Classification of cubic differential systems with a linear center and the line at infinity of maximal multiplicity
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2023-01-08 08:23
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SUBA, Alexandru. Classification of cubic differential systems with a linear center and the line at infinity of maximal multiplicity. In: Conference on Applied and Industrial Mathematics: CAIM 2022, Ed. 29, 25-27 august 2022, Chişinău. Chișinău, Republica Moldova: Casa Editorial-Poligrafică „Bons Offices”, 2022, Ediţia a 29, pp. 74-75. ISBN 978-9975-81-074-6.
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Conference on Applied and Industrial Mathematics
Ediţia a 29, 2022
Conferința "Conference on Applied and Industrial Mathematics"
29, Chişinău, Moldova, 25-27 august 2022

Classification of cubic differential systems with a linear center and the line at infinity of maximal multiplicity


Pag. 74-75

Suba Alexandru
 
Vladimir Andrunachievici Institute of Mathematics and Computer Science
 
 
Disponibil în IBN: 20 decembrie 2022


Rezumat

We consider the real cubic system of differential equations x˙ = y + ax2 + cxy + fy2 + kx3 + mx2y + pxy2 + ry3 ≡ p (x, y) , y˙ = −(x + gx2 + dxy + by2 + sx3 + qx2y + nxy2 + ly3) ≡ q (x, y) , gcd(p, q) = 1, sx4 + (k+q)x3y + (m+n)x2y2 + (l+p)xy3 + ry4 ̸≡ 0 (1) and the homogeneous system associated to the system (1): {x˙ = P (x, y,Z) , y˙ = Q(x, y,Z)}, where P (x, y,Z) = yZ2 + (ax2 + cxy + fy2)Z + kx3 + mx2y + pxy2 + ry3, Q(x, y,Z) = −(xZ2 + (gx2 + dxy + by2)Z + sx3 + qx2y + nxy2+ +ly3). Denote X∞=P (x, y,Z) ∂ ∂x + Q(x, y,Z) ∂ ∂y . The linearized system (1) in critical point (0, 0) has a center in this point, i.e. (0, 0) is a linear center for (1). We say that for (1) the line at infinity Z = 0 has multiplicity ν + 1 if ν is the greatest positive integer such that Zν divides E∞ = PX∞(Q) − QX∞(P))[1]. Theorem 1. In the class of cubic differential systems of the form (1) the maximal multiplicity of the line at infinity is five.Theorem 2. The system (1) has the line at infinity of multiplicity five if and only if its coefficients verify one of the following three set of conditions: C = D = 0, B = −AS3/K3, F = −AS2/K2, G= AS/K, M = S, L = −S4/K3, N = R = −S3/K2, Q= −P = S2/K; (2) A=5F3/B2,C=−6F2/B,D=2F,G=−3F2/B,K=F5/B3,L=BF, M=−3F4/B2,N=−3F2,P=Q=3F3/B,R=−F2,S=−F4/B2, S ̸= 0; (3) A = −K2(2BK − 3FS)/S3,C = −2K(BK − 2FS)/S2,D = 2F, G = K(2FS − BK)/S2,L = S4/K3,M = 3S,N = 3S3/K2, P = 3S2/K,Q = 3S2/K,R = S3/K2, K2(BK − FS)2 + 4S5 = 0, (4) where A = g − b − c + i(a + d − f), C = 2(b + g + i(a + f)), F = c + g − b + i(a − d − f), K = r + s − m − n + i(k − l − p + q), M = n − m − 3(r − s) + i(3(k + l) + p + q), P = m + n + 3(r + s) + i(3(k − l) + p − q), R = m − n − r + s + i(k + l − p − q), B = A, D = C, G = F, L = K, N = M, Q = P, S = R, (5)