#### Abstract

This work deals with the blow-up of solutions for a new class of quasilinear wave equation with variable exponent nonlinearities. To clarify more, we prove in the presence of dispersion term a finite-time blow-up result for the solutions with negative initial energy and also for certain solutions with positive energy. Our results are extension of the recent work (Appl Anal. 2017; 96(9): 1509-1515).

#### 1. Introduction

We study in this paper the following nonlinear wave equation: Here, , be a bounded domain with a smooth boundary , and are constants, and the exponents , , and are given measurable functions on satisfying with

Also, we suppose that , , and satisfy the log-Holder continuity condition:

. In (4), if , the inequality is undefined because is undefined. The inequality is defined for not equal to , but the condition that is completely greater than zero always makes not equal to because . The term is called -Laplacian.

There are many studies that have studied the problem in the case of constant and variable-exponent nonlinearities.

In the case of constant exponent nonlinearity when , , and the explosion term forces the negative-energy solutions to explode in finite time ([1, 2]), whereas when , , and , the dissipation term guarantees the existence of global solutions for any initial data [3].

The problem was first treated by (Levine [2] and Vitillaro [4]) in the case when the both terms are present (the dissipation and source). He debated the case when , and determined the result of blow-up for solutions with negative initial energy. To extend Levine’s results in [5] considered a different method when and discussed the cases when and .

Chen et al. in [6] looked into the nonlinear Laplacian wave equation: when and are given functions. Under suitable conditions on the initial data and the functions , they realized global existence and uniqueness and also discussed the long-time behavior of the solution.

In [7], Benaissa and Mokeddem considered and they achieved an energy decay estimate for the solutions where , , is a positive function and expanded Yang [8] and Messaoudi [9] results. Recently, Mokeddem and Mansour [10] added some modification in the problem of Benaissa and Mokeddem [10] and established the same decay result.

Messaoudi and Houari [11] studied the nonlinear wave equation: where is a bounded domain in , and . They investigated with appropriate conditions imposed on , a global nonexistence result for solutions associated with negative initial energy. In [12], Kafini and Messaoudi treated a nonlinear wave equation with delay term and proved, under appropriate hypotheses on the initial data, that the energy of solutions explodes in a finite time. For more results, see the previous studies ([13–35]). In the case of variable-exponent nonlinearity, Antontsev [36] looked into the problem: when is a nonnegative constant, and are given functions. He discussed the case when and and demonstrated a blow-up result under particular hypothesis on .

Thereafter, Antontsev in [36] considered the same equation and established a local, global existence of weak solutions for specific conditions on and realized a blow-up results for solutions with nonpositive initial energy. In [5], Guo and Gao considered the same problem of Antontsev [13], they picked the constant and realized a blow-up result in finite time, and also, they alleged without any proof the same blow-up result for

Sun et al. in [37] studied the blow-up result for solutions with positive initial energy for the following equation:

They also gave lower and upper bounds for the blow-up time and provided a numerical illustrations for their result.

Lately, Messaoudi and Talahmeh [38] looked into where . They proved a blow-up result for certain solutions with arbitrary positive initial energy. Korpusov [39] generalized this result and established (10), with and are constants.

In this paper, we care to find sufficient conditions on and the initial data for which the blowup happens.

In addition to the introduction, our paper is divided into three sections. The second section deals with variable-exponent Lebesgue and Sobolev spaces and some of their characteristics. We also mention the result of existence, but without demonstration, and the second one deals with the result of blow-up for solutions with negative initial energy.

In the fourth one, we present and demonstrate the theorem of blow-up for certain solutions with positive initial energy.

#### 2. Background and Preliminaries

This section contains some essential concepts and definitions about the Lebesgue and Sobolev spaces with variable exponents which will be useful to us later (see Fan and Zhao [40] and Lars et al. [41], Mezouar and Boulaaras [42]).

Let that is a domain of be a measurable function. We introduce the Lebesgue space with a variable exponent by where

Equipped with the Luxembourg-type norm,

is a Banach space (see Lars et al. [41]).

Now, we introduce the following variable-exponent Sobolev space :

This space is a Banach space with respect to the norm Otherwise, we put to be the closure of in

*Remark 1. *The space is generally defined in a different way for the variable exponent case.

However, the two definitions are equivalent under condition (4) (see Lars et al. [4]). We define as the dual of in the same way as the classical Sobolev spaces, where

Lemma 2. *(Poincare inequality [41]).**Let be a bounded domain and assume that satisfies (4), then
for that is a constant depends only on and . Particularly, determines an equivalent norm on .*

Lemma 3. *(Lars et al. [41]).**If is a measurable function and such that
*

Then, the embedding is continuous and compact.

Lemma 4. *(Hölder’s inequality [41]).**Assume that are measurable functions defined on such that
**If and then with*

Lemma 5. *(Unit ball property [41]).**Assume that is a measurable function on . Then,
*

Lemma 6. *(Lars etal [41]).**If is a measurable function on satisfying (4), then for a.e. , we get
for any *

Proposition 7. *Let and suppose that the exponents satisfy (1) and (2). Then, problem admits a unique weak solution such that
where .*

*Remark 8. *We can achieve the proof of the previous proposition by using the Galerkin method as in [13].

#### 3. Blowing Up for Negative Initial Energy

To introduce and demonstrate our results, we first define our energy as follows:

Theorem 9. *Assume that the assumptions of hold true and suppose that
*

Then, the solution of problem blows up in finite time.

To state and demonstrate our result, In order to prove our result, we give the following Lemma

Lemma 10. *Suppose the conditions of Lemma 3 hold. Then, we have
for any , where is a positive constant which depends on only.*

Corollary 11. *Let the assumptions of Lemma 10 hold. Then, for any , we get
where and are a positive constant.*

Now, we set and use, throughout this paper, to denote a generic positive constant depending on only. From the result of (21) and (23), we give the following Corollary

Corollary 12. *Let the assumptions of Lemma 10 hold. Then, we have
for any and *

Corollary 13. *Let the assumptions of Lemma 10 hold. Then, we have
for any and *

Lemma 14. *Assume that (2) and (4) hold and . Then, the solution of satisfies, for some ,
*

Lemma 15. *Let be the solution of problem and assume that (2) holds. Then,
*

*Remark 16. *We can achieve the proof of the previous Lemmas and Corollaries as in the paper of Messaoudi and Talahmeh [39].

Lemma 17. *Let be the solution of . Then, there exists a constant such that
*

*Proof. *Assume, by contradiction, there exists a sequence such that
Then, Lemmas 3 and 6 give us
This yields
that contrasts with the fact that .

*Proof. *Proof of Theorem 9.

As usual, multiplying by and integrating over in to get
for almost every in since is absolutely continuous; hence, and
for every in , by remembering the condition that . We then introduce
for small to be chosen later and

By taking the derivative of (35) and using (1), we obtain so

Adding and subtracting the term , for , from the right side of (37), we get

So, for small enough, we obtain where

By using Young’s inequality, the last term in (40) yields

Thus, by picking such that for a large constant to be given later, and replacing in (41), we reach to

Combining (40) and (43) yields

Exploiting Lemma 15 and (34) to get

Now, we employ Lemma 10 and (36), and we get And it is easy to see from (46) that So, by using Lemmas 6 and 17, we obtain

Collecting of (45), (47), and (49), we get

In this step, we choose so large that the coefficient

Once is fixed (thus ), we put sufficiently small so that

Subsequently, (50) becomes by virtue of (28). Therefore, we get

Next, we are in the position to obtain an inequality of the form Here, is a positive constant that depends on , (the constant of ).

To achieve (54), we estimate the term(32)Hence,

From Young’s inequality that yields the following estimate where . Putting , we find by (37). Thus, (56) becomes with . We obtain after using

In the end, by noting that and combining it with (51) and (58)), the inequality (54) is achieved.

The proof is completed.

#### 4. Blowing Up for Positive Initial Energy

Now, we are in the position to present and prove one of the main results of this section which is the blow-up for certain solutions with positive energy. For this goal, let be the best constant of the Sobolev embedding and set for that are to be specified later.

We state here the following theorem which will be our main result.

Theorem 18. *Assume that the conditions of Proposition 7 hold true and suppose that
*

Then, the solution of blows up in a finite time.

To demonstrate our theorem, we refer the following two .

Lemma 19. *Let the assumptions in Theorem 18 be fulfilled, and then there exists a constant such that
*

*Proof. *. Exploiting (21), we get
where Let

By noting that , for , we can easily verify that the function is increasing for and decreasing for .

Because , there exists a positive constant such that . So, we get . This means that .

To demonstrate (67), we suppose that , for some . Then, there exists such that . Exploiting the monotonicity of to find which contradicts , for all . Consequently, (67) is determined.

Lemma 20. *Let the assumptions in Theorem 18 be fulfilled, and so we have
*

*Proof. *Exploiting (21), (30), and (64) to get
then from (67), we find

Therefore,

Proof of Theorem 2. It is not hard to determine the proof precisely by repeating the same steps (35) to (58) of the proof of Theorem 1, with the use of Lemma 20.

#### Data Availability

No data were used to support the study.

#### Conflicts of Interest

The author(s) declare(s) that they have no conflicts of interest.

#### Acknowledgments

The third-named author extends his appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through research groups program under grant (R.G.P-2/1/42). The idea and research project in this work have been presented by Prof. Salah Boulaaras to the authors of the paper. The authors also acknowledge to Prof. Salah Boulaaras for first and second revisions and kind comments on this work.