Chinese Journal of Mathematics

Volume 2014, Article ID 379761, 6 pages

http://dx.doi.org/10.1155/2014/379761

## Nontrivial Solutions for a Boundary Value Problem of th-Order Impulsive Differential Equation

^{1}School of Mathematics, Shandong University, Jinan, Shandong 250100, China^{2}Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong 250101, China

Received 3 October 2013; Accepted 28 November 2013; Published 23 January 2014

Academic Editors: H. Lin and D.-B. Wang

Copyright © 2014 Jiafa Xu and Zhongli Wei. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We study the existence of nontrivial solutions for *n*th-order boundary value problem with impulsive effects. We utilize Leray-Schauder degree theory to establish our main results. Furthermore, our nonlinear term *f* is allowed to grow superlinearly and sublinearly.

#### 1. Introduction

In this paper, we consider the existence of nontrivial solutions for the following th-order boundary value problem involving impulsive effects: where , , and denote the Riemann-Stieltjes integral, and and are right continuous on , left continuous at , and nondecreasing on , with .

The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Therefore, many research workers pay their attention to boundary value problems for impulsive differential equations; for example, see [1–10] and the references therein.

In [1], D. Guo utilized fixed point theory to establish some existence theorems of positive solutions for an infinite boundary value problem of th-order nonlinear impulsive singular integrodifferential equations on the half-line in Banach spaces.

In [2, 3], he dealt with multiple positive solutions for the following infinite boundary value problem of first order impulsive superlinear integrodifferential equations on the half-line: In [2], the function is continuous, that is, no singularity, but in [3] the author discussed the singular case, that is, as and . Compare with [2], the key point of [3] is apart from the singular point by some appropriate functions.

Motivated by the works cited above, in particular [1–3], in this paper, we utilize the methods in [11] to study the existence of nontrivial solutions for (1). Nevertheless, our work here improves and extends the corresponding ones in [11]. We first note the impulsive effect as a perturbation to the corresponding problem of (1) without impulse, so we can construct an integral operator for the corresponding boundary value problem without the impulsive terms and find out its first eigenvalue and eigenfunction. Then we establish a special cone associated with the Green function of (1). Finally, by employing Leray-Schauder degree theory, two existence theorems of nontrivial solutions for (1) are obtained.

#### 2. Preliminaries

We first offer some related preliminaries used in the ensuing section. In this work, we always assume that the following two conditions are satisfied:(H1), ,(H2), , where For any , the boundary value problem if and only if can be expressed by , where

Lemma 1 (see [12], Lemma 2.1). * has the following properties:*(i)*, , where ,*(ii)*, , where .**Let , and . Then is a real Banach space and is a cone on . Set , and introduce the following space:
**
with the norm . Clearly, is also a real Banach space.*

Lemma 2 (see [9], Lemma 2.1). *Let (H1) and (H2) hold. Then (1) is equivalent to the following integral equation:
**
where
**Let and
**
Here, is determined by Lemma 1. Clearly, is nonnegative and continuous on .*

*Lemma 3. One has , .*

* Proof . *By Lemma 1 and (8), we obtain , and
Combining these, we can find , . This completes the proof.

*Let
Clearly, is a completely continuous nonlinear operator, and the existence of the solutions of (1) is equivalent to that of fixed points of ; is a completely continuous linear operator, satisfying . Namely, is a positive, completely continuous, linear operator. By the positivity of , , the spectral radius of , denoted by , is positive. The Krein-Rutman theorem [13] then asserts that there are and such that
Denote and .*

*Lemma 4. One has .*

*Proof. *Suppose that , then
for all and thus . This completes the proof.

*Lemma 5 (see [14]). Let be a real Banach space and a bounded open set with . Suppose that is a completely continuous operator. If there is such that , , , then , where stands for the Leray-Schauder topological degree in .*

*3. Main Results*

*Let . We now list our assumptions on .(H3) and uniformly in .(H4)There is a positive number such that
(H5) and uniformly in .(H6)For in (H4), assume that
*

*Two Examples.* Let , , for all . Then , , , , and (H2) holds true.(1)Let , where . Then satisfies (H1). Moreover, we obtain
So, satisfies (H3) and (H4). Also, let . Then satisfies (H1) and
Hence, satisfies (H4). Consequently, (H1), (H3), and (H4) hold true.(2)Let , , where . Then
Moreover,
So, (H1), (H5), (H6) hold.

*Theorem 6. Assume that (H1)–(H4) hold, and (1) has at least one nontrivial solution.*

*Proof. *(H3) implies that there are and such that , and , . Note that if , , and if , . That is, we have
Let , where is given by (12). We claim that is bounded in . Indeed, if , there is such that
Combining this with (20) yields
where **1** stands for the constant function . Multiply (23) by on both sides and integrate over and use (12) to obtain , and thus . Note that (21) holds. Define the Nemytskii operator by . Now (22) is equivalent to
Lemma 4 implies that . Therefore,
Since , the operator has the bounded inverse operator . Therefore, there is such that , . This proves the boundedness of . For each , we have , , , which leads to

On the other hand, (H4) implies that there is a such that , , , . Notice we may choose so that . Thus we have
for all and . We claim that
Suppose the contrary. Then there are and such that . Let . Then and
Multiply (29) by on both sides and integrate over and use (12) to obtain
and then by Lemma 4,
That is, , which contracts to . Consequently, , and thus . This contradicts to . As a result of this, (28) holds true. Hence and are homotopic on . The homotopy invariance of topological degree implies that . This and (26) together imply that . Therefore the operator has at least one point in . Equivalently, (1) has at least one nontrivial solution. This completes the proof.

*Theorem 7. Assume that (H1), (H2), (H5), and (H6) hold; (1) has at least one nontrivial solution.*

*Proof. *(H5) implies there are and such that , , , and , , . Note that if , , and if , . That is, we have
Nowadays, we claim
where is given by (12). Indeed, if the claim is false, then there are and such that
Combining this with (32) yields
Multiply (36) by and integrate over and use (12) to obtain , and thus . Note that (33) holds, and then we find
Lemma 4 implies that . Therefore, . Notice that . We have , contradicting . As a result of this, (34) is true. Invoking Lemma 5 gives

On the other hand, (H6) implies that there is such that

Let . We shall now show that is bounded in . Indeed, if , then for some . In view of (39), let , we have and
Multiply (40) by on both sides and integrate over and use (12) to obtain
and then by Lemma 4,
Hence, . This proves the boundedness of . Choosing and , we have , , . The homotopy invariance of Leray-Schauder topological degree implies that . Combining this with (38) we get . Therefore the operator has at least one point in . Equivalently, (1) has at least one nontrivial solution. This completes the proof.

*Conflict of Interests*

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

*Acknowledgments*

*The authors are supported by the NNSF-China (11371117), Shandong Provincial Natural Science Foundation (ZR2013AM009), GIIFSDU (yzc12063), and IIFSDU (2012TS020).*

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