2 Preliminaries
We first give the definition of an anti-periodic function. Assume that N is a positive integer. Let u
:
R
→
R
N
be a continuous function. We call u
(
t
)
an anti-periodic function on ℝ if u satisfies the following condition:
u
(
t
+
T
2
)
=
−
u
(
t
)
,
for all
t
∈
R
.
Obviously, a
T
2
-anti-periodic function u is a T-periodic function.
Throughout this paper, we will adopt the following notations:
![]()
which is a linear normal space endowed with the norm ∥
⋅
∥
defined by
∥
x
∥
=
max
{
|
x
|
∞
,
|
x
′
|
∞
,
…
,
|
x
(
k
)
|
∞
}
,
for all
x
∈
C
T
k
,
1
2
(
R
,
R
N
)
.
The following lemmas will be useful to prove our main results.
Lemma 2.1 12
If x
∈
C
T
1
(
R
,
R
)
and
∫
0
T
x
(
t
)
d
t
=
0
, then
∫
0
T
|
x
(
t
)
|
2
d
t
⩽
(
T
2
/
4
π
2
)
∫
0
T
|
x
′
(
t
)
|
2
d
t
(Wirtinger inequality) and
|
x
|
∞
2
⩽
(
T
/
12
)
∫
0
T
|
x
′
(
t
)
|
2
d
t
(Sobolev inequality).
Lemma 2.2 Suppose that the following condition holds:
(
H
1
)
(
g
(
t
,
x
1
)
−
g
(
t
,
x
2
)
)
(
x
1
−
x
2
)
<
0
, for all t
,
x
1
,
x
2
∈
R
and
x
1
≠
x
2
.
Then Eq. (1.1) has at most one T-periodic solution.
Proof Assume that
x
1
(
t
)
and
x
2
(
t
)
are two T-periodic solutions of Eq. (1.1). Then we obtain
(
x
i
′
(
t
)
1
+
x
i
′
2
(
t
)
)
′
+
f
(
t
,
x
i
′
(
t
)
)
+
g
(
t
,
x
i
(
t
)
)
=
e
(
t
)
,
i
=
1
,
2
.
It is easy to see that
x
i
(
t
)
∈
C
2
[
0
,
T
]
(i
=
1
,
2
). From (2.1), we know
![]()
Set z
(
t
)
=
x
1
(
t
)
−
x
2
(
t
)
. Now, we prove
z
(
t
)
⩽
0
,
for all
t
∈
R
.
Otherwise, we have
max
t
∈
R
z
(
t
)
=
max
t
∈
[
0
,
T
]
z
(
t
)
>
0
.
Then there exists a
t
∗
∈
[
0
,
T
]
such that
x
1
(
t
∗
)
−
x
2
(
t
∗
)
=
z
(
t
∗
)
=
max
t
∈
R
z
(
t
)
=
max
t
∈
[
0
,
T
]
z
(
t
)
>
0
,
which implies that
z
′
(
t
∗
)
=
x
1
′
(
t
∗
)
−
x
2
′
(
t
∗
)
=
0
,
and
z
″
(
t
∗
)
=
x
1
″
(
t
∗
)
−
x
2
″
(
t
∗
)
⩽
0
.
It follows from (2.2), (2.4) and (2.5) that
g
(
t
∗
,
x
1
(
t
∗
)
)
−
g
(
t
∗
,
x
2
(
t
∗
)
)
=
−
(
x
1
″
(
t
∗
)
(
1
+
x
1
′
2
(
t
∗
)
)
3
−
x
2
″
(
t
∗
)
(
1
+
x
2
′
2
(
t
∗
)
)
3
)
−
(
f
(
t
∗
,
x
1
′
(
t
∗
)
)
−
f
(
t
∗
,
x
2
′
(
t
∗
)
)
)
=
−
1
(
1
+
x
1
′
2
(
t
∗
)
)
3
(
x
1
″
(
t
∗
)
−
x
2
″
(
t
∗
)
)
=
−
1
(
1
+
x
1
′
2
(
t
∗
)
)
3
z
″
(
t
∗
)
⩾
0
.
From (H1), we get
x
1
(
t
∗
)
−
x
2
(
t
∗
)
⩽
0
,
which contradicts (2.3). Thus,
z
(
t
)
⩽
0
,
for all
t
∈
R
.
By using a similar argument, we can also show
z
(
t
)
⩾
0
,
for all
t
∈
R
.
Hence,
x
1
(
t
)
−
x
2
(
t
)
=
z
(
t
)
=
0
,
for all
t
∈
R
.
Therefore, Eq. (1.1) has at most one T-periodic solution. The proof is completed. □
To prove the main result of this paper, we shall use a continuation theorem 1314 as follows.
Lemma 2.3 Let Ω be open bounded in a linear normal space X. Suppose that
f
˜
is a complete continuous field on
Ω
¯
. Moreover, assume that the Leray-Schauder degree
deg
{
f
˜
,
Ω
,
p
}
≠
0
,
for
p
∈
X
∖
f
˜
(
∂
Ω
)
.
Then the equation
f
˜
(
x
)
=
p
has at least one solution in Ω.
3 Main result
In this section, we present and prove our main result concerning the existence and uniqueness of anti-periodic solutions of Eq. (1.1).
Theorem 3.1 Let (H1) hold. Moreover, assume that the following conditions hold:
(
H
2
)
there exists l
>
0
such that
|
g
(
t
,
x
1
)
−
g
(
t
,
x
2
)
|
⩽
l
|
x
1
−
x
2
|
,
for all
t
,
x
1
,
x
2
∈
R
;
(
H
3
)
there exists β
,
γ
>
0
such that
γ
⩽
lim inf
|
x
|
→
∞
f
(
t
,
x
)
x
⩽
lim sup
|
x
|
→
∞
f
(
t
,
x
)
x
⩽
β
,
uniformly in
t
∈
R
;
(
H
4
)
for all t
,
x
∈
R
,
f
(
t
+
T
2
,
−
x
)
=
−
f
(
t
,
x
)
,
g
(
t
+
T
2
,
−
x
)
=
−
g
(
t
,
x
)
,
e
(
t
+
T
2
)
=
−
e
(
t
)
.
Then Eq. (1.1) has a unique anti-periodic solution for l
⋅
T
2
π
<
γ
.
Proof Rewrite Eq. (1.1) in the equivalent form:
{
x
1
′
(
t
)
=
ψ
(
x
2
(
t
)
)
=
x
2
(
t
)
1
−
x
2
2
(
t
)
,
x
2
′
(
t
)
=
−
f
(
t
,
ψ
(
x
2
(
t
)
)
)
−
g
(
t
,
x
1
(
t
)
)
+
e
(
t
)
,
where ψ
(
x
)
=
x
1
−
x
2
. Now, we consider the auxiliary equation of (3.1),
{
x
1
′
(
t
)
=
λ
x
2
(
t
)
1
−
x
2
2
(
t
)
=
λ
ψ
(
x
2
(
t
)
)
,
x
2
′
(
t
)
=
−
λ
f
(
t
,
ψ
(
x
2
(
t
)
)
)
−
λ
g
(
t
,
x
1
(
t
)
)
+
λ
e
(
t
)
,
where λ
∈
(
0
,
1
]
is a parameter. Set
x
(
t
)
=
(
x
1
(
t
)
x
2
(
t
)
)
,
Q
1
(
t
,
x
1
(
t
)
,
x
2
(
t
)
)
=
(
ψ
(
x
2
(
t
)
)
−
f
(
t
,
ψ
(
x
2
(
t
)
)
)
−
g
(
t
,
x
1
(
t
)
)
+
e
(
t
)
)
.
Then Eq. (3.2) can be reduced to the equation as follows:
x
′
(
t
)
=
λ
Q
1
(
t
,
x
1
(
t
)
,
x
2
(
t
)
)
.
By Lemma 2.2 and condition (H1), it is easy to see that Eq. (1.1) has at most one anti-periodic solution. Thus, to prove Theorem 3.1, it suffices to show that Eq. (1.1) has at least one anti-periodic solution. To do this, we shall apply Lemma 2.3. Firstly, we will prove that the set of all possible anti-periodic solutions of Eq. (3.2) is bounded.
Let x
(
t
)
=
(
x
1
(
t
)
,
x
2
(
t
)
)
T
∈
C
T
1
,
1
2
(
R
,
R
2
)
be an arbitrary possible anti-periodic solution of Eq. (3.2). Then
x
1
(
t
)
∈
C
T
1
,
1
2
(
R
,
R
)
. Thus, we have
∫
0
T
x
1
(
t
)
d
t
=
∫
0
T
2
x
1
(
t
)
d
t
+
∫
T
2
T
x
1
(
t
)
d
t
=
∫
0
T
2
x
1
(
t
)
d
t
+
∫
0
T
2
x
1
(
t
+
T
2
)
d
t
=
0
.
It follows from Lemma 2.1 that
|
x
1
|
∞
⩽
T
12
|
x
1
′
|
2
.
Obviously, Eq. (3.2) is equivalent to the following equation:
(
1
λ
x
1
′
(
t
)
1
+
1
λ
2
x
1
′
2
(
t
)
)
′
+
λ
f
(
t
,
1
λ
x
1
′
(
t
)
)
+
λ
g
(
t
,
x
1
(
t
)
)
=
λ
e
(
t
)
.
Multiplying (3.3) by
x
1
′
and integrating from 0 to T, we have
λ
∫
0
T
f
(
t
,
1
λ
x
1
′
(
t
)
)
x
1
′
(
t
)
d
t
+
λ
∫
0
T
g
(
t
,
x
1
(
t
)
)
x
1
′
(
t
)
d
t
=
λ
∫
0
T
e
(
t
)
x
1
′
(
t
)
d
t
.
Since l
⋅
T
2
π
<
γ
, there exists a constant ε
>
0
such that
l
⋅
T
2
π
<
γ
−
ε
.
For such a ε>0, in view of (H3), there exists
M
1
⩾
0
such that for all t,x∈R, x
f
(
t
,
x
)
⩾
(
γ
−
ε
)
x
2
−
M
1
. Hence,
|
λ
∫
0
T
f
(
t
,
1
λ
x
1
′
(
t
)
)
x
1
′
(
t
)
d
t
|
⩾
λ
2
∫
0
T
f
(
t
,
1
λ
x
1
′
(
t
)
)
x
1
′
(
t
)
λ
d
t
⩾
(
γ
−
ε
)
∫
0
T
|
x
1
′
(
t
)
|
2
d
t
−
λ
2
M
1
.
It follows from (3.4) and (3.6) that
(
γ
−
ε
)
∫
0
T
|
x
1
′
(
t
)
|
2
d
t
⩽
|
λ
∫
0
T
f
(
t
,
1
λ
x
1
′
(
t
)
)
x
1
′
(
t
)
d
t
|
+
λ
2
M
1
⩽
|
∫
0
T
g
(
t
,
x
1
(
t
)
)
x
1
′
(
t
)
d
t
|
+
|
∫
0
T
e
(
t
)
x
1
′
(
t
)
d
t
|
+
M
1
⩽
∫
0
T
|
g
(
t
,
x
1
(
t
)
)
−
g
(
t
,
0
)
|
|
x
1
′
(
t
)
|
d
t
+
∫
0
T
(
|
g
(
t
,
0
)
|
+
|
e
(
t
)
|
)
|
x
1
′
(
t
)
|
d
t
+
M
1
⩽
l
∫
0
T
|
x
1
(
t
)
|
|
x
1
′
(
t
)
|
d
t
+
max
t
∈
[
0
,
T
]
{
|
g
(
t
,
0
)
|
+
|
e
(
t
)
|
}
∫
0
T
|
x
1
′
(
t
)
|
d
t
+
M
1
.
For u
,
v
∈
C
(
[
a
,
b
]
,
R
)
, we have the Schwarz inequality
∫
a
b
|
u
(
x
)
|
|
v
(
x
)
|
d
x
⩽
(
∫
a
b
|
u
(
x
)
|
2
d
x
)
1
2
(
∫
a
b
|
v
(
x
)
|
2
d
x
)
1
2
.
Hence,
(
γ
−
ε
)
∫
0
T
|
x
1
′
(
t
)
|
2
d
t
⩽
l
(
∫
0
T
|
x
1
(
t
)
|
2
d
t
)
1
2
⋅
(
∫
0
T
|
x
1
′
(
t
)
|
2
d
t
)
1
2
+
max
t
∈
[
0
,
T
]
{
|
g
(
t
,
0
)
|
+
|
e
(
t
)
|
}
T
(
∫
0
T
|
x
1
′
(
t
)
|
2
d
t
)
1
2
+
M
1
=
l
|
x
1
|
2
|
x
1
′
|
2
+
T
|
x
1
′
|
2
max
t
∈
[
0
,
T
]
{
|
g
(
t
,
0
)
|
+
|
e
(
t
)
|
}
+
M
1
⩽
l
⋅
T
2
π
|
x
1
′
|
2
2
+
T
|
x
1
′
|
2
max
t
∈
[
0
,
T
]
{
|
g
(
t
,
0
)
|
+
|
e
(
t
)
|
}
+
M
1
.
From (3.5) and (3.7), we know that there exists a constant
D
1
>
0
such that
|
x
1
′
|
2
⩽
D
1
,
and
|
x
1
|
∞
⩽
D
1
.
By the first equation of (3.2), we have
∫
0
T
x
2
(
t
)
1
−
x
2
2
(
t
)
d
t
=
0
.
Then there exists η
∈
[
0
,
T
]
such that
x
2
(
η
)
=
0
. It follows that
x
2
(
t
)
=
x
2
(
η
)
+
∫
η
t
x
2
′
(
s
)
d
s
, and so
|
x
2
|
∞
⩽
∫
0
T
|
x
2
′
(
t
)
|
d
t
.
According to (H3), we know there exists
M
2
⩾
0
such that for all t,x∈R,
|
f
(
t
,
x
)
|
⩽
(
β
+
1
)
|
x
|
+
M
2
.
From the second equation of (3.2), we get
∫
0
T
|
x
2
′
(
t
)
|
d
t
⩽
∫
0
T
λ
|
f
(
t
,
1
λ
x
1
′
(
t
)
)
|
d
t
+
∫
0
T
|
g
(
t
,
x
1
(
t
)
)
|
d
t
+
∫
0
T
|
e
(
t
)
|
d
t
⩽
(
β
+
1
)
∫
0
T
|
x
1
′
(
t
)
|
d
t
+
λ
M
2
T
+
∫
0
T
|
g
(
t
,
x
1
(
t
)
)
|
d
t
+
∫
0
T
|
e
(
t
)
|
d
t
⩽
(
β
+
1
)
T
|
x
1
′
|
2
+
∫
0
T
|
g
(
t
,
x
1
(
t
)
)
|
d
t
+
∫
0
T
|
e
(
t
)
|
d
t
+
M
2
T
.
From (3.8), we know that there exists a constant k
>
0
such that
|
g
(
t
,
x
1
(
t
)
|
⩽
k
,
∀
t
∈
[
0
,
T
]
.
Thus,
∫
0
T
|
x
2
′
(
t
)
|
d
t
⩽
(
β
+
1
)
T
D
1
+
k
T
+
T
max
t
∈
[
0
,
T
]
|
e
(
t
)
|
+
M
2
T
,
which implies that there exists a constant
D
2
>
0
such that
|
x
2
|
∞
⩽
D
2
.
Let
M
=
max
{
D
1
,
D
2
}
+
1
.
Set
Ω
=
{
x
∈
C
T
0
,
1
2
(
R
,
R
2
)
=
X
:
∥
x
∥
<
M
}
.
Then Eq. (3.2) has no anti-periodic solution on ∂Ω for λ∈(0,1].
Next, we consider the Fourier series expansions of two functions
x
j
(
t
)
∈
C
T
k
,
1
2
(
R
,
R
)
(j
=
1
,
2
). We have
x
j
(
t
)
=
∑
i
=
0
∞
[
a
2
i
+
1
j
cos
2
π
(
2
i
+
1
)
t
T
+
b
2
i
+
1
j
sin
2
π
(
2
i
+
1
)
t
T
]
.
Define an operator
L
1
:
C
T
k
,
1
2
(
R
,
R
)
→
C
T
k
+
1
,
1
2
(
R
,
R
)
by setting
(
L
1
x
j
)
(
t
)
=
∫
0
t
x
j
(
s
)
d
s
−
T
2
π
∑
i
=
0
∞
b
2
i
+
1
j
2
i
+
1
=
T
2
π
∑
i
=
0
∞
[
a
2
i
+
1
j
2
i
+
1
sin
2
π
(
2
i
+
1
)
t
T
−
b
2
i
+
1
j
2
i
+
1
cos
2
π
(
2
i
+
1
)
t
T
]
.
Then
d
d
t
(
L
1
x
j
)
(
t
)
=
x
j
(
t
)
,
and
|
(
L
1
x
j
)
(
t
)
|
⩽
∫
0
T
|
x
j
(
s
)
|
d
s
+
T
2
π
∑
i
=
0
∞
|
b
2
i
+
1
j
|
2
i
+
1
⩽
T
|
x
j
|
∞
+
T
2
π
(
∑
i
=
0
∞
(
b
2
i
+
1
j
)
2
)
1
2
(
∑
i
=
0
∞
1
(
2
i
+
1
)
2
)
1
2
.
Since
(
∑
i
=
0
∞
1
(
2
i
+
1
)
2
)
1
2
=
π
2
2
,
and
∫
0
T
|
x
j
(
s
)
|
2
d
s
=
T
2
∑
i
=
0
∞
[
(
a
2
i
+
1
j
)
2
+
(
b
2
i
+
1
j
)
2
]
,
we obtain
|
(
L
1
x
j
)
(
t
)
|
⩽
T
|
x
j
|
∞
+
T
4
2
(
∑
i
=
0
∞
[
(
a
2
i
+
1
j
)
2
+
(
b
2
i
+
1
j
)
2
]
)
1
2
⩽
T
|
x
j
|
∞
+
T
4
2
(
2
T
∫
0
T
|
x
j
(
s
)
|
2
d
s
)
1
2
⩽
(
T
+
T
4
)
|
x
j
|
∞
.
Define L
:
C
T
k
,
1
2
(
R
,
R
2
)
→
C
T
k
+
1
,
1
2
(
R
,
R
2
)
by setting
(
L
x
)
(
t
)
=
L
(
x
1
(
t
)
x
2
(
t
)
)
=
(
(
L
1
x
1
)
(
t
)
(
L
1
x
2
)
(
t
)
)
.
Then
|
L
x
|
∞
⩽
(
T
+
T
4
)
|
x
|
∞
, and thus L is continuous.
For any x
(
t
)
∈
C
T
0
,
1
2
(
R
,
R
2
)
, we know from (H4) that
Q
1
(
t
+
T
2
,
x
1
(
t
+
T
2
)
,
x
2
(
t
+
T
2
)
)
=
−
Q
1
(
t
,
x
1
(
t
)
,
x
2
(
t
)
)
.
Therefore,
Q
1
(
t
,
x
1
(
t
)
,
x
2
(
t
)
)
∈
C
T
0
,
1
2
(
R
,
R
2
)
. Define an operator
F
μ
:
Ω
¯
→
C
T
1
,
1
2
(
R
,
R
2
)
⊂
X
by setting
F
μ
(
x
)
=
μ
L
(
Q
1
(
t
,
x
1
(
t
)
,
x
2
(
t
)
)
)
,
μ
∈
[
0
,
1
]
.
It is easy to see that
F
μ
is a compact homotopy, and the fixed point of
F
1
on Ω¯ is the anti-periodic of Eq. (3.1).
Define a homotopic field as follows:
H
μ
(
x
)
:
Ω
¯
×
[
0
,
1
]
→
C
T
0
,
1
2
(
R
,
R
2
)
,
H
μ
(
x
)
=
x
−
F
μ
(
x
)
.
From (3.9), we have
H
μ
(
∂
Ω
)
≠
0
,
μ
∈
[
0
,
1
]
.
Using the homotopy invariance property of degree, we obtain
deg
{
x
−
F
1
(
x
)
,
Ω
,
0
}
=
deg
{
x
,
Ω
,
0
}
≠
0
.
Till now, we have proved that Ω satisfies all the requirements in Lemma 2.3. Consequently, x
−
F
1
(
x
)
=
0
has at least one solution in Ω, i.e., F1 has a fixed point x
(
t
)
=
(
x
1
(
t
)
,
x
2
(
t
)
)
T
on Ω¯. Therefore, Eq. (1.1) has at least one anti-periodic solution x1(t). This completes the proof. □