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.

1. 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 [1, 2].

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 e.g., [3] and references therein), and also in connection to o-minimality [4].

The monodromy problem (determining when the singularity is a center or a focus) was solved by Andreev [5].

A fundamental result concerns the asymptotic form of return maps states that if the singular points of a vector field are algebraically isolated, there exists a semitransversal arc such that the return map admits an asymptotic expansion is positive powers of and logs (with the first term linear), or has its principal part a finite composition of powers and exponentials [6, 7].

In the case when the linear part of the vector field has nonzero eigenvalues there are important results containing the return map [814]. Results are also available for perturbations of Hamiltonians [15, 16] and for perturbations of integrable systems [17]. On the other hand, there are few results available in the general setting [1820]. The recent papers [21, 22] contain methods that can generate general return maps.

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.

2. 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 [23], and for which there are (to the author's knowledge) no methods available to generate the return map.

The main result is the following.

Proposition 2.1. Let with be small enough.
The solution of (2.1) satisfying first returns to the positive -axis at the value satisfying which is a convergent series.
The coefficients can be calculated iteratively. In particular, where with given by where

3. Proof of Proposition 2.1

The proof of Proposition 2.1 also provides an algorithm for calculating iteratively the coefficients .

3.1. Normalization

It is convenient to normalize the variables so that the constant appears as a small parameter in the equation with

the system (2.1) becomes

where is the small parameter

While the initial condition becomes , it is useful to study solutions of (3.2) with the more general initial condition with in a neighborhood of .

Remark 3.1. System (3.2) has the form with , and a small parameter. The recent result [15] gives a formalism for finding the return map for this type of systems; see also [17]. The present construction is concrete, explicit, and suitable for numerical calculations.

3.2. General Behavior of Solutions of (3.2)

Let .

Note the following Lyapunov function for (3.2)

Since the set contains no trajectories besides the origin (which is the only equilibrium point of (3.2)), then the origin is asymptotically stable by the Krasovskii-LaSalle principle.

Consider the solution of (3.2) with the initial condition for some .

Since and then increases and decreases for small . This monotony must change due to (3.4), and this can happen only at some point where or where , whichever comes first. Since increases, then the first occurrence is a point where . At this point so continues to decrease, while so has a maximum, and will continue by decreasing. Again, the monotony must change due to (3.4) and the path cannot cross again the line before the monotony of changes, therefore the next change of monotony happens for , a point where, therefore , denote this value of by .

Denote by the solution for : and . Therefore . We have therefore .

Similar arguments show that the solution of (3.2) continues to turn around the origin, crossing again the positive -axis at a point (and, of course, by (3.4)).

Solutions of (3.2) provide smooth parametrizations for solutions of

Note that following the path one full rotation around the origin corresponds to considering a positive solution of (3.5), followed by a negative one.

3.3. Positive Solutions of (3.5) for

Lemma 3.2 shows that there exists a unique solution of (3.5) so that ; this solution is defined for and establishes an iterative procedure for calculating this solution.

Substituting in (3.5) we obtain

Lemma 3.2. There exists independent of and so that the following holds.
Let . For any with , (3.6) with the condition has a unique solution for . One has for , and is analytic in and   for and .

Remark 3.3. We will use the results of Lemma 3.2 only for such that . In this case (by lowering ) we can take of Lemma 3.2 independent of by taking

Proof of Lemma 3.2. Local analysis shows that solutions of (3.6) satisfying have an expansion in integer and half-integer powers of and we have which inspires the following substitutions.
Denote (with the usual branch of the square root for ) and let
Note that if then necessarily by (3.8).
Equation (3.6) becomes where and the polynomial is given by (2.6). (Note that for .)
Lemma 3.2 follows if we show the following. Lemma 3.4. These exists so that for any with (3.11) has a unique solution for so that .
Moreover, is analytic for with and the terms of its power series can be calculated recursively; in particular, the first terms are (2.4), (2.5).
To prove Lemma 3.4 multiply (3.11) by and integrate; we obtain that is a fixed point () for the operator Let be the Banach space of functions continuous for and analytic on the (complex) disk , continuous on , with the norm
Let be a number with . Let be small enough, so that and .
Let be the ball .
We have . Indeed, note that for we have therefore, since is analytic in , then so is , and therefore so is . Also, if then also because
Moreover, the operator is a contraction on . Indeed, using the estimate we obtain
Therefore the operator has a unique fixed point, which is the solution .
To obtain the power series (3.13) substitute an expansion in (3.11). It follows that with , therefore .
Substitution of (3.13) in followed by expansion in power series in gives where . In particular,
From (3.11) we obtain the recursive system (for and with ), with the only solution with given recursively by
In particular, we have (2.4), (2.5).

The following gathers the conclusions of the present section.

Corollary 3.5. There exists so that for any and with , (3.5) has a unique solution on satisfying and on for .
Moreover, this solution has the form with a solution of (3.11). The map is analytic for .

3.4. Solutions of (3.5) in Other Quadrants and Matching
3.4.1. Solutions in Other Quadrants

We found an expression for the solution of (3.5) for and . In a similar way, expressions in the other quadrants can be found. However, taking advantage of the discrete symmetries of (3.5), these solutions can be immediately written down as follows.

Let , with .

(i)Let be the solution of (3.24), defined for , with and for ,


(ii)the function is also a solution of (3.5), defined for ; we have and for ,(iii)The function is a solution of (3.5), defined for . We have and for ,(iv)The function is a solution of (3.5), defined for and we have and for ,
3.4.2. Matching at the Positive y-Axis

Let and let be the solution of (3.5) as in (i), for and solution as in (ii), for .

The following lemma finds so that , therefore so that is the continuation of

Lemma 3.6. Let with satisfying (3.7). Let so that with small enough so that .
There exists a unique so that and depends analytically on and for for being small enough. One has for some and where with given by (2.4), (2.5).

Proof. Let which is a function analytic in by Lemma 3.2 and relation (3.12).
We have and therefore the implicit equation determines as an analytic function of , for small.
We have therefore
The expansion of in power series of is found as follows. By using (3.24), (3.25) becomes where by substituting (3.13) and followed by power series expansion in we obtain (3.26).

3.4.3. Matching at the Negative y-Axis

Let and given by Lemma 3.6. Consider the solution as in (iii), for , with . Therefore is the continuation of .

Let be a solution of (3.5) as in (iv), for .

The following lemma finds so that therefore so that is the continuation of .

Lemma 3.7. Let and with small enough so that .
There exists a unique so that and depends analytically on and for for small enough. One has for some and where with given by (2.4), (2.5).

Proof. We need to find so that
Note that the function above is the same as (3.27). By Lemma 3.6 the present lemma follows.

3.5. The First Return Map

Let and as in Lemma 3.7. Then given by Lemma 3.7 is the first return to the positive -axis of the solution with and it is analytic in and , therefore, by (3.12), it is analytic in for fixed .

Combining (3.26) and (3.32) we obtain as a convergent power series in , with coefficients dependent on , whose first terms are

To obtain the point where the solution of (2.1) with and first returns to the positive -axis let and multiply (3.34) by (since we have ) and, finally, let . We obtain that is analytic in for small and

where with given by (2.4), (2.5).

Remark 3.8. The first coefficient of the return map (3.34) is, up to a sign, the Melnikov integral of the system (3.2), see[15]; of course, the present results are in agreement with this fact (see the appendix for details).


With the notation and the Melnikov integral of (3.2) is the quantity where . If is the parametrization for the restriction of to the half-line (which means, in the notations used in the present paper, that ) then the return map of (3.2) has the form [15].

For the present system we have

Taking the fourth root in the return map of we obtain (3.34).


The author is grateful to Chris Miller for suggesting the problem.