The Sturm-Liouville boundary-value problem for fourth-order impulsive differential equations is studied. The existence results for one solution and multiple solutions are obtained. The main ideas involve variational methods and three critical points theory.

1. Introduction

The aim of the present paper is to study the following Sturm-Liouville boundary-value problem for the fourth-order impulsive differential equation: where , , , and are real constants, is a positive parameter, , , , , , denote the right (left) limits, respectively, of , at , and , , .

Recently, many authors have studied the existence of solutions for boundary-value problems with impulsive effects [116]. Variational methods are powerful tools for them. We refer the readers to [1719] for related basic information.

In [10], the authors studied the following equation with impulsive effects: By applying critical point theory to (2), several existence results are obtained when is imposed some assumptions and lies in suitable interval. In [8], the authors studied the existence of solutions for the following problem: They essentially proved that when , , and satisfy some conditions, (3) has at least one solution or infinitely many classical solutions via variational methods.

To the best of our knowledge, besides [12, 13] for second-order differential equations, [8] for fourth-order differential equation, limited work has been done in the Sturm-Liouville boundary-value problem, let alone higher order. Motivated by the above facts, we study the existence of solutions for problem (1) by applying variational methods. With the impulse effects and the Sturm-Liouville boundary conditions taken into consideration, the corresponding variational functional will be more complicated than the ones of any fourth-order boundary-value problems before. In our study, some difficulties such as how to prove that the critical points of are just the solutions of problem (1) and how to prove the space and the functional to satisfy the conditions of the related theorems must be overcome. To verify that the weak solution of problem (1) is just the classical solution of (1), we construct a Fundamental Lemma 5, by which we can easily prove that the critical point of the functional is just the solution of problem (1).

This paper is organized as follows. In Section 2, we present some preliminaries and establish the variational structure. In Section 3, we discuss the existence results for one solution and multiple solutions. In Section 4, we discuss the existence results for positive solutions. In Section 5, we will give some examples.

2. Preliminaries and Variational Structure

First we present some theorems that will be needed in the proof of main results.

Theorem 1 (see Theorem 2.2 [19]). Let be a real Banach space and satisfying the Palais-Smale condition (PS). Suppose and(C1)there are constants such that ,(C2)there is an such that . Then possesses a critical value . Moreover, can be characterized as where

Theorem 2 (see Theorem 9.12 [19]). Let be an infinite dimensional Banach space and let be even, satisfy (PS), and . If , where is finite dimensional, and satisfies that(C3)there exist constants such that ,(C4)for each finite dimensional subspace , there is an such that on .Then has unbounded sequence of critical values.

Let be a nonempty set and two functionals. For all , , , we define

Theorem 3 (see Theorem 2.1 [2]). Let be a reflexive real Banach space, a convex, coercive, and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on , and a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact, such that(1) ;(2)for every , such that and one has Assume that there are three positive constants , , with , such that(i) ;(ii) ;(iii) .Then, for each , the functional admits three distinct critical points , , such that , , and .

Let us define the space equipped with the norm We set the functional defined by where , . is differentiable for any and Set the usual norm of , , respectively, as follows:

Lemma 4. The norm is equivalent to the usual norm .

Proof. First, we will show that there exists such that . Since is absolutely continuous in , we have . So which implies where .
Next, we prove that there exists such that .
Obviously, , where . Thus, we have Similar to the above proof, we have By (14) and (15), we have where . By (13) and (16), the proof is complete.

Lemma 5 (Fundamental Lemma). Let . If for every with , satisfying , then there exist such that a.e. on .

Proof. Define by ; we have By the Fubini theorem, we have So In particular, we choose , where By computation, , , , and By (19) and (21), ; that is, , which means . The proof is complete.

Definition 6. A function is said to be a weak solution of (1), if satisfies for all .

Definition 7. A function is said to be a classical solution of problem (1) if satisfies the equation in (1) for a.e. and the impulsive condition and boundary condition in (1). Moreover, is said to be a positive classical solution of problem (1) if , , .

Lemma 8. If is a weak solution of problem (1), then is a classical solution of problem (1).

Proof. By Definition 6, if is a weak solution of (1), then holds for all and hence for all ,   ,   ,   . So . By Lemma 5, we have for a.e. and some . So , a.e. , . Thus satisfies the equation in problem (1) and . By integration by parts for two times, we have that holds for all . Since satisfies the equation of problem (1), (23) becomes for all .
Next we will verify that satisfies impulsive condition in (1). If not, without loss of generality, we assume (24) holds for for , . So , and then . We assume for , . So (24) becomes , which means . Similarly, by choosing particular , we can show that satisfies boundary conditions in problem (1).

Lemma 9. Let ; then , where .

Proof. For any , it follows from the mean-value theorem that for some . Hence, for , using (25) and Hölder’s inequality, we have Similarly, we have . So , which together with (13) yields the results.

Lemma 10. Suppose the following conditions hold.(H1)There exist constants and such that, for , (H2)The impulsive functions satisfy sublinear growth; that is, there exist constants , , and , , such that (H3)The impulsive functions , , are bounded.(H4) , , as , .Then the functional defined by (9) is continuously differentiable. Moreover, it satisfies the Palais-Smale (PS) condition.

Proof. By the continuity of , and , , we know that is continuously differentiable. Next, we will prove that satisfies the Palais-Smale condition. Let be a bounded sequence such that as . Then there exist two constants such that for sufficiently large By (H3), there exists a constant such that Then for sufficiently large, by (H1) (H2) and the definitions of , , we have By Lemma 8, (29), (30), (31), (H2), and (H3), one has So is bounded in , which implies that the sequence weakly converges to .
Next we show that strongly converges to in : Similar to the proof of Proposition 1.2 in [18], the weak convergence implies that uniformly converges to in . Since , converges to in . Thus So as . In other words, converges strongly to in .

Remark 11. By (H1), there exist such that for all , .

3. Existence Results for One Solution and Infinitely Many Solutions

Theorem 12. Suppose that (H1)–(H3) hold. Furthermore, we assume (H4) holds. Then problem (1) has at least one nontrivial solution.

Proof. We will use Theorem 1 to prove the theorem. By Lemma 10, we have known that satisfies the (PS) condition and it is obvious that . By (H4), for any , there exists a such that , which implies for all . Consequently, by Lemma 9, one has, for , Thus as . Therefore, as . Thus (C1) holds.
To verify (C2), we choose , such that , , , are constants. Then by (H1), (H2), Remark 11, Lemma 8, and (30), one has By Hölder’s inequality, we have Substituting (41) into (40), we have as . Hence (C2) holds. Therefore, applying Theorem 1, we deduce that admits a critical value characterized as in the statement of Theorem 1 to , possesses critical value given by where Let be a critical point associated with the critical value of (i.e., ). Condition implies that . Lemma 8 means that IBVP (1) has at least one nontrivial solution.

Theorem 13. Suppose that (H1)–(H4) hold. Moreover, assume that the nonlinearity and impulsive functions , are all odd in . Then IBVP (1) has infinitely many classical solutions.

Proof. We apply Theorem 2 to complete the proof. Clearly is even and . Lemma 10 shows that satisfies (PS) condition. The arguments of Theorem 12 show that satisfies (C3) in Theorem 2. To verify (C4), let be any finite dimensional space in . For any , by (H1), (H2), Remark 11, Lemma 4, and (30), one has For finite dimensional space , the norm is equivalent to .
So there exists satisfying for . Thus, By (45) (46) (47) we have as . That is, there exists such that for . The proof is complete.

4. Existence Result for Three Nonnegative Solutions

In this part, we need the following conditions.(H5)For , , and ,(C5) , , , ;(C6) for almost every and for all .

Lemma 14 (see Lemma 2.2 [12]). For , let . Then the following five properties hold:(i) , ;(ii) ;(iii) ;(iv)if uniformly converges to u in , then uniformly converges to in ;(v) , for a.e. .

Lemma 15. If is a classical solution of problem then for , and hence it is a nonnegative classical solution of (1).

Proof. Since and , we have . If is a classical solution of problem (49), by Lemma 14, (H5) and boundary conditions, we have So for ; that is, . The proof is complete.

Remark 16. By Lemmas 14 and 15, in order to obtain the nonnegative solutions of (1), it is sufficient to show the existence of solutions of (49).
For each , set It is obvious that , , and are differentiable for any . Then we have

Definition 17. A function is said to be a weak solution of (49), if satisfies for all .

Lemma 18. If is a weak solution of (49), then is a classical solution of (49).

Proof. It is similar to the proof of Lemma 8, so we omit it here.

Lemma 19. admits a continuous inverse on .

Proof. First, for every , by (53), we have which means that is coercive. Furthermore, given , one has so is uniformly monotone. By Theorem 26.A(d) of [20], we have that exists and is continuous on . Thus, admits a continuous inverse on . The proof is complete.

Lemma 20. is a continuous and compact operator.

Proof. First we will show that is strongly continuous on . Let as on . By [20], we have converges uniformly to on as . Since is continuous, one has as . Furthermore, , are all continuous. So , which implies that is continuous and that is a compact operator by Proposition 26.2 of [20]. The proof is complete.

Theorem 21. Suppose that the condition (H5) holds. Let , . There exist four positive constants , , , , with such that(H6)(C7) , (C8) , (C9) . Then, for every the problem (1) has at least three distinct nonnegative classical solutions , such that , which means that the problem (1) has at least two distinct positive classical solutions.

Proof. The proof is based on Theorem 3. First, we will prove that and satisfy the hypotheses in Theorem 3. On the one hand, is coercive and its Gâteaux derivative admits a continuous inverse by Lemma 19. On the other hand, is obviously convex. ’s Gâteaux derivative is continuous and compact by Lemma 20. In addition, . By (H5), we have which deduces that for all .
Next, we will verify the conditions (i), (ii), (iii) in Theorem 3. First, we define