1 Introduction and Statement of the Main Results
A planar polynomial differential system is a differential system of the form
(1.1)
{
x
˙
=
P
(
x
,
y
)
,
y
˙
=
Q
(
x
,
y
)
,
where P and Q are real polynomials. We say that the polynomial differential system (1.1) has degreen, if n is the maximum of the degrees of the polynomials P and Q. Usually a polynomial differential system of degree three is denoted simply as a cubic system. The dot in (1.1) denotes the derivative with respect to the independent variable t.
One of the main problems in qualitative theory is the study of the number and configurations of limit cycles for a polynomial vector field, known as the 16th Hilbert problem. For real planar polynomial differential systems, the 16th Hilbert problem restricted to algebraic limit cycles under generic conditions has been solved in [10, 15]. The algebraic limit cycles have been studied by several authors over the last years, see, for instance, [5, 13], as well as their relation with the integrability problem, see [3, 6, 7].
In [11, 14] it is shown that every finite configuration of disjoint closed curves of the plane is topologically realizable as the set of limit cycles of a polynomial vector field, and the realization can be made by algebraic limit cycles. An explicit vector field was also presented exhibiting any given finite configuration of limit cycles. Limit cycles can also be studied by perturbing periodic solutions of centers, see, for instance, [2, 8] and the references therein.
In this paper we want to analyze all cubic polynomial differential systems having two circles as algebraic limit cycles.
In [9] it was proved first that every planar polynomial vector field of degree n with n invariant circles is Darboux integrable without limit cycles, and second that a planar polynomial vector field of degree n has at most n-1 invariant circles as algebraic limit cycles. So, in particular, cubic systems have at most two circles as algebraic limit cycles.
Our first result is to provide a normal form for all cubic polynomial differential systems having two circles as invariant algebraic curves.
Consider two circles on the plane that do not intersect. These circles, after a scaling and a rotation of the coordinates around the origin, can be written, without loss of generality, as one of the following forms:
f1(x,y)=x2+y2-1 and f2(x,y)=x2+y2-r2, with r>1 (in this case both circles are concentric).
f1(x,y)=x2+y2-1 and f2(x,y)=(x-x0)2+y2-r2, with 0<x0<x0+r<1 (this is the case in which one circle, f1=0, contains the other, f2=0, in the bounded region that it delimits, and both circles are not concentric).
f1(x,y)=x2+y2-1 and f2(x,y)=(x-x0)2+y2-r2, with x0>r+1 and r>0 (this is the case in which none of the circles contains the other in the bounded region that they delimit).
Given a polynomial f=f(x,y)∈ℂ[x,y], we say that f=0 is an invariant algebraic curve of system (1.1) if there exists a polynomial K=K(x,y)∈ℂ[x,y], called the cofactor, such that
P
∂
f
∂
x
+
Q
∂
f
∂
y
=
K
f
.
If system (1.1) has degree n, then the cofactor has degree at most n-1.
Theorem 1.1.
A cubic system having the two invariant circles f1=0 and f2=0, as in (I), can be written as
(1.2)
{
x
˙
=
-
y
(
a
0
+
a
1
x
+
a
2
y
+
a
3
x
2
+
a
4
x
y
+
a
5
y
2
)
,
y
˙
=
x
(
a
0
+
a
1
x
+
a
2
y
+
a
3
x
2
+
a
4
x
y
+
a
5
y
2
)
,
where ai∈R for i=0,1,…,5. The cofactors of f1=0 and f2=0 are zero.
Note that system (1.2), after a rescaling of time, can be written as x˙=-y, y˙=x, which has the first integral H=x2+y2. Since the first integral is defined in the whole plane, this system has no limit cycles. So, when the two invariant circles of a cubic system are concentric they cannot be limit cycles.
Theorem 1.2.
A cubic system (x˙,y˙) such that x˙ and y˙ have no common factors and have two invariant circles f1=0 and f2=0, as in (II) or in (III), can be written as
(1.3)
{
x
˙
=
A
1
y
+
A
2
x
y
+
A
3
y
2
+
A
4
x
2
y
+
A
5
x
y
2
+
A
6
y
3
,
y
˙
=
B
0
+
B
1
x
+
B
2
y
+
B
3
x
2
-
A
3
x
y
-
B
0
y
2
+
B
4
x
3
+
B
2
x
2
y
+
B
5
x
y
2
-
B
2
y
3
,
where
A
1
=
-
a
0
-
a
2
(
1
-
r
2
+
x
0
2
)
2
x
0
,
A
2
=
a
2
-
a
1
(
1
-
r
2
+
x
0
2
)
2
x
0
,
A
3
=
-
a
3
(
1
-
r
2
+
x
0
2
)
2
x
0
,
A
4
=
a
0
+
a
1
,
A
5
=
a
3
,
A
6
=
a
0
,
B
0
=
-
a
0
x
0
-
a
2
2
,
B
1
=
a
0
-
a
1
2
+
a
2
(
1
-
r
2
+
x
0
2
)
2
x
0
,
B
2
=
-
a
3
2
,
B
3
=
a
0
x
0
-
a
2
2
+
a
1
(
1
-
r
2
+
x
0
2
)
2
x
0
,
B
4
=
-
a
0
-
a
1
2
,
B
5
=
-
a
0
+
a
1
2
,
with a0,a1,a2,a3∈R. The cofactors of f1=0 and f2=0 are
K
1
=
y
(
a
2
+
2
a
0
x
0
+
a
1
x
+
a
3
y
)
𝑎𝑛𝑑
K
2
=
y
(
a
2
+
a
1
x
+
a
3
y
)
,
respectively.
Note that when a0=0, system (1.3) has the rational first integral f1/f2, and consequently no limit cycles. Therefore, in what follows we consider that a0≠0.
Now we want to characterize when system (1.3) has f1=0 and f2=0 as limit cycles.
The following result shows that the two circles f1=f2=0 are never both limit cycles of system (1.3) when a3=0.
Theorem 1.3.
If a3=0 and a0≠0, then both circles f1=0 and f2=0 are not limit cycles of system (1.3), with x˙ and y˙ not having common factors.
By Theorem 1.3, in order for both f1=0 and f2=0 to be limit cycles of system (1.3), we must have a3≠0. The following result characterizes when the circles f1=0 and f2=0 are limit cycles of system (1.3). This is the main result of this paper because it characterizes when a cubic polynomial differential system has two circles as invariant algebraic limit cycles.
Theorem 1.4.
The two circles f1=0 and f2=0 are limit cycles for system (1.3) without common factors between x˙ and y˙ if and only if
Δ1=(a12+a32)r2-(a2+(2a0+a1)x0)2<0,
either B42+B32+B12≠0 if (II) holds, or B3≠0 and B12-4B0B3>0 if (III) holds with B4=0.
The paper is organized as follows. In Section 2 we prove Theorems 1.1 and 1.2. In Section 3 we give some preliminary results related to system (1.3) that will be used in the proofs of Theorems 1.3 and 1.4. The study of the singular points of system (1.3) is carried out in Section 4. Finally, Theorems 1.3 and 1.4 are proved in Section 5.
2 Proofs of Theorems 1.1 and 1.2
Suppose that a cubic system in the plane has two invariant circles that do not intersect. As pointed out in the introduction, two circles on the plane that do not intersect after a rescaling and a rotation of the coordinates around the origin can be written, without loss of generality, as in (I), (II) or (III).
We write a cubic planar polynomial differential system in the form
(2.1)
x
˙
=
∑
i
+
j
=
0
3
a
i
j
x
i
y
j
,
y
˙
=
∑
i
+
j
=
0
3
b
i
j
x
i
y
j
,
where aij,bij∈ℝ for i+j=0,1,2,3.
We assume that f1=0 and f2=0 are invariant algebraic curves of system (2.1), with cofactors k1 and k2 given, respectively, by
k
1
=
∑
i
+
j
=
0
2
α
i
j
x
i
y
j
and
k
2
=
∑
i
+
j
=
0
2
β
i
j
x
i
y
j
,
where αij,βij∈ℝ for i+j=0,1,2.
We recall that f1=0 and f2=0 are invariant algebraic curves of system (2.1), that is, they satisfy
(2.2)
k
i
f
i
=
∂
f
i
∂
x
x
˙
+
∂
f
i
∂
y
y
˙
=
∑
i
+
j
=
0
3
a
i
j
x
i
y
j
∂
f
i
∂
x
+
∑
i
+
j
=
0
3
b
i
j
x
i
y
j
∂
f
i
∂
y
,
i
=
1
,
2
.
If f1 and f2 are as in (I), from (2.2), we get system (1.2), where we have used the notation a0=a01=b10, a1=a11=b20, a2=a02=b11, a3=a31=b30, a4=a12=b21 and a5=a30=b12. Moreover, in this case, K1=K2=0.
If f1 and f2 are as in case (II) or (III), from (2.2), we get system (1.3) with the corresponding cofactors, where we have used the notation a0=a03, a1=β11, a2=β01 and a3=β02.
3 Preliminary Results on System (1.3)
In this section we introduce some preliminary results on system (1.3) that will be used in the proof of Theorems 1.3 and 1.4.
Lemma 3.1.
For system (1.3), the following hold:
If a3=0, then it is invariant with respect to the change (x,y,t)→(x,-y,-t).
If it has a unique singular point
p
inside the circle
f
1
=
0
(resp. the circle
f
2
=
0
), which is a limit cycle, then this singular point cannot be a center.
Proof.
First we prove statement (a). System (1.3) with a3=0 becomes
{
x
˙
=
A
1
y
+
A
2
x
y
+
A
4
x
2
y
+
A
6
y
3
,
y
˙
=
B
0
+
B
1
x
+
B
3
x
2
-
B
0
y
2
+
B
4
x
3
+
B
5
x
y
2
,
which is obviously invariant with respect to the change (x,y,t)→(x,-y,-t). This concludes the proof of statement (a).
For statement (b), we will prove that if the unique singular point inside f1=0 is a center, then f1=0 is not a limit cycle. The proof for the circle f2 can be given in a similar way.
Consider a Poincaré map defined in a transversal section with endpoints the center and a point of the circle f1=0. This Poincaré map is analytic because the differential system is polynomial, and consequently analytic. We have a continuum of periodic orbits surrounding the center, so the Poincaré map in a neighborhood of the center is the identity, and, by analyticity, it is the identity in all the considered transversal section. Consequently, the circle f1=0 is not a limit cycle. This concludes the proof of statement (b). ∎
We shall need the following result. For the proof of Proposition 3.2, see, for instance, [12].
Proposition 3.2.
Consider a two-dimensional autonomous polynomial differential system having an invariant algebraic curve g=0 with cofactor k. Then all singular points of the differential system are contained in the union of the sets {g=0}∪{k=0}.
Proposition 3.3.
The two circles f1=0 and f2=0 are periodic solutions for system (1.3) with a0≠0 if and only if
(3.1)
Δ
0
=
a
3
2
+
a
1
2
-
a
2
2
<
0
,
Δ
1
=
(
a
1
2
+
a
3
2
)
r
2
-
(
a
2
+
(
2
a
0
+
a
1
)
x
0
)
2
<
0
.
Proof.
Note that f1=0 and f2=0 are invariant algebraic curves of system (1.3), with cofactors K1 and K2, respectively. In view of Proposition 3.2, the singular points of system (1.3) are on {f1=0}∪{K1=0} and on {f2=0}∪{K2=0}.
We first compute the singular points on f1=0. We obtain the four points
z
1
,
2
=
(
x
1
,
2
,
y
1
,
2
)
,
z
3
,
4
=
(
x
3
,
4
,
y
3
,
4
)
,
where
(3.2)
x
1
,
2
=
-
a
2
a
1
±
a
3
Δ
0
a
1
2
+
a
3
2
,
y
1
,
2
=
-
a
2
a
3
∓
a
1
Δ
0
a
1
2
+
a
3
2
and
(3.3)
x
3
=
x
4
=
1
-
r
2
+
x
0
2
2
x
0
,
y
3
,
4
=
∓
1
-
(
1
-
r
2
+
x
0
2
)
2
4
x
0
2
.
Computing the singular points on f2=0, we obtain the points z3=(x3,y3), z4=(x4,y4), given in (3.3), and two additional points
z
5
,
6
=
(
x
5
,
6
,
y
5
,
6
)
,
where
(3.4)
{
x
5
,
6
=
-
a
2
a
1
-
2
a
0
a
1
x
0
+
a
3
2
x
0
∓
a
3
Δ
1
a
1
2
+
a
3
2
,
y
5
,
6
=
-
a
2
a
3
-
2
a
0
a
3
x
0
-
a
1
a
3
x
0
±
a
3
Δ
1
a
1
2
+
a
3
2
.
Note that the two singular points z3 and z4 are on the circles f1=f2=0, and thus they are complex.
By the expressions of K1, K2 given in Theorem 1.2, we have that K1=yk~, with k~=a2+a1x+2a0x0+a3y, and K2=yk^, with k^=a2+a1x+a3y. Computing the singular points of system (1.3) on k~=0, we get the points z5=(x5,y5), z6=(x6,y6) given in (3.4). So, these singular points are on f2=0. Now computing the singular points of (1.3) on k^=0, we get the points z1=(x1,y1) and z2=(x2,y2) given in (3.2). So, these singular points are on f1=0. Finally, on y=0, we have that K1=K2=0. Moreover, computing the singular points (x¯,y¯) of (1.3) on y=0, it follows that they satisfy
F
(
x
)
=
B
0
+
B
1
x
¯
+
B
3
x
¯
2
+
B
4
x
¯
3
=
0
.
In summary, conditions (3.1) together with the condition
(3.5)
F
(
1
)
F
(
-
1
)
F
(
x
0
-
r
)
F
(
x
0
+
r
)
=
(
a
1
2
-
a
2
2
)
(
a
2
+
2
a
0
x
0
+
a
1
(
r
+
x
0
)
)
≠
0
are equivalent to say that there are no singular points on the two circles f1=0 and f2=0 (we recall that f1=f2=0 on y=0 are, respectively, x=±1 and x=x0±r). Note that (3.5) is automatically satisfied since the condition Δ0<0 implies that
a
1
2
-
a
2
2
<
a
3
2
+
a
1
2
-
a
2
2
=
Δ
0
<
0
,
and the condition Δ1<0 implies that
(
a
2
+
2
a
0
x
0
+
a
1
(
r
+
x
0
)
)
(
a
2
+
2
a
0
x
0
+
a
1
x
0
-
a
1
r
)
=
(
a
2
+
2
a
0
x
0
+
a
1
x
0
)
2
-
a
1
2
r
2
>
(
a
2
+
2
a
0
x
0
+
a
1
x
0
)
2
-
(
a
1
2
+
a
3
2
)
r
2
=
-
Δ
1
>
0
.
Consequently, a2+2a0x0+a1(r+x0)≠0. ∎
4 The Singular Points
In this section we study the singular points of system (1.3). For this, we introduce the following notations:
y
˙
|
y
=
0
=
F
(
x
)
=
B
0
+
B
1
x
+
B
3
x
2
+
B
4
x
3
,
Δ
2
=
(
r
-
x
0
+
1
)
(
r
-
x
0
-
1
)
(
r
+
x
0
-
1
)
(
r
+
x
0
+
1
)
,
D
0
=
-
B
1
2
B
3
2
+
4
B
0
B
3
3
+
4
B
1
3
B
4
-
18
B
0
B
1
B
3
B
4
+
27
B
0
2
B
4
2
,
D
1
=
B
3
2
-
3
B
4
B
1
.
First we note that by the definition of the forms of f1 and f2 in cases (II) and (III), we readily have that Δ2≠0. Moreover, under the assumptions Δ0<0 and Δ1<0, we have already proved that the unique singular points of system (1.3) are on y=0 and F(x)=0. Now we will study these points under the same assumptions.
We first recall some important observation that will be used throughout the paper.
Lemma 4.1.
For system (1.3), with Δ0<0, Δ1<0 and a0≠0, the following hold:
Its Jacobian matrix on each of its singular points is of the form
(4.1)
L
(
x
)
=
(
∂
x
˙
∂
x
∂
x
˙
∂
y
∂
y
˙
∂
x
∂
y
˙
∂
y
)
|
y
=
0
=
(
0
A
1
+
A
2
x
+
A
4
x
2
B
1
+
2
B
3
x
+
3
B
4
x
2
B
2
-
A
3
x
+
B
2
x
2
)
,
with A1+A2x+A4x2≠0. Moreover, if a3≠0, then B2-A3x+B2x2≠0.
If
a
3
=
0
and
x
~
is a simple solution of
F
(
x
)
=
0
, then the singular point
(
x
~
,
0
)
is either a saddle or a center. If
a
3
=
0
and
x
~
is a multiple solution of
F
(
x
)
=
0
(either double or triple), then the singular point
(
x
~
,
0
)
will be nilpotent.
Proof.
We first prove statement (a). Compute L(x) as in (4.1). Doing the resultant between A1+A2x+A4x2 and F(x) with respect to x, we get
-
a
0
16
x
0
2
(
a
1
2
-
a
2
2
)
(
a
2
+
2
a
0
x
0
+
a
1
x
0
+
a
1
r
)
(
a
2
+
2
a
0
x
0
+
a
1
x
0
-
a
1
r
)
Δ
2
=
a
0
16
x
0
2
Δ
~
0
Δ
~
1
Δ
2
,
where Δ~0 and Δ~1 are, respectively, Δ0 and Δ1 restricted to a3=0. By the assumptions, we have Δ0<0, Δ1<0, a0≠0, and, by definition, Δ2≠0. So, A1+A2x+A4x2≠0 at the singular points.
If a3≠0, then doing the resultant between B2-A3x+B2x2 and F(x) with respect to x, we get
a
0
2
a
3
2
r
2
Δ
2
8
x
0
2
,
which is different from zero because, by assumption, a0≠0 and, by definition, Δ2x0≠0. So, when a3≠0, B2-A3x+B2x2≠0 at the singular points. This concludes the proof of statement (a).
To prove statement (b), note that if a3=0, then B2=A3=0 and, in view of statement (a), the singular points can be either hyperbolic or nilpotent, but not linearly zero. When a3=0, system (1.3) becomes
(4.2)
{
x
˙
=
A
1
y
+
A
2
x
y
+
A
4
x
2
y
+
A
6
y
3
,
y
˙
=
F
(
x
)
-
B
0
y
2
+
B
5
x
y
2
.
In view of statement (a), the Jacobian matrix at a singular point (x~,0), with F(x~)=0, is of the form (4.1), with A1+A2x~+A4x~2≠0.
If x~ is a simple solution of F(x)=0, then B1+2B3x~+3B4x~2=dF(x)/dx|x=x~≠0. Therefore, the singular point (x~,0) will be either a saddle, a focus or a center. Translating the singular point at the origin, by making the change X=x-x~, Y=y, we see that system (4.2) in the variables (X,Y) is reversible (see Lemma 3.1 (a)), and so the singular point (x~,0) will be either a saddle or a center.
If x~ is a multiple solution of F(x)=0, then B1+2B3x~+3B4x~2=dF(x)/dx|x=x~=0, and the singular point (x~,0) will be nilpotent (see (4.1), taking into account that a3=0). This proves statement (b). ∎
When B4≠0, we will write F(x)=B4G(x) and we will talk about the zeros of G(x)=0.
We shall need the following result which is the part of the Andreev’s theorem [1] for nilpotent singular points, see also [4, Theorem 3.5].
Theorem 4.2.
Let (0,0) be an isolated nilpotent singular point of the vector field X given by
(4.3)
x
˙
=
y
+
A
(
x
,
y
)
,
y
˙
=
B
(
x
,
y
)
,
where A,B are analytic in a neighborhood of (0,0) and also Ax(0,0)=Ay(0,0)=Bx(0,0)=By(0,0)=0. Let y=f(x) be the solution of y+A(x,y)=0 in a neighborhood of (0,0), and consider F(x)=B(x,f(x)) and G(x)=(∂A/∂x+∂B/∂y)(x,f(x)). Assume F(x)=axm+o(xm), with a≠0, and G(x)=0. If m is odd and a>0, then (0,0) is a saddle. If m is odd and a<0, then (0,0) is a focus or a center. If m is even, then (0,0) is a cusp.
Proposition 4.3.
System (1.3), with (x˙,y˙) not having a common factor, a3=0, a0≠0, Δ1<0 and Δ2<0 has the following singular points:
If
B
4
=
0
and
B
3
=
0
, then when
B
1
=
0
, it has no singular points, and when
B
1
≠
0
, it has the unique singular point
(
z
0
,
0
)
, where
(4.4)
z
0
=
-
B
0
B
1
,
and it is either a saddle or a center. We note that
B
0
=
B
1
=
B
3
=
B
4
=
0
is not possible because then
(
x
˙
,
y
˙
)
in system (
1.3
) will have
y
as a common factor.
Assume B4=0 and B3≠0. If (II) holds, or (III) holds with B12-4B0B3>0, then it has two singular points (z1,0) and (z2,0), with
(4.5)
z
1
,
2
=
-
B
1
±
B
1
2
-
4
B
0
B
3
2
B
3
.
Both singular points are either saddles or centers. If (III) holds with B12-4B0B3≤0, then either f1=0, or f2=0 cannot be a limit cycle.
If
B
4
≠
0
and
D
0
=
D
1
=
0
, then it has a unique multiple real solution
z
3
of
G
(
x
)
=
0
. The singular point
(
z
3
,
0
)
is either a saddle or a center.
If B4≠0, D0=0 and D1≠0, then G(x)=0 has two distinct real solutions (one simple z4 and one double z5). The singular point (z4,0) is a saddle or a center, and the singular point (z5,0) is a cusp.
If
B
4
≠
0
and
D
0
<
0
,
then
G
(
x
)
=
0
has three distinct simple real solutions
z
6
,
z
7
,
z
8
. All of them can only be saddles or centers.
If
B
4
≠
0
and
D
0
>
0
, then
G
(
x
)
=
0
has a unique simple real solution
z
9
. The singular point
(
z
9
,
0
)
is either a saddle or a center.
Proof.
The proof follows by direct calculations. Note that when B4=0, equation F(x)=0 becomes the quadratic equation B3x2+B1x+B0=0.
If B3=0, then when B1=0, it has no solutions, and when B1≠0, the solution is as given in (4.4). It is a simple solution of F(x)=0 and, by Lemma 4.1 (b), it is either a saddle or a center. This proves statement (a).
If B4=0 and B3≠0, then F(x)=0 has the two solutions (4.5), where
B
1
2
-
4
B
0
B
3
=
16
a
0
2
r
2
x
0
2
+
a
2
2
Δ
2
4
x
0
2
.
It is easy to see that if we are under assumptions (II), then Δ2>0, and if we are under assumptions (III), then Δ2<0. So, under assumptions (II), the solutions z1 and z2 are simple because B12≠4B0B3>0 and, by Lemma 4.1 (b), both solutions can only be saddles or centers. Under conditions (III), if B12-4B0B3>0, then again z1 and z2 are simple and they are either saddles or centers. On the other hand, if B12-4B0B3≤0, then either we have no singular points or one singular point. But in this case, since none of the circles f1=0 and f2=0 contains the other in the bounded region that they delimit, and since inside one limit cycle there must be a singular point, we conclude that either f1=0, or f2=0 cannot be a limit cycle. This proves statement (b).
If B4≠0 and D0=D1=0, we have a unique triple solution for G(x)=0, that we call z3. This means that F(x)=B4(x-z3)3. With a parametrization of the time, we can rewrite equation (4.2) as
(4.6)
{
x
˙
=
A
~
1
y
+
A
~
2
x
y
+
A
~
4
x
2
y
+
A
~
6
y
3
,
y
˙
=
(
x
-
z
3
)
3
-
B
~
0
y
2
+
B
~
5
x
y
2
,
where A~i=Ai/B4 and B~j=Bj/B4 for i=1,2,4,6 and j=0,5. We introduce the change of variables X=x-z3 and Y=y. Then system (4.6) becomes
(4.7)
{
x
˙
=
A
^
1
Y
+
A
^
2
X
Y
+
A
^
4
X
2
Y
+
A
~
6
Y
3
,
y
˙
=
X
3
-
B
^
0
y
2
+
B
~
5
X
Y
2
,
where A^1=A~1+A~2z3+A~4z32, A^2=A~2+2A~4z3 and B^0=B0-B5z3. Note that, in view of Lemma 4.1 (a), we have that A^1≠0. Now we introduce a scaling of time of the form τ=t/A^1, and system (4.7) becomes
(4.8)
{
x
˙
=
Y
+
A
¯
2
X
Y
+
A
¯
4
X
2
Y
+
A
¯
6
Y
3
,
y
˙
=
1
A
^
1
X
3
-
B
¯
0
Y
2
+
B
¯
5
X
Y
2
,
where
A
¯
2
=
A
^
2
A
^
1
,
A
¯
4
=
A
^
4
A
^
1
,
A
¯
6
=
A
~
6
A
^
1
,
B
¯
0
=
B
^
0
A
^
1
,
B
¯
5
=
B
~
5
A
^
1
.
Note that (0,0) is a nilpotent singular point and that system (4.8) is precisely system (4.3), with
A
(
X
,
Y
)
=
A
¯
2
X
Y
+
A
¯
4
X
2
Y
+
A
¯
6
Y
3
,
B
(
X
,
Y
)
=
1
A
^
1
X
3
-
B
¯
0
Y
2
+
B
¯
5
X
Y
2
.
Under the notation of Theorem 4.2, with (x,y) replaced by (X,Y), we have that the solution Y=f(X) of Y+A(X,Y)=0 is Y=f(X)=0. Then F(X)=B(X,0)=X3/A^1 and G(X)=Y(A¯2+2A¯4X-2B¯0+2B¯5X)Y=0=0. So, we are under the assumptions of Theorem 4.2, with m=3 and a=1/A^1. In view of Theorem 4.2, we have that it is either a saddle, a focus or a center. Since the system is reversible, it can only be a saddle or a center. This proves statement (c).
If B4≠0, D0=0 and D1≠0, then we have a single solution and a double solution for G(x)=0 that we call, respectively, z4 and z5. The solution z4 is simple and thus, by Lemma 4.1 (b), it is either a saddle or a center.
The solution z5 is double. This means that F(x)=B4(x-z4)(x-z5)2. With a parametrization of the time, we can rewrite equation (4.2) as
(4.9)
{
x
˙
=
A
~
1
y
+
A
~
2
x
y
+
A
~
4
x
2
y
+
A
~
6
y
3
,
y
˙
=
(
x
-
z
4
)
(
x
-
z
5
)
2
-
B
~
0
y
2
+
B
~
5
x
y
2
,
where A~i=Ai/B4 and B~j=Bj/B4 for i=1,2,4,6 and j=0,5. We introduce the change of variables X=x-z5 and Y=y. Then system (4.9) becomes
(4.10)
{
x
˙
=
A
^
1
Y
+
A
^
2
X
Y
+
A
^
4
X
2
Y
+
A
~
6
Y
3
,
y
˙
=
(
z
5
-
z
4
)
X
2
+
X
3
-
B
^
0
Y
2
+
B
~
5
X
Y
2
,
where A^1=A~1+A~2z5+A~4z52, A^2=A~2+2A~4z5 and B^0=B0-B5z5. Note that, in view of Lemma 4.1 (a), we have that A^1≠0. Now we introduce a scaling of time of the form τ=t/A^1 and system (4.10) becomes
(4.11)
{
x
˙
=
Y
+
A
¯
2
X
Y
+
A
¯
4
X
2
Y
+
A
¯
6
Y
3
,
y
˙
=
z
5
-
z
4
A
^
1
X
2
+
1
A
^
1
X
3
-
B
¯
0
Y
2
+
B
¯
5
X
Y
2
,
where
A
¯
2
=
A
^
2
A
^
1
,
A
¯
4
=
A
^
4
A
^
1
,
A
¯
6
=
A
~
6
A
^
1
,
B
¯
5
=
B
^
5
A
^
1
,
B
¯
8
=
B
~
8
A
^
1
.
Note that (0,0) is a nilpotent singular point and that system (4.11) is precisely system (4.3) with
A
(
X
,
Y
)
=
A
¯
2
X
Y
+
A
¯
4
X
2
Y
+
A
¯
6
Y
3
,
B
(
X
,
Y
)
=
z
5
-
z
4
A
^
1
X
2
+
1
A
^
1
X
3
-
B
¯
0
Y
2
+
B
¯
5
X
Y
2
.
The solution Y=f(X) of Y+A(X,Y)=0 is Y=f(X)=0. Then
F
(
X
)
=
B
(
X
,
0
)
=
z
5
-
z
4
A
^
1
X
2
+
1
A
^
1
X
3
and G(X)=Y(A¯2+2A¯4X-2B¯0+2B¯5X)Y=0=0. So we are under the assumptions of Theorem 4.2 with m=2 and a=(z5-z4)/A^1. In view of Theorem 4.2, we have that it is a cusp. This proves statement (d).
If B4≠0 and D0<0, then G(x)=0 has three simple real solutions z6, z7 and z8, and so, by Lemma 4.1 (b), they can only be saddles or centers. This proves statement (e).
If B4≠0 and D0>0, then G(x)=0 has a unique simple real solution that we call z9. Again, by Lemma 4.1 (b), it can only be a saddle or a center. This proves statement (f). ∎
Note that, in view of Proposition 4.3, all singular points of system (1.3) with a3=0 have either topological index -1 (and then they are saddles), 0 (and they can only be cusps) or 1 (and they can only be centers). Now we use this information to obtain a restriction on the shape of the singular points when f1=0 and f2=0 are both limit cycles of system (1.3) with a3=0.
To prove Corollary 4.5, we shall need the following result. For a proof, see, for instance, [4, Propositions 6.7 and 6.26, and Example 6.17].
Proposition 4.4.
In the interior of a limit cycle there must be singular points so that the sum of their topological indices is one.
Corollary 4.5.
System (1.3) with a3=0 and having f1=0 and f2=0 as limit cycles can have only three real singular points (-x,0), (0,0) and (x,0). Moreover, (±x,0) are centers and (0,0) is a saddle.
Proof.
In order to prove that f1=0 and f2=0 are limit cycles, we must have that a0≠0, Δ0<0 and Δ1<0, see Proposition 3.3. Also inside the proof of Proposition 3.3, we showed that the singular points of system (1.3) are of the form (z,0), with F(z)=0. In view of Lemma 3.1 (a), system (1.3) with a3=0 is symmetric with respect to the y-axis. So, if system (1.3) has a unique singular point, then this singular point must be the origin (0,0). By Proposition 4.4, it must have topological index one and, by the statements of Proposition 4.3, it must be a center. Consequently, by Lemma 3.1 (b), the circle surrounding this singular point cannot be a limit cycle. So, statements (a), (c) and (f) of Proposition 4.3 cannot hold.
Now assume that there are only two singular points of system (1.3) with a3=0. By symmetry, they are of the form (±x,0), and both of them must have the same index. In view of Proposition 4.3, the only possibility is statement (b) because in statement (d) both singular points have different index. But statement (b) is also not possible, because both points have topological index either 1 or -1, so in view of Proposition 4.4, in at least inside the region delimited by one of the circles f1=0 or f2=0, there is a unique singular point with topological index 1. Therefore, by statement (b) of Proposition 4.3, such a singular point must be a center, and consequently, by Lemma 3.1 (b), that circle cannot be a limit cycle. So statement (b) cannot hold.
In short, the unique possibility is that statement (e) of Proposition 4.3 holds, i.e., there are three singular points of system (1.3) with a3=0. By symmetry, these singular points must be of the form (±x,0) and (0,0). Moreover, (±x,0) have the same index. In view of statement (e) of Proposition 4.3 and the existence of the two invariant circles which are periodic orbits, we must have that (±x,0) are centers and (0,0) is a saddle. This completes the proof of the corollary. ∎
Now we study the singular points of system (1.3) when a3≠0. In this case, in view of Lemma 4.1 (a), the singular points can be either hyperbolic or semi-hyperbolic. More concretely, let x~ be a double zero of F(x). Then B1+2B3x~+3B4x~2=0, and the singular point (x~,0) is either a saddle, a node or a saddle-node. Now we introduce the notation
H
0
(
x
)
=
(
A
1
+
A
2
x
+
A
4
x
2
)
(
B
1
+
2
B
3
x
+
3
B
4
x
2
)
,
H
1
(
x
)
=
(
B
2
-
A
3
x
+
B
2
x
2
)
2
4
+
H
0
(
x
)
.
If x~ is a simple zero of F(x), then the singular point (x~,0) is a saddle if H0(x~)>0, a node if H1(x~)≥0, and a focus if H1(x~)<0.
Proposition 4.6.
System (1.3) with (x˙,y˙) not having a common factor, a3≠0, a0≠0, Δ0<0 and Δ1<0 has the following singular points:
If B4=0 and B3=0, then when B1=0, it has no singular points, and when B1≠0, it has the unique singular point (z0,0), where z0 is given in (4.4). The point z0 is a saddle if H0(z0)>0, a node if H1(z0)≥0, and a focus if H1(z0)<0.
If B4=0 and B3≠0, then it has the two singular points (z1,0) and (z2,0), with z1 and z2 given in (4.5) when either (II) holds, or (III) holds with B12-4B0B3>0. For j=1,2, zj is a saddle if H0(zj)>0, a node if H1(zj)≥0, and a focus if H1(zj)<0. If (III) holds with B12-4B0B3≤0, then either f1=0 or f2=0 cannot be a limit cycle.
If
B
4
≠
0
and
D
0
=
D
1
=
0
, then
F
(
x
)
=
0
has a unique triple solution
z
3
. The singular point
(
z
3
,
0
)
is semi-hyperbolic, and consequently it is either a saddle, a node or a saddle-node.
If B4≠0, D0=0 and D1≠0, then F(x)=0 has two distinct real solutions (one simple z4 and one double z5). The singular point (z4,0) is a saddle if H0(z4)>0, a node if H1(z4)≥0, and a focus if H1(z4)<0. The singular point (z5,0) is semi-hyperbolic.
If B4≠0 and D0>0, then F(x) has three distinct simple real solutions z6,z7,z8. The singular point (zj,0), for j=6,7,8, is a saddle if H0(zj)>0, a node if H1(zj)≥0, and a focus if H1(zj)<0.
If B4≠0 and D0<0, then it has a unique simple real solution z9. The singular point (z9,0) is a saddle if H0(z9)>0, a node if H1(z9)≥0, and a focus if H1(z9)<0.
Proof.
The proof follows by direct calculations. Note that when B4=0, equation F(x)=0 becomes the quadratic equation B3x2+B1x+B0=0. If B3=0, then when B1=0, it has no solutions (note that B0 cannot be zero, otherwise the differential system (x˙,y˙) has the common factor y). If B1≠0, then the solution is the z0 given in (4.4). It is a simple solution of F(x)=0, and so it is hyperbolic. Therefore, by the explanation before the statement of Proposition 4.6, we have that it is a saddle if H0(z0)>0, a node if H1(z0)≥0, and a focus if H1(z0)<0. This proves statement (a).
If B4=0 and B3≠0, then F(x)=0 has the two solutions given in (4.5). Proceeding as in the proof of Proposition 4.3 (b), we conclude that only when condition (III) holds with B12-4B0B3≤0 both solutions collide or disappear, so they are never simple solutions. But, in this case, the same arguments imply that either f1=0, or f2=0 cannot be a limit cycle. On the other cases, z1 and z2 are hyperbolic. Therefore, for j=1,2, zj is a saddle if H0(zj)>0, a node if H1(zj)≥0, and a focus if H1(zj)<0. This proves statement (b).
If B4≠0 and D0=D1=0, we have a unique triple solution z3 for F(x)=0. In this case the singular point (z3,0) is semi-hyperbolic and, by [4, Theorem 2.19], it is either a saddle (if it has index -1), a node (if it has index 1) or a saddle-node (if it has index 0). This proves statement (c).
If B4≠0, D0=0 and D1≠0, we have a simple solution z4 and a double solution z5 for F(x)=0. The singular point (z5,0) is semi-hyperbolic, and thus it is either a saddle, a node or a saddle-node. The singular point (z4,0) is simple and it can be a saddle if H0(z4)>0, a node if H1(z4)≥0, and a focus if H1(z4)<0. This proves statement (d).
The proof of statements (e) and (f) are similar to the previous ones. ∎
,
Jaume Llibre
and
Claudia Valls
