#### Abstract

We construct a variational functional of a class of three-point boundary value problems with impulse. Using the critical points theory, we study the existence of solutions to second-order three-point boundary value problems with impulse.

#### 1. Introduction

In this paper, we study the following three-point boundary value problems with impulse: where , , , and (respectively, ) denote the right limit (respectively, left limit) of at .

The existence of solutions for three-point boundary value problems has been investigated by many authors. See, for example, [1–11] and references cited therein. In [1], Bao et al. studied a class of three-point boundary value problems Using the method upper and lower solutions, some existence results for positive solutions of problems (2) had been obtained. By applying the fixed point theory, many authors have studied the existence of positive solutions for three-point boundary value problems (see [7–11]). In [12], the authors studied Sturm-Liouville boundary value problem of a class of second-order impulsive differential equations: By establishing the corresponding variational principle of problem (3), the existence for solutions of the problems had been obtained. In papers [12–16], the variational methods were applied to impulsive differential equations too. In this paper, we will use the critical point theorem to study three-point boundary value problems (1) with impulse.

The paper is organized as follows: in Section 2, we will construct a variational functional of the boundary value problem (1). In Section 3, using the critical point theory, we will give some sufficient conditions in which problem (1) has solutions. Throughout this paper, we always assume that the following conditions hold.) is measurable in for each , continuous in for almost every .()for any , there exists such that for almost every and for all .

#### 2. Variational Structure

Let be the space of absolutely continuous functions with a weak derivative . We define the operator on by and set and . The following properties are easy consequence of the definition:().().()For each , .

Now, we define the norm over by Then is a Hilbert space and the corresponding inner product is For each , it follows from (6) that and so on .

In order to study problem (1), we define the functional on by where and Under the conditions and , is continuously differentiable, weakly lower semicontinuous on and for all ; see [17].

The following theorem is the main conclusion of this paper.

Theorem 1. *Assume that satisfies the conditions and . If is a critical point of the functional defined by (9), then is a solution of problem (1).*

*Proof. *Let be a critical point of the functional defined by (9). We prove this theorem in three steps.*Step 1.* In this step, we prove that satisfies the equation except at .

We define by
It follows from (11) that for all ,
By the Fubini theorem and (13), we obtain
In particular, we can choose
so that
The theorem of Fourier series implies that
on for some . Integrating (17) over , we obtain
and hence
Similarly, setting
in (14), we have
on . Thus, (19) and (21) imply that satisfies the equation except at .*Step 2.* In this step, we prove that satisfies the boundary value conditions and .

Set
Inserting (22) into (13), we have
It follows form (19) and (23) that
Similarly, setting
equality (13) becomes
and hence
*Step 3.* In this step, we prove that satisfies the conditions .

Inserting
into (13), we have
It follows from (19) and (21) that
This completes the proof of Theorem 1.

#### 3. Solutions of Problem (1)

As applications of Theorem 1, we consider solutions of problem (1). Let and denote the minimum and maximum eigenvalue of the matrix in (9). We have the following theorems.

Theorem 2. *Assume that satisfies and . Assume also that the following conditions hold:*()*.*()*there is a positive constant , with , and a positive function such that
for almost every and for all .**Then, problem (1) has a solution.*

*Proof. *() implies that for each
By (9), (32), and (),
where . It follows that as . Thus, has a critical point and problem (1) has a solution.

Theorem 3. *Assume that , , and are satisfied. Assume also that the following conditions hold:** there is a positive function such that
**
for almost every and for all .*()*, where is in the proof of Theorem 2.**Then, problem (1) has a solution.*

*Proof. *This proof is similar to the proof of Theorem 2.

Theorem 4. *Assume that (A _{1})–(A_{4}) are satisfied. Assume also that the following condition holds:*()

*there are two positive constants and with such that*

*where .*

*Then, problem (1) has at least three solutions.*

In order to prove this theorem, we need the following theorem (see Theorem 2.1 of [18]).

Theorem A. *Let be a reflexive real Banach space; let be a sequentially weakly lower semicontinuous, coercive, and continuously Gâteaux differentiable functional 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 and , such that*(i)(ii)*for each the functional is coercive.**Then, for each the functional has at least three distinct critical points in .*

*Proof of Theorem 4. *Let and define
for each . Then, is a sequentially weakly lower semicontinuous and continuously differentiable functional whose derivative is given by
for all . It follows from (36) that for each ,
so that is coercive. Because is a compact operator, has the continuous inverse if and only if is not the eigenvalues of . If is the eigenvalues of and is a eigenvector of associated with the eigenvalue , then (37) implies that
This is a contradiction, and hence has the continuous inverse. Set for . Then, is a sequentially weakly upper semicontinuous and continuously differentiable functional whose derivative is compact.

Setting and for all , then and
since . Setting , by and (40), we obtain
By (38), implies that for every since . From and (40), we have
and (i) in Theorem A holds.

Since
we can take in Theorem A. Therefore, it is easy to show that
so that is coercive. Using Theorem A, problem (1) has at least three solutions.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

This work is supported by the Natural Science Foundation of Shandong Province of PR China (ZR2011AL007).