On a shock problem involving a nonlinear viscoelastic bar

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bar

Nguyen Thanh Long, Alain Pham Ngoc Dinh, Tran Ngoc Diem

To cite this version:

Nguyen Thanh Long, Alain Pham Ngoc Dinh, Tran Ngoc Diem. On a shock problem involving a

nonlinear viscoelastic bar. Journal of Boundary Value Problems, 2005, 3, pp.337-358. �hal-00008873�

On a shock problem involving a nonlinear viscoelastic bar

Nguyen Thanh Long ^{ 1 } , Alain Pham Ngoc Dinh ^{ 2 } , Tran Ngoc Diem ^{ 1 }

 1  Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University HoChiMinh City, 227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, Vietnam.

E-mail: [email protected]

2 MAPMO, UMR 6628, bâtiment de Mathématiques, Université d’Orléans, BP 6759 Orléans Cedex 2, France.

E-mail: [email protected]

Abstract. We treat an initial boundary value problem for a nonlinear wave equation u tt  u xx  K|u | ^ u |u t | ^ u t  fx, t in the domain 0  x  1, 0  t  T. The boundary condition at the boundary point x  0 of the domain for a solution u involves a time

convolution term of the boundary value of u at x  0 whereas the boundary condition at the other boundary point is of the the form u x 1, t  K 1 u1, t  1 u t 1, t  0 with K 1 and  1 given non-negative constants. We prove existence of a unique solution of such a problem in classical Sobolev spaces. The proof is based on a Galerkin type approximation, various energy

estimates and compactness arguments. In the case of     0, the regularity of solutions is studied also. Finally, we obtain an asymptotic expansion of the solution u, P of this problem up to order N  1 in two small parameters K, .

Keywords: Shock, integral equations, full non-homogeneous conditions, energy-type estimates, compactness, regularity of solutions, asymptotic expansion.

AMS subject classification: 35L70, 35Q72, 35C20.

Address for correspondence: Nguyen Thanh Long.

1. Introduction

Given T  0, we consider the problem to find a pair of functions u, P such that

1. 1

u tt  u xx  Fu, u t   fx,t, 0  x  1, 0  t  T, u x 0, t  Pt,

u x 1, t  K 1 u1, t   1 u t 1, t  0, ux, 0  u 0 x, u t x, 0  u 1 x, where

 Fu, u t   K|u | ^ u  |u t | ^ u t

 u 0 , u 1 , f are given functions

 K, K 1 , , ,  and  1  0 are given constants

and the unknown function ux, t and the unknown boundary value Pt satisfy the following

Cauchy problem for ordinary differential equation

1. 2 P ^// t   ² Pt  hu tt 0, t, 0  t  T, P0  P 0 , P ^/ 0  P 1 ,

where   0, h  0, P 0 , P 1 are given constants. Problem (1.1)-(1.2) describes the shock between a solid body and a nonlinear viscoelastic bar resting on a viscoelastic base with nonlinear elastic constraints at the side, constraints associated with a viscous frictional resistance.

In [1], N.T. An and N.D. Trieu studied a special case of problem 1. 1  1. 2 with     0 and f, u 0 , u 1 and P 0 vanishing, associated with the homogeneous boundary condition

u1, t  0 instead of 1. 1 3 being a mathematical model describing the shock of a rigid body and a linear visoelastic bar resting on a rigid base.

From (1.2), solving the equation ordinary diferential of second order, we get

1. 3 Pt  gt  hu0, t 

t

0

 kt  su0, sds, where

gt  P 0  hu 0 0 cos t  _ ¹ P 1  hu 1 0 sin t, kt  h sin t.

This observation motivates to consider problem (1.1) with a more general boundary term of the form

1. 4 Pt  gt  hu0, t 

t

0

 kt  su0, sds, which we shall do henceforth.

In [8, 9] Dinh and Long studied problem 1. 1 1,2,4 and (1.4) with Dirichlet boundary condition at boundary point x  1 in [9] extending an earlier result of theirs for k  0 in [8].

In [15] Santos has studied the following problem

1. 5

u tt  tu xx  0, 0  x  1, t  0, u0, t  0,

u1, t 

t

0

 Gt  ssu x 1, sds  0, ux, 0  u 0 x, u t x, 0  u 1 x.

The integral in 1. 5 3 is a boundary condition which includes the memory effect. Here, by u we denote the displacement and by G the relaxation function. The function   W _loc ^1,    with t   0  0 and  ^/ t  0 for all t  0. Frictional dissipative boundary condition for the wave equation was studied by several authors, see for example [4, 5, 6, 12, 16, 17, 18, 19]

and the references therein. In these works existence of solutions and exponential stabilization were proved for linear and for nonlinear equations. In contrast with the large literature for frictional dissipative, for boundary condition with memory, we have only a few works as for example [11, 13, 14].

Applying the Volterra’s inverse operator, Santos [15] transformed 1. 5 3 into

1. 6

tu x 1, t  _G0 ¹ Ktu 0 1  _G0 ¹ u t 1, t

 _G ^G

2^/

^0 0 u1, t  _G0 ¹

t

0

 ^{K /} t  su1, sds, where the resolvent kernel satisfies

1. 7 Kt  _G0 ¹

t

0

 G ^/ t  sKsds  _G0 ^1 G ^/ t.

The present paper consists of three main parts. In Part 1 we prove a theorem of global existence and uniqueness of a weak solution u of problem (1.1), (1.4). The proof is based on a Galerkin type approximation in conjunction with various energy estimates, weak convergence compactness arguments. The main difficulty encountered here is the boundary condition at x  1. In order to solve this particular difficulty, stronger assumptions on the initial conditions u 0 and u 1 will be made. We remark that the linearization method in the papers [3,10] cannot be used in [2, 8, 9]. In the case of     0, Part 2 is devoted to the study of the regularity of the solution u. Finally, in Part 3 we obtain an asymptotic expansion of the solution u, P of the problem (1.1), (1.4) up to order N  1 in two small parameters K, . The results obtained here may be considered as generalizations of those in An and Trieu [1] and in Long and Dinh [2, 3, 8, 9, 10].

2. The existence and uniqueness theorem

Put   0, 1, Q T    0, T, T  0. We omit the definitions of the usual function spaces: C ^m , L ^p  and W ^m,p  and denote W ^m,p  W ^m,p , L ^p  W ^0,p  and

H ^m  W ^m,2 , 1  p  , m  IN. The norm in L ² is denoted by . Also, we denote by

,  the scalar product in L ² or the dual pairing between continuous linear functionals and elements of a function space, by  _X the norm of a Banach space X, by X ^/ its dual space, and by L ^p 0, T; X, 1  p   the Banach space of real measurable functions u : 0, T  X such that

u  _L

^p

_0,T;X 

T

0

 ut _X ^p dt

1/p

  for 1  p  , and

u  _L



0,T;X 

0  t  T

ess sup ut _X for p  .

At last, denote ut  ux, t, u ^/ t  u t t  ^ _t ^u x, t, u ^// t  u tt t  ^ _

²

_t

2

^u x, t, u ^r t  ^ _t

^r

^u

r

x,t, u x t  ^ _x ^u x, t, u xx t  ^ _

²

_x ^u

2

x, t.

Further, we make the following assumptions:

H 0    0,   0, K  0,   0,

H 1  h  0, K 1  0, K 1  h  0 and  1  0,

H 2  u 0  H ² and u 1  H ¹ ,

H 3  f, f t  L ² 0, T; L ² ,

H 4  k  H ¹ 0, T  W ^2,1 0, T ,

H 5  g  H ² 0, T .

Then we have the following theorem.

Theorem 1. Let assumptions H 0   H 5  be satisfied. Then there exists a unique weak solution u of problem (1.1), (1.4) such that

2. 1

u  L ^ 0, T; H ² , u t  L ^ 0, T; H ¹ , u tt  L ^ 0, T; L ² , u0,   W ^1, 0, T, u1,   H ² 0, T  W ^1, 0, T, P  W ^{1, } 0, T.

Remark 1. It follows from (2.1) that the component u in the weak solution u, P of problem (1.1), (1.4) satisfies

u  C ⁰ 0, T; H ¹   C ¹ 0, T; L ²   L ^ 0, T; H ² .

Proof of Theorem 1. The proof consists of Steps 1-5.

Step1. Galerkin approximation. Let w j  be an enumeration of a basis of H ² . We find the approximate solution of problem (1.1), (1.4) in the form

u m t  ^m

j1

 ^{c mj tw j ,}

where the coefficient functions c mj satisfy the ordinary differential equation problem

2. 2

u m // t,w j   u mx t, w jx   P m tw j 0  Q m tw j 1

 Fu m t, u m /

t, w j   f t, w j , 1  j  m, P m t  gt  hu m 0, t 

t

0

 kt  s u m 0, sds,

Q m t  K 1 u m 1, t   1 u m / 1, t,

u m 0  u 0m  ^m

j  1

  mj w j  u 0 strongly in H ² ,

u m / 0  u 1m  ^m

j  1

  mj w j  u 1 strongly in H ¹ .

From the assumptions of Theorem 1, this problem has a solution  u m , P m ,Q m  on some interval 0, T m . The following estimates allow one to take T m  T for all m.

Step 2. A priori estimates I. Substituting 2. 2 23 into 2. 2 1 , then multiplying the j ^th equation of 2. 2 1 by c _mj ^/ , summing up with respect to j and afterwards integrating with respect to the time variable from 0 to t, we get

S m t  S m 0  2

t

0

 gsu m / 0, sds

 2

t

0

 ^{u m /} 0, sds

s

0

 ks  u m 0, d  2

t

0

 fs, u m /

sds,

where

2. 3 S m t  u m /

t ²  u mx t ²  _2 ^2K u m t _L 2

2

 hu m 2 0, t

 K 1 u m 2 1, t  2  ₀ ^{t u m /} s

L

^2

 2

ds  2 1  ₀ ^{t |u m /} 1, s| ² ds.

Using assumptions H 4   H 5  and then integrating by parts with respect to the time variable, we get

2. 4

S m t  S m 0  2g0u 0m 0  2gtu m 0, t  2

t

0

 g ^/ su m 0, sds

2u m 0, t

t

0

 kt  u m 0, d  2k0

t

0

 ^u m 2 0, sds

 2

t

0

 ^u m 0, sds

s

0

 ^{k /} s  u m 0, d  2

t

0

 fs, u m / sds.

Then, using 2. 2 45 and (2.3) we get

2. 5 S m 0  2|g0u 0m 0|  C 1 for all m  1,

where C 1 is a constant independent of m. Using the inequality 2ab  a ²  ¹ _ b ² for all a, b   and for all   0, it follows that

2. 6

S m t  C 1  ¹ _ g ² t  u m 2 0, t  ¹ _

t

0

 |g ^/ s| ² ds  

t

0

 u m 2 0, sds

 u m 2 0, t  ¹ _

t

0

 kt  u m 0, d

2

 2|k0|

t

0

 ^u m 2 0, sds



t

0

 u m 2 0, s  ¹ _

s

0

 ^{k /} s  u m 0, d

2

ds

 ¹ _

t

0

 ^{fs 2 ds}  

t

0

 ^{u m /} s ² ds

 C 1  ¹ _ g ² t 

t

0

 ^{|g /} s| ² ds 

t

0

 fs ² ds

2u m 2 0, t  2  |k0|

t

0

 u m 2 0, sds



t

0

 ^{u m /} s ² ds  ¹ _

t

0

 kt  u m 0, d

2

 ¹ _

t

0

 ^{ds s}

0

 ^{k /} s  u m 0, d

2

.

On the other hand, noticing K 1  h  0,

2. 7 v x  ²  hv ² 0  K 1 v ² 1  Cv _H ²

¹

for all v  H ¹ ,

where C  0 is a constant depending only on K 1 and h, and on the other hand, by H ¹   C ⁰ , we have

2. 8 v _C

⁰

_  C 0 v _H

¹

for all v  H ¹ , for some constant C 0  0. Hence it follows from (2.3) that

2. 9 |u m 0, t|  u m t _C

⁰

_  C 0 u m t _H

¹

 ^C

⁰

C

S m t  C 0 S m t .

Now, using the Cauchy-Schwarz inequality, we estimate in the right-hand side of (2.6) the last but one integral as

1  t

0

 kt  u m 0, d

2

 ¹ _

t

0

 ^{k 2} d

t

0

 ^u m 2 0, d

 ^C _

⁰²

t

0

 ^{k 2} d

t

0

 ^{S m} d.

and the last integral as

1  t

0

 ^{ds s}

0

 ^{k /} s  u m 0, d

2

 ¹ _ t

t

0

 ^{|k /} | ² d

t

0

 ^u m 2 0, d

 ^C _

⁰²

t

t

0

 ^{|k /} | ² d

t

0

 ^S m d.

Choosing  so that 0  2C 0

2  ¹ ₂ and using both these estimates, it follows from (2.6) and (2.9) that

S m t  G 1 t  G 2 t  ₀ ^t S m d, where

G 1 t  2C 1  ² _ g ² t 

t

0

 ^{|g /} s| ² ds 

t

0

 fs ² ds ,

G 2 t  2  4C 0

2   |k0|  ^2C _

⁰²

t

0

 k ² d  t

t

0

 |k ^/ | ² d . Since H ¹ 0, T  C ⁰ 0, T, from assumptions H 3   H 5  we deduce that

2. 10 |G i t|  M _T ^{ i } , a. e. on t  0, T, i  1, 2, where the constants M _T ^i are depending on T only. Therefore

2. 11 S m t  M _T ^1  M _T ^2

t

0

 ^{S m} d, 0  t  T m  T, which implies by Gronwall’s lemma

2. 12 S m t  M _T ^1 exptM T

2   M T for all t  0, T.

Step 3. A priori estimates II. Now differentiating 2. 2 1 with respect to t we get

u m /// t, w j   u mx / t, w jx   P m / tw j 0  Q m / tw j 1

 K  1 |u m | ^ u m / t, w j     1 u m / t ^ u m // t,w j 

 f ^/ t, w j , for all 1  j  m.

Multiplying the j ^th equation herein by c _mj ^// , summing up with respect to j and then integrating with respect to the time variable from 0 to t, after some rearrangements we get

2. 13

X m t  X m 0  2

t

0

 ^{g /} su ^{m //} 0, sds

 2

t

0

 k0 u m 0,  

 0

 ^{k /}   su m 0, s ds u ^{m //} 0, d

2 K  1

t

0

 d

1

0

 |u m x, | ^ u m / x, u m // x, dx

2

t

0

 f ^/ s, u m //

sds, where

2. 14

X m t  u m // t ²  u mx / t ²  h u m / 0, t ²  K 1 u m / 1, t ²  2 1 t

0

 ^{|u m //} 1, | ² d

2  1

t

0

 d

1

0

 ^{u m /} x,  ^ u m //

x,  ² dx

 u m // t ²  u mx / t ²  h u m / 0, t ²  K 1 u m / 1, t ²  2 1 t

0

 ^{|u m //} 1, | ² d

 _{2} ^8

2

  1

t

0

 d

1

0

 _d ^d _ ^{u m /} x,  ^{2/2 2} dx.

Integrating by parts in the integrals of the right-hand side of (2.13), we get

2. 15

X m t  X m 0  2g ^/ 0u 1m 0  2g ^/ tu ^{m /} 0, t  2

t

0

 ^{g //} su ^{m /} 0, sds

 2 k0 u m 0, t 

t

0

 ^{k /} t  su m 0, s ds u ^{m /} 0, t  2k0u 0m 0u 1m 0

 2

t

0

 k0 u m / 0,   k ^/ 0 u m 0,  

 0

 ^{k //}   su m 0, s ds u m / 0, d

 2 K  1

t

0

 d

1

0

 |u m x, | ^ u m / x, u m // x, dx  2

t

0

 f ^/ s, u m // sds

 X m 0  2g ^/ 0u 1m 0  2k0u 0m 0u 1m 0  k ^/ 0 u _0m ² 0

k ^/ 0 u m 2 0, t  2g ^/ tu ^{m /} 0, t  2 k0 u m 0, tu ^{m /} 0, t

2

t

0

 ^{g //} su ^{m /} 0, sds  2 k0

t

0

 ^{u m /} 0,  ² d

 2

t

0

 k ^/ t  su m 0, s ds. u m / 0, t  2

t

0

 u m / 0, d



0

 k ^//   su m 0, s ds

2 K  1

t

0

 d

1

0

 ^{|u m} x, | ^ u m /

x, u ^{m //} x, dx  2

t

0

 f ^/ s, u m //

sds.

First, we deduce from 2. 2 3 , (2.14) and asumptions H 4   H 5  that

2. 16 |X m 0  2g ^/ 0u 1m 0  2k0u 0m 0u 1m 0  k ^/ 0 u _0m ² 0|

 C 2  u m //

0 ² ,

where C 2  0 is a constant depending only on u 0 , u 1 , g, k, K, K 1 , h only. But by 2. 2 1  3 we have

u m // 0 ²  u 0mxx , u m // 0  Fu 0m , u 1m , u m // 0  f0, u m // 0.

Therefore

u m // 0  u 0mxx   Fu 0m , u 1m   f0

and by means of 2. 2 4 we deduce that

2. 17 u m //

0  C 3 ,

where C 3  0 is a constant depending on u 0 , u 1 , f, K,  only.

On the other hand, it follows from (2.7)- (2.9) that

2. 18 u m / t _C

0

  C 0 u m / t _H

1

 C 0 X m t . Then, by means of (2.9), (2.12) and (2.18) we deduce that

2. 19

2 K  1

t

0

 d

1

0

 ^|u m x, | ^ u m / x, u m // x, dx

 2 K  1 C 0 M T

 t 0

 u m /  u m //  d

 2 K  1C 0 C 0 M T

 t 0

 ^{X m} d

and from here and (2.13)- (2.17) we obtain

X m t  C 2  C ₃ ²  |k ^/ 0| u m 2 0, t  2 g ^/ tu ^{m /} 0, t

2 k0 u m 0, tu m / 0, t

 2

t

0

 ^{g //} su ^{m /} 0, s ds  2 |k0|

t

0

 ^{u m /} 0,  ² d

2

t

0

 ^{|k /} t  su m 0, s |ds. u m / 0, t

2

t

0

 ^{u m /} 0,  d

 0

 ^{|k //}   su m 0, s| ds

 2 K  1

t

0

 d

1

0

 |u m x,| ^ u m /

x, u ^{m //} x,  dx

2

t

0

 f ^/ s, u m // s ds

 C 2  C ₃ ²  |k ^/ 0|C 0 2

M T  2|g ^/ t| C 0 X m t

 2 |k0|C 0 2

M T X m t  2 C 0 t

0

 |g ^// s| X m s ds

 2 |k0|C 0 2 t

0

 ^{X m} d  2C 0

2 M T

t

0

 ^{|k /} |d X m t

2C 0

2 M T

t

0

 ^{|k //} | d

t

0

 ^{X m}  d

2 K  1C 0 C 0 M T

 t 0

 ^X m d



t

0

 f ^/ s ² ds 

t

0

 X m sds.

We again use the inequality 2ab  a ²  ¹ _ b ² a, b  R,   0 with   ¹ ₄ . Then it

follows that

X m t  C 2  C ₃ ²  |k ^/ 0|C 0

2 M T  2|g ^/ t| C 0 X m t

 4 |k0|C 0

2 M T X m t  2 C 0 t

0

 ^{|g //} s| X m s ds

 2 |k0|C 0 2 t

0

 ^X m d  2C ₀ ² M T t

0

 ^{|k /} |d X m t

C 0 2

M T t

0

 |k ^// | d

t

0

 X m  d

2 K  1C 0 C 0 M T

 t 0

 ^{X m} d



t

0

 f ^/ s ² ds 

t

0

 ^X m sds

 C 2  C ₃ ²  |k ^/ 0|C 0

2 M T  4 |g ^/ t| C 0

2  ¹ ₄ X m t

4 k ² 0C 0

4 M T  ¹ ₄ X m t  C 0 2 t

0

 ^{|g //} s| ² ds



t

0

 ^{X m} sds  2 |k0|C 0 2 t

0

 ^{X m} d

4C 0 4 M T

t

0

 ^{|k } |d

2

 ¹ ₄ X m t

C 0 4

M T t

t

0

 |k ^// | d

2



t

0

 X m d

2 K  1C 0 C 0 M T

 t 0

 ^{X m} d



t

0

 f ^/ s ² ds 

t

0

 ^{X m} sds.

Noting the embedding H ¹ 0, T  C ⁰ 0, T, it follows from assumptions H 3   H 5  that X m t  M T  3   M T  4  t

0

 X m d for all t  0, T, where

M _T ^4  12  8C 0

2 |k0|  8K  1C 0 C 0 M T



and M T  3 

is a constant depending on T, f, g, k, C 2 , C 3 , C 0 and M T only. By Gronwall’s lemma

we deduce that

2. 20 X m t  M T 3 exptM T 4   M T for all t  0, T.

On the other hand, we deduce from 2. 2 23 , (2.3), (2.12), (2.14) and (2.20) that

2. 21

P m  _W

1,

 0,T   M T 5

,

Q m  _H

¹

_0,T  M _T ^{ 6 } , u m / 

u m /

L

^{2}

Q

T



2

^/

 u m /

L

^2

 Q

T



 2

 M _T ^7 ,

t   u m / 2/2



L

²

Q

T



2  X m t  M T ,

2. 22

x   u m / 2/2

 L

²

Q

T



2  ¹ ₄   2 ²

T

0

 ^{dt 1}

0

 ^{u m /} x, t ^ u mx / x, t ² dx

 ¹ ₄   2 ²

T

0

 ^C 0 X m t ^ dt

1

0

 ^{u mx /} x, t ² dx

 ¹ ₄   2 ²

T

0

 C 0 X m t ^ X m tdt

 ¹ ₄   2 ² T C 0 M T



M T  M T  8 

, for all T  0 and   2 ^/    2/  1.

Step 4. Limiting process. From (2.3), (2.12), (2.14), (2.20) and 2. 21 13 we deduce the existence of a subsequence of u m , P m , Q m , still also so denoted, such that

2. 23

u m  u in L ^ 0, T; H ¹  weak , u m /  u ^/ in L ^ 0, T; H ¹  weak , u m /  u ^/ in L ^2 Q T  weakly, u m //  u ^// in L ^ 0, T; L ²  weak , u m 0,   u0,  in W ^1, 0, T weak , u m 1,   u1,  in W ^1, 0, T weak , u m 1,   u1,  in H ² 0, T weakly, P m  P in W ^1, 0, T weak , Q m  Q in H ¹ 0, T weakly, u m / 

u m /

  in L ^{2}

^/

Q T  weakly.

By the compactness lemma of J.L. Lions [7: p.57] we can deduce from 2. 21 4 , 2. 22 and

2. 23 1,2,46 the existence of a subsequence still denoted by u m  such that

2. 24

u m  u strongly in L ² Q T , u m /

 u ^/ strongly in L ² Q T , u m / 2/2

  1 strongly in H ¹ Q T , u m 0,   u0,  strongly in C ⁰ 0, T,

u m 1,   u1,  strongly in H ¹ 0, T, u m / 1,   u ^/ 1,  strongly in C ⁰ 0, T.

From 2. 2 2  3 and 2. 24 4  6 we have

2. 25 P m t  gt  h u0, t 

t

0

 kt  s u0, sds  Pt,

Q m t  K 1 u1, t   1 u ^/ 1, t  Qt

strongly in C ⁰ 0, T from where with 2. 23 89

2. 26 Pt  Pt and Qt  Qt

can be deduced. Using the inequality

2. 27 ||x| ^ x  |y| ^ y |    1R ^ |x  y | x, y  R, R,

for all R  0 and all   0 it follows from (2.9), (2.12) and 2. 24 1 that

2. 28 |u m | ^ u m  |u| ^ u strongly in L ² Q T .

Similarly, we can also obtain from (2.18), (2.20), 2. 24 2 and inequality (2.27) with   , that

u m / 

u m /  |u ^/ | ^ u ^/ strongly in L ² Q T .

Hence, because of (2.28),

2. 29 Fu m , u m /   Fu, u ^/  strongly in L ² Q T .

Passing to the limit in 2. 2 1,45 , by 2. 23 1,2,4 and (2.25)- (2.26) and (2.29) we have u satisfying the problem

u ^// t, v  u x t, v x   Ptv0  Qtv1  Fut,u ^/ t, v  f t, v, u0  u 0 , u ^/ 0  u 1

weak in L ² 0, T weak, for all v  H ¹ . On the other hand, we have from 2. 10  2. 12 and assumption H 3  that

u xx  u ^//  Fu,u ^/   f  L ^ 0, T; L ² 0, 1.

Hence u  L ^ 0, T; H ²  and the existence proof is completed.

Step 5. Uniqueness of the solution. Let u i , P i , i  1, 2 be two weak solutions of problem (1.1), (1.4) such that

u i  L ^ 0, T; H ² , u i

/  L ^ 0, T; H ¹ , u i

//  L ^ 0, T; L ² ,

u i 0,   W ^1, 0, T, u i 1,   H ² 0, T  W ^1, 0, T, P i  W ^1, 0, T .

Then u, P with u  u 1  u 2 and P  P 1  P 2 satisfies the variational problem

u ^// t, v  u x t, v x   Ptv0  Qtv1  K|u 1 | ^ u 1  |u 2 | ^ u 2 , v

  u 1 / 

u 1 /  u 2

/ 

u 2

/ ,v  0  v  H ¹ , u0  u ^/ 0  0,

where

Pt  h u0, t 

t

0

 kt  s u0, sds , Qt  K 1 u1, t   1 u ^/ 1, t.

We take v  u ^/ in 2. 21 1 , afterwards integrating in t, we get

2. 30 Zt  2K

t

0

 |u 1 | ^ u 1  |u 2 | ^ u 2 , u ^/ d  2

t

0

 ^{u /} 0, d

 0

 k  su0, sds,

where

2. 31

Zt  u ^/ t ²  u x t ²  h u ² 0, t  K 1 u ² 1, t

 2 1 t

0

 ^{|u /} 1, s| ² ds  2

t

0

  u ₁ ^{/ } u ₁ ^/  u ₂ ^{/ } u ₂ ^/ ,u ^/ d.

Using inequality (2.27), the first term of the right- hand side of (2.30) can be estimated as

2. 32

2K

t

0

 |u 1 | ^ u 1  |u 2 | ^ u 2 , u ^/ d  2K  1R ^

t

0

 uu ^/  d

 K  1R ^

t

0

 Zd, with R 

i1,2

max u i  _L



0,T;H

¹

 . Using integration by parts in the last integral of (2.30), we get J  2

t

0

 ^{u /} 0, d

 0

 k  su0, sds

 2u0, t

t

0

 kt  su0, sds  2k0

t

0

 ^{u 2} 0, d

2

t

0

 u0, d



0

 k ^/   su0, sds.

On the other hand, it follows from (2.7)- (2.8) and (2.31) that

|u0, t|  ut _C

⁰

_  C 0 ut _H

¹

 ^C

⁰

C

Zt  C 0 Zt .

Thus

2. 33

|J|  2C ₀ ² Zt

t

0

 |kt  s| Zs ds  2|k0|C 0 2 t

0

 Zd

2C 0 2 t

0

 Z d

 0

 |k ^/   s| Zs ds

 ¹ ₂ Zt  2 C 0 4 t

0

 ^{k 2} d

t

0

 Zsds

 2|k0|C 0 2 t

0

 Zd  2C 0

2 t

t

0

 ^{|k /}  | ² d

1/2 t

0

 Zsds can be deduced. It follows from (2.30) and (2.32)- (2.33) that

Zt  m T t

0

 Zsds for all t  0, T, where

m T  2K  1R ^  4 C 0 4 T

0

 ^{k 2} d  4|k0|C 0

2  4C 0

2 T

T

0

 ^{|k /} | ² d

1/2

. By Gronwall’s lemma, we deduce that Z  0 and Theorem 1 is completely proved.

3. Regularity of solutions

In this part, we study the regularity of solution of problem (1.1), (1.4) corresponding to

    0. From here, we assume that h, K, K 1 , ,  1  satisfy assumptions H 0 , H 1 .

Henceforth, we shall impose the following stronger assumptions:

H 1 1  u 0  H ³ and u 1  H ² ,

H 2

1  f, f t , f tt  L ² 0, T; L ²  and f, 0  H ¹ ,

H 3

1  g  H ³ 0, T ,

H 4  1   k  H ² 0, T .

Formally differentiating problem (1.1) with respect to time and letting  u   u t and P  P ^/ we are led to consider the solution  u of problem  Q  :

L  u   u tt   u xx  F u,  u t   

f x, t, x, t  Q T ,

 u x 0, t  Pt,

B 1  u   u x 1, t  K 1  u1, t   1  u t 1, t  0,

 ux, 0   u 0 x,  u t x, 0   u 1 x, Pt   gt  h  u0, t 

t

0

 kt  s u0, sds,

where

Fu, u t   Ku  u t , 

f  f t ,  gt  g ^/ t  ktu 0 0,

 u 0  u 1 ,  u 1  u 0xx  Fu 0 , u 1   fx, 0.

Let u 0 , u 1 , f, g, k satisfy assumptions H 1  1    H 4  1  . Then  u 0 ,  u 1 , 

f ,  g, k satisfy assumptions

H 1   H 4  and by Theorem 1 for problem  Q  there exists a unique weak solution  u, P

such that

3. 1

 u  C ⁰ 0, T; H ¹   C ¹ 0, T; L ²   L ^ 0, T; H ² ,

 u t  L ^ 0, T; H ¹ ,  u tt  L ^ 0, T; L ² ,

 u0,   W ^1, 0, T,  u1,   H ² 0, T  W ^1, 0, T,

P  W ^{1, } 0, T .

Moreover, from the uniqueness of weak solution we have

3. 2  u  u t , P  P ^/ .

It follows from (3.1)-(3.2) that

3. 3

u  C ⁰ 0, T; H ²   C ¹ 0, T; H ¹   C ² 0, T; L ² ,

u t  L ^ 0, T; H ² , u tt  L ^ 0, T; H ¹ , u ttt  L ^ 0, T; L ² , u0,   W ^{2, } 0, T , u1,   H ³ 0, T   W ^{2, } 0, T, P  W ^2, 0, T.

We then have the following theorem.

Theorem 2. Let     0 and let assumptions H 0 , H 1  and H 1

1   H 4

1  hold. Then there exists a unique weak solution u, P of problem (1.1), (1.4) satisfying (3.3).

Similarly, formally differentiating problem (1.1) with respect to time up to order r and letting u ^r  ^ _

^r

_t ^u

r

and P ^r  ^d _{d t}

^r

^P

r

we are led to consider the solution u ^r of problem Q ^r  :

Lu ^r  f ^r x, t, x, t  0, 1  0, T, u x r 0, t  P ^r t,

B 1 u ^{ r }  0,

u ^r x, 0  u ₀ ^r x, u t r x, 0  u ₁ ^r x, P ^r t  g ^r t  hu ^r 0, t 

t

0

 kt  su ^r 0, sds, where the functions u ₀ ^r and u ₁ ^r are defined by the recurrence formulas

u ₀ ^0  u 0 , u ₀ ^r  u ₁ ^r1 , r  1,

u ₁ ^{ 0 }  u 1 , u ₁ ^{ r }  u _0xx ^{ r  1 }  Fu 0  r  1  , u ₁ ^{ r  1 }   _ ^ _t

^r1_r1

^f x, 0, r  1, f ^r  _t ^

^r

^f

r

,

g ^0  g, g ^r  ^d _{d t}

^r

^g

r



0 r1  ^u 0

r1 0 ^d _dt

^

^k , r  1.

Assume that the data u 0 , u 1 , f, g, k satisfy the following conditions:

H 1

r  u 0  H ^r2 and u 1  H ^r1 ,

H 2 r 



^

f

 t

^

 L ² 0, T; L ² , 0    r  1, and ^ _t

^

^f , 0  H ¹ , 0    r  1,

H 3  r   g  H ^{r  2} 0, T ,

H 4 r  k  H ^r1 0, T , r  1.

Then u ₀ ^r , u ₁ ^r , f ^r , g ^r , k satisfy H 1   H 4 . Applying again Theorem 1 for problem Q ^r , there exists a unique weak solution u ^r satisfying (2.1) and the inclusion from Remark1, i.e., such that

3. 4

u ^r  C ⁰ 0, T; H ¹   C ¹ 0, T; L ²   L ^ 0, T; H ² , u t  r   L ^ 0, T; H ¹ , u tt  r   L ^ 0, T; L ² ,

u ^r 0,   W ^1, 0, T , u ^r 1,   H ² 0, T  W ^1, 0, T, P ^r  W ^1, 0, T.

Moreover, from the uniqueness of weak solution we have  u ^r , P ^r    ^ _

^r

_t ^u

r

, ^d _{d t}

^r

^P

r

 . Hence we obtain from (3.4) that

3. 5

u  C ^r1 0, T; H ²   C ^r 0, T; H ¹   C ^r1 0, T; L ² ,

u0,   W ^r1, 0, T, u1,   H ^r2 0, T  W ^r1, 0, T, P  W ^r1, 0, T.

We then have the following theorem.

Theorem 3. Let     0 and let assumptions H 1  and H 1 r   H 4 r  hold. Then there exists a unique weak solution u, P of problem (1.1), (1.4) satisfying (3.5) and



^r

u

 t

^r

 L ^ 0, T; H ² , ^ _{ t}

^r1r1

^u  L ^ 0, T; H ¹ , ^ _

^r2

_t

r2

^u  L ^ 0, T; L ² .  4. Asymptotic expansion of solutions

In this part, we assume that     0 and h, K 1 ,  1 , f, g, k satisfy the assumptions

H 1   H 5 .

We consider the following perturbed problem Q _K, , where K  0,   0 are small parameters:

Q _K, 

Lu  u tt  u xx  Ku  u t  fx, t, 0  x  1, 0  t  T, B 0 u  u x 0, t  Pt,

B 1 u  u x 1, t  K 1 u1, t   1 u t 1, t  0, ux, 0  u 0 x, u t x, 0  u 1 x,

Pt  gt  hu0, t 

t

0

 kt  su0, sds.

Let u 0,0 , P 0,0  be a unique weak solution of problem Q _0,0  as in Theorem 1, corresponding

to K,   0, 0, i. e. ,