Stroh Formalism and Rayleigh Waves

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Kazumi Tanuma

Stroh Formalism and Rayleigh Waves

Previously published in the Journal of Elasticity Volume 89, Issues 1Y3, 2007

Kazumi Tanuma Department of Mathematics Graduate School of Engineering Gunma University Kiryu 376-8515, Japan

Library of Congress Control Number: 2007941543

ISBN-978-1-4020-6388-6

e-ISBN-978-1-4020-6389-3

Printed on acid-free paper. * 2007 Springer Science + Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science + Business Media, LLC., 223 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 987654321 springer.com

Journal of Elasticity Volume 89 I Numbers 1–3 I December 2007

Contents Foreword / Roger Fosdick

1

Preface / Kazumi Tanuma Stroh Formalism and Rayleigh Waves

3 5

Abstract 1 The 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

1.9 1.10 1.11

5

Stroh Formalism for Static Elasticity 6 Basic Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Stroh’s Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Rotational Invariance of Stroh Eigenvector in Reference Plane . . . . . . . 14 Forms of Basic Solutions When Stroh’s Eigenvalue Problem is Degenerate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Rotational Dependence When Stroh’s Eigenvalue Problem is Degenerate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Angular Average of Stroh’s Eigenvalue Problem: Integral Formalism. . . . 25 Surface Impedance Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.8.1 Isotropic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31 1.8.2 Transversely Isotropic Media. . . . . . . . . . . . . . . . . . . . . . . . . . . .35 Justification of the Solutions in the Stroh Formalism . . . . . . . . . . . . . . . 44 Comments and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2 Applications in Static Elasticity 59 2.1 Fundamental Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2.1.1 Fundamental Solution in the Stroh Formalism . . . . . . . . . . . . . . . .59 2.1.2 Formulas for Fundamental Solutions: Examples. . . . . . . . . . . . . . .60 2.2 Piezoelectricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 2.2.1 Basic Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .63 2.2.2 Extension of the Stroh Formalism . . . . . . . . . . . . . . . . . . . . . . . .65 2.2.3 Surface Impedance Tensor of Piezoelectricity . . . . . . . . . . . . . . . .70 2.2.4 Formula for Surface Impedance Tensor of Piezoelectricity: Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71 2.3 Inverse Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 2.3.1 Dirichlet to Neumann map . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74 2.3.2 Reconstruction of Elasticity Tensor . . . . . . . . . . . . . . . . . . . . . . .76 2.3.2.1 Reconstruction of Surface Impedance Tensor from Localized Dirichlet to Neumann Map . . . . . . . . . . . . . . . 76 2.3.2.2 Reconstruction of Elasticity Tensor from Surface Impedance Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

2.4 Comments and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 3 Rayleigh Waves in the Stroh Formalism 95 3.1 The Stroh Formalism for Dynamic Elasticity . . . . . . . . . . . . . . . . . . . . 95 3.2 Basic Theorems and Integral Formalism. . . . . . . . . . . . . . . . . . . . . . . 100 3.3 Rayleigh Waves in Elastic Half-space . . . . . . . . . . . . . . . . . . . . . . . . 105 3.4 Rayleigh Waves in Isotropic Elasticity . . . . . . . . . . . . . . . . . . . . . . . . 111 3.5 Rayleigh Waves in Weakly Anisotropic Elastic Media . . . . . . . . . . . . . 117 3.6 Rayleigh Waves in Anisotropic Elasticity . . . . . . . . . . . . . . . . . . . . . . 127 3.6.1 Limiting Wave Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 3.6.2 Existence Criterion Based on S3 . . . . . . . . . . . . . . . . . . . . . . . . 134 3.6.3 Existence Criterion Based on Z . . . . . . . . . . . . . . . . . . . . . . . . . 141 3.6.4 Existence Criterion Based on Slowness Sections . . . . . . . . . . . . . 145 3.7 Comments and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 3.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 References

151

Index

155

Foreword Roger Fosdick

The Journal of Elasticity: The Physical and Mathematical Science of Solids invites expository articles from time-to-time in order to collect results in areas of research that have made significant impact on the field of solid mechanics and that show continued interest in present-day research and thinking. The Stroh formalism is one such area, especially as it impacts upon the analysis of anisotropic media, that has been generalized from linear elastostatics to cover linear theories of piezoelectro-elasticity and megneto-elasticity as well as elastic wave phenomena. In this work, Kazumi Tanuma presents the essential elements of the Stroh formalism, its major theorems with proofs and applications, in a research-level textbook form. The presentation is self-contained and the development is clear and concise. Professor Tanuma’s efforts to produce a straightforward, accurate and readable account of this subject are clearly evident. He has organized the subject with a common logical thread throughout and he has given basis for not only the importance of this area of research, but, most importantly, for understanding its meaning and significance.

R. Fosdick (*) Department of Aerospace Engineering and Mechanics, University of Minnesota, 110 Union St SE, Minneapolis, MN 55455, USA e-mail: [email protected] Foreword. R. Fosdick (ed.) doi: 10.1007/978-1-4020-6388-6_1, # Springer Science + Business Media B.V. 2007

Preface Kazumi Tanuma

The Stroh formalism is a powerful and elegant mathematical method developed for the analysis of the equations of anisotropic elasticity. The purpose of this exposition is to introduce the essence of this formalism and demonstrate its effectiveness in both static and dynamic elasticity. The Stroh formalism is useful for constructing the solutions to the equations of anisotropic elasticity and for investigating the properties which these solutions possess. The equations of elasticity are complicated, because they constitute a system and, particularly for the anisotropic cases, inherit many parameters from the elasticity tensor. The Stroh formalism reveals simple structures hidden in the equations of anisotropic elasticity and provides a systematic approach to these equations. This fact is not only interesting in itself from a mathematical point of view, but also is important in that the Stroh formalism helps us compute physical quantities explicitly. This exposition is divided into three chapters. Chapter 1 gives a succinct introduction to the Stroh formalism so that the reader could grasp the essentials as quickly as possible. A basic element of the Stroh formalism is Stroh’s eigenvalue problem, each of whose six-dimensional eigenvectors and generalized eigenvectors is composed of a three-dimensional displacement part and a three-dimensional traction part that pertain to two-dimensional deformations in three-dimensional elasticity. The most fundamental parts of this chapter are Section 1.3 and Section 1.5, where rotational invariance and rotational dependence of the Stroh eigenvectors and generalized eigenvectors in the reference plane are proved. On the basis of these results, the Barnett-Lothe integral formalism is developed and the notion of the surface impedance tensor is introduced. In Chapter 2 several important topics in static elasticity, which include fundamental solutions, piezoelectricity, and inverse boundary value problems, are studied on the basis of the Stroh formalism. The surface impedance tensor plays important

K. Tanuma (B) Department of Mathematics, Graduate School of Engineering, Gunma University, Kiryu 376-8515, Japan e-mail: [email protected]

Preface. K.Tanuma doi: 10.1007/978-1-4020-6388-6_2, # Springer Science + Business Media B.V. 2007

4

K. Tanuma

roles in these studies. The Stroh formalism was developed for two-dimensional deformations in three-dimensional elasticity. But the formalism can also be useful in full three-dimensional problems, as we shall see in this chapter. Chapter 3 is devoted to Rayleigh waves, for long a topic of utmost importance in nondestructive evaluation, seismology, and materials science. There we discuss existence, uniqueness, phase velocity, polarization, and perturbation of the phase velocity of Rayleigh waves through the Stroh formalism. We put emphasis on the usefulness of S3 (v), which is one of the three real matrices introduced in the BarnettLothe integral formalism. For instance, the phase velocity of Rayleigh waves can be analyzed by examining only one diagonal component of S3 (v). Chapter 1 is the foundation for both Chapter 2 and Chapter 3. Particularly, quite a number of results in the Stroh formalism for static elasticity apply in a parallel way to dynamic elasticity. Hence even readers who are primarily interested in applications of the Stroh formalism to Rayleigh waves are urged to read Chapter 1 before they proceed to Chapter 3. We give the subjects described above as self-contained a treatment as possible, although we do refer the reader to the literature for the long proofs of several lemmas. We also present practical examples of algebraic calculations as clearly as possible, which will make the Stroh formalism more understandable and attractive. The first draft of several basic sections of this exposition was written as lecture notes for a series of talks I gave, on the invitation of Chi-Sing Man, to graduate students at the University of Kentucky in the spring of 2001. Since then, I have been able to add many subjects, which include joint works with Chi-Sing Man, Gen Nakamura and Masayuki Akamatsu. I owe most to Chi-Sing Man, whose kind suggestion two years ago prompted me to write this exposition. He carefully read the complete manuscript, corrected sentences, and gave me many valuable and important comments and criticism, which not only highly improved this exposition but also let myself learn much on elasticity. Without his help I could not have completed this exposition. I am indebted to Roger Fosdick, who gave me valuable suggestions, constant support, and the chance for this exposition to be published. I would also like to express my thanks to Gen Nakamura, whose valuable guidance has led me to study the equations of elasticity and to learn the Stroh formalism. My thanks go also to Takao Kakita, my adviser, who introduced me to the study on partial differential equations when I was a graduate student. He always gave me valuable suggestions and encouragement. I would also like to thank Yuusuke Iso, who has always shown much interest in my work and has given many helpful suggestions. I am also grateful to my friends and colleagues Masayuki Akamatsu, Ryuichi Ashino, Hiroya Ito, and Mishio Kawashita for valuable discussions with them and helpful advice from them. Lastly I thank Nathalie Jacobs, Anneke Pot, Fiona Routley and Claudia Magpantay of Springer for their kind correspondences with me these two years. This work was partly supported by Grant-in-Aid for Scientific Research (C) (Nos. 16540095 & 19540113), Society for the Promotion of Science, Japan. Kiryu, Japan, March 2007

Stroh Formalism and Rayleigh Waves Kazumi Tanuma

Abstract The Stroh formalism is a powerful and elegant mathematical method developed for the analysis of the equations of anisotropic elasticity. The purpose of this exposition is to introduce the essence of this formalism and demonstrate its effectiveness in both static and dynamic elasticity. The equations of elasticity are complicated, because they constitute a system and, particularly for the anisotropic cases, inherit many parameters from the elasticity tensor. The Stroh formalism reveals simple structures hidden in the equations of anisotropic elasticity and provides a systematic approach to these equations. This exposition is divided into three chapters. Chapter 1 gives a succinct introduction to the Stroh formalism so that the reader could grasp the essentials as quickly as possible. In Chapter 2 several important topics in static elasticity, which include fundamental solutions, piezoelectricity, and inverse boundary value problems, are studied on the basis of the Stroh formalism. Chapter 3 is devoted to Rayleigh waves, for long a topic of utmost importance in nondestructive evaluation, seismology, and materials science. There we discuss existence, uniqueness, phase velocity, polarization, and perturbation of Rayleigh waves through the Stroh formalism. Keywords Anisotropic elasticity · Rayleigh waves · The Stroh formalism · Equations of elasticity · Inverse problems Mathematics Subject Classifications (2000) 74B05 · 74E10 · 74G75 · 74J15

The Table of Contents and Index are also provided as Electronic Supplementary Material for online readers at doi:10.1007/s10659-007-9117-1 K. Tanuma (B) Department of Mathematics, Graduate School of Engineering, Gunma University, Kiryu 376-8515, Japan e-mail: [email protected]

Stroh Formalism and Rayleigh Waves. K.Tanuma doi: 10.1007/978-1-4020-6388-6_3, # Springer Science + Business Media B.V. 2007

6

K. Tanuma

1 The Stroh Formalism for Static Elasticity 1.1 Basic Elasticity Let a Cartesian coordinate system be chosen in space, and let R denote the set of real numbers. Suppose that a region B in R3 with a smooth boundary is occupied by an elastic medium. Let S be a regular oriented surface in the region B and let n = (n1 , n2 , n3 ) be the unit normal to S at x ∈ B. positive The stress tensor σ = σij i, j=1,2,3 at x is defined so that the vector ⎞ ⎛ 3 ⎝ σij n j⎠ j=1

(1.1)

i↓1,2,3

represents the traction or force per unit area at x exerted by the portion of B on the side of S toward which n points on the portion of B on the other side. Let ε = ε(u) = εij i, j=1,2,3 be the infinitesimal strain tensor 1 εij = 2

∂u j ∂ui + , ∂xj ∂ xi

(1.2)

where u = u(x) = (u1 , u2 , u3 ) is the displacement at the place x and (x1 , x2 , x3 ) are the Cartesian coordinates of x. In linearlized elasticity, the stresses σij are related to the strains εij by the generalized Hooke’s law σij =

3

Cijkl εkl ,

i, j = 1, 2, 3.

(1.3)

k,l=1

Here C = C(x) = Cijkl i, j,k,l=1,2,3 is the elasticity tensor. Since the law of balance of angular momentum implies that σij = σ ji and (1.2) implies that εij = ε ji , we may assume that C has the minor symmetries Cijkl = C jikl ,

Cijkl = Cijlk ,

i, j, k, l = 1, 2, 3.

(1.4)

Then (1.3) can be written as σij =

3

Cijkl

k,l=1

∂uk , ∂ xl

i, j = 1, 2, 3.

We also assume the existence of a stored energy function W = W(ε) such that W(0) = 0,

∂ W = σij, ∂εij

(1.5)

Stroh formalism and Rayleigh waves

7

which, combined with (1.3), implies that ∂ ∂εkl

∂ W ∂εij

= Cijkl ,

∂ ∂εij

∂ W ∂εkl

= Cklij.

Hence C has the major symmetry Cijkl = Cklij,

i, j, k, l = 1, 2, 3.

(1.6)

We assume that the stored energy function is positive for any non-zero strain W(ε) > 0 for ε = 0, which implies that C is positive definite or it satisfies the following strong convexity condition: 3

Cijkl εij εkl > 0 for any non-zero 3 × 3 real symmetric matrix (εij).

(1.7)

i, j,k,l=1

In Chapter 1 and Chapter 2 we study static deformations of the elastic medium in B with no body force. Then the displacement u is independent of time and the equations of equilibrium are written as 3 ∂ σij = 0, ∂ xj j=1

i = 1, 2, 3.

(1.8)

By (1.5), the equations of equilibrium can be expressed in terms of the displacement u as 3 j,k,l=1

∂ ∂uk Cijkl = 0, ∂xj ∂ xl

i = 1, 2, 3.

(1.9)

The elasticity tensor C = C(x) = Cijkl i, j,k,l=1,2,3 appearing in (1.3) determines the elastic response of the material point in question. From the symmetries (1.4) and (1.6) it follows that C = C(x) has 21 independent components at each x. We say that the elastic material at x is isotropic if the elasticity tensor C(x) = Cijkl i, j,k,l=1,2,3 satisfies

Cijkl =

3 p,q,r,s=1

Qip Q jq Qkr Qls C pqrs ,

i, j, k, l = 1, 2, 3

(1.10)

8

K. Tanuma

for any orthogonal tensor Q = Qij i, j=1,2,3 .1 For isotropic materials, the components Cijkl (i, j, k, l = 1, 2, 3) are determined by two independent parameters and are given by the representation formula Cijkl = λ δijδkl + μ(δik δ jl + δil δkj),

i, j, k, l = 1, 2, 3,

(1.11)

where δij is Kronecker’s delta symbol and λ = λ(x), μ = μ(x) are scalars called the Lamé moduli. We say that the elastic material at x is anisotropic if it is not isotropic. The largest subgroup G of the orthogonal group such that (1.10) holds for any proper Q = Qij i, j=1,2,3 in G is called the symmetry group of the anisotropic material at x. According to the generators of the subgroup G , the number of independent components in C(x) = Cijkl i, j,k,l=1,2,3 varies and anisotropic elastic materials can be classified (see, for example, Sections 21 and 26 of [33]). To express the components Cijkl (i, j, k, l = 1, 2, 3) conveniently, we introduce the contracted notation called the Voigt notation: We rewrite Cijkl i, j,k,l=1,2,3 using the 6 × 6 matrix as ⎡

C11 C12 C13 ⎢ C22 C23 ⎢ ⎢ C33 Cαβ = ⎢ ⎢ ⎢ ⎣ Sym.

C14 C24 C34 C44

C15 C25 C35 C45 C55

⎤ C16 C26 ⎥ ⎥ C36 ⎥ ⎥, C46 ⎥ ⎥ C56 ⎦ C66

(1.12)

where we have used the rules for replacing the subscript ij (or kl) by α (or β) as follows: ij (or kl) α or β 11 ←→ 1 22 ←→ 2 33 ←→ 3

ij (or kl) α or β 23 or 32 ←→ 4 31 or 13 ←→ 5 12 or 21 ←→ 6.

This replacement is possible because of the minor symmetries (1.4). The symmetry of the matrix (1.12) follows from the major symmetry (1.6). Note that the strong convexity condition (1.7) is equivalent to the assertion that the 6 × 6 matrix (1.12) is positive definite (Exercise 1-1). Now let Rφxi denote the orthogonal tensor corresponding to a right-handed rotation by the angle φ about the i-axis of the Cartesian coordinate system. We give here several typical examples of anisotropic elastic materials with the generators of

we write the generalized Hooke’s law (1.3) as σ = σ (ε), it can be proved that (1.10) is equivalent to

1 If

Q σ (ε) QT = σ ( Q ε QT ), which implies that the transformed σ under the orthogonal tensor Q is related to the transformed ε under the orthogonal tensor Q by the same Hooke’s law. Here σ and ε are interpreted as the 3 × 3 matrices whose (i, j ) components are σij and εij respectively, and the superscript T denotes transposition.


9

their symmetry groups and write the corresponding elasticity tensors in the Voigt notations by using their independent components. (1) Transversely isotropic materials. The generators of G are Rφx3 , 0 < φ < 2π ,2 and then ⎡ C11 C12 C13 0 0 0 ⎢ C C 0 0 0 11 13 ⎢ ⎢ C33 0 0 0 ⎢ Cαβ = ⎢ C 0 0 44 ⎢ ⎣ C44 0 C11 −C12 Sym. 2 (2) Cubic materials. π π π The generators of G are Rx21 , Rx22 , Rx23 , and then ⎡ C11 C12 C12 0 ⎢ C11 C12 0 ⎢ ⎢ C11 0 ⎢ Cαβ = ⎢ C44 ⎢ ⎣ Sym.

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

(1.13)

0 0 0 0 C44

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥. 0 ⎥ ⎥ 0 ⎦ C44

(1.14)

(3) Orthorhombic materials. The generators of G are Rπx2 , Rπx3 , and then ⎡ C11 C12 C13 ⎢ C22 C23 ⎢ ⎢ C33 Cαβ = ⎢ ⎢ ⎢ ⎣ Sym.

0 0 0 C44

0 0 0 0 C55

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥. 0 ⎥ ⎥ 0 ⎦ C66

(1.15)

(4) Monoclinic materials. The generator of G is Rπx3 , and then ⎡ C11 C12 C13 ⎢ C22 C23 ⎢ ⎢ C33 Cαβ = ⎢ ⎢ ⎢ ⎣ Sym.

0 0 0 C44

0 0 0 C45 C55

⎤ C16 C26 ⎥ ⎥ C36 ⎥ ⎥. 0 ⎥ ⎥ 0 ⎦ C66

(1.16)

Note that all these symmetries are, in principle, defined at each point x. When Rθxi belongs to the symmetry group G of the material at x, we say that the i-axis is a 2π -fold symmetry axis at x. For example, for cubic materials described θ above, the 3-axis is a 4-fold symmetry axis.

2 In

this case the 3-axis is called the axis of symmetry. For hexagonal materials, the generator of G π

are Rπx2 and Rx33 . It can be proved that hexagonal materials and transversely isotropic materials have elasticity tensors of the same form.

10

K. Tanuma

Generally, the elasticity tensor C depends on x. However, if C is independent of x, we say that the elastic body is homogeneous. In this case, (1.9) becomes 3

Cijkl

j,k,l=1

∂ 2 uk = 0, ∂ x j∂ xl

i = 1, 2, 3.

In the subsequent sections of this chapter we develop the Stroh formalism, where the elastic body in question is assumed to be homogeneous. 1.2 Stroh’s Eigenvalue Problem Let x = (x1 , x2 , x3 ) be the position vector and let m = (m1 , m2 , m3 ) and n = (n1 , n2 , n3 ) be orthogonal unit vectors in R3 . Let C denote the set of complex numbers. We consider the equations of equilibrium for the homogeneous elastic body in static deformations in the half-space n · x = n1 x1 + n2 x2 + n3 x3 ≤ 0;3 3 j,k,l=1

Cijkl

∂ 2 uk = 0, ∂ x j∂ xl

i = 1, 2, 3.

(1.17)

Here C = Cijkl i, j,k,l=1,2,3 is the elasticity tensor, which has the major symmetry (1.6) and the minor symmetries (1.4) and satisfies the strong convexity condition (1.7). We seek solutions to (1.17) of the form √ −1 ξ(m·x+ p n·x)

u = (u1 , u2 , u3 ) = a e−

∈ C3

(1.18)

in the half-space n · x ≤ 0, each of which satisfies the physical requirement that the amplitude decays with depth below the surface n · x = 0; here a ∈ C3 , p ∈ C, and ξ is a real parameter that is positive.4 A solution of the form (1.18) describes two-dimensional deformations in an elastic body; it depends on the projections of x on the plane spanned by the two orthogonal unit vectors m and n. Such a plane is called the reference plane. Let us determine a ∈ C3 and p ∈ C in (1.18). Substituting (1.18) into (1.17) and noting that √ √ ∂u = − −1(mj + pnj) ξ a e− −1 ξ(m·x+ p n·x) , ∂xj

this exposition, the dot ‘ · ’ denotes the inner product of two vectors in R3 . When denoting the inner product of two vectors in C3 , we use the notation ( , )C3 . 3 Throughout

positive parameter ξ carries the dimension of the reciprocal of length. The rationale for studying this special class of solutions will be given in Section 1.9. See also Remark 1.8.

4 The


11

we get ⎛ ⎝

3

⎞ Cijkl (mj + pnj)(ml + pnl )⎠

j,l=1

⎛ =⎝

3

⎛ Cijkl mjml + p ⎝

j,l=1

+ p2

3

⎞

j,l=1

T=⎝

3 j,l=1

3

Cijkl mjnl +

j,l=1

3

⎞ Cijkl njml ⎠

j,l=1

a = 0.

(1.19)

i↓k→1,2,3

We introduce 3 × 3 real matrices ⎞ ⎛ 3 Cijkl mjml ⎠ Q=⎝ ⎛

i↓k→1,2,3

Cijkl nj nl ⎠

j,l=1

a

⎞

⎛ ,

R=⎝

3

Cijkl mjnl ⎠

j,l=1

i↓k→1,2,3

Cijkl njnl ⎠

⎞ ,

i↓k→1,2,3

.

(1.20)

i↓k→1,2,3

Then it follows from the major symmetry (1.6) that ⎛ ⎞ 3 Cijkl njml ⎠ RT = ⎝ j,l=1

,

i↓k→1,2,3

and hence equation (1.19) can be written as Q + p(R + RT ) + p2 T a = 0,

(1.21)

where the superscript T denotes transposition. For the existence of a non-trivial vector a = 0, we observe that p satisfies the sextic equation det Q + p(R + RT ) + p2 T = 0. (1.22) For the matrices Q, T and the solutions p to (1.22), we have Lemma 1.1 (1) The matrices Q and T are symmetric and positive definite. (2) The characteristic roots pα (1 ≤ α ≤ 6), i.e., the solutions to the sextic equation (1.22), are not real and they occur in complex conjugate pairs. Proof That the matrices Q and T are symmetric follows from definition (1.20) and the major symmetry (1.6). For any non-zero v = (v1 , v2 , v3 )T ∈ R3 , put εij = mi vj + vi m j.

12

K. Tanuma

Then (εij) is a non-zero symmetric matrix. The strong convexity condition (1.7), combined with the minor symmetries (1.4), leads to ⎛ ⎞ 3 3 3 0< Cijkl εij εkl = 4 vi ⎝ Cijkl m jml ⎠ vk = 4(v · Q v), i, j,k,l=1

i,k=1

j,l=1

which implies that Q is positive definite. In the same way, it is proved that T is positive definite. Next we prove (2). The solutions pα (1 ≤ α ≤ 6) to (1.22) are not real if and only if the matrix ⎛ ⎞ 3 ⎝ Cijkl (mj + pnj)(ml + pnl )⎠ is invertible for all p ∈ R. (1.23) j,l=1

i↓k→1,2,3

Since the vector w = (w1 , w2 , w3 ) =

m + pn |m + pn|

( p ∈ R)

moves on the unit circle, it follows from (1) of the lemma that the symmetric matrix ⎞ ⎛ 3 ⎝ Cijkl wjwl ⎠ j,l=1

i↓k→1,2,3

is positive definite, which implies (1.23). Since all the coefficients of the sextic equation (1.22) are real, its roots occur in complex conjugate pairs.

Next we give the traction on the surface n · x = 0 produced by the displacement (1.18). Since the unit outward normal of this surface is the vector n, the traction t on n · x = 0 is given by ⎛ ⎞ 3 t= ⎝ σij n j⎠ j=1 i↓1,2,3 n·x=0

which, by (1.5), becomes

⎛

t= ⎝

3 j,k,l=1

Cijkl

⎞ ∂uk ⎠ nj ∂ xl

.

(1.24)

i↓1,2,3 n·x=0

It follows from (1.18) and (1.20) that ⎞ ⎛ 3 √ Cijkl (ml + pnl ) nj⎠ t = − −1 ξ ⎝ j,l=1 √ √ = − −1 ξ RT + pT a e− −1 ξ m·x .

√ −1 ξ m·x

a e−

i↓k→1,2,3

Hence we define a vector l ∈ C3 as

l = RT + pT a.

(1.25)


13

Then √ √ t = − −1 ξ l e− −1 ξ m·x

(1.26)

is the traction on the surface n · x = 0. Finally we shall see that the relations (1.21) and (1.25) can be recast as a sixdimensional eigenvalue problem, called Stroh’s eigenvalue problem. By (1) of Lemma 1.1, T−1 exists. Hence from (1.25) we get p a = −T−1 RT a + T−1 l

(1.27)

and p l = p RT + p2 T a. The last equation becomes, by (1.21), p l = −[Q + p R] a, and from (1.27) p l = −Q a − R −T−1 RT a + T−1 l = −Q + RT−1 RT a − RT−1 l.

(1.28)

Thus, from (1.27) and (1.28) we obtain a be a column vector in C6 whose first three components consist l of a vector a ∈ C3 that satisfies (1.21) and whose last three components consist of the vector l ∈ C3 given by (1.25). Then the following six-dimensional eigenrelation holds:

Theorem 1.2 Let

a a N =p , l l

(1.29)

where N is the 6 × 6 real matrix defined by N=

−T−1 RT

T−1

−Q + RT−1 RT − RT−1

.

(1.30)

a to (1.29) is called a Stroh eigenvector, where a and l are l its displacement part and traction part, respectively. Lemma 1.1 implies that the eigenvalues pα (1 ≤ α ≤ 6) of N are not real and they occur in complex conjugate pairs. We call the eigenvalue problem (1.29) Stroh’s eigenvalue problem. A solution

14

K. Tanuma

As we shall see, coupling the displacement part and the traction part together in (1.29) leads us to the finding of an elegant result on rotational dependency of these vectors in the reference plane, which could not be proved if we restrict ourselves to consideration of displacements (i.e., the solutions to (1.17)).

1.3 Rotational Invariance of Stroh Eigenvector in Reference Plane The displacement part a ∈ C3 and the traction part l ∈ C3 of a Stroh eigenvector are determined by (1.21) and (1.25). These vectors a and l (and also the characteristic roots p which are the solutions to (1.22)) apparently depend on the two orthogonal unit vectors m and n. In this section we show that a and l do not depend on m and n individually but depend on the vector product m × n. This means that a and l are invariant under the rotations of m and n around m × n, and hence are rotationally invariant in the reference plane. This property is fundamental in the Stroh formalism and is also elegant in itself. We formulate the above more precisely. Let us fix two orthogonal unit vectors e1 and e2 in R3 and take m and n to be those which are obtained by rotating e1 and e2 around e1 × e2 by an angle φ (0 ≤ φ < 2π ) so that m = m(φ) = e1 cos φ + e2 sin φ,

n = n(φ) = −e1 sin φ + e2 cos φ.

(1.31)

Then Q, R, T in (1.20), and hence p in (1.22), can be regarded as functions of φ Q = Q(φ),

R = R(φ),

T = T(φ),

p = p(φ)

(1.32)

and by (1.30), N = N(φ) =

−T(φ)−1 R(φ)T

T(φ)−1

−Q(φ) + R(φ)T(φ)−1 R(φ)T

− R(φ)T(φ)−1

.

(1.33)

Theorem 1.3 Let a and l satisfy (1.21) and (1.25) at φ = 0 with p = p1 being a solution to (1.22) at φ = 0. Then the same a and l satisfy (1.21) and (1.25) for all φ, while p = p(φ) is the solution to the Riccati equation d p = −1 − p2 dφ with p(0) = p1 .

Proof We denote the differentiation

d dφ

m = −e1 sin φ + e2 cos φ = n,

by . From (1.31) we get n = −e1 cos φ − e2 sin φ = −m.

(1.34)


15

Then from (1.20) it follows that ⎞ ⎛ 3 Cijkl mjml ⎠ Q = ⎝ j,l=1

⎛ =⎝

3

i↓k→1,2,3

Cijkl njml +

j,l=1

⎛ R = ⎝

3

=⎝

3

⎞

Cijkl mjnl ⎠

j,l=1

= R + RT ,

i↓k→1,2,3

⎞ Cijkl mjnl ⎠

j,l=1

⎛

3

i↓k→1,2,3

Cijkl njnl −

j,l=1

3

⎞

Cijkl mjml ⎠

j,l=1

=T−Q

i↓k→1,2,3

T

= (R ) , ⎛ ⎞ 3 Cijkl njnl ⎠ T =⎝ j,l=1

⎛ = −⎝

i↓k→1,2,3

3

Cijkl mjnl +

j,l=1

3

⎞ Cijkl njml ⎠

j,l=1

= − R + RT .

(1.35)

i↓k→1,2,3

Put H(φ) = Q + p R + RT + p2 T (φ). From (1.35) we have H (φ) = R + RT + p R + RT + 2 p (T − Q) + 2 p p T − p2 R + RT (φ) = −2 p Q + 1 + p − p2 R + RT + 2 p 1 + p T (φ). Suppose that p(φ) satisfies the Riccati equation (1.34). Then the equation above becomes (1.36) H (φ) = −2 p Q + p(R + RT ) + p2 T (φ) = −2 p(φ) H(φ). Now put h(φ) = Q + p(R + RT ) + p2 T (φ) a = H(φ) a, where a satisfies (1.21) at φ = 0. Then we get h(0) = 0.

(1.37)

h (φ) = −2 p(φ) H(φ) a = −2 p(φ) h(φ).

(1.38)

It follows from (1.36) that

16

K. Tanuma

From (1.37) and the uniqueness of the solution to the initial value problem for the ordinary differential equation (1.38) we obtain h(φ) = Q + p R + RT + p2 T (φ) a = 0 (1.39) for all φ. This implies that a is the solution to (1.21) for all φ. Now put η(φ) = RT + p T (φ) a. Then, by the assumption of the theorem, we have η(0) = l. It follows from (1.34) and (1.35) that η (φ) = T − Q + p T − p R + RT (φ) a = − Q + p R + RT + p2 T (φ) a = −h(φ),

(1.40)

which is equal to zero by (1.39). Hence we obtain η(φ) = RT + p T (φ) a = η(0) = l

(1.41)

for all φ. This implies that l satisfies (1.25) for all φ.

Remark 1.4 The Riccati equation (1.34) can be solved explicitly as p(φ) = tan(φc − φ), where φc is a non-real constant determined by the initial condition p(0) = p1 = tan φc , or √ p(φ) ≡ ± −1. In the proof of Theorem 1.3, we have assumed that p(φ) is the solution to the Riccati equation (1.34). However, it follows that p(φ) must be the solution to (1.34). In fact, (1.39) and (1.41) are to the eigenvalue problem (1.29) with N = equivalent a 6 N(φ) given by (1.33). Then ∈ C is an eigenvector of N(φ) corresponding to the l eigenvalue p(φ). Therefore, p(φ) is unique for each φ and satisfies (1.34). Thus, we can rewrite Theorem 1.3 in terms of Stroh’s eigenvalue problem (1.29) as follows: a Theorem 1.5 Let be an eigenvector of N(0) associated with the eigenvalue l p = p1 . Then it holds that a a N(φ) = p(φ) (1.42) l l for all φ, while the eigenvalue p(φ) of N(φ) satisfies the Riccati equation (1.34) with p(0) = p1 .


17

1.4 Forms of Basic Solutions When Stroh’s Eigenvalue Problem is Degenerate In Section 1.2, considering solutions of the form (1.18) to the equations of equilibrium (1.17) has led us to the six-dimensional Stroh’s eigenvalue problem (1.29). More precisely, the vector a in (1.18) becomes the first three components of an eigenvector of N (1.30), while the vector l in (1.26) becomes the last three components of that eigenvector. On the other hand, when the six-dimensional eigenvalue problem (1.29) does not have six linearly independent eigenvectors, the notions of generalized eigenvectors and Jordan chain will be needed. Then we have to consider solutions of a different form than (1.18). Hence in this section we give the general forms of basic solutions to (1.17) according to the degeneracy of the eigenvalue problem (1.29). But here we restrict ourselves to giving a formal argument for constructing such forms of basic solutions. Later in Section 1.9, these forms will be justified by the theory of Fourier analysis (see also Remark 1.8). Recall that the eigenvalues pα (1 ≤ α ≤ 6) of the six-dimensional eigenvalue problem (1.29) are not real and they occur in complex conjugate pairs (see the paragraph after Theorem 1.2). Henceforth we take Im pα > 0,

α = 1, 2, 3,

where Im p denotes the imaginary part of p, and put pα+3 = pα

(α = 1, 2, 3),

(1.43)

where p denotes the complex conjugate of p. Let m and n be orthogonal unit vectors in R3 . We say that the eigenvalue problem (1.29) is simple if all the six eigenvalues pα (1 ≤ α ≤ 6) are distinct. Then there aα (1 ≤ α ≤ 6) which span C6 . Taking exist six linearly independent eigenvectors lα account of (1.43), we may put

aα+3 lα+3

a = α lα

(α = 1, 2, 3).

(1.44)

In this case, by (1.18) and (1.26), the general form of the basic solution to (1.17) 5 deformations in the half-space n · x ≤ 0, satisfies which describes two-dimensional √ − −1 ξ m·x on n · x = 0 for some complex vector a, and decays to the condition u = a e zero as n · x −→ −∞ is written as u=

3

√ −1 ξ(m·x+ pα n·x)

c α aα e −

,

α=1

5 For

the meaning of “two-dimensional”, see the third paragraph of Section 1.2.

(1.45)

18

K. Tanuma

and the corresponding traction on the surface n · x = 0 is given by 3 √ √ t = − −1 ξ cα lα e− −1 ξ m·x ,

(1.46)

α=1

where cα (1 ≤ α ≤ 3) are complex constants determined from 3α=1 cα aα = a, and ξ > 0. We say that the eigenvalue problem (1.29) is semisimple if there are multiple eigenvalues in the six eigenvalues pα (1 ≤ α ≤ 6) but there remain six linearly aα (1 ≤ α ≤ 6) which span C6 . We also use the conindependent eigenvectors lα ventions (1.43) and (1.44). In this case the general form of the basic solution and the corresponding traction on the surface n · x = 0 are again given by (1.45) and (1.46), respectively. We say that the eigenvalue problem (1.29) is degenerate or non-semisimple if there exist multiple eigenvalues in the six eigenvalues pα and there do not exist six linearly independent eigenvectors any more. The last case is classified into the two cases D1 and D2 according to the Jordan normal form of the 6 × 6 matrix N (1.30). (For the Jordan normal form of a matrix, see, for example, Sections 6.6, 6.9 and 7.7 of [29] and Section 1.4 of [77]). Case D1 There exists a non-singular 6 × 6 matrix P which transforms N to the following Jordan normal form ⎡ ⎢ ⎢ ⎢ P−1 N P = ⎢ ⎢ ⎢ ⎣

⎤

p1 p2 1 p2

⎥ ⎥ ⎥ ⎥, ⎥ p1 ⎥ p2 1 ⎦ p2

where the empty entries in the matrix by zero. De of the right hand side are filled a aα ∈ C6 (1 ≤ α ≤ 6), we see that α (α = 1, 2) noting the α-th column of P by lα lα a3 is a are linearly independent eigenvectors associated with pα (α = 1, 2), and l3 generalized eigenvector which satisfies N

a a a3 − p2 3 = 2 . l3 l3 l2

Likewise we use the convention (1.44). Then all six independent and span C6 .

aα lα

(1.47) (1 ≤ α ≤ 6) are linearly

Lemma 1.6 The general form of the basic solution to (1.17) in Case D1 which describes two-dimensional deformations in the half-space n · x ≤ 0, satisfies the condition


19

√

u = a e− −1 ξ m·x on n · x = 0 for some a ∈ C3 , and decays to zero as n · x −→ −∞ is written as u=

2

√ −1 ξ(m·x+ pα n·x)

c α aα e −

√ √ + c3 a3 − −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x) , (1.48)

α=1

and the corresponding traction on the surface n · x = 0 is given by 3 √ √ cα lα e− −1 ξ m·x , t = − −1 ξ

(1.49)

α=1

where cα (1 ≤ α ≤ 3) are complex constants determined from ξ > 0.

3

α=1 cα

aα = a, and

Proof Looking at the first three rows of the system (1.47), from (1.30) we obtain −T−1 RT a3 + T−1 l3 − p2 a3 = a2 , which gives l3 = RT + p2 T a3 + T a2 .

(1.50)

Looking at the last three rows, we get −Q + RT−1 RT a3 − RT−1 l3 − p2 l3 = l2 . Then substituting (1.50) into the preceding equation, we get Q + p2 R + RT + p22 T a3 + R + p2 T a2 + l2 = 0. Noting that

a2 l2

(1.51)

is an eigenvector of N, which satisfies l2 = RT + p2 T a2 ,

we obtain Q + p2 R + RT + p22 T a3 + R + RT + 2 p2 T a2 = 0.

(1.52)

Hence (1.47) is equivalent to (1.50) and (1.52). To see that (1.48) satisfies (1.17), √ we note that the first two terms 2α=1 cα aα e− −1 ξ(m·x+ pα n·x) in the right hand side of (1.48) observe (1.17), because aα (α = 1, 2) are the same as the displacement part of a Stroh eigenvector in Section 1.2. Hence it remains to prove that the last term √ √ a3 − −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x) satisfies (1.17). Putting √ √ u˜ = (u˜ 1 , u˜ 2 , u˜ 3 ) = a3 − −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x) ,

(1.53)

20

K. Tanuma

we obtain

3 ∂ 2 u˜ k Cijkl ∂ x j∂ xl i↓1,2,3 j,k,l=1 ⎛ ⎛ ⎞ 3 3 3 = − ξ2 ⎝ Cijkl mjml + p2 ⎝ Cijkl mjnl + Cijkl n jml ⎠ j,l=1

+ p22 ⎛ − ξ2 ⎝

3

⎞

j,l=1

Cijkl njnl ⎠

j,l=1

3

j,l=1

a3 −

√

√ −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x)

i↓k→1,2,3

⎞ Cijkl nj(ml + p2 nl )+nl (mj + p2 nj) ⎠

j,l=1

√ −1 ξ(m·x+ p2 n·x)

a2 e −

,

i↓k→1,2,3

and from (1.20) and (1.21) for a = a2 , √ = ξ 2 − Q + p2 R + RT + p22 T a3 − R + RT + 2 p2 T a2 e− −1 ξ(m·x+ p2 n·x) , which is equal to zero by (1.52). √ Now we will show (1.49). By (1.25) and (1.26), the first two terms − −1 ξ √ 2 − −1 ξ m·x in the right hand side of (1.49) are the traction on the surface α=1 cα lα e √ n · x = 0 produced by 2α=1 cα aα e− −1 ξ(m·x+ pα n·x) in (1.48). Hence we will compute the traction on n · x = 0 produced by (1.53). It follows that ⎛ ⎞ 3 ˜k ∂ u ⎝ Cijkl nj ⎠ ∂ xl j,k,l=1

⎛

i↓1,2,3

⎞ 3 √ = − −1 ξ ⎝ Cijkl (ml + p2 nl ) nj⎠ j,l=1

i↓k→1,2,3

√ √ × a3 − −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x) ⎛ ⎞ 3 √ √ − −1 ξ ⎝ Cijkl njnl ⎠ a2 e− −1 ξ(m·x+ p2 n·x) j,l=1

i↓k→1,2,3

√ √ √ = − −1 ξ RT + p2 T a3 − −1 ξ(n · x) a2 e− −1 ξ(m·x+ p2 n·x) −

√

√ −1 ξ(m·x+ p2 n·x)

−1 ξ T a2 e−

.

This is written on the surface n · x = 0 as √ √ − −1 ξ RT + p2 T a3 + T a2 e− −1 ξ m·x , which is equal to √ √ − −1 ξ l3 e− −1 ξ m·x

by (1.50).


21

Now we proceed to the Case D2. Case D2 There exists a non-singular 6 × 6 matrix P which transforms N to the following Jordan normal form ⎡ ⎤ p1 1 ⎢ ⎥ p1 1 ⎢ ⎥ ⎢ ⎥ p1 ⎥. P−1 N P = ⎢ ⎢ ⎥ p1 1 ⎢ ⎥ ⎣ p1 1 ⎦ p1

aα lα

a1 l1

∈ C (1 ≤ α ≤ 6), we see that is aα (α = 2, 3) are an eigenvector associated with the triple eigenvalue p1 , and lα generalized eigenvectors which satisfy Denoting the α-th column of P by

N

a2 a a − p1 2 = 1 , l2 l2 l1

6

N

a3 a a − p1 3 = 2 . l3 l3 l2

a As before, we use the convention (1.44). Then all six α lα independent and span C6 .6

(1.54)

(1 ≤ α ≤ 6) are linearly

Lemma 1.7 The general form of the basic solution to (1.17) in Case D2 which describes two-dimensional deformations in the half-space n · x ≤ 0, satisfies the condition √ u = a e− −1 ξ m·x on n · x = 0 for some a ∈ C3 , and decays to zero as n · x −→ −∞ is written as √ u = c1 a1 + c2 a2 − −1 ξ (n · x) a1 √ √ 1 2 2 e− −1 ξ(m·x+ p1 n·x) , + c3 a3 − −1 ξ (n · x) a2 − ξ (n · x) a1 2

(1.55)

say that an eigenvector v1 and generalized eigenvectors v2 , · · · , vk associated with the multiple eigenvalue λ of the n × n matrix M form a Jordan chain of length k (k ≤ n) if 6 We

Mv1 = λ v1 ,

Mvi − λvi = vi−1 ,

i = 2, 3, · · · , k.

The corresponding Jordan block is a k × k matrix with repeated eigenvalue λ on the diagonal, 1 on the first super diagonal and zero elsewhere. Note that although the eigenvalue problem (1.29) is sixdimensional, the length of the Jordan chains is at most three, because the eigenvalues are not real and occur complex conjugate pairs. in a aα (α = 2, 3) in Case D1 form a Jordan chain of length 2, while α (α = 1, 2, 3) in Case Thus lα lα D2 form a Jordan chain of length 3.

22

K. Tanuma

and the corresponding traction on the surface n · x = 0 is given by 3 √ √ cα lα e− −1 ξ m·x , t = − −1 ξ

(1.56)

α=1

where cα (1 ≤ α ≤ 3) are complex constants determined from ξ > 0.

3

α=1 cα

aα = a, and

This lemma can be proved in a similar way as that for Lemma 1.6. Namely, we write down from the first three rows and the last three rows of the systems (1.54) the four systems in C3 equivalent to (1.54). On the other hand, we substitute (1.55) into the left hand side of (1.17) and (1.24). Then we compare the resulting formulas. The details are left as Exercise 1-4. Remark 1.8 Although the forms of basic solutions (displacements) are different according to the degeneracy of the eigenvalue problem (1.29) (see (1.45), (1.48) and (1.55)), they have the same form on the surface n · x = 0, namely u=

3

√ −1 ξ m·x

c α aα e −

.

(1.57)

α=1

The corresponding tractions on n · x = 0 also assume the same form 3 √ √ t = − −1 ξ cα lα e− −1 ξ m·x α=1

(see (1.46), (1.49) and (1.56)). Looking back on the arguments in this section, we observe that formulas (1.45), (1.48) and (1.55) give, for two-dimensional deformations in the half-space n · x ≤ 0, the forms of the solution for the elementary boundary condition (1.57) on n · x = 0 according to the degeneracy of the eigenvalue problem (1.29). In Section 1.9 we shall see that considering boundary condition (1.57) suffices for the boundary-value problem in the half-space n · x ≤ 0 and the above forms of the basic solution will be justified by the theory of Fourier analysis. 3 From now on, we call the vectors aα ∈ C (α = 1, 2, 3) the displacement parts aα aα 3 6 ∈ C , not only when is an and lα ∈ C (α = 1, 2, 3) the traction parts of lα lα eigenvector of N corresponding to an eigenvalue with positive imaginary part, but aα also when is a generalized eigenvector of N corresponding to an eigenvalue lα with positive imaginary part.

1.5 Rotational Dependence When Stroh’s Eigenvalue Problem is Degenerate In Section 1.3 we have proved that eigenvectors of N are invariant under the rotations of m and n around the vector product m × n (Theorem 1.5). In this section we consider dependence of generalized eigenvectors of N on the rotations of m and n when Stroh’s eigenvalue problem (1.29) is degenerate. As we shall see, generalized


23

eigenvectors do not enjoy such elegant invariance as in Section 1.3, but there is a rule that describes their dependence on the rotations of m and n. As in Section 1.3, we consider here rotations given by (1.31) with rotation angle φ and use notations in (1.32) and (1.33). Corresponding to the Case D1 in Section 1.4, we have

a Theorem 1.9 Let α lα

(α = 1, 2) be linearly independent eigenvectors of N(0) asso a3 ciated with the eigenvalues pα (α = 1, 2), and let be a generalized eigenvector l3 satisfying a a a3 − p2 3 = 2 . (1.58) N(0) l3 l3 l2 Then it follows that

a N(φ) α lα

aα = pα (φ) , lα

α = 1, 2

(1.59)

φ

a3 a3 (φ) = exp 2 p2 (ψ) dψ , l3 (φ) l3 0

(1.60)

a a3 (φ) a3 (φ) − p2 (φ) = 2 N(φ) l3 (φ) l3 (φ) l2

(1.61)

for all φ, and by putting

it follows that

for all φ, where pα (φ) (α = 1, 2) are the eigenvalues of N(φ) and satisfy the Riccati equation (1.34) with pα (0) = pα (α = 1, 2). Proof Equations (1.59) are immediate consequences of Theorem 1.5. Now we will prove (1.61). Recalling that the eigenrelation (1.47) is equivalent to (1.50) and (1.51), we only have to show that (1.62) Q + p2 R + RT + p22 T (φ) a3 (φ) + R + p2 T (φ) a2 + l2 = 0 and

l3 (φ) − RT + p2 T (φ) a3 (φ) − T(φ) a2 = 0

(1.63)

for all φ. Put the left hand sides of (1.62) and (1.63) by g(φ) and ζ (φ), respectively. Then by (1.58), we get g(0) = 0 and ζ (0) = 0.

(1.64)

Since it follows from (1.60) that a3 (φ) = 2 p2 (φ) a3 (φ) and l3 (φ) = 2 p2 (φ) l3 (φ), from (1.36), (1.40), and (1.39) for a = a2 we see that g (φ) = 0,

(1.65)

24

K. Tanuma

which, when combined with (1.64), implies that g(φ) = 0 for all φ. In the same way, from (1.35), (1.40) and (1.65) we obtain ζ (φ) = 2 p2 (φ) l3 (φ) + Q + p2 (R + RT ) + p22 T (φ) a3 (φ) − 2 p2 (φ) RT + p2 T (φ) a3 (φ) + R + RT (φ) a2 and using (1.62) in the second term of the right hand side, we get ζ (φ) = 2 p2 (φ) l3 (φ) − RT + p2 T (φ) a3 (φ) + RT − p2 T (φ) a2 − l2 , and then, from (1.41) with a = a2 , l = l2 we have ζ (φ) = 2 p2 (φ) ζ (φ), which, when combined with (1.64), implies that ζ (φ) = 0

for all φ.

Corresponding to the Case D2, we have a1 be an eigenvector of N(0) associated with the triple eigenTheorem 1.10 Let l1 a value p1 , and let α (α = 2, 3) be generalized eigenvectors satisfying lα a2 a1 a3 a a a2 − p1 = , N(0) − p1 3 = 2 . N(0) l2 l2 l1 l3 l3 l2 Then it follows that

N(φ)

a1 l1

= p1 (φ)

a1 l1

for all φ, and by putting φ

a2 (φ) = exp 2 p1 (ψ) dψ l2 (φ) 0 ψ

φ a a2 exp − 2 p1 (θ) dθ dψ 1 − × l2 l1 0 0 and

a3 (φ) l3 (φ)

it follows that

N(φ)

= exp 0

φ

4 p1 (ψ) dψ

a3 , l3

a a (φ) a2 (φ) − p1 (φ) 2 = 1 l2 (φ) l2 (φ) l1


and

25

a3 (φ) a2 (φ) a3 (φ) N(φ) − p1 (φ) = l3 (φ) l3 (φ) l2 (φ)

for all φ, where p1 (φ) is the eigenvalue of N(φ) that satisfies the Riccati equation (1.34) with p1 (0) = p1 . The proof can be written down in an almost parallel way to that of Theorem 1.9 but will be a little complicated. We suggest an alternative proof of this theorem in Exercise 1-7. Finally, from Theorems 1.5, 1.9 and 1.10 we observe Corollary 1.11 The property that Stroh’s eigenvalue problem (1.29) is simple, semisimple, or degenerate remains invariant under the rotations of m and n around the vector product m × n. Moreover, in the case where the eigenvalue problem (1.29) is degenerate, the structure of the Jordan chains is also invariant under those rotations. Therefore, it is the elasticity tensor C = Cijkl 1≤i, j,k,l≤3 and the unit normal m × n to the reference plane that determine whether Stroh’s eigenvalue problem (1.29) is simple, semisimple, or degenerate and, when Stroh’s eigenvalue problem is degenerate, how long the Jordan chains are. 1.6 Angular Average of Stroh’s Eigenvalue Problem: Integral Formalism We have seen in Sections 1.3 and 1.5 that rotational dependency of eigenvectors and generalized eigenvectors of N is expressed in different forms according to the degeneracy of Stroh’s eigenvalue problem and becomes more complicated as the degeneracy increases. In this section, as a completion of the study on rotational dependency of Stroh’s eigenvalue problem, we take the averages of the formulas in Theorems 1.5, 1.9 and 1.10 with respect to φ over [−π, π ]. Then we arrive at “the angle-averaged Stroh’s eigenvalue problem”, which is valid regardless of whether Stroh’s eigenvalue problem is simple, semisimple or degenerate, and whose form is much simpler than the formulas in Theorems 1.9 and 1.10. Thus we obtain a fundamental theorem in the Barnett-Lothe integral formalism of Stroh’s eigenvalue problem. By this theorem, linear independence of the displacement parts of Stroh eigenvectors or generalized eigenvectors will be proved, and hence we can define the surface impedance tensor in the next section. To begin with, letting N(φ) be given by (1.33), we take the angular average of its eigenvalue p(φ) over [−π, π ]. Lemma 1.12 For each eigenvalue p(φ) of N(φ) with Im p(0) > 0,7 π √ 1 p(φ) dφ = −1. 2π −π

7 Since

(1.66)

p(φ) is non-real and continuous in φ, Im p(0) > 0 is equivalent to Im p(φ) > 0 for any φ.

26

K. Tanuma

Proof According to Remark 1.4, √ √ we have p(φ) = tan(φc − φ) with Im(tan φc ) > 0, or p(φ) ≡ −1. When p(φ) ≡ √−1, we immediately get (1.66). Now let p(φ) = tan(φc − φ) and write φc = φr + −1 φi (φr , φi ∈ R). From Im(tan φc ) > 0 we have φi > 0.8 Then it follows that9 π φ=π 1 1 log cos(φc − φ) p(φ) dφ = φ=−π 2π −π 2π φ=π √ 1 = . log cos(φ − φr ) cosh φi + −1 sin(φ − φr ) sinh φi φ=−π 2π √ The argument of the complex valued cos(φ − φr ) cosh φi + −1 sin(φ − φr ) sinh φi increases by 2π when φ changes from −π to π , because φi > 0 and sinh φi > 0. This proves (1.66).

Definition 1.13 We define the 6 × 6 real matrix S to be the angular average of the 6 × 6 matrix N(φ) over [−π, π ]: π S 1 S2 1 N(φ) dφ, (1.67) S= = 2π −π S3 S1T where S1 , S2 and S3 are 3 × 3 real matrices defined by S1 =

1 2π

S3 =

1 2π

π

−π

π

−π

−T(φ)−1 R(φ)T dφ,

S2 =

1 2π

π

−π

T(φ)−1 dφ,

−Q(φ) + R(φ)T(φ)−1 R(φ)T dφ.

(1.68)

Lemma 1.14 The matrices S2 and S3 are symmetric. Furthermore, S2 is positive definite.10 Proof The matrices S2 and S3 are symmetric, because the integrands T(φ)−1 and −Q(φ) + R(φ)T(φ)−1 R(φ)T are symmetric for all φ. By (1) of Lemma 1.1, the matrix T(φ) is positive definite for all φ, and so is the matrix T(φ)−1 for all φ. Hence its angular average, S2 , is also positive definite.

Now we take the angular average of Stroh’s eigenvalue problem. aα be an eigenvector or a generalized eigenvector of N(0) lα corresponding to the eigenvalues pα (α = 1, 2, 3) with Im pα > 0. Then it follows that √ a aα = −1 α . (1.69) S lα lα

Theorem 1.15 Let

√

√

−1 sinh 2b (a, b ∈ R) has been used. −1 b ) = 12 sin 2a+ cos2 a+sinh2 b √ √ 9 We use the formula cos(a + −1 b ) = cos a cosh b − −1 sin a sinh b (a, b ∈ R). 8 The

10 S 3

formula tan(a +

is negative definite, which we refer to the literature for a proof in Subsection 3.6.2.


27

aα is an eigenvector of N(0), we take the angular averages of lα both sides of (1.42) and use (1.66), which immediately leads to (1.69). Note that aα this consideration applies not only to eigenvectors when Stroh’s eigenvalue lα problem is simple or semisimple, but also to eigenvectors of degenerate Stroh’s eigenvalue problem. Now we consider the cases whereStroh’s eigenvalue problem (1.29) is degenerate, a3 and consider Case D1 first. Let be a generalized eigenvector of N(0) in l3 Theorem 1.9 corresponding to the multiple eigenvalue p2 with Im p2 > 0. Then from (1.60) and (1.61) it follows that φ

a3 a3 a2 N(φ) − p2 (φ) = exp −2 p2 (ψ) dψ , (1.70) l3 l3 l2 0 Proof When

where p2 (φ) is the solution to the Riccati equation (1.34) with p2 (0) = p2 . In order to take the angular average of both sides of this equality, by Remark 1.4, two cases should be considered, i.e., c − φ) with p2 = tan φc and √ the case where p2 (φ) = tan(φ√ the case where p2 (φ) ≡ −1. In the case where p2 (φ) ≡ −1, the right hand side of (1.70) becomes √ a √ a 2 exp −2 −1 φ = (cos 2φ − −1 sin 2φ) 2 , (1.71) l2 l2 whose angular average over [−π, π ] is equal to zero. In the case where p(φ) = tan(φc − φ), the angular average of the right hand side of (1.70) over [−π, π ] becomes 1 2π

π

exp −π

=

1 2π

1 = 2π

π −π π −π

ψ=φ a − 2 log cos(φc − ψ) dφ 2 l2 ψ=0 exp (2 log cos φc − 2 log cos(φc − φ)) dφ

a2 l2

φ=π a −1 cos2 φc a2 2 2 = , (1.72) dφ cos φc tan(φc − φ) l2 l2 φ=−π cos2 (φc − φ) 2π

which is equal to zero, because Im φc > 0 and the function tan(φc − φ) is 2π -periodic in φ ∈ R. In both cases, the angular average of the left hand side of (1.70) over [−π, π ] becomes, by (1.66), √ a a S 3 − −1 α . l3 lα This proves (1.69). By a similar method, the relation (1.69) can be proved for Case D2 on the basis of Theorem 1.10. However, later in Subsection 2.2.2, we shall give a systematic proof of (1.69), which is valid whatever length the Jordan chains in the eigenvalue problem (1.29) have.

28

K. Tanuma

From this theorem we conclude that the angle-averaged Stroh’s eigenvalue problem has the same form (1.69) regardless of whether Stroh’s eigenvalue problem (1.29) is simple, semisimple, or degenerate. 1.7 Surface Impedance Tensor

aα (α = 1, 2, 3) lα be linearly independent eigenvector(s) or generalized eigenvector(s) of Stroh’s eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, 3, Im pα > 0). In this section, on the basis of Theorem 1.15, we first prove that the displacement a parts aα (α = 1, 2, 3) of α above are linearly independent.11 Then we can define lα the surface impedance tensor which, for an elastic half-space, maps the displacements given at the boundary surface to the tractions needed to sustain them. This tensor, which plays a fundamental role in applications of the Stroh formalism, has been used in many studies on forward and inverse problems. In the next chapter we will give several applications of this tensor. Now from Theorem 1.15, the first three rows of the system (1.69) is written, by using notations in (1.68), as √ S1 aα + S2 lα = −1 aα , α = 1, 2, 3. Let m and n be two orthogonal unit vectors in R3 and let

Then we get S2 lα =

√

−1 I − S1 aα ,

α = 1, 2, 3,

where I is the 3 × 3 identity matrix. Since Lemma 1.14 implies that S2 is invertible, multiplying both sides by S−1 2 , we obtain √ −1 −1 S−1 α = 1, 2, 3. (1.73) lα = 2 − S2 S1 a α , Suppose that aα (α = 1, 2, 3) are linearly dependent. Then there exists a set of complex numbers (c1 , c2 , c3 ) = (0, 0, 0) such that 3

cα aα = 0.

α=1

Then from (1.73) it follows that 3

cα lα = 0,

α=1

and therefore 3 α=1

cα

aα lα

= 0.

11 Generally,

if aα (α = 1, 2, 3) are linearly independent in C3 , then

independent in C6 . However, its converse does not always hold.

aα lα

(α = 1, 2, 3) are linearly


29

This contradicts with the assumption that

aα lα

(α = 1, 2, 3) are linearly indepen-

dent. Hence we obtain

aα Theorem 1.16 Let (α = 1, 2, 3) be linearly independent eigenvector(s) or genlα eralized eigenvector(s) of Stroh’s eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, 3, Im pα > 0). Then their displacement parts aα (α = 1, 2, 3) are linearly independent. Thus, we are now in a position to define the surface impedance tensor. Definition 1.17 The surface impedance tensor Z is the 3 × 3 matrix given by √ (1.74) Z = − −1 [l1 , l2 , l3 ][a1 , a2 , a3 ]−1 , where [l1 , l2 , l3 ] and [a1 , a2 , a3 ] denote 3 × 3 matrices which consist of the column vectors lα and aα , respectively. From this definition it follows that √ lα = −1 Z aα ,

α = 1, 2, 3.

(1.75)

Therefore, taking account of Remark 1.8, we see that Z expresses a linear relationship between (i) the displacements given at the surface n · x = 0 of the form (1.57) which pertain to a special class of 2-dimensional deformations of the form (1.45), (1.48) or (1.55) and (ii) the tractions needed to sustain them at that surface. As we shall see soon, Z is a function of m × n. From (1.73) and (1.74) we immediately obtain the integral representation of Z. Theorem 1.18 Z = S−1 2 +

√

−1 S−1 2 S1 ,

(1.76)

where the matrices S1 and S2 are given by (1.68). From (1.31), (1.32), (1.68) and the 2π -periodicity of cos φ and sin φ, we observe Corollary 1.19 The matrices S1 , S2 , and thence Z, are invariant under the rotations of m and n around the vector product m × n. Hence they are functions not of m and n individually but of the vector product m × n. This rotational invariance can also be proved directly from (1.74) and Theorems 1.5, 1.9 and 1.10 (Exercise 1-8). Remark 1.20 There is the arbitrariness of the choice of linearly independent eigenvectors and generalized eigenvectors. However, Theorem 1.18 implies that the arbitrariness is cancelled out in making the product of the two matrices in (1.74), and hence Z is well-defined.

30

K. Tanuma T

Theorem 1.21 The surface impedance tensor Z is Hermitian (i.e., Z = Z ) and positive definite. Here we prove that Z is Hermitian. The positive definiteness of Z is proved in Section 7.D of [22] and [41] indirectly by using the Lagrangian. The direct proof, which is long, is given in [24, 76] and Section 6.6 of [77]. Lemma 1.22 Let S be the angular average of N(φ) given in (1.67). Then S2 = −I,

(1.77)

where I denotes the 6 × 6 identity matrix.

aα (α = 1, 2, 3) be linearly independent eigenvectors or generalized lα eigenvectors of N(0) corresponding to the eigenvalues pα (α = 1, 2, 3) with Im pα > 0. Then from Theorem 1.15 it follows that a a (α = 1, 2, 3). (1.78) S2 α = − α lα lα Proof Let

Recalling (1.44), we take the complex conjugate of both sides of (1.78) to see that (1.78) holds for six linearly independent eigenvectors or generalized eigenvectors of N(0). This proves the lemma.

Proof that Z is Hermitian By the expression of S in (1.67), the (1, 2) blockcomponent of the equality (1.77) turns out to be S1 S2 + S2 S1T = O. Since S2 is symmetric and invertible, it follows that −1 T S−1 , 2 S1 = − S 2 S1 which implies that S−1 2 S1 is anti-symmetric. Hence Z in (1.76) is Hermitian.

Finally, we give one remark. Remark 1.23 Suppose that C = Cijkl i, j,k,l=1,2,3 enjoys the major symmetry (1.6) but does not have the minor symmetries (1.4). We assume, instead of (1.7), the strong ellipticity condition: ⎛ The matrix ⎝

3 j,l=1

⎞ Cijkl ξ jξl ⎠

is positive definite for

i↓k→1,2,3

any non-zero vector ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3 ,

(1.79)


31

and assume that the “traction” on n · x = 0 is given by (1.24). Then all the arguments in Section 1.2 to Section 1.7 except the positive definiteness of Z remain valid.12 Note When the elastic body in question carries residual stress, the elasticity tensor C in equation (1.9) is replaced by a fourth-order tensor B, whose components, sometimes known [74] as “effective elastic coefficients” in the literature, have the major symmetry but not the minor symmetries [34, 46, 47]. 1.8 Examples We give the formulas for the surface impedance tensors for isotropic and for transversely isotropic media. For this purpose, we first investigate the conditions on C = Cijkl i, j,k,l=1,2,3 and on the vector product m × n for the corresponding Stroh’s eigenvalue problem (1.29) to be simple, semisimple, and degenerate (cf. Corollary 1.11). Then we compute the displacement parts and the traction parts of Stroh eigenvectors and generalized eigenvectors. The explicit formulas for the surface impedance tensors obtained here will be used in Chapter 2. 1.8.1 Isotropic Media For isotropic elasticity, the eigenvalue problem (1.29) is degenerate with an eigenvalue p1 of multiplicity three and belongs to Case D1 in Section 1.4 for any orthogonal unit vectors m and n in R3 . We shall observe this in what follows. For the orthogonal unit vectors m = (m1 , m2 , m3 ) and n = (n1 , n2 , n3 ), put yi = mi + p ni ,

i = 1, 2, 3.

(1.80)

Recalling that the components of the elasticity tensor is written as (see (1.11)) Cijkl = λ δijδkl + μ(δik δ jl + δil δkj), we see from (1.19) that

Q+ p R+R

T

⎛

+ p2 T = ⎝

3

⎞ Cijkl (m j + pnj)(ml + pnl )⎠

j,l=1

⎛ =⎝

3 j,l=1

(1.81)

⎞

i↓k→1,2,3

Cijkl yj yl ⎠ i↓k→1,2,3

= (λ + μ)yi yk + μ δik y21 + y22 + y23

i↓k→1,2,3

= μ y21 + y22 + y23 I + (λ + μ) (yi yk )i↓k→1,2,3 ,

(1.82)

12 For the positive definiteness of Z, the strong convexity condition (1.7) is needed (see the references in the comment after Theorem 1.21).

32

K. Tanuma

where I is the 3 × 3 matrix. Then we observe that 3 det Q + p R + RT + p2 T = μ2 (λ + 2μ) y21 + y22 + y23 .13 Note that the strong convexity condition (1.7) for isotropic elasticity is μ > 0,

3λ + 2μ > 0

(1.83)

(Exercise 1-2). Hence λ + 2μ > 0. Since m and n are unit and orthogonal, we have y21 + y22 + y23 = 1 + p2 .

(1.84)

Hence the sextic equation (1.22) becomes 3 det Q + p R + RT + p2 T = μ2 (λ + 2μ) 1 + p2 = 0, which gives the triple characteristic root with a positive imaginary part √ p = −1. This is surely one of the solutions (the singular solution) to the Riccati equation (1.34) (see Remark 1.4). Then by (1.82) and (1.84), the equation for the displacement part a of a Stroh eigenvector, (1.21), becomes ⎡ 2 ⎤ y1 y 1 y 2 y 1 y 3 Q + p R + RT + p2 T a = (λ + μ) ⎣ y1 y2 y22 y2 y3 ⎦ a = 0. (1.85) y1 y3 y2 y3 y23 Write

⎡

⎤ 1 m × n = = ⎣ 2 ⎦ . 3

Then taking account of (1.86) y1 1 + y2 2 + y3 3 = 0 √ and equation (1.84) with p = −1, we have two linearly independent solutions to (1.85): ⎡ ⎤ y1 √ a2 = ⎣ y2 ⎦ = m + −1 n. (1.87) a1 = , y3 p=√−1 It is obvious that the rank of the 3 × 3 matrix in (1.85) is equal to 1, which implies that there exist exactly two linearly independent displacement parts corresponding

13 For

a 3 × 3 matrix M and a scalar q, det(q I + M) = q3 + tr M q2 +

1 (tr M)2 − tr(M2 ) q + det M, 2

where tr denotes the trace of a square matrix. Use this formula by putting q = μ y21 + y22 + y23 and M = (λ + μ)(yi yk )i↓k→1,2,3 .


33

√ to p = −1. Hence, by (1.73), Stroh’s eigenvalue problem (1.29) for isotropic√elasticity has exactly two linearly independent eigenvectors corresponding to p = −1. Therefore, for isotropic elasticity, the eigenvalue problem (1.29) is degenerate and belongs to Case D1. a Now we compute the traction parts l1 and l2 of the Stroh eigenvectors α (α = lα 1, 2) where aα (α = 1, 2) are given by (1.87). By (1.81), (1.25) can be written as ⎛ ⎞ 3 T Cijkl yl n j⎠ a [R + pT] a = ⎝ j,l=1

i↓k→1,2,3

= λ ni yk + μ( p δik + yi nk )

i↓k→1,2,3

a,

(1.88)

where we have used the relation y1 n1 + y2 n2 + y3 n3 = p.

(1.89)

√ Then using the relations (1.84), (1.86), (1.89) with p = −1 and n1 1 + n2 2 + n3 3 = 0, we obtain from (1.87) √ l1 = μ p = −1 μ ,

⎡

⎤ y1 √ = 2 μ(−n + −1 m). l2 = 2 μ p ⎣ y2 ⎦ y3 p=√−1

(1.90)

We compute a3 and l3 which constitute√a generalized eigenvector for (1.29). From (1.87), (1.88), (1.89) and (1.84) with p = −1 we also have ⎡ ⎤ y1 . [R + pT] a2 = λ yi nk + μ( p δik + ni yk ) i↓k→1,2,3 a2 = (λ + μ) p ⎣ y2 ⎦ y3 p=√−1 Then by (1.85) and (1.90), equation (1.51) becomes ⎡ ⎤ ⎤ ⎡ 2 y1 y 1 y1 y2 y1 y3 (λ + μ) ⎣ y1 y2 y22 y2 y3 ⎦ a3 + (λ + 3μ) p ⎣ y2 ⎦ = 0 y1 y3 y2 y3 y23 y3

√ −1 .

p=

(1.91)

Taking account of (1.89), we can solve this for a3 as a3 = −

λ + 3μ n. λ+μ

(1.92)

It follows from (1.81), (1.87) and (1.89) that ⎞ ⎛ 3 Cijkl nj nl ⎠ a2 = λ ni nk + μ (δik + ni nk ) T a2 = ⎝ j,l=1

i↓k→1,2,3

i↓k→1,2,3

⎡

⎤ y1 √ = (λ + μ) p n + μ ⎣ y2 ⎦ = μ m + (λ + 2μ) p n ( p = −1). y3

a2

34

K. Tanuma

Hence by (1.88) and (1.92), (1.50) becomes λ + 3μ μ m + (λ + 2μ) p n + μ m + (λ + 2μ) p n l3 = − λ+μ √ −2μ μ m + −1 (λ + 2μ)n . = λ+μ Using the vectors aα , lα (α = 1, 2, 3) obtained above, we compute the surface impedance tensor for isotropic elasticity. We easily see that det[a1 , a2 , a3 ] = − and

λ + 3μ λ+μ

√ λ + 3μ λ + 3μ , − m, n − −1 m , Cof [a1 , a2 , a3 ] = − λ+μ λ+μ

(1.93)

where Cof [a1 , a2 , a3 ] denotes the cofactor matrix of the matrix [a1 , a2 , a3 ]. Therefore, by simple algebraic calculations of √ √ λ+μ [l1 , l2 , l3 ] Cof [a1 , a2 , a3 ]T , (1.94) Z = − −1 [l1 , l2 , l3 ][a1 , a2 , a3 ]−1 = −1 λ + 3μ we obtain (Exercise 1-10) Theorem 1.24 The surface impedance tensor for isotropic elasticity is given by T Z = Z = Z ij i↓ j→1,2,3 , μ Z ii = 2(λ + 2μ) − (λ + μ) i 2 , i = 1, 2, 3, λ + 3μ √ μ −(λ + μ) i j + −1(−1)k 2μ k , 1 ≤ i < j ≤ 3, (1.95) Z ij = λ + 3μ where (1 , 2 , 3 ) = m × n and the index k is determined by {k} = {1, 2, 3} \ {i, j}. To obtain the surface impedance tensor for isotropic elasticity, the method described above is systematic but is not efficient. We give here an alternative method, which takes full advantage of rotational invariance of the surface impedance tensor in the reference plane (Corollary 1.19) and the rotational symmetry of the elasticity tensor itself. Let us fix the Cartesian coordinate system (x1 , x2 , x3 ). Using spherical coordinates in R3 , we write ⎡ ⎤ ⎡ ⎤ 1 sin α cos β m × n = = ⎣ 2 ⎦ = ⎣ sin α sin β ⎦ (0 ≤ α ≤ π, 0 ≤ β < 2π ). (1.96) 3 cos α Then for

⎡ ⎤ 0 0 = ⎣ 0 ⎦ 1


35

we have = 3 2 0 , where

⎤ cos α 0 sin α 1 0 ⎦ 2 = ⎣ 0 − sin α 0 cos α ⎡

is the transformation matrix pertaining to the rotation of the material around the 2-axis by the angle α and ⎡ ⎤ cos β − sin β 0 3 = ⎣ sin β cos β 0 ⎦ (1.97) 0 0 1 is that pertaining to the rotation of the material around the 3-axis by the angle β. Since the components of the isotropic elasticity tensor (1.81) remain invariant under these transformations, we apply the transformation formula for second order tensors to get14 Z = 3 2 Z0 (3 2 )T ,

(1.99)

where Z0 is the surface impedance tensor for m and n such that m × n = 0 = (0, 0, 1). Hence, by Corollary 1.19, we may take ⎡ ⎤ ⎡ ⎤ 1 0 m = m0 = ⎣ 0 ⎦ , n = n0 = ⎣ 1 ⎦ . 0 0 Then, we see that (1.85), (1.88) and (1.91) assume more simplified forms, and we easily get the corresponding aα , lα (α = 1, 2, 3) to obtain √ ⎡ ⎤ 2(λ + 2μ) −2 −1 μ 0 μ ⎣ √ Z0 = 0 ⎦. 2 −1 μ 2(λ + 2μ) λ + 3μ 0 0 λ + 3μ Finally, from (1.96) and (1.99) we obtain (1.95). 1.8.2 Transversely Isotropic Media For media that are transversely isotropic, the eigenvalue problem (1.29) becomes either simple, semisimple, or degenerate with the appearance of a Jordan chain

that Z is a function of m × n = (Corollary 1.19), we write Z as Z(). Then the transformation formula for second order tensors implies that for any orthogonal tensor Q

14 Recalling

Z∗ ( Q 0 ) = Q Z(0 ) QT , Z∗

is the surface impedance tensor after the material is rotated by Q. where For an isotropic material, Z = Z∗ , and it follows from (1.98) that Z( Q 0 ) = Q Z(0 ) QT . Putting Q = 3 2 , we arrive at (1.99).

(1.98)

36

K. Tanuma

of two. These conditions are determined by the elasticity tensor C = length Cijkl 1≤i, j,k,l≤3 and the vector product m × n of the orthogonal unit vectors m and n (Corollary 1.11). The elasticity tensor of a transversely isotropic medium whose axis of symmetry coincides with the 3-axis is given by (1.13). Hereafter, for simplicity we put C1111 = A,

C1122 = N,

C1133 = F,

C3333 = C,

C2323 = L.

(1.100)

The strong convexity condition (1.7) is equivalent to (Exercise 1-3) L > 0,

1 (A − N) > 0, 2

A + C + N > 0,

(A + N)C > 2F 2 .

(1.101)

As in the previous subsection, we write m × n = = (1 , 2 , 3 ). Then the conditions for the eigenvalue problem (1.29) to be simple, semisimple and degenerate are given, respectively, as follows: Lemma 1.25 (1.29) is degenerate and belongs to Case D1 √ ⇐⇒ 3 = 1 or 3 = −1 or AC − F − 2L = 0 (1.29) is semisimple

√ ⇐⇒ 3 = ±1 and AC − F − 2L = 0 and

2 2L 2L − (AC − F 2 − 2F L) + CL = 0 AL A−N A− N

(1.102)

(1.29) is simple ⇐⇒ all other conditions.

Theorem 1.26 The surface impedance tensor for a transversely isotropic medium whose axis of symmetry coincides with the 3-axis is given by T Z = Z ij i↓ j→1,2,3 = Z , Z 11 Z 22 Z 33

L AG 2L 2 2 2 2 2 1 + 3 + 2 1 + 2 = , K D A− N

L AG 2L = 2 2 + 3 2 + 1 2 1 2 + 2 2 , K D A− N ! 1 = AL − 3 4 − H 3 2 + GHK + (F + L)2 − AC 1 2 + 2 2 3 2 , D


37

√ AG 2L 2 L AGL 2 1− 1 + 2 2 1 2 − −1 A− N− 1 + 2 2 3 , K D A−N D √ −L(F + L) (F + L)K 2 2 2 2 1 + 2 1 3 − −1L −1 + 1 + 2 = 2 , D D √ −L(F + L) 2 (F + L)K 2 = 1 + 2 2 2 3 + −1L −1 + 1 + 2 2 1 , D D

Z 12 = Z 13 Z 23

(1.103) where

" 3 2 +

K=

G=

# $ $ 2AL 2 + (AC − F 2 − 2F L) 2 + 2 + 2√ AL√ 3 1 2 % "

H=

2L 2 1 + 2 2 , A− N

AL

, AL

D = A 3 + L 1 + 2 2

2

2

,

+ AH K − AG 3 2 ,

2 = AL 3 4 + (AC − F 2 − 2F L) 1 2 + 2 2 3 2 + CL 1 2 + 2 2 .

(1.104)

Note In the case 3 = ±1, we have 1 2 + 2 2 = 0. Then K = 1, = AL, G = 2, H = 1 and D = 0. Hence each Z ij in (1.103) has a indeterminate form. In this case, Z is obtained by taking the limit 1 2 + 2 2 −→ 0 in (1.103). In fact, as 1 2 + 2 2 −→ 0 we get 2 2L − A + N 2 K =1+ 1 + 2 2 + O 1 2 + 2 2 , 2(A − N) 2 = AL + (AC − F 2 − 2F L − 2AL) 1 2 + 2 2 + O 1 2 + 2 2 and 2 AC − F 2 − 2F L − 2AL 2 1 + 2 2 + O 1 2 + 2 2 , 2AL 2 AC − F 2 − 2F L − 2AL 2 H =1+ 1 + 2 2 + O 1 2 + 2 2 . 2AL

G=2+

Then D=

2 L(3A − N) 2 1 + 2 2 + O 1 2 + 2 2 . A−N

Substituting these into (1.103), we let 1 2 + 2 2 −→ 0. Then the components of Z in the case 3 = ±1 become Z 11 = Z 22 =

2A(A − N) , 3A − N

√ (A − N)2 3 , Z 12 = − −1 3A − N

Z 33 = L, Z 13 = Z 23 = 0.

(1.105)

38

K. Tanuma

We note that 3 = ±1 implies that the reference plane (i.e., the plane generated by the orthogonal unit vectors m and n) is perpendicular to the axis of symmetry of the transversely isotropic medium in question. Proof of Lemma 1.25 and Theorem 1.26 As in (1.96), we write ⎤ ⎡ ⎤ ⎡ sin α cos β 1 m × n = = ⎣ 2 ⎦ = ⎣ sin α sin β ⎦ (0 ≤ α ≤ π, 0 ≤ β < 2π ). cos α 3 Then for

(1.106)

⎤ sin α 0 = ⎣ 0 ⎦ cos α ⎡

we have = 3 0 , where 3 is the transformation matrix (1.97) related to the rotation of the material around the 3-axis by the angle β. Since the 3-axis has been taken to be the axis of symmetry, the components of the transversely isotropic elasticity tensor (1.100) remain invariant under the transformation by 3 . Hence the transformation formula for the second order tensor can be applied to get15 Z = 3 Z0 3T ,

(1.107)

where Z0 is the surface impedance tensor for m and n such that m × n = 0 = (sin α, 0, cos α). Then, by Corollary 1.19, we may take ⎤ ⎡ ⎡ ⎤ cos α 0 n = n0 = ⎣ 1 ⎦ . (1.108) m = m0 = ⎣ 0 ⎦ , − sin α 0 From Corollary 1.11 it is also possible to consider degeneracy of the eigenvalue problem (1.29) by using (1.108). Let yi (i = 1, 2, 3) be given by (1.80). By (1.100), equation (1.21) becomes ⎛ ⎞ 3 T 2 Q+ p R+R + p T a=⎝ Cijkl yj yl ⎠ a j,l=1

⎡ =⎣

i↓k→1,2,3

Ay21 + 12 (A−N)y22 +Ly23

1 2 (A+N)y1 y2 1 2 2 2 (A−N)y 1 +Ay2 +Ly3 2

Sym. = 0.

15 Again

(F+L)y1 y3 (F+L)y2 y3

⎤ ⎦a

L(y21 +y22 )+Cy23

(1.109)

we use (1.98). For a transversely isotropic material in question, Z = Z∗ for Q = 3 . Then Z( Q 0 ) = Q Z(0 ) QT

for Q = 3 , which implies (1.107).


39

The sextic equation (1.22) can be factorized as

1 (A − N)(y21 + y22 ) + Ly23 det Q + p R + RT + p2 T = 2 2 × AL y21 + y22 + AC − F 2 − 2F L × y21 + y22 y23 + CLy43 = 0.16 Since (1.108) implies that y1 = cos α,

y2 = p,

y3 = − sin α,

(1.110)

we let p1 be the root with Im p1 > 0 of the quadratic equation 1 1 (A − N) y21 + y22 + Ly23 = (A − N) p2 + cos2 α + L sin2 α = 0 2 2

(1.111)

and p2 , p3 be the roots with Im pα > 0 (α = 2, 3) of the quartic equation 2 AL y21 + y22 + AC − F 2 − 2F L y21 + y22 y23 + CLy43 = 0.

(1.112)

Suppose that F + L = 0 and p2 = p3 . Then we can check from (1.111) and (1.112) that three vectors ⎤ ⎡ ⎤ ⎡ (F + L)y1 y3 y2 ⎥ ⎢ + L)y2 y3 a1 = ⎣ −y1 ⎦ , aα = ⎣ & (F (α = 2, 3) (1.113) '⎦ 2 2 2 0 + Ly − A y + y 1 2 3 p= p1 p= pα

2

satisfy (1.109). From (1.112) it follows that A y21 + y2 + Ly23 = 0 at p = p2 and L p = p3 , since otherwise, (1.112) becomes A (F + L)2 y43 = 0. From the supposition and (1.101) we have F + L = 0, A > 0, L > 0. When y3 = 0, (1.112) becomes 2 2 y1 + y22 = 0, which contradicts p2 = p3 . Therefore, a1 and a2 or a1 and a3 are linearly independent even if p1 = p2 or p1 = p3 . Then from (1.73) we see that there exist three linearly independent eigenvectors of the eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, 3). Hence the eigenvalue problem (1.29) is simple or semisimple. By (1.100) and (1.108), (1.25) can be written as ⎞ ⎛ 3 T R + pT a = ⎝ Cijkl yl nj⎠ a j,l=1

⎡1

(A − N)y2 =⎣ Ny1 0 2

16 This

factorization is classically well known [73].

i↓k→1,2,3 1 (A 2

⎤ − N)y1 0 Ay2 Fy3 ⎦ a. Ly3 Ly2

40

K. Tanuma

Then from (1.113) we obtain ⎤ ⎡1 (A − N) y22 − y21 2 ⎦ , l1 = ⎣ (N − A)y1 y2 −Ly1 y3 p= p 1

⎡

&

(A − N)(F + L)y1 y2 y3

'

⎤

⎥ ⎢ ALy22 + ((F + L)N − AF)y21 − F Ly23 y3 ⎥ lα = ⎢ ⎦ ⎣ & ' − A y21 + y22 + Fy23 Ly2

(α = 2, 3). (1.114) p= pα

On the other hand, we easily see from (1.110) and (1.113) that det [a1 , a2 , a3 ] = (F + L) ( p2 − p3 ) sin α × A cos2 α + L sin2 α − Ap2 p3 p1 − A ( p2 + p3 ) cos2 α (1.115) and Cof [a1 , a2 , a3 ] = Aij i↓ j→1,2,3 , where A11 = (F + L)( p2 − p3 ) sin α A cos2 α + L sin2 α − Ap2 p3 , & ' A12 = − A cos2 α + p23 + L sin2 α cos α, & ' A13 = A cos2 α + p22 + L sin2 α cos α, A21 = −A(F + L) p23 − p22 cos α sin α, & ' A22 = − A cos2 α + p23 + L sin2 α p1 , & ' A23 = A cos2 α + p22 + L sin2 α p1 , A31 = (F + L)2 p3 − p2 cos α sin2 α, A32 = (F + L) p1 p3 + cos2 α sin α, A33 = −(F + L) p1 p2 + cos2 α sin α. In the computations of Z0 = Z ij0

i↓ j→1,2,3

√ = − −1

√ = − −1 [l1 , l2 , l3 ] [a1 , a2 , a3 ]−1

1 [l1 , l2 , l3 ] Cof [a1 , a2 , a3 ]T , det [a1 , a2 , a3 ]

(1.116)


41

we need to express p1 , p2 + p3 and p2 p3 in terms of the elastic tensor (1.100). For p1 , we use (1.111) to get p1 =

√

"

√ 2L sin2 α = −1 K. A− N

−1 cos2 α +

For pα (α = 2, 3), we rewrite (1.112) using (1.110) as & ' ALp4 + 2AL cos2 α + (AC − F 2 − 2F L) sin2 α p2 + AL cos4 α + (AC − F 2 − 2F L) cos2 α sin2 α + CL sin4 α = 0, which, from the relations between the roots p22 , p23 and the coefficients of the equation above, implies that p22 + p23 = − p22 p23 =

2AL cos2 α + (AC − F 2 − 2F L) sin2 α , AL

AL cos4 α + (AC − F 2 − 2F L) cos2 α sin2 α + CL sin4 α = . AL AL (1.117)

The numerator can be written as √ √ √ √ L( A cos2 α − C sin2 α)2 + ( AC − F)( AC + F + 2L) cos2 α sin2 α, which is√ positive, because the √ strong convexity condition (1.101) implies that AC − F 2 > 0, AC − F > 0 and AC + F > 0. Taking account of Im pα > 0 (α = 2, 3), we have " p2 p3 = −

= −H. AL

Then from (1.117) it follows that ( p2 + p3 )2 = p22 + p23 + 2 p2 p3 =−

√ √ 2AL cos2 α + (AC − F 2 − 2F L) sin2 α + 2 AL , AL

and since Im( p2 + p3 ) > 0, we have

p2 + p3 = =

√ √

(

√ √ 2AL cos2 α + (AC − F 2 − 2F L) sin2 α + 2 AL −1 AL −1 G.

42

K. Tanuma

0 Let us compute Z 11 . Denoting the (i, j) entry of the matrix [l1 , l2 , l3 ] by Lij, we see from (1.114) and (1.116) that the (1, 1) entry of the matrix [l1 , l2 , l3 ] Cof[a1 , a2 , a3 ]T is

L11 A11 + L12 A12 + L13 A13

p21 − cos2 α A cos2 α + L sin2 α − Ap2 p3 2 + A cos2 α + L sin2 α cos2 α − Ap2 p3 cos2 α

= (A − N)(F + L) ( p2 − p3 ) sin α

= (A − N)(F + L)( p2 − p3 ) sin α

L 2 2 × − cos α − sin α (A cos2 α + L sin2 α + AH) A− N + A cos2 α + L sin2 α cos2 α + AH cos2 α = −(F + L)L( p2 − p3 ) A cos2 α + L sin2 α + AH sin3 α. On the other hand, (1.115) can be written as √ det[a1 , a2 , a3 ] = −1 (F + L)( p2 − p3 ) sin α × (A cos2 α + L sin2 α + AH)K − AG cos2 α √ = −1 (F + L)( p2 − p3 )D sin α. Hence 0 Z 11 =

L sin2 α(A cos2 α + L sin2 α + AH) L sin2 α(D + AG cos2 α) = . D KD

In the same way, by simple but long algebraic calculations, we obtain the other entries: T = Z0 , Z0 = Z ij0 i↓ j→1,2,3

0 = Z 11

L AG sin2 α 1 + cos2 α , K D

0 Z 22 =

L AGK 2 sin α, D

1 AL(− cos4 α − H cos2 α + GHK) D ! + cos2 α sin2 α (F + L)2 − AC ,

√ AGL 2 = − −1 cos α A − N − sin α , D

0 = Z 33

0 Z 12

L(F + L) cos α sin3 α, D

√ (F + L)K 2 = −1 L sin α −1 + sin α . D

0 Z 13 =− 0 Z 23

(1.118)


43

Then from (1.106) and (1.107) we obtain (1.103). This is the generic case. Henceforth, we consider the missing cases, i.e., the case where p2 = p3 and the case where F + L = 0, and we give an outline of the computations. For the details we refer to [69]. First, we consider the case p2 = p3 . By virtue of (1.110), equation (1.112) becomes a quadratic equation in p2 : 2 2 y21 + y22 + Iy23 + Jy43 = p2 + cos2 α + I sin2 α + J sin4 α = 0, (1.119) where AC − F 2 − 2F L , 2AL √ √ −(AC − F 2 )( AC + F + 2L)( AC − F − 2L) . J= 4A2 L2 Hence p2 = p3 is equivalent to I=

J sin4 α = 0, which, by (1.101) and 3 = cos α, is also equivalent to √ 3 = 1 or 3 = −1 or AC − F − 2L = 0.

(1.120)

Suppose that 3 = 1 or 3 = −1. Then we have sin α = 0, cos α = 1 and p1 = p2 = √ p3 = −1. From (1.109) and (1.110) we can take two linearly independent vectors ⎤ ⎡ ⎤ ⎡ 0 √3 a2 = ⎣ −1 ⎦ a1 = ⎣ 0 ⎦ , 1 0 2

and √ we can observe that the rank of the 3 × 3 matrix in (1.109) is one at p = −1. Hence by (1.73), Stroh’s eigenvalue problem (1.29) has√exactly two linearly independent eigenvectors associated with the eigenvalue p = −1 of multiplicity 3. Thus, the eigenvalue problem (1.29) is degenerate and belongs to Case D1. Then from (1.25), (1.50), (1.51) and (1.74) we obtain 0 0 Z 11 = Z 22 =

2A(A − N) , 3A − N

0 = L, Z 33

√ (A − N)2 0 0 = − −1 , Z 12 3 = −Z 21 3A − N

0 0 0 0 Z 13 = Z 23 = Z 31 = Z 32 = 0.

We can check that taking the limit sin2 α −→ 0 in (1.118) gives the above Z ij0 . Thus we obtain (1.105) from (1.107). √ Suppose that 3 = ±1 and AC − F − 2L = 0. Then we get two vectors ⎡ ⎡ ⎤ ⎤ y2 y1 , a2 = ⎣ y2 ⎦ , a1 = ⎣ −y1 ⎦ 0 y 3 p= p p= p 1

2

where p1 (Im p1 > 0) is the root of (1.111) and p2 = p3 (Im p2 = Im p3 > 0) is the root of " C sin2 α = 0. p2 + cos2 α + A

44

K. Tanuma

We can observe that the rank of the 3 × 3 matrix in (1.109) is two at p = p2 when p1 = p2 and is one at p = p1 when p1 = p2 . Hence by (1.73), Stroh’s eigenvalue problem (1.29) has exactly one eigenvector associated with the eigenvalue p2 of multiplicity 2 when p1 = p2 and exactly two linearly independent eigenvectors associated with the eigenvalue p1 of multiplicity 3 when p1 = p2 . Thus, the eigenvalue problem (1.29) is degenerate and belongs to Case D1. Using (1.25), (1.50), (1.51) and (1.74), we can observe that formula (1.118) applies to the present case. Now let p2 = p3 and F + L = 0. Then instead of aα (α = 2, 3) in (1.113) we can take ⎡ ⎤ ⎡ ⎤ 0 y1 a 2 = ⎣ 0 ⎦ , a3 = ⎣ y 2 ⎦ , 1 0 p= p 3

where p2 (Im p2 > 0) is the root of L p2 + cos2 α + C sin2 α = 0 and p3 (Im p3 > 0) is the root of A p2 + cos2 α + L sin2 α = 0. = A, it follows that Since the strong convexity condition (1.101) implies that A−N 2 p1 = p3 . Obviously, a1 in (1.113) and a2 above are linearly independent even when p1 = p2 . Therefore, the eigenvalue problem (1.29) is simple or semisimple. Then using (1.25) and (1.74), we can observe that formula (1.118) also applies to the present case. We note that equality (1.102) holds if and only if (1.111) and (1.112) have a common root, that is, p1 = p2 or p1 = p3 . This makes the eigenvalue problem (1.29) semisimple with an eigenvalue of multiplicity 2, or degenerate with an eigenvalue of multiplicity 3, but the length of the Jordan chain remains the same as before.

Note The eigenvalue problem (1.29) becomes degenerate with an eigenvalue of multiplicity 3 if and only if 3 = 1 or 3 = −1 or ) √ AC − F − 2L = 0 and 2L/(A − N) = C/A. √ The second equality above follows from AC − F − 2L = 0 and (1.102). Isotropic elasticity belongs to this class. 1.9 Justification of the Solutions in the Stroh Formalism We have begun our presentation of the Stroh formalism by seeking solutions of the form (1.18) and have called their superposition (1.45) and its modifications (1.48) and 17 (1.55) the general forms of the solution to (1.17) which describes two-dimensional √ − −1 ξ m·x deformations in an elastic half-space n · x ≤ 0, satisfies the condition u = a e on n · x = 0 for some complex a ∈ C3 , and decays to zero as n · x −→ −∞. In this section, we justify these forms.

17 For

the meaning of “two-dimensional”, see the third paragraph of Section 1.2.


45

We introduce the variables y1 = m · x = m1 x1 + m2 x2 + m3 x3 ,

y 2 = n · x = n1 x 1 + n2 x 2 + n 3 x 3 .

(1.121)

Then the solutions considered are defined in y2 ≤ 0 and depends only on y1 and y2 . Hence, noting that ∂ ∂ ∂ = mj + nj , ∂xj ∂ y1 ∂ y2 we can rewrite (1.17) as ∂2 ∂2 ∂2 + T 2 u(y1 , y2 ) = 0 in Q 2 + (R + RT ) ∂ y1 ∂ y2 ∂ y1 ∂ y2

y2 ≤ 0,

(1.122)

where Q, R and T are matrices defined by (1.20). We impose the boundary condition at y2 = 0 u(y1 , 0) = f (y1 ),

(1.123)

where f (y1 ) is some prescribed boundary data valued in C3 . Applying the Fourier transform with respect to y1 to the boundary-value problem (1.122) and (1.123), we obtain √ √ ∂ ∂2 2 T ( −1 ξ1 ) Q + ( −1 ξ1 )(R + R ) +T 2 * u(ξ1 , y2 ) = 0 in y2 ≤ 0 (1.124) ∂ y2 ∂ y2 and * u(ξ1 , 0) = *f (ξ1 ), where * denotes the Fourier transform with respect to y1 , i.e., √ 1 * u(ξ1 , y2 ) = √ u(y1 , y2 ) e− −1 y1 ξ1 dy1 , 2π R √ 1 *f (ξ1 ) = √ f (y1 ) e− −1 y1 ξ1 dy1 , 2π R and ξ1 is the variable of the Fourier transform. To reduce (1.124) to a first-order differential system in y1 , we introduce τ (y1 , y2 ) such that √ √ ∂ * (1.125) τ (ξ1 , y2 ). ( −1 ξ1 )RT + T u(ξ1 , y2 ) = −1 ξ1 * ∂ y2 The left hand side is the Fourier transform of the traction ⎞ ⎛ 3 ∂u k ⎝ Cijkl n j⎠ ∂ xl j,k,l=1

i↓1,2,3

in (1.24), because it can be written as ∂ ∂ +T u(y1 , y2 ). RT ∂ y1 ∂ y2

46

K. Tanuma

As in (1.27) to (1.30), using the inversibility of T, we see that (1.124) and (1.125) can be recast in a six-dimensional first-order differential system √ T−1 −T−1 RT d * * u u = −1 ξ1 * τ τ dy2 * −Q + RT−1 RT −RT−1 √ * u = −1 ξ1 N . (1.126) * τ * u The value of at y2 = 0 is * τ *f (ξ1 ) * u . (1.127) (ξ1 , 0) = * τ * τ (ξ1 , 0) Hence the solution to (1.126) and (1.127) is obtained as *f (ξ1 ) √ * u −1 ξ1 N y2 . (ξ1 , y2 ) = e * τ * τ (ξ1 , 0)

(1.128)

First we consider the case where the eigenvalue problem (1.29) is simple or aα semisimple, and justifiy the form (1.45). Let ∈ C6 (α = 1, 2, 3) be linearly indelα pendent eigenvectors of N pertaining to the eigenvalues pα (α = 1, 2, 3, Im pα > 0). We use the conventions (1.43) and (1.44). Let ! Pα : C6 −→ v ∈ C6 ; (N − pα I)v = 0 (1 ≤ α ≤ 6) (1.129) be the projection operators on the eigenspaces of N associated with pα , where I is the 6 × 6 identity matrix. Then 6

Pα = I

α=1

and (1.129) implies that N=

6

N Pα =

α=1

6 6 pα Pα . (N − pα I)Pα + pα Pα = α=1

α=1

This is the spectral representation of N. Then since Pα Pβ = δαβ Pα for any α, β (1 ≤ α, β ≤ 6), it follows that k k ∞ √ 6 ∞ √ √ −1 ξ1 y2 −1 ξ1 y2 k −1 ξ1 N y2 e = N = pkα Pα k! k! α=1 18

k=0

=

k=0

6 √ −1 ξ1 pα y2

e

Pα .

α=1

18 When

the eigenvalue problem (1.29) is simple, the spectral representation of N is given in Section 4.A and 4.B of [22] explicitly in terms of eigenvectors of N.


47

Hence (1.128) becomes 6 *f (ξ1 ) √ * u −1 ξ1 pα y2 . (ξ1 , y2 ) = e Pα * τ * τ (ξ1 , 0) α=1

Since Im pα > 0 for α = 1, 2, 3 and Im pα < 0 for α = 4, 5, 6, the boundedness of the solution as y2 −→ −∞ requires that *f (ξ1 ) = 0 for α = 1, 2, 3, ξ1 > 0 =⇒ Pα * τ (ξ1 , 0) which is equivalent to *f (ξ1 ) ∈ P4 C6 ⊕ P5 C6 ⊕ P6 C6 * τ (ξ1 , 0) , + a4 a5 a6 + c5 + c6 ; c4 , c5 , c6 ∈ C for ξ1 > 0, = c4 l4 l5 l6 (1.130) and requires that ξ1 < 0

=⇒

Pα

*f (ξ1 ) * τ (ξ1 , 0)

= 0 for α = 4, 5, 6,

which is equivalent to *f (ξ1 ) ∈ P1 C6 ⊕ P2 C6 ⊕ P3 C6 * τ (ξ1 , 0) + , a1 a2 a3 = c1 + c2 + c3 ; c1 , c2 , c3 ∈ C for ξ1 < 0; l1 l2 l3 (1.131) here ⊕ denotes the direct sum of the vector spaces. To establish (1.130) and (1.131), we first note that there exist scalar functions cα (ξ1 ) (1 ≤ α ≤ 6) such that *f (ξ1 ) = c4 (ξ1 ) a4 + c5 (ξ1 ) a5 + c6 (ξ1 ) a6

for ξ1 > 0

*f (ξ1 ) = c1 (ξ1 ) a1 + c2 (ξ1 ) a2 + c3 (ξ1 ) a3

for ξ1 < 0

and

for any given boundary data f (y1 ), since aα (α = 1, 2, 3) are complete in C3 by Theorem 1.16 and aα+3 = aα (α = 1, 2, 3). Extending cα (ξ1 ) (1 ≤ α ≤ 6) to be functions on R by cα (ξ1 ) = 0 (α = 1, 2, 3) for ξ1 > 0, cα (ξ1 ) = 0 (α = 4, 5, 6) for ξ1 < 0,

(1.132)

48

K. Tanuma

we can write *f (ξ1 ) =

6

cα (ξ1 ) aα

(ξ1 = 0).

α=1

Put

⎛ ⎞ 6 6 √ aβ * u ⎠. e −1 ξ1 pα y2 Pα ⎝ cβ (ξ1 ) (ξ1 , y2 ) = * τ lβ α=1

Then it follows that

β=1

6 √ aα * u −1 ξ1 pα y2 e cα (ξ1 ) . (ξ1 , y2 ) = * τ lα α=1

Substituting the vector * u(ξ1 , y2 ) =

6

e

√ −1 ξ1 pα y2

cα (ξ1 ) aα

(1.133)

α=1

which constitutes the first three components of the above into the left hand side of (1.125), we get 6 √ √ √ ∂ * ( −1 ξ1 )RT + T e −1 ξ1 pα y2 cα (ξ1 ) RT + pα T aα , u(ξ1 , y2 ) = −1 ξ1 ∂ y2 α=1 which, by (1.25), is equal to √

−1 ξ1

6

e

√ −1 ξ1 pα y2

cα (ξ1 ) lα .

α=1

Thus, * τ (ξ1 , 0) =

6

cα (ξ1 ) lα .

α=1

This establishes (1.130) and (1.131). Applying the inverse Fourier transform to (1.133) and using (1.121), we obtain 6 √ 1 e −1 ξ1 ( pα y2 +y1 ) cα (ξ1 ) aα dξ1 u(y1 , y2 ) = √ 2π α=1 R 6 √ 1 = √ e −1 ξ1 (m·x+ pα n·x) cα (ξ1 ) aα dξ1 . 2π α=1 R

Hence u is the Fourier integral with component 6 α=1

cα (ξ1 ) aα e

√ −1 ξ1 (m·x+ pα n·x)

,

(1.134)


49

which becomes for ξ1 < 0, 3

cα (ξ1 ) aα e

√ −1 ξ1 (m·x+ pα n·x)

,

(1.135)

α=1

and becomes for ξ1 > 0, 6

cα (ξ1 ) aα e

√ −1 ξ1 (m·x+ pα n·x)

.

α=4

Putting ξ1 = −ξ in (1.135) gives (1.45). Next we consider the Case D1, where the eigenvalue problem (1.29) is degenerate and there appear Jordan chains of length 2 (see Section 1.4). We justify the form a (1.48). Let α (α = 1, 2) be linearly independent eigenvectors of N associated with lα a the eigenvalues pα (α = 1, 2, Im pα > 0), and let 3 be a generalized eigenvector l3 which satisfies a a a (1.136) N 3 − p2 3 = 2 . l3 l3 l2 Likewise we use the conventions (1.43) for α = 1, 2 and (1.44). Let ! Pα : C6 −→ v ∈ C6 ; (N − pα I)v = 0 , α = 1, 4, ! Pα : C6 −→ v ∈ C6 ; (N − pα I)2 v = 0 , α = 2, 5

(1.137)

be the projection operators on the eigenspaces of N (α = 1, 4) or on the generalized eigenspaces of N (α = 2, 5) associated with pα . Then it follows that Pα = I α=1,2,4,5

and e

√ −1 ξ1 N y2

=

e

√ −1 ξ1 N y2

Pα

α=1,2,4,5

=

e

√ −1 ξ1 pα y2

e

√ −1 ξ1 (N− pα I)y2

α=1,2,4,5

=

α=1,2,4,5

e

∞ √ −1 ξ1 pα y2 k=0

Pα

√ ( −1 ξ1 y2 )k (N − pα I)k Pα . k!

19

Then (1.137) implies that √ √ √ √ e −1 ξ1 N y2 = e −1 ξ1 pα y2 Pα + e −1 ξ1 pα y2 −1 ξ1 y2 (N − pα I) Pα . α=1,2,4,5

α=2,5

19 It is also possible to obtain this formula from the spectral representation of N as before. But here we have adopted a more straightforward method.

50

K. Tanuma

Therefore (1.128) becomes √ * u e −1 ξ1 pα y2 Pα (ξ1 , y2 ) = * τ α=1,2,4,5

+

e

√ −1 ξ1 pα y2

√ −1 ξ1 y2 (N − pα I) Pα

.

*f (ξ1 ) * τ (ξ1 , 0)

α=2,5

.

Since Im pα > 0 for α = 1, 2 and Im pα < 0 for α = 4, 5, the boundedness of the solution as y2 −→ −∞ requires that ξ1 > 0

=⇒

Pα

*f (ξ1 )

= 0 for α = 1, 2,

* τ (ξ1 , 0)

which is equivalent to

*f (ξ1 )

∈ P4 (C6 ) ⊕ P5 (C6 )

* τ (ξ1 , 0)

+ =

c4

a4

+ c5

l4

a5

+ c6

l5

a6

, ; c4 , c5 , c6 ∈ C

l6

for ξ1 > 0, (1.138)

and requires that ξ1 < 0

=⇒

Pα

*f (ξ1 ) * τ (ξ1 , 0)

= 0 for α = 4, 5,

which is equivalent to

*f (ξ1 ) * τ (ξ1 , 0)

∈ P1 (C6 ) ⊕ P2 (C6 )

+ =

c1

a1 l1

+ c2

a2 l2

+ c3

a3 l3

, ; c1 , c2 , c3 ∈ C

for ξ1 < 0. (1.139)

To establish (1.138) and (1.139), we see that there exist scalar functions cα (ξ1 ) (1 ≤ α ≤ 6) such that *f (ξ1 ) = c4 (ξ1 ) a4 + c5 (ξ1 ) a5 + c6 (ξ1 ) a6

for ξ1 > 0

*f (ξ1 ) = c1 (ξ1 ) a1 + c2 (ξ1 ) a2 + c3 (ξ1 ) a3

for ξ1 < 0

and


51

for any given boundary data f (y1 ), since aα (α = 1, 2, 3) are complete in C3 by Theorem 1.16.20 Then extending cα (ξ1 ) (1 ≤ α ≤ 6) by (1.132), we write *f (ξ1 ) =

6

cα (ξ1 ) aα

(ξ1 = 0).

α=1

Put √ * u e −1 ξ1 pα y2 Pα (ξ1 , y2 ) = * τ α=1,2,4,5

+

e

√ −1 ξ1 pα y2

α=2,5

⎞ .⎛ 6 √ aβ ⎠. −1 ξ1 y2 (N − pα I) Pα ⎝ cβ (ξ1 ) l β β=1 (1.140)

Since it follows that ⎛ ⎞ 6 aβ aα aα+1 ⎠ = cα (ξ1 ) cβ (ξ1 ) + cα+1 (ξ1 ) (α = 2, 5) Pα ⎝ lβ lα lα+1 β=1 and from (1.136) that ⎛ (N − pα I) Pα ⎝

6

cβ (ξ1 )

β=1

⎞ a ⎠ = cα+1 (ξ1 ) α (α = 2, 5), lβ lα

aβ

formula (1.140) becomes * u (ξ1 , y2 ) = * τ

e

√ −1 ξ1 pα y2

cα (ξ1 )

+

e

√ −1 ξ1 pα y2

lα -

α=1,2,4,5

aα

cα+1 (ξ1 )

aα+1 lα+1

α=2,5

. √ aα + −1 ξ1 y2 . lα

Substituting the vector

* u(ξ1 , y2 ) =

e

√ −1 ξ1 pα y2

α=1,2,4,5

+

e

√ −1 ξ1 pα y2

cα (ξ1 ) aα √ cα+1 (ξ1 ) aα+1 + −1 ξ1 y2 aα

(1.141)

α=2,5

aα

(α = 1, 2, 3) are linearly independent lα eigenvectors of N but also when they are linearly independent eigenvector(s) and generalized eigenvector(s) of N. 20 Recall

that Theorem 1.16 holds not only when

52

K. Tanuma

which constitutes the first three components of the above into the left hand side of (1.125), we obtain √ ∂ * ( −1 ξ1 )RT + T u(ξ1 , y2 ) ∂ y2 √ √ e −1 ξ1 pα y2 cα (ξ1 ) RT + pα T aα = −1 ξ1 α=1,2,4,5

+

√

−1 ξ1

e

√ −1 ξ1 pα y2

cα+1 (ξ1 )

√ RT + pα T aα+1 + −1 ξ1 y2 aα +T aα ,

α=2,5

which, by (1.25) and (1.50), is equal to √

−1 ξ1 −

e

√ −1 ξ1 pα y2

α=1,2,4,5 √ ξ12 e −1 ξ1 pα y2 α=2,5

cα (ξ1 ) lα +

e

√ −1 ξ1 pα y2

cα+1 (ξ1 ) lα+1

α=2,5

cα+1 (ξ1 ) y2 lα .

Thus, * τ (ξ1 , 0) =

6

cα (ξ1 )lα .

α=1

This establishes (1.138) and (1.139). Applying the inverse Fourier transform to (1.141), we obtain 1 u(y1 , y2 ) = √ 2π

α=1,2,4,5 R

e

√ −1 ξ1 ( pα y2 +y1 )

cα (ξ1 ) aα dξ1

√ √ 1 +√ e −1 ξ1 ( pα y2 +y1 ) cα+1 (ξ1 ) aα+1 + −1 ξ1 y2 aα dξ1 2π α=2,5 R √ 1 = √ e −1 ξ1 (m·x+ pα n·x) cα (ξ1 ) aα dξ1 2π α=1,2,4,5 R √ √ 1 +√ e −1 ξ1 (m·x+ pα n·x) cα+1 (ξ1 ) aα+1 + −1 ξ1 (n · x) aα dξ1 . 2π α=2,5 R (1.142) Hence u is the Fourier integral with component

cα (ξ1 ) aα e

α=1,2,4,5

+

α=2,5

√ −1 ξ1 (m·x+ pα n·x)

√ √ cα+1 (ξ1 ) aα+1 + −1 ξ1 (n · x) aα e −1 ξ1 (m·x+ pα n·x) ,


53

which becomes for ξ1 < 0,

cα (ξ1 ) aα e

√ −1 ξ1 (m·x+ pα n·x)

α=1,2

√ √ + c3 (ξ1 ) a3 + −1 ξ1 (n · x) a2 e −1 ξ1 (m·x+ p2 n·x) ,

(1.143)

and becomes for ξ1 > 0,

cα (ξ1 ) aα e

√ −1 ξ1 (m·x+ pα n·x)

α=4,5

√ √ + c6 (ξ1 ) a6 + −1 ξ1 (n · x) a5 e −1 ξ1 (m·x+ p5 n·x) .

Putting ξ1 = −ξ in (1.143) leads to (1.48). In the Case D2, where the eigenvalue problem (1.29) is degenerate and there appear Jordan chains of length 3, justification of the form (1.55) can be achieved by a similar method using (1.54). The details are left as Exercise 1-13. It can be proved that the solutions u(y1 , y2 ) in (1.134) and (1.142) are real when the boundary data f (y1 ) in (1.123) is a real-valued function (Exercise 1-14). 1.10 Comments and References In Section 1.1 we have summarized briefly linearlized elasticity which is needed in the subsequent sections. For more details we refer to [26, 33]. We have begun our presentation of the Stroh formalism in Section 1.2. The Stroh formalism goes back to Stroh’s original works [65, 66], where he studied dislocations, cracks and elastic waves in a steady state of motion in an anisotropic elastic body. Since then, many researchers extended his study to establish a mathematical theory, which is now called the Stroh formalism. The Stroh formalism reveals simple structures hidden in the equations of anisotropic elasticity and provides a systematic approach to these equations. In this formalism, isotropic elasticity is one example to which this formalism can be applied, although isotropic elasticity had been studied for a long time by using the theory proper only to it. The Stroh formalism is now indispensable to the studies of many areas in anisotropic elasticity, which include elastostatics, elastodynamics, Rayleigh waves, composite materials, cracks, inclusions and inverse problems. In Sections 1.2 and 1.4 we have introduced the Stroh formalism, where we put emphasis on the forms of solutions (i.e., displacements) to the equations of anisotropic elasticity in a half-space and on the forms of the corresponding tractions. We have called these forms of solutions “general forms”, the nomenclature of which is justified in Section 1.9 by the theory of Fourier analysis. These forms will be used in Section 2.3 when we construct approximate solutions pertaining to oscillating Dirichlet data in inverse boundary value problems. In Sections 1.3 and 1.5 we have proved, by an intuitive method, rotational invariance and rotational dependence of the Stroh eigenvectors and generalized eigenvectors in the reference plane. Classically, this rotational dependency was

54

K. Tanuma

proved through the fact that the 6 × 6 matrix N(φ) in (1.33) satisfies the matrix Riccati equation d N(φ) = −I − N(φ)2 , dφ

(1.144)

where I is the 6 × 6 identity matrix [9, 22, 41]. Equation (1.144) can be proved by simple but long computations where we differentiate each 3 × 3 block-component of N(φ) with respect to φ and use the relations (1.35). Comparing the corresponding d block-components of dφ N(φ) and −I − N(φ)2 , we arrive at (1.144). The details are left as Exercise 1-5. We have also suggested a proof of Theorem 1.5 and of Theorem 1.10 in Exercise 1-6 and Exercise 1-7, respectively, which make use of (1.144). It may be difficult to notice without any suggestion, that N(φ) satisfies (1.144). We note that in the case where the eigenvalue problem (1.29) is simple or semisimple Theorem 1.5 suggests (1.144). In fact, formula (1.42) implies that 2 a 2 a N(φ) = p(φ) l l and

d a a = p (φ) N(φ) l l dφ

a is an eigenvector of N(0). Since the eigenvalue p(φ) of N(φ) l satisfies the Riccati equation (1.34), it follows that a d a N(φ) = −I − N(φ)2 . l l dφ a For , we can take six linearly independent eigenvectors of N(0), which proves l (1.144). The Barnett-Lothe integral formalism was originally developed by Barnett and Lothe on the basis of the Stroh formalism ([9, 41]; see also [22]). Theorem 1.15 in Section 1.6, Theorems 1.16, 1.18, 1.21 and Lemma 1.22 in Section 1.7 are the main results in this integral formalism. Among them Theorem 1.15 is the most fundamental. In Subsection 2.2.2, we shall give a rigorous proof of a higher-order dimensional version of this theorem, which can be applied to any degenerate case of Stroh’s eigenvalue problem (cf. [2, 52, 54]). We can see that the results in the integral formalism do not depend on whether the corresponding Stroh’s eigenvalue problem is simple, semisimple or degenerate. Thus, the results in the integral formalism are powerful and useful in the studies on the fundamental solutions in Chapter 2 and on Rayleigh waves in Chapter 3. In Section 1.8, we have computed explicit formulas for the surface impedance tensors on the basis of definition (1.74). Lemma 1.25 and Theorem 1.26 are the results in [69]. To our knowledge, in three-dimensional elasticity, the exact and complete formulas for the surface impedance tensors written in terms of elasticity tensors have been obtained only for isotropic and for transversely isotropic media. The difficulty is caused mainly by the fact that the characteristic equations (1.22) corresponding to the other elastic media can not be factorized. The explicit formulas for the surface for all φ, where


55

impedance tensors given in this section will be used when we derive the explicit forms of the fundamental solutions and prove the identifiability of the elasticity tensor from the Dirichlet to Neumann map in the inverse boundary value problems in Chapter 2. Note that the integral representation (1.76) of the surface impedance tensors is useful for perturbation arguments. On the basis of this representation, the surface impedance tensor is computed for a weakly anisotropic elastic medium by considering the deviation of its elasticity tensor from a comparative isotropic state as a perturbation, and an explicit perturbation formula for the surface impedance tensor is obtained in [72], which is correct to first order in the perturbative part of the elasticity tensor. As we have seen in Section 1.8, for isotropic elasticity, Stroh’s eigenvalue problem (1.29) is degenerate and it always has a Jordan chain of length of exactly two (Case D1). For transversely isotropic materials, the length of the Jordan chain is at most two. Hence there arises the question whether there exists an anisotropic elastic material for which Stroh’s eigenvalue problem has a Jordan chain of length three (i.e, an elastic material which belongs to Case D2, or which is more degenerate than isotropic elasticity). Ting [78] proved that such materials exist. Section 1.9 is devoted to justification of the forms of basic solutions (displacements) introduced in Sections 1.2 and 1.4. This justification, though straightforward, does not seem to have appeared explicitly in the literature before. 1.11 Exercises 1-1 Prove that the strong convexity condition (1.7) is equivalent to the assertion that the 6 × 6 matrix (1.12) in the Voigt notation is positive definite. 1-2 Find the strong convexity condition for the isotropic elasticity tensor (1.11). (Answer: (1.83)) 1-3 Find the strong convexity condition for the elasticity tensor of transversely isotropic materials (1.13). (Answer: (1.101)) 1-4 Prove Lemma 1.7. 1-5 Prove (1.144) by following the suggestions below it. 1-6 Use (1.144) to prove Theorem 1.5. (Suggestion: Put a a k(φ) = N(φ) − p(φ) . l l Then k(0) = 0 from the assumption. Derive the ordinary differential equation for k(φ) and apply the uniqueness of the solution to it as in the proof of Theorem 1.3 to show that k(φ) = 0 for all φ.) 1-7 Prove Theorem 1.10 in the following way: Put a2 (φ) a a2 (φ) f(φ) = N(φ) − p1 (φ) − 1 l2 (φ) l2 (φ) l1 and

χ (φ) = N(φ)

a (φ) a (φ) a3 (φ) − p1 (φ) 3 − 2 . l3 (φ) l3 (φ) l2 (φ)

56

K. Tanuma

Then f(0) = χ (0) = 0 from the assumption. Show that f(φ) = 0,

1-8 1-9 1-10 1-11

χ (φ) = 0

for all φ. (Suggestion: Derive the ordinary differential equations for f(φ) and χ (φ) with the help of (1.144), and apply the uniqueness of the solutions to them.) Prove Corollary 1.19 from Theorems 1.5, 1.9 and 1.10. Establish Remark 1.23. Check (1.93) and complete the calculations of (1.94). Suppose that C = Cijkl i, j,k,l=1,2,3 does not observe any symmetry condition (i.e., it has neither the major symmetry (1.6) nor the minor symmetries (1.4)). We assume, in place of the strong convexity condition (1.7), the following ellipticity condition: ⎞ ⎛ 3 Cijkl ξ jξl ⎠ = 0 det ⎝ j,l=1

i↓k→1,2,3

for any non-zero vector ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3 . In addition to the 3 × 3 matrices Q, R, T in (1.20), we introduce a 3 × 3 matrix ⎞ ⎛ 3 Cijkl njml ⎠ . S=⎝ j,l=1

i↓k→1,2,3

We define the ‘traction’ on n · x = 0, i.e., the Neumann data, by (1.24) and define a vector l by (1.25) with RT replaced by S. Then prove Theorems 1.2, 1.3 and Theorem 1.5 with the matrix RT replaced by the matrix S. 1-12 (Two-dimensional homogeneous elasticity) Consider the plane-strain deformations of orthorhombic materials (1.15) in the (x1 , x2 )-plane. The governing equations become 2 j,k,l=1

Cijkl

∂ 2 uk = 0, ∂ x j∂ xl

i = 1, 2

with C1112 = C2212 = 0. Let m = (m1 , m2 ), n = (n1 , n2 ) be orthogonal unit vectors in R2 . Then, by changing the ranges where the indices i, j, k, l run from {1, 2, 3} to {1, 2}, the arguments in Section 1.2 to Section 1.7 apply in a parallel way. We note that the dimension of the matrices Q, R, T, S1 , S2 , S3 becomes two and the dimension of the matrices N, S and the eigenvalue problem (1.29) becomes four. Hence the length of the Jordan chain is at most two. For simplicity, we put C1111 = A,

C1122 = F,

C2222 = C,

C1212 = L.

(1.145)

Investigate the degeneracy of the eigenvalue problem (1.29) and compute the surface impedance tensor Z, which is a 2 × 2 matrix. Answer: Z 11 Z 12 Z= Z 21 Z 22


with

57

√

Z 11 Z 12

√ C LG , Z 22 = √ , = √ AC + L AC + L √ √ −1 AC − F L , = −Z 21 = ∓ √ AC + L

where

( G=

A LG

AC − F 2 + 2L L

√

AC − F

/ ×/ /= and the ∓ sign corresponds to that of m n = (0, 0, ±1) with m n = (n1 , n2 , 0) (see [53]). (m1 , m2 , 0),/ 1-13 Justify the form of the solution (1.55) from (1.128) in the Case D2, where the eigenvalue problem (1.29) is degenerate and there appear Jordan chains of length 3. 1-14 Following the suggestions below, prove that the solutions u(y1 , y2 ) in (1.134) and (1.142) are real when the boundary data f (y1 ) in (1.123) is a real-valued function. (a) Show that *f (ξ1 ) = *f (−ξ1 ) (ξ1 ∈ R). (b) Show that cα+3 (ξ1 ) = cα (−ξ1 ) (α = 1, 2, 3) for ξ1 > 0. (c) In view of (1.43) and (1.44), recast formulas (1.134) and (1.142) using only quantities with the index α running from 1 to 3.

58

K. Tanuma


59

2 Applications in Static Elasticity 2.1 Fundamental Solutions 2.1.1 Fundamental Solution in the Stroh Formalism Let x ∈ R3 and let Gkm = Gkm (x) be a solution to 3

Cijkl

j,k,l=1

∂2 Gkm + δim δ(x) = 0 ∂ x j∂ xl

in R3 ,

i, m = 1, 2, 3,

(2.1)

where (x1 , x2 , x3 ), δim , and δ(x) are the Cartesian coordinates of x, Kronecker’s delta symbol, and the Dirac delta function, respectively, and C = Cijkl i, j,k,l=1,2,3 is a homogeneous elasticity tensor. Physically, the solution Gkm describes the displacement at the point x in the xk direction due to a unit point force at the origin in the xm direction. We call G = G(x) = (Gkm )k↓m→1,2,3 the fundamental solution to 3 j,k,l=1

Cijkl

∂2 uk (x) + fi (x) = 0 in R3 , ∂ x j∂ xl

it generates the solution

i = 1, 2, 3;

u(x) = (u1 , u2 , u3 ) =

R3

G(x − y)f(y) dy,

where f(x) = ( f1 , f2 , f3 )T . Let x = 0. Take any two orthogonal unit vectors e1 and e2 in R3 such that e 1 × e2 =

x , |x|

where e1 × e2 denotes the vector product. Now we will use the 3 × 3 matrix T(φ) introduced in (1.32): ⎛ T(φ) = ⎝

3

⎞ Cijkl njnl ⎠

j,l=1

,

i↓k→1,2,3

where n = n(φ) = (n1 , n2 , n3 ) = −e1 sin φ + e2 cos φ. The following thoerem is well known (cf. [6, 39, 68]). Theorem 2.1 G(x) = is a fundamental solution.

1 8π 2 |x|

π

−π

T(φ)−1 dφ

(2.2)

60

K. Tanuma

From (2.1), G(x) is given as the inverse Fourier transform of the matrix ⎞ ⎤−1 ⎡⎛ 3 ⎦ , ⎣⎝ Cijkl ξ jξl ⎠ j,l=1

i↓k→1,2,3

where ξi (i = 1, 2, 3) are the variables of the Fourier transform. It can be proved that this inverse Fourier transform can be reduced to a single integral over the unit circle on the plane perpendicular to x. A rigorous proof is given in [54]. Let us write G(x) in terms of the surface impedance tensor Z. By using the notation in (1.68), formula (2.2) can be written as G(x) =

1 S2 . 4π |x|

(2.3)

On the other hand, from the integral representation (1.76) of Z it follows that −1 S2 = Re Z , where Z is the surface impedance tensor defined by (1.74) (or equivalently by (1.76)) with the orthogonal unit vectors m = m(φ) = (m1 , m2 , m3 ) = e1 cos φ + e2 sin φ, n = n(φ) = (n1 , n2 , n3 ) = −e1 sin φ + e2 cos φ, and Re Z denotes the real part of Z. Since it follows that e1 × e2 = m × n, from Corollary 1.19 we obtain Theorem 2.2 G(x) =

−1 1 Re Z 4π |x|

(2.4)

is a fundamental solution, where Z is the surface impedance tensor in (1.74) with m and n satisfying m×n=

x . |x|

2.1.2 Formulas for Fundamental Solutions: Examples In Section 1.8, we have obtained explicit formulas for the surface impedance tensors of isotropic and transversely isotropic materials. In this subsection, applying Theorem 2.2 to these formulas, we derive the formulas for the fundamental solutions pertaining to isotropic and transversely isotropic materials. For x = 0, we write x x2 ,* x3 ). = (* x1 ,* |x| For isotropic materials, from (1.95) and (2.4) we immediately obtain


61

Corollary 2.3 (Fundamental solution for isotropic materials) T GIso (x) = GIso (x) = Gij i↓ j→1,2,3 , Gij =

1 xj . xi * (λ + 3μ)δij + (λ + μ)* 8π μ(λ + 2μ)|x|

For transversely isotropic materials, using the notations (1.100), we obtain Corollary 2.4 (Fundamental solution for transversely isotropic materials) T GTrans (x) = GTrans (x) =

1 Sij i↓ j→1,2,3 , 4π |x|

* * x 21 x 22 0 + S , 22 * * x 21 + * x 21 + * x 22 x 22 * x1* x2 0 0 = S11 − S22 , * x 21 + * x 22

0 S11 = S11

S12

0 S22 = S11

* * x 22 x 21 0 + S , 22 * * x 21 + * x 21 + * x 22 x 22

F+L F+L * * x3 , x3 . x1* x2* S23 = ALGH ALGH 2 1 A* x 23 + H + L * = x1 +* x 22 , ALGH

S13 = S33

(2.5)

where

0 S11

0 S22

2 x 23 x1 +* AL GHK − * x 43 − H * x 23 − AC − (F + L)2 * x 22 * , = x 21 + * AL2 GH * x 22 2 x 23 x1 +* x 22 + AH K − AG* A* x 23 + L * = x 21 + * AGKL * x 22

xi (i = 1, 2, 3), that is, and K, G and H are given by (1.104) with i = * " K=

G=

* x 23 +

# $ √ √ $ 2AL* x 23 + (AC − F 2 − 2F L) * x 21 + * x 22 + 2 AL % "

H=

2L 2 * x1 +* x 22 , A−N

AL , AL

,

2 x 23 + CL * = AL* x 43 + (AC − F 2 − 2F L) * x 21 + * x 22 * x 21 + * x 22 .

62

K. Tanuma

0 0 Note In the case * x3 = ±1, i.e., the case * x 21 + * x 22 = 0, S11 and S22 become indeterTrans minate forms. In this case, as in the note of Theorem 1.26, G (x) is obtained by taking the limit * x 21 + * x 22 −→ 0 in the formula above. In fact, as * x 21 + * x 22 −→ 0 we get

3A − N 3A − N 0 −→ , S22 . 2A(A − N) 2A(A − N) x 21 + * x 22 = 0 become Then the components of Sij i↓ j→1,2,3 in (2.5) in the case * 0 −→ S11

S11 = S22 =

3A − N , 2A(A − N)

S33 =

1 , L

S12 = S13 = S23 = 0.

Recall that in (1.100), the 3-axis is the axis of rotational symmetry. Hence * x3 = ±1 implies that x lies on the axis of rotational symmetry. Outline of Proof of Corollary 2.4 Using spherical coordinates, we write x = (* x1 ,* x2 ,* x3 ) = (sin α cos β, sin α sin β, cos α) |x| (0 ≤ α ≤ π, 0 ≤ β < 2π ).

(2.6)

Instead of applying (2.4) to (1.103) directly, as in Subsection 1.8.2 we make use of the transformation formula for GTrans (x) GTrans (x) = 3 GTrans (x0 ) 3T ,

(2.7)

where |x0 | = |x|,

x0 = (sin α, 0, cos α), |x0 |

and 3 is the transformation matrix (1.97) pertaining to the rotation of the material around the 3-axis by the angle β. Theorem 2.2 implies that GTrans (x0 ) =

−1 1 Re Z0 , 4π |x|

where Z0 is the surface impedance tensor for m × n = (sin α, 0, cos α) and its explicit form is already given by (1.118). Then by simple but long conputations we get T 1 0 GTrans (x0 ) = GTrans (x0 ) = S , 4π |x| ij i↓ j→1,2,3 AL(GHK − cos4 α − H cos2 α) − AC − (F + L)2 cos2 α sin2 α 0 , S11 = AL2 GH sin2 α A cos2 α + L sin2 α + AH K − AG cos2 α S022 = , AGKL sin2 α 1 A(cos2 α + H) + L sin2 α , S033 = ALGH F+L cos α sin α, S023 = 0. S013 = (2.8) S012 = 0, ALGH Thus, from (2.6) and (2.7) we obtain the result.


63

2.2 Piezoelectricity 2.2.1 Basic Theory Piezoelectric materials have been used in many engineering devices because of their intrinsic direct and converse piezoelectric effects that take place between electric fields and mechanical deformations. In piezoelectricity, the mechanical stress tensor σ = (σij)i, j=1,2,3 and the electric displacement D = (D1 , D2 , D3 ) are related to the mechanical displacement u = (u1 , u2 , u3 ) and the electric potential φ by the following constitutive equations: 3

σij =

k,l=1 3

Dj =

∂uk ∂φ + eijl , ∂ xl ∂ xl 3

Cijkl

i, j = 1, 2, 3,

l=1

∂uk ∂φ − jl , ∂ xl ∂ xl 3

eklj

k,l=1

j = 1, 2, 3.

(2.9)

l=1

Here C = (Cijkl )i, j,k,l=1,2,3 is the elasticity tensor, e = (eijl )i, j,l=1,2,3 is the piezoelectric tensor, and = ( jl )i,l=1,2,3 is the dielectric tensor. Hence the elastic and electric fields are coupled through the piezoelectric tensor. These tensors C, e, and satisfy the following symmetry conditions: Cijkl = C jikl = Cklij,

eijl = e jil ,

jl = lj,

i, j, k, l = 1, 2, 3.

(2.10)

We assume that the internal energy function U=

3 3 1 1 Cijkl εij εkl + jl E j El 2 2

i, j,k,l=1

j,l=1

be positive for any non-zero mechanical strain (εij)i, j=1,2,3 and for any non-zero electric field (E1 , E2 , E3 ). This leads to the following strong convexity conditions: 3

Cijkl εij εkl > 0,

i, j,k,l=1

3

jl E j El > 0

(2.11)

j,l=1

for any non-zero 3 × 3 real symmetric matrix (εij) and for any non-zero real vector (E1 , E2 , E3 ) ∈ R3 . The equations of mechanical and electric equilibrium with zero body force and zero free charge are given by 3 ∂σij j=1

respectively.

∂xj

= 0,

i = 1, 2, 3

and

3 ∂ Dj j=1

∂xj

= 0,

(2.12)

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K. Tanuma

To give unified expressions for the constitutive equations (2.9) and for the equilibrium equations (2.12), we introduce the coefficients Bijkl for i, k = 1, 2, 3, 4, j, l = 1, 2, 3 by

Bijkl =

⎧ ⎪ ⎪ Cijkl ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ eijl

i, j, k, l = 1, 2, 3 i, j, l = 1, 2, 3, k = 4 (2.13)

⎪ ⎪ eklj ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ − jl

j, k, l = 1, 2, 3, j, l = 1, 2, 3,

i=4 i=k=4

and define a four-dimensional vector (u˜ k )k=1,2,3,4 by u˜ k =

⎧ ⎨ uk ⎩

k = 1, 2, 3 (2.14)

φ

k=4

and a 4 × 3 tensor (σ˜ ij)i=1,2,3,4, j=1,2,3 by

σ˜ ij =

⎧ ⎨ σij ⎩

i, j = 1, 2, 3 (2.15) i = 4.

Dj

Then (2.9) is written as σ˜ ij =

3 4

Bijkl

k=1 l=1

∂ u˜ k , ∂ xl

i = 1, 2, 3, 4,

j = 1, 2, 3

(2.16)

and (2.12) is written as 3 ∂ σ˜ ij j=1

∂xj

= 0,

i = 1, 2, 3, 4.

In this section we consider homogeneous piezoelectric materials. Then C, e, and

are independent of x = (x1 , x2 , x3 ), and the equilibrium equations become 3 4 k=1 j,l=1

Bijkl

∂ 2 u˜ k = 0, ∂ x j∂ xl

i = 1, 2, 3, 4.

(2.17)

Equations (2.16) and (2.17) are parallel to (1.5) and (1.17), respectively, except that the indices i and k run from 1 to 4. The purpose of this section is to extend the Stroh formalism in Chapter 1 to piezoelectricity.


65

2.2.2 Extension of the Stroh Formalism Let m = (m1 , m2 , m3 ), n = (n1 , n2 , n3 ) be orthogonal unit vectors in R3 . Using Bijkl in (2.13), we define 4 × 4 matrices Q, R and T by ⎛ Q=⎝

3

⎞ Bijkl m jml ⎠

j,l=1

⎛ T=⎝

3

⎛

⎞

⎞ Bijkl m jnl ⎠

j,l=1

i↓k→1,2,3,4

Bijkl n jnl ⎠

j,l=1

R=⎝

,

3

,

i↓k→1,2,3,4

.

(2.18)

i↓k→1,2,3,4

Corresponding to Lemma 1.1, we have Lemma 2.5 (1) The matrices Q and T are symmetric and invertible. (2) The characteristic roots pα (1 ≤ α ≤ 8), i.e., the solutions to the eighth-order equation det Q + p R + RT + p2 T = 0

(2.19)

are not real and they occur in complex conjugate pairs. Proof By (2.13), Bijkl i,k=1,2,3,4, j,l=1,2,3 do not have the minor symmetries any more, but they still enjoy the major symmetry Bijkl = Bklij,

i, k = 1, 2, 3, 4,

j, l = 1, 2, 3.

Hence the matrices Q and T are symmetric. Moreover, we see from (2.13) that the 4 × 4 matrix Q can be written blockwise as Q=

QC qe qeT −Q

,

where the 3 × 3 matrix QC , the three-dimensional column vector qe , and the scalar Q are given by ⎛ QC = ⎝

3

Cijkl m jml ⎠

j,l=1

Q =

3 j,l=1

⎛

⎞

jl m jml .

i↓k→1,2,3

,

qe = ⎝

3 j,l=1

⎞ eijl m jml ⎠ i↓1,2,3

,

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K. Tanuma

Then it follows that Q=

I

0

0T −1

QC qe −qeT Q

,

where I is the 3 × 3 identity matrix and 0 is the three-dimensional column zero vector. Since it follows from (2.11) that QC is positive definite (Lemma 1.1) and that Q is a positive scalar, we see that the 4 × 4 matrix

QC qe

−qeT Q is positive definite, and hence the matrix Q is invertible. In the same way, it is proved that T is invertible. The proof of (2) is exactly the same as that of (2) of Lemma 1.1 except that the indices i and k run from 1 to 4 with Cijkl replaced by Bijkl and that the matrix 3 is invertible.

j,l=1 Bijkl w j wl i↓k→1,2,3,4

Therefore, the arguments in Chapter 1 can be applied to piezoelectricity. Here, the 3 × 3 real matrices Q, R and T in (1.20) are replaced by the 4 × 4 real matrices Q, R and T in (2.18), and the 6 × 6 real matrix N in (1.30) is replaced by the 8 × 8 real matrix N with Q, R and T given by (2.18). Then the eigenvalue problem (1.29) becomes an eight-dimensional eigenrelation; here the vector a becomes a fourdimensional complex vector whose first three components pertain to the mechanical displacement and whose last component to the electric potential (see (2.14)). Also, the vector l becomes a four-dimensional complex vector whose first three components pertain to the mechanical traction on the surface n · x = 0 and whose last component to the normal component of the electric displacement there (consider 3 σ ˜ n , i = 1, 2, 3, 4, where σ˜ ij are given by (2.15)). j=1 ij j When we consider the dependence of eigenvectors and generalized eigenvectors of this extended eight-dimensional eigenvalue problem on the rotations of m and n, we take the 4 × 4 real matrices Q(φ), R(φ), T(φ) and the 8 × 8 real matrix N(φ) to be those which are defined by using Q, R and T in (2.18) with the rotated m = (m1 , m2 , m3 ), n = (n1 , n2 , n3 ) given by (1.31) (cf. (1.32) and (1.33)). Using these matrices, we define the 8 × 8 real matrix S (i.e., the angular average of N(φ)) and its 4 × 4 blocks Si (i = 1, 2, 3) by (1.67) and (1.68), respectively. When establishing the Stroh formalism in piezoelectricity, we shall need to consider the case where the eight-dimensional eigenvalue problem (1.29) has a Jordan chain of length 4. When not only the piezoelectric effect but also the piezomagnetic and magnetoelectric effects are present, the corresponding matrices Q, R and T become 5 × 5 matrices and the matrices N and S become 10 × 10 matrices. Then the eigenvalue problem (1.29) becomes a ten-dimensional eigenrelation and there is a possibility that a Jordan chain of length 5 appears. From now on in this section we prove a higher dimensional version of Theorem 1.15. The proof does not depend on the dimension of the eigenvalue problem (1.29) and is valid whatever length its Jordan chain has. This theorem becomes a basis for the integral formalism in a higher dimension.


Theorem 2.6 Let

a1 l1

67

be an eigenvector of N(0) corresponding to an eigenvalue p1 a with Im p1 > 0, and let α (α = 2, · · · , n) be generalized eigenvectors which form lα a Jordan chain of N(0) of length n so that a a a N(0) α − p1 α = α−1 , α = 2, · · · , n. (2.20) lα lα lα−1 Then it holds that

a S α lα

=

√

aα −1 , lα

α = 1, 2, · · · , n.

(2.21)

To prove the theorem, we need three lemmas. Lemma 2.7 N (φ) = −I − N(φ)2 ,

(2.22)

where I is the identity matrix the dimension of which is equal to that of N(φ) and =

d . dφ

For the proof, see the comments on (1.144). Lemma 2.8 Let p(φ) be the solution to the Riccati equation (1.34) with p(0) = p1 . Moreover, let a1 (φ), . . . , an−1 (φ) be the solutions to the following initial value problems: a1 (φ) = −2 p(φ) a1 (φ),

a1 (0) = 1,

⎧ α−1 ⎪ ⎨ a (φ) = −2 p(φ)a (φ) − a (φ)a (φ), α α−β β α β=1 ⎪ ⎩ aα (0) = 0, α = 2, . . . , n − 1.

(2.23)

(2.24)

Then for α = 2, . . . , n, we have α−1 a a α N(φ) − p(φ)I = aα−β (φ) β lα lβ

(2.25)

β=1

for all φ. Proof We set α−1 a a α − aα−β (φ) β , fα (φ) = N(φ) − p(φ)I lα lβ

α = 2, . . . , n.

β=1

From (2.20) it follows that a a 2 − 1 = 0. f2 (0) = N(0) − p(0)I l2 l1

(2.26)

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K. Tanuma

From Lemma 2.7, (1.34) and (2.23), it follows that a2 a f2 (φ) = −N(φ)2 + p(φ)2 I + 2 p(φ) a1 (φ) 1 , l2 l1 and from (2.26) and Theorem 1.5, f2 (φ) = − N(φ) + p(φ)I f2 (φ). Therefore, the uniqueness of the solution to the initial value problem implies that f2 (φ) = 0 for all φ. This proves (2.25) for α = 2. Suppose that (2.25) holds for α = 2, . . . , m − 1 (m ≥ 3). Then by (2.20), (2.23) and (2.24), a a m fm (0) = N(0) − p(0)I − m−1 = 0. lm lm−1 Moreover from Lemma 2.7, (1.34), (2.23), (2.24) and (2.26), it follows that a m−1 a m f m (φ) = − N(φ) + p(φ)I N(φ) − p(φ)I am−β (φ) β − lm lβ β=1

⎞ m−1 a β ⎠ am−β (φ) = − N(φ) + p(φ)I ⎝fm (φ) + lβ ⎛

+

m−2

β=1

⎛ ⎝2 p(φ)am−β (φ) +

β=1

+ 2 p(φ)a1 (φ)

m−β−1

γ =1

⎞

aβ ⎠ am−β−γ (φ)aγ (φ) lβ

am−1 , lm−1

and regrouping gives m−1 a β am−β (φ) N(φ) − p(φ)I f m (φ) = − N(φ) + p(φ)I fm (φ) − lβ β=1

+

m−2 m−β−1 β=1

γ =1

aβ am−β−γ (φ)aγ (φ) , lβ

and the hypothesis of induction leads us to β−1 m−1 a am−β (φ)aβ−γ (φ) γ f m (φ) = − N(φ) + p(φ)I fm (φ) − lγ β=2 γ =1

+

m−2 m−β−1 β=1

γ =1

aβ am−β−γ (φ)aγ (φ) , lβ


69

where Theorem 1.5 has been used for β = 1 in the second term. Since the last two terms cancel each other out by changing the order of summation in the second term β−1 m−1 β=2 γ =1

we have

=

m−2

m−1

,

γ =1 β=γ +1

f m (φ) = − N(φ) + p(φ)I fm (φ).

Thus, from the uniqueness of the solution to the initial value problem we obtain fm (φ) = 0 for all φ. Now we set

q(φ) = exp −

φ

2 p(ψ)dψ ,

0

r(φ) =

φ

q(ψ)dψ.

(2.27)

0

Lemma 2.9 The solutions a1 (φ), . . . , an−1 (φ) to (2.23) and (2.24) have the following form: aα (φ) = (−1)α−1 r(φ)α−1 r (φ),

α = 1, . . . , n − 1.

(2.28)

Proof Integrating (2.23), we get (2.28) for α = 1. Suppose that (2.28) holds for α = 1, . . . , m − 1. Solving (2.24), we have for m ≥ 2,

φ

am (φ) = −q(φ) 0

q(ψ)−1

m−1

aβ (ψ)am−β (ψ)dψ.

(2.29)

β=1

From the hypothesis of induction and r (ψ) = q(ψ), it follows that q(ψ)−1 aβ (ψ)am−β (ψ) = (−1)m−2 r(ψ)m−2 r (ψ). Thus

am (φ) = −q(φ)

φ

(m − 1)(−1)m−2 r(ψ)m−2 r (ψ)dψ = (−1)m−1 r(φ)m−1 r (φ).

0

Proof of Theorem 2.6 We take the angular average of both sides of (2.25) over [−π, π ]. By Lemma 1.12, the angular average of the left hand side becomes a √ α . S − −1 I lα From (1.71) and (1.72), it follows that r(π ) = r(−π ). Then Lemma 2.9 implies that π α−β φ=π α−β−1 r(φ) aα−β (φ)dφ = (−1) = 0, β = 1, 2, . . . , α − 1. α − β φ=−π −π This proves the theorem.

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K. Tanuma

2.2.3 Surface Impedance Tensor of Piezoelectricity

aα (α = 1, 2, 3, 4) lα be linearly independent eigenvector(s) or generalized eigenvector(s) of the eightdimensional eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, 3, 4, Im pα > 0). Here Q, R and T are taken to be the 4 × 4 matrices in (2.18). Corresponding to Theorem 1.16, we obtain Let m and n be two orthogonal unit vectors in R3 , and let

Theorem 2.10 The vectors aα (α = 1, 2, 3, 4) are linearly independent. Proof On the basis of Theorem 2.6, we can use the proof of Theorem 1.16. We only have to change the range where the index α runs from {1, 2, 3} to {1, 2, 3, 4}. To use the proof of Theorem 1.16 with this new index, it is sufficient to prove the invertibility of the 4 × 4 matrix S2 defined by (1.68) with the 4 × 4 matrix T(φ) given by (2.18) and (1.31). As in the proof of (1) of Lemma 2.5, we can write TC (φ) te (φ) I 0 , T(φ) = −teT (φ) T (φ) 0T −1 where

⎛ TC (φ) = ⎝

3

⎞ Cijkl n jnl ⎠

j,l=1

T (φ) =

3

i↓k→1,2,3

⎛ ,

te (φ) = ⎝

3 j,l=1

⎞ eijl n jnl ⎠

,

i↓1,2,3

jl n jnl

j,l=1

and n = (n1 , n2 , n3 ) is defined in (1.31). Then it follows that −1 π TC (φ) te (φ) I 0 1 . S2 = dφ 2π −π −teT (φ) T (φ) 0T −1 TC (φ) te (φ) By (2.11), the matrix and hence its inverse are positive definite for −teT (φ) T (φ) all the angular average on the right hand side is positive definite. Since φ. Therefore, I 0 is invertible, S2 is also invertible.

0T −1 Thus, we can define the surface impedance tensor of piezoelectricity. Definition 2.11 The surface impedance tensor Z of piezoelectricity is the 4 × 4 matrix given by √ Z = − −1 [l1 , l2 , l3 , l4 ][a1 , a2 , a3 , a4 ]−1 , (2.30) where [l1 , l2 , l3 , l4 ] and [a1 , a2 , a3 , a4 ] denote 4 × 4 matrices which consist of the column vectors lα and aα , respectively.


71

We also have the integral representation of Z. Theorem 2.12 Z = S−1 2 +

√

−1 S−1 2 S1 ,

(2.31)

where the 4 × 4 matrices S1 and S2 are defined by (1.68) with the 4 × 4 matrices Q(φ), R(φ) and T(φ) given by (2.18) and (1.31). Then the arguments in Subsection 2.1.1 can be applied to obtain Theorem 2.13 G(x) =

−1 1 Re Z 4π |x|

(2.32)

is a fundamental solution for piezoelectricity, where Z is the surface impedance tensor defined by (2.30) with m and n satisfying m×n=

x . |x|

2.2.4 Formula for Surface Impedance Tensor of Piezoelectricity: Example Using the algebraic formula (2.30) for the surface impedance tensor, we give the explicit form of Z for piezoelectric materials which have hexagonal symmetry 622. For such materials a Cartesian coordinate system can be chosen so that one coordinate axis is an axis of 6-fold rotational symmetry and the other two axes are 2-fold. Henceforth, we take the 3-axis to be the 6-fold axis and the 1- and π 2- axes to be the 2-fold axes. Then the orthogonal tensors Rx33 and Rπx2 are the generators of the symmetry group G for this piezoelectric material. It follows that for any orthogonal tensor Q = Qij i, j=1,2,3 in G , (1.10) holds for the elasticity tensor C = Cijkl i, j,k,l=1,2,3 , eijl =

3

Qip Q jq Qlr e pqr ,

i, j, l = 1, 2, 3

p,q,r=1

holds for the piezoelectric tensor e = eijl i, j,l=1,2,3 , and jl =

3

Q jp Qlq pq ,

j, l = 1, 2, 3

p,q=1

holds for the dielectric tensor = jl i,l=1,2,3 . Then the elasticity tensor C is given by (1.13), while the non-zero components of the piezoelectric tensor e and the dielectric tensor are proved to be e231 , e132 , 11 , 22 , 33 , and they satisfy e231 = −e132 ,

11 = 22 .

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K. Tanuma

Hereafter, for simplicity, we use the notations (1.100) and put e231 = S,

11 = V,

33 = W.

For the orthogonal unit vectors m and n in R3 , we write m × n = = (1 , 2 , 3 ). Theorem 2.14 The surface impedance tensor for piezoelectric materials of hexagonal symmetry 622 is given by T Z = Z ij i↓ j→1,2,3,4 = Z , ' L& (A − N)M 1 2 + AG 2 2 , D ' L& = (A − N)M 2 2 + AG 1 2 , D & 1 = ALGH − (A − N)M AL 3 2 + H D ' + AC − (F + L)2 1 2 + 2 2 3 2 ,

Z 11 = Z 22 Z 33

Z 44 =

Z 12 =

Z 13 =

Z 23 =

Z 14 = Z 24 = Z 34 =

& 1 −(A − N)V J M + AG (A − N)V 3 2 + J D ' + (A − N)W + 2S2 1 2 + 2 2 3 2 , L {(A − N)M − AG } 1 2 D √ ALGM 2 1 + 2 2 3 , + −1 (A − N) −1 + D −L(F + L)(A − N)M 1 2 + 2 2 1 3 D √ (F + L) 2 − −1 L −1 + 1 + 2 2 2 , D −L(F + L)(A − N)M 2 1 + 2 2 2 3 D √ (F + L) 2 2 1 + 2 + −1 L −1 + 1 , D √ 2ALSG 2 2L 2 1 + 2 2 2 3 + −1 S 1 − 1 + 2 2 1 , D D √ 2ALSG 2L 2 2 2 2 − 2 , 1 + 2 1 3 + −1 S 1 − 1 + 2 D D 2 √ 2L(F + L) 2 −1 S −1 + 3 , (2.33) 1 + 2 2 D


73

where G and H are the same as those in (1.104) and = (A − N)V3 4 + (A − N)W + 2S2 + 2LV 1 2 + 2 2 3 2 2 + 2LW 1 2 + 2 2 , ( J= (

, (A − N)V

2(A − N)V3 4 + (A − N)W + 2S2 + 2LV 1 2 + 2 2 M= (A − N)V √ +2 (A − N)V , = A 3 2 + H + L 1 2 + 2 2 , = (A − N) 3 2 + J + 2L 1 2 + 2 2 , D = − A(A − N)GM 3 2 . To obtain the formula above, it is important to see that the characteristic equation (2.19) can be factorized into

2 2 1 2 2 2 2 2 4 AL y1 + y2 + AC − F − 2F L y1 + y2 y3 + CLy3 2

2 × (A− N)V y21 + y22 + A − N W + 2S2 + 2LV y21 + y22 y23 + 2LWy43 , (2.34) where yi = mi + p ni (i = 1, 2, 3). We note that the first factor of (2.34) is equal to the quartic factor (1.112) of the characteristic equation for transversely isotropic elastic materials, while the second factor of (2.34) depends not only on the elasticity tensor C but also on the piezoelectric tensor e and the dielectric tensor . However, for piezoelectric materials which has hexagonal symmetry 622, it is proved that the conditions for the eight-dimensional eigenvalue problem (1.29) to be simple, semisimple and degenerate are precisely the same as those in Lemma 1.25. Hence the arguments in Subsection 1.8.2 applies in an almost parallel way. For the details of computations, we refer to [2]. When e231 = S = 0, equations (2.17) can be decomposed into the equilibrium equations for transversely isotropic elasticity and the scalar equation of dielectricity. In this case it can be checked by simple computations that the submatrix Z ij i↓ j→1,2,3 of the matrix (2.33) is reduced to the matrix (1.103) in Theorem 1.26. The formula for the fundamental solution can be obtained from Theorem 2.13.

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K. Tanuma

2.3 Inverse Boundary Value Problem 2.3.1 Dirichlet to Neumann Map Let be a bounded domain in R3 occupied by an inhomogeneous, linearly elastic medium. When the displacement f = ( f1 , f2 , f3 ) is given on the boundary ∂, the deformation of the elastic body due to f can be expressed in terms of the displacement u = u(x) = (u1 , u2 , u3 ) as

∂ ∂uk Cijkl = 0, ∂xj ∂ xl j,k,l=1 u = f . 3

i = 1, 2, 3 in ,

(2.35) (2.36)

∂

Here C(x) = Cijkl i, j,k,l=1,2,3 is the elasticity tensor21 and we assume that this tensor now depends on x ∈ . Using the solution u to (2.35) and (2.36), we define the traction t on the boundary ∂ by ⎛ ⎞ 3 ∂uk ⎠ , t=⎝ Cijkl nj (2.37) ∂ xl j,k,l=1

i↓1,2,3 ∂

where n = n(x) = (n1 , n2 , n3 ) is the unit outward normal to ∂ at x ∈ ∂. The boundary displacement f in (2.36) is the Dirichlet data, while the traction t in (2.37) is the Neumann data. We call the map :

f −→ t

(2.38)

the Dirichlet to Neumann map. Physically, ( f ) gives the traction on ∂ produced by the displacement f on ∂. When C(x) is given in and f is prescribed on ∂, the problem of finding the solution u to (2.35) and (2.36) is the forward problem. On the other hand, suppose that the elasticity tensor C(x) is unknown. In this section we are concerned with the following inverse problem, which has received a great deal of attention: Determine the elasticity tensor C(x) from the Dirichlet to Neumann map . This is the problem of determining the elasticity tensor from observations of the displacements on the boundary of the medium and the tractions needed to sustain them on that boundary. Inverse problems where observations are made on the boundary like this are called inverse boundary value problems.

21 Hence

C(x) = Cijkl i, j,k,l=1,2,3 satisfies the symmetry conditions (1.4), (1.6) and the strong convex-

ity condition (1.7) at each x ∈ .


75

We formulate the Dirichlet to Neumann map more precisely using the function spaces. Let ∈ R3 be a bounded domain with a Lipschitz boundary ∂ 22 and let 1 C(x) = Cijkl i, j,k,l=1,2,3 ∈ L∞ (), f ∈ H 2 (∂). Then the Dirichlet problem (2.35), (2.36) admits a unique weak solution u in H 1 ().23 Definition 2.15 The Dirichlet to Neumann map −1

1

H 2 (∂) f −→ ( f ) ∈ H 2 (∂) is defined by < ( f ), g > =

3

i, j,k,l=1

Cijkl

∂uk ∂vi dx ∂ xl ∂ x j

(2.39)

for g = (g1 , g2 , g3 ) ∈ H 1/2 (∂). Here u = (u1 , u2 , u3 ) is the solution to (2.35) and (2.36), v = (v1 , v2 , v3 ) is any function in H 1 () satisfying v|∂ = g and < , > is the bilinear pairing between H 1/2 (∂) and H −1/2 (∂). We note that the right hand side of (2.39) is independent of the choice of the extension v ∈ H 1 () of g. In fact, for another extension v˜ = (v˜1 , v˜2 , v˜3 ) of g we have v − v˜ ∈ H01 (). Then, by the divergence theorem, the difference between the corresponding right hand sides becomes

3

i, j,k,l=1

Cijkl

=−

∂uk ∂(vi − v˜i ) dx ∂ xl ∂xj

3 i, j,k,l=1

∂ ∂uk Cijkl (vi − v˜i ) dx, 24 ∂xj ∂ xl

which, by (2.35), is equal to zero. Hence (2.39) is well-defined.

22

is said to have a Lipschitz boundary ∂ if ∂ can be viewed locally as the graph of a Lipschitz function after an appropriate rotation of the coordinate axes. In the following we will give the function spaces which will be needed in this section. L∞ () : space of measurable functions in which are essentially bounded. α H 1 () : space of functions u in such that ∂∂xαu ∈ L2 () for all multi-index α with |α| ≤ 1. 1 1 H0 () : space of functions u ∈ H () such that u|∂ = 0. This space is equal to the closure in H 1 () α of Cc∞ () with the norm ||u|| H1 () = |α|≤1 || ∂∂xαu || L2 () , where Cc∞ () is the space of C∞ functions with compact support in . 1 H 2 (∂) : the trace of u ∈ H 1 () on ∂. −1 H () : the dual of H01 (). 1

1

H − 2 (∂) : the dual of H 2 (∂).

23 This is proved by Korn’s inequality which is derived from the convexity condition (1.7) and by the Lax-Milgram theorem. For the details we refer to [26, 28, 30]. ∞ ˜ since v − v˜ ∈ H01 (), there exists a sequence {φ n }∞ n=1 ⊂ Cc () such that φ n −→ v − v as n −→ ∞. Hence this right hand side means

3 ∂ ∂uk − lim Cijkl (φ n )i dx, n→∞ ∂xj ∂ xl

24 Precisely,

in

H 1 ()

i, j,k,l=1

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K. Tanuma

When f , g, C(x) and the boundary ∂ have enough regularity, from the divergence theorem and (2.35) again, it follows that < ( f ), g > = =

3 ∂ i, j,k,l=1

∂

Cijkl (x)

∂uk n j vi dσ ∂ xl

t(x) · g dσ,

where dσ denotes the surface element on ∂. This shows that (2.39) provides a weak formulation of (2.38). In this section we consider the problem of reconstructing elasticity tensors C(x) at the boundary ∂ from the localized Dirichlet to Neumann map for isotropic and for transversely isotropic materials. 2.3.2 Reconstruction of Elasticity Tensor Suppose that the elasticity tensor C(x) = Cijkl i, j,k,l=1,2,3 ∈ L∞ () is unknown. We consider the problem of reconstructing C(x) on the boundary ∂ from the localized Dirichlet to Neumann map. More precisely, for x0 ∈ ∂, assuming that ∂ is locally C1 near x0 and that C(x) is continuous around x0 , we input the Dirichlet data f compactly supported in a small neighborhood of x0 on ∂, measure the corresponding Neumann data ( f ) in that neighborhood, and from these observations we reconstruct C(x0 ). This reconstruction problem can be paraphrased as “local reconstruction from local observations”. This reconstruction can be achieved through the following two steps. First, from a countable pairing of f and ( f ) we reconstruct the surface impedance tensor Z in Section 1.7. Secondly, we reconstruct the components of C(x0 ) from Z. As Z expresses a linear relationship between a certain class of displacements given at the surface (Dirichlet data) and the tractions needed to sustain them at that surface (Neumann data) (see the paragraph after Definition 1.17), it is reasonable that Z can be reconstructed from f and ( f ). The explicit forms of Z have been written in terms of the elasticity tensors for isotropic and for transversely isotropic materials in Section 1.8. Hence in this subsection we consider the problem of reconstructing C(x0 ) of isotropic and of transversely isotropic materials. 2.3.2.1 Reconstruction of Surface Impedance Tensor from Localized Dirichlet to Neumann Map For simplicity we set x0 = 0. Let n = (n1 , n2 , n3 ) be the unit outward normal to ∂ at x = 0 ∈ ∂ and we choose an arbitrary unit tangent m = (m1 , m2 , m3 ) to ∂ at x = 0 so that m · n = m1 n1 + m2 n2 + m3 n3 = 0. We write m × n = = (1 , 2 , 3 ),

where

∂ ∂xj

denotes differentiation in the sense of distributions.


77

where m × n denotes the vector product. For x = (x1 , x2 , x3 ) ∈ we put y1 = m · x,

y2 = · x.

(2.40)

Since we have assumed that ∂ is locally C1 near x0 , there exists a C1 function φ(y1 , y2 ) near x = 0 such that , ∂ are given by ∂ = {n · x = φ(y1 , y2 )},

= {n · x < φ(y1 , y2 )}

locally around x = 0 and ∂φ = φ y1 = 0, ∂ y1

∂φ = φ y2 = 0 ∂ y2

at x = 0.

(2.41)

We set y3 = n · x − φ(y1 , y2 )

(2.42)

and introduce the new coordinates y = (y1 , y2 , y3 ) = (y , y3 ). Then , ∂ are given by ∂ = {y3 = 0},

= {y3 < 0}

(2.43)

locally around x = 0. Now we define the Dirichlet data compactly supported in a small neighborhood of x = 0 on ∂. Let η(y ) be a C1 function with compact support in R2 which satisfies η(y )2 dy = 1, supp η ⊂ {|y | < 1}. (2.44) 0 ≤ η ≤ 1, R2

For any vector v in C3 , define the function f N on ∂ by √ √ f N = η N y v e− −1 y1 N ∂ ,

(2.45)

where N is a natural number which is large enough.25 The function f N is localized around x = 0; its support becomes smaller and it oscillates more intensely as N becomes larger. We take f N as the Dirichlet data, which we use to reconstruct the surface impedance tensor Z at x = 0. Theorem 2.16 Assume that ∂ is locally C1 at x = 0 ∈ ∂ and that C(x) is continuous around x = 0. Letting v be an arbitrary vector in C3 , we define f N by (2.45). Then, lim < ( f N ), f N > = Z0 v, v C3 , (2.46) N→∞

where ( , )C3 denotes the inner product of two vectors in C3 3 Z0 v, v C3 = Z ij v j vi i, j=1

25 In the exponent, a parameter with the dimension of the reciprocal of length has been set equal to 1 (see (1.18) and its footnote). Hence N can be interpreted as the number of units with the dimension of the reciprocal of length.

78

K. Tanuma

and Z0 = (Z ij)i, j=1,2,3 is the surface impedance tensor defined by (1.74) with m and n given above and with C = Cijkl i, j,k,l=1,2,3 = C(0). By this theorem, all the components of Z0 = (Z ij)i, j=1,2,3 can be reconstructed from and a suitable choice of v’s in (2.45). Proof Depending on C(0) and vectors m and n, the eigenvalue problem (1.29) becomes simple, semisimple, or degenerate, and when it is degenerate, there are two Cases D1 and D2 (see Section 1.4). Let ζ (y3 ) ∈ C∞ ((−∞, 0]) satisfy 0 ≤ ζ ≤ 1, ζ (y3 ) = 1 for −1/2 ≤ y3 ≤ 0 and 0 for y3 ≤ −1. aα (α = 1, 2, 3) be linearly independent eigenlα vectors of the eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, 3, Im pα > 0). Then Theorem 1.16 implies that their displacement parts aα (α = 1, 2, 3) are linearly independent. Hence, for v in (2.45) there exist cα (α = 1, 2, 3) ∈ C such that

Simple or Semisimple case Let

v=

3

c α aα .

(2.47)

α=1

Taking account of (1.45), we set F N (x) = η

√

3 √ √ N y ζ N y3 cα aα e− −1(y1 + pα y3 )N ,

(2.48)

α=1

which is an extension of f N (y ) and is meant to be an approximate solution to (2.35) and (2.36) with f = f N in a neighborhood of x = 0 as N −→ +∞. From (2.39) it follows that 3 ∂(u N )k ∂ F N i < ( f N ), f N > = Cijkl (x) dx, (2.49) ∂ xl ∂xj i, j,k,l=1

where u N ∈ H 1 () is the solution to

3 ∂(u N )k ∂ Cijkl (x) = 0, i = 1, 2, 3 in ∂xj ∂ xl

,

u N |∂ = f N .

(2.50)

j,k,l=1

Let u N = FN + rN .

(2.51)

Then r N is the solution to

3 3 ∂(r N )k ∂(F N )k ∂ ∂ Cijkl (x) =− Cijkl (x) , ∂xj ∂ xl ∂xj ∂ xl j,k,l=1

j,k,l=1

i = 1, 2, 3 in r N |∂ = 0.

, (2.52)


79

Using (2.51), we recast the right hand side of (2.49) as 3 3 ∂(F N )k ∂ F N i ∂(r N )k ∂ F N i Cijkl (x) dx + Cijkl (x) dx ∂ xl ∂xj ∂ xl ∂xj i, j,k,l=1

i, j,k,l=1

and write < ( f N ), f N > = I + I I,

(2.53)

where I=

3 i, j,k,l=1

II =

3 i, j,k,l=1

Cijkl (x)

∂(F N )k ∂ F N i dx, ∂ xl ∂xj

∂(r N )k ∂ F N i Cijkl (x) dx. ∂ xl ∂xj

It is sufficient to show that lim I = Z0 v, v C3

(2.54)

lim I I = 0.

(2.55)

N→∞

and N→∞

First we note from (2.40), (2.42) and (2.48) that √ √ √ ∂F N = − −1 N η N y ζ N y3 ∂ xl ×

3

√ cα aα ml + pα (nl − φ y1 ml − φ y2 l ) e− −1(y1 + pα y3 )N

α=1

√ √ √ √ + N η y1 N y ml + η y2 N y l ζ N y3 +η ×

3

√

√ N y ζ N y3 (nl − φ y1 ml − φ y2 l )

√ −1(y1 + pα y3 )N

c α aα e −

,

α=1

and we write √ √ √ ∂F N ∼ − −1 N η N y ζ N y3 ∂ xl ×

3

√ cα aα ml + pα (nl − φ y1 ml − φ y2 l ) e− −1(y1 + pα y3 )N .

(2.56)

α=1

Here and hereafter we use the notation ∼ to indicate that we are retaining the dominant terms in N for large N.

80

K. Tanuma

Now we observe that lim I = lim I ,

N→∞

(2.57)

N→∞

where

I =

3

i, j,k,l=1

∂(F N )k ∂ F N i Cijkl (0) dx. ∂ xl ∂xj

In fact, after the change of variables x −→ y, it follows from (2.56) that

I −I∼ N

×

2

|y |≤ √1 N − √1N ≤y3 ≤0

3 α,β=1

3 2 √ 2 √ Ny ζ N y3 Cijkl (y) − Cijkl (0) η i, j,k,l=1

cα cβ (aα )k aβ i ml + pα (nl − φ y1 ml − φ y2 l )

√ × m j + pβ (n j − φ y1 m j − φ y2 j) e− −1( pα − pβ ) y3 N dy, where we have used the fact that the Jacobian related to the change of variables x −→ y equals 1. Then by introducing the scaling transformation zi =

√

N yi

(i = 1, 2),

z 3 = N y3 ,

(2.58)

we get I − I ∼

×

3

√|z |≤1 − N≤z3 ≤0

3 z z3 z3 2 − Cijkl (0) η(z )2 ζ √ Cijkl √ , N N N i, j,k,l=1

√ √ cα cβ (aα )k (aβ )i ml + pα nl − φ y1 (z / N) ml − φ y2 (z / N) l

α,β=1

√ √ √ × m j + pβ n j − φ y1 (z / N) m j − φ y2 (z / N) j e− −1( pα − pβ )z3 dz, where z = (z1 , z2 , z3 ) = (z , z3 ). Since Im pα − pβ > 0,

α, β = 1, 2, 3,

Lebesgue’s dominated convergence theorem and the continuity of C(x) = Cijkl (x) i, j,k,l=1,2,3 at x = 0 imply that the preceding integral tends to zero as N −→ ∞. This proves (2.57). To arrive at (2.54), we will prove lim I = Z0 v, v C3 .

N→∞

(2.59)


81

From the divergence theorem it follows that 3 ∂(F N )k Cijkl (0) F N ν j(x) dσ I = i ∂ xl ∂ i, j,k,l=1

−

3

i, j,k,l=1

Cijkl (0)

∂ 2 (F N )k F N dx. i ∂ x j∂ xl

We write I = I1 − I2 ,

(2.60)

where I1 =

3

∂ i, j,k,l=1

I2 =

3

i, j,k,l=1

Cijkl (0)

Cijkl (0)

∂(F N )k F N ν j(x) dσ, i ∂ xl

∂ 2 (F N )k F N dx, i ∂ x j∂ xl

(2.61)

(2.62)

and ν j(x) is the j-th component of the unit outward normal to ∂ at x ∈ ∂.26 Then from (2.43), (2.48), (2.56) and ζ (0) = 1, it follows that 3 √ √ 2 Cijkl (0) η N y I1 ∼ − −1 N i, j,k,l=1

×

3

|y |≤ √1N

cα cβ (aα )k (aβ )i ml + pα (nl − φ y1 ml − φ y2 l ) ν j(y ) J(y ) dy1 dy2 ,

α,β=1

where we have used the fact that the surface element dσ can be written from (2.40), (2.42) and (2.43) as 4 dσ = J(y ) dy1 dy2 , J(y ) = 1 + φ y1 2 + φ y2 2 . From the scaling transformation (2.58) and regrouping of the summation, it follows that I1 ∼ −

3 3 √ −1 cα cβ Cijkl (0) (aα )k (aβ )i

×

α,β=1

|z |≤1

i, j,k,l=1

√ √ η(z )2 ml + pα nl − φ y1 z / N ml − φ y2 z / N l

√ √ × ν j z / N J z / N dz . 26 Intuitively,

since F N is an approximate solution can expect that I2 will be to (2.35) near x = 0, we ∂(F N )k 3 νj is the corresponding j,k,l=1 Cijkl (0) ∂ x

smaller than I1 for large N. Moreover, since

l

i↓1,2,3

approximate traction on ∂, we can expect that I1 approximates the bilinear pairing of the Neumann and Dirichlet data.

82

K. Tanuma

Recalling (2.41) and (2.44), from Lebesgue’s dominated convergence theorem we get 3 3 √ cα cβ Cijkl (0) (aα )k (aβ )i (ml + pα nl )n j27 lim I1 = − −1

N→∞

α,β=1

√ = − −1

3 α,β=1

i, j,k,l=1

cα cβ [RT + pα T]aα , aβ C3

3 √ = − −1 cα cβ lα , aβ C3 , α,β=1

where we have appealed to (1.20) and (1.25) with C = Cijkl i, j,k,l=1,2,3 = C(0). Hence from (1.75) and (2.47) we obtain 3

lim I1 =

N→∞

α,β=1

cα cβ Z0 aα , aβ C3 = Z0 v, v C3 .

(2.63)

Next we proceed to prove lim I2 = 0.

(2.64)

N→∞

Differentiating (2.56) with respect to x j, we get √ √ ∂ 2FN ∼ −N 2 η N y ζ N y3 ∂ x j∂ xl ×

3

cα aα m j + pα (n j − φ y1 m j − φ y2 j)

α=1

√ × ml + pα (nl − φ y1 ml − φ y2 l ) e− −1(y1 + pα y3 )N .

Then it follows from (2.62) that 3

I2 ∼ −N 2

i, j,k,l=1

×

3

Cijkl (0)

|y |≤ √1 N − √1 ≤y3 ≤0 N

η

√

N y

2 √ 2 ζ N y3

cα cβ (aα )k (aβ )i m j + pα (n j − φ y1 m j − φ y2 j)

α,β=1

√ × ml + pα (nl − φ y1 ml − φ y2 l ) e− −1( pα − pβ )y3 N dy,

that n j denotes the j-th component of the outward unit normal to ∂ at x = 0 ∈ ∂. Hence ν j (x) tends to n j as x → 0.

27 Recall


83

and from the scaling transformation (2.58) and regrouping the summation it follows that

3 3 z3 2 2 I2 ∼ − cα cβ Cijkl (0) (aα )k (aβ )i η(z ) ζ √ √|z |≤1 N − N≤z3 ≤0 α,β=1 i, j,k,l=1 √ √ × m j + pα n j − φ y1 (z / N) m j − φ y2 (z / N) j √ √ √ × ml + pα nl − φ y1 (z / N) ml − φ y2 (z / N) l e− −1( pα − pβ )z3 dz. Recalling (2.41), (2.44) and ζ (0) = 1, from Lebesgue’s dominated convergence theorem we get lim I2 = −

N→∞

3

cα cβ

α,β=1

×

0 −∞

3

Cijkl (0) (aα )k (aβ )i (m j + pα n j)(ml + pα nl )

i, j,k,l=1 √ −1( pα − pβ )z3

e−

dz3 ,

and from (1.20) =−

3 α,β=1

cα cβ [Q + pα (R + RT ) + p2α T]aα , aβ C3

0 −∞

√ −1( pα − pβ )z3

e−

dz3 ,

which, by (1.21), is equal to zero. This proves (2.64). Therefore, from (2.57), (2.60) and (2.63), we obtain (2.54). Formula (2.55) can be proved essentially in a parallel way to [15, 63] (see also [55]). We give an outline of the proof: Since F N is an approximate solution to (2.50), i.e., F N is the dominant term of the solution u N for large N, r N can be taken as the remainder term. Then it can be expected that the integral I I is of less order in N than the integral I for large N. In fact, applying the Lax-Milgram theorem to (2.52), we can estimate ||r N || H01 () from above by the H −1 () norm of the right hand side of (2.52). Then the integral I I can be estimated to get (2.55). For the details we refer to the references above.

aα (α = 1, 2) be linearly independent eigenvectors of lα the eigenvalue problem (1.29) associated with the eigenvalues pα (α = 1, 2, Im pα > 0) a3 be a generalized eigenvector satisfying and l3 a a a3 − p2 3 = 2 . N l3 l3 l2 Degenerate Case D1 Let

Then Theorem 1.16 implies that their displacement parts aα (α = 1, 2, 3) are linearly independent. Hence, for v in (2.45) there exist cα (α = 1, 2, 3) ∈ C such that v=

3 α=1

c α aα .

(2.65)

84

K. Tanuma

Taking account of (1.48), we set

F N (x) = η

√

- 3 √ √ N y ζ N y3 cα aα e− −1(y1 + pα y3 )N

α=1

−

√

−1 c3 y3 N a2 e

√ − −1(y1 + p2 y3 )N

( p2 = p3 ).

(2.66)

Then F N (x) is an extension of f N (y ). Hence equations (2.49) to (2.53) also apply in this case and we will prove (2.54).28 From (2.40), (2.42) and (2.66) we get √ √ √ ∂F N ∼ − −1 N η N y ζ N y3 ∂ xl 3 √ × cα aα ml + pα (nl − φ y1 ml − φ y2 l ) e− −1(y1 + pα y3 )N

(2.67)

α=1

√ + c3 a2 nl − φ y1 ml − φ y2 l − −1 ml + p2 (nl − φ y1 ml − φ y2 l ) Ny3 × e

√ − −1(y1 + p2 y3 )N

.

We note that under the scaling transformation (2.58) the right hand side is of first order in N. Hence (2.57) also applies and we will prove (2.59). We divide I into the two integrals I1 and I2 as in (2.60), (2.61), (2.62) and will prove (2.63) and (2.64). Since it follows from (2.43), (2.66) and (2.67) that

F N |∂ = η

√

N y

3

√ −1 y1 N

c α aα e −

α=1

and 3 √ √ ∂F N ∼ − −1 N η N y cα aα ml + pα (nl − φ y1 ml − φ y2 l ) ∂ xl ∂ α=1 −√−1 y1 N + c3 a2 nl − φ y1 ml − φ y2 l e ,

28 F

N (x)

is meant to be an approximate solution to (2.50) in a neighborhood of x = 0 as N −→ +∞.


85

we get 3 √ I1 ∼ − −1 N Cijkl (0) ⎡ ×⎣

|y

i, j,k,l=1 3

|≤ √1N

η

√

N y

2

cα cβ (aα )k (aβ )i ml + pα (nl − φ y1 ml − φ y2 l )

α,β=1

+ c3

3

⎤ cβ (a2 )k (aβ )i (nl − φ y1 ml − φ y2 l )⎦ ν j(y ) J(y ) dy1 dy2 .

β=1

From the scaling transformation (2.58) and regrouping of the summation, it follows that 3 3 √ I1 ∼ − −1 cα cβ Cijkl (0) (aα )k (aβ )i

×

α,β=1

i, j,k,l=1

|z |≤1

√ √ η(z )2 ml + pα nl − φ y1 (z / N) ml − φ y2 (z / N) l

√ √ × ν j z / N J z / N dz 3 3 √ − −1 c3 cβ Cijkl (0) (a2 )k (aβ )i β=1

×

i, j,k,l=1

|z |≤1

√ √ η(z )2 nl − φ y1 z / N ml − φ y2 z / N l

√ √ × ν j z / N J z / N dz . Then Lebesgue’s dominated convergence theorem implies that 3 3 √ cα cβ Cijkl (0) (aα )k (aβ )i (ml + pα nl )n j lim I1 = − −1

N→∞

α,β=1

i, j,k,l=1

3 3 √ − −1 c3 cβ Cijkl (0) (a2 )k (aβ )i nl n j β=1

i, j,k,l=1

3 √ = − −1 cα cβ [RT + pα T]aα , aβ C3 α,β=1

3 √ − −1 c3 c β T a2 , aβ C 3 β=1

3 √ = − −1 cα cβ lα , aβ C3 , α,β=1

86

K. Tanuma

where we have appealed to (1.20), (1.25) and (1.50) with C = Cijkl i, j,k,l=1,2,3 = C(0). Hence from (1.75) and (2.65) we obtain lim I1 =

N→∞

3 α,β=1

cα cβ Z0 aα , aβ C3 = Z0 v, v C3 .

Next we prove (2.64). From (2.66) we get √ √ ∂ 2FN ∼ − N 2 η N y ζ N y3 ∂ x j∂ xl 3 × cα aα m j + pα (n j − φ y1 m j − φ y2 j) α=1

√ × ml + pα (nl − φ y1 ml − φ y2 l ) e− −1(y1 + pα y3 )N + c3 a2 (n j − φ y1 m j − φ y2 j) ml + p2 (nl − φ y1 ml − φ y2 l )

+ m j + p2 (n j − φ y1 m j − φ y2 j) (nl − φ y1 ml − φ y2 l ) √ − −1 Ny3 m j + p2 (n j − φ y1 m j − φ y2 j)

−√−1(y1 + p2 y3 )N × ml + p2 (nl − φ y1 ml − φ y2 l ) e . Then under the scaling transformation (2.58) it follows from (2.41) and ζ (0) = 1 that 3

Cijkl (0)

j,k,l=1

⎡ ×⎣

3

∂ 2 (F N )k ∼ −N 2 η(z ) ∂ x j∂ xl 3

cα

α=1

√ −1(y1 + pα y3 )N

Cijkl (0) (aα )k (m j + pα n j)(ml + pα nl )e−

j,k,l=1 3

+ c3

Cijkl (0) (a2 )k n j(ml + p2 nl ) + (m j + p2 n j)nl

j,k,l=1

−

√

⎤ √ − −1(y1 + p2 y3 )N ⎦ −1 z3 (m j + p2 n j)(ml + p2 nl ) e -

= −N η(z ) 2

3

√ cα Q + pα (R + RT ) + p2α T aα e− −1(y1 + pα y3 )N

α=1

√ + c3 R+ RT+2 p2 T− −1 z3 Q+ p2 R +RT + p22 T × a2 e

√ − −1(y1 + p2 y3 )N

,


87

where we have used (1.20). From (1.21) for a = a1 , a2 and p2 = p3 , we have 3 j,k,l=1

×

Cijkl (0)

∂ 2 (F N )k ∼ −N 2 η(z ) c3 ∂ x j∂ xl

√ Q + p2 (R + RT ) + p22 T a3 + R + RT + 2 p2 T a2 e− −1(y1 + p2 y3 )N ,

which is equal to zero by (1.52). We apply this to the integrand of I2 as in the simple or semisimple case, which leads to lim I2 = 0.

N→∞

Formula (2.55) can also be proved in the same way as in the simple or semisimple case (see the last paragraph of this case on page 83). aα The degenerate case D2 remains to be considered. In this case, let (α = lα 1, 2, 3) be linearly independent eigenvector and generalized eigenvectors in Lemma 1.7. Then taking account of (1.55), we set √ √ F N (x) = η N y ζ N y3 . - 3 √ √ 1 2 cα aα − −1 y3 N(c2 a1 +c3 a2 )− (y3 N) c3 a1 e− −1(y1 + p1 y3 )N × 2 α=1 (2.68) as an extension of f N (y ), where cα (α = 1, 2, 3) ∈ C are chosen so that v = 3 α=1 cα aα . The proof is parallel to the previous cases, where we need the formulas obtained in the proof of Lemma 1.7. We leave the details as Exercise 2-4.

2.3.2.2 Reconstruction of Elasticity Tensor from Surface Impedance Tensor By Theorem 2.16, we can reconstruct all the components of the surface impedance tensor Z0 at x = 0 ∈ ∂ from the localized Dirichlet to Neumann map . The explicit formulas of Z0 are obtained for isotropic and transversely isotropic elastic materials in Section 1.8. Here we shall restrict our attention to the problem of reconstructing the components of C(0) from Z0 for these materials. In the process of reconstruction we choose some different unit tangents m in (2.45). Theorem 2.17 When the elastic material is isotropic, all the components of C(0), i.e., the two independent parameters λ and μ at x = 0, can be reconstructed algebraically from Z0 . Proof We take m so that m × n = = (1 , 2 , 0), Then from Theorem 1.24 it follows that Z 33 =

2μ(λ + 2μ) λ + 3μ

88

K. Tanuma

and Im Z 13 =

2μ2 2 , λ + 3μ

Im Z 23 =

−2μ2 1 . λ + 3μ

Since it never holds that 1 = 2 = 0 under 3 = 0, the quantity a1 =

2μ2 λ + 3μ

is recovered from Im Z 13 or Im Z 23 . Since λ (λ + 2μ) Z 33 = + 2, = a1 μ μ the quantity a2 =

λ μ

is recovered from Z 33 and a1 . Since Z 33 = 2μ

a2 + 2 , a2 + 3

μ is reconstructed from Z 33 and a2 . Then λ = μ a2

is reconstructed.

Theorem 2.18 Let the elastic material in question be transversely isotropic. Suppose that the normal to ∂ at x = 0 does not coincide with the direction of the axis of rotational symmetry of the material at x = 0. Then all the components of C(0) (5 independent components) can be reconstructed algebraically from Z0 . Proof (First Step) We shall recover the following four quantities: " √ " √ √ L(A − N) AL , a2 = AC − F AC + F + 2L , a1 = √ 2 L + AC √ " L F − AC C . (2.69) a3 = , a4 = √ A L + AC We take m so that m × n = = (1 , 2 , 0). Then from Theorem 1.26, taking account that 21 + 22 = 1, we have Z 11 = a1 21 + a2 22 , Z 12 = (a1 − a2 ) 1 2 ,

Z 22 = a1 22 + a2 21 , √ Z 13 = − −1 a4 2 ,

Z 33 = a2 a3 , √ Z 23 = −1 a4 1 .

We easily see that a1 and a2 can be recovered from Z 11 and Z 22 when 21 = 22 and from Z 11 and Z 12 when 21 = 22 (= 1/2). Furthermore, since the strong convexity


89

condition (1.101) implies that a2 = 0, a3 is recovered from Z 33 and a4 is recovered from Z 13 or Z 23 . Recall that in Theorem 1.26 we have taken the axis of rotational symmetry to be the 3-axis. If the unit outward normal n to ∂ at x = 0 coincides with (0, 0, ±1), we necessarily have 3 = 0, and from the above step only the four quantities ai (i = 1, 2, 3, 4) can be determined from Z0 . There are five independent components A, C, F, L, N to be reconstructed. Hence it is impossible to reconstruct these five independent components at such a point. (Second Step) We reconstruct the parameter L of the elastic tensor. We take m so that 3 is not equal to zero but very close to zero. The formulae for Z 13 and Z 23 in Theorem 1.26 are rewritten as √ √ Z 23 = −b 1 21 + 22 2 3 + −1 b 2 1 , Z 13 = −b 1 21 + 22 1 3 − −1 b 2 2 , where L(F + L) , b1 = D

F+L 2 2 b 2 = L −1 + K 1 + 2 . D

Since b 1 and b 2 are real and 3 = 0, ±1, we see that b 1 and b 2 can be recovered from Z 13 or Z 23 . Moreover, it follows that (2.70) −L + b 1 K 21 + 22 = b 2 . On the other hand, we have K2 = 23 +

2L 2 L2 1 + 22 = 23 + 2 21 + 22 , A− N a1

and combining this with (2.70) we obtain a quadratic equation for L +

1−

b1 a1

,

2 3 2 2 2 1 + 2 L2 + 2b 2 L + b 22 − b 21 21 + 22 23 = 0,

(2.71)

from which L can be reconstructed. (Third Step) We proceed to reconstruct the other parameters A, C, F and N of the elastic tensor. From (2.69) we have (√ " " √ √ a4 AL AC . a2 = AC − F AC + F + 2L = −a4 2 + √ a3 L L + AC √

Since a2 ,√a3 , a4 and L have already been recovered in the previous steps, we can ) recover AC from the above. Using a3 = C/A again, we reconstruct A and C. Finally F and N are reconstructed from a4 and a1 in (2.69).

Thus, by Theorem 2.16, Theorem 2.17 and Theorem 2.18, reconstruction of C(x0 ), x0 ∈ ∂ from the Dirichlet to Neumann map localized around x0 is completed for isotropic and transversely isotropic elastic materials.

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K. Tanuma

2.4 Comments and References The fundamental solution G(x) for general anisotropic elasticity has been studied by many authors and from various approaches, and many expressions of G(x) have been obtained. The attempt to write down the fundamental solution from the standpoint of the Stroh formalism goes back to [45] and can be found in a fairly unified manner in [40]. We emphasize that Theorem 2.2 and Theorem 2.13 can be applied to any class of elastic materials (and also of piezoelectric materials), regardless of whether the corresponding Stroh’s eigenvalue problem is simple, semisimple or degenerate and irrespective of the length of the Jordan chain in question. The word “degenerate” means that Stroh’s eigenvalue problem has multiple eigenvalues, which also implies that the contour integral in the complex plane derived from formula (2.2) has multiple poles in its integrand. Thus degenerate cases need special consideration. Nevertheless, it is important to study degenerate cases because degeneracy can occur very often from additional symmetries that crystals have. The standpoint here is to give the formula of G(x) with a mathematically rigorous proof. Thus, from Theorem 2.2, we obtain by a consistent method for isotropic and transversely isotropic materials explicit forms of G(x) written in terms of their elasticity tensors. It should be pointed out that using Cauchy’s theory of residues, Ting and Lee [84] gave the formula of G(x) in terms of the eigenvalues of Stroh’s eigenvalue problem, which can also be applied to any class of elastic materials. For transversely isotropic materials, several efforts have been made to obtain an explicit form of the fundamental solution (see the references in [49, 54, 84]). The formula in Corollary 2.4 is a result from [54] and is given in a unified expression which is generally valid, regardless of whether the eigenvalue problem (1.29) is degenerate or not.29 On the other hand, as we have stated in the note of Corollary 2.4, for x = 0 such that * x3 = ±1, i.e., * x 21 + * x 22 = 0, we must take the limit * x 21 + * x 22 −→ 0 in (2.5) to obtain GTrans (x). Ting and Lee [84] derived the formula which is uniformly valid regardless of whether * x 21 + * x 22 = 0 or not. The components S 011 and S 022 in [84] are written as L cos2 α + C sin2 α 2 + (A − N)K ALGH - . 2 H + cos2 α 2GH C sin2 α 1 + + , − G(H + GK + K2 ) A− N (A − N)K AL - . 2 H + cos2 α 1 2GH C sin2 α 1 + + + , = AG G(H + GK + K2 ) A− N (A − N)K AL

0 S11 =

0 S22

0 0 and the other S ij0 ’s are the same as those in (2.8). We see that S11 and S22 given in the 2 2 2 x1 +* x 2 −→ 0. expressions above are not indeterminant forms when sin α = * Rudimentary accounts of the theory of piezoelectricity are given, for example, in [62, 75]. Barnett and Lothe [11] extended the Stroh six-dimensional formalism for elasticity to the eight-dimensional formalism for piezoelectricity. Lemma 2.8 and

29 After

the work in [54] we have noticed that in Corollary 2.4 is simpler than that in [54].

D D

=

1 LGH

with D given therein. Hence the formula


91

Lemma 2.9 are the results reported in [2]. These lemmas lead to a mathematically rigorous proof of Theorem 2.6, which is a fundamental theorem in the integral formalism in a higher dimension ([42, 43]). The Stroh formalism has subsequently been extended [4, 5, 25] to cover anisotropic media which are at once piezoelectric, piezomagnetic, and magnetoelectric. It was Calderón [16] who first posed the inverse boundary value problem for the conductivity equation, which can be taken as a mathematical formulation of electrical impedance tomography. Let be a bounded domain in R3 . Given a strictly positive conductivity γ (x) on , we apply a voltage potential f on the boundary ∂ and solve the Dirichlet problem

3 ∂u ∂ γ (x) = 0, in , u∂ = f, ∂ x ∂ x j j j=1 where u = u(x) is the electric voltage potential. The resulting current density distribution on ∂ is 3 ∂u ( f ) = γ (x) n j , ∂xj j=1 ∂

where n = (n1 , n2 , n3 ) is the unit outward normal to ∂. Then ( f ) gives the current distributions on ∂ (the Neumann data) produced by the voltage distributions on ∂ (the Dirichlet data), and we call the Dirichlet to Neumann map for the conductivity equation. The inverse boundary value problem in question is to determine the conductivity from observations of the voltage distributions and the resulting current distributions on ∂. Mathematically this problem can be formulated as follows: Determine γ (x) from . Together with the uniqueness and the stability problems associated with it, the problem of obtaining the formulas for reconstructing γ (x) from has been studied by many authors (for example, [15, 51, 55, 67]). For an overview of these results as well as other works in inverse boundary value problems, we refer to [85]. The inverse boundary value problem considered in Section 2.3 is the analog of Calderón’s problem in elasticity. However, many new difficulties arise because in elasticity the equations of equilibrium constitute an elliptic system. For isotropic elasticity, the problem of determining and reconstructing λ and μ from is studied in [1, 58–60]. For anisotropic elasticity, there are works [53, 57, 60, 61] which use the facts that the Dirichlet to Neumann map (2.38) is a pseudodifferential operator of order 1 and that its full symbol (i.e., all the terms in the asymptotic expansion of the symbol of the pseudodifferential operator) can be written in terms of the surface impedance tensor and its derivatives. In these works, the boundary and the elasticity tensor are assumed to be C∞ at every point. For transversely isotropic materials, Nakamura, Tanuma and Uhlmann [57] proved that the elasticity tensor and all its derivatives at the boundary can be reconstructed from the full symbol of the Dirichlet to Neumann map. Here using the full symbol of the Dirichlet to Neumann map corresponds to making measurements of the tractions on the boundary pertaining to all possible displacements on that boundary. In [57, 61], a layer stripping algorithm was also developed for elastic media, which gives approximation for the elasticity tensor beneath the boundary surface.

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In Section 2.3, we have considered the problem of reconstructing the elasticity tensor at the boundary from the localized Dirichlet to Neumann map for isotropic and transversely isotropic materials, which is part of the results in [56]. We have used explicit Dirichlet data (2.45), which are compactly supported in a neighborhood of a boundary point. The choice of these data is largely inspired by the results in Chapter 1. In fact, their extensions F N ∈ H 1 () in (2.48) and (2.66) have almost √ the same forms as (1.45) and (1.48), respectively, except for the cutoff functions η( N y ) √ and ζ ( N y3 ). It should be noted that the reconstruction method in Subsection 2.3.2 is along the same line as the studies in [15, 63]. In [63], the Lamé coefficient μ and the residual stress in a highly simplified model of elasticity are reconstructed from the localized Dirichlet to Neumann map. For a general anisotropic material, the surface impedance tensor is reconstructed from the Dirichlet to Neumann map by Theorem 2.16, but the surface impedance tensor is not written explicitly in terms of the elasticity tensor. In this case, we may appeal to the numerical approach. Namely, we compute the approximate surface impedance tensors from the estimated values of the elasticity tensor through the integral representation (1.76) and compare them with the surface impedance tensors reconstructed from the Dirichlet to Neumann map. We may repeat this procedure to update the elasticity tensor. Recall that the surface impedance tensor is a function of = m × n. Although a surface impedance tensor, which is Hermitian, provides only 9 equations for the components of the elasticity tensor, we can use as many surface impedance tensors as we please by choosing different unit tangents m at a fixed point on the boundary. The study on reconstruction of the elasticity tensor beneath the boundary surface from the localized Dirichlet to Neumann map is still in progress [56]. As a first step in this approach, reconstruction of the normal derivatives of the elasticity tensor at the boundary from the localized Dirichlet to Neumann map should be considered. There are many other problems for which the Stroh formalism is useful. These are problems of cracks, dislocations, inclusions and holes in anisotropic elastic media, and anisotropic elastic wedges and composites. For a review of these problems we refer to [77, 80].

2.5 Exercises 2-1 Following the suggestions below, derive the formula for GIso (x) directly from the integral in (2.2). (a) For m = m(φ), n = n(φ) given by (1.31), we put

x m × n = = e 1 × e2 = . |x| Show that T(φ) = (λ + 2μ)n ⊗ n + μ(m ⊗ m + ⊗ ),


93

where the tensor product m ⊗ n can be expressed as a 3 × 3 matrix as m ⊗ n = mi n j i↓ j→1,2,3 .30 (b) Write T(φ)−1 in terms of the orthonormal basis {e1 , e2 , } in R3 . (c) Use the equalities π π π cos2 φ dφ = sin2 φ dφ = π, cos φ sin φ dφ = 0, −π

−π

−π

and ⊗=

xi x j |x|2

to derive the formula for GIso (x). 2-2 Is it possible to prove Corollary 1.11 for Jordan chains of arbitrary length by using Lemma 2.8 ? 2-3 (Inverse boundary value problem for two-dimensional elasticity [53]) We consider the inverse boundary value problem for two-dimensional elasticity given in Exercise 1-12. If we change the ranges where the indices i, j, k, l run from {1, 2, 3} to {1, 2}, then the arguments in Subsection 2.3.1 apply in a parallel way. Hence we can define the Dirichlet to Neumann map for two-dimensional elasticity. Let n = (n1 , n2 ) be the unit outward normal to ∂ at x = 0 ∈ ∂ and m = (m1 , m2 ) be a unit tangent there. We put y = m · x = m1 x1 + m2 x2 . Let η(y) be a C1 function with compact support in R which satisfies η(y)2 dy = 1, supp η ⊂ {|y| < 1}. 0 ≤ η ≤ 1, R

For any vector v in C , we define the Dirichlet data f N on ∂ by √ √ f N = η N y v e− −1 yN ∂ , 2

(2.73)

where N is a natural number which is large enough. (a) Show that lim N

N→∞

−1 2

2 < ( f N ), f N > = Z0 v, v C2 = Z ij v j vi , i, j=1

30 Use the following properties of the tensor product in this exercise. Let {v1 , v2 , v3 } be an orthonormal basis in R3 . Then

v1 ⊗ v1 + v2 ⊗ v2 + v3 ⊗ v3 = I (identity matrix). For real a, b , c = 0 −1 1 1 1 a v1 ⊗ v1 + b v2 ⊗ v2 + c v3 ⊗ v3 = v1 ⊗ v1 + v2 ⊗ v2 + v3 ⊗ v3 . a b c

(2.72)

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K. Tanuma

where Z0 = (Z ij)i, j=1,2,3 is the surface impedance tensor obtained in Exercise 1-12 with (1.145) evaluated at x = 0 ∈ ∂. (b) Would it be possible to reconstruct the components of (1.145) from Z0 ? (From this we shall see that the rotational invariance of the surface impedance tensor reduces the information that the Dirichlet to Neumann map pertaining to Dirichlet data (2.73) carries about the elasticity tensor.) (c) How about the case of isotropic elasticity? (Two-dimensional isotropic elasticity is characterized by A = C = λ + 2μ, F = λ, L = μ) 2-4 Use (2.68) to prove (2.54) when the eigenvalue problem (1.29) is degenerate and the length of the Jordan chain is three.


95

3 Rayleigh Waves in the Stroh Formalism 3.1 The Stroh Formalism for Dynamic Elasticity In this Chapter we study dynamic deformations of the elastic medium. In the Cartesian coordinates (x1 , x2 , x3 ), the equations of motion with zero body force can be written, by adding the acceleration term to the right hand side of the equilibrium equations (1.8), as 3 ∂2 ∂ σij = ρ 2 ui , ∂xj ∂t j=1

i = 1, 2, 3,

(3.1)

where σ = σij i, j=1,2,3 is the stress tensor, ρ is the uniform mass density, t is the time, and u = u(x, t) = (u1 , u2 , u3 ) is the displacement at the place x = (x1 , x2 , x3 ) at time t. By (1.5), the equations of motion (3.1) can be written in terms of the displacement u as 3 j,k,l=1

∂2 ∂ ∂uk Cijkl = ρ 2 ui , ∂xj ∂ xl ∂t

i = 1, 2, 3.

(3.2)

Here C = Cijkl i, j,k,l=1,2,3 is the elasticity tensor, which has the minor symmetries (1.4) and the major symmetry (1.6) and satisfies the strong convexity condition (1.7). Throughout this chapter we assume that elastic medium in question is homo the geneous. Then the elasticity tensor C = Cijkl i, j,k,l=1,2,3 is independent of x, so that (3.2) becomes 3 j,k,l=1

Cijkl

∂2 ∂ 2 uk = ρ 2 ui , ∂ x j∂ xl ∂t

i = 1, 2, 3.

(3.3)

Let x = (x1 , x2 , x3 ) be the position vector and let m = (m1 , m2 , m3 ), n = (n1 , n2 , n3 ) be orthogonal unit vectors in R3 . We consider motions of a homogeneous elastic medium which occupies the half-space n · x = n1 x1 + n2 x2 + n3 x3 ≤ 0, and seek solutions to (3.3) of the form √ −1 k(m·x+ p n·x−v t)

u = (u1 , u2 , u3 ) = a e−

∈ C3

(3.4)

in n · x ≤ 0. When Im p, the imaginary part of p ∈ C, is positive, a solution of the preceding form describes a surface wave in n · x ≤ 0; it propagates along the surface n·x = 0 in the direction of m with the wave number k and the phase velocity v > 0, has the polarization defined by a constant vector a, and decays exponentially as n · x −→ −∞. Let us determine the equations that a ∈ C3 and p ∈ C in (3.4) must satisfy. Substituting (3.4) into (3.3) and noting that √ √ ∂u = − −1 k(m j + pn j) a e− −1 k(m·x+ p n·x−v t) ∂xj

96

K. Tanuma

and √ ∂u √ = −1 k v a e− −1 k(m·x+ p n·x−v t) , ∂t

we get

⎛ ⎝

3

⎞ Cijkl (m j + pn j)(ml + pnl ) − ρ v δik ⎠ 2

j,l=1

⎛ =⎝

⎛

3

Cijkl m jml − ρ v 2 δik + p ⎝

j,l=1

+ p2

3

i↓k→1,2,3 3

Cijkl m jnl +

j,l=1

⎞ Cijkl n jnl ⎠

j,l=1

a ⎞

3

Cijkl n jml ⎠

j,l=1

a = 0,

(3.5)

i↓k→1,2,3

where δik is Kronecker’s delta symbol. Now we introduce the dynamic elastic coefficients d = Cijkl − ρ v 2 m jml δik , Cijkl

and the 3 × 3 real matrices ⎞ ⎛ 3 d Cijkl m jml ⎠ Q=⎝ j,l=1

⎛ T=⎝

3

⎛

n jnl ⎠

j,l=1

R=⎝

,

3

(3.6)

⎞ d Cijkl m jnl ⎠

j,l=1

i↓k→1,2,3

⎞ d Cijkl

i, j, k, l = 1, 2, 3

,

i↓k→1,2,3

.

(3.7)

i↓k→1,2,3

Since m and n are orthonormal, we have 3

3

m2j = 1,

j=1

j=1

j,l=1

R=⎝

3 j,l=1

i↓k→1,2,3

(3.8)

,

i↓k→1,2,3

⎞ Cijkl m jnl ⎠

m jn j = 0.

j=1

Then it follows from (3.6) that ⎞ ⎛ 3 Cijkl m jml − ρ v 2 δik ⎠ Q=⎝ ⎛

3

n2j = 1,

,

⎛

T=⎝

(3.9) 3

⎞ Cijkl n jnl ⎠

j,l=1

.

i↓k→1,2,3

Therefore (3.5) can be written as [Q + p(R + RT ) + p2 T]a ⎛ ⎞ 3 d =⎝ Cijkl (m j + pn j)(ml + pnl )⎠ j,l=1

i↓k→1,2,3

a = 0.

(3.10)


97

For the existence of a = 0 that satisfies (3.10), we observe that p must be a solution to the sextic equation det[Q + p(R + RT ) + p2 T] ⎛ ⎞ 3 d = det ⎝ Cijkl (m j + pn j)(ml + pnl )⎠ j,l=1

= 0.

(3.11)

i↓k→1,2,3

Next we give the traction t on the surface n · x = 0 produced by the surface-wave solution (3.4). From (1.24), (3.4) and (3.9) it follows that ⎛ ⎞ 3 ∂u k ⎝ ⎠ t= Cijkl nj ∂ xl j,k,l=1 i↓1,2,3 n·x=0

⎛

⎞ 3 √ = − −1 k ⎝ Cijkl (ml + pnl ) n j⎠ j,l=1

√ −1 k(m·x−v t)

a e−

i↓k→1,2,3

√ √ = − −1 k [RT + pT] a e− −1 k(m·x−v t) .

Hence we define a vector l ∈ C3 as

Then

l = [RT + pT] a.

(3.12)

√ √ t = − −1 k l e− −1 k(m·x−v t)

(3.13)

is the traction on the surface n · x = 0 produced by (3.4). Equations (3.10) and (3.12) have exactly the same form as (1.21) and (1.25) in static elasticity, respectively. Therefore, (3.10) and (3.12) can be recast in Stroh’s sixdimensional eigenrelation a a N =p , (3.14) l l where N is the 6 × 6 real matrix defined by −T−1 RT T−1 N= −Q + RT−1 RT −RT−1

(3.15)

and Q, R and T are defined by (3.7). d When v = 0, it follows that Cijkl = Cijkl and the solutions pα (1 ≤ α ≤ 6) to (3.11), i.e., the eigenvalues of N, are not real (Lemma 1.1). As v increases from v = 0, at some point equation (3.11) ceases to have only complex roots. Since we are concerned with the surface-wave solution (3.4), where Im p > 0, we shall restrict our attention to the range of v for which all the solutions to (3.11) are complex. / = (/ /2 , m /3 ) and / Let m m1 , m n = (/ n1 ,/ n2 ,/ n3 ) be orthogonal unit vectors in R3 which are obtained by rotating the orthogonal unit vectors m and n around their vector product m × n by an angle φ (−π ≤ φ < π ) so that /=m / (φ) = m cos φ + n sin φ, m

/ n =/ n(φ) = −m sin φ + n cos φ.

(3.16)

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K. Tanuma

Let Q(φ), R(φ) and T(φ) be the 3 × 3 real matrices given by ⎛ ⎞ ⎞ ⎛ 3 3 d d / jm /l ⎠ / j/ Q(φ) = ⎝ m m nl ⎠ Cijkl , R(φ) = ⎝ Cijkl j,l=1

⎛ T(φ) = ⎝

3

⎞

i↓k→1,2,3

d / n j/ nl ⎠ Cijkl

j,l=1

j,l=1

,

,

i↓k→1,2,3

(3.17)

i↓k→1,2,3

d (i, j, k, l = 1, 2, 3) are the dynamic elastic coefficients in (3.6).31 Then where Cijkl Q(0), R(0) and T(0) are equal to Q, R and T in (3.7), respectively. We note that d Cijkl no longer have the minor symmetries because of the term ρ v 2 m jml δik , but we see that they still enjoy the major symmetry d d Cijkl = Cklij ,

i, j, k, l = 1, 2, 3.

(3.18)

Hence it follows from (3.17) that Q(φ) and T(φ) are symmetric for all φ. We also note that Q(φ), R(φ) and T(φ) are π -periodic in φ, because they are quadratic in cos φ and sin φ by (3.17). Moreover, it follows that π π π Q φ+ = T(φ), R φ + = −R(φ)T , T φ + = Q(φ). (3.19) 2 2 2 Definition 3.1 The limiting velocity v L = v L (m, n) is the lowest velocity for which the matrices Q(φ) and T(φ) become singular for some angle φ: v L = inf{v > 0 | ∃ φ ; det Q(φ) = 0} = inf{v > 0 | ∃ φ ; det T(φ) = 0}.32

(3.20)

The interval 0 < v < v L is called the subsonic range. It will be observed in the example after Lemma 3.3 and also in Section 3.4 that v L for isotropic elasticity is given by " μ Iso = . (3.21) vL ρ Lemma 3.2 (1) The symmetric matrices Q(φ) and T(φ) are positive definite for all φ if and only if 0 ≤ v < v L . (2) The solutions pα (1 ≤ α ≤ 6) to the sextic equation (3.11), i.e., the eigenvalues of N, are not real and they occur in complex conjugate pairs if and only if 0 ≤ v < vL .

31 We

do not rotate m in (3.6).

32 The

last equality follows from (3.19).


99

Proof To prove (1), we note from (3.6), (3.8), (3.16) and (3.17) that ⎛ Q(φ) = ⎝

3

⎞ /l ⎠ / jm (Cijkl − ρ v m jml δik ) m 2

j,l=1

⎛ =⎝

i↓k→1,2,3

⎞

3

/ jm /l ⎠ Cijkl m

j,l=1

− ρ v 2 cos2 φ I,

(3.22)

i↓k→1,2,3

where I is the 3 × 3 identity matrix. By Lemma 1.1, the real symmetric matrix33 ⎛ ⎝

3

⎞ / jm /l ⎠ Cijkl m

j,l=1

i↓k→1,2,3

is positive definite. Hence it has three positive eigenvalues, which we write as λi (φ) (i = 1, 2, 3),

0 < λ1 (φ) ≤ λ2 (φ) ≤ λ3 (φ).

(3.23)

Then by (3.22), the eigenvalues of Q(φ) are λi (φ) − ρ v 2 cos2 φ,

i = 1, 2, 3.

(3.24)

These are strictly decreasing functions of v if cos φ = 0. Then, as v increases from zero, Q(φ) first becomes singular when v is equal to ( vφ =

λ1 (φ) 1 , ρ | cos φ|

(3.25)

while Q(φ) is positive definite for 0 ≤ v < vφ . Since Q(φ) is π -periodic in φ, we can restrict the range of φ to the interval − π2 < φ < π2 . Thus, we put vL =

min

− π2 0,

α = 1, 2, 3.

3.2 Basic Theorems and Integral Formalism On the basis of Lemma 3.2, we can see that the results in Section 1.3 to Section 1.7 apply in a parallel way to dynamic elasticity when 0 ≤ v < v L . Hence in this section we summarize the fundamental results in the Stroh formalism for dynamic elasticity, which will be needed for the analysis of Rayleigh waves in the subsequent sections. Let m and n be orthogonal unit vectors in R3 . First we give the general forms of the surface-wave solutions to (3.3) in the half-space n · x ≤ 0 which decay exponentially as n · x −→ −∞ and have direction of propagation m, wave number k and phase velocity v on the surface n · x = 0. At the same time we give the corresponding tractions on the surface n · x = 0 produced by the surface-wave solutions. These forms are different according to the degeneracy of Stroh’s eigenvalue problem (3.14). Lemma 3.3 Let 0 ≤ v < v L . aα (α = 1, 2, 3) be linearly independent eigenlα vectors of the eigenvalue problem (3.14) pertaining to the eigenvalues pα (α = 1, 2, 3, Im pα > 0), respectively. As we have seen from (3.4) and (3.13), the surfacewave solution is given by

Simple or Semisimple Case Let

u=

3

√ −1 k(m·x+ pα n·x−v t)

c α aα e −

(3.27)

α=1

and the corresponding traction on the surface n · x = 0 produced by the solution above is given by 3 √ √ cα lα e− −1 k(m·x−v t) , t = − −1 k α=1

where cα (1 ≤ α ≤ 3) are arbitrary complex constants.

(3.28)


101

Degenerate Case D1 Let p1 , p2 = p3 be the eigenvalues of N with positive imagaα inary parts. Let (α = 1, 2) be linearly independent eigenvectors pertaining to lα a pα (α = 1, 2), respectively, and let 3 be a generalized eigenvector satisfying l3 N

a3 a a − p2 3 = 2 . l3 l3 l2

Then the surface-wave solution is given by u=

2

√ −1 k(m·x+ pα n·x−v t)

c α aα e −

α=1

√ √ + c3 a3 − −1 k(n · x) a2 e− −1 k(m·x+ p2 n·x−v t) ,

(3.29)

and the corresponding traction on the surface n · x = 0 produced by the preceding solution is given by 3 √ √ t = − −1 k cα lα e− −1 k(m·x−v t) ,

(3.30)

α=1

where cα (1 ≤ α ≤ 3) are arbitrary complex constants. Degenerate Case D2 Letp1 = p2 = p3 be the triple eigenvalueof N with positive a1 aα imaginary part and let be a corresponding eigenvector. Let (α = 2, 3) be l1 lα generalized eigenvectors satisfying N

a a a2 − p1 2 = 1 , l2 l2 l1

N

a3 a a − p1 3 = 2 . l3 l3 l2

Then the surface-wave solution is given by √ √ u = c1 a1 + c2 a2 − −1 k (n · x) a1 + c3 a3 − −1 k (n · x) a2 −

√ 1 2 k (n · x)2 a1 e− −1 k(m·x+ p1 n·x−v t) , 2

(3.31)

and the corresponding traction on the surface n · x = 0 produced by the solution above is given by 3 √ √ cα lα e− −1 k(m·x−v t) , t = − −1 k α=1

where cα (1 ≤ α ≤ 3) are arbitrary complex constants.

(3.32)

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K. Tanuma

Although the forms of the surface-wave solutions are different according to the degeneracy of the eigenvalue problem (3.14) (see (3.27), (3.29) and (3.31)), they have the same form on the surface n · x = 0: u=

3

√ −1 k(m·x−v t)

c α aα e −

.

(3.33)

α=1

As well, the corresponding tractions on n · x = 0 all have the same form 3 √ √ cα lα e− −1 k(m·x−v t) t = − −1 k

(3.34)

α=1

(see (3.28), (3.30) and (3.32)). Henceforth, we call the vectors aα ∈ C3 (α = 1, 2, 3) 3 given in each case of the lemma the displacement parts and lα∈ C (α = 1, 2, 3) the aα (α = 1, 2, 3) of N. traction parts of eigenvectors or generalized eigenvectors lα Example Isotropic 4 elasticity falls under the semisimple case for any m and n when Iso 0 < v < v L = μρ . In fact, Q + p(R + RT ) + p2 T in (3.10) becomes the matrix which is obtained by adding the term −ρv 2 to the diagonal components of the matrix (1.82), and it can be checked from (3.11) that ( ( √ √ μ − ρv 2 λ + 2μ − ρv 2 p1 = p2 = −1 , p3 = −1 . (3.35) μ λ + 2μ Since the strong convexity4condition (1.83) implies that μ < λ + 2μ, we see from (2) Iso = μρ . The corresponding displacement parts aα (α = 1, 2, 3) of Lemma 3.2 that v L of eigenvectors of N in isotropic elasticity can be taken as a1 = m × n,

a2 = n − p1 m,

a3 = m + p3 n

(3.36)

(Exercise 3-1). The first two vectors are linearly independent. Therefore, the eigenIso . Note that value problem (3.14) for isotropic elasticity is semisimple for 0 < v < v L when v = 0, the eigenvalue problem (3.14) is degenerate and the length of the Jordan chain is two (Subsection 1.8.1). Now for 0 ≤ v < v L , let N(φ) be the 6 × 6 real matrix defined by T(φ)−1 −T(φ)−1 R(φ)T N(φ) = , −Q(φ) + R(φ)T(φ)−1 R(φ)T −R(φ)T(φ)−1

(3.37)

where Q(φ), R(φ) and T(φ) are given by (3.17). Then N(0) equals N in (3.15). To consider how the eigenvalues, eigenvectors and generalized eigenvectors of N(φ) depend on the angle φ, we recall that the major symmetry (3.18) of the dynamic elastic coefficients implies that ⎞ ⎛ 3 d / /l ⎠ n jm Cijkl . R(φ)T = ⎝ j,l=1

i↓k→1,2,3


103

Then the same differentiation formulas as (1.35) hold for Q(φ), R(φ) and T(φ) in (3.17). Hence, when 0 ≤ v < v L , we obtain for dynamic elasticity the same results on rotational dependence as Theorem 1.5, Theorem 1.9 and Theorem 1.10. However, in these results, eigenvectors and generalized eigenvectors of N(φ) no longer have physical meaning when φ = 0,34 because of the term ρ v 2 m jml δik in (3.6). But, we need these results in order to derive the Barnett-Lothe integral formalism for dynamic elasticity, to which we now proceed. Definition 3.4 For 0 ≤ v < v L , we define the 6 × 6 real matrix S = S(v) to be the angular average of the 6 × 6 matrix N(φ) over [−π, π ]: π S 1 S2 1 S= = N(φ) dφ, (3.38) 2π −π S3 S1T where S1 = S1 (v), S2 = S2 (v) and S3 = S3 (v) are 3 × 3 real matrices defined by π π 1 1 −1 T S1 = −T(φ) R(φ) dφ, S2 = T(φ)−1 dφ, 2π −π 2π −π π 1 −Q(φ) + R(φ)T(φ)−1 R(φ)T dφ (3.39) S3 = 2π −π and Q(φ), R(φ) and T(φ) are given by (3.17). From (1) of Lemma 3.2 and from the proof of Lemma 1.14 we immediately arrive at Lemma 3.5 For 0 ≤ v < v L , the matrices S2 and S3 are symmetric. Furthermore, S2 is positive definite. Now we take the angular average of Stroh’s eigenvalue problem as in the proof of Theorem 1.15 to obtain the fundamental theorem in the Barnett-Lothe integral formalism for dynamic elasticity. Thus we have aα be an eigenvector or generalized eigenvector lα of N(0) corresponding to the eigenvalues pα (α = 1, 2, 3) with Im pα > 0. Then for 0 ≤ v < vL , √ a a (3.40) S α = −1 α . lα lα

Theorem 3.6 For 0 ≤ v < v L , let

34 In Chapter 1, eigenvectors

and generalized eigenvectors of N(φ) have the physical significance that they are composed of the displacement parts a(φ) and the traction parts l(φ) on the elastic half-space rotated around m × n by the angle φ.

104

K. Tanuma

Then, as in the proof of Theorem 1.16, from Lemma 3.5 and Theorem 3.6 we obtain aα Corollary 3.7 For 0 ≤ v < v L , let (α = 1, 2, 3) be linearly independent eigenlα vector(s) or generalized eigenvector(s) of N(0) corresponding to the eigenvalues pα (α = 1, 2, 3) with Im pα > 0. Then their displacement parts aα (α = 1, 2, 3) are linearly independent. Thus, we can define the surface impedance matrix for dynamic elasticity. Definition 3.8 For 0 ≤ v < v L , the surface impedance matrix Z = Z(v) is the 3 × 3 matrix given by √ (3.41) Z = − −1 [l1 , l2 , l3 ][a1 , a2 , a3 ]−1 , where [l1 , l2 , l3 ] and [a1 , a2 , a3 ] denote 3 × 3 matrices which consist of the column vectors lα and aα respectively. Therefore, Z expresses a linear relationship between (i) the displacements at the surface n · x = 0 on which surface waves of the form (3.27), (3.29) or (3.31) propagate in the direction of m with the phase velocity v and (ii) the tractions needed to sustain them at that surface. As in the derivation of Theorem 1.18, from (3.38), (3.40) and (3.41) we obtain the integral representation of Z. Theorem 3.9 For 0 ≤ v < v L , Z = S−1 2 +

√

−1 S−1 2 S1 ,

(3.42)

where the matrices S1 and S2 are given by (3.39). As in the proof of Lemma 1.22, we get from Theorem 3.6 that S2 = −I

(0 ≤ v < v L ),

(3.43)

where I denotes the 6 × 6 identity matrix. Then the blockwise expression of (3.43) as obtained from (3.38) gives S21 + S2 S3 = −I,

S1 S2 + S2 S1T = O

(0 ≤ v < v L ).

(3.44)

Since S2 is symmetric and invertible for 0 ≤ v < v L , it follows from the second equality in (3.44) that −1 T (0 ≤ v < v L ), (3.45) S−1 2 S1 = − S 2 S1 which implies that S−1 2 S1 is anti-symmetric. Hence we obtain Corollary 3.10 The surface impedance matrix Z is Hermitian for 0 ≤ v < v L . It is proved that Z is positive definite at v = 0 (see the comment after Theorem 1.21). As we shall see later, the eigenvalues of Z(v) decrease monotonically with v in


105

the interval 0 ≤ v < v L from their positive values at v = 0 (Lemma 3.33), and one of them becomes zero when v equals the Rayleigh-wave velocity v R (Theorem 3.13). 3.3 Rayleigh Waves in Elastic Half-space Rayleigh waves are elastic surface waves which propagate along the traction-free surface with the phase velocity in the subsonic range, and whose amplitude decays exponentially with depth below that surface. Let m and n be orthogonal unit vectors in R3 . Following the setting of Section 3.1, we consider Rayleigh waves which propagate along the surface n · x = 0 in the direction of m with the phase velocity v R satisfying 0 < v R < v L , and whose amplitude decays exponentially as n · x −→ −∞, and which produce no tractions on n · x = 0 (see (3.27), (3.29) and (3.31)). Here v L = v L (m, n) is the limiting velocity in Definition 3.1. Elastic surface waves which propagate along the traction-free surface with the phase velocity greater than the limiting velocity are called supersonic surface waves. But we do not consider them in this chapter.35 aα (α = 1, 2, 3) are linearly We take C3 -vectors aα and lα (α = 1, 2, 3) so that lα independent eigenvector(s) or generalized eigenvector(s) of N = N(0) at v = v R associated with the eigenvalues pα (α = 1, 2, 3, Im pα > 0). Then the existence of Rayleigh waves implies that the corresponding traction on n · x = 0 given by (3.34) vanishes for v = v R . In other words, there exists a set of complex numbers (c1 , c2 , c3 ) = (0, 0, 0) such that 3

cα lα = 0

at v = v R ,

(3.46)

α=1

which is equivalent to det[l1 , l2 , l3 ] = 0

at v = v R .

(3.47)

If the traction parts lα (α = 1, 2, 3) are computed explicitly, it will be straightforward to derive the secular equation for v R from (3.47). For isotropic elasticity, by (3.47), the equation for v R becomes bicubic (Exercise 3-2). The next Lemma provides an alternative method for finding the secular equation for v R , where the integral formalism developed in the last section will be used. Then it is neither necessary to see whether the eigenvalue problem (3.14) is simple, semisimple or degenerate, nor necessary to seek explicit forms of the traction parts lα (α = 1, 2, 3). Lemma 3.11 For given orthogonal unit vectors m and n in R3 , let S3 = S3 (v) be the 3 × 3 real matrix defined by (3.39) for 0 ≤ v < v L . Then a necessary and sufficient condition for the existence of the Rayleigh waves in the half-space n · x ≤ 0 which propagate along the surface n · x = 0 in the direction of m with the phase velocity v R in the subsonic range 0 < v < v L is det S3 = 0

35 For

at v = v R

(0 < v R < v L ).

references on supersonic surface waves, see the first paragraph of Section 3.7.

(3.48)

106

K. Tanuma

Proof Suppose that Rayleigh waves exist. The first three rows of the system (3.40) are written, by using notations in (3.38), as √ S1 aα + S2 lα = −1 aα , α = 1, 2, 3 (3.49) and the last three rows of the system (3.40) are written as √ α = 1, 2, 3. S3 aα + S1T lα = −1 lα ,

(3.50)

Then from (3.46) and (3.50) it follows that for cα in (3.46), S3

3

c α aα = 0

at v = v R .

(3.51)

α=1

Corollary 3.7 implies that 3

c α aα = 0

at v = v R .

α=1

Hence the null space of S3 is not trivial at v = v R , which gives (3.48). Conversely, suppose that there is a speed v R in the subsonic range 0 < v < v L for which (3.48) holds. Let A = [a1 , a2 , a3 ],

L = [l1 , l2 , l3 ]

denote the 3 × 3 matrices which consist of the displacement parts aα and the traction parts lα at v = v R that we have taken at the beginning of this section. Then (3.49) and (3.50) can be rewritten as √ S1 A + S2 L = −1 A and S3 A + S1T L = respectively. Then S2 L =

√

−1 I − S1 A,

√

−1 L,

S3 A =

√

−1 I − S1T L.

Taking the determinants of both sides of these equations, we have √ det S2 det L = det( −1 I − S1 ) det A and

√ det S3 det A = det( −1 I − S1T ) det L.

Hence condition (3.48) implies through (3.53) that √ √ det −1 I − S1T = det −1 I − S1 = 0 or

(3.52)

(3.53)

det L = 0

√ at v = v R . In the case where det( −1 I − S1 ) = 0, since Lemma 3.5 guarantees that det S2 = 0 for 0 ≤ v < v L , from (3.52) we get det L = 0 at v = v R .

Furthermore, we shall see that condition (3.48) can be replaced by a more stringent one.


107

Theorem 3.12 Let m and n be orthogonal unit vectors in R3 . A necessary and sufficient condition for the existence of the Rayleigh waves in the half-space n · x ≤ 0 which propagate along the surface n · x = 0 in the direction of m with the phase velocity v R in the subsonic range 0 < v < v L is rank S3 = 1

at v = v R

(0 < v R < v L ).

(3.54)

+ − Proof We only have to prove 3 necessity. Let a and a be the real part and the imaginary part of the vector α=1 cα aα at v = v R , respectively: 3

c α aα = a+ +

√

−1 a− ,

a+ , a− ∈ R3 ,

α=1

where cα (α = 1, 2, 3) are the coefficients that satisfy (3.46). We will show that a+ and a− are linearly independent. Then, by (3.51), the null space of S3 is two-dimensional, which gives (3.54). From (3.46) and (3.49) it follows that √ √ √ S1 a+ + −1 a− = −1 a+ + −1 a− at v = v R . Since the matrix S1 is real, we get S1 a+ = −a−

and S1 a− = a+ .

(3.55) 3

We have a+ = 0 and a− = 0; otherwise from (3.55) it follows that α=1 cα aα = 0, which contradicts Corollary 3.7. Suppose that there exists a real number k such that a+ = k a− holds. Then by (3.55), −a− = S1 a+ = S1 (ka− ) = k S1 a− = k a+ = k2 a− , which implies k2 = −1. This is a contradiction. Suppose that rank S3 = 0 at v = v R . Then S3 vanishes there and the first equality in (3.44) gives S21 = −I at v = v R . Since S2 is invertible for 0 ≤ v < v L , taking the determinants of both sides of (3.45), we have det S1 = 0 for 0 ≤ v < v L . This is a contradiction.

We give another characterization of the Rayleigh-wave velocity v R in terms of the surface impedance matrix Z = Z(v) in Definition 3.8. Theorem 3.13 A necessary and sufficient condition for the existence of Rayleigh waves in Theorem 3.12 is det Z = 0 at v = v R

(0 < v R < v L ).

Proof Obvious from Corollary 3.7, (3.41) and (3.47).

(3.56)

In the forthcoming Sections 3.4 and 3.5 we shall use Theorem 3.12 to find the secular equations for v R in isotropic elasticity and in weakly anisotropic elastic media. At this point, we prove that v R is unique if it exists. That is, we have

108

K. Tanuma

Theorem 3.14 For given orthogonal unit vectors m and n, the Rayleigh wave is unique if it exists. This theorem is proved by further consideration on the matrix S3 . Lemma 3.15 For 0 ≤ v < v L , let S3 = S3 (v) be the 3 × 3 real symmetric matrix in (3.39). d (1) The symmetric matrix dv S3 is positive definite. (2) The eigenvalues of S3 are monotonic increasing functions of v.

Proof To prove (1), we first differentiate the integrand of S3 , namely −Q(φ) + R(φ)T(φ)−1 R(φ)T ,

(3.57)

with respect to v. From (3.6), (3.8), (3.16) and (3.17) it follows that ⎛ ⎞ ⎞ ⎛ 3 3 / jm /l ⎠ / jm /l ⎠ Cijkl m −V ⎝ m jml δik m Q(φ) = ⎝ j,l=1

⎛ =⎝

3

/ jm /l ⎠ Cijkl m

j,l=1

⎛ R(φ) = ⎝

=⎝

3

3

T(φ) = ⎝

3

⎛

/ j/ nl ⎠ Cijkl m

−V ⎝

=⎝

3

3

⎞ / j/ nl ⎠ m jml δik m

j,l=1

i↓k→1,2,3

i↓k→1,2,3

⎞ / j/ nl ⎠ Cijkl m ⎞

+ V cos φ sin φ I,

i↓k→1,2,3

⎛

−V ⎝

n j/ nl ⎠ Cijkl /

j,l=1

⎛

− V cos2 φ I,

⎞

j,l=1

⎛

i↓k→1,2,3

i↓k→1,2,3

j,l=1

⎛

j,l=1

i↓k→1,2,3

⎞

i↓k→1,2,3

3

⎞ n j/ nl ⎠ m jml δik /

j,l=1

⎞ n j/ nl ⎠ Cijkl /

j,l=1

i↓k→1,2,3

− V sin2 φ I,

i↓k→1,2,3

where V = ρ v . Then we have 2

d Q(φ) = −2ρ v cos2 φ I, dv Also, since

d d R(φ) = 2ρ v cos φ sin φ I = R(φ)T . dv dv

d T(φ) = −2ρ v sin2 φ I dv and d d d T(φ)T(φ)−1 = T(φ) · T(φ)−1 + T(φ) T(φ)−1 = 0, dv dv dv

(3.58)


109

we get d d T(φ)−1 = −T(φ)−1 T(φ) · T(φ)−1 = 2ρ v sin2 φ T(φ)−2 . dv dv Therefore, the derivative of the integrand (3.57) in v becomes d −Q(φ) + R(φ)T(φ)−1 R(φ)T dv = 2ρ v cos2 φ I + cos φ sin φ T(φ)−1 R(φ)T + R(φ)T(φ)−1 + sin2 φ R(φ)T(φ)−2 R(φ)T . We note that the right hand side of the preceding equation can be written as T cos φ I + sin φ T(φ)−1 R(φ)T . 2ρ v cos φ I + sin φ T(φ)−1 R(φ)T Then for any non-zero v ∈ R3 it follows that

π 1 d dS3 v = −Q(φ) + R(φ)T(φ)−1 R(φ)T v dφ v· v· dv 2π −π dv π ρv = cos φ I + sin φ T(φ)−1 R(φ)T v π −π · cos φ I + sin φ T(φ)−1 R(φ)T v dφ, which is positive, because the last integrand is non-negative for all φ and is strictly positive for φ = 0. This proves (1). Let vi = vi (v) ∈ R3 (i = 1, 2, 3) be eigenvectors of the real symmetric matrix S3 associated with the eigenvalues μi = μi (v) ∈ R (i = 1, 2, 3). Differentiating the eigenrelations S3 vi = μi vi ,

i = 1, 2, 3

(3.59)

with respect to v, we have dμi dvi dvi dS3 vi + S3 = vi + μi , dv dv dv dv

i = 1, 2, 3.

Taking the inner products with vi , we get

dμi dS3 dvi dvi vi + vi · S3 = vi · vi + μi vi · , vi · dv dv dv dv Since S3 is symmetric, by (3.59) we have

dvi dvi dvi = (S3 vi ) · = μi vi · , vi · S3 dv dv dv Then it follows that

vi ·

dS3 vi dv

=

dμi vi · vi , dv

i = 1, 2, 3.

i = 1, 2, 3.

i = 1, 2, 3.

110

K. Tanuma

By (1) of the lemma, the left hand side is positive. Also, we have vi · vi > 0 (i = 1, 2, 3). Thus, dμi > 0, dv

i = 1, 2, 3.

This proves (2).

Proof of Theorem 3.14 By Theorem 3.12, the matrix S3 (v R ) has a two-dimensional eigenspace corresponding to the multiple zero eigenvalue. This implies that two of the eigenvalues of S3 (v R ) must vanish together at v = v R . If there were two Rayleigh waves, i.e., there were two v R ’s which satisfy (3.54), then one of the three eigenvalues of S3 (v) would have to vanish twice for 0 ≤ v < v L . This contradicts the monotonicity of the eigenvalues in the last lemma.

Finally, we show that the integral formalism can also be applied to get the formula aα for the polarization vector of the Rayleigh waves at the surface. Let (α = lα 1, 2, 3) be linearly independent eigenvector(s) or generalized eigenvector(s) of N = N(0) at v = v R associated with the eigenvalues pα (α = 1, 2, 3, Im pα > 0). By (3.33), the displacement field of the Rayleigh waves at the surface n · x = 0 is written as u=

3

√ −1 k(m·x−v R t)

c α aα e −

,

(3.60)

α=1

where cα (α = 1, 2, 3) are the coefficients which satisfy (3.46). We call the vector apol =

3

cα aα ∈ C3

α=1

the polarization vector of the Rayleigh waves at the surface. This means that the real part a+ and the imaginary part a− of apol define the plane to which the paths of surface particles are confined, so that their displacements are expressed by taking the √ real part of apol e− −1 k(m·x−v R t) or a+ cos k(m · x − v R t) + a− sin k(m · x − v R t).

(3.61)

√ − −1 k(m·x−v R t)

Note that the imaginary part of apol e differs from the real part (3.61) only by a phase shift of 90 degrees. Now we give an integral expression for the polarization vector apol . Theorem 3.16 Let S1 (v) and S3 (v) be the 3 × 3 real matrices in (3.39), and let e1 and e2 be orthogonal unit vectors in R3 such that S3 (v R )e1 = 0,

(S3 (v R )e1 ) × e2 = 0,

(3.62)

where the symbol × denotes the cross product of vectors. Then the Rayleigh waves described in Theorem 3.12 have at the surface n · x = 0 the polarization vector √ (3.63) apol = (S3 (v R )e1 ) × e2 − −1 S1 (v R ) (S3 (v R )e1 ) × e2 .


111

Proof Formula (3.54) implies that there is a non-zero vector s ∈ R3 such that S3 (v R ) = ± s ⊗ s,

(3.64)

where the symbol ⊗ denotes the tensor product of two vectors.36 Then from (3.51) it follows that S3 (v R )apol = ±(s ⊗ s)apol = ±(apol , s)C3 s = 0. Since s = 0, we have

(3.66)

√ apol , s C3 = a+ · s + −1 a− · s = 0,

and hence a+ · s = a− · s = 0. This implies that s is normal to the plane of polarization. Let e1 and e2 be orthogonal unit vectors in R3 which satisfiy (3.62). Then S3 (v R )e1 , which is equal to (e1 · s)s or −(e1 · s)s, is a non-zero scalar multiple of s, and hence, is normal to the plane of polarization. Therefore, (S3 (v R )e1 ) × e2 is a non-zero real vector lying in the plane of polarization. By taking this vector to be a real part a+ , from (3.55) the imaginary part a− is given by −S1 (v R ) (S3 (v R )e1 ) × e2 .

3.4 Rayleigh Waves in Isotropic Elasticity In this section, on the basis of Theorem 3.12 and Theorem 3.16, we derive the secular Iso equation for the phase velocity v Iso R and compute the polarization vector apol of Rayleigh waves which propagate along the surface of an isotropic elastic half-space. We first compute the integrand of S3 (v), namely −Q(φ) + R(φ)T(φ)−1 R(φ)T , for which we use formulas (3.58). Since the components of the elasticity tensor of an isotropic material is given by Cijkl = λ δijδkl + μ(δik δ jl + δil δkj),

(3.67)

tensor product a ⊗ b of two vectors a = (a1 , a2 , a3 ) and b = (b 1 , b 2 , b 3 ) in R3 (resp. C3 ) can be expressed as a 3 × 3 matrix a ⊗ b = ai b j i↓ j→1,2,3 resp. a ⊗ b = ai b j .

36 The

i↓ j→1,2,3

Then it follows that (a ⊗ b)c = (c · b)a

resp. (a ⊗ b)c = (c, b)C3 a

(3.65)

(resp. c ∈ This will be used in (3.66). Formula (3.64) can be proved by noting from for all c ∈ (3.54) that the 3 × 3 matrix S3 (v R ) has only one linearly independent column vector. R3

C3 ).

112

K. Tanuma

from the identities /·m /= m

3

/ ·/ m n=

/2j = 1, m

j=1

we have

⎛ ⎝

3

⎝

⎞

⎝

j=1

i↓k→1,2,3

i↓k→1,2,3

/⊗m / + μ I, = (λ + μ) m

⎞

/k i↓k→1,2,3 /i/ nk + μ/ = λm ni m

/ j/ nl ⎠ Cijkl m

j,l=1

⎛

/ n j2 = 1,

/k + μ δik + m /k /i m /i m = λm

/ jm /l ⎠ Cijkl m

3

3

/ n ·/ n=

/j/ m n j = 0,

j=1

j,l=1

⎛

3

3

i↓k→1,2,3

/ ⊗/ /, = λm n + μ/ n⊗m

⎞

nk + μ δik + / nk = λ/ ni/ ni/

n j/ nl ⎠ Cijkl /

j,l=1

i↓k→1,2,3

i↓k→1,2,3

= (λ + μ)/ n ⊗/ n + μ I,

(3.68)

where I is the 3 × 3 identity matrix. Hence, /⊗m / + (μ − V cos2 φ) I, Q(φ) = (λ + μ) m

(3.69)

/ ⊗/ / + V cos φ sin φ I, R(φ) = λ m n + μ/ n⊗m

(3.70)

T(φ) = (λ + μ)/ n ⊗/ n + (μ − V sin φ) I,

(3.71)

2

where V = ρ v 2 . Let / ×/ =m n = m × n. Then the triad {/ m,/ n, } forms an orthonormal basis in R3 , and we have /⊗m / +/ I=m n ⊗/ n + ⊗ .

(3.72)

To compute T(φ)−1 , we use (3.72) and rewrite (3.71) as / + ⊗ ) + λ + 2μ − V sin2 φ / T(φ) = μ − V sin2 φ (/ m⊗m n ⊗/ n.37 Then by (2.72) we get T(φ)−1 =

37 It

1 1 / + ⊗ ) + / (/ m⊗m n ⊗/ n. μ − V sin2 φ λ + 2μ − V sin2 φ

(3.73)

follows immediately that det T(φ) = (μ − V sin2 φ)2 (λ + 2μ − V sin2 φ).

From the 4 strong convexity condition (1.83) we get μ < λ + 2μ. Therefore from (3.20) we obtain μ = v Iso L ρ.


113

Note that 0 ≤ V(= ρ v 2 ) < μ < λ + 2μ

(3.74)

Iso . Now from (3.73) and for 0 ≤ v < v L

/ + μm / ⊗/ R(φ)T = λ/ n⊗m n + V cos φ sin φ I, it follows that38 T(φ)−1 R(φ)T =

λ μ / ⊗/ / / m n+ n⊗m 2 μ − V sin φ λ + 2μ − V sin2 φ +

V cos φ sin φ V cos φ sin φ / + ⊗ ) + / (/ m⊗m n ⊗/ n. 2 μ − V sin φ λ + 2μ − V sin2 φ (3.75)

Then from (3.69) and (3.70), − Q(φ) + R(φ)T(φ)−1 R(φ)T /⊗m / − (μ − V cos2 φ) I = −(λ + μ) m λ2 μ2 /⊗m /+ / m n ⊗/ n 2 λ + 2μ − V sin φ μ − V sin2 φ

μ V cos φ sin φ λ V cos φ sin φ /) (/ m ⊗/ n +/ n⊗m + + λ + 2μ − V sin2 φ μ − V sin2 φ +

+

(V cos φ sin φ)2 (V cos φ sin φ)2 / + ⊗ ) + / (/ m⊗m n ⊗/ n. 2 μ − V sin φ λ + 2μ − V sin2 φ

(3.76)

On the other hand, from (3.16) we have /⊗m / = m ⊗ m cos2 φ + (m ⊗ n + n ⊗ m) cos φ sin φ + n ⊗ n sin2 φ, m / ⊗/ m n = (−m ⊗ m + n ⊗ n) cos φ sin φ + m ⊗ n cos2 φ − n ⊗ m sin2 φ, / / = (−m ⊗ m + n ⊗ n) cos φ sin φ − m ⊗ n sin2 φ + n ⊗ m cos2 φ, n⊗m / n ⊗/ n = m ⊗ m sin2 φ − (m ⊗ n + n ⊗ m) cos φ sin φ + n ⊗ n cos2 φ. (3.77) Substituting these into (3.76), we take the angular average of it over [−π, π ]. In this process we use the following formulas of integration: π π 1 1 1 cos2 φ dφ = sin2 φ dφ = , 2π −π 2π −π 2

38 For

any a, b, c, d ∈ R3 , (a ⊗ b)(c ⊗ d) = (c · b)(a ⊗ d).

114

K. Tanuma

and 1 2π 1 2π 1 2π 1 2π 1 2π

π

−π

π

−π

π

−π

π

−π

π

−π

. a−b 1− , a "

1 a −1 + , b a−b √ ) 1 a a − b , 2a − b − 2 2b 2 √ −1 2 3 2 2 − 12ab + 3b − 8 a(a − b ) 8a , 8b 3 ) 1 2 3 8a − 4ab − b 2 − 8a 2 a − b , 3 8b

1 cos2 φ dφ = b a − b sin2 φ sin2 φ dφ = a − b sin2 φ cos2 φ sin2 φ dφ = a − b sin2 φ cos4 φ sin2 φ dφ = a − b sin2 φ cos2 φ sin4 φ dφ = a − b sin2 φ

-

"

(3.78)

where a and b are constants such that 0 < b < a, and we use the fact that the integration of odd functions of φ over [−π, π ] gives zero. Thus, by simple but long computations, we obtain π 1 S3 (v) = −Q(φ) + R(φ)T(φ)−1 R(φ)T dφ 2π −π ( -" . 1 μ λ + 2μ − V 2 2 = (2μ − V) − 4μ m ⊗ m V μ−V λ + 2μ ( -( . 1 λ + 2μ μ − V + (2μ − V)2 − 4μ2 n ⊗ n V λ + 2μ − V μ ) (3.79) − μ(μ − V) ⊗ . Therefore, in terms of the orthonormal basis {m, n, }, S3 (v) can be written componentwise as ⎞ ⎛ R(v) 0 0 4 4 ⎟ ⎜ λ+2μ μ−V ⎟, (3.80) S3 (v) = ⎜ R(v) 0 λ+2μ−V μ ⎠ ⎝ 0 √ 0 0 − μ(μ − V) where 1 R(v) = V

-"

μ (2μ − V)2 − μ−V

(

. λ + 2μ − V 2 4μ , λ + 2μ

V = ρ v2 .

(3.81)

Iso Now we use Theorem 3.12. By (3.74) and4the fact4 that 0 < v Iso R < v L , the (3, 3) √ λ+2μ μ−V in the (2, 2) component component − μ(μ − V) and the multiplier λ+2μ−V μ Iso of (3.80) do not vanish at v = v Iso R . Therefore, (3.54) implies that v R satisfies the equation

R(v) = 0

(3.82)


in the subsonic range 0 < v < "

115

4

μ . ρ

Equation (3.82) can be written as

μ (2μ − V)2 = μ−V

(

λ + 2μ − V 4μ2 . λ + 2μ

Squaring both sides leads to λ + 2μ − V 1 (2μ − V)4 = 16μ3 , μ−V λ + 2μ and hence (λ + 2μ)(2μ − V)4 = (μ − V)(λ + 2μ − V) 16μ3 , which is a quartic equation in V. But the constant term vanishes and we get the cubic equation V 3 − 8μV 2 +

8μ2 (3λ + 4μ) 16μ3 (λ + μ) V− = 0. λ + 2μ λ + 2μ

(3.83)

Proposition 3.17 The velocity of Rayleigh waves which propagate along the surface of Iso an isotropic elastic half-space is uniquely determined. It √ is the simple solution v = v R to (3.83) with V = ρ v 2 in the subsonic range 0 < v < μ/ρ. Proof It is sufficient to prove that in the interval 0 < V < μ, equation (3.83) has only / one solution. Write the left hand side of (3.83) by R(V). Then 3 / = −16μ (λ + μ) R(0) λ + 2μ

which, by (1.83), is negative and / R(μ) = μ3 > 0. / Hence R(V) = 0 has at least one solution in the interval 0 < V < μ. Moreover, it is easy to see that d2 / R(V) = 6V − 16μ < 0 dV 2 / for 0 < V < μ, i.e., R(V) is concave in the interval 0 < V < μ. Thus, the solution in the interval 0 < V < μ is unique and simple.

Next we compute the polarization vector aIso pol of the Rayleigh waves at the surface n · x = 0. To make use of Theorem 3.16, we first compute S1 (v), whose integrand is the negative of (3.75). Substituting (3.77) into (3.75), we take the angular average of

116

K. Tanuma

it over [−π, π ]. In this process, we use the first three integration formulas in (3.78). The result is π 1 T(φ)−1 R(φ)T dφ −S1 (v) = 2π −π ( -( . 1 λ + 2μ μ−V = (2μ − V) − 2μ m ⊗ n V λ + 2μ − V μ -( . " 1 λ + 2μ − V μ 2μ − (2μ − V) n ⊗ m. + V λ + 2μ μ−V Hence, in terms of the orthonormal basis {m, n, }, S1 (v) can be written componentwise as ⎛ ⎞ 0 −S(v) 0 ⎜ 4 μ 4 λ+2μ−V ⎟ S(v) 0 0 ⎟ S1 (v) = ⎜ (3.84) λ+2μ ⎝ μ−V ⎠, 0

0

0

where 1 S(v) = V

-(

λ + 2μ (2μ − V) − λ + 2μ − V

(

. μ−V 2μ . μ

(3.85)

Hereafter, we represent vectors componentwise in terms of the orthonormal basis {m, n, }, too. Now we apply Theorem 3.16 to isotropic elasticity. Take ⎛ ⎞ 0 e1 = ⎝ 0 ⎠ . 1 Then from (3.80) we get ⎛

0 0

⎞

⎟ ⎜ S3 (v Iso R )e1 = ⎝ 4 ⎠, Iso − μ μ − VR Iso Iso 2 where V R = ρ (v Iso R ) . Then by (3.74), S3 (v R )e1 = 0. Next, taking ⎛ ⎞ 1 e2 = ⎝ 0 ⎠ , 0

we have

⎞ 0 4 ⎜ Iso ⎟ S3 (v Iso R )e1 × e2 = ⎝ − μ μ − V R ⎠ = 0, 0 ⎛


117

which is the real part of aIso pol . Then by (3.84), we get ⎛ 4 ⎞ Iso Iso −S(v ) μ μ − V R R ⎟ ⎜ Iso −S1 (v Iso ⎠, R ) S3 (v R )e1 × e2 = ⎝ 0 0 which is the imaginary part of aIso pol . Thus, we obtain ⎛√ ⎞ 4 −1 S v Iso R Iso ⎝ ⎠. aIso 1 pol = − μ μ − V R 0

(3.86)

Therefore, from (3.61) and (3.86) the real displacement field of the Rayleigh waves at the surface n · x = 0 can be written as 4 Iso − μ(μ − V R ) ⎫ ⎧⎛ ⎞ ⎛ Iso ⎞ S vR ⎨ 0 ⎬ Iso Iso (3.87) × ⎝ 1 ⎠ cos k m · x − v R t + ⎝ 0 ⎠ sin k m · x − v R t . ⎭ ⎩ 0 0 This is a componentwise representation relative to the orthonormal basis {m, n, }. Recalling that m is the propagation direction and n is the unit outward normal to the surface, we obtain Proposition 3.18 The longitudinal component and the normal component of the displacements of the Rayleigh waves at the surface of an isotropic elastic half-space have a phase shift of 90 degrees. The polarization ratio, i.e., the ratio of the maximum longitudinal component to the maximum normal component, is 4 ( Iso Iso 2 μ μ − V R 1 λ + 2μ Iso S vR = 2μ − V R , (3.88) = Iso Iso 2μ λ + 2μ − V R 2μ − V R 2 Iso = ρ v Iso and v Iso where V R R R is the velocity of Rayleigh waves in Proposition 3.17. Note The equalities in (3.88) follow from (3.82) and (3.85). 3.5 Rayleigh Waves in Weakly Anisotropic Elastic Media In this section we assume that the elasticity tensor C = Cijkl i, j,k,l=1,2,3 has the form C = CIso + A, where CIso is the isotropic part of C, Iso , CIso = Cijkl i, j,k,l=1,2,3

Iso Cijkl = λ δijδkl + μ δik δ jl + δil δkj

with Lamé constants λ and μ, and A is the perturbative part of C A = aijkl i, j,k,l=1,2,3 .

(3.89)

(3.90)

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K. Tanuma

From the symmetry conditions (1.4) and (1.6) of the elasticity tensor C it follows that aijkl = a jikl = aklij,

i, j, k, l = 1, 2, 3,

but we do not assume any other symmetry for A. Hence the perturbative part A has 21 independent components and we write them in the Voigt notations as ⎡ ⎤ a11 a12 a13 a14 a15 a16 ⎢ a22 a23 a24 a25 a26 ⎥ ⎢ ⎥ ⎢ a33 a34 a35 a36 ⎥ ⎢ ⎥. A = (ars ) = ⎢ (3.91) a44 a45 a46 ⎥ ⎢ ⎥ ⎣ ⎦ a55 a56 Sym. a66 Let m and n be orthogonal unit vectors in R3 . Assuming that the deviation from an isotropic state caused by the perturbative part A is small, we consider Rayleigh waves in the half-space n · x ≤ 0 which propagate along the surface n · x = 0 in the direction of m with the phase velocity v R . In this section we investigate the perturbation of the phase velocity v R of Rayleigh waves, i.e., the shift in v R from its comparative isotropic value v Iso R , caused by the perturbative part A. We present a velocity formula which is correct to within terms linear in the components of A = (ars ). Hereafter, without loss of generality, we take m = (0, 1, 0),

n = (0, 0, 1).

(3.92)

Theorem 3.19 In a weakly anisotropic elastic medium whose elasticity tensor C is given by (3.89), (3.90) and (3.91), the phase velocity of Rayleigh waves which propagate along the surface of the half-space x3 ≤ 0 in the direction of the 2-axis can be written, to within terms linear in the perturbative part A = (ars ), as 1 2ρ v Iso R Iso Iso Iso × γ22 v R a22 + γ23 v Iso R a23 + γ33 v R a33 + γ44 v R a44 ,

v R = v Iso R −

where

(3.93)

(λ + 2μ) −8μ2 (λ + μ) + 2μ(5λ + 6μ)V − (2λ + 3μ)V 2 , γ22 (v) = D(v) 4λ(μ − V) [4μ(λ + μ) − (λ + 2μ)V] , D(v) (λ + 2μ − V) −8μ2 (λ + μ) + 2μ(5λ + 6μ)V − (2λ + 3μ)V 2 γ33 (v) = D(v) V γ22 (v), = 1− λ + 2μ

γ23 (v) =

8μ(λ + 2μ − V) [2μ(λ + μ) − (λ + 2μ)V] , D(v) D(v) = (λ + μ) 8μ2 (3λ + 4μ) − 16μ(λ + 2μ)V + 3(λ + 2μ)V 2 ,

γ44 (v) = −

V = ρ v2

(3.94)


119

and v Iso R is the velocity of Rayleigh waves in the comparative isotropic medium with 2 Iso = ρ v Iso is the unique solution to the cubic equation (3.83) in the C = CIso , i.e., V R R range 0 < V < μ. Remark 3.20 Only four components a22 , a23 , a33 and a44 of the perturbative part A can affect the first-order perturbation of the phase velocity v R . To prove the theorem, we make use of Theorem 3.12. Since we are concerned with the terms in v R up to those linear in the perturbative part A, we first express S3 (v) in (3.39), to within terms linear in A. Then we obtain an approximate secular equation for v R , which contains terms up to those linear in A. Hereafter we denote the matrices Q(φ), R(φ) and T(φ) for an isotropic state given in (3.69), (3.70) and (3.71) by QIso (φ), RIso (φ) and TIso (φ), respectively, and denote by Q(φ), R(φ), T(φ) those of the material whose elasticity tensor is given by (3.89). Then from (3.17) and (3.89) we have Q(φ) = QIso (φ) + QPtb (φ),

R(φ) = RIso (φ) + RPtb (φ),

T(φ) = TIso (φ) + TPtb (φ), where ⎛ QPtb (φ) = ⎝

⎞

3

/ jm /l ⎠ aijkl m

j,l=1

⎛ RPtb (φ) = ⎝

3

⎞

TPtb (φ) = ⎝

3 j,l=1

i↓k→1,2,3

/ j/ nl ⎠ aijkl m

j,l=1

⎛

,

⎞

,

i↓k→1,2,3

n j/ nl ⎠ aijkl /

.

(3.95)

i↓k→1,2,3

Then −1 Iso −1 = T (φ) I + TIso (φ)−1 TPtb (φ) T(φ)−1 = TIso (φ) + TPtb (φ) −1 Iso −1 = I + TIso (φ)−1 TPtb (φ) T (φ) ≈ I − TIso (φ)−1 TPtb (φ) TIso (φ)−1 . Here and hereafter we use the notation ≈ to indicate that we are retaining terms up to those linear in the perturbative part A = (ars ) and are neglecting the higher order terms. Hence T(φ)−1 ≈ TIso (φ)−1 − TIso (φ)−1 TPtb (φ) TIso (φ)−1 .

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K. Tanuma

Then the integrand of S3 (v) becomes − Q(φ) + R(φ)T(φ)−1 RT (φ) ≈ −QIso (φ) − QPtb (φ) + RIso (φ) + RPtb (φ) × TIso (φ)−1 − TIso (φ)−1 TPtb (φ) TIso (φ)−1 RIso (φ)T + RPtb (φ)T ≈ −QIso (φ) + RIso (φ)TIso (φ)−1 RIso (φ)T − QPtb (φ) + RPtb (φ)TIso (φ)−1 RIso (φ)T + RIso (φ)TIso (φ)−1 RPtb (φ)T − RIso (φ)TIso (φ)−1 TPtb (φ)TIso (φ)−1 RIso (φ)T . Therefore, we obtain π 1 Ptb S3 (v) = −Q(φ) + R(φ)T−1 (φ)RT (φ) dφ ≈ SIso 3 (v) + S3 (v); 2π −π here SIso 3 (v)

1 = 2π

π

−π

−QIso (φ) + RIso (φ)TIso (φ)−1 (φ)RIso (φ)T dφ

is of zeroth order in the perturbative part A, which is equal to (3.79), and π 1 −QPtb (φ) + RPtb (φ)TIso (φ)−1 RIso (φ)T (v) = SPtb 3 2π −π + RIso (φ)TIso (φ)−1 RPtb (φ)T

− RIso (φ)TIso (φ)−1 TPtb (φ)TIso (φ)−1 RIso (φ)T dφ, (3.96) which is of first order in the perturbative part A. Now we prove Lemma 3.21 Let sij(v) (i, j = 1, 2, 3) be the (i, j) component of the matrix SPtb 3 (v). Then an approximate secular equation for v R is (v) = 0, where (v) = R(v) + s22 (v), and R(v) is given by (3.81). Proof From (3.79) and (3.92) it follows that ⎞ ⎛ √ 0 − μ(μ − V) 0 ⎟ ⎜ 0 R(v) 0 ⎟. ⎜ SIso 3 (v) = ⎝ ⎠ 4 4 λ+2μ μ−V R(v) 0 0 λ+2μ−V μ

(3.97)


121

Then Ptb S3 (v) ≈ SIso (3.98) 3 (v) + S3 (v) ⎛ √ ⎞ − μ(μ − V) + s11 (v) s12 (v) s13 (v) ⎜ ⎟ R(v) + s22 (v) s23 (v) s12 (v) ⎟. =⎜ ⎝ ⎠ 4 4 λ+2μ μ−V s13 (v) s23 (v) R(v) + s (v) 33 λ+2μ−V μ

By Theorem 3.12, the (3, 3) minor of the preceding matrix, i.e., the determinant of the submatrix formed by striking out the third row and third column, must vanish at v = v R . Since sij(v) (i, j = 1, 2, 3) are linear functions of the perturbative part A, the (3, 3) minor becomes - √ . − μ(μ − V) + s11 (v) s12 (v) det R(v) + s22 (v) s12 (v) ) = − μ(μ − V) + s11 (v) (R(v) + s22 (v)) − s12 (v)2 ) ≈ − μ(μ − V) (R(v) + s22 (v)) + s11 (v)R(v). (3.99) Also, recalling that

R v Iso = 0, R

we have

at v = v R . Since v Iso R is less than

4

s11 (v)R(v) ≈ 0 μ ρ

and v R is close to v Iso R , we get

) μ(μ − V) = 0 at v = v R . Hence (3.99) and Theorem 3.12 imply39 that R(v) + s22 (v) = (v) = 0 at v = v R . This proves the lemma. Lemma 3.22

vR

∂ s22 (v) ∂ ars A=(ars )=0,v=vIso R ≈ v Iso − × ars , R ∂ R ∂v Iso

(3.100)

v=v R

where the summation on the right hand side is taken for the indices of the 21 independent components in (3.91).

39 In the proof, we have used the vanishing of the (3, 3) minor of the matrix SIso (v) + SPtb (v) as an 3 3 approximate secular equation. The vanishing of the other minors may give other approximate secular equations. However, by Theorem 3.14, the Rayleigh wave is unique if it exists. Hence there is no need to consider other secular equations.

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K. Tanuma

Proof From the Taylor expansion of v R around A = (ars ) = 0, we get ∂v R v R ≈ v Iso + ars . R ∂ars (ars )=0

(3.101)

Since we have (v R ) = 0, it follows from the implicit function theorem that ∂ ∂ ∂v R =− ∂ars (ars )=0 ∂ars (ars )=0, v=vIso ∂v R

.−1

(ars )=0, v=v Iso R

.

(3.102)

From (3.97) we get ∂s22 (v) ∂ = ∂ars ∂ars

and

∂ ∂ R(v) ∂s22 (v) = + . ∂v ∂v ∂v

(3.103)

Since s22 (v) is a linear function of A = (ars ), it follows that ∂s22 (v) = 0. ∂v (ars )=0 Hence we have

∂ R(v) ∂ = ∂v (ars )=0, v=vIso ∂v R

v=v Iso R

.

Therefore, from (3.101)–(3.104) we obtain (3.100).40

(3.104)

The next lemma is useful in the computations of the perturbation formula. Lemma 3.23 The effect of the perturbative part A = (ars ) on s22 (v), i.e., on the (2, 2) component of the matrix SPtb 3 (v), comes from a22 , a23 , a33 and a44 . Proof First we prove that the effect of the perturbative part A on the (2, 2) component of the integrand of SPtb 3 (v) (see (3.96)), i.e., on the (2, 2) component of −QPtb (φ) + RPtb (φ)TIso (φ)−1 RIso (φ)T + RIso (φ)TIso (φ)−1 RPtb (φ)T −RIso (φ)TIso (φ)−1 TPtb (φ)TIso (φ)−1 RIso (φ)T ,

(3.105)

comes from a22 , a23 , a33 , a44 , a24 and a34 . From (3.16) and (3.92) we have / = (0, m /3 ) = (0, cos φ, sin φ), /2 , m m / n3 ) = (0, − sin φ, cos φ). n = (0,/ n2 ,/

(3.106)

40 By Proposition 3.17, the unique solution to the cubic equation (3.83) in the range 0 < V < μ is simple. Hence (3.104), i.e., the denominator of (3.102), does not vanish.


123

Then from (3.91) and (3.95) we get ⎡ Q

Ptb

(φ) = ⎣

a66 cos2 φ+2a56 cos φ sin φ + a55 sin2 φ

a26 cos2 φ + (a46 + a25 ) cos φ sin φ+a45 sin2 φ a22 cos2 φ + 2a24 cos φ sin φ + a44 sin2 φ

Sym. ⎤ a46 cos2 φ + (a36 + a45 ) cos φ sin φ + a35 sin2 φ 2 2 a24 cos φ + (a23 + a44 ) cos φ sin φ + a34 sin φ ⎦ , a44 cos2 φ + 2a34 cos φ sin φ + a33 sin2 φ ⎡

R

Ptb

(a55 − a66 ) cos φ sin φ + a56 (cos2 φ − sin2 φ) (φ) = ⎣ (a45 − a26 ) cos φ sin φ + a25 cos2 φ − a46 sin2 φ (a35 − a46 ) cos φ sin φ + a45 cos2 φ − a36 sin2 φ

(3.107)

(a45 − a26 ) cos φ sin φ + a46 cos2 φ − a25 sin2 φ (a44 − a22 ) cos φ sin φ + a24 (cos2 φ − sin2 φ) (a34 − a24 ) cos φ sin φ + a44 cos2 φ − a23 sin2 φ

⎤ (a35 − a46 ) cos φ sin φ + a36 cos2 φ − a45 sin2 φ 2 ⎦, 2 (a34 − a24 ) cos φ sin φ + a23 cos φ − a44 sin φ (a33 − a44 ) cos φ sin φ + a34 (cos2 φ − sin2 φ) ⎡ T

Ptb

(φ) = ⎣

a55 cos2 φ − 2a56 cos φ sin φ + a66 sin2 φ

(3.108)

a45 cos2 φ − (a46 + a25 ) cos φ sin φ + a26 sin2 φ a44 cos2 φ − 2a24 cos φ sin φ + a22 sin2 φ

Sym. ⎤ a35 φ − (a36 + a45 ) cos φ sin φ + a46 sin2 φ a34 cos2 φ − (a23 + a44 ) cos φ sin φ + a24 sin2 φ ⎦ . a33 cos2 φ − 2a34 cos φ sin φ + a44 sin2 φ cos2

(3.109)

From the above we see that the (i, j) components (2 ≤ i, j ≤ 3) of the matrices QPtb (φ), RPtb (φ), TPtb (φ) depend only on a22 , a23 , a33 , a44 , a24 , a34 , whereas the other components depend only on a25 , a26 , a35 , a36 , a45 , a46 , a55 , a56 , a66 . Let us write these matrices symbolically as ⎤ = ⎣ ⎦, ⎡

MPtb

(3.110)

where an entry of ‘’ denotes a component the effect of the perturbative part A on which comes from a25 , a26 , a35 , a36 , a45 , a46 , a55 , a56 , a66 , and that of ‘’ denotes a component the effect of A on which comes from a22 , a23 , a33 , a44 , a24 , a34 . On the other hand, from (3.70), (3.73) and (3.106) we have ⎡

1 0 ⎢ μ − V sin2 φ ⎢ sin2 φ cos2 φ TIso (φ)−1=⎢ ⎢ + ⎣ 2 μ − V sin φ λ + 2μ − V sin2 φ Sym.

0

⎤

⎥ 1 1 − cos φ sin φ ⎥ ⎥ 2 2 μ − V sin φ λ + 2μ − V sin φ ⎥ ⎥ 2 2 ⎦ cos φ sin φ + 2 2 μ − V sin φ λ + 2μ − V sin φ

(3.111)

124

and

K. Tanuma

⎡

V cos φ sin φ 0 ⎢ Iso 0 (V − λ − μ) cos φ sin φ R (φ) = ⎣ 0

μ cos2 φ − λ sin2 φ

0 λ cos2 φ − μ sin2 φ

⎤ ⎥ ⎦ . (3.112)

(V + λ + μ) cos φ sin φ

The (i, 1) and (1, i) components (i = 2, 3) of the matrices TIso (φ)−1 , RIso (φ) vanish, so do those of the matrices TIso (φ)−1 RIso (φ)T , RIso (φ)TIso (φ)−1 . We write these matrices symbolically as ⎤ ⎡ ∗00 (3.113) MIso = ⎣ 0 ∗ ∗ ⎦ , 0∗∗ where ‘∗’ denotes a possibly non-zero component. Now let us evaluate the products of matrices in the integrand (3.105). We see from (3.110) and (3.113) that ⎡ ⎤ MPtb MIso = ⎣ ⎦ , and

⎡

Iso

Ptb

M M

⎤ = ⎣ ⎦.

From this observation, the effect of the perturbative part A = (ars ) on the (2, 2) component of (3.105) comes from a22 , a23 , a33 , a44 , a24 and a34 . Next we prove that the effect of the perturbative part A = (ars ) on s22 (v), i.e., on the (2, 2) component of SPtb 3 (v), comes only from a22 , a23 , a33 and a44 . Since (3.105) is to be integrated over the interval [−π, π ] with respect to φ (see (3.96)), it is sufficient to show that the coefficients of a24 and a34 in the (2, 2) component of (3.105) are odd functions in φ. By (3.111) and (3.112), we can write ⎤ ⎡ even 0 0 TIso (φ)−1 = ⎣ 0 even odd ⎦ 0 odd even and

⎤ odd 0 0 RIso (φ) = ⎣ 0 odd even ⎦ , 0 even odd ⎡

where an entry of ‘even’ (resp. ‘odd’) denotes a component which is an even (resp. odd) function in φ. Then we get ⎡ ⎤ odd 0 0 TIso (φ)−1 RIso (φ)T = ⎣ 0 odd even ⎦ (3.114) 0 even odd


125

and ⎤ odd 0 0 = ⎣ 0 odd even ⎦ . 0 even odd ⎡

RIso (φ)TIso (φ)−1

(3.115)

On the other hand, looking at the coefficients of a24 and a34 in (3.107), (3.108) and (3.109) carefully, we can write ⎡

× QPtb (φ) = ⎣ × × ⎡ × RPtb (φ) = ⎣ × × ⎡ × TPtb (φ) = ⎣ × ×

× a24 -odd a24 , a34 -even × a24 -even a24 , a34 -odd × a24 -odd a24 , a34 -even

⎤ × a24 , a34 -even ⎦ , a34 -odd ⎤ × a24 , a34 -odd ⎦ , a34 -even ⎤ × a24 , a34 -even ⎦ , a34 -odd

(3.116)

(3.117)

(3.118)

where the label ‘aij-odd’ (resp. ‘aij-even’) signifies that the component in question has its aij coefficient being an odd (resp. even) function in φ, and ‘×’ denotes a component which does not depend on a24 and a34 . Thus, we obtain from (3.114) and (3.117), ⎡

Ptb

R

× (φ)T (φ) R (φ) = ⎣ × × Iso

−1

Iso

T

× a24 , a34 -odd a24 , a34 -even

⎤ × a24 , a34 -even ⎦ , a24 , a34 -odd

(3.119)

× a24 , a34 -odd a24 , a34 -even

⎤ × a24 , a34 -even ⎦ , a24 , a34 -odd

(3.120)

and then, ⎡

× RIso (φ)TIso (φ)−1 RPtb (φ)T = ⎣ × × and from (3.114), (3.115) and (3.118),

RIso (φ)TIso (φ)−1 TPtb (φ)TIso (φ)−1 RIso (φ)T ⎤ ⎡ × × × a24 , a34 -even ⎦ . a24 , a34 -odd = ⎣× × a24 , a34 -even a24 , a34 -odd

(3.121)

Then we can observe that the coefficients of a24 and a34 in the (2, 2) components of (3.116), (3.119), (3.120) and (3.121) are odd functions in φ, and so are the coefficients of a24 and a34 in the (2, 2) component of the integrand (3.105). Therefore, the terms linear in a24 and a34 included in the (2, 2) component of SPtb 3 (v) vanish. This completes the proof of the lemma.

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K. Tanuma

Proof of Theorem 3.19 41 Let us start with an orthorhombic elastic half-space x3 ≤ 0 whose elasticity tensor C is given by (1.15) and consider Rayleigh waves that propagate along its free surface. Under the assumption that C is expressed by (3.89), its perturbative part A has the following orthorhombic form: ⎡ ⎤ a11 a12 a13 0 0 0 ⎢ a22 a23 0 0 0 ⎥ ⎢ ⎥ ⎢ a33 0 0 0 ⎥ ⎢ ⎥. A = (ars ) = ⎢ a44 0 0 ⎥ ⎢ ⎥ ⎣ a55 0 ⎦ Sym. a66 Then the phase velocity of Rayleigh waves which propagate along the surface of the half-space x3 ≤ 0 in the direction of the 2-axis satisfies the secular equation42 ROrth (v) = 0,

(3.122)

where 2 2 ROrth (v) = C33 C44 (C22 − V)V 2 − (C44 − V) C33 (C22 − V) − C23

(3.123)

and V = ρv 2 . The components of the perturbative part of A included in ROrth (v) are a22 , a23 , a33 and a44 . By Lemma 3.23, these are exactly the same components that affect the (2, 2) component of SPtb 3 (v). Furthermore, by Lemma 3.22, the effect of the perturbative part A = (ars ) of (3.91) on the perturbation formula comes only from these components. Hence it should also be possible to use (3.122) as a secular equation to obtain the first order perturbation formula for the velocity of Rayleigh waves which propagate along the surface of the half-space of a weakly anisotropic elastic medium whose elasticity tensor is given by (3.89), (3.90) and (3.91). In fact, by the same arguments as in Lemma 3.22, from the Taylor expansion of v R around A = (ars ) = 0 we get ∂v R + ars , (3.124) v R ≈ v Iso R ∂ars (ars )=0 where the summation in the second term of the right hand side is taken for the indices (r, s) = (2, 2), (2, 3), (3, 3) and (4, 4). Moreover, from the implicit function theorem we have .−1 ∂ ROrth ∂v R ∂ ROrth =− . (3.125) ∂aij (ars )=0 ∂aij (ars )=0, v=vIso ∂v (ars )=0, v=vIso R R

41 Of course, it is possible to compute s (v) directly by taking the angular average of the (2, 2) 22 component of (3.105) and then use Lemma 3.22 to obtain the result. But here, we take full advantage of Lemma 3.23 to derive the perturbation formula by a much simpler method. 42 This equation is well known (see, for example, [17, 64]). It is possible to derive (3.122) from (3.47) (see Section 12.10 of [77] and Exercise 3-8).


127

Since ROrth (v) is given by (3.123) explicitly, we can calculate directly its derivatives in (3.125). For example, we get ∂ ROrth ∂ ROrth = 2ρv Iso R ∂v ∂V (ars )=0, v=v Iso R

.

Iso 2 (ars )=0, V=V Iso R =ρ(v R )

Since C22 = C33 = λ + 2μ, C44 = μ and C23 = λ when (ars ) = 0, we have d ∂ ROrth = μ(λ + 2μ)(λ + 2μ − V)V 2 ∂V dV (ars )=0, V=V Iso R 2 − (μ − V) (λ + 2μ)(λ + 2μ − V) − λ2 . Iso V=V R

Then, by a simple computation, this becomes (λ + μ) 8μ2 (3λ + 4μ) − 16μ(λ + 2μ)V + 3(λ + 2μ)V 2 V=V Iso , R

which is equal to D(v) in (3.94). We also have ∂ ROrth ∂a22 (ars )=0, v=vIso R Orth ∂R = ∂C22 (ars )=0, v=vIso R

2 = C33 C44 V 2 − 2C33 (C44 − V) C33 (C22 − V) − C23

(ars )=0, V=V Iso R

= (λ + 2μ) −8μ2 (λ + μ) + 2μ(5λ + 6μ)V − (2λ + 3μ)V 2

V=V Iso R

,

which is equal to the numerator of γ22 (v) in (3.94). A similar method can be applied ROrth ROrth = ∂∂C becomes to get the numerators of the other γij(v)’s in (3.94). Note that ∂ ∂a 44 44 Iso cubic in V. In this case we use (3.83) for V = V R to reduce the power of V. Then we can obtain a quadratic expression in V, which can be factorized and gives the numerator of γ44 (v).

3.6 Rayleigh Waves in Anisotropic Elasticity Let us turn to the general anisotropic material whose elasticity tensor C = Cijkl i, j,k,l=1,2,3 has the symmetries (1.4) and (1.6) and satisfies the strong convexity condition (1.7). Then C has 21 independent components and is expressed under the Voigt notation as (1.12). Let m and n be orthogonal unit vectors in R3 . As in Section 3.3, we consider Rayleigh waves in the half-space n · x ≤ 0 which propagate along the surface n · x = 0 in the direction of m with the phase velocity v R in the subsonic range 0 < v < v L , where v L = v L (m, n) is the limiting velocity in Definition 3.1. In this section we discuss the existence of Rayleigh waves.

128

K. Tanuma

3.6.1 Limiting Wave Solution In Section 3.3, we have obtained several equivalent conditions for the existence of Rayleigh waves which propagate along the surface of an elastic half-space. These conditions are given by the formulas at v = v R , i.e., at the phase velocity of the Rayleigh waves. Hence they are useful in finding the secular equations for the velocity of the Rayleigh waves for isotropic and for weakly anisotropic elastic media (Section 3.4, Section 3.5). However, for general anisotropic elasticity, the existence of Rayleigh waves is not always guaranteed, let alone obtaining their velocity analytically. Hence it is important to have certain criteria for the existence of Rayleigh waves without using their velocity. Thus, in this section we present existence criteria that are based on the limiting velocity. Namely, we give criteria for the existence of Rayleigh waves in an elastic half-space that are expressed in terms of the behavior of S3 (v) and Z(v) in the transonic limit v ↑ v L (m, n), respectively, and in terms of the transonic state of the slowness section, which we shall define later. Before presenting the existence criteria, in this subsection we establish several concepts associated with the limiting velocity v L (m, n). Let us recall from (2) of Lemma 3.2 and Lemma 3.3 that for velocity v such that 0 ≤ v < v L (m, n) there exist surface-wave solutions to (3.3) of the form (3.27) or (3.29) or (3.31) corresponding to the three complex pα (α = 1, 2, 3) with Im pα > 0. When v = v L (m, n), either one, two, or three pα (Im pα > 0) are found to be real. In fact, from (3.17), (3.20) and (3.26) there exists at least one φˆ − π2 < φˆ < π2 such that ˆ det Q(φ)

⎛

= det ⎝

⎞

3

d ˆ ˆ + nl sin φ) ˆ ⎠ Cijkl (m j cos φˆ + n j sin φ)(m l cos φ

j,l=1

= 0.

i↓k→1,2,3

ˆ this becomes Divided by cos6 φ, ⎞ ⎛ 3 d ˆ ˆ ⎠ Cijkl (m j + n j tan φ)(m det ⎝ l + nl tan φ) j,l=1

= 0,

i↓k→1,2,3

which, by (3.11), is equivalent to det[Q + p(R + RT ) + p2 T] = 0 ˆ with real p = tan φ. When v = v L (m, n), solution (3.4) becomes √ −1 k(m·x+tan φˆ n·x−v L t)

u = (u1 , u2 , u3 ) = a e−

,

(3.126)

ˆ This solution represents a plane wave where a = 0 satisfies (3.10) with p = tan φ. called a limiting body wave; it has velocity vL |m + tan φˆ n|

ˆ = v L | cos φ|


129

and has direction of propagation m + tan φˆ n =

m cos φˆ + n sin φˆ . cos φˆ

(3.127)

/ in (3.16) with φ = The direction of the vector (3.127) coincides with that of m φˆ − π2 < φˆ < π2 . Note that the solution (3.126), unlike the surface-wave solution (3.4), no longer decays exponentially with depth beneath the surface n · x = 0. Solution (3.126) is called a limiting wave solution. Limiting wave solutions are given a useful geometric interpretation in terms of the slowness section in the m-n plane. /=m / (φ) = (/ /2 , m /3 ) be the unit vector in (3.16). In an elastic material, Let m m1 , m the solution to (3.3) of the form √ m·x−c(φ) t) −1 k(/

u = (u1 , u2 , u3 ) = a e−

(3.128)

/ , wave number k, represents a body wave which has direction of propagation m velocity c(φ), and polarization a = a(φ). Substituting this into (3.3) leads to ⎛ ⎞ 3 ⎝ / jm /l ⎠ Cijkl m a = ρ c(φ)2 a, (3.129) j,l=1

i↓k→1,2,3

which implies that ρ c(φ) and a(φ) are an eigenvalue and an eigenvector of the real symmetric matrix ⎛ ⎞ 3 ⎝ / jm /l ⎠ Cijkl m , (3.130) 2

j,l=1

i↓k→1,2,3

respectively. We call this matrix the acoustical tensor. As in (3.23), let λi (φ) (i = 1, 2, 3),

0 < λ1 (φ) ≤ λ2 (φ) ≤ λ3 (φ)

be the eigenvalues of the acoustical tensor (3.130). Corresponding to these eigenval/ , velocity ues, there exist three body waves which have direction of propagation m ( ci (φ) =

λi (φ) (i = 1, 2, 3), ρ

0 < c1 (φ) ≤ c2 (φ) ≤ c3 (φ)

(3.131)

and polarizations ai = ai (φ) ∈ R3 (i = 1, 2, 3), which are mutually orthogonal. /=m / (φ) be Definition 3.24 For given orthogonal unit vectors m and n in R3 , let m given by (3.16) and let ci (φ) (i = 1, 2, 3) be the velocity (3.131) of the body wave / . Then the slowness section in the m-n plane which propagates in the direction of m consists of the three closed curves which are generated by the radius vectors 1 /, m ci (φ)

i = 1, 2, 3

(−π < φ ≤ π ).

Since the acoustical tensor (3.130) is π -periodic in φ, the slowness section is centrosymmetric with respect to the origin. The curve corresponding to the slowest

130

K. Tanuma

velocity c1 (φ) defines the silhouette of the slowness section and is called the outer profile. From (3.25), (3.26) and (3.131) it follows that v L = v L (m, n) =

min

− π2

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