On the critical points of the flight return time function of perturbed closed orbits

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On the critical points of the flight return time function of perturbed closed orbits Adriana Buic˘a a , Jaume Giné b , Maite Grau b,∗ a Departamentul de Matematic˘a, Universitatea Babe¸s–Bolyai, Str. Kog˘alniceanu 1, 400084 Cluj-Napoca, Romania b Departament de Matemàtica, Inspires Research Centre, Universitat de Lleida, Avda. Jaume II, 69, 25001 Lleida,

Catalonia, Spain Received 13 April 2018

Abstract We deal here with planar analytic systems x˙ = X(x, ε) which are small perturbations of a period annulus. For each transversal section  to the unperturbed orbits we denote by T  (q, ε) the time needed by a perturbed orbit that starts from q ∈  to return to . We call this the flight return time function. We say that the closed orbit  of x˙ = X(x, 0) is a continuable critical orbit in a family of the form x˙ = X(x, ε) if, for any q ∈  and any  that passes through q, there exists q ε ∈  a critical point of T  ( · , ε) such that q ε → q as ε → 0. In this work we study this new problem of continuability. In particular we prove that a simple critical periodic orbit of x˙ = X(x, 0) is a continuable critical orbit in any family of the form x˙ = X(x, ε). We also give sufficient conditions for the existence of a continuable critical orbit of an isochronous center x˙ = X(x, 0). © 2019 Published by Elsevier Inc. MSC: 34C23; 37G15; 34C05; 34C25; 34C07

1. Introduction In this work we are interested in studying the time needed for a trajectory of a planar differential system to return to a transversal section. We call this time the flight return time function. * Corresponding author.

E-mail addresses: [email protected] (A. Buic˘a), [email protected], [email protected] (M. Grau). https://doi.org/10.1016/j.jde.2018.12.031 0022-0396/© 2019 Published by Elsevier Inc.

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Before presenting the precise definitions and results, this paragraph contains the motivation for the study of this function. In the case that all the orbits are periodic, that is the system has a period annulus, the flight return time function is called the period function and it is independent of the transversal section. The period function has been widely studied by many authors in the last decades, see for instance [9,10,12,14,18–21,24]. The main features of interest of the period function is when it is constant (isochronicity) and its critical points (called critical periods). On the other hand, when the orbits are spirals, the flight return time function strongly depends on the considered transversal section. It turns out, as we will see in Example 2.1, that different transversal sections may give rise to very different qualitative behaviors of the flight return time function. Near singular points or limit cycles, many authors have studied the problem of the existence of a transversal section for which the flight return time function is constant. These transversal sections are called isochrones and their existence and computation are useful in biological models, see [1,7,13,15]. Our interest is to show which features of the period function are maintained for the flight return time function, independently of the transversal section, for a perturbation of a differential system with a period annulus. We will show in Theorem 3 that for a non isochronous period annulus, the critical periods are maintained for the flight return time function of a perturbed differential system. In Theorem 4 we assume that the perturbed system is close enough to an isochronous period annulus (that is, the period annulus of the unperturbed system is isochronous and a finite number of the first Melnikov functions vanish identically, see precise definitions below). We prove that the flight return time function can be defined independently of the transversal section up to a certain order of the perturbation parameter. In this case, the flight return time function may show critical points. We consider a family of planar analytic vector fields of the form x˙ = X(x, ε)

(1)

which depends on the small parameter ε ∈ (R, 0). Here the independent variable is denoted by t , x˙ denotes the derivative with respect to t and x(t) ∈ R2 . We write X(x, ε) =

∞ 

ε k Xk (x)

k=0

and denote by ϕ(t, q, ε) the flow of (1), i.e. for any q ∈ R2 and |ε|  1 the function ϕ(·, q, ε) is the solution of (1) such that ϕ(0, q, ε) = q. Then ϕ(t, q, 0) is the flow of the unperturbed system x˙ = X0 (x) .

(2)

Throughout this work we assume that system (2) has a period annulus P ⊂ R2 , i.e. P is invariant under the flow of (2) and any orbit of (2) in P is a nontrivial closed curve or, equivalently, for any q ∈ P the solution ϕ(·, q, 0) is a nontrivial periodic function. Hence, in particular we have that X0 (x) = 0 for all x ∈ P. We consider also the period function of system (2), T0 : P → R defined such that for each q ∈ P, T0 (q) is the main period of ϕ(·, q, 0). It is known that T0 is analytic, see [24]. When T0 is constant in P we say that the period annulus P is isochronous. When T0 is not constant in P, it is a first integral of system (2). Recall that a first integral H : P → R is a continuous function that it is not locally constant and for any q ∈ P, H (ϕ(t, q, 0)) does not depend on t. If H is of class C 1 , this condition is equivalent to ∇H · X0 = 0 in P.

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Recall that an analytic transversal section  = {γ (s) : s ∈ I } ⊂ P to the flow of (2) in P is defined by an analytic function γ : I →  (parametrization) such that I ⊂ R is a nonempty open interval, γ is one-to-one and onto, and, for any s ∈ I , γ (s) ∧ X0 (γ (s)) = 0 (transversality condition). The wedge product ∧ between two column vectors is the determinant of the matrix which has them as columns. So, the transversality condition means that the vectors γ (s) and X0 (γ (s)) are not collinear. Note also that any analytic transversal section to the flow of (2) in P is also an analytic transversal section to the flow of (1) in P for |ε|  1. Throughout this article, an analytic transversal section to the flow of (2) in P will be simply called transversal section. Take  ⊂  such that the orbit of system (1) (with |ε|  1) that starts in q ∈  returns to . Denote by T  (q, ε) > 0 the time needed, and call the function T  the flight return time to . For simplicity, in the sequel  will be also denoted by . We remark that, given a point q ∗ ∈ P and two different transversal sections  and S such that q ∗ ∈  ∩ S it can happen that T  (q ∗ , ε) = T S (q ∗ , ε). Anyway, for ε = 0, T  (q, 0) = T0 (q) for each q ∈  and T S (q, 0) = T0 (q) for each q ∈ S, where T0 : P → R is the period function of system (2). Given a transversal section  ⊂ P and a parametrization γ : I → , define the function τ γ (s, ε) = T  (γ (s), ε), which is analytic. Given another parametrization γ˜ : I˜ →  and τ γ˜ (σ, ε) = T  (γ˜ (σ ), ε), one can see that τ γ (s, ε) = τ γ˜ (γ˜ −1 ◦ γ (s), ε). Since γ˜ −1 ◦ γ : I → I˜ is a diffeomorphism, one can see that s ε ∈ I is a critical point of τ γ (·, ε) if and only if hε = (γ˜ −1 ◦ γ )(s ε ) ∈ I˜ is a critical point of τ γ˜ (·, ε). Moreover, they correspond to the same point q ε = γ (s ε ) = γ˜ (hε ) ∈ . These comments justify that the following notion does not depend on the parametrization of a transversal section. Definition 1. Let |ε|  1 be fixed. We say that q ε ∈  is a critical point of T  (·, ε) if, for a parametrization γ : I → , the function τ γ ( · , ε) : I → R has an isolated critical point s ε ∈ I such that γ (s ε ) = q ε . We present now the key notion of this work. Definition 2. We say that the closed orbit  of (2) in P is a continuable critical orbit in a family of the form (1) if for any q ∗ ∈ , any transversal section  of (2) in P with q ∗ ∈ , and any |ε|  1, the function T  ( · , ε) has a critical point q ε ∈  such that q ε → q ∗ as ε → 0. The above notion exists in the literature in the particular case that all the orbits of system (1) that start in P are also closed (as the ones of system (2)). In this particular case T  (q, ε) = T (q, ε) for any q ∈ , any  and any |ε|  1, where T (·, ε) is the period function of system (1). Our aim here is to detect the existence of continuable critical orbits in the sense of Definition 2 considering systems (1) whose orbits in P are not necessarily closed. More precisely, we assume that, for an integer m ≥ 0, the first m Melnikov functions vanish identically. In order to explain this condition, we define now the displacement map d γ of (1) on  when considering a parametrization γ on ,   d γ (s, ε) = γ −1 ϕ(T  (γ (s), ε), γ (s), ε) − s,

s ∈ I,

|ε|  1.

(3)

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Note that throughout this paper we will use the symbol O(ε k ) to denote terms of orders higher or equal to k in the Taylor expansion around ε = 0 of an analytic function that may depend on other variables. The displacement map d γ is analytic. In the case that all the orbits of (1) in a neighborhood of P are closed for all |ε|  1, then d γ (s, ε) ≡ 0. Otherwise there exists an integer m ≥ 0 such that γ

d γ (s, ε) = ε m+1 Mm+1 (s) + O(ε m+2 ) ,

s ∈ I,

|ε|  1.

(4)

Let m be the lowest value of m with the property that Mm +1 (s) ≡ 0. The function Mm +1 is called the Melnikov function of order (m + 1) and we say that the first m Melnikov functions vanish identically. It is known that, and proved in Section 2 of this paper (Proposition 6 and Corollary 7), that m does not depend on the transversal section. Before presenting one of the main results of this work we present few more notations and comments. Given a transversal section  ⊂ P, a parametrization γ : I → , and the corresponding flight return time T  , the function s → τ γ (s, ε) = T  (γ (s), ε) is analytic and either is constant on I for any |ε|  1, or there exists an integer m = m(γ ) ≥ 0 such that γ

γ

γ

γ

γ

τ γ (s, ε) = (τ0 + ετ1 + · · · + ε m−1 τm−1 ) + ε m τm (s) + O(ε m+1 ) , γ

γ

s ∈ I,

|ε|  1,

γ

where τ0 , ..., τm−1 are constant and τm (s) is not constant. We have  d  γ γ τ (s, ε) = ε m (τm ) (s) + O(ε m+1 ) , ds

s ∈ I,

|ε|  1.

From the above relation we deduce that, if for each |ε|  1, s ε is a critical point of τ γ (·, ε) such γ that s ε → s ∗ ∈ I as ε → 0, then (τm ) (s ∗ ) = 0. γ γ  = τγ , On the other hand, we have T0 = τ0 and we denote T1 = τ1 , ..., Tm−1 m−1 Tm (q) = τm ◦ γ −1 (q), γ

q ∈ .

 and the function T  (q) do It is clear that the integer m = m() ≥ 0, the constants T1 , ..., Tm−1 m not depend on the parametrization γ . Hence one can write  T  (q, ε) = (T0 + εT1 + · · · + ε m−1 Tm−1 ) + ε m Tm (q) + O(ε m+1 )

(5)

where q ∈  and |ε|  1. We have a given transversal section . Now let  ⊂ P be a periodic orbit of system (2) which is a continuable critical orbit in the sense of Definition 2. Then, any q ∗ ∈  ∩  must be a critical point in the sense of Definition 1 of Tm according to (5). For m ≥ 1, in Theorem 4, we give sufficient conditions so that this is assured for any transversal section . The key results are stated below and proved in Section 2. q ∗ denotes the orbit of the unperturbed system (2) that passes through q ∗ ∈ P.

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Theorem 3. If for a transversal section  = {γ (s) : s ∈ I } ⊂ P of system (2), the function γ τ0 (s) = T0 (γ (s)) has a simple critical point s ∗ ∈ I , then γ (s ∗ ) is a continuable critical orbit in (1). Theorem 4. Assume that there exists m ≥ 1 such that the first m Melnikov functions vanish identically for system (1) and that T0 (q) = T0 is constant. Let T1 , . . . , Tm−1 be some real constants. Assume also that, for a transversal section S we have S T1S (q) = T1 , . . . , Tm−1 (q) = Tm−1 for all q ∈ S.

Then the following statements are valid. (a) For any transversal section ,  T1 (q) = T1 , . . . , Tm−1 (q) = Tm−1 for all q ∈ .

(b) There exists an analytic function Tm in P such that Tm (q) = Tm (q) for all q ∈  and for any . Moreover, if not a constant, Tm is a first integral in P of x˙ = X0 (x). γ (c) If for a transversal section  = {γ (s) : s ∈ I } in P to (2), the function τm (s) = Tm (γ (s)) ∗ has a simple critical point s ∈ I , then γ (s ∗ ) is a continuable critical orbit in (1). Note that one of the hypotheses of Theorem 4 is that T0 (q) = T0 is constant, i.e. the period annulus P of system (2) is isochronous. It is known (see [19]) that, in this case, P must be the period annulus of a nondegenerate singular point of center type. This paper is organized as follows. In section 2 we provide several numerical examples which illustrate the results stated in the Introduction. Section 3 contains some previous results which motivate the proof of the main theorems, given in section 4. In order to apply Theorem 4, we proγ vide several methods to compute the derivative of the function τm in section 5. For the particular case when all the perturbed orbits are closed, formulas to compute the derivative of the period function exist in the literature, see [9,12,14]. Last section 6 contains some analytical examples. 2. Illustrative examples of the main results 2.1. Strong focus We show that the flight return time function defined in a neighborhood of a strong linear focus depends on the transversal section. Take the strong linear focus given by the vector field X = (−y + εx, x + εy), with ε = 0. The flow with initial condition the point q = (x0 , y0 ) is ϕ(t, (x0 , y0 )) = eεt (x0 cos t − y0 sin t, y0 cos t + x0 sin t) . We consider the transversal section given by  = {(s, 0) : s > 0}. It is easy to see that the flight return time function in this case is T  (s, 0) = 2π . We have that  is an isochrone. Now we

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Fig. 1. Flight return time function of the linear strong focus with ε = 1/10 computed on  .

  consider the section  = γ (s) = (s, s 2 ) : s > 0 . In order to have that  is transversal, we   need to check that γ (s) ∧ X(γ (s)) = 0 for s > 0. This gives that s 1 − εs + 2s 2 = 0 for s > 0 √ which implies that |ε| < 8. We assume that ε belongs to this interval for the rest of the example.

In this case, the flight return time function τ γ (s) := T  (s, s 2 ) is defined implicitly as the least 2

positive t such that ϕ(t, (s, s )) ∈  . This is equivalent that t is the least positive solution of

2 eεt (s 2 cos t + s sin t) − e2εt s cos t − s 2 sin t = 0, such that (s 2 cos t + s sin t) > 0. This functional equation cannot be solved analytically. We numerically find the solution for ε = 1/10 and we obtain the graphic shown in Fig. 1. We remark that the behavior of this function is different from the constant value obtained for . If we consider ε = 0, the vector field is the linear center. If we see the vector field X as a perturbation of this isochronous center, neither the hypothesis of Theorem 3 nor Theorem 4 are satisfied. 2.2. A perturbation of a Loud center The classification of the isochronous centers of quadratic systems is due to Loud [17]. The so-called dehomogenized Loud’s systems are the following quadratic systems: x˙ = −y + xy,

y˙ = x + Dx 2 + Fy 2 ,

(6)

where D, F ∈ R. Any isochronous quadratic system can be written in this form for fixed values of (D, F ). Any system of this family has a center at the origin. In [20] a conjectural bifurcation diagram of the period function of the dehomogenized Loud’s systems is provided. It is known that the period function of the vector field defined by (6) when (D, F ) = (−1.5, 2.1) has two critical points. The graphic of the period function in this case (numerically computed) is shown in Fig. 2. In this case the transversal section that we have used is  = {(s, 0) : 0 < s < 0.353}. The boundary of the period annulus of this system corresponds to a value of s greater but very close

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Fig. 2. Period function of system (6) with (D, F ) = (−1.5, 2.1).

to s = 0.353. The maximum of τ (s) = T (s, 0) is at s ≈ 0.312985 and the minimum is at s ≈ 0.35247. We consider the following perturbation of the differential system (6) x˙ = −y + xy + εx 2 ,

y˙ = x + Dx 2 + Fy 2 ,

(7)

where (D, F ) = (−1.5, 2.1) as before. When 0 = |ε|  1, this system has a focus point at the origin since one of its Liapunov constants is V4 = ε(2D − 1)/8 = 0. We take ε = 0.01 and we numerically compute the flight return time function τε (s) = T  (s, ε). The graphic of this function is very similar to Fig. 2 and the maximum of τε (s) is at s ≈ 0.311987 and the minimum is at s ≈ 0.350543. The flight return time function computed for any other transversal section will also have the same qualitative behavior according to Theorem 3. 3. The roots of the key notion and results In this section, for the unperturbed system x˙ = X0 (x),

(2)

we consider a first integral H : P → R such that ∇H (q) = (0, 0) for all q ∈ P. It is known that, in our hypotheses, there exists such a first integral (see [16]). Given a transversal section  ⊂ P to the flow of (2) and taking I = H (), we have that H | :  → I is invertible and I ⊂ R is an open interval. We define a new parametrization on , γ = (H | )−1 . Note that it satisfies that H (γ (h)) = h for all h ∈ I . We say that γ is the parametrization of  with the values of H .

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3.1. A useful Lemma The following result will be useful in this work. As before, q ∗ denotes the orbit of (2) that passes through q ∗ ∈ P. Lemma 5. Let F : P → R be an analytic first integral of system (2). (a) If ∇F (q ∗ ) = (0, 0) for a q ∗ ∈ P then ∇F (q) = (0, 0) for any q ∈ q ∗ . Consider a transversal section  ⊂ P of (2), a parametrization γ : I →  and f γ (s) = F (γ (s)),

s ∈ I.

(b) If (f γ ) (s ∗ ) = 0 for an s ∗ ∈ I then ∇F (γ (s ∗ )) = (0, 0). (c) If ∇F (q ∗ ) = (0, 0) for a q ∗ ∈  then (f γ ) (s ∗ ) = 0 where s ∗ = γ −1 (q ∗ ). Moreover, the multiplicity of s ∗ as critical point of f γ does not depend neither on  nor on γ . Proof. (a) Since F is a first integral of (2) we have F (ϕ(t, q, 0)) = F (q) for all t ∈ R. We take the derivative with respect to q in this last equation ∇F (ϕ(t, q, 0)) · Dϕ(t, q, 0) = ∇F (q). Recall that the matrix Dϕ(t, q, 0) is a fundamental matrix of solutions of the linear system y˙ = DX0 (ϕ(t, q, 0)) · y. Then, by Liouville’s formula we have that det Dϕ(t, q, 0) = 0. Consequently, the conclusion follows. (b) Note that (f γ ) (s) = ∇F (γ (s)) · γ (s). Then ∇F (γ (s ∗ )) · γ (s ∗ ) = 0. But ∇F (γ (s ∗ )) · X0 (γ (s ∗ )) = 0 due to the fact that F is a first integral of (2). Since the vectors γ (s ∗ ) and X0 (γ (s ∗ )) are not collinear, one must have that ∇F (γ (s ∗ )) = (0, 0). (c) First note that (f γ ) (s ∗ ) = ∇F (q ∗ ) · γ (s ∗ ) = 0. Now let S = {γ˜ (σ ) : σ ∈ I˜} be another transversal section and γ˜ be a parametrization of it. Denote f = f γ = F ◦ γ , s ∗ = γ −1 (q ∗ ) and f˜ = f γ˜ = F ◦ γ˜ , σ ∗ = γ˜ −1 (q ∗ ). Hence s ∗ is a critical point of f and σ ∗ is a critical point of f˜. We have to prove that their corresponding multiplicities are equal. It is sufficient to prove that f and f˜ are conjugated up to an analytic diffeomorphism. We consider two cases. First assume that  = S. Then it is easy to see that f˜ ◦ (γ˜ −1 ◦ γ ) = f and γ˜ −1 ◦ γ is an analytic diffeomorphism. Now assume that  = S but both are parameterized with the values of the first integral H (given at the beginning of this section). Then γ = (H | )−1 , γ˜ = (H |S )−1 , and we can write h = s = σ . It is clear that γ (h) and γ˜ (h) are on the same orbit. Then f (h) = f˜(h) because F is a first integral. 2 3.2. When all the orbits are closed: the roots of our key notion In this subsection we assume that, for |ε|  1, all the orbits of system (1) that start in P are closed. For a transversal section  = {γ (s) : s ∈ I } ⊂ P of the unperturbed system (2) denote by

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τ (s, ε) > 0 the main period of ϕ( · , γ (s), ε). The notion of critical periodic orbit is well known in this context. We say that ε = ϕ( · , γ (s ε ), ε) is a critical periodic orbit of system (1) if s ε is an isolated critical point of τ (·, ε). It is worth to mention at this point that this notion does not depend neither on the transversal section  nor on its parametrization. In order to explain this, for any q ∈ P we denote by T (q, ε) > 0 the main period of ϕ( · , q, ε). If not constant, T (·, ε) is an analytic first integral of (1) for which Lemma 5 applies. It is a first integral because it takes a constant value on each periodic orbit. So, Lemma 5 assures that the above notion of critical periodic orbit is well-defined and, moreover, ∇T (q ε , ε) = (0, 0) for q ε = γ (s ε ). Hence q ε is a critical point of T (·, ε) : P → R. Remark that, in this particular case, the flight return function T  (·, ε) defined in the Introduction satisfies T  (q, ε) = T (q, ε) for any q ∈ , for any transversal section  ⊂ P and any |ε|  1. Now it is not difficult to see that the notion of critical point as in Definition 1 reduces, in this context, to the usual notion of critical point. Another well-known notion in this particular context is the following. It is said that the family (depending on ε) of critical periodic orbits ϕ( · , γ (s ε ), ε) of (1) bifurcates from the periodic orbit ϕ( · , γ (s ∗ ), 0) of (2) if s ε → s ∗ as ε → 0. This bifurcation can be checked by finding q ∗ ∈ P and q ε ∈ P such that q ε → q ∗ as ε → 0 and q ε is a critical point of T (·, ε). Now we note that the orbit that passes through q ∗ is a continuable critical orbit in the sense of Definition 2. When dealing with the problem of finding continuable critical orbits in the sense of Definition 2 in the general context defined in the Introduction, we found useful the following remarks made in the particular case when all the orbits in P of (1) are closed. When T (·, ε) is not constant, there exists m ≥ 0 such that T (q, ε) = (T0 + εT1 + · · · + ε m−1 Tm−1 ) + ε m Tm (q) + O(ε m+1 ) , for all q ∈ P , where T0 , ..., Tm−1 are constant and Tm (q) is not constant. Then ∇T (q, ε) = ε m ∇Tm (q) + O(ε m+1 ) , for all q ∈ P. When  is a continuable critical orbit, as we commented above, there exists q ε ∈ P such that q ε → q ∗ ∈  as ε → 0 and ∇T (q ε , ε) = 0. Then ∇Tm (q ε ) + O(ε) = (0, 0), which after taking limits as ε → 0 gives ∇Tm (q ∗ ) = (0, 0). This means that any point of a continuable critical orbit is a critical point of Tm . On the other hand, since 1 [T (q, ε) − (T0 + · · · + ε m−1 Tm−1 )] εm is also an analytic first integral of (1) and Tm (q) is the limit as ε → 0 of this function, we deduce that Tm : P → R is a first integral of the unperturbed system (2). Thus, Lemma 5 applies for Tm and its critical points.

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Coming back to the problem of finding continuable critical orbits as in Definition 2 in the general situation presented in the Introduction, our aim was to find sufficient conditions to assure that the corresponding function Tm has the same order m on any  and that there exists a first integral of (2), Tm : P → R such that Tm is the restriction of Tm to . 3.3. The displacement maps on different transversal sections The next result analyzes the displacement map (as defined in (3)), its corresponding Melnikov functions and how they vary with respect to the transversal section. It is a known result and we present its proof here for completeness. Note that its proof appears also in [3] but, apart from this, we have not found it in the literature. Proposition 6. (a) Let d γ and d γ˜ be the displacement maps of (1) associated to γ : I →  and, respectively, γ˜ : I˜ → . Let k, k˜ ≥ 1 be the orders of the corresponding nonvanishing Melnikov functions and denote ˜

˜

γ˜ k

d γ (s, ε) = ε k Mk (s) + O(ε k+1 ) and d γ˜ (h, ε) = ε k M ˜ (h) + O(ε k+1 ) . γ

Then the corresponding Melnikov functions satisfy k˜ = k

γ˜

Mk (h)γ˜ (h) = Mk (s)γ (s) for s = (γ˜ −1 ◦ γ )(h).

and

γ

(b) We consider now two analytic transversal sections  and S both parameterized with the values of H . Denote the corresponding displacement maps by d(h, ε) and δ(h, ε) and let k, l ≥ 1 be the orders of the corresponding nonvanishing Melnikov functions and denote d(h, ε) = ε k Mk (h) + O(εk+1 ) and

δ(h, ε) = ε l μl (h, ε) + O(ε l+1 ) .

Then the corresponding Melnikov functions satisfy k=l

and Mk (h) = μk (h).

Proof. (a) Let s ∈ I and h = (γ˜ −1 ◦ γ )(s). By the definitions of displacement map and transversal section, we have γ (s + d γ (s, ε)) = γ˜ (h + d γ˜ (h, ε)). Hence ˜

γ

γ˜ k

˜

γ (s + εk Mk (s) + O(ε k+1 )) = γ˜ (h + ε k M ˜ (h) + O(ε k+1 )). After taking the derivative with respect to ε in the above relation we get γ (s + ε k Mk (s) + O(ε k+1 ))(kε k−1 Mk (s) + O(ε k+1 )) γ˜ ˜ γ˜ ˜ ˜ k−1 M ˜ (h) + O(εk+1 )). = γ˜ (h + ε k M ˜ (h) + O(εk+1 ))(kε γ

γ

k

k

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11

γ˜

Assuming that k˜ < k, we obtain that the Melnikov function M ˜ is identically 0, which contradicts k ˜ Because of the symmetry we get k˜ = k. The relation from the conclusion is the definition of k. obtained after dividing by kεk−1 and then taking ε = 0. (b) We consider now the case of two transversal sections. Denote the corresponding Poincaré maps by Pε (h) = P (h, ε) = h + d(h, ε) and, respectively, Rε (h) = R(h, ε) = h + δ(h, ε). Consider also πε (h) = π(ε, h) the transition map between  and S along the flow of (1). Then πε ◦ Pε = Rε ◦ πε , which further gives π(h + εk Mk (h) + O(εk+1 ), ε) = π(h, ε) + ε l μl (π(h, ε)) + O(ε l+1 ) .

(8)

It can be proved that π(h + εk Mk (h) + O(εk+1 ), ε) − π(h, ε) = 0 when l < k. ε→0 εl lim

Assuming that l < k, after dividing by εl relation (8) and taking the limit as ε → 0, we would obtain that the Melnikov function μl (π0 (h)) = μl (h) is identically 0. This contradicts the definition of l. Because of the symmetry it is not necessary to consider the case l > k. Hence l = k. It can be proved that π(h + ε k Mk (h) + O(εk+1 ), ε) − π(h, ε) = Mk (h). ε→0 εk lim

Now, the relation between the Melnikov functions is obtained after dividing by εk relation (8) and taking the limit as ε → 0. 2 A direct consequence of the above result is the following Corollary 7. The constant m ≥ 0 in formula γ

d γ (s, ε) = ε m+1 Mm+1 (s) + O(ε m+2 ) ,

s ∈ I,

(4)

does not depend on γ : I → . Moreover, when an arbitrary transversal section  is parameterized with the values of H , the function Mm+1 does not depend on  and the function Mm+1 : P → R defined by Mm+1 (q) = Mm+1 (H (q)) is an analytic first integral of (2) and satisfies that Mm+1 (h) = Mm+1 (γ (h)). 4. Proofs of the main results We start by presenting the proof of Theorem 3 stated in the Introduction.

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Proof of Theorem 3. By hypothesis, the function τ0 (s) = T0 (γ (s)) has a simple critical point s ∗ ∈ I . Then T0 is not a constant, thus it is a first integral of system (2). Let  = γ (s ∗ ) be the unperturbed closed orbit defined in the statement. Then, by Lemma 5, we have that ∇T0 (q) = (0, 0) for all q ∈ . γ˜ Let S = {γ˜ (σ ) : σ ∈ J } ⊂ P be an arbitrary transversal section and τ0 (σ ) = T0 (γ˜ (σ )). Let γ˜

σ ∗ ∈ J be such that γ˜ (σ ∗ ) ∈  ∩ S. By Lemma 5, we have that σ ∗ is a simple critical point of τ0 . We also have d γ˜ [T S (γ˜ (σ ), ε)] = (τ0 ) (σ ) + O(ε). dσ

Then, the Implicit Function Theorem gives that there exists σ ε ∈ J , a critical point of σ → T S (γ˜ (σ ), ε) such that σ ε → σ ∗ as ε → 0. Thus  fulfills the conditions in Definition 2, so it is a continuable critical orbit. 2 For the proof of Theorem 4 we need the following lemmas. We will work with the Lie bracket of two planar vector fields X and U , [X, U ] = DX U − DU X where the notation DX means the Jacobian matrix of X. The next lemma is written for the planar vector field X0 that satisfies all the hypotheses written in the Introduction, but it works for an arbitrary C 1 planar vector field X0 . Anyway, it recalls some basic properties. Lemma 8. Let U : P → R2 be a C 1 planar vector field. Then (a) [U, X0 ] = 0 if and only if t → U (ϕ(t, q, 0)) is a solution of the variational system y˙ = DX0 (ϕ(t, q, 0))y. (b) [U, X0 ] = 0 if and only if U (ϕ(t, q, 0)) = Dϕ(t, q, 0)U (q). (c) Let T : P → R be a C 1 , nonconstant function. We have that [T X0 , X0 ] = 0 if and only if T is a first integral of x˙ = X0 (x). Proof. (a) We have the following equivalences. The identity [U, X0 ] = 0 is verified if and only if DU (q)X0 (q) = DX0 (q)U (q) for all q ∈ P. And this is true if and only if DU (ϕ(t, q, 0))X0 (ϕ(t, q, 0)) = DX0 (ϕ(t, q, 0))U (ϕ(t, q, 0)) for all q ∈ P. This is equivalent to write d U (ϕ(t, q, 0)) = DX0 (ϕ(t, q, 0))U (ϕ(t, q, 0)). dt The last identity means that U (ϕ(t, q, 0)) is a solution of the variational system y˙ = DX0 (ϕ(t, q, 0))y. (b) This follows immediately from (a) taking into account that Dϕ(t, q, 0) is the principal fundamental matrix solution of the variational system (see [6]). (c) We use the known formula [T X, U ] = T [X, U ] − (∇T · U ) X to obtain [T X0 , X0 ] = −(∇T · X0 ) X0 .

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Hence [T X0 , X0 ] = 0 is equivalent to ∇T · X0 = 0, which means that T is a first integral of x˙ = X0 (x). 2 Consider the functions ϕi (q) such that ϕ(T  (q, ε), q, ε) = q +



ε i ϕi (q) ,

q ∈ .

i≥1

Lemma 9. Assume that the first m Melnikov functions vanish identically for system (1). Then for any transversal section  we have that ϕi (q) = 0, q ∈ , i ∈ {1, ..., m}.

(9)

Proof. From the definition (3) of the displacement map we have γ (s + d γ (s, ε)) = ϕ(T  (γ (s), ε), γ (s), ε).

(10)

On one hand γ

γ (s + d γ (s, ε)) = γ (s + εm+1 Mm+1 (s) + O(ε m+2 )) = γ (s) + O(ε m+1 ), and on the other hand  ϕ(T  (γ (s), ε), γ (s), ε) = γ (s) + εϕ1 (γ (s)) + · · · + ε m ϕm (γ (s)) + O(εm+1 ).

Equating the coefficients of ε, ε 2 , . . . , ε m in both sides of the former relation (10), and replacing q = γ (s), we obtain the conclusion. 2 Remind that the functions Ti (q) are defined such that T  (q, ε) =



ε i Ti (q),

(11)

i≥0

where  is a transversal section and T  (q, ε) is the flight return time to . Of course, the function T0 (q) does not depend on . Lemma 10. Let m ≥ 1 and T1 , . . . , Tm−1 be analytic functions in P. Assume that the first m Melnikov functions vanish identically for system (1) and that  T1 (q) = T1 (q), . . . , Tm−1 (q) = Tm−1 (q) for all q ∈  and for any .

Then there exists an analytic function ψm : P → R2 such that ϕ(T0 + εT1 + ... + ε m−1 Tm−1 , q, ε) = q + ε m ψm (q) + O(ε m+1 ) for all q ∈ P. Moreover,

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ψm (q) = −Tm (q)X0 (q) for all q ∈  and for any  and, consequently, Tm (q) does not depend on . Proof. Denote τm (q, ε) = T0 (q) + εT1 (q) + ... + ε m−1 Tm−1 (q) for all q ∈ P. Fix a q ∈ P and a transversal section  such that q ∈ . Then T  (q, ε) = τm (q, ε) + εm Tm (q) + O(ε m+1 ), 1 m [ϕ(τm (q, ε), q, ε) ε→0 ε

If there exists the limit lim

q ∈ .

− q] then its value is ψm (q), where ψm :

P → R2 is analytic. Our aim now is to prove the existence of this limit for any q ∈ . The following expansions hold ϕ(T  (q, ε), q, ε) = ϕ(ε m Tm (q) + O(ε m+1 ), ϕ(τm (q, ε), q, ε), ε)

(12)

= ϕ(τm (q, ε), q, ε) + εm Tm (q)X0 (q) + O(ε m+1 ). In the first equality above we used a well-known property of the flow, while in the second we used the expansion formula ϕ(ε m Tm (q) + O(ε m+1 ) , p , ε) = p + ε m Tm (q)ϕ(0, ˙ q, 0) + O(εm+1 ) for p = ϕ(τm (q, ε), q, ε) and the fact that ϕ(0, ˙ q, 0) = X0 (q). On the other hand, from Lemma 9 we obtain that ϕ(T  (q, ε), q, ε) = q + O(ε m+1 ).

(13)

Then, equating both expansions (12) and (13) of ϕ(T  (q, ε), q, ε) we obtain, ϕ(τm (q, ε), q, ε) = q − ε m Tm (q)X0 (q) + O(ε m+1 ). Then, indeed there exists the following limit and we have 1 [ϕ(τm (q, ε), q, ε) − q] = −Tm (q)X0 (q). ε→0 ε m

ψm (q) = lim

 0 The above equality holds for all q ∈  and for an arbitrary . Since X0 (q) = from 0 ψm (q) = −Tm (q)X0 (q) we deduce that Tm (q) does not depend on . 2 After applying Lemma 10 successively for m = 1, m = 2, ..., m = m, the next result is obtained.

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Proposition 11. Assume that there exists m ≥ 1 such that the first m Melnikov functions vanish identically for system (1). Then T1 (q), ..., Tm (q) do not depend on the transversal section  that passes through q. Lemma 12. Assume that there exists an integer m ≥ 1 such that the first m Melnikov functions vanish identically for system (1) and that the first m functions T0 , T1 , . . . , Tm−1 are constant in P. Then the function ψm : P → R2 defined like in Lemma 10 by ϕ(T0 + εT1 + ... + εm−1 Tm−1 , q, ε) = q + ε m ψm (q) + O(ε m+1 ) for all q ∈ P satisfies [ψm , X0 ] = 0. Proof. Denote, like in the proof of Lemma 10 τm (ε) = T0 + εT1 + ... + ε m−1 Tm−1 which in our hypotheses does not depend on q. Then we have ϕ(τm (ε), q, ε) = q + ε m ψm (q) + O(ε m+1 ) that, substituting q by ϕ(t, q, ε) gives ϕ(τm (ε), ϕ(t, q, ε), ε) = ϕ(t, q, ε) + εm ψm (ϕ(t, q, ε)) + O(εm+1 ). Using that ψm (ϕ(t, q, ε)) = ψm (ϕ(t, q, 0)) + O(ε), ϕ(τm (ε), ϕ(t, q, ε), ε) = ϕ(t, ϕ(τm (ε), q, ε), ε) and (14) we obtain ϕ(t, q + ε m ψm (q) + O(ε m+1 ), ε) = ϕ(t, q, ε) + ε m ψm (ϕ(t, q, 0)) + O(εm+1 ). On the other hand we have ϕ(t, q + ε m ψm (q) + O(ε m+1 ), ε) = ϕ(t, q, ε) + ε m Dϕ(t, q, 0)ψm (q) + O(ε m+1 ). Hence ψm (ϕ(t, q, 0)) = Dϕ(t, q, 0)ψm (q) which, applying Lemma 8 (b) assures that [ψm , X0 ] = 0.

2

We are ready now to present the proof of Theorem 4 stated in the Introduction.

(14)

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Proof of Theorem 4. We prove (a) and (b) by induction. For m = 1 we have to prove that If the first Melnikov function vanish and T0 is constant, then T1 (q) does not depend on  and, moreover, if not constant, is an analytic first integral of x˙ = X0 (x). Proposition 11 for m = 1 assures that T1 (q) does not depend on . Denote it by T1 (q). The vector field ψ1 (q) defined as in Lemma 10 satisfies ψ1 (q) = −T1 (q)X0 (q) and [ψ1 , X0 ] = 0 (as assured by Lemmas 10 and 12). Thus [T1 X0 , X0 ] = 0, which, by Lemma 8 (c) assures that, if not constant, T1 is a first integral of X0 . Now assume that the statements (a) and (b) are valid for an integer m ≥ 1 and prove them for S ,T S m + 1. Hence we have that the first m + 1 Melnikov functions vanish and that T1S , . . . , Tm−1 m are constant on a section S. The hypothesis of induction assures that T1, . . . , Tm−1 are the same constant on any transversal section and that Tm (q) does not depend on the transversal section and, moreover, it would be a first integral of the unperturbed system x˙ = X0 (x) if not a constant. But since Tm is constant on a transversal section S we deduce that Tm is constant on the whole P. So (a) is proved for m + 1. Since the first m + 1 Melnikov functions vanish, Proposition 11 assures that Tm+1 (q) does not depend on the transversal section. Moreover, since, in addition, the functions T1 , ..., Tm are constant in P, from Lemma 12 we know that [Tm+1 X0 , X0 ] = 0. Then, by Lemma 8 (c) we get that, if not constant, Tm+1 is a first integral of x˙ = X0 (x). So (b) is also proved for m + 1. In conclusion, by induction, (a) and (b) are proved. γ (c) By hypothesis, the function τm (s) = Tm (γ (s)) has a simple critical point s ∗ ∈ I . Thus Tm is not a constant and, by (b), it is a first integral of system (2). Let  = γ (s ∗ ) be the unperturbed closed orbit defined in the statement. Then, by Lemma 5, we have that ∇Tm (q) = (0, 0) for all q ∈ . γ˜ Let S = {γ˜ (σ ) : σ ∈ J } be an arbitrary transversal section and τm (σ ) = Tm (γ˜ (σ )). Let σ ∗ ∈ J be such that γ˜ (σ ∗ ) ∈  ∩ S. By Lemma 5, we have that σ ∗ is a simple critical point γ˜ of τm . We also have d γ˜ [T S (γ˜ (σ ), ε)] = ε m (τm ) (σ ) + O(ε m+1 ). dσ Then, the Implicit Function Theorem gives that there exists a σ ε ∈ J , a critical point of σ → T S (γ˜ (σ ), ε) such that σ ε → σ ∗ as ε → 0. Thus  fulfills the conditions in Definition 2, so it is a continuable critical orbit. 2 5. A formula to compute the derivative of the flight return time function γ

The aim of this section is to give formulas for computing the derivative of the functions τm (s), where m = 0 and m = 1, that appear in Theorem 4. In the particular case that all the perturbed orbits are closed, such formulas exist in the literature. For m = 0 and m = 1 formulas were given by Gasull–Yu in [12]. For arbitrary m ≥ 1 Grau–Villadelprat gave such formulas in [14]. All these authors used the Lie bracket as an important tool. The predecessor of these formulas appeared in [9]. Here we show that similar formulas are valid in the hypotheses of Theorem 4. Note that there are other authors who gave formulas for the successive derivatives of the period function of a plane vector field in the particular case that all the orbits are closed. One of the first articles in this line is the one of Françoise [8], where the method to obtain such formulas relies on the decomposition of a 1-form associated to the relative cohomology of the perturbed Hamiltonian.

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17

For our purpose we present first a result from [4] which extends Theorem 1 in [9], Theorem 2 in [11] and Theorem 1 in [21]. We present also its proof for completeness. We recall that the Lie bracket of two smooth vector fields is the vector field computed with the formula [X, U ] = DX U − DU X. When X ∧ U = 0 in P we have [X, U ] = αX + βU in P for the smooth functions α=

[X, U ] ∧ U X ∧ [X, U ] and β = . X∧U X∧U

Theorem 13. Consider P ⊂ R2 open and X, U ∈ C 1 (P, R2 ) such that X ∧ U = 0 in P. Let α, β ∈ C 1 (P) be such that [X, U ] = αX + βU and let γ : I → P be a solution of x˙ = U (x), that is γ (s) = U (γ (s)), and ϕ(t, q) be the flow of X. Assume that there exists an open interval I ⊂ I such that, for any s ∈ I , each orbit of x˙ = X(x) that starts at γ (s) returns to  = {γ (s) : s ∈ I } and let τ (s) be the time that this orbit needs for the first return to . In addition, let π : I → I be the Poincaré return map to . Then for each s ∈ I we have

τ (s) t τ (s) = α(ϕ(t, γ (s)))e− 0 β(ϕ(v,γ (s)))dv dt

0

π (s) = e−

 τ (s) 0

β(ϕ(t,γ (s)))dt

.

Proof. Note that  is a transversal section to the flow of X since it satisfies γ (s) ∧ X(γ (s)) = 0 for all s ∈ I , where γ (s) = U (γ (s)). The expression of the Poincaré map is π(s) = γ −1 (ϕ(τ (s), γ (s))). Taking the derivative with respect to s in γ (π(s)) = ϕ(τ (s), γ (s)) we obtain γ (π(s)) π (s) =

dϕ (τ (s), γ (s)) τ (s) + Dϕ(τ (s), γ (s)) γ (s). dt

Since ϕ is the flow of X and γ is a solution of U , we have U (γ (π(s))) π (s) = X(γ (π(s))) τ (s) + Dϕ(τ (s), γ (s)) U (γ (s)). Using the notation η(t, q) = Dϕ(t, q)U (q), from the above equation we get η(τ (s), γ (s)) = U (γ (π(s))) π (s) − X(γ (π(s))) τ (s).

(15)

We also have that Dϕ(t, q) is the principal matrix solution of the variational system y˙ = DX(ϕ(t, q))y, thus η(t, q) is the solution of this system that verifies η(0) = U (q). It can be checked that this solution has the expression

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η(t, q) = U (ϕ(t, q))e

−

t 0

t β(ϕ(v,q))dv

− X(ϕ(t, q))

α(ϕ(u, q))e−

u 0

β(ϕ(v,q))dv

du.

0

In order to show the latter assertion we note that η(0, q) = U (q) (thus, the initial condition is satisfied) and we can derive η(t, q) with respect to t . We use that [X, U ] = αX + βU and d dt (ϕ(t, q)) = X(ϕ(t, q)) and we get that d η(t, q) = DX(ϕ(t, q))η(t, q). dt Hence, using also that ϕ(τ (s), γ (s)) = γ (π(s)), we obtain  τ (s)

η(τ (s), γ (s)) = U (γ (π(s))) e− 0 β(ϕ(v,γ (s)))dv u  τ (s) −X(γ (π(s))) 0 α(ϕ(u, γ (s)))e− 0 β(ϕ(v,γ (s)))dv du. The conclusion follows comparing this last relation with (15) and taking into account that the vectors U (γ (π(s))) and X(γ (π(s))) are linearly independent. 2 γ

A genuine approach for finding the derivative of the function τm (that appears in Theorem 4) is to apply Theorem 13 to the vector fields Xε = X(·, ε) and U such that X0 ∧ U = 0 in any point of P. For |ε|  1 we have Xε ∧ U = 0. Then αε =

[Xε , U ] ∧ U , Xε ∧ U

Xε ∧ [Xε , U ] . Xε ∧ U

βε =

Let γ (t) be a solution of x˙ = U (x) and denote by τε and πε the time map and, respectively, the Poincaré map as in Theorem 13. Then, by Theorem 13,

τε (s) =

τ ε (s)

αε (ϕ(t, γ (s), ε))e−

t 0

βε (ϕ(v,γ (s),ε))dv

dt

(16)

0

πε (s) = e−

 τε (s) 0

βε (ϕ(t,γ (s),ε))dt

(17)

.

When the hypotheses of Theorem 4 are fulfilled, we have for an integer m ≥ 0, τε (s) = ε m (τm ) (s) + O(ε m+1 ), γ

πε (s) = 1 + O(εm+1 ).

(18)

Let us expand αε and βε in powers of ε, αε = α0 + εα1 + ε 2 α2 + . . . ,

βε = β0 + εβ1 + ε 2 β2 + . . . .

(19)

Theoretically, to find the expression of (τm ) one has to introduce (18) and (19) in (16) and (17). The final step would be to equate the coefficients of each power of ε in (16) and (17). γ

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Of course, the case m = 0 is the easiest one, and we obtain γ (τ0 ) (s) =

T0

α0 (ϕ(t, γ (s), 0))e−

t 0

β0 (ϕ(v,γ (s),0))dv

dt,

0

where α0 =

[X0 , U ] ∧ U , X0 ∧ U

β0 =

X0 ∧ [X0 , U ] . X0 ∧ U

In the case m = 1, this was the approach in [12] for finding the derivative of the period function. They also took into account that, when the period annulus P of x˙ = X0 (x) is isochronous (thus P surrounds a nondegenerate equilibrium point of center type), then there exists an analytic vector field U in P such that [X0 , U ] = 0 and X0 ∧ U = 0 in P. This result is proved in [22]. Such U is said to be a commutator for X0 . However, in [12] it was noticed that the existence of an analytic vector field U in P such that [X0 , U ] = βU for an analytic function β, and X0 ∧ U = 0 in P, is sufficient to guarantee that the period annulus P of X0 is isochronous. Since, in practice, such a vector field can be easier to find than a commutator, we will use, as in [12], this latter condition for U . In our more general setting, the result follows. Recall that we use the notations X(x, ε) =

∞ 

ε k Xk (x).

k=0

Theorem 14. Assume that the first Melnikov function vanish identically for system (1) and that the period function of (2), T0 , is constant. Let U ∈ C ω (P, R2 ) and β0 ∈ C ω (P) be such that [X0 , U ] = β0 U and X0 ∧ U = 0 in P. Let λ1 , μ1 ∈ C ω (P) be such that X1 = λ1 X0 + μ1 U. γ

Let γ : I → R be a solution of x˙ = U (x) and τ1 (s) = T1 (γ (s)), where the function T1 is like in Theorem 4 II (b) when m = 1. Then γ (τ1 ) (s) = −

T0

∇λ1 (ϕ(t, γ (s), 0)) · U (ϕ(t, γ (s), 0)) e−

t 0

β0 (ϕ(v,γ (s),0))dv

dt.

(20)

0

Proof. We follow the idea of the proof of Theorem 2 in [12], which is presented also in the remarks after Theorem 13. So, first we compute αε and βε . We use that both the Lie bracket and the wedge product are bilinear, and that, by hypothesis, [X0, U ] = β0 U . Thus αε = ε

[X1 , U ] ∧ U X0 ∧ [X1 , U ] + O(ε 2 ) and βε = β0 + ε + O(ε 2 ). X0 ∧ U X0 ∧ U

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Now we use that, by hypothesis, [X0 , U ] = β0 U , X1 = λ1 X0 + μ1 U and the formula [λX, U ] = λ[X, U ] − (∇λ · U )X to compute α1 =

[X1 , U ] ∧ U [λ1 X0 , U ] ∧ U + [μ1 U, U ] ∧ U = = −∇λ1 · U. X0 ∧ U X0 ∧ U

In the sequel we write formulas (16) and (18) for m = 1, αε = εα1 + O(ε 2 ), βε = β0 + O(ε) and we get ετ1 (s) + O(ε 2 ) = ε

T0 +

O(ε)

{α1 (ϕ(t, γ (s), 0)) + O(ε)}e−

t 0

β0 (ϕ(v,γ (s),0))dv+O (ε)

dt.

0

Equating the coefficient of ε in the above equality, and using that α1 = −∇λ1 · U we obtain formula (20). 2 γ

Another approach for finding the derivative of the function τm (that appears in Theorem 4) is to apply Theorem 13 to the vector fields Xε = X(·, ε) and an well-chosen Uε such that Xε ∧ Uε = 0 in P and for any ε with |ε|  1. In this case one has to work with an orbit of x˙ = U (x, ε) as transversal section for the flow of (1) for fixed ε. The result follows. Theorem 15. Assume that there exists m ≥ 1 such that the first m Melnikov functions vanish identically for system (1). Let U0 ∈ C ω (P, R2 ) and β0 ∈ C ω (P) be such that [X0 , U0 ] = β0 U0 and X0 ∧ U0 = 0 in P. Let γ : I → P be a solution of x˙ = U0 (x), that is γ (s) = U0 (γ (s)). Assume also that there exist Uε = U (·, ε) ∈ C ω (P, R2 ) and αm ∈ C ω (P) such that U (x, ·) is analytic in a neighborhood of ε = 0, U (·, 0) = U0 , Uε (γ (s)) = U0 (γ (s)) for all s ∈ I , and [Xε , Uε ] ∧ Uε = ε m (X0 ∧ U0 )αm + O(ε m+1 ).

(21)

Then the functions T0 , T1 , ..., Tm−1 are constant. γ Let τm (s) = Tm (γ (s)), where the function Tm is like in Theorem 4 (b). Then γ (τm ) (s) =

T0

αm (ϕ(t, γ (s), 0))e−

t 0

β0 (ϕ(v,γ (s),0))dv

dt.

(22)

0

Proof. We intend to apply Theorem 13 to the analytic vector fields Xε and Uε when |ε|  1. First note that, since X0 ∧ U0 = 0 in P and Xε ∧ Uε = X0 ∧ U0 + O(ε) we have that, for |ε|  1, Xε ∧ Uε = 0 in P. An important consequence of the hypothesis is that γ : I → R2 is a common solution of x˙ = Uε (x) for all ε ∈ R. Hence the transversal section  can be chosen independent of ε. Define αε =

[Xε , Uε ] ∧ Uε , X ε ∧ Uε

βε =

Xε ∧ [Xε , Uε ] X ε ∧ Uε

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21

such that we have [Xε , Uε ] = αε Xε + βε Uε . All the above assure that the hypotheses of Theorem 13 are fulfilled and, moreover, the situation is similar with the one described just after the proof of Theorem 13. Thus formulas (16), (17) and (18) are valid. Using the above definition of αε and relation (21) from the hypothesis we obtain αε =

ε m (X0 ∧ U0 )αm + O(ε m+1 ) = ε m αm + O(εm+1 ). X ε ∧ Uε

We also have βε = β0 + O(ε). γ

We recall that τj (s) = Tj (γ (s)) with j ≥ 0. After introducing the above expression of αε and γ βε in (16), we deduce that τj (s) is constant for j = 0, 1, . . . , m − 1. By Theorem 4, we deduce that the functions T0 , T1 , . . . , Tm−1 are constant. Now we obtain expression (18) of τε (s) and we have

ε

m

γ (τm ) (s) + O(ε m+1 ) = ε m

T0

αm (ϕ(t, γ (s), 0))e−

t 0

β0 (ϕ(v,γ (s),0))dv

dt + O(εm+1 ).

0

Thus (22) is valid.

2

In the sequel we assure that, in fact, there exists a family of vector fields {Uε } as in Theorem 15. The proof is constructive. Theorem 16. Assume that there exists m ≥ 2 such that the first m Melnikov functions vanish identically, and the functions T0 , T1 , ..., Tm−1 are constant for system (1). Let U0 ∈ C ω (P, R2 ) be such that [X0 , U0 ] = 0 and X0 ∧ U0 = 0 in P. Let γ : I → P be a solution of x˙ = U0 (x). Then there exist δk ∈ C ω (P), 1 ≤ k ≤ m − 1, such that Uε = U0 + (εδ1 + · · · + ε m−1 δm−1 )X0 verifies the hypothesis of Theorem 15. Moreover, the functions δ1 , δ2 , . . . , δm−1 , αm can be found as follows. For each 1 ≤ k ≤ m − 1 there exists a function ak such that [j k (Xε ), j k−1 (Uε )] ∧ j k−1 (Uε ) = ε k (X0 ∧ U0 )ak + O(ε k+1 )

(23)

and the Cauchy Problem for the first order partial differential equation ∇u · X0 = −ak in P,

u| = 0

(24)

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has a unique solution, denoted δk . There exists αm such that [j m (Xε ), j m−1 (Uε )] ∧ j m−1 (Uε ) = ε m (X0 ∧ U0 )αm + O(εm+1 ). Proof. We prove by induction that, for each 1 ≤ k ≤ m − 1 there exists ak and δk like in the statement of the theorem. For k = 1 we have [j 1 (Xε ), j 0 (Uε )] ∧ j 0 (Uε ) = [X0 + εX1 , U0 ] ∧ U0 = ε[X1 , U0 ] ∧ U0 . Hence a1 =

[X1 , U0 ] ∧ U0 . X 0 ∧ U0 γ

Moreover, Theorem 15 relates the derivative of τ1 (s) = T1 with a1 . More precisely, since T1 is a constant, the following relation holds

T0 a1 (ϕ(t, γ (s), 0)) dt = 0 for all s ∈ I.

(25)

0

Relation (25) together to the fact that  is transversal to X0 assures, by the characteristics method (see also [10]) that (24) has a unique solution when k = 1, denoted δ1 . Now let 2 ≤ k ≤ m − 2 be fixed and assume that (induction hypothesis) there exist ai and δi like in the statement of the theorem for all 1 ≤ i ≤ k. We aim to prove that there exist ak+1 and δk+1 like in the statement. Compute [j k+1 (Xε ), j k (Uε )] ∧ j k (Uε )

(26)

= [j k (Xε ) + O(ε k+1 ), j k−1 (Uε ) + ε k δk X0 ] ∧ (j k−1 (Uε ) + ε k δk X0 ) = ε k (X0 ∧ U0 )ak + ε k [X0 , δk X0 ] ∧ U0 + O(ε k+1 ) = ε k (X0 ∧ U0 )(ak + ∇δk · X0 ) + O(ε k+1 ) = ε k · 0 + O(εk+1 ). Then there exists ak+1 such that (23) is valid for k + 1. In order to prove that the Cauchy Problem ∇u · X0 = −ak+1 in P, u| = 0, has a unique solution, as we discussed above, is sufficient to prove that

T0 ak+1 (ϕ(t, γ (s), 0)) dt = 0 for all s ∈ I.

(27)

0

For this we apply Theorem 15 for Xε and U˜ ε = j k (Uε ). Remind that k + 1 ≤ m − 1, hence the Melnikov functions up to order (k + 2) vanish identically and the functions T0 , T1 , . . . , Tk+1 are constant.

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23

First note that U˜ ε (γ (s)) = U0 (γ (s)) for all s ∈ I and all ε ∈ R since, by construction, δi | = 0. Now compute [Xε , U˜ ε ] ∧ U˜ ε = = [j k+1 (Xε ), j k (Uε )] ∧ j k (Uε ) + O(ε k+2 ) = ε k+1 (X0 ∧ U0 )ak+1 + O(ε k+2 ). Then Theorem 15 gives that (27) holds true. The proof by induction is finished. The existence of αm follows by noticing that computations (26) work also for k = m − 1.

2

6. Application to a quadratic perturbation of the linear center In order to apply our results, we look for systems of the form (1) with a focus at the origin for ε = 0. In our example we consider a quadratic perturbation of the linear center X0 = (−y, x). The critical periods of quadratic perturbations of the linear center such that the perturbed family is a center for |ε|  1 have been widely studied, see [20] and the references therein. In this situation, at most two critical periods appear. This maximum value is proved when bifurcating from the inner boundary of the period annulus and also from the period annulus, and it is conjectured when bifurcating from the outer boundary of the period annulus. We consider a quadratic perturbation of the linear center inside a family of the form (1) that may have a focus at the origin. In this case, we prove, under the hypothesis used throughout this paper, that at most two critical points of the flight return time function can appear for a particular example. Theorem 17. Consider system √ 1 (15 + 105) εx 2 + (2a4 ε 4 − c6 ε 6 )xy − (ε + d3 ε 3 )y 2 , 4 y˙ = x + ε 7 λ7 y + a4 ε 4 x 2  √ √ 5 5 + (11 + 105)ε + (17 − 105)d3 ε 3 + A5 ε 5 xy − a4 ε 4 y 2 , 4 46

x˙ = −y + ε 7 λ7 x +

(28)

where λ7 , a4 , c6 , d3 , A5 are real constants and |ε|  1. This system has the first 6 Melnikov functions that vanish identically and M7 (h) can have 0, 1 or 2 simple zeros depending on the values of the parameters. Moreover T0 = 2π , Tj are constant for j = 1, . . . , 5, and the function T6 can have 0, 1 or 2 simple critical points, depending on the values of the parameters, and it is not a constant. Consequently system (28) has at most two limit cycles bifurcating from the period annulus of the linear center and at most two orbits of the linear center are continuable critical orbits in system (28). Proof. System (28) can be embedded in the general Bautin form of a quadratic system x˙ = −y + λx − bx 2 − cxy − dy 2 , y˙ = x + λy + ax 2 + Axy − ay 2 , (29) √ with λ = ε 7 λ7 , b = −(15 + 105)ε/4, c = −(2a4 ε 4 − c6 ε 6 ), d = (ε + d3 ε 3 ), a = a4 ε 4 and √ √ 5 (17 − 105)d3 ε 3 + A5 ε 5 . As its is proved in [2], see also [23] and A = 54 (11 + 105)ε + 46 the references there in, the Poincaré–Liapunov constants of the quadratic system (29) are given by

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V1 V3 V5 V7

= λ, = (2a + c)(b + d), = (b + d)a(A − 2b)(A + 3b + 5d), = (b + d)2 a(A − 2b)(a 2 + 2d 2 + bd).

We expand the Poincaré–Liapunov constants in powers of ε once we substitute the values corresponding to system (28) and we denote by v¯ 2j +1,r the coefficient of ε r in the expansion of V2j +1 , where j = 0, 1, 2, 3 and r ≥ 1. Consider H = (x 2 + y 2 )/2, which is a first integral of the linear center, and the section  = {(x, 0) : x > 0} where only values of x close to 0 for which the section is transversal are considered. The section  is parameterized by the values of H = h. By Theorem 2.2 in [5], there are 4 linearly independent functions h2j +1 B2j +1 (h) which are analytic in [0, +∞) and with B2j +1 (0) a nonzero constant for j = 0, 1, 2, 3, such that the Melnikov function of system (28) writes as Mk (h) =

4 

v¯2j +1,k h2j +1 B2j +1 (h).

(30)

j =0

Since v¯2j +1,r = 0 for r = 1, 2, . . . , 6, we deduce that the first 6 Melnikov functions vanish identically. Moreover, the Melnikov functions of a perturbation of the linear center can be shown to be polynomials in h and after some computations we can find that M7 (h) = 2λ7 h −

√ √ 1 5 (11 + 105)c6 h3 − (2103 + 205 105)a4 h5 . 16 192

It is clear that M7 (h) can have 0, 1 or 2 simple zeros with h > 0 and close enough to h = 0, depending on the values of the parameters. Hence at most two limit cycles can bifurcate from the orbits of the linear center inside our system. Note that when M7 (h) vanishes identically, that is λ7 = c6 = a4 = 0, we have that system (28) is a reversible center by the symmetry (x, y, t) → (−x, y, −t). It is clear that the period of the linear center is T0 = 2π . In order to find the functions Ti (h), for i = 1, 2, . . . , 6, we use Theorems 15 and 16. We have that X0 = (−y, x) and we take U0 = (x, y) such that [X0 , U0 ] = 0 and X0 ∧ U0 = 0 in P = R2 \ {(0, 0)}. We take γ (s) = (es , 0), s ∈ R. We can find the functions δk ∈ C ω (P), 1 ≤ k ≤ 5, such that Uε = U0 + (εδ1 + ε 2 δ2 + ε 3 δ3 + ε 4 δ4 + ε 5 δ5 )X0 satisfies [Xε , Uε ] ∧ Uε = ε 6 (X0 ∧ U0 )α6 + O(ε7 ) where Xε is the vector field associated to (28). We have that √  √ (9 + 105)x 3 1 + (12 + 105) x 2 + y 2 . δ1 (x, y) = −x − 3 3(x 2 + y 2 ) We do not write here the expressions of δk , for k = 2, 3, 4, 5, and α6 because they are very large. By Theorem 15, we deduce that the functions Ti , i = 1, 2, . . . , 5, are constant. Now applying (22), where ϕ(t, γ (s), 0) = (es cos t, es sin t) is the flow of the linear center, we have that

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γ 

τ6 (s)

25

2π =

α6 (ϕ(t, γ (s), 0)) dt. 0

 γ Recall that τ6 = T6 (γ (s)). We obtain that 

γ 

τ6

(s) =

√

√

360 3 105 − 97 d32 − 529 39 + 5 105 A5 12696 √

5 − 15211 + 1481 105 d3 π e2s 368 √

5 − 368921 + 36003 105 π e3s . 192

π es

It is clear that by different values of the parameters d3 and A5 this function can have 0, 1 or 2 zeros and it is never null. 2 Acknowledgments We would like to thank the reviewer of this article for his/her useful comments. The authors are partially supported by a MICINN/FEDER grant number MTM 2017-84383-P and by a AGAUR (Generalitat de Catalunya) grant number 2017 SGR 1276. References [1] A. Algaba, M. Reyes, Isochronous centres and foci via commutators and normal forms, Proc. Roy. Soc. Edinburgh Sect. A 138 (1) (2008) 1–13. [2] N.N. Bautin, On the Number of Limit Cycles Which Appear with the Variation of Coefficients from an Equilibrium Position of Focus or Center Type, Amer. Math. Soc. Transl., vol. 100, 1954, pp. 1–19. [3] A. Buic˘a, On the equivalence of the Melnikov functions method and the averaging method, Qual. Theory Dyn. Syst. 16 (2017) 547–560. [4] A. Buic˘a, Perturbed normalizers and Melnikov functions, J. Math. Anal. Appl. 464 (2018) 266–273. [5] A. Buic˘a, J. Giné, M. Grau, Essential perturbations of polynomial vector fields with a period annulus, Commun. Pure Appl. Anal. 14 (3) (2015) 1073–1095. [6] C. Chicone, Ordinary Differential Equations with Applications, Springer, 2006. [7] C. Chicone, W. Liu, Asymptotic phase revisited, J. Differential Equations 204 (2004) 227–246. [8] J.-P. Françoise, The Successive Derivatives of the Period Function of a Plane Vector Field, J. Differential Equations 146 (1998) 320–335. [9] E. Freire, A. Gasull, A. Guillamon, First derivative of the period function with applications, J. Differential Equations 204 (2004) 139–162. [10] E. Freire, A. Gasull, A. Guillamon, A characterization of isochronous centres in terms of symmetries, Rev. Mat. Iberoam. 20 (2004) 205–222. [11] E. Freire, A. Gasull, A. Guillamon, Limit cycles and Lie symmetries, Bull. Sci. Math. 131 (2007) 501–517. [12] A. Gasull, J. Yu, On the critical periods of perturbed isochronous centers, J. Differential Equations 244 (2008) 696–715. [13] J. Giné, M. Grau, Characterization of isochronous foci for planar analytic differential systems, Proc. Roy. Soc. Edinburgh Sect. A 135 (2005) 985–998. [14] M. Grau, J. Villadelprat, Bifurcation of critical periods from Pleshkan’s isochrones, J. Lond. Math. Soc. 81 (2010) 142–160. [15] A. Guillamon, G. Huguet, A computational and geometric approach to phase resetting curves and surfaces, SIAM J. Appl. Dyn. Syst. 8 (3) (2009) 1005–1042. [16] Y. Ilyashenko, S. Yakovenko, Lectures on Analytic Differential Equations, Amer. Math. Soc., 2008.

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[17] W.S. Loud, Behavior of the period of solutions of certain plane autonomous systems near centers, Contrib. Differ. Equ. 3 (1964) 21–36. [18] F. Mañosas, D. Rojas, J. Villadelprat, Study of the period function of a two-parameter family of centers, J. Math. Anal. Appl. 452 (1) (2017) 188–208. [19] P. Mardeši´c, C. Rousseau, B. Toni, Linearization of isochronous centers, J. Differential Equations 121 (1) (1995) 67–108. [20] P. Mardeši´c, D. Marín, J. Villadelprat, The period function of reversible quadratic centers, J. Differential Equations 224 (1) (2006) 120–171. [21] M. Sabatini, On the period function of planar systems with unknown normalizers, Proc. Amer. Math. Soc. 134 (2006) 531–539. [22] M. Sabatini, Characterizing isochronous centers by Lie brackets, Differ. Equ. Dyn. Syst. 5 (1997) 91–99. [23] D. Schlomiuk, Algebraic and geometric aspects of the theory of polynomial vector fields, in: Bifurcations and Periodic Orbits of Vector Fields, Montreal, PQ, 1992, in: NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 408, Kluwer Acad. Publ., Dordrecht, 1993, pp. 429–467. [24] M. Villarini, Regularity properties of the period function near a centre of a planar vector field, Nonlinear Anal. TMA 19 (1992) 787–803.