- Research Article
- Open Access
Positive Solutions of a Nonlinear Three-Point Integral Boundary Value Problem
© J. Tariboon and T. Sitthiwirattham. 2010
- Received: 4 August 2010
- Accepted: 18 September 2010
- Published: 20 September 2010
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.
- Banach Space
- Partial Differential Equation
- Unique Solution
- Ordinary Differential Equation
- Functional Equation
and so forth.
where . We note that the new three-point boundary conditions are related to the area under the curve of solutions from to .
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 ).
be a completely continuous operator such that
(i) , , and , or
(ii) , , and , .
Then has a fixed point in .
We now state and prove several lemmas before stating our main results.
Let . If and on , then the unique solution of (2.1)-(2.2) satisfies for .
If , then, by the concavity of and the fact that , we have for .
where is the area of triangle under the curve from to for .
By concavity of and , it implies that .
which contradicts the concavity of .
Let . If and for , then (2.1)-(2.2) has no positive solution.
Assume (2.1)-(2.2) has a positive solution .
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.
Set . We divide the proof into three cases.
This completes the proof.
Now we are in the position to establish the main result.
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).
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 )
Thus , .
Hence, , . By the first past of Theorem 1.1, has a fixed point in such that .
Sublinear Case ( and )
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.
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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