摘要
本文证明了：如果ａｋ－ｊ（ｊ＝１，…，ｋ）为多项式，ｄｅｇａｋ－ｊ＝ｎｋ－ｊ，存在某个ａｋ－ｓ（１≤ｓ≤ｋ）满足：当１≤ｊ＜ｓ时，ｎｋ－ｊ／ｊ≤ｎｋ－ｓ／ｓ；当ｓ＜ｊ≤ｋ时，ｎｋ－ｊ＜ｎｋ－ｓ－（ｊ－ｓ）．如果Ｆ０是整函数且满足σ（Ｆ）＝β＜（ｎｋ－ｓ＋ｓ）／ｓ，那么微分方程ｆ（ｋ）＋ａｋ－１ｆ（ｋ－１）＋…＋ａ０ｆ＝Ｆ的解满足λ（ｆ）＝λ（ｆ）＝σ（ｆ）＝（ｎｋ－ｓ＋ｓ）／ｓ。
In this paper, we prove: if a k-j (j=1,…,k) are polynomials, dega k-j =n k-j , there exists some a k-s (1≤s≤k) such that n k-j /j≤n k-s /s if 1≤j<s; and n k-j <n k-s -(j-s) if s<j≤k, if F0 is an entire function satisfying σ(F)=β<(n k-s +s)/s, then a solution of the differential equationf (k) +a k-1 f (k-1) +…+a 0f=Fsatisfies (f)=λ(f)=σ(f)=(n k-s +s)/s, or σ(f)=β.
基金
国家自然科学基金
江西省自然科学基金
关键词
微分方程
零点
复振荡
多项式系数
non homogeneous linear differential equation, entire function, zero sequence, order of growth.
