on a shock problem involving a linear viscoelastic bar

Nonlinear Analysis 63 (2005) 198–224www.elsevier.com/locate/na

On a shock problem involving a linearviscoelastic bar

Nguyen Thanh Longa,∗, Le Van Utb, Nguyen Thi Thao TruccaDepartment of Mathematics and Computer Science, University of Natural Science, Vietnam National University

HoChiMinh City, 227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, VietnambDepartment of Natural Sciences, Can Tho In-serviced University, 256 Nguyen Van Cu Street, Ninh Kieu

District, Can Tho City, VietnamcDepartment of Mathematics, School of Education, Can Tho University, 3-2 Street, Ninh Kieu District,

Can Tho City, Vietnam

Received 7 June 2004; accepted 2 May 2005

Abstract

The paper deals withthe initial-boundary value problem for the linear wave equationutt − �(t)uxx +Ku+ �ut = f (x, t), 0<x <1, 0< t <T,u(0, t)= 0,−�(t)ux(1, t)=Q(t),

u(x,0)= u0(x), ut (x,0)= u1(x),

(1)

whereK, � are given constants andu0, u1, f , � are given functions, the unknown functionu(x, t)and the unknown boundary valueQ(t) satisfy the following linear integral equation

Q(t)=K1(t)u(1, t)+ �1(t)ut (1, t)− g(t)−∫ t

0k(t − s)u(1, s)ds, (2)

whereg, k, K1, �1 are given functions. The paper consists of four parts. In Part 1 we prove a theo-rem of global existence and uniqueness of a weak solution(u,Q) of problem (1)–(2). The proof isbased on a Galerkin type approximation associated to various energy estimates-type bounds, weak-convergence and compactness arguments. In Part 2, it is devoted to the study of the regularity of thesolution(u,Q) with respect to the functions(�, �1, f, g, k,K1). In Part 3 we prove that the solution

∗Corresponding authorE-mail addresses:[email protected](N.T. Long),[email protected](L.V. Ut), [email protected]

(N.T.T. Truc).

0362-546X/$ - see front matter� 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2005.05.007

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N.T. Long et al. / Nonlinear Analysis 63 (2005) 198–224 199

(u,Q) of this problem is stable with respect to the data(K, �, �, �1, f, g, k,K1). Finally, in Part 4 weobtain an asymptotic expansion of the solution(u,Q) of this problem up to orderN + 1 in (K, �),for (K, �) sufficiently small.� 2005 Elsevier Ltd. All rights reserved.

MSC:35L20; 35L70

Keywords:Shock; Integral equations; Energy-type estimates; Compactness; Regularity of solutions; Stability ofthe solutions; Asymptotic expansion

1. Introduction

In this paper,weconsider the followingproblem:Findapair(u,Q)of functions satisfying

utt − �(t)uxx + F(u, ut )= f (x, t), 0<x <1, 0< t <T , (1.1)

u(0, t)= 0, (1.2)

−�(t)ux(1, t)=Q(t), (1.3)

u(x,0)= u0(x), ut (x,0)= u1(x), (1.4)

whereF(u, ut )=Ku+�ut , withK,� are given constants andu0,u1, f,� are given functionssatisfying conditions specified later, and the unknown functionu(x, t) and the unknownboundary valueQ(t) satisfy the following integral equation

Q(t)=K1(t)u(1, t)+ �1(t)ut (1, t)− g(t)−∫ t

0k(t − s)u(1, s)ds, (1.5)

whereg, k,K1, �1 are given functions.In [8] Santos studied the asymptotic behavior of solution of problem (1.1), (1.2), (1.4)

associated with a boundary condition of memory type atx = 1 as follows

u(1, t)+∫ t

0g(t − s)�(s)ux(1, s)ds = 0, t >0. (1.6)

Santos transformed (1.6) into (1.3), (1.5) withK1(t) = g/(0)g(0) , �1(t) = 1

g(0) are positiveconstants.In the case of�1(t) ≡ 0,K1(t)=h�0,�(t)=1, the problem (1.1)–(1.5) is formed from

the problem (1.1)–(1.4) wherein, the unknown functionu(x, t) and the unknown boundaryvalueQ(t) satisfy the following Cauchy problem for ordinary differential equation:{

Q//(t)+ �2Q(t)= hutt (1, t), 0< t <T,Q(0)=Q0, Q/(0)=Q1,

(1.7)

whereh�0,�>0,Q0,Q1 are given constants[6].

200 N.T. Long et al. / Nonlinear Analysis 63 (2005) 198–224

In [1], An and Trieu have studied a special case of problem (1.1)–(1.4), (1.7) withu0 =u1 =Q0 = 0 andF(u, ut )=Ku+ �ut , withK�0, ��0 are given constants. In the latercase the problem (1.1)–(1.4) and (1.7) is a mathematical model describing the shock of arigid body and a linear viscoelastic bar resting on a rigid base[1].From (1.7) we representQ(t) in terms ofQ0,Q1,�, h, utt (1, t) and then by integrating

by parts, we have

Q(t)= hu(1, t)− g(t)−∫ t

0k(t − s)u(1, s)ds, (1.8)

where

g(t)= −(Q0 − hu0(1)) cos�t − 1

�(Q1 − hu1(1)) sin�t , (1.9)

k(t)= h� sin�t . (1.10)

In [2] Bergounioux et al. studied problem (1.1), (1.4) with the mixed boundary conditions(1.2), (1.3) standing for

ux(0, t)= hu(0, t)+ g(t)−∫ t

0k(t − s)u(0, s)ds, (1.11)

ux(1, t)+K1u(1, t)+ �1ut (1, t)= 0, (1.12)

where

g(t)= (Q0 − hu0(0)) cos�t + 1

�(Q1 − hu1(0)) sin�t , (1.13)

k(t)= h� sin�t , (1.14)

whereh�0,�>0,Q0,Q1, K, �,K1, �1 are given constants.The present paper consists of four main parts. In Part 1 we prove a theorem of global

existence and uniqueness of a weak solution(u, Q) of problem (1.1)–(1.5). The proof isbasedonaGalerkin type approximation associated to various energy estimates-type bounds,weak-convergence and compactness arguments. Themain difficulty encountered here is theboundary condition atx=1. In order to solve this particular difficulty, stronger assumptionson the initial conditionsu0 andu1 will be made. We remark that the linearization methodin the papers[3,7] can not be used in[2,5,6]. Part 2 is devoted to the study of the regularityof the solution(u,Q) with respect to the functions(�, �1, f, g, k, K1). In Part 3 we provethat the solution(u,Q) of this problem is stable with respect to the data(K, �, �, �1, f, g,k, K1). Finally, in Part 4 we obtain an asymptotic expansion of the solution(u,Q) of thisproblem up to orderN+1 in (K, �), for (K, �) sufficiently small. The results obtained heremay be considered as generalizations of those in An and Trieu[1] and in Long and Dinh[2,3,5–8].

2. The existence and uniqueness theorem

Put�= (0,1),QT =�× (0, T ), T >0.We omit the definitions of usual function spaces:Cm(�), Lp(�),Wm,p(�).


We denoteWm,p=Wm,p(�),Lp=W0,p(�),Hm=Wm,2(�), 1�p�∞,m=0,1, . . .The norm inL2 is denoted by‖·‖.We also denote by〈·, ·〉 the scalar product inL2 or pair

of dual scalar product of a continuous linear functional with an element of a function space.We denote by‖ ·‖X the norm of a Banach spaceXand byX/ the dual space ofX.We denoteby Lp(0, T ;X), 1�p�∞ for the Banach space of the real functionsu : (0, T ) → X

measurable, such that

‖u‖Lp(0,T ;X) =(∫ T

0‖u(t)‖pX dt

)1/p<∞ for 1�p<∞,

and

‖u‖L∞(0,T ;X) = esssup0<t<T

‖u(t)‖X for p = ∞.

Let u(t), u/(t) = ut (t), u//(t) = utt (t), ux(t), uxx(t) denoteu(x, t),�u�t (x, t),

�2u�t2 (x, t),

�u�x (x, t),

�2u�x2 (x, t), respectively.

We put

V = {v ∈ H 1(0,1) : v(0)= 0}, (2.1)

a(u, v)=∫ 1

0

�u�x

�v�x

dx. (2.2)

V is a closed subspace ofH 1 and onV, ‖v‖H1 and‖v‖V = √a(v, v) = ‖vx‖ are two

equivalent norms.We then have the following lemma.

Lemma 1. The imbeddingV ↪→ C0([0,1]) is compact and

‖v‖C0([0,1])�‖v‖V for all v ∈ V . (2.3)

The proof is straightforward and we omit the details.We make the following assumptions:

(H1) K, � ∈ R,

(H2) u0 ∈ H 2 andu1 ∈ H 1,

(H3) g,K1, �1 ∈ H 1(0, T ), �1(t)��0>0, K1(t)�0,(H4) k ∈ H 1(0, T ),(H5) � ∈ H 2(0, T ), �(t)��0>0,(H6) f, ft ∈ L2(QT ).Then, we have the following theorem.


Theorem 1. Let (H1) − (H6) hold. Then, for everyT >0, there exists a unique weaksolution(u,Q) of problem(1.1)–(1.5)such that{

u ∈ L∞(0, T ;V ∩H 2), ut ∈ L∞(0, T ;H 1), utt ∈ L∞(0, T ;L2),u(1, ·) ∈ H 2(0, T ), Q ∈ H 1(0, T ).

(2.4)

Remark 1. It follows from (2.1), (2.4) that the componentu in the weak solution(u,Q)of problem (1.1)–(1.5) satisfies

u ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2).

Proof of Theorem 1. The proof consists of Steps 1–4.Step1: The Galerkin approximation. Let {wj } be a denumerable base ofV ∩ H 2. We

find the approximate solution of problem (1.1)–(1.5) in the form

um(t)=m∑j=1

cmj (t)wj , (2.5)

where the coefficient functionscmj satisfy the system of ordinary differential equation

〈u//m (t), wj 〉 + �(t)〈umx(t), wjx〉 +Qm(t)wj (1)+ 〈F(um(t), u/m(t)), wj 〉= 〈f (t), wj 〉,1�j�m, (2.6)

Qm(t)=K1(t)um(1, t)+ �1(t)u/m(1, t)−

∫ t

0k(t − s)um(1, s)ds − g(t), (2.7)

um(0)= u0m =

m∑j=1

�mjwj → u0 strongly inH 2,

u/m(0)= u1m =

m∑j=1

�mjwj → u1 strongly inH 1.

(2.8)

From the assumptions of Theorem 1, system (2.6)–(2.8) has solution(um, Qm) on someinterval[0, Tm]. The following estimates allow one to takeTm = T for allm.Step2:A priori estimatesA priori estimatesI. Substituting (2.7) into (2.6), then multiplying thejth equation of

(2.6) byc/mj (t) and summing up with respect toj, we get

S/m(t)= −K

d

dt‖um(t)‖2 − 2�‖u/m(t)‖2 + �/(t)‖umx(t)‖2 +K

/1(t)u

2m(1, t)

+ 2

(∫ t

0k(t − s)um(1, s)ds + g(t)

)u/m(1, t)+ 2〈f (t), u/m(t)〉, (2.9)

where

Sm(t) ≡ ‖u/m(t)‖2 + �(t)‖umx(t)‖2 +K1(t)u2m(1, t)

+ 2∫ t

0�1(s)|u/m(1, s)|2 ds. (2.10)


Integrating by parts with respect to the time variable, we get

Sm(t)= Sm(0)+K‖u0m‖2 − 2g(0)u0m(1)−K‖um(t)‖2 − 2�∫ t

0‖u/m(s)‖2 ds

+∫ t

0�/(s)‖umx(s)‖2 ds +

∫ t

0[K/1(s)− 2k(0)]u2m(1, s)ds

+ 2

(um(1, t)

∫ t

0k(t − s)um(1, s)ds + g(t)um(1, t)

)− 2

∫ t

0um(1, �)d�

(g/(�)+

∫ �

0k/(� − s)um(1, s)ds

)+ 2

∫ t

0〈f (s), u/m(s)〉ds. (2.11)

Using the inequality

2ab��a2 + 1

�b2, ∀a, b ∈ R, ∀�>0, (2.12)

and the following inequalities

Sm(t)�‖u/m(t)‖2 + �0‖umx(t)‖2 + 2�0

∫ t

0

∣∣∣u/m(1, s)∣∣∣2 ds, (2.13)

|um(1, t)|�‖um(t)‖C0(�)�‖umx(t)‖�√Sm(t)

�0, (2.14)

we shall estimate, respectively the following terms on the right-hand side of (2.11)

‖um(t)‖2 =∥∥∥∥um(0)+ ∫ t

0u/m(s)ds

∥∥∥∥2�2‖u0m‖2 + 2t∫ t

0Sm(s)ds, (2.15)

∣∣∣∣∫ t

0�/(s)‖umx(s)‖2 ds

∣∣∣∣ � 1

�0

∫ t

0|�/(s)|Sm(s)ds, (2.16)

∣∣∣∣∫ t

0(K

/1(s)− 2k(0))u2m(1, s)ds

∣∣∣∣ � 1

�0

∫ t

0|K/1(s)− 2k(0)|Sm(s)ds, (2.17)

2

∣∣∣∣um(1, t) ∫ t

0k(t − s)um(1, s)ds

∣∣∣∣ ��Sm(t)+ 1

��20

∫ t

0k2(�)d�

∫ t

0Sm(s)ds,

(2.18)

2|g(t)um(1, t)|��Sm(t)+ 1

��0g2(t), (2.19)

2

∣∣∣∣∫ t

0g/(�)um(1, �)d�

∣∣∣∣ ��∫ t

0Sm(�)d� + 1

��0

∫ t

0|g/(�)|2 d�, (2.20)


2

∣∣∣∣∫ t

0um(1, �)d�

∫ �

0k/(� − s)um(1, s)ds

∣∣∣∣��

∫ t

0Sm(�)d� + 1

��20t

∫ t

0|k/(�)|2 d�

∫ t

0Sm(s)ds, (2.21)

2

∣∣∣∣∫ t

0〈f (s), u/m(s)〉ds

∣∣∣∣ � ∫ t

0‖f (s)‖2 ds +

∫ t

0Sm(s)ds. (2.22)

Combining (2.11), (2.15)–(2.22), we obtain

Sm(t)�Sm(0)+ 3|K|‖u0m‖2 + 2|g(0)u0m(1)|+ 1

��0

(g2(t)+

∫ t

0|g/(s)|2 ds

)+∫ t

0‖f (s)‖2 ds + 1

�0

∫ t

0(|�/(s)| + |K/1(s)− 2k(0)|)Sm(s)ds

+ 1

��20

∫ t

0(k2(�)+ t |k/(�)|2)d�

∫ t

0Sm(s)ds

+ [2� + 1+ 2|�| + 2t |K|]∫ t

0Sm(s)ds + 2�Sm(t). (2.23)

Choosing� = 1/4, it follows from (2.23), that

Sm(t)�2Sm(0)+ 6|K|‖u0m‖2 + 4|g(0)u0m(1)|+ 8

�0

(g2(t)+

∫ t

0|g/(s)|2 ds

)+ 2

∫ t

0‖f (s)‖2 ds

+ 2

�0

∫ t

0(|�/(s)| + |K/1(s)− 2k(0)|)Sm(s)ds

+ 8

�20

∫ t

0(k2(�)+ t |k/(�)|2)d�

∫ t

0Sm(s)ds

+ (3+ 4|�| + 4t |K|)∫ t

0Sm(s)ds. (2.24)

Using assumptions(H1).(H3), (H5), it follows from (2.8), (2.10), that

2Sm(0)+ 6|K|‖u0m‖2 + 4|g(0)u0m(1)|�C1 for all m, (2.25)

whereC1 is a constant depending only on�(0),K1(0), g(0), K, u0, u1.On the other hand, we have

C1 + 8

�0

(g2(t)+

∫ t

0|g/(s)|2 ds

)+ 2

∫ t

0‖f (s)‖2 ds�M(1)

T

for all t ∈ [0, T ], (2.26)

(3+ 4|�| + 4t |K|)+ 8

�20

∫ t

0(k2(�)+ t |k/(�)|2)d��M(2)

T

for all t ∈ [0, T ], (2.27)


whereM(i)T , i = 1,2, is a constant depending only onT. Therefore, it follows from

(2.24)–(2.27), that

Sm(t)�M(1)T +

∫ t

0N(1)T (s)Sm(s)ds, (2.28)

where

N(1)T (s)= 2

�0(|�/(s)| + |K/1(s)− 2k(0)|)+M

(2)T , N

(1)T ∈ L1(0, T ). (2.29)

By Gronwall’s lemma, we obtain

Sm(t)�M(1)T exp

(∫ t

0N(1)T (s)ds

)�MT , for all t ∈ [0, T ]. (2.30)

A priori estimatesII. Now differentiating (2.6) with respect tot we have

〈u///m (t), wj 〉 + �(t)〈u/mx(t), wjx〉 + �/(t)〈umx(t), wjx〉+Q

/m(t)wj (1)+ 〈Ku/m + �u//m (t), wj 〉 = 〈f /(t), wj 〉, (2.31)

for all 1�j�m.Multiplying the jth equation of (2.31) byc//mj (t), summing up with respect toj and then

integrating with respect to the time variable from 0 tot, we have after some rearrangements

Xm(t)=Xm(0)+ 2�/(0)〈u0mx, u1mx〉 +K‖u1m‖2

− 2�/(t)〈umx(t), u/mx(t)〉 −K‖u/m(t)‖2 − 2�∫ t

0‖u//m (s)‖2 ds

− 2∫ t

0[(K1(s)+ �/1(s))u

/m(1, s)+ (K

/1(s)− k(0))um(1, s)]u//m (1, s)ds

+ 2∫ t

0

(∫ s

0k/(s − �)um(1, �)d� + g/(s)

)u//m (1, s)ds

+ 3∫ t

0�/(s)‖u/mx(s)‖2 ds + 2

∫ t

0�//(s)〈umx(s), u/mx(s)〉ds

+ 2∫ t

0〈f /(s), u//m (s)〉ds, (2.32)

where

Xm(t)= ‖u//m (t)‖2 + �(t)‖u/mx(t)‖2 + 2∫ t

0�1(s)|u//m (1, s)|2 ds. (2.33)

Using inequality (2.12) and the following inequalities:

Xm(t)�‖u//m (t)‖2 + �0‖u/mx(t)‖2 + 2�0

∫ t

0|u//m (1, s)|2 ds, (2.34)

|um(1, t)|�‖um(t)‖C0(�)�‖umx(t)‖�√Sm(t)

�0�√MT

�0, (2.35)


|u/m(1, t)|�‖u/m(t)‖C0(�)�‖u/mx(t)‖�√Xm(t)

�0, (2.36)

we estimate without difficulty the following terms in the right-hand side of (2.32) as follows

| − 2�/(t)〈umx(t), u/mx(t)〉|��Xm(t)+ 1

��20MT |�/(t)|2, (2.37)

‖u/m(t)‖2�2‖u1m‖2 + 2t∫ t

0Xm(s)ds, (2.38)

∣∣∣∣−2∫ t

0

(K1(s)+ �/1(s)

)u/m(1, s)u

//m (1, s)ds

∣∣∣∣� 1

��0

∫ t

0|K1(s)+ �/1(s)|2Xm(s)ds + �

1

2�0Xm(t), (2.39)

∣∣∣∣−2∫ t

0(K

/1(s)− k(0))um(1, s)u

//m (1, s)ds

∣∣∣∣�MT

��0

∫ t

0|K/1(s)− k(0)|2 ds + �

1

2�0Xm(t), (2.40)

2

∣∣∣∣∫ t

0u//m (1, s)ds

∫ s

0k/(s − �)um(1, �)d�

∣∣∣∣��

1

2�0Xm(t)+ MT

��0t

(∫ t

0|k/(�)|d�

)2, (2.41)

2

∣∣∣∣∫ t

0g/(s)u

//m (1, s)ds

∣∣∣∣ ��1

2�0Xm(t)+ 1

�

∫ t

0|g/(s)|2 ds, (2.42)

3

∣∣∣∣∫ t

0�/(s)‖u/mx(s)‖2 ds

∣∣∣∣ � 3

�0

∫ t

0|�/(s)|Xm(s)ds, (2.43)

2

∣∣∣∣∫ t

0�//(s)〈umx(s), u/mx(s)〉ds

∣∣∣∣ � 1

�20MT

∫ t

0|�//(s)|2 ds +

∫ t

0Xm(s)ds, (2.44)

2

∣∣∣∣∫ t

0〈f /(s), u//m (s)〉ds

∣∣∣∣ ��∫ t

0Xm(s)ds + 1

�

∫ t

0‖f /(s)‖2 ds. (2.45)


It follows from (2.32), (2.37)–(2.45) that

Xm(t)�Xm(0)+ 2|�/(0)|‖u0m‖V ‖u1m‖V + 3|K|‖u1m‖2

+ 1

��20MT |�/(t)|2 + MT

��0

∫ t

0|K/1(s)− k(0)|2 ds

+ MT

��0t

(∫ t

0|k/(�)|d�

)2+ 1

�20MT

∫ t

0|�//(s)|2 ds

+ 1

�

∫ t

0‖f /(s)‖2 ds + 1

�

∫ t

0|g/(s)|2 ds

+∫ t

0

[2T |K| + 2|�| + 1

��0|K1(s)+ �/1(s)|2 + 3

�0|�/(s)| + � + 1

]Xm(s)ds + �

(1+ 2

�0

)Xm(t). (2.46)

Choosing�>0 with �(1+ 2�0)<1/2, hence, we obtain from (2.46), that

Xm(t)�2Xm(0)+ 4|�/(0)|‖u0m‖V ‖u1m‖V + 6|K|‖u1m‖2 +N(2)T (t)

+∫ t

0N(1)T (s)Xm(s)ds, (2.47)

where

N(2)T (t)= 2

��20MT |�/(t)|2 + 2MT

��0

∫ t

0|K/1(s)− k(0)|2 ds

+ 2MT

��0t

(∫ t

0|k/(�)|d�

)2+ 2

�20MT

∫ t

0|�//(s)|2 ds

+ 2

�

∫ t

0‖f /(s)‖2 ds + 2

�

∫ t

0|g/(s)|2 ds, (2.48)

N(1)T (s)= 4T |K| + 4|�| + 2

��0|K1(s)+ �/1(s)|2 + 6

�0|�/(s)| + 2� + 2,

N(1)T ∈ L1(0, T ). (2.49)

First, we deduce from (2.8) and(H1).(H2), (H5), that

2Xm(0)+ 4|�/(0)|‖u0m‖V ‖u1m‖V + 6|K|‖u1m‖2�2‖u//m (0)‖2 + C2, for all m, (2.50)

whereC2 is a constant depending only on�, u0, u1, K.But by (2.6)–(2.8), we have

‖u//m (0)‖2 − �(0)〈u0mxx, u//m (0)〉 + 〈Ku0m + �u1m, u//m (0)〉 = 〈f (0), u//m (0)〉.

(2.51)


Therefore,

‖u//m (0)‖��(0)‖�u0m‖ + |K|‖u0m‖ + |�|‖u1m‖ + ‖f (0)‖. (2.52)

It follows from (2.8), (2.52), that

‖u//m (0)‖�C3, (2.53)

whereC3 is a constant depending only on�, u0, u1, f, K, �.Noting thatH 1(0, T ) ↪→ C0([0, T ]), it follows from (2.48), that

2C23 + C2 +N

(2)T (t)�M(1)

T , a.e., t ∈ [0, T ], (2.54)

whereM(1)T is a constant depending only onT. Therefore, it follows from (2.47), (2.50),

(2.53), (2.54), that

Xm(t)�M(1)T +

∫ t

0N(1)T (s)Xm(s)ds. (2.55)

By Gronwall’s lemma, we deduce from (2.55), that

Xm(t)�M(1)T exp

(∫ t

0N(1)T (s)ds

)�MT ∀t ∈ [0, T ]. (2.56)

On the other hand, we deduce from (2.7), (2.30), (2.34)–(2.36) and (2.56), that

‖Q/m‖2L2(0,T )�

5

2�0MT ‖�1‖2∞ + 5MT

�0

∫ T

0|K1(s)+ �/1(s)|2 ds

+ 5MT

�0

∫ T

0|K/1(s)− k(0)|2 ds + 5TMT

�0

(∫ T

0|k/(�)|d�

)2+ 5

∫ T

0|g/(s)|2 ds, (2.57)

where‖�1‖∞ = ‖�1‖L∞(0,T ).Hence

‖Qm‖H1(0,T )�MT . (2.58)

Step3: Limiting process. From (2.13), (2.30), (2.34), (2.56) and (2.58), we deduce theexistence of a subsequence of{(um,Qm)} still also so denoted, such that

um → u in L∞(0, T ;V ) weak∗,u/m → u/ in L∞(0, T ;V ) weak∗,u//m → u// in L∞(0, T ;L2) weak∗,um(1, ·) → u(1, ·) in H 2(0, T ) weakly,Qm → Q in H 1(0, T ) weakly.

(2.59)


By the compactness lemma of Lions[4, p. 57]we can deduce from (2.59) the existence ofa subsequence still denoted by{(um, Qm)} such that

um → u strongly inL2(QT ),u/m → u/ strongly inL2(QT ),um(1, ·) → u(1, ·) strongly inH 1(0, T ),u/m(1, ·) → u/(1, ·) strongly inC0[0, T ],Qm → Q strongly inC0[0, T ].

(2.60)

From (2.7) and(2.60)3−4 we have

Qm(t) → K1(t)u(1, t)+ �1(t)u/(1, t)

−∫ t

0k(t − s) u(1, s)ds − g(t) ≡ Q(t) (2.61)

strongly inC0[0, T ].Hence, we deduce from(2.60)5 and (2.61), that

Q(t)= Q(t). (2.62)

Passing to the limit in (2.6)–(2.8) by (2.59)–(2.60) and (2.61)–(2.62) we have(u,Q) satis-fying the problem

〈u//(t), v〉 + �(t)〈ux(t), vx〉 +Q(t)v(1)+ 〈Ku(t)+ �u/(t), v〉= 〈f (t), v〉, ∀v ∈ H 1, (2.63)

u(0)= u0, u/(0)= u1, (2.64)

Q(t)=K1(t)u(1, t)+ �1(t)ut (1, t)−∫ t

0k(t − s)u(1, s)ds − g(t), (2.65)

in L2(0, T ) weakly. On the other hand, we have from (2.63) and assumptions(H5).(H6)

that

uxx = 1

�(t)[u// +Ku(t)+ �u/(t)− f ] ∈ L∞(0, T ;L2). (2.66)

Henceu ∈ L∞(0, T ;V ∩H 2) and the existence proof is completed.Step4:Uniquenessof thesolution. Letu1,u2 be twoweaksolutionsofproblem(1.1)–(1.5)

such thatui ∈ L∞(0, T ;V ∩H 2),

u/i ∈ L∞(0, T ;H 1), u

//i ∈ L∞(0, T ;L2),

ui(1, ·) ∈ H 2(0, T ), Qi ∈ H 1(0, T ), i = 1,2.(2.67)

Then(u,Q) with u= u1 − u2 andQ=Q1 −Q2 satisfy the variational problem{ 〈u//(t), v〉+�(t)〈ux(t), vx〉+Q(t)v(1)+ 〈Ku+ �u/, v〉=0 ∀v ∈ H 1,

u(0)= u/(0)= 0,(2.68)


where

Q(t)=K1(t)u(1, t)+ �1(t)u/(1, t)−∫ t

0k(t − s)u(1, s)ds. (2.69)

We takev = u/ in (2.68)1, afterwards integrating int, we get

S(t)=∫ t

0�/(s)‖ux(s)‖2 ds +

∫ t

0K/1(s)u

2(1, s)ds

+ 2∫ t

0u/(1, s)ds

∫ s

0k(s − �)u(1, �)d� −K‖u(t)‖2

− 2�∫ t

0‖u/(s)‖2 ds, (2.70)

where

S(t)=‖u/(t)‖2+�(t)‖ux(t)‖2+K1(t)u2(1, t)+2

∫ t

0�1(s)|u/(1, s)|2 ds. (2.71)

Noting that

S(t)�‖u/(t)‖2 + �0‖ux(t)‖2 + 2�0

∫ t

0|u/(1, s)|2 ds, (2.72)

|u(1, t)|�‖u(t)‖C0(�)�‖ux(t)‖�

√S(t)

�(t)�√S(t)

�0, (2.73)

‖u(t)‖2�(∫ t

0‖u/(s)‖ds

)2� t

∫ t

0‖u/(s)‖2 ds. (2.74)

We again use inequality (2.12), then, it follows from (2.70)–(2.74), that

S(t)� 1

�0

∫ t

0[|�/(s)| + |K/1(s)|]S(s)ds + (|K|t + 2|�|)

∫ t

0S(s)ds

+ �1

2�0S(t)+ t

��0

∫ t

0k2(�)d�

∫ t

0S(�)d�. (2.75)

Choosing�>0, such that� 12�0

�1/2, we obtain from (2.75), that

S(t)�∫ t

0q1(s)S(s)ds, (2.76)

where{q1(s)= 2

�0(|�/(s)| + |K/1(s)|)+ 2|K|T + 4|�| + 2T

��0

∫ T0 k

2(�)d�,

q1 ∈ L1(0, T ).(2.77)

By Gronwall’s lemma, we deduce thatS ≡ 0 and Theorem 1 is completely proved.�


Remark 2. In the case ofK=K(t),�=�(t), k=k(t, s), we consider the following problemutt − �(t)uxx +K(t)u+ �(t)ut = f (x, t), 0<x <1, 0< t <T , (1.1′)

u(0, t)= 0, (1.2)

−�(t)ux(1, t)=Q(t), (1.3)

u(x,0)= u0(x), ut (x,0)= u1(x), (1.4)

whereK,�,u0,u1, f,� are given functions and the unknown functions(u(x, t),Q(t)) satisfythe following integral equation:

Q(t)=K1(t)u(1, t)+ �1(t)ut (1, t)−∫ t

0k(t, s)u(1, s)ds − g(t), (1.5′)

whereg, k,K1 and�1 are given functions. Now, we assume that

(H1) K ∈ H 1(0, T ), � ∈ H 2(0, T ),

(H4) k,D1k = �k�t

∈ L2(Q∗T ), k∗ ∈ L2(0, T ),

wherek∗(t)= k(t, t) andQ∗T = {(t, s) : 0�s� t�T }.

By an argument analogous to that used to the proof of Theorem 1, we get the followingresult.

Theorem 2. LetT >0. Let (Hi), 2� i�6, i �= 4, (H1) and(H4) hold. Then, there existsa unique weak solution(u,Q) of problem(1.1/), (1.2)–(1.4),(1.5/) such that{

u ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

ut ∈ L∞(0, T ;H 1), utt ∈ L∞(0, T ;L2),u(1, ·) ∈ H 2(0, T ), Q ∈ H 1(0, T ).

(2.78)

3. Regularity of solutions

In this part, we study the regularity of solution of problem (1.1)–(1.5) corresponding tothe data(�, �1,K, �, u0, u1, f, g, k,K1). From here, we assume that(�, �1,K, �, u0, u1, f, g,k,K1) satisfy assumptions(H1).(H6). Henceforth, we shall impose the following strongerassumptions:

(H(1)1 ) K, � ∈ R,

(H(1)2 ) u0 ∈ H 3 andu1 ∈ H 2,

(H(1)3 ) g,K1, �1 ∈ H 2(0, T ), �1(t)��0>0,K1(t)�0,

(H(1)4 ) k ∈ H 2(0, T ),


(H(1)5 ) � ∈ H 3(0, T ), �(t)��0>0,�(t)�/1(t)− �/(t)�1(t)�0,

(H(1)6 ) f, ft , ftt ∈ L2(0, T ;L2) andf (·,0) ∈ H 1.

Formally differentiating problem (1.1) with respect to time and lettingu=ut andQ(t) ≡�(t)(Q(t)�(t) )

/ we are led to consider the solution(u, Q) of problem(P (1)) as follows:

(P (1))

Lu ≡ ut t − �(t )uxx + K(t )u+ �(t )ut = f (x, t), 0<x <1, 0< t <T,u(0, t)= 0,Bu ≡ −�(t )ux(1, t)= Q(t),

u(x,0)= u0(x), ut (x,0)= u1(x),

Q(t)= K1(t )u(1, t)+ �1(t )ut (1, t)−∫ t0 k(t, s) u(1, s)ds − g(t),

whereK(t)=K − �

�/(t)�(t)

, �(t)= � − �/(t)�(t)

,

f (x, t)= �(t)��t

(f

�(t)

)+K

�/(t)�(t)

u,

(3.1)

{u0(x)= u1(x),

u1(x)= �(0)u0xx(x)−Ku0(x)− �u1(x)+ f (x,0),(3.2)

K1(t)=K1(t)+ �(t)(

�1�

)/,

k(t, s)= −�(t)(K1(t)

�(t)

)/− �/(t)

�(t)k1(t − s)+ k(t − s),

with k1(t)=∫ t0 k(�)d�,

g(t)=(

−�(K1

�

)/− �/

�k1(t)+ k(t)

)u0(1)+ �

(g

�

)/.

(3.3)

Let (�, �1,K, �, u0, u1, f, g, k,K1) satisfy assumptions(H(1)1 ).(H (1)

6 ). Then(�, �1, K, �,u0, u1, f , g, k, K1) satisfy assumptions(Hi), 2� i�6, i �= 4, (H1) and(H4). By Theorem2, the problem(P (1)) there exists a unique weak solution(u, Q) such that{

u ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

ut ∈ L∞(0, T ;H 1), utt ∈ L∞(0, T ;L2),u(1, ·) ∈ H 2(0, T ), Q ∈ H 1(0, T ).

(3.4)

Moreover, from the uniqueness of weak solution we have

u= ut , Q(t)= �(t)(Q(t)

�(t)

)/. (3.5)

From the equalitiesQ(t)= �(t)

(Q(t)

�(t)

)/= Q/(t)�(t)−Q(t)�/(t)

�(t),

Q/ = Q+ �/

�Q, Q// = Q/ + �/

�Q/ + �//� − (�/)2

�2Q,

(3.6)


it follows that

Q/,Q// ∈ L2(0, T ), i.e., Q ∈ H 2(0, T ). (3.7)

Then, we deduce from (3.4)–(3.5), (3.7), that{u ∈ C0(0, T ;V ∩H 2) ∩ C1(0, T ;V ∩H 1) ∩ C2(0, T ;L2),ut ∈ L∞(0, T ;V ∩H 2), utt ∈ L∞(0, T ;H 1), uttt ∈ L∞(0, T ;L2),u(1, ·) ∈ H 3(0, T ), Q ∈ H 2(0, T ).

(3.8)

We then have the following theorem.

Theorem 3. Let(�, �1,K, �, u0, u1, f, g, k,K1) satisfy assumptions(H(1)1 ).(H (1)

6 ).Thenthere exists a unique weak solution(u,Q) of problem(1.1)–(1.5)satisfying(3.8).

Similarly, formally differentiating problem (1.1)–(1.5) with respect to time up to order

r, (r�1) and lettingu[r] = �ru�t r andQ

[r](t) = �(t) ddt (Q[r−1](t)

�(t) ) we are led to consider the

solution(u[r],Q[r]) of problem(P (r)):

(P (r))

L[r]u[r] ≡ u[r]t t − �(t)u[r]

xx +K [r](t)u[r] + �[r](t)u[r]t = f [r](x, t),

0<x <1, 0< t <T,u[r](0, t)= 0,Bu[r] ≡ −�(t)u[r]

x (1, t)=Q[r](t),u[r](x,0)= u

[r]0 (x), u

[r]t (x,0)= u

[r]1 (x),

Q[r](t)=K[r]1 (t)u

[r](1, t)+ �1(t)u[r]t (1, t)

− ∫ t0 k[r](t, s) u[r](1, s)ds − g[r](t),

where the functionsK [r], �[r], u[r]0 , u

[r]1 , f

[r], g[r], k[r],K [r]1 , r�1, are defined by the re-

currence formulas

u[0]0 = u0, u

[r]0 = u

[r−1]1 , r�1, (3.9)

u[0]1 = u1,

u[r]1 =

r−1∑j=0

Cjr−1�

(r−1−j)(0)u[j ]0xx − F(u

[r−1]0 , u

[r−1]1 )+ �r−1f

�t r−1 (·,0), r�1,

(3.10)

whereCjr = r!j !(r−j)! ,�[r](t)= �[r−1](t)− �/(t)

�(t), r�1,

�[0](t)= �,(3.11)

K[r](t)=K [r−1](t)+ �(t)

(�[r−1](t)

�(t)

)/, r�1,

K [0](t)=K,

(3.12)


f [r] = �(t)[

��t

(f [r−1]

�(t)

)− u[r−1] �

�t

(K [r−1](t)

�(t)

)], r�1,

f [0] = f,

(3.13)

K [r]1 (t)=K

[r−1]1 (t)+ �(t)

(�1(t)�(t)

)/, r�1,

K[0]1 (t)=K1(t),

(3.14)

k[r](t, s)= k[r−1](t, t)+ k[r−1]2 (t, t)− k

[r−1]2 (t, s)

−�/(t)�(t)

(k[r−1]1 (t, t)− k

[r−1]1 (t, s))− �(t)

(K

[r−1]1 (t)

�(t)

)/, r�2,

k[r−1]1 (t, s)= ∫ s

0 k[r−1](t, �)d�, k

[r−1]2 (t, s)= ∫ s

0�k[r−1]

�t(t, �)d�, r�2,

k[1](t, s)= k(t, s)= −�(t)(K1(t)

�(t)

)/− �/(t)

�(t)

∫ t−s0 k(�)d� + k(t − s),

(3.15)

g[r](t)=k[r−1](t, t)+ k

[r−1]2 (t, t)− �/(t)

�(t)k[r−1]1 (t, t)− �(t)

(K

[r−1]1 (t)

�(t)

)/×u[r−1]

0 (1)+ �(t)(g[r−1](t)

�(t)

)/, r�2,

g[1](t)= g(t)=(

−�(t)(K1(t)

�(t)

)/− �/(t)

�(t)

∫ t0 k(�)d� + k(t)

)u0(1)

+�(t)(g(t)

�(t)

)/.

(3.16)

Assume that the data(�, �1,K, �, u0, u1, f, g, k,K1) satisfy the following assumptions:

(H(r)1 ) K, � ∈ R,

(H(r)2 ) u0 ∈ Hr+2 andu1 ∈ Hr+1, r�1,

(H(r)3 ) g,K1, �1 ∈ Hr+1(0, T ), �1(t)��0>0, K1(t)�0, r�1,

(H(r)4 ) k ∈ Hr+1(0, T ), r�1,

(H(r)5 ) � ∈ Hr+2(0, T ), �(t)��0>0,�(t)�/1(t)− �/(t)�1(t)�0, r�1,

(H(r)6 )

�f

�t ∈ L2(0, T ;L2), (0��r + 1) and ��f

�t� (·,0) ∈ H 1, 0��r − 1.

Then(�, �1,K [r], �[r], u[r]0 , u

[r]1 , f

[r], g[r], k[r],K [r]1 ) satisfy assumptions(Hi), 2� i�6,

i �= 4, (H1) and(H4). Applying again Theorem 2 for problem(P (r)), there exists a uniqueweak solution(u[r],Q[r]) satisfying (2.4) and the inclusion from Remark 1, i.e., such that{

u[r] ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

u[r]t ∈ L∞(0, T ;H 1), u

[r]t t ∈ L∞(0, T ;L2),

u[r](1, ·) ∈ H 2(0, T ), Q[r] ∈ H 1(0, T ).(3.17)


Moreover, from the uniqueness of weak solution we have

(u[r],Q[r])=(

�ru�t r

,�d

dt

(Q[r−1]

�

)). (3.18)

Hence we obtain from (3.17), (3.18) that{u ∈ Cr−1(0, T ;V ∩H 2) ∩ Cr(0, T ;H 1) ∩ Cr+1(0, T ;L2),u(1, ·) ∈ Hr+2(0, T ), Q ∈ Hr+1(0, T ).

(3.19)

We then have the following theorem.

Theorem 4. Let (H (r)1 ).(H (r)

6 ) hold. Then there exists a unique weak solution(u,Q) ofproblem(1.1)–(1.5)satisfying(3.19)and

�ru�t r

∈ L∞(0, T ;V ∩H 2),�r+1u

�t r+1 ∈ L∞(0, T ;H 1),

�r+2u

�t r+2 ∈ L∞(0, T ;L2). (3.20)

4. Stability of the solutions

In this part, we assume that the functionsu0, u1 satisfy(H2). By Theorem 1, problem(1.1)–(1.5) has a unique weak solution(u,Q) depending onK, �, �, �1, f, g, k,K1.

u= u(K, �,�, �1, f, g, k,K1), Q=Q(K, �,�, �1, f, g, k,K1), (4.1)

where(K, �,�, �1, f, g, k,K1) satisfy the assumptions(H1), (H3).(H6) andu0, u1 arefixed functions satisfying(H2).

We put

I(�0,�0)= {(K, �,�, �1, f, g, k,K1) : (K, �,�, �1, f, g, k,K1)

satisfy the assumptions(H1), (H3).(H6)},where�0>0,�0>0 are given constants.Then we have the following theorem.

Theorem 5. LetT >0. Let (H1).(H6) hold. Then, for everyT >0, solutions of the prob-lems(1.1)–(1.5)are stable with respect to the data(K, �,�, �1, f, g, k,K1), i.e.,If (K, �,�, �1, f, g, k,K1), (Kj , �

j ,�j , �j1, fj , gj , kj ,K

j1 ) ∈ I(�0,�0), such that

|Kj −K| + |�j − �| → 0, ‖�j − �‖H2(0,T ) → 0, ‖�j1 − �1‖H1(0,T ) → 0,

‖f j − f ‖L2(0,T ;L2) + ‖f jt − ft‖L2(0,T ;L2) → 0, ‖gj − g‖H1(0,T ) → 0,

‖kj − k‖H1(0,T ) → 0, ‖Kj1 −K1‖H1(0,T ) → 0,

(4.2)

asj → +∞, then

(uj , u/j , uj (1, ·),Qj ) → (u, u/, u(1, ·),Q) (4.3)


in L∞(0, T ;V ) × L∞(0, T ;L2) × H 1(0, T ) × L2(0, T ) strongly, as j → +∞, whereuj = u(Kj , �j ,�j , �j1, f

j , gj , kj ,Kj1 ),Qj =Q(Kj , �j ,�j , �j1, f

j , gj , kj ,Kj1 ).

Proof. First, we note that, if the data(K, �,�, �1, f, g, k,K1) satisfy|K|�K∗, |�|��∗,‖f ‖L2(0,T ;L2) + ‖ft‖L2(0,T ;L2)�f ∗, ‖g‖H1(0,T )�g∗,‖k‖H1(0,T )�k∗, ‖�1‖H1(0,T )��∗

1,

‖�‖H2(0,T )��∗, ‖K1‖H1(0,T )�K∗1 ,

(4.4)

whereK∗, �∗, g∗, k∗, �∗1,�

∗, f ∗,K∗1, are fixed positive constants, then, the a priori esti-

mates of the sequences{um} and{Qm} in the proof of Theorem 1 satisfy

‖u/m(t)‖2 + �0‖umx(t)‖2 + 2�0

∫ t

0|u/m(1, s)|2 ds�MT , ∀t ∈ [0, T ], (4.5)

‖u//m (t)‖2 + �0‖u/mx(t)‖2 + 2�0

∫ t

0|u//m (1, s)|2 ds�MT , ∀t ∈ [0, T ], (4.6)

‖Qm‖H1(0,T )�MT , (4.7)

whereMT is a constant depending only onT, u0, u1, �0, �0,K∗, �∗, f ∗, g∗, �∗, �∗

1, k∗,K∗

1(independent ofK, �, �, �1, f, g, k,K1).Hence, the limit(u,Q) in suitable function spaces of the sequence{(um,Qm)} defined

by (2.6)–(2.8) is a weak solution of problem (1.1)–(1.5) satisfying the a priori estimates(4.5)–(4.7).Now, by (4.2) we can assume that, there exist positive constantsK∗, �∗, f ∗, g∗, �∗, �∗

1,

k∗,K∗1 such that the data(K

j , �j , �j , �j1, fj , gj , kj ,Kj1 ) satisfy (4.4) with (K, �, �, �1,

f, g, k,K1)=(Kj , �j , �j , �j1, f j , gj , kj ,Kj1 ). Then, by the above remark, we have that thesolutions(uj ,Qj) of problem (1.1)–(1.5) corresponding to(K, �, �, �1, f, g, k,K1)=(Kj ,�j , �j , �j1, f

j ,gj , kj ,Kj1 ) satisfy

‖u/j (t)‖2 + �0‖ujx(t)‖2 + 2�0

∫ t

0|u/j (1, s)|2 ds�MT , ∀t ∈ [0, T ], (4.8)

‖u//j (t)‖2 + �0‖u/jx(t)‖2 + 2�0

∫ t

0|u//j (1, s)|2 ds�MT , ∀t ∈ [0, T ], (4.9)

‖Qj‖H1(0,T )�MT , (4.10)

Put Kj =Kj −K, �j = �j − �,�j = �j − �, �1j = �j1 − �1,fj = f j − f, gj = gj − g,

kj = kj − k, K1j =Kj1 −K1.

(4.11)


Then,vj = uj − u andPj =Qj −Q satisfy the variational problem〈v//j (t), v〉 + �(t)〈vjx(t), vx〉 + Pj (t)v(1)+ 〈Kvj + �v/j , v〉

=〈fj , v〉 − �j (t)〈ujx(t), vx〉 − 〈Kjuj + �j u/j , v〉, ∀v ∈ H 1,

vj (0)= v/j (0)= 0,

(4.12)

wherePj (t)=Qj(t)−Q(t)

=K1(t)vj (1, t)+ �1(t)vjt (1, t)−∫ t

0k(t − s)vj (1, s)ds − gj (t), (4.13)

gj (t)=gj (t)−K1j (t)uj (1, t)−�1j (t)ujt (1, t)+∫ t

0kj (t − s)uj (1, s)ds. (4.14)

SubstitutingPj (t) into (4.12), then takingv= v/j in (4.12)1, afterwards integrating int, weobtain

Sj (t)= −K‖vj (t)‖2 − 2�∫ t

0‖v/j (s)‖2 ds +

∫ t

0K/1(s)|vj (1, s)|2 ds

+ 2∫ t

0

(∫ r

0k(r − s)vj (1, s)ds + gj (r)

)v/j (1, r)dr

+∫ t

0�/(s)‖vjx(s)‖2 ds + 2

∫ t

0〈fj , v/j (s)〉ds

− 2∫ t

0�j (s)〈ujx(s), v/jx(s)〉ds − 2

∫ t

0〈Kjuj + �j u

/j , v

/j (s)〉ds, (4.15)

where

Sj (t)= ‖v/j (t)‖2 + �(t)‖vjx(t)‖2 +K1(t)|vj (1, t)|2

+ 2∫ t

0�1(s)|v/j (1, s)|2 ds. (4.16)

Using inequalities (2.12), (4.8), (4.9) and

Sj (t)�‖v/j (t)‖2 + �0‖vjx(t)‖2 + 2�0

∫ t

0|v/j (1, s)|2 ds, (4.17)

then, we can prove the following inequality in a similar manner:

Sj (t)��1

2�0Sj (t)+

[1+ (|K|T + 2|�|)+ 2

��0T

∫ T

0k2(�)d�

] ∫ t

0Sj (s)ds

+ 1

�0

∫ t

0(|K/1(s)| + |�/(s)|)Sj (s)ds

+ 2

�

∫ t

0|gj (r)|2 dr +

∫ T

0‖fj (s)‖2 ds + 4T

MT

�0‖�j‖∞

+ 4TMT

�0(|Kj | + √

�0|�j |), (4.18)

for all �>0 andt ∈ [0, T ].


Choosing�>0 such that� 12�0

�1/2 and denote

Rj = 4

�‖gj‖2L2(0,T ) + 2

∫ T

0‖fj (s)‖2 ds + 8T

MT

�0‖�j‖∞

+ 8TMT

�0(|Kj | + √

�0|�j |), (4.19)

(t)= 2+ 2(|K|T + 2|�|)+ 4

��0T

∫ T

0k2(�)d� + 2

�0(|K/1(t)| + |�/(t)|). (4.20)

Then from (4.18)–(4.20) we have

Sj (t)�Rj +∫ t

0(s)Sj (s)ds. (4.21)

By Gronwall’s lemma, we obtain from (4.21) that

Sj (t)�Rj exp(∫ t

0(s)ds

)�C∗

T Rj , ∀t ∈ [0, T ]. (4.22)

On the other hand, using the embeddingH 1(0, T ) ↪→ C0([0, T ]), it follows from (4.13),(4.14), (4.17), (4.19) and (4.22) that

‖Pj‖L2(0,T )�(‖K1‖∞√

�0+ ‖�1‖∞√

2�0

)√C∗T Rj

+(√

T

�0T

∫ T

0k2(�)d�

)C∗T Rj + ‖gj‖L2(0,T ), (4.23)

Rj �4

�‖gj‖2L2(0,T ) + 2

∫ T

0‖fj (s)‖2 ds + 8T

MT

�0‖�j‖H1(0,T )

+ 8TMT

�0(|Kj | + √

�0|�j |), (4.24)

‖gj‖L2(0,T )�‖gj‖H1(0,T ) +√MT

�0‖K1j‖H1(0,T ) +

√MT

2�0‖�1j‖H1(0,T )

+ √T

√MT

�0‖kj‖H1(0,T ). (4.25)

Finally, by (4.2), (4.11) and the estimates (4.22)–(4.25), we deduce that (4.3) holds.Theorem 5 is completely proved.�

5. Asymptotic expansion of the solution

In thispart,weassume that(u0, u1, f,�, g, k,K1, �1)satisfy theassumptions(H2).(H6).


Weconsider the following perturbed problem,whereK,� are small parameters,|K|�K∗,|�|��∗:

(PK,�)

Au ≡ utt − �(t)uxx = −Ku− �ut + f (x, t), 0<x <1, 0< t <T,u(0, t)= 0,Bu ≡ −�(t)ux(1, t)=Q(t),

u(x,0)= u0(x), ut (x,0)= u1(x),

Q(t)=K1(t)u(1, t)+ �1(t)ut (1, t)− g(t)− ∫ t0 k(t − s)u(1, s)ds.

Let (u0,0,Q0,0) be a unique weak solution of problem(P0,0) (as in Theorem 1) correspond-ing to (K, �)= (0,0), i.e.,

(P0,0)

Au0,0 = P0,0 ≡ f (x, t), 0<x <1, 0< t <T,u0,0(0, t)= 0, Bu0,0 =Q0,0(t),

u0,0(x,0)= u0(x), u/0,0(x,0)= u1(x),

Q0,0(t)=K1(t)u0,0(1, t)+ �1(t)u/0,0(1, t)− g(t)

− ∫ t0 k(t − s)u0,0(1, s)ds,u0,0 ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

u/0,0 ∈ L∞(0, T ;H 1), u

//0,0 ∈ L∞(0, T ;L2),

u0,0(1, ·) ∈ H 2(0, T ), Q0,0 ∈ H 1(0, T ).

Let us consider the sequence of weak solutions(u�1,�2,Q�1,�2), (�1, �2) ∈ Z2+, 1��1 +�2�N , defined by the following problems:

(P�1,�2)

Au�1,�2 = P�1,�2, 0<x <1, 0< t <T,u�1,�2(0, t)= 0, Bu�1,�2 =Q�1,�2(t),

u�1,�2(x,0)= 0, u/�1,�2(x,0)= 0,

Q�1,�2(t)=K1(t)u�1,�2(1, t)+ �1(t)u/�1,�2(1, t)

− ∫ t0 k(t − s)u�1,�2(1, s)ds,u�1,�2 ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

u/�1,�2 ∈ L∞(0, T ;H 1), u

//�1,�2 ∈ L∞(0, T ;L2),

u�1,�2(1, ·) ∈ H 2(0, T ), Q�1,�2 ∈ H 1(0, T ),

where{P0,0 = f (x, t), P1,0 = −u0,0, P0,1 = −u/0,0,P�1,�2 = −u�1−1,�2 − u

/�1,�2−1, (�1, �2) ∈ Z2+, 2��1 + �2�N.

Let (u,Q)= (uK,�,QK,�) be a unique weak solution of problem(PK,�). Then(v, R), with

v = u−∑

0��1+�2�Nu�1,�2K

�1��2, R =Q−∑

0��1+�2�NQ�1,�2K

�1��2,


satisfies the problem

Av = −Kv − �vt + EN(K, �), 0<x <1, 0< t <T,v(0, t)= 0, Bv = R(t),

v(x,0)= vt (x,0)= 0,R(t)=K1(t)v(1, t)+ �1(t)vt (1, t)−

∫ t0 k(t − s)v(1, s)ds,

v ∈ C0(0, T ;V ) ∩ C1(0, T ;L2) ∩ L∞(0, T ;V ∩H 2),

v/ ∈ L∞(0, T ;H 1), v// ∈ L∞(0, T ;L2),v(1, ·) ∈ H 2(0, T ), R ∈ H 1(0, T ),

(5.1)

where

EN(K, �)= −∑

�1+�2=N+1

(u�1−1,�2 + u/�1,�2−1)K

�1��2. (5.2)

Then, we have the following lemma.

Lemma 2. Let (H2).(H6) hold. Then

‖EN(K, �)‖L∞(0,T ;L2)�CN(√

K2 + �2)N+1

, (5.3)

whereCN = ∑�1+�2=N+1(‖u�1−1,�2‖L∞(0,T ;V ) + ‖u/�1,�2−1‖L∞(0,T ;V )) is a constant de-

pendingonlyon theconstants‖u�1−1,�2‖L∞(0,T ;V ),‖u/�1,�2−1‖L∞(0,T ;V ), (�1, �2) ∈ Z2+, �1+�2 =N + 1.

Proof. From (5.2), we have

|EN(K, �)|�∑

�1+�2=N+1

(|u�1−1,�2| + |u/�1,�2−1|)|K�1��2|

�∑

�1+�2=N+1

(‖u�1−1,�2‖V + ‖u/�1,�2−1‖V )|K�1��2|. (5.4)

By the boundedness of the functionsu�1−1,�2, u/�1,�2−1, (�1, �2) ∈ Z2+, �1 + �2 =N + 1 in

the function spaceL∞(0, T ;H 1), we obtain from (5.4), that

‖EN(K, �)‖L∞(0,T ;L2)�

∑�1+�2=N+1

(‖u�1−1,�2‖L∞(0,T ;V ) + ‖u/�1,�2−1‖L∞(0,T ;V ))|K�1��2|. (5.5)


On the other hand, using the Hölder’s inequalityab� 1pap + 1

qbq , 1

p+ 1

q= 1, ∀a, b�0

with a =K2�1N+1 , b = �

2�2N+1 , p = N+1

�1, q = N+1

�2, we obtain

|K�1��2| = |K 2�1N+1�

2�2N+1 |N+1

2 �∣∣∣∣∣ 1N+1�1

(K2�1N+1 )

N+1�1 + 1

N+1�2

(�2�2N+1 )

N+1�2

∣∣∣∣∣N+12

=∣∣∣∣ �1N + 1

K2 + �2N + 1

�2∣∣∣∣N+1

2

�(K2 + �2)N+12 =

(√K2 + �2

)N+1,

(5.6)

for all (�1, �2) ∈ Z2+, �1 + �2 =N + 1.Therefore, it follows from (5.5), (5.6), that

‖EN(K, �)‖L∞(0,T ;L2)�CN(√

K2 + �2)N+1

, (5.7)

whereCN =∑�1+�2=N+1(‖u�1−1,�2‖L∞(0,T ;V ) + ‖u/�1,�2−1‖L∞(0,T ;V )).

The proof of Lemma 2 is complete.�

Next, we obtain the following theorem.

Theorem 6. Let (H2).(H6) hold. Then, for every(K, �) ∈ R2 with |K|�K∗, |�|��∗,problem(PK,�) has a uniqueweak solution(u,Q)=(uK,�,QK,�) satisfying the asymptoticestimations up to orderN + 1 as follows:∥∥∥∥∥∥u/K,� −

∑0��1+�2�N

u/�1,�2K

�1��2

∥∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥∥uK,� −

∑0��1+�2�N

u�1,�2K�1��2

∥∥∥∥∥∥L∞(0,T ;V )

+∥∥∥∥∥∥u/K,�(1, ·)−

∑0��1+�2�N

u/�1,�2(1, ·)K�1��2

∥∥∥∥∥∥L2(0,T )

�D∗N

(√K2 + �2

)N+1

, (5.8)

and ∥∥∥∥∥∥QK,� −∑

0��1+�2�NQ�1,�2K

�1��2

∥∥∥∥∥∥L2(0,T )

�D∗∗N

(√K2 + �2

)N+1

, (5.9)

for all K, � ∈ R, |K|�K∗, |�|��∗, the functions(u�1,�2,Q�1,�2) being the weak solutionsof problems(P�1,�2), (�1, �2) ∈ Z2+, �1 + �2�N.


Proof. By multiplying the two sides of(5.1)1 with v/, and after integration int, we obtain

�(t)=∫ t

0K/1(s)v

2(1, s)ds +∫ t

0�/(s)‖vx(s)‖2 ds −K‖v(t)‖2

− 2�∫ t

0‖v/(s)‖2 ds + 2

∫ t

0v/(1, r)dr

∫ r

0k(r − s)v(1, s)ds

+ 2∫ t

0〈EN(K, �, s), v/(s)〉ds, (5.10)

where

�(t)= ‖v/(t)‖2 + �(t)‖vx(t)‖2 +K1(t)v2(1, t)+ 2

∫ t

0�1(s)|v/(1, s)|2 ds. (5.11)

Noting that{�(t)�‖v/(t)‖2 + �0‖vx(t)‖2 + 2�0

∫ t0 |v/(1, s)|2 ds,

|v(1, t)|�‖v(t)‖C0(�)�‖vx(t)‖�

√�(t)�0,

(5.12)

then, we prove, in a manner similar to the above part, that

�(t)� 1

�0

∫ t

0|K/1(s)|�(s)ds + 1

�0

∫ t

0|�/(s)|�(s)ds

+ (|K|T + 2|�| + 1)∫ t

0�(s)ds + T

�0

∫ T

0k2(�)d�

∫ t

0�(s)ds

+ T ‖EN(K, �, s)‖2L∞(0,T ;L2) + 1

2�0�(t), (5.13)

for all t ∈ [0, T ] and >0.Choosing >0, such that 1

2�0�1/2, we obtain from (5.3), (5.13), that

�(t)�2T C2N(K

2 + �2)N+1 +∫ t

0N1(K, �, s)�(s)ds, (5.14)

where

N1(K, �, s)= 2

�0(|K/1(s)| + |�/(s)|)+ 2(|K|T + 2|�| + 1)

+ 2T

�0

∫ T

0k2(�)d�. (5.15)

By Gronwall’s lemma, it follows from (5.14), (5.15), that

�(t)�2T C2N(K

2 + �2)N+1 exp

(∫ t

0N1(K, �, s)ds

)�2T C2

N(K2 + �2)N+1 exp

(∫ T

0N1(K∗, �∗, s)ds

), (5.16)

for all K, � ∈ R, |K|�K∗, |�|��∗.


It follows that

‖v/(t)‖2 + �0‖vx(t)‖2 + 2�0

∫ t

0|v/(1, s)|2 ds

��(t)�2T C2N(K

2 + �2)N+1 exp

(∫ T

0N1(K∗, �∗, s)ds

)≡ C∗

N(K2 + �2)N+1. (5.17)

Hence

‖v/‖L∞(0,T ;L2) + ‖v‖L∞(0,T ;V ) + ‖v/(1, ·)‖L2(0,T )�D∗N

(√K2 + �2

)N+1

, (5.18)

or ∥∥∥∥∥∥u/K,� −∑

0��1+�2�Nu/�1,�2K

�1��2

∥∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥∥uK,� −

∑0��1+�2�N

u�1,�2K�1��2

∥∥∥∥∥∥L∞(0,T ;V )

+∥∥∥∥∥∥u/K,�(1, ·)−

∑0��1+�2�N

u/�1,�2(1, ·)K�1��2

∥∥∥∥∥∥L2(0,T )

�D∗N

(√K2 + �2

)N+1

, (5.19)

for all K, � ∈ R, |K|�K∗, |�|��∗.On the other hand, it follows from(5.1)4, (5.12), that

‖R‖L2(0,T )�‖K1‖∞‖v‖L∞(0,T ;V ) + ‖�1‖∞‖v/(1, ·)‖L2(0,T )+(1

�0T

∫ T

0k2(�)d�

)1/2(∫ T

0�(s)ds

)1/2�D∗∗

N

(√K2 + �2

)N+1

, (5.20)

hence,∥∥∥∥∥∥QK,� −∑

0��1+�2�NQ�1,�2K

�1��2

∥∥∥∥∥∥L2(0,T )

�D∗∗N

(√K2 + �2

)N+1

. (5.21)

The proof of Theorem 6 is complete.�


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