# Periodic Solutions for Second-Order Ordinary Differential Equations with Linear Nonlinearity

Authors: Xiaohong Hu; Dabin Wang; Changyou Wang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

By using minimax methods in critical point theory, we obtain the existence of periodic solutions for second-order ordinary differential equations with linear nonlinearity.


---

## Body

## 1. Introduction and Main Results

Consider the second-order ordinary differential systems(1)
u
¨
(
t
)
+
m
2
ω
2
u
(
t
)
+
∇
F
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
[
0
,
T
]
,
u
(
0
)
-
u
(
T
)
=
u
˙
(
0
)
-
u
˙
(
T
)
=
0
,
where T
>
0, ω
=
2
π
/
T, m is a nonnegative integer; and F
:
[
0
,
T
]
×
R
N
→
R satisfies the following assumption:(
A
)
F
(
t
,
x
) is measurable in t for every x
∈
R
N and continuously differentiable in x for a
.
e
.
t
∈
[
0
,
T
], and there exist a
∈
C
(
R
+
,
R
+
) and b
∈
L
1
(
[
0
,
T
]
,
R
+
) such that
(2)
|
F
(
t
,
x
)
|
≤
a
(
|
x
|
)
b
(
t
)
,
|
∇
F
(
t
,
x
)
|
≤
a
(
|
x
|
)
b
(
t
)
,
for allx
∈
R
N and a
.
e
.
t
∈
[
0
,
T
], where R
+ is the set of all nonnegative real numbers.In the case ofm
=
0, the existence of periodic solutions for problem (1) is obtained in articles [1–17] with many solvability conditions, such as the coercive type potential condition (see [1]), the convex type potential condition (see [2]), the periodic type potential conditions (see [3]), the even type potential condition (see [4]), the subquadratic potential condition in Rabinowitz’s sense (see [5]), the bounded nonlinearity condition (see [6]), the subadditive condition (see [7]), the sublinear nonlinearity condition (see [9, 15]), and the linear nonlinearity condition (see [13, 14, 16, 17]).In the case ofm
≠
0, Mawhin and Willem [6] prove that problem (1) has at least one solution under the bounded nonlinearity condition; that is, |
∇
F
(
t
,
x
)
|
≤
g
(
t
) for some g
∈
L
1
(
0
,
T
), each x
∈
R
N, and a
.
e
.
t
∈
[
0
,
T
] when
(3)
∫
0
T
‍
F
(
t
,
a
cos
m
ω
t
+
b
sin
m
ω
t
)
d
t
⟶
+
∞
as
|
(
a
,
b
)
|
⟶
∞
in
R
2
N
or
(4)
∫
0
T
‍
F
(
t
,
a
cos
m
ω
t
+
b
sin
m
ω
t
)
d
t
⟶
-
∞
as
|
(
a
,
b
)
|
⟶
∞
in
R
2
N
.Under the sublinear nonlinearity condition, that is, there existf
,
g
∈
L
2
[
0
,
T
] and α
∈
[
0,1
), such that
(5)
|
∇
F
(
t
,
x
)
|
≤
f
(
t
)
|
x
|
α
+
g
(
t
)
,
for x
∈
R
N and a
.
e.  t
∈
[
0
,
T
], Han [18] proves that problem (1) has at least one solution when
(6)
|
(
a
,
b
)
|
-
2
α
∫
0
T
‍
F
(
t
,
a
cos
m
ω
t
+
b
sin
m
ω
t
)
d
t
⟶
+
∞
as
|
(
a
,
b
)
|
⟶
∞
in
R
2
N
or
(7)
|
(
a
,
b
)
|
-
2
α
∫
0
T
‍
F
(
t
,
a
cos
m
ω
t
+
b
sin
m
ω
t
)
d
t
⟶
-
∞
as
|
(
a
,
b
)
|
⟶
∞
in
R
2
N
.Recently, whenm
=
0, Zhao and Wu [13, 14] and Meng and Tang [16, 17] also prove the existence of solutions for problem (1) under linear nonlinearity condition; that is, there exist f
,
g
∈
L
1
(
[
0
,
T
]
,
R
+
) such that
(8)
|
∇
F
(
t
,
x
)
|
≤
f
(
t
)
|
x
|
+
g
(
t
)
.In this paper, motivated by the results mentioned above, we investigate the existence of periodic solutions of problem (1) in the case of m
≥
1.LetH
T
1 be a Hilbert space defined by
(9)
H
T
1
=
{
u
:
[
0
,
T
]
⟶
R
N
∣
u
is
absolutely
continuous
,
u
(
0
)
=
u
(
T
)
and
u
˙
∈
L
2
(
0
,
T
)
}
,
with the norm
(10)
∥
u
∥
=
(
∫
0
T
‍
|
u
(
t
)
|
2
d
t
+
∫
0
T
‍
|
u
˙
(
t
)
|
2
d
t
)
1
/
2
,
for u
∈
H
T
1.Let(11)
H
0
=
{
a
cos
m
ω
t
+
b
sin
m
ω
t
:
a
∈
R
N
,
b
∈
R
N
}
,
H
¯
=
{
∑
k
=
1
m
-
1
‍
a
k
cos
k
ω
t
+
b
k
sin
k
ω
t
:
a
k
∈
R
N
,
b
k
∈
R
N
,
H
¯
=
{
∑
k
=
1
m
-
1
‍
a
k
cos
k
ω
t
+
b
k
si
1
≤
k
≤
m
-
1
∑
k
=
1
m
-
1
‍
}
,
H
~
=
{
u
∈
H
T
1
:
∫
0
T
‍
u
(
t
)
cos
k
ω
t
d
t
=
∫
0
T
‍
u
(
t
)
sin
k
ω
t
d
t
=
0
,
1
≤
k
≤
m
∫
0
T
‍
}
;
then H
T
1
=
H
0
⊕
H
¯
⊕
H
~ ([6]). For all u
∈
H
T
1, we have u
=
u
0
+
u
¯
+
u
~, where u
0
∈
H
0, u
¯
∈
H
¯, and u
~
∈
H
~. It is easy to obtain
(12)
∥
u
-
˙
∥
2
2
≤
(
m
-
1
)
2
ω
2
∥
u
¯
∥
2
2
,
∀
u
¯
∈
H
¯
,
(13)
∥
u
~
˙
∥
2
2
≥
(
m
+
1
)
2
ω
2
∥
u
~
∥
2
2
,
∀
u
~
∈
H
~
.Furthermore, we have∥
u
∥
∞
≤
C
0
∥
u
∥ for some C
0
>
0 and all u
(
t
)
∈
H
T
1 (see, [6, Proposition 1.3]).Our main results are the following theorems.Theorem 1.
Suppose that (A) and (8) hold and
(
i
)
(14)
(
2
+
a
)
C
0
2
∫
0
T
‍
f
(
t
)
d
t
<
min
{
(
2
m
+
1
)
ω
2
1
+
(
m
+
1
)
2
ω
2
,
(
2
m
-
1
)
ω
2
1
+
(
m
-
1
)
2
ω
2
}
,
where a is a parameter and satisfies a
>
1
/
2;
(
ii
)
(15)
lim
u
∈
H
0
,
∥
u
∥
→
∞
inf
∥
u
∥
-
2
∫
0
T
‍
F
(
t
,
u
)
d
t
>
C
0
2
∫
0
T
‍
f
(
t
)
d
t
+
5
C
0
2
2
a
-
1
∫
0
T
‍
f
(
t
)
d
t
+
1
2
a
-
1
.
Then problem (1) has at least one solution.Theorem 2.
Suppose that (A), (8) and (i) hold and
(
iii
)
(16)
lim
u
∈
H
0
,
∥
u
∥
→
∞
sup
∥
u
∥
-
2
∫
0
T
‍
F
(
t
,
u
)
d
t
<
-
[
5
C
0
2
2
a
-
1
∫
0
T
‍
f
(
t
)
d
t
+
C
0
2
∫
0
T
‍
f
(
t
)
d
t
+
m
2
ω
2
2
a
-
1
]
.
Then problem (1) has at least one solution.Remark 3.
(i)  It is worth noting that, in the case ofm
=
0, one solution was obtained by Tang [9] and Han [15] under the sublinear nonlinearity condition.(ii)  It is also worth noting that the sublinear nonlinearity condition in [15, 18] is different from that of [9].


## 2. Proof of Main Results

Let(17)
J
(
u
)
=
1
2
∫
0
T
‍
|
u
˙
(
t
)
|
2
d
t
-
m
2
ω
2
2
×
∫
0
T
‍
|
u
(
t
)
|
2
d
t
-
∫
0
T
‍
F
(
t
,
u
(
t
)
)
d
t
,
for any u
∈
H
T
1. It follows from assumption (
A
) that the functional J on H
T
1 is continuously differentiable; moreover we obtain
(18)
〈
J
′
(
u
)
,
v
〉
=
∫
0
T
‍
(
u
˙
(
t
)
,
v
˙
(
t
)
)
d
t
-
m
2
ω
2
×
∫
0
T
‍
(
u
(
t
)
,
v
(
t
)
)
d
t
-
∫
0
T
‍
(
∇
F
(
t
,
u
(
t
)
)
,
v
(
t
)
)
d
t
for any u
,
v
∈
H
T
1. It is well known that the solutions of problem (1) correspond to the critical points of J (see [6]).For the sake of convenience, we denote(19)
M
1
=
∫
0
T
‍
f
(
t
)
d
t
,
M
2
=
∫
0
T
‍
g
(
t
)
d
t
.Proof of Theorem1.
Firstly, we assert that the functionalJ satisfies (PS) condition. Let {
u
n
} be a sequence in H
T
1 such that {
J
(
u
n
)
}  is bounded and J
′
(
u
n
)
→
0 as n
→
∞. By the proof of [6] Proposition 4.1, we only need to prove that {
u
n
} is bounded. On one hand, we have
(20)
∥
u
¯
n
∥
≥
〈
J
′
(
u
˙
n
)
,
-
u
-
˙
n
〉
=
-
∫
0
T
‍
[
(
u
˙
n
,
u
-
˙
n
)
-
m
2
ω
2
(
u
n
,
u
¯
n
)
-
(
∇
F
(
t
,
u
n
)
,
u
¯
n
)
(
u
˙
n
,
u
-
˙
n
)
-
m
2
ω
2
(
u
n
,
u
¯
n
)
]
d
t
=
-
∫
0
T
‍
|
u
-
˙
n
|
2
d
t
+
m
2
ω
2
×
∫
0
T
‍
|
u
¯
n
|
2
d
t
+
∫
0
T
‍
(
∇
F
(
t
,
u
n
)
,
u
¯
n
)
d
t
≥
[
m
2
-
(
m
-
1
)
2
]
ω
2
×
∫
0
T
‍
|
u
¯
n
|
2
d
t
-
∫
0
T
‍
f
(
t
)
|
u
n
0
+
u
¯
n
+
u
~
n
|
|
u
¯
n
|
d
t
-
∫
0
T
‍
g
(
t
)
|
u
¯
n
|
d
t
≥
(
2
m
-
1
)
ω
2
1
+
(
m
-
1
)
2
ω
2
∥
u
¯
n
∥
2
-
C
0
2
M
1
∥
u
¯
n
∥
2
-
C
0
2
M
1
∥
u
¯
n
∥
∥
u
~
n
∥
-
C
0
2
M
1
∥
u
¯
n
∥
∥
u
n
0
∥
-
C
0
M
2
∥
u
¯
n
∥
≥
(
(
2
m
-
1
)
ω
2
1
+
(
m
-
1
)
2
ω
2
-
2
C
0
2
M
1
)
×
∥
u
¯
n
∥
2
-
C
0
2
M
1
2
∥
u
~
n
∥
2
-
C
0
2
M
1
2
∥
u
n
0
∥
2
-
C
0
M
2
∥
u
¯
n
∥
.
So(21)
C
0
2
M
1
2
(
∥
u
~
n
∥
2
+
∥
u
n
0
∥
2
)
≥
(
(
2
m
-
1
)
ω
2
1
+
(
m
-
1
)
2
ω
2
-
(
2
+
a
)
C
0
2
M
1
)
×
∥
u
¯
n
∥
2
-
(
C
0
M
2
+
1
)
∥
u
¯
n
∥
+
a
C
0
2
M
1
∥
u
¯
n
∥
2
≥
a
C
0
2
M
1
∥
u
¯
n
∥
2
+
C
1
,
where C
1
=
min
s
∈
[
0
,
∞
)
{
(
(
(
2
m
-
1
)
ω
2
/
(
1
+
(
m
-
1
)
2
ω
2
)
)
-
(
2
+
a
)
C
0
2
M
1
)
s
2
-
(
C
0
M
2
+
1
)
s
}.
Since (14), so -
∞
<
C
1
<
0. Then
(22)
∥
u
¯
n
∥
2
≤
∥
u
~
n
∥
2
2
a
+
∥
u
n
0
∥
2
2
a
+
C
2
,
where C
2
=
-
C
1
/
a
C
0
2
M
1
>
0.
On the other hand, we have(23)
∥
u
~
n
∥
≥
〈
J
′
(
u
n
)
,
u
~
n
〉
≥
(
(
2
m
+
1
)
ω
2
1
+
(
m
+
1
)
2
ω
2
-
2
C
0
2
M
1
)
×
∥
u
~
n
∥
2
-
C
0
2
M
1
2
∥
u
¯
n
∥
2
-
C
0
2
M
1
2
∥
u
n
0
∥
2
-
C
0
M
2
∥
u
~
n
∥
.
So(24)
C
0
2
M
1
2
(
∥
u
¯
n
∥
2
+
∥
u
n
0
∥
2
)
≥
(
(
2
m
+
1
)
ω
2
1
+
(
m
+
1
)
2
ω
2
-
(
2
+
a
)
C
0
2
M
1
)
×
∥
u
~
n
∥
2
-
(
C
0
M
2
+
1
)
∥
u
~
n
∥
+
a
C
0
2
M
1
∥
u
~
n
∥
2
≥
a
C
0
2
M
1
∥
u
~
n
∥
2
+
C
3
,
where 0
>
C
3
=
min
s
∈
[
0
,
∞
)
{
(
(
(
2
m
+
1
)
ω
2
/
(
1
+
(
m
+
1
)
2
ω
2
)
)
-
(
2
+
a
)
C
0
2
M
1
)
s
2
-
(
C
0
M
2
+
1
)
s
}.
Then(25)
∥
u
~
n
∥
2
≤
∥
u
¯
n
∥
2
2
a
+
∥
u
n
0
∥
2
2
a
+
C
4
,
where C
4
=
-
C
3
/
a
C
0
2
M
1
>
0.
From (22) and (25), we have
(26)
∥
u
¯
n
∥
2
≤
1
2
a
-
1
∥
u
n
0
∥
2
+
C
5
,
∥
u
~
n
∥
2
≤
1
2
a
-
1
∥
u
n
0
∥
2
+
C
5
,
where C
5
=
max
{
(
4
a
2
C
2
+
2
a
C
4
)
/
(
4
a
2
-
1
)
,
(
4
a
2
C
4
+
2
a
C
2
)
/
(
4
a
2
-
1
)
}.
By (8), (26) we get
(27)
|
∫
0
T
‍
F
(
t
,
u
n
)
-
F
(
t
,
u
n
0
)
d
t
|
=
|
∫
0
T
‍
∫
0
1
‍
∇
F
(
t
,
u
n
0
+
s
(
u
¯
n
+
u
~
n
)
,
u
n
-
u
n
0
)
d
s
d
t
|
≤
∫
0
T
‍
∫
0
1
‍
f
(
t
)
|
u
n
0
+
s
(
u
¯
n
+
u
~
n
)
|
|
u
n
-
u
n
0
|
d
t
+
∫
0
T
‍
∫
0
1
‍
g
(
t
)
|
u
n
-
u
n
0
|
d
t
≤
C
0
2
M
1
∥
u
n
0
∥
2
+
5
2
C
0
2
M
1
∥
u
¯
n
∥
2
+
5
2
C
0
2
M
1
∥
u
~
n
∥
2
+
C
0
M
2
(
∥
u
¯
n
∥
+
∥
u
~
n
∥
)
≤
(
C
0
2
M
1
+
5
C
0
2
M
1
2
a
-
1
)
∥
u
n
0
∥
2
+
2
C
0
M
2
1
2
a
-
1
∥
u
n
0
∥
+
5
C
5
C
0
2
M
1
+
2
C
5
C
0
M
2
.
It follows from (26), (27), and the boundedness of J
(
u
n
) that
(28)
J
(
u
n
)
=
1
2
∫
0
T
‍
|
u
˙
n
|
2
d
t
-
m
2
ω
2
2
∫
0
T
‍
|
u
n
|
2
d
t
-
∫
0
T
‍
F
(
t
,
u
n
)
d
t
≤
1
2
(
∥
u
¯
n
∥
2
+
∥
u
~
n
∥
2
)
-
∫
0
T
‍
F
(
t
,
u
n
)
-
F
(
t
,
u
n
0
)
d
t
-
∫
0
T
‍
F
(
t
,
u
n
0
)
d
t
≤
(
C
0
2
M
1
+
5
C
0
2
M
1
2
a
-
1
+
1
2
a
-
1
)
×
∥
u
n
0
∥
2
+
2
C
0
M
2
1
2
a
-
1
∥
u
n
0
∥
-
∫
0
T
‍
F
(
t
,
u
n
0
)
d
t
+
5
C
5
C
0
2
M
1
+
2
C
5
C
0
M
2
=
∥
u
n
0
∥
2
[
C
0
2
M
1
+
5
C
0
2
M
1
2
a
-
1
+
1
2
a
-
1
+
2
C
0
M
2
1
2
a
-
1
∥
u
n
0
∥
-
1
-
∥
u
n
0
∥
-
2
∫
0
T
‍
F
(
t
,
u
n
0
)
d
t
5
C
0
2
M
1
2
a
-
1
]
+
5
C
5
C
0
2
M
1
+
2
C
5
C
0
M
2
.
The above inequality and (15) imply that {
u
n
0
}  is bounded. Hence {
u
n
} is bounded by (26).
Secondly, we assert that(
J
1
)
J
(
u
)
→
+
∞ as ∥
u
∥
→
∞ in H
~, which implies that inf
u
∈
H
~
J
(
u
)
>
-
∞;
(
J
2
)
J
(
u
)
→
-
∞ as ∥
u
∥
→
∞ in H
0
⊕
H
¯,for all u
∈
H
0
⊕
H
¯; that is, u
=
u
0
+
u
¯; then by (8) and (12) we have
(29)
J
(
u
)
=
1
2
∫
0
T
‍
|
u
˙
(
t
)
|
2
d
t
-
m
2
ω
2
2
∫
0
T
‍
|
u
(
t
)
|
2
d
t
-
∫
0
T
‍
F
(
t
,
u
(
t
)
)
d
t
=
1
2
(
∫
0
T
‍
|
u
¯
˙
(
t
)
|
2
d
t
-
m
2
ω
2
∫
0
T
‍
|
u
¯
(
t
)
|
2
d
t
)
-
∫
0
T
‍
[
F
(
t
,
u
0
+
u
¯
)
-
F
(
t
,
u
0
)
]
d
t
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
≤
1
2
(
1
-
2
m
)
ω
2
∥
u
¯
∥
2
2
-
∫
0
T
‍
∫
0
1
‍
(
∇
F
(
t
,
u
0
+
s
u
¯
)
,
u
¯
)
d
t
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
≤
1
2
(
1
-
2
m
)
ω
2
∥
u
¯
∥
2
2
+
∫
0
T
‍
f
(
t
)
|
u
¯
(
t
)
|
2
d
t
+
∫
0
T
‍
f
(
t
)
|
u
¯
(
t
)
|
|
u
0
|
d
t
+
∫
0
T
‍
g
(
t
)
|
u
¯
(
t
)
|
d
t
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
≤
1
2
(
1
-
2
m
)
ω
2
∥
u
¯
∥
2
2
+
C
0
2
M
1
∥
u
¯
∥
2
+
C
0
2
M
1
∥
u
0
∥
∥
u
¯
∥
+
C
0
M
2
∥
u
¯
∥
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
≤
1
2
(
1
-
2
m
)
ω
2
∥
u
¯
∥
2
2
+
C
0
2
M
1
∥
u
¯
∥
2
+
C
0
2
M
1
2
a
∥
u
0
∥
2
+
a
C
0
2
M
1
2
∥
u
¯
∥
2
+
C
0
M
2
∥
u
¯
∥
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
<
1
2
(
1
-
2
m
)
ω
2
∥
u
¯
∥
2
2
+
(
2
+
a
)
C
0
2
M
1
2
[
1
+
(
m
-
1
)
2
ω
2
]
∥
u
¯
∥
2
2
+
C
0
M
2
[
(
m
-
1
)
ω
+
1
]
∥
u
¯
∥
2
+
C
0
2
M
1
∥
u
0
∥
2
-
∫
0
T
‍
F
(
t
,
u
0
)
d
t
=
{
1
2
(
1
-
2
m
)
ω
2
+
(
2
+
a
)
C
0
2
M
1
2
[
1
+
(
m
-
1
)
2
ω
2
]
}
×
∥
u
¯
∥
2
2
+
C
0
M
2
[
(
m
-
1
)
ω
+
1
]
∥
u
¯
∥
2
+
∥
u
0
∥
2
[
C
0
2
M
1
-
∥
u
0
∥
-
2
∫
0
T
‍
F
(
t
,
u
0
)
d
t
]
,
for∥
u
∥
→
∞ in X if and only if ∥
u
¯
∥
2
→
∞ or ∥
u
0
∥
→
∞. So, by m
≥
1, (14), and (15), we obtain J
(
u
)
→
-
∞ as ∥
u
∥
→
∞ in X.
Letu
∈
H
~; then by (8) and (13), we have
(30)
J
(
u
)
=
1
2
∫
0
T
‍
|
u
˙
(
t
)
|
2
d
t
-
m
2
ω
2
2
∫
0
T
‍
|
u
(
t
)
|
2
d
t
-
∫
0
T
‍
F
(
t
,
u
(
t
)
)
d
t
≥
1
2
(
1
-
m
2
ω
2
(
m
+
1
)
2
ω
2
)
∫
0
T
‍
|
u
~
˙
(
t
)
|
2
d
t
-
∫
0
T
‍
[
F
(
t
,
u
~
)
-
F
(
t
,
0
)
]
d
t
-
∫
0
T
‍
F
(
t
,
0
)
d
t
≥
1
2
2
m
+
1
(
m
+
1
)
2
×
(
m
+
1
)
2
ω
2
1
+
(
m
+
1
)
2
ω
2
∥
u
~
∥
2
-
∫
0
T
‍
∫
0
1
‍
(
∇
F
(
t
,
s
u
~
)
,
u
~
)
d
t
-
∫
0
T
‍
F
(
t
,
0
)
d
t
≥
1
2
(
2
m
+
1
)
ω
2
1
+
(
m
+
1
)
2
ω
2
∥
u
~
∥
2
-
∫
0
T
‍
f
(
t
)
|
u
~
|
2
d
t
-
∫
0
T
‍
g
(
t
)
|
u
~
|
d
t
-
∫
0
T
‍
F
(
t
,
0
)
d
t
≥
(
1
2
(
2
m
+
1
)
ω
2
1
+
(
m
+
1
)
2
ω
2
-
C
0
2
M
1
)
×
∥
u
~
∥
2
-
C
0
M
2
∥
u
~
∥
-
∫
0
T
‍
F
(
t
,
0
)
d
t
.
So, by (14), J is bounded from below on H
~.
Hence, by Rabinowitz’s Saddle point Theorem (see [19, Theorem 4.6]), we obtain that the problem (1) has at least one solution.Proof of Theorem2.
The proof of Theorem2 is similar to the proof of Theorem 1, so we omit it here.



---

*Source: 670287-2014-01-30.xml*