On the nonlinear wave equation with the mixed nonhomogeneous conditions: Linear approximation and asymptotic expansion of solutions

Nonlinear Analysis 71 (2009) 5799–5819

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Nonlinear Analysis

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On the nonlinear wave equation with the mixed nonhomogeneousconditions: Linear approximation and asymptotic expansion of solutionsLe Thi Phuong Ngoc a, Le Khanh Luan b, Tran Minh Thuyet b, Nguyen Thanh Long c,∗a Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Viet Namb Department of Mathematics, University of Economics of HoChiMinh City, 59C Nguyen Dinh Chieu Street, Dist. 3, HoChiMinh City, Viet Namc Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University Ho Chi Minh City, 227 Nguyen Van Cu Street,Dist.5, Ho Chi Minh City, Viet Nam

a r t i c l e i n f o

Article history:Received 3 February 2009Accepted 1 May 2009

MSC:35L2035L7035Q72

Keywords:Faedo–Galerkin methodLinear recurrent sequenceAsymptotic expansion of order N + 1

a b s t r a c t

In this paper, we consider the following nonlinear wave equationutt −

∂

∂x(µ(u)ux) = f (x, t, u, ux, ut), 0 < x < 1, 0 < t < T ,

ux(0, t) = g(t), u(1, t) = 0,u(x, 0) = u0(x), ut(x, 0) = u1(x),

(1)

where u0, u1, µ, f , g are given functions. To problem (1), we associate a linear recursivescheme for which the existence of a local and unique weak solution is proved by applyingthe Faedo–Galerkin method and the weak compact method. In the case of µ ∈ CN+2(R),µ1 ∈ CN+1(R),µ(z) ≥ µ0 > 0,µ1(z) ≥ 0, for all z ∈ R, and g ∈ C3(R+), f ∈ CN+1([0, 1]×R+ × R3), f1 ∈ CN ([0, 1] × R+ × R3), a weak solution uε1,ε2 (x, t) having an asymptoticexpansion of order N + 1 in two small parameters ε1, ε2 is established for the followingequation associated to (1)2,3:

utt −∂

∂x([µ(u)+ ε1µ1(u)]ux) = f (x, t, u, ux, ut)+ ε2f1(x, t, u, ux, ut). (2)

© 2009 Elsevier Ltd. All rights reserved.

1. Introduction

In this paper, we consider the following initial and boundary value problem

utt −∂

∂x(µ(u)ux) = f (x, t, u, ux, ut), 0 < x < 1, 0 < t < T , (1.1)

ux(0, t) = g(t), u(1, t) = 0, (1.2)u(x, 0) = u0(x), ut(x, 0) = u1(x), (1.3)

where u0, u1, µ, f , g are given functions satisfying conditions specified later.Eq. (1.1) constitutes a case, relatively simpler, of a more general equation, namely

utt −∂

∂x(µ(x, t, u)ux) = f (x, t, u, ux, ut), 0 < x < 1, 0 < t < T , (1.4)

∗ Corresponding author.E-mail addresses: [email protected], [email protected] (L.T.P. Ngoc), [email protected] (L.K. Luan), [email protected] (T.M. Thuyet),

[email protected], [email protected] (N.T. Long).

0362-546X/$ – see front matter© 2009 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2009.05.004

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http://dx.doi.org/10.1016/j.na.2009.05.004

5800 L.T.P. Ngoc et al. / Nonlinear Analysis 71 (2009) 5799–5819

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 thenonlinear term f has the simple forms, problem (1.4), with various initial-boundary conditions, has been studied by manyauthors, for example, E. Ortiz, Alain Pham [1], Alain Pham, Long [2,3], Long, Alain Pham [4], Long, Diem [5], Long, Alain Pham,Diem [6,7], Long, Giai [8], Long, Truong [9], Long, Ngoc [10], Long, Tam, Truc [11], Ngoc, Hang, Long [12], Truong, Ngoc, Long[13] and the references therein.In [14], Ficken and Fleishman established the unique global existence and stability of solutions for the equation

uxx − utt − 2αut − βu = εu3 + γ , ε > 0. (1.5)

Rabinowitz [15] proved the existence of periodic solutions for

uxx − utt − 2αut = εf (x, t, u, ux, ut), (1.6)

where ε is a small parameter and f is periodic in time.In a paper of Caughey and Ellison [16], 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 [11], Long, Tam and Truc studied the linear recursive schemes and asymptotic expansion for the nonlinear wave

equation

utt − uxx = f (x, t, u, ux, ut)+ εf1(x, t, u, ux, ut), (1.7)

with the mixed nonhomogeneous conditions

ux(0, t)− h0u(0, t) = g0(t), u(1, t) = g1(t). (1.8)

In the case of g0, g1 ∈ C3(R+), f ∈ CN+1([0, 1] × R+ × R3), f1 ∈ CN([0, 1] × R+ × R3), and some other conditions, anasymptotic expansion of the weak solution uε of order N + 1 in ε is considered.In [12], Ngoc, Hang and Long gave the unique existence, stability and asymptotic expansion of the problem (1.3), (1.4)

with the mixed boundary conditions (1.2) stand forux (0, t) = g0 (t)+

∫ t

0k0 (t − s) u (0, s) ds,

−ux (1, t) = g1 (t)+∫ t

0k1 (t − s) u (1, s) ds,

(1.9)

where µ = µ(x, t), f (x, t, u, ux, ut) = −λut − F(u), with λ is a given constant and µ, F , g0, g1, k0, k1, u0, u1 are givenfunctions.However, by the fact that it is difficult to consider problem (1.4) with some initial-boundary conditions in the case

µ(x, t, u) depending on u, few works were done as far as we know. In order to solve this problem, the linearization methodfor nonlinear term is usually used. Let us present this technique as follows.At first, we note that, for each v = v(x, t) belongs to X being a suitable space of function, we can give some suitable

assumptions to obtain a unique solution u ∈ X of the problem with respect to µ = µ(x, t, v(x, t)) = µ(x, t) andf = f (x, t, v, vx, vt) = f (x, t). It is obvious that u depends on v, so we can suppose that u = A(v). Therefore, the aboveproblem can be reduced to a fixed point problem for operator A : X → X . Based on these ideas, with the first term u0 ischosen, the usual iteration um = A(um−1), m = 1, 2, . . . , is applied to establish a sequence um, which converges to thesolution of the problem, hence the existence result follows.Without loss of generality, we need only consider problem (1.1)–(1.3) instead of problem (1.2)–(1.4) in order to avoid

making the treatment too complicated.The paper consists of four sections. In Section 2, we present some preliminaries. In Section 3, we associate with the

problem (1.1)–(1.3) a linear recurrent sequence which is bounded in a suitable space of functions. The existence of a localsolution and unique weak solution is proved by using the Faedo–Galerkin method and the weak compact method. Notethat the linearization method in this paper and in the papers [3,5,6,11,10,1,13] cannot be used in the papers [2,4,8,9,12].In Section 4, an asymptotic expansion of a weak solution uε1,ε2(x, t) of order N + 1 in two small parameters ε1, ε2 for theequation

utt −∂

∂x([µ(u)+ ε1µ1(u)]ux) = f (x, t, u, ux, ut)+ ε2f1(x, t, u, ux, ut), (1.10)

associated to (1.1)2,3, with µ ∈ CN+2(R), µ1 ∈ CN+1(R), µ(z) ≥ µ0 > 0, µ1(z) ≥ 0 for all z ∈ R, g ∈ C3(R+),f ∈ CN+1([0, 1] × R+ × R3) and f1 ∈ CN([0, 1] × R+ × R3), is established. This result is a relative generalization of[16,2,3,14,4–9,11,10,17,12,1,13].

L.T.P. Ngoc et al. / Nonlinear Analysis 71 (2009) 5799–5819 5801

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). The norm in L2 is denoted by ‖ · ‖. We denote by 〈·, ·〉 the scalar product in L2 or a pair of dualproducts of continuous linear functional with an element of a function space. We denote by ‖ · ‖X the norm of a Banachspace 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 )→ Xmeasurable, such that ‖u‖Lp(0,T ;X) < +∞,with

‖u‖Lp(0,T ;X) =

(∫ T

0‖u(t)‖pXdt

)1/p, if 1 ≤ p <∞,

ess sup0<t<T

‖u(t)‖X , if p = ∞.(2.1)

Let u(t), u′(t) = ut(t) = u(t), u′′(t) = utt(t) = u(t), ux(t) = ∇u(t), uxx(t) = ∆u(t), denote u(x, t), ∂ u∂ t (x, t),∂2 u∂ 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∂wand Dα f = Dα11 D

α22 D

α33 D

α44 D

α55 f , α = (α1, α2, α3, α4, α5) ∈ Z5

+, |α| = α1 + α2 + α3 + α4 + α5 = k,

D(0,0,...,0)f = f .We put

V = v ∈ H1(Ω) : v(1) = 0, (2.2)

a(u, v) =∫ 1

0∇u(x)∇v(x)dx, u, v ∈ V . (2.3)

Then V is a closed subspace of H1 and on V , v 7−→ ‖v‖H1 , v 7−→ ‖vx‖ and v 7−→ ‖v‖V =√a(v, v) = ‖∇v‖ are

equivalent norms.Then we have the following lemmas, the proofs of which are straightforward and are omitted.

Lemma 2.1. The imbedding H1 → C0([0, 1]) is compact and

‖v‖C0(Ω) ≤√2‖v‖H1 for all v ∈ H1. (2.4)

Lemma 2.2. The imbedding V → C0([0, 1]) is compact and

‖v‖C0(Ω) ≤ ‖∇v‖ for all v ∈ V . (2.5)

3. The existence and uniqueness theorem of solution

Wemake the following assumptions:

(H1) u0 ∈ V ∩ H2, u1 ∈ V ,(H2) g ∈ C3(R+),(H3) µ ∈ C2(R), µ (z) ≥ µ0 > 0, ∀z ∈ R,(H4) f ∈ C1(Ω × R+ × R3).

Put ϕ(x, t) = (x− 1)g(t). By the transformation v(x, t) = u(x, t)− ϕ(x, t), we shall reformulate problem (1.1)–(1.3) asa problem with homogeneous boundary conditions as follows

vtt −∂

∂x(µ(v + ϕ)vx) = f (x, t, v, vx, vt), x ∈ (0, 1), 0 < t < T ,

vx(0, t) = v(1, t) = 0,v(x, 0) = v0(x), vt(x, 0) = v1(x),

(3.1)

wheref (x, t, v, vx, vt) = f (x, t, v + ϕ, vx + ϕx, vt + ϕt)− (x− 1)g ′′(t)+ µ′(v + ϕ)(vx + g)g,v0(x) = u0(x)− ϕ(x, 0) = u0(x)− (x− 1)g(0),v1(x) = u1(x)− ϕt(x, 0) = u1(x)− (x− 1)g ′(0),

(3.2)

and g , u0 satisfying the consistency condition g(0) = ux(0, 0) = u′0(0).


Consider T ∗ > 0 fixed, letM > 0, we put

M∗ = sup0≤t≤T∗

(|g(t)| +

∣∣g ′(t)∣∣) , (3.3)

K = K(M, µ) = sup|z|≤M+M∗

(|µ(z)| +

∣∣µ′(z)∣∣+ ∣∣µ′′(z)∣∣) , (3.4)

K0(M, f ) = sup∣∣f (x, t, u, v, w)∣∣ : (x, t, u, v, w) ∈ A∗(M) ,

A∗(M) = (x, t, u, v, w) ∈ [0, 1] × [0, T ∗] × R3 : |u| , |v| , |w| ≤ M,(3.5)

K1 = K1(M, f ) =5∑i=1

K0(M,Dif ). (3.6)

For each T ∈ (0, T ∗] andM > 0, we getW (M, T ) =v ∈ L∞(0, T ; V ∩ H2) : vt ∈ L∞(0, T ; V ) and vtt ∈ L2(QT ),

with ‖v‖L∞(0,T ;V∩H2), ‖vt‖L∞(0,T ;V ), ‖vtt‖L2(QT ) ≤ M,

W1(M, T ) = v ∈ W (M, T ) : vtt ∈ L∞(0, T ; L2),(3.7)

where QT = Ω × (0, T ).We shall choose the first initial term v0 ≡ v0. Suppose that

vm−1 ∈ W1(M, T ), (3.8)

and associate with problem (3.1) the following problem:Find vm ∈ W1(M, T ) (m ≥ 1)which satisfies the following linear variational problem〈v′′m(t), w〉 + 〈µm(t)∇vm(t),∇w〉 = 〈Fm(t), w〉 , ∀w ∈ V ,vm(0) = v0, v′m(0) = v1,

(3.9)

whereµm(t) = µ(ηm(t)), ηm(t) = vm−1(t)+ ϕ(t),Fm(t) = f (·, t, vm−1(t),∇vm−1(t), v′m−1(t)).

(3.10)

Then, we have the following theorem.

Theorem 3.1. Let assumptions (H1)–(H4) hold. Then there exists a constant M > 0 depending on v0, v1, µ, g and a constantT > 0 depending on v0, v1, µ, g, f such that, for v0 ≡ v0, there exists a recurrent sequence vm ⊂W1(M, T ) defined by (3.9)and (3.10).

Proof. The proof consists of several steps.Step 1: The Faedo–Galerkin approximation (introduced by Lions [18]). Consider the basis in V

wj(x) =

√2

1+ λ2jcos(λjx), λj = (2j− 1)

π

2, j ∈ N, (3.11)

constructed by the eigenfunctions of the Laplace operator−∆ = − ∂2

∂x2. Put

v(k)m (t) =k∑j=1

c(k)mj (t)wj, (3.12)

where the coefficients c(k)mj satisfy the system of linear differential equations〈v(k)m (t), wj〉 + 〈µm(t)∇v

(k)m (t),∇wj〉 = 〈Fm(t), wj〉, 1 ≤ j ≤ k,

v(k)m (0) = v0k, v(k)m (0) = v1k,(3.13)

wherev0k =

k∑j=1

α(k)j wj → v0 strongly in V ∩ H2,

v1k =

k∑j=1

β(k)j wj → v1 strongly in V .

(3.14)


Note that by (3.8), and by applying the Banach fixed point theorem, it is not difficult to prove that the system (3.13) hasa unique solution v(k)m on interval [0, T

(k)m ] ⊂ [0, T ], so let us omit the details. The following estimates allow one to take

constant T (k)m = T for allm and k ∈ N (see [19]).Step 2. A priori estimates. Put

X (k)m (t) = ‖v(k)m (t)‖

2+ ‖

√µm(t)∇v(k)m (t)‖

2,

Y (k)m (t) = ‖∇v(k)m (t)‖

2+ ‖

√µm(t)∆v(k)m (t)‖

2,

S(k)m (t) = X(k)m (t)+ Y

(k)m (t)+

∫ t

0‖v(k)m (s)‖

2ds.(3.15)

For all j = 1, 2, . . . , k, multiplying (3.13) by c(k)m (t), summing on j, and integrating by parts with respect to the timevariable from 0 to t , we have

X (k)m (t) = X(k)m (0)+

∫ t

0ds∫ 1

0µ′m(x, s)|∇v

(k)m (x, s)|

2dx+ 2∫ t

0〈Fm(s), v(k)m (s)〉ds

= X (k)m (0)+ I1 + I2. (3.16)

By replacingwj in (3.13) by∆wj, we obtain

a(v(k)m (t), wj)+ a(µm(t)∇v(k)m (t),∇wj

)= −〈Fm(t),∆wj〉, 1 ≤ j ≤ k. (3.17)

Similarly, by multiplying (3.17) by c(k)m (t),we have

Y (k)m (t) = Y(k)m (0)+ 2

⟨∂µm

∂x(0)∇v0k,∆v0k

⟩+ 2 〈Fm(0),∆v0k〉

+

∫ t

0ds∫ 1

0µ′m(x, s)|∆v

(k)m (x, s)|

2dx+ 2∫ t

0

⟨∂

∂s

(∂µm

∂x(s)∇v(k)m (s)

),∆v(k)m (s)

⟩ds

− 2⟨∂µm

∂x(t)∇v(k)m (t),∆v

(k)m (t)

⟩− 2〈Fm(t),∆v(k)m (t)〉 + 2

∫ t

0

⟨∂Fm∂t(s),∆v(k)m (s)

⟩ds

= Y (k)m (0)+ 2⟨∂µm

∂x(0)∇v0k,∆v0k

⟩+ 2〈Fm(0),∆v0k〉 +

7∑j=3

Ij. (3.18)

We shall estimate the integrals on the right-hand sides of (3.16) and (3.18) as follows.First integral I1. From (3.4), (3.8) and (3.10) we have∣∣µ′m(x, t)∣∣ ≤ K(M, µ)(M +M∗). (3.19)

Hence, by (3.15)

I1 =∫ t

0ds∫ 1

0µ′m(x, s)|∇v

(k)m (x, s)|

2dx ≤1µ0K(M, µ)(M +M∗)

∫ t

0X (k)m (s)ds. (3.20)

Second integral I2. From (3.5), (3.8), (3.10) and (3.15), we have

I2 = 2∫ t

0〈Fm(s),

·

v(k)

m (s)〉ds ≤ 2∫ t

0‖Fm(s)‖ ‖v(k)m (s)‖ds ≤ TK

20 (M, f )+

∫ t

0X (k)m (s)ds. (3.21)

Third integral I3. By using and (3.15) and (3.19), we obtain

I3 =∫ t

0ds∫ 1

0µ′m(x, s)|∆v

(k)m (x, s)|

2dx ≤1µ0K(M, µ)(M +M∗)

∫ t

0Y (k)m (s)ds. (3.22)

Fourth integral I4. By the Cauchy–Schwartz inequality, we have

|I4| = 2∣∣∣∣∫ t

0

⟨∂

∂s

(∂µm

∂x(s)∇v(k)m (s)

),∆v(k)m (s)

⟩ds∣∣∣∣ ≤ 2√µ0

∫ t

0I∗4 (s)

√Y (k)m (s)ds, (3.23)

where I∗4 (s) =∥∥∥ ∂∂s ( ∂µm∂x (s)∇v(k)m (s))∥∥∥ .


We continue to estimate the term I∗4 (s) as follows. We have

I∗4 (s) =∥∥∥∥ ∂∂s

(∂µm

∂x(s)∇v(k)m (s)

)∥∥∥∥ = ∥∥∥∥∂µm∂x (s)∇v(k)m (s)+ ∂2µm∂x∂s(s)∇v(k)m (s)

∥∥∥∥≤

∥∥∥∥∂µm∂x (s)∥∥∥∥C0(Ω)

‖∇v(k)m (s)‖ +∥∥∥∥∂2µm∂x∂s

(s)∥∥∥∥ ‖∇v(k)m (s)‖C0(Ω)

≤

(∥∥∥∥∂µm∂x (s)∥∥∥∥C0(Ω)

+1√µ0

∥∥∥∥∂2µm∂x∂s(s)∥∥∥∥)√Y (k)m (s). (3.24)

On the other hand, by ∂µm∂x (x, s) = µ

′(vm−1(x, s)+ ϕ(x, s))(∇vm−1(x, s)+ g(s)), we implies that∥∥∥∥∂µm∂x (s)∥∥∥∥C0(Ω)

≤ K(M, µ)(‖∇vm−1(s)‖C0(Ω) +M

∗

)≤ K(M, µ)(‖∆vm−1(s)‖ +M∗) ≤ K(M, µ)(M +M∗). (3.25)

Similarly, from the following equality

∂2µm

∂x∂s(x, s) = µ′′(vm−1(x, s)+ ϕ(x, s))(v′m−1(x, s)+ ϕ

′(x, s))(∇vm−1(x, s)+ g(s))

+µ′(vm−1(x, s)+ ϕ(x, s))(∇v′m−1(x, s)+ g′(s)),

we obtain that∥∥∥∥∂2µm∂x∂s(s)∥∥∥∥ ≤ K(M, µ) [∥∥v′m−1(s)+ ϕ′(s)∥∥C0(Ω) ‖∇vm−1(s)+ g(s)‖ + ‖∇v′m−1(s)+ g ′(s)‖]≤ K(M, µ)(M +M∗)(1+M +M∗). (3.26)

It follows from (3.24)–(3.26), that

I∗4 (s) =∥∥∥∥ ∂∂s

(∂µm

∂x(s)∇v(k)m (s)

)∥∥∥∥ ≤ K(M, µ)(M +M∗)(1+ 1+M +M∗√µ0

)√Y (k)m (s). (3.27)

Hence we obtain from (3.23) and (3.27), that

|I4| ≤2√µ0K(M, µ)(M +M∗)

(1+

1+M +M∗√µ0

)∫ t

0Y (k)m (s)ds. (3.28)

Fifth integral I5. By the Cauchy–Schwartz inequality, we have

|I5| =∣∣∣∣−2 ⟨∂µm∂x (t)∇v(k)m (t),∆v(k)m (t)

⟩∣∣∣∣ ≤ 1β∥∥∥∥∂µm∂x (t)∇v(k)m (t)

∥∥∥∥2 + β‖∆v(k)m (t)‖2, (3.29)

for all β > 0. On the other hand∥∥∥∥∂µm∂x (t)∇v(k)m (t)∥∥∥∥ = ∥∥∥∥∂µm∂x (0)∇v0k +

∫ t

0

∂

∂s

(∂µm

∂x(s)∇v(k)m (s)

)ds∥∥∥∥

≤

∥∥∥∥∂µm∂x (0)∥∥∥∥C0(Ω)

‖∇v0k‖ +

∫ t

0I∗4 (s)ds. (3.30)

It follows from (3.25), (3.27) and (3.30), that∥∥∥∥∂µm∂x (t)∇v(k)m (t)∥∥∥∥ ≤ ∥∥∥∥∂µm∂x (0)

∥∥∥∥C0(Ω)

‖∇v0k‖

+ K(M, µ)(M +M∗)(1+

1+M +M∗√µ0

)∫ t

0

√Y (k)m (s)ds. (3.31)


Hence we obtain from (3.29), (3.31), that

|I5| ≤β

µ0Y (k)m (t)+

2β

∥∥∥∥∂µm∂x (0)∥∥∥∥2C0(Ω)

‖∇v0k‖2

+2βT K 2(M, µ)(M +M∗)2

(1+

1+M +M∗√µ0

)2 ∫ t

0Y (k)m (s)ds, for all β > 0. (3.32)

Sixth integral I6. By (3.5), (3.10) and (3.15), we obtain

|I6| =∣∣−2〈Fm(t),∆v(k)m (t)〉∣∣ ≤ 1β ‖Fm(t)‖2 + β ∥∥∆v(k)m (t)∥∥2

≤1β

∥∥∥∥Fm(0)+ ∫ t

0

∂Fm∂s(s)ds

∥∥∥∥2 + β

µ0Y (k)m (t)

≤2β‖Fm(0)‖2 +

2βT∫ T

0

∥∥∥∥∂Fm∂s (s)∥∥∥∥2 ds+ β

µ0Y (k)m (t), for all β > 0. (3.33)

It is known that

∂Fm∂t(t) = D2f [vm−1] + D3f [vm−1]v′m−1(t)+ D4f [vm−1]∇v

′

m−1(t)+ D5f [vm−1]v′′

m−1(t), (3.34)

where we have used the notation Dif [vm−1] = Dif (x, t, vm−1(x, t),∇vm−1(x, t), v′m−1(x, t)), i = 2, . . . , 5. So, by (3.6), (3.8)and (3.34), we obtain∥∥∥∥∂Fm∂t (t)

∥∥∥∥ ≤ K1(M, f ) (1+ ∥∥v′m−1(t)∥∥+ ∥∥∇v′m−1(t)∥∥+ ∥∥v′′m−1(t)∥∥)≤ K1(M, f )

(1+ 2M +

∥∥v′′m−1(t)∥∥) . (3.35)

Hence, we deduce from (3.8), (3.33) and (3.35), that

|I6| ≤2β‖Fm(0)‖2 +

2βT2K 21 (M, f )

∫ T

0

[(1+ 2M)2 +

∥∥v′′m−1(s)∥∥2] ds+ β

µ0Y (k)m (t)

≤2β‖Fm(0)‖2 +

4βTK 21 (M, f )

[(1+ 2M)2T +M2

]+β

µ0Y (k)m (t), for all β > 0. (3.36)

Seventh integral I7. By (3.5), (3.8), (3.10), (3.15) and (3.35), we obtain

|I7| = 2∣∣∣∣∫ t

0

⟨∂Fm∂t(s),∆v(k)m (s)

⟩ds∣∣∣∣ ≤ ∫ t

0

∥∥∥∥∂Fm∂t (s)∥∥∥∥ ds+ ∫ t

0

∥∥∥∥∂Fm∂t (s)∥∥∥∥ ‖∆v(k)m (s)‖2ds

≤ K1(M, f )

[(1+ 2M)T +

√T(∫ T

0

∥∥v′′m−1(s)∥∥2 ds)1/2]

+1µ0K1(M, f )

∫ t

0

(1+ 2M +

∥∥v′′m−1(s)∥∥) Y (k)m (s)ds≤ K1(M, f )

[(1+ 2M)T +

√TM

]+1µ0K1(M, f )

∫ t

0

(1+ 2M +

∥∥v′′m−1(s)∥∥) Y (k)m (s)ds. (3.37)

It remains to estimate∫ t0 ‖··

v(k)

m (s)‖2ds.

Eq. (3.13) can be rewritten as follows

〈v(k)m (t), wj〉 −⟨∂

∂x

(µm(t)∇v(k)m (t)

), wj

⟩= 〈Fm(t), wj〉, 1 ≤ j ≤ k. (3.38)

Hence, It follows after replacingwj with··

v(k)

m (t) and integrating that∫ t

0‖v(k)m (s)‖

2ds ≤ 2∫ t

0

∥∥∥∥ ∂∂x (µm(s)∇v(k)m (s))∥∥∥∥2 ds+ 2 ∫ t

0‖Fm(s)‖2ds

≤ 2∫ t

0

∥∥∥∥ ∂∂x (µm(s)∇v(k)m (s))∥∥∥∥2 ds+ 2TK 20 (M, f ). (3.39)


We estimate the term∥∥∥ ∂∂x

(µm(s)∇v

(k)m (s)

)∥∥∥ .By (3.25), we obtain∥∥∥∥ ∂∂x (µm(s)∇v(k)m (s))

∥∥∥∥ = ∥∥∥∥∂µm∂x (s)∇v(k)m (s)+ µm(s)∆v(k)m (s)∥∥∥∥

≤

∥∥∥∥∂µm∂x (s)∥∥∥∥C0(Ω)

∥∥∇v(k)m (s)∥∥+ ‖µm(s)‖C0(Ω) ∥∥∆v(k)m (s)∥∥≤

1√µ0K(M, µ)(M +M∗)

√X (k)m (s)+

1√µ0K(M, µ)

√Y (k)m (s)

≤1√µ0K(M, µ)

√1+ (M +M∗)2

√S(k)m (s). (3.40)

Therefore, from (3.39) and (3.40), we obtain∫ t

0‖v(k)m (s)‖

2ds ≤ 2TK 20 (M, f )+2µ0K 2(M, µ)[1+ (M +M∗)2]

∫ t

0S(k)m (s)ds. (3.41)

Choosing β > 0, with 2βµ0≤12 , it follows from (3.15), (3.16), (3.18), (3.20)–(3.22), (3.28), (3.32), (3.36), (3.37) and (3.41),

that

S(k)m (t) ≤ D0(f , µ, µ′, v0k, v1k, g(0), β)+ D1(M, T , β)

+ 2∫ t

0

(D2(M, T , β)+

1µ0K1(M, f )

∥∥v′′m−1(s)∥∥) S(k)m (s)ds, (3.42)

where

D0(f , µ, µ′, v0k, v1k, g(0), β) = 2s(k)m (0)+ 4〈Fm(0),∆v0k〉

+ 4⟨∂µm

∂x(0)∇v0k,∆v0k

⟩+4β‖Fm(0)‖ +

4β

∥∥∥∥∂µm∂x (0)∥∥∥∥2C0(Ω)

‖∇v0k‖2 ,

D1(M, T , β) = 6TK 20 (M, f )+8β

[(1+ 2M)2T +M2

]TK 21 (M, f )

+ 2((1+ 2M)

√T +M

)√TK1(M, f ),

D2(M, T , β) = 1+2µ0(M +M∗)

(2+√µ0 +M +M∗

)K(M, µ)+

1+ 2Mµ0

K1(M, f )

+ 2

[1β(M +M∗)2

(1+

1+M +M∗√µ0

)2T +

1+ (M +M∗)2

µ0

]K 2(M, µ).

(3.43)

By means of the convergences (3.14) we can deduce the existence of a constantM > 0 independent of k andm such that

D0(f , µ, µ′, v0k, v1k, g(0), β) ≤12M2. (3.44)

Notice that, from (3.5), (3.6), (3.43)2,3 and the assumptions H2–(H4)we have

limT→0+

D1(M, T , β) = limT→0+

T D2(M, T , β) = 0. (3.45)

So, from (3.43)2,3 and (3.45), we can choose T ∈ (0, T ∗], such that(12M2 + D1(M, T , β)

)exp

(2T D2(M, T , β)+

2√T

µ0K1(M, f )M

)≤ M2, (3.46)

and

kT =(1+

1√µ0

)√T(4K 21 (M, f )+ K

2(M, µ)(2+M +M∗)2M2)1/2eT(1+M+M

∗

2µ0K(M,µ)

)< 1. (3.47)


Finally, it follows from (3.42)–(3.44) and (3.46), that

S(k)m (t) ≤ M2 exp

(−2T D2(M, T , β)−

2√T

µ0K1(M, f )M

)

+ 2∫ t

0

(D2(M, T , β)+

1µ0K1(M, f )

∥∥v′′m−1(s)∥∥) S(k)m (s)ds. (3.48)

By using Gronwall’s lemma, we deduce from (3.8), (3.46) and (3.48) that

S(k)m (t) ≤ M2 exp

(−2T D2(M, T , β)−

2√T

µ0K1(M, f )M

)

× exp[2∫ t

0

(D2(M, T , β)+

1µ0K1(M, f )

∥∥v′′m−1(s)∥∥) ds]≤ M2 exp

(−2T D2(M, T , β)−

2√T

µ0K1(M, f )M

)

× exp[2T D2(M, T , β)+

2µ0K1(M, f )

√T∥∥v′′m−1∥∥L2(QT )

]≤ M2, (3.49)

for all t ∈ [0, T (k)m ], for allm and k. So we can take constant T(k)m = T for all k andm. Therefore, we have

v(k)m ∈ W (M, T ), for allm and k. (3.50)

Step 3. Limiting process. From (3.50), we deduce the existence of a subsequence of v(k)m still also so denoted, such thatv(k)m → vm in L∞(0, T ; V ∩ H2)weak*,

v(k)m → v′m in L∞(0, T ; V )weak*,

v(k)m → v′′m in L2(QT )weak,vm ∈ W (M, T ).

(3.51)

Passing to limit in (3.13), we have vm satisfying (3.9), (3.10) in L2(0, T ), weak. On the other hand, it follows from (3.9),(3.10) and (3.51)4 that v′′m =

∂∂x (µm(t)∇vm(t))+ Fm ∈ L

∞(0, T ; L2), hence vm ∈ W1(M, T ) and the proof of Theorem 3.1 iscomplete.

Theorem 3.2. Let assumptions (H1)–(H4) hold. Then:(i) There exist positive constants M and T satisfying (3.44), (3.46) and (3.47) such that the problem (3.1), (3.2) has a unique

weak solution v ∈ W1(M, T ).(ii) On the other hand, the linear recurrent sequence vm defined by (3.9), (3.10) converges to the solution v of problem (3.1),

(3.2) strongly in the space

W1(T ) = w ∈ L∞(0, T ; V ) : w′ ∈ L∞(0, T ; L2). (3.52)

Furthermore, we have also the estimate

‖vm − v‖L∞(0,T ;V ) +∥∥v′m − v′∥∥L∞(0,T ;L2) ≤ CkmT , for all m ∈ N, (3.53)

where the constant kT ∈ (0, 1) is defined as in (3.47) and C is a constant depending only on T , g, v0, v1 and kT .

Proof. (i) Existence of the solution. First, we note thatW1(T ) is a Banach space with respect to the norm (see Lions [18]).

‖w‖W1(T ) = ‖w‖L∞(0,T ;V ) +∥∥w′∥∥L∞(0,T ;L2) . (3.54)

We shall prove that vm is a Cauchy sequence inW1(T ). Letwm = vm+1− vm. Thenwm satisfies the variational problem〈w′′m(t), w〉 + 〈µm+1(t)∇wm(t),∇w〉 + 〈(µm+1(t)− µm(t))∇vm(t),∇w〉= 〈Fm+1(t)− Fm(t), w〉 , ∀w ∈ V ,

wm(0) = w′m(0) = 0.(3.55)


Takingw = w′m in (3.55), after integrating in t , we get

Zm(t) =∫ t

0ds∫ 1

0µ′m+1(x, s) |∇wm(x, s)|

2 dx+ 2∫ t

0

⟨Fm+1(s)− Fm(s), w′m(s)

⟩ds

+ 2∫ t

0

⟨∂

∂x[(µm+1(s)− µm(s))∇vm(s)] , w′m(s)

⟩ds =

3∑i=1

Ji, (3.56)

where

Zm(t) =∥∥w′m(t)∥∥2 + ∥∥∥√µm+1(t)∇wm(t)∥∥∥2 ≥ ∥∥w′m(t)∥∥2 + µ0 ‖∇wm(t)‖2 . (3.57)

and all integrals on the right-hand side of (3.56) are estimated as follows.First integral J1. By (H2) and (H3), we have

|J1| ≤∫ t

0ds∫ 1

0

∣∣µ′m+1(x, s)∣∣ |∇wm(x, s)|2 dx ≤ 1µ0K(M, µ)(M +M∗)

∫ t

0Zm(s)ds. (3.58)

Second integral J2. By (H2) and (H4), we easily have

‖Fm+1(t)− Fm(t)‖ ≤ K1(M, f )[2 ‖∇wm−1(t)‖ +

∥∥w′m−1(t)∥∥] ≤ 2K1(M, f ) ‖wm−1‖W1(T ) . (3.59)

Hence

|J2| ≤ 2∣∣∣∣∫ t

0

⟨Fm+1(s)− Fm(s), w′m(s)

⟩ds∣∣∣∣ ≤ 4TK 21 (M, f ) ‖wm−1‖2W1(T ) + ∫ t

0Zm(s)ds. (3.60)

Third integral J3. We have

∂

∂x[(µm+1(s)− µm(s))∇vm(s)] = (µm+1(s)− µm(s))∆vm(s)+ µ′(ηm+1(s))∇wm−1(s)∇vm(s)

+[µ′(ηm+1(s))− µ′(ηm(s))

]∇ηm(s)∇vm(s), (3.61)

where ηi(s) = vi−1(s)+ ϕ(s), i = m,m+ 1.We also note that

‖∇vm(s)‖C0(Ω) ≤ ‖∆vm(s)‖ ≤ M,

‖∇ηm(s)‖C0(Ω) ≤ ‖∆vm−1(s)‖ + ‖∇ϕ(s)‖C0(Ω) ≤ M +M∗,∥∥µ′(ηm+1(s))∥∥C0(Ω) ≤ K(M, µ),

‖µm+1(s)− µm(s)‖C0(Ω) ≤ K(M, µ) ‖wm−1(s)‖C0(Ω) ≤ K(M, µ) ‖wm−1‖W1(T ) ,∥∥µ′(ηm+1(s))− µ′(ηm(s))∥∥ ≤ K(M, µ) ‖wm−1(s)‖ ≤ K(M, µ) ‖wm−1‖W1(T ) .(3.62)

Therefore, we deduce from (3.61) and (3.62), that∥∥∥∥ ∂∂x [(µm+1(s)− µm(s))∇vm(s)]∥∥∥∥ ≤ ‖(µm+1(s)− µm(s))∆vm(s)‖ + ∥∥µ′(ηm+1(s))∇wm−1(s)∇vm(s)∥∥+∥∥[µ′(ηm+1(s))− µ′(ηm(s))]∇ηm(s)∇vm(s)∥∥

≤ ‖µm+1(s)− µm(s)‖C0(Ω) ‖∆vm(s)‖

+∥∥µ′(ηm+1(s))∥∥C0(Ω) ‖∇wm−1(s)‖ ‖∇vm(s)‖C0(Ω)

+∥∥µ′(ηm+1(s))− µ′(ηm(s))∥∥ ‖∇ηm(s)‖C0(Ω) ‖∇vm(s)‖C0(Ω)

≤ (2+M +M∗)MK(M, µ) ‖wm−1‖W1(T ) . (3.63)

This implies that

|J3| ≤ 2∣∣∣∣∫ t

0

⟨∂

∂x[(µm+1(s)− µm(s))∇vm(s)] , w′m(s)

⟩ds∣∣∣∣

≤ 2K(M, µ)(2+M +M∗)M ‖wm−1‖W1(T )

∫ t

0

√Zm(s)ds

≤ T K 2(M, µ)(2+M +M∗)2M2 ‖wm−1‖2W1(T ) +∫ t

0Zm(s)ds. (3.64)


Combining (3.56), (3.58), (3.60) and (3.64), we obtain

Zm(t) ≤ T(4K 21 (M, f )+ K

2(M, µ)(2+M +M∗)2M2)‖wm−1‖

2W1(T )

+

(2+

M +M∗

µ0K(M, µ)

)∫ t

0Zm(s)ds. (3.65)

Using Gronwall’s lemma, we deduce from (3.65) that

‖wm‖W1(T ) ≤ kT ‖wm−1‖W1(T ) ∀m ∈ N, (3.66)

where kT =(1+ 1

√µ0

)√T(4K 21 (M, f )+ K

2(M, µ)(2+M +M∗)2M2)1/2eT(1+M+M

∗

2µ0K(M,µ)

)< 1 as in (3.47), which

implies that∥∥vm − vm+p∥∥W1(T ) ≤ ‖v0 − v1‖W1(T ) (1− kT )−1kmT ∀m, p ∈ N. (3.67)

It follows that vm is a Cauchy sequence inW1(T ). Then there exists v ∈ W1(T ) such that

vm → v strongly inW1(T ). (3.68)

We also note that vm ∈ W1(M, T ), then from the sequence vmwe can deduce a subsequence vmj such thatvmj → v in L∞(0, T ; V ∩ H2)weak*,v′mj → v′ in L∞(0, T ; V )weak*,v′′mj → v′′ in L2(QT )weak,v ∈ W (M, T ).

(3.69)

Note that∥∥Fm(t)− f (·, t, v(t), vx(t), v′(t))∥∥ ≤ 2K1(M, f ) ‖vm−1 − v‖W1(T ) , (3.70)

|µm(x, t)− µ(v(x, t)+ ϕ(x, t))| ≤ K(M, µ) ‖vm−1 − v‖W1(T ) . (3.71)

Hence, from (3.68), (3.70) and (3.71), we obtain

Fm(t)→ f (·, t, v(t), vx(t), v′(t)) strongly in L∞(0, T ; L2), (3.72)

µm(t)→ µ(v(t)+ ϕ(t)) strongly in L∞(QT ). (3.73)

Finally, passing to limit in (3.9)–(3.10) as m = mj → ∞, it implies from (3.68), (3.69)1,3, (3.72) and (3.73) that thereexists v ∈ W (M, T ) satisfying the equation

〈v′′(t), w〉 + 〈µ(v(t)+ ϕ(t))vx(t),∇w〉 =⟨f (·, t, v(t), vx(t), v′(t)), w

⟩, ∀w ∈ V , (3.74)

and the initial conditions

v(0) = v0, v′(0) = v1. (3.75)

On the other hand, from the assumptions (H3) and (H4)we obtain from (3.69)4, (3.72) and (3.74), that

v′′ = µ′(v + ϕ)(vx + g)vx + µ(v + ϕ)vxx + f (x, t, v, vx, v′) ∈ L∞(0, T ; L2), (3.76)

thus we have v ∈ W1(M, T ). The existence proof is completed.(ii) Uniqueness of the solution.Let v1, v2 ∈ W1(M, T ) be two weak solutions of problem (3.1), (3.2). Then v = v1− v2 satisfies the variational problem〈v′′(t), w〉 + 〈µ1(t)∇v(t),∇w〉 + 〈(µ1(t)− µ2(t))∇v2(t),∇w〉 = 〈F2(t)− F1(t), w〉 , ∀w ∈ V ,v(0) = v′(0) = 0, (3.77)

where

Fi(t) = f (x, t, vi,∇vi, v′i), µi(t) = µ(vi + ϕ), i = 1, 2. (3.78)

We takew = v′ in (3.77)1 and integrate in t to get

Z(t) =∫ t

0ds∫ 1

0µ′1(x, s) |∇v(x, s)|

2 dx+ 2∫ t

0

⟨F1(s)− F2(s), v′(s)

⟩ds

+ 2∫ t

0

⟨∂

∂x[(µ1(s)− µ2(s))∇v2(s)] , v′(s)

⟩ds ≡

3∑i=1

Ji, (3.79)


where

Z(t) =∥∥v′(t)∥∥2 + ∥∥∥√µ1(t)∇v(t)∥∥∥2 ≥ ∥∥v′(t)∥∥2 + µ0 ‖∇v(t)‖2 . (3.80)

We now estimate the terms on the right-hand side of (3.79) as follows.First integral J1. By (H2) and (H3), we have∣∣J1∣∣ ≤ ∫ t

0ds∫ 1

0

∣∣µ′1(x, s)∣∣ |∇v(x, s)|2 dx ≤ 1µ0K(M, µ)(M +M∗)

∫ t

0Z(s)ds. (3.81)

Second integral J2. By (H2) and (H4), we easily have

‖F1(t)− F2(t)‖ ≤ K1(M, f )[2 ‖∇v(t)‖ +

∥∥v′(t)∥∥] . (3.82)

Hence, by (3.80), we obtain∣∣J2∣∣ ≤ 2 ∣∣∣∣∫ t

0

⟨F1(s)− F2(s), v′(s)

⟩ds∣∣∣∣ ≤ 2K1(M, f )(1+ 1

√µ0

)∫ t

0Z(s)ds. (3.83)

Third integral J3. We have∂

∂x[(µ1(s)− µ2(s))∇v2(s)] = (µ1(s)− µ2(s))∆v2(s)+ µ′(η1(s))∇v(s)∇v2(s)

+[µ′(η1(s))− µ′(η2(s))

]∇η2(s)∇v2(s), (3.84)

where ηi(s) = vi(s)+ ϕ(s), i = 1, 2. We also note that

‖∇v2(s)‖C0(Ω) ≤ ‖∆v2(s)‖ ≤ M,

‖∇η2(s)‖C0(Ω) ≤ ‖∆v2(s)‖ + ‖∇ϕ(s)‖C0(Ω) ≤ M +M∗,

‖µ1(s)− µ2(s)‖C0(Ω) ≤ K(M, µ) ‖v(s)‖C0(Ω) ≤ K(M, µ) ‖∇v(s)‖ ,∥∥µ′(η1(s))− µ′(η2(s))∥∥ ≤ K(M, µ) ‖v(s)‖ ≤ K(M, µ) ‖∇v(s)‖ .(3.85)

We deduce from (3.84) and (3.85), that∥∥∥∥ ∂∂x [(µ1(s)− µ2(s))∇v2(s)]∥∥∥∥ ≤ ‖(µ1(s)− µ2(s))∆v2(s)‖ + ∥∥µ′(η1(s))∇v(s)∇v2(s)∥∥+∥∥[µ′(η1(s))− µ′(η2(s))]∇η2(s)∇v2(s)∥∥

≤ ‖µ1(s)− µ2(s)‖C0(Ω) ‖∆v2(s)‖ +∥∥µ′(η1(s))∥∥C0(Ω) ‖∇v2(s)‖C0(Ω) ‖∇v(s)‖

+∥∥µ′(η1(s))− µ′(η2(s))∥∥ ‖∇η2(s)‖C0(Ω) ‖∇v2(s)‖C0(Ω)

≤ (2+M +M∗)MK(M, µ) ‖∇v(s)‖ . (3.86)

This implies that∣∣J3∣∣ ≤ 2 ∣∣∣∣∫ t

0

⟨∂

∂x[(µ1(s)− µ2(s))∇v2(s)] , v′(s)

⟩ds∣∣∣∣

≤ 2∫ t

0

∥∥∥∥ ∂∂x [(µ1(s)− µ2(s))∇v2(s)]∥∥∥∥ ∥∥v′(s)∥∥ ds

≤1√µ0(2+M +M∗)MK(M, µ)

∫ t

0Z(s)ds. (3.87)

Combining (3.79), (3.81), (3.83) and (3.87), we have

Z(t) ≤ K∫ t

0Z(s)ds,

where K = 2K1(M, f )(1+ 1

√µ0

)+

[1µ0(M +M∗)+ M

√µ0(2+M +M∗)

]K(M, µ). Using Gronwall’s lemma, it follows that

Z(t) ≡ 0, ie., v1 ≡ v2.Theorem 3.2 is proved completely.

Remark 3.3. In the case ofµ ≡ 1, f = f (x, t, u, ux, ut)with f ∈ C1([0, 1]×R+×R3), and some other boundary conditionsstanding for (1.2), we have also obtained some results in the papers [3,5,11].


4. Asymptotic expansion of the solution with respect to two small parameters

In this section, let (H1)− (H4) hold. We also make the following assumptions:

(H5) µ1 ∈ C2(R), µ1 ≥ 0,(H6) f1 ∈ C1([0, 1] × R+ × R3).

We consider the following perturbed problem, where ε1, ε2 are small parameters such that, 0 ≤ ε1 ≤ ε1∗ < 1,|ε2| ≤ ε2∗ < 1 :

(Pε1,ε2)

utt −

∂

∂x

(µε1(u)ux

)= Fε2(x, t, u, ux, ut), x ∈ (0, 1), 0 < t < T ,

ux(0, t) = g(t), u(1, t) = 0,u(x, 0) = u0(x), ut(x, 0) = u1(x),µε1(u) = µ(u)+ ε1µ1(u),Fε2(x, t, u, ux, ut) = f (x, t, u, ux, ut)+ ε2f1(x, t, u, ux, ut).

By Theorem 3.2, the problem (Pε1,ε2) ≡ (P−→ε ) has a unique weak solution u depending on−→ε = (ε1, ε2) : u = uε1,ε2 .

When−→ε = (0, 0), (Pε1,ε2) is denoted by (P0,0) ≡ (P0). We shall study the asymptotic expansion of the solution of problem(Pε1,ε2)with respect to ε1, ε2.We use the following notations. For a multi-index γ = (γ1, γ2) ∈ Z2

+, and−→ε = (ε1, ε2) ∈ R+ × R, we put

|γ | = γ1 + γ2, γ ! = γ1!γ2!,∥∥−→ε ∥∥ = √ε21 + ε22,−→ε γ = εγ11 εγ22 ,

α, β ∈ Z2+, α ≤ β ⇐⇒ αi ≤ βi ∀i = 1, 2.

(4.1)

First, we shall need the following lemma.

Lemma 4.1. Let m, N ∈ N and uα ∈ R, α ∈ Z2+, 1 ≤ |α| ≤ N. Then( ∑

1≤|α|≤N

uα−→εα

)m=

∑m≤|α|≤mN

T (m)α [u]−→ε

α, (4.2)

where the coefficients T (m)α [u], m ≤ |α| ≤ mN depending on u = (uα), α ∈ Z2+, 1 ≤ |α| ≤ N defined by the recurrence formulas

T (1)α [u] = uα, 1 ≤ |α| ≤ N,T (m)α [u] =

∑β∈A(m)α

uα−βT(m−1)β [u], m ≤ |α| ≤ mN,m ≥ 2,

A(m)α = β ∈ Z2+: β ≤ α, 1 ≤ |α − β| ≤ N,m− 1 ≤ |β| ≤ (m− 1)N.

(4.3)

The proof of Lemma 4.1 can be found in [17].Now, we assume that

(H7) µ ∈ CN+2(R), µ1 ∈ CN+1(R), µ ≥ µ0 > 0, µ1 ≥ 0,(H8) f ∈ CN+1([0, 1] × R+ × R3), f1 ∈ CN([0, 1] × R+ × R3).

We also use the notation f [u] = f (x, t, u, ux, ut).Let u0 ≡ u0,0 be a unique weak solution of problem

(P0,0

)≡ P0 (as in Theorem 3.2) corresponding to −→ε = (ε1, ε2) =

(0, 0), i.e.,

(P0,0)

u′′0 −

∂

∂x(µ(u0)u0x) = f (x, t, u0, u0x, u′0) ≡ f [u0], x ∈ (0, 1), 0 < t < T ,

u0x(0, t) = g(t), u0(1, t) = 0,u0(x, 0) = u0(x), u′0(x, 0) = u1(x),u0 − ϕ ∈ W1(M, T ).

Let us consider the sequence of weak solutions uγ , γ ∈ Z2+, 1 ≤ |γ | ≤ N , defined by the following problems:

(Pγ )

u′′γ −

∂

∂x

(µ(u0)uγ x

)= Fγ , x ∈ (0, 1), 0 < t < T ,

uγ x(0, t) = uγ (1, t) = 0,uγ (x, 0) = u′γ (x, 0) = 0,uγ ∈ W1(M, T ),


where Fγ , γ ∈ Z2+, 1 ≤ |γ | ≤ N , is defined by the recurrence formulas

Fγ =

f [u0] ≡ f (x, t, u0, u0x, u′0), |γ | = 0,

πν[f ] + π (2)ν [f1] +∑

1≤|ν|≤|γ |,ν≤γ

∂

∂x

[(ρν[µ] + ρ

(1)ν [µ1]

)∇uγ−ν

], 1 ≤ |γ | ≤ N, (4.4)

with ρν[µ] = ρν[µ; uγ γ≤ν], ρ(1)ν [µ] = ρ(1)ν [µ; uγ γ≤ν], πν[f ] = πν[f ; uγ γ≤ν], π (2)ν [f ] = π (2)ν [f ; uγ γ≤ν], |ν| ≤ N ,defined by the recurrence formulas

ρν[µ] =

µ(u0), |ν| = 0,|ν|∑m=1

1m!µ(m)(u0)T (m)ν [u], 1 ≤ |ν| ≤ N, (4.5)

ρ(1)ν [µ] = ρν1−1,ν2 [µ], and ρ(1)ν [µ] = ρ−1,ν2 [µ] = 0, if ν1 = 0, ν = (ν1, ν2) ∈ Z2+, (4.6)

πν[f ] =

f [u0], |ν| = 0,∑1≤|m|≤|ν|

∑(α,β,γ )∈A(m,N)α+β+γ=ν

1m!Dmf [u0]T (m1)α [u]T (m2)β [∇u]T (m3)γ [u′], 1 ≤ |ν| ≤ N, (4.7)

where m = (m1,m2,m3) ∈ Z3+, |m| = m1 + m2 + m3,m! = m1!m2!m3!,Dmf = D

m13 D

m24 D

m35 f , A(m,N) = (α, β, γ ) ∈(

Z2+

)3: m1 ≤ |α| ≤ m1N,m2 ≤ |β| ≤ m2N,m3 ≤ |γ | ≤ m3N, , and

π (2)ν [f ] = πν1,ν2−1[f ], π (2)ν [f ] = πν1,−1[f ] = 0, if ν2 = 0, ν = (ν1, ν2) ∈ Z2+. (4.8)

Then, we have the following lemma.

Lemma 4.2. Let ρν[µ], πν[f ], |ν| ≤ N, be the functions are defined by formulas (4.5) and (4.7). Put h =∑|γ |≤N uγ

−→εγ , then

we have

µ(h) =∑|ν|≤N

ρν[µ]−→ε

ν+∥∥−→ε ∥∥N+1 R(1)N [µ,−→ε ], (4.9)

f [h] =∑|ν|≤N

πν[f ]−→εν+∥∥−→ε ∥∥N+1 R(1)N [f ,−→ε ], (4.10)

where∥∥∥R(1)N [µ,−→ε ]∥∥∥L∞(0,T ;L2) + ∥∥∥R(1)N [f ,−→ε ]∥∥∥L∞(0,T ;L2) ≤ C, with C is a constant depending only on N, T , f , µ, uγ , |γ | ≤ N.

Proof. (i) In the case of N = 1, the proof of (4.9) is easy, hence we omit the details, which we only prove with N ≥ 2. Werewrite as h = u0 +

∑1≤|γ |≤N uγ

−→εγ≡ u0 + h1.

By using Taylor’s expansion of the function µ(h) = µ(u0 + h1) around the point u0 up to order N + 1, we obtain from(4.2), that

µ(u0 + h1) = µ(u0)+N∑m=1

1m!µ(m)(u0)hm1 +

1N!

∫ 1

0(1− θ)Nµ(N+1)(u0 + θh1)hN+11 dθ

= µ(u0)+N∑m=1

1m!µ(m)(u0)

∑m≤|ν|≤mN

T (m)ν [u]−→ε

ν+ R(1)N [µ, h1]

= µ(u0)+N∑m=1

1m!µ(m)(u0)

∑m≤|ν|≤N

T (m)ν [u]−→ε

ν

+

N∑m=1

1m!µ(m)(u0)

∑N+1≤|ν|≤mN

T (m)ν [u]−→ε

ν+ R(1)N [µ, h1], (4.11)

where

R(1)N [µ, h1] =1N!

∫ 1

0(1− θ)Nµ(N+1)(u0 + θh1)hN+11 dθ. (4.12)

We also note that

N∑m=1

1m!µ(m)(u0)

∑m≤|ν|≤N

T (m)ν [u]−→ε

ν=

∑1≤|ν|≤N

(|ν|∑m=1

1m!µ(m)(u0)T (m)ν [u]

)−→ε

ν. (4.13)


On the other hand, if we put

R(1)N [µ,−→ε ] =

∥∥−→ε ∥∥−N−1 ( N∑m=1

1m!µ(m)(u0)

∑N+1≤|ν|≤mN

T (m)ν [u]−→ε

ν+ R(1)N [µ, h1]

), (4.14)

then, by the boundedness of the functions uγ , ∇uγ , u′γ , |γ | ≤ N in the function space L∞(0, T ;H1), we obtain from (4.3),

(4.12), (4.14), that∥∥∥R(1)N [µ,−→ε ]∥∥∥L∞(0,T ;L2) ≤ C,with and C is a constant depending only on N , T , µ, uγ , |γ | ≤ N . Therefore,

we obtain from (4.5), (4.11), (4.13), (4.14), that

µ(u0 + h1) = µ(u0)+∑

1≤|ν|≤N

(|ν|∑m=1

1m!µ(m)(u0)T (m)ν [u]

)−→ε

ν+∥∥−→ε ∥∥N+1 R(1)N [µ,−→ε ]

=

∑|ν|≤N

ρν[µ]−→ε

ν+∥∥−→ε ∥∥N+1 R(1)N [µ,−→ε ]. (4.15)

Hence, the part 1 of Lemma 4.2 is proved.(ii) We only prove (4.10) with N ≥ 2. By using Taylor’s expansion of the function f [u0 + h1] around the point u0 up to

order N + 1,we obtain from (4.2), that

f [u0 + h1] = f [u0] + D3f [u0]h1 + D4f [u0]∇h1 + D5f [u0]h′1

+

∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

1m!Dmf [u0]h

m11 (∇h1)m2 (h′1)

m3 + R(1)N [f , h1]

= f [u0] + D3f [u0]h1 + D4f [u0]∇h1 + D5f [u0]h′1

+

∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑|m|≤|ν|≤|m|N


1m!Dmf [u0]T (m1)α [u]T (m2)β [∇u]T (m3)γ [u′]−→ε ν + R(1)N [f , h1]

= f [u0] + D3f [u0]h1 + D4f [u0]∇h1 + D5f [u0]h′1

+

∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑|m|≤|ν|≤N


1m!Dmf [u0]T (m1)α [u]T (m2)β [∇u]T (m3)γ [u′]−→ε ν

+

∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑N+1≤|ν|≤|m|N


1m!Dmf [u0]T (m1)α [u]T (m2)β [∇u]T (m3)γ [u′]−→ε ν + R(1)N [f , h1],

(4.16)

where

R(1)N [f , h1] =∑|m|=N+1

m=(m1,m2,m3)∈Z3+

N + 1m!

∫ 1

0(1− θ)NDmf [u0 + θh1]h

m11 (∇h1)m2 (h′1)

m3dθ. (4.17)

We also note that

f [u0] + D3f [u0]h1 + D4f [u0]∇h1 + D5f [u0]h′1

+

∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑|m|≤|ν|≤N



= f [u0] +∑1≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑|m|≤|ν|≤N



= f [u0] +∑

1≤|ν|≤N

∑1≤|m|≤|ν|

m=(m1,m2,m3)∈Z3+



=

∑|ν|≤N

πν[µ]−→ε

ν. (4.18)


Similarly, we have also∑2≤|m|≤N

m=(m1,m2,m3)∈Z3+

∑N+1≤|ν|≤|m|N



+ R(1)N [f , h1] =∥∥−→ε ∥∥N+1 R(1)N [f ,−→ε ], (4.19)

where∥∥∥R(1)N [f ,−→ε ]∥∥∥L∞(0,T ;L2) ≤ C , with C is a constant depending only on N , T , f , uγ , |γ | ≤ N.

Then (4.10) holds. Lemma 4.2 is proved.Let u = uε1,ε2 ∈ W1(M, T ) be a unique weak solution of the problem (Pε1,ε2). Then v = u −

∑|γ |≤N uγ

−→εγ≡ u − h

satisfies the problem

v′′ −∂

∂x

(µε1(v + h)vx

)= Fε2 [v + h] − Fε2 [h] +

∂

∂x

([µε1(v + h)− µε1(h)

]hx)

+ E−→ε (x, t), x ∈ (0, 1), 0 < t < T ,vx(0, t) = v(1, t) = 0,v(x, 0) = v′(x, 0) = 0,µε1(v) = µ(v)+ ε1µ1(v),Fε2 [v] = f [v] + ε2f1[v] = f (x, t, v, vx, vt)+ ε2f1(x, t, v, vx, vt),

(4.20)

where

Eε(x, t) = f [h] − f [u0] + ε2f1[h] +∂

∂x([µ(h)− µ(u0)+ ε1µ1(h)] hx)−

∑1≤|γ |≤N

Fγ−→εγ. (4.21)

Then, we have the following lemma.

Lemma 4.3. Let (H1), (H2), (H7), and (H8) hold. Then there exists a constant K such that∥∥E−→ε ∥∥L∞(0,T ;L2) ≤ K ∥∥−→ε ∥∥N+1 , (4.22)

where K is a constant depending only on N, T , f , f1, µ, µ1, uγ , |γ | ≤ N.

Proof. In the case of N = 1, the proof of Lemma 4.3 is easy, hence we omit the details, which we only prove with N ≥ 2.By using formulas (4.9), (4.10) for the functions f1[h] and µ1[h], we obtain

µ1(h) =∑|ν|≤N−1

ρν[µ1]−→ε

ν+∥∥−→ε ∥∥N R (1)N−1[µ1,−→ε ],

f1[h] =∑|ν|≤N−1

πν[f1]−→εν+∥∥−→ε ∥∥N R(1)N−1[f1,−→ε ]. (4.23)

By (4.6), (4.8), we rewrite ε1µ1(h) and ε2f1[h] as follows

ε1µ1(h) =∑|ν|≤N−1

ρν[µ1]ε1−→ε

ν+ ε1

∥∥−→ε ∥∥N R (1)N−1[µ1,−→ε ]=

∑1≤|ν|≤N,ν1≥1

ρν1−1,ν2 [µ1]−→ε

ν+ ε1

∥∥−→ε ∥∥N R (1)N−1[µ1,−→ε ]=

∑1≤|ν|≤N,ν1≥1

ρ(1)ν1−1,ν2

[µ1]−→ε

ν+ ε1

∥∥−→ε ∥∥N R (1)N−1[µ1,−→ε ]=

∑1≤|ν|≤N

ρ(1)ν [µ1]−→ε

ν+ ε1

∥∥−→ε ∥∥N R (1)N−1[µ1,−→ε ]. (4.24)

Similarly

ε2f1[h] =∑

1≤|ν|≤N

π (2)ν [f1]−→ε

ν+ ε2

∥∥−→ε ∥∥N R(1)N−1[f1,−→ε ]. (4.25)

First, we deduce from (4.10) and (4.25), that

f [h] − f [u0] + ε2f1[h] =∑

1≤|ν|≤N

(πν[f ] + π (2)ν [f1]

)−→ε

ν+∥∥−→ε ∥∥N+1 R(1)N [f , f1,−→ε ], (4.26)


where R(1)N [f , f1,−→ε ] = R(1)N [f ,

−→ε ]+ε2‖−→ε ‖R(1)N−1[f1,

−→ε ] is bounded in the function space L∞(0, T ; L2) by a constant dependingonly on N , T , f , f1, uγ , |γ | ≤ N.On the other hand, we deduce from (4.9) and (4.23), that

[µ(h)− µ(u0)+ ε1µ1(h)] hx

=

∑1≤|ν|≤N

(ρν[µ] + ρ

(1)ν [µ1]

)−→ε

ν∑|α|≤N

∇uα−→εα+∥∥−→ε ∥∥N+1 ( ε1∥∥−→ε ∥∥ R (1)N−1[µ1,−→ε ] + R (1)N [µ,−→ε ]

) ∑|α|≤N

∇uα−→εα

=

∑1≤|ν|≤N

∑|α|≤N

(ρν[µ] + ρ

(1)ν [µ1]

)∇uα−→ε

ν+α+∥∥−→ε ∥∥N+1 R (1)N [µ,µ1,−→ε ]

=

∑1≤|ν|≤N,|α|≤N

(ρν[µ] + ρ

(1)ν [µ1]

)∇uα−→ε

ν+α+∥∥−→ε ∥∥N+1 R (1)N [µ,µ1,−→ε ]

=

∑1≤|γ |≤2N

∑1≤|ν|≤N,|γ−ν|≤N

(ρν[µ] + ρ

(1)ν [µ1]

)∇uγ−ν−→ε

γ+∥∥−→ε ∥∥N+1 R (1)N [µ,µ1,−→ε ]

=

∑1≤|γ |≤N

∑1≤|ν|≤N,|γ−ν|≤N

(ρν[µ] + ρ

(1)ν [µ1]


γ

+

∑N+1≤|γ |≤2N

∑1≤|ν|≤N,|γ−ν|≤N

(ρν[µ] + ρ

(1)ν [µ1]


γ+∥∥−→ε ∥∥N+1 R (1)N [µ,µ1,−→ε ]

=

∑1≤|γ |≤N

∑1≤|ν|≤|γ |,ν≤γ

(ρν[µ] + ρ

(1)ν [µ1]


γ+∥∥−→ε ∥∥N+1 R (2)N [µ,µ1,−→ε ], (4.27)

where

R (1)N [µ,µ1,−→ε ] =

(R (1)N [µ,

−→ε ] +ε1∥∥−→ε ∥∥ R (1)N−1[µ1,−→ε ]

) ∑|α|≤N

∇uα−→εα,∥∥−→ε ∥∥N+1 R (2)N [µ,µ1,−→ε ] = ∥∥−→ε ∥∥N+1 R (1)N [µ,µ1,−→ε ]

+

∑N+1≤|γ |≤2N

∑1≤|ν|≤N,|γ−ν|≤N

(ρν[µ]

−→εν+ ρ(1)ν [µ1]


γ.

(4.28)

Hence∂

∂x([µ(h)− µ(u0)+ ε1µ1(h)] hx) =

∑1≤|γ |≤N

∑1≤|ν|≤|γ |,ν≤γ

∂

∂x

[(ρν[µ]

−→εν+ ρ(1)ν [µ1]

)∇uγ−ν

]−→ε

γ

+∥∥−→ε ∥∥N+1 ∂

∂xR (2)N [µ,µ1,

−→ε ]. (4.29)

Combining (4.4), (4.5), (4.7), (4.21), (4.26) and (4.29), we then obtain

E−→ε (x, t) = f [h] − f [u0] + ε2f1[h] +∂

∂x([µ(h)− µ(u0)+ ε1µ1(h)] hx)−

∑1≤|γ |≤N

Fγ−→εγ

=

∑1≤|γ |≤N

πν[f ] + π (2)ν [f1] +

∑1≤|ν|≤|γ |,ν≤γ

∂

∂x

[(ρν[µ] + ρ

(1)ν [µ1]

)∇uγ−ν

]−→ε

γ

−

∑1≤|γ |≤N

Fγ−→εγ+∥∥−→ε ∥∥N+1 [R(1)N [f , f1,−→ε ] + ∂

∂xR (2)N [µ,µ1,

−→ε ]

]

=∥∥−→ε ∥∥N+1 [R(1)N [f , f1,−→ε ] + ∂

∂xR (2)N [µ,µ1,

−→ε ]

]. (4.30)

By the boundedness of the functions uγ ,∇uγ , u′γ , |γ | ≤ N in the function space L∞(0, T ;H1), we obtain from (4.26) and

(4.28), that∥∥E−→ε ∥∥L∞(0,T ;L2) ≤ K ∥∥−→ε ∥∥N+1 , (4.31)

where K is a constant depending only on N , T , f , f1, µ, µ1, uγ , |γ | ≤ N.The proof of Lemma 4.3 is complete.


Now, we consider the sequence of functions vm defined by

v0 ≡ 0,

v′′m −∂

∂x

(µε1(vm−1 + h)vmx

)= Fε2 [vm−1 + h] − Fε2 [h]

+∂

∂x

([µε1(vm−1 + h)− µε1(h)

]hx)+ E−→ε (x, t), x ∈ (0, 1), 0 < t < T ,

vmx(0, t) = vm(1, t) = 0,vm(x, 0) = v′m(x, 0) = 0, m ≥ 1.

(4.32)

Withm = 1, we have the problemv′′1 −

∂

∂x

(µε1(h)v1x

)= E−→ε (x, t), x ∈ (0, 1), 0 < t < T ,

v1x(0, t) = v1(1, t) = 0,v1(x, 0) = v′1(x, 0) = 0.

(4.33)

By multiplying the two sides of (4.33) by v′1, we find without difficulty from (4.22) that

‖v′1(t)‖2+ ‖

√µ1,ε1(t)v1x(t)‖

2= 2

∫ t

0〈E−→ε (s), v

′

1(s)〉ds+∫ t

0ds∫ 1

0µ′1,ε1(x, s)v

21x(x, s)dx

≤ T K 2∥∥−→ε ∥∥2N+2 + ∫ t

0

∥∥v′1(s)∥∥2 ds+ ∫ t

0ds∫ 1

0

∣∣µ′1,ε1(x, s)∣∣ v21x(x, s)dx, (4.34)

where µ1,ε1(x, t) = µε1(h(x, t)). By µ′

1,ε1(x, t) = µ′ε1(h(x, t))h

′(x, t), we have∣∣µ′1,ε1(x, t)∣∣ ≤ (N + 1)M (K(M∗, µ)+ K1(M∗, µ1)) ≡ ζ0,withM∗ = (N + 2)M. (4.35)

It follows from (4.34), (4.35), that

‖v′1(t)‖2+ µ0 ‖v1x(t)‖2 ≤ T K 2

∥∥−→ε ∥∥2N+2 + ∫ t

0

∥∥v′1(s)∥∥2 ds+ ζ0 ∫ t

0‖v1x(s)‖2ds. (4.36)

Using Gronwall’s lemma we obtain from (4.36), that

‖v′1‖L∞(0,T ;L2) + ‖v1‖L∞(0,T ;V ) ≤

(1+

1√µ0

)√T K∥∥−→ε ∥∥N+1 exp [T (1

2+

ζ0

2µ0

)]. (4.37)

We shall prove that there exists a constant CT , independent ofm and−→ε , such that∥∥v′m∥∥L∞(0,T ;L2) + ‖vm‖L∞(0,T ;V ) ≤ CT ∥∥−→ε ∥∥N+1 , with∥∥−→ε ∥∥ ≤ ε∗, for allm. (4.38)

By multiplying the two sides of (4.32) with v′m and after integration in t , we obtain without difficulty from (4.22) that

‖v′m(t)‖2+ µ0‖vmx(t)‖2 ≤ T K 2

∥∥−→ε ∥∥2N+2 + ∫ t

0

∥∥v′m(s)∥∥2 ds+ ∫ t

0ds∫ 1

0

∣∣µ′m,ε1(x, s)∣∣ v2mx(x, s)dx+ 2

∫ t

0

∥∥Fε2 [vm−1 + h] − Fε2 [h]∥∥ ∥∥v′m(s)∥∥ ds+ 2

∫ t

0

∥∥∥∥ ∂∂x ([µε1(vm−1 + h)− µε1(h)] hx)∥∥∥∥ ∥∥v′m(s)∥∥ ds

= T K 2∥∥−→ε ∥∥2N+2 + ∫ t

0

∥∥v′m(s)∥∥2 ds+ I1(t)+ I2(t)+ I3(t), (4.39)

where µm,ε1(t) = µε1(vm−1 + h). We now estimate the integrals on the right-hand side of (4.39) as follows.Estimating I1(t). We have µ′m,ε1(x, t) = µ

′ε1(vm−1 + h)(v′m−1 + h

′), hence∣∣µ′m,ε1(x, t)∣∣ ≤ M∗ (K(M∗, µ)+ K1(M∗, µ1)) ≡ ζ1,withM∗ = (N + 2)M. (4.40)

It follows from (4.40), that

I1(t) =∫ t

0ds∫ 1

0

∣∣µ′m,ε1(x, s)∣∣ v2mx(x, s)dx ≤ ζ1 ∫ t

0‖vmx(s)‖2ds. (4.41)


Estimating I2(t). We also note that ‖f [vm−1 + h] − f [h]‖ ≤ 2K1(M∗, f ) ‖vm−1‖W1(T ) and ‖f1[vm−1 + h] − f1[h]‖ ≤2K1(M∗, f1) ‖vm−1‖W1(T ) , hence, we have∥∥Fε2 [vm−1 + h] − Fε2 [h]∥∥ ≤ ζ2 ‖vm−1‖W1(T ) , (4.42)

where ζ2 = ζ2(M∗, f , f1) = 2 [K1(M∗, f )+ K1(M∗, f1)]. Therefore, we deduce from (4.42), that

I2(t) = 2∫ t

0

∥∥Fε2 [vm−1 + h] − Fε2 [h]∥∥ ∥∥v′m(s)∥∥ ds ≤ Tζ 22 ‖vm−1‖2W1(T ) + ∫ t

0

∥∥v′m(s)∥∥2 ds. (4.43)

Estimating I3(t). First, we need an estimation∥∥ ∂∂x ([µ(vm−1 + h)− µ(h)] hx)

∥∥ .From the equation

∂

∂x([µ(vm−1 + h)− µ(h)] hx) = [µ(vm−1 + h)− µ(h)] hxx +

∂

∂x[µ(vm−1 + h)− µ(h)] hx,

it follows that∥∥∥∥ ∂∂x ([µ(vm−1 + h)− µ(h)] hx)∥∥∥∥ ≤ ‖µ(vm−1 + h)− µ(h)‖C0(Ω) ‖hxx(s)‖

+

∥∥∥∥ ∂∂x [µ(vm−1 + h)− µ(h)]∥∥∥∥ ‖hx(s)‖C0(Ω)

≤√2 ‖h(s)‖H2

[‖µ(vm−1 + h)− µ(h)‖C0(Ω) +

∥∥∥∥ ∂∂x [µ(vm−1 + h)− µ(h)]∥∥∥∥]

≡√2 ‖h(s)‖H2

[I (1)3 (s)+ I (2)3 (s)

]. (4.44)

With I (1)3 (s)we have

I (1)3 (s) = ‖µ(vm−1 + h)− µ(h)‖C0(Ω) ≤ K(M∗, µ) ‖vm−1‖W1(T ) . (4.45)

With I (2)3 (s)we also obtain

I (2)3 (s) =∥∥∥∥ ∂∂x [µ(vm−1 + h)− µ(h)]

∥∥∥∥≤∥∥(µ′(vm−1 + h)∇vm−1)∥∥+ ∥∥[µ′(vm−1 + h)− µ′(h)]∇h∥∥

≤ (1+ ‖∇h‖)K(M∗, µ) ‖∇vm−1‖ ≤ [1+ (N + 1)M] K(M∗, µ) ‖vm−1‖W1(T ) . (4.46)

We deduce from (4.44)–(4.46), that∥∥∥∥ ∂∂x ([µ(vm−1 + h)− µ(h)] hx)∥∥∥∥ ≤ √2(1+M) [2+ (N + 1)M] K(M∗, µ) ‖vm−1‖W1(T ) . (4.47)

Next, by µε1 = µ+ ε1µ1, it follows that∥∥∥∥ ∂∂x ([µε1(vm−1 + h)− µε1(h)] hx)∥∥∥∥ ≤ ζ3 ‖vm−1‖W1(T ) , (4.48)

where

ζ3 = ζ3(M,N, T , µ, µ1) =√2(1+ N)M [2+ (N + 1)M]

(K(M∗, µ)+ K(M∗, µ1)

). (4.49)

By (4.49), we obtain

I3(t) = 2∫ t

0

∥∥∥∥ ∂∂x ([µε1(vm−1 + h)− µε1(h)] hx)∥∥∥∥ ∥∥v′m(s)∥∥ ds ≤ Tζ 23 ‖vm−1‖2W1(T ) + ∫ t

0

∥∥v′m(s)∥∥2 ds. (4.50)

Combining (4.39), (4.41), (4.43), (4.50), we then obtain

‖v′m(t)‖2+ µ0‖vmx(t)‖2 ≤ T K 2

∥∥−→ε ∥∥2N+2 + T (ζ 22 + ζ 23 ) ‖vm−1‖2W1(T )+ 3

∫ t

0

∥∥v′m(s)∥∥2 ds+ ζ1 ∫ t

0‖vmx(s)‖2ds. (4.51)


By using Gronwall’s lemma we deduce from (4.51) that

‖vm‖W1(T ) ≤ σT ‖vm−1‖W1(T ) + δ, for allm ≥ 1, (4.52)

where

σT =

√ζ 22 + ζ

23 ηT , δ = ηT K

∥∥−→ε ∥∥N+1 , ηT =

(1+

1√µ0

)exp

[T(32+

ζ1

2µ0

)]√T . (4.53)

We assume that

σT < 1,with the suitable constant T > 0. (4.54)

We shall now require the following lemma whose proof is immediate.

Lemma 4.4. Let the sequence Ψm satisfy

Ψm ≤ σΨm−1 + δ for all m ≥ 1,Ψ0 = 0, (4.55)

where 0 ≤ σ < 1, δ ≥ 0 are the given constants. Then

Ψm ≤ δ/(1− σ) for all m ≥ 1. (4.56)

Applying Lemma 4.4 with Ψm = ‖vm‖W1(T ) , σ = σT =√ζ 22 + ζ

23 ηT < 1, δ = ηT K

∥∥−→ε ∥∥N+1, it follows from (4.56), that∥∥v′m∥∥L∞(0,T ;L2) + ‖vm‖L∞(0,T ;V ) = ‖vm‖W1(T ) ≤ δ/(1− σT ) = CT ∥∥−→ε ∥∥N+1 , (4.57)

where CT =ηT K

1−√ζ 22+ζ

23 ηT.

On the other hand, the linear recurrent sequence vm defined by (4.32) converges strongly in the space W1(T ) to thesolution v of problem (4.20). Hence, lettingm→+∞ in (4.57) gives∥∥v′∥∥L∞(0,T ;L2) + ‖v‖L∞(0,T ;V ) ≤ CT ∥∥−→ε ∥∥N+1 ,or ∥∥∥∥∥u′ − ∑

|γ |≤N

u′γ−→ε

γ

∥∥∥∥∥L∞(0,T ;L2)

+

∥∥∥∥∥u− ∑|γ |≤N

uγ−→εγ

∥∥∥∥∥L∞(0,T ;V )

≤ CT∥∥−→ε ∥∥N+1 . (4.58)

Thus, we have the following theorem.

Theorem 4.5. Let (H1), (H2), (H7), and (H8) hold. Then there exist constants M > 0 and T > 0 such that, for every −→ε , with∥∥−→ε ∥∥ ≤ ε∗, the problem (P−→ε ) has a unique weak solution u = uε1,ε2 ∈ W1(M, T ) satisfying an asymptotic estimation up toorder N + 1 as in (4.58), the functions uγ , |γ | ≤ N being the weak solutions of problems (Pγ ), |γ | ≤ N, respectively.

Remark 4.6. Typical examples about asymptotic expansion of the solution in a small parameter can be found in theresearches of many authors, such as [3,20,5,6,11,10,12]. However, to our knowledge, in the case of asymptotic expansionin many small parameters, there is only partial results, for example, [7–9,17], . . . concerning asymptotic expansion of thesolution in two or three small parameters.

Acknowledgement

The authors wish to express their sincere thanks to the referee for his/her valuable comments and important remarks.

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On the nonlinear wave equation with the mixed nonhomogeneous conditions: Linear approximation and asymptotic expansion of solutions