Configurations of the Topological Indices of the Planar Polynomial Differential Systems of Degree (2, m)

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 differential systems x˙=P(x,y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{x}=P(x,y)$$\end{document}, y˙=Q(x,y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\dot{y} =Q(x,y)$$\end{document} with degree of P equal to 2 and degree of Q equal to m when these systems have 2m finite singular points.


Introduction and Statement of the Main Results
Consider in R 2 the polynomial differential systeṁ where P (x, y) and Q(x, y) are real polynomials of degrees 2 and m, respectively, 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 [2,5] 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).
We note that it is the first time that Berlinskii's Theorem is extended to polynomial differential systems with arbitrary degree m. In [9] Berlinskii's Theorem was extended to polynomial differential systems degree 3 in the plane R 2 , and in [10] it was extended to polynomial differential of degree 2 in the space R 3 .
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 multiplicities 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 [6] for a proof of this theorem). When all the singular 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 J = ∂P/∂x ∂P/∂y ∂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 a∈A R(a) J(a) = 0.
We denote by A the set of finite singular points of system (1). Given a finite subset B of R 2 , we denote byB its convex hull, by ∂B its boundary, and by #B its cardinal.
We say that A has the configuration (K 0 ; 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 A i ∩ ∂Â i are on the i-th level.
We recall that if we assume that #A = 2m then the Jacobian determinant 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 K i of the i-th level is substituted by the vector (n 1 i +, n 2 i −, . . . , n mi−1 i +, n mi i −), where n j i are positive integers such that j n j i = K i . More precisely, since A i ∩ ∂Â i 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: n 1 i 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 n 2 i and so on. Furthermore, when A i ∩∂Â i is a segment, the 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 where k is the length of the strings 4 (k) , R k and O k ± and is the times that the string S ± is repeated in S ± . Note that the string S ± has length 2 and that S k + = R 2k . We take the convention that O 1 − = − and O 1 + = +. In our notation Berlinskii's Theorem corresponds to the case m = 2 and can be stated as follows.  The proof of Theorem 2 is given in Sect. 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.

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 {p 1 , . . . , p 2m } the set of points of A if there is no information about their indices, by {p + 1 , . . . , p + k } the set of points of A with positive index, and by {p − 1 , . . . , p − 2m−k } the set of points of A with negative index. Also we will denote by L i,j the straight line L i,j (x, y) = 0 through the points p i and p j where i ∈ {1, . . . , 2m}, and by L i a straight line through a point p i ∈ ∂Â such that for all q ∈ A we have L i (A) = 0 and it is zero only at q.
We will also use the following auxiliary result proved in [3].
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.

Proof of Statement (a) of Theorem 2
It was proved in [7] that in the case of polynomial vector fields of degree (2, m) with m odd it holds that a∈A i X (a) = 0. This proves statement (a). We will study each configuration separately. Configuration (2m): Assume that the subscripts of the points in A are in such a way that p 1 , . . . , p 2m are ordered in ∂Â in counterclockwise sense. Also we consider the subscripts in Z/2mZ. Take

Proof of Statement (b) of Theorem 2
Then the Euler Jacobi formula applied to C i yields Since all the points p 1 , . . . , p 2m are in an ellipse, the polynomial C i (x, y) has the same sign on the two points p i+2m and p i+2m+1 . So J(p i+2m )J(p i+2m+1 ) < 0 for all i. Hence the indices of p i and p i+1 are different and the configuration of A must be (2m) = (+, −, . . . , The configuration R 2m can be realized intersecting an ellipse with m parallel straight lines each one having two points in the ellipse. Configuration (m + 1; m − 1) In this case the configuration can only be realized with a conic formed by two straight lines R 1 and R 2 intersecting to a real point p. If the point p is a singular point, then one of the straight lines would have m + 1 singular points on it and in view of Lemma 4 this straight line will be full of singular points, a contradiction. So the point p is not a singular point. In this case in R 1 there are m singular points p 1 , . . . , p m in R 1 (ordered inside R 1 ), and there are m singular points p m+1 , . . . , p 2m in R 2 (ordered inside R 2 ) and without loss of generality we can assume that p 1 , . . . , p m+1 are in the 0-level and p m+2 , . . . , p 2m are in the 1st-level.
Applying the Euler-Jacobi formula with being L k a straight line passing through p k and being parallel to R 1 and taking the subscripts in Z/mZ we get that the points p i and p i+1 for i = m+1, . . . , 2m have opposite index, except for i = m + 2 that then p m+1 and p 2m have the same index. Without loss of generality we can assume that p m+1 has positive index and so the configuration in the 1st-level is R m−1 . Now we study the configuration in the 0-level (and so in R 1 ). Note that since p is not a singular point, the straight line R 2 has points of R 1 (p 1 , . . . , p ) on one side of it and m − points of R 1 (p +1 , . . . , p ) on the other side of it. Applying the Euler-Jacobi formula to 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.
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 p 1 , p 2 the points in one branch of the hyperbola (ordered in clockwise sense) and by p 3 , . . . , p 2m the remaining points which are in the other branch of the hyperbola and ordered in counterclockwise sense. Applying the Euler-Jacobi formula to and taking into account the convex hull of p 1 , p 2 , p 3 , p 4 and the convex hull of p 5 , . . . , p 2m we get that p 1 and p 2 have different indices. Now applying again the Euler-Jacobi formula successively to L 3+j,2m−j , we get that p 1 and p 3 have the same index (because the number of straight lines is m − 2 which is odd) and so p 1 = p + 1 and p 2 = p − 2 . Finally, applying the Euler-Jacobi formula to Applying the Euler-Jacobi formula to we get that p 1 and p 5 have the same index (because the number of straight lines is m − 2 which is odd). In short p 1 = p + 1 and p 4 = p − 4 . Now we apply the Euler-Jacobi formula to with L m+2,m+2 being a straight line through the point p m+2 which separates the points {p 1 , p 2 , p 3 , p 4 } from the points {p 5 , . . . , p m+1 , p m+3 , . . . , p 2m }. Doing so, we get that p 2 and p 5 must have different indices (because the number of straight lines is m − 3 which is even) and so p 2 = p − 2 and p 3 = p + 3 . Finally, applying the Euler-Jacobi formula to for k = 0, . . . , m − 4 we get that p k = p + k for k odd and p k = p − k for k even. with k = 0, . . . , m − 2 we get that p 2+k and p 2m−k have the same index for k = 0, . . . , m − 2. Without loss of generality we can assume that p 2 = p + 2 and so p 2m = p + 2m . Now applying the Euler-Jacobi formula to we get that p 1 and p 2 have opposite indices (because the number of straight lines is m − 1 which is even). So p 1 = p − 1 and p m+1 = p + m+1 . Now applying the Euler-Jacobi formula to we get that p 1 and p 4 have opposite indices (because the number of straight lines is m − 1 which is even). So p 1 = p − 1 and p 3 = p − 3 . Now we apply successively the Euler-Jacobi formula to for k = 0, . . . , m − 4 and we get that p k = p − k for k odd with k = m + 2, and p k = p + k for k even with k = 2. Finally, applying the Euler-Jacobi formula to we get that p m+2 = p − m+2 and p 2 = p + 2 . In short the configuration is (2+, 2−; +, 2−; +, −, . . . , The configuration (Γ, Δ − , R 2m−7 ) can be realized intersecting the hyperbola x 2 − y 2 = 1 with the straight lines x = −2, −1, 1, 2, . . . , m − 2.

Proof of Statement (a) of Theorem 3
It was proved in [7] that in the case of polynomial vector fields of degree (2, m) with m even it holds that a∈A i X (a) = 0 or | a∈A i X (a)| = 2. This proves statement (a).
We will study each one of the configurations separately. 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 p 1 and p 2 have different indices and p 3+j and p 2m−j have also different indices for j = 0, . . . , m − 2. Without loss of generality we can assume that p 3 = p + 3 and p 2m = p − 2m . Moreover, applying the Euler-Jacobi formula to we get that p 1 and p 3 have opposite index (because the number of straight lines is m − 2 which is even) and so p 1 = p − 1 and p 2 = p + 2 . Now proceeding as in the final part of the proof of Theorem 2 we get that p j = p + j for j odd and p j = p − j for j even. Hence, the unique possible configuration is (R 4 ; +, −, . . . , +, −) = (R 4 , R 2m−4 ). Note that here a∈A i(a) = 0. The configuration (R 4 , R 2m−4 ) can be realized intersecting the hyperbola x 2 − y 2 = 1 by the m − 1 straight lines x = −m, 2, 3, . . . , m. 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 p 1 and p 4 have different indices, p 2 and p 3 have different indices and p 5+j and p 2m−j have different indices. Without loss of generality we can assume that p 5 = p + 5 and p 2m = p − 2m . Applying the Euler-Jacobi formula to we get that p 1 and p 2 must have opposite indices and so p 2 = p + 2 and p 3 = p − 3 . Now proceeding as in the last part of the proof of this configuration in Theorem 2 we get that p j = p + j for j odd and p j = p − j for j even. Hence, the unique possible configuration is (R 4 ; R 4 ; +, −, . . . , +, −) = (R 4 ; R 4 ; R 2m−8 ).
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 p 1 and p m+1 have different index, and p 2+j and p 2m−j have the same index for j = 0, . . . , m − 2. Without loss of generality we can assume that p 2 = p + 2 and so p 2m = p + 2m . Moreover, applying the Euler-Jacobi formula to C(x, y) = L m+1,m+2 m−2 j=1 L 2+j,2m−j+1 , we get that p 1 and p 2 have the same indices (because the number of straight lines is m − 1 which is odd). So p 1 = p + 1 and p m+1 = p − m+1 . Now proceeding as in the last part of the proof of Theorem 2 we get that p 2+j = p + 2+j and p 2m−j =