This is a preprint of: “Bifurcations of zeros in translated families of functions and applications”, Pavao Mardesic, David Marı́n, Jordi Villadelprat, J. Dyn. Control Syst., 2020. DOI: [10.1007/s10883-020-09520-3] Bifurcations of zeros in translated families of functions and applications P. Mardešić, D. Marín, and J. Villadelprat Institut de Mathématiques de Bourgogne, UFR Sciences et Techniques, Université de Bourgogne Franche-Comté, UMR 5584, B.P. 47870, 21078 Dijon, France 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 the creation of zeros in a certain type of families of functions. The families studied are given by the difference of two basic functions with a translation made in the argument of one of these functions. The problem is motivated by applications in the 16-th Hilbert problem and its ramifications. Here, we apply the results obtained to the study of bifurcations of critical periods in the Loud family of quadratic centers. 1 Introduction and main results This paper is motivated by the study of bifurcations of critical points of the period function in a neighborhood of a polycycle. A key problem in theses studies is the breaking of separatrices of the polycycle. It appears also in the study of limit cycles corresponding to fixed points of the Poincaré return map of a family of planar vector fields. Contrary to the situation in the study of limit cycles, here by breaking a polycycle it is replaced by a polycycle with less vertices. The simplest situation is when a polycycle with two vertices is broken and a saddle loop polycycle is created. The cyclicity (i.e. number of limit cycles appearing by perturbation) of hyperbolic polycycles has been extensively studied in [1, 20, 24, 16, 17, 18, 15, 4] among others. On the other hand, in our study of critical points of the period function of Loud systems [9, 6, 7, 10, 13] we gave a conjectural bifurcation diagram. We could not prove it in full generality due in part to some phenomena of breaking of separatrices of the polycycle bounding the period annulus. Here we deal with this problem. The simplest setting is the breaking of one separatrix (or two separatrices in the presence of symmetry). This problem leads to the following type of equation ∆(s; ε, µ) := F1 (s; µ) − F2 (s + ε; µ) = 0, (1) where s = 0 corresponds to the polycycle, s ≥ 0 parametrizes the monodromic region, ε ≈ 0 is the parameter controlling the breaking of the separatrix and µ regroups all other parameters of the family, which we study in a neighborhood of a parameter value µ0 . We study the family of functions ∆ν (s) = ∆(s, ε; µ), for the parameter ν = (ε, µ) in a neighborhood of ν0 = (0, µ0 ) and s ≥ 0 close to 0. 1 1.1 General results Given a family of functions the notion of cyclicity has been defined by Roussarie [19]. It counts the maximal number of zeros, counted with multiplicity, born in the family from the origin. A value ν0 ∈ V is a bifurcation value if and only if there exists a sequence (νn , sn ) with sn > 0 and νn ∈ Int V converging to (ν0 , 0) such that ∆νn (sn ) = 0 for all n. Definition 1.1. [19] Let {∆ν }ν∈V be a continuous family of smooth functions on (0, s0 ) and fix ν0 ∈ V ⊂ Rk , i.e. let (s, ν) 7−→ ∆ν (s) be a continuous mapping on (0, s0 ) × V . For each δ, ρ > 0 let N (δ, ρ) be the supremum of the number of isolated zeros counted with multiplicity of ∆ν in (0, δ) for ν ∈ Bρ (ν0 ) ∩ Int V . We define the cyclicity Z(Fν , ν0 ) ∈ N ∪ {0, ∞} as the infimum of N (δ, ρ) varying δ, ρ > 0. We say that a ν0 ∈ V is a regular value of the parameter if Z(∆ν , ν0 ) = 0 and that it is a bifurcation value of the parameter otherwise. In Theorem A, we study the cyclicity of a family of functions of the form (1). The Theorem has two parts. First part is more general and relatively simple. It gives cyclicity Z = 0. We include it here for completenes and because we use it in applications. The second part gives a stronger conclusion (Z = 1), but under more specific hypothesis. It presents the main part of Theorem A. We say that a family of functions F (s; µ) tends to L(µ) as s → 0+ with a uniformly positive (resp. negative) sign at µ0 if for every  > 0, there exists δ > 0 and a neighborhood W of µ0 such that for all µ ∈ W and every 0 < s < δ, we have 0 < F (s; µ) − L(µ) <  (resp. − < F (s; µ) − L(µ) < 0). For example, the family F (s; µ) = s(s − µ) tends uniformly to L(µ) ≡ 0 but not with uniform sign at µ0 = 0. In order to obtain our principal results on higher cyclicity (Z > 0), in the second part of Theorem A we particularize the family of functions (1), by requiring that the functions Fi (s, µ), i = 1, 2, are of the form Fi (s, µ) = fi (s, µ)(ci (µ) + ψi (s, µ)), ci (µ0 ) 6= 0. (2) with (a) ci (µ), fi (s; µ) and ψi (s; µ) continuous functions, with fi and ψi defined on (0, s0 )×U , (b) for each µ ∈ U , ψi ( · ; µ) and fi ( · ; µ) smooth functions on (0, s0 ), (c) fi (s; µ), ψi (s; µ) and Dψi (s; µ) tend to zero as s → 0+ uniformly on compact sets in µ, where D = s∂s is the Euler differential operator. Definition 1.2. We write f1 ≺µ0 f2 (i.e. f1 precedes f2 at µ0 ) if f2 (s; µ) −→ 0 as s → 0+ uniformly on a neighborhood of µ0 . f1 (s; µ) We say that f1 and f2 are orderable at µ0 if f1 ≺µ0 f2 or f2 ≺µ0 f1 . Finally, we write f1 ∼µ0 f2 if f2 (s; µ) −→ 1 as s → 0+ uniformly on a neighborhood of µ0 . f1 (s; µ) For a given µ0 ∈ U , we define the set of admissible functions at µ0 as n o[n o Aµ0 := sλ(µ) : λ ∈ C 0 (U ) with λ(µ0 ) > 0 sω(s; α(µ)) : α ∈ C 0 (U ) with α(µ0 ) = 0 , (3) where recall that ( s−α −1 if α 6= 0, ω(s; α) := α (4) − log s if α = 0, is the Ecalle-Roussarie compensator [20]. Notice also that the admissible functions fi (s; µ) tend to zero as s → 0+ with a uniformly positive sign. 2 Theorem A. Consider a family of functions ∆ε,µ (s) = ∆(s, ε, µ) given by (1), for ν = (ε, µ) ∈ V , where V = [0, ε0 )×U , for some ε0 > 0 and U is an open neighborhood of µ0 ∈ Rk−1 . (i) We assume that the functions Fi (s; µ) tend uniformly on compact neighborhoods of µ0 to continuous real valued functions Li (µ), i = 1, 2, s → 0+ . (a) If L1 (µ0 ) 6= L2 (µ0 ) then ν0 = (0, µ0 ) is not a bifurcation value, i.e. Z(∆ν , ν0 ) = 0. (b) Assume that Fi (s, µ) tend uniformly to Li (µ) as s → 0+ on compact neighborhoods of µ0 with uniform sign. If L1 (µ0 ) = L2 (µ0 ) but Fi − Li is uniformly of the opposite sign, then ν0 = (ε, µ0 ) is not a bifurcation value of ∆ν (s), i.e. Z(∆ν , ν0 ) = 0. (ii) Assume that Fi are given in (2) where fi ∈ Aµ0 (see (3)) and ψi , ci , i = 1, 2, are as above. Suppose that f1 and f2 are orderable at µ0 . Then, the following assertions are equivalent: (a) ν0 = (0, µ0 ) is a bifurcation value for the family {∆ν }ν∈V , (b) f1 ≺µ0 f2 and c1 (µ0 )c2 (µ0 ) > 0, (c) Z(∆ν , ν0 ) = 1. The proof of Theorem A shows the following: Corollary 1.3. Under the assumptions of (ii) in Theorem A, if ν0 = (0, µ0 ) is a bifurcation value for the family {∆ν }ν∈V then there exist δ > 0, a neighbourhood B of ν0 and a continuous function v : B −→ R with v(0, µ0 ) = 0 such that ∆(s; ν) = 0 with s ∈ (0, δ) and ν ∈ B ∩ Int V if and only if s = σ(ν), with   −1 c2 (µ) σ(ε, µ) = σ0 (ε, µ)(1 + v(ε, µ)) and σ0 (ε; µ) := f1 f2 (ε; µ); µ . c1 (µ) The following example is two-fold. It illustrates the appearance of the translated families problem in the context of creation of limit cycles. It is the model example of the displacement function (controling limit cycles) of a generic polycycle with two hyperbolic vertices and two separatrices being broken [17]. It also gives an example where the uniform sign condition is not fulfilled. Example 1.4. Fix r2 > r1 > 0, µ0 = 0 and consider the equation (1) with F1 (s; µ) = sr1 + µ and F2 (s; µ) = sr2 which gives ∆(s; ε, µ) := sr1 + µ − (s + ε)r2 = 0. In that case we have L(µ) = µ which changes sign at µ = 0. Let us sketch the argument showing that Z = 2. Consider the equation given by the derivative ∆0 (s; ε, µ) = r1 sr1 −1 − r2 (s + ε)r2 −1 = 0. We apply Theorem A (ii) to ∆0 (s; ε, µ) = 0 and obtain that there exists a unique solution s = σ(ε) > 0 tending to 0 as ε → 0+ . By Rolle’s theorem, this shows that Z ≤ 2. On the other hand choosing conveniently µ = µ(ε) we obtain a tangency point (σ(ε), σ(ε)r1 + µ(ε)) between the graphs of the functions F1 (s; µ) and F2 (s + ε; µ), showing that Z = 2. 1.2 Critical Periods in Loud systems This work was initially motivated by the study of the bifurcation diagram of the period function of the dehomogenized Loud family of quadratic centers ẋ = −y + xy, (5) ẏ = x + Dx2 + F y 2 . 3 This study was started by Chicone and Jacobs [2]. They focused on the bifurcations of critical periods near the inner boundary (the center itself) of the period annulus by means of the period constants obtained by the Taylor expansion at the origin. In [7], we developed a technique to compute the first coefficient of a uniform asymptotic expansion of the period function at the outer boundary (a polycycle) of the period annulus under the hypothesis that the polycycle only contains linearizable saddles. In [9], we applied this technique to the Loud system obtaining a Jordan curve on which there is bifurcation of critical periods near the outer boundary and an open dense set of regular values. Unfortunately, a union ΓU of straight segments on which the character (bifurcation or regular) remained unknown for different reasons. In particular, along the line {F + D = 0, F 6∈ [0, 1]} a heteroclinic connection between hyperbolic saddles bounding the period annulus occurs. The connection is broken when leaving this line in the space of parameters. We denote by ε the breaking parameter, i.e. the signed distance between the two separatrices measured on a transverse section. The study of the bifurcation of critical periods in that case, leads to the type of bifurcations studied in Theorem A. We prove: Theorem B. Consider the period function of the period annulus of the Loud family (5) containing the origin. Take a parameter (D, F ), with F + D = 0 and F ∈ / {0, 1, 1/2}. (a) If F ∈ / [3/2, 2], then Z = 0. (b) If F ∈ [3/2, 2), then Z = 1. (c) If F = 2, then Z = 2. The following theorem gives a more precise description of the bifurcations occuring at (D, F ) = (−2, 2), for the critical period function T 0 in the Loud Family. Theorem C. (a) There is a curve Γ of double critical periods bifurcating from the polycycle which arrives to the point (D, F ) = (−2, 2). It is contained in the sector {F ≥ 2, F + D ≥ 0} and Γ is given by the graph of a continuous positive function ε = f (F ) with f (2) = 0, where ε = (D + F )U (D, F ) is the breaking parameter, with U (−2, 2) > 0. (b) The curve Γ has a flat tangency with the line F = 2, more precisely, there exist k2 > k1 > 0 such that k2 k1 e− f (F ) < F − 2 < e− f (F ) for F − 2 > 0 small enough. (c) Moreover, crossing the curve Γ from above, two simple critical periods bifurcate from the double critical period. The exponentially flat behavior of the double bifurcation curve Γ explains why it is hard to find numer- ically two critical periods near the point (D, F ) = (−2, 2), see [9, Figure 1]. 2 Proof of Theorem A The proof of (i) is easy. We include it here for completeness and as we use it in applications. The claim (ii) of Theorem A is proved using the implicit function theorem after a convenient blow-up. We use the classical version (Theorem A.1) of the implicit function theorem which requires differentiability only with respect to the variable that we want to isolate. Proposition 2.1. Fix µ0 ∈ U and consider the family {∆ν } in (1) taking f1 , f2 ∈ Aµ0 . Assume that f1 and f2 are orderable at µ0 . If there exists a sequence (νn , sn )n∈N = (εn , µn , sn )n∈N , with εn , sn > 0 and µn ∈ U , converging to (0, µ0 , 0) such that ∆νn (sn ) = 0, for all n, then f1 ≺µ0 f2 and limn→∞ ff21 (ε (sn ;µn ) n ;µn ) = c2 (µ0 ) c1 (µ0 ) . 4 The claim of the proposition is intuitively clear. In ∆ we have a competition between two functions essentially c1 f1 and c2 f2 both tending to zero. One operates a translation by ε in f2 . The only way that the two contributions can cancel is that f2 be smaller than f1 (i.e. f1 ≺ f2 ) and that we take ε > 0 (i.e. s + ε is further away from the origin), so that we increase the contribution of f2 . Proof. By contradiction, suppose that f2 ≺µ0 f1 . Then, since f2 ( · ; µ) is monotonous increasing (see Lemma A.6), we have that 0 6 f2f(s+ε;µ) 1 (s;µ) 6 ff21 (s,µ) (s;µ) , for any ε > 0, which, on account of f2 ≺µ0 f1 , implies that lims→0+ f2f(s+ε;µ) 1 (s;µ) = 0, uniformly on ν = (ε, µ) in a neighbourhood of ν0 = (0, µ0 ). On account of this, by applying Lemma A.2, we can assert that   ∆(sn ; νn ) f1 (sn ; µn ) 0= = (c1 (µn ) + ψ1 (sn ; µn )) − (c2 (µn ) + ψ2 (sn + εn ; µn )) f2 (sn + εn ; µn ) f2 (sn + εn ; µn ) tends to −c2 (µ0 ), as n −→ +∞. Thus c2 (µ0 ) = 0, which contradicts the assumption c1 (µ0 )c2 (µ0 ) 6= 0. Therefore f1 ≺µ0 f2 . At this point we claim that if f1 , f2 ∈ Aµ0 with f1 ≺µ0 f2 , then f2−1 ≺µ0 f1−1 . There are three different cases to consider, namely: (1) f1 (s; µ) = sλ1 (µ) and f2 (s; µ) = sλ2 (µ) , (2) f1 (s; µ) = sλ1 (µ) and f2 (s; µ) = sω(s; α(µ)), (3) f1 (s; µ) = sω(s; α(µ)) and f2 (s; µ) = sλ2 (µ) . The claim is obvious in the first case. In the second case, by (c) in Lemma A.4, the assumption f1 ≺µ0 f2 implies λ1 (µ0 ) < 1. On the other hand, by (c) in Lemma A.6, sκ(α(µ)) α(µ) f2−1 (s) ∼µ0 0 , where α0 (µ) := . ω(s; α (µ)) 1 − α(µ) On account of this, and by applying (c) in Lemma A.4 once again, f2−1 ≺µ0 f1−1 = s1/λ1 (µ) if and only if λ1 (µ0 ) < 1. So the claim follows in the second case. The third case follows exactly the same way. Let us write f1 (sn ; µn ) = rn sin θn and f2 (εn ; µn ) = rn cos θn with rn > 0 and θn ∈ [0, π/2]. Then, due to lims→0+ fi (s; µ) = 0 uniformly on µ ≈ µ0 , by Lemma A.2, we can assert that limn→∞ rn = 0. In addition, since f1 ≺µ0 f2 , on account of 0 6 f1−1 (rn sin θn ) 6 f1−1 (rn ) and the previous claim, we get f1−1 (rn sin θn ) lim = 0. (6) n→∞ f2−1 (rn ) Here and in what follows we omit the dependence of µ when there is no risk of confusion. We write  f2−1 (rn cos θn ) f1−1 (rn sin θn ) f2 (sn + εn ) = f2 f2−1 (rn )An with An := + . f2−1 (rn ) f2−1 (rn ) Suppose at this point that θ? is an accumulation point of the sequence (θn )n∈N . If f2 (s; µ) = sλ(µ) then, taking (6) into account, f2 (sn + εn )(c2 + ψ2 (sn + εn )) 1= f1 (sn )(c1 + ψ1 (sn ))  λ  f −1 (r sin θ ) cos1/λ θn + 1 f −1n(r ) n c2 + ψ2 (f1−1 (rn sin θn ) + f2−1 (rn cos θn )) c2 (µ0 ) cos θ? n = 2 −1  −→ sin θn c1 + ψ1 (f1 (rn sin θn )) c1 (µ0 ) sin θ? 5 as n tends to +∞. Therefore cc21(µ 0 ) cos θ? (µ0 ) sin θ? = 1. Consider now the case f2 (s; µ) = sω(s; α(µ)). Set αn = α(µn ) and αn0 = α0 (µn ) for shortness. Note then the sequences (αn ) and (αn0 ) tend to zero, as n → +∞. From (6) and applying Lemma A.6, rn cos θn ω(rn cos θn ;α0n ) ω(rn ; αn0 ) cos θn lim An = lim rn = lim −α 0 n→∞ n→∞ n→∞ (cos θn ) n ω(r ; α0 ) + ω(cos θ ; α0 ) ω(rn ;α0n ) n n n n 0 0 (cos θn )1+αn (cos θn )1+αn = lim 0 = lim ω(cos θn ;−α0n ) = cos θ? . n→∞ (cos θn )αn ω(cos θn ;α0n ) n→∞ 1+ 1+ ω(rn ;α0n ) ω(rn ;α0n ) If cos θ? 6= 0 then the last equality follows by (b) in Lemma A.4, whereas for cos θ? = 0, it follows easily due to ω(s; α) > 0, for α ≈ 0 and s > 0 small enough. Then, by using Lemma A.6 and limn→+∞ An = cos θ? , we get that f2 (sn + εn )(c2 + ψ2 (sn + εn )) 1= f1 (εn )(c1 + ψ1 (sn ))   1 f2−1 (rn ) c2 + ψ2 (f1−1 (rn sin θn ) + f2−1 (rn cos θn )) c2 (µ0 ) cos θ? = A1−α n n + f 2 (A n ) −→ , sin θn rn c1 + ψ1 (f1−1 (rn sin θn )) c1 (µ0 ) sin θ? as n tends to +∞. Consequently, also in this case, cc21(µ 0 ) cos θ? c1 (µ0 ) (µ0 ) sin θ? = 1. So far we have proved that sin θ? = c2 (µ0 ) , cos θ? which in particular shows that the accumulation point θ? ∈ [0, π/2] is unique. Furthermore, we can also assert that ff21 (ε (sn ;µn ) n ;µn ) sin θn = cos cos θ? c1 (µ0 ) θn −→ sin θ? = c2 (µ0 ) , as n → +∞. Hence, the result is proved. Lemma 2.2. Fix µ0 ∈ U and consider f1 , f2 ∈ Aµ0 and h ∈ C 0 (U ) verifying f1 ≺µ0 f2 and h(µ0 ) > 0. Then s ≺µ0 f1−1 h(µ)f2 (s; µ); µ . Proof. There are three different cases to consider, namely: (1) f1 (s; µ) = sλ1 (µ) and f2 (s; µ) = sλ2 (µ) , (2) f1 (s; µ) = sλ1 (µ) and f2 (s; µ) = sω(s; α(µ)), (3) f1 (s; µ) = sω(s; α(µ)) and f2 (s; µ) = sλ2 (µ) . In the first case, due to f1 ≺µ0 f2 , it is necessary that λ1 (µ0 ) < λ2 (µ0 ), and then f1−1 (h(µ)f2 (s; µ); µ) 1 λ2 (µ) −1 = h(µ) λ1 (µ) s λ1 (µ) −→ 0, as s → 0+ , uniformly on µ ≈ µ0 . s In the second case, by (c) in Lemma A.4, λ1 (µ0 ) < 1, and then f1−1 (h(µ)f2 (s; µ); µ)   λ 1(µ) = h(µ)s1−λ1 (µ) ω(s; α(µ)) −→ 0, as s → 0+ , uniformly on µ ≈ µ0 , 1 s thanks to (c) in Lemma A.4 again. Finally, in the third case, λ2 (µ0 ) > 1 and f1−1 (h(µ)f2 (s; µ); µ) κ(α(µ))h(µ)sλ2 (µ) lim+ = lim s→0 s s→0+ sω(h(µ)sλ2 (µ) ; α(µ)/(1 − α(µ))) 1 1 1− κ(α(µ))h(µ) λ2 (µ) x λ2 (µ) = lim = 0, uniformly on µ ≈ µ0 . x→0 + ω(x; α(µ)/(1 − α(µ))) In the first equality above we use (c) in Lemma A.6, in the second one, Remark A.5 and in the third one, (b) in Lemma A.4. This proves the result. 6 Proof of Theorem A. (i) If L1 (µ0 ) 6= L2 (µ0 ), since Fi (s; µ) tends to Li (µ) as s → 0 uniformly on compact neighborhoods of µ0 and Li (µ) is continuous at µ0 then we get that for any compact neighborhood K of µ0 , there exist E > 0 and δ > 0, such that F1 (s; µ) − F2 (s + ε; µ) is of the sign of L1 (µ0 ) − L2 (µ0 ), for 0 < s < δ, 0 < ε < E, µ ∈ K. Hence ∆ν has no zeros for the above values of (s, ε, µ) and Z(∆ν , (0, µ0 )) = 0. If L1 (µ0 ) = L2 (µ0 ), but Fi − Li are uniformly of the opposite sign, then in a neighborhood of the origin ∆ν corresponds to a sum of two terms of the same sign. Hence, in that neighborhood ∆ν is of the (uniform) sign of F1 (s, µ) and so Z(∆ν , (0, µ0 )) = 0. (ii) Let us prove the equivalence of the statements (a), (b) and (c): That (a) implies (b) follows by applying Proposition 2.1 and using that f1 and f2 are positive functions. The principal step of the proof is to show that (b) implies the statement of Corollary 1.3. To this end, as we explained in Remark A.7, we consider continuous extensions of the families {f2 ( · ; µ)}µ≈µ0 and {f1−1 ( · ; µ)}µ≈µ0 to (−s2 , s2 ) and (−s1 , s1 ), respectively, in order to be able to apply the implicit function theorem. Taking this into account, note that if µ varies in a compact neighbourhood of µ0 , then there exists ε0 > 0 such that   −1 c2 (µ) σ0 (ε; µ) := f1 f2 (ε; µ); µ c1 (µ) defines a continuous family of functions on (−ε0 , ε0 ) with σ0 (0; µ) = 0. In order to study the roots of ∆(s; ε, µ) = 0, see (1), we shall make the generalized blow-up s = σ0 (ε; µ)(1 + v). Our goal is to obtain an equivalent equation, G(ε, v; µ) = 0, with the function G verifying the hypothesis in Theorem A.1. With this end in view, some computations show that c2 (µ) f1 (s; µ)|s=σ0 (ε;µ)(1+v) = f2 (ε; µ)g1 (ε, v; µ) (7) c1 (µ) with   (1 + v)λ1 (µ) , if f1 (s; µ) = sλ1 (µ) , g1 (ε, v; µ) := σ0 (ε;µ)  (1 + v)1−α(µ) + c2 (µ) f1 (1 + v; µ), if f1 (s; µ) = sω(s; α(µ)), f (ε;µ) c1 (µ) 2 where in the second case we use (a) in Lemma A.6. It is to be noted moreover that in this second case the |α|+α function g1 (ε, v; µ) is a priori not defined at ε = 0. However, setting κ1 (α) := |α|−α 2 α 2(1−α) , notice that   σ0 (ε; µ) f1−1 cc12 (µ) (µ) f2 (ε; µ); µ f1−1 (x; µ) κ(α(µ)) lim c (µ) = lim = lim = lim = κ1 (α(µ)) ε→0 2 ε→0 c2 (µ) x→0 x x→0 ω(x; α0 (µ)) c1 (µ) f2 (ε; µ) c1 (µ) f2 (ε; µ) uniformly on µ ≈ µ0 . In the first equality above we use the definition of σ0 , in the second one we take Remark A.5 into account, whereas in the third and forth ones we use (c) in Lemma A.6 and (b) in Lemma A.4. This provides a continuous extension of the function g1 (ε, v; µ) to ε = 0 that, on account of κ1 (0) = 0, verifies g1 (0, 0; µ0 ) = 1. In what follows, by abuse of notation, we refer to this extension as g1 . On the other hand, by Lemma 2.2, σ0 (ε;µ) ε extends to a continuous function σ1 (ε; µ) on ε = 0 such that σ1 (0; µ) = 0. Then some computations show that f2 (s + ε; µ)|s=σ0 (ε;µ)(1+v) = f2 (ε; µ)g2 (ε, v; µ) (8) with ( λ2 (µ) (1 + σ1 (ε; µ)(1 + v)) , if f2 (s; µ) = sλ2 (µ) , g2 (ε, v; µ) := 1−α(µ) (1 + σ1 (ε; µ)(1 + v)) + ε f2 (ε;µ) f2 (1 + σ1 (ε; µ)(1 + v)) , if f2 (s) = sω(s; α(µ)), 7 where, again, in the second case we use (a) in Lemma A.6. Also in this case, thanks to (b) in Lemma A.4, ε 1 |α(µ)| − α(µ) lim = lim = uniformly on µ ≈ µ0 . ε→0 f2 (ε; µ) ε→0 ω(ε; α(µ)) 2 This provides a continuous extension of g2 (ε, v; µ) to ε = 0, verifying g2 (0, 0; µ0 ) = 1. Now, taking (7) and (8) into account, from (1), it follows that ∆(s; ε, µ)|s=σ0 (ε;µ)(1+v) = c2 (µ)f2 (ε; µ)G(ε, v; µ), where ! ! ψ1 σ0 (ε; µ)(1 + v); µ ψ2 ε(1 + σ1 (ε; µ)(1 + v)); µ G(ε, v; µ) := g1 (ε, v; µ) 1 + − g2 (ε, v; µ) 1 + . c1 (µ) c2 (µ) We claim, cf. Definition 1.1, that ∆(ŝ; ε̂, µ̂) = 0 with ŝ ∈ (0, δ), ε̂ ∈ (0, ρ) and kµ̂ − µ0 k < ρ, if and only if G(ε̂, v̂; µ̂) = 0, with ŝ = σ0 (ε̂; µ̂)(1 + v̂), for some v̂ ≈ 0. The sufficiency is obvious due to σ0 (0; µ) = 0. To show the necessity, note first that c2 (µ0 ) 6= 0, by assumption, whereas f2 (ε; µ) = 0, if and only if ε = 0. Accordingly, G(ε̂, v̂; µ̂) = 0, and so it only remains to show that if we take an arbitrary sequence (sn , εn , µn )n∈N , with ∆(sn ; εn , µn ) = 0, for all n, such that limn→+∞ (sn , εn , µn ) = (0, 0, µ0 ), then vn := c2 (µn ) σ0 (εn ;µn ) − 1 tends to zero, as n → +∞. To this end, let us set an := f1 (sn ; µn ) and bn := c1 (µn ) f2 (εn ; µn ), sn f1−1 (an ;µn ) so that limn→∞ an bn = 1, by Proposition 2.1. Then it follows that f1−1 (bn ;µn ) = sn σ0 (εn ;µn ) tends to 1, as n → +∞. This is clear in case that f1 (s; µ) = sλ(µ) , whereas for f1 (s; µ) = sω(s; α(µ)) we have that  −α0n ! f1−1 (an ; µn ) an ω(bn ; αn0 ) ω(an abnn ; αn0 ) bn ω( abnn ; αn0 ) lim = lim = lim = lim + = 1, n→∞ f −1 (bn ; µn ) n→∞ bn ω(an ; αn 0 ) n→∞ ω(an ; αn 0 ) n→∞ an ω(an ; αn0 ) 1 α(µn ) where αn0 := 1−α(µ n) tends to 0, as n → +∞. (In the first and third equalities above we use (c) and (a) in Lemma A.6, respectively, and in the last one (b) in Lemma A.4.) This shows that limn→+∞ vn = 0, as desired, and completes the proof of the claim. Note at this point that (ε, v, µ) 7−→ G(ε, v; µ) is a continuous function in a neighborhood of (ε, v, µ) = (0, 0, µ0 ) with G(0, 0; µ0 ) = 0. Moreover, ∂v gi (ε, v; µ), i = 1, 2, are continuous at ε = 0 as well. In addition, by Lemma A.3, ψi (s; µ) and φi (s; µ) := s∂s ψi (s; µ) form both continuous families of functions on (−s0 , s0 ), with ψi (0; µ) = φi (0; µ) = 0 for i = 1, 2. Therefore the functions  φ1 (σ0 (ε; µ)(1 + v)) ∂v ψ1 σ0 (ε; µ)(1 + v) = 1+v and   σ1 (ε; µ) ∂v ψ2 ε(1 + σ1 (ε; µ)(1 + v)) = φ2 ε(1 + σ1 (ε; µ)(1 + v)) 1 + σ1 (ε; µ)(1 + v) are continuous and vanish at ε = 0. Accordingly    ψ1 (σ0 (ε; µ)(1 + v); µ) g1 (ε, v)∂v ψ1 σ0 (ε; µ)(1 + v); µ ∂v G(ε, v; µ) = ∂v g1 (ε, v; µ) 1 + + c1 (µ) c1 (µ) ! ψ2 (ε(1 + σ1 ε; µ)(1 + v)); µ g2 (ε, v)∂v ψ2 (ε(1 + σ1 (ε; µ)(1 + v); µ) − ∂v g2 (ε, v; µ) 1 + − c2 (µ) c2 (µ) 8 is a continuous function in a neighborhood of (ε, v, µ) = (0, 0, µ0 ) with ∂v G(0, 0; µ0 ) ∈ {λ1 (µ0 ) − λ2 (µ0 ), λ1 (µ0 ) − 1, 1 − λ2 (µ0 )}, which is negative in the corresponding case. Here we use that, by assumption, f1 ≺µ0 f2 , with   sλ1 (µ) , sλ2 (µ) , f1 (s; µ) = and f2 (s; µ) = sω(s; α(µ)), sω(s; α(µ)), and we apply (c) in Lemma A.4. Therefore ∂v G(0, 0; µ0 ) < 0. We can now apply Theorem A.1 to the function (ε, v, µ) 7−→ G(ε, v; µ) at the point (0, 0, µ0 ) in order to conclude that there exists a continuous function v(ε, µ), with v(0, µ0 ) = 0 and such that G(ε1 , v1 ; µ1 ) = 0, if and only  if v1 = v(ε1 , µ1 ), provided (ε1 , v1 , µ1 ) is close enough to (0, 0, µ0 ). Taking σ(ε; µ) := σ0 (ε; µ) 1 + v(ε; µ) , the combination of this with the previous claim proves the statement of Corollary 1.3. In order to prove that (b) implies (c), we must check that the unique zero of ∆(s, ε; µ) which has the form s = σ(ε; µ) obtained before is of multiplicity 1. We have that ∂s ∆ = ∂s f1 (s)(c1 + ψ1 (s)) + f1 (s)∂s ψ1 (s) − ∂s f2 (s + ε)(c2 + ψ2 (s + ε)) − f2 (s + ε)∂s ψ2 (s + ε) We treat three cases: 1. fi (s) = sλi , i = 1, 2, 2. f1 (s) = sλ1 , f2 (s) = sωα (s), with λ1 (µ0 ) < 1, 3. f1 (s) = sωα (s) and f2 (s) = sα 2 (µ0 ), with λ2 (µ0 ) > 1. In the first case, using (7) and (8), we get the evaluation of ∂s ∆ (λ1 −1)sλ1 −1 at s = σ(ε):  ∂s ∆ σ(ε) c2 f2 (ε)g2 =c1 + ψ1 (σ(ε)) + c2 f2 (ε)g1 ∂s ψ1 (σ(ε)) − λ2 (c2 + ψ2 (σ(ε) + ε)) (λ1 − 1)sλ1 −1 s=σ(ε) λ1 c1 f2 (ε)g1 c1 σ(ε) + ε  f2 (σ(ε) + ε) − Dψ2 (σ(ε) + ε) . σ(ε) + ε It tends to c1 (µ) 6= 0, as ε → 0+ , uniformly on µ ≈ µ0 , because (i) gi are bounded, (ii) ψi (s) and s∂s ψi (s) tend to zero, as s → 0+ , uniformly on compact sets of µ, σ(ε) σ(ε)/ε (iii) σ(ε)+ε = σ(ε)/ε+1 → 0, as ε → 0, thanks to the fact σ1 (ε) = σ0 (ε)/ε → 0, as ε → 0,  λ 2 f2 (σ(ε)+ε) σ(ε) (iv) f2 (ε) = ε +1 → 1, as ε → 0. Consider now the second case: f1 (s) = sλ1 and f2 (s) = sωα (s) (with λ1 (µ0 ) < 1). Then  ∂s ∆ σ(ε) c2 λ −1 =c1 + ψ 1 (σ(ε)) + c2 f2 (ε)g1 ∂s ψ1 (σ(ε)) (λ1 − 1)s 1 s=σ(ε) λ1 c1 f2 (ε)g1 c1    f2 (ε)g2 f2 (σ(ε) + ε) − (1 − α) − 1 (c2 + ψ2 (σ(ε) + ε)) − Dψ2 (σ(ε) + ε) . σ(ε) + ε σ(ε) + ε It tends to c1 (µ), as ε → 0+ , uniformly on µ ≈ µ0 , because 9 (i)     σ(ε) f2 (ε)g2 σ(ε) (1 − α)f2 (ε)g2 (σ(ε) + ε) (1 − α) −1 = − → 0, f2 (ε) σ(ε) + ε σ(ε) + ε f2 (ε) f2 (ε) σ(ε) σ(ε)+ε ε(1+ ε ) σ(ε) 1+ ε |α|−α as ε → 0+ , thanks to the fact that f2 (ε) = εωα (ε) = ωα (ε) → 2 ≈ 0, as ε → 0+ , uniformly on µ ≈ µ0 . f2 (σ(ε)+ε) σ(ε)+ε u−α ωα (ε)+ωα (u) (ii) f2 (ε) = σ(ε)+ε ε ωα (ε(1+(1+v(ε))σ1 (ε))) ωα (ε) = ε ωα (ε) → 0, as ε → 0, where u = 1 + (1 + v(ε))σ1 (ε) → 1, as ε → 0. In the third case f1 (s) = sωα (s) and f2 (s) = sλ2 (with λ2 (µ0 ) > 1), we consider ∂s ∆ (1−α)ω−1 . We s=σ(ε) get  ∂s ∆ σ(ε) c2 =c1 + ψ1 (σ(ε)) + f2 (ε)g1 ∂s ψ1 (σ(ε)) (1 − α)ω − 1 s=σ(ε) (1 − α) cc12 f2 (ε)g1 − σ(ε) c1  f2 (ε)g2 f2 (σ(ε) + ε) −λ2 (c2 + ψ2 (σ(ε) + ε)) − Dψ2 (σ(ε) + ε) . σ(ε) + ε σ(ε) + ε It tends to c1 (µ), as ε → 0+ , uniformly on µ ≈ µ0 , because σ(ε) f1−1 (x) |α|−α (i) f2 (ε) ∼ x c ∼ 1 ωα (x) c tends to 2 ≈ 0, as ε → 0+ , uniformly on µ ≈ µ0 . x= c2 f2 (ε) x= c2 f2 (ε) 1 1 λ2 f2 (σ(ε)+ε) (σ(ε)+ε))λ2 (1+ σ(ε) ε ) (ii) (1−α)f2 (ε)−σ(ε) = (1−α)ελ2 −σ(ε) = σ(ε) is bounded, as ε → 0+ , uniformly on µ ≈ µ0 . 1−α− f (ε) 2 By definition (d) implies (a). This completes the proof of the result. 3 Proof of Theorem B Taking into account the symmetry (x, y) → (x, −y) of the Loud system, it suffices to consider half of the period, i.e. the time between local transverse sections Σ± at the outer boundary placed on y = 0 with ±x > 0. The singular point SD = (−1/D, 0) will be allowed to belong to one of the sections Σ± . For convenience we introduce an auxiliary transverse section Σ0 on x = 0 and y > 0. The outer boundary of the period annulus intersects the three transverse sections. We study the time function T and its derivative of orbits from Σ+ to Σ− parametrized by points on Σ0 . Let us explain it in more detail using some facts proved in [9]. The invariant algebraic curves of the Loud system are the line at infinity L∞ , the line L1 = {x = 1} and the conic C = {y 2 /2 = a(D, F )x2 + b(D, F )x + c(D, F )}, where D (D − F + 1) (F − D − 1) a(D, F ) = , b(D, F ) = , c(D, F ) = . 2(1 − F ) (1 − F )(1 − 2F ) 2F (1 − F )(1 − 2F ) When D + F = 0 the conic C becomes degenerated. More precisely, it is the union of the two straight lines y = √F x−1 passing through the point SD , which is a hyperbolic saddle, for D ∈/ [−1, 0]. In that case ± F (F −1) the outer boundary of the period annulus is contained in the triangle C ∪ L1 and C ∪ L∞ , for F < 0 and F > 1 respectively. The other two symmetric vertices Q± are also hyperbolic saddles at finite distance, for F < 0, or at infinity, for F > 1. For D + F 6= 0, the local separatrices of the saddle point SD disconnect from those of Q± , i.e. breaking of the saddle connections occurs, for D + F = 0. If F + D is small and F > 1 (resp. F < 0), then the period annulus is bounded by 10 1. a homoclinic loop through SD (resp. C ∪ L1 ), if F + D < 0, 2. a triangle C0 ∪ L1 3 SD if F + D = 0, 3. C ∪ L∞ (resp. a homoclinic loop through SD ), if F + D > 0. Blowing up the point SD in the Loud family we obtain an exceptional divisor E with two saddles points with hyperbolicity ratios 1/2. In this situation the bifurcation along F + D = 0 becomes the breaking of the heteroclinic connections between the saddle points on E and C ∩ L1 , see Figure 1. We introduce two classes of functions suitable for dealing with the remainder terms in our study. In order to formalize both notions we consider families of smooth functions on s > 0. More concretely, 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 ) if there exist an open  neighbourhood Ω of {s = 0} × U in R N +1 such that (s, µ) → 7 ψ(s; µ) is C K on Ω ∩ (0, +∞) × U . Definition 3.1. Let D := s∂s be the Euler operator and consider some µ̂ ∈ U . We say that ψ(s; µ) ∈ K Cs>0 (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). Definition 3.2. 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 ∈ FL (µ̂), if for each ν = (ν0 , . . . , νN ) ∈ Z≥0 with |ν| = ν0 + · · · + νN 6 K there K N +1 exist a neighbourhood V of µ̂ and C, s0 > 0 such that ∂ |ν| ψ(s; µ) 6 CsL−ν0 for all s ∈ (0, s0 ) and µ ∈ V . ∂sν0 ∂µν11 · · · ∂µνNN T If W is a (not necessarily open) subset of U , then define FLK (W ) := µ̂∈W FLK (µ̂). The first definition is suitable for performing the derivation-division algorithm and the second is well- adapted when derivation with respect to the parameters is needed. The second notion is more general and the precise relationship is given by the following result, see [12, Lemma A.6]: Lemma 3.3. The inclusion FL+ε K (W ) ⊂ sL IK (W ) holds for any ε > 0. Proof of Theorem B. Let us prove (a). By [9, Theorem A], the segment F + D = 0, F ∈ [ 23 , 2] is entirely composed of local bifurcation values at the outer boundary and the segment F + D = 0, F ∈ (0, 1) \ {1/2} is entirely composed of local regular values at the outer boundary. It only remains to show that the points of F + D = 0, with F ∈ (−∞, 0) ∪ (1, 32 ) ∪ (2, +∞) are local regular values at the outer boundary. Taking into account the symmetry (x, y) → (x, −y) of the Loud system, it suffices to consider half of the period, i.e. the time between local transverse sections Σ± at the outer boundary placed on y = 0. At the level D + F = 0 upper half of the boundary of the period annulus is then given by a saddle connection of two saddles P± and the other separatrix of one of them. By convention, we denote the saddle more to the left by P− (see Figure 1 for the case F > 1 and [9, Figure 4] for the case F < 0). Note that one of these saddle points P± is on one of the transverse sections Σ± . For D + F 6= 0, the saddle connection breaks. We denote by S− the local separatrix of P− and S+ the local separatrix of P+ . We also introduce an auxiliary transverse section Σ0 on x = 0 and y > 0. Let us denote by S± ∩ Σ0 = {(0, ζ± )}, σ± (s) = (0, ζ± (1 − s)), σ(s) = (0, ζ(1 − s)), ζ = min(ζ+ , ζ− ), ε := ζ+ − ζ− = (D + F )U (D, F ), with U (D, F ) > 0 for F > 1. 11 D+F >0 D+F =0 D+F >0 Figure 1: In the first row, phase portrait of (5), with D < −1 and F > 1 in the Poincaré disc, where we use thick lines to draw the conic. In the second row, the phase portrait near the period annulus after blowing-up the saddle point SD = (−1/D, 0). Indeed, the Loud systems (5) possesses the Darboux first integral  2  −2F y 2 H(x, y) = (1 − x) − (a(D, F )x + b(D, F )x + c(D, F )) . 2 If F > 1 then {(0, ±ζ+ )} = {H(x, y) = H(−1/D, 0)} ∩ {x = 0} and {(0, ±ζ− )} = {H(x, y) = 0} ∩ {x = 0}. A straightforward computation shows that  −2 F (F − 1) 1 F −1 ε = (D + F ) p + o(D + F ) (9) 2 F (2F − 1) (F − 1) F F is analytic in a neighborhood of any point (D, F ) with D + F = 0. Let T± (s) be the (signed) Dulac time of the trajectory starting at the point σ± (s) ∈ Σ0 ending in Σ± and let T (s) be the period of the orbit passing through σ(s). By definition T+ is negative because we follow the trajectory in opposite sense. Then  1 T− (s) − T+ (s + ε), if ε ≥ 0 T (s) = 2 T− (s − ε) − T+ (s), if ε ≤ 0  0 1 0 T− (s) − T+0 (s + ε), if ε ≥ 0 T (s) = (10) 2 T−0 (s − ε) − T+0 (s), if ε ≤ 0. We will apply (i) of Theorem A to the equation T 0 (s) = 0. We must study the behaviour of the derivative of the Dulac time of a hyperbolic saddle at finite distance or at infinity. By [13, Theorem A] and Lemma 3.3, it follows that the derivative of the Dulac time of a finite saddle of the family has the form 1 τ 0 (s; µ) = [−c0 (µ) + ψ(s; µ)] , with c0 (µ) > 0 (11) s and ψ ∈ I1 , i.e. ψ → 0, sψ 0 → 0 as s → 0+ uniformly on µ. Moreover, the derivative of the Dulac time of the saddle point at infinity of hyperbolicity ratio r(µ) > 0, r(µ) 6= 1, of the Loud family is given by τ 0 (s; µ) = sλ−1 (λc1 (µ) + ψ(s; µ)), (12) 12 with λ = min(1, r(µ)) and ψ ∈ I1 . In the case F < 0, we have P− = (−1/D, 0) ∈ Σ− is a finite saddle point invariant by the symmetry p −y) so that T− is half of the Dulac time of P− between the transverse sections Σ0 and ρ(Σ0 ). ρ(x, y) = (x, P+ = (1, −(D + 1)/F ) is finite saddle point. By (11), T±0 (s) = ±s−1 (c± + ψ± (s)), with c± > 0 and ψ± ∈ I1 . We cannot apply directly (i) of Theorem A. Instead of considering the equation 12 T 0 (s) = 0, we replace it by the equivalent equation 1/T−0 (s) − 1/T+0 (s + ε) = 0, if ε = ζ+ − ζ− ≥ 0 and similarly if ε ≤ 0. Now we can apply assertion (b) in case (i) of Theorem A and conclude that there is no bifurcation. If F > 1, then P− is a saddle at infinity with hyperbolicity ratio r(µ) = 2(F1−1) , whereas P+ = (−1/D, 0) ∈ Σ+ is a finite saddle. By analogous arguments as in the study of T− for F < 0, we have T+0 (s) = s−1 (c+ + ψ(s)), with c+ > 0. The asymptotic expansion for T− depends on r(µ) > 1, or r(µ) < 1, which corresponds to F ∈ (1, 3/2) or F > 3/2 respectively. In the first case, thanks to (12), we have T−0 (s) = c1 (µ) + ψ(s; µ). Hence, lim+ T+0 (s) = ∞ = 6 c1 = lim+ T−0 (s). Using the same trick as in the F < 0 s→0 s→0 case, we conclude by assertion (a) in case (i) of Theorem A that there is no bifurcation. It only remains to study the case F > 3/2 in which 0 < r(µ) < 1. In that case [13, Theorem A] implies that T+0 (s) = 1s [c0 (µ)+ψ+ (s; µ)] and T−0 (s) = sr(µ)−1 (r(µ)c1 (µ)+ψ− (s; µ)), with ci (µ) continuous functions, c0 (µ) > 0 and ψ± ∈ I1 . Thus, lim+ T±0 (s; µ) = ∞. We obtain a new translation family constructed from s→0 1 = s(c0 (µ) + ψ+ (s; µ))−1 = s(c−1 0 (µ) + ψ̂+ (s; µ)) T+0 (s; µ) and 1 = s1−r(µ) (r(µ)c1 (µ) + ψ− (s; µ))−1 = s1−r(µ) (r(µ)−1 c−1 1 (µ) + ψ̂− (s; µ)) T−0 (s; µ) with ψ̂± ∈ I1 . By Lemma A.8, the coefficient c1 (µ) is positive for F ∈ (3/2, 2) and negative for F > 2. Applying Theorem A, we obtain that Z = 1, for F ∈ (3/2, 2) and Z = 0, for F > 2. Let us prove now (b), with F = 3/2. Since the set of bifurcation values is closed, we have Z ≥ 1, for F = 3/2. It remains to show that Z ≤ 1. To prove (b) we take α = 1 − r(µ) ≈ 0. By [13, Theorem A] and Lemma 3.3, ¯ ⇒ T 0 (s) = c1 ω + c2 + sδ I(s), T− (s) = c̄0 + c̄1 sω + c̄2 s + s1+δ I(s) where I¯ ∈ I2 , I ∈ I1 , − T+ (s) = c0 log s + d + sδ I¯+ (s) ⇒ T+0 (s) = s−1 (c0 + sδ I+ (s)), where I¯+ ∈ I2 , I+ ∈ I1 and c0 > 0. Since f1 (s) = ωα (s) ≺ s−1 = f2 (s) it follows as in the beginning of the proof of Proposition 2.1, that a necessary condition for Z ≥ 1 is that ε ≥ 0 in equation (10). Then T 0 (s) = T−0 (s) − T+0 (s + ε) = c1 ωα (s) + c2 + sδ I(s) − (s + ε)−1 (c0 + (s + ε)δ I+ (s + ε)) and it suffices to control the zeros of   (s + ε) 0 c2 sδ I(s) c0 (s + ε)δ ∆(s; ε, α) = T (s) = (s + ε) c1 + + − − I+ (s + ε) ωα (s) ω ω ω ω 1 = εc1 + (εc2 − c0 ) + R, (13) ω where s sδ I(s) (s + ε)δ R = c1 s + c2 + (s + ε) − I+ (s + ε). ω ω ω Since   1 D∆ = (εc2 − c0 )D + DR, ω 13 to prove that Z ≤ 1 it suffices to see that DR  −→0 as (s, ε, α) → (0+ , 0+ , 0). D ω1 We have       s 1 (δ + 1)sδ+1 + δεsδ 1 DR =c1 s + c2 + sD + + (s + ε)sδ D I(s)+ ω ω ω ω    (s + ε)sδ δ(s + ε)δ−1 s δ 1 (s + ε)δ 0 + DI(s) − + (s + ε) D I+ (s + ε) − sI+ (s + ε). ω ω ω ω  Using that D ω1 = s−α ω −2 we obtain that DR 1+α 2   1 =c1 s ω + c2 (s1+α ω + s) + sδ+α ((δ + 1)s + δε)ω + (s + ε)sδ I(s)+ D ω  + sδ+α (s + ε)ωDI(s) − δ(s + ε)δ−1 s1+α ω + (s + ε)δ I+ (s + ε) − (s + ε)δ−1 s1+α ω[(DI+ )(s + ε)]. All the summands, except perhaps the last two, tend to zero, as s → 0+ . The last two summands tend to zero, as (s, ε) → (0+ , 0+ ) and α is small enough. Indeed, I+ (s + ε) and (DI+ )(s + ε) tends to 0 as (s, ε) → (0+ , 0+ ) due to I+ ∈ I1 and on the other hand δ h δ i (s + ε)δ−1 s1+α ωα (s) = s 2 ωα (s) (s + ε)δ−1 s1+α− 2 . The first factor tends to zero, as s → 0+ , and making the substitution s = r cos θ, ε = r sin θ in the second factor, we deduce that h δ i δ δ (s + ε)δ−1 s1+α− 2 = r 2 +α (cos θ + sin θ)δ−1 (cos θ)1+α− 2   tends to zero, as r → 0+ , uniformly in θ ∈ 0, π2 , if α is small enough. Here, we have taken δ ∈ (0, 1). Assertion (c) will be proved in the next section together with the proof of Theorem C. Remark 3.4. From (13), we deduce that, as s → 0+ , the curve   1 Cs = {(ε, α) : ∆(s; ε, α) = 0} = {(ε, α) : εc1 + cα (εc2 − c0 ) + (εc2 − c0 ) − cα + R(s; ε, α) = 0} ωα (s) tends to the bifurcation curve at the outer boundary of the period annulus ΓB = lim Cs = {(ε, α) : εc1 (ε, α) + cα (εc2 (ε, α) − c0 (ε, α)) = 0}. s→0+ Since cα = |α|−α 2 and c0 (ε, α), c1 (ε, α), c2 (ε, α) are C ∞ we obtain a corner in ΓB at the point (−3/2, 3/2). If α > 0, the linear part of the equation defining ΓB is εc1 (0, 0) = 0 and if α < 0 the linear part is εc1 (0, 0) + αc0 (0, 0) = 0. 4 Study of the point (D, F ) = (−2, 2) In this section we will prove assertion (c) in Theorem B and Theorem C. We will use the notations introduced in the proof of Theorem B in the previous section. 14 Proof of assertion (c) in Theorem B and Theorem C. If F ≈ 2 then the hyperbolicity ratio of P− F −2 is r = r(F ) = 2(F1−1) ≈ 12 . Defining ᾱ := 1 − 2r(F ) = F −1 and applying Theorem A of [13] we obtain ∞ T− (s) = T00 + T01 sr + T101 sωᾱ (s) + T100 s + F3/2−δ (s) for every δ > 0. On the other hand, half of the Dulac time of the finite singular point at (−1/D, 0) has the form + + + ∞ T+ (s) = c0 log s + T00 − T101 s log s + T100 s + F2−δ (s) with c0 > 0. Let ε = (D + F )U (D, F ) be the breaking parameter of the connexion, where U (D, F ) > 0 by (9). Since f1 (s) = sr−1 ≺ s−1 = f2 (s) it follows again as in the beginning of the proof of Proposition 2.1, that no bifurcation occurs for ε ≤ 0, see equation (10). Therefore we must study the zeros of the function T 0 (s) = T−0 (s) − T+0 (s + ε), with ε ≥ 0. To this end we define   ¯ D, F ) := s1−r (s + ε)T 0 (s) = (s + ε) rT01 + T101 (1 − ᾱ)s1−r ωᾱ (s) + (T100 − T101 )s1−r + F1−δ ∆(s, ∞ (s)  + + + ∞  + s1−r −c0 + T101 (s + ε) log(s + ε) − (T100 − T101 )(s + ε) + F2−δ (s + ε)  + +  = εrT01 + εT101 (1 − ᾱ)s1−r ωᾱ (s) + −c0 + ε(T100 − T101 − T100 + T101 ) s1−r + 1−r ∞ ∞ + T101 s (s + ε) log(s + ε) + F1−δ (s) + s1−r F2−δ (s + ε). −2 Let us work from now on with the variables z = s1−r , α = 1−r ᾱ = FF− 3 and ε = (D + F )U (D, F ). Note that 2 the map (D, F ) 7→ (ε, α) is a local diffeomorphism at the point (D, F ) = (−2, 2). By Lemma A.8, we have that T01 (D, F ) = −αU01 (ε, α) with U01 (0, 0) > 0 and T101 (−2, 2) > 0. Dividing ∆ ¯ by minus the coefficient 1 in s1−r and taking into account that ωᾱ (s) = ωᾱ (z 1−r ) = 1−r ωα (z) we obtain the function 1 ∆(z, ε, α) := −εαc1 (ε, α) + εc2 (ε, α)zωα (z) − z + c3 (ε, α)zh(z, ε, α) + g(z, ε, α) + zf (z 2−α + ε, ε, α) (14) | {z } R(z,ε,α) which controls the positive zeros of T 0 (s), where c1 (ε, α) = rU01 (ε, α) and c2 (ε, α) = (1−1−r ᾱ)T101 are C ∞ , with c̄1 = c1 (0, 0) and c̄2 = c2 (0, 0) positive, h(z, ε, α) = (z 2−α + ε) log(z 2−α + ε) and f, g ∈ F2−δ . In fact, ∞ c3 (ε, α)h(z, ε, α) + f (z 2−α + ε) = h0 (z 2−α + ε, ε, α) where ∞ h0 (z, ε, α) := c3 (ε, α)z log z + f (z, ε, α) ∈ F1−δ . (15) Let us prove (c) in Theorem B. We claim that the functions 1, zωα (z) and R(z; ε, α) = −z + R(z; ε, α) form an extended complete Chebyshev system, for z ∈ (0, ), see [8]. Since the Wronskian W (1, zωα (z)) > 0 for z > 0 small, it suffices to see that W (R(z; ε, α)) does not vanish on (0, ), where W (ρ) is defined as the Wronskian W (1, zωα (z), ρ) = ∂z (zωα (z))∂z2 ρ − ∂z2 (zωα (z))∂z ρ. Since the Wronskian is linear we compute the contribution of each summand of R separately. (a) Since ∂z (zωα (z)) = (1−α)ωα (z)−1 and ∂z2 (zωα (z)) = (1−α)z −1−α we have that W (−z) = (1−α)z −1−α . (b) Since g ∈ F2−δ ∞ and ωα (z) ∈ F−δ ∞ , we have that ∂zr g ∈ F2−r−δ ∞ , ∞ W (g) ∈ ((1 − α)ωα (z) − 1)F−δ − (1 − α)z −1−α F1−δ ∞ ∞ ⊂ F−2δ W (g) and W (−z) ∞ ∈ F1−3δ tends to zero, as z → 0, uniformly on (ε, α) ≈ (0, 0). 15 (c) Since ∂z (zh0 (z 2−α + ε)) = h0 (z 2−α + ε) + (2 − α)z 2−α (∂z h0 )(z 2−α + ε) and ∂z2 (zh0 (z 2−α + ε)) = (2 − α)(3 − α)z 1−α (∂z h0 )(z 2−α + ε) + (2 − α)2 z 3−2α (∂z2 h0 )(z 2−α + ε), with h0 ∈ F1−δ ∞ , one can check that W (zh0 (z 2−α + ε)) = −h0 (z 2−α + ε) + P1 (ωα (z))z 2 (∂z h0 )(z 2−α + ε) + P2 (ωα (z))z 4−α (∂z2 h0 )(z 2−α + ε), W (−z) Pi (ω) being degree 1 polynomials, and consequently W (zh0 (z 2−α + ε)) |z|2−δ |z|4−2δ + h0 (z 2−α + ε) ≤ C1 2−α + C2 . W (−z) |z + ε|δ |z 2−α + ε|1+δ The limit as (z, ε, α) → (0, ε0 , α0 ) of this upper bound is equal to zero. This is clear for ε0 > 0 and the the case ε0 = 0 follows by using the quasihomogeneous blow-up z = r sin θ, ε = r2−α cos θ, with a θ ∈ [0, π/2] and the fact that the functions (sinbsin θ θ+cos θ)c are bounded in θ ∈ [0, π/2], for a, b, c > 0. On W (zh0 (z 2−α +ε)) account of this and Lemma A.2 we get that W (−z) + h0 (z 2−α + ε) tends to 0 as z → 0 uniformly 2−α on (ε, α) ≈ (0, 0). Consequently, as moreover h0 ∈ F1−δ ∞ we deduce that lim W (zhW 0 (z (−z) +ε)) = −h0 (ε) z→0 uniformly on (ε, α) ≈ (0, 0). Hence W (R) → −1 + h0 (ε) < 0, as z → 0, uniformly on (ε, α) ≈ (0, 0), with h0 (0) = 0. (16) W (z) Consequently, 1, zωα (z) and R(z; ε, α) form a Chebyshev system in a suitable interval (0, ). Hence, F (z; ε, α) has at most 2 zeros in (0, ) and ZT 0 ≤ 2. On the other hand ZT 0 ≥ 2 will follow once we prove assertion (a) in Theorem C, i.e. the existence of the double bifurcation curve Γ arriving to the point (−2, 2). This will complete the proof of assertion (c) in Theorem B. In order to prove assertion (a) in Theorem C we consider the system  ∆(z, ε, α) = 0 (17) ∂z ∆(z, ε, α) = 0 where ∆(z, ε, α) is given in (14) and for convenience we define G(z, ε, α) = ∂z ∆(z, ε, α) = ε(1 − α)c2 (ε, α)ωα (z) − (1 + εc2 (ε, α)) + S(z, ε, α), where S(z, ε, α) = c3 (ε, α)(h(z, ε, α) + z∂z h(z, ε, α)) + zf1 (z 2−α + ε, ε, α) + g1 (z, ε, α). The system (17) will implicitly define a curve in the (ε, α)-plane, which gives the curve Γ, coming back to the (D, F )-plane by the local diffeomorphism (D, F ) 7→ (ε, α). Let us define H(z, ε, α) = (α − 1)∆(z, ε, α) + zG(z, ε, α) = αε(1 − α)c1 (ε, α) − (α + εc2 (ε, α))z + zV0 (z, ε, α), where V0 = (α − 1)R0 + S, R0 (z, ε, α) = c3 (ε, α)h(z, ε, α) + f (z 2−α + ε, ε, α) + g0 (z, ε, α), ∞ f ∈ F2−δ , g0 = g/z ∈ F1−δ ∞ , f1 = (2 − α)∂z f ∈ F1−δ ∞ and g1 = ∂z g ∈ F1−δ ∞ . Hence, V0 (z, ε, α) = c3 (ε, α)(αh(z, ε, α) + z∂z h(z, ε, α)) + f0 (z 2−α + ε, ε, α) + zf1 (z 2−α + ε, ε, α) + f2 (z, ε, α), (18) 16 with f0 = (α − 1)f ∈ F2−δ ∞ and f1 , f2 = (α − 1)g0 + g1 ∈ F1−δ ∞ . We are interested in the solutions of the system ∆(z, ε, α) = G(z, ε, α) = 0, for z > 0. For technical reasons we will rather study the system G(z, ε, |α|) = H(z, ε, |α|) = 0 which is equivalent to the preceding one on α ≥ 0. In order to avoid writing the absolute value of α we will make all the manipulations on α ≥ 0 and they must be extended by parity to α < 0. Note that the functions R0 , S, V0 extend continuously in a neighborhood of (z, ε, α) = (0, 0, 0) thanks to [12, Lemma A.1]. We would like to solve the system G(z, ε, α) = H(z, ε, α) = 0 by applying the implicit function theorem. Unfortunately, the hypothesis are not fulfilled. Therefore, we replace ci (ε, α) by c̄i and S, V by zero and we are able to find explicitly all the solutions of the system G = H = 0 thus obtained. There is a unique solution, which is given by 1 z = z0 (α) := (kα) 1−α ∼ kα, α 1 1/c̄2 k 1−α α 1−α −1/c̄2 ε = ε0 (α) := = α ∼ , (19) (1 − α)ωα (z0 (α)) − 1 c̄2 1 − α − (kα) 1−α 1 + log(kα) where k = c̄2 . c̄1 Here ∼ means that the quotient of the two functions tends to 1 as α → 0. Now, the idea is to find the solutions of the original system G = H = 0 as small perturbations of the above particular solution, i.e. in the following form (1 + v)/c̄2 z = z1 (α, u) := z0 (α)(1 + u), ε = ε1 (α, u, v) := , (20) (1 − α)ωα (z1 (α, u)) − 1 with (u, v) near (0, 0). Notice that z1 (α, u) and ε1 (α, u, v) tend to zero, as α → 0+ , uniformly on u, v ≈ 0, so that they define continuous functions in a neighborhood of (α, u, v) = (0, 0, 0) and z1 (α0 , u) and ε1 (α0 , u, v) are C 1 in u, v for every α0 ≥ 0 small enough. Indeed, z1 (0, u) = 0 and ωα (z1 (α, u)) = ωα (z0 (α)) + z0 (α)−α ωα (1 + u) so that ε1 (0, u, v) = 0. Moreover, z1 (α, u) > 0 and ε1 (α, u, v) > 0, for all u, v ≈ 0 and α > 0 small enough. Setting ci (ε, α) = c̄i (1 + Ci (ε, α)), Ci (0, 0) = 0, (21) we define the functions G1 (α, u, v) := G(z1 (α, u), ε1 (α, u, v), α) = v + (1 + v)C2 (ε1 (α, u, v), α) + S(z1 (α, u), ε1 (α, u, v), α) and 1 H1 (α, u, v) := H(z1 (α, u), ε1 (α, u, v), α) αε1 (α, u, v) 1 α = c̄1 (1 + C1 )(1 − α) − c̄2 (1 + C2 )(1 + u)k 1−α α 1−α (22) z1 (α, u) z1 (α, u)V0 (z1 (α, u), ε1 (α, u, v), α) − + ε1 (α, u, v) αε1 (α, u, v) Putting Φ(α, u, v) = (G1 (α, u, v), H1 (α, u, v)), (23) we will show that the implicit function problem Φ(α, u(α), v(α)) = (0, 0), with (u(0), v(0)) = (0, 0) has a unique solution (u, v) = (u(α), v(α)). By (20), this will define the curve ε(α) = ε1 (α, u(α), v(α)), and ultimately the curve Γ in the (D, F ) parameter space, along which z = z0 (α)(1 + u(α)) is a double critical period of the corresponding Loud system. It remains to verify the hypothesis of the implicit function Theorem A.1, for (23). By (21), it follows that G1 is continuous in a neighborhood of (α, u, v) = (0, 0, 0). In order to prove the continuity at (0, 0, 0) of the second component of Φ, given by H1 , it suffices to show that H1 (α, u, v) tends to 0, for α → 0+ , uniformly on u, v ≈ 0. This has to be verified only for the last two terms in (22). For the before last term, it 17 1−δ (z1 (α,u))1−δ (kα) 1−α (1+u)1−δ follows from Lemma 4.1.a. Indeed, we have in particular, ε1 (α,u,v) = ε1 (α,u,v) also tends to zero, z1 (α,u) 1 α as α → 0+ , uniformly on u, v ≈ 0, if δ is small enough. For the last term, since α = k 1−α α 1−α (1 + u) is bounded for α ≥ 0, the function z1 (α,u) α V0 (z1 (α,u),ε1 (α,u,v)) ε1 (α,u,v) also tends to zero, as α → 0 , uniformly on + u, v ≈ 0, thanks to Lemma 4.1.b. This shows that by putting Φ(0, 0, 0) = 0, and extending by parity in α, the function Φ extends to a continuous function in a neighborhood of (0, 0, 0). It follows by composition of differentiable functions and the fact that and z1 (α0 , u) ε1 (α0 , u, v) are C 1 in u, v, for every α0 ≥ 0 small enough. Finally, we have that G1 (0, u, v) = v, and H1 (0, u, v) = −c̄1 u, with c̄1 6= 0. Hence, ∂(u,v) ∂Φ (0, 0, 0) is surjective. Applying Goursat’s implicit function Theorem A.1, we obtain continuous functions u = u(α) and v = v(α), with u(0) = v(0) = 0, defining thus the curve Γ. Let us prove assertion (b). It can be checked, using (19) and assertion (a) in Lemma A.4, that 1+v lim (− log α)ε1 (α, u, v) = α→0+ c̄2 uniformly on (u, v) in a neighborhood of (0, 0). Consequently, ε1 (α, u(α), v(α)) lim+ −1/c̄2 = 1, (24) α→0 log α which gives us the asymptotic behavior of Γ at the point (−2, 2) with respect to the axis {α = 0} = {F = 2}. Indeed, it follows from (24) that, for any δ > 0, for α > 0 sufficiently close to 0, we have 1 − δ < −c̄2 ε1 log α < 1 + δ. Hence, −1−δ −1+δ e c̄2 ε1 < α < e c̄2 ε1 . In order to prove (c) of Theorem C, we consider the family of functions ∆ε,α (z) given by (14). For parameters (ε, α) on the curve Γ given as a graph of ε = ε(α) let z = z(α) be the corresponding double zero i.e. ∆(z(α), ε(α), α) = 0 and ∂ ∆|(z(α),ε(α),α) = 0. (25) ∂z The idea is simple. First we give the proof modulo two technical claims, whose proof will be given at the end. We consider the function ∆ε(α1 ),α1 , for α1 > 0 small enough in order to satisfy (26) and (28). We know that ∆ε(α1 ),α1 has a double zero at z(α1 ). ∂ 2 ∆ε(α1 ),α1 Claim 1: (z(α1 )) < 0. (26) ∂z 2 Hence, the function ∆ε(α1 ),α1 is strictly concave in a neighborhood of z(α1 ). It follows that it has the value 0 as a strict local maximum at z(α1 ). Moreover, by (26), applying the implicit function theorem to equation ∂ ∆(z, ε, α) = 0 (27) ∂z it follows that (27) has a unique solution z = z(ε, α) in a neighborhood of z(α1 ), for (ε, α) in a neighborhood 2 of (ε(α1 ), α1 ). Moreover, by continuity of ∂∂z∆ 2 , it follows from (26) that for parameters in a small neigh- borhood of (ε(α1 ), α1 ), the solution z = z(ε, α) of the implicit function problem (27) is a local maximum of ∆ε,α . 18 We consider the function ∆ε(α1 ),α , for α in an neighborhood of α1 . Note that here ε = ε(α1 ) is fixed and the point (ε(α1 ), α) leaves the curve Γ. ∂∆ε(α1 ),α (z(ε(α1 ), α)) Claim 2: |α=α1 < 0. (28) ∂α From Claim 2 it will follow that decreasing the value of α from α = α1 close to α1 , the maximal value of ∆ε(α1 ),α will increase and will hence become positive (it is zero for α = α1 ). From local concavity of ∆ε(α1 ),α1 , for δ > 0 arbitrarily small, we have that ∆ε(α1 ),α1 (z(α1 ) ± δ) < 0. The condition ∆ε(α1 ),α (z(α1 ) ± δ) < 0 will be preserved for α < α1 sufficiently close. By continuity of ∆ε(α1 ),α and sign change it follows that there exist two zeros zα± of ∆ε(α1 ),α verifying z(α1 ) − δ < zα− < z(ε(α1 ), α) < zα+ < z(α1 ) + δ. The proof of assertion (b) of Theorem C will be completed, once we prove the two claims (26) and (28). We first prove claim (26). We perform the standard division-derivation procedure. Due to (16) we know that   ∂∆/∂z ∂ ∂(zω α )/∂z < 0, ∂z for z, ε, α sufficiently small strictly positive. Using now the formula for the derivative of a quotient, (25) and the fact that ∂(zω ∂z α) > 0, then Claim 1 (26) follows. Let us now prove Claim 2. We consider the function η(α) = ∆(z(ε(α1 ), α), ε(α1 ), α), for fixed α1 (and dη hence fixed ε(α1 ). We have to prove that dα (α1 ) < 0. Recall that ∂z ∂ ∆(z(ε(α1 ), α1 ), ε(α1 ), α1 ) = 0. Hence, dη by the chain rule it follows that dα (α1 ) = ∂α ∆(z(ε(α1 ), α1 ), ε(α1 ), α)|α=α1 . ∂ Now, using (14) and putting ε1 = ε(α1 ) and z1 = z(α1 ), we have 1 ∂ ∆(z(ε1 , α1 ), ε1 , α)|α=α1 = −c1 (ε1 , α1 ) + c2 (ε1 , α1 )z1 ωα1 (z1 ) ε1 ∂α ∂ωα (z1 ) 1 ∂R(z1 , ε1 , α) + c2 (ε1 , α1 )z1 + |α=α1 . (29) ∂α α=α1 ε1 ∂α Knowing that ε1 > 0, it will be enough to prove that the expression (29) is negative, for α1 sufficienty small. Recall that the first term in the right hand side of (29) tends to −c̄1 < 0. We show that all other terms will tend to zero, for α1 → 0. Recall that z δ ω → 0, for δ > 0, see assertion (c) of Lemma A.4. This solves the second term. For the third term, recalling that | ∂ω ∂α /ω | is bounded (see assertion (d) of Lemma A.4), or that α 2 |∂α ω(z; α)| ≤ Cz −δ using [12, Lemma A.4(b)], it follows equally that the third term tends to zero. Fi- nally, in order to prove that ε11 ∂R(z∂α 1 ,ε1 ,α) |α=α1 tends to zero as α1 → 0, we use (20) for (u, v) = (0, 0) i.e. (19). By continuity, this will be enough in order to show that (29) is negative for α1 > 0 small enough. We have that the growth of 1/ε1 is bounded by C log α1 , see (19) and (20). On the other hand, all the terms of ∂R(z1 ,ε1 ,α) ∂α |α=α1 are bounded by C1 z1 , for some C1 > 0. Indeed, R(z, ε, α) = zh0 (z 2−α + ε, ε, α) + g(z, ε, α) with h0 ∈ F1−δ defined in (15) and g ∈ F2−δ . Then |∂α g| ≤ Cz 2−δ and ∂α (h0 (z 2−α + ε, ε, α)) = z 2−α (− log z)∂z h0 (z 2−α + ε, ε, α) + ∂α h0 (z 2−α + ε, ε, α) is bounded in absolute value by C|z 2−α + ε|−δ z 2−α (− log z) + C 0 |z 2−α + ε|1−δ ≤ C 00 z 1−α |z 2−α + ε|−δ + C 0 |z 2−α + ε|1−δ ≤ C 000 . In order to verify the last inequality it suffices to check that the first summand in the second term tends to 0 as (z, ε) → (0+ , 0+ ) uniformly on α ≈ 0 making the weighted blow-up z = r sin θ, ε = r2−α cos θ, θ ∈ [0, π/2]. Now from z1 ∼ α1 , it follows that ε11 ∂R(z∂α 1 ,ε1 ,α) |α=α1 is bounded by Cα1 log α1 which tends to zero as α1 → 0. 19 α1−δ Lemma 4.1. (a) If δ > 0 is small enough, then ε1 (α,u,v) tends to zero, as α → 0+ uniformly on u, v ≈ 0. (b) 1 ε1 (α,u,v) V0 (z1 (α, u), ε1 (α, u, v), α) tends to zero, as α → 0+ uniformly on u, v ≈ 0. Proof. In order to see assertion (a) we write α1−δ c̄2 α1−δ = ((1 − α)ωα (z0 (α)(1 + u)) − 1) ε1 (α, u, v) 1+v c̄2 1−δ   = α (1 − α)ωα (z0 (α)) + (1 − α)z0 (α)−α ωα (1 + u) − 1 1+v " −α # c̄2 (1 − α) (kα) 1−α − 1 α 1−δ− 1−α −α α1−δ = +α k 1−α ωα (1 + u) − . 1+v αδ 1−α −α α α1−δ The terms c̄2 1+v (1−α) and k 1−α ωα (1 + u) are uniformly bounded in α, u, v ≈ 0. The terms α1−δ− 1−α and 1−α tend to zero as α → 0. To deal with the remaining term we apply L’Hôpital’s rule obtaining that −α   −α 1+2α 1 (kα) 1−α −1 e α − 1−α log(kα) −1 (kα) 1−α (1−α) 2 log(kα) + 1−α lim = lim+ = − lim+ α→0+ αδ α→0 αδ α→0 δαδ−1   −1 α 1−δ− 1−α 1 + 2α 1 = lim α log(kα) + =0 δ α→0+ (1 − α)2 1−α provided δ < 1. Proof of claim (b): According to (18), it suffices to see that (i) α ε1 h(z1 , ε1 , α),   2−α      (z∂z h)(z1 ,ε1 ,α) z12−α z12−α (ii) ε1 = (2 − α) ε1 log(z12−α + ε1 ) + 1 = (2 − α) zε11 ε2−α 1 log ε1 1 + ε1 +1 ,  2−δ f0 (z12−α +ε1 ,ε1 ,α) (z12−α +ε1 )2−δ z12−α (iii) ε1 ≤C ε1 = Cε1−δ 1 1+ ε1 ,  1−δ  δ  1−δ z1 f1 (z12−α +ε1 ,ε1 ,α) z12−α z12−α (iv) ε1 ≤ C zε11 (z12−α + ε1 )1−δ = C zεδ1 1 + ε1 = Cz11−δ z1 ε1 1+ ε1 , 1 f2 (z1 ,ε1 ,α) z11−δ (v) ε1 ≤C ε1 tend to zero, as α → 0+ , uniformly on u, v ≈ 0, using the previous assertion (a) and the fact that f0 ∈ F2−δ ∞ and f1 , f2 ∈ F1−δ . ∞ 5 The state of art of the conjectural bifurcation diagram of the critical periods in Loud system We resume our study of the bifurcation diagram of the period function of the quadratic centers that we initiate in [7]. Let us explain succinctly the results we have obtained so far on this issue. The dehomogeneized Loud’s family (5) of quadratic reversible centers is {Xµ }µ∈R2 , where Xµ := −y(1 − x)∂x + (x + Dx2 + F y 2 )∂y with µ := (D, F ), 20 µ? Γ D = G(F ) Figure 2: Bifurcation diagram of the period function at the polycycle according to [9] and, in colour, the subsequent improvements due  to [5, 10, 11, 25, 26], where µ? = (−F? , F? ) with F? ≈ 2.34. The curve that joins − 23 , 32 and − 12 , 1 is the graphic of an analytic function D = G(F ). The double bifurcation curve Γ of Theorem C appears starting from the point (−2, 2). Following this notation, let ΓU be the union of dotted straight lines in Figure 2, whatever its colour is. Consider also the bold curve ΓB . (Here the subscripts B and U stand for bifurcation and unspecified respectively.) Then [9, Theorem A] shows that the open set R2 \ (ΓB ∪ ΓU ) corresponds to regular values and that the ones in ΓB are bifurcation values (cf.  Definition  1.1). We also conjectured that any parameter in ΓU is regular, except for the segment {0}× 0, 21 in the vertical axis, that should consist of bifurcation parameters. Since the formulation of this conjecture there have been some progress in the study of the parameters in ΓU : • From the results in [5, 26] it follows that the parameters in blue are indeed regular. In these papers the authors determine a region M in the parameter plane for which the corresponding center has a globally monotonic period function. The parameters that we draw in blue are inside the interior of M , which prevents the bifurcation of critical periodic orbits. • By [10, Theorem C] it follows that the parameters in dark green are regular as well. In that paper the authors give an asymptotic expansion of the Dulac time of an unfolding of a saddle-node. The techniques used in [9] enable only to study an unfolding of a hyperbolic saddle.   • Theorem B in [11] shows that the parameters in red, more precisely the segment {0} × 14 , 12 , are bifurcation values of the period function at the polycycle. To this end, after blowing up the polycycle, the authors show that any neighbourhood of a parameter µ̂ ∈ {0}× 14 , 12 contains two parameters, say µ+ and µ− , such that the derivative of the period function near the polycycle is positive for Xµ+ and negative for Xµ− . 21 • More recently, by [25, Corollary B] it follows that the parameters in light green are regular as well. Beyond the dichotomy regular vs bifurcation, a challenging problem is the study of the cyclicity of critical periods Z of the bifurcation parameters, i.e., to compute the exact number of critical periodic orbits that bifurcate from the polycycle. With respect to this problem see the results for the Loud family in [21, 22, 23]. What remains to study in the bifurcation diagram of the period function at the outer boundary is the following: • Along the segment {F = 0, D ∈ [−1, 0]}, which is conjectured to be regular (Z = 0), a saddle- node bifurcation occurs. Unfortunately, we can not apply directly the results in [10] because in the bifurcations studied there the outer boundary of the period annulus had a part of the line at infinity for all values of the parameters. This is not the case for F = 0 as a separatrix bounding the period annulus bifurcates from the line at infinity. • Along the segments {D = 0, F ∈ [0, 41 ]} and {D = −1, F ∈ [0, 1]} bifurcations of degenerate (nilpotent) singularites at the outer boundary of the period annulus occur. Along the first one we conjecture that Z = 2 and that a curve of double critical periods arrives to the point (0, 0). On the other hand, we also conjecture that the second one is regular (Z = 0). We think that, in order to show the regularity in this situation, it is necessary to make higher dimensional blow-ups. A Appendix We put in the appendix some classical results that we need, as well as some specific technicalities A.1 Classical results We start with Goursat’s version of the implicit function theorem which requires continuous differentiability only with respect to the variable that we isolate. Theorem A.1. [3] Let X be an open subset of Rn and let W be an open subset of Rk . Consider (x0 , w0 ) ∈ X × W and Φ : X × W → Rk be such that (a) Φ(x0 , w0 ) = 0; (b) Φ(x, w) is continuous on X × W ; (c) ∂w Φ(x, ·) is continuous on W , for all x ∈ X; (d) ∂w Φ(x0 , w0 ) is surjective. Then there exist a neighborhood X1 × W1 of (x0 , w0 ) and a function φ : X1 → W1 such that φ(x0 ) = w0 and for every (x1 , w1 ) ∈ X1 × W1 we have Φ(x1 , w1 ) = 0 if and only if w1 = φ(x1 ). Moreover, φ is continuous. Lemma A.2. Let {fµ }µ∈U be a continuous family of functions on (0, s0 ) and let K ⊂ U be a compact set. Then lims→0+ fµ (s) = `(µ), uniformly on µ ∈ K, if and only if lim(µ,s)→(µ̂,0+ ) fµ (s) = `(µ̂), for every µ̂ ∈ K. Proof. Assume that lims→0+ fµ (s) = `(µ), uniformly on µ ∈ K. Then µ 7−→ `(µ) is continuous on K. Let us show now that, under the uniformity assumption, fµ (s) tends to `(µ̂), as (s, µ) −→ (0+ , µ̂). Consider a given ε > 0. Then, thanks to the claim, there exists a neighbourhood U of µ̂ such that |`(µ) − `(µ̂)| < ε/2 22 for all µ ∈ U . Furthermore, on account of the uniformity, there exists δ > 0 such that |fµ (s) − `(µ)| < ε/2 for all s ∈ (0, δ) and µ ∈ U . Consequently, |fµ (s) − `(µ̂)| 6 |fµ (s) − `(µ)| + |`(µ) − `(µ̂)| < ε, for all s ∈ (0, δ) and µ ∈ U , and so lim(µ,s)→(µ̂,0+ ) fµ (s) = `(µ̂), as desired. Suppose now that lim(µ,s)→(µ̂,0+ ) fµ (s) = `(µ̂), for every µ̂ ∈ K. Then the map (s, µ) 7−→ fµ (s) extends continuously to [0, s0 /2] × K, which is compact. So the map is uniformly continuous, which clearly implies that lims→0+ fµ (s) = `(µ) is uniform on K. This proves the result. It will be convenient in order to apply the implicit function theorem, to work with functions defined in an open neighborhood of the origin. For that reason, we extend monotone function fˆµ defined on a one-sided neighborhood of the origin to an odd function fˆµ defined in a full neighborhood of the origin. Lemma A.3. Let {fµ }µ∈U be a continuous family of functions on (0, s0 ) with lims→0+ fµ (s) = 0 uniformly on U . For each µ ∈ U , we define    fµ (s), if s ∈ (0, s0 ), ˆ fµ (s) = 0, if s = 0,   −f (−s), if s ∈ (−s , 0). µ 0 Then {fˆµ }µ∈U is a continuous family of functions on (−s0 , s0 ). If in addition s 7−→ fµ (s) is monotonous on (0, s0 ), for all µ ∈ U , then {fˆµ−1 }µ∈U is a continuous family of functions on (−s1 , s1 ), for some s1 > 0. Proof. The continuity of (s, µ) 7−→ fˆµ (s) at some (ŝ, µ̂) ∈ (0, s0 )× U is obvious, for ŝ 6= 0, whereas, for ŝ = 0, it follows by applying Lemma A.2. Suppose additionally that fµ is monotonous on (0, s0 ) for all µ ∈ U. Then fˆµ is monotonous on (−s0 , s0 ) for all µ ∈ U. Accordingly (s, µ) 7−→ (fˆµ (s), µ) is an injective continuous map from the open set (−s0 , s0 )× U ⊂ Rk to Rk . Then, by the Domain Invariance Theorem, it follows that there exists s1 > 0 such that {fˆµ−1 }µ∈U is a continuous family of functions on (−s1 , s1 ). Hence, the result is proved. A.2 Technicalities Recall (4) that ω is a deformation of the logarithmic function. The first claim of the following lemma is the deformation of the fomula for the logarithm of a product for the function ω. Lemma A.4. The following hold: (a) ω(ab; α) = a−α ω(b; α) + ω(a; α), |α|−α 2 , as s → 0+ uniformly on α ≈ 0, 1 (b) ω(s;α) → (c) Let α 7−→ λ(α) be a continuous map at α = 0. Then 1 ≺0 sλ(α) ω(s; α), if and only if λ(0) > 0. ∂α /ω | is bounded. (d) | ∂ω 2 Proof. The equality in (a) is straightforward taking the definition of ω(s; α) into account. The assertion in (b) follows easily from the inequality ω(s; α) > inf(− log s, 1/|α|). Concerning (c), the sufficiency follows −x writing the compensator as ω(s; α) = F (α log s) log s, where F (x) := e x−1 , and using that |F (x)| 6 e|x| for all x ∈ R. To show the necessity we use Lemma A.2, which implies that lim(α,s)→(α̂,0+ ) sλ(α) ω(s; α) = 0 for any α̂ ∈ [−δ, δ] with δ > 0 sufficiently small. Clearly this is not possible if λ(0) 6 0 because then sλ(0) ω(s; 0) = sλ(0) log s tends to −∞ as s −→ 0+ . Thus λ(0) > 0, and so (c) follows. Claim (d) follows from Lemma 4.1.1 in [8]. 23 Remark A.5. If lims→0+ Ψ1 (s; µ) = L(µ) and lims→0+ Ψ2 (s; µ) = 0, with both limits being uniform on µ, then lims→0+ Ψ1 Ψ2 (s; µ); µ = L(µ), uniformly on µ. We shall deal with two types of families of admissible functions, {sλ }λ>0 and {sω(s; α)}α≈0 , both defined in principle, for s > 0. It is clear that each function fλ (s) = sλ in the first family is monotonously increasing and that, by applying Lemma A.3, {fλ }λ>0 and {fλ−1 }λ>0 can be continuously extended to (−s0 , s0 ) for some s0 > 0. It is obvious in addition that fλ−1 (s) = s1/λ . In the following result we show analogous properties for the second family. Lemma A.6. Set fα (s) = sω(s; α). Then the following hold: (a) fα (ab) = a1−α fα (b) + bfα (a), (b) There exists s0 > 0 and ε > 0 such that {fα }α∈(−ε,ε) is a continuous family of monotonous increasing functions on (0, s0 ) with lims→0+ fα (s) = 0, uniformly on α ∈ (−ε, ε). In addition {fα−1 }α∈(−ε,ε) is a continuous family of functions on (0, s0 ), with lims→0+ fα−1 (s) = 0, uniformly on α ∈ (−ε, ε). α+|α| sκ(α) (c) fα−1 (s) ∼0 ω(s;α/(1−α)) , where κ(α) := (1 − α) α 2(1−α) . Proof. The equality in (a) is straightforward taking the definition of ω(s; α) into account. The monotonicity in (b) follows using that, by (b) in Lemma A.4, fα0 (s) = −1 + (1 − α)ω(s; α) tends to +∞ as (α, s) → (0, 0+ ). The fact that fα (s) tends to zero as s → 0+ uniformly on α is a consequence of (c) in Lemma A.4. Taking this into account, the assertion concerning fα−1 follows by applying Lemma A.3. In order to show (c), setting α α0 := 1−α , we first claim that  fα−1 (u) ω sω(s; α); α0 Ψ1 (s; α) := u = ω(u;α0 ) u=f (s) ω(s; α) α tends to κ(α) as s −→ 0+ uniformly on α ≈ 0. Note that (c) will follow once we prove this claim. Indeed, since lims→0+ fα−1 (s) = 0 uniformly on α ≈ 0 by (b), we get the desired conclusion, by applying Remark A.5, with Ψ2 (s; α) = fα−1 (s). α0 In order to prove the claim, we apply Lemma A.2. To this end note that κ(α0 ) = (1 − α0 )α01−α0 , if α0 > 0, κ(0) = 1 and κ(α0 ) = 1 − α0 , if α0 < 0. If α0 6= 0, then, by definition,  −α0  0 s−α −1 ω sω(s; α); α 0 s−α α −1 α = , ω(s; α) s−α − 1 α0 which clearly tends to 1 − α0 , as (α, s) −→ (α0 , 0+ ), in case that α0 < 0. If α0 > 0, then for convenience we write the above equality as  0 0 0 0 0 0 0 ω sω(s; α); α0 s−α +αα (1 − sα )−α − α−α α1+α s−α (1 − sα )−α − α−α 0 = = (1 − α)αα ω(s; α) s−α (1 − sα ) α0 s−α (1 − sα ) 0 0 (1 − sα )−α − α−α sα 0 = α (1 − α)αα , 1−s α0 ω(sω(s;α);α0 ) which tends to (1 − α0 )α01−α0 , as (α, s) −→ (α0 , 0+ ). It only remains to see that ω(s;α) −→ 1 as 24 (α, s) −→ (0, 0+ ). With this aim in view, some manipulations show that   −α0   −α0 −α0 s−α −1 ω sω(s; α); α 0 sω(s; α) −1 α  s α − 1  −1+α= − 1 + α = (1 − α)  − 1 ω(s; α) α0 s−α − 1 s−α − 1   0   0  1−sα −α 1−sα −α − sα −1 ω ω(s; −α); α0 =(1 − α)  α − 1 = (1 − α) α = , (30) 1 − sα 1 − sα ω(s; −α) where we take α0 = α 1−α into account several times. Note that   ω ω(s; −α); α0 ω x; α0 = , with x = ω(s; −α) −→ +∞, as (α, s) −→ (0, 0+ ), (31) ω(s; −α) x e−x −1 due to ω(s; α) > inf(− log s, 1/|α|). Moreover ω(s; α) = F (α log s) log s, where recall that F (x) = x verifies that |F (x)| 6 e|x| , for all x ∈ R. Accordingly, for x > 1 and α0 ∈ [− 12 , 12 ], we can assert that ω(x; α0 ) log x log x 6 1−|α0 | 6 1/2 . x x x ω(x;α0 ) Hence x −→ 0, as x −→ +∞, uniformly on α0 ∈ [− 21 , 12 ]. This, together with (31), implies that  ω ω(s; −α); α0 −→ 0 as (α, s) −→ (0, 0+ ) ω(s; −α) because α0 = α 1−α −→ 0 as α −→ 0. Therefore, on account of (30), we finally obtain   ω sω(s; α); α0 ω ω(s; −α); α0 −1= − α −→ 0 as (α, s) −→ (0, 0+ ), ω(s; α) ω(s; −α) as desired. This proves the claim and so the result follows. Remark A.7. By Lemma A.3, each family {sα }α>0 and {sω(s; α)}α≈0 extends to a continuous family of homeomorphisms {fˆα } on (−s0 , s0 ) with fˆα (0) = 0. Their respective inverses form also a continuous family of functions {fˆα−1 }α on (−s1 , s1 ) for some s1 > 0. In the sequel, by an abuse of notation and when there is no risk of ambiguity, we will denote fˆ and fˆ−1 by f and f −1 , respectively. Lemma A.8. The Dulac time T− (s; F ) of the Loud family  (5) restricted  to the line D + F = 0 between the transverse sections {x = 0} parametrized by σ(s) = 0, √ 1−s and {y = 0, x < 0} satisfies the F (F −1) following: (a) For every F0 > 3/2, there is δ > 0, such that T− (s; F ) = T00 (F ) + T01 (F )sr(F ) + Fr(F ∞ 0 )+δ (F = F0 ) with r(F ) = 2(F −1) and 1 F 1 p √ Γ(1 − r(F )) T01 (F ) = −F − F −1 2 2(F −1) F (F − 1) π 1 . Γ( 2 − r(F )) In particular, T01 (F ) = −(F − 2)U01 (F ) with U01 (2) > 0.   F −2 (b) T− (s; F ) = T00 (F ) + T01 (F )sr(F ) + T101 sω s; F −1 + T100 (F )s + F3/2−δ (F = 2) with T101 (2) > 0. ∞ 25 p y Proof. We perform the projective change of coordinates (u, v) = φ(x, y) = (1 − F (F − 1) 1−F 1 x , 1−F x ) which brings the Loud vector field (x − 1)y∂x + (x + F (y − x ))∂y into XF = v [PF (u, v)u∂u + QF (u, v)v∂v ] 2 2 1 q where PF (u, v) = FF−1 (u − 2)(v − 1) and QF (u, v) = √ 1 (u − 1)((F − 1)v + 1) and the transverse F (F −1) sections x = 0 and y = 0 into v = 1 and u = 1 respectively. The hyperbolicity ratio of the saddle of XF at (u, v) = (0, 0) is r(F ) = 2(F1−1) < 1 for F > 3/2. This gives us the announced expansion in (a). Let us now compute the coefficient T01 (F ). In the coordinate chart (u, v) the parametrization σ(s) translates into σ(s) = (s, 1) and we can take the parametrization τ (s) = (1, s) in the target transverse section. By applying [7, Theorem A], after some tedious but straightforward computations, we obtain that r Z 1 ! F p 1 F u  du T01 (F ) = F − F −1 − F (F − 1) + (1 − )−1−r(F ) − 1 1+r(µ) 2 F −1 0 2 u F 1 p √ Γ(1 − r(F )) = −F − F −1 2 2(F −1) F (F − 1) π 1 Γ( 2 − r(F )) R1  √ 2a−1 πΓ(2−a) thanks to the formula 0 (1 − u2 )−a − 1 udua = (1−a)Γ( 32 −a) 1 − 1−a in which appears the Gamma function Γ. √ Γ(1−r(F )) In particular, Γ( 21 −r(F )) = 2π (F − 2) + O((F − 2) ). 2 To prove (b), it suffices to take F ≈ 2 where the announced asymptotic expansion holds by [13, Theo- rem A]. In fact, we have T− (s) = T00 +T01 sr +T10 s+T02 s2r +F2r0 +δ (r = r0 ) if r0 > 21 and T101 = (1−2r)T02 , T100 = T10 + T02 so that T101 |F =2 = lim1 (1 − 2r)T02 = − lim1 (1 − 2r)T10 . r→ 2 r→ 2 By [14, Theorem A]   σ121 σ111 T10 = −σ120 + B̂1 (1/r − 1, σ120 ) , σ120 Q(0, σ120 ) L1 (σ120 ) with Z   u P (0, z) 1 dz L1 (u) = exp + = (1 + (F − 1)u)2F 0 Q(0, z) r z and p B1 (u) = L1 (u)∂1 Q−1 (0, u) = − F (F − 1)(1 + (F − 1)u)2F −1 . 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