Li and Dong Boundary Value Problems 2011, 2011:59 ppt

RESEARC H Open Access
Multiple positive solutions for a fourth-order
integral boundary value problem on time scales
Yongkun Li
*
and Yanshou Dong
* Correspondence:
cn
Department of Mathematics,
Yunnan University Kunming,
Yunnan 650091 People’s Republic
of China
Abstract
In this article, we investigate the multiplicity of positive solutions for a fourth-order
system of integral boundary value problem on time scales. The existence of multiple
positive solutions for the system is obtained by using the fixed point theorem of
cone expansion and compression type due to Krasnosel’skill. To demonstrate the
applications of our results, an example is also given in the article.
Keywords: positive solutions, fixed points, integral boundary conditions, time scales
1 Introduction
Boundary value problem (BVP) for ordinary differential equations arise in different
areas of applied mathematics and physics and so on, the existence and multiplicity of
positive solutions for such problems have become an important area of investigation in
recent years, lots of significant results have been established by using upper and lower
solution arguments, fixed point indexes, fixed point th eorems and so on (see [1-8] and
the references therein). Especially, the existence of positi ve solutio ns of nonlinear BVP
with integral boundary conditions has been extensively studied by many authors (see
[9-18] and the references therein).
However, the corresponding results for BVP with integral boundary conditions on time
scales are still very few [19-21]. In this article, we discuss the multiple positive solutions
for the following fourth-order system of integral BVP with a parameter on time scales

⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
(
4
)
(t)+λf (t, x(t), x

(t), x

(t), y(t), y

(t), y

(t)) = 0, t ∈ (0, σ (T))
T
,
y
(4)
(t)+μg(t, x(t), x

(t), x


(t), y(t), y

(t), y

(t)) = 0, t ∈ (0, σ (T))
T
,
x(0) = x

(0) = 0,
y(0) = y

(0) = 0,
a
1
x

(0) − b
1
x

(0) =
σ (T)

0
x

(s)A
1
(s)s,

c
1
x

(σ (T)) + d
1
x

(σ (T)) =
σ (T)

0
x

(s)B
1
(s)s,
a
2
y

(0) − b
2
y

(0) =
σ (T)

0
y


(s)A
2
(s)s,
c
2
y

(σ (T)) + d
2
y

(σ (T)) =
σ (T)

0
y

(s)B
2
(s)s,
(1:1)
Li and Dong Boundary Value Problems 2011, 2011:59
/>© 2011 Li and Dong; licensee Springer. This is an Open Access article distr ibuted under the terms of the Creative Commons Attribution
License ( which permits unrestricted use, distribution , and reproduction in any medium,
provided the original work is properly cited.
where a
i
, b
i

, c
i
, d
i
≥ 0, and r
i
= a
i
c
i
s(T)+a
i
d
i
+ b
i
c
i
>0(i =1,2),0< l, μ <+∞, f,
g
∈ C
((
0, σ
(
T
))
T
×
(
R

+
)
6
, R
+
)
, ℝ
+
= [0, +∞), A
i
and B
i
are nonnegative and rd-continu-
ous on
[0, σ
(
T
)
]
T
(
i =1,2
)
.
The main purpose of this article is to establish some sufficient conditions for the exis-
tence of at least two positive solutions for system (1.1) by using the fixed point theorem of
cone expansion and compression type. This article is organized as follows. In Section 2,
some useful lemmas are established. In Section 3, by using the fixed point theorem of
cone expansion and compression type, we establish sufficient conditions for the existence
of at least two positive solutions for system (1.1). An illustrative example is given in

Section 4.
2 Preliminaries
In this section, we will provide several foundational definitions and results from the
calculus on time scales and give some lemmas which a re used in the proof of our
main results.
A time scale
T
is a nonempty closed subset of the real numbers ℝ.
Definition 2.1.[22]For
t
∈ T
, we define the forward jump operator
σ
:
T →
T
by
σ
(
t
)
=inf{τ ∈ T : τ>t
}
, while the backward jump operator
ρ
: T →
T
by
ρ
(

t
)
=sup{τ ∈ T : τ<t
}
.
In this definition, we put
inf∅ =su
p T
and
su
p
∅ =inf
T
,where∅, denotes the empty
set. If s(t) >t,wesaythatt is right-sc attered , while if r(t) <t,wesaythatt is left-
scattered. Also, if
t < su
p T
and s( t)=t,thent is called right-dense, and if
t
> inf
T
and r(t)=t,thent is called left-dense. We also need, below, the set
T
k
,whichis
derived from the time scale
T
as follows: if
T

has a left-scattered maximum m,then
T
k
= T −
m
. Otherwise,
T
k
=
T
.
Definition 2.2.[22]Assume that
x :
T →
R
is a function and let
t
∈
T
k
. Then x is
called di fferentiable at
t
∈
T
if there exists a θ Î ℝ such that for any given ε >0, there is
an open neighborhood U of t such that
|
x
(

σ
(
t
))
− x
(
s
)
− x

(
t
)
|σ
(
t
)
− s|| ≤ε|σ
(
t
)
− s|, s ∈ U
.
In this case, x
Δ
(t) is called the delta derivative of x at t. The second derivative of x(t)
is defined by x
ΔΔ
(t)=(x
Δ

)
Δ
(t).
In a similar way, we can obtain the fourth-order derivative of x(t) is defined by x
(4Δ)
(t) = (((x
Δ
)
Δ
)
Δ
)
Δ
(t).
Definition 2.3. [22]A function
f
: T →
R
is called rd-continuous provided it is contin-
uous at right-dense points in
T
and its left-sided limits exist at left-dense points in
T
.
The set of rd-continuous functions
f
: T →
R
will be denoted by
C

rd
(
T
)
.
Definition 2.4. [22]Afunction
F
: T
→
R
is called a delta-antiderivative of
f
: T →
R
provide F
Δ
(t)=f(t) holds for all
t
∈
T
k
. In this case we define the integral of f by
t

a
f (s)=F(t) − F(a)
.
For convenience, we denote
I =[0,σ
(

T
)
]
T
,
I

=
(
0, σ
(
T
))
T
and for i =1,2,weset
D
1i
=
Q
1i
1 − P
1i
, D
2i
=
P
2i
1 −
Q
2i

, K
1i
=
1
1 − P
1i
, K
2i
=
1
1 −
Q
2i
,
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 2 of 18
where
P
1i
=
σ (T)

0
B
i
(s)
a
i
s + b
i

ρ
i
s, P
2i
=
σ (T)

0
A
i
(s)
a
i
s + b
i
ρ
i
s,
Q
1i
=
σ (T)

0
B
i
(s)
d
i
+ c

i
(σ (T) − s)
ρ
i
s, Q
2i
=
σ (T)

0
A
i
(s)
d
i
+ c
i
(σ (T) − s)
ρ
i
s
.
To establish the existence of multiple positive solutions of system (1.1), let us list th e
following assumptions:
(H
1
)P
j
i
, Q

j
i
∈ [0, 1), D
11
D
21
∈ [0, 1), D
21
D
22
∈ [0, 1), j, i =1,2
.
In order to overco me the difficulty due to the dependence of f, g o n derivatives, we
first consider the following second-order nonlinear system
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
u

(t )+λf (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)=0, t ∈ (0, σ (T))
T
,
v


(t )+μg(t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)=0, t ∈ (0, σ (T))
T
,
a
1
u(0) − b
1
u

(0) =
σ (T)

0
u(s)A
1
(s)s,
c
1

u(σ (T)) + d
1
u

(σ (T)) =
σ (T)

0
u(s)B
1
(s)s,
a
2
v(0) − b
2
v

(0) =
σ (T)

0
v(s)A
2
(s)s,
c
2
v(σ (T)) + d
2
v


(σ (T)) =
σ (T)

0
v(s)B
2
(s)s,
(2:1)
where A
0
is the identity operator, and
A
i
u(t )=
t

0
(t − σ (s))
i−1
u(s)s, A
i
v(t)=
t

0
(t − σ (s))
i−1
v(s)s, i =1,2
.
(2:2)

For the proof of our main results, we will make use of the following lemmas.
Lemma 2.1. The fourth-order system (1.1) has a solution (x, y) if and only if the non-
linear system (2.1) has a solution (u, v).
Proof.If(x, y) is a solution of the fourth-order system (1.1), let u(t)=x
ΔΔ
(t), v(t)=
y
ΔΔ
(t), then it follows from the boundary conditions of system (1.1) that
A
1
u
(
t
)
= x

(
t
)
, A
2
u
(
t
)
= x
(
t
)

, A
1
v
(
t
)
= y

(
t
)
, A
2
v
(
t
)
= y
(
t
).
Thus (u, v)=(x
ΔΔ
(t), y
ΔΔ
(t)) is a solution of the nonlinear system (2.1).
Conversely, if (u, v) is a solution of the nonlinear system (2.1), let x(t)=A
2
u(t), y(t)=
A

2
v(t), then we have
x

(
t
)
= A
1
u
(
t
)
, x

(
t
)
= u
(
t
)
, y

(
t
)
= A
1
v

(
t
)
, y

(
t
)
= v
(
t
),
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 3 of 18
which imply that
x
(
0
)
=0,x

(
0
)
=0,y
(
0
)
=0,y


(
0
)
=0
.
Consequently, (x, y)=(A
2
u(t), A
2
v(t)) is a solution of the fourth-order system (1.1).
This completes the proof.
Lemma 2.2. Assume that D
11
D
21
≠ 1 holds. Then for any h
1
Î C(I’, ℝ
+
), the following
BVP
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪

⎪
⎪
⎪
⎩
u

(t )+h
1
(t )=0, t ∈ (0, σ(T))
T
,
a
1
u(0) − b
1
u

(0) =
σ (T)

0
u(s)A
1
(s)s,
c
1
u(σ (T)) + d
1
u


(σ (T)) =
σ (T)

0
u(s)B
1
(s)
s
(2:3)
has a solution
u
(t )=
σ
(
T
)

0
H
1
(t , s)h
1
(s)s
,
where
H
1
(t , s)=G
1
(t , s)+r

1
(t )
σ (T)

0
B
1
(τ )G
1
(τ , s)τ + r
2
(t )
σ (T)

0
A
1
(τ )G
1
(τ , s)τ
,
G
1
(t , s)=
1
ρ
1

(a
1

σ (s)+b
1
)[d
1
+ c
1
(σ (T) − t)], σ (s) < t,
(a
1
t + b
1
)[d
1
+ c
1
(σ (T) − σ (s))], t ≤ σ (s),
r
11
(t )=
K
11
(a
1
t + b
1
)+K
11
D
21
[d

1
+ c
1
(σ (T) − t)]
ρ
1
(1 − D
11
D
21
)
,
r
21
(t )=
K
21
D
11
(a
1
t + b
1
)+K
21
[d
1
+ c
1
(σ (T) − t)]

ρ
1
(
1 − D
11
D
21
)
.
Proof. First suppose that u is a solution of system (2.3). I t is easy to see by integra-
tion of BVP(2.3) that
u

(t )=u

(0) −
t

0
h
1
(s)s
.
(2:4)
Integrating again, we can obtain
u
(t )=u(0) + tu

(0) −
t


0
(t − σ (s))h
1
(s)s
.
(2:5)
Let t = s(T) in (2.4) and (2.5), we obtain
u

(σ (T)) = u

(0) −
σ (T)

0
h
1
(s)s
,
(2:6)
u
(σ (T)) = u(0) + σ (T)u

(0) −
σ (T)

0
(σ (T) − σ (s))h
1

(s)s
.
(2:7)
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 4 of 18
Substituting (2.6) and (2.7) into the second boundary value condition of system (2.3),
we obtain
c
1
u(0) + (c
1
σ (T)+d
1
)u

(0) =
σ
(
T
)

0
[d
1
+ c
1
(σ (T) − σ (s))]h
1
(s)
s

+
σ (T)

0
u(s)B
1
(s)s.
(2:8)
From (2.8) and the first boundary value condition of system (2.3), we have
u

(0) =
a
1
ρ
1
⎛
⎜
⎝
σ (T)

0
[d
1
+ c
1
(σ (T) − σ (s))]h
1
(s)s +
σ (T)


0
u(s)B
1
(s)
s
(2:9)
−
c
1
a
1
σ (T)

0
u(s)A
1
(s)s
⎞
⎟
⎠
,
u
(0) =
b
1
ρ
1
⎛
⎜

⎝
σ (T)

0
[d
1
+ c
1
(σ (T) − σ (s))]h
1
(s)s +
σ (T)

0
u(s)B
1
(s)
s
−
c
1
a
1
σ (T)

0
u(s)A
1
(s)s
⎞

⎟
⎠
+
1
a
1
σ (T)

0
u(s)A
1
(s)s.
(2:10)
Substituting (2.9) and (2.10) into (2.5), we have
u
(t )=
σ
(
T
)

0
G
1
(t , s)h
1
(s)s +
a
1
t + b

1
ρ
1
σ
(
T
)

0
u(s)B
1
(s)
s
+
d
1
+ c
1
(σ (T) − t)
ρ
1
σ (T)

0
u(s)A
1
(s)s.
(2:11)
By (2.11), we get
σ

(
T
)

0
u(s)B
1
(s)s =
1
1 − P
11
σ
(
T
)

0
B
1
(s)
σ
(
T
)

0
G
1
(s, τ )h
1

(τ )τ
s
+
Q
11
1 − P
11
σ (T)

0
u(s)A
1
(s)s,
(2:12)
σ (T)

0
u(s)A
1
(s)s =
1
1 − Q
21
σ (T)

0
B
1
(s)
σ (T)


0
G
1
(s, τ )h
1
(τ )τ
s
+
P
21
1 − Q
21
σ (T)

0
u(s)B
1
(s)s.
(2:13)
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 5 of 18
By (2.12) and (2.13), we get
σ (T)

0
u(s)A
1
(s)s =
K

11
D
21
1 − D
11
D
21
σ (T)

0
B
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τs
+
K
21
1 − D
11
D
21
σ (T)


0
A
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τs
,
(2:14)
σ
(
T
)

0
u(s)B
1
(s)s =
K
11
1 − D
11
D
21
T


0
B
1
(s)
T

0
G
1
(s, τ )h
1
(τ )τs
+
K
21
D
11
1 − D
11
D
21
σ (T)

0
A
1
(s)
σ (T)


0
G
1
(s, τ )h
1
(τ )τs
.
(2:15)
Substituting (2.14) and (2.15) into (2.11), we have
u
(t)=
σ
(
T
)

0
G
1
(t, s)h
1
(s)s
+
K
11
(a
1
t + b
1
)+K

11
D
21
[d
1
+ c
1
(σ (T) − t)]
ρ
1
(1 − D
11
D
21
)
σ (T)

0
B
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τ s
+

K
21
D
11
(a
1
t + b
1
)+K
21
[d
1
+ c
1
(σ (T) − t)]
ρ
1
(1 − D
11
D
21
)
σ (T)

0
A
1
(s)
σ (T)


0
G
1
(s, τ )h
1
(τ )τ 
s
=
σ (T)

0
G
1
(t, s)h
1
(s)s + r
11
(t)
σ (T)

0
B
1
(s)
σ (T)

0
G
1
(s, τ )h

1
(τ )τ s
+ r
21
(t)
σ (T)

0
A
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τ s
=
σ (T)

0
H
1
(t, s)h
1
(s)s.
(2:16)
Conversely, suppose

u
(t )=

σ
(
T
)
0
H
1
(t , s)h
1
(s)
s
, then
u
(t )=
σ (T)

0
G
1
(t , s)h
1
(s)s + r
11
(t )
σ (T)

0

B
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τ
s
+r
21
(t )
σ (T)

0
A
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ )τs.
(2:17)

Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 6 of 18
Direct differentiation of (2.17) implies
u

(t)=
1
ρ
1
⎛
⎜
⎝
a
1
σ (T)

t
[d
1
+ c
1
(σ (T) − σ (s))]h
1
(s)s − c
1
t

0
(a
1

σ (s)+b
1
)h
1
(s)  s
⎞
⎟
⎠
+
a
1
K
11
− c
1
K
11
D
21
ρ
1
(1 − D
11
D
21
)
σ (T)

0
B

1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ ) τ s
+
a
1
K
21
D
11
− c
1
K
21
ρ
1
(1 − D
11
D
21
)
σ (T)


0
A
1
(s)
σ (T)

0
G
1
(s, τ )h
1
(τ ) τ s
and
u

(
t
)
= −h
1
(
t
),
and it is easy to verify that
a
1
u(0) − b
1
u


(0) =
σ
(
T
)

0
u(s)A
1
(s)s,
c
1
u(σ (T)) + d
1
u

(σ (T)) =
σ (T)

0
u(s)B
1
(s)s
.
This completes the proof.
Lemma 2.3. Assume that D
12
D
22
≠ 1 holds. Then for any h

2
Î C(I’, ℝ
+
), the following
BVP
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
v

(t )+h
2
(t )=0, t ∈ (0, σ(T))
T
,
a
2
v(0) − b
2
v


(0) =
σ (T)

0
v(s)A
2
(s)s,
c
2
v(σ (T)) + d
2
v

(σ (T)) =
σ (T)

0
v(s)B
2
(s)
s
has a solution
v(t)=
σ (T)

0
H
2
(t , s)h

2
(s)s
,
where
H
2
(t , s)=G
2
(t , s)+r
12
(t )
σ (T)

0
B
2
(τ )G
2
(τ , s)τ + r
22
(t )
σ (T)

0
A
2
(τ )G
2
(τ , s)τ
,

G
2
(t , s)=
1
ρ
2

(a
2
σ (s)+b
2
)[d
2
+ c
2
(σ (T) − t)], σ (s) < t ,
(a
2
t + b
2
)[d
2
+ c
2
(σ (T) − σ (s))], t ≤ σ (s),
r
12
(t )=
K
12

(a
2
t + b
2
)+K
12
D
22
[d
2
+ c
2
(σ (T) − t)]
ρ
2
(1 − D
12
D
22
)
,
r
22
(t )=
K
22
D
12
(a
2

t + b
2
)+K
22
[d
2
+ c
2
(σ (T) − t)]
ρ
2
(
1 − D
12
D
22
)
.
Proof. The proof is similar to that of Lemma 2.2 and will omit it here.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 7 of 18
Lemma 2.4. Suppose that (H
1
) is satisfied, for all t, s Î I and i =1,2, we have
(i) G
i
(t, s) >0,H
i
(t, s) >0,
(ii) Lim

i
G
i
(s(s), s) ≤ H
i
(t, s) ≤ M
i
G
i
(s(s), s),
(iii) mG
i
(s(s), s) ≤ H
i
(t, s) ≤ MG
i
(s(s), s),
where
M
i
=1+r
1i
σ
(
T
)

0
B
i

(τ )τ + r
2i
σ
(
T
)

0
A
i
(τ )τ , r
ji
=max
0≤t≤1
r
i
(t),
m
i
=1+r
1i
(t)
σ (T)

0
B
i
(τ )τ + r
2i
(t)

σ (T)

0
A
i
(τ )τ , r
ji
= min
0≤t≤1
r
i
(t),
M =max{M
1
, M
2
}, m = min{L
1
m
1
, L
2
m
2
}, L
i
= min

d
i

d
i
+ c
i
,
b
i
a
i
+ b
i

, i, j =1,2
.
Proof. It is easy to verify that G
i
(t, s) >0, H
i
(t, s) >0andG
i
(t, s) ≤ G
i
(s(s), s), for all t,
s Î I. Since
G
i
(t , s)
G
i
(σ (s), s)

=

d
i
+c
z
(σ (T) −t)
d
i
+c
z
(σ (T) −σ (s))
, σ (s) < t
,
a
i
t+b
i
a
i
σ (s)+b
i
, σ (s) ≥ t.
Thus G
i
(t, s)/G
i
(s(s), s) ≥ L
i
and we have

G
i
(
t, s
)
≥ L
i
G
i
(
σ
(
s
)
, s
).
On the one hand, from the definition of L
i
and m
i
, for all t, s Î I, we have
H
i
(t , s)=G
i
(t , s)+r
1i
(t )
σ
(

T
)

0
B
i
(τ )G
i
(τ , s)τ + r
2i
(t )
σ
(
T
)

0
A
i
(τ )G
i
(τ , s)
τ
≥ L
i
G
i
(σ (s), s)
⎛
⎜

⎝
1+r
1i
(t )
σ (T)

0
B
i
(τ )τ + r
2i
(t )
σ (T)

0
A
i
(τ )τ
⎞
⎟
⎠
≥ L
i
m
i
G
i
(
σ
(

s
)
, s
)
,
and on the other hand, we obtain easily that from the definition of M
i
, for all t, s Î I,
H
i
(t, s) ≤ G
i
(σ (s), s)+r
1i
(t)
σ
(
T
)

0
B
i
(τ ) G
i
(σ (s), s)τ + r
2i
(t)
σ
(

T
)

0
A
i
(τ ) G
i
(σ (s), s)
τ
≤ M
i
G
i
(
σ
(
s
)
, s
)
.
Finally, it is easy to v erify that mG
i
(s(s), s) ≤ H
i
(t, s) ≤ MG
i
(s (s), s). This completes
the proof.

Lemma 2.5. [23]Let E be a Banach space and P be a cone in E. Assume that Ω
1
and
Ω
2
are bounded open subsets of E, such that 0 Î Ω
1
,

1
⊂ 
2
, and let
T : P ∩
(

2
\
1
)
→
P
be a completely continuous operator such that either
(i)||Tu|| ≤ ||u||, ∀u Î P ∩ ∂Ω
1
and ||Tu|| ≥ ||u||, ∀u Î P ∩ ∂Ω
2
,or
(ii)||Tu|| ≥ ||u||, ∀u Î P ∩ ∂ Ω
1

and ||Tu|| ≤ ||u||, ∀u Î P ∩ ∂Ω
2
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 8 of 18
holds. Then T has a fixed point in
P ∩
(

2
\
1
)
.
To obtain the existence of positive solutions for system (2.1), we construct a cone P
in the Banach space Q = C(I, ℝ
+
)×C(I, ℝ
+
)equippedwiththenorm
||(u, v)|| = ||u|| + ||v|| =max
t
∈
I
|u| +max
t
∈
I
|v
|
by

P =

(u, v) ∈ Q|u(t) ≥ 0, v(t) ≥ 0, min
t∈I
(u(t)+v(t)) ≥
m
M
||(u, v)||

.
It is easy to see that P is a cone in Q .
Define two operators T
l
, T
μ
: P ® C(I, ℝ
+
)by
T
λ
(u, v)(t)=λ
T

0
H
1
(t , s)f(t, A
2
u, A
1

u, A
0
u, A
2
v, A
1
v, A
0
v)s, t ∈ I,
T
μ
(u, v)(t)=μ
T

0
H
2
(t , s)g(t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s, t ∈ I

.
Then we can define an operator T : P ® C(I, ℝ
+
)by
T(u, v)=(T
λ
(u, v), T
μ
(u, v)), ∀(u, v) ∈ P
.
Lemma 2.6. Let (H
1
) hold. Then T : P ® P is completely continuous.
Proof. Firstly, we prove that T : P ® P. In fact, for all (u, v) Î P and t Î I, by Lemma
2.4(i) and (H
1
), it is obvious that T
l
(u, v)(t) >0, T
μ
(u, v)(t) >0. In addition, we have
T
λ
(u, v)(t)=λ
σ (T)

0
H
1
(t , s)f(t, A

2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≤ λM
σ (T)

0
G
1
(σ (s), s)f (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0

v)s
,
(2:18)
which implies
|
|T
λ
(u, v)|| ≤ λM

σ (T)
0
G
1
(σ (s), s)f (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)
s
.
And we have
T

λ
(u, v)(t) ≥ λL
1
m
1
σ (T)

0
G
1
(σ (s), s)f (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)
s
≥
m
M
||T
λ
(u, v)||.

In a similar way,
T
μ
(u, v)(t) ≥
m
M
||T
μ
(u, v)||
.
Therefore,
min
t∈I
(T
λ
(u, v)(t)+T
μ
(u, v)(t)) ≥
m
M
||T
λ
(u, v)|| +
m
M
||T
μ
(u, v)|
|
=

m
M
||T
λ
(u, v), T
μ
(u, v)||.
This shows that T : P ® P.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 9 of 18
Secondly, we prove that T is continuous and compact, respectively. Let {(u
k
, v
k
)} Î P
be any sequence of functions with
lim
k
→∞
(u
k
, v
k
)=(u, v) ∈
P
,
|T
λ
(u
k

, v
k
)(t) − T
λ
(u, v)(t)|≤λM
1
sup
t∈I
|f (t, A
2
u
k
, A
1
u
k
, A
0
u
k
, A
2
v
k
, A
1
v
k
, A
0

v
k
)
− f (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)|
σ (T)

0
G
1
(σ (s), s)s
,
from the continuity of f, we know that ||T
l
(u
k
, v
k
)-T

l
(u, v)|| ® 0ask ® ∞. Hence
T
l
is continuous.
T
l
is compact provided that it maps bounded sets into relatively compact sets. L et
¯
f =sup
t
∈
I

|f (t, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)
|
,andletΩ be any bounded subset of P,then
there exists r>0 such that ||(u, v)|| ≤ r for all (u, v) Î Ω.Obviously,from(2.16),we

know that
T
λ
(u, v)(t) ≤ λM
¯
f
σ (T)

0
G
1
(σ (s), s)s
,
so, T
l
Ω is bounded for all (u, v) Î Ω. Moreover, let
L

1
=
λ
¯
f
ρ
1
⎛
⎜
⎝
a
1

σ (T)

0
[d
1
+ c
1
(σ (T) − σ (s))]s + c
1
σ (T)

0
(a
1
σ (s)+b
1
)s
⎞
⎟
⎠
+
λ
¯
f |a
1
K
11
− c
1
K

11
D
21
|
ρ
1
(1 − D
11
D
21
)
σ (T)

0
B
1
(s)
σ (T)

0
G
1
(s, τ )τ s
+
λ
¯
f |a
1
K
21

D
11
− c
1
K
21
|
ρ
1
(1 − D
11
D
21
)
σ (T)

0
A
1
(s)
σ (T)

0
G
1
(s, τ )τ s.
We have
|T
λ
(u, v)


(t)|
≤
λ
ρ
1







a
1
σ (T)

t
[d
1
+ c
1
(σ (T) − σ (s))]f(s, A
2
u, A
1
u, A
0
u, A
2

v, A
1
v, A
0
v)s
−c
1
t

0
(a
1
σ (s)+b
1
)f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s







+
λ|a
1
K
11
− c
1
K
11
D
21
|
ρ
1
(1 − D
11
D
21
)
σ (T)

0
B
1
(s)
σ (T)


0
G
1
(s, τ )f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)τ s
+
λ|a
1
K
21
D
11
− c
1
K
21
|
ρ
1

(1 − D
11
D
21
)
σ (T)

0
A
1
(s)
σ (T)

0
G
1
(s, τ )f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)τ 
s

≤ L

1
.
Thus, f or any (u, v) Î Ω and ∀ε >0, let
δ =
ε
L

1
,thenfort
1
, t
2
Î I,|t
1
- t
2
| < δ,we
have
|
T
λ
(u, v)(t
1
) − T
λ
(u, v)(t
2
)|≤L


1
|t
1
− t
2
| <ε
.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 10 of 18
So, for all (u, v) Î Ω, T
l
Ω is equicontinuous. By Ascoli-Arzela theorem, we obtain
that T
l
: P ® P is completely continuous. In a s imilar way, we can prove that T
μ
: P
® P is completely continuous. Therefore, T : P ® P is completely continuous. This
completes the proof.
3 Main results
In this section, we will give our main results on mul tiplicity of positive solutions of
system (1.1). In the following, for convenience, we set
f
β
= lim
6

i=1
|ϕ

i
|→β
inf
t∈I

f (t, ϕ
1
, , ϕ
6
)
q
1
(t )

6
i=1
|ϕ
i
|
, f
α
= lim
6

i=1
|ϕ
i
|→α
sup
t∈I


f (t, ϕ
1
, , ϕ
6
)
q
2
(t )

6
i=1
|ϕ
i
|
,
g
β
= lim
6

i
=1
|ϕ
i
|→β
inf
t∈I

f (t, ϕ

1
, , ϕ
6
)
q
3
(t )

6
i=1
|ϕ
i
|
, g
α
= lim
6

i
=1
|ϕ
i
|→α
sup
t∈I

f (t, ϕ
1
, , ϕ
6

)
q
4
(t )

6
i=1
|ϕ
i
|
,
where q
i
(t), q
j
(t) Î C
rd
(I’, ℝ
+
) satisfy
0 <
σ (T)

0
G
1
(σ (s), s)q
i
(s)s < +∞ (i =1,2)
,

0 <
σ (T)

0
G
2
(σ (s), s)q
j
(s)s < +∞ (j =3,4).
Theorem 3.1. Assume that (H
1
) holds. Assume further that
(H
2
) thereexistaconstantR>0, and two functions p
i
(t) Î C
rd
(I, R
+
) satisfying
0 <
σ
(
T
)

0
G
i

(σ (s), s)p
i
(s)s < +∞(i =1,2
)
such that
f (t, ϕ
1
, , ϕ
6
) ≤ Rp
1
(t ), ∀t ∈ I,0<
6

i=1
|ϕ
i
|≤R,
g(t, ϕ
1
, , ϕ
6
) ≤ Rp
2
(t ), ∀t ∈ I,0<
6

i
=1
|ϕ

i
|≤R
,
and one of the folloeing conditions is satisfied
(E
1
)
λ ∈ (
M
3
f
0
, M
4
)
,
μ ∈ (
N
3
g
∞
, N
4
)
,
(E
2
)
λ ∈ (
M

3
f
∞
, M
4
)
,
μ ∈ (
N
3
g
0
, N
4
)
,
(E
3
)
λ ∈ (
M
3
F
α
, M
4
)
, μ Î (0, N
4
),

(E
4
) l Î (0, M
4
),
μ ∈ (
N
3
G
α
, N
4
)
,
where
M
3
=
⎛
⎜
⎝
m
2
M
σ (T)

0
G
1
(σ (s), s)q

1
(s)s
⎞
⎟
⎠
−1
, M
4
=
⎛
⎜
⎝
O
1
MN
σ (T)

0
G
1
(σ (s), s)p
1
(s)s
⎞
⎟
⎠
−1
,
N
3

=
⎛
⎜
⎝
m
2
M
σ (T)

0
G
2
(σ (s), s)q
3
(s)s
⎞
⎟
⎠
−1
, N
4
=
⎛
⎜
⎝
o
2
MN
σ (T)


0
G
2
(σ (s), s)p
2
(s)s
⎞
⎟
⎠
−1
,
F
α
= min{
f
0
,
f
∞
}, G
α
= {
g
0
,
g
∞
},
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 11 of 18

O
1
,O
2
satisfy
1
O
1
+
1
O
2
≤ 1, N =1+σ (T)+(σ (T))
2
. Then system (1.1) has at least two
positive solutions.
Proof.Weonlyprovethecaseinwhich(H
2
)and(E
1
) hold, the other case can be
proved similarly. Firstly, from (2.2), we have
2

i
=
0
A
i
u(t)+

2

i
=
0
A
i
v(t) ≤||(u, v)|| + σ (T)||(u, v)|| +(σ(T))
2
||(u, v)|| = N||(u, v)||
.
(3:1)
Take
R
1
=
R
N
,andletΩ
1
={(u, v) Î Q ;||(u , v)|| <R
1
}. For any t Î I,(u, v) Î ∂Ω
1
∩
P, it follows from l <M
4
, μ <N
4
and (H

2
) that
T
λ
(u, v)(t)=λ
σ
(
T
)

0
H
1
(t , s)f(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≤ M
4
M
σ (T)


0
G
1
(σ (s), s)f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)
s
≤ M
4
MR
σ (T)

0
G
1
(σ (s), s)p
1
(s)s
= NM

4
MR
1
σ (T)

0
G
1
(σ (s), s)p
1
(s)s ≤
1
O
1
R
1
and
T
μ
(u, v)(t)=μ
σ (T)

0
H
2
(t , s)g(s, A
2
u, A
1
u, A

0
u, A
2
v, A
1
v, A
0
v)s
≤ N
4
M
σ (T)

0
G
2
(σ (s), s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)
s

≤ N
4
MR
σ (T)

0
G
2
(σ (s), s)p
2
(s)s
= N
4
MNR
1
σ (T)

0
G
2
(σ (s), s)p
2
(s)s ≤
1
O
2
R
1
.
Consequently, for any (u, v) Î ∂Ω

1
∩ P, we have
|
|T(u, v)|| = ||T
λ
(u, v)|| + ||T
μ
(u, v)|| <
1
O
1
R
1
+
1
O
2
R
1
≤ R
1
.
(3:2)
Second, from
λ>
M
3
f
0
, we can choose ε

1
>0 such that lf
0
>M
3
+ ε
1
, then there exists
0 <l
1
<NR
1
such that for any

6
i
=1
|ϕ
i
| < l
1
and t Î I,
f (t, ϕ
1
, , ϕ
6
) ≥
M
3
+ ε

1
λ
q
1
(t )
6

i
=1
|ϕ
i
|
.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 12 of 18
And since
2

i
=
0
A
i
u(t )+
2

i
=
0
A

i
v(t) ≥ u( t)+v(t) ≥
m
M
||(u, v)||
.
(3:3)
Take
R
2
=
l
1
N
< R
1
.
For all (u, v) Î Ω
2
∩ P, where Ω
2
={(u, v) Î Q; ||(u, v)|| <R
2
}, we have
2

i=0
A
i
u(t )+

2

i=0
A
i
v(t) ≥ u( t)+v(t) ≥
m
M
R
2
.
Thus, for all (u, v) Î Ω
2
∩ P, we have
T
λ
(u, v)(t) ≥ λm
σ
(
T
)

0
G
1
(σ (s), s)f (s, A
2
u, A
1
u, A

0
u, A
2
v, A
1
v, A
0
v)s
≥ m( M
3
+ ε
1
)

2

i=0
|A
i
u| +
2

i=0
|A
i
v|

σ (T)

0

G
1
(σ (s), s)q
1
(s)
s
≥ M
3
m
2
M
R
2
σ (T)

0
G
1
(σ (s), s)q
1
(s)s = R
2
.
Consequently, for all (u, v) Î Ω
2
∩ P, we have
|
|T
(
u, v

)
|| ≥ ||T
λ
(
u, v
)
|| ≥ ||
(
u, v
)
||
.
(3:4)
Finally, from μ >N
3
/g
∞
, we can choose ε
2
>0 such that μg
∞
>N
3
+ ε
2
.then,there
exists
l
2
>

m
M
R
such that for any

6
i
=1
|ϕ
i
| < l
2
and t Î I,
g
(t , ϕ
1
, , ϕ
6
) ≥
N
3
+ ε
2
μ
q
3
(t )
6

i

=1
|ϕ
i
|
.
Take
R
3
=

m
M

−1
l
2
> R
1
.
For all (u, v) Î Ω
3
∩ P, where Ω
3
={(u, v) Î Q; ||(u, v)|| <R
3
}, from (3.3), we have
2

i
=

0
A
i
u(t )+
2

i
=
0
A
i
v(t) ≥
m
M
R
3
= l
2
.
(3:5)
Thus, for all (u, v) Î Ω
3
∩ P, we have
T
μ
(u, v)(t) ≥ μm
σ
(
T
)


0
G
2
(σ (s), s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≥ m(N
3
+ ε
2
)

2

i=0
|A
i
u| +
2


i=0
|A
i
v|

σ (T)

0
G
2
(σ (s), s)q
3
(s)s
≥ N
3
m
2
M
R
3
σ (T)

0
G
2
(σ (s), s)q
3
(s)s = R
3

.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 13 of 18
Consequently, for all (u, v) Î Ω
3
∩ P, we have
|
|T(u, v)|| ≥ ||T
μ
(u, v)|| ≥ ||(u, v)||
.
(3:6)
From (3.2), (3.4), and (ii) of Lemma 2.5, it follows that system (2.1) has one positive
solution (u
1
, v
1
) Î P with R
2
≤ ||(u
1
, v
1
)|| ≤ R
1
. Therefore, from Lem ma 2.1, it follows
that system (1.1) has one positive solution (x
1
, y
1

). In the same way, from (3.2), (3.6),
and (i) of Lemma 2.5, it follows that system (2.1) has one positive solution (u
2
, v
2
) Î P
with R
1
≤ ||(u
2
, v
2
)|| ≤ R
3
. Therefore, from Lemma 2.1, it fol lows that system (1.1) has
one positive solution (x
2
, y
2
). Above all, system (1.1) has at least two positive solutions.
This completes the proof.
Theorem 3.2. Assume that (H
1
) holds. Suppose further that
(H
3
) there exist a constant R
0
>0, and two functions w
i

(t) Î C
rd
(I, R
+
)(i =1,2)satis-
fying
0 <

σ (T)
0
G
i
(σ (s), s)w
i
(s)s < +
∞
such that
f (t, ϕ
1
, , ϕ
6
) ≥ R
0
w
1
(t ), ∀t ∈ I,
6

i
=1

|ϕ
i
| > R
0
,
(3:7)
or
g
(t , ϕ
1
, , ϕ
6
) ≥ R
0
w
2
(t ), ∀t ∈ I,
6

i
=1
|ϕ
i
| > R
0
.
(3:8)
Then system (1.1) has at least two positive solutions for each
λ ∈ (M
5

,
M
6
F
0
)
and
μ ∈ (N
5
,
N
6
G
0
)
, where
M
5
=
⎛
⎜
⎝
m
2
M
σ (T)

0
G
1

(σ (s), s)w
1
(s)s
⎞
⎟
⎠
−1
, N
5
=
⎛
⎜
⎝
m
2
M
σ (T)

0
G
2
(σ (s), s)w
2
(s)s
⎞
⎟
⎠
−1
,
M

6
=
⎛
⎜
⎝
O
1
MN
σ (T)

0
G
1
(σ (s), s)q
2
(s)s
⎞
⎟
⎠
−1
, N
6
=
⎛
⎜
⎝
o
2
MN
σ (T)


0
G
2
(σ (s), s)q
4
(s)s
⎞
⎟
⎠
−1
,
F
α
=max{
f
0
,
f
∞
} < ∞, G
α
=max{
g
0
,
g
∞
} < ∞.
Proof. We only prove the case in which (3.7) holds. The other case in which (3.8)

holds can be proved similarly.
Take
R

1
=

m
M

−1
R
0
and let

4
= {(u, v) ∈ Q; ||(u, v)|| < R

1
}
.Foranyt Î I,(u, v) Î ∂Ω
4
∩ P, it follows
from l >M
5
and (H
3
) that
T
λ

(u, v)(t)=λ
σ (T)

0
H
1
(t, s)f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≥ M
5
m
σ (T)

0
G
1
(σ (s), s)f (s, A
2
u, A

1
u, A
0
u, A
2
v, A
1
v, A
0
v)
s
≥ mM
5
R
0
σ (T)

0
G
1
(σ (s), s)w
1
(s)s
= M
5
m
2
M
R


1
σ (T)

0
G
1
(σ (s), s)w
1
(s)s = R

1
.
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 14 of 18
Consequently, for any (u, v) Î ∂Ω
4
∩ P, we have
|
|T
(
u, v
)
|| ≥ ||T
λ
(
u, v
)
|| ≥ ||
(
u, v

)
||
.
(3:9)
From
λ<
M
6
F
α
,
μ<
N
6
G
α
,weknowthat
λ<
M
6
f
0
,
μ<
N
6
g
0
, we can choose ε
3

>0 such that
M
6
- ε
3
>0, N
6
- ε
3
>0andlf
0
<M
6
- ε
3
, μg
0
<N
6
- ε
3
. Then there exists
0 < l
3
< NR
0
< NR

1
such that for any


6
i
=1
|ϕ
i
| < l
3
and t Î I,
f (t, ϕ
1
, , ϕ
6
) ≤
M
6
− ε
3
λ
q
2
(t )
6

i=1
|ϕ
i
|
,
g(t, ϕ

1
, , ϕ
6
) ≤
N
6
− ε
3
μ
q
4
(t )
6

i
=1
|ϕ
i
|.
Take
R

2
=
l
3
N
< R

1

and

5
= {(u, v) ∈ Q; ||(u, v)|| < R

2
}
.Then,forany(u, v) Î Ω
5
∩
P, from(3.1), we have
T
λ
(u, v)(t)=λ
σ
(
T
)

0
H
1
(t , s)f(s, A
2
u, A
1
u, A
0
u, A
2

v, A
1
v, A
0
v)s
≤ λM
σ (T)

0
G
1
(σ (s), s)f (s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≤ M(M
6
− ε
3
)


2

i=0
|A
i
u| +
2

i=0
|A
i
v|

σ (T)

0
G
1
(σ (s), s)q
2
(s)
s
≤ MNM
6
R

2
σ (T)

0

G
1
(σ (s), s)q
2
(s)s =
1
O
1
R

2
and
T
μ
(u, v)(t)=μ
σ (T)

0
H
2
(t , s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1

v, A
0
v)s
≤ μM
σ (T)

0
G
2
(σ (s), s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≤ M( N
6
− ε
3
)

2


i=0
|A
i
u| +
2

i=0
|A
i
v|

σ (T)

0
G
2
(σ (s), s)q
4
(s)
s
≤ MNN
6
R

2
σ (T)

0
G
2

(σ (s), s)q
4
(s)s =
1
O
2
R

2
.
Consequently, for any (u, v) Î ∂Ω
5
∩ P, we have
|
|T(u, v)|| = ||T
λ
(u, v)|| + ||T
μ
(u, v)|| <
1
O
1
R

2
+
1
O
2
R


2
≤ R

2
.
(3:10)
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 15 of 18
From
λ<
M
6
F
α
,
μ<
N
6
G
α
we know that
λ<
M
6
f
∞
,
μ<
N

6
g
∞
, we can choose ε
4
>0 such that
M
6
- ε
4
>0, N
6
- ε
4
>0andlf
∞
<M
6
- ε
4
, μg
∞
<N
6
- ε
4
.Thenthereexists
l
4
>

m
M
R

1
such that for any

6
i
=1
|ϕ
i
| > l
4
and t Î I,
f (t, ϕ
1
, , ϕ
6
) ≤
M
6
− ε
3
λ
q
2
(t )
6


i=1
|ϕ
i
|
,
g(t, ϕ
1
, , ϕ
6
) ≤
N
6
− ε
3
μ
q
4
(t )
6

i
=1
|ϕ
i
|.
Take
R

3
=(

m
M
)
−
1
l
4
> R

1
and let

6
= {(u, v) ∈ Q; ||(u, v)|| < R

3
}
. Then, for any ( u, v)
Î Ω
6
∩ P, we have
T
λ
(u, v)(t)=λ
σ
(
T
)

0

H
1
(t , s)f(s, A
2
u, A
1
u,3A
0
u, A
2
v, A
1
v, A
0
v)s
≤ λM
σ (T)

0
G
1
(σ (s), s)f (s, A
2
u, A
1
u, A
0
u, A
2
v, A

1
v, A
0
v)s
≤ M(M
6
− ε
3
)

2

i=0
|A
i
u| +
2

i=0
|A
i
v|

σ (T)

0
G
1
(σ (s), s)q
2

(s)
s
≤ MNM
6
R

3
σ (T)

0
G
1
(σ (s), s)q
2
(s)s =
1
O
1
R

3
and
T
μ
(u, v)(t)=μ
σ (T)

0
H
2

(t , s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A
0
v)s
≤ μM
σ (T)

0
G
2
(σ (s), s)g(s, A
2
u, A
1
u, A
0
u, A
2
v, A
1
v, A

0
v)s
≤ M( N
6
− ε
3
)

2

i=0
|A
i
u| +
2

i=0
|A
i
v|

σ (T)

0
G
2
(σ (s), s)q
4
(s)
s

≤ MNN
6
R

3
σ (T)

0
G
2
(σ (s), s)q
4
(s)s =
1
O
2
R

3
.
Consequently, for any (u, v) Î ∂Ω
6
∩ P, we have
|
|T(u, v)|| = ||T
λ
(u, v)|| + ||T
μ
(u, v)|| <
1

O
1
R

3
+
1
O
2
R

3
≤ R

3
.
(3:11)
From (3.9), (3.10) and (i) of Lemma 2.5, i t follows that system (2.1) has one positive
solution (u
1
, v
1
) Î P with
R

1
≤||(u
1
, v
1

)|| ≤ R

2
. Therefore, from Lemma 2.1, it follows
that system (1.1) has one positiv e solution (x
1
, y
1
). In the sam e way, from (3.9), (3.11)
and (ii) of Lemma 2.5, it follows that system (2.1) has one positive solution (u
2
, v
2
) Î
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 16 of 18
P with
R

1
≤||(u
2
, v
2
)|| ≤ R

3
.Therefore,fromLemma2.1,it follows that system (1.1)
has one pos itive solution (x
2

, y
2
). Above all, system (1.1) has at least two positive solu-
tions. This completes the proof.
4 A n example
Consider the following BVP with integral boundary conditions:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
(4)
(t )+λf (t, x ( t), x

(t ), x

(t ), y(t), y

(t ), y

(t )) = 0, t ∈ (0, σ (T))
T
,
y
(4)

(t )+μg(t, x(t), x

(t ), x

(t ), y(t), y

(t ), y

(t )) = 0, t ∈ (0, σ (T))
T
,
x(0) = x

(0) = 0,
y(0) = y

(0) = 0,
x

(0) − x

(0) =
σ (T)

0
x

(s)A
1
(s)s,

x

(σ (T)) + x

(σ (T)) =
σ (T)

0
x

(s)B
1
(s)s,
y

(0) − y

(0) =
σ (T)

0
y

(s)A
2
(s)s,
y

(σ (T)) + y


(σ (T)) =
σ (T)

0
(s)B
2
(s)s,
(4:1)
where A
1
(t)=B
1
(t)=t, A
2
(t)=B
2
(t)=t/2 and
f (t, φ
1
, φ
2
, φ
3
, φ
4
, φ
5
, φ
6
)=2t


6

i=1
φ
i

1
2
, t ∈ (0, σ (T))
T
, φ
i
≥ 0, i =1, ,6
,
g(t, φ
1
, φ
2
, φ
3
, φ
4
, φ
5
, φ
6
)=
t
2


6

i=1
φ
i

3
, t ∈ (0, σ (T))
T
, φ
i
≥ 0, i =1, ,6.
we choose O
1
=2,O
2
=4,R =1,p
1
(t)=2t,
p
2
(t )=
t
2
, q
1
(t)=q
3
(t)=1.Itiseasyto

check that f
0
= g
∞
= ∞,(H
1
), (H
2
)and(E
1
) are satisfied. Therefore, by Theorem 3.1,
system (4.1) has at least two positive solutions for each l Î (0, M
4
), μ Î (0, N
4
).
Acknowledgements
This study was supported by the National Natural Sciences Foundation of People’s Republic of China under Grant
10971183.
Authors’ contributions
All authors contributed equally to the manuscript and typed, read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 9 November 2011 Accepted: 29 December 2011 Published: 29 December 2011
References
1. Yuan, C, Jiang, D, O’Regan, D: Existence and uniqueness of positive solutions for fourth-order nonlinear singular
continuous and discrete boundary value problems. Appl Math Comput. 203, 194–201 (2008). doi:10.1016/j.
amc.2008.04.034
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 17 of 18

2. Yang, Z: Existence and uniqueness of positive solutions for a higher order boundary value problem. Comput Math
Appl. 54, 220–228 (2007). doi:10.1016/j.camwa.2007.01.018
3. Zhang, X, Feng, M: Positive solutions for a class of 2nth-order singular boundary value problems. Nonlinear Anal. 69,
1287–1298 (2008). doi:10.1016/j.na.2007.06.029
4. Zhao, J, Ge, W: A necessary and sufficient condition for the existence of positive solutions to a kind of singular three-
point boundary value problem. Nonlinear Anal. 71, 3973–3980 (2009). doi:10.1016/j.na.2009.02.067
5. Lian, H, Pang, H, Ge, W: Triple positive solutions for boundary value problems on infinite intervals. Nonlinear Anal. 67,
2199–2207 (2007). doi:10.1016/j.na.2006.09.016
6. Wei, Z, Zhang, M: Positive solutions of singular sub-linear boundary value problems for fourth-order and second-order
differential equation systems. Appl Math Comput. 197, 135–148 (2008). doi:10.1016/j.amc.2007.07.038
7. Bai, Z: Positive solutions of some nonlocal fourth-order boundary value problem. Appl Math Comput. 215, 4191–4197
(2010). doi:10.1016/j.amc.2009.12.040
8. Zhu, F, Liu, L, Wu, Y: Positive solutions for systems of a nonlinear fourth-order singular semipositone boundary value
problems. Appl Math Comput. 216, 448–457 (2010). doi:10.1016/j.amc.2010.01.038
9. Kang, P, Wei, Z, Xu, J: Positive solutions to fourth-order singular boundary value problems with integral boundary
conditions in abstract spaces. Appl Math Comput. 206, 245–256 (2008). doi:10.1016/j.amc.2008.09.010
10. Feng, M: Existence of symmetric positive solutions for a boundary value problem with integral boundary conditions.
Appl Math Lett. 24, 1419–1427 (2011). doi:10.1016/j.aml.2011.03.023
11. Zhang, X, Ge, W: Positive solutions for a class of boundary-value problems with integral boundary conditions. Comput
Math Appl. 58, 203–215 (2009). doi:10.1016/j.camwa.2009.04.002
12. Feng, M, Ji, D, Ge, W: Positive solutions for a class of boundary-value problem with integral boundary conditions in
Banach spaces. J Comput Appl Math. 222, 351–363 (2008). doi:10.1016/j.cam.2007.11.003
13. Zhao, J, Wang, P, Ge, W: Existence and nonexistence of positive solutions for a class of third order BVP with integral
boundary conditions in Banach spaces. Commun Nonlinear Sci Numer Simulat. 16, 402–413 (2011). doi:10.1016/j.
cnsns.2009.10.011
14. Yang, Z: Existence and uniqueness of positive solutions for an integral boundary value problem. Nonlinear Anal. 69,
3910–3918 (2008). doi:10.1016/j.na.2007.10.026
15. Kong, L: Second order singular boundary value problems with integral boundary conditions. Nonlinear Anal. 72,
2628–2638 (2010). doi:10.1016/j.na.2009.11.010
16. Jankowski, T: Positive solutions for fourth-order differential equations with deviating arguments and integral boundary

conditions. Nonlinear Anal. 73, 1289–1299 (2010). doi:10.1016/j.na.2010.04.055
17. Boucherif, A: Second-order boundary value problems with integral boundary conditions. Nonlinear Anal. 70, 364–371
(2009). doi:10.1016/j.na.2007.12.007
18. Zhang, X, Feng, M, Ge, W: Existence results for nonlinear boundary-value problems with integral boundary conditions in
Banach spaces. Nonlinear Anal.
69, 3310–3321 (2008). doi:10.1016/j.na.2007.09.020
19. Anderson, DR, Tisdell, CC: Third-order nonlocal problems with sign-changing nonlinearity on time scales. Electron J Diff
Equ. 2007(19), 1–12 (2007)
20. Li, YK, Shu, J: Multiple positive solutions for first-order impulsive integral boundary value problems on time scales.
Bound Value Probl. 2011, 12 (2011). doi:10.1186/1687-2770-2011-12
21. Li, YK, Shu, J: Solvability of boundary value problems with Riemann-Stieltjes -integral conditions for second-order
dynamic equations on time scales at resonance. Adv Diff Equ. 2011, 42 (2011). doi:10.1186/1687-1847-2011-42
22. Bohner, M, Peterson, A: Dynamic Equations on Time Scales, An introduction with Applications. Birkhauser, Boston (2001)
23. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, New york (1988)
doi:10.1186/1687-2770-2011-59
Cite this article as: Li and Dong: Multiple positive solutions for a fourth-order integral boundary value problem
on time scales. Boundary Value Problems 2011 2011:59.
Submit your manuscript to a
journal and benefi t from:
7 Convenient online submission
7 Rigorous peer review
7 Immediate publication on acceptance
7 Open access: articles freely available online
7 High visibility within the fi eld
7 Retaining the copyright to your article
Submit your next manuscript at 7 springeropen.com
Li and Dong Boundary Value Problems 2011, 2011:59
/>Page 18 of 18