Arbitrary decays for a viscoelastic equation

Boundary Value Problems20112011:28

DOI: 10.1186/1687-2770-2011-28

Received: 16 February 2011

Accepted: 6 October 2011

Published: 6 October 2011

Abstract

In this paper, we consider the nonlinear viscoelastic equation u t ρ u t t - Δ u - Δ u t t + 0 t g ( t - s ) Δ u ( s ) d s + u p u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq1_HTML.gif, in a bounded domain with initial conditions and Dirichlet boundary conditions. We prove an arbitrary decay result for a class of kernel function g without setting the function g itself to be of exponential (polynomial) type, which is a necessary condition for the exponential (polynomial) decay of the solution energy for the viscoelastic problem. The key ingredient in the proof is based on the idea of Pata (Q Appl Math 64:499-513, 2006) and the work of Tatar (J Math Phys 52:013502, 2010), with necessary modification imposed by our problem.

Mathematical Subject Classification (2010): 35B35, 35B40, 35B60

Keywords

Viscoelastic equation Kernel function Exponential decay Polynomial decay

1 Introduction

It is well known that viscoelastic materials have memory effects. These properties are due to the mechanical response influenced by the history of the materials themselves. As these materials have a wide application in the natural sciences, their dynamics are of great importance and interest. From the mathematical point of view, their memory effects are modeled by an integro-differential equations. Hence, questions related to the behavior of the solutions for the PDE system have attracted considerable attention in recent years. Many authors have focused on this problem for the last two decades and several results concerning existence, decay and blow-up have been obtained, see [128] and the reference therein.

In [3], Cavalcanti et al. studied the following problem
u t ρ u t t - Δ u - Δ u t t + 0 t g ( t - s ) Δ u ( s ) d s - γ Δ u t = 0 , in  Ω × ( 0 , ) , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω , u ( x , t ) = 0 , x Ω , t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ1_HTML.gif
(1.1)
where Ω ⊂ R N , N ≥ 1, is a bounded domain with a smooth boundary ∂Ω, γ ≥ 0, 0 < ρ 2 N - 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq2_HTML.gif if N ≥ 3 or ρ > 0 if N = 1, 2, and the function g: R+R+ is a nonincreasing function. This type of equations usually arise in the theory of viscoelasticity when the material density varies according to the velocity. In that paper, they proved a global existence result of weak solutions for γ ≥ 0 and a uniform decay result for γ > 0. Precisely, they showed that the solutions goes to zero in an exponential rate for γ > 0 and g is a positive bounded C1-function satisfying
1 - 0 g ( s ) d s = 1 - l > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ2_HTML.gif
(1.2)
and
- ξ 1 g ( t ) g ( t ) - ξ 2 g ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ3_HTML.gif
(1.3)

for all t ≥ 0 and some positive constants ξ1 and ξ2. Later, this result was extended by Messaoudi and Tatar [15] to a situation where a nonlinear source term is competing with the dissipation terms induced by both the viscoelasticity and the viscosity. Recently Messaoudi and Tatar [14] studied problem (1.1) for the case of γ = 0, they improved the result in [3] by showing that the solution goes to zero with an exponential or polynomial rate, depending on the decay rate of the relaxation function g.

The assumptions (1.2) and (1.3), on g, are frequently encountered in the linear case (ρ = 0), see [1, 2, 46, 13, 22, 23, 2931]. Lately, these conditions have been weakened by some researchers. For instance, instead of (1.3) Furati and Tatar [8] required the functions e αt g(t) and e αt g'(t) to have sufficiently small L1-norm on (0, ∞) for some α > 0 and they can also have an exponential decay of solutions. In particular, they do not impose a rate of decreasingness for g. Later on Messaoudi and Tatar [21] improved this result further by removing the condition on g'. They established an exponential decay under the conditions g'(t) ≤ 0 and e αt g(t) ∈ L1(0, ∞) for some large α > 0. This last condition was shown to be necessary condition for exponential decay [7]. More recently Tatar [25] investigated the asymptotic behavior to problem (1.1) with ρ = γ = 0 when h(t)g(t) ∈ L1(0, ∞) for some nonnegative function h(t). He generalized earlier works to an arbitrary decay not necessary of exponential or polynomial rate.

Motivated by previous works [21, 25], in this paper, we consider the initial boundary value problem for the following nonlinear viscoelastic equation:
u t ρ u t t - Δ u - Δ u t t + 0 t g ( t - s ) Δ u ( s ) d s + u p u = 0 , in  Ω × ( 0 , ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ4_HTML.gif
(1.4)
with initial conditions
u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ5_HTML.gif
(1.5)
and boundary condition
u ( x , t ) = 0 , x Ω , t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ6_HTML.gif
(1.6)

where Ω ⊂ R N , N ≥ 1, is a bounded domain with a smooth boundary ∂Ω. Here ρ, p > 0 and g represents the kernel of the memory term, with conditions to be stated later [see assumption (A1)-(A3)].

We intend to study the arbitrary decay result for problem (1.4)-(1.6) under the weaker assumption on g, which is not necessarily decaying in an exponential or polynomial fashion. Indeed, our result will be established under the conditions g'(t) ≤ 0 and 0 ξ ( s ) g ( s ) d s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq3_HTML.gif for some nonnegative function ξ(t). Therefore, our result allows a larger class of relaxation functions and improves some earlier results concerning the exponential decay or polynomial decay.

The content of this paper is organized as follows. In Section 2, we give some lemmas and assumptions which will be used later, and we mention the local existence result in Theorem 2.2. In Section 3, we establish the statement and proof of our result related to the arbitrary decay.

2 Preliminary results

In this section, we give some assumptions and lemmas which will be used throughout this work. We use the standard Lebesgue space L p (Ω) and Sobolev space H 0 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq4_HTML.gif with their usual inner products and norms.

Lemma 2.1. (Sobolev-Poincaré inequality) Let 2 p 2 N N - 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq5_HTML.gif, the inequality
u p c s u 2 for  u H 0 1 ( Ω ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equa_HTML.gif

holds with the optimal positive constant c s , where || · || p denotes the norm of L p (Ω).

Assume that ρ satisfies
0 < ρ 2 N - 2 if  N 3 or ρ > 0 if N = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ7_HTML.gif
(2.1)

With regards to the relaxation function g(t), we assume that it verifies

(A1) g(t) ≥ 0, for all t ≥ 0, is a continuous function satisfying
0 < 0 g ( s ) d s = l < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ8_HTML.gif
(2.2)

(A2) g'(t) ≤ 0 for almost all t > 0.

(A3) There exists a positive nondecreasing function ξ(t): [0, ∞) → (0, ∞) such that ξ ( t ) ξ ( t ) = η ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq6_HTML.gif is a decreasing function and
0 ξ ( s ) g ( s ) d s < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ9_HTML.gif
(2.3)

Now, we state, without a proof, the existence result of the problem (1.4)-(1.6) which can be established by Faedo-Galerkin methods, we refer the reader to [3, 5].

Theorem 2.2. Suppose that (2.1) and (A1) hold, and that ( u 0 , u 1 ) H 0 1 ( Ω ) × H 0 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq7_HTML.gif. Assume 0 < ρ 2 N - 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq2_HTML.gif, if N ≥ 3, p > 0, if N = 1, 2. Then there exists at least one global solution u of (1.4)-(1.6) satisfying
u L ( [ 0 , ) ; H 0 1 ( Ω ) ) , u t L ( [ 0 , ) ; H 0 1 ( Ω ) ) , u t t L 2 ( [ 0 , ) ; L 2 ( Ω ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equb_HTML.gif
Next, we introduce the modified energy functional for problem (1.4)-(1.6)
E ( t ) = 1 ρ + 2 u t ρ + 2 ρ + 2 + 1 2 1 - 0 t g ( s ) d s u ( t ) 2 2 + 1 2 u t ( t ) 2 2 + 1 2 ( g u ) ( t ) + 1 p + 2 u ( t ) p + 2 p + 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ10_HTML.gif
(2.4)
where
( g u ) ( t ) = 0 t Ω g ( t - s ) u ( t ) - u ( s ) 2 d x d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ11_HTML.gif
(2.5)
Lemma 2.3. Let u be the solution of (1.4)-(1.6), then the modified energy E(t) satisfies
E ( t ) = 1 2 ( g u ) ( t ) - 1 2 g ( t ) u ( t ) 2 2 1 2 ( g u ) ( t ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ12_HTML.gif
(2.6)

Proof. Multiplying Eq. (1.4) by u t and integrating it over Ω, then using integration by parts and the assumption (A1)-(A2), we obtain (2.6).

Remark. It follows from Lemma 2.3 that the energy is uniformly bounded by E(0) and decreasing in t. Besides, from the definition of E(t) and (2, 2), we note that
( 1 - l ) u 2 2 + u t ( t ) 2 2 + ( g u ) ( t ) 2 E ( 0 ) , t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ13_HTML.gif
(2.7)

3 Decay of the solution energy

In this section, we shall state and prove our main result. For this purpose, we first define the functional
L ( t ) = E ( t ) + i = 1 3 λ i Φ i ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ14_HTML.gif
(3.1)
where λ i are positive constants, i = 1, 2, 3 to be specified later and
Φ 1 ( t ) = 1 ρ + 1 Ω u t ρ u t u d x + Ω u t ( t ) u ( t ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ15_HTML.gif
(3.2)
Φ 2 ( t ) = Ω Δ u t - 1 ρ + 1 u t ρ u t 0 t g ( t - s ) u ( t ) - u ( s ) d s d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ16_HTML.gif
(3.3)
Φ 3 ( t ) = Ω 0 t H ( t - s ) u ( s ) 2 d s d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ17_HTML.gif
(3.4)
here
H ( t ) = ξ ( t ) - 1 t g ( s ) ξ ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equc_HTML.gif

Remark. This functional was first introduced by Tatar [25] for the case of ρ = 0 and without imposing the dispersion term and forcing term as far as (1.4) is concerned.

The following Lemma tells us that L(t) and E(t) + Φ3(t) are equivalent.

Lemma 3.1. There exists two positive constants β1 and β2 such that the relation
β 1 ( E ( t ) + Φ 3 ( t ) ) L ( t ) β 2 ( E ( t ) + Φ 3 ( t ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ18_HTML.gif
(3.5)

holds for all t ≥ 0 and λ i small, i = 1, 2.

Proof. By Hölder inequality Young's inequality Lemma 2.1, (2.7) and (2.2), we deduce that
Ω u t ρ u t u d x 1 2 u t 2 ( ρ + 1 ) 2 ( ρ + 1 ) + 1 2 u 2 2 c s 2 ( ρ + 1 ) 2 u t 2 2 ( ρ + 1 ) + c s 2 2 u 2 2 α 1 2 u t 2 2 + c s 2 2 u 2 2 , Ω u t ( t ) u ( t ) d x 1 2 ( u t 2 2 + u 2 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equd_HTML.gif
Ω u t 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x 1 2 u t 2 2 + 1 2 Ω 0 t g ( t - s ) ( u ( t ) - u ( s ) ) 2 d s d x 1 2 u t 2 2 + l 2 ( g u ) ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Eque_HTML.gif
and
Ω u t ρ u t 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x 1 2 u t 2 ( ρ + 1 ) 2 ( ρ + 1 ) + 1 2 Ω 0 t g ( t - s ) u ( t ) - u ( s ) 2 d s d x α 1 2 u t 2 2 + l c s 2 2 ( g u ) ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equf_HTML.gif
where α 1 = c s 2 ( ρ + 1 ) ( 2 E ( 0 ) ) ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq8_HTML.gif. Therefore, from above estimates, the definition of E(t) by (2.4) and (2.2), we have
L ( t ) = E ( t ) + i = 1 3 λ i Φ i ( t ) E ( t ) + c 1 | | u | | 2 2 + c 2 | | u t | | 2 2 + c 3 g u ( t ) + λ 3 Φ 3 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equg_HTML.gif
and
L ( t ) E ( t ) - c 1 u 2 2 - c 2 u t 2 2 - c 3 g u ( t ) + λ 3 Φ 3 ( t ) 1 ρ + 2 u t ρ + 2 ρ + 2 + 1 2 ( 1 - l ) - c 1 u ( t ) 2 2 + 1 2 - c 2 u t ( t ) 2 2 + 1 2 - c 3 g u ( t ) + 1 p + 2 u ( t ) p + 2 p + 2 + λ 3 Φ 3 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equh_HTML.gif
where c 1 = λ 1 ( c s 2 + ρ + 1 ) 2 ( ρ + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq9_HTML.gif, c 2 = ( λ 1 + λ 2 ) ( α 1 + ρ + 1 ) 2 ( ρ + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq10_HTML.gif, and c 3 = l ( ρ + 1 + c s 2 ) λ 2 2 ( ρ + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq11_HTML.gif. Hence, selecting λ i , i = 1, 2 such that
λ 1 < min ( ρ + 1 ) ( 1 - l ) c s 2 + ρ + 1 , ρ + 1 α 1 + ρ + 1 , λ 2 < min ρ + 1 l ( c s 2 + ρ + 1 ) , ρ + 1 α 1 + ρ + 1 - λ 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equi_HTML.gif
and again from the definition of E(t), there exist two positive constants β1 and β2 such that
β 1 ( E ( t ) + Φ 3 ( t ) ) L ( t ) β 2 ( E ( t ) + Φ 3 ( t ) ) , t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equj_HTML.gif

To obtain a better estimate for Ω u 0 t g ( t - s ) u ( s ) d s d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq12_HTML.gif, we need the following Lemma which repeats Lemma 2 in [25].

Lemma 3.2. For t ≥ 0, we have
Ω u 0 t g ( t - s ) u ( s ) d s d x = 1 2 0 t g ( s ) d s u 2 2 + 1 2 0 t g ( t - s ) u ( s ) 2 2 d s - 1 2 g u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ19_HTML.gif
(3.6)

Proof. Straightforward computations yield this identity.

Now, we are ready to state and prove our result. First, we introduce the following notations as in [24, 25]. For every measurable set AR+, we define the probability measure g ^ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq13_HTML.gif by
g ^ ( A ) = 1 l A g ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equk_HTML.gif
The flatness set and the flatness rate of g are defined by
F g = { s R + g ( s ) > 0 and  g ( s ) = 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ20_HTML.gif
(3.7)
and
R g = g ^ ( F g ) = 1 l F g g ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ21_HTML.gif
(3.8)
Before proceeding, we note that there exists t0 > 0 such that
0 t g ( s ) d s 0 t 0 g ( s ) d s = g * > 0 , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ22_HTML.gif
(3.9)

since g is nonnegative and continuous.

Theorem 3.3. Let ( u 0 , u 1 ) H 0 1 ( Ω ) × H 0 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq7_HTML.gif be given. Suppose that (A1)-(A3), (2, 1) and the hypothesis on p hold. Assume further that R g < g * α l ( 2 α + 2 c s 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq14_HTML.gif, H ( 0 ) < g * ( 4 + l ) - 3 l 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq15_HTML.gif and g * > 3 l 4 + l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq16_HTML.gif with
α = ( g * ( 4 + l ) - 3 l ) 1 - l p 8 c s 2 ( p + 1 ) ( 2 E ( 0 ) ) p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equl_HTML.gif
Then the solution energy of (1.4)-(1.6) satisfies
E ( t ) K ξ ( t ) - μ , t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equm_HTML.gif

where μ and K are positive constants.

Proof. In order to obtain the decay result of E(t), it suffices to prove that of L(t). To this end, we need to estimate the derivative of L(t). It follows from (3.2) and Eq. (1.4) that
Φ 1 ( t ) = 1 ρ + 1 u t ρ + 2 ρ + 2 + u t 2 2 - u 2 2 + Ω u 0 t g ( t - s ) u ( s ) d s d x - u p + 2 p + 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equn_HTML.gif
which together with the identity (3.6) and (2.2) gives
Φ 1 ( t ) 1 ρ + 1 u t ρ + 2 ρ + 2 + u t 2 2 - 1 - l 2 u 2 2 + 1 2 0 t g ( t - s ) u ( s ) 2 2 d s - 1 2 ( g u ) ( t ) - u p + 2 p + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ23_HTML.gif
(3.10)
Next, we would like to estimate Φ 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq17_HTML.gif. Taking a derivative of Φ2 in (3.3) and using Eq. (1.4) to get
Φ 2 ( t ) = 1 - 0 t g ( s ) d s Ω u ( t ) 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x + Ω 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s 2 d x - 1 ρ + 1 0 t g ( s ) d s u t ρ + 2 ρ + 2 - 0 t g ( s ) d s u t 2 2 - Ω u t ( t ) 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x - 1 ρ + 1 Ω u t ρ u t 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x + Ω u p u 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ24_HTML.gif
(3.11)

We now estimate the first two terms on the right-hand side of (3.11) as in [25].

Indeed, for all measure set A and F such that A = R+ - F, we have
Ω u ( t ) 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x = Ω u ( t ) A [ 0 , t ] g ( t - s ) ( u ( t ) - u ( s ) ) d s d x + Ω u ( t ) F [ 0 , t ] g ( t - s ) ( u ( t ) - u ( s ) ) d s d x Ω u ( t ) A [ 0 , t ] g ( t - s ) ( u ( t ) - u ( s ) ) d s d x + F [ 0 , t ] g ( s ) d s u 2 2 - Ω u ( t ) F [ 0 , t ] g ( t - s ) u ( s ) d s d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ25_HTML.gif
(3.12)
To simplify notations, we denote
A t = A [ 0 , t ] and  F t = F [ 0 , t ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equo_HTML.gif
Using Hölder inequality Young's inequality and (2.2), we see that, for δ1 > 0,
Ω u ( t ) A t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x δ 1 u 2 2 + l 4 δ 1 Ω A t g ( t - s ) u ( t ) - u ( s ) 2 d s d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equp_HTML.gif
and
Ω u ( t ) F t g ( t - s ) u ( s ) d s d x 1 2 F t g ( s ) d s u 2 2 + 1 2 F t g ( t - s ) u ( s ) 2 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equq_HTML.gif
Thus, from the definition of g ^ ( F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq18_HTML.gif by (3.8), (3.12) becomes
Ω u ( t ) 0 t g ( t - s ) u ( t ) - u ( s ) d s d x δ 1 + 3 2 l g ^ ( F ) u 2 2 + l 4 δ 1 Ω A t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + 1 2 F t g ( t - s ) u ( s ) 2 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ26_HTML.gif
(3.13)
The second term on the right-hand side of (3.11) can be estimated as follows (see [25]), for δ2 > 0,
Ω 0 t g ( t - s ) u ( t ) - u ( s ) d s 2 d x = Ω A t g ( t - s ) u ( t ) - u ( s ) d s + F t g ( t - s ) u ( t ) - u ( s ) d s 2 d x 1 + 1 δ 2 l Ω A t g ( t - s ) | u ( t ) - u ( s ) | 2 d s d x + ( 1 + δ 2 ) l g ^ ( F ) Ω F t g ( t - s ) | u ( t ) - u ( s ) | 2 d s d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ27_HTML.gif
(3.14)
Using Hölder inequality Young's inequality and (A2) to deal with the fifth term, for δ3 > 0,
Ω u t ( t ) 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x δ 3 u t 2 2 + 1 4 δ 3 Ω 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s 2 d x δ 3 u t 2 2 - g ( 0 ) 4 δ 3 ( g u ) ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ28_HTML.gif
(3.15)
Exploiting Hölder inequality Young's inequality Lemma 2.1 and (A2) to estimate the sixth term, for δ4 > 0,
1 ρ + 1 Ω u t ρ u t 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x 1 ρ + 1 δ 4 u t 2 ( ρ + 1 ) 2 ( ρ + 1 ) + 1 4 δ 4 Ω 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s 2 d x 1 ρ + 1 δ 4 c s 2 ( ρ + 1 ) u t 2 2 ( ρ + 1 ) - g ( 0 ) c s 2 4 δ 4 g u ( t ) 1 ρ + 1 α 1 δ 4 u t 2 2 - g ( 0 ) c s 2 4 δ 4 g u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ29_HTML.gif
(3.16)
For the last term, thanks to Hölder inequality Young's inequality Lemma 2.1, (2.7), (2.2) and (3.8), we have, for δ5 > 0,
Ω u p u 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s d x δ 5 Ω u 2 ( p + 1 ) d x + 1 4 δ 5 Ω 0 t g ( t - s ) ( u ( t ) - u ( s ) ) d s 2 d x δ 5 α 2 u 2 2 + l c s 2 2 δ 5 Ω A t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + l c s 2 2 δ 5 g ^ ( F ) Ω F t g ( t - s ) u ( t ) - u ( s ) 2 d s d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ30_HTML.gif
(3.17)
where α 2 = c s 2 ( p + 1 ) 2 E ( 0 ) 1 - l p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq19_HTML.gif. Thus, gathering these estimates (3.13)-(3.17) and using (3.9), we obtain, for tt0,
Φ 2 ( t ) ( 1 - g * ) δ 1 + 3 2 l g ^ ( F ) + δ 5 α 2 u 2 2 - g ( 0 ) 4 δ 3 + g ( 0 ) c s 4 ( ρ + 1 ) δ 4 ( g u ) ( t ) + 1 - g * 2 F t g ( t - s ) u ( s ) 2 2 d s + δ 3 + α 1 δ 4 ρ + 1 - g * u t 2 2 - g * ρ + 1 u t ρ + 2 ρ + 2 + l 1 + 1 δ 2 + c s 2 2 δ 5 + 1 - g * 4 δ 1 Ω A t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + 1 + δ 2 + c s 2 2 δ 5 l g ^ ( F ) Ω F t g ( t - s ) u ( t ) - u ( s ) 2 d s d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ31_HTML.gif
(3.18)
Further, taking a derivative of Φ3(t), using the fact that ξ ( t ) ξ ( t ) = η ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq6_HTML.gif is a decreasing function and the definition of Φ3(t) by (3.4), we derive that (see [25])
Φ 3 ( t ) = H ( 0 ) u 2 2 - 0 t ξ ( t - s ) ξ ( t - s ) H ( t - s ) u ( s ) 2 2 d s - 0 t g ( t - s ) u ( s ) 2 2 d s H ( 0 ) u 2 2 - η ( t ) 0 t H ( t - s ) u ( s ) 2 2 d s - 0 t g ( t - s ) u ( s ) 2 2 d s = H ( 0 ) u 2 2 - η ( t ) Φ 3 ( t ) - 0 t g ( t - s ) u ( s ) 2 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ32_HTML.gif
(3.19)
Hence, we conclude from (2.6), (3.10), (3.18) and (3.19) that for any tt0 > 0,
L ( t ) = E ( t ) + λ 1 Φ 1 ( t ) + λ 2 Φ 2 ( t ) + λ 3 Φ 3 ( t ) 1 2 - λ 2 g ( 0 ) 4 δ 3 + g ( 0 ) c s 2 4 ( ρ + 1 ) δ 4 ( g u ) ( t ) + 1 ρ + 1 ( λ 1 - λ 2 g * ) u t ρ + 2 ρ + 2 - λ 1 u p + 2 p + 2 - λ 1 2 ( g u ) ( t ) + λ 1 + λ 2 δ 3 + α 1 δ 4 ρ + 1 - g * u t 2 2 + λ 1 2 - λ 3 0 t g ( t - s ) u ( s ) 2 2 d s + λ 2 ( 1 - g * ) δ 1 + 3 2 l g ^ ( F ) + λ 2 δ 5 α 2 - λ 1 ( 1 - l 2 ) + λ 3 H ( 0 ) u 2 2 + λ 2 l 1 + 1 δ 2 + c s 2 2 δ 5 + 1 - g * 4 δ 1 Ω A t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 + δ 2 + c s 2 2 δ 5 l g ^ ( F ) Ω F t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 - g * 2 F t g ( t - s ) u ( s ) 2 2 d s - λ 3 η ( t ) Φ 3 ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ33_HTML.gif
(3.20)
For n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq20_HTML.gif, we consider the sets (see [24, 25])
A n = { s R + n g ( s ) + g ( s ) 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equr_HTML.gif
and observe that
n A n = R + - { F g N g } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equs_HTML.gif
where F g is given in (3.7) and N g is the null set where g' is not defined. In addition, denoting F n = R+ - A n , then
lim n g ^ ( F n ) = g ^ ( F g ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equt_HTML.gif
because A n are increasingly nested. Thus, choosing A = A n , F = F n and λ1 = (g * - ε) λ2 for some ε > 0 in (3.20), we obtain
L ( t ) 1 2 - λ 2 g ( 0 ) 4 δ 3 + g ( 0 ) c s 2 4 ( ρ + 1 ) δ 4 ( g u ) ( t ) - λ 2 ε ρ + 1 u t ρ + 2 ρ + 2 + λ 2 δ 3 + α 1 δ 4 ρ + 1 - ε u t 2 2 - g * - ε λ 2 u p + 2 p + 2 - ( g * - ε ) λ 2 2 g u ( t ) + ( g * - ε ) λ 2 2 - λ 3 0 t g ( t - s ) u ( s ) 2 2 d s + λ 2 ( 1 - g * ) δ 1 + 3 2 l g ^ ( F n ) + λ 2 δ 5 α 2 - λ 2 g * - ε ( 1 - l 2 ) + λ 3 H ( 0 ) u 2 2 + λ 2 l 1 + 1 δ 2 + c s 2 2 δ 5 + 1 - g * 4 δ 1 Ω A n t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 + δ 2 + c s 2 2 δ 5 l g ^ ( F n ) Ω F n t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 - g * 2 F n t g ( t - s ) u ( s ) 2 2 d s - λ 3 η ( t ) Φ 3 ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ34_HTML.gif
(3.21)
At this point, we take δ 3 = δ 4 < ( ρ + 1 ) ε 2 ( ρ + 1 + α 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq21_HTML.gif and select λ2 so that
1 2 - λ 2 g ( 0 ) 4 δ 3 + g ( 0 ) c s 2 4 ( ρ + 1 ) δ 4 1 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equu_HTML.gif
then (3.21) becomes
L ( t ) - λ 2 ε ρ + 1 u t ρ + 2 ρ + 2 - λ 2 ε 2 u t 2 2 - ( g * - ε ) λ 2 u p + 2 p + 2 - ( g * - ε ) λ 2 2 ( g u ) ( t ) + ( g * - ε ) λ 2 2 - λ 3 0 t g ( t - s ) u ( s ) 2 2 d s + λ 2 ( 1 - g * ) δ 1 + 3 2 l g ^ ( F n ) + λ 2 δ 5 α 2 - λ 2 g * - ε 1 - l 2 + λ 3 H ( 0 ) u 2 2 + λ 2 l 1 + 1 δ 2 + c s 2 2 δ 5 + 1 - g * 4 δ 1 - 1 4 n Ω A n t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 + δ 2 + c s 2 2 δ 5 l g ^ ( F n ) Ω F n t g ( t - s ) u ( t ) - u ( s ) 2 d s d x + λ 2 1 - g * 2 F n t g ( t - s ) u ( s ) 2 2 d s - λ 3 η ( t ) Φ 3 ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equv_HTML.gif
For ε, δ2 small enough and large value of n and t0, we see that if
δ 5 = α and  g ^ ( F n ) < g * α l ( 2 α + 2 c s 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equw_HTML.gif
then
( 1 + δ 2 ) + c s 2 2 δ 5 l g ^ ( F n ) - g * - ε 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equx_HTML.gif
and
3 2 ( 1 - g * ) l g ^ ( F n ) < δ ( g * - ε ) 1 - l 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ35_HTML.gif
(3.22)
where
α = g * ( 4 + l ) - 3 l 8 α 2 = ( g * ( 4 + l ) - 3 l ) ( 1 - l ) p 8 c s 2 ( p + 1 ) ( 2 E ( 0 ) ) p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equy_HTML.gif
and
δ = 3 ( 1 - g * ) l g * ( 4 - 2 l ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equz_HTML.gif
Note that α > 0 and 0 < δ < 1 due to g * > 3 l 4 + l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq16_HTML.gif. Furthermore, we require λ2 and λ3 satisfying
λ 2 l 1 + 1 δ 2 + c s 2 2 δ 5 + 1 - g * 4 δ 1 - 1 4 n < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equaa_HTML.gif
and
λ 2 2 < λ 3 < λ 2 ( g * ( 4 + l ) - 3 l ) 8 H ( 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equab_HTML.gif
this is possible because of H ( 0 ) < g * ( 4 + l ) - 3 l 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq15_HTML.gif. Then, letting δ1 be small enough and using (3.22), we see that
λ 2 ( 1 - g * ) δ 1 + 3 2 l g ^ ( F n ) + λ 2 δ 5 α 2 - λ 2 ( g * - ε ) 1 - l 2 + λ 3 H ( 0 ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equac_HTML.gif
Hence, from the definition of E(t) by (2.4), we have, for all tt0,
L ( t ) - c 4 E ( t ) - λ 3 η ( t ) Φ 3 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equad_HTML.gif
for some positive constant c4. As η(t) is decreasing, we have η(t) ≤ c4 after some t*t0. Hence, with the help of the right hand side inequality in (3.5), we find
L ( t ) - c 5 η ( t ) L ( t ) , t t * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equ36_HTML.gif
(3.23)
for some positive constant c5 > 0. An integration of (3.23) over (t*, t) gives
L ( t ) L ( t * ) e - c 5 t * t η ( s ) d s , t t * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equae_HTML.gif
Then using the left hand side inequality in (3.5) leads to
β 1 ( E ( t ) + Φ 3 ( t ) ) L ( t * ) e - c 5 t * t η ( s ) d s , t t * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equaf_HTML.gif
Therefore, by virtue of the continuity and boundedness of E(t) and ξ(t) on the interval [0, t*], we infer that
E ( t ) K ξ ( t ) - μ , t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_Equag_HTML.gif

for some positive constants K and μ.

Similar to those remarks as in [25], we have the following remark.

Remark. Note that there is a wide class of relaxation functions satisfying (A3). More precisely, if ξ(t) = e αt , α > 0, then η(t) = α, this gives the exponential decay estimate E ( t ) c 1 e - c 2 α t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-28/MediaObjects/13661_2011_Article_74_IEq22_HTML.gif, for some positive constants c1 and c2. Similarly, if ξ(t) = (1 + t) α , α > 0, then we obtain the polynomial decay estimate E (t) ≤ c3 (1 + t)-μ, for some positive constants c3 and μ.

Declarations

Acknowledgements

The authors would like to thank very much the anonymous referees for their valuable comments on this work.

Authors’ Affiliations

(1)
General Education Center National Taipei University of Technology

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© Wu; licensee Springer. 2011

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