This is a preprint of: “Center problem for Λ–Ω differential systems”, Jaume Llibre, Rafael Orlando Ramı́rez, Valentı́n Ramı́rez, J. Differential Equations, vol. 267, 6409–6446, 2019. CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS JAUME LLIBRE1 , RAFAEL RAMÍREZ2 AND VALENTÍN RAMÍREZ1 Abstract. The Λ-Ω systems are the real planar polynomial differential equa- tions of degree m ẋ = −y(1 + Λ) + xΩ, ẏ = x(1 + Λ) + yΩ, where Λ = Λ(x, y) and Ω = Ω(x, y) are polynomials of degree at most m − 1 such that Λ(0, 0) = Ω(0, 0) = 0. We study the center problem for these Λ-Ω systems. Any planar vector fields with linear type center can be written as an Λ-Ω systems if and only if the Poincaré-Liapunov first integral is of the form 1 F = (x2 + y 2 )(1 + O(x, y)). These kind of linear type centers are called weak 2 centers, they contain the class of center studied by Alwash and Lloyd [1], and also contain the uniform isochronous centers, and the holomorphic isochronous centers, but they do not coincide with the all class of isochronous centers. The main objective of this paper is to study the center problem for two particular classes of Λ-Ω systems of degree m. First if Λ = µ(a2 x − a1 y), and Ω = a1 x + a2 y + Ωm−1 , where µ, a1 , a2 are constants and Ωm−1 = Ωm−1 (x, y) is a homogenous polynomial of degree m − 1, then we prove the following results. (i) These Λ-Ω systems have a weak center at the origin if and only if (µ + ∫ 2π m − 2)(a21 + a22 ) = 0, and Ωm−1 (cos t, sin t)dt = 0; 0 (ii) If m = 2, 3, 4, 5, 6 and (µ + m − 2)(a21 + a22 ) ̸= 0, then the given Λ–Ω systems have a weak center at the origin if and only if these systems after a linear change of variables (x, y) −→ (X, Y ) are invariant under the transformations (X, Y, t) −→ (−X, Y, −t). . Second if Λ = a1 x + a2 y, and Ω = Ωm−1 , where a1 , a2 are constants and Ωm−1 = Ωm−1 (x, y) is a homogenous polynomial of degree m − 1, then we prove the following results. (i) These Λ-Ω systems have a weak center at the origin if and only if a1 = ∫ 2π a2 = 0, and Ωm−1 (cos t, sin t)dt = 0; 0 (ii) If m = 2, 3, 4, 5 and a21 + a22 ̸= 0, then the given Λ–Ω systems have a weak center at the origin if and only if these systems after a linear change of variables (x, y) −→ (X, Y ) are invariant under the transformations (X, Y, t) −→ (−X, Y, −t). We observe that the main difficulty to prove results (ii) for m > 6 is related with the huge computations necessary for proving them. 2010 Mathematics Subject Classification. 34C05, 34C07. Key words and phrases. linear type center, analytic planar differential system, weak center, reversible system. 1 http://www.gsd.uab.cat 2 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ 1. Introduction ∂ ∂ Let X = P +Q be the real planar polynomial vector field associated to ∂x ∂y the real planar polynomial differential system (1) ẋ = P (x, y), ẏ = Q(x, y), where the dot denotes derivative with respect to an independent variables here called the time t, and P and Q are real coprime polynomials in R[x, y]. We say that polynomial differential system (1) has degree m = max {deg P, deg Q}. In what follows we assume that the origin O := (0, 0) is a singular or equilibrium point, i.e. P (0, 0) = Q(0, 0) = 0. The equilibrium point O is a center if there exists an open neighborhood U of O where all the orbits contained in U \ {O} are periodic. Suppose that the polynomial differential system (1) has a center at the origin. It known that doing a linear change of variables and perhaps a constant scaling of the independent variable, system (1) becomes in one of the next three systems: ẋ = −y + X(x, y), ẏ = x + Y (x, y), (2) ẋ = y + X(x, y), ẏ = Y (x, y), ẋ = X(x, y), ẏ = Y (x, y), where X(x, y) and Y (x, y) are polynomials starting at least with quadratic terms in a neighborhood of the origin. If the origin of system (1) is a center we say that it is of linear type, nilpotent or degenerate if after a linear change of variables and a scaling of the time it can be written as the first, second or third system of (2), respectively. The study of the centers of analytical or polynomial differential systems (1) has a long history. The first works are due to Poincaré [15] and Dulac [6] . Later on were developed by Bendixson [3], Frommer [7], Liapunov [12] and many others. Here we shall work with the differential systems of the form (3) ẋ = −y + X, ẏ = x + Y, where X = X(x, y) and Y = Y (x, y) are polynomials, and the degree of the poly- nomial differential system (3) is m = max {deg X, deg Y } ≥ 2. The center-focus problem asks about conditions on the coefficients of X and Y for distinguish when the origin of system (3) is either a focus or a center. In this paper we shall study the particular case of differential systems (3) of the form (4) ẋ = −y(1 + Λ) + xΩ, ẏ = x(1 + Λ) + yΩ, where Λ = Λ(x, y) and Ω = Ω(x, y) are polynomials such m = max {deg Λ, deg Ω}+ 1. Systems (4) are called Λ–Ω systems. First we observe that if these systems have a center at the origin then they have a first integral of the form 1 2 (5) F = (x + y 2 )(1 + O(x, y)). 2 CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 3 This kind of centers having first integral of the form (5) are called weak centers, they contain the uniform isochronous centers and the holomorphic isochronous centers (for a prof of these results see [11]), but they do not coincide with the all class of isochronous centers (see Remark 19 of [11]). This class of planar polynomial differential systems (4) appear in different fields. If we assume that Λ = λ(x2 + y 2 ) − 1, and Ω = ω(x2 + y 2 ), then system (4) becomes ẋ = −y λ(x2 + y 2 ) + x ω(x2 + y 2 ), ẏ = xλ(x2 + y 2 ) + y ω(x2 + y 2 ), which is called a lambda–omega system (see for instance [8]). On the other hand it is well known the following result (see for instance [2]). Theorem 1. Let X be a polynomial vector field associated to the differential system (3). (i) Then X has either a focus or a center at the origin, and under a formal change of coordinates the differential system associated to X can be written in the Birkhoff normal form ẋ = −y(1 + S2 (x2 + y 2 )) + xS1 (x2 + y 2 ) := P, (6) ẏ = x(1 + S2 (x2 + y 2 )) + yS1 (x2 + y 2 ) := Q, where Sj = Sj (x2 + y 2 ) for j = 1, 2 are formal series in the variable x2 + y 2 and such that S2 (0) = S1 (0) = 0. (i) If X has a center at the origin then the formal system (6) becomes analytic with S1 (x2 +y 2 ) = 0 (see for instance [15, 12]) with the first integral x2 +y 2 . Consequently any linear type center locally is a weak center. Finally, by applying the inverse approach in ordinary differential equations see [9] the following theorem is proved and shows the importance of system (4) in the theory of ordinary differential equations (see Theorem 13 in [11]). Theorem 2. The polynomial differential system (3) has a weak center at the origin if and only if it can be written as (4) with ∑m ( ) j+1 j 3 Λ= Υj−1 + g1 Υj−2 + . . . + gj−2 Υ1 + gj−1 , j=2 2 2 2 1 ∑( ) m Ω= {Υj−1 , H2 } + g1 {Υj−2 , H2 } + . . . + gj−2 {Υ1 , H2 } , 2 j=2 where Υ0 = 1, g0 = 1, gj and Υj are homogenous polynomials of degree j for j ≥ 1 and has the first integral of the form H = H2 Φ = H2 (1 + µ1 Υ1 + . . . + µm−1 Υm−1 ), where H2 = (x + y 2 )/2, and µj = µj (x, y) is a convenient analytic function in the 2 neighborhood of the origin for j = 1, . . . , m − 1. 2. Statement of the main results In this section we give the statements of our main results which will be proved in sections 4, 5 and 6, also we state some conjectures. 4 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ Proposition 3. The polynomial differential system of degree m ẋ = −y(1 + (m − 2)(a1 y − a2 x)) + x(a1 x + a2 y + Ωm−1 ), (7) ẏ = x(1 + (m − 2)a1 y − a2 x)) + y(a1 x + a2 y + Ωm−1 ), where Ωm−1 = Ωm−1 (x, y) is a homogenous polynomial of degree m − 1, has a weak center at the origin if and only if ∫ 2π (8) Ωm−1 (cos t, sin t)dt = 0. 0 Moreover system (7) has the first integral H2 (9) H = ( )2/(m−1) = H2 (1 + h.o.t.), m−1 1+ G(x, y) + Γ(x, y) + Φ(H2 ) m+1 where G is a polynomial of degree m − 1 and such that {H2 , G} = −(m + 1)(a1 x + a2 y + Ωm−1 ), Γ = Γ(x, y) is a convenient polynomial and Φ(H2 ) is a convenient polynomial if m even and a convenient function if m is odd (see the proof for the expressions of Γ and Φ). We observe that if we take a1 = a2 = 0 in Proposition 3 we obtain following corollary. Corollary 4. The polynomial differential system of degree m (10) ẋ = −y + xΩm−1 ẏ = x + yΩm−1 , has a weak uniform center at the origin if and only if (8) holds. Moreover sys- tem (10) has the Poincaré-Liapunov first integral (9) where G is a homogenous polynomial of degree m − 1 and such that {H2 , G} = Ωm−1 . Conti in [5] proved the first part of Corollary4, but the second part providing an explicit expression of the first integral of system (10) is new. Remark 5. The weak centers obtained in Proposition 3 and Corollary 4 in general are not invariant with respect to the transformation (x, y, t) −→ (−x, y, −t). Indeed the polynomial differential system of degree 2k given by formula (7) and (10), when Ω2k−1 is an arbitrary polynomial of degree 2k − 1, in general is not reversible and has the weak center at the origin. We recall that a polynomial differential system has a uniform center at the origin if written in polar coordinates x = r cos θ, y = r sin θ we obtain that θ̇ is constant. Conjecture 6. The polynomial differential system of degree m ẋ = −y(1 + µ(a2 x − a1 y)) + x(a1 x + a2 y + Ωm−1 ), (11) ẏ = x(1 + µ(a2 x − a1 y)) + y(a1 x + a2 y + Ωm−1 ), where (µ+m−2)(a21 +a22 ) ̸= 0, and Ωm−1 = Ωm−1 (x, y) is a homogenous polynomial of degree m − 1 has a weak center at the origin if and only if system (11) after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t). Theorem 7. Conjecture 6 holds for m = 2, 3, 4, 5, 6. CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 5 Conjecture 8. The polynomial differential system of degree m ẋ = −y(1 + a1 x + a2 y) + xΩm−1 , (12) ẏ = x(1 + a1 x + a2 y) + yΩm−1 , where a21 + a22 ̸= 0, and Ωm−1 = Ωm−1 (x, y) is a homogenous polynomial of degree m − 1 has a weak center at the origin if and only if system (12) after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t). Theorem 9. Conjecture 8 holds for m = 2, 3, 4, 5, 6. The only difficulty for proving Conjectures 6 and 8 for the Λ–Ω systems of degree m with m > 6 is the huge number of computations for obtaining the conditions that characterize the centers. 3. Preliminary results In the proofs of the main results of this paper it plays a very important role the following results and notations which we can find in [11] . As usual the Poisson bracket of the functions f (x, y) and g(x, y) is defined as ∂f ∂g ∂f ∂g {f, g} := − . ∂x ∂y ∂y ∂x The following result is a simple consequence of the Liapunov result given in Theorem 1, page 276 of [12] . Corollary 10. Let U = U (x, y) be a homogenous polynomial of degree m. The linear partial differential equation {H2 , V } = U, has a unique homogenous poly- nomial solution V of degree m if m is odd; and if V is a homogenous polynomial solution when m is even then any other homogenous polynomial solution is of the form V + c(x2 + y 2 )m/2 with c ∈ R. Moreover, for m even these solutions exist if ∫ 2π and only if U (x, y)|x=cos t, y=sin t dt = 0. 0 Proposition 11 (see Proposition 6 of [10]). The next relation holds ∫ 2π {H2 , Ψ}|x=cos t, y=sin t dt = 0 0 for an arbitrary C 1 function Ψ = Ψ(x, y) defined in the interval [0, 2π] . Proposition 12 (see Proposition 24 of [10]). Consider the polynomial differential system (1) of degree m which satisfies ∫ 2π ( ) ∂P ∂Q + |x=cos t, y=sin t dt = 0. 0 ∂x ∂y Then there exist polynomials F = F (x, y) and G = G(x, y) of degree m + 1 and m − 1 respectively such that system (1) can be written as (13) ẋ = P = {F, x} + (1 + G){H2 , x}, ẏ = Q = {F, y} + (1 + G){H2 , y}, with G(0, 0) = 0. 6 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ We need the following definitions and notions. A function V = V (x, y) is an ) (1) in an open subset U ⊂ R if V ∈ C (U ), V ̸≡ 2 1 inverse integrating ( factor ) of ( system ∂ P ∂ Q 0 in U and + =0 ∂x V ∂y V Theorem 13 (Reeb ’s criterion). (see for instance [16]) The analytic differential ∞ ∑ ∑∞ system ẋ = −y + Xj , ẏ = x + Yj has a center at the origin if and only if j=2 j=2 there is a local nonzero analytic inverse integrating factor of the form V = 1 + h.o.t. in a neighborhood of the origin. An analytic inverse integrating factor of the form V = 1+h.o.t. in a neighborhood of the origin is called a Reeb inverse integrating factor. The analytic function ∞ ∑ ∑∞ 1 2 H= Hj (x, y) = (x + y 2 ) + Hj (x, y), j=2 2 j=3 where Hj is homogenous polynomials of degree j > 1, is called the Poincaré- Liapunov local first integral if H is constant on the solutions of (3). Theorem 14 (see Theorem 13 and Remark 14 of [11]). Consider the polynomial ∑m ∑m ∂ ∂ vector field X = (−y + Xj ) + (x + Yj ) . Then this vector field has a j=2 ∂x j=2 ∂y Poincaré-Liapunov local first integral H if and only if it has a Reeb inverse inte- grating factor V . Moreover, the differential system associated to the vector field X for which H = (x2 + y 2 )/2 + h.o.t. is a local first integral can be written as ẋ = V {H, x} = {Hm+1 , x} + (1 + g1 ){Hm , x} + . . . + (1 + g1 + . . . + gm−1 ){H2 , x}, (14) ẏ = V {H, y} = {Hm+1 , y} + (1 + g1 ){Hm , y} + . . . + (1 + g1 + . . . + gm−1 ){H2 , y}, and V and H are such that ∞ ∑ V = 1+ gj , j=1 ∑∞ 1 2 2 (15) H= (x + y ) + Hj = τ1 Hm+1 + τ2 Hm + . . . + τm H2 2 j=2 ∫ ( dHm+1 (1 + g1 )dHm (1 + g1 + . . . + gm−1 )dH2 ) = + + ... + , γ V V V where γ is an oriented curve (see for instance [17]), τj = τj (x, y) is a convenient analytic function in the neighborhood of the origin such that τj (0, 0) = 1, and gj = gj (x, y) is an arbitrary homogenous polynomial of degree j which we choose in such a way that V is the inverse Reeb integrating factor which satisfies the first CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 7 order partial differential equation 1 1 + g1 1 + g1 + . . . + gm−1 (16) {Hm+1 , } + {Hm , } + . . . + {H2 , } = 0. V V V Remark 15 (see formula (44) and the proof of Theorem 13 of [10]). From (16),and (15) the following infinite number of equations must hold (17) {Hm+1 , g1 } + {Hm , g2 } + . . . + {H3 , gm−1 } + {H2 , gm } = 0, {Hm+1 , g12 − g2 } + {Hm , g1 g2 − g3 } + . . . + {H3 , g1 gm−1 − gm } + {H2 , g1 gm + gm+1 } = 0, .. .. .. .. . . . . Consequently (18) ∫ 2π ({Hm+1 , g1 } + {Hm , g2 } + . . . + {H3 , gm−1 })|x=cos t,y=sin t dt = 0, 0 ∫ 2π ( ) {Hm+1 , g12 − g2 } + {Hm , g1 g2 − g3 } + . . . + {H3 , g1 gm−1 − gm } x=cos t,y=sin t dt = 0, 0 .. .. .. .. . . . . Conditions (17) and (18) are equivalent to the following relations . (19) {Hm+j+1 , g1 } + {Hm+j , g2 } + . . . + {H3 , gm+j−1 } + {H2 , gm+j } = 0, ∫ 2π ({Hm+j+1 , g1 } + {Hm+j , g2 } + . . . + {H3 , gm+j−1 })|x=cos t,y=sin t dt = 0, 0 for j ≥ 0. Theorem 14 can be applied to determine the Poincaré-Liapunov first integral and Reeb inverse integrating factor for the case when the polynomial differential system is given (see section 8 of [11]. Indeed, given a polynomial vector field X of degree m with a linear type center at the origin of coordinates, using (14) we determine its first integral H and its Reeb inverse integrating factor. Thus, if in m ∑ ∑m (3) X = Xj and Y = Yj with Xj and Yj homogenous polynomials of degree j=2 j=2 j, from (14) equating the terms of the same degree we get {Hj+1 , x} + g1 {Hj , x} + . . . + gj−1 {H2 , x} = Xj {Hj+1 , y} + g1 {Hj , y} + . . . + gj−1 {H2 , y} = Yj , 8 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ for j = 2, . . . , m. Then ∂ H3 = −X2 − yg1 , ∂y ∂ H3 = Y2 − xg1 , ∂x ∂ H4 ∂ H3 = −X3 − g1 − yg2 , ∂y ∂y ∂ H4 ∂ H3 = Y3 − g 1 − xg2 ∂x ∂x ∂ H5 ∂ H4 ∂ H3 = −X4 − g1 − g2 − yg3 , ∂y ∂y ∂y ∂ H5 ∂ H4 ∂ H3 = Y4 − g 1 − g2 − xg3 , ∂x ∂x ∂x (20) .. .. .. .. .. .. . . . . . . ∂ Hm+1 ∂ Hm ∂ H3 ∂ H2 = −Xm − g1 . . . − gm−2 − gm−1 , ∂y ∂y ∂y ∂y ∂ Hm+1 ∂ Hm ∂ H3 ∂ H2 = −Ym − g1 . . . − gm−2 − gm−1 ∂x ∂x ∂x ∂x ∂ Hm+k+1 ∂ Hm+k ∂ H3 ∂ H2 = −g1 . . . − gm+k−2 − gm+k−1 , ∂y ∂y ∂y ∂y ∂ Hm+k+1 ∂ Hm+k ∂ H3 ∂ H2 = −g1 . . . − gm+k−2 − gm+k−1 ∂x ∂x ∂x ∂x .. .. .. .. .. .. . . . . . . for k > 0. From the first two equation of (20) it follows that g1 must satisfy the first order partial differential equation ∂ X2 ∂ Y2 (21) {H2 , g1 } = + , ∂x ∂x which in view of Corollary 10 has a unique solution. Substituting g1 into the first two equations of (20) and using the Eulers Theorem for homogenous polynomial we obtain 1 (22) H3 = (xY2 − yX2 − 2g1 H2 ) . 3 We shall determine g2 as a solution of the first order partial differential equation ∂ X3 ∂ Y3 (23) {H2 , g2 } = + − {H3 , g1 }, ∂x ∂x where g1 is a solution of (21). Then in view of Corollary 10 we get that under the condition ∫ 2π ( ) ∂ X3 ∂ Y3 (24) + − {H3 , g1 } dt = 0, 0 ∂x ∂x x=cos t, y=sin t CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 9 g2 exists and has the form g2 (x, y) = g̃2 (x, y) + cH2 where c is a constant. Hence from the third and fourth equation of (20) we get 1 (25) H4 = (xY3 − yX3 − 3g1 H3 − 2g2 H2 ) . 4 We shall determine g3 as a solution of the first order partial differential equation ∂ X4 ∂ Y4 (26) {H2 , g3 } = + − {H4 , g1 } − {H3 , g2 }, ∂x ∂x where g1 , g2 and H3 , H4 are solutions of the previous differential equations. Then in view of Corollary 10 we get that there exist an unique solution g3 . Hence from the fifth and sixth equation of (20) we get 1 (27) H5 = (xY4 − yX4 − 4g1 H4 − 3g2 H3 − 2g3 H2 ) . 5 By continuing this process we obtain that if gm−1 is a solution of the equation ∂ Xm ∂ Ym (28) {H2 , gm−1 } = + − {Hm , g1 } . . . − {H3 , gm−2 }, ∂x ∂x which exist if and only if ∫ 2π ( ) ∂ Xm ∂ Ym (29) + − {Hm , g1 } . . . − {H3 , gm−2 } dt = 0, 0 ∂x ∂x x=cos t, y=sin t where the homogenous gm−2 , . . . , g1 are solutions of the previous first order differ- entia system, then the homogenous polynomial of degree m + 1 can be calculated as follows 1 Hm+1 = − (xYm − yXm − mg1 Hm − . . . − 2gm−1 H2 ) m+1 Finally under the conditions (30) {H2 , gm+k−1 } = −{Hm+k , g1 } − {Hm+k−1 , g2 } . . . − {H3 , gm+k−2 }, ∫ 2π ( ) 0= − {Hm+k , g1 } − {Hm+k−1 , g2 } . . . − {H3 , gm+k−2 } dt, 0 x=cos t, y=sin t we get that (31) 1 Hm+k+1 = − (−(m + k)g1 Hm+k − (m + k − 1)g2 Hm+k−1 − . . . − 2gm+k−1 H2 ) m+k+1 ∞ ∑ for k > 0. Thus we get the expression of H and of the integrating factor 1 + gj . j=1 Below we need the following results. Let (32) x = κ1 X − κ2 Y, y = κ2 X + κ1 Y, be a non-degenerated linear transformation, i.e. κ21 + κ22 ̸= 0. Then differential system (4) becomes ( ) Ẋ = −Y 1 + Λ̃(X, Y ) + X Ω̃(X, Y ), (33) ( ) Ẏ = X 1 + Λ̃(X, Y ) + Y Ω̃(X, Y ), 10 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ where Λ̃(X, Y ) = Λ(κ1 X − κ2 Y, κ2 X + κ1 Y ) and Ω̃(X, Y ) = Ω(κ1 X − κ2 Y, κ2 X + κ1 Y ). Here we say that system (3) is reversible with respect to a straight line l through the origin if it is invariant with respect to reversion about l and a reversion of time t (see for instance [5]). In particular Poincaré’s Theorem is applied for the case when (3) is invariant under the transformations (x, y, t) −→ (−x, y, −t), or (x, y, t) −→ (x, −y, −t). In the proof of the results which we give later on we need the Poincare’s Theorem (see for instance [14], p.122.) Theorem 16. The origin of system (3) is a center if the system is reversible. Since a rotation with respect to the origin of coordinates is a particular trans- formation of type (32), when a center of system (4) is invariant with respect to a straight line it is not restrictive to assume that such straight line is the x axis. So the center of system (4) will be invariant by the transformation (x, y, t) −→ (−x, y, −t) or (x, y, t) −→ (x, −y, −t). We shall study only the first case, i.e. we shall suppose that the Λ-Ω system is invariant with respect to the transformation (x, y, t) −→ (−x, y, −t). The following proposition is easy to prove (see for instance [13]). Proposition 17. Differential system (33) is invariant under the transformation (X, Y, t) −→ (−X, Y, −t) if and only if it can be written as ( ) Ẋ = −Y 1 + Θ1 (X 2 , Y ) + X 2 Θ2 (X 2 , Y ), (34) ( ) Ẏ = X 1 + Θ1 (X 2 , Y ) + XY Θ2 (X 2 , Y ), Remark 18. Using the notations of (33) and (34) after some computations we can prove that the following relations m−1 ∑ m−1 ∑ ∑ Λ̃(X, Y ) = Λ̃j = λkn X k Y n = Θ1 (X 2 , Y ), j=1 j=1 k+n=j m−1 ∑ m−1 ∑ ∑ Ω̃(X, Y ) = Ω̃j = ωkn X k Y n = XΘ2 (X 2 , Y ), j=1 j=1 k+n=j hold if and only if λ2l−1,j = 0 for l = 1, 2, . . . , [m/2] and j = 0, . . . , m − 2l, ω2l,j = 0 for l = 0, 2, . . . , [(m − 1)/2] and j = 0, . . . , m − 1 − 2l, where [ ] denotes the integer part function. Consequently, [m2 /4] coefficients of Λ̃ must be zero, and [(m2 + 2m − 3)/4] coefficients of Ω̃ must be zero. Corollary 19. Polynomial differential system (34) can be written as ( ) Ẋ = −Y 1 + Θ1 (X 2 , Y ) + X{H2 , Φ} = P (X, Y ), (35) ( ) Ẏ = X 1 + Θ1 (X 2 , Y ) + Y {H2 , Φ} = Q(X, Y ), where Φ = Φ(x, y) is a polynomial of degree at most m − 1 such that {H2 , Φ} = XΘ2 (X 2 , Y ). In particular all uniform isochronous centers which after a linear CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 11 change of variables (x, y) −→ (X, Y ) are invariant under the transformations (X, Y, t) −→ (−X, Y, −t) can be written as (36) Ẋ = −Y + X{H2 , Φ}, Ẏ = X + Y {H2 , Φ}, Proof. By considering that ∫ 2π ∫ 2π 2 XΘ2 (X , Y ) x=cos t, y=sin t dt = Θ2 (1 − sin2 t, sin t) cos t dt = 0. 0 0 By Corollary 10 we get that there exists a polynomial Φ such that XΘ2 (X 2 , Y ) = {H2 , Φ}. Thus we have the formula (35). Since for uniform isochronous centers Θ1 (X 2 , Y ) = 0, the corollary follows.  Proposition 20. Any weak center invariant at the origin of system (35) satisfies that the integral of the divergence on the unit circle is zero. Proof. The weak center of of system (35) satisfies ∂P ∂Q ∂{H2 , Φ} ∂{H2 , Φ} + = 2{H2 , Φ} + X +Y + {H2 , Θ1 } ∂X ∂Y ∂X ∂Y { } ∂Φ ∂Φ = H2 , 2Φ + X +Y + Θ1 . ∂X ∂Y From Proposition 11 satisfy that ∫ 2π ( ) ∂P ∂Q + dt = 0. 0 ∂X ∂Y X = cos t Y = sin t  4. Proofs of Proposition 3 and 4 ∫ 2π Proof of Proposition 3. Sufficiency: If Ωm−1 (cos t, sin t)dt = 0 then (12) holds 0 with P := −y(1 + (m − 2)(a1 y − a2 x)) + x(a1 x + a2 y + Ωm−1 ), Q := x(1 + (m − 2)(a1 y − a2 x)) + y(a1 x + a2 y + Ωm−1 ), because Ωm−1 is a homogenous polynomial of degree m − 1 and from the Euler Theorem for homogenous polynomial we have ∂P ∂Q + = (m + 1) (a1 x + a2 y + Ωm−1 ) . ∂x ∂y Therefore ∫ 2π ( ) ∂P ∂Q + dt = 0. 0 ∂x ∂y x = cos t y = sin t Then in view of Proposition 12 we get that there exist polynomials F and G of degree m + 1 and m − 1 respectively, such that system (7) can be written as ẋ = P = {F, x} + (1 + G){H2 , x}, (37) ˙x = Q = {F, y} + (1 + G){H2 , y}. 12 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ After some computations we obtain {F, H2 } = 2H2 (a1 x + a2 y + Ωm−1 ), {G, H2 } = −(m + 1)(a1 x + a2 y + Ωm−1 ). 2H2 2H2 Therefore {F + G, H2 } = 0. Consequently F = − G + p(H2 ), where m+1 m+1 [(m+1)/2] p(H2 ) = b0 + b1 H2 + . . . + b[m+1)/2] H2 is an arbitrary polynomial of degree ≤ 2[(m + 1)/2], where [ ] denotes the inte- ger part function. Inserting F into the differential system (37) and after some computations we get that ( ) 2H2 m−1 ẋ = P = − {G, x} + 1 + p′ (H2 ) + G {H2 , x}, m+1 m+1 ( ) 2H2 ′ m−1 ẏ = Q = − {G, y} + 1 + p (H2 ) + G {H2 , y}, m+1 m+1 dp where p′ (H2 ) = . In order to have that the linear terms of the previous system dH2 be −y and x respectively, we need that b1 = 0, because G(0, 0) = 0. Therefore 2H2 2H2 Ḣ2 = xẋ + y˙y = − (x{G, x} + y{G, y}) = {H2 , G}, m+1 m+1 ( )( ) ∂G ∂G m−1 ∂G ∂G Ġ = ẋ + ˙y = 1 + p′ (H2 ) + G {H2 , x} + {H2 , y ∂x ∂y m+1 ∂x ∂y ( ) m−1 = {H2 , G} 1 + G + p′ (H2 ) . m+1 Consequently we have the following linear ordinary differential of first order dG (m − 1)G (m + 1)(1 + p′ (H2 )) − = . dH2 2H2 2H2 Thus after the integration we have the existence of a first integral H given by H2 H= ( )2/(m−1) m−1 (38) 1+ G(x, y) + (m − 1)Γ(x, y) + [(m + 1)/2]Φ(H2 ) m+1 = H2 (1 + O(x, y)), hence the origin is a weak center, where 2b2 3b3 ([(m + 1)/2] − 1)b[(m+1)/2]−1 [(m+1)/2]−2 Γ= H2 + H22 + . . . + H , m−3 m−5 [(m1+)/2] − 1 − (m + 1)/2 2 and   b[(m+1)/2] [(m+1)/2]−1  H if m is even, Φ(H2 ) = [(m1+)/2] − (m + 1)/2 2   (m−1)/2 b[(m+1)/2] H2 log H2 if m is odd. We observe that if m is odd then this first integral (38) is non analytic at the origin. CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 13 Necessity: Now we suppose that the origin of system (7) is a center and we must prove that (8) holds. Indeed, if the origin is a center then from Theorem 14 it follows that differential system (7) can be written as (see (14)), where the homogenous polynomials Hk and gl for k = 2, . . . , m + 1 and l = 1, . . . , m − 1 satisfy (39) {Hm+1 , x} + g1 {Hm , x} . . . + gm−1 {H2 , x} := Xm = xΩm−1 , {Hm+1 , y} + g1 {Hm , y} . . . + gm−1 {H2 , y} = Ym = yΩm−1 , {Hm , x} + g1 {Hm−1 , x} . . . + gm−2 {H2 , x} = 0, {Hm , y} + g1 {Hm−1 , y} . . . + gm−2 {H2 , y} = 0, .. .. .. .. . . . . , {H5 , x} + g1 {H4 , x} + g2 {H3 , x} + g3 {H2 , x} = 0, {H5 , y} + g1 {H4 , y} + g2 {H3 , y} + g3 {H2 , y} = 0, {H4 , x} + g1 {H3 , x} + g2 {H2 , x} = 0, {H4 , y} + g1 {H3 , y} + g2 {H2 , y} = 0, {H3 , x} + g1 {H2 , x} := X2 = −y(m − 2)(a1 y − a2 x) + x(a1 x + a2 y), {H3 , y} + g1 {H2 , y} := Y2 = x(m − 2)(a1 y − a2 x) + y(a1 x + a2 y). From the last two equations we obtain (see (21) and (22)) 2 g1 = (m + 1)(a1 y − a2 x), H3 = − H2 g1 . m+1 After from the third and fourth equations starting from the end of system (39) we have (see (23) and (25)) 1 1 c g2 = g 2 + cH2 , H4 = g 2 H2 − H22 . m+1 1 m+1 1 2 Later from the fifth and sixth equations starting from the end of system (39) we get 4 2c 3m + 5 3 c g3 = 2 g13 − g1 H2 , H5 = − 2 g1 H2 + g1 H22 , 3(m + 1) m+1 1(m + 1) 10(m + 1) and so on, where c is an arbitrary constant. Inserting the previous expressions into ∂Xm ∂Ym the homogenous polynomial + we get ∂x ∂y {Hm , g1 } + {Hm−1 , g2 } . . . + {H2 , gm−1 } = (m + 1)Ωm−1 , which we obtain from the compatibility conditions of the first two equations of (39). After some computations it follows that {H2 , Φ} = (m + 1)Ωm−1 , for a convenient polynomial ∫ Φ of degree m − 1. Consequently in view of Proposition 11 we get that 2π Ωm−1 (cos t, sin t)dt = 0. In short the proposition is proved.  0 14 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ Remark 21. Polynomial differential system (7) under the condition (8) can be written as ẋ = −y(1 + (m − 2)(a1 y − a2 x)) + x(a1 x + a2 y + Ωm−1 ) = −y(1 + (m − 2)(a1 y − a2 x)) + x{H2 , Ψ}, ẏ = x(1 + (m − 2)a1 y − a2 x)) + y(a1 x + a2 y + Ωm−1 ) = x(1 + (m − 2)(a1 y − a2 x)) + y{H2 , Ψ}, where {H2 , Ψ} = a1 x + a2 y + Ωm−1 . 5. Proofs of Theorem 7 The proof of Theorem 7 follows from the next propositions. Proposition 22. A quadratic polynomial differential system ẋ = −y(1 + n1 x + n2 y) + x(a1 x + a2 y), (40) ẏ = x(1 + n1 x + n2 y) + y(a1 x + a2 y), has a weak center at the origin if and only if (41) a1 n1 + a2 n2 = 0. Moreover the quadratic differential system (40) satisfying condition (41), after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t). Proof. Sufficiency:. We suppose that (41) holds and we shall show that then the origin of system (40) is a center. Indeed doing the change of variables given in (32) with κ21 + κ22 ̸= 0 to system (??) we obtain the differential system ( ) Ẋ = −Y 1 + (κ1 n1 + κ2 n2 )X + (κ1 n2 − κ2 n1 )Y ( ) +X (κ1 a1 + κ2 a2 )X + (κ1 a2 − κ2 a1 )Y , ( ) Ẏ = X 1 + (κ1 n1 + κ2 n2 )X + (κ1 n2 − κ2 n1 )Y ( ) +Y (κ1 a1 + κ2 a2 )X + (κ1 a2 − κ2 a1 )Y . This system can be written as (34) if and only if (42) κ2 a1 − κ1 a2 = 0, κ1 n1 + κ2 n2 = 0. Clearly system (42) has a nonzero solution κ1 and κ2 if and only if a1 n1 +a2 n2 = 0. Under condition (42) the quadratic system is ( ) Ẋ = −Y 1 + (κ1 n2 − κ2 n1 )Y + (κ1 a1 + κ2 a2 )X 2 , ( ) Ẏ = X 1 + (κ1 n2 − κ2 n1 )Y + (κ1 a1 + κ2 a2 )XY. This system is invariant under the transformation (X, Y, t) −→ (−X, Y, −t), so in view of Theorem 16 the origin is a center, which in view of Theorem 2 is a weak center. CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 15 Necessity: We now suppose that system (40) has a center at the origin. From Theorem 14 we get that quadratic differential system (40) can be written as ẋ = {H3 , x} + (1 + g1 ){H2 , x} = −y(1 + n1 x + n2 y) + x(a1 x + a2 y) := −y + X2 , ẏ = {H3 , y} + (1 + g1 ){H2 , y} = x(1 + n1 x + n2 y) + y(a1 x + a2 y) := x + Y2 . Thus {H3 , x} + g1 {H2 , x} = −y(n1 x + n2 y) + x(a1 x + a2 y) := X2 , {H3 , y} + g1 {H2 , y} = x(n1 x + n2 y) + y(a1 x + a2 y) := Y2 . By determining the homogenous polynomial g1 as the unique solution of the equa- tion (see (21)) ∂ X2 ∂ Y2 {H2 , g1 } = + = (n2 + 3a1 )x + (3a2 − n1 )y, ∂x ∂x we get that (43) g1 = (n1 − 3a2 )x + (n2 + 3a1 )y. In view of homogeneity of H2 and H3 we obtain that (see (22)) 1 H3 = (xY2 − yX2 − 2g1 H2 ) = 2H2 (a2 x − a1 y). 3 On the other hand from conditions (20) since X3 = Y3 = 0 we get that ∂ H4 ∂ H3 ∂ H4 ∂ H3 = −g1 − yg2 , = −g1 − xg2 . ∂y ∂y ∂x ∂x We shall determine g2 as a solution of the first order partial differential equation (see (23)) {H2 , g2 } + {H3 , g1 } = 0, where g1 is given in (43). Then in view of Corollary 10 with V = g2 and U = ∫ 2π −{H3 , g1 }, we get that if {H3 , g1 }|x=cos t, y=sin t dt = 4π(a1 n1 + a2 n2 ) = 0, 0 then g2 exists and has the form g2 (x, y) = g̃2 (x, y) + cH2 where c is a constant. Thus we prove the necessity of the condition (41). In short the proposition is proved.  Proposition 23. A cubic polynomial differential system ẋ = −y(1 + µ(a2 x − a1 y)) +x(a1 x + a2 y + a3 x2 + a4 y 2 + a5 xy), (44) ẏ = x(1 + µ(a2 x − a1 y)) +y(a1 x + a2 y + a3 x2 + a4 y 2 + a5 xy), has a weak center at the origin if and only if (45) a3 + a4 = 0, a1 a2 a5 + (a22 − a21 )a4 = 0, Moreover system (44) under condition (45) and (µ + 1)(a21 + a22 ) ̸= 0, after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t). 16 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ We observe that Proposition 23 when µ = 0 provides Theorem 2.2 of the Collins’ [4]. Proof. Sufficiency: First of all we observe that cubic differential system (44) under the change of variables (32) becomes (46) ( ) Ẋ = −Y 1 + µ(κ1 a2 − κ2 a1 )X − µ(κ2 a2 + κ1 a1 )Y X ( + 2 2 2 (κ21 + κ22 ) ((κ1 a1 + κ2 a2 )X + (κ1 a2 − κ2 a1 )Y ) (κ1 + κ2 ) +(κ21 a3 + κ22 a4 + κ1 κ2 a5 )X 2 ) +(κ22 a3 + κ21 a4 − κ1 κ2 a5 )Y 2 + ((κ21 − κ22 )a5 + 2κ1 κ2 (a4 − a3 ))XY , ( ) Ẏ = X 1 + µ(κ1 a1 − κ2 a2 )X − µ(κ1 a2 − κ2 a1 )Y Y ( + 2 (κ21 + κ22 ) ((κ1 a1 + κ2 a2 )X + (κ1 a2 − κ2 a1 )Y ) (κ1 + κ22 )2 +(κ21 a3 + κ22 a4 + κ1 κ2 a5 )X 2 ) +(κ22 a3 + κ21 a4 − κ1 κ2 a5 )Y 2 + ((κ21 − κ22 )a5 + 2κ1 κ2 (a4 − a3 ))XY , is invariant under the transformation (X, Y, t) −→ (−X, Y, −t) if and only if using Proposition 22 we have κ2 a1 − κ1 a2 = 0, (47) κ21 a3 + κ22 a4 + κ1 κ2 a5 = 0, +κ22 a3 + κ21 a4 − κ1 κ2 a5 − = 0. From these conditions it follows that κ1 and κ2 such that κ21 + κ22 ̸= 0 if and only if a1 κ2 − a2 κ1 = 0 and a3 + a4 = 0, k1 k2 a5 + (k22 − k12 )a4 = 0. We suppose that (45) holds and show that then the origin of system (44) is a center. Clearly if a21 + a22 ̸= 0, then after the change x = a1 X − a2 Y, y = a2 X + a1 Y, we get that system (46) coincide with system (34) when κ1 = a1 and κ2 = a2 , and consequently (47) becomes (45). Since system (46) is invariant under the change (X, Y, t) −→ (−X, Y, −t) i.e. is reversible by Poincaré Theorem its origin is a center, and by Theorem 2 it is a weak center. Thus the sufficiency of proposition is proved. Necessity We suppose that the origin of (44) is a center. We must show that (45) holds. Indeed from Theorem 14 it follows that differential system (44) can be written as {H4 , x} + (1 + g1 ){H3 , x} + (1 + g1 + g2 ){H2 , x} = −y(1 + µ(a2 x − a1 y)) + x(a1 x + a2 y + a3 x2 + a4 y 2 + a5 xy), {H4 , y} + (1 + g1 ){H3 , y} + (1 + g1 + g2 ){H2 , y} = x(1 + µ(a2 x − a1 y)) + y(a1 x + a2 y + a3 x2 + a4 y 2 + a5 xy). CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 17 Consequently taking the homogenous parts of these two previous equalities we obtain (48) ∂H4 ∂H3 ∂ H2 {H4 , x} + g1 {H3 , x} + g2 {H2 , x} = − − g1 − g2 ∂y ∂y ∂y = x(a3 x2 + a4 y 2 + a5 xy) := xΩ2 = X3 , ∂H4 ∂H3 ∂ H2 {H4 , y} + g1 {H3 , y} + g2 {H2 , y} = + g1 + g2 ∂x ∂x ∂x 2 2 = y(a3 x + a4 y + a5 xy) := yΩ2 = Y3 , ∂H3 ∂H2 {H3 , x} + g1 {H2 , x} = − − g1 ∂y ∂y = −µ y(a2 x − a1 y) + x(a1 x + a2 y), ∂H3 ∂H2 {H3 , y} + g1 {H2 , y} = + g1 ∂x ∂x = µ x(a2 x − a1 y) + y(a1 x + a2 y). From the last two equations we have that (see (21) and (22))) g1 = (µ − 3)(a2 x − a1 y), and H3 = 2(a2 x − a1 y)H2 := Υ1 H2 . Inserting g1 and H3 into the two first equations and taking into account the Euler ∂ X3 ∂ Y3 Theorem for homogenous polynomials we get that + = 4Ω2 consequently ∂x ∂x (see (23) and (24)) {H2 , g2 } + {H3 , g1 } − 4Ω2 = 0, and this equation has solution if and only if (see Corollary 10) ∫ 2π ({H3 , g1 } − 4Ω2 )|x=cos t, y=sin t dt = a3 + a4 = 0, 0 Hence the first two equations of system (48) have a solution if and only if a3 +a4 = 0. The solutions of (48) are ( H4 = 9a22 − 3a22 µ − c)x2 + (4a1 a2 µ + 4a4 − 12a1 a2 )yx ) +(6a21 + (2a21 + a22 )µ + 3a22 − 2a2 a5 − c)y 2 H2 /2 := Υ2 H2 , g2 = α1 x2 + (−6a1 a2 + 2a1 µ − 4a4 )yx +((a22 − a21 )(µ − 3) + 2a5 + α1 )y 2 , where c is an arbitrary constant. By inserting a3 + a4 = 0 into (44) we get ẋ = −y(1 − µ(a1 y − a2 x)) + x(a1 x + a2 y − a4 x2 + a4 y 2 + a5 xy), (49) ẏ = x(1 − µ(a2 x − a1 y)) + y(a1 x + a2 y − a4 x2 + a4 y 2 + a5 xy), By calculating g3 as the unique polynomial homogenous of degree four solution of the equation {H4 , g1 } + {H3 , g2 } + {H2 , g3 } = 0, (see (26)) and determining H5 from (26) with X4 = Y4 = 0 we get H5 = −4g1 H4 /5 − 3g2 H3 /5 − 2g3 H2 /5. From equation {H5 , g1 } + {H4 , g2 } + {H3 , g3 } + {H2 , g4 } = 0, 18 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ (see (29) with m = 5 ), and in view of Proposition 11 we get that this equation has a solution for the homogenous polynomial g4 if and only if ∫ 2π 0= ({H5 , g1 } + {H4 , g2 } + {H3 , g3 })|x=cos t, y=sin t 0 = 3π(µ + 1)2 (a4 (a22 − a21 ) + a1 a2 a5 ) = 0. Thus a4 (a22 − a21 ) + a1 a2 a5 = 0. The case when µ + 1 = 0 and m = 3 has been studied in Proposition 3. If a1 = a2 = 0 then system (49) becomes ẋ = −y + x(−a4 x2 + a4 y 2 + a5 xy) := −y + xΩ2 , ẏ = x + y(−a4 x2 + a4 y 2 + a5 xy) = x + yΩ2 , which is a polynomial differential system with homogenous nonlinearities. Conse- quently we have that ∫ 2π ∫ 2π ( ) Ω2 |x=cos t,y=sin t dt = −a4 x2 + a4 y 2 + a5 xy x=cos t,y=sin t dt = 0. 0 0 Hence and in view of Corollary 4 we obtain that the origin is a weak center. In short the proposition is proved.  Proposition 24. A quartic polynomial differential system ẋ = −y(1 + µ(a2 x − a1 y)) +x(a1 x + a2 y + a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ), (50) ẏ = x(1 + µ(a2 x − a1 y)) +y(a1 x + a2 y + a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ), has a weak center at the origin under the assumption (a) a21 + a22 ̸= 0 if and only if (i) a1 a2 ̸= 0 and 1 ( ) a6 + 3 a7 (a31 − 3a22 a1 ) + a9 (a32 − a21 a2 ) = 0, 2a2 (51) 1 ( ) a8 + 2 a7 (3a31 − 3a1 a22 ) + a9 (a32 − 3a21 a2 ) = 0. 2a2 a1 ̸ 0, (ii) a7 = a8 = 0 when a2 = 0 and a1 = (iii) a6 = a9 = 0 when a1 = 0 and a2 ≠ 0. (b) If a1 = a2 = 0. (c) If µ + 2 = 0. Moreover if we assume that (µ + 2)(a21 + a22 ) ̸= 0 then system (50) under the conditions (i) , (ii) and after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t), and in the case (iii) it is invariant under the transformations (X, Y, t) −→ (X, −Y, −t), Proof. Sufficiency: Doing the change of variables (32) system (50) can be written as differential system (33) with m = 4 and Λ := µ(a2 x − a1 y), and Ω := a1 x + a2 y + a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 , CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 19 which is invariant under the transformation (X, Y, t) −→ (−X, Y, −t) if and only if (i) κ1 κ2 ̸= 0 and κ2 a1 − κ1 a2 = 0, 1 ( ) a6 + 3 a7 (κ31 − 3κ22 κ1 ) + a9 (κ32 − κ21 κ2 ) = 0, (52) 2κ2 1 ( ) a8 + 2 a7 (3κ31 − 3κ1 κ22 ) + a9 (κ32 − 3κ21 κ2 ) = 0− 2κ2 κ1 (ii) κ1 = 0 a2 ̸= 0, and a1 = a6 = a9 = 0. (iii) κ2 = 0 a1 ≠ 0 and a2 = a7 = a8 = 0. We suppose that statement (i), (ii) or (iii) hold and we will show that then the origin is a center of system (50). Indeed, if a1 a2 ̸= 0, then after the change x = a1 X − a2 Y, y = a2 X + a1 Y, we get that system (50) coincide with system (34) for m = 4 and with κ1 = a1 and κ2 = a2 and consequently by considering that this system is invariant under the change (X, Y, t) −→ (−X, Y, −t) or (X, Y, t) −→ (X, −Y, −t), i.e. it is reversible. If a2 = a7 = a8 = 0 and a1 ̸= 0, then differential equations (50) become ẋ = −y(1 − µ a1 y)) + x2 (a1 + a6 x2 + a9 y 2 ), ẏ = x(1 − µ a1 y)) + yx(a1 + a6 x2 + +a9 y 2 ). This system is invariant under the change (x, y, t) −→ (−x, y, −t) i.e. it is reversible. If a1 = a6 = a9 = 0 and a2 ̸= 0, then differential system (50) becomes ẋ = −y(1 + µ a2 x)) + xy(a2 + a7 y 2 + a8 x2 ), ẏ = x(1 + µ a2 x)) + y 2 (a2 + a7 y 2 + a8 x2 ). this system is invariant under the change (x, y, t) −→ (x, −y, −t) i.e. it is reversible. Thus in view of the Poincaré Theorem (see Theorem 16) we get that in cases (i), (ii) and (iii) the origin is a center. Furthermore by Theorem 2 this center is a weak center. Necessity: We prove the necessity of the statement (a). We suppose that the origin is a weak center. Indeed, from Theorem 14 it follows that differential system (50) can be written as (53) {H5 , x} + (1 + g1 ){H4 , x} + (1 + g1 + g2 ){H3 , x} + (1 + g1 + g2 + g3 ){H2 , x} = −y(1 + µ(a2 x − a1 y)) + x(a1 x + a2 y + a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ), {H5 , y} + (1 + g1 ){H4 , y} + (1 + g1 + g2 ){H3 , y} + (1 + g1 + g2 + g3 ){H2 , y}, = x(1 + µ(a2 x − a1 y)) + y(a1 x + a2 y + a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ). In view of Corollary 10 and assisted by an algebraic computer we can obtain the solutions of (53), i.e. the homogenous polynomials H5 , H3 , g1 , g3 of degree odd are unique and the homogenous polynomials H4 , g2 of degree even are obtained modulo an arbitrary polynomial of the form c(x2 + y 2 )k where k = 1, 2. Indeed taking the 20 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ homogenous part of these equations we obtain {H3 , x} + g1 {H2 , x} = −y(µ(a2 x − a1 y)) + x(a1 x + a2 y), {H3 , y} + g1 {H2 , y} = x(µ(a2 x − a1 y)) + y(a1 x + a2 y). The solutions of these equations are g1 = (µ − 3)(a2 x − a1 y), H3 = 2H2 (a2 x − a1 y). {H4 , x} + g1 {H3 , x} + g2 {H2 , x} = 0, (54) {H4 , x} + g1 {H3 , x} + g2 {H2 , x} = 0, the compatibility condition of these two last equations becomes of {H3 , g1 } + {H2 , g2 } = 0, and since {H3 , g1 } = {H2 , −(3 − µ)(a2 x − a1 y)2 } we get that {H2 , g2 − (3 − µ)(a2 x − a1 y)2 } = 0. Therefore g2 = (3 − µ)(a2 x − a1y)2 + c2 H2 , where c2 is a constant. Then from system (54) we obtain the solution H2 ( ) H4 = ((µ − 3) (a21 − a22 )(x2 − y 2 ) + 4a1 a2 xy + c1 H22 , 2 where c1 is a constant. Inserting these previous solutions g1 , H3 , g2 and H4 into the partial differential equations {H5 , x} + g1 {H4 , x} + g2 {H3 , x} + g3 {H2 , x} = x(a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ) = xΩ3 := X4 , (55) {H5 , y} + g1 {H4 , y} + g2 {H3 , y} + g3 {H2 , y} = y(a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ) = yΩ3 := Y4 we get that these differential equations have a unique solution. Indeed, in this case (26) becomes (56) {H4 , g1 } + {H3 , g2 } + {H2 , g3 } = 5Ω3 , ∂ X4 ∂ Y4 because + = 5Ω3 , here we have taking into account that Ω3 is a homoge- ∂x ∂x nous polynomial of degree 3. Consequently there exist a unique solution ( ) g3 = a1 (a22 + 2a21 /3)µ2 − a1 (10a21 /3 + 4a22 )µ + a1 (4a21 + 3a23 ) + 5(a9 + 2a6 )/3 y 3 ( ) −a2 (a21 + 2a22 /3)µ2 + a2 (10a22 /3 + 4a21 )µ − (3a21 + 4a22 )a2 − 5/3(a8 + 2a7 ) x3 ( ) + a31 µ2 − 2(2a21 + a22 )a2 µ + 3a1 (a21 + 2a22 ) + 7a6 x2 y ( ) + −a32 µ2 + 2(2a22 + a21 )a2 µ − 3a2 (a21 + 2a22 ) − 5a7 xy 2 +c(1 − µ)(a1 x + a2 y)H2 of (56) and substituting g3 into (55) we get (( ) H5 = −2/3H2 −a32 µ2 + a2 (2a22 − 3a21 )µ + 3a2 (3a21 + a22 ) − a8 − 2a7 x3 ( ) − a31 µ2 + a1 (2a21 − 3a22 )µ + 3a1 (3a22 + a21 ) + a9 + 2a6 y 3 ( ) 3a1 a22 µ2 + 3a1 (a21 − 4a22 )µ + 9a1 (a22 − a21 ) + 3a6 yx2 ( ) − 3a2 a21 µ2 + 3a2 (a21 − 4a22 )µ + 9a2 (a22 − a21 ) − 3a7 y 2 x cH22 (3a1 y − 2a2 x), where c is a constant. CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 21 On the other hand from (19) for m = 4 and j = 0 we get that ∫ 2π ({H5 , g1 } + {H4 , g2 } + {H3 , g3 })|x=cos t,y=sin t dt 0 (57) 3π(µ − 3) =− (a1 (3a7 + a8 ) − a2 (3a6 + a9 )) = 0. 3 Hence we get that (i) If a1 a2 ̸= 0 then by introducing the notations 1 ( ) λ 2 = a6 − 3 a7 (a31 − 3a22 a1 ) + a9 (a32 − a21 a2 ) , 2a2 1 ( ) λ 3 = a8 − 2 a7 (3a31 − 3a1 a22 ) + a9 (a32 − 3a21 a2 ) , 2a2 a1 we get that 3π(µ − 3) − (3a2 λ2 − a1 λ3) = 0. 3 (ii) If a1 = 0 and a2 ̸= 0 then 3a7 + a8 = 0. (iii) If a2 = 0 and a1 ̸= 0 then 3a6 + a9 = 0. Again from (28) with m = 5 we have the differential equation {H5 , g1 } + {H4 , g2 } + {H3 , g3 } + {H2 , g4 } = 0, From this equation we get the homogenous polynomial of degree four g4 which in view of Corollary 10 can be obtained with an arbitrary term of the form c(x2 +y 2 )2 , i.e. g4 is equal to ( ) 1 ( (−9a41 a2 + 11(3a7 + a8 )a21 − a22 (3(11a7 + a8 ) + 54(a7 + a8 )a22 x4 + 18a2 (a4 − a42 )(3a2 µ3 + 6a2 µ2 ) + a42 − a2 a9 )µ − (18(3a7 + a8 )a21 − 36a2 a1 a9 + a8 )a21 +(3a21 a2 ((a21 − 3a22 )µ3 + 2(a22 − a21 )µ2 + a21 (27a32 + 15a7 + 5a8 − 9a21 a2 ) ) −2a2 (a1 a9 + 9a2 a7 )µ + 54a2 (a2 a7 − a1 a9 ) − a22 (3(11a7 + a8 ) x2 y 2 (2 a1 5 ) + a2 a31 (µ − 2)µ2 + ( (7a7 + 10a8 ) + a2 a9 − 2a2 a31 )µ − 12a1 a7 xy 3 3 3 2 (2 7a 4a a2 a9 7 8 + a2 a31 (µ − 2)µ2 + (a1 ( + − 2a32 ) + )µ 3 3 3 3 ) −12a1 a7 − 4a1 a8 + 4a2 a9 x3 y + c(x2 + y 2 )2 , if a2 ̸= 0. In analogous form we can obtain the expression of g4 when a1 ̸= 0. Now we determine the homogenous polynomial H6 from (31) with m = 4 and k = 1 we obtain 5 4 3 2 H6 = − g1 H5 − g2 H4 − g3 H3 − g4 H2 . 6 6 6 6 Since the integral of the homogenous polynomial of degree 5 ∫ 2π ({H6 , g1 } + {H5 , g2 } + {H4 , g3 } + {H3 , g4 })|x=cos t,y=sin t dt ≡ 0, 0 22 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ then we obtain that there is a unique solution for the homogenous polynomial g5 of degree 5 of the equation {H6 , g1 } + {H5 , g2 } + {H4 , g3 } + {H3 , g4 } + {H2 , g5 } = 0. see (30) with m = 4 and k = 2. Calculating the homogenous polynomial H7 of degree 7 from (31) with m = 4 and k = 2 we get 6 5 4 3 2 H7 = − g1 H6 − g2 H5 − g3 H4 − g4 H3 − g5 H2 , 7 7 7 7 7 and inserting these polynomials into the integrand of the following homogenous polynomial of degree 6 we have (58) ∫ 2π ({H7 , g1 } + {H6 , g2 } + {H4 , g3 } + {H3 , g4 })|x=cos t,y=sin t dt = I(a1 , a2 , µ), 0 where I is such that (iii) if a1 = 0 and a2 ̸= 0 then π a31 (µ − 3) ( ) I(a1 , 0, µ) = − (231µ2 + 1230µ + 1776)a7 + 7(5µ + 12)(µ + 2)a9 = 0. 63 (ii) if a2 = 0 and a1 ̸= 0 then 5π a32 (µ − 3) ( ) I(0, a2 , µ) = 21(5µ + 12)(µ + 2)a6 + (49µ2 + 2µ − 432)a9 = 0, 189 (i) If a1 a2 ̸= 0 then π (m − 3) ( ( ) ( I(a1 , a2 , µ) = 21(5µ + 12)(µ + 2)(3a21 + 5a22 ) λ2 + a1 (105a21 + 427a22 )µ2 189 ) ) +(426a21 + 2306a22 )µ + 504a21 + 3384a22 λ3 = 0. By solving equations (57) and (58) and by considering that the three cases can be represented as a linear system Aξ = 0, where ξ = (a6 , a9 )T in the case (iii), ξ = (a7 , a8 )T in the case (ii) and ξ = (λ2 , λ3 )T in the case (i). By considering that the determinant of the matrix A is |A| = (µ − 3)(µ + 2)(21µ2 + 128µ + 212). Then under the condition (µ − 3)(µ + 2) ̸= 0 we obtain the necessity of statement (a) of Proposition 24. Now we study the case when µ = 3. After some computations we get that ∫ 2π ({H5 , g1 } + {H4 , g2 } + {H3 , g3 })|x=cos t,y=sin t dt 0 15π =− (a1 (3a7 + a8 ) − a2 (3a6 + a9 )) = 0. 2 First we shall study the case when a1 = 0, and a2 ̸= 0. From (57) we get that 3a6 + a9 = 0. From (58) we obtain that the integral is identically zero because CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 23 µ = 3 . On the other hand from the relation ∫ 2π ( {H9 , g1 } + {H8 , g2 } + {H7 , g3 } + {H6 , g4 } + {H5 , g5 } 0 ) +{H4 , g6 } + {H3 , g7 } + dt = I(a1 , a2 ) = 0, x=cos t,y=sin t where I is a convenient function in a1 and a2 . If (iii) a1 = 0 and a2 ̸= 0 then I(0, a2 ) = (105a32 a7 − 105a32 a8 − 864a22 c)a6 + (995a32 a7 + 285a22 a8 − 288a22 c)a9 = 0. By considering that a9 = −3a6 we obtain that (3a7 +a8 )a6 = 0. Thus if 3a7 +a8 ̸= 0 then µ−3 = a1 = a6 = a9 = 0. We observe that if a1 = 0 then λ2 = a6 and λ3 = a9 . In analogous form we can study the case when a2 = 0 and a1 ̸= 0. If 3a7 + a8 = 3a6 + a9 = 0 then we obtain differential system ( ) ˙x = −y(1 + µ(a2 x − a1 y) + x a1 x + a2 y + a6 x(x2 − 3y 2 ) + a7 y(y 2 − 3x2 ) , ( ) ˙y = x(1 + µ(a2 x − a1 y) + y a1 x + a2 y + a6 x(x2 − 3y 2 ) + a7 y(y 2 − 3x2 ) , From (29) we have that this system has a weak center at the origin if and only if a2 (3a21 − a22 )a6 + a1 (a21 − 3a22 )a7 = 0. Hence if a1 = 0 then a6 = a9 = 0, and if a2 = 0 then a7 = a8 = 0. In short we prove that the origin is a weak center of (50) in the case when µ = 3. Thus we obtain the necessity of the condition in statement (a). The necessity and sufficiency of statement (c), i.e. when µ + 2 = 0 follows from Proposition 3. Finally we study the necessity and sufficiency of statement (b). Thus when a1 = a2 = 0 we obtain that differential system (50) becomes ẋ = −y + x(a6 x3 + a7 y 3 + a8 xy + a9 xy 2 ) := −y + xΩ3 , ẏ = x + y(a6 x3 + a7 y 3 + a8 xy + a9 xy 2 ) := x + yΩ3 , which is a polynomial differential ∫ 2π system of degree four with homogenous nonlin- earities. Consequently Ω3 |x=cos t,y=sin t dt = 0, then in view of Proposition 4 0 we get that the origin is a weak center. In short the proposition is proved.  Proposition 25. A polynomial differential system of degree five ẋ = −y(1 + µ(a2 x − a1 y)) +x(a1 x + a2 y + a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ), (59) ẏ = x(1 + µ(a2 x − a1 y)) +y(a1 x + a2 y + +a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ), has a weak center at the origin if and only if (a) a12 + 3(a10 + a14 ) = 0, ( ) ( ) (60) 2a31 a32 a13 − a61 + 7(a21 a42 − a42 a21 a10 − a51 a2 − 4a31 a32 + a1 a52 a11 = 0, ( ) 2a21 a22 a14 − a41 − 4a21 a22 + a42 a10 − (a31 a2 − a1 a32 )a11 = 0. Moreover (i) If a1 = 0 and a2 ̸= 0 then a12 = a10 = a14 = 0. 24 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ (ii) If a2 = 0 and a1 = ̸ 0 then a12 = a10 = a14 = 0. (iii) If a1 a2 ̸= 0 then by introducing the notations 1 (( 6 2 4 4 2 6 ) ( 5 3 3 5 ) ) λ1 = a13 − a1 + 7(a 1 a 2 − a 1 a2 − a 2 a 10 − a 1 a2 − 4a 1 a2 + a1 a 2 a 11 , 2a31 a32 1 (( ) ) λ2 = a14 − 3 3 a41 − 4a21 a22 + a42 a10 − (a31 a2 − a1 a32 )a11 , 2a1 a2 we get that λ1 = λ2 = 0. (b) a1 = a1 = 0, and a12 + 3(a10 + a14 ) = 0. (c) µ + 3 = 0. Moreover system (59) under the conditions (60) and (µ+3)(a21 +a22 ) ̸= 0 and after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t), or (X, Y, t) −→ (X, −Y, −t). Proof. Sufficiency: First we write system (59) after the linear change (32). Later we observe that the polynomial differential system obtained is invariant under the transformation (X, Y, t) −→ (−X, Y, −t) if and only if κ1 a2 − κ2 a1 = 0, 2κ21 κ22 a12 + (3κ41 − 6κ22 κ21 + 3κ42 )a10 + (3κ31 κ2 − 3κ32 κ1 )a11 = 0, 2κ31 κ32 a13 − (κ61 − 7κ41 κ22 + 7κ21 κ42 − κ62 )a10 − (κ51 κ2 − 4κ32 κ31 + κ1 κ52 )a11 = 0, 2κ21 κ22 a14 − (κ41 − 4κ21 κ22 + κ42 )a10 + (κ1 κ32 − κ31 κ2 )a11 = 0, From the last three conditions it follows that a12 + 3(a10 + a14 ) = 0. We suppose that (60) holds and then the origin is a center of system (59), and by Theorem 2 is a weak center. We suppose that a21 +a22 ̸= 0. Then after the change x = a1 X − a2 Y, y = a2 X + a1 Y, we get that system (59) coincides with system (34) with κ1 = a1 and κ2 = a2 , and consequently system (59) is invariant under the change (X, Y, t) −→ (−X, Y, −t). Thus in view of the Poincaré Theorem we get that the origin is a center. Again by Theorem 2 this center is weak. We observe that under the assumptions of statement (iii) differential system (59) becomes ẋ = y(1+µ a2 x)+xy(a2 +a11 x3 +a13 xy 2 ), ẏ = x(1+µ a2 x)+y 2 (a2 +a11 x3 +a13 xy 2 ). This system is invariant under the transformation (x, y, t) −→ (x, −y, −t). Under hypothesis of statement (ii) differential system (59) becomes ẋ = −y(1−µ a1 y)+x2 (a1 +a11 x2 y+a13 y 3 ), ẏ = x(1−µ a1 y)+yx(a2 +a11 x2 y+a13 y 2 ). This system is invariant under the transformation (x, y, t) −→ (−x, y, −t). Under the assumptions of statement (i) the conditions λ1 = λ2 = 0 follows immediately. Necessity Now we suppose that the origin is a center of (59) and we prove that (60) holds. Indeed, from Theorem 14 it follows that differential system (59) can be CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 25 written as {H6 , x} + (1 + g1 ){H5 , x} + (1 + g1 + g2 ){H4 , x} + (1 + g1 + g2 + g3 ){H3 , x} +(1 + g1 + g2 + g3 + g4 ){H2 , x} = −y(1 + µ(a2 x − a1 y)) + x(a1 x + a2 y + a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ), {H6 , y} + (1 + g1 ){H5 , y} + (1 + g1 + g2 ){H4 , y} + (1 + g1 + g2 + g3 ){H3 , y} +(1 + g1 + g2 + g3 + g4 ){H2 , y} = x(1 + µ(a2 x − a1 y)) + y(a1 x + a2 y + +a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ), which is equivalent to {H3 , x} + g1 {H2 , x} = −yµ(a2 x − a1 y) + x(a1 x + a2 y), {H3 , y} + g1 {H2 , y} = −yµ(a2 x − a1 y) + y(a1 x + a2 y). These equations have the unique solutions g1 and H3 . Inserting these homogenous polynomials into the equations {H4 , x} + g1 {H3 , x} + g2 {H2 , x} = 0, {H4 , y} + g1 {H3 , y} + g2 {H2 , y} = 0, we get H4 and g2 with arbitrary terms of the form ck (x2 + y 2 )k with ck a constant for k = 1, 2. Inserting g1 , H3 , g2 and H4 into the equations {H5 , x} + g1 {H4 , x} + g2 {H3 , x} + g3 {H2 , x} = 0, {H5 , y} + g1 {H4 , y} + g2 {H3 , y} + g3 {H2 , y} = 0, we have a unique solutions g3 and H5 . Inserting gj and Hj+2 for j = 1, 2, 3 into the equations {H6 , x} + g1 {H5 , x} + g2 {H4 , x} + g3 {H3 , x} + g4 {H2 , x} = x(a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ) = xΩ4 , {H6 , y} + g1 {H5 , y} + g2 {H4 , y} + g3 {H3 , y} + g4 {H2 , y} = y(a10 x4 + a11 x3 y + a12 x2 y 2 + a13 xy 3 + a14 y 4 ) = yΩ4 , we get that this partial differential system has solution if and only if and only if a12 + 3(a10 + a14 ) = 0. Indeed from the computability condition {H5 , g1 } + {H4 , g2 } + {H3 , g3 } + {H2 , g4 } = 6Ω4 , we obtain that if ∫ 2π ({H5 , g1 } + {H4 , g2 } + {H3 , g3 } − 6Ω4 )|x=cos t,y=sin t dt 0 = a12 + 3(a10 + a14 = 0. Then there exist solutions g4 and H6 which we obtain with arbitrary terms of the form dk (x2 + y 2 )k with dk a constant for k = 1, 2. 26 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ Inserting the homogenous polynomials gj and Hj+2 of degree j and j + 2 respec- tively, for j = 1, 2, 3, 4 into (30) for m = 5 and k = 2 we get ∫ 2π ({H7 , g1 } + {H6 , g2 } + {H5 , g3 } + {H4 , g4 } + {H3 , g5 })|x=cos t,y=sin t dt 0 = I1 (a1 , a2 ) = 0. where I1 (a1 , a2 ) is such that I1 (a1 , 0) = 2π a21 (µ + 3)(µ + 2) (a10 − a14 ) = 0, I1 (0, a2 ) = 2π a22 (µ + 3)(µ + 2) (a10 − a14 ) = 0, ( ) I1 (a1 , a2 )|a1 a2 ̸=0 = 2π (µ + 3)(µ + 2) (a21 − a22 )λ1 + a1 a2 λ2 = 0. and ∫ 2π ( {H9 , g1 } + {H8 , g2 } + {H7 , g3 } + {H6 , g4 } + {H5 , g5 } 0 ) +{H4 , g6 } + {H3 , g7 } dt = I2 (a1 , a2 ) = 0, x=cos t,y=sin t where = I2 (a1 , a2 ) is a constant such that π a41 ( ) I2 (a1 , 0)|a10 =a14 = − 2120µ3 + 7161µ2 + 5202µ + 7857 a14 = 0, 864 π a42 ( ) I2 (0, a2 )|a10 =a14 = − 2120µ3 + 7161µ2 + 5202µ + 7857 a14 = 0 = 0, 864 I2 (a1 , a2 )|a1 a2 ̸=0 = ν1 λ1 + ν2 λ2 = 0. where ν1 and ν2 are convenient constants such that the system ( ) 2π (µ + 3)(µ + 2) (a21 − a22 )λ1 + a1 a2 λ2 := σ1 λ1 + σ2 λ2 = 0, ν1 λ1 + ν2 λ2 = 0, is such that ∆ := σ1 ν2 − σ2 ν1 = (µ + 3)(µ + 2)p(µ, a1 , a2 ) with p(µ, a1 , a2 ) a convenient polynomial of degree 3 in the variable µ and p(0, a1 , a2 ) ̸= 0. Consequently if µ is such that ∆ ̸= 0 we obtain that λ1 = λ2 = 0 The case when (µ + 2)p(µ, a1 , a2 ) = 0 could be studied in a similar way to the proof of Proposition 24. The proof of statement (c), i.e. when µ = −3 is studied in Proposition 3. Now we prove statement (b). Thus a1 = a2 = 0 and the differential system (59) under the condition a12 + 3(a10 + a14 ) = 0 becomes ẋ = −y + x(a10 x4 + a11 x3 y − 3(a10 + a14 )x2 y 2 + a13 xy 3 + a14 y 4 ), ẏ = x + y(a10 x4 + a11 x3 y − 3(a10 + a14 )x2 y 2 + a13 xy 3 + a14 y 4 ) Hence we get that Ω4 = a10 x4 + a11 x3 y − 3(a10 + a14 )x2 y 2 + a13 xy 3 + a14 y 4 and ∫ 2π satisfies the condition Ω4 |x=cos t,y=sin t dt = 0, because a12 + 3(a10 + a14 ) = 0 0. Then in view of Corollary 4 we get that the origin is a weak center. Thus the necessity of the statement (b) is proved, and consequently the proposition is proved.  CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 27 Proposition 26. The polynomial differential system of degree six (61) ẋ = −y(1 + µ(a2 x − a1 y)) +x(a1 x + a2 y + a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 xy 4 ), ẏ = x(1 + µ(a2 x − a1 y)) ( ) +y a1 x + a2 y + a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 xy 4 , has a weak center at the origin if and only if (a) the following conditions hold: ( ) ( ) 8a21 a52 a15 + 2a51 a22 − 4a31 a42 + 2a1 a62 a19 − 3a71 − 15a51 a22 + 25a31 a42 − 5a1 a62 a16 ( ) + a72 + 11a41 a32 − 9a21 a52 − 3a61 a2 a20 = 0, 8a31 a42 a17 − (15a71 − 55a51 a22 + 45a31 a42 − 5a1 a62 )a16 − (10a51 a22 − 12a31 a42 + 2a1 a62 )a19 −(−15a61 a2 + 35a41 a32 − 13a21 a52 + a72 )a20 = 0, 2a21 a32 a18 − (5a51 − 10a31 a22 + 5a1 a42 )a16 − (4a31 a22 − 2a1 a42 )a19 − (6a21 a32 − 5a41 a2 − a52 )a20 = 0. Moreover (i) if a1 = 0 and a2 = ̸ 0 then a15 = a17 = 0, (ii) if a2 = 0 and a1 = ̸ 0 then a16 = a18 = 0, (iii) if a1 a2 ̸= 0 then (62) 1 (( 5 2 ) ( ) λ1 = a15 + 2 5 2a1 a2 − 4a31 a42 + 2a1 a62 a19 − 3a71 − 15a51 a22 + 25a31 a42 − 5a1 a62 a16 8a1 a2 ( 7 ) ) + a2 + 11a41 a32 − 9a21 a52 − 3a61 a2 a20 = 0, 1 ( λ2 = a17 − 3 4 (15a71 − 55a51 a22 + 45a31 a42 − 5a1 a62 )a16 + (10a51 a22 − 12a31 a42 + 2a1 a62 )a19 8a1 a2 ) +(−15a61 a2 + 35a41 a32 − 13a21 a52 + a72 )a20 = 0, 1 ( λ3 = a18 + 2 3 − (5a51 − 10a31 a22 + 5a1 a42 )a16 2a1 a2 ) −(4a31 a22 − 2a1 a42 )a19 − (6a21 a32 − 5a41 a2 − a52 )a20 = 0. (b) a1 = a2 = 0. (c) µ + 4 = 0. Moreover, assume that (µ + 4)(a21 + a22 ) ̸= 0 then system (61) under the conditions (i), (ii),(iii) and after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t), or (X, Y, t) −→ (X, −Y, −t). Proof. Sufficiency: First we observe that the polynomial differential system (61) under the conditions (i) becomes ẋ = −y(1 − µ a2 x) + xy(a2 + a16 y 4 + a17 x4 + a19 x2 y 2 ), ẏ = x(1 − µ a2 x) + y 2 (a2 + a16 y 4 + a17 x4 + a19 x2 y 2 ), 28 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ which is invariant under the transformation (x, y, t) −→ (x, −y, −t), and the poly- nomial differential system (61) under the conditions (ii) becomes ẋ = −y(1 + µ a1 y) + x2 (a1 x + a15 x4 + a18 x2 y 2 + a20 y 4 ), ẏ = x(1 + µ a1 y) + xy(a1 x + a15 x4 + a18 x2 y 2 + a20 y 4 ), which is invariant under the transformation (x, y, t) −→ (−x, y, −t) Under the linear change of variables (32) the differential system (61) is invariant under the transformation (X, Y, t) −→ (−X, Y, −t) if and only if ( 5 2 ) ( ) 2κ1 κ2 − 4κ31 κ42 + 2κ1 κ62 a19 − 3κ71 − 155 κ1 κ22 + 25κ31 κ42 − 5κ1 κ62 a16 ( ) + κ72 + 11κ41 κ32 − 9κ21 κ52 − 3κ61 κ2 a20 + 8κ21 κ52 a15 = 0, (15κ71 − 55κ51 κ22 + 45κ31 κ42 − 5κ1 κ62 )a16 + (10κ51 κ22 − 12κ31 κ42 + 2κ1 κ62 )a19 +(−15a61 a2 + 35a41 a32 − 13a21 a52 + a72 )a20 − 8a31 a42 a17 = 0, 2κ21 κ32 a18 − (5κ51 − 10κ31 κ22 + 5κ1 κ42 )a16 −(4κ31 κ22 − 2κ1 κ42 )a19 − (6κ21 κ32 − 5κ41 κ2 − κ52 )a20 = 0. We shall study only the case (iii), i.e. when a1 a2 ̸= 0. We suppose that (62) holds and show that then the origin is a center of system (61). For the case when a21 + a22 ̸= 0, after the change x = a1 X − a2 Y, y = a2 X + a1 Y, we get that this system coincides with system (35) and with κ1 = a1 and κ2 = a2 . Consequently system (35) is invariant under the change (X, Y, t) −→ (−X, Y, −t). Thus in view of the Poincaré Theorem we get that the origin is a center, and by Theorem 2 this center is a weak center. Necessity, Now we suppose that the origin is a center of (61) and we shall prove that (62) holds. Indeed, from Theorem 14 it follows that differential system (61) can be written as {H7 , x} + (1 + g1 ){H6 , x} + (1 + g1 + g2 ){H5 , x} + (1 + g1 + g2 + g3 ){H4 , x} +(1 + g1 + g2 + g3 + g4 ){H3 , x} + (1 + g1 + g2 + g3 + g4 + g5 ){H2 , x} ( = −y(1 + µ(a2 x − a1 y)) + x a1 x + a2 y + a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 ) +a19 x2 y 3 + a20 xy 4 , {H7 , y} + (1 + g1 ){H6 , y} + (1 + g1 + g2 ){H5 , y} + (1 + g1 + g2 + g3 ){H4 , y} +(1 + g1 + g2 + g3 + g4 ){H3 , y} + (1 + g1 + g2 + g3 + g4 + g5 ){H2 , y} ( = x(1 + µ(a2 x − a1 y)) + y a1 x + a2 y + a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 ) +a19 x2 y 3 + a20 xy 4 . CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 29 This partial differential system has solution for arbitrary a1 , a2 , a15 , a16 , a17 , a18 , a19 , a20 .. After some computations we get that (63) ∫ 2π ( ) {H7 , g1 } + {H6 , g2 } + {H5 , g3 } + {H4 , g4 } + {H3 , g5 } dt 0 x=cos t, y=sin t = −π(µ + 4) (a1 λ2 (a1 , a2 ) − 5a2 λ1 (a1 , a2 ) − a2 λ3 (a1 , a2 ))) := ν11 λ1 + ν12 λ2 + ν13 λ3 = 0, where λj = λj (a1 , a2 , µ) for j = 1, 2, 3 are constant defined in (62). By continuing the integration of (see equation (30) with m = 6 and k = 3) we get that ∫ 2π ( {H9 , g1 } + {H8 , g2 } + {H7 , g3 } + {H6 , g4 } + {H5 , g5 } + {H4 , g6 } (64) 0 ) + {H3 , g7 } dt := ν21 λ1 + ν22 λ2 + ν23 λ3 = 0, x=cos t, y=sin t where 25a2 π ( ν21 := (−1062a21 − 558a22 )µ3 + (−8961a21 − 7501a22 )µ2 6912 ) +1344c − 8460a21 − 8796a22 , 5a1 π ( ν22 := − (2070a22 − 354a21 )µ3 + (13897a22 − 2987a21 )µ2 2304 ) +(448c − 4308a22 − 2820a21 ) + (16677a21 − 83367a22 − 1344c , 5a2 ( ν23 := 4014a21 − 2250a22 )µ3 + (27933a21 − 19799a22 )µ2 6912 ) +(−11556a21 − 7764a22 + 1344c) + 99889a22 − 155763a21 − 4032c , where c is an arbitrary constant. Again from equation (30) with m = 6 and k = 5 we get (65) ∫ 2π ( {H11 , g1 } + {H10 , g2 } + {H9 , g3 } + {H8 , g4 } + {H7 , g5 } + {H6 , g6 } 0 ) +{H5 , g7 } + {H4 , g8 }{H3 , g9 } dt := ν31 λ1 + ν32 λ2 + ν33 λ3 = 0, x=cos t, y=sin t where ν3j for j = 1, 2, 3, are convenient constants. From (63), (64) and (65) we obtain the linear system with respect to λ1 , λ2 , λ3 : Aλ = 0, where A = A(µ) is the coefficient matrix with determinant |A| equal to ν11 ν12 ν13 π 3 (a22 a1 )3 (µ + 4) ( ν21 ν22 ν23 =− 12276µ5 + 145467µ4 + 502471µ3 456192 ν31 ν32 ν33 )( ) +480577µ2 − 3775995µ − 2701980 846µ3 + 6149µ2 − 576µ − 34299 ̸= 0. Consequently if |A| ̸= 0 then the unique solution of these linear system is the trivial solution λ1 = λ2 = λ3 = 0. The case when |A| = 0 could be studied in a similar way of Proposition 24. Thus the necessity of statement (a) is proved. To 30 J. LLIBRE, R. RAMÍREZ, V. RAMÍREZ prove statement (c), i.e. when µ = −4 we apply Proposition 3. Finally we prove statement (b), i.e when a1 = a2 = 0. Under these conditions system (61) becomes (66) ẋ = −y + x(a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 x1 y 4 ) := −y + xΩ5 , ẏ = x + y(a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 x1 y 4 := x + yΩ5 , which is a polynomial ∫ 2π differential system of degree six with homogenous nonlinear- ities. Since Ω5 |x=cos t,y=sin t dt = 0, in view of Proposition 4 we obtain that 0 the origin is a weak center of (66). Thus the necessity of the proposition is proved and consequently the proposition is proved.  6. Proof of Theorem 9 The proof of Theorem 9 follows from the next propositions. Proposition 27. A quadratic polynomial differential system ẋ = −y(1 + a1 x + a2 y) + x(a3 x + a4 y), (67) ẏ = x(1 + a1 x + a2 y) + y(a3 x + a4 y), has a weak center at the origin if and only if (68) a1 a3 + a2 a4 = 0. Moreover system (67) under condition (68) and (a21 + a22 ) ̸= 0 and after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t) or transformations (X, Y, t) −→ (X, −Y, −t). Proof. It is analogous to the proof of Proposition 22 with n1 = a1 , n2 = a2 and a1 = a3 , a2 = a4 .  Proposition 28. A cubic polynomial differential system ẋ = −y(1 + a1 x + a2 y) + x(a3 x2 + a4 y 2 + a5 xy), (69) ẏ = x(1 + a1 x + a2 y) + y(a3 x2 + a4 y 2 + a5 xy), has a weak center at the origin if and only if (70) a3 + a4 = 0, a1 a2 a5 + (a22 − a21 )a4 = 0, Moreover system (69), under condition (70) and a21 + a22 ̸= 0, after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformation (X, Y, t) −→ (−X, Y, −t). Proof. It is analogous to the proof of Proposition 23.  Proposition 29. A polynomial differential system of degree four ẋ = −y(1 + a1 x + a2 y) + x(a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ), (71) ẏ = x(1 + a1 x + a2 y) + y(a6 x3 + a7 y 3 + a8 x2 y + a9 xy 2 ), has a weak center at the origin under the assumption (a) a21 + a22 ̸= 0 if and only if CENTER PROBLEM FOR Λ–Ω DIFFERENTIAL SYSTEMS 31 (i) a1 a2 ̸= 0 and 1 ( ) a6 + a7 (a31 − 3a22 a1 ) + a9 (a32 − a21 a2 ) = 0, 2a32 1 ( ) a8 + 2 a7 (3a31 − 3a1 a22 ) + a9 (a32 − 3a21 a2 ) = 0. 2a2 a1 ̸ 0. (ii) a7 = a8 = 0 when a2 = 0 and a1 = (iii) a6 = a9 = 0 when a1 = 0 and a2 ≠ 0. (b) If a1 = a2 = 0. System (71) under the conditions (i) , (ii) and after a linear change of variables (x, y) −→ (X, Y ) it is invariant under the transformations (X, Y, t) −→ (−X, Y, −t), and in the case (iii) it is invariant under the transformations (X, Y, t) −→ (X, −Y, −t), Proof. It is analogous to the proof of Proposition 24.  Proposition 30. A polynomial differential system of degree five (72) ẋ = −y(1 + a1 x + a2 y) + x(a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 xy 4 ), ẏ = x(1 + a1 x + a2 y) + y(a15 x5 + a16 y 5 + a17 x4 y + a18 x3 y 2 + a19 x2 y 3 + a20 xy 4 , has a weak center at the origin under the assumption (a) a21 + a22 ̸= 0 if and only if (i) a1 a2 ̸= 0 and (73) 1 (( ) ( ) 0 = a15 + 2 5 2a51 a22 − 4a31 a42 + 2a1 a62 a19 − 3a71 − 15a51 a22 + 25a31 a42 − 5a1 a62 a16 8a1 a2 ( 7 ) ) + a2 + 11a41 a32 − 9a21 a52 − 3a61 a2 a20 , 1 ( 0 = a17 − 3 4 (15a71 − 55a51 a22 + 45a31 a42 − 5a1 a62 )a16 + (10a51 a22 − 12a31 a42 + 2a1 a62 )a19 8a1 a2 ) +(−15a61 a2 + 35a41 a32 − 13a21 a52 + a72 )a20 , 1 ( 0 = a18 + 2 3 − (5a51 − 10a31 a22 + 5a1 a42 )a16 2a1 a2 ) −(4a31 a22 − 2a1 a42 )a19 − (6a21 a32 − 5a41 a2 − a52 )a20 . ̸ 0. (ii) a1 = a16 = a17 = a19 = 0 when a2 = (iii) a2 = a15 = a18 = a20 = 0 when a1 ≠ 0. (b) If a1 = a2 = 0. 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Reeb, Sur certaines propiétés topologiques des variétés feuilletées , W.T. Wu, G. Reeb (Eds.), Sur les espaces fibrés et les variétés feuilletées, Tome XI, in: Actualités Sci. Indust. vol. 1183, Hermann et Cie, (1952) Paris. [17] S. Sternberg, Lectures on differential geometry, Prentice Hall, 1964. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain. E-mail address: jllibre@mat.uab.cat, valentin.ramirez@e-campus.uab.cat 2Departament d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Vir- gili, Avinguda dels Paı̈sos Catalans 26, 43007 Tarragona, Catalonia, Spain. E-mail address: rafaelorlando.ramirez@urv.cat