# Uniqueness of Limit Cycles for a Class of Cubic

Authors: Xiangdong Xie; Fengde Chen; Qingyi   Zhan
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

A class of cubic systems with two invariant straight linesdx/dt=y(1-x2),dy/dt=-x+δy+nx2+mxy+ly2+bxy2. is studied. It is obtained that the focal quantities of O(0,0) are, W0=δ; if W0=0, then W1=m(n+l); if W0=W1=0, then W2=−nm(b+1); if W0=W1=W2=0, then O is a center, and it has been proved that the above mentioned cubic system has at most one limit cycle surrounding weak focal O(0,0). This paper also aims to solve the remaining issues in the work of Zheng and Xie (2009).


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## Body

## 1. Introduction

The study of the polynomial differential system attracts more and more researchers because of the Hilbert's 16th problem [1–5]. The major problem of the polynomial differential system is to calculate the highest order of focal quantities (also known as focal values, or Lyapunov exponents) at its focal and to decide how many limit cycles surrounding a singular point; the system generated at least under some perturbation of coefficients. All this problem is still open.There are many papers to study the Kukles system, and many achievements are reached, which include the calculation of the focal quantities and decision of the maximum number or limit cycles of the system. Such as paper [5], Hill et al. had studied a class of cubic differential systems and also brought to our attention that a system used to model predator-prey interactions with intratrophic predation could be transformed so that it can be an example of a system of type (1.1).In papers [6, 7], the authors consider a class of cubic Kukles systems E31:(1.1)dxdt=y,dydt=-x+δy+nx2+mxy+ly2+bxy2,
where b≠0. Paper [6] has proved that the focal quantities of O(0,0) in (1.1) are W0=δ,W1=m(n+l),W2=-mnb, if W0=W1=W2=0, O is a center, and if W0=0, system (1.1) has at most one limit cycle surrounding O.Paper [8] considers a class of cubic systems:(1.2)dxdt=y(1+x2),dydt=-x+δy+nx2+mxy+ly2+bxy2,
where b≠0,1, and has proved that the focal quantities of O(0,0) in (1.2) are, W0=δ,W1=m(n+l),W2=-mn(b-1), if W0=W1=W2=0, O is a center, and if W0=0, system (1.2) has at most one limit cycle surrounding O. But in Paper [8], the case b=1 is not considered.Recently, in paper [9], the authors consider a class of cubic systems:(1.3)dxdt=y(1-x),dydt=-x+δy+nx2+mxy+ly2+bxy2,
where b≠0. Paper [9] has proved that the focal quantities of O(0,0) in (1.3) are, W0=δ,W1=m(n+l),W2=-mnb. If W0=W1=W2=0, O is a center, and if W0=0, system (1.3) has at most one limit cycle surrounding O.In this paper, we consider the following cubic system(1.4)dxdt=y(1-x2),dydt=-x+δy+nx2+mxy+ly2+bxy2.
It comes from system (1.1) by adding some invariant straight line, so system (1.4) is said to be the accompany system of (1.1), and it is also said that system (1.1) and (1.4) is part of the accompany system. Paper [10] introduces the concept of accompany system, and studies the qualitative property of some accompany system.Now without loss of generality, we may assume thatm≤0 (if m≥0, let (x,y,t)→(x,-y,-t), then m change its sign), and may assume that l≥0 (if l≤0, let (x,y,t)→(-x,y,-t), then l change its sign, but m does not change its sign). So we study the system (1.4) with m≤0,l≥0.System (1.4) has a critical point O(0,0) and N(1/n,0), if n≠0, and other critical points (if have) lie on the invariant straight line x=±1. Now we transform (1.4) into Lienard equation; note that (1.5)dxdt=-(x2-1)y≡f0(x)-f1(x)y,dydt=x(nx-1)+(mx+δ)y+(l+bx)y2≡g0(x)+g1(x)y+g2(x)y2.Letx=x,ξ=f0(x)-f1(x)y, then (1.5) can be reduced to (1.6)dxdt=ξ,dξdt=-φ0(x)-φ1(x)ξ-φ2(x)ξ2,
where φ0(x)=f1(x)g0(x), φ1(x)=-g1(x), φ2(x)=(g2(x)-f1′(x))/f1(x), and let (1.7)x=x,ξ=ue-∫axψ2(s)ds,x=x,y=u+∫axφ1e∫asφ2(z)dzds,dtdτ=e∫axφ2(s)ds.
System (1.4) can be reduced to (1.8)dxdτ=y-∫axφ1(s)e∫asφ2(z)dzds≡y-F(x),dydτ=-φ0e2∫axφ2(s)ds≡-g(x).


## 2. The Problem of the Center or Focal for Critical PointO(0,0)

In this section, since we will study the problem of center or focus for critical pointO(0,0), we take a=0 in system (1.8). In order to calculate the focal quantities of system (1.4) (or system (1.8)) in O, we need to let δ=0, and only consider, |x|≪1, that is |x-1|=1-x, so (2.1)∫0xφ2(s)ds=∫0xl+bs-2ss2-1ds=b+l2ln|x-1|+b-l2ln|x+1|-ln|x2-1|.So(2.2)F'(x)=f(x)=-(δ+mx)(1-x)(b+l-2)/2(1+x)(b-l-2)/2,g(x)=x(1-nx)(1-x)b+l-1(1+x)b-l-1.We use method of paper [11], so f(x),g(x) need to be written in the power series as follows: (2.3)f(x)≡F'(x)=-mx+mlx2-12m(l2-b+2)x3+16m(l3+8l-3bl)x4+⋯≡b1x+b2x2+b3x3+b4x4⋯,g(x)=x-(2l+n)x2+(2l2-b-n+2nl+1)x3-13(4l3+8l+6nl2+3n-6bl-3nb)x4+⋯≡C0x+C1x2+C2x3+C3x4+⋯.
We use mark in paper [11]; let cn-2=Cn-2/nC0,n=3,4,…, and βm=bm+1/(m+2)-cmb1, then β1=b2/3-c1b1=b2/3-C1b1/3=-m(n+l)/3,g3=2β1=-2m(n+l)/3. If δ=g3=0, then (2.4)β2=b34-c2b1=b34-C2b14=m(3l2-b+4nl)8,β3=b45-c3b1=b45-C3b15=m30(9bl-7l3-8l-12nl2+6nb-6n),
since δ=g3=0, that is, m(n+l)=0, so (2.5)β2=m(nl-b)8,β3=m30(3bl-5nl2-2l),g5=-8c1β2+2β3=-83β2+2β3=2mn15(b+1).
From paper [11], if δ=0,g3>0(g3<0), then O is stable(unstable) weak focal of order one. So for system (1.4), W0=δ; if W0=0, then W1=m(n+l), and if W1>0(W1<0), then O is unstable (stable) weak focal of order one; If W0=W1=0, that is, δ=m(n+l)=0, then W2=-mn(b+1). (note that g3 and W1, g5 and W2 have opposite signs). If W0=W1=W2=0 and b+1≠0, we will prove that O(0,0) is a center of system (1.4). Since W0=W1=W2=0,b+1≠0, that is, δ=m(n+l)=mn=0,b+1≠0, so δ=m=0, or δ=n=l=0. If δ=m=0, then system (1.4) can be reduced to (2.6)dxdτ=y(1-x2)≡P(x,y),dydτ=-x+nx2+ly2+bxy2≡Q(x,y).
This system is symmetry about x-axis because of P(x,-y)=-P(x,y),Q(x,-y)=Q(x,y), so O is a center. If δ=n=l=0, then system (1.4) can be reduced to (2.7)dxdτ=y(1-x2),dydτ=-x+mxy+bxy2.
It is integrable system, so O is a center, hence we have the following theorem.Theorem 2.1.
For system (1.4), let b+1≠0; the focus quantities of O(0,0) are W0=δ; if W0=0, then W1=m(n+l); if W0=W1=0, then W2=-nm(b+1); if W0=W1=W2=0, then O is a center. If W0>0(W0<0), or W0=0,W1>0(W1<0), or W0=W1=0,W2>0(W2<0), then O is an unstable (stable) critical point. If 0<|W0|≪|W1|≪|W2|≪1,W0W1<0,W1W2<0, then system (1.4) has at least two limit cycles surrounding O.


## 3. Nonexistence of Limit Cycle SurroundingO(0,0)

In this section, we study the nonexistence of limit cycle surrounding the weak focalO(0,0). O is weak focal if and only if δ=0, so we let δ=0 in system (1.4).Lemma 3.1.
Ifm=0, system (1.4) has no limit cycles surrounding O.Proof.
Since system (1.4) forms a generalized rotated vector field with respect to parameter δ(refer to paper [12, page 241]), and when δ=m=0, O is a center, so when m=0,δ≠0, system (1.4) has no limit cycles (refer to paper [12, page 244, th. 3.1]).
By Lemma3.1, we let m<0 in the following.
Now we change (1.4) to lienard equation by (1.8):(3.1)dxdτ=y-∫0xφ1(s)e∫0sφ2(z)dzds≡y-F(x),dydτ=-φ0e2∫0xφ2(s)ds≡-g(x).
Since the limit cycle of (1.4) surrounding O(0,0) must lay in 1-x2>0,1-nx>0. Let f(x)≡F'(x), then(3.2)f(x)=-mx1-x2e∫0x((l+bs)/(s2-1))ds,g(x)=x(1-nx)1-x2e2∫0x((l+bs)/(s2-1))ds,fg=-me-∫0x((l+bs)/(s2-1))ds(1-nx),(fg)′=me-∫0x((l+bs)/(s2-1))ds(1-nx)2(1-x2)W(x),
where
(3.3)W(x)=n(b+1)x2+(nl-b)x-(n+l).
Now we define the curveH and L as follows:(3.4)H:F(u)=F(v);L:f(u)g(u)=f(v)g(v),-1<v<0<u<1,nu<1,nv<1.
It easy to see thatL,H are continuously differentiable.
Now Let(3.5)K=K(u,v)=dudv|L-dudv|H=(f(v)g(v))′/(f(u)g(u))′-f(v)f(u)=e-∫0v(l+bs/s2-1)ds(1-nu)2(1-u2)[n(b+1)v2+(nl-b)v-(n+l)]e-∫0u(l+bs/s2-1)ds(1-nv)2(1-v2)[n(b+1)u2+(nl-b)u-(n+l)]-v(1-u2)e∫0v((l+bs)/(s2-1))dsu(1-v2)e∫0u((l+bs)/(s2-1))ds.
IfH and L intersect in P(u,v), then in P, f(u)/g(u)=f(v)/g(v), that is,(3.6)(1-nu)e∫0u((l+bs)/(s2-1))ds=(1-nv)e∫0v((l+bs)/(s2-1))ds,
so in the intersecting point P of H and L,
(3.7)K=K(u,v)=(1-nu)(1-u2)[n(b+1)v2+(nl-b)v-(n+l)](1-nv)(1-v2)[n(b+1)u2+(nl-b)u-(n+l)]-v(1-u2)(1-nu)u(1-v2)(1-nv),
that is,
(3.8)K=K(u,v)=(1-nu)(1-u2)(v-u)[n(b+1)uv+n+l]u(1-nv)(1-v2)[n(b+1)u2+(nl-b)u-(n+l)].Theorem 3.2.
IfW0=0,b+1≠0,andW1W2≥0, then system (1.4) or (3.1) has no limit cycle surrounding O.Proof.
SinceW0=δ,W1W2=-m2n(b+1)(n+l), and we have supposed that m<0, so n(b+1)(n+l)≤0. Now we will prove that system (3.1) has no limit cycles under the conditions of δ=0,b+1≠0,n(b+1)(n+l)≤0.
First supposingn(b+1)>0,n+l<0, we will prove that H and L do not S intersect (S intersect means that H from one side of L astride to another side at intersect point).
(1) If W(x) does not change its sign when x2<1,1-nx>0, then (f/g)'≤0 (equal sign only for some x, the same as below), so f(u)/g(u)<f(v)/g(v), for any -1<v<0<u<1, it means that L does not exist, therefore, L and H do not S intersect.
(2) If W(x) change its sign when x2<1,1-nx>0, and W′(0)>0, then W(x)=0 have one or two real roots in -1<x<0,1-nx>0 (If real roots do not exist, then similar to (1), L does not exist); then the curve y=f(x)/g(x) is shown in Figure 1, and the relative position of curve H and L is shown in Figure 2 (If only part of L exists, it does not influence the proof, the same as below). If H and L have an S intersection point P1(u1,v1) (the first S intersecting point from O), then by Figure 2(3.9)K1≡dudv|L-dudv|H≤0
but since n(b+1)u1v1<0,b+l<0,W(u1)>0(see Figure 1), 1-nu1>0,1-nv1>0, so
(3.10)K1≡K1≡dudv|L-dudv|H=(1-nu1)(1-u12)(v1-u1)[n(b+1)u1v1+n+l]u1(1-nv1)(1-v12)W(u1)>0.
It is a contradiction, so L and H do not S intersect.
(3) If W(x) change its sign when x2<1,1-nx>0, and W'(0)<0, then W(x)=0 have one or two real roots in 0<x<1,1-nx>0, then the curve y=f(x)/g(x) is shown in Figure 3, and the relative position of curves H and L is shown in Figure 4. If H and L have an S intersection point P1(u1,v1)(the first S intersecting point from O), then by Figure 4(3.11)K1≡dudv|L-dudv|H≥0.
Since n(b+1)u1v1<0,b+l<0,1-nu1>0,1-nv1>0, and from the fact that P1(u1,v1) is a intersection point late to O, we have W(u1)<0, so
(3.12)K1≡dudv|L-dudv|H=(1-nu1)(1-u12)(v1-u1)[n(b+1)u1v1+n+l]u1(1-nv1)(1-v12)W(u1)<0.
It is a contradiction, so L and H do not S intersect.
Note that ifW(x) change its sign in x2<1,1-nx>0, then W'(0)≠0, because if W'(0)=0, then W(x)=n(b+1)x2-(n+l)≥0.
Secondly, supposingn(b+1)<0,n+l>0, we will prove that H and L do not S intersect.
(1)′ If W(x) does not change its sign in x2<1,1-nx>0, then (f/g)'≥0,x2<1,1-nx>0, so f(u)/g(u)>f(v)/g(v), for any -1<v<0<u<1, it means that L does not exist, hence L and H do not S intersect.
(2)′ If W(x) change its sign when x2<1,1-nx>0, and W'(0)<0, then W(x)=0 have one or two real roots in -1<x<0,1-nx>0; then the curve y=f(x)/g(x) is shown in Figure 5, and the relative position of curves H and L is shown in Figure 2. If H and L have an S intersection point P1(u1,v1)(the first S intersecting point from O), then by Figure 2(3.13)K1≡dudv|L-dudv|H≤0.
But since n(b+1)u1v1>0,b+l>0,W(u1)<0 (see Figure 5), 1-nu1>0,1-nv1>0, so
(3.14)K1≡dudv|L-dudv|H=(1-nu1)(1-u12)(v1-u1)[n(b+1)u1v1+n+l]u1(1-nv1)(1-v1)W(u1)>0.
It is a contradiction, so L and H do not S intersect.
(3)′ If W(x) change its sign in x2<1,1-nx>0, and W'(0)>0, then the curve y=f(x)/g(x) is shown in Figure 6; then the relative position of curves H and L is similar to Figure 4. If H and L have an S intersecting point P1(u1,v1) (the first S intersecting point from O), then by Figure 4(3.15)K1≡dudv|L-dudv|H≥0
but since W(u1)>0(see Figure 6), and n(b+1)u1v1>0,n+l>0, so
(3.16)K1≡(1-nu1)(1-u1)(v1-u1)[nbu1v1+n+l]u1(1-nv1)(1-v1)W(u1)<0.
It is a contradiction, so L and H do not S intersect.
Note that ifW(x) change its sign in x2<1,1-nx>0, then W'(0)≠0. Because if W'(0)=0, then W(x)=n(b+1)x2-(n+l)≤0.
As far, we have proved thatH and L have no S intersection if W1W2>0. According to the proof that f(u)/g(u)=f(v)/g(v) and F(u)=F(v) has no intersection to F(u)=F(v),G(u)=G(v) has no intersection in paper [13]; it can be extended to that if f(u)/g(u)=f(v)/g(v) and F(u)=F(v) has no intersection, then F(u)=F(v),G(u)=G(v) has no S intersection. Furthermore, based on the proof of paper [14], if F(u)=F(v),G(u)=G(v) has no S intersection, then the corresponding Lienard equation has no limit cycle.
Now, by [12, 15] we have proved that when W0=0,b+1≠0,W1W2>0, system (1.4) or (3.1) has no limit cycles surrounding O.
Finaly, we consider the case:W0=0,b+1≠0,W1W2=0, that is n(n+l)=0. If δ=0,n=0,n+l=0, then O(0,0) is a center of (1.4); If δ=0,n=0,n+l≠0, or δ=0,n≠0,n+l=0, then L≠H(If L≡H, then by (3.8), K(u,v)≡0). If when δ=0,n=0,n+l≠0, or δ=0,n≠0,n+l=0, L and HS have an intersecting point, then when 0<|n|≪1, or 0<|n+l|≪1 this S intersecting point also exists; it is a contradiction to (1)–(3) and (1)′–(3)′, so when W0=0,b+1≠0,W1W2=0, system (1.4) or (3.1) has no limit cycle surrounding O; this completes the proof of Theorem 3.2.Figure 1
The picture ofy=f(x)/g(x) in (2).Figure 2
Relative position ofL and H in (2).Figure 3
The picture ofy=f(x)/g(x) in (3).Figure 4
Relative position ofL and H in (3).Figure 5
The picture ofy=f(x)/g(x) in (2)′.Figure 6
The picture ofy=f(x)/g(x) in (3)′.Lemma 3.3.
IfW0=0,b+1=0,n+l≠0, system (1.4) or (3.1) has no limit cycle surrounding O.Proof.
We consider five cases in the following:
(1) If nl-b=nl+1=0, then by (3.3), W(x)=n+l≠0, so(3.17)(fg)′=me-∫0x((l+bs)/(s2-1))ds(1-nx)2(1-x2)W(x)≠0;
hence f(u)/g(u)≠f(v)/g(v), for any -1<v<0<u<1; it means that L does not exist; therefore, L and H do not S intersect.
(2) If nl+1>0,n+l>0, by (3.3), W(x)=0,x2<1,1-nx>0 has at most one positive real root, then the curve y=f(x)/g(x) is similar to Figure 6 (but L has at most one extremal point, and part of L exist; it does not influence the proof, the same as below), and the relative position of curve H and L is similar to Figure 4 (only below half of L exists, the same as below). If H and L have an S intersection point P1(u1,v1)(the first S intersecting point from O), then by Figure 4(3.18)K1≡dudv|L-dudv|H≥0
but since n(b+1)u1v1=0,b+l>0,W(u1)>0(see Figure 6), 1-nu1>0,1-nv1>0, so
(3.19)K1≡dudv|L-dudv|H=(1-nu1)(1-u12)(v1-u1)[n(b+1)u1v1+n+l]u1(1-nv1)(1-v12)W(u1)<0.
It is a contradiction, so L and H do not S intersect.
(3) The case nl+1<0,n+l<0 is similar to (2); it is easy to prove that L and H do not S intersect.
(4) If nl+1>0,n+l<0, then W(x)=0,x2<1,1-nx>0 has at most one negative real root, then the curve y=f(x)/g(x) is similar to Figure 1, and the relative position of curves H and L is similar to Figure 2. If H and L have an S intersection point P1(u1,v1)(the first S intersecting point from O), then by Figure 2(3.20)K1≡dudv|L-dudv|H≤0
but since n(b+1)u1v1=0,b+l<0,W(u1)>0,1-nu1>0,1-nv1>0, so
(3.21)K1≡dudv|L-dudv|H=(1-nu1)(1-u12)(v1-u1)[n(b+1)u1v1+n+l]u1(1-nv1)(1-v1)W(u1)>0.
It is a contradiction, so L and H do not S intersect.
(5) The case nl+1<0,n+l>0 is similar to (4); it is easy to prove that L and H do not S intersect. This completes the proof of Lemma 3.3.Lemma 3.4.
IfW0=0,W1=0,b+1=0,O(0,0) is a center of system (1.4).Proof.
If whenW0=0,W1=0,b+1=0,O is a weak focal. We let 0<|b+1|≪1, and O change its stability; then there is a limit cycle surrounding O; this means when W0=0,b+1=0,n+l≠0, system (1.4) or (3.1) has a limit cycle surrounding O; it is a contradiction to Lemma 3.3. It is follows that O(0,0) is a center of system (1.4). By Lemma 3.4, Theorem 2.1 can be recension in the following.Theorem 3.5.
For system (1.4), the focal quantities of O(0,0) are W0=δ; if W0=0, then W1=m(n+l); if W0=W1=0, then W2=-nm(b+1); if W0=W1=W2=0,O is a center. If W0>0(W0<0), W0=0,W1>0(W1<0), or W0=W1=0,W2>0(W2<0),O is an unstable (stable) critical point. If 0<|W0|≪|W1|≪|W2|≪1,W0W1<0,W1W2<0, then system (1.4) has at least two limit cycles surrounding O.


## 4. Uniqueness of Limit Cycle SurroundingO(0,0)

In this section, we study the uniqueness of limit cycle surrounding the weak focalO(0,0); O is weak focal if and only if δ=0, so we let δ=0 in system (1.4), we also suppose m<0.Lemma 4.1.
IfW0=0,W1W2<0, H and L have at most one S intersecting point.Proof.
SinceW1W2=-m2n(b+1)(n+l), so we consider the two care of n(b+1)>0,n+l>0 and n(b+1)<0,n+l<0.Care A.
Letn(b+1)>0,n+l>0, so W(x)=n(b+1)x2+(nl-b)x-(n+l)=0 have two real roots :a,k,a<0<k. Let c<d<a<0<k<e<h, such that (see Figure 7)
(4.1)f(c)g(c)=f(k)g(k),f(d)g(d)=f(e)g(e)=f(0)g(0),f(h)g(h)=f(a)g(a).
Without loss of generality, we suppose thath<1,-1<c and nh<1,nc<1(otherwise, only part of L exist; it does not influence the proof); then the graph of y=f/g is shown in Figure 7, and the relative position of H and L is shown in Figure 8, where A(u=0,v=d),B(u=k,v=c),C(u=h,v=a),D(u=e,v=0). Now we suppose H and L intersect in P1(u1,v1)(the first S intersecting point from O, the same as below, and if P1 does not exist, then H and L do not intersect, so the system has no limit cycle surrounding O). We denoted the curve of L from A to B by L(A,B), and L[A,B](A,B∈L[A,B],A,B∈¯L(A,B)) (also define L(B,C),L(C,D),L[B,C],L[C,D] to see the Figure 8), then the following occurs.
(1) If P1∈L(A,B), and If H and L have a second S intersecting point P2(u2,v2), then from Figure 8(4.2)K1=K1(u1,v1)≤0,K2=K2(u2,v2)≥0.
Since P1∈L(A,B), so 0<u1<k, and W(u1)<0, hence from (3.8) n(b+1)u1v1+n+l<0. Since (du/dv)∣H<0, so P2∈L(A,B) and 0<u1<u2<k, hence W(u2)<0. From n(b+1)u2v2+n+l<n(b+1)u1v1+n+l<0, it is follows that K2=K(u2,v2)<0, this is a contradiction, so H and L have no second S intersecting point P2(u2,v2).
(2) If P1∈L[B,C], since in L[B,C], (du/dv)∣L[B,C]>0,(du/dv)∣H<0, so L and H have only one S intersecting point.
(3) If P1∈L(C,D), and If H and L have a second S intersecting point P2(u2,v2), then from Figure 8(4.3)K1=K1(u1,v1)≥0,K2=K2(u2,v2)≤0.
Since P1∈L(C,D), so k<u1, and W(u1)>0; hence from (3.8), n(b+1)u1v1+n+l<0. Since (du/dv)∣H<0, so P2∈L(C,D) and k<u1<u2, hence W(u2)>0. From n(b+1)u2v2+n+l<n(b+1)u1v1+n+l<0, it is follows that K2=K(u2,v2)>0; this is a contradiction, so H and L have no second S interesting point P2(u2,v2).Care B.
Letn(b+1)<0,n+l<0, so W(x)≡n(b+1)x2+(nl-b)x-(n+l)=0 have two real roots :a,k,a<0<k. Let c<d<a<0<k<e<h, such that (see Figure 9)
(4.4)f(c)g(c)=f(k)g(k),f(d)g(d)=f(e)g(e)=f(0)g(0),f(h)g(h)=f(a)g(a).
Without loss of generality, we suppose thath<1,-1<c and nh<1,nc<1, then the graph of y=f/g is shown in Figure 9, and the relative position of H and L is shown in Figure 10, where A(u=0,v=d),B(u=k,v=c),C(u=h,v=a),andD(u=e,v=0). Now we suppose that H and L interest in P1(u1,v1)(the first S interesting point from O, and if P does not exist, then H and L have no intersecting point, so the system has no limit cycle surrounding O), then the following occurs.
(1)′ If P1∈L(A,B), and If H and L have a second S intersecting point P2(u2,v2), then from Figure 10,(4.5)K1=K1(u1,v1)≤0,K2=K2(u2,v2)≥0.
Since P1∈L(A,B), so 0<u1<k, and W(u1)>0, hence from (3.8), n(b+1)u1v1+n+l>0. Since (du/dv)∣H<0, so P2∈L(A,B) and 0<u1<u2<k, so W(u2)>0. From n(b+1)u2v2+n+l>n(b+1)u1v1+n+l>0, it follows that K2=K(u2,v2)<0; this is a contradiction, so H and L have no second S intersecting point P2(u2,v2).
(2)′ If P1∈L[B,C], since in L[B,C], (du/dv)∣L[B,C]>0,(du/dv)∣H<0, so L and H have only one S intersecting point.
(3)′ If P1∈L(C,D), and If H and L have a second S intersecting point P2(u2,v2), then from Figure 10,(4.6)K1=K1(u1,v1)≥0,K2=K2(u2,v2)≤0.
Since P1∈L(C,D), so k<u1, and W(u1)<0; hence n(b+1)u1v1+n+l>0. Since (du/dv)∣H<0, so P2∈L(C,D) and k<u1<u2, so W(u2)<0. From n(b+1)u2v2+n+l>n(b+1)u1v1+n+l>0, it follows that K2=K(u2,v2)>0; this is a contradiction, so H and L have no second S intersecting point P2(u2,v2).
From (1)–(3) and (1)′–(3)′, we have proved that L and H have at most one S intersecting point under the conditions W0=0,W1W2<0.Figure 7
The picture of curvey=f(x)/g(x).Figure 8
Relative position of curvesL and H.Figure 9
The picture of curvey=f(x)/g(x).Figure 10
Relative position of curvesL and H.Lemma 4.2.
LetW0=0;W1W2<0, if n(b+1)>0, then (Ff/g)'≤0,-1<x<0,nx<1; if n(b+1)<0, then (Ff/g)'≥0,0<x<1,nx<1.Proof.
(4.7)(Ffg)′=f2g+F(fg)′=m2x(1-nx)(1-x2)+F(x)me-∫0x((l+bs)/(s2-1))ds(1-nx)2(1-x2)[n(b+1)x2+(nl-b)x-(n+l)]≡m2x(1-nx)(1-x2)+F(x)me-∫0x((l+bs)/(s2-1))ds(1-nx)2(1-x2)W(x)≡me-∫0x((l+bs)/(s2-1))ds(1-nx)2(1-x2)[mx(1-nx)e∫0x((l+bs)/(s2-1))ds+F(x)W(x)].
Now we only have to prove that(4.8)V(x)≡mx(1-nx)e∫0x((l+bs)/(s2-1))ds+F(x)W(x)≥0,
as n(b+1)>0,0<x<1,nx<1(or V(x)≤0, as -1<x<0,n(b+1)<0,nx<1), since
(4.9)V'(x)=m(1-nx)e∫0x((l+bs)/(s2-1))ds+F(x)(2n(b+1)x+(nl-b)),V′′(x)=m(n(b+1)x2+n+l)x2-1e∫0x((l+bs)/(s2-1))ds+F(x)(2n(b+1)).Care A.
n(b+1)>0, according to W1W2<0, we have n+l>0. Since f(x)>0,x>0, and f(x)<0,x<0, so F(x)>0,x≠0,F(0)=0; hence when 0<x<1, V′′(x)>0, that is, V'(x)>V'(0)=-m>0,x>0, so V(x)>V(0)=0,x>0, it follows that V(x)>0, as 0<x<1.Care B.
n(b+1)<0, according to W1W2<0, we have n+l<0, since F(x)>0,x≠0,F(0)=0; hence when -1<x<0, V′′(x)<0, that is, V'(x)>V'(0)=-m>0,-1<x<0, so V(x)<V(0)=0,-1<x<0, it follows that V(x)<0, as -1<x<0.
Sincef/g→-m>0, as x→0, from Lemmas 4.1-4.2 and paper [1], we have the following theorem.Theorem 4.3.
IfW0=0,W1W2<0, then system (1.4) has at most one limit cycle surrounding O(0,0).


## 5. The Remaining Issues in Paper [8]

In this section we will study the remaining issues in paper [8]. Paper [8] considers a class of cubic system (1.2), where b≠0,1, and has proved that the focal quantities of O(0,0) in (1.2) are W0=δ,W1=m(n+l),W2=-mn(b-1), if W0=W1=W2=0, O is a center, and if W0=0, then system (1.2) has at most one limit cycle surrounding O. But in paper [8], the case b=1 is not considered.Lemma 5.1.
Ifδ=0,b=1,n+l≠0, system (1.2) has no limit cycle surrounding O(0,0).Proof.
Paper [8] has proved that when δ=0,W1W2≥0,L and H in paper [8] do not S intersect. Since b-1=0,n+l≠0, so K(u,v)≠0(to see (1.8) in paper [8]), hence L≠H(if L≡H, then K(u,v)≡0). Now if when δ=0,b=-1,n+l≠0, L and H have an S intersecting point P, then under condition: δ=0,0<|b+1|≪1,n+l≠0, this S intersecting point P also exists, this is a contradiction to above. Hence under condition of Lemma 5.1, L and H do not S intersect, and system (1.2) has no limit cycle surrounding O(0,0).Theorem 5.2.
For system (1.2),the focus quantities of O(0,0) are W0=δ; if W0=0, then W1=m(n+l); if W0=W1=0, then W2=-nm(b-1); if W0=W1=W2=0, then O is a center. If W0>0(W0<0), W0=0,W1>0(W1<0), or W0=W1=0,W2>0(W2<0), then O is unstable (stable) critical point.Proof.
By the theorem 1 of paper [8], we only need to prove that under conditions: δ=0,b=1,n+l=0, O is a center of system (1.2).
If whenδ=0,n+l=0,b-1=0, O is a weak focus (not a center). We let 0<|b-1|≪1, and O change its stability, then there is a limit cycle surrounding O; this means that when W0=0,b-1=0,n+l≠0, system (1.2) has a limit cycle surrounding O; this is a contradiction to Lemma 5.1, it follows that O(0,0) is a center of system (1.2).



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