# On a Nonlinear Wave Equation Associated with Dirichlet Conditions: Solvability and Asymptotic Expansion of Solutions in Many Small Parameters

Authors: Le Thi Phuong Ngoc; Le Khanh Luan; Tran Minh Thuyet; Nguyen Thanh Long
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

A Dirichlet problem for a nonlinear wave equation is investigated. Under suitable assumptions, we prove the solvability and the uniqueness of a weak solution of the above problem. On the other hand, a high-order asymptotic expansion of a weak solution in many small parameters is studied. Our approach is based on the Faedo-Galerkin method, the compact imbedding theorems, and the Taylor expansion of a function.


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## Body

## 1. Introduction

In this paper, we consider the following Dirichlet problem:(1.1)utt-∂∂x(μ(x,t,u)ux)=f(x,t,u,ux,ut),0<x<1,0<t<T,(1.2)u(0,t)=u(1,t)=0,(1.3)u(x,0)=ũ0(x),ut(x,0)=ũ1(x),
where ũ0,ũ1,μ, and f are given functions satisfying conditions specified later.In the special cases, when the functionμ(x,t,u) is independent of u,μ(x,t,u)≡1, or μ(x,t,u)=μ(x,t), and the nonlinear term f has the simple forms, the problem (1.1), with various initial-boundary conditions, has been studied by many authors, for example, Ortiz and Dinh [1], Dinh and Long [2, 3], Long and Diem [4], Long et al. [5], Long and Truong [6, 7], Long et al. [8], Ngoc et al. [9], and the references therein.Ficken and Fleishman [10] and Rabinowitz [11] studied the periodic-Dirichlet problem for hyperbolic equations containing a small parameterε, in particular, the differential equation(1.4)utt-uxx=2αut+εf(t,x,u,ut,ux).In [12], Kiguradze has established the existence and uniqueness of a classical solution u∈C2([0,a]×ℝn) of the periodic-Dirichlet problem for the following nonlinear wave equation:(1.5)utt-uxx=g(t,x,u)+g1(u)ut,
under the assumption that g and g1 are continuously differentiable functions (these conditions are sharp and cannot be weakened). Moreover, it is shown that the same results are valid for the equation(1.6)utt-uxx=g(t,x,u)+g1(u)ut+εq(t,x,u,ut,ux),
with sufficiently small ε and continuously differentiable q.In [13], a unified approach to the previous cases was presented discussing the existence unique and asymptotic stability of classical solutions for a class of nonlinear continuous dynamical systems.In [8], Long et al. have studied the linear recursive schemes and asymptotic expansion for the nonlinear wave equation(1.7)utt-uxx=f(x,t,u,ux,ut)+εf1(x,t,u,ux,ut),
with the mixed nonhomogeneous conditions(1.8)ux(0,t)-h0u(0,t)=g0(t),u(1,t)=g1(t).In the case ofg0,g1∈C3(ℝ+),f∈CN+1([0,1]×ℝ+×ℝ3),f1∈CN([0,1]×ℝ+×ℝ3), and some other conditions, an asymptotic expansion of the weak solution uε of order N+1 in ε is considered.This paper consists of four sections. In Section2, we present some preliminaries. Using the Faedo-Galerkin method and the compact imbedding theorems, in Section 3, we prove the solvability and the uniqueness of a weak solution of the problem (1.1)–(1.3). In Section 4, based on the ideals and the techniques used in the above-mentioned papers, we study a high-order asymptotic expansion of a weak solution for  the problem (1.1)–(1.3), where (1.1) has the form of a linear wave equation with nonlinear perturbations containing many small parameters.In order to avoid making the treatment too complicated without losing of generality, at first, an asymptotic expansion of a weak solution u=uε1,ε2(x,t) of order N+1 in two small parameters ε1,ε2 for the following equation:(1.9)utt-∂∂x([μ0(x,t)+ε1μ1(x,t,u)]ux)=f0(x,t)+ε2f1(x,t,u,ux,ut),
associated with (1.2), (1.3), with μ0∈C2([0,1]×ℝ+),μ1∈CN+1([0,1]×ℝ+×ℝ),μ0(x,t)≥μ*>0,μ1(x,t,z)≥0, for all (x,t,z)∈[0,1]×ℝ+×ℝ,f0∈C1([0,1]×ℝ+), and f1∈CN([0,1]×ℝ+×ℝ3) is established. Next, we note that the same results are valid for the equation in p small parameters ε1,…,εp as follows(1.10)utt-∂∂x[(μ0(x,t)+∑i=1pεiμi(x,t,u))ux]=f0(x,t)+∑i=1pεifi(x,t,u,ux,ut),
associated with (1.2), (1.3). The result obtained here is a relative generalization of [5–7, 14], where asymptotic expansion of a weak solution in two or three small parameters is given.


## 2. Preliminaries

PutΩ=(0,1). Let us omit the definitions of usual function spaces that will be used in what follows such as Lp=Lp(Ω),Hm=Hm(Ω),H0m=H0m(Ω). The norm in L2 is denoted by ∥·∥. We denote by 〈·,·〉 the scalar product in L2 or a pair of dual products of continuous linear functional with an element of a function space. We denote by ∥·∥X the norm of a Banach space X and by X′ the dual space of X. We denote Lp(0,T;X),1≤p≤∞, the Banach space of real functions u:(0,T)→X measurable, such that ∥u∥Lp(0,T;X)<+∞, with (2.1)∥u∥Lp(0,T;X)={(∫0T∥u(t)∥Xpdt)1/p,if1≤p<∞,esssup⁡0<t<T∥u(t)∥X,ifp=∞.Letu(t),u′(t)=ut(t)=u̇(t),u′′(t)=utt(t)=ü(t),ux(t)=∇u(t),uxx(t)=Δu(t) denote u(x,t),∂u/∂t(x,t),∂2u/∂t2(x,t),∂u/∂x(x,t),∂2u/∂x2(x,t), respectively. With f∈Ck([0,1]×ℝ+×ℝ3),f=f(x,t,u,v,w), we put D1f=∂f/∂x,D2f=∂f/∂t,D3f=∂f/∂u,D4f=∂f/∂v,D5f=∂f/∂w and Dαf=D1α1D2α2D3α3D4α4D5α5f; α=(α1,α2,α3,α4,α5)∈ℤ+5, |α|=α1+α2+α3+α4+α5=k, D(0,0,…,0)f=f.Similarly, withμ∈Ck([0,1]×ℝ+×ℝ),μ=μ(x,t,z), we put D1μ=∂μ/∂x,D2μ=∂μ/∂t,D3μ=∂μ/∂z and Dβμ=D1β1D2β2D3β3,β=(β1,β2,β3)∈ℤ+3,|β|=β1+β2+β3=k.OnH1, we will use the following norms:(2.2)∥v∥H1=(∥v∥2+∥vx∥2)1/2.Then, we have the following lemma.Lemma 2.1.
The imbeddingH1↪C0(Ω¯) is compact and
(2.3)∥v∥C0(Ω¯)≤2∥v∥H1∀v∈H1.The proof of Lemma2.1 is easy, hence we omit the details.Remark 2.2.
OnH01,v↦∥v∥H1 and v↦∥vx∥ are two equivalent norms. Furthermore, we have the following inequalities:
(2.4)∥v∥C0(Ω¯)≤∥vx∥∀v∈H01.Remark 2.3.
(i) Let us note more that a unique weak solutionu of the problem (1.1)–(1.3) will be obtained in Section 3 (Theorem 3.2) in the following manner.
Findu∈W̃={u∈L∞(0,T;H01∩H2):u′∈L∞(0,T;H01),u′′∈L∞(0,T;L2)} such that u verifies the following variational equation:
(2.5)〈u′′(t),w〉+〈μ(⋅,t,u(t))ux(t),wx〉=〈f(⋅,t,u(t),ux(t),u′(t)),w〉,∀w∈H01,
and the initial conditions
(2.6)u(0)=ũ0,u′(0)=ũ1.
(ii) With the regularity obtained byu∈W̃, it also follows from Theorem 3.2 that the problem (1.1)–(1.3) has a unique strong solution u that satisfies
(2.7)u∈C0(0,T;H1)∩C1(0,T;L2)∩L∞(0,T;H2),ut∈L∞(0,T;H1),utt∈L∞(0,T;L2).On the other hand, byu∈W̃, we can see that u,ux,ut,uxx,uxt,utt∈L∞(0,T;L2)⊂L2(QT).Also, if(u0,u1)∈(H01∩H2)×H01, then the weak solution u of the problem (1.1)–(1.3) belongs to H2(QT). So, the solution is almost classical which is rather natural, since the initial data (u0,u1) do not belong necessarily to C2(Ω¯)×C1(Ω¯).


## 3. The Existence and the Uniqueness of a Weak Solution

We make the following assumptions:(H1)
ũ0∈H01∩H2,ũ1∈H01,(H2)
μ∈C2([0,1]×ℝ+×ℝ),μ(x,t,z)≥μ*>0,forall(x,t,z)∈[0,1]×ℝ+×ℝ,(H3)
f∈C1(Ω¯×ℝ+×ℝ3).Withμ and f satisfying the assumptions (H2) and (H3), respectively, for each T*>0 and M>0 are given, we put the following constants:(3.1)K̃M(μ)=‖μ‖C2(D̃M*),(3.2)KM(f)=‖f‖C1(DM*),
where D̃M*={(x,t,z):0≤x≤1,0≤t≤T*,|z|≤M} and DM*={(x,t,u,v,w)∈ℝ+×ℝ+×ℝ3:0≤x≤1,0≤t≤T*,|u|,|v|,|w|≤M}.For eachT∈(0,T*] and M>0, we get(3.3)W(M,T)={v∈L∞(0,T;H01∩H2):vt∈L∞(0,T;H01),vtt∈L2(QT),with∥v∥L∞(0,T;H01∩H2),∥vt∥L∞(0,T;H01),∥vtt∥L2(QT)≤M},(3.4)W1(M,T)={v∈W(M,T):vtt∈L∞(0,T;L2)},
where QT=Ω×(0,T).We choose the first termu0≡ũ0∈W1(M,T). Suppose that(3.5)um-1∈W1(M,T),m≥1.The problem (1.1)–(1.3) is associated with the following variational problem.Findum∈W1(M,T) such that(3.6)〈um′′(t),v〉+〈μm(t)∇um(t),∇v〉=〈Fm(t),v〉,∀v∈H01,(3.7)um(0)=ũ0,um′(0)=ũ1,
where(3.8)μm(x,t)=μ(x,t,um-1(t)),Fm(x,t)=f(x,t,um-1(x,t),∇um-1(x,t),um-1′(x,t)).Then, we have the following theorem.Theorem 3.1.
Let (H1)–( H3) hold. Then, there exist two constants M>0,T>0 and the linear recurrent sequence  {um}⊂W1(M,T) defined by (3.6)–(3.8).Proof.
The proof consists of three steps.
Step 1.
The Faedo-Galerkin approximation (introduced by Lions [15]).
Consider a special basis{wj} on H01:wj(x)=2sin(jπx),j∈ℕ, formed by the eigenfunctions of the Laplacian -Δ=-∂2/∂x2. Put
(3.9)um(k)(t)=∑j=1kcmj(k)(t)wj,
where the coefficients cmj(k) satisfy the system of linear differential equations
(3.10)〈üm(k)(t),wj〉+〈μm(t)∇um(k)(t),∇wj〉=〈Fm(t),wj〉,1≤j≤k,(3.11)um(k)(0)=ũ0k,u̇m(k)(0)=ũ1k,
where
(3.12)ũ0k=∑j=1kαj(k)wj⟶ũ0stronglyinH01∩H2,ũ1k=∑j=1kβj(k)wj⟶ũ1stronglyinH01.
Note that by (3.5), it is not difficult to prove that the system (3.10), (3.11) has a unique solution um(k)(t) on interval [0,T], so let us omit the details.
Step 2.
A priori estimates. At first, put
(3.13)sm(k)(t)=pm(k)(t)+qm(k)(t)+∫0t‖üm(k)(s)‖2ds,pm(k)(t)=‖u̇m(k)(t)‖2+‖μm(t)∇um(k)(t)‖2,qm(k)(t)=‖∇u̇m(k)(t)‖2+‖μm(t)Δum(k)(t)‖2.
Then, it follows from (3.9)–(3.11), (3.13) that
(3.14)sm(k)(t)=sm(k)(0)+2〈∇μm(0)∇ũ0k,Δũ0k〉+2〈Fm(0),Δũ0k〉+∫0tds∫01μm′(x,s)(|∇um(k)(x,s)|2+|Δum(k)(x,s)|2)dx+2∫0t〈Fm(s),u̇m(k)(s)〉ds+2∫0t〈∂∂s(∇μm(s)∇um(k)(s)),Δum(k)(s)〉ds-2〈∇μm(t)∇um(k)(t),Δum(k)(t)〉-2〈Fm(t),Δum(k)(t)〉+2∫0t〈∂Fm∂t(s),Δum(k)(s)〉ds+∫0t‖üm(k)(s)‖2ds=qm(k)(0)+2〈∇μm(0)∇ũ0k,Δũ0k〉+2〈Fm(0),Δũ0k〉+∑j=17Ij.
Next, we will estimate the termsIj,j=1,2,…,7 on the right-hand side of (3.14) as follows.
First TermI1
We have(3.15)μm′(t)=D2μ(x,t,um-1(t))+D3μ(x,t,um-1(t))um-1′(t).
From (3.1), (3.5), and (3.8), we have
(3.16)|μm′(x,t)|≤(1+M)K̃M(μ).
Hence,(3.17)I1=∫0tds∫01μm′(x,s)(|∇um(k)(x,s)|2+|Δum(k)(x,s)|2)dx≤1+Mμ*K̃M(μ)∫0tsm(k)(s)ds.
Second Term
By using(H3), we obtain from (3.2), (3.5), and (3.13)2 that
(3.18)I2=2∫0t〈Fm(s),u̇m(k)(s)〉ds≤TKM2(f)+∫0tpm(k)(s)ds.Third Term
The Cauchy-Schwartz inequality yields(3.19)|I3|=2|∫0t〈∂∂s(∇μm(s)∇um(k)(s)),Δum(k)(s)〉ds|≤2μ*∫0trm(k)(s)qm(k)(s)ds,
where rm(k)(s)=∥∂/∂s(∇μm(s)∇um(k)(s))∥.
We note(3.20)rm(k)(s)=‖∇μm(s)∇u̇m(k)(s)+∂∂s(∇μm(s))∇um(k)(s)‖≤(‖∇μm(s)‖C0(Ω¯)+1μ*‖∂∂s∇μm(s)‖)sm(k)(s).
On the other hand, by∇μm(x,s)=D1μ(x,s,um-1(x,s))+D3μ(x,s,um-1(x,s))∇um-1(x,s), it is implies that
(3.21)‖∇μm(s)‖C0(Ω¯)≤K̃M(μ)(1+‖∇um-1(s)‖C0(Ω¯))≤2(1+M)K̃M(μ).
Similarly, the following equality(3.22)∂∂s∇μm(x,s)=D1D1μ(x,s,um-1(x,s))+D3D1μ(x,s,um-1(x,s))um-1′(x,s)+[D1D3μ(x,s,um-1(x,s))+D3D3μ(x,s,um-1(x,s))um-1′(x,s)]∇um-1(x,s)+D3μ(x,s,um-1(x,s))∇um-1′(x,s)
gives
(3.23)‖∂∂s∇μm(s)‖≤(1+3M+M2)K̃M(μ).
It follows from (3.20)–(3.23) that
(3.24)rm(k)(s)≤[2(1+M)+1+3M+M2μ*]K̃M(μ)sm(k)(s).
Hence, we obtain from (3.19) and (3.24) that
(3.25)|I3|≤2μ*[2(1+M)+1+3M+M2μ*]K̃M(μ)∫0tsm(k)(s)ds.
Fourth TermI4
By the Cauchy-Schwartz inequality, we have(3.26)|I4|=|-2〈∇μm(t)∇um(k)(t),Δum(k)(t)〉|≤1β‖∇μm(t)∇um(k)(t)‖2+β‖Δum(k)(t)‖2,
for all β>0. On the other hand
(3.27)‖∇μm(t)∇um(k)(t)‖=‖∇μm(0)∇ũ0k+∫0t∂∂s(∇μm(s)∇um(k)(s))ds‖≤‖∇μm(0)‖C0(Ω¯)‖∇ũ0k‖+∫0trm(k)(s)ds.
Hence, we obtain from (3.26), (3.27) that
(3.28)|I4|≤βμ*qm(k)(t)+2β∥∇μm(0)∥C0(Ω¯)2∥∇ũ0k∥2+2βT[2(1+M)+1+3M+M2μ*]2K̃M2(μ)∫0tsm(k)(s)ds,
for all β>0.
Fifth TermI5
By (3.5), (3.8), and (3.13), we obtain
(3.29)|I5|=|-2〈Fm(t),Δum(k)(t)〉|≤1β‖Fm(t)‖2+β‖Δum(k)(t)‖2≤2β‖Fm(0)‖2+2βT∫0T‖∂Fm∂s(s)‖2ds+βμ*sm(k)(t),∀β>0.
Note that(3.30)∂Fm∂t(t)=D2f[um-1]+D3f[um-1]um-1′(t)+D4f[um-1]∇um-1′(t)+D5f[um-1]um-1′′(t),
where we use the notation Dif[um-1]=Dif(x,t,um-1(x,t),∇um-1(x,t),um-1′(x,t)),i=2,…,5. By (3.2), (3.5), and (3.30), we obtain
(3.31)‖∂Fm∂t(t)‖≤KM(f)(1+2M+‖um-1′′(t)‖).
Hence, we deduce from (3.29) and (3.31) that
(3.32)|I5|≤2β∥Fm(0)∥2+4βTKM2(f)[(1+2M)2T+M2]+βμ*sm(k)(t),∀β>0.
Sixth TermI6
By (3.2), (3.5), (3.13)3, and (3.31), we get
(3.33)|I6|=2|∫0t〈∂Fm∂t(s),Δum(k)(s)〉ds|≤∫0t‖∂Fm∂t(s)‖ds+∫0t‖∂Fm∂t(s)‖‖Δum(k)(s)‖2ds≤KM(f)[(1+2M)T+T(∫0T‖um-1′′(s)‖2ds)1/2]+1μ*KM(f)∫0t(1+2M+‖um-1′′(s)‖)qm(k)(s)ds≤KM(f)[(1+2M)T+TM]+1μ*KM(f)∫0t(1+2M+‖um-1′′(s)‖)qm(k)(s)ds.Seventh TermI7
Equation (3.10) is rewritten as follows:
(3.34)〈üm(k)(t),wj〉-〈∂∂x(μm(t)∇um(k)(t)),wj〉=〈Fm(t),wj〉,1≤j≤k.
Hence, by replacingwj with üm(k)(t) and integrating
(3.35)I7=∫0t‖üm(k)(s)‖2ds≤2∫0t‖∂∂x(μm(s)∇um(k)(s))‖2ds+2∫0t‖Fm(s)‖2ds≤2∫0t‖∂∂x(μm(s)∇um(k)(s))‖2ds+2TKM2(f),
we need, estimate ∥∂/∂x(μm(s)∇vm(k)(s))∥.
Combining (3.1), (3.5), and (3.13) yields
(3.36)‖∂∂x(μm(s)∇um(k)(s))‖=‖∇μm(s)∇um(k)(s)+μm(s)Δum(k)(s)‖≤‖∇μm(s)‖C0(Ω¯)‖∇um(k)(s)‖+‖μm(s)‖C0(Ω¯)‖Δum(k)(s)‖≤2μ*(1+M)K̃M(μ)pm(k)(s)+1μ*K̃M(μ)qm(k)(s)≤3μ*(1+M)K̃M(μ)sm(k)(s).
Therefore, from (3.35) and (3.36), we obtain
(3.37)I7≤2TKM2(f)+18μ*(1+M)2K̃M2(μ)∫0tsm(k)(s)ds.
Choosingβ>0, with 2β/μ*≤1/2, it follows from (3.13), (3.14), (3.17), (3.18), (3.25), (3.28), (3.32), (3.33), and (3.37) that
(3.38)sm(k)(t)≤C̃0k+C̃1(M,T)+∫0t(C̃2(M,T)+2μ*KM(f)‖um-1′′(s)‖)sm(k)(s)ds,
where
(3.39)C̃0k=C̃0k(β,f,μ,ũ0,ũ1,ũ0k,ũ1k)=2sm(k)(0)+4〈∇μm(0)∇ũ0k,Δũ0k〉+4〈Fm(0),Δũ0k〉+4β∥∇μm(0)∥C0(Ω¯)2∥∇ũ0k∥2+4β∥Fm(0)∥2,C̃1(M,T)=C̃1(β,f,M,T)=2(3+4β[(1+2M)2T+M2])TKM2(f)+2[M+(1+2M)T]TKM(f),C̃2(M,T)=C̃2(β,f,μ,M,T)=2+2μ0(1+2M)KM(f)+2μ*[(1+4μ*)(1+M)+2(1+3M+M2)]K̃M(μ)+4μ*[1βT(2(1+M)μ*+1+3M+M2)2+9(1+M)2]K̃M2(μ).
By(H1), we deduce from (3.12), (3.39)1 that there exists M>0 independent of m and k, such that
(3.40)C̃0k≤12M2.
Notice that by(H3), we deduce from (3.39)2,3 that
(3.41)lim⁡T→0+C̃1(M,T)=lim⁡T→0+TC̃2(M,T)=0.
So, from (3.39) and (3.41), we can choose T>0 such that
(3.42)(12M2+C̃1(M,T))exp⁡⁡(TC̃2(M,T)+2μ0KM(f)TM)≤M2,(3.43)kT=(1+1μ*)T4KM2(f)+(4+M)2M2K̃M2(μ)eT[1+((1+M)/2μ*)K̃M(μ)]<1.
Finally, it follows from (3.38), (3.40), and (3.42) that
(3.44)sm(k)(t)≤M2exp⁡⁡(-TC̃2(M,T)-2μ0KM(f)TM)+∫0t(C̃2(M,T)+2μ0KM(f)‖um-1′′(s)‖)sm(k)(s)ds.
By using Gronwall's lemma, we deduce from (3.44) that
(3.45)sm(k)(t)≤M2exp⁡(-TC̃2(M,T)-2μ0KM(f)TM)×exp⁡[∫0T(C̃2(M,T)+2μ0KM(f)‖um-1′′(s)‖)ds]≤M2exp⁡(-TC̃2(M,T)-2μ0KM(f)TM)×exp⁡[TC̃2(M,T)+2μ0KM(f)T‖um-1′′‖L2(QT)]≤M2.
Therefore, we have(3.46)um(k)∈W(M,T),∀m,k∈N.
Step 3.
Limiting process.
From (3.46), we can extract from {um(k)} a subsequence still denoted by {um(k)} such that
(3.47)um(k)⟶uminL∞(0,T;H01∩H2)weak*,u̇m(k)⟶um′inL∞(0,T;H01)weak*,üm(k)⟶um′′inL2(QT)weak,
as k→∞, and
(3.48)um∈W(M,T).
Based on (3.47), passing to limit in (3.10), (3.11) as k→∞, we have um satisfying (3.6)–(3.8). On the other hand, it follows from (3.5), (3.6), and (3.47) that
(3.49)um′′=∇μm∇um+μmΔum+f(x,t,um-1,∇um-1,um-1′)∈L∞(0,T;L2).
Hence,um∈W1(M,T), and the proof of Theorem 3.1 is complete.Theorem 3.2.
Let (H1)–( H3) hold. Then, there exist  M>0 and T>0 satisfying   (3.40), (3.42), and  (3.43) such that the problem  (1.1)–(1.3) has a unique weak solution u∈W1(M,T).
Furthermore, the linear recurrent sequence{um} defined by  (3.6)–(3.8) converges to the solution u strongly in the space
(3.50)W1(T)={w∈L∞(0,T;H01):w′∈L∞(0,T;L2)},
with the following estimation:
(3.51)‖um-u‖L∞(0,T;H01)+‖um′-u′‖L∞(0,T;L2)≤CkTm,∀m∈N,
where  kT<1 as in  (3.43) and C is a constant depending only on T,ũ0,ũ1 and kT.Proof.
(i)The existence. First, we note that W1(T) is a Banach space with respect to the norm (see Lions [15])
(3.52)‖w‖W1(T)=‖w‖L∞(0,T;H01)+‖w′‖L∞(0,T;L2).
Next, we prove that{um} is a Cauchy sequence in W1(T). Let vm=um+1-um. Then, vm satisfies the variational problem
(3.53)〈vm′′(t),w〉+〈μm+1(t)∇vm(t),∇w〉=〈∂∂x[(μm+1(t)-μm(t))∇um(t)],w〉+〈Fm+1(t)-Fm(t),w〉,∀w∈H01,vm(0)=vm′(0)=0.
Takingw=vm′ in (3.53)1, after integrating in t, we get
(3.54)Zm(t)=∫0tds∫01μm+1′(x,s)|∇vm(s)|2dx+2∫0t〈Fm+1(s)-Fm(s),vm′(s)〉ds+2∫0t〈∂∂x[(μm+1(s)-μm(s))∇um(s)],vm′(s)〉ds=∑i=13Ji,
in which
(3.55)Zm(t)=‖vm′(t)‖2+‖μm+1(t)∇vm(t)‖2,
and all integrals on the right-hand side of (3.54) are estimated as follows.
First Integral
By (3.16), we obtain
(3.56)|J1|≤|∫0tds∫01μm+1′(x,s)|∇vm(s)|2dx|≤1+Mμ*K̃M(μ)∫0tZm(s)ds.Second Integral
By(H3),
(3.57)‖Fm+1(t)-Fm(t)‖≤2KM(f)[‖∇vm-1(t)‖+‖vm-1′(t)‖]≤2KM(f)‖vm-1‖W1(T),
so
(3.58)|J2|≤2|∫0t〈Fm+1(s)-Fm(s),vm′(s)〉ds|≤4KM(f)‖vm-1‖W1(T)∫0t‖vm′(s)‖ds≤4TKM2(f)‖vm-1‖W1(T)2+∫0tZm(s)ds.Third Integral
Using(H2) again, we get
(3.59)|J3|=2|∫0t〈∂∂x[(μm+1(s)-μm(s))∇um(s)],vm′(s)〉ds|≤∫0t‖∂∂x[(μm+1(s)-μm(s))∇um(s)]‖2ds+∫0tZm(s)ds.
Note that(3.60)∂∂x[(μm+1(s)-μm(s))∇um(s)]=(μm+1(s)-μm(s))Δum(s)+(D1μ[um]-D1μ[um-1])∇um(s)+(D3μ[um]-D3μ[um-1])|∇um(s)|2+D3μ[um-1]∇vm-1(s)∇um(s).
Hence,(3.61)‖∂∂x[(μm+1(s)-μm(s))∇um(s)]‖≤‖μm+1(s)-μm(s)‖C0(Ω¯)‖Δum(s)‖+‖(D1μ[um]-D1μ[um-1])‖C0(Ω¯)‖∇um(s)‖+‖(D1μ[um]-D1μ[um-1])‖C0(Ω¯)‖∇um(t)‖C0(Ω¯)2+‖D3μ[um-1]‖C0(Ω¯)‖∇um(s)‖C0(Ω¯)‖∇vm-1(s)‖.
We also note that(3.62)‖μm+1(s)-μm(s)‖C0(Ω¯)≤K̃M(μ)‖wm-1‖W1(T),‖Diμ[um]-Diμ[um-1]‖C0(Ω¯)≤K̃M(μ)‖wm-1‖W1(T),i=1,3,‖∇um(s)‖C0(Ω¯)≤2‖∇um(s)‖H1≤2∥∇um(s)∥2+‖Δum(s)‖2≤2M,‖D3μ[um]‖C0(Ω¯)≤K̃M(μ),
where we use the notation Diμ[um-1]=Diμ(x,t,um(x,t)),i=1,2,3. Therefore, it implies from (3.61) and (3.62) that
(3.63)‖∂∂x[(μm+1(s)-μm(s))∇um(s)]‖≤(4+M)MK̃M(μ)∥vm-1∥W1(T).
Hence,(3.64)|J3|≤(4+M)2M2TK̃M2(μ)∥vm-1∥W1(T)2+∫0tZm(s)ds.
Combining (3.54)–(3.56), (3.58), and (3.64) yields
(3.65)Zm(t)≤T[4KM2(f)+(4+M)2M2K̃M2(μ)]∥vm-1∥W1(T)2+(2+1+Mμ*K̃M(μ))∫0tZm(s)ds.
Using Gronwall's lemma, (3.65) gives
(3.66)‖vm‖W1(T)≤kT‖vm-1‖W1(T)∀m∈N,
where kT<1 as in (3.43).
Hence, we obtain from (3.66) that
(3.67)‖um+p-um‖W1(T)≤kTm1-kT‖u1-u0‖W1(T)∀m,p∈N,
It follows that{um} is a Cauchy sequence in W1(T). Then, there exists u∈W1(T) such that
(3.68)um⟶ustronglyinW1(T).
On the other hand, from (3.48), we deduce the existence of a subsequence {umj} of {um} such that
(3.69)umj⟶uinL∞(0,T;H01∩H2)weak*,umj′⟶u′inL∞(0,T;H01)weak*,umj′′⟶u′′inL2(QT)weak,(3.70)u∈W(M,T).
Note that(3.71)|μm(x,t)-μ(x,t,u(x,t))|≤K̃M(μ)‖um-1-u‖W1(T),‖Fm(t)-f(⋅,t,u(t),ux(t),u′(t))‖≤2KM(f)‖um-1-u‖W1(T).
Hence, from (3.68) and (3.71), we obtain
(3.72)μm⟶μ(⋅,⋅,u)stronglyinL∞(QT),Fm⟶f(⋅,t,u(t),ux(t),u′(t))stronglyinL∞(0,T;L2).
Finally, passing to limit in (3.6)–(3.8) as m=mj→∞, it implies from (3.68), (3.69), and (3.72) that there exists u∈W(M,T) satisfying the equation
(3.73)〈u′′(t),w〉+〈μ(⋅,t,u(t))ux(t),∇w〉=〈f(⋅,t,u(t),ux(t),u′(t)),w〉,∀w∈H01,u(0)=ũ0,u′(0)=ũ1.
On the other hand, by(H2), we obtain from (3.70), (3.72)2, and (3.73)1 that
(3.74)u′′=D1μ[u]ux+D3μ[u]ux2+μ[u]uxx+f(x,t,u,ux,u′)∈L∞(0,T;L2),
thus u∈W1(M,T), and Step 1 follows.
(ii)The uniqueness of the solution.
Letu1,u2∈W1(M,T) be two weak solutions of the problem (1.1)–(1.3). Then, u=u1-u2 satisfies the variational problem
(3.75)〈u′′(t),w〉+〈μ1(t)ux(t),wx〉=〈∂∂x([μ1(t)-μ2(t)]u2x(t)),w〉+〈F2(t)-F1(t),w〉,∀w∈H01,u(0)=u′(0)=0,μi(t)=μ(x,t,ui(t))≡μ[ui],Fi(t)=f(x,t,ui(t),uix(t),ui′(t)),i=1,2.
We takew=u′ in (3.75)1 and integrate in t to get
(3.76)ρ(t)=∫0tds∫01μ1′(x,s)ux2(x,s)dx+2∫0t〈F1(s)-F2(s),u′(s)〉ds+2∫0t〈∂∂x([μ1(s)-μ2(s)]u2x(s)),u′〉ds≡∑i=13ρi(t),
where
(3.77)ρ(t)=‖u′(t)‖2+‖μ1(t)ux(t)‖2.
We now estimate the terms on the right-hand side of (3.76) as follows:
(3.78)ρ1(t)=∫0tds∫01μ1′(x,s)ux2(x,s)dx≤1μ*(1+M)K̃M(μ)∫0tρ(s)ds≡ρM(1)∫0tρ(s)ds,(3.79)ρ2(t)=2∫0t〈F1(s)-F2(s),u′(s)〉ds≤4KM(f)∫0t(‖ux(s)‖+‖u′(s)‖)‖u′(s)‖ds≤4(1+1μ*)KM(f)∫0tρ(s)ds≡ρM(2)∫0tρ(s)ds,(3.80)ρ3(t)=2∫0t〈∂∂x([μ1(s)-μ2(s)]u2x(s)),u′〉ds≤2∫0t‖∂∂x([μ1(s)-μ2(s)]u2x(s))‖‖u′(s)‖ds.
On the other hand(3.81)∂∂x([μ1(s)-μ2(s)]u2x(s))=[μ1(s)-μ2(s)]u2xx(s)+(D1μ[u1]-D1μ[u2])u2x(s)+(D3μ[u1]-D3μ[u2])u1xu2x+D3μ[u2]uxu2x.
Hence,(3.82)‖∂∂x([μ1(s)-μ2(s)]u2x(s))‖≤‖μ1(s)-μ2(s)‖C0(Ω¯)‖u2xx(s)‖+‖D1μ[u1]-D1μ[u2]‖C0(Ω¯)‖u2x(s)‖+‖D3μ[u1]-D3μ[u2]‖C0(Ω¯)‖u1x(s)‖C0(Ω¯)‖u2x(s)‖C0(Ω¯)+‖D3μ[u2]‖C0(Ω¯)‖ux(s)‖‖u2x(s)‖C0(Ω¯)≤(3+M)MK̃M(μ)‖ux(s)‖.
It follows from (3.80), (3.82) that
(3.83)ρ3(t)≤1μ*(3+M)MK̃M(μ)∫0tρ(s)ds≡ρM(3)∫0tρ(s)ds.
Combining (3.76)–(3.79) and (3.83) yields
(3.84)ρ(t)≤(ρM(1)+ρM(2)+ρM(3))∫0tρ(s)ds.
Using Gronwall's lemma, it follows from (3.84) that ρ≡0 that is, u1≡u2.
Theorem3.2 is proved completely.Remark 3.3.
(i) In the case ofμ≡1,f∈C1(Ω¯×ℝ+×ℝ3) and the boundary condition in [4] standing for (1.2), we obtained some similar results in [4].
(ii) In the case ofμ≡1,f∈C1(Ω¯×ℝ+×ℝ3),f(1,t,u,v,w)=0,forallt≥0,forall(u,v,w)∈ℝ3, and the boundary condition in [8] standing for (1.2), some results as above were given in [8].Remark 3.4.
By Galerkin method, as in Remark2.3, the local existence of a strong solution u∈H2(QT) of the problem (1.1)–(1.3) is proved.In the case ofμ=μ(x,t) and f=f(x,t), obviously, the problem (1.1)–(1.3) is linear. Then, by the same method and applying Banach's theorem [16, Chapter 5, Theorem 17.1], it is not difficult to prove that the problem (1.1)–(1.3) is global solvability. To strengthen some hypotheses, it is possible to prove existence of a classical solution u∈C2(QT)∩C1(Q¯T).


## 4. Asymptotic Expansion of a Weak Solution in Many Small Parameters

In this section, we will study a high-order asymptotic expansion of a weak solution for the problem (1.1)–(1.3), in which (1.1) has the form of a linear wave equation with nonlinear perturbations containing many small parameters.The Problem with Two Small Parameters
At first, we consider the case of the nonlinear perturbations containing two small parameters.
Let(H1) hold. We make the following assumptions: (H4)
μ0∈C2([0,1]×ℝ+),μ1∈CN+1([0,1]×ℝ+×ℝ),μ0≥μ*>0,μ1≥0,(H5)
f0∈C1([0,1]×ℝ+),f1∈CN([0,1]×ℝ+×ℝ3).
We consider the following perturbed problem, whereε1,ε2 are two small parameters such that 0≤εi≤εi*<1,i=1,2:
(Pε⃗)utt-∂∂x(με1(x,t,u)ux)=Fε2(x,t,u,ux,ut),0<x<1,0<t<T,u(0,t)=u(1,t)=0,u(x,0)=ũ0(x),ut(x,0)=ũ1(x),με1(x,t,u)=μ0(x,t)+ε1μ1(x,t,u),Fε2(x,t,u,ux,ut)=f0(x,t)+ε2f1(x,t,u,ux,ut).
By Theorem3.2, the problem (Pε⃗) has a unique weak solution u depending on ε⃗=(ε1,ε2):uε⃗=u(ε1,ε2). When ε⃗=(0,0), (Pε⃗) is denoted by (P0). We will study the asymptotic expansion of uε⃗ with respect to ε1,ε2.
We use the following notations. For a multi-indexα=(α1,α2)∈ℤ+2, and ε⃗=(ε1,ε2)∈ℝ2, we put
(4.1)|α|=α1+α2,α!=α1!α2!,‖ε⃗‖=ε12+ε22,ε⃗α=ε1α1ε2α2,α,β∈Z+2,α≤β⟺αi≤βi∀i=1,2.
We first note the following lemma.Lemma 4.1.
Letm,N∈ℕ and uα∈ℝ,α∈ℤ+2,1≤|α|≤N. Then,
(4.2)(∑1≤|α|≤Nuαε⃗α)m=∑m≤|α|≤mNTα(m)[u]ε⃗α,
where the coefficients Tα(m)[u],m≤|α|≤mN depending on u=(uα),α∈ℤ+2,1≤|α|≤Nare  defined by the recurrent formulas
(4.3)Tα(1)[u]=uα,1≤|α|≤N,Tα(m)[u]=∑β∈Aα(m)uα-βTβ(m-1)[u],m≤|α|≤mN,m≥2,Aα(m)={β∈Z+2:β≤α,1≤|α-β|≤N,m-1≤|β|≤(m-1)N}.The proof of Lemma4.1 can be found in [6].We also use the notationsf1[u]=f1(x,t,u,ux,ut),μ1[u]=μ1(x,t,u).Letu0 be a unique weak solution of the problem (P0) corresponding to ε⃗=(0,0) that is, (P0)u0′′-∂∂x(μ0(x,t)u0x)=f0(x,t),0<x<1,0<t<T,u0(0,t)=u0(1,t)=0,u0(x,0)=ũ0(x),u0′(x,0)=ũ1(x),u0∈W1(M,T).Let us consider the sequence of weak solutionsuγ,γ∈ℤ+2,1≤|γ|≤N, defined by the following problems: (P̃γ)uγ′′-∂∂x(μ0(x,t)uγx)=Fγ,0<x<1,0<t<T,uγ(0,t)=uγ(1,t)=0,uγ(x,0)=uγ′(x,0)=0,uγ∈W1(M,T),
where Fγ,γ∈ℤ+2,1≤|γ|≤N are defined by the recurrent formulas as follows:(4.4)Fγ=πγ(2)[f1]+∑2≤|ν|≤|γ|,ν≤γ∂∂x(ρν(1)[μ1]∇uγ-ν),1≤|γ|≤N,
with ρδ[μ1]=ρδ[μ1;{uγ}γ≤δ], ρδ(1)[μ1]=ρδ(1)[μ1;{uγ}γ≤δ], πδ[f1]=πδ[f1;{uγ}γ≤δ], πδ(2)[f1]=πδ(2)[f1;{uγ}γ≤δ], |δ|≤N-1 defined by(4.5)ρδ[μ1]={μ1[u0],|δ|=0,∑m=1|δ|1m!D3mμ1[u0]Tδ(m)[u],1≤|δ|≤N-1,(4.6)ρδ(1)[μ1]=ρδ1-1,δ2[μ1],δ=(δ1,δ2)∈Z+2,ρδ(1)[μ1]=ρ0,δ2(1)[μ1]=ρ-1,δ2[μ1]=0,ifδ1=0,(4.7)πδ[f1]={f1[u0],|δ|=0,∑1≤|m|≤|δ|m=(m1,m2,m3)∈Z+3∑(α,β,γ)∈A(m,N)α+β+γ=δ1m!Dmf1[u0]Tα(m1)×[u]Tβ(m2)[∇u]Tγ(m3)[u′],1≤|δ|≤N-1,
where m=(m1,m2,m3)∈ℤ+3, |m|=m1+m2+m3, m!=m1!m2!m3!, Dmfj=D3m1D4m2D5m3fj, A(m,N)={(α,β,γ)∈(ℤ+2)3:m1≤|α|≤m1N,m2≤|β|≤m2N,m3≤|γ|≤m3N},(4.8)πδ(2)[f1]=πδ1,δ2-1[f1],δ=(δ1,δ2)∈Z+2,πδ(2)[f1]=πδ1,0(2)[f1]=πδ1,-1[f1]=0,ifδ2=0.Then, we have the following lemma.Lemma 4.2.
Letρν[μ1], πν[f1], |ν|≤N-1 be the functions defined by  (4.5) and  (4.7). Put h=∑|γ|≤Nuγε⃗γ, then one has
(4.9)μ1[h]=∑|ν|≤N-1ρν[μ1]ε⃗ν+‖ε⃗‖NR̃N-1(1)[μ1,ε⃗],(4.10)f1[h]=∑|ν|≤N-1πν[f1]ε⃗ν+‖ε⃗‖NRN-1(1)[f1,ε⃗],
where ∥R̃N-1(1)[μ1,ε⃗]∥L∞(0,T;L2)+∥RN-1(1)[f1,ε⃗]∥L∞(0,T;L2)≤C, with C is a constant depending only on N,T,f1,μ1,uγ,|γ|≤N.Proof.
(i) In the case ofN=1, the proof of (4.9) is easy, hence we omit the details. We only prove with N≥2. We write h=u0+∑1≤|γ|≤Nuγε⃗γ≡u0+h1.
Using Taylor's expansion of the functionμ1[h]=μ1[u0+h1] around the point u0 up to order N, we obtain from (4.2) that
(4.11)μ1[u0+h1]=μ1[u0]+∑m=1N-11m!D3mμ1[u0]h1m+1(N-1)!∫01(1-θ)N-1D3Nμ1[u0+θh1]h1Ndθ=μ1[u0]+∑m=1N-11m!D3mμ1[u0]∑m≤|ν|≤mNTν(m)[u]ε⃗ν+R̃N-1(1)[μ1,h1]=μ1[u0]+∑m=1N-11m!D3mμ1[u0]∑m≤|ν|≤N-1Tν(m)[u]ε⃗ν+∑m=1N-11m!D3mμ1[u0]∑N≤|ν|≤mNTν(m)[u]ε⃗ν+R̃N-1(1)[μ1,h1],
where
(4.12)R̃N-1(1)[μ1,h1]=1(N-1)!∫01(1-θ)N-1D3Nμ1[u0+θh1]h1Ndθ.
We note that(4.13)∑m=1N-11m!D3mμ1[u0]∑m≤|ν|≤N-1Tν(m)[u]ε⃗ν=∑1≤|ν|≤N-1(∑m=1|ν|1m!D3mμ1[u0]Tν(m)[u])ε⃗ν.
On the other hand, if we put(4.14)R̃N-1(1)[μ1,ε⃗]=‖ε⃗‖-N(∑m=1N-11m!D3mμ1[u0]∑N≤|ν|≤mNTν(m)[u]ε⃗ν+R̃N-1(1)[μ1,h1]),
then by the boundedness of the functions uγ,∇uγ,uγ′,|γ|≤N in the function space L∞(0,T;H1), we obtain from (4.3), (4.12), and (4.14) that ∥R̃N-1(1)[μ1,ε⃗]∥L∞(0,T;L2)≤C, with and C is a constant depending only on N,T,μ1,uγ,|γ|≤N. Therefore, we obtain from (4.5), (4.11), (4.13), and (4.14) that
(4.15)μ1[u0+h1]=μ1[u0]+∑1≤|ν|≤N-1(∑m=1|ν|1m!D3mμ1[u0]Tν(m)[u])ε⃗ν+‖ε⃗‖NR̃N-1(1)[μ1,ε⃗]=∑|ν|≤N-1ρν[μ1]ε⃗ν+‖ε⃗‖NR̃N-1(1)[μ1,ε⃗].
Hence, (4.9) in Lemma 4.2 is proved.
(ii) We also only prove (4.10) with N≥2. Using Taylor's expansion of the function f1[u0+h1] around the point u0 up to order N+1, we obtain from (4.2) that
(4.16)f1[u0+h1]=f1[u0]+D3f1[u0]h1+D4f1[u0]∇h1+D5f1[u0]h1′+∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+31m!Dmf1[u0]h1m1(∇h1)m2(h1′)m3+RN-1(1)[f1,h1]=f1[u0]+D3f1[u0]h1+D4f1[u0]∇h1+D5f1[u0]h1′+∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑|m|≤|ν|≤|m|N∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν+RN-1(1)[f1,h1]=f1[u0]+D3f1[u0]h1+D4f1[u0]∇h1+D5f1[u0]h1′+∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑|m|≤|ν|≤N-1∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν+∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑N≤|ν|≤|m|N∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν+RN-1(1)[f1,h1],
where
(4.17)RN-1(1)[f1,h1]=∑|m|=Nm=(m1,m2,m3)∈Z+3Nm!∫01(1-θ)N-1Dmf1[u0+θh1]h1m1(∇h1)m2(h1′)m3dθ.
We also note that(4.18)f1[u0]+D3f1[u0]h1+D4f1[u0]∇h1+D5f1[u0]h1′+∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑|m|≤|ν|≤N-1∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν=f1[u0]+∑1≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑|m|≤|ν|≤N-1∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν=f1[u0]+∑1≤|ν|≤N-1∑1≤|m|≤|ν|m=(m1,m2,m3)∈Z+3∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν=∑|ν|≤N-1πν[f1]ε⃗ν.
Similarly,(4.19)∑2≤|m|≤N-1m=(m1,m2,m3)∈Z+3∑N≤|ν|≤|m|N∑(α,β,γ)∈A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[∇u]Tγ(m3)[u′]ε⃗ν+RN-1(1)[f1,h1]=‖ε⃗‖NRN-1(1)[f1,ε⃗],
where ∥RN-1(1)[f1,ε⃗]∥L∞(0,T;L2)≤C, with C is a constant depending only on N,T,f1,uγ,|γ|≤N.
Then, (4.10) holds. Lemma 4.2 is proved.Remark 4.3.
Lemma4.2 is a generalization of the formula given in [17, page 262, formula (4.38)], and it is useful to obtain Lemma 4.4 below. These lemmas are the key to the asymptotic expansion of a weak solution u=u(ε1,ε2) of order N+1 in two small parameters ε1,ε2.
Byuε⃗=u(ε1,ε2)∈W1(M,T) as a unique weak solution of (Pε⃗), v=uε⃗-∑|γ|≤Nuγε⃗γ≡uε⃗-h satisfies the problem
(4.20)v′′-∂∂x(με1[v+h]vx)=ε2(f1[v+h]-f1[h])+ε1∂∂x[(μ1[v+h]-μ1[h])hx]+Eε⃗(x,t),0<x<1,0<t<T,v(0,t)=v(1,t)=0,v(x,0)=v′(x,0)=0,με1[v]=μ0+ε1μ1[v]=μ0(x,t)+ε1μ1(x,t,v),f1[v]=f1(x,t,v,vx,v′),μ1[v]=μ1(x,t,v),
where
(4.21)Eε⃗(x,t)=ε2f1[h]+ε1∂∂x[(μ1[h]-μ1[u0])hx]-∑1≤|γ|≤NFγε⃗γ.Lemma 4.4.
Let(H1),(H4) and (H5) hold. Then
(4.22)‖Eε⃗‖L∞(0,T;L2)≤E*‖ε⃗‖N+1,
where E* is a constant depending only on N,T,f0,f1,μ0,μ1,uγ,|γ|≤N.Proof.
We only need prove withN≥2.
Using (4.9) for the function μ1[h], we obtain
(4.23)μ1[h]=μ1[u0]+∑1≤|ν|≤N-1ρν[μ1]ε⃗ν+‖ε⃗‖NR̃N-1(1)[μ1,ε⃗].
By (4.6), (4.8), we write
(4.24)ε1(μ1[h]-μ1[u0])=∑1≤|ν|≤N-1ρν[μ1]ε1ε⃗ν+ε1‖ε⃗‖NR̃N-1(1)[μ1,ε⃗]=∑2≤|ν|≤N,ν1≥1ρν1-1,ν2[μ1]ε⃗ν+ε1‖ε⃗‖NR̃N-1(1)[μ1,ε⃗]=∑2≤|ν|≤Nρν(1)[μ1]ε⃗ν+ε1‖ε⃗‖NR̃N-1(1)[μ1,ε⃗].
On the other hand, from (4.24), we compute
(4.25)ε1(μ1[h]-μ1[u0])hx=(∑2≤|ν|≤Nρν(1)[μ1]ε⃗ν+ε1‖ε⃗‖NR̃N-1(1)[μ1,ε⃗])hx=(∑2≤|ν|≤Nρν(1)[μ1]ε⃗ν)∑|α|≤N∇uαε⃗α+ε1‖ε⃗‖NR̃N-1(1)[μ1,ε⃗]hx=∑2≤|ν|≤N,|α|≤Nρν(1)[μ1]∇uαε⃗ν+α+‖ε⃗‖N+1R̃N(1)[μ1,ε⃗]=∑2≤|ν|≤N,|α|≤Nρν(1)[μ1]∇uαε⃗ν+α+‖ε⃗‖N+1R̃N(1)[μ1,ε⃗]=∑2≤|γ|≤2N∑2≤|ν|≤N,|γ-ν|≤Nρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(1)[μ1,ε⃗]=∑2≤|γ|≤N∑2≤|ν|≤N,|γ-ν|≤Nρν(1)[μ1]∇uγ-νε⃗γ+∑N+1≤|γ|≤2N∑2≤|ν|≤N,|γ-ν|≤Nρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(1)[μ1,ε⃗]=∑2≤|γ|≤N∑2≤|ν|≤N,|γ-ν|≤Nρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(2)[μ1,ε⃗]=∑2≤|γ|≤N∑2≤|ν|≤N,ν≤γρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(2)[μ1,ε⃗],
where
(4.26)R̃N(1)[μ1,ε⃗]=ε1‖ε⃗‖R̃N-1(1)[μ1,ε⃗]hx,‖ε⃗‖N+1R̃N(2)[μ1,ε⃗]=∑N+1≤|γ|≤2N∑2≤|ν|≤N,|γ-ν|≤Nρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(1)[μ1,ε⃗].
Hence,(4.27)ε1∂∂x[(μ1[h]-μ1[u0])hx]=∂∂x[∑2≤|γ|≤N∑2≤|ν|≤N,ν≤γρν(1)[μ1]∇uγ-νε⃗γ+‖ε⃗‖N+1R̃N(2)[μ1,ε⃗]]=∑2≤|γ|≤N∑2≤|ν|≤N,ν≤γ∂∂x[ρν(1)[μ1]∇uγ-ν]ε⃗γ+‖ε⃗‖N+1∂∂xR̃N(2)[μ1,ε⃗].
Similarly, we write(4.28)ε2f1[h]=ε2(∑|ν|≤N-1πν[f1]ε⃗ν+‖ε⃗‖NRN-1(1)[f1,ε⃗])=∑1≤|ν|≤Nπν(2)[f1]ε⃗ν+‖ε⃗‖N+1R¯N(1)[f1,ε⃗],
where R¯N(1)[f1,ε⃗]=ε2/∥ε⃗∥RN-1(1)[f1,ε⃗] is bounded in the function space L∞(0,T;L2) by a constant depending only on N,T,f1,uγ,|γ|≤N.
Combining (4.4), (4.21), (4.27), and (4.28) yields
(4.29)Eε⃗(x,t)=ε2f1[h]+ε1∂∂x[(μ1[h]-μ1[u0])hx]-∑1≤|γ|≤NFγε⃗γ=∑1≤|γ|≤N{[πν(2)[f1]+∑2≤|ν|≤N,ν≤γ∂∂x[ρν(1)[μ1]∇uγ-ν]]-Fγ}ε⃗γ+‖ε⃗‖N+1(R¯N(1)[f1,ε⃗]+∂∂xR̃N(2)[μ1,ε⃗])=‖ε⃗‖N+1(R¯N(1)[f1,ε⃗]+∂∂xR̃N(2)[μ1,ε⃗]).
By the boundedness of the functionsuγ,∇uγ,uγ′,|γ|≤N in the function space L∞(0,T;H1), we obtain from (4.26) and (4.29) that
(4.30)‖Eε⃗‖L∞(0,T;L2)≤E*‖ε⃗‖N+1,
where E* is a constant depending only on N,T,f0,f1,μ0,μ1,uγ,|γ|≤N.
The proof of Lemma4.4 is complete.Now, we consider the sequence of functions{vm} defined by(4.31)v0≡0,vm-1′′-∂∂x(με1[vm-1+h]vmx)=ε2(f1[vm-1+h]-fi[h])+ε1∂∂x[(μ1[vm-1+h]-μ1[h])hx]+Eε⃗(x,t),0<x<1,0<t<T,vm(0,t)=vm(1,t)=0,vm(x,0)=vm′(x,0)=0,m≥1.Withm=1, we have the problem(4.32)v1′′-∂∂x(με1[h]v1x)=Eε⃗(x,t),0<x<1,0<t<T,v1(0,t)=v1(1,t)=0,v1(x,0)=v1′(x,0)=0.Multiplying two sides of (4.32)1 by v1′, we compute without difficulty from (4.22) that(4.33)∥v1′(t)∥2+‖μ1,ε1(t)v1x(t)‖2=2∫0t〈Eε⃗(s),v1′(s)〉ds+∫0tds∫01μ1,ε1′(x,s)v1x2(x,s)dx≤TE*2‖ε⃗‖2N+2+∫0t‖v1′(s)‖2ds+∫0tds∫01|μ1,ε1′(x,s)|v1x2(x,s)dx,
where μ1,ε1(x,t)=με1[h(x,t)]=μ0(x,t)+ε1μ1(x,t,h(x,t)). By (4.34)μ1,ε1′(x,t)=μ0′(x,t)+ε1[D2μ1(x,t,h(x,t))+D3μ1(x,t,h(x,t))h′(x,t)],
we get(4.35)|μ1,ε1′(x,t)|≤K̃(μ0)+(1+M*)K̃M*(μ1)≡ζ0,
with M*=(N+1)M,K̃(μ0)=∥μ0∥C1(Q¯T*).It follows from (4.33), (4.35) that(4.36)∥v1′(t)∥2+μ*‖v1x(t)‖2≤TE*2‖ε⃗‖2N+2+∫0t‖v1′(s)‖2ds+ζ0∫0t∥v1x(s)∥2ds.Using Gronwall's lemma, (4.36) gives(4.37)∥v1′∥L∞(0,T;L2)+‖v1x‖L∞(0,T;L2)≤(1+1μ*)TE*‖ε⃗‖N+1exp⁡[(μ*+ζ0)T2μ*].We will prove that there exists a constantCT, independent of m and ε⃗, such that(4.38)‖vm′‖L∞(0,T;L2)+‖vmx‖L∞(0,T;L2)≤CT‖ε⃗‖N+1,with‖ε⃗‖≤ε*<1,∀m.Multiplying two sides of (4.31)1 with vm′ and after integrating in t, we obtain without difficulty from (4.22) that(4.39)∥vm′(t)∥2+μ*∥vmx(t)∥2≤TE*2‖ε⃗‖2N+2+∫0t‖vm′(s)‖2ds+∫0tds∫01|μm,ε1′(x,s)|vmx2(x,s)dx+2ε2∫0t‖f1[vm-1+h]-f1[h]‖‖vm′(s)‖ds+2ε1∫0t‖∂∂x[(μ1[vm-1+h]-μ1[h])hx]‖‖vm′(s)‖ds=TE*2‖ε⃗‖2N+2+∫0t‖vm′(s)‖2ds+Ĵ1(t)+Ĵ2(t)+Ĵ3(t),
where μm,ε1(x,t)=με1[vm-1+h]=μ0(x,t)+ε1μ1(x,t,vm-1(x,t)+h(x,t)). We will estimate the integrals on the right-hand side of (4.39) as follows.First IntegralĴ1(t)
We have(4.40)μm,ε1′(x,t)=μ0′(x,t)+ε1[D2μ1(x,t,vm-1+h)+D3μ1(x,t,vm-1+h)(vm-1′+h′)],
hence
(4.41)|μm,ε1′(x,t)|≤K̃(μ0)+(1+M1*)K̃M1*(μ1)≡χ1,withM1*=(N+2)M.It follows from (4.41) that(4.42)Ĵ1(t)=∫0tds∫01|μm,ε1′(x,s)|vmx2(x,s)dx≤χ1∫0t∥vmx(s)∥2ds.Second IntegralĴ2(t)
We note that(4.43)‖f1[vm-1+h]-f1[h]‖≤2KM1*(f1)‖vm-1‖W1(T).Therefore,(4.44)Ĵ2(t)=2ε2∫0t‖f1[vm-1+h]-f1[h]‖‖vm′(s)‖ds≤Tχ22‖ε⃗‖2‖vm-1‖W1(T)2+∫0t‖vm′(s)‖2ds,
where χ2=χ2(M1*,f1)=2KM1*(f1).Third IntegralĴ3(t)
First, we need to estimate∥∂/∂x[(μ1[vm-1+h]-μ1[h])hx]∥.From the equation(4.45)∂∂x[(μ1[vm-1+h]-μ1[h])hx]=(μ1[vm-1+h]-μ1[h])hxx+∂∂x(μ1[vm-1+h]-μ1[h])hx=(μ1[vm-1+h]-μ1[h])hxx+(D1μ1[vm-1+h]-D1μ1[h])hx+(D3μ1[vm-1+h]-D3μ1[h])(∇vm-1+∇h)hx+D3μ1[h]∇vm-1hx,
it implies that(4.46)‖∂∂x[(μ1[vm-1+h]-μ1[h])hx]‖≤‖μ1[vm-1+h]-μ1[h]‖C0(Ω¯)‖hxx‖+‖D1μ1[vm-1+h]-D1μ1[h]‖C0(Ω¯)‖hx‖+‖D3μ1[vm-1+h]-D3μ1[h]‖C0(Ω¯)‖∇vm-1+∇h‖C0(Ω¯)‖hx‖+‖D3μ1[h]‖C0(Ω¯)‖vm-1‖W1(T)‖hx‖C0(Ω¯).On the other hand, we have(4.47)‖μ1[vm-1+h]-μ1[h]‖C0(Ω¯)≤K̃M1*(μ1)‖vm-1‖W1(T),‖Djμ1[vm-1+h]-Djμ1[h]‖C0(Ω¯)≤K̃M1*(μ1)‖vm-1‖W1(T),j=1,3,‖D3μ1[h]‖C0(Ω¯)≤K̃M1*(μ1).We deduce from (4.46) and (4.47) that(4.48)‖∂∂x[(μ1[vm-1+h]-μ1[h])hx]‖≤(3+2M1*)M1*K̃M1*(μ1)‖vm-1‖W1(T).Next, by (4.48), it follows that(4.49)Ĵ3(t)=2ε1∫0t‖∂∂x[(μ1[vm-1+h]-μ1[h])hx]‖‖vm′(s)‖ds≤Tχ32‖ε⃗‖2‖vm-1‖W1(T)2+∫0t‖vm′(s)‖2ds,
where χ3=χ3(M1*,μ1)=(3+2M1*)M1*K̃M1*(μ1).Combining (4.39), (4.42), (4.44), and (4.49) gives(4.50)∥vm′(t)∥2+μ*∥vmx(t)∥2≤TE*2‖ε⃗‖2N+2+T(χ22+χ32)‖ε⃗‖2‖vm-1‖W1(T)2+3∫0t‖vm′(s)‖2ds+χ1∫0t∥vmx(s)∥2ds≤TE*2‖ε⃗‖2N+2+T(χ22+χ32)‖ε⃗‖2‖vm-1‖W1(T)2+(3+χ1μ*)∫0t(‖vm′(s)‖2+μ*∥vmx(s)∥2)ds.Using Gronwall's lemma, we deduce from (4.50) that(4.51)‖vm‖W1(T)≤σT‖vm-1‖W1(T)+δ,∀m≥1,
where(4.52)σT=χ22+χ32ηT,δ=ηTE*‖ε⃗‖N+1,ηT=(1+1μ*)Texp⁡⁡[T2(3+χ1μ*)].We can assume that(4.53)σT<1,
with sufficiently small T>0.Lemma 4.5.
Let the sequence{ζm} satisfy
(4.54)ζm≤σζm-1+δ∀m≥1,ζ0=0,
where 0≤σ<1,δ≥0 are the given constants. Then,
(4.55)ζm≤δ(1-σ)∀m≥1.This lemma is useful, as it will be said below, and it is easy to prove.Applying Lemma4.5 with ζm=∥vm∥W1(T),σ=σT=χ22+χ32ηT<1,δ=ηTE*∥ε⃗∥N+1, it follows from (4.55) that(4.56)‖vm′‖L∞(0,T;L2)+‖vmx‖L∞(0,T;L2)=‖vm‖W1(T)≤δ(1-σT)≡CT‖ε⃗‖N+1.On the other hand, the linear recurrent sequence{vm} defined by (4.31) converges strongly in the space W1(T) to the solution v of the problem (4.20). Hence, letting m→+∞ in (4.56) yields(4.57)‖v′‖L∞(0,T;L2)+‖vx‖L∞(0,T;L2)≤CT‖ε⃗‖N+1,
it meansthat(4.58)‖u′-∑|γ|≤Nuγ′ε⃗γ‖L∞(0,T;L2)+‖ux-∑|γ|≤Nuγxε⃗γ‖L∞(0,T;L2)≤CT‖ε⃗‖N+1.Consequently, we obtain the following theorem.Theorem 4.6.
Let(H1),(H4) and  (H5) hold. Then there exist  constants  M>0 and  T>0 such that, for every  ε⃗, with  ∥ε⃗∥≤ε*<1, the problem  (Pε⃗) has a unique weak solution  u=uε⃗∈W1(M,T) satisfying an asymptotic expansion up to order  N+1 as in  (4.58), where the functions  uγ,|γ|≤N are the weak solutions of the problems  (P0),  (P̃γ), 1≤|γ|≤N, respectively.The Problem with Many Small Parameters
Next, we note that the results as above still holdfor the problem in p small parameters ε1,…,εp as follows:
(P̂ε⃗)utt-∂∂x[(μ0(x,t)+∑i=1pεiμi(x,t,u))ux]=f0(x,t)+∑i=1pεifi(x,t,u,ux,ut),0<x<1,0<t<T,u(0,t)=u(1,t)=0,u(x,0)=ũ0(x),ut(x,0)=ũ1(x).
For more detail, we also make the following assumptions:(Ĥ4)
μ∈C2([0,1]×ℝ+),μi∈CN+1([0,1]×ℝ+×ℝ),μ0≥μ*>0,μi≥0,i=1,2,…,p,(Ĥ5)
f0∈C1([0,1]×ℝ+),fi∈CN([0,1]×ℝ+×ℝ3),i=1,2,…,p.
For a multi-indexα=(α1,…,αp)∈ℤ+p, and ε⃗=(ε1,…,εp)∈ℝp, we also put
(4.59)|α|=α1+⋯+αp,α!=α1!⋯αp!,‖ε⃗‖=ε12+⋯+εp2,ε⃗α=ε1α1⋯εpαp,α,β∈Z+p,α≤β⟺αi≤βi∀i=1,…,p.
Letu0 be a unique weak solution of the problem (P0), which is (P̂ε⃗) corresponding to ε⃗=(0,…,0).Let the sequence of weak solutions uγ,γ∈ℤ+p,1≤|γ|≤Nbe defined by the problems (P̃γ), in which Fγ,γ∈ℤ+p,1≤|γ|≤N, are defined by suitable recurrent formulas. Then, the following similar theorem holds.Theorem 4.7.
Let(H1),(Ĥ4) and  (Ĥ5) hold. Then there exist constants  M>0 and  T>0 such that, for every  ε⃗, with  ∥ε⃗∥≤ε*<1, the problem  (P̂ε⃗) has a unique weak solution  u=uε⃗∈W1(M,T) satisfying an asymptotic estimation up to order  N+1 as follows:
(4.60)‖u′-∑|γ|≤Nuγ′ε⃗γ‖L∞(0,T;L2)+‖ux-∑|γ|≤Nuγxε⃗γ‖L∞(0,T;L2)≤CT‖ε⃗‖N+1.The proof of Theorem4.7 is similar the one as above let us omit it.Remark 4.8.
Typical examples about asymptotic expansion of solutions in a small parameter can be found in the research of many authors such as [1, 3, 4, 8, 9, 17–19]. However, to our knowledge, in the case of asymptotic expansion in many small parameters, there is only partial results, for example, [5–7, 14], concerning asymptotic expansion of solutions in two or three small parameters.



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