Boundary Value Problems

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Exponential energy decay and blow-up of solutions for a system of nonlinear viscoelastic wave equations with strong damping

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Abstract

In this paper, we consider the system of nonlinear viscoelastic equations

u t t - Δ u + 0 t g 1 ( t - τ ) Δ u ( τ ) d τ - Δ u t = f 1 ( u , v ) , ( x , t ) Ω × ( 0 , T ) , v t t - Δ v + 0 t g 2 ( t - τ ) Δ v ( τ ) d τ - Δ v t = f 2 ( u , v ) , ( x , t ) Ω × ( 0 , T ) Open image in new window

with initial and Dirichlet boundary conditions. We prove that, under suitable assumptions on the functions g i , f i (i = 1, 2) and certain initial data in the stable set, the decay rate of the solution energy is exponential. Conversely, for certain initial data in the unstable set, there are solutions with positive initial energy that blow up in finite time.

2000 Mathematics Subject Classifications: 35L05; 35L55; 35L70.

Keywords

decay blow-up positive initial energy viscoelastic wave equations 

1. Introduction

In this article, we study the following system of viscoelastic equations:
u t t - Δ u + 0 t g 1 ( t - τ ) Δ u ( τ ) d τ - Δ u t = f 1 ( u , v ) , ( x , t ) Ω × ( 0 , T ) , v t t - Δ v + 0 t g 2 ( t - τ ) Δ v ( τ ) d τ - Δ v t = f 2 ( u , v ) , ( x , t ) Ω × ( 0 , T ) , u ( x , t ) = v ( x , t ) = 0 , x Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω , v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x Ω , Open image in new window
(1.1)

where Ω is a bounded domain in ℝ n with a smooth boundary ∂Ω, and g i (·) : ℝ+ → ℝ+, f i (·, ·): ℝ2 → ℝ (i = 1, 2) are given functions to be specified later. Here, u and v denote the transverse displacements of waves. This problem arises in the theory of viscoelastic and describes the interaction of two scalar fields, we can refer to Cavalcanti et al. [1], Messaoudi and Tatar [2], Renardy et al. [3].

To motivate this study, let us recall some results regarding single viscoelastic wave equation. Cavalcanti et al. [4] studied the following equation:
u t t - Δ u + 0 t g ( t - τ ) Δ u ( τ ) d τ + a ( x ) u t + | u | γ u = 0 , i n Ω × ( 0 , ) Open image in new window
for a : Ω → ℝ+, a function, which may be null on a part of the domain Ω. Under the conditions that a(x) ≥ a0 > 0 on Ω1 ⊂ Ω, with Ω1 satisfying some geometry restrictions and
- ξ 1 g ( t ) g ( t ) - ξ 2 g ( t ) , t 0 , Open image in new window
the authors established an exponential rate of decay. This latter result has been improved by Cavalcanti and Oquendo [5] and Berrimi and Messaoudi [6]. In their study, Cavalcanti and Oquendo [5] considered the situation where the internal dissipation acts on a part of Ω and the viscoelastic dissipation acts on the other part. They established both exponential and polynomial decay results under the conditions on g and its derivatives up to the third order, whereas Berrimi and Messaoudi [6] allowed the internal dissipation to be nonlinear. They also showed that the dissipation induced by the integral term is strong enough to stabilize the system and established an exponential decay for the solution energy provided that g satisfies a relation of the form
g ( t ) - ξ g ( t ) , t 0 . Open image in new window
Cavalcanti et al. [1] also studied, in a bounded domain, the following equation:
| u t | ρ u t t - Δ u - Δ u t t + 0 t g 1 ( t - τ ) Δ u ( τ ) d τ - γ Δ u t = 0 , Open image in new window
ρ > 0, and proved a global existence result for γ ≥ 0 and an exponential decay for γ > 0. This result has been extended by Messaoudi and Tatar [2, 7] to the situation where γ = 0 and exponential and polynomial decay results in the absence, as well as in the presence, of a source term have been established. Recently, Messaoudi [8, 9] considered
u t t - Δ u + 0 t g 1 ( t - τ ) Δ u ( τ ) d τ = b | u | γ u , ( x , t ) Ω × ( 0 , ) , Open image in new window

for b = 0 and b = 1 and for a wider class of relaxation functions. He established a more general decay result, for which the usual exponential and polynomial decay results are just special cases.

For the finite time blow-up of a solution, the single viscoelastic wave equation of the form
u t t - Δ u + 0 t g ( t - τ ) Δ u ( τ ) d τ + h ( u t ) = f ( u ) Open image in new window
(1.2)

in Ω × (0, ∞) with initial and boundary conditions has extensively been studied. See in this regard, Kafini and Messaoudi [10], Messaoudi [11, 12], Song and Zhong [13], Wang [14]. For instance, Messaoudi [11] studied (1.2) for h(u t ) = a|u t |m-2u t and f(u) = b|u|p-2u and proved a blow-up result for solutions with negative initial energy if p > m ≥ 2 and a global result for 2 ≤ pm. This result has been later improved by Messaoudi [12] to accommodate certain solutions with positive initial energy. Song and Zhong [13] considered (1.2) for h(u t ) = -Δu t and f(u) = |u|p-2u and proved a blow-up result for solutions with positive initial energy using the ideas of the "potential well'' theory introduced by Payne and Sattinger [15].

This study is also motivated by the research of the well-known Klein-Gordon system
u t t - Δ u + m 1 u + k 1 u v 2 = 0 , v t t - Δ v + m 2 v + k 2 u 2 v = 0 , Open image in new window
which arises in the study of quantum field theory [16]. See also Medeiros and Miranda [17], Zhang [18] for some generalizations of this system and references therein. As far as we know, the problem (1.1) with the viscoelastic effect described by the memory terms has not been well studied. Recently, Han and Wang [19] considered the following problem
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 ) , ( x , t ) 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 ) , ( x , t ) Ω × ( 0 , T ) , u ( x , t ) = v ( x , t ) = 0 , x Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω , v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x Ω , Open image in new window
where Ω is a bounded domain with smooth boundary ∂Ω in ℝ n , n = 1, 2, 3. Under suitable assumptions on the functions g i , f i (i = 1, 2), the initial data and the parameters in the equations, they established several results concerning local existence, global existence, uniqueness, and finite time blow-up (the initial energy E(0) < 0) property. This latter blow-up result has been improved by Messaoudi and Said-Houari [20], to certain solutions with positive initial energy. Liu [21] studied the following system
| u t | ρ u t t - Δ u - γ 1 Δ u t t + 0 t g 1 ( t - τ ) Δ u ( τ ) d τ + f ( u , v ) = 0 , ( x , t ) Ω × ( 0 , T ) , | v t | ρ v t t - Δ v - γ 2 Δ v t t + 0 t g 2 ( t - τ ) Δ v ( τ ) d τ + k ( u , v ) = 0 , ( x , t ) Ω × ( 0 , T ) , u ( x , t ) = v ( x , t ) = 0 , x Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω , v ( x , 0 ) = v 0 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x Ω , Open image in new window

where Ω is a bounded domain with smooth boundary ∂Ω in ℝ n , γ1, γ2 ≥ 0 are constants and ρ is a real number such that 0 < ρ ≤ 2/(n - 2) if n ≥ 3 or ρ > 0 if n = 1, 2. Under suitable assumptions on the functions g(s), h(s), f(u, v), k(u, v), they used the perturbed energy method to show that the dissipations given by the viscoelastic terms are strong enough to ensure exponential or polynomial decay of the solutions energy, depending on the decay rate of the relaxation functions g(s) and h(s). For the problem (1.1) in ℝ n , we mention the work of Kafini and Messaoudi [10].

Motivated by the above research, we consider in this study the coupled system (1.1). We prove that, under suitable assumptions on the functions g i , f i (i = 1, 2) and certain initial data in the stable set, the decay rate of the solution energy is exponential. Conversely, for certain initial data in the unstable set, there are solutions with positive initial energy that blow up in finite time.

This article is organized as follows. In Section 2, we present some assumptions and definitions needed for this study. Section 3 is devoted to the proof of the uniform decay result. In Section 4, we prove the blow-up result.

2. Preliminaries

First, let us introduce some notation used throughout this article. We denote by || · || q the L q (Ω) norm for 1 ≤ q ≤ ∞ and by ||∇ · ||2 the Dirichlet norm in H 0 1 ( Ω ) Open image in new window which is equivalent to the H1(Ω)norm. Moreover, we set
( φ , ψ ) = Ω φ ( x ) ψ ( x ) d x Open image in new window

as the usual L2(Ω) inner product.

Concerning the functions f1(u, v) and f2(u, v), we take
f 1 ( u , v ) = [ a | u + v | 2 ( p + 1 ) ( u + v ) + b | u | p u | v | ( p + 2 ) ] , f 2 ( u , v ) = [ a | u + v | 2 ( p + 1 ) ( u + v ) + b | u | ( p + 2 ) | v | p v ] , Open image in new window
where a, b > 0 are constants and p satisfies
p > - 1 , i f n = 1 , 2 , - 1 < p 1 , i f n = 3 . Open image in new window
(2.1)
One can easily verify that
u f 1 ( u , v ) + v f 2 ( u , v ) = 2 ( p + 2 ) F ( u , v ) , ( u , v ) 2 , Open image in new window
where
F ( u , v ) = 1 2 ( p + 2 ) [ a | u + v | 2 ( p + 2 ) + 2 b | u v | p + 2 ] . Open image in new window

For the relaxation functions g i (t) (i = 1, 2), we assume

(G1) g i (t) : ℝ+ → ℝ+ belong to C1(ℝ+) and satisfy
g i ( t ) 0 , g i ( t ) 0 , f o r t 0 Open image in new window
and
1 - 0 g i ( s ) d s = k i > 0 . Open image in new window

(G2) max 0 g 1 ( s ) d s , 0 g 2 ( s ) d s < 4 ( p + 1 ) ( p + 2 ) 4 ( p + 1 ) ( p + 2 ) + 1 Open image in new window.

We next state the local existence and the uniqueness of the solution of problem (1.1), whose proof can be found in Han and Wang [19] (Theorem 2.1) with slight modification, so we will omit its proof. In the proof, the authors adopted the technique of Agre and Rammaha [22] which consists of constructing approximations by the Faedo-Galerkin procedure without imposing the usual smallness conditions on the initial data to handle the source terms. Unfortunately, due to the strong nonlinearities on f1 and f2, the techniques used by Han and Wang [19] and Agre and Rammaha [22] allowed them to prove the local existence result only for n ≤ 3. We note that the local existence result in the case of n > 3 is still open. For related results, we also refer the reader to Said-Houari and Messaoudi [23] and Messaoudi and Said-Houari [20]. So throughout this article, we have assumed that n ≤ 3.

Theorem 2.1. Assume that (2.1) and (G 1) hold, and that ( u 0 , u 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) Open image in new window, ( v 0 , v 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) Open image in new window. Then problem (1.1) has a unique local solution
u , v C ( [ 0 , T ) ; H 0 1 ( Ω ) ) , u t , v t C ( [ 0 , T ) ; L 2 ( Ω ) ) L 2 ( [ 0 , T ) ; H 0 1 ( Ω ) ) Open image in new window
for some T > 0. If T < ∞, then
lim t T ( k 1 | | u ( t ) | | 2 2 + | | u t ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + | | v t ( t ) | | 2 2 ) = . Open image in new window
(2.2)
Finally, we define
I ( t ) = ( 1 - 0 t g 1 ( τ ) d τ ) | | u ( t ) | | 2 2 + 1 - 0 t g 2 ( τ ) d τ | | v ( t ) | | 2 2 + [ ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ] - 2 ( p + 2 ) Ω F ( u , v ) d x , Open image in new window
(2.3)
J ( t ) = 1 2 1 - 0 t g 1 ( τ ) d τ | | u ( t ) | | 2 2 + 1 - 0 t g 2 ( τ ) d τ | | v ( t ) | | 2 2 + 1 2 [ ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ] - Ω F ( u , v ) d x , Open image in new window
(2.4)
such functionals we could refer to Muñoz Rivera [24, 25]. We also define the energy function as follows
E ( t ) = 1 2 | | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 + J ( t ) , Open image in new window
(2.5)
where
( g i w ) ( t ) = 0 t g i ( t - τ ) | | w ( t ) - w ( τ ) | | 2 2 d τ . Open image in new window

3. Global existence and energy decay

In this section, we deal with the uniform exponential decay of the energy for system (1.1) by using the perturbed energy method. Before we state and prove our main result, we need the following lemmas.

Lemma 3.1. Assume (2.1) and (G 1) hold. Let (u, v) be the solution of the system (1.1), then the energy functional is a decreasing function, that is
E ( t ) = - | | u t ( t ) | | 2 2 - | | v t ( t ) | | 2 2 + 1 2 ( g 1 u ) ( t ) + 1 2 ( g 2 v ) ( t ) - 1 2 g 1 ( t ) | | u ( t ) | | 2 2 - 1 2 g 2 ( t ) | | v ( t ) | | 2 2 0 . Open image in new window
(3.1)
Moreover, the following energy inequality holds:
E ( t ) + s t ( | | u t ( τ ) | | 2 2 + | | v t ( τ ) | | 2 2 ) d τ E ( s ) , f o r 0 s t < T . Open image in new window
(3.2)
Lemma 3.2. Let (2.1) hold. Then, there exists η > 0 such that for any ( u , v ) H 0 1 ( Ω ) × H 0 1 ( Ω ) Open image in new window, we have
| | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | p + 2 p + 2 η ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 ) p + 2 . Open image in new window
(3.3)

Proof. The proof is almost the same that of Said-Houari [26], so we omit it here. □

To prove our result and for the sake of simplicity, we take a = b = 1 and introduce the following:
B = η 1 2 ( p + 2 ) , α * = B - p + 2 P + 1 , E 1 = 1 2 - 1 2 ( p + 2 ) α * 2 , Open image in new window
(3.4)

where η is the optimal constant in (3.3). The following lemma will play an essential role in the proof of our main result, and it is similar to a lemma used first by Vitillaro [27], to study a class of a single wave equation, which introduces a potential well.

Lemma 3.3. Let (2.1) and (G 1) hold. Let (u, v) be the solution of the system (1.1). Assume further that E(0) < E1and
( k 1 | | u 0 | | 2 2 + k 2 | | v 0 | | 2 2 ) 1 2 < α * , Open image in new window
(3.5)
Then
( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ) 1 2 < α * , f o r t [ 0 , T ) . Open image in new window
(3.6)
Proof. We first note that, by (2.5), (3.3) and the definition of B, we have
E ( t ) 1 2 ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ) - 1 2 ( p + 2 ) ( | | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | p + 2 p + 2 ) 1 2 ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ) - B 2 ( p + 2 ) 2 ( p + 2 ) ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 ) p + 2 1 2 α 2 - B 2 ( p + 2 ) 2 ( p + 2 ) α 2 ( p + 2 ) = g ( α ) , Open image in new window
(3.7)
where α = ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ) 1 / 2 Open image in new window. It is not hard to verify that g is increasing for 0 < α < α*, decreasing for α > α*, g(α) → - ∞ as α → +∞, and
g ( α * ) = 1 2 α * 2 - B 2 ( p + 2 ) 2 ( p + 2 ) α * 2 ( p + 2 ) = E 1 , Open image in new window
where α* is given in (3.4). Now we establish (3.6) by contradiction. Suppose (3.6) does not hold, then it follows from the continuity of (u(t), v(t)) that there exists t0 ∈ (0, T) such that
( k 1 | | u ( t 0 ) | | 2 2 + k 2 | | v ( t 0 ) | | 2 2 + ( g 1 u ) ( t 0 ) + ( g 2 v ) ( t 0 ) ) 1 2 = α * . Open image in new window
By (3.7), we observe that
E ( t 0 ) g ( k 1 | | u ( t 0 ) | | 2 2 + k 2 | | v ( t 0 ) | | 2 2 + ( g 1 u ) ( t 0 ) + ( g 2 v ) ( t 0 ) ) 1 2 = g ( α * ) = E 1 . Open image in new window

This is impossible since E(t) ≤ E(0) < E1 for all t ∈ [0, T). Hence (3.6) is established. □

The following integral inequality plays an important role in our proof of the energy decay of the solutions to problem (1.1).

Lemma 3.4. [28]Assume that the function φ : ℝ+ ∪ {0} → ℝ+ ∪ {0} is a non-increasing function and that there exists a constant c > 0 such that
t φ ( s ) d s c φ ( t ) Open image in new window
for every t ∈ [0, ∞). Then
φ ( t ) φ ( 0 ) exp ( 1 - t / c ) Open image in new window

for every tc.

Theorem 3.5. Let (2.1) and (G 1) hold. If the initial data ( u 0 , u 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) Open image in new window, ( v 0 , v 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) Open image in new windowsatisfy E(0) < E1and
( k 1 | | u 0 | | 2 2 + k 2 | | v 0 | | 2 2 ) 1 2 < α * , Open image in new window
(3.8)
where the constants α*, E1are defined in (3.4), then the corresponding solution to (1.1) globally exists, i.e. T = ∞. Moreover, if the initial energy E(0) and k such that
1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) - 5 ( 1 - k ) ( p + 2 ) 2 k ( p + 1 ) > 0 , Open image in new window
where k = min{k1, k2}, then the energy decay is
E ( t ) E ( 0 ) exp ( 1 - a C - 1 t ) Open image in new window

for every taC-1, where C is some positive constant.

Proof. In order to get T = ∞, by (2.2), it suffices to show that
| | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 + k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 Open image in new window
is bounded independently of t. Since E(0) < E1 and
( k 1 | | u 0 | | 2 2 + k 2 | | v 0 | | 2 2 ) 1 2 < α * , Open image in new window
it follows from Lemma 3.3 that
k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) < α * 2 , Open image in new window
which implies that
I ( t ) k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + [ ( g 1 u ) ( t ) + ( g 2 v ) ( t ) ] - 2 ( p + 2 ) Ω F ( u , v ) d x k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 - 2 ( p + 2 ) Ω F ( u , v ) d x = k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 - ( | | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | p + 2 p + 2 ) k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 - η ( k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 ) p + 2 0 , f o r t [ 0 , T ) , (5)  Open image in new window
where we have used (3.3). Furthermore, by (2.3) and (2.4), we get
J ( t ) 1 2 - 1 2 ( p + 2 ) ( 1 - 0 t g 1 ( s ) d s ) | | u ( t ) | | 2 2 + 1 - 0 t g 2 ( s ) d s | | v ( t ) | | 2 2 + 1 2 - 1 2 ( p + 2 ) ( g 1 u ) ( t ) + ( g 2 v ) ( t ) + 1 2 ( p + 2 ) I ( t ) p + 1 2 ( p + 2 ) k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + ( g 1 u ) ( t ) + ( g 2 v ) ( t ) + 1 2 ( p + 2 ) I ( t ) 0 , (4)  Open image in new window
from which, the definition of E(t) and E(t) ≤ E(0), we deduce that
k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 2 ( p + 2 ) ( p + 1 ) J ( t ) 2 ( p + 2 ) ( p + 1 ) E ( t ) 2 ( p + 2 ) ( p + 1 ) E ( 0 ) , Open image in new window
(3.9)
for t ∈ [0, T). So it follows from (16) and Lemma 3.1 that
p + 1 2 ( p + 2 ) k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 + 1 2 ( | | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 ) J ( t ) + 1 2 ( | | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 ) = E ( t ) E ( 0 ) < E 1 , t [ 0 , T ) , (3)  Open image in new window
which implies
| | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 + k 1 | | u ( t ) | | 2 2 + k 2 | | v ( t ) | | 2 2 < C E 1 , Open image in new window

where C is a positive constant depending only on p.

Next we want to derive the decay rate of energy function for problem (1.1). By multiplying the first equation of system (1.1) by u and the second equation of system (1.1) by v, integrating over Ω × [t1, t2] (0 ≤ t1t2), using integration by parts and summing up, we have
Ω u t ( t ) u ( t ) d x | t 1 t 2 - t 1 t 2 | | u t ( t ) | | 2 2 d t + Ω v t ( t ) v ( t ) d x | t 1 t 2 - t 1 t 2 | | v t ( t ) | | 2 2 d t = - t 1 t 2 ( u ( t ) , u t ( t ) ) d t - t 1 t 2 ( v ( t ) , v t ( t ) ) d t - t 1 t 2 | | u ( t ) | | 2 2 d t - t 1 t 2 | | v ( t ) | | 2 2 d t - t 1 t 2 Ω 0 t g 1 ( t - τ ) Δ u ( τ ) d τ u ( t ) d x d t - t 1 t 2 Ω 0 t g 2 ( t - τ ) Δ v ( τ ) d τ v ( t ) d x d t + 2 ( p + 2 ) t 1 t 2 Ω F ( u , v ) d x d t , Open image in new window
which implies
2 t 1 t 2 E ( t ) d t - 2 ( p + 1 ) t 1 t 2 Ω F ( u , v ) d x d t = - Ω u t ( t ) u ( t ) d x | t 1 t 2 - Ω v t ( t ) v ( t ) d x | t 1 t 2 + 2 t 1 t 2 | | u t ( t ) | | 2 2 d t + 2 t 1 t 2 | | v t ( t ) | | 2 2 d t + t 1 t 2 ( g 1 u ) ( t ) d t + t 1 t 2 ( g 2 v ) ( t ) d t - t 1 t 2 0 t g 1 ( τ ) d τ | | u ( t ) | | 2 2 d t - t 1 t 2 0 t g 2 ( τ ) d τ | | v ( t ) | | 2 2 d t - t 1 t 2 ( u ( t ) , u t ( t ) ) d t - t 1 t 2 ( v ( t ) , v t ( t ) ) d t - t 1 t 2 Ω 0 t g 1 ( t - τ ) Δ u ( τ ) d τ u ( t ) d x d t - t 1 t 2 Ω 0 t g 2 ( t - τ ) Δ v ( τ ) d τ v ( t ) d x d t . Open image in new window
(3.10)
For the 11th term on the right-hand side of (3.10), one has
- 2 Ω 0 t g 1 ( t - τ ) Δ u ( τ ) d τ u ( t ) d x = 2 Ω 0 t g 1 ( t - τ ) u ( τ ) u ( t ) d τ d x = 0 t g 1 ( t - τ ) ( | | u ( t ) | | 2 2 + | | u ( τ ) | | 2 2 ) d τ - 0 t g 1 ( t - τ ) ( | | u ( t ) - u ( τ ) | | 2 2 ) d τ . Open image in new window
(3.11)
Similarly,
- 2 Ω 0 t g 2 ( t - τ ) Δ v ( τ ) d τ v ( t ) d x = 0 t g 2 ( t - τ ) ( | | v ( t ) | | 2 2 + | | v ( τ ) | | 2 2 ) d τ - 0 t g 2 ( t - τ ) ( | | v ( t ) - v ( τ ) | | 2 2 ) d τ . Open image in new window
(3.12)
Combining (3.10), (3.11) with (3.12), we have
2 t 1 t 2 E ( t ) d t - 2 ( p + 1 ) t 1 t 2 Ω F ( u , v ) d x d t = - Ω u t ( t ) u ( t ) d x | t 1 t 2 - Ω v t ( t ) v ( t ) d x | t 1 t 2 + 2 t 1 t 2 | | u t ( t ) | | 2 2 d t + 2 t 1 t 2 | | v t ( t ) | | 2 2 d t + 1 2 t 1 t 2 ( g 1 u ) ( t ) d t + 1 2 t 1 t 2 ( g 2 v ) ( t ) d t - 1 2 t 1 t 2 0 t g 1 ( τ ) d τ | | u ( t ) | | 2 2 d t - 1 2 t 1 t 2 0 t g 2 ( τ ) d τ | | v ( t ) | | 2 2 d t - t 1 t 2 ( u ( t ) , u t ( t ) ) d t - t 1 t 2 ( v ( t ) , v t ( t ) ) d t + 1 2 t 1 t 2 0 t g 1 ( t - τ ) | | u ( τ ) | | 2 2 d τ d t + 1 2 t 1 t 2 0 t g 2 ( t - τ ) | | v ( τ ) | | 2 2 d τ d t - Ω u t ( t ) u ( t ) d x | t 1 t 2 - Ω v t ( t ) v ( t ) d x | t 1 t 2 + 2 t 1 t 2 | | u t ( t ) | | 2 2 d t + 2 t 1 t 2 | | v t ( t ) | | 2 2 d t + 1 2 t 1 t 2 ( g 1 u ) ( t ) d t + 1 2 t 1 t 2 ( g 2 v ) ( t ) d t - t 1 t 2 ( u ( t ) , u t ( t ) ) d t - t 1 t 2 ( v ( t ) , v t ( t ) ) d t + 1 2 t 1 t 2 0 t g 1 ( t - τ ) | | u ( τ ) | | 2 2 d τ d t + 1 2 t 1 t 2 0 t g 2 ( t - τ ) | | v ( τ ) | | 2 2 d τ d t . Open image in new window
(3.13)
Now we estimate every term of the right-hand side of the (3.13). First, by Hölder's inequality and Poincaré's inequality
Ω | u ( t ) u t ( t ) | d x + Ω | v ( t ) v t ( t ) | d x 1 2 | | u ( t ) | | 2 2 + 1 2 | | u t ( t ) | | 2 2 + 1 2 | | v ( t ) | | 2 2 + 1 2 | | v t ( t ) | | 2 2 λ 2 | | u ( t ) | | 2 2 + 1 2 | | u t ( t ) | | 2 2 + λ 2 | | v ( t ) | | 2 2 + 1 2 | | v t ( t ) | | 2 2 , (3)  Open image in new window
where λ being the first eigenvalue of the operator - Δ under homogeneous Dirichlet boundary conditions. Then, by (3.9), we see that
Ω | u ( t ) u t ( t ) | d x + Ω | v ( t ) v t ( t ) | d x c 1 E ( t ) , Open image in new window
where c1 is a constant independent on u and v, from which follows that
Ω | u ( t ) u t ( t ) | d x | t 1 t 2 + Ω | v ( t ) v t ( t ) | d x | t 1 t 2 2 c 1 E ( t 1 ) . Open image in new window
(3.14)
Since 0 ≤ J (t) ≤ E (t), from (3.2) we deduce that
t 1 t 2 ( | | u t ( t ) | | 2 2 + | | v t ( t ) | | 2 2 ) d t E ( t 1 ) . Open image in new window
Hence, by Poincaré inequality we get
2 t 1 t 2 | | u t ( t ) | | 2 2 d t + 2 t 1 t 2 | | v t ( t ) | | 2 2 d t 2 c 2 E ( t 1 ) , Open image in new window
(3.15)
where c2 is a constant independent on u and v. In addition, using Young's inequality for convolution ||f * g || q ≤ || f || r ||g|| s with 1/q = 1/r + 1/s - 1 and 1 ≤ q, r, s ≤ ∞, noting that if q = 1, then r = 1 and s = 1, we have
t 1 t 2 0 t g 1 ( t - τ ) | | u ( τ ) | | 2 2 d τ d t = | | g 1 * | | u | | 2 2 | | 1 | | g 1 | | 1 | | | | u | | 2 2 | | 1 = t 1 t 2 g 1 ( t ) d t t 1 t 2 | | u ( t ) | | 2 2 d t ( 1 - k 1 ) t 1 t 2 | | u ( t ) | | 2 2 d t , Open image in new window
(3.16)
and
t 1 t 2 0 t g 2 ( t - τ ) | | v ( τ ) | | 2 2 d τ d t = | | g 2 * | | v | | 2 2 | | 1 | | g 2 | | 1 | | | | v | | 2 2 | | 1 = t 1 t 2 g 2 ( t ) d t t 1 t 2 | | v ( t ) | | 2 2 d t ( 1 - k 2 ) t 1 t 2 | | v ( t ) | | 2 2 d t . Open image in new window
(3.17)
Hence, combining (3.9), (3.16) with (3.17) we then have
t 1 t 2 0 t g 1 ( t - τ ) | | u ( τ ) | | 2 2 d τ d t + t 1 t 2 0 t g 2 ( t - τ ) | | v ( τ ) | | 2 2 d τ d t ( 1 - k 1 ) t 1 t 2 | | u ( t ) | | 2 2 d t + ( 1 - k 2 ) t 1 t 2 | | v ( t ) | | 2 2 d t ( 1 - k ) t 1 t 2 ( | | u ( t ) | | 2 2 + | | v ( t ) | | 2 2 ) d t 2 ( 1 - k ) ( p + 2 ) k ( p + 1 ) t 1 t 2 E ( t ) d t . Open image in new window
(3.18)
From (3.9), we also have
t 1 t 2 0 t g 1 ( t - τ ) | | u ( t ) | | 2 2 d τ d t + t 1 t 2 0 t g 2 ( t - τ ) | | v ( t ) | | 2 2 d τ d t ( 1 - k 1 ) t 1 t 2 | | u ( t ) | | 2 2 d t + ( 1 - k 2 ) t 1 t 2 | | v ( t ) | | 2 2 d t ( 1 - k ) t 1 t 2 ( | | u ( t ) | | 2 2 + | | v ( t ) | | 2 2 ) d t 2 ( 1 - k ) ( p + 2 ) k ( p + 1 ) t 1 t 2 E ( t ) d t . Open image in new window
(3.19)
Combining (3.18) with (3.19), we deduce that
1 2 t 1 t 2 ( g 1 u ) ( t ) d t + 1 2 t 1 t 2 ( g 2 v ) ( t ) d t t 1 t 2 0 t g 1 ( t - τ ) ( | | u ( τ ) | | 2 2 + | | u ( t ) | | 2 2 ) d τ d t + t 1 t 2 0 t g 2 ( t - τ ) ( | | v ( τ ) | | 2 2 + | | v ( t ) | | 2 2 ) d τ d t 4 ( 1 - k ) ( p + 2 ) k ( p + 1 ) t 1 t 2 E ( t ) d t . Open image in new window
(3.20)
Finally, we also have the following estimate
t 1 t 2 ( u ( t ) , u t ( t ) ) d t + t 1 t 2 ( v ( t ) , v t ( t ) ) d t = 1 2 t 1 t 2 d d t | | u ( t ) | | 2 2 d t + 1 2 t 1 t 2 d d t | | v ( t ) | | 2 2 d t = 1 2 ( | | u ( t 2 ) | | 2 2 - | | u ( t 1 ) | | 2 2 ) + 1 2 ( | | v ( t 2 ) | | 2 2 - | | v ( t 1 ) | | 2 2 ) 2 ( p + 2 ) k ( p + 1 ) E ( t 1 ) c 3 E ( t 1 ) . Open image in new window
(3.21)
where c3 is a constant independent on u and v. Combining (3.13)-(3.21), we obtain
2 t 1 t 2 E ( t ) d t - 2 ( p + 1 ) t 1 t 2 Ω F ( u , v ) d x d t C E ( t 1 ) + 5 ( 1 - k ) ( p + 2 ) k ( p + 1 ) t 1 t 2 E ( t ) d t . Open image in new window
(3.22)

where C is a constant independent on u.

On the other hand, from (3.3) and (3.9), we have
2 ( p + 1 ) Ω F ( u , v ) d x = p + 1 p + 2 | | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | ( p + 2 ) ( p + 2 ) p + 1 p + 2 η ( k 1 | | u | | 2 2 + k 2 | | v | | 2 2 ) ( p + 2 ) 2 η 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ( p + 1 ) E ( t ) , Open image in new window
which implies
2 t 1 t 2 E ( t ) d t - 2 ( p + 1 ) t 1 t 2 Ω F ( u , v ) d x d t 2 1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) t 1 t 2 E ( t ) d t . Open image in new window
(3.23)
Note that E(0) < E1, we see that
1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) > 0 . Open image in new window
Thus, combining (3.22) with (3.23), we have
2 1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) t 1 t 2 E ( t ) d t C E ( t 1 ) + 5 ( 1 - k ) ( p + 2 ) k ( p + 1 ) t 1 t 2 E ( t ) d t , Open image in new window
that is
2 1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) - 5 ( 1 - k ) ( p + 2 ) 2 k ( p + 1 ) t 1 t 2 E ( t ) d t C E ( t 1 ) . Open image in new window
(3.24)
Denote
a = 2 1 - η ( 2 ( p + 2 ) ( p + 1 ) E ( 0 ) ) ( p + 1 ) - 5 ( 1 - k ) ( p + 2 ) 2 k ( p + 1 ) . Open image in new window
We rewrite (3.24)
a t E ( τ ) d τ C E ( t ) Open image in new window

for every t ∈ [0, ∞).

Since a > 0 from the assumption conditions, by Lemma 3.4, we obtain the following energy decay for problem (1.1) as
E ( t ) < E ( 0 ) exp ( 1 - a C - 1 t ) Open image in new window

for every tCa-1. □

4. Blow-up of solution

In this section, we deal with the blow-up solutions of the system (1.1). Set
θ i = k i - 1 4 ( p + 2 ) ( p + 1 ) 0 g i ( s ) d s , i = 1 , 2 . Open image in new window
(4.1)

From the assumption (G 2), we have θ i > 0 (i = 1, 2). Similarly Lemma 3.2, we have

Lemma 4.1. Assume (2.1) holds. Then there exists η1> 0 such that for any ( u , v ) H 0 1 ( Ω ) × H 0 1 ( Ω ) Open image in new window, we have
| | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | p + 2 p + 2 η 1 ( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) p + 2 , Open image in new window
(4.2)

where the constants θ i (i = 1, 2) are defined in (4.1).

To prove our result and for the sake of simplicity, we take a = b = 1 and introduce the following:
B 1 = η 1 1 2 ( p + 2 ) , α * = B 1 - p + 2 p + 1 , E 2 = 1 2 - 1 2 ( p + 2 ) α * 2 . Open image in new window
(4.3)

Then we have

Lemma 4.2. Let (G 1), G(2) and (2.1) hold. Let (u, v) be the solution of the system (1.1). Assume further that E(0) < E2and
( θ 1 | | u 0 | | 2 2 + θ 2 | | v 0 | | 2 2 ) 1 2 > α * , Open image in new window
(4.4)
where the constants θ i (i = 1, 2) are defined in (4.1). Then there exists a constant α2 > α*such that
( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) 1 2 α 2 , f o r t ( 0 , T ) . Open image in new window
(4.5)
Proof. We first note that, by (2.5), (4.2) and the definition of B1, we have
E ( t ) 1 2 ( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) - 1 2 ( p + 2 ) ( | | u + v | | 2 ( p + 2 ) 2 ( p + 2 ) + 2 | | u v | | p + 2 p + 2 ) 1 2 ( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) - B 1 2 ( p + 2 ) 2 ( p + 2 ) ( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) p + 2 = 1 2 α 2 - B 1 2 ( p + 2 ) 2 ( p + 2 ) α 2 ( p + 2 ) , (4)  Open image in new window
(4.6)
where α = ( θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 ) 1 / 2 Open image in new window. It is not hard to verify that g is increasing for 0 < α < α * , decreasing for α > α*, g(α) → - ∞ as α → + ∞, and
g ( α * ) = 1 2 α * 2 - B 1 2 ( p + 2 ) 2 ( p + 2 ) α * 2 ( p + 2 ) = E 2 , Open image in new window

where α* is given in (4.3). Since E(0) < E2, there exists α2 > α* such that g(α2) = E(0).

Set α 0 = ( θ 1 | | u 0 | | 2 2 + θ 2 | | v 0 | | 2 2 ) 1 / 2 Open image in new window, then by (4.6) we get g(α0) ≤ E(0) = g (α2), which implies that α0α2. Now, to establish (4.5), we suppose by contradiction that
( θ 1 | | u ( t 0 ) | | 2 2 + θ 2 | | v ( t 0 ) | | 2 2 ) 1 2 < α 2 , Open image in new window
for some t0 > 0. By the continuity of θ 1 | | u ( t ) | | 2 2 + θ 2 | | v ( t ) | | 2 2 Open image in new window we can choose t0 such that
( θ 1 | | u ( t 0 ) | | 2 2 + θ 2 | | v ( t 0 ) | | 2 2 ) 1 2 > α * . Open image in new window
Again, the use of (4.6) leads to
E ( t 0 ) g ( ( θ 1 | | u ( t 0 ) | | 2 2 + θ 2 | | v ( t 0 ) | | 2 2 ) 1 / 2 ) > g ( α 2 ) = E ( 0 ) Open image in new window

This is impossible since E(t) ≤ E(0) for all t ∈ [0, T). Hence (4.5) is established. □

Theorem 4.3. Assume (G 1), (G 2) and (2.1) hold. Then any solution of problem (1.1) with initial data satisfying
( θ 1 | | u 0 | | 2 2 + θ 2 | | v 0 | | 2 2 ) 1 2 > α * a n d E ( 0 ) < E 2 Open image in new window

blows up in finite time, where the constants θ i (i = 1, 2) are defined in (4.1) and α*, E2are defined in (4.3).

Proof. Assume by contradiction that the solution (u, v) is global. Then, for any T > 0 we consider H(t) : [0, T] → ℝ+ defined by
H ( t ) = | | u ( t ) | | 2 2 + | | v ( t ) | | 2 2 + 0 t | | u ( τ ) | | 2 2 d τ + 0 t | | v ( τ ) | | 2 2 d τ + ( T - t ) ( | | u 0 | | 2 2 + | | v 0 | | 2 2 ) + β ( t + s 0 ) 2 , Open image in new window
where β and s0 are positive constants to be determined later. A direct computation yields
H ( t ) = 2 Ω u ( t ) u t ( t ) d x + 2 Ω v ( t ) v t ( t ) d x + | | u ( t ) | | 2 2 + | | v ( t ) | | 2 2 - | | u 0 | | 2 2 - | | v 0 | | 2 2 + 2 β ( t + s 0 ) = 2 Ω u ( t ) u t ( t ) d x + 2 Ω v ( t ) v t ( t ) d x + 2 0 t ( u ( τ ) , u t ( τ ) ) d τ + 2 0 t ( v ( τ ) , v t ( τ ) ) d τ + 2 β ( t + s 0 ) Open image in new window
and
H ( t ) = 2 Ω u ( t ) u t t ( t ) d x + 2 Ω v ( t ) v t t ( t ) d x + 2 | | u t ( t ) | | 2 2 + 2 | | v t ( t ) | | 2 2 + 2 ( u ( t ) , u t ( t ) ) + 2 ( v ( t ) , v t ( t ) ) + 2 β f o r a . e . t [ 0 , T ] . Open image in new window
Multiplying the first equation of system (1.1) by u and the second equation of system (1.1) by v, integrating over Ω, using integration by parts and summing up, we have
( u t t , u ( t ) ) + ( v t t , v ( t ) ) + ( u ( t ) , u t ( t ) ) + ( v ( t ) , v t ( t ) ) = - | | u ( t ) | | 2 2 - | | v ( t ) | | 2 2 - Ω 0 t g 1 ( t - τ ) Δ u ( τ ) d τ u ( t ) d x - Ω 0 t g 2 ( t - τ ) Δ v ( τ ) d τ v ( t ) d x + 2 ( p + 2 ) Ω F ( u , v ) d x , Open image in new window
which implies
H ( t ) = 2 | | u t ( t ) | | 2 2 + 2 | | v t ( t ) | | 2 2 - 2 | | u ( t ) | | 2 2 - 2 | | v ( t ) | | 2 2 + 4 ( p + 2 ) Ω F ( u , v ) d x - 2 Ω 0 t g 1 ( t - τ ) Δ u ( τ ) d τ u ( t ) d x - 2 Ω 0 t g 2 ( t - τ ) Δ v ( τ ) d τ v ( t ) d x + 2 β . (3)  Open image in new window