# Positive Solutions of a Nonlinear Three-Point Integral Boundary Value Problem

Author Affiliations

1 Department of Mathematics, Faculty of Applied Science, King Mongkut's University of Technology North Bangkok, Bangkok 10800, Thailand

2 Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand

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Boundary Value Problems 2010, 2010:519210  doi:10.1155/2010/519210

 Received: 4 August 2010 Accepted: 18 September 2010 Published: 20 September 2010

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.

We study the existence of positive solutions to the three-point integral boundary value problem , , , , where and . We show the existence of at least one positive solution if f is either superlinear or sublinear by applying the fixed point theorem in cones.

### 1. Introduction

The study of the existence of solutions of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by Il'in and Moiseev [1]. Then Gupta [2] studied three-point boundary value problems for nonlinear second-order ordinary differential equations. Since then, nonlinear second-order three-point boundary value problems have also been studied by several authors. We refer the reader to [319] and the references therein. However, all these papers are concerned with problems with three-point boundary condition restrictions on the slope of the solutions and the solutions themselves, for example,

(11)

and so forth.

In this paper, we consider the existence of positive solutions to the equation

(12)

with the three-point integral boundary condition

(13)

where . We note that the new three-point boundary conditions are related to the area under the curve of solutions from to .

The aim of this paper is to give some results for existence of positive solutions to (1.2)-(1.3), assuming that and is either superlinear or sublinear. Set

(14)

Then and correspond to the superlinear case, and and correspond to the sublinear case. By the positive solution of (1.2)-(1.3) we mean that a function is positive on and satisfies the problem (1.2)-(1.3).

Throughout this paper, we suppose the following conditions hold:

;

and there exists such that .

The proof of the main theorem is based upon an application of the following Krasnoselskii's fixed point theorem in a cone.

Theorem 1.1 (see [20]).

Let be a Banach space, and let be a cone. Assume , are open subsets of with , and let

(15)

be a completely continuous operator such that

(i), , and , or

(ii),, and ,.

Then has a fixed point in .

### 2. Preliminaries

We now state and prove several lemmas before stating our main results.

Lemma 2.1.

Let . Then for , the problem

(21)

(22)

has a unique solution

(23)

Proof.

From (2.1), we have

(24)

For , integration from to , gives

(25)

For , integration from to yields that

(26)

that is,

(27)

So,

(28)

Integrating (2.7) from to , where , we have

(29)

From (2.2), we obtain that

(210)

Thus,

(211)

Therefore, (2.1)-(2.2) has a unique solution

(212)

Lemma 2.2.

Let . If and on , then the unique solution of (2.1)-(2.2) satisfies for .

Proof.

If , then, by the concavity of and the fact that , we have for .

Moreover, we know that the graph of is concave down on , we get

(213)

where is the area of triangle under the curve from to for .

Assume that . From (2.2), we have

(214)

By concavity of and , it implies that .

Hence,

(215)

which contradicts the concavity of .

Lemma 2.3.

Let . If and for , then (2.1)-(2.2) has no positive solution.

Proof.

Assume (2.1)-(2.2) has a positive solution .

If , then , it implies that and

(216)

which contradicts the concavity of .

If , then , this is for all . If there exists such that , then , which contradicts the concavity of . Therefore, no positive solutions exist.

In the rest of the paper, we assume that . Moreover, we will work in the Banach space , and only the sup norm is used.

Lemma 2.4.

Let . If and , then the unique solution of the problem (2.1)-(2.2) satisfies

(217)

where

(218)

Proof.

Set . We divide the proof into three cases.

Case 1.

If and , then the concavity of implies that

(219)

Thus,

(220)

Case 2.

If and , then (2.2), (2.13), and the concavity of implies

(221)

Therefore,

(222)

Case 3.

If , then . Using the concavity of and (2.2), (2.13), we have

(223)

This implies that

(224)

This completes the proof.

### 3. Main Results

Now we are in the position to establish the main result.

Theorem 3.1.

Assume and hold. Then the problem (1.2)-(1.3) has at least one positive solution in the case

(i) and (superlinear), or

(ii) and (sublinear).

Proof.

It is known that . From Lemma 2.1, is a solution to the boundary value problem (1.2)-(1.3) if and only if is a fixed point of operator , where is defined by

(31)

Denote that

(32)

where is defined in (2.18).

It is obvious that is a cone in . Moreover, by Lemmas 2.2 and 2.4, . It is also easy to check that is completely continuous.

Superlinear Case ( and )

Since , we may choose so that , for , where satisfies

(33)

Thus, if we let

(34)

then, for , we get

(35)

Thus , .

Further, since , there exists such that , for , where is chosen so that

(36)

Let and . Then implies that

(37)

and so

(38)

Hence, , . By the first past of Theorem 1.1, has a fixed point in such that .

Sublinear Case ( and )

Since , choose such that for , where satisfies

(39)

Let

(310)

then for , we get

(311)

Thus, , . Now, since , there exists so that for , where satisfies

(312)

Choose . Let

(313)

then implies that

(314)

Therefore,

(315)

Thus , . By the second part of Theorem 1.1, has a fixed point in , such that . This completes the sublinear part of the theorem. Therefore, the problem (1.2)-(1.3) has at least one positive solution.

### Acknowledgments

The authors would like to thank the referee for their comments and suggestions on the paper. Especially, the authors would like to thank Dr. Elvin James Moore for valuable advice. This research is supported by the Centre of Excellence in Mathematics, Thailand.

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