# On the Number of Limit Cycles of a Piecewise Quadratic Near-Hamiltonian System

Authors: Jing Tian
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper is concerned with the problem for the maximal number of limit cycles for a quadratic piecewise near-Hamiltonian system. By using the method of the first order Melnikov function, we find that it can have 8 limit cycles.


---

## Body

## 1. Introduction and Main Results

Recently, piecewise smooth systems attracted many researchers’ attention since many real processes and different modern devices can be modeled by them; see [1, 2] and references therein. Due to their nonsmoothness, these systems can have richer dynamical phenomena than the smooth ones (see [3–5] and the references cited therein). For instance, nonsmooth system can have the sliding phenomena and in [6, 7] Giannakopoulos and Pliete have studied the existence of sliding cycles and sliding homoclinic cycles for a planar relay control feedback systems.As we have seen, studying the existence and number of limit cycles is one of the main problems for piecewise smooth systems; see [8–16]. Limit cycles of piecewise smooth linear differential systems defined on two half-planes separated by a straight line x
=
0 or y
=
0 have been studied recently in [8–11], from which one can find that 3 limit cycles can appear for piecewise smooth linear systems. From [12–14], we can know that piecewise linear and quadratic near-Hamiltonian systems can have 2 and 3 limit cycles, respectively. In fact, these papers investigated the problem for limit cycle bifurcations of piecewise linear Hamiltonian systems under piecewise polynomial perturbations of degree n, obtaining some new and interesting results. Furthermore, the authors of [16] studied a piecewise quadratic Hamiltonian system (one side linear and another side quadratic) perturbed inside the class of piecewise polynomial differential systems of degree n and achieved that it can have 2
n
+
[
(
n
+
1
)
/
2
] limit cycles, which implies that piecewise quadratic near-Hamiltonian systems can have 5 limit cycles. The authors of [15] also obtained this result by perturbed quadratic polynomial differential systems containing an isochronous center with piecewise quadratic polynomials of degree 2.In this paper, we mainly study the problem for the maximal number of a quadratic near-Hamiltonian system. That is to say, we consider a system of the form(1)
x
˙
=
y
+
ɛ
p
(
x
,
y
)
,
y
˙
=
-
g
(
x
)
+
ɛ
q
(
x
,
y
)
,
where ɛ
>
0 is a small parameter,
(2)
g
(
x
)
=
{
a
2
x
2
+
a
1
x
+
a
0
,
x
<
0
,
b
2
x
2
+
b
1
x
+
b
0
,
x
>
0
with a
2
b
2
≠
0, and
(3)
p
(
x
,
y
)
=
{
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
,
x
<
0
,
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
,
x
>
0
,
q
(
x
,
y
)
=
{
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
,
x
<
0
,
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
,
x
>
0
.
Then, our main results can be stated as follows.Theorem 1.
There exists a system of form (1) which has 8 limit cycles.In the next section, the proof of Theorem1 is presented.


## 2. Proof of Theorem1

In this section, we divide the proof into two subsections: one section is to present some preliminary knowledge, which is useful to verify our main result; another section is to give the process of the proof of Theorem1.


### 2.1. Preliminaries

Clearly, system (1) has the following two subsystems:(4a)
x
˙
=
y
+
ɛ
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
,
y
˙
=
-
(
a
2
x
2
+
a
1
x
+
a
0
)
+
ɛ
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
,
(4b)
x
˙
=
y
+
ɛ
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
,
y
˙
=
-
(
b
2
x
2
+
b
1
x
+
b
0
)
+
ɛ
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
,which are called left and right subsystems, respectively. And systems (4a)|
ɛ
=
0 and (4b)|
ɛ
=
0 are Hamiltonian with Hamiltonian functions, respectively,
(5)
H
-
(
x
,
y
)
=
1
2
y
2
+
1
3
a
2
x
3
+
1
2
a
1
x
2
+
a
0
x
,
H
+
(
x
,
y
)
=
1
2
y
2
+
1
3
b
2
x
3
+
1
2
b
1
x
2
+
b
0
x
.For systems (4a)|
ɛ
=
0 and (4b)|
ɛ
=
0, we make the following assumption:(H)
a
1
2
>
4
a
2
a
0,  b
1
2
>
4
b
2
b
0.Then, under the condition (H), forɛ
=
0, system (4a) (resp., (4b)) has a hyperbolic saddle S
1
=
(
α
S
1
,
0
) (resp., S
2
=
(
α
S
2
,
0
)) and an elementary center C
1
=
(
α
C
1
,
0
) (resp., C
2
=
(
α
C
2
,
0
)), where
(6)
α
S
1
=
-
a
1
-
a
1
2
-
4
a
2
a
0
2
a
2
,
α
S
2
=
-
b
1
-
b
1
2
-
4
b
2
b
0
2
b
2
,
α
C
1
=
-
a
1
+
a
1
2
-
4
a
2
a
0
2
a
2
,
α
C
2
=
-
b
1
+
b
1
2
-
4
b
2
b
0
2
b
2
.We will assume that(7)
a
2
<
0
,
since, for a
2
>
0, we can make a variable transformation x
→
-
x, y
→
y, together with a time rescaling t
→
-
t such that the resulting system satisfies (7). Further, note that there exist 2 possible cases for level curves of H
(
x
,
y
)
=
(
1
/
2
)
y
2
+
(
1
/
3
)
a
x
3
+
(
1
/
2
)
b
x
2
+
c
x
=
h on the plane, where a
≠
0, b
2
>
4
a
c; see Figure 1.Level curves of(
1
/
2
)
y
2
+
(
1
/
3
)
a
x
3
+
(
1
/
2
)
b
x
2
+
c
x
=
h.
(a)
a
<
0
(b)
a
>
0We remark that, in Figure1, C
=
(
(
-
b
+
b
2
-
4
a
c
)
/
2
a
,
0
), S
=
(
(
-
b
-
b
2
-
4
a
c
)
/
2
a
,
0
). Then, one can obtain 9 possible phase portraits of system (1)|
ɛ
=
0 under (H) and (7) when it has a generalized closed orbit and two hyperbolic saddles; see Figure 2. For system (1)|
ɛ
=
0, a generalized closed orbit passes through the y-axis two times.Possible portraits of (1)|
ɛ
=
0 with two saddles and a generalized closed orbit.
(a)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
>
H
+
(
S
2
)
(b)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
=
H
+
(
S
2
)
(c)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
<
H
+
(
S
2
)
(d)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
>
H
+
(
S
2
)
(e)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
=
H
+
(
S
2
)
(f)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
<
H
+
(
S
2
)
(g)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
>
H
+
(
S
2
)
(h)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
=
H
+
(
S
2
)
(i)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
<
H
+
(
S
2
)Now, we study limit cycle bifurcations on system (1). For simplicity, we consider the case
(8)
α
S
1
<
0
<
α
S
2
<
α
C
2
,
H
-
(
S
1
)
=
H
+
(
S
2
)
≜
h
1
;
that is,
(9)
a
1
2
>
4
a
2
a
0
,
b
1
2
>
4
b
2
b
0
,
a
2
<
0
,
b
2
>
0
,
a
1
<
0
,
b
1
<
0
,
a
0
<
0
,
b
0
>
0
,
a
1
3
-
6
a
0
a
1
a
2
+
(
a
1
2
-
4
a
0
a
2
)
3
/
2
a
2
2
=
b
1
3
-
6
b
0
b
1
b
2
+
(
b
1
2
-
4
b
0
b
2
)
3
/
2
b
2
2
.
Then, the phase portrait of system (1)|
ɛ
=
0 is Figure 2(h), and there are four families of periodic orbits given by
(10)
L
1
h
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
x
0
-
<
x
<
α
S
1
}
,
h
0
-
<
h
<
h
1
,
L
2
h
=
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
α
S
2
<
x
<
x
0
+
}
,
h
0
+
<
h
<
h
1
,
L
3
h
=
L
3
h
-
∪
L
3
h
+
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
α
S
1
<
x
<
0
}
∪
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
0
<
x
<
α
S
2
}
,
mmmmmmmmmmmmmm
0
<
h
<
h
1
,
L
h
=
L
h
-
∪
L
h
+
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
x
<
0
}
∪
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
x
>
0
}
,
h
>
h
1
,
where H
-
(
x
0
-
,
0
)
=
h
1, H
+
(
x
0
+
,
0
)
=
h
1 and H
-
(
C
1
)
=
h
0
-, H
+
(
C
2
)
=
h
0
+. Further, we have
(11)
lim
⁡
h
→
h
1
L
1
h
=
L
1
,
lim
⁡
h
→
h
0
-
L
1
h
=
C
1
,
lim
⁡
h
→
h
1
L
2
h
=
L
2
,
lim
⁡
h
→
h
0
+
L
1
h
=
C
2
,
lim
⁡
h
→
h
1
L
3
h
-
=
L
-
,
lim
⁡
h
→
h
1
L
3
h
+
=
L
+
,
lim
⁡
h
→
0
L
3
h
-
=
lim
⁡
h
→
0
L
3
h
+
=
(
0,0
)
,
lim
⁡
h
→
h
1
L
h
-
=
L
1
∪
L
-
,
lim
⁡
h
→
h
1
L
h
+
=
L
2
∪
L
+
;
see Figure 3. From (11), one can know that, as h approaches h
1, the limit of L
h is L
1
∪
L
-
∪
L
2
∪
L
+ denoted by L, which is a generalized polycycle with two hyperbolic saddles.Figure 3
Phase portrait of system (1)|
ɛ
=
0 under (9).Then, from Theorem 1.1 of [17], for system (1), we have four first order Melnikov functions as follows:
(12)
M
1
(
h
)
=
∮
L
1
h
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
h
∈
(
h
0
-
,
h
1
)
,
M
2
(
h
)
=
∮
L
2
h
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
h
∈
(
h
0
+
,
h
1
)
,
M
3
(
h
)
=
M
3
-
(
h
)
+
M
3
+
(
h
)
=
∫
L
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
+
∫
L
3
h
+
‍
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
m
m
m
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
h
∈
(
0
,
h
1
)
,
M
(
h
)
=
M
-
(
h
)
+
M
+
(
h
)
=
∫
L
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
+
∫
L
h
+
‍
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
h
∈
(
h
1
,
+
∞
)
,
where L
1
h, L
2
h, L
3
h
-, L
3
h
+, L
h
-, and L
h
+ are given in (10).Now, we deduce the expansion ofM
i
(
h
),  i
=
1,2
,
3, and M
(
h
) near h
=
h
1 in the above. First, we have the following.Lemma 2.
The functionM
3
-
(
h
) in (12) has the following expression:
(13)
M
3
-
(
h
)
=
c
0
+
c
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
2
(
h
-
h
1
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
mmmmmmmhmimmmm
0
<
h
1
-
h
≪
1
,
where
(14)
c
0
=
∫
L
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
mmim
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
-
2
a
00
-
2
h
1
-
4
3
h
1
2
h
1
a
02
-
,
c
1
=
-
1
(
a
1
2
-
4
a
0
a
2
)
1
/
4
[
(
a
10
-
+
b
01
-
)
+
α
S
1
(
2
a
20
-
+
b
11
-
)
]
,
c
2
|
c
1
=
0
=
∫
L
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
m
m
m
m
h
m
-
2
h
1
h
1
a
00
-
-
2
2
h
1
a
02
-
,
c
3
=
a
2
2
(
a
1
2
-
4
a
0
a
2
)
5
/
4
(
2
a
20
-
+
b
11
-
)
.Proof.
Denote byU a disk of diameter ϵ
0
>
0 with its center at S
1. Then, for ϵ
0
>
0 small, rewrite M
3
-
(
h
)
=
I
1
(
h
)
+
I
2
(
h
), where
(15)
I
1
(
h
)
=
∫
L
¯
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
I
2
(
h
)
=
∫
L
~
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
with
(16)
L
-
3
h
-
=
L
3
h
-
∩
U
,
L
~
3
h
-
=
L
3
h
-
∖
L
¯
3
h
-
.
Note that I
2
(
h
) is a C
∞ function for h
1
-
h
>
0 small. Thus, it suffices to investigate the expansion of I
1
(
h
) for 0
<
h
1
-
h
≪
1. To do this, we make a variable transformation
(17)
u
=
(
a
1
2
-
4
a
0
a
2
)
1
/
4
(
x
-
α
S
1
)
,
v
=
y
,
τ
=
(
a
1
2
-
4
a
0
a
2
)
1
/
4
t
.
Then system (4a) becomes
(18)
u
˙
=
v
+
ɛ
p
~
-
(
u
,
v
)
,
v
˙
=
u
-
a
2
u
2
(
a
1
2
-
4
a
0
a
2
)
3
/
4
+
ɛ
q
~
-
(
u
,
v
)
,
where
(19)
p
~
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
a
~
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
a
i
j
-
[
u
(
a
1
2
-
4
a
0
a
2
)
1
/
4
+
α
S
1
]
i
v
j
,
(20)
q
~
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
b
~
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
b
i
j
-
(
a
1
2
-
4
a
0
a
2
)
1
/
4
[
u
(
a
1
2
-
4
a
0
a
2
)
1
/
4
+
α
S
1
]
i
v
j
.
For ɛ
=
0, the Hamiltonian function of (18) has the form
(21)
H
~
(
u
,
v
)
=
1
2
(
v
2
-
u
2
)
+
a
2
3
(
a
1
2
-
4
a
0
a
2
)
3
/
4
u
3
+
h
1
and the corresponding first order Melinikov function can be denoted by M
~
(
h
). By Remark 3.1
.
4 of [18], we can know that M
3
-
(
h
)
=
M
~
(
h
), and then, from Corollary 3.2
.
2 and the Theorem 3.2
.
1 of [18], I
1 can be written as
(22)
I
1
(
h
)
=
c
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
where
(23)
c
1
=
-
(
a
~
10
-
+
b
~
01
-
)
=
-
1
(
a
1
2
-
4
a
0
a
2
)
1
/
4
×
[
(
a
10
-
+
b
01
-
)
+
α
S
1
(
2
a
20
-
+
b
11
-
)
]
,
c
3
=
3
2
×
a
2
3
(
a
1
2
-
4
a
0
a
2
)
3
/
4
×
1
(
a
1
2
-
4
a
0
a
2
)
2
/
4
(
2
a
~
20
+
b
~
11
)
=
a
2
2
(
a
1
2
-
4
a
0
a
2
)
5
/
4
(
2
a
20
-
+
b
11
-
)
.
Thus, by the above discussion, one can obtain the expression of (13) and the formulas for c
1 and c
3. Now, we give the expressions of c
0 and c
2 in (13).
On the curveL
3
h intersect y-axis with two points denoted by A
1
(
h
)
=
(
0
,
-
2
h
) and A
(
h
)
=
(
0
,
2
h
). Then, from (12), we have
(24)
M
3
-
(
h
)
=
∫
A
1
(
h
)
A
(
h
)
^
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
=
N
1
(
h
)
+
N
2
(
h
)
,
where
(25)
N
1
(
h
)
=
∫
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
N
2
(
h
)
=
∫
A
(
h
)
A
1
(
h
)
→
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
.
By Green’s formula two times toN
1
(
h
), we obtain
(26)
N
1
(
h
)
=
∫
Int
.
(
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
)
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
m
m
m
m
h
m
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
x
d
y
=
∮
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
‍
m
m
m
m
m
m
m
m
h
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
=
∫
L
3
h
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
m
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
.
Further, one has
(27)
N
2
(
h
)
=
∫
A
(
h
)
A
1
(
h
)
→
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
=
∫
2
h
-
2
h
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
=
-
2
a
00
-
2
h
-
4
3
h
2
h
a
02
-
.
Thus, we have
(28)
M
3
-
(
h
)
=
∫
L
3
h
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
m
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
-
2
a
00
-
2
h
-
4
3
h
2
h
a
02
-
.
Then, let h
→
h
1. Then, by (11), we can obtain the expression of c
0 in (13). Further, differentiating (13) with respect to h, together with Lemma 3.1
.
2 of [18], we obtain
(29)
d
d
h
M
3
-
(
h
)
=
∫
L
3
h
-
‍
(
p
x
+
q
y
)
|
ɛ
=
0
d
t
=
∫
L
3
h
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
-
2
h
h
a
00
-
-
2
2
h
a
02
-
=
c
1
ln
⁡
⁡
(
h
1
-
h
)
+
c
1
+
c
2
+
O
(
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
)
,
which, as c
1
=
0, gives that
(30)
∫
L
3
h
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
-
2
h
h
a
00
-
-
2
2
h
a
02
-
=
c
2
+
O
(
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
)
.
Let h
→
h
1 in the above. Then we can easily obtain the formula for c
2 in (13). This ends the proof.In the same way, one can obtain the expressions ofM
3
+
(
h
).Lemma 3.
ForM
3
-
(
h
) in (12), one has
(31)
M
3
+
(
h
)
=
c
~
0
+
c
~
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
~
2
(
h
-
h
1
)
+
c
~
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
mmmmmmmmmmmmm
0
<
h
1
-
h
≪
1
,
where
(32)
c
~
0
=
∫
L
+
‍
[
1
2
(
a
10
+
+
b
01
+
)
y
+
(
b
11
+
+
2
a
20
+
)
x
y
m
m
m
m
+
1
2
(
a
11
+
+
2
b
02
+
)
y
2
]
d
x
+
2
2
h
1
a
00
+
+
4
3
h
1
2
h
1
a
02
+
,
c
~
1
=
-
1
(
b
1
2
-
4
b
0
b
2
)
1
/
4
[
(
a
10
+
+
b
01
+
)
+
α
S
2
(
2
a
20
+
+
b
11
+
)
]
,
c
~
2
|
c
~
1
=
0
=
∫
L
+
‍
[
(
a
10
+
+
b
01
+
)
+
(
b
11
+
+
2
a
20
+
)
x
+
(
a
11
+
+
2
b
02
+
)
y
]
d
t
+
2
h
1
h
1
a
00
+
+
2
2
h
1
a
02
+
,
c
~
3
=
b
2
2
(
b
1
2
-
4
b
0
b
2
)
5
/
4
(
2
a
20
+
+
b
11
+
)
.By applying Theorem3.2
.
3 of [18], together with the proof of Lemma 2, we have the following.Lemma 4.
The functionsM
i
(
h
), i
=
1,2, given in (12) can be expressed as
(33)
M
1
(
h
)
=
c
01
+
c
1
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
c
21
(
h
-
h
1
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
M
2
(
h
)
=
c
02
+
c
~
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
22
(
h
-
h
1
)
+
c
~
3
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
where c
1, c
3, c
~
1, c
~
3 are given in Lemmas 2 and 3, and
(34)
c
01
=
∮
L
1
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
c
02
=
∮
L
2
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
c
21
|
c
1
=
0
=
∮
L
1
[
(
a
10
-
+
b
01
-
)
+
(
2
a
20
-
+
b
11
-
)
x
‍
mmm
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
|
c
1
=
0
,
c
22
|
c
~
1
=
0
=
∮
L
2
‍
[
(
a
10
+
+
b
01
+
)
+
(
2
a
20
+
+
b
11
+
)
x
mmm
+
(
a
11
+
+
2
b
02
+
)
y
]
d
t
|
c
~
1
=
0
.By above three Lemmas2, 3, and 4, together with (12), one can derive the following.Lemma 5.
ForM
3
(
h
) and M
(
h
), one has
(35)
M
3
(
h
)
=
(
c
0
+
c
~
0
)
+
(
c
1
+
c
~
1
)
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
(
c
2
+
c
~
2
)
(
h
-
h
1
)
+
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
(
O
(
h
1
-
h
)
2
)
,
0
<
h
1
-
h
≪
1
,
M
(
h
)
=
(
c
0
+
c
01
+
c
~
0
+
c
02
)
+
2
(
c
1
+
c
~
1
)
×
(
h
-
h
1
)
ln
⁡
(
h
-
h
1
)
+
(
c
2
+
c
21
+
c
~
2
+
c
22
)
×
(
h
-
h
1
)
+
2
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
-
h
1
)
+
(
O
(
h
1
-
h
)
2
)
,
0
<
h
-
h
1
≪
1
,
where the corresponding coefficients are given in (14), (32), and (34).Then, using Lemmas4 and 5, we can have the following.Theorem 6.
Let (H) and (7) hold. Then if there exists δ
0 such that
(36)
c
01
(
δ
0
)
=
c
02
(
δ
0
)
=
(
c
0
+
c
~
0
)
(
δ
0
)
=
(
c
1
+
c
~
1
)
(
δ
0
)
=
(
c
2
+
c
~
2
)
(
δ
0
)
=
0
,
(
c
3
+
c
~
3
)
(
δ
0
)
<
0
,
c
1
(
δ
0
)
<
0
,
c
~
1
(
δ
0
)
>
0
,
(
c
1
+
c
~
1
)
(
δ
0
)
=
0
,
(
c
21
+
c
~
22
)
(
δ
0
)
=
0
,
rank
⁡
∂
(
c
01
,
c
02
,
c
0
+
c
~
0
,
c
1
+
c
~
1
,
c
2
+
c
~
2
)
∂
δ
=
5
,
then system (1) can have 7 limit cycles near L.Proof.
By the assumption and from the expansions in Lemmas4 and 5 we have
(37)
M
1
(
h
)
=
c
1
(
δ
0
)
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
h
1
-
h
)
<
0
,
mmmmmmmmmmhmhmmm
0
<
h
1
-
h
≪
1
,
M
2
(
h
)
=
c
~
1
(
δ
0
)
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
O
(
h
-
h
1
)
>
0
,
mmmmmmmmmmmmmmm
0
<
h
1
-
h
≪
1
,
M
3
(
h
)
=
(
c
3
+
c
~
3
)
(
δ
0
)
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
(
O
(
h
1
-
h
)
2
)
>
0
,
0
<
h
1
-
h
≪
1
,
M
(
h
)
=
2
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
-
h
1
)
+
(
O
(
h
1
-
h
)
2
)
>
0
,
mmmmmmmmmlmmmmmmmmm
0
<
h
-
h
1
≪
1
.
Note thatc
01, c
02, c
0
+
c
~
0, c
1
+
c
~
1, and c
2
+
c
~
2 can be taken as free parameters. Then we can vary them one by one to change the signs of M
1, M
2, M
3, and M.
First, letc
01
=
c
02
=
c
0
+
c
~
0
=
c
1
+
c
~
1
=
0 and 0
<
-
(
c
2
+
c
~
2
)
≪
1. Then, in this case, we have
(38)
M
1
(
h
)
<
0
,
M
2
(
h
)
>
0
,
M
3
(
h
)
>
0
,
mmmmmmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
,
which means that there exists a zero h
~
1 with 0
<
h
~
1
-
h
1 such that M
(
h
~
1
)
=
0.
Second, fixc
01
=
c
02
=
c
0
+
c
~
0
=
0 and vary 0
≪
-
(
c
1
+
c
~
1
)
≪
-
(
c
2
+
c
~
2
), so that
(39)
M
1
(
h
)
<
0
,
M
2
(
h
)
>
0
,
M
3
(
h
)
<
0
,
mmmmmhmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
>
0
,
for
0
<
h
-
h
1
≪
1
.
Hence, there exist h
1
-
h
~
2
>
0,  h
~
3
∈
(
h
1
,
h
~
1
) such that M
3
(
h
~
2
)
=
0,  M
(
h
~
3
)
=
0.
Third, letc
01
=
c
0
+
c
~
0
=
0,  0
<
-
c
02
≪
-
(
c
1
+
c
~
1
), implying
(40)
M
1
(
h
)
<
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
<
0
,
mmm1mmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
Then M
2
(
h
~
4
)
=
0,  M
(
h
~
5
)
=
0 for some h
~
4
<
h
1,  h
~
5
∈
(
h
1
,
h
~
3
).
Fourth, letc
0
+
c
~
0
=
0 and 0
<
c
01
≪
-
c
02; we have
(41)
M
1
(
h
)
>
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
<
0
,
mmmmmhmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
So there exists h
~
6
<
h
1 such that M
1
(
h
~
6
)
=
0.
Finally, vary0
<
c
0
+
c
~
0
≪
c
01. In this case,
(42)
M
1
(
h
)
>
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
>
0
,
mmmmhmmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
So there exists h
~
7
∈
(
h
~
2
,
h
1
) such that M
3
(
h
~
7
)
=
0.
In a word, we have proved that if(43)
0
<
c
0
+
c
~
0
≪
c
01
≪
-
c
02
≪
-
(
c
1
+
c
~
1
)
≪
-
(
c
2
+
c
~
2
)
≪
1
then M
1, M
2, M
3, and M have one zero, one zero, two zeros, and three zeros which implies that seven limit cycles can appear in the neighbourhood of L.


### 2.2. Proofs

In this section, we will use Theorem6 to prove our main result. For that purpose, let
(44)
a
2
=
-
1
,
a
1
=
-
3
2
,
a
0
=
-
1
2
,
b
2
=
1
,
b
1
=
-
11
3
,
b
0
=
41
48
.
Then, the Hamiltonian functions in (5) can be rewritten as
(45)
H
-
(
x
,
y
)
=
1
2
y
2
-
1
3
x
3
-
3
4
x
2
-
1
2
x
,
H
+
(
x
,
y
)
=
1
2
y
2
+
1
3
x
3
-
11
6
x
2
+
11
48
x
.
And in this case
(46)
S
1
=
(
-
1
2
,
0
)
,
S
2
=
(
1
4
,
0
)
,
C
1
=
(
-
1,0
)
,
C
2
=
(
41
12
,
0
)
,
h
1
=
5
48
.Under (44), it is not hard to obtain the expressions for c
1, c
~
1, c
3, and c
~
3 in Lemmas 2 and 3.Lemma 7.
Under (44), for system (1), one has
(47)
c
1
=
-
2
[
(
a
10
-
+
b
01
-
)
-
1
2
(
2
a
20
-
+
b
11
-
)
]
,
c
3
=
-
2
(
2
a
20
-
+
b
11
-
)
,
c
~
1
=
-
6
19
[
(
a
10
+
+
b
01
+
)
+
1
4
(
2
a
20
+
+
b
11
+
)
]
,
c
~
3
=
1
2
(
6
19
)
5
/
2
(
2
a
20
+
+
b
11
+
)
.For the other expressions of coefficients in Lemmas4 and 5, one has the following.Lemma 8.
Under (44), for system (1), one has
(48)
c
01
=
3
2
20
(
a
10
-
+
b
01
-
)
-
39
2
280
(
2
a
20
-
+
b
11
-
)
,
c
21
=
-
3
2
(
2
a
20
-
+
b
11
-
)
,
c
02
=
361
114
180
(
a
10
+
+
b
01
+
)
+
29963
114
5040
(
2
a
20
+
+
b
11
+
)
,
c
22
=
114
(
2
a
20
+
+
b
11
+
)
,
c
0
=
3
2
20
(
a
10
-
+
b
01
-
)
-
(
39
2
280
-
5
30
168
)
(
2
a
20
-
+
b
11
-
)
-
30
6
a
00
-
-
5
30
432
a
02
-
,
c
~
0
=
(
-
35
9
30
+
361
180
114
)
(
a
10
+
+
b
01
+
)
+
(
29963
5040
114
-
730
63
30
)
(
2
a
20
+
+
b
11
+
)
+
30
6
a
00
+
+
5
30
432
a
02
+
,
c
2
=
(
30
-
3
2
)
(
2
a
20
-
+
b
11
-
)
-
4
30
5
a
00
-
-
30
6
a
02
-
,
c
~
2
=
(
-
2
30
+
114
)
(
2
a
20
+
+
b
11
+
)
+
4
30
5
a
00
+
+
30
6
a
02
+
.Proof.
Note thatc
01
=
∮
L
1
‍
∑
i
+
j
=
0
2
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
a
i
j
-
x
i
y
j
d
y. Then, by Green’s formula two times, we derive
(49)
c
01
=
∫
Int
.
(
L
1
)
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
x
d
y
=
∮
L
1
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
mmm
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
.
On the other hand, along the curveL
1, we have (
1
/
2
)
y
2
-
(
1
/
3
)
x
3
-
(
3
/
4
)
x
2
-
(
1
/
2
)
x
=
5
/
48, for -
(
5
/
4
)
≤
x
≤
-
(
1
/
2
). Then, (49) can be represented as
(50)
c
01
=
2
∫
-
5
/
4
-
1
/
2
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
]
m
m
m
×
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
.
By applying Maple, one can compute that
(51)
∫
-
5
/
4
-
1
/
2
‍
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
3
2
40
,
∫
-
5
/
4
-
1
/
2
‍
x
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
-
39
2
560
.
Inserting the above into (50) gives the expression of c
01.
Recall thatc
1
=
0 if and only if a
10
-
+
b
01
-
=
(
1
/
2
)
(
2
a
20
-
+
b
11
-
). Thus, by (4a)|
ɛ
=
0, we rewrite c
21
|
c
1
=
0 as
(52)
c
21
|
c
1
=
0
=
∮
L
1
[
(
1
/
2
)
(
2
a
20
-
+
b
11
-
)
+
(
2
a
20
-
+
b
11
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
y
d
x
=
1
2
(
2
a
20
-
+
b
11
-
)
∮
L
1
1
+
2
x
y
d
x
=
(
2
a
20
-
+
b
11
-
)
∫
-
5
/
4
-
1
/
2
‍
1
+
2
x
(
5
/
24
)
+
(
2
/
3
)
x
3
+
(
3
/
2
)
x
2
+
x
d
x
,
which, by Maple, yields the conclusion for c
02. Using a similar way, one can derive the formulas for c
02 and c
22. Now, we deduce the formulas for c
0 and c
2.
Note that, on the curveL
-
:
(
1
/
2
)
y
2
-
(
1
/
3
)
x
3
-
(
3
/
4
)
x
2
-
(
1
/
2
)
x
=
(
5
/
48
),  -
(
1
/
2
)
<
x
<
0, we have
(53)
c
0
=
2
∫
-
1
/
2
0
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
]
m
m
m
×
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
-
30
6
a
00
-
-
5
30
432
a
02
-
.
By Maple again, we obtain that
(54)
∫
-
1
/
2
0
‍
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
3
2
40
,
∫
-
1
/
2
0
‍
x
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
-
39
2
560
+
5
30
336
.
Thus, we can obtain the conclusion for c
0.
Whenc
1
=
0, we have (
a
10
-
+
b
01
-
)
=
(
1
/
2
)
(
2
a
20
-
+
b
11
-
). Thus, c
2 in (14) can be rewritten as
(55)
c
2
=
1
2
(
2
a
20
-
+
b
11
-
)
∫
L
-
‍
1
+
2
x
y
d
x
-
2
a
00
-
2
h
1
-
4
a
02
-
3
(
2
h
1
+
h
1
2
h
1
)
|
h
1
=
5
/
48
=
(
2
a
20
-
+
b
11
-
)
∫
-
1
/
2
0
‍
1
+
2
x
(
5
/
24
)
+
(
2
/
3
)
x
3
+
(
3
/
2
)
x
2
+
x
d
x
-
4
5
6
a
00
-
-
30
6
a
02
-
,
which, by Maple, implies the conclusion. Further, we can prove c
~
0 and c
~
2 in the same way to c
0 and c
2, respectively. This finishes the proof.Proof of Theorem1.
Let(56)
δ
1
=
a
10
-
+
b
01
-
,
δ
2
=
2
a
20
-
+
b
11
-
,
δ
3
=
a
10
+
+
b
01
+
,
δ
4
=
2
a
20
+
+
b
11
+
,
δ
5
=
a
00
+
-
a
00
-
,
δ
6
=
a
02
+
-
a
02
-
.
Then, solve the equations
(57)
c
01
=
c
02
=
c
0
+
c
~
0
=
c
1
+
c
~
1
=
c
2
+
c
~
2
=
0
,
which, together with (56), gives that
(58)
δ
1
=
13
57
42
δ
4
,
δ
2
=
57
3
δ
4
,
δ
3
=
-
83
28
δ
4
,
δ
5
=
(
5
57
42
-
5
7
)
δ
4
,
δ
6
=
(
-
18
57
7
+
108
7
)
δ
4
,
where δ
4
>
0. Then, under (58)
(59)
c
1
=
-
114
7
δ
4
<
0
,
c
~
1
=
114
7
δ
4
>
0
,
c
3
+
c
~
3
=
-
6805
20577
114
δ
4
<
0
,
c
1
+
c
~
1
=
0
,
c
21
+
c
~
22
=
0
.
Further,
(60)
rank
⁡
∂
(
c
01
,
c
02
,
c
0
+
c
~
0
,
c
1
+
c
~
1
,
c
2
+
c
~
2
)
∂
(
δ
1
,
δ
2
,
δ
3
,
δ
5
,
δ
6
)
=
5
.
Thus, by Theorem 6, there exist 7 limit cycles near L for ɛ
>
0 small.
Under (58), we have for 0
<
(
5
/
48
)
-
h
≪
1
(61)
M
1
(
h
)
=
-
114
7
δ
4
(
h
-
5
48
)
ln
⁡
⁡
(
5
48
-
h
)
+
O
(
(
h
-
5
48
)
)
<
0
,
M
2
(
h
)
=
114
7
δ
4
(
h
-
5
48
)
ln
⁡
⁡
(
5
48
-
h
)
+
O
(
(
h
-
5
48
)
)
>
0
.
Now, we derive the expansion ofM
1
(
h
). To achieve this, make the transformation
(62)
u
=
2
2
(
x
+
1
)
,
v
=
y
,
together with time scaling τ
=
(
2
/
2
)
t. Then the system becomes
(63)
u
˙
=
v
+
ɛ
p
¯
-
(
u
,
v
)
,
v
˙
=
-
u
-
2
2
u
2
+
ɛ
q
¯
-
(
u
,
v
)
,
where
(64)
p
¯
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
a
¯
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
a
i
j
-
(
2
u
-
1
)
i
v
j
,
q
¯
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
b
¯
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
2
b
i
j
-
(
2
u
-
1
)
i
v
j
.
For ɛ
=
0, the Hamiltonian function of (63) has the form
(65)
H
¯
(
u
,
v
)
=
1
2
(
v
2
+
u
2
)
+
2
2
3
u
3
+
1
12
.
Denote M
¯
1
(
h
) by the first Melnikov function of system (63) near C
1, and note that
(66)
q
-
d
x
-
p
-
d
y
=
(
2
2
q
¯
-
)
d
u
(
2
/
2
)
-
p
¯
-
d
v
=
q
¯
-
d
u
-
p
¯
-
d
v
.
Then we have M
¯
1
(
h
)
=
M
1
(
h
). From Theorem 2.4
.
1 of [18], we can obtain
(67)
M
1
(
h
)
=
b
~
0
(
h
-
1
12
)
+
O
(
(
h
-
1
12
)
2
)
<
0
,
m
m
m
m
m
m
m
m
m
0
<
h
-
1
12
≪
1
,
where, under (58),
(68)
b
~
0
=
2
π
(
p
¯
u
-
+
q
¯
v
-
)
(
0,0
)
=
2
2
π
[
(
a
10
-
+
b
01
-
)
-
(
2
a
20
-
+
b
11
-
)
]
=
2
2
π
(
δ
1
-
δ
2
)
=
-
114
21
π
δ
4
.
Similarly, under (58) we have
(69)
M
2
(
h
)
=
-
114
228
δ
4
(
h
+
1681
324
)
+
O
(
(
h
-
h
+
1681
324
)
2
)
<
0
,
mmmmmmmmmmmmmmmmm
0
<
h
+
1681
324
≪
1
.
This implies that there exists another zero between h
=
-
(
1681
/
324
) and h
=
5
/
48 for M
2
(
h
). This ends the proof.


## 2.1. Preliminaries

Clearly, system (1) has the following two subsystems:(4a)
x
˙
=
y
+
ɛ
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
,
y
˙
=
-
(
a
2
x
2
+
a
1
x
+
a
0
)
+
ɛ
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
,
(4b)
x
˙
=
y
+
ɛ
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
,
y
˙
=
-
(
b
2
x
2
+
b
1
x
+
b
0
)
+
ɛ
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
,which are called left and right subsystems, respectively. And systems (4a)|
ɛ
=
0 and (4b)|
ɛ
=
0 are Hamiltonian with Hamiltonian functions, respectively,
(5)
H
-
(
x
,
y
)
=
1
2
y
2
+
1
3
a
2
x
3
+
1
2
a
1
x
2
+
a
0
x
,
H
+
(
x
,
y
)
=
1
2
y
2
+
1
3
b
2
x
3
+
1
2
b
1
x
2
+
b
0
x
.For systems (4a)|
ɛ
=
0 and (4b)|
ɛ
=
0, we make the following assumption:(H)
a
1
2
>
4
a
2
a
0,  b
1
2
>
4
b
2
b
0.Then, under the condition (H), forɛ
=
0, system (4a) (resp., (4b)) has a hyperbolic saddle S
1
=
(
α
S
1
,
0
) (resp., S
2
=
(
α
S
2
,
0
)) and an elementary center C
1
=
(
α
C
1
,
0
) (resp., C
2
=
(
α
C
2
,
0
)), where
(6)
α
S
1
=
-
a
1
-
a
1
2
-
4
a
2
a
0
2
a
2
,
α
S
2
=
-
b
1
-
b
1
2
-
4
b
2
b
0
2
b
2
,
α
C
1
=
-
a
1
+
a
1
2
-
4
a
2
a
0
2
a
2
,
α
C
2
=
-
b
1
+
b
1
2
-
4
b
2
b
0
2
b
2
.We will assume that(7)
a
2
<
0
,
since, for a
2
>
0, we can make a variable transformation x
→
-
x, y
→
y, together with a time rescaling t
→
-
t such that the resulting system satisfies (7). Further, note that there exist 2 possible cases for level curves of H
(
x
,
y
)
=
(
1
/
2
)
y
2
+
(
1
/
3
)
a
x
3
+
(
1
/
2
)
b
x
2
+
c
x
=
h on the plane, where a
≠
0, b
2
>
4
a
c; see Figure 1.Level curves of(
1
/
2
)
y
2
+
(
1
/
3
)
a
x
3
+
(
1
/
2
)
b
x
2
+
c
x
=
h.
(a)
a
<
0
(b)
a
>
0We remark that, in Figure1, C
=
(
(
-
b
+
b
2
-
4
a
c
)
/
2
a
,
0
), S
=
(
(
-
b
-
b
2
-
4
a
c
)
/
2
a
,
0
). Then, one can obtain 9 possible phase portraits of system (1)|
ɛ
=
0 under (H) and (7) when it has a generalized closed orbit and two hyperbolic saddles; see Figure 2. For system (1)|
ɛ
=
0, a generalized closed orbit passes through the y-axis two times.Possible portraits of (1)|
ɛ
=
0 with two saddles and a generalized closed orbit.
(a)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
>
H
+
(
S
2
)
(b)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
=
H
+
(
S
2
)
(c)
α
S
1
<
0
<
α
C
2
<
α
S
2, H
-
(
S
1
)
<
H
+
(
S
2
)
(d)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
>
H
+
(
S
2
)
(e)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
=
H
+
(
S
2
)
(f)
α
S
1
<
0,  α
C
2
≤
0
<
α
S
2,  H
-
(
S
1
)
<
H
+
(
S
2
)
(g)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
>
H
+
(
S
2
)
(h)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
=
H
+
(
S
2
)
(i)
α
S
1
<
0
<
α
S
2
<
α
C
2, H
-
(
S
1
)
<
H
+
(
S
2
)Now, we study limit cycle bifurcations on system (1). For simplicity, we consider the case
(8)
α
S
1
<
0
<
α
S
2
<
α
C
2
,
H
-
(
S
1
)
=
H
+
(
S
2
)
≜
h
1
;
that is,
(9)
a
1
2
>
4
a
2
a
0
,
b
1
2
>
4
b
2
b
0
,
a
2
<
0
,
b
2
>
0
,
a
1
<
0
,
b
1
<
0
,
a
0
<
0
,
b
0
>
0
,
a
1
3
-
6
a
0
a
1
a
2
+
(
a
1
2
-
4
a
0
a
2
)
3
/
2
a
2
2
=
b
1
3
-
6
b
0
b
1
b
2
+
(
b
1
2
-
4
b
0
b
2
)
3
/
2
b
2
2
.
Then, the phase portrait of system (1)|
ɛ
=
0 is Figure 2(h), and there are four families of periodic orbits given by
(10)
L
1
h
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
x
0
-
<
x
<
α
S
1
}
,
h
0
-
<
h
<
h
1
,
L
2
h
=
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
α
S
2
<
x
<
x
0
+
}
,
h
0
+
<
h
<
h
1
,
L
3
h
=
L
3
h
-
∪
L
3
h
+
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
α
S
1
<
x
<
0
}
∪
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
0
<
x
<
α
S
2
}
,
mmmmmmmmmmmmmm
0
<
h
<
h
1
,
L
h
=
L
h
-
∪
L
h
+
=
{
(
x
,
y
)
∣
H
-
(
x
,
y
)
=
h
,
x
<
0
}
∪
{
(
x
,
y
)
∣
H
+
(
x
,
y
)
=
h
,
x
>
0
}
,
h
>
h
1
,
where H
-
(
x
0
-
,
0
)
=
h
1, H
+
(
x
0
+
,
0
)
=
h
1 and H
-
(
C
1
)
=
h
0
-, H
+
(
C
2
)
=
h
0
+. Further, we have
(11)
lim
⁡
h
→
h
1
L
1
h
=
L
1
,
lim
⁡
h
→
h
0
-
L
1
h
=
C
1
,
lim
⁡
h
→
h
1
L
2
h
=
L
2
,
lim
⁡
h
→
h
0
+
L
1
h
=
C
2
,
lim
⁡
h
→
h
1
L
3
h
-
=
L
-
,
lim
⁡
h
→
h
1
L
3
h
+
=
L
+
,
lim
⁡
h
→
0
L
3
h
-
=
lim
⁡
h
→
0
L
3
h
+
=
(
0,0
)
,
lim
⁡
h
→
h
1
L
h
-
=
L
1
∪
L
-
,
lim
⁡
h
→
h
1
L
h
+
=
L
2
∪
L
+
;
see Figure 3. From (11), one can know that, as h approaches h
1, the limit of L
h is L
1
∪
L
-
∪
L
2
∪
L
+ denoted by L, which is a generalized polycycle with two hyperbolic saddles.Figure 3
Phase portrait of system (1)|
ɛ
=
0 under (9).Then, from Theorem 1.1 of [17], for system (1), we have four first order Melnikov functions as follows:
(12)
M
1
(
h
)
=
∮
L
1
h
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
h
∈
(
h
0
-
,
h
1
)
,
M
2
(
h
)
=
∮
L
2
h
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
h
∈
(
h
0
+
,
h
1
)
,
M
3
(
h
)
=
M
3
-
(
h
)
+
M
3
+
(
h
)
=
∫
L
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
+
∫
L
3
h
+
‍
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
m
m
m
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
h
∈
(
0
,
h
1
)
,
M
(
h
)
=
M
-
(
h
)
+
M
+
(
h
)
=
∫
L
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
+
∫
L
h
+
‍
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
h
∈
(
h
1
,
+
∞
)
,
where L
1
h, L
2
h, L
3
h
-, L
3
h
+, L
h
-, and L
h
+ are given in (10).Now, we deduce the expansion ofM
i
(
h
),  i
=
1,2
,
3, and M
(
h
) near h
=
h
1 in the above. First, we have the following.Lemma 2.
The functionM
3
-
(
h
) in (12) has the following expression:
(13)
M
3
-
(
h
)
=
c
0
+
c
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
2
(
h
-
h
1
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
mmmmmmmhmimmmm
0
<
h
1
-
h
≪
1
,
where
(14)
c
0
=
∫
L
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
mmim
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
-
2
a
00
-
2
h
1
-
4
3
h
1
2
h
1
a
02
-
,
c
1
=
-
1
(
a
1
2
-
4
a
0
a
2
)
1
/
4
[
(
a
10
-
+
b
01
-
)
+
α
S
1
(
2
a
20
-
+
b
11
-
)
]
,
c
2
|
c
1
=
0
=
∫
L
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
m
m
m
m
h
m
-
2
h
1
h
1
a
00
-
-
2
2
h
1
a
02
-
,
c
3
=
a
2
2
(
a
1
2
-
4
a
0
a
2
)
5
/
4
(
2
a
20
-
+
b
11
-
)
.Proof.
Denote byU a disk of diameter ϵ
0
>
0 with its center at S
1. Then, for ϵ
0
>
0 small, rewrite M
3
-
(
h
)
=
I
1
(
h
)
+
I
2
(
h
), where
(15)
I
1
(
h
)
=
∫
L
¯
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
I
2
(
h
)
=
∫
L
~
3
h
-
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
with
(16)
L
-
3
h
-
=
L
3
h
-
∩
U
,
L
~
3
h
-
=
L
3
h
-
∖
L
¯
3
h
-
.
Note that I
2
(
h
) is a C
∞ function for h
1
-
h
>
0 small. Thus, it suffices to investigate the expansion of I
1
(
h
) for 0
<
h
1
-
h
≪
1. To do this, we make a variable transformation
(17)
u
=
(
a
1
2
-
4
a
0
a
2
)
1
/
4
(
x
-
α
S
1
)
,
v
=
y
,
τ
=
(
a
1
2
-
4
a
0
a
2
)
1
/
4
t
.
Then system (4a) becomes
(18)
u
˙
=
v
+
ɛ
p
~
-
(
u
,
v
)
,
v
˙
=
u
-
a
2
u
2
(
a
1
2
-
4
a
0
a
2
)
3
/
4
+
ɛ
q
~
-
(
u
,
v
)
,
where
(19)
p
~
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
a
~
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
a
i
j
-
[
u
(
a
1
2
-
4
a
0
a
2
)
1
/
4
+
α
S
1
]
i
v
j
,
(20)
q
~
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
b
~
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
b
i
j
-
(
a
1
2
-
4
a
0
a
2
)
1
/
4
[
u
(
a
1
2
-
4
a
0
a
2
)
1
/
4
+
α
S
1
]
i
v
j
.
For ɛ
=
0, the Hamiltonian function of (18) has the form
(21)
H
~
(
u
,
v
)
=
1
2
(
v
2
-
u
2
)
+
a
2
3
(
a
1
2
-
4
a
0
a
2
)
3
/
4
u
3
+
h
1
and the corresponding first order Melinikov function can be denoted by M
~
(
h
). By Remark 3.1
.
4 of [18], we can know that M
3
-
(
h
)
=
M
~
(
h
), and then, from Corollary 3.2
.
2 and the Theorem 3.2
.
1 of [18], I
1 can be written as
(22)
I
1
(
h
)
=
c
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
where
(23)
c
1
=
-
(
a
~
10
-
+
b
~
01
-
)
=
-
1
(
a
1
2
-
4
a
0
a
2
)
1
/
4
×
[
(
a
10
-
+
b
01
-
)
+
α
S
1
(
2
a
20
-
+
b
11
-
)
]
,
c
3
=
3
2
×
a
2
3
(
a
1
2
-
4
a
0
a
2
)
3
/
4
×
1
(
a
1
2
-
4
a
0
a
2
)
2
/
4
(
2
a
~
20
+
b
~
11
)
=
a
2
2
(
a
1
2
-
4
a
0
a
2
)
5
/
4
(
2
a
20
-
+
b
11
-
)
.
Thus, by the above discussion, one can obtain the expression of (13) and the formulas for c
1 and c
3. Now, we give the expressions of c
0 and c
2 in (13).
On the curveL
3
h intersect y-axis with two points denoted by A
1
(
h
)
=
(
0
,
-
2
h
) and A
(
h
)
=
(
0
,
2
h
). Then, from (12), we have
(24)
M
3
-
(
h
)
=
∫
A
1
(
h
)
A
(
h
)
^
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
=
N
1
(
h
)
+
N
2
(
h
)
,
where
(25)
N
1
(
h
)
=
∫
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
‍
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
mmm
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
N
2
(
h
)
=
∫
A
(
h
)
A
1
(
h
)
→
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
.
By Green’s formula two times toN
1
(
h
), we obtain
(26)
N
1
(
h
)
=
∫
Int
.
(
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
)
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
m
m
m
m
h
m
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
x
d
y
=
∮
A
1
(
h
)
A
(
h
)
^
∪
A
(
h
)
A
1
(
h
)
→
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
‍
m
m
m
m
m
m
m
m
h
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
=
∫
L
3
h
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
m
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
.
Further, one has
(27)
N
2
(
h
)
=
∫
A
(
h
)
A
1
(
h
)
→
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
=
∫
2
h
-
2
h
‍
(
a
00
-
+
a
01
-
y
+
a
02
-
y
2
)
d
y
=
-
2
a
00
-
2
h
-
4
3
h
2
h
a
02
-
.
Thus, we have
(28)
M
3
-
(
h
)
=
∫
L
3
h
-
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
m
m
m
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
-
2
a
00
-
2
h
-
4
3
h
2
h
a
02
-
.
Then, let h
→
h
1. Then, by (11), we can obtain the expression of c
0 in (13). Further, differentiating (13) with respect to h, together with Lemma 3.1
.
2 of [18], we obtain
(29)
d
d
h
M
3
-
(
h
)
=
∫
L
3
h
-
‍
(
p
x
+
q
y
)
|
ɛ
=
0
d
t
=
∫
L
3
h
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
-
2
h
h
a
00
-
-
2
2
h
a
02
-
=
c
1
ln
⁡
⁡
(
h
1
-
h
)
+
c
1
+
c
2
+
O
(
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
)
,
which, as c
1
=
0, gives that
(30)
∫
L
3
h
-
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
-
2
h
h
a
00
-
-
2
2
h
a
02
-
=
c
2
+
O
(
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
)
.
Let h
→
h
1 in the above. Then we can easily obtain the formula for c
2 in (13). This ends the proof.In the same way, one can obtain the expressions ofM
3
+
(
h
).Lemma 3.
ForM
3
-
(
h
) in (12), one has
(31)
M
3
+
(
h
)
=
c
~
0
+
c
~
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
~
2
(
h
-
h
1
)
+
c
~
3
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
mmmmmmmmmmmmm
0
<
h
1
-
h
≪
1
,
where
(32)
c
~
0
=
∫
L
+
‍
[
1
2
(
a
10
+
+
b
01
+
)
y
+
(
b
11
+
+
2
a
20
+
)
x
y
m
m
m
m
+
1
2
(
a
11
+
+
2
b
02
+
)
y
2
]
d
x
+
2
2
h
1
a
00
+
+
4
3
h
1
2
h
1
a
02
+
,
c
~
1
=
-
1
(
b
1
2
-
4
b
0
b
2
)
1
/
4
[
(
a
10
+
+
b
01
+
)
+
α
S
2
(
2
a
20
+
+
b
11
+
)
]
,
c
~
2
|
c
~
1
=
0
=
∫
L
+
‍
[
(
a
10
+
+
b
01
+
)
+
(
b
11
+
+
2
a
20
+
)
x
+
(
a
11
+
+
2
b
02
+
)
y
]
d
t
+
2
h
1
h
1
a
00
+
+
2
2
h
1
a
02
+
,
c
~
3
=
b
2
2
(
b
1
2
-
4
b
0
b
2
)
5
/
4
(
2
a
20
+
+
b
11
+
)
.By applying Theorem3.2
.
3 of [18], together with the proof of Lemma 2, we have the following.Lemma 4.
The functionsM
i
(
h
), i
=
1,2, given in (12) can be expressed as
(33)
M
1
(
h
)
=
c
01
+
c
1
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
c
21
(
h
-
h
1
)
+
c
3
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
M
2
(
h
)
=
c
02
+
c
~
1
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
c
22
(
h
-
h
1
)
+
c
~
3
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
O
(
(
h
-
h
1
)
2
)
,
0
<
h
1
-
h
≪
1
,
where c
1, c
3, c
~
1, c
~
3 are given in Lemmas 2 and 3, and
(34)
c
01
=
∮
L
1
∑
i
+
j
=
0
2
‍
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
-
x
i
y
j
d
y
,
c
02
=
∮
L
2
∑
i
+
j
=
0
2
‍
b
i
j
+
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
‍
a
i
j
+
x
i
y
j
d
y
,
c
21
|
c
1
=
0
=
∮
L
1
[
(
a
10
-
+
b
01
-
)
+
(
2
a
20
-
+
b
11
-
)
x
‍
mmm
+
(
a
11
-
+
2
b
02
-
)
y
]
d
t
|
c
1
=
0
,
c
22
|
c
~
1
=
0
=
∮
L
2
‍
[
(
a
10
+
+
b
01
+
)
+
(
2
a
20
+
+
b
11
+
)
x
mmm
+
(
a
11
+
+
2
b
02
+
)
y
]
d
t
|
c
~
1
=
0
.By above three Lemmas2, 3, and 4, together with (12), one can derive the following.Lemma 5.
ForM
3
(
h
) and M
(
h
), one has
(35)
M
3
(
h
)
=
(
c
0
+
c
~
0
)
+
(
c
1
+
c
~
1
)
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
(
c
2
+
c
~
2
)
(
h
-
h
1
)
+
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
(
O
(
h
1
-
h
)
2
)
,
0
<
h
1
-
h
≪
1
,
M
(
h
)
=
(
c
0
+
c
01
+
c
~
0
+
c
02
)
+
2
(
c
1
+
c
~
1
)
×
(
h
-
h
1
)
ln
⁡
(
h
-
h
1
)
+
(
c
2
+
c
21
+
c
~
2
+
c
22
)
×
(
h
-
h
1
)
+
2
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
-
h
1
)
+
(
O
(
h
1
-
h
)
2
)
,
0
<
h
-
h
1
≪
1
,
where the corresponding coefficients are given in (14), (32), and (34).Then, using Lemmas4 and 5, we can have the following.Theorem 6.
Let (H) and (7) hold. Then if there exists δ
0 such that
(36)
c
01
(
δ
0
)
=
c
02
(
δ
0
)
=
(
c
0
+
c
~
0
)
(
δ
0
)
=
(
c
1
+
c
~
1
)
(
δ
0
)
=
(
c
2
+
c
~
2
)
(
δ
0
)
=
0
,
(
c
3
+
c
~
3
)
(
δ
0
)
<
0
,
c
1
(
δ
0
)
<
0
,
c
~
1
(
δ
0
)
>
0
,
(
c
1
+
c
~
1
)
(
δ
0
)
=
0
,
(
c
21
+
c
~
22
)
(
δ
0
)
=
0
,
rank
⁡
∂
(
c
01
,
c
02
,
c
0
+
c
~
0
,
c
1
+
c
~
1
,
c
2
+
c
~
2
)
∂
δ
=
5
,
then system (1) can have 7 limit cycles near L.Proof.
By the assumption and from the expansions in Lemmas4 and 5 we have
(37)
M
1
(
h
)
=
c
1
(
δ
0
)
(
h
-
h
1
)
ln
⁡
⁡
(
h
1
-
h
)
+
O
(
h
1
-
h
)
<
0
,
mmmmmmmmmmhmhmmm
0
<
h
1
-
h
≪
1
,
M
2
(
h
)
=
c
~
1
(
δ
0
)
(
h
-
h
1
)
ln
⁡
(
h
1
-
h
)
+
O
(
h
-
h
1
)
>
0
,
mmmmmmmmmmmmmmm
0
<
h
1
-
h
≪
1
,
M
3
(
h
)
=
(
c
3
+
c
~
3
)
(
δ
0
)
(
h
-
h
1
)
2
ln
⁡
(
h
1
-
h
)
+
(
O
(
h
1
-
h
)
2
)
>
0
,
0
<
h
1
-
h
≪
1
,
M
(
h
)
=
2
(
c
3
+
c
~
3
)
(
h
-
h
1
)
2
ln
⁡
⁡
(
h
-
h
1
)
+
(
O
(
h
1
-
h
)
2
)
>
0
,
mmmmmmmmmlmmmmmmmmm
0
<
h
-
h
1
≪
1
.
Note thatc
01, c
02, c
0
+
c
~
0, c
1
+
c
~
1, and c
2
+
c
~
2 can be taken as free parameters. Then we can vary them one by one to change the signs of M
1, M
2, M
3, and M.
First, letc
01
=
c
02
=
c
0
+
c
~
0
=
c
1
+
c
~
1
=
0 and 0
<
-
(
c
2
+
c
~
2
)
≪
1. Then, in this case, we have
(38)
M
1
(
h
)
<
0
,
M
2
(
h
)
>
0
,
M
3
(
h
)
>
0
,
mmmmmmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
,
which means that there exists a zero h
~
1 with 0
<
h
~
1
-
h
1 such that M
(
h
~
1
)
=
0.
Second, fixc
01
=
c
02
=
c
0
+
c
~
0
=
0 and vary 0
≪
-
(
c
1
+
c
~
1
)
≪
-
(
c
2
+
c
~
2
), so that
(39)
M
1
(
h
)
<
0
,
M
2
(
h
)
>
0
,
M
3
(
h
)
<
0
,
mmmmmhmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
>
0
,
for
0
<
h
-
h
1
≪
1
.
Hence, there exist h
1
-
h
~
2
>
0,  h
~
3
∈
(
h
1
,
h
~
1
) such that M
3
(
h
~
2
)
=
0,  M
(
h
~
3
)
=
0.
Third, letc
01
=
c
0
+
c
~
0
=
0,  0
<
-
c
02
≪
-
(
c
1
+
c
~
1
), implying
(40)
M
1
(
h
)
<
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
<
0
,
mmm1mmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
Then M
2
(
h
~
4
)
=
0,  M
(
h
~
5
)
=
0 for some h
~
4
<
h
1,  h
~
5
∈
(
h
1
,
h
~
3
).
Fourth, letc
0
+
c
~
0
=
0 and 0
<
c
01
≪
-
c
02; we have
(41)
M
1
(
h
)
>
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
<
0
,
mmmmmhmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
So there exists h
~
6
<
h
1 such that M
1
(
h
~
6
)
=
0.
Finally, vary0
<
c
0
+
c
~
0
≪
c
01. In this case,
(42)
M
1
(
h
)
>
0
,
M
2
(
h
)
<
0
,
M
3
(
h
)
>
0
,
mmmmhmmmmmm
for
0
<
h
1
-
h
≪
1
,
M
(
h
)
<
0
,
for
0
<
h
-
h
1
≪
1
.
So there exists h
~
7
∈
(
h
~
2
,
h
1
) such that M
3
(
h
~
7
)
=
0.
In a word, we have proved that if(43)
0
<
c
0
+
c
~
0
≪
c
01
≪
-
c
02
≪
-
(
c
1
+
c
~
1
)
≪
-
(
c
2
+
c
~
2
)
≪
1
then M
1, M
2, M
3, and M have one zero, one zero, two zeros, and three zeros which implies that seven limit cycles can appear in the neighbourhood of L.


## 2.2. Proofs

In this section, we will use Theorem6 to prove our main result. For that purpose, let
(44)
a
2
=
-
1
,
a
1
=
-
3
2
,
a
0
=
-
1
2
,
b
2
=
1
,
b
1
=
-
11
3
,
b
0
=
41
48
.
Then, the Hamiltonian functions in (5) can be rewritten as
(45)
H
-
(
x
,
y
)
=
1
2
y
2
-
1
3
x
3
-
3
4
x
2
-
1
2
x
,
H
+
(
x
,
y
)
=
1
2
y
2
+
1
3
x
3
-
11
6
x
2
+
11
48
x
.
And in this case
(46)
S
1
=
(
-
1
2
,
0
)
,
S
2
=
(
1
4
,
0
)
,
C
1
=
(
-
1,0
)
,
C
2
=
(
41
12
,
0
)
,
h
1
=
5
48
.Under (44), it is not hard to obtain the expressions for c
1, c
~
1, c
3, and c
~
3 in Lemmas 2 and 3.Lemma 7.
Under (44), for system (1), one has
(47)
c
1
=
-
2
[
(
a
10
-
+
b
01
-
)
-
1
2
(
2
a
20
-
+
b
11
-
)
]
,
c
3
=
-
2
(
2
a
20
-
+
b
11
-
)
,
c
~
1
=
-
6
19
[
(
a
10
+
+
b
01
+
)
+
1
4
(
2
a
20
+
+
b
11
+
)
]
,
c
~
3
=
1
2
(
6
19
)
5
/
2
(
2
a
20
+
+
b
11
+
)
.For the other expressions of coefficients in Lemmas4 and 5, one has the following.Lemma 8.
Under (44), for system (1), one has
(48)
c
01
=
3
2
20
(
a
10
-
+
b
01
-
)
-
39
2
280
(
2
a
20
-
+
b
11
-
)
,
c
21
=
-
3
2
(
2
a
20
-
+
b
11
-
)
,
c
02
=
361
114
180
(
a
10
+
+
b
01
+
)
+
29963
114
5040
(
2
a
20
+
+
b
11
+
)
,
c
22
=
114
(
2
a
20
+
+
b
11
+
)
,
c
0
=
3
2
20
(
a
10
-
+
b
01
-
)
-
(
39
2
280
-
5
30
168
)
(
2
a
20
-
+
b
11
-
)
-
30
6
a
00
-
-
5
30
432
a
02
-
,
c
~
0
=
(
-
35
9
30
+
361
180
114
)
(
a
10
+
+
b
01
+
)
+
(
29963
5040
114
-
730
63
30
)
(
2
a
20
+
+
b
11
+
)
+
30
6
a
00
+
+
5
30
432
a
02
+
,
c
2
=
(
30
-
3
2
)
(
2
a
20
-
+
b
11
-
)
-
4
30
5
a
00
-
-
30
6
a
02
-
,
c
~
2
=
(
-
2
30
+
114
)
(
2
a
20
+
+
b
11
+
)
+
4
30
5
a
00
+
+
30
6
a
02
+
.Proof.
Note thatc
01
=
∮
L
1
‍
∑
i
+
j
=
0
2
b
i
j
-
x
i
y
j
d
x
-
∑
i
+
j
=
0
2
a
i
j
-
x
i
y
j
d
y. Then, by Green’s formula two times, we derive
(49)
c
01
=
∫
Int
.
(
L
1
)
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
m
m
m
m
+
(
a
11
-
+
2
b
02
-
)
y
]
d
x
d
y
=
∮
L
1
‍
[
1
2
(
a
10
-
+
b
01
-
)
y
+
(
b
11
-
+
2
a
20
-
)
x
y
mmm
+
1
2
(
a
11
-
+
2
b
02
-
)
y
2
]
d
x
.
On the other hand, along the curveL
1, we have (
1
/
2
)
y
2
-
(
1
/
3
)
x
3
-
(
3
/
4
)
x
2
-
(
1
/
2
)
x
=
5
/
48, for -
(
5
/
4
)
≤
x
≤
-
(
1
/
2
). Then, (49) can be represented as
(50)
c
01
=
2
∫
-
5
/
4
-
1
/
2
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
]
m
m
m
×
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
.
By applying Maple, one can compute that
(51)
∫
-
5
/
4
-
1
/
2
‍
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
3
2
40
,
∫
-
5
/
4
-
1
/
2
‍
x
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
-
39
2
560
.
Inserting the above into (50) gives the expression of c
01.
Recall thatc
1
=
0 if and only if a
10
-
+
b
01
-
=
(
1
/
2
)
(
2
a
20
-
+
b
11
-
). Thus, by (4a)|
ɛ
=
0, we rewrite c
21
|
c
1
=
0 as
(52)
c
21
|
c
1
=
0
=
∮
L
1
[
(
1
/
2
)
(
2
a
20
-
+
b
11
-
)
+
(
2
a
20
-
+
b
11
-
)
x
+
(
a
11
-
+
2
b
02
-
)
y
]
y
d
x
=
1
2
(
2
a
20
-
+
b
11
-
)
∮
L
1
1
+
2
x
y
d
x
=
(
2
a
20
-
+
b
11
-
)
∫
-
5
/
4
-
1
/
2
‍
1
+
2
x
(
5
/
24
)
+
(
2
/
3
)
x
3
+
(
3
/
2
)
x
2
+
x
d
x
,
which, by Maple, yields the conclusion for c
02. Using a similar way, one can derive the formulas for c
02 and c
22. Now, we deduce the formulas for c
0 and c
2.
Note that, on the curveL
-
:
(
1
/
2
)
y
2
-
(
1
/
3
)
x
3
-
(
3
/
4
)
x
2
-
(
1
/
2
)
x
=
(
5
/
48
),  -
(
1
/
2
)
<
x
<
0, we have
(53)
c
0
=
2
∫
-
1
/
2
0
‍
[
(
a
10
-
+
b
01
-
)
+
(
b
11
-
+
2
a
20
-
)
x
]
m
m
m
×
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
-
30
6
a
00
-
-
5
30
432
a
02
-
.
By Maple again, we obtain that
(54)
∫
-
1
/
2
0
‍
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
3
2
40
,
∫
-
1
/
2
0
‍
x
5
24
+
2
3
x
3
+
3
2
x
2
+
x
d
x
=
-
39
2
560
+
5
30
336
.
Thus, we can obtain the conclusion for c
0.
Whenc
1
=
0, we have (
a
10
-
+
b
01
-
)
=
(
1
/
2
)
(
2
a
20
-
+
b
11
-
). Thus, c
2 in (14) can be rewritten as
(55)
c
2
=
1
2
(
2
a
20
-
+
b
11
-
)
∫
L
-
‍
1
+
2
x
y
d
x
-
2
a
00
-
2
h
1
-
4
a
02
-
3
(
2
h
1
+
h
1
2
h
1
)
|
h
1
=
5
/
48
=
(
2
a
20
-
+
b
11
-
)
∫
-
1
/
2
0
‍
1
+
2
x
(
5
/
24
)
+
(
2
/
3
)
x
3
+
(
3
/
2
)
x
2
+
x
d
x
-
4
5
6
a
00
-
-
30
6
a
02
-
,
which, by Maple, implies the conclusion. Further, we can prove c
~
0 and c
~
2 in the same way to c
0 and c
2, respectively. This finishes the proof.Proof of Theorem1.
Let(56)
δ
1
=
a
10
-
+
b
01
-
,
δ
2
=
2
a
20
-
+
b
11
-
,
δ
3
=
a
10
+
+
b
01
+
,
δ
4
=
2
a
20
+
+
b
11
+
,
δ
5
=
a
00
+
-
a
00
-
,
δ
6
=
a
02
+
-
a
02
-
.
Then, solve the equations
(57)
c
01
=
c
02
=
c
0
+
c
~
0
=
c
1
+
c
~
1
=
c
2
+
c
~
2
=
0
,
which, together with (56), gives that
(58)
δ
1
=
13
57
42
δ
4
,
δ
2
=
57
3
δ
4
,
δ
3
=
-
83
28
δ
4
,
δ
5
=
(
5
57
42
-
5
7
)
δ
4
,
δ
6
=
(
-
18
57
7
+
108
7
)
δ
4
,
where δ
4
>
0. Then, under (58)
(59)
c
1
=
-
114
7
δ
4
<
0
,
c
~
1
=
114
7
δ
4
>
0
,
c
3
+
c
~
3
=
-
6805
20577
114
δ
4
<
0
,
c
1
+
c
~
1
=
0
,
c
21
+
c
~
22
=
0
.
Further,
(60)
rank
⁡
∂
(
c
01
,
c
02
,
c
0
+
c
~
0
,
c
1
+
c
~
1
,
c
2
+
c
~
2
)
∂
(
δ
1
,
δ
2
,
δ
3
,
δ
5
,
δ
6
)
=
5
.
Thus, by Theorem 6, there exist 7 limit cycles near L for ɛ
>
0 small.
Under (58), we have for 0
<
(
5
/
48
)
-
h
≪
1
(61)
M
1
(
h
)
=
-
114
7
δ
4
(
h
-
5
48
)
ln
⁡
⁡
(
5
48
-
h
)
+
O
(
(
h
-
5
48
)
)
<
0
,
M
2
(
h
)
=
114
7
δ
4
(
h
-
5
48
)
ln
⁡
⁡
(
5
48
-
h
)
+
O
(
(
h
-
5
48
)
)
>
0
.
Now, we derive the expansion ofM
1
(
h
). To achieve this, make the transformation
(62)
u
=
2
2
(
x
+
1
)
,
v
=
y
,
together with time scaling τ
=
(
2
/
2
)
t. Then the system becomes
(63)
u
˙
=
v
+
ɛ
p
¯
-
(
u
,
v
)
,
v
˙
=
-
u
-
2
2
u
2
+
ɛ
q
¯
-
(
u
,
v
)
,
where
(64)
p
¯
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
a
¯
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
a
i
j
-
(
2
u
-
1
)
i
v
j
,
q
¯
-
(
u
,
v
)
=
∑
i
+
j
=
0
2
‍
b
¯
i
j
-
u
i
v
j
=
∑
i
+
j
=
0
2
‍
2
b
i
j
-
(
2
u
-
1
)
i
v
j
.
For ɛ
=
0, the Hamiltonian function of (63) has the form
(65)
H
¯
(
u
,
v
)
=
1
2
(
v
2
+
u
2
)
+
2
2
3
u
3
+
1
12
.
Denote M
¯
1
(
h
) by the first Melnikov function of system (63) near C
1, and note that
(66)
q
-
d
x
-
p
-
d
y
=
(
2
2
q
¯
-
)
d
u
(
2
/
2
)
-
p
¯
-
d
v
=
q
¯
-
d
u
-
p
¯
-
d
v
.
Then we have M
¯
1
(
h
)
=
M
1
(
h
). From Theorem 2.4
.
1 of [18], we can obtain
(67)
M
1
(
h
)
=
b
~
0
(
h
-
1
12
)
+
O
(
(
h
-
1
12
)
2
)
<
0
,
m
m
m
m
m
m
m
m
m
0
<
h
-
1
12
≪
1
,
where, under (58),
(68)
b
~
0
=
2
π
(
p
¯
u
-
+
q
¯
v
-
)
(
0,0
)
=
2
2
π
[
(
a
10
-
+
b
01
-
)
-
(
2
a
20
-
+
b
11
-
)
]
=
2
2
π
(
δ
1
-
δ
2
)
=
-
114
21
π
δ
4
.
Similarly, under (58) we have
(69)
M
2
(
h
)
=
-
114
228
δ
4
(
h
+
1681
324
)
+
O
(
(
h
-
h
+
1681
324
)
2
)
<
0
,
mmmmmmmmmmmmmmmmm
0
<
h
+
1681
324
≪
1
.
This implies that there exists another zero between h
=
-
(
1681
/
324
) and h
=
5
/
48 for M
2
(
h
). This ends the proof.



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*Source: 385103-2014-10-14.xml*