\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 $$\rho _{p}(f)=\limsup_{r\to +\infty } \frac{\log _{p}T(r, f)}{\text{log }r} =\limsup_{r\to +\infty } \frac{\log _{p+1}M(r, f)}{\text{log }r}, \label{e1.1}$$ 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 $$i(f)=\begin{cases} 0,&\text{for f a polynomial}, \\ \min\{ j\in \mathbb{N}: \rho _{j}(f)<+\infty \}, &\text{for f transcendental for which} \\ &\text{there exists j\in \mathbb{N} with \rho _{j}(f)<+\infty} \\ +\infty , &\text{when \rho _{j}(f)=+\infty for all j\in \mathbb{N}}. \end{cases} \label{e1.2}$$ \end{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 $$\tau _{p}(f)=\limsup_{r\to +\infty} \frac{\log _{p-1}T(r, f)}{ r^{\rho_{p}(f)}} =\limsup_{r\to +\infty} \frac{\log _{p}M(r, f)}{ r^{\rho _{p}(f)}} \quad (1\leq p\text{ an integer}). \label{e1.3}$$ \end{definition} For $k\geq 2,$\ we consider the linear differential equation $$f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}(z)f=0, \label{e1.4}$$ 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 $i(A_{s})=p$ $(01$ 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 $m\beta r^{\rho _{p}(f)}. \label{e2.2} \end{lemma} \begin{proof} When$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} 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 $$\log _{p}M(r_{n},f)>(\tau _{p}(f) -\varepsilon )r_{n}^{\rho _{p}(f)}. \label{e2.4}$$ For any given $\beta <\tau _{p}(f)$, there exists a positive integer $n_{1}$ such that for all $n>n_{1}$, we have $$\big(\frac{n}{n+1}\big)^{\rho _{p}(f)}>\frac{\beta }{\tau _{p}(f)-\varepsilon }\,. \label{e2.5}$$ 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{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} Set $E_{2}=\cup_{n=n_{2}}^{\infty } [r_{n},(1+ \frac{1}{n})r_{n}]$, then there holds $$\mathop{\rm lm}(E_{2})=\sum_{n=n_{2}}^\infty \int_{r_{n}}^{(1+\frac{1}{n})r_{n}} \frac{dt}{t} = \sum_{n=n_{2}}^{\infty } \log (1+\frac{1}{n}) =+\infty . \label{e2.7}$$ \end{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 $$\log _{p-1} m (r,f)=\log _{p-1}T(r,f)>\beta r^{\rho _{p}(f)}. \label{e2.8}$$ \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 $$f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_{1}f'+A_{0}f=F, \label{e2.9}$$ 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 $$|A_{0}(z)| \leq \Big|\frac{f^{( k)}}{f}\Big| +|A_{k-1}(z)| \Big|\frac{f^{(k-1)}}{f}\Big| +\dots+| A_{1}(z)| \big|\frac{f'}{f}\big|. \label{e3.1}$$ 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 $$\Big|\frac{f^{(j)}(z)}{f(z)}\Big| \leq B[ T(2r,f)] ^{k+1} \quad (j=1,2,\dots,k). \label{e3.2}$$ 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 $$M(r,A_{0})>\exp _{p}(\alpha _{2}r^{\rho }). \label{e3.5}$$ 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 $$\exp _{p}(\alpha _{2}r^{\rho })\leq kB\exp _{p}(\alpha _{1}r^{\rho })[T(2r,f)] ^{k+1}. \label{e3.6}$$ Since $\alpha _{2}>\alpha _{1}>0$, we get from \eqref{e3.6} that $$\exp ((1-\gamma )\exp _{p-1}(\alpha _{2}r^{\rho }))\leq kB[T(2r,f)] ^{k+1}\text{,} \label{e3.7}$$ 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{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} By Lemma of logarithmic derivative \cite{h1} and \eqref{e4.1}, we obtain \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} holds for all $r$ outside a set $E\subset (0,+\infty )$ with a finite linear measure $m(E)<+\infty$. Noting that \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} For sufficiently large $r$, we have $$O(\log r+\log T(r,f))\leq \frac{1}{2}T(r,f). \label{e4.4}$$ Thus, by \eqref{e4.2}-\eqref{e4.4}, for sufficiently large $r\notin E$, we have $$T(r,A_{s})\leq cT(r,f)+\sum_{j\neq s} T(r,A_{j}), \label{e4.5}$$ 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 $$T(r,A_{j})\leq \exp _{p-1}(r^{\beta _{0}}) \quad (j=0,1,\dots,s-1,s+1,\dots,k-1), \label{e4.6}$$ where $0<\beta _{0}<\rho$. Since $\rho _{p}(A_{s})=\rho$, there exists $\{ r_{n}'\}$ $(r_{n}'\to +\infty )$ such that $$\lim_{r_{n}'\to +\infty } \frac{\log _{p}T(r_{n}',A_{s})}{\log r_{n}'} =\rho . \label{e4.7}$$ 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 $$\frac{\log _{p}T(r_{n},A_{s})}{\log r_{n}} \geq \frac{\log _{p}T(r_{n}',A_{s})}{\log (r_{n}'+\delta +1)} =\frac{\log _{p}T(r_{n}',A_{s})}{\log r_{n}'+\log (1+(\delta +1) /r_{n}')} \label{e4.8}$$ we get $$\lim_{r_{n}\to +\infty }\frac{\log _{p}T( r_{n},A_{s})}{\log r_{n}}=\rho . \label{e4.9}$$ 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 $r_{n}$ $$\exp _{p-1}(r_{n}^{\rho -\varepsilon })\leq cT( r_{n},f)+(k-1)\exp _{p-1}(r_{n}^{\beta _{0}}). \label{e4.12}$$ Therefore, by \eqref{e4.12} we obtain $i(f)\geq p$ and $$\limsup_{r_{n}\to +\infty } \frac{\log _{p}T( r_{n},f)}{\log r_{n}}\geq \rho -\varepsilon \label{e4.13}$$ 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 $$T(r,A_{s})>\exp _{p-1}(\beta _{3}r^{\rho }). \label{e4.16}$$ 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 $$\exp _{p-1}(\beta _{3}r^{\rho })\leq cT(r,f) +(k-1)\exp _{p-1}(\beta _{2}r^{\rho }). \label{e4.17}$$ 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 $$i(0)=i(f^{(k)}+A_{k-1}(z) f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}( z)f)=i(A_{s}) \label{e4.18}$$ and $$\rho _{p}(0)=\rho _{p}(f^{(k)}+A_{k-1}( z)f^{(k-1)}+\dots+A_{1}(z)f'+A_{0}(z)f)=\rho _{p}(A_{s})>0 \label{e4.19}$$ 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 $$M(r,A_{s})>\exp _{p}(\beta _{5}r^{\rho }). \label{e4.22}$$ 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{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} where $d_{1},d_{2}$ are positive constants. By \eqref{e4.23}, we get $$\exp\{ \exp _{p-1}(\beta _{5}r^{\rho })-\exp _{p-1}( \beta _{4}r^{\rho })\} \leq \frac{d_{2}}{d_{1}}r^{s}. \label{e4.24}$$ 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{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} Since $\varphi \not\equiv 0$ and $\rho _{p}(\varphi )<\infty$, we have $W\not\equiv 0$. 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