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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 687595, 12 pages

http://dx.doi.org/10.1155/2013/687595

## On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

^{1}Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China^{2}College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China

Received 31 July 2013; Accepted 8 October 2013

Academic Editor: Zhanbing Bai

Copyright © 2013 Liu Yang et al. 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

Positive solutions for a kind of third-order multipoint boundary value problem under the non-resonant conditions and the resonant conditions are considered. In the nonresonant case, by using Leggett-Williams fixed-point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using Leggett-Williams norm-type theorem due to O’Regan and Zima, existence result of at least one positive solution is established. The results obtained are valid and new for the problem discussed. Two examples are given to illustrate the main results.

#### 1. Introduction

We consider the third-order -point boundary value problem given by where , , , , , , and . For the convenience of writing later, we denote ,, and .

If condition holds, the problem is nonresonant; that is, the associated linear problem has only zero solution, and the differential operator with boundary conditions is invertible. Otherwise, the problem is at resonance.

Third-order differential equations arise in many different areas of applied mathematics and physics, as the varying cross-section or deflection of a curved beam having a constant, three-layer beam and so on [1]. Recently, there have been extensive studies on positive solutions for nonresonant two-point or three-point boundary value problems for nonlinear third-order ordinary differential equations. For examples, Anderson [2] established the existence of at least three positive solutions to problem where is continuous and .

By using the well-known Guo-Krasnosel’skii fixed-point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for the third-order three-point problem:

In another paper [5], Graef and Kong studied the existence of positive solutions for the third-order semipositone boundary value problem: where are constants and is a parameter. Also , and . Moreover are continuous functions. By using the Guo-Krasnosel’skii fixed-point theorem, the author established the existence of positive solutions. For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [6–12] and references therein.

For boundary value problems of second-order or higher-order differential equations at resonance, many existence results of solutions have been established; see [13–25]. In [25], the authors considered the following problem: By using Mawhin continuation theorem [26], the existence results of solutions were obtained under the resonant conditions , , and and , , and , respectively.

It is well known that the problem of existence of positive solution for nonlinear BVP is very difficult when the resonant case is considered. Only few works gave the approach in this area for first- and second-order differential equations [27–34]. To our best knowledge, few paper dealt with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [28–30, 35], we study the positive solution for problem (1) under nonresonant condition and resonant condition , respectively. By using Leggett-Williams fixed-point theorem and its generalization [27, 29], we establish the existence results of positive solutions. The results obtained in this paper are interesting in the following aspects.(1)In the nonresonant case, Green’s function is established and the results obtained are more general than those of earlier work.(2)It is the first time that the positive solution is considered for third-order boundary value problem at resonance.

#### 2. Background Definitions and Lemmas

Let be real Banach spaces. A nonempty closed convex set is said to be a cone provided that , if , and implies .

*Definition 1. *The map is a nonnegative continuous concave functional on if is continuous and

*Definition 2. *Let constants be given and let be a nonnegative continuous concave functional on the cone . Define the convex sets and as follow:

Lemma 3 (Leggett-Williams fixed-point theorem [35]). *Let be a completely continuous operator and let be a nonnegative continuous concave functional on such that for all . Suppose that there exist such that** and for ,** for ,** for with .**Then operator has at least triple fixed-points , and with , and .*

Operator is called a Fredholm operator with index zero, which means that is closed and dim = codim , and there exist continuous projections and such that and . Furthermore, for dim Im codim , there exists an isomorphism . Denote by the restriction of to to and its inverse by , so and the coincidence equation is equivalent to the operator equation: Let be a retraction, which means a continuous mapping such that for all and

Lemma 4 (Leggett-Williams norm-type theorem [28]). *Assume that is a cone in and that and are open bounded subsets of with and . Suppose that is a Fredholm operator of index zero and that** is bounded and continuous and is compact on every bounded subset of ,** for all and ,** maps subsets of into bounded subsets of C,**, where stands for the Brouwer degree,**there exists such that for , where for some and is such that for every ,**,**.**Then the equation has a solution in the set .*

#### 3. Main Results for Nonresonant Case

In this section we consider the positive solution for the nonresonant case with the condition and we always suppose that .

Firstly, we consider the third-order -point boundary value problem given by

Lemma 5. *Suppose . Then problem (11) and (12) is equivalent to
**
where for *

*Proof. *Let be Green’s function of problem with boundary condition (12). We can suppose
where , and are undetermined coefficients.

Considering the properties of Green's function together with boundary condition (12), we have
The explicit expression of Green’s function is obtained by solving the linear function systems.

Lemma 6. *Green’s function satisfies that .*

*Proof. *For , , and ,
which means that is decreasing on variable . Thus
On the other hand, for , , and ,
Thus,
Furthermore,
Thus, for ,
This gives that .

Lemma 7. *If , , and is the solution of problem (11) and (12), then
**
where is a positive constant.*

*Proof. *From
we see that is decreasing on . Considering , we have . Next we claim that . Suppose that, on the contrary, . We have
Thus,
From the concavity of , we have
Multiplying left and right sides by and considering , we have
This completes the proof of Lemma 7.

Let the Banach space be endowed with the maximum norm. We define the cone by

Define the continuous nonnegative concave functional by Define the constants by

Theorem 8. *Suppose that there exist constants such that **, ,**, ,**, .**Then problem (1) has at least three positive solutions , and satisfying
*

*Proof. *The operator is defined by
It is clear that and it is completely continuous.

Next, the conditions of Lemma 3 are checked. If , then and condition implies that
Then
Thus, .

Similar to the proof above, we obtain that . Hence, condition of Lemma 3 is satisfied.

The fact that the constant function implies that . If , from assumption (), Thus which ensures that condition of Lemma 3 is satisfied. Finally we show that condition of Lemma 3 also holds. Suppose that with . Then So, condition of Lemma 3 is satisfied. Thus, an application of Lemma 3 implies that the nonresonant third-order boundary value problem (1) has at least three positive solutions , and satisfying

Here an example is given to illustrate the main results of this section. We consider the following nonresonant three-point boundary value problem: where Here , , , and By a simple computation, we can get that We choose , and . It is easy to check that (1), ,(2), ,(3), .

Thus all conditions of Theorem 8 hold. This ensures that problem (40) has at least three positive solutions , and satisfying

#### 4. Main Results for Resonant Case

In this section the condition is considered. Obviously, problem (1) is at resonance under this condition. The norm-type Leggett-Williams fixed-point theorem will be used to establish the existence results of positive solution.

We define the spaces endowed with the maximum norm. It is well known that and are the Banach spaces.

Define the linear operator and where and the nonlinear operator with It is obvious that . Denote the function , as follow: Define the function as follow:for and .

The function and positive number are given by

Theorem 9. *Suppose that there exists positive constant such that is continuous and satisfies the following conditions:**, for ,** for ,**there exist , and continuous functions
**such that
**
and is nonincreasing on with
**
Then resonant problem (1) has at least one positive solution.*

*Proof. *Firstly we prove that
Indeed, for each , we choose
We can check that
which means . Thus
On the other hand, for each , there exists such that
Integrating both sides on [0, ], we have
Considering the boundary condition together with the resonant condition , we have
Thus,
It is obvious that dim and is closed.

Secondly we see , where
In fact, for each , we have
This induces that . Since , we have . Thus is a Fredholm operator with index zero.

Define two projections and by
Clearly, and . Note that, for , the inverse of is given by
In fact, It is easy to check that

Next we will check that every condition of Lemma 4 is fulfilled. Remark that can be extended continuously on and condition () of Lemma 4 is fulfilled.

Define the set of nonnegative functions and subsets of X by
Remark that and are open and bounded sets. Furthermore

Let the isomorphism and for . Then is a retraction and maps subsets of into bounded subsets of , which ensures that condition () of Lemma 4 is fulfilled.

Then we prove that () of Lemma 4 is fulfilled. For this purpose, suppose that there exist and such that . Then
for all . Thus
Let . The proof is divided into three cases.(1)We show that . Suppose, on the contrary, that achieves maximum value only at . Then the boundary condition in combination with the resonant condition yields that , which is a contradiction.(2)We claim that . Suppose, on the contrary, that achieves maximum value at . From condition , there exists near to zero such that
Then
which contradicts the fact that achieves maximum value at .(3)Thus there exists such that . We may choose nearest to with . From the mean value theory, we claim that there exists such that
However,
Thus
Then, considering assumption , we have
which is a contradiction. Thus () of Lemma 4 is fulfilled.

*Remark 10. *The sign of third-order derivative of a function at point cannot be confirmed when is a maximal value of . Thus the methods in [28] are not applicable directly to this problem.

For , define the projection as follows: where and . Suppose . In view of we obtain Hence implies . Hence, if , we get contradicting . Thus for and . Therefore This ensures Let and . From condition , we see Hence . Moreover, for , we have which means . These ensure that () and () of Lemma 4 hold.

At last, we confirm that () is satisfied. Taking on , we see and we can choose . For , we have Therefore, in view of (), we obtain, for all , So for all , which means () of Lemma 4 holds.

Thus with the application of Lemma 4, we confirm that the equation has a solution , which implies that the resonant problem (1) has at least one positive solution.

Finally an example is given to illustrate the main results of the resonance case. We investigate the resonant third-order three-point boundary value problem: where , , , and Here By a simple computation, we have Choose , , ,, and .

We take Then, It is easy to check that (1), for all ,(2), for all ,(3)