Research Article | Open Access

# Existence and Multiplicity of Solutions to a Boundary Value Problem for Impulsive Differential Equations

**Academic Editor:**Wan-Tong Li

#### Abstract

We investigate the existence and multiplicity of solutions to a boundary value problem for impulsive differential equations. By using critical point theory, some criteria are obtained to guarantee that the impulsive problem has at least one solution, at least two solutions, and infinitely many solutions. Some examples are given to illustrate the effectiveness of our results.

#### 1. Introduction

In this paper, we will investigate the existence and multiplicity of solutions to the boundary value problem for impulsive differential equations: where , , , with ess , ess , and ; ; , , is continuous, are continuous.

Recently, there have been many papers concerned with boundary value problems for impulsive differential equations. Impulsive effects exist 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 (see [1â€“6]).

Impulsive and periodic boundary value problems have been studied extensively in the literature. There have been many approaches to study periodic solutions of differential equations, such as the method of lower and upper solutions, fixed point theory, and coincidence degree theory (see [7â€“10]). However, the study of solutions for impulsive differential equations using variational method has received considerably less attention (see, [11â€“18]). Variational method is, to the best of our knowledge, novel and it may open a new approach to deal with nonlinear problems with some type of discontinuities such as impulses.

Teng and Zhang in [15] studied the existence of solutions to the boundary value problem for impulsive differential equations By using variational methods and iterative methods they showed that there exists a solution for problem (2).

In this paper, we will need the following conditions.(A) is measurable in for every and continuously differentiable in for a.e. and there exist such that for all and a.e. , where .() There exist constants and such that () There exist constants and such that () There exist constants and such that () There exist constants and such that () uniformly, for and . () there exist and such that () there exists such that

We recall some facts which will be used in the proof of our main results. It has been shown, for instance, in [19] that the set of all eigenvalues of the following problem: is given by the sequence of positive numbers where Each eigenvalue is simple with the associated eigenfunction denotes the eigenspace associated to , then .

An outline of this paper is given as follows. In the next section, we present some preliminaries including some basic knowledge and critical point theory. In Section 3, by using the critical point theory, we will establish some sufficient conditions for the existence of solutions of system (1). In Section 4, some examples are given to verify and support the theoretical findings.

#### 2. Preliminaries

In this section, we recall some basic facts which will be used in the proofs of our main results. In order to apply the critical point theory, we make a variational structure. From this variational structure, we can reduce the problem of finding solutions of (1) to the one of seeking the critical points of a corresponding functional.

In [20], the Sobolev space be the endowed with the norm

Throughout the paper, we also consider the norm By PoincarÃ© inequality: where is precisely the largest for which the above inequality holds true. Then while it minimizes and equals to zero exactly on the ray generated by the first eigenfunction .

Let us recall that We denote by the usual -norm. The -dimensional Lebesgue measure of a set is denoted by . By the Sobolev embedding theorem, continuously for , and there exists such that

Lemma 1. *There exist , such that
*

*Proof. *Since ess , ess , , and , we have that ess , ess , where is a positive number and that there exists such that . Thus, by PoincarÃ© inequality, we have
for all . Thereby, for every ,

On the other hand, by PoincarÃ© inequality, one has

Take , , then

The proof is complete.

Lemma 2. *There exists such that if , then
*

*Proof. *If , it follows from the mean value theorem that
for some . Hence, for , by HÃ¶lder inequality and PoincarÃ© inequality
Hence, . The proof is complete.

Take and multiply the two sides of the equality by and integrate it from to , we have Moreover, Combining (28), we have Considering the above, we introduce the following concept solution for problem (1).

*Definition 3. *We say that a function is a solution of problem (1) if the identity
holds for any .

Consider the functional defined by
Using the continuity of and , , one has that . For any , we have
Thus, the solutions of problem (1) are the corresponding critical points of .

*Definition 4. *Let be a normed space. A minimizing sequence for a function is a sequence such that whenever .

*Definition 5. *Let be a Banach space and let . is said to be sequentially weakly lower semicontinuous if as in .

*Definition 6. *Let be a Banach space and let . For any sequence in , if is bounded and as possesses a convergent subsequence, then we say that satisfies the Palais-Smale condition (denoted by condition for short). We say that satisfies the Palais-Smale condition at level (denoted by condition for short) if there exists a sequence in such that and as implies that is a critical value of .

*Definition 7. *Let be a Banach space and let . is said to be coercive if as .

Lemma 8 (see [12]). * If is sequentially weakly lower semicontinuous on a reflexive Banach space and has a bounded minimizing sequence, then has a minimum on . *

*Definition 9. *Let be a real Banach space with a direct sum decomposition . The functional is said to have a local linking at , with respect to , if, for some ,(i),(ii).

If has a local linking at 0, then 0 is critical point (the trivial one). Suppose, furthermore, that there are two sequences of finite dimensional subspaces and such that

*Definition 10 (see [13, Definition 2.2]). * Let . The functional satisfies the condition if every sequence such that is admissible and
contains a subsequence which converges to a critical point of .

Lemma 11 (see [13]). * Suppose that satisfies the following assumptions: *(1)* satisfies the condition,*(2)* has a local linking at 0,*(3)* maps bounded sets into bounded sets,*(4)*for every ,â€‰â€‰ as , . **Then has at least two critical points. *

Lemma 12 (see [14]). *Let E be a Banach space. Let be an even functional which satisfies the condition and . If , where is finite dimensional, and satisfies the following conditions:*(1)*there exist constants such that , where ,*(2)*for each finite-dimensional subspace there is such that , for all with . **Then has an unbounded sequence of critical values. *

#### 3. Existence of Periodic Solutions

Theorem 13. *Assume that ,â€‰â€‰, and are satisfied, then problem (1) has at least one solution. *

*Proof. *Let , . By Lemma 2, we have
for all . This implies that , and is coercive.

On the other hand, we show that is weakly lower semicontinuous. If , , then we have that converges uniformly to on and . Thus

By Lemma 8, has a minimum point on , which is a critical point of . Hence, problem (1) has at least one solution. The proof is complete.

We readily have the following corollary.

Corollary 14. *Assume that ,â€‰â€‰, and are satisfied and and the impulsive functions are bounded. Then problem (1) has at least one solution. *

Lemma 15. *Assume that , , and are satisfied, then satisfies the condition. *

*Proof. *Assume that satisfies that is bounded and as . We will prove that the sequence is bounded.

It follows from , , , and Lemma 2, we have
Hence, is bounded in .

Since is a reflexive Banach space, passing to a subsequence if necessary, we may assume that there is a such that
Notice that
Recalling the following well-known inequality: for any ,
for some constant (Lemmaâ€‰â€‰4.2 in [21]) and using Schwarz inequality, we have
Since
By the assumption , we have
From (39), it follows that in . Thus, satisfies the condition. The proof is complete.

Lemma 16. *Assume that , , , and are satisfied, then satisfies the condition. *

*Proof. *Let be a sequence in such that is admissible and
then there exist a constant such that
for all large . On the other hand, by , there are constants and such that
for all and a.e. . By one has
for all and a.e. . It follows from (47) and (48) that

From and Lemma 2, we have that
for all , where , . Combining (49), (50), and HÃ¶lderâ€™s inequality, we have
for all large , where . On the other hand, by , there exist and such that
for all and a.e. . By ,
for all and a.e. , where . Combining (52) and (53), one has
for all and a.e. . According to , there exists such that
Thus by (46), (54), and (55), we obtain
for all large . From (56), is bounded. If , by HÃ¶lderâ€™s inequality, we have
Since for all , by (51) and (57), is bounded in . If , by Lemma 2, we obtain
Since , , by (51) and (58), is also bounded in . Hence, is also bounded in . Going if necessary to a subsequence, we can assume that in . As the same the proof of Lemma 15, Therefore, in . Hence satisfies the condition.

Theorem 17. *Assume that , , , , , and are satisfied and the following conditions hold.*() *There exist constants , , such that
*()* is nondecreasing. **Then the problem (1) has at least two critical points.*

*Proof. *Let , for . Then . If one has , and if , we have .*Step 1.*â€‰â€‰ has a local linking at 0 with respect to .

For , using we have
It follows from that
Thus, for with , where is small enough.

For , with , we have since . Thus, From and it follows that
Take . We know that has a local linking at 0 with respect to .*Stepâ€‰â€‰2.*â€‰â€‰ maps bounded sets into bounded sets.

Assume , where is a constant. By , one has
which implies that maps bounded sets into bounded sets. *Step 3.* â€‰â€‰For every , as .

By , for any , there exists a constant such that for all . Since is of finite dimension, there exists such that for all , which implies
Choosing , we have as .

Summing up the above, and by Lemma 16, satisfies all the assumptions of Lemma 11. Hence by Lemma 11, problem (1) has at least one nontrivial solution. The proof of Theorem 17 is completed.

Theorem 18. *Assume that , , , , and are satisfied and the following conditions hold.**â€‰. **â€‰ and nondecreasing. **Then the problem (1) has an infinite number of nontrivial solutions. *

*Proof. *, by () and (), is an even functional and .

First, we verify the condition (2) of Lemma 12.
Integrating (65) for from and , respectively, we have
That is,
Combining (67) and (68), we have
where
On the other hand, by the continuity of , is bounded on , there exists such that
Combining (69) and (71), we have
where .

For arbitrary finite-dimensional subspace , and any , there exists such that
By , (72), (73), and Lemma 2, we have