This is a preprint of: “The Euler-Jacobi formula and the planar quadratic-quartic polynomial differential systems”, Jaume Llibre, Clàudia Valls, Proc. Amer. Math. Soc., vol. 149(1), 135–141, 2021. DOI: [10.1090/proc/15257] THE EULER-JACOBI FORMULA AND THE PLANAR QUADRATIC-QUARTIC POLYNOMIAL DIFFERENTIAL SYSTEMS JAUME LLIBRE1 AND CLAUDIA VALLS2 Abstract. The Euler-Jacobi formula provides an algebraic relation be- tween the singular points of a polynomial vector field and their topo- logical indices. Using this formula we obtain the configuration of the singular points together with their topological indices for the planar quadratic–quartic polynomial differential systems when these systems have eight finite singular points. 1. Introduction and statement of the main results Consider the planar polynomial differential system (1) ẋ = P (x, y), ẏ = Q(x, y), in R2 where P (x, y) and Q(x, y) are real polynomials of degree n and m respectively. Assuming that system (1) has nm finite sigular points using the Euler-Jacobi formula we obtain an algebraic relation between the finite singular points of the polynomial differential system (1) and the topological indices of these finite singular points. A proof of the Euler-Jacobi formula can be found in [1]. It also follows from Bezout’s Theorem that in the complex projective plane, and taking into account all the multiplicities of the singular points, if the number of singular points is finite, then it is at most nm. The Euler- Jacobi formula deals with the case in which all the singular points have multiplicity one and are located in the finite part of the projective space. In the two-dimensional case this formula can be enunciated as follows. Consider a system of two real polynomials of degrees n and m respectively in the variables x and y. If the set of zeroes of that system (that we denote by A) contains exactly nm elements, then the Jacobian determinant   ∂P/∂x ∂P/∂y J = det ∂Q/∂x ∂Q/∂y 2010 Mathematics Subject Classification. Primary 34A05. Secondary 34C05, 37C10. Key words and phrases. Euler-Jacobi formula, singular points, topological index, polyno- mial differential systems. 1 2 J. LLIBRE AND C. VALLS evaluated at each zero does not vanish and for any polynomial R of degree less than or equal to n + m − 3 we have X R(a) (2) = 0. J(a) a∈A Using the Euler-Jacobi formula we want to characterize the number and distribution of the singular points of the quadratic–quartic polynomial dif- ferential systems, i.e. of system (1) with n = 2 and m = 4. Consider the polynomial vector field X = (P, Q) associated with the differential system (1) where the degree of P is 2 and the degree of Q is 4. We will call it a quadratic–quartic polynomial differential system. We denote by AX = A the set of {p ∈ R2 : X(p) = 0} of finite singular points. 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 is the natural positive number defined by Ki = #(Ai ∩ ∂ Âi ) We say that A has configuration (K0 ; K1 ; K2 ; . . . ; Kr ; ∗) if we do not specify for the values of Ki for i between r + 1 and q. We are also interested in the study of the (topological) indices of the singular points of X. We say that the singular points of X which belong to Ai ∩ ∂ Âi are on the i-th level. We recall that if we assume that #AX = 8 then the determinant of the Jacobian matrix J is non-zero at any singular point of the vector field X, consequently topological indices of the singular points are ±1, and in this case we substitute the number Ki corresponding to the i-th level by the vec- i −1 j tor (n1i +, n2i −, . . . , nm i +, nm i −) where ni are positive integers such that i P j j ni = Ki . More precisely, when Ai ∩ ∂ Âi is a polygon, these numbers take into account the number of consecutive points with positive and nega- tive indices, viewing the ith level oriented counterclockwise: n1i corresponds to the string with largest number of consecutive points with positive and negative 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 in- dices) is as large as possible. We continue the process for n2i and so on. Furthermore, when Ai ∩ ∂ Âi is a segment, the numbers take into account the number of consecutive points with positive indices, beginning at one of its endpoints. THE EULER-JACOBI FORMULA 3 With this notation we can state the main result of the paper. We denote by iX (a) the index of a singular point a ∈ A of a planar quadratic–quartic polynomial vector field X. Theorem 1. For planar quadratic–quartic polynomial differential systems such that #AX = 8, the following statements hold. P P (a) either a∈A iX (a) = 2 or a∈A iX (a) = 0. P (b) If a∈A iX (a) = 2, then only the following configurations are pos- sible (i) (4; 3; 1) = (4+; 3−; +) or (4−; 3+; −), (ii) (3; 5) = (3+; +, 2−, +, −) or (3−; +, −, +, −, +), andPthere exist examples of such configurations. (c) If a∈A iX (a) = 0, then only the following configurations are possi- ble (i) (8) = (+, −, +, −, +, −, +, −), (ii) (4; 4) = (+, −, +, −; +, −, +, −), and there exist examples of such configurations. The proof of Theorem 1 is given in section 2. We recall that for the planar quadratic-quadratic polynomial differential systems this theorem is the well-known Berinskii’s Theorem proved in [2] and reproved in [4] using the Euler-Jacobi formula. The case quadratic–cubic was also proved in [4], but the case cubic-cubic is much more difficult and is still open with only some partial results. 2. Proof of Theorem 1 In the proof of Theorem 1 we will use the following auxiliary result proved in [3]. Lemma 2. 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 1. First of all we observe that if a configuration exists for quadratic-quartic polynomial vector field X with #AX = 8 then it is possible to construct the same configuration but inter- changing points with index +1 with points with index −1. For doing that it is enough to take Y = (−P, Q) Pinstead of X = (P, Q). So we can restrict ourselves to the cases in which a∈A iX (a) ≥ 0. During the proof we will denote by {p1 , . . . , p8 } 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 , . . . , pl } the set of points of A with 4 J. LLIBRE AND C. VALLS negative index. Also we will denote by Lu,v u,v i,j the straight line Li,j (x, y) = 0 u v through the points pi and pj where u, v ∈ {+, −, ∅}, i ∈ {1, . . . , k}, j ∈ {1, . . . , l}, and by Lui a straight line through a point pui ∈ ∂ Â such that for all q ∈ A we have Lui (A) ≥ 0 and it is zero only at q. It was proved in [6, 5] that in the case of quadratic-quartic polynomial P P vector fields either a∈A iX (a) = 2, or a∈A iX (a) = 0, and by the P above explanation we can only consider that either P a∈A iX (a) = 2, or a∈A iX (a) = 0. This proves statement (a). 2.2. Proof of statement P (b) of Theorem 1. We prove Theorem 1 in the case in which a∈A iX (a) = 2. In this case we have that 5 points have positive index and three points have negative index. First we will show that there are no singular points with index −1 in ∂ Â and so that #(A∩∂ Â) ≤ 5, or equivalently, that the unique possible configurations are (K+, ∗) with K ≤ 5. Indeed, assume first that p− − − − 1 ∈ ∂ Â and denote by p1 , p2 , p3 the + points with negative index and by pi for i = 1, . . . , 5 the points with positive index. Consider the cubic C(x, y) = L− −− 2 1 (x, y)(L2,3 (x, y)) . Since, by the − −− + definitions of L1 and L2,3 we have C(Pi ) ≥ 0 for i = 1, . . . , 5 applying the Euler-Jacobi formula we get X C(P + ) i + =0 i=1,...,5 J(pi ) which is a contradiction because C(p+ + i ) ≥ 0, J(pi ) = 1 for i = 1, . . . , 5, and all of them cannot be zero because it is known that since the maximum degree of P and Q is four, if the five singular points are on the straight line L−− 23 (x, y) = 0 by Lemma 2 this straight line is full of singular points, which is not the case. This contradiction implies that #(A ∩ ∂ Â) ≤ 5 and the configuration of A must be (K+; ∗) with K ≤ 5. We now consider different cases. K = 5: Assume first that K = 5, i.e, #(A ∩ ∂ Â) = 5. Now, applying the Euler-Jacobi formula to C(x, y) = L++ ++ + 1,2 (x, y)L34 (x, y)L5 (x, y) we get to a contradiction. So this configuration is not possible. K = 4: Then #(A ∩ ∂ Â) = 4. Write {p+ + + + 1 , p2 , p3 , p4 } = A ∩ ∂ Â and take a conic C0 (x, y) through them. Since all these points are in the boundary of a convex set, the remaining four singular points are in the same connected component of R2 \{C0 (x, y) = 0}. Assume now that there is a point p+ 5 ∈ A1 , +− − i.e. in the 1st-level of A. Taking L5,k where pk is a point in A1 ∩∂ Â1 contigu- ous with p+ 5 , if the four singular points in A1 are not on a straight line then applying the Euler-Jacobi formula with C(x, y) = C0 (x, y)L+− 5,K , we get a contradiction. On the other hand, if the four singular points are on a straight THE EULER-JACOBI FORMULA 5 line, say L0 (x, y) = 0 then assuming without loss of generality that p+ 1 ∈ ∂ Â, applying the Euler-Jacobi formula to C(x, y) = L0 (x, y)L++ 34 (x, y)L + 1 (x, y) we get a contradiction. So the configuration of A is (4+; 3−; +). Note that the quadratic–quartic system (1) with P (x, y) = y 2 − x2 + 1, ! √ √  √ 4 15 2 + 4 15 x3 y 30 Q(x, y) = −x + √ √ + 1− √ √ x2 y 2 45 2 + 16 15 45 2 + 16 15 √ √  √ 2 4 15 2 + 4 15 xy 3 4 10y 4 4y 3 −x y − √ √  + √ √ + + y 2 + 1, 3 45 2 + 16 15 45 6 + 48 5 3 has the singular points √ √ √ √ √ √ (−2, 3), (−2, − 3), (−1, 0), (1, 0), (2, 3), (2, − 3), (3, −2 2), (4, 15), in the configuration (4+; 3−; +). K = 3: Now assume that #(A ∩ ∂ Â) = 3. Then we have the following possibilities: (3+; 5), (3+; 4; 1) or (3+; 3; 2). Since the polynomial P has degree 2, P (x, y) = 0 is a conic and the eight 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 point, one double straight line or one point) do not allow the configurations (3+; 4; 1) or (3+; 3; 2). Moreover, the configuration (3+; 5) only can be supported by a hyperbola. Now we study the configuration (3+; 5). Since all the singular points lie in a hyperbola and in the 1st-level of A we must have five points, it is clear that one point is in one branch of the hyperbola and the other seven points in the other branch of the hyperbola. Denote by p+ 1 the point in one branch of the hyperbola and by p2 , p3 , p4 , p5 , p6 , p7 , p8 the remaining points which are in the other branch of the hyperbola and ordered. Note that with this notation p2 = p+ + 2 and p8 = p8 . Applying the Euler-Jacobi formula to − C(x, y) = L++ + 28 L37 L46 we get that p1 and p5 have different signs so p5 = p5 . +− ++ Now applying the Euler-Jacobi formula to C(x, y) = L15 L28 L37 we get that p4 and p6 must have the same sign and applying the Euler-Jacobi formula to C(x, y) = L+− ++ 15 L28 L46 we get that p3 and p7 must have the − same sign, and applying the Euler-Jacobi formula to L+ 1 L5 L46 taking into account that p2 = p2 and p8 = p8 we get that p3 = p3 and p7 = p− + + − 7 . Then p4 = p−4 and p 5 = p − 5 and we get the configuration is (3+; +, 2−, +, −). The quadratic–quartic system (1) with P (x, y) = x2 − y 2 − 1, 9x3 y 26xy 3 8y 3 Q(x, y) = −x4 + + x2 y 2 − x2 y − + + y 2 + 1, 11 33 11 6 J. LLIBRE AND C. VALLS has the singular points √ √ √ √ √ √ (−1, 0), (−1, 0), (2, 3), (2, − 3), (3, 2 2), (3, −2 2), (4, 15), (4, − 15), in the configuration (3+; 5). Clearly configurations of the form (2+; ∗) cannot occur because the eight singular points would be on a straight line, and by Lemma 2 this straight line will be full of singular points, a contradiction. Moreover, configurations of the form (1+; ∗) have no meaning. This concludes the proof of statement (b) of Theorem 1. 2.3. ProofP of statement (c) of Theorem 1. We consider now the case in which a∈A iX (a) = 0. In this case we have 4 points with positive index and 4 points with negative one. We separate this study into different cases. Note that configurations (7; 1), (6; 2) and (5; 3) are not possible because any convex hull of seven, six or five 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, by the explanation in the proof of statement (b) we get that the configurations (3; 4; 1), (3; 3; 2), (2; ∗) and (1; ∗) are not possible. In short, the unique possible configurations are (8), (4; 4), (4; 3; 1) and (3; 5). We will study them separately. Configuration (8): Assume that the subscripts of the points in A are in such a way that p1 , p2 , p3 , p4 , p5 , p6 , p7 and p8 are ordered in ∂ Â in counterclockwise sense. Also we consider the subscripts in Z/8Z. Take the cubic Ci (x, y) = Li,i+1 (x, y)Li+2,i+3 (x, y)Li+4,i+5 . Then the Euler Jacobi formula applied to Ci yields Ci (pi+6 ) Ci (pi+7 ) + = 0, J(pi+6 ) J(pi+7 ) so J(pi+6 )J(pi+7 ) < 0 for all i. Hence the indices of Pj and pj+1 are different and the configuration of A must be (8) = (+, −, +, −, +, −, +, −). The quadratic–quartic system (1) with P (x, y) = x2 + y 2 − 1, Q(x, y) = −2x4 + x3 y + x2 y 2 − 3xy 3 + 3y 4 − x3 − x2 y − xy 2 − y 3 + x2 +xy − 4y 2 + x + y + 1, has the singular points 1 1  (0, ±1), (±1, 0), ± √ , ±√ 2 2 in the configuration (+, −, +, −, +, −, +, −). Configuration (4; 4): Denote by p1 , . . . , p4 the points in A ∩ ∂ Â ordered in counterclockwise sense, and by p5 , . . . , p8 the points in A ∩ ∂ Â1 also ordered in counterclockwise sense. Applying the Euler-Jacobi formula to C(x, y) = THE EULER-JACOBI FORMULA 7 L12 L34 L5,6 , C(x, y) = L12 L34 L67 , and C(x, y) = L12 L34 L78 we get that the configuration is (4; +, −, +, −). Moreover we have that in ∂A there are two points with positive index and two points with negative one. We note that the configuration (4; 4) only can be realized in a hyperbola or in a conic formed by two straight lines intersecting in a point. Then applying three times the Euler-Jacobi formula to a cubic formed by the product of three straight lines, two of them defined by two non-contiguous sides of the quadrilateral defined by the boundary of ∂ Â1 and the third straight line defined by one side of the quadrilateral defined by the boundary of ∂ Â, we obtain that the configuration (2+, 2−; +, −, +, −) is not possible. In summary the unique possible configuration is (+, −, +, −; +, −, +, −). The quadratic–quartic system (1) with P (x, y) = x2 − y 2 − 1, 23x2 y 2 Q(x, y) = −x4 + − x2 y − y 4 + y 3 + y 2 + y − 1, 12 √ √ has the eight singular points (±2, ± 3), (±3, ±2 2), in the configuration (+, −, +, −; +, −, +, −). Configuration (4; 3; 1): The arguments used in the proof of case K = 3 of statement (b) of Theorem 1 applied to the configuration (4; 3; 1) show that such a configuration only can be realized if the conic P (x, y) = 0 is a hyperbola. So in the following arguments we take into account that the eight singular points of system (1) are in a hyperbola. Denote by p1 , . . . , p4 the points in A ∩ ∂ Â ordered in counterclockwise sense and by p5 , p6 , p7 the points in the 1st-level of A and p8 in the 2nd-level of A. Applying the Euler- Jacobi formula, iteratively, to C(x, y) = L12 L34 L56 , C(x, y) = L12 L34 L67 and C(x, y) = L12 L34 L57 we get that, without loss of generality, p8 = p+8 and p5 = p−5 , p 6 = p− 6 and p7 = p − 7 . So, we have three points with index + and one point with index − in ∂ Â. By the previous consideration we can assume that ∂ Â is a quadrilateral. Take L++ + k,k+1 where pk are two contiguous points in ∂ Â. Applying the Euler-Jacobi formula to C(x, y) = L−+ −− ++ 5,8 L7,6 Lk,k+1 we reach a contradiction because the remaining points in ∂ Â are in different sides of L++ k,k+1 and have different signs. So, the configuration (4; 3; 1) is not possible. Configuration (3; 5): As in the proof of the case K = 3 of statement (b) of Theorem 1 all the singular points lie in a hyperbola with one point in one branch of the hyperbola and the other seven points in the other branch of the hyperbola. Denote by p1 the point in one branch of the hyperbola and by p2 , p3 , p4 , p5 , p6 , p7 , p8 the remaining points which are in the other branch of the hyperbola and ordered. Applying the Euler-Jacobi formula to C(x, y) = L28 L37 L46 we get that p1 and p5 have different signs so we − can assume without loss of generality that p1 = p+ 1 and p5 = p5 . Now 8 J. LLIBRE AND C. VALLS applying again the Euler-Jacobi formula to C(x, y) = L+− 15 L28 L37 we get that p4 and p6 must have the same sign, applying the Euler-Jacobi formula to C(x, y) = L+−15 L28 L46 we get that p3 and p7 must have the same sign, and finally applying the Euler-Jacobi formula to C(x, y) = L+−15 L37 L46 we get that p2 and p8 must have the same sign, but then the number of positive and negative indices would be odd which is not possible. In short, such a configuration is not possible. This proves statement (c) of Theorem 1 and concludes the proof of the theorem. 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] A. G. Khovanskii, Index of a polynomial vector field, Funktsional Anal. i Prilozhen 13 (1979), 49–58. 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