# Critical Periods of Perturbations of Reversible Rigidly Isochronous Centers

Authors: Jiamei Zhou; Na Li; Maoan Han
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the problem of bifurcation of critical periods of a time-reversible polynomial system of degreen. We first present a new method to find the number of zeros of the period function. Then applying our results, we study the number of critical periods for some polynomial systems and obtain new results.


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## Body

## 1. Introduction

Consider a two-dimensional analytic real differential system of the form(1)x˙=F(x,y),y˙=G(x,y),
where we suppose there is a nondegenerate center at the origin O. Let L(h) denote the orbit passing through the point (h,0) with h>0 of (1) and T(h) denote its period. As we all know, the isolated zeros of derivative of T(h) are named critical periods. An interesting problem is to investigate the number of the critical periods. This is an important problem in the research of period functions and a lot of results have been obtained for polynomial differential systems, for example, monotonicity [1–8], finiteness of critical periods [9, 10] and isochronicity [11–13], and local bifurcation of critical periods [14]. The number of critical periods was also discussed in [15–17] for perturbations of isochronous vector fields.Recently, the authors in [18] studied the following system:
(2)x˙=-y+Fn(x,y)+ε∑i=2m‍P~i(x,y),y˙=x+Kn(x,y)+ε∑i=2m‍Q~i(x,y),
where n,m≥2, 0<ε≪1,
(3)Fn(x,y)=∑j=1[n/2]‍an+1-2j,2j-1xn+1-2jy2j-1,P~i(x,y)=∑j=1[(i+1)/2]‍bi+1-2j,2j-1xi+1-2jy2j-1,Kn(x,y)=∑j=1[n/2]‍an+1-2j,2j-1xn-2jy2j,Q~i(x,y)=∑j=1[(i+2)/2]‍ci+2-2j,2j-2xi+2-2jy2j-2.
Under the conditions above, system (2) is time-reversible and hence has a center at the origin.As shown in [15–20], for 0<ε≪1, the period function T~(h,ε) of system (2) can be written as
(4)T~(h,ε)=T0+∑i=1+∞‍T~i(h)εi,
where T0 is a positive constant. The authors [18] gave expressions of T~1(h) and T~2(h). It is also proved in [18] that one critical period can appear for the case n=m=2 and two critical periods can be found for the case n=m=3.In this paper, we consider the following system:(5)x˙=-y+[Fn(x,y)+λF-l(x,y)]+ε[Pm(x,y)+λP-k(x,y)],y˙=x+[Kn(x,y)+λK-l(x,y)]+ε[Qm(x,y)+λQ-k(x,y)],
where
(6)Fn(x,y)=∑j=1[n/2]‍an+1-2j,2j-1xn-2j+1y2j-1,F-l(x,y)=∑j=1[l/2]‍a-l+1-2j,2j-1xl-2j+1y2j-1,Pm(x,y)=∑i=2m‍∑j=1[(i+1)/2]‍bi+1-2j,2j-1xi+1-2jy2j-1,P-k(x,y)=∑i=2k‍∑j=1[(i+1)/2]‍b-i+1-2j,2j-1xi+1-2jy2j-1,Kn(x,y)=∑j=1[n/2]‍an+1-2j,2j-1xn-2jy2j,K-l(x,y)=∑j=1[l/2]‍a-l+1-2j,2j-1xl-2jy2j,Qm(x,y)=∑i=2m‍∑j=1[(i+2)/2]‍ci+2-2j,2j-2xi+2-2jy2j-2,(7)Q-k(x,y)=∑i=2k‍∑j=1[(i+2)/2]‍c-i+2-2j,2j-2xi+2-2jy2j-2,m, n, l, k≥2, and both ε and λ are small parameters. From the above expressions, we know that the system (5) satisfies the following properties: (i) it is a time-reversible system, and the perturbations with ε are of general form; (ii) for all ε, the origin is a center, and for ε=0, the origin is isochronous since there exist polynomials φn(x,y) and φl(x,y) such that Fn(x,y)=xφn(x,y), Kn(x,y)=yφn(x,y), F-l(x,y)=xφl(x,y), K-l(x,y)=yφl(x,y).LetT(h,ε,λ) denote the period of the periodic orbit of system (5) passing through the point (h,0). Then it has the following expansion:
(8)T(h,ε,λ)=T0+∑i=1+∞‍Ti(h,λ)εi,
where
(9)Ti(h,λ)=Ti0(h)+∑j=1+∞‍Tij(h)λj,
and since (5) becomes (2) as λ=0, we have T(h,ε,0)=T~(h,ε).Next section, we give some preliminary lemmas.


## 2. Preliminary Lemmas

In this section, we cite some results obtained in [18].Lemma 1 (see [18]).
For system (5),
(10)T1(h,0)=T10(h)=-∑i=1m-1‍∫02π‍hiSi(θ)(1+(1-n)hn-1∫0θ‍G0(α)dα)i/(n-1)dθ,(11)T2(h,0)=T20(h)=∫02π‍[(∑i=1m-1‍r0i(θ,h)Si(θ))2-∑i=1m-1‍ir0i-1(θ,h)r1(θ,h)Si(θ)(∑i=1m-1‍r0i(θ,h)Si(θ))2]dθ,
where
(12)r0(θ,h)=[h1-n+(1-n)∫0θ‍G0(α)dα]1/(1-n),(13)G0(θ)=∑j=1[n/2]‍an+1-2j,2j-1cosn-2j(θ)sin2j-1(θ),(14)Si(θ)={∑j=0[(i+1)/2]‍(ci+1-2j,2j-bi+2-2j,2j-1)×cosi+2-2j(θ)sin2j(θ)-b0,i+1sini+2(θ),ieven,∑j=0[(i+1)/2](ci+1-2j,2j-bi+2-2j,2j-1)×cosi+2-2j(θ)sin2j(θ),iodd,(15)r1(θ,h)=∫0θ‍[∑i=1m-1‍r0i+1(α,h)Gi(α)-G0(α)×∑i=1m-1‍r0n+i(α,h)Si(α)]·exp(∫αθ‍nr0n-1(β,h)G0(β)dβ)dα,
moreover, if n is odd, S0(θ)≡0, and
(16)T1(h,0)=-∑k=1[(m-1)/2]‍∫02π‍h2kS2k(θ)(1+(1-n)hn-1∫0θ‍G0(α)dα)2k/(n-1)dθ.Lemma 2 (see [18]).
Letn=m=2 and λ=0 in (5). Then
(17)T1(h,0)=(D02-3D20)π+2(D02-D20)πh2+4(D20-D02)πh+2(1-h)[D20h2+2(D02-D20)h+D20-D02]πh21-2h,
where h∈(0,1/2) and Dij are constants depending on the coefficients appearing in Fn, Kn, Pm, and Qm. Furthermore, for sufficiently small ε,(i)
the centerO preserves the isochronicity when D20=D02=0; period function T(h,ε,0) is increasing (resp., decreasing)  for h∈(0,1/2) when D20=0 and D02>0 (resp., < 0);(ii)
there is at most one critical period in(0,1/2) when D20≠0. Moreover, there is exactly one critical period in (0,1/2) if and only if
(18)r*≔D02D20<-3.Lemma 3 (see [18]).
Letn=m=3 and λ=0 in (5). Then
(19)T1(h,0)=1-1-h2(1+1-h2)1-h2π×[1-h22B03-2D30+(B03+B21-D12-D30)×1-h2+2D30h2],
where h∈(0,1) and Bij, Dij are constants depending on the coefficients appearing in Fn, Kn, Pm, and Qm. Furthermore, for sufficiently small ε, one has the following results.(i)
The centerO preserves the isochronicity when
(20)D20=B11-D02=0,D30=B03=B21-D12=0.(ii)
If (20) do not hold, there are at most two critical periods in (0,1) and the maximum is achievable.


## 3. Main Results

As in [18], one can make the polar coordinates x=rcos(θ), y=rsin(θ), so that system (5) becomes
(21)r˙=rnG0(θ)+λrlG-0(θ)+ε[∑i=1m-1‍ri+1Gi(θ)+λ∑i=1k-1‍ri+1G-i(θ)],θ˙=1+ε[∑i=1m-1‍riSi(θ)+λ∑i=1k-1‍riS-i(θ)],
where G0(θ) and Si(θ) are given by (13) and (14),
(22)G-0(θ)=∑j=1[l/2]a-l+1-2j,2j-1cosl-2j(θ)sin2j-1(θ),(23)Gi(θ)={∑j=1[(i+2)/2]‍(bi+2-2j,2j-1+ci+3-2j,2j-2)×cosi+3-2j(θ)sin2j-1(θ),ieven,∑j=1[(i+2)/2](bi+2-2j,2j-1+ci+3-2j,2j-2)×cosi+3-2j(θ)sin2j-1(θ)+c0,i+1sini+2(θ),iodd,(24)G-i(θ)={∑j=1[(i+2)/2](b-i+2-2j,2j-1+c-i+3-2j,2j-2)×cosi+3-2j(θ)sin2j-1(θ),ieven,∑j=1[(i+2)/2](b-i+2-2j,2j-1+c-i+3-2j,2j-2)×cosi+3-2j(θ)sin2j-1(θ)+c-0,i+1sini+2(θ),iodd,(25)S-i(θ)={∑j=0[(i+1)/2](c-i+1-2j,2j-b-i+2-2j,2j-1)×cosi+2-2j(θ)sin2j(θ)-b-0,i+1sini+2(θ),ieven,∑j=0[(i+1)/2](c-i+1-2j,2j-b-i+2-2j,2j-1)×cosi+2-2j(θ)sin2j(θ),iodd.
Obviously, θ˙≡1 when ε=0, which implies that the unperturbed system (5) |ε=0 has an isochronous center at O which is called a rigidly or uniformly isochronous center. We have the following fundamental result.Theorem 4.
Let (8) and (9) hold. Then, for system (5), one has
(26)T11(h)=-∑i=1m-1‍∫02π‍ir00i-1(θ,h)r01(θ,h)Si(θ)dθ-∑i=1k-1‍∫02π‍r00i(θ,h)S-i(θ)dθ,
where
(27)r00(θ,h)=h[1+(1-n)hn-1∫0θ‍G0(α)dα]1/(n-1),(28)r01(θ,h)=∫0θ‍r00l(α,h)G-0(α)·[exp∫αθ‍nr00n-1(β,h)G0(β)dβ]dα.Proof.
We follow the idea of proving Lemma1 which is Theorem 2.1 given in [18].
From (21), we have
(29)drdθ=([∑i=1m-1‍ri+1Gi(θ)+λ∑i=1k-1ri+1G-i(θ)]rnG0(θ)+λrlG-0(θ)+ε[∑i=1m-1‍ri+1Gi(θ)+λ∑i=1k-1ri+1G-i(θ)])×(1+ε[∑i=1m-1‍riSi(θ)+λ∑i=1k-1‍riS-i(θ)])-1.
Letr(θ,ε,λ,h) be the solution of the above equation satisfying the initial condition r(0,ε,λ,h)=h. It can be written as a series of ε(30)r(θ,ε,λ,h)=∑i=0+∞‍ri(θ,λ,h)εi,
where r0(0,λ,h)=h, ri(0,λ,h)=0, i≥1.
Substituting (30) into differential equation (29) and comparing the coefficients of ε0, we have
(31)ddθr0(θ,λ,h)=r0n(θ,λ,h)G0(θ)+λr0l(θ,λ,h)G-0(θ).
We can write r0(θ,λ,h)=r00(θ,h)+r01(θ,h)λ+⋯, where r00(0,h)=h, r0i(0,h)=0, i≥1. Then, substituting r0(θ,λ,h) into (31), then comparing the coefficients of λ0 and λ, we can obtain
(32)dr00(θ,h)dθ=r00n(θ,h)G0(θ),ddθr01(θ,h)=nr00n-1(θ,h)r01(θ,h)G0(θ)+r00l(θ,h)G-0(θ).
Solving the above ODEs associated with the initial values r00(0,h)=h and r01(0,h)=0, we can get (27) and (28). By (21) we have
(33)T(h,ε,λ)=∫02π‍1θ˙dθ=∫02π‍dθ1+ε[∑i=1m-1riSi(θ)+λ∑i=1k-1riS-i(θ)]=2π-∫02π‍[∑i=1m-1‍r0i(θ,λ,h)Si(θ)2π-∫02π‍+λ∑i=1k-1‍r0i(θ,λ,h)S-i(θ)]dθ·ε+O(ε2).
Therefore,
(34)T1(h,λ)=-∫02π[∑i=1m-1‍r0i(θ,λ,h)Si(θ)+λ∑i=1k-1‍r0i(θ,λ,h)S-i(θ)]dθ=-∑i=1m-1‍∫02π‍r00i(θ,h)Si(θ)dθ-[∑i=1m-1‍∫02π‍ir00i-1(θ,h)r01(θ,h)Si(θ)dθ+∑i=1k-1‍∫02π‍r00i(θ,h)S-i(θ)dθ]λ+O(λ2).
It implies that
(35)T10(h)=-∑i=1m-1‍∫02π‍r00i(θ,h)Si(θ)dθ,T11(h)=-[∑i=1m-1‍∫02π‍ir00i-1(θ,h)r01(θ,h)Si(θ)dθ+∑i=1k-1‍∫02π‍r00i(θ,h)S-i(θ)dθ].
Thus, (26) is proved. This ends the proof of Theorem 4.With the same method as forr01(θ,h), we can compute r0i(θ,h) for i≥2 and give an expression of T1i(h) for i≥2 from the proof of Theorem 4, which are omitted here.Define(36)M0(h)=dT10(h)dh,M1(h)=dT11(h)dh.Another fundamental result is as follows.Theorem 5.
LetT(h,ε,λ) be defined as before with h∈(0,K). Then for 0<|ε|≪|λ|≪1, one has the following.(i)
The period functionT(h,ε,λ) of system (5) is increasing (resp., decreasing) in h∈(0,K) if M0(h)>0 (resp., < 0) for all h∈(0,K).(ii)
IfM0(h) is not identically zero, the number of critical periods of  T(h,ε,λ) in (0,K) is not more than the number of zeros (take the multiplicity into consideration) of M0(h) in (0,K). And there are exactly k critical periods if M0(h) has exactly k simple positive zeros.(iii)
IfM0(h)≡0, the number of critical periods of T(h,ε,λ) in (0,K) is not larger than the number of zeros of M1(h) in (0,K). And k critical periods can appear if M1(h) has k simple zeros. Similarly, the period function T(h,ε,λ) of (5) is increasing (resp., decreasing) in h∈(0,K) if M1(h)>0 (resp., < 0).Proof.
From (8) and (9)
(37)∂T(h,ε,λ)∂h=∑i=1+∞‍∂Ti(h,λ)∂hεi=εΓ(h,ε,λ),Γ(h,ε,λ)=∂T1(h,λ)∂h+∑i=1+∞‍∂Ti+1(h,λ)∂hεi,∂T1(h,λ)∂h=∑i=0+∞‍dT1i(h)dhλi=M0(h)+∑i=1+∞‍dT1i(h)dhλi=Γ1(h,λ).
IfM0(h)>0, h∈(0,K), then there exists a λ0 such that for 0<|λ|<λ0, ∂T1(h,λ)/∂h has the same sign with dT10(h)/dh. For 0<ε≪λ, the sign of the function ∂T(h,ε,λ)/∂h is the same as ∂T1(h,λ)/∂h. Then the conclusion (i) is proved.
Further, suppose thath1,h2,…,hk are the k zeros of M0(h) with the multiplicity m1,m2,…,mk, respectively. We only need to prove that the number of critical periods of T(h,ε,λ) is less than or equal to ∑j=1kmj. For the purpose, it suffices to prove that ∂T(h,ε,λ)/∂h has at most ∑j=1kmj roots. Thus, we only need to prove that ∂T(h,ε,λ)/∂h has at most mj zeros near hj, j=1,2,…,k. At first, we prove that ∂T1(h,λ)/∂h has at most mj zeros near hj for λ small. If it is not the case, then ∂T1(h,λ)/∂h has at least mj+1 zeros near hj; that is, there exists λn→0 such that ∂T1(h,λn)/∂h has at least mj+1 zeros hi,n  (i=1,2,…,mj+1) near hj, where hi,n→hj  (n→∞). By Rolls theorem, ∂2T1(h,λn)/∂h2 has at least mj zeros near hj,…, and  ∂mj+1T1(h,λn)/∂hmj+1 has at least one zero hi,n* near hj. Thus, ∂mj+1T1(hi,n*,λn)/∂hmj+1=0. Letting n→∞, we have ∂mj+1T1(hj,0)/∂hmj+1=∂mjM0(hj)/∂hmj=0. This is a contradiction. Thus, ∂T1(h,λ)/∂h has at most mj zeros near hj. Using the same method, for sufficiently small ε, Γ(h,ε,λ) or ∂T/∂h has at most mj zeros near hj. Thus, T(h,ε,λ) has at most mj critical periods for each hj. Thus, we have proved the first part of (ii).
For the second part of (ii), assume thath1,h2,…,hk are the k simple zeros of M0(h); that is, M0(hj)=0 and M0′(hj)≠0, j=1,2,…,k. Then
(38)Γ1(hj,0)=∂T1(h,λ)∂h|(h,λ)=(hj,0)=0,∂Γ1(h,λ)∂h|(h,λ)=(hj,0)=∂2T1(h,λ)∂h2|(h,λ)=(hj,0)≠0.
By the Implicit Function Theorem, there is a unique function hj(λ) such that Γ1(hj(λ),λ)=0 and hj(0)=hj, j=1,2,…,k. Therefore h1(λ),h2(λ),…,hk(λ) are k simple zeros of ∂T1(h,λ)/∂h. By the same method, there is a unique function hj*(ε,λ) such that Γ(hj*(ε,λ),ε)=0, and hj*(0,λ)=hj(λ). The second part of (ii) is proved.
WhenM0≡0, then
(39)∂T1(h,λ)∂h=Γ1(h,λ)=∑i=1+∞‍dT1i(h)dhλi=dT11(h)dhλ+∑i=2+∞dT1i(h)dhλi=λΓ~1(h,λ),Γ~1(h,λ)=M1(h)+∑i=1+∞‍dT1,i+1(h)dhλi.
Then, we can prove conclusion (iii) in the same way as proving conclusion (ii). The proof is completed.


## 4. Application

In this section, we apply Theorems4 and 5 to the cases n=m=2, l=k=3; n=m=3, l=k=4, and n=l=2, m=k=3, respectively.Case 1 (n=m=2,  l=k=3).
In this case, system (5) becomes
(40)x˙=-y+a11xy+λa-21x2y+ε[b11xy+λ(b-11xy+b-21x2y+b-03y3)],y˙=x+a11y2+λa-21xy2+ε[c20x2+c02y2+λ(c-20x2+c-02y2+c-30x3+c-12xy2)],
where a11≠0. For system (40), O is a center because of the symmetry.
We only need to consider the case ofa11>0 in (40); otherwise, we can use the transformation x→-x,y→-y to change (40) into the same form with opposite signs to the coefficients. Further, suppose a11=1, otherwise, system (40) can be simplified as the form of system (41) by the transformation u=a11x, v=a11y. In this case, system (40) becomes
(41)x˙=-y+xy+λa-21x2y+ε[b11xy+λ(b-11xy+b-21x2y+b-03y3)],y˙=x+y2+λa-21xy2+ε[c20x2+c02y2+λ(c-20x2+c-02y2+c-30x3+c-12xy2)].
It is easy to conclude that system (41) ∣ε=0 has a first integral of the form
(42)H(x,y)=2(1+λa-21)x+(1+λa-21)y2-1(1+λa-21)2[1-(1+λa-21)x]2.Theorem 6.
For system (41), one has(i)
(43)T10(h)=(c02-b11-3c20)π+2(c02-b11-c20)πh2+4(c20-c02+b11)πh+(2(1-h)[c20h2+2(c02-b11-c20)h+c20-c02+b11]π[c20h2+2(c02-b11-c20)h)×(h21-2h)-1,whereh∈(0,1/2). For sufficiently small ε and λ, there is at most one critical period if M0(h) is not identically zero and c20≠0; there is exactly one critical period if and only if ρ≔(c02-b11)/c20<-3. Otherwise, the period function T(h,ε,λ) is increasing for h∈(0,1/2) (resp., decreasing) if c20>0 (resp., < 0);(ii)
ifM0(h)=T10′(h)≡0, then for 0<h≪1,
(44)T11(h)=-[A2h2+A3h3+A4h4+A5h5+A6h6+O(h6)]π,where(45)A2=-14(c-02-b-11)-34c-20+34c-30+14(c-12-b-11)-34b-03,A3=2A2,A4=-238c-20+338c-30-78(c-02-b-11)-218b-03+98(c-12-b-21),A5=-92b-03-112c-20+52(c-12-b-21)+212c-30-32(c-02-b-11),A6=-67564c-20-9916b-03-16564(c-02-b-11)+32316c-30+6916(c-12-b-21).There can appear three critical periods in h∈(0,1/2) if T11(h) is not identically zero.Proof.
For system (41), from (13), (14), (22)–(25), we have
(46)G0(θ)=sin(θ),G-0(θ)=a-21cos(θ)sin(θ),G1(θ)=(c20+b11)cos2(θ)sin(θ)+c02sin3(θ),G-1(θ)=(b-11+c-20)cos2(θ)sin(θ)+c-02sin3(θ),G-2(θ)=(b-21+c-30)cos3(θ)sin(θ)+(b-03+c-12)cos(θ)sin3(θ),S1(θ)=c20cos3(θ)+(c02-b11)cos(θ)sin2(θ),S-1(θ)=c-20cos3(θ)+(c-02-b-11)cos(θ)sin2(θ),S-2(θ)=c-30cos4(θ)+(c-12-b-21)cos2(θ)sin2(θ)-b-03sin4(θ).
From (27), we have
(47)r00(θ,h)=h1-h+hcos(θ).
From (10), we have
(48)T10(h)=-∑i=1m-1‍∫02π‍hiSi(θ)[1+(1-n)hn-1∫0θ‍G0(α)dα]i/(n-1)dθ=-∫02π‍hS1(θ)1-h∫0θ‍sin(α)dαdθ=-∫02π‍h[c20cos3(θ)+(c02-b11)cos(θ)sin2(θ)]1-h+hcos(θ)dθ=-∫02π‍hcos(θ)1-h+hcos(θ)×[c20cos2(θ)+(c02-b11)sin2(θ)]dθ=-∫02π‍(1-1-h1-h+hcos(θ))×[c20cos2(θ)+(c02-b11)sin2(θ)]dθ=(c02-b11-3c20)π+2(c02-b11-c20)πh2+4(c20-c02+b11)πh+2(1-h)[c20h2+2(c02-b11-c20)h+c20+b11-c02]πh21-2h.
Then, from [18], we have
(49)dT10(h)dh=(1-δ)F(δ)πδ3(1-δ)3,
where δ=1-2h and
(50)F(δ)=c20δ4+4c20δ3+(2c20+4c02-4b11)δ2+4c20δ+c20.
Support that c20≠0. By (50) we have four zeros of F(δ) as follows:
(51)δ±=-1-1-ρ±1-ρ-21-ρ,δ-±=-1+1-ρ±1-ρ-21-ρ,
where ρ is given by ρ=(c02-b11)/c20. It is not difficult to examine that δ± is not in (0,1) and δ-+ is not in (0,1). In addition, δ--∈(0,1) if and only if ρ<-3. Thus, there is at most one critical period in (0,1/2). If ρ≥-3, F(δ) has no zero in h∈(0,1/2). From the expression of dT10(h)/dh, we can easily check that dT10(h)/dh>0 (resp., < 0) when c20>0 (resp., < 0). Therefore, for sufficiently small ε and λ, ∂T(h,ε,λ)/∂h>0 (resp., < 0) if c20>0 (resp., < 0). The proof of conclusion (i) is completed.
For conclusion (ii), letM0(h)≡0, which gives c20=c02-b11=0 or S1(θ)=0. From (26), we have that
(52)T11(h)=-∫02π‍r00(θ,h)S-1(θ)dθ-∫02π‍r002(θ,h)S-2(θ)dθ=-∫02πh[c-20cos3(θ)+(c-02-b-11)cos(θ)sin2(θ)]1-h+hcos(θ)dθ-∫02π(((1-h+hcos(θ))2)-1((1-h+hcos(θ))2h2[c-30cos4(θ)+(c-12-b-21)cos2(θ)sin2(θ)-b-03sin4(θ)])(1-h+hcos(θ))-2)dθ.
Then we expandT11(h) at h=0 by letting
(53)T11(h)=-[A2h2+A3h3+A4h4+A5h5+A6h6+O(h6)]π.
By (52), we have
(54)A2=-14(c-02-b-11)-34c-20+34c-30+14(c-12-b-21)-34b-03,A3=2A2,A4=-238c-20+338c-30-78(c-02-b-11)-218b-03+98(c-12-b-21),A5=-92b-03-112c-20+52(c-12-b-21)+212c-30-32(c-02-b-11),A6=-67564c-20-9916b-03-16564(c-02-b-11)+32316c-30+6916(c-12-b-21).
Note that
(55)T11′(h)=-[2A2h+3A3h2+4A4h3+5A5h4+6A6h5+O(h5)]π,
and that
(56)det∂(A2,A4,A5,A6)∂(c-20,c-02-b-11,b-03,c-12-b-21)=|-34-238-34-67564-14-78-218-1656434338-92323161498-99166916|=4117516384≠0.
Therefore, A2,A4,A5,A6 can be taken as free parameters. Hence, we can change the sign of A2,A4,A5,A6, satisfying
(57)0<-A2≪A4≪-A5≪A6,
which ensures that T11′(h) has three positive zeros in h near h=0. Thus, conclusion (ii) is proved.We remark that for cubic system (40), we obtain three critical periods using T11(h), which is one more than the results in Lemma 2.3 obtained in [18].Case 2 (n=m=3, l=k=4).
In this case, system (40) becomes
(58)x˙=-y+a21x2y+λa-31x3y+λa-13xy3+ε[b11xy+b21x2y+b03y3+λ(b-11xy+b-21x2y+b-03y3+b-31x3y+b-13xy3)],y˙=x+a21xy2+λa-31x2y2+λa-13y4+ε[c20x2+c02y2+c30x3+c12xy2+λ(c-20x2+c-02y2+c-30x3+c-12xy2+c-40x4+c-22x2y2+c-04y4)],
where we suppose a21≠0. This is a new system which is not studied in [18]. For the case a21>0, by the rescaling
(59)u=a21x,v=a21y,
system (58) can be written as
(60)u˙=-v+u2v+λA-31u3v+λA-13uv3+ε[B11uv+B21u2v+B03v3+λ(B-11uv+B-21u2v+B-03v3+B-31u3v+B-13uv3)],v˙=u+uv2+λA-31u2v2+λA-13v4+ε[D20u2+D02v2+D30u3+D12uv2+λ(D-20u2+D-02v2+D-30u3+D-12uv2+D-40u4+D-22u2v2+D-04v4)],
where
(61)A-31=a-31a21a21,A-13=a-13a21a21,B11=b11a21,B21=b21a21,B03=b03a21,B-11=b-11a21,B-21=b-21a21,B-03=b-03a21,B-31=b-31a21a21,B-13=b-13a21a21,D20=c20a21,D02=c02a21,D30=c30a21,D12=c12a21,D-20=c-20a21,D-02=c-02a21,D-30=c-30a21,D-12=c-12a21,D-40=c-40a21a21,D-22=c-22a21a21,D-04=c-04a21a21.
If a21<0, system (58) can be simplified as (60) similarly by the change u=-a21x, v=-a21y.Theorem 7.
For system (60), one has(i)
(62)T10(h)=1-1-h2(1+1-h2)1-h2π×[2B03-2D30+(B03+B21-D12-D30)×1-h2+2D30h2],whereh∈(0,1). If M0(h) is not identically zero, there are at most two critical periods in h∈(0,1) for sufficiently small ε and λ, and the maximum can be achievable;(ii)
ifM0(h)≡0, one has
(63)T11(h)=1-1-h2(1+1-h2)1-h2π×[1-h22B-03-2D-30+(B-03+B-21-D-12-D-30)×1-h2+2D-30h2],and there are at most two critical periods inh∈(0,1) if T11(h) is not identically zero.Proof.
From (13), (14), (22)–(25), we have that for system (60)
(64)G0(θ)=cos(θ)sin(θ),G-0(θ)=A-31cos2(θ)sin(θ)+A-13sin3(θ),G1(θ)=(B11+D20)cos2(θ)sin(θ)+D02sin3(θ),G2(θ)=(B21+D30)cos3(θ)sin(θ)+(B03+D12)cos(θ)sin3(θ),G-1(θ)=(B-11+D-20)cos2(θ)sin(θ)+D-02sin3(θ),G-2(θ)=(B-21+D-30)cos3(θ)sin(θ)+(B-03+D-12)cos(θ)sin3(θ),S1(θ)=D20cos3(θ)+(D02-B11)cos(θ)sin2(θ),S2(θ)=D30cos4(θ)+(D12-B21)cos2(θ)sin2(θ)-B03sin4(θ),S-1(θ)=D-20cos3(θ)+(D-02-B-11)cos(θ)sin2(θ),S-2(θ)=D-30cos4(θ)+(D-12-B-21)cos2(θ)sin2(θ)-B-03sin4(θ),S-3(θ)=D-40cos5(θ)+(D-22-B-31)cos3(θ)sin2(θ)+(D-04-B-13)sin4(θ)cos(θ).
From (27) and (28), we have
(65)r00(θ,h)=h1-h2sin2(θ),r01(θ,h)=r003(θ,h)1-1-h2sin2(θ)h.
Hence, from (10),
(66)T10(h)=-∑i=1m-1‍∫02π‍hiSi(θ)[1+(1-n)hn-1∫0θ‍G0(α)dα]i/(n-1)dθ=-∫02π‍hS1(θ)1-h2sin2(θ)dθ-∫02π‍h2S2(θ)1-h2sin2(θ)dθ=1-1-h2(1+1-h2)1-h2π×[2B03-2D30+(B03+B21-D12-D30)1-h2+2D30h2].
Thus,dT10(h)/dh=-(2πF~(δ)/δ3(1+δ)2)1-δ2, where δ=1-h2, F~(δ)=D30δ4+2D30δ3+(D12-B21)δ2-2B03δ-B03. If D30=0, F~(δ)=(D12-B21)δ2-2B03δ-B03, we can control the coefficients D12-B21, -2B03, -B03, which gives two zeros of F~(δ) in (0,1). If D30≠0, according to the expression of F~(δ), we suppose that δ1,δ2,δ3,δ4 are four zeros of F~(δ). From the relationship of root and coefficients, we have δ1+δ2+δ3+δ4=-2; thus not all of the zeros are in (0,1). Then from [18], and it is impossible for F~(δ) to have 3 zeros in (0,1). Thus, the first part of conclusion (i) is proved.
Next, we give an example to show that the zeros in(0,1) can be achievable. We can choose D30=1300, D12-B21=-2553, B03=116, then, F~(δ)=1300δ4+2600δ3-2553δ2-232δ-116=(2δ-1)(5δ-2)(130δ2+377δ+58), so, F~(δ) has two simple zeros 2/5 and 1/2 in (0,1). Thus, T(h,ε,λ) has exactly two critical periods in (0,1). The proof of conclusion (i) is completed.
For the conclusion (ii), ifM0(h)≡0, we get D30=B03=D12-B21=0 or S2(θ)=0.
Hence, from (26), we have
(67)T11(h)=-∫02π‍r01(θ,h)S1(θ)dθ-∫02π‍r00(θ,h)S-1(θ)dθ-∫02π‍r002(θ,h)S-2(θ)dθ-∫02π‍r003(θ,h)S-3(θ)dθ=1-1-h2(1+1-h2)1-h2π×[1-h22B-03-2D-30+(B-03+B-21-D-12-D-30)×1-h2+2D-30h2].
So, dT11(h)/dh=-(2πF~~(δ)/δ3(1+δ)2)1-δ2, where δ=1-h2, F~~(δ)=D-30δ4+2D-30δ3+(D-12-B-21)δ2-2B-03δ-B-03. From the proof above, we can get result (ii). This ends the proof of Theorem 7.Corollary 8.
For system (58) with a21>0, one obtains
(68)T10(h)=1-1-a21h2(1+1-a21h2)1-a21h2π×(1-a21h22b03-2c30+(b03+b21-c12-c30)×1-a21h2+2c30a21h2)(a21)-1,
where h∈(0,1/a21). For sufficiently small ε and λ, there are at most two critical periods for T(h,ε,λ) in h∈(0,1) and the maximum number can be achievable.Proof.
LetT(h,ε,λ) denote the period function of (60) and T-(h-,ε,λ) denote the period function of (58). Then we have h=a21h-, and T(h,ε,λ)∣ε=λ=0=T-(h-,ε,λ)∣ε=λ=0. Hence by Theorem 7 and (61), we have for system (58),
(69)T10(h)=1-1-a21h2(1+1-a21h2)1-a21h2π×(1-a21h22b03-2c30+(b03+b21-c12-c30)×1-a21h2+2c30a21h2)(a21)-1,
which gives the formula of T10(h). Other conclusions in the corollary can also be gotten from results in Theorem 7 similarly. This ends the proof of Corollary 8.Case 3 (n=l=2, m=k=3).
In this case, system (5) becomes
(70)x˙=-y+a11xy+λa-11xy+ε[b11xy+b21x2y+b03y3+λ(b-11xy+b-21x2y+b-03y3)],y˙=x+a11y2+λa-11y2+ε[c20x2+c02y2+c30x3+c12xy2+λ(c-20x2+c-02y2+c-30x3+c-12xy2)].
Without loss of generality, we suppose a11=1, otherwise, system (70) can be simplified as the form of system (71) by the transformation u=a11x, v=a11y. In this case, system (70) becomes
(71)x˙=-y+xy+λa-11xy+ε[b11xy+b21x2y+b03y3+λ(b-11xy+b-21x2y+b-03y3)],y˙=x+y2+λa-11y2+ε[c20x2+c02y2+c30x3+c12xy2+λ(c-20x2+c-02y2+c-30x3+c-12xy2)].Theorem 9.
For system (71) and sufficiently small ε and λ, one has(i)
(72)T10(h)=-π(I1+I2),
where
(73)I1=(c02+b11-3c20)+2(c02+b11-c20)h2+4(c20-c02+b11)h+2(1-h)[c20h2+2(c02+b11-c20)h+c20-c02+b11]h21-2h,I2=(c30-b03-c12+b21)I21+(c12-b21+2b03)I22-b03I23,I21=(-4h6-10h5+78h4-140h3-112h2-7(1-2h)5/2h2+12(1-2h)5/2h-6(1-2h)5/2+6-42h)((1-2h)5/2h2)-1,I22=-22h4+h3-7h2+5h-1-(1-2h)5/2(1-2h)5/2,I23=2h2(4h3-8h2+5h-1)(1-2h)7/2.
There can exist three critical periods for system (71) if T10(h) is not identically zero;(ii)
ifM10(h)=T10′(h)≡0, one has
(74)T11(h)=-π(I3+I4),
where
(75)I3=(c-02-b-11+3c-20)+2(c-02-b-11-c-20)h2+4(c-20-c-02+b-11)h+2(1-h)[c-20h2+2(c-02-b-11-c-20)h+c-20-c-02+b-11](h21-2h)-1,I4=(c-30-b-03-c-12+b-21)I21+(c-12-b-21+2b-03)I22-b-03I23.There can appear three critical periods for system (71) if T11(h) is not identically zero.Proof.
From (13), (14), (22)–(25), we have
(76)G0(θ)=sin(θ),G-0(θ)=a-11sin(θ),G1(θ)=(c20+b11)cos2(θ)sin(θ)+c02sin3(θ),G2(θ)=(b21+c30)cos3(θ)sin(θ)+(b03+c12)cos(θ)sin3(θ),S1(θ)=c20cos3(θ)+(c02-b11)cos(θ)sin2(θ),S2(θ)=c30cos4(θ)+(c12-b21)cos2(θ)sin2(θ)-b03sin4(θ),G-1(θ)=(b-11+c-20)cos2(θ)sin(θ)+c-02sin3(θ),G-2(θ)=(b-21+c-30)cos3(θ)sin(θ)+(b-03+c-12)cos(θ)sin3(θ),S-1(θ)=c-20cos3(θ)+(c-02-b-11)cos(θ)sin2(θ),S-2(θ)=c-30cos4(θ)+(c-12-b-21)cos2(θ)sin2(θ)-b-03sin4(θ).
From (10), we can get
(77)T10(h)=-∫02π‍hS1(θ)1-h+hcos(θ)dθ-∫02π‍h2S2(θ)(1-h+hcos(θ))2dθ=-∫02π((h(c20cos3(θ)+(c02-b11)×cos(θ)sin2(θ)))×(1-h+hcos(θ))-1)dθ-∫02π((h2[c30cos4(θ)+(c12-b21)×cos2(θ)sin2(θ)-b03sin4(θ)])×(1-h+hcos(θ))-2)dθ=-π(I1+I2),
where
(78)I1=(c02-b11-3c20)+2(c02-b11-c20)h2+4(c20-c02+b11)h+2(1-h)[c20h2+2(c02-b11-c20)h+c20-c02+b11[c20h2+2(c02-b11-c20)h](h21-2h)-1,I2=(c30-b03-c12+b21)I21+(c12-b21+2b03)I22-b03I23,I21=(-4h6-10h5+78h4-140h3-112h2-7(1-2h)5/2h2+12(1-2h)5/2h-6(1-2h)5/2+6-42h)×((1-2h)5/2h2)-1,I22=-22h4+h3-7h2+5h-1-(1-2h)5/2(1-2h)5/2,I23=2h2(4h3-8h2+5h-1)(1-2h)7/2.
By the same method as the proof of Theorem 6, we have conclusion (i).
For conclusion (ii), ifM10(h)=T10′(h)≡0, giving S1(θ)=S2(θ)=0, thus from (26),
(79)T11(h)=-∫02π‍r00(θ,h)S-1(θ)dθ-∫02π‍r002(θ,h)S-2(θ)dθ=-∫02π((h[c-20cos3(θ)+(c-02-b-11)cos(θ)sin2(θ)])×(1-h+hcos(θ))-1)dθ-∫02π((h2[c-30cos4(θ)+(c-12-b-21)×cos2(θ)sin2(θ)-b-03sin4(θ)])×(1-h+hcos(θ))-2)dθ=-π(I3+I4),
where
(80)I3=(c-02-b-11+3c-20)+2(c-02-b-11-c-20)h2+4(c-20-c-02+b-11)h+2(1-h)[c-20h2+2(c-02-b-11-c-20)h+c-20-c-02+b-11](h21-2h)-1,I4=(c-30-b-03-c-12+b-21)I21+(c-12-b-21+2b-03)I22-b-03I23.
We can get conclusion (ii) with the same method as the proof of Theorem 6. This ends the proof of Theorem 9.



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*Source: 481501-2013-05-30.xml*