# Positive Solutions for Systems of Nonlinear Higher Order Differential Equations with Integral Boundary Conditions

Authors: Yaohong Li; Xiaoyan Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

By constructing some general type conditions and using fixed point theorem of cone, this paper investigates the existence of at least one and at least two positive solutions for systems of nonlinear higher order differential equations with integral boundary conditions. As application, some examples are given.


---

## Body

## 1. Introduction

In this paper, we consider the following systems of nonlinear mixed higher order differential equations with integral boundary conditions:(1)
u
(
n
1
)
(
t
)
+
a
1
(
t
)
f
1
(
t
,
u
(
t
)
,
v
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
v
(
n
2
)
(
t
)
+
a
2
(
t
)
f
2
(
t
,
u
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
′
(
0
)
=
⋯
=
u
(
n
1
-
2
)
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
n
1
(
t
)
u
(
t
)
d
t
,
v
(
0
)
=
v
′
(
0
)
=
⋯
=
v
(
n
2
-
2
)
(
0
)
=
0
,
v
(
1
)
=
∫
0
1
‍
n
2
(
t
)
v
(
t
)
d
t
,
where f
1
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
,
[
0
,
+
∞
)
), f
2
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
,
[
0
,
+
∞
)
), a
i
∈
C
(
[
0,1
]
,
[
0
,
+
∞
)
), n
i
≥
3, and n
i
(
t
)
∈
L
1
[
0,1
] is nonnegative, i
=
1,2; f
1
(
t
,
0,0
)
≡
f
2
(
t
,
0
)
≡
0.Boundary value problems with integral boundary conditions arise naturally in thermal conduction problems [1], semiconductor problems [2], and hydrodynamic problems [3]. Such problems include two-, three-, and multipoint boundary value problems as special cases and attracted much attention (see [4–12] and the references therein). In particular, we would like to mention the result of Pang et al. [9]. In [9], by applying fixed point index theory, Pang et al. study the expression and properties of Green’s function and obtained the existence of positive solutions for nth-order m-point boundary value problems:
(2)
u
(
n
)
(
t
)
+
a
(
t
)
f
(
u
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
′
(
0
)
=
⋯
=
u
(
n
-
2
)
(
0
)
=
0
,
u
(
1
)
=
∑
i
=
1
m
-
2
‍
α
i
u
(
η
i
)
.
Yang and Wei [10], Feng and Ge [11], and Li and Wei [12] improved and generalized the results of [9] by using different methods.On the other hand, much effort has been devoted to the study of the existence of positive solutions for systems of nonlinear differential equations (see [13–16] and the references therein). In [13], by applying Krasnoselskii fixed point theorem in a cone, Hu and Wang obtained multiple positive solutions of boundary value problems for systems of nonlinear second-order differential equations. In [14], Henderson and Ntouyas extended the results of [13] to systems of nonlinear nth-order three-point boundary value problems:
(3)
u
(
n
)
(
t
)
+
λ
a
(
t
)
f
(
t
,
v
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
v
(
n
)
(
t
)
+
λ
b
(
t
)
h
(
t
,
u
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
′
(
0
)
=
⋯
=
u
(
n
-
2
)
(
0
)
=
0
,
u
(
1
)
=
α
u
(
η
)
,
v
(
0
)
=
v
′
(
0
)
=
⋯
=
v
(
n
-
2
)
(
0
)
=
0
,
v
(
1
)
=
α
v
(
η
)
.
In [15], by using fixed point index theory, Xie and Zhu improved the results of [14]. At the same time, boundary value problems with integral boundary conditions have received attention [16, 17].Motivated by the work of the abovementioned papers, our aim in this paper is to study the existence of positive solutions associated with systems (1) by applying fixed point theorem in cone. Further, we present some general type conditions (H
4)–(H
7) instead of the sublinear or superlinear conditions which are used in [4, 5, 8, 10, 12–14]. Our conditions are applicable for more general functions.


## 2. Several Lemmas

For convenience, we make the following notations. Let(4)
β
i
=
∫
0
1
‍
n
i
(
t
)
t
n
i
-
1
d
t
,
μ
i
=
∫
0
1
‍
K
i
(
s
)
a
i
(
s
)
d
s
,
δ
i
=
∫
a
b
‍
K
i
(
s
)
a
i
(
s
)
d
s
,
i
=
1,2
,
where K
i
(
s
) is defined by Lemma 6 and [
a
,
b
] is some subset of (
0,1
).List the following assumptions:(
H
1)
a
i
∈
C
(
[
0,1
]
,
[
0
,
+
∞
)
), a
i
(
t
) do not vanish identically for t
∈
[
a
,
b
], i
=
1,2;
(H
2)
f
1
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
,
[
0
,
+
∞
)
), f
2
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
,
[
0
,
+
∞
)
);
(H
3)
β
1
,
β
2
∈
[
0,1
);
(H
4)
there existα
∈
(
0,1
], λ
1
>
0 and a sufficiently large M
1
>
1 such that
(1)
f
1
(
t
,
u
,
v
)
≥
λ
1
v
α, for all (
t
,
u
,
v
)
∈
[
0,1
]
×
[
0
,
+
∞
)
×
[
M
1
,
+
∞
),
(2)
f
2
(
t
,
u
)
≥
C
1
u
1
/
α, for all (
t
,
u
)
∈
[
0,1
]
×
[
M
1
,
+
∞
),
whereC
1
=
max
⁡
{
(
γ
δ
2
)
-
1
,
(
γ
δ
2
)
-
1
(
γ
2
λ
1
δ
1
)
-
1
/
α
}; γ is defined by (21).
(H
5)
There existβ
∈
(
0
,
+
∞
), λ
2
>
0 and a sufficiently small ρ
2
∈
(
0,1
) such that
(1)
f
1
(
t
,
u
,
v
)
≤
λ
2
v
β, for all (
t
,
u
,
v
)
∈
[
0,1
]
×
[
0
,
+
∞
)
×
[
0
,
ρ
2
],
(2)
f
2
(
t
,
u
)
≤
C
2
u
1
/
β, for all (
t
,
u
)
∈
[
0,1
]
×
[
0
,
ρ
2
],
whereC
2
=
min
⁡
{
ρ
2
μ
2
-
1
,
μ
2
-
1
/
β
(
μ
1
λ
2
)
-
1
}.
(H
6)
There existp
∈
(
0
,
+
∞
), λ
3
>
0, and M
2
>
0 such that
(1)
f
1
(
t
,
u
,
v
)
≤
λ
3
v
p
+
M
2, for all (
t
,
u
,
v
)
∈
[
0,1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
),
(2)
f
2
(
t
,
u
)
≤
C
3
u
1
/
p
+
M
2, for all (
t
,
u
)
∈
[
0,1
]
×
[
0
,
+
∞
),
whereC
3
=
(
2
μ
1
λ
3
)
-
1
/
p
μ
2
-
1.
(H
7)
There existq
∈
(
0,1
], λ
4
>
0 and a sufficiently small ε
>
0 such that
(1)
f
1
(
t
,
u
,
v
)
≥
λ
4
v
q, for all (
t
,
u
,
v
)
∈
[
0,1
]
×
[
0
,
+
∞
)
×
[
0
,
ε
],
(2)
f
2
(
t
,
u
)
≥
C
4
u
1
/
q, for all (
t
,
u
)
∈
[
0,1
]
×
[
0
,
ε
],
whereC
4
=
γ
-
(
1
/
q
)
(
2
+
q
)
(
λ
4
δ
1
)
-
1
/
q
δ
2
-
1.
(H
8)
f
1
(
t
,
u
,
v
) and f
2
(
t
,
u
) are increasing on u, v and there exists R
>
0 such that
f
1
(
s
,
R
,
∫
0
1
‍
K
2
(
r
)
a
2
(
r
)
f
2
(
r
,
R
)
d
r
)
<
μ
1
-
1
R, for all s
,
r
∈
[
0,1
].Lemma 1.
Ifβ
i
∈
[
0,1
), for any y
(
t
)
∈
C
[
0,1
], higher order differential equations
(5)
w
(
n
i
)
(
t
)
+
y
(
t
)
=
0
,
t
∈
(
0,1
)
,
w
(
0
)
=
w
′
(
0
)
=
⋯
=
w
(
n
i
-
2
)
(
0
)
=
0
,
w
(
1
)
=
∫
0
1
‍
n
i
(
t
)
w
(
t
)
d
t
have a unique solution
(6)
w
(
t
)
=
∫
0
1
‍
K
i
(
t
,
s
)
y
(
s
)
d
s
,
where
(7)
K
i
(
t
,
s
)
=
K
i
1
(
t
,
s
)
+
K
i
2
(
t
,
s
)
,
(8)
K
i
1
(
t
,
s
)
=
1
(
n
i
-
1
)
!
{
t
n
i
-
1
(
1
-
s
)
n
i
-
1
-
(
t
-
s
)
n
i
-
1
,
0
≤
s
≤
t
≤
1
,
t
n
i
-
1
(
1
-
s
)
n
i
-
1
,
0
≤
t
≤
s
≤
1
,
(9)
K
i
2
(
t
,
s
)
=
t
n
i
-
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
K
i
1
(
t
,
s
)
d
t
.Proof.
By Taylor’s formula, we get(10)
w
(
t
)
=
-
1
(
n
i
-
1
)
!
∫
0
t
(
t
-
s
)
n
i
-
1
y
(
s
)
d
s
‍
+
A
(
n
i
-
1
)
!
t
n
i
-
1
.
Letting t
=
1 in (10), we have
(11)
A
=
(
n
i
-
1
)
!
w
(
1
)
+
∫
0
1
(
1
-
s
)
n
i
-
1
y
(
s
)
d
s
.
Substituting w
(
1
)
=
∫
0
1
‍
n
i
(
t
)
w
(
t
)
d
t and (11) into (10), we obtain
(12)
w
(
t
)
=
-
1
(
n
i
-
1
)
!
∫
0
t
(
t
-
s
)
n
i
-
1
y
(
s
)
d
s
‍
+
1
(
n
i
-
1
)
!
∫
0
1
t
n
i
-
1
(
1
-
s
)
n
i
-
1
y
(
s
)
d
s
+
t
n
i
-
1
∫
0
1
n
i
(
s
)
w
(
s
)
d
s
‍
=
1
(
n
i
-
1
)
!
∫
0
t
[
t
n
i
-
1
(
1
-
s
)
n
i
-
1
-
(
t
-
s
)
n
i
-
1
]
‍
y
(
s
)
d
s
+
t
n
i
-
1
∫
0
1
n
i
(
s
)
w
(
s
)
d
s
‍
+
1
(
n
i
-
1
)
!
∫
t
1
t
n
i
-
1
(
1
-
s
)
n
i
-
1
y
(
s
)
d
s
‍
=
∫
0
1
K
i
1
(
t
,
s
)
y
(
s
)
d
s
+
t
n
i
-
1
∫
0
1
n
i
(
s
)
w
(
s
)
d
s
.
Multiplying (12) with n
i
(
t
) and integrating it, we have
(13)
∫
0
1
‍
n
i
(
t
)
w
(
t
)
d
t
=
∫
0
1
‍
n
i
(
t
)
∫
0
1
‍
K
i
1
(
t
,
s
)
y
(
s
)
d
s
d
t
+
∫
0
1
‍
n
i
(
t
)
t
n
i
-
1
d
t
∫
0
1
‍
n
i
(
s
)
w
(
s
)
d
s
,
so
(14)
∫
0
1
‍
n
i
(
t
)
w
(
t
)
d
t
=
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
∫
0
1
‍
K
i
1
(
t
,
s
)
y
(
s
)
d
s
d
t
.
Substituting (14) into (12), we have
(15)
w
(
t
)
=
∫
0
1
‍
K
i
1
(
t
,
s
)
y
(
s
)
d
s
+
t
n
i
-
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
∫
0
1
‍
K
i
1
(
t
,
s
)
y
(
s
)
d
s
d
t
=
∫
0
1
‍
K
i
1
(
t
,
s
)
y
(
s
)
d
s
+
∫
0
1
‍
(
t
n
i
-
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
K
i
1
(
t
,
s
)
d
t
)
y
(
s
)
d
s
=
∫
0
1
‍
[
K
i
1
(
t
,
s
)
+
K
i
2
(
t
,
s
)
]
y
(
s
)
d
s
=
∫
0
1
‍
K
i
(
t
,
s
)
y
(
s
)
d
s
,
where K
i
(
t
,
s
) is defined by (7).Definition 2.
(
u
,
v
)
∈
C
n
1
(
0,1
)
∩
C
[
0,1
]
×
C
n
2
(
0,1
)
∩
C
[
0,1
] is said to be a positive solution of systems (1) if and only if (
u
,
v
) satisfies systems (1) and u
(
t
)
>
0, v
(
t
)
>
0, for any t
∈
[
0,1
].Lemma 3 (see [6]).
Ifβ
i
∈
[
0,1
), the continuous function K
i
1
(
t
,
s
)
,
i
=
1,2, has the following properties: (i)
0
≤
K
i
1
(
t
,
s
)
≤
K
i
1
(
s
), for all t
,
s
∈
[
0,1
], where K
i
1
(
s
)
:
=
s
(
1
-
s
)
n
i
-
1
/
(
n
i
-
2
)
!;
(ii)
K
i
1
(
t
,
s
)
≥
γ
i
(
t
)
K
i
1
(
s
), for all t
,
s
∈
[
0,1
], where γ
i
(
t
)
:
=
(
1
/
(
n
i
-
1
)
)
min
⁡
{
t
n
i
-
1
,
(
1
-
t
)
t
n
i
-
2
}.Remark 4.
Combining (i) and (ii), we can easily see(16)
min
⁡
t
∈
[
a
,
b
]
K
i
1
(
t
,
s
)
≥
γ
i
K
i
1
(
s
)
≥
γ
i
K
i
1
(
t
,
s
)
,
min
⁡
t
∈
[
a
,
b
]
K
i
1
(
t
,
s
)
≥
γ
i
K
i
1
(
s
)
∀
t
,
s
∈
[
0,1
]
,
where γ
i
=
min
⁡
{
γ
i
(
t
)
:
t
∈
[
a
,
b
]
}.Lemma 5.
Ifβ
i
∈
[
0,1
), the continuous function K
i
2
(
t
,
s
) has the following property:
(17)
0
≤
K
i
2
(
t
,
s
)
≤
K
i
2
(
1
,
s
)
≔
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
K
i
1
(
t
,
s
)
d
t
,
0
≤
K
i
2
(
t
,
s
)
≤
K
i
2
(
1
,
)
≔
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
∀
t
,
s
∈
[
0,1
]
.Proof.
From the properties ofK
i
1
(
t
,
s
) and the definition of K
i
2
(
t
,
s
), we can prove easily the results of Lemma 5.Lemma 6.
Ifβ
i
∈
[
0,1
), the continuous function K
i
(
t
,
s
) defined by (7) satisfies (i)
K
i
(
t
,
s
)
≥
0, for all t
,
s
∈
[
0,1
],
(ii)
K
i
(
t
,
s
)
≤
K
i
(
s
) for each t
,
s
∈
[
0,1
], and min
⁡
t
∈
[
a
,
b
]
⁡
K
i
(
t
,
s
)
≥
γ
i
*
K
i
(
s
), for all s
∈
[
0,1
],
whereγ
i
*
=
min
⁡
{
γ
i
,
a
n
i
-
1
}, γ
i is defined in Remark 4 and K
i
(
s
)
=
K
i
1
(
s
)
+
K
i
2
(
1
,
s
).Proof.
(1) From Lemma5 and (i) of Lemma 3, we get the proof of (i) immediately.
(2) From Lemma5 and (i) of Lemma 3, it is obvious that K
i
(
t
,
s
)
≤
K
i
(
s
) for each t
,
s
∈
[
0,1
].
Now, we show that the form (ii) holds. In fact, from (16) and (9), we have
(18)
min
⁡
t
∈
[
a
,
b
]
K
i
(
t
,
s
)
≥
γ
i
K
i
1
(
s
)
+
a
n
i
-
1
1
-
β
i
∫
0
1
‍
n
i
(
t
)
K
i
1
(
t
,
s
)
d
t
≥
γ
i
*
[
K
i
1
(
s
)
+
K
i
2
(
1
,
s
)
]
=
γ
i
*
K
i
(
s
)
,
∀
s
∈
[
0,1
]
.
Then, the proof of Lemma 6 is completed.Remark 7.
From the definition ofγ
i
*, it is obvious that 0
<
γ
i
*
<
1.
It is easy to prove that(
u
,
v
)
∈
C
n
1
(
0,1
)
∩
C
[
0,1
]
×
C
n
2
(
0,1
)
∩
C
[
0,1
] is a positive solution of systems (1) if and only if (
u
,
v
)
∈
C
[
0,1
]
×
C
[
0,1
] is a positive solution of systems of integral equations
(19)
u
(
t
)
=
∫
0
1
‍
K
1
(
t
,
s
)
a
1
(
s
)
f
1
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
,
v
(
t
)
=
∫
0
1
‍
K
2
(
t
,
s
)
a
2
(
s
)
f
2
(
s
,
u
(
s
)
)
d
s
,
where K
i
(
t
,
s
), i
=
1,2, are Green’s functions defined by (7).
It follows from (19) that we can obtain the integral equation:
(20)
u
(
t
)
=
∫
0
1
‍
K
1
(
t
,
s
)
a
1
(
s
)
×
f
1
(
s
,
u
(
s
)
,
∫
0
1
‍
K
2
(
s
,
r
)
a
2
(
r
)
f
2
(
r
,
u
(
r
)
)
d
r
)
d
s
.
In a real Banach spaceC
[
0,1
], the norm is defined by ∥
u
∥
=
max
⁡
t
∈
[
0,1
]
⁡
|
u
(
t
)
|. Set
(21)
P
=
{
min
⁡
t
∈
[
a
,
b
]
u
(
t
)
∈
C
[
0,1
]
∣
u
(
t
)
≥
0
,
t
∈
[
0,1
]
,
min
⁡
t
∈
[
a
,
b
]
u
(
t
)
≥
γ
∥
u
∥
}
,
where γ
=
min
⁡
{
γ
1
*
,
γ
2
*
}. Obviously, P is a positive cone in C
[
0,1
].
Define the operatorT
:
P
→
E by
(22)
T
u
(
t
)
=
∫
0
1
‍
K
1
(
t
,
s
)
a
1
(
s
)
×
f
1
(
s
,
u
(
s
)
,
∫
0
1
‍
K
2
(
s
,
r
)
a
2
(
r
)
f
2
(
r
,
u
(
r
)
)
d
r
)
d
s
,
∀
t
∈
[
0,1
]
.Lemma 8.
Suppose that (H
1)–(H
3) are satisfied; then the operator T
:
P
→
P is completely continuous.Proof.
Letu
∈
P; consider (22); from Lemma 3 and (21), we have
(23)
0
≤
T
u
(
t
)
≤
∥
T
u
∥
≤
∫
0
1
‍
K
1
(
s
)
a
1
(
s
)
×
f
1
(
s
,
u
(
s
)
,
∫
0
1
‍
K
2
(
s
,
r
)
a
2
(
r
)
f
2
(
r
,
u
(
r
)
)
d
r
)
d
s
min
⁡
t
∈
[
a
,
b
]
T
u
(
t
)
≥
γ
∫
0
1
‍
K
1
(
s
)
a
1
(
s
)
×
f
1
(
s
,
u
(
s
)
,
∫
0
1
‍
K
2
(
s
,
r
)
a
2
(
r
)
f
2
(
r
,
u
(
r
)
)
d
r
)
d
s
.
It follows from (23) that we have min
⁡
t
∈
[
a
,
b
]
⁡
T
u
(
t
)
≥
γ
∥
T
u
∥; therefore, operator T
:
P
→
P. It is easy to prove that operator T
:
P
→
P is completely continuous since K
1
(
t
,
s
), K
2
(
t
,
s
), f
1
(
t
,
u
,
v
), f
2
(
t
,
u
), a
1
(
t
), and a
2
(
t
) are continuous.Lemma 9 (see [18]).
SupposeE is a real Banach space and P is cone in E, and let Ω
1, Ω
2 be bounded open sets in E such that θ
∈
Ω
1, Ω
¯
1
⊂
Ω
2. Let operator T
:
P
∩
(
Ω
¯
2
∖
Ω
1
)
→
P be completely continuous. Suppose that one of two conditions holds (i)
∥
T
u
∥
≤
∥
u
∥, for all u
∈
P
∩
∂
Ω
1; ∥
T
u
∥
≥
∥
u
∥, for all u
∈
P
∩
∂
Ω
2;
(ii)
∥
T
u
∥
≥
∥
u
∥, for all u
∈
P
∩
∂
Ω
1; ∥
T
u
∥
≤
∥
u
∥, for all u
∈
P
∩
∂
Ω
2.
Then, operatorT has at least one fixed point in P
∩
(
Ω
¯
2
∖
Ω
1
).Lemma 10 (see [18]).
SupposeE is a real Banach space and P is cone in E, and let Ω
1, Ω
2, and Ω
3 be bounded open sets in E such that θ
∈
Ω
1, Ω
¯
1
⊂
Ω
2, and Ω
¯
2
⊂
Ω
3. Let operator T
:
P
∩
(
Ω
¯
3
∖
Ω
1
)
→
P be completely continuous, such that (1)
∥
T
u
∥
≥
∥
u
∥, for all u
∈
P
∩
∂
Ω
1;
(2)
∥
T
u
∥
≤
∥
u
∥, T
u
≠
u, for all u
∈
P
∩
∂
Ω
2;
(3)
∥
T
u
∥
≥
∥
u
∥, for all u
∈
P
∩
∂
Ω
3.
Then, operatorT has at least two fixed points u
1 and u
2 in P
∩
(
Ω
¯
3
∖
Ω
1
) with u
1
∈
(
Ω
2
∖
Ω
1
) and u
2
∈
(
Ω
¯
3
∖
Ω
¯
2
).


## 3. Main Results

Theorem 11.
Suppose that assumptions (H
1)–(H
5) are satisfied; then systems (1) have at least one positive solution (
u
,
v
) satisfying u
(
t
)
>
0, v
(
t
)
>
0.Proof.
At first, letρ
1
=
M
1
γ
-
1, and set Ω
1
=
{
u
∈
C
[
0,1
]
:
∥
u
∥
<
ρ
1
} and u
∈
P
∩
∂
Ω
1; then min
⁡
t
∈
[
a
,
b
]
⁡
u
(
t
)
≥
γ
∥
u
∥
=
M
1. By Lemma 6 and the assumption (H
4), we have
(24)
v
(
t
)
=
∫
0
1
K
2
(
t
,
s
)
a
2
(
s
)
f
2
(
t
,
u
(
s
)
)
d
s
≥
C
1
∫
0
1
K
2
(
t
,
s
)
a
2
(
s
)
u
1
/
α
(
s
)
d
s
≥
γ
C
1
∫
a
b
K
2
(
s
)
a
2
(
s
)
u
1
/
α
(
s
)
d
s
‍
≥
γ
C
1
∫
a
b
K
2
(
s
)
a
2
(
s
)
d
s
(
γ
∥
u
∥
)
1
/
α
=
γ
C
1
δ
2
M
1
1
/
α
≥
M
1
,
t
∈
[
a
,
b
]
,
min
⁡
t
∈
[
a
,
b
]
⁡
(
T
u
)
(
t
)
≥
γ
∫
0
1
K
1
(
s
)
a
1
(
s
)
f
1
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
≥
γ
λ
1
∫
a
b
K
1
(
s
)
a
1
(
s
)
v
α
(
s
)
d
s
≥
γ
λ
1
δ
1
(
γ
C
1
δ
2
)
α
(
γ
∥
u
∥
)
≥
∥
u
∥
.
Therefore, we have
(25)
∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
1
.
Further, set Ω
2
=
{
u
∈
C
[
0,1
]
:
∥
u
∥
<
ρ
2
}, for u
∈
P
∩
∂
Ω
2; by the assumption (H
5), we have
(26)
v
(
t
)
≤
C
2
∫
0
1
‍
K
2
(
s
)
a
2
(
s
)
u
1
/
β
(
s
)
d
s
≤
C
2
μ
2
∥
u
∥
1
/
β
≤
ρ
2
1
+
(
1
/
β
)
≤
ρ
2
,
t
∈
[
0,1
]
,
(
T
u
)
(
t
)
≤
∫
0
1
K
1
(
s
)
a
1
(
s
)
f
1
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
≤
μ
1
λ
2
∥
v
∥
β
≤
μ
1
λ
2
(
C
2
μ
2
)
β
∥
u
∥
≤
∥
u
∥
.
Therefore, we have
(27)
∥
T
u
∥
≤
∥
u
∥
,
u
∈
P
∩
∂
Ω
2
.
Thus, from (25), (27), Lemma 8, and Lemma 9, operator T has a fixed point u in P
∩
(
Ω
¯
1
∖
Ω
2
). This means that systems (1) have at least one positive solution (
u
,
v
) satisfying u
(
t
)
>
0, v
(
t
)
>
0.Theorem 12.
Suppose that assumptions (H
1)–(H
3) and (H
6)-(H
7) are satisfied; then systems (1) have at least one positive solution (
u
,
v
) satisfying u
(
t
)
>
0, v
(
t
)
>
0.Proof.
At first, it follows from the assumption (H
6) that we have
(28)
(
T
u
)
(
t
)
=
∫
0
1
K
1
(
t
,
s
)
a
1
(
s
)
f
1
(
t
,
u
(
s
)
,
v
(
s
)
)
d
s
‍
≤
∫
0
1
K
1
(
s
)
a
1
(
s
)
(
λ
3
v
p
(
s
)
+
M
2
)
d
s
‍
≤
∫
0
1
K
1
(
s
)
a
1
(
s
)
‍
[
λ
3
(
∫
0
1
K
2
(
s
)
a
2
(
s
)
f
2
(
s
,
u
(
s
)
)
d
s
‍
)
p
+
M
2
(
∫
0
1
K
2
(
s
)
a
2
(
s
)
f
2
(
s
,
u
(
s
)
)
d
s
‍
)
p
]
d
s
≤
μ
1
λ
3
μ
2
p
(
C
3
u
1
/
p
+
M
2
)
p
+
μ
1
M
2
≤
μ
1
λ
3
μ
2
p
(
C
3
∥
u
∥
1
/
p
+
M
2
)
p
+
μ
1
M
2
.
By means of simple calculation, we have
(29)
lim
⁡
u
→
+
∞
⁡
(
μ
1
λ
3
μ
2
p
(
C
3
∥
u
∥
1
/
p
+
M
2
)
p
+
μ
1
M
2
)
∥
u
∥
=
1
2
.
Then, there exists a sufficiently large M
>
0 such that
(30)
μ
1
λ
3
μ
2
p
(
C
3
∥
u
∥
1
/
p
+
M
2
)
p
+
μ
1
M
2
≤
∥
u
∥
.
Set Ω
3
=
{
u
∈
C
[
0,1
]
:
∥
u
∥
<
M
}. For u
∈
P
∩
∂
Ω
3, by (28),(30), we obtain that
(31)
∥
T
u
∥
≤
∥
u
∥
,
u
∈
P
∩
∂
Ω
3
.
Further, since f
2
(
t
,
0
)
≡
0 and f
2
(
t
,
u
) is continuous in [
0,1
]
×
[
0
,
+
∞
), there exists ρ
∈
(
0
,
ε
) such that
(32)
f
2
(
t
,
u
)
<
μ
2
-
1
ρ
,
(
t
,
u
)
∈
[
0,1
]
×
(
0
,
ρ
)
.
Set Ω
4
=
{
u
∈
C
[
0,1
]
:
∥
u
∥
<
ρ
}. For u
∈
P
∩
∂
Ω
4, we have
(33)
v
(
t
)
=
∫
0
1
K
2
(
t
,
s
)
a
2
(
s
)
f
2
(
t
,
u
(
s
)
)
d
s
<
μ
2
-
1
ρ
∫
0
1
K
2
(
s
)
a
2
(
s
)
d
s
=
ρ
.
It follows from the assumption (H
7) and Lemma 6 that we have
(34)
min
⁡
t
∈
[
a
,
b
]
⁡
(
T
u
)
(
t
)
≥
γ
∫
0
1
K
1
(
s
)
a
1
(
s
)
f
1
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
≥
γ
λ
4
∫
a
b
K
1
(
s
)
a
1
(
s
)
d
s
×
(
∫
a
b
K
2
(
s
,
r
)
a
2
(
r
)
f
2
(
r
,
u
(
r
)
)
d
r
)
q
≥
γ
λ
4
δ
1
(
γ
∫
a
b
K
2
(
r
)
a
2
(
r
)
C
4
u
1
/
q
(
r
)
d
r
)
q
≥
γ
2
+
q
λ
4
δ
1
(
C
4
δ
2
)
q
∥
u
∥
≥
∥
u
∥
.
Hence, we have
(35)
∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
4
.
Thus, from (31), (35), Lemmas 8 and 9, operator T has a fixed point u in P
∩
(
Ω
¯
3
∖
Ω
4
). This means that systems (1) have at least one positive solution (
u
,
v
) satisfying u
(
t
)
>
0, v
(
t
)
>
0.Theorem 13.
Suppose that assumptions (H
1)–(H
4) and (H
7)-(H
8) hold. Then, systems (1) have at least two positive solutions (
u
1
,
v
1
) and (
u
2
,
v
2
).Proof.
SetΩ
5
=
{
u
∈
E
:
∥
u
∥
<
R
}. For u
∈
P
∩
∂
Ω
5, from (H
8), we obtain that
(36)
(
T
u
)
(
t
)
≤
∫
0
1
‍
K
1
(
s
)
a
1
(
s
)
f
1
(
s
,
R
,
∫
0
1
‍
K
2
(
r
)
a
2
(
r
)
f
2
(
r
,
R
)
d
r
)
d
s
<
μ
1
-
1
R
∫
0
1
‍
K
1
(
s
)
a
1
(
s
)
d
s
=
R
.
Thus, we have
(37)
∥
T
u
∥
<
∥
u
∥
,
u
∈
P
∩
∂
Ω
5
.
By (H
4) and (H
7), we can get
(38)
∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
1
,
∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
4
.
So, we can choose ρ, R, and ρ
1 such that ρ
<
R
<
ρ
1 and satisfying the above three inequalities. By Lemma 8 and Lemma 10, we guarantee that operator T has two fixed points u
1
∈
P
∩
(
Ω
¯
1
∖
Ω
5
) and u
2
∈
P
∩
(
Ω
¯
5
∖
Ω
4
). This means that systems (1) have at least two positive solutions (
u
1
,
v
1
) and (
u
2
,
v
2
).
In order to illustrate that our assumptions (H
4)–(H
7) are suitable for more general functions, we give some examples.Example 14.
In systems (1), let n
1
=
3, n
2
=
4, a
1
(
t
)
=
a
2
(
t
)
=
1, n
1
(
t
)
=
n
2
(
t
)
=
t, f
1
(
t
,
u
,
v
)
=
(
1
+
t
+
e
-
u
)
v
3
/
2, and f
2
(
t
,
u
)
=
u
5
/
2, so the assumptions (H
1)–(H
3) are satisfied. Choose α
=
1
/
2, β
=
3
/
2; then
(39)
liminf
⁡
u
→
+
∞
f
2
(
t
,
u
)
u
1
/
α
=
+
∞
,
liminf
⁡
v
→
+
∞
f
1
(
t
,
u
,
v
)
v
α
>
0
;
limsup
⁡
u
→
0
+
f
2
(
t
,
u
)
u
1
/
β
=
0
,
limsup
⁡
v
→
0
+
f
1
(
t
,
u
,
v
)
v
β
<
+
∞
uniformly with respect to t
∈
[
0,1
] and (
t
,
u
)
∈
[
0,1
]
×
[
0
,
+
∞
). It is easy to verify that the assumptions (H
4)-(H
5) hold. By Theorem 11, systems (1) have at least one position solution.Example 15.
In systems (1), let n
1
=
3, n
2
=
4, a
1
(
t
)
=
a
2
(
t
)
=
1, n
1
(
t
)
=
n
2
(
t
)
=
t, f
1
(
t
,
u
,
v
)
=
(
1
+
t
+
e
-
u
)
v
1
/
2, and f
2
(
t
,
u
)
=
u
1
/
2, so the assumptions (H
1)–(H
3) are satisfied. Choose p
=
q
=
1
/
2; then
(40)
limsup
⁡
u
→
+
∞
f
2
(
t
,
u
)
u
1
/
p
=
0
,
limsup
⁡
v
→
+
∞
f
1
(
t
,
u
,
v
)
v
p
<
+
∞
,
liminf
⁡
u
→
0
+
f
2
(
t
,
u
)
u
1
/
q
=
+
∞
,
limsup
⁡
v
→
0
+
f
1
(
t
,
u
,
v
)
v
q
>
0
uniformly with respect to t
∈
[
0,1
] and (
t
,
u
)
∈
[
0,1
]
×
[
0
,
+
∞
). It is easy to verify that the assumptions (H
6)-(H
7) hold. By Theorem 12, systems (1) have at least one position solution.



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*Source: 591381-2014-03-10.xml*