General decay of solutions of a viscoelastic equation

J. Math. Anal. Appl. 341 (2008) 1457–1467 www.elsevier.com/locate/jmaa

General decay of solutions of a viscoelastic equation Salim A. Messaoudi King Fahd University of Petroleum and Minerals, Department of Mathematical Sciences, Dhahran 31261, Saudi Arabia Received 15 January 2007

Submitted by M. Nakao

Abstract In this paper we consider the following viscoelastic equation: t utt − u +

g(t − τ )u(τ ) dτ = 0, 0

in a bounded domain, and establish a general decay result which is not necessarily of exponential or polynomial type. This work generalizes and improves earlier results in the literature. © 2007 Elsevier Inc. All rights reserved. Keywords: General decay; Relaxation function; Viscoelastic

1. Introduction In this paper we consider the following problem: ⎧ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ utt (x, t) − u(x, t) + g(t − τ )u(x, τ ) dτ = 0,

(1.1)

0

⎪ ⎪ ⎪ u(x, t) = 0, x ∈ ∂Ω, t  0, ⎪ ⎪ ⎩ u(x, 0) = u0 (x), ut (x, 0) = u1 (x),

in Ω × (0, ∞),

x ∈ Ω,

where Ω is a bounded domain of Rn (n  1) with a smooth boundary ∂Ω and g is a positive nonincreasing function defined on R+ . This type of problems arise in viscoelasticity. For the thermodynamics of materials with fading memory, we refer the reader to the early work of Coleman and Noll [9] and Coleman and Mizel [10] and the references therein. Since the pioneer works of Dafermos [11,12] in 1970, where the general decay was discussed, problems related to viscoelasticity have attracted a great deal of attention and many results of existence and long-time behavior have been established. The first work that dealt with uniform decay was by Dassios and Zafiropoulos [14] in which E-mail address: [email protected]. 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.11.048

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a viscoelastic problem in R3 was studied and a polynomial decay result was proved for exponentially decaying kernels. Also, the uniform stability, for some problems in linear viscoelasticity, has been established in a book by Fabrizio and Morro [15] in 1992. After this, a very important contribution by Rivera was introduced. In 1994, Rivera [22] considered equations for linear isotropic homogeneous viscoelastic solids of integral type which occupy a bounded domain or the whole space Rn , with zero boundary and history data and in the absence of body forces. In the bounded domain case, an exponential decay result was proved for exponentially decaying memory kernels. For the whole space case a polynomial decay result was established and the rate of the decay was given. This result was later generalized to a situation where the kernel is decaying algebraically but not exponentially by Rivera et al. [5]. In their paper, the authors showed that the decay of solutions is also algebraic, at a rate which can be determined by the rate of the decay of the relaxation function. Also, the authors considered both cases the bounded domains and that of a material occupying the entire space. This result was later improved by Barreto et al. in [1], where equations related for linear viscoelastic plates were treated. Precisely, they showed that the solution energy decays at the same decay rate of the relaxation function. For partially viscoelastic materials, Rivera et al. [24] showed that solutions decay exponentially to zero, provided the relaxation function decays in similar fashion, regardless to the size of the viscoelastic part of the material. In [25], a class of abstract viscoelastic systems of the form   utt (t) + Au(t) + βu(t) − g ∗ Aα u (t) = 0, ut (0) = u1 ,

u(0) = u0 ,

(1.2)

for 0  α  1, β  0, were investigated. The main focus was on the case when 0 < α < 1 and the main result was that solutions for (1.2) decay polynomially even if the kernel g decay exponentially. This result is sharp (see [25, Theorem 12]). This result has been improved by Rivera et al. [27], where the authors studied a more general abstract problem than (1.2) and established a necessary and sufficient condition to obtain an exponential decay. In the case of lack of exponential decay, a polynomial decay has been proved. In the latter case they showed that the rate of decay can be improved by taking more regular initial data. Also application to concrete examples was presented. For systems with a localized frictional damping cooperating with the dissipation induced by the viscoelastic term, we mention the work of Cavalcanti et al. [7]. Under the condition −ξ1 g(t)  g  (t)  −ξ2 g(t),

t  0,

with gL1 ((0,∞)) small enough, the authors obtained an exponential rate of decay. Berrimi et al. [3] improved Cavalcanti’s result by showing that the viscoelastic dissipation alone is enough to stabilize the system. To achieve their goal, Berrimi et al. introduced a different functional, which allowed them to weaken the conditions on g as well as on the localized damping. This result has been later extended to a situation, where a source is competing with the viscoelastic dissipation, by Berrimi et al. [4]. Also, Cavalcanti et al. [8] considered t utt − k0 u +



div a(x)g(t − τ )∇u(τ ) dτ + b(x)h(ut ) + f (u) = 0,

0

under similar conditions on the relaxation function g and a(x) + b(x)  δ > 0 and improved the result in [7]. They established an exponential stability when g is decaying exponentially and h is linear and a polynomial stability when g is decaying polynomially and h is nonlinear. Though both results in [3] and [8] improve the earlier one in [7], the approaches are different. A related problem, in a bounded domain, of the form t |ut | utt − u − utt +

g(t − τ )u(τ ) dτ − γ ut = 0,

ρ

(1.3)

0

for ρ > 0, was also studied by Cavalcanti et al. [6]. A global existence result for γ  0, as well as an exponential decay for γ > 0, has been established. This last result has been extended to a situation, where γ = 0, by Messaoudi and Tatar [20,21] and exponential and polynomial decay results have been established in the absence, as well as in the presence, of a source term. For viscoelastic systems with oscillating kernels, Rivera et al. [26] showed that if the kernel satisfies g(0) > 0 and decays exponentially to zero, then the solution decays exponentially to zero. On the other hand, if the kernel decays polynomially, then the corresponding solutions also decays polynomially to zero with the same rate of decay.

S.A. Messaoudi / J. Math. Anal. Appl. 341 (2008) 1457–1467

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For formation of singularities, we quote the early work of Dafermos [13] in 1985. Recently, Messaoudi [18] considered t utt − u + g(t − τ )u(τ ) dτ + aut |ut |m = b|u|γ u, in Ω × (0, ∞), 0

and showed, under suitable conditions on g, that solutions with negative energy blow up in finite time if γ > m and continue to exist if m  γ . This blow-up result has been pushed to some situations, where the initial energy is positive, by Messaoudi [19]. A similar result has been also obtained by Wu [29] using a different method. For more early and recent results related to stability and asymptotic behavior, we refer the reader to the books by Renardy et al. [28] and Liu and Zheng [16], and works by Liu and Liu [17], Rivera and Oquendo [23], and Barreto et al. [2]. In the present work we generalize our earlier decay result to solutions of (1.1). Precisely, we show that the solution energy decays at the same rate of decay of the relaxation function, which is not necessarily decaying in a polynomial or exponential fashion. In fact, our result allows a larger class of relaxation functions (see examples in Remark 2.1). The paper is organized as follows. In Section 2, we present some notations and material needed for our work and state a global existence theorem, which can be proved following exactly the arguments of [7]. Section 3 contains the statement and the proof of our main result. 2. Preliminaries In this section we present some material needed in the proof of our main result. Also, for the sake of completeness we state, without a proof, the global existence result of [7] and [8]. We use the standard Lebesgue space Lp (Ω) and the Sobolev space H01 (Ω) with their usual scalar products and norms. For the relaxation function g we assume (G1) g : R+ → R+ is a differentiable function such that ∞ g(0) > 0,

g(s) ds = l > 0.

1− 0

(G2) There exists a differentiable function ξ satisfying g  (t)  −ξ(t)g(t), t  0,     ξ (t)      ξ(t)   k, ξ(t) > 0, ξ (t)  0, ∀t > 0. Remark 2.1. There are many functions satisfying (G1) and (G2). Examples of such functions are a , ν > 1, g1 (t) = (1 + t)ν g2 (t) = ae−b(t+1) , 0 < p  1, a , ν > 1, g3 (t) = (1 + t)[ln(1 + t)]ν for a and b to be chosen properly. In fact, simple computations show that

(ν+1)/ν −ν , ν > 1. g1 (t) = 1/ν g1 (t) a It is clear as ν approaches 1, the exponent (ν + 1)/ν approaches 2. p

Remark 2.2. Since ξ is nonincreasing then ξ(t)  ξ(0) = M. Remark 2.3. Condition (G1) is necessary to guarantee the hyperbolicity of the system (1.1).

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Proposition. Let (u0 , u1 ) ∈ H01 (Ω) × L2 (Ω) be given. Assume that g satisfies (G1). Then problem (1.1) has a unique global solution     u ∈ C [0, ∞); H01 (Ω) , ut ∈ C [0, ∞); L2 (Ω) . (2.1) We introduce the “modified” energy functional

t  2 1 1 1 E(t) := 1 − g(s) ds ∇u(t)2 + ut 22 + (g ◦ ∇u)(t), 2 2 2

(2.2)

0

where t (g ◦ v)(t) =

2  g(t − τ )v(t) − v(τ )2 dτ.

(2.3)

0

3. Decay of solutions In this section we state and prove our main result. For this purpose we set F (t) := E(t) + ε1 Ψ (t) + ε2 χ(t),

(3.1)

where ε1 and ε2 are positive constants and  Ψ (t) := ξ(t) uut dx, Ω

 χ(t) := −ξ(t)

t ut

Ω

  g(t − τ ) u(t) − u(τ ) dτ dx.

(3.2)

0

Lemma 3.1. If u is a solution of (1.1), then the “modified” energy satisfies 2 1  1 1 E  (t) = (g  ◦ ∇u)(t) − g(t)∇u(t)  (g  ◦ ∇u)(t)  0. 2 2 2

(3.3)

Proof. By multiplying Eq. (1.1) by ut and integrating over Ω, using integration by parts, hypotheses (G1) and (G2) and some manipulations as in [19], we obtain (3.3) for regular solutions. This inequality remains valid for weak solutions by a simple density argument. 2 Lemma 3.2. For u ∈ H01 (Ω), we have  t Ω

  g(t − τ ) u(t) − u(τ ) dτ

2 dx  (1 − l)Cp2 (g ◦ ∇u)(t),

0

where Cp is the Poincaré constant. Proof.  t Ω

0

  g(t − τ ) u(t) − u(τ ) dτ

2 dx =

 t  Ω



  g(t − τ ) g(t − τ ) u(t) − u(τ ) dτ

0

By applying Cauchy–Schwarz inequality and Poincaré’s inequality, we easily see that

2 dx.

S.A. Messaoudi / J. Math. Anal. Appl. 341 (2008) 1457–1467

 t Ω

  g(t − τ ) u(t) − u(τ ) dτ

2

 t

 t dx 

g(t − τ ) dτ Ω

0

0

1461

 2 g(t − τ ) u(t) − u(τ ) dτ dx

0

 (1 − l)Cp2 (g ◦ ∇u)(t).

2

Lemma 3.3. For ε1 and ε2 small enough, we have α1 F (t)  E(t)  α2 F (t)

(3.4)

holds for two positive constants α1 and α2 . Proof. Straightforward computations, using Lemma 3.2, lead to 

 |ut | dx + (ε1 /2)ξ(t)

F (t)  E(t) + (ε1 /2)ξ(t) Ω

Ω

Ω

  g(t − τ ) u(t) − u(τ ) dτ

0

 E(t) + (ε1 + ε2 )/2 M

|ut |2 dx

2

Ω

 t + (ε2 /2)ξ(t)

 |u| dx + (ε2 /2)ξ(t)

2

2 dx



 |ut | dx 2

+ (ε1 /2)Cp2 M

Ω

|∇u|2 dx + (ε2 /2)Cp2 M(1 − l)(g ◦ ∇u)(t) Ω

 α2 E(t).

(3.5)

Similarly,  F (t)  E(t) − (ε1 /2)ξ(t) Ω



1 − M(ε1 + ε2 )/2 2

|∇u|2 dx − (ε2 /2)ξ(t) Ω

− (ε2 /2)ξ(t)Cp2 (1 − l)(g





 |ut |2 dx − (ε1 /2)ξ(t)Cp2



◦ ∇u)(t)



|ut |2 dx +

1 l − M(ε1 /2)Cp2 2

Ω



 1 2 + − M(ε2 /2)Cp (1 − l) (g ◦ ∇u)(t) 2

|ut |2 dx Ω

 |∇u|2 dx Ω

 α1 E(t),

(3.6)

for ε1 and ε1 small enough.

2

Lemma 3.4. Under the assumptions (G1) and (G2), the functional  Ψ (t) := ξ(t)

uut dx Ω

satisfies, along the solution of (1.1), 



Ψ (t)  1 +

k 2 Cp2 l

   l (1 − l) 2 ξ(t) ut dx − ξ(t) |∇u|2 dx + ξ(t)(g ◦ ∇u)(t). 4 2l Ω

Ω

(3.7)

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Proof. By using Eq. (1.1), we easily see that     uutt + u2t dx + ξ  (t) uut dx Ψ  (t) = ξ(t) Ω

Ω

 = ξ(t)

 u2t dx

− ξ(t)

Ω



t

|∇u| dx + ξ(t)

∇u(t).

2

Ω

Ω

g(t − τ )∇u(τ ) dτ dx + ξ  (t)

 uut dx.

(3.8)

Ω

0

We now estimate the third term in the right-hand side of (3.8) as follows: 

t ∇u(t).

Ω

g(t − τ )∇u(τ ) dτ dx 0

1  2



  ∇u(t)2 dx + 1 2

Ω

1  2



 t Ω

  ∇u(t)2 dx + 1 2

Ω

2 dx

0

 t

Ω

  g(t − τ )∇u(τ ) dτ

    g(t − τ ) ∇u(τ ) − ∇u(t) + ∇u(t) dτ

0

We then use Lemma 3.2, Young’s inequality, and the fact that η > 0,  t Ω

    g(t − τ ) ∇u(τ ) − ∇u(t) + ∇u(t) dτ

0

 t

 Ω

  g(t − τ )∇u(τ ) − ∇u(t) dτ

+2 Ω

dx + Ω

0

∞ 0

g(τ ) dτ = 1 − l, to obtain, for any

 t

  g(t − τ )∇u(t) dτ

2 dx

0

  g(t − τ )∇u(t) dτ dx

0

2   t   1    (1 + η) dx + 1 + g(t − τ ) ∇u(τ ) − ∇u(t) dτ dx η Ω Ω 0 0      1 2  1+ (1 − l)(g ◦ ∇u)(t) + (1 + η)(1 − l)2 ∇u(t) dx. η  t

  g(t − τ )∇u(t) dτ

2

g(τ ) dτ 



Ω

By combining (3.8)–(3.10) and using    1 2 2 uut dx  αCp |∇u| dx + u2t dx, 4α Ω

Ω

(3.9)

dx  t

  g(t − τ )∇u(τ ) − ∇u(t) dτ

0

dx.

2

2

0

 t

t

2

α > 0,

Ω

we arrive at

      1 1  ξ  (t)  1 2 ξ(t) Ψ  (t)  1 + 1 + (1 − l)ξ(t)(g ◦ ∇u)(t) u dx + t 4α  ξ(t)  2 η Ω         ξ (t)  2  1 αC ξ(t) ∇u(t)2 dx − 1 − (1 + η)(1 − l)2 − 2 p  2 ξ(t) Ω

(3.10)

S.A. Messaoudi / J. Math. Anal. Appl. 341 (2008) 1457–1467

1463

    1 1 1 2 k ξ(t) ut dx + 1+ (1 − l)ξ(t)(g ◦ ∇u)(t)  1+ 4α 2 η Ω  2 

1 2 2 − 1 − (1 + η)(1 − l) − 2kαCp ξ(t) ∇u(t) dx. 2 

(3.11)

Ω

By choosing η = l/(1 − l) and α

= l/4kCp2 ,

(3.7) is established.

2

Lemma 3.5. Under the assumptions (G1) and (G2), the functional 

t

χ(t) := −ξ(t)

  g(t − τ ) u(t) − u(τ ) dτ dx

ut Ω

0

satisfies, along the solution of (1.1),

χ (t)  δξ(t) 1 + 2(1 − l)2 



  ∇u(t)2 dx +

   Cp2 1 (1 − l) + k ξ(t)(g ◦ ∇u)(t) 2δ + 2δ 4δ

Ω

   t   g(0) 2  + Cp ξ(t) −(g ◦ ∇u) (t) + δ(k + 1) − g(s) ds ξ(t) u2t dx, 4δ

δ > 0.

(3.12)

Ω

0

Proof. Direct computations, using (1.1), yield 



t

χ (t) = −ξ(t)

utt Ω

0

 − ξ(t)

t

   g (t − τ ) u(t) − u(τ ) dτ dx − ξ(t) g(s) ds u2t dx

t



ut Ω

− ξ  (t)

0



0

t ut

Ω

t

∇u(t). Ω

0

 t

− ξ(t)

  g(t − τ ) ∇u(t) − ∇u(τ ) dτ dx

t

g(t − τ )∇u(τ ) dτ . Ω

0

 − ξ(t)

ut

− ξ  (t)

t

   g (t − τ ) u(t) − u(τ ) dτ dx − ξ(t) g(s) ds u2t dx 

0



t ut

Ω

  g(t − τ ) ∇u(t) − ∇u(τ ) dτ dx

0

t

Ω

Ω

  g(t − τ ) u(t) − u(τ ) dτ dx

0

 = ξ(t)

  g(t − τ ) u(t) − u(τ ) dτ dx

0

  g(t − τ ) u(t) − u(τ ) dτ dx.

Ω

(3.13)

0

Similarly to (3.8), we estimates the right-hand side terms of (3.13). So, by using Young’s inequality, the first term gives

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S.A. Messaoudi / J. Math. Anal. Appl. 341 (2008) 1457–1467

t

 −

∇u(t). Ω

   1−l (g ◦ ∇u)(t), g(t − τ ) ∇u(t) − ∇u(τ ) dτ dx  δ |∇u|2 dx + 4δ

∀δ > 0.

(3.14)

Ω

0

Similarly, the second term can be estimated as follows

t

 t   g(t − s)∇u(s) ds . g(t − s) ∇u(t) − ∇u(s) ds dx Ω

0

0

2    t   1    δ  g(t − s)∇u(s) ds  dx +   4δ Ω

Ω

0

 t

0

    g(t − s) ∇u(t) − ∇u(s) + ∇u(t) ds

δ Ω

 t 2       g(t − s) ∇u(t) − ∇u(s) ds  dx  

2

1 dx + 4δ

2    t      g(t − s) ∇u(t) − ∇u(s) ds  dx   Ω

0

0

2    t    1 2    2δ + g(t − s) ∇u(t) − ∇u(s) ds dx + 2δ(1 − l) |∇u|2 dx 4δ Ω Ω 0    1 (1 − l)(g ◦ ∇u)(t) + 2δ(1 − l)2 |∇u|2 dx.  2δ + 4δ

(3.15)

Ω

As for the third and the fourth terms we have   t   g(0) 2  C (g ◦ ∇u)(t) − ut g  (t − τ ) u(t) − u(τ ) dτ dx  δ |ut |2 dx − 4δ p Ω

(3.16)

Ω

0

and t

 ut Ω

  g(t − τ ) u(t) − u(τ ) dτ dx  δ

 |ut |2 dx +

Cp2 4δ

(g ◦ ∇u)(t).

(3.17)

Ω

0

By combining (3.13)–(3.17), the assertion of the lemma is established.

2

Theorem 3.6. Let (u0 , u1 ) ∈ H01 (Ω) × L2 (Ω) be given. Assume that g and ξ satisfy (G1) and (G2). Then, for each t0 > 0, there exist strictly positive constants K and λ such that the solution of (1.1) satisfies E(t)  Ke

−λ

t

t0

ξ(s) ds

,

t  t0 .

(3.18)

Proof. Since g is positive and g(0) > 0 then for any t0 > 0 we have t

t0 g(s) ds 

0

g(s) ds = g0 > 0,

∀t  t0 .

(3.19)

0

By using (3.1), (3.3), (3.7), (3.12) and (3.19), we obtain for t  t0 ,       k 2 Cp2   g(0) 2 1  2 ξ(t) ut dx + − ε2 C M (g  ◦ ∇u)(t) F (t)  − ε2 g0 − δ(1 + k) − ε1 1 + l 2 4δ p Ω     ε1 l − ε2 δ 1 + 2(1 − l)2 ξ(t)∇u22 − 4     Cp2 ε1 (1 − l) 1 + + ε2 2δ + (1 − l) + ε2 k ξ(t)(g ◦ ∇u)(t). 2l 2δ 4δ

(3.20)

S.A. Messaoudi / J. Math. Anal. Appl. 341 (2008) 1457–1467

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At this point we choose δ so small that 1 g0 − δ(1 + k) > g0 , 2

4 1 2 δ 1 + 2(1 − l) < g0 . k2 C 2 l 4(1 + l p ) Whence δ is fixed, the choice of any two positive constants ε1 and ε2 satisfying g0 g0 ε2 < ε1 < ε2 k 2 Cp k2 C 2 4(1 + l ) 2(1 + l p )

(3.21)

will make

  k 2 Cp2   > 0, k1 := ε2 g0 − δ(1 + k) − ε1 1 + l

ε1 l − ε2 δ 1 + 2(1 − l)2 > 0. k2 := 4 We then pick ε1 and ε2 so small that (3.4) and (3.21) remain valid and, further,      Cp2 1 ε1 g(0) 2 1 k3 := − ε2 Cp M − + ε2 2δ + (1 − l) − ε2 k > 0. 2 4δ l 2δ 4δ Hence       Cp2 1 1 g(0) 2 ε1 (1 − l) − ε2 Cp M (g  ◦ ∇u)(t) + + ε2 2δ + (1 − l) + ε2 k ξ(t)(g ◦ ∇u)(t) 2 4δ 2l 2δ 4δ    t 2  1 g(0) 2 − − ε2 Cp M ξ(t − τ )g(t − τ )∇u(τ ) − ∇u(t) dτ dx 2 4δ Ω 0

  Cp2 ε1 (1 − l) 1 + ε2 2δ + k ξ(t)(g ◦ ∇u)(t) (1 − l) + ε2 2l 2δ 4δ  −k3 ξ(t)(g ◦ ∇u)(t), 



+

(3.22)

since ξ is nonincreasing. Therefore, by using (3.4), (3.20), and (3.22), we arrive at F  (t)  −β1 ξ(t)E(t)  −β1 α1 ξ(t)F (t),

∀t  t0 .

(3.23)

A simple integration of (3.23) leads to F (t)  F (t0 )e

−β1 α1

t t0

ξ(s) ds

∀t  t0 .

,

(3.24)

Thus (3.4), (3.24) yield E(t)  α2 F (t0 )e

−β1 α1

This completes the proof.

t t0

ξ(s) ds

= Ke

−λ

t t0

ξ(s) ds

,

∀t  t0 .

(3.25)

2

Corollary 3.7. Under assumptions of Theorem 3.6, there exist strictly positive constants K  and λ such that the solution of (1.1) satisfies E(t)  K  e−λ

t 0

ξ(s) ds

,

t  t0 .

(3.26)

Proof. It suffices to note that E(t)  Ke

−λ

t

 K  e−λ

t0

t 0

ξ(s) ds

 t0 t   = Keλ 0 ξ(s) ds e−λ 0 ξ(s) ds

ξ(s) ds

,

t  t0 .

2

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Remark 3.1. This result generalizes and improves the results of [1,2,4,5], and [6]. In particular, it allows some relaxation functions which satisfy g   −ag ρ , 1  ρ < 2, instead of the usual assumption 1  ρ < 3/2. See g1 in Remark 2.1 Remark 3.2. Note that the exponential and the polynomial decay estimates, given in early works [1,2,4,5], and [6], are only particular cases of (3.25). More precisely, we obtain exponential decay for ξ(t) ≡ a and polynomial decay for ξ(t) = a(1 + t)−1 , where a > 0 is a constant. Remark 3.3. Observe that, similarly to Rivera et al. [23], our result is proved without any condition on g  and g  unlike what was assumed in (2.4) of [8]. We only need g to be differentiable satisfying (G1) and (G2). Remark 3.4. Estimate (3.26) is also true for t ∈ [0, t0 ] by virtue of continuity and boundedness of E(t) and ξ(t). In fact, by virtue of (3.3), we have  t0 t   E(t)  E(0)  E(0)eλ 0 ξ(s) ds e−λ 0 ξ(s) ds ,

∀t ∈ [0, t0 ].

Remark 3.5. A similar result can be established for the semilinear problem ⎧ t ⎪ ⎪ ⎪ p−2 ⎪ ⎪ u(x, t) = 0, ⎨ utt (x, t) − u(x, t) + g(t − τ )u(x, τ ) dτ + b|u|

(3.27)

0

⎪ ⎪ ⎪ u(x, t) = 0, x ∈ ∂Ω, t  0, ⎪ ⎪ ⎩ u(x, 0) = u0 (x), ut (x, 0) = u1 (x),

in Ω × (0, ∞),

x ∈ Ω,

where b > 0 and 2  p  2(n − 1)/(n − 2), if n  3. Acknowledgments The author would like to express his thanks to the referee for his/her valuable suggestions and remarks. This work has been funded by KFUPM under Project # MS/DECAY/334.

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