Open Access

General decay for a system of nonlinear viscoelastic wave equations with weak damping

Boundary Value Problems20122012:146

DOI: 10.1186/1687-2770-2012-146

Received: 19 August 2012

Accepted: 26 November 2012

Published: 13 December 2012

Abstract

In this paper, we are concerned with a system of nonlinear viscoelastic wave equations with initial and Dirichlet boundary conditions in R n ( n = 1 , 2 , 3 ). Under suitable assumptions, we establish a general decay result by multiplier techniques, which extends some existing results for a single equation to the case of a coupled system.

MSC:35L05, 35L55, 35L70.

Keywords

viscoelastic system general decay weak damping

1 Introduction

In this paper, we are concerned with a coupled system of nonlinear viscoelastic wave equations with weak damping
{ u t t Δ u + 0 t g 1 ( t τ ) Δ u ( τ ) d τ + u t = f 1 ( u , v ) , in  Ω × ( 0 , + ) , v t t Δ v + 0 t g 2 ( t τ ) Δ v ( τ ) d τ + v t = f 2 ( u , v ) , in  Ω × ( 0 , + ) , u = v = 0 , on  Ω × ( 0 , + ) , u ( , 0 ) = u 0 , u t ( , 0 ) = u 1 , v ( , 0 ) = v 0 , v t ( , 0 ) = v 1 , in  Ω ,
(1.1)

where Ω R n ( n = 1 , 2 , 3 ) is a bounded domain with smooth boundary Ω, u and v represent the transverse displacements of waves. The functions g 1 and g 2 denote the kernel of a memory, f 1 ( u , v ) and f 2 ( u , v ) are the nonlinearities.

In recent years, many mathematicians have paid their attention to the energy decay and dynamic systems of the nonlinear wave equations, hyperbolic systems and viscoelastic equations.

Firstly, we recall some results concerning single viscoelastic wave equation. Kafini and Tatar [1] considered the following Cauchy problem:
{ u t t Δ u + 0 t g ( t s ) Δ u ( x , s ) d s = 0 , x R n , t > 0 , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x R n .
(1.2)
They established the polynomial decay of the first-order energy of solutions for compactly supported initial data and for a not necessarily decreasing relaxation function. Later Tatar [2] studied the problem (1.2) with the Dirichlet boundary condition and showed that the decay of solutions was an arbitrary decay not necessarily at exponential or polynomial rate. Cavalcanti et al. [3] studied the following equation with Dirichlet boundary condition:
| u t | ρ u t t Δ u Δ u t t + g Δ u γ Δ u t = 0 .

The authors established a global existence result for γ 0 and an exponential decay of energy for γ > 0 . They studied the interaction within the | u t | ρ u t t and the memory term g Δ u . Later on, several other results were published based on [46]. For more results on a single viscoelastic equation, we can refer to [714].

For a coupled system, Agre and Rammaha [15] investigated the following system:
{ u t t Δ u + | u t | m 1 u t = f 1 ( u , v ) , in  Ω × ( 0 , T ) , v t t Δ v + | v t | r 1 v t = f 2 ( u , v ) , in  Ω × ( 0 , T ) ,

where Ω R n ( n = 1 , 2 , 3 ) is a bounded domain with smooth boundary. They considered the following assumptions on f i ( i = 1 , 2 ):

(A1) Let
F ( u , v ) = a | u + v | p + 1 + 2 b | u v | p + 1 2 , f 1 ( u , v ) = F u , f 2 ( u , v ) = F v

with a , b > 0 , p 3 if n = 1 , 2 and p = 3 if n = 3 ; m , r 1 .

(A2) There exist two positive constants c 0 , c 1 such that for all u , v R 2 , F ( u , v ) satisfies
c 0 ( | u | p + 1 + | v | p + 1 ) F ( u , v ) c 1 ( | u | p + 1 + | v | p + 1 ) .
Under the assumptions (A1)-(A2), they established the global existence of weak solutions and the global existence of small weak solutions with initial and Dirichlet boundary conditions. Moreover, they also obtained the blow up of weak solutions. Mustafa [16] studied the following system:
{ u t t Δ u + 0 t g 1 ( t τ ) Δ u ( τ ) d τ + f 1 ( u , v ) = 0 , v t t Δ v + 0 t g 2 ( t τ ) Δ v ( τ ) d τ + f 2 ( u , v ) = 0 ,
(1.3)

in Ω × ( 0 , + ) with initial and Dirichlet boundary conditions, proved the existence and uniqueness to the system by using the classical Faedo-Galerkin method and established a stability result by multiplier techniques. But the author considered the following different assumptions on f i ( i = 1 , 2 ) from (A1)-(A2):

( A 1 ) f i : R 2 R ( i = 1 , 2 ) are C 1 functions and there exists a function F such that
f 1 ( x , y ) = F x , f 2 ( x , y ) = F y , F 0 , x f 1 ( x , y ) + y f 2 ( x , y ) F ( x , y ) ,
( A 2 )
| f i x ( x , y ) | + | f i y ( x , y ) | d ( 1 + | x | β i 1 1 + | y | β i 2 1 ) ,

for all ( x , y ) R 2 , where the constant d > 0 and β i j 1 , ( n 2 ) β i j n for i , j = 1 , 2 .

Han and Wang [17] considered the following coupled nonlinear viscoelastic wave equations with weak damping:
{ u t t Δ u + 0 t g 1 ( t τ ) Δ u ( τ ) d τ + | u t | m 1 u t = f 1 ( u , v ) , in  Ω × ( 0 , T ) , v t t Δ v + 0 t g 2 ( t τ ) Δ v ( τ ) d τ + | v t | r 1 v t = f 2 ( u , v ) , in  Ω × ( 0 , T ) , u = v = 0 , on  Ω × ( 0 , T ) , u ( , 0 ) = u 0 , u t ( , 0 ) = u 1 , v ( , 0 ) = v 0 , v t ( , 0 ) = v 1 , in  Ω ,
(1.4)
where Ω R n is a bounded domain with smooth boundary Ω. Under the assumptions (A1)-(A2) on f i ( i = 1 , 2 ), the initial data and the parameters in the equations, they established the local existence, global existence uniqueness and finite time blow up properties. When the weak damping terms | u t | m 1 u t , | v t | r 1 v t were replaced by the strong damping terms Δ u t , Δ v t , Liang and Gao [18] showed that under certain assumption on initial data in the stable set, the decay rate of the solution energy is exponential when they take
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equg_HTML.gif

a , b > 0 and p > 1 if n = 1 , 2 , 1 < p 1 if n = 3 . Moreover, they obtained that the solutions with positive initial energy blow up in a finite time for certain initial data in the unstable set. For more results on coupled viscoelastic equations, we can refer to [1921].

If we take m = r = 1 in (1.4), the system will be transformed into (1.1). To the best of our knowledge, there is no result on general energy decay for the viscoelastic problem (1.1). Motivated by [16, 17], in this paper, we shall establish the general energy decay for the problem (1.1) by multiplier techniques, which extends some existing results for a single equation to the case of a coupled system. The rest of our paper is organized as follows. In Section 2, we give some preparations for our consideration and our main result. The statement and the proof of our main result will be given in Section 3.

For the reader’s convenience, we denote the norm and the scalar product in L 2 ( Ω ) by and ( , ) , respectively. C 1 denotes a general constant, which may be different in different estimates.

2 Preliminaries and main result

To state our main result, in addition to (A1)-(A2), we need the following assumption.

(A3) g i : R + R + , i = 1 , 2 , are differentiable functions such that
g i ( 0 ) > 0 , 1 0 + g i ( s ) d s = l i > 0 ,
and there exist nonincreasing functions ξ 1 , ξ 2 : R + R + satisfying
g i ( t ) ξ i ( t ) g i ( t ) , t 0 .
Now, we define the energy functional
E ( t ) = 1 2 Ω ( u t 2 + ( 1 0 t g 1 ( s ) d s ) | u | 2 ) d x + 1 2 ( g 1 u ) ( t ) + 1 2 ( g 2 v ) ( t ) + 1 2 Ω ( v t 2 + ( 1 0 t g 2 ( s ) d s ) | v | 2 ) d x Ω F ( u , v ) d x
(2.1)
and the functional
D ( t ) = ( 1 0 t g 1 ( s ) d s ) u ( t ) 2 + ( 1 0 t g 2 ( s ) d s ) v ( t ) 2 + 2 [ ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ] 4 Ω F ( u ( t ) , v ( t ) ) d x ,
(2.2)
where
( g y ) ( t ) = 0 t g ( t s ) y ( t ) y ( s ) 2 d s .

The existence of a global solution to the system (1.1) is established in [17] as follows.

Proposition [17]

Let (A1)-(A3) hold. Assume that D ( 0 ) = u 0 2 + v 0 2 4 Ω F ( u 0 , v 0 ) d x > 0 , 2 p C 0 l ( E ( 0 ) l ) p 1 2 < 1 and that ( u 0 , u 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) , ( v 0 , v 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) , where C 0 is a computable constant and l = min { l 1 , l 2 } . Then the problem (1.1) has a unique global solution ( u ( t ) , v ( t ) ) satisfying
( u ( t ) , u t ( t ) ) C ( R + ; H 0 1 ( Ω ) × L 2 ( Ω ) ) , ( v ( t ) , v t ( t ) ) C ( R + ; H 0 1 ( Ω ) × L 2 ( Ω ) ) .

We are now ready to state our main result.

Theorem 2.1 Let (A1)-(A3) hold. Assume that D ( 0 ) = u 0 2 + v 0 2 4 Ω F ( u 0 , v 0 ) d x > 0 , 2 p C 0 l ( E ( 0 ) l ) p 1 2 < 1 and that ( u 0 , u 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) , ( v 0 , v 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) , where C 0 is a computable constant and l = min { l 1 , l 2 } . Then there exist constants C , η > 0 such that, for t large, the solution of (1.1) satisfies
E ( t ) C e η 0 t ξ ( s ) d s ,
(2.3)
where
ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) } , t 0 .
(2.4)

3 Proof of Theorem 2.1

In this section, we carry out the proof of Theorem 2.1. Firstly, we will estimate several lemmas.

Lemma 3.1 Let u ( t ) , v ( t ) be the solution of (1.1). Then the following energy estimate holds for any t 0 :
E ( t ) = ( u t 2 + v t 2 ) + 1 2 [ ( g 1 u ) + ( g 2 v ) ] 1 2 [ g 1 ( t ) u ( t ) 2 + g 2 ( t ) v ( t ) 2 ] 0 .
(3.1)

Proof Multiplying the first equation of (1.1) by u t and the second equation by v t , respectively, integrating the results over Ω, performing integration by parts and noting that F t ( u , v ) = f 1 ( u , v ) u t + f 2 ( u , v ) v t , we can easily get (3.1). The proof is complete. □

Lemma 3.2 Under the assumption (A3), the following hold:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ10_HTML.gif
(3.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ11_HTML.gif
(3.3)
Proof Using Hölder’s inequality, we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equl_HTML.gif

On the other hand, we repeat the above proof with g , instead of g, we can get (3.3). The proof is now complete. □

Lemma 3.3 Let (A1)-(A3) hold and u ( t ) , v ( t ) be the solution of (1.1). Then the functional I ( t ) defined by
I ( t ) : = Ω ( u u t + v v t ) d x
satisfies
I ( t ) l 1 2 u ( t ) 2 l 2 2 v ( t ) 2 + ( 1 + 1 4 δ ) ( u t 2 + v t 2 ) + C 1 δ ( g 1 u ) + C 1 δ ( g 2 v ) + C 1 Ω F ( u , v ) d x
(3.4)

for all δ > 0 .

Proof By (1.1), a direct differentiation gives
I ( t ) = u t 2 u 2 + Ω u 0 t g 1 ( t τ ) u ( τ ) d τ d x Ω u t u d x + Ω f 1 u d x + v t 2 v 2 + Ω v 0 t g 2 ( t τ ) v ( τ ) d τ d x Ω v t v d x + Ω f 2 v d x .
(3.5)
From the assumptions (A1)-(A2), we derive
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equn_HTML.gif
and
f 1 u + f 2 v = a ( p + 1 ) | u + v | p + 1 + b ( p + 1 ) | u v | p + 1 2 C 1 F ( u , v ) .
(3.6)
By Young’s inequality and (3.2), we deduce for any δ > 0
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ15_HTML.gif
(3.7)
Similarly, we have
Ω v 0 t g 2 ( t τ ) v ( τ ) d τ d x v 2 0 t g 2 ( τ ) d τ + δ v 2 + C 1 4 δ ( g 2 v ) .
(3.8)
Using Young’s inequality and Poincaré’s inequality, we obtain for any δ > 0
Ω u u t d x δ u 2 + 1 4 δ u t 2 δ λ 2 u 2 + 1 4 δ u t 2 ,
(3.9)
where λ is the first eigenvalue of −Δ with the Dirichlet boundary condition. Similarly,
Ω v v t d x δ v 2 + 1 4 δ v t 2 δ λ 2 v 2 + 1 4 δ v t 2 ,
which together with (3.5)-(3.9) gives
I ( t ) ( l 1 δ δ λ 2 ) u 2 ( l 2 δ δ λ 2 ) v 2 + ( 1 + 1 4 δ ) ( u t 2 + v t 2 ) + C 1 4 δ ( g 1 u ) + C 1 4 δ ( g 2 v ) + C 1 Ω F ( u , v ) d x .
(3.10)
Now, we choose δ > 0 so small that
l 1 δ δ λ 2 l 1 2 , l 2 δ δ λ 2 l 2 2 ,

which together with (3.10) gives (3.4). The proof is complete. □

Lemma 3.4 Let (A1)-(A3) hold and u ( t ) , v ( t ) be the solution of (1.1). Then the functional J ( t ) defined by
J ( t ) = J 1 ( t ) + J 2 ( t ) ,
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equr_HTML.gif
satisfies
J ( t ) ( 0 t g 1 ( τ ) 2 δ ) u t 2 + δ C 1 u 2 + C 1 δ ( g 1 u ) C 1 δ ( g 1 u ) ( 0 t g 2 ( τ ) 2 δ ) v t 2 + δ C 1 v 2 + C 1 δ ( g 2 v ) C 1 δ ( g 2 v ) .
(3.11)
Proof A direct differentiation for J 1 ( t ) yields
J 1 ( t ) = Ω u t t 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ Ω u t 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ d x ( 0 t g 1 ( τ ) d τ ) Ω u t 2 d x .
(3.12)
Using the first equation of (1.1) and integrating by parts, we obtain
J 1 ( t ) = ( 1 0 t g 1 ( τ ) d τ ) Ω u 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ d x + Ω ( 0 t g 1 ( t τ ) | u ( t ) u ( τ ) | d τ ) 2 d x + Ω u t 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ d x Ω f 1 ( u , v ) 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ d x Ω u t 0 t g 1 ( t τ ) ( u ( t ) u ( τ ) ) d τ d x ( 0 t g 1 ( τ ) d τ ) Ω u t 2 d x .
(3.13)
From Young’s inequality, Poincaré’s inequality and Lemma 3.2, we derive
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ22_HTML.gif
(3.14)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ23_HTML.gif
(3.15)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ24_HTML.gif
(3.16)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ25_HTML.gif
(3.17)
Now, we estimate the first term on the right-hand side of (3.17). Using the assumptions (A1)-(A2) and Young’s inequality, we arrive at
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ26_HTML.gif
(3.18)
where we used the embedding H 0 1 ( Ω ) L s ( Ω ) for 2 s 2 n / ( n 2 ) if n = 3 or s 2 if n = 1 , 2 and the fact 1 2 ( u t 2 + v t 2 ) + 1 4 l 1 u 2 + 1 4 l 2 v 2 2 E ( 0 ) proved in Lemma 5.1 in [17]. Combining (3.13)-(3.18), we get
J 1 ( t ) ( 0 t g 1 ( τ ) d τ 2 δ ) u t 2 + δ C 1 u 2 + δ C 1 v 2 + C 1 δ ( g 1 u ) C 1 δ ( g 1 u ) .
(3.19)
The same estimate to J 2 ( t ) , we can derive
J 2 ( t ) ( 0 t g 2 ( τ ) d τ 2 δ ) v t 2 + δ C 1 u 2 + δ C 1 v 2 + C 1 δ ( g 2 v ) C 1 δ ( g 2 v ) ,

which together with (3.19) gives (3.11). The proof is now complete. □

Proof of Theorem 2.1 For N 1 , N 2 > 0 , we define the functional K by
K : = N 1 E ( t ) + N 2 J ( t ) + I ( t ) ,
and let
g 0 = min { 0 t 0 g 1 ( s ) d s , 0 t 0 g 2 ( s ) d s }

for some fixed t 0 > 0 .

Using Lemma 3.1 and Lemmas 3.3-3.4, a direct differentiation gives
K ( t ) ( l 2 N 2 δ C 1 ) ( u 2 + v 2 ) + ( C 1 δ + N 2 C 1 δ ) [ ( g 1 u ) + ( g 2 v ) ] ( N 1 + N 2 2 δ 1 1 4 δ ) ( u t 2 + v t 2 ) + C 1 Ω F ( u , v ) d x + ( N 1 2 N 2 C 1 δ ) [ ( g 1 u ) + ( g 2 v ) ] ,
(3.20)

where l = min { l 1 , l 2 } .

Now, we choose δ = 1 4 C 1 N 2 and N 1 , N 2 large enough so that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ29_HTML.gif
(3.21)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ30_HTML.gif
(3.22)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-146/MediaObjects/13661_2012_Article_238_Equ31_HTML.gif
(3.23)
Inserting (3.21)-(3.23) into (3.20), we have
K ( t ) c 1 ( u 2 + v 2 ) c 2 ( u t 2 + v t 2 ) + c 3 [ ( g 1 u ) + ( g 2 v ) ] + ( 4 C 1 2 N 2 l + 4 C 1 2 N 2 l ) [ ( g 1 u ) + ( g 2 v ) ] + C 1 Ω F ( u , v ) d x .
(3.24)
Therefore, for two positive constants ω and C, we obtain
K ( t ) ω E ( t ) + C [ ( g 1 u ) + ( g 2 v ) ] , for all  t t 0 .
(3.25)
On the other hand, we choose N 1 even larger so that K ( t ) is equivalent to E ( t ) , i.e.,
K ( t ) E ( t ) .
(3.26)
Multiplying (3.25) by ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) } and using (A3), we get
ξ ( t ) K ( t ) ω ξ ( t ) E ( t ) + C Ω 0 t ξ 1 ( t τ ) g 1 ( t τ ) | u ( t ) u ( τ ) | 2 d τ d x + C Ω 0 t ξ 2 ( t τ ) g 2 ( t τ ) | v ( t ) v ( τ ) | 2 d τ d x ω ξ ( t ) E ( t ) C Ω 0 t g 1 ( t τ ) | u ( t ) u ( τ ) | 2 d τ d x C Ω 0 t g 2 ( t τ ) | v ( t ) v ( τ ) | 2 d τ d x ω ξ ( t ) E ( t ) C E ( t ) , for all  t t 0 .
(3.27)
By virtue of (A3) and ξ ( t ) 0 , we have
d d t ( ξ ( t ) K ( t ) + C E ( t ) ) ω ξ ( t ) E ( t ) , for all  t t 0 .
(3.28)
Using (3.26), we can easily get
L ( t ) : = ξ ( t ) K ( t ) + C E ( t ) E ( t ) ,
(3.29)
which together with (3.28) yields, for some positive constant η,
L ( t ) η ξ ( t ) L ( t ) , for all  t t 0 .
(3.30)
Integrating (3.30) over ( t 0 , t ) , we arrive at
L ( t ) L ( t 0 ) e η t t 0 ξ ( τ ) d τ C e η t t 0 ξ ( τ ) d τ ,

which together with (3.29) and the boundedness of E and ξ yields (2.3). The proof is now complete. □

Declarations

Acknowledgements

Baowei Feng was supported by the Doctoral Innovational Fund of Donghua University with contract number BC201138, and Yuming Qin was supported by NNSF of China with contract numbers 11031003 and 11271066 and the grant of Shanghai Education Commission (No. 13ZZ048).

Authors’ Affiliations

(1)
College of Information Science and Technology, Donghua University
(2)
Department of Applied Mathematics, Donghua University

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