\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 70, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2009/70\hfil Growth and oscillation of solutions]
{Growth and oscillation of solutions to linear differential
equations with entire coefficients having the same order}
\author[B. Bela\"idi\hfil EJDE-2009/70\hfilneg]
{Benharrat Bela\"idi}
\address{Benharrat Bela\"idi \newline
Department of Mathematics\\
Laboratory of Pure and Applied Mathematics\\
University of Mostaganem\\
B. P. 227 Mostaganem, Algeria}
\email{belaidi@univ-mosta.dz, belaidibenharrat@yahoo.fr}
\thanks{Submitted February 17, 2009. Published June 1, 2009.}
\subjclass[2000]{34M10, 30D35}
\keywords{Linear differential equation; growth of
entire function; iterated order}
\begin{abstract}
In this article, we investigate the growth and
fixed points of solutions of the differential equation
$$
f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_{1}(z)f'+A_{0}(z)f=0,
$$
where $A_{0}(z),\dots$, $A_{k-1}(z)$ are entire functions.
Some estimates are given for the iterated order and iterated
exponent of convergence of fixed points of solutions of the
above equation when most of the coefficients have the same
order with each other.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\section{Introduction and statement of results}
For the definition of the iterated order of an entire function, we
use the same definition as in \cite{k1}, \cite[p. 317]{b6},
\cite[p. 129]{l1}. For all $r\in \mathbb{R}$, we define
$\exp _{1}r:=e^{r}$ and $\exp _{p+1}r:=\exp (\exp_{p}r)$,
$p\in \mathbb{N}$. We also define for all $r$ sufficiently large
$\log _{1}r:=\log r$ and
$\log _{p+1}r:=\log (\log _{p}r)$, $p\in \mathbb{N}$.
Moreover, we denote by $\exp _{0}r:=r$, $\log _{0}r:=r$,
$\log_{-1}r:=\exp _{1}r$ and $\exp _{-1}r:=\log _{1}r$.
\begin{definition} \label{def1.1} \rm
Let $f$ be an entire function. Then the
iterated $p$-order $\rho _{p}(f)$ of $f$ is defined by
\begin{equation}
\rho _{p}(f)=\limsup_{r\to +\infty }
\frac{\log _{p}T(r, f)}{\text{log }r}
=\limsup_{r\to +\infty } \frac{\log _{p+1}M(r,
f)}{\text{log }r},
\label{e1.1}
\end{equation}
where $p$ is an integer, $p\geq 1$, $T(r, f)$ is the Nevanlinna
characteristic function of $f$ and
$M(r,f)=\max_{|z|=r}|f(z)|$;
see \cite{h1,n1}. For $p=1$, this notation is
called order and for $p=2$ hyper-order \cite{y1}.
\end{definition}
\begin{definition}[\cite{b6,l1}] \label{def1.2}\rm
The finiteness degree of the order of an
entire function $f$ is defined by
\begin{equation}
i(f)=\begin{cases}
0,&\text{for $f$ a polynomial}, \\
\min\{ j\in \mathbb{N}: \rho _{j}(f)<+\infty \},
&\text{for $f$ transcendental for which} \\
&\text{there exists $j\in \mathbb{N}$ with $\rho _{j}(f)<+\infty$}
\\
+\infty , &\text{when $\rho _{j}(f)=+\infty$ for all $j\in \mathbb{N}$}.
\end{cases} \label{e1.2}
\end{equation}
\end{definition}
\begin{definition} \label{def1.3}\rm
Let $f$ be an entire function. Then the
iterated $p$-type of an entire function $f$, with iterated $p$-order
$0<\rho_{p}(f)<\infty $ is defined by
\begin{equation}
\tau _{p}(f)=\limsup_{r\to +\infty}
\frac{\log _{p-1}T(r, f)}{ r^{\rho_{p}(f)}}
=\limsup_{r\to +\infty} \frac{\log _{p}M(r, f)}{ r^{\rho _{p}(f)}}
\quad (1\leq p\text{ an integer}).
\label{e1.3}
\end{equation}
\end{definition}
For $k\geq 2,$\ we consider the linear differential equation
\begin{equation}
f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}(z)f=0,
\label{e1.4}
\end{equation}
where $A_{0}(z),\dots, A_{k-1}(z)$ are entire
functions. It is well-known that all solutions of equation
\eqref{e1.4} are entire functions and if some of the coefficients of
\eqref{e1.4} are transcendental, then \eqref{e1.4} has at
least one solution with order $\rho (f)=+\infty $.
A natural question arises: What conditions on
$A_{0}(z),A_{1}(z),\dots ,A_{k-1}(z)$ will guarantee that every solution
$f\not\equiv 0$ 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 in a paper by
Frei \cite{f1}. Extensive work in recent years has been
concerned with the growth of solutions of complex linear differential
equations. Many results have been obtained for the growth and the
oscillation of solutions of the differential equation \eqref{e1.4};
see e.g. \cite{b2,b3,b4,b5,c1,c2,c3,c4,c5,k1,l1,l2,l3,t1,t2,w1}.
Examples of such results are the following two theorems:
\begin{theorem}[\cite{k1}] \label{thmA}
Let $A_{0}(z),\dots$, $A_{k-1}(z)$
be entire functions such that $i(A_{0})=p$ $(0
\max \{ |a_{jn}| :j=1,\dots,k-1\}$,
then every solution $f\not\equiv 0$
of \eqref{e1.4} satisfies $i(f)=p+1$
and $\rho _{p+1}(f)=n$.
\end{corollary}
Replacing the dominant coefficient $A_{0}$ by an arbitrary
coefficient $A_{s}$, where $s\in \{ 1,\dots,k-1\} $, we obtain the
following result which is an extension of Theorems \ref{thmA},
\ref{thmB} and \ref{thm1.1}.
\begin{theorem} \label{thm1.2}
Let $A_{0}(z),\dots, A_{k-1}(z)$ be entire functions. Suppose
that there exists an $A_{s}$ $(1\leq s\leq k-1)$
with $i(A_{s})=p$ $(0
1$ be a given constant. Then there exist a set
$E_{1}\subset $ $(1,+\infty )$ of finite logarithmic measure and
a constant $B>0$ that
depends only on $\alpha $ and $(m,n)$
($m,n$ positive integers with $m\beta r^{\rho _{p}(f)}. \label{e2.2}
\end{equation}
\end{lemma}
\begin{proof} When $p=1$, the Lemma is due to Tu and Yi \cite{t2}.
Thus we assume that $p\geq 2$. By definitions of
iterated order and iterated type, there exists an increasing sequence
$\{ r_{n}\} $, $r_{n}\to \infty $ satisfying
$(1+ \frac{1}{n})r_{n}\beta . \label{e2.3}
\end{equation}
Then there exists a positive integer $n_{0}$ such that for all
$n>n_{0}$ and
for any given $\varepsilon $ $(0<\varepsilon <\tau _{p}(f)
-\beta )$, we have
\begin{equation}
\log _{p}M(r_{n},f)>(\tau _{p}(f)
-\varepsilon )r_{n}^{\rho _{p}(f)}. \label{e2.4}
\end{equation}
For any given $\beta <\tau _{p}(f)$, there exists a positive
integer $n_{1}$ such that for all $n>n_{1}$, we have
\begin{equation}
\big(\frac{n}{n+1}\big)^{\rho _{p}(f)}>\frac{\beta }{\tau
_{p}(f)-\varepsilon }\,. \label{e2.5}
\end{equation}
Taking $n\geq n_{2}=\max\{ n_{0},n_{1}\} $. By
\eqref{e2.4} and \eqref{e2.5} for any $r\in [r_{n},(1+
\frac{1}{n})r_{n}] $, we obtain
\begin{equation}
\begin{aligned}
\log _{p}M(r,f)
&\geq \log _{p}M(r_{n},f)\\
&>(\tau _{p}(f)-\varepsilon )r_{n}^{\rho _{p}(f)}\\
&\geq (\tau _{p}(f)-\varepsilon )\big(\frac{n}{n+1
}r\big)^{\rho _{p}(f)}\\
&>\beta r^{\rho _{p}(f)}.
\end{aligned}\label{e2.6}
\end{equation}
Set $E_{2}=\cup_{n=n_{2}}^{\infty } [r_{n},(1+
\frac{1}{n})r_{n}] $, then there holds
\begin{equation}
\mathop{\rm lm}(E_{2})=\sum_{n=n_{2}}^\infty
\int_{r_{n}}^{(1+\frac{1}{n})r_{n}} \frac{dt}{t}
= \sum_{n=n_{2}}^{\infty } \log (1+\frac{1}{n})
=+\infty . \label{e2.7}
\end{equation}
\end{proof}
Using similar arguments as in the proof of Lemma \ref{lem2.2}, we
obtain the following result.
\begin{lemma} \label{lem2.3}
Let $f(z)$ be an entire function of iterated $p$-order
$0<\rho _{p}(f)<+\infty $ and iterated $p$-type
$0<\tau_{p}(f)<\infty $. Then for any given
$\beta <\tau _{p}(f)$, there exists a set
$E_{3}\subset [1,+\infty )$ that has infinite logarithmic measure, such
that for all $r\in E_{3}$, we have
\begin{equation}
\log _{p-1} m (r,f)=\log _{p-1}T(r,f)>\beta r^{\rho
_{p}(f)}. \label{e2.8}
\end{equation}
\end{lemma}
To avoid some problems caused by the exceptional set we
recall the following Lemmas.
\begin{lemma}[{\cite[p. 68]{b1}}] \label{lem2.4}
Let $g:[0,+\infty )\to \mathbb{R}$ and
$h:[0,+\infty )\to \mathbb{R}$ be monotone non-decreasing functions
such that $g(r)\leq h(r)$ outside of an exceptional set $E_{4}$
of finite linear measure. Then for any $\alpha >1$, there
exists $r_{0}>0$ such that $g(r)\leq h(\alpha r)$ for all $r>r_{0}$.
\end{lemma}
\begin{lemma}[\cite{g2}] \label{lem2.5}
Let $\varphi :[0,+\infty )\to \mathbb{R}$ and
$\psi :[0,+\infty )\to \mathbb{R}$ be monotone non-decreasing functions
such that $\varphi (r)\leq \psi (r)$ for all
$r\notin E_{5}\cup [0,1] $, where $E_{5}\subset (1,+\infty )$
is a set of finite logarithmic measure. Let $\gamma >1$
be a given constant. Then there exists an
$r_{1}=r_{1}(\gamma )>0$ such that
$\varphi (r)\leq \psi (\gamma r)$ for all $r>r_{1}$.
\end{lemma}
\begin{lemma}[\cite{b5,c1}] \label{lem2.6}
Let $A_{0}$, $A_{1},\dots, A_{k-1}$,
$F\not\equiv 0$ be finite iterated $p$-order meromorphic
functions. If $f$ is a meromorphic solution with
$\rho _{p}(f)=+\infty $ and $\rho _{p+1}(f)=\rho <+\infty $
of the equation
\begin{equation}
f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_{1}f'+A_{0}f=F, \label{e2.9}
\end{equation}
then $\overline{\lambda }_{p}(f)=\lambda _{p}(
f)=\rho _{p}(f)=+\infty $ and
$\overline{ \lambda }_{p+1}(f)=\lambda _{p+1}(f)=\rho
_{p+1}(f)=\rho $.
\end{lemma}
\begin{lemma}[\cite{b6,l1}] \label{lem2.7}
Suppose that $k\geq 2$ and $A_{0}( z),\dots,A_{k-1}(z)$
are entire functions of finite iterated $p$-order. If $f(z)$
is a solution of \eqref{e1.4}, then
$i(f)\leq p+1$ and $\rho _{p+1}(f)\leq \max\{
\rho _{p}(A_{j}): j=0,\dots,k-1\} =\rho $.
\end{lemma}
\section{Proof of Theorem \ref{thm1.1}}
Suppose that $f\not\equiv 0$ is a solution of
equation \eqref{e1.4}. From \eqref{e1.4}, we can write
\begin{equation}
|A_{0}(z)| \leq \Big|\frac{f^{(
k)}}{f}\Big| +|A_{k-1}(z)|
\Big|\frac{f^{(k-1)}}{f}\Big| +\dots+|
A_{1}(z)| \big|\frac{f'}{f}\big|. \label{e3.1}
\end{equation}
By Lemma \ref{lem2.1}, there exist a constant $B>0$ and a set
$E_{1}\subset (1,+\infty )$ having finite logarithmic measure
such that for all $z$
satisfying $|z| =r\notin E_{1}\cup [0,1] $,
we have
\begin{equation}
\Big|\frac{f^{(j)}(z)}{f(z)}\Big| \leq B[
T(2r,f)] ^{k+1} \quad (j=1,2,\dots,k). \label{e3.2}
\end{equation}
If $\max \{\rho _{p}(A_{j}):j=1,2,\dots,k-1\}<\rho _{p}(
A_{0})=\rho $, then by Theorem \ref{thmA}, we obtain $i(f)=p+1$
and $\rho _{p+1}(f)=\rho _{p}(A_{0})=\rho $.
If $\max \{\rho _{p}(A_{j}):j=1,2,\dots,k-1\}=\rho
_{p}(A_{0})=\rho$ $(0<\rho <+\infty )$ and
$\max \{\tau _{p}(A_{j}):\rho _{p}(A_{j})=\rho _{p}(A_{0})\}
<\tau _{p}(A_{0}) =\tau $ $(0<\tau <+\infty )$. Then, there exists a set
$I\subseteq\{1,2,\dots,k-1\}$ such that
$\rho _{p}( A_{j})=\rho _{p}(A_{0})=\rho (j\in I)$ and
$\tau _{p}(A_{j})<\tau _{p}(A_{0})$
$(j\in I)$. Thus, we choose $\alpha _{1}$, $\alpha _{2}$
satisfying $\max \{\tau _{p}(A_{j}) : (j\in I)
\}<\alpha _{1}<\alpha _{2}<\tau _{p}(A_{0})=\tau $ such that
for sufficiently large $r$, we have
\begin{gather}
|A_{j}(z)| \leq \exp _{p}(\alpha
_{1}r^{\rho }) \quad (j\in I), \label{e3.3}
\\
|A_{j}(z)| \leq \exp _{p}(r^{\alpha
_{0}}) \quad (j\in\{ 1,2,\dots,k-1\} \backslash
I), \label{e3.4}
\end{gather}
where $0<\alpha _{0}<\rho $. By Lemma \ref{lem2.2}, there exists a set
$E_{2}\subset [1,+\infty )$ with infinite logarithmic measure
such that for all $r\in E_{2}$, we have
\begin{equation}
M(r,A_{0})>\exp _{p}(\alpha _{2}r^{\rho }).
\label{e3.5}
\end{equation}
Hence from \eqref{e3.1}-\eqref{e3.5}, for all $z$ satisfying
$|A_{0}(z)| =M(r,A_{0})$ and
for sufficiently large $|z| =r\in E_{2}\backslash E_{1}\cup [0,1] $,
we have
\begin{equation}
\exp _{p}(\alpha _{2}r^{\rho })\leq kB\exp _{p}(\alpha
_{1}r^{\rho })[T(2r,f)] ^{k+1}. \label{e3.6}
\end{equation}
Since $\alpha _{2}>\alpha _{1}>0$, we get from \eqref{e3.6} that
\begin{equation}
\exp ((1-\gamma )\exp _{p-1}(\alpha _{2}r^{\rho
}))\leq kB[T(2r,f)] ^{k+1}\text{,} \label{e3.7}
\end{equation}
where $\gamma $ $(0<\gamma <1)$ is a real number.
By \eqref{e3.7}, Lemma \ref{lem2.5} and the definition of
iterated order, we have
$i(f)\geq p+1$ and $\rho _{p+1}(f)\geq \rho
_{p}(A_{0})=\rho $.
On the other hand by Lemma \ref{lem2.7}, we have
$i(f)\leq p+1$ and $\rho _{p+1}(f)\leq \rho
_{p}(A_{0})=\rho $, hence $i(f)=p+1$ and $\rho
_{p+1}(f)=\rho _{p}(A_{0})=\rho $.
\section{Proof of Theorem \ref{thm1.2}}
Assume that $f$ is a transcendental solution of \eqref{e1.4}.
It follows from \eqref{e1.4} that
\begin{equation}
\begin{aligned}
A_{s}(z)&=-\Big(\frac{f^{(k)}}{f^{(s)
}}+A_{k-1}(z)\frac{f^{(k-1)}}{f^{(s)}}
+\dots+A_{s+1}(z)\frac{f^{(s+1)}}{f^{(s)}}
\\
&\quad +A_{s-1}(z)\frac{f^{(s-1)}}{f^{(
s)}}+\dots+A_{1}(z)\frac{f'}{f^{(s)}}
+A_{0}(z)\frac{f}{f^{(s)}}\Big)
\\
&=-\frac{f}{f^{(s)}}\Big(\frac{f^{(k)}}{f}
+A_{k-1}(z)\frac{f^{(k-1)}}{f}+\dots+A_{s+1}(
z)\frac{f^{(s+1)}}{f}
\\
&\quad +A_{s-1}(z)\frac{f^{(s-1)}}{f}
+\dots+A_{1}(z)\frac{f'}{f}+A_{0}(z)\Big).
\end{aligned} \label{e4.1}
\end{equation}
By Lemma of logarithmic derivative \cite{h1} and
\eqref{e4.1}, we obtain
\begin{equation}
\begin{aligned}
T(r,A_{s})
&=m(r,A_{s})\\
&\leq m\big(r,\frac{f}{f^{(s)}}\big)
+\sum_{j\neq s} m(r,A_{j}) +O(\log (rT(r,f)))\\
&=m\big(r,\frac{f}{f^{(s)}}\big)+\sum_{j\neq s}
T(r,A_{j})+O(\log (rT(r,f)))
\end{aligned} \label{e4.2}
\end{equation}
holds for all $r$ outside a set $E\subset (0,+\infty )$ with a
finite linear measure $m(E)<+\infty $. Noting that
\begin{equation}
\begin{aligned}
m(r,\frac{f}{f^{(s)}})
&\leq m(r,f) +m\big(r,\frac{1}{f^{(s)}}\big)\\
&\leq T(r,f)+T(r,\frac{1}{f^{(s)}})\\
&=T(r,f)+T(r,f^{(s)})+O(1)\\
&\leq (s+2)T(r,f)+o(T(r,f) )+O(1).
\end{aligned} \label{e4.3}
\end{equation}
For sufficiently large $r$, we have
\begin{equation}
O(\log r+\log T(r,f))\leq \frac{1}{2}T(r,f). \label{e4.4}
\end{equation}
Thus, by \eqref{e4.2}-\eqref{e4.4}, for sufficiently large
$r\notin E$, we have
\begin{equation}
T(r,A_{s})\leq cT(r,f)+\sum_{j\neq s}
T(r,A_{j}), \label{e4.5}
\end{equation}
where $c$ is a positive constant.
If $\max \{\rho _{p}(A_{j}):j=0,1,\dots,s-1,s+1,\dots,k-1\}
<\rho _{p}(A_{s})=\rho $, then for
sufficiently large $r$, we have
\begin{equation}
T(r,A_{j})\leq \exp _{p-1}(r^{\beta _{0}}) \quad
(j=0,1,\dots,s-1,s+1,\dots,k-1), \label{e4.6}
\end{equation}
where $0<\beta _{0}<\rho $. Since $\rho _{p}(A_{s})=\rho $,
there exists $\{ r_{n}'\}$ $(r_{n}'\to +\infty )$ such that
\begin{equation}
\lim_{r_{n}'\to +\infty } \frac{\log _{p}T(r_{n}',A_{s})}{\log r_{n}'}
=\rho .
\label{e4.7}
\end{equation}
Set the linear measure of $E$, $m(E)=\delta <+\infty $, then
there exists a point $r_{n}\in [r_{n}',r_{n}'+\delta +1] -E$. From
\begin{equation}
\frac{\log _{p}T(r_{n},A_{s})}{\log r_{n}}
\geq \frac{\log _{p}T(r_{n}',A_{s})}{\log (r_{n}'+\delta +1)}
=\frac{\log _{p}T(r_{n}',A_{s})}{\log r_{n}'+\log (1+(\delta +1)
/r_{n}')} \label{e4.8}
\end{equation}
we get
\begin{equation}
\lim_{r_{n}\to +\infty }\frac{\log _{p}T(
r_{n},A_{s})}{\log r_{n}}=\rho . \label{e4.9}
\end{equation}
So for any given $\varepsilon $
$(0<\varepsilon <\rho -\beta_{0})$, and for $j\neq s$,
\begin{gather}
T(r_{n},A_{j})\leq \exp _{p-1}(r_{n}^{\beta _{0}}) \label{e4.10}
\\
T(r_{n},A_{s})\geq \exp _{p-1}(r_{n}^{\rho -\varepsilon}) \label{e4.11}
\end{gather}
hold for sufficiently large $r_{n}$. By \eqref{e4.5},
\eqref{e4.10}, \eqref{e4.11} and Lemma \ref{lem2.4}, we obtain for
sufficiently large $r_{n}$
\begin{equation}
\exp _{p-1}(r_{n}^{\rho -\varepsilon })\leq cT(
r_{n},f)+(k-1)\exp _{p-1}(r_{n}^{\beta
_{0}}). \label{e4.12}
\end{equation}
Therefore, by \eqref{e4.12} we obtain $i(f)\geq p$ and
\begin{equation}
\limsup_{r_{n}\to +\infty }
\frac{\log _{p}T( r_{n},f)}{\log r_{n}}\geq \rho -\varepsilon \label{e4.13}
\end{equation}
and since $\varepsilon >0$ is arbitrary, we get
$\rho _{p}(f) \geq \rho _{p}(A_{s})=\rho $. On the other hand
by Lemma \ref{lem2.7},
we have $i(f)\leq p+1$ and
$\rho _{p+1}(f)\leq \rho _{p}(A_{s})$. Hence, we obtain
$p\leq i(f) \leq p+1$ and $\rho _{p+1}(f)\leq \rho _{p}(A_{s})
\leq \rho _{p}(f)$.
If $\max \{\rho _{p}(A_{j}):j=0,1,\dots,s-1,s+1,\dots,k-1\}
=\rho _{p}(A_{s})=\rho$ $(0<\rho <+\infty )$ and
$\max \{\tau _{p}(A_{j}):\rho _{p}(A_{j})=\rho _{p}(A_{s})
\}<\tau _{p}(A_{s})=\tau $ $(0<\tau <+\infty )$.
Then, there exists a set $J\subseteq\{ 0,1,\dots,s-1,s+1,\dots,k-1\}$
such that $\rho _{p}(A_{j})=\rho _{p}(A_{s})=\rho$
$(j\in J)$ and $\tau _{p}(A_{j})<\tau _{p}(A_{s})$ $(j\in J)$.
Thus, we choose $\beta _{2}$, $\beta _{3}$ satisfying
$\max \{\tau _{p}(A_{j}):(j\in J)\}<\beta _{2}<\beta _{3}<\tau _{p}(A_{s})
=\tau $ such that for sufficiently large $r$, we have
\begin{gather}
T(r,A_{j})\leq \exp _{p-1}(\beta _{2}r^{\rho }) \quad (j\in J),
\label{e4.14}\\
T(r,A_{j})\leq \exp _{p-1}(r^{\beta _{1}})\quad
(j\in\{ 0,1,\dots,s-1,s+1,\dots,k-1\} \backslash J),
\label{e4.15}
\end{gather}
where $0<\beta _{1}<\rho $. By Lemma \ref{lem2.3}, there exists a set
$E_{3}\subset [1,+\infty )$ with infinite logarithmic measure
such that for all $r\in E_{3}$, we have
\begin{equation}
T(r,A_{s})>\exp _{p-1}(\beta _{3}r^{\rho }).
\label{e4.16}
\end{equation}
Hence from \eqref{e4.5}, \eqref{e4.14}, \eqref{e4.15},
\eqref{e4.16} and Lemma \ref{lem2.4}, for sufficiently large
$|z| =r\in E_{3}$, we have
\begin{equation}
\exp _{p-1}(\beta _{3}r^{\rho })\leq cT(r,f)
+(k-1)\exp _{p-1}(\beta _{2}r^{\rho }). \label{e4.17}
\end{equation}
By this inequality and the definition of iterated order,
we have $i(f)\geq p$ and $\rho _{p}(f)\geq \rho _{p}(
A_{s})=\rho $. On the other hand by Lemma \ref{lem2.7},
we have $i(f)\leq p+1$ and $\rho _{p+1}(f)\leq \rho _{p}(
A_{s})$. Hence, we obtain $p\leq i(f)\leq p+1$ and
$\rho_{p+1}(f)\leq \rho _{p}(A_{s})\leq \rho _{p}(f)$.
Suppose that $f$ is a polynomial of $\deg f=m\geq s$. If
$\max \{\rho _{p}(A_{j}):j=0,1,\dots,s-1,s+1,\dots,k-1\}<\rho
_{p}(A_{s})=\rho $, then
\begin{equation}
i(0)=i(f^{(k)}+A_{k-1}(z)
f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}(
z)f)=i(A_{s}) \label{e4.18}
\end{equation}
and
\begin{equation}
\rho _{p}(0)=\rho _{p}(f^{(k)}+A_{k-1}(
z)f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}(z)f)=\rho _{p}(A_{s})>0 \label{e4.19}
\end{equation}
this is a contradiction by \eqref{e1.4}.
If $\max \{\rho _{p}(A_{j}):j=0,1,\dots,s-1,s+1,\dots,k-1\}
=\rho _{p}(A_{s})=\rho$
$(0<\rho <+\infty )$ and
$\max \{\tau _{p}( A_{j}):\rho _{p}(A_{j})=\rho _{p}(A_{s})
\}<\tau _{p}(A_{s})=\tau $ $(0<\tau <+\infty )$.
Then, there exists a set
$K\subseteq\{ 0,1,\dots,s-1,s+1,\dots,k-1\}$ such that
$\rho _{p}(A_{j})=\rho _{p}(A_{s})=\rho \mathit{\ }(j\in K)$
and $\tau _{p}(A_{j}) <\tau _{p}(A_{s})$ $(j\in K)$. Thus, we choose
$\beta _{4}$, $\beta _{5}$ satisfying
$\max \{\tau _{p}(A_{j}): (j\in K)\}<\beta _{4}<\beta _{5}<\tau _{p}(
A_{s})=\tau $ such that for sufficiently large $r$, we have
\begin{gather}
|A_{j}(z)| \leq \exp _{p}(\beta_{4}r^{\rho })
\quad (j\in K), \label{e4.20} \\
|A_{j}(z)| \leq \exp _{p}(r^{\beta
_{6}}) \quad (j\in\{ 0,1,\dots,s-1,s+1,\dots,k-1\}
\backslash K), \label{e4.21}
\end{gather}
where $0<\beta _{6}<\rho $. By Lemma \ref{lem2.2}, there exists a set
$E_{2}\subset [1,+\infty )$ with infinite logarithmic measure
such that for all $r\in E_{2}$, we have
\begin{equation}
M(r,A_{s})>\exp _{p}(\beta _{5}r^{\rho }).
\label{e4.22}
\end{equation}
Hence from \eqref{e1.4}, \eqref{e4.20}-\eqref{e4.22},
for all $z$ satisfying $|A_{s}(z)| =M(r,A_{s})$ and for sufficiently
large $|z| =r\in E_{2}$, we have
\begin{equation}
\begin{aligned}
d_{1}r^{m-s}\exp _{p}(\beta _{5}r^{\rho })
&\leq | A_{s}(re^{i\theta })f^{(s)}(re^{i\theta })|\\
&\leq \sum_{j\neq s}|A_{j}(re^{i\theta })f^{(j)}(re^{i\theta })| \\
&\leq d_{2}r^{m}\exp _{p}(\beta _{4}r^{\rho }),
\end{aligned} \label{e4.23}
\end{equation}
where $d_{1},d_{2}$ are positive constants. By
\eqref{e4.23}, we get
\begin{equation}
\exp\{ \exp _{p-1}(\beta _{5}r^{\rho })-\exp _{p-1}(
\beta _{4}r^{\rho })\} \leq \frac{d_{2}}{d_{1}}r^{s}.
\label{e4.24}
\end{equation}
Hence by $\beta _{5}>\beta _{4}>0$, from \eqref{e4.24} we obtain a
contradiction. Therefore, if $f$ is a non-transcendental solution, then it
must be a polynomial of $\deg f\leq s-1$. This proves
Theorem \ref{thm1.2}.
\section{Proof of Theorem \ref{thm1.3}}
Suppose that $f\not\equiv 0$ is a solution of
equation \eqref{e1.4}. Then by Theorem \ref{thm1.1}, we have
$\rho _{p}( f)=\infty $ and
$\rho _{p+1}(f)=\rho _{p}( A_{0})=\rho $. Set
$w=f-\varphi $. Since $\rho _{p}(\varphi)<\infty $, then we
have $\rho _{p}(w)=\rho _{p}(f-\varphi )=\rho _{p}(f)=\infty $
and $\rho _{p+1}(w)=\rho _{p+1}(f-\varphi )=\rho _{p+1}(f)
=\rho _{p}(A_{0})=\rho $. Substituting $f=w+\varphi $ into
equation \eqref{e1.4}, we obtain
\begin{equation}
\begin{aligned}
w^{(k)}+A_{k-1}(z)w^{(k-1)}+\dots+A_{0}(z)w
&=-(\varphi^{(k)}+A_{k-1}(z)\varphi
^{(k-1)}+\dots+A_{0}(z)\varphi)\\ &=W.
\end{aligned} \label{e5.1}
\end{equation}
Since $\varphi \not\equiv 0$ and $\rho _{p}(\varphi )<\infty$,
we have $W\not\equiv 0$. Then by Lemma \ref{lem2.6}, we obtain
$\overline{\lambda }_{p}(w)=\lambda _{p}(w)
=\rho _{p}(w)=\infty $ and
$\overline{\lambda }_{p+1}(w)=\lambda_{p+1}(w)
=\rho _{p+1}(w)=\rho _{p}(A_{0})=\rho $; i.e.,
$\overline{\lambda }_{p}(f-\varphi)=\lambda _{p}(f-\varphi )
=\rho _{p}(f) =\infty $ and $\overline{\lambda }_{p+1}(f-\varphi )
=\lambda_{p+1}(f-\varphi )=\rho _{p+1}(f)=\rho _{p}(A_{0})=\rho $.
\subsection*{Acknowledgements}
The author would like to thank the anonymous
referees for their helpful remarks and suggestions to improve
this article.
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\end{document}