\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 43, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/43\hfil Growth of solutions] {Growth of solutions to linear differential equations with entire coefficients \\ of slow growth} \author[C.-Y. Zhang, J. Tu\hfil EJDE-2010/43\hfilneg] {Cui-Yan Zhang, Jin Tu} \address{Cui-Yan Zhang \newline College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, 330022, China} \email{zhcy0618@163.com} \address{Jin Tu \newline College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, 330022, China} \email{tujin2008@sina.com} \thanks{Submitted February 8, 2010. Published March 26, 2010.} \thanks{Supported by grants GJJ09463 from the Youth Foundation of Education Bureau, \hfill\break\indent and 2009GQS0013 from the Natural Science Foundation of Jiangxi Province in China} \subjclass[2000]{30D35, 34M10} \keywords{Linear differential equations; hyper order, hyper lower order} \begin{abstract} In this article, we investigate the hyper order of solutions of higher-order linear differential equations with entire coefficients of slow growth. We assume that the lower order of the dominant coefficient in the high-order linear equations is less than $1/2$, and obtain some results which extend the results in \cite{c2,h4,h5,t1}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction and statement of results} We shall assume that readers are familiar with the fundamental results and the standard notation of the Nevanlinna's theory of meromorphic functions \cite{h1,l1}. We use $\sigma(f)$ and $\mu(f)$ to denote the order and low order of meromorphic function $f(z)$ respectively. We use $\sigma_{2}(f)$ and $\mu_2(f)$ to denote the hyper order and hyper lower order of $f(z)$, which are defined as \cite{y1} \begin{gather*} \sigma_{2}(f)=\limsup_{r\to \infty}\frac{\log\log T(r,f)}{\log r} =\limsup_{r\to \infty}\frac{\log_{2}T(r,f)}{\log r},\\ \mu_2(f)=\liminf_{r\to \infty} \frac{\log_{2}T(r,f)}{\log r}. \end{gather*} The hyper exponent of convergence of zeros and distinct zeros of $f(z)$ are respectively defined to be (see \cite{c1}) $$ \lambda_{2}(f)=\limsup_{r\to \infty}\frac{\log_{2}N(r,f)}{\log r},\quad \overline{\lambda_{2}}(f)=\limsup_{r\to \infty} \frac{\log_{2}\overline{N}(r,f)}{\log r}. $$ It is easy to see that $\sigma(f)=\infty$ if $\sigma_{2}(f)>0$. We denote the linear measure of a set $E\subset[1,\infty)$ by $m E=\int_{E}dt$ and the logarithmic measure of $E$ by $m_{l}E=\int_{E}\frac{dt}{t}$. The upper and lower logarithmic density of $E$ are defined by (see \cite{b3}) \begin{gather*} \overline{\mathop{\rm log\,dens}}(E) =\limsup_{r\to \infty}\frac{m_{l}(E\cap[1,r])}{\log r}, \\ \underline{\mathop{\rm log\,dens}} (E)=\liminf_{r\to \infty}\frac{m_{l}(E\cap[1,r])}{\log r}. \end{gather*} Nevanlinna's value distribution theory has become a very useful tool in investigating the growth of solutions of linear differential equations. By the definition of hyper order, the growth of infinite order solutions of the differential equations can be estimated more precisely. In recent years, many papers began to investigate the hyper order of the infinite order solutions of the linear differential equations (see e.g. \cite{c1,c2,t1}). For the second order linear differential equation \begin{equation} f''+A(z)f'+B(z)f=0, \label{e1.1} \end{equation} where $A(z),B(z)\not\equiv 0$ are entire functions, it is well known that every nonconstant solution $f$ of \eqref{e1.1} has infinite order if $\sigma(A)<\sigma(B)$ or $A(z)$ is a polynomial and $B(z)$ is transcendental. Gundersen \cite{g2} proved the following result. \begin{theorem}[\cite{g2}] \label{thmA} Let $A(z)$ and $B(z)$ be entire functions such that \begin{itemize} \item[(i)] $\sigma(B)<\sigma(A)<1/2$ or \item[(ii)] $A(z)$ is transcendental with $\sigma(A)=0$ and $\sigma(B)$ is a polynomial. \end{itemize} Then every nonconstant solution of \eqref{e1.1} has infinite order. \end{theorem} In 1991, Hellerstein, Miles and Rossi improved Theorem \ref{thmA} by proving the following result. \begin{theorem}[\cite{h4}] \label{thmB} If $A(z)$ and $B(z)$ are entire functions with $\sigma(B)<\sigma(A)\leq\frac{1}{2}$, then any nonconstant solution of \eqref{e1.1} has infinite order. \end{theorem} In 1992, Hellerstein, Miles and Rossi improved Theorem \ref{thmB} and proved the following result. \begin{theorem}[\cite{h5}] \label{thmC} $A_0,\dots,A_{k-1},F$ be entire functions. Suppose that there exists an $A_s$ $(0\leq s\leq k-1)$ such that $$ b=\max\{\sigma(F),\sigma(A_{j})(j\neq s)\}<\sigma(A_{s})\leq\frac{1}{2}. $$ Then every solution of the equation \begin{equation} f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_sf^{(s)}+\dots+A_0f=F \label{e1.2} \end{equation} is either a polynomial or an entire function of infinite order. \end{theorem} In 2000, Chen and Yang \cite{c2} gave a more precise estimate of the growth of the solutions of \eqref{e1.2} and its homogeneous differential equation and obtained the following results. \begin{theorem}[\cite{c2}] \label{thmD} Let $A_0,\dots,A_{k-1},F\not\equiv0$ be entire functions, such that there exists an $A_s$ $(0\leq s\leq k-1)$ satisfying $$ b=\max\{\sigma(F),\sigma(A_{j})(j\neq s)\}<\sigma(A_{s})<1/2. $$ Then every transcendental solution of \eqref{e1.2} satisfies $\sigma_{2}(f)=\sigma(A_{s })$. \end{theorem} \begin{theorem}[\cite{c2}] \label{thmE} Let $A_j(z)$ $(j=1,\dots,k-1)$ be entire functions such that $\max\{\sigma(A_j), j=1,\dots,k-1\}<\sigma(A_0)<\infty$, then every nontrivial solution $f$ of \begin{equation} f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_sf^{(s)}+\dots+A_0f=0 \label{e1.3} \end{equation} satisfies $\sigma_{2}(f)=\sigma(A_{0})$. \end{theorem} In 2008, Tu and Deng \cite{t1} investigated the growth of solutions of \eqref{e1.3} and obtained the following results. \begin{theorem}[\cite{t1}] \label{thmF} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions. Suppose that there exists some $s\in\{1,\dots,k-1\}$ such that $\max\{\sigma(A_j):j\not=0,s\}<\sigma(A_0)\leq1/2$ and that $A_s(z)$ has a finite deficient value, then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\sigma(A_0)\leq\sigma_2(f)\leq\sigma(A_s)$. \end{theorem} \begin{theorem}[\cite{t1}] \label{thmG} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions. Suppose that there exists some $s\in\{1,\dots,k-1\}$ such that $\max\{\sigma(A_j):j\not=0,s\}<\sigma(A_0)<1/2$. Suppose that $A_s(z)$ is an entire function of genus $q\geq1$, and that all the zeros of $A_s(z)$ lie in the angular sector $\theta_1\leq\arg z\leq\theta_2$ satisfying $$ \theta_2-\theta_1\leq\frac{\pi}{q+1}. $$ Then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\sigma(A_0)\leq\sigma_2(f)\leq\sigma(A_s)$. \end{theorem} Most of the above theorems are related to the problem: Does every transcendental solution of \eqref{e1.1}-\eqref{e1.3} have infinite order when the order of any one of the middle coefficients is greater than others? From Theorems \ref{thmA}--\ref{thmD}, we know that the answer is affirmative when the fastest growing entire coefficient satisfies $\sigma(A_{s})\leq1/2$ $(s\in \{1,\dots ,k-1\})$. It is mentioned that Theorem \ref{thmB} and Theorem \ref{thmC} also hold under the hypothesis $\sigma(B)<\mu(A)\leq 1/2$ in \eqref{e1.1} or $\max\{\sigma(F),\sigma(A_{j})(j\neq s)\}<\mu(A_{s})\leq 1/2$ in \eqref{e1.2} (see \cite{h4,h5}). However the proof of the case $\sigma(A_{s})=\frac{1}{2}$ or $\mu(A_{s})=1/2$ is more complicated. In the Theorem \ref{thm1.1} below, we estimate the hyper order of the transcendental solutions of \eqref{e1.1}-\eqref{e1.3} under the assumption that the lower order of the dominant coefficient in \eqref{e1.1}-\eqref{e1.3} is less than $1/2$. \begin{theorem} \label{thm1.1} Let $A_0,\dots,A_{k-1},F$ be entire functions such that there exists an $A_s$ $(1\leq s\leq k-1)$ satisfying $$ b=\max\{\sigma(F),\sigma(A_{j})(j\neq s)\}<\mu(A_{s})<1/2. $$ Then every transcendental solution of \eqref{e1.2} satisfies $\mu(A_{s})\leq\sigma_{2}(f)\leq\sigma(A_{s })$. Furthermore, if $F\not\equiv0$, then every transcendental solution of \eqref{e1.2} satisfies \[ \mu(A_{s})\leq\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\sigma_{2}(f) \leq\sigma(A_{s}). \] \end{theorem} \begin{corollary} \label{coro1.1} If $s=1$, then every non-constant solution $f$ of \eqref{e1.2} satisfies $\mu(A_{1})\leq\sigma_{2}(f)\leq\sigma(A_{1})$. Furthermore, if $F\not\equiv0$, then every non-constant solution $f$ of \eqref{e1.2} satisfies $\mu(A_{1})\leq\overline{\lambda}_{2}(f)=\lambda_{2}(f) =\sigma_{2}(f)\leq\sigma(A_{1})$. \end{corollary} \begin{corollary} \label{coro1.2} If $A(z),B(z)$ are entire functions with $\sigma(B)<\mu(A)<1/2$, then every solution $f\not\equiv0$ of \eqref{e1.1} satisfies $\mu(A)\leq\sigma_{2}(f)\leq\sigma(A)$. \end{corollary} \begin{corollary} \label{coro1.3} Under the hypotheses of Theorem \ref{thm1.1}, if $\varphi(z)$ is a transcendental entire function with $\sigma(\varphi)<\infty$, then every transcendental solution $f$ of \eqref{e1.2} or \eqref{e1.3} satisfies \[ \mu(A_{s})\leq\overline{\lambda}_{2}(f-\varphi) =\lambda_{2}(f-\varphi)=\sigma_{2}(f)\leq\sigma(A_{s}). \] \end{corollary} \begin{remark} \label{rmk1.1}\rm Theorem \ref{thm1.1} is an extension of Theorems \ref{thmB}--\ref{thmD}. If $\mu(A_{s})=\sigma(A_{s})<1/2$, by Theorem \ref{thmD}, we have that every transcendental solution of \eqref{e1.2} satisfies $\sigma_{2}(f)=\mu(A_{s})=\sigma(A_{s})$, then Theorem \ref{thm1.1} holds. Therefore, we only need to prove that Theorem \ref{thm1.1} holds in the case $\mu(A_{s})<1/2$ and $\mu(A_{s})<\sigma(A_{s})$. In Theorem \ref{thm1.1}, if $s=0$, we remove the restriction $\mu(A_0)<1/2$ and have the following result. \end{remark} \begin{theorem} \label{thm1.2} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions satisfying \[ \max\{\sigma(A_j):j\not=0\}<\mu(A_0)\leq\sigma(A_0)<\infty, \] then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu(A_0)=\mu_2(f)\leq\sigma_2(f)=\sigma(A_0)$. \end{theorem} \begin{corollary} \label{coro1.4} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions satisfying \[ \max\{\sigma(A_j):j\not=0\}<\mu(A_0)=\sigma(A_0)<\infty, \] then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu_2(f)=\sigma_2(f)=\sigma(A_0)$. \end{corollary} The following theorem studies the case when there are two dominant coefficients in \eqref{e1.3}. \begin{theorem} \label{thm1.3} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions. Suppose that there exists some $s\in\{1,\dots,k-1\}$ such that $\max\{\sigma(A_j):j\not=0,s\}<\mu(A_0)<1/2$ and that $A_s(z)$ has a finite deficient value, then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu(A_0)\leq\sigma_2(f)\leq\max\{\sigma(A_0),\sigma(A_s)\}$. \end{theorem} \begin{corollary} \label{coro1.5} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions. Suppose that there exists some $s\in\{1,\dots,k-1\}$ such that $\max\{\sigma(A_j):j\not=0,s\}<\mu(A_0)<1/2$. Suppose that $A_s(z)$ is an entire function of genus $q\geq1$, and that all the zeros of $A_s(z)$ lie in the angular sector $\theta_1\leq\arg z\leq\theta_2$ satisfying $$ \theta_2-\theta_1\leq\frac{\pi}{q+1}. $$ Then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu(A_0)\leq\sigma_2(f)\leq\max\{\sigma(A_0),\sigma(A_s)\}$. \end{corollary} \begin{corollary} \label{coro1.6} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions. Suppose that there exists some $s\in\{1,\dots,k-1\}$ such that $\max\{\sigma(A_j):j\not=0,s\}<\mu(A_0)<1/2$ and $A_s(z)=h_s(z)e^{a_sz}$, where $\sigma(h_s)<1$ and $a_s\neq0$ is a complex number, then every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu(A_0)\leq\sigma_2(f)\leq \max\{\sigma(A_0),1\}$. \end{corollary} \begin{remark} \label{rmk1.2} \rm Theorem \ref{thm1.2} extends Theorem \ref{thmE}. The meaning of Corollary \ref{coro1.4} is that all solutions of \eqref{e1.3} are regular growing when the dominant coefficient $A_0$ is regular growing. However, we can not give any information about $\mu_2(f)$ in Theorem \ref{thm1.1} and Theorem \ref{thm1.3}. Theorem \ref{thm1.3} is a supplement to Theorems \ref{thm1.1}--\ref{thm1.2} and an improvement of Theorem \ref{thmF}. Corollaries \ref{coro1.5}--\ref{coro1.6} are the immediate conclusions of Theorem \ref{thm1.3}, since $A_s(z)$ in Corollary \ref{coro1.5} has zero as a finite deficient value \cite{k2} and $A_s(z)=h_s(z)e^{a_sz}$ in Corollary \ref{coro1.6} also has zero as a finite deficient value. If $A_s(z)$ in Theorem \ref{thm1.3} has a finite deficient value, then $\sigma(A_s)>1/2$ \cite{t1}. \end{remark} \subsection*{Open problems} Do Theorems \ref{thm1.1} and \ref{thm1.3} hold under the hypothesis $\mu(A_{s})=\sigma(A_{s})=\frac{1}{2}$? Does Theorem \ref{thmD} hold under the hypothesis $\sigma(A_{s})=\frac{1}{2}$? \section{Lemmas for the proofs of main results} \begin{lemma}[\cite{g1}] \label{lem2.1} Let $f(z)$ be a transcendental meromorphic function and $\alpha>1$ be a given constant, for any given $\varepsilon>0$, (i) there exist a set $E_{1}\subset[1,\infty)$ that has finite logarithmic measure and a constant $B>0$ that depends only on $\alpha$ and $(m,n) (m,n\in\{0,\dots,k\}$ with $m0$ that depends only on $\alpha$ and $(m,n) (m,n\in\{0,\dots,k\}$ with $m(\cos\pi\alpha)B(r)\} >1-\frac{\sigma}{\alpha}.\label{e2.2} \end{equation} \end{lemma} \begin{lemma}[\cite{b2}] \label{lem2.3} Let $f(z)$ be entire with $\mu(f)=\mu<1/2$ and $\mu<\sigma=\sigma(f)$. If $\mu\leq\delta<\min(\sigma,\frac{1}{2})$ and $\delta<\alpha<1/2$, then \begin{equation} \overline{\mathop{\rm log\,dens}}\{r:A(r) >(\cos\pi\alpha)B(r)>r^{\delta}\}>C(\sigma,\delta,\alpha),\label{e2.3} \end{equation} where $C(\sigma,\delta,\alpha)$ is a positive constant depending only on $\sigma,\delta$ and $\alpha$. \end{lemma} \begin{lemma}[\cite{c2}] \label{lem2.4} Let $f(z)$ be a transcendental entire function. Then there is a set $E_{2}\subset(1,+\infty)$ having finite logarithmic measure such that when we take a point $z$ satisfying $|z|=r\not\in E_{2}$ and $|f(z)|=M(r,f)$, we have \begin{equation} |\frac{f(z)}{f^{(s)}(z)}|\leq 2r^{s},\quad (s\in N).\label{e2.4} \end{equation} \end{lemma} \begin{lemma}[\cite{h2,j1}] \label{lem2.5} 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 $E_{3}$ of $r$ of finite logarithmic measure, we have \begin{equation} \frac{f^{(k)}(z)}{f(z)}=\left(\frac{\nu_{f}(r)}{z}\right)^{k}(1+o(1)), \quad (k\in N, r\notin E_{3}), \label{e2.5} \end{equation} where $\nu_{f}(r)$ is the central index of $f$. \end{lemma} \begin{lemma}[\cite{c1,t2}] \label{lem2.6} Let $f(z)$ be an entire function of infinite order satisfying $\sigma_{2}(f)=\sigma$ and $\mu_{2}(f)=\mu$. Then \begin{equation} \limsup_{r\to \infty}\frac{\log_{2}\nu_{f}(r)}{\log r}=\sigma,\quad \liminf_{r\to \infty}\frac{\log_{2}\nu_{f}(r)}{\log r}=\mu, \label{e2.6} \end{equation} where $\nu_{f}(r)$ is the central index of $f$. \end{lemma} \begin{lemma}[\cite{l1}] \label{lem2.7} Let $g:(0,+\infty)\to \mathbb{R}$,$ h: (0,+\infty) \to \mathbb{R}$ be monotone increasing functions such that (i) $g(r)\leq h(r)$ 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 $r>r_0$. (ii) $g(r)\leq h(r)$ outside of an exceptional set $E_4$ of finite logarithmic measure. Then, for any $\alpha>1$, there exists $r_0>0$ such that $g(r)\leq h(r^{\alpha})$ for all $r>r_0$. \end{lemma} \begin{lemma} \label{lem2.8} Let $f(z)$ be an entire function with $\mu(f)<\infty$. Then for any given $\varepsilon>0$, there exists a set $E_{5}\subset(0,+\infty)$ having infinite logarithmic measure such that for all $r\in E_5$, we have \begin{equation} M(r,f)<\exp\{r^{\mu(f)+\varepsilon}\}.\label{e2.7} \end{equation} \end{lemma} \begin{proof} By the definition of 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 $n\geq n_1$, we have \begin{equation} M(r_n,f)\leq \exp\{r_n^{\mu(f)+\frac{\varepsilon}{2}}\}.\label{e2.8} \end{equation} Let $E_5=\bigcup_{n=n_1}^{\infty}[(\frac{n}{n+1})r_n,r_n]$, then for any ${r}\in E_5$, we have \begin{equation} M(r,f)\leq M(r_n,f)\leq \exp\{r_n^{\mu(f)+\frac{\varepsilon}{2}}\}\leq \exp\{[(1+\frac{1}{n})r]^{\mu(f)+\frac{\varepsilon}{2}}\}\leq \exp\{r^{\mu(f)+\varepsilon}\}.\label{e2.9} \end{equation} and $m_l{E_5}=\sum^{\infty}_{n=n_1}\int_{\frac{n}{n+1}r_n}^{r_n}\frac{dt}{t} =\sum^{\infty}_{n=n_1}\log(1+\frac{1}{n})=\infty$. The proof is complete. \end{proof} \begin{lemma} \label{lem2.9} Let $A_j$ $(j=0,\dots,k-1)$ be entire functions of finite order, then all solutions of \eqref{e1.3} satisfies $\mu_2(f)\leq\max\{\mu(A_0),\sigma(A_j):j=1,\dots,k-1\}$. \end{lemma} \begin{proof} From \eqref{e1.3}, we have \begin{equation} |\frac{f^{(k)}(z)}{f(z)}|\leq |A_{k-1}||\frac{f^{(k-1)}(z)}{f(z)}| +\dots+|A_s||\frac{f^{(s)}(z)}{f(z)}|+\dots+|A_0|.\label{e2.10} \end{equation} By Lemma \ref{lem2.5}, there exists a set $E_{3}\subset(1,+\infty)$ having finite logarithmic measure such that for all $z$ satisfying $|z|=r\not\in E_{3}$ and $|f(z)|=M(r,f)$, we have \begin{equation} |\frac{f^{(j)}(z)}{f(z)}|=\Big(\frac{\nu_{f}(r)}{r}\Big)^j(1+o(1)) \quad (j=1,\dots,k-1).\label{e2.11} \end{equation} Set $\max\{\mu(A_0),\sigma(A_j):j=1,\dots,k-1\}=a$, then for any given $\varepsilon>0$ and for sufficiently large $r$, we have \begin{equation} |A_{j}(z)|<\exp\{r^{a+\varepsilon}\}\quad (j=1,\dots,k-1).\label{e2.12} \end{equation} By Lemma \ref{lem2.8}, for any given $\varepsilon>0$, there exists a set $E_{5}\subset(1,+\infty)$ having infinite logarithmic measure such that for all $|z|=r\in E_{5}$, we have \begin{equation} |A_{0}(z)|<\exp\{r^{a+\varepsilon}\}.\label{e2.13} \end{equation} Substituting \eqref{e2.11}-\eqref{e2.13} into \eqref{e2.10}, for any given $\varepsilon>0$ and for sufficiently large $r\in E_{5}\backslash E_3$ and $|f(z)|=M(r,f)$, we have \begin{equation} \Big(\frac{\nu_{f}(r)}{r}\Big)^k|1+o(1)| \leq k\exp\{r^{a+\varepsilon}\}\Big(\frac{\nu_{f}(r)}{r}\Big)^{k-1} |1+o(1)|,\label{e2.14} \end{equation} then \begin{equation} \nu_{f}(r)\leq kr\exp\{r^{a+\varepsilon}\}.\label{e2.15} \end{equation} Then by Lemma \ref{lem2.6} we have $\mu_2(f)\leq a$. Thus, the proof is complete. \end{proof} \begin{lemma}[\cite{f1}] \label{lem2.10} Let $f(z)$ be a meromorphic function of finite order $\sigma$. For any given $\zeta>0$ and $l$, $00$ and $\alpha>0$ are constants, not always the same at each occurrence. Since $0<\mu(A_s)<1/2, \mu(A_s)<\sigma(A_s)$, we choose $\varepsilon,\delta$ such that \begin{equation} b+\varepsilon<\mu(A_{s})\leq\delta<\min\{\sigma(A_{s}), \frac{1}{2}\},\label{e3.4} \end{equation} where $b=\max\{\sigma(F),\sigma(A_{j})(j\neq s)\}$. For sufficiently large $r$, we have \begin{gather} |A_{j}(z)|\leq\exp\{r^{b+\varepsilon}\},\quad (j=0,\dots ,s-1,s+1,\dots,k-1),\label{e3.5} \\ |F(z)|\leq\exp\{r^{b+\varepsilon}\}. \label{e3.6} \end{gather} Since $M(r,f)>1$ for sufficiently large $r$, by \eqref{e3.6}, we have \begin{equation} \frac{|F(z)|}{M(r,f)}\leq|F(z)|\leq\exp\{r^{b+\varepsilon}\}.\label{e3.7} \end{equation} By Lemma \ref{lem2.3} ($\mu(A_{s})<\sigma(A_{s})$), there exists a set $H_1\subset(1,+\infty)$ having infinite logarithmic measure such that for all $z$ satisfying $|z|=r\in H_1$, we have \begin{equation} |A_{s}(z)|>\exp\{r^{\delta}\}.\label{e3.8} \end{equation} By Lemma \ref{lem2.4}, there is a set $E_{2}\subset(1,+\infty)$ of finite logarithmic measure such that for a point $z$ satisfying $|z|=r\notin [0,1)\cup E_{2}$ and $|f(z)|=M(r,f)$, we have \begin{equation} |\frac{f(z)}{f^{(s)}(z)}|\leq 2r^{s}.\label{e3.9} \end{equation} By \eqref{e3.1}-\eqref{e3.3} and \eqref{e3.5}-\eqref{e3.9}, for all $z$ satisfying $|z|=r\in H_1-([0,1]\cup E_{1}\cup E_{2})$ and $|f(z)|=M(r,f)$, we have \begin{equation} \exp\{r^{\delta}\}\leq M\cdot \exp\{r^{b+\varepsilon}\} \cdot r^{\alpha}\cdot [T(2r,f)]^{2k}.\label{e3.10} \end{equation} Again by \eqref{e3.4} and \eqref{e3.10}, we see that for a point $z$ satisfying$|z|=r\in H_1-([0,1]\cup E_{1}\cup E_{2})$ and $|f(z)|=M(r,f)$, we have \begin{equation} \exp\{r^{\delta}(1+o(1))\}\leq[T(2r,f)]^{2k}.\label{e3.11} \end{equation} Since $\delta$ is arbitrarily close to $\mu(A_{s})$, from \eqref{e3.11}, we obtain \begin{equation} \limsup_{r\to \infty}\frac{\log_{2}T(r,f)}{\log r}\geq\mu(A_{s}). \label{e3.12} \end{equation} On the other hand, from Lemma \ref{lem2.5}, there is a set $E_{3}\subset(1,+\infty)$ having finite logarithmic measure such that for all $z$ satisfying $|z|=r\notin [0,1]\cup E_{3}$ and $|f(z)|=M(r,f)$, we have \begin{equation} \frac{f^{(j)}(z)}{f(z)}=\Big(\frac{\nu_{f}(r)}{z}\Big)^{j}(1+o(1)), \quad (j=1,\dots,k).\label{e3.13} \end{equation} For any given $\varepsilon>0$ and for sufficiently large $r$, we have \begin{equation} |A_{s}(z)|\leq\exp\{r^{\sigma(A_{s})+\varepsilon}\}. \label{e3.14} \end{equation} Now we take a point $z$ satisfying $|z|=r\notin [0,1]\cup E_{3}$ and $|f(z)|=M(r,f)$ and substitute \eqref{e3.5}-\eqref{e3.7},\eqref{e3.13}-\eqref{e3.14} into \eqref{e1.2}, then we obtain \begin{equation} \Big(\frac{\nu_{f}(r)}{|z|}\Big)^{k}|1+o(1)|\leq(k+1) \Big(\frac{\nu_{f}(r)}{|z|}\Big)^{k-1}|1+o(1)|\exp\{r^{\sigma(A_{s}) +\varepsilon}\}.\label{e3.15} \end{equation} This gives \begin{equation} \limsup_{r\to \infty}\frac{\log_{2}\nu_{f}(r)}{\log r}\leq\sigma(A_{s})+\varepsilon.\label{e3.16} \end{equation} Since $\varepsilon$ is arbitrary, by Lemma \ref{lem2.6} and \eqref{e3.16}, we have $\sigma_{2}(f)\leq\sigma(A_{s})$. Combining this and \eqref{e3.12}, we obtain $$ \mu(A_{s})\leq\sigma_{2}(f)\leq\sigma(A_{s }). $$ Assume that if $f$ is a transcendental solution of \eqref{e1.2}, then $\sigma(f)=\infty$ by \eqref{e3.12}. We next show that $\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\sigma_{2}(f)$ if $F\not\equiv0$. By \eqref{e1.2}, it is easy to see that if $f$ has a zero at $z_{0}$ of order more than $k$, then $F$ must has a zero at $z_{0}$. Hence we have \begin{equation} N(r,\frac{1}{f})\leq k\overline{N}(r,\frac{1}{f})+N(r,\frac{1}{F}).\label{e3.17} \end{equation} 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{e3.18} \end{equation} Hence \begin{equation} m(r,\frac{1}{f})\leq\sum_{j=1}^{k}m(r,\frac{f^{(j)}}{f}) +\sum_{j=0}^{k-1}m(r,A_{j})+m(r,\frac{1}{F}).\label{e3.19} \end{equation} By \eqref{e3.17} and \eqref{e3.19}, we obtain that \begin{equation} T(r,f)\leq k\overline{N}(r,\frac{1}{f})+M(\log(rT(r,f)))+T(r,F) +\sum_{j=0}^{k-1}T(r,A_{j}),\quad (r\notin E_{4}), \label{e3.20} \end{equation} where $E_{4}\subset(0,+\infty)$ is a set having finite linear measure. For sufficiently large $r$, we have \begin{gather} M(\log(rT(r,f)))\leq\frac{1}{2}T(r,f),\label{e3.21}\\ \sum_{j=0}^{k-1}T(r,A_{j})+T(r,F)\leq(k+1)r^{\sigma(A_{s})+\varepsilon}. \label{e3.22} \end{gather} By \eqref{e3.20}-\eqref{e3.22}, we have \begin{equation} T(r,f)\leq2k\overline{N}(r,\frac{1}{f})+2(k+1)r^{\sigma(A_{s}) +\varepsilon},\label{e3.23} \end{equation} hence $\sigma_{2}(f)\leq\overline{\lambda}_{2}(f)$ by \eqref{e3.23}. Therefore, $\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\sigma_{2}(f)$. By $\mu(A_{s})\leq\sigma_{2}(f)\leq\sigma(A_{s})$, we have $\mu(A_{s})\leq\overline{\lambda}_{2}(f)=\lambda_{2}(f) =\sigma_{2}(f)\leq\sigma (A_{s})$. \subsection{Proof of Corollaries} Using the similar proof in Theorem \ref{thm1.1}, we can easily obtain the Corollaries \ref{coro1.1}--\ref{coro1.2}. Assume that $f(z)$ is a transcendental solution of \eqref{e1.2} or \eqref{e1.3}, then we have $\mu(A_{s})\leq\sigma_{2}(f)\leq\sigma(A_{s})$. Set $g(z)=f(z)-\varphi$, then we have $\sigma_2(g)=\sigma_2(f)$, and $\overline{\lambda}_{2}(g)=\overline{\lambda}_{2}(f-\varphi)$. Substituting $f=g+\varphi$ into \eqref{e1.2} or \eqref{e1.3}, we obtain \begin{equation} g^{(k)}+A_{k-1}g^{(k-1)}+\dots+A_{0}g =F-(\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots +A_{0}\varphi)\label{e3.24} \end{equation} or \begin{equation} g^{(k)}+A_{k-1}g^{(k-1)}+\dots+A_{0}g =-(\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_{0}\varphi).\label{e3.25} \end{equation} If $F-(\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_{0}\varphi)\equiv0$ or $\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_{0}\varphi\equiv0$, by Theorem \ref{thm1.1}, we have $\sigma(\varphi)=\infty$. This is a contradiction. Therefore, $\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_{0}\varphi\not\equiv0$ and $F-(\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_{0}\varphi)\not\equiv0$. Using the similar proof in \eqref{e3.17}-\eqref{e3.23}, we can easily obtain $\overline{\lambda}_{2}(g)=\lambda_{2}(g)=\sigma_{2}(g)$, therefore Corollary \ref{coro1.3} holds. \section{Proof of Theorem \ref{thm1.2}} From Theorem \ref{thmE}, we know that every nontrivial solution $f$ of \eqref{e1.3} satisfies $\sigma_{2}(f)=\sigma(A_{0})$. Then we only need to prove that every nontrivial solution $f$ of \eqref{e1.3} satisfies $\mu_{2}(f)=\mu(A_{0})$. From \eqref{e1.3}, we have \begin{equation} -A_0=\frac{f^{(k)}}{f}+\dots+A_1\frac{f'}{f}.\label{e4.1} \end{equation} By this equality and the logarithmic derivative lemma, we have \begin{equation} m(r,A_0)\leq\sum_{j=1}^{k}m\left(r,\frac{f^{(k)}}{f}\right) +\sum_{j=1}^{k-1}m(r,A_{j})\leq O\{\log rT(r,f)\} + \sum_{j=1}^{k-1}m(r,A_{j}),\label{e4.2} \end{equation} where $r\not\in E_7$, $E_7\subset (1,\infty)$ is a set having finite linear measure, not necessarily the same at each occurrence. Set $\max\{\sigma(A_j):j\not=0\}=c$, then for any given $\varepsilon(0<2\varepsilon<\mu(A_0)-c)$ and for sufficiently large $r$, we have \begin{equation} m(r,A_0)>r^{\mu(A_0)-\varepsilon},\quad m(r,A_j)0$ at $a\in C$ as stated in the hypothesis. Then it follows from the definition of deficiency that for all sufficiently large $r$, we have $$ m\Big(r,\frac{1}{A_s-a}\Big)\geq dT(r,A_s). $$ Hence, for any sufficiently large $r$, there exists a point $z_r=re^{i\theta_r}$ such that \begin{equation} \log|A_s(z_r)-a|\leq-dT(r,A_s).\label{e5.1} \end{equation} Assume first that $A_s(z)$ has zero as a deficient value, that is, $a=0$. By Lemma \ref{lem2.10}, for any given $l(00$, there exists a set $E_\zeta\subset[0,\infty)$ of lower logarithmic density greater than $1-\zeta$ such that for all $r\in E_\zeta$ and for all $\theta\in[\theta_r-l,\theta_r+l]$, then we have $$ \log|A_s(re^{i\theta})|\leq0. $$ In fact, if we choose $l$ sufficiently small in \eqref{e2.16}, we have \begin{align*} \log|A_s(re^{i\theta})| &=\log|A_s(re^{i\theta_r})|+ \int^{\theta}_{\theta_r}\frac{d}{dt}\log|A_s(re^{it})|dt\\ &\leq-d T(r,A_s)+r\int^{\theta}_{\theta_r}|\frac{A_s'(re^{it})} {A_s(re^{it})}||dt|\\ &\leq(-d+\varepsilon_1)T(r,A_s)\leq 0, \end{align*} where $0<\varepsilon_10$ and for sufficiently small $l>0$, there exists a set $E_\zeta\subset[1,\infty)$ of lower logarithmic density greater than $1-\zeta$ such that for all $r\in E_\zeta$ and for all $\theta\in[\theta_r-l,\theta_r+l]$, we have $$ \log|A_s(re^{i\theta})-a|\leq0. $$ Thus for these $r$ and $\theta$, we have \begin{equation} |A_s(re^{i\theta})|\leq|a|+1.\label{e5.2} \end{equation} Since $0<\mu(A_0)<1/2$, we divide the proof into two cases: (i) $0<\mu(A_0)=\sigma(A_0)<1/2$; (ii) $0<\mu(A_0)<1/2, \mu(A_0)<\sigma(A_0)$.\par Case (i): $0<\mu(A_0)=\sigma(A_0)<1/2$. By Lemma \ref{lem2.2}, there exists a set $H_1\subset[1,\infty)$ of lower logarithmic density greater than $0$ such that for any given $\varepsilon_2>0$ and for all $r\in H_1$, we have \begin{equation} |A_0(z)|>\exp\{r^{\mu(A_0)-\varepsilon_2}\}.\label{e5.3} \end{equation} Let $f\not\equiv0$ be a solution of \eqref{e1.3}. From \eqref{e1.3}, we obtain \begin{equation} |A_0(z)|\leq|\frac{f^{(k)}(z)}{f(z)}|+\dots+ |A_s(z)\frac{f^{(s)}(z)}{f(z)}| +\dots+|A_1(z)\frac{f'{(z)}}{f(z)}|.\label{e5.4} \end{equation} By Lemma \ref{lem2.1}(ii), there exists a set $E_1\subset[0,2\pi)$ with linear measure zero such that for all $z=re^{i\theta}$ satisfying $\arg z=\theta\in[0,2\pi)\backslash E_1$ and for all sufficiently large $r$, we have \begin{equation} |\frac{f^{(j)}(re^{i\theta})}{f(re^{i\theta})}|\leq r[T(2r,f)]^k, \quad (j=1,\dots,k).\label{e5.5} \end{equation} Furthermore, choosing $\varepsilon_2$ small enough such that $\max\{\sigma(A_j),j\not=0,s\}=\beta<\mu(A_0)-2\varepsilon_2$, then for sufficiently large $r$, we have \begin{equation} |A_j(z)|<\exp\{r^{\beta+\varepsilon_2}\},\quad (j\not=0,s). \label{e5.6} \end{equation} Hence by \eqref{e5.2}-\eqref{e5.6}, for all sufficiently large $r\in E_\zeta\bigcap H_1$ and for all $\theta\in[\theta_r-l,\theta_r+l]\backslash E_1$, we have \begin{equation} \exp\{r^{\mu(A_0)-\varepsilon_2}\}\leq kr\exp\{r^{\beta+\varepsilon_2}\} [T(2r,f)]^k.\label{e5.7} \end{equation} Since $\varepsilon_2$ is arbitrarily small and $\beta+\varepsilon_2<\mu(A_0)-\varepsilon_2$, by \eqref{e5.7}, we have $\sigma_2(f)\geq\mu(A_0)$. On the other hand, by Lemma \ref{lem2.11}, $\sigma_2(f)\leq \sigma(A_s)=\max\{\sigma(A_0),\sigma(A_s)\}$. Therefore, every solution $f\not\equiv0$ of \eqref{e1.3} satisfies $\mu(A_0)\leq\sigma_2(f)\leq\sigma(A_s) =\max\{\sigma(A_0),\sigma(A_s)\}$. Case (ii) $0<\mu(A_0)<1/2, \mu(A_0)<\sigma(A_0)$. By Lemma \ref{lem2.3}, there exists a set $H_2\subset[1,\infty)$ of upper logarithmic density greater than $0$ such that for any given $\delta(\mu(A_0)\leq\delta<\min(\sigma,\frac{1}{2}))$ and for all $r\in H_2$, we have \begin{equation} |A_0(z)|>\exp\{r^\delta\}.\label{e5.8} \end{equation} Note that the set $E_{\zeta}\cap H_2$ has a positive upper logarithmic density. In fact, without loss of generality, set $\overline{\mathop{\rm log\,dens}}(H_2)=2\zeta>0$, we have $$ \zeta\leq\overline{\mathop{\rm log\,dens}}(H_2) +\underline{\mathop{\rm log\,dens}} (E_{\zeta})-\overline{\mathop{\rm log\,dens}}(E_{\zeta}\cup H_2)\leq \overline{\mathop{\rm log\,dens}}(E_{\zeta}\cap H_2). $$ By the same reasoning, we know that the set $E_{\zeta}\cap H_1$ in \eqref{e5.7} also has a positive upper logarithmic density. Hence from \eqref{e5.4}-\eqref{e5.6} and \eqref{e5.8}, for all sufficiently large $r$ in $E_{\zeta}\cap H_2$ and for all $\theta\in[\theta_r-l,\theta_r+l]\backslash E_1$, we have \begin{equation} \exp\{r^{\delta}\}\leq kr\exp\{r^{\beta+\varepsilon_2}\} [T(2r,f)]^k,\label{e5.9} \end{equation} where $0<\varepsilon_2<\delta-\beta$. By \eqref{e5.9}, we get $\sigma_2(f)\geq \delta$, since $\delta$ is arbitrarily close to $\mu(A_0)$, we have $\sigma_2(f)\geq\mu(A_0)$. 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