Abstract

Via the Fountain theorem, we obtain the existence of infinitely many solutions of the following superlinear elliptic boundary value problem: in on , where is a bounded domain with smooth boundary and is odd in and continuous. There is no assumption near zero on the behavior of the nonlinearity , and does not satisfy the Ambrosetti-Rabinowitz type technical condition near infinity.

1. Introduction

Consider the following nonlinear problem: which has been receiving much attention during the last several decades. Here is a bounded smooth domain and is a continuous function on and odd in . We make the following assumptions on : there exist constants and such that , for all , and where ; there exists a constant such that

Note that Costa and Magalhães in [1] introduced a condition similar to , which also appeared in [2].

In this paper, we will study the existence of infinitely many nontrivial solutions of (1) via a variant of Fountain theorems established by Zou in [3]. Fountain theorems and their dual form were established by Bartsch in [4] and by Bartsch and Willem in [5], respectively. They are effective tools for studying the existence of infinitely many large or small energy solutions. It should be noted that the P.S. condition and its variants play an important role for these theorems and their applications.

We state our main result as follows.

Theorem 1. Assume that hold and is odd in . Then problem (1) possesses infinitely many solutions.

Problem (1) was studied widely under various conditions on ; see, for example, [610]. In 2007, Rabinowitz et al. [6] studied the problem where is a bounded smooth domain, and assumed, , , ,, , s.t. , for all and ,  and for small.

They got the existence of at least three nontrivial solutions. was given by Ambrosetti and Rabinowitz [11] to ensure that some compactness and the Mountain Pass setting hold.

However, there are many functions which are superlinear but do not necessarily need to satisfy . For example, It is easy to check that does not hold. On the other hand, in order to verify , it is usually an annoying task to compute the primitive function of and sometimes it is almost impossible, for example, More examples are presented in Remark 2.

We recall that implies a weaker condition In [12], Willem and Zou gave one weaker condition Note that is much weaker than the above conditions.

In [13], Schechter and Zou proved that under the hypotheses that problem (1) has a nontrivial weak solution.

Recently, Miyagaki and Souto in [2] proved that problem (1) has a nontrivial solution via the Mountain Pass theorem under the following conditions: and they adapted some monotonicity arguments used by Schechter and Zou [13]. This approach is interesting, but many powerful variational tools such as the Fountain theorem and Morse theory are not directly applicable. In addition, the monotonicity assumption on is weaker than the monotonicity assumption on .

As to the case in the current paper, we make some concluding remarks as follows.

Remark 2. To show that our assumptions and are weaker than , we give two examples:(1),(2),
which do not satisfy . Example can be found in [3]. So the case considered here cannot be covered by the cases mentioned in [6, 11].

Remark 3. Compared with papers [11, 12], we do not assume any superlinear conditions near zero. Compared with paper [2], we do not impose any kind of monotonic conditions. In addition, although we do not assume holds, we are able to check the boundedness of P.S. (or P.S.*) sequences. So, our result is different from those in the literature.

Our argument is variational and close to that in [2, 3, 13, 14]. The paper is arranged as follows. In Section 2 we formulate the variational setting and recall some critical point theorems required. We then in Section 3 complete the proof of Theorem 1.

2. Variational Setting

In this section, we will first recall some related preliminaries and establish the variational setting for our problem. Throughout this paper, we work on the space equipped with the norm

Lemma 4. embeds continuously into , for all , and compactly into , for all ; hence there exists such that where .

Define the Euler-Lagrange functional associated to problem (1), given by where . Note that () implies that

In view of (16) and Sobolev embedding theorem, and are well defined. Furthermore, we have the following.

Lemma 5 (see [15] or [16]). Suppose that is satisfied. Then and is compact and hence . Moreover for all , and critical points of on are solutions of (1).

Lemma 6 (see [17]). Assume that , and . Then for every , and the operator is continuous.

Let be a Banach space equipped with the norm and , where for any . Set and . Consider the following functional defined by The following variant of the Fountain theorems was established in [3].

Theorem 7 (see [3, Theorem 2.1]). Assume that the functional defined above satisfies the following: maps bounded sets to bounded sets uniformly for ; furthermore, for all ; for all ; moreover, or as ; there exists such that Then where and . Moreover, for a.e. , there exists a sequence such that

In order to apply the above theorem to prove our main results, we define the functionals , , and on our working space by for all and . Note that , where is the functional defined in (15).

From Lemma 5, we know that , for all . It is known that is a selfadjoint operator with a sequence of eigenvalues (counted with multiplicity) and the corresponding system of eigenfunctions forming an orthogonal basis in . Let , for all .

3. Proof of Theorem 1

Lemma 8. Assume that ()-() hold. Then there exists a positive integer and two sequences as such that where and for all .

Proof
Step  1. We first prove (25).
By (16) and (23), for all and , we have where is the constant in (16). Let Then since is compactly embedded into . Combining (14), (27), and (28), we have By (29), there exists a positive integer such that since . Evidently, Combining (30) and (31), direct computation shows Step  2. We then verify (26).
We claim that for any finite-dimensional subspace , there exists a constant such that Here and in the sequel, always denotes the Lebesgue measure in .
If not, for any , there exists such that Let , for all . Then , for all , and Passing to a subsequence if necessary, we may assume , for some , since is of finite dimension. Evidently, . In view of Lemma 4 and the equivalence of any two norms on , we have and .
By the definition of norm , there exists a constant such that For any , let Set . Then for large enough, by (36) and (38), we have Consequently, for large enough, there holds This is in contradiction to (37). Therefore (34) holds.
Consequently, for any , there exists a constant such that where , for all , and . By (), for any , there exists a constant such that Combining (23), (42), (43), and (), for any and , we have with . Now for any , if we choose then (44) implies ending the proof.

Proof of Theorem 1. It follows from (16), (23), and Lemma 5 that maps bounded sets to bounded sets uniformly for . In view of the evenness of in , it holds that for all . Thus the condition () of Theorem 7 holds. Besides, as and since . Thus the condition () of Theorem 7 holds. And Lemma 8 shows that the condition () holds for all . Therefore, by Theorem 7, for any and a.e. , there exists a sequence such that as , where with and .
Furthermore, it follows from the proof of Lemma 8 that where and by (32).
Claim  1.   possesses a strong convergent subsequence in , for and .
In fact, by the boundedness of the , passing to a subsequence, as ,  we may assume By the Sobolev embedding theorem, Lemma 6 implies that Observe that By (47), it is clear that It follows from the Hölder inequality, (51), and (52) that as . Thus by (53), (54), and (55), we have proved that that is, in .
Thus, for each , we can choose such that the sequence obtained a convergent subsequence; passing again to a subsequence, we may assume This together with (47) and (49) yields
Claim 2.    is bounded in for all .
For notational simplicity, we will set for all throughout this paragraph. If is unbounded in , we define . Since , without loss of generality we suppose that there is such that Let . If , from   it follows that On the other hand, after a simple calculation, we have We conclude that has zero measure and .
Moreover, from (49) and (58) By (), which contradicts (62). Hence is bounded.
Claim 3.   possesses a convergent subsequence with the limit for all .
In fact, by Claim 2, without loss of generality, we have assume By virtue of the Riesz Representation theorem, and can be viewed as and , respectively, where is the dual space of . Note that that is By Lemma 5, is also compact. Due to the compactness of and (64), the right-hand side of (66) converges strongly in and hence in .
Now for each , by (58), the limit is just a critical point of with . Since as in (49), we get infinitely many nontrivial critical points of . Therefore (1) possesses infinitely many nontrivial solutions by Lemma 5.

Acknowledgments

The authors would like to thank the referee for valuable comments and helpful suggestions. The first author would like to acknowledge the hospitality of Professor Y. Ding of the AMSS of the Chinese Academy of Sciences, where this paper was written during his visit. Anmin Mao was supported by NSFC (11101237) and ZR2012AM006.