APPMATHISRN Applied Mathematics2090-55722090-5564International Scholarly Research Network62590810.5402/2011/625908625908Research ArticleOn a Nonlinear Wave Equation Associated with Dirichlet Conditions: Solvability and Asymptotic Expansion of Solutions in Many Small ParametersNgocLe Thi Phuong1LuanLe Khanh2ThuyetTran Minh2LongNguyen Thanh3JauberteauF.1Nhatrang Educational College01 Nguyen Chanh StreetNhatrang CityVietnam2Department of MathematicsUniversity of Economics of Ho Chi Minh City 59C Nguyen Dinh Chieu Street, District 3Ho Chi Minh CityVietnamueh.edu.vn3Department of Mathematics and Computer ScienceUniversity of Natural ScienceVietnam National University Ho Chi Minh City227 Nguyen Van Cu Street, District 5Ho Chi Minh CityVietnamvnuhcm.edu.vn201130052011201109032011120420112011Copyright © 2011 Le Thi Phuong Ngoc et al.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.

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.

1. Introduction

In this paper, we consider the following Dirichlet problem: utt-x(μ(x,t,u)ux)=f(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), where ũ0,ũ1,μ, and f are given functions satisfying conditions specified later.

In the special cases, when the function μ(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 , Dinh and Long [2, 3], Long and Diem , Long et al. , Long and Truong [6, 7], Long et al. , Ngoc et al. , and the references therein.

Ficken and Fleishman  and Rabinowitz  studied the periodic-Dirichlet problem for hyperbolic equations containing a small parameter    ε, in particular, the differential equationutt-uxx=2αut+εf(t,x,u,ut,ux).

In , Kiguradze has established the existence and uniqueness of a classical solution uC2([0,a]×n) of the periodic-Dirichlet problem for the following nonlinear wave equation: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 equationutt-uxx=g(t,x,u)+g1(u)ut+εq(t,x,u,ut,ux), with sufficiently small ε and continuously differentiable q.

In , 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 , Long et al. have studied the linear recursive schemes and asymptotic expansion for the nonlinear wave equationutt-uxx=f(x,t,u,ux,ut)+εf1(x,t,u,ux,ut), with the mixed nonhomogeneous conditionsux(0,t)-h0u(0,t)=g0(t),u(1,t)=g1(t).

In the case of g0,g1C3(+),fCN+1([0,1]×+×3),f1CN([0,1]×+×3), 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 Section 2, 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: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 μ0C2([0,1]×+),μ1CN+1([0,1]×+×),μ0(x,t)μ*>0,μ1(x,t,z)0, for all (x,t,z)[0,1]×+×,f0C1([0,1]×+), and f1CN([0,1]×+×3) is established. Next, we note that the same results are valid for the equation in p small parameters ε1,,εp as followsutt-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 [57, 14], where asymptotic expansion of a weak solution in two or three small parameters is given.

2. Preliminaries

Put Ω=(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),1p, the Banach space of real functions u:(0,T)X measurable, such that uLp(0,T;X)<+, with uLp(0,T;X)={(0Tu(t)Xpdt)1/p,if1p<,esssup0<t<Tu(t)X,ifp=.

Let u(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 fCk([0,1]×+×3),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)+5, |α|=α1+α2+α3+α4+α5=k, D(0,0,,0)f=f.

Similarly, with μCk([0,1]×+×),μ=μ(x,t,z), we put D1μ=μ/x,D2μ=μ/t,D3μ=μ/z and Dβμ=D1β1D2β2D3β3,β=(β1,β2,β3)+3,|β|=β1+β2+β3=k.

On H1, we will use the following norms:vH1=(v2+vx2)1/2.

Then, we have the following lemma.

Lemma 2.1.

The imbedding H1C0(Ω¯) is compact and vC0(Ω¯)2vH1vH1.

The proof of Lemma 2.1 is easy, hence we omit the details.

Remark 2.2.

On H01,vvH1 and vvx are two equivalent norms. Furthermore, we have the following inequalities: vC0(Ω¯)vxvH01.

Remark 2.3.

(i) Let us note more that a unique weak solution u of the problem (1.1)–(1.3) will be obtained in Section 3 (Theorem 3.2) in the following manner.

Find uW̃={uL(0,T;H01H2):uL(0,T;H01),u′′L(0,T;L2)} such that u verifies the following variational equation: u′′(t),w+μ(,t,u(t))ux(t),wx=f(,t,u(t),ux(t),u(t)),w,wH01, and the initial conditions u(0)=ũ0,u(0)=ũ1.

(ii) With the regularity obtained by uW̃, it also follows from Theorem 3.2 that the problem (1.1)–(1.3) has a unique strong solution u that satisfies uC0(0,T;H1)C1(0,T;L2)L(0,T;H2),utL(0,T;H1),uttL(0,T;L2).

On the other hand, by uW̃, we can see that u,ux,ut,uxx,uxt,uttL(0,T;L2)L2(QT).

Also, if (u0,u1)(H01H2)×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:

ũ0H01H2,ũ1H01,

μC2([0,1]×+×),μ(x,t,z)μ*>0,forall  (x,t,z)[0,1]×+×,

fC1(Ω¯×+×3).

With μ and f satisfying the assumptions (H2) and (H3), respectively, for each T*>0 and M>0 are given, we put the following constants:K̃M(μ)=μC2(D̃M*),KM(f)=fC1(DM*), where D̃M*={(x,t,z):0x1,0tT*,|z|M} and DM*={(x,t,u,v,w)+×+×3:0x1,0tT*,|u|,|v|,|w|M}.

For each T(0,T*] and M>0, we getW(M,T)={vL(0,T;H01H2):vtL(0,T;H01),vttL2(QT),withvL(0,T;H01H2),vtL(0,T;H01),vttL2(QT)M},W1(M,T)={vW(M,T):vttL(0,T;L2)}, where QT=Ω×(0,T).

We choose the first term u0ũ0W1(M,T). Suppose thatum-1W1(M,T),m1.

The problem (1.1)–(1.3) is associated with the following variational problem.

Find umW1(M,T) such thatum′′(t),v+μm(t)um(t),v=Fm(t),v,vH01,um(0)=ũ0,um(0)=ũ1, whereμm(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 ).

Consider a special basis {wj} on H01:wj(x)=2sin(jπx),j, formed by the eigenfunctions of the Laplacian -Δ=-2/x2. Put um(k)(t)=j=1kcmj(k)(t)wj, where the coefficients cmj(k) satisfy the system of linear differential equations üm(k)(t),wj+μm(t)um(k)(t),wj=Fm(t),wj,1jk,um(k)(0)=ũ0k,u̇m(k)(0)=ũ1k, where ũ0k=j=1kαj(k)wjũ0stronglyinH01H2,ũ1k=j=1kβj(k)wjũ1stronglyinH01.

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 sm(k)(t)=pm(k)(t)+qm(k)(t)+0tüm(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 sm(k)(t)=sm(k)(0)+2μm(0)ũ0k,Δũ0k+2Fm(0),Δũ0k+0tds01μm(x,s)(|um(k)(x,s)|2+|Δum(k)(x,s)|2)dx+20tFm(s),u̇m(k)(s)ds+20ts(μm(s)um(k)(s)),Δum(k)(s)ds-2μm(t)um(k)(t),Δum(k)(t)-2Fm(t),Δum(k)(t)+20tFmt(s),Δum(k)(s)ds+0tüm(k)(s)2ds=qm(k)(0)+2μm(0)ũ0k,Δũ0k+2Fm(0),Δũ0k+j=17Ij.

Next, we will estimate the terms Ij,j=1,2,,7 on the right-hand side of (3.14) as follows.

First Term <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M126"><mml:mrow><mml:msub><mml:mrow><mml:mi>I</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

We have 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 |μm(x,t)|(1+M)K̃M(μ).

Hence, I1=0tds01μm(x,s)(|um(k)(x,s)|2+|Δum(k)(x,s)|2)dx1+Mμ*K̃M(μ)0tsm(k)(s)ds.

Second Term

By using (H3), we obtain from (3.2), (3.5), and (3.13)2 that I2=20tFm(s),u̇m(k)(s)dsTKM2(f)+0tpm(k)(s)ds.

Third Term

The Cauchy-Schwartz inequality yields |I3|=2|0ts(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 rm(k)(s)=μm(s)u̇m(k)(s)+s(μm(s))um(k)(s)(μm(s)C0(Ω¯)+1μ*sμm(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 μm(s)C0(Ω¯)K̃M(μ)(1+um-1(s)C0(Ω¯))2(1+M)K̃M(μ).

Similarly, the following equality sμm(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 sμm(s)(1+3M+M2)K̃M(μ).

It follows from (3.20)–(3.23) that rm(k)(s)[2(1+M)+1+3M+M2μ*]K̃M(μ)sm(k)(s).

Hence, we obtain from (3.19) and (3.24) that |I3|2μ*[2(1+M)+1+3M+M2μ*]K̃M(μ)0tsm(k)(s)ds.

Fourth Term <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M142"><mml:mrow><mml:msub><mml:mrow><mml:mi>I</mml:mi></mml:mrow><mml:mrow><mml:mn>4</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

By the Cauchy-Schwartz inequality, we have |I4|=|-2μm(t)um(k)(t),Δum(k)(t)|1βμm(t)um(k)(t)2+βΔum(k)(t)2, for all β>0. On the other hand μm(t)um(k)(t)=μm(0)ũ0k+0ts(μm(s)um(k)(s))dsμm(0)C0(Ω¯)ũ0k+0trm(k)(s)ds.

Hence, we obtain from (3.26), (3.27) that |I4|βμ*qm(k)(t)+2βμm(0)C0(Ω¯)2ũ0k2+2βT[2(1+M)+1+3M+M2μ*]2K̃M2(μ)0tsm(k)(s)ds, for all β>0.

Fifth Term <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M148"><mml:mrow><mml:msub><mml:mrow><mml:mi>I</mml:mi></mml:mrow><mml:mrow><mml:mn>5</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

By (3.5), (3.8), and (3.13), we obtain |I5|=|-2Fm(t),Δum(k)(t)|1βFm(t)2+βΔum(k)(t)22βFm(0)2+2βT0TFms(s)2ds+βμ*sm(k)(t),β>0.

Note that Fmt(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 Fmt(t)KM(f)(1+2M+um-1′′(t)).

Hence, we deduce from (3.29) and (3.31) that |I5|2βFm(0)2+4βTKM2(f)[(1+2M)2T+M2]+βμ*sm(k)(t),β>0.

Sixth Term <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M154"><mml:mrow><mml:msub><mml:mrow><mml:mi>I</mml:mi></mml:mrow><mml:mrow><mml:mn>6</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

By (3.2), (3.5), (3.13)3, and (3.31), we get |I6|=2|0tFmt(s),Δum(k)(s)ds|0tFmt(s)ds+0tFmt(s)Δum(k)(s)2dsKM(f)[(1+2M)T+T(0Tum-1′′(s)2ds)1/2]+1μ*KM(f)0t(1+2M+um-1′′(s))qm(k)(s)dsKM(f)[(1+2M)T+TM]+1μ*KM(f)0t(1+2M+um-1′′(s))qm(k)(s)ds.

Seventh Term <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M157"><mml:mrow><mml:msub><mml:mrow><mml:mi>I</mml:mi></mml:mrow><mml:mrow><mml:mn>7</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

Equation (3.10) is rewritten as follows: üm(k)(t),wj-x(μm(t)um(k)(t)),wj=Fm(t),wj,1jk.

Hence, by replacing wj with üm(k)(t) and integrating I7=0tüm(k)(s)2ds20tx(μm(s)um(k)(s))2ds+20tFm(s)2ds20tx(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 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 I72TKM2(f)+18μ*(1+M)2K̃M2(μ)0tsm(k)(s)ds.

Choosing β>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 sm(k)(t)C̃0k+C̃1(M,T)+0t(C̃2(M,T)+2μ*KM(f)um-1′′(s))sm(k)(s)ds, where C̃0k=C̃0k(β,f,μ,ũ0,ũ1,ũ0k,ũ1k)=2sm(k)(0)+4μm(0)ũ0k,Δũ0k+4Fm(0),Δũ0k+4βμm(0)C0(Ω¯)2ũ0k2+4βFm(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 C̃0k12M2.

Notice that by (H3), we deduce from (3.39)2,3 that limT0+C̃1(M,T)=limT0+TC̃2(M,T)=0.

So, from (3.39) and (3.41), we can choose T>0 such that (12M2+C̃1(M,T))exp(TC̃2(M,T)+2μ0KM(f)TM)M2,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 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 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)Tum-1′′L2(QT)]M2.

Therefore, we have um(k)W(M,T),m,kN.

Step 3.

Limiting process.

From (3.46), we can extract from {um(k)} a subsequence still denoted by {um(k)} such that um(k)uminL(0,T;H01H2)weak*,u̇m(k)uminL(0,T;H01)weak*,üm(k)um′′inL2(QT)weak, as k, and umW(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 um′′=μmum+μmΔum+f(x,t,um-1,um-1,um-1)L(0,T;L2).

Hence, umW1(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 uW1(M,T).

Furthermore, the linear recurrent sequence {um} defined by (3.6)–(3.8) converges to the solution u strongly in the space W1(T)={wL(0,T;H01):wL(0,T;L2)}, with the following estimation: um-uL(0,T;H01)+um-uL(0,T;L2)CkTm,mN, 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 ) wW1(T)=wL(0,T;H01)+wL(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 vm′′(t),w+μm+1(t)vm(t),w=x[(μm+1(t)-μm(t))um(t)],w+Fm+1(t)-Fm(t),w,wH01,vm(0)=vm(0)=0.

Taking w=vm in (3.53)1, after integrating in t, we get Zm(t)=0tds01μm+1(x,s)|vm(s)|2dx+20tFm+1(s)-Fm(s),vm(s)ds+20tx[(μm+1(s)-μm(s))um(s)],vm(s)ds=i=13Ji, in which 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 |J1||0tds01μm+1(x,s)|vm(s)|2dx|1+Mμ*K̃M(μ)0tZm(s)ds.

Second Integral

By (H3), Fm+1(t)-Fm(t)2KM(f)[vm-1(t)+vm-1(t)]2KM(f)vm-1W1(T), so |J2|2|0tFm+1(s)-Fm(s),vm(s)ds|4KM(f)vm-1W1(T)0tvm(s)ds4TKM2(f)vm-1W1(T)2+0tZm(s)ds.

Third Integral

Using (H2) again, we get |J3|=2|0tx[(μm+1(s)-μm(s))um(s)],vm(s)ds|0tx[(μm+1(s)-μm(s))um(s)]2ds+0tZm(s)ds.

Note that 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, 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 μm+1(s)-μm(s)C0(Ω¯)K̃M(μ)wm-1W1(T),Diμ[um]-Diμ[um-1]C0(Ω¯)K̃M(μ)wm-1W1(T),i=1,3,um(s)C0(Ω¯)2um(s)H12um(s)2+Δum(s)22M,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 x[(μm+1(s)-μm(s))um(s)](4+M)MK̃M(μ)vm-1W1(T).

Hence, |J3|(4+M)2M2TK̃M2(μ)vm-1W1(T)2+0tZm(s)ds.

Combining (3.54)–(3.56), (3.58), and (3.64) yields Zm(t)T[4KM2(f)+(4+M)2M2K̃M2(μ)]vm-1W1(T)2+(2+1+Mμ*K̃M(μ))0tZm(s)ds.

Using Gronwall's lemma, (3.65) gives vmW1(T)kTvm-1W1(T)mN, where kT<1 as in (3.43).

Hence, we obtain from (3.66) that um+p-umW1(T)kTm1-kTu1-u0W1(T)m,pN,

It follows that {um} is a Cauchy sequence in W1(T). Then, there exists uW1(T) such that umustrongly  in  W1(T).

On the other hand, from (3.48), we deduce the existence of a subsequence {umj} of {um} such that umjuinL(0,T;H01H2)weak*,umjuinL(0,T;H01)weak*,umj′′u′′inL2(QT)weak,uW(M,T).

Note that |μm(x,t)-μ(x,t,u(x,t))|K̃M(μ)um-1-uW1(T),Fm(t)-f(,t,u(t),ux(t),u(t))2KM(f)um-1-uW1(T).

Hence, from (3.68) and (3.71), we obtain μmμ(,,u)stronglyinL(QT),Fmf(,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 uW(M,T) satisfying the equation u′′(t),w+μ(,t,u(t))ux(t),w=f(,t,u(t),ux(t),u(t)),w,wH01,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 u′′=D1μ[u]ux+D3μ[u]ux2+μ[u]uxx+f(x,t,u,ux,u)L(0,T;L2), thus uW1(M,T), and Step 1 follows.

(ii) The uniqueness of the solution.

Let u1,u2W1(M,T) be two weak solutions of the problem (1.1)–(1.3). Then, u=u1-u2 satisfies the variational problem u′′(t),w+μ1(t)ux(t),wx=x([μ1(t)-μ2(t)]u2x(t)),w+F2(t)-F1(t),w,wH01,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 take w=u in (3.75)1 and integrate in t to get ρ(t)=0tds01μ1(x,s)ux2(x,s)dx+20tF1(s)-F2(s),u(s)ds+20tx([μ1(s)-μ2(s)]u2x(s)),udsi=13ρi(t), where 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: ρ1(t)=0tds01μ1(x,s)ux2(x,s)dx1μ*(1+M)K̃M(μ)0tρ(s)dsρM(1)0tρ(s)ds,ρ2(t)=20tF1(s)-F2(s),u(s)ds4KM(f)0t(ux(s)+u(s))u(s)ds4(1+1μ*)KM(f)0tρ(s)dsρM(2)0tρ(s)ds,ρ3(t)=20tx([μ1(s)-μ2(s)]u2x(s)),uds20tx([μ1(s)-μ2(s)]u2x(s))u(s)ds.

On the other hand 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, 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 ρ3(t)1μ*(3+M)MK̃M(μ)0tρ(s)dsρM(3)0tρ(s)ds.

Combining (3.76)–(3.79) and (3.83) yields ρ(t)(ρM(1)+ρM(2)+ρM(3))0tρ(s)ds.

Using Gronwall's lemma, it follows from (3.84) that ρ0 that is, u1u2.

Theorem 3.2 is proved completely.

Remark 3.3.

(i) In the case of μ1,fC1(Ω¯×+×3) and the boundary condition in  standing for (1.2), we obtained some similar results in .

(ii) In the case of μ1,fC1(Ω¯×+×3),f(1,t,u,v,w)=0,forallt0,forall(u,v,w)3, and the boundary condition in  standing for (1.2), some results as above were given in .

Remark 3.4.

By Galerkin method, as in Remark 2.3, the local existence of a strong solution uH2(QT) of the problem (1.1)–(1.3) is proved.

In the case of μ=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 uC2(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:

μ0C2([0,1]×+),  μ1CN+1([0,1]×+×),  μ0μ*>0,  μ10,

f0C1([0,1]×+),  f1CN([0,1]×+×3).

We consider the following perturbed problem, where ε1,ε2 are two small parameters such that 0εiεi*<1,  i=1,2: 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 Theorem 3.2, the problem (Pε) has a unique weak solution u depending on ε=(ε1,ε2):uε=u(ε1,ε2). When ε=(0,0), (Pε) is denoted by (P0). We will study the asymptotic expansion of uε with respect to ε1,ε2.

We use the following notations. For a multi-index α=(α1,α2)+2, and ε=(ε1,ε2)2, we put |α|=α1+α2,α!=α1!α2!,ε=ε12+ε22,εα=ε1α1ε2α2,α,βZ+2,αβαiβii=1,2.

We first note the following lemma.

Lemma 4.1.

Let m,N and uα,α+2,1|α|N. Then, (1|α|Nuαεα)m=m|α|mNTα(m)[u]εα, where the coefficients Tα(m)[u],m|α|mN depending on u=(uα),α+2,1|α|Nare defined by the recurrent formulas Tα(1)[u]=uα,1|α|N,Tα(m)[u]=βAα(m)uα-βTβ(m-1)[u],m|α|mN,m2,Aα(m)={βZ+2:βα,1|α-β|N,m-1|β|(m-1)N}.

The proof of Lemma 4.1 can be found in .

We also use the notations f1[u]=f1(x,t,u,ux,ut),μ1[u]=μ1(x,t,u).

Let u0 be a unique weak solution of the problem (P0) corresponding to ε=(0,0) that is, 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),u0W1(M,T).

Let us consider the sequence of weak solutions uγ,γ+2,1|γ|N, defined by the following problems: 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,uγW1(M,T), where Fγ,γ+2,1|γ|N are defined by the recurrent formulas as follows:Fγ=πγ(2)[f1]+2|ν||γ|,νγx(ρν(1)[μ1]uγ-ν),1|γ|N, with ρδ[μ1]=ρδ[μ1;{uγ}γδ], ρδ(1)[μ1]=ρδ(1)[μ1;{uγ}γδ], πδ[f1]=πδ[f1;{uγ}γδ], πδ(2)[f1]=πδ(2)[f1;{uγ}γδ], |δ|N-1 defined byρδ[μ1]={μ1[u0],|δ|=0,m=1|δ|1m!D3mμ1[u0]Tδ(m)[u],1|δ|N-1,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,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)+3, |m|=m1+m2+m3, m!=m1!m2!m3!, Dmfj=D3m1D4m2D5m3fj, A(m,N)={(α,β,γ)(+2)3:m1|α|m1N,m2|β|m2N,m3|γ|m3N},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.

Let ρν[μ1], πν[f1], |ν|N-1 be the functions defined by (4.5) and (4.7). Put h=|γ|Nuγεγ, then one has μ1[h]=|ν|N-1ρν[μ1]εν+εNR̃N-1(1)[μ1,ε],f1[h]=|ν|N-1πν[f1]εν+εNRN-1(1)[f1,ε], where R̃N-1(1)[μ1,ε]L(0,T;L2)+RN-1(1)[f1,ε]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 of N=1, the proof of (4.9) is easy, hence we omit the details. We only prove with N2. We write h=u0+1|γ|Nuγεγu0+h1.

Using Taylor's expansion of the function μ1[h]=μ1[u0+h1] around the point u0 up to order N, we obtain from (4.2) that 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]εν+R̃N-1(1)[μ1,h1]=μ1[u0]+m=1N-11m!D3mμ1[u0]m|ν|N-1Tν(m)[u]εν+m=1N-11m!D3mμ1[u0]N|ν|mNTν(m)[u]εν+R̃N-1(1)[μ1,h1], where R̃N-1(1)[μ1,h1]=1(N-1)!01(1-θ)N-1D3Nμ1[u0+θh1]h1Ndθ.

We note that m=1N-11m!D3mμ1[u0]m|ν|N-1Tν(m)[u]εν=1|ν|N-1(m=1|ν|1m!D3mμ1[u0]Tν(m)[u])εν.

On the other hand, if we put R̃N-1(1)[μ1,ε]=ε-N(m=1N-11m!D3mμ1[u0]N|ν|mNTν(m)[u]εν+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,ε]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 μ1[u0+h1]=μ1[u0]+1|ν|N-1(m=1|ν|1m!D3mμ1[u0]Tν(m)[u])εν+εNR̃N-1(1)[μ1,ε]=|ν|N-1ρν[μ1]εν+εNR̃N-1(1)[μ1,ε].

Hence, (4.9) in Lemma 4.2 is proved.

(ii) We also only prove (4.10) with N2. 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 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]εν+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]εν+2|m|N-1m=(m1,m2,m3)Z+3N|ν||m|N(α,β,γ)A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[u]Tγ(m3)[u]εν+RN-1(1)[f1,h1], where 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 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]εν=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]εν=f1[u0]+1|ν|N-11|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]εν=|ν|N-1πν[f1]εν.

Similarly, 2|m|N-1m=(m1,m2,m3)Z+3N|ν||m|N(α,β,γ)A(m,N)α+β+γ=ν1m!Dmf1[u0]Tα(m1)[u]Tβ(m2)[u]Tγ(m3)[u]εν+RN-1(1)[f1,h1]=εNRN-1(1)[f1,ε], where RN-1(1)[f1,ε]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.

Lemma 4.2 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.

By uε=u(ε1,ε2)W1(M,T) as a unique weak solution of (Pε), v=uε-|γ|Nuγεγuε-h satisfies the problem v′′-x(με1[v+h]vx)=ε2(f1[v+h]-f1[h])+ε1x[(μ1[v+h]-μ1[h])hx]+Eε(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 Eε(x,t)=ε2f1[h]+ε1x[(μ1[h]-μ1[u0])hx]-1|γ|NFγεγ.

Lemma 4.4.

Let (H1),(H4) and (H5) hold. Then EεL(0,T;L2)E*εN+1, where E* is a constant depending only on N,T,f0,f1,μ0,μ1,uγ,|γ|N.

Proof.

We only need prove with N2.

Using (4.9) for the function μ1[h], we obtain μ1[h]=μ1[u0]+1|ν|N-1ρν[μ1]εν+εNR̃N-1(1)[μ1,ε].

By (4.6), (4.8), we write ε1(μ1[h]-μ1[u0])=1|ν|N-1ρν[μ1]ε1εν+ε1εNR̃N-1(1)[μ1,ε]=2|ν|N,ν11ρν1-1,ν2[μ1]εν+ε1εNR̃N-1(1)[μ1,ε]=2|ν|Nρν(1)[μ1]εν+ε1εNR̃N-1(1)[μ1,ε].

On the other hand, from (4.24), we compute ε1(μ1[h]-μ1[u0])hx=(2|ν|Nρν(1)[μ1]  εν+ε1εNR̃N-1(1)[μ1,ε])hx=(2|ν|Nρν(1)[μ1]εν)|α|Nuαεα+ε1εNR̃N-1(1)[μ1,ε]hx=2|ν|N,|α|Nρν(1)[μ1]uαεν+α+εN+1R̃N(1)[μ1,ε]=2|ν|N,|α|Nρν(1)[μ1]uαεν+α+εN+1R̃N(1)[μ1,ε]=2|γ|2N2|ν|N,|γ-ν|Nρν(1)[μ1]uγ-νεγ+εN+1R̃N(1)[μ1,ε]=2|γ|N2|ν|N,|γ-ν|Nρν(1)[μ1]uγ-νεγ+N+1|γ|2N2|ν|N,|γ-ν|Nρν(1)[μ1]uγ-νεγ+εN+1R̃N(1)[μ1,ε]  =2|γ|N2|ν|N,|γ-ν|Nρν(1)[μ1]uγ-νεγ+εN+1R̃N(2)[μ1,ε]=2|γ|N2|ν|N,νγρν(1)[μ1]uγ-νεγ+εN+1R̃N(2)[μ1,ε], where R̃N(1)[μ1,ε]=ε1εR̃N-1(1)[μ1,ε]hx,εN+1R̃N(2)[μ1,ε]=N+1|γ|2N2|ν|N,|γ-ν|Nρν(1)[μ1]uγ-νεγ+εN+1R̃N(1)[μ1,ε].

Hence, ε1x[(μ1[h]-μ1[u0])hx]=x[2|γ|N2|ν|N,νγρν(1)[μ1]uγ-νεγ+εN+1R̃N(2)[μ1,ε]]=2|γ|N2|ν|N,νγx[ρν(1)[μ1]uγ-ν]εγ+εN+1xR̃N(2)[μ1,ε].

Similarly, we write ε2f1[h]=ε2(|ν|N-1πν[f1]εν+εNRN-1(1)[f1,ε])=1|ν|Nπν(2)[f1]εν+εN+1R¯N(1)[f1,ε], where R¯N(1)[f1,ε]=ε2/εRN-1(1)[f1,ε] 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 Eε(x,t)=ε2f1[h]+ε1x[(μ1[h]-μ1[u0])hx]-1|γ|NFγεγ=1|γ|N{[πν(2)[f1]+2|ν|N,νγx[ρν(1)[μ1]uγ-ν]]-Fγ}εγ+εN+1(R¯N(1)[f1,ε]+xR̃N(2)[μ1,ε])=εN+1(R¯N(1)[f1,ε]+xR̃N(2)[μ1,ε]).

By the boundedness of the functions uγ,uγ,uγ,|γ|N in the function space L(0,T;H1), we obtain from (4.26) and (4.29) that EεL(0,T;L2)E*εN+1, where E* is a constant depending only on N,T,f0,f1,μ0,μ1,uγ,|γ|N.

The proof of Lemma 4.4 is complete.

Now, we consider the sequence of functions {vm} defined byv00,vm-1′′-x(με1[vm-1+h]vmx)=ε2(f1[vm-1+h]-fi[h])+ε1x[(μ1[vm-1+h]-μ1[h])hx]+Eε(x,t),0<x<1,  0<t<T,vm(0,t)=vm(1,t)=0,vm(x,0)=vm(x,0)=0,m1.

With m=1, we have the problemv1′′-x(με1[h]v1x)=Eε(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) thatv1(t)2+μ1,ε1(t)v1x(t)2=20tEε(s),v1(s)ds+0tds01μ1,ε1(x,s)v1x2(x,s)dxTE*2ε2N+2+0tv1(s)2ds+0tds01|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 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|μ1,ε1(x,t)|K̃(μ0)+(1+M*)K̃M*(μ1)ζ0, with M*=(N+1)M,K̃(μ0)=μ0C1(Q¯T*).

It follows from (4.33), (4.35) thatv1(t)2+μ*v1x(t)2TE*2ε2N+2+0tv1(s)2ds+ζ00tv1x(s)2ds.

Using Gronwall's lemma, (4.36) givesv1L(0,T;L2)+v1xL(0,T;L2)(1+1μ*)TE*εN+1exp[(μ*+ζ0)T2μ*].

We will prove that there exists a constant CT, independent of m and ε, such thatvmL(0,T;L2)+vmxL(0,T;L2)CTεN+1,withεε*<1,m.

Multiplying two sides of (4.31)1 with vm and after integrating in t, we obtain without difficulty from (4.22) thatvm(t)2+μ*vmx(t)2TE*2ε2N+2+0tvm(s)2ds+0tds01|μm,ε1(x,s)|vmx2(x,s)dx+2ε20tf1[vm-1+h]-f1[h]vm(s)ds+2ε10tx[(μ1[vm-1+h]-μ1[h])hx]vm(s)ds=TE*2ε2N+2+0tvm(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 Integral <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M414"><mml:msub><mml:mrow><mml:mover accent="true"><mml:mrow><mml:mi>J</mml:mi></mml:mrow><mml:mo>̂</mml:mo></mml:mover></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

We have μm,ε1(x,t)=μ0(x,t)+ε1[D2μ1(x,t,vm-1+h)+D3μ1(x,t,vm-1+h)(vm-1+h)], hence |μm,ε1(x,t)|K̃(μ0)+(1+M1*)K̃M1*(μ1)χ1,withM1*=(N+2)M.

It follows from (4.41) thatĴ1(t)=0tds01|μm,ε1(x,s)|vmx2(x,s)dxχ10tvmx(s)2ds.

Second Integral <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M418"><mml:msub><mml:mrow><mml:mover accent="true"><mml:mrow><mml:mi>J</mml:mi></mml:mrow><mml:mo>̂</mml:mo></mml:mover></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

We note that f1[vm-1+h]-f1[h]2KM1*(f1)vm-1W1(T).

Therefore,Ĵ2(t)=2ε20tf1[vm-1+h]-f1[h]vm(s)dsTχ22ε2vm-1W1(T)2+0tvm(s)2ds, where χ2=χ2(M1*,f1)=2KM1*(f1).

Third Integral <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M422"><mml:msub><mml:mrow><mml:mover accent="true"><mml:mrow><mml:mi>J</mml:mi></mml:mrow><mml:mo>̂</mml:mo></mml:mover></mml:mrow><mml:mrow><mml:mn>3</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>t</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

First, we need to estimate /x[(μ1[vm-1+h]-μ1[h])hx].

From the equationx[(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 thatx[(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+hC0(Ω¯)hx+D3μ1[h]C0(Ω¯)vm-1W1(T)hxC0(Ω¯).

On the other hand, we haveμ1[vm-1+h]-μ1[h]C0(Ω¯)K̃M1*(μ1)vm-1W1(T),Djμ1[vm-1+h]-Djμ1[h]C0(Ω¯)K̃M1*(μ1)vm-1W1(T),j=1,3,D3μ1[h]C0(Ω¯)K̃M1*(μ1).

We deduce from (4.46) and (4.47) thatx[(μ1[vm-1+h]-μ1[h])hx](3+2M1*)M1*K̃M1*(μ1)vm-1W1(T).

Next, by (4.48), it follows thatĴ3(t)=2ε10tx[(μ1[vm-1+h]-μ1[h])hx]vm(s)dsTχ32ε2vm-1W1(T)2+0tvm(s)2ds, where χ3=χ3(M1*,μ1)=(3+2M1*)M1*K̃M1*(μ1).

Combining (4.39), (4.42), (4.44), and (4.49) givesvm(t)2+μ*vmx(t)2TE*2ε2N+2+T(χ22+χ32)ε2vm-1W1(T)2+30tvm(s)2ds+χ10tvmx(s)2dsTE*2ε2N+2+T(χ22+χ32)ε2vm-1W1(T)2+(3+χ1μ*)0t(vm(s)2+μ*vmx(s)2)ds.

Using Gronwall's lemma, we deduce from (4.50) thatvmW1(T)σTvm-1W1(T)+δ,m1, whereσT=χ22+χ32ηT,δ=ηTE*εN+1,ηT=(1+1μ*)Texp[T2(3+χ1μ*)].

We can assume thatσT<1, with sufficiently small T>0.

Lemma 4.5.

Let the sequence {ζm} satisfy ζmσζm-1+δm1,ζ0=0, where 0σ<1,δ0 are the given constants. Then, ζmδ(1-σ)m1.

This lemma is useful, as it will be said below, and it is easy to prove.

Applying Lemma 4.5 with ζm=vmW1(T),σ=σT=χ22+χ32ηT<1,δ=ηTE*εN+1, it follows from (4.55) thatvmL(0,T;L2)+vmxL(0,T;L2)=vmW1(T)δ(1-σT)CTεN+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) yieldsvL(0,T;L2)+vxL(0,T;L2)CTεN+1, it means    thatu-|γ|NuγεγL(0,T;L2)+ux-|γ|NuγxεγL(0,T;L2)CTεN+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 ε, with εε*<1, the problem (Pε) has a unique weak solution u=uεW1(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 hold    for the problem in p small parameters ε1,,εp as follows: 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:

μC2([0,1]×+),μiCN+1([0,1]×+×),μ0μ*>0,μi0,i=1,2,,p,

f0C1([0,1]×+),fiCN([0,1]×+×3),i=1,2,,p.

For a multi-index α=(α1,,αp)+p, and ε=(ε1,,εp)p, we also put |α|=α1++αp,α!=α1!αp!,ε=ε12++εp2,εα=ε1α1εpαp,α,βZ+p,αβαiβii=1,,p.

Let u0 be a unique weak solution of the problem (P0), which is (P̂ε) corresponding to ε=(0,,0).    Let the sequence of weak solutions uγ,γ+p,1|γ|N    be defined by the problems (P̃γ), in which Fγ,γ+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 ε, with εε*<1, the problem (P̂ε) has a unique weak solution u=uεW1(M,T) satisfying an asymptotic estimation up to order N+1 as follows: u-|γ|NuγεγL(0,T;L2)+ux-|γ|NuγxεγL(0,T;L2)CTεN+1.

The proof of Theorem 4.7 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, 1719]. However, to our knowledge, in the case of asymptotic expansion in many small parameters, there is only partial results, for example, [57, 14], concerning asymptotic expansion of solutions in two or three small parameters.

Acknowledgments

The authors wish to express their sincere thanks to the referees for the suggestions and valuable comments. The authors are also extremely grateful for the support given by Vietnam's National Foundation for Science and Technology Development (NAFOSTED) under Project no. 101.01-2010.15.

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