\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 125, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/125\hfil Growth of meromorphic solutions]
{Growth of meromorphic solutions of higher order
linear differential equations}

\author[L. Wang, H. Liu \hfil EJDE-2014/125\hfilneg]
{Lijun Wang, Huifang Liu}  % in alphabetical order

\address{Lijun Wang \newline
College of Mathematics and Information Science,
Jiangxi Normal University, Nanchang 330022, China}
\email{lijunwangz@163.com}

\address{Huifang Liu \newline
College of Mathematics and Information Science,
Jiangxi Normal University, Nanchang 330022, China}
\email{liuhuifang73@sina.com, 925268196@qq.com}

\thanks{Submitted January 27, 2014. Published May 14, 2014.}
\subjclass[2000]{30D35, 39B12}
\keywords{Meromorphic function; differential equations; growth; order}

\begin{abstract}
 In this article, we investigate the growth of
 meromorphic solutions of the differential equations
 \begin{gather*}
 f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=0,\\
 f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=F,
 \end{gather*}
 where $A_j, F$ $(j=0,\dots,k-1)$ are meromorphic functions. When there
 exists one dominant coefficient with lower order less than $1/2$, we
 obtain some estimations of the hyper order and the hyper convergence
 exponent of zeros of meromorphic solutions of the above  equations.
\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

\section{Introduction and statement of results}

It is assumed that the reader is familiar with the standard
notations and the fundamental results of the Nevanlinna theory
\cite{b10,b13,b16}. Let $f(z)$ be a nonconstant
meromorphic function in the complex plane. We use the symbols
$\sigma(f)$ and $\mu(f)$ to denote the order and the lower order of
$f$ respectively, and use $ \lambda(f)$ and $\lambda(1/f)$
to denote the convergence exponent of zeros and poles of $f$,
respectively. In order to estimate the rate of growth of meromorphic
function of infinite order more precisely, we recall the following
definitions.

\begin{definition}[\cite{b13}]
Let $f(z)$ be a
nonconstant meromorphic function in the complex plane. Its hyper
order $\sigma_2(f)$ is defined by
$$\sigma_{2}(f)=\limsup_{r\to+\infty}\frac{\log^+\log ^+T(r,f)}{\log
r}.$$
\end{definition}


\begin{definition}[\cite{b13}] \rm
Let $f(z)$ be a nonconstant meromorphic function in the complex plane.
Its hyper convergence exponent of zeros and distinct zeros of f(z) are
respectively defined by
$$
\lambda_{2}(f)=\limsup_{r\to+\infty}\frac{\log^+\log^+ N(r,f)}{\log r},\quad
\overline{\lambda_2}(f)=\limsup_{r\to+\infty}\frac{\log^+\log^+
\overline{N}(r,f)}{\log r}.
$$
\end{definition}

 Consider the second-order linear differential equation
\begin{equation}
f''+A(z)f'+B(z)f=0, \label{e1.1}
\end{equation}
where $A(z), B(z)$ are entire functions. It is well known that every
nonconstant solution $f$ of  \eqref{e1.1} has infinite order if
$\sigma(A)<\sigma(B)$. When the order of the coefficients of
\eqref{e1.1} are less than $1/2$, Gundersen \cite{b8} proved the
following result.

\begin{theorem} \label{thmA}
Suppose that $A(z), B(z)$ are entire functions. If $\sigma(B)<\sigma(A)<1/2$,
or $A(z)$ is transcendental and $\sigma(A)=0$, $B(z)$ is a polynomial,
then every nonconstant solution $f$ of \eqref{e1.1} satisfies $\sigma(f)=\infty$.
\end{theorem}

Hellerstein, Miles and Rossi  \cite{b11} investigated the case
$\sigma(B)<\sigma(A)\leq 1/2$, and also obtained that every
nonconstant solution $f$ of \eqref{e1.1} satisfies
$\sigma(f)=\infty$, which improved Theorem \ref{thmA}. Meanwhile, in
\cite{b12} they also considered the linear differential
equation
\begin{equation}
f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=F, \label{e1.2}
\end{equation}
and obtained the following result.

\begin{theorem} \label{thmB}
 Suppose that $A_0, A_1,\dots, A_{k-1}, F$
are entire functions. If there exists some $s\in\{0, 1, \dots,
k-1\}$ such that $\max\{\sigma(F), \sigma(A_{j}):j\neq
s\}<\sigma(A_s)\leq  1/2$, then every solution $f$ of  \eqref{e1.2}
is either a polynomial  or a transcendental entire function of infinite order.
\end{theorem}

When the coefficients $A_0, A_1,\dots, A_{k-1}$ and $F$ are
meromorphic functions, many authors investigated the value
distribution of solutions of \eqref{e1.2} and its corresponding
homogeneous differential equation
\begin{equation}
f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=0 \label{e1.3}
\end{equation}
(see \cite{b1,b2,b5,b6,b7,b15}).
Especially, Chen \cite{b5} obtained the following result.

\begin{theorem} \label{thmC}
Suppose that $A_0,A_1,\dots,A_{k-1}$ are
meromorphic functions. If there exists some $A_s(0 \leq s \leq k-1)$ satisfying
\[
\max\Big\{\sigma(A_{j})(j\neq
s),\lambda(\frac{1}{A_s})\Big\}<\mu(A_s)\leq\sigma(A_s)<1/2,
\]
then every transcendental meromorphic solution $f$ whose poles are
of uniformly bounded multiplicities,  of  \eqref{e1.3} satisfies
$\sigma_{2}(f)=\sigma(A_s)$, and every nontranscendental
meromorphic solutions $f$ is a polynomial with degree
$\deg f\leq s-1$.
\end{theorem}

In \cite{b7}, the authors pointed out that the condition that the
multiplicity of poles of the solution $f$ is uniformly bounded in
Theorem \ref{thmC} is necessary (see Remark 3.1). The above obtained results
are related to the question: what conditions on coefficients
$A_j(j=0, \dots, k-1)$ will guarantee that every transcendental
solution of  \eqref{e1.2} or \eqref{e1.3} is of infinite order?
From Theorems \ref{thmA}--\ref{thmC}, we know that the answer is affirmative, if there
exists one  dominant coefficient $A_s$ such that $\mu(A_s)\leq
\sigma(A_s)<1/2$. In this paper, we continue to investigate the
above question. In the following results, we estimate the hyper
order,  the hyper convergence exponent of zeros of transcendental
meromorphic solutions of  \eqref{e1.2} or \eqref{e1.3} under the
condition that the dominant coefficient $A_s$ satisfying
$\mu(A_s)<1/2$.


\begin{theorem} \label{thm1.1}
Suppose that $A_0,A_1,\dots, A_{k-1}, F$ are meromorphic
functions of finite order. If there exists some
$A_s(0 \leq s \leq k-1)$ such that
 \begin{equation}
 b=\max\Big\{\sigma(F), \sigma(A_{j})(j\neq
s),\lambda\Big(\frac{1}{A_s}\Big)\Big\}<\mu(A_s)
<1/2, \label{e1.4}
\end{equation}
Then
\begin{itemize}
\item[(i)]
Every transcendental meromorphic solution $f$ whose poles are of
uniformly bounded multiplicities,  of \eqref{e1.2} satisfies
$\mu(A_s)\leq\sigma_2(f)\leq \sigma(A_s)$. Furthermore, if
$F\not\equiv 0$, then we have
$\mu(A_s)\leq\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)\leq
\sigma(A_s)$.
\item[(ii)]
If $s\geq 2$, then every nontranscendental meromorphic solution $f$
of \eqref{e1.2} is a polynomial with degree $\deg f\leq s-1$. If
$s=0$ or $ 1$, then every nonconstant solution of \eqref{e1.2}
is transcendental.
\end{itemize}
\end{theorem}

\begin{corollary} \label{coro1.1}
Suppose that $A_0,A_1,\dots, A_{k-1}, F(\not\equiv 0)$ are
meromorphic functions. If there exists some $A_s(0 \leq s \leq
k-1)$ such that
\begin{gather*}
\max\Big\{\sigma(F), \sigma(A_{j})(j\neq
s),\lambda\Big(\frac{1}{A_s}\Big)\Big\}<\mu(A_s)=\sigma(A_s)
<1/2,
\end{gather*}
then every transcendental meromorphic solution $f$ whose poles are
of uniformly bounded multiplicities,  of \eqref{e1.2} satisfies
$\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)=\sigma(A_s)$.
\end{corollary}


\begin{corollary}  \label{coro1.2}
Suppose that $A_0,A_1,\dots, A_{k-1}, F(\not\equiv 0)$ are entire functions.
If there exists some $A_s(0 \leq s \leq k-1)$ such that
\[
\max\{\sigma(F), \sigma(A_{j})(j\neq s)\}<\mu(A_s)=\sigma(A_s) <1/2,
\]
then every transcendental solution $f$ of \eqref{e1.2} satisfies
$\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)=\sigma(A_s)$, and
every nontranscendental solution $f$ is a polynomial with degree
$\deg f\leq s-1$.
\end{corollary}


\begin{remark} \rm
From Corollary \ref{coro1.2}, we obtain the precise estimation of the growth of
transcendental solutions in Theorem \ref{thmB} when $\mu(A_s)=\sigma(A_s) < 1/2$.
\end{remark}


When $F$ is of infinite order, we obtain the following results.

\begin{theorem} \label{thm1.2}
Suppose that $A_0,A_1,\dots, A_{k-1}$, $Q(\not\equiv 0)$ are
meromorphic functions of finite order, $P$ is a transcendental
entire function, such that
\begin{equation}
\max\Big\{\sigma(P),\sigma(Q), \sigma(A_{j})(1\leq j\leq k-1),
\lambda\Big(\frac{1}{A_0}\Big)\Big\}<\mu(A_0) <
\frac{1}{2}.\label{e1.5}
\end{equation}
Then every solution $f$ of the equation
\begin{equation}
f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=Qe^P\label{e1.6}
\end{equation}
is transcendental, and every transcendental meromorphic solution $f$
whose poles are of uniformly bounded multiplicities, of \eqref{e1.6} satisfies
$\mu(A_0)\leq\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)\leq
\sigma(A_0)$.
\end{theorem}



\begin{corollary} \label{coro1.3}
Suppose that $A_0,A_1,\dots, A_{k-1}$, $Q(\not\equiv 0)$ are
entire functions of finite order, $P$ is a transcendental entire
function, such that
\[
\max\{\sigma(P),\sigma(Q), \sigma(A_{j})(1\leq j\leq k-1)
\}<\mu(A_0) < \frac{1}{2}.
\]
Then every solution $f$ of \eqref{e1.6} satisfies
$\mu(A_0)\leq\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)\leq
\sigma(A_0)$.
\end{corollary}



\begin{corollary} \label{coro1.4}
Suppose that $A_0,A_1,\dots, A_{k-1}$, $Q(\not\equiv
0)$ are entire functions of finite order, $P$ is a transcendental
entire function, such that
\[
\max\{\sigma(P),\sigma(Q), \sigma(A_{j})(1\leq j\leq k-1)
\}<\mu(A_0)=\sigma(A_0)< \frac{1}{2}.
\]
Then every solution
 $f$ of \eqref{e1.6} satisfies $\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)=
\sigma(A_0)$.
\end{corollary}


\section{Preliminaries}


\begin{lemma}[\cite{b9}]  \label{lem2.1}
Let $f(z)$ be a transcendental meromorphic function, and let
$\Gamma=\{(k_1,j_1),\dots,(k_{m},j_{m})\}$ be a finite set of
distinct pairs of integers such that
$k_{i}>j_{i}\geq0$ $(i=1,\dots,m)$. Let $\alpha>1$ be a given
constant. Then there exists a set $E\subset(1,+\infty)$ that has a
finite logarithmic measure, and a constant $B>0$ depending only on
$\alpha$ and $\Gamma$, such that for all $z$ with
$|z|=r\not\in[0,1]\cup E$ and $(k,j)\in\Gamma$, we have
\[
\big|\frac{f^{(k)}(z)}{f^{(j)}(z)}\big|\leq B\Big(\frac{T(\alpha
r,f)}{r}{\log^{\alpha}r}\cdot\log T(\alpha r,f)\Big)^{k-j}.
\]
\end{lemma}

It is known that the Wiman-Valiron theory (see \cite{b13}) is an
important tool while considering the value distribution theory of
entire solutions of differential equations. In \cite{b5}, using the
Wiman-Valiron theory for entire functions and the Hadamard
factorization theorem for meromorphic functions, the author obtained
the following Wiman-Valiron theory for meromorphic functions, which
is a generalization of \cite[Lemma 5]{b14}. Their proofs are quite
parallel.


\begin{lemma} [\cite{b5}] \label{lem2.2}
Let $f(z)=g(z)/d(z)$ be a meromorphic function, where $g(z)$ and $d(z)$
are entire functions, such that
\begin{gather*}
\mu(g)=\mu(f)=\mu \leq \sigma(g)=\sigma(f) \leq+\infty,\quad
\lambda(d)=\sigma(d)=\lambda\Big(\frac{1}{f}\Big)< \mu.
\end{gather*}
Then there exists a set $E\subset(1,+\infty)$ that has a finite
logarithmic measure, such that for point $z$ with
$|z|=r\not\in[0,1]\cup E$ and $|g(z)|=M(r,g)$, we have
\[
\frac{f^{(n)}(z)}{f(z)}=\Big(\frac{v_g(r)}{z}\Big)^n(1+o(1)), (n\geq 1),
\]
where $v_g(r)$ denotes the central index of
$g(z)$.
\end{lemma}


\begin{lemma}[\cite{b5}] \label{lem2.3}
Let $f(z)=g(z)/d(z)$ be a meromorphic function, where $g(z)$ and $d(z)$
are entire functions, such that
\begin{gather*}
\mu(g)=\mu(f)=\mu \leq \sigma(g)=\sigma(f) \leq+\infty,\quad
\lambda(d)=\sigma(d)=\lambda\Big(\frac{1}{f}\Big)< \mu.
\end{gather*}
Then there exists a set $E\subset(1,+\infty)$ that has a finite
logarithmic measure, such that for point $z$ with
$|z|=r\not\in[0,1]\cup E$ and $|g(z)|=M(r,g)$, we have
\[
\big|\frac{f(z)}{f^{(s)}(z)}\big|\leq r^{2s},
\]
where $s$ is a positive integer.
\end{lemma}


Since \cite[Lemma 2.3]{b5} is published in Chinese, for the
convenience of the non-Chinese readers, we show the following proof
of \cite[Lemma 2.3]{b5}.


\begin{proof}
By Lemma \ref{lem2.2}, there exists a set $E_1\subset(1,+\infty)$ that has a
finite logarithmic measure, such that for point $z$ with
$|z|=r\not\in[0,1]\cup E_1$ and $|g(z)|=M(r,g)$, we have
\begin{equation} \label{e2.1}
\frac{f^{(s)}(z)}{f(z)}=\Big(\frac{v_g(r)}{z}\Big)^s(1+o(1)).
\end{equation}
Since $\mu(g)=\liminf_{r\to \infty}\frac{\log^+
v_g(r)}{\log r}$, for any given $\varepsilon>0$, there exists $R>0$
such that
\begin{equation} \label{e2.2}
 v_g(r) >r^{\mu(g)-\varepsilon}
\end{equation}
holds for $r>R$. If $\mu(g)=\infty$, then we replace
$\mu(g)-\varepsilon$ by a sufficiently large positive constant $M$.
Let $E=E_1\bigcup[1, R]$, then $E$ has a finite logarithmic measure,
and by \eqref{e2.1} and \eqref{e2.2}, we obtain the result in
 Lemma \ref{lem2.3}.
\end{proof}


\begin{lemma}[\cite{b4, b6}] \label{lem2.4}
 Let $g(z)$ be a meromorphic function of finite order.
Then for any given $\varepsilon > 0$,
there exists a set $E\subset(1,+\infty)$ that has a finite
logarithmic measure, such that for all $z$ with
$|z|=r\not\in[0,1]\cup E$, we have
$|g(z)|\leq\exp\{r^{\sigma(g)+\varepsilon}\}$.
\end{lemma}

\begin{lemma}[\cite{b3,b15}] \label{lem2.5}
Let $g(z)$ be an entire function with $0\leq\mu(g)<1$. Then for
every $\alpha\in(\mu(g),1)$, there exists a set $E\subset[0,\infty)$
such that \begin{gather*} \overline{\log \rm dens}E\geq{1-\frac{\mu(g)}{\alpha}},
\end{gather*}
where $E=\{r\in[0,\infty):m(r)>M(r)\cos\pi\alpha\},
~m(r)=\inf_{|z|=r}\log|g(z)|, ~M(r)=\sup_{|z|=r}\log|g(z)|$.
\end{lemma}


\begin{lemma} [\cite{b13}] \label{lem2.6}
Let $g:(0, \infty)\to R $ and $h:(0, \infty)\to R $
be monotone nondecreasing functions such that $g(r)\leq h(r)$
outside of an exceptional set $H$ of finite logarithmic measure.
Then for any $\alpha>1$, there exists $r_0>0$ such that $g(r)\leq
h(\alpha r)$ holds for all $r>r_0$.
\end{lemma}

\begin{lemma} \label{lem2.7}
Let $f(z)$ be a meromorphic function such that
$\lambda(1/f) < \mu(f) < 1/2$. Then for any given
$\varepsilon$ $(0 <2\varepsilon <\mu(f)-\lambda(1/f))$, there exists a set
$E\subset(1,+\infty)$ with $\overline{\log \rm dens}E> 0$, such that for all
$z$ satisfying $|z|=r\in E$, we have
\[
|f(z)|\geq\exp\{(1-o(1))r^{\mu(f)-\varepsilon}\}.
\]
\end{lemma}

The above result might be known, but we give the proof for the
convenience of the readers.

\begin{proof}
From the Hadamard factorization theorem, we obtain
\begin{equation} \label{e2.3}
f(z)=\frac{g(z)}{d(z)},
\end{equation}
where $g(z)$ is an entire function, $d(z)$ is the canonical product
of $f(z)$ formed with its poles such that
\begin{equation} \label{e2.4}
\lambda(d)=\sigma(d)=\lambda\Big(\frac{1}{f}\Big)< \mu(f).
\end{equation}
By \eqref{e2.3} we obtain $T(r,g) \leq T(r,f)+T(r,d)$. Then combining with
\eqref{e2.4}, we obtain $\mu(g)\leq \mu(f)$. On the other hand, take a
sequence $\{r_{n}\}$ such that 
\[
\lim_{r_n \to \infty}\frac{\log T(r_n, g)}{\log r_n}=\mu(g),
\]
 hence we have
\begin{equation} \label{e2.5}
\liminf_{r_n\to \infty}\frac{\log T(r_n, f)}{\log r_n}\geq\mu(f).
\end{equation}
By \eqref{e2.4} and \eqref{e2.5}, for any given
$\varepsilon(0 < 2\varepsilon < \mu(f)-\sigma(d))$, there exists a
subsequence of $\{r_{n}\}$, for
convenience, we also denote it as $\{r_{n}\}$, such that for
sufficiently large $r_{n}$, we have
\[
T(r_{n},f) > r_{n}^{\mu(f)-\varepsilon},\quad
T(r_{n},d) < r_{n}^{\sigma(d)+\varepsilon}.
\]
Then combining with $T(r,f)\leq T(r,g)+T(r,d)+O(1)$, we obtain
$\mu(f)\leq \mu(g)$. Hence we have
\[
 \mu(g)= \mu(f)<1/2.
\]
By Lemma \ref{lem2.5}, set $\alpha_0=\frac{\frac{1}{2}+\mu(g)}{2}$, then
there exists a set $E_1$ with
$\overline{\log \rm dens}E_1\geq{1-\frac{\mu(g)}{\alpha_0}}$, such that for all $z$
with $|z|=r\in E_1$, we have
\begin{equation} \label{e2.6}
\log|g(z)| \geq \cos(\pi\alpha_0)\log M(r,g).
\end{equation}
By the definition of $\mu(g)$, for any given $\varepsilon(0 <
2\varepsilon <\mu(f)-\lambda(\frac{1}{f}))$, there exists $r_1 >
0$ such that
\begin{equation} \label{e2.7}
\log M(r,g) \geq r^{\mu(g)-\frac{\varepsilon}{2}}
\end{equation}
holds for $r> r_1$. Since
\begin{equation} \label{e2.8}
\frac{\cos(\pi\alpha_0) r^{\mu(g)-\frac{\varepsilon}{2}}}
{r^{\mu(g)-\varepsilon}}\to+\infty, (r\to+\infty),
\end{equation}
by \eqref{e2.6}--\eqref{e2.8}, there exists $r_{2}( \geq r_1)$ such that for all
$z$ with $|z|=r\in E_1\setminus[0,r_{2}]$, we have
\begin{equation} \label{e2.9}
|g(z)| \geq \exp\{\cos(\pi\alpha_0)
r^{\mu(g)-\frac{\varepsilon}{2}}\} \geq
\exp\{r^{\mu(g)-\varepsilon}\}.
\end{equation}
On the other hand, there exists $R
> 0$ such that for $r > R$, we have
\begin{equation} \label{e2.10}
|d(z)| \leq \exp\{r^{\sigma(d)+\varepsilon}\}.
\end{equation}
Set $E=E_1\cap[R,+\infty]\cap[r_{2},+\infty]$, then
$\overline{\log \rm dens} E>0$. By \eqref{e2.3},\eqref{e2.9} and \eqref{e2.10}, we obtain
that
\[
|f(z)|\geq
\exp\{r^{\mu(g)-\varepsilon}-r^{\sigma(d)+\varepsilon}\}
=\exp\{(1-o(1))r^{\mu(f)-\varepsilon}\}
\]
holds for $|z|=r\in E$.
\end{proof}

\section{Proofs of main results}

In the sequel, we use the symbols $E$ and $E_1$ to denote any set of
finite logarithmic measure and any set of finite linear measure, not
necessarily the same at each occurrence. We also use $M$ to denote
any positive constant, not necessarily the same at each occurrence.


\begin{proof}[Proof of Theorem \ref{thm1.1}]
Firstly, suppose that $f(z)$ is a transcendental meromorphic
solution whose poles are of uniformly bounded multiplicities of \eqref{e1.2}.
From \eqref{e1.2}, we know that the poles of
$f(z)$ can only occur at the poles of $A_0, A_1, \dots, A_{k-1}$,
$F$. Note that the multiplicities of poles of $f$ are uniformly
bounded, and thus we have
\begin{equation} \label{e3.1}
n(r,f) \leq
O\Big\{{\sum^{k-1}_{j=0}}\overline{n}(r,A_{j})+\overline{n}(r,
F)\Big\}.
\end{equation}
Then by \eqref{e1.4} and \eqref{e3.1} we obtain
\begin{equation} \label{e3.2}
\lambda\Big(\frac{1}{f}\Big) \leq b.
\end{equation}
From \eqref{e1.2} we obtain
\begin{equation} \label{e3.3}
-A_s=\frac{f^{(k)}}{f^{(s)}}+\dots+A_{s+1}\cdot\frac{f^{(s+1)}}{f^{(s)}}
+\Big[A_{s-1}\cdot\frac{f^{(s-1)}}{f}+
\dots+A_0\Big]\cdot\frac{f}{f^{(s)}}-\frac{F}{f}\frac{f}{f^{(s)}}.
\end{equation}
By the lemma of the logarithmic derivative and \eqref{e3.3}, we obtain
\begin{equation} \label{e3.4}
\begin{aligned}
T(r,A_s) &\leq  N(r,A_s)+\sum_{j \neq s} m(r,A_{j})+m(r,\frac{F}{f})+2m(r,\frac{f}{f^{(s)}})+O(\log rT(r,f))\\
         &\leq  N(r,A_s)+\sum_{j \neq s} T(r,A_{j})+T(r,F)+T(r,\frac{1}{f})\\
         &\quad+2N(r,f^{(s)})+2N(r,\frac{1}{f})+O(\log rT(r,f))\\
         &\leq  N(r,A_s)+\sum_{j \neq s} T(r,A_{j})+T(r,F)+2(s+1)N(r,f)\\
         &\quad +3T(r,f)+O(\log rT(r,f))\\
         &\leq  N(r,A_s)+\sum_{j \neq s} T(r,A_{j})+T(r,F)+2(s+1)N(r,f)\\
         &\quad +4T(r,f),(r\not\in E_1).
\end{aligned}
\end{equation}
By \eqref{e1.4}, \eqref{e3.2}, \eqref{e3.4} and Lemma \ref{lem2.6}, we obtain
\begin{equation} \label{e3.5}
\mu(A_s)\leq\mu(f).
\end{equation}
From the Hadamard factorization theorem, we obtain
\begin{equation} \label{e3.6}
f(z)=\frac{g(z)}{d(z)},
\end{equation}
where $g(z)$ is an entire function, $d(z)$ is the canonical product
of $f(z)$ formed with its poles such that
$\lambda(d)=\sigma(d)=\lambda(\frac{1}{f})$. By \eqref{e1.4},\eqref{e3.2} and
\eqref{e3.5}, we obtain
\begin{equation} \label{e3.7}
\lambda(d)=\sigma(d)=\lambda(\frac{1}{f})<\mu(f).
\end{equation}
By the definition of order, for any given
$\varepsilon(0 < 2\varepsilon < \sigma(f)-\sigma(d))$, there exists a sequence
$\{r_{n}\}$ such that for sufficiently large $r_{n}$, we have
\begin{equation} \label{e3.8}
T(r_{n},f) > r_{n}^{\sigma(f)-\varepsilon},\quad T(r_{n},d) <
r_{n}^{\sigma(d)+\varepsilon}.
\end{equation}
By \eqref{e3.6}, we obtain
\begin{equation} \label{e3.9}
T(r,f)\leq T(r,g)+T(r,d)+O(1).
\end{equation}
Hence by \eqref{e3.8} and \eqref{e3.9}, we obtain $\sigma(f)\leq \sigma(g)$. On the
other hand, by \eqref{e3.6} we obtain $T(r,g) \leq T(r,f)+T(r,d)$. Then
combining with \eqref{e3.7}, we obtain $\sigma(g)\leq \sigma(f)$. Hence we
have
\begin{equation} \label{e3.10}
\sigma(g)=\sigma(f).
\end{equation}
Using the similar proof to that of Lemma \ref{lem2.7}, we obtain
\begin{equation} \label{e3.11}
 \mu(g)= \mu(f).
\end{equation}
So by \eqref{e3.7}, \eqref{e3.10},\eqref{e3.11} and Lemma \ref{lem2.3}, there exists a set
$E\subset(1,+\infty)$ of finite logarithmic measure, such that for
all $z$ satisfying $|z|=r\not\in[0,1]\cup E$ and $|g(z)|=M(r,g)$, we
have
\begin{equation} \label{e3.12}
\big|\frac{f(z)}{f^{(s)}(z)}\big|\leq r^{2s}.
\end{equation}
By Lemma \ref{lem2.1}, there exists a set $E\subset(1,\infty)$ of finite
logarithmic measure and $B>0$, such that for all $z$ satisfying
$|z|=r\not\in[0,1]\cup E$, we have
\begin{gather} \label{e3.13}
|\frac{f^{(j)}(z)}{f^{(s)}(z)}|\leq
Br[T(2r,f)]^{j-s+1},\quad (j=s+1,\dots,k) , \\
 \label{e3.14}
|\frac{f^{(j)}(z)}{f(z)}|\leq
Br[T(2r,f)]^{j+1},\quad (j=1,\dots,s-1).
\end{gather}
By \eqref{e1.4}, \eqref{e3.2}, \eqref{e3.7} and Lemma \ref{lem2.4},
for any given $\varepsilon$ $(0<2\varepsilon<\mu(A_s)-b)$, there exists a set
$E\subset(1,+\infty)$ of finite logarithmic measure, such that for
all $z$ satisfying $|z|=r\not\in[0,1]\cup E$, we have
\begin{equation} \label{e3.15}
|A_{j}(z)|\leq\exp\{r^{b+\varepsilon}\}(j \neq s), \quad
|F(z)|\leq\exp\{r^{b+\varepsilon/2}\}, \quad
|d(z)|\leq\exp\{r^{b+\varepsilon/2}\}.
\end{equation}
Hence for all $z$ satisfying $|z|=r\not\in[0,1]\cup E$ and
$|g(z)|=M(r,g)$, we have
\begin{equation} \label{e3.16}
\Big|\frac{F(z)}{f(z)}\Big|=\frac{|F(z)d(z)|}{M(r,g)}\leq
\exp\{r^{b+\varepsilon}\}.
\end{equation}
By Lemma \ref{lem2.7}, there exists a set $H_0\subset(1,+\infty)$ with
$\overline{\log \rm dens}H_0
> 0$, such that for all $z$ satisfying $|z|=r\in H_0$, we have
\begin{equation} \label{e3.17}
|A_s(z)|\geq\exp\{(1-o(1))r^{\mu(A_s)-\varepsilon}\}.
\end{equation}
Let $H=H_0-([0,1]\cup E)$, then we have
$\overline{\log \rm dens}H > 0$, and for all $z$ satisfying $|z|=r\in H$
and $|g(z)|=M(r,g)$,
by \eqref{e3.3},\eqref{e3.12}--\eqref{e3.17}, we have
\begin{equation} \label{e3.18}
\begin{aligned}
\exp\{(1-o(1))r^{\mu(A_s)-\varepsilon}\}
&\leq |A_s(z)|\\
&\leq (k-s)\cdot\exp\{r^{b+\varepsilon}\}Br\cdot[T(2r,f)]^{k-s+1}
+s\cdot\exp\{r^{b+\varepsilon}\}\\
&\quad \times Br\cdot[T(2r,f)]^{s}\cdot r^{2s}
+\exp\{r^{b+\varepsilon}\}\cdot r^{2s}\\
&\leq (k+1)Br\cdot\exp\{r^{b+\varepsilon}\}\cdot[T(2r,f)]^{k+1}\cdot
r^{2s}.
 \end{aligned}
 \end{equation}
Hence by \eqref{e3.18}, we obtain $\sigma_{2}(f)\geq \mu(A_s)$. Now we prove
that $\sigma_{2}(f)\leq \sigma(A_s)$. By \eqref{e3.7},
\eqref{e3.10},\eqref{e3.11} and
Lemma \ref{lem2.2}, there exists a set $E\subset(1,+\infty)$ of finite
logarithmic measure, such that for all $z$ satisfying
$|z|=r\not\in[0,1]\cup E$ and $|g(z)|=M(r,g)$, we have
\begin{equation} \label{e3.19}
\frac{f^{(j)}(z)}{f(z)}=\Big(\frac{v_g(r)}{z}\Big)^j(1+o(1)), (j=1,
\dots, k).
\end{equation}
By Lemma \ref{lem2.4}, for any given $\varepsilon>0$, there exists a set
$E\subset(1,+\infty)$ of finite logarithmic measure, such that for
all $z$ satisfying $|z|=r\not\in[0,1]\cup E$, we have
\begin{equation} \label{e3.20}
|A_s(z)|\leq\exp\{r^{\sigma(A_s)+\varepsilon}\}.
\end{equation}
Then by \eqref{e1.2}, \eqref{e3.16}, \eqref{e3.19} and \eqref{e3.20},
for all $z$ satisfying
$|z|=r\not\in[0,1]\cup E$ and $|g(z)|=M(r,g)$, we have
\begin{equation} \label{e3.21}
v_g(r)\leq Mr\exp\{r^{\sigma(A_s)+\varepsilon}\}.
\end{equation}
Hence by \eqref{e3.10}, \eqref{e3.21} and Lemma \ref{lem2.6}, we
obtain $\sigma_{2}(f)\leq \sigma(A_s)$.

Let $F\not\equiv 0$. Next we prove that
$\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)$.
By \eqref{e1.2} we obtain
\begin{equation} \label{e3.22}
\frac{1}{f}=\frac{1}{F}\Big(\frac{f^{(k)}}{f}+A_{k-1}\frac{f^{(k-1)}}{f}
+\dots+A_0\Big).
\end{equation}
Suppose that $z_0$ is a zero of $f$ with order $\alpha(>k)$, if
$z_0$ is not a pole of $A_j$  $(j=0,\dots, k-1)$, then $z_0$ must be a
zero of $F$ with order $\alpha-k$. Hence
\begin{equation} \label{e3.23}
N\Big(r, \frac{1}{f}\Big)\leq k\overline{N}\Big(r,
\frac{1}{f}\Big)+N\Big(r, \frac{1}{F}\Big) +\sum_{j=0}^{k-1}
N(r,A_{j}).
\end{equation}
By \eqref{e3.22} we obtain
\begin{equation} \label{e3.24}
m\Big(r, \frac{1}{f}\Big)\leq m\Big(r, \frac{1}{F}\Big)
+\sum_{j=0}^{k-1} m(r,A_{j})+O(\log rT(r,f)), (r\not\in E_1).
\end{equation}
Combining \eqref{e3.23} and \eqref{e3.24}, we obtain
\begin{equation} \label{e3.25}
T(r, f)\leq k\overline{N}\Big(r, \frac{1}{f}\Big)+T(r,
F)+\sum_{j=0}^{k-1} T(r,A_{j})+O(\log rT(r,f)), (r\not\in E_1).
\end{equation}
Take a sequence $\{r_n'\}$ satisfying
$\lim_{r_n'\to\infty}\frac{\log\log T(r_n', f)}{\log
r_n'}=\sigma_2(f)$, set $\operatorname{meas} E_1=\delta$, then there exists
$r_n\in[r_n', r_n'+\delta+1]$ such that
\begin{equation} \label{e3.26}
\liminf_{r_n\to\infty}\frac{\log\log T(r_n, f)}{\log
r_n}\geq\lim_{r_n'\to\infty}\frac{\log\log T(r_n', f)}{\log
(r_n'+\delta+1)}=\sigma_2(f).
\end{equation}
Hence by \eqref{e1.4}, \eqref{e3.26} and $\sigma_2(f)\geq \mu(A_s)$, for
sufficiently large $r_n$, we have
\begin{equation} \label{e3.27}
T(r_n, F)=o(T(r_n, f)),~T(r_n, A_j)=o(T(r_n, f)),\quad (0\leq j\leq k-1).
\end{equation}
Then by \eqref{e3.25} and \eqref{e3.27}, we obtain
 $\sigma_2(f)\leq \overline{\lambda_2}(f)$. Since
$\overline{\lambda_2}(f)\leq\sigma_2(f)$, we obtain
$\mu(A_s)\leq\overline{\lambda_2}(f)=\lambda_2(f)=\sigma_2(f)\leq\sigma(A_s)$.

Secondly, suppose that $f$ is a nonconstant rational solution of
\eqref{e1.2}. When $s\geq 2$, if $z_0$ is a pole of $f$ with order
$m(\geq 1)$, or $f$ is a polynomial with degree more than $s-1$, then
$f^{(s)}\not\equiv 0$. Hence by \eqref{e1.2}, \eqref{e3.15} and \eqref{e3.17},
for all $z$ satisfying $|z|=r\in H_0-([0,1]\cup E)$, we have
\[
\exp \{(1-o(1))r^{\mu(A_s)-\varepsilon}\}\leq|A_s(z)|\leq r^M\exp
\{r^{b+\varepsilon} \}.
\]
This is impossible. So every nontranscendental solution $f$ of
\eqref{e1.2} is a polynomial with degree $\deg f\leq s-1$. By the similar
argument, we obtain that every nonconstant solution of  \eqref{e1.2} is
transcendental when $s=0$ or $1$ʱ.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
From the hypothesis  we know that every meromorphic
solution of  \eqref{e1.6} is of infinite order. So every meromorphic
solution of  \eqref{e1.6} is transcendental. Suppose that $f(z)$ is a
transcendental meromorphic solution whose poles are of uniformly
bounded multiplicities. Set $f=ge^P$, then we have
\begin{equation} \label{e3.28}
\overline{\lambda_2}(g)=\overline{\lambda_2}(f),\quad
\lambda_2(g)=\lambda_2(f).
\end{equation}
Substituting $f=ge^P$ into \eqref{e1.6}, we obtain
\begin{equation} \label{e3.29}
g^{(k)}+B_{k-1}g^{(k-1)}+\dots+B_0g=Q,
\end{equation}
where
\begin{gather} \label{e3.30}
B_{k-1}=A_{k-1}+kP', \\
 \label{e3.31}
\begin{aligned}
B_{k-j}&=A_{k-j}+(k-j+1)A_{k-j+1}P'
 +\sum_{m=2}^j A_{k-j+m}
\Big[\binom{m}{k-j+m}(P')^m \\
&\quad +D_{m-1}(P')\Big],\quad 
 j=2, 3,\dots, k; \; A_k\equiv 1. 
\end{aligned}
\end{gather}
Here $D_{m-1}(P')$ is a differential polynomial in $P'$ of degree
$m-1$, its coefficients are constants. By \eqref{e1.5},\eqref{e3.30}
and \eqref{e3.31}, we obtain
\begin{equation} \label{e3.32}
\mu(B_0)=\mu(A_0), \quad
\lambda\Big(\frac{1}{B_0}\Big)<\mu(A_0),\quad \sigma(B_j)<\mu(A_0),
(1\leq j\leq k-1).
\end{equation}
Hence by \eqref{e1.5}, \eqref{e3.29},\eqref{e3.32} and Theorem \ref{thm1.1},
we obtain
\begin{equation} \label{e3.33}
\mu(A_0)\leq\overline{\lambda_2}(g)=\lambda_2(g)=\sigma_2(g)\leq\sigma(A_0).
\end{equation}
Since $\sigma_2(e^P)=\sigma(P)<\mu(A_0)\leq \sigma_2(g)$, we obtain
$\sigma_2(f)=\sigma_2(g)$. Then combining \eqref{e3.28} and \eqref{e3.33},
we obtain $\mu(A_0)\leq\overline{\lambda_2}(f)
=\lambda_2(f)=\sigma_2(f)\leq\sigma(A_0)$.
\end{proof}

\subsection*{Acknowledgments}
This work is supported by the National Natural Science Foundation of
China (No. 11201195), the Natural Science Foundation of Jiangxi,
China (No. 20122BAB201012, 20132BAB201008).
The authors want to thank the anonymous referees for their valuable 
suggestions and comments.


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\end{document}