Open Access

Asymptotic behaviour of solution for multidimensional viscoelasticity equation with nonlinear source term

Boundary Value Problems20132013:42

DOI: 10.1186/1687-2770-2013-42

Received: 24 April 2012

Accepted: 11 February 2013

Published: 1 March 2013

Abstract

In this paper we study the initial-boundary value problem of the multidimensional viscoelasticity equation with nonlinear source term u t t Δ u t i = 1 N x i σ i ( u x i ) = f ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq1_HTML.gif. By using the potential well method, we first prove the global existence. Then we prove that when time t + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq2_HTML.gif, the solution decays to zero exponentially under some assumptions on nonlinear functions and the initial data.

1 Introduction

This paper considers the initial-boundary value problem (IBVP) of the multidimensional viscoelasticity equation with nonlinear source term
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ1_HTML.gif
(1.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ2_HTML.gif
(1.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ3_HTML.gif
(1.3)

where u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq3_HTML.gif is the unknown function with respect to the spacial variable x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq4_HTML.gif and the time variable t , Ω R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq5_HTML.gif is a bounded domain.

The viscoelasticity equation
u t t u x x t = σ ( u x ) x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ4_HTML.gif
(1.4)

was suggested and studied by Greenberg et al. [1, 2] from viscoelasticity mechanics in 1968. Under the condition σ ( s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq6_HTML.gif and higher smooth conditions on σ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq7_HTML.gif and the initial data, they obtained the global existence of classical solutions for the initial-boundary value problem of Eq. (1.4).

After that many authors [311] studied the global well-posedness of IBVP for Eq. (1.4). In [310] the global existence, uniqueness and stability of solution were studied thoroughly. And in [11] the blow up of solution was discussed. Furthermore, in [1214] the global existence of solution for IBVP of some multidimensional viscoelasticity equation was considered. And in [11] the blow up of solution for IBVP of the multidimensional generalisation of Eq. (1.4) was proved. Recently, in [15] and [16], the IBVP of the multidimensional viscoelasticity equation with nonlinear damping and source terms
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ5_HTML.gif
(1.5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ6_HTML.gif
(1.6)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ7_HTML.gif
(1.7)

was studied, and by using the potential well method, the global existence of weak solution was proved under some assumptions on nonlinear functions σ i ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq8_HTML.gif, f ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq9_HTML.gif, g ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq10_HTML.gif and the initial data. But we do not know how the global solution behaves as the time goes to infinity, namely the asymptotic behaviour of problem (1.1)-(1.3) is still open up to now. In the present paper, we try to study this problem by the multiplier method [1722].

The main purpose of present paper is to consider the asymptotic behaviour of solution for problem (1.1)-(1.3). Since in the proof of the asymptotic behaviour of solution the global existence theory is required, it is necessary to give the proof of global existence of solution for problem (1.1)-(1.3).

In this paper, suppose that σ ( s ) = ( σ 1 ( s ) , , σ N ( s ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq11_HTML.gif and f ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq9_HTML.gif satisfy the following assumptions:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equa_HTML.gif
where the constants in (H1) and (H2) are all positive and satisfy
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equb_HTML.gif

In this paper, we first give some definitions and lemmas (Section 2). Then we prove the global existence of solution (Section 3). Finally, we prove the asymptotic behaviour of solution (Section 4).

In this paper, we denote L p ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq12_HTML.gif by p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq13_HTML.gif, = L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq14_HTML.gif and ( u , v ) = Ω u v d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq15_HTML.gif.

2 Preliminaries

In this section, we will give some definitions and prove some lemmas for problem (1.1)-(1.3).

For problem (1.1)-(1.3), we define
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equc_HTML.gif

Remark 2.1 Note that the definitions of J ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq16_HTML.gif and I ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq17_HTML.gif in the present paper are different from those in [11] and [15]. The definitions given in this paper will be shown more natural and rational because they are a part of the total energy E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq18_HTML.gif.

Lemma 2.2 Let (H1) and (H2) hold. Set
σ i ¯ ( s ) = σ i ( s ) a s , G i ¯ ( s ) = 0 s σ i ¯ ( s ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equd_HTML.gif
Then the following hold:
  1. (i)

    σ i ¯ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq19_HTML.gif is increasing and s σ i ¯ ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq20_HTML.gif s R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq21_HTML.gif;

     
  2. (ii)

    0 G i ¯ ( s ) s σ i ¯ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq22_HTML.gif s R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq21_HTML.gif.

     

Proof This lemma follows from σ i ¯ ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq23_HTML.gif and σ ¯ i ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq24_HTML.gif. □

Lemma 2.3 Let (H1) and (H2) hold, u W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq25_HTML.gif. Then the following hold:
  1. (i)

    If 0 < u m + 1 < r ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq26_HTML.gif, then I δ ( u ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq27_HTML.gif;

     
  2. (ii)

    If I δ ( u ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq28_HTML.gif, then u m + 1 > r ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq29_HTML.gif;

     
  3. (iii)

    If I δ ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq30_HTML.gif, then u m + 1 r ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq31_HTML.gif,

     
where
r ( δ ) = ( A δ b C q + 1 ) 1 q m , C = sup u W 0 1 , m + 1 ( Ω ) / 0 u q + 1 u m + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Eque_HTML.gif
Proof
  1. (i)
    If 0 < u m + 1 < r ( δ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq26_HTML.gif, then we have
    Ω u f ( u ) d x Ω | u f ( u ) | d x b Ω | u | q + 1 d x = b u q + 1 q + 1 b C q + 1 u m + 1 q + 1 = b C q + 1 A u m + 1 q m A u m + 1 m + 1 < δ i = 1 N Ω u x i σ i ( u x i ) d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equf_HTML.gif
     
which gives I δ ( u ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq27_HTML.gif.
  1. (ii)
    If I δ ( u ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq28_HTML.gif, then we have
    δ A u m + 1 m + 1 δ i = 1 N Ω u x i σ i ( u x i ) d x < Ω u f ( u ) d x b C q + 1 u q m m + 1 u m + 1 m + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equg_HTML.gif
     
which gives
u m + 1 > r ( δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equh_HTML.gif
  1. (iii)
    If I δ ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq30_HTML.gif and u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq32_HTML.gif, then by
    δ A u m + 1 m + 1 δ i = 1 N Ω u x i σ i ( u x i ) d x < Ω u f ( u ) d x b C q + 1 u m + 1 q m u m + 1 m + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equi_HTML.gif
     
we get
u m + 1 r ( δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equj_HTML.gif

 □

Lemma 2.4 Let (H1) and (H2) hold. Then the following holds:
d d 0 = ( p l ) A ( p + 1 ) ( l + 1 ) ( A b C q + 1 ) m + 1 q m . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ8_HTML.gif
(2.1)
Proof For any u N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq33_HTML.gif, by Lemma 2.3, we have u m + 1 r ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq34_HTML.gif and
J ( u ) = i = 1 N Ω G i ( u x i ) d x Ω F ( u ) d x 1 l + 1 i = 1 N Ω u x i σ i ( u x i ) d x 1 p + 1 Ω u f ( u ) d x = ( 1 l + 1 1 p + 1 ) i = 1 N Ω u x i σ i ( u x i ) d x + 1 p + 1 I ( u ) = p l ( p + 1 ) ( l + 1 ) i = 1 N Ω u x i σ i ( u x i ) d x p l ( p + 1 ) ( l + 1 ) A u m + 1 m + 1 p l ( p + 1 ) ( l + 1 ) A r m + 1 ( 1 ) = ( p l ) A ( p + 1 ) ( l + 1 ) ( A b C q + 1 ) m + 1 q m , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equk_HTML.gif

which gives (2.1). □

Now, for problem (1.1)-(1.3), we define
W = { u W 0 1 , m + 1 ( Ω ) I ( u ) > 0 } { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equl_HTML.gif

3 Global existence of solution

In this section, we prove the global existence of weak solution for problem (1.1)-(1.3).

Definition 3.1 We call u = u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq35_HTML.gif a weak solution of problem (1.1)-(1.3) on Ω × [ 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq36_HTML.gif if u L ( 0 , T ; W 0 1 , m + 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq37_HTML.gif, u t L ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq38_HTML.gif satisfying
  1. (i)
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equm_HTML.gif
     
  2. (ii)
    u ( x , 0 ) = u 0 ( x ) in  W 0 1 , m + 1 ( Ω ) ; u t ( x , 0 ) = u 1 ( x ) in  L 2 ( Ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equn_HTML.gif
     

Theorem 3.2 Let (H1) and (H2) hold, u 0 ( x ) W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq39_HTML.gif, u 1 ( x ) L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq40_HTML.gif. Assume that E ( 0 ) < d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq41_HTML.gif, u 0 ( x ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq42_HTML.gif. Then problem (1.1)-(1.3) admits a global weak solution u L ( 0 , ; W 0 1 , m + 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq43_HTML.gif and u t L ( 0 , ; L 2 ( Ω ) ) L 2 ( 0 , ; H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq44_HTML.gif.

Proof Let { w j ( x ) } j = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq45_HTML.gif be a system of base functions in W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq46_HTML.gif. Construct the approximate solutions of problem (1.1)-(1.3)
u n ( x , t ) = j = 1 n g j n ( t ) w j ( x ) , n = 1 , 2 , , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equo_HTML.gif
satisfying
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ9_HTML.gif
(3.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ10_HTML.gif
(3.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ11_HTML.gif
(3.3)
Multiplying (3.1) by g s n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq47_HTML.gif and summing for s, we get
d d t E n ( t ) + u n t 2 = 0 , 0 t < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ12_HTML.gif
(3.4)
and
E n ( t ) + 0 t u n τ 2 d τ = E n ( 0 ) , 0 t < , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ13_HTML.gif
(3.5)
where
E n ( t ) = 1 2 u n t 2 + J ( u n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equp_HTML.gif
From (3.2) and (3.3), we have E n ( 0 ) E ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq48_HTML.gif as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq49_HTML.gif. Hence, for sufficiently large n, we have E n ( 0 ) < d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq50_HTML.gif and
1 2 u n t 2 + J ( u n ) + 0 t u n τ 2 d τ < d , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ14_HTML.gif
(3.6)

On the other hand, since W is an open set in W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq46_HTML.gif, Eq. (3.2) implies that for sufficiently large n, we have u n ( 0 ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq51_HTML.gif. Next, we prove that u n ( t ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq52_HTML.gif for 0 < t < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq53_HTML.gif and sufficiently large n. If it is false, then there exists a t 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq54_HTML.gif such that u n ( t 0 ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq55_HTML.gif, i.e. I ( u n ( t 0 ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq56_HTML.gif and u n ( t 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq57_HTML.gif, i.e. u n ( t 0 ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq58_HTML.gif. So, by the definition of d, we get J ( u n ( t 0 ) ) d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq59_HTML.gif, which contradicts (3.6).

From (3.6) we have
1 2 u n t 2 + i = 1 N Ω G i ( u n x i ) d x Ω F ( u n ) d x + 0 t u n τ 2 d τ < d , 0 t < , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equq_HTML.gif
which gives
1 2 u n t 2 + 1 l + 1 i = 1 N Ω u n x i σ i ( u n x i ) d x 1 p + 1 Ω u n f ( u n ) d x + 0 t u n τ 2 d τ < d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equr_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equs_HTML.gif
which together with u n ( t ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq52_HTML.gif gives
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x + 0 t u n τ 2 d τ < d , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equt_HTML.gif
and
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) A u n m + 1 m + 1 + 0 t u n τ 2 d τ < d , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ15_HTML.gif
(3.7)
From (3.7) we can get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ16_HTML.gif
(3.8)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ17_HTML.gif
(3.9)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ18_HTML.gif
(3.10)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ19_HTML.gif
(3.11)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ20_HTML.gif
(3.12)

Hence there exist u, χ = ( χ 1 , χ 2 , , χ N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq60_HTML.gif, η and a subsequence { u ν } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq61_HTML.gif of { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq62_HTML.gif such that as ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq63_HTML.gif, u ν u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq64_HTML.gif in u L ( 0 , ; W 0 1 , m + 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq65_HTML.gif weak-star, and a.e. in Q = Ω × [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq66_HTML.gif, u ν t u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq67_HTML.gif in L ( 0 , ; L 2 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq68_HTML.gif weak-star and in L 2 ( 0 , ; H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq69_HTML.gif weakly, σ i ( u ν x i ) χ i = σ i ( u x i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq70_HTML.gif in L ( 0 , ; L ( m + 1 ) ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq71_HTML.gif weak-star, ( m + 1 ) = m + 1 m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq72_HTML.gif, f ( u v ) η = f ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq73_HTML.gif in L ( 0 , ; L ( q + 1 ) ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq74_HTML.gif weak-star, ( q + 1 ) = q + 1 q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq75_HTML.gif.

Integrating (3.1) with respect to t, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ21_HTML.gif
(3.13)
Letting n = ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq76_HTML.gif in (3.13), we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equu_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equv_HTML.gif

On the other hand, from (3.2) and (3.3), we get u ( x , 0 ) = u 0 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq77_HTML.gif in W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq46_HTML.gif, u t ( x , 0 ) = u 1 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq78_HTML.gif in L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq79_HTML.gif. Therefore u is a global weak solution of problem (1.1)-(1.3). □

4 Asymptotic behaviour of solution

In this section, we prove the main conclusion of this paper - the asymptotic behaviour of solution for problem (1.1)-(1.3).

Lemma 4.1 Let (H1) and (H2) hold, u 0 ( x ) W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq39_HTML.gif, u 1 ( x ) L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq40_HTML.gif. Then, for the approximate solutions u n ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq80_HTML.gif of problem (1.1)-(1.3) constructed in the proof of Theorem 3.2, the following hold:
  1. (i)
    I ( u n ) = u n t 2 d d t ( ( u n t , u n ) + 1 2 u n 2 ) ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ22_HTML.gif
    (4.1)
     
  2. (ii)
    Furthermore, if E ( 0 ) < d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq81_HTML.gif and u 0 ( x ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq42_HTML.gif, then for sufficiently large n, there exists a δ 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq82_HTML.gif such that
    I ( u n ) ( 1 δ 1 ) i = 1 N ( σ i ( u n x i ) , u n x i ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ23_HTML.gif
    (4.2)
     
Proof (i) Multiplying (3.1) by g s n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq83_HTML.gif and summing for s, we get (4.1).
  1. (ii)
    From
    E ( 0 ) < d 0 = p l ( p + 1 ) ( l + 1 ) A ( A b C q + 1 ) m + 1 q m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equw_HTML.gif
     
it follows that there exists a δ 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq82_HTML.gif such that
E ( 0 ) < p l ( p + 1 ) ( l + 1 ) A ( A δ 1 b C q + 1 ) m + 1 q m d ( δ 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ24_HTML.gif
(4.3)
From (3.2), (3.3) and (4.3), it follows that E m ( 0 ) < d ( δ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq84_HTML.gif for sufficiently large n. Hence from (3.5) we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equx_HTML.gif
and
i = 1 N Ω G i ( u n x i ) d x Ω F ( u n ) d x < d ( δ 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equy_HTML.gif
which gives
1 l + 1 i = 1 N Ω u n x i σ i ( u n x i ) d x 1 p + 1 Ω u n f ( u n ) d x < d ( δ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equz_HTML.gif
and
p l ( p + 1 ) ( l + 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x + 1 p + 1 I ( u n ) < d ( δ 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equaa_HTML.gif
which together with u n W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq85_HTML.gif for sufficiently large n gives
p l ( p + 1 ) ( l + 1 ) A u n m + 1 m + 1 < p l ( p + 1 ) ( l + 1 ) A ( A δ 1 b C q + 1 ) m + 1 q m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equab_HTML.gif
and
u n m + 1 < ( A δ 1 b C q + 1 ) 1 q m = r ( δ 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equac_HTML.gif
Hence, by Lemma 2.3, we have I δ 1 ( u n ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq86_HTML.gif or u n = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq87_HTML.gif. So, we have
I ( u n ) = i = 1 N Ω u n x i σ i ( u n x i ) d x Ω u n f ( u n ) d x = ( 1 δ 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x + I δ 1 ( u n ) ( 1 δ 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equad_HTML.gif

 □

Theorem 4.2 Let (H1) and (H2) hold, u 0 ( x ) W 0 1 , m + 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq39_HTML.gif, u 1 ( x ) L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq40_HTML.gif. Assume that E ( 0 ) < d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq81_HTML.gif, u 0 ( x ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq42_HTML.gif. Then, for the global weak solution u given in Theorem 3.2, there exist positive constants C and λ such that
u t 2 + u m + 1 m + 1 C e λ t , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ25_HTML.gif
(4.4)
Proof Let { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq62_HTML.gif be the approximate solutions of problem (1.1)-(1.3) in the proof of Theorem 3.2, then (3.4) holds. Multiplying (3.4) by e α t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq88_HTML.gif ( α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq89_HTML.gif), we get
d d t ( e α t E n ( t ) ) + e α t u n t 2 = α e α t E n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equae_HTML.gif
and
e α t E n ( t ) + 0 t e α τ u n τ 2 d τ = E n ( 0 ) + α 0 t e α τ E n ( τ ) d τ , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ26_HTML.gif
(4.5)
From (H2), Lemma 2.2 and Lemma 4.1, we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equaf_HTML.gif
where
C ( r , δ 1 ) = 1 1 δ 1 r r + 1 + 1 r + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equag_HTML.gif
Hence we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ27_HTML.gif
(4.6)
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ28_HTML.gif
(4.7)
From
1 2 u n t 2 + i = 1 N Ω G i ( u n x i ) d x Ω F ( u n ) d x = E n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equah_HTML.gif
we get
1 2 u n t 2 + 1 l + 1 i = 1 N Ω u n x i σ i ( u n x i ) d x 1 p + 1 Ω u n f ( u n ) d x E n ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equai_HTML.gif
and
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x + 1 p + 1 I ( u n ) E n ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equaj_HTML.gif
which together with u n W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq85_HTML.gif for sufficiently large n gives
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) i = 1 N Ω u n x i σ i ( u n x i ) d x E n ( t ) , 1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) ( a u n 2 + i = 1 N Ω u n x i σ ¯ i ( u n x i ) d x ) E n ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ29_HTML.gif
(4.8)
and
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) a u n 2 E n ( t ) , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ30_HTML.gif
(4.9)
From (4.8) and the Poincaré inequality u 2 λ 1 u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq90_HTML.gif, it follows that there exists a constant C 0 = C 0 ( p , l , a , λ 1 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq91_HTML.gif such that
( u n t 2 + u n 2 + u n 2 ) C 0 E n ( t ) , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ31_HTML.gif
(4.10)
From (4.5)-(4.10) it follows that there exists a C 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq92_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ32_HTML.gif
(4.11)
Choose α such that
0 < α < min { 1 2 C 0 , λ 1 1 2 + C ( r , δ 1 ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equak_HTML.gif
Then from (4.11) we get
e α t E n ( t ) 2 ( C 0 α + 1 ) E n ( 0 ) + 2 α 2 C 0 0 t e α τ E n ( τ ) d τ 2 ( C 0 α + 1 ) d 0 + 2 α 2 C 0 0 t e α τ E n ( τ ) d τ 3 d 0 + 2 α C 0 0 t e α τ E n ( τ ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equal_HTML.gif
From this and the Gronwall inequality, we get
e α t E n ( t ) 3 d 0 e 2 α 2 C 0 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equam_HTML.gif
and
E n ( t ) 3 d 0 e λ t , 0 t < , λ = α ( 1 2 C 0 α ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ33_HTML.gif
(4.12)
On the other hand, from (4.8) we get
1 2 u n t 2 + p l ( p + 1 ) ( l + 1 ) A u n m + 1 m + 1 E n ( t ) , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equan_HTML.gif
Hence, there exists a C 1 = C 1 ( p , l , A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq93_HTML.gif such that
u n t 2 + u n m + 1 m + 1 C 1 E n ( t ) , 0 t < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equ34_HTML.gif
(4.13)
Let { u ν } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq61_HTML.gif be the subsequence of { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq62_HTML.gif in the proof of Theorem 3.2. Then from (4.13) and(4.12), we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_Equao_HTML.gif

which gives (4.4), where C = 3 d 0 C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq94_HTML.gif, λ = α ( 1 2 C 0 α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-42/MediaObjects/13661_2012_Article_305_IEq95_HTML.gif. □

Declarations

Acknowledgements

We thank the referees for their valuable suggestions which helped us improve the paper so much. This work was supported by the National Natural Science Foundation of China (11101102), Ph.D. Programs Foundation of the Ministry of Education of China (20102304120022), the Support Plan for the Young College Academic Backbone of Heilongjiang Province (1252G020), the Natural Science Foundation of Heilongjiang Province (A201014), Science and Technology Research Project of the Department of Education of Heilongjiang Province (12521401), Foundational Science Foundation of Harbin Engineering University and Fundamental Research Funds for the Central Universities.

Authors’ Affiliations

(1)
College of Science, Harbin Engineering University
(2)
College of Automation, Harbin Engineering University
(3)
Department of Mathematics, Cape Breton University

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