#### Abstract

This present work deals with the blow up of the coupled Klein-Gordon system with strong damping, distributed delay, and source terms, under suitable conditions.

#### 1. Introduction

In the present paper, we consider the following system:whereand , and are the time delay with , and are a *L*^{∞} functions, and are differentiable functions.

Viscous materials are the opposite of flexible materials that have dissipate mechanical energy and the ability to store. The mechanical properties of viscous materials are so important that we find them in many applications of natural sciences. Many authors have been concerned with this problem in recent decades.

If there is only one equation and if , that is, for absence of , and . Our problem (1) has been studied by Berrimi and Messaoudi [1]. Using Galerkin’s method they proved the result of local existence. They also made it clear that the local solution is global in time under suitable conditions and at the same rate of decaying (exponential or polynomial) of the kernel . In addition, the authors themselves demonstrated that the dissipation can be deduced by the term viscous integral and that it is strong enough to stabilize the solution oscillations. Their results were also obtained under weaker conditions than those used by Cavalcanti et al. [2].

In [3], the authors considered the following problem:where the authors proved the exponential decay result. This subsequent result was improved by Berrimi et al. in [1], as they showed that the viscosity elastic dissipation alone is strong enough to stabilize the problem even with the exponential rate with respect to the kernel assumptions. In the case , in problem (1), Kafini and Messaoudi in [4] proved a blow up result for the following problem:where satisfies

The initial data was backed by negative energy as

In [5], Song and Xue considered the following problem:where the authors showed that there were solutions of (7) with initial energy according to suitable assumptions on . Moreover, they showed the blow up in a finite time. Then, the same authors in [6] continued to prove that there were solutions of (7) with positive initial energy that blow up in finite time. In [7], the author studied the following problem:where they proved the exponential growth result under suitable assumptions. The authors in [8] studied the following problem:where they showed a blow up result if and established the global existence. In the coupled equation case, the authors in [9] studied the following system:with and nonlinear functions satisfying appropriate conditions. According to certain restrictions imposed on the initial data and parameters, they obtained numerous results on the existence of weak solutions. They obtained many results on the presence of weak solutions. In addition, by using the same techniques similar to that in [10] with negative initial, energy blows up for a finite period of time.

In [11], the authors have proved the solution of the problem:where under some restrictions on positive initial energy for certain conditions on the functions and , the authors proved the blows up in finite time of solution.

The result of [11] has been extended by the authors in [12], where they studied the following system:they proved that the solutions of a system of wave equations with degenerate damping, viscoelastic term and strong nonlinear sources acting in both equations at the same time are globally nonexisting provided that the initial data are sufficiently large in a bounded domain of .

As complement to these works, we are working to prove the blow up result with distributed delay of problem (1), under appropriate assumptions, and we prove these results using the energy method. In the following, let ,.

The present paper is organized as follows. In Section 2, we give some necessarily assumptions for the main result. In Section 3, we prove the blow up result.

#### 2. Assumptions

We consider the following suitable assumptions.(A1) are differentiable and decreasing functions such that(A2)There exists a constants such that(A3) are a functions so that

#### 3. Blow up

In this section, we obtain the proof of the blow up result of the solution of problem (1). First, of all in [13], we introduce the new variablesthen, we obtain

Let us denote by

Therefore, problem (1) get the following form:with initial and boundary conditionswhere

Theorem 1. *Assume (14), (16), and (17) hold. Let*

For any initial data,wherethen, problem (22) has a unique solutionfor some .

Lemma 2. *There exists a function such thatwherewe take for convenience.*

Lemma 3. (see [12]). *There exist two positive constants and such that*

We define the energy functional (see, e.g., [14–16] and reference therein).

Lemma 4. *Assume (14), (16), (17), and (25) hold, let be a solution of (22), then is nonincreasing, that is,satisfieswhere*

*Proof. *By multiplying (3.4)_{1}, (3.4)_{2} by , and integrating over , we getand, from (3.4)_{3}, (3.4)_{4}, we havethen, we get

And by using Young’s inequality, (14), (16), and (17) in (38), we obtain (33).

Now, we define the functional

Theorem 5. *Assume (14)–(17) and (25) hold. Assume further that , then the solution of problem (22) blow up in finite time.*

*Proof. *From (32), we haveTherefore,hence,We setwhere to be assigned later and

By multiplying (3.4)_{1}, (3.4)_{2} by and with a derivative of (45), we get

Using Young’s inequality, we getand we have

We obtain, from (47),

Therefore, using (43) and by setting *δ*_{1}, *δ*_{1} so that, and , substituting in (50), we get

For , from (39),substituting in (51), we get

Since (25) hold, we obtain by using (44) and (46)for some positive constants *c*_{4}, *c*_{5}. By using (46) and the algebraic inequality,we have, where . Also, sincewe concludesubstituting (58) and (61) in (55), we get

Combining (53) and (63), using (31), we getwhere , .

In this case, we take small enough, thenassumingwe havechoose so large thatfix and *a*, we appoint small enough so that

Then, for , we estimate (64) and it becomes

By (31), for , we get

Using Holder’s and Young’s inequalities, we havewhere put , to get

Subsequently, for and by using (39), we get

Therefore,