Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 185701, 14 doc
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Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2010, Article ID 185701, 14 pages
doi:10.1155/2010/185701
Research Article
Positive Solutions for Impulsive Equations of
Third Order in Banach Space
Jingjing Cai
Department of Mathematics, Tongji University, Shanghai 200092, China
Correspondence should be addressed to Jingjing Cai,
Received 4 September 2010; Accepted 30 November 2010
Academic Editor: John Graef
Copyright q 2010 Jingjing Cai. This is an o pen 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.
Using the fixed-point theorem, this paper is devoted t o study the multiple and single positive
solutions of third-order boundary value problems for impulsive differential equations in ordered
Banach spaces. The arguments are based on a specially constructed cone. At last, an example is
given to illustrate the main results.
1. Introduction
The purpose of this paper is to establish the existence of positive solutions for the following
third-order three-point boundary value problems BVP, for short in Banach space E
−x
t
λf
1
t, x
t
,y
t
,t∈
0, 1
\
{
t
1
,t
2
, ,t
m
}
,
−y
t
μf
2
t, x
t
,y
t
,t∈
0, 1
\
{
t
1
,t
2
, ,t
m
}
,
Δx
t
k
−I
1,k
x
t
k
, Δy
t
k
−I
2,k
y
t
k
,k 1, 2, ,m,
x
0
x
0
θ, x
1
− αx
η
θ, y
0
y
0
θ, y
1
− αy
η
θ,
1.1
where f
i
∈ C0, 1 × P × P, P, I
i,k
∈ CP, P, i 1, 2, k 1, 2, ,m. Δx
t
k
x
t
k
− x
t
−
k
,
Δy
t
k
y
t
k
− y
t
−
k
,μ>0, λ>0. θ is the zero element of E.
Recently, third-order boundary value problems cf. 1–9 have attracted many authors
attention due to their wide range of applications in applied mathematics, physics, and
engineering, especially in the bridge issue. To our knowledge, most papers in literature
2AdvancesinDifference Equations
concern mainly about the existence of positive solutions for the cases in which the spaces
are real and the equations have no parameters. And many authors consider nonlinear term
have same linearity. In this paper, we consider the existence of solutions when the nonlinear
terms have different properties, the space is abstract and the equations have two different
parameters.
In 3, Guo et al. studied the following nonlinear three-point boundary value problem:
u
t
a
t
f
u
t
0,
u
0
u
0
0,u
1
αu
η
,
1.2
where a ∈ C0, 1, 0, ∞, f ∈ C0, ∞, 0, ∞. The authors obtained at least one
positive solutions of BVP 1.2 by using fixed-point theorem when f is sublinear or suplinear.
In 8, Yao and Feng used the upper and lower solutions method proved some
existence results for the following third-order two-point boundary value problem
u
t
f
t, u
t
0, 0 ≤ t ≤ 1,
u
0
u
0
u
1
0.
1.3
Inspired by the above work, the aim of this paper is to establish some simple criteria
for the existence of nontrivial solutions for BVP 1.1 under some weaker conditions. The new
features of this paper mainly include the following aspects. Firstly, we consider the system
1.1 in abstract space while 3, 8 talk about equations in real space E R.Secondly,we
obtained the positive solutions when the two parameters have different ranges. Thirdly, f
1
and f
2
in system 1.1 may have different properties. Fourthly, f
i
i 1, 2 in system 1.1 not
only contains x, y but also t, which is much more complicated. Finally, the main technique
used here is the fixed-point theory and a special cone is constructed to study the existence of
nontrivial solutions.
We recall some basic facts about ordered Banach spaces E. The cone P in E induces
a partial order on E,thatis,x ≤ y if and only if y − x ∈ P, P is said to be normal if
there exists a positive constant N such that θ ≤ x ≤ y implies x≤Ny, without loss
of generality, suppose, in present paper, the normal constant N 1. α· denotes the measure
of noncompactness cf. 10.
Some preliminaries and a number of lemmas to the derivation of the main results are
given in Section 2, then the proofs of the theorems are given in Section 3, followed by an
example, in Section 4, to demonstrate the validity of our main results.
2. Preliminaries and Lemmas
In this paper we will consider the Banach space E, ·,denoteJ 0 , 1 and PC
2
J, E{x |
x
∈ CJ, E, x
is continuous at t
/
t
k
and x
is left continuous at t t
k
, the right limit x
t
k
exists, k 1, 2, ,m}.Foranyx ∈ PC
2
J, E we define x
1
sup
t∈J
xt and x, y
2
x
1
y
1
for x, y ∈ PC
2
J, E × PC
2
J, E.
For convenience, let us list the following assumption.
Advances in Difference Equations 3
A f
i
∈ C0, 1 × P × P, P, I
i,k
∈ CP, P, i 1, 2, k 1, 2, ,m.Foranyt ∈ 0, 1
and r>0, ft, P
r
,P
r
{ft, u, v : u, v ∈ P
r
} is relatively compact in E,where
P
r
{x ∈ P |x≤r}.
Lemma 2.1. Assume that αη
/
1, then for any y ∈ C0, 1, the following boundary value problem:
−u
t
y
t
,t∈
0, 1
\
{
t
1
,t
2
, ,t
m
}
,
Δu
t
k
−I
k
u
t
k
,k 1, 2, ,m,
u
0
u
0
θ, u
1
− αu
η
θ
2.1
has a unique solution
u
t
1
0
G
t, s
y
s
ds
m
k1
G
t, t
k
I
k
u
t
k
,
2.2
where
G
t, s
1
2
1 − αη
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
2ts − s
2
1 − αη
t
2
s
α − 1
,s≤ min
η, t
,
t
2
1 − αη
t
2
s
α − 1
,t≤ s ≤ η,
2ts − s
2
1 − αη
t
2
αη − s
,η≤ s ≤ t,
t
2
1 − s
, max
η, t
≤ s.
2.3
Proof. The proof is similar to Lemma 2.2 in 3,weomitit.
Lemma 2.2 see 3. Assume that 0 <η<1 and 1 <α<1/η.Then0 ≤ Gt, s ≤ gs for any
t, s ∈ 0, 1 × 0, 1,wheregs1 α/1 − αηs1 − s, s ∈ 0, 1.
Lemma 2.3 see 3. Let 0 <η<1 and 1 <α<1/η, then for any t, s ∈ η/α, η × 0
, 1,
Gt, s ≥ σgs,where
0 <σ
η
2
2α
2
1 α
min
{
α − 1, 1
}
< 1.
2.4
Inthepaper,wedefineconeK as follows:
K
x ∈ PC
2
J, E
| x
t
≥ θ, x
t
≥ σx
s
,t∈
η
α
,η
,s∈
0, 1
. 2.5
Lemma 2.4 see 10. Let E be a Banach space and K ⊂ E be a cone. Suppose Ω
1
and Ω
2
∈ E are
bounded open sets, θ ∈ Ω
1
, Ω
1
⊂ Ω
2
, A : K ∩ Ω
2
\ Ω
1
→ K is completely continuous such that
4AdvancesinDifference Equations
either
i Au≤u for any u ∈ K ∩ ∂Ω
1
and Au≥u for any u ∈ K ∩ ∂Ω
2
or
ii Au≥u for any u ∈ K ∩ ∂Ω
1
and Au≤u for any u ∈ K ∩ ∂Ω
2
.
Then A has a fixed-point in K ∩
Ω
2
\ Ω
1
.
Lemma 2.5. The vector x, y ∈ PC
2
J, E × PC
2
J, E is a solution of differential systems 1.1 if
and only if x, y ∈ PC
2
J, E is the solution of the following integral systems:
x
t
λ
1
0
G
t, s
f
1
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
1,k
x
t
k
,
y
t
μ
1
0
G
t, s
f
2
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
2,k
y
t
k
.
2.6
Define operators T
1
: K → K, T
2
: K → K and T : K × K → K × K as follows:
T
1
x, y
λ
1
0
G
t, s
f
1
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
1,k
x
t
k
,
T
2
x, y
μ
1
0
G
t, s
f
2
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
2,k
y
t
k
,
T
x, y
t
T
1
x, y
,T
2
x, y
t
.
2.7
As we know, BVP 1.1 has a positive solution x, y ifandonlyifx, y ∈ K × K is the fixed-point
of T.
Lemma 2.6. T : K × K → K × K is completely continuous.
Proof. By condition A we get T
1
x, yt ≥ θ, T
2
x, yt ≥ θ,forallx, y ∈ K.Forany
t ∈ η/α, η,wehave
T
1
x, y
t
1
0
G
t, s
f
1
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
1,k
x
t
k
≥ σ
1
0
g
s
f
1
s, x
s
,y
s
ds σ
m
k1
g
t
k
I
1,k
x
t
k
≥ σ
1
0
G
u, s
f
1
s, x
s
,y
s
ds σ
m
k1
G
u, t
k
I
1,k
x
t
k
σT
1
x, y
u
,u∈
0, 1
.
2.8
Advances in Difference Equations 5
Similarly
T
2
x, y
t
≥ σT
2
x, y
u
,u∈
0, 1
. 2.9
So T : K × K → K × K.
Next, we prove T : K × K → K × K is completely continuous. We first prove that T
1
is
continuous. Let x
n
,y
n
∈ Kn 1, 2, and x
0
,y
0
∈ K such that x
n
,y
n
− x
0
,y
0
2
→
0 n →∞.Letr sup
n
x
n
,y
n
2
,then
x
0
,y
0
2
≤ r,
x
0
1
≤ r,
y
0
1
≤ r,
x
n
1
≤ r,
y
n
1
≤ r. 2.10
By A,weobtain
f
i
t, x
n
t
,y
n
t
−→ f
i
t, x
0
t
,y
0
t
,
n −→ ∞
, for any t ∈
0, 1
,i 1, 2,
I
1,k
x
n
t
k
−→ I
1,k
x
0
t
k
,
n −→ ∞
,k 1, 2, ,m,
I
2,k
y
n
t
k
−→ I
2,k
y
0
t
k
,
n −→ ∞
,k 1, 2, ,m.
2.11
Hence
T
1
x
n
,y
n
t
− T
1
x
0
,y
0
t
1
0
G
t, s
f
1
s, x
n
s
,y
n
s
ds
m
k1
G
t, t
k
I
1,k
x
n
t
k
−
1
0
G
t, s
f
1
s, x
0
s
,y
0
s
ds −
m
k1
G
t, t
k
I
1,k
x
0
t
k
≤
1
0
G
t, s
f
1
s, x
n
s
,y
n
s
− f
1
s, x
0
s
,y
0
s
ds
m
k1
G
t, t
k
I
1,k
x
n
t
k
− I
1,k
x
0
t
k
≤
1
0
g
s
f
1
s, x
n
s
,y
n
s
− f
1
s, x
0
s
,y
0
s
ds
m
k1
g
t
k
I
1,k
x
n
t
k
− I
1,k
x
0
t
k
.
2.12
Since
T
1
x
n
,y
n
− T
1
x
0
,y
0
1
sup
t∈0,1
T
1
x
n
,y
n
t
− T
1
x
0
,y
0
t
.
2.13
6AdvancesinDifference Equations
By 2.11–2.13 and Lebesgue-dominated convergence theorem
T
1
x
n
,y
n
− T
1
x
0
,y
0
1
−→ 0
n −→ ∞
. 2.14
So T
1
is continuous. Similarly, T
2
is continuous. It follows that T is continuous.
Next we prove T is compact. Let V {x
n
,y
n
}⊂K × K be bounded, V
1
{x
n
} and
V
2
{y
n
}.Letx
n
,y
n
2
≤ r for some r>0, then x
n
1
≤ r, y
n
1
≤ r. It is easy to see that
{T
1
x
n
,y
n
t} is equicontinuous. By condition A we have
α
T
1
V
t
α
1
0
G
t, s
f
1
s, x
n
s
,y
n
s
ds
m
k1
G
t, t
k
I
1,k
x
n
t
k
: x
n
∈ V
1
,y
n
∈ V
2
≤ 2
1
0
α
G
t, s
f
1
s, V
1
s
,V
2
s
m
k1
α
G
t, t
k
I
1,k
V
1
t
k
0
2.15
which implies that αT
1
V 0. So, αTV0, it follows that T is compact. The lemma is
proved.
In this paper, denote
f
β
i
lim sup
x
y
→ β
max
t∈0,1
f
t, x, y
x
y
,f
i,β
lim inf
x
y
→ β
min
t∈η/α,η
f
t, x, y
x
y
,
ψf
i
β
lim sup
x
y
→ β
max
t∈0,1
ψ
f
t, x, y
x
y
,
ψf
i
β
lim inf
x
y
→ β
min
t∈η/α,η
ψ
f
t, x, y
x
y
.
I
i,β
k
lim inf
x
→ β
I
i,k
x
x
,I
β
i
k
lim sup
x
→ β
I
i,k
x
x
,k 1, 2, ,m.
2.16
where β 0orβ ∞, ψ ∈ P
∗
{ψ ∈ E
∗
: ψx ≥ θ, ∀x ∈ P} and ψ 1. P
∗
is a dual cone
of P.
We list the assumptions:
H
1
ψf
1
0
>m
1
, ψf
2
∞
>m
2
,wherem
1
,m
2
∈ 0, ∞;
H
2
f
0
i
<m
3
, ψf
1
∞
>m
4
,I
i,0
k0, i 1, 2, where m
3
,m
4
> 0andm
3
m
4
;
H
3
ψf
1
0
>m
5
, f
∞
i
<m
6
, I
∞
i
k0, i 1, 2, where m
5
,m
6
> 0andm
6
m
5
.
Advances in Difference Equations 7
For convenience, denote
a
1
1
4
m
3
1
0
gsds
−1
,α
2
m
4
σ
η
α/η
Gη, sds
−1
,
a
3
m
5
σ
η
η/α
Gη, sds
−1
,α
4
1
4
m
6
1
0
gsds
−1
.
2.17
3. Main Results
Theorem 3.1. Assume that (A), (H
1
) and the following condition H
hold, then BVP 1.1 has at
least two positive solution while λ ∈ 0, 1/4M
1
1
0
gsds and μ ∈ 0 , 1/4M
2
1
0
gsds.
H
: m
1
λσ
η
η/α
Gη, sds ≥ 1; m
2
μσ
η
η/α
Gη, sds ≥ 1;
2
i1
m
k1
gt
k
M
i,k
< 1/2,where
M
i
max
t∈0,1, 0≤uv≤1
f
i
t, u, v > 0, M
i,k
max
0≤u≤1
{I
i,k
u}.
Proof. Let Ω
1
{x, y ∈ K × K : x, y
2
< 1},thenforx, y ∈ ∂Ω
1
,wehave
T
1
x, y
t
≤
λ
1
0
g
s
f
1
x
s
,y
s
ds
m
k1
g
t
k
I
1,k
x
t
k
≤ λM
1
1
0
g
s
ds
m
k1
g
t
k
M
1,k
,
3.1
that is,
T
1
x, y
1
≤ λM
1
1
0
g
s
ds
m
k1
g
t
k
M
1,k
,
3.2
Similarly
T
2
x, y
1
≤ M
2
μ
1
0
g
s
ds
m
k1
g
t
k
M
2,k
.
3.3
So
Tx, y
2
≤
λM
1
μM
2
1
0
g
s
ds
2
i1
m
k1
g
t
k
M
i,k
< 1
x, y
2
.
3.4
Hence
Tx, y
2
<
x, y
2
, for any
x, y
∈ ∂Ω
1
. 3.5
8AdvancesinDifference Equations
Since ψf
1
0
>m
1
,thereexistε
1
> 0and0<R
1
< 1suchthatψf
1
t, u, v ≥ m
1
ε
1
uv for 0 ≤uv≤R
1
and t ∈ η/α, η.LetΩ
2
{x, y ∈ K×K : x, y
2
<R
1
}.
Then for any x, y ∈ ∂Ω
2
,byH
1
and the definition of ψ,weobtain
T
1
x, y
1
≥ ψ
T
1
x, y
η
≥ λ
η
η/α
G
η, s
ψ
f
1
t, x
s
,y
s
ds
≥
m
1
ε
1
λσ
η
η/α
G
η, s
x
1
y
1
ds
R
1
m
1
ε
1
λσ
η
η/α
G
η, s
ds.
3.6
By 3.6 and H
Tx, y
2
≥
T
1
x, y
1
≥ R
1
m
1
ε
1
λσ
η
η/α
G
η, s
ds
>R
1
x, y
2
,
x, y
∈ ∂Ω
2
,
3.7
Similarly, by ψf
2
∞
>m
2
,thereexistε
2
> 0andR
2
> 1suchthatψf
2
t, u, v ≥
m
2
ε
2
u v for t ∈ η/α, η and u, v ∈ P with 0 ≤u v≤R
2
.LetΩ
3
{x, y ∈
K × K : x, y
2
<R
2
}.Thenforanyx, y ∈ ∂Ω
3
,
T
2
x, y
1
≥ R
2
μ
m
2
ε
2
σ
η
η/α
G
η, s
ds.
3.8
So we have by 3.8 and H
Tx, y
2
≥
T
2
x, y
1
>R
2
x, y
2
, for any
x, y
∈ ∂Ω
3
. 3.9
By 3.5, 3.7, 3.9 and Lemma 2.4 wegetthatBVP1.1 has at least two positive solutions
with x
1
,y
1
2
< 1 < x
2
,y
2
2
.
Corollary 3.2. Assume that (A) and the following condition h old, then the conclusion of Theorem 3.1
also holds.
ψf
1
0
>m
1
,ψ
f
2
∞
>m
2
, where m
1
,m
2
∈
0, ∞
. 3.10
Theorem 3.3. Assume that (A) and (H
2
)hold,thenBVP1.1 has at least one positive solution when
λ ∈ a
2
,a
1
and μ ∈ 0,a
1
.
Proof. By Lemma 2.6,weseethatT : K × K → K × K is completely continuous. By H
2
,
there exists r
1
> 0, ε
3
> 0, ε>0suchthatfori 1, 2,
f
i
t, x
t
,y
t
≤
m
3
− ε
3
x
t
y
t
,
I
i,k
x
t
k
≤ ε
x
t
k
, 3.11
Advances in Difference Equations 9
for any x, y ∈ K with 0 ≤x
1
y
1
≤ r
1
,wherem
3
− ε
3
> 0, ε>0suchthat
ε
m
k1
g
t
k
≤
1
2
.
3.12
Let Ω
4
{x, y ∈ K × K : x, y
2
<r
1
}.Thenforanyx, y ∈ ∂Ω
4
,weobtain
T
1
x, y
t
λ
1
0
G
t, s
f
1
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
1,k
x
t
k
≤ λ
1
0
g
s
f
1
s, x
s
,y
s
ds
m
k1
g
t
k
I
1,k
x
t
k
≤ λ
m
3
− ε
3
1
0
g
s
ds
x
t
y
t
ε
m
k1
g
t
k
x
t
k
≤ λ
m
3
− ε
3
1
0
g
s
ds
x
1
y
1
ε
m
k1
g
t
k
x
1
λ
m
3
− ε
3
r
1
1
0
g
s
ds ε
m
k1
g
t
k
x
1
≤
1
4
r
1
ε
m
k1
g
t
k
x
1
,
3.13
Similarly
T
2
x, y
t
≤ μ
m
3
− ε
3
r
1
1
0
g
s
ds ε
m
k1
g
t
k
y
1
≤
1
4
r
1
ε
m
k1
g
t
k
y
1
.
3.14
It follows that
Tx, y
2
T
1
x, y
1
T
2
x, y
1
≤ r
1
x, y
2
, 3.15
which implies
Tx, y
2
≤
x, y
2
, for any
x, y
∈ ∂Ω
4
. 3.16
On the other hand, by ψf
1
∞
>m
4
,thereexistsR>0, ε
4
> 0suchthatψf
1
t, xt,yt ≥
m
4
ε
4
xt yt for x
1
y
1
>Rand t ∈ η/α, η.LetR
1
max{2r
1
,R/σ},
Ω
5
{x, y ∈ K × K : x, y
2
<R
1
}.Foranyx, y ∈ ∂Ω
5
,wehave
x
t
≥ σx
s
,y
t
≥ σy
s
,
x
t
≥ σ
x
s
,
y
t
≥ σ
y
s
,t∈
η
α
,η
,s∈
0, 1
.
3.17
10 Advances in Difference Equations
By the definition of T
1
we get
T
1
x, y
1
≥ ψ
T
1
x, y
η
≥ λ
η
η/α
G
η, s
ψ
f
1
t, x
s
,y
s
ds
≥ λ
m
4
ε
4
η
η/α
G
η, s
x
s
y
s
ds
≥ λ
m
4
ε
4
σ
η
η/α
G
η, s
x
u
y
u
ds.
3.18
So
T
1
x, y
1
≥ λ
m
4
ε
4
σ
η
η/α
G
η, s
x
1
y
1
ds R
1
λ
m
4
ε
4
σ
η
η/α
G
η, s
ds.
3.19
Hence
Tx, y
2
≥
T
1
x, y
1
≥ R
1
λ
m
4
ε
4
σ
η
η/α
G
η, s
ds ≥ R
1
x, y
2
.
3.20
Therefore
Tx, y
2
≥
x, y
2
, ∀
x, y
∈ ∂Ω
5
. 3.21
By 3.16, 3.21 and Lemma 2.4, it is easily seen that T has a fixed-point x
∗
,y
∗
∈ Ω
5
\
Ω
4
.
Corollary 3.4. Let (A) and the following conditions hold, then BVP 1.1 has at least one positive
solution while μ ∈ a
2
,a
1
and λ ∈ 0,a
1
.
f
0
i
<m
3
,
ψf
2
∞
>m
4
,I
i,0
k
0,i 1, 2.
3.22
Theorem 3.5. Let (A) and (H
3
)hold,thenBVP1.1 has at least one positive solution while λ ∈
a
3
,a
4
and μ ∈ 0,a
4
.
Advances in Difference Equations 11
Proof. Since ψf
1
0
>m
5
, we choose R
3
> 0, ε
5
> 0suchthatψf
i
t, u, v ≥ m
5
ε
5
uv
for 0 ≤u v≤R
3
and t ∈ η/α, η.LetΩ
6
{x, y ∈ K × K : x, y
2
<R
3
}.Thenfor
any x, y ∈ ∂Ω
6
,
T
1
x, y
1
≥ ψ
T
1
x, y
η
≥ λ
η
η/α
G
η, s
ψ
f
1
t, x
s
,y
s
ds
≥ λ
m
5
ε
5
σ
η
η/α
G
η, s
x
1
y
1
ds
R
2
λ
m
5
ε
5
σ
η
η/α
G
η, s
ds.
3.23
So
Tx, y
2
T
1
x, y
1
T
2
x, y
1
≥
T
1
x, y
1
≥ R
3
λ
m
5
ε
5
σ
η
η/α
G
η, s
ds
≥ R
3
,
3.24
which implies
Tx, y
2
≥
x, y
2
, ∀
x, y
∈ ∂Ω
6
. 3.25
On the other hand, by f
∞
i
<m
6
and I
∞
i
k0 i 1, 2,thereexistM>0, ε
6
> 0, ε>0such
that m
6
− ε
6
> 0and
f
i
t, x
t
,y
t
≤
m
6
− ε
6
x
t
y
t
,
I
i,k
x
t
k
≤ ε
x
t
k
,
for any
x
1
y
1
x, y
2
≥ M, t ∈
0, 1
,
3.26
where ε satisfies
ε
m
k1
g
t
k
≤
1
2
.
3.27
12 Advances in Difference Equations
Let R
4
max{M, 2R
3
} and Ω
7
{x, y | x, y ∈ K × K : x, y
2
<R
4
}.Thenforany
x, y ∈ ∂Ω
7
,wehave
T
1
x, y
t
λ
1
0
G
t, s
f
1
s, x
s
,y
s
ds
m
k1
G
t, t
k
I
1,k
x
t
k
≤ λ
1
0
g
s
f
1
s, x
s
,y
s
ds
m
k1
g
t
k
I
1,k
x
t
k
≤ λ
m
6
− ε
6
1
0
g
s
ds
x
t
y
t
ε
m
k1
g
t
k
x
t
k
≤ λ
m
6
− ε
6
1
0
g
s
ds
x
1
y
1
ε
m
k1
g
t
k
x
1
λ
m
6
− ε
6
R
4
1
0
g
s
ds ε
m
k1
g
t
k
x
1
≤
1
4
R
4
ε
m
k1
g
t
k
x
1
.
3.28
Similarly
T
2
x, y
t
≤ μ
m
6
− ε
6
R
4
1
0
g
s
ds ε
m
k1
g
t
k
y
1
≤
1
4
R
4
ε
m
k1
g
t
k
y
1
.
3.29
Hence
Tx, y
2
≤
1
2
R
4
εR
4
m
k1
g
t
k
≤ R
4
x, y
2
.
3.30
So
Tx, y
2
≤
x, y
2
, ∀
x, y
∈ ∂Ω
7
. 3.31
By 3.25, 3.31,andLemma 2.4, T has a fixed-point x
∗
,y
∗
∈ Ω
7
\ Ω
6
.
Corollary 3.6. Assume that (A) and the following conditions hold, then BVP 1.1 has at least one
positive solution while μ ∈ a
3
,a
4
and λ ∈ 0,a
4
.
ψf
2
0
>m
5
,f
∞
i
<m
6
,I
∞
i
k
0,i 1, 2. 3.32
Advances in Difference Equations 13
4. An Example
In this section, we construct an example to demonstrate the application of our main results
obtained in Section 3. Consider the following third-order boundary value problem:
−x
n
t
λt
2
e
−t
x
n
t
y
n
t
2
,
−y
n
t
μt
2
e
−t
x
n
t
y
n
t
2
,
Δx
n
1
3
−x
4
n
1
3
, Δy
n
1
3
−y
4
n
1
3
,
Δx
n
0
x
n
0
θ, x
n
1
− 2x
n
2
5
θ,
y
n
0
y
n
0
θ, y
n
1
− 2y
n
2
5
θ.
4.1
Conclusion 1. BVP4.1 has at least one positive solution.
Proof. E R
m
{x x
1
,x
2
, ,x
m
, x
n
∈ R, n 1, 2, ,m}.Definex max
1≤n≤m
|x
n
|.
P {x ∈ E : x
i
> 0, i 1, 2, ,m}. x x
1
,x
2
, ,x
m
, f f
1
,f
2
, ,f
m
. g
n
f
n
t
2
e
−t
x
n
ty
n
t
2
, we know that P
∗
P,letψ 1, 1, ,1,thenforanyx ∈ P,
ψft, x, y
m
k1
f
n
t, x, y. It is easy to see that A is satisfied. On the other hand,
ψ
f
t, x, y
x
y
≥
f
t, x, y
x
y
∞,f
0
lim sup
x
y
→ 0
max
t∈0,1
f
t, x, y
x
y
0,
4.2
that is, ψf
∞
∞. Similarly, g
0
0, it is easy to see that I
i,0
k0, where k 1, i 1, 2. In
this example, α 2, η 2/5, σ η
2
/2α
2
1 α 25/96 and
G
t, s
5
2
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
2ts − s
2
5
t
2
s, s ≤
2
5
,t
,
t
2
t
2
s
5
,t≤ s ≤
2
5
,
2ts − s
2
5
t
2
4
5
− s
,
2
5
≤ s ≤ t,
t
2
1 − s
, max
2
5
,t
≤ s,
4.3
and gs1 αs1 − s/1 − αη15s1 − s.
Let m
1
5,m
2
3000. By computing, we get
a
1
1
4
1
0
5gsds
−1
8
25
,a
2
4
3125
2/5
1/5
G
2
5
,s
ds
−1
.
4.4
14 Advances in Difference Equations
Above all, the conditions of Theorem 3.3 are satisfied. Then for any λ ∈ a
2
, ∞ and μ ∈
0,a
1
,BVP4.1 has at least one positive solution.
Acknowledgments
The author thanks Professor Liu and Professor Lou for many useful discussions and helpful
suggestions. The work was partially supported by NSFC 10971155 and Innovation program
of Shanghai Municipal Education Commission 09ZZ33.
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