# Poincaré Bifurcations of Two Classes of Polynomial Systems

Authors: Jing Wang; Shuliang Shui
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

Using bifurcation methods and the Abelian integral, we investigate the number of the limit cycles that bifurcate from the period annulus of the singular point when we perturb the planar ordinary differential equations of the formẋ=-yC(x,y), ẏ=xC(x,y) with an arbitrary polynomial vector field, where C(x,y)=1-x3 or C(x,y)=1-x4.


---

## Body

## 1. Introduction and Main Results

In the qualitative theory of real planar differential systems, one of the main problems is to determine the existence and number of the limit cycles of the polynomial differential system. In the general case, this is a very difficult task. Therefore, the researchers consider the weak Hilbert 16th problem. In addition, the existence of invariant algebraic curves in polynomial systems may influence the number of limit cycles. For example, the planar quadratic systems with one invariant line or conic curve or cubic curve can have at most one limit cycle [1–4]. In [3], the authors proved that the cubic systems with four invariant lines have at most one limit cycle. In [5, 6], the authors proved that a real polynomial system of degree m with irreducible invariant algebraic curves has at most 1+(m-1)(m-2)/2 limit cycles if m is even and (m-1)(m-2)/2 limit cycles if m is odd.In this paper, we consider the weak Hilbert 16th problem that the unperturbed systems have a linear center and an invariant algebraic curve(1)x˙=-yC(x,y)+ɛP(x,y),y˙=xC(x,y)+ɛQ(x,y),
where P(x,y) and Q(x,y) are polynomials of degree n in ℝ2, the algebraic curve C(x,y) satisfies C(0,0)≠0, and ɛ∈ℝ is a sufficient small parameter.It is obvious that, on the regionΩ={(x,y)∣C(x,y)≠0}, the system (1) is equivalent to the following form:
(2)x˙=-y+ɛP(x,y)C(x,y),y˙=x+ɛQ(x,y)C(x,y).
When ɛ=0, system (2) is a Hamilton system with a family of ovals
(3)γh={(x,y)∈ℝ2∣H(x,y)=x2+y2=h,h>0}.Define the Abelian integral(4)Φ(h)=∮γh‍P(x,y)dy-Q(x,y)dxC(x,y),
which is also called first-order Melnikov function of (2). According to the Poincaré-Pontryagin theorem [7], the number of isolated real zeros of Φ(h) controls the number of limit cycles of system (1) that bifurcate from the periodic annulus of the perturbed system (1) with ɛ=0. That is to say, when the Φ(h) does not vanish exactly, the maximum number of the isolated real zeros of Φ(h) is corresponding to the upper bound of the number of limit cycles which bifurcate from periodic annulus of unperturbed systems.For the arbitrary polynomialsP(x,y), Q(x,y) of given degree n, the number of limit cycles of (2) depends on the different choices of C(x,y). At present, several works have figured out this problem for the particular choices of C(x,y). In [8], the authors studied the system (1) with C(x,y)=1+x and proved that the number of limit cycles that bifurcate from the period orbits is at most n. The authors in [9, 10] studied the number of limit cycles which bifurcate from (1) when ɛ=0 with C(x,y)=1+Bx+Ax2 and C(x,y)=1+ax+by+cx(x2+y2), respectively. The authors in [11] studied the number of limit cycles of system (1) with C(x,y)=y2+Ax2+Bx+C. In [12], the authors studied the number of limit cycles of system (1) with C(x,y)=(x+a)(x+b) and obtain that the system can have at most 3[(n-1)/2]+2 limit cycles if a≠b and 2[(n-1)/2]+1 if a=b, respectively. In [13], the authors studied the case the curves C(x,y)=0 are three lines, two of them parallel and one perpendicular, and [14, 15] studied the case the curves are k(k>3) lines, and any two of them are parallel or perpendicular directions. The authors in [16] studied the case the curves are consistent by k nonzero points. The authors in [17] considered system (1) with C(x,y)=1+x4 and proved that 3[(n+1)/2]-2 limit cycles can at most bifurcate from the periodic orbits of the unperturbed system. In [18], the authors proved that the system (1) with C(x,y)=(1-x)m has at most n+m-1 limit cycles.The aim of this paper is to investigate the upper bound of the number of limit cycles bifurcate from the periodic annulus of the center of the unperturbed system (1) (ɛ=0) with the perturbed polynomials P(x,y), Q(x,y) of given degree n, and C(x,y)=1-x3 or C(x,y)=1-x4.Consider the planar differential system(5)x˙=-yC(x,y)+ɛ∑0≤k+j≤n‍ak,jxkyj,y˙=xC(x,y)+ɛ∑0≤k+j≤n‍bk,jxkyj,
where ɛ∈ℝ is a sufficient small parameter. Applying the Abelian integral, we obtain the following two main theorems.Theorem 1.
IfC(x,y)=1-x3, the lower bound of the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ=0 is 4[(n+1)/2]-2.Theorem 2.
IfC(x,y)=1-x4, the upper bound of the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ=0 is 3[(n+1)/2]-2.Our primary purpose is to calculate the concrete expression ofΦ(h); then we can obtain the number of limit cycles of the perturbed system (5) by determining the isolated real zeros of Abelian integral Φ(h). In Sections 2 and 3, we prove these two theorems with the different methods, respectively.


## 2. The proof of Theorem1

Taking the change of variablex=hcosθ, y=hsinθ(0<h<1), we have
(6)Φ(h)=∮γh‍∑0≤k+j≤nak,jxk+1yj+∑0≤k+j≤nbk,jxkyj+11-x3dθ=∑0≤k+j≤n‍h(k+j+1)/2∫02π‍ak,jcosk+1θsinjθ+bk,jcoskθsinj+1θ1-h3/2cos3θdθ=∑0≤k+j≤n‍(ak,jΦk+1,j(h)+bk,jΦk,j+1(h)),
where
(7)Φk,j=h(k+j)/2∫02π‍coskθsinjθ1-h3/2cos3θdθ.Firstly, we have the following obvious result.Lemma 3.
Ifj is odd, Φk,j=0(k≥0).According to Lemma3, we can rewrite the Φ(h) as follows:
(8)Φ(h)=∑s=1n‍as,0Φs+1,0+∑s=1n‍∑k+j=s‍(ak,j+bk+1,j-1)Φk+1,j+∑s=0soddnb0,sΦ0,s+1.
Denote l=[(n+1)/2] and ak,j*=ak,j+bk+1,j-1(j≥1); then
(9)∑s=0n‍as,0Φs+1,0=∑s=1l‍(a2s-1,0Φ2s,0+a2s-2,0Φ2s-1,0),∑s=0soddnb0,sΦ0,s+1=∑s=1l‍b0,2s-1Φ0,2s,∑s=0n‍∑k+j=s‍(ak,j+bk+1,j-1)Φk+1,j=∑s=1l‍(∑m=1s-1‍a2s-2m-1,2m*Φ2s-2m,2m+∑m=1s-1‍a2s-2m-2,2m*Φ2s-2m-1,2m).
Therefore, we have
(10)Φ(h)=∑s=1l‍(a2s-1,0Φ2s,0+∑m=1s-1‍a2s-2m-1,2m*Φ2s-2m,2m+b0,2s-1Φ0,2s+a2s-2,0Φ2s-1,0+∑m=1s-1‍a2s-2m-2,2m*Φ2s-2m-1,2m).
Let
(11)Γ0=2π,Γ2i=∫02π‍cos2iθdθ=2π(2i-1)!!2i!!(i≥1),Ak,j=∑r=0k‍(-1)rCkrΓ2j+2r,(12)Bi,j=a2i-1,0Γ2i+6j+∑m=1i-1‍a2i-2m-1,2m*Am,i-m+3j+b0,2i-1Ai,3j,(13)Ci,j=a2i-2,0Γ2i+2+6j+∑m=1i-1‍a2i-2m-2,2m*Am,i-m+3j+1,
where ai,j*=0 if i<0. Then, we have the following lemma.Lemma 4.
Abelian integralΦ(h) has an expansion in the form
(14)Φ(h)=h(p0+p1h+p2h2+⋯+p3lh3l+p3l+1h3l+1+p3l+2h3l+2+o(h3l+2)),h∈(0,1),
where
(15)pj=∑s=0[(j+1)/3]‍(Bj-3s+1,s+Cj-3s,s),ifj<l,pj=∑s=0[l/3]‍(Bl-3s,u+s+Cl-3s-1,u+s),ifj=l+3u-1(u=1,2,3,…),pj=∑s=0[(l-2)/3]‍Bl-3s-2,u+s+1+∑s=0[l/3]‍Cj-3s,u+s,ifj=l+3u(u=0,1,2,…),pj=∑s=0[(l-1)/3]‍(Bl-3s-1,u+s+Cl-3s-2,u+s),ifj=l+3u-2(u=1,2,3,…).Proof.
Firstly, we have(16)Φ2s,0=hs∫02π‍cos2sθ1-h3/2cos3θdθ=hs∫02π‍(cos2sθ+h3cos2s+6θ+⋯+h3icos2s+6iθ+⋯)dθ=hs(Γ2s+h3Γ2s+6+⋯+h3iΓ2s+6i+⋯),Φ2s-2m,2m=hs∫02π‍cos2s-2mθsin2mθ1-h3/2cos3θdθ=hs∑r=0m‍(-1)rCmr∫02π‍(cos2s-2m+2rθ+h3cos2s-2m+6+2rθ+⋯+h3icos2s-2m+2r+6iθ+⋯)dθ=hs(Am,s-m+h3Am,s-m+3+⋯+h3iAm,s-m+3i+⋯),Φ0,2s=hs∫02π‍sin2sθ1-h3/2cos3θdθ=hs∑r=0s‍(-1)rCsr∫02π‍(cos2rθ+h3cos2r+6θ+⋯+h3icos2r+6iθ+⋯)dθ=hs(As,0+h3As,3+⋯+h6As,3i+⋯),Φ2s-1,0=hs-1/2∫02π‍cos2s-1θ1-h3/2cos3θdθ=hs-1/2∫02π‍(h3/2cos2s+2θ+h9/2cos2s+8θ+⋯+h3/2+3icos2s+2+6iθ⋯)dθ=hs∫02π‍(hcos2s+2θ+h4cos2s+8θ+⋯+h1+3icos2s+2+6iθ⋯)dθ=hs(hΓ2s+2+h4Γ2s+8+⋯+h1+3iΓ2s+2+6i+⋯),Φ2s-2m-1,2m=hs-1/2∫02π‍cos2s-2m-1θsin2mθ1-h3/2cos3θdθ=hs-1/2×∑r=0m‍(-1)rCmr∫02π‍(h3/2cos2s-2m+2+2rθ+h9/2cos2s-2m+8+2rθ+⋯+h3/2+3icos2s-2m+2+2r+6i+⋯)dθ=hs(hAm,s-m+1+h4Am,s-m+4+⋯+h1+3iAm,s-m+1+3i+⋯).
Substituting the previous formulas into (10), we have
(17)Φ(h)=∑s=1l‍hs×[(a2s-1,0Γ2s+∑m=1s-1‍a2s-2m-1,2m*Am,s-m+b0,2s-1As,0∑m=1s‍)+h3(a2s-1,0Γ2s+6+∑m=1s-1‍a2s-2m-1,2m*Am,s-m+3+b0,2s-1As,3∑m=1s‍)+⋯+h3i(a2s-1,0Γ2s+6i+∑m=1s-1‍a2s-2m-1,2m*Am,s-m+3i+b0,2s-1As,3i∑m=1s‍)+⋯+h(a2s-2,0Γ2s+2+∑m=1s-1‍a2s-2m-2,2m*Am,s-m+1)+h4(a2s-2,0Γ2s+8+∑m=1s-1‍a2s-2m-2,2m*Am,s-m+4)+⋯+h1+3i(∑m=1s-1‍a2s-2,0Γ2s+6i+2+∑m=1s-1‍a2s-2m-2,2m*Am,s-m+3i+1)+⋯∑m=1s‍].
Using (12) and (13), (17) can be written as follows:
(18)Φ(h)=∑s=1l‍hs(Bs,0+h3Bs,1+⋯+h3iBs,i+⋯+hCs,0+h4Cs,1+⋯+h1+3iCs,i+⋯)=h[(B1,0+h3B1,1+⋯+h3iB1,i+⋯+hC1,0+h4C1,1+⋯+h1+3iC1,i+⋯)+h(B2,0+h3B2,1+⋯+h3iB2,i+⋯+hC2,0+h4C2,1+⋯+h1+3iC2,i+⋯)+h2(B3,0+h3B3,1+⋯+h3iB3,i+⋯+hC3,0+h4C3,1+⋯+h1+3iC3,i+⋯)+⋯+hl-1(Bl,0+h3Bl,1+⋯+h3iBl,i+⋯+hCl,0+h4Cl,1+⋯+h1+3iCl,i+⋯)].
Thus, we haveifj<l,
(19)pj=Bj+1,0+Bj-2,1+⋯+Bj-3i+1,i+⋯+Cj,0+Cj-3,1+⋯+Cj-3i,i+⋯,ifj=l+3u-1(u=1,2,3,…),
(20)pj=Bl,u+Bl-3,u+1+⋯+Bl-3i,u+i+⋯+Cl-1,u+Cl-4,u+1+⋯+Cl-3i-1,u+i+⋯,ifj=l+3u(u=0,1,2,⋯),
(21)pj=Bl-2,u+1+Bl-5,u+2+⋯+Bl-3i-2,u+i+1+⋯+Cl,u+Cl-3,u+1+⋯+Cl-3i,u+i+⋯,ifj=l+3u-2(u=1,2,3,…),
(22)pj=Bl-1,u+Bl-4,u+2+⋯+Bl-3i-1,u+i+⋯+Cl-2,u+Cl-5,u+1+⋯+Cl-3i-2,u+i+⋯, where i=0,1,2,… and Bi,j=Ci,j=0 if i<0. The proof is completed.By Lemma4, we regard a2j-1,0, a2j-3,2*,…, a1,2j-2*, b0,2j-1, a2j-2,0, a2j-4,2*,…, a2,2j-4*, a0,2j-2*(j≥1) as free parameters, and denote vectors
(23)αj=(a2j-1,0,a2j-3,2*,…,a1,2j-2*,b0,2j-1),βj=(a2j-2,0,a2j-4,2*,…,a2,2j-4*,a0,2j-2*)
and Jacobian matrices
(24)Jj=∂(pj-1,pj+2,pj+5,…,p4j-1)∂αj,(25)Hj=∂(pj,pj+3,pj+6,…,p4j-3)∂βj,(26)Il=∂(p0,p1,p2,…,pl-2,pl-1,pl,…,p4l-5,p4l-4,p4l-3,p4l-1)∂(α1,β1,α2,β2,α3,β3,…,αl,βl),
where j, l≥1. Then, Jj is (j+2)×(j+2) matrix, Hj is j×j matrix, and Il is (4l-1)×(l2+2l) matrix.For matricesJj, Hj and Il, we have Lemmas 5 and 6, respectively.Lemma 5.
Forj≥1, det(Jj)≠0 and det(Hj)≠0. That is, rank(Jj)=j+1, rank(Hj)=j.Proof.
According to Lemma4 and (24), we can obtain that
(27)Jj=(Γ2jA1,j-1A2,j-2⋯Aj-2,2Aj-1,1Aj,0Γ2j+6A1,j+2A2,j+1⋯Aj-2,5Aj-1,4Aj,3Γ2j+12A1,j+5A2,j+4⋯Aj-2,8Aj-1,7Aj,6Γ2j+18A1,j+8A2,j+7⋯Aj-2,11Aj-1,10Aj,9⋮⋮⋮⋮⋮⋮Γ8j-6A1,4j-4A2,4j-5⋯Aj-2,3j-1Aj-1,3j-2Aj,3j-3Γ8jA1,4j-1A2,4j-2⋯Aj-2,3j+2Aj-1,3j+1Aj,3j).
Define(j+1)×(j+1) matrix
(28)ithDi=(1(-1)i-2Ci-1i-11(-1)i-3Ci-1i-2⋱⋮1(-1)1Ci-121(-1)0Ci-1111⋱1).
Then, we have
(29)Jj′≡JjD2D3⋯Dj+1=(Γ2jΓ2j-2Γ2j-4⋯Γ4Γ2Γ0Γ2j+6Γ2j+4Γ2j+2⋯Γ10Γ8Γ6Γ2j+12Γ2j+10Γ2j+8⋯Γ16Γ14Γ12Γ2j+18Γ2j+16Γ2j+14⋯Γ22Γ20Γ18⋮⋮⋮⋮⋮⋮Γ8j-6Γ8j-8Γ8j-10⋯Γ6j-2Γ6j-4Γ6j-6Γ8jΓ8j-2Γ8j-4⋯Γ6j+4Γ6j+2Γ6j).
Let Γij=j!!/i!!; then Γj=2πΓjj-1 (j≥1). Denote
(30)Tj-k=(Γ2j+6k2j+4k-1Γ2j+6k-22j+4k-3⋯Γ8k+46k+3Γ8k+26k+1Γ8k6k-1Γ2j+6k+62j+4k+5Γ2j+6k+42j+4k+3⋯Γ8k+106k+9Γ8k+86k+7Γ8k+66k+5⋮⋮⋮⋮⋮Γ8j-68j-2k-7Γ8j-88j-2k-9⋯Γ6j+2k-26j-3Γ6j+2k-46j-5Γ6j+2k-66j-7Γ8j8j-2k-1Γ8j-28j-2k-3⋯Γ6j+2k+46j+3Γ6j+2k+26j+1Γ6j+2k6j-1)(0≤k≤j),
where Tj-k is (j-k+1)×(j-k+1) matrix. Then, det(Jj)=det(Jj′)=(2π)j+1det(Tj).
We add entries of thei+1th(0<i≤j-k+1) column which times -(2(j+2-i)-1)/2(j+2-i) to the ith column and obtain(31)Tj-k⟶Tj-k1≡(00⋯0Γ8k6k-16k+3j+3kΓ2j+6k+62j+4k+36k+3j+3k-1Γ2j+6k+42j+4k+1⋯6k+34k+1Γ8k+86k+5Γ8k+66k+512k+6j+3kΓ2j+6k+122j+4k+912k+6j+3k-1Γ2j+6k+102j+4k+7⋯12k+64k+1Γ8k+146k+11Γ8k+126k+11⋮⋮⋮⋮(6k+3)(j-k-1)j+3kΓ8j-68j-2k-9(6k+3)(j-k-1)j+3k-1Γ8j-88j-2k-11⋯(6k+3)(j-k-1)4k+1Γ6j+2k-46j-7Γ6j+2k-66j-7(6k+3)(j-k)j+3kΓ8j8j-2k-3(6k+3)(j-k)j+3k-1Γ8j-28j-2k-5⋯(6k+3)(j-k)4k+1Γ6j+2k+26j-1Γ6j+2k6j-1)≡(01×(j-k)Γ8k6k-1Tj-k-10ak),
whereak=(Γ8k+66k+5,Γ8k+126k+11,…,Γ6j+2k-66j-7,Γ6j+2k6j-1)T. We can write Tj-k-10 as follows:
(32)Tj-k-10=(6k+312k+6⋱(6k+3)(j-k))×Tj-k-1(1j+3k1j+3k-1⋱14k+1);
then det(Tj-k)=(-1)j-k+2Γ8k6k-1(6k+3)j-k(j-k)!((4k)!/(j+3k)!)det(Tj-k-1).
Summarizing above results, we have(33)det(Jj)=(2π)j+1(-1)(j+2)+(j+1)+⋯+33j9j-115j-2⋯(6j-9)2(6j-3)j!(j-1)!⋯2!×Γ85Γ1611Γ2417⋯Γ8j6j-11j!4!(j+3)!8!(j+6)!⋯(4j-4)!(4j-3)!≠0;
therefore,rank(Jj)=j+1.
According to Lemma4 and (25), we have
(34)Hj-1=(Γ2jA1,j-1A2,j-2⋯Aj-3,3Aj-2,2Γ2j+6A1,j+2A2,j+1⋯Aj-3,6Aj-2,5Γ2j+12A1,j+5A2,j+4⋯Aj-3,9Aj-2,8⋮⋮⋮⋮⋮Γ8j-18A1,4j-10A2,4j-11⋯Aj-3,3j-6Aj-2,3j-7Γ8j-12A1,4j-7A2,4j-8⋯Aj-3,3j-3Aj-2,3j-4).
By the above proof procedure, we can obtainrank(Hj-1)=j-1 in a similar way. That is rank(Hj)=j. The proof is completed.Lemma 6.
Forl≥1, rank(Il)=4l-1.Proof.
Firstly, ifl=1,
(35)I1=∂(p0,p1,p3)∂(α1,β1)=(Γ2A1,0000Γ4Γ8A1,30)⟶(Γ2A1,00Γ8A1,3000Γ4),
it is easy to know that rank(I1)=3.
Ifl=2, we have
(36)I2=∂(p0,p1,p2,p3,p4,p5,p7)∂(α1,β1,α2,β2)=(Γ2A1,000000000Γ4A1,1A2,0Γ400000000Γ6A1,2Γ8A1,300000000Γ10A1,4A2,3Γ1000000000Γ12A1,500Γ16A1,7A2,6Γ1600),⟶(Γ2A1,0000000Γ8A1,300000000Γ4A1,1A2,000000Γ10A1,4A2,300000Γ16A1,7A2,6000000000Γ6A1,2000000Γ12A1,5)⟶(J1J2H202×1).
Then, by Lemma 5, rank(I2)=rank(J1)+rank(J2)+rank(H2)=7.
Forl≥3, let γl1=(p0,p1,…,pl-3), γl2=(pl-2,pl+1,…,p4l-5), γl3=(pl-1,pl+2,…,p4l-1), γl4=(pl,pl+3,…,p4l-3); we have
(37)Il⟶Il*=∂(γl1,γl2,γl3,γl4)∂(α1,β1,α2,β2,…,αl,βl).
According to Lemma 4 and the definitions of Jj and Hj, simplifying it by elementary transformation of matrix, we obtain
(38)Il*⟶(El-2Jl-1JlHl0l×(l2-2l+1)).
Therefore, rank(Il)=rank(El-2)+rank(Jl-1)+rank(Jl)+rank(Hl)=4l-1. The proof is completed.By Lemma6, it is obvious that p0,p1,p2,…,pl-2,pl-1,pl,…,p4l-4,p4l-3,p4l-1 are independent. Now, we have the following lemma.Lemma 7.
Forl≥1, one can write pj as follows:
(39)pl+3u-1=x1pl-1+x2pl+2+⋯+xl+1p4l-1,u>l,pl+3u-2=y1pl-2+y2pl+1+⋯+ylp4l-5,u>l-1,pl+3u=z1pl+1+z2pl+4+⋯+zlp4l-3,u>l-1,
where (x1,x2,…,xl+1), (y1,y2,…,yl) and (z1,z2,…,zl) are nonzero vectors.Proof.
According to Lemma4, if j=l+3u-1(u=1,2,3,…),
(40)pj=∑s=0[l/3]‍(Bl-3s,u+s+Cl-3s-1,u+s).
Substituting pl-1,pl+2,…,p4l-1 into the first equation, we obtain a linear equation
(41)dl+3u-1=JlTX(u>l),
where X=(x1,x2,…,xl+1)T, dl+3u-1=(Γ2l+6u,A1,l+3u,A2,l+3u-2,…,Al-1,3u+1,Al,3u)T. According to Lemma 5, det(JlT)≠0, (41) has unique solution X=(x1′,x2′,…,xl+1′)T≠0. That is, the first formula holds.
Ifj=l+3u-2(u=1,2,3,…), in a similar way, we can prove that the second formula holds.
Ifj=l+3u(u=0,1,2,…), we have
(42)pj=∑s=0[(l-2)/3]‍Bl-3s-2,u+s+1+∑s=0[l/3]‍Cj-3s,u+s,
substituting pl+1,pl+4,…,p4l-3 into the third equation, we obtain a linear equation
(43)cl+3u=HlTZ(u>l-1),
where Z=(z1,z2,…,zl)T, cl+3u=(Γ2l+6u+2,A1,l+3u,A2,l+3u-1,…,Al-2,3u+3,Al-1,3u+2)T. According to Lemma 5, det(HlT)≠0, (43) has unique solution Z=(z1′,z2′,…,zl′)T≠0. That is, the third formula also holds. The proof is completed.Now, we prove Theorem1.Proof.
Forh∈(0,1), Abelian integral Φ(h) has an expansion of the following form:
(44)Φ(h)=h(p0+p1h+p2h2+⋯+p4l-3h4l-3+p4l-2h4l-2+p4l-1h4l-1+o(h4l-1)).
According to Lemma 6, for l≥1, p0,p1,p2,…,pl-2,pl-1,pl,…,p4l-4,p4l-3,p4l-1 are independent.
Letp0=p1=p2=⋯=p4l-1=p4l-3=0 and p4u-1=1; then, by Lemma 7, p4l-2=0. Thus (44) becomes Φ(h)=h4l+o(h4l), and Φ(h)>0 if h∈(0,1). Furthermore, we take p0=p1=p2=⋯=p4l-5=p4l-4=0, then p4l-2=0 still holds. Choosing proper p4l-3∈(-1,0) such that Φ(h)=p4l-3h4l-2+h4l+o(h4l)<0, by Descartes’ rule of signs, (44) has a root h1 on interval (0,1).
Letp0=p1=p2=⋯=p4l-6=p4l-5=0, and choose proper p4l-4(p4l-4∈(0,1)) so that Φ(h)=p4l-4h4l-3+p4l-3h4l-2+h4l+o(h4l)>0; then (44) has the second root h2 on interval (0, 1). In a similar way, we take proper pi(i=4l-6,4l-7,…,2,1,0) in turn such that pipi-1<0 and |pi|∈(0,1). According to Descartes’ rule of signs, we can obtain 4l-4 zeros h3,h4,…,h4l-3,h4l-2 on interval (0,1).
Applying the Poincaré-Pontryagin theorem, the system (5) with C(x,y)=1-x3 can have at least 4[(n+1)/2]-2 limit cycles for suitable ak,j and bk,j(0≤k+j≤n). The proof of Theorem 1 is completed.


## 3. The proof of Theorem2

In this section, we will prove Theorem2. At first, all the primary computations to express the Abelian integral Φ(h) and some concerned lemmas are presented.Taking the change of variablex=hcosθ, y=hsinθ(0<h<1), then by (4) we have
(45)Φ(h)=∮γh‍∑0≤k+j≤nak,jxk+1yj+∑0≤k+j≤nbk,jxkyj+11-x4dθ=12(Φ1+Φ2),
where
(46)Φ1=∮γh‍∑0≤k+j≤nak,jxk+1yj+∑0≤k+j≤nbk,jxkyj+11-x2dθ,Φ2=∮γh‍∑0≤k+j≤nak,jxk+1yj+∑0≤k+j≤nbk,jxkyj+11+x2dθ.
Denote
(47)Φk,j1=∮γh‍xkyj1-x2dθ=∮γh‍xkyj(1-x)(1+x)dθ,Φk,j2=∮γh‍xkyj1+x2dθ=∮γh‍xkyj(1-ix)(1+ix)dθ,wherei2=-1; then we have
(48)Φ1=∑0≤k+j≤n‍(ak,jΦk+1,j1+bk,jΦk,j+11),Φ2=∑0≤k+j≤n‍(ak,jΦk+1,j2+bk,jΦk,j+12).Lemma 8.
Letω1=1, ω2=-1, ω3=i, ω4=-i; then
(49)Ψs=∮γh‍dθ1-ωsx=2π1-ωs2h,
wheres=1, 2, 3, 4. And
(50)Φ0,01=∮γh‍dθ(1-x)(1+x)=2π1-h,Φ1,01=∮γh‍xdθ(1-x)(1+x)=0,Φ0,02=∮γh‍dθ(1-ix)(1+ix)=2π1+h,Φ1,02=∮γh‍xdθ(1-ix)(1+ix)=0.Proof.
We use the residue theorem to compute theΨs(s=1,2,3,4).
Whenx=hcosθ, we have
(51)Ψs=∮γh‍dθ1-ωsx=∫02π‍dθ1-ωshcosθ.
Leteiθ=z; then cosθ=(z2+1)/2z, dθ=dz/iz. The previous formula becomes(52)Ψs=∮|z|=1‍11-ωsh·((z2+1)/2z)dziz=-2i∮|z|=1‍dzωshz2-2z+ωsh=-4πRes[1ωshz2-2z+ωsh,z1=1-1-ωs2hωsh]=-4πlimz→z1z-z1ωshz2-2z+ωsh=2π1-ωs2h;
hence (49) holds. For the first formula of (50),
(53)Φ0,01=12(∮γh‍dθ1-ω1x+∮γh‍dθ1+ω2x)=12(2π1-h+2π1-h)=2π1-h,
and the others can be proved in a similar way. The proof is completed.Lemma 9.
Ifj is odd, the integrands in Φk,j1 and Φk,j2 are odd functions with respect to the variable θ; therefore, Φk,j1=0 and Φk,j2=0.
Define(54)Φk,j=∮γh‍ak,jxk+1yj+bk,jxkyj+11-x4dθ;
then, according to Lemmas 8 and 9 and the definition of Φ(h), one knows that Φ0,0=0.Lemma 10.
Ifj is even, then
(55)Φk,j1=∑s=0j/2‍(-1)shj/(2-s)Cj/2sΦk+2s,01,(56)Φk,j2=∑s=0j/2‍(-1)shj/(2-s)Cj/2sΦk+2s,02.Proof.
Ifj is even, then
(57)Φk,j1=∮γh‍xkyj1-x2dθ=∮γh‍xk(h-x2)j/21-x2dθ=∮γh‍∑s=0j/2‍(-1)shj/2-sCj/2sxk+2s1-x2dθ=∑s=0j/2‍(-1)shj/2-sCj/2sΦk+2s,01.
Similarly, (56) also holds. The proof is completed.Lemma 11.
(i)  Ifk is odd, Φk,01=0.
(ii)  Ifk is even, then
(58)Φk,01=2π1-h×[1-∑l=1l≡k(mod2)k2l-kh(k-l)/21-hCk-l(k-l)/2].Proof.
Whenk>1,
(59)Φk,01=∮γh‍xk1-x2dθ=12(∮γh‍xk1-xdθ+∮γh‍xk1+xdθ)≡12(M1+M2).
We use the residue theorem to compute the integralsM1 and M2. Denote eiθ=z; thus, cosθ=(z2+1)/2z, dθ=dz/iz. We have
(60)M1=∮γh‍xk1-xdθ=∮|z|=1‍(h·((z2+1)/2z))k1-h·((z2+1)/2z)·1izdz=-21-k·hk/2i∮γh‍(z2+1)kzk·(hz2-2z+h)dz=-22-kπhk/2[Res(M1,0)+Res(M1,z1=1-1-hh)].
Since z1=(1-1-h)/h is the first-order zero of the equation hz2-2z+h=0, the residue of M1 at z1 is
(61)Res(M1,z1)=limz→z1(z2+1)kzk·(hz2-2z+h)·(z-z1)=-2k-1hk/2·1-h.
For the residue at z=0, we have the expansion of M1 in the form
(62)M1=(z2+1)kzk·(hz2-2z+h)=12z·((z2+1)/z)k(h/2)·((z2+1)/z)-1=∑l=1∞‍2l-1h-l/2z-1(z2+1z)k-l;
the coefficient of z-1 is corresponding to the residue of M1 at z=0; therefore,
(63)Res(M1,0)=∑l=1l≡k(mod2)k2l-1h-l/2Ck-l(k-l)/2.
Substituting them into (60), we have
(64)M1=-22-kπhk/2[-2k-1hk/2·1-h+∑l=1l≡k(mod2)k2l-1h-l/2Ck-l(k-l)/2]=2π1-h[1-∑l=1l≡k(mod2)k2l-kh(k-l)/21-hCk-l(k-l)/2].
We can compute M2 in a similar way and obtain
(65)M2=2π1-h×[(-1)k+∑l=1l≡k(mod2)k(-1)l-12l-kh(k-l)/21-hCk-l(k-l)/2].
By the formulas (64) and (65), Φk,01 becomes
(66)Φk,01=π1-h×[1-∑l=1l≡k(mod2)k2l-kh(k-l)/21-hCk-l(k-l)/2+(-1)k+∑l=1l≡k(mod2)k(-1)l-12l-kh(k-l)/21-hCk-l(k-l)/2],
from the previous formula, it is easy to know that, if k is odd, Φk,01=0, and if k is even, formula (58) is obtained. The proof is completed.In a similar way, we can prove the following lemma.Lemma 12.
(i) Ifk is odd, Φk,02=0.
(ii) Ifk is even, then
(67)Φk,02=2πik1+h×[1-∑l=1l≡k(mod2)k2l-kh(k-l)/2i-k-l1+hCk-l(k-l)/2].Using Lemmas11 and 12, it is easy to see that, if k is even, Φk,0=0. Therefore, by Lemma 10 and the definition of Φk,j, if n is even, Σk+j=nΦk,j=0.Lemma 13.
Consider a function of the form(68)F(x)=P0(x)+P1(x)1+x+P2(x)1-x,
where Pj(x)(j=1,2) are real polynomials of degree n and the degree of P0 is n0. Then the number 𝒵(F) of real zeros of F(x) in U=[0,1), taking into account their multiplicities, satisfies 𝒵(F)≤2n+n0+2; here deg(0)=-1.To prove Lemma13, we need the following lemma and a known principle, the Derivation-division algorithm.Lemma 14.
For anyn≥0, m≥1 and the real constants α,
(69)𝒟(pn(x)(x+1)α)=qn(x)(x+1)α-1,𝒟(pn(x)(1+x1-x)α)=qn+1(x)(1+x)α-1(1-x)α+1,𝒟m(pn(x)(1+x1-x)α)=qn+m(x)(1+x)α-m(1-x)α+m.
In particular, when m=n+1, formula (69) becomes in the following form:
(70)𝒟n+1(pn(x)(1+x1-x)α)=qn(x)(1+x)α-(n+1)(1-x)α+(n+1),
where pn(x), qn(x) are polynomials of degree n, 𝒟n=dn/dxn with n≥1.The previous lemma has been proved in [14] and [19]. Now, we will prove Lemma 13.Proof.
DifferentiatingF(x) in formula (68) n0+1 times, 𝒟n0+1P0(x)=0. According to Lemma 14 and dividing the expression (1+x)-1/2-(n0+1), which does not vanish in U=[0,1), let a=-1/2-(n0+1); we can obtain
(71)F1(x)≡(1+x)1/2+n0+1𝒟n0+1F(x)=P11(x)+P12(x)(1-x1+x)a,
where P11(x), P12(x) are suitable polynomials of degree n. Applying Rolle’s theorem, it follows that 𝒵(F)≤𝒵(F1)+n0+1.
DifferentiatingF1(x) in formula (71) n+1 times and applying Lemma 14 again and dividing the expression (1-x)a-(n+1)/(1+x)a+(n+1), which does not vanish in interval U=[0,1), we have
(72)F2(x)≡(1-x)-a+n+1(1+x)a+n+1𝒟n+1F1(x),
where F2(x) is a polynomials of degree n. Therefore according to Rolle’s theorem, taking into account their multiplicities, the total number 𝒵(F) of real zeros of F(x) in interval U=[0,1) satisfies 𝒵(F)≤2n+n0+2. The proof is completed.Now, we prove Theorem2.Proof.
From Lemma9, we have
(73)Φ(h)=12[∑0≤k+j≤n‍ak,j(Φk+1,j1+Φk+1,j2)+∑0≤k+j≤n‍bk,j(Φk,j+11+Φk,j+12)]=12[∑0≤k+j≤njevenak,j(Φk+1,j1+Φk+1,j2)+∑0≤k+j≤njoddbk,j(Φk,j+11+Φk,j+12)].
By Lemmas11 and Lemma 12, the previous formula becomes
(74)Φ(h)=12[∑0≤k+j≤nkoddjeven‍ak,j(Φk+1,j1+Φk+1,j2)+∑0≤k+j≤nkevenjodd‍bk,j(Φk,j+11+Φk,j+12)].
According to Lemma10, we have that, if j is even,
(75)Φk+1,jσ=∑s=0j/2‍(-1)shj/2-sCj/2sΦk+1+2s,0σ;
ifj is odd,
(76)Φk,j+1σ=∑s=0(j+1)/2‍(-1)sh(j+1)/2-sC(j+1)/2sΦk+2s,0σ,
where σ=1,2.
According to (75) and (76), we can obtain that, if k is odd, j is even,
(77)12(Φk+1,j1+Φk+1,j2)=∑s=0j/2‍(-1)shj/2-sCj/2s×[π1-h+πik+1+2s1+h-∑l=1l≡k+1+2s(mod2)k+1+2s2l-k-1-2sπ·h(k+1+2s-l)/2Ck+1+2s-l(k+1+2s-l)/2-∑l=1l≡k+1+2s(mod2)k+1+2s2l-k-1-2sπ·h(k+1+2s-l)/2i-k-1-2s-lCk+1+2s-l(k+1+2s-l)/2πik+1+2s1+h],
and if k is even, j is odd,(78)12(Φk,j+11+Φk,j+12)=∑s=0(j+1)/2‍(-1)sh(j+1)/2-sC(j+1)/2s×[π1-h+πik+2s1+h-∑l=1l≡k+2s(mod2)k+2s2l-k-2sπ·h(k+2s-l)/2Ck+2s-l(k+2s-l)/2-∑l=1l≡k+2s(mod2)k+2s2l-k-2sπ·h(k+2s-l)/2i-k-2s-lCk+2s-l(k+2s-l)/2πik+2s1+h].
From (77) and (78), we have, if k is odd, j is even,
(79)12[ak,j(Φk+1,j1+Φk+1,j2)+bk+1,j-1(Φk+1,j1+Φk+1,j2)]=(ak,j+bk+1,j-1)×∑s=0j/2‍(-1)shj/2-sCj/2s×[π1-h+πik+1+2s1+h-∑l=1l≡k+1+2s(mod2)k+1+2s2l-k-1-2sπ·h(k+1+2s-l)/2Ck+1+2s-l(k+1+2s-l)/2-∑l=1l≡k+1+2s(mod2)k+1+2s2l-k-1-2sπ·h(k+1+2s-l)/2i-k-1-2s-lCk+1+2s-l(k+1+2s-l)/2πik+2s1+h].
Letm=[(n+1)/2] and
(80)Ψk,j=12[ak,j(Φk+1,j1+Φk+1,j2)+bk+1,j-1(Φk+1,j1+Φk+1,j2)],
where k is odd, j is even.
According to (79), the Abelian integral Φ(h) of system (5) with C(x,y)=1-x4 has the following form:
(81)Φ(h)=∑0≤k+j≤n‍Φk,j=∑u=1m‍∑k+j=2u-1koddΨk,j+12∑j=1joddnb0,j(Φ0,j+11+Φ0,j+12)=∑u=0m‍(Pu-1(h)+Qu-1(h)1-h+Ru-1(h)1+h)≡Pm-1(h)+Qm-1(h)1-h+Rm-1(h)1+h,
where Pm(h), Qm(h), and Rm(h) denote a polynomial of variable h of degree m, whose coefficients are linear combinations of ak,j, bk,j(0≤k+j≤n).
According to Lemma13, taking into account their multiplicities, the maximum number of real zeros of (81) in interval U=[0,1) is 3m-1. From (73), (75), and (76), we know that Φ(0)=0. Hence, Φ(h) has at least 3m-2 real zeros in the open interval (0,1).
Applying the Poincaré-Pontryagin theorem, the upper bound of number of limit cycles for the system (5) with C(x,y)=1-x4 is 3m-2. That is, the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ=0 is 3[(n+1)/2]-2. The proof of Theorem 2 is completed.



---

*Source: 861329-2013-08-05.xml*