This is a preprint of: “Configurations of the topological indices of the planar polynomial differential systems of degree (2,m)”, Jaume Llibre, Clàudia Valls, RESULTS MATH, vol. 76, 21, 2021. DOI: [10.1007/s00025-020-01322-0] CONFIGURATIONS OF THE TOPOLOGICAL INDICES OF THE PLANAR POLYNOMIAL DIFFERENTIAL SYSTEMS OF DEGREE (2, m) JAUME LLIBRE1 AND CLAUDIA VALLS2 Abstract. Using the Euler-Jacobi formula there is a relation between the singular points of a polynomial vector field and their topological indices. Using this formula we obtain the configuration of the singular points together with their topological indices for the polynomial differ- ential systems ẋ = P (x, y), ẏ = Q(x, y) with degree of P equal to 2 and degree of Q equal to m when these systems have 2m finite singular points. 1. Introduction and statement of the main results Consider in R2 the polynomial differential system (1) ẋ = P (x, y), ẏ = Q(x, y), where P (x, y) and Q(x, y) are real polynomials of degrees 2 and m, respec- tively, or simply of degree (2, m). The motivation of our paper comes from the fact that for the planar quadratic polynomial differential systems (i.e. the ones of degree (2, 2)) the characterization of all configurations of the indices of the singular points of all systems that have four singular points is the well-known Berlinskii’s Theorem proved in [2, 6] and reproved in [4] using the Euler-Jacobi formula. More precisely, the Berlinskii’s Theorem can be stated as follows: Assume that a real quadratic system has exactly four real singular points. In this case if the quadrilateral formed by these points is convex, then two opposite singular points are anti-saddles (i.e. nodes, foci or centers) and the other two are saddles. If this quadrilateral is not convex, then either the three exterior vertices are saddles and the interior vertex is an anti-saddle or the exterior vertices are anti-saddles and the interior vertex is a saddle. We want to extend the Berlinskii’s Theorem from degree (2, 2) to degree (2, 2m) for all m ≥ 2, i.e., we shall obtain all configurations of the polynomial differential system of degree (2, m). 2010 Mathematics Subject Classification. Primary 34A05. Secondary 34C05, 37C10. Key words and phrases. Euler-Jacobi formula, singular points, topological index, polyno- mial differential systems, Berlinskii’s Theorem. 1 2 J. LLIBRE AND C. VALLS Assuming that the differential system (1) has 2m finite singular points, then using the Euler-Jacobi formula we obtain a relation between the finite singular points of the polynomial differential system (1) and the topological indices of their singular points. In the complex projective plane, and taking into account all the multiplic- ities of the singular points of a polynomial differential system, if the number of singular points is finite, then it is at most 2m, see for more details the Bézout’s Theorem (see [7] for a proof of this theorem). When all the sin- gular points have multiplicity one and are located in the finite part of the projective space, we can apply the Euler-Jacobi formula (see [1] for a proof of such formula). For system (1) if the set of zeroes contains exactly 2m elements, then the Jacobian determinant ∂P/∂x ∂P/∂y J= ∂Q/∂x ∂Q/∂y evaluated at each zero does not vanish, and for any polynomial R of degree less than or equal to m − 1 we have X R(a) (2) = 0. J(a) a∈A We denote by A the set of finite singular points of system (1). Given a finite subset B of R2 , we denote by B̂ its convex hull, by ∂B its boundary, and by #B its cardinal. Set A0 = A and for i ≥ 1 Ai = Ai−1 \ (Ai−1 ∩ ∂ Âi−1 ). There is an integer q such that Aq+1 = ∅. We say that A has the configuration (K0 ; K1 ; K2 ; . . . ; Kq ) where Ki = #(Ai ∩ ∂ Âi ). We are also interested in the study of the (topological) indices of the singular points of system (1). We say that the singular points of system (1) which belong to Ai ∩ ∂ Âi are on the i-th level. We recall that if we assume that #A = 2m then the Jacobian determi- nant J is non-zero at any singular point of system (1), consequently the topological indices of the singular points are ±1, and then the number Ki i −1 of the i-th level is substituted by the vector (n1i +, n2i −, . . . , nm i +, nm i −), i j P j where ni are positive integers such that j ni = Ki . More precisely, since Ai ∩∂ Âi is a convex polygon, these numbers take into account the number of consecutive points with positive and negative indices, viewing the ith level oriented counterclockwise: n1i corresponds to the string with largest number of consecutive points with positive indices. If there are several strings with the same number of points we choose one such that the next string (that has points with negative indices) is as large as possible. We continue the process with n2i and so on. Furthermore, when Ai ∩ ∂ Âi is a segment, the THE EULER-JACOBI FORMULA 3 numbers take into account the number of consecutive points with positive indices, beginning at one of its endpoints. We denote by i(a) the index of a singular point a ∈ A of system (1). To introduce the main results of the paper we introduce some notations 4(k) = (4; . . . ; 4), ∆− = (+, 2−), ∆+ = (2+, −), Γ = (2+, 2−), Rk = (+, −, . . . , +, −), k O+ = (2+, −, +, −, +, . . . , +, −), k O− = (+, 2−, +, −, +, . . . , +, −), where k is the length of the strings 4(k) , Rk and O± k . We take the convention 1 = − and O 1 = +. that O− + In our notation Berlinskii’s Theorem corresponds to the case m = 2 and can be stated as follows. Theorem 1 ( Berlinskii’s Theorem). For planar quadratic polynomial dif- ferential systems such that #A = 4 the following statements hold: P P (a) i(a) = 0 or | a∈A i(a)| = 2. P a∈A (b) If a∈A i(a) = 0, only the configuration (4) = R4 is possible and there P are examples of such configuration. 1 ) and (3) = (c) If | a∈A i(a)| = 2, only the configurations (3) = (3+, O− 1 (3−, O+ ) are possible and there are examples of such configurations. With the notation introduced above we can state the first main theorem in the paper that deals with m ≥ 3 odd. Theorem 2. For planar polynomial differential systems of degree (2, m) such that #A = 2m with m ≥ 3 odd, the following statements hold. P (a) a∈A i(a) = 0. (b) Only the following configurations are possible (i) (2m) = R2m ; (ii) (4(j) ; 2m − 4j) = (Γ1 ; . . . ; Γj ; R2m−4j ) for j = 1, . . . , (m − 3)/2 where Γk = Γ for k = 1, . . . , j; (iii) (4(j) ; 3; 2m − 4j − 3) = (Γ1 ; . . . ; Γj ; ∆+ , R2m−4j−3 ) for j even and (4(j) ; 3; 2m − 4j − 3) = (Γ1 ; . . . ; Γj ; ∆− , R2m−4j−3 ) for j odd, where j = 0, 1, . . . , (m − 3)/2 and Γk = Γ for k = 1, . . . , j. There exist examples of all these configurations. The proof of Theorem 2 is given in section 2. Note that the particular case m = 3 was proved in [4]. Now we introduce the second main result of the paper which deals with the case m ≥ 2 even. Theorem 3. For planar polynomial differential systems of degree (2, m) such that #A = 2m with m ≥ 2 even, the following statements hold. 4 J. LLIBRE AND C. VALLS P P (a) a∈A i(a) = 0 or | a∈A i(a)| = 2. P (b) If a∈A i(a) = 0, only the following configurations are possible (i) (2m) = R2m ; (ii) (4(j) ; 2m − 4j) = (R14 ; . . . ; Rj4 ; R2m−4j ) for j = 1, . . . , (m − 3)/2 where Rk4 = R for k = 1, . . . , j. There P exist examples of such configurations. (c) If | a∈A i(a)| = 2, only the following configurations are possible 2m−4j−3 (iii) (4(j) ; 3; 2m − 4j − 3) = (S1 +; S2 −; . . . ; Sj δj ; 3(−δj ); O− ) (j) 2m−4j−3 or (4 ; 3; 2m−4j−3) = (S1 −; S2 +; . . . ; Sj (−δj ); 3δj ; O+ ) for j = 1, . . . , (m − 3)/2, where Sk = 4 for k = 1, . . . , j and δj = + if j is odd and δj = − if j is even. There exist examples of such configurations. The proof of Theorem 3 is given in section 3. The particular cases m = 2 is the well-known Berinskii’s Theorem and m = 4 was proved in [9]. 2. Proof of Theorem 2 First of all we observe that if a configuration exists for a polynomial vector field X with degrees (2, m) and #A = 2m, then it is possible to construct the same configuration but interchanging the indices of the singular point, i.e. the points with index +1 become with index −1 and vice versa. For doing that it is enough to take Y = (−P, Q) instead of X = (P, Q). In the proof of Theorem 2 we will denote by {p1 , . . . , p2m } the set of points of A if there is no information about their indices, by {p+ + 1 , . . . , pk } − − the set of points of A with positive index, and by {p1 , . . . , p2m−k } the set of points of A with negative index. Also we will denote by Li,j the straight line Li,j (x, y) = 0 through the points pi and pj where i ∈ {1, . . . , 2m}, and by Li a straight line through a point pi ∈ ∂ Â such that for all q ∈ A we have Li (A) 6= 0 and it is zero only at q. We will also use the following auxiliary result proved in [3]. Lemma 4. Let X = (P, Q) be a polynomial vector field with max(deg P, deg Q) = n. If X has n singular points on a straight line L(x, y) = 0, this line is an isocline. If X has n + 1 singular points on L(x, y) = 0 then this line is full of singular points. 2.1. Proof of statement (a) of Theorem 2. It was proved in [8] that in thePcase of polynomial vector fields of degree (2, m) with m odd it holds that a∈A iX (a) = 0. This proves statement (a). 2.2. Proof of statement (b) of Theorem 2. By statement (a) we have m points with positive index and m points with negative one. Note that configurations (2m − 1; 1), (2m − 2; 2), (2m − 3; 3), (2m − 4; 4), (2m − 4; 3; 1), THE EULER-JACOBI FORMULA 5 (2m−5, ∗), . . . , (5; ∗) are not possible because any convex hull of 2m−1, 2m− 2, . . . , 5 points on a conic has at most four points in the boundary of the convex hull except for the ellipse, but in the case of the ellipse cannot be points in the 1–level. Furthermore, the unique possible configurations of the form (3; ∗) are (3; 2m − 3). Indeed, since the polynomial P has degree 2, P (x, y) = 0 is a conic and the 2m finite singular points of system (1) are on this conic. Therefore any real conic (ellipse, parabola, hyperbola, two parallel straight lines, two straight lines intersecting in a real point, one double straight line or one point) do not allow the configuration of the form (3; ∗; ∗). In a similar manner a configuration of the form (4; 5; ∗), or (4; 4; 5; ∗) or (4; 3; ∗; ∗). Clearly configurations of the form (2; ∗) cannot occur because the 2m singular points would be on a straight line, and by Lemma 4 this straight line will be full of singular points, a contradiction. Moreover, configurations of the form (1; ∗) have no meaning. In short, the unique possible configurations are (2m), (4(j) ; 2m − 4j − 4), (3; 2m − 3) and (4(j) ; 3; 2m − 4j − 3) for j = 1, . . . , (m − 3)/2. Moreover we will show that each configuration mentioned above is realizable. We will study each configuration separately. Configuration (2m): Assume that the subscripts of the points in A are in such a way that p1 , . . . , p2m are ordered in ∂ Â in counterclockwise sense. Also we consider the subscripts in Z/2mZ. Take m−2 Y Ci (x, y) = Li+2j,i+2j+1 , for j = 1, 2, . . . , 2m. j=0 Then the Euler Jacobi formula applied to Ci yields Ci (pi+2m ) Ci (pi+2m+1 ) + = 0. J(pi+2m ) J(pi+2m+1 ) Since all the points p1 , . . . , p2m are in an ellipse, the polynomial Ci (x, y) has the same sign on the two points pi+2m and pi+2m+1 . So J(pi+2m )J(pi+2m+1 ) < 0 for all i. Hence the indices of pi and pi+1 are different and the configuration of A must be (2m) = (+, −, . . . , +, −) = R2m . The configuration R2m can be realized intersecting an ellipse with m parallel straight lines each one having two points in the ellipse. Configurations (4(j) ; 2m − 4j − 4) for j = 0, 1, . . . , (m − 3)/2. We will only prove the cases j = 0 and j = 1 since the other cases are done in a similar manner. So, we study the cases (4; 2m − 4) and (4; 4; 2m − 8). Configuration (4; 2m − 4). Clearly the configuration (4; 2m − 4) only can be realizable being P (x, y) = 0 a hyperbola or a conic formed by two straight lines which intersect. 6 J. LLIBRE AND C. VALLS First we prove it for the case when P (x, y) = 0 is a hyperbola. Since all the singular points lie in a hyperbola and in the 1st-level of A we must have 2m − 4 points, it is clear that two points are in one branch of the hyperbola and the other 2m−2 points in the other branch of the hyperbola. Denote by p1 , p2 the points in one branch of the hyperbola (ordered in clockwise sense) and by p3 , . . . , p2m the remaining points which are in the other branch of the hyperbola and ordered in counterclockwise sense. Applying the Euler-Jacobi formula to m−2 Y C(x, y) = L3+j,2m , j=0 and taking into account the convex hull of p1 , p2 , p3 , p4 and the convex hull of p5 , . . . , p2m we get that p1 and p2 have different indices. Now applying again the Euler-Jacobi formula successively to m−2 Y Ck (x, y) = L1,2 L3+j,2m−j , j=0,j6=k with k = 0, . . . , m − 2 we get that p3+k and p2m−k have different indices for k = 0, . . . , m − 2. Without loss of generality we can assume that p3 = p+ 3 and p2m = p− 2m . Moreover, applying the Euler-Jacobi formula to m−2 Y C(x, y) = L2,2m L3+j,2m−j , j=1 we get that p1 and p3 have the same index (because the number of straight − lines is m − 2 which is odd) and so p1 = p+ 1 and p2 = p2 . Finally, applying the Euler-Jacobi formula to k−1 Y m−k−3 Y Ck (x, y) = L1,2 L3+j,2m−j L5+j+k,2m−j−k , j=0 j=0 − for k = 0, . . . , m − 3 we get that pk = p+ k for k odd and pk = pk for k even. Hence the unique possible configuration is (2+, 2−; +, −, . . . , +, −) = (Γ, R2m−4 ). If the conic is the product of two intersecting straight lines at the real point p, by Lemma 4 on each straight line we must have exactly m points, and the point p on every straight line separates 1 point from the other m − 1 points of the same straight line. In this configuration we can repeat the arguments done on the hyperbola for obtaining the configuration (Γ, R2m−4 ). The configuration (Γ, R2m−4 ) can be realized intersecting the hyperbola x2 − y 2 = 1 by the m − 1 straight lines x = −m, 2, 3, . . . , m. Configuration (4; 4; 2m − 8). The configuration (4; 4; 2m − 8) only can be realized being P (x, y) = 0 a hyperbola or a conic formed by two straight lines which intersect. Since the proof in both cases follow the same lines, THE EULER-JACOBI FORMULA 7 we only do it for the hyperbola. In that case, observe that since all the singular points lie in a hyperbola and in the 1st-level and 2nd level of A we must have four points, it is clear that four points are in one branch of the hyperbola and the other 2m − 4 points in the other branch of the hyperbola. Denote by p1 , p2 , p3 , p4 the points in one branch of the hyperbola ordered in clockwise sense and by p5 , . . . , p2m the remaining points which are in the other branch of the hyperbola and ordered in counterclockwise sense. Applying the Euler-Jacobi formula to m−3 Y m−3 Y C(x, y) = L1,4 L5+j,2m−j and C(x, y) = L2,3 L5+j,2m−j , j=0 j=0 we get that p2 and p3 have different indices and p1 and p4 have different indices. Moreover, applying the Euler-Jacobi formula to m−3 Y C(x, y) = L1,4 L2,3 L5+j,2m−j j=0,j6=k with k = 0, . . . , m − 3 we get that p5+k and p2m−k have different indices. − Without loss of generality we can assume that p5 = p+ 5 and p2m = p2m . Applying the Euler-Jacobi formula to m−3 Y C(x, y) = L4,2m L2,3 L5+j,2m−j , j=1 we get that p1 and p5 have the same index (because the number of straight − lines is m − 2 which is odd). In short p1 = p+ 1 and p4 = p4 . Now we apply the Euler-Jacobi formula to m−3 Y C(x, y) = L4,2m L3,2m−1 L5+j,2m−j−1 , j=1 with Lm+2,m+2 being a straight line through the point pm+2 which separates the points {p1 , p2 , p3 , p4 } from the points {p5 , . . . , pm+1 , pm+3 , . . . , p2m }. Do- ing so, we get that p2 and p5 must have different indices (because the number of straight lines is m−3 which is even) and so p2 = p− + 2 and p3 = p3 . Finally, applying the Euler-Jacobi formula to k−1 Y m−k−4 Y Ck (x, y) = L1,4 L2,3 L5+j,2m−j L7+j+k,2m−j−k , j=0 j=0 − for k = 0, . . . , m − 4 we get that pk = p+ k for k odd and pk = pk for k even. Hence, the unique possible configuration is (2+, 2−; 2+, 2−; +, −, . . . , +, −) = (Γ; Γ; R2m−8 ). The configuration (Γ; Γ; R2m−8 ) can be realized intersecting the hyperbola x2 − y 2 = 1 with the straight lines x = −3, −2, 2, 3, . . . , m − 1. 8 J. LLIBRE AND C. VALLS Configurations (4(j) ; 3; 2m − 4j − 3) for j = 0, 1, . . . , (m − 3)/2. We will only prove the cases j = 0, j = 1 and j = 2 since the other cases are done in a similar manner. So, we study the cases (3; 2m − 3), (4; 3; 2m − 7) and (4, 4, 3, 2m − 11). Configuration (3; 2m − 3). In this case it is clear that this configuration only can be realized being P (x, y) = 0 a hyperbola. Since all the singular points lie in a hyperbola and in the 1st-level of A we must have 2m − 3 points, it is clear that one point is in one branch of the hyperbola and the other 2m − 1 points in the other branch of the hyperbola. Denote by p1 the point alone in one branch of the hyperbola and by p2 , . . . , p2m the remaining points which are in the other branch of the hyperbola ordered in counterclockwise sense. Applying the Euler-Jacobi formula to m−2 Y C(x, y) = L2+j,2m−j j=0 we get that p1 and pm+1 have different indices. Now applying again the Euler-Jacobi formula successively to m−2 Y Ck (x, y) = L1,m+1 L2+j,2m−j , j=0,j6=k with k = 0, . . . , m − 2 we get that p2+k and p2m−k have the same index for k = 0, . . . , m − 2. Without loss of generality we can assume that p2 = p+ 2 and so p2m = p+ 2m . Now applying the Euler-Jacobi formula to m−2 Y C(x, y) = Lm+1,m+2 L2+j,2m−j+1 , j=1 we get that p1 and p2 have opposite indices (because the number of straight lines is m − 1 which is even). So p1 = p− + 1 and pm+1 = pm+1 . Now applying the Euler-Jacobi formula to k−1 Y m−k−2 Y Ck (x, y) = L1,m+1 L2+j,2m−j L4+j+k,2m−j−k , j=0 j=0 for k = 0, . . . , m − 2, where Lm+2,m+2 is a straight line through the point pm+2 that separates the point p1 from {p2 , . . . , pm+1 , pm+3 , . . . , p2m }, we − get that p2+k = p+ + 2+k and p2m−k = p2m−k for k even and p2+k = p2+k and p2m−k = p− 2m−k for k odd. Hence, the unique possible configuration is (2+, −; +, −, . . . , +, −) = (∆+ , R2m−3 ). The configuration (∆+ , R2m−3 ) can be realized intersecting the hyperbola x2 − y 2 = 1 with the straight lines x = −1, 1, 2, . . . , m − 1. Configuration (4; 3; 2m − 7). In this case it is clear that the configuration (4; 3; 2m − 7) only can be realized being P (x, y) = 0 a hyperbola. Since all THE EULER-JACOBI FORMULA 9 the singular points lie in a hyperbola and in the 0-level of A we must have four points and in the first level we must have three points, it is clear that three points are in one branch of the hyperbola and the other 2m − 3 points in the other branch of the hyperbola. Denote by p1 , p2 , p3 the points in one branch of the hyperbola ordered in clockwise sense and by p4 , . . . , p2m the remaining points which are in the other branch of the hyperbola and ordered in counterclockwise sense. Applying the Euler-Jacobi formula to m−3 Y C(x, y) = L1,3 L4+j,2m−j , j=0 we get that p2 and pm+2 have different indices. Now applying again the Euler-Jacobi formula to m−3 Y m−3 Y C(x, y) = L2,m+2 L4+j,2m−j and Ck (x, y) = L2,m+2 L1,3 L4+j,2m−j , j=0 j=0,j6=k with k = 0, . . . , m − 3 we get that p1 and p3 have the same index and p4+k and p2m−k have the same indices for k = 0, . . . , m − 3. Without loss of generality we can assume that p4 = p+ + 4 and so p2m = p2m . Moreover, applying the Euler-Jacobi formula to m−3 Y C(x, y) = L2,m+2 L3,2m L4+j,2m−j , j=1 we get that p1 and p4 have opposite indices (because the number of straight lines is m − 1 which is even). So p1 = p− − 1 and p3 = p3 . Now we apply successively the Euler-Jacobi formula to k−1 Y m−k−4 Y Ck (x, y) = L1,3 L2,m+3 L4+j,2m−j L6+j+k,2m−j−k , j=0 j=0 for k = 0, . . . , m − 4 and we get that pk = p−k for k odd with k 6= m + 2, and pk = p+k for k even with k 6 = 2. Finally, applying the Euler-Jacobi formula to m−4 Y C(x, y) = L1,3 L2,m+1 L4+j,2m−j , j=0 we get that pm+2 = p− + m+2 and p2 = p2 . In short the configuration is (2+, 2−; +, 2−; +, −, . . . , +, −) = (Γ, ∆− , R2m−7 ). The configuration (Γ, ∆− , R2m−7 ) can be realized intersecting the hyper- bola x2 − y 2 = 1 with the straight lines x = −2, −1, 1, 2, . . . , m − 2. Configuration (4; 4; 3; 2m − 11). As before, in this case it is clear that the configuration (4; 4; 3; 2m − 11) only can be realized being P (x, y) = 0 a hyperbola. Since all the singular points lie in a hyperbola and in the 0-level and 1-st level of A we must have four points and in the 2nd level of A we 10 J. LLIBRE AND C. VALLS must have three points, it is clear that five points are in one branch of the hyperbola and the other 2m − 5 points in the other branch of the hyperbola. Denote by p1 , p2 , p3 , p4 , p5 the points in one branch of the hyperbola ordered in clockwise sense and by p6 , . . . , p2m the remaining points which are in the other branch of the hyperbola and ordered in counterclockwise sense. Applying the Euler-Jacobi formula to m−4 Y C(x, y) = L1,5 L2,4 L6+j,2m−j , j=0 we get that p3 and pm+3 have different indices. Now applying again the Euler-Jacobi formula to m−4 Y m−4 Y C(x, y) = L3,m+3 L1,5 L6+j,2m−j , C(x, y) = L3,m+3 L2,4 L6+j,2m−j j=0 j=0 and m−4 Y Ck (x, y) = L3,m+3 L1,5 L2,4 L6+j,2m−j , j=0,j6=k with k = 0, . . . , m − 4 we get that p2 and p4 have the same indices, p1 and p5 have the same indices and p6+k and p2m−k have the same indices for k = 0, . . . , m − 4. Without loss of generality we can assume that p6 = p+ 6 and so p2m = p+ 2m . Moreover, applying the Euler-Jacobi formula to m−4 Y C(x, y) = L3,m+3 L5,2m L2,4 L6+j,2m−j , j=1 we get that p1 and p6 have opposite indices (because the number of straight lines is m − 3 which is even). So p1 = p− − 1 and p5 = p5 . Now we apply the Euler-Jacobi formula to m−5 Y C(x, y) = L5,2m L4,2m−1 L3,m+2 L6+j,2m−j−2 , j=0 and we get that p1 and p2 have opposite indices. So p2 = p+ + 2 and p4 = p4 . Now we apply the Euler-Jacobi formula to k−1 Y m−k−5 Y C(x, y) = L1,5 L2,4 L3,m+4 L6+j,2m−j L8+j+k,2m−j−k , j=0 j=0 and we get that p3 = p− − + 3 , pk = pk for k ≥ 7 odd with k 6= m + 4 and pk = pk for k ≥ 6 even. Finally, applying the Euler-Jacobi formula to m−5 Y C(x, y) = L1,5 L2,4 L3,m+2 L6+j,2m−j , j=0 THE EULER-JACOBI FORMULA 11 we obtain that pm+4 and pm+3 have opposite indices, and so pm+4 = p− m+4 . In short the unique possible configuration is (2+, 2−; 2+, 2−; 2+, −; +, −, . . . , +, −) = (Γ, Γ, ∆+ , R2m−11 ). The configuration (Γ, Γ, ∆+ , R2m−11 ) can be realized intersecting the hy- perbola x2 − y 2 = 1 with the straight lines x = −3, −2, −1, 1, 2, . . . , m − 3. 3. Proof of Theorem 3 3.1. Proof of statement (a) of Theorem 3. It was proved in [8] that in the case P of polynomial vector P fields of degree (2, m) with m even it holds that a∈A iX (a) = 0 or | a∈A iX (a)| = 2. This proves statement (a). 3.2. Proof of statements (b) and (c) of Theorem 3. Proceeding as in the same manner as in the proof of Theorem 2 we have that the only possible configurations are (2m), (4(j) ; 2m−4j−4), (3; 2m−3) and (4(j) ; 3; 2m−4j−3) for j = 1, . . . , (m − 3)/2. We will study each one of the configurations separately. Configuration (2m): Proceeding exactly in the same way as in the proof of Theorem 2 we get that the unique possible configuration is (+, −, . . . , +, −) = R2m . Again, the configuration R2m can be realized intersecting an ellipse with m parallel straight lines each one having two points in the ellipse. Configurations (4(j) ; 2m − 4j) for j = 1, . . . , (m − 3)/2. We will only prove the cases j = 1 and j = 2 since the other cases are done in a similar manner. Configuration (4; 2m − 4). We take the same notation as in the proof of the configuration (4; 2m − 4) in Theorem 2 and proceeding exactly as in that proof we get that p1 and p2 have different indices and p3+j and p2m−j have also different indices for j = 0, . . . , m − 2. Without loss of generality we can − assume that p3 = p+ 3 and p2m = p2m . Moreover, applying the Euler-Jacobi formula to m−2 Y C(x, y) = L2,2m L3+j,2m−j , j=1 we get that p1 and p3 have opposite index (because the number of straight lines is m − 2 which is even) and so p1 = p− + 1 and p2 = p2 . Now proceeding as in the final part of the proof of Theorem 2 we get that pj = p+ j for j − odd and pj = pj for j even. Hence, the unique possible configuration is (R4 ; +, −, . . . , +, −) = (R4 , R2m−4 ). P Note that here a∈A i(a) = 0. The configuration (R4 , R2m−4 ) can be realized intersecting the hyperbola x2 − y 2 = 1 by the m − 1 straight lines x = −m, 2, 3, . . . , m. 12 J. LLIBRE AND C. VALLS Configuration (4; 4; 2m − 8). We take the same notation as the proof of the configuration (4; 4; 2m − 8) in Theorem 2. Proceeding as there we get that p1 and p4 have different indices, p2 and p3 have different indices and p5+j and p2m−j have different indices. Without loss of generality we can assume − that p5 = p+5 and p2m = p2m . Applying the Euler-Jacobi formula to m−3 Y C(x, y) = L4,2m L2,3 L5+j,2m−j , j=1 we get that p1 and p5 have opposite index (because the number of straight lines is m − 2 which is even). In short p1 = p− + 1 and p4 = p4 . Now applying the Euler-Jacobi formula to m−4 Y C(x, y) = L4,2m L3,2m−1 L5+j,2m−j−2 , j=0 − we get that p1 and p2 must have opposite indices and so p2 = p+ 2 and p3 = p3 . Now proceeding as in the last part of the proof of this configuration in − Theorem 2 we get that pj = p+ j for j odd and pj = pj for j even. Hence, the unique possible configuration is (R4 ; R4 ; +, −, . . . , +, −) = (R4 ; R4 ; R2m−8 ). P Note that here a∈A i(a) = 0. The configuration (R4 ; R4 ; R2m−8 ) can be realized intersecting the hyperbola x2 − y 2 = 1 with the straight lines x = −3, −2, 2, 3, . . . , m − 1. This proves statement (b) of the Theorem. Configurations (4(j) ; 3; 2m − 4j − 3) for j = 0, 1, . . . , (m − 3)/2. We will only prove the cases j = 0, j = 1 and j = 2 since the other cases are done in a similar manner. So, we study the cases (3; 2m − 3), (4; 3; 2m − 7) and (4, 4, 3, 2m − 11). Configuration (3; 2m − 3): We take the same notation as in the proof of the configuration (3; 2m − 3) in Theorem 2 and proceeding as in the first part of that proof we get that p1 and pm+1 have different index, and p2+j and p2m−j have the same index for j = 0, . . . , m − 2. Without loss of generality we can assume that p2 = p+ + 2 and so p2m = p2m . Moreover, applying the Euler-Jacobi formula to m−2 Y C(x, y) = Lm+1,m+2 L2+j,2m−j+1 , j=1 we get that p1 and p2 have the same indices (because the number of straight − lines is m − 1 which is odd). So p1 = p+ 1 and pm+1 = pm+1 . Now proceeding as in the last part of the proof of Theorem 2 we get that p2+j = p+ 2+j and + − − p2m−j = p2m−j for j even and p2+j = p2+j and p2m−j = p2m−j for j odd. Hence, the unique possible configuration is (3+; +, 2−, +, −, . . . , +, −) = 2m−3 (3+; O− ). THE EULER-JACOBI FORMULA 13 P P Note that here a∈A i(a) = 2, so when a∈A i(a) = −2 we would 2m−3 2m−3 have the configuration (3−; O+ ) . The configuration (3+; O− ) can be realized intersecting the hyperbola x2 − y 2 = 1 with the straight lines x = −1, 1, 2, . . . , m − 1. Configuration (4; 3; 2m − 7). We use the notation in the configuration (4; 3; 2m − 7) in the proof of Theorem 2. Proceeding as there we get that p2 and pm+2 have different indices, p1 and p3 have the same index and p4+j and p2m−j have the same index for j = 0, . . . , m − 2. Without loss of generality we can assume that p4 = p+ + 4 and so p2m = p2m . Moreover, applying the Euler-Jacobi formula to m−3 Y C(x, y) = L2,m+2 L3,2m L4+j,2m−j , j=1 we get that p1 and p4 have the same indices (because the number of straight lines is m − 1 which is odd). So p1 = p+ + 1 and p3 = p3 . Now proceeding as in the last part of the proof of this configuration in Theorem 2 we get that − p4+j = p+ + 4+j and p2m−j = p2m−j for j even and p4+j = p4+j and p2m−j = p−2m−j for j odd. In short the configuration is (4+; 3−; 2+, −, +, −, . . . , +, −) = 2m−7 (4+, 3−, O+ ). P P Note that here a∈A i(a) = 2, so when a∈A i(a) = −2 we would have 2m−7 2m−7 the configuration (4−, 3+, O− ). The configuration (4+, 3−, O+ ) can 2 2 be realized intersecting the hyperbola x − y = 1 with the straight lines x = −2, −1, 1, 2, . . . , m − 2. Configuration (4; 4; 3; 2m − 11). We use the notation in the configuration (4; 3; 2m − 7) in the proof of Theorem 2. Proceeding as there we get that p3 and pm+3 have different indices, p1 and p5 have the same index, p2 and p4 have the same index and p6+j and p2m−j have the same index for j = 0, . . . , m − 2. Without loss of generality we can assume that p6 = p+ 6 and so p2m = p+ 2m . Moreover, applying the Euler-Jacobi formula to m−4 Y C(x, y) = L3,m+3 L5,2m L2,4 L6+j,2m−j , j=1 we get that p1 and p6 have the same index (because the number of straight lines is m − 2 which is even). So p1 = p+ + 1 and p5 = p5 . Now, proceeding as in the last part of the proof of configuration (4; 4; 3; 2m − 11) in Theorem 2 we get that p2 = p− − + + 2 , p4 = p4 , p6+j = p6+j and p2m−j = p2m−j for j even and p6+j = p6+j and p2m−j = p2m−j for j odd, p3 =3 and pm+3 = p− − − + m+3 . In 2m−11 short the unique possible configuration is (4+; 4−; 3+; O− ). P P Note that here a∈A i(a) = 2, so when a∈A i(a) = −2 we would 2m−11 have the configuration (4−; 4+; 3−; O+ ). Moreover, the configuration 14 J. LLIBRE AND C. VALLS 2m−11 (4+; 4−; 3+; O+ ) can be realized intersecting the hyperbola x2 − y 2 = 1 with the straight lines x = −3, −2, −1, 1, 2, . . . , m − 3. This proves statement (c) of the theorem and concludes the proof of it. Acknowledgements The first author is supported by the Ministerio de Economı́a, Indus- tria y Competitividad, Agencia Estatal de Investigación grant MTM2016- 77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Re- cerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The second author is partially supported by FCT/Portugal through UID/MAT/04459/2013. References [1] V. Arnold, A. Varchenko and S. Goussein-Zude, Singularités des applications différentialbes, Mir, Moscow, 1982. [2] A. N. Berlinskii, On the behavior of the integral curves of a differential equation, Izv. Vyssh. Uchebn. Zaved. Mat. 2 (1960), 3–18. [3] C. Chicone and T. Jinghuang, On general properties of quadratic systems, Amer. Math. Monthly 81 (1982), 167–178. [4] A. Cima, A. Gasull and F. Manõsas, Some applications of the Euler-Jacobi for- mula to differential equations, Proc. Amer. Math. 118 (1993), 151–163. [5] A. Cima and J. Llibre, Configurations of fans and nests of limit cycles for polyno- mial vector fields in the plane, J. Differential Equations 82 (1989), 71–97. [6] W.A. Coppel, A survey of quadratic systems, J. Differential Equations 2 (1966), 293–304. [7] W. Fulton, Algebraic curves, Mathematic Lecture Note Series, Benjamin, 1974. [8] A. G. Khovanskii, Index of a polynomial vector field, Funktsional Anal. i Prilozhen 13 (1979), 49–58. [9] J. Llibre and C. Valls, The Euler-Jacobi formula and the planar quadratic-quartic polynomial differential systems, preprint, 2019. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat 2 Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, Av. Rovisco Pais 1049–001, Lisboa, Portugal Email address: cvalls@math.ist.utl.pt