# The Number of Limit Cycles of a Polynomial System on the Plane

Authors: Chao Liu; Maoan Han
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We perturb the vector fieldx˙=-yC(x,y), y˙=xC(x,y) with a polynomial perturbation of degree n, where C(x,y)=(1-y2)m, and study the number of limit cycles bifurcating from the period annulus surrounding the origin.


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## Body

## 1. Introduction and Main Result

The main task in the qualitative theory of real plane differential systems is to determine the number of limit cycles, which is related to Hilbert's 16th problem as well as weakened Hilbert's 16th problem, posed by Arnold in [1].Consider a planar system of the form(1)x˙=-yC(x,y)+εP(x,y),y˙=xC(x,y)+εQ(x,y),whereP, Q, and C are real polynomials, C(0,0)≠0, and ε is a small real parameter. It is well known that the number of zeros of the Abelian integral
(2)M(ρ)=∫γρ‍Q(x,y)dx-P(x,y)dyC(x,y),whereγρ={(x,y):x2+y2=ρ2}, controls the number of limit cycles of (1) that bifurcate from the periodic orbits of the unperturbed system (1) with ε=0; see [2].The problem of finding lower and upper bounds for the number of zeros ofM(ρ), when P and Q are arbitrary polynomials of a given degree, say n, and C is a particular polynomial, has been faced in several recent papers. In the case of perturbing the linear center by arbitrary polynomials P and Q of degree n, that is, considering x˙=-y+εp(x,y), y˙=x+εq(x,y), there are at most [(n-1)/2] limit cycles up to first order in ε, see [3], where [·] denotes the integer part function. Also it is known that perturbing the quadratic center x˙=-y(1+x), y˙=x(1+x) inside the polynomial systems of degree n we can obtain at most n limit cycles up to first order in ε (see [4]). The authors of [5] studied the perturbation of the cubic center x˙=-y(1+x)(2+x), y˙=x(1+x)(2+x) inside the polynomial differential systems of degree n, and they obtained that 2n+2-(-1)n is an upped bound for the number of limit cycles up to first order in ε. In [6], the authors studied the perturbations of x˙=-y(a+x)(b+y), y˙=x(a+x)(b+y), and they obtained that 3[(n-1)/2]+4 if a≠b and, respectively, 2[(n-1)/2]+2 if a=b, up to first order in ε, are upper bounds for the number of the limit cycles. In [7] the authors studied the maximum number σ of limit cycles which can bifurcate from the periodic orbits of the quartic center x˙=-yf(x,y), y˙=xf(x,y) with f(x,y)=(x+a)(y+b)(x+c) and abc≠0 by perturbing it inside the class of polynomial vector fields of degree n. They proved that 4[(n-1)/2]+4≤σ≤5[(n-1)/2]+14. In [8] the authors studied the bifurcation of limit cycles of the system x˙=-y(x2-a2)(y2-b2)+εP(x,y), y˙=x(x2-a2)(y2-b2)+εQ(x,y) for ε sufficiently small, where a,b∈R-0 and P, Q are polynomials of degree n. They obtained that up to first order in ε the upper bound for the number of limit cycles that bifurcate from the period annulus of the quintic center given by ε=0 is (3/2)(n+sin2(nπ/2))+1 if a≠b or n-1 if a=b. More results can be found in [9, 10].But few of these algebraic curves have a multiple factor. In [11], the authors took C(x,y)=(1-y)m and proved that an upper bound for the number of zeros of M(ρ) on (0,1) is n+m-1 and that this bound is reached when m=1. The approach of [11] is mainly based on the explicit computation of M(ρ). In [12], the authors obtained the maximum of zeros of M(ρ), taking into account their multiplicities, is [(m+n)/2]-1 when n<m-1 and n when n≥m-1. In [12], the authors improve the upper bound given in [11] and provide the optimal upper bound for the zeros of M(ρ). In this paper, motivated by [12] we take C(x,y)=(1-y2)m and obtain the following theorem.Theorem 1.
Consider (1) with C(x,y)=(1-y2)m. Let
(3)M(ρ)=∫γρ‍Q(x,y)dx-P(x,y)dy(1-y2)m,
whereγρ={(x,y):x2+y2=ρ2}, ρ∈(0,1), and P and Q are polynomials of degree n. Then the maximum number of zeros of M(ρ), taking into account their multiplicities, is m+[(n-1)/2]-1 when n<2m-1 and 2[(n+1)/2]-1 when n≥2m-1. Moreover, when m=1,2 and n=2, the corresponding maximum number, which is 1, can be reached by taking suitable P and Q.


## 2. Preliminary Results

To study the property ofM(ρ), we need to make some preliminaries. First we introduce a function of the form
(4)Ij(ρ)=∫02π‍1(1-ρ2sin2θ)jdθ,whereρ∈[0,1) and j is an integer.This section contains some preliminary computations to express the Abelian integralM(ρ) given in (3) in terms of polynomials.Lemma 2.
LetRl(x) be a polynomial of degree l in x. Then for ρ∈[0,1) and m≥0 it holds that
(5)∫02π‍Rl(ρ2sin2θ)(1-ρ2sin2θ)mdθ=∑j=0l‍αm-jIm-j(ρ),
for someαj∈R.Proof.
Note that, forj≥0,
(6)∫02π‍ρ2jsin2jθ(1-ρ2sin2θ)mdθ=∫02π‍(1-(1-ρ2sin2θ))j(1-ρ2sin2θ)mdθ=∑k=0j‍(jk)(-1)k∫02π‍(1-ρ2sin2θ)k(1-ρ2sin2θ)mdθ=∑k=0j‍(jk)(-1)kIm-k(ρ).
The result follows by applying the above formula to each term of(7)Rl(ρ2sin2θ)=∑j=0lrj(ρ2sin2θ)j.Lemma 3.
LetIj be the functions introduced in (4). Then, for 1≠j∈Z,
(8)(ρ2-1)Ij(ρ)=3-2j2j-2(2-ρ2)Ij-1(ρ)+j-2j-1Ij-2(ρ).Proof.
Note that(9)Ij-1(ρ)=∫02π‍1-ρ2sin2θ(1-ρ2sin2θ)jdθIj-1(ρ)=Ij(ρ)-ρ2∫02π‍sin2θ(1-ρ2sin2θ)jdθ.
Using integration by parts in the last integral, we obtain(10)ρ2∫02π‍sin2θ(1-ρ2sin2θ)jdθ∫02π‍=ρ2∫02π‍sinθ(1-ρ2sin2θ)jd(-cosθ)∫02π‍=-ρ2sinθcosθ(1-ρ2sin2θ)j|02π∫02π‍=+∫02π‍(((1-ρ2sin2θ)2j)-1(ρ2cos2θ(1-ρ2sin2θ)j+2jρ4sin2θcos2θ(1-ρ2sin2θ)j-1)×((1-ρ2sin2θ)2j)-1)dθ∫02π‍=∫02π‍ρ2cos2θ(1-ρ2sin2θ)jdθ+2j∫02π‍ρ4sin2θcos2θ(1-ρ2sin2θ)j+1dθ∫02π‍=ρ2Ij+∫02π‍1-ρ2sin2θ-1(1-ρ2sin2θ)jdθ∫02π=‍+2j∫02π‍ρ4sin2θ(1-sin2θ)(1-ρ2sin2θ)j+1dθ.
Note that(11)∫02π‍ρ4sin2θ(1-sin2θ)(1-ρ2sin2θ)j+1dθ=ρ2∫02π‍ρ2sin2θ(1-ρ2sin2θ)j+1dθ-∫02π‍ρ4sin4θ(1-ρ2sin2θ)j+1dθ=ρ2∫02π‍1-(1-ρ2sin2θ)(1-ρ2sin2θ)j+1dθ-∫02π‍[1-(1-ρ2sin2θ)]2(1-ρ2sin2θ)j+1dθ=ρ2(Ij+1-Ij)-(Ij+1-2Ij+Ij-1).
Then(12)ρ2∫02π‍sin2θ(1-ρ2sin2θ)jdθ=ρ2Ij+(Ij-1-Ij)+2j[ρ2(Ij+1-Ij)-Ij+1+2Ij-Ij-1].
Substituting the formula above into (9), we find
(13)Ij-1(ρ)=(1-2j)(2-ρ2)Ij(ρ)+(2j-1)Ij-1(ρ)Ij-1(ρ)=+2j(1-ρ2)Ij+1(ρ).
Replacingj by j-1, we can obtain the conclusion.Lemma 4.
The functions in (4) satisfy
(14)Ij(ρ)=1(1-ρ2)j-1/2I1-j(ρ),j∈Z.
Moreover,(15)Ij(ρ)=R¯(-j)(ρ2),j≤0,Ij(ρ)=R¯j-1(ρ2)(1-ρ2)j-1/2,j≥1,
whereR¯l denotes a polynomial of (exact) degree l.Proof.
It is easy to check that equality (14) is true for j=0,1. Now we prove that it is true for any j≥2 by induction.
Suppose that it is true forj and j+1; that is,
(16)(1-ρ2)j-1/2Ij(ρ)=I1-j(ρ),(1-ρ2)j+1/2Ij+1(ρ)=I-j(ρ).
We need to prove that(17)(1-ρ2)j+3/2Ij+2(ρ)=I-(j+1)(ρ).
By Lemma3, we have
(18)(1-ρ2)Ij+2(ρ)=-2j+12j+2(ρ2-2)Ij+1(ρ)-jj+1Ij(ρ).
Hence,(19)(1-ρ2)j+3/2Ij+2(ρ)=-2j+12j+2(ρ2-2)(1-ρ2)j+1/2Ij+1(ρ)-jj+1(1-ρ2)j+1/2Ij(ρ).
Then it follows from assumption (16) that
(20)(1-ρ2)j+3/2Ij+2(ρ)=-2j+12j+2(ρ2-2)I-j(ρ)-jj+1(1-ρ2)I1-j(ρ).
By Lemma3 again we have
(21)(1-ρ2)I1-j(ρ)=-2j+12j(ρ2-2)I-j(ρ)-j+1jI-(j+1)(ρ).
Substituting it into (20) we obtain
(22)(1-ρ2)j+3/2Ij+2(ρ)=-2j+12j+2(ρ2-2)I-j(ρ)-jj+1[-2j+12j(ρ2-2)I-j(ρ)-j+1jI-(j+1)(ρ)]=I-(j+1)(ρ),
which gives (17). Hence equality (14) holds for j≥0.
Ifj≤-1, then j~=1-j≥2. By applying (14) for j~≥2, it is easy to see that (14) holds for all j≤-1.
The first formula in (15) for the case of j≤0 follows directly from (4). The second one follows from the first one together with (14). This completes the proof.Some explicit expressions ofIj(ρ) are
(23)I1(ρ)=2π(1-ρ2)1/2,I2(ρ)=(2-ρ2)π(1-ρ2)3/2,I3(ρ)=(3ρ4-8ρ2+8)π4(1-ρ2)5/2,I0(ρ)=2π,I-1(ρ)=(2-ρ2)π,I-2(ρ)=(3ρ4-8ρ2+8)π4.Lemma 5.
For any nonnegative integer numbersp, q, and m, one has
(24)∫02π‍sin2p+1θcosqθ(1-ρ2sin2θ)mdθ=0,(25)∫02π‍sinpθcos2q+1θ(1-ρ2sin2θ)mdθ=0.Proof.
Since the integrand is an odd function ofθ, (24) follows. Further
(26)∫02π‍sinpθcos2q+1θ(1-ρ2sin2θ)jdθ=∫02π‍sinpθ(1-sin2θ)q(1-ρ2sin2θ)jdsinθ=0.
Then (25) follows, and the proof is ended.Lemma 6.
LetM(ρ) be the Abelian integral given in (3). Then, there exist polynomials R~l of degree l, l=0,1,…,[(n+1)/2], such that, for ρ∈[0,1),
(27)M(ρ)=∑j=1[(n+1)/2]+1‍R~[(n+1)/2]+1-j(ρ2)Im+1-j(ρ).Proof.
In polar coordinates,x=ρcosθ and y=ρsinθ, the integral M(ρ) writes as
(28)M(ρ)=-∫02π‍(((1-ρ2sin2θ)m)-1(Q(ρcosθ,ρsinθ)ρsinθ+P(ρcosθ,ρsinθ)ρcosθ)×((1-ρ2sin2θ)m)-1)dθ.
Note that(29)∫02π‍sinpθcos2qθ(1-ρ2sin2θ)mdθ=∫02π‍sinpθ(1-sin2θ)q(1-ρ2sin2θ)mdθ.
Then by Lemma5, we have
(30)M(ρ)=∑j=1n‍Rj(cosθ,sinθ)ρj(1-ρ2sin2θ)mdθM(ρ)=∑j=1[(n+1)/2]‍Sj(sin2θ)ρ2j(1-ρ2sin2θ)mdθ,
whereRj(x,y) denotes a homogeneous polynomial of degree j in (x,y), and Sj(x) denotes a polynomial of degree j in x having the form
(31)Sj(sin2θ)=∑l=0j‍s2j,2l(sin2θ)l.
By the above formula, we get(32)∑j=1[(n+1)/2]‍Sj(sin2θ)ρ2j=s2,2sin2θρ2+s4,4sin4θρ4+⋯+s2[(n+1)/2],2[(n+1)/2]sin2[(n+1)/2]θρ2[(n+1)/2]+ρ2(s2,0+s4,2sin2θρ2+⋯+s2[(n+1)/2],2[(n+1)/2]-2×sin2[(n+1)/2]-2θρ2[(n+1)/2]-2)+⋯+ρ2[(n+1)/2]-2(s2[(n+1)/2]-2,0+s2[(n+1)/2],2sin2θρ2)+ρ2[(n+1)/2](s2[(n+1)/2],0)=T[(n+1)/2](ρ2sin2θ)+ρ2T[(n+1)/2]-1(ρ2sin2θ)+ρ4T[(n+1)/2]-2(ρ2sin2θ)+⋯+ρ2[(n+1)/2]T0(ρ2sin2θ)=∑j=0[(n+1)/2]‍ρ2jT[(n+1)/2]-j(ρ2sin2θ),
whereTi is a polynomial of degree i in ρ2sin2θ, i=0,1,…,[(n+1)/2].
Hence by (30) and Lemma 2,
(33)M(ρ)=∑j=1[(n+1)/2]Sj(sin2θ)ρ2j(1-ρ2sin2θ)mdθ=∑j=0[(n+1)/2]‍ρ2j∫02π‍T[(n+1)/2]-j(ρ2sin2θ)(1-ρ2sin2θ)mdθ=∑j=0[(n+1)/2]‍ρ2j(Im-[n+1/2]+jαmIm(ρ)+αm-1Im-1(ρ)+⋯+αm-[(n+1)/2]+jIm-[(n+1)/2]+j(ρ))=∑j=1[(n+1)/2]+1‍R~([(n+1)/2]+1-j)(ρ2)Im+1-j(ρ).
This ends the proof.Assumeδ=1-ρ2, ρ∈[0,1). By Lemma 4, we have
(34)Ij*(δ)=R-j*(δ2),j≤0,Ij*(δ)=Rj-1*(δ2)δ2j-1,j≥1,where(35)R-j*(δ2)=R-j(1-δ2),Ij*(δ)=Ij(1-δ2),whereRl* denotes a polynomial of degree l.Lemma 7.
LetM(ρ) be the Abelian integral given in (3). Then for δ∈(0,1](36)M(ρ)=Rm+[(n-1)/2]*(δ2)δ2m-1,n<2m-1,M(ρ)=R2[(n+1)/2]*(δ)δ,m=1(n≥2m-1),M(ρ)=Rm+[(n+1)/2]-1*(δ2)δ2m-1+R[(n+1)/2]-m*(δ2),=Rm+[(n+1)/2]-1*(δ2)δ2m-15111m≠1(n≥2m-1).Proof.
We have the following cases.
Case  1 (m+1-([(n+1)/2]+1)>0). In this case we have [(n+1)/2]<m, or n<2m-1.
By Lemmas4 and 6, we have
(37)M(ρ)=R~[(n+1)/2](ρ2)Im(ρ)M(ρ)=+R~[(n+1)/2]-1(ρ2)Im-1(ρ)M(ρ)=+⋯+R~0(ρ2)Im-[(n+1)/2](ρ).
LetM(ρ)=M*(δ). Then
(38)M*(δ)=R[(n+1)/2]*(δ2)Im*(δ)+R[(n+1)/2]-1*(δ2)Im-1*(δ)+⋯+R0*(δ2)Im-[(n+1)/2]*(δ)=R[(n+1)/2]*(δ2)Rm-1*(δ2)δ2m-1+R[(n+1)/2]-1*(δ2)Rm-2*(δ2)δ2m-3+⋯+R0*(δ2)Rm-[(n+1)/2]-1*(δ2)δ2m-2[(n+1)/2]-1=Rm+[(n-1)/2]*(δ2)δ2m-1,
where(39)R[(n+1)/2]+m-1*(δ2)=R[(n+1)/2]*(δ2)Rm-1*(δ2)+δ2R[(n+1)/2]-1*(δ2)Rm-2*(δ2)+⋯+δ(2[(n+1)/2])R0*(δ2)Rm-[(n+1)/2]-1*(δ2),
which is a polynomial of degree[(n+1)/2]+m-1   (=[(n-1)/2]+m) in δ2.
Hence,(40)M(ρ)=M*(δ)=R[(n-1)/2]+m*(δ2)δ2m-1.
Case  2 (m+1-([(n+1)/2]+1)≤0). In this case we have [(n+1)/2]≥m or n≥2m-1.(i) Whenm=1, we have
(41)M(ρ)=∑j=1[(n+1)/2]+1‍R~[(n+1)/2]+1-j(ρ2)I2-j(ρ)M(ρ)=R~[(n+1)/2])(ρ2)I1(ρ)+R~[(n+1)/2]-1(ρ2)I0(ρ)M(ρ)=+R~[(n+1)/2]-2(ρ2)I-1(ρ)M(ρ)=+⋯+R~0(ρ2)I1-[(n+1)/2](ρ).Thus,(42)M*(δ)=R[(n+1)/2]*(δ2)R0*(δ2)δ+R[(n+1)/2]-1*(δ2)·2π+R[(n+1)/2]-2*(δ2)R1*(δ2)+⋯+R0*(δ2)R[(n+1)/2]-1*(δ2)=R2[(n+1)/2]*(δ)δ,where(43)R2[(n+1)/2]*(δ)=R[(n+1)/2]*(δ2)R0*(δ2)+2πδR[(n+1)/2]-1*(δ2)+δR[(n+1)/2]-2*(δ2)R1*(δ2)+⋯+δR0*(δ2)R[(n+1)/2]-1*(δ2).Then(44)M*(δ)=R2[(n+1)/2]*(δ)δ.(ii) Whenm≠1, by Lemmas 4 and 6, we have
(45)M(ρ)=R~[(n+1)/2]+m-1(ρ2)(1-ρ2)m-1/2M(ρ)+∑j=0[(n+1)/2]-m‍R~[(n+1)/2]-m-j(ρ2)I-j(ρ)M(ρ)=R~[(n+1)/2]+m-1(ρ2)(1-ρ2)m-1/2+R~[(n+1)/2]-m(ρ2)I0(ρ)M(ρ)+R~[(n+1)/2]-m-1(ρ2)I-1(ρ)M(ρ)+⋯+R~0(ρ2)I-([(n+1)/2]-m)(ρ).Thus,(46)M*(δ)=R[(n+1)/2]+m-1*(δ2)δ2m-1+R[(n+1)/2]-m*(δ2)R0*(δ2)+R[(n+1)/2]-m-1*(δ2)R1*(δ2)+···+R0*(δ2)R[(n+1)/2]-m*(δ2)=R[(n+1)/2]+m-1*(δ2)δ2m-1+R[(n+1)/2]-m*(δ2).This ends the proof.The following lemma can be found in [5].Lemma 8.
Consider the family of functions(47)F(x)=Ai(x)+Bj(x)(a-x)α,
defined on(-∞,a), where Ai and Bj are polynomials of degrees i and j, respectively, and α∉Z. Then each nontrivial function of the form has at most i+j+1 real zeros, taking into account their multiplicities. Moreover, there exist polynomials Ai and Bj such that the corresponding function has exactly this number of zeros on (-∞,a).


## 3. Proof of Theorem1

Proof.
By Lemma7, when n<2m-1 we have
(48)M(ρ)=Rm+[(n-1)/2]*(δ2)δ2m-1.
As the numerator of the above expression is a polynomial inδ2 of degree m+[(n-1)/2], the maximum number of zeros of M(ρ) in [0,1) is m+[(n-1)/2].
Whenm=1 (n≥2m-1), then
(49)M(ρ)=R2[(n+1)/2]*(δ)δ.
As before, the maximum number of zeros ofM(ρ) in [0,1) is 2[(n+1)/2].
Whenm≠1(n≥2m-1), we have
(50)M(ρ)=M~(δ2)=Rm+[(n+1)/2]-1*(δ2)δ2m-1+R[(n+1)/2]-m*(δ2).
Then, using Lemma8, M~(δ2) has at most (m+[(n+1)/2]-1)+([(n+1)/2]-m)+1=2[(n+1)/2] zeros in δ2∈(0,1].
Finally, for alln and m we know that M(0)=M~(1)=0. Then the maximum number of zeros of M(ρ), taking into account their multiplicities, is m+[(n-1)/2]-1 when n<2m-1 and 2[(n+1)/2]-1 when n≥2m-1. The proof is completed.


## 4. Two Illustration Examples on the Maximum Number

Consider the system(51)x˙=-y(1-y2)+εP(x,y),y˙=x(1-y2)+εQ(x,y),where(52)P(x,y)=∑i+j≤2‍pijxiyj,Q(x,y)=∑i+j≤2‍qijxiyjandε is a small real parameter.Let(53)M(ρ)=∫γρ‍Q(x,y)dx-P(x,y)dy1-y2,whereγρ={(x,y):x2+y2=ρ2}, ρ∈(0,1). Assuming x=ρcosθ,y=ρsinθ, we have
(54)M(ρ)=-∫02π‍p10ρ2+(q01-p10)ρ2sin2θ1-ρ2sin2θdθ=[-p10ρ2-(q01-p10)]I1+(q01-p10)I0,where(55)Ij(ρ)=∫02π‍1(1-ρ2sin2θ)jdθ,ρ∈[0,1),Letδ=1-ρ2, ρ∈[0,1). Then
(56)M~(δ)=M(ρ)=(δ-1)(p10δ+q01)δ.ObviouslyM~(δ)=0 for δ∈(0,1) if and only if δ=-q01/p10. Thus for system (51) the function M(ρ) can have 1 simple zero in ρ∈(0,1).Now we consider the system(57)x˙=-y(1-y2)2+εP¯(x,y),y˙=x(1-y2)2+εQ¯(x,y),where(58)P¯(x,y)=∑i+j≤2‍p¯ijxiyj,(59)Q¯(x,y)=∑i+j≤2‍q¯ijxiyj,andε is a small real parameter.Let(60)M(ρ)=∫γρ‍Q¯(x,y)dx-P¯(x,y)dy(1-y2)2M(ρ)=[-p¯10ρ2-(q¯01-p¯10)]I2+(q¯01-p¯10)I1,whereγρ={(x,y):x2+y2=ρ2}, ρ∈(0,1).As before, letδ=1-ρ2, ρ∈[0,1). Then
(61)M¯(δ)=M(ρ)=(1-δ2)(-p¯10δ2-q¯01)δ3.It follows thatM¯(δ)=0 for ∈(0,1) if and only if δ=-q¯01/p¯10. Thus for system (57) the function M(ρ) can have 1 simple zero in ρ∈(0,1).



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