Kuilin WU Haihua LIANG
The second part of Hilbert’s 16th problem(see[6])asks about the maximum number and the location of limit cycles of a planar polynomial vector fields of degreen.If the quadratic centers belong to the Hamiltonian class,then the study of the number of limit cycles bifurcating from a period annulus or annuli(i.e.,the weak Hilbert’s 16th problem forn=2)is finished,and the study of the number of limit cycles bifurcating from a singular loop,or from infinity is partially finished(see[3–4,8–9,12,15,17]and the references therein).If the quadratic centers belong to the reversible class(and do not belong to the Hamiltonian class),then the study seems very difficult,and the known results are very limited.
A weaker version of this problem is proposed by Arnold(see[1])to study the zeros of Abelian integrals,that is the weak Hilbert’s 16th problem or infinitesimal Hilbert’s 16th problem.The problem is related to in the following way.Consider the perturbed system of a Hamiltonian vector fieldXε=XH+εY,where
A closed connected component of a level curve{H=h}is denoted byγhand called an oval ofH.The Abelian integral is
Therefore the number ofisolated zeros ofI(h),counted with multiplicities,provides an upper bound for the number of ovals of{H=h}that generates limit cycles ofXεforε≈0(see[5,12]for details).If the unperturbed system is integrable and non-Hamiltonian,then one has to consider pseudo-Abelian integrals(see[1]for details).As far as we know,most of the papers investigate the Hamiltonian centers and few papers study the non-Hamiltonian centers(see[3,8,17]for instance).Recently,Zhao[16]proved that the cyclicity of the period annulus ofQ4is less than or equal to five.
It is well-known by Iliev[8]that any quadratic polynomial reversible Lotka-Volterra system can be written in the complex form
or in the real form
wherebis a real parameter.Arnold[2]declared that the in finitesimal problem for system(1.1)is still open.For the system(1.1)of genus one,Gautier et al.[4]classify this kind of systems into 6 cases(rlv1)–(rlv6).Until now,[7,11]have studied the cases(rlv1)and(rlv2),respectively,and[5,14]have studied the cases(rlv3)and(rlv4),respectively.
The cyclicity of system(1.1)under quadratic perturbations forb=0,,was studied in[3–4],respectively.In this paper we study the caseb=,that is,the number of limit cycles bifurcates from the period annulus of the following quadratic reversible Lotka-Volterra system:
System(1.2)has a first integral
with the integrating factor
Note that system(1.1)forb=can be reduced to system(1.2)by using a linear transformation.There is a period annulus surrounding the center at(x,y)=(0,0)bounded by three straight lines
corresponding toThe intersection pointsandof the three straight lines are three saddles of systemTherefore,the period annulus can be expressed bywhere the periodic orbith}.
The next result is a particular case of Theorem 3 in[8].For convenience,we state it in the present paper.
Lemma 1.1The exact upper bound for the number of limit cycles produced by the period annulus of system(1.2)under quadratic perturbations is equal to the maximal number of zerosin(counting multiplicities)of the Abelian integral
where a,b and c are arbitrary constant.
The main result of this paper is the following theorems.
Theorem 1.1The cyclicity of the period annulus of system(1.2)under quadratic perturbations is two.
This paper is organized in the following way.In Section 2 we introduce the definitions and the notations that we shall use.In Section 3 we give the proof of Theorem 1.1 by applying Theorem B in[5].We shall rewrite the Abelian integral(1.3)as a linear combination of{I0(h),I1(h),I2(h)}and prove that(I0(h),I1(h),I2(h))forms an extended complete Chebychev system.Due to Theorem B in[5],we turn the problem of the number of zeros of the Abelian integral into a pure algebraic problem,namely,counting zeros of a polynomial.To solve the latter problem we shall use the notion of a resultant.The interested reader is referred to the appendix in[5]for details.
In order to prove the main results,some definitions and lemmas are needed.
Definition 2.1Let f0(x),f1(x),···,fn?1(x)be analytic functions on an open interval L ofR.Then we have the following:
(a)(f0(x),f1(x),···,fn?1(x))is a Chebyshev system(in short,T-system)on L if any nontrivial linear combination
has at most n?1isolated zeros on L.
(b)(f0(x),f1(x),···,fn?1(x))is a complete Chebyshev system(in short,CT-system)on L if(f0(x),f1(x),···,fk?1(x))is a T-system for all k=1,2,···,n.
(c)(f0(x),f1(x),···,fn?1(x))is an extended complete Chebyshev system(in short,ECT-system)on L if,for all k=1,2,···,n,any nontrivial linear combination
has at most k?1isolated zeros on L counted with multiplicities.
It is clear that if(f0(x),f1(x),···,fn?1(x))is an ECT-system onL,then(f0(x),f1(x),···,fn?1(x))is a CT-system onL.However,the reverse implication is not true.Moreover,if(f0(x),f1(x),···,fn?1(x))is a T-system onL,andf(x)is an analytic function and has a constant sign onL,then(f(x)f0(x),f(x)f1(x),···,f(x)fn?1(x))is a T-system onL.
Remark 2.1If(f0(x),f1(x),···,fn?1(x))is an ECT-system onL,then,for eachk=0,1,2,···,n?1,there exists a linear combination with exactlyksimple zeros onL(see[10,13]for instance).
Definition 2.2Let f0(x),f1(x),···,fn?1(x)be analytic functions on an open interval L of.The Wronskian of(f0(x),f1(x),···,fn?1(x))at x∈L is given by
The following result is well-known(see[10,13]for instance).
Lemma 2.1The system(f0(x),f1(x),···,fn?1(x))is an ECT-system on L if and only if,for each k=1,2,···,n,
Recall thatH(x,y)=A(x)+B(x)y2with
It is clear thatH(x,y)has a local minimum at(x,y)=(0,0),B(x)>0 andA(x)have a local minimum atx=0.Denote the period ann(ulus ass)ociated to the center origin byand the projection ofon thex-axis byThen there exists a unique analytic involution functionσ(x),such thatfor allThe next result is a particular case of Theorem B in[5].For convenience,we state it in the present paper.
Theorem 3.1(see[5])Let us consider the Abelian integrals
where,for each h∈(0,h0),γhis the oval surrounding the origin inside the level curve{A(x)+B(x)y2=h}.Let σ be the involution associated to A(x)and we define
Then(I0,I1,···,In?1)is an ECT-system on(0,h0)if s>n?2and(l0,l1,···,ln?1)is a CT-system on(0,xr).
Recall that a mappingσis an involution ifσ?σ=Id andσ≠Id.An involutionσis a diffeomorphism with a unique fixed point.Noting thatli(x)=?li(σ(x)),we have that(l0,l1,···,ln?1)is a CT-system on(0,xr)if and only if(l0,l1,···,ln?1)is a CT-system on(xl,0).
We rewrite the Abelian integral(1.3)asI(h)=aI0(h)+bI1(h)+cI2(h),where
The following lemma,proved in[5],establishes a formula to change the integrand of an Abelian integral into other Abelian integrals that we want.
Lemma 3.1(see[5])Let γhbe an oval inside the level curve{A(x)+B(x)y2=h},and consider a function F(x)such thatis analytic at x=0.Then,for any k∈N,
where
In what follows,we shall apply Theorem 3.1 to prove that(I0(h),I1(h),I2(h))is an ECT-system.By Lemma 3.1,it yields that
However,we discover thatn=3 ands=1 in the integrand of(I0(h),I1(h),I2(h)),so that the conditions>n?2 is not fulfilled.Therefore we must takes=3 and apply Lemma 3.1 to overcome the shortcomings.Applying Lemma 3.1,we have that
where
Exactly in the same way we obtain
where
It is clear that(I0(h),I1(h),I2(h))is an ECT-system in the intervalif and only ifis an ECT-system in the interval
Setting
then
Denote byσthe involution associated toA(x),i.e.,A(x)=A(σ(x)).In order to compute Wronskians,we setz=σ(x)and
Thenli(x)=Li(x,z).We only need to prove that system(L0(x,z),L1(x,z),L0(x,z))is an ECT-system onOn account of
it turns out thatz=σ(x)is defined by
and
We shall depend on Wolfram Mathematica to compute three Wronskians and the resultant between two polynomials to show the nonexistence of zeros of a polynomial on the interval.In the following,we show the following lemma.
Lemma 3.2Systemis an ECT-system on the intervali.e.,systemis an ECT-system on the intervalwhere z is defined by(3.2).
ProofBy Lemma 2.1,we split the proof into three cases to show that the three Wronskians have no zeros on
First,note thatW[L0(x,z)]=L0(x,z).By the common denominator and the factorization,we have
whereα0(x,z)is a polynomial of degree 15 in(x,z)with a very long expression.It follows from direct computations that the resultant with respect tozbetweenα0(x,z)andq(x,z)is(i.e.,eliminatingzfromα0(x,z)=0 andq(x,z)=0),where
Note that=36,andhas a local minimum 35.0652 atx≈0.486258 on the interval.Therefore,α0(x,z)=0 andq(x,z)=0 have no common roots for any,which implies thatW[L0(x,z)]≠0 for anyorz∈(?1,0).
Secondly,by the definition ofW[L0(x,z),L1(x,z)],it follows that
whereα1(x,z)is a polynomial of degree 20 in(x,z)with a very long expression.It follows from direct computations that the resultant with respect tozbetweenα1(x,z)andq(x,z)is 131072(1+x)22(?1+2x)10ζ1(z),where
By calculating,we have thatζ1(0)=2799360,=6561,andζ1(x)has a local minimum 6128.2996 atx≈0.487723 on the intervalTherefore,α1(x,z)=0 andq(x,z)=0 have no common roots on the interval,andW[L0(x,z),L1(x,z)]≠0 for any
Finally,let us compute the third Wronskian and we have that
whereα2(x,z)is a polynomial of degree 50 in(x,z)with a very long expression.The resultant with respect tozbetweenα2(x,z)andq(x,z)is 618475290624(1+x)37(?1+2x)19ζ2(x),where
ζ2(0)=314424115200,=76527504,andhas a local minimum 7.07214×107atx≈0.490262 on the intervalHence,we can assert thatW[L0,L1,L2]≠0 for anyBy Lemma 2.1,the proof of the result is completed.
Proof of Theorem 1.1By Lemma 1.1,Lemma 3.2 and Theorem 3.1,we obtain Theorem 1.1.
Remark 3.1The proof depends on the symbolic computations by Wolfram mathematica and some very long expressions are omitted for the sake of briefness,while the derivative process can be done precisely.
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Chinese Annals of Mathematics,Series B2014年1期