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Discrete Dynamics in Nature and Society
Volume 2013 (2013), Article ID 809262, 5 pages
http://dx.doi.org/10.1155/2013/809262
Research Article

Existence of Multiple Solutions for a Class of Biharmonic Equations

Department of Mathematics, Qilu Normal University, Jinan 250013, China

Received 17 July 2013; Revised 28 October 2013; Accepted 11 November 2013

Academic Editor: Cengiz Çinar

Copyright © 2013 Chunhan Liu and Jianguo Wang. 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

By a symmetric Mountain Pass Theorem, a class of biharmonic equations with Navier type boundary value at the resonant and nonresonant case are discussed, and infinitely many solutions of the equations are obtained.

1. Introduction and Main Results

In this paper, we study the following fourth-order elliptic equation: where is the biharmonic operator, is a constant, is a bounded smooth domain, , is a parameter, , , , is a continuous function on .

This fourth-order semilinear elliptic problem can be considered as an analogue of a class of second-order problems which have been studied by many authors. A main tool of seeking solutions of the problem is the Mountain Pass Theorem (see [13]). In [4], Pei studied the following problem: and obtained at least three nontrivial solutions by using the minimax method and Morse theory.

Problem (1) has been studied extensively in recent years; we refer the reader to [511] and the references therein. In [12], the author showed that the problem (1) admits at least three (or four or five) nontrivial solutions by using the minimax method and Mountain Pass Theory.

However, to the best of author's knowledge, there have been very few results dealing with (1) using a symmetric Mountain Pass Theorem. This paper will make some contribution in the research field. In this paper, we study the problem (1) by a symmetric Mountain Pass Theorem at the resonant and nonresonant case and obtain infinitely many solutions of the equation.

Consider eigenvalue problem Let us denote that are the eigenvalues and are the corresponding eigenfunctions of the eigenvalue problem (3). It is well known that , and the first eigenfunction , .

It is easy to see that , , are eigenvalues of the problem and are still the corresponding eigenfunctions.

Let be the Hilbert space equipped with the inner product and the deduced norm

Suppose that . Let us define a norm of the space as follows: It is easy to verify that the norm is equivalent to the norm on , and for all , the following Poincaré inequality holds: where .

Throughout this paper, the weak solutions of (1) are the critical points of the associated functional where .

Obviously , and for all , , In order to establish solutions for problem (1), we make the following assumptions:, ;, , for all , ;  , uniformly for a.e. , where , or ;   uniformly in .

Definition 1 (see [13]). Let , we say that satisfies the Cerami condition at the level , if any sequence with possesses a convergent subsequence; satisfies the condition if satisfies for all .

Lemma 2. Let ; ; then

Proof. It is similar to the proof of Lemma 2.5 in [13].

Theorem 3 (see [14] a symmetric Mountain Pass Theorem). Suppose that is even, , and(i)there exist , , and a finite dimensional linear subspace such that (ii)there exist a sequence of linear subspace , , and such that If satisfies condition, then possesses infinitely many distinct critical points corresponding to positive critical values.

Remark 4. If satisfies the condition, Theorem 3 still holds.

The main results of this paper are as follows.

Theorem 5. Assume that hold, and , , is not an eigenvalue of (4); then there exist such that for , (1) has infinitely many solutions.

Theorem 6. Assume that hold, and , ; then there exist such that for , (1) has infinitely many solutions.

2. Proofs of Theorems

Proof of Theorem 5. (i) Assume that is a sequence, that is, We claim that is bounded. Assume as a contradiction that . Setting , then . Without loss of generality, we assume From (15) we know that Next we consider the two possible cases: (a), (b).
In case (a), from we derive For in , we have In case (b), we have Then By (17) and , we have We can easily see that . In fact, if , then , which contradicts . Hence, is an eigenvalue of the problem (4); this contradicts our assumption. Then is bounded; there exist a subsequence of (we can also denote by ) and , such that in . By Lemma 2 and (15), we have where , and as . Hence in . We verify satisfies condition.
(ii) There exists some such that , where .
By and , taking , for any given , there exists such that where
Taking such that , combining Lemma 2, Poincaré inequality, and Sobolev embedding, we have where are constant.
Let ; we claim that there exists such that
It is easy to see that has a minimum at ; substituting in , we have where .
We take , then there exists such that , where .
(iii) There exists , such that For , implies that for any , there exists such that Taking such that . By Lemma 2 and , we have Hence there exists such that Summing up the above proofs, satisfies all the conditions of Theorem 3 and Remark 4. Then the problem (1) has infinitely many solutions.

Proof of Theorem 6. Similar to the proof of Theorem 5(i), we have We can easily see that . In fact, if , then , which contradicts . Then is a corresponding eigenfunction of the eigenvalue ; hence , a.e. .
By , we have It follows from Fatous Lemma that
On the other hand, (15) implies that where , which contradicts (36); hence is bounded. Similar to the proof of Theorem 5(i), we have in . Hence we verify satisfies condition.
Similar to [15], let , and ; then It follows that for every , there exists such that For , we have over the interval , we have Letting , we see that , , , a.e. . In a similar way, we have , , , a.e. . Hence By Lemma 2 and , we have Hence there exists such that Summing up the above proofs and Theorem 5(ii), satisfies all the conditions of Theorem 3 and Remark 4, then the problem (1) has infinitely many solutions.

Acknowledgments

This work was supported by the National Nature Science Foundation of China (10971179) and the Project of Shandong Province Higher Educational Science and Technology Program (J09LA55, J12LI53).

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