#### Abstract

In this article, we study the existence of solutions for nonlocal -biharmonic Kirchhoff-type problem with Navier boundary conditions. By different variational methods, we determine intervals of parameters for which this problem admits at least one nontrivial solution.

#### 1. Introduction

We consider the problem with Navier boundary conditions. where is a bounded domain in with smooth boundary , , and is the -biharmonic operator defined by , where We denoted by

We assume that the weight and the Kirchhoff function satisfy the following conditions:

(m). , with such that where

(M1). is a continuous function verifying where are real numbers such that and

Example 1. A typical example of () satisfying the conditions ()-() is given by where such that , , and

Put . Then, we have , where is the conjugate of

Furthermore, Indeed,

Problem () is related to the stationary problem of a model introduced by Kirchhoff . To be more precise, Kirchhoff established a model given by the equation where are constants, which extends the classical D’Alambert’s wave equation, by considering the effects of the changes in the length of the strings during the vibrations. In two dimensions, Kirchhoff equations model the oscillations of thin plates and the most usual plate operator is the biharmonic operator .

Fourth-order equations have various applications in many domains like microelectromechanical systems, surface diffusion on solids, thin film theory, and interface dynamics; for recent contributions concerning this type of equations, we refer to . In recent years, the study of variational problems with variable exponent has received considerable attention; these problems arises from nonlinear electrorheological fluids, elastic mechanics, image restoration, and mathematical biology (see ). The interplay between the fourth-order equation and the variable exponent equation goes to the -biharmonic problems. The -biharmonic operator possesses more complicated structure than the -biharmonic operator , where is a real constant; for example, it is not homogeneous. A study on -biharmonic problems with Navier boundary condition was treated by many authors (see, for example, ). The authors in [21, 22] proved the existence and multiplicity of weak solutions for the -biharmonic problems under Navier boundary conditions. Their approach is of variational nature and does not require any symmetry of the nonlinearities. In , a similar problem to ours has been investigated in the case of -Laplacian and with weight 1. In , the authors examined a -biharmonic Kirchhoff-type problem but in the case where the weight is bounded and without parameter . Motivated by the above papers and the results in [24, 25], we determine by different variational methods intervals of parameters for which this problem admits at least one nontrivial solution.

#### 2. Preliminaries

We state some definitions and basic properties of variable exponent Lebesgue-Sobolev spaces. We refer the reader to  for details.

For any , we define the variable exponent Lebesgue space by with the norm

Proposition 1 (see ). The space is separable, uniformly convex, and reflexive and its conjugate space is , where for all For and , we have

Proposition 2 (see ). Let be the modular of the space. For , we have

For , we define the variable exponent Sobolev space where , where is a multi-index and . The space equipped with the norm becomes a separable, reflexive, and uniformly convex Banach space.

Proposition 3 (). For such that for all , there is a continuous embedding If we replace with , the embedding is compact.

We denote where is the closure of in

For , we define

endowed with the above norm is a separable and reflexive Banach space.

Remark 4. From , the norms and are equivalent in

Let be a measurable real function a.e. . We define the weighted variable exponent Lebesgue space

equipped with the norm is a Banach space which has similar properties with the usual variable exponent Lebesgue spaces. The modular of this space is defined by

Proposition 5 (). For , we have (1)(2)(3)(4)(5)

In the same way as in , we show the following proposition.

Proposition 6. Assume that the boundary of possesses the cone property and Suppose that
If and then there is a compact embedding

Denote the operator defined by for all

Proposition 7 (). The operator satisfies the following assertions: (i) is continuous, bounded, and strictly monotone(ii) is a mapping of type, namely, and , which imply (iii) is a homeomorphism

#### 3. The Main Result

We say that is a weak solution of () if for every

For any , the energy functional corresponding to problem () is defined as where

Standard arguments imply that and for any Hence, we can infer that critical points of functional are the weak solutions for problem ()

In the sequel, we use fountain theorem to study the existence of multiple solutions of () We obtain the following result.

Theorem 8. Assume that () (), and ; then, for every , problem () has a sequence of weak solutions such that as

Before proving Theorem 8, we give some preliminary results.

Since is a reflexive and separable Banach space, then is too. There exist (see ) and such that where denote the duality product between and . We define

Theorem 9 (fountain theorem, see ). is a Banach space; is an even functional. If for every there exist such that (a)(b)(c) satisfies the condition for every Then, has a sequence of critical values tending to .

Lemma 10. If , then under condition ()

Proof. For every , there exist such that There exists a subsequence of such that For every and for every , we obtain and we conclude that ; hence, On the other hand, by condition () and Proposition 5, there is a compact embedding ; hence, in By relation (27) we conclude that

Proof of Theorem 8. , and is even. According to Lemmas 17 and 18, satisfies condition for every . We will prove that if is large enough, then there exist such that and hold.
(a) By condition (), we have for any such that For large enough, choose such that and we take Then, we have By Lemma 10 and the fact that , the assertion is verified.
() For any with and , using , we have By and dim , it is easy to see that as for

Now, a nontrivial solution of () is given by using the coercivity and the weakly lower semicontinuity of .

Theorem 11. If we assume that (), (), and hold, then there exists such that for any , problem () possesses a nontrivial weak solution.

Lemma 12. Assume (), (), and Then, the functional is coercive on

Proof. By Proposition 5 and the compact imbedding , there exists such that Therefore, for and under condition (), we obtain Since , we infer that as

Proof of Theorem 11. is a coercive functional and weakly lower semicontinuous on Then, there exists a global minimizer of (cf. Theorem 1.2 ). Thus, is a weak solution of () Now, we prove that is a nontrivial solution for large enough. Letting be a constant and be an open subset of with , we assume that is such that for any and in There exists such that , Hence, for , is a nontrivial weak solution of ().

In the following theorem, we apply Ekeland variational principle  to get a nontrivial solution to problem ().

Theorem 13. Assume that the conditions (), (), and are satisfied. Then, there exists such that for any , problem () has at last one nontrivial weak solution.

Lemma 14. Assume (), (), and There exists and two positive real numbers such that for any for any with

Proof. By using the condition () and Proposition 6, the embedding from to is compact. Then, there exists such that for all ,

Consider such that . Then, we get

Furthermore, Proposition 5 yields and we conclude that

For , using (), (39), and Proposition 2, we get

By the above inequality, we remark that if we define then for any and any with , there exists such that , which end the proof of Lemma 14.

Lemma 15. Assume () and There exists such that , and for small enough.

Proof. Since , there exists such that . On the other hand, we have ; then, there exists an open ball such that for all Thus, we conclude that for all Let be such supp(), for all and for all Then, using the above information and , for all , we obtain for all with

Proof of Theorem 13. Let be defined as in (41) and . By Lemma 14, it follows that on the boundary of the ball centered at the origin and of radius in , denoted by , we have On the other hand, by Lemma 15, there exists such that for all small enough. Moreover, by relations (53), we have for any It follows that Choose Applying the Ekeland variational principle  to the functional , we find such that The fact that implies that We deduce that there exists a sequence such that It is clear that is bounded in Thus, there exists and a subsequence still denoted by such that in Moreover, by the Hölder inequality, we get where is the conjugate exponent of , i.e., for all
By the compact embedding , in Using Proposition 5, we obtain and we deduce that Hence, (50) and (51) imply that as
Using (48), we infer that From () and assertion (ii) of Proposition 7, it follows that in Thus, and

If in addition we have the following condition on

(). There exists such that .

Then by help of the Mountain Pass theorem, we obtain the following.

Theorem 16. Assume that (), (), (), and hold. For every , problem () has a nontrivial weak solution.

To prove Theorem 16, we need the two following lemmas.

Lemma 17. For and under conditions (), (), and , there exist and such that for any with

Proof. Let Under conditions (), (), and as in Lemma 14, we show the existence of a constant such that for all with , we have Since , then we can choose such that and we have for every

Lemma 18. For and under conditions () and , there exists with where is given in Lemma 17, such that

Proof. Let such that and and By (), we have Since , we obtain Then, for large enough, we can take such that and

Proof of Theorem 16. By Lemmas 17 and 18 and the mountain pass theorem of Ambrosetti and Rabinowitz , we deduce the existence of a sequence and positive real number such that Firstly, we prove that is bounded in Arguing by contradiction and passing to a subsequence, we have as Considering , for large enough and using (), (), (48), and Proposition 2, we have But, this cannot hold true since and Hence, is bounded in The fact that is reflexive implies that there exists a subsequence, still denoted by , and such that in Actually, with similar arguments as those used in the proof of Theorem 13, we can show that in Thus, and

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.