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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 287865, 24 pages
doi:10.1155/2012/287865
Research Article
A Justification of Two-Dimensional Nonlinear
Viscoelastic Shells Model
Fushan Li
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
Correspondence should be addressed to Fushan Li, fushan99@163.com
Received 19 October 2011; Accepted 23 November 2011
Academic Editor: Wing-Sum Cheung
Copyright q 2012 Fushan Li. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
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.
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 timedependent 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. Section 2 begins with some preliminaries and then
gives the main result. In Section 3, we give the proof of the main theorem.
2
Abstract and Applied Analysis
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.
Let ω ⊂ 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 : ∂α θy are linear independent at all points
y ∈ ω. They form the covariant basis of the tangent plane to the surface S θω at the point
θy; 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 θy.
One then defines the first fundamental form, also known as metric tensor aαβ or
β
aαβ , the second fundamental form, also known as the curvature tensor bαβ or bα , and the
Christoffel symbols Γσαβ of the surface S by setting whenever no confusion should arise, we
henceforth drop the explicit dependence on the variable y ∈ ω
aαβ : aα · aβ ,
bαβ : a3 · ∂β aα ,
aαβ : aα · aβ ,
β
bα : aβσ bσα ,
Γσαβ : aσ · ∂β aα .
2.1
√
Note the symmetries aαβ aβα , bαβ bβα , and Γσαβ Γσβα . The area element along S is ady,
where a : detaαβ . All the functions defined above are at least continuous over the set ω. In
particular, there exists a constant a0 > 0 such that ay ≥ a0 , for all y ∈ ω.
In addition, let the covariant derivatives bβσ |α and the covariant components cαβ of the
third form of the surface S be defined by
bβσ |α : ∂α bβσ Γσατ bβτ − Γταβ bτσ ,
2.2
cαβ : bασ bσβ .
For each ε > 0, we consider a shell with thickness 2ε and middle surface S, whose lamé
relaxation modules λt ≥ 0 and μt > 0 t ≥ 0 are independent of ε. We define the sets
Ωε : ω × −ε, ε,
Γε
ω × {ε},
Γε− ω × {−ε},
Γε0 γ0 × −ε, ε,
2.3
where γ0 ⊂ γ and γ0 /
∅. Note that Γε
∪ Γε− ∪ Γε0 constitutes a partition of the boundary of the
ε
set Ωε . Let xε xiε denote a generic point in the set Ω , and let ∂εi : ∂/∂xiε ; hence xαε yα
and ∂εα ∂α .
ε
We then define a mapping Θ : Ω → R3 by
Θxε : θy x3ε a3 y,
ε
∀xε y, x3ε ∈ Ω ,
2.4
Abstract and Applied Analysis
3
ε
then there exists ε0 > 0 such that for all 0 < ε ≤ ε0 the mapping Θ : Ω → R3 is an
injective mapping and the three vectors gεi xε : ∂εi Θxε ∂εα ∂/∂xα , ∂ε3 ∂/∂x3ε are linear
ε
ε
independent for each xε ∈ Ω . The injectivity of the mapping Θ : Ω → R3 ensures that the
physical problem described below is meaningful.
The three vectors gεi xε form the covariant basis at the point Θxε , and the three
j
vectors gi,ε xε defined by gj,ε xε · gεi xε δi form the contravariant. We define the metric
ε
tensor gijε or g ij,ε and the Christoffel symbols of the manifold ΘΩ by setting we omit
the explicit dependence on xε gijε gεi · gεj ,
k,ε
Γk,ε
· ∂εi gεj .
ij : g
g ij,ε : gi,ε · gj,ε ,
2.5
Note the symmetries
gijε gjiε ,
g ij,ε g ji,ε ,
ε
The volume element in the set ΘΩ is
k,ε
Γk,ε
ij Γji .
2.6
ε ε
g dx , where g ε : detgijε .
ε : ΘΩε is the reference configuration of a viscoelastic
For each 0 < ε ≤ ε0 , the set Ω
shell with middle surface S θω and thickness 2ε. We assume that the material constituting
ε
the shell is homogeneous isotropic and ΘΩ is of a nature state, so that the material is
characterized by its two lamé relaxation modules λt ≥ 0 and μt > 0 t ≥ 0. Under the
action of forces, the shell undergoes a displacement field.
ε t uεi tgi,ε in terms of curvilinear coordinates xε of the reference configuration
Let u
ε
ΘΩ . Then, the covariant displacement field uε t uεi t satisfies the following threedimensional equations c.f. 1, 10:
ε
ε
ε
1,4
ε
ε
ε
uε t ∈ L∞
c −∞, T ; WΩ with WΩ : v ∈ W Ω , v 0 on Γ0 ,
ε
ε
uεitt tviε g ij,ε g ε dxε Aijkl,ε 0Ekl
uε tFij
uε t, vε g ε dxε
Ωε
Ωε
t
Ωε
−∞
Ωε
A
f i,ε tviε
ijkl,ε
2.7
ε
ε
t − τEkl
uε τFij
uε τ, vε g ε dτ dxε
g ε dxε Γε
∪Γε−
hi,ε tviε
g ε dΓε ,
∀vε ∈ WΩε ,
∞
where the symbol L∞
c denotes the subspace of L > 0 such that there exists a constant T such
that the functions vanish as s < −T . And,
Aijkl,ε t : λtg ij,ε g kl,ε μt g ik,ε g jl,ε g il,ε g jk,ε
2.8
designate the contravariant components of the three-dimensional elasticity tensor,
ε
Eij
vε :
1 ε
ε
ε
ε
vij vji
,
g mn,ε vmi
vnj
2
p,ε
ε
where vij
: ∂εj viε − Γij vpε ,
2.9
4
Abstract and Applied Analysis
designate the strains in the curvilinear coordinates associated with an arbitrary displacement
field viε g i,ε of the manifold ΘΩ,
ε
Fij
uε t, vε :
1 ε
ε
ε
ε
,
vij vji
g mn,ε uεmi tvnj
uεnj tvmi
2
2.10
and, finally, f i,ε ∈ L∞ 0, T ; L2 Ωε and hi,ε ∈ L∞ 0, T ; Γε
∪ Γε− denote the contravariant
components of the applied body and surface force densities, respectively, applied to the
interior ΘΩε of the shell and to its “uper” and “lower” faces ΘΓε
and ΘΓε− , and
designate the area element along ∂Ωε . We thus assume that there are no surface forces applied
to the portion Θγ − γ0 × −ε, ε of the lateral face of the shell.
We record in passing the symmetries
Aijkl,ε Ajikl,ε Aklij,ε
2.11
and the relation
p,ε
Γ3,ε
α3 Γ33 0,
Aαβσ3,ε Aα333,ε 0
ε
in Ω .
2.12
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 to ε 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 tai y : ω →
R3 of the middle surface S θω. In other words, ζiε t, yai y is the displacement of the
point θy ∈ S.
“Asymptotic analysis” means that our objective is to study the behavior of the
ε
displacement field uεi tgi,ε : Ω → R3 as ε → 0, an endeavour that will be a behavior as
ε
ε → 0 of the covariant components uεi t : Ω → R of the displacement field, that is, the
ε
behavior of the unknown uε t uεi t : Ω → R3 of the three-dimensional shell problem.
ε
Since these fields are defined on sets Ω 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.
Let Ω : 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
ε
through the bijection π ε : x x1 , x2 , x3 ∈ Ω → xε xiε x1 , x2 , εx3 ∈ Ω ; we thus
ε
have ∂εα ∂α and ∂ε3 1/ε∂3 . Let uε uεi , Γεij , g ε , Aijkl,ε : Ω → R and the vector fields
vε viε appearing in the three-dimensional problem 2.7 be associated with the functions
p
Γij ε, gε, Aijkl ε : Ω → R and the scaled vector fields v vi defined by
ui εt, x uεi t, xε ,
p
p,ε
Γij εx Γij xε ,
vi x viε xε gεx g ε xε ,
ε
∀xε ∈ Ω ,
Aijkl εx Aijkl,ε xε ∀xε ∈ Ωε .
2.13
Abstract and Applied Analysis
5
Functions f i εt : Ω → R and hi tε : Γ
∪ Γ− → R are defined by setting
∀xε ∈ Ωε ,
f i εt, x f i,ε t, xε 2.14
∀xε ∈ Γε
∪ Γε− .
hi εt, x hi,ε t, xε Then the scaled unknown uεt defined above satisfies c.f. 1
1,4
uεt ∈ L∞
c −∞, T ; WΩ with WΩ : v ∈ W Ω, v 0 on Γ0 ,
uitt εtvi gεdx Aijkl ε0Ekl ε; uεtFij ε; uεt, v gεdx
Ω
t
Ω
−∞
Ω
A
ijkl
εt − τEkl ε; uετFij ε; uετ, v gεdτ dx
1
f εtvi gεdx hi εtvi gεdΓ,
ε Γ
∪Γ−
Ω
i
2.15
∀v ∈ WΩ,
where
1
uij εt uji εt g mn εumi εtunj εt ,
2
1
Fij ε; uεt, v :
vij ε vji ε g mn ε umi εtvnj ε unj εtvmi ε ,
2
Eij ε; uεt :
p
uβα εt : ∂α uβ εt − Γαβ εup εt,
u3α εt : ∂α u3 εt − Γσα3 εtuσ εt,
uα3 εt :
p
vβα ε ∂α vβ − Γaβ εvp ,
v3α ε : ∂α v3 − Γσα3 εvσ ,
1
1
∂3 uα εt − Γσα3 εuσ εt,
vα3 ε : ∂3 vα − Γσα3 εvσ ,
ε
ε
1
1
v33 ε : ∂3 v3 .
u33 εt : ∂3 u3 εt,
ε
ε
2.16
The functions Aijkl ε are called the contravariant components of the scaled threedimensional elasticity tensor of the shell. The functions Eij ε; uεt are called the scaled
strains in the curvilinear coordinates because they satisfy
ε
∀xε ∈ Ω .
ε
Eij ε; uεtx Eij
uε txε 2.17
Note that the above definitions likewise imply that
ε
Fij ε; uεt, vx Fij
uε t, vε xε ,
ε
vij εx vij
xε uij εtx uεij txε ,
ε
∀xε ∈ Ω .
2.18
For notational brevity, the point x of some functions is suppressed where no confusion
can arise.
6
Abstract and Applied Analysis
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
unknown uεt, 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 unknown uε satisfying problem 2.15 admits a formal
asymptotic expansion of the form
uεt u0 t εu1 t ε2 u2 t · · ·
2.19
with u0 t ∈ L∞ 0, T ; WΩ 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
f i,ε t, xε f i,0 t, x,
x ε ∈ Ωε ,
hi,ε t, xε εhi,1 t, x,
xε ∈ Γε
∪ Γε− ,
2.20
where the functions f i,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 term u0 t is independent of the transverse variable x3 and
1
0
ζ t 1/2 −1 u0 tdx3 satisfies the following two-dimensional variation problem:
1,4
ζ0 t ∈ L∞
c −∞, T ; Wω with Wω : η ∈ W ω; η 0 on γ0 ,
√
√
0
ij
0
0
ζitt tηj a ady aαβστ 0Eστ
t, η ady
tFαβ
ω
t
ω
−∞
ω
√
0
0
a αβστ t − τEστ
τ, η ady dτ
τFαβ
√
pi,0 tηi ady,
2.21
∀η ∈ Wω, a.e. − ∞ < t ≤ T,
ω
where (recall that amn am · an :
1 0
0
0
0
ζαβ t ζβα
t amn ζmα
tζnβ
t ,
2
1
0
0
0
Fαβ t, η :
ηαβ ηβα amn ζmα
tηnβ ζnβ
tηmα ,
2
ηαβ ∂β ηα − Γσαβ ησ − bαβ η3 ,
η3β : ∂β η3 bβσ ησ ,
aαβστ t : ataαβ aστ μt aασ aβτ aατ aβσ ,
μs
2λs
2λ0μ0
−1
at : L
,
with a0 λ0
2μ0
λs 2
μs
1
1
i,0
i,0
i,1
i,1
p t :
f tdx3 h ·, 1 h ·, −1 ,
2
−1
0
Eαβ
t :
2.22
Abstract and Applied Analysis
7
λs,
μ
s denote Laplace transformation of λt, μt, respectively, and L−1 denotes the inverse
Laplace transformation.
Lemma 2.2. For small ε > 0, it is not difficult to verify the following relations:
g ij ε aij εx3 g ij,1 O ε2 ,
2.23
where
aij : ai · aj ,
β
g i3,1 0,
p
p,0
p,1
Γij ε Γij εx3 Γij O ε2 ,
g αβ,1 : 2aασ bσ ,
2.24
where
Γσ,0
: Γσαβ ,
αβ
Γσ,1
: −bβσ |α ,
αβ
Γ3,0
: bαβ ,
αβ
β,0
β
p,0
Γ3,0
α3 Γ33 : 0,
Γα3 : −bα ,
Γ3,1
: −bασ bσβ ,
αβ
τ σ
Γσ,1
α3 : −bα bτ ,
p,1
Γ3,1
α3 Γ33 : 0,
2.25
√
Aijkl εt gε Aijkl t a εBijkl,1 t ε2 Bijkl,2 t o ε2 ,
where
Aαβστ t : λtaαβ aστ μt aασ aβτ aατ aβσ ,
Aαβ33 t : λtaαβ ,
A3333 t : λt 2μt,
Aα3σ3 t : μtaασ ,
2.26
Aαβσ3 t Aα333 t : 0.
Lemma 2.3 see 1. Let ω 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
∂α aβ Γσαβ aσ bαβ a3 ,
β
β
∂α aβ −Γασ aσ bα a3
2.27
and Weingarten
∂α a3 ∂α a3 −bαβ aβ −bασ aσ .
Lemma 2.4. Let ω ∈ L2 Ω be a function such that
on γ × −1, 1. Then, ω 0.
Proof. Thanks to Theorem 3.4-1 in 1.
Ω
2.28
ω∂3 υ 0 for all υ ∈ C∞ Ω satisfing υ 0
8
Abstract and Applied Analysis
Lemma 2.5. Assume that the scaled unknown satisfying 2.15 admits for each 0 < ε < ε0 a formal
asymptotic expansion of the form
uεt 1
1 −N
u t N−1 u−N
1 t · · ·
N
ε
ε
2.29
with
u−N t, u−N
1 t ∈ L∞ 0, T ; WΩ,
WΩ v ∈ W1,4 Ω, v 0 on Γ0 , u−N t /
0,
2.30
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 functions Aijkl 0 be defined as in Lemma 2.2. Then, for any symmetric
matrices skl and tij ,
Aijkl 0skl tij λ0aαβ aστ μ0 aασ aβτ aατ aβσ sστ tαβ 4μ0aασ sα3 tσ3
λ0a s33 tαβ λ0a sστ t33 λ0 2μ0 s33 t33 .
αβ
στ
2.31
This formula, which immediately follows from the definitions, will be constantly put
to use in the ensuing arguments.
ii Let aij : ai · aj . Then, for any y ∈ ω and any matrix tij ,
aij y amn y tim tjn ≥ 0,
aij y amn y tim tjn 0 ⇐⇒ tij 0.
2.32
Given any y ∈ ω and any matrix tij , let ti y : tim am y and let ti yq denote the qth
Cartesian component of the vector ti y. We thus have
aij y amn y tim tjn aij y
tim am y · tjn an y
aij y ti y · tj y
p p p i p j aij y ti y
ti y a y · tj y a y
tj y
3 p i 2
ti y a y .
2.33
p1
Hence, aij yamn ytim tjn ≥ 0 and
p aij amn y tim tjn 0 ⇒ ti y ai y 0, p 1, 2, 3,
⇒ ti y tim am y 0, i 1, 2, 3,
⇒ tim 0,
for the three vectors ai y are linear independent.
i, m 1, 2, 3,
2.34
Abstract and Applied Analysis
9
iii Assume that the formal asymptotic expansion of the scaled unknown is of the
form
uεt 1 −N
1
u t N−1 u−N
1 t · · ·
εN
ε
for some integer N ≥ 0,
2.35
with u−N ∈ L∞ 0, T ; WΩ and u−N
1 ∈ L∞ 0, T ; WΩ.
p
Together with the asymptotic behavior of the functions g ij ε and Γij ε as ε → 0,
such an expansion induces specific formal asymptotic expansions of the various functions
appearing in the formulation of problem 2.15
1 −N
1
1 −N
umα t · · · ,
um3 εt N
1 u−N−1
m3 t N um3 t · · · ,
N
ε
ε
ε
1 −2N
1
−2N−1
Eα3 ε; uεt 2N
1 Eα3
Eαβ ε; uεt 2N Eαβ t · · · ,
t · · · ,
ε
ε
1
1 −N
−2N−2
Fαβ ε; uεt, v N Fαβ
E33 ε; uεt 2N
2 E33
t · · · ,
t, v · · · ,
ε
ε
1
1
−N−1
−N−2
F33 ε; uεt, v N
2 F33
Fα3 ε; uεt, v N
1 Fα3
t, v · · · ,
t, v · · · ,
ε
ε
2.36
umα εt q
q
q
where, by definition, uij t, Eij t, and Fij t, v designate for each q ∈ Z the coefficient of εq
in the induced expansions of uij εt, Eij ε; uεt, and Fij ε; uεt, v.
Note in passing that, while the functions factorizing the powers of ε 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,
p,0
−N
−N
u−N
mα t ∂α um t − Γαm up t,
p,0
−N
1
u−N
t − Γm3 u−N
p t,
m3 t ∂3 um
2.37
that is, the factor of 1/εN in umα εt depends on u−N t but the one in um3 εt depends
also on u−N
1 t.
Likewise, it should be remembered that the expression of some factor may differ
according to which value of N is considered, for instance,
−2N
Eαβ
t ⎧1
−N
mn −N
⎪
⎪
⎨ 2 a umα tunβ t
if N ≥ 1,
⎪
⎪
⎩ 1 u0 t u0 t amn u0 tu0 t
if N 0,
αβ
βα
mα
nβ
2
⎧1
−N
mn
⎪
u−N
if N ≥ 1,
u
tv
tv
⎪
nβ
mα
⎨2a
mα
nβ
−N
Fαβ
t, v ⎪
⎪
⎩ 1 v v amn u0 tv u0 tv
if N 0,
αβ
βα
nβ
mα
mα
nβ
2
2.38
10
Abstract and Applied Analysis
where
p,0
vmα : ∂α vm − Γαm vp .
2.39
We are now in a position to start the cancellation of the factors of the successive powers
of ε found in the variational equations of problem 2.15 when uεt is replaced by its formal
expansion. In what follows, Lr designates for any integer r ≥ −3N − 4 the linear form defined
by
L t, v :
r
√
f tvi adx i,r
Ω
Γ
∪Γ−
√
hi,r
1 tvi adΓ.
2.40
iv Assume that N ≥ 0. Since the lowest power of ε in the left-hand side is ε−3N−4 , we
are naturally led to first try
f i εt 1
f i,−3N−4 t,
ε−3N−4
hi εt 1
hi,−3N−3 t.
ε3N
3
2.41
Comparing the coefficients of ε−3N−4 in 2.15 and using Lemma 2.2 and 2.36, we get
the equations
√
−2N−2
−N−2
λ0 2μ0 E33
tF33
t, v adx
Ω
t
−∞
Ω
√
−2N−2
−N−2
λ t − τ 2μ t − τ E33
τF33
τ, v adx dτ L−3N−4 t, v
2.42
for all v ∈ WΩ. Since
−2N−2
E33
t 1 mn
−N
a ∂3 u−N
m t∂3 un t,
2
−N−2
F33
t, v amn ∂3 u−N
m t∂3 vn ,
2.43
we must have
−3N−4
L
t, v Ω
f
i,−3N−4
√
tvi adx Γ
∪Γ−
√
hi,−3N−3 tvi adΓ 0
2.44
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
f i,−3N−4 t 0,
hi,−3N−3 t 0.
2.45
Abstract and Applied Analysis
11
By recalling 2.42–2.45, we have
1
2
λ0 2μ0
Ω
1
2
t
−N
amn ∂3 u−N
m t∂3 un t
amn ∂3 u−N
m t∂3 vn
√
adx
√
−N
amn ∂3 u−N
adx dτ 0,
λ t − τ 2μ t − τ amn ∂3 u−N
m τ∂3 un τ
m τ∂3 vn
−∞
Ω
2.46
that is,
d
dt
t −∞
Ω
λt − τ 2μt − τ
−N
amn ∂3 u−N
m τ∂3 un τ
amn ∂3 u−N
m τ∂3 vn
dx dτ 0.
2.47
Therefore,
t −∞
Ω
λt − τ 2μt − τ
−N
amn ∂3 u−N
m τ∂3 un τ
amn ∂3 u−N
m τ∂3 vn
dx dτ const,
2.48
which implies
t −∞
Ω
λt − τ 2μt − τ
−N
amn ∂3 u−N
m τ∂3 un τ
amn ∂3 u−N
m τ∂3 vn
dx dτ 0.
2.49
Letting v u−N τ in 2.49 shows that
t −∞
Ω
2
−N
λt − τ 2μt − τ amn ∂3 u−N
m τ∂3 un τ dx dτ 0.
2.50
Since the symmetric aij is positive definite, we conclude that
∂3 u−N t ∂3 u−N
m t 0
in Ω,
2.51
that is, u−N t is independent of x3 . Inserting 2.51 into 2.43 yields
−2N−2
E33
t 0,
−N−2
F33
t, v 0 ∀v ∈ WΩ.
2.52
12
Abstract and Applied Analysis
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
u−N t ∈ L∞ 0, T ; Wω,
Wω : η ∈ W1,4 ω; η 0 on γ0 ,
−2N−2
Eij
t 0 in Ω,
−N−2
Fij
t, v 0 ∀v ∈ WΩ
2.53
2.54
Noting 2.36 and 2.51, we also have
−2N−1
Eij
t 0
in Ω.
2.55
−2N−1
Since Eαβ
t 0 the leading term in the formal expansion of Eαβ ε; uεt is order
−2N−1
t 0 since ∂3 u−N t 0, each factor of 1/ε2N
1 in the expansion of
of −2N and Ei3
Eα3 ε; uεt vanishes because it contains some derivative ∂3 u−N
m and the leading term in
the expansion of E33 ε; uεt is of order strictly higher than −2N − 1, our next try is thus
f i εt 1
f i,−3N−3 t,
ε3N
3
hi εt 1
hi,−3N−2 t.
ε3N
2
2.56
Comparing the coefficient of ε−3N−3 in 2.15 then yields equations the functions Aijkl 0 are
defined in Lemma 2.2
Ω
√
−2N−1
−N−2
Aijkl 0Ekl
tFij
t, v adx
t
−∞
Ω
A
ijkl
−2N−1
−N−2
0Ekl
τFij
τ, v
√
2.57
adx dτ L
−3N−3
t, v
for all v ∈ WΩ. But since 2.55, we must let f i,−3N−3 t 0 and hi,−3N−2 t 0 first
requirement and accordingly try
f i εt 1
f i,−3N−2 t,
ε3N
2
hi εt 1
hi,−3N−1 t.
ε3N
2
2.58
In which case the cancellation of the coefficient of ε−3N−2 in 2.15 yields the equations
Ω
√
−2N
−N−2
Aijkl 0Ekl
tFij
t, v adx
t
Ω
−∞
A
ijkl
√
−2N
−N−2
t − τEkl
τFij
τ, v adx dτ L−3N−2 t, v
2.59
for all v ∈ WΩ. But since 2.54, we must let f i,−3N−2 t 0 and hi,−3N−1 t 0 first
requirement.
Abstract and Applied Analysis
13
v Assume that N ≥ 1. Our next try being thus
f i εt 1
ε3N
1
f i,−3N−1 ,
hi ε 1
ε3N
2
2.60
hi,−3N ,
the cancellation of the coefficient of ε−3N−1 in the variational equations of problem 2.15 then
yields the equations
Ω
√
−2N
−N−1
Aijkl 0Ekl
tFij
t, v adx
t
−∞
Ω
A
ijkl
t −
−2N
−N−1
τEkl
τFij
τ, v
√
2.61
adx dτ L
−3N−1
t, v
for all v ∈ WΩ, where
−2N
Eαβ
t −2N
Eα3
t 1 mn −N
a umα tu−N
nβ t,
2
1 mn −N
a umα tu−N
n3 t,
2
−2N
E33
t 1 mn −N
a um3 tu−N
n3 t,
2
−N−1
Fαβ
t, v 0,
−N−1
Fα3
t, v 1 mn −N
a umα t∂3 vn ,
2
2.62
−N−1
F33
t, v amn u−N
m3 t∂3 vn ,
t being those defined in 2.36.
the functions u−N
mi
Letting v ∈ WΩ be independent of x3 then shows that we must let f i,−3N−1 t 0 and
i,−3N
0; hence,
h
Ω
√
−2N
−N−1
Aijkl 0Ekl
tFij
t, v adx
t
Ω
−∞
A
ijkl
t −
−2N
−N−1
τEkl
τFij
τ, v
√
2.63
adx dτ 0
N
for all v ∈ WΩ. Let the field wN wm
be defined for all y, x3 ∈ Ω by
p,0
N
wm
t : u−N
1
t − 1 x3 Γm3 u−N
m
p t.
2.64
Then, wN t ∈ L∞ 0, T ; WΩ because both u−N t and u−N
1 t are assumed to be in the
space L∞ 0, T ; WΩ.
N
t u−N
t, so that
Furthermore, ∂3 wm
m3
−N−1
−2N
Fα3
t, wN t Eα3
t,
−N−1
−2N
t, wN t 2E33
F33
t.
2.65
14
Abstract and Applied Analysis
Using Lemma 2.2 and 2.65, we get
−2N
−N−1
−2N
−2N
−2N
−2N
t, wN 2λ0amn Emn
Aijkl 0Ekl
tFij
tE33
t 4μ0amn Em3
tEn3
t. 2.66
Since
−2N
amn Emn
t 1 ij mn −N
a a uim tu−N
jn t ≥ 0 in Ω,
2
2.67
by ii
−2N
E33
t 1 mn −N
a um3 tu−N
n3 t ≥ 0,
2
−2N
−2N
amn Em3
tEn3
t ≥ 0 in Ω
2.68
the matrix amn is positive definite, in a similar way as in iv, we can obtain from 2.63
that
t −∞
Ω
−2N
−N−1
Aijkl t − τEkl
τFij
τ, vdx dτ 0
2.69
for all v ∈ WΩ.
Letting v wN τ in 2.69 and noting 2.66–2.68, we conclude that
−2N
−2N
amn Em3
tEn3
t 0 in Ω;
2.70
hence the matrix amn is positive definite
−2N
Em3
t 0 in Ω.
2.71
−2N
In particular then, E33
t 1/2amn u−N
tu−N
t 0 by 2.62 and thus the
m3
n3
mn
matrix a is positive definite
u−N
m3 t 0.
2.72
vi Assume that N ≥ 2 the case N 1 is considered separately, c.f. viii. Our next
try being thus
f i εt 1
ε3N
f i,−3N t,
hi εt 1
ε3N−1
hi,−3N
1 t,
2.73
Abstract and Applied Analysis
15
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 gε, c.f. Lemma 2.2
Ω
√
−2N
−N
−2N
1
−N−1
Aijkl 0 Ekl
tFij
t, v Ekl
tFij
t, v adx
Ω
t
−∞
Ω
−2N
−N−1
Bijkl,1 Ekl
tFij
t, vdx
Ω
B
A
ijkl,1
ijkl
√
−2N
−N
−2N
1
−N−1
t − τ Ekl
τFij
τ, v Ekl
τFij
τ, v adx dτ
2.74
−2N
−N−1
t − τEkl
τFij
τ, vdx dτ L−3N t, v
for all v ∈ WΩ, where by 2.62 and 2.72
1 mn −N
−2N
a umα tu−N
Ei3
t 0,
mβ t,
2
1
−N−1
−N−1
−N−1
F33
Fα3
Fαβ
t, v 0,
t, v amn u−N
t, v 0,
mα t∂3 vn ,
2
1
−N
−N
F33
amn u−N
1
Fαβ
t, v amn u−N
tvnβ u−N
tvmα ,
mα
nβ
m3 t∂3 vn ,
2
−2N
Eαβ
t 2.75
−N
−N
the last expression of F33
t, v being valid only if N ≥ 2 the expressions of Fα3
t, v are not
−2N
α3στ
0 0 by Lemma 2.2 and Eα3 t 0 by 2.71.
needed since A
−N−1
−N
v F33
0 if ∂3 v 0, we thus conclude that the variational
Noting that Fα||3
equations 2.74 reduce to
Ω
√
−2N
−N
Aαβστ 0Eστ
tFαβ
t, v adx
t
Ω
−∞
A
αβστ
t −
−2N
−N
τEστ
τFαβ
τ, v
√
2.76
adx dτ L
−3N
t, v
for all v ∈ WΩ that are independent of x3 .
−2N
−N
Since each term in the sum Aαβστ 0Eστ
tFαβ
t, v is cubic with respect to the
t, the linearization trick second requirement implies that L−3N t, v 0 for
functions u−N
mα
all v ∈ WΩ that are independent of x3 . Hence, we must let f i,−3N t 0 and hi,−3N
1 t 0.
Hence
Ω
√
−2N
−N
Aαβστ 0Eστ
tFαβ
t, v adx t
Ω
−∞
A
αβστ
√
−2N
−N
t − τEστ
τFαβ
τ, v adx dτ 0.
2.77
16
Abstract and Applied Analysis
In a similar way as in iv, we can obtain from 2.77 that
t −∞
Ω
−2N
−N
Aαβστ t − τEστ
τFαβ
τ, vdx dτ 0.
2.78
Recalling that u−N t is independent of x3 by 2.51, we may let v u−N τ in 2.78.
This gives
t −∞
Ω
−2N
−2N
Aαβστ t − τEστ
τEαβ
τdx dτ 0
2.79
−N
−2N
τ, u−N τ 2Eαβ
t by 2.75. But see Lemma 2.2
since Fαβ
Aαβστ 0 λ0aαβ aστ μ0 aασ aβτ aατ aβσ
2.80
and thus the matrix amn is positive definite
−2N
Eαβ
t 1 mn −N
a umα tu−N
nβ t 0 in Ω
2
2.81
to reach this conclusion, observe that aασ aβτ tστ tαβ ≥ 0 and that aασ aβτ tστ tαβ 0 only if tαβ 0
by ii; these relations in turn imply that
u−N
mα t 0.
2.82
By definition see 2.36 and Lemma 2.2,
p,0
−N
−N
−N
σ −N
−N
u−N
βα t ∂α uβ t − Γαβ up t ∂α uβ t − Γαβ uσ t − bαβ u3 t,
u−N
3α t
∂α u−N
3 t
−
p,0
Γα3 u−N
p t
∂α u−N
3 t
bασ u−N
σ t.
2.83
−N
∞
∞
Let ζi t u−N
i t|x3 0 . Then, ζi t ∈ L 0, T ; Wω since ui t ∈ L 0, T ; WΩ and
−N
−N
∂3 ui t 0 in Ω and ζi t 0 on γ0 since ui t 0 on Γ0 . The above relations combined
with the Gauss and Weingarten formulas Lemma 2.3 then imply that ∂α ζi tai 0 in ω
and hence that ζi t 0. We have thus shown that
u−N t 0
∀N ≥ 2.
2.84
vii Finally, assume that N 1. The only difference from vi is that now
−1
F33
t, v ∂3 v3 amn u0m3 t∂3 vn .
2.85
Abstract and Applied Analysis
17
But since the arguments that led in vi to the conclusion that u−N 0 for N ≥ 2 only required
−1
t, v 0,
that consideration of functions v ∈ W that are independent of x3 , in which case F33
they can be reproduced verbatim for N 1, thus showing that
u−1 t 0.
2.86
The proof is complete.
3. The Proof of the Main Result
Proof. The proof comprises three parts.
i Using Lemma 2.5, uεt can be expanded as
uεt u−N t u−N
1 t · · · ,
3.1
with u−N ∈ L∞ 0, T ; WΩ. Letting N 0, we thus infer that ∂3 u0 t 0 in Ω, that
ζ0 t :
1
2
1
−1
Wω η ∈ W1,4 ω; η 0 on γ0 ,
u0 tdx3 ∈ L∞ 0, T ; Wω,
3.2
and also that see 2.54 and 2.55
q
Eij t 0
∀ integers q ≤ −1,
3.3
q
Fij t, v 0 ∀ integers q ≤ −2 ∀v ∈ WΩ,
and, finally, that we must let f i,−2 t 0 and hi,−1 t 0.
ii Our next try is thus
f i εt 1 i,−1
f t,
ε
hi εt hi,0 t,
3.4
where it is understood as in the proof of Lemma 2.5 that each function f i,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
L t, v :
r
Ω
√
f tvi adx i,r
Γ
∪Γ−
√
hi,r
1 tvi adΓ,
r ≥ −1.
3.5
The cancellation of the coefficient of ε−1 in the variational equations of problem 2.15
then yields the equations:
Ω
√
0
−1
Aijkl 0Ekl
tFij
t, v adx t
Ω
−∞
A
ijkl
√
0
−1
t − τEkl
τFij
τ, v adx dt L−1 t, v
3.6
18
Abstract and Applied Analysis
for all v ∈ WΩ, where
1 0
uαβ t u0βα t amn u0mα tu0nβ t ,
2
1 0
0
0
Eα3
uα3 t u03α t amn u0mα tun3 t ,
t 2
1
0
0
0
0
E33
t u33 t amn um3 tun3 t,
2
1
−1
−1
∂3 vα amn u0mα t∂3 vn ,
Fα3
Fαβ
t, v 0,
t, v 2
0
Eαβ
t 3.7
0
−1
F33
t, v ∂3 v3 amn um3 t∂3 vn ,
0
p,0
u0mα t : ∂α u0m t − Γαm u0p t,
p,0
um3 t : ∂3 u1m t − Γm3 u0p t.
3.8
0
The special notation um3 t emphasizes that, by contrast with the functions u0mα t,
0
which only depend on u0 t, the functions um3 t also depend on u1 t.
−1
t, v imply that L−1 t, v 0 for all for all
The expressions of the functions Fij
v ∈ WΩ that are independent of x3 . Hence, we must let f i,−1 t 0 and hi,0 t 0 first
requirement, so that we are left with the equations
Ω
√
0
−1
Aijkl 0Ekl
tFij
t, v adx t
Ω
−∞
A
ijkl
√
0
−1
t − τEkl
τFij
τ, v adx dt 0
3.9
−1
for all v ∈ WΩ. When the functions Fij
t, v are replaced by their expression given in 3.7,
the integrand in 3.9 takes the form wτ ∂3 vτ w3 ∂3 v3 .
Then, Lemma 2.4 shows that the functions wτ and w3 vanish in Ω, that is,
0
0
0
0
λ0aαβ Eαβ
t λ0 2μ0 E33
t aστ uσ3 t 2μ0Eα3
t aατ aασ aβτ u0βσ t
t 0
0
0
λ t − saαβ Eαβ
s λ t − s 2μ t − s E33
s aστ uσ3 s
−∞
0
2μ t − sEα3
s aατ aασ aβτ u0βσ s ds 0 in Ω, τ 1, 2,
0
0
0
0
λ0aαβ Eαβ
t λ0 2μ0 E33
t 1 u33 t 2μ0aασ Eα3
tu03σ t
t 0
0
0
λ t − saαβ Eαβ
s λ t − s 2μ t − s E33
s 1 u33 s
−∞
0
2μ t − saασ Eα3
su03σ s ds 0
in Ω,
3.10
Abstract and Applied Analysis
19
that is,
t −∞
t −∞
0
0
0
λt − τaαβ Eαβ
τ λt − s 2μt − s E33
s aστ uσ3 s
0
2μt − sEα3
s aατ aασ aβτ u0βσ s ds 0 in Ω, τ 1, 2,
λt − sa
αβ
0
Eαβ
s
0
0
λt − s 2μt − s E33
s 1 u33 s
0
2μt − saασ Eα3
su03σ s ds 0
3.11
in Ω.
Under the conditions of integral mean value theorem, one obvious solution to this
system of three equations is
0
Eα3
t 0 in Ω,
αβ 0
0
Eαβ s λs
λsa
2
μs E33
s 0 in Ω,
3.12
0
0
s denote the Laplace transformation of λt, μt, and Eαβ
t c.f. 11.
λs,
μ
s, and Eαβ
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
lin
0
0
Eα3
eα3
t
s,
lin
αβ 0
0
αβ 0
0
s
λsa
eαβ s λs
Eαβ s λs
λsa
2μt − s E33
2
μs e33
s,
3.13
0
0
by definition of the functions Ei3
t and ei3
t as the coefficient of ε0 in the formal expansions
of the functions Ei3 ε, uεt and ei3 ε, uεt, the latter being precisely the linear part in
2.
Since it was found in the linear case see 2 that
0
eα||3
t 0 in Ω,
αβ 0
0
eαβ s λs
λsa
2
μs e33
s 0
in Ω,
3.14
the linearization trick second requirement suggests that we only retain the “obvious”
solution found above.
iii Our next try is thus
f i εt f i,0 t,
hi εt εhi,1 t.
3.15
20
Abstract and Applied Analysis
The cancellation of the coefficient of ε0 in the variational equations of problem 2.15
then leads to the equations
Ω
√
u0itt tvj aij adx t
Ω
Ω
−∞
B
A
ijkl,1
ijkl
Ω
√
0
0
1
−1
Aijkl 0 Ekl
Fij
tFij
t, v Ekl
t, v adx
√
0
0
1
−1
t − τ Ekl
τFij
τ, v Ekl
τFij
τ, v adx dτ
0
−1
0Ekl
tFij
t, vdx
t
Ω
−∞
B
ijkl,1
0
−1
t − τEkl
τFij
τ, vdx dτ L0 t, v
3.16
0
1
for all v ∈ WΩ, where the functions Eij
t and Fij
t are defined by means of formal
expansions
0
1
Eij ε; uεt Eij
t εEij
t · · · ,
Fij ε; uεt, v 1 −1
0
F t, v Fij
t, v · · · .
ε ij
3.17
−1
0
0
Note that, while the functions Eij
t, Fij
t, v, and Fij
t, v depend only on u0 t and
1
u1 t, the functions Eij
t depend also on u2 t but not on u3 t; each term involving u3 t
vanishes because it contains some derivative ∂3 u0m t as a factor. For this reason the formal
asymptotic expansion of uεt must be “at least” of the form
uεt 2
εq uq t · · · .
3.18
q0
In particular then, we must have by 3.7
Ω
√
u0itt tvj aij adx t
Ω
−∞
A
ijkl
Ω
t −
√
0
0
Aijkl 0Ekl
tFij
t, v adx
0
0
τEkl
τFij
τ, v
√
3.19
adx dτ L t, v
0
−1
t, v 0 for such functions; equivalently,
for all v ∈ WΩ that are independent of x3 since Fij
after performing the usual identification, we must have
Ω
u0itt tηj aij
√
adx t
Ω
−∞
A
ijkl
t −
Ω
√
0
0
Aijkl 0Ekl
tFij
t, η adx
0
0
τEkl
τFij
τ, η
for all η ∈ Wω {η ∈ W1,4 ω; η 0 on γ0 }.
√
3.20
adx dτ L t, v
0
Abstract and Applied Analysis
21
Using Lemma 2.2 and 3.12, 3.20 can be written as
Ω
√
u0itt ηj aij adx √
0
0
0
Aαβστ 0Eστ
t Aαβ33 0E33
t Fαβ
t, η adx
Ω
t √
αβστ
αβ33
0
0
0
A
t − τEστ
τ A
t − τE33
τ Fαβ
τ, η adτ dx
Ω
−∞
√
0
0
0
A33στ 0Eστ
t A3333 0E33
t F33
t, η adx
Ω
3.21
t √
33στ
3333
0
0
0
A
t − τEστ
τ A t − τE33
τ F33
τ, η adτ dx
Ω
−∞
Ω
√
f tηi adx i,0
Γ
∪Γ−
√
hi,1 tηi adΓ
for all η ∈ Wω {η ∈ W1,4 ω; η 0 on γ0 }, that is,
Ω
√
u0itt ηj aij adx
Ω
∂
∂t
Ω
∂
∂t
−∞
t
−∞
√
αβστ
0
αβ33
0
0
A
adx
t − τEστ τ A
t − τE33 τ Fαβ τ, ηdτ
√
33στ
0
3333
0
0
A
adx
t − τEστ τ A t − τE33 τ F33 τ, ηdτ
√
f tηi adx i,0
Ω
t
Γ
∪Γ−
3.22
√
hi,1 tηi adΓ.
Setting
Ct :
t 0
0
0
Aαβστ t − τEστ
τ Aαβ33 t − τE33
τ Fαβ
τ, ηdτ
−∞
t 0
0
0
A33στ t − τEστ
τ A3333 t − τE33
τ F33
τ, ηdτ,
3.23
−∞
we have
αβ33 sE0 s ∗ F0 s, η
αβστ sE0 s ∗ F0 s, η A
Cs
A
στ
αβ
33
αβ
33στ sE0 s ∗ F0 s, η A
3333 sE0 ∗ F0 s, η,
A
στ
33
33
33
3.24
22
Abstract and Applied Analysis
where ∗ denotes convolution. Substituting 3.12 into 3.24, we get
αβστ sE0 s ∗ F0 s, η −
Cs
A
στ
αβ
λs
λs
2
μs
33στ sE0 s ∗ F0 s, η −
A
στ
33
αβ33 saστ E0 s ∗ F0 s, η
A
στ
αβ
λs
λs
2
μs
3333 saστ E0 s ∗ F0 s, η
A
στ
33
2
μsλs
0
0
0
0
aαβ aστ Eστ
μ
s aασ aβτ aατ aβσ Eστ
s ∗ Fαβ
s, η s ∗ Fαβ
s, η.
λs
2
μs
3.25
Applying the inverse Laplace transformation to 3.25, we obtain
Ct t
−∞
0
0
aαβστ t − τEστ
τFαβ
τ, ηdτ,
3.26
where
at : L
−1
aαβστ t : ataαβ aστ
μs
2λs
,
λs
2
μs
μt aασ aβτ aατ aβσ .
3.27
Inserting 3.26 into 3.22, we get the equation in Theorem 2.1.
Since u0 t is independent of x3 , it may be identified with a function ζ0 t ∈
L∞ 0, T ; WΩ. Consequently, the functions
1 0
uβα t u0βα t amn u0mα tu0nβ t ∈ L∞ 0, T ; L2 Ω ,
2
1
0
Fαβ
∈ L∞ 0, T ; L2 Ω ,
ηαβ ηβα amn u0mα tηnβ u0nβ tηmα
t, η :
2
0
Eαβ
t :
3.28
which are thus also independent of x3 , may be likewise identified with functions denoted
for convenience by the same symbols
1 0
0
0
0
ζαβ t ζβα
t amn ζmα
tζnβ
t ∈ L∞ 0, T ; L2 ω ,
2
1
0
0
0
Fαβ
ηαβ ηβα amn ζmα
∈ L∞ 0, T ; L2 ω ,
t, η :
tηnβ ζnβ
tηmα
2
0
Eαβ
t :
3.29
where
ηαβ ∂β ηα − Γσαβ ησ − bαβ η3 ,
η3η : ∂3 η3 bβσ ησ ,
for all η ∈ Wω. The last variational problem is thus indeed two-dimensional.
3.30
Abstract and Applied Analysis
23
The definition of at implies
μs.
as λs
2
μs 2λs
3.31
Applying the inverse Laplace transformation to 3.31, we get
t
at − τ λτ 2μτ dτ 0
t
2λt − τμτdτ.
3.32
0
Therefore
a0 λt 2μt t
a t − τ λτ 2μτ dτ 2λ0μt 0
t
2λ t − τμτdτ.
3.33
0
Letting t 0 in 3.33, we obtain immediately that
a0 2λ0μ0
.
λ0 2μ0
3.34
Acknowledgments
This paper is supported by the National Natural Science Foundation of China 10871116, the
Natural Science Foundation of Shandong Province of China ZR2011AM008, ZR2011AQ006,
the China Postdoctoral Science Foundation 20090451291, the Postdoctoral Science Foundation of Shandong Province of China 20093043, and STPF of university in Shandong Province
of China J09LA04.
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Abstract and Applied Analysis
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