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ISRN Applied Mathematics
Volume 2011 (2011), Article ID 625908, 33 pages
http://dx.doi.org/10.5402/2011/625908
Research Article

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

1Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Vietnam
2Department of Mathematics, University of Economics of Ho Chi Minh City, 59C Nguyen Dinh Chieu Street, District 3, Ho Chi Minh City, Vietnam
3Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University Ho Chi Minh City, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Vietnam

Received 9 March 2011; Accepted 12 April 2011

Academic Editor: F. Jauberteau

Copyright © 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.

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.

1. Introduction

In this paper, we consider the following Dirichlet problem: 𝑢𝑡𝑡𝜕𝜕𝑥𝜇(𝑥,𝑡,𝑢)𝑢𝑥=𝑓𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡,0<𝑥<1,0<𝑡<𝑇,(1.1)𝑢(0,𝑡)=𝑢(1,𝑡)=0,(1.2)𝑢(𝑥,0)=̃𝑢0(𝑥),𝑢𝑡(𝑥,0)=̃𝑢1(𝑥),(1.3) where ̃𝑢0,̃𝑢1,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 [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 equation𝑢𝑡𝑡𝑢𝑥𝑥=2𝛼𝑢𝑡+𝜀𝑓𝑡,𝑥,𝑢,𝑢𝑡,𝑢𝑥.(1.4)

In [12], 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:𝑢𝑡𝑡𝑢𝑥𝑥=𝑔(𝑡,𝑥,𝑢)+𝑔1(𝑢)𝑢𝑡,(1.5) 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 equation𝑢𝑡𝑡𝑢𝑥𝑥=𝑔(𝑡,𝑥,𝑢)+𝑔1(𝑢)𝑢𝑡+𝜀𝑞𝑡,𝑥,𝑢,𝑢𝑡,𝑢𝑥,(1.6) 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 equation𝑢𝑡𝑡𝑢𝑥𝑥=𝑓𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡+𝜀𝑓1𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡,(1.7) with the mixed nonhomogeneous conditions𝑢𝑥(0,𝑡)0𝑢(0,𝑡)=𝑔0(𝑡),𝑢(1,𝑡)=𝑔1(𝑡).(1.8)

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𝜕𝜇𝜕𝑥0(𝑥,𝑡)+𝜀1𝜇1𝑢(𝑥,𝑡,𝑢)𝑥=𝑓0(𝑥,𝑡)+𝜀2𝑓1𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡,(1.9) 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𝜕𝜇𝜕𝑥0(𝑥,𝑡)+𝑝𝑖=1𝜀𝑖𝜇𝑖𝑢(𝑥,𝑡,𝑢)𝑥=𝑓0(𝑥,𝑡)+𝑝𝑖=1𝜀𝑖𝑓𝑖𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡,(1.10) 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),𝐻𝑚0=𝐻𝑚0(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)=𝑇0𝑢(𝑡)𝑝𝑋𝑑𝑡1/𝑝,if1𝑝<,esssup0<𝑡<𝑇𝑢(𝑡)𝑋,if𝑝=.(2.1)

Let 𝑢(𝑡),𝑢(𝑡)=𝑢𝑡(𝑡)=̇𝑢(𝑡),𝑢(𝑡)=𝑢𝑡𝑡(𝑡)=̈𝑢(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=𝐷𝛼11𝐷𝛼22𝐷𝛼33𝐷𝛼44𝐷𝛼55𝑓; 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=𝐷𝛽11𝐷𝛽22𝐷𝛽33,𝛽=(𝛽1,𝛽2,𝛽3)3+,|𝛽|=𝛽1+𝛽2+𝛽3=𝑘.

On 𝐻1, we will use the following norms:𝑣𝐻1=𝑣2+𝑣𝑥21/2.(2.2)

Then, we have the following lemma.

Lemma 2.1. The imbedding 𝐻1𝐶0(Ω) is compact and 𝑣𝐶0(Ω)2𝑣𝐻1𝑣𝐻1.(2.3)

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

Remark 2.2. On 𝐻10,𝑣𝑣𝐻1 and 𝑣𝑣𝑥 are two equivalent norms. Furthermore, we have the following inequalities: 𝑣𝐶0(Ω)𝑣𝑥𝑣𝐻10.(2.4)

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 𝑢𝑊={𝑢𝐿(0,𝑇;𝐻10𝐻2)𝑢𝐿(0,𝑇;𝐻10),𝑢𝐿(0,𝑇;𝐿2)} such that 𝑢 verifies the following variational equation: 𝑢(𝑡),𝑤+𝜇(,𝑡,𝑢(𝑡))𝑢𝑥(𝑡),𝑤𝑥𝑓=,𝑡,𝑢(𝑡),𝑢𝑥(𝑡),𝑢(𝑡),𝑤,𝑤𝐻10,(2.5) and the initial conditions 𝑢(0)=̃𝑢0,𝑢(0)=̃𝑢1.(2.6)
(ii) With the regularity obtained by 𝑊𝑢, it also follows from Theorem 3.2 that the problem (1.1)–(1.3) has a unique strong solution 𝑢 that satisfies 𝑢𝐶00,𝑇;𝐻1𝐶10,𝑇;𝐿2𝐿0,𝑇;𝐻2,𝑢𝑡𝐿0,𝑇;𝐻1,𝑢𝑡𝑡𝐿0,𝑇;𝐿2.(2.7)

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

Also, if (𝑢0,𝑢1)(𝐻10𝐻2)×𝐻10, 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)̃𝑢0𝐻10𝐻2,̃𝑢1𝐻10, (𝐻2)𝜇𝐶2([0,1]×+×),𝜇(𝑥,𝑡,𝑧)𝜇>0,forall(𝑥,𝑡,𝑧)[0,1]×+×, (𝐻3)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:𝐾𝑀(𝜇)=𝜇𝐶2(𝐷𝑀),𝐾(3.1)𝑀(𝑓)=𝑓𝐶1(𝐷𝑀),(3.2) where 𝐷𝑀={(𝑥,𝑡,𝑧)0𝑥1,0𝑡𝑇,|𝑧|𝑀} and 𝐷𝑀={(𝑥,𝑡,𝑢,𝑣,𝑤)+×+×30𝑥1,0𝑡𝑇,|𝑢|,|𝑣|,|𝑤|𝑀}.

For each 𝑇(0,𝑇] and 𝑀>0, we get𝑊(𝑀,𝑇)=𝑣𝐿0,𝑇;𝐻10𝐻2𝑣𝑡𝐿0,𝑇;𝐻10,𝑣𝑡𝑡𝐿2𝑄𝑇,with𝑣𝐿(0,𝑇;𝐻10𝐻2),𝑣𝑡𝐿(0,𝑇;𝐻10),𝑣𝑡𝑡𝐿2(𝑄𝑇),𝑊𝑀(3.3)1(𝑀,𝑇)=𝑣𝑊(𝑀,𝑇)𝑣𝑡𝑡𝐿0,𝑇;𝐿2,(3.4) where 𝑄𝑇=Ω×(0,𝑇).

We choose the first term 𝑢0̃𝑢0𝑊1(𝑀,𝑇). Suppose that𝑢𝑚1𝑊1(𝑀,𝑇),𝑚1.(3.5)

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,𝑣𝐻10,𝑢(3.6)𝑚(0)=̃𝑢0,𝑢𝑚(0)=̃𝑢1,(3.7) where𝜇𝑚(𝑥,𝑡)=𝜇𝑥,𝑡,𝑢𝑚1(𝑡),𝐹𝑚(𝑥,𝑡)=𝑓𝑥,𝑡,𝑢𝑚1(𝑥,𝑡),𝑢𝑚1(𝑥,𝑡),𝑢𝑚1(𝑥,𝑡).(3.8)

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 𝐻10𝑤𝑗(𝑥)=2sin(𝑗𝜋𝑥),𝑗, formed by the eigenfunctions of the Laplacian Δ=𝜕2/𝜕𝑥2. Put 𝑢𝑚(𝑘)(𝑡)=𝑘𝑗=1𝑐(𝑘)𝑚𝑗(𝑡)𝑤𝑗,(3.9) where the coefficients 𝑐(𝑘)𝑚𝑗 satisfy the system of linear differential equations ̈𝑢𝑚(𝑘)(𝑡),𝑤𝑗+𝜇𝑚(𝑡)𝑢𝑚(𝑘)(𝑡),𝑤𝑗=𝐹𝑚(𝑡),𝑤𝑗𝑢,1𝑗𝑘,(3.10)𝑚(𝑘)(0)=̃𝑢0𝑘,̇𝑢𝑚(𝑘)(0)=̃𝑢1𝑘,(3.11) where ̃𝑢0𝑘=𝑘𝑗=1𝛼𝑗(𝑘)𝑤𝑗̃𝑢0stronglyin𝐻10𝐻2,̃𝑢1𝑘=𝑘𝑗=1𝛽𝑗(𝑘)𝑤𝑗̃𝑢1stronglyin𝐻10.(3.12)
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)+𝑡0̈𝑢𝑚(𝑘)(𝑠)2𝑝𝑑𝑠,𝑚(𝑘)(𝑡)=̇𝑢𝑚(𝑘)(𝑡)2+𝜇𝑚(𝑡)𝑢𝑚(𝑘)(𝑡)2,𝑞𝑚(𝑘)(𝑡)=̇𝑢𝑚(𝑘)(𝑡)2+𝜇𝑚(𝑡)Δ𝑢𝑚(𝑘)(𝑡)2.(3.13)
Then, it follows from (3.9)–(3.11), (3.13) that 𝑠𝑚(𝑘)(𝑡)=𝑠𝑚(𝑘)(0)+2𝜇𝑚(0)̃𝑢0𝑘,Δ̃𝑢0𝑘+2𝐹𝑚(0),Δ̃𝑢0𝑘+𝑡0𝑑𝑠10𝜇𝑚||(𝑥,𝑠)𝑢𝑚(𝑘)||(𝑥,𝑠)2+||Δ𝑢𝑚(𝑘)||(𝑥,𝑠)2𝑑𝑥+2𝑡0𝐹𝑚(𝑠),̇𝑢𝑚(𝑘)(𝑠)𝑑𝑠+2𝑡0𝜕𝜕𝑠𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠),Δ𝑢𝑚(𝑘)(𝑠)𝑑𝑠2𝜇𝑚(𝑡)𝑢𝑚(𝑘)(𝑡),Δ𝑢𝑚(𝑘)𝐹(𝑡)2𝑚(𝑡),Δ𝑢𝑚(𝑘)(𝑡)+2𝑡0𝜕𝐹𝑚𝜕𝑡(𝑠),Δ𝑢𝑚(𝑘)(𝑠)𝑑𝑠+𝑡0̈𝑢𝑚(𝑘)(𝑠)2𝑑𝑠=𝑞𝑚(𝑘)(0)+2𝜇𝑚(0)̃𝑢0𝑘,Δ̃𝑢0𝑘+2𝐹𝑚(0),Δ̃𝑢0𝑘+7𝑗=1𝐼𝑗.(3.14)
Next, we will estimate the terms 𝐼𝑗,𝑗=1,2,,7 on the right-hand side of (3.14) as follows.
First Term 𝐼1
We have 𝜇𝑚(𝑡)=𝐷2𝜇𝑥,𝑡,𝑢𝑚1(𝑡)+𝐷3𝜇𝑥,𝑡,𝑢𝑚1𝑢(𝑡)𝑚1(𝑡).(3.15)

From (3.1), (3.5), and (3.8), we have ||𝜇𝑚||𝐾(𝑥,𝑡)(1+𝑀)𝑀(𝜇).(3.16)
Hence, 𝐼1=𝑡0𝑑𝑠10𝜇𝑚||(𝑥,𝑠)𝑢𝑚(𝑘)||(𝑥,𝑠)2+||Δ𝑢𝑚(𝑘)||(𝑥,𝑠)2𝑑𝑥1+𝑀𝜇𝐾𝑀(𝜇)𝑡0𝑠𝑚(𝑘)(𝑠)𝑑𝑠.(3.17)
Second Term
By using (𝐻3), we obtain from (3.2), (3.5), and (3.13)2 that 𝐼2=2𝑡0𝐹𝑚(𝑠),̇𝑢𝑚(𝑘)(𝑠)𝑑𝑠𝑇𝐾2𝑀(𝑓)+𝑡0𝑝𝑚(𝑘)(𝑠)𝑑𝑠.(3.18)
Third Term
The Cauchy-Schwartz inequality yields ||𝐼3||||||=2𝑡0𝜕𝜕𝑠𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠),Δ𝑢𝑚(𝑘)||||2(𝑠)𝑑𝑠𝜇𝑡0𝑟𝑚(𝑘)(𝑠)𝑞𝑚(𝑘)(𝑠)𝑑𝑠,(3.19) where 𝑟𝑚(𝑘)(𝑠)=𝜕/𝜕𝑠(𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)).

We note 𝑟𝑚(𝑘)(𝑠)=𝜇𝑚(𝑠)̇𝑢𝑚(𝑘)𝜕(𝑠)+𝜕𝑠𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)𝜇𝑚(𝑠)𝐶0(Ω)+1𝜇𝜕𝜕𝑠𝜇𝑚(𝑠)𝑠𝑚(𝑘)(𝑠).(3.20)
On the other hand, by 𝜇𝑚(𝑥,𝑠)=𝐷1𝜇(𝑥,𝑠,𝑢𝑚1(𝑥,𝑠))+𝐷3𝜇(𝑥,𝑠,𝑢𝑚1(𝑥,𝑠))𝑢𝑚1(𝑥,𝑠), it is implies that 𝜇𝑚(𝑠)𝐶0(Ω)𝐾𝑀(𝜇)1+𝑢𝑚1(𝑠)𝐶0(Ω)𝐾2(1+𝑀)𝑀(𝜇).(3.21)
Similarly, the following equality 𝜕𝜕𝑠𝜇𝑚(𝑥,𝑠)=𝐷1𝐷1𝜇𝑥,𝑠,𝑢𝑚1(𝑥,𝑠)+𝐷3𝐷1𝜇𝑥,𝑠,𝑢𝑚1𝑢(𝑥,𝑠)𝑚1+𝐷(𝑥,𝑠)1𝐷3𝜇𝑥,𝑠,𝑢𝑚1(𝑥,𝑠)+𝐷3𝐷3𝜇𝑥,𝑠,𝑢𝑚1𝑢(𝑥,𝑠)𝑚1(𝑥,𝑠)𝑢𝑚1(𝑥,𝑠)+𝐷3𝜇𝑥,𝑠,𝑢𝑚1(𝑥,𝑠)𝑢𝑚1(𝑥,𝑠)(3.22) gives 𝜕𝜕𝑠𝜇𝑚(𝑠)1+3𝑀+𝑀2𝐾𝑀(𝜇).(3.23)
It follows from (3.20)–(3.23) that 𝑟𝑚(𝑘)(𝑠)2(1+𝑀)+1+3𝑀+𝑀2𝜇𝐾𝑀(𝜇)𝑠𝑚(𝑘)(𝑠).(3.24)
Hence, we obtain from (3.19) and (3.24) that ||𝐼3||2𝜇2(1+𝑀)+1+3𝑀+𝑀2𝜇𝐾𝑀(𝜇)𝑡0𝑠𝑚(𝑘)(𝑠)𝑑𝑠.(3.25)
Fourth Term 𝐼4
By the Cauchy-Schwartz inequality, we have ||𝐼4||=|||2𝜇𝑚(𝑡)𝑢𝑚(𝑘)(𝑡),Δ𝑢𝑚(𝑘)|||1(𝑡)𝛽𝜇𝑚(𝑡)𝑢𝑚(𝑘)(𝑡)2+𝛽Δ𝑢𝑚(𝑘)(𝑡)2,(3.26) for all 𝛽>0. On the other hand 𝜇𝑚(𝑡)𝑢𝑚(𝑘)(=𝑡)𝜇𝑚(0)̃𝑢0𝑘+𝑡0𝜕𝜕𝑠𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)𝑑𝑠𝜇𝑚(0)𝐶0(Ω)̃𝑢0𝑘+𝑡0𝑟𝑚(𝑘)(𝑠)𝑑𝑠.(3.27)

Hence, we obtain from (3.26), (3.27) that ||𝐼4||𝛽𝜇𝑞𝑚(𝑘)2(𝑡)+𝛽𝜇𝑚(0)2𝐶0Ω̃𝑢0𝑘2+2𝛽𝑇2(1+𝑀)+1+3𝑀+𝑀2𝜇2𝐾2𝑀(𝜇)𝑡0𝑠𝑚(𝑘)(𝑠)𝑑𝑠,(3.28) for all 𝛽>0.
Fifth Term 𝐼5
By (3.5), (3.8), and (3.13), we obtain ||𝐼5||=|||𝐹2𝑚(𝑡),Δ𝑢𝑚(𝑘)|||1(𝑡)𝛽𝐹𝑚(𝑡)2+𝛽Δ𝑢𝑚(𝑘)(𝑡)22𝛽𝐹𝑚(0)2+2𝛽𝑇𝑇0𝜕𝐹𝑚𝜕𝑠(𝑠)2𝛽𝑑𝑠+𝜇𝑠𝑚(𝑘)(𝑡),𝛽>0.(3.29)

Note that 𝜕𝐹𝑚𝜕𝑡(𝑡)=𝐷2𝑓𝑢𝑚1+𝐷3𝑓𝑢𝑚1𝑢𝑚1(𝑡)+𝐷4𝑓𝑢𝑚1𝑢𝑚1(𝑡)+𝐷5𝑓𝑢𝑚1𝑢𝑚1(𝑡),(3.30) where we use the notation 𝐷𝑖𝑓[𝑢𝑚1]=𝐷𝑖𝑓(𝑥,𝑡,𝑢𝑚1(𝑥,𝑡),𝑢𝑚1(𝑥,𝑡),𝑢𝑚1(𝑥,𝑡)),𝑖=2,,5. By (3.2), (3.5), and (3.30), we obtain 𝜕𝐹𝑚𝜕𝑡(𝑡)𝐾𝑀𝑢(𝑓)1+2𝑀+𝑚1.(𝑡)(3.31)
Hence, we deduce from (3.29) and (3.31) that ||𝐼5||2𝛽𝐹𝑚(0)2+4𝛽𝑇𝐾2𝑀(𝑓)(1+2𝑀)2𝑇+𝑀2+𝛽𝜇𝑠𝑚(𝑘)(𝑡),𝛽>0.(3.32)
Sixth Term 𝐼6
By (3.2), (3.5), (3.13)3, and (3.31), we get ||𝐼6||||||=2𝑡0𝜕𝐹𝑚𝜕𝑡(𝑠),Δ𝑢𝑚(𝑘)||||(𝑠)𝑑𝑠𝑡0𝜕𝐹𝑚𝜕𝑡(𝑠)𝑑𝑠+𝑡0𝜕𝐹𝑚𝜕𝑡(𝑠)Δ𝑢𝑚(𝑘)(𝑠)2𝑑𝑠𝐾𝑀(𝑓)(1+2𝑀)𝑇+𝑇𝑇0𝑢𝑚1(𝑠)2𝑑𝑠1/2+1𝜇𝐾𝑀(𝑓)𝑡0𝑢1+2𝑀+𝑚1𝑞(𝑠)𝑚(𝑘)(𝑠)𝑑𝑠𝐾𝑀(𝑓)(1+2𝑀)𝑇++1𝑇𝑀𝜇𝐾𝑀(𝑓)𝑡0𝑢1+2𝑀+𝑚1𝑞(𝑠)𝑚(𝑘)(𝑠)𝑑𝑠.(3.33)
Seventh Term 𝐼7
Equation (3.10) is rewritten as follows: ̈𝑢𝑚(𝑘)(𝑡),𝑤𝑗𝜕𝜇𝜕𝑥𝑚(𝑡)𝑢𝑚(𝑘)(𝑡),𝑤𝑗=𝐹𝑚(𝑡),𝑤𝑗,1𝑗𝑘.(3.34)

Hence, by replacing 𝑤𝑗 with ̈𝑢𝑚(𝑘)(𝑡) and integrating 𝐼7=𝑡0̈𝑢𝑚(𝑘)(𝑠)2𝑑𝑠2𝑡0𝜕𝜇𝜕𝑥𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)2𝑑𝑠+2𝑡0𝐹𝑚(𝑠)2𝑑𝑠2𝑡0𝜕𝜇𝜕𝑥𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)2𝑑𝑠+2𝑇𝐾2𝑀(𝑓),(3.35) we need, estimate 𝜕/𝜕𝑥(𝜇𝑚(𝑠)𝑣𝑚(𝑘)(𝑠)).
Combining (3.1), (3.5), and (3.13) yields 𝜕𝜇𝜕𝑥𝑚(𝑠)𝑢𝑚(𝑘)=(𝑠)𝜇𝑚(𝑠)𝑢𝑚(𝑘)(𝑠)+𝜇𝑚(𝑠)Δ𝑢𝑚(𝑘)(𝑠)𝜇𝑚(𝑠)𝐶0(Ω)𝑢𝑚(𝑘)+𝜇(𝑠)𝑚(𝑠)𝐶0(Ω)Δ𝑢𝑚(𝑘)2(𝑠)𝜇𝐾(1+𝑀)𝑀(𝜇)𝑝𝑚(𝑘)1(𝑠)+𝜇𝐾𝑀(𝜇)𝑞𝑚(𝑘)3(𝑠)𝜇𝐾(1+𝑀)𝑀(𝜇)𝑠𝑚(𝑘)(𝑠).(3.36)
Therefore, from (3.35) and (3.36), we obtain 𝐼72𝑇𝐾2𝑀(𝑓)+18𝜇(1+𝑀)2𝐾2𝑀(𝜇)𝑡0𝑠𝑚(𝑘)(𝑠)𝑑𝑠.(3.37)
Choosing 𝛽>0, with 2𝛽/𝜇1/2, 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)(𝐶𝑡)0𝑘+𝐶1(𝑀,𝑇)+𝑡0𝐶2(2𝑀,𝑇)+𝜇𝐾𝑀(𝑢𝑓)𝑚1(𝑠𝑠)𝑚(𝑘)(𝑠)𝑑𝑠,(3.38) where 𝐶0𝑘=𝐶0𝑘𝛽,𝑓,𝜇,̃𝑢0,̃𝑢1,̃𝑢0𝑘,̃𝑢1𝑘=2𝑠𝑚(𝑘)(0)+4𝜇𝑚(0)̃𝑢0𝑘,Δ̃𝑢0𝑘+4𝐹𝑚(0),Δ̃𝑢0𝑘+4𝛽𝜇𝑚(0)2𝐶0Ω̃𝑢0𝑘2+4𝛽𝐹𝑚(0)2,𝐶1𝐶(𝑀,𝑇)=14(𝛽,𝑓,𝑀,𝑇)=23+𝛽(1+2𝑀)2𝑇+𝑀2𝑇𝐾2𝑀(𝑓)+2𝑀+(1+2𝑀)𝑇𝑇𝐾𝑀(𝐶𝑓),2𝐶(𝑀,𝑇)=22(𝛽,𝑓,𝜇,𝑀,𝑇)=2+𝜇0(1+2𝑀)𝐾𝑀+2(𝑓)𝜇1+4𝜇(1+𝑀)+21+3𝑀+𝑀2𝐾𝑀+4(𝜇)𝜇1𝛽𝑇2(1+𝑀)𝜇+1+3𝑀+𝑀22+9(1+𝑀)2𝐾2𝑀(𝜇).(3.39)
By (𝐻1), we deduce from (3.12), (3.39)1 that there exists 𝑀>0 independent of 𝑚 and 𝑘, such that 𝐶0𝑘12𝑀2.(3.40)
Notice that by (𝐻3), we deduce from (3.39)2,3 that lim𝑇0+𝐶1(𝑀,𝑇)=lim𝑇0+𝑇𝐶2(𝑀,𝑇)=0.(3.41)
So, from (3.39) and (3.41), we can choose 𝑇>0 such that 12𝑀2+𝐶1𝑇𝐶(𝑀,𝑇)exp22(𝑀,𝑇)+𝜇0𝐾𝑀(𝑓)𝑇𝑀𝑀2𝑘,(3.42)𝑇=11+𝜇𝑇4𝐾2𝑀(𝑓)+(4+𝑀)2𝑀2𝐾2𝑀(𝜇)𝑒𝑇[1+((1+𝑀)/2𝜇)𝐾𝑀(𝜇)]<1.(3.43)
Finally, it follows from (3.38), (3.40), and (3.42) that 𝑠𝑚(𝑘)(𝑡)𝑀2𝐶exp𝑇22(𝑀,𝑇)𝜇0𝐾𝑀(𝑓)+𝑇𝑀𝑡0𝐶22(𝑀,𝑇)+𝜇0𝐾𝑀𝑢(𝑓)𝑚1𝑠(𝑠)𝑚(𝑘)(𝑠)𝑑𝑠.(3.44)
By using Gronwall's lemma, we deduce from (3.44) that 𝑠𝑚(𝑘)(𝑡)𝑀2𝐶exp𝑇22(𝑀,𝑇)𝜇0𝐾𝑀(𝑓)𝑇𝑀×exp𝑇0𝐶22(𝑀,𝑇)+𝜇0𝐾𝑀𝑢(𝑓)𝑚1(𝑠)𝑑𝑠𝑀2𝐶exp𝑇22(𝑀,𝑇)𝜇0𝐾𝑀(𝑓)𝑇𝐶𝑇𝑀×exp22(𝑀,𝑇)+𝜇0𝐾𝑀(𝑓)𝑇𝑢𝑚1𝐿2(𝑄𝑇)𝑀2.(3.45)
Therefore, we have 𝑢𝑚(𝑘)𝑊(𝑀,𝑇),𝑚,𝑘.(3.46)
Step 3. Limiting process.
From (3.46), we can extract from {𝑢𝑚(𝑘)} a subsequence still denoted by {𝑢𝑚(𝑘)} such that 𝑢𝑚(𝑘)𝑢𝑚in𝐿0,𝑇;𝐻10𝐻2weak,̇𝑢𝑚(𝑘)𝑢𝑚in𝐿0,𝑇;𝐻10weak,̈𝑢𝑚(𝑘)𝑢𝑚in𝐿2𝑄𝑇weak,(3.47) as 𝑘, and 𝑢𝑚𝑊(𝑀,𝑇).(3.48)
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+𝑓𝑥,𝑡,𝑢𝑚1,𝑢𝑚1,𝑢𝑚1𝐿0,𝑇;𝐿2.(3.49)
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)=𝑤𝐿0,𝑇;𝐻10𝑤𝐿0,𝑇;𝐿2,(3.50) with the following estimation: 𝑢𝑚𝑢𝐿(0,𝑇;𝐻10)+𝑢𝑚𝑢𝐿(0,𝑇;𝐿2)𝐶𝑘𝑚𝑇,𝑚,(3.51) where 𝑘𝑇<1 as in (3.43) and 𝐶 is a constant depending only on 𝑇,̃𝑢0,̃𝑢1 and 𝑘𝑇.

Proof. (i) The existence. First, we note that 𝑊1(𝑇) is a Banach space with respect to the norm (see Lions [15]) 𝑤𝑊1(𝑇)=𝑤𝐿(0,𝑇;𝐻10)+𝑤𝐿(0,𝑇;𝐿2).(3.52)
Next, we prove that {𝑢𝑚} is a Cauchy sequence in 𝑊1(𝑇). Let 𝑣𝑚=𝑢𝑚+1𝑢𝑚. Then, 𝑣𝑚 satisfies the variational problem 𝑣𝑚+𝜇(𝑡),𝑤𝑚+1(𝑡)𝑣𝑚=𝜕(𝑡),𝑤𝜇𝜕𝑥𝑚+1(𝑡)𝜇𝑚(𝑡)𝑢𝑚+𝐹(𝑡),𝑤𝑚+1(𝑡)𝐹𝑚(𝑡),𝑤,𝑤𝐻10,𝑣𝑚(0)=𝑣𝑚(0)=0.(3.53)
Taking 𝑤=𝑣𝑚 in (3.53)1, after integrating in 𝑡, we get 𝑍𝑚(𝑡)=𝑡0𝑑𝑠10𝜇𝑚+1||(𝑥,𝑠)𝑣𝑚||(𝑠)2𝑑𝑥+2𝑡0𝐹𝑚+1(𝑠)𝐹𝑚(𝑠),𝑣𝑚(𝑠)𝑑𝑠+2𝑡0𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚(𝑠),𝑣𝑚(𝑠)𝑑𝑠=3𝑖=1𝐽𝑖,(3.54) in which 𝑍𝑚𝑣(𝑡)=𝑚(𝑡)2+𝜇𝑚+1(𝑡)𝑣𝑚(𝑡)2,(3.55) and all integrals on the right-hand side of (3.54) are estimated as follows.
First Integral
By (3.16), we obtain ||𝐽1||||||𝑡0𝑑𝑠10𝜇𝑚+1||(𝑥,𝑠)𝑣𝑚||(𝑠)2||||𝑑𝑥1+𝑀𝜇𝐾𝑀(𝜇)𝑡0𝑍𝑚(𝑠)𝑑𝑠.(3.56)
Second Integral
By (𝐻3), 𝐹𝑚+1(𝑡)𝐹𝑚(𝑡)2𝐾𝑀(𝑓)𝑣𝑚1+𝑣(𝑡)𝑚1(𝑡)2𝐾𝑀𝑣(𝑓)𝑚1𝑊1(𝑇),(3.57) so ||𝐽2||||||2𝑡0𝐹𝑚+1(𝑠)𝐹𝑚(𝑠),𝑣𝑚||||(𝑠)𝑑𝑠4𝐾𝑀𝑣(𝑓)𝑚1𝑊1(𝑇)𝑡0𝑣𝑚(𝑠)𝑑𝑠4𝑇𝐾2𝑀𝑣(𝑓)𝑚12𝑊1(𝑇)+𝑡0𝑍𝑚(𝑠)𝑑𝑠.(3.58)
Third Integral
Using (𝐻2) again, we get ||𝐽3||||||=2𝑡0𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚(𝑠),𝑣𝑚||||(𝑠)𝑑𝑠𝑡0𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚(𝑠)2𝑑𝑠+𝑡0𝑍𝑚(𝑠)𝑑𝑠.(3.59)

Note that 𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚=𝜇(𝑠)𝑚+1(𝑠)𝜇𝑚(𝑠)Δ𝑢𝑚+𝐷(𝑠)1𝜇𝑢𝑚𝐷1𝜇𝑢𝑚1𝑢𝑚𝐷(𝑠)+3𝜇𝑢𝑚𝐷3𝜇𝑢𝑚1||𝑢𝑚||(𝑠)2+𝐷3𝜇𝑢𝑚1𝑣𝑚1(𝑠)𝑢𝑚(𝑠).(3.60)
Hence, 𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚𝜇(𝑠)𝑚+1(𝑠)𝜇𝑚(𝑠)𝐶0(Ω)Δ𝑢𝑚+𝐷(𝑠)1𝜇𝑢𝑚𝐷1𝜇𝑢𝑚1𝐶0(Ω)𝑢𝑚+𝐷(𝑠)1𝜇𝑢𝑚𝐷1𝜇𝑢𝑚1𝐶0(Ω)𝑢𝑚(𝑡)2𝐶0Ω+𝐷3𝜇𝑢𝑚1𝐶0(Ω)𝑢𝑚(𝑠)𝐶0(Ω)𝑣𝑚1.(𝑠)(3.61)
We also note that 𝜇𝑚+1(𝑠)𝜇𝑚(𝑠)𝐶0(Ω)𝐾𝑀𝑤(𝜇)𝑚1𝑊1(𝑇),𝐷𝑖𝜇[𝑢𝑚]𝐷𝑖𝜇[𝑢𝑚1]𝐶0(Ω)𝐾𝑀𝑤(𝜇)𝑚1𝑊1(T),𝑖=1,3,𝑢𝑚(𝑠)𝐶0(Ω)2𝑢𝑚(𝑠)𝐻12𝑢𝑚(𝑠)2+Δ𝑢𝑚(𝑠)2𝐷2𝑀,3𝜇[𝑢𝑚]𝐶0(Ω)𝐾𝑀(𝜇),(3.62) where we use the notation 𝐷𝑖𝜇[𝑢𝑚1]=𝐷𝑖𝜇(𝑥,𝑡,𝑢𝑚(𝑥,𝑡)),𝑖=1,2,3. Therefore, it implies from (3.61) and (3.62) that 𝜕𝜇𝜕𝑥𝑚+1(𝑠)𝜇𝑚(𝑠)𝑢𝑚𝐾(𝑠)(4+𝑀)𝑀𝑀(𝜇)𝑣𝑚1𝑊1(𝑇).(3.63)
Hence, ||𝐽3||(4+𝑀)2𝑀2𝑇𝐾2𝑀(𝜇)𝑣𝑚12𝑊1(𝑇)+𝑡0𝑍𝑚(𝑠)𝑑𝑠.(3.64)
Combining (3.54)–(3.56), (3.58), and (3.64) yields 𝑍𝑚(𝑡)𝑇4𝐾2𝑀(𝑓)+(4+𝑀)2𝑀2𝐾2𝑀(𝜇)𝑣𝑚12𝑊1(𝑇)+2+1+𝑀𝜇𝐾𝑀(𝜇)𝑡0𝑍𝑚(𝑠)𝑑𝑠.(3.65)
Using Gronwall's lemma, (3.65) gives 𝑣𝑚𝑊1(𝑇)𝑘𝑇𝑣𝑚1𝑊1(𝑇)𝑚,(3.66) where 𝑘𝑇<1 as in (3.43).
Hence, we obtain from (3.66) that 𝑢𝑚+𝑝𝑢𝑚𝑊1(𝑇)𝑘𝑚𝑇1𝑘𝑇𝑢1𝑢0𝑊1(𝑇)𝑚,𝑝,(3.67)
It follows that {𝑢𝑚} is a Cauchy sequence in 𝑊1(𝑇). Then, there exists 𝑢𝑊1(𝑇) such that 𝑢𝑚𝑢stronglyin𝑊1(𝑇).(3.68)
On the other hand, from (3.48), we deduce the existence of a subsequence {𝑢𝑚𝑗} of {𝑢𝑚} such that 𝑢𝑚𝑗𝑢in𝐿0,𝑇;𝐻10𝐻2weak,𝑢𝑚𝑗𝑢in𝐿0,𝑇;𝐻10weak,𝑢𝑚𝑗𝑢in𝐿2𝑄𝑇weak,(3.69)𝑢𝑊(𝑀,𝑇).(3.70)
Note that ||𝜇𝑚||𝐾(𝑥,𝑡)𝜇(𝑥,𝑡,𝑢(𝑥,𝑡))𝑀𝑢(𝜇)𝑚1𝑢𝑊1(𝑇),𝐹𝑚(𝑡)𝑓,𝑡,𝑢(𝑡),𝑢𝑥(𝑡),𝑢(𝑡)2𝐾𝑀𝑢(𝑓)𝑚1𝑢𝑊1(𝑇).(3.71)
Hence, from (3.68) and (3.71), we obtain 𝜇𝑚𝜇(,,𝑢)stronglyin𝐿𝑄𝑇,𝐹𝑚𝑓,𝑡,𝑢(𝑡),𝑢𝑥(𝑡),𝑢(𝑡)stronglyin𝐿0,𝑇;𝐿2.(3.72)
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))𝑢𝑥𝑓(𝑡),𝑤=,𝑡,𝑢(𝑡),𝑢𝑥(𝑡),𝑢(𝑡),𝑤,𝑤𝐻10,𝑢(0)=̃𝑢0,𝑢(0)=̃𝑢1.(3.73)
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]𝑢2𝑥[𝑢]𝑢+𝜇𝑥𝑥+𝑓𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝐿0,𝑇;𝐿2,(3.74) 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=𝑢1𝑢2 satisfies the variational problem 𝑢(𝑡),𝑤+𝜇1(𝑡)𝑢𝑥(𝑡),𝑤𝑥𝜕=𝜇𝜕𝑥1(𝑡)𝜇2𝑢(𝑡)2𝑥(𝑡),𝑤+𝐹2(𝑡)𝐹1(𝑡),𝑤,𝑤𝐻10,𝑢(0)=𝑢𝜇(0)=0,𝑖(𝑡)=𝜇𝑥,𝑡,𝑢𝑖𝑢(𝑡)𝜇𝑖,𝐹𝑖(𝑡)=𝑓𝑥,𝑡,𝑢𝑖(𝑡),𝑢𝑖𝑥(𝑡),𝑢𝑖(𝑡),𝑖=1,2.(3.75)
We take 𝑤=𝑢 in (3.75)1 and integrate in 𝑡 to get 𝜌(𝑡)=𝑡0𝑑𝑠10𝜇1(𝑥,𝑠)𝑢2𝑥(𝑥,𝑠)𝑑𝑥+2𝑡0𝐹1(𝑠)𝐹2(𝑠),𝑢(𝑠)𝑑𝑠+2𝑡0𝜕𝜇𝜕𝑥1(𝑠)𝜇2𝑢(𝑠)2𝑥(𝑠),𝑢𝑑𝑠3𝑖=1𝜌𝑖(𝑡),(3.76) where 𝑢𝜌(𝑡)=(𝑡)2+𝜇1(𝑡)𝑢𝑥(𝑡)2.(3.77)
We now estimate the terms on the right-hand side of (3.76) as follows: 𝜌1(𝑡)=𝑡0𝑑𝑠10𝜇1(𝑥,𝑠)𝑢2𝑥1(𝑥,𝑠)𝑑𝑥𝜇𝐾(1+𝑀)𝑀(𝜇)𝑡0𝜌(𝑠)𝑑𝑠𝜌𝑀(1)𝑡0𝜌𝜌(𝑠)𝑑𝑠,(3.78)2(𝑡)=2𝑡0𝐹1(𝑠)𝐹2(𝑠),𝑢(𝑠)𝑑𝑠4𝐾𝑀(𝑓)𝑡0𝑢𝑥(+𝑢𝑠)(𝑢𝑠)(1𝑠)𝑑𝑠41+𝜇𝐾𝑀(𝑓)𝑡0𝜌(𝑠)𝑑𝑠𝜌𝑀(2)𝑡0𝜌𝜌(𝑠)𝑑𝑠,(3.79)3(𝑡)=2𝑡0𝜕𝜇𝜕𝑥1(𝑠)𝜇2(𝑢𝑠)2𝑥(𝑠),𝑢𝑑𝑠2𝑡0𝜕𝜇𝜕𝑥1(𝑠)𝜇2(𝑢𝑠)2𝑥(𝑢𝑠)(𝑠)𝑑𝑠.(3.80)
On the other hand 𝜕𝜇𝜕𝑥1(𝑠)𝜇2𝑢(𝑠)2𝑥=𝜇(𝑠)1(𝑠)𝜇2𝑢(𝑠)2𝑥𝑥𝐷(𝑠)+1𝜇𝑢1𝐷1𝜇𝑢2𝑢2𝑥+𝐷(𝑠)3𝜇𝑢1𝐷3𝜇𝑢2𝑢1𝑥𝑢2𝑥+D3𝜇𝑢2𝑢𝑥𝑢2𝑥.(3.81)
Hence, 𝜕𝜇𝜕𝑥1(𝑠)𝜇2𝑢(𝑠)2𝑥𝜇(𝑠)1(𝑠)𝜇2(𝑠)𝐶0(Ω)𝑢2𝑥𝑥+𝐷(𝑠)1𝜇[𝑢1]𝐷1𝜇[𝑢2]𝐶0(Ω)𝑢2𝑥+𝐷(𝑠)3𝜇[𝑢1]𝐷3𝜇[𝑢2]𝐶0(Ω)𝑢1𝑥(𝑠)𝐶0(Ω)𝑢2𝑥(𝑠)𝐶0(Ω)+𝐷3𝜇[𝑢2]𝐶0(Ω)𝑢𝑥𝑢(𝑠)2𝑥(𝑠)𝐶0(Ω)𝐾(3+𝑀)𝑀𝑀𝑢(𝜇)𝑥.(𝑠)(3.82)
It follows from (3.80), (3.82) that 𝜌3(1𝑡)𝜇(𝐾3+𝑀)𝑀𝑀(𝜇)𝑡0𝜌(𝑠)𝑑𝑠𝜌𝑀(3)𝑡0𝜌(𝑠)𝑑𝑠.(3.83)
Combining (3.76)–(3.79) and (3.83) yields 𝜌𝜌(𝑡)𝑀(1)+𝜌𝑀(2)+𝜌𝑀(3)𝑡0𝜌(𝑠)𝑑𝑠.(3.84)
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 [4] standing for (1.2), we obtained some similar results in [4].
(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 [8] standing for (1.2), some results as above were given in [8].

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(𝑄𝑇).

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)𝜇0𝐶2([0,1]×+),𝜇1𝐶𝑁+1([0,1]×+×),𝜇0𝜇>0,𝜇10, (𝐻5)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𝜕𝜇𝜕𝑥𝜀1(𝑥,𝑡,𝑢)𝑢𝑥=𝐹𝜀2𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡𝑢,0<𝑥<1,0<𝑡<𝑇,𝑢(0,𝑡)=𝑢(1,𝑡)=0,(𝑥,0)=̃𝑢0(𝑥),𝑢𝑡(𝑥,0)=̃𝑢1𝜇(𝑥),𝜀1(𝑥,𝑡,𝑢)=𝜇0(𝑥,𝑡)+𝜀1𝜇1𝐹(𝑥,𝑡,𝑢),𝜀2𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡=𝑓0(𝑥,𝑡)+𝜀2𝑓1𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡.(𝑃𝜀)
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=!,𝜀𝜀21+𝜀22,𝜀𝛼=𝜀𝛼11𝜀𝛼22,𝛼,𝛽2+,𝛼𝛽𝛼𝑖𝛽𝑖𝑖=1,2.(4.1)
We first note the following lemma.

Lemma 4.1. Let 𝑚,𝑁 and 𝑢𝛼,𝛼2+,1|𝛼|𝑁. Then, 1|𝛼|𝑁𝑢𝛼𝜀𝛼𝑚=𝑚|𝛼|𝑚𝑁𝑇𝛼(𝑚)[𝑢]𝜀𝛼,(4.2) where the coefficients 𝑇𝛼(𝑚)[𝑢],𝑚|𝛼|𝑚𝑁 depending on 𝑢=(𝑢𝛼),𝛼2+,1|𝛼|𝑁are defined by the recurrent formulas 𝑇𝛼(1)[𝑢]=𝑢𝛼𝑇,1|𝛼|𝑁,𝛼(𝑚)[𝑢]=𝛽𝐴𝛼(𝑚)𝑢𝛼𝛽𝑇𝛽(𝑚1)[𝑢]𝐴,𝑚|𝛼|𝑚𝑁,𝑚2,𝛼(𝑚)=𝛽2+||||||𝛽||.𝛽𝛼,1𝛼𝛽𝑁,𝑚1(𝑚1)𝑁(4.3)

The proof of Lemma 4.1 can be found in [6].

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𝜕𝜇𝜕𝑥0(𝑥,𝑡)𝑢0𝑥=𝑓0𝑢(𝑥,𝑡),0<𝑥<1,0<𝑡<𝑇,0(0,𝑡)=𝑢0𝑢(1,𝑡)=0,0(𝑥,0)=̃𝑢0(𝑥),𝑢0(𝑥,0)=̃𝑢1𝑢(𝑥),0𝑊1(𝑀,𝑇).(𝑃0)

Let us consider the sequence of weak solutions 𝑢𝛾,𝛾2+,1|𝛾|𝑁, defined by the following problems: 𝑢𝛾𝜕𝜇𝜕𝑥0(𝑥,𝑡)𝑢𝛾𝑥=𝐹𝛾𝑢,0<𝑥<1,0<𝑡<𝑇,𝛾(0,𝑡)=𝑢𝛾𝑢(1,𝑡)=0,𝛾(𝑥,0)=𝑢𝛾𝑢(𝑥,0)=0,𝛾𝑊1(𝑃(𝑀,𝑇),𝛾) where 𝐹𝛾,𝛾2+,1|𝛾|𝑁 are defined by the recurrent formulas as follows:𝐹𝛾=𝜋𝛾(2)𝑓1+||𝛾||2|𝜈|,𝜈𝛾𝜕𝜌𝜕𝑥𝜈(1)𝜇1𝑢𝛾𝜈||𝛾||,1𝑁,(4.4) with 𝜌𝛿[𝜇1]=𝜌𝛿[𝜇1;{𝑢𝛾}𝛾𝛿], 𝜌𝛿(1)[𝜇1