The cyclicity of the period annulus of a reversible quadratic system

We prove that perturbing the periodic annulus of the reversible quadratic polynomial differential system $\dot x=y+ax^2$, $\dot y=-x$ with a ≠ 0 inside the class of all quadratic polynomial differential systems we can obtain at most two limit cycles, including their multiplicities. Since the first integral of the unperturbed system contains an exponential function, the traditional methods cannot be applied, except in Figuerasa, Tucker and Villadelprat (2013, J. Diff. Equ., 254, 3647–3663) a computer-assisted method was used. In this paper, we provide a method for studying the problem. This is also the first purely mathematical proof of the conjecture formulated by Dumortier and Roussarie (2009, Discrete Contin. Dyn. Syst., 2, 723–781) for q ⩽ 2. The method may be used in other problems.


Introduction and statement of the main results
We recall that a centre of a planar differential system is a singular point p of the system having a neighbourhood filled up of periodic orbits with the unique exception of the point p. The period annulus of a centre is the maximal region filled up with the periodic orbits surrounding the centre.
There is a big programme whose objective is to find the exact upper bound for the number of limit cycles that can bifurcate from the periodic orbits of the period annuli of the quadratic polynomial differential systems under quadratic perturbations, see for instance the second part of the book of Christhopher and Li [3]. This upper bound is called the cyclicity of the period annulus. This programme started with Arnold [1,2] and has produced more than 100 articles, see for instance the references of [3].
Here, we contribute to this programme determining this upper bound for the period annulus of the centre of the quadratic polynomial differential systemṡ with a = 0. We note that to study the cyclicity of the period annulus of system (1.1) is equivalent to study the cyclicity of the period annulus of the systeṁ Indeed, doing the change of variables (X, Y ) → (x, y) where X = 4x/a and Y = 4y/a system (1.1) becomes system (1.2). System (1.2) has the first integral and the corresponding integrating factor R(y) = 8e 8y . The phase portrait of system (1.2) in the Poincaré disc is shown in figure 1. This phase portrait has a unique finite singular point, the centre localized at the origin of coordinates. It has two pairs of infinite singular points localized at the endpoints of the x and y axes. At the endpoint of the positive x-half-axis there is a hyperbolic stable node, at the endpoint of the negative x-half-axis there is a hyperbolic unstable node, at the endpoints of the y-axis there is a nilpotent saddle, having a hyperbolic sector at the endpoint of the positive y-half-axis and three hyperbolic sectors at the endpoint of the negative y-axis. For the definitions of first integral and integrating factor see chapter 8, for the definition of the Poincaré disc see chapter 5 and for the definitions of hyperbolic and nilpotent singular points see chapters 2 and 3 respectively of [4]. The boundary of the period annulus of the centre of system (1.2) localized at the origin of coordinates is the parabola y = −4x 2 + 1/8. Then the period annulus can be expressed by In what follows, we will say simply quadratic system instead of quadratic polynomial differential system. It is known, see for instance [7], that any reversible quadratic system can be written in the complex formż = −iz + az 2 + 2|z| 2 + bz 2 where z = x + iy, or in the real forṁ where a and b are real parameters. When a = b = 1 the reversible quadratic system (2.1) becomes system (1.2). Our main result is the following one.  [5]. Dumortier and Roussarie formulated a conjecture on p. 726 of [5], that {J 1 (h), J 3 (h), . . . ,J 2q+1 (h)} forms a strict Chebyshev system for h ∈ (0, 1/2) and for any integer q 0. This conjecture is obviously true for q = 0. We give a positive answer to this conjecture for q = 1 in lemma 2.4 and for q = 2 in lemmas 2.5-2.7, by using purely mathematical method. Note also that Figuerasa, Tucker and Villadelprat in [6] gave a proof of this conjecture for q 2 by using theoretical analysis and computations by computer, that are based on computer-assisted techniques. For example, the computations for the proof of a lemma take six and a half hours on a desktop computer with a 2.8 GHz CPU, see remark 4.11 of [6].

Proof of theorem 1.1
We first state a result by Iliev, see statement (ii).
First, using integration by parts we have γ h e 8y dx = −I 1 (h).
If α < 0, we denote x 0 the positive root of 3x 2 + α = 0. Suppose that the intersection points of the curve γ h and the axis {(x, z) |z = 0} are (±x M (h), 0), the most left and the most right points of γ h , then by (2.9) γ h has two branches z = z i (x, h) with (2.14) Note that γ h tends to the origin as h → −1/8 + , monotonically expands as h increases from −1/8, and tends to infinity in ±x direction as  z-axis, so we will only consider the side x 0. We divide the integral form of M (h)/2 into two parts as follows: we use the last quality of (2.11) in the first integral to change dz to dx; in the second integral, Making one more derivative with respect to h by using (2.10), we have (2.14)), h < 0, and 3x 2 2 − α > 0. Besides, 3x 2 0 + α = 0 and ∂z2(x0,h) ∂h − ∂z1(x0,h) ∂h is bounded by (2.10), so the above middle term is zero and we will directly omit the similar terms in further calculations.
If α 3 = 0 in (2.2), without loss of generality we consider where α and β are arbitrary constants. From (2.7) we have Proof. When α 0, it is obvious that M (h) < 0, because dz/x < 0 along γ h . Hence we suppose α < 0 in the rest part, and denote the only positive root of F (x) = 0 by x 0 , and If h ∈ (h 0 , 0), then x M (h) > x 0 , the proof is completely similar to the proof of lemma 2.4, so we only list some different computations. We first rewrite M (h) as Then by (2.10) we obtain Proof. In this case, F (x) = 0 has exactly one positive root x 0 . Similar to the proof of lemma 2.5, we denote If h ∈ (h 0 , 0), then x M (h) > x 0 , we get the same expressions (2.18) and (2.19). Now β < 0, α 0, hence F (x) < 0 on (0, x 0 ); when x x 0 , we have F (x) 0, then G(x) = 3F (x) − 6βx 2 − 4α > 0, hence M (h) < 0. In any case we obtain that M (h) has at most one zero for h ∈ (−1/8, 0).
Finally we consider the most complicated case β < 0 and α > 0. Thus, (−h)M (h), hence M (h), has at most one zero on (h 1 , 0). Since we have proved that M (h) > 0 on (h 0 , h 1 ], we get that M (h) has at most one zero on (h 0 , 0), hence M (h) has at most two zeros on (h 0 , 0). assertion 2 is proved.
Supping up the results in assertions 1 and 2, we obtain that M (h) has at most two zeros on (−1/8, 0). All multiplicities of the zeros are taken into account.