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 form ẋ=-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.
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 region Ω={(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 polynomials P(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.
If C(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.
If C(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 of Φ(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 Theorem 1
Taking the change of variable x=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.
If j is odd, Φk,j=0(k≥0).
According to Lemma 3, 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 integral Φ(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 have
if j<l,
(19)pj=Bj+1,0+Bj-2,1+⋯+Bj-3i+1,i+⋯+Cj,0+Cj-3,1+⋯+Cj-3i,i+⋯,
if j=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+⋯,
if j=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+⋯,
if j=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 Lemma 4, 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 matrices Jj, Hj and Il, we have Lemmas 5 and 6, respectively.
Lemma 5.
For j≥1, det(Jj)≠0 and det(Hj)≠0. That is, rank(Jj)=j+1, rank(Hj)=j.
Proof.
According to Lemma 4 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 the i+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),
where ak=(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 Lemma 4 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 obtain rank(Hj-1)=j-1 in a similar way. That is rank(Hj)=j. The proof is completed.
Lemma 6.
For l≥1, rank(Il)=4l-1.
Proof.
Firstly, if l=1,
(35)I1=∂(p0,p1,p3)∂(α1,β1)=(Γ2A1,0000Γ4Γ8A1,30)⟶(Γ2A1,00Γ8A1,3000Γ4),
it is easy to know that rank(I1)=3.
If l=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.
For l≥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 Lemma 6, 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.
For l≥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 Lemma 4, 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.
If j=l+3u-2(u=1,2,3,…), in a similar way, we can prove that the second formula holds.
If j=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 Theorem 1.
Proof.
For h∈(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.
Let p0=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).
Let p0=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 Theorem 2
In this section, we will prove Theorem 2. At first, all the primary computations to express the Abelian integral Φ(h) and some concerned lemmas are presented.
Taking the change of variable x=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,
where i2=-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.
Let ω1=1, ω2=-1, ω3=i, ω4=-i; then
(49)Ψs=∮γh
dθ1-ωsx=2π1-ωs2h,
where s=1, 2, 3, 4. And
(50)Φ0,01=∮γh
dθ(1-x)(1+x)=2π1-h,Φ1,01=∮γhxdθ(1-x)(1+x)=0,Φ0,02=∮γh
dθ(1-ix)(1+ix)=2π1+h,Φ1,02=∮γhxdθ(1-ix)(1+ix)=0.
Proof.
We use the residue theorem to compute the Ψs(s=1,2,3,4).
When x=hcosθ, we have
(51)Ψs=∮γhdθ1-ωsx=∫02πdθ1-ωshcosθ.
Let eiθ=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.
If j 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-x4
d
θ;
then, according to Lemmas 8 and 9 and the definition of Φ(h), one knows that Φ0,0=0.
Lemma 10.
If j 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.
If j 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) If k is odd, Φk,01=0.
(ii) If k is even, then
(58)Φk,01=2π1-h×[1-∑l=1l≡k(mod2)k2l-kh(k-l)/21-hCk-l(k-l)/2].
Proof.
When k>1,
(59)Φk,01=∮γhxk1-x2dθ=12(∮γhxk1-xdθ+∮γhxk1+xdθ)≡12(M1+M2).
We use the residue theorem to compute the integrals M1 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(mod2)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(mod2)k2l-1h-l/2Ck-l(k-l)/2]=2π1-h[1-∑l=1l≡k(mod2)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(mod2)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(mod2)k2l-kh(k-l)/21-hCk-l(k-l)/2+(-1)k+∑l=1l≡k(mod2)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) If k is odd, Φk,02=0.
(ii) If k is even, then
(67)Φk,02=2πik1+h×[1-∑l=1l≡k(mod2)k2l-kh(k-l)/2i-k-l1+hCk-l(k-l)/2].
Using Lemmas 11 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 Lemma 13, we need the following lemma and a known principle, the Derivation-division algorithm.
Lemma 14.
For any n≥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=
dn/
dxn with n≥1.
The previous lemma has been proved in [14] and [19]. Now, we will prove Lemma 13.
Proof.
Differentiating F(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.
Differentiating F1(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 Theorem 2.
Proof.
From Lemma 9, 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 Lemmas 11 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 Lemma 10, we have that, if j is even,
(75)Φk+1,jσ=∑s=0j/2(-1)shj/2-sCj/2sΦk+1+2s,0σ;
if j 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(mod2)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(mod2)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(mod2)k+2s2l-k-2sπ·h(k+2s-l)/2Ck+2s-l(k+2s-l)/2-∑l=1l≡k+2s(mod2)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(mod2)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(mod2)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].
Let m=[(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-1k
odd
Ψk,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 Lemma 13, 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.
Acknowledgments
The second author is partially supported by the National Natural Science Foundation of China, Grants no. 11171309 and no. 11172269, and the Zhejiang Provincial Natural Science Foundation of China Grant no. Y6110195.
CoppelW. A.Some quadratic systems with at most one limit cycle19892Chichester, UKWiley6188MR1000976XieX. D.CaiS. L.The planar quadratic system with an invariant parabola has at most one limit cycle19931715401542ZegelingA.KooijR. E.Uniqueness of limit cycles in polynomial systems with algebraic invariants199449172010.1017/S0004972700016026MR1262668ShuiS. L.The planar quadratic system with an invariant cubic curve has at most one limit cycle2001244590595MR1894454LlibreJ.RamírezR.SadovskaiaN.On the 16th Hilbert problem for algebraic limit cycles201024861401140910.1016/j.jde.2009.11.023MR2593047ZhangX.The 16th Hilbert problem on algebraic limit cycles201125171778178910.1016/j.jde.2011.06.008MR2823674ChristopherC.LiC.2007Basel, SwitzerlandBirkhäuserviii+171Advanced Courses in Mathematics. CRM BarcelonaMR2325099LlibreJ.Pérez del RíoJ. S.RodríguezJ. A.Averaging analysis of a perturbated quadratic center2001461455110.1016/S0362-546X(99)00444-7MR1845576XiangG.HanM.Global bifurcation of limit cycles in a family of polynomial systems2004295263364410.1016/j.jmaa.2004.03.047MR2072038XiangG.HanM.ZhangT.The number of limit cycles for a family of polynomial systems20054911-121669167810.1016/j.camwa.2005.02.007MR2154676LiS.ZhaoY.LiJ.On the number of limit cycles of a perturbed cubic polynomial differential center2013404221222010.1016/j.jmaa.2013.03.010MR3045167BuicăA.LlibreJ.Limit cycles of a perturbed cubic polynomial differential center20073231059106910.1016/j.chaos.2005.11.060MR2286546CollB.LlibreJ.ProhensR.Limit cycles bifurcating from a perturbed quartic center2011444-531733410.1016/j.chaos.2011.02.009MR2795937GasullA.LázaroJ. T.TorregrosaJ.Upper bounds for the number of zeroes for some Abelian integrals201275135169517910.1016/j.na.2012.04.033MR2927579AtabaigiA.NyamoradiN.ZangenehH. R. Z.The number of limit cycles of a quintic polynomial system200957467768410.1016/j.camwa.2008.10.079MR2489090GasullA.ProhensR.TorregrosaJ.Bifurcation of limit cycles from a polynomial non-global center200820494596010.1007/s10884-008-9112-7MR2448220YaoH.HanM.The number of limit cycles of a class of polynomial differential systems201275134135710.1016/j.na.2011.08.037MR2846805XiangG.HanM.Global bifurcation of limit cycles in a family of multiparameter system20041493325333510.1142/S0218127404011144MR2099173GasullA.LázaroJ. T.TorregrosaJ.On the Chebyshev property for a new family of functions2012387263164410.1016/j.jmaa.2011.09.019MR2853132