The Return Map for a Planar Vector Field with Nilpotent Linear Part: A Direct and Explicit Derivation

Abstract

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 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 [6, 7].

In the case when the linear part of the vector field has nonzero eigenvalues there are important results containing the return map [8–14]. 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 [18–20]. 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

𝑋̇=−𝑌,𝑌̇=𝑋³−𝑌³, (2.1) 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 𝜖 with 0<𝜖<𝜖₀ be small enough.
The solution of (2.1) satisfying 𝑋(0)=𝜖,𝑌(0)=0 first returns to the positive 𝑋-axis at the value 𝑋 satisfying 𝑋=𝜖+^∞_𝑛=1𝑋_𝑛𝜖^3𝑛+1, (2.2) which is a convergent series.
The coefficients 𝑋_𝑛 can be calculated iteratively. In particular, 𝑋₁=−2^3/2𝑐₁,𝑋₂=16𝑐²₁,𝑋₃=−2^3/2(20𝑐₂𝑐₁+8𝑐₃+49𝑐³₁), (2.3) where 𝑐_𝑛=𝑣_𝑛(1) with 𝑣_𝑛 given by 𝑣₁(𝜉)=𝜉^−2𝜉₀𝑡⁴𝑝(𝑡)^3/2𝑑𝑡,𝑣₂3(𝜉)=−2𝜉^−2𝜉₀𝑡⁴𝑝(𝑡)^1/2𝑣₁𝑣(𝑡)𝑑𝑡,₃(𝜉)=𝜉^−2𝜉₀𝑡⁴3[−2𝑝(𝑡)^1/2𝑣₂3(𝑡)+8𝑣₁²(𝑡)𝑝(𝑡)^1/2]𝑑𝑡, (2.4)(2.5) where 𝑝(𝑡)=4−6𝑡²+4𝑡⁴−𝑡⁶. (2.6)

3. Proof of Proposition 2.1

The proof of Proposition 2.1 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 𝜖 appears as a small parameter in the equation with

𝑋=𝜖𝑥,𝑌=2^−1/2𝜖²𝑦,𝜏=2^−3/2𝜖𝑡, (3.1) the system (2.1) becomes

𝑑𝑥𝑑𝜏=−2𝑦,𝑑𝑦𝑑𝜏=4𝑥³−𝛼𝑦³, (3.2) where 𝛼 is the small parameter

𝛼=2^1/2𝜖³. (3.3)

While the initial condition 𝑋(0)=𝜖 becomes 𝑥(0)=1, it is useful to study solutions of (3.2) with the more general initial condition 𝑥(0)=𝜂 with 𝜂 in a neighborhood of 1.

Remark 3.1. System (3.2) has the form 𝑑𝐻+𝛼𝜔=0 with 𝐻=𝑦²+𝑥⁴, 𝜔=𝑦³𝑑𝑥 and 𝛼 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 𝛼>0.

Note the following Lyapunov function for (3.2)

𝐿(𝑥,𝑦)=𝑦²+𝑥⁴𝑑,with𝑑𝜏𝐿(𝑥(𝜏),𝑦(𝜏))=−2𝛼𝑦(𝜏)⁴≤0. (3.4)

Since the set {𝑦=0} 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 𝑥(0)=𝜂,𝑦(0)=0 for some 𝜂∈[1/2,3/2].

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 (3.4), and this can happen only at some point where 𝑦=0 or where 4𝑥³=𝛼𝑦³, whichever comes first. Since 𝑦 increases, then the first occurrence is a point where 4𝑥³=𝛼𝑦³. At this point 𝑥^′<0 so 𝑥 continues to decrease, while 𝑦^′′=−24𝑥²𝑦<0 so 𝑦 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 4^1/3𝑥=𝛼^1/3𝑦 before the monotony of 𝑥 changes, therefore the next change of monotony happens for 𝑦=0, a point where, therefore 𝑥<0, denote this value of 𝑥 by −𝜂.

Denote by 𝑥₀(𝜏),𝑦₀(𝜏) the solution for 𝛼=0: 𝑥^′₀=−2𝑦,𝑦^′₀=4𝑥³₀ and 𝑥₀(0)=𝜂,𝑦₀(0)=0. Therefore 𝑥⁴₀+𝑦²₀=𝜂⁴. We have 𝑥(𝜏)=𝑥₀(𝜏)+𝑂(𝛼),𝑦(𝜏)=𝑦₀(𝜏)+𝑂(𝛼) therefore 𝜂=𝜂+𝑂(𝛼).

Similar arguments show that the solution 𝑥(𝜏),𝑦(𝜏) of (3.2) continues to turn around the origin, crossing again the positive 𝑥-axis at a point 𝜂=𝜂+𝑂(𝛼) (and, of course, 𝜂<𝜂 by (3.4)).

Solutions (𝑥(𝜏),𝑦(𝜏)) of (3.2) provide smooth parametrizations for solutions 𝑦(𝑥) of

𝑑𝑑𝑥(𝑦²)=−4𝑥³+𝛼𝑦³. (3.5)

Note that following the path (𝑥(𝜏),𝑦(𝜏)) 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 𝑥>0

Lemma 3.2 shows that there exists a unique solution 𝑦≥0 of (3.5) so that 𝑦(𝜂)=0; this solution is defined for 𝑥∈[0,𝜂] and establishes an iterative procedure for calculating this solution.

Substituting 𝑦=𝑢^1/2 in (3.5) we obtain

𝑑𝑢𝑑𝑥=−4𝑥³+𝛼𝑢^3/2. (3.6)

Lemma 3.2. There exists 𝛿₀>0 independent of 𝜂 and 𝛼 so that the following holds.
Let 𝜂>0. For any 𝛼 with |𝛼|<𝛼₀≤𝛿₀𝜂⁻³/2, (3.6) with the condition 𝑢(𝜂)=0 has a unique solution 𝑢=𝑢(𝑥;𝛼,𝜂) for 𝑥∈[0,𝜂]. One has 𝑢(𝑥;𝛼,𝜂)>0 for 𝑥∈[0,𝜂), 𝛼>0 and 𝑢(𝑥;𝛼,𝜂) is analytic in 𝛼 and 𝜂  for |𝛼|<𝛼₀ and 𝜂>0.

Remark 3.3. We will use the results of Lemma 3.2 only for 𝜂 such that 1/2≤𝜂≤3/2. In this case (by lowering 𝛼₀) we can take 𝛼₀ of Lemma 3.2 independent of 𝜂 by taking 𝛼₀≤𝛿₀𝜂2min⁻³=4³/27³𝛿₀2≡𝑐₀𝛿₀. (3.7)

Proof of Lemma 3.2. Local analysis shows that solutions of (3.6) satisfying 𝑢(𝜂)=0 have an expansion in integer and half-integer powers of 𝜂−𝑥 and we have 𝑢(𝑥)=𝜂⁴−𝑥⁴−165𝛼𝜂^9/2(𝜂−𝑥)^5/2(1+𝑜(1)),(𝑥→𝜂−), (3.8) which inspires the following substitutions.
Denote 𝜉=𝑥1−𝜂, (3.9) (with the usual branch of the square root for 𝑥/𝜂<1) and let 𝑢(𝑥)=𝜂⁴−𝑥⁴−𝜂³(𝜂−𝑥)𝑣(𝜉). (3.10)
Note that if 𝑢(𝜂)=0 then necessarily 𝑣(0)=0 by (3.8).
Equation (3.6) becomes 𝜉𝑑𝑑𝜉𝑣(𝜉)+2𝑣(𝜉)=𝛿𝜉³[𝑝(𝜉)−𝑣(𝜉)]^3/2, (3.11) where 𝛿=2𝜂³𝛼, (3.12) and the polynomial 𝑝 is given by (2.6). (Note that 𝑝(𝜉)∈[1,4] for 𝜉∈[0,1].)
Lemma 3.2 follows if we show the following. Lemma 3.4. These exists 𝛿₀>0 so that for any 𝛿 with |𝛿|<𝛿₀, (3.11) has a unique solution 𝑣=𝑣(𝜉;𝛿) for 𝜉∈[0,1] so that 𝑣(0)=0.
Moreover, 𝑣(𝜉;⋅) is analytic for 𝛿∈ℂ with |𝛿|<𝛿₀ and the terms of its power series 𝑣(𝜉;𝛿)=_𝑛≥1𝛿^𝑛𝑣_𝑛(𝜉) (3.13) can be calculated recursively; in particular, the first terms are (2.4), (2.5).To prove Lemma 3.4 multiply (3.11) by 𝜉 and integrate; we obtain that 𝑣 is a fixed point (𝑣=𝒥[𝑣]) for the operator 𝒥𝑣(𝜉)=𝛿𝜉^−2𝜉₀𝑡⁴[𝑝(𝑡)−𝑣(𝑡)]^3/2𝑑𝑡=𝛿𝜉³¹₀𝑠⁴[𝑝(𝜉𝑠)−𝑣(𝜉𝑠)]^3/2𝑑𝑠. (3.14) Let ℬ be the Banach space of functions 𝑓(𝜉;𝛿) continuous for 𝜉∈[0,1] and analytic on the (complex) disk |𝛿|<𝛿₀, continuous on |𝛿|≤𝛿₀, with the norm ‖𝑓‖=sup_𝜉∈[0,1]sup_{|𝛿|≤𝛿0}|𝑓(𝜉;𝛿)|. (3.15)
Let 𝑚 be a number with 0<𝑚<1. Let 𝛿₀>0 be small enough, so that 𝛿₀<𝑚/√5 and 𝛿₀<10/3/√5.
Let ℬ_𝑚 be the ball ℬ_𝑚={𝑓∈ℬ_𝑚;‖𝑓‖≤𝑚}.
We have 𝒥∶ℬ_𝑚→ℬ_𝑚. Indeed, note that for 𝑓∈ℬ_𝑚 we have |𝑝(𝑡)−𝑓(𝑡;𝛿)|≥|𝑝(𝑡)|−|𝑓(𝑡;𝛿)|≥1−𝑚>0, (3.16) therefore, since 𝑓 is analytic in 𝛿, then so is [𝑝(𝑡)−𝑓(𝑡;𝛿)]^3/2, and therefore so is 𝒥𝑓. Also, if 𝑓∈ℬ_𝑚 then also 𝒥𝑓∈ℬ_𝑚 because |𝒥𝑓(𝜉;𝛿)|≤|𝛿|¹₀𝑠⁴[|𝑝(𝜉𝑠)|+|𝑓(𝜉𝑠;𝛿)|]^3/2𝛿𝑑𝑠≤⁰(4+𝑚)^3/25<𝛿₀√5<𝑚. (3.17)
Moreover, the operator 𝒥 is a contraction on ℬ_𝑚. Indeed, using the estimate |(𝑝−𝑓₁)^3/2−(𝑝−𝑓₂)^3/2|≤|𝑓₁−𝑓₂|32sup_{|𝑓|≤𝑚}|𝑝−𝑓|^1/2≤|𝑓₁−𝑓₂|3(4+𝑚)^1/22 (3.18) we obtain |𝒥𝑓₁−𝒥𝑓₂|≤𝑐‖𝑓₁−𝑓₂‖with𝑐=𝛿₀3√510<1. (3.19)
Therefore the operator 𝒥 has a unique fixed point, which is the solution 𝑣(𝜉;𝛿).
To obtain the power series (3.13) substitute an expansion 𝑣(𝜉;𝛿)=𝑣₀(𝜉)+∑_𝑛≥1𝛿^𝑛𝑣_𝑛(𝜉) in (3.11). It follows that 𝜉𝑣^′₀+2𝑣₀=0 with 𝑣₀(0)=0, therefore 𝑣₀(𝜉)≡0.
Substitution of (3.13) in (𝑝−𝑣)^3/2 followed by expansion in power series in 𝛿 gives (𝑝−𝑣)^3/2=𝑝^3/2(1−_𝑛≥1𝛿^𝑛𝑣_𝑛𝑝)^3/2≡𝑝^3/2(1+_𝑛≥1𝛿^𝑛𝑅_𝑛), (3.20) where 𝑅_𝑛=𝑅_𝑛[𝑣₁,…,𝑣_𝑛,𝑝]. In particular, 𝑅₁3=−2𝑣₁𝑝,𝑅₂3=(−2𝑣₂𝑝+38𝑣₁²𝑝²). (3.21)
From (3.11) we obtain the recursive system 𝜉𝑑𝑣𝑑𝜉^𝑛+2𝑣_𝑛=𝜉³𝑝^3/2𝑅_𝑛−1, (3.22) (for 𝑛≥1 and with 𝑅₀=1), with the only solution with 𝑣_𝑛(0)=0 given recursively by 𝑣_𝑛(𝜉)=𝜉^−2𝜉₀𝑡⁴𝑝(𝑡)^3/2𝑅_𝑛−1(𝑡)𝑑𝑡. (3.23)
In particular, we have (2.4), (2.5).

The following gathers the conclusions of the present section.

Corollary 3.5. There exists 𝛼₀>0 so that for any 𝜂∈[1/2,3/2] and 𝛼 with |𝛼|<𝛼₀, (3.5) has a unique solution 𝑦(𝑥) on [0,𝜂] satisfying 𝑦(𝜂)=0 and 𝑦>0 on [0,𝜂) for 𝛼>0.
Moreover, this solution has the form 𝑦=𝜙(𝑥;𝛼,𝜂)=[𝜂⁴−𝑥⁴−𝜂³(𝜂−𝑥)𝑣(𝑥1−𝜂;2𝜂³𝛼)]^1/2, (3.24) with 𝑣=𝑣(𝜉,𝛿) a solution of (3.11). The map 𝛼↦𝑣(𝜉;2𝜂³𝛼) is analytic for |𝛼|<𝛼₀.

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 𝑦(𝑥) of (3.5) 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 (3.5), these solutions can be immediately written down as follows.

Let 𝜂∈[1/2,3/2], 𝛼 with |𝛼|<𝛼₀.

(i)Let 𝑦₁=𝜙(𝑥;𝛼,𝜂) be the solution of (3.24), defined for 𝑥∈[0,𝜂], with 𝑦₁(𝜂)=0 and 𝑦₁>0 for 𝛼>0,

Then:

(ii)the function 𝑦₂=𝜙(−𝑥;−𝛼,𝜂) is also a solution of (3.5), defined for 𝑥∈[−𝜂,0]; we have 𝑦₂(−𝜂)=0 and 𝑦₂≥0 for 𝛼>0,(iii)The function 𝑦₃=−𝜙(−𝑥;𝛼,𝜂) is a solution of (3.5), defined for 𝑥∈[−𝜂,0]. We have 𝑦₃(−𝜂)=0 and 𝑦₃≤0 for 𝛼>0,(iv)The function 𝑦₄=−𝜙(𝑥;−𝛼,𝜂) is a solution of (3.5), defined for 𝑥∈[0,𝜂] and we have 𝑦₄(𝜂)=0 and 𝑦₄≤0 for 𝛼>0,

3.4.2. Matching at the Positive y-Axis

Let 𝜂,𝜂∈[1/2,3/4] and let 𝑦₁(𝑥)=𝜙(𝑥;𝛼,𝜂) be the solution of (3.5) as in (i), for 𝑥∈[0,𝜂] and 𝑦₂(𝑥)=𝜙(−𝑥;−𝛼,𝜂) solution as in (ii), for 𝑥∈[−𝜂,0].

The following lemma finds 𝜂 so that 𝑦₁(0)=𝑦₂(0), therefore so that 𝑦₁ is the continuation of 𝑦₂

Lemma 3.6. Let |𝛼|<𝛼₀ with 𝛼₀ satisfying (3.7). Let 𝜂 so that |𝜂−1|<𝑐₁|𝛼| with 𝛼₀ small enough so that 𝑐₁𝛼₀≤1/2.
There exists a unique 𝜂=𝜂+𝑂(𝛼) so that 𝜙(0;𝛼,𝜂)=𝜙(0;−𝛼,𝜂), (3.25) and 𝜂 depends analytically on 𝜂 and 𝛼 for |𝛼|<𝛼₁ for 𝛼₁ being small enough. One has |𝜂−1|≤𝑐₂|𝛼| for some 𝑐₂>0 and 𝜂=𝜂−𝜂⁴𝑐₁𝛼+2𝛼²𝜂⁷𝑐²₁−𝛼³𝜂¹⁰(4𝑐₃+212𝑐³₁+10𝑐₂𝑐₁)+𝑂(𝛼⁴), (3.26) where 𝑐_𝑛=𝑣_𝑛(1) with 𝑣_𝑛(𝜉) given by (2.4), (2.5).

Proof. Let 𝐹(𝜂,𝜂,𝛼)=𝜙(0;−𝛼,𝜂)−𝜙(0;𝛼,𝜂), (3.27) which is a function analytic in (𝜂,𝜂,𝛼) by Lemma 3.2 and relation (3.12).
We have 𝐹(𝜂,𝜂,0)=0 and 𝜕𝐹𝜕𝜂(𝜂,𝜂,0)=−4𝜂³[1−𝑣(1;0)]=−4𝜂³≠0, (3.28) therefore the implicit equation 𝐹(𝜂,𝜂,𝛼)=0 determines 𝜂=𝜂(𝜂,𝛼) as an analytic function of 𝛼, 𝜂 for 𝛼 small.
We have |𝜂−𝜂|≤|𝛼|sup_{|𝛼|<𝛼1,|𝜂|<𝑐1𝛼1}|𝜕𝜂𝜕𝛼|=𝑐^′₁|𝛼|, (3.29) therefore |𝜂−1|≤|𝜂−𝜂|+|𝜂−1|≤(𝑐₁+𝑐^′₁)|𝛼|≡𝑐₂|𝛼|. (3.30)
The expansion of 𝜂 in power series of 𝛼 is found as follows. By using (3.24), (3.25) becomes 𝜂⁴−𝜂⁴𝑣(1;−2𝜂³𝛼)=𝜂⁴−𝜂⁴𝑣(1;2𝜂³𝛼), (3.31) where by substituting (3.13) and 𝜂=∑_𝑛≥0𝛿^𝑛𝜂_𝑛 followed by power series expansion in 𝛼 we obtain (3.26).

3.4.3. Matching at the Negative y-Axis

Let 𝜂,𝜂∈[1/2,3/4] and 𝜂 given by Lemma 3.6. Consider the solution 𝑦₃=−𝜙(−𝑥;𝛼,𝜂) as in (iii), for 𝑥∈[−𝜂,0], with 𝑦₃(−𝜂)=0. Therefore 𝑦₃ is the continuation of 𝑦₂.

Let 𝑦₄(𝑥)=−𝜙(−𝑥;−𝛼,𝜂) be a solution of (3.5) as in (iv), for 𝑥∈[0,𝜂].

The following lemma finds 𝜂 so that 𝑦₃(0)=𝑦₄(0) therefore so that 𝑦₄ is the continuation of 𝑦₃.

Lemma 3.7. Let |𝛼|<𝛼₁ and |𝜂−1|≤𝑐₂|𝛼| with 𝛼₁ small enough so that 𝜂∈[1/2,3/2].
There exists a unique 𝜂>0 so that 𝜙(0;−𝛼,𝜂)=𝜙(0;𝛼,𝜂) and 𝜂 depends analytically on 𝜂 and 𝛼 for |𝛼|<𝛼₂ for 𝛼₂ small enough. One has |𝜂−1|≤𝑐₃|𝛼| for some 𝑐₃>0 and 𝜂=𝜂−𝜂⁴𝑐₁𝛼+2𝛼²𝜂⁷𝑐²₁−𝛼³𝜂¹⁰(4𝑐₃+212𝑐³₁+10𝑐₂𝑐₁)+𝑂(𝛼⁴), (3.32) where 𝑐_𝑛=𝑣_𝑛(1) with 𝑣_𝑛(𝜉) given by (2.4), (2.5).

Proof. We need to find 𝜂=𝜂(𝜂,𝛼) so that 𝐹(𝜂,𝜂,𝛼)=𝜙(0;−𝛼,𝜂)−𝜙(0;𝛼,𝜂)=0. (3.33)
Note that the function 𝐹 above is the same as (3.27). By Lemma 3.6 the present lemma follows.

3.5. The First Return Map

Let 𝜂∈[1/2,3/2] and 𝛼₂ as in Lemma 3.7. Then 𝜂 given by Lemma 3.7 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 (3.12), it is analytic in 𝜖 for fixed 𝜂.

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

𝜂=𝜂−2𝜂⁴𝑐₁𝛼+8𝜂⁷𝑐²₁𝛼²−𝜂¹⁰(20𝑐₂𝑐₁+8𝑐₃+49𝑐³₁)𝛼³+𝑂(𝛼⁴). (3.34)

To obtain the point 𝑋 where the solution of (2.1) with 𝑋(0)=𝜖>0 and 𝑌(0)=0 first returns to the positive 𝑋-axis let 𝛼=2^1/2𝜖³ and multiply (3.34) by 𝜖 (since we have 𝑋=𝜖𝑥) and, finally, let 𝜂=1. We obtain that 𝑋 is analytic in 𝜖 for small 𝜖 and

𝑋=𝜖−2^3/2𝑐₁𝜖⁴+16𝑐²₁𝜖⁷−2^3/2(20𝑐₂𝑐₁+8𝑐₃+49𝑐³₁)𝜖¹⁰+𝑂(𝜖¹³), (3.35) where 𝑐_𝑛=𝑣_𝑛(1) with 𝑣_𝑛 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).

Appendix

With the notation 𝐻=𝑦²+𝑥⁴ and 𝜔=𝑦³𝑑𝑥 the Melnikov integral of (3.2) is the quantity 𝑀(𝑇)=∫_𝐻=𝑇𝜔 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=𝜂⁴) then the return map of (3.2) has the form 𝑇↦𝑇−𝛼𝑀(𝑇)+𝑂(𝛼²) [15].

For the present system we have

𝑀(𝑇)=_{𝑦2+𝑥4=𝑇}𝑦³𝑑𝑥=4^𝑇1/4₀(𝑇−𝑥⁴)^3/27𝑑𝑥=4𝑇4¹₀(1−𝑠⁴)^3/2𝑑𝑠=8𝑇^7/4𝑐₁. (A.1) Taking the fourth root in the return map of 𝑇 we obtain (3.34).

Acknowledgment

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