# A Note on Small Amplitude Limit Cycles of Liénard Equations Theory

Authors: Yassine Bouattia; Djalil Boudjehem; Ammar Makhlouf; Sulima Ahmed Zubair; Sahar Ahmed Idris
Journal: Mathematical Problems in Engineering
 (2021)

---

## Abstract

In this paper, we demonstrate using a counterexample for a theorem of the small amplitude limit cycles in some Liénard systems and show that that there will be no solutions unless we add an extra condition. A new condition is derived for some specific Liénard systems where a violation of the small amplitude limit cycles theorem takes place.


---

## Body

## 1. Introduction

A lot of previous works consider studies on a limit cycles’ existence for Liénard systems [1–3]. It represents a very important class of nonlinear systems due to its appearance in some branches of science and engineering as well as in some ecological models, planar physical models, and even in some chemical models, where using a suitable transformation can change these systems into nonlinear Liénard systems. However, an extensive attention has been also devoted to the question of its uniqueness [4–6]; this uniqueness can be verified using different ways of methods based on Poincare–Bendixson theorem. In [4], Zhou etal. proposed a set of theorems for the limit cycles’ uniqueness for the Liénard systems; the proposed theorems represent a guarantee to complete the proof of some previous works’ propositions. In [7], Sabatini and Gabriele studied the uniqueness of limit cycles for a class of planar dynamical systems taking into account those which are equivalent to Liénard systems, and they have also proved a theorem for limit cycles of a class of plane differential systems. In the paper proposed by Li and Llibre [8], the authors proved that for any classical Liénard differential equation of degree four, there exists at most one hyperbolic limit cycle. In [9], a sufficient condition for the existence and the uniqueness of limit cycles for Liénard systems has been proposed for some applications.In the theory of small amplitude limit cycles, Liénard systems haven solutions, However, in this paper, we use a counterexample to demonstrate that the existence of n solutions for some systems is not true unless we add an extra new condition.We consider in our study the systems given by the following form:(1)x˙=y−Fx,y˙=−gx,where F and g are polynomials of order n of x and y. For several classes of such systems and in cases where the critical point is under perturbation of the coefficients in F and g, the maximum number of limit cycles that can bifurcate out can be formulated in terms of the degree of F and g [10–12].


## 2. Bendixon Criterion

We consider the following autonomous system:(2)x˙=Px,y,y˙=Qx,y.LetX=P,Q be the vector field and divX=∂P/∂x+∂Q/∂y.Theorem 1.
LetD be a simply connected open subset of ℝ. If divX=∂P/∂x+∂Q/∂y is of constant sign and not identically zero in D, then the system defined by 2 has no periodic orbit lying entirely in the region D.Proof.
Ifγ is a periodic orbit in D, then Px,ydy−Qx,ydx=0 on γ. Since the interior U of γ is simply connected, we can apply Greenâs theorem to obtain the following:(3)0=∮Px,ydy−Qx,ydx=∬∂P∂xx,y+∂Q∂yx,ydxdy.
This is a contradiction since our hypothesis implies that the integral on the right cannot be zero.Proof.
If we suppose that the system given by 2 has a periodic solution of a periodT, then it has a closed orbit Γ in D. Let G be the interior of Γ, we can apply Greens theorem to obtain the following:(4)I=∬G∂P∂xx,y+∂Q∂yx,ydxdy=∮ΓPx,ydy−Qx,ydx,=∫0TPxt,ytdydt−Qxt,ytdxdtdt=∫0TPxt,ytQxt,yt−Qxt,ytPxt,ytdt=0.
SincedivX is either >0 or <0, then ∬GdivXdxdy will not be zero; therefore, there are no periodic solutions.


## 3. A Note on Liénard Equations Theory

We consider the following system:(5)x˙=y+c2y2+⋯+cLyL−a1x+a3x3+⋯+a2n+1x2n+1,y˙=−x+b2x2+⋯+bNxN,where c2,c3,…,cL,a1,a3,…,a2n+1,b2,b3,… and bN are real coefficients.Theorem 2 (see [1]).
For the system of form (2), there are at most n small-amplitudes limit cycles. If a1,a3,…,a2n+1 are so chosen that(6)a1≪a3≪⋯≪a2n+1,a2j−1a2j+1<0j=1,…,n,then there are exactly n small-amplitudes limit cycles.Proof.
(counterexample).
We suppose the following system:(7)x˙=X=ϕy−Fx,y˙=Y=−gx,where(8)Fx=∑j=0n−1j2j+11010n−jx2j+1,n=2m,gx=x+b2x2+⋯+bNxN,ϕy=y+c2y2+⋯+cLyL.
By puttinga2j+1=−1j/2j+11010n−j, we obtain(9)a2j−1a2j+1=1/2j−11010n−j+11/2j+11010n−j=2j+12j−110−10.
As(10)a2j−1a2j+1=2j+12j−110−10≤3×10−10≪1,forj=1,…,n,then(11)a1≪a3≪⋯≪a2n+1,a2j−1a2j+1<0,forj=1,…,n.
However,(12)divX,Y=∂ϕy−∑j=0na2j+1x2j+1∂x+∂−gx∂y,=−∑j=0n−1j1010n−jx2j=−∏j=0n/2−1x2−11010cos2j+1n+1π2+11020sin22j+1n+1π<0,because(13)fx=∑j=0n−1j1010n−jx2j,=∏j=0n/2−1x−10−5ei2j+1π/2n+2x−10−5e−i2j+1π/2n+2x+10−5ei2j+1π/2n+2x+10−5e−i2j+1π/2n+2=∏j=0n/2−1x2−11010cos2j+1n+1π2+11020sin22j+1n+1π.Theorem 3 (see [2]).
We consider the following equation:(14)r˙=rv0+v1r2+v2r4+⋯+vnr2n.If the focus valuesvj given in equation (3) satisfy the following conditions:(15)vjvj+1<0,andvj≪vj+1≪1,forj=0,1,2,…,n−1,then the polynomial equation given by r˙=0 in equation (3) has n positive real roots for r2.Proof.
(counterexample).
We consider the following equation:(16)r˙=fr=r∑j=0n−1j1010n−j+10r2j.
By puttingvj=−1j/1010n−j+10, we obtain(17)vjvj+1<0,andvj≪vj+1≪1,forj=0,1,2,…,n−1,because(18)vjvj+1=1010n−j−1+101010n−j+10=10−10≪1,forj=0,1,2,…,n−1,andvn=10−10≪1.
However,(19)frr=∑j=0n−1j1010n−j+10r2j=10−10∏j=0n/2−1r2−11010cos2j+1n+1π2+11020sin22j+1n+1π≠0,∀r∈ℝ.


## 4. Examples

In this section, by using the counterexample, we can demonstrate that Theorems2 and 3 are not true. However, the previous theorems will be true if we add the following condition: a0/a2≪a2/a4≪a4/a6≪a6/a8≪⋯≪a2j−2/a2j, j=1,…,n.Example 1.
We consider the following equation:(20)r˙=fr,orfr=∑j=04vjr2j+1=r11050−11040r2+11030r4−11020r6+11010r8.
We have(21)vjvj+1<0,andvj≪vj+1≪1,forj=0,1,…,3,because(22)vjvj+1=10−10≪1,forj=0,1,…,3,andv4=10−10≪1.
However,(23)frr=10−10r−1105cosπ10+isinπ10r−1105cos3π10+isin3π10,r−1105cos7π10+isin7π10r−1105cos9π10+isin9π10r−1105cos11π10+isin11π10r−1105cos13π10+isin13π10r−1105cos17π10+isin17π10r−1105cos19π10+isin19π10≠0,∀r∈ℝ,where the system roots are given by(24)r1=1105cosπ10+isinπ10=9.5106×10−6+3.0902×10−6i,r2=1105cos3π10+isin3π10=5.8779×10−6+8.0902×10−6i,r3=1105cos7π10+isin7π10=−5.8779×10−6+8.0902×10−6i,r4=1105cos9π10+isin9π10=−9.5106×10−6+3.0902×10−6i,r5=1105cos11π10+isin11π10=−9.5106×10−6−3.0902×10−6i,r6=1105cos13π10+isin13π10=−5.8779×10−6−8.0902×10−6i,r7=1105cos17π10+isin17π10=5.8779×10−6−8.0902×10−6i,r8=1105cos19π10+isin19π10=9.5106×10−6−3.0902×10−6i.Example 2.
Let us consider now the following system:(25)r˙=fr=r∑j=04−1j1044−j+1r2j=r1101024−110256r2+11064r4−11016r6+1104r8,with positive roots such as(26)r1=10−6r2=10−24r3=10−96r4=10−384,because(27)vjvj+1<0,andvj≪vj+1≪1,forj=0,1,…,3,vjvj+1≪vj+1vj+2≪1,forj=0,…,2.Example 3.
We suppose the following system:(28)x˙=y−εb1x+b3x3+b5x5+b7x7+b9x9,y˙=−x,orb1=210−196,b3=−8310−192+10−152+10−132+10−112,b5=16510−148+10−128+10−110+10−88+10−68+10−48,b7=−1283510−4+10−44+10−64+10−84,b9=25663.
By puttinga2j+1=εb2j+1,j=0,4¯, we obtain(29)x˙=y−a1x+a3x3+a5x5+a7x7+a9x9,y˙=−x,and by applying the first-order averaging method [13, 14] on (14), we obtain(30)f0r=rr8−10−4+10−44+10−64+10−84r6+10−148+10−128+10−110+10−88+10−68+10−48r4−10−192+10−152+10−132+10−112r2+10−196.
f0r=0 implied r1=10−2, r2=10−22, r3=10−32, and r4=10−42, then there are exactly 4 small-amplitudes limit cycles ri, i=1,4¯
Note thata2j+1/a2j+3=b2j+1/b2j+3≪1 for j=0,…,3 and b2j+1/b2j+3≪b2j+3/b2j+5 for j=0,…,2.


## 5. Conclusion

In this work, by using a counterexample for a theorem of the small amplitude limit cycles in some Liénard systems, we have shown that that there will be no solutions unless an extra condition is added. In addition, a new condition is derived for some specific Liénard systems where a violation of the small amplitude limit cycles theorem takes place. However, these theorems will be true if we add the following condition:a0/a2≪a2/a4≪a4/a6≪a6/a8≪⋯≪a2j−2/a2j, j=1,…,n.



---

*Source: 3101000-2021-09-29.xml*