#### Abstract

We use variational methods to investigate the solutions of damped impulsive differential equations with mixed boundary conditions. The conditions for the multiplicity of solutions are established. The main results are also demonstrated with examples.

#### 1. Introduction

Impulsive effect exists widely in many evolution processes in which their states are changed abruptly at certain moments of time. The theory of impulsive differential systems has been developed by numerous mathematicians [16]. Applications of impulsive differential equations with or without delays occur in biology, medicine, mechanics, engineering, chaos theory, and so on [711].

In this paper, we consider the following second-order damped impulsive differential equations with mixed boundary conditions: where , , is continuous, , are continuous, and for .

The characteristic of (1) is the presence of the damped term . Most of the results concerning the existence of solutions of these equations are obtained using upper and lower solutions methods, coincidence degree theory, and fixed point theorems [1215]. On the other hand, when there is no presence of the damped term, some researchers have used variational methods to study the existence of solutions for these problems [1621]. However, to the best of our knowledge, there are few papers concerned with the existence of solutions for impulsive boundary value problems like problem (1) by using variational methods.

For this nonlinear damped mixed boundary problem (1), the variational structure due to the presence of the damped term is not apparent. However, inspired by the work [22, 23], we will be able to transform it into a variational formulation. In this paper, our aim is to study the existence of distinct pairs of nontrivial solutions of problem (1). Our main results extend the study made in [22, 23], in the sense that we deal with a class of problems that is not considered in those papers.

#### 2. Preliminaries and Statements

Let , , , . We transform (1) into the following equivalent form:

Obviously, the solutions of (2) are solutions of (1).

Define the space . It is easy to see that and is a closed subset of . So is a Hilbert space with the usual inner product in .

Consider the Hilbert spaces with the inner product inducing the norm We also consider the inner product inducing the norm

Consider the problem As is well known, (7) possesses a sequence of eigenvalues with The corresponding eigenfunctions are normalized so that ; here

Now multiply (2) by and integrate on the interval : Then, a weak solution of (2) is a critical point of the following functional: where .

We say that is a classical solution of IBVP (1) if it satisfies the following conditions: satisfies the first equation of (1) a.e. on ; the limits , , exist and impulsive condition of (1) holds; satisfies the boundary condition of (1).

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

Proof. If is a weak solution of (1), then is a weak solution of (2), so holds for all ; that is, By integrating by part, we have Thus holds for all . Without loss of generality, for any and with , for every , then substituting into (14), we get Hence satisfies the first equation of (2). Therefore, by (14) we have
Next we will show that satisfies the impulsive and the boundary condition in (2). If the impulsive condition in (2) does not hold, without loss of generality, we assume that there exists such that Let ; then which contradicts (16). So satisfies the impulsive condition in (2) and (16) implies If , pick ; one has which contradicts (19), so satisfies the boundary condition. Therefore, is a solution of (1).

Lemma 2. Let . Then there exists a constant , such that where .

Proof. By Hölder inequality, for ,

Lemma 3 (see [24, Theorem 9.1]). Let be a real Banach space, with even, bounded from below, and satisfying P.S. condition. Suppose ; there is a set such that is homeomorphic to by an odd map and . Then possesses at least distinct pairs of critical points.

#### 3. Main Results

Theorem 4. Suppose that the following conditions hold.(H1)There exist , which is the kth eigenvalue of (7) such that (H2)There exist and such that (H3) and are odd about .(H4), , as , .Then, for , problem (1) has at least distinct pairs of solutions.

Proof. Set Consider
Next, we will verify that the solutions of problem (26) are solutions of problem (1).
In fact, let be the solution of problem (26). If , then there exists an interval such that When , by (H1), we have That is, is nondecreasing in . By and , we have That is, for any . Since , then . So, there exists a constant such that , which contradicts (27). Then . Similarly, we can prove that .
Therefore, any solution of (26) is a solution of (1). Hence to prove Theorem 4, it suffices to produce at least distinct pairs of critical points of where .
We will apply Lemma 3 to finish the proof.
By (30) and (H3), is even and .
Next, we will show that is bounded from below.
Let , . By (H1) and (H3), we have for ; thus So, we have for any . Therefore, is bounded from below.
In the following we will show that satisfies the P.S. condition. Let such that is a bounded sequence and ; then there exists such that By (32), we have So is bounded in . From the reflexivity of , we may extract a weakly convergent subsequence that, for simplicity, we call in . In the following we will verify that strongly converges to : By in , we see that uniformly converges to in . So So we obtain , as . That is, strongly converges to in , which means that satisfies the P.S. condition.
Now set , where is defined in (9). It is clear that is homeomorphic to by an odd map for any . In the following we verify that if is sufficiently small.
For any . By (H4) and (30), we have for small . Since , and the proof is complete.

Theorem 5. Suppose that the following conditions hold.(H1)There exist , which is the kth eigenvalue of (7) such that (H2) for any .(H3) and are odd about .(H4), , as , .Then, for , problem (1) has at least distinct pairs of solutions.

Proof. The proof is similar to the proof of Theorem 4, and therefore we omit it.

Theorem 6. Suppose that the following conditions hold.(H1)There exist , which is the kth eigenvalue of (7) such that (H2) and are odd about .(H3), , as , .Then, for , problem (1) has at least distinct pairs of solutions.

Proof. Set Consider
Next, we will verify that the solutions of problem (41) are solutions of problem (1).
In fact, let . By the definitions of and , (41) is reduced to The solution of (42) satisfies . So and .
Let . By the definitions of and , (41) is reduced to The solution of (43) satisfies , . So and .
Therefore, the solutions of (41) are solutions of (1). Hence to prove Theorem 6, it suffices to produce at least distinct pairs of critical points of where .
We will apply Lemma 3 to finish the proof.
By (44) and (H2), is even and .
Next, we will show that is bounded from below.
By (H1) and (H2), we have and for ; thus So, we have for any . Therefore, is bounded from below.
In the following we will show that satisfies the P.S. condition. Let such that is a bounded sequence and ; then there exists such that By (46), we have So is bounded in . From the reflexivity of , we may extract a weakly convergent subsequence that, for simplicity, we call in . In the following we will verify that strongly converges to : By in , we see that uniformly converges to in . So So we obtain , as . That is, strongly converges to in , which means satisfies the P.S. condition.
Now set , where is defined in (9). It is clear that is homeomorphic to by an odd map for any . In the following we verify that if is sufficiently small.
For any , . By (H3) and (44), we have for small . Since , and the proof is complete.

#### 4. Example

To illustrate how our main results can be used in practice we present the following example.

Example 1. Let , and consider the following problem:
Compared with (1), , . Obviously (H2), (H3), and (H4) are satisfied. Let , ; then (H1) is satisfied. By Theorem 4, for , , problem (52) has at least distinct pairs of solutions.

Example 2. Let , and consider the following problem:
Compared with (1), , . Obviously (H2), (H3), and (H4) are satisfied. Let , ; then (H1) is satisfied. By Theorem 5, for , , problem (53) has at least distinct pairs of solutions.

Example 3. Let , and consider the following problem:
Compared with (1), , . Obviously (H2) and (H3) are satisfied. Let ; then (H1) is satisfied. By Theorem 6, for , , problem (54) has at least distinct pairs of solutions.

#### Conflict of Interests

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

#### Acknowledgments

This work is partially supported by the National Natural Science Foundation of China (no. 71201013) and the Innovation Platform Open Funds for Universities in Hunan Province (no. 13K059).