modeling and numerical simulation of wave propagation in elastic wave … · 2014. 1. 15. · we...

Unterschrift des Betreuers

DIPLOMARBEIT

Modeling and Numerical Simulationof Wave Propagation in Elastic

Wave Guides

Ausgefuhrt am Institut fur

Analysis und Scientific Computing

unter der Anleitung von

Univ.Prof. Dipl.-Ing. Dr.techn. Joachim Schoberl

Univ.Ass. Dipl.-Math. Dr.rer.nat. Lothar Nannen

durch

Martin Halla BSc

Enenkelstraße 11-13/38, 1160 Wien

10. Dezember 2012

Martin Halla

Die approbierte Originalversion dieser Diplom-/Masterarbeit ist an der Hauptbibliothek der Technischen Universität Wien aufgestellt (http://www.ub.tuwien.ac.at). The approved original version of this diploma or master thesis is available at the main library of the Vienna University of Technology (http://www.ub.tuwien.ac.at/englweb/).

i

Danksagung

Meinem Betreuer Dr. Lothar Nannen mochte ich dafur danken, dass er mir dieses Diplo-marbeitsthema vorgeschlagen und mich bei der Bearbeitung fruchtvoll angeleitet hat.Weiters danke ich Prof.Dr. Joachim Schoberl dafur, dass er die offizielle Betreuung dieserDiplomarbeit ubernommen hat.Meinen Eltern mochte ich dafur danken, dass sie mich wahrend meines Studiums im-mer unterstutzt haben. Mag. Verena Berger danke ich fur das sprachliche Korrek-turlesen dieser Dipomarbeit. Schlussendlich mochte ich auch meinen Studienkollegen furdas gemeinsame studentische Leben danken. Ohne sie ware das Studium nicht dasselbegewesen.

Contents

Chapter 1. Introduction and Outline 51.1. Outline 51.2. Linear Elasticity Equations 61.3. Geometry used 7

Chapter 2. Modeling 92.1. Properties of Wave Numbers 112.2. Properties of Lamb Modes 192.3. Limiting Absorption Principle 20

Chapter 3. Numerical Concepts 273.1. Acoustic Wave Propagation 273.2. Wave Numbers and Dispersion Curves 293.3. Complex Scaling 343.4. Hardy Space Method 37

Chapter 4. A Model Problem 414.1. Formulation of the Problem 414.2. Collapsing Wave Numbers 434.3. Galerkin Formulation 444.4. Numerical Tests 49

Chapter 5. Summary and Outlook 53

Appendix A 55Matlab Code of solve modelproblem.m 55Matlab Code of ext Matrices.m 58

Bibliography 63

iii

CHAPTER 1

Introduction and Outline

In this diploma thesis we want to discuss wave propagation in elastic wave guides.Apart from electro magnetic waves, every wave needs a medium to be transmitted in.If the shape and form of this medium is clearly defined, we speak of a wave guide.Due to scaling reasons studying of unbounded wave guides is necessary to understandwave propagation in bounded wave guides. Hence we focus on unbounded wave guides.As we are only dealing with elastic waves in this thesis, a reader without physicalbackground knowledge can imagine an elastic wave as some body that oscillates. Aswas unfortunately seen on November 7th 1940 bridges are also elastic wave guides:The Tacoma Narrows Bridge in the U.S. state of Washington dramatically collapseddue to a physical phenomenon called aeroelastic flutter, which causes rapid periodicmotion, see [3] and [12]. This bridge made history as Galloping Gertie. However, inour context the wave guide will always have the simple geometry of a beam, plate or pipe.

1.1. Outline

To complete the picture we give a brief introduction into elasticity theory in Section 1.2.The advanced reader might take a short look into this section to check the notation used.As at the moment research is not yet ready to deal with arbitrary geometries of waveguides, we restrict our work on a simple beam. We explain this and some additionalscaling arguments in Section 1.3.Chapter 2 deals with constructing physically meaningful solutions of the elasticity equa-tions equipped with suitable boundary and radiation conditions. We follow [1] to con-struct solutions of the elasticity equations analytically, which fulfill the boundary condi-tions on one part of the boundary. In Section 2.1 we show some new results about theasymptotic properties of the wave numbers kn. We use these properties to justify severalassumptions in Section 2.2, which we need in Section 2.3. There we use the limitingabsorption principle to justify our choice of the wave numbers kn.In Chapter 3 we present the problem of guided acoustic wave propagation and discussdifferent concepts for numerical methods, namely complex scaling aka perfectly matchedlayer and the Hardy space method. Complex scaling reaches back to the sixties. A his-torical synopsis can be found in [9]. The concept was redeveloped as perfectly matchedlayers (PML) in [2]. First convergence results where published in [5]. In [18] PML wasapplied to guided elastic wave propagation by projecting onto eigen modes. We show whyPML cannot be used or adapted to elastic wave propagation without such a projection

5

6 1. INTRODUCTION AND OUTLINE

onto the eigen modes. In contrast we show why and how the Hardy space method canbe adapted to the elastic case without a projection onto the eigen modes.In Chapter 4 we discuss an artificial model problem, which exemplifies several propertiesand difficulties of the Hardy space method for elastic waves, in order to prepare the Hardyspace method for the elastic case.

1.2. Linear Elasticity Equations

As we are analysing elastic waves it is appropriate to give a brief summary of linearelastic continuum mechanics. We closely adhere to [8] and [1]. But first we define ournotation. Let us mention that we use Einstein’s summation convention in this section,but when we switch to a simpler geometry in the next section, we drop it. The notationwill be used as follows:

• let Ω ⊂ R3 be an open simple connected set, which describes our body of interest,

• let x =

xyz

∈ R3 be our position vector,

• let t ∈ R+ be our time variable,• let u : Ω × R+ → R3 be our displacement vector, i.e. u(x, t) describes theposition of the point x ∈ Ω at time t,

• let ǫ = (ǫij)i,j∈1,2,3 be our (small) strain tensor,• let τ = (τij)i,j∈1,2,3 be our stress tensor,• let ρ ∈ R+ be the density constant,• let f : Ω → R3 be the external force per unit,• let λ and µ be Lame’s elasticity constants for which 0 < µ <∞, 0 < 2µ+3λ <∞hold.

Now the linear elastic model consists of the conservation of linear and angular momentum,and a constitutive relation between force and deformation, the density is approximatedas a constant ρ. Due to the conservation of angular momentum, the stress tensor τ issymmetric, i.e. τij = τji. The following partial differential equation can be derived fromthe conservation of linear momentum :

∂kτki + ρfi = ρ∂t∂tui.

As we are only treating the linear model the deformation of our body Ω can be describedvia the (small) strain tensor

ǫij = (∂iuj + ∂jui)/2.

For a homogeneous, isotropic, linearly elastic solid, stress and strain are related by

τij = λǫkkδij + 2µǫij,

where µ and λ are material constants. If we plug the above equations into another wegain

(λ+ µ)∂i∂kuk + µ∂j∂jui + ρfi = ρ∂t∂tui.

1.3. GEOMETRY USED 7

In vectorial notation this reads

µ∆u+ (λ+ µ)∇ div u+ ρf = ρ∂t∂tu. (1.1)

This equation can be better understood, if we apply a Hemlholtz decomposition to u.Thus let curl be the standard curl -operator: Be ψ : Ω → R3 be smooth enough, then

curl ψ = ∇× ψ =

∂yψ3 − ∂zψ2

∂zψ1 − ∂xψ3

∂xψ2 − ∂yψ1

.

We say φ : Ω → R, ψ : Ω → R3 is Helmholtz decomposition of u, if

u = ∇φ+ curl ψ.

Often the additional constraint div ψ = 0 is demanded. For the existence of generalHelmholtz decompositions see [17]. For Helmholtz decompositions specified to the elas-ticity equations we refer to [1]. We will see later why a Helmholtz decomposition is souseful in Chapter 2.

1.3. Geometry used

We assume u to be time harmonic. I.e. from now on u is only a function of x and wewrite for some positive frequency ω ∈ R+

u(x, t) = u(x)e−iωt,

and so

∂t∂tu = −ω2u.

The physical quantity is described by Re u. As research on numerical methods for thisparticular case of guided elastic waves is comparatively young, we will focus on a verysimple geometry. Thus we take a two dimensional beam Ω := (−M,M) × (−h, h) withh,M ∈ R+ and suppose there is motion in the (x, y)-plane only. Hence

u3 = 0,

∂z( ) = 0.

This two dimensional model, gained by reducing one dimension, is called plain strain. Itis not the only choice possible: We could also reduce the dimension by one, but allowmovement in the z-dimension, the model in this case is called plain stress. For a moredetailed description, we refer to [4]. This thesis focuses on plain strain.We suppose the beam to be homogeneous with a small local anomaly in the center, i.e.a hole χ (or possibly some other disturbance). Due to the stated size proportion, it issufficient to rescale and assume the beam to be infinitely long, i.e. Ω := R× (−h, h) \ χ.We also omit exterior force f , but assume that we can split our displacement vector u

into a given incoming and a reflected outgoing wave, i.e. u = uin + uout. Since u andour incoming wave uin fulfill Equation (1.1), our outgoing wave to be calculated uout also

8 1. INTRODUCTION AND OUTLINE

Figure 1. A decomposition of the rescaled beam.

fulfills Equation (1.1).Summarizing the above statements, our problem reads

• (Ω = R× (−h, h)) \ χ.• We search for uout : Ω → R2 such that

* µ∆uout + (λ+ µ)∇ div uout = −ρω2uout in Ω,* ∂nuout = −∂nuin on ∂χ,* uout is stress free on the outer boundary, i.e.

τ21 = µ(∂xuout,2 + ∂yuout,1) = 0,τ22 = λ(∂xuout,1 + ∂yuout,2) + 2µ∂yuout,2 = 0

on ∂ (Ω ∪ χ).

When we speak of u in the following chapters, we always mean uout, but drop the indexfor the sake of simplicity.From the numerical point of view it is useful to split the domain Ω into Ωleft,Ωcenter

and Ωright as seen in Figure 1. On Ωcenter some standard numerical scheme like a finiteelement method or a boundary element method can be used and coupled through theboundary with numerical schemes for Ωleft and Ωright. As Ωleft and Ωright are infinite, thedevelopment of numerical schemes for these domains is rather difficult. However sinceΩleft is just an inversion of Ωright, we only have to deal with Ωright, which we will do inthe following chapters.

CHAPTER 2

Modeling

We are interested in wave propagation in elastic wave guides. In particular we investigatetime harmonic wave propagation in a half open two dimensional stripe Ω := R+×(−h, h).I.e. for a given positive frequency ω ∈ R+ and boundary data g, we look for the displace-ment function u : Ω → R2, such that

Figure 1. Our domain of interest.

µ∆u+ (λ+ µ)∇ div u = −ρω2u in R+ × (−h, h), (2.1)

u = g on 0 × (−h, h), (2.2)

with natural stress free boundary conditions, i.e.

τ21 = µ(∂xu2 + ∂yu1) = 0 on R+ × ±h, (2.3)

τ22 = λ(∂xu1 + ∂yu2) + 2µ∂yu2 = 0 on R+ × ±h. (2.4)

We take a Helmholtz decomposition of u and consider u to be two dimensional, i.e. we

associate ψ with

00ψ

. This reads as

u = ∇φ+ curl ψ ⇔ u1 = ∂xφ+ ∂yψ, u2 = ∂yφ− ∂xψ,

Equation (2.1) then reads

∇[

(λ+ 2µ)∆φ+ ρω2φ]

+ curl[

µ∆ψ + ρω2ψ]

= 0. (2.5)

9

10 2. MODELING

We define

cL :=

√

ρ

λ+ 2µ,

cT :=

√

ρ

µ.

Because of 0 < µ <∞, 0 < 2µ+ 3λ we have

cL, cT > 0,

andc2Tc2L

=2µ+ λ

µ= 2 +

λ

µ

≥ 2− 2

3=

4

3.

(2.6)

Equation (2.5) is clearly fulfilled if

∆φ = −ω2

c2Lφ, (2.7)

∆ψ = −ω2

c2Tψ. (2.8)

We use the following product ansatz

φ(x, y) = eikx (A1 sin(αy) + A2 cos(αy)) ,

ψ(x, y) = eikx (B1 sin(βy) +B2 cos(βy)) ,

with k, α, β ∈ C. We see that equations (2.7) and (2.8) are satisfied iff

α2 =ω2

c2L− k2, (2.9)

β2 =ω2

c2T− k2. (2.10)

An inspection of φ and ψ shows, that we can split the modes into symmetric, i.e. A1 =B2 = 0, and antisymmetric ones, i.e. A2 = B1 = 0. Hence we get the symmetric Lambmodes as

u1 = (ikA2 cos(αy) + βB1 cos(βy)) eikx,

u2 = (−αA2 sin(αy)− ikB1 sin(βy)) eikx.

(2.11)

If we insert these into the boundary conditions at y = ±h

τ21 = τ22 = 0, (2.12)

2.1. PROPERTIES OF WAVE NUMBERS 11

this leads to a system of two homogeneous linear equations for the constants A2, B1.Since the system is homogeneous, the determinant of the coefficients has to be zero. Asin [1, page 223] calculated

0 = 4k2αβ sin(αh) cos(βh) + (k2 − β2)2 cos(αh) sin(βh). (2.13)

If α, β, k fulfill equations (2.9),(2.10) and (2.13) we can solve Equation (2.12) and gain

A2 =(

k2 − β2)2

sin(βh),

B1 = 2ikα sin(αh).(2.14)

We only analyze the symmetric modes, since the case for the antisymmetric ones issimilar. Hence, we further analyze Equation (2.13). We observe that for every solutionk, −k also solves the equation. For k /∈ R, we choose k such that Im k > 0 in order toattain u ∈ L∞(R+ × (−h, h)). For k ∈ R we choose k so that d

dωk(ω) > 0. If k ∈ R and

ddωk(ω) = 0, we choose k so that d2

dω2k(ω) < 0 and so on. The objective of this chapteris to justify this decision using the limiting absorption principle. First let us furtherinvestigate Equation (2.13). We are looking for solutions of

F (k) = 0 (2.15)

withF (k) = 4k2αβ sin(αh) cos(βh) + (k2 − β2)2 cos(αh) sin(βh),

α =

√

ω2

c2L− k2,

β =

√

ω2

c2T− k2.

(2.16)

Remark 2.0.1. Since F is symmetric in α, antisymmetric in β and we are looking forsolutions of F (k) = 0, how we define the root in the definitions of α and β is of noconsequence.

2.1. Properties of Wave Numbers

Lemma 2.1.1. There exist at most N roots kn of F and the only cluster point of⋃

n∈Nkn can be at infinity.

Proof. We can split R× iR+0 in R+

0 × iR+0 and R−

0 × iR+0 . On each of these sets the

root of k2−ω2 can be defined analytically and so F . Obviously F is not constant. Hencewith [16] the claim follows.

Theorem 2.1.2. If there exists an infinite number of kn, then there holds for large n

kn,±1,±2 ≈ ±1 1

2harccosh

(

2√2nπ +

√2π ±2

√2π

4

)

+i

h

(

2nπ + π ±2 π

4

)

(2.17)

12 2. MODELING

Proof. First we set

c1 :=ω2

2c2L,

c2 :=ω2

2c2T.

Due to Lemma 2.1.1 we may assume |k| to be big enough, such that linearising is justifi-able. Thus

α = ik

√

1− 1

k2

(

ω

cL

)2

≈ ik − i

k

ω2

2c2L,

β = ik

√

1− 1

k2

(

ω

cT

)2

≈ ik − i

k

ω2

2c2T,

sin(icjh

k) ≈ icjh

k, j = 1, 2,

cos(icjh

k) ≈ 1, j = 1, 2.

The linearisation of Equation (2.15) is

0 =k4(

1− c1k2

)(

1− c2k2

)

[

sin(ikh) cos(ikh)− ic1h

k2cos2(ikh)

+ic2h

k2sin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]

− k4(

1− c2k2

)2[


kcos2(ikh)

+ic1h

ksin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]

.

(2.18)

As we may assume |k| to be big enough, we have k, β 6= 0. Equation (2.18) is equivalentto

0 =

[


kcos2(ikh) +

ic2h

ksin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]

− c1k2

[


kcos2(ikh) +

ic2h

ksin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]

−[


kcos2(ikh) +

ic1h

ksin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]

+c2k2

[


kcos2(ikh) +

ic1h

ksin2(ikh)− c1c2h

2

k2sin(ikh) cos(ikh)

]


The first and the third line cancel each other to (c2−c1) ihk . The terms with sin2 and cos2

of the second and forth line cancel each other to (c21 − c22)ihk3cos2(ikh). The remaining

terms can be written as sin(ikh) cos(ikh)k2

[

−c1 + c21c2h2

k2+ c2 − c1c22h

2

k2

]

. Thus we have

0 = (c2 − c1)ih

k+ (c21 − c22)

ih

k3cos2(ikh)

+sin(ikh) cos(ikh)

k2

[

−c1 +c21c2h

2

k2+ c2 −

c1c22h

2

k2

]

.(2.19)

Due to (2.6) and ω 6= 0 there holds c1 6= c2. Equation (2.19) is divided by (c2 − c1)ihk

which yields

0 = 1− (c1 + c2)cos2(ikh)

k2+

sin(ikh) cos(ikh)

ikh

(

1− c1c2h2

k2

)

.

This equation can be reformulated as

1 = (c1 + c2)cos2(ikh)

k2− sin(ikh) cos(ikh)

ikh

(

1− c1c2h2

k2

)

. (2.20)

We want to show that for k → ∞ the equation yields no solution if cos2(ikh)k2

or sin(ikh) cos(ikh)ik3h

does not converge to zero.

If we assume | cos2(ikh)||k|2 ≥ |k|−1/2 for k big enough, then

1 ≥∣

∣

∣

∣

∣

∣

∣

∣

sin(ikh) cos(ikh)

ikh

∣

∣

∣

∣

(

1

2− c1c2h

2

|k|2)

+

∣

∣

∣

∣

cos2(ikh)

k2

∣

∣

∣

∣

( |k|2h

| tan(ikh)| − (c1 + c2)

)∣

∣

∣

∣

≥∣

∣|k|1/2| tan(ikh)| − |k|−1/2(c1 + c2)∣

∣ .

Let δ be positive. For k /∈⋃

z∈Zk : |ikh − zπ| ≤ δ the right hand side is unbounded,which is a contradiction. But for k ∈

⋃

z∈Zk : |ikh − zπ| ≤ δ we have | cos(ikh)| ≈ 1.

Hence cos2(ikh)k2

≥ |k|−1/2 cannot hold. Thus for big k, solutions kn fulfill | cos2(iknh)||kn|2 ≤

|kn|−1/2.

If we assume | sin(ikh) cos(ikh)||ik3h| ≥ |k|−1/2 for k big enough, then

1 ≥∣

∣

∣

∣

∣

∣

∣

∣

sin(ikh) cos(ikh)

ikh

∣

∣

∣

∣

−∣

∣

∣

∣

sin(ikh) cos(ikh)c1c2h2

ik3h

∣

∣

∣

∣

−∣

∣

∣

∣

cos2(ikh)

k2

∣

∣

∣

∣

(c1 + c2)

∣

∣

∣

∣

≥∣

∣

∣

∣

∣

∣

∣

∣

sin(ikh) cos(ikh)

ik3h

∣

∣

∣

∣

(

|k|2 − c1c2h2 − (c1 + c2)| cot(ikh)ih|

)

∣

∣

∣

∣

≥∣

∣|k|3/2 − c1c2h2|k|−1/2 − (c1 + c2)h| cot(ikh)||k|1/2

∣

∣ .

Let δ be positive. For k /∈⋃

z∈Zk : |ikh − zπ| ≤ δ the right hand side is unbounded,which is a contradiction. But for k ∈ ⋃z∈Zk : |ikh−zπ| ≤ δ we have | cos(ikh)| ≈ 1 and

sin(ikh) ≈ ikh. Hence | sin(iknh) cos(iknh)||ik3nh|

≥ |k|−1/2 cannot hold. Thus for big k, solutions

14 2. MODELING

kn fulfill | sin(iknh) cos(iknh)||ik3nh|

≤ |kn|−1/2.

Dropping the small terms in Equation (2.20) yields

ikh = − sin(ikh) cos(ikh). (2.21)

Setting ikh = x+ iy with x ∈ R−0 , y ∈ R, Equation (2.21) becomes

x+ iy = − sin(x+ iy) cos(x+ iy)

= − (sin(x) cosh(y) + i cos(x) sinh(y)) (cos(x) cosh(y)− i sin(x) sinh(y))

= − sin(x) cos(x)(

cosh2(y) + sinh2(y))

− i(

cos2(x)− sin2(x))

sinh(y) cosh(y)

= −1

2sin(2x) cosh(2y)− i

1

2

(

2 cos2(x)− 1))

sinh(2y).

If the equation is split into the real and imaginary part, this results in

x = −1

2sin(2x) cosh(2y), (2.22)

y = −1

2

(

2 cos2(x)− 1))

sinh(2y). (2.23)

If 2 cos2(x) − 1 6= 0, Equation (2.23) can only hold for bounded y. Due to this andEquation (2.22) x is bounded, which contradicts the assumption that |k| is big. Thus2 cos2(x)− 1 = 0 has to hold, i.e. x = −π

2(n + 1

2). Due to x < 0 Equation (2.22) is only

solvable if sin(2x) ≥ 0. Hence

x = −2nπ − π ± π

4.

Solving Equation (2.22) yields

y = ±1 1

2arccosh

(

2√2nπ +

√2π ±2

√2π

4

)

.

Due to ikh = x+ iy, i.e. k = yh− ix

hwe have

k = ±1 1

2harccosh

(

2√2nπ +

√2π ±2

√2π

4

)

+i

h

(

2nπ + π ±2 π

4

)

and the claim is proven.

Lemma 2.1.3. If there exists an infinite number of kn, then there exist N ∈ N, C > 0such that for all n ≥ N kn is analytic at ω and |k′n(ω)| ≤ C.

Proof. First we want to show, that for |k| big enough, we can apply the analyticimplicit function theorem to F . For every k ∈ (R× iR+

0 ) \ ω, the root can be defined


analytically in a neighbourhood of k× ω and thus F . ∂kF 6= 0 remains to be shown. Wehave

∂kF = 4k2αβ cos(αh) cos(βh)−hkα

(2.24a)

+ 4k2αβ sin(αh) sin(βh)hk

β(2.24b)

+ (k2 − β2)2 sin(αh) sin(βh)hk

α(2.24c)

+ (k2 − β2)2 cos(αh) cos(βh)−hkβ

(2.24d)

+ 16k3(1− c2k2

) cos(αh) sin(βh) (2.24e)

+ 8kαβ sin(αh) cos(βh) (2.24f)

+ 4k2−kβα

sin(αh) cos(βh) (2.24g)

+ 4k2−kαβ

sin(αh) cos(βh). (2.24h)

Again for the sake of simplicity we set

c1 :=ω2

2c2L,

c2 :=ω2

2c2T.

We assume |k| to be big enough, so that linearising α, β is justifiable, i.e.

α = ik

√

1− 1

k2

(

ω

cL

)2

≈ ik − ic1k,

β = ik

√

1− 1

k2

(

ω

cT

)2

≈ ik − ic2k.

16 2. MODELING

We take a closer look at the first four terms of ∂kF :

(2.24a) + (2.24b) + (2.24c) + (2.24d) ≈− i4hk4(1− c2

k2) cos(αh) cos(βh)

+ i4hk4(1− c1k2

) sin(αh) sin(βh)

− i4hk4(1− c2k2

)2(1− c1k2

)−1 sin(αh) sin(βh)

+ i4hk4(1− c2k2

)2(1− c2k2

)−1 cos(αh) cos(βh)

= i4hk4 sin(αh) sin(βh)(1− c1k2

)

(

1− (1− c2k2)2

(1− c1k2)2

)

= i4hk4 sin(αh) sin(βh)(1− c1k2

)

(

2(c2−c1)k2

+c21−c22k4

(1− c1k2)2

)

= i8h(c2 − c1)k2 sin(αh) sin(βh) +O(|k0 sin(αh) sin(βh)|).

Now we inspect the second four terms of ∂kF :

(2.24e) + (2.24f) + (2.24g) + (2.24h) ≈

+ 16k3(1− c2k2

) cos(αh) sin(βh)

− 8k3(1− c1k2

)(1− c2k2

) sin(αh) cos(βh)

− 4k3(1− c1k2

)−1(1− c2k2

) sin(αh) cos(βh)

− 4k3(1− c1k2

)(1− c2k2

)−1 sin(αh) cos(βh)

= 16k3(1− c2k2

) cos(αh) sin(βh)

− k3(1− c2k2

) sin(αh) cos(βh)

·(

8(1− c1k2

) + 4(1− c2k2

)−1 + 4(1− c1k2

)(1− c2k2

)−2)

= k3(1− c2k2

) sin(αh) cos(βh)

·(

16 + 16(tan(βh)

tan(αh)− 1)− 16 +O(|k|−2)

)

.

Hence we examine

tan(βh)

tan(αh)− 1 =

tan(βh)− tan(αh)

tan(αh).


Due to Theorem 2.1.2 we know Im kn → ∞, hence | tan(αnh)| ≥ c for n big enough. Weexamine further

tan(βh)− tan(αh) ≈ tan(ikh) + (1 + tan2(ikh))−ic1k

− tan(ikh)− (1 + tan2(ikh))−ic2k

= (1 + tan2(ikh))i(c2 − c1)

k.

There holds

tan(x+ iy) =sin(2x)

cos(2x) + cosh(2y)+ i

sinh(2y)

cos(2x) + cosh(2y).

Due to Theorem 2.1.2 we have

tan(ikn,±1,±2h) ≈ ∓21

π√2(2n+ 1±2 1

4)+ i

√

(π√2(2n + 1±2 1

4))2 − 1

π√2(2n+ 1±2 1

4)

.

Hence

(1 + tan2(ikn,±1,±2h)) ≈ O(1

n) = O(|kn|−1).

Thus

(2.24e) + (2.24f) + (2.24g) + (2.24h) = O(|kn sin(αnh) sin(βnh)|).

This yields

∂kF (ω, kn) ≈ i8h(c2 − c1)k2n sin(αnh) sin(βnh)

+O(|kn sin(αnh) sin(βnh)|).

Therefore for n big enough we have F (ω, kn) 6= 0. Hence for n big enough kn is analytic.Now we can calculate k′n as k′n = −∂ωF

∂kF. If we show

∂ωF (ω, kn) ≈ O(|k2n sin(αnh) sin(βnh)|),

18 2. MODELING

we obtain k′n bounded. We have

∂ωF = 4k2β

α

ω

c2Lsin(αh) cos(βh) (2.25a)

+ 4k2α

β

ω

c2Tsin(αh) cos(βh) (2.25b)

− 4(k2 − ω2

c2T)ω

c2Tcos(αh) sin(βh) (2.25c)

+ 4k2βω

c2Lcos(αh) cos(βh) (2.25d)

− 4k2αω

c2Tsin(αh) sin(βh) (2.25e)

− (2k2 − ω2

c2T)2

ω

αc2Lsin(αh) sin(βh) (2.25f)

+ (2k2 − ω2

c2T)2

ω

βc2Tcos(αh) cos(βh). (2.25g)

The first three terms (2.25a)-(2.25c) of ∂ωF are of the desired order. For the last fourterms we obtain

(2.25d) + (2.25e) + (2.25f) + (2.25g) ≈i8(c1 − c2)k

3 sin(αh) sin(βh) + i8(c1 − c2)k3 cos(αh) cos(βh)

+O(|k sin(αh) sin(βh)|)= i8(c1 − c2)k

3 cos(αh) cos(βh) (1 + tan(αh) tan(βh))

+O(|k sin(αh) sin(βh)|)

through linearizing α, β. Yet another linearization shows

tan(αnh) = tan(iknh)− (1 + tan2(iknh))ic1kn

+O(|kn|−2)

= tan(iknh) +O(|kn|−2)

= i+O(|kn|−1).

Thus

1 + tan(αh) tan(βh) = O(|kn|−1).

Hence

∂ωF (ω, kn) = O(|k2n sin(αnh) sin(βnh)|)

and k′n is bounded.

2.2. PROPERTIES OF LAMB MODES 19

2.2. Properties of Lamb Modes

For every solution kn with Im kn > 0 if kn /∈ R and k′n > 0 if kn ∈ R we associate a modeun:

un,1 = (iknAn,2 cos(αny) + βnB1 cos(βny)) eiknx,

un,2 = (−αAn,2 sin(αny)− iknB1 sin(βny)) eiknx.

(2.26)

From now on we assume

Assumption 2.2.1. There exist N kn and for every symmetric function g ∈ [L2(−h, h)]2,there exists a unique sequence cn ∈ l∞ such that

g(y) =∑

n∈Ncn

un(0, y)

‖un(0, .)‖L2(−h,h)

. (2.27)

Then one (not the unique) solution of problem (2.1)-(2.4) with g symmetric is

u(x, y) =∑

n∈Ncn

un(x, y)

‖un(0, .)‖L2(−h,h)

. (2.28)

Remark 2.2.2. Due to Theorem 2.1.2 it is plausible that there exist N solutions kn ofEquation (2.13). We proof the linear independence of the functions un: Let any linear

combination of zero∑M

m=1 cmunm(0, y) = 0 with cm ∈ C be given. Then there exists al ∈ 1, . . . ,M such that either

|αnl| ≥ |αnm |, |βnm| ∀m ∈ 1, . . . ,M,

or

|βnl| ≥ |αnm|, |βnm| ∀m ∈ 1, . . . ,M.

We assume that the first case holds. The second one can be treated in a similar way.Since

M∑

m=1

cmunm(0, y) = 0,

there holdsM∑

m=1

cm (−αnmAnm,2 sin(αnmy)− iknmBnm,1 sin(βnmy)) = 0, ∀y ∈ C.

We multiply with e−|αnl|t and set y = iαnl

t. Hence

M∑

m=1

cme−|αnl

|t (−αnmAnm,2 sin(αnmiαnlt)− iknmBnm,1 sin(βnmiαnl

t)) = 0.

Taking the limit t→ +∞ yields cl = 0.We also want to show that Assumption 2.2.1 is plausible in the sense, that the system

20 2. MODELING

un is complete. To do this, we take a look at a well known orthogonal basis of thesymmetric function space on the interval (-h,h):

(

cos(πnhy)

0

)

,

(

0sin(πn

hy)

)

, n ∈ N.

Due to

kn,±1,±2 ≈ ±1 1

2harccosh

(

2√2nπ +

√2π ±2

√2π

4

)

+i

h

(

2nπ + π ±2 π

4

)

,

we have

αn ∼ βn ∼ −1

h

(

2nπ + π ±2 π

4

)

±1 i · Cn,

and

un,±1,±2 ∼(

Dn cos((

2nπ + π ±2 π4

)

yh±1 i · Cn · y

h

)

En sin((

2nπ + π ±2 π4

)

yh±1 i · Cn · y

h

)

)

with Cn ∈ R for large n. When we compare the two systems, we see that the sin and costerms are split in the first system, but not in the second one. However, for every n weget two functions in the first system, one sin and one cos, which we can associate withthe two functions in the second system generated by ±1 for every n. We also see thatfrequency in the first system is πn

h. In the second system the frequency increases faster,

i.e. 2πnh. Yet due to ±2 we still have two functions for each n. Thus for this concept of

counting, we have to divide 2πnh

by 2, which yields same frequency growing rate as in thefirst system.Hence these heuristic arguments state that Assumption 2.2.1 is plausible. In order toprove this in a rigorous way one could consider the best approximation of a given g inthe finite sub spaces span unNn=1 with N ∈ N. Writing g in the previous orthogonalbasis and calculating the approximation error would then lead to the spectral norm of theinverse of a explicit given N ×N matrix. If one showed that this term converges to zerofor N → ∞, the claim would be proven.

2.3. Limiting Absorption Principle

Now we show that the given u is indeed the physical solution, due to the correct choice ofthe wave numbers kn. In natural physical processes one always observes the absorption ofenergy. But this absorption can be considered sufficiently small. Thus, it is adequate toset the absorption parameter in the physical/mathematical model to zero. Nevertheless,one has to revert to the damped model, because the change to the undamped modelproduces a loss of information. I.e. we add a damping parameter iǫ to the frequency ωand choose the sign of the wave numbers so that the solution uǫ of the problem convergesto zero for x → ∞. Afterwards we let ǫ tend to zero and calculate limǫ→0 u

ǫ =: u. An

2.3. LIMITING ABSORPTION PRINCIPLE 21

analysis of u then shows that for real k, k′(ω) > 0 is the correct criterion for deciding onthe sign of the wave number.This approach is called the limiting absorption principle. We now apply it and carry outthe mathematically rigorous details. The formulation of the damped problem is: Findu ∈ L∞ (R+ × (−h, h)), so that

µ∆u+ (λ+ µ)∇ div u = ρ(ω + iǫ)2u in R+ × (−h, h), (2.29)

u = g on 0 × (−h, h), (2.30)

τ21 = µ(∂xu2 + ∂yu1) = 0 on R+ × ±h, (2.31)

τ22 = λ(∂xu1 + ∂yu2) = 0 on R+ × ±h. (2.32)

Finding the solution uǫ of problem (2.29)-(2.32) and taking the limit u := limǫ→0 uǫ then

gives us the physical solution. Similarly to the undamped problem

uǫn,1 =(

ikǫnAǫn,2 cos(α

ǫny) + βǫ

nBǫn,1 cos(β

ǫny))

eikǫnx,

uǫn,2 =(

−αǫnA

ǫn,2 sin(α

ǫny)− ikǫnB

ǫn,1 sin(β

ǫny))

eikǫnx,

(2.33)

solve equations (2.29), (2.31)-(2.32), with

αǫn =

√

(ω + iǫ)2

c2L− (kǫn)

2,

βǫn =

√

(ω + iǫ)2

c2T− (kǫn)

2,

Aǫn,2 =

(

(kǫn)2 − (βǫ

n)2)2

sin(βǫnh),

Bǫn,1 = 2ikǫnα

ǫn sin(α

ǫnh),

(2.34)

where kǫn := kn(ω + iǫ) are the solutions of

4(kǫ)2αǫβǫ sin(αǫh) cos(βǫh) + ((kǫ)2 − (βǫ)2)2 cos(αǫh) sin(βǫh) = 0. (2.35)

Assumption 2.3.1. For every n ∈ N there holds: ǫ 7→ kn(ω+ iǫ) is continuous at ǫ = 0.For every n ∈ N with kn(ω) 6= 0, ǫ 7→ kn(ω + iǫ) is differentiable at ǫ = 0.

For big n Lemma 2.1.3 already yields the previous assumption. For the remaining finitelymany small n we simply have to assume the property.

Remark 2.3.2. For k(ω) = 0 we cannot expect k to be differentiable. If we considerF (ω, k) = 0 with k = 0 we gain

ω4

c4Tcos

(

ωh

cL

)

sin

(

ωh

cT

)

= 0.

22 2. MODELING

Solving this equations yields

ωn,1 =cLh(nπ − π

2),

ωn,2 =cThnπ.

Because of F = 0 we have ∂ωF + ω′∂kF = 0 and thus

ω′ = −∂kF∂ωF

.

Due to k = 0 we have

∂kF (ωn,j, 0) = 0, j = 1, 2

∂ωF (ωn,j, 0) 6= 0, j = 1, 2.

Hence ω′(0) = 0 and therefore ∄k′(ω) for k = 0. If we take a look at the dispersion curvesin Section 3.2, we see that for certain ω there indeed exist certain kn(ω) with kn(ω) 6= 0and ∄k′n(ω).

In order to ensure u ∈ L∞ we choose kǫn so that Im kǫn ≥ 0. Due to Assumption 2.3.1for kn /∈ R the sign of Im kn determines the sign of Im kǫn for ǫ small enough. Due toLemma 2.1.3 this holds uniformly in n. If kn ∈ R we have

kǫn = kn(ω + iǫ) ≈ kn(ω) + iǫk′n(ω).

Thus we choose kn(ω + iǫ) so that k′n(ω) > 0. In the case that k′n vanishes we have toassume ǫ 7→ kn(ω + iǫ) ∈ C3 in a neighbourhood of 0 (again Lemma 2.1.3 helps us) andgain

kǫn = kn(ω + iǫ) ≈ kn(ω)−ǫ2

2!k′′n(ω)−

iǫ3

3!k′′′n (ω),

and thus take kn(ω + iǫ) with k′′′n (ω) < 0 and so on. Similar to the undamped problemwe have to assume

Assumption 2.3.3. There exists ǫ0 > 0 such that for all ǫ ∈ [0, ǫ0) there holds: For

every symmetric function g ∈ [L2(−h, h)]2, there exists a unique sequence cǫn ∈ l∞ suchthat

g(y) =∑

n∈Ncǫn

uǫn(0, y)

‖uǫn(0, .)‖L2(−h,h)

. (2.36)

Hence we can write the solution of the damped problem as

uǫ(x, y) =∑

n∈Ncǫn

uǫn(x, y)

‖uǫn(0, .)‖L2(−h,h)

. (2.37)

Now we want to show that uǫ converges to u in some norms of interest. In order to dothat, we require two more plausible assumptions:

Assumption 2.3.4. For every fixed n ∈ N there holds cǫnǫ→0−−→ cn.


Assumption 2.3.5. There exists C > 0 such that |cǫn| ≤ C for all n ∈ N, ǫ ∈ [0, ǫ0).

Remark 2.3.6. Due to Lemma 2.1.3 and Assumption 2.3.1 we have kǫn → kn uniformly

in n. If we consider the background of the problem to solve, we can also show uǫnL2(−h,h)−−−−−→

un uniformly in n for fixed x = 0: Originally we were investigating the elastic behaviourof an infinite beam. The left boundary of our half infinite beam is just set for analyticalreasons. Thus it is acceptable to assume the half beam begins already at x = −η < 0 andhence

uǫn,1 =(

ikǫnAǫn,2 cos(α

ǫny) + βǫ

nBǫn,1 cos(β

ǫny))

eikǫn(x+η),

uǫn,2 =(

−αǫnA

ǫn,2 sin(α

ǫny)− ikǫnB

ǫn,1 sin(β

ǫny))

eikǫn(x+η).

Due to the uniform convergence of kǫn and the exponentially damped term eikǫnη, we have

uǫn(0, y) → un(0, y) uniformly in L∞(−h, h). The coefficients of a function u with respectto an orthonormal basis are linear continuous functionals of u. We expect this to besimilar for the ”nonorthogonal basis” un(0, y). Therefore the above observations suggestcǫn → cn in l∞, which would yield assumptions 2.3.4 and 2.3.5.

Theorem 2.3.7. Let uǫ be the solution of the damped problem with damping parameter ǫand u be the stated solution of the undamped problem. Then for every x ∈ R+ there holds

limǫ→0‖u− uǫ‖[L2(x×(−h,h))]2 = 0.

Proof. Due to Lebesgue’s dominated convergence theorem it suffices to show that

(1) |uǫ| ≤ C ∀y ∈ (−h, h), ∀0 ≤ ǫ ≤ ǫ0,(2) for almost every y ∈ (−h, h), uǫ converges pointwise to u.

Due to Assumption 2.3.3, we can write

u =∑

n∈Ncn

un(0, y)

‖un(0, .)‖L2

eiknx,

uǫ =∑

n∈Ncǫn

uǫn(0, y)

‖uǫn(0, .)‖L2

eikǫnx.

First we show that

‖uǫn(0, .)‖L∞(−h,h)

‖uǫn(0, y)‖L2(−h,h)

|eikǫnx| ≤ Ce−π4h

nx ∀n ∈ N, ǫ ∈ [0, ǫ0).

We know that for n big enough we have Im kn,±1,±2 ≈ 2πhn. Thus Im kn & π

2hn. Due

to |k′n| ≤ C we have Im kǫn ≥ π2hn − C. A straight forward calculation yields that

‖uǫn(0,y)‖L∞(−h,h)

‖uǫn(0,y)‖L2(−h,h)

is bounded by a polynomial in |kǫn|. Additionally due to Theorem 2.1.2

we have |Re kǫn| ≤ | Im kǫn|+ C. Thus‖uǫ

n(0,y)‖L∞(−h,h)

‖uǫn(0,y)‖L2(−h,h)

|eikǫnx

2 | is bounded.

24 2. MODELING

Now using Assumption 2.3.5 we obtain

|∑

n∈Ncǫn

uǫn(0, y)

‖uǫn(0, .)‖L2(−h, h)eikǫnx| ≤ C

∑

n∈Ne−

π4h

nx

≤ C.

Now we obtain the pointwise convergence of uǫ:

|∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2

eikǫnx − cn

un(0, y)

‖un(0, .)‖L2

eiknx)

| ≤∑

n∈N|cǫn − cn|

|uǫn(0, y)|‖uǫn(0, .)‖L2

|eikǫnx|

+∑

n∈N|cn||

uǫn(0, y)

‖uǫn(0, .)‖L2

eikǫnx − un(0, y)

‖un(0, .)‖L2

eiknx|

For any given δ > 0 we choose N ∈ N such that

4C∞∑

n=N

e−π4h

nx ≤ δ.

Since∑N

n=1 |cǫn − cn| |uǫn(0,y)|

‖uǫn(0,.)‖L2

|eikǫnx|+∑Nn=1 |cn||

uǫn(0,y)

‖uǫn(0,.)‖L2

eikǫnx − un(0,y)

‖un(0,.)‖L2eiknx| is contin-

uous in ǫ, we have

lim supǫ→0

|∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2


‖un(0, .)‖L2

eiknx)

| ≤ δ.

As δ was chosen arbitrarily, we have

limǫ→0

|∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2


‖un(0, .)‖L2

eiknx)

| = 0.

Theorem 2.3.8. Let uǫ be the solution of the damped problem with damping parameterǫ and u be the stated solution of the undamped problem. Then

limǫ→0‖u− uǫ‖[L∞((η,D)×(−h,h))]2 = 0

holds for every 0 < η < D <∞.


Proof. We use the same main ideas used in the proof of Theorem 2.3.7. For everyy ∈ (−h, h), x ∈ (η,D) we have

∣

∣

∣

∣

∣

∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2

eikǫnx −cn

un(0, y)

‖un(0, .)‖L2

eiknx)∣

∣

∣

∣

≤∣

∣

∣

∣

∣

∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2

− cnun(0, y)

‖un(0, .)‖L2

)

eikǫnx

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∑

n∈Ncn

un(0, y)

‖un(0, .)‖L2

(

eikǫnx − eiknx

)

∣

∣

∣

∣

∣

.

We can estimate the second summand as∣

∣

∣

∣

∣

∑

n∈Ncn

un(0, y)

‖un(0, .)‖L2

(

eikǫnx − eiknx

)

∣

∣

∣

∣

∣

≤∑

n∈N|eiknx|‖un(0, .)‖L∞

‖un(0, .)‖L2

∣

∣1− ei(kǫn−kn)x

∣

∣ .

As in the proof of the previous theorem we have

∑

n∈N|eiknx|‖un(0, .)‖L∞

‖un(0, .)‖L2

∣


∣

∣ ≤ C∑

n∈Ne−

πnη4h

∣


∣

∣ .

If ǫ ∈ [0, ǫ0) and using the explicit formula of exp, we conclude

∑

n∈Ne−

πnη4h

∣


∣

∣ ≤ C∑

n∈Ne−

πnη4h |kǫn − kn|D.

This estimate is uniform in x and y and converges to zero for ǫ→ 0 as well. To deal withthe first summand we start with a simple estimate

∣

∣

∣

∣

∣

∑

n∈N

(

cǫnuǫn(0, y)

‖uǫn(0, .)‖L2

− cnun(0, y)

‖un(0, .)‖L2

)

eikǫnx

∣

∣

∣

∣

≤

C∑

n∈Ne−

ηπn8h sup

y∈(−h,h)

∣

∣

∣

∣

cǫne− ηπn

8huǫn(0, y)

‖uǫn(0, .)‖L2

− cne− ηπn

8hun(0, y)

‖un(0, .)‖L2

∣

∣

∣

∣

.

Now if we can show that for all n ∈ N supy∈(−h,h)

∣

∣

∣cǫne

− ηπn8h

uǫn(0,y)

‖uǫn(0,.)‖L2

− cne− ηπn

8hun(0,y)

‖un(0,.)‖L2

∣

∣

∣

converges to zero for ǫ → 0, as well as that

supy∈(−h,h)

∣

∣

∣cǫne

− ηπn8h

uǫn(0,y)

‖uǫn(0,.)‖L2

− cne− ηπn

8hun(0,y)

‖un(0,.)‖L2

∣

∣

∣stays bounded in n ∈ N, ǫ ∈ [0, ǫ0), the

stated theorem is proven. The boundedness holds due to the term e−ηπn8h and the same

arguments as in Theorem 2.3.7. To prove the convergence we show uǫn(0, y)L∞

−−→ un(0, y).

26 2. MODELING

The rest follows easily. Now

| cos(αny)− cos(αǫny)| =

∣

∣

∣

∣

∫ αǫn

αn

y sin(γy)dγ

∣

∣

∣

∣

≤ |αǫn − αn|h sup

y∈(−h,h),γ∈Uδ(αn)

| sin(γy)|.

From this the stated follows.

Remark 2.3.9. Due to the exponential decreasing terms eiknx ∼ e−πnx4h convergence in

additional norms could be shown, if wanted to.

CHAPTER 3

Numerical Concepts

In this chapter we present some numerical methods for acoustic wave propagation andstudy why, why not and how these concepts can be adapted to the elastic case. Henceit is necessary to give a short introduction to acoustic wave propagation, which we do inSection 3.1, where we adhere very closely to [14].In Section 3.2 we elaborate on the qualitative difference between acoustic and elasticwave propagation, which is the behaviour of the wave numbers and the dispersion curves.In particular this means that elastic dispersion curves can have negative slopes, whichentails that phase and group velocity have different signs.The first numerical concept which we present in Section 3.3, is complex scaling aka per-fectly matched layers. However, we will not give a full discussion of this method, butonly sketch its main ideas. We will further show that without a projection onto eigenmodes this concept cannot be successfully adapted to our specific elastic problem.Hence in Section 3.4 we will introduce another numerical concept for acoustic wave prop-agation, i.e. the Hardy space method. Again we do not give rigorous proofs for thisnumerical concept, but we elaborate on why and how it can be successfully adapted toelastic wave propagation.

3.1. Acoustic Wave Propagation

To describe acoustic waves we require the following variables:

• The space variable x ∈ Rd, where d = 1, 2, 3 is the space dimension,• the time variable t ∈ R+

0 ,• the velocity field v(x, t) ∈ Rd,• the pressure field p(x, t) ∈ R,• the density field ρ(x, t) ∈ R.

Additionally we assume that

• p0 = const,• v0 = 0,• ρ0 = ρ0(x)

describe the stationary state. The relationship between v, p and ρ can be described bythe linearised Euler equation

ρ0∂tv +∇p = 0, (3.1)

27

28 3. NUMERICAL CONCEPTS

the linearised continuity equation

∂tρ+ div (ρ0v) = 0 (3.2)

and

p = c2ρ, (3.3)

where c = c(x) describes the speed of sound. If we plug Equation (3.3) into Equation (3.2)we gain

1

c2∂tp+ div (ρ0v) = 0. (3.4)

Now taking the time derivative of Equation (3.4) and subtracting it from the divergenceof Equation (3.1) yields

1

c2∂t∂tp = ∆p.

If we consider time harmonic waves of the form

p(x, t) = u(x)e−iωt,

with the amplitude u and the positive frequency ω, we gain

∆u+ω2

c2u = 0.

In the case of a homogeneous medium, i.e. c = const, we obtain the Helmholtz equation

∆u+ κ2u = 0,

with κ = ωc. Now we consider an acoustic wave guide with the same geometry as described

in Section 1.3. Hence we gain the problem (for given boundary g):Find u ∈ H2

loc(R+ × (−h, h);C) such that

∆u+ κ2u = 0 for (x, y) ∈ R+ × (−h, h), (3.5)

∂yu = 0 for x ∈ R+, y = ±h, (3.6)

u = g for x = 0, y ∈ (−h, h). (3.7)⋃

n∈Ncos(πhny) are the eigen functions to the Sturm-Liouville problem ∂2xu = 0 withNeumann boundary conditions ∂xu(±h) = 0. Thus

⋃

n∈Ncos(πhny) is an orthogonalbasis of L2(−h, h). Additionally, every basis function fulfills the boundary conditions aty = ±h. Hence we can write

g(y) =∑

n∈Ncn cos(

π

hny)

with

cn =1

π

(

g(y) cos(π

hny))

L2(−h,h).

3.2. WAVE NUMBERS AND DISPERSION CURVES 29

As in the elastic problem we use a product and series ansatz

u(x, y) =∑

n∈Ncn cos(

π

hny)e±iknx (3.8)

with wave numbers

kn :=

√

κ2 − π2

h2n2, (3.9)

which solves equations (3.5)-(3.7). As in the elastic case for each n we can choose a sign

of i√

κ2 − π2

h2n2. If κ < πhn we choose the sign, such that e±i

√

κ2−π2

h2n2x stays bounded. If

κ > πhn we choose the sign, such that the imaginary part of i

√

κ2 − π2

h2n2 is positive. This

choice can be justified by the limiting absorption principle as was done for the elasticwave numbers. This is much easier to be shown for acoustic wave guides, thus we leaveit to the reader. One way to implement this condition on the wave numbers into theset of Equations (3.5)-(3.7) is the so called pole condition, see [11, 15, 10]. If we haveu(x, y) =

∑

n∈N cn cos(πhny)eiknx, we can easily calculate the Laplace transformation of u

in x:

u(s, y) =

∫ ∞

0

u(x, y)e−sxdx

=∑

n∈Ncn cos(

π

hny)

∫ ∞

0

e(ikn−s)xdx

=∑

n∈Ncn cos(

π

hny)

1

ikn − s.

Since we demanded Im kneiπ/4 ≥ 0, which is equivalent to Re ikne

iπ/4 ≤ 0, u(s) has ananalytic extension to Re seiπ/4 > 0. Thus one way to complete our formulation of theproblem is:Find u ∈ H2

loc(R+ × (−h, h);C) such that

∆u+ κ2u = 0 for (x, y) ∈ R+ × (−h, h), (3.10)

∂yu = 0 for x ∈ R+, y = ±h, (3.11)

u = g for x = 0, y ∈ (−h, h), (3.12)

u(s) has an analytic extension on the half plane Re seiπ/4 > 0. (3.13)

3.2. Wave Numbers and Dispersion Curves

3.2.1. Material Parameters. Up to now we have described our material in theelastic problem with the help of Lame’s parameters: Lame’s first parameter λ and theshear modulus µ. However, for some purposes it is more practical to use Young’s modulus


E and Poisson’s ration ν to describe the material, as done in [1]:

E =µ(3λ+ 2µ)

λ+ µ,

ν =λ

2(λ+ µ),

λ =Eν

(1 + ν)(1− 2ν),

µ =E

2(1 + ν).

(3.14)

If we express our wave speeds cT and cL with E and ν we get

cT =

√

µ

ρ=

√

E

ρ(1 + ν)

1

2,

cL =

√

λ+ 2µ

ρ=

√

E

ρ(1 + ν)

(1− ν)

(1− 2ν).

(3.15)

We take a look at the relation equation for k and ω:

F (ω, k) = 4k2αβ sin(αh) cos(βh) + (k2 − β2)2 cos(αh) sin(βh),

α =

√

ω2

c2L− k2,

β =

√

ω2

c2T− k2.

We see that ω either occurs as ω2

c2Tor as ω2

c2L. Since we have

c2T =E

ρ(1 + ν)

1

2,

c2L =E

ρ(1 + ν)

(1− ν)

(1− 2ν),

(3.16)

there either occurs ω2Eρ

1(1+ν)2

or ω2Eρ

1−ν(1+ν)(1−2ν)

. We see that the factor Eρonly scales ω,

whereas the size of ν makes a qualitative difference in the solutions of the equation. Thatis why we fixed E, ρ and h in figures 2-3 and only varied the Poission’s ratio ν.

3.2.2. Dispersion Curves. A dispersion curve is a curve (ω, kn(ω)) :ω ∈ (ωn,∞), where ωn is the root of kn. In the acoustic case we have the wavenumbers explicitly given as

kn(ω) =

√

ω2

c2− π2

h2n2,


0 5 10 15 20 25 300

5

10

15

20

k

ω

Figure 1. Acoustic dispersion curves with c = h = 1.

with the root

ωn = cπ

hn.

In the elastic case we also expect that if ω < ωn, kn(ω) is imaginary and if ω > ωn, kn(ω)is real. The roots ωn of the even elastic dispersion curves can be easily calculated bysolving

ω4

c4Tcos

(

ωh

cL

)

sin

(

ωh

cT

)

= 0.

Hence

ωn,1 =cLh(nπ − π

2),

ωn,2 =cThnπ.

For the odd elastic dispersion curves we calculate

ω4

c4Tcos

(

ωh

cT

)

sin

(

ωh

cL

)

= 0.

Hence

ωn,1 =cTh(nπ − π

2),

ωn,2 =cLhnπ.

Although it may seem simple, finding and calculating all dispersion curves is rathertricky. One knows the points where dispersion curves start, but one does not know howmany different curves have the same root. We take for example ωn = 0: at least two


Figure 2. Elastic dispersion curves with h = 1, cT = 1, cL =√

2(1−ν)1−2ν

, ν =

0.3663, compare with [18]. Even curves are red, odd curves are blue.

Figure 3. Elastic dispersion curves for concrete: h = 1, cT = 1, cL =√

2(1−ν)1−2ν

, ν = 0.2. Even curves are red, odd curves are blue.

linear curves, i.e. k = ωcT

and k = ωcL, have roots there. But plugging these equations

into u shows, that these curves correspond to u = 0.

In many text books, as in [1], in addition to the phase velocity ω the group velocity isdefined as ∂kω. If we set the material parameters E, ν and h in the elastic case or cand h in the acoustic case and choose a frequency ω, we can read from the figure of thedispersion curves how many real wave numbers exist. Moreover one can see which sign ofk is the physically correct choice by regarding the slope of the curves. We see that in the


acoustic case group and phase velocity always have the same sign. Yet, when regardingthe given figures of the elastic dispersion curves, one can see that there actually existwave numbers with different sign of group and phase velocity, which justifies our effortinto developing a numerical scheme.

3.2.3. Wave Numbers. In the acoustic case we can calculate the wave numbersexplicitly as

kn(ω) =

√

ω2

c2− π2

h2n2, n ∈ N0 : kn ∈ R,

kn(ω) = i

√

π2

h2n2 − ω2

c2, n ∈ N0 : kn /∈ R.

Figure 4 shows them in the complex plane. The wave numbers which are the correct

−50 −40 −30 −20 −10 0 10 20 30 40 50−60

−40

−20

0

20

40

60

real axis

imag

inar

y ax

is

Figure 4. Acoustic wave numbers in the complex plane.

physical choice are marked red, the non-physical ones are marked blue. We see that thecorrect wave numbers lie in the rotated upper half of the complex plane. In contrast,in the elastic case we only know the asymptotic behaviour of the wave numbers due toChapter 2, i.e. for the symmetric wave number there holds for large n:

kn,±1,±2 ≈ ±1 1

2harccosh

(

2√2nπ +

√2π ±2

√2π

4

)

+i

h

(

2nπ + π ±2 π

4

)

.

Additionally, as we pointed out in the previous subsection, the real valued wave numbersdo not necessary lie in one half plane, but can have different signs. Thus we can expect


the elastic wave numbers to look similar to Figure 5.This general knowledge about the wave numbers can be used to develop and justify

−3 −2 −1 0 1 2 3−100

−80

−60

−40

−20

0

20

40

60

80

100

real axis

imaginary

axis

Figure 5. Expected appearance of elastic wave numbers in the complex plane.

different numerical methods, which do not depend on the wave numbers. Two of themwill be introduced in the next sections.

3.3. Complex Scaling

Complex scaling reaches back to the sixties. A historical synopsis can be foundin [9]. The concept was redeveloped as perfectly matched layers (PML) in [2]. Firstconvergence results where published in [5]. In [18] PML was applied to guided elasticwave propagation by projecting onto eigen modes.

We set Ω = (−h, h)×R+. The idea of complex scaling follows a very classical approach.We reformulate the problem in a weak sense, i.e. we search for u ∈ H such that

∫

Ω

∇u · ∇v +∫

Ω

κ2uv = 0 ∀v ∈ H0,

u = g for x = 0, y ∈ (−h, h),u ”is outgoing”,

(3.17)

with some Hilbert space H ⊂ H1(Ω) and H0 = v ∈ H : v|0×(−h,h) = 0, but sincethis implies u ∈ H1(Ω), which does not hold, this clearly does not work. Hence wemodify u in such a way, that it becomes an element of H1(Ω). The modification used iscomplex scaling. As we want to explain the expression PML we do not restrict ourselvesto the exterior domain but split Ω into an interior domain Ωinn := (0, x0)× (−h, h) and

3.3. COMPLEX SCALING 35

an exterior domain Ωext := (x0,∞) × (−h, h) with x0 ∈ R+. We define an anisotropicscaling γ, such that x is transformed to a complex variable on the exterior domain:

γ(x) =

x x ≤ x0

x+ i∫ x−x0

0σ(r)dr x0 < x

. (3.18)

We demand σ to be a continuous function σ : R+0 → R+ and exp(−

∫ x

0σ(r)dr) ∈ H1(R+).

Then we define

uscal := u γ. (3.19)

Now we set Vg := v ∈ H1(Ω) : v = g on 0 × (−h, h) and we define the variationalproblem: Find uscal ∈ Vg such that

a(uscal, v) + b(uscal, v) = 0, ∀v ∈ V0, (3.20)

with

a(u, v) :=

∫

Ω

1

γ′∂xu∂xv + γ′∂yu∂yv, (3.21)

and

b(u, v) :=

∫

Ω

γ′uv. (3.22)

If u is a solution of the strong formulation (3.10)-(3.13), we calculate that uscal solves thevariational formulation (3.21):

0 = ∂x∂xu+ ∂y∂yu+ κ2u

0 = (∂x∂xu) γ + (∂y∂yu) γ + κ2u γ

=1

γ′∂x((∂xu) γ) + ∂y∂y(u γ) + κ2u γ

=1

γ′∂x

1

γ′∂x(u γ) + ∂y∂yuscal + κ2uscal

=1

γ′∂x

1

γ′∂xuscal + ∂y∂yuscal + κ2uscal

= ∂x1

γ′∂xuscal + γ′∂y∂yuscal + γ′κ2uscal.

We further have

uscal(x, y) =∑

n∈Ncn cos(

π

hny)eiknγ(x) ∈ H1(Ω).


Hence for v ∈ V0 we have

0 =

∫

Ω

(∂x1

γ′∂xuscal + γ′∂y∂yuscal + γ′κ2uscal)v

=

∫

Ω

1

γ′∂xuscal∂xv + γ′∂yuscal∂yv + γ′κ2uscalv

= a(uscal, v) + b(uscal, v).

On the other hand, if uscal solves the variational problem (3.21), we have

1

γ′∂x

1

γ′∂xuscal + ∂y∂yuscal + κ2uscal = 0 on Ω, (3.23)

∂yuscal = 0 on y = ±h, (3.24)

uscal = g on x = 0, (3.25)

uscal ∈ H1(Ω). (3.26)

If we solve Equations (3.23)-(3.25) and consider condition (3.26) we get

uscal(x, y) =∑

n∈Ncn cos(

π

hny)eiknγ(x) ∈ H1(Ω).

Hence uscal|Ωinn= u|Ωinn

. Now that we know that the restriction on Ωinn of the solutionof the variational formulation (3.21) gives us a restriction on Ωinn of the solutionof (3.10)-(3.13), it suggests itself to perform a Galerkin Method on the weak formulation.Proofs of convergence for finite element methods on complex scaling can be foundhere [14].

In the implementation of the method, the calculation of the interior and the exterior partis split. The name perfectly matched layer is earned by the fact that at the interfacebetween interior and exterior domain, the bilinear forms produce the same Neumannboundary terms:∫

Ωext

1

γ′∂xuscal∂xv + γ′∂yuscal∂yv = −

∫

Ωext

∂x1

γ′∂xuscalv + γ′∂2yuscalv +

∫

x0×(−h,h)

1

γ′∂xuscalv

= −∫

Ωext

∂x1

γ′∂xuscalv + γ′∂2yuscalv +

∫

x0×(−h,h)

∂xuv.

The core of the mentioned method is Re iknγ(x)x→∞−−−→ −∞ for all n ∈ N. This is

possible because all wave numbers kn lie in one half plane. In elasticity this is notnecessarily the case: If there exist −kn, km > 0, there would have to hold Re iknγ(x) =

−kn∫ x−x0

0σ(r)dr

x→∞−−−→ −∞ and Re ikmγ(x) = −km∫ x−x0

0σ(r)dr

x→∞−−−→ −∞, which canclearly not both be satisfied. Thus to perform PML for elasticity one would have toseparate the modes with negative group velocity, as done in [18].

3.4. HARDY SPACE METHOD 37

3.4. Hardy Space Method

The Hardy space method continues with the observation that u(s, y) has in s an analyticextension on the half plane Re seiπ/4 > 0, but before we proceed, we need some analyticalbackground. The Hardy space was first introduced by Godfrey Harold Hardy in [7]. Wetake the stated definitions and results from [13].

Definition 3.4.1.

(1) The Hardy space H±(S1) consists of all functions f ∈ L2(S1) for which thereexist g with the following properties

• g is analytic on z ∈ C : |z±| < 1,• there exists C > 0 such that

∫

S1 |g(r±z)|2dz ≤ C for all 0 < r < 1,

• f is the L2-trace of g, i.e.∫

S1 |g(r±z)− f(z)|2dz rր1−−→ 0.(2) The Hardy space H±(R) consists of all functions f ∈ L2(R) for which there exist

g with the following properties• g is analytic on C± := z ∈ C : Im ±z > 0,• there exists C > 0 such that

∫

R |g(z ± iǫ)|2dz ≤ C for all 0 < ǫ,

• f is the L2-trace of g, i.e.∫

R|g(z ± iǫ)− f(z)|2dz ǫց0−−→ 0.

(3) Let κ0 ∈ C with Re κ0 > 0. Then H±(κ0R) := f : f(κ0.) ∈ H±(R).Lemma 3.4.2.

(1) The Hardy space H±(S1) equipped with the standard L2(S1) product is a Hilbertspace.

(2) The monomials z±n, n ∈ N0 are a orthogonal basis of H±(S1).(3) If f ∈ H±(S1), then the function g from Definition 3.4.1 is uniquely given by

∑∞n=0(f, z

n)z±n.

Lemma 3.4.3. Let Re κ0 > 0 and

φ(z) := iκ0z + 1

z − 1, z ∈ S1

be a family of maps from S1 to κ0R. Then the Mobius transformations M

(Mf)(z) := (f φ)(z) 1

z − 1, z ∈ S1, f ∈ H−(κ0R)

are a family of unitary maps from H−(κ0R) to H+(S1) up to the factor√

2|κ0|.Proposition 3.4.4. Let Re κ0 > 0. The Hardy spaces H±(κoR) are Hilbert spaces withthe standard L2 product.

Lemma 3.4.5. Let M ≥ 0, κ0 > 0 and f, g : R+ → C be two functions, so that f := Lfexists for s ∈ C : Re s ≥ M and g := Lg exists for s ∈ C : Re s ≥ −M − ǫwith some ǫ > 0. Further let f and g have an analytic extension onto a neighbourhood


of E(M,κ0) := s ∈ C : Re is/κ0 ≥ 0 ∨ Re s ≥ M and let f and g be bounded onE(M,κ0). Then there holds

∫

R+

f(r)g(r)dr = − i

2π

∫

κ0R+

f(s)g(−s)ds = −κ0π

∫

S1

Mf(z)Mg(z)dz.

Lemma 3.4.6. Let f : R+0 → C be Laplace transformable. Further let F := MLf ∈

H+(S1). Then

f(0) = lims→∞

s(Lf)(s) = 2iκ0F (1)

holds.

Now we develop a numerical scheme for

∆u+ κ2u = 0 for (x, y) ∈ R+ × (−h, h),∂yu = 0 for x ∈ R+, y = ±h,u = g for x = 0, y ∈ (−h, h),u ”outgoing”.

(3.27)

The following integrals and transformation have to be understood only in a formal sense.For a mathematical rigorous justification we refer to [13]. We multiply with a testfunction v and integrate by parts:

∫ ∞

0

∫ h

−h

∂xu∂xv + ∂yu∂yv − κ2uv dydx =

∫ h

−h

∂xuv dy.

We use a Galerkin method with tensor product functions, where we define our basisfunctions later:

u(x, y) =∑

j,m

cj,mbj,x(x)bm,y(y).

We gain the integrals∫ h

−h

∂ybm,y(y)∂ybn,y(y) dy,

∫ h

−h

bm,y(y)bn,y(y) dy,

with bounded integration domains and the integrals∫ ∞

0

∂xbj,x(x)∂xbl,x(x) dx,

∫ ∞

0

bj,x(x)bl,x(x) dx.

3.4. HARDY SPACE METHOD 39

with unbounded integration domains. The latter we can draw back to H+(S1) with helpof Lemma 3.4.5:

∫ ∞

0

∂xbj,x(x)∂xbl,x(x) dx = −κ0π

∫

S1

(ML∂xbj,x)(s)ML∂xbl,x)(−s)ds∫ ∞

0

bj,x(x)bl,x(x) dx = −κ0π

∫

S1

(MLbj,x)(s)MLbl,x)(−s)ds.

At S1 we have an orthogonal basis zj , j ∈ N0 at hand. This choice of basis functions may

−3 −2 −1 0 1 2 3−100

−80

−60

−40

−20

0

20

40

60

80

100

real axis

imag

inar

y ax

is

Figure 6. We need a transformation from the upper part to S1.

seem practical, because of it’s orthogonality, but it is difficult to implement the Dirichletboundary conditions with it. Due to Lemma 3.4.6 it is convenient to set

Mbj,x(z) :=1

2iκ0, j = 0

Mbj,x(z) := (z − 1)zj−1, j > 0.

With this choice of basis functions we simply have

u(0, y) =∑

c0,mbm,y(y).

We gain

bj,x(x) =

j∑

k=0

(

jk

)

(2iκ0x)k+1

(k + 1)!

by transforming back. For bm,y we can use some standard finite element basis.


We obtained u to be outgoing by demanding MLu ∈ H+(S1), which is equivalent toLu ∈ H+(κ0R). Now in the case of an elastic wave guide our wave numbers do not liein one half plane and therefore u has no analytic extension to any half plane. Thus wedefine a −iΓ like in Figure 6. By using a Schwarz-Christoffel mapping, see [6], we candefine H±(Γ) analogously to H±(κ0R). Now we can implement the pole condition bydemanding that u has an analytic extension to H+(Γ).

CHAPTER 4

A Model Problem

In order to understand the problems arising in the Hardy space method by switchingfrom the acoustic case to the elastic case, we study a model problem. It captures themain structure, properties and difficulties of the general problem, but unlike most modelproblems it is not the simplest case of some general problem, i.e. all parameters are con-stants and either 1 or 0, neither does it have a physical meaning. This artificial problem isdesigned only for the purpose of studying analytical properties of and numerical conceptsfor guided elastic wave propagation.As we have the explicit solution and more importantly the explicit wave numbers for ourmodel problem at hand, we can elaborate on and study degenerated cases of the modelproblem, i.e. collapsing wave numbers. We expect the same behaviour for guided elasticwave propagation, although we have no rigorous proof for this conclusion. Nevertheless,if we develop a numerical scheme capable of handling the degenerated cases of our modelproblem, we can hope and expect the scheme also to work successfully for guided elasticwave propagation.In the last part of this chapter we consider only one branch of the model problem andcarry out numerical tests. We conclude that the approximation results are satisfying. Thearising matrix of the linear equation system is not stable, but the condition is suitablesmall for a practical number of degrees of freedom.

4.1. Formulation of the Problem

We work on the same domain as in the previous chapters, i.e. Ω := R+ × (−h, h) withh ∈ R+. For the sake of simplicity we set h := π/2. Let a, b, c ∈ R+ be parameters. Inthe last chapters we studied the time harmonic two dimensional linear elastic equations,i.e. a two dimensional second order system in space. Hence it is consistent to start withour model problem as a one dimensional fourth order system. Let

H :=(

−a2∂2x + b2∂2y)2

+ c2∂4y . (4.1)

We are looking for solutions of

Hu = −∂2t u. (4.2)

We are interested only in time harmonic solutions. Hence for a given frequency ω ∈ R+

we write

u(x, y, t) = u(x, y)e−itω. (4.3)

41

42 4. A MODEL PROBLEM

Thus we gain the equation

Hu = ω2u. (4.4)

With the help of

D1 := −a2∂2x + b2∂2y − ic∂2y , (4.5)

D2 := −a2∂2x + b2∂2y + ic∂2y , (4.6)

we can factorize H as

H := D2D1. (4.7)

This and defining

v :=1

ωD1u, (4.8)

allows us to reformulate the equation as a two dimensional second order system:

D1u− ωv = 0,

−ωu+D2v = 0.(4.9)

We add homogeneous Dirichlet boundary conditions at the unbounded part of the bound-ary:

u = v = 0 at y = ±π/2. (4.10)

A product ansatz u = eikx cos(ny) yields the solutions

u = eikx cos(ny),

v =1

ω(a2k2 − b2n2 + icn2)eikx cos(ny),

(4.11)

withn ∈ N0,

k±1,±2,n = ±1 1

a

√

b2n2 ±2√ω2 − c2n4.

(4.12)

The indices of ± indicate that the choices of the signs are independently. We only allowoutgoing wave numbers k. As in the previous chapters we call non real wave numbersoutgoing if Im k > 0 and incoming if not. We call real wave numbers for which ∂ωk existsoutgoing if ∂ωk > 0 and incoming if not. We call real wave numbers degenerated if ∂ωkdoes not exist and treat this special cases in the next section.Up to the end of this section we assume ω such that there do not exist any degeneratedwave numbers. If we add further Neumann boundary conditions on x = 0 we see that

4.2. COLLAPSING WAVE NUMBERS 43

0 50 100 1500

2

4

6

8

10

12

14

16

18

20

ω

k

Figure 1. Dispersion curves of the model problem.

the problem, find u, v ∈ H2(Ω), so that

D1u− ωv = 0,

−ωu+D2v = 0,

∂xu = f, ∂xv = g at x = 0,

u, v = 0 at y = ±π/2,u, v outgoing,

(4.13)

has a unique solution. Observing elastic dispersion curves no intersection between dis-persion curves with different signs of group and phase velocity seem to occur. Hence wechoose our parameters in such a way, that this holds for our model problem. b = 1 andc ≥ 2 satisfy our demand. Thus we can describe the possible domain of the outgoingwave numbers as follows: there exists ξ ∈ R+ such that for every outgoing k there holdsk ∈ z ∈ C : Im z > 0 ∪ (−ξ, 0) ∪ (ξ,∞). For the incoming wave numbers there ac-cordingly holds k ∈ z ∈ C : Im z < 0 ∪ (−∞,−ξ) ∪ (0, ξ). Hence we can force u to beoutgoing by the pole condition: u, v have an extension in H−(Γ) with Γ as in Figure 6.

4.2. Collapsing Wave Numbers

We speak of collapsing wave numbers if two different wave numbers have the same value.This either happens if two different dispersion curves intersect or if a dispersion curvehas a turning point, i.e. k′(ω) = ∞. In the first case we can distinguish two situations:slopes of the two dispersion curves have the same sign or they do not. Since in elasticitywe never observed the latter and we have already chosen the parameters of our modelproblem accordingly, we only discus the situation with the same sign of slopes. The casek′(ω) = ∞ always occurs at roots k = 0 and can happen if the dispersion curve has an


additional turning point k 6= 0.

Intersecting dispersion curves in the model problem: The to k corresponding eigen space

in y is now two dimensional, i.e. u = eikx cos(ny) and u = eikx cos(my) with n 6= mare both solutions, but this gives neither rise to analytical nor to numerical problems,because the dispersion curves have the same sign of group velocity.

Turning points of dispersion curves in the model problem: In the model problem turning

points exist at every root k = 0, ω2 = (c2 + b4)n4 of a dispersion curve. All butthe first dispersion curve have a turning point at k = bn, ω2 = c2n4. If k = 0 thesolutions u = e±ikx cos(ny) collapse to cos(ny), x cos(ny). If ω2 = c2n4 the solutions

e1a

√b2n2±2

√ω2−c2n4

cos(ny) collapse to e1abn cos(ny), xe

1abn cos(ny). The same happens for

the solutions corresponding to −k. Due to continuity and boundedness arguments, wecall solutions with a xeikx-term ”non-physical”.

The case of intersecting dispersion curves should not create numerical problems. The caseof turning points creates problems, because in this case wave numbers lie on our usualseparation path Γ. We could modify Γ in such a way that 0, respectively both ±k lie inthe allowed pole domain, but if the elasticity problem behaves like the model problemadditional non-physical modes would arise. An extra condition to the formulation of theproblem would be needed to eliminate those non-physical modes. If those non-physicalmodes of the elasticity problem behaved like the ones of the model problem, the conditionu ∈ L∞ would be analytically sufficient. However, the question how to implement thisinto a numerical scheme arises. We refer to Chapter 5 for some ideas.

4.3. Galerkin Formulation

We want to break the model problem down to an even simpler problem. We take themodel problem with the parameters a = b = c = 1 and solve it only for the secondbranch:

(−∂2x − 1 + i)u− ωv = 0,

−ωu+ (−∂2x − 1− i)v = 0.(4.14)

If we multiply these equations with appropriate test functions p, q, we gain formally theweak formulation

∫ ∞

0

u′p′ + (−1 + i)up− ωvpdx = u′0p(0),

∫ ∞

0

−ωuq + v′q′ + (−1− i)vqdx = v′0q(0).

(4.15)

4.3. GALERKIN FORMULATION 45

Analogously to the standard Hardy space method we are able to derive a variationalformulation in H−(Γ):

−i2π

∫

Γ

L(u′)(s)L(p′)(−s) + ((i− 1)L(u)(s)− ωL(v)(s))L(p)(−s)ds = u′0p(0),

−i2π

∫

Γ

L(v′)(s)L(q′)(−s) + (−ωL(u)(s)− (i+ 1)L(v)(s))L(q)(−s)ds = v′0q(0).

(4.16)

In the standard Hardy space method the integrals are mapped back using the Mobiustransformation onto the unit circle, where we choose

Ψ−1(z) :=1

2iκ0, Ψj(z) :=

z − 1

2iκ0zj , z ∈ S1, j = 0, . . . , N, (4.17)

as basis functions. The corresponding basis of H−(κ0R) (p0 := iκ0) is

Ψp0−1(s) :=

1

s− p0,Ψp0

j (s) :=2p0

(s− p0)2

(

s+ p0s− p0

)j

, j = 0, . . . , N. (4.18)

It is also useful to state

ψp0−1(s) := (L∂xL−1Ψp0

−1)(s) =p0

s− p0,

ψp0j (s) := (L∂xL−1Ψp0

j )(s) =2sp0

(s− p0)2

(

s+ p0s− p0

)j

, j = 0, . . . , N.

(4.19)

The modification, which seems useful, is the usage of two poles p0 and p1 in order toilluminate the edges of Γ. Using the basis functions (4.18) with p0 and p1 leads to

Ψp0−1(s) :=

1

s− p0,

Ψp0j (s) :=

2p0(s− p0)2

(

s+ p0s− p0

)j

, j = 0, . . . , N,

Ψp1j (s) :=

2p1(s− p1)2

(

s+ p1s− p1

)j

, j = 0, . . . , N.

(4.20)

For the integrals we can use Cauchy’s integral theorem to rewrite the integration over Γas an integration over a line. The line is chosen so that the distance to the poles of the


integrand is minimized. Hence we are left with four kinds of integrals:

M11jk :=

−i2π

∫

−ip0R

Ψp0j (s)Ψp0

k (−s)ds, S11jk :=

−i2π

∫

−ip0R

ψp0j (s)ψp0

j (−s)ds

M22jk :=

−i2π

∫

−ip1R

Ψp1j (s)Ψp1

k (−s)ds, S22jk :=

−i2π

∫

−ip1R

ψp1j (s)ψp1

j (−s)ds

M12jk :=

−i2π

∫

−i(p0+p1)R+p0−p1

2

Ψp0j (s)Ψp1

k (−s)ds, S12jk :=

−i2π

∫

−i(p0+p1)R+p0−p1

2

ψp0j (s)ψp1

j (−s)ds

M21jk :=

−i2π

∫

−i(p1+p0)R+p1−p0

2

Ψp1j (s)Ψp0

k (−s)ds, S21jk :=

−i2π

∫

−i(p1+p0)R+p1−p0

2

ψp1j (s)ψp0

j (−s)ds.

We define

M :=

(

M11 M12

M21 M22

)

,

S :=

(

S11 S12

S21 S22

)

.

(4.21)

With these two matrices we can build the discretization matrix of (4.16):

A :=

(

S + (−1 + i)M −ωM−ωM S + (−1− i)M

)

. (4.22)

We want to point out that if we apply this method to an elastic wave guide, the givenmatrices M and S will be important. In elasticity also mixed integrals

∫∞0u∂xvdx arise.

Therefore we state

D11jk :=

−i2π

∫

−ip0R

Ψp0j (s)ψp0

k (−s)ds,

D22jk :=

−i2π

∫

−ip1R

Ψp1j (s)ψp1

k (−s)ds,

D12jk :=

−i2π

∫

−i(p0+p1)R+p0−p1

2

Ψp0j (s)ψp1

k (−s)ds,

D21jk :=

−i2π

∫

−i(p1+p0)R+p1−p0

2

Ψp1j (s)ψp0

k (−s)ds,

D :=

(

D11 D12

D21 D22

)

.

(4.23)

The exterior discretization matrix of the elastic problem is build with the very samematrices M,S and D. Hence we can learn much by studying these. We observe that thematrix blocks Sij behave like the matrix blocks M ij . Therefore it is sufficient to focuson the matrix M . M11 and M22 are matrices of the standard Hardy space method andhence are tridiagonal. It can be seen in the numerical results, that the the second twokinds of integrals and thus the entries of M12,M21 become large. This results in a bad

4.3. GALERKIN FORMULATION 47

condition number of A. Therefore a slightly different method is implemented using theansatz and test functions:

Ψp0−1(s) :=

1

s− p0,

Ψp0j (s) :=

2p0(s− p0)2

(

s+ p1s− p0

)j

, j = 0, . . . , N,

Ψp1j (s) :=

2p1(s− p1)2

(

s+ p0s− p1

)j

, j = 0, . . . , N.

(4.24)

Now M12,M21 become tridiagonal and the first two kinds of integrals become large. Wecan also use a mixture of the two variants (4.18) and (4.24). Let w : N0 → N0 such thatw(n+ 1)− w(n) ∈ 0, 1 ∀n ∈ N0. Then

Ψp0−1(s) :=

1

s− p0,

Ψp0,p1j (s) :=

2p0(s− p0)2

(

s+ p0s− p0

)w(j)(s + p1s− p0

)j−w(j)

, j = 0, . . . , N,

Ψp1,p0j (s) :=

2p1(s− p1)2

(

s+ p1s− p1

)w(j)(s + p0s− p1

)j−w(j)

, j = 0, . . . , N.

(4.25)

If for example w(n+1)−w(n) := nmod2, we expect the same structure ofM11,M22,M12

and M21. To see this we calculate∫

−i(p0+p1)R+p0−p1

2

ψp0,p1j (s)ψp1,p0

k (−s)ds

=

∫

4s2p0p1(s− p0)2(s+ p1)2

(

s+ p0s− p0

)w(j)(s+ p1s− p0

)j−w(j)(s− p1s+ p1

)w(k)(s− p0s+ p1

)k−w(k)

ds

=

∫

4s2p0p1(s− p0)2+j+w(k)−k(s+ p1)2+k+w(j)−j

(s+ p0)w(j)(s− p1)

w(k)ds.

Because the order of the polynomial in the denominator minus the order of the polynomialin the nominator is greater or equal two, the integral is 2πi times the sum of residuals inthe upper part of the half space bounded by the integration path. Now if 2+j+w(k)−k ≤0, there exist no residuals in that domain and the integral vanishes. If 2+k+w(j)−j ≤ 0we calculate the integral with the residuals in the other part of C, which do not exist inthis case. The other integrals can be calculated in a similar way.We now focus on the matrices obtained by the ansatz and test functions (4.24). Since wecalculate integrals of meromorphic functions with different poles, we expect and observethat the different matrix blocks grow at a different rate in N . Taking the limit N → ∞


10 20 30 40 50 60

−60

−50

−40

−30

−20

−10

−15

−10

−5

0

5

10

15

20

(a) Basis (4.20)

10 20 30 40 50 60

−60

−50

−40

−30

−20

−10

−15

−10

−5

0

5

(b) Basis (4.24)

10 20 30 40 50 60

−60

−50

−40

−30

−20

−10

−15

−10

−5

0

5

10

(c) Basis (4.25) with w(n+1)−w(n) := nmod 2

Figure 2. log10 of the absolute value of matrix A with different basis.

this results in a ”matrix” with the structure

0 0 0 00 1 0 10 0 0 00 1 0 1

.

A matrix with such a structure is singular. We observe in numerical computations

M11jk ≈ C0p

j+k0 + C1p

j+k1 + w(p0, p1)

pj+k+10

,

M22jk ≈ C0p

j+k0 + C1p

j+k1 + w(p0, p1)

pj+k+11

,

(4.26)

with a polynomial w of degree j + k − 1. Thus it seems appropriate to rescale:

4.4. NUMERICAL TESTS 49

10 20 30 40 50 60 70 80 90 100

−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

−15

−10

−5

0

5

10

(a) Unscaled Basis (4.24)

10 20 30 40 50 60 70 80 90 100

−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

−20

−18

−16

−14

−12

−10

−8

−6

−4

−2

0

(b) Scaled Basis (4.27)

Figure 3. log10 of the absolute value of the matrix A.

Ψp0−1(s) :=

1

s− p0,

Ψp0,p1j (s) :=

2p0(s− p0)2

(

s+ p1s− p0

· |p0|max(|p0|, |p1|)

)j

, j = 0, . . . , N,

Ψp1,p0j (s) :=

2p1(s− p1)2

(

s+ p0s− p1

· |p1|max(|p0|, |p1|)

)j

, j = 0, . . . , N.

(4.27)

Numerical tests show that this scaling performs well, f.e. it reduces the condition numberof A with N = 15, i.e. 60 degrees of freedom, from 1013 to 108. The still high conditionnumber seems to be generated by the numerical calculation of the integrals: In the matlabcode the on the unit circle uniformly distributed integration nodes are mapped to thegiven line using the Mobius transformation. M12 andM21 should be tridiagonal matrices.If we set the non-tridiagonal entries to zero, the condition number of A (again N = 15)reduces to 105. However, the error becomes high. The cause of this might be that onlyparts ofM are corrected, but notM11 andM22. Hence we expect a stable and convergentmethod if we improve the numerical integration of the unbounded integrals. This couldbe done by calculating the integrals analytically: We know the integrals are equal to theresidues of the integrands at p0 respectively p1. If f is a meromorphic function with apole of nth order at p, we can calculate the residue as

res(f, p) =

(

1

(n− 1)!∂n−1x (x− p)nf(x)

)∣

∣

∣

∣

x=p

. (4.28)

4.4. Numerical Tests

For ω ∈ (0, 1) we have two imaginary wave numbers. For ω ∈ (1,√2) we have two real

wave numbers and for ω ∈ (√2,∞) we have one real and one imaginary wave number.


At ω = 1,√2 the wave numbers are degenerated. In elasticity we have infinitely many

wave numbers and hence all the above non-generated cases will occur at once. Thus wewant the method to work for all mentioned cases.

−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4

−1.5

−1

−0.5

0

0.5

1

1.5

2

real part

imag

inar

y pa

rt

Figure 4. Green/blue: poles of the outgoing solution for ω ∈ [0, 2].

We want to experimentally verify the convergence of the method. Thus the questionarises in which norm the error has to be measured. The reason for studying this exteriorspace problem is to couple it with the interior space problem, see Section 1.3. Thus theappropriate error has to be measured in the Dirichlet data calculated by this method.Since our domain is R+, the error norm is defined as the absolute value at 0. We take

uexact(x) := C1eik1(ω)x + C2e

ik2(ω)x,

vexact(x) :=1

ω((−∂2x − 1 + i)uexact)(x),

(4.29)

and calculate numerically unum, vnum with Neumann boundary data u′exact(0), v′exact(0).

Then we gain the error as

erroru := |uexact(0)− unum(0)|,errorv := |vexact(0)− vnum(0)|.

(4.30)

We use the method with Basis 4.27. We notice a strong influence of the chosen polesp0, p1 on the approximation results. If we take a look at Figure 5, we observe convergenceeverywhere except in the region of the degenerated points, i.e. ω = 1,

√2. If we change

now p1 from −0.75i−0.2 to −0.5i−0.2, we notice in Figure 6 that for ω < 1 the methoddoes not converge any more. This is due to bad approximation properties of this basisfor ω in that region. As we can see in Figure 7(d) the slow convergence is dominated bythe rising condition number. In Figure 7 we see that the condition of A is bounded withthe very high constant 1022. If we couple the exterior problem with the interior problem,we expect the interior error to be in a range of 10−5. Thus we state that ω outside thedegenerated area, an error of 10−5 with a condition 106 is reached with N = 10, i.e. 40degrees of freedom of A.

4.4. NUMERICAL TESTS 51

0 1 210

−15

10−10

10−5

100

ω

rel. erroru

rel. errorv

(a) N = 20

0 1 210

−20

10−15

10−10

10−5

100

ω

rel. erroru

rel. errorv

(b) N = 30

0 1 210

−20

10−15

10−10

10−5

100

ω

rel. erroru

rel. errorv

(c) N = 40

Figure 5. Error for ω ∈ [0, 2]. To analyze convergence, plots are given fordifferent N . Parameters are C1 = 1, C2 = 1, p0 = 1.3i − 0.4, p1 =−0.75i− 0.2.

0 1 210

−15

10−10

10−5

100

105

ω

rel. erroru

rel. errorv

(a) N = 20

0 1 210

−20

10−15

10−10

10−5

100

105

ω

rel. erroru

rel. errorv

(b) N = 30

0 1 210

−15

10−10

10−5

100

105

ω

rel. erroru

rel. errorv

(c) N = 40

Figure 6. Error for ω ∈ [0, 2]. To analyze convergence, plots are given fordifferent N . Parameters are C1 = 1, C2 = 1, p0 = 1.3i − 0.4, p1 =−0.5i− 0.2.


0 10 20 30 40 50 10

−20

10−10

100

1010

1020

1030

ω

rel. erroru

rel. errorv

Condition

(a) p1 = −0.75i− 0.2, ω = 0.5

0 10 20 30 40 50 10

−20

10−10

100

1010

1020

1030

ω

rel. erroru

rel. errorv

Condition

(b) p1 = −0.75i− 0.2, ω = 1.2

0 10 20 30 40 50 10

−20

10−10

100

1010

1020

1030

ω

rel. erroru

rel. errorv

Condition

(c) p1 = −0.75i− 0.2, ω = 1.75

0 10 20 30 40 50 10

−10

100

1010

1020

1030

ω

rel. erroru

rel. errorv

Condition

(d) p1 = −0.55i− 0.2, ω = 0.5

Figure 7. Error of u, v and condition number of matrix A. Parametersare C1 = 1, C2 = 1, p0 = 1.3i− 0.4.

CHAPTER 5

Summary and Outlook

In Chapter 1 we presented the problem of guided time harmonic elastic wave propagation.In the first part of Chapter 2 we derived an analytical description of the solution using theLamb modes. In Section 2.1 we presented new asymptotic results of the wave numbers,i.e. for large n there hold

kn,±1,±2 ≈ ±1 1

2harccosh

(

2√2nπ +

√2π ±2

√2π

4

)

+i

h

(

2nπ + π ±2 π

4

)

,

and

|k′n(ω)| ≤ C.

In Section 2.2 we further outlined properties of the Lamb modes. Wave numbers kn aredefined to be outgoing if Im kn > 0 or k′n > 0 if kn ∈ R. As far as the author knows,there exists no literature showing a mathematical justification for this criterion in thisspecific case of an elastic wave guide. With the results of the previous sections wewere able to give a rigorous proof of this physical decision criterion using the limitingabsorption principle.In Chapter 3 we gave a brief introduction to guided acoustic wave propagation andcarried out the similarities/differences between acoustic and elasticity. We presented twopopular numerical methods for acoustic wave propagation, i.e. perfectly matched layersand the Hardy space method, and analyzed how suitable these concepts are for guidedelastic wave propagation.In Chapter 4 we formulated a model problem. We analyzed degenerated wave numbersof the model problem, in order to understand those in elasticity. In Section 4.3we formulated a numerical scheme for the model problem and carried out differentmodifications in order to get a stable method. In Section 4.4 we presented numericalresults.

To continue the work of this thesis one would have to carry out the given ideas forstabilization of the numerical method in full detail. The next step would be to makenumerical experiments with the method for the full elastic problem. We also give an ideahow to handle degenerated wave numbers, i.e. k′n(ω) = ∞, which we call pole swapping.The case kn=0, i.e. the case where the two different real wavenumbers ±kn first both

53

54 5. SUMMARY AND OUTLOOK

become zero and then complex, could be treated numerically in the following way: Leta ∈ C with Re a < 0. We define

L(U)(s) := s

s+ aL(u)(s). (5.1)

We observe U(0) = u(0) and U ′(0) = u′(0). We know that we can write the analyticalsolution as

L(u)(s) =∑ an

s− pn. (5.2)

Thus we calculate

L(U ′) = −U(0) + sL(U)(s) = −u(0) + s2

s+ aL(u)(s)

=∑

an

(

−1 +s2

s+ a· 1

s− pn

)

=∑ an

s− pn

(

pn +s2

s+ a− s

)

= L(u′)(s)− sa

s+ aL(u)(s)

= L(u′)(s)− aL(U)(s).Hence

L(u′) = L(U ′) + aL(U). (5.3)

We define L(V ) in an analogous way

L(V )(s) := s

s+ aL(v)(s). (5.4)

Now we can formulate the problem for the variables U and V .

−i2π

∫

Γ

(L(U ′)(s) +aL(U)(s))L(p′)(−s) + (i− 1)s+ a

sL(U)(s)L(p)(−s)

− ωs+ a

sL(v)(s)L(p)(−s)ds = u′0p(0),

−i2π

∫

Γ

(L(V ′)(s) +aL(V ))L(q′)(−s)− (1 + i)s+ a

sL(V )(s)L(q)(−s)

− ωs+ a

sL(U)(s)L(q)(−s)ds = v′0q(0).

(5.5)

L(U) is holomorph at 0 and the integrands have a pole at 0. Thus Γ has to be chosensuch that 0 lies on the right hand side of Γ. In the case that ±ξ are both physical wavenumbers, we set

L(U)(s) := s(s+ ξ2)

(s+ a)(s+ b)(s + c)L(u)(s). (5.6)

Appendix A

The matlab code used in Chapter 4 is listed in this appendix.

Matlab Code of solve modelproblem.m

This script solves the model problem.

1 clear a l l ;

2 close a l l ;3

4 pol1 =1.32 i −0.4 ;

5 pol2=−0.75 i −0.2 ;6 dimH=10;

7 ba s i s =4;8

9 [ S ,M]= ext Mat r i c e s (dimH , [ pol1 , po l2 ] , b a s i s ) ;10 e r r o r u = [ ] ;

11 e r r o r v = [ ] ;12 ab s e r r o r u = [ ] ;

13 ab s e r r o r v = [ ] ;14 CondA= [ ] ;

15 k1vec = [ ] ;16 k2vec = [ ] ;

17 omega vec =0 . 0 1 : 0 . 0 5 : 2 ;18

19 for omega=omega vec

20 C1=1;21 C2=1;

22

23 i f omega<1

24 i f real ( sqrt (1+sqrt (1−omegaˆ2)∗ i )∗ i )>025 k1=−sqrt (1+sqrt (1−omegaˆ2)∗ i )∗ i ;26 else

27 k1=sqrt (1+sqrt (1−omegaˆ2)∗ i ) ∗ i ;55

56 APPENDIX A

28 end

29 i f real (−sqrt (1−sqrt (1−omegaˆ2)∗ i ) ∗ i )>030 k2=sqrt (1−sqrt (1−omegaˆ2)∗ i ) ∗ i ;31 else

32 k2=−sqrt (1−sqrt (1−omegaˆ2)∗ i )∗ i ;33 end

34 e l s e i f omega<sqrt ( 2 )

35 k1=sqrt (1+sqrt ( omegaˆ2−1) )∗ i ;36 k2=−sqrt (1−sqrt ( omegaˆ2−1) ) ∗ i ;37 else

38 k1=sqrt (1+sqrt ( omegaˆ2−1) )∗ i ;39 i f real (−sqrt (1−sqrt ( omegaˆ2−1) ) ∗ i )>040 k2=sqrt (1−sqrt ( omegaˆ2−1) ) ∗ i ;41 else

42 k2=−sqrt (1−sqrt ( omegaˆ2−1) ) ∗ i ;43 end

44 end

45

46 k1vec=[k1vec , k1 ] ;47 k2vec=[k2vec , k2 ] ;

48 exact u = @(x ) C1∗exp ( k1∗x ) + C2∗exp ( k2∗x ) ;49 exact u d1 = @(x ) C1∗k1∗exp( k1∗x ) + C2∗k2∗exp( k2∗x ) ;50 exact u d2 = @(x ) C1∗k1ˆ2∗exp( k1∗x ) + C2∗k2ˆ2∗exp( k2∗x ) ;51 exact u d3 = @(x ) C1∗k1ˆ3∗exp( k1∗x ) + C2∗k2ˆ3∗exp( k2∗x ) ;52

53 exact v = @(x ) 1/omega∗(− exact u d2 (x ) + (−1+ i )∗ exact u (x ) ) ;

54 exact v d1 = @(x ) 1/omega∗(− exact u d3 (x ) + (−1+ i ) ∗exact u d1 (x ) ) ;

55

56 A=[S+(−1+i )∗M , −omega∗M ; −omega∗M , S+(−1− i ) ∗M] ;

57 rhs=[−exact u d1 (0 ) ; zeros (dimH , 1 ) ;− exact u d1 (0 ) ; zeros (dimH , 1 ) ; . . .

58 −exact v d1 (0 ) ; zeros (dimH, 1 ) ;− exact v d1 (0 ) ; zeros (dimH

, 1 ) ] ;59

60 s o l=A \ rhs ;61 e r r o r u =[ e r ro r u , abs ( s o l ( 1 )+s o l (dimH+2)−exact u (0 ) ) /abs (

exact u (0 ) ) ] ;

MATLAB CODE OF SOLVE MODELPROBLEM.M 57

62 e r r o r v =[ e r ro r v , abs ( s o l (2∗dimH+3)+so l (3∗dimH+4)−exact v (0 ) )/abs ( exact v (0 ) ) ] ;

63 ab s e r r o r u =[ abs e r ro r u , abs ( s o l ( 1 )+s o l (dimH+2)−exact u (0 ) )

] ;64 ab s e r r o r v =[ abs e r ro r v , abs ( s o l (2∗dimH+3)+so l (3∗dimH+4)−

exact v (0 ) ) ] ;65 CondA=[CondA, condest (A) ] ;

66 end

67

68 f igure ( 1 ) ;69

70 subplot ( 2 , 2 , 1 ) ;71 semilogy ( omega vec , e r r o r u , omega vec , e r r o r v ) ;

72 t i t l e ( ’ r e l . e r r o r s in u 0 and v 0 ’ ) ;73 xlabel ( ’ omega ’ ) ;

74 legend ( ’ r e l . e r r o r u 0 ’ , ’ r e l . e r r o r v 0 ’ ) ;75

76 subplot ( 2 , 2 , 2 ) ;

77 semilogy ( omega vec , CondA) ;78 t i t l e ( ’ Matrix Condit ion ’ ) ;

79 xlabel ( ’ omega ’ ) ;80 ylabel ( ’CondA ’ ) ;

81

82 subplot ( 2 , 2 , 3 ) ;

83 t i t l e ( ’modes ’ ) ;84 plot ( real ( k1vec ) , imag( k1vec ) , ’ b ’ ) ;

85 hold on ;86 plot ( real ( k2vec ) , imag( k2vec ) , ’ g ’ ) ;

87 plot ( real ( [ po l1 pol2 ] ) , imag ( [ po l1 pol2 ] ) , ’ r . ’ , ’ MarkerSize ’ , 20) ;88 l ine ( [ −0.1 , 0 . 1 ] , [ 1 , 1 ] , ’ Color ’ , ’ k ’ , ’ LineWidth ’ ,2 ) ;

89 l ine ( [ −0.1 , 0 . 1 ] , [ 1 , 1 ] , ’ Color ’ , ’ k ’ , ’ LineWidth ’ ,2 ) ;90 l ine ( [ −0.1 , 0 . 1 ] , [ 0 , 0 ] , ’ Color ’ , ’ k ’ , ’ LineWidth ’ ,1 , ’ L ineSty l e

’ , ’−− ’ ) ;

91 plot ( real ( k1vec (1 ) ) , imag( k1vec (1 ) ) , ’ b . ’ , ’ MarkerSize ’ , 15) ;92 plot ( real ( k2vec (1 ) ) , imag( k2vec (1 ) ) , ’ g . ’ , ’ MarkerSize ’ , 15) ;

93 axis ( [ real ( k2vec (end) )−0.5 0 . 5 imag( k2vec (1 ) )−0.5 imag( k1vec (end) ) +0 .5 ] )

94 xlabel ( ’ r e a l part ’ ) ;95 ylabel ( ’ imaginary part ’ ) ;

58 APPENDIX A

96 legend ( ’ k1 ’ , ’ k2 ’ , ’ po l e s ’ , ’ s q r t ( x i ) ’ , ’ Locat ion ’ , ’ NorthWest ’ ) ;97

98 subplot ( 2 , 2 , 4 ) ;

99 X=1: s ize (A, 1 ) ;100 pcolor (X,−X, log10 (abs (A) ) ) ; colorbar ;

101 t i t l e ( [ ’ cond (A)=’ num2str (max(CondA) , ’%0.1e ’ ) ] ) ;102

103 f igure ( 2 ) ;104 pcolor (X,−X, log10 (abs (myA) ) ) ; colorbar ;

105 t i t l e ( ’ l og10 o f abso lu t e va lue o f matrix A ’ ) ;106

107 f igure ( 1 ) ;108

109 fpr int f ( ’=======================\n ’ ) ;110 fpr int f ( ’ po l1=%e\n ’ , po l1 ) ;

111 fpr int f ( ’ po l2=%e\n ’ , po l2 ) ;112 fpr int f ( ’ b a s i s=%e\n ’ , b a s i s ) ;

113 fpr int f ( ’dimH=%e\n ’ , dimH) ;

114 fpr int f ( ’ e r r o r u (0 ) : %e\n ’ ,max( e r r o r u ) ) ;115 fpr int f ( ’ e r r o r v (0 ) : %e\n ’ ,max( e r r o r v ) ) ;

116 fpr int f ( ’ cond (A) : %e\n ’ , max(CondA) ) ;

Matlab Code of ext Matrices.m

This function returns the matrices M and S.

1 function [ S , M]= ext Mat r i c e s (dimH , po les , b a s i s )2

3 pol1=po l e s (1 ) ;4 pol2=po l e s (2 ) ;

5 n ipo i n t s=10∗dimH+10;6 theta=linspace (0 ,2∗pi , n i p o i n t s +1) ;

7 theta=theta ( 1 : end−1)+theta (2 ) /2 ;

8 z=exp( theta ∗1 i ) ;9 weights=2∗pi /( n i po i n t s ) ∗ones (1 , n i po i n t s ) ;

10

11 %========== 11 ======================

12 k0=pol1 ;13 m = @( z ) k0 ∗( z+1) . / ( z−1) ;

14 s=m( z )−0.00 i ;15 we ight s s=(−2∗k0 ) . / (2∗ pi ) ∗weights . / abs ( z−1) . ˆ 2 ;

MATLAB CODE OF EXT MATRICES.M 59

16 [ va l s1 , d e r i v s 1 ]= b a s i s e v a l ( s , dimH , [ pol1 , po l2 ] , b a s i s ) ;17 [ va l s2 , d e r i v s 2 ]= b a s i s e v a l (−s , dimH , [ pol1 , po l2 ] , b a s i s ) ;

18 M 11=( va l s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ va l s 2 . ’ ;

19 S 11=( d e r i v s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ d e r i v s 2 . ’ ;20

21 %========== 22 ======================22 k0=pol2 ;

23 m = @( z ) k0 ∗( z+1) . / ( z−1) ;24 s=m( z )+0.00 i ;

25 we ight s s=(−2∗k0 ) . / (2∗ pi ) ∗weights . / abs ( z−1) . ˆ 2 ;26 [ va l s1 , d e r i v s 1 ]= b a s i s e v a l ( s , dimH , [ pol2 , po l1 ] , b a s i s ) ;

27 [ va l s2 , d e r i v s 2 ]= b a s i s e v a l (−s , dimH , [ pol2 , po l1 ] , b a s i s ) ;28 M 22=( va l s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ va l s 2 . ’ ;


31 %========== 12 ======================32 k0=pol1+pol2 ;

33 m = @( z ) k0 ∗( z+1) . / ( z−1) ;

34 s=m( z )+(pol1−pol2 ) /2−0.00 i ;35 we ight s s=(−2∗k0 ) . / (2∗ pi ) ∗weights . / abs ( z−1) . ˆ 2 ;


38 M 12=( va l s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ va l s 2 . ’ ;39 S 12=( d e r i v s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ d e r i v s 2 . ’ ;

40

41 %========== 21 ======================

42 k0=pol2+pol1 ;43 m = @( z ) k0 ∗( z+1) . / ( z−1) ;

44 s=m( z )+(pol2−pol1 ) /2+0.00 i ;45 we ight s s=(−2∗k0 ) . / (2∗ pi ) ∗weights . / abs ( z−1) . ˆ 2 ;


48 M 21=( va l s 1 .∗ repmat ( we ights s , dimH+1 ,1) )∗ va l s 2 . ’ ;


51 %========== tog e t h e r ======================52 S=[S 11 , S 12 ; S 21 , S 22 ] ;

53 M=[M 11 , M 12 ; M 21 , M 22 ] ;54

60 APPENDIX A

55 %========== eva l ua t e b a s i s ===============56 function [ val , d e r i v ]= b a s i s e v a l ( po ints , n , po les , b a s i s )

57

58 pol1=po l e s (1 ) ;59 pol2=po l e s (2 ) ;

60

61 po in t s=reshape ( po ints , 1 , length ( po in t s ) ) ;

62 va l =1./( po ints−pol1 ) ;63 der i v=pol1 . / ( po ints−pol1 ) ;

64

65 i f n>0

66 va l=[ va l ; va l (end , : ) .∗2∗ pol1 . / ( po ints−pol1 ) ] ;67 der i v =[ de r i v ; de r i v (end , : ) . ∗ 2 . ∗ po in t s . / ( po ints−pol1 ) ] ;

68

69 switch ba s i s

70 case 171 for k=2:n

72 va l=[ va l ; va l (end , : ) . ∗ ( po in t s+pol1 ) . / ( po ints−pol1

) ] ;73 der i v =[ de r i v ; de r i v (end , : ) . ∗ ( po in t s+pol1 ) . / (

po ints−pol1 ) ] ;74 end

75 case 276 for k=2:n

77 va l=[ va l ; va l (end , : ) . ∗ ( po in t s+pol2 ) . / ( po ints−pol1) ] ;

78 der i v =[ de r i v ; de r i v (end , : ) . ∗ ( po in t s+pol2 ) . / (po ints−pol1 ) ] ;

79 end

80 case 3

81 m = 2 ;82 o th e rba s i s = [ 1 ] ;

83 for k=2:n

84 i f any(mod(k ,m)==othe rba s i s )85 va l=[ va l ; va l (end , : ) . ∗ ( po in t s+pol2 ) . / ( po ints−

pol1 ) ] ;86 der i v =[ de r i v ; de r i v (end , : ) . ∗ ( po in t s+pol2 ) . / (

po ints−pol1 ) ] ;87 else

MATLAB CODE OF EXT MATRICES.M 61

88 va l=[ va l ; va l (end , : ) . ∗ ( po in t s+pol1 ) . / ( po ints−pol1 ) ] ;

89 der i v =[ de r i v ; de r i v (end , : ) . ∗ ( po in t s+pol1 ) . / (

po ints−pol1 ) ] ;90 end

91 end

92 case 4

93 damp = abs ( po l1 /max(abs ( po l1 ) ,abs ( po l2 ) ) ) ;94 for k=2:n

95 va l=[ va l ; va l (end , : ) . ∗ ( po in t s+pol2 ) . / ( po ints−pol1)∗damp ] ;

96 der i v =[ de r i v ; de r i v (end , : ) . ∗ ( po in t s+pol2 ) . / (po ints−pol1 )∗damp ] ;

97 end

98 end

99 end

Bibliography

[1] J. Achenbach. Wave propagation in elastic solids. North-Holland Publishing Company, 1975.[2] J.-P. Berenger. Three-dimensional perfectly matched layer for the absorption of electromagnetic

waves. J. Comput. Phys., 127(2):363–379, 1996.[3] Y. K. Billah and R. H. Scanlan. Resonance, tacoma narrows bridge failure, and undergraduate

physics textbooks. American Journal of Physics, 59(2):118–124, 1991.[4] D. Braess. Finite Elemente. Springer-Verlag Berlin Heidelberg, 2007.[5] F. Collino and P. Monk. The perfectly matched layer in curvilinear coordinates. SIAM J. Sci.

Comput., 19(6):2061–2090 (electronic), 1998.[6] T. A. Driscoll and L. N. Trefethen. Schwarz-Christoffel Mapping. Cambridge Monographs on Applied

and Computational Mathematics, 2002.[7] G. H. Hardy. The mean value of the modulus of an analytic function. Proc. London Math. Soc. 2,

14:269–277, 1915.[8] J. Harris. Linear Elastic Waves. Cambridge University Press, 2001.[9] P. D. Hislop and I. M. Sigal. Introduction to spectral theory, volume 113 of Applied Mathematical

Sciences. Springer-Verlag, New York, 1996. With applications to Schrodinger operators.[10] T. Hohage and L. Nannen. Hardy space infinite elements for scattering and resonance problems.

SIAM J. Numer. Anal., 47(2):972–996, 2009.[11] T. Hohage, F. Schmidt, and L. Zschiedrich. Solving time-harmonic scattering problems based on

the pole condition. I. Theory. SIAM J. Math. Anal., 35(1):183–210 (electronic), 2003.[12] A. Larsen. Aerodynamics of the tacoma narrows bridge - 60 years later. Structural Engineering

International, 10(4):243–248, 2000.[13] L. Nannen. Hardy-Raum Methoden zur numerischen Losung von Streu- und Resonanzproblemen

auf unbeschrankten Gebieten. Dissertation, Georg-August-Universitat zu Gottingen, 2008. Online:http://num.math.uni-goettingen.de/picap/pdf/E630.pdf.

[14] L. Nannen. Streu- und Resonanzprobleme, June 2012. Online:http://www.asc.tuwien.ac.at/~lnannen/Lectures/ScatteringProblems/ScatteringProblems.pdf.

[15] L. Nannen and A. Schadle. Hardy space infinite elements for Helmholtz-type problems with un-bounded inhomogeneities. Wave Motion, 48(2):116–129, 2011.

[16] R. Remmert. Funktionentheorie. Springer-Verlag Berlin Heidelberg, 1992.[17] J. Schoberl. Numerical methods for maxwell equations, April 2009. Online:

http://www.asc.tuwien.ac.at/~schoeberl/wiki/lva/notes/maxwell.pdf.[18] E. A. Skelton, S. D. M. Adams, and R. V. Craster. Guided elastic waves and perfectly matched

layers. Wave Motion, 44(7-8):573–592, 2007.

63

http://num.math.uni-goettingen.de/picap/pdf/E630.pdf

http://www.asc.tuwien.ac.at/~lnannen/Lectures/ScatteringProblems/ScatteringProblems.pdf

http://www.asc.tuwien.ac.at/~schoeberl/wiki/lva/notes/maxwell.pdf

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