#### Abstract

This paper is devoted to investigating the eigenvalue problems of a class of nonlinear impulsive singular boundary value problem in Banach spaces: ; where denotes the zero element of Banach space, , , , is a parameter, and may be singular at and . The arguments are mainly based upon the theory of fixed point index, measure of noncompactness, and the special cone, which is constructed to overcome the singularity.

#### 1. Introduction

Consider the following eigenvalue problems of singular boundary value problem (SBVP) with impulse in Banach space : where denotes the zero element of Banach space , , , , is a parameter, may be singular at and .

During the last few decades, the existence of positive solution of nonlinear singular boundary value problems has gained considerable popularity (see [1â€“12] and references therein). In recent years, there were also a lot of papers which dealt with eigenvalue problems (see [13â€“27]). Some of them considered singular case (see, for instance, [13â€“16, 18, 19], etc.).

On the other hand, as we know, the theory of impulsive differential equations has found its extensive applications in realistic mathematical modeling of a wide variety of practical situations such as physics, chemical technology, population dynamics, biotechnology, and economics. It has emerged as an important area of investigation in recent years (see [28â€“36] and references therein).

To the best of our knowledge, there is no paper studying the eigenvalue problems of the impulsive singular boundary value problem in Banach spaces. The main purpose of this paper is to fill this gap. By using the theory of fixed point index, measure of noncompactness, and the special cone which is constructed to overcome the singularity, we investigate the existence of eigenvalues of (1).

The main features of the present paper are as follows. By virtue of a special transformation, we first convert (1) into another solvable form such that the associated operator can be used to overcome the influence of impulse and parameter . Then a special cone is constructed to deal with the singularity of (1).

This paper is organized as follows. In Section 2, we provide some basic definitions, preliminaries facts, and lemmas. Meanwhile, some transformations are introduced to convert (1) into another solvable form. In Section 3, the main results are presented and proved. Finally, an example is worked out to demonstrate the application of the main result.

#### 2. Preliminaries and Conversion of (1)

Let be a normal solid cone of real Banach space . Without loss of generality, suppose the normal constant is 1. Let denote the dual cone of , , and , . Denoted by the Banach space of all continuous functions with norm .

We define is a map from into such that is continuous at and left continuous at and its right limit at denoted by ) exists for , and is continuous at and left continuous at and the right limit at denoted by ) exists for .

Let , and *âˆ¶*. It is well known that is a Banach space with the norm =, .

Evidently, and are Banach spaces with norm and , , respectively. Furthermore, is a closed subspace of .

For each , let ; it is easy to see . Conversely, for each , let ; then .

We convert (1) into another solvable form first.

Lemma 1. *If is a solution of impulsive SBVP (1), then satisfies SBVP:
**
Conversely, if is a solution of SBVP (2), then is a solution of impulsive SBVP (1).*

Lemma 2. * is a solution of SBVP (2) if and only if is a solution of the integral equation:
**
where
*

By Lemmas 1 and 2, we can obtain the following.

Lemma 3. * is a solution of impulsive SBVP (1) if and only if is a solution of the integral equation:
**
where
*

Let and donated the Kuratowski noncompactness measure of bounded sets in and , respectively. If nontrivial function satisfies problem (5) for some , then is called an eigenvalue and is called an eigenfunction of problem (1) corresponding to the eigenvalue .

It is well known that the following conclusions hold.

Lemma 4 (see [17]). *Let be a cone in Banach space , , and . Let operator be a strict set contraction. If and for , then .*

Lemma 5 (see [17]). * is relatively compact if and only if is equicontinuous and for any is a relatively compact set in .*

#### 3. Main Results

For convenience, let

For the forthcoming analysis, we list the following assumptions:, where satisfy where .There exists a constant such that There exist such that where .There exist and such that for and uniformly for , and .There exist and such that for and uniformly for , and .

Define It is easy to check that is a cone in space and . Let . As in [2], we can prove that where is defined in Lemma 2. So for all , we have

To solve eigenvalues of impulsive SBVP (1), we first consider operator associated with (5) and defined by For , from the definition of , we have By , we know So the operator is well defined in .

For the sake of overcoming the singularity, choose with and consider the approximate problem of (17): For any , Hence .

Lemma 6. *Let conditions and be satisfied; then for any , the operator is a strict set contraction from into .*

*Proof. *Obviously . Now we prove that is continuous. Let as ; we have
So and the dominated convergence theorem imply that
We now show that
In fact, if (24) is not true, then there exist a positive number and a sequence such that
Since is relatively compact, there is a subsequence of which converges to some . Without loss of generality, we may assume that itself converges to ; that is,
By virtue of (23) and (26), we have , and so (26) contradicts with (25). Hence (24) holds, and the continuity of is proved.

By , it is easy to see that is bounded from into .

Now we will prove that the operator is a strict set contraction from into . Let
By , as , we have
So for any bounded set , we have
Hence
where denotes the Hausdorff metric, which implies
For any , by , we have
It is easy to show that . Therefore, we can have . Consequently, the operator is a strict set contraction from into . The proof is thus completed.

Theorem 7. *Let conditions , and be satisfied; then there exists a positive number such that for any given , the impulsive SBVP (1) has a nontrivial eigenfunction with corresponding to eigenvalue .*

*Proof. *By Lemma 6, the operator is a strict set contraction from into . Observing , if , we can choose a with
and there is a such that
Choose and sufficiently large such that ; we have
So
By virtue of Lemma 4, we have . Since , it follows from the homotopy invariance of fixed point index for strict set contraction that there exist and such that
where .

Let . Obviously, is uniformly bounded. We shall show that is equicontinuous. By virtue of boundary condition, we need to consider only the following eight cases.*Case 1. *, the boundary value condition is . So we need only to show that uniformly converge to with respect to as , and is equicontinuous on any closed subinterval of .

By , let
Using the absolute continuity of integration, for any , there exists a , such that
as . Since
we have
To prove that uniformly converge to with respect to as , we need only to show
Notice that
This together with (39) implies that (42) holds. For in (39), choose
Then for , (39) implies that
Hence (43) holds. Very similarly, we can obtain that uniformly converge to with respect to as . Now, we show that is equicontinuous on for any . Notice that

So for any , we have
This together with (39) implies that is equicontinuous on .*Case 2. *, the boundary value condition is . We need to show that uniformly converge to with respect to as . This can be obtained by the similar way in Case 1, so we omit it. Next, we show is equicontinuous on for any . In fact, for any , we have
This together with (39) implies that is equicontinuous on .*Case 3. *, the boundary value condition is . The equicontinuity of on of this case can be obtained by the same way of Case 2, so we omit it.*Case 4. *, the boundary value condition is . Similarly, by Case 1, we can get that uniformly converge to with respect to as . For any , we need to show that is equicontinuous on . In fact, for any , we have
This together with (39) implies that is equicontinuous on .*Case 5. *, the boundary value condition is