#### Abstract

The authors discuss multiple
solutions for the *n*th-order singular boundary value problems of
nonlinear integrodifferential equations in Banach spaces by means
of the fixed point theorem of cone expansion and compression. An
example for infinite system of scalar third-order singular nonlinear
integrodifferential equations is offered.

#### 1. Introduction

Singular nonlinear boundary value problems of the ordinary differential equations appeared frequently in applications. With Taliaferro [1] treating the general problem, Callegari and Nachman [2] considered existence questions in boundary layer theory, and Luning and Perry [3] obtained constructive results for generalized Emden-Fowler problems. Results have also been obtained for singular boundary value problems arising in reaction-diffusion theory and in non-Newtonian fluid theory [4]. Singular nonlinear boundary value problems of the ordinary differential equations have made great progress in recent years (please see [5–8]).

In the above papers, singular problems are studied in scalar case. In Chen [9], the boundary value problems of a class of th-order nonlinear integrodifferential equations of mixed type in Banach space are considered, and the existence of three solutions is obtained by using the fixed point index theory. But such equations do not have singular nonlinear terms. As much as we know, there are a few papers ([10–18]) to consider the singular problems in abstract Banach spaces. In Liu [10], the following singular problems in Banach spaces were investigated by constructing a special convex closed set and using Mönch fixed point theorem, where denotes the zero element of . In [10], (1) under certain conditions, there is at least one solution. And, in the methods, under normal circumstances, to investigate the singular problems, at first, one needs to consider the approximation problems which have no singularities. However, in the study of integrodifferential equations in infinite dimensional Banach space, this method is very complicated and difficult.

In this paper, not considering approximative problems, informed by the characteristic of nonlinear term, we construct a new cone, and through the cone we create a new special cone. Moreover, through finding the relations from to ( belongs to the special cone), we triumphantly overcome the singularity and use the fixed point theorem of cone expansion and compression directly to obtain the existence of multiple solutions for singular boundary value problems of nonlinear integrodifferential equations in Banach spaces. Finally, an example of scalar third-order singular nonlinear integrodifferential equations for an infinite system is offered. With the previous methods, one can not get the results in this paper.

Let be a cone in Banach space which defines a partial ordering in by if and only if . Let . is said to be normal if there exists a positive constant such that implies , where denotes the zero element of , and the smallest is called the normal constant of . For convenience, in the following, we set as a normal cone and . Let , in which and . Obviously, is a normal cone of , and the normal constant of also is 1. Cone is the key to overcome the singular nonlinear term (please see the last example).

We consider the following singular boundary value problem (SBVP for short) for an th-order nonlinear integrodifferential equations in Banach spaces : where , , with ( denotes the set of all nonnegative real numbers).

is singular at , , andor if , ,

Let . A map is called a solution of SBVP (2) if it satisfies (2).

#### 2. Preliminaries and Several Lemmas

Denote is a map from into and is continuous on . The norm of is defined by whereObviously is a Banach space.

Let Obviously, is a cone of . For , we write and .

Let be continuous. We call the abstract generalized integral convergence if exists. Analogously, we can define the convergence of other kinds of abstract generalized integrals.

We will use to denote the Kuratowski measure of noncompactness of set in space . For details of the Kuratowski measure of noncompactness, please see [19].

Lemma 1 (see [19]). *Let be a bounded set of Suppose that is equicontinuous. Then,where *

Lemma 2 (see [19]). *If is bounded and equicontinuous, then is continuous on . Moreover, *

Lemma 3 (see [19]). *Let be a bounded set of Then, where is defined by Lemma 1.*

Lemma 4 (the fixed point theorem of cone expansion and compression [see [20]]). * is a cone of real Banach space . Let Suppose that is a strict set contraction such that one of the following two conditions is satisfied: *(i)*, and , ;*(ii)*, and , .**Then, has at least a fixed point in .*

#### 3. Main Results and an Example

To continue, let us formulate some conditions.

There exist , , , and , , such thatwhere is nonincreasing and and are nondecreasing.

For any , And, there exists a such that where , , , and are defined as in condition , and

For any , , is uniformly continuous on and there exist , such that

*Remark 5. *Obviously, condition is satisfied automatically when is finite dimensional.

There exist , and denotes the dual cone of such that for . At the same time, one of the conditions is satisfied uniformly in , with .

*Remark 6. *Because is singular at , condition is easy to be satisfied. And only one of the conditions is satisfied.

There exist , and denotes the dual cone of such that for . At the same time, one of the following conditions is satisfied: uniformly in , with .

*Remark 7. *In condition , only one of the conditions is satisfied.

To avoid singularity, let Obviously, is a normal cone in , and the normal constant of is 1.

Lemma 8. *Suppose . Then, *

*Proof. *For any , that is, , Therefore, which implies that Because of , and the normal characters of , it is easy to get Hence,It follows from (28) and (29) and the normal characters of that By (27) and (30), the conclusion holds.

*Remark 9. *Formula (23) implies that the norm of is decided by th-order derivative .

*Remark 10. *Inequality (24) implies that controls distance between and . This is one of the keys to apart from the singularities of the nonlinear term .

Lemma 11. *For , the following conclusion holds: where *

*Proof. *In fact, for , , we get ConsiderIt follows from (33) and (34) that that is,On the other hand, for , it is easy to get It follows from (36) and (37) that Since for (38) holds, we get (31).

Lemma 12. *Suppose conditions and are satisfied. Then, for any , , in which the operator is defined bywhere *

*Proof. *At first, we show that the operator defined by (39) is reasonable for with any . In fact, for with , by and , which implies that defined by (39) is reasonable for .

Next, we show that . For ,which implies and . This together withgives Therefore, holds.

Finally, by Lemma 11 and (39) and (40), one can see It follows from (44) and (45) that .

Lemma 13. *Let cone be normal and let conditions and be satisfied. Then, is a fixed point of operator if and only if is a solution for SBVP (2).*

*Proof. *By Lemma 12, . For , Taylor’s formula with the integral remainder term gives Substituting into (46), we get Let be the solution of SBVP (2). Then, (48) implies Comparing this with (39) and (40), we have , which means is the fixed point of the operator in .

On the other hand, let be the fixed point of the operator . By (39) and (40), where . It follows by taking and in (50) that that is,Then, (51)-(52) imply that is the solution for SBVP (2) in .

Lemma 14. *Suppose conditions are satisfied. Letwith , . Then,with .*

*Proof. *Apart from the singularities, let By conditions and , for any , , one can see thatBy virtue of absolute continuity of the Lebesgue integrable function and (56), it is easy to see that in which denotes the Hausdorff distance between and . Therefore, Now, we check that For , it is easy to see that is bounded, which implies that are bounded. Sincewe havewhere , , , and . Take , , such that Obviously, and ; moreover, both of them are equacontinuous on . It follows from (62), , and Lemmas 1 and 2 that Analogously, it is easy to get It follows from , (61), (63), and (64) that Hence, by (58), we know (59) is true, and the conclusion holds.

Lemma 15. *Let conditions , , and be satisfied. Suppose that , in which . Then, is a strict set contraction from into .*

*Proof. *By Lemma 12, , and , it is easy to see that and is a bounded operator. We check that is continuous. In fact, let , . For , it is easy to get For , by (31) and (50),