Abstract

Using three critical points theorems, we prove the existence of at least three solutions for a quasilinear biharmonic equation.

1. Introduction

In this paper, we show the existence of at least three weak solutions for the Navier boundary value problem where is a nonempty bounded open set with a sufficient smooth boundary , , is an -Carathéodory function, and is a Lipschitz continuous function with Lipschitz constant ; that is, for every and .

Motivated by the fact that such problems are used to describe a large class of physical phenomena, many authors looked for existence and multiplicity of solutions for forth-order nonlinear equations. For an overview on this subject, we cite the papers [123]. For instance, when , in [22], Liu and Li, using Ricceri’s three critical points theorem [24], established the existence of at least three weak solutions for the following problem: where , are real constants and is a positive parameter and is a -Carathéodory function. Later, some authors generalized this type of equation (see [15, 10, 12, 15, 19]). When , Liu and Su [21] also used Ricceri’s three critical points theorem [24] to established the existence of at least three weak solutions for the following problem: where is a nonempty bounded open set with a sufficient smooth boundary , , , and is an -Carathéodory function. After that some authors used different critical point theorems to get one nontrivial, at least three, and infinitely many solutions (see [68, 16, 18]). Elliptic systems were also considered by [9, 11, 13, 14, 17, 20, 23].

The goal of the present paper is to establish some new criteria for (1) to have at least three weak solutions (Theorems 4 and 5). Our analysis is mainly based on recent critical point theorems that are contained in Theorems 2 and 3. In fact, employing rather different three critical points theorems, under different assumptions on the nonlinear term , we obtain the exact collections of for which (1) admits at least three weak solutions in the space .

A special case of our main results is the following theorem.

Theorem 1. Let be a Lipschitz continuous function with the Lipschitz constant and such that , where is defined by (17). Let be a continuous function and put for each . Assume that for some and in and Then, there is such that for each the problem admits at least three weak solutions.

2. Preliminaries

First we here recall for the reader’s convenience our main tools to prove the results. The first result has been obtained in [25] and the second one in [26].

Theorem 2 (see [25, Theorem 3.1]). Let be a separable and reflexive real Banach space, a nonnegative continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on , and a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Assume that there exists such that and that Further, assume that there are , , such that and here denotes the closure of in the weak topology. Then, for each the equation has at least three solutions in and, moreover, for each , there exists an open interval and a positive real number such that, for each , (10) has at least three solutions in whose norms are less than .

Theorem 3 (see [26, Theorem 3.6]). Let be a reflexive real Banach space; let be a sequentially weakly lower semicontinuous, coercive, and continuously Gâteaux differentiable whose Gâteaux derivative admits a continuous inverse on , and let be a sequentially weakly upper semicontinuous and continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Assume that there exist and with , such that (A1);(A2)for each the functional is coercive.Then, for each , the functional has at least three distinct critical points in .

Let be an -Carathéodory function and let be a Lipschitz continuous function with the Lipschitz constant ; that is, for every , and .

Put for all and . Denote the usual norm in is defined by Note that is a separable and reflexive real Banach space.

We say that a function is a weak solution of problem (1) if for all .

It is well known that is embedded in and for all . Since , one has .

Suppose that the Lipschitz constant of the function satisfies . For other basic notations and definitions, we refer the reader to [2729].

3. Main Results

Our main results are the following theorems.

Theorem 4. Assume that there exist a function , a positive function , and two positive constants and with such that(A1);(A2);(A3) for a.e. and for all .Then, for each in problem (1) admits at least three weak solutions in and, moreover, for each , there exists an open interval and a positive real number such that, for each , the problem (1) admits at least three weak solutions in whose norms are less than .

Theorem 5. Assume that there exists a function and a positive constant such that (B1);(B2);(B3).Then, for each problem (1) admits at least three weak solutions.

Let us give particular consequences of Theorems 4 and 5 for a fixed test function . Now, fix and pick with such that where denotes the ball with center at and radius of . Put

where , , and denotes the volume of .

Corollary 6. Assume that there exist a positive function and three positive constants , , and with and such that Assumption (A3) in Theorem 4 holds. Furthermore, suppose that (A4) for a.e. and all ;(A5) .Then, for each in problem (1) admits at least three weak solutions in and, moreover, for each , there exist an open interval and a positive real number such that, for each , problem (1) admits at least three weak solutions in whose norms are less than .

Proof. We claim that all the assumptions of Theorem 4 are fulfilled with given by and . It is easy to verify that , and, in particular, one has and consequently from (22) we see that Thus, (A1) holds. Since for each , the condition (A4) ensures that so, from (A5), and thus (A2) holds. Next notice that In addition note that Finally note that and, taking into account that and , we have the desired conclusion directly from Theorem 4.

Corollary 7. Assume that there exist two positive constants and with such that the assumption (A4) in Corollary 6 holds. Furthermore, suppose that (B4);(B5).Then, for each problem (1) admits at least three weak solutions.

Proof. All the assumptions of Theorem 5 are fulfilled by choosing as given in (25) and and bearing in mind that and recalling Hence, by applying Theorem 5, we have the conclusion.

Proof of Theorem 1. Fix for some . Since there is such that and In fact, one has where . Hence, there is such that and . From Corollary 7 we have the desired conclusion.

4. Proofs

Proof of Theorem 4. Our aim is to apply Theorem 2 to our problem. To this end, for each , we let the functionals be defined by and put The functionals and satisfy the regularity assumptions of Theorem 2. Indeed, by standard arguments, we have that is Gâteaux differentiable and sequentially weakly lower semicontinuous and its Gâteaux derivative at the point is the functional , given by for every . Furthermore, the differential is a Lipschitzian operator. Indeed, for any , there holds Recalling that is Lipschitz continuous and the embedding is compact, the claim is true. In particular, we derive that is continuously differentiable. The inequality (17) yields for any the estimate By the assumption , it turns out that is a strongly monotone operator. So, by applying Minty-Browder theorem [29, Theorem ],     admits a Lipschitz continuous inverse. On the other hand, the fact that is compactly embedded into implies that the functional is well defined, continuously Gâteaux differentiable, and with compact derivative, whose Gateaux derivative at the point is given by for every . Note that the weak solutions of (1) are exactly the critical points of . Also, since is Lipschitz continuous and satisfies , we have from (17) that for all , and so is coercive.
Furthermore from (A3) for any fixed , using (46), taking (17) into account, we have and so Also according to (A1) we achieve . From the definition of and by using (46) we have So, we obtain Therefore, from (A2) and (46), we have Now, we can apply Theorem 2. Note that, for each , Note also that (A2) implies Also, From (A2) it follows that since (note so and now apply (A2)). Now with and from Theorem 2 (note ) it follows that, for each , the problem (1) admits at least three weak solutions and there exist an open interval and a real positive number such that, for each , the problem (1) admits at least three weak solutions whose norms in are less than . Thus, the conclusion is achieved.

Proof of Theorem 5. To apply Theorem 3 to our problem, we take the functionals as given in the proof of Theorem 4. Let us prove that the functionals and satisfy the conditions required in Theorem 3. The regularity assumptions on and , as requested in Theorem 3, hold. According to (B1) we deduce . From the definition of we have and it follows that Therefore, due to assumption (B2), we have Furthermore, from (B3) there exist two constants with such that for all and all . Fix . Then for all . Now, to prove the coercivity of the functional , first we assume that . So, for any fixed using (61), we have and thus On the other hand, if , clearly we obtain . Both cases lead to the coercivity of functional .
So, assumptions (A1) and (A2) in Theorem 3 are satisfied. Hence, by using Theorem 3, the problem (1) admits at least three distinct weak solutions in .

Conflict of Interests

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

Acknowledgments

This work was supported by the Fundamental Research Funds for the Central Universities (no. XDJK2013D007), the Scientific Research Fund of SUSE (no. 2011KY03), and the Scientific Research Fund of Sichuan Provincial Education Department (no. 12ZB081).