# Periodic Solutions of a Stage-Structured Plant-Hare Model with Toxin-Determined Functional Responses

Authors: Huicheng Wang; Patricia J. Y. Wong; Yonghui Xia
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The purpose of this paper is to obtain some sufficient conditions for the global existence of multiple positive periodic solutions of a delayed stage-structured plant-hare model with a toxin-determined functional response. Some novel estimation techniques to construct two open subsets for a priori bounds are employed.


---

## Body

## 1. Introduction

A lot of classical predator-prey models have been well studied (e.g., see [1–12]). Recently, Gao et al. [13] considered a nonautonomous plant-hare dynamical system with a toxin-determined functional response given by
(1)
N
˙
(
t
)
=
r
(
t
)
N
(
t
)
[
1
-
N
(
t
)
K
]
-
C
(
N
(
t
)
)
P
(
t
)
,
P
˙
(
t
)
=
B
(
t
)
C
(
N
(
t
)
)
P
(
t
)
-
d
(
t
)
P
(
t
)
,
(2)
C
(
N
(
t
)
)
=
f
(
N
(
t
)
)
[
1
-
f
(
N
(
t
)
)
4
G
]
,
f
(
N
(
t
)
)
=
e
δ
N
(
t
)
1
+
h
e
δ
N
(
t
)
,
where N
(
t
) denotes the density of plant at time t and P
(
t
) denotes the herbivore biomass at time t.On the other hand, many experts argued that the predator-prey models should be modified to fit the more realistic environment. They suggested that one should take the stage structure factor into consideration. Because it is very unrealistic to assume that each individual predator admits the same ability of attacking in the classical predator-prey models. They divided the individuals into two stages in life history, namely, immature and mature stages, where the rate of the immature predator attacking the prey and the reproductive rate can be ignored, while the mature predators are responsible for the prey. For example, one can refer to [14, 15] and the references cited therein. To discuss the effects of Holling type IV functional responses on a stage-structured model, the authors in [16] proposed the following delayed system:
(3)
d
x
(
t
)
d
t
=
x
(
t
)
[
r
1
(
t
)
-
a
1
(
t
)
∫
-
∞
t
‍
iiiiiiii
×
∫
-
∞
t
‍
K
(
t
-
s
)
x
(
s
)
d
s
iiiiiiii
∫
-
∞
t
‍
-
a
2
(
t
)
y
2
(
t
)
x
2
(
t
)
/
m
+
x
(
t
)
+
a
]
,
d
y
1
(
t
)
d
t
=
b
1
(
t
)
x
(
t
)
y
2
(
t
)
x
2
(
t
)
/
m
+
x
(
t
)
+
a
-
β
(
t
)
y
1
(
t
)
-
b
1
(
t
-
τ
)
exp
⁡
(
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
)
×
x
(
t
-
τ
)
y
2
(
t
-
τ
)
x
2
(
t
-
τ
)
/
m
+
x
(
t
-
τ
)
+
a
,
d
y
2
(
t
)
d
t
=
b
1
(
t
-
τ
)
exp
⁡
(
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
)
×
x
(
t
-
τ
)
y
2
(
t
-
τ
)
x
2
(
t
-
τ
)
/
m
+
x
(
t
-
τ
)
+
a
-
r
2
(
t
)
y
2
(
t
)
.
However, Holling IV type functional response is not appropriate for the plant-hare model if we explore the impact of plant toxicity on the dynamics of plant-hare interactions. Because such kind of plant can produce toxicity to protect itself. Therefore, in the present paper, we discuss the stage-structured plant-hare model with toxin-determined functional response as follows:
(4)
d
x
(
t
)
d
t
=
x
(
t
)
[
r
1
(
t
)
-
a
(
t
)
x
(
t
)
e
2
δ
2
(
h
δ
(
t
)
)
2
iiiiiii
-
4
G
e
δ
y
2
(
t
)
+
(
4
G
h
-
1
)
e
2
δ
2
x
(
t
)
y
2
(
t
)
4
G
(
1
+
h
e
δ
x
(
t
)
)
2
]
,
d
y
1
(
t
)
d
t
=
-
β
(
t
)
y
1
(
t
)
+
(
(
b
1
(
t
)
4
G
e
δ
B
x
(
t
)
y
2
(
t
)
δ
2
(
(
h
δ
(
t
)
)
2
)
-
1
+
(
4
G
h
-
1
)
e
2
δ
2
B
x
2
(
t
)
y
2
(
t
)
)
×
(
4
G
(
1
+
h
e
δ
x
(
t
)
)
2
)
-
1
)
-
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
(
(
4
G
e
δ
B
x
(
t
-
τ
)
y
2
(
t
-
τ
)
δ
2
(
(
h
δ
(
t
)
)
2
)
-
1
+
(
4
G
h
-
1
)
e
2
δ
2
B
x
2
(
t
-
τ
)
y
2
(
t
-
τ
)
)
×
(
4
G
(
1
+
h
e
δ
x
(
t
-
τ
)
)
2
)
-
1
)
,
d
y
2
(
t
)
d
t
=
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
(
(
4
G
e
δ
B
x
(
t
-
τ
)
y
2
(
t
-
τ
)
δ
2
(
(
h
δ
(
t
)
)
2
)
-
1
+
(
4
G
h
-
1
)
e
2
δ
2
B
x
2
(
t
-
τ
)
y
2
(
t
-
τ
)
)
×
(
4
G
(
1
+
h
e
δ
x
(
t
-
τ
)
)
2
)
-
1
)
-
r
2
(
t
)
y
2
(
t
)
,
where x
(
t
) denotes the density of the plant at time t, y
1
(
t
) is the density of immature individual hares at time t, and y
2
(
t
) denotes the density of mature individual hares at time t, respectively; r
1
(
t
), r
2
(
t
), a
(
t
), b
1
(
t
), and β
(
t
) are continuously positive periodic functions with period ω. B is the conversion rate, e is the encounter rate per hare, δ is the fraction of food items encountered that the hares ingest, G measures the toxicity level, and h is the time for handing one unit of plant. e, δ, G, and h are positive real constants. r
1
(
t
) is the intrinsic growth rate of the prey, a
(
t
) is the density-dependent coefficient of the plant, and r
2
(
t
) is the death rate of the mature hares.For any continuousω-periodic function f
(
t
), we always adopt the following notations throughout this paper:
(5)
f
-
=
1
ω
∫
0
ω
‍
f
(
t
)
d
t
,
f
L
=
min
⁡
t
∈
[
0
,
ω
]
f
(
t
)
,
f
M
=
max
⁡
[
0
,
ω
]
f
(
t
)
,
where f is a continuous ω-periodic function.The purpose of this paper is to obtain some sufficient conditions for the global existence of multiple positive periodic solutions of system (4). Our method is based on Mawhin’s coincidence degree and novel estimation techniques for a priori bounds of unknown solutions to L
x
=
λ
N
x. To the best of our knowledge, it is the first time that a delayed stage-structured plant-hare dynamical system with a toxin-determined functional response has been proposed and studied by using this method.Remark 1.
The termexp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
(
x
(
t
-
τ
)
y
2
(
t
-
τ
)
/
(
x
2
(
t
-
τ
)
/
m
+
x
(
t
-
τ
)
+
a
)
) in the third equation of (4) involves x
(
t
-
τ
)
y
2
(
t
-
τ
) instead of x
(
t
-
τ
)
y
2
(
t
); the method used in [13] cannot be applied to system (4) directly. Thus, novel estimation techniques must be employed for a priori bounds of unknown solutions to the operator equation L
x
=
λ
N
x. More specifically, integrating the second equation of system (1) over [
0
,
ω
], the authors in [13] obtained
(6)
∫
0
ω
(
(
4
G
e
δ
B
(
t
)
exp
⁡
⁡
{
u
1
(
t
)
}
δ
2
(
(
exp
⁡
⁡
{
u
1
}
)
2
)
-
1
‍
iiiiiii
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
t
)
exp
⁡
⁡
{
2
u
1
(
t
)
}
)
iiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
d
t
=
d
¯
ω
.
It follows that
(7)
d
¯
ω
≥
∫
0
ω
‍
4
G
e
δ
B
(
t
)
exp
⁡
{
u
1
(
t
)
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
)
}
)
2
d
t
.
By some arguments, this inequality then leads them to
(8)
d
¯
h
2
e
2
δ
2
exp
⁡
⁡
(
2
u
1
(
η
1
)
)
-
(
e
δ
B
¯
exp
⁡
{
-
2
r
¯
ω
}
-
2
h
e
δ
d
¯
)
×
exp
⁡
⁡
{
u
1
(
η
1
)
}
+
d
¯
>
0
,
which implies that
(9)
x
1
(
η
1
)
<
ln
⁡
h
-
0
,
x
1
(
η
1
)
>
ln
⁡
h
+
0
,
where
(10)
h
±
0
=
(
e
δ
B
¯
exp
⁡
{
-
2
r
¯
ω
}
-
2
h
e
δ
d
¯
)
±
Δ
1
2
d
¯
h
2
e
2
δ
2
,
Δ
1
=
[
e
δ
B
¯
exp
⁡
{
-
2
r
¯
ω
}
-
2
h
e
δ
d
¯
]
2
-
4
d
¯
2
h
2
e
2
δ
2
.
It should be noted that it is possible to construct two open subsets Ω
1 and Ω
2 due to (9). The essential reason to obtain (9) is the inequality (7). In inequality (7), there is no variable u
2
(
t
) and only one variable u
1
(
t
).
However, since the termx
(
t
-
τ
)
y
2
(
t
-
τ
) is in the third equation of (4), by same arguments in [13], we will see that
(11)
∫
0
ω
‍
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
-
τ
)
-
u
2
(
t
)
}
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
-
τ
)
}
δ
2
(
(
exp
⁡
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
iiiiiiiiiiiiii
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
(
t
-
τ
)
}
)
iiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
)
d
t
=
r
¯
2
ω
.
Note that both u
1 and u
2 appear simultaneously in the above equality. If we were to use the same ideas in [13], then the above equality does not lead us anywhere. Thus, some new arguments should be employed to obtain a priori bounds for u
1. To see how to overcome this difficulty, the reader can refer to (33)–(56) in Section 2.Remark 2.
It should be noted that the standard estimation techniques used in [16] are not applicable to the system (4) either, due to the term C
(
N
(
t
)
). If we were to use the standard arguments in [16], we can not obtain two positive roots of exp
⁡
(
u
1
(
ξ
1
)
). Consequently, we can not construct two open subsets. Thus, we can not obtain two positive solutions in these two open subsets.


## 2. Existence of Multiple Positive Periodic Solutions

In this section, we will study the existence of multiple periodic solutions of (4). We recall a few concepts and results from [17].Lemma 3 (see [17]).
LetΩ
⊂
X be an open bounded set. Let L be a Fredholm mapping of index zero and N
L-compact on Ω
¯. Assume(a)
for eachλ
∈
(
0,1
), x
∈
∂
Ω
∩
Dom
⁡
L, L
x
≠
λ
N
x;
(b)
for eachx
∈
∂
Ω
∩
Ker
⁡
L, Q
N
x
≠
0;
(c)
deg
⁡
{
J
Q
N
,
Ω
∩
Ker
⁡
L
,
0
}
≠
0.Then L
x
=
N
x has at least one solution in Ω
¯
∩
Dom
⁡
L.Lemma 4.
Ifβ
(
t
) and g
(
t
) are ω-periodic functions, then the system
(12)
d
y
(
t
)
d
t
=
β
(
t
)
y
(
t
)
+
g
(
t
)
has a unique ω-periodic solution which can be represented as y
(
t
)
=
∫
-
∞
t
‍
exp
⁡
(
∫
s
t
‍
β
(
σ
)
d
σ
)
g
(
s
)
d
s.Throughout, we assume the following:(A
1)
1
/
4
h
<
G
<
1
/
3
h;
(A
2)
4
h
r
2
M
exp
⁡
{
τ
β
M
}
exp
⁡
{
2
r
¯
1
ω
}
/
b
1
L
<
B
<
4
G
r
2
L
h
2
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
/
(
4
G
h
-
1
).We further introduce six positive numbers which will be used later as follows:(13)
h
±
=
(
(
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
exp
⁡
{
-
2
r
¯
1
ω
}
-
2
h
e
δ
r
2
M
)
Δ
1
iiiii
±
Δ
1
)
×
(
2
r
2
M
h
2
e
2
δ
2
)
-
1
,
l
±
=
(
[
4
G
h
2
e
δ
B
exp
⁡
{
2
r
¯
1
ω
}
(
b
1
M
)
-
1
Δ
2
iii
-
2
h
e
δ
(
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
)
]
ii
±
Δ
2
)
×
(
2
h
2
e
2
δ
2
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
⁡
{
τ
β
L
}
iiiiiiiiiiiiiiiiiiii
[
(
b
1
M
)
-
1
(
b
1
M
)
-
1
-
(
4
G
h
-
1
)
B
]
)
-
1
,
u
±
=
(
4
G
e
δ
B
-
8
G
h
e
δ
r
¯
2
b
¯
-
1
)
±
Δ
3
2
[
4
G
r
¯
2
b
¯
-
1
h
2
e
2
δ
2
-
(
4
G
h
-
1
)
e
2
δ
2
B
]
,
where
(14)
Δ
1
=
[
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
exp
⁡
{
-
2
r
¯
1
ω
}
-
2
h
e
δ
r
2
M
]
2
-
4
(
r
2
M
)
2
h
2
e
2
δ
2
,
Δ
2
=
[
4
G
h
2
e
δ
B
exp
⁡
{
2
r
¯
1
ω
}
(
(
b
1
M
)
-
1
)
iii
-
2
h
e
δ
(
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
)
]
2
-
4
h
2
e
2
δ
2
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
2
,
Δ
3
=
(
4
G
e
δ
B
-
8
G
h
e
δ
r
¯
2
b
¯
-
1
)
2
-
16
G
r
¯
2
b
¯
-
1
[
4
G
r
¯
2
b
¯
-
1
h
2
e
2
δ
2
-
(
4
G
h
-
1
)
e
2
δ
2
B
]
,
b
¯
=
(
1
/
ω
)
∫
0
ω
‍
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
}
d
s
d
t. Under assumptions (
A
1
) and (
A
2
), it is not difficult to show that
(15)
l
-
<
u
-
<
h
-
<
h
+
<
u
+
<
l
+
.Theorem 5.
In addition to (A
1) and (A
2), suppose that(A
3)
r
¯
1
-
a
¯
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
ω
}
>
0.Then system (4) has at least two positive ω-periodic solutions.Proof.
Note that the first equation and the third equation of (4) can be separated from the whole system. Consider the following subsystem:
(16)
d
x
(
t
)
d
t
=
x
(
t
)
[
r
1
(
t
)
-
a
(
t
)
x
(
t
)
δ
2
(
h
(
t
)
)
2
iiiiii
-
4
G
e
δ
y
2
(
t
)
+
(
4
G
h
-
1
)
e
2
δ
2
x
(
t
)
y
2
(
t
)
4
G
(
1
+
h
e
δ
x
(
t
)
)
2
]
,
d
y
2
(
t
)
d
t
=
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
(
(
4
G
e
δ
B
x
(
t
-
τ
)
y
2
(
t
-
τ
)
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
h
(
t
)
)
2
)
-
1
iiiiiii
×
B
(
t
)
x
2
(
t
-
τ
)
y
2
(
t
-
τ
)
)
iiii
×
(
4
G
(
1
+
h
e
δ
x
(
t
-
τ
)
)
2
)
-
1
)
-
r
2
(
t
)
y
2
(
t
)
.
Make the change of variables
(17)
x
(
t
)
=
exp
⁡
{
u
1
(
t
)
}
,
y
(
t
)
=
exp
⁡
{
u
2
(
t
)
}
;
then system (16) can be rewritten as
(18)
u
˙
1
(
t
)
=
r
1
(
t
)
-
a
(
t
)
exp
⁡
{
u
1
(
t
)
}
-
(
(
4
G
e
δ
exp
⁡
{
u
2
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
h
(
t
)
)
2
)
-
1
iiiiii
e
2
δ
2
×
exp
⁡
{
u
1
(
t
)
+
u
2
(
t
)
}
)
iiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
∶
=
f
1
(
t
,
u
)
,
u
˙
2
(
t
)
=
-
r
2
(
t
)
+
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
-
τ
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
9
iiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
t
-
τ
)
}
)
iiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
)
×
exp
⁡
{
u
2
(
t
-
τ
)
-
u
2
(
t
)
}
∶
=
f
2
(
t
,
u
)
.
Take(19)
X
=
Y
=
{
x
=
(
u
1
,
u
2
)
T
∈
C
(
ℝ
,
ℝ
2
)
∣
x
(
t
+
ω
)
=
x
(
t
)
}
and define
(20)
∥
x
∥
=
max
⁡
t
∈
[
0
,
ω
]
|
u
1
(
t
)
|
+
max
⁡
t
∈
[
0
,
ω
]
|
u
2
(
t
)
|
,
iiiiiiiiiiiiiiiiiii
x
=
(
u
1
,
u
2
)
T
∈
X
or
Y
;
here |
·
| denotes the Euclidean norm. Then X and Y are Banach spaces with the norm ∥
·
∥. Set
(21)
L
:
Dom
⁡
L
∩
X
,
L
(
u
1
(
t
)
,
u
2
(
t
)
)
T
=
(
d
u
1
(
t
)
d
t
,
d
u
2
(
t
)
d
t
)
T
,
where Dom
⁡
L
=
{
(
u
1
(
t
)
,
u
2
(
t
)
)
T
∈
C
1
(
ℝ
,
ℝ
2
)
}. Further, N
:
X
→
X is defined by
(22)
N
(
u
1
u
2
)
=
(
f
1
(
t
,
u
)
f
2
(
t
,
u
)
)
.
Define
(23)
P
(
u
1
u
2
)
=
Q
(
u
1
u
2
)
=
(
1
ω
∫
0
ω
‍
u
1
(
t
)
d
t
1
ω
∫
0
ω
‍
u
2
(
t
)
d
t
)
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
(
u
1
u
2
)
∈
X
=
Y
.
It is not difficult to show that L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) K
p
:
Im
⁡
L
→
Dom
⁡
L
∩
Ker
⁡
P exists. Standard arguments show that N is L-compact on Ω
¯ for any open bounded set Ω
⊂
X.
Now, we will search for two appropriate open bounded subsets in order to apply the continuation theorem.
Corresponding to the operator equationL
x
=
λ
N
x, λ
∈
(
0,1
), we have
(24)
u
˙
1
(
t
)
=
λ
r
1
(
t
)
-
λ
[
a
(
t
)
exp
⁡
{
u
1
(
t
)
}
(
(
{
u
1
(
t
)
}
)
2
)
-
1
)
iiiiiii
-
(
(
4
G
e
δ
exp
⁡
{
u
2
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
u
1
(
t
)
+
u
2
(
t
)
}
)
iiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
]
,
(25)
u
˙
2
(
t
)
=
-
λ
r
2
(
t
)
+
λ
[
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
iiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
-
τ
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
t
-
τ
)
}
)
iiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
)
iiiiiii
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
-
τ
)
-
u
2
(
t
)
}
]
.
Supposex
=
(
u
1
(
t
)
,
u
2
(
t
)
)
T
∈
X is a solution of (24) and (25) for a certain λ
∈
(
0,1
). Integrating (24), (25) over the interval [
0
,
ω
], we obtain
(26)
∫
0
ω
‍
a
(
t
)
exp
⁡
{
u
1
(
t
)
}
d
t
+
∫
0
ω
‍
(
(
4
G
e
δ
exp
⁡
{
u
2
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
u
1
(
t
)
+
u
2
(
t
)
}
)
iiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
d
t
=
r
¯
1
ω
,
(27)
∫
0
ω
‍
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
-
τ
)
-
u
2
(
t
)
}
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
-
τ
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
(
{
u
1
(
t
)
}
)
2
)
-
1
)
9
iiiiiiiiiiiiiiiii
e
2
δ
2
×
exp
⁡
{
2
u
1
(
t
-
τ
)
}
)
iiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
)
d
t
=
r
¯
2
ω
.
It follows from (
A
1
), (24), and (26) that
(28)
∫
0
ω
‍
|
u
˙
1
(
t
)
|
d
t
=
λ
∫
0
ω
‍
|
r
1
(
t
)
-
a
(
t
)
exp
⁡
{
u
1
(
t
)
}
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiii
-
(
(
4
G
e
δ
exp
⁡
{
u
2
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
(
{
u
1
(
t
)
}
)
2
)
-
1
)
9
iiiiiiiiiiiiiiiiiiiiiiiiiii
e
2
δ
2
×
exp
⁡
{
u
1
(
t
)
+
u
2
(
t
)
}
)
iiiiiiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
|
d
t
<
∫
0
ω
‍
r
1
(
t
)
d
t
+
∫
0
ω
‍
a
(
t
)
exp
⁡
⁡
{
u
1
(
t
)
}
d
t
+
∫
0
ω
‍
(
(
4
G
e
δ
exp
⁡
⁡
{
u
2
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
⁡
{
u
1
(
t
)
+
u
2
(
t
)
}
)
iiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
d
t
=
∫
0
ω
‍
r
1
(
t
)
d
t
+
r
¯
1
ω
=
2
r
¯
1
ω
;
that is,
(29)
∫
0
ω
‍
|
u
˙
1
(
t
)
|
d
t
<
2
r
¯
1
ω
.
Similarly, it follows from (
A
1
), (25), and (27) that
(30)
∫
0
ω
‍
|
u
˙
2
(
t
)
|
d
t
<
2
r
¯
2
ω
.
Since (
u
1
(
t
)
,
u
2
(
t
)
)
T
∈
X, there exist ξ
i
,
η
i
∈
[
0
,
ω
] such that
(31)
u
i
(
ξ
i
)
=
min
⁡
t
∈
[
0
,
ω
]
u
i
(
t
)
,
u
i
(
η
i
)
=
max
⁡
t
∈
[
0
,
ω
]
u
i
(
t
)
,
i
=
1,2
.
Multiplying (25) by exp
⁡
{
u
2
(
t
)
} and integrating over [
0
,
ω
], we obtain
(32)
∫
0
ω
‍
r
2
(
t
)
exp
⁡
{
u
2
(
t
)
}
d
t
=
∫
0
ω
‍
[
b
1
(
t
-
τ
)
exp
⁡
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
iiiiiiiiiiiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
-
τ
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
t
-
τ
)
}
)
iiiiiiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
-
τ
)
}
)
2
)
-
1
)
iiiiiiiiiiiiiii
{
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
-
τ
)
}
]
d
t
=
∫
-
τ
ω
-
τ
‍
[
b
1
(
σ
)
exp
⁡
{
-
∫
σ
σ
+
τ
‍
β
(
s
)
d
s
}
iiiiiiiiiiiiiiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
σ
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
σ
)
}
)
iiiiiiiiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
σ
)
}
)
2
)
-
1
)
iiiiiiiiiiiiiiiiiii
{
-
∫
σ
σ
+
τ
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
σ
)
}
]
d
t
=
∫
0
ω
‍
[
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
iiiiiiiiiiiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
t
)
}
)
iiiiiiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
iiiiiiiiiiiiiiii
{
-
∫
σ
σ
+
τ
‍
β
(
s
)
d
s
}
×
exp
⁡
⁡
{
u
2
(
t
)
}
]
d
t
;
that is,
(33)
∫
0
ω
‍
r
2
(
t
)
exp
⁡
(
u
2
(
t
)
)
d
t
=
∫
0
ω
‍
[
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
iiiiiiiiiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiii
×
e
2
δ
2
exp
⁡
{
2
u
1
(
t
)
}
)
iiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
iiiiiiiiiiiiii
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
)
}
]
d
t
.
It follows from (27), (33), and (
A
1
); we see that
(34)
r
2
L
∫
0
ω
‍
exp
⁡
{
u
2
(
t
)
}
d
t
≤
∫
0
ω
‍
r
2
(
t
)
exp
⁡
{
u
2
(
t
)
}
d
t
=
∫
0
ω
‍
[
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
iiiiiiiiiiiiiii
×
(
(
4
G
e
δ
B
exp
⁡
{
u
1
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
(
{
u
1
(
t
)
}
)
2
)
-
1
iiiiiiiiiiiiiiiiiiiiiiiiii
e
2
δ
2
×
exp
⁡
⁡
{
2
u
1
(
t
)
}
)
iiiiiiiiiiiiiiiiiiiiiii
×
(
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
t
)
}
)
2
)
-
1
)
iiiiiiiiiiiiiiii
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
×
exp
⁡
{
u
2
(
t
)
}
]
d
t
≤
∫
0
ω
‍
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
×
[
4
G
e
δ
B
exp
⁡
{
u
1
(
t
)
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
)
}
)
2
iiiiiiiiiiiii
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
(
t
)
}
4
G
h
2
e
2
δ
2
exp
⁡
{
2
u
1
(
t
)
}
]
×
exp
⁡
{
u
2
(
t
)
}
d
t
≤
b
1
M
exp
⁡
{
-
τ
β
L
}
×
[
e
δ
B
exp
⁡
{
u
1
(
η
1
)
}
(
1
+
h
e
δ
exp
⁡
{
u
1
(
ξ
1
)
}
)
2
iiiiiiiiiiii
+
(
4
G
h
-
1
)
B
4
G
h
2
]
∫
0
ω
‍
exp
⁡
{
u
2
(
t
)
}
d
t
,
which implies
(35)
r
2
L
≤
b
1
M
exp
⁡
{
-
τ
β
L
}
×
[
e
δ
B
exp
⁡
{
u
1
(
η
1
)
}
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
ξ
1
)
}
)
2
+
(
4
G
h
-
1
)
B
4
G
h
2
]
.
So
(36)
u
1
(
η
1
)
≥
ln
⁡
(
(
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
iiiiiiii
×
[
(
b
1
M
)
-
1
]
(
1
+
h
e
δ
exp
⁡
{
u
1
(
ξ
1
)
}
)
2
)
iiiiiii
×
(
4
G
h
2
e
δ
B
)
-
1
)
.
This, combined with (29), gives
(37)
u
1
(
t
)
≥
u
1
(
η
1
)
-
∫
0
ω
‍
|
u
˙
1
(
t
)
|
d
t
>
ln
⁡
(
(
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
iiiiiiiiiii
×
[
(
b
1
M
)
-
1
]
(
1
+
h
e
δ
exp
⁡
⁡
{
u
1
(
ξ
1
)
}
)
2
)
iiiiiiiiii
×
(
4
G
h
2
e
δ
B
)
-
1
)
-
2
r
¯
1
ω
.
In particular, we have
(38)
u
1
(
ξ
1
)
>
ln
⁡
(
(
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
iiiiiiiiiii
×
[
(
b
1
M
)
-
1
]
(
1
+
h
e
δ
exp
⁡
{
u
1
(
ξ
1
)
}
)
2
)
iiiiiiiiii
×
(
4
G
h
2
e
δ
B
)
-
1
)
-
2
r
¯
1
ω
,
or
(39)
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
h
2
e
2
δ
2
×
exp
⁡
⁡
{
2
u
1
(
ξ
1
)
}
-
[
4
G
h
2
e
δ
B
exp
⁡
{
2
r
¯
ω
}
[
(
b
1
M
)
-
1
]
iiiiiiii
-
2
h
e
δ
(
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
)
]
×
exp
⁡
{
u
1
(
ξ
1
)
}
+
[
4
G
h
2
r
2
L
(
b
1
M
)
-
1
exp
⁡
⁡
{
τ
β
L
}
-
(
4
G
h
-
1
)
B
]
<
0
.
In view of (
A
2
), we have
(40)
ln
⁡
l
-
<
u
1
(
ξ
1
)
<
ln
⁡
l
+
.
Similarly, it follows from (33) that
(41)
b
1
L
exp
⁡
{
-
τ
β
M
}
×
4
G
e
δ
B
exp
⁡
{
u
1
(
ξ
1
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
(
ξ
1
)
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
×
∫
0
ω
‍
exp
⁡
{
u
2
(
t
)
}
d
t
≤
∫
0
ω
‍
b
1
(
t
)
exp
⁡
{
-
∫
t
t
+
τ
‍
β
(
s
)
d
s
}
×
4
G
e
δ
B
exp
⁡
{
u
1
(
t
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
(
t
)
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
t
)
}
)
2
×
exp
⁡
{
u
2
(
t
)
}
d
t
=
∫
0
ω
‍
r
2
(
t
)
exp
⁡
{
u
2
(
t
)
}
d
t
≤
r
2
M
∫
0
ω
‍
exp
⁡
{
u
2
(
t
)
}
d
t
,
which implies
(42)
b
1
L
exp
⁡
{
-
τ
β
M
}
×
4
G
e
δ
B
exp
⁡
{
u
1
(
ξ
1
)
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
(
ξ
1
)
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
≤
r
2
M
;
that is,
(43)
b
1
L
exp
⁡
{
-
τ
β
M
}
d
s
e
δ
B
exp
⁡
{
u
1
(
ξ
1
)
}
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
≤
r
2
M
.
So
(44)
u
1
(
ξ
1
)
≤
ln
⁡
r
2
M
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
.
This, combined with (29), gives
(45)
u
1
(
t
)
≤
u
1
(
ξ
1
)
+
∫
0
ω
‍
|
u
˙
1
(
t
)
|
d
t
<
ln
⁡
r
2
M
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
+
2
r
¯
1
ω
.
In particular, we have
(46)
u
1
(
η
1
)
<
ln
⁡
r
2
M
(
1
+
h
e
δ
exp
⁡
{
u
1
(
η
1
)
}
)
2
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
+
2
r
¯
1
ω
,
or
(47)
r
2
M
h
2
e
2
δ
2
exp
⁡
{
2
u
1
(
η
1
)
}
-
[
b
1
L
e
δ
B
exp
⁡
{
-
τ
β
M
}
exp
⁡
{
-
2
r
¯
1
ω
}
-
2
h
e
δ
r
2
M
]
×
exp
⁡
{
u
1
(
η
1
)
}
+
r
2
M
>
0
.
It follows from (
A
2
) that
(48)
u
1
(
η
1
)
<
ln
⁡
h
-
or
u
1
(
η
1
)
>
ln
⁡
h
+
.
From (29) and (40), we find
(49)
u
1
(
t
)
≤
u
1
(
ξ
1
)
+
∫
0
ω
‍
|
u
˙
1
(
t
)
|
d
t
<
ln
⁡
l
+
+
2
r
¯
1
ω
≜
H
11
.
On the other hand, it follows from (
A
1
), (26), and (49) that
(50)
r
¯
1
ω
≥
∫
0
ω
‍
4
G
e
δ
exp
⁡
⁡
{
u
2
(
ξ
2
)
}
4
G
(
1
+
h
e
δ
exp
⁡
⁡
{
ln
⁡
l
+
+
2
r
¯
ω
}
)
2
d
t
,
(51)
r
¯
1
ω
≤
∫
0
ω
‍
a
(
t
)
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
1
ω
}
d
t
+
∫
0
ω
‍
e
δ
exp
⁡
{
u
2
(
η
2
)
}
d
t
+
∫
0
ω
‍
e
δ
exp
⁡
{
u
2
(
η
2
)
}
2
d
t
.
It follows from (50) that
(52)
u
2
(
ξ
2
)
≤
ln
⁡
r
¯
1
(
1
+
h
e
δ
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
1
ω
}
)
2
e
δ
.
This, combined with (30), gives
(53)
u
2
(
t
)
≤
u
2
(
ξ
2
)
+
∫
0
ω
‍
|
u
˙
2
(
t
)
|
d
t
<
ln
⁡
r
¯
1
(
1
+
h
e
δ
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
1
ω
}
)
2
e
δ
+
2
r
¯
2
ω
≜
H
21
.
Moreover, because of (
A
3
), it follows from (51) that
(54)
u
2
(
η
2
)
≥
ln
⁡
2
(
r
¯
1
-
a
¯
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
ω
}
)
3
e
δ
.
This, combined with (30) again, gives
(55)
u
2
(
t
)
≥
u
2
(
η
2
)
-
∫
0
ω
‍
|
u
˙
2
(
t
)
|
d
t
>
ln
⁡
2
(
r
¯
1
-
a
¯
exp
⁡
{
ln
⁡
l
+
+
2
r
¯
ω
}
)
3
e
δ
-
2
r
¯
2
ω
≜
H
22
.
It follows from (53) and (55) that
(56)
max
⁡
t
∈
[
0
,
ω
]
u
2
(
t
)
<
max
⁡
{
|
H
21
|
,
|
H
22
|
}
≜
H
2
.
Now, let us considerQ
N
x with
(57)
Q
N
(
u
1
,
u
2
)
T
=
(
r
¯
1
ω
-
a
¯
ω
exp
⁡
{
u
1
}
-
w
e
2
δ
2
(
{
u
1
}
)
2
iiiiiiiii
×
4
G
e
δ
exp
⁡
{
u
2
}
+
(
4
G
h
-
1
)
e
2
δ
2
exp
⁡
{
u
1
+
u
2
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
}
)
2
,
iiiiiiiii
-
r
¯
2
ω
+
b
¯
ω
iiiiiiiii
×
4
G
e
δ
B
exp
⁡
{
u
1
}
+
(
4
G
h
-
1
)
e
2
δ
2
B
exp
⁡
{
2
u
1
}
4
G
(
1
+
h
e
δ
exp
⁡
{
u
1
}
)
2
)
T
.
In view of (
A
1
)
,
(
A
2
), and (
A
3
), we can show that the equation Q
N
(
u
1
,
u
2
)
T
=
0 has two distinct solutions
(58)
u
~
=
(
ln
⁡
u
-
,
ln
⁡
4
G
(
r
¯
1
-
a
¯
u
-
)
(
1
+
h
e
δ
u
-
)
2
4
G
e
δ
+
(
4
G
h
-
1
)
e
2
δ
2
u
-
)
,
u
^
=
(
ln
⁡
u
+
,
ln
⁡
4
G
(
r
¯
1
-
a
¯
u
+
)
(
1
+
h
e
δ
u
+
)
2
4
G
e
δ
+
(
4
G
h
-
1
)
e
2
δ
2
u
+
)
.
Choose C
>
0 such that
(59)
C
>
max
⁡
{
|
ln
⁡
4
G
(
r
¯
1
-
a
¯
u
-
)
(
1
+
h
e
δ
u
-
)
2
4
G
e
δ
+
(
4
G
h
-
1
)
e
2
δ
2
u
-
|
,
i
i
i
i
i
i
i
i
i
|
ln
⁡
4
G
(
r
¯
1
-
a
¯
u
+
)
(
1
+
h
e
δ
u
+
)
2
4
G
e
δ
+
(
4
G
h
-
1
)
e
2
δ
2
u
+
|
}
.
We define two open bounded subsets. Let
(60)
Ω
1
=
{
x
=
(
u
1
,
u
2
)
T
∈
X
∣
u
1
(
t
)
∈
(
ln
⁡
l
-
,
ln
⁡
h
-
)
,
max
⁡
t
∈
[
0
,
ω
]
|
u
2
(
t
)
|
i
i
i
i
i
max
⁡
t
∈
[
0
,
ω
]
|
u
2
(
t
)
|
<
H
2
+
C
}
,
Ω
2
=
{
x
=
(
u
1
,
u
2
)
T
∈
X
∣
min
⁡
t
∈
[
0
,
ω
]
u
1
(
t
)
∈
(
ln
⁡
l
-
,
ln
⁡
l
+
)
,
i
i
i
i
i
max
⁡
t
∈
[
0
,
ω
]
u
1
(
t
)
∈
(
ln
⁡
h
+
,
H
11
)
,
max
⁡
t
∈
[
0
,
ω
]
|
u
2
(
t
)
|
<
H
2
+
C
}
.
Then both Ω
1 and Ω
2 are bounded open subsets of X. It follows from (16) and (59) that u
~
∈
Ω
1 and u
^
∈
Ω
2. With the help of (16), (40), (48), (49), (56), and (59), it is easy to see that Ω
1
∩
Ω
2
=
ϕ, and Ω
i satisfies the requirement (a) in Lemma 3 for i
=
1,2. Moreover, Q
N
x
≠
0 for x
∈
∂
Ω
∩
Ker
⁡
L
=
∂
Ω
∩
ℝ
2. A direct computation gives deg
⁡
{
J
Q
N
,
Ω
i
∩
Ker
⁡
L
,
0
}
≠
0. Here, J is taken as the identity mapping since Im
⁡
Q
=
Ker
⁡
L. So far we have proved that Ω
i satisfies all the assumptions in Lemma 3. Hence, (16) has at least two ω-periodic solutions u
*
(
t
) and u
+
(
t
) with u
*
∈
Dom
⁡
L
∩
Ω
¯
1 and u
+
∈
Dom
⁡
L
∩
Ω
¯
2. Obviously, u
* and u
+ are different. Let x
*
(
t
)
=
exp
⁡
(
u
1
*
(
t
)
), y
2
*
(
t
)
=
exp
⁡
(
u
2
*
(
t
)
) and x
+
(
t
)
=
exp
⁡
(
u
1
+
(
t
)
), y
2
+
(
t
)
=
exp
⁡
(
u
2
+
(
t
)
). Then, by (18), (
x
*
(
t
)
,
y
2
*
(
t
)
) and (
x
+
(
t
)
,
y
2
+
(
t
)
) are two different positive ω-periodic solutions of (4). By the periodicity of the coefficients of system (4), it is not difficult to verify that
(61)
g
*
(
t
)
=
-
β
(
t
)
y
1
(
t
)
+
(
(
b
1
(
t
)
4
G
e
δ
B
x
*
(
t
)
y
2
*
(
t
)
(
x
*
)
2
(
(
x
*
(
t
)
)
2
)
-
1
i
i
i
i
i
i
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
x
*
)
2
(
t
)
y
2
*
(
t
)
)
i
i
i
i
i
×
(
4
G
(
1
+
h
e
δ
x
*
(
t
)
)
2
)
-
1
)
-
b
1
(
t
-
τ
)
exp
⁡
(
-
∫
t
-
τ
t
‍
β
(
s
)
d
s
)
×
(
(
4
G
e
δ
B
x
*
(
t
-
τ
)
y
2
*
(
t
-
τ
)
(
x
*
)
2
(
(
x
*
(
t
)
)
2
)
-
1
i
i
i
i
i
i
+
(
4
G
h
-
1
)
e
2
δ
2
B
(
x
*
)
2
(
t
-
τ
)
y
2
*
(
t
-
τ
)
)
i
i
i
i
i
×
(
4
G
(
1
+
h
e
δ
x
*
(
t
-
τ
)
)
2
)
-
1
)
is also ω-periodic. Then, from Lemma 4, we know that
(62)
d
y
1
(
t
)
d
t
=
-
β
(
t
)
y
1
(
t
)
+
g
*
(
t
)
has a unique ω-periodic solution denoted by y
1
*
(
t
). And
(63)
d
y
2
(
t
)
d
t
=
-
β
(
t
)
y
1
(
t
)
+
g
+
(
t
)
has a unique ω-periodic solution denoted by y
1
+
(
t
). Therefore, (
x
*
(
t
)
,
y
1
*
(
t
)
,
y
2
*
(
t
)
) and (
x
+
(
t
)
,
y
1
+
(
t
)
,
y
2
+
(
t
)
) are two different ω-periodic solutions of system (3). This completes the proof of Theorem 5.



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*Source: 357534-2014-05-08.xml*