The first return map for planar vector fields with nilpotent linear part with a center or a focus
aa r X i v : . [ m a t h . C A ] M a y THE FIRST RETURN MAP FOR PLANAR VECTORFIELDS WITH NILPOTENT LINEAR PART WITH ACENTER OR A FOCUS
RODICA D. COSTIN
Abstract.
The return map for planar vector fields with nilpotentlinear part (having a center or a focus and under an assumptiongenerically satisfied) is found as a convergent power series whoseterms can be calculated iteratively. The first nontrivial coefficientis the value of an Abelian integral, and the following ones areexplicitly given as iterated integrals built with algebraic functions. Introduction
The study of planar vector fields has been the subject of intense in-vestigation, due to their importance in applications, and in connectionto Hilbert’s 16th Problem [21]. Significant progress has been made inthe geometric theory of these fields, as well as in bifurcation theory,normal forms, foliations, and the study of Abelian integrals (see therecent book [7]).First return maps have been studied in relation to existence of closedorbits; more generally, return maps are important in a large arrayof applications (see [13] and references therein) and also in logic, inconnection to o-minimality [19].A fundamental result regarding the asymptotic form of return mapsstates that if the singular points of a C ∞ vector field are algebraicallyisolated, there exists a semitransversal arc such that the return mapadmits an asymptotic expansion is positive powers of x and logs (withthe first term linear), or has its principal part a finite composition ofpowers and exponentials [14], [20].In the case when the linear part of the vector field has non-zero eigen-values there is a good understanding of the return map [2], [3], [4], [5], [8],[9], as well as for perturbations of Hamiltonians [10], [11], or for per-turbations of integrable systems [12]. In the general setting, however,there are few results available [1], [6], [17], [18].The present paper establishes an iterative procedure for calculatingthe return map as an integer power series for generic vector fields withnilpotent linear part in the case of a center or a focus. The first few coefficients in these series are explicitly given. Using an algorithm givenhere these coefficients can be found explicitly up to any (finite) orderin terms of iterated integrals involving algebraic functions.A fundamental result concerning planar vector fields analytic near astationary point (0 ,
0) states that if the linear part is not zero, havingboth eigenvalues zero, then such fields have the normal form near (0 , x = y, ˙ y = a ( x ) + yb ( x )with a ( x ) , b ( x ) analytic at 0, with a (respectively, b ) having a zero oforder at least two (respectively, one) at x = 0 [15].Furthermore, if the origin is a center or a focus, and if the linear partof this system is not radial (i.e. does not have the form ˙ x = λx, ˙ y = λy ),then after an analytic change of variables (1) can be written as [16](2) ˙ z = − w f ( z ) + z l +1 g ( z ) , ˙ w = k z k − f ( z ) + k w z l g ( z )where(3) 1 ≤ k ≤ l + 1 and f (0) = 0The present paper studies the first return map for (2) under the sup-plementary assumption :(4) k = l + 12. Main result
The main result is the following:
Theorem 1.
Consider the system (2) with f ( z ) , g ( z ) analytic at and satisfying(3), (4), and its transversal T : f ( z ) w = z l +1 g ( z ) ( z > , small).Then the first return to T of the solution with z (0) = ǫ (with ǫ smallenough) is at the point with z = Z ( ǫ ) analytic in ǫ , with Z ( ǫ ) = ǫ + X n ≥ l − k +2 ǫ n Z n where Z n can be calculated iteratively in terms of iterated integrals in-volving algebraic functions. The proof of Theorem 1 shows that T is indeed transversal. Also,the proof contains, and it is built around, an algorithm for calculatingiteratively the coefficients Z n ; in particular, the first two of them arecalculated: (61), (62) with the notations (9), (7), (27), (29), (52).Necessary conditions for a system to have closed orbits (thus a cen-ter) are discussed in § IRST RETURN MAP 3 Proof of Theorem 1
Normalization.
Eliminating the time in (2) we obtain(5) dwdz = kz k − f ( z ) + kwz l g ( z ) − wf ( z ) + z l +1 g ( z )Solutions of (2) provide smooth parametrizations for solutions of (5).There are points where the graph of a solution w ( z ) of (5) has verticaltangents; at such a point the parametrization (2) prescribes that w ( z )is continued with another solution through the same point, and havinga vertical tangent as well.It is convenient to straighten the curve T using a substitution: let(6) w = z l +1 F ( z ) + w with(7) F ( z ) = g ( z ) f ( z )(note that F ( z ) is analytic at 0 since f (0) = 0). Equation (5) becomes(8) d ( w ) dz = − kz k − A ( z ) − w z p + k − B ( z )with the notations(9) p = l − k + 1 , p ≥ A ( z ) = 1 + z p F ( z ) ≡ A F ( z )(11) B ( z ) = 2 z dFdz + 2( p + 2 k ) F ( z ) ≡ B F ( z )The problem translates into the study of the first return to the posi-tive z -axis of solutions of (8) which start close enough to the origin, i.e.which satisfy the initial condition w ( ǫ ) = 0 with ǫ > ǫ in the equation.With the notation(12) z = ǫx, w = ǫ k y equation (8) becomes(13) d ( y ) dx = − k x k − A ( ǫx ) − ǫ p y x p + k − B ( ǫx )While the initial condition w ( ǫ ) = 0 becomes y (1) = 0, it is usefulto study solutions with the more general initial condition y ( η ) = 0 with η in a neighborhood of 1. RODICA D. COSTIN
It is assumed that(14) η ∈ (1 − η , η ) for some fixed η ∈ (0 , General behavior of solutions of (13).
In parametrized formequation (13) is(15) ˙ x = − y, ˙ y = 2 k x k − A ( ǫx ) + ǫ p y x p + k − B ( ǫx )Consider the solution of (15) with the initial condition x (0) = η, y (0) =0 for some η >
0. Let ( x , y ) be the solution of the system (15) for ǫ = 0: ˙ x = − y , ˙ y = 2 k x k − , x (0) = η, y (0) = 0This solution satisfies(16) y + x k = η k and clearly(17) x ( t ) = x ( t ) + O ( ǫ ) , y ( t ) = y ( t ) + O ( ǫ ) ( ǫ → Positive solutions of (13) for x > . Lemma 2 shows that thereexists a unique solution of (13) so that y ( η ) = 0 and y ≥ ǫ = 0.It also shows that this solution is defined for x ∈ [0 , η ] and establishesan iterative procedure for calculating its power series expansion in ǫ .With the substitution y = u / equation (13) becomes(18) dudx = − k x k − A ( ǫx ) − ǫ p u / x p + k − B ( ǫx ) Lemma 2.
There exists ǫ > so that so that the following holds.Let η with (14). For any ǫ with | ǫ | < ǫ equation (18) with thecondition u ( η ) = 0 has a unique solution u = u ( x ; ǫ, η ) for x ∈ [0 , η ] .We have u ( x ; 0 , η ) > for x ∈ [0 , η ) , and u ( x ; ǫ, η ) is analytic in ǫ and η . Proof of Lemma 2. Denote(19) ˜ P ( x ; η, ǫ ) = 1 η − x Z xη (cid:2) − kt k − A ( ǫt ) (cid:3) dt = η k − − x/η Z x/η kσ k − A ( ǫησ ) dσ Heuristics.
A particular vector field was studied in [6], unveilingthe main ideas involved. In the present general case a few heuristicconsiderations are the following.
IRST RETURN MAP 5
Looking for solutions of (13) with y ( η ) = 0 we obtain that y ( x ) ∼ c ( η − x ) / (1 + o (1)) ( x → η ) with c = 2 kη k − A ( ǫη )Note that A ( ǫη ) = 1 + O ( ǫ ) > ǫ small enough, in view of (10).Therefore(20) u = y = ( η − x ) ˜ P ( x ; η, ǫ ) + ∆( x )with ∆ = o ( η − x ) as x → η . In fact, substitution of (20) in (18) gives(21) ∆ ∼ const ( η − x ) / These observations motivate the following substitutions.Denote ξ = r − xη (with the usual branch of the square root for x/η < δ = ǫη Note that (19) can be written ˜ P ( x ; η, ǫ ) ≡ η k − P ( ξ ; δ ) where(23) P ( ξ ; δ ) = ξ − Z − ξ k σ k − A ( δσ ) dσ = 2 k Z (1 − ξ s ) k − A (cid:0) δ (1 − ξ s ) (cid:1) ds With the substitution(24) u ( x ) = ( η − x ) η k − [ P ( ξ ; δ ) + v ( ξ ; δ )]equation (18) becomes(25) ξ dvdξ + 2 v = 2 δ p ξ (1 − ξ ) p + k − ( P + v ) / B (cid:0) δ (1 − ξ ) (cid:1) Note that y ( η ) = 0 implies u ( η ) = 0 and then necessarily v (0) = 0by (19), (20), (21).Lemma 2 follows with ǫ = δ / (1 − η ) if we show the following: Lemma 3.
These exists δ > so that for any δ with | δ | < δ equation(25) has a unique solution v = v ( ξ ; δ ) so that v (0) = 0 and this solutionis defined for ξ ∈ [0 , .Moreover, v ( ξ ; · ) is analytic for δ ∈ C with | δ | < δ .The terms of its power series (26) v ( ξ ; δ ) = X n ≥ p δ n v n ( ξ ) RODICA D. COSTIN can be calculated recursively. In particular, the first terms are as fol-lows.Using the notations (27) Φ p,k ( ξ ) = ξ − Z − ξ s p + k − (1 − s k ) / ds (28) Ψ k ( ξ ) = ξ − Z − ξ ds s k (1 − s k ) / Z s dσ σ k (1 − σ k ) / and (29) B = 2( p +2 k ) F (0) ≡ B F ;0 , B = 2 ( p +2 k +1) F ′ (0) ≡ B F ;1 we have: (30) v p ( ξ ) = B Φ p,k ( ξ ) and, moreover, for p ≥ we have (31) v ( ξ ; δ ) = δ p v p ( ξ ) + δ p +1 B Φ p + k +1 ,k ( ξ ) + O (cid:0) δ p +2 (cid:1) while for p = 1 we have (32) v ( ξ ; δ ) = δ v ( ξ ) + δ (cid:20) B Φ ,k ( ξ ) + 12 B Ψ k ( ξ ) (cid:21) + O (cid:0) δ (cid:1) Proof of Lemma 3.
Multiplying (25) by ξ and integrating we obtain: v ( ξ ; δ ) = ξ − (cid:20) C + 2 δ p Z ξ t (1 − t ) p + k − [ P ( t ; δ ) − v ( t ; δ )] / B (cid:0) δ (1 − t ) (cid:1) dt (cid:21) Since v (0) = 0 then the constant C must vanish. It follows that v is afixed point ( v = J [ v ]) for the operator(33) J [ v ] ( ξ ; δ ) = 2 δ p ξ − Z ξ t (1 − t ) p + k − [ P ( t ; δ ) + v ( t ; δ )] / B (cid:0) δ (1 − t ) (cid:1) dt = 2 δ p ξ Z s (1 − ξ s ) p + k − [ P ( ξs ; δ ) + v ( ξs ; δ )] / B (cid:0) δ (1 − ξ s ) (cid:1) ds Estimates for P ( t ; δ ) : Let r > F is analytic in a neighborhoodof | z | ≤ r , and so that | A ( z ) − | ≤ c for all | z | ≤ r for some c with0 < c < / (2 k ) (see (10)).Then from (23), for ξ ∈ [0 ,
1] and | δ | ≤ r we have the upper estimate(34) | P ( ξ ; δ ) | ≤ k (1 + c ) Φ p,k ( ξ ) is expressible in terms of the incomplete beta function. IRST RETURN MAP 7 and the lower estimate(35) | P ( ξ ; δ ) | ≥ k Z (1 − ξ s ) k − ds − (cid:12)(cid:12) k Z (1 − ξ s ) k − (cid:2) A (cid:0) δ (1 − ξ s ) (cid:1) − (cid:3) ds (cid:12)(cid:12) ≥ k Z (1 − s ) k − ds − k Z (1 − ξ s ) k − (cid:12)(cid:12) A (cid:0) δ (1 − ξ s ) (cid:1) − (cid:12)(cid:12) ds ≥ − kc > The operator J is contractive: Let M = sup | z |≤ r | B ( z ) | .Let µ be a number with 0 < µ < − kc .Let δ > δ p
23 [2 k (1 + c ) + µ ] / M < µ, δ p M − kc − µ ) / < B be the Banach space of functions f ( ξ ; δ ) continuous for ξ ∈ [0 ,
1] and analytic on the (complex) disk | δ | < δ , continuous on | δ | ≤ δ , with the norm k f k = sup ξ ∈ [0 , sup | δ |≤ δ | f ( ξ ; δ ) | Let B µ be the ball B µ = { f ∈ B ; k f k ≤ µ } .The operator J defined by (33) is defined on B µ . Indeed, for f ∈ B µ we have(37) | P ( t ; δ ) + f ( t ; δ ) | ≥ | P ( t ; δ ) | − | f ( t ; δ ) | ≥ − kc − µ > f and P are analytic in δ , then so is ( P + f ) / , andtherefore so is J [ f ].Also, J B µ ⊂ B µ since for f ∈ B µ , using (33), (34), (36) we have (cid:12)(cid:12) J f ( ξ ; δ ) (cid:12)(cid:12) ≤ | δ | p
23 [2 k (1 + c ) + µ ] / M < µ
Moreover, the operator J is a contraction on B µ . Indeed, using theestimate (cid:12)(cid:12) ( P + f ) / − ( P + f ) / (cid:12)(cid:12) ≤ | f − f |
12 sup f ∈B µ | P + f | − / ≤ | f − f | − kc − µ ) / (by (37)) we obtain (cid:12)(cid:12) J f − J f (cid:12)(cid:12) ≤ c k f − f k with c = δ p M − kc − µ ) / < J has a unique fixed point in B µ , which isthe solution v ( ξ ; δ ), analytic for δ ∈ C with | δ | < δ . RODICA D. COSTIN
A recursive algorithm for calculating the power series in ǫ : To obtain the power series (26) substitute an expansion v ( ξ ; δ ) =+ P n ≥ δ n v n ( ξ ) in (25). It follows that for n < p we have ξv ′ n + 2 v n = 0with v n (0) = 0, therefore v n ( ξ ) ≡ n < p .Using (23), (10) we obtain for P ( ξ ; δ ) a power series in δ , with coef-ficients polynomials in ξ :(38) P ( ξ ; δ ) = P ( ξ ) + X m ≥ δ p + m P p + m ( ξ )where(39) P ( ξ ) = 1 − (1 − ξ ) k ξ and(40) P p + m ( ξ ) = F ,m k p + 2 k + m − (1 − ξ ) k +2 p + m ξ with the notation(41) F ( z ) ≡ X m ≥ F ,m z m and in particular(42) F , = F (0) Substitution of (26), (38), followed by expansion in power series in δ give(43) ( P ( ξ ; δ ) + v ( ξ ; δ )) / B (cid:0) δ (1 − ξ ) (cid:1) ≡ X n ≥ δ n R n ( ξ )where R n = R n [ v p , . . . , v n , ξ ].From (25) and (43) we obtain the recursive system ξ dv n dξ + 2 v n = 2 ξ (1 − ξ ) p + k − R n − p , n ≥ p with the only solution with v n (0) = 0 given recursively by(44) v n ( ξ ) = 2 ξ − Z ξ t (1 − t ) p + k − R n − p ( t ) dt The first terms:
To calculate the first few R n note that(45) ( P ( ξ ; δ ) + v ( ξ ; δ )) / B (cid:0) δ (1 − ξ ) (cid:1) = B P ( ξ ) / + δB P ( ξ ) / (1 − ξ ) + δ p B v p ( ξ )2 P ( ξ ) / + O (cid:0) δ (cid:1) IRST RETURN MAP 9 where B , B are coefficients in the expansion B ( z ) = B + zB + O ( z ),and in view of (11), they are (29).In particular(46) v p ( ξ ) = 2 B ξ − Z ξ t (1 − t ) p + k − P ( t ) / dt = B ξ − Z − ξ s p + k − (1 − s k ) / ds which equals (30).For p ≥ P + v ) / B (cid:0) δ (1 − ξ ) (cid:1) = B P ( ξ ) / + δB P ( ξ ) / (1 − ξ ) + O (cid:0) δ (cid:1) and using (45) we obtain (31).For p = 1 we have O ( δ p ) = O ( δ p +1 ) and there is one more term inthe second nontrivial coefficient of v ( ξ ; δ ). The calculation is straight-forward: using relation (45) for p = 1 and (44) we obtain (32). (cid:3) The following Corollary gathers the conclusions of the present sec-tion. Many quantities depend on the function F , see (7), (10), (11),(23), and we add the subscript F for them: Corollary 4.
There exists ǫ > so that for any η with (14) and ǫ with | ǫ | < ǫ equation (13) has a unique solution y ( x ) on [0 , η ] satisfying y ( η ) = 0 and y > on [0 , η ) for ǫ = 0 .Moreover, this solution has the form (48) y ( x ) = φ F ( x ; ǫ, η ) ≡ ( η − x ) / η k − / (cid:20) P F (cid:18)r − xη ; ǫη (cid:19) + v F (cid:18)r − xη ; ǫη (cid:19)(cid:21) / with P F ( ξ ; δ ) ≡ P ( ξ ; δ ) given by (23) and v F ( ξ ; δ ) solution of (25) with v F (0; δ ) = 0 and v F ( ξ ; 0) = 0 .The map ( ǫ, η ) v F ( ξ ; ǫη ) is analytic for | ǫ | < ǫ and η as in (14). (cid:3) Solutions of (13) in other quadrants and matching.
Solutions in the four quadrants.
Corollary 4 gives an expressionfor the solution y ( x ) of (13) for x > y >
0. Solutions in the otherquadrants are found in the following way.Let ǫ with | ǫ | < ǫ with ǫ given by Lemma 2. (i) Let y ( x ) = φ F ( x ; ǫ, η ) the solution (48) of (13), defined for x ∈ [0 , η ], with y ( η ) = 0. We have y ( x ) = ( η k − x k ) / + O ( ǫ ) in view of(16), (17), and we will refer to y as a ”solution in the first quadrant”.Solutions ”in the other quadrants” are obtained as follows.For a function F ( z ) denote by J i F the following functions:(49) ( J F )( x ) = ( − p + k F ( − x ) , ( J F )( x, y ) = ( − p + k − F ( − x ) , ( J F )( x ) = − F ( x )Note that J J = J and A J F = A F . (ii) It is easy to check that the function y ( x ) = φ J F ( − x ; ǫ, η ) isa solution of (13). It is obviously defined for x ∈ [ − η, y ( − η ) = 0 and y ( x ) = ( η k − x k ) / + O ( ǫ ). (iii) Similarly, the function y ( x ) = − φ J F ( − x ; ǫ, η ) is a solutionof (13) defined for x ∈ [ − η, y ( − η ) = 0 and y ( x ) = − ( η k − x k ) / + O ( ǫ ). (iv) The function y ( x ) = − φ J F ( x ; ǫ, η )( x ; ǫ, η ) is a solution of (13)defined for x ∈ [0 , η ]; we have y ( η ) = 0 and y ( x ) = − ( η k − x k ) / + O ( ǫ ).3.4.2. Matching at the positive y -axis. Let η, ˜ η satisfying (14) and let y ( x ) = φ F ( x ; ǫ, η ) solution as in (i) , for x ∈ [0 , η ] and ˜ y ( x ) = φ J F ( − x ; ǫ, ˜ η ) solution as in (ii) , for x ∈ [ − ˜ η, η so that y (0) = ˜ y (0), therefore so that y is the continuation of ˜ y : Lemma 5.
Let | ǫ | < ǫ . Let η so that | η − | ≤ c | ǫ | with c smallenough so that c ǫ < η .There exists a unique ˜ η = η + O ( ǫ ) so that (50) φ F (0; ǫ, η ) = φ J F (0; ǫ, ˜ η ) Denote this (51) ˜ η = N F ( η, ǫ ) Moreover, N F ( η, ǫ ) depends analytically on η and ǫ for | ǫ | < ǫ for ǫ small enough. We have | ˜ η − | ≤ c | ǫ | for some c > .Furthermore we have, in the notations (27), (28), (29), and with (52) θ p = ( − p + k − IRST RETURN MAP 11 that for p ≥ η = N F ( η, ǫ ) = η + ǫ p η p +1 B Φ p,k (1)2 k (1 + θ p )+ ǫ p +1 η p +2 B Φ p +1 ,k (1)2 k (1 − θ p ) + O ( ǫ p +2 ) and for p = 1(54) ˜ η = N F ( η, ǫ ) = η + ǫ η B Φ ,k (1)2 k (1 + θ )+ ǫ η (1 − θ ) B Φ ,k (1) + (1 + θ ) /k B Φ ,k (1) k + O ( ǫ ) Proof of Lemma 5.
Using (48) equation (50) is equivalent to solving the implicit equation(55) G (˜ η, η, ǫ ) = 0 where G (˜ η, η, ǫ ) = ˜ η k [ P J F (1; ǫ ˜ η ) + v J F (1; ǫ ˜ η )] − η k [ P F (1; ǫη ) + v F (1; ǫη )]which is a function analytic in (˜ η, η, ǫ ) by Lemma 2.We have G ( η, η,
0) = 0 (by (38), (39), (26)) and ∂G∂ ˜ η ( η, η,
0) = 2 kη k − = 0therefore by the implicit function theorem and using the compactness ofthe interval | η − | ≤ c ǫ , equation G (˜ η, η, ǫ ) = 0 determines ˜ η = ˜ η ( η, ǫ )as an analytic function if | ǫ | ≤ ǫ for ǫ small enough.Since ˜ η ( η,
0) = η we have, for | ǫ | ≤ ǫ and | η − | ≤ c ǫ , | ˜ η ( η, ǫ ) − η | ≤ | ǫ | sup | ǫ |≤ ǫ , | η − |≤ c ǫ (cid:12)(cid:12) ∂ ˜ η∂ǫ (cid:12)(cid:12) = c ′ | ǫ | therefore | ˜ η ( η, ǫ ) − | ≤ | ˜ η ( η, ǫ ) − η | + | η − | ≤ ( c + c ′ ) | ǫ | ≡ c | ǫ | We can assume c ǫ < η by lowering ǫ .The expansion of ˜ η ( η, ǫ ) in power series of ǫ is found by introducingthe expansions (38), (26), (44) in (55) and noting that P (1) = 1, P J F ;2 (1) = P (1) (see (39), (40), (42)), and that the coefficients of B J F in (29) are B J F ;0 = ( − p + k B and B J F ;1 = ( − p + k +1 B . (cid:3) Matching at the negative y -axis. Let η, ˜˜ η satisfying (15) and | η − | ≤ c | ǫ | as in Lemma 5. Let ˜ η given by Lemma 5. Considerthe solution ˜ y ( x ) = − φ J F ( − x ; ǫ, ˜ η ) as in (iii) , for x ∈ [ − ˜ η, y ( − ˜ η ) = 0. Therefore ˜ y is the continuation of ˜ y .Let ˜˜ y ( x ) = − φ J F ( x ; ǫ, ˜˜ η ) be a solution of (13) as in (iv) , for x ∈ [0 , ˜˜ η ].The following Lemma finds ˜˜ η so that ˜ y (0) = ˜˜ y (0), therefore so that˜˜ y is the continuation of ˜ y : Lemma 6.
Let | ǫ | < ǫ and | ˜ η − | ≤ c | ǫ | with ǫ small enough so that ˜ η satisfies (15).There exists a unique ˜˜ η > so that (56) − φ J F (0; ǫ, ˜˜ η ) = − φ J F (0; ǫ, ˜ η ) Moreover, we have (57) ˜˜ η = N J F (˜ η, ǫ ) where N F denotes the function (51) of Lemma 5.Therefore ˜˜ η depends analytically on ˜ η and ǫ for | ǫ | < ǫ for ǫ smallenough. We have | ˜˜ η − | ≤ c | ǫ | for some c > .Furthermore, the first coefficients of the ǫ series of N J F ( η, ǫ ) and N F ( η, ǫ ) coincide: (58) N J F ( η, ǫ ) = N F ( η, ǫ ) + O ( ǫ p +2 ) Proof.
Note that the equation (56) for ˜˜ η = ˜˜ η (˜ η, ǫ ) is the same as the equa-tion (55) for ˜ η = ˜ η ( η, ǫ ), only with F replaced by J F . Noting that P J F ;2 (1) = P (1) and that B J F ;0 = θ p B , B J F ;1 = − θ p B Lemma 6follows from Lemma 5. (cid:3)
The first return map.
Let η satisfying (14) and ǫ as in Lemma 6.Then ˜˜ η given by Lemma 6 is the first return to the positive x -axis ofthe solution with x (0) = η, y (0) = 0 and it is analytic in ǫ and η :(59) ˜˜ η = N J F ( N F ( η, ǫ ) , ǫ )At this point it is enough to take η = 1. Going back through the sub-stitutions (12), (6), we obtain that the first return map to the transver-sal w = z l +1 F ( z ) (with z > z (0) = ǫ , w (0) = ǫ l +1 F ( ǫ ) is attained for(60) z = ǫ N J F ( N F (1 , ǫ ) , ǫ ) ≡ ǫ + X n ≥ p +1 Z n ǫ n IRST RETURN MAP 13 and the terms of this convergent power series in ǫ can be calculatedrecursively.In particular, the first terms are obtained from (53), (54), (58): for p ≥ z = ǫ + 2 ǫ p +1 B Φ p,k (1)2 k (1 + θ p )+ 2 ǫ p +2 B Φ p +1 ,k (1)2 k (1 − θ p ) + O ( ǫ p +3 )and for p = 1(62) z = ǫ + 2 ǫ B Φ ,k (1)2 k (1 + θ )+ 2 ǫ (cid:20) (1 − θ ) B Φ ,k (1) + (1 + θ ) /k B Φ ,k (1) k + (cid:18) B Φ ,k (1)2 k (1 + θ ) (cid:19) + O ( ǫ )3.6. Closed trajectories.
Relation (60) together with the construc-tive method presented allows to calculate, recursively, the coefficients Z n , and therefore to decide whether the origin is a center or a focus:the origin is a center if and only if all Z n vanish.As a practical matter, for each value of p = l − k + 1 all the powerseries should first be properly ordered (the order in which the termsappear does depend on the value of p ) and then the series can becalculated (at least in principle) to any order n . The conditions Z n = 0for n ≤ n can be written in terms of the function F , and they arenecessary conditions for the fixed point to be a center.The first such conditions follow from the first two nontrivial termscalculated here: using (61), (62) for the origin to be a center we musthave Z p +1 = Z p +2 = 0 which implies B (1 + θ p ) = 0 and B (1 − θ p ) =0, which in turn means that if p + k − F ′ (0) = 0, while if p + k − F (0) = 0. Acknowledgements.
The author is grateful to Chris Miller for sug-gesting the problem, to Barbara Keyfitz for very interesting discussions,and to Christiane Rousseau for illuminating e-mail correspondence.
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