Abstract and Applied Analysis

Volume 2016, Article ID 2349172, 8 pages

http://dx.doi.org/10.1155/2016/2349172

## Existence of Solutions for a Robin Problem Involving the -Laplace Operator

Department of Mathematics, Faculty of sciences, Department of Mathematics, Mohamed I University, 60000 Oujda, Morocco

Received 27 January 2016; Accepted 9 May 2016

Academic Editor: Patricia J. Y. Wong

Copyright © 2016 Mostafa Allaoui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In this article we study the nonlinear Robin boundary-value problem , on . Using the variational method, under appropriate assumptions on , we obtain results on existence and multiplicity of solutions.

#### 1. Introduction

The aim of this article is to analyze the existence of solutions of the following problem:where is a bounded smooth domain, is the outer unit normal derivative on , is a continuous function on with , and with and is a continuous function. The main interest in studying such problems arises from the presence of the -Laplace operator , which is a natural extension of the classical -Laplace operator obtained in the case when is a positive constant. However, such generalizations are not trivial since the -Laplace operator possesses a more complicated structure than -Laplace operator; for example, it is inhomogeneous.

We make the following assumptions on the function :(): is a continuous function and there exist two constants such that where and for all .():the following limit holds uniformly for a.e :(): as and uniformly for .(): there exist two positive constants and such that where and .

By the famous Mountain Pass lemma we state the first result.

Theorem 1. *Suppose that the conditions ()–() with hold. Then problem (1) has at least a nontrivial weak solution.*

Assume the following hypotheses:(): ():, and there exists such that for every and where .():, for , .

We are now in the position to state our second theorem.

Theorem 2. *Suppose that is Lipschitz continuous function. Under the assumptions () and ()–(), problem (1) has a sequence of weak solutions such that as .*

For the next theorem we assume that satisfies the following conditions:(): for all and with .(): is nondecreasing with respect to , .

Theorem 3. *Suppose that the conditions (), (), and () with hold. Then problem (1) has a positive solution.*

Nonlinear boundary-value problems with variable exponent have received considerable attention in recent years. This is partly due to their frequent appearance in applications such as the modeling of electrorheological fluids [1–4] and image processing [5], but these problems are very interesting from a purely mathematical point of view as well. Many results have been obtained on this kind of problems; see for example [6–13]. In [9], the authors have studied the case ; they proved the existence of infinitely many eigenvalue sequences. Unlike the -Laplacian case, for a variable exponent ( constant), there does not exist a principal eigenvalue and the set of all eigenvalues is not closed under some assumptions. Finally, they presented some sufficient conditions that the infimum of all eigenvalues is zero and positive, respectively.

In [14], the authors obtained results on existence and multiplicity of solutions for problem (1) in the case , under () and the following Ambrosetti-Rabinowitz type condition:Here, we notice that () is much weaker than the condition in the constant exponent case.

Very recently, the authors in [15] studied the following problem:where is a positive parameter, is locally Lipschitz function in the -variable integrand, and is the subdifferential with respect to the -variable in the sense of Clarke. They claim that problem (6) admits at least two nontrivial solutions.

In the first result, we consider problem (1) when the nonlinear term is superlinear at infinity but does not satisfy the type condition, used in [14, 16], which is necessary to ensure the boundedness of the Palais-Smale (PS) type sequences of the associated functional. To overcome these difficulties, we will use the Mountain Pass Theorem [17] with Cerami condition () which is weaker than Palais-Smale (PS) condition.

In the second result, a distinguishing feature is that we have assumed some conditions only at zero; however, there are no conditions imposed on at infinity, which is necessary in many works. Finally, in Theorem 3, applying the subsuper solution method we get a positive solution of problem (1).

This article is organized as follows. First, we will introduce some basic preliminary results and lemmas in Section 2. In Section 3, we will give the proofs of our main results.

#### 2. Preliminaries

For completeness, we first recall some facts on the variable exponent spaces and . For more details, see [18, 19]. Suppose that is a bounded open domain of with smooth boundary and , where Denote by and . Define the variable exponent Lebesgue space by with the norm Define the variable exponent Sobolev space by with the norm We refer the reader to [11, 18] for the basic properties of the variable exponent Lebesgue and Sobolev spaces.

Lemma 4 (see [19]). *Both and are separable and uniformly convex Banach spaces.*

Lemma 5 (see [19]). *Hölder inequality holds, namely, where .*

Lemma 6 (see [18]). *Assume that the boundary of possesses the cone property and and for , then there is a compact embedding , where *

Now, we introduce a norm, which will be used later.

Let with and, for , define Then, by Theorem in [16], is also a norm on which is equivalent to .

An important role in manipulating the generalized Lebesgue-Sobolev spaces is played by the mapping defined by the following.

Lemma 7 (see [16]). *Let with . For one has * ***;* ***;* ***;* ***.*

*We recall the definition of the following condition (), see [20].*

*Definition 8 (see [20]). *Let be a Banach space and . Given , one says that satisfies the Cerami condition (one denotes condition ()) if (i)any bounded sequence such that and has a convergent subsequence;(ii)there exist constants such that If satisfies condition () for every , one says that satisfies condition ().

*Note that condition () is weaker than the (PS) condition. However, it was shown in [17] that from condition () it is possible to obtain a deformation lemma, which is fundamental in order to get some min-max theorems.*

*Theorem 9 (see [17]). Let a Banach space, , , and , such that and If satisfies the condition with Then is a critical value of .*

*Here, problem (1) is stated in the framework of the generalized Sobolev space .*

*The Euler-Lagrange functional associated with (1) is defined as in One says that is a weak solution of (1) if for all .*

*Standard arguments imply that and for all . Thus, the weak solutions of (1) coincide with the critical points of .*

*3. Proof of Main Results*

*For simplicity, we use , to denote the general positive constants whose exact values may change from line to line.*

*Noting that is the sum of () type map and a weakly-strongly continuous map, so is of () type. To see that Cerami condition () holds, it is enough to verify that any Cerami sequence is bounded.*

*Proof of Theorem 1. *We check the assumption of compactness of the Mountain Pass Theorem as in the following lemma.

Lemma 10.* Suppose that (**)–(**) hold. If **, then any ** sequence of ** is bounded.**Proof*. Let be a sequence of . If is unbounded, up to a subsequence we may assume thatLet , then is bounded in ; up to a subsequence we have If , we have ; that is,Dividing (23) by , we get On the other side, using () and lemma of Fatou we obtain we obtain a contradiction.

If , since in and , by the continuity of the Nemitskii operator, we see that in as ; therefore,We choose a sequence such that Given , since for large enough we have , using (26) with , we obtainThat is, , but , ; we see that and It yields Therefore,so we get Appropriately, we have From (), there exist two constants and such thatHence, , which is impossible and thus is bounded in .

We will show that possesses the Mountain Pass geometry.

Lemma .* Under the conditions (**)–(**), there exist ** and ** such that ** when *.*Proof*. In view of () and (), there exists such that Therefore, for we haveSince , the function is strictly positive in a neighborhood of zero. It follows that there exist and such thatTo apply the Mountain Pass Theorem, it suffices to show thatfor a certain .

Let ; by (), we can choose a constant , such that Let be large enough; we havewhere is a constant, which implies thatIt follows that there exists such that and . According to the Mountain Pass Theorem, admits a critical value which is characterized bywhereThis completes the proof.

*Proof of Theorem 2. *The main idea (developed by Wang [21]) is to extend to an appropriate function in order to prove for the associated modified functional the existence of a sequence of weak solutions tending to zero in norm. Therefore, it is worth recalling the following proposition.

Proposition (see [22]).* Let **, where ** is a Banach space. Assume that ** satisfies the (**) condition and is even and bounded from below, and **. If for any **, there exists a **-dimensional subspace ** and ** such that**where **, then ** has a sequence of critical values ** satisfying ** as **.*

We need to state the following results. *Claim 1*. When , then .

Indeed, suppose that . Thus Then we obtainwhich contradicts the assumption (). *Claim 2*. There exist and such that is odd andwhere .

In fact, let us define where is a positive constant and is a cut-off function presented as follows:For , (46) easily holds.

On the other hand, we have It is easy to check that for we have Hence, (48) is satisfied. In the rest, from (), we can choose small enough to get when and the formula (47) holds since .*Claim 3*. The associated modified functional satisfies the Palais-Smale condition.

In fact, by Claim , it is easy to see that is even and . For , we haveBecause with is a positive constant, is coercive, that is, as . Hence, to verify that satisfies () condition on , it is enough to verify that any () sequence is bounded. Hence, by the coercivity of , any () sequence is bounded in .

Next, we modify and extend to get satisfying the assertions of Proposition .

For any we have independent smooth functions for , and define the subspace .

From Claim , for we can obtainBy (53) and as it is well known that all norms in are equivalent, for sufficiently small and suitable positive constant we obtain As a consequence of this fact, we observe that the conditions of Proposition hold and thus there exists a sequence of negative critical values for the functional such that as .

Afterwards, for any satisfying and , is sequence of . Passing, if necessary, to a subsequence still denoted by , we may suppose that has a limit.

From Claims and it is clear that 0 is the only critical point when the energy is zero and thus converges to 0. It follows from [23, 24] thatSo in view of Claim , we have . Thereby, the sequences are solutions of problem (1).

*Proof of Theorem 3. *Firstly, we recall the definition of subsupersolution of problem (1) as follows. We call a subsolution (resp. supersolution) of (1) if, for every with , Lemma .* Let **. Suppose that ** satisfies the subcritical growth condition (**) and the function ** is nondecreasing in **. If there exist a subsolution ** and a supersolution ** of (1) such that **, then (1) has a minimal solution ** and a maximal solution ** in the order interval ** (i.e., **).*

The proof of Lemma is built on the fixed point theory for the increasing operator on the order interval (see e.g., [25]) and is similar to that given in [26] for the -Laplacian case.

According to Proposition in [15], the mapping such that for all , in ; is a strictly monotone, bounded homeomorphism, and consequently we have the following.

Proposition .* Let ** with ** for **, then for ** (or **), the problem**has a unique solution ** in **.*

Let us consider the following problem:with . By Proposition , the strong maximum principle [27] and the result of regularity in [28], problem has a unique positive solution such that for each .

Taking , for any with we haveHence, is a positive supersolution of problem (1).

Obviously 0 is a subsolution of (1). By Lemma , (1) has a solution .

*Competing Interests*

*The author declares that he has no competing interests.*

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