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On the modulus of solutions of a first order differential equation

Complex Variables 2026-01-16 v6

Abstract

Let P(z)=zn+an2zn2++a0P(z)=z^{n}+a_{n-2}z^{n-2}+\cdots+a_0 be a nonconstant polynomial and S(z)S(z) be a nonzero rational function and denote h(z)=S(z)eP(z)h(z)=S(z)e^{P(z)}. Let θ(0,π/2n)\theta\in(0,\pi/2n) be a constant and ε>0\varepsilon>0 be a small constant. It is shown that if f(z)f(z) is a solution of the first order differential equation f(z)=h(z)f(z)+1f'(z)=h(z)f(z)+1, then there is a sequence {rk}\{r_{k}\} such that the set E=l=0[r2l,r2l+1]E=\cup_{l=0}^{\infty}[r_{2l},r_{2l+1}] has infinite logarithmic measure and for all rEr\in E, \begin{equation}\tag{\dag} \begin{split} |f(re^{i\theta})|\geq (1-\varepsilon)\frac{\sqrt[n]{\sin n\theta}}{n}r\exp\left(e^{(1-\varepsilon)r^n\cos n\theta}\sin\varepsilon\right). \end{split} \end{equation} When h(z)=ezh(z)=e^{z}, we also give a lower bound for f(reiθ)|f(re^{i\theta})| for other values of rr. The estimate in ()(\dag) yields that the hyper-order ς(f)\varsigma(f) of f(z)f(z) is equal to nn, giving a partial answer to Br\"{u}ck'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.

Keywords

Cite

@article{arxiv.2407.00580,
  title  = {On the modulus of solutions of a first order differential equation},
  author = {Yueyang Zhang},
  journal= {arXiv preprint arXiv:2407.00580},
  year   = {2026}
}

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30 pages