Centers for Trigonometric Abel Equations

In this paper we introduce the notion of strongly persistent centers, together with the condition of the annulation of some generalized moments, for Abel differential equations with trigonometric coefficients as a natural candidate to characterize the centers of composition type for these equations. We also recall several related concepts and discuss the differences between the trigonometric and the polynomial cases.

where P and Q are analytic functions starting with second order terms. The problem of determining necessary and sufficient conditions on P and Q for system (1) to have a center at the origin is known as the center-focus problem. From the works of Poincaré and Lyapunov it is well-known that Eq. (1) has a center at the origin when an infinite sequence of polynomial conditions among the coefficients of the Taylor expansions of P and Q at the origin are satisfied. These conditions are given by the vanishing of the so called Lyapunov constants. Moreover, when P and Q are polynomials of a given degree this set of conditions is finite due to the Hilbert's basis Theorem. Nevertheless from these results it is not easy to obtain explicit conditions on P and Q that force the origin to be a center.
Therefore it is natural to consider the following problem: Center-focus problem for Abel equations. Given the equatioṅ defined on the cylinder (r, θ) ∈ R × R/(2π Z), where A and B are trigonometric polynomials, give necessary and sufficient conditions on A and B to ensure that all the solutions r = r (θ, r 0 ), with initial condition r (0, r 0 ) = r 0 and |r 0 | small enough are 2π -periodic, i.e. r (0, r 0 ) = r (2π, r 0 ). For short, if this property holds we will say the the Abel equation has a center (at r = 0). It is well known that also for the above problem the existence of a center is guaranteed if some polynomials computed from the coefficients of A(θ ) and B(θ ) vanish. These quantities are given by the return map between {θ = 0} and {θ = 2π }, near r = 0, and can be thought as the Lyapunov constants for the Abel equation, see for instance [3,13]. The first two center conditions are In [3] the authors introduce a simple condition called composition condition, for short CC, which ensures that the corresponding Abel equation has a center. Roughly speaking the composition condition says that the primitives of the functions A and B depend functionally on a new 2π -periodic function. See Definition 1 in next section for the precise statement of the result. When an Abel equation has a center because A and B satisfy the CC we will say that the equation has a CC-center. In [1] it was shown that this condition is not necessary to have a center.
For planar differential equations (1) the simplest explicit conditions that imply that the origin is a center are either that the vector field is Hamiltonian, or that the vector field is reversible with respect to a straight line. The notion of CC-centers for Abel equations can also be seen as the simplest explicit condition for these equations to have a center. Moreover it is not difficult to prove that the Hamiltonian or reversible (with respect to a straight line) centers for planar differential equations (1) with homogeneous nonlinearities correspond, via the Cherkas transformation, to CC-centers for the corresponding Abel equations.
On the other hand several authors try to characterize the so-called persistent centers, see for instance [2] and the references therein and to relate them with the CC-centers. It is said that the Abel equatioṅ has a persistent center if it has a center for all small enough. In next section we give some equivalent formulations. It is easy to see that CC-centers are persistent centers but the converse of this implication is an open question.
Persistent centers satisfy the so-called moments condition, see for instance [2] or Theorem 8 for a stronger result. This last condition says that Summarizing, associated to trigonometric Abel equations we have CC-centers, persistent centers and moments condition. The current interest is to relate these three concepts.
Motivated by the above problem many authors have faced the equivalent question when the functions A and B, instead of being trigonometric polynomials are usual polynomials, see [5][6][7][8]10,12,17]. To fix the problem, in this situation the question is to give necessary and sufficient conditions on the real polynomials A(t) and B(t) to ensure that the solutions of the equation , for some given a and b and for initial conditions close enough to w = 0. Notice that contrary to what happens in the trigonometric case with a = 0 and b = 2π, such condition does not imply the global periodicity of w(t). There are many significative advances for this polynomial case. In [16] an example of a polynomial Abel equation satisfying the moments condition and not satisfying the composition condition is given. Later on, in [15] a full algebraic characterization of the moments condition in the polynomial case is done. In Sect. 4 we recall this result and prove that a natural trigonometric analogous to it does not hold. In this paper we consider the following natural extension of the Abel equation (3) on the cylinder,ṙ where A(θ ) and B(θ ) are trigonometric polynomials. Notice that the case m = 1 can be transformed into a Riccati equation which can be explicitly solved. In this case it is easy to see that (6) has a center if and only if the two conditions given in (4) hold.
Our purpose is to point out some relations between, persistent centers, CC-centers and moments condition for Eq. (6). In Sect. 2 we give several relations between these three concepts and in Sect. 3 we present some examples which show that some of them are not equivalent. Also we introduce several new classes, symmetric centers, degreepersistent centers and strongly persistent centers. Related to this last class we also consider some generalized moments conditions, which can be useful to have a better understanding of the situation. Finally, in Sect. 4, we present a diagram with the known implications among the different classes of centers and conditions considered in the paper.

Persistent Centers, Moment Conditions and the Composition Condition
In all the paper given any function C(θ ) we will denote by C(θ ) = θ 0 C(t)dt. In [3] the authors give the following sufficient condition for Eq. (6) to have a center, named the composition condition.

Definition 1
The functions A(θ ) and B(θ ) satisfy the composition condition (CC) if there exist C 1 -functions u, A 1 and B 1 , with u being 2π -periodic, and such that To see that the CC implies the existence of a center one can consider the differential equation Let R = R(u) one of its solutions. It is easy to see that then r (θ ) = R(u(θ )) is a solution of (6). Therefore, if A and B satisfy the CC given in (7), then the corresponding differential equation (6) has center for all n, m ∈ N, because r (2π) = R(u(2π)) = R(u(0)) = r (0). As we have already explained in the introduction there are centers for (6) which are no CC-centers, see [1,2] and also Proposition 18.
Another important notion is that of the persistent center. Before introducing it we prove some preliminary results. (6) and let r (θ, ρ) be the solution such that r (0, ρ) = ρ. Set

Lemma 2 Consider the Abel equation
Then I is and open interval containing 0. If in addition Eq. (6) has a center then I coincides with the set of initial conditions for which the solution is defined for all time. Moreover, in this case, it also coincides with the set of initial conditions that correspond to periodic orbits.
Proof First we prove that I is and open interval containing 0. It is open because of the continuous dependence of the flow on the initial conditions. It is an interval because if 0 < ρ 1 / ∈ I then r (θ, ρ 1 ) goes to infinity at some point θ 1 ∈ (0, 2π) and this fact implies that for all ρ > ρ 1 the corresponding solution goes to infinity as well, at some point 0 < θ ρ ≤ θ 1 . Hence ρ / ∈ I for all ρ > ρ 1 . A similar argument can be used when 0 > ρ 1 / ∈ I. So the return map ρ → r (2π, ρ) is well defined on I and, if the equation has a center, it is the identity for ρ small enough. Since it is analytic it follows that the return map is the identity on I. This ends the proof of the lemma.
The proof of the next lemma is straightforward.

Moreover, this differential equation has a center if and only if C(2π)
The proof of the following result follows the same ideas that the one of Proposition 2.1 of [4].

has a center. Then
be the solution of (6) such that r (0, ρ) = ρ. If the equation has a center then u k (2π) = 0, for all k ≥ 1. Plugging the above expression in (6) we obtain that Thus the condition V m = B(2π) = 0 holds. Therefore, from Lemma 3 we have that the equationṙ = B(θ )r m has also a center. Let r = R(θ, ρ) be its solution satisfying and plugging it again in (6) we obtain that Then and A(2π) = 0 as we wanted to prove. Proposition 5 Set n > m ≥ 1. The following three conditions are equivalent: has a center for all small enough. (ii) The differential equationṙ has a center for all ∈ R. (iii) The differential equationṙ has a center for all λ, μ ∈ R. Moreover, if m > 1, all the above conditions are also equivalent to: (iv) The differential equationṙ has a center for all small enough.
Set 0 = ∞ if S is unbounded and 0 = sup(S) otherwise. We will denote by r = r (θ, , ρ 0 ) the solution of (9) such that r (0, , ρ 0 ) = ρ 0 . We will show that Then, by the continuity of solutions with respect to parameters, the same holds for r (θ, , ρ 0 ) and r (θ, , Since it is analytic and, also from Lemma 2, restricted . Thus Eq. (10) has a center for all ∈ ( 0 − δ, 0 + δ) which gives a contradiction with the definition of 0 .
ii)⇒ iii). Assume that ii) holds. By Lemma 3,ṙ = μ B(θ ) r m has a center if and only if B(2π) = 0. This last condition is guaranteed by Lemma (4). So (11) has a center when λ = 0. Otherwise, observe that the change R = αr transforms Eq. (11) intoṘ If n is even choosing α = λ 1 n−1 we get that Eq. (11) is conjugated tȯ which has a center. This finish the proof in the case n even. If n is odd and λ > 0 the same argument holds. So in this case we have proved that Eq. (11) has a center for all λ ≥ 0 and for all μ ∈ R. To obtain the desired result for all λ ∈ R we fix μ ∈ R and consider λ 0 = inf{λ ∈ R : Eq. (11) has a center for all λ > λ} and use the same arguments than in the proof of the previous implication to show that It is obvious that (iii) ⇒ (i). Lastly to see that (iv)⇔ (ii) we repeat the same arguments used in the proof of (ii)⇒ (iii), choosing now α = μ 1−m . Note that this last equivalence does not hold when m = 1.
The above result motivates the following definition: (6) has a persistent center iḟ

Definition 6 Equation
has a center for all λ, μ ∈ R.
We want to remark that most authors use the definition of persistent center given in item (iv) of Proposition 5. Other also refer to the notion of persistent at infinity using the definition of item (i). Our result shows that all are equivalent. We have chosen the above one because it is more symmetric.
Notice that if A(θ ) and B(θ ) satisfy the CC then the same is true for λA(θ ) and μB(θ ). So, each center which satisfies the CC is a persistent center. Next result proves that if a center is persistent, then the moments of A with respect to B and the moments of B with respect to A must be zero. As far as we know the second fact is a new result.

Theorem 8 If (6) has a persistent center then
Proof We prove first the new set of conditions (14). From the hypothesis and Proposition 5 we know that equationṙ = A(θ )r n + B(θ )r m has a center for all ∈ R. From Lemma 2 we also know that there exist ρ and such that the above equation has the 2π -periodic solution r (θ, , ρ) for all |ρ| < ρ and | | < . Then, for 0 < ρ < ρ : r (θ, , ρ)) n−m dθ.
To prove (13) we can proceed similarly starting from (12) and the following equality:

Remark 9
The above result also holds for polynomial Abel equations.
Note that from the above theorem we know that if an equation satisfies the composition condition then all the moments vanish. From this result it is natural to formulate the following two questions. (13) and (14). Consider its associated Abel equation (6). Then:

Question 10 Assume that the functions A and B are trigonometric polynomials such that all the moments of A with respect to B and the ones of B with respect to A vanish, see
• Does it have a center? If yes: • Is the center persistent?
• Is the center a CC-center?
In Sect. 3 we give examples which answer negatively all the above questions. It is worth to comment that the question of whether one of the moments condition implies or not the composition conjecture, for the case of polynomials (not trigonometric polynomials) has attracted during the last years a wide interest. This question has been known as the Composition Conjecture. Finally it was proved to be false and in [16] the author gave a pair of polynomials such that the moments of one of them respect to other vanish but they do not satisfy the composition condition. For the sake of completeness we reproduce here this example: Example 11 [16] Take A(t) = T 2 (t)+T 3 (t) and B(t) = T 6 (t) where T i denotes the ith Chebyshev polynomial and T i its derivative. Thus T 2 (t) = 2t 2 − 1 and T 3 (t) = 4t 3 − 3t. It is well known that T 6 Thus we get where P and Q are some suitable polynomials.
However A and B do not satisfy the composition condition. An easy way to see this is to show that some moment of B respect to A does not vanish. This is the case because .
Since for the above example not all the the moments of B respect to A vanish, for the polynomial case, remains the following question, which seems to us much more natural that the so called composition conjecture:

Question 12 Given an interval [a, b], let A and B be polynomials such that all the moments of A with respect to B and the ones of B with respect to A vanish. Is it true that A and B satisfy the composition condition?
The tools developed in [15] will surely be very useful to answer the above question. As we have said, in next the section we will see that the answer to this question is "no" in the case of trigonometric polynomials.
All the above results and comments lead us to believe that the composition condition for trigonometric polynomials must be related with a stronger condition than the moments conditions (13) and (14). We introduce now the following definition.

Definition 13
We say that Eq. (6) has a strongly persistent center if has a center for all α, β, γ , δ ∈ R.
Clearly CC-centers are strongly persistent. Moreover strongly persistent centers satisfy that some "generalized moments" have to be equal to zero, as we will see in the next lemma. Brudnyi [8][9][10] also introduced iterated integrals and some generalized moments to express the Taylor expansion of the return map of generalized Abel equations.

Lemma 14
If (6) has a strongly persistent center then the following generalized moments vanish: for all p, q ∈ N.
Proof From Theorem 8 we know that Taking β = 0 and α = 1 : for some a i = 0. Notice that for i = 0 and i = k the corresponding integrals are equal to zero. Hence: for all γ = 0 and δ = 0. Taking the limits when γ and δ tend to zero, we get, respectively Reasoning inductively we see that for all i ∈ N, and hence that 2π 0 A(θ ) A p (θ ) B q (θ ) dθ = 0 for all p, q ∈ N. Starting with β = 1 and α = 0 we obtain the other set of conditions.

Hence it is natural to introduce the following open questions:
Question 15 Assume that the functions A and B are trigonometric polynomials such that all the generalized moments given in (17) vanish. Consider its associated Abel equation (6). Then:

• Does it have a center? If yes:
• Is the center strongly persistent or persistent?
• Is the center a CC-center?
We want to stress that the example that we will construct in the next section, which answers negatively the items stated in Question 10, gives no information about Question 15 because as we will see it has at least one non-zero generalized moment.
A smaller (or equal) class than strongly persistent centers is introduced in the following definition: Definition 16 We say that Eq. (6) has a symmetric center if it has a center and has also a center.
Clearly CC-centers are symmetric centers. As far as we know, no example is known of symmetric center not being a CC-center.
Finally we introduce the class of degree-persistent centers, as follows: Definition 17 We say that Eq. (6) has a degree-persistent center if has a center for all p and q in N.
As in the previous case, CC-centers are degree-persistent centers and it would be nice to study the converse implication. Notice that degree-persistent centers form, in principle, a smaller class that the one of the symmetric centers.

The Example
Systems (1), with P and Q of degree 2, can be written asż = i z + Az 2 + Bzz + Cz 2 , where z = x + iy and A, B, C are complex numbers and are usually called quadratic systems. Quadratic systems having a center are classified in four families. The family given by the single condition B = 0 is known as the Lotka-Volterra class. When C = 1/4, the corresponding Abel equation obtained using the Cherkas transformation is (6) with A(θ ) = a cos(2θ) + b sin(2θ) + (1/32) sin(6θ) and B(θ ) = cos(3θ), where a and b are arbitrary real parameters. Motivated for this equation we consider the trigonometric polynomials A and B given in next proposition.

Then:
(i) For all k ≥ 0, (ii) For a = 0 or a 2 = 3b 2 , the equation dr/dθ = A(θ )r n + B(θ )r m , with n > m has a CC-center. Proof (i) To see that for all a, b, c it holds that I k = 0 we will prove that each one of the three following integrals and cos r (2 θ) = (1 − 2 sin 2 (θ )) r we get that there exists a polynomial Q(x) ∈ R[x] such that K = 2π 0 cos(θ ) Q(sin 2 (θ )) dθ and hence K is zero. Lastly, equality (19) is obtained by direct computations.
In order to prove assertion (ii) assume that a = 0. Since we see that B(θ ) is a function of u(θ ) = sin θ. Using that cos(2 θ) and cos(6 θ) are also polynomials in sin θ we deduce that A(θ ) also depends on u(θ ) = sin θ. So, the CC is satisfied and equation dr/dθ = A(θ )r n + B(θ )r m has a center for all n, m. If a 2 = 3 b 2 then it can be seen that A(θ ) and B(θ ) are polynomials in cos(θ + π/6) and the CC is also satisfied. In order to prove (iii) and (iv) we compute, following [3] or [13,Prop. 3.1], the first coefficients V i of the return map of the Abel equation. We obtain that V i = 0, for i = 2, . . . , 10, and V 11 = a(3b 2 − a 2 )(1 − 32c) 4320 π.
Thus a necessary condition to have a center is that either a = 0 or a 2 = 3b 2 or c = 1/32. The fact that these conditions are sufficient comes from (ii) and from the fact that when c = 1/32, these cases corresponds to a quadratic center in the plane with a center at the origin. So if a = 0 and a 2 = 3b 2 a center exists if and only if c = 1/32. This implies that such a center is not persistent. Note also that from (i) these centers do not satisfy that all generalized moments associated to A and B vanish.
Remark 19 (i) Observe that the family presented in the above proposition answers negatively the questions stated in Question 10. Notice also that from this proposition conditions I k = 0, J k = 0, k ≥ 0 does not imply the existence of center. (ii) Note that item (v) of the above proposition shows that the Abel equation considered has neither a degree-persistent center nor a symmetric center. Moreover, computing also several Lyapunov constants, it can also be seen that for all a, b with a = 0, a 2 = 3b 2 , the equation dr/dθ = A(θ )r 4 + B(θ )r 2 has neither a center.
In [15] the following characterization of the moments condition in the polynomial case was proved.
Notice that the above result gives the reason for which a couple of polynomials A and B satisfying the moment condition does not necessarily satisfy the CC. The point is the existence of at least two essentially different functions w 1 and w 2 in the above decompositions. This is the case for the example obtained in [16], recalled in Example 11.
The following result shows that the natural translation of the above result to the trigonometric case does not hold.
Proof We know from Proposition 18 that  w(θ )). There are only three possibilities: either B 1 has degree 6 and w is linear or B 1 has degree 3 and w is quadratic or B 1 has degree 2 and and w is cubic. We examine with detail the case when B 1 has degree 6 and w is linear. In this case we would get: . Assume for instance that p 6 = −1. Doing the Fourier series of the right term of this equality, looking at its coefficients of cos(6 θ) and sin(6 θ) and equaling them to the corresponding ones of B(θ ) we obtain: (r 2 − s 2 )((s 2 + r 2 ) 2 − 16r 2 s 2 ) = 0 and rs(−3s 2 + r 2 )(−s 2 + 3r 2 ) = 16, which has a finite number of solutions all of them satisfying that rs = 0.
This system has no solution with rs = 0. This ends the proof of the non-existence of a decomposition of this type in this case. The proof in the other cases follows by similar computations. On the other hand it is clear that A(θ ) = sin(3 θ) is not a polynomial function of B(θ ) = sin(2 θ) − cos(6 θ).

Conclusions
We have seen that the study of the CC-centers for trigonometric Abel equations and other problems surrounding them, like the characterization of the persistent centers or the relation with the moments condition, is quite different to the same questions for polynomial Abel equations. We have also introduced in Definition 13 the notion of strongly persistent centers and related with them the generalized moments condition, see (17) in Lemma 14. These generalized moments seem a natural candidate to characterize persistent and CC-centers. Other possible characterization of CC-centers are given by the classes of symmetric centers or the one degree-persistent centers.
Next diagram shows the known relations among the concepts appearing in this paper. A crossed dotted implication means that the implication does not hold. We believe that the implications not given in the diagram suggest interesting problems to be studied. Perhaps the more important ones are: Are all the persistents centers, CC-centers? Do the generalized moments conditions imply the existence of a strongly persistent center?
Finally notice that many of the above concepts, problems and results can be extended to general equations of the formṙ = k≥2 A k (θ )r k , having either a finite or an infinite sum, see [8][9][10].