On the number of limit cycles bifurcating from a non-global degenerated center

ON THE NUMBER OF LIMIT CYCLES BIFURCATING FROM A NON-GLOBAL DEGENERATED CENTER ARMENGOL GASULL(1), CHENGZHI LI(2) AND CHANGJIAN LIU(2) Abstract. We give an upper bound for the number of zeros of an Abelian integral. This integral controls the number of limit cycles that bifurcate, by a polynomial perturbation of arbitrary degree, from the periodic orbits of the integrable system (1 + x)dH = 0,where His the quasi-homogeneous Hamiltonian H(x, y) = x2k/(2k) + y2/2.The main tool used in the proof is the Argument Principle applied to a suitable complex extension of the Abelian integral. 1. Introduction and Main Result Consider the perturbed integrable system      ˙x=−∂H(x, y) ∂y R(x, y) + εP(x, y), ˙y=∂H(x, y) ∂x R(x, y) + εQ(x, y), (1)ε and associated to it define the Abelian integral I(h) = ZΓh P(x, y)dy −Q(x, y)dx R(x, y),(2) taken along the real ovals of the level curves {H(x, y) = h}.It is well known that each isolated zero h∗of I(h) gives rise, for εsmall enough, to a limit cycle of the perturbed system (1)ε,which tends to the oval {H(x, y) = h∗} when εgoes to zero, see for instance [8]. (1)Department of Matem`atiques, Universitat Aut`onoma de Barcelona, 08193 Bellaterra, Barcelona, Spain. (2)LMAM and School of Mathematical Sciences, Peking University, Beijing 100871, China. The first author is partially supported by the DGES grant number BFM2002-04236- C02-2 and CONACIT grant number 2001SGR00173. The second author is partially supported by the grant NSFC–10231020 and RFDP of China. 1 2 ARMENGOL GASULL, CHENGZHI LI AND CHANGJIAN LIU A challenging problem consists in, fixed Hand R, consider Pand Q polynomials of degree at most nand give upper bounds for the maximum number of isolated zeroes of I(h) in terms of n. This question is related with the known as weak or infinitesimal Hilbert’s sixteenth problem. Also, considered as a method for obtaining limit cycles for polynomial vector fields it is strongly related with the Hilbert’s sixteenth problem. The most studied cases are the ones for which system (1)0is Hamiltonian (i.e. those for which R(x, y)≡1) and more concretely the ones having H either a cubic polynomial or a function of hyperelliptic type, H(x, y) = y2+V(x). This paper is devoted to study the above problem when H(x, y) is the quasi-homogeneous hyperelliptic function x2k/(2k) + y2/2, k ≥1 and R(x, y) = 1 + x. Notice that in this case the phase portrait of (1)0is a non-global center, which phase portrait consists of a global center (degenerated if k > 1) cut by the line of critical points {x=−1}.Our main result is the following theorem: Theorem A. Let P(x, y)and Q(x, y)be real polynomials of degree at most n∈ {2`, 2`+ 1}, ` ≥1.Then the maximum number of isolated zeros of the Abelian integral I(h) = ZΓh P(x, y)dy −Q(x, y)dx 1 + x,(3) where Γh={(x, y) : x2k/(2k) + y2/2 = h, 0< h < 1/(2k)},is k(`+ 1) + 2`+ 3 −sin2¡kπ 2¢`. The above result has been already proved in [4] when k= 1.Indeed in that paper it is proved that the maximum number of isolated zeros of I(h) is nand that this number can be attained. Notice that our upper bound, 2`+4,is slightly bigger. This small difference can be easily corrected following carefully our proof of Theorem A when k= 1. As it will be seen along the proof of our theorem, the expression of I(h) includes some polynomials in h1 kand the function J(h) = √1−2kh Zh 0 t1−k 2kpk−1(t1 k) (1 −2kt)3/2dt, where pk−1a real polynomial of degree k−1.In the particular case k= 1, J(h) = 2π£√1−2h−1¤and the problem of the study of the zeros of I(h) can be reduced to the study of the zeros of a polynomial in a new variable w=√1−2h. For k > 1 the expression of I(h) is more complicated and our study relies on the study of an extension of I(h) in a suitable complex domain. Theorem A will be proved in Section 3 by applying the Argument Principle to this extension of I(h). As far as we know this method to LIMIT CYCLES BIFURCATING FROM A DEGENERATED CENTER 3 estimate the number of zeros of Abelian integrals was introduced by Petrov in [5, 6, 7]. It is also worth to mention that in [2] a related question is considered by using a slightly different approach. In that paper by using the Poincar´e- Lyapunov polar coordinates, see [1, 3], a lower bound for the number of zeros of I(h),when k=`, is obtained. This result is studied there to get a lower bound for the number of nested limit cycles of a polynomial system of degree n∈ {2`, 2`+ 1}.For instance when n= 2`the lower bound given in that paper is (`2+ 3`)/2.Notice that by applying Theorem A we get that an upper bound for the zeros of I(h) is `2+ 3`+ 3 −sin2¡`π 2¢`. The gap between these two results remains to be studied. 2. Preliminary Results Along this paper any polynomial in R[x] of degree mis denoted by pm(x) although its coefficients may vary from one expression to another. If it is necessary we will also use indistinctly qm(x).Notice that in particular the constants will be denoted by p0(x)≡p0, p0 m(x) = pm−1(x), pm(x)pn(x) = pn+m(x),and so on. Define Ij(h) = ZΓh xjydx , j = 0,1,2, . . . ; and J(h) = ZΓh y 1 + xdx . (4) We give several preliminary results on the above functions which, as we will see, generate the Abelian integral I(h). Lemma 1. The functions Ij(h)introduced in (4) are I2j+1(h)≡0and I2j(h) = p0h2j+k+1 2k, j = 0,1,2, . . . where p06= 0 is a constant independent on h. Proof. The oval of H(x, y) = his given by x2k 2k+y2 2=h , (5) or, equivalently, by y=y±(x, h) = ±r2h−x2k k.(6) Hence Ij(h) = 2 Zx∗ −x∗ xjr2h−x2k kdx , 4 ARMENGOL GASULL, CHENGZHI LI AND CHANGJIAN LIU where x∗=2k √2kh. Thus, by the symmetry of the integrator it is clear that I2m+1 = 0 for all m. In case of even subindex we have that I2m(h) = 4 Zx∗ 0 x2mr2h−x2k kdx . By making the substitution x=2k √2kh sin2θ, we obtain I2m(h) = Ck,mh2m+k+1 2k, where p0:= Ck,m =4√2(2k)2m+1 2k kZπ 2 0 (sin θ)2m+1−k kcos2θ dθ 6= 0 , as we wanted to prove. ¤ The proof of next result is straightforward. Lemma 2. Consider the functions Ij(h)and J(h)introduced in (4). Their derivatives can be expressed as I0 j(h) = ZΓh xj ydx , j = 0,1,2, . . . ;J0(h) = ZΓh 1 y(1 + x)dx . By using previous lemma and Lemma 1 we get: Lemma 3. Write m= 2ror m= 2r+ 1.The following equalities hold: ZΓh ypm(x)dx =hk+1 2kpr(h1 k),and ZΓh pm(x) ydx =h1−k 2kpr(h1 k). Lemma 4. (i) The function J(h)introduced in (4) is also given by the expression J(h) = √1−2kh Zh 0 s1−k 2kpk−1(s1 k) (1 −2ks)3/2ds . (7) where pk−1(x)is a polynomial of degree exactly k−1.Furthermore, limh→(1 2k)−J(h)exists, and it is finite and non-zero. (ii) When k= 1 then J(h) = 2π£√1−2h−1¤. LIMIT CYCLES BIFURCATING FROM A DEGENERATED CENTER 5 Proof. By Lemmas 2, 3 and equations (4) and (5) we have J(h) = ZΓh y2 y(1 + x)dx =ZΓh 2h−x2k k y(1 + x)dx = 2hJ0(h)−1 kZΓh x2k y(x+ 1) dx = 2hJ0(h)−1 kZΓh (x+ 1)p2k−1(x)+1 y(x+ 1) dx =µ2h−1 k¶J0(h) + ZΓh p2k−1(x) ydx =µ2h−1 k¶J0(h) + h1−k 2kpk−1(h1 k). Notice that from the construction of p2k−1,all its coefficients are non null. From the computations used to prove Lemma 3 it is easy to see that the same property happens for the coefficients of pk−1.In particular it is exactly of degree k−1. By definition of J(h), we have J(0) = 0. By integration of the above ordinary differential equation for J(h) we obtain (7). From (6) we have that near x=−1, y±(x, 1 2k) = ±√1 + x³√2 + O(x+ 1)´1/2. Hence lim h→(1 2k)− J(h) = 2 Z1 −1 y+(x, 1 2k) 1 + xdx < ∞, and is different from zero, as we wanted to prove. To end the proof of the lemma let us study J(h) when k= 1.In this case notice that the integral appearing in (7) can be computed explicitly. By using that p0(h) = −2π, the expression for J(h) follows. ¤ Proposition 5. Assume that n= 2`or n= 2`+ 1.Then the Abelian integral I(h)given in (2) can be expressed in the form I(h) = 1 2h−1/k ³hk+1 2kpk`+k−1(h1 k) + q`+1(h)J(h)´,(8) where as usual pk`+k−1(h)and q`+1(h)are real polynomials in hof degree given by their respective subindexes. Proof. We assume n= 2`+ 1, the proof for the case n= 2`is similar. Let us start by studying the integral RΓh Q(x,y) 1+xdx. The polynomial Q(x, y) can be decomposed as the sum of its even and odd parts with respect to the variable y, i.e. Q(x, y) = Qe(x, y) + Qo(x, y),where Qe(x, −y) = Qe(x, y) 6 ARMENGOL GASULL, CHENGZHI LI AND CHANGJIAN LIU and Qo(x, −y) = −Qo(x, y).In particular Qe(x, y) = ˜ Q(x, y2) being ˜ Q another suitable polynomial. Notice that ZΓh Qe(x, y) 1 + xdx =ZΓh ˜ Q(x, y2) 1 + xdx =ZΓh ˜ Q(x, 2h−x2k/(2k)) 1 + xdx ≡0. Thus RΓh Q(x,y) 1+xdx =RΓh Qo(x,y) 1+xdx. Write Qo(x, y) = (9) =pn−1(x)y+pn−3(x)y3+···+p2(x)y2`−1+p0(x)y2`+1 =y³pn−1(x)+pn−3(x)£2h−x2k/(2k)¤2+···+p0(x)£2h−x2k/(2k)¤`´ =y¡p2k`(x) + hp2k(`−1)(x) + h2p2k(`−2)(x) + ···+h`p0(x)¢. Hence Qo(x,y) 1+x= =y¡p2k`−1(x) + hp2k(`−1)−1(x) + h2p2k(`−2)−1(x) + ···+h`−1p2k−1(x)¢+ q`(h)y 1+x. By using Lemma 3 we get ZΓh Qo(x, y) 1 + xdx =hk+1 2k³pk`−1(h1 k)+hpk`−k−1(h1 k)+h2pk`−2k−1(h1 k)+··· +h`−1pk−1(h1 k)´+q`(h)J(h) =hk+1 2kpk`−1(h1 k) + q`(h)J(h). Let us study the term RΓh P(x,y) 1+xdy in the expression of I(h).By using (5) we have that over Γh,dy dx =−x2k−1 y, hence ZΓh P(x, y) 1 + xdy =−ZΓh x2k−1P(x, y) y(1 + x)dx. As in the previous case we decompose P(x, y) = Pe(x, y) + Po(x, y),being Peand Pothe even and odd parts, respectively, with respect to the variable y. Thus ZΓh x2k−1P(x, y) y(1 + x)dx =ZΓh x2k−1Pe(x, y) y(1 + x)dx = ZΓh x2k−1p2`+1(x) y(1 + x)dx +ZΓh ¡p2`−1(x)+p2`−3(x)y2+···+p1(x)y2`−2¢x2k−1y 1+xdx. LIMIT CYCLES BIFURCATING FROM A DEGENERATED CENTER 7 Notice that the numerator of the last integral above, when the points (x, y) are on Γhcan be written as ¡p2`−1(x) + p2`−3(x)y2+···+p1(x)y2`−2¢x2k−1y= =Ãp2`−1(x) + p2`−3(x)·2h−x2k 2k¸+···+p1(x)·2h−x2k 2k¸`−1!x2k−1y= =¡p2k`(x) + hp2k(`−1)(x) + h2p2k(`−2)(x) + ···+h`−1p2k(x)¢y. This last expression coincides with the one of (9). Thus following the same steps that in that case we obtain that ZΓh x2k−1P(x, y) y(1 + x)dx =ZΓh x2k−1p2`+1(x) y(1 + x)dx +hk+1 2kpk`−1(h1 k) + q`(h)J(h). (10) To end the proof we study the integral appearing in the righthand side of the above expression. We have ZΓh x2k−1p2`+1(x) y(1 + x)dx =ZΓh p2(`+k)(x) y(1 + x)dx = =ZΓh p2(`+k)−1(x) ydx +ZΓh p0 y(1 + x)dx = =h1−k 2kp`+k−1(h1 k) + p0J0(h), where we have used Lemmas 2 and 3. Finally, from the proof of Lemma 4 we have that J0(h) = 1 2h−1/k ³J(h) + h1−k 2kpk−1(h1 k)´. Collecting all the results we get I(h) = hk+1 2kpk`−1(h1 k) + q`(h)J(h) + p0 2h−1/k ³J(h) + h1−k 2kpk−1(h1 k)´ =1 2h−1/k ³hk+1 2kpk`+k−1(h1 k) + q`+1(h)J(h)´, as we wanted to prove. ¤ From previous proposition to study the zeros of I(h) in the real interval (0,1 2k) it suffices to study a function of the form M(h) := hk+1 2kpk`+k−1(h1 k) + q`+1(h)J(h).(11) One of the standard tools to give an upper bound of its number of zeros consists in extending the function to a suitable subset of the complex plane, 8 ARMENGOL GASULL, CHENGZHI LI AND CHANGJIAN LIU and later to apply the Argument Principle to this extended function. In this case this can be done in the set D=C\µ{h∈R:h≤0}∪½h∈R:h≥1 2k¾¶. Essentially the problems to get a full extension to all Ccome from the functions h1 kand (1 −2kh)3/2. Lemma 6. The function M(h)given in (11) and defined for h∈(0,1 2k), can be extended to Das a single-valued analytic function of h. We denote this extension also by M(h). Proof. We first rewrite the real function J(h) as J(h) = √1−2kh "ak+Zh 1 4k s1−k 2kpk−1(s1 k) (1 −2ks)3/2ds#, where h∈(0,1 2k) and ak=R1 4k 0 s1−k 2kpk−1(s1 k) (1−2ks)3/2ds is a constant with parameter k. Then it can be extended to J(h) = √1−2kh "ak+Zγh s1−k 2kpk−1(s1 k) (1 −2ks)3/2ds#,(12) where γhis any path contained in Djoining 1 4kand h. Thus, the function J(h) in form (12) is single-valued and analytic in D. On the other hand, the functions h1 kand hk+1 2kare obviously single-valued and analytic for h∈D. Hence, M(h) can be extended to Das a single-valued analytic function of h.¤ In order to use the Argument Principle to M(h), we define G=GR,ε ⊂D (a simply connected region) with ∂G =C=CR,ε a simple closed curve, CR,ε =CR∪C1 ε∪C2 ε∪L1 ±(R, ε)∪L2 ±(R, ε), where CR={z∈C,|z|=RÀ1},C1 ε={z∈C,|z−1 2k|=ε¿1}, C2 ε={z∈C,|z|=ε¿1},L1 ±(R, ε) = {z∈R,1 2k+ε≤z≤R}, L2 ±(R, ε) = {z∈R,−R≤z≤ −ε}. See Figure 1. LIMIT CYCLES BIFURCATING FROM A DEGENERATED CENTER 9 Figure 1. Domain G=GR,ε ⊂D. Lemma 7. Let M(h)and J(h)be the extensions to D⊂Cof the functions given in Lemma 6 and (12). The following statements hold: (i) Im(M(h)) has at most `+ 1 zeros for h∈L1 ±. (ii) Im(M(h)) has at most k` +k−1−sin2¡kπ 2¢`zeros for h∈L2 ±. (iii) J(h)∼0as h→0and J(h)∼p0(constant) as h→1 2k. (iv) There exist constants Band C6= 0 such that J(h)∼Bh1 2+Ch1 2−1 2k as |h| → ∞. Proof. (i) Write J(h) = √1−2khK(h) where K(h) = µak+Rγh s1−k 2kpk−1(s1 k) (1−2ks)3/2ds¶, being γhany path contained in Djoining 1 4k and h. Notice that both functions √1−2kh and K(h) are holomorphic in D. To get more information on K(h) at L1 ±(R, ε) we consider in each case a concrete path joining 1 4kand h. Let us fix h∈L1 +(R, ε); the another case can be similarly studied. Take the path Γhformed by a curve contained in the first quadrant between 1 4kand 1 2k+ 2ε∈R,followed by the segment {z∈R,:1 2k+ 2ε≤z≤h}.The function K(h) can be computed by using Γh.In particular K(h)−K(1 2k+ 2ε)= Zh 1 2k+2ε s1−k 2kpk−1(s1 k) (1 −2ks)3/2ds =iZh 1 2k+2ε t1−k 2kpk−1(t1 k) (2kt −1)3/2dt ∈iR.