This is a preprint of: “Highest weak focus order for trigonometric Liénard equations”, Armengol Gasull, Jaume Giné, Clàudia Valls, Ann. Mat. Pur. Appl., 2019. DOI: [10.1007/s10231-019-00936-8] HIGHEST WEAK FOCUS ORDER FOR TRIGONOMETRIC LIÉNARD EQUATIONS ARMENGOL GASULL, JAUME GINÉ, AND CLAUDIA VALLS Abstract. Given a planar analytic differential equation with a critical point which is a weak focus of order k, it is well-known that at most k limit cycles can bifurcate from it. Moreover, in case of analytic Liénard differential equations this order can be computed as one half of the multiplicity of an associated planar analytic map. By using this approach, we can give an upper bound of the maximum order of the weak focus of pure trigonometric Liénard equations only in terms of the degrees of the involved trigonometric polynomials. Our result extends to this trigonometric Liénard case a similar result known for polynomial Liénard equations. 1. Introduction and main results Recall that a critical point of a planar analytic vector field is called a focus if the eigenvalues of its linear approximation at the point are not real, i.e. α ± iβ, β 6= 0. Moreover, when α 6= 0 the point is called a strong focus and, otherwise, it is called a weak focus. The complex Poincaré’s normal form of its associated differential equation at this weak focus point is  X∞  ż = iz β + cj (z z̄)2j , cj = αj + iβj ∈ C. j=1 The values αj give the so called Lyapunov quantities and can be computed in many other ways, see for instance [1, 4, 12, 16, 17, 18, 19, 23]. When all αj = 0 the weak focus is a center, otherwise, if αk 6= 0, is the first non-zero αj then it is said that the origin is a weak focus of order k. It is well known that k is the maximum number of limit cycles (isolated periodic orbits) that bifurcate from this type of points, and that this amount of limit cycles is attained for some analytic perturbations. Therefore, given an analytic family, F, of planar analytic differential equations depending on finitely many parameters, it is very interesting to know which is the maximum order of the weak focus inside this family, σ(F), see [24]. This number is known to exist when the Lyapunov quantities are polynomials on the parameters of the system, because of the Hilbert’s basis Theorem. Unfortunately Hilbert’s result is not constructive and, in general, an explicit bound of the number of needed Lyapunov quantities is not known. In fact, even for cubic vector fields this number is nowadays unknown. The most important family F of systems of arbitrary degree for which an explicit upper bound of σ(F) is known is the family of polynomial Liénard equations. This 2010 Mathematics Subject Classification. Primary 34C07. Secondary 13H15; 34C25; 37C27. Key words and phrases. Trigonometric Liénard equation, weak focus, cyclicity. 1 2 A. GASULL, J. GINÉ, AND C. VALLS family is the one formed for the planar vector fields with a weak focus at the origin  ẋ = y, ẏ = g(x) + yf (x), where f and g are polynomials with given degrees, satisfying f (0) = g(0) = 0 and g 0 (0) < 0. This upper bound (which is not sharp in general) depends on these degrees and it is given in [6, 7, 8, 9, 15]. This proof relies on two main facts: a relation between the order of a weak focus for analytic Liénard equations with the multiplicity at the origin of a planar polynomial map, and on Bezout’s Theorem. The same tools can be applied to solve the same question for trigonometric Liénard systems. Notice that, as well as polynomial systems, trigonometric systems are differen- tial systems of current interest, see for instance [3, 13, 14, 20, 21, 22, 25]. We will say that ( θ̇ = y, (1) ẏ = G0 (θ) + yF 0 (θ), is a pure trigonometric Liénard system if F and G are 2π-trigonometric polynomials satisfying F (0) = F 0 (0) = 0, G(0) = G0 (0) = 0, G00 (0) < 0. We denote by Lm,n the family of all pure trigonometric Liénard systems where F and G satisfy the above properties and their degrees are at most m and n, respectively. Because of the lack of symmetry between F and G we also introduce the subclass formed by the F ’s that also satisfy F 00 (0) < 0. We denote it by L∗m,n ⊂ L.m,n Recall that if a real 2π-trigonometric polynomial p is such that its Fourier series satisfies X̀ p(θ) = ak ekiθ , a−k = ak , with a` 6= 0, (2) k=−` then its degree is `. n\m 1 2 3 4 1 0 1 2 3 2 1 2 6 7 3 2 6 7 - 4 3 7 - - Table 1. Some values of σ(Lm,n ). Our main result is: Theorem 1. Let σ(Lm,n ) (resp. σ(L∗m,n )) be highest weak focus order for systems (1) inside family Lm,n (resp. σ(L∗m,n )). Then, (i) For all positive m and n, σ(L1,n ) = n−1 and σ(Lm,1 ) = m−1. Similarly, σ(L∗1,n ) = n − 1 and σ(L∗m,1 ) = m − 1. (ii) For all positive m ≥ n, σ(Lm,n ) = σ(L∗m,n ) = σ(L∗n,m ) ≤ σ(Ln,m ). TRIGONOMETRIC LIÉNARD EQUATIONS 3 (iii) For n and m ≥ 2, m + n − 2 ≤ σ(Lm,n ) ≤ 6mn − 3(m + n) + 1. (iv) For small m and n, some values of σ(Lm,n ) are given in Table 1. Although, comparing with the results of Table 1, the upper bounds given in item (iii) of the theorem are not sharp, the more important fact is that they are explicit. If in system (1), instead of F and G we consider F (θ) = Fe(θ) + αθ and G(θ) = Fe(θ) + βθ, with Fe and G e trigonometric polynomials, we will say that (1) is a (non pure) trigonometric Liénard system. In Section 4 we give some partial results for this case. 2. Preliminary results Let (P, Q) : R2 → R2 be an analytic function at (0, 0). As usual we denote by µ0 [P, Q] its multiplicity at this point. Recall that when (P (0, 0), Q(0, 0)) 6= (0, 0) then µ0 [P, Q] = 0 and, otherwise, µ0 [P, Q] is the number of (P, Q)-complex preimages around (0, 0) of any regular point near the origin, see [2]. When (0, 0) is a non isolated zero then it is said that the multiplicity is infinity. Notice also that µ0 [P, Q] = µ0 [Q, P ]. In text proposition we collect some useful properties to compute multiplicities, see again [2]. As usual, O(k) denotes terms of order at least k. Proposition 2. Let (P, Q) : R2 → R2 be an analytic map at the origin with finite mul- tiplicity and let R : R2 → R and S : R → R be also analytic at the origin. Then: (a) The multiplicity of (P, Q) at the origin does not depend on the choice of coordi- nates; (b) It holds that µ0 [RP, Q] = µ0 [R, Q] + µ0 [P, Q]. In particular, if R(0, 0) 6= 0 then µ0 [RP, Q] = µ0 [P, Q]; (c) Write P = Pj + O(j + 1) and Q = Qk + O(k + 1), with Pj and Qk homogeneous with respective degrees j and k. Then µ0 [P, Q] ≥ jk, and the equality holds if and only if the system Pj = 0, Qk = 0 has only the trivial solution (0, 0) in C2 . (d) It holds that µ0 [P + RQ, Q] = µ0 [P, Q]. (e) If P (x, y) = (y − S(x))R(x, y), with S(0) = 0 and R(0, 0) 6= 0, and Q(x, S(x)) = axk + O(k + 1), with a 6= 0, then µ0 [P, Q] = k. In [5] the authors proved the following nice theorem which is based on previous results of [6]. Our results will be strongly based on it. Theorem 3. ([5]) Consider the Liénard system  ẋ = y, (3) ẏ = g(x) + yf (x), where f and g are analytic at the origin and satisfy f (0) = g(0) = 0 and g 0 (0) < 0. Define   F (x) − F (y) G(x) − G(y) µ(f, g) = µ0 , , x−y x−y Rx Rx where F (x) = 0 f (s) ds and G(x) = 0 g(s) ds. Then (i) if µ(f, g) = ∞, the origin of (3) is a center, 4 A. GASULL, J. GINÉ, AND C. VALLS (ii) if µ(f, g) < ∞, the origin of (3) is a weak focus of order µ(f, g)/2. We will use a different characterization of trigonometrical polynomials of degree `. Consider the ring of trigonometric polynomials, Rt [θ] = R[sin θ, cos θ]. It is well known that its quotient field, Rt (θ), is isomorphic to R(x) by means of the morphism Φ : Rt (θ) → R(x) defined by 2x 1 − x2 Φ(sin θ) = and Φ(cos θ) = . (4) 1 + x2 1 + x2 Note that  sin θ   2x  Φ(tan(θ/2)) = Φ = x and Φ(θ) = arctan . 1 + cos θ 1 − x2 Observe also that Φ is a well-defined change of variables around the origin. If p is a trigonometric polynomial of degree `, as in (2), it holds that P (x) Φ(p(θ)) = , with P ∈ R[x], deg(P ) ≤ 2` and gcd(P (x), 1 + x2 ) = 1. (1 + x2 )` (5) Moreover, the converse is also true: for each P under the above hypotheses, there exists P (x) a trigonometric polynomial, p, of degree at most `, such that Φ(p(θ)) = (1+x 2 )` , see [10, 11]. Consider now a pair f, g of 2π-periodic trigonometric polynomials of degrees m and n, respectively and define Z x Z x F (x) = f (θ) dθ, G(x) = g(θ) dθ. (6) 0 0 Then F (θ) = αθ + Fe(θ) and G(θ) = βθ + G(θ) e where α = F (2π) 2π and β == G(2π)2π and Fe and Ge are also trigonometric polynomials of degrees n and m, respectively. Using the change of variables Φ given in (4) we have that   2x   2x  M (x) F (θ) = F arctan 2 = α arctan 2 + , 1−x 1−x (1 + x2 )m   2x   2x  N (x) G(θ) = G arctan 2 = β arctan 2 + , (7) 1−x 1−x (1 + x2 )n where M is the polynomial of degree less than or equal to 2m associated to Fe and N is the polynomial of degree less than or equal to 2n associated to G.e As a corollary of the previous results we prove the following proposition. Proposition 4. Let f and g be 2π-trigonometric polynomials with degrees m and n, and such that f (0) = g(0) = 0. Then, following the notation introduced in (6) and (7), it holds that   F (θ) − F (ψ) G(θ) − G(ψ) µ0 , = µ0 [Pα (x, y), Qβ (x, y)] , θ−ψ θ−ψ where  Pα (x, y) = α∆(x, y)(1 + x2 )m (1 + y 2 )m + M (x)(1 + y 2 )m − M (y)(1 + x2 )m /(x − y),  Qβ (x, y) = β∆(x, y)(1 + x2 )n (1 + y 2 )n + N (x)(1 + y 2 )n − N (y)(1 + x2 )n /(x − y) TRIGONOMETRIC LIÉNARD EQUATIONS 5  and ∆(x, y) = 2 arctan(x) − arctan(y) . Proof. Notice that given any analytic map r it holds that R(x, y) = (r(x) − r(y)/(x − y) is analytic as well at (0, 0). Moreover at the origin R(0, 0) = r0 (0). Thus, taking r(θ) = tan(θ/2) we have that R(0, 0) = 1/2 and R and 1/R are analytic at zero. Hence, by using property (b) of Proposition 2, we get that     F (θ) − F (ψ) G(θ) − G(ψ) F (θ) − F (ψ) 1 G(θ) − G(ψ) 1 µ0 , = µ0 , θ−ψ θ−ψ θ−ψ R(θ, ψ) θ−ψ R(θ, ψ)   F (θ) − F (ψ) G(θ) − G(ψ) = µ0 , tan(θ/2) − tan(ψ/2) tan(θ/2) − tan(ψ/2) Notice that by (4), if we take (x, y) = (tan(θ/2), tan(ψ/2)) it holds that F (θ) − F (ψ) 1  e M (x) M (y)  = α∆(x, y) + − , tan(θ/2) − tan(ψ/2) x−y (1 + x2 )m (1 + y 2 )m G(θ) − G(ψ) 1  e N (x) N (y)  = α∆(x, y) + − , tan(θ/2) − tan(ψ/2) x−y (1 + x2 )n (1 + y 2 )n where  2x   2y  e ∆(x, y) = arctan − arctan . 1 − x2 1 − y2 Observe also that for |x| < 1 and |y| < 1,  2x   2y   e ∆(x, y) = arctan − arctan = 2 arctan(x) − arctan(y) = ∆(x, y). 1 − x2 1 − y2 Hence, by property (a) of Proposition 2 we obtain that   F (θ) − F (ψ) G(θ) − G(ψ) µ0 , = tan(θ/2) − tan(ψ/2) tan(θ/2) − tan(ψ/2)   Pα (x, y) Qα (x, y) µ0 , . (1 + x2 )m (1 + y 2 )m (1 + x2 )n (1 + y 2 )n Finally, by using again property (b) of the same proposition, we can remove in each component the factor (1 + x2 )−m (1 + y 2 )−m and the factor (1 + x2 )−n (1 + y 2 )−n , without changing the multiplicity, because they do not vanish at (0, 0), giving the desired result.  3. Proof of Theorem 1 Proof of Theorem 1. (i) We will prove that σ(Lm,1 ) = m − 1. The proof that σ(L1,n ) = n − 1 follows similarly. We will use Theorem 3 and Proposition 4. Note that following the notation of Proposition 4, µ(F 0 , G0 ) = µ0 [P0 (x, y), Q0 (x, y)] , because α = β = 0, where M (x)(1 + y 2 )m − M (y)(1 + x2 )m P0 (x, y) = , x−y 6 A. GASULL, J. GINÉ, AND C. VALLS and bx2 (1 + y 2 ) − by 2 (1 + x2 ) Q0 (x, y) = = b(x + y). x−y By property (e) of Proposition 2, to know the multiplicity µ0 [P0 , Q0 ] is suffices to study the function M (x) − M (−x) M odd (x) H(x) = P0 (x, −x) = (1 + x2 )m = (1 + x2 )m , 2x x where M odd is the odd part of M. Clearly, when H 6= 0, its highest order term at the origin is x2(m−1) . Hence, by Theorem 3, σ(Lm,1 ) = m − 1, as we wanted to prove. The proofs for σ(L∗m,1 ) and σ(L∗1,n ) are similar. (ii) Simply because L∗n,m ⊂ Ln,m it holds that σ(L∗n,m ) ≤ σ(Ln,m ). Moreover, by Theorem 3, and due to the symmetry between F and G in L∗n,m it holds that σ(L∗m,n ) = σ(L∗n,m ). Hence, we only need to prove that σ(Lm,n ) = σ(L∗m,n ). To prove this equality, notice that if the trigonometric polynomials F and G that give the maximum weak focus order correspond to an F such that for it corresponding expression as in (7) it holds that M (x) = a2 x2 + 0(3), with a2 6= 0, then it is clear that the maximum highest order in Lm,n is also taken for an element that is in L∗m,n and the equality follows. Otherwise, the maximum highest order is reached for an F such that its corresponding M satisfies M (x) = 0(3). Let us prove in this situation how to construct another F such that the order of the origin is the same but this new M, say M c, is such that M c(x) = b2 x2 + 0(3), with b2 6= 0. Let M and N such that m ≥ n and 2σ(Lm,n ) = µ0 [P0 , Q0 ], where recall that   M (x)(1 + y 2 )m − M (y)(1 + x2 )m N (x)(1 + y 2 )n − N (y)(1 + x2 )n (P0 , Q0 ) = , . x−y x−y Consider, as new F, a trigonometric polynomial of degree m, Fb, such that its correspond- ing M according to (5) is the polynomial of degree 2m, M c(x) = N (x)(1+x2 )m−n +M (x). Notice that Mc(x)(1 + y 2 )m − Mc(y)(1 + x2 )m Pb0 (x, y) = x−y   N (x)(1 + x2 )m−n + M (x) (1 + y 2 )m − N (y)(1 + y 2 )m−n + M (y) (1 + x2 )m = x−y 2 m−n 2 m−n = (1 + x ) (1 + y ) Q0 (x, y) + P0 (x, y). Hence, by property (d) of Proposition 2, h i µ0 Pb0 , Q0 = µ0 [P0 , Q0 ] = 2σ(Lm,n ) and M c(x) = N (x)(1 + x2 )m−n + M (x) = x2 + 0(3), as we wanted to prove. (iii) We start with the upper bound. Recall that if a polynomial map (P, Q) has only isolated (real or complex) singularities and Z denotes the set formed for all of them, TRIGONOMETRIC LIÉNARD EQUATIONS 7 then by Bezout’s theorem X µz [P, Q] ≤ deg(P ) deg(Q). z∈Z Recall that in our situation, α = β = 0 and by Proposition 4,   0 0 F (θ) − F (ψ) G(θ) − G(ψ) µ(F , G ) = µ0 , = µ0 [P0 (x, y), Q0 (x, y)] , θ−ψ θ−ψ where last two functions are polynomials and deg (P0 (x, y)) = 4m − 2 and deg (Q0 (x, y)) = 4n − 2, because the term of degree 4m−1 (resp. 4n−1) of P0 (resp. Q0 ) vanishes. Moreover, the four points (±i, ±i) are also singularities of (P0 , Q0 ). By using property (c) of Theorem 2 it is not difficult to prove that µ(i,i) [P0 , Q0 ] = µ(−i,−i) [P0 , Q0 ] ≥ (m − 1)(n − 1), µ(i,−i) [P0 , Q0 ] = µ(−i,i) [P0 , Q0 ] ≥ mn. Hence, by the above inequalities, X µ(F 0 , G0 ) ≤ deg(P0 ) deg(Q0 ) − µz [P0 , Q0 ] z∈Z\{(0,0)} ≤ 4(2m − 1)(2n − 1) − 2mn − 2(m − 1)(n − 1). Finally, by Theorem 3, σ(Lm,n ) ≤ 2(2m − 1)(2n − 1) − mn − (m − 1)(n − 1) = 6mn − 3(m + n) + 1. Now we compute the lower bound. We consider F 0 and G0 such that their correspond- ing expressions, as rational functions following (4) and (5), are x2m x2 + x2n−1 M (x) = and N (x) = . (1 + x2 )m (1 + x2 )n For each i, ` ∈ N introduce the following polynomials xi (1 + y 2 )` − y i (1 + x2 )` x2i (1 + y 2 )` − y 2i (1 + x2 )` Ri,` (x, y) = and S2i,` (x, y) = . x−y x2 − y 2 Then, x2m (1 + y 2 )m − y 2m (1 + x2 )m P0 (x, y) = = S2m,m (x, y)(x + y), x−y (x2 + x2n−1 )(1 + y 2 )n − (y 2 + y 2n−1 )(1 + x2 )n Q0 (x, y) = x−y = S2,n (x, y)(x + y) + R2n−1,n (x, y), where P0 and Q0 are the polynomials appearing in Proposition 4. Notice that x2m − y 2m S2,n (x, y) = 1 + O(1) and S2m,m (x, y) = + O(2m − 1). x2 − y 2 8 A. GASULL, J. GINÉ, AND C. VALLS By Proposition 4 and Theorem 3, µ(F 0 , G0 ) =µ0 [S2m,m (x, y)(x + y), S2,n (x, y)(x + y) + R2n−1,n (x, y)] =µ0 [S2m,m (x, y), S2,n (x, y)(x + y) + R2n−1,n (x, y)] + µ0 [x + y, S2,n (x, y)(x + y) + R2n−1,n (x, y)] , where in the last equality we have used property (d) of Proposition 2. We consider separately each of the terms of the sum. By using again the properties of Proposition 4, since S2,n (0, 0) 6= 0, µ0 [S2m,m (x, y), S2,n (x, y)(x + y) + R2n−1,n (x, y)] = µ0 [S2m,m (x, y), x + y + O(2)] =  2m   2m  x − y 2m x − y 2m µ0 + O(2m − 1), x + y + O(2) = µ0 , x + y = 2(m − 1). x2 − y 2 x2 − y 2 Similarly, by property (e) of Proposition 4, the second term coincides with the degree of the lowest term at the origin of R2n−1,n (x, −x) = (1 + x2 )n x2n−2 . Hence, µ0 [x + y, S2,n (x, y)(x + y) + R2n−1,n (x, y)] = 2(n − 1). Putting all the results together µ(F 0 , G0 ) = 2(m + n − 2). Hence, by Theorem 3, the order of the corresponding weak focus in m + n − 2 and the lower bound of the theorem follows. (iv) We only will give the full details of two cases of Table 1, (m, n) ∈ {(2, 3), (3, 2)}. The others follow similarly. When (m, n) = (2, 3), M (x)(1 + y 2 )2 − M (y)(1 + x2 )2 N (x)(1 + y 2 )3 − N (y)(1 + x2 )3 P0 (x, y) = , Q0 (x, y) = , x−y x−y where M (x) = b2 x2 + b3 x3 + b4 x4 and N (x) = x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6 because G(0) = G0 (0) = 0 and G00 (0) 6= 0 (the coefficient of x2 is normalized to one for the sake of simplicity). Since Q0 (x, y) = x + y + O(2), we have that ∂Q0 (0, 0)/∂y 6= 0, and by the Weier- strass Preparation Theorem it holds that Q0 (x, y) = (y − S(x))R(x,P y) for some analytic functions such that R(0, 0) 6= 0 and S(x) = −x + O(2) = −x + ∞ i i=2 ai x . Moreover, Q0 (x, S(x)) ≡ 0. Hence, by property (e) of Proposition 2, we can compute the maxi- mum multiplicity by taking P0 (x, S(x)) and vanishing this power series to the highest possible order by using the free parameters bi of M and ci of N. The first non-zero term is −4(c3 b2 − b3 )x2 which forces b3 = c3 b2 to have order bigger than 2. The next order is 4(−c3 b2 + 2c3 c4 b2 − c5 b2 − 2c3 b4 )x4 . It can be seen that if c3 = 0 we obtain lower vanishing order. So, we assume that c3 6= 0 we take b4 = (−c3 b2 + 2c3 c4 b2 − c5 b2 )/(2c3 ) to go on. The next power is 1 (3c43 − 2c23 c4 + 4c3 c5 − 3c33 c5 − 4c3 c4 c5 + 2c25 + 6c23 c6 )b2 x6 . 2c3 TRIGONOMETRIC LIÉNARD EQUATIONS 9 From here we have that c6 = (−3c43 + 2c23 c4 − 4c3 c5 + 3c33 c5 + 4c3 c4 c5 − 2c25 )/(6c23 ) to arrive to order bigger than 6. The next power is 1 − c3 (−21c3 + 22c3 c4 − 11c5 )(c3 − c5 )b2 x8 . 6 If (c3 −c5 )b2 = 0 we have that multiplicity infinity (or in other words that the correspond- ing Liénard system has a center at the origin). Then we must take c5 = c3 (−21+22c4 )/11. The next power is 2  c33 (−4 + 143c23 )(−16 + 11c4 )b2 x10 . 1331 The case √ (−16 + 11c4 )b2 = 0 gives again a case of multiplicity infinity. Hence we take c3 = ±2/ 143 and we have that the next power is 65536 ± √ ((−16 + 11c4 )b2 )x12 . (8) 353829047 143 Therefore the highest multiplicity is 12 which implies that σ(L2,3 ) = 6. To get σ(L3,2 ), notice first that by item (ii), σ(L3,2 ) ≤ σ(L2,3 ) = 6. Moreover, since the cases giving rise to order of the weak focus 6 satisfy that b2 6= 0, see (8), taking one of them and as a new M as M/b2 , we have that σ(L3,2 ) ≥ 6. Thus σ(L3,2 ) = 6, as we wanted to show.  4. Some results for the non pure trigonometric case For the case of non pure trigonometric polynomials a table similar to Table 1, but for the values σ(Lα,β m,n ), can be done. We present some cases in Table 2, where the numbers with a star mean a lower bound for the highest weak focus order and simply correspond to the values σ(Lm,n ). n\m 1 2 3 4 1 1 3 5 7 2 3 4 6∗ 7∗ 3 5 6∗ 7∗ - 4 7 7∗ - - Table 2. Some values of σ(Lα,β m,n ). We only will give some details for the case (m, n) = (2, 2). For these values 2α (arctan(x) − arctan(y)) (1 + x2 )2 (1 + y 2 )2 + M (x)(1 + y 2 )2 − M (y)(1 + x2 )2 Pα = , x−y 2β (arctan(x) − arctan(y)) (1 + x2 )2 (1 + y 2 )2 + N (x)(1 + y 2 )2 − N (y)(1 + x2 )2 Qβ = , x−y where M (x) = −2αx+b2 x2 +b3 x3 +b4 x4 and N (x) = −2βx+x2 +c3 x3 +c4 x4 . We proceed as in the proof of item (iv) of Theorem 1 by using property (e) of Proposition 2. Hence we have to find the highest order at zero of Pα (x, S(x)) where S is the analytic function that satisfies S(0) = 0 and Qβ (x, S(x)) ≡ 0. The first non-zero order is (−3c3 b2 + 3b3 − 10 A. GASULL, J. GINÉ, AND C. VALLS 10b2 β + 10α)x2 /3 which yields b3 = (3c3 b2 + 10b2 β − 10α)/3 if we want to arrive to higher order. The next power is 2 (15c3 c4 b2 − 15c3 b4 − 8b2 β + 50c4 b2 β − 50b4 β + 8α)x4 . 15 From it, to go on, we impose that b4 = (15c3 c4 b2 − 8b2 β + 50c4 b2 β + 8α)/(5(3c3 + 10β)) because, otherwise, if we take 3c3 + 10β = 0 it can be seen that we arrive to a lower vanishing order. The next power is 8(α − b2 β) − 720c3 + 945c33 + 1260c3 c4 − 3072β + 9450c23 β + 4200c4 β 1575(3c3 + 10β)  + 31500c3 β 2 + 35000β 3 x6 . If α − b2 β = 0 we have that multiplicity infinity, that is, we obtain that Pα (x, S(x)) ≡ 0. Then we must take 720c3 − 945c33 + 3072β − 9450c23 β − 31500c3 β 2 − 35000β 3 c4 = . 420(3c3 + 10β) The next power is 4(α − b2 β) (432 + 7(3c3 + 10β)2 (360 + 77(3c3 + 10β)2 ))x8 59535 Hence we must impose that 432 + 7(3c3 + 10β)2 (360 + 77(3c3 + 10β)2 ) = 0 to obtain higher multiplicity. Doing the reparametrization 3c3 + 10β = k1 , this second term does not vanish because it corresponds to 432 + 2520k12 + 539k14 , which has no real roots. Hence the maximum multiplicity is 8 and by Theorem 3, σ(Lα,β 2,2 ) = 4. Remark 5. In general, in this work we have not addressed the question of knowing if the highest order cyclicity gives rise to the corresponding number of limit cycles inside the Liénard trigonometric family. In general, the easiest way to ensure that this happens is to prove that the gradients of the Lyapunov quantities (the coefficient of the even orders in the above procedure) have the maximum rank at zero. For instance, it can be seen that this is the case in the above example. We end this section with a particular result refereed to a subfamily of non pure trigono- metric Liénard systems. Proposition 6. Let σ(Lα,0 m,1 ) be highest weak focus order for systems (1) inside the family non pure trigonometric Liénard systems with α ∈ R and β = 0. Then σ(Lα,0 m,1 ) = σ(Lm,1 ) + 1 = m. Proof. Arguing as in the proof of item (i) of Theorem 1 we know that µ(F 0 , G0 ) = µ0 [Pα (x, y), Q0 (x, y)] , where Q0 (x, y) = b(x + y), b 6= 0, and Pα (x, y) is as in the state- ment of Proposition 4. Hence, by property (e) of Proposition 2, to know the above multiplicity it suffices to know the highest order at the origin of the map  2 m  2 m 2α(1 + x ) arctan(x) M odd (x) K(x) = Pα (x, −x) = (1 + x ) + , x x TRIGONOMETRIC LIÉNARD EQUATIONS 11 where M odd is the odd part of M, which recall that is a polynomial of degree at most 2m. Clearly the coefficients of M can be chosen in such a way that K starts at the origin with terms of order at least 2m. Hence, to prove that the maximum order of K 6= 0 at the origin is 2m we need to prove that the coefficient of order 2m + 1 at the origin of the function (1 + x2 )m arctan (x) is not null. With this aim, we fix m and consider X ∞ 2 m H(x) = (1 + x ) arctan (x) = hk x2k+1 , for |x| < 1, k=0 where, for the sake of simplicity, we have removed the dependence on m of the function and on its Taylor series. We will prove that for all k ∈ N, hk 6= 0. It is not difficult to check that (1 + x2 )H 0 (x) = 2m xH(x) + (1 + x2 )m . As a consequence, h0 = 1, and equating the terms with x2k in the above expression, we obtain that for all k ≥ 1,  (2(m − k) + 1)hk−1 + m k hk = , 2k + 1  where m k = 0 for k > m. The above recurrence implies that hk > 0 for all k ≤ m and k−m (−1) hk > 0 for all k > m, because 2(m − k) + 1 is positive for k ≤ m and negative otherwise. Hence hk 6= 0 for all k ≥ 0, and in particular hm , the coefficient of x2m+1 , is not zero, as we wanted to prove. Thus, H[F 0 , G0 ] ≤ 2m and this upper bound is attained. Hence, by Theorem 3 the upper bound of the order of the weak focus for the family considered is m and this values is reached.  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Series A: Monographs and Treatises, 30. World Scientific Publishing Co., Inc., River Edge, NJ, 1998. [23] J. M. Pearson, N. G. Lloyd and C. J. Christopher, Algorithmic derivation of centre conditions SIAM Rev. 38 (1996), 619–636. [24] R. Roussarie, Bifurcation of planar vector fields and Hilbert’s sixteenth problem, Birkhäuser Verlag, Basel, 1998. [25] J. A. Sanders and R. Cushman, Limit cycles in the Josephson equation, SIAM J. Math. Anal. 17 (1986), 495–511. Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellater- ra, Barcelona, Catalonia, Spain Email address: gasull@mat.uab.cat 1 Departament de Matemàtica, Universitat de Lleida, Avda. Jaume II, 69; 25001 Lleida, Catalonia, Spain Email address: gine@matematica.udl.cat Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, Av. Rovisco Pais 1049–001, Lisboa, Portugal Email address: cvalls@math.ist.utl.pt