# On the Limit Cycles of a Class of Generalized Kukles Polynomial Differential Systems via Averaging Theory

Authors: Amar Makhlouf; Amor Menaceur
Journal: International Journal of Differential Equations
 (2015)

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## Abstract

We apply the averaging theory of first and second order to a class of generalized Kukles polynomial differential systems to study the maximum number of limit cycles of these systems.


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## Body

## 1. Introduction

One of the main topics in the theory of ordinary differential equations is the study of limit cycles: their existence, their number, and their stability. A limit cycle of a differential equation is an isolated periodic orbit in the set of all periodic orbits of the differential equation. The second part of the 16th Hilbert’s problem [1] is related to the least upper bound on the number of limit cycles of polynomial vector fields having a fixed degree. This problem and the Riemann conjecture are the only two problems on the list of Hilbert which have not been solved. Here we consider a very particular case of the sixteenth Hilbert problem. We study the upper bound of the generalized Kukles polynomial system (1)x˙=-y,y˙=Qx,y,where Q(x,y) is a polynomial with real coefficients of degree n.Kukles [2], in 1944, introduced the differential system (2)x˙=-y,y˙=x+a1x2+a2xy+a3y2+a4x3+a5x2y+a6xy2+a7y3,and he gives the necessary and sufficient conditions in order that this system has a center at the origin. This cubic system without the term y3 was also studied in [3] and the authors called it reduced. In [4] a description of the bifurcation of its critical period appears, and [5] presents the existence of reduced Kukles systems with five limit cycles. In the paper [6], the author studied the class of reduced Kukles systems under the cubic perturbation (3)x˙=-y,y˙=x-a+1ax2+2-a1-ay2+1ax3+ɛgx,y,g(x,y)=b01y+b11xy+b02y2+b12xy2+b21x2y+b03y3,where ɛ>0 is small and a>2 is a constant.In [7] the author proves that some cubic systems of form (1) can have seven limit cycles. In [8], Chavarriga et al. studied the maximum number of small amplitude limit cycles for Kukles systems which can coexist with some invariant algebraic curves. Also they give a family of cubic Kukles systems (4)x˙=-y,y˙=x+λy+abx2-a(a2+3c-bλ)xy-2aby2y˙-b2cx3-b(b2-a2c-2c2+bcλ)x2yy˙-b2(-a2-3c+bλ)xy2+b3y3,with an invariant hyperbola h(x,y)=1+abx-b2cx2-b3xy=0 with b≠0, which coexist with one or two small amplitude limit cycles. In [9] the author studied the maximum number of limit cycles of the generalized polynomial Liénard differential equations by using the first and second averaging method. In [10], Llibre and Mereu studied the maximum number of limit cycles of the Kukles polynomial differential systems(5)x˙=y,y˙=-x-∑k≥1‍εkfnk(x)+gmk(x)y+hlk(x)y2+d0ky3,where for every k the polynomials fnk(x),  gmk(x), and hlk(x) have degrees n,  m, and l, respectively, d0k≠0 is a real number, and ɛ is a small parameter.In this work we study the maximum number of limit cycles given by averaging theory of first and second order, which can bifurcate from the periodic orbits of the linear centerx˙=y,  y˙=-x perturbed inside the following class of generalized Kukles polynomial differential systems:(6)x˙=y,y˙=-x-∑k≥1‍εkfnk(x)+gmk(x)y+hlk(x)y2+gmk(x)y3,where for every k the polynomials fnk(x),  gmk(x), and hlk(x) have degrees n,  m, and l, respectively, and ɛ is a small parameter. We have considered the same polynomial gmk(x) as the coefficient of y and y3. With this choice, we can apply the first and second order of the averaging method. If we consider the coefficients of yi for i=0,1,2,3 as arbitrary polynomials, it is difficult to apply the second order averaging method, because to pass from the first order to the second order averaging method, we must put the averaged function of the first order F10(r) (see (8)) identically null. In this case the calculations of the averaged function of the second order F20(r) (see (9)) become difficult. If we replace d0k by hlk(x) in the coefficient of y3 of the differential systems (5), we can apply the first order averaging method but it is not easy to apply the second averaging method. To apply the second averaging method, we must put F10(r)=0 which is equivalent to cancel all coefficients of the polynomial F10(r). The conditions on the coefficient of F10(r) make the calculations of F20(r) difficult. We have used the averaging method for looking for the limit cycles of many classes of Liénard systems. Here we do the same for Kukles differential systems. Comparatively, with the results of the paper [10], we obtained more limit cycles than the results of this paper. More precisely our main result is the following.Theorem 1.
Assume that fork=1,2 the polynomials fnk(x),  gmk(x), and hlk(x) have degrees n,  m, and l, respectively, with n,m,l≥1. Then for ɛ sufficiently small the maximum number of limit cycles of the Kukles polynomial differential systems (6) bifurcating from the periodic orbits of the linear centre x˙=y,  y˙=-x, using averaging theory (a)
of first order is
(i)
no limit cycle form=1,(ii)
(m+2)/2 limit cycles for m≥2,(b)
of second order ismax⁡⁡n/2+(m+1)/2,l/2+(m+3)/2,(m+2)/2,
where· denotes the integer part function.


## 2. First and Second Order Averaging Method

In proof of our main result we use the averaging theory as it is presented in [11]. Consider the differential system(7)x˙(t)=ɛF1(x,t)+ε2F2(x,t)+ε3R(x,t,ɛ),where F1,F2:D×R→Rn,  R:D×R×-εf,εf→Rn are continuous functions, T-periodic in the first variable, and D is an open subset of Rn. Assume that the following hypotheses (i) and (ii) hold.(i)F1(·,t)∈C1(D) for all t∈R, F1, F2, R, and DxF1 are locally Lipschitz with respect to x, and R is differentiable with respect to ɛ. We define(8)F10(z)=1T∫0T‍F1(z,s)ds,(9)F20z=1T∫0T‍DzF1z,s·∫0s‍F1z,tdt+F2z,sds.(ii) ForV⊂D an open and bounded set and for each ɛ∈-εf,εf∖{0}, there exists aɛ∈V such that F10(aɛ)+F20aɛ=0 and dB(F10+ɛF20,V,aɛ)≠0.Then, forɛ>0 sufficiently small there exists aT-periodic solution φ(·,ɛ) of system (7) such that φ(0,ɛ)=aɛ.The expressiondB(F10+ɛF20,V,aɛ)≠0 means that the Brouwer degree of the function F10+ɛF20:V→Rn at the fixed point aɛ is not zero. A sufficient condition for the inequality to be true is that the Jacobian of the function F10+ɛF20 at aɛ is not zero.IfF10 is not identically zero, then the zeros of F10+ɛF20 become mainly the zeros of F10 for ɛ sufficiently small. In this case the previous result provides the averaging theory of first order.IfF10 is identically zero and F20 is not identically zero, then the zeros of F10+ɛF20 are mainly the zeros of F20 for ɛ sufficiently small. In this case the previous result provides the averaging theory of second order. For more information about the averaging theory see [12, 13].


## 3. Proof of Theorem1

### 3.1. Proof of Statement (a) of Theorem1

In order to apply the first order averaging method we write system (6) with k=1, in polar coordinates (r,θ) where x=rcos⁡θ, y=rsinθ, r>0. If we take fn1(x)=∑i=0n‍aixi, gm1(x)=∑i=0m‍bixi, and hl1(x)=∑i=0l‍cixi, system (6) can be written as follows: (10)r˙=-ɛsinθPr,θ,θ˙=-1-ɛrcos⁡θPr,θ,where (11)Pr,θ=∑i=0n‍airicos⁡iθ+∑i=0m‍biri+1cos⁡iθsinθ+∑i=0m‍biri+3cos⁡iθsin3θ+∑i=0l‍ciri+2cos⁡iθsin2θ.If we take θ as a new independent variable, system (10) becomes (12)drdθ=ɛsinθPr,θ+oε2=ɛF1r,θ+oε2.By using the notation introduced in Section 2 we have that (13)F10(r)=12π∫02π‍sinθP(r,θ)dθ.We know that(14)∫02π‍cos⁡iθsinjθdθ=0ifioddorjisoddIi,j≠0ifiisevenandjeven.Letk be a positive integer. We define ev(k) as the largest even integer less than or equal to k, and od(k) as the largest odd integer less than or equal to k.Hence(15)F10(r)=12π∑i=0ievenevm‍biIi,2ri+1+12π∑i=0ievenev(m)‍biIi,4ri+3.We obtain(16)F10r=r2πb0I0,2+b2I2,2+b0I0,4r2+b4I4,2+b2I2,4r4bevmrevm+2pppipp+⋯+bev(m)Iev(m),2+bev(m)-2Iev(m)-2,4rev(m)ppiippb2I2,2+b0I0,4r2+bevmrevm+2.For the casem=1, we obtain that (17)F10r=b0r2πI0,2+I0,4r2=b0r21+34r2.There is no positive root forF10(r).Form≥2, the polynomial F10(r) has at most [(m+2)/2] positive roots. Hence (a) of Theorem 1 is proved.


### 3.2. Proof of Statement (b) of Theorem1

For proving statement (b) of Theorem1 we will use the second-order averaging theory. If we write(18)fn1(x)=∑i=0n‍aixi,fn2(x)=∑i=0n‍a~ixi,gm1x=∑i=0m‍bixi,gm2x=∑i=0m‍b~ixi,hl1(x)=∑i=0l‍cixi,hl2(x)=∑i=0l‍c~ixithen system (6) with k=2 in polar coordinates (r,θ), r>0, becomes(19)r˙=-ɛsinθPr,θ-ε2sinθKr,θ,θ˙=-1-ɛrcos⁡θPr,θ-ε2rcos⁡θKr,θ,where(20)Kr,θ=∑i=0n‍a~iricos⁡iθ+∑i=0m‍b~iri+1cos⁡iθsinθ+∑i=0m‍b~iri+3cos⁡iθsin3θ+∑i=0l‍c~iri+2cos⁡iθsin2θ.Takingθ as the new independent variable system, (19) can be written as (21)drdθ=ɛF1(r,θ)+ε2F2(r,θ)+o(ε3),where F1(r,θ)=sinθP(r,θ) and F2(r,θ)=sinθKr,θ-(cos⁡θsinθ/r)P2(r,θ).In order to apply the averaging theory of second order,F10 must be identically zero. Therefore from (16), F10 is identically zero if and only if bi=0 for i even.Now we determine the corresponding function(22)F20=12π∫02π‍ddrF1r,θ·∫0θ‍F1(r,ϕ)dϕ+F2(r,θ)dθ.For this we compute(23)ddrF1r,θ=∑i=1n‍iairi-1cos⁡iθsinθ+∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ+∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ+∑i=0l‍i+2ciri+1cos⁡iθsin3θ,∫0θ‍F1r,ϕdϕ=∑j=0n‍ajrj∫0θ‍cos⁡jϕsinϕdϕ+∑j=1joddod(m)‍bjrj+1∫0θ‍cos⁡jϕsin2ϕdϕ+∑j=1joddod(m)‍bjrj+3∫0θ‍cos⁡jϕsin4ϕdϕ+∑j=0l‍cjrj+2∫0θ‍cos⁡jϕsin3ϕdϕ=∑j=0n‍ajrj1j+1(1-cos⁡j+1θ)+∑j=1joddod(m)‍bjrj+1∑k=1(j+3)/2‍αk,jsin2k-1θ+∑j=1joddod(m)‍bjrj+3∑k=1(j+5)/2‍βk,jsin2k-1θ+∑j=0jevenevl‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θ+∑j=1joddodl‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θ,where αk,j, βk,j, γk,j, and δk,j are constants.The integral∫02π‍d/drF1(r,θ)∫0θ‍F1(r,ϕ)dϕdθ will be given in several lemmas.Lemma 2.
The integral(24)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(25)∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij,where Aij and Bij are real constants.Proof.
We have that(a1)
(26)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=0n‍ajrj1j+11-cos⁡j+1θdθ=0,(b1)
(27)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij,whereAij=∫02π‍cos⁡iθsinθ∑k=1(j+3)/2‍αk,jsin2k-1θdθ≠0 for i⩾2 even and j⩾1 odd,(c1)
(28)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=0,(d1)
(29)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e1)
(30)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrkj+3βk,jsin2k-1θdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij,whereBij=∫02π‍cos⁡iθsinθ∑k=1(j+5)/2‍βk,jsin2k-1θdθ≠0 for i⩾2 even and j⩾1 odd.
We have that the sum of the integrals (a1)–(e1) is polynomial (3.6). This ends the proof of the lemma.Lemma 3.
The integral(31)∫02π‍∑i=0m‍i+1biricos⁡iθsin2θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(32)-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij,where Ii+j+1,2 and Dij are real constants.Proof.
We have that(a2)
(33)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=0n‍ajrj1j+1(1-cos⁡j+1θ)dθ=-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2,whereIi+j+1,2=∫02π‍cos⁡i+j+1θsin2θdθ≠0 for i odd and j even.(b2)
(34)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=0,(c2)
(35)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij,whereDij=∫02π‍cos⁡iθsin2θγ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ≠0 for i⩾1 odd and j⩾0 even,(d2)
(36)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e2)
(37)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrj+3βk,jsin2k-1θdθ=0.
We have that the sum of the integrals (a2)–(e2) is polynomial (32). This ends the proof of the lemma.Lemma 4.
The integral(38)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(39)∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij,where Eij and Fij are real constants.Proof.
We have that(a3)
(40)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=0n‍ajrj1J+11-cos⁡j+1θdθ=0,(b3)
(41)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij,whereEij=∫02π‍cos⁡iθsin3θ∑k=1(j+3)/2‍αk,jsin2k-1θ≠0 for i⩾0 even and j⩾1 odd,(c3)
(42)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=0,(d3)
(43)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e3)
(44)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrj+3βkjsin2k-1θdθ=∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij,whereFij=∫02π‍cos⁡iθsin3θ∑k=1(j+5)/2‍βk,jsin2k-1θ≠0 for i⩾0 even and j⩾1 odd.
We have that the sum of the integrals (a3)–(e3) is polynomial (39). This completes the proof of the lemma.Lemma 5.
The integral(45)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(46)-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij,where Ii+j+1,4 and Hij are real constants.Proof.
We have that(a4)
(47)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=0n‍ajrj1J+1(1-cos⁡j+1θ)dθ=-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4,whereIi+j+1,4=∫02π‍cos⁡i+j+1θsin4θdθ≠0 for i odd and j even,(b4)
(48)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=0,(c4)
(49)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij,whereHij=∫02π‍cos⁡iθsin4θγ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ≠0 for i⩾1 odd and j⩾0 even,(d4)
(50)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e4)
(51)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+3βk,jsin2k-1θdθ=0.
We have that the sum of the integrals (a4)–(e4) is polynomial (46). This ends the proof of the lemma.From Lemmas2–5 we have that(52)∫02π‍ddrF1(r,θ)∫0θ‍F1(r,ϕ)dϕdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij.Now we calculate the integral ∫02π‍F2(r,θ)dθ.(53)∫02π‍F2r,θdθ=∫02π‍sinθKr,θdθ-∫02π‍sinθcos⁡θrP2(r,θ)dθ.First we calculate∫02π‍sinθK(r,θ)dθ. Noting that K(r,θ) is given by (20), we have(54)∫02π‍sinθKr,θdθ=∑i=0n‍a~iri∫02π‍cos⁡iθsinθdθ+∑i=0m‍b~iri+1∫02π‍cos⁡iθsin2θdθ+∑i=0m‍b~iri+3∫02π‍cos⁡iθsin4θdθ+∑i=0l‍c~iri+2∫02π‍cos⁡iθsin3θdθ.Hence(55)∫02π‍sinθKr,θdθ=∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4.Now noting that P(r,θ) is given by (11), we compute(56)∫02π‍sinθcos⁡θrP2(r,θ)dθ=∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθ2dθ+∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθ2dθ+∫02π‍sinθcos⁡θr∑i=0l‍ciri+2cos⁡iθsin2θ2dθ+∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+3cos⁡iθsin3θ2dθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=1joddod(m)‍bjrj+1cos⁡jθsinθdθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=1joddod(m)‍bjrj+3cos⁡jθsin3θdθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθpppppppp·∑j=1joddod(m)‍bjrj+3cos⁡jθsin3θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+3cos⁡iθsin3θpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ.Hence(57)∫02π‍sinθcos⁡θrP2r,θdθ=2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2+2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4+2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4+2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.From (55) and (57) we obtain(58)∫02π‍F2r,θdθ=∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.ThenF20 is the polynomial(59)F20(r)=12π∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍F20(r)=ooo12π+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2BijF20(r)=ooo12π-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2F20(r)=ooo12π+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2DijF20(r)=ooo12π+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2EijF20(r)=ooo12π+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4FijF20(r)=ooo12π-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4F20(r)=ooo12π+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4HijF20(r)=ooo12π+∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4F20(r)=ooo12π-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2F20(r)=ooo12π-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4F20(r)=ooo12π-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4F20(r)=ooo12π-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.Note that in order to find the positive roots ofF20 after dividing by r, we must find the zeros of a polynomial in the variable r2 of degree equal to the max{(ev(n)+od(m)+2-1)/2,(ev(l)+od(m)+4-1)/2,(ev(m)+3-1)/2}, we conclude that F20 has at most max⁡⁡n/2+(m+1)/2,l/2+(m+3)/2,(m+2)/2 positive roots. Hence (b) of Theorem 1 is proved.


## 3.1. Proof of Statement (a) of Theorem1

In order to apply the first order averaging method we write system (6) with k=1, in polar coordinates (r,θ) where x=rcos⁡θ, y=rsinθ, r>0. If we take fn1(x)=∑i=0n‍aixi, gm1(x)=∑i=0m‍bixi, and hl1(x)=∑i=0l‍cixi, system (6) can be written as follows: (10)r˙=-ɛsinθPr,θ,θ˙=-1-ɛrcos⁡θPr,θ,where (11)Pr,θ=∑i=0n‍airicos⁡iθ+∑i=0m‍biri+1cos⁡iθsinθ+∑i=0m‍biri+3cos⁡iθsin3θ+∑i=0l‍ciri+2cos⁡iθsin2θ.If we take θ as a new independent variable, system (10) becomes (12)drdθ=ɛsinθPr,θ+oε2=ɛF1r,θ+oε2.By using the notation introduced in Section 2 we have that (13)F10(r)=12π∫02π‍sinθP(r,θ)dθ.We know that(14)∫02π‍cos⁡iθsinjθdθ=0ifioddorjisoddIi,j≠0ifiisevenandjeven.Letk be a positive integer. We define ev(k) as the largest even integer less than or equal to k, and od(k) as the largest odd integer less than or equal to k.Hence(15)F10(r)=12π∑i=0ievenevm‍biIi,2ri+1+12π∑i=0ievenev(m)‍biIi,4ri+3.We obtain(16)F10r=r2πb0I0,2+b2I2,2+b0I0,4r2+b4I4,2+b2I2,4r4bevmrevm+2pppipp+⋯+bev(m)Iev(m),2+bev(m)-2Iev(m)-2,4rev(m)ppiippb2I2,2+b0I0,4r2+bevmrevm+2.For the casem=1, we obtain that (17)F10r=b0r2πI0,2+I0,4r2=b0r21+34r2.There is no positive root forF10(r).Form≥2, the polynomial F10(r) has at most [(m+2)/2] positive roots. Hence (a) of Theorem 1 is proved.


## 3.2. Proof of Statement (b) of Theorem1

For proving statement (b) of Theorem1 we will use the second-order averaging theory. If we write(18)fn1(x)=∑i=0n‍aixi,fn2(x)=∑i=0n‍a~ixi,gm1x=∑i=0m‍bixi,gm2x=∑i=0m‍b~ixi,hl1(x)=∑i=0l‍cixi,hl2(x)=∑i=0l‍c~ixithen system (6) with k=2 in polar coordinates (r,θ), r>0, becomes(19)r˙=-ɛsinθPr,θ-ε2sinθKr,θ,θ˙=-1-ɛrcos⁡θPr,θ-ε2rcos⁡θKr,θ,where(20)Kr,θ=∑i=0n‍a~iricos⁡iθ+∑i=0m‍b~iri+1cos⁡iθsinθ+∑i=0m‍b~iri+3cos⁡iθsin3θ+∑i=0l‍c~iri+2cos⁡iθsin2θ.Takingθ as the new independent variable system, (19) can be written as (21)drdθ=ɛF1(r,θ)+ε2F2(r,θ)+o(ε3),where F1(r,θ)=sinθP(r,θ) and F2(r,θ)=sinθKr,θ-(cos⁡θsinθ/r)P2(r,θ).In order to apply the averaging theory of second order,F10 must be identically zero. Therefore from (16), F10 is identically zero if and only if bi=0 for i even.Now we determine the corresponding function(22)F20=12π∫02π‍ddrF1r,θ·∫0θ‍F1(r,ϕ)dϕ+F2(r,θ)dθ.For this we compute(23)ddrF1r,θ=∑i=1n‍iairi-1cos⁡iθsinθ+∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ+∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ+∑i=0l‍i+2ciri+1cos⁡iθsin3θ,∫0θ‍F1r,ϕdϕ=∑j=0n‍ajrj∫0θ‍cos⁡jϕsinϕdϕ+∑j=1joddod(m)‍bjrj+1∫0θ‍cos⁡jϕsin2ϕdϕ+∑j=1joddod(m)‍bjrj+3∫0θ‍cos⁡jϕsin4ϕdϕ+∑j=0l‍cjrj+2∫0θ‍cos⁡jϕsin3ϕdϕ=∑j=0n‍ajrj1j+1(1-cos⁡j+1θ)+∑j=1joddod(m)‍bjrj+1∑k=1(j+3)/2‍αk,jsin2k-1θ+∑j=1joddod(m)‍bjrj+3∑k=1(j+5)/2‍βk,jsin2k-1θ+∑j=0jevenevl‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θ+∑j=1joddodl‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θ,where αk,j, βk,j, γk,j, and δk,j are constants.The integral∫02π‍d/drF1(r,θ)∫0θ‍F1(r,ϕ)dϕdθ will be given in several lemmas.Lemma 2.
The integral(24)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(25)∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij,where Aij and Bij are real constants.Proof.
We have that(a1)
(26)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=0n‍ajrj1j+11-cos⁡j+1θdθ=0,(b1)
(27)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij,whereAij=∫02π‍cos⁡iθsinθ∑k=1(j+3)/2‍αk,jsin2k-1θdθ≠0 for i⩾2 even and j⩾1 odd,(c1)
(28)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=0,(d1)
(29)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e1)
(30)∫02π‍∑i=1n‍iairi-1cos⁡iθsinθ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrkj+3βk,jsin2k-1θdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij,whereBij=∫02π‍cos⁡iθsinθ∑k=1(j+5)/2‍βk,jsin2k-1θdθ≠0 for i⩾2 even and j⩾1 odd.
We have that the sum of the integrals (a1)–(e1) is polynomial (3.6). This ends the proof of the lemma.Lemma 3.
The integral(31)∫02π‍∑i=0m‍i+1biricos⁡iθsin2θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(32)-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij,where Ii+j+1,2 and Dij are real constants.Proof.
We have that(a2)
(33)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=0n‍ajrj1j+1(1-cos⁡j+1θ)dθ=-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2,whereIi+j+1,2=∫02π‍cos⁡i+j+1θsin2θdθ≠0 for i odd and j even.(b2)
(34)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=0,(c2)
(35)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij,whereDij=∫02π‍cos⁡iθsin2θγ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ≠0 for i⩾1 odd and j⩾0 even,(d2)
(36)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e2)
(37)∫02π‍∑i=1ioddod(m)‍i+1biricos⁡iθsin2θ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrj+3βk,jsin2k-1θdθ=0.
We have that the sum of the integrals (a2)–(e2) is polynomial (32). This ends the proof of the lemma.Lemma 4.
The integral(38)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(39)∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij,where Eij and Fij are real constants.Proof.
We have that(a3)
(40)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=0n‍ajrj1J+11-cos⁡j+1θdθ=0,(b3)
(41)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij,whereEij=∫02π‍cos⁡iθsin3θ∑k=1(j+3)/2‍αk,jsin2k-1θ≠0 for i⩾0 even and j⩾1 odd,(c3)
(42)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=0,(d3)
(43)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e3)
(44)∫02π‍∑i=0l‍i+2ciri+1cos⁡iθsin3θ·∑j=1joddod(m)‍∑k=1(j+5)/2‍bjrj+3βkjsin2k-1θdθ=∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij,whereFij=∫02π‍cos⁡iθsin3θ∑k=1(j+5)/2‍βk,jsin2k-1θ≠0 for i⩾0 even and j⩾1 odd.
We have that the sum of the integrals (a3)–(e3) is polynomial (39). This completes the proof of the lemma.Lemma 5.
The integral(45)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ∫0θ‍F1(r,ϕ)dϕdθis, in the variable r, the polynomial(46)-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij,where Ii+j+1,4 and Hij are real constants.Proof.
We have that(a4)
(47)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=0n‍ajrj1J+1(1-cos⁡j+1θ)dθ=-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4,whereIi+j+1,4=∫02π‍cos⁡i+j+1θsin4θdθ≠0 for i odd and j even,(b4)
(48)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+1αk,jsin2k-1θdθ=0,(c4)
(49)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=0jevenev(l)‍cjrj+2γ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ=∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij,whereHij=∫02π‍cos⁡iθsin4θγ1,j+∑k=2(j+6)/2‍γk,jcos⁡2k-3θdθ≠0 for i⩾1 odd and j⩾0 even,(d4)
(50)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(l)‍cjrj+2δ1,j+∑k=2(j+5)/2‍δk,jcos⁡2k-2θdθ=0,(e4)
(51)∫02π‍∑i=1ioddod(m)‍i+3biri+2cos⁡iθsin4θ·∑j=1joddod(m)‍∑k=1(j+3)/2‍bjrj+3βk,jsin2k-1θdθ=0.
We have that the sum of the integrals (a4)–(e4) is polynomial (46). This ends the proof of the lemma.From Lemmas2–5 we have that(52)∫02π‍ddrF1(r,θ)∫0θ‍F1(r,ϕ)dϕdθ=∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2Bij-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2Dij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2Eij+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4Fij-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4Hij.Now we calculate the integral ∫02π‍F2(r,θ)dθ.(53)∫02π‍F2r,θdθ=∫02π‍sinθKr,θdθ-∫02π‍sinθcos⁡θrP2(r,θ)dθ.First we calculate∫02π‍sinθK(r,θ)dθ. Noting that K(r,θ) is given by (20), we have(54)∫02π‍sinθKr,θdθ=∑i=0n‍a~iri∫02π‍cos⁡iθsinθdθ+∑i=0m‍b~iri+1∫02π‍cos⁡iθsin2θdθ+∑i=0m‍b~iri+3∫02π‍cos⁡iθsin4θdθ+∑i=0l‍c~iri+2∫02π‍cos⁡iθsin3θdθ.Hence(55)∫02π‍sinθKr,θdθ=∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4.Now noting that P(r,θ) is given by (11), we compute(56)∫02π‍sinθcos⁡θrP2(r,θ)dθ=∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθ2dθ+∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθ2dθ+∫02π‍sinθcos⁡θr∑i=0l‍ciri+2cos⁡iθsin2θ2dθ+∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+3cos⁡iθsin3θ2dθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=1joddod(m)‍bjrj+1cos⁡jθsinθdθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=1joddod(m)‍bjrj+3cos⁡jθsin3θdθ+2∫02π‍sinθcos⁡θr∑i=0n‍airicos⁡iθpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθpppppppp·∑j=1joddod(m)‍bjrj+3cos⁡jθsin3θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+1cos⁡iθsinθpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ+2∫02π‍sinθcos⁡θr∑i=1ioddod(m)‍biri+3cos⁡iθsin3θpppppppp·∑j=0l‍cjrj+2cos⁡jθsin2θdθ.Hence(57)∫02π‍sinθcos⁡θrP2r,θdθ=2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2+2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4+2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4+2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.From (55) and (57) we obtain(58)∫02π‍F2r,θdθ=∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.ThenF20 is the polynomial(59)F20(r)=12π∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+jAij∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍F20(r)=ooo12π+∑i=2ievenev(n)‍∑j=1joddod(m)‍iaibjri+j+2BijF20(r)=ooo12π-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+1j+1ajbiri+jIi+j+1,2F20(r)=ooo12π+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+2DijF20(r)=ooo12π+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+2EijF20(r)=ooo12π+∑i=0ievenev(l)‍∑j=1joddod(m)‍i+2cibjri+j+4FijF20(r)=ooo12π-∑1≤i≤od(m)0≤j≤ev(n)ioddandjeven‍i+3j+1ajbiri+j+2Ii+j+1,4F20(r)=ooo12π+∑i=1ioddod(m)‍∑j=0jevenev(l)‍i+1bicjri+j+4HijF20(r)=ooo12π+∑i=0ievenev(m)‍b~iri+1Ii,2+∑i=0ievenev(m)‍b~iri+3Ii,4F20(r)=ooo12π-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+jIi+j+1,2F20(r)=ooo12π-2∑0≤i≤ev(n)1≤j≤od(m)ievenandjodd‍aibjri+j+2Ii+j+1,4F20(r)=ooo12π-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+2Ii+j+1,4F20(r)=ooo12π-2∑1≤i≤od(m)0≤j≤ev(l)ioddandjeven‍bicjri+j+4Ii+j+1,6.Note that in order to find the positive roots ofF20 after dividing by r, we must find the zeros of a polynomial in the variable r2 of degree equal to the max{(ev(n)+od(m)+2-1)/2,(ev(l)+od(m)+4-1)/2,(ev(m)+3-1)/2}, we conclude that F20 has at most max⁡⁡n/2+(m+1)/2,l/2+(m+3)/2,(m+2)/2 positive roots. Hence (b) of Theorem 1 is proved.



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*Source: 325102-2015-03-15.xml*