# Fifteen Limit Cycles Bifurcating from a Perturbed Cubic Center

Authors: Amor Menaceur; Mufda Alrawashdeh; Sahar Ahmed Idris; Hala Abd-Elmageed
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

In this work, we study the bifurcation of limit cycles from the period annulus surrounding the origin of a class of cubic polynomial differential systems; when they are perturbed inside the class of all polynomial differential systems of degree six, we obtain at most fifteenth limit cycles by using the averaging theory of first order.


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## Body

## 1. Introduction and Statement of the Main Result

Hilbert in 1900 was interested in the maximum number of the limit cycles that a polynomial differential system of a given degree can have. This problem is the well-known 16th Hilbert problem, which together with the Riemann conjecture are the two problems of the famous list of 23 problems of Hilbert which remain open. See for more details [1, 2].A classical way to produce limit cycles is by perturbing a system which has a center, in such a way that limit cycles bifurcate in the perturbed system from some of the periodic orbits of the period annulus of the center of the unperturbed system [3–7].In [8], the authors improved the result of the maximum number of limit cycles for a class of polynomial differential systems which bifurcate from the period annulus surrounding the origin of the system:(1)u˙=v−vu−y+au+v+a,v˙=−u+uu−v+au+v+a,where u−y+au+v+a=1 is a conic, a2≠1, and a≤2 by using the first order of the averaging theory method.In [9], the authors improved the result of the maximum number of limit cycles of sixth polynomial differential systems which bifurcate from the period annulus surrounding the origin of the system:(2)u˙=−vu−v2−a2,v˙=uu−v2−a2,where u−v2−a=0 is a conic and a≠0, by using the first order of the averaging theory method.In this work, we perturb the cubic systems equation (1). Thus, we consider these classes of all polynomial differential systems of degree n, i.e.,(3)u˙=v−vu−y+au+v+a+εPu,v,v˙=−u+uu−v+au+v+a+εQu,v,where u−v+au+v+a=1 is a conic, a>2, Pu,v and Qu,v are the real polynomials of degree n≥3, and ε is a small parameter. Main result of this study is the following Theorem 1.Theorem 1.
For the sufficiently smallε and the polynomials Pu,v and Qu,v having degree 6, suppose that a>2, system equation (3) has at most 15 limit cycles bifurcating from the period annulus surrounding the origin of cubic polynomial differential system equation (1) using averaging theory of first order (Figures 1 and 2).Figure 1
Phase portrait of the cubic system equation (1) with a=2.Figure 2
Phase portrait of the cubic system equation (1) with a=2.


## 2. The Averaging Theory of First Order

Theorem 2.
Consider the following two initial value problems:(4)x˙=εRt,x+ε2Gt,x,ε,x0=x0,and(5)y˙=εf0y,y0=x0,where x,y, and x0∈D is an open domain of ℝ, t∈0,∞, ε∈0,ε0, R and G are the periodic functions with their period T with its variable t, and f0y is the average function of Rt,y with respect to t, i.e.,(6)f0y=1T∫0TFt,ydt.Assume that(i)
R, ∂R/∂x, ∂2R/∂x2, G, and ∂G/∂x are well defined, continuous, and bounded by a constant independent by ε∈0,ε0 in 0,∞×D(ii)
T is a constant independent of ε(iii)
yt belongs to D on the time scale 1/ε. Then, the following statements hold.(a)
On the time scale1/ε, we have(7)xt−yt=Oε,asε⟶0.(b)
Ifp is an equilibrium point of the averaged system equation (5), such that(8)∂f0∂yy=p≠0.Then, system equation (4) has a T-periodic solution ϕt,ε⟶p as ε⟶0.(c)
If equation (8) is a negative, the corresponding periodic solution ϕt,ε of equation (4) according to t,x is asymptotically stable for all ε sufficiently small, and if equation (8) is a positive, then it is unstable.For more details on the averaging method, see [10, 11].


## 3. Proof of Theorem1

Fora>2, the cubic system equation (1) has a unique period annulus:(9)A=u,v:0<u2+v2<a2−22.According to Figures1 and 2, this proof is based on the first order of the averaging theory method, in polar coordinates r,θ, where u=rcosθ, v=rsinθ, and r>0. We take(10)Pu,v=∑k=1n∑i+j=kpijuivj,Qu,v=∑k=1n∑i+j=kqijuivj.Equation (3) can be written as follows:(11)r˙=ε∑k=1ncosθMkcosθ,sinθ+sinθNkcosθ,sinθrk,θ˙=Sr,θ+ε∑k=1ncosθNkcosθ,sinθ−sinθMkcosθ,sinθrk−1,where(12)Mkcosθ,sinθ=∑i+j=kpijicosiθsinjθ,Nkcosx,sinx=∑i+j=kqijcosiθsinjθ,and(13)Sr,θ=rcosθ−sinθ+arcosθ+sinθ+a−1.Therefore, we have(14)drdθ=ε∑k=1ncosθMkcosθ,sinθ+sinθNkcosθ,sinθrkSr,θ+ε2Rr,θ,ε.The averaged function of equation (14) is(15)f0r=12π∑k=1nrk∫02πcosθMkcosθ,sinθ+sinθNkcosθ,sinθSr,θdθ.Forn=6, we get(16)f0r=∑k=16∑i+j=k6pijAi+1,jr+qijAi,j+1rrk,where(17)Ap,q=12π∫02πcospθsinqθSr,θdθ.According to Theorem2, every simple zero of the average function f0r provides a limit cycle of system equation (3). Now, we prove Theorem 1; in the first step, we compute the integral f0r, and in the second step, the number of its simple zeros is studied.Lemma 1.
From the above, we have(18)A0,0=−G1−G22SH1H2andA1,0=aG1−G24rSH1H2+G1+G24rH1H2,where(19)H1=a−12−r2,H2=a+12−r2,G1=−2r2−2aS+2,G2=−2r2+2aS+2,with(20)S=2r2−a2+2≠0.Proof.
Assume thatz=eiθ and C is the circle z=1; we get(21)2π0dθSr,θ≡2ir2Czz−z1z−z2z−z3z−z4dz,2π0cosθdθSr,θ≡1ir2Cz2+1z−z1z−z2z−z3z−z4dz,whose poles are(22)z1,2=−a−2r2−a2+2±−2r2+2+2a2r2−a2+22r,z3,4=−a+2r2−a2+2±−2r2+2−2a2r2−a2+22r.
By applying the residue theorem, fora>2, we obtain z1<1, z3<1,C encloses the two singular points of the integrand, so(23)A0,0=12π02πdθSr,θ,=1πir2zz−z1z−z2z−z3z−z4dz,=2r2z1z1−z2z1−z3z1−z4+z3z3−z1z3−z2z3−z4.
Therefore, we have(24)A0,0=−−2r2−2a2r2−a2+2+2−−2r2+2a2r2−a2+2+222r2−a2+2a−12−r2a+12−r2,A1,0=12π02πcosθdθSr,θ,=12πir2z2+1z−z1z−z2z−z3z−z4dz,=1r2z12+1z1−z2z1−z3z1−z4+z32+1z3−z1z3−z2z3−z4.
Therefore, we get(25)A1,0=a−2r2−2a2r2−a2+2+2−−2r2+2a2r2−a2+2+24r2r2+2−a2a−12−r2a+12−r2+−2r2−2a2r2−a2+2+2+−2r2+2a2r2−a2+2+24ra−12−r2a+12−r2.
This completes the proof.Lemma 2.
Under the previous notations, we have(a)
A2k+1,0=−a/rA2k,0+r2+1−a2/2r2A2k−1,0(b)
A2k,0=−a/rA2k−1,0+r2+1−a2/2r2A2k−2,0+λk−1/2kk−1!r2,Whereλk−1=3.5….2k−3(c)
A0,2k=∑i=0k−1iCkiA2i,0(d)
An−2m+1,2m=∑i=0m−1iCmiAn−2m+2i+1,0, m=1,…,n+1/2.Proof.
Puttingu=rcosθ, v=rsinθ, we get for (a) and (b)(26)cosnθSr,θ=cosnθa2−2r2−2/2+2rcosθ+a/22,=−2arcosn−1θ+r2+1−a2cosn−2θ+Sr,θcosn−2θ2r2Sr,θ,=−arcosn−1θSr,θ+r2+1−a22r2cosn−2θSr,θ+12r2cosn−2θ.
Thus,(27)An,0=−a2πr∫02πcosn−1θSr,θdθ+r2+1−a24πr2∫02πcosn−2θSr,θdθ+14πr2∫02πcosn−2θdθ.
Therefore,(28)A2k+1,0=−arA2k,0+r2+1−a22r2A2k−1,0,A2k,0=−arA2k−1,0+r2+1−a22r2A2k−2,0+14πr2∫02πcos2k−2θdθ,and(29)∫02πcos2k−1θdθ=πλk−12k−2k−1!,where λk−1=3.5….2k−3,λk=2k−1λk−1 ([12]).
Thus,(30)A2k,0=−arA2k−1,0+r2+1−a22r2A2k−2,0+λk−12kk−1!r2,(31)csin2kθSr,θ=1−cos2θkSr,θ=∑i=0k−1iCkicos2iθSr,θ,then(32)A0,2k=∑i=0k−1iCkiA2i,0,(33)dcosn−2m+1θsin2mθSr,θ=cosn−2m+1θ1−cos2θmSr,θ=∑i=0m−1iCmicosn−2m+2i+1θSr,θ,m=1,…,n+12.
Thus,(34)An−2m+1,2m=∑mi=0−1iCmiAn−2m+2i+1,0,m=1,…,n+12.
This completes the proof.Remark 1.
Ap,2k+1=0.
By Lemmas1 and 2, we have(35)A2,0=14πr2∫02πdθ−arA1,0+r2+1−a22r2A0,0=12r2−araG1−G24rSH1H2+G1+G24rH1H2+r2+1−a22r2−G1−G22SH1H2=−14SH1H2r2−2SH1H2+aSG1+G2+r2+1G1−G2.
Then,(36)A0,2=A0,0−A2,0=−G1−G22SH1H2−−14SH1H2r2−2SH1H2+aSG1+G2+r2+1G1−G2=−2SH1H2+aSG1+G2−r2−1G1−G24SH1H2r2,and we also have(37)A3,0=−4aSH1H2+r2+a2+1SG1+G2+a3r2−a2+3G1−G28SH1H2r3,A1,2=−−4aSH1H2+−r2+a2+1SG1+G2+ar2−a2+3G1−G28SH1H2r3,A4,0=−18SH1H2r4−4r2−2a2+1SH1H2+2ar2+1SG1+G2+r4+2a2+1r2+−a4+2a2+1G1−G2,A2,2=18SH1H2r42−a2−1SH1H2+2aSG1+G2+−r4+2a2r2+−a4+2a2+1G1−G2,and(38)A0,4=−18SH1H2r422r2−a2−1SH1H2+2a−r2+1SG1+G2+r4+2a2−1r2+−a4+2a2+1G1−G2,A5,0=116SH1H2r54a−3r2−2SH1H2+r4+22a2+1r2+−a4+4a2+1SG1+G2+a5r4+10r2−a4+5G1−G2,A3,2=−116SH1H2r54a−r2−2SH1H2+−r4+2a2r2+−a4+4a2+1SG1+G2+−ar4+2aa2+2r2+a−a4+5G1−G2,A1,4=116SH1H2r54ar2−2SH1H2+r4−2r2+−a4+4a2+1SG1+G2+−3ar4+2a2a2−1r2+a5−a4G1−G2,A6,0=−116SH1H2r6−7r4+2−5a2−3r2+2a4−4a2−1SH1H2+a3r4+2a2+3r2+−a4+2a2+3SG1+G2+r6+32a2+1r4+34a2−a4+1r2+6a2−3a4+1G1−G2,A4,2=116SH1H2r6r4+−6a2−2r2+2a4−8a2−2SH1H2+−ar4+2a3+2ar2+2a3−a5+3aSG1+G2+−r6+2a2−1r4+8a2−a4+1r2+6a2−3a4+1G1−G2.
In addition, we have(39)A2,4=−116SH1H2r6r4+21−a2r2+2a4−4a2−1SH1H2+−ar4+2aa2−1r2+a−a4+2a2+3SG1+G2+r6+−2a2−1r4+a4+4a2−1r2+6a2−3a4+1G1−G2,A0,6=116SH1H2r6−7r4+2a2+3r2+2a4−4a2−1SH1H2+3ar4+2aa2−3r2+a−a4+2a2+3SG1+G2+−r6+3−2a2+1r4+3a4−1r2+−3a4+6a2+1G1−G2,A7,0=132SH1H2r74−3a−7r4+47a−10a2−8r2+8a4−4a2+3a−1SH1H2+r6+4a2+12a−1r4+−a4+8a3−8a2+24a−5r2+−4a5+3a4+8a3−12a2+12a−3SG1+G2+5a+4r6+24a2−5a+12r4+−a5−12a4+48a2−25a+12r2+3a5−12a4+24a2−15a+4G1−G2,A5,2=−132SH1H2r743a−7r4+411a−10a2−8r2+8a4−4a2+3a−1SH1H2+−r6+−4a2+12a−5r4+a4+8a3−16a2+24a−7r2+−4a5+3a4+8a3−12a2+12a−3SG1+G2+−5a+4r6+24a2−25a+12r4+a5−12a4+48a2−35a+12r2+3a5−12a4+24a2−15a+4G1−G2,A3,4=132SH1H2r745a−7r4+415a−10a2−7r2+8a4−4a2+3a−1SH1H2+r6+−8a2+12a−5r4+3a4+8a3−24a2+24a−9r2+−4a5+3a4+8a3−12a2+12a−3SG1+G2+−3a+4r6+−4a3+24a2−33a+12r4+3a5−12a4+48a2−45a+12r2+3a5−12a4+24a2−15a+4G1−G2,A1,6=−132SH1H2r743a−7r4+419a−10a2−7r2+8a4−4a2+3a−1SH1H2+−r6+−8a2+12a−1r4+5a4+8a3−32a2+24a−11r2+−4a5+3a4+8a3−12a2+12a−3SG1+G2+3a+4r6+−12a3+24a2−29a+12r4+5a5−12a4+48a2.−55a+12r2+3a5−12a4+24a2−15a+4G1−G2.
Using equation (15), we get(40)f0r=1rSH1H2XSH1H2+YSG1+G2+ZG1−G2,where(41)X=x4r4+x2r2+x0,Y=y6r6+y4r4+y2r2+y0,and(42)Z=z6r6+z4r4+z2r2+a2−1x0+2ay02,with the coefficients xi,yi, and zi the polynomials in the coefficients of a,pi,j, and qi,j.
In fact, there are only ten independent parameters betweenxi, yi, and zi with respect to pij,qij, and a. In order to bound the zeros number of numerator of f0r, it is sufficient to bound the zeros number of(43)Kr=X2S2H12H22−YSG1+G2+ZG1−G22.
Since(44)G1+G2G1−G2=−4aS,G1+G22=41−r2+2G1G2,and(45)G1−G22=41−r2−2G1G2,we have(46)Kr=X2S2H12H22−4Y2S21−r2−4Z21−r2+8aYZS2−2Y2S2−Z2G1G2.
Finally, in order to bound the zeros number of the above expression, we should bound the zeros of the following polynomial:(47)Hr=X2S2H12H22−4Y2S21−r2−4Z21−r2+8aYZS22−4Y2S2−Z22G12G22.
We have(48)Hr=S216Y4a2S4+4Y2r2−1+8YZa+X2a−12−r2a+12−r22−4Y42r2−22−32Y2Z2a2S2+16Z4a2+8Z2r2−14Y2r2−1+8YZa+X2a−12−r2a+12−r2+8Y2Z22r2−22.
Therefore, we get(49)Hr=S2d30r30+⋯+d2r2+d0,where di are the polynomials in a,xi,yi, and zi. We conclude that f0r has at most 15 simple zeros. Hence, Theorem 1 is proved.


## 4. Conclusion

As we know, the limit cycles and a polynomial differential system is the well-known 16th Hilbert problem, which together with the Riemann conjecture are the two problems of the famous list of 23 problems of Hilbert which remain open. In addition, a classical way to produce limit cycles is by perturbing a system which has a center, in such a way that limit cycles bifurcate in the perturbed system from some of the periodic orbits of the period annulus of the center of the unperturbed system; in this work, by using the averaging theory of first order, we study the bifurcation of limit cycles from the period annulus surrounding the origin of a class of cubic polynomial differential systems; when they are perturbed inside the class of all polynomial differential systems of degree six, we have obtained at most 15th limit cycles for this kind of the problem; in the next study, we will try to extend the same tools but for higher degrees.



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