Eigenvalues for a Neumann Boundary Problem Involving the -Laplacian

Abstract

We study the existence of weak solutions to the following Neumann problem involving the -Laplacian operator:  , , , . Under some appropriate conditions on the functions ,  ,  , and  , we prove that there exists such that any is an eigenvalue of the above problem. Our analysis mainly relies on variational arguments based on Ekeland’s variational principle.

1. Introduction

This paper is considered with the existence of weak solutions to the following Neumann problem: where is a bounded domain with a smooth boundary, with on , and is a parameter. is nonnegative, , and for some .

The operator is called a -Laplacian. If is a constant, then operator is the well-known -Laplacian, and (1) is the usual -Laplacian equation. As a matter of fact, the -Laplacian has more complicated nonlinearities than the -Laplacian. For example, it is not homogeneous. The study of differential equations with -growth conditions is an interesting and attractive topic and has been the object of considerable attention in recent years [1–6].

In [3], by applying a variational principle due to B. Ricceri and the theory of the variable exponent Sobolev spaces, the author considered the Neumann problem of -Laplacian:

In [5], based on the three critical points theorem due to B. Ricceri, the author studied the problem

In [6], the author established the existence of at least three solutions of the problem

However, in [3, 5, 6], the function assumed that . In this paper, by using Ekeland’s variational principle, we show that when , there exists such that any ; problem (1) has at least one nontrivial weak solution. For more applications of Ekeland’s variational principle to other problems, see, for example, [7–10]. Our result is partly motivated by these nice papers.

This paper is organized as follows. In Section 2, we recall some basic facts about the variable exponent Lebesgue and Sobolev spaces and also give some propositions that will be used later. In Section 3, we obtain the existence of weak solutions of problem (1).

2. Preliminaries

In order to deal with the -Laplacian problem, we need some theories on spaces , and properties of -Laplacian which will be used later. Let Through this paper, for any , we use the notations denote We can introduce the norm on by and becomes a Banach space; we call it a generalized Lebesgue space.

The variable exponent Sobolev space is defined by and it can be equipped with the norm The spaces , are separable, reflexive Banach spaces [11].

From now on, we denote by the space . For any , define then, it is easy to see that is a norm on and equivalent to . In the following, we will use instead of on .

Proposition 1 (see [11, 12]). The conjugate space of is , where . For any and , one has the following Hölder-type inequality: Moreover, if with , then, for , , , one has

Proposition 2 (see [11]). Put , ; then(i);(ii); ;(iii); .

From Proposition 2, the following inequalities hold:

Proposition 3 (see [13]). Let be such that for a.e.  . Let . Then, one has the following:(i)if , then ;(ii)if , then .

For any , let

Proposition 4 (see [11, 14]). Assume that satisfy on ; then the imbedding from to is compact and continuous.

We define by

Proposition 5 (see [15]). (i) is weakly lower semicontinuous, , and
(ii) is a mapping of type ; that is, if and , then ;
(iii) is a homeomorphism.

3. Main Result

We say that is a weak solution of problem (1) if

It follows that we can seek for weak solutions of problem (1). We need the following assumptions.

(A) There exist , , and such that

(B) Consider and there exists a subset with such that for , where denotes the Lebesgue measure.

Remark 6. Regarding condition (A), we have where . Thus, by Proposition 4, the embeddings and , are continuous and compact.

By a standard argument, we have that is weakly lower semicontinuous, , and

Remark 7. is weakly lower semicontinuous, , and Thus, is weak solution of problem (1) if and only if is a critical point of .

Note from Remark 6 that the embedding is continuous; then, there exists a positive constant such that

Lemma 8. For any , there exist and such that for any and with .

Proof. Let be fixed. Then, , and, from (26), we know that From Propositions 1 and 3, (15), (21), and (26), it follows that Hence, if we let then, for any and with , there exists such that .

Lemma 9. There exists such that , , and for small enough.

Proof. From condition (B) and (20), there exists a subset , and on . If we let and , then there exists such that . Moreover, since , there exists an open set , , and for . Thus, in .
Let be nontrivial such that , and in . From (22), then for , we have Since , in fact, if this is not true, ; by Proposition 2, we have and so in , which is a contradiction. Hence, for

In this paper, we study problem (1) by using Ekeland’s variation principle, which is recalled below.

Lemma 10 (see [16]). Let be a complete metric space and let be a lower semicontinuous functional, bounded from below, and not identically equal to . Let be given and be such that Then, there exists such that and for any in , where denotes the distance between two elements in .

We now state our main theorem.

Theorem 11. Assume that conditions (A) and (B) hold. Then, there exists such that any ; problem (1) has at least one nontrivial weak solution.

Proof. Let be defined by (29), and . By Lemma 8, we have where is the ball in and is the boundary of . For any , by an argument similar to those used in Lemma 8, we can obtain that Note from Lemma 9 that there exists such that for small enough. Then, from (35) and (36), Let Applying Lemma 10, we see that there exists such that Since , we have .
Now, define a functional by By (40), is a minimum point of , and for small enough and all , we have then when , we have Together with (39), there exists a sequence such that By the reflexivity of , there exists such that . Note that Then, from (44), we have From (21) and Propositions 1 and 3, we have where is defined in Remark 6; by the continuous and compact embedding of and , we can get Now, we conclude that Thus, by Proposition 5, we have . Hence, Therefore, is a nontrivial weak solution of problem (1).

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are thankful to the referees for their helpful suggestions. Project is supported by the Applied Basic Research Projects of Yunnan Province (no. 2013FD031) and the National Natural Science Foundation of China (no. 11461083).