Abstract and Applied Analysis

Volume 2014 (2014), Article ID 793639, 7 pages

http://dx.doi.org/10.1155/2014/793639

Research Article

Solvability of Third-Order Three-Point Boundary Value Problems

Dongyuan Liu and Zigen Ouyang

School of Mathematics and Physics, University of South China, Hengyang 421001, China

Received 28 October 2013; Accepted 9 April 2014; Published 29 April 2014

Academic Editor: Patricia J. Y. Wong

Copyright © 2014 Dongyuan Liu and Zigen Ouyang. 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.

We are interested in the existence theorems for a third-order three-point boundary value problem. In the nonresonant case, using the Krasnosel’skii fixed point theorem, we obtain some sufficient conditions for the existence of the positive solutions. In addition, we focus on the resonant case, the boundary value problem being transformed into an integral equation with an undetermined parameter, and the existence conditions being obtained by the Intermediate Value Theorem.

1. Introduction

Consider the following third-order nonlinear differential equations:

subject to the following boundary value conditions:

where , , and .

Recently, the existence of solutions for boundary value problem has been investigated by many authors [1–9]. Further, many authors focused on the existence of solutions or positive solutions for higher order differential equations with boundary value problems [10–16].

Sun [15] has investigated the following three-order three-point boundary value problem:

with

where , is a positive parameter, , and . Using the Krasnosel’skii fixed point theorem, they obtained some existence conditions for the positive solutions of the problem (3)-(4).

Usually, by constructing Green’s function , the authors transform the problem (1)-(2) () into the following integral equation:

Therefore, some fixed point theorems can be used to prove the existence of the positive solutions of the problem (1)-(2).

However, if we consider the problem (1)-(2) when , the respective integral equation of the problem (1)-(2) has not the form of (5). Thus, we cannot prove the existence of solutions of (1)-(2) only by fixed point theorems.

Despite the success in the study of (1)-(2) and (3)-(4), it has been recognized that for the resonant cases, that is, (1)-(2) with , no much work has been known except recent work [12, 17, 18] and the references therein where the so-called coincidence-degree method was employed.

Lu and Ge [12] considered the following higher-order -point boundary value problem:

with

where and . Using the coincidence-degree method, they obtained a sufficient condition for the existence of solutions for the problem (6)-(7).

More recently, Ouyang and Li [19] have discussed a class of fractional order differential equations of the following three-point boundary value problem with resonance:

Using a new method, they obtained some sufficient conditions for the existence of solutions for the fractional order boundary value problem (8).

The purpose of this paper is to study the problem (1)-(2) for the nonresonant case and the resonant case. In the nonresonant case, the Krasnosel’skii fixed point theorem is used to prove the existence of positive solutions for the problem (1)-(2). In the resonant case, a completely new method is incorporated; we transform the problem into an integral equation with an undetermined parameter. The Intermediate Value Theorem is applied to determine the particular value of the parameter so that true solutions exist. Not only the existence conditions of the solutions, but also the prove of the main results are more simple than [12].

We introduce two lemmas as follows.

Lemma 1 (the Krasnosel’skii fixed point theorem [20]). Let be a cone in a Banach space . Assume that and are open subsets of with and . Suppose that is a completely continuous operator so that either

(i) for and for

or

(ii) for and for . Then has a fixed point .

Lemma 2 (see [21]). Let be a closed and convex subset of a Banach space . Assume that is a relatively open subset of with and is completely continuous. Then at least one of the following two properties holds:

(i) has a fixed point in ; (ii) there exist and with .

The paper is arranged as follows. In Section 2, we discuss the existence of the positive solutions of the problem (1)-(2) in nonresonant case. Section 3 is devoted to the existence of the solutions of the problem (1)-(2) in resonant case. Finally, we give some examples to illustrate our results.

For convenience, we set

To present our result, we assume that

 (H₁), , , and , , or , ; (H₂); there exist nonnegative functions so that

where ; (H₃)for any , as , and as .

2. The Nonresonant Case ( )

Throughout this paper, we suppose that is a space of continuous functions in , equipped with the norm

In this section, we consider the nonresonant case, that is, (1) with the boundary value problem (2) with . We have the following theorem.

Theorem 3. Assume that holds and . Then problem (1)-(2) has at least one positive solution.

Proof of Theorem 3. Suppose that is a solution of the problem (1)-(2). Integrating both sides of (1) three times from to , one gets

Imposing the first two boundary conditions in (2), we have

Imposing the last boundary condition in (2), we obtain

Substituting (13) and (14) into (12), the problem (1)-(2) is equivalent to the following integral equation:

where is defined by

When , the function is decreasing with respect to , and is independent on the parameter when . So Green’s function satisfies that . Further, for any , Green’s function satisfies

and the solution of (1)-(2) is nonnegative and satisfies

for any and .

In fact, since , from () and (15), . Noticing that

thus, is decreasing; it follows from the boundary value condition that

Hence is concave; that is, for any and ,

Noticing that , then , and is nonincreasing; thus is the maximum point of in ; then . For any , one gets

Therefore, (18) holds.

Suppose that is a given number and satisfies . We denote a cone as follows:

and define an operator on by

Now, we prove the existence of the positive solution of (1)-(2). For convenience, we only prove the case that , , the prove of the case that , is similar, and we omit it here.

Since , then for , there exists a such that

for any . We define the set by

Then, for any , we have

which implies that

From , for the given and

there exists a so that

Let

Then, for any , we obtain from (17) that

which follows

It is easy to show that is a completely continuous function. Combining (28)-(33) and Lemma 1, the operator has at least one fixed point in , which is a positive solution of (15). The proof is completed.

Suppose that

: , , , and , , or , .

From the proof of Theorem 3, we have the following.

Corollary 4. Assume that holds and . Then problem (1)-(2) has at least one positive solution.

3. The Resonant Case ()

In this case, the boundary value condition (2) can be rewritten as

We have the following main theorem.

Theorem 5. Assume that and hold. Further, suppose that

where

Then problem (1)–(34) has at least one solution.

Proof of Theorem 5. Using a similar method as in the proof of Theorem 3, the problem (1)–(34) is equivalent to the following integral equation:

where is defined by (36).

It is obvious that Green’s function is decreasing with respect to when , and it is independent on the parameter when ; thus, the function is not increasing with respect to and .

Let

Given any value , satisfies the Hammerstein integral equation by (37):

To obtain the solvability of (39), we replace by a real constant ; that is,

Define the Hammerstein operator:

for any real number . From the condition , it is easy to know that

here, we have made use of

for any . By (35), that is,

we then see that the operator maps the ball onto itself, where

It is easy to show that the operator is a compact operator. From Lemma 2 and using a similar method of Theorem 3.6 in [22], we obtain that the operator has a fixed point for any real number . Let be the fixed point of with a given parameter ; that is, . For the solvability of (40), we need to find a so that ; that is,

Define

We only need to claim by assumption that

It is obvious that is continuously dependent on the parameter ; this would help us to show the existence of such that .

Now, we show that as .

On the contrary, assume that there exists a sequence such that

We now claim that for all is unbounded from below, which is dependent on and . In fact, suppose that is bounded from below by a constant ; by assumption , is bounded from below for and ; that is, there exists an so that

Replacing in (50) by , one gets

for and , which implies that the fixed point of satisfies

which implies that as . From the condition , as for all . Noticing that for , we have

as , which contradicts ; thus, our claim is true, and

Since is a fixed point of the operator , that is, the function satisfies

From (54) and (55), it is impossible that

as is sufficiently large.

Now, we define

Then, is not empty, and . In fact, assume that is empty; then, is bounded from below for any and , which follows from (55) and that is bounded from below. This contradicts (54).

From (50), there exists a positive number so that

Notice that for any and , and for any .

Therefore, we have from that

From , we have from (58) and (59) that

Taking the minimal values of both sides of the above, it yields

This contradicts (54).

Therefore, we have proved that . By a similar method, we can prove that and the detail is omitted.

Since the function is continuous with respect to the parameter . From the Intermediate Value Theorem, there exists a so that .

Let

Then, . From (62) and (39), it is obvious that solves

which implies that is a solution of the problem (1)–(34). This completes the proof.

4. Examples

In this section, we give two examples to illustrate our main result.

Example 1. Consider the following boundary value problem:

where

From Theorem 3, the problem (64) has a positive solution if .

Example 2. Consider the following boundary value problem:

where

Here . Thus, with

By applying our theorems, the problem (66) has a solution if .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work was supported by the NSF of Hunan Province (no. 13JJ3074), Major Project of Department of Education of Hunan Province (no. 13A088), the Scientific Research Foundation of Hengyang City (no. J1), and the Construct Program in USC.

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