Abstract and Applied Analysis

Volume 2014, Article ID 968792, 6 pages

http://dx.doi.org/10.1155/2014/968792

## Parameter Dependence of Positive Solutions for Second-Order Singular Neumann Boundary Value Problems with Impulsive Effects

Department of Mathematics and Physics, North China Electric Power University, Beijing 102206, China

Received 21 March 2014; Accepted 7 May 2014; Published 19 May 2014

Academic Editor: Juan J. Nieto

Copyright © 2014 Xuemei Zhang. 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

The author considers the Neumann boundary value problem = and establishes the dependence results of the solution on the parameter , which cover equations without impulsive effects and are compared with some recent results by Nieto and O’Regan.

#### 1. Introduction

Impulsive effects exist widely in many evolution processes in which their states are changed abruptly at certain moments of time. The theory and applications of impulsive differential equations have been emerging as an important area of investigation in recent years [1–6]. There is a vast literature on the existence of solutions by using different methods such as bifurcation theory [7, 8], fixed point theorems in cones [9–12], the method of lower and upper solutions [13, 14], and the theory of critical point theory and variational methods [15–19]. We remark that on second-order impulsive differential equations with parameter only a few results have been obtained; see, for instance, [9, 20–22]. To the best of our knowledge, these papers only consider the existence of positive solutions. However, the corresponding results for the dependence of the solution on the parameter for second-order impulsive differential equations are not investigated until now. In this paper, we try to solve this kind of problem.

Consider the Neumann boundary value problems
where is a constant, is a parameter, , is a nonnegative measurable function on , on any open subinterval in , which may be singular at and/or , (where is fixed positive integer) are fixed points with , and , where and represent the right-hand limit and left-hand limit of at , respectively. In addition, , , and satisfy(H_{1});(H_{2}), , where , .

Some special cases of (1) have been investigated. For example, Nieto and O’Regan [17] studied problem (1) with and for . By using variational methods and critical point theory, the authors proved the existence of solutions of problem (1).

For ease of exposition, we set

Our main results are as follows.

Theorem 1. *Assume that and hold. Then the following two conclusions hold:*(H_{3})*if , , and , , , then for every problem (1) has a positive solution satisfying ;*(H_{4})*if , , and , , , then for every problem (1) has a positive solution satisfying .*

*Remark 2. *Assume that and hold. Furthermore, suppose that or , , in or or , , in . Then the conclusions of Theorem 1 also hold.

*Remark 3. *It follows from the conditions of Theorem 1 that we develop some ideas of Guo and Lakshmikantham essentially; for detail, see Theorem in [23].

*Remark 4. *For simplicity we only consider Neumann boundary conditions since all the results obtained in this paper can also be adapted with minor changes to the other boundary conditions.

#### 2. Preliminaries

Let , , and Then is a real Banach space with norm where , .

A function is called a solution of problem (1) if it satisfies (1).

In our main results, we will make use of the following lemmas.

Lemma 5. *If and hold, then problem (1) has a unique solution given by
**
where
*

*Proof. *The proof is similar to that of Lemma 2.4 in [24].

By (6), we can prove that has the following property:

Define a cone in by where It is easy to see is a closed convex cone of .

Define an operator by

From (10), we know that is a solution of problem (1) if and only if is a fixed point of operator .

Lemma 6. *Suppose that and hold. Then and is completely continuous.*

*Proof. *For , it follows from (7) and (10) that
It follows from (7), (10), and (11) that
Thus, .

Next, by similar arguments of Lemmas 5 and 6 [12] one can prove that is completely continuous. So it is omitted, and the theorem is proved.

To obtain positive solutions of problem (1), the following fixed point theorem in cones is fundamental, which can be found in [23, page 94].

Lemma 7. *Let be a cone in a real Banach space . Assume , are bounded open sets in with , . If
**
is completely continuous such that either *(a)*, , and , , or*(b)*, , and , ,**then has at least one fixed point in .*

#### 3. Proofs of the Main Results

For convenience we introduce some notations where , , and is a constant.

*Proof of Theorem 1. *We need only prove this theorem under condition since the proof is similar when holds, provided the proper adjustments are made.

If , , then there exist and such that
where satisfies
Then for we have
where
It follows from (17) that
If , , then there exist and such that
where satisfies

Let . Thus, when we have
and then we get
This yields

Hence, for given condition of Lemma 7 is satisfied of operator , which implies that has a fixed point in .

It remains to prove as . In fact, if not, there would exist a number and a sequence such that
Furthermore, the sequence contains a subsequence that converges into a number , where . For simplicity, suppose that itself converges into .

If , then for sufficiently large (), and therefore
which contradicts .

If , then for sufficiently large (), and therefore it follows from that for any there exists such that
and hence it follows from (10) that
Since is arbitrary, we have in contradiction to . Therefore, as and the proof is complete.

*Remark 8. *Comparing with Nieto and O’Regan [17], the main features of this paper are as follows.(i)The parameter is considered.(ii)The parameter dependence of the solution is available.(iii), not for .

#### 4. An Example

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

*Example 9. *Consider the following boundary value problems
Evidently, is the trivial solution of problem (29).

*Conclusion*. Problem (29) has at least one positive solution for any .

*Proof. *Problem (29) can be regarded as a problem of the form (1), where

It follows from the definition of , , and that and hold, and is singular at and . By calculating, we have
Then, the condition of Theorem 1 holds. Hence, by Theorem 1, the conclusion follows, and the proof is complete.

#### Conflict of Interests

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

#### Acknowledgments

The authors are indebted to the referee’s suggestions. These have greatly improved this paper. This work is sponsored by the Project NSFC (11301178) and the Fundamental Research Funds for the Central Universities (2014MS58).

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