\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 226, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/226\hfil Growth of solutions] {Growth of solutions to linear differential equations with entire coefficients} \author[ H. Hu, X.-M. Zheng \hfil EJDE-2012/226\hfilneg] {Hui Hu, Xiu-Min Zheng} % in alphabetical order \address{Hui Hu \newline Institute of Mathematics and Information Science, Jiangxi Normal University, 330022, China} \email{h\_h87\_6@hotmail.com} \address{Xiu-Min Zheng (Corresponding author)\newline Institute of Mathematics and Information Science, Jiangxi Normal University, 330022, China} \email{zhengxiumin2008@sina.com} \thanks{Submitted May 8, 2012. Published December 17, 2012.} \subjclass[2000]{30D35, 34M05} \keywords{Iterated $p$-order; iterated $p$-lower order; iterated $p$-lower type; \hfill\break\indent iterated exponent of convergence of distinct zeros} \begin{abstract} In this article, we study the growth of solutions of linear differential equations with some dominant entire coefficients. Especially, we obtain some results on the iterated $p$-lower order of these solutions, which extend previous results. Moreover, we investigate the iterated exponent of convergence of distinct zeros of $f^{(j)}(z)-\varphi(z)$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \def\R{\mathbb{R}} \section{Introduction} We shall assume that readers are familiar with the fundamental results and the standard notations of Nevanlinna's theory; see e.g. \cite{h1,l1,v1}. Let us define inductively for $r\in[0,+\infty)$, $\exp_1 r=e^r$ and $\exp_{p+1}r=\exp(\exp_p r)$, $p\in \mathbb{N}$. For all sufficiently large $r$, we define $\log_1 r=\log r$ and $\log_{p+1}r=\log(\log_p r)$, $p\in \mathbb{N}$. We also denote $\exp_0 r=r=\log_0 r$ and $\exp_{-1}r=\log_1 r$. We recall the following definitions of finite iterated order; see e.g. \cite{b2,c1,l1,k1,t1,t3}. \begin{definition} \label{def1.1}\rm The iterated $p$-order $\sigma_p(f)$ of a meromorphic function $f(z)$ is defined as $$\sigma_p(f)=\limsup_{r\to\infty} \frac{\log_pT(r,f)}{\log r}\quad (p\in \mathbb{N}).$$ \end{definition} \begin{remark} \label{rmk1.2} \rm If $f(z)$ is an entire function, then $$\sigma_p(f)=\limsup_{r\to\infty} \frac{\log_pT(r,f)}{\log r} =\limsup_{r\to\infty} \frac{\log_{p+1}M(r,f)}{\log r} =\limsup_{r\to\infty} \frac{\log_p{\nu_f(r)}}{\log r},$$ where $p\in \mathbb{N}$, $\nu_f(r)$ is the central index of $f(z)$. \end{remark} \begin{definition} \label{def1.3} \rm The iterated $p$-lower order $\mu_p(f)$ of a meromorphic function $f(z)$ is defined by $$\mu_p(f)=\liminf_{r\to\infty} \frac{\log_pT(r,f)}{\log r}\quad (p\in \mathbb{N}).$$ \end{definition} \begin{remark} \label{rmk1.4} \rm The iterated $p$-lower order $\mu_p(f)$ of an entire function $f(z)$ is defined by $$\mu_p(f)=\liminf_{r\to\infty} \frac{\log_pT(r,f)}{\log r} =\liminf_{r\to\infty} \frac{\log_{p+1}M(r,f)}{\log r} =\liminf_{r\to\infty} \frac{\log_p{\nu_f(r)}}{\log r}\quad (p\in \mathbb{N}).$$ \end{remark} \begin{definition} \label{def1.5} \rm The finiteness degree of the order of a meromorphic function $f(z)$ is defined by \begin{equation*} {i}(f)=\begin{cases} 0, & \text{if $f$ is rational}; \\ \min\{j\in \mathbb{N}:\sigma _{j}(f)<\infty\}, & \text{if $f$ is transcendental with}\\ &\sigma_{j}(f)<\infty \text{ for some }j\in\mathbb{N}; \\ \infty, & \text{if } \sigma _{j}(f)=\infty \text{ for all }j\in \mathbb{N}. \end{cases} \end{equation*} \end{definition} \begin{definition} \label{def1.6}\rm The iterated convergence exponent of the sequence of $a$-points of a meromorphic function $f(z)$ is defined by $$\lambda_p(f-a)=\lambda_p(f,a)=\limsup_{r\to\infty} \frac{\log_p N(r,\frac{1}{f-a})}{\log r}\quad (p\in \mathbb{N}),$$ and the iterated convergence exponent of the sequence of distinct $a$-points of a meromorphic function $f(z)$ is defined by $$\overline{\lambda}_p(f-a)=\overline{\lambda}_p(f,a)=\limsup_{r\to\infty} \frac{\log_p \overline{N}(r,\frac{1}{f-a})}{\log r}\quad (p\in \mathbb{N}).$$ If $a=0$, the iterated convergence exponent of the zeros or the iterated convergence exponent of the distinct zeros is defined respectively by $$\lambda_p(f)=\lambda_p(f,0)=\limsup_{r\to\infty} \frac{\log_p N(r,\frac{1}{f})}{\log r}~~(p\in \mathbb{N}),$$ or $$\overline{\lambda}_p(f)=\overline{\lambda}_p(f,0) =\limsup_{r\to\infty} \frac{\log_p \overline{N}(r,\frac{1}{f})}{\log r}\quad (p\in \mathbb{N}).$$ If $a=\infty$, the iterated convergence exponent of the poles or the iterated convergence exponent of the distinct poles is defined respectively by $$\lambda_p(\frac{1}{f})=\limsup_{r\to\infty} \frac{\log_p N(r,f)}{\log r}\quad (p\in \mathbb{N}),$$ or $$\overline{\lambda}_p(\frac{1}{f}) =\limsup_{r\to\infty} \frac{\log_p \overline{N}(r,f)}{\log r}\quad (p\in \mathbb{N}).$$ Furthermore, we can get the definitions of $\lambda_p(f-\varphi)$ and $\overline{\lambda}_p(f-\varphi)$, when $a$ is replaced by a meromorphic function $\varphi$. \end{definition} \begin{definition} \label{def1.7}\rm Let $f(z)$ be an entire function. Then the iterated $p$-type of an entire function $f(z)$, with iterated $p$-order $0<\sigma_p(f)<\infty$ is defined by $$\tau_p(f)=\limsup_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\sigma_p(f)}} =\limsup_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\sigma_p(f)}}\quad (p\in \mathbb{N}\backslash\{1\}).$$ \end{definition} We definite the iterated $p$-lower type of $f(z)$ as follows. \begin{definition} \label{def1.8} \rm Let $f(z)$ be an entire function. Then the iterated $p$-lower type of an entire function $f(z)$, with iterated $p$-lower order $0<\mu_p(f)<\infty$, is defined by $$\underline\tau_p(f)=\liminf_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\mu_p(f)}}=\liminf_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\mu_p(f)}}\quad(p\in \mathbb{N}\backslash\{1\}).$$ \end{definition} \begin{remark} \label{rmk1.9} \rm If $p=1$, then the equalities \begin{gather*} \limsup_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\sigma_p(f)}} =\limsup_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\sigma_p(f)}},\\ \liminf_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\mu_p(f)}} =\liminf_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\mu_p(f)}} \end{gather*} in Definitions \ref{def1.7} and \ref{def1.8} respectively fail to hold. For example, for the function $f(z)=e^z$, we have $\lim_{r\to\infty}\frac{T(r,f)}{r}=\frac{1}{\pi}\neq1 =\lim_{r\to\infty}\frac{\log M(r,f)}{r}$. Therefore, we assume $p\in\mathbb{N}\backslash\{1\}$ in the following. \end{remark} We denote the linear measure and the logarithmic measure of a set $E\subset[0,+\infty)$ by $mE=\int_E dt$ and $m_l E=\int_E dt/t$ respectively (see e.g. \cite{h2}). \section{Main Results} In 1998, Kinnunen investigated complex oscillation properties of the solutions of the higher order linear differential equations $$f^{(n)}+A_{n-1}(z)f^{(n-1)}+\dots+A_1(z)f'+A_0(z)f=0 \label{e2.1}$$ and $$f^{(n)}+A_{n-1}(z)f^{(n-1)}+\dots+A_1(z)f'+A_0(z)f=F(z), \label{e2.2}$$ with entire coefficients of finite iterated order and obtained the following result in \cite{k1}. \begin{theorem} \label{thmA} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions and let $i(A_0)=p$, $00)$ and $\max\{\tau_p(A_j):\sigma_p(A_j)=\sigma_p(A_0)\}<\tau_p(A_0) =\tau(0<\tau<\infty)$. Then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $i(f)=p+1$ and $\sigma_{p+1}(f)=\sigma_p(A_0)$. \end{theorem} Theorems \ref{thmA} and \ref{thmB} investigated the iterated order of solutions of \eqref{e2.1}, when there is some dominating coefficient with iterated order. Another question is: If there is some dominating coefficient with iterated lower order, what can we say about the growth of solutions of \eqref{e2.1}. For the special case $p=2$, Zhang-Tu in \cite{z1} discussed it and obtained the following result. \begin{theorem} \label{thmC} Let $A_0(z), \dots,A_{n-1}(z)$ be entire functions satisfying $\max\{\sigma(A_j)$, $j=1,\dots,{n-1}\}<\mu(A_0)\leq\sigma(A_0)<\infty$, then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $$\mu(A_0)=\mu_2(f)\leq\sigma_2(f)=\sigma(A_0).$$ \end{theorem} In this paper, we investigate the above problems. Moreover, we investigate the iterated exponent of convergence of distinct zeros of $f^{(j)}(z)-\varphi(z)$. Firstly, we extend Theorem \ref{thmC} into a general case and obtain the same result. \begin{theorem} \label{thm2.1} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions of finite iterated order satisfying $\max\{\sigma_p(A_j),j=1,\dots,n-1\}<\mu_p(A_0)\leq\sigma_p(A_0)<\infty$, then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $$\mu_p(A_0)=\mu_{p+1}(f)\leq\sigma_{p+1}(f)=\sigma_p(A_0).\label{e2.3}$$ \end{theorem} Secondly, when there are some coefficients with iterated order equal to $\mu_p(A_0)$, we obtain the following two results. \begin{theorem} \label{thm2.2} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions, and let $i(A_0)=p$. Assume that $\max\{\sigma_p(A_j):j\neq0\}\leq\mu_p(A_0)\leq\sigma_p(A_0)$ and $\tau_1=\max\{\tau_p(A_j):\sigma_p(A_j)=\mu_p(A_0)\}<\underline\tau_p(A_0) =\tau(0<\tau<\infty)$. Then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $$\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f) =\lambda_{p+1}(f-\varphi)=\overline{\lambda}_{p+1}(f-\varphi),\label{e2.4}$$ where $\varphi(z)\not\equiv 0$ is an entire function satisfying $\sigma_{p+1}(\varphi)<\mu_p(A_0)$. \end{theorem} \begin{theorem} \label{thm2.3} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions of finite iterated order satisfying $\max\{\sigma_p(A_j),j\neq0\}\leq\mu_p(A_0)=\mu$ and $$\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<1.$$ Then every non-trivial solution $f(z)$ of \eqref{e2.1} satisfies $$\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f) =\lambda_{p+1}(f-\varphi)=\overline{\lambda}_{p+1}(f-\varphi),\label{e2.5}$$ where $\varphi(z)\not\equiv 0$ is an entire function satisfying $\sigma_{p+1}(\varphi)<\mu_p(A_0)$. \end{theorem} \begin{remark} \label{rmk0}\rm All solutions of \eqref{e2.1} in Theorems \ref{thm2.1}, \ref{thm2.2}, \ref{thm2.3} are of regular growth $\mu_{p+1}(f)=\sigma_{p+1}(f)$, when the coefficient $A_0(z)$ is of regular growth $\mu_p(A_0)=\sigma_p(A_0)$. \end{remark} \begin{theorem} \label{thm2.4} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be meromorphic functions of finite iterated order satisfying $$\max\{\lambda_p(\frac{1}{A_0}),\sigma_p(A_j),j=1,\dots,n-1\} <\mu_p(A_0)\leq\sigma_p(A_0)<\infty,$$ if $f(z)\not\equiv 0$ is a meromorphic solution of \eqref{e2.1} satisfying $\frac{N(r,f)}{{\overline N}(r,f)}<\exp_{p-1}\{r^b\}$, $(b<\mu_p(A_0))$, then we have $$\sigma_p(A_0)=\sigma_{p+1}(f)={\lambda}_{p+1}(f^{(j)}-\varphi) =\overline{\lambda}_{p+1}(f^{(j)}-\varphi),\ (j=0,1,\dots),\label{e2.6}$$ where $\varphi(z)\not\equiv 0$ is a meromorphic function satisfying $\sigma_{p+1}(\varphi)<\sigma_p(A_0)$. \end{theorem} \begin{corollary} \label{coro2.1} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ satisfying the hypotheses of Theorem \ref{thm2.1}, then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $$\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f) =\lambda_{p+1}(f^{(j)}-\varphi)=\overline{\lambda}_{p+1}(f^{(j)}-\varphi),$$ where $\varphi(z)\not\equiv 0$ is an entire function satisfying $\sigma_{p+1}(\varphi)<\sigma_p(A_0)$. \end{corollary} \section{Lemmas for the proofs of main results} \begin{lemma} \label{lem3.1} Let $f(z)$ be a transcendental entire function. There exists a set $E_1$ of $r$ of finite logarithmic measure, such that for all $z$ satisfying $|z|=r\not\in E_1$ and $|f(z)|=M(r,f)$, we have $$\frac{f^{(k)}(z)}{f(z)}=\big(\frac{\nu_f(r)}{z}\big)^k(1+o(1)),\quad (k\in \mathbb{N}, r\not\in E_1),$$ where $\nu_f(r)$ is the central index of $f(z)$. \end{lemma} \begin{lemma}[\cite{h2,l1}] \label{lem3.2} Let $g:[0,+\infty)\to\mathbb{R}$ and $h:[0,+\infty)\to\mathbb{R}$ be monotone increasing functions. If (i) $g(r)\leq h(r)$ outside of an exceptional set of finite linear measure, or (ii) $g(r)\leq h(r)$, $r\not\in E_2\cup(0,1]$, where $E_2\subset [1,\infty)$ is a set of finite logarithmic measure, then for any constant $\alpha>1$, there exists $r_0=r_0(\alpha)>0$ such that $g(r)\leq h(\alpha r)$ for all $r>r_0$. \end{lemma} \begin{lemma}[\cite{t2}] \label{lem3.3} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be meromorphic functions of finite iterated order satisfying $\max\{\sigma_p(A_j),j=1,\dots,n-1\}<\mu_p(A_0)\leq\sigma_p(A_0)<\infty$, if $f(z)\not\equiv 0$ is a meromorphic solution of \eqref{e2.1} satisfying $\frac{N(r,f)}{{\overline N}(r,f)}<\exp_{p-1}\{r^b\}$, $(b<\mu_p(A_0))$, then $\sigma_{p+1}(f)=\sigma_p(A_0)$. \end{lemma} \begin{lemma} \label{lem3.4} Let $f(z)$ be an entire function with $\mu_p(f)<\infty$, then for any given $\varepsilon>0$, there exists a set $E_4\subset(1,+\infty)$ having infinite logarithmic measure such that for all $r\in E_4$, we have $$\mu_p(f)=\lim_{r\to\infty,\, r\in E_4}\frac{\log_pT(r,f)}{\log r} =\lim_{r\to\infty,\, r\in E_4} \frac{\log_{p+1}M(r,f)}{\log r} =\lim_{r\to\infty,\, r\in E_4}\frac{\log_p{\nu_f(r)}}{\log r},$$ and $$M(r,f)<\exp_p\{r^{\mu_p(f)+\varepsilon}\}.$$ \end{lemma} \begin{proof} We use a similar proof as \cite[Lemma 3.8]{t2}. By the definition of iterated $p$-lower order, there exists a sequence $\{r_n\}_{n=1}^{\infty}$ tending to $\infty$ satisfying $(1+\frac{1}{n})r_n0$, there exists an $n_1$ such that for $n\geq n_1$ and any $r\in[r_n,(1+\frac{1}{n})r_n]$, we have $${\frac{\log_{p+1}M(r_n,f)}{\log(1+\frac{1}{n})r_n}} \leq\frac{\log_{p+1}M(r,f)}{\log r} \leq\frac{\log_{p+1}M((1+\frac{1}{n})r_n,f)}{\log r_n}.$$ Let $E_4={\cup_{n=n_1}^{\infty}}[r_n,(1+\frac{1}{n})r_n]$, then for any $r\in E_4$, we have $$\lim_{r\to\infty,\, r\in E_4} \frac{\log_{p+1}M(r,f)}{\log r} =\lim_{r_n \to\infty}{\frac{\log_{p+1}M(r_n,f)}{\log r_n}}=\mu_p(f),$$ and $$m_l E={\sum_{n=n_1}^{\infty}}{\int_{r_n}^{(1+\frac{1}{n})r_n}}{\frac{dt}{t}} ={\sum_{n=n_1}^{\infty}}\log(1+\frac{1}{n})=\infty.$$ It is easy to see $$\lim_{r\to\infty,\, r\in E_4} \frac{\log_pT(r,f)}{\log r} =\lim_{r\to\infty,\, r\in E_4}\frac{\log_{p+1}M(r,f)}{\log r} =\lim_{r\to\infty,\, r\in E_4}\frac{\log_p{\nu_f(r)}}{\log r}.$$ The proof is complete. \end{proof} \begin{lemma}[\cite{t2}] \label{lem3.5} Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions of finite iterated order satisfying $i(A_0)=p,~\sigma_p(A_0)=\sigma$ and $$\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<1,$$ then every non-trivial solution $f(z)$ of \eqref{e2.1} satisfies $\sigma_{p+1}(f)=\sigma_p(A_0)=\sigma$. \end{lemma} \begin{lemma}[\cite{g1}] \label{lem3.6} Let $f(z)$ be a transcendental meromorphic function. Let $\alpha>1$ be a constant, and $k$ and $j$ be integers satisfying $k>j\geq0$. Then the following two statements hold: (a) There exists a set $E_6\subset[1,\infty)$ which has finite logarithmic measure, and a constant $C>0$, such that for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, we have $$\big|\frac{f^{(k)}(z)}{f^{(j)}(z)}\big| \leq C\big[\frac{T(\alpha r,f)}{r}(\log r)^{\alpha}\log T(\alpha r,f)\big]^{k-j}. \label{e3.1}$$ (b) There exists a set $E_6'\subset[0,2\pi)$ which has linear measure zero, such that if $\theta\in[0,2\pi)-E_6'$, then there is a constant $R=R(\theta)>0$ such that \eqref{e3.1} holds for all $z$ satisfying $\arg z=\theta$ and $|z|\geq R$. \end{lemma} \begin{lemma}[\cite{t1}] \label{lem3.7} Let $A_0(z), A_1(z), \dots,A_{n-1}(z), F(z)\not\equiv 0$ be meromorphic functions and let $f(z)$ be a meromorphic solution of \eqref{e2.2} satisfying one of the following two conditions \begin{itemize} \item[(i)] $\max\{i(F)=q, i(A_j), j=0,1,\dots,n-1\}0$ and sufficiently large $r$. By the hypotheses of Lemma \ref{lem3.10}, there exists a set $E_9$ having infinite logarithmic measure such that for all $|z|=r\in E_9$, we have $$m(r,B_0)>\exp_{p-1}\{r^{\beta_0-\varepsilon}\}.\label{e3.15}$$ By \eqref{e3.4},\eqref{e3.14} and \eqref{e3.15}, we have $$\exp_{p-1}\{r^{\beta_0-\varepsilon}\}\leq O\left(\log(rT(r,f))\right)+(n-1)\exp_{p-1} \{r^{\beta_1+\varepsilon}\},\label{e3.16}$$ for any given $\varepsilon(0<2\varepsilon<\beta_0-\beta_1)$, where $r\in E_9\backslash E, r\to\infty$, and $E$ is a set of $r$ of finite linear measure. By \eqref{e3.16}, we have $\sigma_{p+1}(f)\geq\beta_0$. \end{proof} \section{Proofs of main theorems} \begin{proof}[Proof of Theorem \ref{thm2.1}] By Theorem \ref{thmA}, we know that every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies $\sigma_{p+1}(f)=\sigma_p(A_0)$. Then we only need to prove that every solution $f(z)$ of \eqref{e2.1} satisfies $\mu_{p+1}(f)=\mu_p(A_0)$. We rewrite \eqref{e2.1} as $$|A_0(z)|\leq\big|\frac{f^{(n)}(z)}{f(z)}\big|+|A_{n-1}(z)| \big|\frac{f^{(n-1)}(z)}{f(z)}\big|+\dots+|A_1(z)|\big|\frac{f'(z)}{f(z)}\big|. \label{e4.1}$$ Set $\max\{\sigma_p(A_j):j\neq0\}=c$, then for any given $\varepsilon(0<2\varepsilon<\mu_p(A_0)-c)$ and for sufficiently large $r$, we have $$M(r,A_0)\geq\exp_{p}\{r^{\mu_p(A_0)-\varepsilon}\},\label{e4.2}$$ and $$M(r,A_j)\leq\exp_{p}\{r^{c+\varepsilon}\},\quad (j=1,2,\dots,n-1).\label{e4.3}$$ By Lemma \ref{lem3.6}, there exists a set $E_6$ having finite logarithmic measure and a constant $C>0$ such that for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, we have $$\big|\frac{f^{(k)}(z)}{f(z)}\big|\leq C(T(2r,f))^{k+1},\ (k\geq1).\label{e4.4}$$ Substituting \eqref{e4.2}-\eqref{e4.4} into \eqref{e4.1}, for the above $\varepsilon>0$, we have $$\exp_{p}\{r^{\mu_p(A_0)-\varepsilon}\}\leq Cn\exp_{p}\{r^{c+\varepsilon}\}\left(T(2r,f)\right)^{n+1},\label{e4.5}$$ for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, $r\to\infty$ and $|A_0(z)|=M(r,A_0)$. By Lemma \ref{lem3.2} and \eqref{e4.5}, we have $\mu_{p+1}(f)\geq\mu_p(A_0)-\varepsilon$. Since $\varepsilon>0$ is arbitrary, we obtain $$\mu_{p+1}(f)\geq\mu_p(A_0).\label{e4.6}$$ By \eqref{e2.1}, we have $$\big|{\frac{f^{(n)}(z)}{f(z)}}\big| \leq|A_{n-1}(z)|\big|{\frac{f^{(n-1)}(z)}{f(z)}}\big|+\dots +|A_1(z)|\big|{\frac{f'(z)}{f(z)}}\big|+|A_0(z)|.\label{e4.7}$$ By Lemma \ref{lem3.1}, there exists a set $E_1\subset(1,+\infty)$ having finite logarithmic measure such that for all $z$ satisfying $|z|=r\not\in E_1$, and $|f(z)|=M(r,f)$, we have $$\big|\frac{f^{(j)}(z)}{f(z)}\big| =\big|\frac{\nu_f(r)}{z}\big|^j|1+o(1)|,\quad (j=1,\dots,n).\label{e4.8}$$ By Lemma \ref{lem3.4}, there exists a set $E_4\subset(1,+\infty)$ having infinite logarithmic measure such that for all $|z|=r\in E_4\backslash E_1$, we have $$|A_0(z)|\leq M(r,A_0)\leq\exp_p\{r^{\mu_p(A_0)+\varepsilon}\}. \label{e4.9}$$ Hence, by \eqref{e4.3},\eqref{e4.7}-\eqref{e4.9}, we have $$|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{r^{\mu_p(A_0)+\varepsilon}\} r^n|\nu_f(r)|^{n-1}|1+o(1)|,$$ then we obtain $$|\nu_f(r)||1+o(1)|\leq nr^n\exp_p\{r^{\mu_p(A_0)+\varepsilon}\},\quad (r\in E_4\backslash E_1).\label{e4.10}$$ By the definition of iterated $p$-lower order and \eqref{e4.10}, we have $\mu_{p+1}(f)\leq\mu_p(A_0)+\varepsilon$. Since $\varepsilon>0$ is arbitrary, we have $$\mu_{p+1}(f)\leq\mu_p(A_0).\label{e4.11}$$ By \eqref{e4.6} and \eqref{e4.11}, we obtain $\mu_{p+1}(f)=\mu_p(A_0)$. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.2}] By Theorem \ref{thmB}, we have $\sigma_{p+1}(f)=\sigma_p(A_0)$. Now we need to prove (1) $\mu_{p+1}(f)=\mu_p(A_0)$ and (2) $\sigma_{p+1}(f)=\overline{\lambda}_{p+1}{(f-\varphi)}$. (1) On the one hand, we set $b=\max\{\sigma_p(A_j),\sigma_p(A_j)<\mu_p(A_0)\}$. If $\sigma_p(A_j)<\mu_p(A_0)$, then for any given $\varepsilon(0<2\varepsilon<\min\{\mu_p(A_0)-b,\tau-\tau_1\})$ and for sufficiently large $r$, we have $$M(r,A_j)\leq\exp_{p}\{r^{b+\varepsilon}\}\leq\exp_{p} \{r^{\mu_p(A_0)-\varepsilon}\}.\label{e4.12}$$ If $\sigma_p(A_j)=\mu_p(A_0),~\tau_p(A_j)\leq\tau_1<\tau =\underline{\tau}_p(A_0)$, then for sufficiently large $r$, we have \begin{gather} M(r,A_j)\leq\exp_{p}\{(\tau_1+\varepsilon)r^{\mu_p(A_0)}\},\label{e4.13} \\ M(r,A_0)\geq\exp_{p}\{(\tau-\varepsilon)r^{\mu_p(A_0)}\}.\label{e4.14} \end{gather} By \eqref{e4.12}, \eqref{e4.13}, \eqref{e4.14}, \eqref{e4.1} and \eqref{e4.4}, we obtain $$\exp_{p}\{(\tau-\varepsilon)r^{\mu_p(A_0)}\}\leq n\exp_{p}\{(\tau_1+\varepsilon)r^{\mu_p(A_0)}\}CT(r,f)^{n+1},\label{e4.15}$$ where $C>0$ is a constant, for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, $r\to\infty$ and $|A_0(z)|=M(r,A_0)$. By Lemma \ref{lem3.2} and \eqref{e4.15}, we have $\mu_{p+1}(f)\geq\mu_p(A_0)$. On the other hand, by Lemma \ref{lem3.4}, there exists a set $E_4$ having infinite logarithmic measure such that for all $r\in E_4$, we have $$|A_0(z)|\leq M(r,A_0)\leq\exp_p\{(\tau+\varepsilon)r^{\mu_p(A_0)}\}.\label{e4.16}$$ By \eqref{e4.7}, \eqref{e4.8}, \eqref{e4.12}, \eqref{e4.13} and \eqref{e4.16}, we have $$|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{(\tau+\varepsilon)r^{\mu_p(A_0)}\}r^n| \nu_f(r)|^{n-1}|1+o(1)|,\label{e4.17}$$ where $r\in E_4\backslash E_1$, $r\to\infty$. By the definition of iterated $p$-lower order and \eqref{e4.17}, we obtain $\mu_{p+1}(f)\leq\mu_p(A_0)$. Thus, we have $\mu_{p+1}(f)=\mu_p(A_0)$. (2) We prove that $\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Assume that $f(z)\not\equiv 0$ is a solution of \eqref{e2.1}, then $\sigma_{p+1}(f)=\sigma_p(A_0)$. Set $g=f-\varphi$, since $\sigma_{p+1}(\varphi)<\mu_p(A_0)\leq\sigma_p(A_0)$, then $\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$, $\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$. Substituting $f=g+\varphi, f'=g'+\varphi',\dots, f^{(n)}=g^{(n)}+\varphi^{(n)}$, into \eqref{e2.1}, we obtain $$g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g =-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi]. \label{e4.18}$$ If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv 0$, then by Lemma \ref{lem3.8}, we have $\sigma_{p+1}(\varphi)\geq\mu_p(A_0)$, which is a contradiction. Since $F(z)\not\equiv 0$ and $\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$. By Lemma \ref{lem3.7} and \eqref{e4.18}, we have $\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$. Therefore, $\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi) =\sigma_{p+1}(f)=\sigma_p(A_0)$. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.3}] By Lemma \ref{lem3.5}, we have $\sigma_{p+1}(f)=\sigma_p(A_0)$. Now we need to prove (1) $\mu_{p+1}(f)=\mu_p(A_0)$ and (2) $\sigma_{p+1}(f)=\overline{\lambda}_{p+1}{(f-\varphi)}$. (1) On the one hand, by \eqref{e4.1} and the logarithmic derivative lemma, we have $$m(r,A_0)\leq \sum_{j=1}^{n-1}m(r,A_j)+O(\log(rT(r,f))),\quad (r\not\in E),\label{e4.19}$$ where $E$ is a set of $r$ of finite linear measure. Setting $\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<\beta<1$, for sufficiently large $r$, we have $$\sum_{j=1}^{n-1}m(r,A_j)<\beta m(r,A_0).\label{e4.20}$$ By \eqref{e4.19} and \eqref{e4.20}, we have $$(1-\beta)m(r,A_0)\leq O(\log(rT(r,f))),\quad (r\not\in E).\label{e4.21}$$ By $\mu_p(A_0)=\mu$, for any given $\varepsilon>0$ and sufficiently large $r$, we have $$m(r,A_0)\geq\exp_{p-1}\{r^{\mu-\varepsilon}\}.\label{e4.22}$$ By \eqref{e4.21} and \eqref{e4.22}, for the above $\varepsilon>0,r\not\in E,r\to\infty$, we have $$(1-\beta)\exp_{p-1}\{r^{\mu-\varepsilon}\}\leq O\big(\log(rT(r,f))\big). \label{e4.23}$$ By Lemma \ref{lem3.2} and \eqref{e4.23}, we have $\mu-\varepsilon\leq\mu_{p+1}(f)$. Since $\varepsilon>0$ is arbitrary, we have $\mu_p(A_0)=\mu\leq\mu_{p+1}(f)$. On the other hand, since $\max\{\sigma_p(A_j),j\neq0\}\leq\mu_p(A_0)=\mu$, for any given $\varepsilon>0$ and sufficiently large $r$, we have $$|A_j(z)|\leq\exp_p\{r^{\mu+\varepsilon}\},\quad (j=1,\dots,n-1).\label{e4.24}$$ By Lemma \ref{lem3.4}, there exists a set of $E_2$ having infinite logarithmic measure such that for all $r\in E_2$, we have $$|A_0(z)|\leq\exp_p\{r^{\mu+\varepsilon}\}.\label{e4.25}$$ By \eqref{e4.7}, \eqref{e4.8}, \eqref{e4.24} and \eqref{e4.25}, we have $$|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{r^{\mu+\varepsilon}\}r^n |\nu_f(r)|^{n-1}|1+o(1)|.\label{e4.26}$$ By \eqref{e4.26}, for the above $\varepsilon>0$, we obtain $$|\nu_f(r)||1+o(1)|\leq nr^n\exp_p\{r^{\mu+\varepsilon}\},\label{e4.27}$$ where $|z|=r\in E_2\backslash E_1,r\to\infty,|f(z)|=M(r,f)$. By \eqref{e4.27}, we obtain $\mu_{p+1}(f)\leq\mu+\varepsilon$. Since $\varepsilon>0$ is arbitrary, we have $\mu_{p+1}(f)\leq\mu$. Thus, we have $\mu_{p+1}(f)=\mu_p(A_0)$. (2) We prove that $\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Setting $g=f-\varphi$, since $\sigma_{p+1}(\varphi)<\mu_p(A_0)$, we have $\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$, $\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$. Substituting $f=g+\varphi,f'=g'+\varphi',\dots, f^{(n)}=g^{(n)}+\varphi^{(n)}$ into \eqref{e2.1}, we obtain $$g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g =-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi].\label{e4.28}$$ If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv 0$, then by part (1), we have $\sigma_{p+1}(\varphi)\geq\mu_p(A_0)$, which is a contradiction. Since $F(z)\not\equiv 0$ and $\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$. By Lemma \ref{lem3.7} and \eqref{e4.28}, we have $\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$. Therefore, $\mu_p(A_0)=\mu_{p+1}(f)\leq\sigma_{p+1}(f)=\sigma_p(A_0) =\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi)$. The proof is complete. \end{proof} \section{Proof of Theorem \ref{thm2.4}} By Lemma \ref{lem3.3}, we have $\sigma_{p+1}(f)=\sigma_p(A_0)$. Now we prove that $\overline{\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$. (1) We prove the $\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Setting $g=f-\varphi$, since $\sigma_{p+1}(\varphi)<\sigma_p(A_0)$, we have $\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$, $\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$. Substituting $f=g+\varphi, f'=g'+\varphi',\dots, f^{(n)}=g^{(n)}+\varphi^{(n)}$ into \eqref{e2.1}, we obtain $$g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g =-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi].\label{e5.1}$$ Since $\lambda_p(\frac{1}{A_0})<\mu_p(A_0)$, we have $N(r,A_0)=o(T(r,A_0)),r\to\infty$. Therefore, by Lemma \ref{lem3.9}, we have $$\begin{split} \sigma_p(A_0)&=\limsup_{r\to\infty} \frac{\log_p T(r,A_0)}{\log r} =\lim_{r\to\infty,\, r\in E_8}\frac{\log_p T(r,A_0)}{\log r}\\ &=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}, \end{split}\label{e5.2}$$ where $E_8$ is a subset of $r$ of infinite logarithmic measure. Combining the assumption and \eqref{e5.2}, we have $$\limsup_{r\to\infty} \frac{\log_p m(r,A_j)}{\log r}<\lim_{r\to\infty,\, r\in E_8}\frac{\log_p m(r,A_0)}{\log r}=\sigma_p(A_0),\ j=1,\dots,n.\label{e5.3}$$ If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv 0$, then by Lemma \ref{lem3.10}, we have $\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction. Since $F(z)\not\equiv 0$ and $\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$, by Lemma \ref{lem3.7} and \eqref{e5.1}, we have $\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$. Therefore, $\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi) =\sigma_{p+1}(f)=\sigma_p(A_0)$. (2) We prove that $\overline{\lambda}_{p+1}(f'-\varphi)=\sigma_{p+1}(f)$. Setting $g_1=f'-\varphi$, we have $\sigma_{p+1}(g_1)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and $$f'=g_1+\varphi,\dots,f^{(n+1)}=g_1^{(n)}+\varphi^{(n)}.\label{e5.4}$$ By \eqref{e2.1}, we have $$f(z)=-\frac{1}{A_0(z)}\left(f^{(n)}+\dots+A_1(z)f'\right).\label{e5.5}$$ The derivative of \eqref{e2.1} is $$f^{(n+1)}+A_{n-1}f^{(n)}+(A_{n-1}'+A_{n-2})f^{(n-1)}+\dots +(A_1'+A_0)f'+A_0'f=0.\label{e5.6}$$ Substituting \eqref{e5.4} and \eqref{e5.5} into \eqref{e5.6}, we obtain \begin{align*} &g_1^{(n)}+(A_{n-1}-\frac{A_0'}{A_0})g_1^{(n-1)}+(A_{n-2}+A_{n-1}' -\frac{A_{n-1}A_0'}{A_0})g_1^{(n-2)} +\dots\\ & +(A_0+A_1'-\frac{A_1A_0'}{A_0})g_1\\ &=-\big[\varphi^{(n)}+(A_{n-1}-\frac{A_0'}{A_0})\varphi^{(n-1)}+\dots +(A_0+A_1'-\frac{A_1A_0'}{A_0})\varphi\big]. \end{align*} Setting $$\begin{gathered} B_{n-1}=A_{n-1}-\frac{A_0'}{A_0},\quad B_{n-2}=A_{n-2}+A_{n-1}'-\frac{A_{n-1}A_0'}{A_0},\\ \dots,\quad B_0=A_0+A_1'-\frac{A_1A_0'}{A_0}, \end{gathered}\label{e5.7}$$ we have $$g_1^{(n)}+B_{n-1}g_1^{(n-1)}+B_{n-2}g_1^{(n-2)}+\dots+B_0g_1 =-\big[\varphi^{(n)}+B_{n-1}\varphi^{(n-1)}+\dots+B_0\varphi\big].\label{e5.8}$$ By \eqref{e5.3} and \eqref{e5.7}, we have $$\limsup_{r\to\infty} \frac{\log_pm(r,B_j)}{\log r}<\lim_{r\to\infty,\, r\in E_8} \frac{\log_pm(r,A_0)}{\log r}=\sigma_p(A_0),\quad (j\neq0),\label{e5.9}$$ and $$\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r} =\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,B_0)}{\log r},\label{e5.10}$$ where $E_8$ is a subset of infinite logarithmic measure $r$. Let $F_1(z)=\varphi^{(n)}+B_{n-1}\varphi^{(n-1)}+\dots+B_0\varphi$. We affirm $F_1(z)\not\equiv 0$. If $F_1(z)\equiv 0$, then by \eqref{e5.9}, \eqref{e5.10} and Lemma \ref{lem3.10}, we obtain $\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction. Since $F_1(z)\not\equiv 0$, and $\sigma_{p+1}(F_1)<\sigma_{p+1}(g_1)=\sigma_p(A_0)$. By Lemma \ref{lem3.7} and \eqref{e5.8}, we obtain $$\overline{\lambda}_{p+1}(f'-\varphi)={\lambda}_{p+1}(f'-\varphi)=\sigma_{p+1}(f).$$ (3) We prove that $\overline{\lambda}_{p+1}(f''-\varphi)=\sigma_{p+1}(f)$. Setting $g_2=f''-\varphi$, we have $\sigma_{p+1}(g_2)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and $$f''=g_2+\varphi,\dots,f^{(n+2)}=g_2^{(n)}+\varphi^{(n)}.\label{e5.11}$$ Substituting \eqref{e5.5} into \eqref{e5.6}, we have $$\begin{split} &f^{(n+1)}+(A_{n-1}-\frac{A_0'}{A_0})f^{(n)}+(A_{n-2}+A_{n-1}' -\frac{A_{n-1}A_0'}{A_0})f^{(n-1)}+\dots\\ &+(A_0+A_1'-\frac{A_1A_0'}{A_0})f'=0. \end{split}\label{e5.12}$$ The derivative of \eqref{e5.12} is $$\begin{split} &f^{(n+2)}+(A_{n-1}-\frac{A_0'}{A_0})f^{(n+1)} +\big[(A_{n-1}-\frac{A_0'}{A_0})'+(A_{n-2}+A_{n-1}'-\frac{A_{n-1}A_0'}{A_0})\big] f^{(n)}\\ & +\dots+(A_0+A_1'-\frac{A_1A_0'}{A_0})'f'=0. \end{split}\label{e5.13}$$ By \eqref{e5.12}, we have $$f'=-\big[\frac{1}{A_0+A_1'-\frac{A_1A_0'}{A_0}}f^{(n+1)} +\frac{A_{n-1}-\frac{A_0'}{A_0}}{A_0+A_1' -\frac{A_1A_0'}{A_0}}f^{(n)}+\dots+\frac{A_1+A_2' -\frac{A_2A_0'}{A_0}}{A_0+A_1'-\frac{A_1A_0'}{A_0}}f''\big].\label{e5.14}$$ Substituting \eqref{e5.14} into \eqref{e5.13}, we have \begin{align*} &f^{(n+2)}+\big[(A_{n-1}-\frac{A_0'}{A_0})-\frac{(A_0+A_1' -\frac{A_1A_0'}{A_0})'}{A_0+A_1'-\frac{A_1A_0'}{A_0}}\big]f^{(n+1)}+\dots\\ & +\big[(A_0+A_1'-\frac{A_1A_0'}{A_0})+(A_1+A_2'-\frac{A_2A_0'}{A_0})'\\ &-\frac{(A_1+A_2'-\frac{A_2A_0'}{A_0})(A_0+A_1'-\frac{A_1A_0'}{A_0})'}{A_0+A_1' -\frac{A_1A_0'}{A_0}}\big]f'' =0. \end{align*} Setting $$\begin{gathered} C_{n-1}=B_{n-1}-\frac{B_0'}{B_0},\quad C_{n-2}=B_{n-2}+B_{n-1}'-\frac{B_{n-1}B_0'}{B_0},\\ \dots,\quad C_0=B_0+B_1'-\frac{B_1B_0'}{B_0}, \end{gathered}\label{e5.15}$$ we obtain $$f^{(n+2)}+C_{n-1}(z)f^{(n+1)}+\dots+C_0(z)f''=0.\label{e5.16}$$ Substituting \eqref{e5.11} into \eqref{e5.16}, we obtain $$g_2^{(n)}+C_{n-1}(z)g_2^{(n-1)}+\dots+C_0(z)g_2 =-\big[\varphi^{(n)}+C_{n-1}(z)\varphi^{(n-1)}+\dots+C_0(z)\varphi\big]. \label{e5.17}$$ By \eqref{e5.2}, \eqref{e5.9}, \eqref{e5.10} and \eqref{e5.15}, we have $$\limsup_{r\to\infty} \frac{\log_pm(r,C_j)}{\log r}<\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}=\sigma_p(A_0),(j\neq0),\label{e5.18}$$ and $$\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r} =\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,C_0)}{\log r},\label{e5.19}$$ where $E_8$ is a subset of $r$ of infinite logarithmic measure. If $F_2(z)\equiv\varphi^{(n)}+C_{n-1}(z)\varphi^{(n-1)}+\dots+C_0(z)\varphi\equiv 0$, then by \eqref{e5.18}, \eqref{e5.19} and Lemma \ref{lem3.10}, we have $\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction. Therefore, $F_2(z)\not\equiv 0$. Since $\sigma_{p+1}(F_2)<\sigma_{p+1}(g_2)=\sigma_p(A_0)$, by Lemma \ref{lem3.7} and \eqref{e5.17}, we have $$\overline{\lambda}_{p+1}(f''-\varphi)={\lambda}_{p+1}(f''-\varphi) =\sigma_{p+1}(f).$$ (4) We prove that $\overline{\lambda}_{p+1}(f'''-\varphi)=\sigma_{p+1}(f)$. Setting $g_3=f'''-\varphi$, then $\sigma_{p+1}(g_3)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and $$f'''=g_3+\varphi,\quad \dots,\quad f^{(n+3)}=g_3^{(n)}+\varphi^{(n)}.\label{e5.20}$$ The derivative of \eqref{e5.16} is $$f^{(n+3)}+C_{n-1}f^{(n+2)}+(C_{n-1}'+C_{n-2})f^{(n+1)}+\dots +(C_1'+C_0)f'''+C_0'f''=0.\label{e5.21}$$ By \eqref{e5.16}, we have $$f''=-\big[\frac{1}{C_0}f^{(n+2)}+\frac{C_{n-1}}{C_0}f^{(n+1)} +\dots+\frac{C_1}{C_0}f'''\big].\label{e5.22}$$ Substituting \eqref{e5.22} into \eqref{e5.21}, we have \begin{aligned} &f^{(n+3)}+\Big(C_{n-1}-\frac{C_0'}{C_0}\Big)f^{(n+2)} +\Big(C_{n-2}+C_{n-1}'-\frac{C_{n-1}C_0'}{C_0}\Big)f^{(n+1)}\\ &+\dots+\Big(C_0+C_1'-\frac{C_1C_0'}{C_0}\Big)f'''=0. \end{aligned} \label{e5.23} Setting $$\begin{gathered} D_{n-1}=C_{n-1}-\frac{C_0'}{C_0},\quad D_{n-2}=C_{n-2}+C_{n-1}'-\frac{C_{n-1}C_0'}{C_0},\\ \dots,\quad D_0=C_0+C_1'-\frac{C_1C_0'}{C_0}, \end{gathered}\label{e5.24}$$ we have $$f^{(n+3)}+D_{n-1}(z)f^{(n+2)}+\dots+D_0(z)f'''=0.\label{e5.25}$$ Substituting \eqref{e5.20} into \eqref{e5.25}, we obtain $$g_3^{(n)}+D_{n-1}(z)g_3^{(n-1)}+\dots+D_0(z)g_3 =-[\varphi^{(n)}+D_{n-1}(z)\varphi^{(n-1)}+\dots+D_0(z)\varphi].\label{e5.26}$$ By \eqref{e5.18}, \eqref{e5.19} and \eqref{e5.24}, we have $$\limsup_{r\to\infty} \frac{\log_pm(r,D_j)}{\log r} <\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r} =\sigma_p(A_0),\ (j\neq0),\label{e5.27}$$ and $$\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r} =\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,D_0)}{\log r},\label{e5.28}$$ where $E_8$ is a subset of $r$ of infinite logarithmic measure. Let $F_3(z)=\varphi^{(n)}+D_{n-1}(z)\varphi^{(n-1)}+\dots+D_0(z)\varphi\equiv 0$, by \eqref{e5.27}, \eqref{e5.28} and Lemma \ref{lem3.10}, we have $F_3(z)\not\equiv 0$. Since $\sigma_{p+1}(F_3)<\sigma_{p+1}(g_3)=\sigma_p(A_0)$, by Lemma \ref{lem3.7} and \eqref{e5.26}, we have $$\overline{\lambda}_{p+1}(f'''-\varphi)={\lambda}_{p+1}(f'''-\varphi) =\sigma_{p+1}(f).$$ (5) We prove that $\overline{\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$, $(j>3)$. Setting $g_j=f^{(j)}-\varphi$, $(j>3)$, then $\sigma_{p+1}(g_j)=\sigma_{p+1}(f^{(j)})=\sigma_p(A_0)$ and $$f^{(j+1)}=g_{j}'+\varphi',\quad \dots,f^{(n)}=g_{j}^{(n-j)}+\varphi^{(n-j)},\quad (j>3).\label{e5.29}$$ By successive derivation on \eqref{e5.25}, we also get an equation which has similar form with \eqref{e5.23}. Furthermore, combining \eqref{e5.29}, we can get \begin{aligned} &g_{j}^{(n)}+(H_{n-1}-\frac{H_0'}{H_0})g_{j}^{(n-1)}+\dots +(H_0+H_1'-\frac{H_1H_0'}{H_0})g_j \\ &=-\big[\varphi^{(n)}+\dots+(H_0+H_1'-\frac{H_1H_0'}{H_0})\varphi\big], \end{aligned} \label{e5.30} where $H_j(z)$, $(j=0,1,\dots,n-1)$ are meromorphic functions which have the same form as $D_j(z),(j=1,\dots,n-1)$. Setting $G_{n-1}=H_{n-1}-\frac{H_0'}{H_0}$, \dots, $G_0=H_0+H_1'-\frac{H_1H_0'}{H_0}$, we have $$\limsup_{r\to\infty} \frac{\log_pm(r,G_j)}{\log r} <\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r} =\sigma_p(A_0),\ (j\neq 0),$$ and $$\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8} \frac{\log_pm(r,A_0)}{\log r} =\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,G_0)}{\log r},$$ where $E_8$ is a subset of $r$ of infinite logarithmic measure. By Lemmas \ref{lem3.7} and \ref{lem3.10}, we can get $\overline{\lambda}_{p+1}(g_j)={\lambda}_{p+1}(g_j)=\sigma_{p+1}(g_j)$; i.e., $\overline{\lambda}_{p+1}(f^{(j)}-\varphi) ={\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$. the proof of Theorem \ref{thm2.4} is complete. \subsection*{Acknowledgements} The authors are grateful to the referees and editors for their valuable comments which lead to the improvement of this paper. This project is supported by the National Natural Science Foundation of China (11126145. 11171119), and the Natural Science Foundation of Jiangxi Province in China (20114BAB211003, 20122BAB211005). \begin{thebibliography}{00} \bibitem{b1} B. Bela\"idi; \emph{Growth and Oscillation of Solutions to Linear Differential Equations with Entire Coefficients having the same Order}, Elec. J. Diff. Eqns., 2009 (2009), no. 70, 1-10. \bibitem{b2} L. G. 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