# A Justification of Two-Dimensional Nonlinear Viscoelastic Shells Model

Authors: Fushan Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By applying formal asymptotic analysis and Laplace transformation, we obtain two-dimensional
nonlinear viscoelastic shells model satisfied by the leading term of asymptotic expansion of the solution
to the three-dimensional equations.


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## Body

## 1. Introduction

In the case of pure nonlinear elasticity, Ciarlet and his collaborators have studied membrane shells, flexural shell and Koiter shell (see [1] and the references therein). The linear viscoelasticity was studied in [2–5], and Li [6–8] studied the global existence and uniqueness of weak solution, uniform rates of decay, and limit behavior of the solution to nonlinear viscoelastic Marguerre-von Kármán shallow shells. Xiao studied the time-dependent nonlinear elastic shells by the method of asymptotic analysis (see [9]).Motivated by the above work, we deal with nonlinear viscoelastic shells and give the identification of two-dimensional variation problem satisfied by the leading term of of asymptotic expansion of the solution to the three-dimensional equations. The main contributions of this paper are the following: (a) the problem considered in this paper is nonlinear viscoelastic shells, to our knowledge this model has not been considered; (b) applying Laplace transformation, we overcome the difficulties caused by the integral term in the model; (c) the calculation and derivation are precise.This paper is organized as follows. Section2 begins with some preliminaries and then gives the main result. In Section 3, we give the proof of the main theorem.


## 2. Preliminaries and Main Results

We use the following conventions and notations throughout this work:Greek indices and exponents (except ε) belong to the set {1,2}, Latin indices and exponents (except when otherwise indicated, as, e.g., when they are used to index sequences) belong to the set {1,2,3}, and the summation convention with respect to the repeated indices and exponents is systematically used. The sign := indicates that the right-hand side defines the left-hand side.Letω⊂R2 be a bounded connected open set with a Lipschitz boundary γ, let y=(yα) denote a generic point in the set ω¯, and let ∂α:=∂/∂yα. Let θ:ω¯→R3 be an injective mapping of C3 such that the two vectors aα(y):=∂αθ(y) are linear independent at all points y∈ω¯. They form the covariant basis of the tangent plane to the surface S=θ(ω¯) at the point θ(y); the two vectors aα(y) of the same tangent plane defined by the relations aα(y)·aβ(y):=δβα constitute its contravariant basis. We also define the unit vector a3(y)=a3(y)=a1(y)×a2(y)/|a1(y)×a2(y)| which is normal to the S at the point θ(y).One then defines the first fundamental form, also known as metric tensor(aαβ) or (aαβ), the second fundamental form, also known as the curvature tensor (bαβ) or (bαβ), and the Christoffel symbols Γαβσ of the surface S by setting (whenever no confusion should arise, we henceforth drop the explicit dependence on the variable y∈ω¯)
(2.1)aαβ:=aα⋅aβ,aαβ:=aα⋅aβ,bαβ:=a3⋅∂βaα,bαβ:=aβσbσα,Γαβσ:=aσ⋅∂βaα.
Note the symmetries aαβ=aβα,bαβ=bβα, and Γαβσ=Γβασ. The area element along S is ady, where a:=det(aαβ). All the functions defined above are at least continuous over the set ω¯. In particular, there exists a constant a0>0 such that a(y)≥a0,forally∈ω¯.In addition, let the covariant derivativesbβσ|α and the covariant components cαβ of the third form of the surface S be defined by
(2.2)bβσ|α:=∂αbβσ+Γατσbβτ-Γαβτbτσ,cαβ:=bασbσβ.For eachε>0, we consider a shell with thickness 2ε and middle surface S, whose lamé relaxation modules λ(t)≥0 and μ(t)>0(t≥0) are independent of ε. We define the sets
(2.3)Ωε:=ω×(-ε,+ε),Γ+ε=ω×{ε},Γ-ε=ω×{-ε},Γ0ε=γ0×[-ε,ε],
where γ0⊂γ and γ0≠∅. Note that Γ+ε∪Γ-ε∪Γ0ε constitutes a partition of the boundary of the set Ωε. Let xε=(xiε) denote a generic point in the set Ω¯ε, and let ∂iε:=∂/∂xiε; hence xαε=yα and ∂αε=∂α.We then define a mappingΘ:Ω¯ε→R3 by
(2.4)Θ(xε):=θ(y)+x3εa3(y),∀xε=(y,x3ε)∈Ω¯ε,
then there exists ε0>0 such that for all 0<ε≤ε0 the mapping Θ:Ω¯ε→R3 is an injective mapping and the three vectors giε(xε):=∂iεΘ(xε)(∂αε=∂/∂xα,∂3ε=∂/∂x3ε)are linear independent for each xε∈Ω¯ε. The injectivity of the mapping Θ:Ω¯ε→R3 ensures that the physical problem described below is meaningful.The three vectorsgiε(xε) form the covariant basis at the point Θ(xε), and the three vectors gi,ε(xε) defined by gj,ε(xε)·giε(xε)=δij form the contravariant. We define the metric tensor (gijε) or (gij,ε) and the Christoffel symbols of the manifold Θ(Ω¯ε) by setting (we omit the explicit dependence on xε)
(2.5)gijε=giε⋅gjε,gij,ε:=gi,ε⋅gj,ε,Γijk,ε:=gk,ε⋅∂iεgjε.
Note the symmetries
(2.6)gijε=gjiε,gij,ε=gji,ε,Γijk,ε=Γjik,ε.
The volume element in the set Θ(Ω¯ε) is gεdxε, where gε:=det(gijε).For each0<ε≤ε0, the set Ω̂ε¯:=Θ(Ω¯ε) is the reference configuration of a viscoelastic shell with middle surface S=θ(ω¯) and thickness 2ε. We assume that the material constituting the shell is homogeneous isotropic and Θ(Ω¯ε) is of a nature state, so that the material is characterized by its two lamé relaxation modules λ(t)≥0 and μ(t)>0(t≥0). Under the action of forces, the shell undergoes a displacement field.Letûε(t)=uiε(t)gi,ε in terms of curvilinear coordinates xε of the reference configuration Θ(Ω¯ε). Then, the covariant displacement field uε(t)=(uiε)(t) satisfies the following three-dimensional equations (c.f. [1, 10]):
(2.7)uε(t)∈Lc∞(-∞,T;W(Ωε))withW(Ωε):={vε∈W1,4(Ωε),vε=0on⁡Γ0ε},∫Ωεuittε(t)viεgij,εgεdxε+∫ΩεAijkl,ε(0)Ek∥lε(uε(t))Fi∥jε(uε(t),vε)gεdxε+∫Ωε∫-∞tA′ijkl,ε(t-τ)Ek∥lε(uε(τ))Fi∥jε(uε(τ),vε)gεdτdxε=∫Ωεfi,ε(t)viεgεdxε+∫Γ+ε∪Γ-εhi,ε(t)viεgεdΓε,∀vε∈W(Ωε),
where the symbol Lc∞ denotes the subspace of L∞>0 such that there exists a constant T such that the functions vanish as s<-T. And,
(2.8)Aijkl,ε(t):=λ(t)gij,εgkl,ε+μ(t)(gik,εgjl,ε+gil,εgjk,ε)
designate the contravariant components of the three-dimensional elasticity tensor,
(2.9)Ei∥jε(vε):=12(vi∥jε+vj∥iε+gmn,εvm∥iεvn∥jε),wherevi∥jε:=∂jεviε-Γijp,εvpε,
designate the strains in the curvilinear coordinates associated with an arbitrary displacement field viεgi,ε of the manifold Θ(Ω¯),
(2.10)Fi∥jε(uε(t),vε):=12(vi∥jε+vj∥iε+gmn,ε{um∥iε(t)vn∥jε+un∥jε(t)vm∥iε}),
and, finally, fi,ε∈L∞(0,T;L2(Ωε)) and hi,ε∈L∞(0,T;(Γ+ε∪Γ-ε)) denote the contravariant components of the applied body and surface force densities, respectively, applied to the interior Θ(Ωε) of the shell and to its “uper” and “lower” faces Θ(Γ+ε) and Θ(Γ-ε), and designate the area element along ∂Ωε. We thus assume that there are no surface forces applied to the portion Θ((γ-γ0)×[-ε,ε]) of the lateral face of the shell.We record in passing the symmetries(2.11)Aijkl,ε=Ajikl,ε=Aklij,ε
and the relation
(2.12)Γα33,ε=Γ33p,ε=0,Aαβσ3,ε=Aα333,ε=0inΩ¯ε.Our final objective consists in showing, by means of the method of formal asymptotic expansions that, if the data are of an appropriate order with respect toε as ε→0, the above three-dimensional problems are “asymptotically equivalent” to a “two-dimensional problem posed over the middle surface of the shell.” This means that the new unknown should be ζε(t)=(ζiε(t)), where ζiε(t) are the covariant components of the displacement ζiε(t)ai(y):ω¯→R3 of the middle surface S=θ(ω¯). In other words, ζiε(t,y)ai(y) is the displacement of the point θ(y)∈S.“Asymptotic analysis” means that our objective is to study the behavior of the displacement fielduiε(t)gi,ε:Ω¯ε→R3 as ε→0, an endeavour that will be a behavior as ε→0 of the covariant components uiε(t):Ω¯ε→R of the displacement field, that is, the behavior of the unknown uε(t)=(uiε(t)):Ω¯ε→R3 of the three-dimensional shell problem.Since these fields are defined on setsΩ¯ε that themselves vary with ε, our first task naturally consists in transforming the three-dimensional problems into problems posed over a set that does not depend on ε.Furthermore, we transform problem (2.7) into an equivalent problem independent of ε, posed over the domain.LetΩ:=ω×(-1,+1),Γ0=γ0×[-1,1],Γ+:=ω×{+1}, and Γ-:=ω×{-1}, and let x=(xi) denote a generic point in Ω¯. With each point x∈Ω¯, we associate the point xε through the bijection πε:x=(x1,x2,x3)∈Ω¯→xε=(xiε)=(x1,x2,εx3)∈Ω¯ε; we thus have ∂αε=∂α and ∂3ε=(1/ε)∂3. Let uε=(uiε),Γijε,gε,Aijkl,ε:Ω¯ε→R and the vector fields vε=(viε) appearing in the three-dimensional problem (2.7) be associated with the functions Γijp(ε),g(ε),Aijkl(ε):Ω¯→R and the scaled vector fields v=(vi) defined by
(2.13)ui(ε)(t,x)=uiε(t,xε),vi(x)=viε(xε)∀xε∈Ω¯ε,Γijp(ε)(x)=Γijp,ε(xε),g(ε)(x)=gε(xε),Aijkl(ε)(x)=Aijkl,ε(xε)∀xε∈Ωε.
Functions fi(ε)(t):Ω→R and hi(t)(ε):Γ+∪Γ-→R are defined by setting
(2.14)fi(ε)(t,x)=fi,ε(t,xε)∀xε∈Ωε,hi(ε)(t,x)=hi,ε(t,xε)∀xε∈Γ+ε∪Γ-ε.
Then the scaled unknown u(ε)(t) defined above satisfies (c.f. [1])
(2.15)u(ε)(t)∈Lc∞(-∞,T;W(Ω))withW(Ω):={v∈W1,4(Ω),v=0on⁡Γ0},∫Ωuitt(ε)(t)vig(ε)dx+∫ΩAijkl(ε)(0)Ek∥l(ε;u(ε)(t))Fi∥j(ε;u(ε)(t),v)g(ε)dx+∫Ω∫-∞tA′ijkl(ε)(t-τ)Ek∥l(ε;u(ε)(τ))Fi∥j(ε;u(ε)(τ),v)g(ε)dτdx=∫Ωfi(ε)(t)vig(ε)dx+1ε∫Γ+∪Γ-hi(ε)(t)vig(ε)dΓ,∀v∈W(Ω),
where
(2.16)Ei∥j(ε;u(ε)(t)):=12(ui∥j(ε)(t)+uj∥i(ε)(t)+gmn(ε)um∥i(ε)(t)un∥j(ε)(t)),Fi∥j(ε;u(ε)(t),v):=12(vi∥j(ε)+vj∥i(ε)+gmn(ε){um∥i(ε)(t)vn∥j(ε)+un∥j(ε)(t)vm∥i(ε)}),uβ∥α(ε)(t):=∂αuβ(ε)(t)-Γαβp(ε)up(ε)(t),vβ∥α(ε)=∂αvβ-Γaβp(ε)vp,u3∥α(ε)(t):=∂αu3(ε)(t)-Γα3σ(ε)(t)uσ(ε)(t),v3∥α(ε):=∂αv3-Γα3σ(ε)vσ,uα∥3(ε)(t):=1ε∂3uα(ε)(t)-Γα3σ(ε)uσ(ε)(t),vα∥3(ε):=1ε∂3vα-Γα3σ(ε)vσ,u3∥3(ε)(t):=1ε∂3u3(ε)(t),v3∥3(ε):=1ε∂3v3.The functionsAijkl(ε) are called the contravariant components of the scaled three-dimensional elasticity tensor of the shell. The functions Ei∥j(ε;u(ε)(t)) are called the scaled strains in the curvilinear coordinates because they satisfy
(2.17)Ei∥j(ε;u(ε)(t))(x)=Ei∥jε(uε(t))(xε)∀xε∈Ω¯ε.
Note that the above definitions likewise imply that
(2.18)Fi∥j(ε;u(ε)(t),v)(x)=Fi∥jε(uε(t),vε)(xε),ui∥j(ε)(t)(x)=ui∥jε(t)(xε),vi∥j(ε)(x)=vi∥jε(xε)∀xε∈Ω¯ε.For notational brevity, the pointx of some functions is suppressed where no confusion can arise.The following two requirements constantly guide the procedures of the formal asymptotic analysis. The first requirement asserts that no restriction should be imposed on the applied forces entering the right-hand side of the equations used for determining the leading term. The second requirement asserts that, by retaining only the linear terms in any relation satisfied by terms of arbitrary order in the formal asymptotic expansion of the scaled unknownu(ε)(t), a relation of linear theory should be recovered. For brevity, we will call it “linearization trick” (see [1]).Theorem 2.1.
Assume that the scaled unknownu(ε) satisfying problem (2.15) admits a formal asymptotic expansion of the form
(2.19)u(ε)(t)=u0(t)+εu1(t)+ε2u2(t)+⋯
with u0(t)∈L∞(0,T;W(Ω)) and u1(t),u2(t)∈L∞(0,T;W1,4(Ω)). Then in order that no restriction be put on the applied forces and that the linearization be satisfied, the components of the applied forces must be of the form
(2.20)fi,ε(t,xε)=fi,0(t,x),xε∈Ωε,hi,ε(t,xε)=εhi,1(t,x),xε∈Γ+ε∪Γ-ε,
where the functions fi,0∈L∞(0,T;L2(Ω)) and hi,1∈Ł∞(0,T;L2(Γ+∪Γ-)) are independent of ε.
This being the case, the leading termu0(t) is independent of the transverse variable x3 and ζ0(t)=(1/2)∫-11u0(t)dx3 satisfies the following two-dimensional variation problem:
(2.21)ζ0(t)∈Lc∞((-∞,T];W(ω))withW(ω):={η∈W1,4(ω);η=0onγ0},∫ωζitt0(t)ηjaijady+∫ωaαβστ(0)Eσ∥τ0(t)Fα∥β0(t,η)ady+∫ω∫-∞ta′αβστ(t-τ)Eσ∥τ0(τ)Fα∥β0(τ,η)adydτ=∫ωpi,0(t)ηiady,∀η∈W(ω),a.e.-∞<t≤T,
where (recall that amn=am·an):
(2.22)Eα∥β0(t):=12(ζα∥β0(t)+ζβ∥α0(t)+amnζm∥α0(t)ζn∥β0(t)),Fα∥β0(t,η):=12(ηα∥β+ηβ∥α+amn{ζm∥α0(t)ηn∥β+ζn∥β0(t)ηm∥α}),ηα∥β=∂βηα-Γαβσησ-bαβη3,η3∥β:=∂βη3+bβσησ,aαβστ(t):=a(t)aαβaστ+μ(t)(aασaβτ+aατaβσ),a(t):=L-1{2λ̂(s)μ̂(s)λ̂(s)+2μ̂(s)}witha(0)=2λ(0)μ(0)λ(0)+2μ(0),pi,0(t):=12(∫-1+1fi,0(t)dx3+hi,1(⋅,1)+hi,1(⋅,-1)),λ̂(s),μ̂(s) denote Laplace transformation of λ(t),μ(t), respectively, and L-1 denotes the inverse Laplace transformation.Lemma 2.2.
For smallε>0,it is not difficult to verify the following relations:
(2.23)gij(ε)=aij+εx3gij,1+O(ε2),
where
(2.24)aij:=ai⋅aj,gαβ,1:=2aασbσβ,gi3,1=0,Γijp(ε)=Γijp,0+εx3Γijp,1+O(ε2),
where
(2.25)Γαβσ,0:=Γαβσ,Γαβ3,0:=bαβ,Γα3β,0:=-bαβ,Γα33,0=Γ33p,0:=0,Γαβσ,1:=-bβσ|α,Γαβ3,1:=-bασbσβ,Γα3σ,1:=-bατbτσ,Γα33,1=Γ33p,1:=0,Aijkl(ε)(t)g(ε)=Aijkl(t)a+εBijkl,1(t)+ε2Bijkl,2(t)+o(ε2),
where
(2.26)Aαβστ(t):=λ(t)aαβaστ+μ(t)(aασaβτ+aατaβσ),Aαβ33(t):=λ(t)aαβ,Aα3σ3(t):=μ(t)aασ,A3333(t):=λ(t)+2μ(t),Aαβσ3(t)=Aα333(t):=0.Lemma 2.3 (see [1]).
Letω be a domain in R2, and let θ∈C2(ω¯;R3) be an injective mapping such that the two vectors aα are linear independent at all points of ω¯. The derivatives of the vectors of the covariant and contravariant basis are given by the formulas of Gauss
(2.27)∂αaβ=Γαβσaσ+bαβa3,∂αaβ=-Γασβaσ+bαβa3
and Weingarten
(2.28)∂αa3=∂αa3=-bαβaβ=-bασaσ.Lemma 2.4.
Letω∈L2(Ω) be a function such that ∫Ωω∂3υ=0 for all υ∈C∞(Ω¯) satisfing υ=0 on γ×[-1,+1]. Then, ω=0.Proof.
Thanks to Theorem  3.4-1 in [1].Lemma 2.5.
Assume that the scaled unknown satisfying (2.15) admits for each 0<ε<ε0 a formal asymptotic expansion of the form
(2.29)u(ε)(t)=1εNu-N(t)+1εN-1u-N+1(t)+⋯
with
(2.30)u-N(t),u-N+1(t)∈L∞(0,T;W(Ω)),W(Ω)={v∈W1,4(Ω),v=0onΓ0},u-N(t)≠0,
for some integer N∈Z. Then, N=0.Proof.
The proof is broken into seven parts. Before beginning the proper induction in (iv), we record several useful preliminaries.
(i) Let the functionsAijkl(0) be defined as in Lemma 2.2. Then, for any symmetric matrices (skl) and (tij),
(2.31)Aijkl(0)skltij=(λ(0)aαβaστ+μ(0)(aασaβτ+aατaβσ))sστtαβ+4μ(0)aασsα3tσ3+λ(0)aαβs33tαβ+λ(0)aστsστt33+(λ(0)+2μ(0))s33t33.
This formula, which immediately follows from the definitions, will be constantly put to use in the ensuing arguments.
(ii) Letaij:=ai·aj. Then, for any y∈ω¯ and any matrix (tij),
(2.32)aij(y)amn(y)timtjn≥0,aij(y)amn(y)timtjn=0⟺tij=0.
Given any y∈ω¯ and any matrix (tij), let ti(y):=timam(y) and let [ti(y)]q denote the qth Cartesian component of the vector ti(y). We thus have
(2.33)aij(y)amn(y)timtjn=aij(y){(timam(y))⋅(tjnan(y))}=aij(y){ti(y)⋅tj(y)}=aij(y)[ti(y)]p[tj(y)]p=([ti(y)]pai(y))⋅([tj(y)]paj(y))=∑p=13|[ti(y)]pai(y)|2.
Hence, aij(y)amn(y)timtjn≥0 and
(2.34)aijamn(y)timtjn=0⟹[ti(y)]pai(y)=0,p=1,2,3,⟹ti(y)=timam(y)=0,i=1,2,3,⟹tim=0,i,m=1,2,3,
for the three vectors ai(y) are linear independent.
(iii) Assume that the formal asymptotic expansion of the scaled unknown is of the form(2.35)u(ε)(t)=1εNu-N(t)+1εN-1u-N+1(t)+⋯forsomeintegerN≥0,
with u-N∈L∞(0,T;W(Ω)) and u-N+1∈L∞(0,T;W(Ω)).
Together with the asymptotic behavior of the functionsgij(ε) and Γijp(ε) as ε→0, such an expansion induces specific formal asymptotic expansions of the various functions appearing in the formulation of problem (2.15)
(2.36)um∥α(ε)(t)=1εNum∥α-N(t)+⋯,um∥3(ε)(t)=1εN+1um∥3-N-1(t)+1εNum∥3-N(t)+⋯,Eα∥β(ε;u(ε)(t))=1ε2NEα∥β-2N(t)+⋯,Eα∥3(ε;u(ε)(t))=1ε2N+1Eα∥3-2N-1(t)+⋯,E3∥3(ε;u(ε)(t))=1ε2N+2E3∥3-2N-2(t)+⋯,Fα∥β(ε;u(ε)(t),v)=1εNFα∥β-N(t,v)+⋯,Fα∥3(ε;u(ε)(t),v)=1εN+1Fα∥3-N-1(t,v)+⋯,F3∥3(ε;u(ε)(t),v)=1εN+2F3∥3-N-2(t,v)+⋯,
where, by definition, ui∥jq(t),Ei∥jq(t), and Fi∥jq(t,v) designate for each q∈Z the coefficient of εq in the induced expansions of ui∥j(ε)(t),Ei∥j(ε;u(ε)(t)), and Fi∥j(ε;u(ε)(t),v).
Note in passing that, while the functions factorizing the powers ofε are by definition independent of ε, they are dependent on one or several terms uq(t),q≥-N. In this respect, particular caution should be exercised as regards this dependence. For instance,
(2.37)um∥α-N(t)=∂αum-N(t)-Γαmp,0up-N(t),um∥3-N(t)=∂3um-N+1(t)-Γm3p,0up-N(t),
that is, the factor of 1/εN in um∥α(ε)(t) depends on u-N(t) but the one in um∥3(ε)(t) depends also on u-N+1(t).
Likewise, it should be remembered that the expression of some factor may differ according to which value ofN is considered, for instance,
(2.38)Eα∥β-2N(t)={12amnum∥α-N(t)un∥β-N(t)ifN≥1,12(uα∥β0(t)+uβ∥α0(t)+amnum∥α0(t)un∥β0(t))ifN=0,Fα∥β-N(t,v)={12amn{um∥α-N(t)vn∥β+un∥β-N(t)vm∥α}ifN≥1,12(vα∥β+vβ∥α+amn{um∥α0(t)vn∥β+un∥β0(t)vm∥α})ifN=0,
where
(2.39)vm∥α:=∂αvm-Γαmp,0vp.
We are now in a position to start the cancellation of the factors of the successive powers ofε found in the variational equations of problem (2.15) when u(ε)(t) is replaced by its formal expansion. In what follows, Lr designates for any integer r≥-3N-4 the linear form defined by
(2.40)Lr(t,v):=∫Ωfi,r(t)viadx+∫Γ+∪Γ-hi,r+1(t)viadΓ.
(iv) Assume thatN≥0. Since the lowest power of ε in the left-hand side is ε-3N-4, we are naturally led to first try
(2.41)fi(ε)(t)=1ε-3N-4fi,-3N-4(t),hi(ε)(t)=1ε3N+3hi,-3N-3(t).
Comparing the coefficients ofε-3N-4 in (2.15) and using Lemma 2.2 and (2.36), we get the equations
(2.42)∫Ω(λ(0)+2μ(0))E3∥3-2N-2(t)F3∥3-N-2(t,v)adx+∫Ω∫-∞t(λ′(t-τ)+2μ′(t-τ))E3∥3-2N-2(τ)F3∥3-N-2(τ,v)adxdτ=L-3N-4(t,v)
for all v∈W(Ω). Since
(2.43)E3∥3-2N-2(t)=12amn∂3um-N(t)∂3un-N(t),F3∥3-N-2(t,v)=amn∂3um-N(t)∂3vn,
we must have
(2.44)L-3N-4(t,v)=∫Ωfi,-3N-4(t)viadx+∫Γ+∪Γ-hi,-3N-3(t)viadΓ=0
for all v∈W that are independent of x3. Consequently, the first requirement (that there be no restriction on the applied forces) implies that we must let
(2.45)fi,-3N-4(t)=0,hi,-3N-3(t)=0.
By recalling (2.42)–(2.45), we have
(2.46)12∫Ω(λ(0)+2μ(0))[(amn∂3um-N(t)∂3un-N(t))(amn∂3um-N(t)∂3vn)]adx+12∫Ω∫-∞t(λ′(t-τ)+2μ′(t-τ))[(amn∂3um-N(τ)∂3un-N(τ))(amn∂3um-N(τ)∂3vn)]adxdτ=0,
that is,
(2.47)ddt∫-∞t∫Ω(λ(t-τ)+2μ(t-τ))[(amn∂3um-N(τ)∂3un-N(τ))(amn∂3um-N(τ)∂3vn)]dxdτ=0.
Therefore,
(2.48)∫-∞t∫Ω(λ(t-τ)+2μ(t-τ))[(amn∂3um-N(τ)∂3un-N(τ))(amn∂3um-N(τ)∂3vn)]dxdτ=const,
which implies
(2.49)∫-∞t∫Ω(λ(t-τ)+2μ(t-τ))[(amn∂3um-N(τ)∂3un-N(τ))(amn∂3um-N(τ)∂3vn)]dxdτ=0.
Lettingv=u-N(τ) in (2.49) shows that
(2.50)∫-∞t∫Ω(λ(t-τ)+2μ(t-τ))[amn∂3um-N(τ)∂3un-N(τ)]2dxdτ=0.
Since the symmetric(aij) is positive definite, we conclude that
(2.51)∂3u-N(t)=(∂3um-N(t))=0inΩ,
that is, u-N(t) is independent of x3. Inserting (2.51) into (2.43) yields
(2.52)E3∥3-2N-2(t)=0,F3∥3-N-2(t,v)=0∀v∈W(Ω).
A usual, any function defined on Ω¯ that is independent of x3 is identified with a function defined on ω¯, and (2.36) and (2.52) imply
(2.53)u-N(t)∈L∞(0,T;W(ω)),W(ω):={η∈W1,4(ω);η=0on⁡γ0},(2.54)Ei∥j-2N-2(t)=0inΩ,Fi∥j-N-2(t,v)=0∀v∈W(Ω)
Noting (2.36) and (2.51), we also have
(2.55)Ei∥j-2N-1(t)=0inΩ.
Since Eα∥β-2N-1(t)=0 (the leading term in the formal expansion of Eα∥β(ε;u(ε)(t)) is order of -2N) and Ei∥3-2N-1(t)=0 (since ∂3u-N(t)=0, each factor of 1/ε2N+1 in the expansion of Eα∥3(ε;u(ε)(t)) vanishes because it contains some derivative ∂3um-N and the leading term in the expansion of E3∥3(ε;u(ε)(t)) is of order strictly higher than (-2N-1)), our next try is thus
(2.56)fi(ε)(t)=1ε3N+3fi,-3N-3(t),hi(ε)(t)=1ε3N+2hi,-3N-2(t).
Comparing the coefficient of ε-3N-3 in (2.15) then yields equations (the functions Aijkl(0) are defined in Lemma 2.2)
(2.57)∫ΩAijkl(0)Ek∥l-2N-1(t)Fi∥j-N-2(t,v)adx+∫Ω∫-∞tAijkl(0)Ek∥l-2N-1(τ)Fi∥j-N-2(τ,v)adxdτ=L-3N-3(t,v)
for all v∈W(Ω). But since (2.55), we must let fi,-3N-3(t)=0 and hi,-3N-2(t)=0 (first requirement) and accordingly try
(2.58)fi(ε)(t)=1ε3N+2fi,-3N-2(t),hi(ε)(t)=1ε3N+2hi,-3N-1(t).
In which case the cancellation of the coefficient of ε-3N-2 in (2.15) yields the equations
(2.59)∫ΩAijkl(0)Ek∥l-2N(t)Fi∥j-N-2(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l-2N(τ)Fi∥j-N-2(τ,v)adxdτ=L-3N-2(t,v)
for all v∈W(Ω). But since (2.54), we must let fi,-3N-2(t)=0 and hi,-3N-1(t)=0 (first requirement).
(v) Assume thatN≥1. Our next try being thus
(2.60)fi(ε)(t)=1ε3N+1fi,-3N-1,hi(ε)=1ε3N+2hi,-3N,
the cancellation of the coefficient of ε-3N-1 in the variational equations of problem (2.15) then yields the equations
(2.61)∫ΩAijkl(0)Ek∥l-2N(t)Fi∥j-N-1(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l-2N(τ)Fi∥j-N-1(τ,v)adxdτ=L-3N-1(t,v)
for all v∈W(Ω), where
(2.62)Eα∥β-2N(t)=12amnum∥α-N(t)un∥β-N(t),Fα∥β-N-1(t,v)=0,Eα∥3-2N(t)=12amnum∥α-N(t)un∥3-N(t),Fα∥3-N-1(t,v)=12amnum∥α-N(t)∂3vn,E3∥3-2N(t)=12amnum∥3-N(t)un∥3-N(t),F3∥3-N-1(t,v)=amnum∥3-N(t)∂3vn,
the functions um∥i-N(t) being those defined in (2.36).
Lettingv∈W(Ω) be independent of x3 then shows that we must let fi,-3N-1(t)=0 and hi,-3N=0; hence,
(2.63)∫ΩAijkl(0)Ek∥l-2N(t)Fi∥j-N-1(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l-2N(τ)Fi∥j-N-1(τ,v)adxdτ=0
for all v∈W(Ω). Let the field wN=(wmN) be defined for all (y,x3)∈Ω by
(2.64)wmN(t):=um-N+1(t)-(1+x3)Γm3p,0up-N(t).
Then, wN(t)∈L∞(0,T;W(Ω)) because both u-N(t) and u-N+1(t) are assumed to be in the space L∞(0,T;W(Ω)).
Furthermore,∂3wmN(t)=um∥3-N(t), so that
(2.65)Fα∥3-N-1(t,wN(t))=Eα∥3-2N(t),F3∥3-N-1(t,wN(t))=2E3∥3-2N(t).
Using Lemma2.2 and (2.65), we get
(2.66)Aijkl(0)Ek∥l-2N(t)Fi∥j-N-1(t,wN)=2λ(0)amnEm∥n-2N(t)E3∥3-2N(t)+4μ(0)amnEm∥3-2N(t)En∥3-2N(t).
Since(2.67)amnEm∥n-2N(t)=12aijamnui∥m-N(t)uj∥n-N(t)≥0inΩ,
(by (ii))
(2.68)E3∥3-2N(t)=12amnum∥3-N(t)un∥3-N(t)≥0,amnEm∥3-2N(t)En∥3-2N(t)≥0inΩ
(the matrix (amn) is positive definite), in a similar way as in (iv), we can obtain from (2.63) that
(2.69)∫-∞t∫ΩAijkl(t-τ)Ek∥l-2N(τ)Fi∥j-N-1(τ,v)dxdτ=0
for all v∈W(Ω).
Lettingv=wN(τ) in (2.69) and noting (2.66)–(2.68), we conclude that
(2.70)amnEm∥3-2N(t)En∥3-2N(t)=0inΩ;
hence (the matrix (amn) is positive definite)
(2.71)Em∥3-2N(t)=0inΩ.
In particular then,E3∥3-2N(t)=(1/2)amnum∥3-N(t)un∥3-N(t)=0 (by (2.62)) and thus (the matrix (amn) is positive definite)
(2.72)um∥3-N(t)=0.
(vi) Assume thatN≥2 (the case N=1 is considered separately, c.f. (viii)). Our next try being thus
(2.73)fi(ε)(t)=1ε3Nfi,-3N(t),hi(ε)(t)=1ε3N-1hi,-3N+1(t),
the cancellation of the coefficient of ε-3N in the variational equations of problem (2.15) then yields the equations (note that two terms are needed here from the expansions of the functions Aijkl(ε)g(ε), c.f. Lemma 2.2))
(2.74)∫ΩAijkl(0){Ek∥l-2N(t)Fi∥j-N(t,v)+Ek∥l-2N+1(t)Fi∥j-N-1(t,v)}adx+∫ΩBijkl,1Ek∥l-2N(t)Fi∥j-N-1(t,v)dx+∫Ω∫-∞tA′ijkl(t-τ){Ek∥l-2N(τ)Fi∥j-N(τ,v)+Ek∥l-2N+1(τ)Fi∥j-N-1(τ,v)}adxdτ+∫ΩB′ijkl,1(t-τ)Ek∥l-2N(τ)Fi∥j-N-1(τ,v)dxdτ=L-3N(t,v)
for all v∈W(Ω), where (by (2.62) and (2.72))
(2.75)Eα∥β-2N(t)=12amnum∥α-N(t)um∥β-N(t),Ei∥3-2N(t)=0,Fα∥β-N-1(t,v)=0,Fα∥3-N-1(t,v)=12amnum∥α-N(t)∂3vn,F3∥3-N-1(t,v)=0,Fα∥β-N(t,v)=12amn(um∥α-N(t)vn∥β+un∥β-N(t)vm∥α),F3∥3-N=amnum∥3-N+1(t)∂3vn,
the last expression of F3∥3-N(t,v) being valid only if N≥2 (the expressions of Fα∥3-N(t,v) are not needed since Aα3στ(0)=0 by Lemma 2.2 and Eα∥3-2N(t)=0 by (2.71)).
Noting thatFα||3-N-1(v)=F3∥3-N=0 if ∂3v=0, we thus conclude that the variational equations (2.74) reduce to
(2.76)∫ΩAαβστ(0)Eσ∥τ-2N(t)Fα∥β-N(t,v)adx+∫Ω∫-∞tA′αβστ(t-τ)Eσ∥τ-2N(τ)Fα∥β-N(τ,v)adxdτ=L-3N(t,v)
for all v∈W(Ω) that are independent of x3.
Since each term in the sumAαβστ(0)Eσ∥τ-2N(t)Fα∥β-N(t,v) is cubic with respect to the functions um∥α-N(t), the linearization trick (second requirement) implies that L-3N(t,v)=0 for all v∈W(Ω) that are independent of x3. Hence, we must let fi,-3N(t)=0 and hi,-3N+1(t)=0. Hence
(2.77)∫ΩAαβστ(0)Eσ∥τ-2N(t)Fα∥β-N(t,v)adx+∫Ω∫-∞tA′αβστ(t-τ)Eσ∥τ-2N(τ)Fα∥β-N(τ,v)adxdτ=0.
In a similar way as in (iv), we can obtain from (2.77) that
(2.78)∫-∞t∫ΩAαβστ(t-τ)Eσ∥τ-2N(τ)Fα∥β-N(τ,v)dxdτ=0.
Recalling thatu-N(t) is independent of x3 by (2.51), we may let v=u-N(τ) in (2.78). This gives
(2.79)∫-∞t∫ΩAαβστ(t-τ)Eσ∥τ-2N(τ)Eα∥β-2N(τ)dxdτ=0
since Fα∥β-N(τ,u-N(τ))=2Eα∥β-2N(t) (by (2.75)). But (see Lemma 2.2)
(2.80)Aαβστ(0)=λ(0)aαβaστ+μ(0)(aασaβτ+aατaβσ)
and thus (the matrix (amn) is positive definite)
(2.81)Eα∥β-2N(t)=12amnum∥α-N(t)un∥β-N(t)=0inΩ
(to reach this conclusion, observe that aασaβτtστtαβ≥0 and that aασaβτtστtαβ=0 only if tαβ=0 by (ii)); these relations in turn imply that
(2.82)um∥α-N(t)=0.
By definition (see (2.36) and Lemma 2.2),
(2.83)uβ∥α-N(t)=∂αuβ-N(t)-Γαβp,0up-N(t)=∂αuβ-N(t)-Γαβσuσ-N(t)-bαβu3-N(t),u3∥α-N(t)=∂αu3-N(t)-Γα3p,0up-N(t)=∂αu3-N(t)+bασuσ-N(t).
Let ζi(t)=ui-N(t)|x3=0. Then, ζi(t)∈L∞(0,T;W(ω)) since ui-N(t)∈L∞(0,T;W(Ω)) and ∂3ui-N(t)=0 in Ω and ζi(t)=0 on γ0 since ui-N(t)=0 on Γ0. The above relations combined with the Gauss and Weingarten formulas (Lemma 2.3) then imply that ∂α(ζi(t)ai)=0 in ω and hence that ζi(t)=0. We have thus shown that
(2.84)u-N(t)=0∀N≥2.
(vii) Finally, assume thatN=1. The only difference from (vi) is that now
(2.85)F3∥3-1(t,v)=∂3v3+amnum∥30(t)∂3vn.
But since the arguments that led in (vi) to the conclusion that u-N=0 for N≥2 only required that consideration of functions v∈W that are independent of x3, in which case F3∥3-1(t,v)=0, they can be reproduced verbatim for N=1, thus showing that
(2.86)u-1(t)=0.
The proof is complete.


## 3. The Proof of the Main Result

Proof.
The proof comprises three parts.
(i) Using Lemma2.5, u(ε)(t) can be expanded as
(3.1)u(ε)(t)=u-N(t)+u-N+1(t)+⋯,
with u-N∈L∞(0,T;W(Ω)). Letting N=0, we thus infer that ∂3u0(t)=0 in Ω, that
(3.2)ζ0(t):=12∫-11u0(t)dx3∈L∞(0,T;W(ω)),W(ω)={η∈W1,4(ω);η=0on⁡γ0},
and also that (see (2.54) and (2.55))
(3.3)Ei∥jq(t)=0∀integersq≤-1,Fi∥jq(t,v)=0∀integersq≤-2∀v∈W(Ω),
and, finally, that we must let fi,-2(t)=0 and hi,-1(t)=0.
(ii) Our next try is thus(3.4)fi(ε)(t)=1εfi,-1(t),hi(ε)(t)=hi,0(t),
where it is understood as in the proof of Lemma 2.5 that each function fi,r(t)∈L∞(0,T;L2(Ω)) and each function hi,r+1(t)∈L∞(0,T;L2(Γ+∪Γ-)), r≥-1, appearing here and subsequently is independent of ε; likewise, we again let
(3.5)Lr(t,v):=∫Ωfi,r(t)viadx+∫Γ+∪Γ-hi,r+1(t)viadΓ,r≥-1.
The cancellation of the coefficient ofε-1 in the variational equations of problem (2.15) then yields the equations:
(3.6)∫ΩAijkl(0)Ek∥l0(t)Fi∥j-1(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l0(τ)Fi∥j-1(τ,v)adxdt=L-1(t,v)
for all v∈W(Ω), where
(3.7)Eα∥β0(t)=12(uα∥β0(t)+uβ∥α0(t)+amnum∥α0(t)un∥β0(t)),Eα∥30(t)=12(uα∥3(0)(t)+u3∥α0(t)+amnum∥α0(t)un∥3(0)(t)),E3∥30(t)=u3∥3(0)(t)+12amnum∥3(0)(t)un∥3(0)(t),Fα∥β-1(t,v)=0,Fα∥3-1(t,v)=12(∂3vα+amnum∥α0(t)∂3vn),F3∥3-1(t,v)=∂3v3+amnum∥3(0)(t)∂3vn,(3.8)um∥α0(t):=∂αum0(t)-Γαmp,0up0(t),um∥3(0)(t):=∂3um1(t)-Γm3p,0up0(t).
The special notationum∥3(0)(t) emphasizes that, by contrast with the functions um∥α0(t), which only depend on u0(t), the functions um∥3(0)(t) also depend on u1(t).
The expressions of the functionsFi∥j-1(t,v) imply that L-1(t,v)=0 for all for all v∈W(Ω) that are independent of x3. Hence, we must let fi,-1(t)=0 and hi,0(t)=0 (first requirement), so that we are left with the equations
(3.9)∫ΩAijkl(0)Ek∥l0(t)Fi∥j-1(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l0(τ)Fi∥j-1(τ,v)adxdt=0
for all v∈W(Ω). When the functions Fi∥j-1(t,v) are replaced by their expression given in (3.7), the integrand in (3.9) takes the form (wτ∂3vτ+w3∂3v3).
Then, Lemma2.4 shows that the functions wτ and w3 vanish in Ω, that is,
(3.10)(λ(0)aαβEα∥β0(t)+(λ(0)+2μ(0))E3∥30(t))aστuσ∥3(0)(t)+2μ(0)Eα∥30(t)(aατ+aασaβτuβ∥σ0(t))+∫-∞t[(λ′(t-s)aαβEα∥β0(s)+(λ′(t-s)+2μ′(t-s))E3∥30(s))aστuσ∥3(0)(s)+2μ′(t-s)Eα∥30(s)(aατ+aασaβτuβ∥σ0(s))]ds=0inΩ,τ=1,2,(λ(0)aαβEα∥β0(t)+(λ(0)+2μ(0))E3∥30(t))(1+u3∥3(0)(t))+2μ(0)aασEα∥30(t)u3∥σ0(t)+∫-∞t[(λ′(t-s)aαβEα∥β0(s)+(λ′(t-s)+2μ′(t-s))E3∥30(s))(1+u3∥3(0)(s))+2μ′(t-s)aασEα∥30(s)u3∥σ0(s)]ds=0inΩ,
that is,
(3.11)∫-∞t[(λ(t-τ)aαβEα∥β0(τ)+(λ(t-s)+2μ(t-s))E3∥30(s))aστuσ∥3(0)(s)+2μ(t-s)Eα∥30(s)(aατ+aασaβτuβ∥σ0(s))]ds=0inΩ,τ=1,2,∫-∞t[(λ(t-s)aαβEα∥β0(s)+(λ(t-s)+2μ(t-s))E3∥30(s))(1+u3∥3(0)(s))+2μ(t-s)aασEα∥30(s)u3∥σ0(s)]ds=0inΩ.
Under the conditions of integral mean value theorem, one obvious solution to this system of three equations is(3.12)Eα∥30(t)=0inΩ,λ̂(s)aαβÊα∥β0(s)+(λ̂(s)+2μ̂(s))Ê3∥30(s)=0inΩ,λ̂(s),μ̂(s),and Êα∥β0(s) denote the Laplace transformation of λ(t),μ(t), and Eα∥β0(t) (c.f. [11]). But there may be other solutions to this nonlinear system. Denoting by [⋯]lin the linear part with respect to (any component of) u0(t) or u1(t) in the expression [⋯], we have
(3.13)[Eα∥30(t)]lin⁡=eα∥30(s),[λ̂(s)aαβÊα∥β0(s)+(λ̂(s)+2μ(t-s))Ê3∥30(s)]lin⁡=λ̂(s)aαβêα∥β0(s)+(λ̂(s)+2μ̂(s))ê3∥30(s),
by definition of the functions Ei∥30(t) and ei∥30(t) as the coefficient of ε0 in the formal expansions of the functions Ei∥3(ε,u(ε))(t) and ei∥3(ε,u(ε))(t), the latter being precisely the linear part in [2].
Since it was found in the linear case (see [2]) that
(3.14)eα||30(t)=0inΩ,λ̂(s)aαβêα∥β0(s)+(λ̂(s)+2μ̂(s))ê3∥30(s)=0inΩ,
the linearization trick (second requirement) suggests that we only retain the “obvious” solution found above.
(iii) Our next try is thus(3.15)fi(ε)(t)=fi,0(t),hi(ε)(t)=εhi,1(t).
The cancellation of the coefficient ofε0 in the variational equations of problem (2.15) then leads to the equations
(3.16)∫Ωuitt0(t)vjaijadx+∫ΩAijkl(0){Ek∥l0(t)Fi∥j0(t,v)+Ek∥l1Fi∥j-1(t,v)}adx+∫Ω∫-∞tA′ijkl(t-τ){Ek∥l0(τ)Fi∥j0(τ,v)+Ek∥l1(τ)Fi∥j-1(τ,v)}adxdτ+∫ΩBijkl,1(0)Ek∥l0(t)Fi∥j-1(t,v)dx+∫Ω∫-∞tB′ijkl,1(t-τ)Ek∥l0(τ)Fi∥j-1(τ,v)dxdτ=L0(t,v)
for all v∈W(Ω), where the functions Ei∥j1(t) and Fi∥j0(t) are defined by means of formal expansions
(3.17)Ei∥j(ε;u(ε)(t))=Ei∥j0(t)+εEi∥j1(t)+⋯,Fi∥j(ε;u(ε)(t),v)=1εFi∥j-1(t,v)+Fi∥j0(t,v)+⋯.
Note that, while the functionsEi∥j0(t), Fi∥j-1(t,v), and Fi∥j0(t,v) depend only on u0(t) and u1(t), the functions Ei∥j1(t) depend also on u2(t) (but not on u3(t); each term involving u3(t) vanishes because it contains some derivative ∂3um0(t) as a factor). For this reason the formal asymptotic expansion of u(ε)(t) must be “at least” of the form
(3.18)u(ε)(t)=∑q=02εquq(t)+⋯.
In particular then, we must have (by (3.7))
(3.19)∫Ωuitt0(t)vjaijadx+∫ΩAijkl(0)Ek∥l0(t)Fi∥j0(t,v)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l0(τ)Fi∥j0(τ,v)adxdτ=L0(t,v)
for all v∈W(Ω) that are independent of x3 since Fi∥j-1(t,v)=0 for such functions; equivalently, after performing the usual identification, we must have
(3.20)∫Ωuitt0(t)ηjaijadx+∫ΩAijkl(0)Ek∥l0(t)Fi∥j0(t,η)adx+∫Ω∫-∞tA′ijkl(t-τ)Ek∥l0(τ)Fi∥j0(τ,η)adxdτ=L0(t,v)
for all η∈W(ω)={η∈W1,4(ω);η=0onγ0}.
Using Lemma2.2 and (3.12), (3.20) can be written as
(3.21)∫Ωuitt0ηjaijadx+∫Ω{Aαβστ(0)Eσ∥τ0(t)+Aαβ33(0)E3∥30(t)}Fα∥β0(t,η)adx+∫Ω∫-∞t{A′αβστ(t-τ)Eσ∥τ0(τ)+A′αβ33(t-τ)E3∥30(τ)}Fα∥β0(τ,η)adτdx+∫Ω{A33στ(0)Eσ∥τ0(t)+A3333(0)E3∥30(t)}F3∥30(t,η)adx+∫Ω∫-∞t{A′33στ(t-τ)Eσ∥τ0(τ)+A′3333(t-τ)E3∥30(τ)}F3∥30(τ,η)adτdx=∫Ωfi,0(t)ηiadx+∫Γ+∪Γ-hi,1(t)ηiadΓ
for all η∈W(ω)={η∈W1,4(ω);η=0onγ0}, that is,
(3.22)∫Ωuitt0ηjaijadx+∫Ω∂∂t(∫-∞t{Aαβστ(t-τ)Eσ∥τ0(τ)+Aαβ33(t-τ)E3∥30(τ)}Fα∥β0(τ,η)dτ)adx+∫Ω∂∂t(∫-∞t{A33στ(t-τ)Eσ∥τ0(τ)+A3333(t-τ)E3∥30(τ)}F3∥30(τ,η)dτ)adx=∫Ωfi,0(t)ηiadx+∫Γ+∪Γ-hi,1(t)ηiadΓ.
Setting(3.23)C(t):=∫-∞t{Aαβστ(t-τ)Eσ∥τ0(τ)+Aαβ33(t-τ)E3∥30(τ)}Fα∥β0(τ,η)dτ+∫-∞t{A33στ(t-τ)Eσ∥τ0(τ)+A3333(t-τ)E3∥30(τ)}F3∥30(τ,η)dτ,
we have
(3.24)Ĉ(s)=Âαβστ(s)Êσ∥τ0(s)*F̂α∥β0(s,η)+Âαβ33(s)Ê3∥30(s)*F̂α∥β0(s,η)+Â33στ(s)Êσ∥τ0(s)*F̂3∥30(s,η)+Â3333(s)Ê3∥30*F̂3∥30(s,η),
where * denotes convolution. Substituting (3.12) into (3.24), we get
(3.25)Ĉ(s)=Âαβστ(s)Êσ∥τ0(s)*F̂α∥β0(s,η)-λ̂(s)λ̂(s)+2μ̂(s)Âαβ33(s)aστÊσ∥τ0(s)*F̂α∥β0(s,η)+Â33στ(s)Êσ∥τ0(s)*F̂3∥30(s,η)-λ̂(s)λ̂(s)+2μ̂(s)Â3333(s)aστÊσ∥τ0(s)*F̂3∥30(s,η)=μ̂(s)(aασaβτ+aατaβσ)Êσ∥τ0(s)*F̂α∥β0(s,η)+2μ̂(s)λ̂(s)λ̂(s)+2μ̂(s)aαβaστÊσ∥τ0(s)*F̂α∥β0(s,η).
Applying the inverse Laplace transformation to (3.25), we obtain
(3.26)C(t)=∫-∞taαβστ(t-τ)Eσ∥τ0(τ)Fα∥β0(τ,η)dτ,
where
(3.27)a(t):=L-1{2λ̂(s)μ̂(s)λ̂(s)+2μ̂(s)},aαβστ(t):=a(t)aαβaστ+μ(t)(aασaβτ+aατaβσ).
Inserting (3.26) into (3.22), we get the equation in Theorem 2.1.
Sinceu0(t) is independent of x3, it may be identified with a function ζ0(t)∈L∞(0,T;W(Ω)). Consequently, the functions
(3.28)Eα∥β0(t):=12(uβ∥α0(t)+uβ∥α0(t)+amnum∥α0(t)un∥β0(t))∈L∞(0,T;L2(Ω)),Fα∥β0(t,η):=12(ηα∥β+ηβ∥α+amn{um∥α0(t)ηn∥β+un∥β0(t)ηm∥α})∈L∞(0,T;L2(Ω)),
which are thus also independent of x3, may be likewise identified with functions (denoted for convenience by the same symbols)
(3.29)Eα∥β0(t):=12(ζα∥β0(t)+ζβ∥α0(t)+amnζm∥α0(t)ζn∥β0(t))∈L∞(0,T;L2(ω)),Fα∥β0(t,η):=12(ηα∥β+ηβ∥α+amn{ζm∥α0(t)ηn∥β+ζn∥β0(t)ηm∥α})∈L∞(0,T;L2(ω)),
where
(3.30)ηα∥β=∂βηα-Γαβσησ-bαβη3,η3∥η:=∂3η3+bβσησ,
for all η∈W(ω). The last variational problem is thus indeed two-dimensional.
The definition ofa(t) implies
(3.31)â(s)(λ̂(s)+2μ̂(s))=2λ̂(s)μ̂(s).
Applying the inverse Laplace transformation to (3.31), we get
(3.32)∫0ta(t-τ)(λ(τ)+2μ(τ))dτ=∫0t2λ(t-τ)μ(τ)dτ.
Therefore
(3.33)a(0)(λ(t)+2μ(t))+∫0ta′(t-τ)(λ(τ)+2μ(τ))dτ=2λ(0)μ(t)+∫0t2λ′(t-τ)μ(τ)dτ.
Lettingt=0 in (3.33), we obtain immediately that
(3.34)a(0)=2λ(0)μ(0)λ(0)+2μ(0).



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*Source: 287865-2012-02-19.xml*