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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2011, Article ID 456426, 19 pages
doi:10.1155/2011/456426
Research Article
Positive Solutions of nth-Order Nonlinear
Impulsive Differential Equation with Nonlocal
Boundary Conditions
Meiqiang Feng,
1
Xuemei Zhang,
2
and Xiaozhong Yang
2
1
School of Science, Beijing Information Science & Technology University, Beijing 100192, China
2
Department of Mathematics and Physics, North China E lectric Power University, Beijing 102206, China
Correspondence should be addressed to Meiqiang Feng,
Received 25 March 2010; Accepted 9 May 2010
Academic Editor: Feliz Manuel Minh
´
os
Copyright q 2011 Meiqiang Feng et al. 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.
This paper is devoted to study the existence, nonexistence, and multiplicity of positive solutions
for the nth-order nonlocal boundary value problem with impulse effects. The arguments are based
upon fixed point theorems in a cone. An example is worked out to demonstrate the main results.
1. Introduction
The theory of impulsive differential equations describes processes which experience a sudden


change of their state at certain moments. Processes with such a character arise naturally
and often, especially in phenomena studied in physics, chemical technology, population
dynamics, biotechnology, and economics. For an introduction of the basic theory of impulsive
differential equations, see Lakshmikantham et al. 1; for an overview of existing results and
of recent research areas of impulsive differential equations, see Benchohra et al. 2.The
theory of impulsive differential equations has become an important area of investigation in
the recent years and is much richer than the corresponding theory of differential equations;
see, for instance, 3–14 and their references.
At the same time, a class of boundary value problems with integral boundary
conditions arise naturally in thermal conduction problems 15, semiconductor problems
16, hydrodynamic problems 17. Such problems include two, three, and multipoint
boundary value problems as special cases and attract much attention; see, for instance,
7, 8, 11, 18–44 and references cited therein. In particular, we would like to mention some
results of Eloe and Ahmad 19 and Pang et al. 22.In19, by applying the fixed point
2 Boundary Value Problems
theorem in cones due to the work of Krasnosel’kii and Guo, Eloe and Ahmad established the
existence of positive solutions of the following nth boundary value problem:
x
n

t

 a

t

f

t, x


t

 0,t∈

0, 1

,
x

0

 x


0

 ···  x
n−2

0

 0,
x

1

 αx

η


.
1.1
In 22, Pang et al. considered the expression and properties of Green’s function for
the nth-order m-point boundary value problem
x
n

t

 a

t

f

x

t

 0, 0 <t<1,
x

0

 x


0

 ···  x

n−2

0

 0,
x

1


m−2

i1
β
i
x

ξ
i

,
1.2
where 0 <ξ
1

2
< ··· <ξ
m−2
< 1,β
i

> 0,

m−2
i1
β
i
ξ
m−1
i
< 1. Furthermore, they obtained the
existence of positive solutions by means of fixed point index theory.
Recently, Yang and Wei 23 and the author of 24 improved and generalized the
results of Pang et al. 22 by using different methods, respectively.
On the other hand, it is well known that fixed point theorem of cone expansion and
compression of norm type has been applied to various boundary value problems to show the
existence of positive solutions; for example, see 7, 8, 11, 19, 23, 24. However, there are few
papers investigating the existence of positive solutions of nth impulsive differential equations
by using the fixed point theorem of cone expansion and compression. The objective of the
present paper is to fill this gap. Being directly inspired by 19, 22, using of the fixed point
theorem of cone expansion and compression, this paper is devoted to study a class of nonlocal
BVPs for nth-order impulsive differential equations with fixed moments.
Consider the following nth-order impulsive differential equations with integral
boundary conditions:
x
n

t

 f


t, x

t

 0,t∈ J, t
/
 t
k
,
−Δx
n−1
|
tt
k
 I
k

x

t
k

,k 1, 2, ,m,
x

0

 x



0

 ···  x
n−2

0

 0,x

1



1
0
h

t

x

t

dt.
1.3
Here J 0, 1,f ∈ CJ × R

,R

,I

k
∈ CR

,R

, and R

0, ∞,t
k
k  1, 2, ,m
where m is fixed positive integer are fixed points with 0 <t
1
<t
2
< ··· <t
k
< ··· <t
m
<
1, Δx
n−1
|
tt
k
 x
n−1
t

k
 − x

n−1
t

k
, where x
n−1
t

k
 and x
n−1
t

k
 represent the right-hand
limit and left-hand limit of x
n−1
t at t  t
k
, respectively, h ∈ L
1
0, 1 is nonnegative.
For the case of h ≡ 0, problem 1.3 reduces to the problem studied by Samo
˘
ılenko and
Perestyuk in 4. By using the fixed point index theory in cones, the authors obtained some
Boundary Value Problems 3
sufficient conditions for the existence of at least one or two positive solutions to the two-point
BVPs.
Motivated by the work above, in this paper we will extend the results of 4, 19 , 22–

24 to problem 1.3. On the other hand, it is also interesting and important to discuss the
existence of positive solutions for problem 1.3 when I
k
/
 0 k  1, 2, ,m,,n≥ 2, and
h
/
≡ 0. Many difficulties occur when we deal with them; for example, the construction of
cone and operator. So we need to introduce some new tools and methods to investigate the
existence of positive solutions for problem 1.3. Our argument is based on fixed point theory
in cones 45.
To obtain positive solutions of 1.3, the following fixed point theorem in cones is
fundamental which can be found in 45, page 93.
Lemma 1.1. Let Ω
1
and Ω
2
be two bounded open sets in Banach space E, such that 0 ∈ Ω
1
and
Ω
1
⊂ Ω
2
.LetP be a cone in E and let operator A : P ∩ Ω
2
\ Ω
1
 → P be completely continuous.
Suppose that one of the following two conditions is satisfied:

i Ax
/
≥x, ∀x ∈ P ∩ ∂Ω
1
; Ax
/
≤x, ∀x ∈ P ∩ ∂Ω
2
;
ii Ax
/
≤x, ∀x ∈ P ∩ ∂Ω
1
; Ax
/
≥x, ∀x ∈ P ∩ ∂Ω
2
.
Then, A has at least one fixed point in P ∩ Ω
2
\ Ω
1
.
2. Preliminaries
In order to define the solution of problem 1.3, we will consider the following space.
Let J

 J \{t
1
,t

2
, ,t
n
},and
PC
n−1

0, 1



x ∈ C

0, 1

: x
n−1
|
t
k
,t
k1

∈ C

t
k
,t
k1


,
x
n−1

t

k

 x
n−1

t
k

, ∃ x
n−1

t

k


,k 1, 2, ,m.
2.1
Then PC
n−1
0, 1 is a real Banach space with norm

x


pc
n−1
 max


x


,


x




,


x




, ,



x
n−1






,
2.2
where x
n−1


 sup
t∈J
|x
n−1
t|,n 1, 2,
A function x ∈ PC
n−1
0, 1 ∩ C
n
J

 is called a solution of problem 1.3 if it satisfies
1.3.
To establish the existence of multiple positive solutions in PC
n−1
0, 1 ∩ C
n
J


 of
problem 1.3, let us list the following assumptions:
H
1
 f ∈ CJ × R

,R

,I
k
∈ CR

,R

;
H
2
 μ ∈ 0, 1, where μ 

1
0
htt
n−1
dt.
4 Boundary Value Problems
Lemma 2.1. Assume that H
1
 and H
2
 hold. Then x ∈ PC

n−1
0, 1 ∩ C
n
J

 is a solution of
problem 1.3 if and only if x is a solution of the following impulsive integral equation:
x

t



1
0
H

t, s

f

s, x

s

ds 
m

k1
H


t, t
k

I
k

x

t
k

, 2.3
where
H

t, s

 G
1

t, s

 G
2

t, s

, 2.4
G

1

t, s


1

n − 1

!



t
n−1

1 − s

n−1


t − s

n−1
, 0 ≤ s ≤ t ≤ 1,
t
n−1

1 − s


n−1
, 0 ≤ t ≤ s ≤ 1,
2.5
G
2

t, s


t
n−1
1 −

1
0
h

t

t
n−1
dt

1
0
h

t

G

1

t, s

dt.
2.6
Proof. First suppose that x ∈ PC
n−1
0, 1 ∩ C
n
J

 is a solution of problem 1.3.Itiseasyto
see by integration of 1.3 that
x
n−1

t

 x
n−1

0



t
0
f


s, x

s

ds 

0<t
k
<t

x
n−1

t

k

− x
n−1

t
k


 x
n−1

0




t
0
f

s, x

s

ds −

0<t
k
<t
I
k

x

t
k

.
2.7
Integrating again and by boundary conditions, we can get
x
n−2

t


 x
n−1

0

t −

t
0

t − s

f

s, x

s

ds −

0<t
k
<t
I
k

x

t
k


t − t
k

.
2.8
Similarly, we get
x

t

 −
1

n − 1

!

t
0

t − s

n−1
f

s, x

s


ds  x
n−1

0

t
n−1

n − 1

!


t
k
<t
I
k

x

t
k

t − t
k

n−1

n − 1


!
. 2.9
Letting t  1in2.9,wefind
x
n−1

0



n − 1

!x

1



1
0

1 − s

n−1
f

s, x

s


ds


t
k
<1
I
k

x

t
k

1 − t
k

n−1
.
2.10
Boundary Value Problems 5
Substituting x1

1
0
htxtdt and 2.10 into 2.9,weobtain
x

t


 −
1

n − 1

!

t
0

t − s

n−1
f

s, x

s

ds 
t
n−1

n − 1

!


n − 1


!

1
0
h

t

x

t

dt


1
0

1 − s

n−1
f

s, x

s

ds 


t
k
<1
I
k

x

t
k

1 − t
k

n−1



t
k
<t
I
k

x

t
k

t − t

k

n−1

n − 1

!


1
0
G
1

t, s

f

s, x

s

ds 
m

k1
G
1

t, t

k

I
k

x

t
k

 t
n−1

1
0
h

t

x

t

dt.
2.11
Multiplying 2.11 with ht and integrating it, we have

1
0
h


t

x

t

dt 

1
0
h

t


1
0
G
1

t, s

f

s, x

s

ds dt 


1
0
h

t

m

k1
G
1

t, t
k

I
k

x

t
k

dt


1
0
h


t

t
n−1
dt

1
0
h

t

x

t

dt,
2.12
that is,

1
0
h

t

x

t


dt 
1
1 −

1
0
h

t

t
n−1
dt


1
0
h

t


1
0
G
1

t, s


f

s, x

s

ds dt


1
0
h

t

m

k1
G
1

t, t
k

I
k

x

t

k

dt

.
2.13
Then we have
x

t



1
0
G
1

t, s

f

s, x

s

ds 
m

k1

G
1

t, t
k

I
k

x

t
k


t
n−1
1 −

1
0
h

t

t
n−1
dt



1
0
h

t


1
0
G
1

t, s

f

s, x

s

ds dt 

1
0
h

t

m


k1
G
1

t, t
k

I
k

x

t
k

dt

.
2.14
Then, the proof of sufficient is complete.
6 Boundary Value Problems
Conversely, if x is a solution of 2.3,directdifferentiation of 2.3 implies that, for
t
/
 t
k
,
x



t


1

n − 2

!

t
0

t
n−2

1 − s

n−1


t − s

n−2

f

s, x

s


ds

1

n − 2

!

1
t
t
n−2

1 − s

n−1
f

s, x

s

ds

1

n − 2

!


t
k
<t

t
n−2

1 − t
k

n−1


t − t
k

n−2

I
k

x

t
k


1

n − 2


!

t
k
≥t
t
n−2

1 − t
k

n−1
I
k

x

t
k



n − 1

t
n−2
1 −

1

0
h

t

t
n−1
dt


1
0
h

t


1
0
G
1

t, s

f

s, x

s


ds dt


1
0
h

t

m

k1
G
1

t, t
k

I
k

x

t
k

dt

,
.

.
.
x
n−1

t



1
0

1 − s

n−1
f

s, x

s

ds −

t
0
f

s, x

s


ds 

t
k
<1

1 − t
k

n−1
I
k

x

t
k



t
k
<t
I
k

x

t

k



n − 1

!
1 −

1
0
h

t

t
n−1
dt


1
0
h

t


1
0
G

1

t, s

f

s, x

s

ds dt


1
0
h

t

m

k1
G
1

t, t
k

I
k


x

t
k

dt

.
2.15
Evidently,
Δx
n−1
|
tt
k
 −I
k

x

t
k

,

k  1, 2, ,m

,
2.16

x
n

t

 −f

t, x

t

.
2.17
So x ∈ C
n
J

 and Δx
n−1
|
tt
k
 −I
k
xt
k
, k  1, 2, ,m, and it is easy to verify that
x0x

0··· x

n−2
00,x1

1
0
htxtdt, and the lemma is proved.
Similar to the proof of that from 22, we can prove that Ht, s,G
1
t, s,andG
2
t, s
have the following properties.
Proposition 2.2. The function G
1
t, s defined by 2.5 satisfyong G
1
t, s ≥ 0 is continuous for all
t, s ∈ 0, 1,G
1
t, s > 0, ∀t, s ∈ 0, 1.
Boundary Value Problems 7
Proposition 2.3. There exists γ>0 such that
min
t∈t
m
,1
G
1

t, s


≥ γG
1

τ

s

,s

, ∀s ∈

0, 1

,
2.18
where τs is defined in 2.20.
Proposition 2.4. If μ ∈ 0, 1, then one has
i G
2
t, s ≥ 0 is continuous for all t, s ∈ 0, 1,G
2
t, s > 0, ∀t, s ∈ 0, 1;
ii G
2
t, s ≤ 1/1 − μ

1
0
htG

1
t, sdt, ∀t ∈ 0, 1,s∈ 0, 1.
Proof. From the properties of G
1
t, s and the definition of G
2
t, s, we can prove that the
results of Proposition 2.4 hold.
Proposition 2.5. If μ ∈ 0, 1, the function Ht, s defined by 2.4 satisfies
i Ht, s ≥ 0 is continuous for all t, s ∈ 0, 1,Ht, s > 0, ∀t, s ∈ 0, 1;
ii Ht, s ≤ Hs ≤ H
0
for each t, s ∈ 0, 1, and
min
t∈t
m
,1
H

t, s

≥ γ

H

s

, ∀s ∈

0, 1


,
2.19
where γ

 min{γ,t
n−1
m
}, and
H

s

 G
1

τ

s

,s

 G
2

1,s



s



s
1 −

1 − s

11/n−2
,H
0
 max
s∈J
H

s

,
2.20
γ is defined in Proposition 2.3.
Proof. i From Propositions 2.2 and 2.4,weobtainthatHt, s ≥ 0 is continuous for all t, s ∈
0, 1,andHt, s > 0, ∀t, s ∈ 0, 1.
ii From ii of Proposition 2.2 and ii of Proposition 2.4, we have Ht, s ≤ Hs for
each t, s ∈ 0, 1.
Now, we show that 2.19 holds.
In fact, from Proposition 2.3, we have
min
t∈t
m
,1
H


t, s

≥ γG
1

τ

s

,s


t
n−1
m
1 − μ

1
0
h

t

G
1

t, s

dt

≥ γ


G
1

τ

s

,s


1
1 − μ

1
0
h

t

G
1

t, s

dt

 γ


H

s

, ∀s ∈

0, 1

.
2.21
Then the proof of Proposition 2.5 is completed.
Remark 2.6. From the definition of γ

, it is clear that 0 <γ

< 1.
8 Boundary Value Problems
Lemma 2.7. Assume that H
1
 and H
2
 hold. Then, the solution x of problem 1.3 satisfies xt ≥
0, ∀t ∈ J.
Proof. It is an immediate subsequence of the facts that Ht, s ≥ 0on0, 1 × 0, 1.
Remark 2.8. From ii of Proposition 2.5, one can find that
γ

H


s

≤ H

t, s

≤ H

s

,t∈

t
m
, 1

,s∈ J. 2.22
For the sake of applying Lemma 1.1, we construct a cone K in PC
n−1
0, 1 by
K 

x ∈ PC
n−1

0, 1

: x ≥ 0,x

t


≥ γ

x

s

,t∈

t
m
, 1

,s∈ J

. 2.23
Define T : K → K by

Tx

t



1
0
H

t, s


f

s, x

s

ds 
m

k1
H

t, t
k

I
k

x

t
k

. 2.24
Lemma 2.9. Assume that H
1
 and H
2
 hold. Then, TK ⊂ K, and T : K → K is completely
continuous.

Proof. From Proposition 2.5 and 2.24, we have
min
t∈t
m
,1

Tx

t

 min
t∈

t
m
,1


1
0
H

t, s

f

s, x

s


ds 
m

k1
H

t, t
k

I
k

x

t
k



1
0
min
t∈

t
m
,1

H


t, s

f

s, x

s

ds 
m

k1
min
t∈

t
m
,1

H

t, t
k

I
k

x

t

k

≥ γ



1
0
H

s

f

s, x

s

ds 
m

k1
H

t
k

I
k


x

t
k


≥ γ



1
0
max
t∈

0,1

H

t, s

f

s, x

s

ds 
m


k1
max
t∈

0,1

H

t, t
k

I
k

x

t
k


≥ γ

max
t∈0,1


1
0
H


t, s

f

s, x

s

ds 
m

k1
H

t, t
k

I
k

x

t
k


 γ


Tx


, ∀x ∈ K.
2.25
Thus, TK ⊂ K.
Next, by similar arguments to those in 8 one can prove that T : K → K is completely
continuous. So it is omitted, and the lemma is proved.
Boundary Value Problems 9
3. Main Results
Write
f
β
 lim sup
x → β
max
t∈J
f

t, x

x
,f
β
 lim inf
x → β
min
t∈J
f

t, x


x
,
I
β

k

 lim inf
x → β
I
k

x

x
,I
β

k

 lim sup
x → β
I
k

x

x
,
3.1

where β denotes 0

or ∞.
In this section, we apply Lemma 1.1 to establish the existence of positive solutions for
BVP 1.3.
Theorem 3.1. Assume that H
1
 and H
2
 hold. In addition, letting f and I
k
satisfy the following
conditions:
H
3
 f
0
 0 and I
0
k0,k 1, 2, ,m;
H
4
 f

 ∞ or I

k∞,k 1, 2, ,m,
BVP 1.3 has at least one positive solution.
Proof. Considering H
3

, there exists η>0 such that
f

t, x

≤ εx, I
k

x

≤ ε
k
x, k  1, 2, ,m, ∀0 ≤ x ≤ η, t ∈ J, 3.2
where ε, ε
k
> 0satisfy
max
{
H
0
, 1  G
0
}

ε 
m

k1
ε
k


< 1;
3.3
here
G
0
 max

G
1
0
,G
2
0
, ,G
n−1
0

,
G
1
0
 max
t,s∈J,t
/
 s
G

2t


t, s

 max
t,s∈J,t
/
 s

n − 1

t
n−2
1 − μ

1
0
h

t

G
1

t, s

dt,
G
2
0
 max
t,s∈J,t

/
 s
G

2t

t, s

 max
t,s∈J,t
/
 s

n − 1

n − 2

t
n−3
1 − μ

1
0
h

t

G
1


t, s

dt,
.
.
.
G
n−1
0
 max
t,s∈J,t
/
 s
G
n−1
2t

t, s

 max
t,s∈J,t
/
 s

n − 1

!
1 − μ

1

0
h

t

G
1

t, s

dt.
3.4
10 Boundary Value Problems
Now, for 0 <r<η, we prove that
Tx
/
≥x, x ∈ K,

x

pc
n−1
 r.
3.5
In fact, if there exists x
1
∈ K, x
1

pc

n−1
 r such that Tx
1
≥ x
1
. Noticing 3.2, then we have
0 ≤ x
1

t



1
0
H

t, s

f

s, x
1

s

ds 
m

k1

H

t, t
k

I
k

x
1

t
k

≤ εr

1
0
H

s

ds  r
m

k1
H

t
k


ε
k
≤ rH
0

ε 
m

k1
ε
k

<r

x
1

pc
n−1
,


x

1

t






1
0


H

t

t, s



f

s, x
1

s

ds 
m

k1


H


t

t, t
k



I
k

x
1

t
k



1
0



G

1t

t, s







G

2t

t, s




f

s, x
1

s

ds

m

k1



G


1t

t, t
k






G

2t

t, t
k




I
k

x
1

t
k




1
0

1  G
1
0

f

s, x
1

s

ds 
m

k1

1  G
1
0

I
k

x
1


t
k

≤ r

1  G
1
0


ε 
m

k1
ε
k

<r

x
1

pc
n−1
,


x

1


t





1
0


H

t

t, s



f

s, x
1

s

ds 
m

k1



H

t

t, t
k



I
k

x
1

t
k



1
0



G

1t


t, s






G

2t

t, s




f

s, x
1

s

ds

m

k1




G

1t

t, t
k






G

2t

t, t
k




I
k

x
1


t
k

Boundary Value Problems 11


1
0

1  G
2
0

f

s, x
1

s

ds 
m

k1

1  G
2
0

I

k

x
1

t
k

≤ r

1  G
2
0


ε 
m

k1
ε
k

<r

x
1

pc
n−1
,

.
.
.



x
n−1
1

t






1
0



H
n−1
t

t, s





f

s, x
1

s

ds 
m

k1



H
n−1
t

t, t
k




I
k

x
1


t
k



1
0




G
n−1
1t

t, s








G
n−1
2t

t, s






f

s, x
1

s

ds

m

k1




G
n−1
1t

t, t
k









G
n−1
2t

t, t
k





I
k

x
1

t
k



1
0


1  G
n
0

f

s, x
1

s

ds 
m

k1

1  G
n
0

I
k

x
1

t
k

≤ r


1  G
n
0


ε 
m

k1
ε
k

<r

x
1

pc
n−1
, 3.6
where
G

1t

t, s


1


n − 2

!



t
n−2

1 − s

n−1


t − s

n−2
if 0 ≤ s ≤ t ≤ 1,
t
n−2

1 − s

n−1
if 0 ≤ t ≤ s ≤ 1,
G

1t


t, s


1

n − 3

!



t
n−3

1 − s

n−1


t − s

n−3
if 0 ≤ s ≤ t ≤ 1,
t
n−3

1 − s

n−1
if 0 ≤ t ≤ s ≤ 1,

.
.
.
G
n−1
1t

t, s






1 − s

n−1
− 1if0≤ s ≤ t ≤ 1,

1 − s

n−1
if 0 ≤ t ≤ s ≤ 1,
max
t,s∈J,t
/
 s




G
N
1t

t, s




 1,N 1, 2, ,n− 1.
3.7
Therefore, x
1

pc
n−1
< x
1

pc
n−1
, which is a contraction. Hence, 3.2 holds.
12 Boundary Value Problems
Next, turning to H
4
.Case1. f

 ∞. There exists τ>0 such that
f


t, x

≥ Mx, t ∈ J, x ≥ τ, 3.8
where M>γ

H
0
1 − t
m

−1
. Choose
R>max

r, τ

γ


−1

. 3.9
We show that
Tx
/
≤x, x ∈ K,

x

pc

n−1
 R.
3.10
In fact, if there exists x
0
∈ K, x
0

pc
n−1
 R such that Tx
0
≤ x
0
, then
x
0

t

≥ γ

x
0

s

,t∈

t

m
, 1

,s∈ J. 3.11
This and 3.9 imply that
min
t∈t
m
,1
x
0

t

≥ γ


x
0

pc
n−1
 γ

R>τ.
3.12
So, we have
t ∈ J ⇒ x
0


t



Tx
0

t

≥ min
t∈

t
m
,1


1
t
m
H

t, s

f

s, x
0

s


ds ≥ γ

H
0
M

1
t
m
x
0

s

ds,
3.13
that is,

1
t
m
x
0

t

dt ≥ γ

H

0
M

1 − t
m


1
t
m
x
0

s

ds.
3.14
It is easy to see that

1
t
m
x
0

s

ds > 0.
3.15
In fact, if


1
t
m
x
0
sds  0, then x
0
t0, for t ∈ t
m
, 1. Since x
0
∈ K, x
0
s0, ∀s ∈ J.
Hence, x
0

pc
n−1
 x
n−1
0


 x
0


 0, which contracts x

0

pc
n−1
 R. So, 3.15 holds.
Therefore, M ≤ γ

H
0
1 − t
m

−1
, this is also a contraction. Hence, 3.10 holds.
Case 2. I

k∞,k 1, 2, ,m. There exists τ
1
> 0 such that
I
k

x

≥ M
k
x, x ≥ τ
1
, 3.16
Boundary Value Problems 13

where M
k
> γ

H
0

−1
,k  1, 2, ,m. If we define M

 min{M
k
: k  1, 2, ,m}, then
M

> γ

H
0

−1
. Choose
R>max

r, τ
1

γ



−1

. 3.17
We prove that 3.10 holds.
In fact, if there exists x
00
∈ K, x
00

pc
n−1
 R such that Tx
00
≤ x
00
, then
x
00

t

≥ γ

x
00

s

,t∈


t
m
, 1

,s∈ J. 3.18
This and 3.17 imply that
min
t∈t
m
,1
x
00

t

≥ γ


x
00

pc
n−1
 γ

R>τ
1
.
3.19
So, we have

t ∈ J ⇒ x
00

t



Tx
00

t

≥ min
t∈t
m
,1
m

k1
H

t, t
k

I
k

x
00


t
k

≥ γ

H
0
m

k1
M
k
x
00

t
k

≥ γ

H
0
M

m

k1
x
00


t
k

.
3.20
From 3.20,weobtainthat
x
00

t
1

≥ γ

H
0
M

m

k1
x
00

t
k

,
x
00


t
2

≥ γ

H
0
M

m

k1
x
00

t
k

,
.
.
.
x
00

t
k

≥ γ


H
0
M

m

k1
x
00

t
k

.
3.21
So, we have
m

k1
x
00

t
k

≥ mγ

H
0

M

m

k1
x
00

t
k

.
3.22
14 Boundary Value Problems
From the definition of M

, we can find that
m

k1
x
00

t
k

>m
m

k1

x
00

t
k

,x
00
∈ K,

x
00

pc
n−1
 R.
3.23
Similar to the proof in case 1, we can show that

m
k1
x
00
t
k
 > 0. Then, from 3.23,
we have m<1, which is a contraction. Hence, 3.10 holds.
Applying i of Lemma 1.1 to 3.2 and 3.10 yields that T has a fixed point x ∈
K
r,R


{x : r ≤x
pc
n−1
≤ R}. T hus, it follows that BVP 1.3 has at least one positive solution, and
the theorem is proved.
Theorem 3.2. Assume that H
1
 and H
2
 hold. In addition, letting f and I
k
satisfy the following
conditions:
H
5
 f

 0 and I

k0,k 1, 2, ,m;
H
6
 f
0
 ∞ or I
0
k∞,k 1, 2, ,m,
BVP 1.3 has at least one positive solution.
Proof. Considering H

5
, there exists r>0 such that ft, x ≤ εr, I
k
x ≤ ε
k
r, and k 
1, 2, ,m,forx ≥
r, t ∈ J, where ε, ε
k
> 0 satisfy max{H
0
, 1  G
0
}ε 

m
k1
ε
k
 < 1.
Similar to the proof of 3.2, we can show that
Tx
/
≥x, x ∈ K,

x

pc
1


r
.
3.24
Next, turning to H
6
. Under condition H
6
, similar to the proof of 3.10, we can also
show that
Tx
/
≤x, x ∈ K,

x

pc
1

R
.
3.25
Applying i of Lemma 1.1 to 3.24 and 3.25 yields that T has a fixed point
x ∈
K
r,R
 {x : r ≤x
pc
n−1
≤ R}. Thus, it follows that BVP 1.3 has one positive solution, and
the theorem is proved.

Theorem 3.3. Assume that H
1
, H
2
, H
3
, and H
5
 hold. In addition, letting f and I
k
satisfy
the following condition:
H
7
 there is a ς>0 such that γ

ς ≤ x ≤ ς and t ∈ J implies
f

t, x

≥ τς, I
k

x

≥ τ
k
ς, k  1, 2, , 3.26
where τ, τ

k
≥ 0 satisfy τ 

m
k1
τ
k
> 0,τ

1
t
m
H1/2,sds 

m
k1
τ
k
H1/2,t
k
 > 1,BVP
1.3 has at least two positive solutions x

and x
∗∗
with 0 < x


pc
n−1

<ς<x
∗∗

pc
n−1
.
Proof. We choose ρ, ξ with 0 <ρ<ς<ξ.IfH
3
 holds, similar to the proof of 3.2, we can
prove that
Tx
/
≥x, x ∈ K,

x

pc
1
 ρ.
3.27
Boundary Value Problems 15
If H
5
 holds, similar to the proof of 3.24, we have
Tx
/
≥x, x ∈ K,

x


pc
n−1
 ξ.
3.28
Finally, we show that
Tx
/
≤x, x ∈ K,

x

pc
n−1
 ς.
3.29
In fact, if there exists x
2
∈ K with x
2

pc
n−1
 ς, then by 2.23, we have
x
2

t

≥ γ



x
2

pc
n−1
 γ

ς,
3.30
and it follows from H
7
 that
x
2

t



1
t
m
H

1
2
,s

f


s, x
2

s

ds 
m

k1
H

1
2
,t
k

I
k

x
2

t
k

≥ ς

τ


1
t
m
H

1
2
,s

ds 
m

k1
τ
k
H

1
2
,t
k


>ς

x
2

pc
n−1

,
3.31
that is, x
2

pc
n−1
> x
2

pc
n−1
, which is a contraction. Hence, 3.29 holds.
Applying Lemma 1.1 to 3.27, 3.28,and3.29 yields that T has two fixed points
x

,x
∗∗
with x

∈ K
ρ,ς
,x
∗∗
∈ K
ς,ξ
. Thus it follows that BVP 1.3 has two positive solutions
x

,x

∗∗
with 0 < x


pc
n−1
<ς<x
∗∗

pc
n−1
. The proof is complete.
Our last results corresponds to the case when problem 1.3 has no positive solution.
Write
ΔH
0

1  m

. 3.32
Theorem 3.4. Assume H
1
, H
2
,ft, x < Δ
−1
x, t ∈ J, x > 0, and I
k
x < Δ
−1

x, ∀x>0,then
problem 1.3 has no positive solution.
16 Boundary Value Problems
Proof. Assume to the contrary that problem 1.3 has a positive solution, that is, T has a fixed
point y. Then y ∈ K, y > 0fort ∈ 0, 1,and


y





1
0
H

s

f

s, y

s


ds 
m

k1

H

t
k

I
k

y

t
k


<

1
0
H

s

Δ
−1
y

s

ds 
m


k1
H

t
k

Δ
−1


y



≤ H
0
Δ
−1


y




m

k1
H

0
Δ
−1


y



 H
0
Δ
−1

1  m



y






y



,

3.33
which is a contradiction, and this completes the proof.
To illustrate how our main results can be used in practice we present an example.
Example 3.5. Consider the following boundary value problem:
−x
4

t


3

t
5
 1x
5
tanh x, t ∈ J, t
/

1
2
,
−Δx
3
|
t
1
1/2
 x
3


1
2

,
x

0

 x


0

 x


0

 0,x

1



1
0
tx

t


dt.
3.34
Conclusion. BVP 3.34 has at least one positive solution.
Proof. BVP 3.34 can be regarded as a BVP of the form 1.3, where
h

t

 t, μ 

1
0
t · t
3
dt 
1
5
,t
1

1
2
,f

t, x


3


t
5
 1x
5
tanh x, I
1

x

 x
3
,
G
1

t, s


1
6



t
3

1 − s

3



t − s

3
, 0 ≤ s ≤ t ≤ 1,
t
3

1 − s

3
, 0 ≤ t ≤ s ≤ 1,
G
2

t, s


1
24
t
3

3
4
s − 2s
2

3
2

s
3

1
4
s
5

.
3.35
Boundary Value Problems 17
It is not difficult to see that conditions H
1
 and H
2
 hold. In addition,
f
0
 lim sup
x → 0
max
t∈J
f

t, x

x
 0,I
0


k

 lim sup
x → 0
I
k

x

x
 0,
f

 lim inf
x →∞
min
t∈J
f

t, x

x
 ∞.
3.36
Then, conditions H
3
 and H
4
 of Theorem 3.1 hold. Hence, by Theorem 3.1,the
conclusion follows, and the proof is complete.

Acknowledgment
This work is supported by the National Natural Science Foundation of China 10771065,
the Natural Sciences Foundation of Heibei Province A2007001027, the Funding Project
for Academic Human Resources Development in Institutions of Higher Learning Under
the Jurisdiction of Beijing Municipality PHR201008430, the Scientific Research Common
Program of Beijing Municipal Commission of EducationKM201010772018 and Beijing
Municipal Education Commission71D0911003. T he authors thank the referee for his/her
careful reading of the paper and useful suggestions.
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