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

Abstract

We study the existence of nontrivial solutions for nth-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).