The center problem for a 1:−4 resonant quadratic system

J. Math. Anal. Appl. 420 (2014) 1568–1591

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The center problem for a 1 : −4 resonant quadratic system Brigita Ferčec a , Jaume Giné b , Matej Mencinger c,d,∗ , Regilene Oliveira e a Center for Applied Mathematics and Theoretical Physics, University of Maribor, Krekova 2, SI-2000 Maribor, Slovenia b Departament de Matemàtica, Universitat de Lleida, Av. Jaume II, 69, 25001 Lleida, Spain c Faculty of Civil Engineering, University of Maribor, Smetanova 17, SI-2000 Maribor, Slovenia d Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubljana, Slovenia e Departamento de Matemática, ICMC–USP, Caixa Postal 668, São Carlos, Brazil

a r t i c l e

i n f o

Article history: Received 18 October 2013 Available online 25 June 2014 Submitted by D. O’Regan Keywords: Quadratic system Resonant center Saddle point Center variety Decomposition of affine varieties Modular arithmetic

a b s t r a c t The main objective of this paper is to find necessary and sufficient conditions for a 1 : −4 resonant system of the form x˙ = x − a10 x2 − a01 xy − a12 y 2 ,

y˙ = −4y + b2,−1 x2 + b10 xy + b01 y 2

to have a center at the origin. Since applying a linear change of variables any system of this form can be transformed either to system with a10 = 1 or a10 = 0, these are the two cases considered here. When a10 = 1 there appear 46 resonant center conditions and for a10 = 0 there are 9 center conditions. To obtain necessary conditions for integrability the computation of the resonant saddle quantities (focus quantities) and the decomposition of the variety of the ideal generated by an initial string of them were used. The theory of Darboux first integrals and some other methods, as the monodromy arguments for instance, are used to show the sufficiency. Since decomposition of the variety mentioned above was performed using modular computations the obtained conditions of integrability represent the complete list of the integrability conditions only with very high probability and there remains an open problem to verify this statement. © 2014 Elsevier Inc. All rights reserved.

1. Introduction In this paper we consider differential autonomous systems in C2 of the form x˙ = px + P (x, y),

y˙ = −qy + Q(x, y),

* Corresponding author. E-mail addresses: [email protected] (B. Ferčec), [email protected] (J. Giné), [email protected] (M. Mencinger), [email protected] (R. Oliveira). http://dx.doi.org/10.1016/j.jmaa.2014.06.060 0022-247X/© 2014 Elsevier Inc. All rights reserved.

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where p, q ∈ N and P and Q are complex polynomials. In particular, we consider the case p = 1, q = 4 and P , Q being quadratic polynomials. The main goal of the paper is to determine when the elementary singular point located at the origin is a resonant center. Definition 1.1. A p : −q resonant elementary singular point of analytic system (1) is a resonant center if there exists a local analytic first integral of the form Φ(x, y) = xq y p +



φj−q,k−p xj y k .

(2)

j+k>p+q+1

The p : −q resonant center is a generalization of the concept of a real center to systems of differential equations in C2 of the form (1), it was given by Żoładek in [37]. This definition is a generalization of the concept of 1 : −1 resonant center introduced by Dulac in [10]. The classical real center problem (which originates from the work of Lyapunov and Poincaré [25,30]) was studied by many authors (see e.g. [3,10,19] and references therein). Some generalizations of the problem are discussed in [12] and [29]. Note that the 1 : −1 resonant center in C2 is a complexification of the classical real planar center (see, for instance [10] for more details). The saddle-node and the node case were studied in [35,37]. The study of p : −q resonant centers is relatively new (see e.g. [2,8,20,26,27,32,37] and the references given there). Two p : −q resonant systems for which the center problem has been solved completely are the 1 : −1 resonant cases with quadratic and homogeneous cubic nonlinearities [5,10,34]. The resonant center problem for the 1 : −2 resonance case was investigated in [16] where 20 cases for integrability were given. In [17] the sufficiency of these cases are proven, so this case is almost finished due to the use of modular arithmetics. Recently, the case 1 : −3 was considered also using modular arithmetics, see [9]. Then the 1 : −4 resonance presented in this paper is in the actual limit of the current computational facilities. The objective is to have the classification of lower resonances in order to use them for the study of particular systems with higher resonances via blow-up transformations. The most studied resonant centers are the 1 : −λ quadratic Lotka–Volterra systems, i.e. systems of the form x˙ = x(1 + ax + by), y˙ = y(−λ + cx + dy), see [6]. In [23] certain sufficient conditions for integrability and linearizability for general λ ∈ N and necessary and sufficient conditions for λ = p2 and for λ = p2 (with p ∈ N) are presented. The cases 3 : −4 and 3 : −5 of resonances have been studied in [27]. Recently, the integrability of the Lotka–Volterra type systems (1) were studied in [15,21,26]. Inside the 1 : −1 resonant centers, the case when P and Q are homogeneous quintic nonlinearities has been studied in [14,22] but not finished. Concerning the generalized p : −q resonant center many papers are considering 1 : −q and 2 : −q resonant centers (with mostly homogeneous quadratic, cubic or quartic nonlinearities added); see e.g. [2,8]. In [37] for the general p : −q resonant quadratic case there are exhibited fifteen independent sufficient conditions for existence of a center, along with the corresponding first integrals for each one. In [2] the 1 : −q resonant center problem for certain cubic Lotka–Volterra system was studied for integers q ≤ 9. For odd q ≤ 9 the authors obtained necessary and sufficient conditions for existence of a center, whilst for even q < 9 only necessary conditions were obtained. This may indicate that the analysis of a 1 : −q resonant center might be more difficult for q being even. The computational difficulties which occur when performing computations with Computer Algebra Systems (CAS) Singular and Mathematica for the present paper may definitely confirm this idea. By the definition any nonconstant differentiable complex function which is constant on trajectories of (1) is a first integral of (1):  ∂Ψ   ∂Ψ  px + P (x, y) + −qy + Q(x, y) ≡ 0. Ψ˙ := ∂x ∂y Throughout the work P and Q are assumed to be quadratic polynomials.

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Thus, we can write system (1) in the form x˙ = px −

n−1 

y˙ = −qy +

ajk xj+1 y k ,

j+k=1 j≥−1

n−1 

bkj xk y j+1 .

(4)

j+k=1 j≥−1

According to Definition 1.1 to find conditions for existence of a resonant center at a p : −q resonant elementary singular point of system (4) we look for a formal series in the form (2) satisfying (3). In other words, the approach is based on computing the saddle quantities [2,5,8,17,32,37], gkq,kp , which are polynomials in the coefficients ai,j , bi,j of (4) with the property that a formal first integral (3) exists if and only if every saddle quantity gkq,kp vanishes on the coefficients of (4), that is, if and only if the coefficients ai,j , bi,j of (4) lie in the variety V(B) in C(n+4)(n−1) of the (Bautin) ideal B = gkq,kp : k ∈ N in the polynomial ring C[a10 , a01 , a−12 , . . . , b2,−1 , b10 , b01 ] which we abbreviate to C[a, b]. Therefore, to start the computational process we write down the initial string of (2) up to order 2N + 1 Ψ2N +1 (x, y) = xq y p +

2N +1 

φj−q,k−p xj y k .

(5)

j+k=p+q+1

Then for each i = p + q + 1, . . . , 2N + 1 we equate coefficients of terms of order i in the expression  ∂Ψ2N +1   ∂Ψ2N +1  px + P (x, y) + −qy + Q(x, y) . ∂x ∂y

(6)

Now let denote the coefficients of xk1 +q y k2 +p in (6) by gk1 ,k2 and set them to be zero for k1 + k2 ≤ 0. For k1 + k2 ≥ 1 they are given [33, p. 117] recursively by: k1 +k 2 −1 

gk1 ,k2 = (pk1 − qk2 )φk1 ,k2 −



 (i1 + q)ak1 −i1 ,k2 −i2 − (i2 + p)bk1 −i1 ,k2 −i2 φi1 ,i2 .

(7)

i1 +i2 =0 i1 ≥−q,i2 ≥−p

The coefficients gk1 ,k2 defined in (7) can obviously be set to zero, as long as k1 and k2 satisfy the conditions k1 = kq and k2 = kp (note that setting gk1 ,k2 to zero determines the coefficient φk1 ,k2 by the previous ones, φi1 ,i2 , as defined in (7) – using initial condition φ0,0 = 1). However, the coefficients gkq,kp (i.e. gk1 ,k2 for k1 = kq and k2 = kp) cannot be set to zero simply by choosing a certain value for φi1 ,i2 . Therefore, the corresponding polynomials for gkq,kp are defined to be the saddle quantities:

gkq,kp =

kq+kp−1 

  (i1 + q)akq−i1 ,kp−i2 − (i2 + p)bkq−i1 ,kp−i2 φi1 ,i2 .

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i1 +i2 =0 i1 ≥−q,i2 ≥−p

Obviously, if for a fixed system (4) with the coefficients (a∗ , b∗ ), we have gkq,kp (a∗ , b∗ ) = 0 for all k ∈ N, then we obtain a formal first integral Ψ defined in (5). Therefore to find necessary conditions for existence of formal first integral we need to find the set of all parameters (a, b) where all polynomials gkq,kp vanish, i.e. to find the variety of the Bautin ideal B. By the Hilbert Basis Theorem (see e.g. [33, Theorem 1.1.6]) the ideal B is finitely generated, i.e. there exists K ∈ N such that B = BK , where BK = gkq,kp : 1 ≤ k ≤ K. Therefore, we have to find first few saddle quantities

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and then to compute the variety of the ideal they generate, where the variety of the ideal generated by polynomials f1 , . . . , fs is the set of common solutions of polynomial system f1 = 0, . . . , fs = 0, i.e.     V f1 , . . . , fs  = a = (a1 , . . . , an ) ∈ kn : fi (a) = 0, for every i = 1, . . . , s . The variety of the ideal B, V(B), is called the center variety. In the following two sections we consider necessary and sufficient conditions for the existence of a 1 : −4 resonant center for system (4) with quadratic nonlinearities. 2. Statement of the main results Let us consider system (4) for p = 1, q = 4 and P , Q being quadratic polynomials: x˙ = x − a10 x2 − a01 xy − a−12 y 2 , y˙ = −4y + b2,−1 x2 + b10 xy + b01 y 2 ,

(9)

where x, y, aij , bji ∈ C. First note that using a linear transformation every system (9) can be transformed to either a system (9) with a10 = 1 or a10 = 0. In both cases the (minimal primary) decomposition of the corresponding ideal BK cannot be performed in polynomial rings of characteristic zero, i.e. in the ring Q[a, b] (where [a, b] denotes the coefficients of system (9)). However, it becomes possible in the polynomial ring of a proper prime characteristics, i.e. in Zp [a, b]. See more details in the proof of Theorem 2.1, in Section 3. The sufficient conditions of integrability of system (9) with a10 = 1 and a10 = 0 are given in the following two theorems. Theorem 2.1. System (9) with a10 = 1 has a 1 : −4 resonant center at the origin if one of the following 46 conditions holds: a01 = a−12 = 6b210 − 18b10 + b01 b2,−1 + 12 = 0; a01 = a−12 = 4b210 − 12b10 + 9b01 b2,−1 = 0; 5b10 − 4 = 4a01 + 7b01 = a−12 = 25b01 b2,−1 + 24 = 0; 19b10 − 8 = 4a01 + 7b01 = a−12 = 361b01 b2,−1 + 168 = 0; 13b10 − 16 = 4a01 + 7b01 = a−12 = 169b01 b2,−1 + 72 = 0; 7b10 + 1 = 4a01 + 7b01 = a−12 = 49b01 b2,−1 + 12 = 0; b2,−1 = a−12 = −4a01 − 3b01 + b01 b10 = 0; b10 − 1 = a01 − b01 = a−12 = 49b01 b2,−1 − 24 = 0; 23b10 − 14 = a01 − b01 = a−12 = 529b01 b2,−1 − 168 = 0; b10 = 4a01 + 3b01 = a−12 = 0; 19b10 − 16 = 4a01 + 11b01 = a−12 = 361b01 b2,−1 + 168 = 0; 23b10 − 22 = a01 − 2b01 = a−12 = 529b01 b2,−1 − 168 = 0; b10 − 5 = 2a01 − b01 = a−12 = 0; b2,−1 = b10 − 1 = 0; b2,−1 = b10 = 0; b2,−1 = b10 − 2 = 0; 13b10 − 16 = 1872a201 + 1092b01 a01 + 78b201 + 600a−12 = 0, 338a01 b2,−1 + 169b01 b2,−1 − 180 = 108a01 + 9b01 + 65a−12 b2,−1 = 0; 18) −15b10 + 19a01 b2,−1 + 11b01 b2,−1 + 6 = 120 − 300b10 + 114b210 + 49b01 b2,−1 = 18a01 b10 − 48a01 − 18b01 + 12b01 b10 − 7a−12 b2,−1 = 0,

1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17)

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19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42)

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72 − 2232b10 + 1329b01 b2,−1 + 570b01 b10 b2,−1 − 2527a−12 b22,−1 = 0, 3192a01 + 1674b01 − 1221b01 b10 + 38a−12 b2,−1 + 105b201 b2,−1 + 1083a−12 b10 b2,−1 = 504a201 − 144a−12 + 42a01 b01 −45b201 +306a−12 b10 +30b201 b10 −133a−12 b01 b2,−1 = 72a301 +96a01 a−12 +54a201 b01 +36a−12 b01 + 9a01 b201 − 15a−12 b01 b10 + 17a2−12 b2,−1 = 0; b2,−1 = b10 − 8 = 4a01 − 5b01 = 0; b2,−1 = b10 − 3 = a01 = 0; b01 = a01 = a−12 = 0; 61b10 − 92 = 7a01 + b01 = 3721b01 b2,−1 − 4116 = 0, 427a−12 b2,−1 − 27b01 = 549b201 − 9604a−12 = 0; 131b10 − 112 = 36a01 − 97b01 = 0, 17 161b01 b2,−1 − 4536 = 131a−12 b2,−1 − 56b01 = 131b201 − 81a−12 = 0; 8b10 + 9 = 19a01 + 12b01 = 32b01 b2,−1 − 57 = 0, 152a−12 b2,−1 − 27b01 = 36b201 − 361a−12 = 0; 22b10 − 9 = 53a01 + 54b01 = 121b01 b2,−1 + 159 = 0, 583a−12 b2,−1 − 108b01 = 396b201 + 2809a−12 = 0; 191b10 − 72 = 4a01 + 177b01 = 36 481b01 b2,−1 + 504 = 0, 1134b01 + 191a−12 b2,−1 = 1719b201 − 4a−12 = 0; b10 − 12 = 2a01 − 9b01 = 2b01 b2,−1 + 3 = 0, 2a−12 b2,−1 − 27b01 = 9b201 + a−12 = 0; 5b10 − 4 = 2a01 − 9b01 = 50b01 b2,−1 − 9 = 0, 10a−12 b2,−1 − 9b01 = 5b201 − a−12 = 0; 23b10 − 31 = 3a01 + 4b01 = 529b01 b2,−1 + 216 = 0, 23a−12 b2,−1 − 48b01 = 46b201 + 9a−12 = 0; 38b10 − 81 = 107a01 + 6b01 = 361b01 b2,−1 − 321 = 0, 2033a−12 b2,−1 − 108b01 = 684b201 − 11 449a−12 = 0; 13b10 − 6 = 2a01 − 9b01 = 1352b01 b2,−1 − 147 = 0, 52a−12 b2,−1 − 27b01 = 234b201 − 49a−12 = 0; 19b10 − 28 = 31a01 + 23b01 = 361b01 b2,−1 + 2604 = 0, 589a−12 b2,−1 − 189b01 = 171b201 + 3844a−12 = 0; b10 − 12 = 38a01 − 21b01 = b01 b2,−1 + 114 = 0, 38a−12 b2,−1 − 63b01 = 21b201 + 1444a−12 = 0; 4b10 − 3 = 148a01 − 81b01 = 256b01 b2,−1 − 111 = 0, 27b01 + 148a−12 b2,−1 = 576b201 + 1369a−12 = 0; 11b10 − 27 = a01 + 3b01 = 121b01 b2,−1 + 24 = 0, 11a−12 b2,−1 − 36b01 = 33b201 + 2a−12 = 0; 121b10 − 97 = a01 + 28b01 = 14 641b01 b2,−1 + 504 = 0, 504b01 + 121a−12 b2,−1 = 121b201 − a−12 = 0; 4b10 − 3 = 3a01 − 41b01 = 392b01 b2,−1 − 27 = 0, 14a−12 b2,−1 − 39b01 = 364b201 − 9a−12 = 0; 71b10 + 8 = 22a01 + 31b01 = 5041b01 b2,−1 + 1584 = 0, 781a−12 b2,−1 − 81b01 = 639b201 + 1936a−12 = 0; 57b10 − 44 = a01 − 37b01 = 1083b01 b2,−1 − 28 = 0, 19a−12 b2,−1 − 119b01 = 969b201 − 4a−12 = 0; 89b10 − 248 = 58a01 − b01 = 7921b01 b2,−1 − 4176 = 0, 2581a−12 b2,−1 − 81b01 = 801b201 − 13 456a−12 = 0; 11b10 + 8 = 7a01 + b01 = 121b01 b2,−1 + 84 = 0, 77a−12 b2,−1 − 27b01 = 99b201 + 196a−12 = 0; b2,−1 = 2b10 + 1 = b01 = a01 = 0;

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b2,−1 = 2b10 − 1 = b01 = a01 = 0; b2,−1 = 2b10 − 5 = b01 = a01 = 0; b2,−1 = 2b10 − 3 = b01 = a01 = 0; −18 346b10 + 17 857b01 b2,−1 + 7100 = 10 027b10 + 17 857a01 b2,−1 − 14 708 = −36 738a01 − 25 857b01 + 262 753a−12 b2,−1 = 2551b210 − 5224b10 + 1594 = −7966a01 − 440 739b01 + 262 753b01 b10 = 0, 525 506b10 a01 − 194 666a01 + 60 473b01 = 819b201 + 3606a−12 − 9670a−12 b10 = 788b10 a−12 − 209a−12 + 819a01 b01 = 0, 1638a201 − 9355a−12 + 5469a−12 b10 = 0.

Theorem 2.2. System (9) with a10 = 0 has a 1 : −4 resonant center at the origin if and only if one of the following 9 conditions holds: 1) 2) 3) 4) 5) 6) 7) 8) 9)

b2,−1 = b10 = 0; b2,−1 = b01 = a−12 = 0; a01 = a−12 = 6b210 + b01 b2,−1 = 0; a01 = a−12 = 4b210 + 9b01 b2,−1 = 0; b01 = a01 = a−12 = 0; b10 = 2a01 − b01 = a−12 = 0; b10 = 4a01 + 3b01 = a−12 = 0; b10 = 2a01 + b01 = a−12 = 0; 19a01 + 11b01 = 114b210 + 49b01 b2,−1 = 0, 30b01 b10 − 133a−12 b2,−1 = 35b201 + 361a−12 b10 = 0;

3. Proofs of the main results In this section we prove the main Theorems 2.1 and 2.2 of the paper. Since the Darboux method of integration is the main tool for proving the sufficiency of the conditions in Theorems 2.1 and 2.2 let first recall some definitions concerning this method which is commonly used as a tool for investigating the center problem [8,12,33]. In this section we use it to prove the existence of first integrals in most cases of polynomial systems (9) listed in Theorems 2.1 and 2.2. The method is described in detail in [33]. A good survey of recent applications of Darboux method to polynomial systems in Rn and Cn is also [28]. We consider the system (9) written as x˙ = P (x, y),

y), y˙ = Q(x,

(10)

are polynomials that have no nonconstant common factor, and m = where x, y ∈ C, P and Q max(deg(P ), deg(Q)). We define the algebraic partial integral or Darboux factor of system (10) to be a polynomial f (x, y) such that ∂f ∂f P+ Q = Kf, ∂x ∂y where K(x, y) is called a cofactor. It turns out that K(x, y) is a polynomial of degree at most m − 1. An integrating factor on an open set Ω for system (10) is a differentiable function μ(x, y) on Ω such that ∂μ ∂μ y ) P+ Q = −μ(P x + Q ∂x ∂y y stands for the divergence of (P , Q). holds throughout on Ω, where P x + Q

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It is easily seen that if there are the Darboux factors f1 , f2 , . . . , fk with the cofactors K1 , K2 , . . . , Kk satisfying k 

αi Ki = 0,

(11)

i=1

then H = f1α1 · · · fkαk is a first integral of (10), and if k 

y = 0 αi Ki + P x + Q

(12)

i=1

then the equation admits the (Darboux) integrating factor μ = f1α1 · · · fkαk .

(13)

3.1. Proof of Theorem 2.1 Following the approach described in Section 3 of [33] the saddle quantities were computed. In this case we compute the first 6 saddle quantities g4,1 , . . . , g24,6 . The first saddle quantity is g4,1 = −1512b01 b410 + 6048a01 a10 b310 + 9072a10 b01 b310 + 684a−12 b2,−1 b310 − 18 144a01 a210 b210 − 16 632a210 b01 b210 + 9072a201 b2,−1 b210 − 3654b201 b2,−1 b210 + 7452a10 a−12 b2,−1 b210 − 2916a01 b01 b2,−1 b210 + 12 096a01 a310 b10 + 10 476a01 a−12 b22,−1 b10 − 1086a−12 b01 b22,−1 b10 + 9072a310 b01 b10 + 10 962a10 b201 b2,−1 b10 − 6480a201 a10 b2,−1 b10 − 19 800a210 a−12 b2,−1 b10 + 15 516a01 a10 b01 b2,−1 b10 + 1120a2−12 b32,−1 + 3024a301 b22,−1 − 567b301 b22,−1 − 756a01 b201 b22,−1 − 5088a01 a10 a−12 b22,−1 + 2268a201 b01 b22,−1 + 4464a10 a−12 b01 b22,−1 − 4896a201 a210 b2,−1 − 6804a210 b201 b2,−1 + 8640a310 a−12 b2,−1 − 12 744a01 a210 b01 b2,−1 . The other saddle quantities are too large to be presented here. Then, we set in the obtained polynomials a10 = 1. The procedure is to compute a set of necessary conditions g4i,i = 0, i = 1, . . . , M for M large enough. Usually these necessary conditions are polynomials in the parameters of the system with long rational coefficients. Therefore, it is an extremely difficult computational problem to determine the irreducible components of the variety V = V (g4i,i : i = 1, . . . , M ). We have followed the algorithm described in [31] which makes use of modular arithmetics, that is, we replace the ring Q[A] by the ring Zp [A], where A denotes the coefficients of the system and p is a prime number. The most difficult part of this algorithm is the last one where we need to ensure that all the points of the variety V have been found. That is, we know that all the encountered points belong to the decomposition of V but we do not know whether the given decomposition is complete. We remark that, nevertheless, it is practically sure that the given list is complete, see for instance [1]. Therefore, in the following we provide sufficient conditions for (9) to have a resonant center. It turns out that they are practically also necessary conditions. Although the proof of the completeness of the decomposition is not given exactly (over the field of characteristic zero) the probability of the opposite event is very low, see the estimation given in [17] for the 1 : −2 resonance where the Faugére method [13] is used.

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In this paper, to obtain the necessary conditions for integrability we find the minimal decomposition of the variety of the ideal B6 = g4,1 , . . . , g24,6 . The computational tool which we use is the routine minAssGTZ [7] of the computer algebra system Singular [24] which is based on the Gianni–Trager–Zacharias algorithm [18]. Since computations are too laborious they cannot be completed in the field of rational numbers. Therefore, we choose the approach based on making use of modular computations [1,31,36]. We choose some primes p = 32 003, 104 729, 4 256 233, 7 368 787, 15 485 863, 179 595 127, 433 494 437 and 479 001 599 and compute the decomposition over the field Zp for each p listed above. First note that during the computations over the fields of finite characteristic p listed above we arrived at two different problems. The first one is that in some cases the obtained decomposition was defined by different polynomials (for different characteristics of Zp ). Secondly we observed that when computing the decomposition over the field Zp we obtained 46 components for some values of p while for some other values of p we obtained 47 components of the center variety. For each decomposition we performed rational reconstruction algorithm to obtain ideals in Q[a, b]. For √ instance, computing the decomposition of affine variety V ( B6 ) over the field of characteristic 32 003 we find that one of the obtained component is b10 + 4572 = a01 − 7999b01 = a−1,2 = b01 b2,−1 − 1306 = 0. Performing rational reconstruction using Mathematica we obtain 4572 ≡ 1/7 mod 32 003, 7999 ≡ −7/4 mod 32 003 and 1306 ≡ −12/49 mod 32 003. Therefore the corresponding component over characteristic zero is b10 + 1/7 =√a01 + 7/4b01 = a−1,2 = b01 b2,−1 + 12/49 = 0. If we use another prime p the corresponding component of V ( B 6 ) looks similar but with different parameters. However, after rational reconstruction we obtain that both components coincide and we arrive at the component 6) of Theorem 2.1. Then we checked by a direct computation using CAS Mathematica, if the 6 saddle quantities g4,1 , . . . , g24,6 are equal to zero under the obtained conditions. However, in cases 17–18, 22–41 and 46 it turned out that (after the decomposition and rational reconstruction) the obtained conditions were actually not the center variety conditions since they did not yield the saddle quantities to be zero. We noted that in all these “problematic” cases the saddle quantities did not vanish because of the presence of a particular polynomial of the form fp = b01 b2,−1 + K. Actually it was the value of K in fp which was the essence of the problem. Such kind of errors probably appear because of using computations over the fields of finite characteristic. Usually such problems are solved by choosing another prime p in order to compute the decomposition again and improve the result. But in this cases we tried eight different primes p listed above but the problem remained the same. Therefore, we tried to “correct” the value for constant K in the polynomial fp in all these problematic cases. To this end in the problematic cases we used all the other polynomials (i.e. conditions) to compute the “right” value for K. For b01 and b2,−1 satisfying the other (i.e. “non-problematic”) conditions we solved the equation fp = 0 for K. However, note that the problematic polynomials fp in the cases 17–18, 22–41 and 46 cannot be omitted (they imply also that b01 b2,−1 = 0). The second problem was the fact that the decomposition over some fields of finite characteristic contained 46 cases for some values of p and 47 cases for some other values of p. We noted that for those values of p for which 46 components were obtained there was always one large component which seemed to be decomposed into two smaller ones similar to those two from the decompositions consisting of 47 components (which was obtained for some other value of p). Since the large component under no chosen prime p was lying in the center variety after rational reconstruction we compared its decomposition for different primes and picked out from them those polynomials which were in common. Next we computed in Q[a, b] the decomposition of the variety of the ideal generated by these common polynomials. Then, after checking that the saddle quantities g4,1 , . . . , g24,6 are equal to zero we obtained the “critical” 46-th component listed in Theorem 2.1. Furthermore, we noted that in the field C[a, b] this “critical” 46-th component can be decomposed into two components whose polynomials contain some square roots, which possibly explains why in the decompositions with 47 components we always obtain (the last) two components containing

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quotients of big numbers in numerators and denominators (i.e. these quotients were the approximations for these square roots). Checking if some conditions in computations of the decomposition over the field of finite characteristic were lost is a very difficult computational problem which can again not be performed (completely) over

46 rational numbers. We first compute intersection P = i=1 Pi over the field of characteristic zero, where Pi denotes the √ component i from Theorem 2.1. We obtain 112 polynomials p1 , . . . , p112 . We would like to check √ if B6 = P . Computing over the field of characteristic 0 Gröbner basis of each ideal 1 − wg4k,k , P : k = 1, . . . , 6, with √ w a new parameter and g4k,k a saddle quantity we find that they are all {1}. This implies √ that B6 ⊂ P . √ √ To check the opposite inclusion, P ⊂ B6 we must use computations with modular arithmetics. We choose prime p = 179 595 127 and after computing Gröbner basis of each ideal 1 − wpk , B6 : k = 1, . . . , 112, and pk in P over the field Zp we find that they are all {1}. Then we repeat computations over prime 32 003 and find that the Gröbner basis of ideal wpk , B6 : k = 1, . . . , 112 is {1} for each k = 1, . . . , 112 and pk √ 1 −√ in B6 . We can conclude that equality P = B6 holds with high probability. 3.2. Necessity of the cases of Theorem 2.2 We use the same saddle quantities g4,1 , . . . , g24,6 as in the proof of Theorem 2.1 and set a10 = 0. Using again the routine MinAssGTZ of CAS system Singular we compute the decomposition of the variety V(g4,1 , . . . , g24,6 ). Computations cannot be completed in the field of rational numbers, therefore we choose prime p = 32 003 and compute the decomposition in the finite field Zp [a10 , a01 , a−12 , b2,−1 , b10 , b01 ]. We obtain nine components and after applying the rational reconstruction algorithm we obtain nine components listed in Theorem 2.2. Following the decomposition algorithm [31] we first check that all saddle quantities are zero under each condition. Then we check that no condition is lost. We denote by Pi, i = 1, . . . , 9, the components from Theorem 2.2 and we compute the intersection P =

9 

Pi .

i=1

We obtain nine polynomials q1 , . . . , q9 in P . Now we compute over the field of characteristic zero Gröbner bases of ideals 1 − wqi , B6 : i = 1, . . . , 9 over the field of characteristic 0 and Gröbner bases of ideals 1 − wg4i,i , P : i = 1, . . . , 6 (over the field of characteristic 0) and find that they are all {1}. Therefore, no condition is lost. 3.3. Sufficiency of the cases of Theorem 2.2 There are nine cases here. In all nine cases we applied the Darboux method to prove the existence of the first integral. Below is the case-by-case analysis: Case 1. In this case system (9) is written as x˙ = x − a−12 y 2 − a01 xy,

y˙ = −4y + b01 y 2 ,

and we find two invariant lines l1 = 1 − b01 y/4 and l2 = y yielding the Darboux integrating factor of the (a −5b )/(4b01 ) −3/4 form μ = l1 01 01 l2 . By Theorem 4.13 of [4] there exists a first integral of the form (14). Case 2. Here the corresponding system (9) is x˙ = x − a01 xy, and it admits the Darboux first integral Ψ = x4 y(1 −

y˙ = −4y + b01 y 2 , b01 y −(b01 −4a01 )/(b01 ) . 4 )

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Case 3. In this case system (9) takes the form y˙ = −4y −

x˙ = x,

6b210 x2 + b10 yx + b01 y 2 , b01

and it has two invariant curves 5b210 x2 5 b01 y − 2b10 x + b01 b10 yx − , 4 12 4 15b210 x2 25 25b310 x3 5 b01 y + − b01 b210 yx2 + 3b10 x − b01 b10 yx − , l2 = 1 + 12 4 24 6 4

l1 = 1 +

and one invariant line l3 = x. We are able to compute the Darboux integrating factor of the form μ = (l1 l2 )−1 l33 and after integration we obtain a first integral of the form (14). Case 4. Here the system is of the form y˙ = −4y −

x˙ = x,

We compute two invariant curves l1 = 1 − l2 = 1 +

b10 x 3

−

4b210 x2 + b10 yx + b01 y 2 . 9b01

b01 y 4 ,

25b310 x3 125 25 5 125b410 x4 5b2 x2 4b10 x b01 y + − b01 b310 yx3 + 10 − b01 b210 yx2 + − b01 b10 yx − , 1944 81 648 6 72 3 12 4

and one invariant line l3 = x yielding the Darboux integrating factor μ = (l1 l2 )−1 l33 and a first integral of the form (14). Case 5. In this case we compute the Darboux integrating factor

−1 x4 yb610 x4 b410 x3 b310 x2 b210 x5 b510 + + + + xb10 μ = x3 1 + + 120b2,−1 120 24 6 2 and after integration we obtain a first integral of the required form. Case 6. The system under conditions of this case admits the first integral Ψ =−

6yx4 3b01 y 2 x4 + x6 + . b2,−1 2b2,−1

Case 7. In this case system (9) is written as x˙ = x +

3b01 yx , 4

y˙ = −4y + b2,−1 x2 + b01 y 2 .

We have found two invariant curves b01 y 1 b2 y 2 , and l1 = 1 + b01 b2,−1 x2 + 01 − 4 16 2 3b2 y 2 3b01 y 3 2 2 3 3 b3 y 3 b01 b2,−1 x4 + b01 b2,−1 x2 − b201 b2,−1 yx2 − 01 + 01 − , l2 = 1 + 128 8 32 64 16 4 −3/2

and compute the Darboux first integral Ψ = l1 l2 . Case 8. Here the corresponding system is of the form x˙ = x +

b01 yx , 2

y˙ = −4y + b2,−1 x2 + b01 y 2

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and it has two invariant curves l1 = 1 + 18 b01 b2,−1 x2 − l2 =

b01 y 4 ,

and

3b2 y 2 3b01 y 3 2 2 3 3 3 3 b3 y 3 b01 b2,−1 x4 + b01 b2,−1 y 2 x2 + b01 b2,−1 x2 − b201 b2,−1 yx2 − 01 + 01 − , 64 128 8 16 64 16 4

yielding the Darboux first integral Ψ = l1−3 l2 . Case 9. In this case system (9) is written as x˙ = x +

35b201 y 2 11b01 xy , + 361b10 19

y˙ = −4y −

114b210 x2 + b10 yx + b01 y 2 . 49b01

In this case we found three invariant curves l1 = 1 − l3 =

6b10 x 6b01 y − , 7 19

l2 = −

19b10 x2 7b01 y 2 7y + yx − − , 7b01 76b10 b10

6 021 872 768b610 x6 24 463 858 120b510 x5 950 822 016b510 yx5 + − 6 6 16 807b01 7203b01 2401b501 +

645 234 257 915b410 x4 62 554 080b410 y 2 x4 3 070 362 760b410 yx4 + − 49 392b601 343b401 1029b501

+

317 418 559 107b310 x3 2 194 880b310 y 3 x3 153 778 780b310 y 2 x3 − + 6 3 12 544b01 49b01 147b401

−

60 676 805 995b310 yx3 43 320b210 y 4 x2 548 720b210 y 3 x2 + − 5 2 7056b01 7b01 3b301

+

5 504 368 077b210 x2 2 779 095 325b210 y 2 x2 179 390 896 451b210 yx2 + − 224b601 1344b401 16 128b501

−

456b10 y 5 x 95 665b10 y 4 x 61 353 755b10 y 3 x 3 322 273 253b10 y 2 x + − + b01 6b201 288b301 2304b401

+

611 596 453b10 x 1 384 139 341b10 yx 3325y 5 4 485 425y 4 6 − + 14y − + 64b601 256b501 6b01 576b201

−

108 269 315y 3 59 296 055y 2 + , 2304b301 576b401 −2/3 −5/6 −1/3 l2 l3 .

yielding the following integrating factor μ = l1 integral of the required form.

By Theorem 4.13 of [4] there exists a first

3.4. Sufficiency of the cases of Theorem 2.1 In this subsection we prove the sufficiency of all 46 conditions of Theorem 2.1 by doing a case-by-case analysis of all 46 cases. Note that 3 distinct methods are used for proving the existence of a first integral for each case. The Darboux method, the monodromy argument (as in case 17 on page 1582, see [6] for more details) and the inductive method, i.e. for the (formal) first integral we set a trial solution of the ∞ form Ψ (x, y) = k=1 fk (x)y k , where functions fk are determined recursively by some first order differential equations and finally using mathematical induction we prove that Ψ actually takes the desired form yielding the existence of the first integral; see cases 14–16 on page 1581 for details. The monodromy argument is described in detail in [6, Theorem 9]. Roughly speaking, if all the singular points on an invariant curve except the origin are integrable and if all of them but one have identity monodromy map corresponding to the invariant curve then the origin is also integrable. In the simple case all the singular points are linearizable nodes, then the saddle at the origin is integrable.

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Below is the case-by-case analysis of all 46 cases. Case 1. In this case system (9) takes the form x˙ = x − x2 ,

y˙ = −4y −

6b210 − 18b10 + 12 2 x + b10 xy + b01 y 2 . b01

This system has three algebraic curves: l1 = x, l2 = 1 − x, and l3 = 1 +

11b10 x2 3x2 5 5b210 x2 b01 yx b01 y − + − 2b10 x − + b01 b10 yx + 2x − , 4 4 2 2 12 4

which allow us to construct a Darboux integrating factor of the form μ = l13 l2−9+5b10 l3−2 . Integration yields a first integral, whose series expansion is of the form Ψ = x4 y + h.o.t.

(14)

Case 2. The corresponding system for this case is x˙ = x − x2 ,

y˙ = −4y −

4b210 − 12b10 2 x + b10 xy + b01 y 2 . 9b01

It has three algebraic curves: l1 = x, l2 = 1 − x and l3 = 1 −

b10 x b01 y − 3 4 5(−3+b10 )/3 −2 l3

which allow us to construct a Darboux integrating factor of the form μ = l13 l2 integral of the form (14). Case 3. Here the corresponding system is x˙ = x − x2 +

7b01 xy , 4

y˙ = −4y −

This system admits three algebraic curves: l1 = x, l2 = 1 +

x 5

and a first

24x2 4xy + b01 y 2 . + 25b01 5 −

b01 y 4 ,

and

8 3 16x4 16x3 2 1 1 2b01 yx − b01 yx3 + + b201 y 2 x2 − b01 yx2 − b301 y 3 x + b201 y 2 x + 625 125 125 50 25 40 10 5 b301 y 3 3b201 y 2 8x b401 y 4 + − + − b01 y, − 5 256 16 8

l3 = 1 +

−5/2

yielding the Darboux integrating factor μ = l13 l21 l3 and a first integral of the form (14). b201 y 2 b01 y 7b01 yx 49x2 Case 4. In this case we have three algebraic curves: l1 = 2x − 26x 19 − 4 + 1, l2 = 361 − 38 19 + 16 − b01 y −2 −7/2 3 l3 . After integration we obtain a first integral of 2 + 1 and l3 = x yielding integrating factor μ = l1 l2 the form (14). 2 b201 y 2 9b01 yx b01 y Case 5. Here we find following invariant curves: l1 = 81x − 22x 169 − 26 13 + 16 − 2 + 1, l2 = b2 y 2

−5

b01 y 01 yx 2 −2 3 01 − 29b156 + 3x 13 + 16 − 2 + 1 and l3 = x. We obtain the Darboux integrating factor μ = l1 l2 l3 , which yields a first integral of the form (14). 6x2 169

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Case 6. Under these conditions system (9) becomes x˙ = x − x2 +

7b01 yx , 4

y˙ = −4y −

12x2 yx + b01 y 2 . − 49b01 7

b2 y 2

2

b01 y 01 It admits two invariant curves: l1 = 1 + x49 − b0114yx − 8x and l2 = x yielding the Darboux 7 + 16 − 2 −9/2 3 l2 and a first integral of the form (14). integrating factor μ = l1 Case 7. Here we have three invariant curves: l1 = 1 − x − b014 y , l2 = x and l3 = y. We can construct the (−4+b10) 4 Darboux first integral H = l1 l2 l3 which is of the form (14). Case 8. The corresponding system here is

x˙ = x − x2 − b01 yx,

y˙ = −4y +

24x2 + yx + b01 y 2 . 49b01 4

3

9x 6 This system admits three algebraic invariant curves: l1 = 1 − x7 − b014 y , l2 = 1 + 2401 + 343 b01 yx3 − 12x 343 + 2 −5/2 2 3 4b yx 1 2 2 2 6 30x 12x 2 01 − 7 and l3 = x bringing the Darboux integrating factor μ = l1 l2 l3 49 b01 y x − 49 b01 yx + 49 + 7 and a first integral of the form (14). 2 10b01 yx b01 y 2x Case 9. In this case we have three algebraic curves: l1 = 1 + 105x − 30x 529 + 23 23 , l2 = − 23 − 4 and −7/2 2 3 l3 = x, which allow us to construct the Darboux integrating factor of the form μ = l1 l2 l3 . Integration brings a first integral of the form (14). Case 10. In this case system (9) takes the form

3b01 yx , y˙ = −4y + b2,−1 x2 + b01 y 2 4  = 1 + 12 (± 1 − b01 b2,−1 − 1)x − b014 y and l3 = x yielding the Darboux

x˙ = x − x2 + and it has three algebraic curves: l1,2

integrating factor μ = l1a1 l2a2 l3b , where a1,2 = −

5(

 1−b b ±1)  01 2,−1 2 1−b01 b2,−1

and b = 3. By [4, Theorem 4.13] there exists

a first integral of the form (14). Case 11. The corresponding system is x˙ = x − x2 +

11b01 yx , 4

It has three algebraic curves l1 = 1 +

2x 19

−

y˙ = −4y −

168x2 16yx + b01 y 2 . + 361b01 19

b01 y 4 ,

729b01 yx5 2754x5 1215b201 y 2 x4 2079b01 yx4 2655x4 135b301 y 3 x3 729x6 − + + − + − 47 045 881 4 952 198 2 476 099 2 085 136 260 642 130 321 109 744 2 2 3 3 3 4 4 2 3 3 2 2 2 2 645b01 yx 636x 135b01 y x 183b01 y x 485b01 y x 39 621b01 y x − + + − + − b01 yx2 + 27 436 6859 6859 92 416 5776 2888 361 9b5 y 5 x 53b401 y 4 x 45 3 3 3b5 y 5 39x2 21 b01 yx 30x b601 y 6 − 01 + − b01 y x + b201 y 2 x + − + − 01 + 361 9728 2432 304 76 2 19 4096 512 4 4 3 3 2 2 5b y 15b01 y 3b01 y 15b01 y − 01 + − , + 256 16 16 2

l2 = 1 +

−5/2 3 l3

and l3 = x yielding the Darboux integrating factor μ = l12 l2 the form (14). Case 12. For this case system (9) takes the form x˙ = x − x2 − 2b01 yx,

y˙ = −4y +

and integration brings a first integral of

168x2 22yx + b01 y 2 , + 529b01 23

B. Ferčec et al. / J. Math. Anal. Appl. 420 (2014) 1568–1591

and it has the following algebraic curves l1 = 1 −

2x 23

−

1581

b01 y 4 ,

1372b01 yx5 4074x5 294b201 y 2 x4 1856b01 yx4 105x4 28b301 y 3 x3 2401x6 + − + − + + 148 035 889 6 436 343 6 436 343 279 841 279 841 12 167 12 167 2 2 3 3 3 840b01 yx 1036x 1 4 4 2 280b01 y x 14 3 3 2 70 2 2 2 224 + − + b y x − b y x + b y x − b01 yx2 − 12 167 12 167 12 167 529 01 529 01 529 01 529 4 399x2 28b01 yx 42x − b201 y 2 x + − , + 529 23 23 23

l2 = 1 +

−5/2 3 l3

and l3 = x. We can construct the Darboux integrating factor of the form μ = l16 l2 brings a first integral of the form (14). Case 13. Under these conditions system (9) is written as x˙ = x − x2 −

b01 yx , 2

and integration

y˙ = −4y + b2,−1 x2 + 5yx + b01 y 2 ,

and it has the Darboux integrating factor μ = x3 which yields a first integral Ψ = x4 y −

b2,−1 x6 1 − yx5 − b01 y 2 x4 . 6 4

Case 14 – Case 16. In case 14 and case 16 we find only f1 = y and in case 15 we find f1 = y and f2 = 1 − b014 y . However, in all three cases we cannot construct the Darboux first integral or the Darboux integrating factor. Noting that conditions in these cases are b2,−1 = 0 and b10 = A, where A = 1, 0 and 2, respectively, system (9) is written as x˙ = x − x2 − a01 xy − a−12 y 2 ,

y˙ = −4y + Axy + b01 y 2 .

We look for a formal first integral in the form Ψ (x, y) = recursively by the differential equation

∞ k=1

fk (x)y k . The functions fk are determined

  −a−12 fk−2 (x) − a01 xfk−1 (x) + (k − 1)b01 fk−1 + (Ax − 4)kfk (x) + x(1 − x)fk (x) = 0.

If k = 1, 2, 3, 4 and setting the integration constant equal to 1 we observe that f1 (x) =

−x4 , (x − 1)4−A

f3 (x) =

x12 + · · · , (x − 1)3(4−A)

f2 (x) =

x8 + · · · , (x − 1)2(4−A)

f4 (x) =

x16 + · · · . (x − 1)4(4−A)

p4k (x) So, suppose by induction that fk (x) = (x−1) k(4−A) , where pi (x) denotes a polynomial of degree at most i and k = 1, . . . , n − 1. Let check the form of fk (x) for k = n. In order to complete this task we solve the differential equation

fn (x) =

  a01 xfn−1 + a−12 fn−2 − b01 (n − 1)fn−1 (4 − Ax)n fn (x) + , x(1 − x) x(1 − x)

(15)

using the induction assumption about fn−1 and fn−2 . As the general solution of linear differential equation of the form f  (x) = g(x)f (x) + h(x) is 

f (x) = Ce

g(x)dx



+e

 g(x)dx

e−



g(x)dx

h(x)dx,

(16)

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and, in our case we have g(x) =

(4−Ax)n x(1−x)

and

h(x) = it follows that e



g(x)dx

e− Rewriting e−



g(x)dx



x4n (x−1)n(4−A)

=

g(x)dx

h(x) =

p4n−3 (x) , x(x − 1)(n−1)(4−A)+2

and

(x − 1)n(4−A) p4n−3 (x) p4n−A−1 (x) · = . x4n x4n+1 x(x − 1)(n−1)(4−A)+2

h(x) as

a0 + a1 x + · · · + a4n−A−1 x4n−A−1 a0 a1 a4n−A−1 p4n−A−1 (x) = = 4n+1 + 4n + · · · + , 4n+1 4n+1 x x x x xA+2 and integrating, yields 

e−



g(x)dx

h(x)dx =

a0 a1 a4n−A−1 + 4n−1 + · · · + 4n x x xA+1

for some a0 , a1 , . . . , a4n−A−1 . Therefore, using (16), choosing integration constant C = 1 at each step i of computing functions fi (x) for i = 1, 2, . . . , we obtain the solution of (15) fn (x) = =

  x4n x4n a0 a1 a4n−A−1 + + + · · · + x4n−1 xA+1 (x − 1)n(4−n) (x − 1)n(4−n) x4n [1 · x4n + a4n−A−1 x4n−A−1 + · · · + a0 ] p4n = , n(4−A) (x − 1) (x − 1)n(4−A)

where p4n denotes a polynomial of degree at most 4n. For choices A = 1, 0 and 2 satisfying cases 14, 15 ∞ and 16, respectively it follows that there exists an analytic first integral of the form Ψ (x, y) = k=1 fk (x)y k whose power series expansion is of the form (14). Case 17. Under conditions of this case system (9) is written as 13b201 y 2 91 78a201 y 2 + + a01 b01 y 2 , 25 100 50 16yx 180x2 + + b01 y 2 . y˙ = −4y + 169(2a01 + b01 ) 13

x˙ = x − x2 − a01 yx +

(17)

169 2 In this case we are able to find one invariant curve f1 = x2 + 13 5 (2a01 + b01 )yx + 100 (4a01 + 4b01 a01 + b201 )y 2 − 169 30 (2a01 + b01 )y, which is not enough to construct the Darboux first integral or the Darboux integrating factor. Therefore, we use the monodromy arguments (see e.g. [6] for details). In order to avoid some difficulties which appear in the computations of the singular points in this case we reparameterize coefficients a01 and b01 by a01 = (−7s21 + s22 )/30 and b01 = (22s21 − s22 )/15, which yields the system

  13 2  2 1  s1 4s1 − s22 y 2 + xy 7s21 − s22 , 300 30 16xy (22s21 − s22 )y 2 180x2 + , + y˙ = −4y + 169s21 13 15

x˙ = x − x2 −

(18)

and the invariant curve has the form f1 = x2 +13s21 xy/5+169s21 y(3s21 y−10)/300. We compute the eigenvalues of all singular points of system (18) on the curve f1 = 0. The ratios of eigenvalues are 3 and 3 for finite

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points and 2 for the point at the infinity. We recall that here the sum of the ratio of eigenvalues on an invariant conic must be 4. Thus, all singular points on f1 = 0 except the saddle point at the origin are linearizable nodes whose monodromy is the identity and therefore, the monodromy of the saddle at the origin is also the identity. Hence the saddle at the origin is integrable. Case 18. In order to avoid some difficulties which appear in computations with square roots we reparam1 eterize coefficients a01 and b01 by a01 = 16 (9s21 − s22 ) and b01 = 14 (−5s21 + s22 ). Thus, in this case system (9) becomes 21s41 + 43s31 s2 + 27s21 s22 + 5s1 s32 2 s22 − 9s21 xy + y , 16 64 −y(5ys21 − s22 y + 16)(7s1 + 5s2 )2 + 16s2 (7s1 + 5s2 )xy + 384x2 . y˙ = 4(7s1 + 5s2 )2

x˙ = x − x2 +

(19)

1 1 −s2 )x 2 4 2 3 We can find two invariant curves f1 = 1 + 2(s 7s1 +5s2 + 16 (s1 − s2 )(3s1 + s2 )y and f2 = 49y s1 + 168s2 y s1 + 214s22 y 2 s21 + 224xys21 − 784ys21 + 120s32 y 2 s1 − 1120s2 ys1 + 384s2 xys1 + 256x2 + 25s42 y 2 − 400s22 y + 160s22 xy which turns out to be not enough to construct the Darboux integrating factor or the Darboux first integral. Therefore, in this case we also use the monodromy arguments (see e.g. [6]). We compute the eigenvalues of all singular points of system (9) on the curve f2 = 0. The ratios of eigenvalues are 2 and 4 for finite points and 2 for the point at the infinity. Hence, as in the previous case, all singular points on f2 = 0 except the saddle point at the origin are linearizable nodes whose monodromy is the identity and therefore, the monodromy of the saddle at the origin is also the identity. Hence the saddle at the origin is integrable. Case 19. In this case the corresponding system (9) is of the form

x˙ = x − x2 −

5b01 yx − a−12 y 2 , 4

It admits two invariant curves l1 = y and l2 = b 4

l1 l2b

4x2 b01

y˙ = −4y + b01 y 2 + 8xy.

+ yx −

4x b01

+

4a−12 y 2 9b01

which yields the Darboux first

4

integral of the form Ψ = = x y + h.o.t. Case 20. Here the corresponding system is x˙ = x − x2 − a−12 y 2 ,

y˙ = −4y + b01 y 2 + 3xy

and it has three invariant lines l1,2 = 1 − x + 18 (−b01 ± integrating factor l1a1 l2a2 l3b , where a1,2

 b201 − 16a−12 )y and l3 = y yielding the Darboux

 5(±b01 + b201 − 16a−12 )  =− 8 b201 − 16a−12

3 and b = − . 4

By [4, Theorem 4.13] there exists a first integral of the form (14). Case 21. Under conditions of this case system (9) becomes x˙ = x − x2 ,

y˙ = −4y + b2,−1 x2 + b10 yx.

It has two invariant curves l1 = 1 − x and l2 = x. Using constructed Darboux integrating factor of the form μ = l1−5+b10 l23 we obtain a first integral of the form (14). In cases 22–41 all conditions are of the form: b10 = A, a01 = Bb01 , a−12 = Cb201 and b2,−1 = bD . 01 Therefore, the system (9) with coefficients A, B, C, and D corresponding to one of cases 22–41 is of the form x˙ = x − x2 − Bb01 xy − Cb201 y 2 ,

y˙ = −4y +

D 2 x + Axy + b01 y 2 . b01

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Case 22. In this case we obtain three invariant curves 40x 10b01 y 61b01 yx 61x 3721b201 y 2 − , l 2 = x2 + − + , 61 49 343 49 470 596 125b01 yx 110x 125b201 y 2 5b01 y 6125x2 + − + − , l3 = 1 + 7442 854 61 19 208 28 l1 = 1 −

−7/2

which bring the Darboux integrating factor μ = l1 l23 l3−2 and using it we obtain a first integral of the required form. Case 23. In this case we find three invariant curves 5b01 y 17 161b201 y 2 5x 1703b01 yx − , l2 = 14x2 + − 131x + , 131 36 36 729 279 936x3 16 848yx2 979 776x2 288y 2 x 300 672yx 66 024y 2 169 776y + − + − + 131y 3 + + , l3 = 3 2 3 2 17 161b01 131b01 81 875b01 b01 625b01 625b01 625b201 l1 = 1 −

−1/3 −1/3 −5/6

and we construct the integrating factor of the form μ = l1 l2 l3 . By [4, Theorem 4.13] there exists a first integral of the form (14). Case 24. In this case we construct the Darboux first integral of the form Ψ = l1−10 l24 l3 , where 16b201 y 2 5x 5b01 y 8b01 yx − , l 2 = x2 + − 4x + , 8 38 19 361 64b01 y 8b01 yx 16b201 y 2 + − . l3 = x2 + 19 361 19

l1 = 1 −

Case 25. In this case the system has three invariant curves 15x 20b01 y 5x 45b01 y − , l2 = 1 + − , 11 53 22 212 484b201 y 2 44b01 yx + 11x + , l 3 = x2 − 53 2809

l1 = 1 −

−7/2

yielding the Darboux integrating factor μ = l1 l2−2 l33 and a first integral of the form (14). Case 26. In this case we find two invariant curves 2483b01 yx 191x 36 481b201 y 2 − + , 36 9 36 625b01 yx3 14 500x3 1875b201 y 2 x2 9000b01 yx2 6450x2 625b301 y 3 x 625x4 − + + − + − l2 = 1 330 863 361 6 967 871 6 967 871 291 848 36 481 36 481 3056 875b301 y 3 975b201 y 2 1475b01 yx 260x 625b401 y 4 7125 2 2 b y x− − + − + − 10b01 y, + 764 01 191 191 256 8 8 l1 = x2 −

−7/2

and we construct the Darboux integrating factor μ = l13 l2 Case 27. Here we have two invariant curves l 1 = x2 +

b01 yx 2x 2b201 y 2 − − , 3 3 3

. After integration we obtain a first integral (14).

l2 = x2 − 8b01 yx − 9b201 y 2 + 4b01 y,

yielding the Darboux first integral of the form Ψ = l14 l2 = x4 y + P6 (x, y) + P7 (x, y) + P8 (x, y) + P9 (x, y) + P10 (x, y), where Pk (x, y) denotes homogeneous polynomial of degree k.

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Case 28. Here we find two invariant curves passing through the origin 50b201 y 2 , 9 2yx3 84x3 12y 2 x2 192yx2 12x2 6y 3 x 132y 2 x 144yx + 3 − + − + + − + 3 + 5y 4 5b01 25b401 5b201 5b301 5b401 b01 b201 b01

l1 = x2 + 5b01 yx − 10x + l2 =

3x4 125b401 −

120y 3 60y 2 80y − 2 − 3 , b01 b01 b01 −1/3 −5/6 l2 .

consequently, we can construct the Darboux integrating factor of the form μ = l1 of [4] there exists a first integral of required form. Case 29. In this case we find two invariant curves l1 = 1 −

5x 5b01 y + , 23 12

and l2 = 18x2 −

By Theorem 4.13

529b01 y 253b01 yx 529b201 y 2 + + . 2 3 2

They are not enough to construct the Darboux first integral or the Darboux integrating factor. Therefore, we attempt to use the monodromy argument (see e.g. [6]). We compute the eigenvalues of all singular points of system (9) corresponding to case 29 on the curve l2 = 0. The ratios of eigenvalues are: −4 (for (0, 0)) and 23 3 for finite singular point ( 25 , − 25b301 ) and 3 and 2 for singular points at the infinity. Thus, all singular points on l2 = 0 except the saddle point at the origin are linearizable nodes whose monodromy is the identity and therefore, the monodromy of the saddle at the origin is also the identity. Hence the saddle at the origin is integrable. −9/2 −2 3 Case 30. Here we find the Darboux integrating factor of the form μ = l1 l2 l3 , where l1 , l2 and l3 are algebraic invariant curves of the form 15x 20b01 y 35x 35b01 y − , l2 = 1 − − , 19 107 38 428 19b01 yx 19x 361b201 y 2 − + . l 3 = x2 + 107 16 45 796 l1 = 1 −

After integration using μ we obtain a first integral of the form (14). Case 31. In this case the system has two invariant curves 425b01 yx 35x 225b201 y 2 5b01 y 525x2 + − + + , 2704 364 26 196 7 26b01 yx 52x 1352b201 y 2 − + , l 2 = x2 + 3 3 147 l1 = 1 +

−7/2

which yields the Darboux integrating factor μ = l1 l23 and a first integral of the required form. Case 32. Here we find three invariant curves of the form 40x 10b01 y 19b01 yx 19x 361b201 y 2 − , l2 = x2 − + + , 19 31 93 9 34 596 5 984 416 080x4 128 697 120yx3 1 595 844 288x3 1 037 880y 2 x2 4 289 904yx2 l3 = − + + − 130 321b401 6859b301 6859b401 361b201 361b301 l1 = 1 −

−

206 868 704x2 3720y 3 x 92 256y 2 x 22 879 488yx 372y 3 149 916y 2 476 656y − − + + 5y 4 + + − . 4 2 3 1805b01 19b01 19b01 95b01 b01 5b201 5b301

We construct the Darboux integrating factor of the form μ = l1 (l2 l3 )−1 and by Theorem 4.13 of [4] there exists a first integral of the form (14). 1/2

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Case 33. Using two invariant curves of the form l1 = 1 +

50b01 yx 75b201 y 2 5b01 y 75x2 + − 10x − − 2 19 2888 38

and

b01 y 2b01 yx 21b201 y 2 − + 19 1444 19

l 2 = x2 −

1

− 34

we can construct the Darboux integrating factor μ = l14 l2

and a first integral of the form (14). −9 −7

Case 34. Using the Darboux integrating factor of the form μ = l1 2 l2 2 l33 , where l1 = 1 −

5x 45b01 y − , 32 148

l3 = x2 +

l2 = 1 −

and

232b01 yx 32x 2048b201 y 2 − − 111 3 4107

we compute a first integral of required form. Case 35. Here we obtain three invariant curves l1 = 1 − l2 = 1 −

35x 35b01 y + 32 148

15x 5b01 y − , 11 2

and l3 = x2 −

5x 11

+

5b01 y 4 ,

121b01 y 77b01 yx 363b201 y 2 + + . 4 8 4

We compute the Darboux integrating factor μ = l1 (l2 l3 )−3/4 yielding the first integral of the form (14). y 5x Case 36. Using three invariant curves of the form l1 = 1 + 121 − 5b01 4 , 1/4

l 2 = x2 − l3 = −

242b01 yx 121x 14 641b201 y 2 − + , 9 9 81

and

14 580x3 6885yx2 84x2 90y 2 x 222yx 6y 2 y + + − − + 45y 3 − + 2 , 3 2 3 2 1 771 561b01 14 641b01 14 641b01 11b01 121b01 b01 b01 −1/2 −1/3 −5/6

we construct the Darboux integrating factor μ = l1 l2 l3 . By Theorem 4.13 of [4] there exists a first integral of the form (14). Case 37. In this case the system admits two invariant curves

l1 =

27x3 27yx2 27x2 9y 2 x 99yx 27x 364y 2 + + + + − 3 + 7y 3 + , 3 2 3 2 3136b01 112b01 70b01 4b01 20b01 5b01 15b01

and l2 =

27x3 27yx2 27x2 9y 2 x 153yx 231y 2 21y + − + − + 7y 3 + + . 3 2 3 2 3136b01 112b01 4480b01 4b01 160b01 160b01 40b201 −1/3 −5/6 l2

We compute the Darboux integrating factor μ = l1 form (14) is ensured by Theorem 4.13 of [4].

and the existence of the first integral of the −9

Case 38. Here we find a first integral of the form (14) using the Darboux integrating factor μ = l1 2 l23 , where l1 = 1 −

80x 5b01 y − 71 11

and l2 = x2 −

5041b201 y 2 71b01 yx + 71x + . 22 1936

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Case 39. Here we find three invariant lines of the form 578 125b01 yx4 896 875x4 250 000b201 y 2 x3 678 125b01 yx3 43 750x5 + + + + 1 805 076 171 190 008 018 126 672 012 1 666 737 1 111 158 15 625b301 y 3 x2 4 446 875b201 y 2 x2 4375b01 yx2 1625x2 171 875b401 y 4 x 8750x3 + + − + + + 555 579 4332 233 928 12 996 3249 4104 1 034 375b301 y 3 x 23 125 2 2 690 625b401 y 4 500b01 yx 100x 53 125b501 y 5 + + b y x+ − + + 4104 684 01 171 57 288 576 2125b201 y 2 25b01 y 74 375b301 y 3 + + , + 144 16 2 16x4 32yx3 64x3 8y 2 x2 176yx2 224x2 8y 3 x l2 = + + + + − + 3 518 667b401 61 731b301 48 735b401 361b201 1805b301 243 675b401 19b01 l1 = 1 +

+

224y 2 x 1664yx 92y 3 124y 2 16y 4 − + 3y + + + , 95b201 4275b301 5b01 75b201 75b301

and l3 =

16x4 32yx3 48x3 8y 2 x2 568yx2 288x2 8y 3 x + + + + + + 4 3 4 2 3 4 3 518 667b01 61 731b01 34 295b01 361b01 5415b01 9025b01 19b01 +

244y 2 x 504yx 288x 102y 3 408y 2 + − + 3y 4 + + . 2 3 4 95b01 475b01 475b01 5b01 25b201

We compute the Darboux first integral Ψ = l1−3 l2 l34 whose series expansion is of the form (14). Case 40. In this case we find two invariant curves l1 = 1 −

80x 5b01 y − , 89 29

and l2 = 27x2 +

267b01 yx 89x 7921b201 y 2 − + , 58 3 40 368

− 11

and we construct the Darboux integrating factor μ = l1 2 l23 and then a first integral of the form (14). Case 41. In this case we compute the Darboux first integral Ψ = l1−4 l24 l3 , where 25b01 yx 10x 25b201 y 2 5b01 y 25x2 + − − − , 242 154 11 392 28 121b201 y 2 11b01 yx + 11x + , l 2 = x2 − 7 196 121b01 y 11b01 yx 121b201 y 2 + + . l 3 = x2 − 7 196 14 l1 = −

Case 42. The corresponding system (9) in this case is of the form x˙ = x − x2 − a−12 y 2 ,

y˙ = −4y − xy/2.

We find two invariant curves for this system l1 = 1 − x + a−12 y 2 /8 and l2 = y which allow us to construct −17/8 −3/4 the Darboux integrating factor of the form μ = l1 l2 and the existence of the first integral of the form (14) is assured by Theorem 4.13 of [4]. Case 43 – Case 45. For all three cases we have b01 = 0, b2,−1 = 0, a01 = 0 and b10 = A, where A is 1/2, 5/2 and 3/2 for cases 43–45, respectively. Therefore, the corresponding system is of the form x˙ = x − x2 − a−12 y 2 ,

y˙ = −4y + Axy.

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In each of these three cases we are able to find only one invariant line l1 = y which is not enough to construct the Darboux first integral or the Darboux integrating factor. Therefore we look for a formal first integral ∞ expressed in the form Ψ (x, y) = k=1 fk (x)y k . The functions fk are determined recursively by the first order differential equation  −a−12 fk−2 (x) − 4kfk (x) + kAxfk (x) + x(1 − x)fk (x) = 0.

Let pi (x) denote a polynomial of degree at most i. Using induction on k we show that fk (x) = For the basis of induction, by setting the integration constant to one, we have f1 (x) = assuming that the induction hypothesis holds for j < k, we have fj (x) = order linear differential equation fk (x) =

p4j (x) . (x−1)(4−A)j

p4k (x) . (x−1)(4−A)k 4

x . (x−1)(4−A)

Next,

Then we have the first

 (x) a−12 fk−2 (4 − Ax)k fk (x) + . x(1 − x) x(1 − x)

(20)

Recall that the solution of the linear differential equation f  (x) = g(x)f (x) + h(x) is (16). Noting that p fk−2 = (x−1)4(k−2) (4−A)(k−2) and applying similar computational procedure as in cases 14–16 we find that the solution for (20) is  x4k x4k p4k−2A−2 (x) fk (x) = + dx (4−A)k (4−A)k x4k+1 (x − 1) (x − 1)

 a0 x4k a1 a2 a4k−2A−2 1 + dx + + + · · · + x4k+1 x4k x4k−1 x2A+3 (x − 1)(4−A)k

a0 a1 a2 a4k−2A−2 x4k 1 + + + + · · · + x4k x4k−1 x4k−2 x2A+2 (x − 1)(4−A)k =

p4k (x) (x − 1)(4−A)k

where p4k (x) denote polynomials of degree at most 4k. This proves the induction and the proof of cases 43–45 is completed for A = 12 , 52 , 32 . Case 46. After computing the coefficients from the conditions of this case we find that this case can be decomposed into two subcases: 675 a) b2,−1 = − 3(−5396504+250 507 601a01 √

√

2612+525 10 , 2551 √ 3(539 504+250 675 10) b2,−1 = , 6 507 601a 01 √ 2612−525 10 . 2551

10)

, a−12 =

b10 =

b)

a−12 = −

√ 234(410 175 10a201 +1 368 511a201 ) , 74 632 321

√ 234(410 175 10a201 −1 368 511a201 ) , 74 632 321

b01 = − 2(7725

b01 =

√

10a01 +24 529a01 ) , 8639

√ 2(7725 10a01 −24 529a01 ) , 8639

b10 =

In both cases we find two invariant curves. Let la,b and fa,b denote invariant curves for both subcases. Index a corresponds to invariant curve for system (9) filled with conditions from a) and index b corresponds to invariant curve for system (9) filled with conditions from b). Using this notation we have la,b

√ √ √ (44 132 822 ∓ 706 860 10)x3 (238 ∓ 2970 10)yx2 7(∓51 980 720 + 16 414 566 10)x2 = + ± 78 091 212a201 5102a01 382 650a201 √ √ √ 41(∓10 780 + 3414 10)yx (5 972 420 ∓ 1 888 566 10)x 2(2970 10a01 ± 238a01 )y 3 2 + 2y x ± + ∓ 300a01 300a201 25 917 √ 13 (∓5720 + 1806 10)y 2 , ± 150

B. Ferčec et al. / J. Math. Anal. Appl. 420 (2014) 1568–1591

fa,b

1589

√ √ (3 861 805 637 ∓ 1 162 343 360 10)x6 (6 625 166 ∓ 3 041 960 10)yx5 = + 94 002 296 445a201 36 849 195a01 √ √ 2(∓853 872 612 205 + 270 026 378 041 10)x5 2(∓259 106 075 + 81 917 729 10)yx4 2 4 ± +y x ± 921 229 875a201 72 225a01 √ √ 4 3(440 609 747 ∓ 139 334 678 10)x 8(±3 599 117 + 2 318 725 10)a01 y 3 x3 + ∓ 66 875a201 24 958 071 √ √ 2 3 8(16 131 771 463 ∓ 5 100 608 812 10)yx3 32(∓4 893 737 + 1 549 508 10)y x + ± 14 445 1 805 625a01 √ √ 3 64(∓6 451 604 305 + 2 040 198 883 10)x 8(±82 712 334 587 + 19 372 813 330 10)a201 y 4 x2 ± ± 15 046 875a201 215 612 775 369 √ √ 4(∓9 118 925 659 + 2 886 803 191 10)y 2 x2 4(∓1 253 404 102 870 + 394 018 564 231 10)a01 y 3 x2 ∓ ± 623 951 775 601 875 √ √ 2 4(194 210 144 462 ∓ 61 415 401 673 10)x2 4(∓3 095 596 649 260 + 978 840 105 391 10)yx + ± 45 140 625a01 75 234 375a201 √ 32(±169 898 542 620 406 + 62 274 561 197 195 10)a301 y 5 x ∓ 3 104 464 610 687 985 √ 16(∓107 163 377 491 130 + 34 634 486 590 247 10)a201 y 4 x ± 598 924 376 025 √ √ 8(∓736 346 688 059 + 229 612 484 741 10)a01 y 3 x 8(∓106 721 196 235 + 33 582 210 361 10)y 2 x ∓ ∓ 577 733 125 45 140 625 √ 8(8 528 758 152 668 ∓ 2 696 860 731 977 10)yx + 225 703 125a01 √ 16(±296 870 402 740 081 997 + 86 796 109 377 846 580 10)a401 y 6 ± 8 939 823 257 244 500 805 √ 16(∓685 577 182 006 925 980 + 202 494 657 574 882 111 10)a301 y 5 ± 25 870 538 422 399 875 √ 4(∓10 191 711 380 951 723 + 3 116 773 245 583 502 10)a201 y 4 ± 14 973 109 400 625 √ 16(∓419 435 752 079 855 + 131 894 867 253 953 10)a01 y 3 ± 389 969 859 375 √ √ 8(∓2 846 157 446 809 + 899 975 292 211 10)y 4(∓2 381 664 790 141 + 752 392 059 784 10)y 2 ± . ± 225 703 125 225 703 125a01 1/5 −7/10

In both cases we can construct the Darboux integrating factor of the form μa,b = la,b fa,b . By [4, Theorem 4.13] there exists a first integral of the form (14) for system of cases a) and b), respectively. 4. Conclusions In order to obtain the necessary conditions for the existence of a 1 : −4 resonant center for (9) the decomposition of the center variety was found. However, we arrived at huge computational difficulties when computing the decomposition. The decomposition gives 46 cases (of necessary conditions) for Theorem 2.1 and 9 cases for Theorem 2.2. The computations for the decomposition in Theorem 2.1 were still extremely laborious. The corresponding computational problem was solved by using the routine minAssGTZ [7] of CAS Singular which is based on the Gianni–Trager–Zacharias algorithm [18]. Like e.g. in [8] also in the case of (9) which was the main problem of the present paper the decomposition of varieties of ideals generated

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by saddle quantities (8) could not be performed in polynomial ring of characteristic zero, but it became possible in the rings of finite characteristics Zp [a, b]. The modular approach was for the first time successfully applied in p : −q resonant center problems in [11,17]. Preparing this paper the authors have used modular approach for the following prime numbers: p = 32 003, 104 729, 4 256 233, 7368 787, 15 485 863, 179 595 127, 433 494 437 and 479 001 599. The computational problems are described in the proof of Theorem 2.1. To perform the rational reconstruction we used the following Mathematica code [31,36]: RATCONVERT[c,m]:=Block[u={1,0,m},v={0,1,c},r], While[Sqrt[m/2] ≤ v[[3]],r=u-Quotient[u[[3]],v[[3]]] v; u=v; v=r]; If[Abs[v[[2]]] ≥ Sqrt[m/2], err, v[[3]]/v[[2]]] Given an integer number c and a natural number m the function produces a couple of integer numbers v2 and v3 such that v3 /v2 ≡ c mod m and |v2 |, |v3 | ≤ m/2. Finally to prove the existence of the first integral in the 55 cases of Theorems 2.1 and 2.2 we have used the Darboux method, the inductive method (see [2,8,31]), as well as the monodromy arguments [6]. From the results presented in this work we have the following conjecture: Conjecture 4.1. The conditions 1–46 of Theorem 2.1 are the necessary and sufficient conditions for integrability of system (9). In order to verify this conjecture we have to check that over the field of characteristic 0 all Gröbner bases

46 of 1 − wpk , B6  for k = 1, . . . , 112 are equal to {1}, where pk are polynomials in P = i=1 Pi and Pi denotes the component given by the item i) in Theorem 2.1. Unfortunately the authors were not able to complete computations over the field of rational numbers. Acknowledgments The authors would like to thank Prof. Valery G. Romanovski for fruitful discussions and comments about the work. The authors are grateful to the referees for their valuable instructions and suggestions to improve this paper. The first and the third authors want to thank the Slovenian Research Agency for support (grant P1-0288). The second author is partially supported by a MINECO/FEDER grant number MTM2011-22877 and by an AGAUR (Generalitat de Catalunya) grant number 2014SGR 1204. The first, third and fourth authors want to acknowledge the support of the project FP7-PEOPLE-2012-IRSES-316338. The fourth author is partially supported by CNPq fellowship “Projeto Universal” 472796/2013-5. References [1] E.A. Arnold, Modular algorithms for computing Gröbner bases, J. Symbolic Comput. 35 (2003) 403–419. [2] X. Chen, J. Giné, V.G. Romanovski, D.S. Shafer, The 1 : −q resonant center problem for certain cubic Lotka–Volterra systems, Appl. Math. Comput. 218 (2012) 11620–11633. [3] C. Christopher, C. Li, Limit Cycles of Differential Equations, Birkhäuser Verlag, Basel, 2007. [4] C. Christopher, P. Mardešić, C. Rousseau, Normalizable, integrable and linearizable saddle points for complex quadratic systems in C2 , J. Dyn. Control Syst. 9 (2003) 311–363. [5] C. Christopher, C. Rousseau, Nondegenerate linearizable centers of complex planar quadratic and symmetric cubic systems in C2 , Publ. Math. 45 (1) (2001) 95–123. [6] C. Christopher, C. Rousseau, Normalizable, integrable and linearizable saddle points in the Lotka–Volterra system, Qual. Theory Dyn. Syst. 5 (1) (2004) 11–61. [7] W. Decker, S. Laplagne, G. Pfister, H.A. Schonemann, SINGULAR 3-1 library for computing the prime decomposition and radical of ideals, primdec.lib, 2010. [8] D. Dolićanin, J. Giné, R. Oliveira, V.G. Romanovski, The center problem for a 2 : −3 resonant cubic Lotka–Volterra system, Appl. Math. Comput. 220 (2013) 12–19.

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