Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2011, Article ID 894135, 20 pages
doi:10.1155/2011/894135
Research Article
Solutions to a Three-Point Boundary Value Problem
Jin Liang
1
and Zhi-Wei Lv
2, 3
1
Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
2
Department of Mathematics and Physics, Anyang Institute of Technology, Anyang, Henan 455000, China
3
Department of Mathematics, University of Science a nd 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
Copyright q 2011 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 problem u
tatft, ut,u
t
0, 0 <t<1; u0u
00; u
t
f
t, u
t
,u
t
0, 0 <t<1,
u
0
u
0
0,u
1
y
,x,y∈ Y, t ∈
0, 1
. 1.2
Assume that
H
a ∈ C
0, 1
,
0, ∞
, 0 <
1
0
1 − s
sa
s
ds < ∞,
v
,
min f
0
lim
v → 0
min
t∈0,1
inf
u∈0,∞
f
t, u, v
v
,
max f
∞
lim
v → ∞
max
t∈0,1
sup
u∈0,∞
f
t, u, v
v
u
,
u
, 2.1
where
u
max
t∈0,1
|
u
t
|
,
u
s
f
s, u
s
,u
s
ds
αt
2
2
1 − αη
1
0
G
1
η, s
1
2
⎧
⎨
⎩
2t − t
2
− s
s, s ≤ t,
1 − s
t
2
, t ≤ s,
G
1
t, s
∂G
t, s
∂t
u
1
u
.
Proof. Let u ∈ C
1
0, 1 be the unique solution of 1.1.Thenitisobviousthatut is
nonnegative. By Lemmas 2.1 and 2.2, we have the following.
4AdvancesinDifference Equations
i For t ≤ η,
u
t
1
2
1 − αη
t
0
2ts − s
2
t
2
1 − αη
t
2
s
α − 1
a
s
f
s, u
s
,u
s
ds
t
1
2
1 − αη
t
0
2s
1 − αη
2ts
α − 1
a
s
f
s, u
s
,u
s
ds
1
η
2t
1 − s
a
s
f
s, u
s
2
1 − αη
t
2
s
α − 1
a
s
f
s, u
s
,u
s
ds
ds
1
t
t
2
1 − s
a
s
f
s, u
s
,u
s
ds λt
2
f
s, u
s
,u
s
ds
t
η
2s
1 − αη
2t
αη − s
a
s
s
,u
s
ds 2λt
.
2.6
Advances in Difference Equations 5
On the other hand, for η ≤ s ≤ t,wehave
2s
1 − αη
2t
αη − s
−
2ts − s
2
1 − αη
s
1 − t
2 − t
.
2.7
Since α ∈ 1/2η, 1/η,
2s
1 − αη
2t
αη − s
≥
2ts − s
2
1 − αη
t
2
αη − s
The proof is completed.
Lemma 2.4. Let u ∈ C
1
0, 1 be the unique solution of 1.1.Then
min
t∈1/4,3/4
u
t
≥
1
4
u
1
.
2.10
Proof. From 2.3, it follows that
u
t
1
0
η, s
a
s
f
s, u
s
,u
s
ds
λt
1 − αη
≤
1
0
1 − s
sa
f
s, u
s
,u
s
ds
λ
1 − αη
.
2.11
Hence,
u
1
u
G
1
η, s
a
s
f
s, u
s
,u
s
ds
λ
1 − αη
.
2.12
6AdvancesinDifference Equations
By Lemmas 2.2 and 2.3,weget,foranyt ∈ 1/4, 3/4,
min
t∈1/4,3/4
ds
αt
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
,u
s
ds
α
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
,u
s
ds
λ
≥
1
4
u
1
.
2.15
Set
K
r
{
u ∈ K :
u
1
<r
}
,∂K
r
{
u ∈ K :
u
1
≤ s
,s>r>0.
2.16
Define an operator T by
Tu
t
1
0
G
t, s
a
s
f
s, u
s
,u
,u
s
ds
λt
2
2
1 − αη
.
2.17
Lemma 2.1 implies that 1.1 has a solution u ut if and only if u is a fixed point of T.
From Lemmas 2.1 and 2.2 and the Ascoli-Arzela theorem, the following follow.
Lemma 2.5. The operator defined in 2.17 is completely continuous and satisfies TK ⊂ K.
Advances in Difference Equations 7
Theorem 2.6 see 10. Let E be a real Banach Space, let K ⊂ E be a c one, and Ω
r
{u ∈ K : u≤
r}. Let operator T : K∩Ω
r
→ K be completely continuous and satisfy Tx
/
x, for all x ∈ ∂Ω
r
.Then
i if Tx≤x, for all x ∈ ∂Ω
: βx >a
0
}
/
∅ and βTx >a
0
for x ∈ Pβ, a
0
,b
0
,
b Tx <d
0
for x≤d
0
,
c βTx >a
0
for x ∈ Pβ, a
0
,c with Tx >b
0
.
Then, T has at least three fixed points x
1
,x
2
, and x
3
in P
3. Main Results
In this section, we give new existence theorem about two positive solutions or three positive
solutions for 1.1.
Write
Λ
1
1
0
1 − s
sa
s
ds
α
1 − αη
1
0
G
1
η, s
a
η, s
a
s
ds
−1
.
3.1
Theorem 3.1. Assume that
H
1
min f
0
min f
∞
∞;
H
2
there exists a constant ρ
1
> 0 such that ft, u, v ≤ 1/2Λ
1
ρ
1
,fort ∈ 0, 1, u ∈ 0,ρ
1
t∈0,1
inf
u∈0,∞
f
t, u, v
v
∞,
3.3
8AdvancesinDifference Equations
there is ρ
0
∈ 0,ρ
1
such that
f
t, u, v
≥ 8Λ
2
v, for t ∈
0, 1
,u∈
0, ∞
1
2
1
0
G
1
1
2
,s
a
s
f
s, u
s
,u
s
ds
λ
2
1 − αη
≥
1
0
G
1
1
2
,s
a
s
f
s, u
s
,u
s
ds
≥
1
2
3/4
1/4
1 − s
sa
s
8Λ
2
u
s
ds
α
2
sa
s
ds
α
2
1 − αη
3/4
1/4
G
1
η, s
a
s
ds
1
4
u
1
1
, ∀u ∈ ∂Ω
ρ
0
. 3.8
By Theorem 2.6,wehave
i
T, Ω
ρ
0
,K
0. 3.9
On the other hand, since
min f
∞
lim
v → ∞
min
t∈0,1
inf
u∈0,∞
f
t, u, v
v
∞,
0
.
3.11
Let Ω
ρ
∗
0
{u ∈ K : u
1
<ρ
∗
0
}. Then, by a argument similar to that above, we obtain
Tu
1
≥
u
1
, ∀u ∈ ∂Ω
ρ
∗
0
.
3.12
By Theorem 2.6,
i
1
0
G
1
t, s
a
s
f
s, u
s
,u
s
ds
αt
1 − αη
1
1 − s
sa
s
1
2
Λ
1
ρ
1
ds
α
1 − αη
1
0
G
1
η, s
a
s
1
2
Λ
η, s
a
s
ds
ρ
1
1/2
1 − αη
ρ
1
1 − αη
1
2
ρ
1
1
2
ρ
0
<ρ
1
<ρ
∗
0
, it follows that
i
T, Ω
ρ
∗
0
\ Ω
ρ
1
,K
−1,i
T, Ω
ρ
1
\ Ω
ρ
0
,K
1. 3.16
10 Advances in Difference Equations
1
<
u
2
1
. 3.17
The proof of Theorem 3.1 is completed.
Theorem 3.2. Assume that
H
3
max f
0
max f
∞
0;
H
4
there exists a constant ρ
2
> 0 such that ft, u, v ≥ 2Λ
2
ρ
2
,fort ∈ 0, 1,u ∈ 0,ρ
2
and
v ∈ 1/4ρ
2
t∈0,1
sup
u∈0,∞
f
t, u, v
v
0,
3.19
we see that there exists ρ
∗
∈ 0,ρ
2
such that
f
t, u, v
≤
1
2
Λ
1
v, for t ∈
0, 1
,u∈
ρ
∗
.
3.22
Then Lemmas 2.2 and 2.3 and 3.20 implies that for any u ∈ ∂Ω
ρ
∗
,
Tu
t
1
0
G
1
t, s
a
s
f
s, u
,u
s
ds
λt
1 − αη
Advances in Difference Equations 11
≤
1
0
1 − s
sa
s
1
2
Λ
1
u
s
2
Λ
1
1
0
1 − s
sa
s
ds
α
1 − αη
1
0
G
1
η, s
a
s
ds
1
.
3.23
So Tu
≤u
1
.Hence,Tu
1
≤u
1
,forallu ∈ ∂Ω
ρ
∗
.
Applying Theorem 2.6,wehave
i
T, Ω
ρ
∗
,K
1. 3.24
Next, by
max f
∞
lim
v → ∞
0, ∞
,v≥ r
0
.
3.26
Case 1. max
t∈0,1
ft, u, v is unbounded.
Define a function f
∗
: 0, ∞ → 0, ∞ by
f
∗
ρ
max
f
t, u, v
: t ∈
0, 1
,u,v∈
}, it follows from 3.26–3.28 that
f
t, u, v
≤ f
∗
ρ
∗
≤
1
2
Λ
1
ρ
∗
, for t ∈
0, 1
,u,v∈
0,ρ
∗
.
3.29
12 Advances in Difference Equations
ds
αt
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
,u
s
ds
a
s
1
2
Λ
1
ρ
∗
ds
λ
1 − αη
≤
1
2
Λ
1
1
0
1 − s
sa
s
ds
,andthenTu
1
≤ ρ
∗
.
Case 2. max
t∈0,1
ft, u, v is bounded.
In this case, there exists an M>0suchthat
max
t∈0,1
f
t, u, v
≤ M, for t ∈
0, 1
,u,v∈
0, ∞
.
3.31
Choosing ρ
∗
≥ max{2ρ
2
, 2M/Λ
ds
αt
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
,u
s
ds
ρ
∗
2
≤
ρ
∗
2
ρ
∗
2
ρ
∗
,
3.32
which implies Tu
≤ρ
∗
,andthenTu
1
≤ ρ
∗
.
Therefore, in both cases, taking
Ω
ρ
T, Ω
ρ
∗
,K
1. 3.35
Finally, put Ω
ρ
2
{u ∈ K : u
1
<ρ
2
}.ThenH
4
implies that
Tu
1
2
1
0
G
1
1
1
η, s
a
s
f
s, u
s
,u
s
ds
λ
2
1 − αη
≥
3/4
1/4
2Λ
2
ρ
2
ds
≥ Λ
2
ρ
2
3/4
1/4
1 − s
sa
s
ds
α
1 − αη
3/4
1/4
G
1
0. 3.37
From 3.24, 3.35, 3.37,andρ
∗
<ρ
2
<ρ
∗
, it follows that
i
T, Ω
ρ
∗
\ Ω
ρ
2
,K
1,i
T, Ω
ρ
2
\ Ω
ρ
∗
,K
−1. 3.38
<
u
2
1
. 3.39
The proof of Theorem 3.2 is completed.
Theorem 3.3. Let there exist d
0
, a
0
, b
0
,andc with
0 <d
0
<a
0
< 32a
0
<b
0
≤ c, 3.40
14 Advances in Difference Equations
such that
f
t, u, v
0
,t∈
0, 1
,u∈
a
0
,b
0
,v∈
a
0
,b
0
, 3.42
f
t, u, v
≤
1
2
Λ
1
c, t ∈
u
2
,
u
3
1
>d
0
,β
u
3
<a
0
, 3.44
for λ ≤ 1/21 − αηd
0
.
Proof. Let
β
u
min
t∈1/4,3/4
s
f
s, u
s
,u
s
ds
αt
1 − αη
1
0
G
1
η, s
a
s
ds
α
1 − αη
1
0
G
1
η, s
a
s
ds
c
1/2
1 − αη
c
1 − αη
c.
0
∈
u ∈ K
β, a
0
,b
0
: β
u
>a
0
/
∅. 3.48
Advances in Difference Equations 15
By 3.42,wehave,foranyu ∈ Kβ, a
0
,b
0
,
β
Tu
min
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
,u
s
ds
λt
2
s
ds
αt
2
2
1 − αη
1
0
G
1
η, s
a
s
f
s, u
s
a
s
35Λ
2
a
0
ds
α
2
1 − αη
1
4
2
3/4
1/4
G
1
η, s
a
s
1
t, s
a
s
f
s, u
s
,u
s
ds
αt
1 − αη
1
0
G
1
s
1
4
Λ
1
d
0
ds
α
1 − αη
1
0
G
1
η, s
a
s
1
4
Λ
1
d
0
ds
0
,wehave
β
Tu
min
t∈1/4,3/4
1
0
G
t, s
a
s
f
s, u
s
,u
s
s
ds
λt
2
2
1 − αη
16 Advances in Difference Equations
≥ min
t∈1/4,3/4
1
0
1
2
t
2
1 − s
sa
s
s
f
s, u
s
,u
s
ds
λt
2
2
1 − αη
≥
1
2
×
1
16
η, s
a
s
f
s, u
s
,u
s
ds
λ
1 − αη
≥
1
32
1
0
η, s
a
s
f
s, u
s
,u
s
ds
λ
1 − αη
≥
1
32
1
0
η, s
a
s
f
s, u
s
,u
s
ds
λt
1 − αη
1
32
Tu
t
3
∈ K
c
satisfying
u
1
1
<d
0
,a
0
<β
u
2
,
u
3
1
>d
0
,β
u
3
u
t
2
0, 0 <t<1,
u
0
u
0
0,u
1
− u
1
2
u
t
1/2
u
t
2
.
4.2
Then,
min f
0
min f
∞
∞. 4.3
So, the condition H
1
is satisfied. Observe
Λ
1
1
0
1 − s
sa
s
ds
α
1 − αη
η
0
1 − η
sa
s
ds
1
η
1/2
0
1 −
1
2
s
1
10
ds
1
1/2
1
2
1 − s
1
10
ds
−1
24.
4.4
Taking
ρ
1
2
Λ
1
ρ
1
12ρ
1
.
4.6
Thus, condition H
2
is satisfied.
Therefore, by Theorem 3.1,theproblem4.1 has at least two positive solutions u
1
and
u
2
such that
0 <
u
1
1
< 4 <
u
2
1
t
u
t
2
5
−u
t
0, 0 <t<1,
u
0
u
0
0,u
1
t
2
5
−u
t
.
4.10
Then,
max f
0
max f
∞
0, 4.11
that is, the condition H
3
is satisfied. Moreover,
Λ
2
3/4
1/4
1 − s
sa
1
1 −
1/2
1/2
1/4
1 −
1
2
s2ds
3/4
1/2
1
2
1 − s
2ds
−1
48
29
.
8
2
5
−8
8
2
8ρ
2
> 2Λ
2
ρ
2
96
29
ρ
2
.
4.14
Thus, condition H
4
is satisfied.
Consequently, by Theorem 3.2,weseethatfor
0 <λ≤
1
2
1 − αη
ρ
6
5
, Λ
2
48
29
.
4.17
Let
d
0
1,a
0
2,b
0
99,c
1425
6
,
f
t, u, v
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1 − t
10
t
2
u
10
tv
10
,t∈
0, 1
,u∈
0, 1
,v∈
1, 2
,
1 − t
10
t
2
10
t
10
140,t∈
0, 1
,u∈
2, 99
,v∈
2, 99
,
1 − t
10
0, 1
,u∈
99, ∞
,v∈
99, ∞
.
4.18
Then,
f
t, u, v
1 − t
10
t
2
u
10
tv
10
≤
1
f
t, u, v
1 − t
10
t
2
10
t
10
140 ≥ 140 > 35Λ
2
a
0
35 ×
48
29
× 2,t∈
0, 1
,u∈
2, 99
,v∈
1 t
2
u
2
≤
3
10
140 2
1423
10
<
1
2
Λ
1
c
1
2
×
6
5
×
1425
6
,t∈
,v∈
0,c
.
4.20
20 Advances in Difference Equations
That is, the conditions of Theorem 3.3 are satisfied. Consequently, the problem 1.1 has at
least three positive solutions u
1
,u
2
,u
3
∈ K
c
for
λ ≤
1
2
1 − αη
d
0
1
4
4.21
2002.
3 J. Henderson, Ed., Boundary Value Problems for Functional-Differential Equations, World Scientific, River
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