# Positive Solutions for Coupled Nonlinear Fractional Differential Equations

Authors: Wenning Liu; Xingjie Yan; Wei Qi
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We consider the existence of positive solutions for a coupled system of nonlinear fractional differential equations with integral boundary values. Assume the nonlinear term is superlinear in one equation and sublinear in the other equation. By constructing two conesK
1, K
2 and computing the fixed point index in product cone K
1
×
K
2, we obtain that the system has a pair of positive solutions. It is remarkable that it is established on the Cartesian product of two cones, in which the feature of two equations can be opposite.


---

## Body

## 1. Introduction

Fractional calculus is a very old concept dating back to 17th century; it involves fractional integration and fractional differentiation. At the first stage, fractional calculus theory is mainly focused on pure mathematical fields. In the last few decades, fractional differential equations and fractional integration equations have found many applications in various fields, such as science and engineering, physics, chemistry, biology, economics, and signal and image processing; for details, see [1–5]. In recent years, fractional differential equations have attracted increasing interests for their extensive applications, which leads to intensive development of the theory of fractional calculus. And more and more results about the existence and uniqueness of solutions appear. We can refer to [6] for the latest studies of fractional calculus.The existence theory for initial value problems has been paid considerable attentions by many authors; see [3, 7, 8] and the references cited therein. Also the existence of positive solutions for boundary value problem of nonlinear fractional differential equation has attracted attentions from many researches; see [9–14]. By using the cone extension method Feng et al. [15] studied the existence of solutions for higher-order nonlinear fractional differential equation with integral boundary conditions:
(1)
D
0
+
α
u
(
t
)
+
g
(
t
)
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
u
′
(
0
)
=
·
·
·
=
u
(
n
-
2
)
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
h
(
t
)
u
(
t
)
d
t
,
where D
0
+
α is the standard Riemann-Liouville fractional derivative of order n
-
1
<
α
<
n, n
≥
3, g
∈
𝒞
(
(
0,1
)
,
(
0
,
+
∞
)
) and g may be singular at t
=
0 or t
=
1, h
∈
L
1
(
0,1
) is nonnegative, and f
∈
𝒞
(
(
0,1
)
×
(
0
,
+
∞
)
,
(
0
,
+
∞
)
).In recent years, many authors have established the existence and uniqueness for solutions of some systems of nonlinear fractional differential equations; readers can see [16–21]. For example, by using the alternative of Leray-Schauder theorem method Wang et al. [22] obtained the existence and uniqueness of positive solution of nonzero boundary values problem for a coupled system of nonlinear fractional differential equations:
(2)
D
0
+
α
u
(
t
)
=
f
(
t
,
v
(
t
)
)
,
0
<
t
<
1
,
D
0
+
β
v
(
t
)
=
g
(
t
,
u
(
t
)
)
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
a
u
(
ξ
)
,
v
(
0
)
=
0
,
v
(
1
)
=
b
v
(
ξ
)
,
where 1
<
α, β
<
2, 0
≤
a
≤
b
≤
1, 0
<
ξ
<
1, f
,
g
:
(
0,1
)
×
ℝ
×
ℝ
→
ℝ are given continuous functions and D
0
+
α is the standard Riemann-Liouville fractional derivative. By using the cone extension and cone compression Zhao et al. [23] studied the existence and nonexistence of positive solutions for a class of third order boundary value problem with integral boundary conditions in Banach spaces:
(3)
x
′′′
+
f
(
t
,
x
(
t
)
)
=
0
,
t
∈
J
,
x
(
0
)
=
0
,
x
′′
(
0
)
=
0
,
x
(
1
)
=
∫
0
1
‍
g
(
t
)
x
(
t
)
d
t
.
Or
(4)
x
′′′
+
f
(
t
,
x
(
t
)
)
=
0
,
t
∈
J
,
x
(
0
)
=
∫
0
1
‍
g
(
t
)
x
(
t
)
d
t
,
x
′′
(
0
)
=
0
,
x
(
1
)
=
0
,
where J
=
(
0,1
), f
∈
𝒞
(
(
0,1
)
×
P
,
P
), g
∈
L
1
(
0,1
) is nonnegative and D
0
+
α is the standard Riemann-Liouville fractional derivative; P is a cone.By using the cone extension and cone compression method Yang et al. [24] established sufficient conditions for the existence and nonexistence of positive solutions for a general class with integral boundary value problem for a coupled system of fractional equations:
(5)
D
0
+
α
u
(
t
)
+
a
(
t
)
f
(
t
,
v
(
t
)
)
=
0
,
0
<
t
<
1
,
D
0
+
β
v
(
t
)
+
b
(
t
)
g
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
ϕ
(
t
)
u
(
t
)
d
t
,
v
(
0
)
=
0
,
v
(
1
)
=
∫
0
1
‍
φ
(
t
)
v
(
t
)
d
t
,
where 1
<
α, β
≤
2, a
,
b
∈
𝒞
(
(
0,1
)
,
(
0
,
+
∞
)
), f
,
g
∈
𝒞
(
(
0,1
)
×
(
0
,
+
∞
)
,
(
0
,
+
∞
)
), and ϕ
,
φ
∈
L
1
(
0,1
) are nonnegative and D
0
+
α is the standard Riemann-Liouville fractional derivative.However, all these works are obtained by getting a fixed point of a completely continuous operator in a cone of product Banach spacesX
1
×
X
2. Thus, there is no essential difference between we consider the positive solution of coupled equations and positive solution of single equation. Furthermore, the authors cannot guarantee the obtained pair of positive solutions are all positive; maybe one is positive and another is zero. The features of nonlinear term are the same, which means the growth properties of nonlinear term are similar.Motivated by the above mentioned works, in this paper, we consider the existence of positive solution to boundary value problem for a coupled system of nonlinear fractional differential equations as follows:(6)
D
0
+
α
u
(
t
)
+
f
(
t
,
u
(
t
)
,
v
(
t
)
)
=
0
,
0
<
t
<
1
,
D
0
+
β
v
(
t
)
+
g
(
t
,
u
(
t
)
,
v
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
ϕ
(
t
)
u
(
t
)
d
t
,
v
(
0
)
=
0
,
v
(
1
)
=
∫
0
1
‍
φ
(
t
)
v
(
t
)
d
t
,
where 1
<
α, β
≤
2, f
,
g
∈
𝒞
(
(
0,1
)
×
(
0
,
+
∞
)
×
(
0
,
+
∞
)
,
(
0
,
+
∞
)
), and ϕ
,
φ
∈
L
1
(
0,1
) are nonnegative and D
0
+
α is the standard Riemann-Liouville fractional derivative. f and g have different features. It is very difficult to deal with our problem if we directly use the usual method because the features of nonlinear term are different. In order to overcome the difficulties, by using the ideals in [25–27], we need to consider our problem on the Cartesian product of two cones in the space 𝒞
(
0,1
); thus we can better exploit the feature of two equations. We choose a cone K
1
×
K
2 in the Cartesian product of two cones in 𝒞
(
0,1
)
×
𝒞
(
0,1
). By computing the fixed point index in K
1
×
K
2, we obtain the existence of positive solution (
u
,
v
) of problem (6) such that u
>
0, v
>
0.


## 2. Preliminary

In this section, we review some basic facts which are used throughout this paper. We can see [10, 16, 25–29] for more details.Definition 1.
For a functionf
:
[
a
,
+
∞
)
→
ℝ, the Riemann-Liouville fractional integral of order α
>
0 is defined by
(7)
I
a
+
α
f
(
t
)
=
1
Γ
(
α
)
∫
a
t
‍
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,and the right-hand side is pointwise defined on [
a
,
+
∞
), where a
∈
ℝ and Γ is the gamma function.Definition 2.
For a functionf
:
[
a
,
+
∞
)
→
ℝ, the Riemann-Liouville fractional derivative of order α
>
0 is given by
(8)
D
a
+
α
f
(
t
)
=
1
Γ
(
α
)
d
n
d
t
n
-
1
∫
a
t
‍
(
t
-
s
)
n
-
α
-
1
f
(
s
)
d
s
,
and the right-hand side is pointwise defined on [
a
,
+
∞
), where a
∈
ℝ, n
=
[
α
]
+
1 and Γ is the gamma function.LetE be a real Banach space with norm ∥
∥ and let A
:
E
→
E be an operator. A nonempty, closed, convex set P
⊂
E is said to be a cone provided the following: (i) if x
∈
P and λ
≥
0, then λ
x
∈
P; (ii) if x
∈
P and -
x
∈
P, then x
=
0 are satisfied. u is said to be a fixed point of A, if u
=
A
u. Ω is a bounded open subset of E with boundary ∂
Ω. Denote i
(
A
,
P
∩
Ω
,
P
) the fixed point index of A.Lemma 3 (see [25–27]).
LetE be a Banach space and let K
i
⊂
E
(
i
=
1,2
) be a closed convex cone in E. For r
i
>
0
(
i
=
1,2
), denote K
r
i
=
{
u
∈
K
i
∣
∥
u
∥
<
r
i
}, ∂
K
i
=
{
u
∈
K
i
∣
∥
u
∥
=
r
i
}. Suppose A
i
:
K
i
→
K
i is completely continuous. If u
i
≠
A
i
u
i, ∀
u
i
∈
∂
K
i, then
(9)
i
(
A
,
K
r
1
×
K
r
2
,
K
1
×
K
2
)
=
i
(
A
1
,
K
r
1
,
K
1
)
·
i
(
A
2
,
K
r
2
,
K
2
)
,
where A
(
u
,
v
)
=
(
A
1
u
,
A
2
v
) and for all (
u
,
v
)
∈
(
K
1
×
K
2
).In order to prove the existence of solution for problem (6), we can translate it into obtaining a fixed point of operator. Then we use the topological method to deal with it.Theorem 4.
LetE be a Banach space and let K
1
,
K
2
⊂
E be a closed convex cone in E. Denote K
r
i
=
{
u
∈
K
i
∣
∥
u
∥
<
r
i
}, ∂
K
i
=
{
u
∈
K
i
∥
u
∥
=
r
i
}, where r
>
0, i
=
1,2. Let D
=
(
K
R
1
∖
K
r
1
¯
)
×
(
K
R
2
∖
K
r
2
¯
). Suppose T
:
K
1
×
K
2
→
K
1
×
K
2 is completely continuous and T
(
u
,
v
)
=
(
T
1
(
u
,
v
)
,
T
2
(
u
,
v
)
). For λ
∈
[
0,1
], let
(10)
H
λ
,
1
(
u
,
v
)
=
λ
T
1
(
u
,
v
)
+
(
1
-
λ
)
T
1
(
u
,
0
)
,
H
λ
,
2
(
u
,
v
)
=
λ
T
2
(
u
,
v
)
+
(
1
-
λ
)
T
2
(
0
,
v
)
.
If H
λ
,
1, H
λ
,
2 satisfy the following conditions: (i)
H
λ
,
1 is uniformly cone extension about v
∈
K
R
2 at K
R
1
∖
K
r
1
¯, that is to say,
(11)
∥
H
λ
,
1
∥
<
∥
u
∥
,
∀
u
∈
∂
K
r
1
,
∥
H
λ
,
1
∥
>
∥
u
∥
,
∀
u
∈
∂
K
R
1
;
(ii)
H
λ
,
2 is uniformly cone compression about u
∈
K
R
1 at K
R
2
∖
K
r
2
¯, that is to say,
(12)
∥
H
λ
,
2
∥
>
∥
v
∥
,
∀
v
∈
∂
K
r
2
,
∥
H
λ
,
2
∥
<
∥
v
∥
,
∀
v
∈
∂
K
R
2
,then i
(
T
,
D
,
K
1
×
K
2
)
=
-
1. Thus the mapping T in the region D has a fixed point.Proof.
Following a similar procedure to the method given in [25–27], we present a proof. We suppose that T
1
(
u
,
0
)
=
A
1
u and T
2
(
0
,
v
)
=
A
2
v. Let A
(
u
,
v
)
=
(
A
1
u
,
A
2
v
). For λ
∈
[
0,1
], for any (
u
,
v
)
∈
K
1
×
K
2, set
(13)
H
(
(
u
,
v
)
,
λ
)
=
λ
T
(
u
,
v
)
+
(
1
-
λ
)
A
(
u
,
v
)
.
Since T is completely continuous, hence the mapping H is also completely continuous. By Lemma 3 and homotopy invariance of fixed point index (see [7, 28]), we have
(14)
i
(
T
,
D
,
K
1
×
K
2
)
=
i
(
A
1
,
K
R
1
∖
K
r
1
¯
,
K
1
)
·
i
(
A
2
,
K
R
2
∖
K
r
2
¯
,
K
2
)
.
Next, we computei
(
T
,
D
,
K
1
×
K
2
). By assumptions (i) and (ii), also using the properties of fixed point index (see [7, 28]), we have
(15)
i
(
A
1
,
K
r
1
,
K
1
)
=
i
(
A
2
,
K
R
2
,
K
2
)
=
1
,
i
(
A
1
,
K
R
1
,
K
1
)
=
i
(
A
2
,
K
r
2
,
K
2
)
=
0
.
Thanks to (15), we obtain
(16)
i
(
T
,
D
,
K
1
×
K
2
)
=
i
(
A
1
,
K
R
1
∖
K
r
1
¯
,
K
1
)
·
i
(
A
2
,
K
R
2
∖
K
r
2
¯
,
K
2
)
=
(
i
(
A
1
,
K
R
1
,
K
1
)
-
i
(
A
1
,
K
r
1
¯
,
K
1
)
)
·
(
i
(
A
2
,
K
R
2
,
K
2
)
-
i
(
A
2
,
K
r
2
¯
,
K
2
)
)
=
(
0
-
1
)
·
(
1
-
0
)
=
-
1
.
The proof is complete.Remark 5.
IfH
λ
,
1, H
λ
,
2 were all cone extension or cone compression, we can obtain the same conclusions.The following lemmas are necessary to prove the existence of positive solution for problem (6).Lemma 6 (see [15]).
Assume that∫
0
1
‍
ϕ
(
t
)
u
(
t
)
d
t
≠
1. Then for any ψ
∈
𝒞
(
0,1
), the solution of boundary value problem
(17)
D
0
+
α
u
(
t
)
+
ψ
(
t
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
ϕ
(
t
)
u
(
t
)
d
t
,
is given by u
(
t
)
=
∫
0
1
‍
G
1
α
(
t
,
s
)
ψ
(
s
)
d
s, where G
1
α
(
t
,
s
)
=
G
2
α
(
t
,
s
)
+
G
3
α
(
t
,
s
),
(18)
G
2
α
(
t
,
s
)
=
{
t
α
-
1
(
1
-
s
)
α
-
1
-
(
t
-
s
)
α
-
1
Γ
(
α
)
,
0
≤
s
≤
t
≤
1
.
t
α
-
1
(
1
-
s
)
α
-
1
Γ
(
α
)
,
0
≤
t
≤
s
≤
1
.
G
3
α
(
t
,
s
)
=
t
α
-
1
1
-
∫
0
1
‍
ϕ
(
t
)
t
α
-
1
d
t
∫
0
1
‍
ϕ
(
t
)
G
2
α
(
t
,
s
)
d
t
.Lemma 7 (see [15]).
If∫
0
1
‍
ϕ
(
t
)
t
α
-
1
d
t
∈
[
0,1
), then Green’s function G
1
α
(
t
,
s
) has the following properties: (i)
G
1
α
(
t
,
s
)
≥
0 is continuous for any t
,
s
∈
[
0,1
], G
1
α
(
t
,
s
)
>
0 for any t
,
s
∈
(
0,1
);
(ii)
G
1
α
(
t
,
s
)
≤
G
1
α
(
s
) for any t
,
s
∈
[
0,1
] and min
⁡
t
∈
[
θ
,
1
-
θ
]
G
1
α
(
t
,
s
)
≥
γ
θ
G
1
α
(
s
), where G
1
α
(
s
)
=
G
2
α
(
s
,
s
)
+
G
3
α
(
1
,
s
), γ
θ
=
θ
α
-
1, θ
∈
(
0
,
1
/
2
).


## 3. Main Results

In this section, we will discuss the existence of positive solutions for problem (6). For a coupled system, many researchers transformed the problem into computing the fixed point index of the composition operator on the single cone in the Cartesian product space and the nonlinear terms in two equations have similar features. However it is very difficult to deal with our problem directly. Denote 𝒞
+
[
0,1
] the continuous function in [
0,1
] with nonnegative value. In order to make full use of the different nature of the two equations, we construct single cone K
i and open set K
R
i
∖
K
r
i
¯, i
=
1,2 in 𝒞
+
[
0,1
]. Thus we transform the problem into computing the fixed point index of operator that is defined in Cartesian product of two cones in open set D
=
(
K
R
1
∖
K
r
1
¯
)
×
(
K
R
2
∖
K
r
2
¯
) of product cone K
1
×
K
2.


### 3.1. The Transformation of the Problem

Firstly, we construct the relevant coneK in Banach space 𝒞
+
[
0,1
]. Then we transform the problem into computing the fixed point index of Cartesian product of cone with the nature of analytical property of Green’s function.Following, we introduce notations:(19)
𝒞
+
[
0,1
]
=
{
u
∈
𝒞
[
0,1
]
∣
u
(
t
)
≥
0
,
∀
t
∈
[
0,1
]
}
,
with the norm ∥
u
∥
=
max
⁡
t
∈
[
0,1
]
{
|
u
(
t
)
|
}. Clearly, it is a Banach space. Define
(20)
K
1
=
{
u
∈
𝒞
+
[
0,1
]
∣
u
(
t
)
≥
(
1
2
)
α
-
1
∥
u
(
t
)
∥
,
11111111111111111
∀
t
∈
[
θ
,
1
-
θ
]
,
θ
∈
(
0
,
1
2
)
u
(
t
)
≥
(
1
2
)
α
-
1
∥
u
(
t
)
∥
,
}
,
K
2
=
{
u
∈
𝒞
+
[
0,1
]
∣
u
(
t
)
≥
(
1
2
)
β
-
1
∥
u
(
t
)
∥
,
11111111111111111
∀
t
∈
[
θ
,
1
-
θ
]
,
θ
∈
(
0
,
1
2
)
u
(
t
)
≥
(
1
2
)
β
-
1
∥
u
(
t
)
∥
,
}
.Forλ
∈
[
0,1
]
,
u
,
v
∈
𝒞
+
[
0,1
], we define the operators
(21)
T
λ
,
1
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
,
T
λ
,
2
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
,
T
λ
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
,
as follows:
(22)
T
λ
,
1
(
u
,
v
)
=
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
1111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
,
T
λ
,
2
(
u
,
v
)
=
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
1111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
,
T
λ
(
u
,
v
)
(
t
)
=
(
T
λ
,
1
(
u
,
v
)
(
t
)
,
T
λ
,
2
(
u
,
v
)
(
t
)
)
.Then we translate that problem (6) possesses a pair of solutions (
u
,
v
) if and only if T
λ
(
u
,
v
)
(
t
) have a fixed point in K
1
×
K
2.Lemma 8.
The mappingT
λ
:
K
1
×
K
2
→
K
1
×
K
2 as mentioned above is completely continuous.Proof.
For any(
u
,
v
)
∈
K
1
×
K
2, we want to prove T
λ
(
u
,
v
)
∈
K
1
×
K
2, that is, to prove T
λ
,
1
(
u
,
v
)
∈
K
1 and T
λ
,
2
(
u
,
v
)
∈
K
2. By the preceding definition and the nature of Green's function, we have for t
∈
[
θ
,
1
-
θ
], θ
∈
(
0
,
1
/
2
)
(23)
T
λ
,
1
(
u
,
v
)
=
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
∫
0
1
‍
γ
θ
G
1
α
(
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
=
γ
θ
∫
0
1
‍
G
1
α
(
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
(
1
2
)
α
-
1
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
(
1
2
)
α
-
1
∥
T
λ
,
1
(
u
,
v
)
∥
.
Similarly,T
λ
,
2
(
u
,
v
)
≥
(
1
/
2
)
β
-
1
∥
T
λ
,
2
(
u
,
v
)
∥, t
∈
[
θ
,
1
-
θ
], θ
∈
(
0
,
1
/
2
). Thus, T
λ
,
1
(
u
,
v
)
∈
K
1, T
λ
,
2
(
u
,
v
)
∈
K
2, and T
λ
(
K
1
×
K
2
)
⊂
K
1
×
K
2.
It is easy to prove thatT
λ
(
u
,
v
) is continuous in K
1
×
K
2.
So as Arzela-Ascoli theorem, we prove thatT
λ
:
K
1
×
K
2
→
K
1
×
K
2 is completely continuous.


### 3.2. The Main Result and Its Proof

Now, we present our main result.Theorem 9.
Assume that the nonlinear terms of the problem (6) f and g meet the following conditions:
(24)
(
H
1
)
{
(
a
1
)
lim
⁡
u
→
0
+
sup
⁡
max
⁡
t
∈
[
0,1
]
f
(
t
,
u
,
v
)
u
=
0
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
ℝ
+
,
(
b
1
)
lim
⁡
u
→
+
∞
inf
⁡
min
⁡
t
∈
[
0,1
]
f
(
t
,
u
,
v
)
u
=
+
∞
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
ℝ
+
,
(
H
2
)
{
(
a
2
)
lim
⁡
v
→
0
+
inf
⁡
min
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
v
=
+
∞
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
u
∈
ℝ
+
,
(
b
2
)
lim
⁡
v
→
+
∞
sup
⁡
max
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
v
=
0
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
u
∈
ℝ
+
,
(
H
3
)
lim
⁡
u
→
+
∞
sup
⁡
max
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
=
h
(
v
)
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
[
0
,
M
]
,
where h
(
v
) is continuous in ℝ
+ and M is a constant.
Then the problem (6) has at least a pair of positive solutions.Proof.
We select an appropriate open setD
=
(
K
R
1
∖
K
r
1
¯
)
×
(
K
R
2
∖
K
r
2
¯
) in the product of cone K
1
×
K
2; then we will verify that the solution operator meets the conditions of Theorem 4.
In the following, we separate our proof into four steps.
Step  1. Select r
1
>
0, in such a fashion that ∥
T
λ
,
1
(
u
,
v
)
∥
<
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
1
×
K
1. Based on assuming (
H
1
)
(
a
1
), for ɛ
1
=
(
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists r
1
>
0, such that f
(
t
,
u
,
v
)
<
ɛ
1
u, for all t
∈
[
0,1
], 0
≤
u
≤
r
1, v
≥
0. Then when λ
∈
[
0,1
]
,
(
u
,
v
)
∈
∂
K
r
1
×
K
1, we have for t
∈
[
0,1
]
(25)
∥
T
λ
,
1
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
∫
0
1
‍
|
<
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
α
(
s
)
|
λ
ɛ
1
u
(
s
)
+
(
1
-
λ
)
ɛ
1
u
(
s
)
|
d
s
≤
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
α
(
s
)
ɛ
1
|
u
(
s
)
|
d
s
≤
ɛ
1
∫
0
1
‍
G
1
α
(
s
)
max
⁡
t
∈
[
0,1
]
|
u
(
s
)
|
d
s
≤
ɛ
1
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
u
∥
=
∥
u
∥
.
That is to say, ∥
T
λ
,
1
(
u
,
v
)
∥
<
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
1
×
K
1.
Step  2. Select R
1
>
r
1; we claim that ∥
T
λ
,
1
(
u
,
v
)
∥
>
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
1
×
K
1. In view of the assumption (
H
1
)
(
b
1
), for N
1
=
(
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists R
0
>
0, such that f
(
t
,
u
,
v
)
<
N
1
u, for all t
∈
[
0,1
], u
≥
R
0, v
≥
0. Let R
1
>
max
⁡
{
r
1
,
γ
θ
R
0
}. Then when λ
∈
[
0,1
]
,
(
u
,
v
)
∈
∂
K
R
1
×
K
1, we have t
∈
[
0,1
]
(26)
∥
T
λ
,
1
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
∫
0
1
‍
|
>
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
γ
θ
G
1
α
(
s
)
|
λ
N
1
u
(
s
)
+
(
1
-
λ
)
N
1
u
(
s
)
|
d
s
≥
max
⁡
t
∈
[
0,1
]
γ
θ
∫
0
1
‍
G
1
α
(
s
)
N
1
|
u
(
s
)
|
d
s
≥
N
1
γ
θ
∫
0
1
‍
G
1
α
(
s
)
max
⁡
t
∈
(
0,1
)
|
u
(
s
)
|
d
s
≥
N
1
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
u
∥
=
∥
u
∥
.
That is to say, ∥
T
λ
,
1
(
u
,
v
)
∥
>
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
1
×
K
1.
Step  3. Select r
2
>
0, such that ∥
T
λ
,
2
(
u
,
v
)
∥
>
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2. In view of the assumption (
H
2
)
(
a
2
), for N
2
=
(
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists r
2
>
0, such that g
(
t
,
u
,
v
)
>
N
2
v, for all t
∈
[
0,1
], 0
≤
v
≤
r
2, u
≥
0. Hence when λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2, we have for t
∈
[
0,1
]
(27)
∥
T
λ
,
2
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
∫
0
1
‍
|
>
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
γ
θ
G
1
β
(
s
)
|
λ
N
2
v
(
s
)
+
(
1
-
λ
)
N
2
v
(
s
)
|
d
s
≥
max
⁡
t
∈
[
0,1
]
γ
θ
∫
0
1
‍
G
1
β
(
s
)
N
2
|
v
(
s
)
|
d
s
≥
N
2
γ
θ
∫
0
1
‍
G
1
β
(
s
)
max
⁡
t
∈
(
0,1
)
|
v
(
s
)
|
d
s
≥
N
2
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
v
∥
=
∥
v
∥
.
That is to say, ∥
T
λ
,
2
(
u
,
v
)
∥
>
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2.
Step  4. Select R
2
>
r
2, such that ∥
T
λ
,
2
(
u
,
v
)
∥
<
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2. Based on the condition (
H
2
)
(
b
2
), for ɛ
2
=
(
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists R
′
>
0, such that g
(
t
,
u
,
v
)
<
ɛ
2
v, for all t
∈
[
0,1
], v
≥
R
′, u
≥
0. Furthermore, paying attention to the assumptions (
H
3
), we can select R
2
>
0
,
R
2 which is sufficiently large and
(28)
max
⁡
t
∈
[
0,1
]
{
g
(
t
,
u
,
v
)
∣
0
≤
v
≤
R
′
,
u
≥
0
}
<
ɛ
2
R
2
.
Thus g
(
t
,
u
,
v
)
<
ɛ
2
R
2, where t
∈
[
0,1
], 0
≤
v
≤
R
2, u
≥
0. Hence when λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2, we have for t
∈
[
0,1
]
(29)
∥
T
λ
,
2
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
∫
0
1
‍
|
<
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
β
(
s
)
(
λ
ɛ
2
R
2
+
(
1
-
λ
)
ɛ
2
R
2
)
d
s
≤
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
β
(
s
)
ɛ
2
R
2
d
s
≤
ɛ
2
R
2
∫
0
1
‍
G
1
β
(
s
)
max
⁡
t
∈
[
0,1
]
v
(
s
)
d
s
≤
ɛ
2
R
2
∫
0
1
‍
G
1
α
(
s
)
d
s
=
∥
v
∥
.
That is to say, ∥
T
λ
,
2
(
u
,
v
)
∥
<
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2.
Combining with (25)–(29), we know the conditions of Theorem 4 are all established. Thus the completely continuous operator T has a fixed point in D; that is to say, at least a pair of positive solutions exist for the problem (6). The proof is complete.Remark 10.
Binding (27), (29) and the assumption (
H
1
)
(
a
1
), we can get a nonnegative solution (
0
,
v
) of the original equations (6). In fact, by (27) and (29), we know the fixed point index i
(
T
0,2
(
u
,
v
)
,
K
R
2
∖
K
r
2
¯
,
K
2
)
=
-
1; that is to say, the second equation of the original equations has a solution (
0
,
v
). Further, by hypothesis (
H
1
)
(
a
1
), we have f
(
t
,
0
,
v
(
t
)
)
=
0, so the solution of the second equation is also the solution of the first equation.


## 3.1. The Transformation of the Problem

Firstly, we construct the relevant coneK in Banach space 𝒞
+
[
0,1
]. Then we transform the problem into computing the fixed point index of Cartesian product of cone with the nature of analytical property of Green’s function.Following, we introduce notations:(19)
𝒞
+
[
0,1
]
=
{
u
∈
𝒞
[
0,1
]
∣
u
(
t
)
≥
0
,
∀
t
∈
[
0,1
]
}
,
with the norm ∥
u
∥
=
max
⁡
t
∈
[
0,1
]
{
|
u
(
t
)
|
}. Clearly, it is a Banach space. Define
(20)
K
1
=
{
u
∈
𝒞
+
[
0,1
]
∣
u
(
t
)
≥
(
1
2
)
α
-
1
∥
u
(
t
)
∥
,
11111111111111111
∀
t
∈
[
θ
,
1
-
θ
]
,
θ
∈
(
0
,
1
2
)
u
(
t
)
≥
(
1
2
)
α
-
1
∥
u
(
t
)
∥
,
}
,
K
2
=
{
u
∈
𝒞
+
[
0,1
]
∣
u
(
t
)
≥
(
1
2
)
β
-
1
∥
u
(
t
)
∥
,
11111111111111111
∀
t
∈
[
θ
,
1
-
θ
]
,
θ
∈
(
0
,
1
2
)
u
(
t
)
≥
(
1
2
)
β
-
1
∥
u
(
t
)
∥
,
}
.Forλ
∈
[
0,1
]
,
u
,
v
∈
𝒞
+
[
0,1
], we define the operators
(21)
T
λ
,
1
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
,
T
λ
,
2
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
,
T
λ
(
u
,
v
)
:
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
⟶
𝒞
+
[
0,1
]
×
𝒞
+
[
0,1
]
,
as follows:
(22)
T
λ
,
1
(
u
,
v
)
=
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
1111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
,
T
λ
,
2
(
u
,
v
)
=
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
1111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
,
T
λ
(
u
,
v
)
(
t
)
=
(
T
λ
,
1
(
u
,
v
)
(
t
)
,
T
λ
,
2
(
u
,
v
)
(
t
)
)
.Then we translate that problem (6) possesses a pair of solutions (
u
,
v
) if and only if T
λ
(
u
,
v
)
(
t
) have a fixed point in K
1
×
K
2.Lemma 8.
The mappingT
λ
:
K
1
×
K
2
→
K
1
×
K
2 as mentioned above is completely continuous.Proof.
For any(
u
,
v
)
∈
K
1
×
K
2, we want to prove T
λ
(
u
,
v
)
∈
K
1
×
K
2, that is, to prove T
λ
,
1
(
u
,
v
)
∈
K
1 and T
λ
,
2
(
u
,
v
)
∈
K
2. By the preceding definition and the nature of Green's function, we have for t
∈
[
θ
,
1
-
θ
], θ
∈
(
0
,
1
/
2
)
(23)
T
λ
,
1
(
u
,
v
)
=
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
∫
0
1
‍
γ
θ
G
1
α
(
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
=
γ
θ
∫
0
1
‍
G
1
α
(
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
(
1
2
)
α
-
1
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
111111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
≥
(
1
2
)
α
-
1
∥
T
λ
,
1
(
u
,
v
)
∥
.
Similarly,T
λ
,
2
(
u
,
v
)
≥
(
1
/
2
)
β
-
1
∥
T
λ
,
2
(
u
,
v
)
∥, t
∈
[
θ
,
1
-
θ
], θ
∈
(
0
,
1
/
2
). Thus, T
λ
,
1
(
u
,
v
)
∈
K
1, T
λ
,
2
(
u
,
v
)
∈
K
2, and T
λ
(
K
1
×
K
2
)
⊂
K
1
×
K
2.
It is easy to prove thatT
λ
(
u
,
v
) is continuous in K
1
×
K
2.
So as Arzela-Ascoli theorem, we prove thatT
λ
:
K
1
×
K
2
→
K
1
×
K
2 is completely continuous.


## 3.2. The Main Result and Its Proof

Now, we present our main result.Theorem 9.
Assume that the nonlinear terms of the problem (6) f and g meet the following conditions:
(24)
(
H
1
)
{
(
a
1
)
lim
⁡
u
→
0
+
sup
⁡
max
⁡
t
∈
[
0,1
]
f
(
t
,
u
,
v
)
u
=
0
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
ℝ
+
,
(
b
1
)
lim
⁡
u
→
+
∞
inf
⁡
min
⁡
t
∈
[
0,1
]
f
(
t
,
u
,
v
)
u
=
+
∞
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
ℝ
+
,
(
H
2
)
{
(
a
2
)
lim
⁡
v
→
0
+
inf
⁡
min
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
v
=
+
∞
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
u
∈
ℝ
+
,
(
b
2
)
lim
⁡
v
→
+
∞
sup
⁡
max
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
v
=
0
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
u
∈
ℝ
+
,
(
H
3
)
lim
⁡
u
→
+
∞
sup
⁡
max
⁡
t
∈
[
0,1
]
g
(
t
,
u
,
v
)
=
h
(
v
)
i
s
e
s
t
a
b
l
i
s
h
e
d
c
o
n
s
i
s
t
e
n
t
a
b
o
u
t
v
∈
[
0
,
M
]
,
where h
(
v
) is continuous in ℝ
+ and M is a constant.
Then the problem (6) has at least a pair of positive solutions.Proof.
We select an appropriate open setD
=
(
K
R
1
∖
K
r
1
¯
)
×
(
K
R
2
∖
K
r
2
¯
) in the product of cone K
1
×
K
2; then we will verify that the solution operator meets the conditions of Theorem 4.
In the following, we separate our proof into four steps.
Step  1. Select r
1
>
0, in such a fashion that ∥
T
λ
,
1
(
u
,
v
)
∥
<
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
1
×
K
1. Based on assuming (
H
1
)
(
a
1
), for ɛ
1
=
(
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists r
1
>
0, such that f
(
t
,
u
,
v
)
<
ɛ
1
u, for all t
∈
[
0,1
], 0
≤
u
≤
r
1, v
≥
0. Then when λ
∈
[
0,1
]
,
(
u
,
v
)
∈
∂
K
r
1
×
K
1, we have for t
∈
[
0,1
]
(25)
∥
T
λ
,
1
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
∫
0
1
‍
|
<
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
α
(
s
)
|
λ
ɛ
1
u
(
s
)
+
(
1
-
λ
)
ɛ
1
u
(
s
)
|
d
s
≤
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
α
(
s
)
ɛ
1
|
u
(
s
)
|
d
s
≤
ɛ
1
∫
0
1
‍
G
1
α
(
s
)
max
⁡
t
∈
[
0,1
]
|
u
(
s
)
|
d
s
≤
ɛ
1
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
u
∥
=
∥
u
∥
.
That is to say, ∥
T
λ
,
1
(
u
,
v
)
∥
<
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
1
×
K
1.
Step  2. Select R
1
>
r
1; we claim that ∥
T
λ
,
1
(
u
,
v
)
∥
>
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
1
×
K
1. In view of the assumption (
H
1
)
(
b
1
), for N
1
=
(
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists R
0
>
0, such that f
(
t
,
u
,
v
)
<
N
1
u, for all t
∈
[
0,1
], u
≥
R
0, v
≥
0. Let R
1
>
max
⁡
{
r
1
,
γ
θ
R
0
}. Then when λ
∈
[
0,1
]
,
(
u
,
v
)
∈
∂
K
R
1
×
K
1, we have t
∈
[
0,1
]
(26)
∥
T
λ
,
1
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
α
(
t
,
s
)
(
λ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
f
(
s
,
u
(
s
)
,
0
)
)
d
s
∫
0
1
‍
|
>
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
γ
θ
G
1
α
(
s
)
|
λ
N
1
u
(
s
)
+
(
1
-
λ
)
N
1
u
(
s
)
|
d
s
≥
max
⁡
t
∈
[
0,1
]
γ
θ
∫
0
1
‍
G
1
α
(
s
)
N
1
|
u
(
s
)
|
d
s
≥
N
1
γ
θ
∫
0
1
‍
G
1
α
(
s
)
max
⁡
t
∈
(
0,1
)
|
u
(
s
)
|
d
s
≥
N
1
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
u
∥
=
∥
u
∥
.
That is to say, ∥
T
λ
,
1
(
u
,
v
)
∥
>
∥
u
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
1
×
K
1.
Step  3. Select r
2
>
0, such that ∥
T
λ
,
2
(
u
,
v
)
∥
>
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2. In view of the assumption (
H
2
)
(
a
2
), for N
2
=
(
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists r
2
>
0, such that g
(
t
,
u
,
v
)
>
N
2
v, for all t
∈
[
0,1
], 0
≤
v
≤
r
2, u
≥
0. Hence when λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2, we have for t
∈
[
0,1
]
(27)
∥
T
λ
,
2
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
∫
0
1
‍
|
>
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
γ
θ
G
1
β
(
s
)
|
λ
N
2
v
(
s
)
+
(
1
-
λ
)
N
2
v
(
s
)
|
d
s
≥
max
⁡
t
∈
[
0,1
]
γ
θ
∫
0
1
‍
G
1
β
(
s
)
N
2
|
v
(
s
)
|
d
s
≥
N
2
γ
θ
∫
0
1
‍
G
1
β
(
s
)
max
⁡
t
∈
(
0,1
)
|
v
(
s
)
|
d
s
≥
N
2
γ
θ
∫
0
1
‍
G
1
α
(
s
)
d
s
∥
v
∥
=
∥
v
∥
.
That is to say, ∥
T
λ
,
2
(
u
,
v
)
∥
>
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
r
2
×
K
2.
Step  4. Select R
2
>
r
2, such that ∥
T
λ
,
2
(
u
,
v
)
∥
<
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2. Based on the condition (
H
2
)
(
b
2
), for ɛ
2
=
(
∫
0
1
‍
G
1
α
(
s
)
d
s
)
-
1
>
0, there exists R
′
>
0, such that g
(
t
,
u
,
v
)
<
ɛ
2
v, for all t
∈
[
0,1
], v
≥
R
′, u
≥
0. Furthermore, paying attention to the assumptions (
H
3
), we can select R
2
>
0
,
R
2 which is sufficiently large and
(28)
max
⁡
t
∈
[
0,1
]
{
g
(
t
,
u
,
v
)
∣
0
≤
v
≤
R
′
,
u
≥
0
}
<
ɛ
2
R
2
.
Thus g
(
t
,
u
,
v
)
<
ɛ
2
R
2, where t
∈
[
0,1
], 0
≤
v
≤
R
2, u
≥
0. Hence when λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2, we have for t
∈
[
0,1
]
(29)
∥
T
λ
,
2
(
u
,
v
)
∥
=
max
⁡
t
∈
[
0,1
]
|
∫
0
1
‍
G
1
β
(
t
,
s
)
(
λ
g
(
s
,
u
(
s
)
,
v
(
s
)
)
11111111111111111111
+
(
1
-
λ
)
g
(
s
,
0
,
v
(
s
)
)
)
d
s
∫
0
1
‍
|
<
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
β
(
s
)
(
λ
ɛ
2
R
2
+
(
1
-
λ
)
ɛ
2
R
2
)
d
s
≤
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
1
β
(
s
)
ɛ
2
R
2
d
s
≤
ɛ
2
R
2
∫
0
1
‍
G
1
β
(
s
)
max
⁡
t
∈
[
0,1
]
v
(
s
)
d
s
≤
ɛ
2
R
2
∫
0
1
‍
G
1
α
(
s
)
d
s
=
∥
v
∥
.
That is to say, ∥
T
λ
,
2
(
u
,
v
)
∥
<
∥
v
∥, where λ
∈
[
0,1
], (
u
,
v
)
∈
∂
K
R
2
×
K
2.
Combining with (25)–(29), we know the conditions of Theorem 4 are all established. Thus the completely continuous operator T has a fixed point in D; that is to say, at least a pair of positive solutions exist for the problem (6). The proof is complete.Remark 10.
Binding (27), (29) and the assumption (
H
1
)
(
a
1
), we can get a nonnegative solution (
0
,
v
) of the original equations (6). In fact, by (27) and (29), we know the fixed point index i
(
T
0,2
(
u
,
v
)
,
K
R
2
∖
K
r
2
¯
,
K
2
)
=
-
1; that is to say, the second equation of the original equations has a solution (
0
,
v
). Further, by hypothesis (
H
1
)
(
a
1
), we have f
(
t
,
0
,
v
(
t
)
)
=
0, so the solution of the second equation is also the solution of the first equation.



---

*Source: 790862-2014-05-12.xml*