Qualitative Theory of Dynamical Systems (2022) 21:77 https://doi.org/10.1007/s12346-022-00609-7 Abelian Integrals and Non-generic Turning Points Renato Huzak1 · David Rojas2 Received: 26 January 2022 / Accepted: 29 April 2022 © The Author(s) 2022 Abstract In this paper we initiate the study of the Chebyshev property of Abelian integrals generated by a non-generic turning point in planar slow-fast systems. Such Abelian integrals generalize the Abelian integrals produced by a slow-fast Hopf point (or generic turning point), introduced in Dumortier et al. (Discrete Contin Dyn Syst Ser S 2(4):723–781, 2009), and play an important role in studying the number of limit cycles born from the non-generic turning point. Keywords Abelian integrals · Chebyshev systems · Planar turning points Mathematics Subject Classification 34E15 · 34E17 1 Introduction In this work we are concerned with the study of the Abelian integrals  I j (h) = e−2n ȳ x̄ 2n+2 j+1 d ȳ, n ∈ N, j = −n, 0, 1, 2 . . . γh and with their derivatives  d 2n + 2 j + 1 ¯ I j (h) = − I j (h) where I¯j (h) = x̄ 2 j+1 d ȳ. (1) dh 4n 2 γh A proof of (1) is given in Sect. 2. Here γh , with h ∈]0, 1[, denotes the oval surrounding the origin described by the level of energy {H (x̄, ȳ) = h} where B David Rojas david.rojas@udg.edu 1 Hasselt University, Campus Diepenbeek, Agoralaan Gebouw D, 3590 Diepenbeek, Belgium 2 Departament d’Informàtica, Matemàtica Aplicada i Estadística, Universitat de Girona, 17003 Girona, Catalonia, Spain 0123456789().: V,-vol 77 Page 2 of 18 R. Huzak, D. Rojas Fig. 1 The blow-up locus and indication of the period annulus formed by {γh }h∈]0,1[ and its boundary (see Sect. 3) H (x̄, ȳ) = 2ne−2n ȳ ( ȳ − x̄ 2n + 2n 1 ). We assume that the ovals are oriented clock- wise. The boundary of the period annulus formed by {γh }h∈]0,1[ consists of the origin (x̄, ȳ) = (0, 0) and the curve ȳ = x̄ 2n − 2n 1 (see Fig. 1). They correspond to the level sets h = 1 and h = 0, respectively. When n = 1, the above Abelian integrals arise from a slow-fast Hopf point at the origin (x, y) = (0, 0) in slow-fast family of Liénard systems   ∂ ∂ y − x 2 + O(x 3 ) +  2 (α − x) , (2) ∂x ∂y where  ≥ 0 is a small singular perturbation parameter and α ∼ 0 is a regular parameter. This has been observed in [5] where the cyclicity of (x, y) = (0, 0) (i.e., the maximum number of limit cycles in an (, α)-uniform neighborhood of (x, y) = (0, 0)) in the Liénard family (2) has been studied using so-called family blow-up. If the Liénard family is analytic or C ∞ -smooth with finite codimension, then the cyclicity of the origin is finite (see [5,Theorem 7.3]). To find a good upper bound for the cyclicity, the following conjecture formulated in [5] has to be solved: Conjecture Let n = 1. For each m ≥ 0, the functions I¯j j = −1, 0, 1, . . . , m − 1, defined in (1), form an extended complete Chebyshev system on [h 0 , 1] for any h 0 ∈ ]0, 1[. Extended complete Chebyshev systems (shortly, ECT-systems) are defined in Sect. 2. For m ≤ 2, the conjecture is proved in [7] or [13]. The conjecture is also true near the center h = 1, i.e. for each m ≥ 0 there exists  > 0 such that ( I¯−1 , I¯0 , . . . , I¯m−1 ) is an ECT-system on [1 − , 1] (see [7,Corollary 3.5]). Further- more, in [14,Theorem A] it has been shown that for each m ≥ 0 there exists  > 0 Abelian Integrals and Non-generic Turning Points Page 3 of 18 77 such that ( I¯−1 , I¯0 , . . . , I¯m−1 ) is an ECT-system on ]0, ]. The conjecture has been solved recently by Chengzhi Li and Changjian Liu in the paper [12]. The main purpose of this paper is to study the Chebyshev property of ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ), with m = 0, 1, . . . , on the interval ]0, 1[ for any fixed integer n > 1. When m = 0, the set of functions is only formed by I¯−n . Our motivation is the following generalization of system (2):   ∂ ∂ y − x 2n + O(x 2n+1 ) +  2n ( 2n−1 α − x 2n−1 ) , (3) ∂x ∂y where n > 1,  ≥ 0 is the singular perturbation parameter and α ∼ 0. We say that slow- fast Liénard family (3) has a non-generic turning point at the origin (x, y) = (0, 0). Like in the generic case (n = 1), to study the cyclicity of (x, y) = (0, 0) inside (3), one typically uses a family blow-up at (x, y, ) = (0, 0, 0). After desingular- ization of (3) near the origin, one has to deal with the Chebyshev property of the integrals ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ). For more details about the connection between the non-generic turning point (3) and the Abelian integrals ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ), we refer the reader to Sect. 3. Large canard limit cycles of (3), of size O(1) in the (x, y)-phase space, have been treated in [2]. As far as we know, the cyclicity of (x, y) = (0, 0) in (3) has not been studied. For the study of the cyclicity of some other slow-fast points, different from the Liénard systems (2) and (3), see e.g. [3, 8, 9, 11]. We stress that the goal of this paper is not to prove finite cyclicity of non-generic turning points, which is a result that needs further research to approach and it is beyond the scope of this manuscript. We mostly focus on the ECT-property of the ordered set ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ). We prove that a result similar to [7,Corollary 3.5] and [14,Theorem A] is true for each fixed n > 1 (see Theorems 1 and 2 in Sect. 2). The main difference between the statement of [7,Corollary 3.5] (n = 1) and Theorem 1 (n > 1) is that for n > 1 the boundary point h = 1 is not included in the interval on which the ECT-property holds. The function I¯−n goes to infinity as h → 1− (see Lemma 2). Applying methods from [13] to the case where n > 1 it can be seen that, for any fixed integer n > 1, ( I¯−n , I¯0 ) is an ECT-system on [, 1 − ]. The proof is analogous to the proof for n = 1 and we therefore omit it for the sake of brevity. We did not succeed in using [13] to prove the same for the ordered set ( I¯−n , I¯0 , I¯1 ) when n > 1. This is a topic of further study. Further, we prove the monotonicity property of the quotient I0 /I−n on the interval ]0, 1[ for each n > 1 (see Theorem 3). Theorem 3 naturally generalizes [4,Theorem 18] which covers the case of n = 1. Theorem 3 can be used to prove existence and uniqueness of limit cycles of planar systems obtained after desingularization of non- generic turning points. For more details see Theorem 4 in Sect. 3. In Sect. 2 we recall the definition of ECT-systems and state the main results of this paper. As already mentioned above, in Sect. 3 we motivate our study of the Abelian integrals. We prove the main results in Sect. 4. 77 Page 4 of 18 R. Huzak, D. Rojas 2 Definitions and Statement of Results Definition 1 Let f 0 , f 1 , . . . , f n−1 be analytic functions on a real interval with nonempty interior I . The ordered set of functions ( f 0 , f 1 , . . . , f n−1 ) is an extended complete Chebyshev system (in short, ECT-system) on I if, for all k = 1, 2, . . . n, any nontrivial linear combination α0 f 0 (x) + α1 f 1 (x) + · · · + αk−1 f k−1 (x) has at most k − 1 isolated zeros on I counted with multiplicity. (Notice that in this abbreviation "T" stands for Tchebycheff, which in some sources is the transcription of the Russian name Chebyshev.) One can prove (see [7,Lemma 3.7]) that ( f 0 , f 1 , . . . , f n−1 ) is an ECT-system on I if and only if the sequence F 1 , . . . , F n−1 , with F k = { f kk , f k+1 k , . . . , f k }, can be n−1 constructed such that: fi ( f ik ) 1. Defining f i1 := f0 , i = 1, . . . , n − 1, the functions f ik+1 := , for k = ( f kk ) 1, . . . , n − 2 and i = k + 1, . . . , n − 1, are analytic on I and 2. f 0 and ( f kk ) , k = 1, . . . , n − 1, are nowhere zero on I . This equivalent definition of ECT-system has been used in [5] with I being a closed interval [a, b]. If I = [a, b], then we have the following stability property of ECT- system (see [5,Proposition 7.6]): If ( f 0 , f 1 , . . . , f n−1 ) is an ECT-system on [a, b] and if gi is an analytic function sufficiently close to f i in the C n−1 -topology, for i = 0, . . . , n − 1, then (g0 , g1 , . . . , gn−1 ) is also an ECT-system on [a, b]. We will use this stability property in Sect. 3. The following results show that the number of zeros of any nontrivial linear combi- nation of the Abelian integrals are bounded locally near the endpoints of the interval ]0, 1[. We note that in the following statements, when m = 0 the set of functions is only formed by I¯−n . Theorem 1 For each m ≥ 0 there exists  > 0 such that ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ) is an ECT-system on [1 − , 1[. Theorem 1 will be proved in Sect. 4.1. Theorem 2 For each m ≥ 0 there exists  > 0 such that ( I¯−n , I¯0 , I¯1 , . . . , I¯m−1 ) is an ECT-system on ]0, ]. We prove Theorem 2 in Sect. 4.2. I0 (h) Theorem 3 Let P(h) = I−n (h) . Then P(h) > 0 and P  (h) < 0, for all h ∈]0, 1[, and limh→1− P(h) = 0. For n = 1, Theorem 3 has been proved by Chengzhi Li in [4] using the method from [1]. We use the same technique to prove it in the case where n > 1 (see Sect. 4.3). Lemma 1 The formula in (1) holds for every positive integer n and j = −n, 0, 1, 2 . . . . Abelian Integrals and Non-generic Turning Points Page 5 of 18 77 Fig. 2 Dynamics of X 0,α,a in a neighborhood of the contact point Proof Notice that the oval γh = {H (x̄, ȳ) = h}, where H is defined in Sect. 1, has two components: x̄ = x̄− ( ȳ, h) < 0 and x̄ = x̄+ ( ȳ, h) > 0. We have x̄+ = −x̄− , due to the symmetry of γh , and ∂∂hx̄± 2n ȳ = − 2e 2n−1 . Now, it easily follows that 4n x̄±    d e2n ȳ I j (h) =(2n + 2 j + 1) e−2n ȳ x̄ 2n+2 j − 2 2n−1 d ȳ dh γh 4n x̄  2n + 2 j + 1 =− x̄ 2 j+1 d ȳ 4n 2 γh where we use the notation x̄ for x̄± . This gives (1). 3 Motivation Consider slow-fast polynomial Liénard equations   ẋ = y − x 2n + mk=1 ak x 2n+k X ,α,a : (4) ẏ =   2n 2n−1 α−x 2n−1 , where m, n ≥ 1, a := (a1 , . . . , am ) is kept in a compact set K ⊂ Rm ,  ≥ 0 is a small singular perturbation parameter and α ∼ 0 is a small regular parameter. To study the number and configurations of limit cycles of (4), we can use geometric singular perturbation theory. The theory is essentially composed of two parts, one of which, called Fenichel theory [6], describes the dynamics of X ,α,a near normally hyperbolic manifolds. The other part is family blow-up [4, 11]; it is used to desingularize X ,α,a for example near the origin (x, y) = (0, 0) where the normal hyperbolicity is lost. More precisely, the fast subsystem X 0,α,a has the curve of singularities S = {(x, y) ∈ R2 |y = x 2n + m k=1 ak x 2n+k } and horizontal fast movements. The crit- ical curve S contains near the origin (x, y) = (0, 0) a normally repelling part x < 0, a normally attracting part x > 0 and a nilpotent contact point x = 0 which separates them (see Fig. 2). The contact point is generic (resp. non-generic) when n = 1 (resp. n > 1). For  > 0 and  ∼ 0, the dynamics of X ,α,a , uniformly away from S, can be 77 Page 6 of 18 R. Huzak, D. Rojas described using regular horizontal orbits of the fast subsystem X 0,α,a . Near the nor- mally hyperbolic parts of S, the dynamics of X ,α,a is given by the slow flow (often called slow dynamics) x 2n−1 x = − ∂F ∂ x (x, a) where F(x, a) := x 2n + m k=1 ak x 2n+k . When x ∼ 0 and x = 0, the slow dynamics points from the attracting part of S to the repelling part of S (note that x  = − 2n 1 + O(x) < 0). Thus, we call the contact point (x, y) = (0, 0) a turning point. To see how the Abelian integrals defined in Sect. 1 come into play, we blow up the ∂ origin (x, y, ) = (0, 0, 0) in X ,α,a + 0 ∂ using the following “singular" coordinate change (see [2, 10]) (x, y, ) = (r x̄, r 2n ȳ, r ¯ ), r ≥ 0, ¯ ≥ 0, (x̄, ȳ, ) ¯ ∈ S2 . (5) We work with different charts in (5). Family directional chart {¯ = 1}. In this chart we have (x, y) = ( x̄,  2n ȳ), with  = r , where we keep (x̄, ȳ) in a large compact set in R2 . In these new coordinates the system X ,α,a , defined in (4), becomes (after division by  2n−1 > 0)   x̄˙ = ȳ − x̄ 2n + m k=1 ak  x̄ k 2n+k F X ,α,a : (6) ȳ˙ = α − x̄ 2n−1 . When  = α = 0, system (6) is given by  x̄˙ = ȳ − x̄ 2n (7) ȳ˙ = −x̄ 2n−1 . System (7) has a center at the origin (x̄, ȳ) = (0, 0), with   −2n ȳ 1 H (x̄, ȳ) = 2ne ȳ − x̄ 2n + 2n as a first integral, and the invariant curve { ȳ = x̄ 2n − 2n 1 } is the boundary of the period annulus (Fig. 1). The phase directional charts {x̄ = ±1, ȳ = ±1}. The most interesting phase directional chart is the chart { ȳ = +1}. In this chart we find two semi-hyperbolic singularities p± located on the equator of the blow up locus (they are the end points of the invariant curve { ȳ = x̄ 2n − 2n 1 }). For a detailed study of X ,α,a in the { ȳ = +1}- direction see e.g. [10]. The other phase directional charts ({x̄ = ±1, ȳ = −1}) are not relevant when we study limit cycles of X ,α,a near the origin in the (x, y)-space. To find the cyclicity of (x, y) = (0, 0) in X ,α,a , we have to study three differ- ent types of limit periodic sets, for  = α = 0, which may produce limit cycles after perturbations: the center γ1 represented by {H (x̄, ȳ) = 1}, closed orbits of (7) Abelian Integrals and Non-generic Turning Points Page 7 of 18 77 surrounding the origin, denoted by γh = {H (x̄, ȳ) = h} where h ∈]0, 1[, and the polycycle γ0 consisting of the singularities p± and heteroclinic orbits between them. Let ρ > 0 be arbitrarily small and fixed. Limit cycles bifurcating from ∪h∈[ρ,1] γh can F be studied inside X ,α,a using the family chart {¯ = 1}. In order to treat γ0 , the charts {¯ = 1} and { ȳ = +1} have to be combined. In this paper we don’t study limit cycles produced by γ0 (see Remark 2). Define a section  = {x̄ = 0, ȳ ≥ 0} parametrized by h ∈]0, 1] by means of the relation H (0, ȳ) = h and a section 0 ⊂ {x̄ = 0, ȳ > 0} parametrized by h ∈ [ρ, 1 − ρ]. For  ≥ 0 small enough we define the Poincaré map P(h, , α, a) of X ,α,a F from 0 ⊂  to . Notice that we focus on the return map P with (, α) ∼ (0, 0),  ≥ 0 and a ∈ K , defined uniformly away from γ0 and γ1 . We have Proposition 1 The Poincaré map P(h, , α, a) of X ,α,a F can be written as    P(h, , α, a) = h + 4n 2 α − e−2n ȳ d x̄ + U−n (h, , α, a) γh l    −2n ȳ 2(n+ j)+1 + 4n 2 a2 j+1  2 j+1 − e x̄ d ȳ + U j (h, , α, a) j=0 γh (8) where l is the largest integer with the property 2l + 1 ≤ m and U−n , U j are analytic functions identically zero when (, α) = (0, 0): U−n (h, 0, 0, a) = 0 and U j (h, 0, 0, a) = 0 for j = 0, . . . , l. Proof We first study the Poincaré map P(h, α, A) of   x̄˙ = ȳ − x̄ 2n + m k=1 Ak x̄ 2n+k (9) ȳ˙ = α − x̄ 2n−1 where α ∼ 0 and A = (A1 , . . . , Am ) ∼ (0, . . . , 0). (If Ak = ak  k , then sys- F tem (9) becomes X ,α,a and P(h, , α, a1 , . . . , am ) = P(h, α, a1 , . . . , am  m ).) For (α, A) = (0, 0), the function H is the Hamiltonian of the (Hamiltonian) vector field (9), multiplied by 4n 2 e−2n ȳ . If we denote by  the dual 1−form of (9), then we have m 4n 2 e−2n ȳ  = d H + 4n 2 αe−2n ȳ d x̄ + 4n 2 Ak e−2n ȳ x̄ 2n+k d ȳ. k=1 Since γh,α,A 4n 2 e−2n ȳ  = 0 and γh,α,A d H = P(h, α, A) − h, where γh,α,A is a part of the orbit of (9) (multiplied by 4n 2 e−2n ȳ ) between h ∈ 0 and the next intersection P(h, α, A) ∈  in forward time, we obtain 77 Page 8 of 18 R. Huzak, D. Rojas    −2n ȳ P(h, α, A) = h + 4n α − e 2 d x̄ + V−n (h, α, A) γh m    −2n ȳ 2n+k + 4n 2 Ak − e x̄ d ȳ + Vk (h, α, A) , (10) k=1 γh where V−n , Vk are analytic functions, V−n (h, 0, 0) = 0 and Vk (h, 0, 0) = 0 for k = 1, . . . , m. Here we used the fact that γh,α,A converges uniformly to the oval γh when (α, A) → (0, 0), where the section 0 is parametrized by h ∈ [ρ, 1 − ρ]. On the other hand, P(h, α, A) = h for α = A2 j+1 = 0, j = 0, . . . , l, due to the symmetry (x, t) → (−x, −t). This, together with (10) and a Taylor expansion, implies    −2n ȳ P(h, α, A) = h + 4n α − 2 e d x̄ + V−n (h, α, A) γh l    + 4n 2 A2 j+1 − e−2n ȳ x̄ 2n+2 j+1 d ȳ + V j (h, α, A) j=0 γh for some new analytic functions V−n , V j with V−n (h, 0, 0) = 0 and V j (h, 0, 0) = 0 for j = 0, . . . , l. This implies (8). Theorem 4 Let U be any compact set in the interior of the period annulus of (7) and let κ > 0 be arbitrary and fixed. Then there exist sufficiently small 0 > 0 and α0 > 0 such that system X ,α,a F –given in (6)–has at most one periodic orbit in U, for all (, α, a) ∈ [0, 0 ] × [−α0 , α0 ] × K , (, α) = (0, 0) and |a1 | ≥ κ. Proof Let ρ > 0 be small enough such that U ⊂ ∪h∈[ρ,1−ρ] γh and let κ > 0 be small and fixed. If |a1 | ≥ κ, then the Poincaré map given in (8) can be written as    −2n ȳ P(h, , α, a) =h + 4n α − e 2 d x̄ + U−n (h, , α, a) γ  h  + 4n 2 a1  − e−2n ȳ x̄ 2n+1 d ȳ + U0 (h, , α, a) (11) γh where h ∈ [ρ, 1 − ρ] and U−n , U0 are analytic functions and are equal to zero when (, α) = (0, 0). Following Theorem 3, (I−n , I0 ), with I−n (h) = 2n 1 −2n ȳ d x̄ since  −2n ȳ γh e γh d e x̄ = 0 and I0 (h) = γh e−2n ȳ x̄ 2n+1 d ȳ, is an ECT-system on [ρ, 1 − ρ]. This implies that (−2n I−n , −I0 ) is an ECT-system on [ρ, 1 − ρ]. Now, using the stability property (Sect. 2) of (−2n I−n , −I0 ) on the segment [ρ, 1 − ρ], we have that (−2n I−n + U−n , −I0 + U0 ) is an ECT-system on [ρ, 1 − ρ], for each (, α) ∼ (0, 0), a ∈ K and |a1 | ≥ κ (note that U−n , U0 are equal to zero when (, α) = (0, 0)). Now, using (11), the equation {P(h, , α, a) − h = 0} has at most one solution in [ρ, 1 − ρ] counted with multiplicity, for each (, α) ∼ (0, 0), (, α) = (0, 0), a ∈ K and |a1 | ≥ κ. This solution corresponds to a periodic orbit of X ,α,aF . Abelian Integrals and Non-generic Turning Points Page 9 of 18 77 Remark 1 Theorem 4 gives the uniqueness of limit cycles of X ,α,a F in the compact set U, for |a1 | ≥ κ. Using (11) and The Implicit Function Theorem we see that X ,α,a F has a limit cycle near γh , with h ∈ [ρ, 1 − ρ], for   a1  I0 α=− (h) + o(1) 2n I−n where o(1) → 0 as  → 0. Remark 2 To find an optimal upper bound for the number of limit cycles of (4) in a fixed neighborhood of (x, y) = (0, 0), independent of  → 0, it is more suitable to study the Chebyshev property of ( I¯−n , I¯0 , I¯1 , . . . , I¯l ). These integrals appear in the expression for the derivative of P given in (8). The reason for this comes from [5] where the same has been done for n = 1. In fact, the Chebyshev property of the derivatives is relevant in a gluing process with the polycycle γ0 (see for example [5,Proposition 7.17] or [8]). As already observed in [14] for n = 1, we recall that the Chebyshev property of ( I¯−n , I¯0 , I¯1 , . . . , I¯l ) in the limit h → 0, obtained in Theorem 2, does not say anything about the number of limit cycles produced by γ0 for n > 1. One has to use different techniques to study the cyclicity of γ0 (see [5]). This topic is therefore not a subject of the present paper. 4 Proofs of Theorem 1–Theorem 3 4.1 Proof of Theorem 1 2 j+1+n Lemma 2 For each j ∈ {−n, 0, 1, 2, . . . }, I¯j (h) = (1 − h) 2n (a j + g j (h)) where a j = 0 and g j is an analytic function at h = 1 with g j (1) = 0. Proof When n = 1, this has been proved in [7,Lemma 3.4]. When n > 1, Lemma 2 can be proved in similar fashion. However, for the sake of completeness, we will give a sketch of the proof of this lemma. We know that γh can be described by {C( ȳ) + D( ȳ)x̄ 2n = 1 − h}, with C( ȳ) = 1 − 2ne−2n ȳ ( ȳ + 2n 1 ) and D( ȳ) = 2ne−2n ȳ . The oval γh , with h ∈]0, 1[, intersects the ȳ-axis at − 2n < ȳ− (h) < 0 < ȳ+ (h) < +∞. 1 Clearly, C( ȳ± (h)) =√ 1 − h. Since C( ȳ) = 2n ȳ (1 + O( ȳ)) near1 ȳ = 0, it follows 2 2 that g( ȳ) = sgn( ȳ) C( ȳ) is an analytic diffeomorphism on ] − 2n , +∞[. We have √ ȳ± (h) = g −1 (± 1 − h) and   √   2 j+1 g −1 ( 1−h) 1 − h − C( ȳ) 2n I¯j (h) = x̄ d ȳ = −2 2 j+1 √ d ȳ γh g −1 (− 1−h) D( ȳ)  1 √ 2 j+1+n (g −1 ) ( 1 − hs) 2 j+1 = −2(1 − h) 2n 2 j+1 √ (1 − s 2 ) 2n ds, −1 (D 2n )(g −1 ( 1 − hs)) 77 Page 10 of 18 R. Huzak, D. Rojas √ j = −n, 0, 1, . . . . In2 the last step we use the change of coordinates g( ȳ) = with 1 − hs and C( ȳ) = g( ȳ) . Note that the function (g −1 ) (z) z → 2 j+1 (D 2n )(g −1 (z)) is analytic at z = 0, and thus can be written as k≥0 bk z k. We obtain now    2 j+1+n 1 √ 2 j+1 I¯j (h) = −2(1 − h) 2n bk ( 1 − hs)k (1 − s 2 ) 2n ds −1 k≥0  1 2 j+1+n 2 j+1 = −2(1 − h) 2n b2k (1 − h)k s 2k (1 − s 2 ) 2n ds. k≥0 −1 2 j+1+n This implies the analyticity of I¯j (h)/(1−h) 2n at h = 1, for each j = −n, 0, 1, . . . . As (g −1 ) (0) = √1 and D(0) = 2n, we have 2n  √ I¯j (h) 1 2 j+1 −2 2π ( 2 j+1+2n 2n ) lim 2 j+1+n = −2b0 (1 − s ) 2 2n ds = 2 j+1+2n =: a j h→1− (1 − h) 2n −1 (2n) 2n ( 2 j+1+3n ) 2n where is the Gamma function. Proof of Theorem 1 For each α ∈ Q we say that a function f belongs to the set Rα if there exists  > 0 such that f (h) = (1 − h)α F(h) for all h ∈ [1 − , 1[, where F is an analytic function at h = 1 satisfying F(1) = 0. We notice that if f ∈ Rα and g ∈ Rβ with α > β then ( gf ) ∈ Rα−β−1 . Let us fix 0 ≤ k ≤ m − 1 and consider any function in the linear span of I¯−n , I¯0 , . . . , I¯k , that is (h) = η−n I¯−n (h) + η0 I¯0 (h) + · · · + ηk I¯k (h) where η−n , η0 , . . . , ηk ∈ R. We can consider ηk = 0 since otherwise (h) belongs to the linear span with lesser index k. For the sake of compactness we rewrite the previous equality as 0 = η−n I¯−n 0 + η0 I¯00 + · · · + ηk I¯k0 . From Lemma 2 we know that I¯−n 0 ∈R 1−n . In particular, there exists 0 > 0 such that 2n I¯−n (h) = 0 for all h ∈ [1 − 0 , 1[. This allows the first step of the division-derivation 0 algorithm, producing the function         0 I¯00 I¯10 I¯k0 1 := = η0 + η1 + · · · + ηk . I¯−n 0 I¯−n 0 I¯−n 0 I¯−n 0 Abelian Integrals and Non-generic Turning Points Page 11 of 18 77 For an analytic function f , let us denote by Z( f , ) the number of zeros of f on the interval [1 − , 1[ counted with multiplicity. By Rolle’s Theorem, we have that I¯0 Z( 0 , 0 ) ≤ Z( 1 , 0 ) + 1. Let us denote I¯1 := ( 0i ) for all i = 0, 1, . . . , k. We i I¯−n point out that, from Lemma 2 and the remark at the beginning of the proof, I¯i1 ∈ R i . n Therefore I¯01 ∈ R0 and so there exists 0 < 1 < 0 such that I¯01 (h) = 0 for all h ∈ [1 − 1 , 1[. We can then perform the second step of the division-derivation algorithm, obtaining         1 I¯11 I¯21 I¯k1 2 := = η1 + η2 + · · · + ηk . I¯01 I¯1 0 I¯1 0 I¯1 0 ¯1 Therefore we can ensure that Z( 0 , 1 ) ≤ Z ( 2 , 1 ) + 2. Now, denoting I¯i2 := ( I¯i1 ) I 0 we have that I¯i2 ∈ R i−n for i = 1, 2, . . . , k. In particular, I¯12 ∈ R 1−n and we can n n perform the next step in the division-derivation algorithm. Following this procedure, the ( j + 1)-step in the algorithm is ⎛ ⎞ ⎛ ⎞ ⎞ ⎛ I¯j j I¯k j j j+1 := ⎝ ⎠ = ηj ⎝ ⎠ + · · · + ηk ⎝ ⎠ I¯j−1 I¯j−1 I¯ j j j j−1 ¯j with I¯i := ( ¯ ji ) for i = j, j + 1, . . . , k and I¯j ∈ R 1−n . Thus we can perform j+1 I j+1 I j−1 n it since the (k + 1)-step, obtaining     k I¯kk k+1 := = ηk , I¯k−1 k I¯kk−1 ¯k with I¯kk+1 := ( ¯kk ) ∈ R 1−n and ηk = 0. In particular I¯kk+1 does not vanish on I Ik−1 n [1−k+1 , 1[ for some 0 < k+1 < k < · · · < 0 and Z( 0 , k ) ≤ Z ( k+1 , k )+k +1. Since the sequence of 0 , 1 , . . . , k+1 does not depend on but only on the Abelian integrals I¯−n , I¯0 , . . . , I¯k , this shows that, by taking  = k+1 , any function in the linear span of I¯−n , I¯0 , . . . , I¯k has at most k + 1 zeros, counted with multiplicity, on the interval [1 − , 1[. 4.2 Proof of Theorem 2 The oval γh intersects the ȳ-axis in two points (0, ȳ± (h)) for each h ∈]0, 1[, where ȳ− (h) < 0 < ȳ+ (h) are the two solutions of   1 f ( ȳ) := H (0, ȳ) = 2ne−2n ȳ ȳ + = h. 2n 77 Page 12 of 18 R. Huzak, D. Rojas From the definition of γh , we have that any (x, y) ∈ γh satisfies the equality x̄ 2n = ȳ + 2n 1 − 2n h 2n ȳ e . Therefore we can write the Abelian integrals (1) as  0  ȳ+ (h) h 2n ȳ 2 2n j+1 h 2n ȳ 2 2n j+1 I¯j (h) = 2 [ ȳ + 1 2n − 2n e ] d ȳ +2 [ ȳ + 1 2n − 2n e ] d ȳ ȳ− (h) 0 for j = −n, 0, 1, 2, . . . Let us denote by I¯− ¯+ j (h) (resp. I j (h)) the left-hand side integral (resp. right-hand side integral) of the splitting of I¯j (h). We point out that the functions ȳ± (h) are analytic on ]0, 1[. Moreover, since f  (− 2n 1 ) = 0, the function ȳ− (h) can also be extended analytically to h = 0 by ȳ− (0) := − 2n 1 . In consequence, the functions I¯j (h) are analytic at h = 0 since 2n > −1 for j ∈ {−n, 0, 1, 2, . . . }. − 2 j+1 Following the steps in [14] we perform the change of variable h = f ( 2ns 1 ) = e−1/s ( 1s + 1) and we define Iˆj (s) = I¯j ( f ( 2ns 1 )). We also define Iˆ± j (s) accordingly. 2n+2 j+1 Lemma 3 For each j ∈ {−n, 0, 1, 2, . . . }, Iˆ+ j (s) = (2ns) − 2n ψ 2 j+1 (s) for all 2n s > 0, where ψα (s) := 1 −t/s ) − t]α dt, 0 [(1 + s)(1 − e for α > −1. Proof From the fact that ȳ+ (h) is the positive solution of f ( ȳ) = h we have that ȳ+ ( f (u)) = u. Then, performing the change of variable ȳ = 1−t 2ns we get  1 2ns  2 j+1 Iˆ+ ¯+ j (s) = I j ( f ( 2ns )) = 1 [ ȳ + 1 2n − 1 2n f 1 2ns e2n ȳ ] 2n d ȳ 0  1 2n+2 j+1 2 j+1 = (2ns)− 2n [(1 + s)(1 − e−t/s ) − t] 2n dt 0 2 j+1 with 2n > −1 for all j = −n, 0, 1, 2, . . . This proves the result. Let us define the function  1−e1− s 1 Jα (s) := [x + s ln(1 − x)]α d x, for s > 0. 0 Next two results are proved in [14,Lemma 2.3] and [14,Lemma 2.4] respectively. Lemma 4 ψα (s) = − α+1 1 [(1 + s)(1 − e−1/s ) − 1]α+1 + (1 + s)α+1 Jα ( 1+s s ). Lemma 5 For every α > −1, lims→0+ Jα (s) = 1 1+α . Moreover, for every ∈ N, lims→0+ s ∂s Jα (s) = 0. Abelian Integrals and Non-generic Turning Points Page 13 of 18 77 Proof of Theorem 2 For each α ∈ Q we say that a function f belongs to the set Sα if there exists  > 0 such that f (s) = s α F(s) for all s ∈]0, [, with  a if = 0, lim s ∂s F(s) = s→0 + 0 if > 0, for some a = 0. We claim at this point that if f ∈ Sα and g ∈ Sβ with α = β then ( f /g) ∈ Sα−β−1 . Indeed, deriving the quotient we obtain      f (s) α−β−1 F(s) F(s)  =s (α − β) +s . g(s) G(s) G(s) Let us denote Q 0 (s) = F(s) G(s) and Q(s) = (α − β)Q 0 (s) + s Q 0 (s). A simple compu- tation shows that Q (k) (s) = (k + α − β)Q (k) (k+1) 0 (s) + s Q 0 (s), and so (k) (k+1) s k Q (k) (s) = (k + α − β)s k Q 0 (s) + s k+1 Q 0 (s). (k) Let us show that lims→0+ s k Q 0 (s) = 0 for all k > 0 by induction. For k = 1 we have   F(s)  s F  (s)G(s) − s F(s)G  (s) s = . G(s) G(s)2 Since lims→0+ s F  (s) = lims→0+ sG  (s) = 0 and lims→0+ G(s) is a non-zero con- stant, lims→0+ s Q 0 (s) = 0. Assuming the property to be true up to k − 1 we use the following recursive expression for the derivative of the quotient of two functions proposed in [15], ⎛  F(s) ( j−1) ⎞  (k) k F(s) 1 ⎝ (k) G (k+1− j) (s) G(s) ⎠. = F (s) − k! G(s) G(s) (k + 1 − j)! ( j − 1)! j=1 Then multiplying by the term s k the property follows by lims→0+ s k F (k) = 0, ( j−1) lims→0+ s k+1− j G (k+1− j) (s) = 0 and lims→0+ s j−1 Q 0 (s) = 0 for j = 1, . . . , k. (k) This shows lims→0+ s k Q 0 (s) = 0 for all k > 0 and so lims→0+ s k Q (k) (s) = 0 for all k > 0. Moreover, F(s) lim Q(s) = lim (α − β) =a s→0+ s→0+ G(s) for some a = 0. This proves the claim. 77 Page 14 of 18 R. Huzak, D. Rojas The fact that the map s → f ( 2ns 1 ) is an analytic diffeomorphism from ]0, +∞[ to ]0, 1[ and lims→0+ f ( 2ns ) = 0 imply the equivalence between proving Theorem 2 and 1 showing that the set ( Iˆ−n , Iˆ0 , Iˆ1 , . . . , Iˆm−1 ) is an ECT-system on ]0, ]. According to the notation introduced at the beginning of the section, Iˆj (s) = 2 Iˆ− ˆ+ ˆ± ¯± j (s) + 2 I j (s), where I j = I j ( f (1/2ns)). Since I¯− j (h) can be extended analytically at h = 0 and the map s  → f ( 2ns ) is flat at 1 s = 0 the function Iˆ− ˆ− j (s) can be written as I j (s) = a j + g j (s), where a j is a constant and g j is a flat function at s = 0. In consequence, by Lemma 3, 2n+2 j+1  2n+2 j+1 Iˆj (s) = 2(2ns)− 2n (2ns) 2n (a j + g j (s)) + ψ 2 j+1 (s) , 2n and by Lemma 4, 2n+2 j+1  2n+2 j+1 2n+2 j+1 Iˆj (s) = 2(2ns)− 2n s 2n (ã j + g̃ j (s)) + (1 + s) 2n J 2 j+1 ( 1+s s ) , 2n where ã j is a constant and g̃ j is a flat function at s = 0. The last equality together 2n+2 j+1 with Lemma 5 allows to write Iˆj (s) = 2(2ns)− 2n L j (s) with  2n if = 0, lim s ∂s L j (s) = 2n+2 j+1 s→0+ 0 if > 0, for each j ∈ {−n, 0, 1, 2, . . . }. Let us fix 0 ≤ k ≤ m − 1 and consider any function in the linear span of Iˆ−n , Iˆ0 , Iˆ1 , . . . , Iˆk , (s) = η−n Iˆ−n (s) + η0 Iˆ0 (s) + · · · + ηk Iˆk (s), where η−n , η0 , . . . , ηk ∈ R. We can consider ηk = 0 since otherwise (s) belongs to the linear span with lesser index k. For the sake of compactness we rewrite 0 = η−n Iˆ−n 0 + η0 Iˆ00 + · · · + ηk Iˆk0 . From the previous discussion, we have Iˆ−n0 ∈S −1/2n . In particular, there exists 0 > 0 such that Iˆ−n (s) = 0 for all s ∈]0, 0 ] and the first step of the division-derivation 0 algorithm can be performed, producing       0 Iˆ00 Iˆk0 1 = = η0 + · · · + ηk . Iˆ−n 0 Iˆ−n 0 Iˆ−n 0  Iˆ0j  Let us denote Iˆ1j := ˆI−n 0 for all j ∈ {0, 1, 2, . . . }. From the claim before Iˆ1j ∈ S− 2n+ j . In particular, Iˆ01 ∈ S−2 and there exists 0 < 1 < 0 such that Iˆ01 (s) = 0 for n Abelian Integrals and Non-generic Turning Points Page 15 of 18 77 all s ∈]0, 1 ] and a second step of the division-derivation algorithm can be performed. Following this procedure, the ( j + 1)-step in the algorithm is ⎛ ⎞ ⎛ ⎞ ⎞ ⎛ Iˆj j Iˆk j j j+1 := ⎝ ⎠ = ηj ⎝ ⎠ + · · · + ηk ⎝ ⎠ Iˆj−1 Iˆj−1 Iˆ j j j j−1 ˆj with Iˆi := ( ˆ ji ) with Iˆj ∈ R− 1+n . Thus we can perform it since the (k +1)-step j+1 I j+1 I j−1 n to obtain     k Iˆkk k+1 := = ηk , Iˆk−1 k Iˆk k−1 ˆk with Iˆkk+1 := ( ˆkk ) ∈ R− 1+n and ηk = 0. In particular Iˆkk+1 does not vanish I Ik−1 n on ]0, k+1 ] for some 0 < k+1 < k < · · · < 0 and, if we denote by Z( f , ) the number of zeros of an analytic function f on the interval ]0, ] counted with multiplicity, Z( 0 , k ) ≤ Z ( k+1 , k ) + k + 1. Since the sequence of 0 , 1 , . . . , k+1 does not depend on but only on the Abelian integrals Iˆ−n , Iˆ0 , . . . , Iˆk , this shows that, by taking  = k+1 , any function in the linear span of Iˆ−n , Iˆ0 , . . . , Iˆk has at most k + 1 zeros, counted with multiplicity, on the interval ]0, ]. This proves that ( Iˆ−n , Iˆ0 , Iˆ1 , . . . , Iˆm−1 ) is an ECT-system on ]0, ] and thus the result. 4.3 Proof of Theorem 3 0 (h) Let P(h) := II−n (h) for h ∈]0, 1[. It is clear that I−n (h) < 0 and I0 (h) < 0 (and hence P(h) > 0) for all h ∈]0, 1[ due to the chosen orientation of the oval γh . Near h = 1, the function P can be written as P(h) = (1 − h) (c + o(1)), (12) where c is a positive constant and o(1) is an analytic function at h = 1 and equal to zero for h = 1. From (12) follows that P  (h) < 0, for h ∼ 1 and h < 1, and that limh→1− P(h) = 0. In the rest of the proof we show that P  (h) < 0 for all h ∈]0, 1[. Suppose that this is not true on ]0, 1[. Then there exists the largest h 0 < 1 such that P  (h 0 ) = 0. We prove that P(h) > P(h 0 ) for h > h 0 and h ∼ h 0 (this will produce a contradiction with the largest h 0 such that P  (h 0 ) = 0). If we define q(h) := I−n (h 0 )I0 (h) − I−n (h)I0 (h 0 ), then we have q(h 0 ) = 0 and, for h > h 0 , 77 Page 16 of 18 R. Huzak, D. Rojas Fig. 3 The ovals γh 0 and γh , with h > h 0 and h ∼ h 0 , and the region  = 1 ∪ 2 ∪ 3 ∪ 4 with the orientation on the boundary I0 (h) I0 (h 0 ) P(h) − P(h 0 ) = − I−n (h) I−n (h 0 ) q(h) q(h) − q(h 0 ) = = I−n (h)I−n (h 0 ) I−n (h)I−n (h 0 ) h − h0 q  (h̃) = · I−n (h) I−n (h 0 ) h − h0     = · I0 (h̃) − P(h 0 )I−n (h̃) (13) I−n (h) where h̃ ∈]h 0 , h[. In the last step we differentiate q and use the definition of P. If we write Q(h) := I0 (h) − P(h 0 )I−n  (h), then, using (13) and the fact that I −n < 0, it suffices to prove that Q(h) < 0 for h > h 0 and h ∼ h 0 . We have Q(h 0 ) = 0 because P  (h 0 ) = 0. Since I−n  (h) = − 1 I¯ (h) and 4n 2 −n I0 (h) = − 2n+12 ¯0 (h), with I¯−n and I¯0 defined in (1), we get I 4n    2n + 1 1 Q(h) = − x̄d ȳ − P(h 0 ) − x̄ 1−2n d ȳ 4n 2 γh 4n 2 γh  1 (2n + 1)x̄ 2n − P(h 0 ) =− 2 d ȳ. (14) 4n γh x̄ 2n−1  0) Consider two vertical lines x̄ = ± 2n P(h 2n+1 . These two vertical lines intersect the oval γh 0 through the interior (Fig. 3) since Q(h 0 ) = 0. See also (14). Let’s denote by  the region between γh and γh 0 , with h > h 0 and h ∼ h 0 . We can write  = Abelian Integrals and Non-generic Turning Points Page 17 of 18 77 1 ∪ 2 ∪ 3 ∪ 4 (see Fig. 3). Then we have Q(h) =Q(h) − Q(h 0 )    1 (2n + 1)x̄ 2n − P(h 0 ) (2n + 1)x̄ 2n − P(h 0 ) =− 2 d ȳ − d ȳ 4n γh x̄ 2n−1 γh 0 x̄ 2n−1  1 (2n + 1)x̄ 2n − P(h 0 ) =− 2 d ȳ 4n ∂ x̄ 2n−1    1 (2n + 1)x̄ 2n − P(h 0 ) =− 2 + d ȳ 4n ∂1 ∪∂3 ∂2 ∪∂4 x̄ 2n−1  1 (2n + 1)x̄ 2n + P(h 0 )(2n − 1) =− 2 d x̄d ȳ 4n 1 ∪3 x̄ 2n  1 P(h 0 ) − (2n + 1)x̄ 2n − 2 d x̄d ȳ. (15) 4n 2 ∪4 (x̄ 2n − ȳ)2 In the last step we used Green’s Theorem and   (2n + 1)x̄ 2n − P(h 0 ) (2n + 1)x̄ 2n − P(h 0 ) d ȳ = d x̄, ∂2 ∪∂4 x̄ 2n−1 ∂2 ∪∂4 x̄ 2n − ȳ using d H (x̄, ȳ) = 0 and the factthat the numerator in the above integrals vanishes 0) along the vertical lines x̄ = ± 2n P(h2n+1 . It is clear that the first integral in (15) is positive. Since the curve {x̄ 2n − ȳ = 0} does not intersect 2 and 4 (more precisely, the curve {x̄ 2n − ȳ = 0} intersects γh 0 in its right-most point and its left-most point), we have that the second integral in (15) is also positive. We conclude that Q(h) < 0 for h > h 0 and h ∼ h 0 . This completes the proof of Theorem 3. Acknowledgements D.R. have been partially supported by the Ministerio de Ciéncia e Innovación grant PID2020-118281GB-C31, and he is member of the Consolidated Research Group 2017 SGR 1617 funded by the Generalitat de Catalunya. D.R. is a Serra Húnter Fellow. Funding Open Access funding provided thanks to the CRUE-CSIC agreement with Springer Nature. Data Availibility Statement All data generated or analysed during this study are included in this published article. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 77 Page 18 of 18 R. Huzak, D. Rojas References 1. 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