On the Conditions of Kukles for the Existence of a Centre

For a class of cubic differential systems, Kukles established necessary and sufficient conditions for the existence of a centre. This paper presents a particular system that shows that either Kukles' conditions are incomplete, or the multiplicity of the origin as a focus is greater than or equal to 9. The second author described in [6] a mechanical procedure based on Poincare's classic method, and implemented a differential equations manipulation system DEMS for computing the local Liapunov functions and Liapunov constants for differential systems of centre and focus type. The applications of Wu's characteristic set method and Buchberger's Grobner basis method to the problems concerning the relations among the Liapunov constants and other conditions are described in [7]. In this paper, we consider a class of cubic differential systems. The relations between the Liapunov constants computed by DEMS, and the conditions given by Kukles for the existence of a centre, are studied by using the methods proposed in [7]. We thus present a particular cubic differential system that indicates that either the Kukles conditions [4] are incomplete or the multiplicity of the origin as a focus is greater than or equal to 9. For a class of differential systems of centre and focus type dx dy _, . , ~~x =y f = x + Q(xy) (DE) in which Q(x, y) is a polynomial of degree n beginning with terms of degree greater than 1, Kukles [3] established certain criteria for the existence of a centre. He applied his criteria to the cases when Q(x,y) is a polynomial of the third and fifth degrees and got the necessary and sufficient conditions for the existence of a centre. If Q(x,y) is of degree three and we write Q(x, y) = a20 x 2 + an xy + a02 y 2 + a30 x* + a21 x y + a12 xy 2 + a03 y \ Kukles [4] stated that the origin is a centre if and only if one of the following conditions holds: (1) •••,v'11 = vlx\a _0 consist of 3, 15, 53, 136 and 298 terms, respectively. With the order of variables as a02 < au < a20 < n "< 3o "< 2i> t n e Grobner basis (GB'), consisting of 11 polynomials, of (PS) = {v'z,...,v'n) is computed. The first one in (GB') may be factorized as g = a02 au(a20 + a02f (a12 a20 + 2a12 a02 + a20 a 2 02 + aj2). Computing the Grobner bases of the polynomial sets obtained from (GB') in replacing g by an, aQ2 and a20 + a02, respectively, we find out that all zeros of (GB') in these cases are (i) «2i = «n = 0, (ii) aQ2 = a12 = azo = a21 + a20 an = 0, (iii) a02 = a20 = a21 = 0 and (iv) a20 + aQ2 = a12 = a3Q = a2l = 0 . Computing then the Grobner basis (GB") of the polynomial set obtained from (GB') in replacing g by a12 a20 + 2al2 a02 + a20 a\2 + a\2, we find that it consists of 5 polynomials of which 4 polynomials have the factors an or a02. We compute further the Grobner basis of the polynomial set obtained from (GB") by removing the factors axx or a02 of the polynomials in it. This Grobner basis consists of three polynomials, as follows: ft = ^ ^ + 2a a + a20 a02 + a03, It is the same as the Grobner basis of (KS2) by removing some factors alx or a02. Note that the above (i), (iii) are the Kukles conditions (3), (4) and (ii), (iv) are special cases of the condition (2). Summing up these results, we obtain the following. ON THE CONDITIONS OF KUKLES FOR THE EXISTENCE OF A CENTRE 3 THEOREM 2. If a03 = 0, then v3 = • • • = vn = 0 if and only if one of the Kukles conditions (l)-(4) holds. The above theorem is also proved by the method of characteristic sets and the detailed discussion is omitted here. In what follows, we check the Kukles condition (1). Due to the restriction of our computer working space, we are unable to compute either the characteristic series or the Grobner basis of the polynomial set consisting of v3,...,vn. For verifying the correctness of the Kukles condition (1), we consider now the particular case a n = 0. Then the characteristic series is computed. We find in this case that i>3 = • • • = vlx = 0 does not imply that one of (l)-(4) holds; that is, there exist certain real numbers a{j such that v3 = • •• = vn = 0 but none of the conditions (l)-(4) holds. In particular, let tfn = 0, a12 = 0, a21 = -3« 0 3 , a20 = -\a02, a30 = ^ 0 2 Then the first two Liapunov constants v3 and vh are zero and the next three are 7 = ~i68oo3(288a03 — a02),