dðx� aÞ ¼ ddx

Hðx� aÞ ð1:2:7Þ

This relation will be understood from the graphical discussion on delta and stepfunctions.

(3) Heisenberg’s delta function: d�ðxÞIn many applications, we have to express a semi-infinite distribution of a

physical quantity, such as the uniform load over the surface. In order to treat thesemi-infinite distribution, another delta function is defined. That is Heisenberg’sdelta function which is the Fourier transform of the semi-infinite distribution.

Let us consider Fourier transform of Heaviside’s unit step function,

HðxÞ ¼ 1; 0\ x\ þ10; �1\ x\ 0

�ð1:2:8Þ

Its formal Fourier transform is

�hðnÞ ¼Z1�1

HðxÞ expðþinxÞdx ¼Z10

expðþinxÞdx ð1:2:9Þ

This integral cannot be evaluated in the regular sense of calculus. Instead, we shalllook for the transformed image function �hðnÞ so that its Fourier inversion integralresults in the step function.

Remember the delta function whose Fourier transform is a constant, and assumethat the image function is the sum of delta function and a newly introducedunknown function �h1ðnÞ,

�hðnÞ ¼ pdðnÞ þ �h1ðnÞ ð1:2:10Þ

Let us apply the Fourier inversion integral to the above function and look for thesuitable form of the unknown function �h1ðnÞ. The inversion integral is processed as


HðxÞ ¼ 12p

Z1�1

�hðnÞ expð�inxÞdn ¼ 12p

Z1�1

pdðnÞ þ �h1ðnÞ� �

expð�inxÞdn

¼ 12þ 12p

Z1�1

�h1ðnÞ expð�inxÞdn )must be 1; x[ 0

0; x\ 0

�ð1:2:11Þ

From the last line in the above equation, we learn that the Fourier transform(inversion) integral of the unknown function �h1ðnÞ must be þ1=2 in x[ 0 and�1=2 in x\ 0, i.e.

12p

Z1�1

�h1ðnÞ expð�inxÞdn ¼þ1=2 ; x[ 0

�1=2; x\ 0

(ð1:2:12Þ

We remember the integration formula that gives �1=2 in each semi-infinitex-region (Erdélyi 1954, vol I, pp. 64, 3), that is

1p

Z10

sinðxnÞn

dn ¼ 12p

Z1�1

inexpð�ixnÞdn ¼

þ1=2 ; x[ 0

�1=2 ; x\ 0

(ð1:2:13Þ

Comparing Eq. (1.2.12) with Eq. (1.2.13), we learn that the whole integrand inEq. (1.2.12) must be the function

sinðxnÞn

ð1:2:14Þ

and thus we can guess that the suitable functional form for �h1ðnÞ is

�h1ðnÞ ¼ in

ð1:2:15Þ

We shall now examine whether this form is really suitable or not. Substitute theabove Eq. (1.2.15) into the last line in Eq. (1.2.11) and split the integral into realand imaginary parts, we get

12þ 12p

Z1�1

inexpð�inxÞdx ¼ 1

2þ 12p

Z1�1

1nsinðnxÞdxþ i

2p

Z1�1

1ncosðnxÞdx

ð1:2:16Þ

1.2 Distributions and Their Integration Formulas 9

The third term on the right hand side vanishes due to the anti-symmetric nature ofthe integrand. The second term is evaluated with the aid of formula (1.2.13) as

12þ 12p

Z1�1

sinðnxÞn

dn ¼ 12þ 1p

Z10

sinðnxÞn

dn ¼ 12þ 12

þ1 ; x[ 0�1 ; x\ 0

�

¼ 1 ; x[ 00 ; x\ 0

�ð1:2:17Þ

Then, we learn that the supposed function �h1ðnÞ in Eq. (1.2.15) is correct and thatthe Fourier transform of the step function is given by

�hðnÞ ¼ pdðnÞ þ in

ð1:2:18Þ

This function is called “Heisenberg’s delta function” and its transform pair is

dþðnÞ ¼ pdðnÞ þ in¼

Z10

expðþinxÞdx; HðxÞ ¼ 12p

Z1�1

dþðnÞ expð�inxÞdn

ð1:2:19Þ

Heisenberg’s delta function dþðnÞ is the Fourier transform of the step functionHðxÞ. If we have the another step function Hð�xÞ, its transform pair is also given by

d�ðnÞ ¼ pdðnÞ � in¼

Z0

�1expðþinxÞdx; Hð�xÞ ¼ 1

2p

Z1�1

d�ðnÞ expð�inxÞdn

ð1:2:20Þ

Then, Heisenberg’s delta function has two definitions as

d�ðnÞ ¼ pdðnÞ � in; Hð�xÞ ¼ 1

2p

Z1�1

d�ðnÞ expð�inxÞdn ð1:2:21Þ

As a byproduct, we can obtain a new formula. Two operations, addition andsubtraction of Eqs. (1.2.19) and (1.2.20) give us the formulas as

Z10

cosðnxÞdx ¼ pdðnÞ;Z10

sinðnxÞdx ¼ 1n

ð1:2:22Þ

The formula on the left is the same as the integral in Eq. (1.2.6), but with c ¼ 0.


1.3 Branch Cut and Argument of Square Root Functions

When we apply the multiple integral transform to the dynamic problem, some root(radical) functions of the complex variable appear. The square root function is one ofmulti-valued functions and we have to introduce branch cuts in order to make the rootfunction single-valued. Many textbooks for the complex analysis describe the branchcut/line for very simple multi-valued functions. However, in the practical applica-tions, we have to consider the branch cut for a little bit complicated function. In thissubsection, we show how to introduce the branch cut for the fundamental square rootfunction following the method of Ewing et al. (1957, Sect. 2.5, pp. 44–49). Thismethod is very simple and is easily understood even for college students. Afterintroducing the branch cut, we also show the argument of the root function along thebranch cut and that our introduction of the branch cut satisfies the regular conver-gence condition in the complex plane. This condition is crucial for the inversionintegral.

We shall discuss the branch cut and the argument of two square root (radical)functions: gðzÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffi

z� z0p

;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pwhere z0 is a complex constant. The branch

cut is introduced under the conditional: the real part of the root function is positivein the complex z-plane, i.e. Re gðzÞf g� 0. Discussions on these two root functionsare carried out, separately.

1.3.1 Square Root Function 1: gðzÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiz� z0

p

Let us consider the introduction of the branch cut/line for a simple square rootfunction which is defined by

gðzÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiz� z0

p ð1:3:1Þ

where zð¼ xþ iyÞ is the complex variable and z0ð¼ aþ ibÞ is an arbitrary complexconstant. We impose the conditional

Reffiffiffiffiffiffiffiffiffiffiffiffiz� z0

pð Þ� 0 ð1:3:2Þ

on the root function.It is easily found that the branch point is z ¼ z0 which is derived from gðzÞ ¼ 0.

In order to discuss the branch cut, we introduce the capital letters, X and Y, whichdenote the real and imaginary parts of the root function, respectively. The rootfunction is rewritten in the detailed form,

X þ iY ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ iy� ðaþ ibÞ

pð1:3:3Þ

1.3 Branch Cut and Argument of Square Root Functions 11

Squaring both sides, and equating the real and imaginary parts, we have thesimultaneous equations for two unknowns, X and Y, as

X2 � Y2 ¼ x� a

XY ¼ ðy� bÞ=2

(ð1:3:4Þ

Since the conditional is equivalent to

X ¼ Reffiffiffiffiffiffiffiffiffiffiffiffiz� z0

pð Þ� 0 ð1:3:5Þ

the border for this conditional is X ¼ 0 and then, from the second equation in(1.3.4), we have

y� b ¼ 0 ð1:3:6Þ

On the other hand, substituting X ¼ 0 into the first equation in (1.3.4), we have

�Y2 ¼ x� a ð1:3:7Þ

Since �Y2 \ 0; x must be smaller than a,

x\ a ð1:3:8Þ

The two conditions, Eqs. (1.3.6) and (1.3.8), give one semi-infinite lineðx\ a; y ¼ bÞ in the complex z-plane. This semi-infinite horizontal line with theedge ðx ¼ a; y ¼ bÞ is called the branch cut or line. The conditional Re ffiffiffiffiffiffiffiffiffiffiffiffi

z� z0p� � ¼

0 is satisfied on this branch cut which is shown by the thick line PQ in Fig. 1.3.Now, we shall examine whether the conditional Re

ffiffiffiffiffiffiffiffiffiffiffiffiz� z0

p� �� 0 is satisfied ornot in the whole complex z-plane. We introduce the polar coordinates ðr; hÞ for thecomplex variable as

z� z0 ¼ r expðþihÞ ð1:3:9Þ

and measure its argument h from the horizontal line PC as shown in Fig. 1.3. Thenthe root function is expressed in the polar form as


p ¼ ffiffir

pexpðþih=2Þ ð1:3:10Þ

If the argument h is defined within �p h þ p, the argument of the root functionis in the range, �p=2 h=2 þ p=2, and thus the real part of the root function ispositive in the whole complex z-plane. Based on this argument definition, we canspecify the argument of the root function on the line along the branch cut. Theargument of the root function on the upper horizontal line AB in Fig. 1.3 ismeasured from the horizontal line PC and is arg


p� � ¼ þp=2. On the lower


line A0B0, the argument is argffiffiffiffiffiffiffiffiffiffiffiffiz� z0

p� � ¼ �p=2. Thus, the argument of the rootfunction is in the range

�p=2 argffiffiffiffiffiffiffiffiffiffiffiffiz� z0

pð Þ þ p=2 ð1:3:11Þ

Therefore, the real part of the root function is positive in the whole complexz-plane.

In the above discussion, the argument of the variable z� z0 is restricted in therange �p h þ p. However, if we take the positive argument only, i.e. 0 h, wehave to introduce two Riemann sheets. The two sheets are connected along thebranch cut as shown in Fig. 1.4 and the argument of the variable z� z0 is measuredfrom the horizontal line as that in Fig. 1.3, but, when the argument exceeds þp(cross the cut), we measure the argument in the lower sheet. After rounding oneanti-clockwise rotation in the lower sheet, the argument is then measured in theupper sheet. The variation of the argument is shown by the solid and dotted thincurves in Fig. 1.4, schematically. Due to this definition, the argument of the variableis argðz� z0Þ ¼ þp on the upper line AB, and argðz� z0Þ ¼ þ3p on the lower lineA0B0. Thus, the argument of the square root function is positive in the upperRiemann sheet, but, unfortunately, it is not continuous across the horizontal line PCin Fig. 1.3, i.e. h ¼ 0 and h ¼ þ4p in Fig. 1.4. Due to this discontinuous nature ofthe argument, the former definition is preferable for the argument of the rootfunction as defined in Fig. 1.3.

x α=

y β=

PQ

Im( )y z=

Re( )x z=

0iz z re θ+− =

θθ π= +

θ π= −

0θ =A

A'

B

B'C

( )0arg / 2z z π− = +

( )0arg / 2z z π− = −

00 arg / 2z z π< − <

00 arg / 2z z π> − > −

Fig. 1.3 Definition of the arguments of the variable z� z0 and its square root functionffiffiffiffiffiffiffiffiffiffiffiffiz� z0

p


1.3.2 Square Root Function 2: gðzÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

q

This square root function appears frequently in many dynamic problems. The

conditional, Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� � 0 is also employed in order to guarantee the con-

vergence at infinity. We shall discuss the introduction of the branch cut under thisconditional.(1) Branch cut

The branch cut and the argument of a little bit complicated square root function,gðzÞ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p, is discussed. The conditional imposed on the function is similar

to that in the previous subsection, i.e. Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� � 0. It should be understood

that the variable is z ¼ xþ iy and the complex constant is z0 ¼ aþ ib where a andb are positive. It is easily found that two branch points of this root function are atz ¼ �ðaþ ibÞ. In order to discuss the branch cut, we introduce the capital X and

4θ π= +0 or 4θ π=2θ π= +

θ π= +

3θ π= +

upper Riemann sheet

lower Riemann sheet

branch cut/line

branch point

0z z= 0θ =

Fig. 1.4 Two Riemann sheets and argument variation of the variable z� z0 (upper sheet: solidcurve, lower sheet: dotted curve)


Y for the real and imaginary parts of the function and rewrite the root function in theexplicit form,

X þ iY ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðxþ iyÞ2 � ðaþ ibÞ2

qð1:3:12Þ

Squaring the both sides, and equating the real and imaginary parts, we have thesimultaneous equations for the unknowns, X and Y,

X2 � Y2 ¼ x2 � y2 � ða2 � b2ÞXY ¼ xy� ab

�ð1:3:13Þ

Since the conditional is X ¼ Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� � 0, its border is X ¼ 0. Substituting

X ¼ 0 into the second of Eq. (1.3.13), we have the simple equation for thehyperbola in the complex z-plane,

y ¼ abx

ð1:3:14Þ

This is the border curve for the conditional, but its range must be specified. So, wesubstitute X ¼ 0 into the first of Eq. (1.3.13),

�Y2 ¼ x2 � y2 � ða2 � b2Þ ð1:3:15Þ

As the left side of the above equation is negative, the right side also must benegative,

x2 � y2 � ða2 � b2Þ 0 ð1:3:16Þ

Substituting the Eq. (1.3.14) into the above equation, we have the quadraticinequality for x2,

x4 � ða2 � b2Þx2 � a2b2 ¼ ðx2 � a2Þðx2 þ b2Þ 0 ð1:3:17Þ

Solving for x2, we have

�b2 x2 þ a2 ð1:3:18Þ

Then, the admissible range for the variable x in the complex z-plane is

0 jxj a ð1:3:19Þ

The combination of two Eqs. (1.3.14) and (1.3.19) determines the branch cut which

satisfies the conditional Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� ¼ 0. This shows two semi-infinite hyper-

bolic curves and their edges are corresponding to the branch points as shown inFig. 1.5.


When the real or imaginary part of the constant z0 vanishes, the branch cutbecomes a simple line. In the case of the pure real constant, z0 ¼ a, the branch cutfor the root function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

pis the line along the real and imaginary axes as

shown in Fig. 1.6a, where the cut along the real axis is slightly shifted from the axisfor the visual explanation. On the other hand, in the case of the pure imaginary

constant z0 ¼ þib, the branch cut for the root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

pis the semi-infinite

straight line along the imaginary axis as shown in Fig. 1.6b.It is easily anticipated that the branch cut/curve depends on the sign of the

imaginary part of the complex constant z0. The former discussion is carried out forthe positive imaginary part of the constant. So, we shall reconsider the branch cut inthe case of the negative imaginary part of the constant, i.e. z0 ¼ a� ib. Employingthe same conditional and procedure as those in the former, the branch points for theroot function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pare z ¼ �ða� ibÞ and the branch cuts are given by

y ¼ � abx; 0 jxj a ð1:3:20Þ

This is the semi-hyperbolic curve shown in Fig. 1.7. Two cuts are introduced in thesecond and the fourth quadrants in the complex z-plane and the edge of the cut iscorresponding to each branch point, z ¼ �ða� ibÞ. When the imaginary part of theconstant vanishes, z0 ¼ a; b ¼ 0, the branch cut for the root function


p

α

β

β−

α−

0branch point: z z=

0branch point: z z= −

branch cut/line yx

αβ=

yx

αβ=

branch cut/line

Re( )x z=

Im( )y z=2 2

0

0

root function

z z

z iα β−

= +

Fig. 1.5 Branch cut/line for the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pwith z0 ¼ aþ ib


becomes the line on the real and imaginary axes as shown in Fig. 1.8a. On the otherhand, when the real part vanishes, z0 ¼ �ib; a ¼ 0, the branch cut for the root

functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

pbecomes the two semi-infinite lines on the imaginary axis as

shown in Fig. 1.8b.Comparing Fig. 1.8a with Fig. 1.6a, we learn that the branch cut on the real axis

is different in spite of the same form of the root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

p. This difference

x α= +x α= −

branch cut

branch cut

2 2root function: z α−

Im( )y z=

Re( )x z=

y β= +

y β= −

branch cut

branch cut

2 2root function: z β+

Im( )y z=

Re( )x z=

(a)

(b)

Fig. 1.6 a Branch cut for the root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

p. b Branch cut for the root function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

p


comes from the approaching direction of the zero imaginary part of the constant z0.When the imaginary part approaches to zero from above, the branch cut for the rootfunction


pis given by the lines in Fig. 1.6a. But, when the imaginary part

approaches to zero from below, the branch cut is given by the lines in Fig. 1.8a. Inthe practical applications, the real constant a expresses the frequency x of the time-harmonic vibration and the root function is given as

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � x2

p. Therefore, we are

puzzled which is the suitable branch cut. This puzzle is solved by the sign of theexponential time factor expð�ixtÞ and physical conditions. However the answerand verification are given in the later chapters, a brief conclusion is stated here. Thebranch cut in Fig. 1.6a is employed when the time factor is negative, i.e. expð�ixtÞ,and the cut in Fig. 1.8a is employed when the time factor is positive, expðþixtÞ.

When the real part of the complex constant vanishes, a ! 0, the branch cut layson the imaginary axis as shown in Figs. 1.6b and 1.8b. These two figures are same.

Thus, the branch cut for the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

pis unchanged regardless

of the sign of the imaginary part b.

(2) Argument

We have just determined the branch cut for the square root function under the

conditional Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� ¼ 0. In this subsection, we examine whether the real

part of the root function is positive or not in the whole complex plane. In order to

αβ

β−α−

0branch point: z z=

0branch point: z z= −

branch cut

yx

αβ= −

yx

αβ= −

branch cut/line

Re( )x z=

Im( )y z=2 2

0

0

root function

z z

z iα β−

= −

Fig. 1.7 Branch cut/line for the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pwith z0 ¼ a� ib


y β= +

y β= −

branch cut

branch cut

2 2root function: z β+

Im( )y z=

Re( )x z=

x α= +

x α= −

branch cut

branch cut

2 2root function: z α−

Im( )y z=

Re( )x z=

(a)

(b)

Fig. 1.8 a Branch cut for the root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

p. b Branch cut for the root function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

p


discuss the argument of the root function, we factorize the function as the productof two simple root functions asffiffiffiffiffiffiffiffiffiffiffiffiffiffi

z2 � z20

q¼ ffiffiffiffiffiffiffiffiffiffiffiffi

z� z0p ffiffiffiffiffiffiffiffiffiffiffiffi

zþ z0p ð1:3:21Þ

and introduce the polar coordinate ðrj; hjÞ; j ¼ 1; 2 for each variable as

z� z0 ¼ r1 expðþih1Þ; zþ z0 ¼ r2 expðþih2Þ ð1:3:22Þ

where rj is the radial arm/distance from each branch point, and the arguments hj ismeasured from the horizontal line parallel to the positive real axis. Then, the rootfunction is rewritten in the polar form,ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

z2 � z20

q¼ ffiffiffiffiffiffiffiffi

r1r2p

expfþiðh1 þ h2Þ=2g ð1:3:23Þ

and the discussion for the argument of the square root function is carried out on itspolar angle ðh1 þ h2Þ=2. We assume that the real and imaginary parts of the con-stant a and b are positive. Therefore, the branch points and cuts are in the first andthird quadrants in the complex z-plane as shown in Fig. 1.5. The argument of theroot function is discussed on three paths: (1) path along the branch cut, (2) path onthe non-branch cut hyperbola, and (3) the line which connects two branch points.We shall discuss their arguments, separately.

(2:1) Right side of the upper branch cut

Firstly, we consider the argument on the right side of the upper branch cut inFig. 1.9a, whose branch cuts are the same as those in Fig. 1.5. The mark “P1”denotes an arbitrary point on the upper branch cut and its coordinate is ðxP; yPÞ inthe complex plane. The arguments h1 of the radial arm r1 ¼ O1P1 is measured fromthe horizontal line O1A1. Since the branch point is at ða; bÞ, the complementaryangle /1 of the radial arm r1 is introduced as

tan/1 ¼yP � ba� xP

ð1:3:24Þ

and thus the argument h1 is expressed by

h1 ¼ p� /1 ð1:3:25Þ

On the other hand, the argument h2 of the radial arm r2 ¼ O2P1 is measured fromthe horizontal line O2A2. Since the lower branch point is at ð�a;�bÞ, the argumentof the radial arm, i.e. h2, is given by

tan h2 ¼ bþ yPaþ xP

ð1:3:26Þ


1P ( , )P Px y

( , )α β

( , )α β− −

1O

2O

1θ

2θ

1A1θ

2A

1B

2B

Re( )x z=

Im( )y z=

1r

2r

1φ

( , )α β

( , )α β− −

1O

2O

1θ

2θ( , )Q Qx y

1Q

1A

2A

1B

2B

Re( )x z=

Im( )y z=

1r

2r

( , )α β

( , )α β− −

1O

2O

1 ( )θ π ψ= − −

2θ ψ=

1A

1r

1B

2B

R

Im( )y z=

Re( )x z=

1r

2r

ψ

(a)

(b)

(c)

Fig. 1.9 Arguments, h1 and h2, a along the upper branch cut, b on the upper non-branch cuthyperbola, c on the line connecting two branch points, d along the lower branch cut and e onthe lower non-branch cut hyperbola


Two Eqs. (1.3.24) and (1.3.26) are looks different, but, the point P1 is on thehyperbolic curve at the limit. The hyperbolic curve gives the relation between xPand yP,

yP ¼ abxP

ð1:3:27Þ

Substituting the above relation into the two Eqs. (1.3.24) and (1.3.26),

tan h2 ¼ tan/1 ¼bxP

ð1:3:28Þ

2P( , )P Px y− −

( , )α β

( , )α β− −

1O

2O

2θ

1θ

2θ

1A

2A

1B

2B

Re( )x z=

Im( )y z=

1r

2r

1ψ

2φ

( , )α β

( , )α β− −

1O

2O2θ

( , )Q Qx y− −

2Q

1A

1θ

2A

1B

2B

Re( )x z=

Im( )y z=

1r

2r2ψ

1ψ

(d)

(e)

Fig. 1.9 (continued)


we find that h2 ¼ /1. The argument of the root function at the right side of theupper branch cut is thus given by

ðh1 þ h2Þ=2 ¼ ðp� /1 þ /1Þ=2 ¼ þp=2 ð1:3:29Þ

This shows that the value of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pat the right side of

the upper branch cut is positive imaginary.

(2:2) Left side of the upper branch cut

We now consider the argument of the root function at the left side of the upperbranch cut at P1. The argument h1 must be measured in the clockwise directionfrom the horizontal line O1A1, since the arm r1 cannot cross its own branch cut.However, the argument h2 can cross the cut, since the cut is not for the lower branchpoint. Thus, from the geometry in Fig. 1.9a, we have

h1 ¼ �ðpþ /1Þ; h2 ¼ /1 ð1:3:30Þ

and the argument of the root function is

ðh1 þ h2Þ=2 ¼ �p=2 ð1:3:31Þ

Then, the value of the root function at the left side of the upper branch cut isnegative imaginary.

(2:3) On the upper non-branch cut hyperbola

One arbitrary point Q1 is taken on the non-branch cut hyperbola as shown inFig. 1.9b and we assume that its coordinate is ðxQ; yQÞ in the complex plane. Thearm r1 from the upper branch point is denoted by the line O1Q1 and the arm r2 fromthe lower branch point is O2Q1. They are shown in Fig. 1.9b. The argument of thearm r1 is measured from the horizontal line O1A1 in the clockwise direction andthat of the arm r2 is measured from the horizontal line O2A2 in the anti-clockwisedirection. A simple analytical geometry shows that the arguments h1 and h2 are

tan h1 ¼ � b� yQxQ � a

; tan h2 ¼ bþ yQaþ xQ

ð1:3:32Þ

Substituting the equation for the hyperbola yQ ¼ ab=xQ into the above equations,we have

tan h1 ¼ � bxQ

; tan h2 ¼ bxQ

ð1:3:33Þ


Thus, h2 ¼ �h1, and the argument of the root function on the non-branch cuthyperbola is

ðh1 þ h2Þ=2 ¼ 0 ð1:3:34Þ

This shows the value of the root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

pis positive real.

(2:4) On the line connecting two branch points

We take an arbitrary point R on the line O1O2, which connects two branchpoints, as shown in Fig. 1.9c. Since the argument of each arm from the branch pointis unchanged on this line, it is no need to specify the location of the point R.Introducing the complementary angle w for the acute angle \B1O1O2, the argumenth2 for the lower arm r2 is expressed by

h2 ¼ w ð1:3:35Þ

and that for the upper arm r1 is

h1 ¼ �ðp� wÞ ð1:3:36Þ

Then, the argument of the root function is given by

ðh1 þ h2Þ=2 ¼ �ðp=2� wÞ ð1:3:37Þ

This shows that the argument of the root function is negative on the connecting line,since 0\w\ p=2. Thus, the real part of the square root function is positive.

(2:5) Right side of the lower branch cut

We take an arbitrary point P2 on the lower branch cut as shown in Fig. 1.9d, andconsider the argument of the root function on the right side of the cut. Assuming thelocation of the point P2 as ð�xP;�yPÞ, the complementary angle w1ð¼\B1O1P2Þ isgiven by

tanw1 ¼bþ yPaþ xP

ð1:3:38Þ

Then, the argument of the radial arm r1 is measured in the clockwise direction,

h1 ¼ �ðp� w1Þ ð1:3:39Þ

The argument h2 of the arm r2ð¼ O2P2Þ is also measured in the clockwise direction.It is given by

tan h2 ¼ � yP � ba� xP

ð1:3:40Þ


Substituting the equation for the hyperbola, yP ¼ ab=xP, into Eqs. (1.3.38) and(1.3.40), we have

tanw1 ¼bxP

; tan h2 ¼ � bxP

ð1:3:41Þ

Then,

h2 ¼ �w1 ð1:3:42Þ

Two Eqs. (1.3.39) and (1.3.42) give the argument of the root function as

ðh1 þ h2Þ=2 ¼ �p=2 ð1:3:43Þ

This shows that the value of the root function on the right side of the lower branchcut is negative imaginary.

(2:6) Left side of the lower branch cut

When we consider the argument on the left side of the lower branch cut, anattention must be paid to the argument h2 since the arm r2 from the lower branchpoint cannot cross its own branch cut. As shown in Fig. 1.9d, the argument h2 ismeasured from the horizontal line O2A2 in the counterclockwise direction. Intro-ducing the complementary angle /2 as shown in Fig. 1.9d, the argument h2 is givenby

h2 ¼ 2p� /2 ð1:3:44Þ

On the other hand, the arm r1 can cross the lower branch cut and its argument is

h1 ¼ �ðp� w1Þ ð1:3:45Þ

The two complementary angles are given by

tanw1 ¼bþ yPaþ xP

; tan/2 ¼yP � ba� xP

ð1:3:46Þ

Substituting the equation for the hyperbola, yP ¼ ab=xP, into the above, we findthat w1 ¼ /2. Thus, the argument of the root function is

ðh1 þ h2Þ=2 ¼ þp=2 ð1:3:47Þ

This shows that the value of the root function on the left side of the lower branchcut is positive imaginary.


(2:7) On the lower non-branch cut hyperbola

We consider the argument on the lower non-branch cut hyperbola and take anarbitrary point Q2 on it, as shown in Fig. 1.9e. Assuming the point Q2 is atð�xQ;�yQÞ, we introduce two complementary angles w1ð¼\B1O1Q2Þ andw2ð¼\B2O2Q2Þ. They are

tanw1 ¼bþ yQaþ xQ

; tanw2 ¼b� yQxQ � a

ð1:3:48Þ

Since the point Q2 is on the hyperbola, we substitute yQ ¼ ab=xQ and then find thatw1 ¼ w2. The two arguments are

h1 ¼ �ðp� w1Þ; h2 ¼ p� w2 ð1:3:49Þ

Thus, the argument of the root function is

ðh1 þ h2Þ=2 ¼ 0 ð1:3:50Þ

This shows the value of the root function on the non-branch cut hyperbola ispositive real.

arg( ) / 2g π= +

arg( ) / 2g π= +

1O

2O

arg( ) 0g =

arg( ) 0g =

2 2

square root function

( ) ( )g z z iα β= − +

/ 2 arg( ) 0gπ− < <

0 arg( ) / 2g π< < +

Im( )y z=

Re( )x z=

0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

Fig. 1.10 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � ðaþ ibÞ2

q(The imaginary part of the

constant is positive)


Summarizing the above discussions in (2.1)–(2.7), the argument of the squareroot function in the whole complex plane is shown schematically in Fig. 1.10.However the argument in the non-specific point/region was not discussed, we caneasily guess the argument from our result. It is shown in the square box in thefigure. When the imaginary part of the complex constant vanishes, b ! 0, Fig. 1.10is reduced to Fig. 1.11. On the other hand, when the real part of the complexconstant vanishes, a ! 0, Fig. 1.10 is reduced to Fig. 1.12. In any case, theargument of the square root function in the whole complex z-plane is in the range

�p=2ðh1 þ h2Þ=2 þ p=2 ð1:3:51Þ

This shows that the conditional Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � z20

p� � 0 is satisfied in the whole com-

plex plane and thus we have introduced the branch cuts and defined the argument ofthe square root function, correctly.

If we employ the complex constant with the negative imaginary part,z0 ¼ a� ib, the branch cuts are in the second and the fourth quadrants in thecomplex z-plane as shown in Fig. 1.7. The discussion in the above subsections canbe applied to the argument of the root function. Although the detailed discussionsfor the argument are not repeated here, the final results are schematically shown inFigs. 1.13, 1.14 and 1.15.

arg( ) / 2g π= +

arg( ) / 2g π= +

α+

α−

arg( ) 0g =arg( ) 0g =

2 2


( )g z z α= −

/ 2 arg( ) 0gπ− < <0 arg( ) / 2g π< < +

Im( )y z=

Re( )x z=

0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

/ 2 arg( ) 0gπ− < <

Fig. 1.11 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

p(The positive imaginary part of the

complex constant vanishes)


One comment should be repeated here. Comparing Fig. 1.11 with Fig. 1.14 in thecase of vanishing imaginary part of the complex constant, the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

pon the real axis x ¼ ReðzÞ is the positive or negative purely imaginary as

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 � a2

p¼ þi

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p�i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p�

; 0 jxj\ a ð1:3:52Þ

The sign depends on the vanishing direction of the imaginary part of the complexconstant z0. Thus, as was mentioned before, we have to pay special attention to thesquare root function with the real constant


p.

1.4 Comments on Inversion Techniques and IntegrationFormulas

The integral transform technique is a powerful tool for solving linear differentialequations. However, its success is up to the evaluation of the inversion integral. Sofar, many integration formulas have been found and published. The most com-prehensive formula books are Erdélyi (1954), Gradshteyn and Ryzhik (1980),

arg( ) / 2g π= +

arg( ) / 2g π= +

iβ+

iβ−

arg( ) 0g =arg( ) 0g =

2 2


( )g z z β= +

/ 2 arg( ) 0gπ− < <0 arg( ) / 2g π< < +

Im( )y z=

Re( )x z=

0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

/ 2 arg( ) 0gπ− < <

arg( ) 0g =

Fig. 1.12 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

pin the complex plane (The real part of

the complex constant vanishes)


Magnus et al. (1966), Watson (1966), Moriguchi et al. (1972) and Titchmarsh(1948).

The books from Erdélyi (1954) to Watson (1966) are well-known and it is notdifficult to obtain access to them. The book Watson (1966) deals solely with Besselfunctions and not with integral transforms, but it gives many integration formulasfor the Hankel transform. The book Moriguchi et al. (1972), written in Japanese, isvery compact and is separated into three small handy books. In spite of its com-pactness, the principal formulas which are included in Erdélyi (1954), Gradshteynand Ryzhik (1980), Magnus et al. (1966) and Watson (1966) are cited. The authorbelieves that the three handy books are most convenient as a “first aid.” The lastbook Titchmarsh (1948) describes the mathematics of the theory of Fourier trans-form. When someone needs more detailed mathematics for the integral transform,this book will give proper answers.

If a desired formula cannot be found in these books, the complex integral isemployed to evaluate the inversion integral. The complex integral based on Cau-chy’s integral theorem is the most useful evaluation technique and the discussion on

arg( ) / 2g π= +

arg( ) / 2g π= +

1O

2O

arg( ) 0g =

arg( ) 0g =

2 2


( ) ( )g z z iα β= − −

/ 2 arg( ) 0gπ− < <

/ 2 arg( ) 0gπ− < <

Im( )y z=

Re( )x z=0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

Fig. 1.13 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � ða� ibÞ2

q(The imaginary part of the

complex constant is negative)

1.4 Comments on Inversion Techniques and Integration Formulas 29

the branch cut in the previous subsections will be helpful. If the complex integraldoes not give any compact result, the inversion integral is left in its definition formor is converted to the numerically tractable form by the complex integral.

Exercises

(1:1) Apply the finite Fourier transform, complex Fourier series defined byEq. (1.1.6), to Dirac’s delta function which is defined in a finite regionð�p;þpÞ as

dðx� aÞ ; �p\ a\ þ p ðaÞ

and then show the series form of the delta function,

dðx� aÞ ¼ 12p

1þ 2X1n¼1

cosfnðx� aÞg" #

ðbÞ

arg( ) / 2g π= +

arg( ) / 2g π= +

α+

α−arg( ) 0g =

arg( ) 0g =

2 2


( )g z z α= −

/ 2 arg( ) 0gπ− < <0 arg( ) / 2g π< < +

Im( )y z=

Re( )x z=

0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

/ 2 arg( ) 0gπ− < <

Fig. 1.14 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 � a2

pin the complex plane (The negative

imaginary part of the complex constant vanishes)


ia+

ia−

1θ

2θ

Re( )z

Im( )z1r

2r

Fig. 1.16 Two branch cutsfor the root function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ a2

p

arg( ) / 2g π= +

arg( ) / 2g π= +

β+

β−

arg( ) 0g =arg( ) 0g =

2 2


( )g z z β= +

/ 2 arg( ) 0gπ− < <0 arg( ) / 2g π< < +

Im( )y z=

Re( )x z=

0 arg( ) / 2g π< < +

arg( ) / 2g π= −

arg( ) / 2g π= −

/ 2 arg( ) 0gπ− < <

arg( ) 0g =

Fig. 1.15 Argument of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ b2

pin the complex plane (The real part of

the complex constant vanishes)

1.4 Comments on Inversion Techniques and Integration Formulas 31

(1:2) If we expand the x-range to ð�1;þ1Þ, the Fourier series in the aboveequation (b) gives an infinite sequence of delta function, i.e.

12p

1þ 2X1n¼1

cosfnðx� aÞg" #

¼Xþ1

m¼�1dðx� a� 2mpÞ ðcÞ

Explain why the infinite sequence?(1:3) For the root function

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiz2 þ a2

p, where a is a positive real constant, if we

introduce two branch cuts as shown in Fig. 1.16, how is the argument of theroot function along the branch cut?

References

Erdélyi A (ed) (1954) Tables of integral transforms, vol I and II. McGraw-Hill, New-YorkEwing WM, Jardetzky WS, Press F (1957) Elastic waves in layered media. McGraw-Hill,

New YorkGradshteyn IS, Ryzhik IM (Jefferey A, ed) (1980) Table of integrals, series, and products, 5th edn.

Academic Press, San DiegoMagnus W, Oberhettinger F, Soni RP (1966) Formulas and theorems for the special functions of

mathematical physics. Springer, New YorkMoriguchi S, Udagawa K, Ichimatsu S (1972) Mathematical formulas, vols I–III (in Japanese).

Iwanami, TokyoSneddon IN (1951) Fourier transforms. McGraw-Hill, New YorkTitchmarsh EC (1948) Introduction to the theory of Fourier integrals, 2nd edn. Oxford, LondonWatson GN (1966) A treatise on the theory of Bessel functions. Cambridge University Press,

Cambridge


Chapter 2Green’s Functions for Laplaceand Wave Equations

This chapter shows the solution method for Green’s functions of 1, 2 and 3DLaplace and wave equations. Lengthy and detailed explanations are given in orderto instruct the basic technique of the integral transform. Especially, Fourierinversion integral for the time-harmonic Green’s function is discussed in detail, andthree evaluation techniques are introduced in Sect. 2.5.

2.1 1D Impulsive Source

We start from the simplest wave equation that has two variables: a single spacevariable x and the time t,

@2/@x2

¼ 1c2

@2/@t2

� PdðxÞdðtÞ ð2:1:1Þ

The nonhomogeneous term represents a wave source with magnitude P. The twoDirac’s delta functions, dðxÞ and dðtÞ, show the location and the impulsive nature ofthe source. The Green’s function is a particular solution of the differential equationcorresponding to the impulsive source. The Green’s function is sought under thequiescent condition at an initial time,

/jt¼0¼@/@t

��t¼0

¼ 0 ð2:1:2Þ

and the convergence condition at infinity,

/jx!�1¼ @/@x

��x!�1

¼ 0 ð2:1:3Þ

To obtain the particular solution of the wave equation (2.1.1), we apply theintegral transforms and reduce the differential equation to an algebraic equation inthe transformed domain. Since the unknown function /ðx; tÞ has two variables, we


33

apply the double transform: Laplace transform with respect to the time variable t:0� t\þ1,

f �ðsÞ ¼ L f ðtÞ½ � ¼Zþ1

0


and Fourier transform with respect to the space variable x: �1\ x\ þ 1,

�f ðnÞ ¼Zþ1

�1f ðxÞ expðþinxÞdx; f ðxÞ ¼ 1

2p

Zþ1

�1


Firstly, we multiply the kernel of the Laplace transform expð�stÞ to both sides ofthe differential equation (2.1.1),

Z10

@2/@x2

¼ 1c2

@2/@t2

� PdðxÞdðtÞ� �

expð�stÞdt ð2:1:6Þ

and perform the Laplace transform integral term by term. The order of integrationand differentiation are interchanged for the first term in the left hand side of theequation. The first term in the right hand side is left in its order and then integratedby parts. The last nonhomogeneous term, which has the delta function, is evaluatedby using the formula (1.2.3). Then, Eq. (2.1.6) is rewritten as

d2

dx2

Z10

/ expð�stÞdt ¼ 1c2

Z10

@2/@t2

expð�stÞdt � PdðxÞZ10

dðtÞ expð�stÞdt ð2:1:7Þ

We define the Laplace transform of the unknown function as

/� ¼Z10

/ expð�stÞdt ð2:1:8Þ

The time-derivative term is integrated by parts,

Z10

@2/@t2

expð�stÞdt ¼ @/@t

expð�stÞ� �t!1

t¼0þ s/ expð�stÞ½ �t!1

t¼0 þ s2Z10

/ expð�stÞdt ¼ s2/�

ð2:1:9Þ

34 2 Green’s Functions for Laplace and Wave Equations

http://dx.doi.org/10.1007/978-3-319-17455-6_1

where the quiescent condition at the initial time is incorporated. To the last non-homogeneous term, the simple integration formula for the delta function,

Z10

dðtÞ expð�stÞdt ¼ 1 ð2:1:10Þ

is applied. Then, we have the ordinary differential equation for the function /� inthe transformed domain,

d2/�

dx2¼ ðs=cÞ2/� � PdðxÞ ð2:1:11Þ

It is possible to obtain the exact solution for the above ordinary differential equationby the elementary method. However, in order to demonstrate the integral transformtechnique, we further apply Fourier transform to the ordinary differential equa-tion (2.1.11). The Fourier transform defined by Eq. (2.1.5) is applied toEq. (2.1.11),

Zþ1

�1

d2/�

dx2¼ ðs=cÞ2/� � PdðxÞ

� �expðþinxÞdx ð2:1:12Þ

The Fourier transform integral is applied to each term as

Zþ1

�1

d2/�

dx2expðþinxÞdx ¼ ðs=cÞ2

Zþ1

�1/� expðþinxÞdx� P

Zþ1

�1dðxÞ expðþinxÞdx

ð2:1:13Þ

The convergence condition is also transformed, as

Z10

/jx!�1 ¼ @/@x

��x!�1

¼ 0� �

expð�stÞdt ) /�jx!�1 ¼ d/�

dx

��x!�1

¼ 0

ð2:1:14Þ

Defining the Fourier transform of the Laplace transformed unknown function,

�/� ¼Zþ1

�1/� expðþinxÞdx ð2:1:15Þ

2.1 1D Impulsive Source 35

the transform of the space-derivative in Eq. (2.1.13) is carried out with the aid of theconvergence condition,

Zþ1

�1

d2/�

dx2expðþinxÞdx¼ d/�

dxexpðþinxÞ

� �x!þ1

x!�1�in /� expðþinxÞ½ �x!þ1

x!�1�n2Zþ1

�1/� expðþinxÞdx

¼ �n2~/�

ð2:1:16Þ

The integration formula for the delta function

Zþ1

�1dðxÞ expðþinxÞdx ¼ 1 ð2:1:17Þ

is also used for evaluating the last term. The Fourier transform of the ordinarydifferential equation (2.1.12) becomes a simple algebraic equation for the doubletransformed unknown function �/�,

�n2�/� ¼ s2

c2�/� � P ð2:1:18Þ

Then we have the exact expression for the double transformed function �/�,

�/� ¼ P

n2 þ ðs=cÞ2 ð2:1:19Þ

The unknown function has just been determined explicitly in the transformeddomain. We shall carry out two inverse transforms successively. As the firstinversion, we apply the Fourier inversion integral which is defined by the second ofEq. (2.1.5). The formal Fourier inversion is given by the integral

/� ¼ 12p

Zþ1

�1

P

n2 þ ðs=cÞ2 expð�inxÞdn ð2:1:20Þ

Due to the symmetric nature of the integrand, the integral is reduced to the simplersemi-infinite integral,

/� ¼ Pp

Z10

1

n2 þ ðs=cÞ2 cosðnxÞdn ð2:1:21Þ


This is a simple integral and the integration formula (Erdélyi 1954, vol. I, pp. 8, 11),

Zþ1

0

1x2 þ a2

cosðxyÞdx ¼ p2a

expð�ajyjÞ ð2:1:22Þ

can be applied. Then, Eq. (2.1.21) gives

/� ¼ P2ðs=cÞ expð�sjxj=cÞ ð2:1:23Þ

The next step is to carry out the Laplace inversion. The symbolical form of theLaplace inversion is given by

/ ¼ L�1 /�½ � ¼ cP2L�1 1

sexpð�sjxj=cÞ

� �ð2:1:24Þ

Fortunately, we have the Laplace inversion formula (Erdélyi 1954, vol. I, pp. 241, 1),

L�1 1sexpð�asÞ

� �¼ 0; t\a

1; t[ a

�¼ Hðt � aÞ ð2:1:25Þ

where Hð:Þ is Heaviside’s unit step function. Applying this formula to Eq. (2.1.24),the solution for the non-homogeneous 1D wave equation is obtained as

/ ¼ cP2Hðt � jxj=cÞ ¼ cP

2Hðct � jxjÞ ð2:1:26Þ

This solution shows an expanding (or out-going) 1D wave with the uniformamplitude cP/2 and with the velocity c as shown in Fig. 2.1. Consequently, we getthe Green’s function for the 1D wave equation,

@2/@x2

¼ 1c2

@2/@t2

� PdðxÞdðtÞ ) /ðx; tÞ ¼ cP2Hðct � jxjÞ ð2:1:27Þ

/ 2cP

ct

x

Fig. 2.1 1D expanding wavefrom a source point


2.2 1D Time-Harmonic Source

When the source is time-harmonic, the nonhomogeneous term in Eq. (2.1.1) isreplaced with a harmonic function, but the source location is unchanged. Thus, thewave equation with a time-harmonic source is given by

@2/@x2

¼ 1c2

@2/@t2

� QdðxÞ expðþixtÞ ð2:2:1Þ

where Q is the source magnitude and x the frequency of the time-harmonicvibration. We assume that its solution satisfies the convergence condition at infinity,i.e.

/jx!�1 ¼ @/@x

��x!�1

¼ 0 ð2:2:2Þ

As the first step of the solution method, we assume that the solution is also time-harmonic,

/ðx; tÞ ¼ /#ðxÞ expðþixtÞ ð2:2:3Þ

where /# is called the “amplitude function.” Due to this assumption, the conver-gence condition (2.2.2) is rewritten for the amplitude function,

/#��x!�1¼ d/#

dx

��x!�1

¼ 0 ð2:2:4Þ

Substituting the time-harmonic assumption of Eq. (2.2.3) into the wave equa-tion (2.2.1), we have the ordinary differential equation for the amplitude function,

d2/#

dx2þ ðx=cÞ2/# ¼ �QdðxÞ ð2:2:5Þ

The exact solution of this ordinary differential equation can also be obtained by theelementary method. However, in order to demonstrate the integral transformtechnique, we apply the Fourier transform, which is defined by Eq. (2.1.5), to theordinary differential equation (2.2.5),

Zþ1

�1

d2/#

dx2þ ðx=cÞ2/# ¼ �QdðxÞ

� �expðþinxÞdx ð2:2:6Þ


Defining the Fourier transform of the amplitude function as

�/# ¼Zþ1

�1/# expðþinxÞdx ð2:2:7Þ

the space derivative term in Eq. (2.2.6) is integrated by parts as

Zþ1

�1

d2/#

dx2expðþinxÞdx ¼ d/#

dxexpðþinxÞ

� �x!þ1

x!�1�in /# expðþinxÞ� x!þ1

x!�1

� n2Zþ1

�1/# expðþinxÞdx

ð2:2:8Þ

We apply the convergence condition of Eq. (2.2.4) and the definition of the Fouriertransform (2.2.7) to the above equation. The Fourier transform of the doublederivative is then reduced to

Zþ1

�1

d2/#

dx2expðþinxÞdx ¼ �n2�/# ð2:2:9Þ

The nonhomogeneous term is evaluated as

�QZþ1

�1dðxÞ expðþinxÞdx ¼ �Q ð2:2:10Þ

Then, Eq. (2.2.6) gives a simple algebraic equation for the Fourier transformedamplitude function,

�n2�/# þ ðx=cÞ2�/# ¼ �Q ð2:2:11Þ

The Fourier transformed amplitude is determined completely,

�/# ¼ Q

n2 � ðx=cÞ2 ð2:2:12Þ

Our next task is to invert the transformed amplitude. Applying the formal Fourierinversion integral to Eq. (2.2.12), we get

/# ¼ 12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ð2:2:13Þ

2.2 1D Time-Harmonic Source 39

Inspecting the integrand, we see that it has two simple poles at n ¼ �ðx=cÞ, that isto say, the poles are located on the integration path (real axis in the complexn-plane). Since the integration cannot be performed through these singular points,we have to distort the integration path around the poles. There are two ways ofdeforming the path. One is through an upper semi-circle, another is through a lowersemi-circle as shown in Fig. 2.2. We have to determine which semi-circle is suit-able. Discussing the nature of the initial wave equation (2.2.1), we learn that thewave will expand to the outer region from the source point, i.e. wave radiation fromthe source. Therefore, we have to choose the path so that the inversion integralresults in a radiating (out-going) wave from the source.

Still, it is somewhat complicated to explain the path selection. To aid theunderstanding, two integrals with complex frequency are considered. After dis-cussing the wave nature derived from each integral, we will determine andunderstand the path distortion.

Let us introduce and add a small imaginary number e to the frequency inEq. (2.2.13) so that the poles are shifted from the real axis and are not on theintegration path. The complex frequency is considered in two ways, positive andnegative imaginary parts, x ! x� ie. Employing the complex frequency -, weconsider the complex integral,

U ¼ 12p

ZC

P

n2 � ð-=cÞ2 expð�inxÞdn ð2:2:14Þ

where the integrand is the same as that in the Fourier inversion integral (2.2.13), butthe frequency is complex, - ¼ x� ie. The integration loop C for the two cases ofcomplex frequency, with positive and negative imaginary parts, is discussedseparately.

(1) Small positive imaginary part: - ¼ xþ ie

When the frequency has a small positive imaginary part, the poles are shiftedfrom the real axis. The integration path C is chosen so that the integrand vanishes

/ cω−

/ cω+Re( )ξ

Im( )ξ

? ?

Fig. 2.2 Possible deformations of the integration path around the pole


on the large semi-circle with infinite radius. We employ the lower closed loop Cð�Þ

in the case of positive x and the upper loop CðþÞ in that of negative x as shown inFig. 2.3a.

( ) /i cω ε− +

( ) /i cω ε+

Re( )ξ

Im( )ξ

0x <

0x >

( )C +

( )C −

( ) /i cω ε− −

( ) /i cω ε−

Re( )ξ

Im( )ξ

0x <

0x >

( )C +

( )C −

(a)

(b)

Fig. 2.3 Integration loop forthe complex integral U in thecase of a the positiveimaginary part of thefrequency, b the negativeimaginary part of thefrequency


Applying Cauchy’s integral theorem (Jordan’s lemma) to the complex integral Uwith the loop Cð�Þ in Fig. 2.3a, the integral along the real axis is evaluated as theresidue at the lower pole n ¼ �-=c ¼ �ðxþ ieÞ=c,

� 12p

Zþ1

�1

P

n2 � ð-=cÞ2 expð�inxÞdn ¼ 2pi2p

Pnþ -=c

n2 � ð-=cÞ2 expð�inxÞ" #

n¼�-=c

ð2:2:15Þ

Rewriting the above equation, and taking the limit e ! 0, we have in the case ofpositive x,

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ iP2ðx=cÞ expðþixx=cÞ; x[ 0 ð2:2:16Þ

On the other hand, when x\ 0, we employ the upper loop CðþÞ for the complexintegral U and have

12p

Zþ1

�1

P


Pn� -=c


n¼-=c

ð2:2:17Þ

Then, we take the limit e ! 0,

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ iP2ðx=cÞ expð�ixx=cÞ; x\0 ð2:2:18Þ

Unifying the two Eqs. (2.2.16) and (2.2.18), we have for the Fourier inversionintegral, where the positive imaginary part of the complex frequency tends to zero,i.e.

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ iP2ðx=cÞ expðþixjxj=cÞ; xþ ieje!0 ð2:2:19Þ

(2) Small negative imaginary part: - ¼ x� ie

When the imaginary part of the complex frequency is negative, the poles are alsoshifted from the real axis as shown in Fig. 2.3b. In order to guarantee the con-vergence on the large semi-circle, the lower loop Cð�Þ is employed in the case ofpositive x, and the upper CðþÞ in that of negative x.


When x[ 0, we employ the loop Cð�Þ and apply Jordan’s lemma to the complexintegral U in Eq. (2.2.14). The integral along the real axis is converted to the residueat the lower pole, n ¼ -ð¼ x� ieÞ=c,

� 12p

Zþ1

�1

P


Pn� -=c


n¼-=c

ð2:2:20Þ

Rewriting the above and taking the limit e ! 0, we have in the case of positive x,

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ � iP2ðx=cÞ expð�ixx=cÞ; x[ 0 ð2:2:21Þ

Similarly, when we employ the upper loop CðþÞ in the case of x\0,

12p

Zþ1

�1

P


Pnþ -=c


n¼�-=c

ð2:2:22Þ

Taking the limit e ! 0, the Fourier inversion integral is evaluated as

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ � iP2ðx=cÞ expðþixx=cÞ; x\0 ð2:2:23Þ

Unifying two Eqs. (2.2.21) and (2.2.23), we have the unified expression when thenegative imaginary part of the frequency approached to zero,

12p

Zþ1

�1

P

n2 � ðx=cÞ2 expð�inxÞdn ¼ � iP2ðx=cÞ expð�ixjxj=cÞ; x� ieje!0 ð2:2:24Þ

(3) Selection of the imaginary part

Two expressions are obtained for the single Fourier inversion integral. They areEqs. (2.2.19) and (2.2.24) and are summarized as

/# ¼� iP

2ðx=cÞ expð�ixjxj=cÞ; x� ieje!0

þ iP2ðx=cÞ expðþixjxj=cÞ; xþ ieje!0

8<: ð2:2:25Þ


The wave nature of the two expressions is discussed by multiplying the time factor,

/ ¼� iP

2ðx=cÞ expfþixðt � jxj=cÞg; x� ieje!0

þ iP2ðx=cÞ expfþixðt þ jxj=cÞg; xþ ieje!0

8<: ð2:2:26Þ

Inspecting the argument of the exponential function in the above equation, theupper solution shows an out-going (radiation) wave from the source, since theargument of the exponential function is t � jxj=c. On the other hand, the lowerincludes t þ jxj=c and shows an incoming wave, coming from infinity. Since thesuitable solution must be the out-going wave, we employ the upper formula inEqs. (2.2.25) and (2.2.26). Thus, the complex frequency with the negative imagi-nary part is the right assumption. Then, the suitable integration contour for theFourier inversion integral is that of Fig. 2.3b and the final result for the time-harmonic response is given by

@2/@x2

¼ 1c2

@2/@t2

� QdðxÞ expðþixtÞ ) / ¼ � iQ2ðx=cÞ expfþixðt � jxj=cÞg

ð2:2:27Þ

We have just learned that the right selection for the complex frequency is- ¼ x� ie, i.e. negative imaginary part, and the suitable integration loop is Cð�Þ inFig. 2.3b. If we do not introduce the small imaginary part and keep the integrationpath on the real axis, the poles is on the real axis and the integration path around thepole must be deformed by a small semi-circle shown in Fig. 2.4a. This deformationis valid only in the case of the positive time factor, expðþixtÞ. If we assume thenegative time factor expð�ixtÞ, the integration path on the real axis is deformed asthat shown in Fig. 2.4b. Thus, the selection of the deformed path around the poledepends on the sign of the imaginary part of the complex frequency. Therefore, wecould answer to the initial question about the deformation of the integration path forthe Fourier inversion integral.

2.3 2D Static Source

Let us consider Green’s function for a typical partial differential equation, the so-called Laplace equation. The Laplace equation with a source S is the nonhomo-geneous differential equation,

@2/@x2

þ @2/@y2

¼ �SdðxÞdðyÞ ð2:3:1Þ


The product of two delta functions in the nonhomogeneous term shows the locationof the source, i.e. the source S is placed at the coordinate origin (0, 0) in (x, y)-plane.The convergence condition at infinity,

/j ffiffiffiffiffiffiffiffiffix2þy2

p!1¼ @/

@x

�� ffiffiffiffiffiffiffiffiffix2þy2

p!1

¼ @/@y


p!1

¼ 0 ð2:3:2Þ

is also imposed.Now, we apply the integral transforms to Eq. (2.3.1). Since the unknown

function / has two space variables, we apply the double Fourier transform definedby

�/ðnÞ ¼Zþ1

�1/ðxÞ expðþinxÞdx; /ðxÞ ¼ 1

2p

Zþ1

�1

�/ðnÞ expð�inxÞdn ð2:3:3Þ

~/ðgÞ ¼Zþ1

�1/ðyÞ expðþigyÞdy; /ðyÞ ¼ 1

2p

Zþ1

�1

~/ðgÞ expð�igyÞdg ð2:3:4Þ

/ cω−

/ cω+Re( )ξ

Im( )ξNegative time factor

exp( )i tω−

/ cω−

/ cω+Re( )ξ

Im( )ξPositive time factor

exp( )i tω+

(a)

(b)

Fig. 2.4 Path deformation for the inversion integral. a Positive time factor, b Negative time factor

2.3 2D Static Source 45

to the differential equation (2.3.1) successively, as

Zþ1

�1

Zþ1

�1

@2/@x2

þ @2/@y2

¼ �SdðxÞdðyÞ� �

expðþinxÞdx0@

1A expðþigyÞdy ð2:3:5Þ

With the aid of the convergence condition (2.3.2), each term is transformed asfollows:

Zþ1

�1

Zþ1

�1

@2/@x2

expðþinxÞdx0@

1A expðþigyÞdy ¼ �n2~�/

Zþ1

�1

Zþ1

�1

@2/@y2

expðþinxÞdx0@

1A expðþigyÞdy ¼ �g2~�/

Zþ1

�1

Zþ1

�1SdðxÞdðyÞ expðþinxÞdx

0@

1A expðþigyÞdy ¼ S

ð2:3:6Þ

Then, we have the simple algebraic equation �ðn2 þ g2Þ~�/ ¼ �S for the doubletransformed function and its solution is given by

~�/ ¼ S

n2 þ g2ð2:3:7Þ

The reader will find that the partial differential equation (2.3.1) is transformed tothe simple algebraic equation. There is thus no need of solving the differentialequation directly. The subsequent inversion process is however crucial for thesolution in the actual space. The formal Fourier inversion integral with respect tothe parameter g,

�/ ¼ 12p

Zþ1

�1

S

n2 þ g2expð�igyÞdg ¼ S

p

Z10

1

n2 þ g2cosðgyÞdg ð2:3:8Þ

is evaluated with the aid of the formula (2.1.22) and yields

�/ ¼ S2jnj expð�jnjjyjÞ ð2:3:9Þ


The next inversion is to evaluate the Fourier inversion integral with respect to theparameter n,

/ ¼ 12p

Zþ1

�1

S2jnj expð�jnjjyjÞ expð�inxÞdn ð2:3:10Þ

Inspecting the integrand, the singular point at n ¼ 0 lies on the real axis, i.e. on theintegration path. It is impossible to evaluate the integral in the regular sense. So, wehave to deform the integration path around the pole as that in the previous section.But it was somewhat complicated to determine the path deformation. In order toavoid this troublesome work, we employ a simpler way for evaluating the integral.

Since the trouble stems from the singular point at n ¼ 0, in order to avoid thetrouble, we differentiate equation (2.3.10) with respect to the space variables, x andy, respectively,

@/@x

¼ � S4p

Zþ1

�1

injnj expð�jnjjyjÞ expð�inxÞdn ð2:3:11Þ

@/@y

¼ � S4p

sgnðyÞZþ1

�1expð�jnjjyjÞ expð�inxÞdn ð2:3:12Þ

where sgn(.) is the sign function defined by

sgnðyÞ ¼ þ1; y[ 0�1; y\0

�ð2:3:13Þ

Using the symmetry of the integrand in Eqs. (2.3.11) and (2.3.12), the integrals arereduced to the real valued semi-infinite integrals,

@/@x

¼ � S2p

Z10

expð�njyjÞ sinðnxÞdn ð2:3:14Þ

@/@y

¼ � S2p

sgnðyÞZ10

expð�njyjÞ cosðnxÞdn ð2:3:15Þ


The two integrals in the above equations are well-known from Calculus and wehave the formulas,

Z10

expð�nyÞ sinðnxÞdn ¼ xx2 þ y2

;

Z10

expð�nyÞ cosðnxÞdn ¼ yx2 þ y2

ð2:3:16Þ

Then, Eqs. (2.3.14) and (2.3.15) are evaluated as

@/@x

¼ � S2p

xx2 þ y2

;@/@y

¼ � S2p

yx2 þ y2

ð2:3:17Þ

The derivative of / is completely determined. We integrate the above two equationswith respect to x and y, respectively. This integration leads to two expressions forthe single function / as

@/@x

) / ¼ � S4p

log x2 þ y2� �þ c0ðyÞ

@/@y

) / ¼ � S4p

log x2 þ y2� �þ c00ðxÞ

ð2:3:18Þ

Since the above two equations must be equal, two constant terms should beidentical,

c0ðyÞ ¼ c00ðxÞ ð2:3:19Þ

This condition is satisfied only when the two terms are pure constant and do notinclude any space variable:

c0ðyÞ ¼ c00ðxÞ ¼ constant ð2:3:20Þ

Then, we have Green’s function for the Laplace equation,

@2/@x2

þ @2/@y2

¼ �SdðxÞdðyÞ ) / ¼ � S4p

log x2 þ y2� �þ arbitrary constant

ð2:3:21Þ

This Green’s function does not satisfy the convergence condition at infinity, since ithas the constant. This is because we could not carry out the Fourier inversionintegral of Eq. (2.3.10) directly.



The 2D wave equation with an impulsive source is given by

@2/@x2

þ @2/@y2

¼ 1c2

@2/@t2

� PdðxÞdðyÞdðtÞ ð2:4:1Þ

The nonhomogeneous term states that the source with magnitude P is placed at thecoordinate origin and is impulsive in time. The quiescent condition at an initialtime,

/jt¼0¼@/@t

��t¼0

¼ 0 ð2:4:2Þ



p!1¼ @/

@x


p!1

¼ @/@y


p!1

¼ 0 ð2:4:3Þ

are also employed. As the wave equation (2.4.1) has two space variables x and y,and one time variable t, the triple integral transform is applied to the differentialequation (2.4.1): the Laplace transform with respect to the time variable,

/�ðsÞ ¼Z10

/ðtÞ expð�stÞdt ð2:4:4Þ

and the double Fourier transform with respect to the space variables,

�/ðnÞ ¼Zþ1


2p

Zþ1

�1


~/ðgÞ ¼Zþ1


2p

Zþ1

�1


Applying this triple integral transform with the quiescent and convergence condi-tions, the original differential equation (2.4.1) is transformed to the simple algebraic

equation for the unknown function ~�/�,

�n2~�/� � g2~�/

� ¼ ðs=cÞ2~�/� � P ð2:4:7Þ


Then, the exact expression for ~�/�in the transformed domain is given by

~�/� ¼ P

n2 þ g2 þ ðs=cÞ2 ð2:4:8Þ

For the inversion, three inversion integrals must be carried out successively. Thefirst one is the Fourier inversion integral with respect to the parameter g. This isreduced to the semi-infinite integral as

�/� ¼ 12p

Zþ1

�1

P

n2 þ g2 þ ðs=cÞ2 expð�igyÞdg ¼ Pp

Z10

cosðgyÞg2 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cÞ2

q 2 dg

ð2:4:9Þ

The integral on the far right side is easily evaluated by applying the formula(2.1.22). Then, the first Fourier inversion integral in Eq. (2.4.9) yields

�/� ¼ P

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cÞ2

q exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cÞ2

q� ð2:4:10Þ

Secondly, we apply the inversion integral with respect to the parameter n,

/� ¼ 12p

Zþ1

�1

P



q� expð�inxÞdn ð2:4:11Þ

The above integral is also reduced to the semi-infinite integral, as

/� ¼ P2p

Z10



q� cosðnxÞdn ð2:4:12Þ

and we apply the integration formula (Erdélyi 1954, vol. I, pp. 17, 27)

Z10

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2

p exp �cffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2

p� �cosðbxÞdx ¼ K0 a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 þ c2

p� �ð2:4:13Þ

where K0ð:Þ is the zeroth order modified Bessel function of the second kind. Then,Eq. (2.4.12) takes the compact form

/� ¼ P2p

K0sc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p� �ð2:4:14Þ


The last inversion is the Laplace inversion. The Laplace inversion is symboli-cally expressed as

/ ¼ P2p

L�1 K0sc


p� �h ið2:4:15Þ

We have the suitable inversion formula (Erdélyi 1954, vol. I, pp. 277, 8),

L�1 K0 asð Þ½ � ¼ Hðt � aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � a2

p ¼ 0; t\a1ffiffiffiffiffiffiffiffiffi

t2�a2p ; t[ a

�ð2:4:16Þ

Applying this formula to Eq. (2.4.15), the simple expression for / is obtained as

/ ¼ cP2p

H ct � rð ÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðctÞ2 � r2

q ¼ cP2p

0; ct\r1ffiffiffiffiffiffiffiffiffiffiffiffi

ðctÞ2�r2p ; ct[ r

(ð2:4:17Þ

where Hð:Þ is Heaviside’s unit step function and the radial distance r from thesource is defined by

r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

pð2:4:18Þ

Consequently, we have the exact expression for Green’s function of the 2D waveequation as

@2/@x2

þ @2/@y2

¼ 1c2

@2/@t2

� PdðxÞdðyÞdðtÞ ) / ¼ cP2p

Hðct � rÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðctÞ2 � r2

q ð2:4:19Þ


When the source is vibrating harmonically, the nonhomogeneous 2D wave equationis given by

@2/@x2

þ @2/@y2

¼ 1c2

@2/@t2

� QdðxÞdðyÞ expðþixtÞ ð2:5:1Þ

where Q and x are the magnitude and the frequency of the source, respectively. Weassume that its Green’s function satisfies the convergence condition at infinity,


p!1¼ @/

@x


p!1

¼ @/@y


p!1

¼ 0 ð2:5:2Þ


As the standard technique for the time-harmonic response, the Green’s functionis assumed as the product

/ðx; y; tÞ ¼ /#ðx; yÞ expðþixtÞ ð2:5:3Þ

where /#ðx; yÞ is an amplitude function to be determined. Substituting thisassumption into the nonhomogeneous wave equation (2.5.1), we have the reducedwave equation (so-called Helmholtz equation) for the amplitude function /#ðx; yÞ,

@2/#

@x2þ @2/#

@y2þ ðx=cÞ2/# ¼ �QdðxÞdðyÞ ð2:5:4Þ

The convergence condition (2.5.2) is also rewritten for the amplitude function as

/#�� ffiffiffiffiffiffiffiffiffi

x2þy2p

!1¼ @/#

@x


p!1

¼ @/#

@y


p!1

¼ 0 ð2:5:5Þ

The double Fourier transform with respect to two space variables as defined byEqs. (2.4.5) and (2.4.6) is applied to the nonhomogeneous Helmholtzequation (2.5.4),

Zþ1

�1

Zþ1

�1

@2/#

@x2þ @2/#

@y2¼ �ðx=cÞ2/# � QdðxÞdðyÞ

� �expðþinxÞdx

24

35 expðþigyÞdy

ð2:5:6Þ

Defining the double transform of the amplitude function as

~�/# ¼

Zþ1

�1

Zþ1

�1/#ðx; yÞ expðþinxÞdx

24

35 expðþigyÞdy ð2:5:7Þ

Equation (2.5.6) is transformed into the algebraic equation for the unknown

amplitude function ~�/#,

�n2~�/# � g2~�/

# ¼ �ðx=cÞ2~�/# � Q ð2:5:8Þ

Then, the explicit expression for the amplitude function in the transformed domainis given by

~�/# ¼ Q

n2 þ g2 � ðx=cÞ2 ð2:5:9Þ


As the first inversion, we apply the Fourier inversion integral with respect to theparameter g. Its formal inversion integral is simplified as

�/# ¼ 12p

Zþ1

�1

Q

n2 þ g2 � ðx=cÞ2 expð�igyÞdg ¼ Qp

Z10

cosðgyÞg2 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q 2 dg

ð2:5:10Þ

The far right integral is evaluated with the aid of the formula (2.1.22) and yields

�/# ¼ Q

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� ð2:5:11Þ

where the real part of the square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

qmust be positive in

order to guarantee the convergence at infinity jyj ! 1, i.e.

Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� �� 0 ð2:5:12Þ

The second Fourier inversion integral with respect to the parameter n is given by

/# ¼ Q4p

Zþ1

�1


q exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� expð�inxÞdn ð2:5:13Þ

The integrand in the above inversion integral has two branch points at n ¼ �x=cwhich are on the integration path, and the square root function (radical)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 � ðx=cÞ2q

changes its sign around the branch point. Thus, we have to discuss

the path deformation around the branch point as was done in the case of the 1Dtime-harmonic problem in Sect. 2.2. However, the singular point for the 1Dproblem was the simple pole, but that for the present 2D problem is the branch pointfor the square root function. Therefore, we have to discuss the introduction of thebranch cut in the complex plane. As the detailed discussion for the introduction ofbranch cuts has been carried out in Sect. 1.3.2 in Chap. 1, we do not repeat, but, wecite here its result. If the reader needs more about the introduction of the branch cut,please return to Sect. 1.3 in Chap. 1.

In order to evaluate the inversion integral of Eq. (2.5.13), we define the integral I as

I ¼ 12p

Zþ1

�1


q exp �inx� jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� dn ð2:5:14Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

and consider the complex integral having the same integrand as that in the aboveequation,

U ¼ 12p

ZC

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

q exp �ifx� jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

q� df ð2:5:15Þ

In the subsequent analysis, it should be understood that the real and imaginary partsof the complex variable f are n and g, i.e. f ¼ nþ ig. The square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

f2 � ðx=cÞ2q

has two branch points at f ¼ �x=c, and the corresponding two

branch cuts must be introduced in the complex f-plane, in order to make the radicalbe single-valued and to guarantee the convergence condition,

Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

q� �� 0 ð2:5:16Þ

The discussion on the branch cut for the square root function in Sect. 1.3.2shows two ways to consider the complex frequency: one is the complex frequencywith a small positive imaginary part and the other with a small negative imaginarypart. As the shape of the branch cut depends on the sign of the imaginary part, weevaluate the integral I for these two cases of the complex frequency, separately, andthen determine the suitable imaginary part.

(1) Complex frequency with the negative imaginary part

If we add a small negative imaginary part to the frequency, the branch points areshifted from the real axis and then two branch cuts are introduced along thehyperbola in the second and the fourth quadrants in the complex f-plane as shownin Figs. 1.7 and 1.8 in Chap. 1. When the imaginary part of the frequency tends tozero as the limit, the branch cut lies on the real and imaginary axes as shown by thethick lines in Fig. 1.8a. Therefore, we take the closed loop C for our complexintegral of Eq. (2.5.15) as shown in Fig. 2.5. We have two closed loops for C: one isthe upper closed loop CðþÞ and the other is the lower loop Cð�Þ. The selection of thetwo loops depends on the convergence of the integrands at the infinity jfj ! 1.The lower loop is employed when the parameter (space variable) x is positive, andthe upper loop is employed when x\0. The following evaluation is carried out forthe case of the positive space variable, x[ 0.

When x[ 0, the loop for the complex integral is Cð�Þ as shown in Fig. 2.5

where the integration path is the straight line AOB��!

on the real axis. In order toexclude the lower branch cut, the path along the branch cut is introduced. The path

is composed of two lines, CDE and FGH along the branch cut, and a small circle EF\

around the branch point. It should be understood that the radius of the small circle

vanishes as the limit. Two edges of the straight line AOB��!

are connected to the edges


http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

of the path CDE and FGH through two quarter circles, AC\

and BH\, respectively. It

is also understood that the radius of the quarter circle tends to infinity as the limit.Then, the closed loop Cð�Þ is the sum of the integrals along these paths. In theclockwise direction, the closed loop is

Cð�Þ : AOB��!þ BH

\ þHGF��!þ FE

\ þEDC��!þ CA

\ð2:5:17Þ

The complex integral of Eq. (2.5.15) is decomposed into the integrals along thepath segments as

U ¼ 12p

ZAOB��! þ

ZBH\

þZ

HGF��! þ

ZFE\

þZ

EDC��! þ

ZCA\

0BB@

1CCA 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

f2 � ðx=cÞ2q exp �ifx� jyj

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

q� df

ð2:5:18Þ

The integral along the small circle FE\

around the branch point vanishes as its radius

tends to zero. The two integrals along the large arcs BH\

and CA\

also vanish as theradius tends to infinity due to the convergence condition of Eq. (2.5.16). Thus, sinceno singular point is included in the closed loop and the total value of the complex

/ cω+

/ cω−

Im( )η ζ=

Re( )ξ ζ=

BO

C

D E

FG

H

A

( )Loop

0

C

x

+

<

( )Loop

0

C

x

−

>

Fig. 2.5 Closed loop C for the complex integral U in Eq. (2.5.15) (the negative imaginary part ofthe complex frequency vanished)


integral vanishes, i.e. U ¼ 0, the integral along the real axis AOB��!

is converted tothe integrals along the branch cut,

12p

ZAOB��!



q� df

¼ 12p

ZCDE��! þ

ZFGH��!

0BB@

1CCA 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

f2 � ðx=cÞ2q exp �ifx� jyj


q� df

ð2:5:19Þ

where the directions of paths CDE��!

and FGH��!

are inverse of the paths EDC��!

and HGF��!

,respectively. Equation (2.5.19) states that it is enough to consider the integrals

along the branch cut to evaluate the integral on the real axis. The line paths CDE��!

and FGH��!

are further decomposed into the line segments along the real and imag-inary axes. They are

CDE��! ¼ CD

�!þ DE�!

; FGH��! ¼ FG

�!þ GH�! ð2:5:20Þ

The argument and value of the square root function have already been determinedin Sect. 1.3.2(2) in Chap. 1 and are tabulated in Table 2.1 with the integrationvariable along each line segment. Applying these results, the integral along eachpath is given by

12p

ZAOB��!



q� df

¼ 12p

Z1�1



q� dn

ð2:5:21Þ

12p

ZCD�!



q� df

¼ 12p

Z10

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q exp �gx� ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

ð2:5:22Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

12p

ZDE�!



q� df

¼ � i2p

Zx=c0

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q exp �inx� ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

ð2:5:23Þ

12p

ZFG�!



q� df

¼ � i2p

Zx=c0


q exp �inxþ ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

ð2:5:24Þ

12p

ZGH�!



q� df

¼ 12p

Z10


q exp �gxþ ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

ð2:5:25Þ

Table 2.1 Value andargument of the square rootfunction on the integrationpath in Fig. 2.5

Pathffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

qVariable f and its range

AOB! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 � ðx=cÞ2q

f ¼ n

�1\n\þ1CD!

þiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

qf ¼ �ig

0\g\1DE!

þiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

qf ¼ þn

0\n\x=c

FG!

�iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

qf ¼ þn

0\n\x=c

GH!

�iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

qn ¼ �ig

0\g\1n and g are positive real, except BOA

!


Substituting the above equations into Eq. (2.5.19) with the path decomposition ofEq. (2.5.20), we have for the integral I,

I ¼ 12p

Z1�1



q� dn

¼ 1p

Z10


q expð�gxÞ cos yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

� 1p

Zx=c0


q sinðnxÞ cos yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

� ip

Zx=c0


q cosðnxÞ cos yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

ð2:5:26Þ

The Fourier inversion integral has just been converted to three real-valued integrals.Fortunately, we can evaluate these integrals with the aid of integration formulas.Two integration formulas (Gradshteyn and Ryzhik 1980, pp. 755, 6.677, No. 4)*

Z10

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ x2

p cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ x2

p� �expð�cxÞdx�

Za

0

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p� �sinðcxÞdx

¼ � p2Y0 a


p� � ð2:5:27Þ

and (Erdélyi 1954, vol. I, pp. 28, 42)

Za

0

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p� �cosðcxÞdx ¼ p

2J0 a


p� �ð2:5:28Þ

are applied to the right hand side of Eq. (2.5.26) and the Hankel function (Watson1966, p. 73) is introduced. Then, Eq. (2.5.26) is simplified as

12p

Zþ1

�1



q� dn

¼ � 12Y0

xc


p� �� i2J0

xc


p� �¼ � i

2Hð2Þ

0xc


p� �ð2:5:29Þ


where J0ð:Þ and Y0ð:Þ are Bessel functions of the first and second kind, respectively,

and Hankel function of the second kind is defined by Hð2Þ0 ð:Þ ¼ J0ð:Þ � iY0ð:Þ.

Finally, we have the simple expression for the Fourier inversion integral I. That is

12p

Zþ1

�1



q� dn

¼ � i2Hð2Þ

0xc


p� �ð2:5:30Þ

It is easily understood that the above equation is also valid for the negative spacevariable, x\0, since Eq. (2.5.30) is the even function of the space variable x. So, ifwe perform the complex integral with the upper closed loop CðþÞ in Fig. 2.5, we canobtain the same expression for the negative space variable. However, the result inEq. (2.5.30) is valid only when the negative imaginary part of the complex fre-quency tends to zero.

(2) Complex frequency with the positive imaginary part

If we assume that the imaginary part of the complex frequency approaches tozero from the positive, the branch cuts corresponding to two branch points areintroduced in the first and the third quadrants in the complex plane as was discussedin Sect. 1.3.2 in Chap. 1 and are shown in Fig. 1.11 as its zero limit. Thus, the

/ cω+

/ cω−

Im( )η ζ=

Re( )ξ ζ=

BO

C´

D´E´

F´G´

H´

A

( )Loop

0

C

x

+

<

( )Loop

0

C

x

−

>

Fig. 2.6 Closed loop C for the complex integral U in Eq. (2.5.15) (the positive imaginary part ofthe complex frequency vanished)


http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

closed loop for the complex integral U is taken as Cð�Þ in the complex f-plane asshown in Fig. 2.6 and the value and argument of the square root function are listedin Table 2.2. We can apply the same evaluation procedure as that in the previoussubsection. The analysis is carried out for the case of the positive space variable,x[ 0. The integral on the real axis, i.e. the integral I, is given by

I ¼ 12p

ZAOB��!



q� df

¼ 12p

Z1�1



q� dn

ð2:5:31Þ

and the integrals along the branch cut are

12p

ZC0D0��!



q� df

¼ 12p

Z10


q exp �gx� ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

ð2:5:32Þ

12p

ZD0E0��!



q� df

¼ i2p

Zx=c0


q exp þinx� ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

ð2:5:33Þ

Table 2.2 Value andargument of the root functionalong the integration path inFig. 2.6

Pathffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

qVariable f and its range

AOB! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 � ðx=cÞ2q

f ¼ n

�1\n\þ1C0D0!

þiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

qf ¼ �ig

0\g\1D0E0!


qf ¼ �n

0\n\x=c

F0G0!�i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

qf ¼ �n

0\n\x=c

G0H0!�i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

qn ¼ �ig

0\g\1n and g are positive real, except BOA

!


12p

ZF0G0��!



q� df

¼ i2p

Zx=c0


q exp þinxþ ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

ð2:5:34Þ

12p

ZG0H0��!



q� df

¼ 12p

Z10


q exp �gxþ ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

ð2:5:35Þ

Since no singular point is included in the closed loop, U ¼ 0. Thus, the integral I onthe real axis is given by

I ¼ 12p

Z1�1



q� dn

¼ 1p

Z10


q expð�gxÞ cos yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� dg

� 1p

Zx=c0


q sinðnxÞ cos yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

þ ip

Zx=c0



q� dn

ð2:5:36Þ

The above equation can be simplified by applying the formulas (2.5.27) and(2.5.28). We have

12p

Z1�1



q� dn

¼ þ i2Hð1Þ

0xc


p� �ð2:5:37Þ

where Hankel function of the first kind is defined by Hð1Þ0 ð:Þ ¼ J0ð:Þ þ iY0ð:Þ.

*Derivation of formula (2.5.27).


We have the integration formula (Gradshteyn and Ryzhik 1980, pp. 755, 6.677),

Z10

Y0 affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ z2

p� �cosðbxÞdx ¼

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � b2

p sin zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � b2

q� �; b\a

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � a2

p exp �zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � a2

q� �; b[ a

8>>><>>>:

ð2:5:38Þ

If we consider the above as the Fourier cosine transform with respect to thetransform parameter b, its inverse cosine transform is given by

p2Y0 a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ z2

p� �¼

Za

0



q� �cosðbxÞdb

�Z1a

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � a2

p exp �zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � a2

q� �cosðbxÞdb

ð2:5:39Þ

Making the change of variable, u ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � a2

p, for the second integral in the right

hand side, the formula (2.5.27) is obtained as

p2Y0 a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ z2

p� �¼

Za

0



q� �cosðbxÞdb

�Z10

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu2 þ a2

p exp �zuð Þ cos xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu2 þ a2

p� �du

ð2:5:40Þ

(3) Selection of the branch cut and the integration loop

The Fourier inversion integral I has just been evaluated in two ways of thebranch cut and we get two expressions for the single integral I as

12p

Zþ1

�1



q� dn

¼� i2Hð2Þ

0 ðxr=cÞ; x � x� ieje!0

þ i2Hð1Þ

0 ðxr=cÞ; x � xþ ieje!0

8><>:

ð2:5:41Þ

where the radial distance r from the source point is defined by



pð2:5:42Þ

In order to determine the correct evaluation, we multiply the time factor expðþixtÞand replace the Hankel functions with their asymptotic forms (Watson 1966, p. 198)as

Hð1Þ0 ðzÞ

ffiffiffiffiffi2pz

rexpf�iðz� p=4Þg

Hð2Þ0 ðzÞ

ffiffiffiffiffi2pz

rexpfþiðz� p=4Þg

; z ! 1 ð2:5:43Þ

The asymptotic form of the integral I yields to

expðþixtÞ2p

Zþ1

�1



q� dn

r!1

� i2

ffiffiffiffiffiffiffiffiffi2cpxr

rexpðþpi=4Þ expfþixðt � r=cÞg; x � x� ieje!0

þ i2

ffiffiffiffiffiffiffiffiffi2cpxr

rexpð�pi=4Þ expfþixðt þ r=cÞg; x � xþ ieje!0

8>>><>>>:

ð2:5:44Þ

It is clear that the asymptotic form in the upper line shows the out-going wave fromthe source point r = 0, and that in the lower line does the in-coming wave from theinfinity. Thus, the right evaluation for the integral I is

12p

Zþ1

�1



q� dn ¼ � i

2Hð2Þ

0 ðxr=cÞ

ð2:5:45Þ

and the correct introduction of the branch cut and the integration loop are shown inFig. 2.5 for our positive time factor expðþixtÞ. However, if we employ the negativetime factor expð�ixtÞ, the branch cut and the loop in Fig. 2.6 are the right choices,needless to say.

Anyway, we could evaluate the infinite integral of the Fourier inversion andobtained the amplitude function in the closed form as

/# ¼ � iQ4Hð2Þ

0 ðxr=cÞ; r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

pð2:5:46Þ

Thus, the 2D Green’s function for the time-harmonic source is given by


@2/@x2

þ @2/@x2

¼ 1c2

@2/@t2

� QdðxÞdðyÞ expðþixtÞ ) / ¼ � iQ4Hð2Þ

0 ðxr=cÞ expðþixtÞð2:5:47Þ

When the time factor is negative, we have

@2/@x2

þ @2/@x2

¼ 1c2

@2/@t2

� QdðxÞdðyÞ expð�ixtÞ ) / ¼ þ iQ4Hð1Þ

0 ðxr=cÞ expð�ixtÞð2:5:48Þ

(4) Convolution integral

The solution of Eq. (2.5.47) can be obtained directly by applying the convolu-tion integral of the impulsive solution which is given by Eq. (2.4.19). Let usconsider its convolution. The convolution integral for the Laplace transform isdefined by

/ðx; y; tÞ ¼Z t

0

/ðimplseÞðx; y; t0Þ expfþixðt � t0Þgdt0 ð2:5:49Þ

where /ðimplseÞ is the solution for the impulsive source. We substitute the impulsiveGreen’s function of Eq. (2.4.19) with the source magnitude Q into Eq. (2.5.49),

/ðx; y; tÞ ¼Z t

0

cQ2p

Hðct0 � rÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðct0Þ2 � r2

q expfþixðt � t0Þgdt0 ð2:5:50Þ

and examine the supporting region for the step function. We have

/ðx; y; tÞ ¼ cQ2p

expðþixtÞHðt � r=cÞZ t

r=c

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðct0Þ2 � r2

q expð�ixt0Þdt0 ð2:5:51Þ

As we are considering the steady-state response, the time in the upper integrationlimit and that in the step function can be set to be infinite as

/ðx; y; tÞ ¼ cQ2p

expðþixtÞ limt!1 Hðt � r=cÞ

Z t

r=c

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðct0Þ2 � r2

q expð�ixt0Þdt0264

375 ð2:5:52Þ


Taking the limit for the time, we have

/ðx; y; tÞ ¼ Q2p

expðþixtÞZ1r

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu2 � r2

p expð�ixu=cÞdu ð2:5:53Þ

This is just the integral representation for the Hankel function of the second kind(Watson 1966, p. 170),

Hð2Þ0 ðxÞ ¼ 2i

p

Z11

1ffiffiffiffiffiffiffiffiffiffiffiffiffiu2 � 1

p expð�ixuÞdu ð2:5:54Þ

Thus, we have

/ðx; y; tÞ ¼ � iQ4expðþixtÞHð2Þ

0 ðrx=cÞ ð2:5:55Þ

(5) Direct evaluation

In the former subsection (1) and (2), the Fourier inversion is carried out byapplying the Cauchy complex integral. It was lengthy troublesome work. Withoutapplying the Cauchy theorem, we can evaluate the integral I,

I ¼ 12p

Zþ1

�1



q� dn ð2:5:56Þ

Firstly, we reduce the infinite integral to the semi-infinite integral,

I ¼ 1p

Zþ1

0


q cosðnxÞ exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� dn ð2:5:57Þ

In the complex n-plane, the integration path for this integral is just on the real axis

and two branch cuts for the radicalffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

qare introduced on the real and

imaginary axes as was discussed in Sect. 1.3.2 in Chap. 1. The slightly shiftedbranch cuts are shown in Fig. 2.7 where the integration path for the semi-infiniteintegral is OP. The integration path in the region 0\n\x=c is slightly up from the


http://dx.doi.org/10.1007/978-3-319-17455-6_1

lower branch cut. When the cut approaches to the real axis as was discussed inSect. 1.3.2 (2), the argument of the square root function in the integrand yields toþp=2, i.e.


q¼ þi


q; jnj\x=c ð2:5:58Þ

In the other positive region n[x=c on the real axis, the root function is positivereal. Then, the semi-infinite integral is decomposed into two integrals in the regions,0\n\x=c and n[x=c, as

I ¼ 1p

Zx=c0

1

ðþiÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q cosðnxÞ exp �ijyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

þ 1p

Zþ1

x=c



q� dn

ð2:5:59Þ

The above integral I is further decomposed into the real and imaginary parts,

Re( )ξ

/ cω+

/ cω−

Im( )ξ

O PQ

branch cut

branch cut

Fig. 2.7 Branch cuts and integration path for the inversion integral of Eq. (2.5.57)


http://dx.doi.org/10.1007/978-3-319-17455-6_1

I ¼ � 1p

Zx=c0


q cosðnxÞ sin jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q� dn

þ 1p

Zþ1

x=c



q� dn

� ip

Zx=c0



q� dn

ð2:5:60Þ

We make the change of variable for the first and second integrals in the aboveequation. The changes are defined by

g ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

q; n ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � g2

q; dn ¼ � gdgffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðx=cÞ2 � g2q ð2:5:61Þ

for the first integral, and

g ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q; n ¼


q; dn ¼ gdgffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

g2 þ ðx=cÞ2q ð2:5:62Þ

for the second integral. Equation (2.5.60) yields to

I ¼ � 1p

Zx=c0

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � g2

q cos xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � g2

q� sinðjyjgÞdg

þ 1p

Z10


q cos xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q� expð�jyjgÞdg

� ip

Zx=c0



q� dn

ð2:5:63Þ

Then, we apply the formula of Eq. (2.5.27) to the first two integrals and the formulaof Eq. (2.5.28) to the third integral. The integral I is exactly evaluated as

I ¼ � 12Y0

xc


p� �� i2J0

xc


p� �¼ � i

2Hð2Þ

0xc


p� �ð2:5:64Þ

This is just the same as Eqs. (2.5.30) and (2.5.45) for the integral I.



The static 3D Green’s function for the Laplace equation is a particular solution ofthe nonhomogeneous differential equation,

@2/@x2

þ @2/@y2

þ @2/@z2

¼ �SdðxÞdðyÞdðzÞ ð2:6:1Þ

where S is the magnitude of the source which is placed at the coordinate origin(0, 0, 0). The convergence condition at infinity,

/j ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2

p!1¼ @/

@x

�� ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2

p!1

¼ @/@y


p!1

¼ @/@z


p!1

¼ 0

ð2:6:2Þ

is applied to the Green’s function /.The Green’s function is obtained by the method of integral transform. For three

space variables, the triple Fourier transform defined by

�/ðnÞ ¼Zþ1


2p

Zþ1

�1


~/ðgÞ ¼Zþ1


2p

Zþ1

�1


/ðfÞ ¼Zþ1

�1/ðzÞ expðþifzÞdz; /ðzÞ ¼ 1

2p

Zþ1

�1/ðfÞ expð�ifzÞdf ð2:6:5Þ

is applied to Eq. (2.6.1),

Zþ1

�1

Zþ1

�1

Zþ1

�1

@2/@x2

þ @2/@y2

þ @2/@z2

¼ �SdðxÞdðyÞdðzÞ�

expðþinxÞdx24

35 expðþigyÞdy

* +expðþifzÞdz

ð2:6:6Þ

Applying the convergence condition (2.6.2), the above Eq. (2.6.6) is transformed to

the algebraic equation for the triple transformed function ~�/,

�ðn2 þ g2 þ f2Þ ~�/ ¼ �S ð2:6:7Þ


The transformed function is determined explicitly,

~�/ ¼ S

n2 þ g2 þ f2ð2:6:8Þ

The first Fourier inversion integral with respect to the parameter f is

~�/ ¼ 12p

Zþ1

�1

S

n2 þ g2 þ f2expð�ifzÞdf ¼ S

p

Z10

1

f2 þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2

p 2 cosðfzÞdf ð2:6:9Þ

The above integral is easily evaluated with the aid of the formula (2.1.22). It yields

~�/ ¼ S

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2

p exp �jzjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2

q� �ð2:6:10Þ

The second inversion integral with respect to the parameter g is reduced to the semi-infinite integral as

�/ ¼ 12p

Zþ1

�1

S



q� �expð�igyÞdg

¼ S2p

Z10



q� �cosðgyÞdg

ð2:6:11Þ

The latter semi-infinite integral can be evaluated by using the formula (2.4.13). Itfollows that

�/ ¼ S2p

K0 jnjffiffiffiffiffiffiffiffiffiffiffiffiffiffiy2 þ z2

p� �ð2:6:12Þ

The last inversion integral with respect to the parameter n is also reduced to thesemi-infinite integral,

/ ¼ S4p2

Zþ1

�1K0 jnj

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiy2 þ z2

p� �expð�inxÞdn

¼ S2p2

Z10

K0 nffiffiffiffiffiffiffiffiffiffiffiffiffiffiy2 þ z2

p� �cosðnxÞdn

ð2:6:13Þ


Fortunately, we have the suitable integration formula, (Erdélyi 1954, vol. I, pp. 49, 40)

Z10

K0 anð Þ cosðbnÞdn ¼ p

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ b2

p ð2:6:14Þ

Applying this formula to Eq. (2.6.13), the last inversion integral is exactly evaluatedas

/ ¼ S

4pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2 þ z2

p ð2:6:15Þ

Consequently, we have the static 3D Green’s function for the Laplace equation,

@2/@x2

þ @2/@y2

þ @2/@z2

¼ �SdðxÞdðyÞdðzÞ ) / ¼ S4pR

ð2:6:16Þ

where R is the 3D radial distance from the source,

R ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2 þ z2

pð2:6:17Þ


Green’s function for the 3D wave equation is discussed. The wave equation with animpulsive point source located at the coordinate origin is given by

@2/@x2

þ @2/@y2

þ @2/@z2

¼ 1c2

@2/@t2

� PdðxÞdðyÞdðzÞdðtÞ ð2:7:1Þ

where P is the magnitude of the source. The quiescent condition at an initial time,

/jt¼0¼@/@t

��t¼0

¼ 0 ð2:7:2Þ



p!1¼ @/

@x


p!1

¼ @/@y


p!1

¼ @/@z


p!1

¼ 0 ð2:7:3Þ

are also imposed.Our dynamic Green’s function has four variables: three space and one time

variables. Laplace transform with respect to the time,


/�ðsÞ ¼Zþ1

0

/ðtÞ expð�stÞdt ð2:7:4Þ

and the triple Fourier transform with respect to three space variables,

�/ðnÞ ¼Zþ1


2p

Zþ1

�1


~/ðgÞ ¼Zþ1


2p

Zþ1

�1


/ðfÞ ¼Zþ1

�1/ðzÞ expðþifzÞdz; /ðzÞ ¼ 1

2p

Zþ1

�1/ðfÞ expð�ifzÞdf ð2:7:7Þ

are applied to the nonhomogeneous wave equation (2.7.1). With the aid of thequiescent and convergence conditions, the wave equation is transformed to the

simple algebraic equation for the multi-transformed unknown function ~�/�,

�fn2 þ g2 þ f2 þ ðs=cÞ2g ~�/� ¼ �P ð2:7:8Þ

The inversion starts from the transformed function,

~�/� ¼ P

n2 þ g2 þ f2 þ ðs=cÞ2 ð2:7:9Þ

As the first inversion, the Fourier inversion integral with respect to the parameter f

~�/� ¼ 1

2p

Zþ1

�1

P

n2 þ g2 þ f2 þ ðs=cÞ2 expð�ifzÞdz

¼ Pp

Z10

cosðfzÞf2 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cÞ2

q 2 dz

ð2:7:10Þ


is carried out by applying the formula (2.1.22). It follows that

~�/� ¼ P

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cÞ2

q exp �jzjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cÞ2

q� ð2:7:11Þ

The second inversion is the integral with respect to the parameter g,

�/� ¼ 12p

Zþ1

�1

P



q� expð�igyÞdg

¼ P2p

Z10



q� cosðgyÞdg

ð2:7:12Þ

The latter semi-infinite integral is also evaluated by applying the formula (2.4.13). Ityields

�/� ¼ P2p

K0


q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiy2 þ z2

p� �ð2:7:13Þ

The third inversion integral with respect to the parameter n is

/� ¼ P2p

12p

Zþ1

�1K0



p� �expð�inxÞdn

¼ P2p2

Z10

K0



p� �cosðnxÞdn

ð2:7:14Þ

The integration formula (Erdélyi 1954, vol. I, pp. 56, 43)

Z10

K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

q� �cosðcnÞdn ¼ p

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p� �ð2:7:15Þ

is very helpful for our task. Then, applying the above formula to the last integral inEq. (2.7.14), we have

/� ¼ P


p exp � sc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2 þ z2

p� �ð2:7:16Þ


The last is the Laplace inversion. Its symbolical form is

/ ¼ P


p L�1 exp � sc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2 þ z2

p� �h ið2:7:17Þ

The transform parameter “s” is included only in the argument of the exponentialfunction. We remember the simple Laplace inversion formula for the delta function,

L�1 expð�asÞ½ � ¼ dðt � aÞ ð2:7:18Þ

Thus, applying this inversion formula, Eq. (2.7.17) is fully inverted as

/ ¼ P4pR

d t � R=cð Þ ð2:7:19Þ

where the radial distance R from the source is defined by Eq. (2.6.17). Finally, wehave the 3D Green’s function for the wave equation with the impulsive pointsource,

@2/@x2

þ @2/@y2

þ @2/@z2

¼ 1c2

@2/@t2

� PdðxÞdðyÞdðzÞdðtÞ ) / ¼ P4pR

d t � R=cð Þð2:7:20Þ

This is the very simple expression in spite of the 3D nature!


This section derives the 3D Green’s function for a time-harmonic source. It is theconvolution integral of the impulsive Green’s function obtained in the previoussection. The wave equation with the time-harmonic source is given by

@2/@x2

þ @2/@y2

þ @2/@z2

¼ 1c2

@2/@t2

� QdðxÞdðyÞdðzÞ expðþixtÞ ð2:8:1Þ

where Q is the magnitude of the source and x the frequency of the time-harmonicvibration. The convergence condition at infinity,


p!1¼ @/

@x


p!1

¼ @/@y


p!1

¼ @/@z


p!1

¼ 0 ð2:8:2Þ

is also imposed.


The standard multiple integral transform technique is available for getting thetime-harmonic Green’s function. However, as shown in the case of 2D Green’sfunction in Sect. 2.5 (4), we take a very simple way, i.e. the convolution integral ofthe impulsive Green’s function.

In the previous section we get the Green’s function for the impulsive source.Replacing the source magnitude with unit 1, the Green function is given by

@2/@x2

þ @2/@y2

þ @2/@z2

¼ 1c2

@2/@t2

� dðxÞdðyÞdðzÞdðtÞ ) / ¼ 14pR

d t � R=cð Þ ð2:8:3Þ

Employing the time-harmonic source with the frequency x and the magnitude Q

Q expfþixðt � t0Þg ð2:8:4Þ

the convolution integral for the Laplace transform is given by

/ ¼Z t

0

Q4pR

d t0 � R=cð Þ expfþixðt � t0Þgdt0 ð2:8:5Þ

It is very easy to evaluate the above integral, since the integrand includes Dirac’sdelta function and we can apply the simple integration formula (1.2.3) in Sect. 1.2,

Zb

a

f ðxÞd x� cð Þdx ¼ f ðcÞ; a\c\b0; c\a or b\c

�ð2:8:6Þ

Then, we can evaluate the integral in Eq. (2.8.5) and have for /

/ ¼ Q4pR

H t � R=cð Þ expfþixðt � R=cÞg ð2:8:7Þ

The step function ahead of the equation means that Eq. (2.8.7) is the transientresponse to the time-harmonic source and the disturbance starts from the wavearrival t = R/c. When sufficient long time has passed and the response becomessteady, the step function is meaningless. Then, we have the steady-state time-harmonic response as

/ ¼ Q4pR

expfþixðt � R=cÞg ð2:8:8Þ

Therefore, the 3D Green’s function for the wave equation with the time-harmonicsource is given by


http://dx.doi.org/10.1007/978-3-319-17455-6_1

Tab

le2.3

Green’s

functio

nforLaplace

andWaveequatio

ns

Differentialequatio

nSo

urce

Green’s

functio

n

1D@2/

@x2¼

1 c2@2/

@t2�So

urce

PdðxÞd

ðtÞ/ðx;

tÞ¼

cP 2Hðct

�jxjÞ

QdðxÞexpð�i

xtÞ

/ðx;

tÞ¼

�iQ

2ðx=cÞexpf

�ixðt�jxj=cÞg

2D@2/

@x2þ

@2/

@y2¼

�Sou

rce

SdðxÞ

dðyÞ

/ðx;

yÞ¼

�S 4plogðrÞ

þarbitraryconstant;

r¼


x2þy2

p@2/

@x2þ

@2/

@x2¼

1 c2@2/

@t2�So

urce

PdðxÞd

ðyÞdðtÞ

/ðx;

y;tÞ¼

cP

2pffiffiffiffiffiffiffiffi

ffiffiffiffiðct

Þ2 �r2

pH

ct�r

ðÞ;

r¼


x2þy2

pQdðxÞd

ðyÞexpðþi

xtÞ

/ðx;

y;tÞ¼

�iQ 4H

ð2Þ

0ðx

r=cÞexpðþi

xtÞ;

r¼


x2þy2

pQdðxÞd

ðyÞexpð�i

xtÞ

/ðx;

y;tÞ¼

þiQ 4H

ð1Þ

0ðx

r=cÞexpð�i

xtÞ;

r¼


x2þy2

p3D

@2/

@x2þ

@2/

@y2þ

@2/

@z2¼

�Sou

rce

SdðxÞ

dðyÞd

ðzÞ/ðx;

y;zÞ

¼S

4pR;

R¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiffiffiffix2

þy2

þz2

p@2/

@x2þ

@2/

@y2þ

@2/

@z2¼

1 c2@2/

@t2�So

urce

PdðxÞd

ðyÞdðzÞd

ðtÞ/ðx;

y;z;tÞ¼

P 4pRdt�

R=c

ðÞ;

R¼



þy2

þz2

pQdðxÞd

ðyÞdðzÞexpð�i

xtÞ

/ðx;

y;z;tÞ¼

Q 4pRexpf

�ixðt�R=cÞg

;R¼



þy2

þz2

p


@2/@x2

þ @2/@y2

þ @2/@z2

¼ 1c2

@2/@t2

� QdðxÞdðyÞdðzÞ expðþixtÞ

) / ¼ Q4pR

expfþixðt � R=cÞgð2:8:9Þ

Exercise

(2:1) Using the impulsive response (2.1.27) for 1D wave equation, derive thetime-harmonic Green’s function through the convolution integral andcompare it with the time-harmonic Green’s function (2.2.27).

Appendix

See Table 2.3.

References

Erdélyi A (ed) (1954) Tables of integral transforms, vols I, II, McGraw-Hill, New YorkGradshteyn IS, Ryzhik IM (1980) In: Jefferey A (ed) Table of integrals, series, and products, 5th

edn. Academic Press, San DiegoWatson GN (1966) A treatise on the theory of Bessel functions, Cambridge University Press,

Cambridge


Chapter 3Green’s Dyadic for an Isotropic ElasticSolid

The present chapter shows how to derive an exact closed form solution, the so-called Green’s dyadic, for elasticity equations. Introducing the Cartesian coordinatesystem (xi) ≡ (x, y, z), we employ the notation u � ðuiÞ for the displacement,e � ðeijÞ for the strain, r � ðrijÞ for the stress, B � ðBiÞ for the body force and ρ forthe density. The governing equations for the deformation of an isotropic elasticsolid are constituted by the following equations of motion,

@rxx@x

þ @ryx@y

þ @rzx@z

þ qBx ¼ q@2ux@t2

@rxy@x

þ @ryy@y

þ @rzy@z

þ qBy ¼ q@2uy@t2

@rxz@x

þ @ryz@y

þ @rzz@z

þ qBz ¼ q@2uz@t2

ð3:1Þ

and by the stress-strain relation, the so-called Hooke’s law,

rxxryyrzzrxzryzrxy

26666664

37777775¼

kþ 2l k k 0 0 0kþ 2l k 0 0 0

kþ 2l 0 0 02l 0 0

sym: 2l 02l

26666664

37777775

exxeyyezzexzeyzexy

26666664

37777775

ð3:2Þ

Here (λ, μ) are Lame’s constants which are expressed by Young’s modulus E andPoisson ratio ν as

k ¼ mEð1þ mÞð1� 2mÞ ; l ¼ E

2ð1þ mÞ ð3:3Þ


77

The strains in Eq. (3.2) are defined by the displacement gradient,

exx ¼ @ux@x

; eyy ¼ @uy@y

; ezz ¼ @uz@z

;

exy ¼ 12

@ux@y

þ @uy@x

� �; eyz ¼ 1

2@uy@z

þ @uz@y

� �; ezx ¼ 1

2@ux@z

þ @uz@x

� � ð3:4Þ

It should be noticed that the symmetry relations for the stress and strain componentsare satisfied:

rij ¼ rji; eij ¼ eji; i; j ¼ x; y; z ð3:5Þ

The elasticity equations constitute a set of coupled partial differential equationswith 15 unknowns. In order to reduce the differential equations to more compactforms, the strain in Hooke’s law is replaced with the displacement gradient and thenthe stress is substituted into the equations of motion. We then get a set of dis-placement equations, the so-called Navier equations, with only three unknownfunctions (displacement components),

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

þ @uz@z

� �þ @2ux

@x2þ @2ux

@y2þ @2ux

@z2¼ 1

c2s

@2ux@t2

� 1c2s

Bx

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

þ @uz@z

� �þ @2uy

@x2þ @2uy

@y2þ @2uy

@z2¼ 1

c2s

@2uy@t2

� 1c2s

By

ðc2 � 1Þ @@z

@ux@x

þ @uy@y

þ @uz@z

� �þ @2uz

@x2þ @2uz

@y2þ @2uz

@z2¼ 1

c2s

@2uz@t2

� 1c2s

Bz

ð3:6Þ

where cs and cd are the velocities of shear and dilatational waves, respectively, andare defined by

cs ¼ffiffiffilq

r; cd ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffikþ 2l

q

sð3:7Þ

The velocity ratio γ is defined and expressed by Poisson ratio ν,

c ¼ cdcs

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffikþ 2l

l

s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ð1� mÞ1� 2m

rð3:8Þ

Then, in this chapter we shall show a solution method for the coupled dis-placement Eqs. (3.6) with nonhomogeneous body force terms Bi. The solutionmethod that we employ is the integral transforms: Fourier transforms with respect tothe space variables and Laplace transform with respect the time variable.

78 3 Green’s Dyadic for an Isotropic Elastic Solid

A particular solution corresponding to a point body force is called a Green’sfunction. However, each displacement component produced by one source com-ponent is called “Green’s dyadic.” Thus, the present chapter shows the solutionmethod for the Green’s dyadic.


We assume here the deformation of plane-strain. Take the 2D coordinate system (x,y) in an infinite elastic solid and neglect the anti-plane deformation produced by theanti-plane displacement uz. The in-plane displacement components (ux, uy) arefunctions of two space variables (x, y) and the time t. As a wave source, we assumean impulsive point body force with magnitude Pi placed at the coordinate origin.Under these assumptions, the displacement equations (3.6) is reduced to the simplerform

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

� �þ @2ux

@x2þ @2ux

@y2¼ 1

c2s

@2ux@t2

� Px

c2sdðxÞdðyÞdðtÞ

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

� �þ @2uy

@x2þ @2uy

@y2¼ 1

c2s

@2uy@t2

� Py

c2sdðxÞdðyÞdðtÞ

ð3:1:1Þ

Now, we consider these coupled differential equations. The solutions are twodisplacement components (ux, uy) corresponding to the nonhomogeneous bodyforce Pi. Our solution strategy is very simple, namely, to transform the differentialequations into the simultaneous algebraic equations in the transformed domain. Weemploy Laplace transform with respect to the time, defined by

u�j ¼Z10

uj expð�stÞdt ð3:1:2Þ

and the double Fourier transform with respect to the two space variables, defined by

�uj ¼Zþ1

�1uj expðþinxÞdx; uj ¼ 1

2p

Zþ1

�1�uj expð�inxÞdn

~uj ¼Zþ1

�1uj expðþigyÞdy; uj ¼ 1

2p

Zþ1

�1~uj expð�igyÞdg

ð3:1:3Þ

3 Green’s Dyadic for an Isotropic Elastic Solid 79

where the subscript j stands for x and y. The quiescent condition at an initial time,

uj��t¼0 ¼

@uj@t

��t¼0

¼ 0 ð3:1:4Þ

and the convergence condition at infinity

uj�� ffiffiffiffiffiffiffiffiffi

x2þy2p

!1 ¼ @uj@x


p!1

¼ @uj@y


p!1

¼ 0 ð3:1:5Þ

are assumed.Applying the triple integral transform to the displacement equations (3.1.1), we

have the algebraic equations for the transformed displacement components ~�u�j ,

�inðc2 � 1Þ �in~�u�x � ig~�u�y� �

� n2 þ g2�

~�u�x ¼ ðs=csÞ2~�u�x � Px=c2s

�igðc2 � 1Þ �in~�u�x � ig~�u�y� �

� n2 þ g2�

~�u�y ¼ ðs=csÞ2~�u�y � Py=c2s

ð3:1:6Þ

Solving for the displacement, we have the exact expressions in the transformeddomain (ξ, η, s),

~�u�x ¼Px

c2s

1a2s

� n2Px þ ngPy� 1

s21a2d

� 1a2s

� �ð3:1:7aÞ

~�u�y ¼Py

c2s

1a2s

� ngPx þ g2Py� 1

s21a2d

� 1a2s

� �ð3:1:7bÞ

where two radicals are defined by

ad ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cdÞ2

q; as ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=csÞ2

qð3:1:8Þ

Our main task is to invert the transformed displacement to the original space(x, y, t). Examining the Eqs. (3.1.7), we learn that the inversion of the fourtransformed functions is enough for evaluating the displacement. These fourfundamental functions are

~�I�0 ¼1

c2sa2s

ð3:1:9Þ

~�I�xx ¼ � n2

s21a2d

� 1a2s

� �ð3:1:10Þ


~�I�xy ¼ � ngs2

1a2d

� 1a2s

� �ð3:1:11Þ

~�I�yy ¼ � g2

s21a2d

� 1a2s

� �ð3:1:12Þ

If we could have the four inversions, the displacement is expressed as

ux ¼ PxðI0 þ IxxÞ þ PyIxy; uy ¼ PxIxy þ PyðI0 þ IyyÞ ð3:1:13Þ

Then, we shall consider the inversion of the four fundamentals, successively.

(1) Inversion of ~�I�0Rewriting ~�I�0 as

~�I�0 ¼1c2s

1

n2 þ g2 þ ðs=csÞ2; ð3:1:14Þ

we can find that this equation is the same as Eq. (2.4.8) for the transformed Green’s

function ~�/�with the replacement P → 1/cs

2 and c → cs. Then, we apply the samemathematics as that in Sect. 2.4 and have the inversion,

I0ðx; y; tÞ ¼ 12pcs

Hðcct � rÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � r2

q : ð3:1:15Þ

where H(.) is Heaviside’s unit step function.

(2) Inversion of ~�I�xxAs we have no previous result for the second inversion ~�I�xx, the inversion inte-

grals are carried out successively. The Fourier inversion integral with respect to theparameter η is given by

�I�xx ¼ � n2

2ps2

Z1�1

1a2d

� 1a2s

� �expð�igyÞdg

¼ � n2

ps2

Z10

1

n2 þ g2 þ ðs=cdÞ2� 1

n2 þ g2 þ ðs=csÞ2( )

cosðgyÞdgð3:1:16Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2

The semi-infinite integral is evaluated with the use of the formula (2.1.22) and isarranged so that each term has the same radical,

�I�xx ¼ � 12s2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cdÞ2

q� ðs=cdÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 þ ðs=cdÞ2q

8><>:

9>=>; exp �jyj


q �

þ 12s2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

q� ðs=csÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 þ ðs=csÞ2q

8><>:

9>=>; exp �jyj


q �

ð3:1:17Þ

The next Fourier inversion integral with respect to the parameter ξ is reduced to thesemi-infinite integral as

I�xx ¼ � 12ps2

Z10


q� ðs=cdÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n2 þ ðs=cdÞ2q

8><>:

9>=>; exp �jyj


q �cosðnxÞdn

þ 12ps2

Z10



n2 þ ðs=csÞ2q

8><>:

9>=>; exp �jyj


q �cosðnxÞdn

ð3:1:18Þ

We have already had the integration formula (2.4.13). Here, it is recited again,

Z10





p� �ð3:1:19Þ

where K0(.) is the zeroth order modified Bessel function of the second kind. Thisformula is applicable only for the second term in the bracket; another formula isnecessary for the first term. Differentiating the formula (3.1.19) twice with respectto the parameter “c,” we have the new formula for our use,

Z10

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2

pexp �c

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2

p� �cosðbxÞdx

¼ a2c2

b2 þ c2K0 a


p� �� affiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 þ c2p b2 � c2

b2 þ c2K1 a


p� � ð3:1:20Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2

Applying these two formulas to the integrals in Eq. (3.1.18), we have

I�xx ¼ þ 12pc2d

x2

r2K0 rs=cdð Þ þ 1

2psrcd

x2 � y2

r2K1 rs=cdð Þ

� 12pc2s

x2

r2K0 rs=csð Þ � 1

2psrcs

x2 � y2

r2K1 rs=csð Þ

ð3:1:21Þ

where the radial distance from the source is defined by


pð3:1:22Þ

The last inversion is the Laplace inversion. Its inversion is expressed symboli-cally as

Ixx ¼ þ 12pc2d

x2

r2L�1 K0 rs=cdð Þ½ � þ 1

2prcd

x2 � y2

r2L�1 1

sK1 rs=cdð Þ

�

� 12pc2s

x2

r2L�1 K0 rs=csð Þ½ � � 1

2prcs

x2 � y2

r2L�1 1

sK1 rs=csð Þ

� ð3:1:23Þ

We have already used the inversion formula (2.4.16),

L�1 K0ðasÞ½ � ¼ Hðt � aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � a2

p ð3:1:24Þ

One more inversion formula that includes the modified Bessel function of thesecond kind is that (Erdélyi 1954, vol. I, pp. 277, 11),

L�1 1sK1ðasÞ

� ¼ Hðt � aÞ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � a2

p

að3:1:25Þ

Applying these two inversion formulas to Eq. (3.1.23), we have the final form forIxx,

Ixxðx; y; tÞ ¼ þHðt � r=cdÞ2pcd

x2

r21ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðcdtÞ2 � r2q þ x2 � y2

r4

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � r2

q8><>:

9>=>;

� Hðt � r=csÞ2pcs

x2


ðcstÞ2 � r2q þ x2 � y2

r4

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � r2

q8><>:

9>=>; ð3:1:26Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

(3) Inversion of ~�I�yyThe same inversion procedure as that for ~�I�xx is applied to the inversion of

~�I�yy, andits final form is

Iyyðx; y; tÞ ¼ þHðt � r=cdÞ2pcd

y2


ðcdtÞ2 � r2q þ y2 � x2

r4


q8><>:

9>=>;


y2


ðcstÞ2 � r2q þ y2 � x2

r4


q8><>:

9>=>; ð3:1:27Þ

The difference between Ixx in Eq. (3.1.26) and Iyy in Eq. (3.1.27) should benoticed. The difference is in the space variables only. If we replace x with y in Ixx ofEq. (3.1.26), we could have Iyy of Eq. (3.1.27). This is easily anticipated from thecomparison of the transformed forms in Eqs. (3.1.10) and (3.1.12). In theseequations, each transform parameter corresponds to the original space variable. Ifwe exchange ξ with η in Eq. (3.1.10), it yields Eq. (3.1.12). Similarly, in the originalspace we can exchange the space variables.

(4) Inversion of ~�I�xyThe inversion technique for ~�I�xy is essentially the same as that for the former two

cases, but the integration formulas are slightly different. The Fourier inversionintegral with respect to the parameter η is reduced to the semi-infinite integral

�I�xy ¼ � 12ps2

Z1�1

ng1a2d

� 1a2s

� �expð�igyÞdg ¼ in

ps2

Z10

ga2d

� ga2s

� �sinðgyÞdg

ð3:1:28Þ

The integration formula (Erdélyi 1954, vol. I, pp. 65, 15)

Z10

xx2 þ a2

sinðbxÞdx ¼ p2sgnðbÞ expð�ajbjÞ; ð3:1:29Þ

is applied to Eq. (3.1.28). It then follows that

�I�xy ¼in2s2

sgnðyÞ exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cdÞ2

q �� exp �jyj


q �� ð3:1:30Þ


where sgnð:Þ is the sign function defined by

sgnðxÞ ¼ þ1 ; x[ 0�1 ; x\ 0

ð3:1:31Þ

The next Fourier inversion integral with respect to the parameter ξ

I�xy ¼sgnðyÞ2s2

12p

Zþ1

�1ðþinÞ exp �jyj


q �� exp �jyj


q �� e�inxdn

ð3:1:32Þ

is also reduced to the semi-infinite integral

I�xy ¼sgnðyÞ2ps2

Z10

n exp �jyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cdÞ2

q �� exp �jyj


q �� sinðnxÞdn

ð3:1:33Þ

Applying the integration formula (Erdélyi 1954, vol. I, pp. 75, 35)

Z10

x exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2

p� �sinðcxÞdx ¼ a2bc

b2 þ c2K2 a


p� �ð3:1:34Þ

to the integral in Eq. (3.1.33), we have for Ixy*

I�xy ¼xy

2pr21c2d

K2 rs=cdð Þ � 1c2s

K2 rs=csð Þ �

: ð3:1:35Þ

Lastly, the Laplace inversion formula (Erdélyi 1954, vol. I, pp. 277, 12)

L�1 K2ðbsÞ½ � ¼ Hðt � bÞ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � b2

p þ 2b2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � b2

p� �ð3:1:36Þ

is applied to Eq. (3.1.35). The final form of Ixy is given by

Ixy ¼ þ xy2pr4

ðr=cdÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðr=cdÞ2

q þ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðr=cdÞ2

q8><>:

9>=>;Hðt � r=cdÞ

� xy2pr4

ðr=csÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðr=csÞ2

q þ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðr=csÞ2

q8><>:

9>=>;Hðt � r=csÞ:

ð3:1:37Þ


(5) Green’s dyadic

We have obtained the exact expressions for the three fundamentals, Ixx, Ixy andIyy, and the unified expression for these is given by

Iijðx; y; tÞ ¼ þHðt � r=cdÞ2pcd

xixjr2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdt2Þ � r2p þ 2xixj

r2� dij

� � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdt2Þ � r2p

r2

( )


xixjr2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � r2

q þ 2xixjr2

� dij

� � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � r2

qr2

8><>:

9>=>;

ð3:1:38Þ

where the subscripts i and j stand for x and y, and it should be understood thatxx � x; xy � y. Further, δij is Kronecker’s delta defined by

dij ¼ 1; i ¼ j0; i 6¼ j

ð3:1:39Þ

Thus, we have just obtained the exact expressions for the four fundamental func-tions. Substituting Eqs. (3.1.15) and (3.1.38) into Eq. (3.1.13), the displacement inthe actual space (x, y, t) is expressed as

uiðx; y; tÞ ¼Xj¼x;y

PjHðt � r=cdÞ

2pcd

xixjr2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � r2

q þ 2xixjr2

� dij

� � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � r2

qr2

8><>:

9>=>;

264

�Hðt � r=csÞ2pcs

xixjr2

� dij� � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðcstÞ2 � r2q þ 2xixj

r2� dij


qr2

8><>:

9>=>;375

ð3:1:40Þ

In this equation, the operation of Heaviside’s unit step function determines thedisturbed region. H(t − r/cd) shows a circular cylindrical region disturbed by thedilatational wave which expands with velocity cd, while H(t − r/cs) does that by theshear wave with cs. These two waves are the basic disturbances in the 2D dynamicdeformation and are shown in Fig. 3.1.

Now, we rewrite the above Eq. (3.1.40) as


PjGijðx; y; tÞ ð3:1:41Þ


The function Gij(x, y, t) is called “Green’s dyadic” and it expresses the displacementcomponent in the i-axis direction due to the unit body force in the j-axis direction.The explicit form of the dyadic Gij is given by

Gijðx; y; tÞ ¼ Hðt � r=cdÞ2pcd

xixjr2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � r2

q þ 2xixjr2

� dij

� � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � r2

qr2

8><>:

9>=>;


xixjr2

� dij� � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðcstÞ2 � r2q þ 2xixj

r2� dij


qr2

8><>:

9>=>;

ð3:1:42Þ


When the source is harmonically vibrating with frequency ω, the non-homogeneousbody force term in the displacement equation (3.1.1) is replaced with the time-harmonic source, i.e.

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

� �þ @2ux

@x2þ @2ux

@y2¼ 1

c2s

@2ux@t2

� Qx

c2sdðxÞdðyÞ expðixtÞ

ð3:2:1aÞ

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

� �þ @2uy

@x2þ @2uy

@y2¼ 1

c2s

@2uy@t2

� Qy

c2sdðxÞdðyÞ expðixtÞ

ð3:2:1bÞ

where Qi is the magnitude of the source component.

dc t

sc t

x

yFig. 3.1 Radiation of P andSV waves from a point source


It is possible to apply the integral transform method for obtaining the time-harmonic Green’s dyadic. However, we do not employ the method here since thecumbersome complex integration must be discussed as that for the time-harmonicGreen’s function in Sect. 2.5. We employ a simpler way, i.e. the convolutionintegral. The time-harmonic response can be obtained by the convolution integral ofthe impulsive response as

uiðx; y; tÞ ¼ limt!1

Z t

0

uðimpulseÞi ðx; y; t0Þ expfþixðt � t0Þgdt0 ð3:2:2Þ

where uðimpulseÞi ðx; y; t0Þ is the impulsive response given by Eq. (3.1.41) with(3.1.42), where the source magnitude Pi for the impulsive solution is replaced withQi. In this convolution integral, the time-harmonic function, exp(+iωt), is excludedfrom the limit, but we keep the upper limit of the integral be infinite, since thesteady-state response takes place long time after the initial disturbance. Then, theconvolution integral for the time-harmonic response takes the form

uiðx; y; tÞ ¼ expðþixtÞZt!1

0

uðimpulseÞi ðx; y; t0Þ expð�ixt0Þdt0 ð3:2:3Þ

Substituting Eqs. (3.1.41) with (3.1.42) into Eq. (3.2.3), we have

uiðx; y; tÞ ¼ expðþixtÞXj¼x;y

Qj1

2pcd

xixjr2R1

r=cd

expð�ixt0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcd t0Þ2 � r2

p dt0

þ 2xixjr2 � dij

� �1r2R1

r=cd


qexpð�ixt0Þdt0

8>>><>>>:

9>>>=>>>;

266664

� 12pcs

xixjr2 � dij� R1

r=cs

expð�ixt0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcst0Þ2 � r2

p dt0

þ 2xixjr2 � dij

� �1r2R1

r=cs


qexpð�ixt0Þdt0

8>>><>>>:

9>>>=>>>;

377775

ð3:2:4Þ

Two integrals in the above equation are the integral representations of Hankelfunction of the second kind (Watson 1966, p. 169). They are

Z1a

expð�ixzÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 � a2

p dx ¼ � ip2Hð2Þ

0 ðazÞ

Z1a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 � a2

pexpð�ixzÞdx ¼ þ ipa

2zHð2Þ

1 ðazÞð3:2:5Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

Applying these formulas to the integrals in Eq. (3.2.4), the time-harmonic responseis expressed as


Qjgijðx; yÞ expðþixtÞ ð3:2:6Þ

where gij(x, y) is the Green’s dyadic for the time-harmonic response and is given by

gijðx; yÞ ¼ � i4

1c2d

xixjr2

Hð2Þ0 ðrx=cdÞ þ dij � 2xixj

r2

� �cdrx

Hð2Þ1 ðrx=cdÞ

��

� 1c2s

xixjr2

� dij� �

Hð2Þ0 ðrx=csÞ þ dij � 2xixj

r2

� �csrx

Hð2Þ1 ðrx=csÞ

� ð3:2:7Þ

Then, Eq. (3.2.6) with Eq. (3.2.7) gives the particular solution of the displacementequation (3.2.1).

If the frequency is negative, ω → −ω( =ωe−πi), and the time factor is exp(−iωt),we apply the formula (Watson 1966, p. 75)

Hð2Þn ðxe�piÞ ¼ ð�1Þnþ1Hð1Þ

n ðxÞ ð3:2:8Þ

to the Hankel function in Eq. (3.2.7), and then the displacement is expressed as


Qjgijðx; yÞ expð�ixtÞ ð3:2:9Þ

and its dyadic gij (x, y) is given by

gijðx; yÞ ¼ þ i4

1c2d

xixjr2

Hð1Þ0 ðrx=cdÞ þ dij � 2xixj

r2

� �cdrx

Hð1Þ1 ðrx=cdÞ

��

� 1c2s

xixjr2

� dij� �

Hð1Þ0 ðrx=csÞ þ dij � 2xixj

r2

� �csrx

Hð1Þ1 ðrx=csÞ

� ð3:2:10Þ

where Hn(1)(.) is the n-th order Hankel function of the first kind and defined by

Hð1Þn ðxÞ ¼ JnðxÞ þ iYnðxÞ ð3:2:11Þ


If we consider to derive the static solution from the time-harmonic solution, wehave to take the limit ω → 0 in the dyadic given by Eq. (3.2.7) or (3.2.10).However, the limit does not exist since the static displacement in the 2D plane


deformation does not vanish at infinity. We have to start from the original differ-ential equation and solve it. The static source is a non-time-dependent point bodyforce and the displacement equation which has no inertia term is given by

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

� �þ @2ux

@x2þ @2ux

@y2¼ � Sx

c2sdðxÞdðyÞ

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

� �þ @2uy

@x2þ @2uy

@y2¼ � Sy

c2sdðxÞdðyÞ

ð3:3:1Þ

where the static body force is placed at the coordinate origin and its magnitude is Si.For these coupled differential equations, the convergence condition at infinity

uj�� ffiffiffiffiffiffiffiffiffi

x2þy2p

!1¼ @uj@x


p!1

¼ @uj@y


p!1

¼ 0 ð3:3:2Þ

is assumed in order to apply the Fourier transform (but, the final solution does notsatisfy the first condition in the above equation). We should understand that theconvergence condition guarantees the application of the Fourier transform.

In order to obtain the static Green’s dyadic, we apply the double Fouriertransform with respect to two space variables. Applying the double Fourier trans-form defined by Eq. (3.1.3) to the displacement equations (3.3.1), we have thealgebraic equations for the transformed displacement components ~�uj,

�inðc2 � 1Þ �in~�ux � ig~�uy� � n2 þ g2

� ~�ux ¼ �Sx=c

2s

�igðc2 � 1Þ �in~�ux � ig~�uy� � n2 þ g2

� ~�uy ¼ �Sy=c

2s

ð3:3:3Þ

and their solutions are

~�ux ¼ Sxc2s

1

n2 þ g2� c2 � 1

c2n2

ðn2 þ g2Þ2( )

� Syc2s

c2 � 1c2

ng

ðn2 þ g2Þ2

~�uy ¼ � Sxc2s

c2 � 1c2

ng

ðn2 þ g2Þ2 þSyc2s

1

n2 þ g2� c2 � 1

c2g2

ðn2 þ g2Þ2( ) ð3:3:4Þ

Examining the above equations, we learn that four inversions are needed for the fullFourier inversion since the displacement is given by

~�ux ¼ Sxc2s

~�I0 � c2 � 1c2

~�Ixx

� �� Sy

c2s

c2 � 1c2

~�Ixy

~�uy ¼ � Sxc2s

c2 � 1c2

~�Ixy þ Syc2s

~�I0 � c2 � 1c2

~�Iyy

� � ð3:3:5Þ


where the four fundamentals to be inverted are

~�I0 ¼ 1

n2 þ g2; ~�Ixx ¼ n2

ðn2 þ g2Þ2 ;~�Iyy ¼ g2

ðn2 þ g2Þ2 ;~�Ixy ¼ ng

ðn2 þ g2Þ2ð3:3:6Þ

The inversion for each fundamental is carried out in the following subsections.

(1) Inversion of ~�I0

We have already inverted this function in Sect. 2.3. The result is given byEq. (2.3.21), i.e.

~�/ ¼ S

n2 þ g2) / ¼ � S

4plog ðx2 þ y2Þ þ const: ð3:3:7Þ

Applying this result, we have the inversion for ~�I0 as

I0 ¼ � 14p

log ðx2 þ y2Þ þ const: ð3:3:8Þ

(2) Inversion of ~�Ixx

The Fourier inversion integral with respect to the parameter η is reduced to thesemi-infinite integral,

�Ixx ¼ 12p

Zþ1

�1

n2

ðn2 þ g2Þ2 expð�igyÞdg ¼ n2

p

Z10

1

ðn2 þ g2Þ2 cosðgyÞdg ð3:3:9Þ

The integration formula (Gradshteyn and Ryzhik 1980, pp. 449, 3.729 1)

Z10

1

ðx2 þ b2Þ2 cosðaxÞdx ¼p4b3

ð1þ abÞ expð�abÞ; a[ 0; b[ 0 ð3:3:10Þ

is applied to the far right integral in Eq. (3.3.9). We have for �Ixx

�Ixx ¼ n2

p

Z10

1

ðn2 þ g2Þ2 cosðgyÞdg ¼ 14jnj þ

jyj4

� �expð�jnjjyjÞ ð3:3:11Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2

The last Fourier inversion with respect to the parameter ξ is given by the sum of twointegrals,

Ixx ¼ 12p

Zþ1

�1

14jnj þ

jyj4

� �expð�jnjjyjÞ expð�inxÞdn

¼ 14p

Z10

1nexpð�njyjÞ cosðnxÞdnþ jyj

4p

Z10

expð�njyjÞ cosðnxÞdnð3:3:12Þ

Inspecting the two integrals in the last equation, the second integral can be eval-uated by applying the formula (2.3.16) in Chap. 2. The first integral, however, hasno formula since its integrand has the first order singularity at ξ = 0 and it isimpossible to evaluate the integral in this form. We extract the first integral

Ið1Þxx ¼ 14p

Z10

1nexpð�njyjÞ cosðnxÞdn ð3:3:13Þ

and consider its derivative with respect to each space variable as

@Ið1Þxx

@x¼ � 1

4p

Z10

expð�njyjÞ sinðnxÞdn

@Ið1Þxx

@jyj ¼ � 14p

Z10

expð�njyjÞ cosðnxÞdnð3:3:14Þ

These two integrals are easily evaluated by the formulas (2.3.16) to yield

@Ið1Þxx

@x¼ � 1

4px

x2 þ y2;

@Ið1Þxx

@jyj ¼ � 14p

jyjx2 þ y2

ð3:3:15Þ

Then, we return the two derivatives to the original one by the integration withrespect to each space variable. Two expressions for the single Ixx

(1) are obtained as

Ið1Þxx ¼ � 14p

Zx

x2 þ y2dx ¼ � 1

8plogðx2 þ y2Þ þ C1ðyÞ

Ið1Þxx ¼ � 14p

Z jyjx2 þ y2

djyj ¼ � 18p

logðx2 þ y2Þ þ C2ðxÞ ð3:3:16Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

http://dx.doi.org/10.1007/978-3-319-17455-6_2

These two equations must be equal. This condition is satisfied by settingC1(y) = C2(x) and thus the two functions must be a pure constant,

C1ðyÞ ¼ C2ðxÞ ¼ const: ð3:3:17Þ

Then, we have for Ixx(1),

Ið1Þxx ¼ � 18p

logðx2 þ y2Þ þ const: ð3:3:18Þ

The second integral in Eq. (3.3.12) is easily evaluated by the formula (2.3.16)and it yields

Ið2Þxx ¼ jyj4p

Z10

expð�njyjÞ cosðnxÞdn ¼ 14p

y2

x2 þ y2ð3:3:19Þ

Finally, substituting Eqs. (3.3.18) and (3.3.19) into (3.3.12), we have the inversion

Ixx ¼ � 18p

logðx2 þ y2Þ þ 14p

y2

x2 þ y2þ const: ð3:3:20Þ

Further, since the constant in the above equation is arbitrary, we can rewrite Ixx as

Ixx ¼ � 18p

logðx2 þ y2Þ � 14p

x2


(3) Inversion of ~�Iyy

Rewriting the transformed equation in Eq. (3.3.5), we learn that this inversioncan be decomposed into the sum of the former two functions, as

~�Iyy ¼ 1

n2 þ g2� n2

ðn2 þ g2Þ2 ¼~�I0 � ~�Ixx ð3:3:22Þ

Then, we can apply the former results, Eqs. (3.3.8) and (3.3.21), to the aboveequation,

Iyy ¼ I0 � Ixx

¼ � 14p

logðx2 þ y2Þ þ const:þ 18p


y2

x2 þ y2þ const:

ð3:3:23Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

The final form of Iyy is given by

Iyy ¼ � 18p


y2


This can be derived from Ixx in Eq. (3.3.21) by the exchange of variables. Changingx with y in Eq. (3.3.21), we have the inversion for Iyy.

(4) Inversion of ~�Ixy

The Fourier inversion integral with respect to the parameter η is reduced to thesemi-infinite integral,

�Ixy ¼ 12p

Zþ1

�1

ng

ðn2 þ g2Þ2 expð�igyÞdg ¼ � inp

Z10

g

ðn2 þ g2Þ2 sinðgyÞdg ð3:3:25Þ

The integration formula (Gradshteyn and Ryzhik 1980, pp. 449, 3.729 2),

Z10

x

ðx2 þ b2Þ2 sinðaxÞdx ¼pa4b

expð�abÞ ð3:3:26Þ

is applied to the last integral in Eq. (3.3.25). We have

�Ixy ¼ � iy4sgnðnÞ expð�jnjjyjÞ ð3:3:27Þ

The next Fourier inversion integral with respect to the parameter ξ is reduced to thesimple integral,

Ixy ¼ � iy4

12p

Zþ1

�1sgnðnÞ expð�jnjjyjÞ expð�inxÞdn

¼ � y4p

Z10

expð�njyjÞ sinðnxÞdnð3:3:28Þ

Applying the formula (2.3.16), we have for Ixy,

Ixy ¼ � 14p

xyx2 þ y2

ð3:3:29Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

(5) 2D Kelvin’s solution

The four inversions in Eq. (3.3.5) have thus been completed. Since the dis-placement in the original space is given by

ux ¼ Sxc2s

I0 � c2 � 1c2

Ixx

� �� Sy

c2s

c2 � 1c2

Ixy

uy ¼ � Sxc2s

c2 � 1c2

Ixy þ Syc2s

I0 � c2 � 1c2

Iyy

� � ð3:3:30Þ

we substitute I0 and Iij given by Eqs. (3.3.8), (3.3.21), (3.3.24) and (3.3.29) into theabove Eq. (3.3.30). The final form for the displacement is given by

ux ¼ Sx4pc2d

�ðc2 þ 1Þ logðrÞ þ ðc2 � 1Þ xr

� �2 �þ Sy4pc2d

ðc2 � 1Þ xyr2

uy ¼ Sx4pc2d

ðc2 � 1Þ xyr2

þ Sy4pc2d

�ðc2 þ 1Þ logðrÞ þ ðc2 � 1Þ yr

� �2 � ð3:3:31Þ

where cd = γcs defined by Eq. (3.8) is used. The constant term is omitted since itgives a simple rigid motion with no strain. The reader should notice that theconstant term in the displacement breaks the applicability of the Fourier transform;thus the last inversion integral with respect to the parameter ξ has the singular point.

The expression for the displacement is rewritten in terms of the dyadic,

ux ¼ SxgðstaticÞxx ðx; yÞ þ Syg

ðstaticÞxy ðx; yÞ

uy ¼ SxgðstaticÞxy ðx; yÞ þ Syg

ðstaticÞyy ðx; yÞ

ð3:3:32Þ

where the dyadic for the static source is given by

gðstaticÞxx ðx; yÞ ¼ 14pc2d


� �2 �

gðstaticÞxy ðx; yÞ ¼ 14pc2d

ðc2 � 1Þ xyr2

gðstaticÞyy ðx; yÞ ¼ 14pc2d


� �2 � ð3:3:33Þ

This static Green’s dyadic for the plane deformation is called the “two dimensionalKelvin’s solution.”



An impulsive point source placed at the coordinate origin is expressed by the bodyforce

Bx

By

Bz

0@

1A ¼

Px

Py

Pz

0@

1AdðxÞdðyÞdðzÞdðtÞ; ð3:4:1Þ

where Pi is the magnitude in the i-direction. The displacement equation with thisimpulsive source is given by

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

þ @uz@z

� �þ @2ux

@x2þ @2ux

@y2þ @2ux

@z2¼ 1

c2s

@2ux@t2

� Px

c2sdðxÞdðyÞdðzÞdðtÞ

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

þ @uz@z

� �þ @2uy

@x2þ @2uy

@y2þ @2uy

@z2¼ 1

c2s

@2uy@t2

� Py


ðc2 � 1Þ @@z

@ux@x

þ @uy@y

þ @uz@z

� �þ @2uz

@x2þ @2uz

@y2þ @2uz

@z2¼ 1

c2s

@2uz@t2

� Pz


ð3:4:2Þ

where cs is the shear wave velocity and the velocity ratio γ is defined by Eq. (3.8).On these displacement equations, we impose the quiescent condition at an initialtime,

uijt¼0 ¼@ui@t

jt¼0 ¼ 0 ð3:4:3Þ


uij ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2p

!1 ¼ @ui@x

j ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2

p!1 ¼ @ui

@yj ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2þy2þz2p

!1 ¼ @ui@z

j ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2

p!1 ¼ 0

ð3:4:4Þ

Laplace transform with respect to the time variable,

u�j ¼Z10

uj expð�stÞdt ð3:4:5Þ


and the triple Fourier transform with respect to three space variables,

�uj ¼Zþ1

�1uj expðþinxÞdx; uj ¼ 1

2p

Zþ1

�1�uj expð�inxÞdn

~uj ¼Zþ1

�1uj expðþigyÞdy; uj ¼ 1

2p

Zþ1

�1~uj expð�igyÞdg

uj ¼Zþ1

�1uj expðþifzÞdz; uj ¼ 1

2p

Zþ1

�1uj expð�ifzÞdf

ð3:4:6Þ

are applied to the displacement equations (3.4.2). The simple algebraic equationsfor the transformed displacement are

ðc2 � 1Þ �n2~�u�x � ng~�u�y � nf~�u�z� �

� ðn2 þ g2 þ f2Þ~�u�x ¼scs

� �2

~�u�x �Px

c2s

ðc2 � 1Þ �ng~�u�x � g2~�u�y � gf~�u�z� �

� ðn2 þ g2 þ f2Þ~�u�y ¼scs

� �2

~�u�y �Py

c2s

ðc2 � 1Þ �nf~�u�x � gf~�u�y � f2~�u�z� �

� ðn2 þ g2 þ f2Þ~�u�z ¼scs

� �2

~�u�z �Pz

c2s

ð3:4:7Þ

and the displacement components in the transformed domain are given by

~�u�x ¼Px

c2sb2s

� n2Px þ ngPy þ nfPz� 1

s21

b2d� 1

b2s

!

~�u�y ¼Py

c2sb2s

� ngPx þ g2Py þ gfPz� 1

s21

b2d� 1

b2s

!

~�u�z ¼Pz

c2sb2s

� nfPx þ gfPy þ f2Pz� 1

s21

b2d� 1

b2d

!ð3:4:8Þ

where the radicals are defined by

bd ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ f2 þ ðs=cdÞ2

q; bs ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ f2 þ ðs=csÞ2

qð3:4:9Þ

In order to explore the convenient way for the inversion of the transformeddisplacement, we inspect the expression of Eq. (3.4.8). It is found that seveninversion formulas are necessary.


They are

~�I�0ðn; g; 1; sÞ ¼

1

c2sb2s

; ð3:4:10Þ

~�I�11ðn; g; f; sÞ ¼

n2

s21

b2d� 1

b2s

!; ~�I

�12ðn; g; f; sÞ ¼

g2

s21

b2d� 1

b2s

!;

~�I�13ðn; g; f; sÞ ¼

f2

s21

b2d� 1

b2s

! ð3:4:11Þ

~�I�21ðn; g; f; sÞ ¼

ngs2

1

b2d� 1

b2s

!; ~�I

�22ðn; g; f; sÞ ¼

gfs2

1

b2d� 1

b2s

!;

~�I�23ðn; g; f; sÞ ¼

nfs2

1

b2d� 1

b2s

! ð3:4:12Þ

The first inversion for ~�I�0 is the same as that of the 3D wave equation in Sect. 2.7.

Applying that result to our inversion, we have

I0ðx; y; z; tÞ ¼ 14pc2sR

dðt � R=csÞ ð3:4:13Þ

where the 3D radial distance is defined by


pð3:4:14Þ

Subsequently, we consider the inversion of ~�I�1j in Eq. (3.4.11). The three Fourier

transforms are the same in the form of definition integral and thus the exchange oftwo space variables is equivalent to the exchange of two integration parameters.That is

~�I�12ðn; g; f; sÞ ¼ ~�I

�11ðg; n; f; sÞ ) I12ðx; y; z; tÞ ¼ I11ðy; x; z; tÞ

~�I�13ðn; g; f; sÞ ¼ ~�I

�11ðf; g; n; sÞ ) I13ðx; y; z; tÞ ¼ I11ðz; y; x; tÞ

ð3:4:15Þ

Due to this exchangeability, if we could obtain the inversion for ~�I�11 as F(x, y, z, t), the

inversion of ~�I�12 is given by the exchange of the space variables as F(y, x, z, t) and the


http://dx.doi.org/10.1007/978-3-319-17455-6_2

inversion of ~�I�13 is also given by the variable exchange as F(z, y, x, t). Thus, we can

draw a schematic inversion diagram as

~�I�11ðn; g; f; sÞ ¼

n2

s21

b2d� 1

b2s

!) I11ðx; y; z; tÞ ¼ Fðx; y; z; tÞ

# exchange n with g # exchange x with y

~�I�12ðn; g; f; sÞ ¼

g2

s21

b2d� 1

b2s

!) I12ðx; y; z; tÞ ¼ Fðy; x; z; tÞ

ð3:4:16Þ

Similarly, the third inversion ~�I�13 is also obtained by the exchange of the space

variables as

~�I�11ðn; g; f; sÞ ¼

n2

s21

b2d� 1

b2s

!) I11ðx; y; z; tÞ ¼ Fðx; y; z; tÞ

# exchange n with f # exchange x with z

~�I�13ðn; g; f; sÞ ¼

f2

s21

b2d� 1

b2s

!) I13ðx; y; z; tÞ ¼ Fðz; y; x; tÞ

ð3:4:17Þ

Then, the necessary inversion formula is reduced to only one, that is

~�I�11ðn; g; f; sÞ ¼

n2

s21

b2d� 1

b2s

!ð3:4:18Þ

As for the second group of the inversion in Eq. (3.4.12), we learn that thenecessary inversion formula is

~�I�21ðn; g; f; sÞ ¼

ngs2

1

b2d� 1

b2s

!ð3:4:19Þ

If we could get the inversion as G(x, y, z, t), the other two are given by the exchangeof the space variables. They are

~�I�21ðn; g; f; sÞ ¼

ngs2

1

b2d� 1

b2s

!) I21ðx; y; z; tÞ ¼ Gðx; y; z; tÞ

# exchange n with f # exchange x with z

~�I�22ðn; g; f; sÞ ¼

gfs2

1

b2d� 1

b2s

!) I22ðx; y; z; tÞ ¼ Gðz; y; x; tÞ

ð3:4:20Þ


and

~�I�21ðn; g; f; sÞ ¼

ngs2

1

b2d� 1

b2s

!) I21ðx; y; z; tÞ ¼ Gðx; y; z; tÞ

# exchange g with f # exchange y with z

~�I�23ðn; g; f; sÞ ¼

nfs2

1

b2d� 1

b2s

!) I23ðx; y; z; tÞ ¼ Gðx; z; y; tÞ

ð3:4:21Þ

Consequently, it is sufficient to develop the inversion formulas only for ~�I�11 and

~�I�21.

(1) Inversion of ~�I�11

The formal Fourier inversion integral with respect to the parameter f is given by

~�I�11ðn; g; z; sÞ ¼12p

Zþ1

�1

n2

s21

b2d� 1

b2s

!expð�ifzÞdf

¼ n2

ps2

Z10

1

f2 þ a2d� 1

f2 þ a2s

!cosðfzÞdf

ð3:4:22Þ

where

aj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cjÞ2

q; j ¼ d; s ð3:4:23Þ

Applying the simple integration formula (2.1.22) to the integral in Eq. (3.4.22), wehave

~�I�11ðn; g; z; sÞ ¼n2

2s21ad

expð�adjzjÞ � 1asexpð�asjzjÞ

�ð3:4:24Þ

The second inversion integral with respect to the parameter η is given by

�I�11ðn; y; z; sÞ ¼12p

Zþ1

�1

n2

2s21ad


�expð�igyÞdg

¼ n2

2ps2

Z10

exp �jzjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cdÞ2

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cdÞ2

q �exp �jzj

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=csÞ2

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=csÞ2

q2664

3775 cosðgyÞdg

ð3:4:25Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

The integration formula (2.4.13) in Chap. 2,

Z10





p� �ð3:4:26Þ

is applied to the integral in Eq. (3.4.25). We have for �I�11

�I�11ðn; y; z; sÞ ¼n2

2ps2K0 �r


q� �� K0 �r


q� � �ð3:4:27Þ

where K0(.) is the zeroth order modified Bessel function of the second kind and

�r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiy2 þ z2

pð3:4:28Þ

The last Fourier inversion integral with respect to the parameter ξ is given by

I�11ðx; y; z; sÞ ¼12p

Zþ1

�1

n2

2ps2K0 �r


q� �� K0 �r


q� � �expð�inxÞdn

¼ 12p2s2

Z10

n2 K0 �rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cdÞ2

q� �� K0 �r


q� � �cosðnxÞdn

ð3:4:29Þ

The last integral, which includes the modified Bessel function, has not yet beentabulated in reference books. However, we have one formula which resembles ourintegral, i.e.x

I ¼Z10

K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ c2

p� �cosðbxÞdx ¼ p

2exp �c

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ b2

p� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ b2

p ð3:4:30�Þ

*Notice: the above equation is just the Fourier cosine inversion of Eq. (3.4.26)!If we differentiate this integration formula with respect to the parameter “b”

twice, it yields to our necessary formula,

� @2I@b2

¼Z10

x2K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ c2

p� �cosðbxÞdx ¼ � p

2@2

@b2exp �c


p� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ b2

p( )

ð3:4:31Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

That is,

Z10

x2K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ c2

p� �cosðbxÞdx

¼ p2

1a2 þ b2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ b2

p þ c

� �1� 3b2

a2 þ b2

� �� b2c2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ b2p

�exp �c


p� �ð3:4:32Þ

Applying this new formula (3.4.32) to the integral in Eq. (3.4.29), we have

I�11ðx; y; z; sÞ ¼1

4pR2 1� 3x2

R2

� �1Rs2

þ 1� 3x2

R2

� �1cds

� x2

c2dR

�expf�sðR=cdÞg

�

� 1� 3x2

R2

� �1Rs2

þ 1� 3x2

R2

� �1css

� x2

c2s R

�expf�sðR=csÞg

ð3:4:33Þ

where the 3D radial distance R is defined by Eq. (3.4.14).The last Laplace inversion is carried out by applying the inversion formulas,


L�1 1sexpð�asÞ

� ¼ Hðt � aÞ ¼ 1; t[ a

0; t\ a

ð3:4:35Þ

L�1 1s2expð�asÞ

� ¼ Hðt � aÞðt � aÞ ¼ t � a; t[ a

0; t\ a

ð3:4:36Þ

where δ(.) and H(.) are Dirac’s delta and Heaviside’s unit step functions defined inSect. 1.2. Finally, the application of Laplace inversion formulas leads to the explicitexpression for the inversion:

I11ðx; y; z; tÞ ¼ 14pR

� x2

c2dR2dðt � R=cdÞ þ x2

c2s R2 dðt � R=csÞ

þ 1� 3x2

R2

� �tR2 Hðt � R=cdÞHðR=cs � tÞ

� ð3:4:37Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

The other inversions for ~�I�12 and

~�I�13 can be obtained by the simple exchange of the

space variables as shown in Eqs. (3.4.16) and (3.4.17). They are


� y2

c2dR2dðt � R=cdÞ þ y2


þ 1� 3y2

R2


� ð3:4:38Þ


� z2

c2dR2dðt � R=cdÞ þ z2


þ 1� 3z2

R2


� ð3:4:39Þ

(2) Inversion of ~�I�21

The Fourier inversion integral with respect to the parameter ζ

~�I�21ðn; g; z; sÞ ¼12p

Zþ1

�1

ngs2

1

b2d� 1

b2s

!expð�ifzÞdf

¼ ngps2

Z10

1

f2 þ a2d� 1

f2 þ a2s

!cosðfzÞdf

ð3:4:40Þ

is evaluated by applying the formula (2.1.22) in Chap. 2. It is

~�I�21ðn; g; z; sÞ ¼ng2s2

1ad


�ð3:4:41Þ

The second inversion integral with respect to the parameter η is given by

�I�21ðn; y; z; sÞ ¼12p

Zþ1

�1

ng2s2

1ad


�expð�igyÞdg

¼ �in2ps2

Z10

gffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ c2d

p exp �jzjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ c2d

q� �� gffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

g2 þ c2sp exp �jzj

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ c2s

q� �( )sinðgyÞdg

ð3:4:42Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_2

where

cj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cjÞ2

q; j ¼ d; s ð3:4:43Þ

We apply the integration formula (Erdélyi 1954, vol. I, pp. 75, 36),

Z10

xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ a2


p� �sinðbxÞdx ¼ abffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 þ c2p K1 a


p� �

ð3:4:44Þ

to the integral in Eq. (3.4.42). It follows that

�I�21ðn; y; z; sÞ ¼y

2p�rs2�in


qK1 �r


q� �

þ inffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

qK1 �r


q� �� ð3:4:45Þ

where �r has already been defined by Eq. (3.4.28). The last Fourier inversion integralwith respect to the parameter ξ is given by

I�21ðn; y; z; sÞ ¼y

2p�rs212p

Zþ1

�1�in


qK1 �r


q� �

þ inffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

qK1 �r


q� ��expð�inxÞdn

¼ � y2p2�rs2

Z10

nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cdÞ2

qK1 �r


q� �

� nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

qK1 �r


q� ��sinðnxÞdn

ð3:4:46Þ

We have the formula (Erdélyi 1954, vol. I, pp. 113, 45)

Z10

xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ b2

pK1 a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ b2

p� �sinðcxÞdx

¼ p2

ab2c

ða2 þ c2Þ3=21þ 3

bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p þ 3b2ða2 þ c2Þ

�exp �b

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p� � ð3:4:47Þ


The integrals in Eq. (3.4.46) are evaluated and we have

I�21ðx; y; z; sÞ ¼ � xy4pR3

1c2d

1þ 3cdsR

þ 3c2ds2R2

�expf�sðR=cdÞg

�

� 1c2s

1þ 3cssR

þ 3c2ss2R2

�expf�sðR=csÞg

ð3:4:48Þ

The Laplace inversion is carried out by applying the formulas (3.4.34)–(3.4.36).Then, the final form of I21(x, y, z, t) is given by

I21ðx; y; z; tÞ ¼ � xy4pR3

1c2d

dðt � R=cdÞ � 1c2s

dðt � R=csÞ þ 3tR2 Hðt � R=cdÞHðR=cs � tÞ

�

ð3:4:49Þ

We have just obtained the exact expressions for the two principal parts: I11 andI21. Other two inversions can be obtained by the exchange of the space variables asshown in Eqs. (3.4.20) and (3.4.21). They are

I22ðx; y; z; tÞ ¼ � yz4pR3

1c2d



�

ð3:4:50Þ

I23ðx; y; z; tÞ ¼ � xz4pR3

1c2d



�

ð3:4:51Þ


We have just obtained the exact inversion of the fundamentals. InspectingEq. (3.4.8), we learn that the displacement components can be expressed in terms ofIij(x, y, z, t) as

ux ¼ Px I0ðx; y; z; tÞ � I11ðx; y; z; tÞf g þ Py �I21ðx; y; z; tÞf g þ Pz �I23ðx; y; z; tÞf guy ¼ Px �I21ðx; y; z; tÞf g þ Py I0ðx; y; z; tÞ � I12ðx; y; z; tÞf g þ Pz �I22ðx; y; z; tÞf guz ¼ Px �I23ðx; y; z; tÞf g þ Py �I22ðx; y; z; tÞf g þ Pz I0ðx; y; z; tÞ � I13ðx; y; z; tÞf g

ð3:4:52Þ


Substituting Eqs. (3.4.13), (3.4.37–3.4.39) and (3.4.49–3.4.51) into Eq. (3.4.52), wehave the following explicit expressions for the displacement components:

ux ¼ Px

4pR

x2

c2dR2dðt � R=cdÞ þ 1

c2s1� x2

R2

� �dðt � R=csÞ

� tR2 1� 3x2

R2

� �Hðt � R=cdÞHðR=cs � tÞ

8>>><>>>:

9>>>=>>>;

þ Py

4pRxyc2dR

2dðt � R=cdÞ � xy

c2s R2 dðt � R=csÞ þ 3xyt

R4 Hðt � R=cdÞHðR=cs � tÞ �

þ Pz

4pRxz

c2dR2dðt � R=cdÞ � xz

c2s R2 dðt � R=csÞ þ 3xzt


ð3:4:53aÞ

uy ¼ Px

4pRxyc2dR

2dðt � R=cdÞ � xy

c2sR2 dðt � R=csÞ þ 3xyt


þ Py

4pR

y2


c2s1� y2

R2


� tR2 1� 3y2

R2


8>>><>>>:

9>>>=>>>;

þ Pz

4pRyz

c2dR2dðt � R=cdÞ � yz

c2sR2 dðt � R=csÞ þ 3yzt


ð3:4:53bÞ

uz ¼ Px

4pRxz

c2dR2dðt � R=cdÞ � xz

c2sR2 dðt � R=csÞ þ 3xzt


þ Py

4pRyz

c2dR2dðt � R=cdÞ � yz

c2sR2 dðt � R=csÞ þ 3yzt


þ Pz

4pR

z2


c2s1� z2

R2


� tR2 1� 3z2

R2


8>>><>>>:

9>>>=>>>;

ð3:4:53cÞ

In terms of the Green’s dyadic, we can rewrite the displacement components as

ui ¼Xj¼x;y;z

PjGijðx; y; z; tÞ; i ¼ x; y; z ð3:4:54Þ


where the dyadic Gijðx; y; z; tÞ is given by

Gijðx; y; z; tÞ ¼ 14pR

xixjc2dR

2dðt � R=cdÞ þ 1

c2sdij � xixj

R2


� tR2 dij � 3xixj

R2


8>><>>:

9>>=>>; ð3:4:55Þ

It is clear that there are two spherical waves. Their fronts are given bycdt = R and cst = R for dilatational and shear waves, respectively. We also find thatthe displacement has the singularity of the delta function (i.e. the first order sin-gularity) at their wave fronts.


When the source is a time-harmonic vibration, the body force with magnitude Qi isgiven by

Bx

By

Bz

0@

1A ¼

Qx

Qy

Qz

0@

1AdðxÞdðyÞdðzÞ expðþixtÞ; ð3:5:1Þ

and the displacement equation is

ðc2 � 1Þ @

@x@ux@x

þ @uy@y

þ @uz@z

� �þ @2ux

@x2þ @2ux

@y2þ @2ux

@z2

¼ 1c2s

@2ux@t2

� Qx

c2sdðxÞdðyÞdðzÞ expðixtÞ

ðc2 � 1Þ @

@y@ux@x

þ @uy@y

þ @uz@z

� �þ @2uy

@x2þ @2uy

@y2þ @2uy

@z2

¼ 1c2s

@2uy@t2

� Qy


ðc2 � 1Þ @@z

@ux@x

þ @uy@y

þ @uz@z

� �þ @2uz

@x2þ @2uz

@y2þ @2uz

@z2

¼ 1c2s

@2uz@t2

� Qz


ð3:5:2Þ

In this section we do not employ the method of the integral transform for solvingthe displacement equations. Instead, we take a short-cut. That is the convolution


integral with the impulsive Green’s dyadicobtained in the previous section. Whenwe express the displacement in terms of the time-harmonic Green’s dyadic,

ui ¼Xj¼x;y;z

Qjgijðx; y; z; tÞ; i ¼ x; y; z ð3:5:3Þ

the dyadic for the time-harmonic response can be derived by the convolutionintegral of the impulsive dyadic given by Eq. (3.4.55). The convolution integral isevaluated as follows

gijðx; y; z; tÞ ¼ limt!1

Z t

0

Gijðx; y; z; t0Þ expfþixðt � t0Þgdt0

¼ expðþixtÞZt!1

0

Gijðx; y; z; t0Þ expð�ixt0Þdt0ð3:5:4Þ

Substituting Eq. (3.4.55) into the last integral in the above equation, we can easilyperform the integration and obtain the time-harmonic Green’s dyadic gij as

gijðx; y; z; tÞ

¼ 14pR

xixjc2dR

2þ 1

ðxRÞ2 1þ ixRcd

� �dij � 3xixj

R2

� �( )expfixðt � R=cdÞg

þ 14pR

1c2s

dij � xixjR2

� �� 1

ðxRÞ2 1þ ixRcs

� �dij � 3xixj

R2

� �( )expfixðt � R=csÞg

ð3:5:5Þ


When a static point force is placed at the coordinate origin, the body force withmagnitude Si is given by

Bx

By

Bz

0@

1A ¼

SxSySz

0@

1AdðxÞdðyÞdðzÞ ð3:6:1Þ


The static deformation can be considered as the time-harmonic response with zero-frequency. Taking the limit ω → 0 in Eq. (3.5.4), the displacement can beexpressed in terms of the static dyadic as

ui ¼Xj¼x;y;z

SjGðKÞij ðx; y; zÞ; i ¼ x; y; z ð3:6:2Þ

where the source magnitude Qi is replaced with Si and the static dyadic GðKÞij ðx; y; zÞ

is derived from the limit and is given by

GðKÞij ðx; y; zÞ ¼ lim

x!0gijðx; y; z; tÞ ¼ 1

8pc2dRðc2 þ 1Þdij þ ðc2 � 1Þ xixj

R2

n oð3:6:3Þ

Since the velocity ratio can be replaced with the simple function of Poisson ratio νas

c2 ¼ 2ð1� mÞ1� 2m

ð3:6:4Þ

the dyadic can also be rewritten as

GðKÞij ðx; y; zÞ ¼ 1

8pð1� 2mÞc2dRð3� 4mÞdij þ xixj

R2

n oð3:6:5Þ

This dyadic is called “Kelvin’s solution” for the static deformation.

3.7 Torsional Source

This section shows two impulsive Green’s functions for the axisymmetric torsionproblem. The first one is corresponding to a circular ring force, and the second to apoint torque. We take the cylindrical coordinate system (r, θ, z) in an infinite elasticsolid and assume that the symmetry axis of the deformation/torsion is the z-axis.The ring force lies on the z = 0 plane and its symmetry axis is also the z-axis. On theother hand, the point torque is place on the coordinate origin. Both produce theaxisymmetric torsional deformation.

Due to the axisymmetric nature of the deformation, the non-vanishing dis-placement component is the circumferential displacement uθ only, and the gov-erning equation for the axisymmetric torsional deformation is given by

@2uh@r2

þ 1r@uh@r

� uhr2

þ @2uh@z2

� 1c2s

@2uh@t2

¼ � 1c2s

Bhðr; z; tÞ ð3:7:1Þ


where cs is the shear (torsional) wave velocity and Bθ is the circumferential bodyforce which represents the ring or point source. When the source is an impulsivering force with radius a, the body force is defined by

Bhðr; z; tÞ ¼ Qdðr � aÞ

rdðzÞdðtÞ ð3:7:2Þ

where δ(·) is Dirac’s delta function and Q is the magnitude of the source. On theother hand, when the source is an impulsive point torque, the body force is definedby the limit of the ring force as

Bhðr; z; tÞ ¼ T2pq

lima!0

dðr � aÞar

� dðzÞdðtÞ ð3:7:3Þ

where T is the magnitude of the torque. As we have no suitable mathematicalformula for the limit in the square bracket, the limiting form for the point torquesource is retained. In this section, we shall discuss Green’s functions for these twosources, separately.

3.7.1 Ring Source

In order to solve the differential equation (3.7.1) with the source (3.7.2), we applythe triple integral transform which is defined in Chap. 1: Hankel transform withrespect to the radial variable r, Fourier transform with respect to the axial variablez and Laplace transform with respect to the time t. The triple transform is expressedas

Z10

expð�stÞdtZ1�1

expðþifzÞdzZ10

@2uh@r2

þ 1r@uh@r

� uhr2

þ @2uh@z2

� 1c2s

@2uh@t2

¼ � 1c2s

Qdðr � aÞ

rdðzÞdðtÞ

8>>><>>>:

9>>>=>>>;rJ1ðnrÞdr

ð3:7:4Þ

The convergence condition at infinity

uhj ffiffiffiffiffiffiffiffiffir2þz2p !1 ¼ @uh

@r

�� ffiffiffiffiffiffiffiffiffir2þz2

p !1¼ @uh

@z


p !1¼ 0 ð3:7:5Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

and the quiescent condition at the initial time

uhjt¼0¼@uh@t

��t¼0

¼ 0 ð3:7:6Þ

are employed, and the transformed displacement is defined as

~u�h ¼Z10


expðþifzÞdzZ10

ruhJ1ðnrÞdr ð3:7:7Þ

The application of the triple integral transform yields to the simple algebraicequation for the transformed displacement,

� n2 þ f2 þ ðs=csÞ2n o

~u�h ¼ � Qc2s

J1ðnaÞ ð3:7:8Þ

Then, we consider the inversion of the displacement,

~u�h ¼Qc2s

J1ðnaÞn2 þ f2 þ ðs=csÞ2

ð3:7:9Þ

As the first step, the Fourier inversion integral defined by Eq. (1.1.21c) in Chap. 1 isapplied and it is reduced to the semi-infinite integral as

~u�h ¼Qpc2s

J1ðnaÞZ10

1

n2 þ f2 þ ðs=csÞ2cosðfzÞdf ð3:7:10Þ

The simple integration formula (2.1.22) in Chap. 2 is applied to the above integral.It follows that

~u�h ¼Q2c2s

J1ðnaÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

q exp � zj jffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

q �ð3:7:11Þ

The second step is to apply the Hankel inversion with n = 1 defined by Eq. (1.1.19)in Chap. 1. The Hankel inversion integral is given by

u�h ¼Q2c2s

Z10

nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2


q �J1ðnaÞJ1ðnrÞdn ð3:7:12Þ

3.7 Torsional Source 111

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

The product of two Bessel functions is replaced with the integral form of the singleBessel function (Watson 1966, p. 361),

J1ðnaÞJ1ðnrÞ ¼ 1p

Zp0

J0 nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosu

p� �cosudu ð3:7:13Þ

and the order of integration is exchanged. Equation (3.7.12) yields

u�h ¼Q

2pc2s

Zp0

cosuduZ10



q �J0ðnZÞdn

ð3:7:14Þ

where

Z ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosu

pð3:7:15Þ

Fortunately, we have the nice integration formula (Erdélyi 1954, vol. II, pp. 9, 24)for the inner integral, i.e.

Z10



q �J0ðnZÞdn

¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZ2 þ z2

p exp �ðs=csÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZ2 þ z2

pn o ð3:7:16Þ

Applying this integration formula to Eq. (3.7.14), and recalling the definition of Z,we have the single finite integral for the Laplace transformed displacement,

u�h ¼Q

2pc2s

Zp0

exp �ðs=csÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ z2

pn offiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ z2

p cosudu ð3:7:17Þ

Before going to the Laplace inversion, the integral is simplified by the change ofvariable, φ → u,

u ¼ 1cs

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ z2

pð3:7:18Þ


Equation (3.7.17) is simplified as

u�h ¼Q

4parcs

ZR2=cs

R1=cs

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR22 � ðcsuÞ2

ðcsuÞ2 � R21

�s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðcsuÞ2 � R21

R22 � ðcsuÞ2

s( )expð�suÞdu ð3:7:19Þ

where

R1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr � aÞ2 þ z2

q; R2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr þ aÞ2 þ z2

qð3:7:20Þ

Examining the integrand, we learn that the Laplace transform parameter s isincluded only in the argument of the exponential function. So, the simple Laplaceinversion formula for the delta function


is applied to Eq. (3.7.19). It yields

uh ¼ Q4parcs

ZR2=cs

R1=cs

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR22 � ðcsuÞ2

ðcsuÞ2 � R21

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcsuÞ2 � R2

1

R22 � ðcsuÞ2

s( )dðt � uÞdu ð3:7:22Þ

In order to evaluate the integral, we examine the supporting region for the deltafunction and applied the simple formula (1.2.3) in Chap. 1,

Zba

f ðxÞdðx� cÞdx ¼ f ðcÞ; a\ c\ b0; c\ a or b\ c

ð3:7:23Þ

We have the simple form for the torsional displacement, i.e. Green’s function forthe torsional deformation,

uh ¼ Q4pcsar

HðR2 � cstÞHðcst � R1ÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR22 � ðcstÞ2

ðcstÞ2 � R21

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � R2

1

R22 � ðcstÞ2

s( )ð3:7:24Þ

where H(.) is Heaviside’s unit step function defined in Sect. 1.2.

3.7.2 Point Torque Source

This subsection considers the Green’s function corresponding to the point torquesource defined by Eq. (3.7.3). As was employed in the previous subsection, we also


http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

apply the triple integral transform to the governing Eq. (3.7.1) with the point torquesource,

Z10


expðþifzÞdzZ10

@2uh@r2 þ 1

r@uh@r � uh

r2 þ @2uh@z2 � 1

c2s@2uh@t2

¼ � T2pl lim

a!0

dðr�aÞar

� dðzÞdðtÞ

8><>:

9>=>;rJ1ðnrÞdr

ð3:7:25Þ

The limit in the source term is taken after the Hankel transform. Its procedure is

Z10

lima!0

dðr � aÞar

rJ1ðnrÞdr ¼ lima!0

Z10

dðr � aÞar

rJ1ðnrÞdr ¼ lima!0

J1ðnaÞa

¼ n2

ð3:7:26Þ

The transformed governing equation yields to the simple algebraic equation,

� n2 þ f2 þ ðs=csÞ2n o

~u�h ¼ � T4pl

n ð3:7:27Þ

Thus, the triple transformed displacement is given by

~u�h ¼T4pl

n

n2 þ f2 þ ðs=csÞ2ð3:7:28Þ

Now, we consider the inversion. As was done for the inversion integral ofEq. (3.7.10), the Fourier inversion integral with respect to the parameter ζ is easilyevaluated as

~u�h ¼T8pl



q �ð3:7:29Þ

The second inversion, i.e. Hankel inversion integral, is separated into two terms as

u�h ¼T8pl

Z10

n2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2


q �J1ðnrÞdn

¼ T8pl

Z10



n2 þ ðs=csÞ2q

8><>:

9>=>; exp � zj j


q �J1ðnrÞdn

ð3:7:30Þ


The second term in the last line can be evaluated by applying the formula (Erdélyi1954, vol. II, pp. 19, 10)

Z10

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

p exp �affiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

q �J1ðnyÞdn

¼ 1by

expð�abÞ � exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ y2

p� �n oð3:7:31Þ

As to the first term in the Hankel inversion integral (3.7.30), we derive a suitableintegration formula from the formula (3.7.31). Differentiating Eq. (3.7.31) twicewith respect to the parameter α, we have

Z10

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

qexp �a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

q �J1ðnyÞdn

¼ byexpð�abÞ þ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ y2p y

a2 þ y2� a2b

yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ y2

p !

exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ y2

p� �ð3:7:32Þ

Applying these two formulas to Eq. (3.7.30), we have the Laplace transformeddisplacement as

u�h¼ T

8plrR2

scsþ 1R

� �expð�sR=csÞ ð3:7:33Þ

where R is the distance from the source point (coordinate origin),

R ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ z2

pð3:7:34Þ

The last Laplace inversion is very simple. Fortunately, we have two simpleinversion formulas,

L�1 expð�asÞf g ¼ dðt � aÞ; L�1 s expð�asÞf g ¼ d0ðt � aÞ ð3:7:35Þ

where d0(.) is the derivative of the delta function as

d0ðtÞ ¼ ddðtÞdt

ð3:7:36Þ


Applying these two inversion formulas to Eq. (3.7.33), we have the final form of thetorsional displacement as

uh ¼ T8pl

rR2

1csd0ðt � R=csÞ þ 1

Rdðt � R=csÞ

�ð3:7:37Þ

The above equation shows only one spherical wave and its front is R = cst. It shouldbe noticed that disturbance is only on the wave front and no disturbance its inside,R < cst, due to the nature of the delta function.

Exercises

(3:1) From the unified expression for the Green’s dyadic (3.1.42), derive theexplicit expressions for Gxx, Gxy, Gyx,Gyy and show that Gxy = Gyx.

(3:2) Show that the explicit expression of the Green’s dyadic (3.4.55) is the sameas the corresponding one in the displacement equation (3.4.53).

(3:3) When the body force Bθ in the governing equation (3.7.1) is a suddenlyapplied point torque,

Bhðr; z; tÞ ¼ T02pq

lima!0

dðr � aÞar

� dðzÞHðtÞ ð3:7:39Þ

where H(t) is Heaviside’s unit step function and T0 is the magnitude of the torque,show that the corresponding Green’s function is given by

uh ¼ T08pl

rR2

1csdðt � R=csÞ þ 1

RHðt � R=csÞ

�ð3:7:40Þ

Appendix

See Table 3.1.


Tab

le3.1

Green’s

dyadic

forelastody

namic

equatio

ns

Displacem

entequatio

nsSo

urce

Bi

Green’s

dyadic

2D2D

inplanedeform

ation(plane

strain)

ðc2�1Þ

@ @x

@u x @xþ@u y @y

�� þ

@2u x

@x2

þ@2u x

@y2

¼1 c2 s

@2u x

@t2

�1 c2 sBxðx

;y;tÞ

ðc2�1Þ

@ @y

@u x @xþ@u y @y

�� þ

@2u y

@x2

þ@2u y

@y2

¼1 c2 s

@2u y

@t2

�1 c2 sByðx

;y;tÞ

PidðxÞ

dðyÞd

ðtÞu iðx;

y;tÞ¼X j¼x;y

PjG

ijðx;

y;tÞ;

i;j¼

x;y

Gijðx;

y;tÞ¼

Hðt�r=c dÞ

2pc d

x ix j r2

1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiffiðc d

tÞ2�r2

qþ

2xix

j

r2�d ij

��


ffiffiffiffiffiffiðc d

tÞ2�r2

qr2

8 > < > :9 > = > ;

�Hðt�r=c sÞ

2pc s

x ix j r2�d ij

��

1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiðc s

tÞ2�r2

qþ

2xix

j

r2�d ij

��


ffiffiffiffiffiðc s

tÞ2�r2

qr2

8 > < > :9 > = > ;

QidðxÞ

dðyÞexpðþi

xtÞ

u iðx;

y;tÞ¼X j¼x;y

Qjg

ijðx;

yÞexpðþi

xtÞ;

i;j¼

x;y

g ijðx

;yÞ¼

�i 4

1 c2 d

x ix j r2H

ð2Þ

0ðrx

=c dÞþ

d ij�2x

ixj

r2

�� c d rx

Hð2Þ

1ðrx

=c dÞ

�

�

�1 c2 s

x ix j r2�d ij

�� H

ð2Þ

0ðrx

=c sÞþ

d ij�2x

ixj

r2

�� c

s

rxH

ð2Þ

1ðrx

=c sÞ

�

S idðxÞd

ðyÞu i

¼X j¼x;y

S jgð

staticÞ

ijðx;

yÞ;i;j¼

x;y

gðstaticÞ

xxðx;

yÞ¼

14p

c2 d

�ðc2

þ1Þ

logðrÞ

þðc2

�1Þ

x r��2

�

gðstaticÞ

xyðx;

yÞ¼

14p

c2 d

ðc2�1Þ

xy r2

gðstaticÞ

yyðx;

yÞ¼

14p

c2 d

�ðc2

þ1Þ

logðrÞ

þðc2

�1Þ

y r��2

�

3Dðc2

�1Þ

@ @x

@u x @xþ@u y @yþ@u z @z

�� þ

@2u x

@x2

þ@2u x

@y2

þ@2u x

@z2

¼1 c2 s

@2u x

@t2

�1 c2 sBxðx

;y;z;tÞ

ðc2�1Þ

@ @y


�� þ

@2u y

@x2

þ@2u y

@y2

þ@2u y

@z2

¼1 c2 s

@2u y

@t2

�1 c2 sByðx

;y;z;tÞ

ðc2�1Þ

@ @z


�� þ

@2u z

@x2

þ@2u z

@y2

þ@2u z

@z2

¼1 c2 s

@2u z

@t2

�1 c2 sBzðx

;y;z;tÞ

PidðxÞ

dðyÞd

ðzÞdðtÞ

u i¼X j¼x;y;z

PjG

ijðx;

y;z;tÞ;

i;j¼

x;y;z

Gijðx;

y;z;tÞ¼

14p

R

x ix j

c2 dR2dðt�

R=c dÞþ

1 c2 sd ij�x ix j R2

�� dð

t�R=c sÞ

�t R2

d ij�3x

ixj

R2

�� H

ðt�R=c dÞH

ðR=c s

�tÞ

8 > > > < > > > :

9 > > > = > > > ;Q

idðxÞ

dðyÞd

ðzÞexpðþi

xtÞ

u i¼X j¼x;y;z

Qjg

ijðx;

y;z;tÞ;

i;j¼

x;y;z

g ijðx

;y;z;tÞ¼

14p

Rx ix j

c2 dR2þ

1

ðxRÞ2

1þix

R c d

�� d i

j�3x

ixj

R2

��

() ex

pfixðt�R=c dÞg

þ1

4pR

1 c2 sd ij�x ix j R2

��

1

ðxRÞ2

1þix

R c s

�� d i

j�3x

ixj

R2

��

() ex

pfixðt�R=c sÞg

(con

tinued)

Appendix 117

Tab

le3.1

(con

tinued)

Displacem

entequatio

nsSo

urce

Bi

Green’s

dyadic

S idðxÞd

ðyÞdðzÞ

u i¼X j¼x;y;z

S jG

ðKÞ

ijðx;

y;zÞ;

i;j¼

x;y;z

GðK

Þij

ðx;y;zÞ

¼1

8pð1

�2mÞc2 d

Rð3

�4mÞd

ijþx ix j R2

no

Torsional

ring

source

@2u h

@r2

þ1 r@u h @r�

u h r2þ

@2u h

@z2

�1 c2 s

@2u h

@t2

¼�

1 c2 sQ

dðr�

aÞr

dðzÞd

ðtÞu h

¼Q

4pc sar

HR2�c st

ðÞH

c st�

R1

ðÞ


ffiffiffiffiffiffiR2 2�ðc s

tÞ2ðc s

tÞ2�R2 1

s�


ffiffiffiffiffiffiðc s

tÞ2�R2 1

R2 2�ðc s

tÞ2

s(

)

R1¼


ffiffiffiffiffiffiffiffiffiffi

ðr�aÞ

2þz2

q;

R2¼


ffiffiffiffiffiffiffiffiffiffi

ðrþaÞ

2þz2

qPo

inttorque

source

@2u h

@r2

þ1 r@u h @r�

u h r2þ

@2u h

@z2

�1 c2 s

@2u h

@t2

¼�

T 2pl

lim a!0

dðr�

aÞar

� dð

zÞdðtÞ

u h¼

T 8pl

r R2

1 c sd0ðt�R=c sÞþ

1 Rdðt�

R=c sÞ

�

d0ðtÞ¼

ddðtÞ dt

;R¼


r2þz2

pSu

ddenly

appliedpointtorque

@2u h

@r2

þ1 r@u h @r�

u h r2þ

@2u h

@z2

�1 c2 s

@2u h

@t2

¼�

T 0 2pl

lim a!0

dðr�

aÞar

� dð

zÞHðtÞ

u h¼

T 0 8pl

r R2

1 c sdðt�

R=c sÞþ

1 RHðt�R=c sÞ

�

R¼


r2þz2

p


References

Erdélyi A (ed) (1954) Tables of integral transforms, vol I and II. MacGraw-Hill, New-YorkGradshteyn IS, Ryzhik IM (Jefferey A (ed)) (1980) Table of integrals, series, and products, 5th

edn. Academic Press, San DiegoWatson GN (1966) A treatise on the theory of bessel functions. Cambridge University Press,

Cambridge (1966)

References 119

Chapter 4Acoustic Wave in a Uniform Flow

Traditionally, when we use the word “wave,” we think about acoustic waves orwater waves. The acoustic wave is much more familiar to our daily life and weexperience many wave phenomena such as reflection, refraction, diffraction, theDoppler effects, etc. The governing equations for the acoustic wave are rigorouslyderived from the fluid equations and Green’s function for the acoustic wave in aflowing fluid is discussed by applying the method of integral transform.

4.1 Compressive Viscous Fluid

Motions and disturbances in fluids such as water, oil and gas are governed by fourgroups of equations:

(1) Equations of motion,

@rxx@x

þ @ryx@y

þ @rzx@z

þ qBx ¼ q@vx@t

þ vx@vx@x

þ vy@vx@y

þ vz@vx@z

� �@rxy@x

þ @ryy@y

þ @rzy@z

þ qBy ¼ q@vy@t

þ vx@vy@x

þ vy@vy@y

þ vz@vy@z

� �@rxz@x

þ @ryz@y

þ @rzz@z

þ qBz ¼ q@vz@t

þ vx@vz@x

þ vy@vz@y

þ vz@vz@z

� � ð4:1:1Þ

(2) Constitutive equations for the linear Newtonian fluid,

rxx ¼ �p� 2l3

@vx@x

þ @vy@y

þ @vz@z

� �þ 2l

@vx@x

ryy ¼ �p� 2l3

@vx@x

þ @vy@y

þ @vz@z

� �þ 2l

@vy@y

rzz ¼ �p� 2l3

@vx@x

þ @vy@y

þ @vz@z

� �þ 2l

@vz@z

rxy ¼ l@vx@y

þ @vy@x

� �; ryz ¼ l

@vy@z

þ @vz@y

� �; rzx ¼ l

@vx@z

þ @vz@x

� �ð4:1:2Þ


121

(3) Continuity equation

@q@t

þ @ðqvxÞ@x

þ @ðqvyÞ@y

þ @ðqvzÞ@z

¼ 0 ð4:1:3Þ

(4) Equation of state for the acoustic medium (adiabatic change)

pp0

¼ qq0

� �j

ð4:1:4Þ

In above equations, the viscosity l is a known/given constant that specifies thenature of the fluid. The body force Bi is assumed to be a source and its functionalform must be specified. The stress rij, the particle velocity vi, the hydro-staticpressure p, and the density q are unknown functions to be determined. The sub-script “0” stands for the quantities at a reference state. The constant parameter j isthe ratio of two specific heats. Thus, the governing equation for the linear fluid is aset of strongly coupled partial differential equations with 11 unknowns.

Substituting the constitutive Eq. (4.1.2) into the equations of motion (4.1.1), weobtain the well-known Navier-Stokes equations,

� @p@x

þ l3@

@x@vx@x

þ @vy@y

þ @vz@z

� �þ l

@2vx@x2

þ @2vx@y2

þ @2vx@z2

� �þ qBx

¼ q vx@vx@x

þ vy@vx@y

þ vz@vx@z

þ @vx@t

� � ð4:1:5aÞ

� @p@y

þ l3@

@y@vx@x

þ @vy@y

þ @vz@z

� �þ l

@2vy@x2

þ @2vy@y2

þ @2vy@z2

� �þ qBy

¼ q vx@vy@x

þ vy@vy@y

þ vz@vy@z

þ @vy@t

� � ð4:1:5bÞ

� @p@z

þ l3@

@z@vx@x

þ @vy@y

þ @vz@z

� �þ l

@2vz@x2

þ @2vz@y2

þ @2vz@z2

� �þ qBz

¼ q vx@vz@x

þ vy@vz@y

þ vz@vz@z

þ @vz@t

� � ð4:1:5cÞ

When the fluid is incompressible and no density change takes place, the con-tinuity equation is simplified and the equation of state is unnecessary. Then, theincompressible fluid is governed by a simplified form of the Navier-Stokes equa-tions. However, when we discuss acoustic waves, the two reminder Eqs. (4.1.3) and(4.1.4) must be fully incorporated since acoustic waves express the propagation ofdensity changes.

122 4 Acoustic Wave in a Uniform Flow

4.2 Linearization

The nonlinearity stems from the acceleration terms in Navier-Stokes equations(4.1.5), the product ðqviÞ in the continuity equation (4.1.3) and the power of densityin the equation of state (4.1.4). Thus, the governing equations for acoustic wavesare fully nonlinear, coupled partial differential equations. In order to reduce thedifferential equations to tractable ones, we introduce the assumption that theacoustic wave is an infinitesimal small disturbance superimposed on a referencenonlinear fluid motion. This assumption states that the infinitesimal small distur-bance is a deviation from the nonlinear reference state. We introduce a smallparameter e, and represent the disturbance with ð�vi; �p; �qÞ. The five unknownfunctions, the velocity component vi, the static hydro-pressure p, and the density q,are approximated as the power of the small parameter e,

vi ¼ Vi þ e�vi þ Oðe2Þ; p ¼ p0 þ e�pþ Oðe2Þ; q ¼ q0 þ e�qþ Oðe2Þ ð4:2:1Þ

We also assume that the wave source described by the body force is a smallquantity of order e1,

Bi ¼ e�Bi þ Oðe2Þ ð4:2:2Þ

Substituting Eqs. (4.2.1–4.2.2) into Eq. (4.1.5a), and neglecting terms of higherorder than Oðe1Þ, the equation is reduced to two parts, the zeroth and first orderterms. The zeroth order term is

� @p0@x

þ l3@

@x@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vx

@x2þ @2Vx

@y2þ @2Vx

@z2

� �

¼ þ q0@Vx

@tþ Vx

@Vx

@xþ Vy

@Vx

@yþ Vz

@Vx

@z

� � ð4:2:3Þ

and the first order term is

e � @�p@x

þ l3@

@x@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vx@x2

þ @2�vx@y2

þ @2�vx@z2

� �þ q�Bx

�

¼ þ �q Vx@Vx

@xþ Vy

@Vx

@yþ Vz

@Vx

@z

� �

þ q0@�vx@t

þ q0 Vx@�vx@x

þ �vx@Vx

@x

� �þ q0 Vy

@�vx@y

þ �vy@Vx

@y

� �þ q0 Vz

@�vx@z

þ �vz@Vx

@z

� ��ð4:2:4Þ

Similarly, we substitute Eqs. (4.2.1–4.2.2) into the reminder of Eq. (4.1.5). FromEq. (4.1.5b), we have the zeroth order term

4.2 Linearization 123

� @p0@y

þ l3@

@y@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vy

@x2þ @2Vy

@y2þ @2Vy

@z2

� �

¼ þ q0@Vy

@tþ Vx

@Vy

@xþ Vy

@Vy

@yþ Vz

@Vy

@z

� � ð4:2:5Þ

and the first order term

e � @�p@y

þ l3@

@y@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vy@x2

þ @2�vy@y2

þ @2�vy@z2

� �þ q�By

�

¼ þ �q Vx@Vy

@xþ Vy

@Vy

@yþ Vz

@Vy

@z

� �

þ q0@�vy@t

þ q0 Vx@�vy@x

þ �vx@Vy

@x

� �þ q0 Vy

@�vy@y

þ �vy@Vy

@y

� �þ q0 Vz

@�vy@z

þ �vz@Vy

@z

� ��

ð4:2:6Þ

From Eq. (4.1.5c), we also have for the zeroth order term

� @p0@z

þ l3@

@z@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vz

@x2þ @2Vz

@y2þ @2Vz

@z2

� �

¼ þ q0@Vz

@tþ Vx

@Vz

@xþ Vy

@Vz

@yþ Vz

@Vz

@z

� � ð4:2:7Þ

and for the first order term

e � @�p@z

þ l3@

@z@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vz@x2

þ @2�vz@y2

þ @2�vz@z2

� �þ q�Bz

�

¼ þ �q Vx@Vz

@xþ Vy

@Vz

@yþ Vz

@Vz

@z

� �

þ q0@�vz@t

þ q0 Vx@�vz@x

þ �vx@Vz

@x

� �þ q0 Vy

@�vz@y

þ �vy@Vz

@y

� �þ q0 Vz

@�vz@z

þ �vz@Vz

@z

� ��

ð4:2:8Þ

Substituting Eq. (4.2.1) into Eq. (4.1.3), the continuity equation is decomposedinto the two parts,

@ðq0 þ e�qÞ@t

þ @ðq0 þ e�qÞðVx þ e�vxÞ@x

þ @ðq0 þ e�qÞðVy þ e�vyÞ@y

þ @ðq0 þ e�qÞðVz þ e�vzÞ@z

¼ @q0@t

þ @ðq0VxÞ@x

þ @ðq0VyÞ@y

þ @ðq0VzÞ@z

þ e@�q@t

þ @ðq0�vxÞ@x

þ @ðq0�vyÞ@y

þ @ðq0�vzÞ@z

þ @ðVx�qÞ@x

þ @ðVy�qÞ@y

þ @ðVz�qÞ@z

� �þ Oðe2Þ ¼ 0

ð4:2:9Þ


The equation of state for the adiabatic change in Eq. (4.1.4) is approximated as

p0 þ e�pp0

¼ q0 þ e�qq0

� �j

) 1þ e�pp0

¼ 1þ ej�qq0

þ Oðe2Þ ) �pp0

¼ j�qq0

ð4:2:10Þ

Thus, we have the linear relation between the pressure and density deviations.Consequently, we have the equations for the reference state as the Oðe0Þ part,

� @p0@x

þ l3@

@x@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vx

@x2þ @2Vx

@y2þ @2Vx

@z2

� �

¼ þ q0@Vx

@tþ Vx

@Vx

@xþ Vy

@Vx

@yþ Vz

@Vx

@z

� � ð4:2:11aÞ

� @p0@y

þ l3@

@y@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vy

@x2þ @2Vy

@y2þ @2Vy

@z2

� �

¼ þ q0@Vy

@tþ Vx

@Vy

@xþ Vy

@Vy

@yþ Vz

@Vy

@z

� � ð4:2:11bÞ

� @p0@z

þ l3@

@z@Vx

@xþ @Vy

@yþ @Vz

@z

� �þ l

@2Vz

@x2þ @2Vz

@y2þ @2Vz

@z2

� �

¼ þ q0@Vz

@tþ Vx

@Vz

@xþ Vy

@Vz

@yþ Vz

@Vz

@z

� � ð4:2:11cÞ

@q0@t

þ @ðq0VxÞ@x

þ @ðq0VyÞ@y

þ @ðq0VzÞ@z

¼ 0 ð4:2:12Þ

The disturbance of Oðe1Þ is governed by the equations

� @�p@x

þ l3@

@x@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vx@x2

þ @2�vx@y2

þ @2�vx@z2

� �þ q0�Bx

¼ þ �q Vx@Vx

@xþ Vy

@Vx

@yþ Vz

@Vx

@z

� �

þ q0@�vx@t

þ q0 Vx@�vx@x

þ �vx@Vx

@x

� �þ q0 Vy

@�vx@y

þ �vy@Vx

@y

� �þ q0 Vz

@�vx@z

þ �vz@Vx

@z

� �

ð4:2:13aÞ

� @�p@y

þ l3@

@y@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vy@x2

þ @2�vy@y2

þ @2�vy@z2

� �þ q0�By

¼ þ �q Vx@Vy

@xþ Vy

@Vy

@yþ Vz

@Vy

@z

� �

þ q0@�vy@t

þ q0 Vx@�vy@x

þ �vx@Vy

@x

� �þ q0 Vy

@�vy@y

þ �vy@Vy

@y

� �þ q0 Vz

@�vy@z

þ �vz@Vy

@z

� �ð4:2:13bÞ

4.2 Linearization 125

� @�p@z

þ l3@

@z@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vz@x2

þ @2�vz@y2

þ @2�vz@z2

� �þ q0�Bz

¼ þ �q Vx@Vz

@xþ Vy

@Vz

@yþ Vz

@Vz

@z

� �

þ q0@�vz@t

þ q0 Vx@�vz@x

þ �vx@Vz

@x

� �þ q0 Vy

@�vz@y

þ �vy@Vz

@y

� �þ q0 Vz

@�vz@z

þ �vz@Vz

@z

� �ð4:2:13cÞ

@�q@t

þ @ðq0�vxÞ@x

þ @ðq0�vyÞ@y

þ @ðq0�vzÞ@z

þ @ðVx�qÞ@x

þ @ðVy�qÞ@y

þ @ðVz�qÞ@z

¼ 0 ð4:2:14Þ

�q ¼ q0jp0

�p ð4:2:15Þ

We should understand that the zeroth order equations, which govern the referencefluid flow, are already satisfied by the reference quantities ðVi; p0; q0Þ. On the otherhand, the first order equations, which include the change/deviation from the ref-erence state, are the coupled differential equations for the 5 unknowns ð�vi; �p; �qÞ.

4.3 Viscous Acoustic Fluid

When the reference state is a uniform flow with viscosity l, we can assumeVi; p0; q0 ¼ const:, and then the first order Eq. (4.2.13) are reduced to the simplerforms

� @�p@x

þ l3@

@x@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vx@x2

þ @2�vx@y2

þ @2�vx@z2

� �þ q0�Bx

¼ þ q0@�vx@t

þ q0 Vx@�vx@x

þ Vy@�vx@y

þ Vz@�vx@z

� � ð4:3:1aÞ

� @�p@y

þ l3@

@y@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vy@x2

þ @2�vy@y2

þ @2�vy@z2

� �þ q0�By

¼ þ q0@�vy@t

þ q0 Vx@�vy@x

þ Vy@�vy@y

þ Vz@�vy@z

� � ð4:3:1bÞ

� @�p@z

þ l3@

@z@�vx@x

þ @�vy@y

þ @�vz@z

� �þ l

@2�vz@x2

þ @2�vz@y2

þ @2�vz@z2

� �þ q0�Bz

¼ þ q0@�vz@t

þ q0 Vx@�vz@x

þ Vy@�vz@y

þ Vz@�vz@z

� � ð4:3:1cÞ


With use of the linear relation between the pressure and density deviations, thecontinuity equation (4.2.14) can be rewritten in terms of the pressure and velocitygradients as

@�p@t

þ Vx@�p@x

þ Vy@�p@y

þ Vz@�p@z

þ jp0@�vx@x

þ @�vy@y

þ @�vz@z

� �¼ 0 ð4:3:2Þ

The above four Eqs. (4.3.1) and (4.3.2) are the governing equations for linear-ized acoustic waves in the uniformly flowing viscous fluid. The coupled differentialequations, which have four unknowns ð�vi; �pÞ, can be reduced to a single differentialequation by the introduction of a velocity potential. We assume that the velocitycomponents ð�viÞ can be derived from a single velocity potential /ðx; y; z; tÞ as

�vx ¼ @/@x

; �vy ¼ @/@y

; �vz ¼ @/@z

ð4:3:3Þ

As to the body force, we also assume that there exists a body force potential B(x, y,z, t) and each body force component is derived as,

�Bx ¼ @B@x

; �By ¼ @B@y

; �Bz ¼ @B@z

ð4:3:4Þ

Substituting Eqs. (4.3.3) and (4.3.4) into (4.3.1a), we obtain

@�p@x

þ q0@

@x@/@t

þ Vx@/@x

þ Vy@/@y

þ Vz@/@z

� �� 4l

3@

@x@2/@x2

þ @2/@y2

þ @2/@z2

� �� q0

@B@x

¼ 0

ð4:3:5Þ

Since all terms in the above equation are derivatives with respect to the spacevariable x, we integrate it with respect to the variable x,

�p ¼ �q0@/@t

þ Vx@/@x

þ Vy@/@y

þ Vz@/@z

� �þ 4l

3@2/@x2

þ @2/@y2

þ @2/@z2

� �þ q0B

ð4:3:6Þ

where the constant term is neglected since it is one of the reference quantities.Substitution into Eqs. (4.3.1b) and (4.3.1c) leads to the same equation and thus theequations of motion in all three directions are reduced to the single Eq. (4.3.6).

The velocity potential defined by Eq. (4.3.3) is also substituted into the conti-nuity equation (4.3.2),

@�p@t

þ Vx@�p@x

þ Vy@�p@y

þ Vz@�p@z

þ jp0@2/@x2

þ @2/@y2

þ @2/@z2

� �¼ 0 ð4:3:7Þ

4.3 Viscous Acoustic Fluid 127

Then, we have the coupled differential Eqs. (4.3.6) and (4.3.7) in terms of only twounknowns, the velocity potential and the pressure deviation. Further, if we sub-stitute Eq. (4.3.6) into Eq. (4.3.7), we have just the single differential equation forthe velocity potential /,

r2/þ 4l3jp0

@

@tþ Vx

@

@xþ Vy

@

@yþ Vz

@

@z

� �r2/

¼ q0jp0

@

@tþ Vx

@

@xþ Vy

@

@yþ Vz

@

@z

� �2

/� q0jp0

@B@t

þ Vx@B@x

þ Vy@B@y

þ Vz@B@z

� �ð4:3:8Þ

where r2 is the Laplacian operator defined by

r2 ¼ @2

@x2þ @2

@y2þ @2

@z2ð4:3:9Þ

We can also derive the differential equation for the pressure deviation. We applythe Laplacian to both sides of Eq. (4.3.6),

r2�p ¼ �q0@

@tþ Vx

@

@xþ Vy

@

@yþ Vz

@

@z

� �r2/þ 4l

3r2 r2/� �þ q0r2B

ð4:3:10Þ

and derive the Laplacian of the velocity potential from Eq. (4.3.7) as

r2/ ¼ � 1jp0

@�p@t

þ Vx@�p@x

þ Vy@�p@y

þ Vz@�p@z

� �ð4:3:11Þ

The Laplacian of the velocity potential in Eq. (4.3.10) is replaced with the abovepressure gradient. We therefore have the single differential equation for the pressuredeviation as

r2�pþ 4l3jp0

@

@tþ Vx

@

@xþ Vy

@

@yþ Vz

@

@z

� �r2�p

¼ q0jp0

@

@tþ Vx

@

@xþ Vy

@

@yþ Vz

@

@z

� �2

�pþ q0@2B@x2

þ @2B@y2

þ @2B@z2

� � ð4:3:12Þ

Two differential Eqs. (4.3.8) and (4.3.12) for the potential and the pressure aremathematically the same, but the non-homogeneous terms of the body force are


little bit different. We easily see that both equations are wave equations and theirwave velocity is

c ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðjp0Þ=q0

pð4:3:13Þ

If we introduce a reference length l and define Reynolds number,

Re ¼ ðq0clÞ=l ð4:3:14Þ

and Mach numbers,

Mj ¼ Vj=c; j ¼ x; y; z ð4:3:15Þ

the governing equation for the viscous acoustic media is simplified to

r2/þ 4l3Re

1c@

@tþMx

@

@xþMy

@

@yþMz

@

@z

� �r2/

¼ 1c@

@tþMx

@

@xþMy

@

@yþMz

@

@z

� �2

/� 1c

1c@B@t

þMx@B@x

þMy@B@y

þMz@B@z

� �ð4:3:16Þ

for the velocity potential, and to

r2�pþ 4l3Re

1c@

@tþMx

@

@xþMy

@

@yþMz

@

@z

� �r2�p

¼ 1c@

@tþMx

@

@xþMy

@

@yþMz

@

@z

� �2

�pþ q0@2B@x2

þ @2B@y2

þ @2B@z2

� � ð4:3:17Þ

for the pressure deviation. These two differential equations have only two param-eters, Reynolds and Mach numbers. Thus, the nature of acoustic waves in theflowing viscous fluid is characterized by these two numbers.

4.4 Wave Radiation in a Uniform Flow

This section discusses a 2D wave propagation problem in a uniform flow. Take the(x, y) plane and assume that all quantities are independent of z, and assume that thenon-viscous acoustic fluid (l ¼ 0) is flowing along the x-axis with the uniformvelocity Vx (see Fig. 4.1). Further, we also assume the Laplacian of the body forcepotential as the wave source Q,

4.3 Viscous Acoustic Fluid 129

q0@2B@x2

þ @2B@y2

� �¼ QdðxÞdðyÞdðtÞ ð4:4:1Þ

Then, we substitute Re ¼ 1; My ¼ Mz ¼ 0 into Eq. (4.3.17). The governingequation for the pressure deviation is then reduced to the simple form

r2p ¼ 1c@

@tþMx

@

@x

� �2

pþ QdðxÞdðyÞdðtÞ ð4:4:2Þ

where c is the acoustic wave velocity defined by Eq. (4.3.13), and Mach number Mx

is defined by Eq. (4.3.15). Furthermore, the pressure in this equation is the deviationfrom the reference state and the correct notation is �p (over bar). However, we do notuse the over bar for the pressure deviation since we apply the double Fouriertransform and one of the transforms is classified by the over bar. So, in order toavoid the confusion, the over bar is dropped from the pressure deviation.

Since our acoustic field is of infinite extent, it is enough to obtain a particularsolution of Eq. (4.4.2). Laplace transform with respect to the time t and doubleFourier transform with respect to two space variables x and y are applied. The tripleintegral transform is defined as

Laplace transform: f �ðsÞ ¼Z10


Fourier transform: �f ðnÞ ¼Z10


Z10

�f ðnÞ expð�inxÞdn

ð4:4:4Þ

Fourier transform: ~f ðgÞ ¼Z10


Z10

~f ðgÞ expð�igyÞdg

ð4:4:5Þ

xV

( )( ) ( )

i t

tQ x y

e ω

δδ δ

⎧⎪⎨⎪⎩

x

y

Fig. 4.1 A point source in a uniformly flowing fluid


Applying the triple transform to Eq. (4.4.2), the simple algebraic equation for thetransformed pressure is obtained as

�ðn2 þ g2Þ~�p� ¼ ðs=c� inMxÞ2~�p� þ Q ð4:4:6Þ

The pressure in the transformed domain is thus given by

~�p� ¼ � Q

n2 þ g2 þ s=c� inMxð Þ2 ¼ � Qg2 þ q2

ð4:4:7Þ

where q is written as

q2 ¼ n2 þ s=c� inMxð Þ2

¼ 1�M2x

� �n� iMx

1�M2xðs=cÞ

� �2

þ s=c1�M2

x

� �2( )

¼ 1�M2x

� �n� iaMxð Þ2þa2

n oð4:4:8Þ

The Laplace transform parameter s is included in the parameter a which is definedby

a ¼ s=c1�M2

xð4:4:9Þ

As to the first inversion, we apply the Fourier inversion with respect to theparameter g, and reduce it to the semi-infinite integral,

�p� ¼ � 12p

Zþ1

�1

Qg2 þ q2

expð�igyÞdg ¼ �Qp

Z10

1g2 þ q2

cosðgyÞdg ð4:4:10Þ

The last integral is easily evaluated by the formula (2.1.22) in Chap. 2 and yields

�p� ¼ � Q2q

expð�qjyjÞ ð4:4:11Þ

The next step is to apply the Fourier inversion integral with respect to the parameter n,

p� ¼ �Q2

12p

Zþ1

�1

1qexpð�qjyjÞ expð�inxÞdn ð4:4:12Þ

4.4 Wave Radiation in a Uniform Flow 131

http://dx.doi.org/10.1007/978-3-319-17455-6_2

Recalling q in Eq. (4.4.8), the integral in the above equation is rewritten in theexplicit form,

p� ¼ � Q

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

x

p 12p

Zþ1

�1

exp �jyj ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

x

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin� iaMxð Þ2þa2

q� inx

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin� iaMxð Þ2þa2

q dn

ð4:4:13Þ

Introducing the variable transform, n ! 1, as

1 ¼ n� iaMx ð4:4:14Þ

we obtain the simpler integrand, but the integration path is shifted to the complexplane, not on the real axis, i.e.

p� ¼ �Q2exp þaMxxð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

p 12p

Zþ1�iaMx

�1�iaMx


x

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ a2

p� i1x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ a2

p d1

ð4:4:15Þ

The integration path for this integral is slightly shifted from the real axis in thecomplex 1-plane and is shown by the line CD in Fig. 4.2. In order to transform theintegral to that along the real axis, we consider the complex integral U with the

closed loop LðCDBAC��!Þ in the figure,

U ¼ �Q2exp þaMxxð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

p 12p

ZL


x


p� i1x


p d1 ð4:4:16Þ

The integrand has two branch points, at 1 ¼ �ia. Two branch cuts are introducedalong the imaginary axis, as shown in the figure. Fortunately, if we assume subsonic

iα+

iα−

Branch cut

A B

C D

Re( )ς

Im( )ς

xi Mα−

Fig. 4.2 Transform ofintegration path


motion of the flow, Mxð¼Vx=cÞ\1, these branch cuts do not cross the integrationline CD and no singular point are then included in the closed loop L. In addition, thetwo integrals along the line AC and BD, with an infinite real part, vanish since

Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ a2

p [ 0. Then, applying Cauchy’s integral theorem to the complex

integral U in Eq. (4.4.16), the integral along the complex path CD is converted tothat along the real axis AB in the 1-plane. That is

p� ¼ �Q2exp þaMxxð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

p 12p

Zþ1

�1


x


p� i1x


p d1 ð4:4:17Þ

The above integral can be further reduced to the semi-infinite integral,

p� ¼ �Q2expðþaMxxÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

p 1p

Z10


x



p cosð1xÞd1 ð4:4:18Þ

Applying the integration identity, which is the integral representation of the mod-ified Bessel function (Erdélyi 1954, vol. I, pp. 17, 27; Watson 1966, p. 172),

K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p ¼Z10

exp �yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ a2


p cosð1xÞd1

¼Z11

exp �auffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiu2 � 1

p du ð4:4:19Þ

to Eq. (4.4.18), we have the other integral form for the pressure,

p� ¼ � Q2p

exp þaMxxð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

x

p Z11

1ffiffiffiffiffiffiffiffiffiffiffiffiffiu2 � 1

p exp �auffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q� du ð4:4:20Þ

Further, we recall the definition of a which includes the Laplace transformparameter s. The above Eq. (4.4.20) can be rewritten as

p� ¼ � Q2p


x

p Z11

exp � s=c1�M2

xuffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q�Mxx

n offiffiffiffiffiffiffiffiffiffiffiffiffiu2 � 1

p du ð4:4:21Þ


Introducing the variable transform, u ! t, as defined by

t ¼uffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q�Mxx

c 1�M2x

� � ; u ¼ 1�M2x

� �ct þMxxffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 1�M2x

� �y2

q ;

du ¼ c 1�M2x

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q dt

ð4:4:22Þ

Equation (4.4.21) is converted to the form of the Laplace transform integral,

p� ¼ � cQ2p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

x

q Z1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þ 1�M2

xð Þy2p

�Mxx

c 1�M2xð Þ

expð�stÞdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

x

� �ct þMxx

� �2� x2 þ 1�M2x

� �y2

� �q

ð4:4:23Þ

The above equation is the Laplace transformed pressure, but it is just in the formof the Laplace transform integral. Thus, the original pressure is its integrand with ashifted starting time since the lower limit is not zero. Inspecting Eq. (4.4.23), wecan find the original pressure as

p ¼ � cQ2p


x

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

� �ct þMxx

� �2� x2 þ 1�M2x

� �y2

� �q ð4:4:24Þ

where its valid time range is

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q�Mxx

c 1�M2x

� � \t\1 ð4:4:25Þ

In order to obtain a more compact expression, the argument in the radical and thetime range are rewritten as

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ 1�M2

x

� �y2

q\ct 1�M2

x

� �þMxx ) x� Vxtð Þ2þy2\ðctÞ2 ð4:4:26Þ

1�M2x

� �ct þMxx

� �2� x2 þ 1�M2x

� �y2

� � ¼ 1�M2x

� � ðctÞ2 � x� Vxtð Þ2�y2n o

ð4:4:27Þ


Finally, the pressure deviation in the flowing fluid is expressed in the simple form,

p ¼ � Q2p

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � x� Vxtð Þ2þy2

n o=c2

r H ct �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� Vxtð Þ2þy2

q� �ð4:4:28Þ

We have just arrived at the final form of the solution that gives the pressurefluctuations in the flowing fluid with the stationary impulsive point source. Thedisturbed region is easily derived from the argument of the step function: it is acircle but its center is moving with the flow velocity,

x� Vxtð Þ2þy2 ¼ ctð Þ2 ð4:4:29Þ

Figure 4.3 shows the typical wave front and the circular disturbed region in theflowing fluid.

4.5 Time-Harmonic Wave in a Uniform Flow

The steady-state acoustic response produced by a stationary time-harmonic sourceis governed by

r2p� 1c@

@tþMx

@

@x

� �2

p ¼ QdðxÞdðyÞ expðþixtÞ ð4:5:1Þ

ct

xV t

x

y

source

Flow xV

Fig. 4.3 Disturbed circularregion in a uniform flow


A solution to this equation can be obtained by the convolution integral of theimpulsive response as

p ¼ limt!1

Z t

0

pðimpulseÞðx; y; t0Þ exp ixðt � t0Þf gdt0 ð4:5:2Þ

Substituting the impulsive response of Eq. (4.4.24), not the final one, we have

p ¼ � cQ2p

expðixtÞZ1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þ 1�M2

xð Þy2�Mx

px

c 1�M2xð Þ


x

pexpð�ixt0Þdt0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1�M2x

� �ct0 þMxx

� �2� x2 þ 1�M2x

� �y2

� �q

ð4:5:3Þ

where the source magnitude Q is not the same as that of the impulsive source. Forthis integral, we make the change of variable, t0 ! u, as

u ¼ 1�M2x

� �ct0 þMxx; t0 ¼ u�Mxx

1�M2x

� �c; dt0 ¼ du

1�M2x

� �c

ð4:5:4Þ

Equation (4.5.3) is now rewritten as

p ¼ � Q2p


x

p exp ix t þ Mx

1�M2x

xc

� �� Z1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þ 1�M2

xð Þy2pexp � ixu

1�M2xð Þc

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu2 � x2 þ ð1�M2

x Þy2� �q du

ð4:5:5Þ

Since the integral in the above equation is just the definition integral of the Hankelfunction of the second kind (which is Eq. (3.2.5) in Sect. 3.2), we replace theintegral with the Hankel function. The final form for the time-harmonic response isthen given by

p ¼ iQ


x

p Hð2Þ0

xc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ ð1�M2

x Þy2p

1�M2x

!exp ix t þ Mx

1�M2x

xc

� �� ð4:5:6Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_3

Exercises

(4:1) When a wave source is moving with uniform velocity V along the x-axis,how do you change the nonhomogeneous wave source term in Eq. (4.4.1)?

(4:2) Using the asymptotic formula for the Hankel function (Watson 1966, p. 198),

Hð2Þm ðzÞ�

ffiffiffiffiffi2pz

rexp �i z� mp=2� p=4ð Þf g ðaÞ

show that the time-harmonic response (4.5.6) is the out-going wave.

References

Erdélyi A (ed) (1954) Tables of integral transforms, vols I and II. McGraw-Hill, New YorkWatson GN (1966) A treatise on the theory of bessel functions. Cambridge University Press,

Cambridge

4.5 Time-Harmonic Wave in a Uniform Flow 137

Chapter 5Green’s Functions for Beam and Plate

This chapter presents dynamic Green’s functions for the elastic beam and plate. TheGreen’s function is the deflection response produced by a point load. The deflectionequation with the nonhomogeneous term of the applied load is discussed byapplying the method of integral transform.

5.1 An Impulsive Load on a Beam

We shall obtain Green’s function for an elastic beam. The deflection of the beam,w(x, t), is governed by the well-known partial differential equation, so called beam/deflection equation,

EI@4w@x4

þ qA@2w@t2

¼ pðx; tÞ ð5:1:1Þ

where EI is bending rigidity, ρA mass per unit length and p(x, t) the load on thebeam. An impulsive point loadwith magnitude P is assumed by

pðx; tÞ ¼ PdðxÞdðtÞ ð5:1:2Þ

where dð:Þ is Dirac’s delta function. The nonhomogeneous deflection equation(5.1.1) is rewritten as

a4@4w@x4

þ @2w@t2

¼ QdðxÞdðtÞ ð5:1:3Þ

where

a4 ¼ EIqA

; Q ¼ PqA

ð5:1:4Þ


139

For solving the deflection equation, we employ the quiescent condition at an initialtime,

wjt¼0 ¼@w@x

��t¼0

¼ 0 ð5:1:5Þ


wjjxj!1¼ @w@x

��jxj!1

¼ @2w@x2

��jxj!1

¼ @3w@x3

��jxj!1

¼ 0 ð5:1:6Þ

The higher derivatives in the above equation mean the vanishing of the moment andshear force at infinity.

In order to solve the nonhomogeneous deflection equation, we employ themethod of integral transform. Since the deflection depends on two variables, weapply the following double transform: Laplace transform with respect to the timevariable,

f �ðsÞ ¼Z10


and Fourier transform with respect to the space variable,

�f ðnÞ ¼Zþ1


2p

Zþ1

�1


Applying the double transform to the deflection equation (5.1.3), we have thealgebraic equation for the transformed deflection and then the deflection in thetransformed domain is given by

�w� ¼ Q

ðanÞ4 þ s2ð5:1:9Þ

The Laplace inversion is applied firstly. The Laplace inversion formula (Erdélyi1954, vol. I, pp. 150, 1),

L�1 1s2 þ a2

� �¼ 1

asinðatÞ ð5:1:10Þ

is applied to Eq. (5.1.9) and thus we have for the Fourier transformed deflection,

140 5 Green’s Functions for Beam and Plate

�w ¼ Q

ðanÞ2 sinða2tn2Þ ð5:1:11Þ

The Fourier inversion integral of the above equation is reduced to the semi-infiniteintegral, as

w ¼ 12p

Zþ1

�1

Q

ðanÞ2 sinða2tn2Þ expð�inxÞdn ¼ Q

pa2

Z10

1

n2sinða2tn2Þ cosðxnÞdn

ð5:1:12Þ

Fortunately, we have a convenient formula for the inversion integral. That is theformula (Erdélyi 1954, vol. I, pp. 23, 3)

Z10

1x2

sinðax2Þ cosðxyÞdx ¼ p2y S

yffiffiffiffiffiffiffiffi2pa

p� �

� Cyffiffiffiffiffiffiffiffi2pa

p� ��

þ ffiffiffiffiffiffipa

psin

y2

4aþ p

4

� �

ð5:1:13Þ

where C(x) and S(x) are Fresnel integrals/functions (Erdélyi 1954, vol. II, p. 431)defined by

CðxÞSðxÞ

�¼ 1ffiffiffiffiffiffi

2pp

Zx

0

1ffiffiffiu

p sinðuÞcosðuÞ

� �du ð5:1:14Þ

Applying this formula to the last integral in Eq. (5.1.12) and rewriting theexpression, we obtain the Green’s function for the dynamic deflection of the beamas

wðx; tÞ ¼ Qa

ffiffiffitp

rpzffiffiffi2

p SðzÞ � CðzÞf g þ sinp4ð2z2 þ 1Þ

n o� ð5:1:15Þ

where the dimensionless variable z is defined by

z ¼ x

affiffiffiffiffiffiffi2pt

p ð5:1:16Þ

Note that the Green’s function given by Eq. (5.1.15) is the solution for theimpulsive response; however, it does not show any wave nature since the deflectionwave is dispersive and its wave velocity depends on the frequency. The impulsivesource includes an infinite frequency, i.e. the initial disturbance spreads all over thebeam at once without showing any wave nature.

5.1 An Impulsive Load on a Beam 141

5.2 A Moving Time-Harmonic Load on a Beam

When a time-harmonic load with frequency ω is moving with the uniform velocityV, the deflection equation for the beam is given by,

EI@4w@x4

þ qA@2w@t2

¼ Pdðx� VtÞ expðþixtÞ ð5:2:1Þ

where the load location is expressed by the delta function and is at x ¼ Vt. Theconvergence condition given by Eq. (5.1.6) is also assumed.

Since the time variable t is included not only in the argument of the exponentialfunction, but also in that of delta function, it is not good to assume a simple time-harmonic vibration such as wðx; tÞ ¼ w#ðxÞ expðþixtÞ. We directly apply theFourier transform defined by Eq. (5.1.8) to the deflection equation (5.2.1),

d2�wdt2

þ ðanÞ4�w ¼ Q expfþiðVnþ xÞtg ð5:2:2Þ

where α and Q are defined by Eq. (5.1.4). The particular solution corresponding tothe nonhomogeneous loading term is easily obtained as

�w ¼ Q

ðanÞ4 � ðnV þ xÞ2 expfþiðnV þ xÞtg ð5:2:3Þ

Factorizing the denominator, we get

ðanÞ4 � ðnV þ xÞ2 ¼ a4ðn� nþ1 Þðn� n�1 Þðn� nþ2 Þðn� n�2 Þ ð5:2:4Þ

where the eigenvalues are given by

n�1 ¼ 1a

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ V

2a

� �2s

þ V2a

8<:

9=;; n�2 ¼ 1

a�i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� V

2a

� �2s

� V2a

8<:

9=; ð5:2:5Þ

the Fourier inversion integral with respect to the parameter n is expressed as

w ¼ Q2pa4

expðþixtÞZþ1

�1

expf�inðx� VtÞgðn� nþ1 Þðn� n�1 Þðn� nþ2 Þðn� n�2 Þ

dn ð5:2:6Þ

In order to evaluate the integral in Eq. (5.2.6), we apply the theory of complexintegrals (the residue theorem). Following the discussion in Sect. 2.2, the complexfrequency with a small negative imaginary part is assumed. Due to this assumption,all singular points do not lie on the real axis. Two singular points, n�1 and nþ2 , are inthe upper complex n-plane and the other two, nþ1 and n�2 , are in the lower plane.Then, the complex integral


http://dx.doi.org/10.1007/978-3-319-17455-6_2

U ¼ 12p

ZC�


dn ð5:2:7Þ

is to be discussed. Two closed loops C� are shown in Fig. 5.1. We employ the circuitCþ in the case of x� Vt\ 0 and C� in that of x� Vt [ 0 in order to guarantee theconvergence on the large semi-circle. Jordan’s lemma is applied to the complexintegral Φ and after somewhat lengthy calculation we arrive at the expressions,

12p

Zþ1

�1


dn

¼ � ia3

4

exp �ði=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

qþ ðV=2aÞ

� �ðx� VtÞ

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

qþ ðV=2aÞ

� �2

� a3

4

exp ð1=aÞ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

qþ iðV=2aÞ

� �ðx� VtÞ

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

q� iðV=2aÞ

� �2 ; x� Vt[ 0

¼ � ia3

4

exp ði=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

q� ðV=2aÞ

� �ðx� VtÞ

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ ðV=2aÞ2

q� ðV=2aÞ

� �2

� a3

4

exp ð1=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

qþ iðV=2aÞ

� �ðx� VtÞ

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� ðV=2aÞ2

qþ iðV=2aÞ

� �2 ; x� Vt\0

ð5:2:8Þ

Then, the deflection produced by the moving time-harmonic load is given by

wðx; tÞ ¼ � iQ4a

exp �ði=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ X2

pþ X

�ðx� VtÞ þ ixt

n offiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ X2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2

pþ X

�2

� Q4a

exp ð1=aÞ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2

pþ iX


n offiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2


p� iX

�2 ; x� Vt[ 0

ð5:2:9aÞ

5.2 A Moving Time-Harmonic Load on a Beam 143

wðx; tÞ ¼ � iQ4a

exp ði=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ X2

p� X


n offiffiffiffiffiffiffiffiffiffiffiffiffiffiffixþ X2


p� X

�2

þ Q4a

exp ð1=aÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2

pþ iX


n offiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2


pþ iX

�2 ; x� Vt\0

ð5:2:9bÞ

where

X ¼ V2a

ð5:2:10Þ

Inspecting the above equations, the radicalffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix� X2

pin the denominator gives a

critical velocity

Vcr ¼ 2affiffiffiffix

p ð5:2:11Þ

at which the deflection divergence takes place.When the load is stationary at the coordinate origin, the simple time-harmonic

response is obtained by taking the limit X ! 0 ðV ! 0Þ in Eq. (5.2.10). It yields

wðx; tÞ ¼ � Q

4affiffiffiffix

p 3 i exp þi xt �ffiffiffiffix

pa

jxj� ��

þ exp �ffiffiffiffix

pa

jxj þ ixt

� ��

ð5:2:12Þ

2

1( / 2 ) ( / 2 )c cα ω α

ξα

+ + +=

2

1( / 2 ) ( / 2 )c cα ω α

ξα

− − +=

2

2( / 2 ) ( / 2 )c i cα ω α

ξα

+ − + −=

2

2( / 2 ) ( / 2 )c i cα ω α

ξα

− − − −=

Re( )ξ

Re( )ξ

Im( )ξ Im( )ξ

C−C+

Fig. 5.1 Closed loops C� for the complex integral Φ


This is the particular solution of the deflection equation with the time-harmonicstationary load,

a4@4w@x4

þ @2w@t2

¼ QdðxÞ expðþixtÞ ð5:2:13Þ

5.3 An Impulsive Load on a Plate

Let us consider an impulsive response of an infinite elastic plate. Taking the(x, y) coordinate on the neutral plane of the plate, the deflection equation for theplate with an impulsive point loadis given by

D@4w@x4

þ 2@4w

@x2@y2þ @4w

@y4

� �þ qh

@2w@t2

¼ PdðxÞdðyÞdðtÞ ð5:3:1Þ

where D is bending rigidity, ρh density per unit area and P magnitude of the load.The initial condition,

wjt¼0¼@w@t

��t¼0

¼ 0 ð5:3:2Þ

and the convergence condition,

wj ffiffiffiffiffiffiffiffiffix2þy2

p!1¼ @w

@x


p!1

¼ @2w@x2


p!1

¼ @3w@x3


p!1

¼ 0

@w@y


p!1

¼ @2w@y2


p!1

¼ @3w@y3


p!1

¼ @2w@x@y


p!1

¼ 0

ð5:3:3Þ

are applied to the deflection equation.In order to obtain the particular solution corresponding to the non-homogeneous

loading term, we apply the triple integral transform to Eq. (5.3.1). The Laplacetransform with respect to the time t,

f �ðsÞ ¼Z10


and the double Fourier transform with respect to the space variables (x, y),

5.2 A Moving Time-Harmonic Load on a Beam 145

�f ðnÞ ¼Zþ1


2p

Zþ1

�1


~f ðgÞ ¼Zþ1

�1f ðyÞ expðþigyÞdy; f ðyÞ ¼ 1

2p

Zþ1

�1

~f ðgÞ expð�igyÞdg ð5:3:6Þ

are applied to Eq. (5.3.1). The exact solution in the transformed domain is easilyobtained as

~�w� ¼ Q

s2 þ fb2ðn2 þ g2Þg2 ð5:3:7Þ

where

b4 ¼ Dqh

; Q ¼ Pqh

ð5:3:8Þ

Now, let us consider the inversion. The Laplace inversion of Eq. (5.3.7) is easilyperformed with use of the formula (5.1.10). It follows that

~�w ¼ Q

b2ðn2 þ g2Þ sinfb2ðn2 þ g2Þtg; t[ 0 ð5:3:9Þ

The double Fourier inversion is expressed by the double integral as

w ¼ 1

ð2pÞ2Zþ1

�1

Zþ1

�1

Q

b2ðn2 þ g2Þ sinfb2ðn2 þ g2Þtg expf�iðnxþ gyÞgdndg

ð5:3:10Þ

If we introduce the polar coordinates (r, θ) and the polar integration variables(u, φ) as

x ¼ r cos h; y ¼ r sin h ð5:3:11Þ

n ¼ u sinu; g ¼ u cosu; dndg ¼ ududu ð5:3:12Þ

the double integral can be rewritten as

w ¼ Q

ð2pÞ2b2Z10

Z2p0

1usinðb2tu2Þ expf�iur sinðuþ hÞgdudu ð5:3:13Þ


The exponential function in the integrand is the generating function of Besselfunction (Watson 1966, p. 22),

expf�iur sinðuþ hÞg ¼Xþ1

�1JnðurÞ expf�inðuþ hÞg ð5:3:14Þ

where Jnð:Þ is the n-th order Bessel function of the first kind. We substitute theabove into Eq. (5.3.13) and exchange the order of the summation and integration as

w ¼ Q

ð2pÞ2b2Z10

Z2p0

1usinðb2tu2Þ

Xþ1

�1JnðurÞ expf�inðuþ hÞgdudu

¼ Q

ð2pÞ2b2Xþ1

�1

Z10

1usinðb2tu2ÞJnðurÞdu

Z2p0

expf�inðuþ hÞgduð5:3:15Þ

The inner integral with respect to the angle variable φ is easily evaluated as

Z2p0

expf�inðuþ hÞgdu ¼ 2p; n ¼ 00; n 6¼ 0

�ð5:3:16Þ

Then, we have the single integral for the deflection

w ¼ Q

2pb2

Z10

1usinðb2tu2ÞJ0ðurÞdu ð5:3:17Þ

Lastly, the integration formula (Erdélyi 1954, vol. II, pp. 11, 39),

Z10

1xsinðax2ÞJ0ðbxÞdx ¼ 1

2si

b2

4a

� �ð5:3:18Þ

where si(x) is one of sine integrals (Erdélyi 1954, vol. II, p. 430) defined by

si(xÞ ¼ �Z1x

sinðuÞu

du ð5:3:19Þ

5.3 An Impulsive Load on a Plate 147

is applied to Eq. (5.3.17). The Green’s function for the dynamic plate deflection isthus given by

w ¼ Q

4pb2si

r2

4b2t

� �; t[ 0; r ¼


pð5:3:20Þ

5.4 A Time-Harmonic Load on a Plate

Let us consider an elastic plate on which a time-harmonic point load is applied. Thedynamic deflectionof the plate is governed by the deflection equation with thenonhomogeneous loading term,

D@4w@x4

þ 2@4w

@x2@y2þ @4w

@y4

� �þ qh

@2w@t2

¼ PdðxÞdðyÞ expðþixtÞ ð5:4:1Þ

where P is the magnitude of the load (but is not of the same dimension as that of theimpulsive load in the previous section). The convergence condition at infinity,

wj ffiffiffiffiffiffiffiffiffix2þy2

p!1¼ @w

@x


p!1

¼ @2w@x2


p!1

¼ @3w@x3


p!1

¼ 0

@w@y


p!1

¼ @2w@y2


p!1

¼ @3w@y3


p!1

¼ @2w@x@y


p!1

¼ 0

ð5:4:2Þ

is also assumed. The double Fourier transform defined by Eqs. (5.3.5) and (5.3.6) isapplied to the deflection equation (5.4.1). We then have the simple differentialequation,

d2~�wdt2

þ b4ðn2 þ g2Þ2~�w ¼ Q expðþixtÞ ð5:4:3Þ

and its particular solution is given by

~�w ¼ Q

b4ðn2 þ g2Þ2 � x2expðþixtÞ ð5:4:4Þ

where β and Q are defined by Eq. (5.3.8).The formal Fourier inversion is given by the double integral

w ¼ Q

ð2pÞ2 expðþixtÞZþ1

�1

Zþ1

�1

expð�inx� igyÞb4ðn2 þ g2Þ2 � x2

dndg ð5:4:5Þ


Introducing the polar coordinates defined by Eqs. (5.3.11) and (5.3.12) in theprevious section, the double Fourier inversion integral can be rewritten as

w ¼ Q

ð2pÞ2 expðþixtÞZ10

Z2p0

u

ðbuÞ4 � x2expf�iur sinðuþ hÞgdudu ð5:4:6Þ

The inner integral with respect to the angular variable φ is evaluated with the aid ofthe formulas (5.3.14) and (5.3.16), and the above Eq. (5.4.6) is reduced to the singleintegral,

w ¼ Q

2pb4expðþixtÞ

Z10

u

u4 � ffiffiffiffix

p=bð Þ4

J0ðurÞdu ð5:4:7Þ

where the radial distance r isffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p.

The last integral with respect to the variable u is evaluated by applying thecomplex integral theory. As was done in the former time-harmonic problem inSect. 5.2 and earlier too, we introduce a small negative imaginary part to thefrequency, and then consider the complex integral

U ¼ZC

z

z4 � ffiffiffiffix

p=bð Þ4

Hð1Þ0 ðzrÞdz ð5:4:8Þ

where Hð1Þ0 ð:Þ is Hankel function of the first kind. The closed loop C is shown in

Fig. 5.2. Since the Hankel function has a logarithmic singularity and its branchpoint is at the origin, one branch cut along the negative real axis is introduced. Theintegrand has four poles, whereof two are in the upper z-plane due to the intro-duction of the imaginary part to the frequency. They are marked in the figure andthese two poles are specified with argument

z ¼ �ffiffiffiffix

pb

¼ffiffiffiffix

pb

expðþpiÞ; z ¼ þi

ffiffiffiffix

pb

¼ffiffiffiffix

pb

expðþpi=2Þ ð5:4:9Þ

It should be understood that the argument of z along the positive real axis is zero,but that along the negative real axis is þp. Then, we apply Cauchy’s theorem to thecomplex integral Φ in Eq. (5.4.8). The integral along the large semi-circle vanishesas its radius tends to infinity, and that along the small semi-circle also vanishes asits radius tends to zero. Thus, the integral along the real axis is converted to the sumof two residues at the poles. That is

5.4 A Time-Harmonic Load on a Plate 149

ZR!1

e!0

u


p=bð Þ4

Hð1Þ0 ðurÞ � Hð1Þ

0 ðurepiÞn o

du

¼ 2pi � b2

4xHð1Þ

0r

ffiffiffiffix

pb

epi=2� �

þ b2

4xHð1Þ

0r

ffiffiffiffix

pb

epi� �� ð5:4:10Þ

The formulas for the Hankel function (Watson 1966, pp. 75 and 78),

Hð1Þ0 ðxepiÞ ¼ �Hð2Þ

0 ðxÞ; Hð1Þ0 ðxepi=2Þ ¼ � 2i

pK0ðxÞ ð5:4:11Þ

are very useful for arranging Eq. (5.4.10). Then, we have the integration formula forour use,

Z10

u


p=bð Þ4

J0ðruÞdu ¼ � b2

2xpi2Hð2Þ

0r

ffiffiffiffix

pb

� �þ K0

rffiffiffiffix

pb

� �� ð5:4:12Þ

Applying this formula to Eq. (5.4.7), the time-harmonic response of the plate, i.e.the time-harmonic Green’s function, is given by

wðx; y; tÞ ¼ � Q

4pb2x

pi2Hð2Þ

0r

ffiffiffiffix

pb

� �þ K0

rffiffiffiffix

pb

� �� expðþixtÞ ð5:4:13Þ

where r is the radial distance from the load.

/z ω β= −

/z i ω β= +arg( )z π= +

arg( ) 0z =branch cut

A B

C

D E F

Im( )z

Re( )z

R → ∞

0ε →

Fig. 5.2 A closed loop C for the complex integral Φ


Tab

le5.1

Green’s

functio

nsforbeam

andplate

Deflectio

nequatio

nSo

urce

p(x,

t)Green’s

functio

n

1Dbeam

EI@

4w

@x4þqA

@2w

@t2¼

pðx;tÞ

PdðxÞd

ðtÞwðx;

tÞ¼

Q a

ffiffi t pppz ffiffi 2p

SðzÞ�CðzÞ

fgþ

sin

p 4ð2z2

þ1Þ

�

hi ;

z¼

xaffiffiffiffiffi 2p

tp

a¼

EI

qA�

1=4;

Q¼

P qA

PdðxÞexpðþi

xtÞ

wðx;

tÞ¼

�Q

4affiffiffi xp 3

iexp

þixt�

ffiffiffi xp ajxj

�

no þ

exp

�ffiffiffi xp ajxj

þixt

�

hi

2Dplate

D@4w

@x4

þ2

@4w

@x2@y2

þ@4w

@y4

��

þqh@2w

@t2

¼pðx;y;tÞ

PdðxÞd

ðyÞdðtÞ

wðx;

y;tÞ¼

Q4p

b2si

r2 4b2t

� ;

r¼


x2þy2

pPdðxÞd

ðyÞexpðþi

xtÞ

wðx;

y;tÞ¼

�Q

4pb2x

pi 2H

ð2Þ

0rffiffiffi xp b� þ

K0

rffiffiffi xp b�

no ex

pðþi

xtÞ;

r¼


x2þy2

pb¼

D qh�1=

4;

Q¼

P qh

5.4 A Time-Harmonic Load on a Plate 151

The reader will find that the deflection has two components: One is a time-

harmonic wave which is given by the product Hð2Þ0 r

ffiffiffiffix

p=bð Þ expðþixtÞ. The other

is a simple time-harmonic vibration K0 rffiffiffiffix

p=bð Þ expðþixtÞ whose amplitude

decays exponentially with the distance. The same nature can be found in the 1Dbeam response in Eq. (5.2.12). Table 5.1 shows the summary for the deflectionGreen’s functions.

Exercises

(5:1) Derive the wave velocity from the time-harmonic responses (5.2.9), (5.2.12)and (5.4.13) and show that all velocities depend on the frequency.

(5:2) Using the generating function for Bessel function (5.3.14), derive the integralrepresentation for the Bessel function as

JnðzÞ ¼ 12p

Zþp

�p

expð�iz sinuþ inuÞdu ¼ 1p

Zþp

0

cosðnu� z sinuÞdu ðaÞ

(5:3) The asymptotic formula for the Hankel function is given by

Hð1Þ0 ðzÞ�

ffiffiffiffiffi2pz

rexpfþiðz� p=4Þg; z ! 1 ðbÞ

Explain why the integral along the semi-circle BCD in Fig. 5.2 vanishes forthe complex integral (5.4.8).

Appendix

See Table 5.1.

References

Erdélyi A (ed) (1954) Tables of integral transforms, vol I and II. McGraw-Hill, New-YorkWatson GN (1966) A treatise on the theory of bessel functions. Cambridge University Press,

Cambridge


Chapter 6Cagniard-de Hoop Technique

The success of the integral transform method hinges on the evaluation of inversionintegrals. It is not always easy to find a suitable integration formula. If we cannotfind any suitable formula, the inversion is left in its integral form and somenumerical techniques must be applied for the evaluation. However, in the case ofdouble inversion, such as Laplace and Fourier inversions, if we could convert thefirst Fourier inversion integral to a definition integral of Laplace transform, the nextLaplace inversion can be carried out by inspection without evaluating its Laplaceinversion integral. For example, when we have the double transformed functionFðn; sÞ and its Fourier inversion integral

f �ðx; sÞ ¼Zþ1

�1Fðn; sÞ expð�inxÞdn ð6:1Þ

if we could apply some mathematical techniques to the integral and convert theintegral to the definition form of Laplace transform,

f �ðx; sÞ ¼Z1a

Gðx; tÞ expð�stÞdt ð6:2Þ

i.e. the Laplace transform integral, the Laplace inversion is easily carried out byinspection and its inversion is given by

f ðx; tÞ ¼ Gðx; tÞ; t[ a0; t\a

�¼ Hðt � aÞGðx; tÞ ð6:3Þ

The most important and substantial part of this technique is to convert theFourier inversion integral to the form of the Laplace transform,

Zþ1

�1Fðn; sÞ expð�inxÞdn ¼

Z1a

Gðx; tÞ expð�stÞdt ð6:4Þ


153

This conversion is carried out with aids of the theory of complex integrals and iscalled “Cagniard-de Hoop technique.” Cagniard (1962) is the author of this tech-nique, but his original technique is to map the inversion integral onto the integral inanother complex t-plane. De Hoop (1961) has modified the technique to use the samecomplex plane without mapping. The present chapter explains the technique devel-oped by de Hoop, but uses the name “Cagniard-de Hoop technique.” The essentialidea is to convert the first inversion integral to the definition integral of the secondintegral transform. This technique is applicable to other combinations of two integraltransforms, not limited to Laplace-Fourier transforms. In addition, it will be worth tocite textbooks (Achenbach 1973; Fung 1970; Graff 1975; Miklowitz 1978), whichtreat elastic waves and explain the Cagniard-de Hoop technique in some depths.

The first section in the present chapter treats a very simple problem in order todemonstrate the Cagniard-de Hoop technique in details. In the following sections,2D and 3D Green’s functions (dyadic) for the coupled elastodynamic equations arediscussed and some additional explanations for the Cagniard-de-Hoop techniqueare described.

6.1 2D Anti-plane Deformation

As the first example of the Cagniard-de Hoop technique, the simplest elastody-namic problem is discussed. We consider an elastic half space and take theCartesian coordinate (x, y, z) as shown in Fig. 6.1 where the z-axis is normal to thepaper plane. The surface of the solid is denoted by y ¼ 0 and its interior y[ 0. Animpulsive anti-plane shear load directed to the negative z-direction is applied on thesurface and is expressed by

ryz��y¼0 ¼ PzdðxÞdðtÞ ð6:1:1Þ

x

y

0( ) ( )yz zy

P x tσ δ δ=

=0y =

,μ ρ

Fig. 6.1 An impulsive anti-plane shear load on thesurface of a semi-infiniteelastic solid

154 6 Cagniard-de Hoop Technique

where Pz is the load magnitude. Since the load of infinite length is directed to theanti-plane z-direction, an anti-plane deformation is induced. The equation of motionis given by

@rxz@x

þ @ryz@y

¼ q@2uz@t2

ð6:1:2Þ

where q is the density. Hooke’s law for two shear stresses is given by

rxz ¼ l@uz@x

; ryz ¼ l@uz@y

ð6:1:3Þ

where l is the shear rigidity. We also employ the quiescent condition at an initialtime,

uzjt¼0 ¼@uz@t

��t¼0

¼ 0 ð6:1:4Þ


uzj ffiffiffiffiffiffiffiffiffix2þy2p

!1 ¼ @uz@x


p!1

¼ @uz@y


p!1

¼ 0 ð6:1:5Þ

Equations (6.1.1)–(6.1.5) constitute the present elastodynamic problem.Substituting Hooke’s law of Eq. (6.1.3) into the equation of motion (6.1.2), weobtain the typical wave equation for the anti-plane displacement uz,

@2uz@x2

þ @2uz@y2

¼ 1c2s

@2uz@t2

ð6:1:6Þ

where the shear wave velocity cs is defined by

cs ¼ffiffiffiffiffiffiffiffil=q

pð6:1:7Þ

In order to solve our elastodynamic problem, Laplace and Fourier transforms areemployed; Laplace transform with respect to the time t,

f �ðsÞ ¼Z10


and Fourier transform with respect to the space variable x,

6.1 2D Anti-plane Deformation 155

�f ðnÞ ¼Zþ1


2p

Zþ1

�1


With the aid of the quiescent and convergence conditions, the displacementequation (6.1.6) and Hooke’s law (6.1.3) are transformed to

d2�u�zdy2

� n2 þ ðs=csÞ2n o

�u�z ¼ 0 ð6:1:10Þ

�r�xz ¼ �inl�u�z ; �r�yz ¼ ld�u�zdy

ð6:1:11Þ

After solving the transformed displacement equation (6.1.10), the displacement andstress, which satisfy the convergence condition at infinity, are given by

�u�z ¼ Aðn; sÞ expð�asyÞ; �r�yz ¼ �lasAðn; sÞ expð�asyÞ ð6:1:12Þ

Here Aðn; sÞ is an unknown coefficient to be determined by the loading condition,and the radical is defined by

as ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=csÞ2

q; ReðasÞ[ 0 ð6:1:13Þ

The boundary condition is also transformed, to

�r�yz��y¼0

¼ Pz ð6:1:14Þ

Substituting the second equation in Eq. (6.1.12) into the above transformedboundary condition, the coefficient is determined as

Aðn; sÞ ¼ � Pz

lasð6:1:15Þ

and thus, the displacement in the transformed domain is explicitly determined as

�u�z ¼ � Pz

lasexpð�asyÞ ð6:1:16Þ

Now, we shall consider the inversion. The formal Fourier inversion for thedisplacement is given by the infinite integral


u�z ¼ � Pz

2pl

Zþ1

�1

1asexpð�asy� inxÞdn ð6:1:17Þ

The above Eq. (6.1.17) is the Laplace transformed displacement, but in the formof Fourier inversion integral. If we could convert the integral into the definitionform of Laplace transform integral, its Laplace inversion can be performed byinspection (i.e. the Cagniard-de Hoop technique).

Let us start to apply the technique. First of all, in order to eliminate the Laplacetransform parameter s, leaving it only in the argument of the exponential function inthe integrand, a variable transform from n to 1 is introduced as

n ¼ s1 ð6:1:18Þ

Here we assume that the Laplace transform parameter s is a positive real con-stant. The integral in Eq. (6.1.17) is rewritten as

u�z ¼ � Pz

2pl

Zþ1

�1

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

p exp �s yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

qþ i1x

� �n od1 ð6:1:19Þ

where the radical must satisfy the radiation condition,

Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

q� �[ 0 ð6:1:20Þ

Examining the integral in Eq. (6.1.19), the Laplace transform parameter isincluded only in the argument of the exponential function as a simple multiplier. Ifwe could change the argument to a simple product, such as st where t is a newvariable, the integral will have the form of the definition integral of the Laplacetransform. So, the variable transform from 1 to the new variable t is introduced as

t ¼ yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

qþ i1x ð6:1:21Þ

However, its inverse gives multiple values for 1

1ð�Þs ¼ �itx� y

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðx2 þ y2Þ=c2s

px2 þ y2

ð6:1:22Þ

Due to the multi-valuedness, we are puzzled as to which one is the suitableinversion for 1. In order to solve this puzzle, we examine the inverse as a functionof the real parameter t. We separate the real and imaginary parts of the inverse as

1ð�Þs ¼ �X � iY , where


X ¼ yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðx2 þ y2Þ=c2s

px2 þ y2

; Y ¼ txx2 þ y2

ð6:1:23Þ

and eliminate the parameter t. An equation for the hyperbolic curve in (X, Y) planeis derived as

Xy

� �2

� Yx

� �2

¼ � 1c2s

ð6:1:24Þ

Thus, the inverse 1ð�Þs has the form of two semi-hyperbolic curves in the complex 1-

plane. When x[ 0, 1ð�Þs is the left half and 1ðþÞ

s is the right half of the hyperbola inthe lower 1-plane, as shown in Fig. 6.2. On the other hand, when x\0, the

hyperbolic curves 1ð�Þs are in the upper 1-plane. The connected two semi-hyperbolic

curves CAB shown in the figure is called Cagniard’s path.If we could connect the real line with the edges of two semi-hyperbolas in the

complex plane, a closed loop for the complex integral is formed; and then Cauchy’sintegral theorem can be applied. Thus, we consider the complex integral U, whoseintegrand is the same as that of the Fourier inversion integral in Eq. (6.1.19),

U ¼ 12p

IC



qþ i1x

� �n od1 ð6:1:25Þ

Fig. 6.2 Cagniard’s path


The integrand has two branch points at

1 ¼ �i=cs ð6:1:26Þ

In order to guarantee the radiation condition of Eq. (6.1.20), two branch cuts, whichare discussed in Sect. 1.3.2 in Chap. 1, are introduced along the imaginary axis inthe complex 1-plane as shown in Fig. 6.3. When the real parameter t varies fromffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ y2p

=cs to þ1, the inverse 1ðþÞs moves on the semi-hyperbola from A to B,

and the other inverse 1ð�Þs moves on the other semi-hyperbola from A to C. These

two semi-hyperbolas are connected at the saddle point A,

1saddle ¼ � ix

csffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p at t ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cs ð6:1:27Þ

This saddle point is always smaller in magnitude than the branch point 1 ¼ �i=cs.The other edges at the infinite t must be connected with the real line. The edge B isconnected to the line at the positive infinity F with the large arc. The edge C is alsoconnected to the line at the negative infinity D with the large arc, as shown inFig. 6.3. Then, the closed loop for the complex integral is formed by the loopCABFEDC. The closed loop C(−) in the lower 1-plane is employed in the case ofpositive x and the loop C(+) in the upper plane is in the case of negative x. In either

Fig. 6.3 Closed loop Cð�Þ for the complex integral U


http://dx.doi.org/10.1007/978-3-319-17455-6_1

case, no singular point is included in the closed loop. The integral along the largearc vanishes as its radius tends to infinity, since the radiation condition ofEq. (6.1.20) is kept. Thus, the integral along the real axis is converted to that alongthe two semi-hyperbolas, CA and AB, due to Cauchy’s integral theorem.

The integral along the hyperbola is one of the parametric integrals and itsparameter is the variable t. The identity between two integrals along the real axisand along the hyperbolas is thus given by

12p

Zþ1

�1



qþ i1x

� �n od1

¼ 12p

Z1ffiffiffiffiffiffiffiffiffix2þy2

p=cs


p d1dt

��1¼1ðþÞ

s

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

p d1dt

��1¼1ð�Þ

s

8<:

9=; expð�stÞdt

ð6:1:28Þ

The integrand in the right hand side of Eq. (6.1.28) can be simplified. Explicitexpressions for the radical and the gradient are derived from the definition ofEq. (6.1.21) and its inverse Eq. (6.1.22). They are

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1ð�Þs

� �2þ 1=c2s

r ¼ x2 þ y2

yt � ixffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðx2 þ y2Þ=c2s

p

d1ð�Þs

dt¼ � yt � ix

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðx2 þ y2Þ=c2s

pðx2 þ y2Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

t2 � ðx2 þ y2Þ=c2sp

ð6:1:29Þ

Substituting these into the integral in the right hand side of Eq. (6.1.28), theparametric integral along the hyperbola is simplified as

12p


p=cs


p d1dt

��1¼1ðþÞ

s

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1=c2s

p d1dt

��1¼1ð�Þ

s

8<:

9=; expð�stÞdt

¼ 1p


p=cs

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðx2 þ y2Þ=c2s

p expð�stÞdt ð6:1:30Þ


Thus, we have converted the Fourier inversion integral to the integral form of theLaplace transform,

12p

Zþ1

�1



qþ i1x

� �n od1

¼ 1p


p=cs


p expð�stÞdtð6:1:31Þ

Substituting the right hand side of the above equation into the Laplace transformeddisplacement in Eq. (6.1.19), we have the Laplace transformed displacement in theform of Laplace transform integral,

u�z ðx; y; sÞ ¼ � Pz

pl


p=cs


p expð�stÞdt ð6:1:32Þ

This integral is just the form of Laplace transform, but its lower limit is not zero.The integral states that the original function is vanishing before t ¼


p=cs

and after this time the function has the form of the integrand. Then, we can easilyanticipate the original displacement function before Laplace transform, i.e. Laplaceinversion of Eq. (6.1.32),

uzðx; y; tÞ ¼ L�1 u�z ðx; y; sÞ

¼ � Pz

pl

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2�ðx2þy2Þ=c2s

p ; t[ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cs

0; t\ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cs

8<: ð6:1:33Þ

Utilizing Heaviside’s unit step function, the displacement is expressed in thecompact form,

uzðx; y; tÞ ¼ � Pz

pl1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

t2 � ðx2 þ y2Þ=c2sp H cst �


p� �ð6:1:34Þ

This is our final result for the double inversion. The reader should notice that we didnot use any integration formula for Laplace and Fourier inversions but we did theinversion. This is the Cagniard-de Hoop technique!

The conditional for the step function gives the circular region disturbed by thetransient SH-wave,

ðcstÞ2 [ x2 þ y2 ð6:1:35Þ


and its edge is a circular (cylindrical in 3D) wave front with the center at the sourcepoint,

ðcstÞ2 ¼ x2 þ y2 ð6:1:36Þ

6.2 2D In-plane Deformation

We consider a 2D transient response of an semi-infinite elastic solid. Take thecoordinate system (x, y) so that the surface of the solid is at y ¼ 0 and its interior iny[ 0 as shown in Fig. 6.4. An impulsive point load is applied on the surface and isexpressed by

ryx��y¼0 ¼ PxdðxÞdðtÞ; ryy

��y¼0 ¼ PydðxÞdðtÞ ð6:2:1Þ

where Pj; j ¼ x; y are the components of the load. The in-plane deformation of theelastic solid is governed by the equations of motion,

@rxx@x

þ @ryx@y

¼ q@2ux@t2

@rxy@x

þ @ryy@y

¼ q@2uy@t2

ð6:2:2Þ

where q is density. Hooke’s law for the plane strain is given by

x

y

0( ) ( )yy yy

P x tσ δ δ=

=

0( ) ( )yx xy

P x tσ δ δ=

=0y =

, ,λ μ ρ

Fig. 6.4 An impulsive loadon the surface of a semi-infinite elastic solid


rxx ¼ k@ux@x

þ @uy@y

� �þ 2l

@ux@x

ryy ¼ k@ux@x

þ @uy@y

� �þ 2l

@uy@y

rxy ¼ ryx ¼ l@ux@y

þ @uy@x

� � ð6:2:3Þ

where k and l are Lame’s constants. We employ the quiescent condition at aninitial time,

uijt¼0 ¼@ui@t

��t¼0

¼ 0; i ¼ x; y ð6:2:4Þ

and the convergence (radiation) condition at infinity

uij ffiffiffiffiffiffiffiffiffix2þy2p

!1¼ @ui@x


p!1

¼ @ui@y


p!1

¼ 0; i ¼ x; y ð6:2:5Þ

Equations from (6.2.1) to (6.2.5) constitute the impulsive Lamb’s problem(Fung 1970) for the 2D semi-infinite elastic solid. Substituting Hooke’s law into theequations of motion, the displacement equations are obtained as

c2@2ux@x2

þ @2ux@y2

þ ðc2 � 1Þ @2uy

@x@y¼ 1

c2s

@2ux@t2

ðc2 � 1Þ @2ux

@x@yþ @2uy

@x2þ c2

@2uy@y2

¼ 1c2s

@2uy@t2

ð6:2:6Þ

where c is the velocity ratio defined by Eq. (3.8) in Chap. 3. We apply the doubleintegral transform: Laplace transform with respect to the time t,

f �ðsÞ ¼Z10


and Fourier transform with respect to the space variable x,

�f ðnÞ ¼Zþ1


2p

Zþ1

�1


With use of the quiescent and radiation conditions, the displacement equations aretransformed to the coupled ordinary differential equations with constantcoefficients,

6.2 2D In-plane Deformation 163

http://dx.doi.org/10.1007/978-3-319-17455-6_3

d2�u�xdy2

� fc2n2 þ ðs=csÞ2g�u�x � inðc2 � 1Þ d�u�y

dy¼ 0

�inðc2 � 1Þ d�u�x

dyþ c2

d2�u�ydy2

� fn2 þ ðs=csÞ2g�u�y ¼ 0

ð6:2:9Þ

Hooke’s law and the boundary condition are also transformed to

1l�r�xy ¼

1l�r�yx ¼

d�u�xdy

� in�u�y

1l�r�yy ¼ �inðc2 � 2Þ�u�x þ c2

d�u�ydy

ð6:2:10Þ

and

�r�yx��y¼0

¼ Px; �r�yy��y¼0

¼ Py ð6:2:11Þ

The solution for the displacement equation (6.2.9) is given by

�u�x ¼inad

Aðn; sÞ expð�adyÞ þ Bðn; sÞ expð�asyÞ

�u�y ¼ Aðn; sÞ expð�adyÞ � inasBðn; sÞ expð�asyÞ

ð6:2:12Þ

where Aðn; sÞ and Bðn; sÞ are unknown coefficients, and the two radicals are definedby

aj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðs=cjÞ2

q; j ¼ d; s ð6:2:13Þ

with the radiation condition ReðajÞ[ 0. Substituting Eq. (6.2.12) into Eq. (6.2.10),the stress components are obtained as

1l�r�yx ¼ �2inAðn; sÞ expð�adyÞ � a2s þ n2

asBðn; sÞ expð�asyÞ

1l�r�yy ¼ � a2s þ n2

adAðn; sÞ expð�adyÞ þ 2inBðn; sÞ expð�asyÞ

ð6:2:14Þ

Applying the boundary condition of Eq. (6.2.11) to the above stresses, we obtainthe following simple algebraic equations for the unknown coefficients,


�2inAðn; sÞ � a2s þ n2

asBðn; sÞ ¼ Px=l

� a2s þ n2

adAðn; sÞ þ 2inBðn; sÞ ¼ Py=l

ð6:2:15Þ

Their solutions are given by

Aðn; sÞ ¼ � 2inadasðPx=lÞ þ adða2s þ n2ÞðPy=lÞRðn; sÞ

Bðn; sÞ ¼ � asða2s þ n2ÞðPx=lÞ � 2inadasðPy=lÞRðn; sÞ

ð6:2:16Þ

where Rðn; sÞ is Rayleigh equation, defined by

Rðn; sÞ ¼ ða2s þ n2Þ2 � 4adasn2 ð6:2:17Þ

Substitution of Eq. (6.2.16) into Eq. (6.2.12) gives the displacement in the doubletransformed domain. The transformed displacement can also be expressed as thesum of the dilatational and shear wave contributions,

�u�x ¼Px

l�IðdÞ�xx ðn; y; sÞ þ �IðsÞ�xx ðn; y; sÞn o

þ Py

l�IðdÞ�xy ðn; y; sÞ þ �IðsÞ�xy ðn; y; sÞn o

�u�y ¼Px

l�IðdÞ�yx ðn; y; sÞ þ �IðsÞ�yx ðn; y; sÞn o

þ Py

l�IðdÞ�yy ðn; y; sÞ þ �IðsÞ�yy ðn; y; sÞn o

ð6:2:18Þ

where the superscripts “d” and “s” for k in �IðkÞ�ij stand for the dilatational and shearwave contributions, respectively. The first subscript i stands for the direction of thedisplacement component and the second j for the component of the load. Moreexplicit expressions for each of the contributions are

�IðdÞ�xx ðn; y; sÞ ¼ 2n2asRðn; sÞ expð�adyÞ

�IðsÞ�xx ðn; y; sÞ ¼ � asða2s þ n2ÞRðn; sÞ expð�asyÞ

�IðdÞ�xy ðn; y; sÞ ¼ � inða2s þ n2ÞRðn; sÞ expð�adyÞ

�IðsÞ�xy ðn; y; sÞ ¼ 2inadasRðn; sÞ expð�asyÞ

ð6:2:19Þ


for �u�x , and

�IðdÞ�yx ðn; y; sÞ ¼ � 2inadasRðn; sÞ expð�adyÞ

�IðsÞ�yx ðn; y; sÞ ¼ inða2s þ n2ÞRðn; sÞ expð�asyÞ

�IðdÞ�yy ðn; y; sÞ ¼ � adða2s þ n2ÞRðn; sÞ expð�adyÞ

�IðsÞ�yy ðn; y; sÞ ¼ 2n2adRðn; sÞ expð�asyÞ

ð6:2:20Þ

for �u�y .We have obtained explicit expressions for the transformed displacement com-

ponents. Our next task is to invert the displacement into the real domain ðx; y; tÞ.However, it might be impossible to invert the displacement as a whole. Theinversion must be carried out for each wave contribution one by one. As all dila-

tational wave contributions given by �IðdÞ�ij have the same exponential function

expð�adyÞ and the shear wave contributions �IðsÞ�ij does the function expð�asyÞ, weinvert these two wave contributions separately.

(1) Inversion of the dilatational wave contribution

As a representative for the dilatational wave contribution, we consider the

inversion of �IðdÞ�xx . The formal Fourier inversion integral with respect to theparameter n is given by

IðdÞ�xx ðx; y; sÞ ¼ 12p

Zþ1

�1

2n2asRðn; sÞ expð�ady� inxÞdn ð6:2:21Þ

In order to eliminate the Laplace transform parameter s in the integrand exceptthe exponential function, the variable transform from n to the new variable 1 isintroduced as

n ¼ ðs=cdÞ1 ð6:2:22Þ

The inversion integral is rewritten as


Zþ1

�1FðdÞxx ð1Þ exp �ðs=cdÞ y

ffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pþ ix1

� �n od1 ð6:2:23Þ


where

FðdÞxx ð1Þ ¼ 212

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ c2

pRð1Þ ð6:2:24Þ

and its denominator, which is called the Rayleigh equation/function, is redefined by

Rð1Þ ¼ ð212 þ c2Þ2 � 412ffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ c2

pð6:2:25Þ

Note that the Rayleigh equation Rð1Þ differs from Rðn; sÞ in Eq. (6.2.17).Subsequently, we consider the complex integral U whose integrand is the same

as that in Eq. (6.2.23),

U ¼ 12p

ICð�Þ

FðdÞxx ð1Þ exp �ðs=cdÞ y


pþ ix1

� �n od1 ð6:2:26Þ

The closed loops C� are discussed here. The two radicalsffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pand


pin the integrand have each two branch points at 1 ¼ �i; �ic, respectively. Thus,four branch cuts are introduced along the imaginary axis in the complex 1-plane asshown in Fig. 6.5. Only the lower half plane is shown since the path and singularpoint are symmetric about the real axis. The Rayleigh equation has two pureimaginary roots, 1 ¼ �icR, where cR is the velocity ratio of the dilatational wave cdto the Rayleigh wave cR,

cR ¼ cd=cR ð6:2:27Þ

This velocity ration cR is always greater than c, since the following inequality holdsfor the isotropic media:

cR\cs\cd ð6:2:28Þ

As to the argument of the exponential function, we hope to transform it to theproduct of the new (time) variable t and the transform parameter s. So, we introducethe new variable t as

t ¼ yffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pþ ix1

� �=cd ð6:2:29Þ

Solving for the variable 1, we have two solutions, the so-called “Cagniard’s path”,

1ð�Þd ¼

�ixðcdtÞ � yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � ðx2 þ y2Þ

qx2 þ y2

ð6:2:30Þ


These solutions 1ð�Þd take the form of two symmetrical semi-hyperbolas in the

complex 1-plane as the real parameter t varies fromffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cd to infinity. The

two semi-hyperbolas are connected to each other at the saddle point

1ðdÞsaddle ¼ � ixffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p at t ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cd ð6:2:31Þ

This connected hyperbola is called Cagniard’s path as shown in Fig. 6.5. Fortu-nately, the saddle point is not on any branch cut. Then, we can determine the closedloop in the complex 1-plane.

When the Cagniard’s path is in the lower plane, the closed loop is composed ofan infinite line along the real axis, Caginiard’s path and two large arcs whichconnect the line to the Cagniard’s path. When the Caginiard’s path is in the upperplane, the closed loop is composed of the similar ones in the lower plane and issymmetric with the lower loop about the real axis. The lower loop C(−), and thebranch cuts and points are shown in Fig. 6.5. The closed loop C(−) is employedwhen the space variable x is positive, and the loop C(+) when it is negative, due tothe convergence at the large arc. In any case, positive or negative x, i.e. the upper orlower circuit in the complex plane, the closed loop does not include any singular

Im( )ς

Re( )ς

iς = −

iς γ= −

Riς γ= −

+

( )dς −

( )dς +

( )C − 0x >

2 2

ix

x yς −=

+

Fig. 6.5 Closed loop C(−) and Cagniard’s path for the dilatational wave contribution


point. Then we can apply Cauchy’s integral theorem to the complex integral U.Since the integral along the large arc vanishes as its radius tends to infinity, we canconvert the integral along the real axis to that along the Cagniard’s path. That is


Zþ1

�1FðdÞxx ð1Þ exp �ðs=cdÞ y


pþ ix1

� �n od1

¼ 12p

Zþ1

ffiffiffiffiffiffiffiffiffix2þy2

p=cd

FðdÞxx ð1ðþÞ

d Þ d1ðþÞd

dt� FðdÞ

xx ð1ð�Þd Þ d1

ð�Þd

dt

( )expð�stÞdt

ð6:2:32Þ

If we understand that the integration variable t as the real time variable, the integralhas just the form of Laplace transform integral, but with the shifted starting timeffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ y2p

=cd . As the Laplace inversion of IðdÞ�xx ðx; y; sÞ is the original functionbefore Laplace transform, the inversion is carried out by inspection on the last lineof Eq. (6.2.32). Utilizing the step function, the original function is given by

L�1 IðdÞ�xx ðx; y; sÞh i

¼ 12p

FðdÞxx ð1ðþÞ

d Þ d1ðþÞd

dt� FðdÞ


ð�Þd

dt

( )H t �


p=cd

� �ð6:2:33Þ

Consequently, the double inversion for �IðdÞ�xx ðn; y; sÞ is given by

IðdÞxx ðx; y; tÞ ¼ 12p

FðdÞxx ð1ðþÞ

d Þ d1ðþÞd

dt� FðdÞ


ð�Þd

dt

( )H t �


p=cd

� �ð6:2:34Þ

The same Cagniard-de Hoop technique can be applied to the other dilatational wave

contributions �IðdÞ�ij . Thus, we have the unified expression for the dilatational wavecontributions as

IðdÞij ðx; y; tÞ ¼ 12p

FðdÞij ð1ðþÞ

d Þ d1ðþÞd

dt� FðdÞ

ij ð1ð�Þd Þ d1

ð�Þd

dt

( )H t �


p=cd

� �ð6:2:35Þ


where FðdÞij ð1Þ are given by

FðdÞxx ð1Þ ¼ 212


pRð1Þ ; FðdÞ

xy ð1Þ ¼ � i1ð212 þ c2ÞRð1Þ

FðdÞyx ð1Þ ¼ � 2i1



pRð1Þ ; FðdÞ

yy ð1Þ ¼ � ð212 þ c2Þffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pRð1Þ

ð6:2:36Þ

(2) Inversion of the shear wave contribution

We consider the double inversion for the shear wave contribution, which has theexponential function expð�asyÞ. As a representative, we consider the inversion of

�IðsÞ�xx ðn; y; sÞ ¼ � asða2s þ n2ÞRðn; sÞ expð�asyÞ ð6:2:37Þ

Its formal Fourier inversion is given by

IðsÞ�xx ðx; y; sÞ ¼ 12p

Zþ1

�1� asða2s þ n2Þ

Rðn; sÞ� �

expð�asy� inxÞdn ð6:2:38Þ

Introducing the variable transform defined by Eq. (6.2.22), the inversion integral isrewritten as


Zþ1

�1FðsÞxx ð1Þ exp �ðs=cdÞ y


pþ i1x

� �n od1 ð6:2:39Þ

where

FðsÞxx ð1Þ ¼ � ð212 þ c2Þ


pRð1Þ ð6:2:40Þ

and the Rayleigh equation Rð1Þ is defined by Eq. (6.2.25).We shall consider the complex integral U whose integrand is the same as that in

Eq. (6.2.39),

U ¼ 12p

ICð�Þ

FðsÞxx ð1Þ exp �ðs=cdÞ y


pþ i1x

� �n od1 ð6:2:41Þ

The closed loop Cð�Þ is discussed here. Four branch points at 1 ¼ �i; �ic arefound from the two radicals, and corresponding branch cuts are introduced along


the imaginary axis in the 1-plane. The denominator, i.e. the Rayleigh equation, hastwo symmetric zeros at 1 ¼ �icR. These zeros give the simple poles for the inte-grand and the poles are on the imaginary axis (on the branch cut), but is greater inmagnitude than any of the branch points. The branch points, cuts and poles areshown in Fig. 6.6a, b.

As to the Cagniard’s path, we introduce a new parameter t, defined by

t ¼ yffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ c2

pþ i1x

� �=cd ð6:2:42Þ

Its inversion gives the Cagniard’s path in the complex 1-plane,

1ð�Þs ¼

�ixðcdtÞ � yffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcdtÞ2 � c2ðx2 þ y2Þ

qx2 þ y2

ð6:2:43Þ

The saddle point at t ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cs is

1ðsÞsaddle ¼ � icxffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p ð6:2:44Þ

Comparing the magnitude of the saddle point with those of two branch points, wefind that the saddle point is always smaller than the branch points 1 ¼ �ic of theshear wave, but greater than the branch points 1 ¼ �i of the dilatational wave, if theinequality

cjxjffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p [ 1 ) y\jxjffiffiffiffiffiffiffiffiffiffiffiffiffic2 � 1

pð6:2:45Þ

holds. Thus, we have two different closed loops for the complex integral.When the saddle point is smaller in magnitude than the branch points 1 ¼ �i, the

Cagniard’s path does not cross any branch cut; the closed loop is then composed of theline along the real axis, the Cagniard’s path, and two large arcs which connect the linewith the Cagniard’s path. This closed loop is similar to that in the case of the dilata-tional wave contribution and the Cagniard’s path is denoted by the path I in Fig. 6.6a.

On the other hand, when the saddle point is greater in magnitude than the branchpoint of the dilatational wave, i.e. the inequality of Eq. (6.2.45) holds, theCagniard’s path crosses the branch cut. The Cagniard’s path has to be deformedalong the imaginary axis, so that the closed loop does not include any branch pointsand cuts. This deformed Cagniard’s path is shown as the path II in Fig. 6.6b, wherethe regular Cagniard’s path is deformed by two short lines along the dilatationalbranch cut and a small circle around the dilatational branch point. Thus, the closedloop for this case is composed of the line along the real axis, the deformedCagniard’s path II, and two large arcs. Then, two closed loops are considered forthe complex integral Φ and its choice depends on the inequality in Eq. (6.2.45).


When the inequality of Eq. (6.2.45) does not hold, we employ the closed loopwith the Cagniard’s path I, and the Cauchy’s integral theorem is applied to thecomplex integral Φ in Eq. (6.2.41). The integral along the real axis is converted tothat along the Cagniard’s path I, since no singular point is included in the loop andthe integral along the large arc vanishes. That is,

Im( )ς

Re( )ς

iς = −

iς γ= −

Riς γ= −

( )sς −

( )sς +

2 21

x

x y

γ <+

0x >

0ε →

Im( )ς Re( )ς

iς = −

iς γ= −

Riς γ= −

( )sς − ( )

sς +

2 21

x

x y

γ >+

0x >

(a)

(b)

Fig. 6.6 Two closed loops for the 2D Cagniard’s technique. a Loop with Cagniard’s path I.b Loop with deformed Cagniard’s path II



Zþ1



pþ i1x

� �n od1

¼ 12p


p=cs

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ

xx ð1ð�Þs Þ d1

ð�Þs

dt

( )expð�stÞdt

ð6:2:46Þ

Then, the Laplace inversion is carried out by inspection and the double inversion

for �IðsÞ�xx is given by

IðsÞxx ðx; y; tÞ

¼ 12p

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ


ð�Þs

dt

( )H t �


p=cs

� �H 1� c xj jffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ y2p

!

ð6:2:47Þ

where the last step function is the conditional for the inequality.When the inequality of Eq. (6.2.45) holds, we have to employ the loop with the

Cagniard’s path II. In addition to the regular hyperbolic path, the deformedCagniard’s path II has one small circle around the branch point 1 ¼ �i and twoshort lines along the branch cut. We perform the complex integral along the closedloop shown in Fig. 6.6b. The integral around the small circle vanishes as its radiustends to zero, but the line integral along the branch cut does not vanish and must bediscussed. As was discussed in Sect. 1.3.2 in Chap. 1, the radical for the dilatationalwave,


p, has different arguments, depending on the side of the cut. How-

ever, the radicalffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ c2

pfor the shear wave does not change its argument since

the saddle point is smaller in magnitude than the branch points �ic. The sign andargument of the two radicals on the deformed Cagniard’s path II are summarized inTable 6.1, where the new radicals are introduced as

bd ¼ffiffiffiffiffiffiffiffiffiffiffiffiffig2 � 1

p; bs ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffic2 � g2

pð6:2:48Þ

Using the new notations, the complex integral U in Eq. (6.2.45) is carried outalong the closed loop with the deformed Cagniard’s path II. Since no singular pointis included in the loop and the integral along the large arc vanishes, the line integral

Table 6.1 Argument andmagnitude of the radical andRayleigh functions

Radicals Path AB: 1 ¼ �ig Path DE: 1 ¼ �igffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pþi

ffiffiffiffiffiffiffiffiffiffiffiffiffig2 � 1

p¼ þibd �i

ffiffiffiffiffiffiffiffiffiffiffiffiffig2 � 1

p¼ �ibdffiffiffiffiffiffiffiffiffiffiffiffiffiffi

12 þ c2p

þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffic2 � g2

p¼ þbs þ


p¼ þbs

Rð1Þ ðc2 � 2g2Þ2 þ 4ig2bdbs ðc2 � 2g2Þ2 � 4ig2bdbs

bd ¼ffiffiffiffiffiffiffiffiffiffiffiffiffig2 � 1

p; bs ¼


p


http://dx.doi.org/10.1007/978-3-319-17455-6_1

along the real axis is converted to that along the Cagniard’s path II. The deformedCagniard path is composed of two lines along the imaginary axis with differentargument for the radicals and the regular Cagniard’s path. Since the integral along

the small circle BCD\

vanishes as its radius tends to zero, the Fourier inversionintegral, i.e. the integral along the real axis, is converted to the three integrals


Zþ1



pþ i1x

� �n od1

¼ 12p

Zþ1


p=cs

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þ 12p

ZAB



pþ i1x

� �n od1

þ 12p

ZDE



pþ i1x

� �n od1

ð6:2:49Þ

The last two integrals along the branch line are rearranged further. With the use ofTable 6.1, the two branch line integrals are unified as

12p

ZAB



pþ i1x

� �n od1

þ 12p

ZDE



pþ i1x

� �n od1

¼ 12p

Zcxffiffiffiffiffiffiffix2þy2

p

1

f ðsÞxx ðgÞ exp �ðs=cdÞðbsyþ xgÞf gdg

ð6:2:50Þ

where the new integrand is

f ðsÞxx ðgÞ ¼ � 8g2ðc2 � 2g2Þbdb2sðc2 � 2g2Þ4 þ 16g4b2db

2s

ð6:2:51Þ

In order to transform the last integral in Eq. (6.2.50) to the form of Laplace transformintegral, we introduce the variable transform from g to the new variable t as

t ¼ bsyþ xgð Þ=cd ð6:2:52Þ


Solving for g, we have

gH ¼xðcdtÞ � y

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffic2ðx2 þ y2Þ � ðcdtÞ2

qx2 þ y2

ð6:2:53Þ

Then, the last integral in Eq. (6.2.50) is transformed to that of the Laplacetransform,

12p

Zcxffiffiffiffiffiffiffix2þy2p

1

f ðsÞxx ðgÞ exp �ðs=cdÞðbsyþ xgÞf gdg

¼ 12p

Zffiffiffiffiffiffiffiffiffix2þy2p

=cs

xþyffiffiffiffiffiffiffiffic2�1

p� =cd

f ðsÞxx ðgHÞdgHdt

expð�stÞdt

ð6:2:54Þ

Finally, we have converted the Fourier inversion integral to the form of Laplacetransform integral as


Zþ1



pþ i1x

� �n od1

¼ 12p

Zþ1


p=cs

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þ 12p

Zffiffiffiffiffiffiffiffiffix2þy2p

=cs

xþyffiffiffiffiffiffiffiffic2�1

p� =cd


expð�stÞdt

ð6:2:55Þ

The Laplace inversion is carried out by inspection. The branch line integral ofEq. (6.2.54) appears only when the conditional of Eq. (6.2.45) holds. Using the stepfunction for this conditional, the Laplace inversion is finally expressed by


¼ Hcxffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ y2p � 1

!12p

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ


ð�Þs

dt

( )H t �


p=cs

� �"

þ 12p


Hffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p=cs � t

� �H t � xþ y

ffiffiffiffiffiffiffiffiffiffiffiffiffic2 � 1

pcd

!#

ð6:2:56Þ


Comparing this expression with that of Eq. (6.2.47), we see that the aboveequation includes the full conditional. Thus, the final result for the double inver-sions of the shear wave contribution is given by


¼ 12p

FðsÞxx ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ


ð�Þs

dt

( )H t �


p=cs

� �

þ 12p


Hcxffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ y2p � 1

!H


p=cs � t

� �H t � xþ y


pcd

!

ð6:2:57Þ

The same inversion technique is applied to another double inversion for the shearwave contribution and we have the unified expression for the shear wave contri-bution as

IðsÞij ðx; y; tÞ

¼ 12p

FðsÞij ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ

ij ð1ð�Þs Þ d1

ð�Þs

dt

( )H cst �


p� �

þ 12p

f ðsÞij ðgHÞdgHdt

Hcjxjffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p � 1

!H


p� cst

� �H cdt � jxj � y


p� �ð6:2:58Þ

where the integrands are given by

FðsÞxx ð1Þ ¼ � ð212 þ c2Þ


pRð1Þ ; FðsÞ

xy ð1Þ ¼2i1



pRð1Þ

FðsÞyx ð1Þ ¼

i1ð212 þ c2ÞRð1Þ ; FðsÞ

yy ð1Þ ¼212


pRð1Þ

ð6:2:59Þ

f ðsÞxx ðgÞ ¼ � 8g2ðc2 � 2g2Þbdb2sRHðgÞ ; f ðsÞxy ðgÞ ¼ � 4gðc2 � 2g2Þ2bdbs

RHðgÞ

f ðsÞyx ðgÞ ¼8g3ðc2 � 2g2Þbdbs

RHðgÞ ; f ðsÞyy ðgÞ ¼ 4g2ðc2 � 2g2Þ2bdRHðgÞ

ð6:2:60Þ

and

RHðgÞ ¼ ðc2 � 2g2Þ4 þ 16g4b2db2s ð6:2:61Þ



We have just inverted the two wave contributions and thus the displacementresponse is given by

ux ¼ Px

lIðdÞxx ðx; y; tÞ þ IðsÞxx ðx; y; tÞn o

þ Py

lIðdÞxy ðx; y; tÞ þ IðsÞxy ðx; y; tÞn o

uy ¼ Px

lIðdÞyx ðx; y; tÞ þ IðsÞyx ðx; y; tÞn o

þ Py

lIðdÞyy ðx; y; tÞ þ IðsÞyy ðx; y; tÞn o ð6:2:62Þ

When we express the displacement in terms of Green’s dyadic Gijðx; y; tÞ,

ux ¼ PxGxxðx; y; tÞ þ PyGxxðx; y; tÞuy ¼ PxGyxðx; y; tÞ þ PyGyyðx; y; tÞ

ð6:2:63Þ

the dyadic is given by

Gijðx; y; tÞ ¼ 1l

IðdÞij ðx; y; tÞ þ IðsÞij ðx; y; tÞn o

ð6:2:64Þ

A more detailed expression is

Gijðx; y; tÞ

¼ 12pl

FðdÞxx ð1ðþÞ

d Þ d1ðþÞd

dt� FðdÞ


ð�Þd

dt

( )Hðcdt � rÞ

"

þ FðsÞij ð1ðþÞ

s Þ d1ðþÞs

dt� FðsÞ

ij ð1ð�Þs Þ d1

ð�Þs

dt

( )Hðcst � rÞ

þ f ðsÞij ðgHÞdgHdt

Hðcjxj � rÞHðr � cstÞH cdt � jxj � yffiffiffiffiffiffiffiffiffiffiffiffiffic2 � 1

p� ��ð6:2:65Þ

where the radial distance r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

pis introduced.

It should be noticed that the step function Hðcdt � rÞ shows the disturbed cir-cular region produced by the dilatational wave and Hðcst � rÞ does the regiondisturbed by the shear wave. These two circular waves are emanating from thesource point. The triple product of the step functions

Hðcjxj � rÞHðr � cstÞH cdt � jxj � yffiffiffiffiffiffiffiffiffiffiffiffiffic2 � 1

p� �

gives the region disturbed by the shear wave which is induced by the precursordilatational wave. We call this region “von Schmidt wave”. Its wave region isshown in Fig. 6.7.


6.3 3D Dynamic Lamb’s Problem

The transient response of a fully 3D elastic half space is discussed here. We take the3D coordinate system (x, y, z). The surface is denoted by z ¼ 0 and the interior byz[ 0 as shown in Fig. 6.8. The 3D deformation of the isotropic elastic solid isgoverned by equations of motion,

@rxx@x

þ @ryx@y

þ @rzx@z

¼ q@2ux@t2

@rxy@x

þ @ryy@y

þ @rzy@z

¼ q@2uy@t2

@rxz@x

þ @ryz@y

þ @rzz@z

¼ q@2uz@t2

ð6:3:1Þ

and Hooke’s law for the isotropic elastic solid,

rxx ¼ k@ux@x

þ @uy@y

þ @uz@z

� �þ 2l

@ux@x

ryy ¼ k@ux@x

þ @uy@y

þ @uz@z

� �þ 2l

@uy@y

rzz ¼ k@ux@x

þ @uy@y

þ @uz@z

� �þ 2l

@uz@z

x

2 1 | |

( | | )

y x

x r

γγ

= −=

2

| |

1

dc t xy

γ

−=

−

sc t r=

dc t r=

y

Fig. 6.7 von Schmidt wave region



þ @uy@x

� �

ryz ¼ rzy ¼ l@uz@y

þ @uy@z

� �

rzx ¼ rxz ¼ l@ux@z

þ @uz@x

� �ð6:3:2Þ

where k and l are Lame’s constants.We assume that a point load is suddenly applied at the coordinate origin on the

surface z ¼ 0, i.e.

rzxjz¼0 ¼ PxdðxÞdðyÞHðtÞrzy��z¼0 ¼ PydðxÞdðyÞHðtÞ

rzzjz¼0 ¼ PzdðxÞdðyÞHðtÞð6:3:3Þ

where Pj, j = x, y, z are components of the load. Note that the loading time function isHeaviside’s unit step function, notDirac’s delta function. This is because thatwe get thesimpler solution for the dynamic response. The quiescent condition at an initial time

uijt¼0 ¼@ui@t

��t¼0

¼ 0; i ¼ x; y; z ð6:3:4Þ


uij ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2þy2þz2p

!1¼ @ui@x


p!1

¼ @ui@y


p!1

¼ @ui@z


p!1

¼ 0; i ¼ x; y; z

ð6:3:5Þ

are employed.

x

y

0( ) ( ) ( )zz zz

P x y H tσ δ δ=

=

0( ) ( ) ( )zx xz

P x y H tσ δ δ=

=

0z =

, ,λ μ ρ

z

0( ) ( ) ( )zy yz

P x y H tσ δ δ=

=

Fig. 6.8 A suddenly appliedload Pi on the surface of asemi-infinite elastic solid

6.3 3D Dynamic Lamb’s Problem 179

Substituting Hooke’s law of Eq. (6.3.2) into the equation of motion (6.3.1), thedisplacement equations are given by

c2@2ux@x2

þ @2ux@y2

þ @2ux@z2

þ ðc2 � 1Þ @2uy@x@y

þ @2uz@x@z

� �¼ 1

c2s

@2ux@t2

@2uy@x2

þ c2@2uy@y2

þ @2uy@z2

þ ðc2 � 1Þ @2ux@x@y

þ @2uz@y@z

� �¼ 1

c2s

@2uy@t2

@2uz@x2

þ @2uz@y2

þ c2@2uz@z2

þ ðc2 � 1Þ @2ux@x@z

þ @2uy@y@z

� �¼ 1

c2s

@2uz@t2

ð6:3:6Þ

The present 3D elastodynamic problem is also discussed by the integral trans-form method. With use of the quiescent and convergence conditions, we applyLaplace transform with respect to the time, defined by

f �ðsÞ ¼Z10


and the double Fourier transform with respect to two space variables x and y,defined by

~�f ðn; gÞ ¼Zþ1

�1

Zþ1

�1f ðx; yÞ expðþinxþ igyÞdxdy;

f ðx; yÞ ¼ 1

ð2pÞ2Zþ1

�1

Zþ1

�1

~�f ðn; gÞ expð�inx� igyÞdndgð6:3:8Þ

to the displacement equations (6.3.6) and Hooke’s law of Eq. (6.3.2). The dis-placement equations are transformed to the coupled ordinary differential equations,

d2~�u�xdz2

� fc2n2 þ g2 þ ðs=csÞ2g~�u�x � ngðc2 � 1Þ~�u�y � inðc2 � 1Þ d~�u�zdz

¼ 0

d2~�u�ydz2

� fn2 þ c2g2 þ ðs=csÞ2g~�u�y � ngðc2 � 1Þ~�u�x � igðc2 � 1Þ d~�u�zdz

¼ 0

c2d2~�u�zdz2

� fn2 þ g2 þ ðs=csÞ2g~�u�z � inðc2 � 1Þ d~�u�x

dz� igðc2 � 1Þ d

~�u�ydz

¼ 0

ð6:3:9Þ


The stress components to be used for the boundary condition are transformed asfollows:

1l~�r�zx ¼

d~�u�xdz

� in~�u�z

1l~�r�zy ¼

d~�u�ydz

� ig~�u�z

1l~�r�zz ¼ c2

d~�u�zdz

� inðc2 � 2Þ~�u�x � igðc2 � 2Þ~�u�y

ð6:3:10Þ

The boundary condition on the surface is also transformed to

1l~�r�zj��z¼0

¼ Pj

ls; j ¼ x; y; z ð6:3:11Þ

The general solution for the displacement equations (6.3.9) is obtained as

~�u�x ¼inad

Aðn; g; sÞ expð�adzÞ þ Bðn; g; sÞ expð�aszÞ

~�u�y ¼igad

Aðn; g; sÞ expð�adzÞ þ Cðn; g; sÞ expð�aszÞ

~�u�z ¼ Aðn; g; sÞ expð�adzÞ � ias

nBðn; g; sÞ þ gCðn; g; sÞf g expð�aszÞ

ð6:3:12Þ

The stress components are

1l~�r�zx ¼ �2inAðn; g; sÞ expð�adzÞ � 1

asða2s þ n2ÞBðn; g; sÞ þ ngCðn; g; sÞ

expð�aszÞ1l~�r�zy ¼ �2igAðn; g; sÞ expð�adzÞ � 1

asngBðn; g; sÞ þ ða2s þ n2ÞCðn; g; sÞ

expð�aszÞ

1l~�r�zz ¼ � a2s þ n2 þ g2

adAðn; g; sÞ expð�adzÞ þ 2i nBðn; g; sÞ þ gCðn; g; sÞf g expð�aszÞ

ð6:3:13Þ

where Aðn; g; sÞ; Bðn; g; sÞ and Cðn; g; sÞ are unknown coefficients to be determinedby the boundary condition. The radicals are defined by

aj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ g2 þ ðs=cjÞ2

q; ReðajÞ[ 0; j ¼ d; s ð6:3:14Þ

where the conditional ReðajÞ[ 0 guarantees the convergence condition at infinity.Applying the boundary condition (6.3.11) to the stresses of Eq. (6.3.13), we obtainthe following simple algebraic equations for the coefficients, Aðn; g; sÞ; Bðn; g; sÞand Cðn; g; sÞ,


�2inAðn; g; sÞ � 1as

ða2s þ n2ÞBðn; g; sÞ þ ngCðn; g; sÞ ¼ Px

ls

�2igAðn; g; sÞ � 1as

ngBðn; g; sÞ þ ða2s þ n2ÞCðn; g; sÞ ¼ Py

ls

� a2s þ n2 þ g2

adAðn; g; sÞ þ 2i nBðn; g; sÞ þ gCðn; g; sÞf g ¼ Pz

ls

ð6:3:15Þ

The coefficients are determined as

Aðn; g; sÞ ¼ � 2inadassRðn; g; sÞ

Px

l� 2igadassRðn; g; sÞ

Py

l� adða2s þ n2 þ g2Þ

sRðn; g; sÞPz

lð6:3:16Þ

Bðn; g; sÞ ¼ n2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ � 1

sas

� �Px

l

þ ngða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ

Py

lþ 2inadassRðn; g; sÞ

Pz

l

ð6:3:17Þ

Cðn; g; sÞ ¼ ngða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ

Px

l

þ g2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ � 1

sas

� �Py

lþ 2igadassRðn; g; sÞ

Pz

l

ð6:3:18Þ

where Rayleigh equation is defined by

Rðn; g; sÞ ¼ ðn2 þ g2 þ a2s Þ2 � 4adasðn2 þ g2Þ ð6:3:19Þ

Then, the transformed displacement components are given by

~�u�x ¼Px

l~�I�ðdÞxx ðn; g; z; sÞ þ ~�I�ðsÞxx ðn; g; z; sÞ þ ~�I�ðSHÞ

xx ðn; g; z; sÞn o

þ Py

l~�I�ðdÞxy ðn; g; z; sÞ þ ~�I�ðsÞxy ðn; g; z; sÞn o

þ Pz

l~�I�ðdÞxz ðn; g; z; sÞ þ ~�I�ðsÞxz ðn; g; z; sÞn o ð6:3:20aÞ

~�u�y ¼Px

l~�I�ðdÞyx ðn; g; z; sÞ þ ~�I�ðsÞyx ðn; g; z; sÞn o

þ Py

l~�I�ðdÞyy ðn; g; z; sÞ þ ~�I�ðsÞyy ðn; g; z; sÞ þ ~�I�ðSHÞ

yy ðn; g; z; sÞn o

þ Pz

l~�I�ðdÞyz ðn; g; z; sÞ þ ~�I�ðsÞyz ðn; g; z; sÞn o ð6:3:20bÞ


~�u�z ¼Px

l~�I�ðdÞzx ðn; g; z; sÞ þ ~�I�ðsÞzx ðn; g; z; sÞn o

þ Py

l~�I�ðdÞzy ðn; g; z; sÞ þ ~�I�ðsÞzy ðn; g; z; sÞn o

þ Pz

l~�I�ðdÞzz ðn; g; z; sÞ þ ~�I�ðsÞzz ðn; g; z; sÞn o ð6:3:20cÞ

where

~�I�ðdÞxx ðn; g; z; sÞ ¼ 2n2assRðn; g; sÞ expð�adzÞ

~�I�ðsÞxx ðn; g; z; sÞ ¼ n2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�aszÞ

~�I�ðSHÞxx ðn; g; z; sÞ ¼ � 1

sasexpð�aszÞ

~�I�ðdÞxy ðn; g; z; sÞ ¼ 2ngassRðn; g; sÞ expð�adzÞ

~�I�ðsÞxy ðn; g; z; sÞ ¼ ngða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�aszÞ

~�I�ðdÞxz ðn; g; z; sÞ ¼ � inða2s þ n2 þ g2ÞsRðn; g; sÞ expð�adzÞ

~�I�ðsÞxz ðn; g; z; sÞ ¼ 2inadassRðn; g; sÞ expð�aszÞ

ð6:3:21aÞ

~�I�ðdÞyx ðn; g; z; sÞ ¼ 2ngassRðn; g; sÞ expð�adzÞ

~�I�ðsÞyx ðn; g; z; sÞ ¼ ngða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�aszÞ

~�I�ðdÞyy ðn; g; z; sÞ ¼ 2g2assRðn; g; sÞ expð�adzÞ

~�I�ðsÞyy ðn; g; z; sÞ ¼ g2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�aszÞ

~�I�ðSHÞyy ðn; g; z; sÞ ¼ � 1

sasexpð�aszÞ

~�I�ðdÞyz ðn; g; z; sÞ ¼ � igða2s þ n2 þ g2ÞsRðn; g; sÞ expð�adzÞ

~�I�ðsÞyz ðn; g; z; sÞ ¼ 2igadassRðn; g; sÞ expð�aszÞ

ð6:3:21bÞ


~�I�ðdÞzx ðn; g; z; sÞ ¼ � 2inadassRðn; g; sÞ expð�adzÞ

~�I�ðsÞzx ðn; g; z; sÞ ¼ inða2s þ n2 þ g2ÞsRðn; g; sÞ expð�aszÞ

~�I�ðdÞzy ðn; g; z; sÞ ¼ � 2igadassRðn; g; sÞ expð�adzÞ

~�I�ðsÞzy ðn; g; z; sÞ ¼ igða2s þ n2 þ g2ÞsRðn; g; sÞ expð�aszÞ

~�I�ðdÞzz ðn; g; z; sÞ ¼ � adða2s þ n2 þ g2ÞsRðn; g; sÞ expð�adzÞ

~�I�ðsÞzz ðn; g; z; sÞ ¼ 2adðn2 þ g2ÞsRðn; g; sÞ expð�aszÞ

ð6:3:21cÞ

In the above equations, the superscript k in ~�I�ðkÞij indicates the wave type. Thesuperscript “d” stands for the dilatational wave (P-wave), “s” for the verticallypolarized shear wave (SV-wave), and “SH” for the horizontally polarized shearwave (SH-wave). The meaning of the subscripts i and j are the same as those in the2D problem. Note that the SH-wave contribution appears only in the horizontalcomponents and that no SH-wave contribution is included in the vertical compo-nent which is normal to the surface.

Now, we shall invert each wave component separately.

(1) Inversion of P wave contribution: ~�I�ðdÞij ðn; g; z; sÞ

As a representative, we consider the inversion of ~�I�ðdÞxx ðn; g; z; sÞ. Its formalFourier double inversion with respect to the parameters n and g is given by

I�ðdÞxx ðx; y; z; sÞ ¼ 1

ð2pÞ2Zþ1

�1

Zþ1

�1

2n2assRðn; g; sÞ expð�adz� inx� igyÞdndg ð6:3:22Þ

We introduce the variable transform from ðn; gÞ to the new variables (p, q) definedby

n ¼ ðs=rÞðxp� yqÞ; g ¼ ðs=rÞðypþ xqÞ ð6:3:23Þ

where the horizontal distance defined by r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

pis introduced. This variable

transform corresponds to the rotation of the axes ðn; gÞ. The equivalent area elementis given by dndg ¼ s2dpdq. The double integral in Eq. (6.3.22) is converted toanother double integral,


I�ðdÞxx ðx; y; z; sÞ ¼ 1

ð2prÞ2Zþ1

�1

Zþ1

�1

2ðx2p2 þ y2q2 � 2xypqÞbsRðp; qÞ exp �sðbdzþ irpÞf gdpdq

ð6:3:24Þ

where

Rðp; qÞ ¼ ðb2s þ p2 þ q2Þ2 � 4bdbsðp2 þ q2Þ ð6:3:25Þ

bj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2j

q; ReðbjÞ[ 0; j ¼ d; s ð6:3:26Þ

Note that the Rayleigh equation R(p, q) is different from the former Rayleighequation Rðn; g; sÞ.

The integrand in Eq. (6.3.24) includes both odd and even functions of the newvariable q. As the integration with respect to q is carried out over the whole rangefrom �1 to þ1, the integral whose integrand has an odd power (or the anti-symmetric function of q) vanishes. Equation (6.3.24) is then slightly simplified, to

I�ðdÞxx ðx; y; z; sÞ ¼Z10

dqZþ1

�1FðdÞxx ðp; qÞ exp �sðbdzþ irpÞf gdp ð6:3:27Þ

where the new notation for the integrand is introduced as

FðdÞxx ðp; qÞ ¼ ðx2p2 þ y2q2Þbs

ðprÞ2Rðp; qÞ ð6:3:28Þ

The inner integral in Eq. (6.3.27) is in a convenient form for the application ofCagniard-de Hoop technique; thus we apply the technique to the inner integral withrespect to the variable p.

Now, we shall consider the complex integral U whose integrand is the same asthat of the inner integral in Eq. (6.3.27),

U ¼ZC

FðdÞxx ðp; qÞ exp �sðbdzþ irpÞf gdp ð6:3:29Þ

In order to transform the argument of the exponential function to the simple productof the new variable t and the Laplace transform parameter s, the new variable t isintroduced as

t ¼ bdzþ irp ð6:3:30Þ


Solving for p, the Cagniard’s path is given by

pð�Þd ¼ �irt � z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðq2 þ 1=c2dÞR2

pR2 ð6:3:31Þ

and its gradient is

dpð�Þd

dt¼ � zt � ir


pR2


p ð6:3:32Þ

where the 3D radial distance R is defined by


pð6:3:33Þ

The integrand has four branch points at

p ¼ �iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2d

q; �i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2s

qð6:3:34Þ

From these branch points, four branch cuts are introduced along the imaginary axisin the complex p-plane as was discussed in Sect. 1.3.2 in Chap. 1. Since the saddlepoint of the Cagniard’s path at t ¼ R

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2d

p,

pðdÞsaddle ¼ � irR


qð6:3:35Þ

is smaller in magnitude than any of the branch points, the Cagniard’s path does notcross any cut. In addition, the Rayleigh poles p ¼ �i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2R

p, which are

derived from Rðp; qÞ ¼ 0, are on the imaginary axis and is greater in magnitudethan the saddle point and all branch points. Then, the closed loop C is composed ofthe infinite line along the real p-axis, the Cagniard’s path, and two large arcs whichconnect the Cagniard’s path to the line on the real axis.

The closed loop C is shown in Fig. 6.9. The complex integral U in Eq. (6.3.29) iscarried out along the closed loop. Since no singular point is included in the loop andthe integrals along the large arc vanish, the integral along the real axis is convertedto that along the Cagniard’s path, i.e.

Zþ1

�1FðdÞxx ðp; qÞ exp �sðbdzþ irpÞf gdp

¼Z1

Rffiffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2d

pFðdÞxx ðpðþÞ

d ; qÞ dpðþÞd

dt� FðdÞ

xx ðpð�Þd ; qÞ dp

ð�Þd

dt

( )expð�stÞdt

ð6:3:36Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

Substituting the above integral into the inner integral in Eq. (6.3.27), we have


dqZþ1


pFðdÞxx ðpðþÞ

d ; qÞ dpðþÞd

dt� FðdÞ


ð�Þd

dt

( )expð�stÞdt

ð6:3:37Þ

As the inner integral is just in the form of Laplace transform integral, we exchangethe order of integration so that the outer integral can be in the form of Laplacetransform. In discussing the supporting region for the double integral as shown inFig. 6.10, the exchange is symbolically carried out as

Z10

dqZ1


pgðt; qÞ expð�stÞdt ¼

Z1R=cd

expð�stÞdtZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2d

p

0

gðt; qÞdq

ð6:3:38Þ

Im( )p

Re( )p

2 21/ di q c− +

2 21/ si q c− +

2 21/ Ri q c− +

( )dp−

( )dp+

Closed loop C and Cagniard’s path

Fig. 6.9 Closed loop and Cagniard’s path for the complex integral U


where g(t, q) is an arbitrary non-singular function. Then, the double integral inEq. (6.3.37) is converted to the Laplace transform integral as


R=cd

expð�stÞdtZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2d

p

0

FðdÞxx ðpðþÞ

d ; qÞ dpðþÞd

dt� FðdÞ


ð�Þd

dt

( )dq

ð6:3:39Þ

We have just arrived at the form of Laplace transform integral, and its integrand isin the form of the finite integral with respect to the variable q. The Laplaceinversion is carried out by inspection and it results in the following integral:

IðdÞxx ðx; y; z; tÞ ¼ Hðt � R=cdÞZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2d

p

0

FðdÞxx ðpðþÞ

d ; qÞ dpðþÞd

dt� FðdÞ


ð�Þd

dt

( )dq

ð6:3:40Þ

We can easily find that the disturbed region by the dilatational wave is given by theoperation of the step function Hðt � R=cdÞ and it is the semi-sphere R� cdt withradius cdt.

The triple inversion for the dilatational wave contribution has been carried outsuccessfully. The other dilatational wave contributions can be inverted by the sametechnique developed here. The unified expression for the dilatational wave con-tribution is given by

IðdÞij ðx; y; z; tÞ ¼ Hðt � R=cdÞZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2d

p

0

FðdÞij ðpðþÞ

d ; qÞ dpðþÞd

dt� FðdÞ

ij ðpð�Þd ; qÞ dp

ð�Þd

dt

( )dq

ð6:3:41Þ

t

t

2 2( / ) 1/ dq t R c= −

2 21/ dt R q c= +

/ dR c

q

Fig. 6.10 Supporting regionfor the double integral inEq. (6.3.38)


where pð�Þd are defined by Eq. (6.3.31). The integrands are given by

FðdÞxx ðp; qÞ ¼ ðx2p2 þ y2q2Þbs

ðprÞ2Rðp; qÞFðdÞxy ðp; qÞ ¼ xyðp2 � q2Þbs

ðprÞ2Rðp; qÞ

FðdÞxz ðp; qÞ ¼ � ixpðb2s þ p2 þ q2Þ

2p2rRðp; qÞ

ð6:3:42aÞ

FðdÞyx ðp; qÞ ¼ xyðp2 � q2Þbs

ðprÞ2Rðp; qÞ

FðdÞyy ðp; qÞ ¼ ðy2p2 þ x2q2Þbs

ðprÞ2Rðp; qÞ

FðdÞyz ðp; qÞ ¼ � iypðb2s þ p2 þ q2Þ

2p2rRðp; qÞ

ð6:3:42bÞ

FðdÞzx ðp; qÞ ¼ � ixpbdbs

p2rRðp; qÞFðdÞzy ðp; qÞ ¼ � iypbdbs

p2rRðp; qÞ

FðdÞzz ðp; qÞ ¼ � ðb2s þ p2 þ q2Þbd

2p2Rðp; qÞ

ð6:3:42cÞ

(2) Inversion of SV wave contribution: ~�I�ðsÞij ðn; g; z; sÞWe consider the inversion of the typical SV-wave contribution,

~�I�ðsÞxx ðn; g; z; sÞ ¼ n2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�aszÞ ð6:3:43Þ

The formal double Fourier inversion is given by

I�ðsÞxx ðx; y; z; sÞ ¼ 1

ð2pÞ2Zþ1

�1

Zþ1

�1

n2ða2s þ n2 þ g2 � 4adasÞsasRðn; g; sÞ expð�asz� inx� igyÞdndg

ð6:3:44Þ


The variable transform defined by Eq. (6.3.23) is introduced. Examining the oddand non-symmetric nature with respect to the variable q, we obtain the followingsimpler form for the double integral,

I�ðsÞxx ðx; y; z; sÞ ¼Z10

dqZþ1

�1FðsÞxx ðp; qÞ expf�sðbszþ irpÞgdp ð6:3:45Þ

where the integrand is given by

FðsÞxx ðp; qÞ ¼

ðx2p2 þ y2q2Þðb2s þ p2 þ q2 � 4bdbsÞ2ðprÞ2bsRðp; qÞ

ð6:3:46Þ

Now, we consider the complex integral U whose integrand is the same as that ofthe inner integral in Eq. (6.3.45),

U ¼ZC

FðsÞxx ðp; qÞ expf�sðbszþ irpÞgdp ð6:3:47Þ

In order to determine the closed loopC, the Cagniard’s path is examined. Introducingthe time variable t as

t ¼ bszþ irp ð6:3:48Þ

the Cagniard’s path is given by

pð�Þs ¼ �irt � z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðq2 þ 1=c2s ÞR2

pR2 ð6:3:49Þ

and its saddle point at the time t ¼ Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2s

pis

pðsÞsaddle ¼ �irR


qð6:3:50Þ

The integrand has the four branch points given by Eq. (6.3.34) and branch cutsare introduced along the imaginary axis as shown in Fig. 6.11a, b. Comparing themagnitude of the branch points with that of the saddle point, we see that there aretwo cases: the saddle point is on the branch cut or it is not; in other words, theCagniard’s path crosses the branch cut or it does not.

When the saddle point is smaller in magnitude than the branch point of thedilatational wave,

rR


q\


qð6:3:51Þ


the Cagniard’s path does not cross any branch cut. Then, the closed loop C iscomposed of the infinite line along the real axis, the Cagniard’s path, and two largearcs which connect the line to the Cagniard’s path. This closed loop is similar to

Im( )p

Re( )p

2 21/ di q c− +

2 21/ si q c− +

2 21/ Ri q c− +

( )sp−

( )sp+

2 2 2 21/ 1/s dr

q c q cR

+ < +

0ε →

Im( )p Re( )p

2 21/ di q c− +

2 21/ si q c− +

2 21/ Ri q c− +

( )sp−

( )sp+

2 2

| |1

x

x y

γ >+

(a)

(b)

Fig. 6.11 Two closed loops for the 3D Cagniard’s technique. a Loop with Cagniard’s path I.b Loop with deformed Cagniard’s path II


that in the case of the dilatational wave and is shown in Fig. 6.11a. Employing thisclosed loop, we apply Cauchy’s theorem to the complex integral U in Eq. (6.3.47).Since the loop does not include any singular point, the integral along the real axis isconverted to that along the Cagniard’s path, i.e.

Zþ1

�1FðsÞxx ðp; qÞ expf�sðbszþ irpÞgdp

¼Z1

Rffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

pFðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ

xx ðpð�Þs ; qÞ dp

ð�Þs

dt

( )expð�stÞdt

ð6:3:52Þ

When the saddle point is larger in magnitude than the branch point of thedilatational wave,

rR


q[


q) q\

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr=csÞ2 � ðR=cdÞ2

q=z ð6:3:53Þ

the Cagniard’s path crosses the branch cut. In order to avoid the crossing, the path isdeformed along the cut, such as the Cagniard’s path II in Sect. 6.2. Our deformedCagniard’s path is composed of two symmetric semi-hyperbolas, two short linesalong the cut, and a small circle around the the branch point of the dilatationalwave. We name the deformed path the Cagniard path II, too. Figure 6.11b showsthe Cagniard path II and the closed loop C. The closed loop C has two large arcswhich connect the line with the Cagniard path II. Then, we apply Cauchy’s theoremto the complex integral U with this closed loop.

No singular point is included in the loop. The integral along the small circle

(BCD\

) vanishes as its radius tends to zero, and those along the large arcs (IJ\and

FG\) also vanish as the radius tends to infinity. Then, the integral along the real axis

(IHG��!

) is converted to the sum of two integrals. One is that along the regular

Cagniard’s path ( AJ�!

and EF�!

) and the other is the sum of two line integrals along

the branch cut (AB�!

and DE�!

). Thus the integral along the real axis is converted tothe sum of two integrations as

Zþ1


¼Z1


pFðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þZAB�! þ

ZDE�!

0BB@

1CCAFðsÞ

xx ðp; qÞ expf�sðbszþ irpÞgdp ¼ 0

ð6:3:54Þ


Following the method in Sect. 1.3.2 in Chap. 1, we examine the argument of the

radicals along the branch cut, AB�!

and DE�!

in Fig. 6.11b. Due to the radiation

condition ReðbjÞ[ 0, the argument of the radical bd on AB�!

, which is the left side

of the cut, is þp=2 and that on the right side DE�!

is �p=2. The argument of theradicals and the integration variable along the cut are summarized as follows:

On AB�!

: p ¼ �i1;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2d

q\ 1\

rR


qbd ¼ þi

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 � ðq2 þ 1=c2dÞ

q; bs ¼ þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2s � 12

q ð6:3:55Þ

On DE�!

: p ¼ �i1;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq2 þ 1=c2d

q\ 1\

rR


qbd ¼ �i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 � ðq2 þ 1=c2dÞ

q; bs ¼ þ


q ð6:3:56Þ

The two line integrals along the branch cut are calculated and are unified as follows:

ZAB�! þ

ZDE�!

0BB@

1CCAFðsÞ

xx ðp; qÞ expf�sðbszþ irpÞgdp

¼ � Rðr=RÞffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p

ffiffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2d

p FðsÞxx ðp; qÞ

�� p¼�i1

bd¼þicdbs¼cs

expf�sðcszþ 1rÞgð�id1Þ

þ Rðr=RÞffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


p FðsÞxx ðp; qÞ

�� p¼�i1bd¼�icdbs¼cs

expf�sðcszþ 1rÞgð�id1Þ

¼ Rðr=RÞffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


p f ðsÞxx ð1; qÞ expf�sðcszþ r1Þgd1

ð6:3:57Þ

where a new notation for the integrand is introduced as

f ðsÞxx ð1; qÞ ¼ ð�iÞ FðsÞxx ðp; qÞ

�� p¼�i1bd¼�icdbs¼cs

�FðsÞxx ðp; qÞ

�� p¼�i1bd¼þicdbs¼cs

8<:

9=; ð6:3:58Þ

and

cd ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 � ðq2 þ 1=c2dÞ

q; cs ¼


qð6:3:59Þ

In order to reduce the last integral in Eq. (6.3.57) to the form of Laplacetransform, the variable transform from 1 to the time t is introduced as


http://dx.doi.org/10.1007/978-3-319-17455-6_1

t ¼ cszþ 1r ð6:3:60Þ

Its inverse is

1H ¼ rt � zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðq2 þ 1=c2s ÞR2 � t2

pR2 ð6:3:61Þ

The integral along the branch cut is thus converted to that of Laplace transform,

Zðr=RÞffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


pf ðsÞxx ð1; qÞ expf�sðcszþ r1Þgd1

¼ZRffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p

zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s�1=c2d

pþr


pf ðsÞxx ð1H ; qÞ d1Hdt� �

expð�stÞdt

ð6:3:62Þ

Substituting the above equation into Eq. (6.3.54), we have the inner integral in theform of Laplace transform integral,

Zþ1


¼Z1


pFðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þZRffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


pþr


pf ðsÞxx ð1H ; qÞ

d1Hdt

� �expð�stÞdt

ð6:3:63Þ

Comparing this equation with Eq. (6.3.52), we readily see that the last integral inthe above equation appears only when the conditional of Eq. (6.3.53) holds. Usingthe step function, we can express the inner integral in the unified form as

Zþ1


¼Z1


pFðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þ Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr=csÞ2 � ðR=cdÞ2

q� qz

� � ZRffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


pþr



d1Hdt


ð6:3:64Þ


We substitute the above equation into Eq. (6.3.45) to get


dqZ1


pFðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )expð�stÞdt

þZ10

Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr=csÞ2 � ðR=cdÞ2

q� qz

� �dq

ZRffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


pþr



d1Hdt


ð6:3:65Þ

To perform the Laplace inversion, the order of integration must be interchanged.Discussing the supporting region for the double integral such as in Fig. 6.12a, weexchange the order of the integrations in the first double integral as

Z10

dqZ1


pgðt; qÞdt¼

Z1R=cs

dtZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2s

p

0

gðt; qÞdq ð6:3:66Þ

The supporting region for the second double integral is also shown in Fig. 6.12b.After some examinations, we obtain the exchange formula

Z10

Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr=csÞ2 � ðR=cdÞ2

q� qz

� �dq

ZRffiffiffiffiffiffiffiffiffiffiffiffiq2þ1=c2s

p


pþr


phðt; qÞdt

¼ZR=cs

r=cdþzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s�1=c2d

pdt

Z1r

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit�z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s�1=c2d

pð Þ2�ðr=cdÞ2q

0

hðt; qÞdq

þZR2

z


p

R=cs

dtZ1

r




ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2s

phðt; qÞdq

ð6:3:67Þ

where g(t, q) and h(t, q) in Eqs. (6.3.66) and (6.3.67) are arbitrary non-singularintegrands.


t

t

2 2( / ) 1/ sq t R c= −

2 21/ st R q c= +

/ sR c

q

22 21/ 1/s d

Rc c

z−

/ sR c

2 2/ 1/ 1/d s dr c z c c+ −

0q =2 21

( / ) ( / )s dq r c R cz

= −

q

t

2 21/ st R q c= +

2 2/ 1/ 1/d s dt r c z c c= + −

2 21 ( / ) 1/ sq t R c= − ( )2 2 2

21

1/ 1/ ( / )s d dq t z c c r cr

= − − −

1q 2q

t

t

(a)

(b)

Fig. 6.12 Supporting region for the double integral in a Eq. (6.3.66) and b Eq. (6.3.67)


Applying the exchange formulas to the double integral in Eq. (6.3.65), we havethe Laplace transform integral for the Laplace transformed component,


R=cs

expð�stÞdtZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2s

p

0

FðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )dq

þZR=cs

r=cdþzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s�1=c2d

pexpð�stÞdt

Z1r




0

f ðsÞxx ð1H ; qÞd1Hdt

� �dq

þZR2

z


p

R=cs

expð�stÞdtZ1

r





pf ðsÞxx ð1H ; qÞ d1Hdt� �

dq

ð6:3:68Þ

Finally, the Laplace inversion is carried out by inspection, and the SV-wave con-tribution is given by

IðsÞxx ðx; y; z; tÞ

¼ Hðt � R=csÞZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2s

p

0

FðsÞxx ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ


ð�Þs

dt

( )dq

þ H t � r=cd � zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s � 1=c2d

q� �HðR=cs � tÞ

Z1r




0

f ðsÞxx ð1H ; qÞd1Hdt

� �dq

þ H t � R=csð ÞH R2

z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s � 1=c2d

q� t

� � Z1r






d1Hdt

� �dq

ð6:3:69Þ

Examining the operation of the step functions ahead of the integral, the first termwith Hðt � R=csÞ shows a semi-spherical region disturbed by SV-wave. The secondterm which has the product of the two step functions,

H t � r=cd � zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s � 1=c2d

p� �HðR=cs � tÞ, shows a region of von-Schmidt wave.

The last term with H t � R=csð ÞH R2

z


p � t� �

does not show the wave

nature, but appears only in the case of the 3D deformation (this non-wave front isdiscussed by Gakenheimer and Miklowitz (1969)). These wave regions are denotedby regions A and B in Fig. 6.13.


To the other SV-wave contributions, the same inversion technique is applied andwe obtain the unified expression for SV-wave contributions as

IðsÞij ðx; y; z; tÞ

¼ Hðt � R=csÞZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðt=RÞ2�1=c2s

p

0

FðsÞij ðpðþÞ

s ; qÞ dpðþÞs

dt� FðsÞ

ij ðpð�Þs ; qÞ dp

ð�Þs

dt

( )dq

þ H t � r=cd � zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=c2s � 1=c2d

q� �HðR=cs � tÞ

Z1r




0

f ðsÞij ð1H ; qÞ d1Hdt� �

dq

þ H t � R=csð ÞH R2

z


q� t

� � Z1r





pf ðsÞij ð1H ; qÞ d1Hdt� �

dq

ð6:3:70Þ

z

r

sc t R=

dc t R=

dc t r z= +

2 2 1 dR c t zγ − = ⋅

2 1z r γ= −

2 1γ −

Fig. 6.13 Wave fronts and disturbed regions A and B


where pð�Þs and 1H are given by Eqs. (6.3.49) and (6.3.61) respectively and

FðsÞxx ðp; qÞ ¼

ðx2p2 þ y2q2Þðb2s þ p2 þ q2 � 4bdbsÞ2ðprÞ2bsRðp; qÞ

FðsÞxy ðp; qÞ ¼

xy

2ðprÞ2ðp2 � q2Þðb2s þ p2 þ q2 � 4bdbsÞ

bsRðp; qÞ

FðsÞxz ðp; qÞ ¼

xp2r

ipbdbsRðp; qÞ

ð6:3:71aÞ

FðsÞyx ðp; qÞ ¼

xy

2ðprÞ2ðp2 � q2Þðb2s þ p2 þ q2 � 4bdbsÞ

bsRðp; qÞ

FðsÞyy ðp; qÞ ¼

ðy2p2 þ x2q2Þðb2s þ p2 þ q2 � 4bdbsÞ2ðprÞ2bsRðp; qÞ

FðsÞyz ðp; qÞ ¼

yp2r

ipbdbsRðp; qÞ

ð6:3:71bÞ

FðsÞzx ðp; qÞ ¼

x2p2r

ipðb2s þ p2 þ q2ÞRðp; qÞ

FðsÞzy ðp; qÞ ¼

y2p2r

ipðb2s þ p2 þ q2ÞRðp; qÞ

FðsÞzz ðp; qÞ ¼

bdðp2 þ q2Þp2Rðp; qÞ

ð6:3:71cÞ

f ðsÞxx ð1; qÞ ¼ ð�iÞ ðx1Þ2 � ðyqÞ2

2ðprÞ2cs

� c2s � 12 þ q2 � 4icdcsðc2s � 12 þ q2Þ2 þ 4icdcsð12 � q2Þ �

c2s � 12 þ q2 þ 4icdcsðc2s � 12 þ q2Þ2 � 4icdcsð12 � q2Þ

( )

f ðsÞxy ð1; qÞ ¼ ð�iÞ xy

2ðprÞ212 þ q2

cs



( )

f ðsÞxz ð1; qÞ ¼ � xp2r

1cdcs

� 1

ðc2s � 12 þ q2Þ2 þ 4icdcsð12 � q2Þ þ1

ðc2s � 12 þ q2Þ2 � 4icdcsð12 � q2Þ

( )

ð6:3:72aÞ


f ðsÞyx ð1; qÞ ¼ ð�iÞ xy

2ðprÞ212 þ q2

cs



( )

f ðsÞyy ð1; qÞ ¼ ð�iÞ xy

2ðprÞ2ðy1Þ2 � ðxqÞ2

cs



( )

f ðsÞyz ð1; qÞ ¼ � yp2r

1cdcs

� 1



( )

ð6:3:72bÞ

f ðsÞzx ð1; qÞ ¼ ðþiÞ x2p2r

1ðc2s � 12 þ q2Þ

� 1

ðc2s � 12 þ q2Þ2 þ 4icdcsð12 � q2Þ �1


( )

f ðsÞzy ð1; qÞ ¼ ðþiÞ y2p2r

1ðc2s � 12 þ q2Þ

� 1

ðc2s � 12 þ q2Þ2 þ 4icdcsð12 � q2Þ �1


( )

f ðsÞzz ð1; qÞ ¼ cdð12 � q2Þp2

� 1



( )

ð6:3:72cÞ

(3) Inversion of SH wave contribution: ~�I�ðSHÞii ðn; g; z; sÞ

Two SH-wave contributions are identical

~�I�ðSHÞxx ðn; g; z; sÞ ¼ ~�I�ðSHÞ

yy ðn; g; z; sÞ ¼ � 1sas

expð�aszÞ ð6:3:73Þ

The formal double Fourier inversion is given by

I�ðSHÞii ðx; y; z; sÞ ¼ 1

ð2pÞ2Zþ1

�1

Zþ1

�1� 1sas

� �expð�asz� inx� igyÞdndg ð6:3:74Þ


Now, we introduce the variable transform defined by Eq. (6.3.23),

I�ðSHÞii ðx; y; z; sÞ ¼ 1

2p2

Z10

dqZþ1

�1� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

p2 þ q2 þ 1=c2sp

!exp �s z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

qþ irp

� �n odp

ð6:3:75Þ

The Cagniard-de Hoop technique is applied to the inner integral. Introducing thevariable transform from p to the new variable t,

t ¼ zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

qþ irp ð6:3:76Þ

the Cagniard’s path is given by

pð�Þs ¼ �irt � z


pR2 ð6:3:77Þ

We have exact expressions for the radical and the gradient,

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

q ��p¼pð�Þ

s

¼ tz� irffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � ðq2 þ 1=c2s ÞR2

pR2 ð6:3:78Þ

dpð�Þs

dt¼ � tz� ir


pR2


p ð6:3:79Þ

Following the former discussion for the Cagniard-de Hoop technique, we applyCauchy’s integral theorem to the complex integral,

U ¼ 12p2

Zc

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

p !

exp �s zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

qþ irp

� �n odp

ð6:3:80Þ

and find that the integral along the real axis is converted to that along the Cagn-iard’s path. Since

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

p dpdt p¼pðþÞ

Ps

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 þ q2 þ 1=c2s

p dpdt

��p¼pð�Þ

s

¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � q2 þ 1ð ÞR2

pð6:3:81Þ


the double inversion integral in Eq. (6.3.75) is transformed to

I�ðSHÞii x; y; z; sð Þ ¼ 1

p2

Z10

dqZ1


p� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

t2 � q2 þ 1=c2s�

R2q

0B@

1CA exp �stð Þdt

ð6:3:82Þ

The order of integration is also exchanged as

I�ðSHÞii x; y; z; sð Þ ¼ 1

p2

Z1R=cs

exp �stð ÞdtZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit=Rð Þ2�1=c2s

p

0

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � q2 þ 1=c2s

� R2

q0B@

1CAdq

ð6:3:83Þ

It is very lucky that the inner integral with respect to the variable q can be evaluatedexactly as

Zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit=Rð Þ2�1=c2sp

0

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � q2 þ 1=c2s

� R2

q0B@

1CAdq ¼ p

2Rð6:3:84Þ

Thus, the SH-wave contribution is reduced to the simple Laplace transform integral,

I�ðSHÞii x; y; z; sð Þ ¼ � 1

2pR

Z1R=Cs

expð�stÞdt ð6:3:85Þ

The Laplace inversion is carried out by inspection and we have the final result,

I�ðSHÞii x; y; z; tð Þ ¼ � 1

2pRH t � R=Csð Þ; i ¼ x; y ð6:3:86Þ

Consequently, we have obtained the SH-wave contribution in the closed form.


The three wave contributions have just been inverted by the Cagniard-de Hooptechnique. The displacement response is expressed in terms of the Green’s dyadicGij(x, y, z, t) as


ux ¼ PxGxx x; y; z; tð Þ þ PyGxy x; y; z; tð Þ þ PzGxz x; y; z; tð Þuy ¼ PxGyx x; y; z; tð Þ þ PyGyy x; y; z; tð Þ þ PzGyz x; y; z; tð Þuz ¼ PxGzx x; y; z; tð Þ þ PyGzy x; y; z; tð Þ þ PzGzz x; y; z; tð Þ

ð6:3:87Þ

where the dyadic components are given by

Gxx x; y; z; tð Þ ¼ 1l

IðdÞxx x; y; z; tð Þ þ IðsÞxx x; y; z; tð Þ þ ISHxx x; y; z; tð Þn o

Gxy x; y; z; tð Þ ¼ 1l

IðdÞxy x; y; z; tð Þ þ IðsÞxy x; y; z; tð Þn o

Gxz x; y; z; tð Þ ¼ 1l

IðdÞxz x; y; z; tð Þ þ IðsÞxz x; y; z; tð Þn o ð6:3:88aÞ

Gyx x; y; z; tð Þ ¼ 1l

IðdÞyx x; y; z; tð Þ þ IðsÞyx x; y; z; tð Þn o

Gyy x; y; z; tð Þ ¼ 1l

IðdÞyx x; y; z; tð Þ þ IðsÞyx x; y; z; tð Þ þ IðSHÞyy x; y; zð Þ

n oGyz x; y; z; tð Þ ¼ 1

lIdyz x; y; z; tð Þ þ ISyz x; y; z; tð Þn o ð6:3:88bÞ

Gzx x; y; z; tð Þ ¼ 1l

IðdÞzx x; y; z; tð Þ þ IðsÞzx x; y; z; tð Þn o

Gzy x; y; z; tð Þ ¼ 1l

IðdÞzy x; y; z; tð Þ þ IðsÞzy x; y; z; tð Þn o

Gzy x; y; z; tð Þ ¼ 1l

IðdÞzz x; y; z; tð Þ þ IðsÞzy x; y; z; tð Þn o ð6:3:88cÞ

Exercises

(6:1) In Sect. 6.2, if the applied load is a semi-infinite distribution such as

ryz��y¼0 ¼ dðtÞ p0; x[ 0

0; x\0

�ðaÞ

the transformed boundary condition of Eq. (6.1.14) is replaced with

�r�yz��y¼0

¼ p0dþ nð Þ ðbÞ

where dþð�Þ is Heisenberg’s delta function. Verify the above equation (b).(6:2) In Sect. 6.3, when the load is distributed uniformly in a quarter region

x[ 0; y[ 0ð Þ on the surface, how do you change the mathematicalexpression for the boundary condition (6.3.3)?


References

Achenbach JD (1973) Wave propagation in elastic solids. North-Holland, New YorkCagniard L (1962) Reflection and refraction of progressive seismic waves (translated by Flinn ED,

Dix CH). McGraw-Hill, New YorkDe-Hoop AT (1961) The surface line source problem. Appl Sci Res B 8:349–356Fung YC (1970) Foundation of solid mechanics (Japanese translation by Ohashi Y et al (1965)

Bai-Fu-Kan, Tokyo). Prentice-Hall, New JerseyGakenheimer DC, Miklowitz J (1969) Transient excitation of an elastic half space by a point load

traveling on the surface. J Appl Mech (Trans ASME Ser E) 36:505–515Graff KF (1975) Wave motion in elastic solids. Clarendon Press, OxfordMiklowitz J (1978) The theory of elastic waves and waveguides. North-Holland, New York


Chapter 7Miscellaneous Green’s Functions

This last chapter presents five Green’s functions and one application technique ofthe complex integral. The first and second sections consider the 2D static Green’sdyadic for an orthotropic elastic solid, and for an inhomogeneous elastic solid. Thethird section discusses the Green’s function for torsional waves in an anisotropicsolid. The fourth and fifth sections are concerned with Green’s function for SHwaves. One is wave reflection at a moving boundary and the other is wave scat-tering by a rigid inclusion in an inhomogeneous solid. All the Green’s functions areobtained in the closed form by applying the method of integral transform.Especially, in the fourth section which discusses wave reflection, a conversionformula between two different Laplace inversion integrals is developed so that wecan treat the moving boundary problem. In the fifth section which discusses thewave scattering, a summation formula is derived for the Fourier series of theSchlömlich type. It enables us to obtain the closed expression for the wave scat-tering problem. The last section shows an excellent application technique of thecomplex integral. It reduces a semi-infinite integral, which includes the product oftwo Bessel functions, to a finite integral that is very suitable for numericalcomputations.

7.1 2D Static Green’s Dyadic for an OrthotropicElastic Solid

An anisotropic elastic solid whose orthogonal Young’s moduli differ from eachother is called an “orthotropic” solid. In 2D in-plane deformation, Hooke’s law forthe orthotropic elastic solid is given by

rxxryyrxy

24

35 ¼

C11 C12 0C12 C22 00 0 C66

24

35 exx

eyyexy

24

35 ð7:1:1Þ


205

where Cij are elastic moduli. The strain components in the state of plane strain aredefined by

exx ¼ @ux@x

; eyy ¼ @uy@y

; exy ¼ 12

@ux@y

þ @uy@x

� �ð7:1:2Þ

The equilibrium equations are

@rxx@x

þ @rxy@y

¼ �QxdðxÞdðyÞ@rxy@x

þ @ryy@y

¼ �QydðxÞdðyÞð7:1:3Þ

where Qi is the magnitude of a body force.Substituting Hooke’s law into the equilibrium equations, we obtain the dis-

placement equations as

C11@2ux@x2

þ C66

2@2ux@y2

þ C12 þ C66

2

� �@2uy@x@y

¼ �QxdðxÞdðyÞ

C12 þ C66

2

� �@2ux@x@y

þ C66

2@2uy@x2

þ C22@2uy@y2

¼ �QydðxÞdðyÞð7:1:4Þ

Since we are concerned with Green’s dyadic, a particular solution corresponding tothe body force is explored. As the elastic medium is of infinite extent, the con-vergence condition at infinity is assumed as


!1¼ @ui@x


p!1

¼ @ui@y


p!1

¼ 0; i ¼ x; y ð7:1:5Þ

We employ the double Fourier transform with respect to two space variables asdefined by

�f ðnÞ ¼Z1�1


Z1�1


~f ðgÞ ¼Z1�1


Z1�1


206 7 Miscellaneous Green’s Functions

The displacement equations (7.1.4) are transformed to the algebraic equations forthe transformed displacement components,

fC11n2 þ ðC66=2Þg2g~�ux þ ðC12 þ C66=2Þng~�uy ¼ Qx

ðC12 þ C66=2Þng~�ux þ fðC66=2Þn2 þ C22g2g~�uy ¼ Qy

ð7:1:8Þ

Explicit expressions for the displacement components are obtained in the trans-formed domain,

~�ux ¼ þ Qx

C22

n2 þ bg2

ðg2 þ p21n2Þðg2 þ p22n

2Þ �Qy

C22

ðcþ 1Þngðg2 þ p21n

2Þðg2 þ p22n2Þ

~�uy ¼ � Qx

C22

ðcþ 1Þngðg2 þ p21n

2Þðg2 þ p22n2Þ þ

Qy

C22

an2 þ g2


2Þ

ð7:1:9Þ

where the eigenvalues, pj ðj ¼ 1; 2Þ, are

p1p2

� �¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiab� cðcþ 2Þ þ 2

ffiffiffiffiffibc

p4b

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiab� cðcþ 2Þ � 2

ffiffiffiffiffibc

p4b

sð7:1:10Þ

and the elastic modulus parameters, a; b; c, are introduced as

a ¼ C11

C66=2; b ¼ C22

C66=2; c ¼ C12

C66=2ð7:1:11Þ

Inspecting the transformed displacement in Eq. (7.1.9), we find that twoinversion formulas are necessary for the full inversion. They are

~�I1ðn; g; a; bÞ ¼ an2 þ bg2


2Þ ;~�I2ðn; gÞ ¼ ng


2Þð7:1:12Þ

If we get the inversions for these two expressions, the displacement in the actualspace can be expressed as

ux ¼ þ Qx

C22I1ðx; y; 1; bÞ � Qy

C22ðcþ 1ÞI2ðx; yÞ

uy ¼ � Qx

C22ðcþ 1ÞI2ðx; yÞ þ Qy

C22I1ðx; y; a; 1Þ

ð7:1:13Þ

7.1 2D Static Green’s Dyadic for an Orthotropic Elastic Solid 207

(1) Fourier inversion with respect to the parameter g

The formal Fourier inversion with respect to the parameter g is given by theintegrals

�I1ðn; y; a; bÞ ¼ 12p

Zþ1

�1

an2 þ bg2


2Þ expð�igyÞdg

�I2ðn; yÞ ¼ 12p

Zþ1

�1

ng


2Þ expð�igyÞdgð7:1:14Þ

These two integrals can be evaluated by the application of some formulas, but, inorder to show the application of Jordan’s lemma, we consider the complex integralsin the g-plane.

Let us consider the complex integral whose integrand is the same as that of thecorresponding integral in Eq. (7.1.14),

U1 ¼ 12p

IL

an2 þ bg2


2Þ expð�igyÞdg

U2 ¼ 12p

IL

ng


2Þ expð�igyÞdgð7:1:15Þ

The integrand has the four simple poles at g ¼ �ipjjnj; ðj ¼ 1; 2Þ as shown inFig. 7.1. The closed loop L is composed of a straight line along the real axis and a

Fig. 7.1 Closed loop L forFourier inversion integral


semi-circle with infinite radius. Due to the convergence on the semi-circle, we haveto choose the lower loop when y > 0 and the upper loop when y < 0. Since no othersingular point exists in the loop, the integral along the real axis is converted to thesum of the residues at the two poles. Thus, we have for �Ij; j ¼ 1; 2

�I1ðn; y; a; bÞ ¼ � a� bp212p1ðp21 � p22Þ

1jnj exp �p1jnjjyjð Þ þ a� bp22

2p2ðp21 � p22Þ1jnj exp �p2jnjjyjð Þ

�I2ðn; yÞ ¼ þ sgnðyÞ2ðp21 � p22Þ

inexp �p1jnjjyjð Þ � i

nexp �p2jnjjyjð Þ

� �ð7:1:16Þ

where the sign function is defined by

sgnðyÞ ¼ þ1; y[ 0�1; y\0

�ð7:1:17Þ

(2) Fourier inversion with respect to the parameter n

Applying the formal Fourier inversion integral with respect to the parameter n toEq. (7.1.16), we have

I1ðx; y; a; bÞ ¼ � a� bp212p1ðp21 � p22Þ

12p

Zþ1

�1

1jnj exp �p1jnjjyj � inxð Þdn

þ a� bp222p2ðp21 � p22Þ

12p

Zþ1

�1

1jnj exp �p2jnjjyj � inxð Þdn

I2ðx; yÞ ¼ þ sgnðyÞ2ðp21 � p22Þ

12p

Zþ1

�1

inexp �p1jnjjyj � inxð Þdn

8<:

� 12p

Zþ1

�1

inexp �p2jnjjyj � inxð Þdn

9=;

ð7:1:18Þ

Inspecting the above equations, the two necessary inversion integrals are extracted as

I1j ¼ 12p

Zþ1

�1

1jnj exp �pjjnjjyj

� expð�inxÞdn;

I2j ¼ 12p

Zþ1

�1

inexp �pjjnjjyj�

expð�inxÞdn;ð7:1:19Þ


The second integral I2j is reduced to a semi-infinite integral and can be evaluated bythe application of the formula (Erdélyi 1954, vol. I, pp. 72, 2),

Z10

1xexpð�axÞ sinðbxÞdx ¼ tan�1 b

a

� �ð7:1:20Þ

That is

I2j ¼ 12p

Zþ1

�1


expð�inxÞdn

¼ 1p

Z10

1nexp �pjnjyj�

sinðnxÞdn

¼ 1ptan�1 x

pjjyj� �

ð7:1:21Þ

The first integral I1j has the first order singularity at n ¼ 0, and thus this integralcannot be evaluated in this form. The singular behavior is the same as that in the 2Dstatic plane problem in Chap. 3. We differentiate the integral with respect to thevariable x and jyj, respectively, to obtain

@I1j@x

¼ 12p

Zþ1

�1

�injnj exp �pjjnjjyj

� expð�inxÞdn

@I1j@jyj ¼

12p

Zþ1

�1

�pjjnjjnj exp �pjjnjjyj

� expð�inxÞdn

ð7:1:22Þ

Reducing the above integrals to semi-infinite ones, we get

@I1j@x

¼ � 1p

Z10

exp �pjnjyj�

sinðnxÞdn

@I1j@jyj ¼ � pj

p

Z10

exp �pjnjyj�

cosðnxÞdnð7:1:23Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_3

and then we apply the formulas (Erdélyi 1954, vol. I, pp. 14, 72), to get

Z10

exp �pjnjyj�

cosðnxÞdn ¼ pjjyjx2 þ p2j y2

Z10

exp �pjnjyj�

sinðnxÞdn ¼ xx2 þ p2j y2

ð7:1:24Þ

The two integrals with different derivative are expressed in terms of algebraicfunctions,

@I1j@x

¼ � 1p

xx2 þ p2j y2

;@I1j@jyj ¼

�p2jp

jyjx2 þ p2j y2

ð7:1:25Þ

In order to obtain the formula for I1j, we integrate the derivatives with respect toeach space variable,

I1j ¼ � 12p

log x2 þ p2j y2

�þ CxðyÞ; I1j ¼ � 1

2plog x2 þ p2j y

2 �

þ CyðxÞð7:1:26Þ

Since the two expressions for I1j must be same, the integration “constant” functionsCx and Cy also must be a simple constant, without any space variable. Thus, wehave for I1j

I1j ¼ � 12p

log x2 þ p2j y2

�þ const: ð7:1:27Þ

Summarizing the above discussion for the inversion integrals, we obtain thesimple formulas,

I1j ¼ 12p

Zþ1

�1

1jnj exp �pjjnjjyj

� expð�inxÞdn ¼ � 1

2plog x2 þ p2j y

2 �

I2j ¼ 12p

Zþ1

�1


expð�inxÞdn ¼ 1ptan�1 x

pjjyj� � ð7:1:28Þ

where the arbitrary constant for the integral I1j is neglected since it does not produceany stress and strain. Substituting these formulas into Eq. (7.1.18), we obtain theclosed expressions for the double Fourier inversions I1 and I2 as


I1ðx; y; a; bÞ ¼ þ a� bp214pp1ðp21 � p22Þ

log x2 þ p21y2� � a� bp22

4pp2ðp21 � p22Þlog x2 þ p22y

2� I2ðx; yÞ ¼ þ sgnðyÞ

2pðp21 � p22Þtan�1 x

p1jyj� �

� tan�1 xp2jyj� ��

ð7:1:29Þ

where the sign function sgn(.) is defined by Eq. (7.1.17). Thus, the displacementcomponents in Eq. (7.1.13) are expressed in the closed form as

uxðx; yÞ ¼ þ Qx

4pC22ðp21 � p22Þ1� bp21

p1log x2 þ p21y

2� � 1� bp22p2

log x2 þ p22y2� � �

� Qy

2pC22ðp21 � p22Þðcþ 1Þ tan�1 x

p1y

� �� tan�1 x

p2y

� ��

uyðx; yÞ ¼ � Qx

2pC22ðp21 � p22Þðcþ 1Þ tan�1 x

p1y

� �� tan�1 x

p2y

� ��

þ Qy

4pC22ðp21 � p22Þa� p21p1

log x2 þ p21y2� � a� p22

p2log x2 þ p22y

2� � �ð7:1:30Þ

Furthermore, if we introduce the notation of the Green’s dyadic Gij, the displace-ment can be rewritten as

uxðx; yÞ ¼ QxGxxðx; yÞ þ QyGxyðx; yÞuyðx; yÞ ¼ QxGyxðx; yÞ þ QyGyyðx; yÞ

ð7:1:31Þ

where the dyadic components are given by

Gxxðx; yÞ ¼ þ 14pC22ðp21 � p22Þ

1� bp21p1

log x2 þ p21y2� � 1� bp22

p2log x2 þ p22y

2� � �

Gxyðx; yÞ ¼ Gyxðx; yÞ ¼ � 12pC22ðp21 � p22Þ

ðcþ 1Þ tan�1 xp1y

� �� tan�1 x

p2y

� ��

Gyyðx; yÞ ¼ þ 14pC22ðp21 � p22Þ

a� p21p1

log x2 þ p21y2� � a� p22

p2log x2 þ p22y

2� � �ð7:1:32Þ

It should be noticed that the symmetry Gxy ¼ Gyx is held in spite of the ortho-tropic nature. The logarithmic singularity at the source point and at the infinity isthe same as that for isotropic media in Chap. 3. These singular behaviors areinherent in 2D plane elasticity.


http://dx.doi.org/10.1007/978-3-319-17455-6_3

7.2 2D Static Green’s Dyadic for an InhomogeneousElastic Solid

Materials whose material parameters, such as elastic moduli and density, arevarying with the space point are called “inhomogeneous” materials. Many mathe-matical models for the inhomogeneity are proposed. The simplest model for theinhomogeneity is a uniaxial exponential function for the material parameters. Thismodel is mathematically simple and tractable, but not fully correct since thematerial parameters vanish or diverge at infinity. In spite of the unrealistic nature ofthe exponential model, this type of the inhomogeneity is frequently used for stressanalysis due to its simplicity.

Let us consider the 2D plane-strain deformation of an inhomogeneous isotropicelastic solid. Hooke’s law for an inhomogeneous solid is the same as that for ahomogeneous one, but its elastic moduli ðk;lÞ are functions of the space variables,

rxx ¼ ðkþ 2lÞ @ux@x

þ k@uy@y

ryy ¼ k@ux@x

þ ðkþ 2lÞ @uy@y


þ @uy@x

� � ð7:2:1Þ

As we are concerned with Green’s dyadic corresponding to a point source, the 2Dequilibrium equations with a point body force are given by

@rxx@x

þ @ryx@y

¼ �BxdðxÞdðyÞ@rxy@x

þ @ryy@y

¼ �BydðxÞdðyÞð7:2:2Þ

where the body force is placed at the coordinate origin ð0; 0Þ. The inhomogeneitywhich we adopt here is a uniaxial exponential function and thus, Lame’s constantsðk; lÞ are assumed to be exponential functions with one space variable y,

kðyÞ ¼ k0 expfkðy=hÞg; lðyÞ ¼ l0 expfkðy=hÞg ð7:2:3Þ

where ðk0; l0Þ are moduli at the coordinate origin, k is an inhomogeneity parameter,and h is a reference length. Since two Lame’s constants have the same exponentialfunction, Poisson ratio v is constant throughout the medium.

Substituting Hooke’s law of Eq. (7.2.1) with the inhomogeneity of Eq. (7.2.3)into the equilibrium equations (7.2.2), we obtain displacement equations withconstant coefficients,

7.2 2D Static Green’s Dyadic for an Inhomogeneous Elastic Solid 213

c2@2ux@x2

þ @2ux@y2

þ ðc2 � 1Þ @2uy

@x@yþ kh

@ux@y

þ @uy@x

� �¼ � Bx

lðyÞ dðxÞdðyÞ

ðc2 � 1Þ @2ux

@x@yþ @2uy

@x2þ c2

@2uy@y2

þ kh

ðc2 � 2Þ @ux@x

þ c2@uy@y

� �¼ � By

lðyÞ dðxÞdðyÞ

ð7:2:4Þ

where the constant material parameter c is defined by

c ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffikþ 2l

l

s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik0 þ 2l0

l0

s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ð1� mÞ1� 2m

rð7:2:5Þ

In order to obtain a particular solution corresponding to the nonhomogeneous bodyforce term, we apply the double Fourier transform with respect to two spacevariables,

�f ðnÞ ¼Zþ1


2p

Zþ1

�1

�f ðnÞ expð�inxÞdn

~f ðgÞ ¼Zþ1


2p

Zþ1

�1

~f ðgÞ expð�igyÞdgð7:2:6Þ

The convergence condition at infinity is also imposed on the displacement,


!1¼ @ui@x


p!1

¼ @ui@y


p!1

¼ 0; i ¼ x; y ð7:2:7Þ

With the aid of the integration formulas which include the delta function,

Zþ1

�1dðxÞ expðþinxÞdx ¼ 1;

Zþ1

�1

dðyÞlðyÞ expðþigyÞdy ¼ 1

lð0Þ ð7:2:8Þ

the displacement equations (7.2.4) are transformed to the algebraic equations for thedisplacement components,

fc2n2 þ g2 þ igðk=hÞg~�ux þ fðc2 � 1Þngþ inðk=hÞg~�uy ¼ Bx

lð0Þfðc2 � 1Þngþ inðc2 � 2Þðk=hÞg~�ux þ fn2 þ c2g2 þ igc2ðk=hÞg~�uy ¼ By

lð0Þð7:2:9Þ


where

lð0Þ ¼ l0 ð7:2:10Þ

Solving Eq. (7.2.9) for the transformed displacement components, we get

~�ux ¼ þ Bx

lð0Þn2 þ c2ð�g2 þ j2Þ

c2Dðn; �gÞ � By

lð0Þðc2 � 1Þ�g� ijðc2 � 3Þ

c2Dðn; �gÞ n ð7:2:11aÞ

~�uy ¼ � Bx

lð0Þðc2 � 1Þ�gþ ijðc2 � 3Þ

c2Dðn; �gÞ nþ By

lð0Þc2n2 þ �g2 þ j2

c2Dðn; �gÞ ð7:2:11bÞ

where the new inhomogeneity parameter j is introduced as

j ¼ k=ð2hÞ ð7:2:12Þ

and the denominator Dðn; �gÞ and �g are given by

Dðn; �gÞ ¼ ðn2 þ �g2 þ j2Þ2 þ 4ð1� 2=c2Þj2n2 ð7:2:13Þ

�g ¼ gþ ij ð7:2:14Þ

In order to apply the inversion integral, we factorize the denominator as

Dðn; �gÞ ¼ ðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ ð7:2:15Þ

where the eigenvalues xj; j ¼ 1; 2 are

x1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p� jp; x2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

pþ jp ð7:2:16Þ

and

p ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 2=c2

p; q ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ð1� 1=c2Þ

p; q[ p ð7:2:17Þ

The formal Fourier inversion with respect to the parameter ξ is given by theintegrals,

~ux ¼ þ Bx

c2lð0Þ12p

Zþ1

�1

n2 þ c2ð�g2 þ j2Þðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ expð�inxÞdn

� By

c2lð0Þ12p

Zþ1

�1

ðc2 � 1Þ�g� ijðc2 � 3Þðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ n expð�inxÞdn

ð7:2:18aÞ


~uy ¼ � Bx

c2lð0Þ12p

Zþ1

�1

ðc2 � 1Þ�gþ ijðc2 � 3Þðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ n expð�inxÞdn

þ By

c2lð0Þ12p

Zþ1

�1

c2n2 þ �g2 þ j2

ðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ expð�inxÞdn

ð7:2:18bÞ

The above integrals can be evaluated by the complex integral theory. So, we applyCauchy’s theorem (Jordan’s lemma) to the integral in the complex n-plane. As anexample, the first integral in Eq. (7.2.18a) is discussed. We represent it by thecomplex integral Uxx,

Uxx ¼ 12p

ZC

n2 þ c2ð�g2 þ j2Þðn� ix1Þðnþ ix1Þðn� ix2Þðnþ ix2Þ expð�inxÞdn ð7:2:19Þ

where the closed loop C is composed of a semi-circle with infinite radius and astraight line along the real axis (see Fig. 7.2). In order to guarantee the convergenceon the large semi-circle, we have to employ the lower loop for x > 0, and the upperfor x < 0. The integrand has four poles; two of them are in the upper plane and othertwo are in the lower plane. Each closed loop includes two poles. Applying theJordan’s lemma to the complex integral Uxx in Eq. (7.2.19), the integral along thereal axis is converted to the sum of two residues. When we employ the lower loopfor x[ 0, the complex integral yields

Uxx ¼ � 12p

Zþ1

�1


¼ 2piResð�ix1Þ þ 2piResð�ix2Þð7:2:20Þ

The detailed calculation for the residues is

2piResð�ix1Þ þ 2piResð�ix2Þ

¼ 2pi2p

n2 þ c2ð�g2 þ j2Þðn� ix1Þðn� ix2Þðnþ ix2Þ expð�inxÞ

��n¼�ix1

þ 2pi2p

n2 þ c2ð�g2 þ j2Þðn� ix1Þðn� ix2Þðnþ ix1Þ expð�inxÞ

��n¼�ix2

ð7:2:21Þ


Arranging the above equation, the inversion integral along the real axis is evaluatedexactly as

12p

Zþ1

�1


¼ c2 � 14jp

sinhðjpxÞ þ c2 þ 1

4ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p coshðjpxÞ( )

e�xffiffiffiffiffiffiffiffiffiffiffiffiffi�g2þj2q2

pð7:2:22Þ

Similarly, the other integrals in Eq. (7.2.18) can be evaluated and we have

~ux ¼ þ Bx

4c2lð0Þc2 þ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p coshðjpxÞ þ c2 � 1jp

sinhðjpjxjÞ( )

exp �jxjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p �

þ By

4c2lð0Þigðc2 � 1Þ þ jðc2 � 3Þ

jpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p sinh jpjxjð Þ exp �jxjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p �ð7:2:23aÞ

~uy ¼ þ Bx

4c2lð0Þigðc2 � 1Þ � jðc2 � 3Þ

jpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p sinhðjpjxjÞ exp �jxjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p �

þ By

4c2lð0Þc2 þ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p coshðjpxÞ � c2 � 1jp

sinhðjpjxjÞ( )


p �ð7:2:23bÞ

2iω+

1iω+

2iω−

1iω−

Re( )ξ

Im( )ξ

x a<

x a>

+∞

+∞

−∞

−∞

Fig. 7.2 Closed loop C forthe complex integral Uxx


Our next task is to evaluate the inversion integral with respect to the parameter g.The formal inversion of the displacement is expressed in terms of the fundamentalinversion integrals Iiðx; yÞ; i ¼ 0; 1; 2,

ux ¼ þ Bx

4lð0Þc2 � 1c2

expð�jyÞ c2 þ 1c2 � 1

coshðjpxÞI0ðx; yÞ þ sinhðjpjxjÞjp

I1ðx; yÞ� �

þ By

4lð0Þc2 � 1c2

sinhðjpjxjÞjp

expð�jyÞ I2ðx; yÞ þ jc2 � 3c2 � 1

I0ðx; yÞ� �

ð7:2:24aÞ

uy ¼ þ Bx

4lð0Þc2 � 1c2

sinhðjpjxjÞjp

expð�jyÞ I2ðx; yÞ � jc2 � 3c2 � 1

I0ðx; yÞ� �

þ By

4lð0Þc2 � 1c2

expð�jyÞ c2 þ 1c2 � 1

coshðjpxÞI0ðx; yÞ � sinhðjpjxjÞjp

I1ðx; yÞ� �

ð7:2:24bÞ

where the fundamental integrals Ii are defined by

I0ðx; yÞ ¼ 12p

Zþ1

�1


p� i�gy

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p dg ð7:2:25Þ

I1ðx; yÞ ¼ 12p

Zþ1

�1exp �jxj


p� i�gy

�dg ð7:2:26Þ

I2ðx; yÞ ¼ 12p

Zþ1

�1i�gexp �jxj


p� i�gy

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�g2 þ j2q2

p dg ð7:2:27Þ

and the variable �g is defined by Eq. (7.2.14).Observing and examining the above three Eqs. (7.2.25)–(7.2.27), we find the

relation among the fundamental integrals as

I1ðx; yÞ ¼ � @I0ðx; yÞ@jxj ; I2ðx; yÞ ¼ � @I0ðx; yÞ

@yð7:2:28Þ

Since the two integrals I1 and I2 can be derived from I0, it is enough to evaluate theonly one integral I0. For the integral I0 in Eq. (7.2.25), we introduce the variabletransform from g to the new variable 1,

1 ¼ �g ¼ gþ ij ð7:2:29Þ


Equation (7.2.25) is transformed to the integral along the complex line from �1þij to þ1þ ij,

I0ðx; yÞ ¼ 12p

Zþ1þij

�1þij

exp �jxjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ j2q2

p� i1y

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ j2q2

p d1 ð7:2:30Þ

The above integral resembles the formula (Erdélyi 1954, p. 17, (27)),

Zþ1

�1

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ c2

p exp �affiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ c2

p �expð�ibxÞdx ¼ 2K0 c


p �ð7:2:31Þ

where K0ð:Þ is the modified Bessel function of zeroth order. However, this formulacannot be applied directly since the integration path for our integral of Eq. (7.2.30)is not real. In order to apply the formula, we need to transform the integral to onealong the real path.

Let us consider the complex integral U along the rectangular closed loopABCDA shown in Fig. 7.3,

U ¼ 12p

ZABCDA


p� i1y


p d1 ð7:2:32Þ

A

BC

iκ+

D

i qκ+

i qκ−branch cut

branch cut

iκ−∞ + iκ+∞ +

−∞ +∞

{ }2 2Re ( ) 0qς κ+ >

Re( )ς

Im( )ς

Fig. 7.3 Transform of integration path from the complex line to the real line


The integrand has two branch points at 1 ¼ �ijq. Two branch cuts along theimaginary axis are also introduced as shown in the figure. Fortunately, neither of

two branch cuts cross the integration lines DA�!

and BC�!

, especially the line BC�!

since qð¼1=ð1� mÞÞ[ 1. No other singular point is included in the closed loopABCDA and the integrals along the vertical straight lines, AB and CD, vanish atinfinity, since the real part of the radical is positive. Thus, the integral along the

complex line CB�!

is converted to one along the real axis DA�!

. That is

12p

ZCB�!


p� i1y


p d1 ¼ 12p

ZDA�!


p� i1y


p d1

ð7:2:33Þ

Consequently, the integral (7.2.30) along the complex line is converted to that alongthe real path,

I0ðx; yÞ ¼ 12p

Zþ1

�1


p� i1y


p d1 ð7:2:34Þ

Therefore, we can apply the integration formula Eq. (7.2.31) to the above integraland have the exact expression for the integral I0,

I0ðx; yÞ ¼ 1pK0 jjjq


p �ð7:2:35Þ

Substituting this Eq. (7.2.35) into Eq. (7.2.28), the other two integrals, I1 and I2, areexactly evaluated as

I1 ¼ jjjqjxjpffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p K1 jjjqffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p �

I2 ¼ jjjqypffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p K1 jjjqffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p � ð7:2:36Þ

Finally, we substitute Eqs. (7.2.35) and (7.2.36) into Eq. (7.2.24) and rewrite thedisplacement in terms of Green’s dyadic Gij,

ux ¼ BxGxxðx; yÞ þ ByGxyðx; yÞuy ¼ BxGyxðx; yÞ þ ByGyyðx; yÞ

ð7:2:37Þ


where the exact expressions for the dyadic are

Gxxðx; yÞ ¼ c2 � 14pc2lð0Þ expð�jyÞ

� c2 þ 1c2 � 1

coshðjpxÞK0 jjjqrð Þ þ jjjq xrsinhðjpxÞ

jpK1 jjjqrð Þ

� �ð7:2:38aÞ

Gxyðx; yÞ ¼ c2 � 14pc2lð0Þ expð�jyÞ sinhðjpxÞ

jpjjjq y

rK1 jjjqrð Þ þ j

c2 � 3c2 � 1

K0 jjjqrð Þ� �

ð7:2:38bÞ

Gyxðx; yÞ ¼ c2 � 14pc2lð0Þ expð�jyÞ sinhðjpxÞ

jpjjjq y

rK1 jjjqrð Þ � j

c2 � 3c2 � 1

K0 jjjqrð Þ� �

ð7:2:38cÞ

Gyyðx; yÞ ¼ c2 � 14pc2lð0Þ expð�jyÞ

� c2 þ 1c2 � 1

coshðjpxÞK0 jjjqrð Þ � jjjq xrsinhðjpxÞ

jpK1 jjjqrð Þ

� �ð7:2:38dÞ

Here the radial distance r from the source is defined by


pð7:2:39Þ

7.2.1 2D Kelvin’s Solution for Homogeneous Media

Inspecting the dyadic in Eq. (7.2.38), we can find that there is only one inhomo-geneity parameter, i.e. jð¼j=ð2hÞÞ. When this parameter vanishes, the medium ishomogeneous. Thus, taking the limit j ! 0 as

sinhðjpxÞjp

� x; cosh ðjpxÞ� 1 ð7:2:40Þ

K0ðjjjqrÞ� � logðjjjqrÞ; jjjqK1ðjjjqrÞ� 1r

ð7:2:41Þ


we have the dyadic for the homogeneous medium, i.e. the 2D Kelvin’s solution,

Gxxðx; yÞ ¼ 14pc2l0


�2� �ð7:2:42aÞ

Gxyðx; yÞ ¼ Gyxðx; yÞ ¼ c2 � 14pc2l0

xyr2

ð7:2:42bÞ

Gyyðx; yÞ ¼ 14pc2l0


�2� �ð7:2:42cÞ

where l0 is a uniform rigidity throughout the medium. The reader will find that thisdyadic is the same as the 2D static dyadic of Eq. (3.3.33) in Sect. 3.3.

7.3 Green’s Function for Torsional Waves in a MonoclinicMaterial

The axisymmetric torsional deformation of an elastic solid is governed by theequation of motion

1r2@ðr2rrhÞ

@rþ @rzh

@z¼ q

@2uh@t2

� qBh ð7:3:1Þ

and Hooke’s law

rrhrzh

�¼ C44 C45

C54 C55

�erhezh

�; C45 ¼ C54 ð7:3:2Þ

where uh � uhðr; z; tÞ is the torsional/circumferential displacement, rrh and rzh aretorsional stresses and erh and ezh are strains. When C44 ¼ C55 ¼ l andC45 ¼ C54 ¼ 0, the elastic solid is isotropic. On the other hand, when C44 6¼ C55

and C45 ¼ C54 6¼ 0, the solid is anisotropic and is called “monoclinic.” Themonoclinic material is one of anisotropic solids. As Green’s function for the iso-tropic solid has been obtained in Sect. 3.7 in Chap. 3, we shall obtain the Green’sfunction for the monoclinic material.

Two strains are defined by

erh ¼ 12

@uh@r

� uhr

� �; ezh ¼ 1

2@uh@z

ð7:3:3Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_3

http://dx.doi.org/10.1007/978-3-319-17455-6_3

We employ the same circumferential ring body force as that in Sect. 3.7,

Bh ¼ Qdðr � aÞ

rdðzÞdðtÞ ð7:3:4Þ

where Q is the magnitude of the body force and dð:Þ is Dirac’s delta function.Substituting Eq. (7.3.2) into the equation of motion with Eq. (7.3.4), we have thedisplacement equation,

@2uh@r2

þ 1r@uh@r

� uhr2

þ 2b@2uh@r@z

þ br@uh@z

þ a2@2uh@z2

� 1c2

@2uh@t2

¼ � Qc2

dðr � aÞr

dðzÞdðtÞ ð7:3:5Þ

where

a ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC44=C55

p; b ¼ C45=C55; c ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC55=ð2qÞ

pð7:3:6Þ

Our task in the present section is to obtain a particular solution corresponding tothe nonhomogeneous term of the ring force. In order to reduce the partial differ-ential equation to the ordinary one, we apply the double integral transform: Laplacetransform with respect to the time t

f �ðsÞ ¼Z10


and Fourier transform with respect to the axial length z,

f ðfÞ ¼Zþ1

�1f ðzÞ expðþifzÞdz , f ðzÞ ¼ 1

2p

Zþ1

�1f ðfÞ expð�ifzÞdf ð7:3:8Þ

Assuming the quiescent condition at an initial time,

uhjt¼0 ¼@uh@t

��t¼0

¼ 0 ð7:3:9Þ


uhj ffiffiffiffiffiffiffiffiffir2þz2p !1¼ @uh

@r


p !1¼ @uh

@z


p !1¼ 0 ð7:3:10Þ

7.3 Green’s Function for Torsional Waves in a Monoclinic Material 223

http://dx.doi.org/10.1007/978-3-319-17455-6_3

we apply the double transform to the displacement equation (7.3.5). It follows that

d2u�hdr2

þ 1r� 2ibf

� �du�hdr

� 1r2

þ ibfr

þ ðafÞ2 þ ðs=cÞ2� �

u�h ¼ � Qc2

dðr � aÞr

ð7:3:11Þ

In order to obtain the simpler Bessel equation, we assume the solution as a productof two unknown functions. They are assumed as

u�h ¼ gðrÞUðrÞ ð7:3:12Þ

Substituting the above equation into Eq. (7.3.11), we have

g U00 þ 1rU0 � 1

r2U

� �þ 2ðg0 � ibfgÞU0 þ g0 � ibfg

rþ g00 � 2ibfg0

� �U

� gfðafÞ2 þ ðs=cÞ2gU ¼ � Qc2

dðr � aÞr

ð7:3:13Þ

where the prime (.)′ denotes the differentiation with respect to the radial variabler. Inspecting the above equation, if we assume

g0 � ibfg ¼ 0 ) gðrÞ ¼ expðþibfrÞ ð7:3:14Þ

we have the nonhomogeneous Bessel equation for another unknown U(r),

U00 þ 1rU0 � 1

rU � fðcfÞ2 þ ðs=cÞ2gU ¼ � Q

c2dðr � aÞ

rexpð�ibfrÞ ð7:3:15Þ

where the parameter γ is the combination of two anisotropy parameters,

c ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � b2

qð7:3:16Þ

Equation (7.3.15) is in the form of Bessel differential equation for the unknown U(r). In order to obtain the particular solution, we apply the Hankel transform withthe order 1,

~f ðnÞ ¼Z10

rf ðrÞJ1ðnrÞdr , f ðrÞ ¼Z10

n~f ðnÞJ1ðnrÞdn ð7:3:17Þ


to Eq. (7.3.15), and then have the simple algebraic equation for the transformedfunction. The transformed function ~UðnÞ is given by

~UðnÞ ¼ Qc2

expð�ibfaÞ J1ðanÞn2 þ ðcfÞ2 þ ðs=cÞ2 ð7:3:18Þ

Its formal Hankel inversion is given by

UðrÞ ¼ Qc2

expð�ibfaÞZ10

n

n2 þ ðcfÞ2 þ ðs=cÞ2 J1ðanÞJ1ðrnÞdn ð7:3:19Þ

The product of two Bessel functions is replaced with its integral form. From theaddition theorem of the Bessel function (Watson 1966, p. 358), we can easily derivethe integral representation for the product as

J1ðanÞJ1ðrnÞ ¼ 1p

Zp0

J0 nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosu

p �cosudu ð7:3:20Þ

This equation is the same as Eq. (3.7.13) in Chap. 3. We substitute the aboveintegral representation into Eq. (7.3.19) and exchange the order of integration. Itfollows that

UðrÞ ¼ Qpc2

expð�ibfaÞZp0

cosuduZ10

nJ0 n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosu

p �n2 þ ðcfÞ2 þ ðs=cÞ2 dn

ð7:3:21Þ

The simple integration formula (Erdélyi 1954, vol. II, pp. 23, 12)

Z10

n

n2 þ a2J0ðbnÞdn ¼ K0ðabÞ ð7:3:22Þ

where K0ð:Þ is the modified Bessel function of the first kind, is applied toEq. (7.3.21). We have the single integral representation for UðrÞ,

UðrÞ ¼ Qpc2

expð�ibfaÞZp0

K0


p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcfÞ2 þ ðs=cÞ2

q� �cosudu

ð7:3:23Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_3

The product of Eq. (7.3.12) has just been determined by Eqs. (7.3.14) and (7.3.23).Thus the double transformed torsional displacement is given by

u�h ¼Qpc2

expfþibfðr � aÞgZp0

K0



q� �cosudu

ð7:3:24Þ

Let us start for the inversion of the double transform. Firstly, we apply theFourier inversion integral with respect to the parameter f to the above equation andexchange the order of integration. It yields

u�h ¼Qpc2

Zp0

cosudu

� 1p

Z10

K0



q� �cos ffz� bðr � aÞg½ df

ð7:3:25Þ

Applying the integration formula (Erdélyi 1954, p. 56, (43))

Z10

K0 affiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ b2

p �cosðcxÞdx ¼ p

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 þ c2

p �ð7:3:26Þ

to Eq. (7.3.25), we have

u�h ¼Q

2pcc2

Zp0

exp �ðs=cÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ fz� bðr � aÞg2

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ fz� bðr � aÞg2

q cosudu

ð7:3:27Þ

Before going to the Laplace inversion, we make the change of variable for theintegral, u ! t,

t ¼ 1c

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffir2 þ a2 � 2ar cosuþ fz� bðr � aÞg2

qð7:3:28Þ

The new variable t is a simple variable, not the real time at the present stage. But, atthe next stage we understand it as the real time. With the change of the variable,Eq. (7.3.27) is simplified as


u�h ¼Q

4pcc1ar

ZR2=c

R1=c

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR22 � ðctÞ2

ðctÞ2 � R21

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðctÞ2 � R2

1

R22 � ðctÞ2

s( )expð�stÞdt ð7:3:29Þ

where

R1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr � aÞ2 þ fz� bðr � aÞg2

q; R2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðr þ aÞ2 þ fz� bðr � aÞg2

qð7:3:30Þ

Equation (7.3.29) is the Laplace transformed displacement and the integral is just inthe form of the Laplace transform. Therefore, the original function of the dis-placement is its integrand and the Laplace inversion can be carried out byinspection. Finally, we have the exact closed form for the displacement. That is theGreen’s function for the torsional ring source in the monoclinic material,

uh ¼ Q4pcc

Hðct � R1ÞHðR2 � ctÞ 1ar

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR22 � ðctÞ2

ðctÞ2 � R21

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðctÞ2 � R2

1

R22 � ðctÞ2

s( )ð7:3:31Þ

The reader will find that this equation can be reduced to Eq. (3.7.26) in Chap. 3when the material is isotropic, a ! 1; b ! 0 ðc ! 1Þ; c ! cs.

7.4 Reflection of a Transient SH-Wave at a MovingBoundary

As an example of a moving boundary problem, wave reflection at a moving edge isdiscussed. The simplest single SH-wave is considered and a conversion formulabetween two Laplace transforms is developed so that the integral transform methodcan be applicable to the moving boundary problem.

Let us consider a semi-infinite elastic solid and define Cartesian coordinates(x, y) as shown in Fig. 7.4. The initial position of a moving edge of the half space isat x ¼ l and the edge moves toward the positive x-direction with velocity V. A pointand stationary source of SH-wave is placed at the coordinate origin and is assumedas an impulsive body force with magnitude Q. The SH-wave field produced by theimpulsive source is governed by the nonhomogeneous displacement equation,

@2uz@x2

þ @2uz@y2

¼ 1c2s

@2uz@t2

� Qc2s

dðxÞdðyÞdðtÞ ð7:4:1Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_3

where uz is the anti-plane displacement and cs is SH-wave velocity. The stresscomponents follow Hooke’s law as

rzx ¼ l@uz@x

; rzy ¼ l@uz@y

ð7:4:2Þ

where l is rigidity.We assume the stress-free condition at the moving edge x ¼ Vt þ l,

rxzjx¼Vtþl ¼ 0 ð7:4:3Þ

The radiation condition at infinity

uzj ffiffiffiffiffiffiffiffiffix2þy2p

!1¼ @uz@x


p!1

¼ @uz@y


p!1

¼ 0 ð7:4:4Þ

and the quiescent condition at an initial time

uzjt¼0¼@uz@t

��t¼0

¼ 0 ð7:4:5Þ

are also assumed. Equations (7.4.1)–(7.4.5) constitute the present elastodynamicproblem.

The wave field can be decomposed into two parts: the incident wave uinc and thereflected wave urefl:. The total displacement (wave field) is the sum of the two,

uzðx; y; tÞ ¼ uincðx; y; tÞ þ ureflðx; y; tÞ ð7:4:6Þ

Vt

x l=0x =

x

y

SH-source

Q

Moving boundary

0t = 0t >Fig. 7.4 Moving edge of asemi-infinite elastic solid


where uinc and urefl are a particular and the general solution of the non-homoge-neous wave equation (7.4.1), respectively.

(1) Incident wave

The incident wave is generated by the impulsive source and radiates from thesource point. The incident wave is given as the particular solution of the nonho-mogeneous wave equation for the displacement,

@2uinc@x2

þ @2uinc@y2

¼ 1c2s

@2uinc@t2

� Qc2s

dðxÞdðyÞdðtÞ ð7:4:7Þ

The particular solution has already been obtained in Sect. 2.4, i.e.

uincðx; y; tÞ ¼ Q2pcs

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðcstÞ2 � ðx2 þ y2Þ

q H cst �ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p �ð7:4:8Þ

However, we do not use this solution for the latter analysis, instead, the Fouriertransformed solution is employed. We apply the triple integral transform toEq. (7.4.7): Laplace transform with respect to the time,

f �ðsÞ ¼Z10

f ðtÞ expð�stÞdt; f ðtÞ ¼ 12pi

ZBrðsÞ

f �ðsÞ expðþstÞds ð7:4:9Þ

where “Br(s)” in the inverse transform denotes the Bromwich line in the complexs-plane, and the double Fourier transform with respect to the space variables x and y,

�f ðnÞ ¼Zþ1


2p

Zþ1

�1


~f ðgÞ ¼Zþ1


2p

Zþ1

�1


The nonhomogeneous wave equation (7.3.7) is reduced to the simple algebraicequation for the transformed displacement,

�fn2 þ g2 þ ðs=csÞ2g~�u�inc ¼ � Qc2s

ð7:4:12Þ

7.4 Reflection of a Transient SH-Wave at a Moving Boundary 229

http://dx.doi.org/10.1007/978-3-319-17455-6_2

and the particular solution in the transformed domain is then given by

~�u�inc ¼Qc2s

1

n2 þ g2 þ ðs=csÞ2ð7:4:13Þ

Its formal Fourier inversion with respect to the parameter n is given by

~u�inc ¼Q

2pc2s

Zþ1

�1

1

n2 þ g2 þ ðs=csÞ2expð�inxÞdn ð7:4:14Þ

and is easily evaluated by the application of the formula (2.1.22), to give

~u�inc ¼Q

2c2sasexpð�asjxjÞ ð7:4:15Þ

where the radical as is defined by

as ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðs=csÞ2

q; ReðasÞ[ 0 ð7:4:16Þ

Furthermore, the formal Laplace inversion for the incident wave of Eq. (7.4.15) isgiven by the Bromwich integral,

~uinc ¼ Q2c2s

12pi

ZBrðsÞ

1asexpð�asjxj þ stÞds ð7:4:17Þ

where BrðsÞ denotes the Bromwich line from cs � i1 to cs þ i1 in the complex

s-plane. The transformed stress ~rðincÞxz produced by the incident wave is alsoobtained in the form of Laplace inversion integral,

~rðincÞxz ¼ � Q2c2s

sgnðxÞ 12pi

ZBrðsÞ

expð�asjxj þ stÞds ð7:4:18Þ

where sgn(x) is the sign function defined by Eq. (7.1.17).In the subsequent discussion for the stress-free condition, the integral form of

Eq. (7.4.18) is convenient since we develop the conversion formula for two dif-ferent Laplace transforms. Therefore, we shall stop the discussion for the incidentwave.


http://dx.doi.org/10.1007/978-3-319-17455-6_2

(2) Reflected wave

The reflected wave is the general solution of the homogeneous displacementequation,

@2urefl@x2

þ @2urefl@y2

¼ 1c2s

@2urefl@t2

ð7:4:19Þ

The reflected wave is generated at the moving edge and runs toward the negative x-direction. So, we transform the Eq. (7.4.19) to the moving coordinate systemðz; y; tÞ where the moving coordinate z is defined by

z ¼ Vt þ l� x ð7:4:20Þ

Due to this coordinate transform, the displacement of the reflected wave has thedifferent set of three variables, as

urefl:ðx; y; tÞ � urefl:ðz; y; tÞ ð7:4:21Þ

The wave equation (7.3.19) is transformed to

m2 @2urefl@z2

� 2Mc2

@2urefl@z@t

� 1c2s

@2urefl@t2

þ @2urefl@y2

¼ 0 ð7:4:22Þ

where Mach number M and its related parameter m are defined by

M ¼ V=cs; m ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1�M2

pð7:4:23Þ

In order to solve the differential equation (7.4.22), we apply the double integraltransform: Laplace transform with respect to the time variable in the movingcoordinate system. It is defined by

fðpÞ ¼Z10

f ðtÞ expð�ptÞdt; f ðtÞ ¼ 12pi

ZBrðpÞ

fðpÞ expðptÞdp ð7:4:24Þ

Fourier transform with respect to the space variable y is defined by

~f ðgÞ ¼Zþ1


2p

Zþ1

�1


It should be noted that the Laplace transform is not the same as that in Eq. (7.4.9),but that the Fourier transform is the same as that in Eq. (7.4.11). This is because wehave introduced the moving coordinate system along the x-axis, not along the


y-axis. Thus, it should be understood that the time variable t in the moving coor-dinate system is slightly different from the time in the fixed coordinate system sincethe variable z includes the time variable.

Using the quiescent condition at the initial time and the convergence condition atinfinity, y ! �1, the wave equation (7.4.22) is transformed to the ordinary dif-ferential equation,

m2 d2~urefldz2

� 2ðp=csÞM d~urefldz

� g2 þ ðp=csÞ2n o

~urefl ¼ 0 ð7:4:26Þ

This solution which satisfies the convergence condition at z ! þ1 is given by

~urefl ¼ Aðg; pÞ exp �bs þMpmcs

� �zm

� �ð7:4:27Þ

where Aðg; pÞ is an unknown coefficient and the radical bs is defined by

bs ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ p

mcs

� �2s

; ReðbsÞ� 0 ð7:4:28Þ

Then, the formal Laplace inversion of the reflected wave is given by the Bromwichintegral,

~urefl ¼ 12pi

ZBrðpÞ

Aðg; pÞ exp �bs þMpmcs

� �zm

� �expðptÞdp ð7:4:29Þ

where BrðpÞ denotes the Bromwich line from cp � i1 to cp þ i1 in the complexp-plane. In order to obtain the stress, we change the moving coordinate z back to theoriginal one, z ¼ Vt þ l� x, to get

~urefl ¼ 12pi

ZBrðpÞ

Aðg; pÞ exp �bs þMpmcs

� �Vt þ l� x

mþ pt

� �dp ð7:4:30Þ

We then substitute this expression into the first of Eq. (7.4.2). The stress producedby the reflected wave is given by

~rðreflÞxz ¼ 12pi

ZBrðpÞ

1m

bs �Mpmcs

� �Aðg; pÞ exp 1

m�bs þ

Mpmcs

� �ðVt þ l� xÞ þ pt

� �dp

ð7:4:31Þ

The unknown coefficient is determined by the stress-free condition on themoving edge. The stress is also the sum of two wave contributions, the incident and


the reflected waves. Thus, the Fourier transformed boundary condition is given bythe sum of each wave contribution,

~rxz ¼ ~rðincÞxz þ ~rðreflÞxz ¼ 0; at x ¼ Vt þ l ð7:4:32Þ

Substituting Eqs. (7.4.18) and (7.4.31) into the above condition, we obtain

� Q2c2s

12pi

ZBrðsÞ

expf�asðVt þ lÞ þ stgds

þ 12pi

ZBrðpÞ

1m

bs �Mpmcs

� �Aðg; pÞ expðptÞdp ¼ 0 ð7:4:33Þ

This is the integral equation for the unknown coefficient Aðg; pÞ. The secondintegral on the left-hand side is in the form of Laplace inversion integral withrespect to the parameter p, however, the first integral is not (in the form of Laplaceinversion integral). We thus have two different Laplace inversion integrals in asingle equation and it is not possible to apply any one of the Laplace transforms toreduce it to an algebraic equation for the unknown. Thus, the inversion integral inthe first term must be converted to that with respect to the parameter p, i.e. thematching of the inversion integral.

Let us consider the complex integral UA whose integrand is the same as that inthe first integral in Eq. (7.4.33),

UA ¼ 12pi

ZC

expð�aslÞ expfðs� asVÞtgds ð7:4:34Þ

where the radical as is defined by Eq. (7.4.16). The integration loop C is discussednow. The integrand has two branch points s ¼ �icsg and corresponding two branchcuts are introduced along the imaginary axis in the complex s-plane. These are shownin Fig. 7.5. If we introduce the variable transform from s to the new variable p as

p ¼ s� asV ð7:4:35Þ

its inverse is given by

s=cs ¼ 1m2 p=cs �M

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðmgÞ2 þ ðp=csÞ2

q� �ð7:4:36Þ

In order to determine the multiple sign ð�Þ, we examine Eq. (7.4.35). We have thefollowing asymptotic relation between two variables:

s� p1�M

; jpj ! 1 ð7:4:37Þ


This relation must be hold for the inverse of the variable transform, i.e. we have tochoose the positive sign in Eq. (7.4.36). Then, the suitable inversion for the variabletransform is

s=cs ¼ 1m2 p=cs þM

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðmgÞ2 þ ðp=csÞ2

q� �ð7:4:38Þ

When the new variable p varies along the Bromwich line BrðpÞ in the complex p-plane, i.e.

BrðpÞ: cp � i1 ! cp þ i1 ð7:4:39Þ

where cp is a proper constant, the inverse s given by Eq. (7.4.38) moves along thebumped curve ACB in Fig. 7.6, but it is almost straight at infinity. At infinity, theinverse function takes the asymptotic form

s�cp þ i11�M

; p ! cp þ i1cp � i11�M

; p ! cp � i1

8><>: ð7:4:40Þ

Re( )s

Im( )s

A

C

B

sγ

pp iγ= + ∞s iγ + ∞

pp iγ= − ∞s iγ − ∞

( )s p

sic η

sic η−

Fig. 7.5 Closed loop C forthe complex integral UA inEq. (7.4.34)


So, if we take the real part of the Bromwich line BrðsÞ from cs � i1 to cs þ i1 as

cs ¼cp

1�Mð7:4:41Þ

the bumped curve can be connected to the edge of the Bromwich line BrðsÞ andthen, these two curves constitute a closed loop ACBA as shown in the figure. Thus,we employ this closed loop C for the complex integral UA in Eq. (7.4.34).

Fortunately, the real part of the two edges is sufficiently large and the twoBromwich lines

BrðpÞ: cp � i1\p\cp þ i1BrðsÞ: cs � i1\s\cs þ i1 ð7:4:42Þ

have sufficiently large real parts so that all singular points of the integrand are in theleft side of the Bromwich line BrðsÞ. Then, Cauchy’s integral theorem can beapplied to the complex integral. Since the closed loop C(ACBA) does not includeany singular point, the integral along BrðsÞ is converted to a parametric integralalong the bumped curve. That is,

12pi

Zcsþi1

cs�i1expð�aslÞ expfðs� asVÞtgds ¼ 1

2pi

Zcpþi1

cp�i1expð�aslÞ dsdp �

s¼sðpÞexpðptÞdp

ð7:4:43Þ

where sðpÞ is the inverse given by Eq. (7.4.38).

Fig. 7.6 Closed loop C for the complex integral UB


Consequently, we could convert the Laplace inversion integral with respect tothe parameter s to that with respect to the parameter p and thus, Eq. (7.4.33) isrewritten in the form of the single Laplace inversion integral with respect to theparameter p, i.e.

12pi

ZBrðpÞ

1m

bs �Mpmcs

� �Aðg; pÞ expðptÞdp

¼ Q2c2s

12pi

ZBrðpÞ

expð�aslÞ dsdp �

s¼sðpÞexpðptÞdp ð7:4:44Þ

Both sides of the above equation have the same form of Laplace inversion integralwith respect to the single parameter p. Applying the Laplace transform with respectto the time t to both sides, we obtain the simple algebraic equation for the unknowncoefficient,

1m

bs �Mpmcs

� �Aðg; pÞ ¼ Q

2c2sexpð�aslÞ dsdp �

s¼sðpÞð7:4:45Þ

Thus, the coefficient is determined as

Aðg; pÞ ¼ Q2c2s

mbs � ðMpÞ=ðmcsÞ expð�aslÞ dsdp

�s¼sðpÞ

ð7:4:46Þ

Using the relations derived from Eqs. (7.4.35) and (7.4.38), the radical and thegradient are given by

as ¼ 1m2 Mðp=csÞ þ mbsf g ð7:4:47Þ

dsðpÞdp

¼ Mðp=mcsÞ þ bsm2bs

ð7:4:48Þ

The coefficient is then rewritten in the explicit form,

Aðg; pÞ ¼ Q2c2s

1mbs

bs þMðp=mcsÞbs �Mðp=mcsÞ exp �Mðp=csÞ þ mbs

m2 l

� �ð7:4:49Þ

Lastly, substituting Eq. (7.4.49) into Eq. (7.4.27), the double transformed dis-placement produced by the reflected wave is given by

~urefl ¼Q2c2s

1mbs

bs þMðp=mcsÞbs �Mðp=mcsÞ exp �Mðp=csÞ þ mbs

m2 l

� �exp �bs þ

Mpmcs

� �zm

� �ð7:4:50Þ


(3) Cagniard’s technique

We have obtained the displacement produced by the reflected wave, but in thetransformed domain. As the incident wave has already been obtained in the explicitform by Eq. (7.4.8), our next task is to perform the double inversion for thedisplacement in Eq. (7.4.50). The double inversion is carried out by applying theCagniard-de Hoop technique. The formal Fourier inversion with respect to theparameter g is given by

urefl ¼Q

4pc2s

Zþ1

�1

1mbs

bs þMðp=mcsÞbs �Mðp=mcsÞ exp �Mðp=csÞðl� zÞ

m2 � bsðzþ lÞm

� igy

� �dg

ð7:4:51Þ

Since the radical bs ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðp=mcsÞ2

qhas the Laplace transform parameter p, we

introduce the variable transform from g to the new variable 1 as

g ¼ ðp=mcsÞ1 ð7:4:52Þ

Equation (7.4.51) is rewritten as

urefl ¼Q

4pmc2s

Zþ1

�1

1ffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1

pþMffiffiffiffiffiffiffiffiffiffiffiffiffi

12 þ 1p

�Mexp �ðp=mcsÞ Mðl� zÞ

mþ zþ l

m


pþ i1y

� � �d1

ð7:4:53Þ

Now, let us consider the complex integral UB whose integrand is the same as thatof Eq. (7.4.53),

UB ¼ 12p

ZC




12 þ 1p


mþ zþ l

m


pþ i1y

� � �d1

ð7:4:54Þ

The closed loop C is discussed here. The integrand has two branch points at 1 ¼ �i,and two branch cuts are thus introduced along the imaginary axis in the complex 1-plane. The vanishing point of the denominator,


p�M ¼ 0, gives two poles

at 1 ¼ �im on the imaginary axis; but they are smaller in magnitude than the branchpoints.


The Cagniard’s path is determined by the variable transform from 1 to the newvariable t as

t ¼ 1mcs

Mðl� zÞm

þ zþ lm


pþ i1y

� �ð7:4:55Þ

Solving for 1, the Cagniard’s path is given by

1ð�Þs ¼

�imyfm2cst �Mðl� zÞg � ðzþ lÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifm2cst �Mðl� zÞg2 � fðzþ lÞ2 þ ðmyÞ2g

qðzþ lÞ2 þ ðmyÞ2

ð7:4:56Þ

The saddle point of the Cagniard’s path is

1saddle ¼ �imyffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðzþ lÞ2 þ ðmyÞ2q ð7:4:57Þ

at

t ¼ 1m2cs

Mðl� zÞ þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðzþ lÞ2 þ ðmyÞ2

q� �ð7:4:58Þ

If we take the closed loop composed of the Cagniard’s path, the straight line alongthe real axis and two large arcs which connect the straight line with the Cagniard’spath, the pole 1 ¼ �im is inside of the loop when

m\mjyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðzþ lÞ2 þ ðmyÞ2q ) jyj[ zþ l

Mð7:4:59Þ

but outside when

m[mjyjffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðzþ lÞ2 þ ðmyÞ2q ) jyj\ zþ l

Mð7:4:60Þ

The location of the pole depends on this inequality. The contribution from the poleat 1 ¼ �im is included only when the inequality of Eq. (7.4.59) holds. Thus, wedetermine the closed loop C as ABCDOEA shown in Fig. 7.6. Applying Cauchy’stheorem to the complex integral UB in Eq. (7.4.54), the line integral along the realaxis is converted to the sum of the integral along the Cagniard’s path and theresidue at the pole. That is


12p

Zþ1

�1




12 þ 1p


mþ zþ l

m


pþ i1y

� � �d1

¼ 12p

Z1Mðl�zÞþ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðzþlÞ2þðmyÞ2

pm2cs




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1



12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

24

35e�ptdt

� 2pi2p

1þ imffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ 1



12 þ 1p


mþ zþ l

m


pþ i1y

� � �" #1¼�im

H jyj � zþ lM

� �

ð7:4:61Þ

Rearranging the second residue term in the right hand side, we obtain the somewhatsimpler form,

12p

Zþ1

�1




12 þ 1p


mþ zþ l

m


pþ i1y

� � �d1

¼ 12p

Z1Mðl�zÞþ


pm2cs




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

24

35e�ptdt

þ 2Mm

exp � pmcs

2Mlm

þ my

� �� H jyj � zþ l

M

� �

ð7:4:62Þ

Substituting this equation into Eq. (7.4.53), we obtain the Laplace transformeddisplacement as the sum of the Laplace transform integral and the simple expo-nential function,

urefl ¼Q

4pmc2s

Z1Mðl�zÞþ


pm2cs




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

26666664

37777775e�ptdt

þ MQm2c2s

exp � pmcs

2Mlm

þ my


M

� �ð7:4:63Þ

The first term is just in the form of Laplace transform integral and is easily invertedby inspection. The second term is the exponential function and the simple Laplaceinversion formula,

L�1 expð�apÞ½ ¼ dðt � aÞ ð7:4:64Þ


is applied. Thus, the Laplace inversion is easily carried out and we get the final formfor the displacement produced by the reflected wave as

urefl ¼ Q4pmc2s

H t �Mðl� zÞ þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðzþ lÞ2 þ ðmyÞ2

qm2cs

0@

1A




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

24

35

� MQm2c2s

d t � 1mcs

2Mlm

þ my


M

� �ð7:4:65Þ

Fortunately, we can simplify the above equation and get an exact closed formsolution. Substituting Eq. (7.4.56) into Eq. (7.4.55), we have for the radical


p¼

ðzþ lÞfm2cst �Mðl� zÞg � iðmyÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifm2cst �Mðl� zÞg2 � fðzþ lÞ2 þ ðmyÞ2g

qðzþ lÞ2 þ ðmyÞ2n o

ð7:4:66Þ

and thus we have the simple expression for the complex valued function as




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s




12 þ 1p

�M

d1dt

( )1¼1ðþÞ

s

¼ 2m2csffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiT2 � R2

p T2 �M2R2 � ðmyÞ2T2 � 2Mðzþ lÞT þM2R2 � ðmyÞ2

ð7:4:67Þ

where

T ¼ m2cst �Mðl� zÞ; R ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðzþ lÞ2 þ ðmyÞ2

qð7:4:68Þ

Then, Eq. (7.4.65) is simplified as the exact closed form solution for the reflectedwave,

csQurefl ¼ m

2pH T � Rð Þ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

T2 � R2p T2 �M2R2 � ðmyÞ2

T2 � 2Mðzþ lÞT þM2R2 � ðmyÞ2

�Mmd mðcstÞ � 2Ml

mþ my


M

� � ð7:4:69Þ


(4) Wave fronts

Disturbed regions and wave fronts are easily obtained. They are given by theconditionals of the step and delta functions in Eq. (7.4.69). Returning to the originalfixed coordinate from the moving coordinate, we find that the step function attachedto the first term in Eq. (7.4.69) gives the disturbed region by the reflected wave:

Wrefl: : H t �Mðl� zÞ þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðzþ lÞ2 þ ðmyÞ2

qm2cs

0@

1A) cst � 2lM

m2

� �2

� x� 2lm2

� �2

þ y2

ð7:4:70Þ

This is a circular region, but not a full circle since the edge is moving and theconditional for x is imposed as x�Mcst þ l. This reflected wave is shown in Fig. 7.7.The second term includes the product of delta and step functions.

2

21s

lc t

M m⎛ ⎞= +⎜ ⎟⎝ ⎠

O

A

B C

Q

y

x22 /l m

2edge 2 (1 1/ )x l m= +

3 /r l M=

sc t

M

Fig. 7.7 Wave fronts for incident and reflected waves at the time cst ¼ lM

2m2 þ 1�


Wflat: d t � 1mcs

2Mlm

þ my


M

� �

) d cst � 2Mlm2 � y

� �H Mjyj � x�Mcst � 2lð Þ

ð7:4:71Þ

Examining the product, we find that the delta function gives a line singular waveand the step function restricts its line segment BC (in the figure). Thus, the wave isgiven as

Flat wave:Wflat: cst ¼ yþ 2Mlm2 ð7:4:72Þ

Supporting region: jyj[ � x� 2lM

þ cst ð7:4:73Þ

The incident wave has already been given by Eq. (7.4.8) and it is the circularregion. Its front is given by

Winc: H cst �ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p �) ðcstÞ2 � x2 þ y2; x\MðcstÞ þ l ð7:4:74Þ

Figure 7.7 shows the typical wave front shape for the incident and reflected waves.

7.5 Wave Scattering by a Rigid Inclusionin an Inhomogeneous Elastic Solid

The problem of wave scattering is one of the traditional elastodynamic problems.However, as far as the author knows, no exact closed form solution for transient(impulsive) waves is known for isotropic homogeneous elastic and acousticmaterials (Friedlander 1958; Pao and Mow 1973). Only one exact closed formsolution (Green’s function) was obtained for SH-waves in an inhomogeneouselastic solid (Watanabe 1982). The present section shows the exact closed form ofGreen’s function for the wave scattering problem in a radially inhomogeneouselastic solid.

Let us consider a radially inhomogeneous elastic solid with a borehole and takethe cylindrical coordinate ðr; h; zÞ. We assume that the shear modulus l and thedensity q are functions of the radial distance r from the coordinate origin as

l � l0ðr=aÞk; q � q0ðr=aÞk�2 ð7:5:1Þ

where l0 and q0 are the reference values at r ¼ a and k is an inhomogeneityparameter. Upon this assumption, the shear wave velocity is the linear function ofthe radial distance as


cs ¼ffiffiffiffiffiffiffiffil=q

p¼ cs0ðr=aÞ; cs0 ¼

ffiffiffiffiffiffiffiffiffiffiffiffil0=q0

pð7:5:2Þ

A rigid circular (cylindrical) inclusion with radius a is placed at the coordinateorigin and SH-wave source is also placed at ðr ¼ b; h ¼ 0Þ as shown in Fig. 7.8.Assuming the anti-plane deformation, i.e. SH-wave problem, the equation ofmotion and Hooke’s law are given by

@rrz@r

þ 1rrrz þ 1

r@rhz@h

þ qBz ¼ q@2uz@t2

ð7:5:3Þ

rrz ¼ l@uz@r

; rhz ¼ l1r@uz@h

ð7:5:4Þ

where uz and rrz; rhz are the anti-plane displacement and its associated stresscomponents. The body force Bz is assumed as the point/line wave source,

Bz ¼ B0dðr � bÞ

rdðhÞdðtÞ ð7:5:5Þ

where B0 is the magnitude of the source and dð:Þ is Dirac’s delta function.As the rigid circular inclusion is inserted in the hole, the elastic solid occupies its

outer region a\r\1 and the anti-plane displacement is fixed at the boundary, i.e.the fixed boundary condition for the elastic solid, is

uzjr¼a ¼ 0; 0� t\1 ð7:5:6Þ

In addition to the above boundary condition, we employ the radiation condition atinfinity as

uzjr!1¼ @uz@r

��r!1

¼ 0 ð7:5:7Þ

a

b

0θ =θ π= ±

SH-wave source

Rigid inclusion ( , )r θ

Fig. 7.8 A circular rigid inclusion and a SH-wave source in an inhomogeneous elastic solid

7.5 Wave Scattering by a Rigid Inclusion … 243

The above equations from (7.5.1) to (7.5.7) constitute our wave scatteringproblem. Substituting Eqs. (7.5.4) and (7.5.5) into the equation of motion (7.5.3),we have the single displacement equation as

@2uz@r2

þ l0

lþ 1

r

� �@uz@r

þ 1r2@2uz@h2

� ql@2uz@t2

¼ � qlB0

dðr � bÞr

dðhÞdðtÞ ð7:5:8Þ

where the prime denotes the differentiation with respect to the radial variable as

l0 ¼ dldr

ð7:5:9Þ

Recalling Eqs. (7.5.1) and (7.5.2), the displacement Eq. (7.5.8) is rewritten as

@2uz@r2

þ kþ 1r

@uz@r

þ 1r2@2uz@h2

� 1c2s0

ar

�2 @2uz@t2

¼ � B0

c2s0

ar

�2 dðr � bÞr

dðhÞdðtÞ

ð7:5:10Þ

Now, we start to solve the above differential equation. The variable transform forthe radial variable

z ¼ logðr=aÞ , r ¼ a expðzÞ ¼ aez ð7:5:11Þ

is introduced. Equation (7.5.10) is reduced to the nonhomogeneous differentialequation with constant coefficients as

@2uz@z2

þ k@uz@z

þ @2uz@h2

� acs0

� �2 @2uz@t2

¼ �B0acs0

� �2 dðaez � bÞaez

dðhÞdðtÞ ð7:5:12Þ

Then, we apply the Laplace transform with respect to the time t and the Fourierfinite transform with respect to the circumferential angle θ. They are defined inChap. 1 as

Laplace transform: f �ðsÞ ¼Z10


Fourier finite transform: fn ¼Zp�p

f ðhÞ expðþinhÞdh , f ðhÞ ¼ 12p

Xþ1

n¼�1fn expð�inhÞ

ð7:5:14Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

Applying the double transform with the quiescent condition at an initial time,

uzjt¼0 ¼@uz@t

��t¼0

¼ 0 ð7:5:15Þ

Equation (7.5.12) is reduced to the simple ordinary differential equation,

d2u�zndz2

þ kdu�zndz

� fn2 þ ðas=cs0Þ2gu�zn ¼ �B0acs0

� �2 dðaez � bÞaez

ð7:5:16Þ

The homogeneous solution which satisfies the radiation condition (7.5.7) is easilyobtained as

u�ðhÞzn ¼ CnðsÞ exp �zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðas=cs0Þ2 þ ðk=2Þ2

qþ k=2

� � �ð7:5:17Þ

where CnðsÞ is an unknown coefficient to be determined by the boundary condition.In order to obtain a particular solution corresponding to the wave source B0, we

apply the Fourier transform with respect to the variable z to the ordinary differentialequation (7.5.16). The transform is defined by

f ðfÞ ¼Z1�1

f ðzÞ expðþifzÞdz , f ðzÞ ¼ 12p

Z1�1

f ðfÞ expð�ifzÞdf ð7:5:18Þ

We have the simple algebraic equation for the triple transformed displacement,

f2 þ ikfþ n2 þ ðas=cs0Þ2n o

u�ðpÞzn ¼ B0a

bcs0

� �2

exp þif logðb=aÞf g ð7:5:19Þ

where the source (nonhomogeneous) term is evaluated as

B0acs0

� �2 Z1�1

dðaez � bÞaez

expðþifzÞdz

¼ B0acs0

� �2Z10

dðr � bÞr

exp þif logðr=aÞf g drr

¼ B0a

bcs0

� �2

exp þif logðb=aÞf g

ð7:5:20Þ


Thus, we have the displacement in the triple transformed domain,

u�ðpÞzn ¼ B0a

bcs0

� �2 exp þif logðb=aÞf gðfþ ik=2Þ2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 ð7:5:21Þ

Now, let us consider the inversion. Firstly, we apply the Fourier inversionintegral with respect to the parameter f defined by the second of Eq. (7.5.18). It is

u�ðpÞzn ¼ B0a

bcs0

� �2 12p

Z1�1

exp½�iffz� logðb=aÞgðfþ ik=2Þ2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 df ð7:5:22Þ

If we make the change of variable from f to g as g ¼ fþ ik=2, the integrand inEq. (7.5.22) is simplified, but, its path of integration is the complex line from�1þ ik=2 to þ1þ ik=2, not a real path.

u�ðpÞzn ¼ B0a

bcs0

� �2

exp½�ðk=2Þfz� logðb=aÞg

� 12p

Z1þik=2

�1þik=2

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg

ð7:5:23Þ

In order to have a real line integral, we consider the complex integral whoseintegrand is the same as that in Eq. (7.5.23). That is

U ¼ 12p

ZL

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg ð7:5:24Þ

The integration loop L is a rectangular form, as shown in Fig. 7.9. It is composed of

two parallel horizontal lines AB�!

and CD�!

, and two finite vertical lines DA�!

and BC�!

.The edges of the horizontal line tend to infinite, i.e. �1þ ik=2. Examining thesingularity of the integrand, we find two poles at

g ¼ �iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðas=cs0Þ2 þ ðk=2Þ2

qð7:5:25Þ

which are outside of the loop L and the integrand tends to zero at the infinityg ! �1þ ik=2. Then, we apply the Cauchy theorem to the complex integral U.

Since the integrals along the two vertical lines DA�!

and BC�!

vanish and no singular

point is included in the loop, the integral along the upper horizontal line AB�!

is

converted to that along the lower horizontal line DC�!

, i.e. the real axis in theg-plane. Thus we have the conversion formula for our integral as


12p

Z1þik=2

�1þik=2

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg

¼ 12p

Z1�1

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg ð7:5:26Þ

The infinite integral in the right hand side of the above equation is reduced to thesemi-infinite integral

12p

Z1�1

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg ¼ 1

p

Z10

cos½gfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg

ð7:5:27Þ

and the simple integration formula (Erdélyi 1954, vol. I, p. 8, (11))

Z10

1x2 þ a2

cosðxyÞdx ¼ p2a

expð�a yj jÞ ð7:5:28Þ

/ 2iλ+

( )niq s−

/ 2iλ−

( )niq s+

Im( )η

Re( )η

/ 2iλ−∞ + / 2iλ+∞ +

2 2 20( ) ( / ) ( / 2)snq s n as c λ= + +

Fig. 7.9 Integration loop L for the complex integral U


is applied. Thus, we can evaluate our integral as

12p

Z1þik=2

�1þik=2

exp½�igfz� logðb=aÞgg2 þ n2 þ ðas=cs0Þ2 þ ðk=2Þ2 dg ¼ 1

2qnðsÞ exp �qnðsÞ z� logðb=aÞj jf g

ð7:5:29Þ

where

qnðsÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðas=cs0Þ2 þ ðk=2Þ2

qð7:5:30Þ

Substituting Eq. (7.5.29) into Eq. (7.5.23), the double transformed particularsolution is given by

u�ðpÞzn ¼ B0

2a

bcs0

� �2 1qnðsÞ exp½�ðk=2Þfz� logðb=aÞg exp �qnðsÞ z� logðb=aÞj jf g

ð7:5:31Þ

Summing the homogeneous and particular solutions, Eqs. (7.5.17) and (7.5.31), wehave the full solution for the differential equation (7.5.16). That is

u�zn ¼ u�ðhÞzn þ u�ðpÞzn

¼ CnðsÞ exp �z qnðsÞ þ k=2f g½

þ B0

2a

bcs0

� �2 1qnðsÞ exp½�ðk=2Þfz� logðb=aÞg exp �qnðsÞ z� logðb=aÞj jf g

ð7:5:32Þ

In order to determine the coefficient CnðsÞ, we apply the boundary condition,Eq. (7.5.6). The double transformed boundary condition with the change of variableis given by

u�zn��z¼0 ¼ 0 ð7:5:33Þ

Substituting Eq. (7.5.32) into the above equation, we have for the coefficient,

CnðsÞ ¼ �B0

2a

bcs0

� �2 1qnðsÞ exp½fðk=2Þ � qnðsÞg logðb=aÞg ð7:5:34Þ


and thus the displacement in the double transformed domain is completelydetermined,

u�zn ¼B0

2a

bcs0

� �2

exp �ðk=2Þ z� logðb=aÞf g½

� 1qnðsÞ exp �qnðsÞ z� logðb=aÞj jf g � 1

qnðsÞ exp �qnðsÞ zþ logðb=aÞf g½ � �

ð7:5:35Þ

For the simplification in the subsequent analysis, we return the variable z to theoriginal radial distance r. The transformed solution is rewritten as

u�zn ¼B0

2a

bcs0

� �2 br

� �k=2 exp �qnðsÞ logðr=bÞj jf gqnðsÞ � exp �qnðsÞ logðbr=a2Þ

� �qnðsÞ

�ð7:5:36Þ

We have just obtained the solution in the double transformed domain. Oursubsequent work is to invert the solution into the actual ðr; h; tÞ space. As the firstinversion, we consider its Laplace inversion. The Laplace inversion is symbolicallywritten as

uzn ¼ B0

2a

bcs0

� �2 br

� �k=2

L�1 exp �qnðsÞ logðr=bÞj jf gqnðsÞ

�� L�1 exp �qnðsÞ logðbr=a2Þ

� �qnðsÞ

��

ð7:5:37Þ

Fortunately, since qnðsÞ is the simple square root function given by Eq. (7.5.30), wecan apply the inversion formula (Erdélyi 1954, vol, I, p. 248, (24)),

L�1 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 þ a2

p exp �bffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 þ a2

p � �¼ Hðt � bÞJ0 a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 � b2

p �ð7:5:38Þ

where H(.) is Heaviside’s unit step function and J0(.) is Bessel function of thezeroth order. It follows that

uzn ¼ B0

2a

b2cs0

br

� �k=2

H s� logðr=bÞj jð ÞJ0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðk=2Þ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ

q� �

� H s� logðbr=a2Þ� J0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ ðk=2Þ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ

q� ��ð7:5:39Þ

where the dimensionless time, τ = cs0t/a, is introduced. The next is the inversion of theFourier finite transform, i.e. Fourier series. Applying the inversion formula inEqs. (7.5.14)–(7.5.39), we have for the displacement


uz ¼ B0

2a

b2cs0

br

� �k=2

� H s� logðr=bÞj jð Þ 12p

Xþ1

n¼�1J0



q� �expð�inhÞ

"

� H s� logðbr=a2Þ� 12p

Xþ1

n¼�1J0



q� �expð�inhÞ

#

ð7:5:40Þ

Since both coefficients in the two series include only even power of the index n, theFourier series is reduced to the semi-infinite sum as

uz ¼ B0

2pa

b2cs0

br

� �k=2

� H s� logðr=bÞj jð ÞX1n¼0

enJ0



q� �cosðnhÞ

"

� H s� logðbr=a2Þ� X1n¼0

enJ0



q� �cosðnhÞ

#

ð7:5:41Þ

where

en ¼ 1=2; n ¼ 01; n� 1

�ð7:5:42Þ

Formally we have just arrived at the exact solution for the scattering problem,however, this solution does not show wave propagation phenomena clearly. Inorder to obtain the solution which shows the wave phenomena, i.e. wave frontshape, we have to discuss the Fourier series. Let us consider the Fourier seriesdefined by

S ¼X1n¼0

J0 affiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 þ b2

q� �cosðnhÞ ð7:5:43Þ

where a and b are positive constants. We replace the Bessel function with itsintegral form (Erdélyi 1954, vol. I, p. 28, (42)),


q� �¼ 2

p

Za0

cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � n2

p �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � n2

p cosðnnÞdn ð7:5:44Þ


and change the order of the summation and integration. Then, expanding theintegration range, two summations are unified as

S ¼ 1p

Za0



p dnX1n¼0

en cosfnðhþ nÞg þ cosfnðh� nÞg½

¼ 1p

Zþa

�a



p dnX1n¼0

en cosfnðn� hÞgð7:5:45Þ

The Fourier series in the integrand is just the form of Fourier expansion of aninfinite periodic array of the delta function (see Exercises (1.1) and (1.2)). That is

X1n¼0

en cosfnðn� hÞg ¼ pXþ1

m¼�1dðn� hþ 2mpÞ ð7:5:46Þ

We substitute the above equation into the last line in Eq. (7.5.45) and change theorder of integration and summation. It follows that

S ¼Xþ1

m¼�1

Zþa

�a



p dðn� hþ 2mpÞdn ð7:5:47Þ

Examining the supporting region for the delta function, we can evaluate the integralwith use of the simple formula (1.2.3) in Chap. 1. It results

S ¼Xþ1

m¼�1H a� h� 2mpj jð Þ

cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � ðh� 2mpÞ2

q� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � ðh� 2mpÞ2

q ð7:5:48Þ

The above equation looks like the sum of the infinite terms. But, due to theoperation of the step function, this summation is truncated to the finite sum. Thus,we have just reduced the infinite sum of the Fourier series to the finite sum ofsimple algebraic function. Further, if we expand the range of the circumferential/polar angle h from the finite range ð�p\h\þ pÞ to the infinite rangeð�1\h\þ1Þ and rewrite the angle

h � h� 2mp ð7:5:49Þ


http://dx.doi.org/10.1007/978-3-319-17455-6_1

The summation with the index m in Eq. (7.5.48) is reduced to the single term. That is

S ¼ Hða� hÞcos b

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � h2

p �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � h2

p ; �1\h\þ1 ð7:5:50Þ

Consequently, we have obtained the very simple expression for the Fourierseries. That is the summation formula,

X1n¼0


q� �cosðnhÞ ¼ Hða� hÞ

cos bffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � h2

p �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � h2

p ; �1\ h\þ1

ð7:5:51Þ

Substituting this formula into Eq. (7.5.41), we have the closed form for thedisplacement

uz ¼ B0

2pa

b2cs0

br

� �k=2

� H s� logðr=bÞj jð ÞHðffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ

q� hj jÞ

cos ðk=2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ � h2

q� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ � h2

q2664

� H s� logðbr=a2Þ� Hð

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ

q� hj jÞ

cos ðk=2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ � h2

q� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ � h2

q3775

ð7:5:52Þ

The product of two step functions in the above equation can be unified and then wehave the final form for the displacement, i.e. the Green function for the scatteringproblem,

uz ¼ B0

2pa

b2cs0

br

� �k=2

H s�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffilog2ðr=bÞ þ h2

q� � cos ðk=2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ � h2

q� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðr=bÞ � h2

q2664

� H s�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffilog2ðbr=a2Þ þ h2

q� � cos ðk=2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ � h2

q� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � log2ðbr=a2Þ � h2

q3775

ð7:5:53Þ


The first term in the bracket shows the direct wave from the source and its wavefront is given by

s ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffilog2ðr=bÞ þ h2

q) r=b ¼ exp �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � h2

p �ð7:5:54Þ

The second term shows the scattering wave and its front is given by

s ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffilog2ðbr=a2Þ þ h2

q) r=a ¼ ða=bÞ exp �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 � h2

p �ð7:5:55Þ

The both equations show the spiral curves in the ðr; hÞ-plane and the time-devel-opment of these wave fronts can be found in the paper (Watanabe 1982).

7.6 An Excellent Application of Cauchy Complex Integral

The present section does not show any Green’s function, but a very excellentapplication of the Cauchy complex integral is described. This application techniqueis originated by Mal (1970), when he considered the wave scattering by a crack.The dynamic crack problem is reduced to the Fredholm integral equation whosekernel is in the form of semi-infinite integral with the product of two Besselfunctions,

Kðx; yÞ ¼Z10


q� n

� �J0ðxnÞJ0ðynÞdn; x\ y ð7:6:1Þ

where J0ð:Þ is the Bessel function of the first kind. As we have no suitable inte-gration formula for this semi-infinite integral, we have to evaluate it numerically.However, the integrand shows very slow convergence and its numerical evaluationtakes much CPU time for sufficient accuracy. If we could transform the semi-infinite integral to finite integrals, we can evaluate the kernel without regard to theconvergence and accuracy. Mal (1970) has reduced the integral to the finite onesuch as

Kðx; yÞ ¼ þiZx=c0


qJ0ðxnÞHð2Þ

0 ðynÞdn; x\ y ð7:6:2Þ

where Hð2Þ0 ð:Þ is Hankel function of the second kind. After Mal, many researchers

have employed his evaluation technique for the dynamic crack and punch problems.But, they (including Mal) did not describe its mathematics in detail. Thus, applying


our former discussion on the branch cut in Sect. 1.3.2 in Chap. 1, his technique isexplained in some depths and then one can learn the power of the Cauchy complexintegral.

As the kernel function of Eq. (7.6.1) is the same in the form, regardless of thepositive and negative time factors, we assume the positive time factor expðþixtÞand the radiation/convergence condition at infinity

Reffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

q� �� 0 ð7:6:3Þ

Due to these assumptions, the branch cuts in the complex n-plane are introduced aswas discussed in Sect. 1.3.2 in Chap. 1. The cut and integration path for the semi-infinite integral of the kernel are shown in Fig. 7.10, where the cut is the bold line

and the integration path is the semi-infinite line OAB��!

. Since the integration path ison the positive real axis, we decompose the integral into two parts: one is the finite

integral on OA�!

0� n\x=cð Þ and the other on AB�!

x=c\n\1ð Þ. As was discussedin Sect. 1.3.2, the square root function on the path OA

�!takes the positive imaginary

values as


q¼ þi


q; 0� n\x=c ð7:6:4Þ

and that on the path AB�!

does the positive real value as


q¼ þ


q; x=c\n\1 ð7:6:5Þ

/ cω+/ cω−

Im( )ξ

Re( )ξ

O A B

integration pathbranch cut

branch cut

Fig. 7.10 Integration path in the complex ξ-plane


http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

http://dx.doi.org/10.1007/978-3-319-17455-6_1

So, the kernel K(x, y) of Eq. (7.6.1) is decomposed into two integrals as

Kðx; yÞ ¼Zx=c0


q� n

� �J0ðxnÞJ0ðynÞdn

þZ1x=c


q� n


ð7:6:6Þ

The first integral in the above equation is in the form of the finite integral, but the

second one on the path AB�!

is still in that of the semi-infinite integral. We have totransform the latter integral to the finite ones. This is the main purpose of thepresent section.

Since the relation between Bessel and Hankel functions exists,

J0ðZÞ ¼ 12

Hð1Þ0 ðZÞ þ Hð2Þ

0 ðZÞn o

; Hð1=2Þ0 ðZÞ ¼ J0ðZÞ � iY0ðZÞ ð7:6:7Þ

the latter integral in Eq. (7.6.6) is separated into two semi-infinite integrals as

I ¼Z1x=c


q� n


¼ 12

Z1x=c


q� n

� �J0ðxnÞHð1Þ

0 ðynÞdn

þ 12

Z1x=c


q� n


0 ðynÞdn

ð7:6:8Þ

We consider two complex integrals whose integrands are the same as those in theabove equation, respectively, and are defined by

U1 ¼ 12

IC1


q� f

� �J0ðxfÞHð1Þ

0 ðyfÞdf ð7:6:9Þ

U2 ¼ 12

IC2


q� f


0 ðyfÞdf ð7:6:10Þ

We assume the real and imaginary parts of the complex variable f are n and g,respectively, i.e. f ¼ nþ ig. Both integrands have two branch points at f ¼ �x=cfor the square root function, and one branch point at the origin f ¼ 0 for the

7.6 An Excellent Application of Cauchy Complex Integral 255

logarithmic nature of the Hankel function. In order to examine the convergence atinfinity, we apply the asymptotic expression of the Bessel function (Watson 1966,pp. 198–199) to the product of Bessel and Hankel functions and have

J0ðxfÞHð1Þ0 ðyfÞ

��jfj!1

� 1pf

ffiffiffiffiffixy

p expfþiðy� xÞfg � i expfþiðyþ xÞfg½ ð7:6:11Þ

J0ðxfÞHð2Þ0 ðyfÞ

��jfj!1

� 1pf

ffiffiffiffiffixy

p expf�iðy� xÞfg þ i expf�iðyþ xÞfg½ ð7:6:12Þ

We learn that the product of Bessel and Hankel functions shows the exponential

decay at infinity. Since y > x, the product J0ðxfÞHð1Þ0 ðyfÞ converges in the upper

ζ-plane, i.e. Im(ζ) > 0, and J0ðxfÞHð2Þ0 ðyfÞ does in the lower ζ-plane, i.e. Im(ζ) < 0.

Two branch cuts for the square root function are the same as those in Fig. 7.10, andthe branch cut for the Hankel function is introduced along the negative real axisstarting from the origin, i.e. the dotted line OP in Fig. 7.11. So, we take the closedloop Γ1 in the first quadrant and Γ2 in the fourth quadrant as shown in the figure.These two loops are composed of line paths and circular arcs: They are

/ cω+/ cω−

Im( )η ζ=

Re( )ξ ζ=1A 1B 1C 1D

1E

1F

2A 2B 2C 2D

2E

2F

OP

1Loop Γ

2Loop Γ

Fig. 7.11 Integration contours Γ1 and Γ1 for the complex integrals Φ1 and Φ2


C1:A1B1��!þ B1C1

\þC1D1��!þ D1E1

\ þE1F1��!þ F1A1

\ð7:6:13Þ

C2:A2B2��!þ B2C2

\þC2D2��!þ D2E2

\ þE2F2��!þ F2A2

\ð7:6:14Þ

The complex integral is decomposed into the integrals along these path segments.

U1 ¼ 12

ZA1B1��! þ

ZB1C1

\

þZ

C1D1��! þ

ZD1E1

\

þZ

E1F1��! þ

ZF1A1

\

0BBB@

1CCCA


q� f


0 ðyfÞdf

ð7:6:15Þ

U2 ¼ 12

ZA2B2��! þ

ZB2C2

\

þZ

C2D2��! þ

ZD2E2

\

þZ

E2F2��! þ

ZF2A2

\

0BBB@

1CCCA


q� f


0 ðyfÞdf

ð7:6:16Þ

The radii of the quarter- and semi-circular paths, BjCj

\and FjAj

\; j = 1, 2, are very

small and tend to zero as the limit. Thus, the integrals on these small circular pathsvanish when we take their zero limit of the radius. On the other hand, due to theexponential decay of the product of Bessel and Hankel functions, the integral on the

quarter-circle DjEj

\; j ¼ 1; 2 vanishes as the radius of the quarter circle tends to

infinity. Consequently, since no singular point such as the pole is included in eachloop, the complex integral around the closed loop vanishes, i.e. Uj ¼ 0, and thus the

integral on the line paths CjDj��!

; j ¼ 1; 2 is transformed to the sum of two integrals on

the paths AjBj��!

and EjFj��!

,

12

ZC1D1��!


q� f


0 ðyfÞdf

¼ � 12

ZA1B1��! þ

ZE1F1��!

0BBB@

1CCCA


q� f


0 ðyfÞdfð7:6:17Þ


12

ZC2D2��!


q� f


0 ðyfÞdf

¼ � 12

ZA2B2��! þ

ZE2F2��!

0BBB@

1CCCA


q� f


0 ðyfÞdfð7:6:18Þ

The values of the square root function and integration variable on each line path aretabulated in Table 7.1, where the product of Bessel and Hankel functions is derivedby using the formulas

J0ð�ixÞ ¼ I0ðxÞ; Hð1Þ0 ð�ixÞ ¼ � 2i

pK0ðxÞ; Hð2Þ

0 ð�ixÞ ¼ þ 2ipK0ðxÞ ð7:6:19Þ

where I0(.) and K0(.) are the modified Bessel function of the first and second kinds.Then, we can rewrite Eqs. (7.6.17) and (7.6.18). They are

12

Z1x=c


q� n


0 ðynÞdn

¼ � 12

Zx=c0


q� n


0 ðynÞdn

þ ip

Z10


q� g

� �I0ðxgÞK0ðygÞdg

ð7:6:20Þ

Table 7.1 Variable transforms on the line path

Path Integration variable fand its range

Square root functionffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif2 � ðx=cÞ2

q Product of Bessel and Hankel

functions J0ðxfÞHð1=2Þ0 ðyfÞ

A1B1��! f ¼ n

0\n\x=cþi


qJ0ðxnÞHð1Þ

0 ðynÞ

C1D1��! f ¼ n

x=c\n\1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

qJ0ðxnÞHð1Þ

0 ðynÞ

E1F1��! f ¼ þig; þ i ¼ eþpi=2

0\g\1 þiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q � 2ip I0ðxgÞK0ðygÞ

A2B2��! f ¼ n

0\n\x=c�i


qJ0ðxnÞHð2Þ

0 ðynÞ

C2D2��! f ¼ n

x=c\n\1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � ðx=cÞ2

qJ0ðxnÞHð2Þ

0 ðynÞ

E2F2��! f ¼ �ig; �i ¼ e�pi=2

0\g\1 �iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig2 þ ðx=cÞ2

q þ 2ip I0ðxgÞK0ðygÞ


12

Z1x=c


q� n


0 ðynÞdn

¼ þ 12

Zx=c0

iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

qþ n


0 ðynÞdn

� ip

Z10


q� g

� �I0ðxgÞK0ðygÞdg

ð7:6:21Þ

Adding two Eqs. (7.6.20) and (7.6.20), we have two finite integrals for the semi-infinite integral in Eq. (7.6.8) as

Z1x=c


q� n


¼ � 12

Zx=c0


q� n


0 ðynÞdn

þ 12

Zx=c0


qþ n


0 ðynÞdn

ð7:6:22Þ

This equation is simplified by using the formula (7.6.7) for the Hankel function. Ityields

Z1x=c


q� n


¼Zx=c0

nJ0ðynÞ þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðx=cÞ2 � n2

qY0ðynÞ

� �J0ðxnÞdn

ð7:6:23Þ

The semi-infinite integral is just transformed to the finite integral and thus we canreplace the second integral in the right hand side of Eq. (7.6.6) with the above finiteintegral. Therefore, the kernel function of the semi-infinite integral is reduced to thefinite one, i.e.

Kðx; yÞ ¼ þiZx=c0


qJ0ðxnÞHð2Þ

0 ðynÞdn; y[ x ð7:6:24Þ

This is the same as Eq. (7.6.2).


When we evaluate the kernel function, the change of variable from n to u,

n ¼ ðx=cÞu ð7:6:25Þ

is introduced and the numerical computation is carried out for the simplifiedintegral,

Kðx; yÞ ¼ þiðx=cÞZ10

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� u2

pJ0ðxxu=cÞHð2Þ

0 ðxyu=cÞdu; y[ x ð7:6:26Þ

As we have the good subroutines for Bessel functions, the numerical integration forthe unit interval is a very easy task. This completes the application of the complexintegral.

Exercise

(7:1) Decompose the integrand in Sect. 7.1, as

n2

g2 þ p21n2�

g2 þ p22n2� ¼ � 1

p21 � p22� 1

g2 þ p21n2 �

1

g2 þ p22n2

!

g2

g2 þ p21n2�

g2 þ p22n2� ¼ þ 1

p21 � p22� p21

g2 þ p21n2 �

p22g2 þ p22n

2

! ðaÞ

and carry out the inversion integral (7.1.14) with the aid of the tabulatedintegration formulas.

References

Erdélyi A (ed) (1954) Tables of integral transforms, vols I and II. McGraw-Hill, New YorkFriedlander FG (1958) Sound pulses. Cambridge University Press, CambridgeMal AK (1970) Interaction of elastic waves with a Griffith crack. Int J Eng Sci 8:763–776Pao Y-H, Mow C-C (1973) Diffraction of elastic waves and dynamic stress concentrations. Crane

Russak, New YorkWatanabe K (1982) Scattering of SH-wave by a cylindrical discontinuity in an inhomogeneous

elastic medium. Bull JSME 25(205):1055–1060Watson GN (1966) A treatise on the theory of Bessel functions, vol 2. Cambridge University

Press, Cambridge


Index

AAchenbach, J.D., 154Acoustic fluid, 126, 129Acoustic wave, 121–123, 127, 129, 130Adiabatic change, 122, 125Amplitude function, 38, 39, 52, 63Anisotropic elastic solid, 205Anti-plane

— deformation, 79, 154, 155— displacement, 228, 243— shear load, 154

BBeam

1D —, 152Bending rigidity, 139, 145Bessel function, 5, 29, 50, 59, 82, 83, 101, 112,

133, 147, 152, 205, 219, 225, 249, 250,253, 256, 258, 260

Body force— potential, 127, 129

Boundary condition, 156, 164, 181, 203, 243,245, 248

Branch cut, 1, 11–21, 23–27, 30, 32, 53, 54,56, 59, 60, 62, 63, 65, 66, 132, 149,159, 167, 168, 170, 171, 173, 186,190–192, 220, 233, 237, 254, 256

Branch line, 174Branch point, 11, 14–16, 20, 23–25, 53–55, 59,

132, 149, 159, 167, 170, 171, 173, 186,190, 192, 220, 233, 237, 255

Bromwich integral, 230, 232Bromwich line, 2, 229, 230, 232, 234, 235Bumped curve, 234, 235

CCagniard, L., 154Cagniard’s

— path, 158, 167–169, 171, 174, 186, 187,190–192, 201, 238

— path I, 171, 172, 191— path II, 171–173, 191, 192

Cagniard-de Hoop technique, 154, 157, 161,169, 185, 201, 202, 237

Calculus, 8, 48Cartesian coordinate

— system, 77Cauchy’s (integral) theorem, 149, 192, 216,

238Complex frequency, 40, 42, 44, 54, 55, 59, 142Complex integral, 1, 29, 40, 42, 43, 54–56, 59,

60, 65, 132, 133, 142–144, 149, 150,154, 158, 159, 167, 169–173, 185–187,190, 192, 201, 205, 208, 216, 217, 219,233–235, 237, 238, 246, 247, 253–257,260

Continuity equation, 122–124, 127Convergence condition, 5, 11, 33, 35, 36, 38,

39, 45, 46, 48, 49, 51, 52, 54, 55, 68,70, 71, 73, 80, 90, 96, 110, 140, 142,145, 148, 155, 156, 179–181, 206, 214,223, 232, 254

Convolution integral, 64, 73, 74, 76, 88, 108,136

Critical velocity, 144

DDeflection

— divergence, 144

© Springer International Publishing Switzerland 2015K. Watanabe, Integral Transform Techniques for Green’s Function,Lecture Notes in Applied and Computational Mechanics 76,DOI 10.1007/978-3-319-17455-6

261

Deflection (cont.)dynamic —, 141, 148dynamic plate —, 148— equation, 139, 140, 142, 145, 148— response, 139

De-Hoop, A.T., 154Delta function, 7, 8, 30, 34–36, 45, 73, 107,

113, 115, 116, 142, 214, 241, 242, 251Denominator, 142, 144, 167, 171, 215, 237Density, 77, 122, 123, 125, 127, 145, 155, 162,

213, 242Dilatational wave, 78, 86, 166–169, 171, 173,

177, 184, 188, 190–192Dirac’s delta function, 6–8, 30, 74, 110, 139,

179, 223, 243Dispersive, 141Displacement

— equation, 78–80, 87, 89, 90, 96, 97, 107,116, 156, 163, 164, 180, 181, 206, 207,213, 214, 223, 224, 227, 231, 244

Distribution, 6–8Doppler effects, 121Double transform, 34, 52, 140, 224, 226, 245Dyadic

Green’s —, 77, 79, 86–90, 95, 105, 106,108, 116, 177, 202, 205, 206, 212, 213,220

EEigenvalues, 142, 207, 215Elastic

— beam, 139— moduli, 206, 213— modulus, 207— solid, 77, 79, 109, 154, 162, 163, 178,

179, 205, 213, 222, 228, 242–244Elasticity equations, 77, 78Elastodynamic

— equations, 154— problem, 154, 155, 180, 228, 2423D —, 180

Equation of motion, 155, 180, 222, 223, 243,244

Equation of state, 122, 123, 125Equilibrium equation

2D —, 213Erd’elyi, A., 9, 28, 29, 37, 50, 51, 58, 70, 72,

83–85, 104, 112, 115, 140, 141, 147,210, 211, 219, 225, 226, 247, 250

Ewing, W.M., 11

FFixed condition, 243Fixed coordinate system, 232

Fourier— cosine series, 3— cosine transform, 5, 62— inversion, 4, 8, 33, 36, 39, 40, 42–44,

46–48, 50, 53, 58, 59, 62, 63, 65, 69,71, 81, 82, 84, 85, 90–92, 94, 100, 101,103, 104, 111, 114, 131, 141, 142, 146,148, 149, 153, 156, 157, 161, 166, 170,174, 175, 189, 200, 208, 209, 211, 215,226, 230, 237, 246

— sine series, 3— sine transform, 4, 5

Fourier transformcomplex —, 3double —, 45, 49, 52, 79, 90, 130, 145,

148, 180, 206, 214, 229finite —, 3, 30triple —, 68, 71, 97

Fredholm integral equation, 253Fresnel integral, 141Fundamental functions, 80, 86Fung, Y.C., 154, 163

GGakenheimer, D.C., 197Gradshteyn, I.S., 28, 29, 58, 62, 91, 94Graff, K.F., 154Green’s dyadic, 77, 79, 86–90, 95, 105, 106,

108, 116, 177, 202, 205, 206, 212–220Green’s function

2D —, 63, 743D —, 68, 70, 73, 74, 154

HHankel function, 58, 59, 61, 63, 65, 88, 89,

136, 149, 150, 152, 253, 255–259Hankel transform, 5, 29, 224Heaviside’s (unit), 6, 37, 86, 113, 161, 179,

249Heisenberg’s delta function, 6, 8, 10, 203Helmholtz equation, 52Hooke’s law, 77, 78, 155, 164, 213, 243Hydro-static pressure, 122Hyperbola

semi —, 15, 16, 158, 168Hyperbolic curves

semi —, 15, 16, 158

IIchimatsu, 29Impulsive

— body force, 79, 96, 110— point load, 139, 145, 162— (point) source, 40

262 Index

— response, 88, 136, 1452D response, 49

Incident wave, 228, 229, 237, 242Inclusion, 205, 242, 243Incoming wave, 44Inhomogeneity, 213, 221, 242Inhomogeneous (isotropic elastic) solid, 205,

213Inhomogeneous media, 221Initial condition, 145In-plane

— deformation, 79, 90, 162, 205— displacement, 79

Isotropic, 178, 212

JJardetzky, W.S., 11Jordan’s lemma, 42, 43, 208, 216

KKelvin’s solution

2D —, 95, 221Kernel, 1, 34, 253, 260Kronecker’s delta, 86

LLamb’s problem, 163, 178Lame’s constants, 77, 163, 213Laplace

— equation, 44, 48, 68, 70— inversion, 2, 153— transform, 2, 5

Laplacian— operator, 128

Logarithmic singularity, 149, 212

MMach numbers, 129, 231Magnus, 29Mal, A.K., 253Miklowitz, J., 154Monoclinic, 222, 227Moriguchi, 29Moving

— boundary, 205, 227— coordinate, 231, 232— edge, 232

Mow, C.-C., 242

NNavier-Stokes equations, 122Navier equations, 78Newtonian fluid, 121

OOrthotropic, 205, 212Out-going wave, 44, 63

PPao, Y.-H., 242Parametric integral, 160, 235Particle velocity, 122Plane strain, 162Plate, 145, 148Poisson ratio, 77, 78, 109, 213Poles, 40, 218, 257Press, F., 11Pulse, 7

QQuiescent condition, 33, 35, 70, 96, 140, 179,

228, 245

RRadiation condition, 157, 159, 163, 228, 245Radiation wave, 40, 44Rayleigh

— equation/function, 167— poles, 186— wave, 167

Reference— length, 129, 213— state, 122, 123, 126, 130

Reflected wave, 228, 231, 232, 237, 241Reflection, 121, 205, 227Refraction, 121Residue, 42, 142, 238, 239Reynolds number, 129Riemann sheets, 13, 14Rigid motion, 95Ring source, 110, 227Ryzhik, I.M., 28

SSaddle point, 159, 168, 173, 190, 192, 238Scattering, 205, 242SH-wave, 161, 184, 202, 227, 228, 243Shear rigidity, 157Shear wave, 86, 96, 155, 165, 173, 242Sign function, 47, 85, 212Sine integrals, 147Specific heats, 122Static hydro-pressure, 123Step functions, 1, 6, 8, 10, 74, 135, 177, 252Strain, 77, 78, 95, 211Stress, 77, 122, 181, 211, 230, 243Stress-free, 228, 232

Index 263

Subsonic, 132SV-wave, 184, 189, 197

TTime-harmonic

— function, 88— Green’s function, 33— (point) load, 148— response, 44— solution, 89— source, 38, 51, 73, 87— vibration, 18, 38— wave, 135, 151

Titchmarsh, E.C., 29Torque

point —, 109, 110— source, 113

Torsion problem, 109Transient response

2D —, 162Triple integral transform, 49, 80, 111, 145, 229

UUdagawa, K., 29Uniform flow, 129, 135

VVelocity

— gradients, 127— potential, 127, 128— ratio, 78, 109

Viscosity, 122, 126Viscous fluid, 121, 127von Schmidt wave, 177, 178

WWater waves, 121Watson, G.N., 29Wave equation

2D —, 49, 513D —, 70, 98

Wave fronts, 107, 135, 197, 198, 241, 253Wave nature, 40, 141, 197Wave reflection, 205, 227Wave source, 33, 79, 129, 243, 245

YYoung’s moduli, 205Young’s modulus, 77

264 Index

integral transform techniques for green's function

Documents