GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES

## WAVE PROPAGATION IN COMPOSITE

## MATERIALS

by Demet ERSOY

July, 2008 ˙IZM˙IR

## MATERIALS

A Thesis Submitted to the

Graduate School of Natural and Applied Sciences of Dokuz Eylül University In Partial Fulfillment of the Requirements for the Degree of Master of Science in

Mathematics

by Demet ERSOY

July, 2008 ˙IZM˙IR

We have read the thesis entitled ”WAVE PROPAGATION IN COMPOSITE MATE-RIALS” completed by Demet ERSOY under supervision of PROF. DR. VALERY G. YAKHNO and we certify that in our opinion it is fully adequate, in scope and in quality, as a thesis for the degree of Master of Science.

————————————– PROF. DR. VALERY G. YAKHNO

Supervisor

————————————– ————————————–

(Jury Member) (Jury Member)

————————————– Prof. Dr. Cahit HELVACI

Director

Graduate School of Natural and Applied Sciences

I would like to express my deepest gratitude to my supervisor Prof. Dr. Valery G. Yakhno for his continual presence, invaluable guidance, help and never ending patience throughout the course of this work.

I would like to thank to Prof. Dr. ˙Ismihan Bayramoˇglu for his understanding, encourage-ments and valuable suggestions. Special thanks to all members of Izmir University of Eco-nomics Department of Mathematics and to all members of Dokuz Eylül University Department of Mathematics.

I would like to express my gratitude to TÜB˙ITAK for their generous financial support.

I am also grateful to my fiance Murat ÖZDEK and to my family for their continual presence, understanding, endless supporting and confidence to me throughout my life.

Finally, it is a pleasure to express my gratitude to all my close friends who have always cared about my work, and increased my motivation which I have strongly needed.

Demet ERSOY

ABSTRACT

The system of anisotropic elasticity with piecewise constant coefficients is considered in this thesis. The main object of the thesis is to model an initial value problem (IVP) and an initial boundary value problem (IBVP) for the considered system. The main results are explicit formulae for solutions of initial value problem and initial boundary value problem. Using these formulae the simulation of elastic waves have been obtained. Results of the simulations have clear physical interpretation of wave propagation in layered medium from the point source.

The method of characteristics has been used for constructing explicit formulae and MAT-LAB codes has been successfully applied for the simulation of the waves.

Keywords: anisotropic elastic system, elastic layered medium, initial value problem, initial boundary value problem, modeling, simulation, wave propagation.

ÖZ

Bu tezde parçalı sabit katsayılı, anizotropik elastik sistem çalı¸sıldı. Bu tezdeki ana hedef çalı¸sılan sistemin ba¸slangıç deˇger problemine (BDP) ve ba¸slangıç sınır deˇger problemine (BSDP) modellenmesidir. Bu ba¸slangıç deˇger ve ba¸slangıç sınır deˇger problemlerinin temel sonucu formüllerle belirtilen çözümleridir. Bu formüller kullanılarak elastik dalgaların simulasyonları elde edilmi¸s ve sonuçları katmanlı elastik ortamlarda olu¸san dalga yayılımının fiziksel yorum-larıyla uyum göstermi¸stir.

Çözümleri elde edebilmek için karakteristikler metodu kullanılmı¸s ve dalgaların simulasy-onları için MATLAB kodları ba¸sarılı bir ¸sekilde uygulanmı¸stır.

Anahtar Sözcükler: Anizotropik elastik sistem, elastik katmanlı ortam, ba¸slangıç deˇger prob-lemi, ba¸slangıç sınır deˇger probprob-lemi, modelleme, simulasyon, dalga yayılımı.

Page

THESIS EXAMINATION RESULT FORM ... ii

ACKNOWLEDGEMENTS ... iii

ABSTRACT... iv

ÖZ ... v

CHAPTER ONE – INTRODUCTION... 1

1.1 Equations of Anisotropic Elasticity ... 1

1.2 Problems and Methods for Equations of Anisotropic Elasticity ... 2

1.2.1 Plane Wave Formalism-Stroh Formalism ... 2

1.2.2 Green’s Functions Method... 3

1.2.3 Finite Element Method ... 3

1.2.4 Polynomial Solution Method ... 4

1.3 Plan of the Thesis... 4

CHAPTER TWO – INITIAL BOUNDARY VALUE PROBLEM OF ANISOTROPIC LAY-ERED ELASTIC HALF SPACE... 6

2.1 Statement of the Problem... 6

2.2 Assumptions ... 7

2.3 Reduction to IBVP for Wave Equations in Two Layered Half Space... 8

2.4 IBVP Of Isotropic Elastic Half Space ... 11

2.5 Construction of the Solution ... 12

2.6 Zero Step ... 12

2.6.1 The Region R1... 13

2.6.2 The Region R2... 14

2.7 The First Step... 15

2.7.1 The Region R3... 17

2.7.2 The Region R4... 18

2.7.3 The Region R5... 19

2.7.4 Matching Conditions Between R4 and R5... 20

2.8 General Case... 21

2.8.1 The Region R(4n-2) ... 24 vi

2.8.3 The Region R(4n)... 28

2.8.4 The Region R(4n+1) ... 29

2.8.5 Matching Conditions Between R(4n) and R(4n+1)... 29

2.9 Examples of Simulations of Wave Propagation in Two Layered Medium... 30

*2.9.1 Example 1 - The Pulse Point Source is Between the Boundaries x*3= 0
*and x*_{3}*= ` ... 30*

*2.9.2 Example 2 - The Pulse Point Source is between ` and ∞ ... 34*

2.10 Conclusion of Chapter Two ... 38

CHAPTER THREE – INITIAL VALUE PROBLEM IN THREE LAYERED MEDIUM... 39

3.1 IVP of Wave Equations in Three Layered Medium ... 40

3.2 Construction of the Solution ... 41

3.3 Zero Step ... 41

3.4 The First Step... 44

3.4.1 The Region R4... 45

3.4.2 The Region R5... 46

3.4.3 Matching Conditions Between R4 and R5... 47

3.4.4 The Region R6... 48

3.4.5 The Region R7... 49

3.4.6 Matching Conditions Between R6 and R7... 50

3.5 General Case... 50

*3.5.1 The Region R(5n − 2) ... 54*

*3.5.2 The Region R(5n − 1) ... 55*

*3.5.3 Matching Conditions Between R(5n − 2) and R(5n − 1)... 56*

*3.5.4 The Region R(5n)... 57*

3.5.5 The Region R(5n+1) ... 60

*3.5.6 The Region R(5n + 2) ... 61*

*3.5.7 Matching Conditions Between R(5n + 1) and R(5n + 2)... 62*

3.6 Examples of Simulations of Wave Propagation in Three Layered Medium... 62

*3.6.1 Example 1 - The Pulse Point Source is Between −∞ and x = 0... 63*

*3.6.2 Example 2 - The Pulse Point Source is Between the Boundaries x = 0 and*
*x = `... 67*

3.6.3 Example 3 - The Pulse Point Source is Between 0 and ∞... 70

CHAPTER FOUR – INITIAL VALUE PROBLEMS WITH ONE BOUNDARY ... 76

4.1 IVP-I... 76

4.1.1 The Region R1 and R2 ... 79

4.1.2 The Region R3... 80

4.1.3 The Region R4... 81

4.1.4 Matching Conditions Between R3 and R4... 82

4.2 Examples of Simulations of Wave Propagation ... 87

*4.2.1 Example 1 - The Pulse Point Source is Between −∞ and 0 ... 87*

4.2.1.1 Commands of Matlab for Example 1 ... 89

*4.2.1.2 Results of Simulations by the Formula (4.2.2) ... 91*

4.2.2 Example 2 - The Pulse Point Source is Between 0 and ∞... 92

4.2.2.1 Commands of Matlab for Example 2 ... 94

*4.2.2.2 Results of Simulations by the Formula (4.2.2) ... 96*

4.3 Conclusion of Chapter Four ... 98

CHAPTER FIVE – CONCLUSION... 99

REFERENCES ... 100

INTRODUCTION

Anisotropic elasticity has been mostly studied in different applied sciences such as engi-neering sciences, geophysics, solids and structures sciences etc. for the last thirty years due to its applications to composite materials. [(Ting, 2000), (Yahkno & Akmaz, 2005)]

The propagation of elastic waves in anisotropic media is governed by a system of second order partial differential equations.[see, for example, (Dieulesaint and Royer, 1980), (Fedorov, 1968), (Ting, 1996), (Ting & Barnet & Wu, 1990)] Here, we formulate shortly the problems which are considered in this thesis.

1.1 Equations of Anisotropic Elasticity

*Let x = (x*1*, x*2*, x*3*) ∈ R*2*× [0, ∞) and t ∈ R be variables. The displacement of the point x is*

*the vector u(x,t) = (u*1*, u*2*, u*3) with components

*u(x,t) = u _{j}(x,t),*

*for each j = 1, 2, 3.*

Initial value problem (IVP) of anisotropic elastic layered medium is described by the fol-lowing differential equations,

*ρ(x*_{3})∂
2_{u}*j*
*∂t*2 =
3

### ∑

*k=1*3

### ∑

*`=1*3

### ∑

*m=1*∂

*∂ xk*³

*cjk`m(x*3)

*∂ uj*

*∂ xm*´

*,*

*0 < x*3

*< `,*

*` < x*3

*< ∞,*

*t ∈ R,*

*j = 1, 2, 3,*(1.1.1)

with initial data

*uj(x, 0) =*ϕ*j(x) ,*
*∂ u _{j}*

*∂t*

*(x,t)*¯ ¯ ¯

*t=0*=ψ

*j(x),*

*0 < x*3

*< `,*

*` < x*3

*< ∞,*

*j = 1, 2, 3,*(1.1.2)

and matching conditions

*u _{j}(x*

_{3}

*,t)*¯ ¯ ¯

*x*3

*=`−0*

*= u*

_{j}(x_{3}

*,t)*¯ ¯ ¯

*x*3

*=`+0*

*,*(1.1.3) 3

### ∑

*`=1*3

### ∑

*m=1*

*c*

_{j3`m}(x_{3})

*∂ uj*

*∂ x*¯ ¯ ¯

_{m}*x*3

*=`−0*=

_{∑}

3
*`=1*3

### ∑

*m=1*

*c*

_{j3`m}(x_{3})

*∂ uj*

*∂ x*¯ ¯ ¯

_{m}*x*3

*=`+0*

*,*(1.1.4)

*where ` is given number,*
n

*cjk`m(x*3)

o_{3}

*jk`m=1* are the elastic moduli of the medium; *ρ(x*3*) > 0*

is the density of the elastic medium; ϕ* _{j},*ψ

_{j}and F_{j}are smooth functions for each j = 1, 2, 3.For initial boundary value problem (IBVP) of anisotropic elastic layered medium, we add
*the following boundary condition to the system (1.1.1) − (1.1.4), the boundary condition*

3

### ∑

*`=1*3

### ∑

*m=1*

*cj3`m*

*∂ u*

_{`}*∂ xm*¯ ¯ ¯

*x*3=0

*= Fj(t),*

*t ∈ R.*(1.1.5)

The elastic moduli of the medium is positive definite and satisfy the symmetry property

*cjk`m(x*3*) = c`m jk(x*3*) = ck j`m(x*3)

so that the system of anisotropic elasticity can be written as Cauchy problem of second order partial differential equations (Yahkno & Akmaz, 2005). The assumptions and detailed expla-nations can be found in the Chapter 2.

1.2 Problems and Methods for Equations of Anisotropic Elasticity

In the recent years, there exists substantially modern methods for solving initial and bound-ary value problems [(Boyce & DiPrima, 1992), (Dieulesaint & Royer, 1980), (Courant & Hilbert, 1989), (Cohen & Heikkola & Joly & Neittaan, 2003)] so that many researchers get a great chance to study more about the phenomena of the elastic wave propagation. And the developments of computer facilities-applications of analytical methods [(Rand & Rovenski, 2005), (Pavlovic, 2003)], special softwares such as Mathematica, Maple, Matlab etc.-provide better understanding of invisible elastic waves.

In this section, we mention some approaches for constructing solutions of IVPs and IBVPS.

*1.2.1 Plane Wave Formalism-Stroh Formalism*

Stroh formalism (Stroh, 1958) is a well-known approach for the system of elasticity in material sciences, applied mathematics and Physics community (Ting, 2000). In the method of plane wave approach, the system of elasticity is considered in a unbounded domain and the

solution of the systems have the form

*u(x,t) = a f (x.n − ct).* (1.2.1)

*where n, a, c are values to be determined. Substitution of (1.2.1) into the system, gives us*

*(Λ −λ I)a = 0,* (1.2.2)

where*λ = c*2 and Λ is second-order tensor with components

Λ* _{jl}*=

_{∑}

3
*k,m=1*

*c _{jklm}n_{k}nm*

*for all nk* *and nm. The construction of a solution is reduced to eigenvalues and eigenfunctions*

*problem for Λ.*

*1.2.2 Green’s Functions Method*

A different method to obtain the solution of the system is Green’s functions method. The
main idea of applying this method is Fourier transforms. The system is firstly solved in the
transformed domain. Then the solution of the system is derived by using
Fourier-inverse transform (Yang, 2004). In the article of Yang (2004), after applying 2-D Fourier
*trans-form with the variables (k*1*, k*2), the solution in Fourier-transformed domain is the following

*˜ui(k*1*, k*2*, y*3) =

Z Z

*u(y*1*, y*2*, y*3*)eik*α*y*α*dy*1*dy*2*,*

*where e stands for exponential function, i is the imaginary number for both variables y*1*, y*2*.*

Fourier-inverse transform yield the solution of the system in the domain.

*1.2.3 Finite Element Method*

Besides the analytical approaches, the numerical methods can be applied to solve the sys-tems. Finite element and finite difference methods are mostly used for some problems de-scribed by partial differential equations including system of elasticity. This approach is based on converting partial differential equations into an approximating system of ordinary differen-tial equations.

*1.2.4 Polynomial Solution Method*

Polynomial Solution method (PS-method) is an analytical method for constructing solution
of partial differential problems with the special form of initial data and inhomogeneous term
[(Yakhno & Akmaz, 2005), (Yakhno & Akmaz, 2007)]. In the article of Yakhno & Akmaz
(2005), it is proved that if the initial data are polynomials with respect to the lateral variables
*(x*1*, x*2*), then the solution of the problem which has coefficient functions depending on the other*

*variable x*3*, is in the form of polynomials depending of the same variables. The system in the*

article (Yakhno & Akmaz, 2005) can be written as follows

ρ∂
2* _{u}*γ

*j*

*∂t*2 = 3

### ∑

*k=1*∂ σγ

_{jk}*∂ x*

_{k}*,*

*j = 1, 2, 3, x ∈ R*3

_{, t > 0}*u*γ

*ϕγ*

_{j}(x, 0) =*(x),*

*j = 1, 2, 3, x ∈ R*3

*∂ u*γ

_{j}*∂t*

*(x,t)*¯ ¯ ¯

*t=0*=ψ γ

_{(x),}*3 where*

_{j = 1, 2, 3, x ∈ R}*u*γ

_{j}*= D*γ

*uj,*ϕγ

_{j}*= D*γϕ

*j,*ψγ

_{j}*= D*γψ

*j,*σγ

*=*

_{jk}### ∑

*`,m=1*

*Cjk`m*ε

*γ*

_{`m}*,*ε

*γ =1*

_{`m}_{2}µ

*∂ u*γ

_{`}*∂ xm*+

*∂ u*γ

*m*

*∂ x*¶

_{`}*.*

By applying Polynomial Solution method (PS-method), the solution can be written in the
form
*uj(x*1*, x*2*, x*3*,t) =*
∞

### ∑

*k=0*∞

### ∑

*s=0*

*Us,k*

_{j}*(x*3

*,t)xs*1

*xk*2 where

*Us,k*

_{j}*(x*3

*,t) =*1 ∂

_{s!k!}*s+k*

*∂ xs*1

*xk*2

*uj(x*1

*, x*2

*, x*3

*,t)*¯ ¯ ¯

*x*1

*=x*2=0

*,*

*j = 1, 2, 3;*

*s, k = 0, 1, 2.*

1.3 Plan of the Thesis

The system of anisotropic elasticity with piecewise constant coefficients is a mathematical
model of elastic wave propagation in layered media (composite elastic materials). The main
*goal of the thesis is to construct explicit formulae for the solutions of the considered problems*

*and using these formulae to obtain the simulation of the elastic waves. The thesis is organized*

In Chapter 1, we describe initial value problem (IVP) and initial boundary value problem (IBVP) of anisotropic elastic layered medium. We mention about other studies and approaches for solving the system of anisotropic elasticity and the way of finding solutions. In addition, the main goal of this thesis is given.

In Chapter 2, we reformulate initial boundary value problem of anisotropic elasticity in
two layered half space. The following section deals with the reduction of the system to the
Cauchy problem of the wave equation. For solving this problem, we separate the half space
*into different subregions. By using the method of characteristics, the solution of IBVP*
is investigated in these subregions. The explicit formula of a solution is constructed. The
simulations of wave propagation are obtained and analyzed.

Chapter 3 starts with the formulations of initial value problem (IVP) of the wave equation with piecewise constant coefficients. IBVP in Chapter 2 is reformulated as IVP in three lay-ered medium. Similarly, we separate the space into different subregions and the solution of the problem is investigated independently. By using the explicit formula of the solution, the simulations of wave propagation are obtained and analyzed.

Chapter 4 starts with initial value problem (IVP) that is formulated in Chapter 3 with two layered space. The techniques of finding solution is described in detail. Analysis of the formu-lations and the results of the simuformu-lations are dealed extensively. In addition, the Matlab codes of IVP in two layered medium are given.

INITIAL BOUNDARY VALUE PROBLEM OF ANISOTROPIC LAYERED ELASTIC HALF SPACE

*Let x = (x*1*, x*2*, x*3*) ∈ R*3*,* *t ∈ R* and let

*•* ϕ = (ϕ1*,*ϕ2*,*ϕ3) and ψ = (ψ1*,*ψ2*,*ψ3) be given vector functions

*depending on x;*

*• F = (F*1*, F*2*, F*3*) be given vector function depending on t;*

*• u = (u*1*, u*2*, u*3*) be unknown vector function depending on x and t.*

2.1 Statement of the Problem

Initial boundary value problem of anisotropic elastic half space is to find unknown function

*u = (u*1*, u*2*, u*3) satisfying the following system of differential equations

*ρ(x*3)
∂2_{u}*j*
*∂t*2 =
3

### ∑

*k=1*3

### ∑

*`=1*3

### ∑

*m=1*∂

*∂ xk*³

*cjk`m(x*3)

*∂ u*

_{j}*∂ xm*´

*, 0 < x*3

*< `, ` < x*3

*< ∞, t ∈ R*(2.1.1)

with initial data

*uj(x, 0) =*ϕ*j(x) ,*
*∂ u _{j}*

*∂t*

*(x,t)*¯ ¯ ¯

*t=0*=ψ

*j(x), 0 < x*3

*< `, ` < x*3

*< ∞,*(2.1.2)

the boundary condition

3

### ∑

*`=1*3

### ∑

*m=1*

*c*

_{j3`m}∂ u`*∂ x*¯ ¯ ¯

_{m}*x*3=0

*= F*

_{j}(t),*t ∈ R*(2.1.3)

and matching conditions

*uj(x*3*,t)*
¯
¯
¯
*x*3*=`−0*
*= uj(x*3*,t)*
¯
¯
¯
*x*3*=`+0*
(2.1.4)
3

### ∑

*`=1*3

### ∑

*m=1*

*cj3`m(x*3)

*∂ uj*

*∂ xm*¯ ¯ ¯

*x*3

*=`−0*= 3

### ∑

*`=1*3

### ∑

*m=1*

*cj3`m(x*3)

*∂ uj*

*∂ xm*¯ ¯ ¯

*x*3

*=`+0*(2.1.5)

*where ` is given number, (x*1*, x*2*) ∈ R*2 *and for each j = 1, 2, 3 uj(x,t) is jth component of*

*the displacement vector u(x,t) = (u*1*(x,t), u*2*(x,t), u*3*(x,t)) ;* *ρ(x*3) is the density of the elastic

medium and n

*c _{jk`m}(x*3)

o_{3}

*jk`m=1* are the elastic moduli of the medium.

2.2 Assumptions

*The elastic moduli c _{jk`m}(x*

_{3}) satisfy the symmetry properties

*cjk`m(x*3*) = c`m jk(x*3*) = ck j`m(x*3)

*and also cjk`m(x*3*) is positive definite for each j, k, `, m = 1, 2, 3 i.e. there exists a positive*

*constant M such that*

3

### ∑

*j,k,`,m=1*

*cjk`m(x*3)ε

*jk*ε

*`m*

*> M ·*3

### ∑

*j,k,`,m=1*ε2

*jk*

for allε* _{jk}* such that ε

*=ε*

_{jk}*.*

_{k j}*There exists a real, symmetric, positive definite 6x6 matrix C = (c*γσ*(x*3))*6x6* which includes

*cjk`m(x*3*) as its entries by relating the pair ( j, k) of indices j, k = 1, 2, 3 to a single index*

*γ = 1, 2, . . . , 6 and the pair (`, m) of indices `, m = 1, 2, 3 to a single index σ = 1, 2, . . . , 6 .*

*(1, 1) ↔ 1 ,* *(2, 3), (3, 2) ↔ 4 ,*
*(2, 2) ↔ 2 ,* *(1, 3), (3, 1) ↔ 5 ,*
*(3, 3) ↔ 3 ,* *(1, 2), (2, 1) ↔ 6 .*

(2.2.1)

*due to the symmetry properties. Then the matrix C is the following,*

*C(x*_{3}) =
*c*11*(x*3*) c*12*(x*3*) c*13*(x*3*) c*14*(x*3*) c*15*(x*3*) c*16*(x*3)
*c*21*(x*3*) c*22*(x*3*) c*23*(x*3*) c*24*(x*3*) c*25*(x*3*) c*26*(x*3)
*c*31*(x*3*) c*32*(x*3*) c*33*(x*3*) c*34*(x*3*) c*35*(x*3*) c*36*(x*3)
*c*41*(x*3*) c*42*(x*3*) c*43*(x*3*) c*44*(x*3*) c*45*(x*3*) c*46*(x*3)
*c*_{51}*(x*3*) c*52*(x*3*) c*53*(x*3*) c*54*(x*3*) c*55*(x*3*) c*56*(x*3)
*c*61*(x*3*) c*62*(x*3*) c*63*(x*3*) c*64*(x*3*) c*65*(x*3*) c*66*(x*3)
*= (c*_{γσ}*(x*_{3}))_{6×6}

In this work, we assume that
*c*_{γσ}*(x*_{3}) =
*c*1
γσ*,* *0 < x*3*< `;*
*c*2
γσ*,* *` < x*3*< ∞.*
*ρ(x*_{3}) =
ρ1_{,}* _{0 < x}*
3

*< `;*ρ2

_{,}*3*

_{` < x}*< ∞.*(2.2.2)

*where c*1

γσ*, c*2γσ*,* ρ1*> 0 and*ρ2*> 0 are given constants.*

2.3 Reduction to IBVP for Wave Equations in Two Layered Half Space

*Under these assumptions, the equations (2.1.1) − (2.1.3) can be written as follows*

ρ∂2*u*
*∂t*2 *= A*33
∂2_{u}*∂ x*32+
3

### ∑

*i= j6=3 ; i, j=1*

*Ai j*∂2

_{u}*∂ xi∂ xj,*

*0 < x*3

*< `,*

*` < x*3

*< ∞*(2.3.1)

*u(x, 0) =ϕ(x) ,*

*∂ u*

*∂t(x,t)*¯ ¯ ¯

*t=0*=

*ψ(x) ,*

*0 < x*3

*< `,*

*` < x*3

*< ∞*(2.3.2)

*A*33

*3 ¯ ¯ ¯*

_{∂ x}∂ u*x*3=0 + 2

### ∑

*i=1*

*Ai*

_{∂ x}∂ u*i*¯ ¯ ¯

*x*3=0

*= F(t),*

*t ∈ R*(2.3.3)

*where u is the vector u(x,t) =*
³

*u*_{1}*(x,t), u*_{2}*(x,t), u*_{3}*(x,t)*
´

under the assumption that u does
*not depend on the variables x*1 *and x*2 *i.e. u(x,t) = u(x*3*,t) . And where the matrices are*

as follow,
*A*11*(x*3) =
*c*11*(x*3*) c*16*(x*3*) c*15*(x*3)
*c*16*(x*3*) c*66*(x*3*) c*56*(x*3)
*c*_{15}*(x*3*) c*56*(x*3*) c*55*(x*3)
*,*
*A*12*(x*3) =1
2
*2c*16*(x*3) *c*12*(x*3*) + c*66*(x*3*) c*14*(x*3*) + c*56*(x*3)
*c*_{66}*(x*3*) + c*12*(x*3) *2c*26*(x*3) *c*46*(x*3*) + c*25*(x*3)
*c*56*(x*3*) + c*14*(x*3*) c*25*(x*3*) + c*46*(x*3) *2c*45*(x*3)
*,*
*A*22*(x*3) =
*c*_{66}*(x*_{3}*) c*_{26}*(x*_{3}*) c*_{46}*(x*_{3})
*c*26*(x*3*) c*22*(x*3*) c*24*(x*3)
*c*_{46}*(x*_{3}*) c*_{24}*(x*_{3}*) c*_{44}*(x*_{3})
*,*

*A*13*(x*3) =
*2c*_{15}*(x*3) *c*14*(x*3*) + c*56*(x*3*) c*13*(x*3*) + c*55*(x*3)
*c*56*(x*3*) + c*14*(x*3) *2c*46*(x*3) *c*36*(x*3*) + c*45*(x*3)
*c*55*(x*3*) + c*13*(x*3*) c*45*(x*3*) + c*36*(x*3) *2c*35*(x*3)
*,*
*A*33*(x*3) =
*c*55*(x*3*) c*45*(x*3*) c*35*(x*3)
*c*_{45}*(x*3*) c*44*(x*3*) c*34*(x*3)
*c*35*(x*3*) c*34*(x*3*) c*33*(x*3)
*,*
*A*23*(x*3) = 1_{2}
*2c*_{56}*(x*3) *c*46*(x*3*) + c*25*(x*3*) c*36*(x*3*) + c*45*(x*3)
*c*25*(x*3*) + c*46*(x*3) *2c*24*(x*3) *c*23*(x*3*) + c*44*(x*3)
*c*_{45}*(x*3*) + c*36*(x*3*) c*44*(x*3*) + c*23*(x*3) *2c*34*(x*3)
*,*
*A*1*(x*3) =
*c*15*(x*3*) c*56*(x*3*) c*55*(x*3)
*c*14*(x*3*) c*46*(x*3*) c*45*(x*3)
*c*13*(x*3*) c*36*(x*3*) c*35*(x*3)
*, A*2*(x*3) =
*c*56*(x*3*) c*25*(x*3*) c*45*(x*3)
*c*_{46}*(x*3*) c*24*(x*3*) c*44*(x*3)
*c*36*(x*3*) c*23*(x*3*) c*34*(x*3)
*.*
We assume that
*c*45*(x*3*) = 0, c*35*(x*3*) = 0, c*34*(x*3*) = 0,*
*c*_{54}*(x*_{3}*) = 0, c*_{53}*(x*_{3}*) = 0, c*_{43}*(x*_{3}*) = 0.*
*Under these assumptions, A*33 has diagonal form,

*A*33*(x*3) =
*c*55*(x*3) 0 0
0 *c*44*(x*3) 0
0 0 *c*33*(x*3)
(2.3.4)

*Then the equations (2.3.1) − (2.3.3) can be written as*

∂2_{u}*∂t*2 *= Λ(x*3)
∂2_{u}*∂ x*_{3}2 *,* *0 < x*3*< ` ,* *` < x*3*< ∞, t ∈ R,*
*u(x, 0) =ϕ(x) ,* *∂ u*
*∂t(x,t)*
¯
¯
¯
*t=0*=*ψ(x) ,* *0 < x*3*< ` ,* *` < x*3*< ∞*
*Λ(x*3)* _{∂ x}∂ u*
3
¯
¯
¯

*x*3=0

*= F(t),*

*t ∈ R*

*where Λ(x*3) =* _{ρ(x}*1

3)*A*33*(x*3*),* *ρ(x*3*) > 0.*

*Consider the matching conditions (2.1.4) − (2.1.5), The equation (2.1.4) is obvious. *
*Un-der the above assumptions and notations in (2.2.1) , the equation (2.1.5) has the form,*

*A*_{33}*(x*_{3})*∂ u*
*∂ x*_{3}
¯
¯
¯
*x*3*=`−0*
+

_{∑}

2
*i=1*

*A*

_{i}(x_{3})

*∂ u*

*∂ x*¯ ¯ ¯

_{i}*x*3

*=`−0*

*=A*

_{33}

*(x*

_{3})

*∂ u*

*∂ x*

_{3}¯ ¯ ¯

*x*3

*=`+0*+

_{∑}

2
*i=1*

*A*

_{i}(x_{3})

*∂ u*

*∂ x*¯ ¯ ¯

_{i}*x*3

*=`+0*

*Since there is no dependence on x*1 *and x*2*. So the equation (2.2.1) has the following form,*

*Λ(x*3)* _{∂ x}∂ u*
3
¯
¯
¯

*x*3

*=`−0*

*= Λ(x*3)

*3 ¯ ¯ ¯*

_{∂ x}∂ u*x*3

*=`+0*

*where Λ(x*

_{3}) = 1

*ρ(x*

_{3})

*A*33

*(x*3

*) ,*

*ρ(x*3

*) > 0 where A*33

*(x*3

*) is defined in (2.3.4).*

Notice that the matrix Λ is diagonal. Since the matrix C is positive definite and*ρ(x*_{3}*) > 0,*
then
Λ =
*d*2
11*(x*3) 0 0
0 *d*2
22*(x*3) 0
0 0 *d*2
33*(x*3)
(2.3.5)
*where d*2_{11}=*c*55*(x*3)
ρ_{(}*x*3) *, d*
2
22=
*c*44*(x*3)
ρ_{(}*x*3) *, d*
2
33=
*c*33*(x*3)
ρ_{(}*x*3) *.*

*The initial and boundary value problem of anisotropic elastic half space is for each k = 1, 2, 3,*

∂2_{U}*k*
*∂t*2 *= dkk*2*(x*3)∂
2_{U}*k*
*∂ x*_{3}2 *,* *0 < x*3*< ` , ` < x*3*< ∞,* *t ∈ R,* (2.3.6)

with initial and boundary conditions,

*U _{k}(x, 0) = Φ_{k}(x) ,*

*∂Uk*

*∂t*

*(x,t)*¯ ¯ ¯

*t=0*= Ψ

*k(x)*

*0 < x*3

*< ` , ` < x*3

*< ∞*(2.3.7)

*d*2

_{kk}*(x*3)

*∂U*

_{k}*∂ x*3 ¯ ¯ ¯

*x*3=0

*= Fk(t) ,*

*t ∈ R*(2.3.8)

and the matching conditions,

*Uk(x*3*,t)*
¯
¯
¯
*x*3*=`−0*
*= Uk(x*3*,t)*
¯
¯
¯
*x*3*=`+0*
(2.3.9)
*d _{kk}*2

*(x*3)

*∂U*3

_{∂ x}k*(x*3

*,t)*¯ ¯ ¯

*x*3

*=`−0*

*= d*2

_{kk}*(x*3)

*∂U*3

_{∂ x}k*(x*3

*,t)*¯ ¯ ¯

*x*3

*=`+0*(2.3.10)

2.4 IBVP Of Isotropic Elastic Half Space

Let Φ*k(x*3*), Ψk(x*3*) and dkk(x*3*) for k = 1, 2, 3 are in the form,*

*d _{kk}(x*

_{3}) = α

*3*

_{k}, 0 < x*< `;*β

*3*

_{k}, ` < x*< ∞.*(2.4.1) Φ

*k(x*3) = ϕ

*k(x*3

*), 0 < x*3

*< `;*

*w*3

_{k}(x*), ` < x*3

*< ∞.*Ψ

*k(x*3) = ψ

*k(x*3

*), 0 < x*3

*< `;*φ

*3*

_{k}(x*),*

*` < x*3

*< ∞.*(2.4.2)

In our further consideration, we consider the scalar equation with fixed k together with initial data and boundary condition. We will omit the index k for simplicity writing.

Figure 2.1 The Regions for n=2,3,4,. . .

*Initial boundary value problem (2.3.6) − (2.3.10) may be written in the form of*

*Uk(x*3*,t) =*
*uk(x*3*,t), 0 < x*3*< `;*
*vk(x*3*,t), ` < x*3*< ∞.*
as follows,
∂2_{u}*k*
*∂t*2 =α
2
*k*
∂2_{u}*k*
*∂ x*_{3}2 *,* *0 < x*3*< ` ,* *t ∈ R,* (2.4.3)
∂2_{v}*k*
*∂t*2 =β*k*2
∂2_{v}*k*
*∂ x*32 *,* *` < x*3*< ∞ ,* *t ∈ R,* (2.4.4)

with initial and boundary data,
*u _{k}(x*

_{3}

*, 0) =*ϕ

_{k}(x_{3}

*),*

*∂ uk*

*∂t*

*(x*3

*,t)*¯ ¯ ¯

*t=0*=ψ

*k(x*3

*),*

*0 < x*3

*< `,*(2.4.5)

*vk(x*3

*, 0) = wk(x*3

*),*

*∂ v*3

_{∂t}k(x*,t)*¯ ¯ ¯

*t=0*=φ

*k(x*3

*),*

*` < x*3

*< ∞,*(2.4.6) α

*2*

_{k}*∂ uk*

*∂ x*

_{3}¯ ¯ ¯

*x*3=0

*= F*

_{k}(t) ,*for k = 1, 2, 3.*(2.4.7)

and the matching conditions,

*uk*
¯
¯
¯
*x*3*=`−0*
*= vk*
¯
¯
¯
*x*3*=`+0*
(2.4.8)
α* _{k}*2

*∂ uk*

*∂ x*3 ¯ ¯ ¯

*x*3

*=`−0*=β

*2*

_{k}*∂ vk*

*∂ x*3 ¯ ¯ ¯

*x*3

*=`+0*(2.4.9)

2.5 Construction of the Solution

To find the solution, we separate half space into subregions and the formulation of the
*solution of the problem (2.4.3) − (2.4.9) is constructed for each subregions, independently by*
*using the method of characteristics.*

*uk(x*3*,t) =*

n

*ukm(x*3*,t),* *if (x*3*,t) ∈ Rm* (2.5.1)

*Here, k denotes the the component of the matrix u(x*3*,t) and m denotes the index of *

subre-gion.

2.6 Zero Step

Zero step includes the regions R1 and R2 (see, Figure 2.1) Let us consider the problem
*(2.4.3) − (2.4.9) for zero step. Notice that in this step there is no*

boundary, so we use only initial conditions.

*Theorem 2.6.1. Let* ϕ*k(x*3*),*ψ*k(x*3*), wkand*φ*k(x*3*) be given continuous*

*functions depending on x*3*; uk(x*3*,t) is unknown function in the form (2.5.1). Then the solution*

*Uk(x*3*,t) =*
1
2[ϕ*k(x*3+α*kt) +*ϕ*k(x*3*−*α*kt)]*
+ 1
2α* _{k}*
Z

*3+α*

_{x}*kt*

*x*3

*−*α

*kt*ψ

*(γ)dγ,*

_{k}*if (x*3

*,t) ∈ R1;*1 2

*[wk(x*3+β

*kt) − w(x*3

*−*β

*kt)]*+ 1 2β

*k*Z

_{x}_{3}

_{+}

_{β}

_{k}_{t}*x*3

*−*β

*kt*φ

*k*(ν)dν ,

*if (x*3

*,t) ∈ R2.*(2.6.1)

*where*

*R1 =*½

*(x*3

*,t)*¯ ¯

*¯ 0 < x*3

*< ` ,*

*t <*

_{α}

*x*3

*k*

*∧ t <*

*` − x*3 α

*¾*

_{k}*R2 =*½

*(x*3

*,t)*¯ ¯

*¯ ` < x*3

*< ∞ ,*

*t <*

*x*3

_{β}

*− `*

*k*¾

*for each k = 1, 2, 3.*

*Proof. Let us consider the problem (2.4.3) − (2.4.4) with initial conditions*

*(2.4.5) − (2.4.6) in the regions R1 and R2, respectively.*

*2.6.1 The Region R1*

*Let us consider the problem (2.4.3) − (2.4.9) in the region R1,*

*R1 =*
½
*(x*3*,t)*
¯
¯
*¯ 0 < x*3*< ` ,* *t <*_{α}*x*3
*k* *∧ t <*
*` − x*3
α*k*
¾
*for k = 1, 2, 3.*

*The equation (2.4.3) can be written*

*∂ qk*
*∂t* *−*α*k*
*∂ qk*
*∂ x*_{3} *= 0,* *(x*3*,t) ∈ R1,* (2.6.2)
*∂ uk*
*∂t* +α*k*
*∂ uk*
*∂ x*3 *= qk(x*3*,t),* *(x*3*,t) ∈ R1.* (2.6.3)

For the solution of the problem, we use the method of characteristics. So, the characteristics of
*the equations (2.6.2) − (2.6.3) are respectively,*

*d*ξ

*d*τ *= −αk,* *ξ (t) = x*3 ; *ξ = −αkτ + x*3+α*kt,*

*d*ξ

*d*τ =α*k,* *ξ (t) = x*3 ; ξ = α*kτ + x*3*−*α*kt.*
By integrating along the characteristics, we get the following

*qk(x*3*,t) =*ψ*k(x*3+α*kt) +*α*k*ϕ*k0(x*3+α*kt)*
and
Z * _{t}*
0
∂
∂ τ
h

*uk(x*3

*−*α

*k(t −τ), τ)*i

*d*τ = Z

*0 ψ*

_{t}*k(x*3

*−*α

*kt + 2*α

*kτ)dτ*+α

*Z*

_{k}*0 ϕ*

_{t}*0*

*k(x*3

*−*α

*kt + 2*α

*kτ)dτ*Let

*x*3

*−*α

*kt + 2*α

*kτ = γ ,*2α

*kdτ = dγ*γ

*3*

_{low}= x*−*α

*kt ,*γ

*up= x*3+α

*kt*So, we get

*uk(x*3

*,t) − uk(x*3

*−*α

*kt, 0) =*1

_{2}[ϕ

*k(x*3+α

*kt) −*ϕ

*k(x*3

*−*α

*kt)]*+ 1 2α

*Z*

_{k}*3+α*

_{x}*kt*

*x*3

*−*α

*kt*ψ

*(γ)dγ*

_{k}*By substituting the initial conditions (2.4.5), we have the solution*

*uk(x*3*,t) =* 1_{2}[ϕ*k(x*3+α*kt) +*ϕ*k(x*3*−*α*kt)] +*_{2α}1
*k*
Z _{x}_{3}_{+}_{α}_{k}_{t}*x*3*−*α*kt*
ψ* _{k}*(γ)dγ, (x

_{3}

*,t) ∈ R1.*

*2.6.2 The Region R2*

*Let us consider the problem (2.4.3) − (2.4.9) in the region R2, for each k = 1, 2, 3.*
*The equation (2.4.4) can be written*

*∂ q _{k}*

*∂t*

*−*β

*k*

*∂ q*

_{k}*∂ x*

_{3}

*= 0 ,*

*(x*3

*,t) ∈ R2,*(2.6.4)

*∂ vk*

*∂t*+β

*k*

*∂ vk*

*∂ x*

*= qk(x*3

*,t) ,*

*(x*3

*,t) ∈ R2.*(2.6.5)

*The characteristic of the equation (2.6.4) − (2.6.5) are respectively,*

*d*ξ

*d*τ *= −βk,* *ξ (t) = x*3 ; *ξ = −βkτ + x*3+β*kt,*

*d*ξ

*d*τ =β*k,* *ξ (x*3*) = t* ; ξ = β*kτ + x*3*−*β*kt.*
Then by the same argument, we integrate along the characteristics so we get,

*v _{k}(x*

_{3}

*,t) =*1 2

*[wk(x*3+β

*kt) − w(x*3

*−*β

*kt)] +*1 2β

*Z*

_{k}*3+β*

_{x}*kt*

*x*3

*−*β

*kt*φ

*(ν)dν, (x*

_{k}_{3}

*,t) ∈ R2.*

2.7 The First Step

The first step includes the regions R3, R4 and R5 (see, Figure 2.1). In this step, we consider
*initial boundary data and also matching conditions defined on the boundary x = `.*

Before finding the solution for the first step, we must define the following
functions,
*uk(0,t) = gk(t),* *uk(`,t) = fk(t) and* *∂ u _{∂ x}k*
3
¯
¯
¯

*x*3

*=`*

*= Gk(t) .*(2.7.1)

We must construct these functions by initial and boundary data and also by the matching conditions.

*Theorem 2.7.1. Let* ϕ* _{k}(x*3

*),*ψ

*k(x*3

*), wkand*φ

*k(x*3

*) be given continuous*

*functions depending on x*3*; Fk(t) be given continuous function depending on t; uk(x*3*,t) is*

*unknown function in the form (2.5.1). Then the solution of the problem (2.4.3) − (2.4.9) for*
*the first step is the following,*

*Uk(x*3*,t) =*
*gk*
µ
*t −x*3
α* _{k}*
¶
+1
2[ϕ

*k(x*3+α

*kt) −*ϕ

*k(−x*3+α

*kt)]*+ 1 2α

*Z*

_{k}*3+α*

_{x}*kt*

*−x*3+α

*kt*ψ

*(µ)dµ,*

_{k}*if (x*

_{3}

*,t) ∈ R3;*

*fk*µ

*t +x*3

*− `*α

*k*¶ +1 2[ϕ

*k(x*3

*−*α

*kt) −*ϕ

*k(−x*3

*−*α

*kt + 2`)]*

*−*1 2α

*Z*

_{k}

_{x}_{3}

_{−}_{α}

*kt*

*−x*3

*−*α

*kt+2`*ψ

*k*(µ)dµ,

*if (x*3

*,t) ∈ R4;*

*f*µ

_{k}*t −x*3

*− `*β

*¶ +1 2*

_{k}*[wk(x*3+β

*kt) − wk(−x*3+β

*kt + 2`)]*+ 1 2β

*k*Z

*3+β*

_{x}*kt*

*−x*3+β

*kt+2`*φ

*(ν)dν,*

_{k}*if (x*

_{3}

*,t) ∈ R5.*(2.7.2)

*where*

*R3 =*½

*(x*3

*,t)*¯ ¯

*¯ 0 < x*3

*< ` ,*

_{α}

*x*3

*k*

*< t <` − x*3 α

*¾*

_{k}*,*

*R4 =*½

*(x*3

*,t)*¯ ¯

*¯ 0 < x*3

*< ` ,*

*` − x*

_{α}3

*k*

*< t <*

*x*3 α

*k*¾

*,*

*R5 =*½

*(x*3

*,t)*¯ ¯

*¯ ` < x*3

*< ∞ ,*

*x*3

_{β}

*− `*

*k*

*< t <*

*x*3

*− `*β

*k*+

*`*α

*k*¾

*,*

*and the functions defined in (2.7.1) are constructed by initial-boundary data and the matching*
*conditions as follows*
*gk(t) = (ϕk*(α*kt) −*ϕ*k*(0)) +
Z * _{t}*
0 ψ

*k*(α

*kτ)dτ −*1 α

*k*Z

*0*

_{t}*Fk*(τ)dτ (2.7.3)

*Gk(t) =*1 α

*k*

*f*

*0*

*k(t) +*ϕ

*k0(` −*α

*kt) −*1 α

*k*ψ

*α*

_{k}(` −*(2.7.4)*

_{k}t)*fk(t) =*

_{α}α

*k*

*k*+β

*k*[ϕ

*k(` −*α

*kt) −*ϕ

*k(`)] −*

_{α}1

*k*+β

*k*Z

_{`−}_{α}

*kt*

*`*ψ

*k(s)ds*+ β

*k*α

*+β*

_{k}*β*

_{k}[wk(` +*kt) − w(`)] +*1 α

*+β*

_{k}*Z*

_{k}

_{`+}_{β}

*kt*

*`*φ

*k(z)dz*(2.7.5)

*for each k = 1, 2, 3.*

*Proof. Let us consider the problem (2.4.3)−(2.4.4) with initial-boundary data (2.4.5)−(2.4.7)*

Now, we analyze the regions, independently.

*2.7.1 The Region R3*

*Let us consider the problem (2.4.3) − (2.4.9) in the region R3 (see, Figure 2.1), for k =*
*1, 2, 3.*
*R3 =*
½
*(x*_{3}*,t)*
¯
¯
*¯ 0 < x*3*< ` ,*
*x*3
α_{k}*< t <*
*` − x*3
α* _{k}*
¾

*The equation (2.4.3) can be written as in the form,*

*∂ q _{k}*

*∂t*

*−*α

*k*

*∂ q*

_{k}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R3,*(2.7.6)

*∂ u*

_{k}*∂t*+α

*k*

*∂ u*

_{k}*∂ x*

_{3}

*= qk(x*3

*,t) ,*

*(x*3

*,t) ∈ R3.*(2.7.7)

*The characteristic of the equation (2.7.6) − (2.7.7) are respectively,*

*d*ξ
*d*τ *= −αk,* *ξ (t) = x*3 ; *ξ = −αkτ + x*3+α*kt ,*
*d*ξ
*d*τ =α*k,* *ξ (t) = x*3 ; ξ = α*kτ + x*3*−*α*kt* and if ξ = 0 ; *τ = t −*
*x*3
α*k* *.*

By integrating along the characteristics,

*qk(x*3*,t) =*ψ*k(x*3+α*kt) +*α*k*ϕ*k0(x*3+α*kt)*

Then by integrating along the characteristic,

*u _{k}(x*3

*,t) − uk*µ

*0,t −x*3 α

*¶ = Z*

_{k}

_{t}*t−*

_{α}

*x3*

*k*ψ

*3*

_{k}(x*−*α

*kt + 2*α

*kτ)dτ*+α

*Z*

_{k}

_{t}*t−*

_{α}

*x3*

*k*ϕ

_{k}0(x_{3}

*−*α

*α*

_{k}t + 2*Let*

_{k}τ)dτ ,*x*3

*−*α

*kt + 2*α

*kτ = ν ,*2α

*kdτ = dν*ν

*low= −x*3+α

*kt ,*ν

*up= x*3+α

*kt*

*By substituting the initial conditions (2.4.5), we have the solution*
*uk(x*3*,t) = gk*
µ
*t −x*3
α*k*
¶
+1
2[ϕ*k(x*3+α*kt) −*ϕ*k(−x*3+α*kt)]*
+ 1
2α*k*
Z * _{x}*
3+α

*kt*

*−x*3+α

*kt*ψ

*(µ)dµ ,*

_{k}*(x*3

*,t) ∈ R3,*

*and the function gk(t) defined in (2.7.1) is the following,*

*gk(t) = (ϕk*(α*kt) −*ϕ*k*(0)) +
Z * _{t}*
0 ψ

*k*(α

*kτ)dτ −*1 α

*k*Z

*0*

_{t}*Fk*(τ)dτ .

*2.7.2 The Region R4*

*Let us consider the problem (2.4.3) − (2.4.9) in the region R4 (see, Figure 2.1), for k =*
*1, 2, 3.*
*R4 =*
½
*(x*3*,t)*
¯
¯
*¯ 0 < x*3*< ` ,* *` − x*_{α} 3
*k* *< t <*
*x*3
α*k*
¾

*The equation (2.4.3) can be written as in the form,*

*∂ q _{k}*

*∂t*+α

*k*

*∂ q*

_{k}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R4 ,*(2.7.8)

*∂ u*

_{k}*∂t*

*−*α

*k*

*∂ u*

_{k}*∂ x*

_{3}

*= qk(x*3

*,t) ,*

*(x*3

*,t) ∈ R4 .*(2.7.9)

*The characteristics of the equations (2.7.8) − ((2.7.9)) are respectively,*

*d*ξ
*d*τ =α*k,* *ξ (t) = x*3 ; ξ = α*kτ + x*3*−*α*kt ,*
*d*ξ
*d*τ *= −αk,* *ξ = −αkτ + x*3+α*kt , whenξ = ` ;* *τ = t +*
*x*3*− `*
α*k* *.*

By integrating along the characteristics, we get

*Similarly, we integrate along the characteristic and by using the boundary data (2.4.7), we get*
the following formula

*uk(x*3*,t) = fk*
µ
*t +x*3*− `*
α* _{k}*
¶
+1
2[ϕ

*k(x*3

*−*α

*kt) −*ϕ

*k(−x*3

*−*α

*kt + 2`)]*

*−*1 2α

*Z*

_{k}

_{x}_{3}

_{−}_{α}

*kt*

*−x*3

*−*α

*kt+2`*ψ

*k*(µ)dµ ,

*(x*3

*,t) ∈ R4.*

*2.7.3 The Region R5*

*Let us consider the problem (2.4.3) − (2.4.9) in the region R5 (see, Figure 2.1), for k =*
*1, 2, 3.*
*R5 =*
½
*(x*3*,t)*
¯
¯
*¯ ` < x*3*< ∞ ,* *x*3_{β}*− `*
*k*
*< t <* *x*3*− `*
β* _{k}* +

*`*α

*¾*

_{k}*The equation (2.4.4) can be written as in the form,*

*∂ q _{k}*

*∂t*

*−*β

*k*

*∂ q*

_{k}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R5,*(2.7.10)

*∂ v*

_{k}*∂t*+β

*k*

*∂ v*

_{k}*∂ x*

_{3}

*= qk(x*3

*,t) ,*

*(x*3

*,t) ∈ R5.*(2.7.11)

*The characteristics of the equations (2.7.10) − ((2.7.11)) are respectively,*

*d*ξ
*d*τ *= −βk,* *ξ (t) = x*3 ; *ξ = −βkτ + x*3+β*kt,*
*d*ξ
*d*τ =β*k,* *ξ (t) = x*3; ξ = β*kτ + x*3*−*β*kt, when* *ξ = `;* *τ = t −*
*x*3*− `*
β*k* *.*
So
*qk(x*3*,t) =*φ*k(x*3+β*kt) +*β*kw0k(x*3+β*kt) ,*

Similarly, by integrating along the characteristics and by using initial conditions, we get the
following formula
*vk(x*3*,t) = hk*
µ
*t −x*3*− `*
β*k*
¶
+1
2*[wk(x*3+β*kt) − wk(−x*3+β*kt + 2`)]*

+ 1
2β*k*

Z _{x}_{3}_{+}_{β}_{k}_{t}

*−x*3+β*kt+2`*

φ* _{k}*(ν)dν ,

*(x*3

*,t) ∈ R5.*

*To find the functions fk(t) and Gk(t) defined in (2.7.1), we must use the matching conditions*

*in (2.4.8) − (2.4.9).*

*2.7.4 Matching Conditions Between R4 and R5*

The formula for the region R4 is in the form,

*uk(x*3*,t) = fk*
µ
*t −` − x*3
α*k*
¶
+1
2[ϕ*k(x*3*−*α*kt) −*ϕ*k(−x*3*−*α*kt + 2`)]*
*−* 1
2α*k*
Z _{x}_{3}_{−}_{α}_{k}_{t}*−x*3*−*α*kt+2`*
ψ*k*(ν)dν ,

and the formula for the region R5 is in the form,

*vk(x*3*,t) = hk*
µ
*t +` − x*3
β* _{k}*
¶
+1
2

*[wk(x*3+β

*kt) − wk(−x*3+β

*kt + 2`)]*+ 1 2β

*Z*

_{k}

_{x}_{3}

_{+}

_{β}

_{k}_{t}*−x*3+β

*kt+2`*φ

*k*(ν)dν .

*By the first matching condition (2.4.8), we have,*

*u _{k}(` − 0,t) = v_{k}(` + 0,t) = f_{k}(t)*

*To use the second matching condition (2.4.9), we must differentiate the formulas for the regions*
*R4 and R5, and substitute x = `. Then we get the function Gk(t) defined in (2.7.1),*

*Gk(t) =* *∂ u _{∂ x}k*
3
¯
¯
¯

*x*3

*=`−0*= 1 α

_{k}f*0*

*k(t) +*ϕ

*k0(` −*α

*kt) −*

_{α}1

*k*ψ

*k(` −*α

*kt)*

*By using the second matching condition (2.4.9) and by integrating the resulting formula from*
*0 to t, we get the function fk(t) defined in (2.7.1) as follows,*

*fk(t) =*
α* _{k}*
α

*k*+β

*k*[ϕ

*k(` −*α

*kt) −*ϕ

*k(`)] −*1 α

*k*+β

*k*Z

_{`−}_{α}

_{k}_{t}*`*ψ

*k(s)ds*+ β

*k*α

*+β*

_{k}*β*

_{k}[wk(` +*kt) − w(`)] +*1 α

*+β*

_{k}*Z*

_{k}

_{`+}_{β}

*kt*

*`*φ

*k(z)dz*

2.8 General Case

*In zero and the first step, we have constructed the formulations of uk(x*3*,t), vk(x*3*,t) and*

*the functions gk(t), fk(t), Gk(t) defined in (2.7.1) for n = 0 and n = 1. After the first step, we*

*generalize the number of the step with index n, for n = 2, 3, . . . So, we reformulate the initial*
boundary value problem.

*Initial boundary value problem is to find u _{nk}(x*3

*,t) in the form*

*Unk(x*3*,t) =*
*u _{nk}(x*

_{3}

*,t), 0 < x*

_{3}

*< `;*

*vnk(x*3

*,t), ` < x*3

*< ∞.*

*for each k = 1, 2, 3 and n = 2, 3, . . . satisfying*

∂2_{u}*nk*
*∂t*2 =α*k*2
∂2_{u}*nk*
*∂ x*32 *,* *0 < x*3*< ` ,* *t ∈ R,* (2.8.1)
∂2_{v}*nk*
*∂t*2 =β*k*2
∂2_{v}*nk*
*∂ x*32 *,* *` < x*3*< ∞ ,* *t ∈ R,* (2.8.2)

with initial and boundary data,

*unk(x, 0) =*ϕ*k(x*3*) ,*
*∂ u _{nk}*

*∂t*

*(x,t)*¯ ¯ ¯

*t=0*=ψ

*k(x*3

*) ,*

*0 < x*3

*< ` ,*(2.8.3)

*vnk(x, 0) = wk(x*3

*) ,*

*∂ v*¯ ¯ ¯

_{∂t}nk(x,t)*t=0*=φ

*k(x*3

*) ,*

*` < x*3

*< ∞ ,*(2.8.4) α

*2*

_{k}*∂ unk*

*∂ x*

_{3}¯ ¯ ¯

*x*3=0

*= F*

_{k}(t) ,*t ∈ R*(2.8.5)

and the matching conditions,

*u _{nk}*
¯
¯
¯

*x*3

*=`−0*

*= v*¯ ¯ ¯

_{nk}*x*3

*=`+0*(2.8.6) α2

*k*

*∂ u*

_{nk}*∂ x*3 ¯ ¯ ¯

*x*3

*=`−0*=β2

*k*

*∂ v*

_{nk}*∂ x*3 ¯ ¯ ¯

*x*3

*=`+0*(2.8.7)

*The General case includes the regions R(4n − 2), R(4n − 1), R(4n) and R(4n + 1) (see, *
Fig-ure 2.1). Notice that, unlike in the first step, in the general case we have an additional subregion,
namely the region R(4n-2).

*also matching conditions defined on the boundary x = `.*

Before finding the solution for the general case, we must define the following
functions,
*unk(0,t) = gnk(t),* *unk(`,t) = fnk(t) and*
*∂ u _{nk}*

*∂ x*3 ¯ ¯ ¯

*x*3

*=`*

*= Gnk(t)*(2.8.8)

We must construct these functions by initial-boundary data and also by the matching conditions.
*Similar to (2.5.1), the solution of the problem (2.8.1) − (2.8.7) for the general case will be*
*found in the following form by using the method of characteristics.*

*ukn(x*3*,t) =*

n

*uknm(x*3*,t),* *if (x*3*,t) ∈ Rm* (2.8.9)

*Here, the index k denotes the component of the vector function u(x*3*,t), the index n denotes*

the number of the step and the index m denotes the number of subregion.

*Theorem 2.8.1. Let* ϕ* _{k}(x*3

*),*ψ

*k(x*3

*), wkand*φ

*k(x*3

*) be given continuous*

*functions depending on x*3*; Fk(t) be given continuous function depending on t; uk(x*3*,t) is*

*unknown function in the form (2.5.1). Then the solution of the problem (2.8.1) − (2.8.7) for*
*the general case is the following,*

*Uk(x*3*,t) =*
*g _{(n−1)k}*
µ

*t −x*3 α

*k*¶ +1 2

*f(n−1)k*µ

*t +x*3

*− `*α

*k*¶

*−*1 2

*f(n−1)k*µ

*t −x*3

*+ `*α

*k*¶ +α

*k*2 Z

*α*

_{t+}x3−`*k*

*t−x3+`*

_{α}

_{k}*G(n−1)k*(µ)dµ,

*if (x*3

*,t) ∈ R(4n − 2);*

*g*µ

_{nk}*t −x*3 α

*¶ +1 2 ·*

_{k}*f*µ

_{(n−1)k}*t +x*3

*− `*α

*¶*

_{k}*− f*µ

_{(n−1)k}*t −x*3

*+ `*α

*¶¸ +α*

_{k}*k*2 Z

*α*

_{t+}x3−`*k*

*t−x3+`*

_{α}

*k*

*G*(µ)dµ,

_{(n−1)k}*if (x*3

*,t) ∈ R(4n − 1);*

*fnk*µ

*t +x*3

*− `*α

*¶ +1 2 ·*

_{k}*g*³

_{(n−1)k}*t −x*3 α

*´*

_{k}*− g*³

_{(n−1)k}*t +x*3

*− 2`*α

*´¸*

_{k}*−*α

*k*2 Z

*α*

_{t−}x3*k*

*t+x3−2`*

_{α}

*k*

*F*(γ)dγ ,

_{(n−1)k}*if (x*3

*,t) ∈ R(4n);*

*fnk*µ

*t −x*3

*− `*β

*k*¶ +1 2

*[wk(x*3+β

*kt) − wk(−x*3+β

*kt + 2`)]*+ 1 2β

*k*Z

_{x}_{3}

_{+}

_{β}

_{k}_{t}*−x*3+β

*kt+2`*φ

*k*(ν)dν,

*if (x*3

*,t) ∈ R(4n + 1).*(2.8.10)

*where*
*R(4n − 2) =*
½
*(x*3*,t)*
¯
¯
*¯ 0 < x*3*< ` ,*
*(n − 2)`*
α*k* *< t −*
*x*_{3}
α*k* *<*
*(n − 1)`*
α*k* *and*
*(n − 1)`*
α_{k}*< t +*
*x*_{3}
α_{k}*<*
*n`*
α* _{k}*
¾

*R(4n − 1) =*½

*(x*3

*,t)*¯ ¯

*¯ 0 < x*3

*< ` ,*

*x*3

*+ (n − 1)`*

_{α}

*k*

*< t <n` − x*3 α

*¾*

_{k}*R(4n) =*½

*(x*3

*,t)*¯ ¯

*¯ 0 < x*3

*< ` ,*

*n` − x*

_{α}3

*k*

*< t <x*3

*+ (n − 1)`*α

*¾*

_{k}*R(4n + 1) =*½

*(x*3

*,t)*¯ ¯

*¯ ` < x*3

*< ∞ ,*

*(n − 1)`*

_{α}

*k*

*< t −x*3

*− `*β

_{k}*<*

*n`*α

*¾*

_{k}*for each n = 2, 3, . . . and the functions defined in (2.8.8) are constructed by initial-boundary*
*data and the matching conditions as follows*

*Gnk(t) =* _{α}1
*kf*
*0*
*nk(t) −*
1
α*kg*
*0*
*(n−1)k*
µ
*t −* *`*
α*k*
¶
*+ F _{(n−1)k}*
µ

*t −*

*`*α

*k*¶

*,*(2.8.11)

*gnk(t) =*·

*f*µ

_{(n−1)k}*t −*

*`*α

*k*¶

*− f*µ

_{(n−1)k}*−*

*`*α

*k*¶¸ +α

*Z*

_{k}*α*

_{t−}`*k*

*−`*α

*k*

*G*(γ)dγ

_{(n−1)k}*−*1 α

*k*Z

*0*

_{t}*Fnk*(τ)dτ, (2.8.12)

*fnk(t) =*

_{α}α

*k*

*k*+β

*k[g(n−1)k*µ

*t −*

*`*α

*k*¶

*− g*µ

_{(n−1)k}*−*

*`*α

*k*¶ ] + β

*k*α

*k*+β

*kwk(` +*β

*kt)*

*−*β

*k*α

*k*+β

*kwk(`) −*α2

*k*α

*k*+β

*k*Z

_{t−}*`*α

*k*

*−`*α

*k*

*F*+ 1 α

_{(n−1)k}(s)ds*+β*

_{k}*Z*

_{k}

_{`+}_{β}

*kt*

*`*φ

*k(z)dz*(2.8.13)

*for each k = 1, 2, 3 and n = 2, 3, . . .*

*Proof. If we notice the subregions in 0 < x*3*< `, namely the regions R(4n − 2),*

*R(4n − 1) and R(4n) (see, Figure 2.1), we do not use the initial conditions. Instead, we use*

*the functions, defined in (2.8.8). As a result of this situation, the formulation of the defined*
*functions (2.8.11) − (2.8.13) is in the form of recurrence relations.*

*2.8.1 The Region R(4n-2)*

*The region R(4n − 2) has a different form (see, Figure 2.2). We use the boundary*
*condition Fk(t) and the functions f(n−1)k(t), g(n−1)kand G(n−1)k* which we must find in the

previous step.

*In this region, we assume that there is a jump at x =* *`*

2*. We will apply the following matching*
conditions when the speeds are the same.

*unk*(
*`*
2*− 0,t) = unk*(
*`*
2*+ 0,t)* (2.8.14)
(α2
*k*)
*∂ u _{nk}*

*∂ x*3 ¯ ¯ ¯

*x*3=

*`*2

*−0*= (α2

*k*)

*∂ u*

_{nk}*∂ x*3 ¯ ¯ ¯

*x*3=

*`*2+0 (2.8.15)

Figure 2.2 The Region R(4n-2)

*Let us consider the problem (2.8.1) − (2.8.7) in the region R(4n-2), for k = 1, 2, 3. and*

*n = 2, 3, . . .*
*R(4n − 2) =*
½
*(x*_{3}*,t)*
¯
¯
*¯ 0 < x*3*< ` ,*
*(n − 2)`*
α_{k}*< t −*
*x*_{3}
α_{k}*<*
*(n − 1)`*
α* _{k}* and

*(n − 1)`*α

_{k}*< t +*

*x*3 α

_{k}*<*

*n`*α

*¾*

_{k}*The equation (2.8.1) can be written as in the form,*
*∂ q _{nk}*

*∂t*+α

*k*

*∂ q*

_{nk}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R(4n − 2),*(2.8.16)

*∂ u*

_{nk}*∂t*

*−*α

*k*

*∂ u*

_{nk}*∂ x*

_{3}

*= qnk(x*3

*,t) ,*

*(x*3

*,t) ∈ R(4n − 2).*(2.8.17)

*The characteristics of the equation (2.8.16) − (2.8.17) are the following,*

*d*ξ
*d*τ =α*k,* *ξ (t) = x*3; ξ = α*kτ + x*3*−*α*kt , when*ξ = 0 ; *τ = t −*
*x*3
α_{k}*,*
*d*ξ
*d*τ *= −αk,* *ξ (t) = x*3; *ξ = −αkτ + x*3+α*kt , when*ξ =
*`*
2; *τ = t +*
*2x*3*− `*
2α_{k}*.*

By integrating along the characteristic *ξ = x*3*−*α*k(t −τ), from t −*_{α}*x*3
*k* *to t,*

*qnk(x*3*,t) = g(n−1)k(t −*

*x*_{3}

α*k*)

Then by integrating along the characteristic *ξ = x*_{3}+α* _{k}(t −τ), from t +2x*3

*− `*2α

_{k}*to t,*Z

_{t}*t+2x3−`*

_{2}

_{α}

*k*∂ ∂ τ

*[unk(x*3+α

*k(t −τ), τ)]dτ =*Z

_{t}*t+2x3−`*

_{2}

_{α}

*k*

*g0*µ 2τ − t −

*x*3 α

*¶*

_{k}*d*τ Let 2τ − t −

*x*3 α

*k*=

*µ ,*

*2dτ = dµ*µ

*low= t +x*3

_{α}

*− `*

*k*

*,*µ

*up= t −*

*x*3 α

*k*

*By letting unk*(

*`*

_{2}

*,t) = mnk(t), we get*

*u*3

_{nk}(x*,t) = mnk*µ

*t +2x*3

*− `*2α

*¶ +1 2 ·*

_{k}*g*µ

_{(n−1)k}*t −x*3 α

*¶*

_{k}*− g*µ

_{(n−1)k}*t +x*3

*− `*α

*¶¸*

_{k}*.*(2.8.18)

*Similarly, the equation (2.8.1) can be written as in the form,*

*∂ q _{nk}*

*∂t*

*−*α

*k*

*∂ q*

_{nk}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R(4n − 2) ,*(2.8.19)

*∂ u*

_{nk}*∂t*+α

*k*

*∂ u*

_{nk}*∂ x*

_{3}

*= qnk(x*3

*,t) ,*

*(x*3

*,t) ∈ R(4n − 2) .*(2.8.20)

*The characteristics of the equation (2.8.18) − (2.8.19) are the following,*
*d*ξ
*d*τ *= −αk,* *ξ (t) = x*3; *ξ = −αkτ + x*3+α*kt , whenξ = ` ;* *τ = t +*
*x*_{3}*− `*
α*k* *,*
*d*ξ
*d*τ =α*k,* *ξ (t) = x*3; ξ = α*kτ + x*3*−*α*kt, when*ξ =
*`*
2 ; *τ = t +*
*` − 2x*3
2α*k* *.*

By integrating along the characteristic *ξ = x*3+α*k(t −τ), from t +*

*x*3*− `*
α_{k}*to t,*
*qnk(x*3*,t) = f(n−1)k0*
µ
*t +x*3*− `*
α*k*
¶
+α* _{k}G_{(n−1)k}*
µ

*t +x*3

*− `*α

*k*¶

*Similarly, by letting unk*(*`*_{2}*+ 0,t) = r(t) and integrating along the characteristic*

*ξ = x*_{3}*−*α* _{k}(t −τ), from t +` − 2x*3
2α

*k*

*to t, we get*

*unk(x*3

*,t) = r*µ

*t +` − 2x*3 2α

*¶ +1 2 ·*

_{k}*f*µ

_{(n−1)k}*t +x*3

*− `*α

*¶*

_{k}*− f*µ

_{(n−1)k}*t −x*3 α

*¶¸ +α*

_{k}*k*2 Z

*α*

_{t+}x3−`

_{k}*t−x3*

_{α}

_{k}*G(n−1)k(z)dz*(2.8.21)

*If we use the first matching condition (2.8.14), we get*

*m(t) = r(t)*

*By using the second matching condition (2.8.15), we get*

*m(t) =* 1
2
·
*g*
µ
*t −* *`*
2α*k*
¶
*− g*
µ
*−* *`*
2α*k*
¶¸
+α*k*
2
Z _{t−}*`*
2α*k*
*−* *`*
2α*k*
*G _{(n−1)k}*(µ)dµ
+1
2
·

*f*µ

*t −*

*`*2α

*¶*

_{k}*− f*µ

*−*

*`*2α

*¶¸*

_{k}*,*

*If we substitute the function m(t) into the formulation (2.8.21), we get*

*u _{nk}(x*

_{3}

*,t) = g*µ

_{(n−1)k}*t −x*3 α

*k*¶ +1 2

*f(n−1)k*µ

*t +x*3

*− `*α

*k*¶

*−*1 2

*f(n−1)k*µ

*t −x*3

*+ `*α

*k*¶ +α

*k*2 Z

*α*

_{t+}x3−`*k*

*t−x3+`*

_{α}

*k*

*G*(µ)dµ,

_{(n−1)k}*(x*

_{3}

*,t) ∈ R(4n − 2).*

*2.8.2 The Region R(4n-1)*

*Let us consider the problem (2.8.1) − (2.8.7) in the region R(4n-1), for k = 1, 2, 3. and*

*n = 2, 3, . . .*
*R(4n − 1) =*
½
*(x*3*,t)*
¯
¯
*¯ 0 < x*3*< ` ,* *x*3*+ (n − 1)`*_{α}
*k* *< t <*
*n` − x*3
α*k*
¾

*The equation (2.8.1) can be written in the form,*

*∂ q _{nk}*

*∂t*

*−*α

*k*

*∂ q*

_{nk}*∂ x*3

*= 0 ,*

*(x*3

*,t) ∈ R(4n − 1),*(2.8.22)

*∂ u*

_{nk}*∂t*+α

*k*

*∂ u*

_{nk}*∂ x*

_{3}

*= qnk(x*3

*,t) ,*

*(x*3

*,t) ∈ R(4n − 1).*(2.8.23)

*The characteristics of the equation (2.8.22) − (2.8.23) are the following,*

*d*ξ
*d*τ *= −αk,* *ξ (t) = x*3; *ξ = −αkτ + x*3+α*kt , whenξ = ` ;* *τ = t +*
*x*_{3}*− `*
α*k* *,*
*d*ξ
*d*τ =α*k,* *ξ (t) = x*3 ; ξ = α*kτ + x*3*−*α*kt* and if ξ = 0 ; *τ = t −*
*x*3
α*k* *.*

So, by integrating along the characteristic*ξ = x*_{3}+α* _{k}(t −τ) from t +x*3

*− `*α

*k*

*, to t,*

*qnk(x*3

*,t) = f(n−1)k0*µ

*t +x*3

*− `*α

*¶ +α*

_{k}*kG(n−1)k*µ

*t +x*3

*− `*α

*¶*

_{k}*,*

By integrating along the characteristic,*ξ = x*_{3}*−*α* _{k}(t −*τ) we get the solution

*unk(x*3*,t) = gnk*
µ
*t −x*3
α*k*
¶
=1
2
·
*f _{(n−1)k}*
µ

*t +x*3

*− `*α

*k*¶

*− f*µ

_{(n−1)k}*t −x*3

*+ `*α

*k*¶¸ +α

*k*2 Z

*α*

_{t+}x3−`*k*

*t−x3+`*

_{α}

*k*

*G*(µ)dµ ,

_{(n−1)k}*(x*3

*,t) ∈ R(4n − 1).*

*And the function gnk* *defined in (2.8.8) is in the form,*

*gnk(t) =*
·
*f _{(n−1)k}*
µ

*t −*

*`*α

*¶*

_{k}*− f*µ

_{(n−1)k}*−*

*`*α

*¶¸ +α*

_{k}*k*Z

*α*

_{t−}`*k*

*−`*α

_{k}*G*(γ)dγ

_{(n−1)k}*−*1 α

*Z*

_{k}*0*

_{t}*Fnk*(τ)dτ .

*2.8.3 The Region R(4n)*

*Let us consider the problem (2.8.1) − (2.8.7) in the region R(4n), for k = 1, 2, 3. and*

*n = 2, 3, . . .*
*R(4n) =*
½
*(x*3*,t)*
¯
¯
*¯ 0 < x*3*< ` ,* *n` − x*_{α} 3
*k* *< t <*
*x*3*+ (n − 1)`*
α*k*
¾

*The equation (2.8.1) can be written as in the form,*

*∂ q _{nk}*

*∂t*+α

*k*

*∂ q*

_{nk}*∂ x*

_{3}

*= 0 ,*

*(x*3

*,t) ∈ R(4n),*(2.8.24)

*∂ unk*

*∂t*

*−*α

*k*

*∂ unk*

*∂ x*

_{3}

*= qnk(x*3

*,t) ,*

*(x*3

*,t) ∈ R(4n).*(2.8.25)

*The characteristics of the equations (2.8.24) − (2.8.25) are the following*

*d*ξ
*d*τ =α*k,* *ξ (t) = x*3 ; ξ = α*kτ + x*3*−*α*kt* and if ξ = 0 ; *τ = t −*
*x*3
α*k* *,*
*d*ξ
*d*τ *= −αk,* *ξ (t) = x*3 *ξ = −αkτ + x*3+α*kt ,* when*ξ = ` ;* *τ = t +*
*x*3*− `*
α_{k}*.*
So, by integrating along the characteristic*ξ = x*_{3}*−*α_{k}(t −τ) from t −*x*3

α*k* to t,
*qnk(x*3*,t) = g0 _{(n−1)k}*
µ

*t −x*3 α

*¶*

_{k}*−*α

*kF(n−1)k*µ

*t −x*3 α

*¶*

_{k}*,*

Similarly, by integrating along the characteristic*ξ = −α _{k}τ + x*

_{3}+α

*3*

_{k}t from t +x*− `*

α*k* to t, we
get
*unk(x*3*,t) = fnk*
µ
*t +x*3*− `*
α*k*
¶
+1
2
·
*g _{(n−1)k}*
³

*t −x*3 α

*k*´

*− g*³

_{(n−1)k}*t +x*3

*− 2`*α

*k*´¸

*−*α

*k*2 Z

*α*

_{t−}x3*k*

*t+x3−2`*

_{α}

_{k}*F(n−1)k*(γ)dγ ,

*(x*3

*,t) ∈ R(4n).*(2.8.26)

*2.8.4 The Region R(4n+1)*

*Let us consider the problem (2.8.1)−(2.8.7) in the region R(4n+1), for k = 1, 2, 3. and n =*
*2, 3, . . .*
*R(4n + 1) =*
½
*(x*3*,t)*
¯
¯
*¯ ` < x*3*< ∞ ,* *(n − 1)`*_{α}
*k*
*< t −x*3*− `*
β_{k}*<*
*n`*
α* _{k}*
¾

*The equation (2.8.2) can be written as in the form,*

*∂ qnk*
*∂t* *−*β*k*
*∂ qnk*
*∂ x*3 *= 0 ,* *(x*3*,t) ∈ R(4n + 1),* (2.8.27)
*∂ v _{nk}*

*∂t*+β

*k*

*∂ v*

_{nk}*∂ x*3

*= qnk(x*3

*,t) ,*

*(x*3

*,t) ∈ R(4n + 1).*(2.8.28)

*The characteristics of the equation (2.8.27) − (2.8.28) are the following,*

*d*ξ
*d*τ *= −βk,* *ξ (x*3*) = t* ; *ξ = −βkτ + x*3+β*kt ,*
*d*ξ
*d*τ =β*k,* *ξ (x*3*) = t* ; ξ = β*kτ + x*3*−*β*kt ,* when*ξ = ` ;* *τ = t −*
*x*_{3}*− `*
β*k* *.*

So, by integrating alongξ = β_{k}τ + x_{3}*−*β_{k}t from 0 to t,

*qnk(x*3*,t) =*φ*k(x*3+β*kt) +*β*kw0k(x*3+β*kt),*

Similarly, by integrating along the characteristicξ = β*kτ + x*3*−*β*kt fromτ = t −x*3_{β}*− `*
*k* to t,
we get
*vnk(x*3*,t) = fnk*
µ
*t −x*3*− `*
β*k*
¶
+1
2*[wk(x*3+β*kt) − wk(−x*3+β*kt + 2`)]*
+ 1
2β*k*
Z _{x}_{3}_{+}_{β}_{k}_{t}*−x*3+β*kt+2`*
φ* _{k}*(ν)dν , (x

_{3}

*,t) ∈ R(4n + 1).*(2.8.29)

*2.8.5 Matching Conditions Between R(4n) and R(4n+1)*

*Consider the formulations in (2.8.26) − (2.8.29) for the regions R(4n) and R(4n + 1).*
*By the first matching condition (2.8.6), we get the relation*

*To apply the second matching condition (2.8.7), we must differentiate the formulations in*
*(2.8.26) − (2.8.29), then by substituting x*3*= `, we get the function Gnk(t) defined in (2.8.8)*

*Gnk(t) =* *∂ u _{∂ x}nk*
3
¯
¯
¯

*x*3

*=`−0*= 1 α

_{k}*f*

*0*

*nk(t) −*1 α

_{k}g*0*

*(n−1)k*µ

*t −*

*`*α

*¶*

_{k}*+ F*µ

_{(n−1)k}*t −*

*`*α

*¶*

_{k}*and by the second matching condition (2.8.7), we get the function f _{nk}(t),*

*f _{nk}(t) =* α

*k*α

*+β*

_{k}*µ*

_{k}[g(n−1)k*t −*

*`*α

*¶*

_{k}*− g*µ

_{(n−1)k}*−*

*`*α

*¶ ] + β*

_{k}*k*α

*+β*

_{k}*β*

_{k}[wk(` +*kt) − wk(`)] +*1 α

*+β*

_{k}*Z*

_{k}

_{`+}_{β}

*kt*

*`*φ

*k(z)dz*

*−*α

*k*2 α

*+β*

_{k}*Z*

_{k}*α*

_{t−}`*k*

*−*

*`*α

*k*

*F*

_{(n−1)k}(s)ds .2.9 Examples of Simulations of Wave Propagation in Two Layered Medium

In this section, we deal with examples of simulations of wave propagation in two layered elastic half space. As the mathematical model of wave propagation, we study IBVP of wave equations in two layered medium.,

We took a pulse point source in different positions in half space: Between the boundaries

*x*3*= 0 and x*3*= `, outside the boundary x*3*= `. In each case, the half space has two layers*

with different speed. The speed of the first layer isα = 1 and the speed of the second layer
is*β = 2. We considered the matching conditions (2.4.8)−(2.4.9) only on the boundary x*_{3}*= `.*

For all examples, we omit the index k for simplicity writing.

*2.9.1 Example 1 - The Pulse Point Source is Between the Boundaries x*3*= 0 and x*3*= `*

*Let us consider initial boundary value problem (2.4.3) − (2.4.9) with its general form*
*(2.8.1) − (2.8.7) for k = 1 and n = 2, 3, . . . The initial conditions* *ϕ(x*_{3}*),ψ(x*_{3}*), w(x*3*),φ (x*3)

have the following form

*w(x*_{3}*) = 0,* *φ (x*_{3}*) = 0.*

where*δ (x*3*) is Dirac delta function, the boundary ` = 40, the point source is located at x*03= 10

and the boundary condition

*F(x*3*) = 0.*

By the properties of Dirac delta function and the assumptions, the solution of IBVP can be
written as follows:
*U(x*3*,t) =*
1
2
£
*δ (x*3+*αt − x*03) +*δ (x*3*−αt − x*03)
¤
*, if (x*3*,t) ∈ R1;*
*0,* *if (x*3*,t) ∈ R2.*
*U(x*3*,t) =*
*g*
³
*t −x*3
α
´
+1
2*·δ (x*3+*αt − x*
0
3)
*−*1
2*·δ (−x*3+*αt − x*
0
3*),* *if (x*3*,t) ∈ R3;*
*f*
µ
*t +x*3*− `*
α
¶
+1
2*·δ (x*3*−αt − x*
0
3)
*−*1
2*·δ (−x*3*−αt + 2` − x*
0
3*),* *if (x*3*,t) ∈ R4;*
*f*
µ
*t −x*3*− `*
β
¶
*,* *if (x*3*,t) ∈ R5.*

*Here, the function g(t), f (t) and G(t), constructed in Theorem 2.7.1, can be also written as*

*g(t) =δ (αt − x*0_{3}*),*
*G(t) = −* β
α(α + β )*·*
∂
*∂t*
£
*δ (` − αt − x*0_{3})¤
*f (t) =* α
α + β *·δ (` − αt − x*
0
3)
*For n = 2, 3, . . .*