doi:10.1155/2011/894135
Research Article
Solutions to a Three-Point Boundary Value Problem
Jin Liang
1and Zhi-Wei Lv
2, 31Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
2Department of Mathematics and Physics, Anyang Institute of Technology, Anyang, Henan 455000, China 3Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China
Correspondence should be addressed to Jin Liang,[email protected]
Received 25 November 2010; Accepted 19 January 2011
Academic Editor: Toka Diagana
Copyrightq2011 J. Liang and Z.-W. Lv. 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.
By using the fixed-point index theory and Leggett-Williams fixed-point theorem, we study the exis-tence of multiple solutions to the three-point boundary value problemut atft, ut, ut 0, 0< t <1;u0 u0 0;u1−αuη λ, whereη∈0,1/2,α∈1/2η,1/ηare constants, λ∈0,∞is a parameter, anda,fare given functions. New existence theorems are obtained, which extend and complement some existing results. Examples are also given to illustrate our results.
1. Introduction
It is known that when differential equations are required to satisfy boundary conditions at more than one value of the independent variable, the resulting problem is called a multipoint boundary value problem, and a typical distinction between initial value problems and multipoint boundary value problems is that in the former case one is able to obtain the solutions depend only on the initial values, while in the latter case, the boundary conditions at the starting point do not determine a unique solution to start with, and some random choices among the solutions that satisfy these starting boundary conditions are normally not to satisfy the boundary conditions at the other specified points. As it is noticed elsewhere see, e.g., Agarwal 1, Bisplinghoff and Ashley 2, and Henderson3, multi point boundary value problem has deep physical and engineering background as well as realistic mathematical model. For the development of the research of multi point boundary value problems for differential equations in last decade, we refer the readers to, for example,
In this paper, we study the existence of multiple solutions to the following three-point boundary value problem for a class of third-order differential equations with inhomogeneous three-point boundary values,
ut atft, ut, ut0, 0< t <1,
u0 u0 0, u1−αuηλ,
1.1
whereη ∈0,1/2,α∈1/2η,1/η,λ∈0,∞, anda,fare given functions. To the authors’ knowledge, few results on third-order differential equations with inhomogeneous three-point boundary values can be found in the literature. Our purpose is to establish new existence theorems for1.1which extend and complement some existing results.
Let X be an Banach space, and let Y be a cone in X. A mapping β is said to be a nonnegative continuous concave functional onY ifβ:Y → 0,∞is continuous and
βtx 1−ty≥tβx 1−tβy, x, y∈Y, t∈0,1. 1.2
Assume that
H
a∈C0,1,0,∞, 0<
1
0
1−ssasds <∞,
f∈C0,1×0,∞×0,∞,0,∞.
1.3
Define
maxf0 lim
v→0tmax∈0,1u∈sup0,∞
ft, u, v v ,
minf0 lim
v→0tmin∈0,1u∈inf0,∞
ft, u, v v ,
maxf∞vlim→∞max t∈0,1u∈sup0,∞
ft, u, v v ,
minf∞vlim→∞min t∈0,1u∈inf0,∞
ft, u, v v .
1.4
2. Lemmas
LetEC10,1be a Banach Space with norm
u1max
u,u, 2.1
where
u max
t∈0,1|ut|, u
max
t∈0,1u
t.
2.2
It is not hard to see Lemmas2.1and2.2.
Lemma 2.1. Letu∈C10,1be the unique solution of 1.1. Then
ut
1
0
Gt, sasfs, us, usds
αt2
21−αη
1
0
G1
η, sasfs, us, usds λt
2
21−αη,
2.3
where
Gt, s 1
2 ⎧ ⎨ ⎩
2t−t2−ss, s≤t,
1−st2, t≤s,
G1t, s ∂Gt, s
∂t
⎧ ⎨ ⎩
1−ts, s≤t,
1−st, t≤s.
2.4
Lemma 2.2. One has the following.
i0≤G1t, s≤1−ss, 1/2t21−ss≤Gt, s≤G1, s 1/21−ss.
iiG1t, s≥1/4G1s, s 1/41−ss, fort∈1/4,3/4, s∈0,1.
iii G11/2, s≥1/21−ss, fors∈0,1.
Lemma 2.3. Letu∈C10,1be the unique solution of 1.1. Thenutis nonnegative and satisfies u1u.
iFort≤η,
ut 1
21−αη
t
0
2ts−s21−αηt2sα−1asfs, us, usds
η
t
t21−αηt2sα−1asfs, us, usds
1
η
t21−sasfs, us, usdsλt2
,
ut 1
21−αη
t
0
2s1−αη2tsα−1asfs, us, usds
η
t
2t1−αη2tsα−1asfs, us, usds
1
η
2t1−sasfs, us, usds2λt
,
2.5
that is,ut≤ut.
iiFort≥η,
ut 1
21−αη
η
0
2ts−s21−αηt2sα−1asfs, us, usds
t
η
2ts−s21−αηt2αη−sasfs, us, usds
1
t
t21−sasfs, us, usdsλt2
,
ut 1
21−αη
η
0
2s1−αη2tsα−1asfs, us, usds
t
η
2s1−αη2tαη−sasfs, us, usds
1
t
2t1−sasfs, us, usds2λt
.
On the other hand, forη≤s≤t, we have
2s1−αη2tαη−s−2ts−s21−αηt2αη−s
αηt−s2s−t s1−t2−t ss−t
αηt−s
2s−t− s αη
s1−t2−t.
2.7
Sinceα∈1/2η,1/η,
2s1−αη2tαη−s≥2ts−s21−αηt2αη−s. 2.8
So,ut≤ut. Therefore,ut≤ut, which means
u ≤u, u1u. 2.9
The proof is completed.
Lemma 2.4. Letu∈C10,1be the unique solution of 1.1. Then
min
t∈1/4,3/4u
t≥ 1
4u1. 2.10
Proof. From2.3, it follows that
ut
1
0
G1t, sasf
s, us, usds
αt
1−αη
1
0
G1
η, sasfs, us, usds λt
1−αη
≤ 1
0
1−ssasfs, us, usds
α
1−αη
1
0
G1
η, sasfs, us, usds λ
1−αη.
2.11
Hence,
u1u≤ 1
0
1−ssasfs, us, usds
α
1−αη
1
0
G1
η, sasfs, us, usds λ 1−αη.
By Lemmas2.2and2.3, we get, for anyt∈1/4,3/4,
min
t∈1/4,3/4u
t min
t∈1/4,3/4 1
0
G1t, sasf
s, us, usds
αt
1−αη
1
0
G1
η, sasfs, us, usds λt
1−αη
≥ 1 4
1
0
1−ssasfs, us, usds
α
1−αη
1
0
G1
η, sasfs, us, usds λ
1−αη
.
2.13
Thus,
min
t∈1/4,3/4u
t≥ 1
4u1. 2.14
Define a cone by
K
u∈E:u≥0, min
t∈1/4,3/4u
t≥ 1
4u1
. 2.15
Set
Kr {u∈K:u1< r}, ∂Kr{u∈K:u1r}, r >0,
Kr {u∈K:u1≤r}, K
β, r, su∈K:r≤βu, u1≤s, s > r >0.
2.16
Define an operatorTby
Tut
1
0
Gt, sasfs, us, usds
αt2
21−αη
1
0
G1
η, sasfs, us, usds λt
2
21−αη.
2.17
Lemma2.1implies that1.1has a solutionuutif and only ifuis a fixed point ofT.
From Lemmas2.1and2.2and the Ascoli-Arzela theorem, the following follow.
Theorem 2.6see10. LetEbe a real Banach Space, letK⊂Ebe a cone, andΩr {u∈K:u ≤
r}. Let operatorT :K∩Ωr → Kbe completely continuous and satisfyTx /x, for allx∈∂Ωr. Then iifTx ≤ x, for all x∈∂Ωr, theniT,Ωr, K 1,
iiifTx ≥ x, for all x∈∂Ωr, theniT,Ωr, K 0.
Theorem 2.7see8. LetT :Pc → Pcbe a completely continuous operator andβa nonnegative continuous concave functional onP such thatβx ≤ x for allx ∈ Pc. Suppose that there exist
0< d0< a0 < b0≤csuch that
a{x∈Pβ, a0, b0:βx> a0}/∅andβTx> a0forx∈Pβ, a0, b0,
bTx< d0forx ≤d0,
cβTx> a0forx∈Pβ, a0, cwithTx> b0.
Then,T has at least three fixed pointsx1, x2, and x3inPcsatisfying
x1< d0, a0< βx2, x3> d0, βx3< a0. 2.18
3. Main Results
In this section, we give new existence theorem about two positive solutions or three positive solutions for1.1.
Write
Λ1
1
0
1−ssasds α
1−αη
1
0
G1
η, sasds
−1
,
Λ2
3/4
1/4
1−ssasds α
1−αη
3/4
1/4
G1
η, sasds
−1
.
3.1
Theorem 3.1. Assume that
H1minf0minf∞ ∞;
H2there exists a constantρ1 >0such thatft, u, v≤1/2Λ1ρ1, fort∈0,1,u∈0, ρ1 andv∈0, ρ1.
Then, the problem1.1has at least two positive solutionsu1andu2such that
0<u11< ρ1<u21, 3.2
forλsmall enough.
Proof. Since
minf0 lim
v→0tmin∈0,1u∈inf0,∞
ft, u, v
there isρ0∈0, ρ1such that
By Theorem2.6, we have
iT,Ωρ0, K
0. 3.9
On the other hand, since
minf∞vlim→∞min t∈0,1u∈inf0,∞
ft, u, v
there existρ0∗,ρ∗0> ρ1such that
ft, u, v≥8Λ2v, fort∈0,1, u∈0,∞, v≥ 1 4ρ
∗
0. 3.11
LetΩρ∗0{u∈K:u1 < ρ∗0}. Then, by a argument similar to that above, we obtain
Tu1≥ u1, ∀u∈∂Ωρ0∗. 3.12
By Theorem2.6,
iT,Ωρ∗0, K
0. 3.13
Finally, letΩρ1 {u∈K :u1 < ρ1}, and letλsatisfy 0< λ≤1/21−αηρ1for any
u∈∂Ωρ1. Then,H2implies
Tut
1
0
G1t, sasf
s, us, usds
αt
1−αη
1
0
G1
η, sasfs, us, usds λt
1−αη
≤ 1
0
1−ssas1
2Λ1ρ1ds
α
1−αη
1
0
G1
η, sas1
2Λ1ρ1ds
λ
1−αη
≤ 1 2Λ1
1
0
1−ssasds α
1−αη
1
0
G1
η, sasds
ρ1
1/21−αηρ1 1−αη
1
2ρ1 1 2ρ1
u1,
3.14
which means thatTu ≤ u1. Thus,Tu1 ≤ u1, for allu∈∂Ωρ1. Using Theorem2.6, we get
iT,Ωρ1, K
1. 3.15
From3.9–3.15andρ0< ρ1 < ρ∗0, it follows that
iT,Ωρ∗0\Ωρ1, K
−1, iT,Ωρ1\Ωρ0, K
Therefore,Thas fixed pointu1 ∈Ωρ1\Ωρ0and fixed pointu2 ∈Ωρ∗0\Ωρ1. Clearly,u1,u2are both positive solutions of the problem1.1and
0<u11< ρ1<u21. 3.17
The proof of Theorem3.1is completed.
Theorem 3.2. Assume that
H3maxf0maxf∞0;
H4there exists a constantρ2 >0such thatft, u, v≥2Λ2ρ2, fort∈0,1, u∈0, ρ2and
v∈1/4ρ2, ρ2.
Then, the problem1.1has at least two positive solutionsu1andu2such that
0<u11< ρ2<u21 3.18
forλsmall enough.
Proof. By
maxf0 lim
v→0tmax∈0,1u∈sup0,∞
ft, u, v
v 0, 3.19
we see that there existsρ∗∈0, ρ2such that
ft, u, v≤ 1
2Λ1v, fort∈0,1, u∈0,∞, v∈
0, ρ∗. 3.20
Put
Ωρ∗
u∈K:u1 < ρ∗, 3.21
and letλsatisfy
0< λ≤ 1
2
1−αηρ∗. 3.22
Then Lemmas2.2and2.3and3.20implies that for anyu∈∂Ωρ∗,
Tut
1
0
G1t, sasf
s, us, usds
αt
1−αη
1
0
G1
η, sasfs, us, usds λt
≤ 1
0
1−ssas1
2Λ1u
sds α
1−αη
1
0
G1
η, sas1
2Λ1u
sds λ
1−αη
≤ 1 2Λ1
1
0
1−ssasds α
1−αη
1
0
G1
η, sasds
u1
1−αηρ∗
21−αη
≤ 1 2u1
1 2ρ∗
u1.
3.23
SoTu ≤ u1. Hence,Tu1≤ u1, for allu∈∂Ωρ∗.
Applying Theorem2.6, we have
iT,Ωρ∗, K
1. 3.24
Next, by
maxf∞vlim→∞max t∈0,1u∈sup0,∞
ft, u, v
v 0, 3.25
we know that there existsr0 > ρ2such that
ft, u, v≤ 1
2Λ1v, fort∈0,1, u∈0,∞, v≥r0. 3.26
Case 1. maxt∈0,1ft, u, vis unbounded.
Define a functionf∗:0,∞ → 0,∞by
f∗ρmaxft, u, v:t∈0,1, u, v∈0, ρ. 3.27
Clearly,f∗is nondecreasing and limρ→∞f∗ρ/ρ0, and
f∗ρ≤ 1
2Λ1ρ, forρ > r0. 3.28
Takingρ∗≥max{2r0,2λ/1−αη,2ρ2}, it follows from3.26–3.28that
ft, u, v≤f∗ρ∗≤ 1
2Λ1ρ
By Lemmas2.2and2.3and3.28, we have
In this case, there exists anM >0 such that
max Therefore, in both cases, taking
we get
Tu1≤ u1, ∀u∈∂Ωρ∗. 3.34
By Theorem2.6, we have
iT,Ωρ∗, K
1. 3.35
Finally, putΩρ2{u∈K:u1< ρ2}. ThenH4implies that
Tu
1 2
1
0
G1
1 2, s
asfs, us, usds
α
21−αη
1
0
G1
η, sasfs, us, usds λ
21−αη
≥ 3/4
1/4 1
21−ssas2Λ2ρ2ds
α
21−αη
3/4
1/4
G1
η, sas2Λ2ρ2ds
≥Λ2ρ2 3/4
1/4
1−ssasds α
1−αη
3/4
1/4
G1
η, sasds
ρ2,
3.36
that is,Tu ≥ρ2, and thenTu1≥ u1, for allu∈∂Ωρ2. By virtue of Theorem2.6, we have
iT,Ωρ2, K
0. 3.37
From3.24,3.35,3.37, andρ∗< ρ2< ρ∗, it follows that
iT,Ωρ∗\Ωρ2, K
1, iT,Ωρ2\Ωρ∗, K
−1. 3.38
Hence,T has fixed pointu1 ∈Ωρ2\Ωρ∗and fixed pointu2∈Ωρ∗\Ωρ2. Obviously,u1,u2are both positive solutions of the problem1.1and
0<u11< ρ2<u21. 3.39
The proof of Theorem3.2is completed.
Theorem 3.3. Let there existd0,a0,b0, andcwith
such that
ft, u, v≤ 1
4Λ1d0, t∈0,1, u∈0, d0, v∈0, d0, 3.41
ft, u, v≥35Λ2a0, t∈0,1, u∈a0, b0, v∈a0, b0, 3.42
ft, u, v≤1
2Λ1c, t∈0,1, u∈0, c, v∈0, c. 3.43
Then problem1.1has at least three positive solutionsu1,u2,u3satisfying
u11< d0, a0< βu2, u31> d0, βu3< a0, 3.44
forλ≤1/21−αηd0.
Proof. Let
βu min
t∈1/4,3/4|ut|, u∈K. 3.45
Then,βis a nonnegative continuous concave functional onKandβu≤ u1for eachu∈K. Letube inKc. Equation3.43implies that
Tut
1
0
G1t, sasf
s, us, usds
αt
1−αη
1
0
G1
η, sasfs, us, usds λt
1−αη
≤ 1 2Λ1
1
0
1−ssasds α
1−αη
1
0
G1
η, sasds
c1/2
1−αηc
1−αη
c.
3.46
Hence,Tu1≤c. This means thatT :Kc → Kc.
Take
u0t 1
2a0b0, t∈0,1. 3.47
Then,
u0∈
u∈Kβ, a0, b0
:βu> a0
/
By3.42, we have, for anyu∈Kβ, a0, b0,
Therefore,ain Theorem2.7holds.
≥ min
So,cin Theorem2.7holds. Thus, by Theorem2.7, we know that the operatorT has at least three positive fixed pointsu1, u2, u3∈Kcsatisfying
u11< d0, a0< βu2, u31> d0, βu3< a0. 3.53
4. Examples
In this section, we give three examples to illustrate our results.
Example 4.1. Consider the problem
whereη1/2,α1. Set
So, the conditionH1is satisfied. Observe
Λ1
Thus, conditionH2is satisfied.
Therefore, by Theorem3.1, the problem4.1has at least two positive solutionsu1and
Example 4.2. Consider the problem
ut 2×581t2sinutut25−ut0, 0< t <1,
u0 u0 0, u1−u
1 2
λ,
4.9
whereη1/2,α1. Set
at 2, ft, u, v 581t2sinutut25−ut. 4.10
Then,
maxf0maxf∞0, 4.11
that is, the conditionH3is satisfied. Moreover,
Λ2
3/4
1/4
1−ssasds α
1−αη
3/4
1/4
G1
η, sasds
−1
3/4
1/4
1−ss2ds 1
1−1/2 1/2
1/4
1−1 2
s2ds
3/4
1/2 1
21−s2ds −1
48
29.
4.12
Taking
ρ28, fort∈0,1, u∈
0, ρ2
, v∈
1 4ρ2, ρ2
, 4.13
we get
ft, u, v≥58825−8828ρ2>2Λ2ρ2 96
29ρ2. 4.14
Thus, conditionH4is satisfied.
Consequently, by Theorem3.2, we see that for
0< λ≤ 1
2
1−αηρ∗≤ 14ρ2, 4.15
the problem4.9has at least two positive solutionsu1andu2such that
That is, the conditions of Theorem3.3are satisfied. Consequently, the problem1.1has at least three positive solutionsu1, u2, u3∈Kcfor
λ≤ 1
2
1−αηd0 1
4 4.21
satisfying
u11<1, 2< βu2, u31>1, βu3<2. 4.22
Acknowledgments
This paper was supported partially by the NSF of China 10771202and the Specialized Research Fund for the Doctoral Program of Higher Education of China2007035805.
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