#### Abstract

By using the variant fountain theorem, we study the existence of multiple solutions for a class of superquadratic fourth-order elliptic problem with Navier boundary value condition.

#### 1. Introduction

Consider the following fourth-order boundary value problem: where denotes the biharmonic operator, is a bounded domain with smooth boundary, and .

The fourth-order elliptic equations which contain a biharmonic operator can describe the static form change of beam or the motion of rigid body. Thus the fourth-order elliptic equations are widely applied in physics, oceanics, aerospace engineering and other engineering. In [1], Lazer and Mckenna considered the biharmonic problem: where and . They pointed out that this type of nonlinearity furnishes a model to study traveling waves in suspension bridges. Afterwards, in [2], they have proved the existence of solutions when and ( is the sequence of the eigenvalues of in ) by the global bifurcation method. In [3] the existence of a negative solution of (2) was proved when by using the Leray-Schauder degree. In particular, in [1, 4] the authors observed that problem (2) was interesting also when the nonlinearity was replaced by a somewhat more general function . In [5], Micheletti and Pistoia used a variational linking theorem to investigate the existence of two solutions for a more general nonlinearity . Moreover, by using a variational result, they and Saccon also showed the existence of three solutions for some special (see [6]). Next year, in [7], Micheletti and Saccon obtained two results about the existence of two nontrivial solutions and four nontrivial solutions by the similar variational approach, depending on the position of a suitable parameter with respect to the eigenvalues of the linear part. In recent years, more researchers have used variational approach to investigate the fourth-order elliptic equations. In [8], Xu and Zhang studied the existence of positive solutions of problem (1) when satisfied the local superlinearity and sublinearity condition and by the classical mountain pass theorem. Recently, in [9], Pu et al. used the least action principle, the Ekeland variational principle, and the mountain pass theorem to prove the existence and multiplicity of solutions of (1) when . For other related results, see [8–14] and the references therein. Here, we emphasize that most authors considered the case .

The variant fountain theorems established in [15] have been used in the study of a class of semilinear elliptic equations (see [16, 17]) and the investigation of the Hamiltonian system (see [18, 19]). Inspired by [9, 17], we will use the variant fountain theorem to investigate the problem (1). More precisely, we make the following assumptions.There exist constants and such that for all and Here, is the primitive of the nonlinearity .There exist constants , and such that Our main result is the following theorem.

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

*Remark 2. *In Theorem 1, we do not assume , which is widely used in the investigation of the fourth-order equations. As is known, the so-called global Ambrosetti-Rabinowitz condition (AR-condition for short) is introduced by Ambrosetti and Rabinowitz in [20] and wildly used to the existence of infinitely many solutions for superquadratic situation: there is a constant such that, for all and , the nonlinearity is assumed to satisfy
In fact, if we choose
where , , and for all . Then it is easy to see that satisfies the conditions in Theorem 1 with , , , , and , but does not satisfy the AR-condition (6).

*Remark 3. *By , we can obtain that there exists a constant such that
And by , there exists a constant such that

#### 2. Preliminaries

In this section, we will establish the variational setting for our problem and state a variant fountain theorem.

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

A weak solution of problem (1) is a such that for any . Here and in the sequel, always denotes the standard inner product in . Let be the functional defined by It is well known that a critical point of the functional in corresponds to a weak solution of problem (1).

Let be the eigenvalues of in . Then the eigenvalue problem has infinitely many eigenvalues , .

Define a selfadjoint linear operator by with domain . Here, denotes the inner product in and in the sequel is simply denoted by . Then the sequence of eigenvalues of is just . Denote the corresponding system of eigenfunctions by ; it forms an orthogonal basis in .

Denote Here, denotes the cardinal of a set. Let Decompose as Then also possesses the orthogonal decomposition with

We define on a new inner product and the associated norm by Therefore, can be written as where for all . Then and are continuously differentiable.

Direct computation shows that for all with and , respectively. It is known that is compact.

Denote by the usual norm of for all ; then by the Sobolev embedding theorem, there exists a such that Noting that the constants and appeared in and satisfies

To prove our main result Theorem 1, we need an abstract critical point theorem found in [15].

Let be a Banach space with the norm and with for any . Set and . Consider the following -functional defined by

Theorem 4 (see [15, Theorem 2.1]). *Assume that the functional defined above satisfies the following:** maps bounded sets to bounded sets for , and for all ;** for all ; moreover, or as ;**there exist such that
**Then
**
where and . Moreover, for a.e. , there exists a sequence such that
*

In order to apply this theorem to prove our main result, we define the functionals , , and on our working space as follows: for all and . Then for all and Let , . Note that is just equal to the functional defined in (22).

#### 3. Proof of Theorem 1

In this section we firstly establish the following two lemmas and then give the proof of Theorem 1.

Lemma 5. *Assume that and hold. Then for all . Furthermore, or as .*

*Proof. *Since , by (30), it is obvious that for all .

By the similar method used in the proof of Lemma 2.6 of [17], for any finite-dimensional subspace , there exists a constant such that
where is the Lebesgue measure in .

Now for the finite-dimensional subspace , there exists a constant corresponding to the one in (33). Let
Then . By , there exist positive constants and such that
Note that
for any with . Combining (35) and (36), for any with , we have
which implies that
Combining this with and (30), we have
The proof is completed.

Lemma 6. *Let , be satisfied. Then there exist a positive integer and two sequences as such that
**
where and for all .*

*Proof. **Step **1.* We first prove (40).

By virtue of (8) and (31), for any
where , are the constants in (8). Let
Then
since is compactly embedded into . Note that
where is the integer given in (16). Combining (24), (42), (43), and (45), for , we have
where is the constant given in (24). By (44), there exists a positive integer such that
since . Clearly,
Combining (46) and (47), direct computation shows
*Step **2.* We then prove (41).

Note that for any , is of finite dimension, so we can choose sufficiently large such that
By and (8), for the former , there exists a such that

Consequently, by (50) and (51), we have
for all . Now for any , if we choose
then (52) implies
The proof is completed.

Now we prove our main result Theorem 1.

*Proof of Theorem 1. *In view of (8), (24), and (31), maps bounded sets to bounded sets uniformly for . By virtue of the evenness of in , it holds that for all . Therefore the condition of Theorem 4 holds. Lemma 5 shows that the condition holds, whereas Lemma 6 implies that condition holds for all , where is given in Lemma 6. Thus, by Theorem 4, for each and a.e. , there exists a sequence such tha
where
with and .

Moreover, by the proof of Lemma 6, we have
where and as by (48).

Since the sequence obtained by (55) is bounded, it is clear that, for each , we can choose such that the sequence has a strong convergent subsequence.

In fact, without loss of generality, assume that
for some since .

Note that
That is,
In view of (55), (58), (59), and the compactness of , the right-hand side of (61) converges strongly in and hence in . Together with (58), has a strong convergent subsequence in .

Without loss of generality, we assume
This together with (55) and (57) yields
Now we claim that the sequence in (63) is bounded in and possesses a strong convergent subsequence with the limit for each . For the sake of notational simplicity, throughout the remaining proof of Theorem 1 we always denote .

Now we claim that is bounded in . Otherwise, going to a subsequence if necessary, we can assume that as . By (9), we have
which yields that
Write . It follows from , (24), (25), (32), and the Hölder inequality that
for any . Here and in the sequel, we denote for different positive constants. Since and , we have . So, by (65) we get
Similarly, we have
By , there also exist constants and such that
So we get
keeping in mind that and (24). Hence, by (67) and (68), we get
Then we arrive at
which is a contradiction. Thus, is bounded in . Then the proof that has a strong convergent subsequence is the same as the preceding proof of .

Now for each , by (63), the limit is just a critical point of with . Since as in (57), we get infinitely many nontrivial critical points of . Therefore, system (1) possesses infinitely many nontrivial solutions.

#### Conflict of Interests

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

#### Acknowledgments

This study was supported by the Fundamental Research Funds for the Central Universities and the National Natural Science Foundation of China (no. 61001139).