On the number of critical periods for planar polynomial ... - Unesp

Nonlinear Analysis 69 (2008) 1889–1903www.elsevier.com/locate/naOn the number of critical periods for planar polynomial systemsAnna Cima a , Armengol Gasull a,∗ , Paulo R. da Silva ba Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spainb IBILCE–UNESP, CEP 15054–000 S. J. Rio Preto, São Paulo, BrazilReceived 11 April 2007; accepted 20 July 2007AbstractIn this paper we get some lower bounds for the number of critical periods of families of centers which are perturbations of thelinear one. We give a method which lets us prove that there are planar polynomial centers of degree l with at least 2[(l − 2)/2]critical periods as well as study concrete families of potential, reversible and Liénard centers. This last case is studied in more detailand we prove that the number of critical periods obtained with our approach does not increases with the order of the perturbation.c○ 2007 Elsevier Ltd. All rights reserved.MSC: 34C23; 37C10; 37C27Keywords: Period function; Critical periods; Perturbations; Potential systems; Reversible centers; Hamiltonian centers; Liénard centers1. IntroductionConsider the set V l of all the polynomial vector fields of the forml∑ẋ = −y + P(x, y) = −y + P n (x, y),n=2l∑ẏ = x + Q(x, y) = x + Q n (x, y),n=2(1)having a center at the origin, where P n and Q n are homogeneous polynomials of degree n. Given a vector field X ∈ V llet P be the period annulus of the center, i.e. the open subset of the phase plane formed by all the periodic orbits ofX surrounding the origin. The period function T : P −→ R + associates with any point (x, y) ∈ P the period of theperiodic orbit passing through (x, y). Since all the points belonging to the same periodic orbit γ have the same periodwe may denote by T (γ ) the period of the periodic orbit. We say that T is an increasing (resp. decreasing) functionif for any couple of periodic orbits γ 0 and γ 1 in P with γ 0 contained in the region surrounded by γ 1 , we have thatT (γ 1 ) − T (γ 0 ) > 0 (resp. T (γ 1 ) − T (γ 0 ) < 0). The local maximum or minimum of the period function are calledcritical periods.∗ Corresponding author. Tel.: +34 93 581 2909; fax: +34 93 581 2790.E-mail addresses: cima@mat.uab.cat (A. Cima), gasull@mat.uab.cat (A. Gasull), prs@ibilce.unesp.br (P.R. da Silva).0362-546X/$ - see front matter c○ 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2007.07.031

1892 A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903Table 1Values of N l k (pot)k/l 2 3 4 5 6 7 81 0 0 0 1 1 2 22 0 0 2 2 4 4 63 0 0 1 1 2 3 44 0 1 2 2 4 5 65 0 0 1 1 3 3 46 0 0 2 2 4 4 67 0 0 2 2 3 – –8 0 1 – ≥3 – – –Proposition 3. With the above notation,[ ] l − 3N1 l = N 1 l (Ham) = N 1 l (rev) = N 1 l (pot) = ,2[ ] l − 2N2 l ≥ N 2 l (Ham) ≥ N 2 l (pot) = 2 .2Notice that from the previous results it follows that N 3 > N 3 2 (pot).Proposition 4. The numbers Nk l (pot) for k ≤ 8 and l ≤ 8 are given in Table 1.We make some comments about the results of Table 1:– The results of the first two columns are not surprising because, as we have already said, it is proved in [5] thatN 2 (pot) = 0 and in [11] that N 3 (pot) = 1.– The results of our study give for the numbers N 4 (pot) and N 5 (pot) the same lower bounds as are obtained in [5]by studying the period constants at the origin, namely 2 and 3, respectively.– The easiest open problem suggested by the above table is that of whether N 4 (pot) is 2 or greater.– The empty places in the last two rows correspond to values of k and l for which we have not been able to completeall the computations involved.The proof of Proposition 3, as well as some examples showing that N8 5(pot) ≥ 3 and N 4 7 (pot) ≥ 5, are given inSection 4.Liénard centers are studied in Section 5. Recall that N 2 (Lie) = 0, see [2]. We prove the following result:Theorem 5. Fix l even. The following hold:(i) For any j ∈ N, N2 l j−1 (Lie) = 0 and N 2 l j (Lie) = N 2 l (Lie) ≤ l − 2.(ii) For l = 2, 4, 6 and 8 we have that N2 l (Lie) = l − 2.Note that the above results say that the maximum number of critical periods obtained by using our method doesnot increase when more T j (ρ) are considered. For instance this does not happen when we study potential systems;see Table 1. This property makes it natural to wonder whether N2 l(Lie) is or is not equal to N l (Lie), as happens forl = 2. Notice also that a proof that N2 l (Lie) = l − 2 would give a new proof of Theorem 1 in the case l even.Finally, reversible centers are studied in Section 6. In particular we have studied the case l = 2 and the maximumnumber of critical periods that we have found is 2, giving new support to the conjecture stated in [3] that N 2 (rev) = 2.2. Proof of Theorem 2In this section we prove Theorem 2. (a) For the sake of simplicity, we only give the details for the computation ofT k (ρ), k = 1, 2, 3. The T k (ρ) for k ≥ 4 can be obtained similarly. We start by finding the first terms of the solution

A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903 1893r(θ, ε; ρ) = ∑ ∞i=0 r i (θ; ρ)ε i of (4) with the initial condition r(0, ε) = ρ. By replacing this expression in (4) weobtain thatr 0 (θ) ≡ ρ, r 1 (θ; ρ) =r 2 (θ; ρ) =∫ θNote that for any k ≥ 1,0∫ θ0A 1 (ρ, ψ)dψ,(A 2 (ρ, ψ) − A 1 (ρ, ψ)B 1 (ρ, ψ) + ∂ A )1∂r (ρ, ψ)r 1(ψ; ρ) dψ.B k (r(θ, ε; ρ), θ) = B k (ρ, θ) + ∂ B k∂r (ρ, θ)r 1(θ; ρ)ε + O(ε 2 )and that1∣1 + B 1 (r, θ)ε + B 2 (r, θ)ε 2 + B 3 (r, θ)ε 3 + O(ε 4 )= 1 + (−B 1 (r, θ))ε + (−B 2 (r, θ) + B1 2 (r, θ))ε2∣r=r(θ,ε;ρ)+ (−B 3 (r, θ) + 2B 1 (r, θ)B 2 (r, θ) − B1 3 (r, θ))ε3 + O(ε 4 )| r=r(θ,ε;ρ)(= 1 − εB 1 (ρ, θ) + ε 2 B1 2 (ρ, θ) − B 2(ρ, θ) − ∂ B )1(ρ, θ)r 1 (θ; ρ)∂r(+ ε 3 − ∂ B 1(ρ, θ)r 2 (θ; ρ) − ∂ B 2(ρ, θ)r 1 (θ; ρ) − 1 ∂ 2 B 1 (ρ, θ)∂r∂r2 ∂r 2 r1 2 (θ; ρ))∂ B 1 (ρ, θ)+ 2B 1 r 1 (θ; ρ) − B 3 (ρ, θ) +2B 1 (ρ, θ)B 2 (ρ, θ) − B1 3 ∂r(ρ, θ) + O(ε 4 ).From (3) we know that∫ 2π ∫dθ 2πT (ρ, ε) =0˙θ = 10 1 + B 1 (r, θ)ε + B 2 (r, θ)ε 2 + · · · ∣ dθr=r(θ,ε;ρ)∫ 2π∫ 2π(= 2π − ε B 1 (ρ, θ)dθ + ε 2 B1 2 (ρ, θ) − B 2(ρ, θ) − ∂ B )1(ρ, θ)r 1 (θ; ρ) dθ00∂r∫ 2π(+ ε 3 − ∂ B 1(ρ, θ)r 2 (θ; ρ) − ∂ B 2(ρ, θ)r 1 (θ; ρ) − 1 ∂ 2 B 1 (ρ, θ)0 ∂r∂r2 ∂r 2 r1 2 (θ; ρ))∂ B 1 (ρ, θ)+ 2B 1 r 1 (θ; ρ) −B 3 (ρ, θ) + 2B 1 (ρ, θ)B 2 (ρ, θ) − B1 3 ∂r(ρ, θ) dθ + O(ε 4 ),as we wanted to prove.(b) The result follows by applying the implicit function theorem to the equation0 = 1 ε k ∂T (ρ, ε)∂ρ= T k ′ (ρ) + O(ε). □3. Proof of Theorem 1It is clear that Theorem 1 is a consequence of Proposition 3. We prove this in the sequel.Proof of Proposition 3. We start by proving that N1lεP(x, y), ẏ = x + εQ(x, y). ThusT 1 (ρ) = −∫ 2π0B 1 (ρ, θ)dθ = −∫ 2π0≤ [l−32]. Take any center in V l of the form ẋ = −y +∣x Q(x, y) − y P(x, y) ∣∣∣(x,y)=(ρρ 2dθ.cos θ,ρ sin θ)

andMoreover∫ 2π0B 2 (ρ, ψ)dψ =l∑(j=2A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903 1895b 2 j−1The integrals ∫ 2π0B1 2dθ and ∫ 2π0l∑n=2n=2)(2 j)!4 j ( j!) 2 2π ρ 2 j−2 .( )∑n−1(2n + 1)!2πa 2m a 2n−2m(2n + 2)4m=1 n (n!) 2 ρ 2n−2∂ B 1∂r r 1dθ are given respectively by(l∑ ∑n−1() ) (2n + 1)!2πa 2m a 2n−2m (2n − 2m − 1) −(2m + 1)(2n + 2)4m=1n (n!) 2 ρ 2n−2 .Collecting all the above results, by using Theorem 2, we get that T 1 = 0 and the expression (7) for T 2 . Take k suchthat l = 2k or l = 2k + 1. In order to see that N2 l(pot) ≤ 2k − 2 we observe that T 2(ρ) is an even polynomial withouta constant term and we claim that its degree is 4k −2. If l = 2k, then 2l−2 = 4k −2. If l = 2k +1, then 2l−2 = 4k,but the coefficient of ρ 4k is zero. This is so because b 4k+1 = 0 (b i runs from i = 2 until 2k + 1) and a 2m a 2l−2m = 0for all m = 1, 2, . . . , l − 1 (a 2i takes the values a 2 until a 2k ). Hence, T2 ′ (ρ) has at most 2k − 2 positive zeros and wehave that N2 l (pot) ≤ 2k − 2.To prove that the equality holds we need to choose the values a 2 , a 4 , . . . , a 2k and b 3 , b 5 , . . . , b 2k−1 such that T2 ′(ρ)has arbitrary coefficients. Indeed this will be a consequence of the following formula, which shows a kind of triangularstructure for the coefficients of the polynomial:T 2 (ρ) = −k∑{J 2 (n)b 2n−1 + f n (a 2 , a 4 , . . . , a 2k )} ρ 2n−2n=2{ (−1) l − 1+2+2k−1 ∑n=k+2}J 2 (k + 1)b 2k+1 + [J 1 (2, 2k) + J 1 (2k, 2)] a 2 a 2k + f k+1 (a 4 , a 6 , . . . , a 2k ) ρ 2k{[J 1 (2n − 2k, 2k) + J 1 (2k, 2n − 2k)] a 2n−2k a 2k+ f n (a 2n−2k+2 , a 2n−2k+4 , . . . , a 2k )} ρ 2n−2 + J 1 (2k, 2k)a 2 2k ρ4k−2 .Here the functions f n are polynomials in their variables. Write T 2 (ρ) = ∑ 2k−1i=1 t 2iρ 2i . Consider a new polynomialτ(ρ) = ∑ 2k−1i=1 τ 2iρ 2i with τ 4k−2 positive and such that τ ′ (ρ) has 2k − 2 simple positive roots. Let us see how tochoose a 2 , a 4 , . . . , a 2k and b 3 , b 5 , . . . , b 2k−1 such that T 2 (ρ) = τ(ρ). We start by fixing a 2k such that t 4k−2 = τ 4k−2 .Notice that this is possible for the triangular structure of the formula for the coefficient of ρ 4k−4 . We can continuewith the same procedure until we obtain t 2k . At this point we have fixed a 2k , a 2k−2 , . . . , a 4 , a 2 and b 2k+1 (if it exists).The rest of the coefficients can be easily obtained by choosing b 2k−1 , b 2k−3 , . . . , b 3 . □4. Potential systemsIn this section we give some details about the proof of Proposition 4. To obtain the results given in Table 1 weapply the method developed in Theorem 2 to potential centers. For the sake of brevity we omit the statement of theresult for k = 7. The lower bound for two representative examples, N4 7(pot) and N 8 5 (pot), is given in more detail atthe end of the section.The following five sets of conditions, depending on the coefficients appearing in (8): a 2 , . . . , a 8 , b 2 , . . . , b 8 ,c 2 , . . . , c 8 , d 2 , . . . , d 8 , w 2 , . . . , w 8 , will be used in our analysis.(C1) a 3 = a 5 = a 7 = 0.(C2) a 6 = a 8(= 0, b 3 = 10 9 a2 2 , b 5)= 14 5 a 2a 4 , b 7 = 3625 a2 4 .(C3) c 7 =3564 6340a 4 b 4 + 15 8 a 2b 6 , c 5 = 8 ( 745a 4 b 2 + 7 4 a )2b 4 , c3 = 209 a 2b 2 .

A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903 1897T 4,6 = 6340 a 4c 4 + 15 8 b 2b 6 − 3564 d 7 + 6380 b2 4 − 25936 a3 2 a 4 + 15 8 a 2c 6 ,T 4,8 = 9964 b 4b 6 + 385192 a 2c 8 − 100180 a2 2 a2 4 + 9964 a 4c 6 + 385192 b 2b 8 ,T 4,9 = 1172 a 2(35a 2 b 8 + 27a 4 b 6 ),T 4,10 = 1001640 a 4c 8 −T 4,11 = 11320 a 4(72a 4 b 6 + 245a 2 b 8 ),T 4,12 = 715512 b 6b 8 −69 0698000 a 2a 3 4 + 12871792 b2 6 + 1001640 b 4b 8 ,33 03316 000 a4 4 , T 4,13 = 1001320 a2 4 b 8, T 4,14 =12 15518 432 b2 8 .(e) If the coefficient values satisfy conditions (C1)–(C4) then T 1 ≡ T 2 ≡ T 3 ≡ 0, T 4 (ρ) = π 385192 a 2c 8 ρ 8 andT 5 (ρ) = π ∑ 10l=1 T 5,lρ l with T 5,1 = T 5,3 = T 5,5 = T 5,7 = 0 andT 5,2 = 5 3 a 2d 2 − 3 4 w 3 + 5 3 b 2c 2 ,T 5,4 = − 5 8 w 5 + 7 4 b 2c 4 + 7 4 a 2d 4 − 14027 a3 2 b 2 + 7 4 b 4c 2 ,T 5,6 = 6340 b 4c 4 + 15 8 a 2d 6 − 3564 w 7 − 25936 a3 2 b 4 + 15 8 b 2c 6 ,T 5,8 = 9964 b 4c 6 + 385192 a 2d 8 + 385192 b 2c 8 ,T 5,9 = 38572 a2 2 c 8, T 5,10 = 1001640 b 4c 8 .(f) If the coefficient values satisfy conditions (C1)–(C5) then T 1 ≡ T 2 ≡ T 3 ≡ 0, T 4 (ρ) = π 385192 a 2c 8 ρ 8 ,∑T 5 (ρ) = π[T 5,8 ρ 8 + T 5,9 ρ 9 + T 5,10 ρ 10 14] and T 6 (ρ) = π T 6,l ρ lwith T 6,1 = T 6,3 = T 6,5 = T 6,7 = T 6,13 = 0 andT 6,2 = 5 6 c2 2 − 3 4 v 3 + 5 3 a 2w 2 + 5 3 b 2d 2 ,T 6,4 = − 5 8 v 5 + 7 4 a 2w 4 + 7 4 c 2c 4 − 14027 a3 2 c 2 + 7 4 b 4d 2 − 709 a2 2 b2 2 + 7 4 b 2d 4 ,T 6,6 = 6340 b 4d 4 + 15 8 c 2c 6 − 3564 v 7 + 6380 c2 4 − 25936 a3 2 c 4 + 158 a 2w 6 + 1855324 a6 2 − 25912 a2 2 b 2b 4 + 158 b 2d 6 ,T 6,8 = 9964 b 4d 6 + 385192 a 2w 8 − 100180 a2 2 b2 4 + 9964 c 4c 6 + 385192 c 2c 8 − 16516 a3 2 c 6 + 385192 b 2d 8 ,T 6,9 = 1172 a 2(35a 2 d 8 + 27b 4 c 6 + 70b 2 c 8 ),T 6,10 = 1001640 c 4c 8 + 12871792 c2 6 + 1001640 b 4d 8 +38 8853456 a3 2 c 8,T 6,11 = 59364 a 2b 4 c 8 , T 6,12 = 715512 c 6c 8 , T 6,14 =l=112 15518 432 c2 8 .An example showing that N 7 4(pot) ≥ 5: Consider a system like in Proposition 6 satisfying conditions (C1)–(C3),a 4 = 1, b 6 = 0 and, of course, b 8 = c 8 = d 8 = w 8 = v 8 = y 8 = z 8 = 0. For this choice of parameters we have thatT 1 = T 2 = T 3 = 0 and

ThenwhereT 1 (ρ) ≡ 0 and T 2 (ρ) =A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903 18992k∑n=2( )∑n−1a 2m a 2n−2m I (n, m) ρ 2n−2m=1I (n, m) = 2(n − 2m + 2n2 − 2nm)(2n)!4 n π, (11)(n + 1)(2n − 2m + 1)(n!) 2and a j = 0 for j 2k + 2.Proof. These systems are given in polar coordinates bywithWe haveṙ = ε A 1 (r, θ) + ε 2 A 2 (r, θ), ˙θ = 1 + εB 1 (r, θ) + ε 2 B 2 (r, θ)A 1 =k∑a 2m r 2m cos 2m+1 θ, A 2 =m=1B 1 = − sin θT 1 (ρ) =∫ 2π0k∑a 2m r 2m−1 cos 2m θ,m=1(sin ψk∑b 2m r 2m cos 2m+1 θ,m=1B 2 = − sin θ)k∑a 2m r 2m−1 cos 2m ψ dψ = 0m=1k∑b 2m r 2m−1 cos 2m θ.because the integrand is an odd function. The derivative of B 1 with respect to r at (ρ, θ) is∂ B 1∂r= − sin θk∑(2m − 1)a 2m ρ 2m−2 cos 2m θ.m=1To compute T 2 we first need to know r 1 (θ):r 1 (θ) =∫ θ0A 1 (ρ, ψ)dψ =∫ θ0cos(ψ)m=1k∑a 2m ρ 2m cos 2m ψdψm=1( )k∑ m∑= (−1) l a 2mC m,l2l + 1 sin2l+1 θ ρ 2m .m=1 l=0The square B1 2 is given by( )2k∑ ∑n−1a 2m a 2n−2m sin 2 θ cos 2n θn=2 m=1ρ 2n−2with a j = 0 for j 2k + 2.The product − ∂ B 1∂r r 1 is given by( ( ))2k∑ ∑n−1m∑D m,l sin 2l+2 θ a 2m a 2n−2m cos 2n−2m θ ρ 2n−2 ,n=2 m=1 l=0where D m,l = (−1) l C m,l2n−2m−12l+1.Moreover∫ 2π0B 2 (ρ, ψ)dψ = 0,because it is a sum of integrals of products of the kind sin ψ cos l ψ.

1900 A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903withThenT 2 =2k∑n=2I 1 (n) =( )∑n−1a 2m a 2n−2m I 1 (n) ρ 2n−2 +m=1∫ 2π02k∑n=2sin 2 θ cos 2n θdθ, I 2 (n, m, l) =( ( ))∑n−1m∑D m,l a 2m a 2n−2m I 2 (n, m, l) ρ 2n−2 (12)m=1 l=0∫ 2πThe recurrent integrals I 1 and I 2 which appear in (12) are respectivelyI 1 =0sin 2l+2 θ cos 2n−2m θdθ.(2n)!4 n n!(n + 1)! π, I (2n − 2m)!(2l + 1)!2 =4 n−m+l (n − m)!(l)!(l + 1 + n − m)! π. □Remark 8. Fix some l = 2k. Note that for all s = 1, 2, . . . , k,I (2s, s) =2(4s)!16 s (2s + 1) 2 ((2s − 1)!) 2 π ≠ 0.Let T 2 (ρ) be given by Proposition 7. Then the condition T 2 (ρ) ≡ 0 is equivalent to the condition a 2 = a 4 = · · · =a 2k = 0.By using the same tools as in the proof of Theorem 2(a) it is not difficult to prove the following result.Proposition 9. Fix j ∈ N. Assume that for any small ε, equationṙ = ∑ i≥ jA i (r, θ)ε i ,˙θ = 1 + ∑ B i (r, θ)ε i , (13)i≥ jhas a center at the origin. Let T (ρ, ε) = 2π + ∑ ∞i=1 T i (ρ)ε i be the period of the solution of (13) that passes throughthe point θ = 0 and r = ρ. Then T 1 = T 2 = · · · = T j−1 = 0,T m (ρ) = −T 2 j (ρ) =∫ 2π0∫ 2π0where r 1 (θ; ρ) = ∫ θ0 A j(ρ, ψ)dψ.(B m (ρ, θ)dθ, for m = j, j + 1, . . . , 2 j − 1,B 2 j (ρ, θ) − B 2 j(ρ, θ) − ∂ B )j(ρ, θ)r 1 (θ; ρ) dθ,∂rBy using the same arguments as in the proof of Proposition 7 we get:Proposition 10. Fix l = 2k and j ≥ 1. Considerẋ = −y +∞∑i= j( )k∑ε i a2m i x2m , ẏ = x, (14)m=1and the period function T (ρ, ε) = 2π + ∑ ∞i=1 T i (ρ)ε i of the solution of (14) that passes through the point θ = 0 andr = ρ. Then T 1 = T 2 = · · · = T 2 j−1 = 0 and( )2k∑ ∑n−1T 2 j (ρ) = a j 2m a j 2n−2m I (n, m) ρ 2n−2 .n=2 m=1Theorem 11. Fix an even number l. Then for any j ∈ N, N l 2 j−1 (Lie) = 0, N l 2 j (Lie) = N l 2 (Lie) and N l 2 (Lie) ≤l − 2.

A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903 1901Proof. The first two assertions follow from the above three propositions and Remark 8. To see the third one, fromProposition 7 we get that( )l∑ ∑n−1T 2 ′ (ρ) = ρ (2n − 2) a 2m a 2n−2m I (n, m) ρ 2n−4n=2m=1and, hence, the maximum number of positive solutions of T2 ′ is l − 2. □To end this section we fix small even values of l and study the numbers N l 2 (Lie). Recall that N 2 (Lie) = 0, see [2],so we take l ≥ 4. We proveProposition 12. N2 l (Lie) = l − 2, for l = 4, 6, 8.Proof. Assume that a 4 ≠ 0. By using Proposition 7 we have that( 1T 2 (ρ) = π3 a2 2 ρ2 + 2 ( 53 a 2a 4 ρ 4 +8 a 2a 6 + 7 )20 a2 4 ρ 6( 27+40 a 4a 6 + 7 ) ( 29712 a 2a 8 ρ 8 +896 a2 6 + 77 )120 a 4a 8 ρ 10 + 143224 a 6a 8 ρ 12 + 715 )2304 a2 8 ρ14 .Thus we have that(T 2 ′ (ρ) = πa2 2 ρ 23 + 8 3Note that+(1485 a62448 a22(a 4ρ 2 21+a 2 10)+ 77 a 4 a 812a 2 2a 2 4a 2 2ρ 8 + 42956( 2T 2 ′ (ρ) = πa2 2 ρ 3 + 8 ( 213 R + 10 + 15 )4 b+) (a 6ρ 4 +a 2a 2 2a 2 2+ 15 27 a 4 a 645 a22 + 14 3)a 6 a 8ρ 10 + 5005 a82 ρ 12 .1152( 27R 2 +5 b + 14 )3 c R 3)a 8ρ 6a 2( 1485448 b2 + 77 )12 c R 4 + 42956 bcR5 + 5005 )1152 c2 R 6 := πa2 2 ρ P(b, c, R),where R := a 4 ρ 2 /a 2 , b := a 2 a 6 /a4 2 and c := a2 2 a 8/a4 3.It is not difficult to check that the polynomial equations in the variable R, P(0, 0, R) = 0, P(0.1, 0, R) = 0 andP(0.1, 0.0009, R) = 0 have two, four and six simple negative solutions, respectively. In the first two cases we cansolve exactly the equations, while in the third one we use the Sturm sequence of P(0.1, 0.0009, R) in the interval[−150, 0]. By taking suitable a 2 , a 4 , a 6 and a 8 the result follows. □Theorem 5 follows from Theorem 11 and Proposition 12.Remark 13. (i) We know that N2 l(Lie) ≤ l − 2. It seems natural to believe that N 2 l (Lie) is indeed this upper boundbut we have only proved this fact for l = 2, 4, 6 and 8. Our approach can also be used to study bigger values of l.(ii) In [1] it is proved that all the period constants of Liénard centers are positive or zero. So the standard tools usedto bifurcate critical periods from the origin do not work in this case. On the other hand our approach can be usedto obtain critical periods bifurcating from periodic orbits far from the origin.6. Reversible systemsIn this section we give an example that proves that N 2 6(rev) ≥ 2. Indeed our construction of the example havingtwo critical periods shows that with our method no more critical periods can be obtained, unless we go to higher orderperturbations. In fact, if the conjecture given in [3] is true, even going to higher order perturbations the maximumnumber of expected critical points is 2. Unfortunately we have not been able to perform all the computations for amore general perturbation.

1902 A. Cima et al. / Nonlinear Analysis 69 (2008) 1889–1903Considerẋ = −y + ε(a 1 x 2 + a 2 y 2 ) + ε 2 (b 1 x 2 + b 2 y 2 ) + ε 3 (c 1 x 2 + c 2 y 2 ) + ε 4 (d 1 x 2 + d 2 y 2 ),ẏ = x + ε(a 3 xy) + ε 2 (b 3 xy) + ε 3 (c 3 xy) + ε 4 (d 3 xy).By using Theorem 2 we obtain thatT 2 (ρ) = T 2,2 (a)ρ 2 = π )(4a1 2 12+ 10a2 2 + 10a 1a 2 − a 2 a 3 + a3 2 − 5a 1a 3 ρ 2 ,(15)andT 4 (ρ) = T 4,2 (a, b, c)ρ 2 + T 4,4 (a)ρ 4 ,where T 4,2 is a homogeneous polynomial of degree 2 in a, b, c andT 4,4 (a) =π (168a 3 a2 2 1152a 1 + a3 4 + 3080a3 2 a 1 − 2a 2 a3 3 + 1540a4 2 + 400a4 1 + 369a2 3 a2 1 − 712a 3a13)+ 1424a 2 a1 3 − 58a3 3 a 1 +700a 3 a2 3 + 21a2 3 a2 2 − 828a 3a 2 a1 2 + 126a2 3 a 2a 1 + 2772a2 2 a2 1 .A solution of T 2,2 (a) = T 4,4 (a) = 0 isa 3 = 1, a 2 = s/5, a 1 = (33 − 18s)/(10s + 155),where s = (−11 + √ 105)/4.Now we consider system (15) fixing the above values of a. Moreover we take all the other parameters equal to zeroexcept c 3 = 1 and c 1 = (−55 + 13 √ 105)/320.For these values of the parameters we have that: T k (ρ) ≡ 0, k = 2, 3, 4, 5 and T 6 (ρ) isπ(2279 − 221 √ 10510 240ρ 2 + 1701 − 167√ 10548 000)ρ 4 + 371 − 33√ 105ρ 6 .3 600 000The equation T6 ′ (ρ) = 0 has two positive simple roots and therefore, according Theorem 2, the system (15), for εsufficiently small, has at least two critical periods, as we wanted to prove. □After this paper was finished a related paper appeared; see [9]. There the authors prove that for any l even thereare Liénard equations having at least l − 2 critical periods; see our Theorem 5.AcknowledgementsThe authors are supported by the joint project CAPES–MECD grant HBP2003–0017. The first two authors arealso supported by grants MTM2005-06098-C02-1 and 2005SGR-00550. The third author is partially supported by aFAPESP-BRAZIL grant 10246-2.References[1] A. Cima, A. Gasull, V. Mañosa, F. Mañosas, Algebraic properties of the Liapunov and period constants, Rocky Mountain J. Math. 27 (1997)471–501.[2] C. Chicone, The monotonicity of the period function for planar Hamiltonian vector fields, J. Differential Equations 69 (1987) 310–321.[3] C. Chicone, Review of reference [8] in Math. Sci. Net., ref 94h:58072.[4] C. Chicone, F. Dumortier, Finiteness for critical periods of planar analytic vector fields, Nonlinear Anal. 20 (1993) 315–335.[5] C. Chicone, M. Jacobs, Bifurcations of critical periods for plane vector fields, Trans. Amer. Math. Soc. 102 (1998) 706–710.[6] C.J. Christopher, N.G. Lloyd, Polynomial systems: A lower bound for the Hilbert numbers, Proc. Roy. Soc. London Ser. A 450 (1995)219–224.[7] C.B. Collins, The period function of some polynomial systems of arbitrary degree, Differential Integral Equations 9 (1996) 251–266.[8] W.A. Coppel, L. Gavrilov, The period function of a Hamiltonian quadratic system, Differential Integral Equations 6 (1993) 1357–1365.[9] P. De Maesschalck, F. Dumortier, The period function of classical Liénard equations, J. Differential Equations 233 (2007) 380–403.[10] A. Gasull, A. Guillamon, V. Mañosa, F. Mañosas, The period function for Hamiltonian systems with homogeneous nonlinearities,J. Differential Equations 139 (1997) 237–260.[11] L. Gavrilov, Remark on the number of critical points of the period, J. Differential Equations 101 (1993) 58–65.

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