Journal of Function Spaces and Applications

Volume 2013 (2013), Article ID 567970, 6 pages

http://dx.doi.org/10.1155/2013/567970

Research Article

Positive Solutions of a Singular Third-Order -Point Boundary Value Problem

Shaolin Zhou and Xiaoling Han

College of Mathematics and Information Science, Northwest Normal University, Lanzhou 730070, China

Received 4 December 2012; Accepted 16 January 2013

Academic Editor: To Ma

Copyright © 2013 Shaolin Zhou and Xiaoling Han. 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.

This paper is concerned with the existence and nonexistence of positive solutions to the singular third-order -point boundary value problem ,  , where    are constants, is a parameter, is continuous and is allowed to be singular at and . The results here essentially extend and improve some known results.

1. Introduction and the Main Results

Singular boundary value problems for nonlinear ordinary differential equations arise in a variety of areas of applied mathematics, physics, chemistry, and so on. For earlier works, see [1–3]. Nonsingular third-order multipoint boundary value problems have been studied by many authors by using different type of techniques, see, for example, [4–9] and the references therein. In recent years, singular third-order multipoint boundary value problems have also received much attention, see [10–12].

Very recently, motivated by Ma [13], Sun [14] considered the third-order three-point boundary value problem

where ,   are constants and is a parameter. Under the following assumptions:

and ;; is superlinear, that is, ,  ; is sublinear, that is, ,  ,

where ,  .

By using Guo-Krasnosel’skii fixed point theorem, the author established the following results.

Theorem A (see [14, Theorem 3.1]). Suppose that (), (), and () hold. Then the problem has at least one positive solution for small enough and has no positive solution for large enough.

Theorem B (see [14, Theorem 3.2]). Suppose that (), (), and () hold. If nondecreasing, then there exists a positive constant such that the problem has at least one positive solution for and has no positive solution for .

Theorem C (see [14, Theorem 3.3]). Suppose that (), (), and () hold. Then the problem has at least one positive solution for any .

Being directly inspired by the previously mentioned works, we will consider the existence and nonexistence of positive solutions to the following third-order -point BVP:

where ,   are constants and is a parameter, is allowed to be singular at and . Here, the solution of BVP of (1), is called positive solution if is positive on and satisfies (1) and the boundary conditions .

We assume that (), () hold and make the following additional assumptions:

and ; and ,

where

From , we know that there exists such that . Let satisfy .

Our main results are the following.

Theorem 1. Let ()–() hold. Then there exists a positive number such that BVP of (1), has at least one positive solution for and none for .

Theorem 2. Let ()–() and () hold. Then BVP of (1), has at least one positive solution for any .

The proof of previous theorems is based on the Schauder fixed-point theorem.

Remark 3. BVP is a special case of (1), with ,  , and ,  .

Remark 4. allows but do not require the nonlinearity to be sublinear at zero and infinity; allows but do not require the nonlinearity to be sublinear at zero and infinity.

Remark 5. We do not assume any monotonicity condition on the nonlinearity as in [14]. We find that the nondecreasing condition of can be removed from Theorem 3.2 in [14], and the same result is obtained in Theorem 1.

Remark 6. It is obvious that Theorem 1 is an extension and complement of Theorems 3.1 and 3.2; furthermore, Theorem 2 is also an extension of Theorem 3.3 in [14].

2. Preliminary Lemmas

In this section, we present some notation and preliminary lemmas.

Let equipped with the norm .

Lemma 7 (see [15, Lemma 2.1]). Suppose that .

(i) Then ,   and

(ii) Let . Then

for every with , and

for every with .

Lemma 8. Suppose that ()–() hold, then BVP

has a unique nonnegative solution which can be represented as

where

and

Proof. The proof of the uniqueness is standard and hence is omitted here. Now we prove the existence of the solution.

From ()–() and Lemma 7, we conclude that the integration in (7) is well defined. Let ; then BVP (6) may be reduced to boundary value problems

We claim that (9), (10) have a nonnegative solution which can be represented as

In fact, from ()–() and Lemma 7, for each , and . Combining the continuity of and , we have

Thus . Moreover

Similarity, . From (15), we get ,  .

By Lemma 7, we have from (13) that

Again applying (13), we have

This together with (13) implies that . The claim is proved.

By Lemma 7, we obtain from (11), (12), and (13) that

It is easy to see that , and moreover, is a nonnegative solution of the BVP (6).

The proof is complete.

Lemma 9 (see [14, Lemmas 2.2 and 2.3]). For any , one has

(i), where ,(ii), and .

Lemma 10. Suppose that ()–() hold; then the unique nonnegative solution of (6) satisfies

The proof is similar to Lemma 2.4 in [14].

Lemma 11. Suppose that ()–() hold. Let and . Then BVP

has a unique solution satisfying on .

Proof. From (20) and (21), we obtain . Again applying (21), we have

Now set ; then we have that

This together with the fact that implies that . Thus on . The proof is complete.

3. Proof of the Main Results

In this section, we will prove our main results.

Proof of Theorem 1. We divide the proof into three steps.

Step 1. We first prove the existence of positive solutions to (1), for sufficiently small .

Let be the unique solution of

Then . Let ; then is a positive solution of BVP (1), if and only if is a nonnegative solution of BVP

Let . Since ; then there exists a positive number such that

Define a closed convex subset in by

and an operator by

where Modeling the proof of Lemma 2.3 in [10], we can show that is a completely continuous operator. From Lemma 8, we know that is a nonnegative solution of (25) if and only if is a fixed point of .

Suppose that ; we claim that .

In fact, from Lemma 9 and (28), we have

The claim is proved. Using the Schauder fixed point theorem, we conclude that has a fixed point in , and then is a positive solution of (1), .

Step 2. We verify that BVP of (1), has no positive solutions for large enough.

Suppose to the contrary that BVP of (1), has at least one positive solution for any . Then there exist , with , such that for any positive integer , BVP of (1), has a positive solution . Thus is a nonnegative solution to (25). On the one hand, we have

On the other hand, since , there exists such that , for any . Let be large enough that . By Lemma 10, we have

This implies that

Thus

which is a contradiction.

Step 3. Let   BVP of (1), has at least one positive solution} and ; then . We show that (1), have positive solution for any . From the definition of , we know that, for any , there exists such that (1), have positive solution .

Now we consider the following third-order -point boundary value problem:

where

Since is bounded, by Schauder fixed point theorem, the problem (34) has a solution .

By Lemma 8, satisfies

thus .

Let and , where ,  . We will show that . Noticing that, if holds for any , combining with , we get . Thus we prove that ; then we have .

Suppose to the contrary that .

If , then, from , and the continuity of , there exists such that . Moreover, in . Thus in . This contradicts with the fact that in .

If , we claim that there exists such that .

In fact, from the fact that and , we have that Thus, there exists such that .

Let such that ; then, we only need to deal with the following four cases.

Case 1. in . In this case, we have

We easily verify that

This contradicts with the fact that .

Case 2. There exists such that and in . In this case, we have

We easily obtain in , a contradiction again.

Case 3. There exists such that and in . In this case, if , then, for any , we have . Thus

If , then and satisfies

By Lemma 11, we also have in , a contradiction again.

Case 4. There exists such that and in . The same as Case 2, we can lead to a contradiction.

Summarizing the previous discussion, we assert that ; thus . Up to now, the problem (1), has a solution .

Proof of Theorem 2. Since , there exists , and such that ,  . Next we consider two cases: is bounded or is unbounded.

Case 1. Suppose that is bounded, that is, , for all . By Schauder fixed point theorem the problem of (1), has a positive solution.

Case 2. If is unbounded. Since , there exists a positive number such that , for . Since is unbounded, for any , we are able to choose

such that

Defining a closed convex subset in by

For each , we have . By (28) and Lemma 9, we obtain that

That is, . By using the Schauder fixed point theorem, we assert that has a fixed point , and then is a positive solution of BVP of (1), .

4. Example

Consider the boundary value problem

where ,  ,  ,  , and ,  . Obviously , and , and hold. By calculating, we have ,  ,  ,  . Let , then . We easily verify that

that is, is satisfied. Therefore, Theorem 1 now guarantees that there exists a positive number such that BVP of (46), has at least one positive solution for and none for .

But we cannot apply Theorem B [14, Theorem 3.2]. In fact, does not satisfy monotonicity condition. Moreover, condition of Theorem B does not hold.

Acknowledgments

The authors are grateful to the anonymous referee for his or her constructive comments and suggestions which led to improvement of the original paper. X. Han is supported by the NNSF of China (no.11101335).

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