关于一阶微分方程解的模
On the modulus of solutions of a first order differential equation
AI总结：
本文研究了一阶微分方程解的模估计，证明了解的超阶为n，并给出了二阶代数微分方程解的增长阶描述。
摘要：
设$P(z)=z^{n}+a_{n-2}z^{n-2}+\cdots+a_0$为非常数多项式，$S(z)$为非零有理函数，令$h(z)=S(z)e^{P(z)}$。令$θ∈(0,π/2n)$为常数，$ε>0$为小常数。证明如果$f(z)$是一阶微分方程$f'(z)=h(z)f(z)+1$的解，则存在一个序列$\{r_{k}\}$，使得集合$E=\cup_{l=0}^\infty[r_{2l},r_{2l+1}]$有无限对数测度，并且对于所有$r∈E$，有$$|f(re^{iθ})|\geq (1-\varepsilon)\frac{\sqrt[n]{\sin nθ}}{n}r\exp\left(e^{(1-\varepsilon)r^n\cos nθ}\sin\varepsilon\right).$$当$h(z)=e^{z}$时，我们还给出了对于其他$r值的$|f(re^{iθ})|$的下界。估计$(†)$表明$f(z)$的超阶$ς(f)$等于$n$，给出了Brück在全纯函数唯一性理论中的猜想的部分答案。该方法的扩展还给出了Hayman自主情况下二阶代数微分方程的整个解的增长阶的完整描述。
英文摘要：
Let $P(z)=z^{n}+a_{n-2}z^{n-2}+\cdots+a_0$ be a nonconstant polynomial and $S(z)$ be a nonzero rational function and denote $h(z)=S(z)e^{P(z)}$. Let $θ\in(0,π/2n)$ be a constant and $\varepsilon>0$ be a small constant. It is shown that if $f(z)$ is a solution of the first order differential equation $f'(z)=h(z)f(z)+1$, then there is a sequence $\{r_{k}\}$ such that the set $E=\cup_{l=0}^{\infty}[r_{2l},r_{2l+1}]$ has infinite logarithmic measure and for all $r\in E$, \begin{equation}\tag† \begin{split} |f(re^{iθ})|\geq (1-\varepsilon)\frac{\sqrt[n]{\sin nθ}}{n}r\exp\left(e^{(1-\varepsilon)r^n\cos nθ}\sin\varepsilon\right). \end{split} \end{equation} When $h(z)=e^{z}$, we also give a lower bound for $|f(re^{iθ})|$ for other values of $r$. The estimate in $(†)$ yields that the hyper-order $ς(f)$ of $f(z)$ is equal to $n$, giving a partial answer to Brück's conjecture in uniqueness theory of meromorphic functions. An extension of the method also yields a complete description on the order of growth of entire solutions of a second order algebraic differential equation of Hayman in the autonomous case.
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