Volume 2009, Article ID 590856,11pages doi:10.1155/2009/590856
Research Article
The Return Map for a Planar Vector
Field with Nilpotent Linear Part: A Direct and
Explicit Derivation
Rodica D. Costin
Ohio State University, Columbus, OH 43210, USA
Correspondence should be addressed to Rodica D. Costin,[email protected]
Received 18 May 2009; Revised 4 July 2009; Accepted 24 August 2009
Recommended by Manfred H. Moller
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.
Copyrightq2009 Rodica D. Costin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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 integrals1,2.
The Poincar´e 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.,3and references therein, and also in connection to o-minimality4.
The monodromy problemdetermining when the singularity is a center or a focus was solved by Andreev5.
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 ofxand logswith the first term linear, or has its principal part a finite composition of powers and exponentials6,7.
the other hand, there are few results available in the general setting18–20. The recent papers
21,22contain 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
˙
X−Y, Y˙ 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 class23, and for which there areto the author’s knowledge no methods available to generate the return map.
The main result is the following.
Proposition 2.1. Letwith 0< < 0be small enough.
The solution of 2.1satisfyingX0 , Y0 0 first returns to the positiveX-axis at the valueXsatisfying
X
∞
n1
Xn3n1, 2.2
which is a convergent series.
The coefficientsXncan be calculated iteratively. In particular,
X1 −23/2c1, X2 16c21, X3−23/2
20c2c18c349c31
, 2.3
wherecnvn1withvngiven by
v1ξ ξ−2
ξ
0
t4pt3/2 dt, v2ξ − 3 2ξ
−2
ξ
0
t4pt1/2v1tdt, 2.4
v3ξ ξ−2
ξ
0
t4
−3 2 pt
1/2
v2t 3 8
v12t
pt1/2
dt, 2.5
where
pt 4−6t24t4−t6. 2.6
3. Proof of
Proposition 2.1
3.1. Normalization
It is convenient to normalize the variablesX, Y, t so that the constantappears as a small parameter in the equation with
Xx, Y 2−1/22y, τ 2−3/2t, 3.1
the system2.1becomes
dx dτ −2y,
dy dτ 4x
3−αy3, 3.2
whereαis the small parameter
α21/23. 3.3
While the initial conditionX0 becomesx0 1, it is useful to study solutions of
3.2with the more general initial conditionx0 ηwithηin a neighborhood of 1.
Remark 3.1. System 3.2has the formdHαω 0 withH y2 x4,ω y3dxandαa small parameter. The recent result15gives a formalism for finding the return map for this type of systems; see also17. 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 for3.2
L x, yy2x4, with d
dτL xτ, yτ
−2αyτ4≤0. 3.4
Since the set {y 0} contains no trajectories besides the origin which is the only equilibrium point of3.2, then the origin is asymptotically stable by the Krasovskii-LaSalle principle.
Consider the solution of3.2with the initial conditionx0 η, y0 0 for some
η∈1/2,3/2.
Sincey0 > 0 andx0 0, x0 < 0 thenyincreases andxdecreases for small
τ > 0. This monotony must change due to3.4, and this can happen only at some point where y 0 or where 4x3 αy3, whichever comes first. Sincey increases, then the first occurrence is a point where 4x3 αy3. At this pointx<0 soxcontinues to decrease, while
Denote by x0τ, y0τ the solution for α 0: x0 −2y, y0 4x30 and x00
η, y00 0. Thereforex40y20 η4. We have xτ x0τ Oα, yτ y0τ Oα thereforeηηOα.
Similar arguments show that the solutionxτ, yτof3.2continues to turn around the origin, crossing again the positivex-axis at a pointηηOα and, of course,η < ηby
3.4.
Solutionsxτ, yτof3.2provide smooth parametrizations for solutionsyxof
d dx
y2−4x3αy3. 3.5
Note that following the path xτ, yτ one full rotation around the origin corresponds to considering a positive solution of3.5, followed by a negative one.
3.3. Positive Solutions of
3.5
for
x >
0
Lemma 3.2shows that there exists a unique solution y ≥ 0 of 3.5 so thatyη 0; this solution is defined forx ∈ 0, ηand establishes an iterative procedure for calculating this solution.
Substitutingyu1/2in3.5we obtain
du dx −4x
3αu3/2. 3.6
Lemma 3.2. There existsδ0>0 independent ofηandαso that the following holds.
Letη >0. For anyαwith|α|< α0≤δ0η−3/2,3.6with the conditionuη 0 has a unique solutionuux;α, ηforx∈0, η. One hasux;α, η>0 forx∈0, η,α >0 andux;α, ηis analytic inαandη for|α|< α0andη >0.
Remark 3.3. We will use the results ofLemma 3.2only forηsuch that 1/2≤η ≤3/2. In this caseby loweringα0we can takeα0ofLemma 3.2independent ofηby taking
α0≤ δ0 2 minη
−3 43/273δ0
2 ≡c0δ0. 3.7
Proof ofLemma 3.2. Local analysis shows that solutions of3.6satisfyinguη 0 have an expansion in integer and half-integer powers ofη−xand we have
ux η4−x4−16
5 αη
9/2 η−x5/2
1o1, x−→η−, 3.8
which inspires the following substitutions. Denote
ξ
1−x
with the usual branch of the square root forx/η <1and let
ux η4−x4−η3 η−xvξ. 3.10
Note that ifuη 0 then necessarilyv0 0 by3.8. Equation3.6becomes
ξ d
dξvξ 2vξ δξ
3pξ−vξ3/2
, 3.11
where
δ2η3α, 3.12
and the polynomialpis given by2.6.Note thatpξ∈1,4forξ∈0,1. Lemma 3.2follows 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
vvξ;δforξ∈0,1so thatv0 0.
Moreover,vξ;·is analytic forδ∈Cwith|δ|< δ0and the terms of its power series
vξ;δ
n≥1
δnvnξ 3.13
can be calculated recursively; in particular, the first terms are2.4,2.5.
To proveLemma 3.4multiply3.11byξand integrate; we obtain thatvis a fixed point
vJvfor the operator
Jvξ δξ−2 ξ
0
t4pt−vt3/2dtδξ3 1
0
s4pξs−vξs3/2ds. 3.14
LetBbe the Banach space of functionsfξ;δcontinuous forξ ∈0,1and analytic on the
complexdisk|δ|< δ0, continuous on|δ| ≤δ0, with the norm
f sup
ξ∈0,1 sup
|δ|≤δ0
fξ;δ. 3.15
Letmbe a number with 0< m <1. Letδ0>0 be small enough, so thatδ0< m/ √
5 and
δ0<10/3/ √
5.
LetBmbe the ballBm{f ∈ Bm;f ≤m}.
We haveJ:Bm → Bm. Indeed, note that forf ∈ Bmwe have
therefore, sincef is analytic inδ, then so ispt−ft;δ3/2, and therefore so isJf. Also, if
f∈ Bmthen alsoJf ∈ Bmbecause
Jfξ;δ≤ |δ| 1
0
s4|pξs||fξs;δ|3/2ds≤ δ04m
3/2
5 < δ0 √
5< m. 3.17
Moreover, the operatorJis a contraction onBm. Indeed, using the estimate
p−f1
3/2−
p−f2
3/2≤
f1−f2 3
2 |f|≤msupp−f 1/2≤
f1−f2
34m1/2
2 3.18
we obtain
Jf1− Jf2≤cf1−f2 withcδ03√5
10 <1. 3.19
Therefore the operatorJhas a unique fixed point, which is the solutionvξ;δ. To obtain the power series3.13substitute an expansionvξ;δ v0ξ
n≥1δnvnξ
in3.11. It follows thatξv0 2v00 withv00 0, thereforev0ξ≡0.
Substitution of3.13inp−v3/2followed by expansion in power series inδgives
p−v3/2 p3/2
1−
n≥1
δnvn p
3/2 ≡p3/2
1
n≥1
δnRn
, 3.20
whereRnRnv1, . . . , vn, p. In particular,
R1− 3 2
v1
p, R2 −3 2 v2 p 3 8
v12
p2
. 3.21
From3.11we obtain the recursive system
ξd
dξvn2vnξ
3p3/2R
n−1, 3.22
forn≥1 and withR01, with the only solution withvn0 0 given recursively by
vnξ ξ−2
ξ
0
t4pt3/2Rn−1tdt. 3.23
The following gathers the conclusions of the present section.
Corollary 3.5. There existsα0 >0 so that for anyη ∈ 1/2,3/2andαwith|α|< α0,3.5has a unique solutionyxon0, ηsatisfyingyη 0 andy >0 on0, ηforα >0.
Moreover, this solution has the form
yφ x;α, η
η4−x4−η3 η−xv
1−x
η; 2η
3α
1/2
, 3.24
withvvξ, δa solution of3.11. The mapα→vξ; 2η3αis analytic for|α|< α 0.
3.4. Solutions of
3.5
in Other Quadrants and Matching
3.4.1. Solutions in Other Quadrants
We found an expression for the solutionyxof3.5forx > 0 andy >0. In a similar way, expressions in the other quadrants can be found. However, taking advantage of the discrete symmetries of3.5, these solutions can be immediately written down as follows.
Letη∈1/2,3/2,αwith|α|< α0.
iLety1 φx;α, ηbe the solution of3.24, defined forx ∈ 0, η, withy1η 0 andy1>0 forα >0,
Then:
iithe functiony2 φ−x;−α, ηis also a solution of3.5, defined forx∈−η,0; we havey2−η 0 andy2≥0 forα >0,
iiiThe functiony3−φ−x;α, ηis a solution of3.5, defined forx∈−η,0. We have
y3−η 0 andy3≤0 forα >0,
ivThe functiony4 −φx;−α, ηis a solution of3.5, defined forx ∈0, ηand we havey4η 0 andy4≤0 forα >0,
3.4.2. Matching at the Positive y-Axis
Letη,η∈1/2,3/4and lety1x φx;α, ηbe the solution of3.5as ini, forx∈0, η andy2x φ−x;−α,ηsolution as inii, forx∈−η, 0.
The following lemma finds η so that y10 y20, therefore so that y1 is the continuation ofy2
Lemma 3.6. Let|α|< α0withα0satisfying3.7. Letηso that|η−1|< c1|α|withα0small enough so thatc1α0≤1/2.
There exists a uniqueηηOαso that
andηdepends analytically onηandαfor|α|< α1forα1being small enough. One has|η−1| ≤c2|α| for somec2>0 and
ηη−η4c1α2α2η7c21−α3η10
4c321 2 c
3
110c2c1
Oα4, 3.26
wherecnvn1withvnξgiven by2.4,2.5.
Proof. Let
F η, η, α φ 0;−α,η−φ 0;α, η, 3.27
which is a function analytic inη, η, α byLemma 3.2and relation3.12. We haveFη, η,0 0 and
∂F ∂η η, η,0
−4η31−v1; 0 −4η3/0, 3.28
therefore the implicit equationFη, η, α 0 determinesηηη, αas an analytic function of
α,ηforαsmall. We have
η−η≤ |α| sup
|α|<α1,|η|<c1α1
∂∂αηc1|α|, 3.29
therefore
η−1≤η−ηη−1≤ c1c1
|α| ≡c2|α|. 3.30
The expansion ofηin power series of αis found as follows. By using3.24,3.25 becomes
η4−η4v1;−2η3αη4−η4v1; 2η3α, 3.31
where by substituting3.13andηn≥0δnηnfollowed by power series expansion inαwe
obtain3.26.
3.4.3. Matching at the Negative y-Axis
Letη,η∈1/2,3/4andηgiven byLemma 3.6. Consider the solutiony3 −φ−x;α,ηas in
iii, forx∈−η, 0, withy3−η 0. Thereforey3is the continuation ofy2. Lety4x −φ−x;−α,ηbe a solution of3.5as iniv, forx∈0,η.
Lemma 3.7. Let|α|< α1and|η−1| ≤c2|α|withα1small enough so thatη∈1/2,3/2.
There exists a uniqueη >0 so thatφ0;−α,η φ0;α,ηandηdepends analytically onη
andαfor|α|< α2forα2small enough. One has|η−1| ≤c3|α|for somec3>0 and
ηη−η4c1α2α2η7c21−α3η10
4c3 21
2 c 3
110c2c1
Oα4, 3.32
wherecnvn1withvnξgiven by2.4,2.5.
Proof. We need to findηηη, α so that
Fη,η, α φ0;−α,η−φ 0;α,η0. 3.33
Note that the functionFabove is the same as3.27. ByLemma 3.6the present lemma follows.
3.5. The First Return Map
Letη∈1/2,3/2andα2as inLemma 3.7. Thenηgiven byLemma 3.7is the first return to the positivex-axis of the solution withx0 η, y0 0 and it is analytic inαandη, therefore, by3.12, it is analytic infor fixedη.
Combining 3.26 and 3.32 we obtain η as a convergent power series in η, with coefficients dependent onη, whose first terms are
ηη−2η4c1α8η7c21α2−η10
20c2c18c349c31
α3Oα4. 3.34
To obtain the pointX where the solution of2.1withX0 >0 andY0 0 first returns to the positiveX-axis letα21/23and multiply3.34bysince we haveXx
and, finally, letη1. We obtain thatXis analytic infor smalland
X−23/2 c1416c217−23/2
20c2c18c349c31
10O13, 3.35
wherecnvn1withvngiven by2.4,2.5.
Remark 3.8. The first coefficient of the return map3.34is, up to a sign, the Melnikov integral of the system3.2, see15; of course, the present results are in agreement with this factsee the appendix for details.
Appendix
With the notationH y2x4 andω y3dxthe Melnikov integral of3.2is the quantity
x >0which means, in the notations used in the present paper, thatT η4then the return map of3.2has the formT →T−αMT Oα2 15.
For the present system we have
MT
y2x4Ty 3dx4
T1/4
0
T−x43/2dx4T
7 4
1
0
1−s43/2ds8T7/4c1. A.1
Taking the fourth root in the return map ofTwe obtain3.34.
Acknowledgment
The author is grateful to Chris Miller for suggesting the problem.
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