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

Authors: Rodica D. Costin
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## 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.


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## Body

## 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 aC∞ 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(2.1)Ẋ=-Y,Ẏ=X3-Y3,
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.
Letϵ 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 X̃̃ satisfying
(2.2)X̃̃=ϵ+∑n=1∞Xnϵ3n+1,
which is a convergent series.
The coefficientsXn can be calculated iteratively. In particular,
(2.3)X1=-23/2c1,X2=16c12,X3=-23/2(20c2c1+8c3+49c13),
where cn=vn(1) with vn given by
(2.4)v1(ξ)=ξ-2∫0ξt4p(t)3/2dt,v2(ξ)=-32ξ-2∫0ξt4p(t)1/2v1(t)dt,(2.5)v3(ξ)=ξ-2∫0ξt4[-32p(t)1/2v2(t)+38v12(t)p(t)1/2]dt,
where
(2.6)p(t)=4-6t2+4t4-t6.


## 3. Proof of Proposition2.1

The proof of Proposition2.1 also provides an algorithm for calculating iteratively the coefficients Xn.


### 3.1. Normalization

It is convenient to normalize the variablesX,Y,t so that the constant ϵ appears as a small parameter in the equation with(3.1)X=ϵx,Y=2-1/2ϵ2y,τ=2-3/2ϵt,
the system (2.1) becomes(3.2)dxdτ=-2y,dydτ=4x3-αy3,
where α is the small parameter(3.3)α=21/2ϵ3.While the initial conditionX(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)

Letα>0.Note the following Lyapunov function for (3.2)(3.4)L(x,y)=y2+x4,withddτL(x(τ),y(τ))=-2αy(τ)4≤0.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].Sincey′(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 41/3x=α1/3y 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 -η̃.Denote byx0(τ),y0(τ) the solution for α=0: x0′=-2y,y0′=4x03 and x0(0)=η,y0(0)=0. Therefore x04+y02=η4. We have x(τ)=x0(τ)+O(α),y(τ)=y0(τ)+O(α) therefore η̃=η+O(α).Similar arguments show that the solutionx(τ),y(τ) of (3.2) continues to turn around the origin, crossing again the positive x-axis at a point η̃̃=η+O(α) (and, of course, η̃̃<η by (3.4)).Solutions(x(τ),y(τ)) of (3.2) provide smooth parametrizations for solutions y(x) of(3.5)ddx(y2)=-4x3+αy3.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

Lemma3.2 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.Substitutingy=u1/2 in (3.5) we obtain(3.6)dudx=-4x3+αu3/2.Lemma 3.2.
There existsδ0>0 independent of η and α so that the following holds.
Letη>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 Lemma3.2 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
(3.7)α0≤δ02minη-3=43/273δ02≡c0δ0.Proof of Lemma3.2.
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
(3.8)u(x)=η4-x4-165αη9/2(η-x)5/2(1+o(1)),(x→η-),
which inspires the following substitutions.
Denote(3.9)ξ=1-xη,
(with the usual branch of the square root for x/η<1) and let
(3.10)u(x)=η4-x4-η3(η-x)v(ξ).
Note that ifu(η)=0 then necessarily v(0)=0 by (3.8).
Equation (3.6) becomes
(3.11)ξddξv(ξ)+2v(ξ)=δξ3[p(ξ)-v(ξ)]3/2,
where
(3.12)δ=2η3α,
and the polynomial p is given by (2.6). (Note that p(ξ)∈[1,4] for ξ∈[0,1].)
Lemma3.2 follows if we show the following. Lemma 3.4.
These existsδ0>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
(3.13)v(ξ;δ)=∑n≥1δnvn(ξ)
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
(3.14)𝒥v(ξ)=δξ-2∫0ξt4[p(t)-v(t)]3/2dt=δξ3∫01s4[p(ξs)-v(ξs)]3/2ds.
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
(3.15)∥f∥=supξ∈[0,1]sup|δ|≤δ0|f(ξ;δ)|.
Letm be a number with 0<m<1. Let δ0>0 be small enough, so that δ0<m/5 and δ0<10/3/5.
Letℬm be the ball ℬm={f∈ℬm;∥f∥≤m}.
We have𝒥:ℬm→ℬm. Indeed, note that for f∈ℬm we have
(3.16)|p(t)-f(t;δ)|≥|p(t)|-|f(t;δ)|≥1-m>0,
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∈ℬm then also 𝒥f∈ℬm because
(3.17)|𝒥f(ξ;δ)|≤|δ|∫01s4[|p(ξs)|+|f(ξs;δ)|]3/2ds≤δ0(4+m)3/25<δ05<m.
Moreover, the operator𝒥 is a contraction on ℬm. Indeed, using the estimate
(3.18)|(p-f1)3/2-(p-f2)3/2|≤|f1-f2|32sup|f|≤m|p-f|1/2≤|f1-f2|3(4+m)1/22
we obtain
(3.19)|𝒥f1-𝒥f2|≤c∥f1-f2∥withc=δ03510<1.
Therefore the operator𝒥 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≥1δnvn(ξ) in (3.11). It follows that ξv0′+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
(3.20)(p-v)3/2=p3/2(1-∑n≥1δnvnp)3/2≡p3/2(1+∑n≥1δnRn),
where Rn=Rn[v1,…,vn,p]. In particular,
(3.21)R1=-32v1p,R2=(-32v2p+38v12p2).
From (3.11) we obtain the recursive system
(3.22)ξddξvn+2vn=ξ3p3/2Rn-1,
(for n≥1 and with R0=1), with the only solution with vn(0)=0 given recursively by
(3.23)vn(ξ)=ξ-2∫0ξt4p(t)3/2Rn-1(t)dt.
In particular, we have (2.4), (2.5).The following gathers the conclusions of the present section.Corollary 3.5.
There existsα0>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(3.24)y=ϕ(x;α,η)=[η4-x4-η3(η-x)v(1-xη;2η3α)]1/2,
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 solutiony(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.Letη∈[1/2,3/2], α with |α|<α0.(i)
Lety1=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 functiony2=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 functiony3=-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 functiony4=-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

Letη,η̃∈[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 ỹ2(x)=ϕ(-x;-α,η̃) solution as in (ii), for x∈[-η̃,0].The following lemma findsη̃ so that y1(0)=ỹ2(0), therefore so that y1 is the continuation of ỹ2Lemma 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 uniqueη̃=η+O(α) so that
(3.25)ϕ(0;α,η)=ϕ(0;-α,η̃),
and η̃ depends analytically on η and α for |α|<α1 for α1 being small enough. One has |η̃-1|≤c2|α| for some c2>0 and
(3.26)η̃=η-η4c1α+2α2η7c12-α3η10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
Let(3.27)F(η̃,η,α)=ϕ(0;-α,η̃)-ϕ(0;α,η),
which is a function analytic in (η̃,η,α) by Lemma 3.2 and relation (3.12).
We haveF(η,η,0)=0 and
(3.28)∂F∂η̃(η,η,0)=-4η3[1-v(1;0)]=-4η3≠0,
therefore the implicit equation F(η̃,η,α)=0 determines η̃=η̃(η,α) as an analytic function of α, η for α small.
We have(3.29)|η̃-η|≤|α|sup|α|<α1,|η|<c1α1|∂η̃∂α|=c1′|α|,
therefore
(3.30)|η̃-1|≤|η̃-η|+|η-1|≤(c1+c1′)|α|≡c2|α|.
The expansion ofη̃ in power series of α is found as follows. By using (3.24), (3.25) becomes
(3.31)η̃4-η̃4v(1;-2η̃3α)=η4-η4v(1;2η3α),
where by substituting (3.13) and η̃=∑n≥0δnη̃n followed by power series expansion in α we obtain (3.26).


#### 3.4.3. Matching at the Negative y-Axis

Letη,η̃̃∈[1/2,3/4] and η̃ given by Lemma 3.6. Consider the solution ỹ3=-ϕ(-x;α,η̃) as in (iii), for x∈[-η̃,0], with ỹ3(-η̃)=0. Therefore ỹ3 is the continuation of ỹ2.Letỹ̃4(x)=-ϕ(-x;-α,η̃̃) be a solution of (3.5) as in (iv), for x∈[0,η̃̃].The following lemma findsη̃̃ so that ỹ3(0)=ỹ̃4(0) therefore so that ỹ̃4 is the continuation of ỹ3.Lemma 3.7.
Let|α|<α1 and |η̃-1|≤c2|α| with α1 small enough so that η̃∈[1/2,3/2].
There exists a uniqueη̃̃>0 so that ϕ(0;-α,η̃̃)=ϕ(0;α,η̃) and η̃̃ depends analytically on η̃ and α for |α|<α2 for α2 small enough. One has |η̃̃-1|≤c3|α| for some c3>0 and
(3.32)η̃̃=η̃-η̃4c1α+2α2η̃7c12-α3η̃10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
We need to findη̃̃=η̃̃(η̃,α) so that
(3.33)F(η̃̃,η̃,α)=ϕ(0;-α,η̃̃)-ϕ(0;α,η̃)=0.
Note that the functionF above is the same as (3.27). By Lemma 3.6 the present lemma follows.


### 3.5. The First Return Map

Letη∈[1/2,3/2] and α2 as in Lemma 3.7. Then η̃̃ 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 η̃̃ as a convergent power series in η, with coefficients dependent on η, whose first terms are(3.34)η̃̃=η-2η4c1α+8η7c12α2-η10(20c2c1+8c3+49c13)α3+O(α4).To obtain the pointX̃̃ where the solution of (2.1) with X(0)=ϵ>0 and Y(0)=0 first returns to the positive X-axis let α=21/2ϵ3 and multiply (3.34) by ϵ (since we have X=ϵx) and, finally, let η=1. We obtain that X̃̃ is analytic in ϵ for small ϵ and(3.35)X̃̃=ϵ-23/2c1ϵ4+16c12ϵ7-23/2(20c2c1+8c3+49c13)ϵ10+O(ϵ13),
where cn=vn(1) with vn 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).


## 3.1. Normalization

It is convenient to normalize the variablesX,Y,t so that the constant ϵ appears as a small parameter in the equation with(3.1)X=ϵx,Y=2-1/2ϵ2y,τ=2-3/2ϵt,
the system (2.1) becomes(3.2)dxdτ=-2y,dydτ=4x3-αy3,
where α is the small parameter(3.3)α=21/2ϵ3.While the initial conditionX(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)

Letα>0.Note the following Lyapunov function for (3.2)(3.4)L(x,y)=y2+x4,withddτL(x(τ),y(τ))=-2αy(τ)4≤0.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].Sincey′(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 41/3x=α1/3y 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 -η̃.Denote byx0(τ),y0(τ) the solution for α=0: x0′=-2y,y0′=4x03 and x0(0)=η,y0(0)=0. Therefore x04+y02=η4. We have x(τ)=x0(τ)+O(α),y(τ)=y0(τ)+O(α) therefore η̃=η+O(α).Similar arguments show that the solutionx(τ),y(τ) of (3.2) continues to turn around the origin, crossing again the positive x-axis at a point η̃̃=η+O(α) (and, of course, η̃̃<η by (3.4)).Solutions(x(τ),y(τ)) of (3.2) provide smooth parametrizations for solutions y(x) of(3.5)ddx(y2)=-4x3+αy3.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

Lemma3.2 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.Substitutingy=u1/2 in (3.5) we obtain(3.6)dudx=-4x3+αu3/2.Lemma 3.2.
There existsδ0>0 independent of η and α so that the following holds.
Letη>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 Lemma3.2 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
(3.7)α0≤δ02minη-3=43/273δ02≡c0δ0.Proof of Lemma3.2.
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
(3.8)u(x)=η4-x4-165αη9/2(η-x)5/2(1+o(1)),(x→η-),
which inspires the following substitutions.
Denote(3.9)ξ=1-xη,
(with the usual branch of the square root for x/η<1) and let
(3.10)u(x)=η4-x4-η3(η-x)v(ξ).
Note that ifu(η)=0 then necessarily v(0)=0 by (3.8).
Equation (3.6) becomes
(3.11)ξddξv(ξ)+2v(ξ)=δξ3[p(ξ)-v(ξ)]3/2,
where
(3.12)δ=2η3α,
and the polynomial p is given by (2.6). (Note that p(ξ)∈[1,4] for ξ∈[0,1].)
Lemma3.2 follows if we show the following. Lemma 3.4.
These existsδ0>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
(3.13)v(ξ;δ)=∑n≥1δnvn(ξ)
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
(3.14)𝒥v(ξ)=δξ-2∫0ξt4[p(t)-v(t)]3/2dt=δξ3∫01s4[p(ξs)-v(ξs)]3/2ds.
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
(3.15)∥f∥=supξ∈[0,1]sup|δ|≤δ0|f(ξ;δ)|.
Letm be a number with 0<m<1. Let δ0>0 be small enough, so that δ0<m/5 and δ0<10/3/5.
Letℬm be the ball ℬm={f∈ℬm;∥f∥≤m}.
We have𝒥:ℬm→ℬm. Indeed, note that for f∈ℬm we have
(3.16)|p(t)-f(t;δ)|≥|p(t)|-|f(t;δ)|≥1-m>0,
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∈ℬm then also 𝒥f∈ℬm because
(3.17)|𝒥f(ξ;δ)|≤|δ|∫01s4[|p(ξs)|+|f(ξs;δ)|]3/2ds≤δ0(4+m)3/25<δ05<m.
Moreover, the operator𝒥 is a contraction on ℬm. Indeed, using the estimate
(3.18)|(p-f1)3/2-(p-f2)3/2|≤|f1-f2|32sup|f|≤m|p-f|1/2≤|f1-f2|3(4+m)1/22
we obtain
(3.19)|𝒥f1-𝒥f2|≤c∥f1-f2∥withc=δ03510<1.
Therefore the operator𝒥 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≥1δnvn(ξ) in (3.11). It follows that ξv0′+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
(3.20)(p-v)3/2=p3/2(1-∑n≥1δnvnp)3/2≡p3/2(1+∑n≥1δnRn),
where Rn=Rn[v1,…,vn,p]. In particular,
(3.21)R1=-32v1p,R2=(-32v2p+38v12p2).
From (3.11) we obtain the recursive system
(3.22)ξddξvn+2vn=ξ3p3/2Rn-1,
(for n≥1 and with R0=1), with the only solution with vn(0)=0 given recursively by
(3.23)vn(ξ)=ξ-2∫0ξt4p(t)3/2Rn-1(t)dt.
In particular, we have (2.4), (2.5).The following gathers the conclusions of the present section.Corollary 3.5.
There existsα0>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(3.24)y=ϕ(x;α,η)=[η4-x4-η3(η-x)v(1-xη;2η3α)]1/2,
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 solutiony(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.Letη∈[1/2,3/2], α with |α|<α0.(i)
Lety1=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 functiony2=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 functiony3=-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 functiony4=-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

Letη,η̃∈[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 ỹ2(x)=ϕ(-x;-α,η̃) solution as in (ii), for x∈[-η̃,0].The following lemma findsη̃ so that y1(0)=ỹ2(0), therefore so that y1 is the continuation of ỹ2Lemma 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 uniqueη̃=η+O(α) so that
(3.25)ϕ(0;α,η)=ϕ(0;-α,η̃),
and η̃ depends analytically on η and α for |α|<α1 for α1 being small enough. One has |η̃-1|≤c2|α| for some c2>0 and
(3.26)η̃=η-η4c1α+2α2η7c12-α3η10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
Let(3.27)F(η̃,η,α)=ϕ(0;-α,η̃)-ϕ(0;α,η),
which is a function analytic in (η̃,η,α) by Lemma 3.2 and relation (3.12).
We haveF(η,η,0)=0 and
(3.28)∂F∂η̃(η,η,0)=-4η3[1-v(1;0)]=-4η3≠0,
therefore the implicit equation F(η̃,η,α)=0 determines η̃=η̃(η,α) as an analytic function of α, η for α small.
We have(3.29)|η̃-η|≤|α|sup|α|<α1,|η|<c1α1|∂η̃∂α|=c1′|α|,
therefore
(3.30)|η̃-1|≤|η̃-η|+|η-1|≤(c1+c1′)|α|≡c2|α|.
The expansion ofη̃ in power series of α is found as follows. By using (3.24), (3.25) becomes
(3.31)η̃4-η̃4v(1;-2η̃3α)=η4-η4v(1;2η3α),
where by substituting (3.13) and η̃=∑n≥0δnη̃n followed by power series expansion in α we obtain (3.26).


### 3.4.3. Matching at the Negative y-Axis

Letη,η̃̃∈[1/2,3/4] and η̃ given by Lemma 3.6. Consider the solution ỹ3=-ϕ(-x;α,η̃) as in (iii), for x∈[-η̃,0], with ỹ3(-η̃)=0. Therefore ỹ3 is the continuation of ỹ2.Letỹ̃4(x)=-ϕ(-x;-α,η̃̃) be a solution of (3.5) as in (iv), for x∈[0,η̃̃].The following lemma findsη̃̃ so that ỹ3(0)=ỹ̃4(0) therefore so that ỹ̃4 is the continuation of ỹ3.Lemma 3.7.
Let|α|<α1 and |η̃-1|≤c2|α| with α1 small enough so that η̃∈[1/2,3/2].
There exists a uniqueη̃̃>0 so that ϕ(0;-α,η̃̃)=ϕ(0;α,η̃) and η̃̃ depends analytically on η̃ and α for |α|<α2 for α2 small enough. One has |η̃̃-1|≤c3|α| for some c3>0 and
(3.32)η̃̃=η̃-η̃4c1α+2α2η̃7c12-α3η̃10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
We need to findη̃̃=η̃̃(η̃,α) so that
(3.33)F(η̃̃,η̃,α)=ϕ(0;-α,η̃̃)-ϕ(0;α,η̃)=0.
Note that the functionF above is the same as (3.27). By Lemma 3.6 the present lemma follows.


## 3.4.1. Solutions in Other Quadrants

We found an expression for the solutiony(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.Letη∈[1/2,3/2], α with |α|<α0.(i)
Lety1=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 functiony2=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 functiony3=-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 functiony4=-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

Letη,η̃∈[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 ỹ2(x)=ϕ(-x;-α,η̃) solution as in (ii), for x∈[-η̃,0].The following lemma findsη̃ so that y1(0)=ỹ2(0), therefore so that y1 is the continuation of ỹ2Lemma 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 uniqueη̃=η+O(α) so that
(3.25)ϕ(0;α,η)=ϕ(0;-α,η̃),
and η̃ depends analytically on η and α for |α|<α1 for α1 being small enough. One has |η̃-1|≤c2|α| for some c2>0 and
(3.26)η̃=η-η4c1α+2α2η7c12-α3η10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
Let(3.27)F(η̃,η,α)=ϕ(0;-α,η̃)-ϕ(0;α,η),
which is a function analytic in (η̃,η,α) by Lemma 3.2 and relation (3.12).
We haveF(η,η,0)=0 and
(3.28)∂F∂η̃(η,η,0)=-4η3[1-v(1;0)]=-4η3≠0,
therefore the implicit equation F(η̃,η,α)=0 determines η̃=η̃(η,α) as an analytic function of α, η for α small.
We have(3.29)|η̃-η|≤|α|sup|α|<α1,|η|<c1α1|∂η̃∂α|=c1′|α|,
therefore
(3.30)|η̃-1|≤|η̃-η|+|η-1|≤(c1+c1′)|α|≡c2|α|.
The expansion ofη̃ in power series of α is found as follows. By using (3.24), (3.25) becomes
(3.31)η̃4-η̃4v(1;-2η̃3α)=η4-η4v(1;2η3α),
where by substituting (3.13) and η̃=∑n≥0δnη̃n followed by power series expansion in α we obtain (3.26).


## 3.4.3. Matching at the Negative y-Axis

Letη,η̃̃∈[1/2,3/4] and η̃ given by Lemma 3.6. Consider the solution ỹ3=-ϕ(-x;α,η̃) as in (iii), for x∈[-η̃,0], with ỹ3(-η̃)=0. Therefore ỹ3 is the continuation of ỹ2.Letỹ̃4(x)=-ϕ(-x;-α,η̃̃) be a solution of (3.5) as in (iv), for x∈[0,η̃̃].The following lemma findsη̃̃ so that ỹ3(0)=ỹ̃4(0) therefore so that ỹ̃4 is the continuation of ỹ3.Lemma 3.7.
Let|α|<α1 and |η̃-1|≤c2|α| with α1 small enough so that η̃∈[1/2,3/2].
There exists a uniqueη̃̃>0 so that ϕ(0;-α,η̃̃)=ϕ(0;α,η̃) and η̃̃ depends analytically on η̃ and α for |α|<α2 for α2 small enough. One has |η̃̃-1|≤c3|α| for some c3>0 and
(3.32)η̃̃=η̃-η̃4c1α+2α2η̃7c12-α3η̃10(4c3+212c13+10c2c1)+O(α4),
where cn=vn(1) with vn(ξ) given by (2.4), (2.5).Proof.
We need to findη̃̃=η̃̃(η̃,α) so that
(3.33)F(η̃̃,η̃,α)=ϕ(0;-α,η̃̃)-ϕ(0;α,η̃)=0.
Note that the functionF above is the same as (3.27). By Lemma 3.6 the present lemma follows.


## 3.5. The First Return Map

Letη∈[1/2,3/2] and α2 as in Lemma 3.7. Then η̃̃ 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 η̃̃ as a convergent power series in η, with coefficients dependent on η, whose first terms are(3.34)η̃̃=η-2η4c1α+8η7c12α2-η10(20c2c1+8c3+49c13)α3+O(α4).To obtain the pointX̃̃ where the solution of (2.1) with X(0)=ϵ>0 and Y(0)=0 first returns to the positive X-axis let α=21/2ϵ3 and multiply (3.34) by ϵ (since we have X=ϵx) and, finally, let η=1. We obtain that X̃̃ is analytic in ϵ for small ϵ and(3.35)X̃̃=ϵ-23/2c1ϵ4+16c12ϵ7-23/2(20c2c1+8c3+49c13)ϵ10+O(ϵ13),
where cn=vn(1) with vn 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).



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*Source: 590856-2009-10-18.xml*