The return map for a planar vector field with nilpotent linear part: a direct and explicit derivation

Using a direct approach the return map near a focus of a planar vector field with nilpotent linear part is found as a convergent power series which is a perturbation of the identity and whose terms can be calculated iteratively. The first nontrivial coefficient is the value of an Abelian integral, and the following ones are explicitly given as iterated integrals.


Introduction
The study of planar vector fields has been the subject of intense research, particularly in connection to Hilbert's 16th Problem. Significant progress has been made in the geometric theory of these fields, as well as in bifurcation theory, normal forms, foliations, and the study of Abelian integrals [6], [20].
The Poincaré first return maps have been studied in view of their relevance for establishing the existence of closed orbits, and also due to their large number of applications (see for example [12] and references therein), and also in connection to o-minimality [16].
A fundamental result concerns the asymptotic form of return maps states that if the singular points of a C ∞ vector field are algebraically isolated, there exists a semitransversal arc such that the return map admits an asymptotic expansion is positive powers of x and logs (with the first term linear), or has its principal part a finite composition of powers and exponentials [13], [17].
The present paper studies an example of a field with nilpotent linear part, near a focus. The main goal is to establish techniques that allow to deduce the return map as a suitable series which can be calculated algorithmically and can be used in numerical calculations.

Main result
The paper studies the return map for the system which has a nilpotent linear part (both eigenvalues are zero). This is one of the simplest examples of systems in this class [19], and for which there are (to the author's knowledge) no methods available to generate the return map.
The main result is the following: The solution of (1) satisfying X(0) = ǫ, Y (0) = 0 first returns to the positive X-axis at the valueX satisfying which is a convergent series.
The coefficients X n can be calculated iteratively. In particular, where c n = v n (1) with v n given by

Proof of Proposition 1
The proof of Proposition 1 also provides an algorithm for calculating iteratively the coefficients X n .
3.1. Normalization. It is convenient to normalize the variables X, Y, t so that the constant ǫ appears as a small parameter in the equation: with X = ǫx, Y = 2 −1/2 ǫ 2 y, τ = 2 −3/2 ǫt the system (1) becomes where α is the small parameter: While the initial condition X(0) = ǫ becomes x(0) = 1, it is useful to study solutions of (5) with the more general initial condition x(0) = η with η in a neighborhood of 1.
Remark. The system (5) has the form dH + αω = 0 with H = y 2 + x 4 , ω = y 3 dx and α a small parameter. The recent result [9] gives a formalism for finding the return map for this type of systems; see also [11]. The present construction is concrete, explicit, and suitable for numerical calculations.
Note the following Lyapunov function for (5): Since the set {y = 0} contains no trajectories besides the origin (which is the only equilibrium point of (5)), then the origin is asymptotically stable by the Krasovskii-LaSalle principle.
Since y ′ (0) > 0 and x ′ (0) = 0, x ′′ (0) < 0 then y increases and x decreases for small τ > 0. This monotony must change due to (7), and this can happen only at some point where y = 0 or where 4x 3 = αy 3 , whichever comes first. Since y increases, then the first occurrence is a point where 4x 3 = αy 3 . At this point x ′ < 0 so x continues to decrease, while y ′′ = −24x 2 y < 0 so y has a maximum, and will continue by decreasing. Again, the monotony must change due to (7) and the path cannot cross again the line 4 1/3 x = α 1/3 y before the monotony of x changes, therefore the the next change of monotony happens for y = 0, a point where, therefore x < 0; denote this value of x by −η.
Solutions (x(τ ), y(τ )) of (5) provide smooth parametrizations for solutions y(x) of Note that following the path (x(τ ), y(τ )) one full rotation around the origin corresponds to considering a positive solution of (8), followed by a negative one.
3.3. Positive solutions of (8) for x > 0. Lemma 2 shows that there exists a unique solution y ≥ 0 of (8) so that y(η) = 0, that this solution is defined for x ∈ [0, η] and establishes an iterative procedure for calculating this solution.
To prove Lemma 3 multiply (12) by ξ and integrate; we obtain that v is a fixed point Moreover, the operator J is a contraction on B m . Indeed, using the estimate Therefore the operator J has a unique fixed point, which is the solution v(ξ; δ).
Substitution of (14) in (p − v) 3/2 followed by expansion in power series in δ give In particular, From (12) we obtain the recursive system (for n ≥ 1 and with R 0 = 1), with the only solution with v n (0) = 0 given recursively by In particular, we have (2), (3).
The following gathers the conclusions of the present section.

Solutions of (8) in other quadrants and matching.
3.4.1. Solutions in other quadrants. We found an expression for the solution y(x) of (8) for x > 0 and y > 0. In a similar way, expressions in the other quadrants can be found. However, taking advantage of the discrete symmetries of equation (8), these solutions can be immediately written down as follows. (16), defined for x ∈ [0, η], with y 1 (η) = 0 and y 1 > 0 for α > 0. Then: is also a solution of (8), defined for x ∈ [−η, 0]; we have y 2 (−η) = 0 and y 2 ≥ 0 for α > 0.

Proof.
We need to findη =η(η, α) so that Note that the function F above is the same as (19). By Lemma 5 the present Lemma follows.
3.5. The first return map. Let η ∈ [1/2, 3/2] and α 2 as in Lemma 6. Thenη given by Lemma 6 is the first return to the positive x-axis of the solution with x(0) = η, y(0) = 0 and it is analytic in α and η, therefore, by (13), it is analytic in ǫ for fixed η.
Remark. The first coefficient of the return map (21) is, up to a sign, the Melnikov integral of the system (5), see [9]; of course, the present results are in agreement with this fact (see §4 for details).
Acknowledgement. The author is grateful to Chris Miller for suggesting the problem.

Appendix
With the notation H = y 2 + x 4 and ω = y 3 dx the Melnikov integral of (5) is the quantity M(T ) = H=T ω where T > 0. If T is the parametrization for the restriction of H to the half-line x > 0 (which means, in the notations used in the present paper, that T = η 4 ) then the return map of (5) has the form T → T − αM(T ) + O(α 2 ) [9]. For the present system we have Taking the fourth root in the return map of T we obtain (21).