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The Scientific World Journal
Volume 2014, Article ID 194310, 5 pages
Research Article

Infinitely Many Weak Solutions of the -Laplacian Equation with Nonlinear Boundary Conditions

1Xingyi Normal University for Nationalities, Xingyi, Guizhou 562400, China
2Human Resources and Social Security Bureau, Buyi and Miao Autonomous Prefecture in Southwest Guizhou, Guizhou 562400, China

Received 24 August 2013; Accepted 27 October 2013; Published 14 January 2014

Academic Editors: E. A. Abdel-Salam and P. Candito

Copyright © 2014 Feng-Yun Lu and Gui-Qian Deng. 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.


We study the following -Laplacian equation with nonlinear boundary conditions: and where is a bounded domain in with smooth boundary . We prove that the equation has infinitely many weak solutions by using the variant fountain theorem due to Zou (2001) and do not need to satisfy the or condition.

1. Introduction

In this paper, we study the following-Laplacian equation: whereis a bounded domain inwith smooth boundaryandis the outer normal derivative,is the-Laplacian with,is real parameter, and The perturbation functionssatisfy the following conditions:(F1) are odd in;(F2)there exist , , , such that (F3)There exists (where ) such that for a.e.and. Moreover, uniformly for.(F4)Assume that one of the following conditions hold:(1) uniformly for;(2) uniformly for; furthermore, and are decreasing inforis large enough;(3) uniformly for; is increasing inforis large enough; moreover, there exists such that where.

Remark 1. The above conditions were given in Zou [1] for the semilinear case.

Remark 2. A simple example which satisfies (F1)–(F4) is where.

Equation (1) is posed in the framework of the Sobolev space with the norm

The corresponding energy functional of (1) is defined by for, whereandis the measure on the boundary. It is easy to see that and for all. It is well-known that the weak solution of (1) corresponds to the critical point of the energy functional on.

Remark 3. Under condition (2), it is easy to check that norm (7) is equivalent to the usual one, that is, the norm defined in (7) with .

In [2], the author researched (1) and obtained the existence of infinitely many weak solutions. Moreover, the existence of three solutions for (1) was researched in [3] by using a three-critical-point theorem due to Ricceri [4]. Also, some authors researched and obtained the existence of infinitely many weak solution without requiring any symmetric conditions and also with discontinuous nonlinearities; see [5, 6]. Recently, this equation was studied by J.-H. Zhao and P.-H. Zhao [7] via Bartsch’s dual fountain theorem in [8] and obtained the existence of infinitely many weak solutions for (1) under the case of Remark 2. They obtained the following theorem.

Theorem A. Let, where. Then there exists a constantsuch that, for any, (1)for any, (1) has a sequence of solutionssuch thatas;(2)for any, (1) has a sequence of solutionssuch thatas.

The main ingredient for the proof of the above theorem is a dual fountain theorem in [8]. It should be noted that theorcondition and its variants play an important role in this theorem and its application. While the variant fountain theorem in Zou [1] does not need not theorcondition, we obtain the following generalized result by using Zou’s theorem.

Theorem 4. Assume that (F1)–(F4) hold; then there exists a constantsuch that, for any, (1) has infinitely many weak solutionssatisfying

This paper is organized as follows. In Section 2, we recall some preliminary theorems and lemmas. In Section 3, we give the proof of Theorem 4.

2. Preliminaries

In what follows, we make use of the following notations:(or) denotes Banach space with the norm;denotes the conjugate space for;denotes Lebesgue space with the norm ; is the dual pairing of the spaces and; we denote by(resp., ) the strong (resp., weak) convergence; denote (possibly different) positive constants.

For completeness, we first recall the variant fountain theorem in Zou [1]. Letbe a Banach space with norm and with for any. Set, .

Theorem 5 (see [1, Theorem 2.2]). The -functional defined by ,, satisfies(A1) maps bounded sets to bounded sets uniformly for; furthermore, for all .(A2) for all; as on any finite dimensional subspace of.(A3)There exists such that for all,

Then there exist , , such that

Particularly, if has a convergent subsequence for every, thenhas infinitely many nontrivial critical points satisfying as .

Remark 6. Obviously, the sequenceis asequence.

For our working space , is a reflexive and separable Banach space; then there areand such that

We write; thencan be defined as that in the beginning of Theorem 5. Consider defined by

Thenfor all;ason any finite dimensional subspace of;for allWe need the following lemmas.

Lemma 7 (see [7, Lemma 3.5]). If, then one has

3. Proof of Theorem 4

First, we check the condition of Theorem 5.

Lemma 8. Assume (F1)–(F3); then (A1)–(A3) hold.

Proof. (A1) and (A2) are obvious. Let ; we assume that and define
Observe that for any . Note that ; there exists a constant such that By the Sobolev trace imbedding inequality, we have Then we takesuch that, for all, By (F3), for any, there existssuch that Then, by (F1)–(F3) and (18)–(21), we obtain Note that ; we may choose and sufficiently small that holds true for anywith. So we have for any with . Choosing by Lemma 7, for,as, it follows thatas, so there existssuch that when . Thus, for , , and , we have ; then for all .
On the other hand, if with being small enough, since all the norms are equivalent on the finite dimensional space and, thenfor all.
Furthermore, if with , , we see that Therefore, as . Thus, (A3) holds.

By Theorem 5, we have the following lemma.

Lemma 9. There exist and such that

In order to complete our proof of Theorem 4, by a standard argument (see the proof of Lemma 3.4 in Zhao [7]), we only need to show that is bounded.

Lemma 10. is bounded in.

Proof. Since , then
We can choose and if such that . If, up to a subsequence, as , then, by (F2), for is large enough. Obviously, it is a condition if (F4)(1) holds.
Otherwise, we set ; then, up to a subsequence, If in and in (F4)(2), then, for is large enough, by Fatou’s Lemma, we have that this is a contradiction. It is similar if in (F4)(3). Thus, .
Let such that For anylarge enough, and , for is large enough, we have that which implies that . Obviously,
It follows that
If (F4)(2) holds, we have that and are decreasing infor is large enough. Therefore, for all and ; it is a contradiction.
If (F4)(3) holds, then we have that which implies
On the other hand, by the property of , for is large enough, since , we have that this implies that is bounded, which contradicts (39).
By the above arguments, we have that is bounded.

Remark 11. In fact, our result still holds if we consider a weaker condition than (F4)(2); that is, there is such that for all or ,, where and .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.


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