\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 54, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/54\hfil Growth of solutions] {Growth of solutions of complex differential equations with coefficients of finite iterated order} \author[J. Tu, Z. Chen, X. Zheng\hfil EJDE-2006/54\hfilneg] {Jin Tu, Zongxuan Chen, Xiumin Zheng} % in alphabetical order \address{Jin Tu \newline School of Mathematical Sciences, Beijing Normal University, Beijing, 100875, China} \email{tujin2008@sina.com} \address{Zongxuan Chen \newline Department of Mathematics, South China Normal University, Guangzhou, 510631, China} \email{chzx@sina.com} \address{Xiumin Zheng \newline Institute of Mathematics and Information, Jiangxi Normal University, Nanchang, 330027, China} \email{xiaogui88@sohu.com} \date{} \thanks{Submitted January 17, 2006. Published April 28, 2006.} \thanks{Supported by grant 10371009 from the Natural Science Foundation of China} \subjclass[2000]{30D35, 34M10} \keywords{Differential equations; growth of solutions; iterated order} \begin{abstract} In this paper, we investigate the growth of solutions to the differential equation $$ f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_0(z)f=F(z), $$ where the coefficients are of finite iterated order. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \section{Introduction} It is well known that all solutions of the complex differential equations \begin{gather} f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_0(z)f=0, \label{e1.1}\\ f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_0(z)f=F(z) \label{e1.2} \end{gather} are entire functions, provided that the coefficients $A_0(z),A_1(z),\dots,A_{k-1}(z),F(z)$ are entire functions with $A_0(z)\not\equiv 0$. A natural question arises: What conditions on $A_0(z),A_1(z),\dots,A_{k-1}(z),F(z)$ will guarantee that every solution $f\not\equiv0$ has infinite order? Also: For solutions of infinite order, how to express the growth of them explicitly, it is a very important problem. Partial results have been available since a paper of Frei \cite{f1}. For high order differential equations, the following results have been obtained. \begin{theorem}[{\cite[Theorem 2.1]{b3}}] \label{thm1.1} Let $A_0(z),A_1(z),\dots,A_{k-1}(z)$ be entire functions with $A_0(z)\not\equiv0$, such that for some real constants $\alpha,\beta,\mu,\theta_1,\theta_2$, with $0\leq\beta<\alpha, \mu>0, \theta_1<\theta_2$, we have \begin{gather} |A_0(z)|\geq e^{\alpha|z|^{\mu}},\label{e1.3}\\ |A_j(z)|\leq e^{\beta|z|^{\mu}},\quad j=1,\dots,k-1,\label{e1.4} \end{gather} as $z\to\infty$ with $\theta_1\leq \arg z\leq\theta_2$. Then every solution $f\not\equiv0$ of \eqref{e1.1} has infinite order. \end{theorem} \begin{theorem}[{\cite[Theorem 1]{b1}}] \label{thm1.2} Let $H$ be a set of complex numbers satisfying $\overline{\mathop{\rm dens}}\{|z|:z\in H\}>0$, and let $A_0(z),A_1(z),\dots,A_{k-1}(z)$ be entire functions and satisfy \eqref{e1.3} and \eqref{e1.4} as $z\to\infty$ for $z\in H$. Then every solution $f\not\equiv0$ of \eqref{e1.1} satisfies $\sigma(f)=\infty$ and $\sigma_2(f)\geq\mu$. \end{theorem} \begin{theorem}[{\cite[Theorem 2]{b1}}] \label{thm1.3} Let $H$ be a set of complex numbers satisfying $\overline{\mathop{\rm dens}}\{|z|:z\in H\}>0$, and let $A_0(z),A_1(z),\dots,A_{k-1}(z)$ be entire functions with $\max\{\sigma (A_j):j=1,\dots,k-1\}\leq\sigma(A_0)=\sigma<+\infty$ such that for some constants $0\leq\beta<\alpha$ and for any $\varepsilon>0$, we have \begin{gather} |A_0(z)|\geq e^{\alpha|z|^{\sigma-\varepsilon}} ,\label{e1.5}\\ |A_j(z)|\leq e^{\beta|z|^{\sigma-\varepsilon}},\quad j=1,\dots,k-1,\label{e1.6} \end{gather} as $z\to\infty$ for $z\in H$. Then every solution $f\not\equiv0$ of \eqref{e1.1} satisfies $\sigma(f)=\infty$ and $\sigma_2(f)=\sigma(A_0)$. \end{theorem} \begin{theorem}[{\cite[Theorem 1.1]{b2}}] \label{thm1.4} Let $H,A_0(z),A_1(z),\dots,A_{k-1}(z)$ satisfy the hypotheses of Theorem \ref{thm1.3}, and let $F\not\equiv0$ be an entire function with $\sigma(F)<+\infty$. Then every solution $f(z)$ of \eqref{e1.2} satisfies $\overline{\lambda}_2(f)=\sigma_2(f)=\sigma$, with at most one exceptional solution $f_0$ satisfying $\sigma_2(f_0)<\sigma$. \end{theorem} \section{Notation and results} In this section, we prove some results concerning the above questions when the coefficients of \eqref{e1.1} and \eqref{e1.2} are of finite iterated order. For $r\in[0,\infty)$, we define $\exp_{1}r=e^{r}$ and $\exp_{i+1}r=\exp(\exp_{i}r)$ ($i\in \mathbb{N}$). For $r$ sufficiently large, we define $\log_{1}r=\log r$, $\log_{i+1}r=\log(\log_{i}r)$ ($i\in \mathbb{N}$). To express the rate of growth of entire function of infinite order, we introduce the notion of iterated order \cite{k1}. \begin{definition} \label{def2.1} \rm The iterated $i$-order of an entire function $f$ is defined by \begin{equation} \sigma_{i}(f)=\limsup_{r\to\infty} \frac{\log_{i+1}M(r,f)}{\log r} =\limsup_{r\to\infty} \frac{\log_{i}T(r,f)}{\log r}\quad(i\in \mathbb{N}). \label{e2.1} \end{equation} \end{definition} \begin{definition} \label{def2.2} \rm The finiteness degree of the order of an entire function $f$ is defined by \begin{equation} i(f)=\begin{cases} 0 & \text{if $f$ is a polynomial}, \\ \min\{j\in \mathbb{N}:\sigma _{j}(f)<\infty\} & \text{if $f$ is transcendental with}\\ &\quad \sigma_{j}(f)<\infty \text{ for some }j\in\mathbb{N}, \\ \infty & \text{if } \sigma _{j}(f)=\infty\; \forall j\in \mathbb{N}. \end{cases} \label{e2.2} \end{equation} \end{definition} \begin{definition} \label{def2.3} \rm The iterated convergence exponent of the sequence of zeros of an entire function $f$ is defined by \begin{equation} \lambda_{i}(f)=\limsup_{r\to \infty} \frac{\log_{i}n(r,1/f)}{\log r}\quad(i\in \mathbb{N}).\label{e2.3} \end{equation} \end{definition} The linear measure of a set $E\subset[0,+\infty)$ is defined as $m(E)=\int^{+\infty}_0\chi_E(t)\,dt$. The logarithmic measure of a set $E\subset[1,+\infty)$ is defined by $lm(E)=\int^{+\infty}_1 \chi_E(t)/t\, dt$, where $\chi_E(t)$ is the characteristic function of $E$. The upper and lower densities of $E$ are \begin{equation} \overline{\mathop{\rm dens}}E =\limsup_{r\to\infty} \frac{m(E\cap[0,r])}{r},\quad \underline{\mathop{\rm dens}}E=\liminf_{r\to\infty} \frac{m(E\cap[0,r])}{r}.\label{e2.4} \end{equation} In this paper, we improve the results of Bela\"{\i}di \cite{b1,b2,b3}, and we obtain the following results: \begin{theorem} \label{thm2.1} Let $A_0(z),A_1(z),\dots,A_{k-1}(z)$ be entire functions with $A_0(z)\not\equiv0$ such that for real constants $\alpha,\beta,\mu,\theta_1,\theta_2$ and positive integer $p$ with $0\leq\beta<\alpha,\mu>0,\theta_1<\theta_2,1\leq p<\infty$, we have \begin{gather} |A_0(z)|\geq \exp_p\{\alpha|z|^{\mu}\}, \label{e2.5} \\ |A_j(z)|\leq \exp_p\{\beta|z|^{\mu}\},\quad j=1,\dots,k-1,\label{e2.6} \end{gather} as $z\to\infty$ with $\theta_1\leq argz\leq\theta_2$. Then $\sigma_{p+1}(f)\geq\mu$ holds for all non-trivial solutions of \eqref{e1.1}. \end{theorem} \begin{theorem} \label{thm2.2} Let $H$ be a set of complex numbers satisfying $\overline{\mathop{\rm dens}}\{|z|:z\in H\}>0$, and let $A_0(z)$, $A_1(z),\dots,A_{k-1}(z)$ be entire functions and satisfy \eqref{e2.5} and \eqref{e2.6} as $z\to\infty$ for $z\in H$, where $0\leq\beta<\alpha,\mu>0$, $1\leq p<\infty$. Then every solution $f\not\equiv0$ of \eqref{e1.1} satisfies $\sigma_{p+1}(f)\geq\mu$. \end{theorem} \begin{theorem} \label{thm2.3} Let $H$ be a set of complex numbers satisfying $\overline{\mathop{\rm dens}}\{|z|:z\in H\}>0$, and let $A_0(z),A_1(z),\dots,A_{k-1}(z)$ be entire functions of iterated order with $\max\{\sigma_p(A_j):j=1,\dots,k-1\}\leq\sigma_p(A_0)=\sigma<+\infty$, $1\leq p<\infty$ such that for some constants $0\leq\beta<\alpha$ and for any given $\varepsilon>0$, we have \begin{gather} |A_0(z)|\geq \exp_p\{\alpha|z|^{\sigma-\varepsilon}\}\label{e2.7} \\ |A_j(z)|\leq \exp_p\{\beta|z|^{\sigma-\varepsilon}\},\quad j=1,\dots,k-1,\label{e2.8} \end{gather} as $z\to\infty$ for $z\in H$. Then every solution $f\not\equiv0$ of \eqref{e1.1} satisfies $\sigma_{p+1}(f)=\sigma_{p}(A_0)=\sigma$. \end{theorem} \begin{theorem} \label{thm2.4} Let $H,A_0(z)$, $A_1(z),\dots,A_{k-1}(z)$ satisfy the hypotheses of Theorem \ref{thm2.3}, and let $F\not\equiv0$ be an entire function of iterated order with $i(F)=q$. \begin{itemize} \item[(i)] If $qp+1$ or $q=p+1, \sigma_p(A_0)<\sigma_{p+1}(F)<+\infty$, then every solution $f(z)$ of \eqref{e1.2} satisfies $i(f)=q$ and $\sigma_{q}(f)=\sigma_q(F)$. \end{itemize} \end{theorem} \section{Preliminaries for proving the main results} To prove the above theorems, we need the following lemmas: \begin{lemma}[\cite{g1}] \label{lem3.1} Let $f(z)$ be a nontrivial entire function, and let $\alpha>1$ and $\varepsilon>0$ be given constants. Then there exist a constant $c>0$ and a set $E_1\subset[0,\infty)$ having finite linear measure such that for all $z$ satisfying $|z|=r\not\in E_1$, we have \begin{equation} \big|\frac{f^{(k)}(z)}{f(z)}\big| \leq c[T(\alpha r,f)r^\varepsilon\log T(\alpha r,f)]^k\quad (k\in \mathbb{N}).\label{e3.1} \end{equation} \end{lemma} \begin{lemma}[Wiman-Valiron \cite{h1,v1}] \label{lem3.2} Let $f(z)$ be a transcendental entire function, and let $z$ be a point with $|z|=r$ at which $|f(z)|=M(r,f)$. Then for all $|z|$ outside a set $E_2$ of $r$ of finite logarithmic measure, we have \begin{equation} \frac{f^{(k)}(z)}{f(z)}=\Big(\frac{\nu_{f}(r)}{z}\Big)^{k} (1+o(1)) \quad(k\in \mathbb{N},r\not\in E_2).\label{e3.2} \end{equation} where $\nu_f(r)$ is the central index of $f$. \end{lemma} \begin{lemma}[\cite{h2}] \label{lem3.3} Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be an entire function, $\mu(r)$ be the maximum term, i.e. $\mu(r)=\max\{|a_n|r^n; n=0,1,\dots\}$, and let $\nu_f(r)$ be the central index of $f$. Then \begin{itemize} \item[(i)] For $|a_{0}|\neq 0$, \begin{equation} \log\mu(r)=\log|a_{0}|+\int_{0}^{r}\frac{\nu_{f}(t)}{t}dt,\label{e3.3} \end{equation} \item[(ii)] For $r0$ is a constant. By \eqref{e2.1} and \eqref{e3.7}, \begin{equation} \limsup_{r\to\infty} \frac{\log_{p+1}\nu_{f}(r)}{\log r}\leq\limsup_{r\to\infty} \frac{\log_{p+2}M(r,f)}{\log r}=\sigma.\label{e3.8} \end{equation} On the other hand, from \eqref{e3.4}, we have \begin{equation} M(r,f)<\mu(r)\{\nu_{f}(2r)+2\} =|a_{\nu_{f}(r)}|r^{\nu_{f}(r)}\{\nu_{f}(2r)+2\}, \label{e3.9} \end{equation} Since $\{|a_n|\}$ is a bounded sequence, we have \begin{equation} \log_{p+2}M(r,f)\leq\log_{p+1}\nu_{f}(2r) \big[1+\frac{\log_{p+2}\nu_{f}(2r)} {\log_{p+1}\nu_{f}(2r)}\big]+\log_{p+2}r+c_{2},\label{e3.10} \end{equation} where $c_2>0$ is a constant. Hence \begin{equation} \sigma=\limsup_{r\to\infty} \frac{\log_{p+2}M(r,f)}{\log r} \leq\limsup_{r\to\infty} \frac{\log_{p+1}\nu_{f}(2r)}{\log2r}= \limsup_{r\to\infty} \frac{\log_{p+1}\nu_{f}(r)}{\log r}. \label{e3.11} \end{equation} From \eqref{e3.8} and \eqref{e3.11}, we obtain the conclusion \eqref{e3.5}. \end{proof} \begin{lemma}[\cite{k1}] \label{lem3.5} Let $f(z)$ be an entire function with $i(f)=p+1$, then \begin{equation} \sigma_{p+1}(f)=\sigma_{p+1}(f'). \label{e3.12} \end{equation} \end{lemma} \begin{lemma} \label{lem3.6} Let $A_0(z),\dots,A_{k-1}(z)$ be entire functions, with $,F\not\equiv0$ and let $f(z)$ be a solution of \eqref{e1.2} satisfying one of the following conditions: \begin{itemize} \item[(i)] $\max\{i(F)=q,i(A_{j})(j=0,\dots,k-1)\}< i(f)=p+1$ ($1\leq p<\infty$), \item[(ii)] $\max\{\sigma_{p}(F),\sigma_{p}(A_{j}) (j=0,\ldots,k-1)\}<\sigma_{p+1}(f)=\sigma$. \end{itemize} Then $\overline{\lambda}_{p+1}(f)=\lambda_{p+1}(f)=\sigma_{p+1}(f)=\sigma$. \end{lemma} \begin{proof} From \eqref{e1.2}, we have \begin{equation} \frac{1}{f}=\frac{1}{F}\Big(\frac{f^{(k)}}{f}+A_{k-1}\frac{f^{(k-1)}}{f} +\dots+A_{0}\Big),\label{e3.13} \end{equation} it is easy to see that if $f$ has a zero at $z_0$ of order $\alpha(>k)$, then $F$ must have a zero at $z_0$ of order $\alpha-k$, hence \begin{gather} n(r,\frac{1}{f})\leq k\overline{n}(r,\frac{1}{f})+n(r,\frac{1}{F}), \label{e3.14} \\ N(r,\frac{1}{f})\leq k\overline{N}(r,\frac{1}{f})+N(r,\frac{1}{F}). \label{e3.15} \end{gather} By \eqref{e3.13}, we have \begin{equation} m(r,\frac{1}{f})\leq m(r,\frac{1}{F})+\sum_{j=0}^{k-1}m(r,A_{j}) +O\left(\log T(r,f)+\log r\right) (r\not\in E_3),\label{e3.16} \end{equation} where $E_3$ is a subset of $r$ of finite linear measure. By \eqref{e3.15} and \eqref{e3.16}, for $r\not\in E_3$, we get \begin{equation} T(r,f)=T(r,\frac{1}{f})+O(1) \leq k \overline{N}(r,\frac{1}{f})+T(r,F)+\sum_ {j=0}^{k-1}T(r,A_{j})+O\{\log(r T(r,f))\}.\label{e3.17} \end{equation} For sufficiently large $r$, we have \begin{gather} O\{\log r +\log T(r,f)\}\leq\frac{1}{2}T(r,f),\label{e3.18}\\ T(r,A_{0})+\dots+T(r,A_{k-1})\leq k\exp_{p-1}\{r^{\sigma+\varepsilon}\}, \label{e3.19}\\ T(r,F)\leq\exp_{p-1}\{r^{\sigma(F)+\varepsilon}\}.\label{e3.20} \end{gather} Thus, by \eqref{e3.17}-\eqref{e3.20}, for $r\not\in E_3$, we have \begin{equation} T(r,f)\leq 2k\overline{N}(r,\frac{1}{f}) +2k\exp_{p-1}\{r^{\sigma+\varepsilon}\} +2\exp_{p-1}\{r^{\sigma(F)+\varepsilon}\} .\label{e3.21} \end{equation} Hence for any $f$ with $\sigma_{p+1}(f)=\sigma$, by \eqref{e3.21}, we have $\sigma_{p+1}(f)\leq\overline{\lambda}_{p+1}(f)$. Therefore, $\overline{\lambda}_{p+1}(f)=\lambda_{p+1}(f)=\sigma_{p+1}(f)=\sigma$. \end{proof} \section{Proofs of theorems} \begin{proof}[Proof of Theorem \ref{thm2.1}] Let $f$ be a solution of \eqref{e1.1}, and rewritten \eqref{e1.1} as \begin{equation} A_{0}=-\Big(\frac{f^{(k)}}{f}+A_{k-1}\frac{f^{(k-1)}}{f} +\dots+A_{1}\frac{f'}{f}\Big). \label{e4.1} \end{equation} By Lemma \ref{lem3.1}, there exist a constant $c>0$ and a set $E_1\subset[0,\infty)$ having finite linear measure such that $|z|=r\not\in E_1$ for all $z=re^{i\theta}$. Then we have \begin{equation} \Big|\frac{f^{(j)}(z)}{f(z)}\Big|\leq c[rT(2r,f)]^{2k},\quad j=1,\dots,k-1.\label{e4.2} \end{equation} By \eqref{e4.1}, \eqref{e4.2} and the hypothesis of Theorem \ref{thm2.1}, we get \begin{equation} \exp_p\{\alpha|z|^{\mu}\}\leq|A_0(z)|\leq k\exp_p\{\beta|z|^{\mu}\} c[rT(2r,f)]^{2k}\label{e4.3} \end{equation} as $z\to\infty$ with $|z|=r\not\in E_1, \theta_1\leq\arg z=\theta\leq\theta_2$. By \eqref{e4.3} and \eqref{e2.1}, we have $\sigma_{p+1}(f)\geq\mu$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.2}] From \eqref{e1.1}, it follows that \begin{equation} |A_0(z)|\leq \big|\frac{f^{(k)}(z)}{f(z)}\big| +|A_{k-1}(z)|\big|\frac{f^{(k-1)}(z)}{f(z)}\big| +\dots+|A_1(z)|\big|\frac{f^{\prime}(z)}{f(z)}\big|. \label{e4.4} \end{equation} By the hypotheses of Theorem \ref{thm2.2}, there exists a set $H$ with $\overline{\mathop{\rm dens}}\{|z|:z\in H\}>0$ such that for all $z$ satisfying $z\in H$, we have \begin{gather} |A_0(z)|\geq \exp_p\{\alpha|z|^\mu\}, \label{e4.5} \\ |A_j(z)|\leq \exp_p\{\beta|z|^\mu\},\quad j=1,\dots,k-1, \label{e4.6} \end{gather} as $z\to\infty$. Hence from \eqref{e4.2}, \eqref{e4.4}-\eqref{e4.6}, it follows that for all $z$ satisfying $z\in H$ and $z\not\in E_1$, we have \begin{equation} \exp_p\{\alpha|z|^{\mu}\}\leq k\exp_p\{\beta|z|^{\mu}\}c[rT(2r,f)]^{2k} \label{e4.7} \end{equation} as $z\to\infty$. Thus, there exists a set $H_1=H\setminus E_1$ with $\overline{\mathop{\rm dens}}\{|z|:z\in H_1\}>0$ such that \begin{equation} \exp_p\{(\alpha-\beta)|z|^{\mu}\}\leq kc[rT(2r,f)]^{2k} \label{e4.8} \end{equation} as $z\to\infty$. Therefore, by \eqref{e4.8} and Definition \ref{def2.1}, we obtain $\sigma_{p+1}(f)\geq\mu$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.3}] By Theorem \ref{thm2.2}, we have $\sigma_{p+1}(f)\geq\sigma-\varepsilon$, since $\varepsilon$ is arbitrary, we get $\sigma_{p+1}(f)\geq\sigma_p(A_0)=\sigma$. On the other hand, by Lemma \ref{lem3.2}, there exists a set $E_2\subset[1,\infty)$ having finite logarithmic measure such that \eqref{e3.2} holds for all $z$ satisfying $|z|=r\not\in [0,1]\bigcup E_2$ and $|f(z)|=M(r,f)$. By Definition \ref{def2.1}, for any given $\varepsilon>0$ and for sufficiently large $r$, we have \begin{equation} |A_j(z)|\leq \exp_p\{r^{\sigma+\varepsilon}\},\quad j=0,1,\dots,k-1.\label{e4.9} \end{equation} Substituting \eqref{e3.2} and \eqref{e4.9} in \eqref{e1.1}, for all $z$ satisfying $|z|=r\not\in [0,1]\bigcup E_2$ and $|f(z)|=M(r,f)$, we have \begin{equation} \big(\frac{\nu_{f}(r)}{|z|}\big)^{k}|1+o(1)| \leq k\big(\frac{\nu_{f}(r)}{|z|} \big)^{k-1}|1+o(1)| \exp_{p}\{r^{\sigma+\varepsilon}\}. \label{e4.10} \end{equation} By \eqref{e4.10}, we get \begin{equation} \limsup_{r\to\infty} \frac{\log_{p+1}\nu_{f}(r)}{\log r}\leq \sigma +\varepsilon.\label{e4.11} \end{equation} Since $\varepsilon$ is arbitrary, by \eqref{e4.11} and Lemma \ref{lem3.4}, we obtain $\sigma_{p+1}(f)\leq\sigma$. This and the fact that $\sigma_{p+1}(f)\geq\sigma$ yield $\sigma_{p+1}(f)=\sigma$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.4}] (i) First, we show that \eqref{e1.2} can possess at most one exceptional solution $f_0$ satisfying $\sigma_{p+1}(f_0)\leq\sigma$ or $i(f_0)