一类四次Hamilton函数Abel积分零点个数的估计

ESTIMATION OF THE NUMBER OF ZEROS OF ABELIAN INTEGRAL FOR A KIND OF QUARTIC HAMILTONIANS

证明了Abel积分I(h)=∮_(Γ_h)Q(x,y)dx-P(x,y)dy的零点个数的最小上界B(2n+2)= B(2n+1)≤3[n/2]+12[(n-1)/2]+4([p]表示p的整数部分),这里Γ_h是代数曲线H(x,y)= x^2±x^4+y^4=h的连通闭分支,h∈∑(Γ_h存在的最大开区间),P(x,y),Q(x,y)是关于x,y的次数不超过2n+2或2n+1的实多项式.

It is proved that the supremum of the number of zeros of Abelian integral I（h） = ∮Гh Q（x,y）dx - P（x,y）dy satisfies B（2n ＋ 2） = B（2n ＋ 1） 〈 3[n/2] ＋ 12[n-1/4] ＋ 4 （[p] denotes the entire part of p ）, where Гh is a compact component of H（x, y） = x^2 ± x^4 ± y^4 = h, h ∈∑ （a maximal open interval of Fh）, P（x, y）, Q（x, y） are real polynomials of x, y with degree not greater than 2n ＋ 2 or 2n ＋ 1.