On a Singular Second-Order Multipoint Boundary Value Problem at Resonance

Abstract

The aim of this paper is to derive existence results for a second-order singular multipoint boundary value problem at resonance using coincidence degree arguments.

1. Introduction

In this paper we derive existence results for the second-order singular multipoint boundary value problem of the form where is Caratheodory’s function (i.e., for each the function is measurable on ; for a.e. , the function is continuous on ). Let , , , , and , where and have singularity at .

In [1] Gupta et al. studied the above equation when and have no singularity and . They obtained existence of a solution by utilising the Leray-Schauder continuation principle. In [2] Ma and O’Regan derived existence results for the same equation when and have a singularity at and . They also utilised the Leray-Schauder continuation method. These results correspond to the nonresonance case. The purpose of this article is therefore to derive existence results for (1) when (the resonance case) and when and have a singularity at . We shall employ coincidence degree arguments in obtaining our results. In this case, the methods used in [1, 2] are not valid.

Research on singular differential equations is important because singular differential equations are useful in the modeling of many problems in the physical and engineering sciences; see [3].

In general singular boundary value problems can be difficult to solve because they may blow up near the singularity. The existence and multiplicity of solutions for second-order nonsingular boundary value problems have been extensively studied by many researchers. However to the best of our knowledge the corresponding problem for second-order differential equations at resonance and with a singularity had not received much attention in the literature. For recent results in these directions see [1, 2, 4–9] and references therein.

The rest of this paper is organised as follows. In Section 2, we present some definitions, lemmas, and theorems necessary for obtaining our main results. In Section 3, we derive some lemmas and the main theorem. In what follows we shall utilise the following assumptions:For , , and .There exist with and , a.e., , . is such that .

2. Preliminaries

In this section we state some definitions, theorems, and lemmas that will be used in the subsequent section.

Definition 1. Let and be real Banach spaces. One says that the linear operator is a Fredholm mapping of index zero if and are of finite dimension, where denotes the image of .

As a result of Definition 1, we will require the continuous projections , such that , , , , and is an isomorphism.

Definition 2. Let be a Fredholm mapping of index zero and a bounded open subset of such that . The map is called -compact on , if the map is bounded and is compact, where one denotes by the generalised inverse of . In addition is -completely continuous if it is -compact on every bounded .

Theorem 3 (see [10]). Let be a Fredholm operator of index zero and let be -compact on . Assume that the following conditions are satisfied: (i) for every .(ii), for every .(iii), with being a continuous projection such that . Then the equation has at least one solution in .

In what follows, we shall make use of the following classical spaces, , , , and . Let denote the space of all absolute continuous functions on , = , = .

.

Let be the Banach space defined by with the norm Let be the Banach space with the normWe denote the norm in by . We define the linear operator bywhere and is defined byThen boundary value problem (1) can be written as

Lemma 4 (see [2]). Let . Then (i).(ii).

Lemma 5. If then (i);(ii);(iii) is a Fredholm operator of index zero and the continuous operator can be defined by where .(iv)The linear operator can be defined as (v) for all .

Proof. (i) It is obvious that(ii) We show that To do this, we consider the problemand we show that (13) has a solution satisfying , if and only ifSuppose (13) has a solution satisfying , ; then we obtain from (13) that and applying the boundary conditions we get since , and using (i) of Lemma 4 we get On the other hand if (14) holds, let ; then , where and . Then from Lemma 4   and = . Hence(iii) For , we define the projection as where .
We show that is well defined and bounded.In addition it is easily verified that We therefore conclude that is a projection. If , then from (14) . Hence . Let ; that is, . Then Thus, and therefore and hence . It follows that since , then . ThereforeThis implies that is Fredholm mapping of index zero.
(iv) We define by and clearly is continuous and linear and and . We now show that the generalised inverse of is given by For we have and for we know that since , , and .
This shows that .
(v) We conclude that

Lemma 6. The operator defined by is -completely continuous.

Proof. Suppose is an open bounded subset of . Let . From condition (A1) and each we have We can deduce from (A1) and (A2) that : This shows that is bounded in and is continuous by using the Lebesgue Dominated Convergence Theorem. Next we show that is compact.
By using (31) we derive This indicates that the sequence is uniformly bounded in . Also for Hence the sequence is bounded in and . Thus is bounded in .
Next we show that the sequence is equicontinuous. Let , ; then for every . By (i) of Lemma 4  . Thus the sequence is equicontinuous on and by Arzela-Ascoli Theorem is convergent. Next we prove that the sequence is also equicontinuous on . We have for Using (i) of Lemma 4 and the fact that and are in we conclude that . Therefore The sequence is therefore equicontinuous on and therefore converges to some with , .
We then conclude that is relatively compact and since is bounded we conclude from Definition 2 that is -compact on every bounded subset of and hence is -completely continuous.

3. Main Result

In this section we will state and prove the main existence results for problem (1).

Theorem 7. Assume that the following conditions are satisfied: There exists a positive constant such that, for each , if for all then There exists a positive constant such that for and either or . Then (1) has at least one solution in provided

To prove Theorem 7, we first establish some lemmas.

Lemma 8. Let then is bounded in .

Proof. Let . We let , . Since it is clear that ; hence for all . Therefore by assumption (H1) there exist such that . Now We note that :From (41) and (42) we get From the definition of we obtain From (43) and (44) we get Since we obtain thatTherefore is bounded in .

Lemma 9. The set is a bounded subset of .

Proof. Let with , . Then implies . We therefore derive from (H2) that

Lemma 10. The sets and are bounded in provided (H2)(i) and (H2)(ii) are satisfied simultaneously.

Proof. If then, for with , , we have If , it follows from (48) that ; that is, , and therefore by Lemma 9. we have . However if and then using assumption (H2)(i) we obtain the contradiction Thus . Hence is bounded in . We can use the same argument to prove that is also bounded in .

Proof of Theorem 7. We show that the conditions of Theorem 3 are satisfied where is an open and bounded set such that . It is easily seen that conditions (i) and (ii) of Theorem 3 are satisfied by using Lemmas 8 and 9. To verify the third condition we set . We choose the isomorphism defined by , . By Lemma 10, we derive that for all . Hence Therefore problem (1) has at least one solution in .

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work is supported by the Covenant University Centre for Research, Innovation and Discovery (CUCRID).