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]=𝜌𝛿(1)[𝜇1;{𝑢𝛾}𝛾𝛿], 𝜋𝛿[𝑓1]=𝜋𝛿[𝑓1;{𝑢𝛾}𝛾𝛿], 𝜋𝛿(2)[𝑓1]=𝜋𝛿(2)[𝑓1;{𝑢𝛾}𝛾𝛿], |𝛿|𝑁1 defined by𝜌𝛿𝜇1=𝜇1𝑢0,||𝛿||=0,||𝛿||𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑇𝛿(𝑚)[𝑢]||𝛿||𝜌,1𝑁1,(4.5)𝛿(1)𝜇1=𝜌𝛿11,𝛿2𝜇1𝛿,𝛿=1,𝛿22+,𝜌𝛿(1)𝜇1=𝜌(1)0,𝛿2𝜇1=𝜌1,𝛿2𝜇1=0,if𝛿1𝜋=0,(4.6)𝛿𝑓1=𝑓1𝑢0,||𝛿||=0,||𝛿||1|𝑚|𝑚=(𝑚1,𝑚2,𝑚3)3+(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝛿1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼×[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢||𝛿||,1𝑁1,(4.7) where 𝑚=(𝑚1,𝑚2,𝑚3)3+, |𝑚|=𝑚1+𝑚2+𝑚3, 𝑚!=𝑚1!𝑚2!𝑚3!, 𝐷𝑚𝑓𝑗=𝐷𝑚13𝐷𝑚24𝐷𝑚35𝑓𝑗, 𝐴(𝑚,𝑁)={(𝛼,𝛽,𝛾)(2+)3𝑚1|𝛼|𝑚1𝑁,𝑚2|𝛽|𝑚2𝑁,𝑚3|𝛾|𝑚3𝑁},𝜋𝛿(2)𝑓1=𝜋𝛿1,𝛿21𝑓1𝛿,𝛿=1,𝛿22+,𝜋𝛿(2)𝑓1=𝜋𝛿(2)1,0𝑓1=𝜋𝛿1,1𝑓1=0,if𝛿2=0.(4.8)

Then, we have the following lemma.

Lemma 4.2. Let 𝜌𝜈[𝜇1], 𝜋𝜈[𝑓1], |𝜈|𝑁1 be the functions defined by (4.5) and (4.7). Put =|𝛾|𝑁𝑢𝛾𝜀𝛾, then one has 𝜇1[]=|𝜈|𝑁1𝜌𝜈𝜇1𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝜇1,𝑓,𝜀(4.9)1[]=|𝜈|𝑁1𝜋𝜈𝑓1𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝑓1,,𝜀(4.10) where 𝑅(1)𝑁1[𝜇1,𝜀]𝐿(0,𝑇;𝐿2)+𝑅(1)𝑁1[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 =𝑢0+1|𝛾|𝑁𝑢𝛾𝜀𝛾𝑢0+1.
Using Taylor's expansion of the function 𝜇1[]=𝜇1[𝑢0+1] around the point 𝑢0 up to order 𝑁, we obtain from (4.2) that 𝜇1𝑢0+1=𝜇1𝑢0+𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑚1+1(𝑁1)!10(1𝜃)𝑁1𝐷𝑁3𝜇1𝑢0+𝜃1𝑁1𝑑𝜃=𝜇1𝑢0+𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑚|𝜈|𝑚𝑁𝑇𝜈(𝑚)[𝑢]𝜀𝜈+𝑅(1)𝑁1𝜇1,1=𝜇1𝑢0+𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑚|𝜈|𝑁1𝑇𝜈(𝑚)[𝑢]𝜀𝜈+𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑁|𝜈|𝑚𝑁𝑇𝜈(𝑚)[𝑢]𝜀𝜈+𝑅(1)𝑁1𝜇1,1,(4.11) where 𝑅(1)𝑁1𝜇1,1=1(𝑁1)!10(1𝜃)𝑁1𝐷𝑁3𝜇1𝑢0+𝜃1𝑁1𝑑𝜃.(4.12)
We note that 𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑚|𝜈|𝑁1𝑇𝜈(𝑚)[𝑢]𝜀𝜈=1|𝜈|𝑁1|𝜈|𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑇𝜈(𝑚)[𝑢]𝜀𝜈.(4.13)
On the other hand, if we put 𝑅(1)𝑁1𝜇1=,𝜀𝜀𝑁𝑁1𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑁|𝜈|𝑚𝑁𝑇𝜈(𝑚)[𝑢]𝜀𝜈+𝑅(1)𝑁1𝜇1,1,(4.14) 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 𝑅(1)𝑁1[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 𝜇1𝑢0+1=𝜇1𝑢0+1|𝜈|𝑁1|𝜈|𝑚=11𝐷𝑚!𝑚3𝜇1𝑢0𝑇𝜈(𝑚)[𝑢]𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝜇1=,𝜀|𝜈|𝑁1𝜌𝜈𝜇1𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝜇1.,𝜀(4.15)
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+1] around the point 𝑢0 up to order 𝑁+1, we obtain from (4.2) that 𝑓1𝑢0+1=𝑓1𝑢0+𝐷3𝑓1𝑢01+𝐷4𝑓1𝑢01+𝐷5𝑓1𝑢01+𝑚2|𝑚|𝑁1𝑚=1,𝑚2,𝑚33+1𝐷𝑚!𝑚𝑓1𝑢0𝑚111𝑚21𝑚3+𝑅(1)𝑁1𝑓1,1=𝑓1𝑢0+𝐷3𝑓1𝑢01+𝐷4𝑓1𝑢01+𝐷5𝑓1𝑢01+2|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+|𝑚||𝜈||𝑚|𝑁(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈+𝑅(1)𝑁1𝑓1,1=𝑓1𝑢0+𝐷3𝑓1𝑢01+𝐷4𝑓1𝑢01+𝐷5𝑓1𝑢01+2|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+|𝑚||𝜈|𝑁1(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈+2|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+𝑁|𝜈||𝑚|𝑁(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈+𝑅(1)𝑁1𝑓1,1,(4.16) where 𝑅(1)𝑁1𝑓1,1=|𝑚|=𝑁𝑚=(𝑚1,𝑚2,𝑚3)3+𝑁𝑚!10(1𝜃)𝑁1𝐷𝑚𝑓1𝑢0+𝜃1𝑚111𝑚21𝑚3𝑑𝜃.(4.17)
We also note that 𝑓1𝑢0+𝐷3𝑓1𝑢01+𝐷4𝑓1𝑢01+𝐷5𝑓1𝑢01+2|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+|𝑚||𝜈|𝑁1(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈=𝑓1𝑢0+1|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+|𝑚||𝜈|𝑁1(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈=𝑓1𝑢0+1|𝜈|𝑁11|𝑚||𝜈|𝑚=(𝑚1,𝑚2,𝑚3)3+(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈=|𝜈|𝑁1𝜋𝜈𝑓1𝜀𝜈.(4.18)
Similarly, 2|𝑚|𝑁1𝑚=(𝑚1,𝑚2,𝑚3)3+𝑁|𝜈||𝑚|𝑁(𝛼,𝛽,𝛾)𝐴(𝑚,𝑁)𝛼+𝛽+𝛾=𝜈1𝐷𝑚!𝑚𝑓1𝑢0𝑇(𝑚1)𝛼[𝑢]𝑇(𝑚2)𝛽[]𝑇𝑢(𝑚3)𝛾𝑢𝜀𝜈+𝑅(1)𝑁1𝑓1,1=𝜀𝑁𝑅(1)𝑁1𝑓1,,𝜀(4.19) where 𝑅(1)𝑁1[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 satisfies the problem 𝑣𝜕𝜇𝜕𝑥𝜀1[]𝑣𝑣+𝑥=𝜀2𝑓1[]𝑣+𝑓1[]+𝜀1𝜕𝜇𝜕𝑥1[]𝑣+𝜇1[]𝑥+𝐸𝜀𝑣(𝑥,𝑡),0<𝑥<1,0<𝑡<𝑇,(0,𝑡)=𝑣(1,𝑡)=0,𝑣(𝑥,0)=𝑣𝜇(𝑥,0)=0,𝜀1[𝑣]=𝜇0+𝜀1𝜇1[𝑣]=𝜇0(𝑥,𝑡)+𝜀1𝜇1𝑓(𝑥,𝑡,𝑣),1[𝑣]=𝑓1𝑥,𝑡,𝑣,𝑣𝑥,𝑣,𝜇1[𝑣]=𝜇1(𝑥,𝑡,𝑣),(4.20) where 𝐸𝜀(𝑥,𝑡)=𝜀2𝑓1[]+𝜀1𝜕𝜇𝜕𝑥1[]𝜇1𝑢0𝑥||𝛾||1𝑁𝐹𝛾𝜀𝛾.(4.21)

Lemma 4.4. Let (𝐻1),(𝐻4) and (𝐻5) hold. Then 𝐸𝜀𝐿(0,𝑇;𝐿2)𝐸𝜀𝑁+1,(4.22) 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[], we obtain 𝜇1[]=𝜇1𝑢0+1|𝜈|𝑁1𝜌𝜈𝜇1𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝜇1.,𝜀(4.23)
By (4.6), (4.8), we write 𝜀1𝜇1[]𝜇1𝑢0=1|𝜈|𝑁1𝜌𝜈𝜇1𝜀1𝜀𝜈+𝜀1𝜀𝑁𝑅(1)𝑁1𝜇1=,𝜀2|𝜈|𝑁,𝜈11𝜌𝜈11,𝜈2𝜇1𝜀𝜈+𝜀1𝜀𝑁𝑅(1)𝑁1𝜇1=,𝜀2|𝜈|𝑁𝜌𝜈(1)𝜇1𝜀𝜈+𝜀1𝜀𝑁𝑅(1)𝑁1𝜇1.,𝜀(4.24)
On the other hand, from (4.24), we compute 𝜀1𝜇1[]𝜇1𝑢0𝑥=2|𝜈|𝑁𝜌𝜈(1)𝜇1𝜀𝜈+𝜀1𝜀𝑁𝑅(1)𝑁1𝜇1,𝜀𝑥=2|𝜈|𝑁𝜌𝜈(1)𝜇1𝜀𝜈|𝛼|𝑁𝑢𝛼𝜀𝛼+𝜀1𝜀𝑁𝑅(1)𝑁1𝜇1,𝜀𝑥=2|𝜈|𝑁,|𝛼|𝑁𝜌𝜈(1)𝜇1𝑢𝛼𝜀𝜈+𝛼+𝜀𝑁+1𝑅𝑁(1)𝜇1=,𝜀2|𝜈|𝑁,|𝛼|𝑁𝜌𝜈(1)𝜇1𝑢𝛼𝜀𝜈+𝛼+𝜀𝑁+1𝑅𝑁(1)𝜇1=,𝜀||𝛾||22𝑁||||2|𝜈|𝑁,𝛾𝜈𝑁𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(1)𝜇1=,𝜀||𝛾||2𝑁||||2|𝜈|𝑁,𝛾𝜈𝑁𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+||𝛾||𝑁+12𝑁||||2|𝜈|𝑁,𝛾𝜈𝑁𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(1)𝜇1=,𝜀||𝛾||2𝑁||||2|𝜈|𝑁,𝛾𝜈𝑁𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(2)𝜇1=,𝜀||𝛾||2𝑁2|𝜈|𝑁,𝜈𝛾𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(2)𝜇1,,𝜀(4.25) where 𝑅𝑁(1)𝜇1=𝜀,𝜀1𝑅𝜀(1)𝑁1𝜇1,𝜀𝑥,𝜀𝑁+1𝑅𝑁(2)𝜇1=,𝜀||𝛾||𝑁+12𝑁||||2|𝜈|𝑁,𝛾𝜈𝑁𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(1)𝜇1.,𝜀(4.26)
Hence, 𝜀1𝜕𝜇𝜕𝑥1[]𝜇1𝑢0𝑥=𝜕𝜕𝑥||𝛾||2𝑁2|𝜈|𝑁,𝜈𝛾𝜌𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝑅𝑁(2)𝜇1=,𝜀||𝛾||2𝑁2|𝜈|𝑁,𝜈𝛾𝜕𝜌𝜕𝑥𝜈(1)𝜇1𝑢𝛾𝜈𝜀𝛾+𝜀𝑁+1𝜕𝑅𝜕𝑥𝑁(2)𝜇1.,𝜀(4.27)
Similarly, we write 𝜀2𝑓1[]=𝜀2|𝜈|𝑁1𝜋𝜈𝑓1𝜀𝜈+𝜀𝑁𝑅(1)𝑁1𝑓1=,𝜀1|𝜈|𝑁𝜋𝜈(2)𝑓1𝜀𝜈+𝜀𝑁+1𝑅𝑁(1)𝑓1,,𝜀(4.28) where 𝑅𝑁(1)[𝑓1,𝜀]=𝜀2/𝜀𝑅(1)𝑁1[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[]+𝜀1𝜕𝜇𝜕𝑥1[]𝜇1𝑢0𝑥1|𝛾|𝑁𝐹𝛾𝜀𝛾=||𝛾||1𝑁𝜋𝜈(2)𝑓1+2|𝜈|𝑁,𝜈𝛾𝜕𝜌𝜕𝑥𝜈(1)𝜇1𝑢𝛾𝜈𝐹𝛾𝜀𝛾+𝜀𝑁+1𝑅𝑁(1)𝑓1+𝜕,𝜀𝑅𝜕𝑥𝑁(2)𝜇1=,𝜀𝜀𝑁+1𝑅𝑁(1)𝑓1+𝜕,𝜀𝑅𝜕𝑥𝑁(2)𝜇1.,𝜀(4.29)
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)𝐸𝜀𝑁+1,(4.30) 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 by𝑣0𝑣0,𝑚1𝜕𝜇𝜕𝑥𝜀1𝑣𝑚1𝑣+𝑚𝑥=𝜀2𝑓1𝑣𝑚1+𝑓𝑖[]+𝜀1𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥+𝐸𝜀𝑣(𝑥,𝑡),0<𝑥<1,0<𝑡<𝑇,𝑚(0,𝑡)=𝑣𝑚𝑣(1,𝑡)=0,𝑚(𝑥,0)=𝑣𝑚(𝑥,0)=0,𝑚1.(4.31)

With 𝑚=1, we have the problem𝑣1𝜕𝜇𝜕𝑥𝜀1[]𝑣1𝑥=𝐸𝜀𝑣(𝑥,𝑡),0<𝑥<1,0<𝑡<𝑇,1(0,𝑡)=𝑣1𝑣(1,𝑡)=0,1(𝑥,0)=𝑣1(𝑥,0)=0.(4.32)

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=2𝑡0𝐸𝜀(𝑠),𝑣1(𝑠)𝑑𝑠+𝑡0𝑑𝑠10𝜇1,𝜀1(𝑥,𝑠)𝑣21𝑥(𝑥,𝑠)𝑑𝑥𝑇𝐸2𝜀2𝑁+2+𝑡0𝑣1(𝑠)2𝑑𝑠+𝑡0𝑑𝑠10|||𝜇1,𝜀1|||𝑣(𝑥,𝑠)21𝑥(𝑥,𝑠)𝑑𝑥,(4.33) where 𝜇1,𝜀1(𝑥,𝑡)=𝜇𝜀1[(𝑥,𝑡)]=𝜇0(𝑥,𝑡)+𝜀1𝜇1(𝑥,𝑡,(𝑥,𝑡)). By 𝜇1,𝜀1(𝑥,𝑡)=𝜇0(𝑥,𝑡)+𝜀1𝐷2𝜇1(𝑥,𝑡,(𝑥,𝑡))+𝐷3𝜇1(𝑥,𝑡,(𝑥,𝑡)),(𝑥,𝑡)(4.34) we get|||𝜇1,𝜀1|||𝐾𝜇(𝑥,𝑡)0+1+𝑀𝐾𝑀𝜇1𝜁0,(4.35) with 𝑀=(𝑁+1)𝑀,𝐾(𝜇0)=𝜇0𝐶1(𝑄𝑇).

It follows from (4.33), (4.35) that𝑣1(𝑡)2+𝜇𝑣1𝑥(𝑡)2𝑇𝐸2𝜀2𝑁+2+𝑡0𝑣1(𝑠)2𝑑𝑠+𝜁0𝑡0𝑣1𝑥(𝑠)2𝑑𝑠.(4.36)

Using Gronwall's lemma, (4.36) gives𝑣1𝐿(0,𝑇;𝐿2)+𝑣1𝑥𝐿(0,𝑇;𝐿2)11+𝜇𝑇𝐸𝜀𝑁+1𝜇exp+𝜁0𝑇2𝜇.(4.37)

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.(4.38)

Multiplying two sides of (4.31)1 with 𝑣𝑚 and after integrating in 𝑡, we obtain without difficulty from (4.22) that𝑣𝑚(𝑡)2+𝜇𝑣𝑚𝑥(𝑡)2𝑇𝐸2𝜀2𝑁+2+𝑡0𝑣𝑚(𝑠)2𝑑𝑠+𝑡0𝑑𝑠10||𝜇𝑚,𝜀1||𝑣(𝑥,𝑠)2𝑚𝑥(𝑥,𝑠)𝑑𝑥+2𝜀2𝑡0𝑓1𝑣𝑚1+𝑓1[]𝑣𝑚(𝑠)𝑑𝑠+2𝜀1𝑡0𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥𝑣𝑚(𝑠)𝑑𝑠=𝑇𝐸2𝜀2𝑁+2+𝑡0𝑣𝑚(𝑠)2𝐽𝑑𝑠+1𝐽(𝑡)+2𝐽(𝑡)+3(𝑡),(4.39) where 𝜇𝑚,𝜀1(𝑥,𝑡)=𝜇𝜀1[𝑣𝑚1+]=𝜇0(𝑥,𝑡)+𝜀1𝜇1(𝑥,𝑡,𝑣𝑚1(𝑥,𝑡)+(𝑥,𝑡)). We will estimate the integrals on the right-hand side of (4.39) as follows.

First Integral 𝐽1(𝑡)
We have 𝜇𝑚,𝜀1(𝑥,𝑡)=𝜇0(𝑥,𝑡)+𝜀1𝐷2𝜇1𝑥,𝑡,𝑣𝑚1++𝐷3𝜇1𝑥,𝑡,𝑣𝑚1𝑣+𝑚1+,(4.40) hence ||𝜇𝑚,𝜀1||𝐾𝜇(𝑥,𝑡)0+1+𝑀1𝐾𝑀1𝜇1𝜒1,with𝑀1=(𝑁+2)𝑀.(4.41)

It follows from (4.41) that𝐽1(𝑡)=𝑡0𝑑𝑠10||𝜇𝑚,𝜀1||𝑣(𝑥,𝑠)2𝑚𝑥(𝑥,𝑠)𝑑𝑥𝜒1𝑡0𝑣𝑚𝑥(𝑠)2𝑑𝑠.(4.42)

Second Integral 𝐽2(𝑡)
We note that 𝑓1𝑣𝑚1+𝑓1[]2𝐾𝑀1𝑓1𝑣𝑚1𝑊1(𝑇).(4.43)

Therefore,𝐽2(𝑡)=2𝜀2𝑡0𝑓1𝑣𝑚1+𝑓1[]𝑣𝑚(𝑠)𝑑𝑠𝑇𝜒22𝜀2𝑣𝑚12𝑊1(𝑇)+𝑡0𝑣𝑚(𝑠)2𝑑𝑠,(4.44) where 𝜒2=𝜒2(𝑀1,𝑓1)=2𝐾𝑀1(𝑓1).

Third Integral 𝐽3(𝑡)
First, we need to estimate 𝜕/𝜕𝑥[(𝜇1[𝑣𝑚1+]𝜇1[])𝑥].

From the equation𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥=𝜇1𝑣𝑚1+𝜇1[]𝑥𝑥+𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥=𝜇1𝑣𝑚1+𝜇1[]𝑥𝑥+𝐷1𝜇1𝑣𝑚1+𝐷1𝜇1[]𝑥+𝐷3𝜇1𝑣𝑚1+𝐷3𝜇1[]𝑣𝑚1+𝑥+𝐷3𝜇1[]𝑣𝑚1𝑥,(4.45) it implies that𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥𝜇1[𝑣𝑚1+]𝜇1[]𝐶0(Ω)𝑥𝑥+𝐷1𝜇1[𝑣𝑚1+]𝐷1𝜇1[]𝐶0(Ω)𝑥+𝐷3𝜇1𝑣𝑚1+𝐷3𝜇1[]𝐶0(Ω)𝑣𝑚1+𝐶0(Ω)𝑥+𝐷3𝜇1[]𝐶0(Ω)𝑣𝑚1𝑊1(𝑇)𝑥𝐶0(Ω).(4.46)

On the other hand, we have𝜇1𝑣𝑚1+𝜇1[]𝐶0(Ω)𝐾𝑀1𝜇1𝑣𝑚1𝑊1(𝑇),𝐷𝑗𝜇1𝑣𝑚1+𝐷𝑗𝜇1[]𝐶0(Ω)𝐾𝑀1𝜇1𝑣𝑚1𝑊1(𝑇)𝐷,𝑗=1,3,3𝜇1[]𝐶0(Ω)𝐾𝑀1𝜇1.(4.47)

We deduce from (4.46) and (4.47) that𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥3+2𝑀1𝑀1𝐾𝑀1𝜇1𝑣𝑚1𝑊1(𝑇).(4.48)

Next, by (4.48), it follows that𝐽3(𝑡)=2𝜀1𝑡0𝜕𝜇𝜕𝑥1𝑣𝑚1+𝜇1[]𝑥𝑣𝑚(𝑠)𝑑𝑠𝑇𝜒23𝜀2𝑣𝑚12𝑊1(𝑇)+𝑡0𝑣𝑚(𝑠)2𝑑𝑠,(4.49) where 𝜒3=𝜒3(𝑀1,𝜇1)=(3+2𝑀1)𝑀1𝐾𝑀1(𝜇1).

Combining (4.39), (4.42), (4.44), and (4.49) gives𝑣𝑚(𝑡)2+𝜇𝑣𝑚𝑥(𝑡)2𝑇𝐸2𝜀2𝑁+2𝜒+𝑇22+𝜒23𝜀2𝑣𝑚12𝑊1(𝑇)+3𝑡0𝑣𝑚(𝑠)2𝑑𝑠+𝜒1𝑡0𝑣𝑚𝑥(𝑠)2𝑑𝑠𝑇𝐸2𝜀2𝑁+2𝜒+𝑇22+𝜒23𝜀2𝑣𝑚12𝑊1(𝑇)+𝜒3+1𝜇𝑡0𝑣𝑚(𝑠)2+𝜇𝑣𝑚𝑥(𝑠)2𝑑𝑠.(4.50)

Using Gronwall's lemma, we deduce from (4.50) that𝑣𝑚𝑊1(𝑇)𝜎𝑇𝑣𝑚1𝑊1(𝑇)+𝛿,𝑚1,(4.51) where𝜎𝑇=𝜒22+𝜒23𝜂𝑇,𝛿=𝜂𝑇𝐸𝜀𝑁+1,𝜂𝑇=11+𝜇𝑇𝑇exp2𝜒3+1𝜇.(4.52)

We can assume that𝜎𝑇<1,(4.53) with sufficiently small 𝑇>0.

Lemma 4.5. Let the sequence {𝜁𝑚} satisfy 𝜁𝑚𝜎𝜁𝑚1+𝛿𝑚1,𝜁0=0,(4.54) where 0𝜎<1,𝛿0 are the given constants. Then, 𝜁𝑚𝛿(1𝜎)𝑚1.(4.55)

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+𝜒23𝜂𝑇<1,𝛿=𝜂𝑇𝐸𝜀𝑁+1, it follows from (4.55) that𝑣𝑚𝐿(0,𝑇;𝐿2)+𝑣𝑚𝑥𝐿(0,𝑇;𝐿2)=𝑣𝑚𝑊1(𝑇)𝛿1𝜎𝑇𝐶𝑇𝜀𝑁+1.(4.56)

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,(4.57) it meansthat𝑢||𝛾||𝑁𝑢𝛾𝜀𝛾𝐿(0,𝑇;𝐿2)+𝑢𝑥||𝛾||𝑁𝑢𝛾𝑥𝜀𝛾𝐿(0,𝑇;𝐿2)𝐶𝑇𝜀𝑁+1.(4.58)

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), (𝑃𝛾), 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: 𝑢𝑡𝑡𝜕𝜇𝜕𝑥0(𝑥,𝑡)+𝑝𝑖=1𝜀𝑖𝜇𝑖𝑢(𝑥,𝑡,𝑢)𝑥=𝑓0(𝑥,𝑡)+𝑝𝑖=1𝜀𝑖𝑓𝑖𝑥,𝑡,𝑢,𝑢𝑥,𝑢𝑡𝑢,0<𝑥<1,0<𝑡<𝑇,(0,𝑡)=𝑢(1,𝑡)=0,𝑢(𝑥,0)=̃𝑢0(𝑥),𝑢𝑡(𝑥,0)=̃𝑢1(𝑃(𝑥).𝜀)
For more detail, we also make the following assumptions: (𝐻4)𝜇𝐶2([0,1]×+),𝜇𝑖𝐶𝑁+1([0,1]×+×),𝜇0𝜇>0,𝜇𝑖0,𝑖=1,2,,𝑝, (𝐻5)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=!,𝜀𝜀21++𝜀2𝑝,𝜀𝛼=𝜀𝛼11𝜀𝛼𝑝𝑝,𝛼,𝛽𝑝+,𝛼𝛽𝛼𝑖𝛽𝑖𝑖=1,,𝑝.(4.59)
Let 𝑢0 be a unique weak solution of the problem (𝑃0), which is (𝑃𝜀) corresponding to 𝜀=(0,,0).Let the sequence of weak solutions 𝑢𝛾,𝛾𝑝+,1|𝛾|𝑁be defined by the problems (𝑃𝛾), in which 𝐹𝛾,𝛾𝑝+,1|𝛾|𝑁, are defined by suitable recurrent formulas. Then, the following similar theorem holds.

Theorem 4.7. 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 estimation up to order 𝑁+1 as follows: 𝑢||𝛾||𝑁𝑢𝛾𝜀𝛾𝐿(0,𝑇;𝐿2)+𝑢𝑥||𝛾||𝑁𝑢𝛾𝑥𝜀𝛾𝐿(0,𝑇;𝐿2)𝐶𝑇𝜀𝑁+1.(4.60)

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.