We study the existence of positive solutions to the threepoint 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 secondorder ordinary differential equations was initiated by Il'in and Moiseev [1]. Then Gupta [2] studied threepoint boundary value problems for nonlinear secondorder ordinary differential equations. Since then, nonlinear secondorder threepoint boundary value problems have also been studied by several authors. We refer the reader to [3–19] and the references therein. However, all these papers are concerned with problems with threepoint boundary condition restrictions on the slope of the solutions and the solutions themselves, for example,
and so forth.
In this paper, we consider the existence of positive solutions to the equation
with the threepoint integral boundary condition
where . We note that the new threepoint 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
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
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
has a unique solution
Proof.
From (2.1), we have
For , integration from to , gives
For , integration from to yields that
that is,
So,
Integrating (2.7) from to , where , we have
From (2.2), we obtain that
Thus,
Therefore, (2.1)(2.2) has a unique solution
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
where is the area of triangle under the curve from to for .
Assume that . From (2.2), we have
By concavity of and , it implies that .
Hence,
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
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
where
Proof.
Set . We divide the proof into three cases.
Case 1.
If and , then the concavity of implies that
Thus,
Case 2.
If and , then (2.2), (2.13), and the concavity of implies
Therefore,
Case 3.
If , then . Using the concavity of and (2.2), (2.13), we have
This implies that
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
Denote that
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
Thus, if we let
then, for , we get
Thus , .
Further, since , there exists such that , for , where is chosen so that
Let and . Then implies that
and so
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
Let
then for , we get
Thus, , . Now, since , there exists so that for , where satisfies
Choose . Let
then implies that
Therefore,
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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