Volume 2010, Article ID 519210,11pages doi:10.1155/2010/519210
Research Article
Positive Solutions of a Nonlinear Three-Point
Integral Boundary Value Problem
Jessada Tariboon
1, 2and Thanin Sitthiwirattham
11Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
2Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand
Correspondence should be addressed to Jessada Tariboon,jsdtrb@hotmail.com
Received 4 August 2010; Accepted 18 September 2010
Academic Editor: Raul F. Manasevich
Copyrightq2010 J. Tariboon and T. Sitthiwirattham. 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
uatfu 0,t∈0,1,u0 0,α0ηusdsu1, where 0< η <1 and 0< α <2/η2. We show the existence of at least one positive solution iffis 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 Gupta2studied 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 to3–19and 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,
u0 0, αuηu1, u0 βuη, αuηu1, u0 0, αuηu1, u0−βu0 0, αuηu1, αu0−βu0 0, uηu1 0,
1.1
In this paper, we consider the existence of positive solutions to the equation
uatfu 0, t∈0,1, 1.2
with the three-point integral boundary condition
u0 0, α
η
0
usdsu1, 1.3
where 0< η <1. We note that the new three-point boundary conditions are related to the area under the curve of solutionsutfromt0 totη.
The aim of this paper is to give some results for existence of positive solutions to1.2
1.3, assuming that 0< α <2/η2andfis either superlinear or sublinear. Set
f0 lim
u→0 fu
u , f∞ulim→ ∞
fu
u . 1.4
Then f0 0 and f∞ ∞ correspond to the superlinear case, and f0 ∞ and f∞ 0
correspond to the sublinear case. By the positive solution of 1.2-1.3 we mean that a functionutis positive on 0< t <1 and satisfies the problem1.2-1.3.
Throughout this paper, we suppose the following conditions hold:
H1f ∈C0,∞,0,∞;
H2a∈C0,1,0,∞and there existst0∈η,1such thatat0>0.
The proof of the main theorem is based upon an application of the following Krasnoselskii’s fixed point theorem in a cone.
Theorem 1.1see20. LetEbe a Banach space, and letK⊂Ebe a cone. AssumeΩ1,Ω2are open subsets ofEwith0∈Ω1, Ω1⊂Ω2, and let
A:K∩Ω1\Ω2
−→K 1.5
be a completely continuous operator such that
iAuu,u∈K∩∂Ω1, andAuu,u∈K∩∂Ω2;or
iiAuu,u∈K∩∂Ω1, andAuu,u∈K∩∂Ω2.
2. Preliminaries
We now state and prove several lemmas before stating our main results.
Lemma 2.1. Letαη2
/
2. Then fory∈C0,1, the problem
uyt 0, t∈0,1, 2.1
u0 0, α
η
0
usdsu1, 2.2
has a unique solution
ut 2t
2−αη2 1
0
1−sysds− αt
2−αη2 η
0
η−s2ysds−
t
0
t−sysds. 2.3
Proof. From2.1, we have
ut −yt. 2.4
Fort∈0,1, integration from 0 tot, gives
ut u0−
t
0
ysds. 2.5
Fort∈0,1, integration from 0 totyields that
ut u0t−
t
0 x
0
ysds dx, 2.6
that is,
ut u0t−
t
0
t−sysds. 2.7
So,
u1 u0−
1
0
Integrating2.7from 0 toη, whereη∈0,1, we have
η
0
usdsu0η
2
2 −
η
0 x
0
x−sysds dx
u0η
2
2 − 1 2
η
0
η−s2ysds.
2.9
From2.2, we obtain that
u0−
1
0
1−sysdsu0αη
2
2 −
α
2
η
0
η−s2ysds. 2.10
Thus,
u0 2 2−αη2
1
0
1−sysds− α
2−αη2 η
0
η−s2ysds. 2.11
Therefore,2.1-2.2has a unique solution
ut 2t
2−αη2 1
0
1−sysds− αt
2−αη2 η
0
η−s2ysds−
t
0
t−sysds. 2.12
Lemma 2.2. Let0< α < 2/η2. Ify∈C0,1andyt0on0,1, then the unique solutionuof 2.1-2.2satisfiesu0fort∈0,1.
Proof. Ifu10, then, by the concavity ofuand the fact thatu0 0, we haveut0 for
t∈0,1.
Moreover, we know that the graph ofutis concave down on0,1, we get
η
0
usds 1
2ηu
η, 2.13
where1/2ηuηis the area of triangle under the curveutfromt0 totηforη∈0,1. Assume thatu1<0. From2.2, we have
η
0
usds <0. 2.14
Hence,
u1 α
η
0
usds αη
2 u
η> u
η
η , 2.15
which contradicts the concavity ofu.
Lemma 2.3. Letαη2>2. Ify∈C0,1andyt0fort∈0,1, then2.1-2.2has no positive solution.
Proof. Assume2.1-2.2has a positive solutionu.
Ifu1>0, then0ηusds >0, it implies thatuη>0 and
u1 1 α
η
0
usds αη
2 u
η αη
2
2
uη η >
uη
η , 2.16
which contradicts the concavity ofu.
Ifu1 0, then0ηusds0, this isut≡0 for allt∈0, η. If there existsτ ∈η,1 such thatuτ>0, thenu0 uη< uτ, which contradicts the concavity ofu. Therefore, no positive solutions exist.
In the rest of the paper, we assume that 0< αη2 < 2. Moreover, we will work in the
Banach spaceC0,1, and only the sup norm is used.
Lemma 2.4. Let0< α <2/η2. Ify∈C0,1andy0, then the unique solutionuof the problem 2.1-2.2satisfies
inf
t∈η,1utγu, 2.17
where
γ:min
η,αη
2
2 ,
αη1−η
2−αη2
. 2.18
Proof. Setuτ u. We divide the proof into three cases.
Case 1. Ifητ 1 and inft∈η,1ut uη, then the concavity ofuimplies that
uη
η
uτ
τ uτ. 2.19
Thus,
inf
Case 2. Ifητ 1 and inft∈η,1ut u1, then2.2,2.13, and the concavity ofuimplies
u1 α
η
0
usdsαη
2
2
uη η
αη2
2
uτ
τ
αη2
2 uτ. 2.21
Therefore,
inf
t∈η,1ut
αη2
2 u. 2.22
Case 3. Ifτ η < 1, then inft∈η,1ut u1. Using the concavity ofuand2.2,2.13, we
have
uσu1 u1−u
η
1−η 0−1
u1
1−1−2/αη 1−η
u1 2−αη
2
αη1−η.
2.23
This implies that
inf
t∈η,1ut
αη1−η
2−αη2 u. 2.24
This completes the proof.
3. Main Results
Now we are in the position to establish the main result.
Theorem 3.1. AssumeH1andH2hold. Then the problem1.2-1.3has at least one positive solution in the case
if00andf∞∞(superlinear), or
Proof. It is known that 0< α <2/η2. FromLemma 2.1,uis a solution to the boundary value
problem1.2-1.3if and only ifuis a fixed point of operatorA, whereAis defined by
Aut 2t
2−αη2 1
0
1−sasfusds− αt
2−αη2 η
0
η−s2asfusds
− t
0
t−sasfusds.
3.1
Denote that
K
u|u∈C0,1, u0, inf
ηt1utγu
, 3.2
whereγis defined in2.18.
It is obvious thatKis a cone inC0,1. Moreover, by Lemmas2.2and2.4,AK ⊂K. It is also easy to check thatA:K → Kis completely continuous.
Superlinear Case (
f
00
and
f
∞∞
).
Sincef00, we may chooseH1>0 so thatfu u, for 0< uH1, where >0 satisfies
2 2−αη2
1
0
1−sasds1. 3.3
Thus, if we let
Ω1{u∈C0,1| u< H1}, 3.4
then, foru∈K∩∂Ω1, we get
Aut 2t
2−αη2 1
0
1−sasfusds
2t
2−αη2 1
0
1−sasusds
2
2−αη2 1
0
1−sasdsu
u.
3.5
Further, sincef∞ ∞, there existsH2 > 0 such thatfu ρu, foru H2, where
ρ >0 is chosen so that
ργ 2η
2−αη2 1
η
1−sasds1. 3.6
LetH2 max{2H1,H2/γ}andΩ2{u∈C0,1| u< H2}. Thenu∈K∩∂Ω2implies that
inf
ηt1utγuγH2
H2, 3.7
and so
Auη 2η
2−αη2 1
0
1−sasfusds− αη
2−αη2 η
0
η−s2asfusds
− η
0
η−sasfusds
2η
2−αη2 1
0
1−sasfusds− αη
2−αη2 η
0
η2−2ηss2asfusds
− 1
2−αη2 η
0
2−αη2η−sasfusds
2η
2−αη2 1
0
1−sasfusds αη
2
2−αη2 η
0
sasfusds
− αη
2−αη2 η
0
s2asfusds− 2η
2−αη2 η
0
asfusds
2
2−αη2 η
0
sasfusds
2η
2−αη2 1
η
1−sasfusds 2
1−η
2−αη2 η
0
sasfusds
αη
2−αη2 η
0
sη−sasfusds
2η
2−αη2 1
η
1−sasfusds
2ηρ
2−αη2 1
η
1−sasusds 2ηργ
2−αη2 1
η
1−sasdsuu.
3.8
Hence,Au u,u ∈ K∩∂Ω2. By the first past of Theorem 1.1,Ahas a fixed point in
Sublinear Case (
f
0∞
and
f
∞0
).
Sincef0∞, chooseH3 >0 such thatfuMufor 0< uH3, whereM >0 satisfies
2ηγM
2−αη2 1
η
1−sasds1. 3.9
Let
Ω3{u∈C0,1| u< H3}, 3.10
then foru∈K∩∂Ω3, we get
Auη 2η
2−αη2 1
0
1−sasfusds− αη
2−αη2 η
0
η−s2asfusds
− η
0
η−sasfusds
2η
2−αη2 1
η
1−sasfusds
2ηγM
2−αη2 1
η
1−sasdsuu.
3.11
Thus,Auu,u∈K∩∂Ω3. Now, sincef∞0, there existsH4>0 so thatfuλufor
uH4, whereλ >0 satisfies
2λ
2−αη2 1
0
1−sasds1. 3.12
ChooseH4 max{2H3,H4/γ}. Let
Ω4{u∈C0,1| u< H4}, 3.13
thenu∈K∩∂Ω4implies that
inf
Therefore,
Aut 2t
2−αη2 1
0
1−sasfusds− αt
2−αη2 η
0
η−s2asfusds
− t
0
t−sasfusds
2t
2−αη2 1
0
1−sasfusds
2λu
2−αη2 1
0
1−sasdsu.
3.15
ThusAu u,u∈K∩∂Ω4. By the second part ofTheorem 1.1,Ahas a fixed pointuin
K∩Ω4\Ω3, such thatH3 u H4. This completes the sublinear part of the theorem.
Therefore, the problem1.2-1.3has 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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