2 Preliminaries
In this section, we will provide several foundational definitions and results from the calculus on time scales and give some lemmas which are 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
∅
=
sup
T
and sup
∅
=
inf
T
, where ∅, denotes the empty set. If σ(t) > t, we say that t is right-scattered, while if ρ(t) < t, we say that t is left-scattered. Also, if t
<
sup
T
and σ(t) = t, then t is called right-dense, and if t
>
inf
T
and ρ(t) = t, then t is called left-dense. We also need, below, the set
T
k
, which is 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 differentiable at t
∈
T
if there exists a θ ∈ ℝ such that for any given ε > 0, there is an open neighborhood U of t such that
|
x
(
σ
(
t
)
)
-
x
(
s
)
-
x
Δ
(
t
)
|
σ
(
t
)
-
s
|
|
≤
ε
|
σ
(
t
)
-
s
|
,
s
∈
U
.
In this case, xΔ(t) is called the delta derivative of x at t. The second derivative of x(t) is defined by xΔΔ(t) = (xΔ)Δ(t).
In a similar way, we can obtain the fourth-order derivative of x(t) is defined by x(4Δ)(t) = (((xΔ)Δ)Δ)Δ(t).
Definition 2.3. 22 A function f
:
T
→
R
is called rd-continuous provided it is continuous 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
r
d
(
T
)
.
Definition 2.4. 22 A function F
:
T
→
ℝ
is called a delta-antiderivative of f
:
T
→
R
provide FΔ(t) = f(t) holds for all t
∈
T
k
. In this case we define the integral of f by
∫
a
t
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, we set
D
1
i
=
Q
1
i
1
-
P
1
i
,
D
2
i
=
P
2
i
1
-
Q
2
i
,
K
1
i
=
1
1
-
P
1
i
,
K
2
i
=
1
1
-
Q
2
i
,
where
P
1
i
=
∫
0
σ
(
T
)
B
i
(
s
)
a
i
s
+
b
i
ρ
i
Δ
s
,
P
2
i
=
∫
0
σ
(
T
)
A
i
(
s
)
a
i
s
+
b
i
ρ
i
Δ
s
,
Q
1
i
=
∫
0
σ
(
T
)
B
i
(
s
)
d
i
+
c
i
(
σ
(
T
)
-
s
)
ρ
i
Δ
s
,
Q
2
i
=
∫
0
σ
(
T
)
A
i
(
s
)
d
i
+
c
i
(
σ
(
T
)
-
s
)
ρ
i
Δ
s
.
To establish the existence of multiple positive solutions of system (1.1), let us list the 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 overcome the difficulty due to the dependence of f, g on derivatives, we first consider the following second-order nonlinear system
u
Δ
Δ
(
t
)
+
λ
f
(
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
Δ
Δ
(
t
)
+
μ
g
(
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
Δ
(
0
)
=
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
,
c
1
u
(
σ
(
T
)
)
+
d
1
u
Δ
(
σ
(
T
)
)
=
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
,
a
2
v
(
0
)
-
b
2
v
Δ
(
0
)
=
∫
0
σ
(
T
)
v
(
s
)
A
2
(
s
)
Δ
s
,
c
2
v
(
σ
(
T
)
)
+
d
2
v
Δ
(
σ
(
T
)
)
=
∫
0
σ
(
T
)
v
(
s
)
B
2
(
s
)
Δ
s
,
where A0 is the identity operator, and
A
i
u
(
t
)
=
∫
0
t
(
t
-
σ
(
s
)
)
i
-
1
u
(
s
)
Δ
s
,
A
i
v
(
t
)
=
∫
0
t
(
t
-
σ
(
s
)
)
i
-
1
v
(
s
)
Δ
s
,
i
=
1
,
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 nonlinear 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) = xΔΔ(t), v(t) = yΔΔ(t), then it follows from the boundary conditions of system (1.1) that
A
1
u
(
t
)
=
x
Δ
(
t
)
,
A
2
u
(
t
)
=
x
(
t
)
,
A
1
v
(
t
)
=
y
Δ
(
t
)
,
A
2
v
(
t
)
=
y
(
t
)
.
Thus (u, v) = (xΔΔ(t), yΔΔ(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) = A2u(t), y(t) = A2v(t), then we have
x
Δ
(
t
)
=
A
1
u
(
t
)
,
x
Δ
Δ
(
t
)
=
u
(
t
)
,
y
Δ
(
t
)
=
A
1
v
(
t
)
,
y
Δ
Δ
(
t
)
=
v
(
t
)
,
which imply that
x
(
0
)
=
0
,
x
Δ
(
0
)
=
0
,
y
(
0
)
=
0
,
y
Δ
(
0
)
=
0
.
Consequently, (x, y) = (A2u(t), A2v(t)) is a solution of the fourth-order system (1.1). This completes the proof.
Lemma 2.2. Assume that D11D21 ≠ 1 holds. Then for any h1 ∈ C(I', ℝ+), the following BVP
u
Δ
Δ
(
t
)
+
h
1
(
t
)
=
0
,
t
∈
(
0
,
σ
(
T
)
)
T
,
a
1
u
(
0
)
-
b
1
u
Δ
(
0
)
=
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
,
c
1
u
(
σ
(
T
)
)
+
d
1
u
Δ
(
σ
(
T
)
)
=
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
has a solution
u
(
t
)
=
∫
0
σ
(
T
)
H
1
(
t
,
s
)
h
1
(
s
)
Δ
s
,
where
H
1
(
t
,
s
)
=
G
1
(
t
,
s
)
+
r
1
(
t
)
∫
0
σ
(
T
)
B
1
(
τ
)
G
1
(
τ
,
s
)
Δ
τ
+
r
2
(
t
)
∫
0
σ
(
T
)
A
1
(
τ
)
G
1
(
τ
,
s
)
Δ
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). It is easy to see by integration of BVP(2.3) that
u
Δ
(
t
)
=
u
Δ
(
0
)
-
∫
0
t
h
1
(
s
)
Δ
s
.
Integrating again, we can obtain
u
(
t
)
=
u
(
0
)
+
t
u
Δ
(
0
)
-
∫
0
t
(
t
-
σ
(
s
)
)
h
1
(
s
)
Δ
s
.
Let t = σ(T) in (2.4) and (2.5), we obtain
u
Δ
(
σ
(
T
)
)
=
u
Δ
(
0
)
-
∫
0
σ
(
T
)
h
1
(
s
)
Δ
s
,
u
(
σ
(
T
)
)
=
u
(
0
)
+
σ
(
T
)
u
Δ
(
0
)
-
∫
0
σ
(
T
)
(
σ
(
T
)
-
σ
(
s
)
)
h
1
(
s
)
Δ
s
.
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
Δ
(
0
)
=
∫
0
σ
(
T
)
[
d
1
+
c
1
(
σ
(
T
)
-
σ
(
s
)
)
]
h
1
(
s
)
Δ
s
+
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
.
From (2.8) and the first boundary value condition of system (2.3), we have
u
Δ
(
0
)
=
a
1
ρ
1
∫
0
σ
(
T
)
[
d
1
+
c
1
(
σ
(
T
)
-
σ
(
s
)
)
]
h
1
(
s
)
Δ
s
+
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
-
c
1
a
1
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
,
u
(
0
)
=
b
1
ρ
1
∫
0
σ
(
T
)
[
d
1
+
c
1
(
σ
(
T
)
-
σ
(
s
)
)
]
h
1
(
s
)
Δ
s
+
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
-
c
1
a
1
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
+
1
a
1
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
.
Substituting (2.9) and (2.10) into (2.5), we have
u
(
t
)
=
∫
0
σ
(
T
)
G
1
(
t
,
s
)
h
1
(
s
)
Δ
s
+
a
1
t
+
b
1
ρ
1
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
+
d
1
+
c
1
(
σ
(
T
)
-
t
)
ρ
1
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
.
By (2.11), we get
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
=
1
1
-
P
11
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
Q
11
1
-
P
11
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
,
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
=
1
1
-
Q
21
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
P
21
1
-
Q
21
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
.
By (2.12) and (2.13), we get
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
=
K
11
D
21
1
-
D
11
D
21
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
K
21
1
-
D
11
D
21
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
,
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
=
K
11
1
-
D
11
D
21
∫
0
T
B
1
(
s
)
∫
0
T
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
K
21
D
11
1
-
D
11
D
21
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
.
Substituting (2.14) and (2.15) into (2.11), we have
u
(
t
)
=
∫
0
σ
(
T
)
G
1
(
t
,
s
)
h
1
(
s
)
Δ
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
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
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
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
=
∫
0
σ
(
T
)
G
1
(
t
,
s
)
h
1
(
s
)
Δ
s
+
r
11
(
t
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
r
21
(
t
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
=
∫
0
σ
(
T
)
H
1
(
t
,
s
)
h
1
(
s
)
Δ
s
.
Conversely, suppose u
(
t
)
=
∫
0
σ
(
T
)
H
1
(
t
,
s
)
h
1
(
s
)
Δ
s
, then
u
(
t
)
=
∫
0
σ
(
T
)
G
1
(
t
,
s
)
h
1
(
s
)
Δ
s
+
r
11
(
t
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
r
21
(
t
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
.
Direct differentiation of (2.17) implies
u
Δ
(
t
)
=
1
ρ
1
a
1
∫
t
σ
(
T
)
[
d
1
+
c
1
(
σ
(
T
)
-
σ
(
s
)
)
]
h
1
(
s
)
Δ
s
-
c
1
∫
0
t
(
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
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
+
a
1
K
21
D
11
-
c
1
K
21
ρ
1
(
1
-
D
11
D
21
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
h
1
(
τ
)
Δ
τ
Δ
s
and
u
Δ
Δ
(
t
)
=
-
h
1
(
t
)
,
and it is easy to verify that
a
1
u
(
0
)
-
b
1
u
Δ
(
0
)
=
∫
0
σ
(
T
)
u
(
s
)
A
1
(
s
)
Δ
s
,
c
1
u
(
σ
(
T
)
)
+
d
1
u
Δ
(
σ
(
T
)
)
=
∫
0
σ
(
T
)
u
(
s
)
B
1
(
s
)
Δ
s
.
This completes the proof.
Lemma 2.3. Assume that D12D22 ≠ 1 holds. Then for any h2 ∈ C(I', ℝ+), the following BVP
v
Δ
Δ
(
t
)
+
h
2
(
t
)
=
0
,
t
∈
(
0
,
σ
(
T
)
)
T
,
a
2
v
(
0
)
-
b
2
v
Δ
(
0
)
=
∫
0
σ
(
T
)
v
(
s
)
A
2
(
s
)
Δ
s
,
c
2
v
(
σ
(
T
)
)
+
d
2
v
Δ
(
σ
(
T
)
)
=
∫
0
σ
(
T
)
v
(
s
)
B
2
(
s
)
Δ
s
has a solution
v
(
t
)
=
∫
0
σ
(
T
)
H
2
(
t
,
s
)
h
2
(
s
)
Δ
s
,
where
H
2
(
t
,
s
)
=
G
2
(
t
,
s
)
+
r
12
(
t
)
∫
0
σ
(
T
)
B
2
(
τ
)
G
2
(
τ
,
s
)
Δ
τ
+
r
22
(
t
)
∫
0
σ
(
T
)
A
2
(
τ
)
G
2
(
τ
,
s
)
Δ
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.
Lemma 2.4. Suppose that (H1) is satisfied, for all t, s ∈ I and i = 1, 2, we have
(i) Gi(t, s) > 0, Hi(t, s) > 0,
(ii) LimiGi(σ(s), s) ≤ Hi(t, s) ≤ MiGi(σ(s), s),
(iii) mGi(σ(s), s) ≤ Hi(t, s) ≤ MGi(σ(s), s),
where
M
i
=
1
+
r
1
i
¯
∫
0
σ
(
T
)
B
i
(
τ
)
Δ
τ
+
r
2
i
¯
∫
0
σ
(
T
)
A
i
(
τ
)
Δ
τ
,
r
j
i
¯
=
max
0
≤
t
≤
1
r
i
(
t
)
,
m
i
=
1
+
r
1
i
¯
(
t
)
∫
0
σ
(
T
)
B
i
(
τ
)
Δ
τ
+
r
2
i
¯
(
t
)
∫
0
σ
(
T
)
A
i
(
τ
)
Δ
τ
,
r
j
i
¯
=
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 Gi(t, s) > 0, Hi(t, s) > 0 and Gi(t, s) ≤ Gi(σ(s), s), for all t, s ∈ 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 Gi(t, s)/Gi(σ(s), s) ≥ Li and we have
G
i
(
t
,
s
)
≥
L
i
G
i
(
σ
(
s
)
,
s
)
.
On the one hand, from the definition of Li and mi, for all t, s ∈ I, we have
H
i
(
t
,
s
)
=
G
i
(
t
,
s
)
+
r
1
i
(
t
)
∫
0
σ
(
T
)
B
i
(
τ
)
G
i
(
τ
,
s
)
Δ
τ
+
r
2
i
(
t
)
∫
0
σ
(
T
)
A
i
(
τ
)
G
i
(
τ
,
s
)
Δ
τ
≥
L
i
G
i
(
σ
(
s
)
,
s
)
1
+
r
1
i
(
t
)
∫
0
σ
(
T
)
B
i
(
τ
)
Δ
τ
+
r
2
i
(
t
)
∫
0
σ
(
T
)
A
i
(
τ
)
Δ
τ
≥
L
i
m
i
G
i
(
σ
(
s
)
,
s
)
,
and on the other hand, we obtain easily that from the definition of Mi, for all t, s ∈ I,
H
i
(
t
,
s
)
≤
G
i
(
σ
(
s
)
,
s
)
+
r
1
i
(
t
)
∫
0
σ
(
T
)
B
i
(
τ
)
G
i
(
σ
(
s
)
,
s
)
Δ
τ
+
r
2
i
(
t
)
∫
0
σ
(
T
)
A
i
(
τ
)
G
i
(
σ
(
s
)
,
s
)
Δ
τ
≤
M
i
G
i
(
σ
(
s
)
,
s
)
.
Finally, it is easy to verify that mGi(σ(s), s) ≤ Hi(t, s) ≤ MGi(σ(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 Ω1 and Ω2 are bounded open subsets of E, such that 0 ∈ Ω1,
Ω
1
¯
⊂
Ω
2
, and let T
:
P
∩
(
Ω
2
¯
\
Ω
1
)
→
P
be a completely continuous operator such that either
(i) ||Tu|| ≤ ||u||, ∀u ∈ P ∩ ∂Ω1 and ||Tu|| ≥ ||u||, ∀u ∈ P ∩ ∂Ω2, or
(ii) ||Tu|| ≥ ||u||, ∀u ∈ P ∩ ∂Ω1 and ||Tu|| ≤ ||u||, ∀u ∈ P ∩ ∂Ω2
holds. Then T has a fixed point in
P
∩
(
Ω
2
¯
\
Ω
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+) equipped with the norm |
|
(
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 Tλ, Tμ : P → C(I, ℝ+) by
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
T
H
1
(
t
,
s
)
f
(
t
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
,
t
∈
I
,
T
μ
(
u
,
v
)
(
t
)
=
μ
∫
0
T
H
2
(
t
,
s
)
g
(
t
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
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 (H1) hold. Then T : P → P is completely continuous.
Proof. Firstly, we prove that T : P → P. In fact, for all (u, v) ∈ P and t ∈ I, by Lemma 2.4(i) and (H1), it is obvious that Tλ(u, v)(t) > 0, Tμ(u, v)(t) > 0. In addition, we have
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
σ
(
T
)
H
1
(
t
,
s
)
f
(
t
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
λ
M
∫
0
σ
(
T
)
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
)
Δ
s
,
which implies |
|
T
λ
(
u
,
v
)
|
|
≤
λ
M
∫
0
σ
(
T
)
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
)
Δ
s
. And we have
T
λ
(
u
,
v
)
(
t
)
≥
λ
L
1
m
1
∫
0
σ
(
T
)
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
)
Δ
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.
Secondly, we prove that T is continuous and compact, respectively. Let {(uk, vk)} ∈ 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
)
|
≤
λ
M
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
)
|
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
Δ
s
,
from the continuity of f, we know that ||Tλ(uk, vk) - Tλ(u, v)|| → 0 as k → ∞. Hence Tλ is continuous.
Tλ is compact provided that it maps bounded sets into relatively compact sets. Let
f
̄
=
sup
t
∈
I
′
|
f
(
t
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
|
, and let Ω be any bounded subset of P, then there exists r > 0 such that ||(u, v)|| ≤ r for all (u, v) ∈ Ω. Obviously, from (2.16), we know that
T
λ
(
u
,
v
)
(
t
)
≤
λ
M
f
̄
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
Δ
s
,
so, TλΩ is bounded for all (u, v) ∈ Ω. Moreover, let
L
′
1
=
λ
f
̄
ρ
1
a
1
∫
0
σ
(
T
)
[
d
1
+
c
1
(
σ
(
T
)
-
σ
(
s
)
)
]
Δ
s
+
c
1
∫
0
σ
(
T
)
(
a
1
σ
(
s
)
+
b
1
)
Δ
s
+
λ
f
̄
|
a
1
K
11
-
c
1
K
11
D
21
|
ρ
1
(
1
-
D
11
D
21
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
Δ
τ
Δ
s
+
λ
f
̄
|
a
1
K
21
D
11
-
c
1
K
21
|
ρ
1
(
1
-
D
11
D
21
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
Δ
τ
Δ
s
.
We have
|
T
λ
(
u
,
v
)
Δ
(
t
)
|
≤
λ
ρ
1
a
1
∫
t
σ
(
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
)
Δ
s
-
c
1
∫
0
t
(
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
)
Δ
s
+
λ
|
a
1
K
11
-
c
1
K
11
D
21
|
ρ
1
(
1
-
D
11
D
21
)
∫
0
σ
(
T
)
B
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
f
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
τ
Δ
s
+
λ
|
a
1
K
21
D
11
-
c
1
K
21
|
ρ
1
(
1
-
D
11
D
21
)
∫
0
σ
(
T
)
A
1
(
s
)
∫
0
σ
(
T
)
G
1
(
s
,
τ
)
f
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
τ
Δ
s
≤
L
′
1
.
Thus, for any (u, v) ∈ Ω and ∀ε > 0, let δ
=
ε
L
′
1
, then for t1, t2 ∈ I, |t1 - t2| < δ, we have
|
T
λ
(
u
,
v
)
(
t
1
)
-
T
λ
(
u
,
v
)
(
t
2
)
|
≤
L
1
′
|
t
1
-
t
2
|
<
ε
.
So, for all (u, v) ∈ Ω, TλΩ is equicontinuous. By Ascoli-Arzela theorem, we obtain that Tλ : P → P is completely continuous. In a similar way, we can prove that Tμ : 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 multiplicity of positive solutions of system (1.1). In the following, for convenience, we set
f
β
=
lim
∑
i
=
1
6
|
φ
i
|
→
β
inf
t
∈
I
′
f
(
t
,
φ
1
,
…
,
φ
6
)
q
1
(
t
)
∑
i
=
1
6
|
φ
i
|
,
f
α
=
lim
∑
i
=
1
6
|
φ
i
|
→
α
sup
t
∈
I
′
f
(
t
,
φ
1
,
…
,
φ
6
)
q
2
(
t
)
∑
i
=
1
6
|
φ
i
|
,
g
β
=
lim
∑
i
=
1
6
|
φ
i
|
→
β
inf
t
∈
I
′
f
(
t
,
φ
1
,
…
,
φ
6
)
q
3
(
t
)
∑
i
=
1
6
|
φ
i
|
,
g
α
=
lim
∑
i
=
1
6
|
φ
i
|
→
α
sup
t
∈
I
′
f
(
t
,
φ
1
,
…
,
φ
6
)
q
4
(
t
)
∑
i
=
1
6
|
φ
i
|
,
where qi(t), qj(t) ∈ Crd(I', ℝ+) satisfy
0
<
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
i
(
s
)
Δ
s
<
+
∞
(
i
=
1
,
2
)
,
0
<
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
j
(
s
)
Δ
s
<
+
∞
(
j
=
3
,
4
)
.
Theorem 3.1. Assume that (H1) holds. Assume further that
(H2) there exist a constant R > 0, and two functions pi(t) ∈ Crd(I, R+) satisfying 0
<
∫
0
σ
(
T
)
G
i
(
σ
(
s
)
,
s
)
p
i
(
s
)
Δ
s
<
+
∞
(
i
=
1
,
2
)
such that
f
(
t
,
φ
1
,
…
,
φ
6
)
≤
R
p
1
(
t
)
,
∀
t
∈
I
,
0
<
∑
i
=
1
6
|
φ
i
|
≤
R
,
g
(
t
,
φ
1
,
…
,
φ
6
)
≤
R
p
2
(
t
)
,
∀
t
∈
I
,
0
<
∑
i
=
1
6
|
φ
i
|
≤
R
,
and one of the folloeing conditions is satisfied
(E1) λ
∈
(
M
3
f
0
,
M
4
)
, μ
∈
(
N
3
g
∞
,
N
4
)
,
(E2) λ
∈
(
M
3
f
∞
,
M
4
)
, μ
∈
(
N
3
g
0
,
N
4
)
,
(E3) λ
∈
(
M
3
F
α
,
M
4
)
, μ ∈ (0, N4),
(E4) λ ∈ (0, M4), μ
∈
(
N
3
G
α
,
N
4
)
,
where
M
3
=
m
2
M
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
1
(
s
)
Δ
s
-
1
,
M
4
=
O
1
M
N
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
p
1
(
s
)
Δ
s
-
1
,
N
3
=
m
2
M
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
3
(
s
)
Δ
s
-
1
,
N
4
=
o
2
M
N
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
p
2
(
s
)
Δ
s
-
1
,
F
α
=
min
{
f
0
,
f
∞
}
,
G
α
=
{
g
0
,
g
∞
}
,
O1, O2 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. We only prove the case in which (H2) and (E1) hold, the other case can be proved similarly. Firstly, from (2.2), we have
∑
i
=
0
2
A
i
u
(
t
)
+
∑
i
=
0
2
A
i
v
(
t
)
≤
|
|
(
u
,
v
)
|
|
+
σ
(
T
)
|
|
(
u
,
v
)
|
|
+
(
σ
(
T
)
)
2
|
|
(
u
,
v
)
|
|
=
N
|
|
(
u
,
v
)
|
|
.
Take
R
1
=
R
N
, and let Ω1 = {(u, v) ∈ Q; ||(u, v)|| < R1}. For any t ∈ I, (u, v) ∈ ∂Ω1 ∩ P, it follows from λ < M4, μ < N4 and (H2) that
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
σ
(
T
)
H
1
(
t
,
s
)
f
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
M
4
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
M
4
M
R
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
p
1
(
s
)
Δ
s
=
N
M
4
M
R
1
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
p
1
(
s
)
Δ
s
≤
1
O
1
R
1
and
T
μ
(
u
,
v
)
(
t
)
=
μ
∫
0
σ
(
T
)
H
2
(
t
,
s
)
g
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
N
4
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
N
4
M
R
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
p
2
(
s
)
Δ
s
=
N
4
M
N
R
1
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
p
2
(
s
)
Δ
s
≤
1
O
2
R
1
.
Consequently, for any (u, v) ∈ ∂Ω1 ∩ 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
.
Second, from λ
>
M
3
f
0
, we can choose ε1 > 0 such that λf0 > M3 + ε1, then there exists 0 < l1 < NR1 such that for any
∑
i
=
1
6
|
φ
i
|
<
l
1
and t ∈ I,
f
(
t
,
φ
1
,
…
,
φ
6
)
≥
M
3
+
ε
1
λ
q
1
(
t
)
∑
i
=
1
6
|
φ
i
|
.
And since
∑
i
=
0
2
A
i
u
(
t
)
+
∑
i
=
0
2
A
i
v
(
t
)
≥
u
(
t
)
+
v
(
t
)
≥
m
M
|
|
(
u
,
v
)
|
|
.
Take
R
2
=
l
1
N
<
R
1
.
For all (u, v) ∈ Ω2 ∩ P, where Ω2 = {(u, v) ∈ Q; ||(u, v)|| < R2}, we have
∑
i
=
0
2
A
i
u
(
t
)
+
∑
i
=
0
2
A
i
v
(
t
)
≥
u
(
t
)
+
v
(
t
)
≥
m
M
R
2
.
Thus, for all (u, v) ∈ Ω2 ∩ P, we have
T
λ
(
u
,
v
)
(
t
)
≥
λ
m
∫
0
σ
(
T
)
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
)
Δ
s
≥
m
(
M
3
+
ε
1
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
1
(
s
)
Δ
s
≥
M
3
m
2
M
R
2
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
1
(
s
)
Δ
s
=
R
2
.
Consequently, for all (u, v) ∈ Ω2 ∩ P, we have
|
|
T
(
u
,
v
)
|
|
≥
|
|
T
λ
(
u
,
v
)
|
|
≥
|
|
(
u
,
v
)
|
|
.
Finally, from μ > N3/g∞, we can choose ε2 > 0 such that μg∞ > N3 + ε2. then, there exists
l
2
>
m
M
R
such that for any
∑
i
=
1
6
|
φ
i
|
<
l
2
and t ∈ I,
g
(
t
,
φ
1
,
…
,
φ
6
)
≥
N
3
+
ε
2
μ
q
3
(
t
)
∑
i
=
1
6
|
φ
i
|
.
Take
R
3
=
m
M
-
1
l
2
>
R
1
.
For all (u, v) ∈ Ω3 ∩ P, where Ω3 = {(u, v) ∈ Q; ||(u, v)|| < R3}, from (3.3), we have
∑
i
=
0
2
A
i
u
(
t
)
+
∑
i
=
0
2
A
i
v
(
t
)
≥
m
M
R
3
=
l
2
.
Thus, for all (u, v) ∈ Ω3 ∩ P, we have
T
μ
(
u
,
v
)
(
t
)
≥
μ
m
∫
0
σ
(
T
)
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
)
Δ
s
≥
m
(
N
3
+
ε
2
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
3
(
s
)
Δ
s
≥
N
3
m
2
M
R
3
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
3
(
s
)
Δ
s
=
R
3
.
Consequently, for all (u, v) ∈ Ω3 ∩ P, we have
|
|
T
(
u
,
v
)
|
|
≥
|
|
T
μ
(
u
,
v
)
|
|
≥
|
|
(
u
,
v
)
|
|
.
From (3.2), (3.4), and (ii) of Lemma 2.5, it follows that system (2.1) has one positive solution (u1, v1) ∈ P with R2 ≤ ||(u1, v1)|| ≤ R1. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x1, y1). 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 (u2, v2) ∈ P with R1 ≤ ||(u2, v2)|| ≤ R3. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x2, y2). Above all, system (1.1) has at least two positive solutions. This completes the proof.
Theorem 3.2. Assume that (H1) holds. Suppose further that
(H3) there exist a constant R0 > 0, and two functions wi(t) ∈ Crd(I, R+) (i = 1, 2) satisfying 0
<
∫
0
σ
(
T
)
G
i
(
σ
(
s
)
,
s
)
w
i
(
s
)
Δ
s
<
+
∞
such that
f
(
t
,
φ
1
,
…
,
φ
6
)
≥
R
0
w
1
(
t
)
,
∀
t
∈
I
,
∑
i
=
1
6
|
φ
i
|
>
R
0
,
or
g
(
t
,
φ
1
,
…
,
φ
6
)
≥
R
0
w
2
(
t
)
,
∀
t
∈
I
,
∑
i
=
1
6
|
φ
i
|
>
R
0
.
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
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
w
1
(
s
)
Δ
s
-
1
,
N
5
=
m
2
M
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
w
2
(
s
)
Δ
s
-
1
,
M
6
=
O
1
M
N
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
2
(
s
)
Δ
s
-
1
,
N
6
=
o
2
M
N
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
4
(
s
)
Δ
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
Ω
4
=
{
(
u
,
v
)
∈
Q
;
|
|
(
u
,
v
)
|
|
<
R
1
′
}
. For any t ∈ I, (u, v) ∈ ∂Ω4 ∩ P, it follows from λ > M5 and (H3) that
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
σ
(
T
)
H
1
(
t
,
s
)
f
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≥
M
5
m
∫
0
σ
(
T
)
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
)
Δ
s
≥
m
M
5
R
0
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
w
1
(
s
)
Δ
s
=
M
5
m
2
M
R
′
1
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
w
1
(
s
)
Δ
s
=
R
′
1
.
Consequently, for any (u, v) ∈ ∂Ω4 ∩ P, we have
|
|
T
(
u
,
v
)
|
|
≥
|
|
T
λ
(
u
,
v
)
|
|
≥
|
|
(
u
,
v
)
|
|
.
From λ
<
M
6
F
α
, μ
<
N
6
G
α
, we know that λ
<
M
6
f
0
, μ
<
N
6
g
0
, we can choose ε3 > 0 such that M6 - ε3 > 0, N6 - ε3 > 0 and λf0 < M6 - ε3, μg0 < N6 - ε3. Then there exists 0
<
l
3
<
N
R
0
<
N
R
1
′
such that for any
∑
i
=
1
6
|
φ
i
|
<
l
3
and t ∈ I,
f
(
t
,
φ
1
,
…
,
φ
6
)
≤
M
6
-
ε
3
λ
q
2
(
t
)
∑
i
=
1
6
|
φ
i
|
,
g
(
t
,
φ
1
,
…
,
φ
6
)
≤
N
6
-
ε
3
μ
q
4
(
t
)
∑
i
=
1
6
|
φ
i
|
.
Take
R
2
′
=
l
3
N
<
R
1
′
and
Ω
5
=
{
(
u
,
v
)
∈
Q
;
|
|
(
u
,
v
)
|
|
<
R
2
′
}
. Then, for any (u, v) ∈ Ω5 ∩ P, from(3.1), we have
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
σ
(
T
)
H
1
(
t
,
s
)
f
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
λ
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
M
(
M
6
-
ε
3
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
2
(
s
)
Δ
s
≤
M
N
M
6
R
′
2
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
2
(
s
)
Δ
s
=
1
O
1
R
′
2
and
T
μ
(
u
,
v
)
(
t
)
=
μ
∫
0
σ
(
T
)
H
2
(
t
,
s
)
g
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
μ
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
M
(
N
6
-
ε
3
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
4
(
s
)
Δ
s
≤
M
N
N
6
R
′
2
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
4
(
s
)
Δ
s
=
1
O
2
R
′
2
.
Consequently, for any (u, v) ∈ ∂Ω5 ∩ 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
′
.
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 ε4 > 0 such that M6 - ε4 > 0, N6 - ε4 > 0 and λf∞< M6 - ε4, μg∞ < N6 - ε4. Then there exists
l
4
>
m
M
R
1
′
such that for any
∑
i
=
1
6
|
φ
i
|
>
l
4
and t ∈ I,
f
(
t
,
φ
1
,
…
,
φ
6
)
≤
M
6
-
ε
3
λ
q
2
(
t
)
∑
i
=
1
6
|
φ
i
|
,
g
(
t
,
φ
1
,
…
,
φ
6
)
≤
N
6
-
ε
3
μ
q
4
(
t
)
∑
i
=
1
6
|
φ
i
|
.
Take
R
3
′
=
(
m
M
)
-
1
l
4
>
R
1
′
and let
Ω
6
=
{
(
u
,
v
)
∈
Q
;
|
|
(
u
,
v
)
|
|
<
R
3
′
}
. Then, for any (u, v) ∈ Ω6 ∩ P, we have
T
λ
(
u
,
v
)
(
t
)
=
λ
∫
0
σ
(
T
)
H
1
(
t
,
s
)
f
(
s
,
A
2
u
,
A
1
u
,
3
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
λ
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
M
(
M
6
-
ε
3
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
2
(
s
)
Δ
s
≤
M
N
M
6
R
′
3
∫
0
σ
(
T
)
G
1
(
σ
(
s
)
,
s
)
q
2
(
s
)
Δ
s
=
1
O
1
R
′
3
and
T
μ
(
u
,
v
)
(
t
)
=
μ
∫
0
σ
(
T
)
H
2
(
t
,
s
)
g
(
s
,
A
2
u
,
A
1
u
,
A
0
u
,
A
2
v
,
A
1
v
,
A
0
v
)
Δ
s
≤
μ
M
∫
0
σ
(
T
)
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
)
Δ
s
≤
M
(
N
6
-
ε
3
)
∑
i
=
0
2
|
A
i
u
|
+
∑
i
=
0
2
|
A
i
v
|
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
4
(
s
)
Δ
s
≤
M
N
N
6
R
′
3
∫
0
σ
(
T
)
G
2
(
σ
(
s
)
,
s
)
q
4
(
s
)
Δ
s
=
1
O
2
R
′
3
.
Consequently, for any (u, v) ∈ ∂Ω6 ∩ 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
′
.
From (3.9), (3.10) and (i) of Lemma 2.5, it follows that system (2.1) has one positive solution (u1, v1) ∈ P with
R
1
′
≤
|
|
(
u
1
,
v
1
)
|
|
≤
R
2
′
. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x1, y1). In the same way, from (3.9), (3.11) and (ii) of Lemma 2.5, it follows that system (2.1) has one positive solution (u2, v2) ∈ P with
R
1
′
≤
|
|
(
u
2
,
v
2
)
|
|
≤
R
3
′
. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x2, y2). Above all, system (1.1) has at least two positive solutions. This completes the proof.