Abstract
This paper is concerned with the blow-up of certain solutions with positive initial energy to the following quasilinear wave equation: . This work generalizes the blow-up result of solutions with negative initial energy.
1. Introduction
Let be an open bounded Lipschitz domain in . We consider the following nonlinear hyperbolic equation:
Here, is a Lipschitz continuous boundary. The initial conditions meet the following:
The Kirchhoff function is continuous and has the standard form:
The elliptic nonhomogeneous -Laplacian operator is defined by where is the vectorial divergence and is the gradient of . The functional is the naturally associated -Dirichlet energy integral. The term with a variable exponent plays the role of a source, and the dissipative term with a variable exponent is a strong damping term.
The coefficients and are continuous in and satisfy where is a constant defined in (38). We assume that the Kirchhoff function , defined by (3), satisfies the following hypotheses: (i)For , there exist such that(ii)For all , it holds that
The exponents , , and are continuous and satisfy where the constants and are given in (10) and
Also, we can define by
We also assume that , , and satisfy the log-Hölder continuity condition for .
Problem (1) models several physical and biological systems such as viscoelastic fluids, filtration processes through a porous medium, and fluids with viscosity dependent on temperature. In the intention of problem (1), we can see that it is linked to the following equation presented by Kirchhoff and Hensel [1] in 1883:
The parameters , , , , and represent, respectively, the length of the string, the area of the cross-section, Young’s modulus of the material, the mass density, and the initial tension. This equation is an extension of the classic d’Alembert’s wave equation by looking at the effects of changes in the length of the string during the vibrations. As for this problem, it has been studied. More precisely, for , the global existence and nonexistence results can be found in [2, 3], and for , the main results of existence and nonexistence are in the paper [4]. In recent years, hyperbolic problems with a constant exponent have been studied by many authors; we refer to interesting works [5–7]. However, only a little research has been done regarding hyperbolic problems with nonlinearities of the variable exponent type; some interesting works can be found in [8–13].
Recently, in [14], Piskin studied the following wave equation with variable exponent nonlinearities:
The author proved, by using the modified energy functional method, the existence of solutions. We have also looked at the asymptotic behavior of the Kirchhoff wave equation problems. We can say that the investigation into the determination of the type, as well as the rate of decay, was the focus of attention of many researchers whose work was represented in [15, 16]. Motivated by previous studies, in this work, we consider problem (1), which is more interesting and applicable in the real approach of sciences, so a finite-time blow-up for certain solutions with positive and also negative initial energy has been proved. More precisely, our aim here is to find sufficient conditions on the variable exponents , , and and the initial data for which the blow-up occurs. This paper is organized as follows. After the introduction in the first section, we will give some preliminaries in Section 2. Then, in Section 3, we state the main results which will be proved in Sections 4 and 5.
2. Preliminaries
Regarding some definitions and basic properties of the generalized Lebesgue-Sobolev spaces and , where is an open subset of , we refer to the book of Musielak [17] and the papers [18, 19]. Let
For any , we write
Then, for any , we define the variable exponent Lebesgue space as follows: where is the modular of , and it is defined by
It is equipped with the following so-called Luxemburg norm on this space defined by the formula
Variable exponent Lebesgue spaces resemble classical Lebesgue spaces in many aspects: they are Banach spaces, the Hölder inequality holds, they are reflexive if and only if , and their continuous functions are dense if .
Lemma 1. Suppose that satisfies (15); then, where is a constant that depends only on , , and .
Lemma 2. If and are measurable functions such that then the embedding is continuous and compact.
Lemma 3. Let . The spaces and are separable, uniformly convex, and reflexive Banach spaces. The conjugate space of is , where For and , we have
Lemma 4. If is a measurable function on and , then and are equivalent. For , we have
Lemma 5. If is a measurable function on , then for all .
3. Main Results
Now, we state without proof the following existence result.
Proposition 6. Assume that (2) holds and the coefficients , , , and satisfy (3) and (9) and the exponents , , and satisfy (12). Then, problem (1) has a unique weak solution such that where is the conjugate exponent of .
Remark 7. The proof can be established by employing the Galerkin method as in the work of Antontsev [8].
We first define the energy function. Let
where
In order to investigate the properties of , the following lemma is necessary.
Lemma 8. Suppose that is a solution of problem (1) that satisfies (29); then, we have
Proof. By using the energy function (30) and problem (1), we directly deduce (32).
We also introduce the following lemma.
Lemma 9. Suppose that the conditions of Lemmas 1–5 hold. Then, there exists a constant , which is a generic constant that depends on only, such that for any and .
Proof. If , then
If , then we deduce by Lemma 4 that . Then, Lemmas 2 and 5 imply
Let be the best constant of the Sobolev embedding
We set
Now, the main results of the blow-up for certain solutions with positive/negative initial energy are given by the following theorems.
Theorem 10. Let the assumptions of Proposition 6 be satisfied, and assume that
Then, the solutions of (1) blow up in finite time:
Theorem 11. Let the assumptions of Proposition 6 be satisfied, and assume that Then, the solution of (1) blows up in finite time (42).
4. Proof of Theorem 10
To prove Theorem 10, we need the following lemmas.
Lemma 12. Let the assumptions of Theorem 10 hold; then, there exists such that for any , there exist a constant such that
Proof. By using the hypothesis (10) and the function (30), we obtain
where . Let the function
be defined by
Notice that , for . It is easy to check that the function is increasing for and decreasing for . On the other hand, by (38), we deduce that, for any , since , there exists a positive constant such that . Then, we have . This implies that .
Now, we suppose on the contrary that for some . Then, there exists such that . Using the monotonicity of , we have
which contradicts , for all .
Lemma 13. Let the assumptions of Theorem 10 hold. Then, in light of Lemma 12, we have
Proof. By using (30), we get
Let
Lemma 14. Let the assumptions of Theorem 10 be satisfied; then, we have
Proof. Using (30), (32), and (51), we obtain Then, the use of (10) gives Now, recalling in (38), we have On the other hand, we use (9) to get Combining (55) with (56) gives (52).
Corollary 15. Under the assumptions of Lemma 9, we have (i)(ii)(iii)for any and .
Lemma 16. Assume that (12) and (15) hold. Then, the solution of (1) satisfies for some .
Proof. Let We have This implies Now, given (52), (60) leads to Thus, (57) follows.
Lemma 17. Suppose that (12) holds, and is a solution of (1). Then,
Proof.
We set for small, which will be specified later, and for
Now, we are in a position to prove Theorem 10.
Proof. We differentiate (64) and use the equation in (1) to get
Adding and subtracting the term , for , from the right-hand side of (56), by using (49) and (10), we get
Then, for small enough, we have
where
Recall Young’s inequality
where , such that . Applying (70) to estimate the term , we get
where
where is a large constant to be specified later.
Now, by using (32) and (49), we get
Combining (73) and (71) yields
Substituting (74) in (68), we obtain
To estimate the last term in (75), we use (62) and (52) to get
Then, we use (65) and Lemma 9, for
to deduce from (76) that
By exploiting Lemmas 5 and 12, we get
Combining (75), (78), and (79) leads to
Now, we pick large enough and so small such that
Then, by using (57), (80) takes the form
Therefore, we get
On the other hand, the application of the Hölder inequality yields
and from (70), we get
for . Setting , we get by virtue of (65). Therefore, (85) takes the form
where . By recalling Corollary 15, we get
Now, (87) and the following Minkowski’s inequality
will give
By combining (82) and (89), we obtain
where is a positive constant. A simple integration of (90) over yields
which implies that the solution blows up in finite time , such that
This completes the proof of Theorem 10.
5. Proof of Theorem 11
We set
To prove our main result, we first establish the following lemma.
Lemma 18. Let be the solution of (1). Then, there exists a constant such that
Proof. Suppose that, by contradiction, there exists a sequence such that Then, by using Lemmas 2 and 5, we get which contradicts the fact that .
Using (93) and (94) and applying the same procedures used to prove Theorem 10 will give the proof of Theorem 11.
Data Availability
No data were used in this study.
Conflicts of Interest
The authors declare that they have no competing interests.
Authors’ Contributions
The authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.