This is a preprint of: “Asymptotic expansion of the Dulac map and time for unfoldings of hyperbolic saddles: local setting”, David Marı́n, Jordi Villadelprat, J. Differential Equations, vol. 269(10), 8425–8467, 2020. DOI: [10.1016/j.jde.2020.06.024] Asymptotic expansion of the Dulac map and time for unfoldings of hyperbolic saddles: local setting D. Marín and J. Villadelprat BGSMath and Departament de Matemàtiques, Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain Departament d’Enginyeria Informàtica i Matemàtiques, ETSE, Universitat Rovira i Virgili, 43007 Tarragona, Spain Abstract. In this paper we study unfoldings of planar vector fields in a neighbourhood of a hyperbolic resonant saddle. We give a structure theorem for the asymptotic expansion of the local Dulac time (as well as the local Dulac map) with the remainder uniformly flat with respect to the unfolding parameters. Here local means close enough to the saddle in order that the normalizing coordinates provided by a suitable normal form can be used. The principal part of the asymptotic expansion is given in a monomial scale containing a deformation of the logarithm, the so-called Roussarie-Ecalle compensator. Especial attention is paid to the remainder’s properties concerning the derivation with respect to the unfolding parameters. Contents 1 Introduction and statements of the results 1 2 Further results on Roussarie’s series expansion 7 3 Dulac map 16 4 Dulac time 19 A Results about the class FLK (W ) 25 B Differentiation formulas and integration of series 32 1 Introduction and statements of the results In this paper we study unfoldings of planar vector fields in a neighbourhood of a hyperbolic resonant saddle. It can be viewed as the continuation of a previous paper where we give a C K normal form for the unfolding with respect to the conjugacy relation, see [10, Theorem A]. By means of this normal form in that paper 2010 AMS Subject Classification: 34C07; 34C20; 34C23. Key words and phrases: Dulac map, Dulac time, asymptotic expansion, uniform flatness. This work has been partially funded by the Ministry of Science, Innovation and Universities of Spain through the grants MTM2015-66165-P and MTM2017-86795-C3-2-P, the Agency for Management of University and Research Grants of Catalonia through the grants 2017SGR1725 and 2017SGR1617, and by the “María de Maeztu” Programme for Units of Excellence in R&D (MDM-2014-0445). 1 Σ1 1 P1 φ 1 Σ01 Σ02 D0 P2 Σ2 Figure 1: Auxiliary transverse sections in the decomposition of the Dulac map D = P2 ◦ D0 ◦ P1 and the Dulac time T = T1 + T0 ◦ P1 + T2 ◦ D0 ◦ P1 , where P1 (respectively, P2 ) is the Poincaré map from Σ1 to Σ01 (respectively, Σ2 to Σ02 ) and T1 (respectively T2 ) is the time that spends the flow to do this transition. Here the local Dulac map is D0 and the local Dulac time is T0 . we also determine an asymptotic expansion, uniform with respect to the parameters, for the local Dulac time of a resonant saddle, see [10, Theorem B]. The Dulac map of a saddle is the transition map from a transverse section Σ1 in the stable separatrix to a transverse section Σ2 in the unstable separatrix, whereas the Dulac time is the time that spends the flow to do this transition, see Figure 1. By local we mean that Σ1 and Σ2 cannot be at arbitrary distance from the saddle but close enough in order that we can use the normalizing coordinates provided by the normal form. In other words, and more precisely, the local Dulac map (respectively, local Dulac time) is the Dulac map (respectively, Dulac time) of the normal form. The asymptotic expansion of the Dulac map (see [15, Chapter 5] and references therein) is a key tool to study the cyclicity of a polycycle Γ (i.e., the maximum number of limit cycles that bifurcate from Γ) and to this end the remainder in the asymptotic expansion must be uniformly flat. In this respect recall that the second part of Hilbert’s 16th problem asks for the maximum number of limit cycles, called H(n), of a polynomial vector field P (x, y)∂x + Q(x, y)∂y as a function of n = max(deg(P ), deg(Q)). It is still unknown whether H(n) is finite. In case that the polycycle is monodromic and its return map is the identity then there is an annulus foliated by periodic orbits where the period function (i.e., the time of the return map) is defined. In this context the object of study are the so-called critical periodic orbits, which are the critical points of the period function. Similarly as with Hilbert’s 16th problem, it arises the notion of criticality of a polycycle Γ, i.e., the maximum number of critical periodic orbits that bifurcate from Γ, see [7, 9]. In the same way as for the cyclicity, an asymptotic expansion of the Dulac time with remainder uniformly flat constitutes a key tool to study the criticality of a polycycle. Both asymptotic expansions are of similar nature, they are given in a monomial scale containing the so-called Roussarie-Ecalle compensator, which is deformation of the logarithm. The asymptotic expansion of the local Dulac map (respectively, time) is a basic building block for establishing an asymptotic expansion of the Dulac map (respectively, time) and, in its turn, the latter is essential to study the cyclicity (respectively, criticality) of polycycles. In the present paper we focus on the local setting. Our main result for the local Dulac time is an asymptotic expansion that improves the one we previously obtained, see [10, Theorem B], in two aspects. Firstly because it gives a more precise description of the monomials appearing in the principal part. And secondly, more important, it shows that the remainder can be smoothly extended also with respect to the unfolding parameters. This was in fact our initial motivation to tackle the problem. In order to state our main theorems some results concerning 2 normal forms are needed. Let V be an open subset of RN and consider a C ∞ unfolding {Xµ }µ∈V of a hyperbolic saddle point at the origin. More precisely, Xµ = A(x, y; µ)x∂x + B(x, y; µ)y∂y , where A, B ∈ C ∞ (U × V ) for some open neighbourhood U of (0, 0) ∈ R2 and A(0, 0; µ)B(0, 0; µ) < 0 for all µ ∈ V. The hyperbolicity ratio of the saddle is B(0, 0; µ) λ = λ(µ) = − . A(0, 0; µ) Given m, n ∈ Z we also consider the collinear family 1 Yµ = Xµ . xm y n The reason why we permit this “polar” factor is because, when dealing with polynomial vector fields, a special attention must be paid to the study of those polycycles with vertices at infinity in the Poincaré disc. The factor can come from the line at infinity in a saddle at infinity or, more generally, appear in a divisor after desingularizing more general singular points at infinity of a polycycle. The case of lines of zeros in at least one of the separatrices is also allowed as it can appear after desingularizing a degenerate singular point at finite distance. It is important to remark that (by means of a reparametrization of time) this factor can be neglected to study the Dulac map but, on the contrary, this cannot be done when dealing with the Dulac time. For the same reason, to study the Dulac time we need normal forms with respect to the conjugacy relation rather than the equivalence relation. We recall at this point Theorem A in [10], which generalizes well-known orbital normal forms with respect to the equivalence relation (see [6, 15] and references therein). To this end let us fix µ0 ∈ V and denote λ0 = λ(µ0 ) for shortness. If λ0 ∈ Q, say λ0 = p/q with (p, q) = 1, then that result shows that for any k ∈ N the family {Yµ }µ∈V is C k conjugated, by means of a diffeomorphism Φ(x, y, µ) = (φ(x, y, µ), µ) defined in a neighbourhood of (0, 0, µ0 ) ∈ R2 × V , to the normal form 1    YµN F = x∂ x + − λ(µ) + P (u; µ) y∂y , η(µ)xm y n + u` Q(u; µ) where η is a C ∞ function, P and Q are polynomials in the resonant monomial u = xp y q with the coefficients being also C ∞ functions in µ, and     max m n p, q if mp − nq 6= 0, ` :=   (1)  max m , n + 1 if mp − nq = 0. p q Finally, if λ0 ∈/ Q then the result shows that we can take P = Q = 0. (In this paper we use the common notation b · c and d · e for the floor and ceiling functions respectively.) As we already explained, our aim in this paper is to study the Dulac time (as well as the Dulac map) associated to YµN F . (Note in this respect that the only interesting case is the resonant one, i.e., λ0 ∈ Q, because otherwise both maps can be computed explicitly.) More generally, we consider the polynomial normal family 1 Yα,β := PM Xα (2) β0 xm y n + u` i=1 βi ui−1 where  PN −1  Xα := x∂x + 1 q −p + i=0 αi+1 ui y∂y . (3) 3 In this way, setting α = (α1 , . . . , αN ) ∈ RN and β = (β0 , . . . , βM ) ∈ RM +1 , we thus consider the coefficients of the polynomials P ( · ; µ) and Q( · ; µ) in the normal form YµN F as independent parameters. Naturally we work with α1 ≈ 0 because p − α1 λ = λ(α1 ) := . q Note also that, with regard to the Dulac map, we can ignore the time and take Xα instead of Yα,β . That being said, we denote the Dulac map between (0, 1) × {1} and {1} × (0, 1) by D( · ; α). Similarly, the Dulac time between the same sections is denoted by T ( · ; α, β). More explicitly, let ϕ(t; s, α) be the solution of Xα passing through (s, 1) ∈ R2 with s > 0 at t = 0. Then, since this solution reaches {y = 1} at time t = − ln s due to ϕ1 (t; s, α) = set , it turns out that D(s; α) = ϕ2 (− ln s; s, α). Likewise, if φ(t; s, α, β) is the solution of Yα,β passing through (s, 1) ∈ R2 with s > 0 then the Dulac time is the function T ( · ; α, β) verifying φ1 (t; s, α, β)|t=T (s;α,β) = 1 for all s > 0 small enough. The present paper has two main results, namely: Theorem A, devoted to the Dulac map D(s; α), and Theorem B, addressed to the Dulac time T (s; α, β). The idea behind the proof, and also the aim of the result, is the same for both theorems. We show firstly that we can write the function as an infinite series for s > 0 and α1 small enough. Secondly, that we can truncate this series in order that the tail is uniformly flat at s = 0. And, thirdly, that the finite truncation can be expressed in terms of a polynomial in sp and sp ω(s; α1 ), where ω is a deformation of the logarithm (see Definition 1.3), the so-called Ecalle-Roussarie compensator. In this paper we use a more general notion of flatness (see Definition 1.2), which constitutes the key point in our approach as well as the main motivation to tackle the problem. Let us advance that it has better properties with respect to parameters and that this enables us to elucidate a delicate point which we think did not received the required attention in the literature (see Remark 1.4). Definition 1.1. Consider K ∈ Z≥0 ∪ {+∞} and an open subset U of RN . We say that a function ψ(s; µ) belongs to the class Cs>0 K (U ), respectively Cs=0 K (U ), if there exist an open neighbourhood V of {(s, µ) ∈ RN +1 ; s = 0, µ ∈ U } = {0} × U  in RN +1 such that (s, µ) 7→ ψ(s; µ) is C K on V ∩ (0, +∞) × U , respectively V.  More formally, the definition of Cs>0 K (U ) and Cs=0 K (U ) must be thought in terms of germs with respect to relative neighborhoods of {0} × U in (0, +∞) × U . In doing so these sets become rings and we have the inclusions C K (U ) ⊂ Cs=0 K K (U ) ⊂ Cs>0 (U ). These facts are implicitly used in Lemma A.3. We can now introduce the notion of (finitely) flatness that we shall use in the sequel. Definition 1.2. Consider K ∈ Z≥0 ∪ {+∞} and an open subset U of RN . Given some L ∈ R and µ̂ ∈ U , we say that ψ(s; µ) ∈ Cs>0 K (U ) is (L, K)-flat with respect to s at µ̂, and we write ψ ∈ FLK (µ̂), if for each ν = (ν0 , . . . , νN ) ∈ Z≥0 with |ν| = ν0 + · · · + νN 6 K there exist a neighbourhood V of µ̂ and C, s0 > 0 N +1 such that ∂ |ν| ψ(s; µ) 6 CsL−ν0 for all s ∈ (0, s0 ) and µ ∈ V . (4) ∂sν0 ∂µν11 · · · ∂µνNN T If W is a (not necessarily open) subset of U then define FLK (W ) := µ̂∈W FLK (µ̂).  The class FLK (W ) consists in those functions ψ(s; µ) that are (finitely) flat along {0} × W. The usual notion of (finitely) flatness is addressed to functions ψ that are smooth at s = 0 and not depending on parameters. In that context one simply requires the s derivatives of ψ to vanish at s = 0 up to order K − 1. When dealing with functions that are not smooth at s = 0, the natural and common definition is to require 4 the estimates in (4). In this non-smooth context, and when the function depends on parameters, one can alternatively require (4) to hold for all µ ∈ V but only for derivation with respect to s. This is precisely the notion of flatness used in [14, 15] for the remainder of the asymptotic expansion of the Dulac map (cf. Remark 1.4). For instance the function (s, µ) 7→ sµ is obviously L-flat at any µ̂ > L according to this alternative notion whereas to show that (s, µ) 7→ sµ belongs to FL∞ ({µ > L}) requires some computations (see Lemma A.4). Coming again to Definition 1.2, note that the case L < K is not excluded (and so it may occur that L − ν0 is negative) and that the case L = K corresponds to the usal notion of (finitely) flatness. The principal part of D( · ; α) and T ( · ; α, β) will be expressed in terms of the following deformation of the logarithm. Definition 1.3. The function defined for s > 0 and κ ∈ R by means of ( −κ s −1 κ if κ 6= 0, ω(s; κ) = − ln s if κ = 0, is called the Ecalle-Roussarie compensator.  Lemma A.4 gives several properties of the Ecalle-Roussarie compensator in relation with the class FLK (W ) as introduced in Definition 1.2. It shows in particular that (s, κ) 7→ ω(s; κ) belongs to F−ε ∞ ({κ < ε}) for all ε > 0. With regard to the parameter space of the family of vector fields in (3), hereafter we denote U0 := {α ∈ RN ; α1 = 0} = {0} × RN −1 . We can now state our two main results. In both statements we set ω = ω(s; α1 ) and λ = λ(α1 ) for the sake of shortness. The first one is a structure theorem for the asymptotic expansion of the local Dulac map. Theorem A. Let D( · ; α) be the Dulac map of the vector field Xα in (3) between the sections (0, 1) × {1} and {1} × (0, 1). Then for each L ∈ R there exists a unique ∆(z, w; α) ∈ Q[z, w, α], with deg(z,w) ∆ 6 Lp − 1q , and DL ∈ FL∞ (U0 ) such that D(s; α) = sλ ∆(sp , sp ω; α) + DL (s; α). Moreover, ∆(0, 0; α) = 1 in case that L > p q and ∆ ≡ 0 otherwise. This result has strong connections with the seminal works on the structure of the local Dulac map by A. Mourtada and R. Roussarie. Indeed, we write the principal part along the same lines as Mourtada, see [13, Proposition 2], in the sense that it is the Dulac map of the linear vector field x∂x − λy∂y , i.e., s 7→ sλ , multiplied by a unity (that we show is polynomial in sp and sp ω). Roussarie (see [14, Theorem F] or [15, Chapter 5]) writes the principal part in a different way and it is difficult to compare since he considers the case p = q = 1 only, which does not fit very well for q 6= 1. Next we make some further comments about it. Remark 1.4. The proof of Theorem A (and also the forthcoming Theorem B) relies on some previous results by R. Roussarie in [14] (see also [15, Chapter 5]) that we gather in Lemma 2.1 and constitute our starting point. In that paper the author studies the cyclicity of a saddle loop and to this aim he proves Theorem F, which describes the structure of the local Dulac map D(s; α). That result is very similar to our Theorem A, but important differences exist. Firstly his result is addressed to the case p = q = 1 because at that time it was already well-known that the cyclicity of a saddle loop with λ0 6= 1 is at most one. Secondly his result is more precise in the description of the principal part, i.e., D − DL , since he divides it in the ideal generated by the coefficients α1 , α2 , . . . , αN . And, thirdly, his proof concerning the remainder consists in showing that it verifies ∂sk DL (s; α) 6 CsL−k . This kind of estimate, similar to (4) but without derivation with respect to parameters, behaves well through the so-called derivation-division algorithm that yields to the main result in [14] on the cyclicity of the saddle 5 loop (which in our opinion is perfectly right and, what is more, correctly proved). However it does not enable to assert that DL extends to a C L function in (s, α) at s = 0 (see Example A.2 for a counterexample). Sadly enough, this is precisely what the author states in Theorem F with regard to the remainder DL (see also Theorem 14 in [15, page 103]). This inexactness yields to a crucial gap in a subsequent paper by the same author [16]. Indeed, in that paper he studies the smoothness property of the bifurcation diagram of a generic saddle loop unfolding of codimension 2, and to prove the main result he appeals to this (unproved) claim in Theorem F. To be more precise, by taking advantage of the smoothness with respect to parameters of the remainder, he is able to apply an ad hoc implicit function theorem to prove Proposition 2.1. In our Theorem A we show that DL ∈ FL∞ (U0 ), i.e., that the above bound holds for derivation with respect parameters as well, and on the other hand we prove (see Lemma A.1) that any function in FLK (U0 ) with L > K extends to a C K function in a neighbourhood of {0} × U0 in RN +1 . We can thus fill the gap between the proof of [14, Theorem F] and its statement. This shows in particular the validity of the proof of [16, Proposition 2.1], which constitutes a key step to show the main result in that paper.  Next result provides  the structure  of the asymptotic expansion of the local Dulac time and in its statement we assume that max m n p, q > 0. Let us point out however that we do not need this assumption in any of the previous auxiliary results. In this regard note that this hypothesis is satisfied if m and n are not both negative. From the point of view of the bifurcation of critical periodic orbits, the most interesting situation comes from the Dulac time associated to a saddle placed in the line at infinity, and in this case either m > 0 or n > 0. Theorem B. Let T ( · ; α, β) be the Dulac  time ofthe vector field Yα,β in (2) between the sections (0, 1)×{1} and {1} × (0, 1). Suppose that κ := max m , p q n > 0. Then for each L ∈ R we can write T (s; α, β = T L (s; α, β) + TL (s; α, β), where (1) the principal part is given by T L (s; α, β) := τ0 (β) ln s + sλn τ1 (sp , sp ω; α, β) − sm τ1 (sp , 0; α, β) + sκp τ2 (sp , sp ω; α, β), with τ0 (β) ∈ Q[β], τ1 (z, w; α, β) ∈ Q(α1 )[z, w, α2 , . . . , αN , β] and τ2 (z, w; α, β) ∈ Q[z, w, α, β], (2) and the remainder TL (s; α, β) belongs to FL∞ (U0 × RM +1 ). Moreover the principal part verifies the following: (a) τ1 is linear in β and without poles along α1 = 0. (b) τ2 is linear in β and τ2 (z, 0; α, β) = 0. (c) τ1 = 0 if mp − nq = 0. (d) τ0 = −β0 if (m, n) = (0, 0) and τ0 = −β1 if ` = 0, whereas τ0 = 0 in any other case. In a previous paper we already give a structure theorem for the asymptotic expansion of the local Dulac time, see [10, Theorem B]. The main difference between both results is that we can now guarantee that the remainder TL is flat along s = 0, not only for the derivation with respect to s, but also with respect to α and β (cf. Definition 1.2). Consequently, as we explain in Remark 1.4, by applying Lemma A.1 we can assert that if K < L then the remainder TL (s; α, β) extends to a C K function in a neighbourhood of {0} × U0 × RM +1 in R × RN × RM +1 . We are convinced that this regularity of the remainder will be crucial in future applications, for instance to have a better understanding of the bifurcation diagram of the critical periodic orbits of the Loud’s centers, see [9]. In fact this kind of property has already been used to study the 6 period of the limit cycle appearing in one-parameter saddle loop bifurcations (see [4, Theorem 16]). To this end the authors prove Proposition 23, which corresponds to Theorem B particularized to m = n = 0 and K = L = 1. (As a matter of fact while trying to extend it we realized that their proof contains a bridgeable mistake that we correct here, see Remark 2.4.) Coming back to our previous result in [10], let us note that the principal part T L that we provide here is more precise than the one given there. The paper is organized as follows. In Section 2, taking Roussarie’s results in [15, Chapter 5] as starting point, we consider the solution ϕ(t; s, α) of Xα passing through (s, 1) ∈ R2 at t = 0 and we expand ϕ2 (t; s, α) as a power series in s for each fixed t and α. We obtain sharp uniform estimates for the radius of convergence of this series (see Lemma 2.5) and also for the derivatives of its coefficients (see Lemma 2.7). Next, on account of D(s; α) = ϕ2 (− ln s; s, α), in Section 3 we use these results to prove Theorem A. Section 4 is devoted to the proof of Theorem B and to this end, see (2), we take advantage of the previous results thanks to the identity ! Z − ln s MX +`−1 T (s; α, β) = β0 x m y n + βi+1−` (xp y q )i dt. 0 i=` {(x,y)=ϕ(t;s,α)} Some technical but crucial issues about the sets FLK (W ) are treated in Appendix A. Among other properties we show that any g(s; µ) ∈ FLK (W ) with L > K extends to a C K -function (on s and the parameter µ) along s = 0. (This applies in particular to the remainder DL in Theorem A, as well as to TL in Theorem B.) Finally, in Appendix B we recall some specific results from analysis and calculus, in particular the multivariate Faa di Bruno formula for higher-order derivatives of a composite function (see Theorem B.1) that we use repeatedly all over the paper. 2 Further results on Roussarie’s series expansion Observe that performing the singular change of variables {u = xp y q , x = x}, the differential equation given by the vector field Xα in (3) is brought to the following form: ( ẋ = x, PN u̇ = P (u; α) := i=1 αi ui . The first equation gives x(t, x0 ) = x0 et and we denote by u(t, u0 ; α) the solution of the second one with initial condition u(0, u0 ; α) = u0 . For each fixed t and α, we expand it as a power series in u0 , +∞ X u(t, u0 ; α) = gi (t; α)ui0 . (5) i=1 In what follows, for any given δ > 0 we define Uδ := {α = (α1 , . . . , αN ) ∈ RN ; |α1 | < δ}. Following this notation, Roussarie [15, §5.1.2] shows the next result with regard to the series in (5). Lemma 2.1. The following assertions hold: eα1 t −1 (a) For all i ∈ N, gi (t; α) = eα1 t ḡi−1 (t; α) with ḡi (t; α) ∈ Q[α, Ω] where Ω := α1 and degΩ ḡi 6 i. (b) For each compact set C ⊂ Uδ with δ ∈ (0, 12 ] there exist K0 , C0 > 0 such that if t > 0, |u0 | < C0−1 e−δt and α ∈ C then the series (5) is absolutely convergent and |u(t, u0 ; α)| 6 K0 |u0 |eδt . (c) For all i ∈ N, t > 0 and α ∈ C, |gi (t; α)| 6 K0 C0−1 (C0 eδt )i . 7 Proof. Assertion (a) is proved in Proposition 10, whereas (b) follows from the proofs of Lemmas 18 and 19 because, using the author’s notation, +∞ X +∞ X |u(t, u0 ; α)| 6 |gi (t; α)||u0 |i 6 Gi (t)|u0 |i = U (t, |u0 |). i=1 i=1 Finally (c) follows from (b) by applying the Cauchy’s estimates (see for instance [18, Theorem 10.26]). Corollary 2.2. For each compact set C ⊂ Uδ with δ ∈ (0, 21 ] there exist C0 > 0 such that the function u(t, u0 ; α) is analytic on an open set containing {(t, u0 , α) ∈ RN +2 ; t > 0, |u0 | < C0−1 e−δt , α ∈ C}. Proof. Recall that u(t, u0 ; α) is the solution of u̇ = P (u; α) with initial condition u(0, u0 ; α) = u0 . Let us denote its maximal interval of existence by (ω− , ω+ ), where ω± = ω± (u0 , α). Since P is analytic on RN +1 , D = {(t, u0 , α) ∈ RN +2 ; ω− (u0 , α) < t < ω+ (u0 , α)} is an open set in RN +2 and u(t, u0 ; α) is analytic in D (see [5, Theorem 1.1] and [19, page 34]). Moreover, for the same reason, if ω+ is finite then |u(t, u0 ; α)| tends to +∞ as t % ω+ (see [1, Theorem 1.263] or [19, page 17] for instance). Note on the other hand that, by Lemma 2.1, if |u0 | < C0−1 e−δt then |u(t, u0 ; α)| 6 K0 C0−1 for all t ∈ (0, ω+ ) and α ∈ C. Arguing by contradiction this implies that ω+ > − 1δ ln(C0 |u0 |) and concludes the proof of the result. Given ν = (ν0 , ν1 , . . . , νN ) ∈ ZN +1 ≥0 , we write ν ∂ |ν| ∂t,α = ∂tν0 ∂α1ν1 νN · · · ∂αN and, following this notation, we expand ∂ νt,α u(t, u0 ; α) as a power series in u0 , +∞ X ν ∂t,α u(t, u0 ; α) = hi (t, α)ui0 . (6) i=1 Similarly as in Lemma 2.1, we want to estimate the functions hi and the convergence of the above series in terms of t and α. This is the aim of the next result, where we also write ν = (ν0 , ν̄) with ν̄ = (ν1 , . . . , νN ) for the sake of convenience. Theorem 2.3. For all ν ∈ ZN ≥0 +1 there exists a real number ρν̄ , satisfying 1 6 ρν̄ 6 max(|ν̄|, 1) and independent from ν0 , such that for each compact set C ⊂ Uδ with δ ∈ (0, 12 ] there exist Cν̄ > 0, independent from ν0 , and Kν > 0 such that if t > 0, α ∈ C and |u0 | < Cν̄−1 e−δt then (i) |∂ νt,α u(t, u0 ; α)| 6 Kν |u0 |eρν̄ δt , and (ii) the series in (6) is absolutely convergent. Moreover, for all i ∈ N, α ∈ C and t > 0, hi (t, α) = ∂t,α ν gi (t; α) and i−1 ν ∂t,α gi (t; α) 6 Kν eρν̄ δt Cν̄ eδt . Finally there exists M > 0 such that if |u0 | < (2C0 )−1 e−δt then eα1 t −1 |∂u20 u(t, u0 ; α)| 6 M eα1 t Ω(t, α1 ) where Ω(t, α1 ) := α1 . for all t > 0 and α ∈ C. 8 Proof. We begin by proving assertions (i) and (ii) in case that ν0 = 0, i.e., only derivation with respect to parameters. To this end for the sake of shortness we use the compact notation (0,ν1 ,...,νN ) ∂ |ν̄| ∂t,α = ∂αν̄ = νN where ν̄ = (ν1 , . . . , νN ). ∂α1ν1 · · · ∂αN The proof follows by induction on |ν̄|. The base case |ν̄| = 0 is (b) in Lemma 2.1. To show the induction step we first perform the partial derivation ∂αν̄ on both sides of the equality ∂t u = P (u; α) and then apply Theorem B.1 to obtain ∂t ∂αν̄ u(t, u0 ; α) =∂αν̄ P (u(t, u0 ; α); α) X X Yq ` QN ` λ (∂αj u)kj0 i=1 (∂αj αi )kji = ∂ P (u; α) ν̄! . (7) j=1 kj !(`j !)|kj | 16|λ|6|ν̄| p(ν̄,λ) u=u(t,u0 ;α) ∂ |λ| P (u;α) Here, for λ = (λ0 , . . . , λN ) ∈ ZN +1 ≥0 , we use the notation ∂ P (u; α) = λ λ λ and, for kj ∈ ZN +1 ≥0 , ∂uλ0 ∂α1 1 ···∂αnN we write kj = (kj0 , . . . , kjN ). Note also that both summations Pq are multidimensional Pq and the second one is subject to the coupling conditions given in (37), namely i=1 ki = λ and i=1 |ki |`i = ν̄. In this respect we observe the following: ` (a) The only summand in (7) that contains a factor ∂αj u with |`j | = |ν̄| is ∂u P (u; α)∂αν̄ u. Indeed, this is so because if `j = ν̄ and kj0 6= 0 then |ki | = 0 for i 6= j and λ = kj = (1, 0, . . . , 0). (b) If kj0 = 0 for all j then λ0 = 0. Consequently any summand in (7) not containing a factor ∂ ` u with PN |`| > 0 has the factor ∂ (0,`) P (u; α) = i=1 (∂α` αi )ui , which is a polynomial vanishing at u = 0. Accordingly we can split the right hand side of the equation (7) so that it writes as ∂t ∂αν̄ u(t, u0 ; α) = ∂u P (u(t, u0 ; α); α)∂αν̄ u(t, u0 ; α) + Rν̄1 (t, u0 , α) + Rν̄2 (t, u0 , α), where we define Rν̄1 to be the sum of those summands with kj0 = 0 for all j = 1, 2, . . . , q while Rν̄2 is the sum of the remaining summands. Note then that Rν̄1 (t, u0 , α) = uS(u; α)|u=u(t,u0 ;α) for some polynomial S(u; α) with degu S = N − 2. The above equality is a first order linear differential equation for ∂αν̄ u(t, u0 ; α) that, setting Rν̄ = Rν̄1 + Rν̄2 and Z t  B(t, u0 , α) := exp ∂u P (u(s, u0 ; α); α)ds (8) 0 for the sake of shortness, yields to Z t Rν̄ (s, u0 , α) ∂αν̄ u(t, u0 ; α) = B(t, u0 , α) ds. (9) 0 B(s, u0 , α) P∞ Note that we can write ∂u P (u(s, u0 ; α); α) = i=0 pi (s, α)ui0 with the same radius of convergence as (5) because ∂u P ( · ; α) is polynomial. In addition we have p0 (s; α) ≡ α1 . Thus, applying (b) in Lemma 2.1 and setting K1 := sup{|∂u P (u; α) − α1 |; |u| 6 K0 C0−1 , α ∈ C}, if |u0 | < C0−1 e−δt then Z t ∞ Z t X ∞ Z t X (∂u P (u(s, u0 ; α); α) − α1 ) ds 6 |pi (s; α)||u0 |i ds 6 K1 (C0−1 eδs )i |u0 |i ds 0 i=1 0 i=1 0 ∞ X X∞ eδit − 1 K1 = K1 C0i |u0 |i 6 (C0 |u0 |eδt )i < +∞, i=1 δi δ i=1 9 where in the second inequality we use Cauchy’s estimates (see [18, Theorem 10.26]). Thus, by Lemma B.5, Z t +∞ Z X t  ∂u P (u(s, u0 ; α); α)ds = pi (s; α)ds ui0 0 i=0 0 and the series converges absolutely for |u0 | < C0−1 e−δt . Furthermore, setting K2 := δ , K1 if |u0 | 6 (2C0 )−1 e−δt then Z t α1 t − K2 6 ∂u P (u(s, u0 ; α); α)ds 6 α1 t + K2 for all t > 0 and α ∈ C. 0 Consequently, recall (8), if |u0 | 6 (2C0 )−1 e−δt then e−K2 eα1 t 6 B(t, u0 , α) 6 eK2 eα1 t for all t > 0 and α ∈ C. (10) On the other hand, since x 7→ e±x are entire functions, the Taylor series of B(t, u0 , α) and 1/B(t, u0 , α) at u0 = 0 converge absolutely for all t > 0 and α ∈ C provided that |u0 | < C0−1 e−δt . Therefore, from (9) and taking the previous bounds into account, we get that if |u0 | 6 (2C0 )−1 e−δt then Z t |∂αν̄ u(t, u0 ; α)| 6 e2K2 eα1 t |Rν̄ (s, u0 , α)| e−α1 s ds for all t > 0 and α ∈ C. (11) 0 We are now in position to prove the validity of assertions (i) and (ii) for the case ν0 = 0 and to this end recall that Rν̄ = Rν̄1 + Rν̄2 . Let us begin with the study of Rν̄2 by noting that in each one of its summands we have that (∂ `j u(t, u0 ; α))kj0 verifies |`j | < |ν̄| for j = 1, 2, . . . , q and that there is at least one exponent kj0 strictly positive. Accordingly, thanks to the induction hypothesis, for each j = 1, 2, . . . , q we know that if |u0 | < C`−1 j e−δt then |(∂ `j u(t, u0 ; α))kj0 | 6 (K`j |u0 |eρ`j δt )kj0 for all t > 0 and α ∈ C. We define [ p? (ν̄) := {(k1 , . . . , kq ; `1 , . . . , `q ) ∈ p(ν̄, λ) ; |`j | < |ν̄|} , 16|λ|6|ν̄| which is nonempty if and only if |ν̄| > 1. Taking this into account, if |ν̄| > 1 then we set   Cν̄ := max 2C0 , max C`j ; (k1 , . . . , kq ; `1 , . . . , `q ) ∈ p? (ν̄) ,  P  q ρν̄ := max 1, max j=1 kj0 ρ`j ; (k1 , . . . , kq ; `1 , . . . , `q ) ∈ p? (ν̄) and Q  q kj0 K3 := max j=1 (K`j ) ; (k1 , . . . , kq ; `1 , . . . , `q ) ∈ p? (ν̄) , whereas if |ν̄| = 1 then we define Cν̄ = 2C0 , ρν̄ = 1 and K3 = 1. Furthermore we define  K4 := sup |∂ λ P (u; α)|; |u| 6 K0 C0−1 , α ∈ C, 1 6 |λ| 6 |ν̄| and nP P Qq QN `j o i=1 (∂ αi )kji K5 := sup 16|λ|6|ν̄| p(ν̄,λ) ν! j=1 |kj | ;α∈C kj !(`j !) Pq Pq Note moreover that |u0 | j=1 kj0 6 |u0 | due to j=1 kj0 > 1 and |u0 | 6 1. On account of these definitions and applying (b) in Lemma 2.1, from (7) it follows that if |u0 | < Cν̄−1 e−δt then |Rν̄2 (t, u0 , α)| 6 K6 |u0 |eρν̄ δt for all t > 0 and α ∈ C, where we set K6 := K3 K4 K5 . Let us proceed next with the study of Rν̄1 . In this case, due to Rν̄1 (t, u0 , α) = uS(u; α)|u=u(t,u0 ;α) , we define  K7 := sup |S(u; α)|; |u0 | 6 K0 C0−1 , α ∈ C . 10 Thus, by applying Lemma 2.1, if |u0 | < C0−1 e−δt then |Rν̄1 (t, u0 , α)| 6 K0 K7 |u0 |eδt for all t > 0 and α ∈ C. Finally, taking Cν̄ > C0 into account, we can assert that if |u0 | < Cν̄−1 e−δt then |Rν̄ (t, u0 , α)| 6 |Rν̄1 (t, u0 , α)| + |Rν̄2 (t, u0 , α)| 6 K0 K7 |u0 |eδt + K6 |u0 |eρν̄ δt 6 K8 |u0 |eρν̄ δt (12) for all t > 0 and α ∈ C, where we set K8 := max(K6 , K0 K7 ) and we use that ρν̄ > 1. We can now plug this inequality in (11) to obtain that if |u0 | < Cν̄−1 e−δt then Z t e(ρν̄ δ−α1 )t − 1 |∂αν̄ u(t, u0 ; α)| 6 K8 e2K2 |u0 |eα1 t e(ρν̄ δ−α1 )s ds = K8 e2K2 |u0 |eα1 t 6 Kν |u0 |eρν̄ δt 0 ρν̄ δ − α1 2K2 for all t > 0 and α ∈ C, where Kν := K8Ke 9 with K9 := inf{ρν̄ δ − α1 ; α ∈ C}, which is strictly positive because |α1 | < δ 6 ρν̄ δ. This proves the inductive step with regard to assertion (i). Let us turn now to assertion (ii). Since Rν̄ is a polynomial of ∂ ` u with 0 6 |`| < |ν̄| on account of property (a), by the induction hypothesis we get that the Taylor series of Rν̄ (t, u0 , α) at u0 = 0 is absolutely convergent for all t > 0 and α ∈ C provided that |u0 | < Cν̄−1 e−δt . Furthermore, from (12), if |u0 | < Cν̄−1 e−δt then |Rν̄ (t, u0 , α)| 6 K8 Cν̄−1 e(ρν̄ −1)δt for all t > 0 and α ∈ C. Recall on the other hand that the Taylor series of B(t,u1 0 ,α) at u0 = 0 is absolutely convergent for all t > 0 and α ∈ C provided that |u0 | < C0−1 e−δt . In addition (10) shows that |B(t,u1 0 ,α)| 6 eK2 e−α1 t for this range P∞ of values. Hence, due to C0 < 2C0 6 Cν̄ , if |u0 | < Cν̄−1 e−δt then the series RB(t,u ν̄ (t,u0 ,α) 0 ,α) = i=0 ri (t; α)ui0 is absolutely convergent and the upper bound Rν̄ (t, u0 , α) 6 K8 Cν̄−1 eK2 e((ρν̄ −1)δ−α1 )t 6 K8 Cν̄−1 eK2 eρν̄ δt B(t, u0 , α) holds for all t > 0 and α ∈ C due to |α1 | < δ. Note also that r0 (t; α) ≡ 0. As we did before, the Cauchy’s estimates show that |ri (t; α)| 6 K8 Cν̄−1 eK2 Cν̄i e(ρν̄ +i)δt for all i ∈ N, t > 0 and α ∈ C. Consequently, for all i ∈ N, t > 0 and α ∈ C, we get Z t e(ρν̄ +i)δt − 1 |ri (s; α)|ds 6 K8 Cν̄−1 eK2 Cν̄i 6 K9 Cν̄i e(ρν̄ +i)δt , 0 (ρν̄ + i)δ K8 Cν̄−1 eK2 where K9 = ρν̄ δ . Thanks to these estimates we can assert that if |u0 | < Cν̄−1 e−δt then Z t X +∞ Z t +∞ X Rν̄ (s, u0 , α) i ds 6 |u0 |i |ri (s; α)|ds 6 K9 eρν̄ δt Cν̄ |u0 |eδt <∞ 0 B(s, u0 , α) i=1 0 i=1 for all t > 0 and α ∈ C. Therefore, by Lemma B.5, Z t X +∞ Z t  Rν̄ (s, u0 , α) ds = ri (s; α)ds ui0 0 B(s, u0 , α) i=1 0 and the series converges absolutely for |u0 | < Cν̄−1 e−δt . Since we already prove this fact for the Taylor series of B(t, u0 , α) at u0 = 0, from (9) it follows that the Taylor series of ∂αν̄ u(t, u0 ; α) at u0 = 0 converges absolutely for all t > 0 and α ∈ C provided that |u0 | < Cν̄−1 e−δt . This shows the inductive step concerning assertion (ii) for the case ν0 = 0. 11 Let us turn now to the proof of the case ν0 > 0. Since ∂t u = P (u; α) we deduce that ∂tn u = Pn (u; α) for all n ∈ N, where Pn := P ∂u Pn−1 with P0 (u; α) := u. The application of Faa di Bruno formula given by Theorem B.1 yields to (ν ,ν1 ,...,νN ) ∂t,α0 u(t, u0 ; α) = ∂αν̄ Pν0 (u(t, u0 ; α)) X X Yq ` QN ` (∂αj u)kj0 i=1 (∂αj αi )kji = ∂ λ Pν0 (u; α) ν̄! (13) j=1 kj !(`j !)|kj | 16|λ|6|ν̄| p(ν̄,λ) u=u(t,u0 ;α) = R̂ν1 (t, u0 , α) + R̂ν2 (t, u0 , α), where R̂ν1 consists in all the summands with kj0 = 0 for all j. Accordingly, due to Pν0 (0; α) ≡ 0, exactly as we did to show (b), we can write R̂ν1 (t, u0 , α) = uSν (u; α)|u=u(t,u0 ;α) for some polynomial Sν . Thus, setting  K10 := sup |Sν (u; α)|; |u| 6 K0 C0−1 , α ∈ C and applying (b) in Lemma 2.1, if |u0 | < C0−1 e−δt then |Rν1 (t, u0 , α)| 6 K0 K10 |u0 |eδt for all t > 0 and α ∈ C. On the other hand, since we have already proved the validity of (i) for the particular case ν0 = 0, it follows ` that for each j there exist K`j , C`j > 0 and ρ`j > 1 such that |∂αj u(t, u0 ; α)| 6 K`j |u0 |etρ`j for all t > 0 and α ∈ C provided that |u0 | 6 C`−1 j e−δt . Thus, taking upper bounds in (13) as we did before with Rν̄2 , it follows that there exists Kν > 0 (which we take satisfying Kν > 2K0 K10 for convenience) such that if |u0 | < Cν̄−1 e−δt then |R̂ν2 (t, u0 , α)| 6 21 Kν |u0 |eρν̄ δt for all t > 0 and α ∈ C. (Here we remark that ρν̄ > 1 and Cν̄ > C0 are the ones previously defined when we tackle the case ν0 = 0.) Hence, if |u0 | < Cν̄−1 e−δt then ν̄ |∂t,α u(t, u0 ; α)| 6 |R̂ν1 (t, u0 , α)| + |R̂ν2 (t, u0 , α)| 6 K0 K10 |u0 |eδt + 21 Kν |u0 |eρν̄ δt 6 Kν |u0 |eρν δt for all t > 0 and α ∈ C. Finally the fact that the Taylor series of ∂t,α ν u(t, u0 ; α) is absolutely convergent for ` all t > 0 and α ∈ C provided that |u0 | < Cν̄−1 e−δt follows from (13) using that this is true for ∂αj u(t, u0 ; α) for all j and that ∂ λ Pν0 (u; α) is polynomial in u. So far we have proved assertions (i) and (ii) except for the validity of the upper bound ρν̄ 6 max(|ν̄|, 1). Lemma 2.1 shows that this is true for |ν̄| = 0 because we can take ρ0 = 1. The proof for |ν̄| > 1 follows by induction taking into account that  P  q ρν̄ := max 1, max j=1 kj0 ρ ` j ; (k1 , . . . , k q ; `1 , . . . , ` q ) ∈ p ? (ν̄) . The base case is also true by definition because ρν̄ = 1 for |ν̄| = 1 (recall that in this case the set p? (ν̄) is empty). The inductive step follows by noting that, due to the definition of p(ν̄, λ), q X q X q X q X kj0 ρ`j 6 kj0 |`j | 6 (kj0 + · · · + kjN )|`j | = |kj |(`j1 + · · · + `jN ) = |ν̄|, j=1 j=1 j=1 j=1 Pq where in the first inequality we use the inductive step and in the last equality take j=1 |kj |`j = ν̄ into account. Let us prove next the statement concerning the coefficients hi (t, α) in the series (6). To this aim observe that, by assertion (ii), this series converges absolutely for all t > 0 and α ∈ C provided that |u0 | < Cν̄−1 e−δt . This implies that, for each fixed t > 0 and α ∈ C, 1 i ν hi (t, α) = ∂ ∂ u(t, u0 ; α) for all i ∈ N. (14) i! u0 t,α u0 =0 On the other hand, thanks to assertion (i), if |u0 | < Cν̄−1 e−δt then |∂ νt,α u(t, u0 ; α)| 6 Kν |u0 |eρν̄ δt < Kν Cν̄−1 e(ρν̄ −1)δt . 12 Therefore, by applying the Cauchy’s estimates, i i−1 |hi (t; α)| 6 Kν Cν̄−1 e(ρν̄ −1)δt Cν̄ eδt = Kν̄ eρν̄ δt Cν̄ eδt for all i > 1, α ∈ C and t > 0. Recall in addition that, by (b) in Lemma 2.1, if |u0 | < C0−1 e−δt then P+∞ u(t, u0 ; α) = i=1 gi (t, α)ui0 converges absolutely for all t > 0 and α ∈ C. In particular, for each fixed t > 0 and α ∈ C, we can assert that gi (t, α) = i!1 ∂ui 0 u(t, u0 ; α) u0 =0 holds for all i ∈ N. Consequently 1 i ν 1 ν i ν hi (t, α) = ∂ ∂ u(t, u0 ; α) = ∂ ∂ u(t, u0 ; α) = ∂t,α gi (t, α), i! u0 t,α u0 =0 i! t,α u0 u0 =0 where in the first equality we use (14) and in the second one Corollary 2.2. It only remains to be proved the upper bound for |∂u20 u(t, u0 , α)|. To this end we observe that Z t   ∂u0 u(t, u0 ; α) = exp ∂u P u(s, u0 ; α); α ds = B(t, u0 , α), 0 where we use ∂t ∂u0 u = ∂u P (u; α)∂u0 u in the first equality and (8) in the second one. Therefore Z t  ∂u20 u(t, u0 ; α) = B(t, u0 , α) ∂u2 P u(s, u0 ; α); α B(s, u0 , α)ds. 0  Setting K11 := sup |∂u2 P (u; α)|; |u| 6 K0 C0−1 , α ∈ C , (b) in Lemma 2.1 and the inequalities in (10) show that if |u0 | < (2C0 )−1 e−δt then Z t |∂u20 u(t, u0 ; α)| 6 K11 e 2K2 α1 t e eα1 s ds = M eα1 t Ω(t, α1 ), 0 where we take M = K11 e2K2 . This completes the proof of the result. Remark 2.4. Let us mention that Theorem 2.3 corrects a mistake in the proof of [4, Propositon 23]. The authors of that paper split the proof into two intermediate claims. The second one is a particular case of assertion (i) in Theorem 2.3 (it corresponds to |ν̄| = 1 andPN = 1) but the proof given there is not right. +∞ Indeed, they consider in page 283 the series p(u(ξ, u0 )) = i=1 pi (ξ)ui0 but the summation index should run from i = 0. This may seem a typo but it has important consequences in order to bound the derivative with respect to parameters because, transferred to our proof, it yields to the factor eα1 t in (10). That being said, except for this bridgeable mistake in the proof of [4, Propositon 23], the main result in that paper with regard to the period of the limit cycle emerging from a saddle loop bifurcation is perfectly correct.  At this point let us denote by t 7→ (x(t, p0 ; α), y(t, p0 ; α)) the solution of the differential system given by the vector field Xα in (3) passing through p0 = (x0 , y0 ) ∈ R2 . It is clear that x(t, p0 ; α) = x0 et . We are interested in the analytical properties of y(t, p0 ; α) with the initial condition p0 = (s, 1). This is the reason why we first studied u = xp y q and in this respect, by Lemma 2.1, we know that +∞ X +∞ X u(t, u0 ; α) = gi (t; α)ui0 = u0 eα1 t ḡi (t; α)ui0 , i=1 i=0 where the series converge absolutely and we use that gi (t; α) = eα1 t ḡi−1 (t; α). Thus, since x(t, p0 ; α) = x0 et , +∞ +∞ ! q X X y(t, p0 ; α) = e−pt y0q eα1 t ḡi (t; α)ui0 = y0q e(α1 −p)t 1 + ḡi (t; α)ui0 . (15) i=0 i=1 13 P+∞ η  j Since (1 + z)η = k=0 k z k for |z| < 1, with the aim of computing y(t, x0 , y0 ; α) for any j ∈ Z we set k  X  X j/q ψ0j := 1 and, for k ∈ N, ψkj := ḡi1 · · · ḡir . (16) r=1 r i1 +···+ir =k Our next task is to prove the following result. Lemma 2.5. For each compact set C ⊂ Uδ with δ ∈ (0, 12 ] there exist C0 , M > 0 such that the identity +∞ X j y(t, s, 1; α) = e−λjt ψkj (t; α)skp k=0  holds for all j ∈ Z, t > 0, α ∈ C and s > 0 with sp max M Ω(t, α1 ), 4C0 eδt < 2, Moreover under these conditions the series is absolutely convergent. Proof. Since gi (t; α) = eα1 t ḡi−1 (t; α) and ḡ0 = 1, from (5) we get +∞ X u(t, u0 ; α) − u0 eα1 t ∂u20 u(t, ξ; α)u20 ḡi (t; α)ui0 = = for some ξ ∈ [0, u0 ], i=1 u0 eα1 t 2u0 eα1 t where in the second equality we apply Taylor’s theorem taking u(t, 0; α) = 0 and ∂u0 u(t, 0; α) = eα1 t into account. By applying Theorem 2.3, there exist C0 , M > 0 such that if |u0 | < (2C0 )−1 e−δt then |∂u20 u(t, u0 ; α)| 6 M eα1 t Ω(t, α1 ) for all t > 0 and α ∈ C. P+∞ Hence if |u0 | < (2C0 )−1 e−δt then ḡi (t; α)ui0 6 M 2 Ω(t, α1 )|u0 | for all t > 0 and α ∈ C. Therefore, if   i=1 P +∞ , (2C0 )−1 e−δt then i=1 ḡi (t; α)u0 < 1 for all t > 0 and α ∈ C. Accordingly, since 2 i |u0 | < min M Ω(t,α 1) P+∞  (1 + z)j/q = k=0 j/q k z for |z| < 1, from (15) and (16) it follows that k +∞ !j/q +∞ j X X y(t, x0 , y0 ; α) = y0j ej(α1 −p)t/q 1+ ḡi (t; α)ui0 = y0j ej(α1 −p)t/q ψkj (t; α)uk0 i=1 k=0  for all t > 0 and α ∈ C provided that |u0 | < 1/ max M 2 Ω(t, α1 ), 2C0 e . Furthermore the second series δt converges absolutely because so it does the first one thanks to Lemma 2.1. Finally, since u0 = xp0 y0q and λ = p−αq , the result follows taking (x0 , y0 ) = (s, 1). 1 The following is a technical lemma that will be used in the proof of our last result in this section. Lemma 2.6. For each m, n ∈ Z≥0 there exist Pmn , Qmn ∈ Z[x, y] with degx Pmn = degx Qmn = m and degy Pmn = degy Qmn = n such that for any a ∈ R, ∂xn ∂ym eaxy = am eaxy Pmn (x, ay) and ∂xn ∂ym xy = xy−n Qmn (ln x, y). In particular, there exist M1 , M2 > 0 such that ∂xn ∂ym eaxy 6 M1 max(1, |x|, |ay|)m+n |a|m eaxy and ∂xn ∂ym xy 6 M2 max(1, | ln x|, |y|)m+n xy−n .   Proof. Note that ∂xn ∂ym eaxy = ∂xn eaxy (ax)m and ∂xn ∂ym xy = ∂xn xy (ln x)m . From here the proof follows by induction on n. To this end we set Pm0 (x, y) = Qm0 (x, y) = xm . Then the inductive step follows by taking Pm,n+1 = yPmn + ∂x Pmn and Qm,n+1 = (y − n)Qmn + ∂x Qmn . 14 Lemma 2.7. For each compact set C ⊂ Uδ with δ ∈ (0, 21 ], j ∈ Z and ν ∈ ZN ≥0 +1 there exist Cjν , Kν > 0 such that ν |∂t,α ψkj (t; α)| 6 Kν (k + 1)3|ν| (Cjν )k max(1, t)|ν| e(3k+|ν|)δt for all k ∈ Z≥0 , t > 0 and α ∈ C. Proof. Note first that, on account of the definition in (16), k  X  X j/q ∂ ν ψkj = ∂ ν (ḡi1 · · · ḡir ), r=1 r i1 +···+ir =k where, due to gi (t; α) = eα1 t ḡi−1 (t; α) and applying Theorem B.2, X ∂ ν (ḡi1 · · · ḡir ) = a`0 ,...,`r ∂ `0 (e−α1 rt )∂ `1 gi1 +1 · · · ∂ `r gir +1 . `0 +...+`r =ν ν  We remark that this summation is multidimensional with `0 , . . . , `r ∈ ZN ≥0 +1 and a`0 ,...,`r = `0 ,...,` r are the generalized multinomial coefficients (cf. Remark B.3). Setting `0 = (`00 , . . . , `0N ) then, by Lemma 2.6, ∂ `0 (e−α1 rt ) = ∂t`00 ∂α`01 1 (e−α1 rt ) = (−r)`01 e−α1 rt P`01 `00 (t, −α1 r) if `02 = . . . = `0N = 0 and zero otherwise. In addition |∂ (e`0 −α1 rt )| 6 M`0 max(1, |t|, |rα1 |)|`0 | r`01 e−α1 rt . On the other hand, by Theorem 2.3, ` |∂t,α gi+1 (t; α)| 6 K` eρ` δt (C` eδt )i for all i ∈ N, α ∈ C and t > 0. Thus, if we set M̂ν := max(M` ; ` 6 ν), Ĉν := max(C` ; ` 6 ν) and K̂ν := max(K` ; ` 6 ν), then X |∂ ν (ḡi1 · · · ḡir ) (t; α)| 6 a`0 ,...,`r M̂ν (K̂ν )r max(1, |t|)|ν| r2|ν| e(r+ρ`1 +...+ρ`r )δt (Ĉν eδt )i1 +...+ir . `0 +...+`r =ν Here we use |α1 | 6 δ < 1, `0 6 ν and r > 1, which implies max(1, |t|, |rα1 |)|`0 | r`01 6 max(1, |t|, |r|)|ν| rν1 6 max(1, |t|)|ν| r2|ν| . Hence, since ρν 6 max(|ν|, 1) 6 1 + |ν| thanks to Theorem 2.3 and, on the other hand, |`0 | + . . . + |`r | = |ν| and r 6 k = i1 + · · · + ir , we obtain X |∂ ν (ḡi1 · · · ḡir ) (t; α)| 6 M̂ν (K̂ν Ĉν )k k 2|ν| max(1, t)|ν| e(3k+|ν|)δt a`0 ,...,`r . `0 +...+`r =ν P Thus, since `0 +...+`r =ν a`0 ,...,`r = (r + 1)|ν| 6 (k + 1)|ν| thanks to Remark B.3, we get |∂ ν (ḡi1 · · · ḡir ) (t; α)| 6 M̂ν (K̂ν Ĉν )k (k + 1)3|ν| max(1, t)|ν| e(3k+|ν|)δt .  Accordingly, since | j/q r | 6 max(|j/q|, 1)r 6 max(|j|, 1)k for all j ∈ Z, k X X ∂ ν ψkj (t; α) 6 M̂ν (K̂ν Ĉν ) (k + 1) k 3|ν| max(1, t) e |ν| (3k+|ν|)δt max(|j|, 1)k 1 r=1 i1 +···+ir =k  k = M̂ν p(k) max(|j|, 1)K̂ν Ĉν (k + 1)3|ν| max(1, t)|ν| e(3k+|ν|)δt ,  where p(k) is the number of partitions of k and it is easy to see that p(k) 6 2k−1 k 6 22k−1 6 4k . Hence, setting Cjν = 4 max(|j|, 1)K̂ν Ĉν and Kν = M̂ν , the result follows. 15 3 Dulac map This section is entirely devoted to prove Theorem A, that will follow almost immediately from Theorem 3.3. In the proof of this result, and the forthcoming Proposition 4.2, we will use the following lemma together with this easy observation: Remark 3.1. The function φ(s) = sα (− ln s)m is monotonous increasing on the interval (0, 1e ) provided that α > m > 0 because ∂s φ(s) = −sα−1 (− ln(s))m−1 (m + α ln s).  P Lemma 3.2. For every ρ ∈ (0, 1) and n ∈ Z+ there exists A > 0 such that k>K k n rk 6 AK n rK for all K ∈ N and 0 6 r 6 ρ. n  P+∞ Pn Proof. Setting c` := ` i=0 in−` ρi and A := `=0 c` we obtain +∞ X +∞ X n   X +∞ X n X n kn rk = (i + K)n ri+K = rK K` in−` ri 6 rK c` K ` 6 AK n rK , i=0 ` i=0 k=K `=0 `=0 where in the last inequality we take K > 1 into account. Theorem 3.3. Consider the family of vector fields {Xα }α∈Uδ defined in (3) and let D( · ; α) be the Dulac map of Xα between the transversal sections {y = 1} and {x = 1}. Then the following holds: (a) For each compact set C ⊂ Uδ with δ ∈ (0, 21 ] there exists s0 > 0 such that +∞ X D(s; α) = ψk1 (− ln s; α)skp+λ , for all s ∈ (0, s0 ) and α ∈ C, k=0 and the series is absolutely convergent. Moreover, for each K ∈ N there exists ∆(z, w; α) ∈ Q[z, w, α] with degz,w (∆) < K and ∆(0, 0; α) = 1 such that K−1 X ψk1 (− ln s; α)skp+λ = sλ ∆(sp , sp ω; α), where ω = ω(s; α1 ). k=0 (b) Finally, for each L ∈ R there exists KL ∈ Z≥0 such that +∞ X ψk1 (− ln s; α)skp+λ ∈ FL∞ (U0 ). (17) k=KL Proof. The solution x(t, x0 , y0 ; α) = x0 et of Xα with initial condition (x0 , y0 ) = (s, 1) intersects the transversal section {x = 1} at t = − ln s. Hence the Dulac map is given by D(s; α) = y(t, s, 1; α)|t=− ln s . On account of this, the first assertion in (a) will follow by applying Lemma 2.5 once we show that we can take s0 > 0 small enough such that  sp max M Ω(t, α1 ), 4C0 eδt t=− ln s < 2 for all s ∈ (0, s0 ). In this respect observe that, by applying (b) in Lemma A.4, sp Ω(− ln s, α1 ) = sp ω(s; α1 ) tends to 0 as s → 0+ uniformly in α1 ∈ (−δ, δ), and this is also true for sp−δ because p − δ > p − 21 > 0. Consequently it is clear that there exists s0 > 0 small enough such that the above inequality holds and so the first assertion is true. With regard to second one, from (a) in Lemma 2.1 and (16) it follows that ψk1 (− ln s; α) = ηk (ω; α) where ηk ∈ Q[ω, α] with degω (ηk ) 6 k. Then it is clear that, for each k = 0, 1, . . . , K −1, there exists a homogeneous 16 polynomial η̂k ∈ Q[z, w, α] with degw (η̂k ) 6 k such that we can write ψk1 (− ln s; α)skp = η̂k (sp , sp ω; α). Since η̂0 ≡ 1, this shows the validity of the second assertion in (a). In order to prove (b) we claim that for each ν ∈ ZN +1 ≥0 there exists s0 > 0 small enough such that the P series k>0 ∂s,α (ψk (− ln s; α)s ν 1 pk+λ ) converges uniformly on (0, s0 ) × C, where C is any compact set in Uδ that we hereafter. By the Weierstrass P M -test, to this end it suffices to show that there exists a sequence of positive numbers {Mk }k∈N with k>1 Mk < ∞ such that, for some kν ∈ N large enough, ν |∂s,α (ψk1 (− ln s; α)skp+λ )| 6 Mk , for all k > kν , s ∈ (0, s0 ) and α ∈ C. By applying Theorem B.2 we have that X ∂ ν (ψk1 (− ln s; α)skp+λ ) = a`1 `2 ∂ `1 (ψk1 (− ln s; α))∂ `2 (spk+λ ) (18) `1 +`2 =ν  with a`1 `2 = ν `1 ,`2 . Setting `ˆ1 = (0, `11 , . . . , `1N ) it turns out that, for each fixed s and α, ˆ `1 |∂s,α (ψk1 (− ln s; α))| = |∂s`10 (∂α`1 ψk1 )(− ln s; α)| 6 C`10 s−`10 max |(∂ (j,`11 ,...,`1N ) ψk1 )(− ln s, α)|, j∈{0,...,`10 } where C`10 > 0 depends only on `10 . The above inequality is clear in case that `10 = 0, whereas for `10 > 1 it follows easily by applying the one-dimensional Faa di Bruno formula n X X n Y (∂ i g(s))ki ∂sn (f (g(s))) = (∂sj f )(g(s)) n! j=1 i=1 (ki !)(i!)ki p(n,j) ˆ taking Pn n = `10 , f = ∂α`1 ψk1 and g = − ln s and noting that, in doing so, ∂ i g(s) = (−1)i (i − 1)!s−i and i=1 iki = n. Thus by applying Lemma 2.7 we deduce that, for all s ∈ (0, 1/e) and α inside a compact subset C of Uδ with δ ∈ (0, 21 ], `1 |∂s,α (ψk1 (− ln s; α))| 6 K̂`1 (k + 1)3|`1 | (Ĉ`1 )k (− ln s)|`1 | s−(3k+|`1 |)δ−`10 , (19) where, following the notation in that result, K̂`1 = C`10 max(K(j,`11 ,...,`1N ) ; j = 0, . . . , `10 ) and Ĉ`1 = max(C1,(j,`11 ,...,`1N ) ; j = 0, . . . , `10 ) p−α1 and we use that max(1, − ln s) = − ln s for s ∈ (0, 1/e). In addition, since λ = q , Lemma 2.6 shows that |∂ `2 (spk+λ )| = |∂s`20 ∂α`21 1 (spk+λ )| 6 M2 max(− ln s, pk + λ)|`2 | spk+λ−`20 q −`21 6 C`2 (k + 1)|`2 | (− ln s)|`2 | spk+λ−`20 , because pk + λ 6 p(k + 1) + 1 6 2p(k + 1) due to p, q > 1, |α1 | 6 δ < 1 and we set C`2 = (2p)|`2 | q −`21 M2 . Here we also use that max(x, y) 6 xy when x, y > 1. Using this inequality and the one in (19), from (18) we obtain |∂ ν (ψk1 (− ln s; α)skp+λ )| 6 K̄ν (C̄ν )k (k + 1)3|ν| (− ln s)|ν| s(p−3δ)k+λ−|ν|δ−ν0 , (20) P where we set C̄ν := max(Ĉ`1 ; `1 6 ν) and, on account of `1 +`2 =ν a`1 `2 = 2|ν| , K̄ν := 2|ν| max(K̂`1 C`2 ; `1 + `2 = ν). Let us remark that the above estimate holds for all s ∈ (0, 1/e) and α ∈ C ⊂ Uδ with δ ∈ (0, 21 ]. As a matter of fact, at this point we shrink it so that δ ∈ (0, 14 ), which in particular implies p − 3δ > 14 . Consequently, using also the fact that λ > 0, from (20) we get |∂ ν (ψk1 (− ln s; α)skp+λ )| 6 K̄ν (C̄ν )k (k + 1)3|ν| (− ln s)|ν| s(k−|ν|)/4−ν0 =: mk (s). (21) 17 On account of Remark 3.1 it easily follows that a sufficient condition for s 7→ mk (s) to be monotonous increasing on (0, 1/e) is that k > 9|ν| =: kν . Note on the other hand that mk (s) = K̄ν (C̄ν s1/4 )k (k + 1)3|ν| (− ln s)|ν| s−|ν|/4−ν0 .  Thus if we take s0 := min 1e , (2C̄ν )−4 then the series with general term Mk := mk (s0 ) is summable and, additionally, from (21) and the monotonicity of mk (s) on (0, 1/e), |∂ ν (ψk1 (− ln s; α)skp+λ )| 6 mk (s) < Mk for all s ∈ (0, s0 ) and k > kν . This proves the validity of the claim and consequently, by applying Lemma B.4 recursively, if s ∈ (0, s0 ) and α ∈ Uδ then ! X+∞ +∞ X ν 1 pk+λ ν  ∂s,α ψk (− ln s; α)s = ∂s,α ψk1 (− ln s; α)spk+λ (22) k=KL k=KL for all ν ∈ZN ≥0 +1 and KL ∈ N. (We stress that the above identity is valid regardless of KL > kν and this is crucial in what follows because kν depends on ν.) We are now in position to finish the proof. We will show that (17) holds taking KL := max(0, d4Le + 4). To this end, recall Definition 1.1, we fix any ν ∈ ZN ≥0 +1 and α? = (0, α2 , . . . , αN ) ∈ U0 = {0} × RN −1 , and we take a relatively compact neighbourhood V of α? contained in Uδ with δ = min( 41 , |ν| 1 ). Then, from (22) and using the upper bound in (20), for each s ∈ (0, s0 ) and α ∈ V we have +∞ ! +∞ X X ν ∂s,α 1 ψk (− ln s; α)spk+λ 6 ν ∂s,α (ψk1 (− ln s; α)spk+λ ) k=KL k=KL +∞ X 6 K̄ν (− ln s)|ν| sλ−|ν|δ−ν0 (k + 1)3|ν| (C̄ν sp−3δ )k k=KL 6 K̄ν Mν s −|ν|δ−ν0 A(KL + 1)3|ν| (C̄ν )KL s(p−3δ)KL 1 6 K̄ν Mν A(KL + 1)3|ν| (C̄ν )KL s 4 KL −1−ν0 6 CsL−ν0 . In the third inequality above we apply Lemma 3.2 and set Mν := sup{sλ (− ln s)|ν| ; s ∈ (0, s0 ), |α1 | 6 δ}. Next, in the fourth inequality, we take δ = min( 14 , |ν| 1 ) into account. Finally in the last inequality we set C := K̄ν Mν A(KL + 1)3|ν| (C̄ν )KL and use that KL > 4(L + 1). This completes the proof of the result. Proof of Theorem A. By Theorem 3.3, for each compact set C ⊂ Uδ with δ ∈ (0, 21 ] there exists s0 > 0 such that +∞ X D(s; α) = ψk1 (− ln s; α)skp+λ for all s ∈ (0, s0 ) and α ∈ C. k=0 In addition, for each L ∈ R there exists KL ∈ Z≥0 such that +∞ X DL (s; α) := ψk1 (− ln s; α)skp+λ ∈ FL∞ (U0 ). k=KL If KL = 0 then the result follows taking ∆ ≡ 0. If, on the contrary, KL ∈ N then by Theorem 3.3 we know ˆ that there exists ∆(z, ˆ 0; α) = 1 such that w; α) ∈ Q[z, w, α] with ∆(0, L −1 KX ˆ p , sp ω; α), ψk1 (− ln s; α)skp+λ = sλ ∆(s k=0 18 where ω = ω(s; α1 ). By gathering the homogenous part of ∆ ˆ of i-th degree, for i = 0, 1, . . . , dˆ:= deg(z,w) ∆, ˆ P ˆ ˆ p , sp ω) = d sλ+pi pi (ω; α) where pi (w; α) ∈ Q[w, α] with degw pi 6 i. it turns out that we can write sλ ∆(s i=0 p−α1 Then, due to λ = q and by (d) in Lemma A.4, note that sλ+pi pi (ω; α) ∈ FL∞ (U0 ) provided that i > Lp − 1q . Consequently if L > pq then there exists a unique polynomial ∆(z, w; α) ∈ Q[z, w, α] with ∆(0, 0; α) = 1 and deg(z,w) ∆ 6 b Lp − 1q c =: d, such that d X dˆ X ∆(z, w; α) = pi s pi (ω; α) and sλ+pi pi (ω; α) ∈ FL∞ (U0 ), i=0 i=d+1 where, in case that dˆ P 6 d, the second summation is void and we set pi ≡ 0 for i > d. ˆ Hence the result +∞ ˆ and DL respectively. Observe on the other follows taking ∆ and k=d+1 ψk1 (− ln s; α)skp+λ instead of ∆ ˆ p , sp ω) = Pd sλ+pi pi (ω; α) ∈ F ∞ (U0 ) and so in this case the result follows hand that if L < pq then sλ ∆(s i=0 L ˆ This concludes the proof of the result since the uniqueness of the polynomial ∆ taking ∆ ≡ 0 instead of ∆. / FL∞ (U0 ) if i 6 d. in the statement follows from the fact that sλ+pi ω ` ∈ 4 Dulac time In this section we will prove Theorem B. To this aim, for the sake of convenience, we begin by introducing Z − ln s Tijk (s; α) := e(i−λj)t ψkj (t; α)dt, for i, j ∈ Z and k ∈ N, (23) 0 and in its regard we prove the following result. Lemma 4.1. For each i, j ∈ Z and δ ∈ (0, 12 ] there exists k0 ∈ Z≥0 such that for all ν ∈ ZN ≥0 +1 and compact set C ⊂ Uδ there exist Cν , Kν > 0 so that the upper bound |∂ ν Tijk (s; α)| 6 Kν (k + 1)3|ν| (Cν )k (− ln s)|ν| sλj−i−ν0 −(3k+|ν|)δ holds for all k > k0 , s ∈ (0, 1/e) and α ∈ C. Proof. The result follows by applying Lemma 2.7 to the given compact set C ⊂ Uδ and ν ∈ ZN +1 ≥0 . Denote ν = (ν0 , ν1 , . . . , νN ) and suppose first that ν0 = 0. In this case if s ∈ (0, 1/e) and α ∈ C then X  Z − ln s ν |∂ (Tijk (s; α)| 6 ν ∂α`1 (e(i−λj)t )∂α`2 ψkj (t; α) dt `1 , `2 0 `1 +`2 =ν Z − ln s 6 2|ν| K̂ν (k + 1)3|ν| (Cν )k e(i−λj)t (jt/q)`11 max(1, t)|`2 | e(3k+|`2 |)δt dt 0 Z − ln s 6 Kν (k + 1)3|ν| (Cν )k (− ln s)|ν| e(i−λj+(3k+|ν|)δ)t dt 0 6 Kν (k + 1) 3|ν| k (Cν ) (− ln s) s |ν| λj−i−(3k+|ν|)δ , where in the first inequality we apply Theorem B.2, in the second one Lemma 2.7 and Remark B.3, in the third one we set Kν := 2|ν| (j/q)ν1 K̂ν and we use that max(1, t) 6 − ln s for all t ∈ (0, − ln s) due to s ∈ (0, 1/e), and in the last one we take  l  m k > k0 := max 0, 3δ1 1 − i + p+δ q |j| (24) 19   in order that i − λj + (3k + |ν|)δ) > 1 holds for all α ∈ Uδ and k > k0 . Here we use that λ ∈ p−δ p+δ q , q due p−α1 to |α1 | < δ and λ = q . We stress, and this is crucial, that k0 is independent from ν and C. This proves the result for ν0 = 0. Let us consider next the case ν0 > 1 and to this end we denote ν 0 := (ν0 −1, ν1 , . . . , νN ). Thus, from (23) and Theorem B.2, 0   X  ν0  ∂ ν Tijk (s; α) = −s−1 ∂ ν sλj−i ψkj (− ln s; α) = −s−1 ∂ `1 (sλj−i )∂ `2 (ψkj (− ln s; α)). 0 ` 1 , `2 `1 +`2 =ν Then the application of Lemma 2.6 and Lemma 2.7 show respectively ∂ `1 (sλj−i ) 6 M`1 (− ln s)|`1 | max(1, |λj − i|)|`1 | sλj−i−`10 (|j|/q)`10 and, since max(1, − ln s) = − ln s due to s ∈ (0, 1/e), ∂ `2 (ψkj (− ln s; α)) 6 K̂`2 (k + 1)3|`2 | (Ĉj`2 )k (− ln s)|`2 | s−(3k+|`2 |)δ . n o Setting K̄ν := sup M`1 K̂`2 max(1, |λj − i|)|`1 | (|j|/q)`10 ; α ∈ C, `1 + `2 = ν 0 and Cν := max(Ĉj`2 ; `2 6 ν 0 ), we can assert that if s ∈ (0, 1/e) and α ∈ C then |∂ ν Tijk (s; α)| 6 2|ν|−1 K̄ν (k + 1)3|ν| (Cν )k (− ln s)|ν| sλj−i−ν0 −(3k+|ν|)δ . Here we also take `1 + `2 = ν 0 = ν − (1, 0, . . . , 0) and Remark B.3 into account. Consequently, setting k0 := 0 and Kν := 2|ν|−1 K̄ν , the result follows in case that ν0 > 1. Recall at this point, see (2), that Theorem B concerns with the Dulac time associated to 1 Yα,β = PM Xα , β0 xm y n + u` i=1 βi ui−1 where m, n, ` ∈ Z and u = xp y q with p, q ∈ N. For this reason, as an intermediate step, we next consider the Dulac time Tij ( · ; α) of xi1yj Xα for any i, j ∈ Z. In its regard the next statement explains the convenience of introducing Tijk , see (23). Proposition 4.2. For each compact set C ⊂ Uδ with δ ∈ (0, 14 ] there exists s0 > 0 such that the Dulac time Tij ( · ; α) of the vector field xi1yj Xα , where i, j ∈ Z, writes as +∞ X Tij (s; α) = si+pk Tijk (s; α) for all s ∈ (0, s0 ) and α ∈ C (25) k=0 and the series is absolutely convergent. Moreover, for each L ∈ R there exists KL ∈ Z≥0 such that +∞ X si+pk Tijk (s; α) ∈ FL∞ (U0 ). (26) k=KL  Proof. Let t 7→ x(t, p0 ; α), y(t, p0 ; α) be the solution of Xα passing through p0 ∈ R2 at t = 0. Note that if p0 = (s, 1) with s > 0 then x(t, p0 ; α) = set intersects the transversal section {x = 1} at t = − ln s. Thus the time Tij (s; α) that spends the solution of xi1yj Xα starting at (s, 1) with s > 0 to reach the transversal section {x = 1} is given by Z − ln s Z − ln s +∞ X Tij (s; α) = i (x(t, s, 1; α)) (y(t, s, 1; α)) dt = j si e(i−λj)t ψkj (t; α)skp dt, 0 0 k=0 20 where in the second equality we apply Lemma 2.5. In this respect observe that, due to ∂t Ω(t, α1 ) = eα1 t > 0, for all t ∈ (0, − ln s) we have    sp max M 2 Ω(t, α1 ), 2C0 e δt < sp max M 2 Ω(t, α1 ), 2C0 e δt t=− ln s = sp max M 2 ω(s; α1 ), 2C0 s −δ < 1, provided that s > 0 is small enough because lims→0+ sp−δ = 0 and, by (b) in Lemma A.4, sp ω(s; α1 ) tends to zero as s → 0+ uniformly on Uδ . Consequently, recall the definition in (23), the first assertion in the statement will follow by applying Lemma B.5 once we show that for each compact set C ⊂ Uδ with δ ∈ (0, 41 ] there exists s0 > 0 such that +∞ Z X − ln s si+kp e(i−λj)t ψkj (t; α) dt < +∞ for all s ∈ (0, s0 ) and α ∈ C. (27) k=0 0 With this aim let us note that, by applying Lemma 2.7 with |ν| = 0, Z − ln s Z − ln s si+kp (i−λj)t e ψkj (t; α) dt 6 K0 (Cj0 ) s k i+kp e(i−λj+3kδ)t dt 0 0 s−(i−λj+3kδ) − 1 3 = K0 (Cj0 )k si+kp 6 K0 (Cj0 )k sk(p− 4 )+λj , i − λj + 3kδ where in the last inequality we use that p − 3δ > 41 , due to δ ∈ (0, 14 ), and we take k large enough so that 3 i − λj + 3kδ > 1. Thus the above upper bound readily shows the validity of (27) taking s0 = (Cj0 )−1/(p− 4 ) 3 because it guarantees that Cj0 sp− 4 < 1 for all s ∈ (0, s0 ). With regard to the last assertion in the statement let us first note that, by applying Theorem B.2,  X  ν  ν i+pk ∂ s Tijk (s; α) = ∂ `1 (si+pk )∂ `2 (Tijk (s; α)). `1 , `2 `1 +`2 =ν Accordingly, by Lemma 4.1, there exists k0 ∈ Z≥0 such that, for all ν ∈ ZN ≥0 +1 and compact set C ⊂ Uδ , X  ν  ν |∂ (s i+pk Tijk (s; α))| 6 K`2 |i + kp|`10 (k + 1)3|`2 | (C`2 )k (− ln s)|`2 | sλj+pk−`10 −`20 −(3k+|ν|)δ `1 , `2 `1 +`2 =ν provided that k > k0 , s ∈ (0, 1e ) and α ∈ C. Since |i + pk| 6 (k + 1)(|i| + p), setting Ĉν = max(C`2 ; `2 6 ν) and K̂ν := 2|ν| max(K`2 (|i| + p)`10 ; `1 + `2 = ν), we can assert that if k > k0 , s ∈ (0, 1/e) and α ∈ C then |∂ ν (si+pk Tijk (s; α))| 6 K̂ν (k + 1)4|ν| (Ĉν )k (− ln s)|ν| sλj+(p−3δ)k−ν0 −|ν|δ (28) 6 K̂ν (k + 1)4|ν| (Ĉν )k (− ln s)|ν| sγ+(k−|ν|)/4−ν0 =: mk (s), where in the first inequality we also take `1 + `2 = ν and Remark B.3 into account, and in the second one we use that δ ∈ (0, 14 ), p > 1 and λj > − p+δq |j| =: γ. (Let us remark, it will be important later on when we use the previous inequalities, that k0 is independent from ν and C.) On account of Remark 3.1, a sufficient condition for s 7→ mk (s) to be monotonous increasing on (0, 1/e) is that k > 9|ν| + 4(ν0 − γ), and for this reason we set k̄ν := max (d5|ν| + 4(ν0 − γ)e , k0 ) . Note on the other hand that, due to mk (s) = K̂ν (Ĉν s1/4 )k (k + 1)4|ν| (− ln s)|ν| sγ−|ν|/4−ν0 , 21 if we set s0 := min(1/e, (2Ĉν )−4 ) then the series with general term Mk := mk (s0 ) is summable and, moreover, thanks to the monotonicity of mk (s) on (0, 1/e), |∂ ν (si+pk Tijk (s; α))| 6 mk (s) 6 Mk for all s ∈ (0, s0 ), α ∈ C and k > k̄ν . P∞ Hence, thanks to the Weierstrass M-test, for each ν ∈ ZN ≥0 +1 the series k=0 ∂ ν (si+pk Tijk (s; α)) converges uniformly for s ∈ (0, s0 ) and α ∈ C. Consequently, by applying recursively Lemma B.4 starting from (25), we have that for each compact set C ⊂ Uδ and ν ∈ ZN ≥0 +1 there exists s0 > 0 small enough such that if s ∈ (0, s0 ) and α ∈ C then +∞ ! +∞ X X  ν ν i+pk ∂ Tij (s; α) = ∂ s Tijk (s; α) = ∂ ν si+pk Tijk (s; α) . (29) k=0 k=0 We are now in position to finish the proof. Indeed, we claim that (26) holds taking  l m  KL := max k0 , 4L + 4p+1 q |j| + 8 . (Recall that k0 is the nonnegative integer given by Lemma 4.1, see (24), which is relevant for our purpose because it guarantees the upper bound (28) for k > k0 .) We point out that KL is independent from ν and C. In order to show (26), recall Definition 1.1, we fix any ν ∈ ZN +1 ≥0 and α = (0, α2 , . . . , αN ) ∈ U0 = {0}×R ? N −1 , and we take a relatively compact neighbourhood V of α contained in Uδ with δ = min( 4 , |ν| ). Then, ? 1 1 from (29) and using the upper bound in (28), for each s ∈ (0, s0 ) and α ∈ V we have +∞ ! +∞ X X ν ∂s,α (si+pk Tijk (s; α) 6 ν ∂s,α (si+pk Tijk (s; α)) k=KL k=KL +∞ X 6 K̂ν (− ln s)|ν| sλj−|ν|δ−ν0 (k + 1)4|ν| (Ĉν sp−3δ )k k=KL 6 K̂ν M̂ν s λj−|ν|δ−ν0 −1 A(KL + 1)4|ν| (Ĉν )KL s(p−3δ)KL 1 6 K̂ν M̂ν A(KL + 1)4|ν| (Ĉν )KL sλj+ 4 KL −2−ν0 6 CsL−ν0 . In the third inequality above we apply Lemma 3.2 and set M̂ν := sup{s(− ln s)|ν| ; s ∈ (0, s0 )}. Next, in the fourth inequality, we take δ = min( 41 , |ν| 1 ) and p > 1 into account. Finally in the last inequality we use the definition of KL , which implies λj+ 14 KL −2 > L due to λ < p+δ q , and we set C := K̂ν M̂ν A(KL +1)4|ν| (Ĉν )KL . This completes the proof of the result. Finally, and this will be the last ingredient for the proof of Theorem B, we next study the finite truncation of the series given in (25). We will show that it can be written in terms of polynomials in sp and sp ω. Lemma 4.3. Consider i, j ∈ Z and K ∈ N and define K−1 X TijK (s; α) := si+pk Tijk (s; α). k=0 Then, setting ω = ω(s; α1 ), the following holds: (a) If iq − jp 6= 0 then there exists τij K (z, w; α) ∈ Q(α1 )[z, w, α2 , . . . , αN ], with degz,w (τij K ) < K and not having poles along α1 = 0, such that Tij (s) = s τij (s , s ω; α) − s τij (s , 0; α). K λj K p p i K p (b) If (i, j) = ν(p, q) with ν ∈ N then there exists %K ij (z, w; α) ∈ Q[z, w, α], with degz,w (%ij ) < K + ν and K ij (z, 0; α) = 0, such that Tij (s) = %ij (s , s ω; α). %K K K p p 22 Proof. By applying (a) in Lemma 2.1, from the definition in (16) we get the existence of a polynomial Rkj (z; α) ∈ Q[z, α] with degz (Rkj ) 6 k such that eα1 t −1 ψkj (t; α) = Rkj (Ω(t; α1 ); α), where Ω(t; α) = α1 . Accordingly, from the definition in (23) and by performing the coordinate change w = Ω(t; α1 ), we get Z − ln s Z ω(s;α1 ) i−λj α1 −1 Tijk (s; α) = e(i−λj)t Rkj (Ω(t; α1 ); α) dt = (1 + α1 w) Rkj (w; α)dw, (30) 0 0 where we use that Ω(− ln s; α1 ) = ω(s; α1 ) by definition. If i − λj|α1 =0 6= 0, which is equivalent to pj−qi 6= 0, after integrating by parts k times we obtain i−λj (1 + α1 w) α1 ∂w Rkj (w; α)(1 + α1 w) Tijk (s; α) = Rkj (w; α) − i − λj i − λj + α1 ! ω(s;α1 ) 2 j k j ∂w Rk (w; α)(1 + α1 w)2 (−1)k ∂w Rk (w; α)(1 + α1 w)k + + ··· + . (i − λj + α1 )(i − λj + 2α1 ) (i − λj + α1 ) · · · (i − λj + kα1 ) 0 It is clear then that there exists a polynomial τijk (w; α) ∈ Q(α1 )[w, α2 , . . . , αN ], not having poles along α1 = 0 and with degw (τijk ) 6 k, such that we can write  i−λj  si+kp Tijk (s; α) = si+kp (1 + α1 ω) α1 τijk (ω; α) − τijk (0; α) = sλj+kp τijk (ω; α) − τijk (0; α)si+kp , where we set ω = ω(s; α1 ) for shortness and in the second equality we use that 1 + α1 ω = s−α1 . On account of this there exists τ̂ijk (z, w; α) ∈ Q(α1 )[z, w, α2 , . . . , αN ], which is homogenous of degree k in z and w, such that si+kp Tijk (s; α) = sλj τ̂ijk (sp , sp ω; α) − si τ̂ijk (sp , 0; α). PK−1 In view of this it is clear that the assertion in (a) follows taking τij K := k=0 τ̂ijk . With regard to the one in (b) we note that, since (p, q) = 1, the equality pj − qi = 0 holds if and only if there exists ν ∈ Z such that (i, j) = ν(p, q). In this case, from (30), we deduce that Z ω(s;α1 ) Tijk (s; α) = (1 + α1 w)ν−1 Rkj (w; α)dw. 0 If ν ∈ N then Tijk (s; α) = %ijk (ω; α) − %ijk (0; α), where %ijk (z; α) ∈ Q[z, α] with degz (%ijk ) 6 k + ν. Hence there exists %̂ijk (z, w; α) ∈ Q[z, w, α], homogeneous of degree k + ν in z and w, such that si+kp Tijk (s; α) = sp(ν+k) Tijk (s; α) = %̂ijk (sp , sp ω; α) − %̂ijk (sp , 0; α). PK−1 Since TijK (s; α) = k=0 si+pk Tijk (s; α), this shows that (b) follows taking K−1 X %K ij (z, w; α) := (%ijk (z, w; α) − %ijk (z, 0; α)) , k=0 which concludes the proof of the result. We are now in position to prove our second main result. 23 Proof of Theorem B. Recall that the family of vector fields under consideration is given by 1 Yα,β = PM +`−1 Xα , β0 xm y n + i=` βi+1−` ui where ` ∈ Z is defined in (1), u = xp y q , (p, q) = 1 and  PN −1  Xα = x∂x + 1q −p + i=0 αi+1 ui y∂y .  Let us denote the solution of Xα passing through p0 ∈ R2 at t = 0 by t 7→ x(t, p0 ; α), y(t, p0 ; α) . Then, if p0 = (s, 1) with s > 0, x(t, p0 ; α) = set intersects the transversal section {x = 1} at t = − ln s. Consequently the time T (s; α, β) that spends the solution of Yα,β starting at (s, 1) with s > 0 to reach the transversal section {x = 1} is given by Z MX +`−1 ! − ln s T (s; α, β) = β0 x m y n + βi+1−` (xp y q )i dt 0 i=` {x=x(t,s,1;α),y=y(t,s,1;α)} MX +`−1 = β0 Tmn (s; α) + βi+1−` Tip,iq (s; α), i=` where Tij ( · ; α) is the Dulac time of xi1yj Xα , which is precisely our concern in Proposition 4.2 and Lemma 4.3. It is clear then that, by applying Proposition 4.2, for each compact set C ⊂ Uδ with δ ∈ (0, 41 ] there exists s0 > 0 such that +∞ MX+`−1 ! X kp m ip T (s; α, β) = s β0 s Tmnk (s; α) + βi+1−` s Tip,iq,k (s; α) k=0 i=` for all s ∈ (0, s0 ) and α ∈ C and the series is absolutely convergent. Furthermore we can assert that, for the given L ∈ R, there exists KL ∈ Z≥0 such that +∞ MX +`−1 ! X TL (s; α, β) := s kp m β0 s Tmnk (s; α) + ip βi+1−` s Tip,iq,k (s; α) ∈ FL∞ (U0 × RM +1 ), k=KL i=` where U0 × RM +1 stands for the set {(α, β) ∈ RM +N +1 ; α1 = 0}. This assertion follows by taking (26) into account and applying, in this order, (c), (b) (g) and (e) in Lemma A.3. On the other hand, by Lemma 4.3, there exist τ0 (z, w; α) ∈ Q(α1 )[z, w, α2 , . . . , αN ] without poles along α1 = 0 and %i (z, w; α) ∈ Q[z, w, α] with %i (z, 0; α) = 0, i = 0, 1, . . . , M , such that such that setting  λn  s τ0 (s , s ω; α) − s τ0 (s , 0; α) if mq − np 6= 0, p p m p  L0 (s; α) := sκp %0 (sp , sp ω; α) if mq − np = 0 and (m, n) 6= (0, 0),   − ln s if (m, n) = (0, 0), ( s`p %1 (sp , sp ω; α) if ` > 0, L1 (s; α) := − ln s if ` = 0, and Li (s; α) :=s(`+i−1)p %i (sp , sp ω; α), for i = 2, 3, . . . , M, 24 then ! L −1 KX MX +`−1 T L (s; α, β) := skp β0 sm Tmnk (s; α) + βi+1−` sip Tip,iq,k (s; α) k=0 i=` M X =β0 L0 (s; α) + β1 L1 (s; α) + βi Li (s; α). (31) i=2 With regard to the cases considered in the definition of L0 , let us note that if mq − np = 0 then,  due to (p, q) = 1, there exists η ∈ Z such that (m, n) = η(p, q). Thus, by assumption, η = κ := max m , p q n >0 and hence, on account of Definition 1, ` = η + 1 > 0. (In particular, if mq − np = 0 and (m, n) 6= (0, 0) then η = κ ∈ N, and so the assertion with respect to L0 follows by (b) in Lemma 4.3.) If, on the contrary, mq − np 6= 0 then, by Definition 1 again, ` = κ > 0. Note also that if (i, j) = (0, 0) then Tij (s; α) = − ln s, which yields to the subcases (m, n) = (0, 0) and ` = 0 in L0 and L1 , respectively. In this respect, L0 (s; α) = − ln s in case that (m, n) = (0, 0), which implies ` > 1, and then, L1 (s; α) 6= − ln s. On the other hand, L1 (s; α) = − ln s in case that ` = 0, which implies mp − nq 6= 0 due to (1) and the assumption κ > 0. Accordingly, in this case, L0 (s; α) 6= − ln s. Taking the previous considerations into account, the assertions with respect to T L follow from (31). This concludes the proof of the result. A Results about the class FLK (W ) The present section is devoted to show a number of general properties about the class FLK (W ). We first prove that any g(s; µ) ∈ FLK (W ) extends to a finitely smooth function (on s and the parameter µ) along s = 0. (This applies in particular to the remainder DL in Theorem A, as well as to TL in Theorem B.) On the contrary, we will provide an example showing that a function g(s; µ) verifying the estimates in (4) but only with respect to the s derivative (i.e., with ν1 = . . . = νN = 0) may not have an extension along s = 0 which is C L on s and the parameter µ (see Example A.2). Lemma A.1. Let U be an open set of RN , K ∈ Z≥0 and g(s; µ) ∈ Cs>0 K (U ) such that, for some W ⊂ U and L ∈ R, g(s; µ) ∈ FLK (W ). If L > K then g extends to a C K -function ĝ, defined in some open neighbourhood of {0} × W in RN +1 , and satisfying ∂ ν ĝ(0; µ) = 0 for all µ ∈ W and ν ∈ ZN +1 ≥0 with |ν| 6 K. Proof. Due to g(s; µ) ∈ Cs>0 K (U ), by definition there exists an  open neighbourhood V of {0} × U in R N +1 such that (s, µ) 7→ g(s; µ) is C on V+ := V ∩ (0, +∞) × U . Then the function K  g(|s|; µ) if s 6= 0 and (|s|, µ) ∈ V+ , ĝ(s; µ) := 0 if s = 0 and µ ∈ U ,   is well defined on (s, µ) ∈ RN +1 ; (|s|, µ) ∈ V+ ∪ {0} × U , which is an open neighbourhood of {0} × U in RN +1 . Moreover, for ν = (ν0 , ν1 , . . . , νN ) ∈ ZN≥0 +1 with |ν| 6 K, it is easy to show (by induction on ν0 ) that ∂ ν ĝ(s; µ) = sgn(s)ν0 ∂ ν g(|s|; µ) for s 6= 0 with (|s|, µ) ∈ V+ . (32) Next we fix any µ̂ ∈ W . Then, due to g(s; µ) ∈ FLK (µ̂), by definition there exist s0 , ε, C > 0 such that, for each ν ∈ ZN ≥0 +1 with |ν| 6 K, |∂ ν g(s; µ)| 6 CsL−ν0 for all s ∈ (0, s0 ) and kµ − µ̂k < ε. (33) ≥0 with |ν| 6 K. Since µ̂ We claim that ĝ(s; µ) is of class C K at (0, µ̂) and that ∂ ν ĝ(0; µ̂) = 0 for all ν ∈ ZN +1  is arbitrary and, on account of (32), ĝ is C K on (s, µ) ∈ RN +1 ; (|s|, µ) ∈ V+ , the result will follow once 25 we prove the claim. To prove it we will show by induction on ν0 that if |ν| 6 K then |∂ ν ĝ(s; µ)| 6 C|s|L−ν0 for all (s, µ) with s ∈ (−s0 , s0 ) and kµ − µ̂k < ε. (This will imply that ∂ ν ĝ is continuous and vanishes at any (0, µ) with kµ − µ̂k < ε.) Denote ν̄ = (ν1 , ν2 , . . . , νN ) ∈ ZN ≥0 for shortness so that ν = (ν0 , ν̄). The base case ν0 = 0 is clear because, taking (32) and ĝ(0; µ) = 0 into account,  (0;ν̄) (0;ν̄) ∂ g(|s|; µ) if s 6= 0, ∂ ĝ(s; µ) = 0 if s = 0, that has absolute value smaller than C|s|L if kµ− µ̂k < ε and s ∈ (−s0 , s0 ) thanks to (33) and ∂ (0;ν̄) ĝ(0; µ) = 0. Let us take next any ν0 > 1 and show the inductive step. Then, by using (32) and that ∂ (ν0 −1,ν̄) ĝ(0; µ) = 0 due to the induction hypothesis, we get   sgn(s)ν0 ∂ ν g(|s|; µ) if s 6= 0, ν ∂ ĝ(s; µ) = (ν −1,ν̄)  lim ∂ 0 ĝ(z,µ) if s = 0. z→0z Therefore |∂ ν ĝ(s; µ)| = |sgn(s)ν0 ∂ ν g(|s|; µ)| 6 C|s|L−ν0 in case that 0 < |s| < s0 , thanks to (33), whereas ∂ (ν0 −1,ν̄) ĝ(z,µ) C|z|L−ν0 +1 ∂ ν ĝ(0; µ) = 0 because the induction hypothesis implies z 6 |z| = C|z|L−ν0 , which tends to zero as z → 0 due to L > K > |ν| > ν0 . Accordingly |∂ ĝ(s; µ)| 6 C|s| ν for all (s, µ) with L−ν0 s ∈ (−s0 , s0 ) and kµ − µ̂k < ε, and this proves the induction step. Consequently the claim is true and the result follows. Example A.2. With regard to the previous result we now exhibit a C ∞ function g(s; µ) on (0, +∞) × R verifying |∂si g(s; µ)| 6 CsL−i for all s > 0, µ ∈ R and i = 0, 1, . . . , L, but such that ∂µ g(s; µ) does not have a continuous extension along s = 0. Let us begin by taking a C ∞ bump function ϕ : R −→ [0, +∞) defined by ϕ(x) = exp(−x2 /(x2 − 1)2 ) if |x| 6 1 and zero otherwise. Let us fix besides any α ∈ (0, 1) and define β = 1+α 2 . Then, for each k ∈ Z≥0 , define Ek := {(s, µ) ∈ R2 ; pk (s, µ) 6 1} where  2  2 2(s − βαk ) µ pk (s, µ) := + . αk (1 − β) α(L+1)k The sets Ek , k ∈ Z≥0 , are pairwise disjoint and, furthermore, every (s, µ) 6= (0, 0) has an open neighbourhood that intersects at most one Ek . This shows that +∞ X g(s; µ) := αLk ϕ(pk (s, µ)) k=0 is a well defined C ∞ function on R2 \{(0, 0)}. For the same reason we can commute derivation and summation and then, by applying Theorem B.1, +∞ n n  ri X X  X Y ∂si pk (s, µ) n Lk (j) ∂s g(s; µ) = α ϕ pk (s; µ) n! , for all (s, µ) 6= (0, 0), j=1 r ,...,r i=1 ri !(i!)ri k=0 1 n Pn Pn where the third summation is subject to the coupling conditions i=1 ri = j and i=1 iri = n. Observe  k 2−i  i that ∂si pk (s, µ) = 2 2(s−βα αk (1−β) ) 2 αk (1−β) for i = 1, 2 and zero for i > 3. Thus, if (s, µ) ∈ Ek then  i |∂si pk (s, µ)| 6 2 αk (1−β) 2 for all i ∈ N. Consequently, if (s, µ) ∈ Ek0 and n ∈ N then we get n Y |∂sn g(s, µ)| 6 C 0 αLk0 (α−k0 )iri = C 0 αk0 (L−n) 6 Cα(k0 +1)(L−n) 6 CsL−n , i=1 26 where C 0 is a positive constant (depending on n, α and kϕ(j) k, j = 1, 2, . . . , n), C := C 0 αn−L and we use use that s ∈ [αk0 +1 , αk0 ]. The same inequality is valid for n = 0 since |g(s; µ)| 6 αLk0 = α−L αL(k0 +1) 6 α−L sL . Accordingly g verifies the desired bounds with respect to the s derivatives. The sequence of points (si , µi ) := (βαi , 2−1/2 αLi ) ∈ Ei tends to (0, 0) as i → ∞ and, on the other hand, an easy computations gives |∂µ g(si ; µi )| = αLi |∂µ ϕ(pi (si , µi ))| = α−i |ϕ0 (1/2)|, which tends to +∞ as i → ∞. This shows that ∂µ g(s; µ) does not have a continuous extension at (0, 0).  Next result gathers some general properties with regard to operations between functions in FLK (W ) with K ∈ Z≥0 ∪ {∞} and L ∈ R. 0 Lemma A.3. Let U and U 0 be open sets of RN and RN respectively and consider W ⊂ U and W 0 ⊂ U 0 . Then the following holds: T S (a) FLK (W ) ⊂ FLK (Ŵ ) for any Ŵ ⊂ W and n FLK (Wn ) = FLK ( n Wn ). (b) FLK (W ) ⊂ FLK (W × W 0 ). (c) C K (U ) ⊂ Cs=0 K (U ) ⊂ F0K (W ). 0 (d) If K > K 0 and L > L0 then FLK (W ) ⊂ FLK0 (W ). (e) FLK (W ) is closed under addition. K−|ν| (f ) If f ∈ FLK (W ) and ν ∈ ZN ≥0 +1 with |ν| 6 K then ∂ ν f ∈ FL−ν0 (W ). K (g) FLK (W ) · FLK0 (W ) ⊂ FL+L 0 (W ). (h) Assume that φ : U 0 −→ U is a C K function with φ(W 0 ) ⊂ W and let us take g ∈ FLK0 (W 0 ) with L0 > 0 and verifying g(s; η) > 0 for all η ∈ W 0 and s > 0 small enough. Consider also any f ∈ FLK (W ). Then h(s; η) := f (g(s; η); φ(η)) is a well-defined function that belongs to FLL K 0 0 (W ). Proof. Let us begin by showing (g) since the previous assertions are straightforward. Take f (s; µ) ∈ FLK (W ) and g(s; µ) ∈ FLK0 (W ) and fix µ̂ ∈ W and ν̂ ∈ ZN ≥0 +1 with |ν̂| 6 K. Then, by definition, it follows that there 0 exist a neighbourhood V of µ̂ and C, s0 > 0 such that |∂ ν f (s; µ)| 6 CsL−ν0 and |∂ ν g(s; µ)| 6 CsL −ν0 for all µ ∈ V , s ∈ (0, s0 ) and ν ∈ Z≥0 with |ν| 6 |ν̂|. Thus, by applying Leibniz’s rule (see Theorem B.2), if N +1 µ ∈ V and s ∈ (0, s0 ) then  X  ν̂  0 ∂ f (s; µ)g(s; µ) 6 ν̂ |∂ ν1 f (s; µ)| |∂ ν2 g(s; µ)| 6 ĈsL+L −ν̂0 , ν1 , ν2 ν1 +ν2 =ν̂ P ν̂  where we use that ν10 + ν20 = ν̂0 and set Ĉ := C 2 ν1 +ν2 =ν̂ ν1 ,ν2 = C 2 2|ν̂| . Thus f g ∈ FL+L K 0 (W ). 0 0 Let us turn next to show the assertion in (h). To this end fix any ν̂ ∈ ZN ≥0 +1 and η̂ ∈ U 0 ⊂ RN . Then, by 0 definition, it follows that there exist a neighbourhood V 0 of η̂ and C 0 , s1 > 0 such that |∂ ν g(s; η)| 6 C 0 sL −ν0 0 for all η ∈ V 0 , s ∈ (0, s1 ) and ν ∈ ZN ≥0 +1 with |ν| 6 |ν̂|. On the other hand, there exist a neighbourhood V of µ̂ := φ(η̂) ∈ U ⊂ R and C, s2 > 0 such that |∂ ν f (s; µ)| 6 CsL−ν0 for all µ ∈ V , s ∈ (0, s2 ) and N ν ∈ ZN +1 ≥0 with |ν| 6 |ν̂|. Consider now a relatively compact neighbourhood V 00 of η̂ with V 00 ⊂ V 0 and φ(V ) ⊂ V. Then, on account of L0 > 0, there exists s3 ∈ (0, s1 ) such that g(s; η) ∈ (0, s2 ) for all s ∈ (0, s3 ) 00 27 and η ∈ V 00 . The application of Faà di Bruno formula (see Theorem B.1) to compute the derivative of h(s; η) = f (g(s; η); φ(η)) yields X X q Y N Y ν̂ λ ∂ h(s; η) = ∂ f (u; µ) {u=g(s;η),µ=φ(η)} (ν̂!) Cki `i (∂ `i g(s, η))ki0 (∂ `i φi (η))kij . 16|λ|6|ν̂| p(ν̂,λ) i=1 j=1 QN 0 Here we set Cki `i := 1 ki !(`i !)|ki | + 1) for shortness. Note that the vectors λ, ki ∈ ZN and q := −1 + i=0 (ν̂i ≥0 +1 N +10 P q P q and `i ∈ Z≥0 are subject to the coupling conditions i=1 ki = λ and i=1 |ki |`i = ν̂. So if we define Q nQ Q o q q N 1 Ck` := i=1 Cki `i and Ck` 2 := sup i=1 `i j=1 |∂ φi (η)| kij ; η ∈ V 00 and we take any s ∈ (0, s3 ) and η ∈ V 00 , X X q Y 0 |∂ ν̂ h(s; η)| 6 Cg(s; η)L−λ0 1 (ν̂!)Ck` 2 Ck` (C 0 sL −`i0 )ki0 16|λ|6|ν̂| p(ν̂,λ) i=1 X X Pq 0 = Cg(s; η) L−λ0 Ck` s i=1 −`i0 )ki0 3 (L 16|λ|6|ν̂| p(ν̂,λ) X X 0 L0 L−λ0 0 6 C(C s ) 3 L λ0 −ν̂0 Ck` s 16|λ|6|ν̂| p(ν̂,λ) Qq Pq where we set Ck` 3 := (ν̂!)Ck`1 C2 (C 0 )ki0 = (ν̂!)Ck` 1 2 Ck` (C 0 )λ0 and we use that i=1 ki0 `i0 6 ν̂0 . Con- P k` i=1 0 L−λ0 P LL0 −ν̂0 sequently, setting Ĉ := 16|λ|6|ν̂| C(C ) p(ν̂,λ) Ck` , this shows that |∂ h(s; η)| 6 Ĉs for all 3 ν̂ s ∈ (0, s3 ) and η ∈ V 00 , which proves the validity of (h). This completes the proof of the result. Next result gathers some interesting properties of the Ecalle-Roussarie compensator that will be used in this (and a subsequent) paper. In the statement we use the notation x+ := max(x, 0) and x− := max(−x, 0) for, respectively, the positive and negative part of a given x ∈ R. Note in particular that then x = x+ − x− and |x| = x+ + x− . Lemma A.4. The following assertions hold: (a) For each compact set I ⊂ R and ν ∈ Z2≥0 there exists a constant C > 0 such that + |∂ ν ω(s; α)| 6 Cs−α −ν0 | ln s||ν|+1 for all α ∈ I and s ∈ (0, 1/e). Moreover lims→0+ 1 ω(s;α) = α− uniformly on α ∈ R so that, in particular, lim(s,α)→(0+ ,0) 1 ω(s;α) = 0. (b) For each ε > 0, (s, α) 7→ ω(s; α) belongs to F−ε ∞ ({α < ε}) and (s; α) 7→ 1 ω(s;α) belongs to F−ε ∞ (R). (c) For each L ∈ R and ` ∈ Z, (s, α, β) 7→ sβ ω ` (s; α) belongs to FL∞ ({(α, β) ∈ R2 ; β > L + `+ α+ }). (d) If p(z; µ) ∈ C K (U )[z, z −1 ], where U is some open set of RN , then the function (s, α, β, µ) 7→ sβ p(ω(s; α); µ) belongs to FLK ({(α, β, µ) ∈ R2 × U ; α = 0, β > L}). Proof. For the sake of convenience we prove first the assertion (c) for ` = 0. To this end we apply Lemma 2.6, which shows that for each i, j ∈ Z≥0 there exists M > 0 so that, for every s ∈ (0, 1/e), |∂si ∂βj sβ | 6 M sβ−i max(| ln s|, |β|)i+j = M sL−i sβ−L max(| ln s|, |β|)i+j . (34) Let us fix β̂  ∈ R with β̂ > L and take a compact neighborhood I of β̂ such that β − L > 0 for all β ∈ I. Thus C := M sup sβ−L max(| ln s|, |β|)i+j ; β ∈ I, s ∈ (0, 1/ε) is finite and so, from (34), |∂si ∂βj sβ | 6 CsL−i for all 28 s ∈ (0, 1/e) and β ∈ I. Hence sβ belongs to FL∞ ({β > L}), which is a subset of FL∞ ({(α, β) ∈ R2 ; β > L}) by (b) in Lemma A.3. We show next the validity of the inequality in (a). Take ν = (ν0 , ν1 ) ∈ Z2≥0 and a compact set I of R and let us consider first the case ν0 > 0. Then, if α ∈ I and s ∈ (0, 1/e), |∂ ν ω(s; α)| = |∂ (ν0 −1,ν1 ) s−α−1 | 6 M s−α−ν0 max(| ln s|, |α + 1|)|ν|−1 + 6 Cs−α−ν0 | ln s||ν|−1 6 Cs−α −ν0 | ln s||ν|+1 , where the first inequality follows from (34) taking i = ν0 − 1, j = ν1 and β = −α − 1, and the second one setting C := M max(1, sup{|α + 1|; α ∈ I})|ν|−1 and using previously that max(a, b) 6 a max(1, b) for any a > 1 and b > 0. In order to prove the same inequality for ν0 = 0 note that ω(s; α) = F (α ln s) ln s with −x F (x) := e x−1 and so, in this case, ∂ ν ω(s; α) = ∂αν1 (F (α ln s) ln s) = (ln s)ν1 +1 F (ν1 ) (α ln s). We claim that − |F (n) (x)| 6 ex for all x ∈ R and n ∈ Z≥0 . + In this respect observe that, due to x− |x=α ln s = max(−α ln s, 0) = − ln s max(α, 0) = ln(s−α ), the claim + + will imply |∂ ν ω(s; α)| 6 s−α | ln s|ν1 +1 = s−α −ν0 | ln s||ν|+1 for all s ∈ (0, 1/e) and α ∈ R and, consequently, the validity of the inequality in (a) for ν0 = 0 as well. To prove the claim we note that F is an entire function which, differentiating term by term its Taylor’s series at x = 0, verifies +∞ X +∞ X (−1)n (−x)r−n (−x)k F (n) (x) = − = (−1)n+1 for all x ∈ R. r=n (r − n)!(r + 1) k!(k + n + 1) k=0 Hence, on account of k+n+1 1 6 1, we get |F (n) (x)| 6 e|x| for all x ∈ R. In its turn this implies the claim for x 6 0 because,in thiscase, x− = |x|. The proof of the claim for x > 0 is a little more involved. We must e−x −1 show that ∂xn −x 6 1 for all x > 0, and it is clear that this will follow once we prove that   ex − 1 0 < ∂xn 6 1 for all x 6 0. (35) x To prove these two inequalities we first check by induction on n ∈ Z≥0 that   +∞ X ex − 1 (−x)k ∂xn = ex n! , x (k + n + 1)! k=0 x which is valid for all x ∈ R because x 7→ e x−1 is an entire function. Hence, for any n ∈ Z≥0 , we can assert x  x  x  that ∂xn e x−1 > 0 for all x 6 0. In particular this implies ∂xn−1 e x−1 6 ∂xn−1 e x−1 x=0 = n1 6 1 for all x 6 0 and n ∈ N. Thus both inequalities in (35) are true and so the claim follows for x > 0 as well. Let us prove now that lims→0+ ω(s;α) 1 = α− uniformly on α ∈ R. By distinguishing the cases α < 0, α = 0 and α > 0, one can check that ω(s;α)1 1 − α− = ω(s;|α|) , which is strictly positive in case that s ∈ (0, 1) R 1 −α−1 due to ω(s; α) = s x dx. Accordingly, for each given ε > 0 we must find s0 ∈ (0, 1) small enough such that if s ∈ (0, s0 ) then 1 1 − α− = < ε for all α ∈ R. (36) ω(s; α) ω(s; |α|) |α| −1/|α| If α 6= 0 then ω(s;|α|) 1 = s−|α| −1 . So, in this case, the above inequality holds if and only if s < (1 + |α|/ε) . In this regard note that, for every ε > 0 and α ∈ R,  − |α| 1  − |α| 1 − 1ε |α| |α| e = lim 1+ 6 1+ α→0 ε ε 29 −1/x because the function x 7→ 1 + xε is increasing on (0, +∞) for every ε > 0. Hence this shows that, for α 6= 0, the inequality in (36) follows taking s0 = e−1/ε . This is also true for α = 0 because in this case the inequality in (36) simply writes as − ln1s < ε. Thus lims→0+ ω(s;α) 1 = α− uniformly on α ∈ R, as desired. We turn next to the proof of the two assertions in (b). To show the first one we consider the given ε > 0 and any α̂ < ε, and we take a compact neighbourhood I of α̂ such that α < ε for all α ∈ I. Then, by applying (a), for each ν ∈ Z2≥0 there exists C > 0 such that + |∂ ν ω(s; α)| 6 Cs−ε−ν0 sε−α | ln s||ν|+1 for all α ∈ I and s ∈ (0, 1/e). + Thus, since α+ < ε if and only if α < ε, taking Ĉ := C sup{sε−α | ln s||ν|+1 ; s ∈ (0, 1/e), α ∈ I}, from the previous estimate we get |∂ ν ω(s; α)| 6 Ĉs−ε−ν0 for all s ∈ (0, 1/e) and α ∈ I. This proves that ∞ ω(s; α) ∈ F−ε ({α < ε}), as desired. Let us prove next that ω(s;α) 1 ∞ ∈ F−ε (R) for all ε > 0. So consider any α̂ ∈ R and take a compact neighbourhood I of α̂. Theorem B.1 shows that, for any ν ∈ Z2≥0 with |ν| > 1,   |ν| X X q Y 1 ∂ ν = (−1)n n!(ω(s; α))−1−n ν! Cki `i (∂ `i ω(s; α))ki , ω(s; α) n=1 i=1 p(ν,n) with Cki `i = 1 and q = (ν0 +1)(ν1 +1)−1, and where the second summation is multidimensional and ki !(`i )|ki | Pq Pq subject to the coupling conditions i=1 ki = n and i=1 `i ki = ν. On account of this and the inequality Q q + in (a) there exists C 0 > 0 such that i=1 |∂ `i ω(s; α)|ki 6 C 0 s−nα −ν0 | ln s|n+|ν| for all α ∈ I and s ∈ (0, 1/ε). + Consequently, taking s−α = max(1, s−α ) = max(1, 1 + αω(s; α)) also into account, we can assert that there exist suitable positive constants C 00 and C such that if s ∈ (0, 1/e) and α ∈ I then   X|ν| 1 max(1, 1 + αω(s; α))n ε ∂ ν 6 C 00 s−ε−ν0 n+1 s | ln s|n+|ν| 6 Cs−ε−ν0 , ω(s; α) n=1 ω(s; α) where in the second inequality we also use that, by applying (a), lims→0+ 1 ω(s;α) = α− uniformly on α ∈ R. ε Observe that, by the same reason, lims→0+ ω(s;α)s = 0 uniformly for α ∈ I, which implies that the above inequality holds for |ν| = 0 as well. This proves that the function ω(s;α) 1 belongs to F−ε ∞ (R) for any ε > 0. With regard to the assertion in (c) recall that the case ` = 0 is already proved. Here, for the sake of shortness in the exposition, we shall use the Heaviside step function H(`), which is defined by H(`) = 0 if ` < 0 and H(`) = 1 if ` > 0. By applying (b) together with Lemma A.3, and distinguishing the cases ` < 0 and ` > 0, it can be easily checked that ∞  ∞  ω ` (s; α) ∈ F−|`|ε {α ∈ R; H(`)α < ε} ⊂ F−|`|ε {(α, β) ∈ R2 ; H(`)α < ε, β > L} . Similarly, but applying (c) with ` = 0, we get   sβ ∈ FL∞ {β > L} ⊂ FL∞ {(α, β) ∈ R2 ; H(`)α < ε, β > L} . Consequently, by (g) in Lemma A.3, ∞  sβ ω ` (s; α) ∈ FL−|`|ε {(α, β) ∈ R2 ; H(`)α < ε, β > L} for all L ∈ R and ε > 0.  Hence sβ ω ` (s; α) ∈ FL∞ {(α, β) ∈ R2 ; H(`)α < ε, β > L + |`|ε} for all L ∈ R and ε > 0. Thus, by (a) in Lemma A.3, the function (s, α, β) 7→ sβ ω ` (s; α) belongs to ! \  [n o ∞ 2 ∞ 2 FL {(α, β) ∈ R ; H(`)α < ε, β > L + |`|ε} = FL (α, β) ∈ R ; H(`)α < ε, β > L + |`|ε ε>0 ε>0  = FL∞ {(α, β) ∈ R2 ; β > L + `+ α+ } , 30 where once again the second equality follows by distinguishing the cases ` > 0 and ` < 0. This proves assertion (c) for ` 6= 0. Finally assertion (d) follows by applying (c) in the present result and, in this order, (c), (b), (g) and (e) in Lemma A.3. This concludes the proof of the result. Next we introduce the set of functions IK (W ) that we previously used in [8, 9, 11, 12] to describe the properties of the remainder TL of the Dulac time. In this respect let us quote that Mourtada uses essentially the same definition in his study of the cyclicity of the hyperbolic polycycles (see for instance [13]). This set of functions is not used in the present paper and our aim is only to relate it with the set FK L (W ) for completeness and reader’s convenience. Definition A.5. Consider K ∈ Z≥0 ∪ {+∞} and an open subset U of RN . Let D := s∂s be the Euler operator and consider some µ̂ ∈ U . We say that ψ(s; µ) ∈ Cs>0 K (U ) belongs to the class IK (µ̂) if for each k = 0, 1, . . . , K there exists a neighbourhood V of µ̂ such that lim D k ψ(s; µ) = 0 uniformly on µ ∈ V . s→0+ T If W is a (not necesarily open) subset of U then we define IK (W ) = µ̂∈W IK (µ̂).  The following result shows that the remainder DL in Theorem A and TL in Theorem B can be written in terms of the class Ik (W ), which is more suitable in order to perform the derivation-division algorithm. Lemma A.6. Let U be an open set of RN , W ⊂ U, L ∈ R, K ∈ Z≥0 ∪ {+∞} and ε > 0. Then the inclusion K FL+ε (W ) ⊂ sL IK (W ) holds. Proof. Clearly it suffices to show that FL+ε K (µ̂) ⊂ sL IK (µ̂) for any µ̂ ∈ W because then, by definition, \ \ \ K K FL+ε (W ) = FL+ε (µ̂) ⊂ sL IK (µ̂) ⊂ sL IK (µ̂) = sL IK (W ). µ̂∈W µ̂∈W µ̂∈W So fix µ̂ ∈ W and let us show that FL+ε K (µ̂) ⊂ sL IK (µ̂). To this end we note that one can easily verify by induction that for all k ∈ Z≥0 there exist ηik ∈ Z≥0 , i = 0, 1, . . . , k, such that the identity k X D k g(s; µ) = ηik si ∂si g(s; µ) i=0 holds for any C k -function g. On the other hand, if ψ ∈ FL+ε K (µ̂) then for each i = 0, 1, . . . , K there exist a neighbourhood Vi of µ̂ and Ci , si > 0 such that |∂s ψ(s; µ)| 6 Ci sL+ε−i for all s ∈ (0, si ) and µ ∈ Vi . Thus, i Pk setting V̄k := ∩ki=0 Vi , ŝk := min(si ; i = 0, . . . , k) and Ĉk := i=0 ηik Ci , by applying the above identity we get that if k = 0, 1, . . . , K then k k ! X X |D k ψ(s; µ)| 6 ηik si |∂si ψ(s; µ)| 6 ηik Ci sL+ε = Ĉk sL+ε for all s ∈ (0, ŝk ) and µ ∈ V̄k . i=0 i=0 Taking this into account, since i   i    X i X i D i s−L ψ(s; µ) = D i−k (s−L )D k ψ(s; µ) = (−L)i−k s−L D k ψ(s; µ), k k k=0 k=0 we can assert that i X    i D i −L s ψ(s; µ) 6 |L|i−k Ĉk sε = C̃i sε for all s ∈ (0, s̃i ) and µ ∈ Ṽi , k k=0 31 Pi  where Ṽi := ∩ik=0 V̄k , s̃i := min(ŝk ; k = 0, . . . , i) and C̃i := k=0 ki |L|i−k Ĉk . It is clear that the above upper  bound implies that lims→0+ D i s−L ψ(s; µ) = 0 uniformly on µ ∈ Ṽi for i = 0, 1, . . . , K, which implies s−L ψ(s; µ) ∈ IK (µ̂), as desired. This proves the validity of the result.  Corollary A.7. For each ` ∈ Z, (s, α, β) 7→ sβ ω ` (s; α) belongs to sL I∞ {(α, β) ∈ R2 ; β > L + `+ α+ } for all L ∈ R. Proof. The result follows by noting that \  \ L  ∞ sβ ω ` (s; α) ∈ FL+ε {(α, β) ∈ R2 ; β > L + ε + `+ α+ } ⊂ s I∞ {(α, β) ∈ R2 ; β > L + ε + `+ α+ } ε>0 ε>0 ! [n o  = sL I∞ (α, β) ∈ R2 ; β > L + ε + `+ α+ = sL I∞ {(α, β) ∈ R2 ; β > L + `+ α+ } , ε>0 where we apply firstly Lemma A.4 and secondly Lemma A.6. B Differentiation formulas and integration of series In this section, for reader’s convenience, we state some specific results from analysis and calculus that we use all along. To begin with, since we use several times the multivariate Faa di Bruno formula to calculate the derivative of a composition of functions, we provide its explicit expression according to [3, Theorem 2.1]. To this end some notation is needed. If ν = (ν1 , . . . , νd ) ∈ Zd≥0 and x = (x1 , . . . , xd ) ∈ Rd then we define d X d Y Y d ∂ |ν| |ν| = νi , ν! = (νi !), ∂xν = ν1 ν νd and x = xiνi . i=1 i=1 ∂x1 · · · ∂xd i=1 Moreover, if ` = (`1 , . . . , `d ) ∈ Zd≥0 , we write ` 6 ν provided `i 6 νi for i = 1, . . . , d. Let f (y1 , . . . , ym ) and g (x1 , . . . , xd ), . . . , g (x1 , . . . , xd ) be real-valued functions and set (1) (m)   h(x1 , . . . , xd ) = f g (1) (x1 , . . . , xd ), . . . , g (m) (x1 , . . . , xd ) . Theorem B.1 (Multivariate Faa di Bruno formula). Let ν = (ν1 , . . . , νd ) ∈ Zd≥0 with |ν| > 0 and x0 ∈ Rd be given. Suppose that all the partial derivatives ∂x` with ` 6 ν of g 1 , . . . , g m exist and are continuous in a neighbourhood of x0 . Assume moreover that all the partial derivatives ∂yλ f (y), with λ ∈ Zm ≥0 and  |λ| 6 |ν|, exist and are continuous in a neighbourhood of g (x ), . . . , g (x ) ∈ R . Then ∂x h(x) exits in 1 0 m 0 m ν a neighbourhood of x0 and it is given by X X Yq (g`i (x))ki hν (x) = fλ (g(x)) (ν!) , i=1 (ki !)(`i !)|ki | 16|λ|6|ν| p(ν,λ) where ( ) q X q X p(ν, λ) = (k1 , . . . , kq ; `1 , . . . , `q ) : ki = λ and |ki |`i = ν . (37) i=1 i=1 In the statement `1 , . . . , `q ∈ Zd≥0 is a complete listing of all vectors ` 6 ν with |`| > 0, k1 , . . . , kq ∈ Zm ≥0 Qd   (1) (m) and q = −1+ i=1 (νi +1). We also set hν (x) = ∂xν h(x), fλ (y) = ∂yλ f (y) and g` (x) = g` (x), . . . , g` (x) (i) where g` (x) = ∂x` g (i) (x). We will also appeal to the following Leibniz formula for the partial derivatives of a product of functions (see for instance [2, Theorem C, p. 132]). 32 Theorem B.2. If f1 , . . . , fr ∈ C ∞ (U ) for some open subset U of Rd and ν ∈ Zd≥0 then r Y X  Yr ν ν ∂ fi = ∂ `i fi , i=1 `1 , . . . , `r i=1 `1 +...+`r =ν ν  Q d νi ! where `1 , . . . , `r ∈ Zd≥0 and `1 ,...,`r := ν! `1 !···`r ! = `1i !···`ri ! . i=1 ν  Remark B.3. The generalized multinomial coefficients `1 ,...,`r satisfy  νi ! d Y Xr d Y X X d Y X    νi ! νi ! ν r |ν| = 1 = = = i=1 j=1 i=1 `1i ! · · · `ri ! ` ! · · · `ri ! i=1 1i `1 , . . . , `r `1 +...+`r =ν `1 +...+`r =ν `1 +...+`r =ν thanks to the multinomial identity (see [2, Theorem B, p 28]) m !n X X n! xi = xa1 1 · · · xamm , i=1 a 1 ! · · · am ! where the summation takes place over all (a1 , . . . , am ) ∈ Zm ≥0 such that a1 + . . . + am = n.  The following result is also well-known (see [17, Theorem 7.17] for instance). Lemma B.4. Suppose that {fn } is a sequence of functions, differentiable on [a, b] and such that {fn (x0 )} converges for some point x0 ∈ [a, b]. If {fn0 } converges uniformly on [a, b], then {fn } converges uniformly on [a, b] to a function f such that f 0 (x) = lim fn0 (x) for all x ∈ [a, b]. n→∞ Lemma P R B.5. Let E be a measurable R P set of R and consider P R a sequence of measurable functions {fn }n∈N . If n>1 E |f n (x)|dx < +∞ then E n>1 f n (x)dx = n>1 E fn (x)dx. Proof. The problem is to show that Z Z k X lim ψk (x)dx = lim ψk (x)dx, where ψk (x) := fn (x) for each k ∈ N, k→+∞ E E k→+∞ n=1 and this follows by the Lebesgue’s dominated convergence theorem (see [17, Theorem 11.32]) because k X +∞ X |ψk (x)| 6 |fn (x)| 6 |fn (x)| =: Ψ(x) for all k ∈ N n=1 n=1 R and, on the other hand, EP R < +∞ by hypothesis. Ψ(x)dx R P In this regard let us remark that, due to |fn | > 0 for all n ∈ N, the equality n>1 E |fn (x)|dx = E n>1 |fn (x)|dx holds (see [17, Theorem 11.30]). References [1] C. Chicone, “Ordinary differential equations with applications”, Texts in Applied Mathematics, 34. Springer, New York, 2006. 33 [2] L. Comtet, “Advanced combinatorics. The art of finite and infinite expansions", D. Reidel Publishing Co., Dordrecht, 1974. [3] G. M. Constantine and T. H. Savits, A multivariate Faà di Bruno formula with applications, Trans. Amer. Math. Soc. 348 (1996) 503–520. [4] A. 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