# Existence of Positive Periodic Solutions forn-Dimensional Nonautonomous System

Authors: Youjun Liu; Huanhuan Zhao; Jurang Yan
Journal: Discrete Dynamics in Nature and Society
 (2014)

---

## Abstract

In this paper we consider the existence, multiplicity, and nonexistence of positive periodic solutions forn-dimensional nonautonomous functional differential system x
'
(
t
)
=
H
(
t
,
x
(
t
)
)
-
λ
B
(
t
)
F
(
x
(
t
-
τ
(
t
)
)
), where h
i are ω-periodic in t and there exist ω-periodic functions α
i
,
β
i
∈
C
(
R
,
R
+
) such that α
i
(
t
)
≤
(
h
i
(
t
,
x
)
/
x
i
)
≤
β
i
(
t
)
,
∫
0
ω
‍
α
i
(
t
)
d
t
>
0
, for x
∈
R
+
n all  with x
i
>
0
, and t
∈
R, l
i
m
x
i
→
0
+
(
h
i
(
t
,
x
)
/
x
i
) exist for t
∈
R; b
i
∈
C
(
R
,
R
+
) are ω-periodic functions and ∫
0
ω
‍
b
i
(
t
)
d
t
>
0
;
f
i
∈
C
(
R
+
n
,
R
+
), f
i
(
x
)
>
0 for x
>
0; τ
∈
(
R
,
R
) is an ω-periodic function. We show that the system has multiple or no positive ω-periodic solutions for sufficiently large or small λ
>
0, respectively.


---

## Body

## 1. Introduction

In this paper, we consider the first-ordern-dimensional nonautonomous functional differential system(1)
x
′
(
t
)
=
H
(
t
,
x
(
t
)
)
-
λ
B
(
t
)
F
(
x
(
t
-
τ
(
t
)
)
)
,
where λ
>
0 is a parameter;
(2)
x
=
[
x
1
,
x
2
…
,
x
n
]
⊤
,
B
(
t
)
=
diag
⁡
[
b
1
(
t
)
,
b
2
(
t
)
,
…
,
b
n
(
t
)
]
;
H
(
t
,
x
)
=
[
h
1
(
t
,
x
)
,
h
2
(
t
,
x
)
,
…
,
h
n
(
t
,
x
)
]
⊤
,
F
(
x
)
=
[
f
1
(
x
)
,
f
2
(
x
)
,
…
,
f
n
(
x
)
]
⊤
.Let(3)
R
=
(
-
∞
,
+
∞
)
,
R
+
=
[
0
,
+
∞
)
,
R
+
n
=
∏
i
=
1
n
‍
R
+
,
and for any x
=
[
x
1
,
x
2
…
,
x
n
]
⊤
∈
R
+
n, the norm of x is defined as ∥
x
∥
=
max
⁡
1
≤
i
≤
n
|
x
i
|.Throughout this paper, we usei
=
1,2
,
…
,
n, unless otherwise stated.For the system (1), we assume that(
H
1
)
τ
∈
(
R
,
R
) is an ω-periodic function, b
i
∈
C
(
R
,
R
+
) are ω-periodic functions, and
(4)
∫
0
ω
‍
b
i
(
t
)
d
t
>
0
;
(
H
2
)
f
i
∈
C
(
R
+
n
,
R
+
), f
i
(
x
)
>
0 for ∥
x
∥
>
0; h
i
∈
C
(
R
×
R
+
n
,
R
+
)
,
h
i are ω-periodic in t and there exist ω-periodic functions α
i
,
β
i
∈
C
(
R
,
R
+
) such that
(5)
α
i
(
t
)
≤
h
i
(
t
,
x
)
x
i
≤
β
i
(
t
)
,
∫
0
ω
‍
α
i
(
t
)
d
t
>
0
,
h
h
h
h
h
h
h
h
h
h
h
∀
x
∈
R
+
n
with
x
i
>
0
,
t
∈
R
.In addition, lim
⁡
x
i
→
0
+
(
h
i
(
t
,
x
)
/
x
i
) exist for t
∈
R.We note that in (1) F
(
x
) may have a singularity near x
=
0; that is,
(6)
lim
⁡
x
→
0
+
f
i
(
x
)
=
∞
.As we well know, the system (1) is sufficiently general to include particular mathematical models which describe multiple population dynamics. Recently, due to the theoretical and practical significance, the existence of positive periodic solution of some particular cases of periodic system (1) has been extensively studied; see, for example, [1–15]. Cheng and Zhang [1], Kang and Cheng [2], Kang et al. [3], Kang and Zhang [4], and Liu et al. [5] studied the existence, multiplicity, and nonexistence of positive periodic solutions. The existence of positive periodic solutions of the scalar functional differential equation(7)
x
′
(
t
)
=
a
(
t
)
g
(
x
(
t
)
)
x
(
t
)
-
λ
b
(
t
)
f
(
x
(
t
-
τ
(
t
)
)
)
has been studied by Wang [6]. By employing behaviours of the quotient f
(
x
)
/
x as x
→
0
+ and x
→
∞, several interesting results on the existence and nonexistence of positive periodic solutions of (7) have been obtained. In [7], Weng and Sun studied more general scalar periodic functional differential equation
(8)
x
′
(
t
)
=
h
(
t
,
x
)
-
λ
b
(
t
)
f
(
x
(
t
-
τ
(
t
)
)
)
,
where the existence theorems of positive periodic solutions of (8) are obtained by employing the behaviours of f
(
x
)
/
x at any point x
∈
(
0
,
+
∞
) and x
→
0
+
,
x
→
∞. The result in [7] generalized and improved those in [6]. O’Regan and Wang [8] investigated then-dimensional periodic system
(9)
x
′
(
t
)
=
A
(
t
)
g
(
x
(
t
)
)
-
λ
B
(
t
)
F
(
x
(
t
-
τ
(
t
)
)
)
.
By employing behaviours of f
(
x
)
/
∥
x
∥ as ∥
x
∥
→
0
+ and ∥
x
∥
→
∞, under quite general conditions, several existence theorems of positive periodic solutions are proved.A solutionx
=
[
x
1
,
x
2
…
,
x
n
]
⊤
,
t
∈
R of (1) is said to be positive if its all components x
i
(
t
) are positive; x is said to be ω-periodic (
ω
>
0
) if x
i
(
t
)
=
x
i
(
t
+
ω
)
,
t
∈
R.


## 2. Preliminary

Lemma 1 (see [9]).
LetE be a Banach space and K a cone in E. For r
>
0, define K
r
=
{
u
∈
K
:
∥
x
∥
<
r
}. Assume that T
:
K
¯
r
→
K is completely continuous such that T
x
≠
x for x
∈
∂
K
r
=
{
u
∈
K
:
∥
x
∥
=
r
}. (i)
If∥
T
x
∥
≥
∥
x
∥ for x
∈
∂
K
r, then
(10)
i
(
T
,
K
r
,
K
)
=
0
.
(ii)
If∥
T
x
∥
≤
∥
x
∥ for x
∈
∂
K
r, then
(11)
i
(
T
,
K
r
,
K
)
=
1
.Lemma 2 (see [9, 10]).
LetX be a Banach space and K a cone in X. Assume Ω
1
,
Ω
2 are open subsets of X with 0
∈
Ω
1
,
Ω
¯
1
⊂
Ω
2. Let
(12)
T
:
K
∩
(
Ω
¯
2
∖
Ω
1
)
⟶
K
be a completely continuous operator such that one of the following conditions is satisfied: (a)
∥
T
y
∥
≤
∥
y
∥ for y
∈
K
∩
∂
Ω
1 and ∥
T
y
∥
≥
∥
y
∥ for y
∈
K
∩
∂
Ω
2;
(b)
∥
T
y
∥
≥
∥
y
∥ for y
∈
K
∩
∂
Ω
1 and ∥
T
y
∥
≤
∥
y
∥ for y
∈
K
∩
∂
Ω
2.Then T has at least one fixed point in K
∩
(
Ω
¯
2
∖
Ω
1
).In order to apply Lemmas1 and 2 to system (1), we take(13)
X
=
{
x
(
t
)
:
x
(
t
)
∈
C
(
R
,
R
n
)
,
x
(
t
+
ω
)
=
x
(
t
)
,
t
∈
R
}
,
endowed with the norm ∥
x
∥
=
max
⁡
1
≤
i
≤
n
|
x
i
|
0, where |
x
i
|
0
=
sup
⁡
t
∈
[
0
,
ω
]
|
x
i
(
t
)
|; then X is a Banach space.Define the operator(14)
T
λ
:
X
⟶
X
by
(15)
(
T
λ
x
)
(
t
)
=
(
(
T
λ
x
)
1
(
t
)
,
(
T
λ
x
)
2
(
t
)
,
…
,
(
T
λ
x
)
n
(
t
)
)
⊤
,
where
(16)
(
T
λ
x
)
i
(
t
)
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
,
(17)
G
i
(
t
,
s
)
=
exp
⁡
(
-
∫
t
s
(
h
i
(
θ
,
x
(
θ
)
)
/
x
i
(
θ
)
)
d
θ
)
1
-
exp
⁡
⁡
(
-
∫
0
ω
‍
(
h
i
(
θ
,
x
(
θ
)
)
/
x
i
(
θ
)
)
d
θ
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
≤
s
≤
t
+
ω
.
Let m
=
min
⁡
1
≤
i
≤
n
min
⁡
t
,
s
∈
[
0
,
ω
]
G
i
(
t
,
s
) and M
=
max
⁡
1
≤
i
≤
n
max
⁡
t
,
s
∈
[
0
,
ω
]
G
i
(
t
,
s
), clearly;
(18)
0
<
m
≤
G
i
(
t
,
s
)
≤
M
,
t
≤
s
≤
t
+
ω
.
Define a set by
(19)
K
=
{
x
(
t
)
=
(
x
1
,
x
2
…
,
x
n
)
⊤
∈
X
:
x
i
(
t
)
≥
σ
∥
x
∥
,
t
∈
[
0
,
ω
]
}
,
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
where
σ
=
m
M
.We use the following notations.Letr
>
0 be a constant, and x
∈
K, defining
(20)
Ω
r
=
{
x
∈
X
:
∥
x
∥
<
r
}
,
∂
Ω
r
=
{
x
∈
X
:
∥
x
∥
=
r
}
,
ρ
μ
(
r
)
∶
=
max
⁡
1
≤
i
≤
n
sup
⁡
∥
x
∥
=
r
f
i
(
x
)
∥
x
∥
,
ρ
l
(
r
)
∶
=
min
⁡
1
≤
i
≤
n
inf
⁡
∥
x
∥
=
r
f
i
(
x
)
∥
x
∥
,
f
i
0
=
lim
⁡
∥
x
∥
→
0
+
f
i
(
x
)
∥
x
∥
,
f
i
∞
=
lim
⁡
∥
x
∥
→
∞
f
i
(
x
)
∥
x
∥
,
F
0
=
max
⁡
1
≤
i
≤
n
{
f
i
0
}
,
F
∞
=
max
⁡
1
≤
i
≤
n
{
f
i
∞
}
,
I
0
=
number
of
zeros
in
the
set
{
F
0
,
F
∞
}
,
I
∞
=
number
of
infinities
in
the
set
{
F
0
,
F
∞
}
,
b
μ
∶
=
max
⁡
1
≤
i
≤
n
∫
0
ω
‍
b
i
(
t
)
d
t
,
b
l
∶
=
min
⁡
1
≤
i
≤
n
∫
0
ω
‍
b
i
(
t
)
d
t
.Lemma 3.
Assume that(
H
1
)-(
H
2
) hold; then T
λ
(
K
)
⊂
K and T
λ
:
K
→
K is continuous and completely continuous.Proof.
In view of the definition ofK, for x
∈
K, we have
(21)
(
T
λ
x
)
i
(
t
+
ω
)
=
λ
∫
t
+
ω
t
+
2
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
=
λ
∫
t
t
+
ω
‍
G
i
(
t
+
ω
,
s
+
ω
)
b
i
(
s
+
ω
)
×
f
i
(
x
(
s
+
ω
-
τ
(
s
+
ω
)
)
)
d
s
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
=
(
T
λ
x
)
i
(
t
)
.
It is easy to see that ∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s is a constant because of the periodicity of b
i
(
t
)
f
i
(
x
(
t
-
τ
(
t
)
)
).
Notice that, forx
∈
K and t
∈
[
0
,
ω
],
(22)
(
T
λ
x
)
i
(
t
)
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
λ
m
∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
=
λ
m
∫
0
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
=
m
M
λ
M
∫
0
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
m
M
max
⁡
t
∈
[
0
,
ω
]
⁡
|
(
T
λ
x
)
i
(
t
)
|
.
Thus T
λ
(
K
)
⊂
K and it is easy to show that T
λ
:
K
→
K is continuous and completely continuous.Lemma 4.
Assume that(
H
1
)-(
H
2
) hold; then a function x
(
t
)
∈
K is a positive ω-periodic solution of (1) if and only if T
λ
x
=
x, x
∈
K.Proof.
Ifx
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
K and T
λ
x
=
x, then
(23)
x
i
′
(
t
)
=
d
d
t
(
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
)
≥
λ
G
i
(
t
,
t
+
ω
)
b
i
(
t
+
ω
)
f
i
(
x
(
t
+
ω
-
τ
(
t
+
ω
)
)
)
-
λ
G
i
(
t
,
t
)
b
i
(
t
)
f
i
(
x
(
t
-
τ
(
t
)
)
)
+
h
i
(
t
,
x
)
x
i
(
t
)
(
T
λ
x
)
i
(
t
)
=
λ
[
G
i
(
t
,
t
+
ω
)
-
G
i
(
t
,
t
)
]
b
i
(
t
)
f
i
(
x
(
t
-
τ
(
t
)
)
)
+
h
i
(
t
,
x
)
=
h
i
(
t
,
x
)
-
λ
b
i
(
t
)
f
i
(
x
(
t
-
τ
(
t
)
)
)
.
Thus x is a positive ω-periodic solution of (1). On the other hand, if x
=
(
x
1
,
x
2
,
…
,
x
n
) is a positive ω-periodic solution of (1), then x
i
′
(
t
)
=
h
i
(
t
,
x
)
-
λ
b
i
(
t
)
f
i
(
x
(
t
-
τ
(
t
)
)
) and
(24)
(
T
λ
x
)
i
(
t
)
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
=
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
(
h
i
(
s
,
x
)
-
x
i
′
(
s
)
)
d
s
=
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
h
i
(
s
,
x
)
d
s
-
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
x
i
′
(
s
)
d
s
=
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
h
i
(
s
,
x
)
d
s
-
G
i
(
t
,
s
)
x
i
(
s
)
|
t
t
+
ω
-
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
h
i
(
s
,
x
)
d
s
=
x
i
(
t
)
.
Thus, T
λ
x
=
x; furthermore, in view of the proof of Lemma 3, we also have x
i
(
t
)
≥
(
m
/
M
)
max
⁡
t
∈
[
0
,
ω
]
|
(
T
λ
x
)
i
(
t
)
| for t
∈
[
0
,
ω
]. That is, x is a fixed point of T
λ in K.Lemma 5.
Assume that(
H
1
)-(
H
2
) hold; for any η
>
0 and x
∈
K, if there exists a component f
i of F such that f
i
(
x
(
t
)
)
≥
x
i
(
t
)
η, then
(25)
∥
T
λ
x
∥
≥
λ
η
m
2
b
l
M
∥
x
∥
.Proof.
Sincex
∈
K and f
i
(
x
(
t
)
)
≥
x
i
(
t
)
η, we have
(26)
(
T
λ
x
)
i
(
t
)
=
λ
∫
0
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
λ
m
∫
0
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
λ
η
m
∫
0
ω
‍
b
i
(
s
)
x
i
(
s
-
τ
(
s
)
)
d
s
≥
λ
η
m
2
b
l
M
∥
x
∥
.
Thus ∥
T
λ
x
∥
≥
(
λ
η
m
2
b
l
/
M
)
∥
x
∥. The proof is complete.For eachi
=
1,2
,
…
,
n, let f
^
(
r
)
:
R
+
→
R
+ be the function given by
(27)
f
^
i
(
r
)
=
max
⁡
{
f
i
(
x
)
:
x
∈
R
+
n
,
∥
x
∥
≤
r
,
t
∈
[
0
,
ω
]
}
.
Let f
^
i
0
=
lim
⁡
r
→
0
(
f
^
i
(
r
)
/
r
) and f
^
i
∞
=
lim
⁡
r
→
∞
(
f
^
i
(
r
)
/
r
).Lemma 6 (see [11]).
Assume(
H
2
) holds. Then f
^
i
0
=
f
i
0 and f
^
i
∞
=
f
i
∞.Lemma 7.
Assume that(
H
1
)-(
H
2
) hold and let r
>
0. If x
∈
∂
Ω
r and there exists an ϵ
>
0, such that f
^
i
(
r
)
≤
ϵ
r for t
∈
[
0
,
ω
], then
(28)
∥
T
λ
x
∥
≤
λ
ϵ
M
b
μ
∥
x
∥
.Proof.
From the definition ofT
λ, for x
∈
∂
Ω
r, we have
(29)
(
T
λ
x
)
i
(
t
)
=
∫
0
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≤
λ
M
∫
0
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≤
λ
M
∫
0
ω
‍
b
i
(
s
)
f
^
i
(
r
)
d
s
≤
λ
ϵ
M
b
μ
∥
x
∥
.
That is, ∥
T
λ
x
∥
≤
λ
ϵ
M
b
μ
∥
x
∥. The proof is complete.The following two lemmas are weak forms of Lemmas5 and 7.Lemma 8.
Assume that(
H
1
)-(
H
2
) hold. If x
∈
∂
Ω
r
,
r
>
0, then
(30)
∥
T
λ
x
∥
≥
λ
m
b
l
m
^
r
,
where m
^
r
=
min
⁡
{
f
i
(
x
)
:
x
∈
R
+
n
,
m
r
/
M
≤
∥
x
∥
≤
r
}.Lemma 9.
Assume that(
H
1
)-(
H
2
) hold. If x
∈
∂
Ω
r
,
r
>
0, then
(31)
∥
T
λ
x
∥
≤
λ
M
b
μ
M
^
(
r
)
,
where M
^
r
=
max
⁡
{
f
i
(
x
)
:
x
∈
R
+
n
,
∥
x
∥
≤
r
}.


## 3. The Main Results

Theorem 10.
Assume that(
H
1
)-(
H
2
) hold and there exist positive constants r
1 and r
2 with r
1
<
r
2 such that
(32)
m
b
l
ρ
l
(
r
2
)
>
M
b
μ
ρ
μ
(
r
1
)
;
then for
(33)
1
m
b
l
ρ
l
(
r
2
)
≤
λ
≤
1
M
b
μ
ρ
μ
(
r
1
)
,
the system (1) has at least a positive ω-periodic solution x
(
t
) satisfying r
1
≤
∥
x
∥
≤
r
2.Proof.
From (32) for λ satisfying (33) we have that
(34)
λ
m
b
l
ρ
l
(
r
2
)
≥
1
,
λ
M
b
μ
ρ
μ
(
r
1
)
<
1
,
or
(35)
λ
m
b
l
ρ
l
(
r
2
)
>
1
,
λ
M
b
u
ρ
μ
(
r
1
)
≤
1
.
Let x
∈
K and ∥
x
∥
=
r
1; by (16) and (34), we have
(36)
(
T
λ
x
)
i
(
t
)
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≤
λ
M
∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≤
λ
M
∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
∥
x
∥
∥
x
∥
d
s
≤
λ
M
b
μ
ρ
μ
(
r
1
)
∥
x
∥
≤
∥
x
∥
.
That is, ∥
T
λ
x
i
∥
≤
∥
x
∥. This implies that ∥
T
λ
x
∥
≤
∥
x
∥ for x
∈
K
∩
∂
Ω
r
1.
Ifx
∈
K and ∥
x
∥
=
r
2, by (16) and (35), we have(37)
(
T
λ
x
)
i
(
t
)
=
λ
∫
t
t
+
ω
‍
G
i
(
t
,
s
)
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
λ
m
∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
d
s
≥
λ
m
∫
t
t
+
ω
‍
b
i
(
s
)
f
i
(
x
(
s
-
τ
(
s
)
)
)
∥
x
∥
∥
x
∥
d
s
≥
λ
m
b
l
ρ
l
(
r
2
)
∥
x
∥
≥
∥
x
∥
.
This implies that ∥
T
λ
x
∥
≥
∥
x
∥ for x
∈
K
∩
∂
Ω
r
2. By Lemma 2(a), T
λ has a fixed point in K
∩
(
Ω
¯
r
2
∖
Ω
r
1
). It follows from Lemma 2 that (1) has an ω-periodic solution x with r
1
≤
∥
x
∥
≤
r
2. The proof is complete.Theorem 11.
Assume that(
H
1
)-(
H
2
) hold and there exist positive constants r
1 and r
2 with r
1
<
r
2 such that
(38)
m
b
l
ρ
l
(
r
1
)
>
M
b
μ
ρ
μ
(
r
2
)
;
then for
(39)
1
m
b
l
ρ
l
(
r
1
)
≤
λ
≤
1
M
b
μ
ρ
μ
(
r
2
)
,
the system (1) has at least a positive ω-periodic solution x
(
t
) satisfying r
1
≤
∥
x
∥
≤
r
2.Proof.
The proof of Theorem11 is similar to that of Theorem 10, so we omit it. The proof is complete.Theorem 12.
Assume that(
H
1
)-(
H
2
) hold.(a)
IfI
0
=
1 or I
0
=
2, then (1) has I
0 positive ω-periodic solution(s) for λ
>
1
/
m
b
l
m
^
1
>
0.
(b)
IfI
∞
=
1 or I
∞
=
2, then (1) has I
∞ positive ω-periodic solution(s) for 0
<
λ
<
1
/
M
b
μ
M
^
1.
(c)
IfI
0
=
0 or I
∞
=
0, then (1) has no positive ω-periodic solution for sufficiently large or small λ
>
0, respectively.Proof.
(a) Choose a numberr
1
=
1. By Lemma 8 we infer that there exists λ
0
=
1
/
m
b
l
m
^
1
>
0 such that
(40)
∥
T
λ
(
x
)
∥
≥
∥
x
∥
for
x
∈
∂
Ω
r
1
,
λ
>
λ
0
.
IfF
0
=
0, this implies that f
i
0
=
0. It follows from Lemma 6 that f
^
i
0
=
0; therefore, we choose 0
<
r
2
<
r
1 so that f
^
i
(
r
2
)
≤
ϵ
r
2, where the constant ϵ
>
0 satisfies λ
ϵ
M
b
μ
<
1. By Lemma 7, it follows that
(41)
∥
T
λ
x
∥
≤
λ
ϵ
M
b
μ
∥
x
∥
<
∥
x
∥
,
for
x
∈
∂
Ω
r
2
,
t
∈
[
0
,
ω
]
;
it follows from Lemma 1 that
(42)
i
(
T
λ
,
Ω
r
1
,
K
)
=
0
,
i
(
T
λ
,
Ω
r
2
,
K
)
=
1
.
Thus i
(
T
λ
,
Ω
r
1
∖
Ω
¯
r
2
,
K
)
=
-
1, which implies T
λ has a fixed point in Ω
r
1
∖
Ω
¯
r
2, which is positive ω-periodic solution of (1) for λ
>
λ
0.
IfF
∞
=
0, then f
i
∞
=
0. It follows from Lemma 6 that f
^
i
∞
=
0. Therefore, we choose r
3
>
2
r
1 such that f
^
i
(
t
,
r
3
)
≤
ϵ
r
3, where the constant ϵ
>
0 satisfies λ
ϵ
M
b
μ
<
1. By Lemma 7, it follows that
(43)
∥
T
λ
x
∥
≤
λ
ϵ
M
b
μ
∥
x
∥
<
∥
x
∥
,
for
x
∈
∂
Ω
r
3
,
t
∈
[
0
,
ω
]
;
it follows from Lemma 1 that
(44)
i
(
T
λ
,
Ω
r
1
,
K
)
=
0
,
i
(
T
λ
,
Ω
r
3
,
K
)
=
1
;
thus i
(
T
λ
,
Ω
r
3
∖
Ω
¯
r
1
,
K
)
=
1, which implies T
λ has a fixed point in Ω
r
3
∖
Ω
¯
r
1, which is positive ω-periodic solution of (1) for λ
>
λ
0.
IfF
0
=
F
∞
=
0, it is easy to see from the above proof that T
λ has fixed points x
1 in Ω
r
1
∖
Ω
¯
r
2 and x
2 in Ω
r
3
∖
Ω
¯
r
1 such that
(45)
r
2
<
∥
x
1
∥
<
r
1
<
∥
x
2
∥
<
r
3
.
Consequently, (1) has two positive ω-periodic solutions for λ
>
λ
0.
(b) Choose a numberr
1
=
1; by Lemma 9 we infer that there exists a λ
0
=
1
/
M
b
μ
M
^
1
>
0 such that
(46)
∥
T
λ
(
x
)
∥
<
∥
x
∥
for
x
∈
∂
Ω
r
1
,
0
<
λ
<
λ
0
.
IfF
0
=
∞, there exists a component f
i such that f
i
0
=
∞. Therefore there is a positive number r
2
<
r
1 such that
(47)
f
i
(
x
)
≥
η
∥
x
∥
,
for
∥
x
∥
≤
r
2
,
t
∈
[
0
,
ω
]
,
where the constant η
>
0 satisfies λ
η
m
2
b
l
M
>
1. Lemma 5 implies that
(48)
∥
T
λ
x
∥
≥
λ
η
M
b
μ
∥
x
∥
>
∥
x
∥
,
for
x
∈
∂
Ω
r
2
,
t
∈
[
0
,
ω
]
.
It follows from Lemma 1 that
(49)
i
(
T
λ
,
Ω
r
1
,
K
)
=
1
,
i
(
T
λ
,
Ω
r
2
,
K
)
=
0
.
Thus i
(
T
λ
,
Ω
r
1
∖
Ω
¯
r
2
,
K
)
=
1 which implies T
λ has a fixed point in Ω
r
1
∖
Ω
¯
r
2, which is positive ω-periodic solution of (1) for 0
<
λ
<
λ
0.
IfF
∞
=
∞, there exists a component f
i such that f
i
∞
=
∞. Therefore there is a positive number H
^ such that
(50)
f
i
(
x
)
≥
η
∥
x
∥
,
for
∥
x
∥
≥
H
^
,
where the constant η
>
0 satisfies λ
η
m
2
b
l
M
>
1. Let r
3
=
max
⁡
{
2
r
1
,
M
H
^
/
m
}; if x
∈
∂
Ω
r
3, then
(51)
min
⁡
t
∈
[
0
,
ω
]
x
i
(
t
)
≥
m
M
∥
x
∥
=
m
M
r
3
≥
H
^
.
Hence,
(52)
f
i
(
x
)
≥
η
x
i
(
t
)
.
Again, it follows from Lemma 5 that
(53)
∥
T
λ
x
∥
≥
λ
η
M
b
μ
∥
x
∥
>
∥
x
∥
,
for
x
∈
∂
Ω
r
3
,
t
∈
[
0
,
ω
]
.
It follows from Lemma 1 that
(54)
i
(
T
λ
,
Ω
r
1
,
K
)
=
1
,
i
(
T
λ
,
Ω
r
3
,
K
)
=
0
.
Thus i
(
T
λ
,
Ω
r
3
∖
Ω
¯
r
1
,
K
)
=
-
1, which implies T
λ has a fixed point in Ω
r
3
∖
Ω
¯
r
1, which is positive ω-periodic solution of (1) for 0
<
λ
<
λ
0.
IfF
0
=
F
∞
=
∞, it is easy to see from the above proof that T
λ has fixed points x
1 in Ω
r
1
∖
Ω
¯
r
2 and x
2 in Ω
r
3
∖
Ω
¯
r
1 such that
(55)
r
2
<
∥
x
1
∥
<
r
1
<
∥
x
2
∥
<
r
3
.
Consequently, (1) has two positive ω-periodic solutions for 0
<
λ
<
λ
0.
(c) IfI
0
=
0, then F
0
>
0 and F
∞
>
0; there exist two components f
i and f
j such that
(56)
f
i
0
>
0
,
f
i
∞
>
0
.
It is easy to show (see [11]) that positive numbers η
,
r
1 exist such that
(57)
f
i
(
x
)
≥
η
∥
x
∥
,
∥
x
∥
≤
r
1
,
(58)
f
j
(
x
)
≥
η
∥
x
∥
,
∥
x
∥
≥
m
M
r
1
.
Assumey
(
t
)
=
[
y
1
,
y
2
,
…
,
y
n
] is a positive ω-periodic solution of (1), we will show that this leads to a contradiction for λ
>
λ
0
=
M
/
η
m
2
b
l. In fact, if ∥
y
∥
≤
r
1, (57) implies that
(59)
f
i
(
y
)
≥
η
y
i
(
t
)
,
for
t
∈
[
0
,
ω
]
.
On the other hand, if ∥
y
∥
>
r
1, then
(60)
min
⁡
t
∈
[
0
,
ω
]
y
i
(
t
)
≥
σ
∥
y
∥
≥
σ
r
1
,
which together with (58), implies that
(61)
f
j
(
y
)
≥
η
y
j
(
t
)
,
for
t
∈
[
0
,
ω
]
.
Since T
λ
(
y
)
=
y
(
t
), for t
∈
[
0
,
ω
], it follows from Lemma 5 that, for λ
>
λ
0,
(62)
∥
y
∥
=
∥
T
λ
(
y
)
∥
≥
λ
η
m
2
b
l
M
∥
y
∥
>
∥
y
∥
,
which is a contradiction.
IfI
∞
=
0, then F
0
<
∞ and F
∞
<
∞; there exist two components f
i and f
j such that f
i
0
<
∞ and f
j
∞
<
∞. It is easy to show (see [11]) that positive numbers ϵ exist such that
(63)
f
i
(
x
)
≤
ϵ
∥
x
∥
.
Assume that y
(
t
)
=
[
y
1
,
y
2
,
…
,
y
n
] is a positive ω-periodic solution of (1); we will show that this leads to a contradiction for 0
<
λ
<
λ
0
=
1
/
ϵ
M
b
μ. In fact, for 0
<
λ
<
λ
0, since T
λ
(
y
)
=
y
(
t
), we find
(64)
∥
y
∥
=
∥
T
λ
(
y
)
∥
≤
λ
ϵ
M
b
μ
∥
y
∥
<
∥
y
∥
,
which is a contradiction. The proof is complete.Theorem13 is a direct consequence of the proof of Theorem 12(c). Under the conditions of Theorem 13, we are able to give explicit intervals of λ such that (1) has no positive ω-periodic solution.Theorem 13.
Assume that(
H
1
)-(
H
2
) hold.(a)
If there is ac
1
>
0 such that f
i
(
x
)
≥
c
1
∥
x
∥,  x
∈
R
+
n, then there exists a λ
0
=
M
/
m
2
c
1
b
l such that, for all λ
>
λ
0, (1) has no positive ω-periodic solution.
(b)
If there is ac
2
>
0 such that f
i
(
x
)
≤
c
2
∥
x
∥,  x
∈
R
+
n, then there exists a λ
0
=
1
/
M
c
2
b
μ such that, for all 0
<
λ
<
λ
0, (1) has no positive ω-periodic solution.Theorem 14.
Assume that(
H
1
)-(
H
2
) hold and I
0
=
I
∞
=
0. If
(65)
M
max
⁡
{
F
0
,
F
∞
}
m
2
b
l
<
λ
<
1
min
⁡
{
F
0
,
F
∞
}
M
b
μ
,
then (1) has a positive ω-periodic solution.Proof.
(a) IfF
∞
>
F
0, then there exist two components f
i and f
j such that f
i
0
<
f
j
∞. It is easy to see that there exists an 0
<
ϵ
<
f
j
∞ such that
(66)
M
m
2
b
l
(
f
j
∞
-
ϵ
)
<
λ
<
1
M
b
μ
(
f
i
0
+
ϵ
)
.
Now, turning to f
i
0 and f
j
∞, there is an r
1
>
0 such that
(67)
f
i
(
x
)
≤
(
f
i
0
+
ϵ
)
∥
x
∥
,
for
0
<
∥
x
∥
<
r
1
.
Thus
(68)
f
i
(
x
)
≤
(
f
i
0
+
ϵ
)
∥
x
∥
,
for
x
∈
∂
Ω
r
1
,
t
∈
[
0
,
ω
]
.
We have by Lemma 7 that
(69)
∥
T
λ
x
∥
≤
λ
(
f
i
0
+
ϵ
)
M
b
μ
∥
x
∥
<
∥
x
∥
,
hhhhhhh
for
x
∈
∂
Ω
r
1
,
t
∈
[
0
,
ω
]
.
On the other hand, there is an H
^
>
r
1 such that
(70)
f
j
(
x
)
≥
(
f
j
∞
-
ϵ
)
∥
x
∥
,
for
∥
x
∥
≥
H
^
.
Let r
2
=
max
⁡
{
2
r
1
,
m
H
^
/
M
}. It follows that
(71)
x
i
(
t
)
≥
m
M
∥
x
∥
≥
H
^
,
for
x
∈
∂
Ω
r
2
,
t
∈
[
0
,
ω
]
.
Thus f
i
(
x
)
≤
(
f
j
∞
-
ϵ
)
∥
x
∥, for x
∈
∂
Ω
r
2 and t
∈
[
0
,
ω
]. In view of Lemma 7, we have
(72)
∥
T
λ
x
∥
≥
λ
(
f
j
∞
-
ϵ
)
m
2
b
l
M
∥
x
∥
>
∥
x
∥
,
hhhhhhhh
for
x
∈
∂
Ω
r
2
,
t
∈
[
0
,
ω
]
.
It follows from Lemma 1 that
(73)
i
(
T
λ
,
Ω
r
1
,
K
)
=
1
,
i
(
T
λ
,
Ω
r
2
,
K
)
=
0
.
Thus i
(
T
λ
,
Ω
r
2
∖
Ω
¯
r
1
,
K
)
=
-
1. Hence, T
λ has a fixed point in Ω
r
2
∖
Ω
¯
r
1. Consequently, (1) has a positive ω-periodic solution.
(b) IfF
∞
>
F
0. The remaining part of the proof is similar to that of Theorem 14(a); therefore it is omitted. The proof is complete.


## 4. Remarks

Remark 15.
Based on the conditionH
2 of [8], we may obtain the inequality sequence of
(74)
l
a
i
(
t
)
≤
A
(
t
)
G
(
x
)
≤
L
a
i
(
t
)
,
i
=
1,2
,
…
,
n
.
Comparatively, in terms of condition of H
1 presented in this paper, we have the inequality sequence of
(75)
α
i
(
t
)
x
i
≤
h
i
(
t
,
x
)
≤
β
i
(
t
)
x
i
,
i
=
1,2
,
…
,
n
;
it is clear that result of this paper can be applied to even more wider domain. Additionally, an extra requirement of
(76)
0
<
l
<
G
(
x
)
<
L
,
included in H
2 of [8], is not demanded here.
For example, when letting(77)
H
(
t
,
x
)
=
A
(
t
)
G
(
x
)
=
[
a
1
(
t
)
0
0
a
2
(
t
)
]
[
(
2
+
sin
x
1
)
x
1
(
2
+
cos
⁡
x
2
)
x
2
]
and then
(78)
G
(
x
)
=
[
(
2
+
sin
x
1
)
x
1
(
2
+
cos
⁡
x
2
)
x
2
]
,
we can obtain the formula of
(79)
h
1
(
t
,
x
)
=
a
1
(
t
)
(
2
+
sin
x
1
)
x
1
,
h
2
(
t
,
x
)
=
a
2
(
t
)
(
2
+
cos
⁡
x
2
)
x
2
.
According to condition H
1, the formulas would be further achieved as
(80)
a
1
(
t
)
≤
h
1
(
t
,
x
)
x
1
=
a
1
(
t
)
(
2
+
sin
x
1
)
x
1
x
1
≤
3
a
1
(
t
)
,
a
2
(
t
)
≤
h
2
(
t
,
x
)
x
2
=
a
2
(
t
)
(
2
+
cos
⁡
x
2
)
x
2
x
2
≤
3
a
2
(
t
)
.
However, there do not exist l
>
0 such that
(81)
0
<
l
<
g
1
(
x
1
)
=
(
2
+
sin
x
1
)
x
1
.
Therefore, function of H
(
t
,
x
)
=
A
(
t
)
G
(
x
) cannot satisfy the conditions proposed in [8].Remark 16.
The technique of proof of theorems in this paper differs from that of theorems in [8]. What is more, the condition of Theorems 10 and 11 is obviously weaker here than proposed in [8].Remark 17.
In this paper the application scope of Theorems12–14 is relatively wider than that of Theorems 1.1
–
1.3 in [6], respectively.



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*Source: 268418-2014-08-17.xml*