This is a preprint of: “Centers of planar generalized Abel equations”, Jaume Llibre, Clàudia Valls, J. Differential Equations, vol. 268, 6481–6487, 2020. DOI: [10.1016/j.jde.2019.11.046] CENTERS OF PLANAR GENERALIZED ABEL EQUATIONS JAUME LLIBRE1 AND CLÀUDIA VALLS2 Abstract. We deal with the differential equation dr ṙ = = a(θ)rn + b(θ)rm , dθ where (r, θ) are the polar coordinates in the plane R2 , m and n are integers such that m > n ≥ 2, and a, b are C 1 functions. Note that when n = 2 and m = 3 we have an Abel differential equation. For this class of generalized Abel equations we characterize a new family of centers. 1. Introduction and statement of the results Consider the generalized Abel equation dr (1) ṙ = = a(θ)rn + b(θ)rm , dθ defined in the plane (r, θ) ∈ [0, +∞) × S1 in polar coordinates where S1 = R/(2πZ). Here m and n are integers such that m > n ≥ 2, θ ∈ [−π, π] and a, b are C 1 - functions. The origin of the plane is a center for the differential equation (1) if there is a neighborhood of it where all the solutions are periodic except the equilibrium point at the origin. When n = 2 and m = 3 the differential equation (1) is a particular family of Abel equations. In fact the Abel equations are of the form ṙ = a(θ) + b(θ)r + c(θ)r2 + d(θ)r3 , and they appeared by the first time in the works of Niels Henryk Abel, see [7]. Today there are more than 1400 papers in MathSciNet with the name “Abel equation” in their tittle, see for instance the papers [1, 2, 3, 5, 6, 8] for results on centers in the Abel equations and the references quoted therein. The main objective of this work is to provide a new family of centers in the generalized Abel equation (1). Thus, our main result is the following. Theorem 1. If a(θ) and b(θ) are C 1 odd functions and m ≥ 2n−1, then the origin r = 0 of the differential equation (1) is a center. The proof of Theorem 1 is given in the next section. We note that Theorem 1 in the particular case n = 2 and m = 3, i.e. when equation (1) is an Abel equation, already was obtained in [4] by Araujo, Lemos and Alves. 2010 Mathematics Subject Classification. 34C25, 34A34. Key words and phrases. Centers, generalized Abel equations. 1 http://www.gsd.uab.cat 2 J. LLIBRE AND C. VALLS 2. Proof of Theorem 1 In order to prove Theorem 1 we introduce some auxiliary results that will be used in its proof. Proposition 2. The origin r = 0 of equation (1) is a center if and only if ∫ π ∫ π (2) a(θ) dθ = 0 and b(θ)r(θ; ρ)m−n dθ = 0, −π −π for |ρ| < ρ0 with ρ0 sufficiently small where r(θ; ρ) is the solution of equation (1) such that r(−π; ρ) = ρ. Proof. First we prove sufficiency. Note that dividing equation (1) by rn and inte- grating it, we obtain ∫ θ ∫ θ 1 1 − r1−n (θ; ρ) = a(s) ds + b(s)r(s; ρ)m−n ds − , n−1 −π −π (n − 1)ρn−1 where ρ is the initial condition. So we have ρn−1 rn−1 (θ; ρ) = ∫θ ∫θ , 1 − (n − 1)ρn−1 [ −π a(s) ds + −π b(s)r(s; ρ)m−n ds] which yields ρ (3) r(θ; ρ) = ( ∫θ ∫θ ) n−1 1 . 1 − (n − 1)ρn−1 [ −π a(s) ds + −π b(s)r(s; ρ)m−n ds] Taking θ = π in the previous equation we have ρ r(π; ρ) = ( ∫π ∫π ) 1 . 1 − (n − 1)ρn−1 [ −π a(s) ds + −π b(s)r(s; ρ)m−n ds] n−1 If (2) holds, then x(π; ρ) = ρ, and the sufficiency in the theorem follows. Assume now that equation (1) has a center on r = 0. We first note that any solution of equation (1), r(θ; ρ) can be expanded in power series in ρ for |ρ| < ρ0 with ρ0 sufficiently small and θ ∈ [−π, π] in the form (4) r(θ; ρ) = r0 (θ) + r1 (θ)ρ + r2 (θ)ρ2 + . . . Clearly r0 (θ) = 0 because r(θ; 0) = 0. Substituting (4) into equation (1) we get ṙ1 (θ) = 0 because n ≥ 2, and since r(π; ρ) = ρ we must have r1 (θ) = 1. Hence we have (5) r(θ; ρ) = ρ + r2 (θ)ρ2 + . . . Note that a sufficient and necessary condition for r = 0 to be a center of (1) is that ri (π; ρ) = 0 for i = 2, 3, . . .. Substituting (4) into (1) and computing the coefficients in ρj for j ≥ 2 we get that ṙi (θ; ρ) = 0 for i = 2, . . . , n − 1. Since ri (π; ρ) = 0 we must have ri (θ; ρ) = 0 for i = 2, . . . , n − 1. On the other hand ∫ θ ṙn (θ; ρ) = a(θ) that is rn (θ; ρ) = a(θ) dθ. −π CENTERS OF PLANAR GENERALIZED ABEL EQUATIONS 3 ∫π Since rn (π; ρ) = 0 we obtain that −π a(s) ds = 0. But then, from (3) we readily ∫π get that −π b(s)r(s; ρ)m−n ds = 0, which proves the necessity. This concludes the proof of the proposition.  Note that by equation (3) a solution of equation (1) is equivalent to a solution of the integral equation ρ (6) r(θ; ρ) = ( ∫θ ) 1 , 1 − (n − 1)ρ n−1 (a(s) + b(s)r(s; ρ)m−n ) ds n−1 −π for θ ∈ [−π, π], where r(−π; ρ) = ρ. We denote m−n (7) y(θ) = rn−1 (θ; ρ) and ℓ= . n−1 Then equation (6) becomes ρn−1 (8) y(θ) = ∫θ . (1 − (n − 1)ρn−1 −π (a(s) + b(s)y(s)ℓ ) ds We define the operator T : BM → C([−π, π]) by ρn−1 T (y)(θ) = ∫θ , 1 − (n − 1)ρn−1 −π (a(s) + b(s)y(s)ℓ ) ds for θ ∈ [−π, π], where BM = {y ∈ E : ∥y∥∞ ≤ M }, being E the closed subspace of C([−π, π]) defined by E = {y ∈ C([−π, π]) : y is an even function}. Note that if the operator T has a unique fixed point, i.e., a unique y ∈ BM such that T (y)(θ) = y then (8) has an even solution. As usual ∥r∥∞ = maxθ∈[−π,π] |r(θ)|. We take the notation ρn−1 Jθ (y) = ∫θ . 1 − (n − 1)ρn−1 −π (a(s) + b(s)y(s)ℓ ) ds We define A = max |a(θ)|, B = max |b(θ)|. θ∈[−π,π] θ∈[−π,π] Proposition 3. For { ( M )1/(n−1) } 1 (9) 0 ≤ ρ < min ( )1/(n−1) , , 4π(n − 1)(A + BM ℓ ) 2 the operator T : BM → C([−π, π]) is continuous and compact. Moreover T (BM ) ⊂ BM . Proof. For each y1 , y2 ∈ C([−π, π]), we have ∫ θ |T (y1 )(θ) − T (y2 )(θ)| = (n − 1)Jθ (y1 )Jθ (y2 ) b(s)(y1 (s)ℓ − y2 (s)ℓ ) ds , −π 4 J. LLIBRE AND C. VALLS for θ ∈ [−π, π]. Note that since 1 0≤ρ≤ ( )1/(n−1) , 4π(n − 1)(A + BM ℓ ) we have ∫ θ n−1 1 |(n − 1)ρ (a(s) + b(s)y(s)ℓ ) ds| ≤ 2π(n − 1)ρn−1 (A + BM ℓ ) < . −π 2 Therefore ∫ θ (10) 1 − (n − 1)ρn−1 (a(s) + b(s)y(s)ℓ ) ds > 1/2, −π which yields Jθ (y) ≤ 2ρn−1 . Moreover, since m ≥ 2n − 1 we have that ℓ ≥ 1 and by the Mean Value theorem we get |y1 (s)ℓ − y2 (s)ℓ | ≤ ℓM ℓ−1 ∥y1 − y2 ∥∞ . Hence ∫ θ 2n−2 |T (y1 )(θ) − T (y2 )(θ)| ≤ 4(n − 1)ρ b(s)(y1 (s)ℓ − y2 (s)ℓ ) ds −π ≤ 8π(n − 1)ρ2n−2 BℓM ℓ−1 ∥y1 − y2 ∥∞ , for each θ ∈ [−π, π]. So, ∥T (y1 ) − T (y2 )∥∞ ≤ 8π(n − 1)ρ2n−2 BℓM ℓ−1 ∥y1 − y2 ∥∞ for all y1 , y2 ∈ C([−π, π]). Hence the operator T is continuous. For proving that the operator T is compact, we shall see that T is bounded and equicontinuous. Now we prove that it is bounded. Indeed, by (10) and the condition in ρ in (9) we have |T (y)(θ : −π)| ≤ 2ρn−1 , for all y ∈ BM and θ ∈ [−π, π] and so (11) ∥T (y)∥∞ ≤ 2ρn−1 for all y ∈ BM , proving that the operator T is bounded. Now we show that T is equicontinuous. For each θ1 , θ2 ∈ [−π, π] (that without loss of generality we can assume that θ2 > θ1 ), and any y ∈ BM , we have |T (y)(θ1 ) − T (y)(θ2 )| = (n − 1)Jθ1 (y)Jθ2 (y) × ∫ θ1 ∫ θ2 (a(s) + b(s)y(s)ℓ ) ds − (a(s) + b(s)y(s)ℓ ) ds −π −π ∫ θ2 ≤ 4(n − 1)ρ2n−2 |a(s) + b(s)y(s)ℓ | ds θ1 ≤ 4(n − 1)ρ2n−2 (A + BM ℓ )|θ2 − θ1 |. Therefore T (BM ) is an equicontinuous subset of C([−π, π]). By Ascoli-Arzela The- orem (see for instance [9]) we have that T : BM → C([−π, π]) is compact. Since the functions a(θ) and b(θ) are odd and by assumptions y(θ) is an even function (and so also y(θ)ℓ is an even function), we have that a(s) + b(s)y(s)ℓ is odd. Taking into account that the integral of an odd function is an even function CENTERS OF PLANAR GENERALIZED ABEL EQUATIONS 5 ∫θ we conclude that −π (a(s) + b(s)y(s)ℓ ) ds is an even function in θ. Hence, for each y ∈ E we have that T (y)(θ) = T (y)(−θ) for all θ ∈ [−π, π]. Therefore, T (y) ∈ E for every y ∈ E. Moreover, by (9) and (11) we have that ∥T (y)∥∞ ≤ 2ρn−1 < M for all y ∈ BM . So T : BM → BM is well defined. This concludes the proof of the proposition.  Proposition 4. Under the assumptions of Theorem 1 there are infinitely many closed even solutions r(θ; ρ) of system (1) for ρ satisfying (9). Proof. It follows from Proposition 3 that the operator T : BM → BM is well defined, continuous and compact. By the Schauder fixed point Theorem, see [9], the operator T has a fixed point y satisfying ρ T (y)(θ) = y(θ) = ∫θ 1 − (n − 1)ρ n−1 (a(s) + b(s)y(s)ℓ ) ds −π and y(−π) = ρn−1 for each ρ satisfying (9). From (7) there exists r(θ; ρ) such that ρ r(θ; ρ) = ( ∫θ ) 1 1 − (n − 1)ρ n−1 (a(s) + b(s)r(s; ρ)m−n ) ds n−1 −π and r(−π; ρ) = ρ. Note that r(−θ; ρ) = r(θ; ρ) and so the solution is closed and even. In short, there are many closed even solutions of system (1) near the origin.  Proof of Theorem 1. To prove Theorem 1 we first show that if r̄(θ; ρ) is a solution of equation (1) that satisfies r̄(−π; ρ) = ρ with ρ satisfying (9), then r̄(θ; ρ) is closed and even. Indeed, by Proposition 4 there is r(θ; ρ) a closed even solution of system (1) such that r(−π; ρ) = ρ, and by the uniqueness of solutions of an ordinary differential equation, we obtain that r̄(θ; ρ) = r(θ; ρ). Hence if a and b are odd functions in the variable θ, then each solution of equation (1) with initial condition ρ satisfying (9) is a closed even solution. Hence, for any ρ satisfying (9) we have ∫ π ∫ π a(s) ds = 0 and b(s)r(s; ρ)m−n ds = 0. −π −π Therefore it follows from Proposition 2 that r = 0 is a center for equation (1).  Acknowledgements The first author is partially supported by a FEDER-MINECO grant MTM2016- 77278-P, a MINECO grant MTM2013-40998-P, and an AGAUR grant number 2014SGR-568. The second author is partially supported by FCT/Portugal through UID/MAT/04459/2013. References [1] M.J. Alvarez, J.L. 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Houzel, The work of Niels Henryk Abel, The Legacy of Niels Henryk Abel, The Abel Bicentennial, Oslo, 2002. [8] F. Pakovich, Solution of the parametric center problem for the Abel differential equation, J. Eur. Math. Soc. (JEMS) 19 (2017), 2343–2369. [9] E. Zeidler, Nonlinear Functional Analysis and its Applications, Springer-Verlag, New York, 1986. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat 2 Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, Av. Rovisco Pais 1049–001, Lisboa, Portugal E-mail address: cvalls@math.ist.utl.pt