Bifurcation of limit cycles and integrability conditions for 6-parameter families of polynomial...

Nonlinear Analysis 69 (2008) 256–275www.elsevier.com/locate/na

Bifurcation of limit cycles and integrability conditions for6-parameter families of polynomial vector fields of arbitrary degree

Claudia Valls

Departamento de Matematica, Instituto Superior Tecnico, Av. Rovisco Pais 1049-001, Lisboa, Portugal

Received 11 December 2006; accepted 22 May 2007

Abstract

We characterize the center variety and the maximum order of any focus of the 6-parameter family of real planar polynomialvectors given, in complex notation, by z = iz + Azn1 z j1 + Bzn2 z j2 + Czn3 z j3 , where A, B, C ∈ C \ {0}, (n1, j1) 6= (n2, j2) 6=

(n3, j3), nk + jk > 1 for k = 1, 2, 3 n1 + j1 = n2 + j2 = n3 + j3, |1 − n3 + j3| = |1 − n2 + j2| 6= |1 − n1 + j1| and(1 − n1 + j1)(1 − n2 + j2) 6= 0.c© 2008 Elsevier Ltd. All rights reserved.

Keywords: Center manifolds; Integrability condition; Quadratic polynomial vector fields

1. Introduction and statement of the main results

A planar real analytic differential system in complex notation, z = x + iy, can be written as

z = iz + F(z, z). (1)

We are concerned mainly with the conditions under which the origin of (1) is a center, and also with the question ofdetermining the number of limit cycles which can bifurcate from the origin. These are very related issues and theirknowledge will provide the dynamical properties of (1). In its real formulation, systems like (1) are used to modelvarious phenomena like the crystallization of agates (see [19,24]).

A usual method of looking for nondegenerate centers (i.e. having purely imaginary eigenvalues) of planarpolynomial vector fields such as that in (1) is to calculate the successive coefficients v2i+1, called the Liapunovconstants of the polynomial vector field, of the return map of the vector field about the origin. When all the v2i+1vanish, then the origin is a center (notion first introduced by Poincare in [17]). The set of coefficients for which allthe vi vanish is called the center variety of the family of polynomial vector fields. By the Hilbert Basis Theorem, thecenter variety is an algebraic set. On the other hand, if there is an integer N satisfying v2k+1 = 0 for k < N butv2N+1 6= 0, then the origin is a fine focus of order N . The order of a focus is an invariant of the system. At mostN limit cycles can bifurcate from a fine focus of order N under perturbation of the coefficients. The stability of theorigin is determined by the sign of the first non-vanishing Liapunov coefficient.

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C. Valls / Nonlinear Analysis 69 (2008) 256–275 257

In general, is very difficult to study the centers and the focus, since to do it requires a good knowledge not onlyof the common zeros of the polynomials vi , but also of the finite generated ideal that they generate in the ring ofpolynomials, taking as variables the coefficients of the polynomial vector field. Furthermore, in general the calculationof the Liapunov constants is not easy, and the computational complexity of finding their common zeros grows veryquickly. A number of algorithms have been developed to compute them automatically up to a certain order (see [2,5,6,8,12–15] and the references therein). For analytical computation of the Liapunov constants, we are using the resultsof [3]. First, for each m, we will see which are the admissible monomials that appear in v2m+1. After this, we willcompute the coefficients of the polynomial v2m+1. For doing that, we will use the knowledge of some types of centersfor the systems (2).

The classification of centers in polynomial vector fields started with the quadratic ones with the works of Dulac,Kapteyn, Bautin, Zoładek,... see [20] for references. It continued with the symmetric cubic systems (those withoutquadratic terms) and projective quadratic systems [21,25,16]. Liapunov constants are also well known for Lienardsystems [26,1,3].

Our goal is to study the centers and the maximum order of any focus of a 6-parameter family of linear polynomialvector fields with arbitrary homogeneous nonlinearities. Despite the fact that linear systems with quadratic and cubicnonlinearities are well understood (see [15,18,25]), very few families of centers of arbitrary degree are known; seefor instance [4,9] for the case of 4-parameter families of polynomial vector fields of arbitrary degree, [10] for thecase of 6-parameter families of polynomial vector fields, [23] for the case of an 8-parameter particular families ofpolynomial vector fields and [22] for the case of particular families of polynomial vector fields of arbitrarily numbersof parameters (greater or equal 6) and arbitrary degrees.

We consider the family of real polynomial differential equations in (x, y) ∈ R2 that in complex notation can bewritten as

z = iz + Azn1 z j1 + Bzn2 z j2 + Czn3 z j3 , (2)

with A, B, C ∈ C \ {0}, and where n1, j1, n2, j2, n3 and j3 are non-negative integers such that (n1, j1) 6= (n2, j2) 6=

(n3, j3), nk + jk > 1 for k = 1, 2, 3, n1+ j1 = n2+ j2 = n3+ j3 = n, and |1−n3+ j3| = |1−n2+ j2| 6= |1−n1+ j1|.Note that this implies that system (2) is a linear system with homogeneous nonlinearities of degree n1 + j1, and

1 − n3 + j3 = −(1 − n2 + j2) = n2 + j2 − 1. (3)

Set N1 = |1 − n1 + j1|, K1 = |1 − n2 + j2|, L1 = |1 − n3 + j3|, M1 = gcd{N1, K1}, N1 = M1 N2, K1 = M1 K2.We consider the case in which the parameters A, B, C are all different from zero, since the study of the center

manifold variety in the case when at least one of the parameters is zero has already been studied in [9] (in themore general case of the nonlinearity of the vector field not necessarily being homogeneous). For an homogeneousnonlinearity, the results in [9] are:

Proposition 1. For system (2) with C = B = 0 the following holds:

(a) if j1 6= n1 − 1 then the center manifold variety of system (2) is the set {A ∈ C \ {0}},(b) if j1 = n1 − 1 then the center manifold variety of system (2) is the set {A ∈ C \ {0} : Re(A) = 0}.

The cases A = B = 0 or A = C = 0 are symmetric.

We set MK2,N2 = (n − 1)(K2 + N2)/2 and d = (1 − n1 − j1)(1 − n2 − j2). We note that since N2, K2 ≥ 1 andgcd{N2, K2} = 1, N2 + K2 ≥ 3. Furthermore, if n1 + j1 − 1 is odd, then N1 and K1 are odd, and thus, N2 and K2 arealso odd. Hence, in this case, N2 + K2 is even. In short, MK2,N2 is always an integer number.

Proposition 2. For system (2) with C = 0 the following holds:

(a) if j1 = n1 − 1 and j2 = n2 − 1, then the center manifold variety of system (2) is the set {(A, B) ∈ (C \ {0})2:

Re(A) = Re(B) = 0},(b) if j1 = n1 − 1 and j2 6= n2 − 1, then the center manifold variety of system (2) is the set {(A, B) ∈ (C \ {0})2

:

Re(A) = 0}. The case j1 6= n1 − 1 and j2 = n2 − 1 is symmetric,(c) if j1 6= n1 − 1 and j2 6= n2 − 1, we have

(c.1) v2m+1 = 0 for m = 1, 2, . . . , MK2,N2 − 1,

258 C. Valls / Nonlinear Analysis 69 (2008) 256–275

(c.2) there exists a constant Kn1, j1,n2, j2 such that

v2m+1 =

Kn1, j1,n2, j2 Im( AK2 B N2) if d > 0 and K2 + N2 evenKn1, j1,n2, j2 Re( AK2 B N2) if d > 0 and K2 + N2 oddKn1, j1,n2, j2 Im(AK2 B N2) if d < 0 and K2 + N2 evenKn1, j1,n2, j2 Re(AK2 B N2) if d < 0 and K2 + N2 odd,

(c.3) if the constant Kn1, j1,n2, j2 introduced in statement (c.2) is different from zero, then the center manifoldvariety of system (2) is:

{(A, B) ∈ (C \ {0})2: Im( AK2 B N2) = 0} if d > 0 and K2 + N2 even

{(A, B) ∈ (C \ {0})2: Re( AK2 B N2) = 0} if d > 0 and K2 + N2 odd

{(A, B) ∈ (C \ {0})2: Im(AK2 B N2) = 0} if d < 0 and K2 + N2 even

{(A, B) ∈ (C \ {0})2: Re(AK2 B N2) = 0} if d < 0 and K2 + N2 odd

(c.4) for K2 + N2 = 2, Kn1, j1,n2, j2 = −2π and for K2 + N2 = 3,

Kn1, j1,n2, j2 =

3π(n1 − 1)/(1 + j1 − n1) if d < 0 and K2 = 13π(n1 − 1 − 3 j1)/(4(1 + j1 − n1)) if d < 0 and K2 = 2jπ/(1 + j1 − n1) if d > 0 and K2 = 1(3 j1 − n1 + 1)π/(4(1 + j1 − n1)) if d > 0 and K2 = 2.

Furthermore, we also consider the case in which n1 6= 1 + j1 and n2 6= 1 + j2, since the case in which eithern1 = 1 + j1 or n2 = 1 + j2 is studied in [10] (in the more general case of the nonlinearity of the vector field beingnot necessarily homogeneous). For homogeneous nonlinearities, the results in [10] are

Proposition 3. For system (2) with A, B, C ∈ C \ {0}, and j3 = n3 − 1 (note that then n must be odd), the followingholds:

(a) Re(C) = 0 is a necessary condition to have a center at the origin and v2k+1 = 0 for k = 1, . . . , MK2,N2 ,(b) there exists a constant Kn1, j1,n2, j2 such that

v2m+1 =

Kn1, j1,n2, j2 Im( AK2 B N2) if d > 0 and K2 + N2 evenKn1, j1,n2, j2 Re( AK2 B N2) if d > 0 and K2 + N2 oddKn1, j1,n2, j2 Im(AK2 B N2) if d < 0 and K2 + N2 evenKn1, j1,n2, j2 Re(AK2 B N2) if d < 0 and K2 + N2 odd,

(c) if the constants Kn1, j1,n2, j2 introduced in statement (b) are different from zero, then the center manifold variety ofsystem (2) is the subset of {(A, B, C) ∈ (C \ {0})3

} such that

Re(C) = Im( AK2 B N2) = 0 if d > 0 and K2 + N2 even

Re(C) = Re( AK2 B N2) = 0 if d > 0 and K2 + N2 odd

Re(C) = Im(AK2 B N2) = 0 if d < 0 and K2 + N2 even

Re(C) = Re(AK2 B N2) = 0 if d < 0 and K2 + N2 odd

(d) if Re(C) = 0, then for K2 + N2 = 2 we have Kn1, j1,n2, j2 = −2π , and for K2 + N2 = 3 we have

Kn1, j1,n2, j2 =

3π(n1 − 1)/(1 + j1 − n1) if d < 0 and K2 = 13π(n1 − 1 − 3 j1)/(4(1 + j1 − n1)) if d < 0 and K2 = 2jπ/(1 + j1 − n1) if d > 0 and K2 = 1(3 j1 − n1 + 1)π/(4(1 + j1 − n1)) if d > 0 and K2 = 2.

From (3), we can always assume that (1 − n1 + j1)(1 − n2 + j2) > 0. The main results in this paper are thefollowing.

Theorem 4. We consider the family of polynomial differential systems (2). Then, the following statements hold:

(a) vk = 0 for any k such that 2k/(n1 + j1 − 1) 6∈ N.(b) Im(BC) = 0 is a necessary condition for system (2) to have a center at the origin.


Theorem 5. We consider the family of polynomial differential systems (2) with Im(BC) = 0. Then, v2k+1 = 0 fork = 2, 3, . . . , m with 2m < (n1 + j1 − 1)(K2 + N2).

We set

LK2,N2A,C = AK2C N2 − (−1)K2+N2 AK2C N2 . (4)

Theorem 6. We consider the family of polynomial differential systems (2) with Im(BC) = 0. Then, there exist realconstants Lk , k = 0, . . . , N2, such that for 2m = (n1 + j1 − 1)(K2 + N2),

v2m+1 = LK2,N2A,C

N2∑k=0

Lk

(B

C

)k

.

Furthermore, if n1 = 0, then

v2m+1 =1

n2CLK2,N2

A,C (n2 B + n3C)

N2∑k=1

Lk

k−1∑j=0

(B

C

)k−1− j (−n3

n2

) j

.

Theorem 7. We consider the family of polynomial differential systems (2) with Im(BC) = 0. If for some k =

0, . . . , N2, the constants Lk in Theorem 5 are different from zero, and LK2,N2A,C = 0, then the origin of (2) is a center.

Theorem 8. We consider the family of polynomial differential systems (2) with Im(BC) = 0 and n1 + j1 − 1 even.Then, the following holds:

(a) if 2m = (n1 + j1 − 1)(K2 + N2 + 1), then v2m+1 = 0.(b) if 2m = (n1 + j1 − 1)(K2 + N2 + 3) and N2 + K2 ≥ 4, then v2m+1 = 0.

We recall that as pointed out before, if n1 + j1 − 1 is odd then K2 + N2 is even, and the conditions 2m =

(n1 + j1 − 1)(K2 + N2 + 1) and 2m = (n1 + j1 − 1)(K2 + N2 + 3) are never satisfied for any m ∈ N and in thesetwo cases v2m+1 = 0.

With the constants Lk introduced in Theorem 6, we introduce the sets

S1 =

{C, B ∈ C \ {0} :

N2∑k=0

Lk

(B

C

)k

= 0

}, if n1 6= 0

and

S1 =

{C, B ∈ C \ {0} :

N2∑k=1

Lk

k−1∑j=0

(B

C

)k−1− j (−n3

n2

) j

= 0

}, if n1 = 0.

Theorem 9. We consider the family of polynomial differential systems (2) with Im(BC) = 0. If there existsk1 ∈ {0, . . . , N2} such that the constants Lk in Theorem 6 satisfy Lk1 6= 0, and B, C ∈ S1, then there exist constantsD1, D2 ∈ C such that

v2m+1 = (D1 AA + D2CC)LK2,N2A,C , for 2m = (n1 + j1 − 1)(K2 + N2 + 2).

Given A ∈ C \ {0} and D1, D2, the constants in Theorem 9, we introduce the set

S2 = {C ∈ C \ {0} : D2CC = −D1 AA}.

Theorem 10. Under the same hypothesis of Theorem 9, if N2 + K2 ≥ 5, D1, D2 6= 0 and C ∈ S2, then there existsa constant L such that for 2m = (n1 + j1 − 1)(N2 + K2 + 4) we have

v2m+1 = L(AA)2LK2,N2A,C .


We note that if in Theorem 10 L 6= 0, then system (2) has a fine focus at the origin of order [(n1 + j1 − 1)(N2 +

K2 + 4)]/2, and this is the maximum order of any focus around the origin for system (2).In the following theorem, we provide infinitely many examples for which the constants Lk in Theorem 6 and the

constants D1, D2 in Theorem 9 are different from zero.The proofs of Theorems 5, 6 and 8–10 are strongly based on the algebraic properties of the Liapunov constants of

the polynomial vector fields studied by Cima, Gasull, Manosa and Manosas in [3].The paper is organized as follows. In Section 2, we illustrate our results with three examples: a linear system

with homogeneous quartic nonlinearities, a linear system with homogeneous cubic nonlinearities, a linear systemwith homogeneous quartic nonlinearities, and finally a family of homogeneous nonlinearities of degree n even. InSection 3, we present some basic results. In Section 4, we prove Theorem 4, and in Section 5 we provide the proof ofTheorem 5. Theorem 6 is proved in Section 6 and Theorems 7 and 8 are proved in Section 7. Finally, in Section 8, weprove Theorem 9, and in Section 9 we prove Theorem 10.

2. Examples

2.1. First example

We consider the following linear systems with quadratic homogeneous nonlinearities:

z = iz + Az2+ Bzz + Cz2, (5)

with A, B, C ∈ C\ {0}. We refer to this known system, since it serves as a test for our results. Computing numericallythe Liapunov constants, after removing the non-zero multiplicative factors, we obtain

v1 = 0,

v2 = BC − BC,

v3 = 0,

v4 = 0,

v5 = (B + 2C)(C3 A − C3 A)(C − 2B),

v6 = 0,

(6)

and if B = C/2, then

v7 = (C3 A − C3 A)(CC − 4AA). (7)

Furthermore, from the results of Dulac, Kapteyn, Bautin, Zoładek, (see [20,25]), we have that

Theorem 11. System (5) is a center if and only if

(1) B + 2C = 0(2) BC − BC = 0, C3 A − C3 A = 0(3) B − 2C = 0 and CC − AA = 0.

In the notation of (2), we have that n1 = 0, j1 = 2, n2 = 1, j2 = 1 and n3 = 2, j3 = 0. Furthermore, N1 = 3,K1 = 1 and thus, M1 = 1, N2 = 3 and K2 = 1.

Then, since n1 + j1 − 1 = 1, Theorems 4–6 imply that if BC = BC = 0, then v1 = v2 = v3 = v4 = 0. Thismatches perfectly well with the computed values of v1, v2, v3 and v4 in (6).

Since system (5) has n1 = 0, by Theorem 6, after simplifying by B/C3 (note that C 6= 0) and sinceL1,2

A,C = C3 A − C3 A, we obtain

v5 = (B + 2C)(C3 A − C3 A)((L1 − 2L2 + 4L3)C2+ (L2 − 2L3)(BC + L3 B2)).

This matches with v4 in (6) taking L1 = −3, L2 = −2 and L3 = 0 or L1 = 14, L2 = −3 and L3 = −2.


Furthermore, by Theorem 8(a), we obtain v6 = 0. In this example S1 = {B, C ∈ C : B = C/2} and hence byTheorem 9 applied with B = C/2, we have that

v7 = (D1 AA + D2CC)(C3 A − C3 A),

which matches with v7 in (7) taking D1 = −4 and D2 = 1.Note that by the expression of v3 in (6), Lemma 14(b) and Theorem 7, we have that system (5) has a center at the

origin if B + 2C = 0 or Im(BC) = 0 and C3 A = C3 A, which matches with the first two statements in Theorem 11.

2.2. Second example

We consider the following linear systems with cubic homogeneous nonlinearities:

z = iz + Az3+ Bzz2

+ Cz3, (8)

with A, B, C ∈ C \ {0}. We refer to this known system, since it serves as a test for our results.Computing numerically the Liapunov constants after removing the non-zero multiplicative factors, we obtain

(see [15] for a more detailed explanation of the numerical method to compute the Liapunov constants)

v1 = 0,

v2 = BC − BC,

v3 = (3C + B)(B2 A + B2 A)(3B − C),

v4 = 0,

(9)

and if B = C/3, after multiplying by 9, we have

v5 = (B2 A + B2 A)(4CC − 9AA). (10)

In the notation of (2), we have that n1 = 0, j1 = 3, n2 = 1, j2 = 2 and n3 = 3, j3 = 0. Furthermore, N1 = 4,K1 = 2 and thus, M1 = 2, N2 = 2 and K2 = 1.

Then, since n1 + j1 − 1 = 2, Theorems 4 and 5 imply that if BC = BC = 0, then v1 = v2 = 0. This matchesperfectly well with the computed values of v1 and v2 in (9). Since system (8) has n1 = 0, by Theorem 6, using thatC 6= 0, L1,2

A,C = C2 A + C2 A, BC = BC , and simplifying by 1/B2, we obtain

v3 = (3C + B)(B2 A + B2 A)(C(L1 − 3L2) + L2 B).

This matches with v3 in (9) taking L1 = 8 and L2 = 3. Furthermore, by Theorem 8(a) we obtain v4 = 0. In thisexample, S1 = {B, C ∈ C : B = C/3}, and hence by Theorem 9 applied with C = 3B, and using that BC = BC andsimplifying by C2/B2, we have that

v5 = (D1 AA + D2CC)(C2 A + C2 A),

which matches with v5 in (10) taking D1 = −9 and D2 = 4.Furthermore, from the results of Lloyd, Pearson, Popa, Sibirskii and Vulpe (see [15,18,25]) we have that

Theorem 12. System (8) is a center if and only if

(1) 3B = −C(2) BC = BC, C2 A = −C2 A(3) B = 3C and 4CC = 9AA.

Note that by the expression of v3 in (9) and Lemma 14(b) and Theorem 7, we have that system (8) has a centerat the origin if 3B + C = 0 or Im(BC) = 0 and C2 A = −C2 A, which matches with the first two statements ofTheorem 12.


2.3. Third example

We consider the following linear systems with quartic homogeneous nonlinearities:

z = iz + (α1 + iα2)z4+ (α3 + iα4)zz3

+ (α5 + iα6)z4, (11)

with α j ∈ R for j = 1, . . . , 6. We refer to this known system, since it also serves as a test for our results.Computing numerically the Liapunov constants, we obtain (see [7] for a more detailed explanation of the

computation of the Liapunov constants in this example)

vk = 0 if k is not divisible by 3,

v3 = −2(α1α4 + α2α3),

v3k = 0 for k = 2, 3.

(12)

Choosing α1 = α3 = 0, we have

v12 = −4

525α5(α

25 − 3α2

6)(4α2 − α4)(α2 + 2α4)(α2 − 2α4)(59α4 + 30α2)(4α4 + α2). (13)

Then, making α2 = 2α4,

v15 = −7

320α5

4α5(α25 − 3α2

6)(52 278α24 − 20 723(α2

5 + α26)). (14)

Now taking α25 + α2

6 = 1 and α4 =

√20 72352 278 , we get

v18 = −189333165503483774277911

381799174273650079066137600

√1083356994α5(α

25 − 3α2

6). (15)

Therefore, if α5(α25 − 3α2

6) 6= 0, system (11) has a fine focus at the origin of order 16.In the notation of (2), we have that A = α5 + iα6, n1 = 0, j1 = 4, B = α3 + iα4, n2 = 1, j2 = 3 and C = α1 + iα2,

n3 = 4, j3 = 0. Then, N1 = 5, K1 = 3 and thus, M1 = 1, N2 = 5 and K2 = 3.Then, since n1 + j1 − 1 = 3, Theorem 4(a) implies that vk = 0 for k not divisible by 3. Furthermore, Theorem 4

together with Theorem 5 imply that if Im(BC) = 0, then v1 = · · · = v11 = 0. This matches perfectly well with thecomputed values of v1, . . . , v11 in (12).

Now, we make the restriction α1 = α3 = 0, that is, B = iα4 6= 0 and C = iα2 6= 0. Then,

n2 B + n3C = −i(4α2 − α4), L3,5A,C = A3C5

− A3C5=

C5

B5 ( A3 B5− A3 B5) = −2iα5

2α5(α25 − 3α2

6). (16)

Since system (11) has n1 = 0, Theorem 6 together with (16) imply

v12 = −2iα5(α25 − 3α2

6)(4α2 − α4)

5∑k=1

(−1)k−1Lk

k−1∑j=0

4 jα5+ j−k2 α

k−1− j4 .

This matches perfectly well with (13) by taking L1, . . . , L5 such that

5∑k=1

(−1)k−1Lk

k−1∑j=0

4 jα5+ j−k2 α

k−1− j4 = −i

2048525

(α2 + 2α4)(α2 − 2α4)(59α4 + 30α2)(4α4 + α2).

Furthermore, by setting α4 = α2/2 (note that in this case B, C ∈ S1) and by Theorem 9, taking into account (16), thatB = iα2/2, AA = α2

5 + α26 and CC = α2

2 , we get that

v15 = 2iα52α5(α

25 − 3α2

6)(D1(α25 + α6)

2+ D2α

22),


which matches with (14) by taking D1 = −145 061

640 i and D2 =365 946

640 i. Then, since D1, D2 6= 0, setting

AA = α25 + α2

6 = 1 and α2 =

√20 72352 278 (note that in this case C ∈ S2), by Theorem 10, we get

v18 = 2i(

20 72352 278

)5/2

Lα5(α25 − 3α2

6),

which matches perfectly well with (15).

2.4. Fourth example

We consider the following linear systems with homogeneous nonlinearities of order n even greater than or equal tofour:

z = iz + Azn−2

2 zn+2

2 + Bzn2 z

n2 + Cz

n+22 z

n−22 , (17)

with n ≥ 4 even and A, B, C ∈ C \ {0}. We refer to this system, since it serves as a test for our results for infinitelymany examples. Computing numerically the Liapunov constants (see [11] for a more detailed explanation of thecomputation of the LiapunovLiapunov constants in this example), and after removing the non-zero multiplicativefactors, we obtain

vk = 0 if 1 ≤ k < 2n − 2,

v2n−2 = BC − BC,

vk = 0, if 2n − 1 ≤ k < 4n − 3

v4n−3 =

(2 +

B

C

)(C3 A − C3 A)

((4 − n) − (n + 2)

B

C

)((n − 2) + n

B

C

).

(18)

If n = 4 and B = −(n − 2)C/n, then

vk = 0, if 4n − 2 ≤ k < 6n − 5

v6n−5 = (AC3− AC3)(5|C |

2− 6|A|

2).(19)

If n ≥ 6 and B = (4 − n)C/(2 + n), then

vk = 0, if 4n − 2 ≤ k < 6n − 5

v6n−5 = (AC3− AC3)

((n2

− 12 − 40)|A|2−

240(n − 1)2

(n + 2)2 |C |2)

.(20)

If d ≥ 6 and B = −(n − 2)C/n, then

vk = 0, if 4n − 2 ≤ k < 6n − 5

v6n−5 = (AC3− AC3)

((n2

− 12 − 40)|A|2−

240(n − 1)2(n − 2)2

(n − 4)2n2 |C |2)

.(21)

In the notation of (2), we have that n1 = (n − 2)/2, j1 = (n + 2)/2, n2 = n/2, j2 = n/2 and n3 = (n + 2)/2,j3 = (n − 2)/2. Furthermore, N1 = 3, K1 = 1 and thus, M1 = 1, N2 = 3 and K2 = 1.

Then, since n1 + j1 − 1 = n − 1, Theorems 4–6 imply that if BC = BC = 0, then vk = 0 for k < 4n − 3. Thismatches perfectly well with the computed values of vk in (18).

Since system (17) has n1 6= 0 and L1,2A,C = C3 A − C3 A, by Theorem 6, we obtain

v4n−3 = (C3 A − C3 A)

(L0 + L1

B

C+ L2

(B

C

)2

+ L3

(B

C

)3)

.

This matches with v4n−3 in (18), taking L0 = (8−2n)(n −2), L1 = 14n −5n2, L2 = 4(1−n2) and L3 = −n(2+n).


Furthermore, by Theorem 4(a) and 8(a), we obtain vk = 0 for 4n − 2 ≤ k < 6n − 5. In this example, if n = 4 then

S1 = {B, C ∈ C : B = −(n − 2)C/n} ∪ {B, C ∈ C : B = −2C}.

By Theorem 9 applied with B = −(n − 2)C/n, we have that

v6n−5 = (D1 AA + D2CC)(C3 A − C3 A), (22)

which matches with (19) taking D1 = −6 and D2 = 5.Finally, if n is even with n ≥ 6, then

S1 = {B, C ∈ C : B = (4 − n)C/(2 + n)} ∪ {B, C ∈ C : B = −(n − 2)C/n} ∪ {B, C ∈ C : B = −2C}.

Theorems 4(a) and 9 applied with B = (4 − n)C/(2 + n) match with (20) taking D1 = n2− 12n − 40 and

D2 = −240(n −1)2/(n +2)2 (see (22)). Furthermore, if we apply Theorem 9 with B = −(n −2)C/n, we have again(22), which matches with (21) taking D1 = (n − 4)2(n2

− 12n − 40) and D2 = −240(n − 1)2(n − 2)2/n2.

3. Preliminary results

Using polar coordinates r2= zz, θ = arctan(Im(z)/Re(z)), in a neighborhood of the origin system, (1) becomes

dr

dθ=

r2 P2(θ) + r3 P3(θ) + · · ·

1 + r Q2(θ) + r2 Q3(θ) + · · ·(23)

where

Pk(θ) = Re(e−iθ Fk(eiθ , e−iθ )), Qk(θ) = Im(e−iθ Fk(eiθ , e−iθ )),

and

F(z, z) =

∑k≥2

Fk(z, z) =

∑k≥2

Fk(eiθ , e−iθ )rk .

Here Fk denotes the homogeneous part of degree k of F in the variables z and z.Consider the solution r(θ, x) of (23) that takes the value x at θ = 0. Then,

r(θ, x) = u1(θ)x + u2(θ)x2+ · · · ,

with u1(0) = 1 and uk(0) = 0 for k ≥ 2. Then h(x) = r(2π, x) is the return map. Assume that system (23) satisfiesuk(2π) = 0 for k = 2, 3, . . . , 2m and u2m+1(2π) 6= 0. Therefore v2m+1 = u2m+1(2π) is the m-th Liapunov constant.

It is known that the Liapunov constant v2m+1 is a polynomial with its variables being the coefficients of Fi ,i = 2, 3, . . . , 2m + 1 and their conjugates. We put F(z, z) =

∑k+l≥2 Akl zk zl , and we will use the notation

v2m+1 = v2m+1(F2, F3, . . . , F2m+1) = v2m+1(Akl , Akl).

The algebraic properties of the Liapunov constant are established in the following result, see [3].

Proposition 13. Let v2m+1 be the m-th Liapunov constant of system (1). Then, it satisfies the following properties:

(a) v2m+1(λ1−k+l Akl , λ

−(1−k+l) Akl) = v2m+1(Akl , Akl).(b) v2m+1(λ

k+l−1 Akl , λk+l−1 Akl) = λ2mv2m+1(Akl , Akl).

Now there is an easy way for listing the monomials that appear in v2m+1. For K ∈ R, let

M = K

(r∏

i=1

Amiki li

)(r+s∏

i=r+1

Amiki li

)(24)

be a monomial of v2m+1. Then, Proposition 13(a) implies

r∑i=1

(1 − ki + li )mi =

r+s∑i=r+1

(1 − ki + li )mi , (25)


and Proposition 13(b) implies

r+s∑i=1

(ki + li − 1)mi = 2m. (26)

These last two equalities will be very useful in order to carry out the effective computation of the Liapunov constantsof system (2).

We say that a system (1) is reversible with respect to a straight line if it is invariant under the change of variablesw = e−iγ z, τ = −t and say that a system (1) is Hamiltonian if it satisfies Re(∂ F/∂z) = 0. For systems (2), we havethe following result.

Lemma 14. The following holds:

(a) System (2) is reversible if and only if A = − Ae−i(1−n1+ j1)γ , B = −Be−i(1−n2+ j2)γ and C = −Ce−i(1−n3+ j3)γ ,for some γ ∈ R.

(b) If n1 6= 0, system (2) is never Hamiltonian. Furthermore, if n1 = 0, then system (2) is Hamiltonian if and only ifn2 B + n3C = 0, and in this situation, the origin of system (2) is a center.

Furthermore, in those two situations, the origin of system (2) is a center.

Proof. The proof of Lemma 14(a) follows directly from its definition, and the proof of Lemma 14(b) also followsdirectly from its definition, taking into account that A 6= 0. �

From (24) with An1 j1 = A, An2 j2 = B and An3 j3 = C , we have that the monomials which appear in v2m+1 are ofthe form

M = K Am1 Bm2Cm3 Am4 Bm5Cm6 ,

with K ∈ R, where m1, m2, m3, m4, m5 and m6 are non-negative integers satisfying (25) and (26); that is,

(1 − n1 + j1)(m1 − m4) + (1 − n2 + j2)(m2 − m5) + (1 − n3 − j3)(m3 − m6) = 0 (27)

and

(n1 + j1 − 1)(m1 + m4) + (n2 + j2 − 1)(m2 + m5) + (n3 + j3 − 1)(m3 + m6) = 2m. (28)

From (3) and (27), we get

σ1 N1(m1 − m4) + σ2 K1(m2 − m5 + m6 − m3) = 0, (29)

where for k = 1, 2,

σk =

{1 if 1 − nk + jk > 0,

−1 if 1 − nk + jk < 0.(30)

Then, since N1 = M1 N2, K1 = M1 K2, after dividing by M1, (29) becomes

σ1 N2(m1 − m4) + σ2 K2(m2 − m5 + m6 − m3) = 0. (31)

Let

s1 = m2 + m6, s2 = m3 + m5. (32)

Then, from (31) we have

σ1 N2(m1 − m4) = σ2 K2(s2 − s1). (33)

Now, since gcd{N2, K2} = 1, from (33), we get that there exists r ∈ Z ∪ {0} such that

m1 = m4 + σ2 K2r and s2 = s1 + σ1 N2r. (34)


4. Proof of Theorem 4

4.1. Proof of Theorem 4(a)

By (32), since n1 + j1 = n2 + j2 = n3 + j3, (28) becomes

(n1 + j1 − 1)(m1 + m4 + s1 + s2) = 2m. (35)

This implies that vm = 0 if 2m is not divisible by n1 + j1 − 1. This finishes the proof of the statement a. �

4.2. Proof of Theorem 4(b)

By Theorem 4(a), we can assume that 2m is divisible by n1 + j1 − 1, i.e., 2m = (n1 + j1 − 1)m, for some apositive integer m. Let m = 2. We compute v2m+1 = v2(n1+ j1)−1, by means of the Poincare section. Recall thatn1 − j1 − 1 = σ1 N1, n2 − j2 − 1 = σ2 K1 and n3 − j3 − 1 = −(n2 − j2 − 1) = −σ2 K1 and thus, if we introduce thenotation

FA(θ) = Re(A) cos(σ1 N1θ) − Im(A) sin(σ1 N1θ),

G B,C (θ) = (Re(B) + Re(C)) cos(σ2 N2θ) − (Im(B) − Im(C)) sin(σ2 N2θ),

HA(θ) = Re(A) sin(σ1 N1θ) + Im(A) cos(σ1 N1θ),

IB,C (θ) = (Re(B) − Re(C)) sin(σ2 N2θ) + (Im(B) + Im(C)) cos(σ2 N2θ),

then system (2) in polar coordinates takes the form

r = rn1+ j1(FA(θ) + G B,C (θ)), θ = 1 + rn1+ j1−1(HA(θ) + IB,C (θ))

and thus,

dr

dθ=

rn1+ j1(FA(θ) + G B,C (θ))

1 + rn1+ j1−1(HA(θ) + IB,C (θ))

and thus, in a neighbourhood of r = 0, dr/dθ can be written as:

dr

dθ= rn1+ j1(FA(θ) + G B,C (θ)) − r2(n1+ j1)−1(FA(θ) + G B,C (θ))(HA(θ) + IB,C (θ)) + O(r3(n1+ j1)−2). (36)

We write, by r(θ, x), the solution of (36) such that at θ = 0 is equal to x , and write it as

r(θ, x) = u1(θ)x + un1+ j1(θ)xn1+ j1 + · · · .

Computing in (36) the terms with xn1+ j1 , we get that

dun1+ j1(θ)

dθ= FA(θ) + G B,C (θ) (37)

and thus since un1+ j1(0) = 0 we get

un1+ j1(θ) =Re(A)

σ1 N1sin(σ1 N1θ) +

Im(A)

σ1 N1cos(σ1 N1θ) +

Re(B) + Re(C)

σ2 N2sin(σ2 N2θ)

+Im(B) − Im(C)

σ2 N2cos(σ2 N2θ) −

Im(A)

σ1 N1−

Im(B) − Im(C)

σ2 N2. (38)

Now, writing h(x) = r(2π, x), we have from (37) that

h(x) := v1x + vn1+ j1 xn1+ j1 + · · · = u1(2π)x + un1+ j1(2π)xn1+ j1 + · · ·

= x + O(xn1+ j1+1).

Now, we compute u2(n1+ j1)−1(θ). From (36) and (38) computing the coefficients of x2(n1+ j1)−1 we obtain that

du2(n1+ j1)−1

dθ= (n1 + j1)un1+ j1(θ)(FA(θ) + G B,C (θ)) − (FA(θ) + G B,C (θ))(HA(θ) + IB,C (θ))


and since u2(n1+ j1)−1(0) = 0, we obtain

v2(n1+ j1)−1 = u2(n1+ j1)−1(2π) =

∫ 2π

0

du(2n1+ j1)−1

dθdθ

= Re(B)Im(C) + Im(B)Re(C) = Im(BC).

Therefore, in order that v2(n1+ j1)−1 = 0, it must hold that Im(BC) = 0, which yields that Im(BC) = 0 is a necessarycondition in order that system (2) has a center at the origin. This finishes the proof of the theorem. �


By Theorem 4(a), we can assume that 2m is divisible by n1 + j1 − 1, i.e., 2m = (n1 + j1 − 1)m, for some positiveinteger m. Thus, (35), together with (34), imply that

2m4 + 2s1 + (σ2 K2 + σ1 N2)r = m. (39)

We consider two different cases.

Case 1: 1 + j1 > n1 and 1 + j2 > n2. In this case, if m < K2 + N2, then the unique solution of (34), and (39) withσ1 = σ2 = 1, is r = 0. Indeed, since m < K2 + N2 and m4, s1 are positive integers, from (39) with σ1 = σ2 = 1, weget that r ≤ 0. Similarly, from (34) and (39) with σ1 = σ2 = 1, we obtain that 2m1+2s2−(K2+N2)r = m < K2+N2.Now using that m1 and s2 are positive integers, we get that r ≥ 0. Thus, r = 0, and from (34) and (39), we have thatm1 = m4, s2 = s1 and 2(m4 + s1) = m. Thus, m must be divisible by 2, and then v2m+1 = 0 if m = (n1 + j1 − 1)mwith m < K2 + N2 odd.

Let m = 2m with m a positive integer. Then, the expression of v2m+1 with m = (n1 + j1 − 1)m, becomes

v2m+1 =

m∑m4=0

s1∑m2=0

s1∑m5=0

βm4,s1,m2,m5(AA)m4 Bm2 Bm5Cs1−m5Cs1−m2 (40)

with βm4,s1,m2,m5 real constants and s1 = m − m4.Let Γ1 = ei(1−n1+ j1)γ and Γ2 = ei(1−n2+ j2)γ . From Lemma 14(a), if A = −AΓ1, B = −BΓ2, and C = −CΓ−1

2 ;then, since for this system we have v2m+1 = 0, we conclude that

0 = (−1)mm∑

m4=0

A2m4Γm41 Γ s1

2

s1∑m2=0

s1∑m5=0

(−1)m5−m2βm4,s1,m2,m5 Bm2+m5C2s1−(m2+m5)Γm2+m52 . (41)

Now, we note that (41) is valid for any value of the complex numbers A, B and C . Thus, computing the differentpowers of A, we have that for every m4 = 0, . . . , m,

s1∑m2=0

s1∑m5=0

(−1)m5−m2βm4,s1,m2,m5(Γ2 B)m2+m5C2s1−(m2+m5) = 0. (42)

Let r = m2 + m5, then (42) becomes(s1∑

r=0

r∑m2=0

+

2s1∑r=s1+1

s1∑m2=r−s1

)(−1)rβm4,s1,m2,r−m2 Br C2s1−rΓ r

2 = 0, (43)

where we have used the notation(s1∑

r=0

r∑m2=0

+

2s1∑r=s1+1

s1∑m2=r−s1

)ar,m2 =

s1∑r=0

r∑m2=0

ar,m2 +

2s1∑r=s1+1

s1∑m2=r−s1

ar,m2 .

Note that Br C2s1−r has degree 2s1. Then, computing the different degrees 2s1 of (43) and for each degree 2s1,computing the different powers of B and C we get that for r = 0, . . . , s1,


r∑m2=0

βm4,s1,m2,r−m2 = 0 ⇒ βm4,s1,0,r = −

r∑m2=1

βm4,s1,m2,r−m2 (44)

and for r = s1 + 1, . . . , 2s1, we get that

s1∑m2=r−s1

βm4,s1,m2,r−m2 = 0 ⇒ βm4,s1,r−s1,s1 = −

s1∑m2=r−s1+1

βm4,s1,m2,r−m2 . (45)

Now, we write v2m+1 in (40) as

v2m+1 =

m∑m4=0

(AA)m4

(s1∑

r=0

r∑m2=0

+

2s1∑r=s1+1

s1∑m2=s1−r

)βm4,s1,m2,r−m2 Bm2 Br−m2Cs1−r+m2Cs1−m2 . (46)

Then, (46) together with (44) and (45), imply

v2m+1 =

m∑m4=0

(AA)m4

s1∑r=0

r∑m2=1

βm4,s1,m2,r−m2 [−Br Cs1−r Cs1 + Bm2 Br−m2Cs1−r+m2Cs1−m2 ]

+

m∑m4=0

(AA)m4

2s1∑r=s1+1

s1∑m2=s1−r

βm4,s1,m2,r−m2 [−Br−s1 Bs1C2s1−r+ Bm2 Br−m2Cs1−r+m2Cs1−m2 ].

(47)

Clearly, we can rewrite v2m+1 in (47) as

v2m+1 =

m∑m4=0

(AA)m4

s1∑r=0

r∑m2=1

βm4,s1,m2,r−m2 Br−m2Cs1−r Cs1−m2 [Bm2Cm2 − Bm2Cm2 ]

+

m∑m4=0

(AA)m4

2s1∑r=s1+1

s1∑m2=s1−r

βm4,s1,m2,r−m2

× Br−s1 Br−m2Cs1−m2 [Bs1−r+m2Cs1−r+m2 − Bs1−r+m2Cs1−r+m2 ]. (48)

Since by hypothesis Im(BC) = 0, i.e., BC = BC from (48), we conclude that v2m+1 = 0 for any m = 2(n1+ j1−1)mwith 2m < N2 + K2. This finishes the proof of the theorem.

Case 2: 1 + j1 < n1 and 1 + j2 < n2. In this case, 1 − n1 + j1 = −N1 and 1 − n2 + j2 = −K1. The same argumentsare used as in Case 1, but working with (34) and (39) with σ1 = −1, σ2 = −1, give the proof of the theorem in thiscase. �


Let 2m = (n1 + j1 − 1)(K2 + N2). We consider two different cases.

Case 1: 1 + j1 > n1 and 1 + j2 > n2. In this case, we first note that 1 − n1 + j1 = N1, 1 − n2 + j2 = K1, and ifm = K2 + N2, then the unique solution of (34), and (39) with σ1 = σ2 = 1, is r ∈ {−1, 0, 1}. Indeed, from (39),we get that 2m4 + 2s1 + (K2 + N2)r = K2 + N2. Thus, since m4 and s1 are positive integers, we get that r ≤ 1.Furthermore, (34) and (39) imply that 2m1 + 2s2 − (K2 + N2)r = K2 + N2. Thus, since m1 and s2 are positiveintegers, we obtain that r ≥ −1. Thus, r ∈ {−1, 0, 1}.

If r = 1, then (39) implies that 2m4 + 2s1 = 0, which yields m4 = s1 = 0. Therefore, (34) with σ1 = σ2 = 1 and(32) yield m2 = m4 = m6 = 0 and m1 = K2, s2 = m3 + m5 = N2.

If r = 0, proceeding as in the proof of Theorem 4, we get that m1 = m4, s2 = s1 and 2m4 + 2s1 = N2 + K2. Thus,we have two possibilities: if N2+K2 is odd, then m1 = m4 = s1 = s2 = 0, i.e m1 = m2 = m3 = m4 = m5 = m6 = 0,and if N2 + K2 is even, then m4 + s1 = (N2 + K2)/2.

Finally, if r = −1, then (39) implies that 2m1 + 2s2 = 0 which yields m1 = s2 = 0. Therefore, (34) withσ1 = σ2 = 1 and (32) yield m1 = m3 = m5 = 0 and m4 = K2, s1 = m2 + m6 = N2.


We now consider two different subcases.Subcase 1.1: N2 + K2 is odd. In this case, as pointed out before, only the terms with r = 1 and r = −1 appear inv2m+1, that is,

v2m+1 =

N2∑k=0

[AK2αk BkC N2−k+ AK2γk BkC N2−k

], (49)

where αk and γk are real constants. From Lemma 14(a), if A = −AΓ1, B = −BΓ2 and C = −CΓ−12 , then since for

this system we have v2m+1 = 0, we conclude that

0 = AK2

N2∑k=0

[αk(−1)kΓ k2 + γk(−1)K2+N2−kΓ K2

1 Γ−N2+k2 ]BkC N2−k .

Now, using that Γ K21 Γ−N2

2 = 1, we obtain

0 = AK2

N2∑k=0

[αk + (−1)K2+N2γk](−1)kΓ k2 BkC N2−k . (50)

Since this equation must hold for every A, B, C ∈ C, computing the different powers in the variables A, B and C , weobtain

αk + (−1)K2+N2γk = 0 for k = 0, . . . , N2

that is,

αk = (−1)K2+N2+1γk for k = 0, . . . , N2. (51)

Thus, from (49) and (51) we get

v2m+1 =

N2∑k=0

[AK2(−1)K2+N2+1γk BkC N2−k+ AK2γk BkC N2−k

]

=

N2∑k=0

γk[ AK2 BkC N2−k− (−1)K2+N2 AK2 BkC N2−k

].

Now, since Im(BC) = 0, i.e., BC = BC with the notation introduced in (4), we get that

v2m+1 = LK2,N2A,C

N2∑k=0

γk

(B

C

)k

, (52)

which finishes the proof of the first statement of the theorem in this case.Now, assume n1 = 0. By Lemma 14(b) with n1 = 0, if n2 B + n3C = 0, we have that v2m+1 = 0 and thus, from

(52),

LK2,N2A,C

N2∑k=0

γk

(−n3

n2

)k

= 0, that is γ0 = −

N2∑k=1

γk

(−n3

n2

)k

. (53)

Therefore, inserting (53) into (52), we obtain

v2m+1 = LK2,N2A,C

[γ0 +

N2∑k=1

γk

(B

C

)k]

= LK2,N2A,C

N2∑k=1

γk

[(B

C

)k

−

(−n3

n2

)k]

=1

n2CLK2,N2

A,C (n2 B + n3C)

N2∑k=1

γk

k−1∑j=0

(B

C

)k−1− j (−n3

n2

) j

.

This finishes the proof of the theorem in this case.


Subcase 1.2: N2 + K2 is even. In this case, since K2 and N2 are coprime, it must hold that K2 and N2 are both odd.Furthermore, as pointed out before, the terms with r = 1, r = 0 and r = −1 appear in v2m+1, that is,

v2m+1 =

N2∑m=0


]

+

m∑m4=0

s1∑m2=0

s1∑m5=0

βm4,s1,m2,m5(AA)m4 Bm2 Bm5Cs1−m5Cs1−m2 (54)

where m = (N2 + K2)/2, s1 = m − m4 and αk , γk , βm4,s1,m2,m5 are real constants.From Lemma 14(a), if A = −AΓ1, B = −BΓ2, and C = −CΓ−1

2 , then since for this system we have v2m+1 = 0,from (54), we conclude that

0 = AK2

N2∑m=0

[αk + (−1)K2+N2γk](−1)kΓ k2 BkC N2−k

+ (−1)mm∑

m4=0

A2m4Γm41 Γ s1

2

s1∑m2=0

s1∑m5=0

× (−1)m5−m2βm4,s1,m2,m5 Bm2+m5C2s1−(m2+m5)Γm2+m52 . (55)

Now, we note that (55) must hold for every value of A, B, C ∈ C; then computing the even powers of A and since K2is odd, we get that

0 = (−1)mm∑

m4=0

A2m4Γm41 Γ s1

2

s1∑m2=0

s1∑m5=0

(−1)m5−m2βm4,s1,m2,m5 Bm2+m5C2s1−(m2+m5)Γm2+m52 .

Now, since this is the same equation as (41), proceeding as in the proof of Theorem 5 using that Im(BC) = 0, we getthat βm4,s1,m2,m5 = 0 for m4 = 0, . . . , m, s1 = m − m4, m2 = 0, . . . , s1 and m5 = 0, . . . , s1. Thus, (55) becomesexactly the same as (50), and proceeding as in Subcase 1.1.1, the theorem holds in this case.

Case 2: 1 + j1 < n1 and 1 + j2 < n2. In this case, 1 − n1 + j1 = −N1 and 1 − n2 + j2 = −K1. The same argumentsused in Case 1, but working with (34) and (39) with σ1 = −1 and σ2 = −1 (see (30)), imply the proof of the theoremin this case. �

7. Proof of Theorems 7 and 8

7.1. Proof of Theorem 7

From Lemma 14(a), we know that system (2) has a reversible center at the origin if and only if for some α, wehave A = − Ae−i(1−n1+ j1)α , B = −Be−i(1−n2+ j2)α and C = −Cei(1−n2+ j2)α . To prove Theorem 7, we just need toshow that AK2C N2 = (−1)K2+N2 AK2C N2 and Im(BC) = 0 is a sufficient condition for the center variety. So, sinceBC = BC , we have that(

− A

A

)K2

=

(−C

C

)N2

=

(−B

B

)N2

. (56)

Now let θ1 and θ2 be such that eiθ1 = − A/A and eiθ2 = −B/B. Then, using (56) together with the fact thatN2 = N1/M1 and K2 = K1/M1, we obtain

K1θ1 = N1θ2 (mod 2π). (57)

Now, using that N1 = |1 − n1 + j1|, K1 = |1 − n2 + j2| and that sign (1 − n1 + j1)(1 − n2 + j2) > 0, we get that(57) reads as

(1 − n1 + j1)θ1 = (1 − n2 + j2)θ2 (mod 2π). (58)

Set α = θ1/(1 − n1 + j1). Then, using (58), we get


ei(1−n1+ j1)α = eiθ1 = −A

Aand ei(1−n2+ j2)α = −B/B. (59)

Consequently, the theorem follows from (59) and the fact that Im(BC) = 0. �

7.2. Proof of Theorem 8(a)

By (39) with m = N2 + K2 + 1, we have that

N2 + K2 + 1 =

2m4 + 2s1 + (N2 + K2)r, if σ1 = σ2 = 12m1 + 2s2 + (N2 + K2)r, if σ1 = σ2 = −12m1 + 2s1 + (N2 + K2)r, if σ1 = −1, σ2 = 12m4 + 2s2 + (N2 + K2)r, if σ1 = 1, σ2 = −1.

(60)

We consider two different cases.

Case 1: N2 + K2 is even. In this case, the left-hand side of (60) is odd while the right-hand side is even. Thus, thiscase is not possible, and v2m+1 with 2m = (n1 + j1 − 1)(N2 + K2 + 1) is zero.

Case 1: N2 + K2 is odd. In this case, since K2 + N2 ≥ 3, the unique solution of (60) is given by r = 0.Then, proceeding exactly in the same way as we did in the proof of Theorem 5, we obtain that v2m+1 = 0 for2m = (n1 + j1 − 1)(N2 + K2 + 1). �

7.3. Proof of Theorem 8(b)

Let 2m = (n1 + j1 − 1)(K2 + N2 + 3). By (39) with m = N2 + K2 + 3, we obtain (60) with the right-hand sidereplaced by N2 + K2 + 3. Then, using that N2 + K2 > 3, and proceeding exactly in the same way as in the proof ofTheorem 8(a), we obtain that v2m+1 = 0 for 2m = (n1 + j1 − 1)(N2 + K2 + 3). �


Let 2m = (n1 + j1 − 1)(K2 + N2 + 2). We will only consider the case 1 + j1 > n1 and 1 + j2 > n2, since theproof of the case 1 + j1 < n1 and 1 + j2 < n − 2 can be carried out in the same way working with (34) and (39)with σ1 = −1 and σ2 = −1. When 1 + j1 > n1 and 1 + j2 > n2, 1 − n1 + j1 = N1, 1 − n2 + j2 = K1 and ifm = K2 + N2, then the unique solution of (34) and (39) with σ1 = σ2 = 1, is r ∈ {−1, 0, 1}. Indeed, from (39), weget that 2m4 + 2s1 + (K2 + N2)r = K2 + N2 + 2. Thus, since m4 and s1 are positive integers and N2 + K2 ≥ 3, weget that r ≤ 1. Furthermore, (34) and (39) imply that 2m1 + 2s2 − (K2 + N2)r = K2 + N2. Thus, since m1 and s2 arepositive integers and N2 + K2 ≥ 3, we obtain that r ≥ −1. Thus, r ∈ {−1, 0, 1}.

If r = 1, then (39) implies that 2m4 + 2s1 = 2, which yields m4 + s1 = 1. Therefore, (34) with σ1 = σ2 = 1 and(32) yield m4 = 1, m2 = m6 = 0 and m1 = K2 + 1, s2 = m3 + m5 = N2 or m4 = 0, m2 + m6 = 1 and m1 = K2,s2 = m3 + m5 = N2 + 1.

If r = 0, proceeding as in the proof of Theorem 4, we get that m1 = m4, s2 = s1 and 2m4 + 2s1 = N2 + K2 + 2.Thus, we have two possibilities: if N2 + K2 is odd, then m1 = m4 = s1 = s2 = 0; that is, m1 = m2 = m3 = m4 =

m5 = m6 = 0, and if N2 + K2 is even, then m4 + s1 = (N2 + K2 + 2)/2.Finally, if r = −1, then (39) implies that 2m1 + 2s2 = 2, which yields m1 + s2 = 1. Therefore, (34) with

σ1 = σ2 = 1 and (32) yield m1 = 1, m3 = m5 = 0 and m4 = K2 + 1, s1 = m2 + m6 = N2 or m1 = 0, m3 + m5 = 1and m4 = K2, s1 = m2 + m6 = N2 + 1.

We will only prove the case N2 + K2 odd, since as in the proof of Theorem 6, the case N2 + K2 even can bereduced to this one. When N2 + K2 is odd, only the terms with r = 1 and r = −1 appear in v2m+1; that is, if we writev2m+1 = v2m+1 + v2m+1, then

v2m+1 = AAN2∑

k=0


]


where αk and γk are real constants and

v2m+1 =

N2+1∑k=0

[AK2βk BkCC N2+1−k] +

N2+1∑k=0

[AK2δk B BkC N2+1−k]

+

N2+1∑k=0

[ AK2µk BkCC N2+1−k] +

N2+1∑k=0

[ AK2νk B BkC N2+1−k],

where βk , δk , µk and νk are real constants.

First, note that the degree of AA in v2m+1 is K2, while the degree of AA in v2m+1 is K2 + 2. Hence, we can treatv2m+1 and v2m+1 separately. Proceeding for v2m+1 exactly in the same way as in the proof of Theorem 6, we obtainthat

v2m+1 = (AA)LK2,N2A,C

N2∑k=0

γk

(B

C

)k

. (61)

Now we rewrite v2m+1 as

v2m+1 = AK2β0CC N2+1+ AK2µ0CC N2+1

+ AK2δN2+1 B B N2+1+ AK2νN2+1 B B N2+1

+

N2∑k=0

[AK2βk+1 Bk+1CC N2−k] +

N2∑k=0

[ AK2µk+1 Bk+1CC N2−k]

+

N2∑k=0

[AK2δk B BkC N2+1−k] +

N2∑k=0

[ AK2νk B BkC N2+1−k], (62)

where in the fifth and sixth sums, we have introduced the change k − 1 → k. From Lemma 14(a), if A = −AΓ1,B = −BΓ2 and C = −CΓ−1

2 , then v2m+1 = 0 and thus, also taking into account that Γ K21 Γ−N2

2 = 1, we conclude,after dividing by AK2 , that

0 = −C N2+2Γ−12 [β0 + µ0(−1)K2+N2 ] + B N2+2(−1)N2+1Γ N2+1

2 [δN2+1 + (−1)K2+N2νN2+1]

+

N2∑k=0

Bk+1C N2+1−k(−1)kΓ k2 [βk+1 + δk + (−1)K2+N2(µk+1 + νk)]. (63)

Clearly, from (63), by computing the different powers in the variables B and C , we obtain

β0 = (−1)K2+N2+1µ0, δN2+1 = (−1)K2+N2+1νN2+1,

βk+1 = −[δk + (−1)K2+N2(νk + µk+1)], for k = 0, . . . , N2.(64)

Thus, from (62) and (64), we get

v2m+1 = µ0CC[ AK2C N2 − (−1)K2+N2 AK2C N2 ] + νN2+1 B B[ AK2 B N2 − (−1)K2+N2 AK2 B N2 ]

+

N2∑k=0

δk AK2 BkC N2−k[BC − BC] +

N2∑k=0

[µk+1 Ak2 Bk+1CC N2−k]

+

N2∑k=0

[(−1)K2+N2+1µk+1 AK2 Bk+1CC N2−k]

+

N2∑k=0

[νk Ak2 B BkC N2+1−k] +

N2∑k=0

[(−1)K2+N2+1νk AK2 Bk+1CC N2−k].


Now, since Im(BC) = 0, i.e., BC = BC , we get that

v2m+1 = LK2,N2A,C

[µ0CC + νN2+1 B B

(B

C

)N2

+ CCN2∑

k=0

(µk+1 + νk)

(B

C

)k+1]

. (65)

By hypothesis, B, C ∈ S1, and thus, there exists E ∈ C such that B = EC . Then, by (61) and (65),we have

v2m+1 = (D1 AA + D2CC)LK2,N2A,C

where

D1 =

N2∑k=0

γk Ek, D2 = µ0 + νN2+1 E N2+1 E +

N2∑k=0

(µk+1 + νk)Ek+1.

This concludes the proof of the theorem. �


Let 2m = (n1 + j1 − 1)(K2 + N2 + 4). We will only prove the case 1 + j1 > n1 and 1 + j2 > n2, since theother can be proved analogously. Then, proceeding exactly in the same was as in the proof of Theorem 9 taking intoaccount that by hypotheses K2 + N2 > 4, we get that the solution of (34) and (39) is r ∈ {−1, 0, 1}.

If r = 1, then (39) implies that 2m4 + 2s1 = 4, which yields m4 + s1 = 2. Therefore, (34) with σ1 = σ2 = 1 and(32) yield m4 = 2, m2 = m6 = 0 and m1 = K2 +2, s2 = m3 +m5 = N2 or m4 = 1, m2 = m6 = 1 and m1 = K2 +1,s2 = m3 + m5 = N2 + 1 or m4 = 0, m2 + m6 = 2 and m1 = K2, s2 = m3 + m5 = N2 + 2.

If r = 0, proceeding as in the proof of Theorem 4, we get that m1 = m4, s2 = s1 and 2m4 + 2s1 = N2 + K2 + 4.Thus, we have two possibilities: if N2 + K2 is odd, then m1 = m4 = s1 = s2 = 0; that is, m1 = m2 = m3 = m4 =

m5 = m6 = 0, and if N2 + K2 is even, then m4 + s1 = (N2 + K2 + 4)/2.Finally, if r = −1 then (39) implies that 2m1 + 2s2 = 4, which yields m1 + s2 = 2. Therefore, (34) with

σ1 = σ2 = 1 and (32) yield m1 = 2, m3 = m5 = 0 and m4 = K2 + 2, s1 = m2 + m6 = N2 or m1 = 1, m3 = m5 = 1and m4 = K2 + 1, s1 = m2 + m6 = N2 + 1 or m1 = 0, m3 + m5 = 2 and m4 = K2, s1 = m2 + m6 = N2 + 2.

We will only prove the case N2 + K2 odd, since the proof of the case N2 + K2 even can be reduced to this one.In this case, as pointed out before, only the terms with r = 1 and r = −1 appear in v2m+1; that is, if we writev2m+1 = v2m+1 + v2m+1, then

v2m+1 = (AA)2N2∑

k=0


]

+ (AA)

N2+1∑k=0

[AK2βk BkCC N2+1−k] + (AA)

N2+1∑k=0

[AK2δk B BkC N2+1−k]

+ (AA)

N2+1∑k=0

[ AK2µk BkCC N2+1−k] + (AA)

N2+1∑k=0

[ AK2νk B BkC N2+1−k], (66)

where αk , βk , γk , δk , µk and νk are real constants, and

v2m+1 =

N2+2∑k=0

[AK2τk BkC2C N2+2−k] +

N2+2∑k=0

[AK2θk BC BkC N2+2−k]

+

N2+2∑k=0

[AK2λk B2 BkC N2+2−k] +

N2+2∑k=0

[ AK2ζk B2 BkC N2+2−k]

+

N2+2∑k=0

[ AK2ρk BC BkC N2+2−k] +

N2+2∑k=0

[ AK2ωkC2 BkC N2+2−k], (67)

where τk , θk , λk , ζk , ρk and ωk are real constants.


First, we note that the degree of the variable AA in v2m+1 given in (67) is K2, while in v2m+1 given in (66) isgreater than K2. Then, we can treat separately v2m+1 that v2m+1. Proceeding for v2m+1 as in the proof of Theorem 9,we obtain that there exist two constants D3 and D4 such that

v2m+1 = AA(D3 AA + D4CC)LK2,N2A,C . (68)

We rewrite v2m+1 in (67) as

v2m+1 = AK2τ0C2C N2+2+ AK2ω0C2C N2+2

+ AK2λN2+2 B2 B N2+2

+ AK2ζN2+2 B2 B N2+2+ AK2τ1 BC2C N2+1

+ AK2ω1C2 BC N2+1

+ AK2θ0 BCC N2+2+ AK2ρ0 BCC N2+2

+ AK2λN2+1 B2 B N2+1C

+ AK2ζN2+1 B2 B N2+1C + AK2θN2+2 BC B N2+2+ AK2ρN2+2 BC B N2+2

+

N2+1∑k=1

[AK2τk+1 Bk+1C2C N2+1−k] +

N2+1∑k=1

[AK2θk BC BkC N2+2−k]

+

N2+1∑k=1

[AK2λk−1 B2 Bk−1C N2+3−k] +

N2+1∑k=1

[ AK2ζk−1 B2 Bk−1C N2+3−k]

+

N2+1∑k=1

[ AK2ρk BC BkC N2+2−k] +

N2+1∑k=1

[ AK2ωk+1C2 Bk+1C N2+1−k].

From Lemma 14(a), if A = −AΓ1, B = −BΓ2 and C = −CΓ−12 , v2m+1 = 0, and also taking into account that

Γ K21 Γ−N2

2 = 1, after dividing by AK2 , we conclude that

0 = C N2+1Γ−22 [τ0 + (−1)K2+N2ω0] + B N2+2Γ N2+2

2 (−1)N2 [λN2+2 + (−1)K2+N2ζN2+2]

− BC N2+3Γ−12 [τ1 + θ0 + (−1)K2+N2(ω1 + ρ0)]

+ C B N2+3Γ N2+12 (−1)N2+1

[λN2+1 + θN2+2 + (−1)K2+N2(ζN2+1 + ρN2+2)]

+

N2+1∑k=1

Bk+1C N2+3−k(−1)k+1Γ k−12 (τk+1 + θk + λk−1 + (−1)K2+N2(ζk−1 + ρk + ωk+1)].

Now computing the different powers of B and C , we get

τ0 = (−1)K2+N2+1ω0, λN2+2 = (−1)K2+N2+1ζN2+2

θ0 = −[τ1 + (−1)K2+N2(ω1 + ρ0)]

θN2+2 = −[λN2+1 + (−1)K2+N2(ζN2+1 + ρN2+2)]

θk = −[τk+1 + λk−1 + (−1)K2+N2(ζk−1 + ρk + ωk+1)],

(69)

for k = 1, . . . , N2 + 1. Therefore, using that BC = BC , we have

AK2 B N2 − (−1)K2+N2 AK2 B N2 =

(B

C

)N2

[ AK2C N2 − (−1)K2+N2 AK2C N2 ].

We obtain from (69) that v2m+1 = LK2,N2A,C V2m+1 where

V2m+1 = CC

[CCω0 + C B

(ω1 + ρ0 +

N2+1∑k=1

ρk

(B

C

)k)]

+ B B

(B

C

)N2 [B BζN2+1 + BC(ζN2+1 + ρN2+1)

]+ (BC)2

N2+1∑k=1

(ζk−1 + ωk+1)

(B

C

)k−1

. (70)


By hypothesis B, C ∈ S1 and C ∈ S2. Thus, there exist E ∈ C and E1 ∈ R such that B = EC and CC = E1 AA.Hence, B B = E E E1 AA and BC = E E1 AA. Then, from (68) and (70), we obtain that

v2m+1 = v2m+1 + v2m+1 = (AA)2LK2,N2A,C V2m+1

with

V2m+1 = D3 + E1

[D4 + E1ω0 + E E1

(ω1 + ρ0 +

N2+1∑k=1

ρk Ek

)]

+ E N2+1 E E1[E E E1ζN2+1 + E E1(ζN2+1 + ρN2+1)

]+ E2 E2

1

N2+1∑k=1

(ζk−1 + ωk+1)Ek−1.

Thus, the theorem is proved with L = V2m+1. �

Acknowledgments

The author wishes to thank the referee for his comments on a previous version of the paper. The author has beensupported by the Center for Mathematical Analysis, Geometry, and Dynamical Systems, through FCT by ProgramPOCTI/FEDER and the grant SFRH/BPD 26464/2006.

References

[1] T.R. Blows, N.G. Lloyd, The number of small-amplitude limit cycles of Lienard equations, Math. Proc. Cambridge Philos. Soc. 95 (1984)751–758.

[2] J. Chavarriga, Integrable systems in the plane with a centre linear part, Appl. Math. 22 (1994) 285–309.[3] A. Cima, A. Gasull, V. Manosa, F. Manosas, Algebraic properties of the Liapunov and period constants, Rocky Mountain J. Math. 27 (1997)

471–501.[4] A. Cima, J. Llibre, J.C. Medrado, New family of centers for polynomial vector fields of arbitrary degree, 2005, preprint.[5] J. Franoise, R. Pons, Une approche algorithmique du probleme du centre pour des perturbations homogenes, Bull. Sci. Math. 120 (1996)

1–17.[6] A. Gasull, A. Guillamon, V. Manosa, An analytical-numerical method of computation of the Liapunov and period constants derived from their

algebraic structure, Universitat Autonoma de Barcelona, Preprint No 6.[7] J. Huang, F. Wang, L. Wang, J. Yang, A quartic system and a quintic system with fine focus or order 18, 2006, preprint.[8] V. Kertesz, R. Kooij, Degenerate Hopf bifurcation in two dimensions, Nonlinear Anal. 17 (1991) 267–283.[9] J. Llibre, C. Valls, Centers for polynomial vector fields of arbitrary degree, 2006, preprint.

[10] J. Llibre, C. Valls, 6-parameter families of centers for polynomial vector fields of arbitrary degree, 2006, preprint.[11] J. Llibre, C. Valls, Classification of the centers for the generalized quadratic systems, 2007, preprint.[12] N. Lloyd, J. Pearson, Computing centre conditions for certain cubic systems, J. Comput. Appl. Math. 40 (1992) 323–336.[13] N. Lloyd, J. Pearson, V. Romanovsky, Computing integrability conditions for a cubic differential system, Comput. Math. Appl. 32 (1996)

99–107.[14] N. Lloyd, J. Pearson, Five limit cycles for a simple cubic system, Publ. Math. 41 (1997) 199–208.[15] N.G. Lloyd, J.M. Pearson, Bifurcation of limit cycles and integrability of planar dynamical systems in complex form, J. Phys. A: Math. Gen.

32 (1999) 1973–1984.[16] N.G. Lloyd, C.J. Christopher, J. Devlin, J.M. Pearson, N. Yasmin, Quadratic-like cubic systems. Planar nonlinear dynamical systems (Delft,

1995), Differential Equations Dynam. Systems 5 (1997) 329–345.[17] H. Poincare, Memoire sur les courbes definies par les equations differentielles, in: Oeuvreus de Henri Poincare, vol. I, Gauthiers-Villars, Paris,

1951, pp. 95–114.[18] M.N. Popa, K.S. Sibirskii, Focal cyclicity of critical points of a differential system, J. Differential Equations 22 (1986) 1539–1545.[19] V. Sheplev, N. Bryxina, M.G. Slin’ko, The algorithm of calculation of Liapunov’s coefficients by analysis of chemical self-sustained

oscillations, Dokl. Akad. Nauk 359 (1998) 789–792.[20] D. Schlomiuk, Algebraic particular integrals, integrability and the problem of the center, Trans. Amer. Math. Soc. 338 (1993) 799–841.[21] K.S. Sibirskii, On the number of limit cycles arising from a singular point of focus or center type, Dokl. Akad. Nauk SSSR 161 (1965)

304–307 (in Russian); Soviet Math. Dokl. 6 (1965) 428–431.[22] C. Valls, New arbitrary-parameter families of centers for polynomial vector fields of arbitrary degree, 2006, preprint.[23] C. Valls, Center manifold variety for 8-parameter families of polynomial vector fields of arbitrary degree, 2006, preprint.[24] Y. Wang, J. Merino, Self-organisational origin of agates: banding, fibre twisting, composition and dynamic crystallization model, Geochim.

Cosmochim. Acta 54 (1990) 1627–1638.[25] H. Zoładek, Quadratic systems with center and their perturbations, J. Differential Equations 109 (1994) 223–273.[26] C. Zuppa, Order of cyclicity of the singular point of Lienard’s polynomial vector field, Bol. Soc. Bras. Mat. 12 (1981) 105–111.

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