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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 73, pp. 1--14.\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-2014/73\hfil Oscillation of meromorphic solutions]
{Oscillation of meromorphic solutions to linear differential equations
with coefficients of $[p,q]$-order}

\author[H.-Y. Xu, J. Tu \hfil EJDE-2014/73\hfilneg]
{Hong-Yan Xu, Jin Tu}

\address{Hong-Yan Xu \newline
Department of Informatics and Engineering,
Jingdezhen Ceramic Institute,
Jingdezhen, Jiangxi 333403, China}
\email{xhyhhh@126.com}

\address{Jin Tu \newline
Institute of Mathematics and informatics, 
Jiangxi Normal University,
Nanchang, Jiangxi 330022, China}
\email{tujin2008@sina.com}

\thanks{Submitted June 20, 2013. Published March 16, 2014.}
\subjclass[2000]{34M10, 30D35}
\keywords{Linear differential equation; oscillation; small function;
$[p,q]$-order}

\begin{abstract}
 We study the relationship between ``small functions'' and
 the derivative of  solutions to the higher order linear differential equation
 $$
 f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=0,\quad (k\geq 2)
 $$
 Here $A_j(z)$ $(j=0,1,\dots,k-1)$ are entire functions or meromorphic
 functions of $[p,q]$-order.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction and statement of main results}

 The study of oscillation theory for linear differential
equations in the complex plane $\mathbb{C}$ was started by Bank and
Laine \cite{2,3}. After their well-known work, many important
results have been obtained, see for example \cite{16, 17}.

We assume that the reader knows the standard notations and the
fundamental results of the Nevanlinna value distribution theory of
meromorphic functions \cite{11,14}. In addition, we use
$\sigma(f)$, $\lambda(f)$ and $\overline{\lambda}(f)$ to denote
the order, the exponent of convergence of the zero-sequence,  and
the exponent of convergence of the nonzero zero sequence of a meromorphic
function $f(z)$, respectively. We also denote by  $\tau(f)$
 the type of an entire function $f(z)$ with
$0<\sigma(f)=\sigma<+\infty$ (see \cite{14}).


We use $mE=\int_Edt$ and $m_l E=\int_E\frac{dt}{t}$ to denote the linear measure
 and the logarithmic measure of a set
$E\subset[1,+\infty)$, respectively. We denote by $S(r,f)$ any
quantity satisfying $S(r,f)=o(T(r,f)),$ as $r\to+\infty$,
possibly outside of a set with finite linear measure. A meromorphic
function $\psi(z)$ is called a small function with respect to $f$ if
$T(r,\psi)=S(r,f)$.

For results on the growth of solutions to
equations of the form
\begin{equation} \label{e1}
  f''+A(z)f'+B(z)f=0,
\end{equation}
with $A(z)$ and $B(z)(\not\equiv0)$ are entire functions, the reader is referred to
\cite{1,6,7,10,13}.

In 1996, Kwon \cite{15} investigated the hyper-order of the
solutions of \eqref{e1} and obtained the following result.

\begin{theorem}[\cite{15}] \label{thmA}
Let $A(z)$ and $B(z)$ be
entire functions such that $\sigma(A)<\sigma(B)$ or
$\sigma(B)<\sigma(A)< 1/2$. Then every solution
$f\not\equiv0$ of \eqref{e1} satisfies
$\sigma_2(f)\geq \max\{\sigma(A),\sigma(B)\}$.
\end{theorem}

In 2006, Chen and Shon \cite{8} investigated the zeros concerning
small functions and fixed points of solutions of second order linear
differential equations and obtained the following results.

\begin{theorem}[\cite{8}] \label{thmB}
Let $A_{j}(z)\not\equiv0 $ $(j=1,2)$ be entire functions with $\sigma(A_{j})<1$,
 suppose that $a,b$ are complex numbers that satisfy $ab\neq0$ and
$\arg a\neq\arg b$ or $a=cb$ $(0<c<1)$. If $\varphi(z)\not\equiv0$ is an entire
function of finite order, then every non-trivial solution $f$ of
equation
$$
f''+A_{1}(z)e^{az}f'+A_{2}(z)e^{bz}f=0
$$
satisfies $\overline{\lambda}(f-\varphi)=\overline{\lambda}(f'-\varphi)
=\overline{\lambda}(f''-\varphi)=\infty$.
\end{theorem}

\begin{theorem}\cite{8} \label{thmC}
Let $A_{1}(z)\not\equiv0$, $\varphi(z)\not\equiv0$, $Q(z)$ be entire
functions with $\sigma(A_{1})<1$, $1<\sigma(Q)<\infty$ and
$\sigma(\varphi)<\infty$, then every non-trivial solution $f$ of
equation
$$
f''+A_{1}(z)e^{az}f'+Q(z)f=0
$$
satisfies $\overline{\lambda}(f-\varphi)=\overline{\lambda}(f'-\varphi)
=\overline{\lambda}(f''-\varphi)=\infty$,
where $a\neq0$ is a complex number.
\end{theorem}

In 2012, Wu and Chen \cite{wc} investigate the problem on the
fixed-points of solutions of some second order differential equation
with transcendental entire function coefficients and obtained the
following theorems.

\begin{theorem}[{\cite[Theorem 1]{wc}}] \label{thmD}
Let $A_{j}(z)\not\equiv0 (j=0,1)$ be entire functions, $P(z)$ be a
polynomial satisfying $\sigma(A_1)<\deg P(z)$ and
$0<\sigma(A_0)<1/2$, and let $\varphi(z)(\not\equiv0)$ be an
entire function of finite order. Then every non-trivial solution $f$
of equation
$$
f''+A_{1}(z)e^{P(z)}f'+A_0(z)f=0
$$
satisfies $\overline{\lambda}(f-\varphi)=\infty$.
\end{theorem}

\begin{theorem}[{\cite[Theorem 2]{wc}}] \label{thmE}
Under the assumptions of Theorem \ref{thmD}, every non-trivial solution
$f$ of the equation
$$
f''+A_{1}(z)e^{P(z)}f'+A_0(z)f=0
$$
satisfies
\begin{itemize}
\item[(i)]
$\overline{\lambda}(f-z)=\overline{\lambda}(f'-z)
=\overline{\lambda}(f''-z)=\sigma(f)=\infty$;

\item[(ii)] $g(z)$ has infinitely many fixed points and
$\overline{\lambda}(g-z)=\infty$, where
$g(z)=d_0f(z)+d_1f'(z)+d_2f''(z)$, $d_0d_2\neq0$.
\end{itemize}
\end{theorem}

An interesting subject arises naturally about the problems of the
zeros concerning small function and fixed points of solutions of
differential equations
\begin{equation} \label{e2}
  f^{(k)}+A_{k-1}f^{(k-1)}+\dots+A_0f=0,\quad (k\geq2)
\end{equation}
where $A_j(z)$ $(j=0,1,\dots,k-1)$ are entire functions.

In 2000s, Bela\"{\i}di \cite{4}, Bela\"{\i}di and El Farissi
\cite{5}  (see also \cite{fb,18,19}) investigated the fixed
points and the relationship between small functions and differential
polynomials of solutions of \eqref{e2} and obtained some results
which improve Theorem \ref{thmC}.

 Recently, the growth of solutions of higher order linear
 differential equation with meromorphic coefficients of $[p,q]$-order
 was studied and some results were obtained in \cite{Bela,Liu-Tu}.

 In this article, we  study the zeros of small
functions and the fixed points of solutions to equation \eqref{e2} with entire
or meromorphic coefficients of $[p,q]$-order  and obtain some
results that   extend the work of Chen and  Bela\"{\i}di.

Before stating our theorems, we introduce the concepts of
entire functions of $[p,q]$-order (see \cite{Jun-Kap1, Jun-Kap2, Liu-Tu}).
Juneja and co-authors \cite{Jun-Kap1, Jun-Kap2}
 introduced the concept of entire functions of
$[p, q]$-order, and studied some of their properties for
$p,q$ integers satisfying $p>q\geq1$.

\begin{definition} \label{def1.1} \rm
  If $f(z)$ is a transcendental entire function, the $[p,q]$-order
  of $f(z)$ is defined by
  $$
  \sigma_{[p,q]}(f)=\limsup_{r\to\infty}
\frac{\log_{p+1}M(r,f)}{\log_q r}
=\limsup_{r\to\infty}\frac{\log_pT(r,f)}{\log_q r},
$$
  where $p,q$ are two integers and $p>q\geq1$.
  \end{definition}

  \begin{remark} \label{rmk1.1}\rm
 For sufficiently large $r\in[1,\infty)$, we define
$\log_{i+1} r=\log_{i}(\log r)$ $(i\in \mathbb{N})$
 and $\exp_{i+1} r=\exp(\exp_{i} r)$ $(i\in \mathbb{N})$ and
 $\exp_{0} r=r=\log_{0} r$,
$\exp_{-1} r=\log r$.
  \end{remark}

  \begin{definition} \label{def1.2} \rm
    The $[p,q]$-type of an entire function $f$ of $[p,q]$-order
    $\sigma$ $(0<\sigma<\infty)$ is defined by
    $$
    \tau_{[p,q]}=\tau_{[p,q]}(f)
=\limsup_{r\to\infty}\frac{\log_{p}M(r,f)}{(\log_{q-1}  r)^\sigma}.
$$
  And the $[p,q]$ exponent of convergence of the zero sequence of
  $f$ is defined by
$$
 \lambda_{[p,q]}=\lambda_{[p,q]}(f)=\limsup_{r\to\infty}
\frac{\log_{p}n(r,\frac{1}{f})}{\log_q   r}
=\limsup_{r\to\infty}\frac{\log_pN(r,\frac{1}{f})}{\log_q   r},
$$
  and the $[p,q]$ exponent of convergence of the distinct zero
  sequence of $f$ is defined by
$$
 \overline{\lambda}_{[p,q]}=\overline{\lambda}_{[p,q]}(f)
 =\limsup_{r\to\infty}\frac{\log_{p}\overline{n}(r,\frac{1}{f})}{\log_q
  r}=\limsup_{r\to\infty}\frac{\log_p\overline{N}(r,\frac{1}{f})}{\log_q
  r}.
$$
  \end{definition}

Let $\varphi(z)$ be an entire function with
$\sigma_{[p,q]}(\varphi)<\sigma_{[p,q]}(f)$, the $[p,q]$ exponent of
convergence of zeros and distinct zeros of $f(z)-\varphi(z)$ are
defined to be
$$
\lambda_{[p,q]}(f-\varphi)=\limsup_{r\to\infty}
\frac{\log_pN(r,\frac{1}{f-\varphi})}{\log_q r},
\overline{\lambda}_{[p,q]}(f-\varphi)=\limsup_{r\to\infty}
\frac{\log_pN(r,\frac{1}{f-\varphi})}{\log_q r},
$$
especially if $\varphi(z)=z$, we use $\lambda_{[p,q]}(f-z)$
and $\overline{\lambda}_{[p,q]}(f-z)$ to denote the $[p,q]$ exponent
of convergence of fixed points and distinct fixed points of $f(z)$,
respectively.

 Next we state our main results.

 \begin{theorem} \label{thm1.1}
It  $A_j(z)$ $(j=0,1,\dots,k-1)$ are entire functions and satisfy one
of the following two conditions:
\begin{itemize}
\item[(i)] $\max\{\sigma_{[p,q]}(A_j):
j=1,2,\dots,k-1\}<\sigma_{[p,q]}(A_0)<\infty$;

\item[(ii)] $\max\{\sigma_{[p,q]}(A_j):
j=1,2,\dots,k-1\}\leq\sigma_{[p,q]}(A_0)<\infty$ and\\
$\max\{\tau_{[p,q]}$ $(A_j)|\sigma_{[p,q]}(A_j)
=\sigma_{[p,q]}(A_0)>0\}=\tau_1<\tau_{[p,q]}(A_0)=\tau$,
\end{itemize}
then for every solution $f\not\equiv0$ of \eqref{e2} and for any entire function
$\varphi(z)\not\equiv0$ satisfying
$\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$. Moreover
\begin{align*}
\overline{\lambda}_{[p+1,q]}(f-\varphi)
&=\overline{\lambda}_{[p+1,q]}(f'-\varphi)
=\overline{\lambda}_{[p+1,q]}(f''-\varphi)\\
&=\overline{\lambda}_{[p+1,q]}(f^{(i)}-\varphi)=\sigma_{[p+1,q]}(f)\\
&=\sigma(A_0), \quad (i\in \mathbb{N}).
\end{align*}
\end{theorem}

 Throughout this paper we assume  that $A_0$ does not vanish
identically.

\begin{theorem} \label{thm1.2}
If $A_j(z)$, $j=0,1,\dots,k-1$ are meromorphic functions satisfying
  $\max\{\sigma_{[p,q]}$ $(A_j): j=1,2,\dots,k-1\}<\sigma_{[p,q]}(A_0)$ and
$ \delta(\infty,A_0)>0$, then for every meromorphic solution $f\not\equiv0$
of \eqref{e2}  and for any meromorphic function $\varphi(z)\not\equiv0$
   satisfying $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$,
   we have
\[
\overline{\lambda}_{[p+1,q]}(f^{(i)}-\varphi)
=\lambda_{[p+1,q]}(f^{(i)}-\varphi)\geq
   \sigma_{[p,q]}(A_0) (i=0,1,\dots),
\]
   where $f^{(0)}=f$.
\end{theorem}

\begin{example} \label{examp1.1} \rm
  For the equation
\begin{equation}
  f''+\frac{e^{2z}+e^z-1}{1-e^z}f'+\frac{-e^{2z}}{1-e^z}f=0,
  \label{esharp}
\end{equation}
  we can easily see that this equation  has a solution
  $f(z)=e^{e^z}+e^z$.  The functions
$\frac{e^{2z}+e^z-1}{1-e^z},\frac{-e^{2z}}{1-e^z}$
are meromorphic and satisfy $\sigma(\frac{e^{2z}+e^z-1}{1-e^z})=
  \sigma(\frac{-e^{2z}}{1-e^z})=1$ and
 $\delta\big(\infty,\frac{-e^{2z}}{1-e^z}\big)=\frac{1}{2}$.
 Taking $\varphi(z)=e^z$, then
 $\sigma_{[2,1]}(\varphi)<\sigma_{[1,1]}(\frac{-e^{2z}}{1-e^z})$. Thus,
  we get that $\overline{\lambda}_{[2,1]}(f'-\varphi)
=\overline{\lambda}_{[2,1]}(e^{e^z}e^z)
=0\neq1=\sigma_{[1,1]}(\frac{-e^{2z}}{1-e^z})$.
\end{example}

For $p>q\geq1$, we have the following example.

\begin{example} \label{examp1.2} \rm
 Consider the equation
 $$
 f''+A_1f'+A_0f=0, \label{esharp1}
$$
where
 $$
A_1=-\frac{1+e^z-2e^{e^z}+e^{2e^z}-2e^{e^z}e^z+e^{3e^z}e^z}{(1-e^{e^z})^2}, \quad
A_0=\frac{e^{3e^z}e^{2z}-e^{2e^z}e^{2z}}{(1-e^{e^z})^2}.
$$
 Obviously, $A_0, A_1$ are meromorphic functions,
$\sigma_{[2,1]}(A_1)=\sigma_{[2,1]}(A_0)=1$ and
$\delta(\infty,A_0)>0$. By calculating, the equation \eqref{esharp1}
has a solution $f(z)=e^{e^{e^z}}+e^{e^z}$. Taking $\varphi(z)=e^{e^z}e^z$,
then $\sigma_{[3,1]}(\varphi)<\sigma_{[2,1]}(A_0)$.
Thus, we can get that
 $\overline{\lambda}_{[3,1]}(f'-\varphi)
=\overline{\lambda}_{[3,1]}(e^{e^{e^z}}e^{e^z}e^z)
 =0\neq 1=\sigma_{[2,1]}(A_0)$.
\end{example}

From Theorems \ref{thm1.1} and \ref{thm1.2}, we obtain the following
corollaries.

\begin{corollary} \label{coro1.1}
  Under the assumptions of Theorem \ref{thm1.1}, if $\varphi(z)=z$, for every
  solution $f\not\equiv0$ of \eqref{e2}, we have
\begin{align*}
  \overline{\lambda}_{[p+1,q]}(f-z)
&=\overline{\lambda}_{[p+1,q]}(f'-z)
=\overline{\lambda}_{[p+1,q]}(f''-z)\\
&=\overline{\lambda}_{[p+1,q]}(f^{(i)}-z)=\sigma_{[p+1,q]}(f)\\
&=\sigma_{[p,q]}(A_0), \quad (i\in \mathbb{N}).
\end{align*}
\end{corollary}

\begin{corollary} \label{coro1.2}
 Under the assumptions of Theorem \ref{thm1.2}, if $\varphi(z)=z$, for every
meromorphic solution $f\not\equiv0$ of \eqref{e2}, we have
$\overline{\lambda}_{[p+1,q]}(f^{(i)}-z)=\lambda_{[p+1,q]}(f^{(i)}-z)\geq
   \sigma_{[p+1,q]}(A_0), (i=0,1,\dots)$,
   where $f^{(0)}=f$.
\end{corollary}

\section{Preliminary results}

To prove our theorems, we require the following lemmas.

\begin{lemma}[{\cite[Lemma 2.1]{xtz}}] \label{lem2.1}
Assume $f\not\equiv0$ is a solution of  \eqref{e2}, set
$g=f-\varphi$, then $g$ satisfies the equation
\begin{equation} \label{e3}
  g^{(k)}+A_{k-1}g^{(k-1)}+\dots+A_0g
  =-[\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_0\varphi].
\end{equation}
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{xtz}}] \label{lem2.2}
  Assume $f\not\equiv0$ is a solution of equation \eqref{e2}, set
$g_1=f'-\varphi$ , then $g_1$ satisfies the equation
\begin{equation} \label{e4}
  g_1^{(k)}+U^1_{k-1}g_1^{(k-1)}+\dots+U^1_0g_1
  =-[\varphi^{(k)}+U^1_{k-1}\varphi^{(k-1)}+\dots+U^1_0\varphi],
\end{equation}
where $U_j^1=A'_{j+1}+A_j-\frac{A'_0}{A_0}A_{j+1}$,
$j=0,1,2,\dots,k-1$ and $A_k\equiv1$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.5]{xtz}}] \label{lem2.3}
 Assume $f\not\equiv0$ is a solution of equation \eqref{e2}, set
$g_i=f^{(i)}-\varphi$, then $g_i$ satisfies the equation
\begin{equation} \label{e5}
  g_i^{(k)}+U^i_{k-1}g_i^{(k-1)}+\dots+U^i_0g_i
  =-[\varphi^{(k)}+U^i_{k-1}\varphi^{(k-1)}+\dots+U^i_0\varphi],
\end{equation}
where
$U_j^i={U^{i-1}_{j+1}}'+U^{i-1}_j-\frac{{U^{i-1}_0}'}{U^{i-1}_0}U^{i-1}_{j+1}$,
$j=0,1,2,\dots,k-1$, $U^{i-1}_k\equiv1$ and $ i\in \mathbb{N}$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 3.9]{Liu-Tu}}] \label{lem2.4}
Let $f(z)$ be an entire function of
$[p,q]$-order, then $\sigma_{[p,q]}(f)=\sigma_{[p,q]}(f')$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 3.10]{Liu-Tu}}] \label{lem2.5}
Let $f(z)$ be an entire function of $[p,q]$-order satisfying
  $\sigma_{[p,q]}(f)=\sigma_2$, then there exists a set
  $E\subset[1,+\infty)$ with infinite logarithmic measure such that
  for all $r\in E$, we have
  $$
  \lim_{r\to\infty}\frac{\log_p T(r,f)}{\log_q r}=\sigma_2, \quad
  r\in E.
$$
\end{lemma}

\begin{lemma} \label{lem2.6}
Let  $A_0(z), A_1(z), \dots, A_{k-1}(z)$ be entire functions with
 $[p,q]$-order and satisfy $\max\{\sigma_{[p,q]}(A_j):
  j=1,2,\dots,k-1\}=\sigma_1<\sigma_{[p,q]}(A_0)<\infty$, and set
$$
U_j^1=A'_{j+1}+A_j-\frac{A'_0}{A_0}A_{j+1}
$$
and
$$
U_j^i={U^{i-1}_{j+1}}'+U^{i-1}_j-\frac{{U^{i-1}_0}'}{U^{i-1}_0}U^{i-1}_{j+1},
$$
where $j=0,1,2,\dots,k-1$, $A_k\equiv1$, $U^{i-1}_k\equiv1$ and
$i\in \mathbb{N}$. Then there exists a set $E$ with infinite
logarithmic measure such that for $r\in E$, we have
\begin{equation} \label{e6}
\begin{aligned}
\sigma_{[p,q]}(A_0)
&=\lim_{r\to\infty}\frac{\log_p   m(r,U_0^i)}{\log_q   r}\\
&>\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m(r,
  U_j^i)\}}{\log_q r}=\sigma_1.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We will use the inductive method to prove it.

First, when $i=1$, it follows that
$U_j^1=A'_{j+1}+A_j-\frac{A'_0}{A_0}A_{j+1}$ for
$j=0,1,2,\dots,k-1$ and $A_k\equiv1$. When $j=0$, that is,
$U_0^1=A'_{1}+A_0-\frac{A'_0}{A_0}A_{1}$. Then, we have
\begin{equation} \label{e7}
  m(r,U_0^1)\leq
  m(r,A_1)+m(r,A_0)+m\big(r,\frac{A'_1}{A_1}\big)+m\big(r,\frac{A'_0}{A_0}\big)+O(1).
\end{equation}
From $A_0=-A'_1+U_0^1+\frac{A'_0}{A_0}A_{1}$, we have
\begin{equation} \label{e8}
 m(r,A_0) \leq
  m(r,A_1)+m(r,U_0^1)+m\big(r,\frac{A'_1}{A_1}\big)
+m\big(r,\frac{A'_0}{A_0}\big)+O(1).
\end{equation}
When $j\neq0$, from the definitions of $U_j^1(j=1\dots,k)$, we have
\begin{equation} \label{e9}
\begin{aligned}
  m\left(r,U_j^1\right)
&\leq   m(r,A_{j+1})+m(r,A_j)+m\big(r,\frac{A'_0}{A_0}\big)\\
&\quad +m\big(r,\frac{A'_{j+1}}{A_{j+1}}\big)+O(1),\quad
  j=1,2,\dots,k-1.
\end{aligned}
\end{equation}
Since $A_0(z),\dots, A_{k-1}(z)$ are entire functions with
$\max\{\sigma_{[p,q]}(A_j):
j=1,2,\dots,k-1\}<\sigma_{[p,q]}(A_0)<\infty$ and \eqref{e9}, we have
\begin{equation} \label{e10}
\begin{aligned}
&\max_{1\leq j\leq k-1}\{m(r,U_j^1)t\}\\
&\leq \max_{1\leq j\leq   k-1}\{m(r,A_j)+o(m(r,A_0)+
  O(\log(rT(r,f)))\}+O(1),
\end{aligned}
\end{equation}
holds for all $r\in E_1-E_2$ (where $E_1$ is a set of infinite
logarithmic measure and $E_2$ is a set of finite linear measure).
From \eqref{e7}, \eqref{e8}, \eqref{e10} and  Lemma \ref{lem2.5},
there exists a set
$E\subset[1,+\infty)$ with infinite logarithmic measure such that
\begin{equation} \label{e11}
\begin{aligned}
 \sigma_{[p,q]}(A_0)
&= \lim_{r\to\infty}\frac{\log_p  m(r,U_0^1)}{\log_q   r}\\
&>\sigma_1=\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}
\{\log_p m(r,   A_j)\}}{\log_q r}\\
&\geq\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m
 (r,  U_j^1)\}}{\log_q r}, ~r\in E.
\end{aligned}
\end{equation}
Now, suppose that \eqref{e6} holds for $i\leq n(n\in \mathbb{N})$, thus,
there exists a set $E$ with infinite logarithmic measure such that
\begin{equation} \label{e12}
\begin{aligned}
 \sigma_{[p,q]}(A_0)
&= \lim_{r\to\infty}\frac{\log_p   m(r,U_0^n)}{\log_q   r}\\
&>\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m(r,
  U_j^n)\}}{\log_q r}=\sigma_1.
\end{aligned}
\end{equation}
Next, we prove that \eqref{e6} holds for $i=n+1$. From the assumptions of
this lemma, we have
$U_j^{n+1}={U^{n}_{j+1}}'+U^{n}_j-\frac{{U^{n}_0}'}{U^{n}_0}U^{n}_{j+1}$,
$(j=0,1,2,\dots,k-1)$ and $U^{n}_k\equiv1$ for $i=n+1$. Thus, when
$j=0$, it follows  that
 $ U_0^{n+1}={U^{n}_{1}}'+U^{n}_0-\frac{{U^{n}_0}'}{U^{n}_0}U^{n}_{1}$.
 Then, we have
\begin{equation} \label{e13}
  m\left(r,U_0^{n+1}\right)\leq
  m\left(r,U^n_0\right)+m(r,U_1^n)+m(r,\frac{{U_0^n}'}{U_0^n})
  +m(r,\frac{{U_1^n}'}{U_1^n})+O(1).
\end{equation}
And since
$U^{n}_0=-{U^{n}_{1}}'+U_0^{n+1}+\frac{{U^{n}_0}'}{U^{n}_0}U^{n}_{1}$,
we have
\begin{equation} \label{e14}
 m(r,U^n_0)  \leq
 m(r,U_0^{n+1})+m(r,U_1^n)+m(r,\frac{{U_0^n}'}{U_0^n})
  +m(r,\frac{{U_1^n}'}{U_1^n})+O(1).
\end{equation}
When $j\neq0$, it follows from the definitions of
$U_j^{n+1}$ $(j=1,2,\dots,k-1)$ and $U_k^{n}\equiv1$ that
\begin{equation} \label{e15}
 m(r,U_j^{n+1})
\leq  m(r,U_{j+1}^n)+m(r,U_j^n)
 +m(r,\frac{{U_{j+1}^n}'}{U_{j+1}^n})+m(r,\frac{{U_0^n}'}{U_0^n})+O(1).
\end{equation}
From \eqref{e12}--\eqref{e15}, there exists a set $E$ with infinite logarithmic
measure such that
\begin{equation} \label{e16}
\begin{aligned}
 \lim_{r\to\infty}\frac{\log_p
 m\left(r,U_0^{n+1}\right)}{\log_q   r}
&=\lim_{r\to\infty}\frac{\log_p
  m\left(r,U_0^n\right)}{\log_q   r}=\sigma_{[p,q]}(A_0)\\
&>\sigma_1=\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m\left(r,
  U_j^n\right)\}}{\log_q r}\\
&=\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m(r,
  U_j^{n+1})\}}{\log_q r},\quad  r\in E.
\end{aligned}
\end{equation}
Thus, the proof  is complete.
\end{proof}

\begin{lemma} \label{lem2.7}
  Let $H_j(z)$ $(j=0,1,\dots,k-1)$ be meromorphic functions of finite
  $[p,q]$-order. If
  $$
\limsup_{r\to\infty}\frac{\max_{1\leq j\leq k-1}\{\log_p m(r,H_j)\}}{\log_q r}
=\beta_1
$$
and there exists a set $E_1$ with infinite logarithmic measure
such that
$$
  \lim_{r\to\infty}\frac{\log_p m(r,H_0)}{\log_q   r}=\beta_2>\beta_1
$$
holds for all $r\in E_1$, then every meromorphic solution $f\not\equiv0$ of
\begin{equation} \label{e17}
    f^{(k)}+H_{k-1}f^{(k-1)}+\dots+H_1f'+H_0f=0
\end{equation}
satisfies $\sigma_{[p+1,q]}(f)\geq \beta_2$.
  \end{lemma}

\begin{proof}
Assume that $f(z)\not\equiv0$ is a meromorphic solution of \eqref{e17}.
From \eqref{e17}, we have
\begin{equation} \label{e18}
m(r,H_0)\leq m(r,\frac{f^{(k)}}{f}) +\dots+m(r,\frac{f'}{f})
+\sum_{j=1}^{k-1}m(r,H_j)+O(1).
\end{equation}
By the logarithmic derivative lemma and \eqref{e18}, we have
\begin{equation} \label{e19}
  m(r,H_0)\leq O\{\log rT(r,f)\}+\sum_{j=1}^{k-1}m(r,H_j),\quad
  r\not\in E_2,
\end{equation}
where $E_2\subset[1,+\infty)$ is a set with finite linear measure.
From the assumptions of Lemma \ref{lem2.7}, there exists a set $E_1$ with
infinite logarithmic measure such that for all $|z|=r\in E_1-E_2$,
we have
\begin{equation} \label{e20}
 \exp_p\{(\beta_2-\varepsilon)\log_q r\}\leq O\{\log
  rT(r,f)\}+(k-1)\exp_p\{(\beta_1+\varepsilon)\log_qr\},
\end{equation}
where $0<2\varepsilon<\beta_2-\beta_1$. From (20), we have
$\sigma_{[p+1,q]}(f)\geq \beta_2$.
\end{proof}

\begin{lemma}[\cite{12}] \label{lem2.8}
 Let $f(z)$ be a transcendental meromorphic function and  $\alpha>1$  
be a given constant. Then for any given $\varepsilon>0$, there exists a set
$E_{7}\subset[1,\infty)$ that has finite logarithmic measure and a
constant $M>0$ that depends only on $\alpha$ and $(m,n)
(m,n\in\{0,\dots,k\}$ with $m<n$) such that for all $z$ satisfying
$|z|=r\not\in [0,1]\cup E_{7}$, we have
$$
\big|\frac{f^{(n)}(z)}{f^{(m)}(z)}\big|
\leq M \Big(\frac{T(\alpha r,f)}{r}
(\log^{\alpha} r)\log T(\alpha r,f)\Big)^{n-m}. 
$$
\end{lemma}

\begin{lemma}[{\cite[Lemma 3.13]{Liu-Tu}}] \label{lem2.9}
Let $f(z)$ be an entire function of $[p,q]$-order satisfying 
$\sigma_{[p,q]}(f)=\sigma$, $\tau_{[p,q]}(f)=\tau$, $0<\sigma<\infty$,
 $0<\tau<\infty$, then for any given $\beta<\tau$, there exists a set 
$E_{4}\subset[1,+\infty)$ that has infinite logarithmic measure such that 
for all $r\in E_{4}$, we have
$$
\log_p M(r,f)>\beta (\log_{q-1}r)^{\sigma}.
$$
\end{lemma}

\begin{lemma} \label{lem2.10}
 Let $A_0(z), A_1(z), \dots, A_{k-1}(z)$ be entire functions with
  finite $[p,q]$-order and satisfy 
$\max\{\sigma_{[p,q]}(A_{j}): j=1,2,\dots,k-1\}
\leq\sigma_{[p,q]}(A_0)=\sigma_2<\infty$ and 
$\max\{\tau_{[p,q]}(A_j)$ $|\sigma_{[p,q]}$
$(A_{j})=\sigma_{[p,q]}(A_0)>0\}=\tau_1<\tau_{[p,q]}(A_0)=\tau$,
and let $U_j^1, U_j^i$ be as stated  in Lemma \ref{lem2.6}.
Then for any given $\varepsilon(0<2\varepsilon<\tau-\tau_1)$, 
there exists a set $E_5$ with infinite logarithmic measure such that
\begin{equation} \label{e21}
 |U_j^i|\leq
  \exp_p\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\},\quad
 |U_0^i| \geq\exp_p\{(\tau-\varepsilon)(\log_{q-1}r)^{\sigma_2}\},
\end{equation}
   where $i\in \mathbb{N}$ and $j=1,2,\dots,k-1$.
\end{lemma}

\begin{proof}
We will use the induction method for this proof.

(i) First, we prove that $U_j^i(j=0,1,\dots,k-1)$ satisfy \eqref{e21} when
$i=1$.
  From the definition $U_j^1=A'_{j+1}+A_j-\frac{A'_0}{A_0}A_{j+1} (j\neq0)$ and
  $U_0^1=A'_{1}+A_0-\frac{A'_0}{A_0}A_{1}$, we have
 \begin{equation} \label{e22}
   |U_0^1|\geq-|A_{1}|\left(|\frac{A_{1}'}{A_{1}}|
    +|\frac{A_0'}{A_0}|\right)+
    |A_0|
  \end{equation}
  and  
\begin{equation} \label{e23}
|U_j^1|\leq|A_{j+1}|\big(|\frac{A_{j+1}'}{A_{j+1}}|+|\frac{A_0'}{A_0}|\big)+|A_j|,
    \quad j=1,2,\dots,k-1;\, A_k\equiv1.
  \end{equation}
From  Lemma \ref{lem2.8}, Lemma \ref{lem2.9} and \eqref{e22}--\eqref{e23}, 
for any $\varepsilon(0<4\varepsilon<\tau-\tau_1)$, there exists a 
set $E_5$ with infinite   logarithmic measure such that
\begin{equation} \label{e24}
   \begin{aligned}
|U_0^1|
&\geq-2M\exp_p\{(\tau_1+\frac{\varepsilon}{8})
(\log_{q-1}r)^{\sigma_2}\}(T(2r,A_0))^2\\
&\quad +\exp_p\{(\tau-\frac{\varepsilon}{4})(\log_{q-1}r)^{\sigma_2}\}
\\
&\geq -2M\exp_p\{(\tau_1+\frac{\varepsilon}{8})(\log_{q-1}r)^{\sigma_2}\}
     \big(\exp_p\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\}\big)^2
     \\
&\quad +\exp_p\{\left(\tau-\frac{\varepsilon}{4}\right)(\log_{q-1}r)^{\sigma_2}\}
     \\
&\geq\exp_p\{\left(\tau-\frac{\varepsilon}{2}\right)(\log_{q-1}r)^{\sigma_2}\}
   \end{aligned}
\end{equation}
   and
\begin{equation} \label{e25}
\begin{aligned}
    |U_j^1|
&\leq 2M\exp_p\{(\tau_1+\frac{\varepsilon}{4})(\log_{q-1}r)^{\sigma_2}\}
    (T(2r,A_0))^2\\
&\quad +\exp_p\{(\tau_1+\frac{\varepsilon}{4})(\log_{q-1}r)^{\sigma_2}\}\\
&\leq 2M\exp_p\{(\tau_1+\frac{\varepsilon}{4})(\log_{q-1}r)^{\sigma_2}\}
\big(\exp_p\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\}\big)^2
\\
&\quad +\exp_p\{(\tau_1+\frac{\varepsilon}{4})(\log_{q-1}r)^{\sigma_2}\}
\\
&\leq \exp_p\{(\tau_1+\frac{\varepsilon}{2})(\log_{q-1}r)^{\sigma_2}\},\quad
    j\neq0,
\end{aligned}
\end{equation}
where $M>0$ is a constant, not necessarily the same at each
occurrence.

  (ii) Next, we show that $U_j^i$ $(j=0,1,2,\dots,k-1)$ satisfy \eqref{e21}
   when $i=2$.
From $U_0^2={U_{1}^1}'+U_0^1-\frac{{U_0^1}'}{U_0^1}U_{1}^1$ and
   $U_j^2={U_{j+1}^1}'+U_j^1-\frac{{U_0^1}'}{U_0^1}U_{j+1}^1$
$(j=0,1,\dots,k-1)$ and $U_k^1\equiv1$, we have
\begin{equation}  \label{e26}
|U_0^2|\geq
  |U_0^1|-|U_{1}^1|\Big(|\frac{{U_{1}^1}'}{U_{1}^1}|
  +|\frac{{U_{0}^1}'}{U_{0}^1}|\Big)
\end{equation}
and
\begin{equation} \label{e27}
  |U_j^2|\leq
  |U_j^1|+|U_{j+1}^1|\Big(|\frac{{U_{j+1}^1}'}{U_{j+1}^1}|
  +|\frac{{U_{0}^1}'}{U_{0}^1}|\Big), \quad
  j=1,2,\dots,k-1.
\end{equation}
By the conclusions in (i), Lemma \ref{lem2.8} and Lemma \ref{lem2.9}, 
\eqref{e24}--\eqref{e25}, for
all $|z|=r\in E_5$, we have
\begin{equation} \label{e28}
\begin{aligned}
|U_0^2|&\geq-2M\exp_p\big\{(\tau_1+\frac{\varepsilon}{2})
(\log_{q-1}r)^{\sigma_2}\big\}
     \big(\exp_p\big\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\big\}\big)^2
\\
&\quad +\exp_p\{\left(\tau-\frac{\varepsilon}{2}\right)(\log_{q-1}r)^{\sigma_2}\}
\\
&\geq  \exp_p\{(\tau-\varepsilon)(\log_{q-1}r)^{\sigma_2}\}
   \end{aligned}
\end{equation}
   and
\begin{equation} \label{e29}
  \begin{aligned}
    |U_j^2|
&\leq2M\exp_p\{(\tau_1+\frac{\varepsilon}{2})(\log_{q-1}r)^{\sigma_2}\}
    \exp_p\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\}
\\
&\quad +\exp_p\{(\tau_1+\frac{\varepsilon}{2})(\log_{q-1}r)^{\sigma_2}\}
    \\
&\leq  \exp_p\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\},\quad j\neq0.
   \end{aligned}
\end{equation}

  (iii) Now, suppose that \eqref{e21} holds for $i\leq n (n\in \mathbb{N})$.
 Thus, for any given
   $\varepsilon(0<4\varepsilon<\tau-\tau_1)$, there exists a set $E_5$ with infinite
   logarithmic measure such that
\begin{equation} \label{e30}
 |U_j^i|\leq
  \exp_p\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\},
  |U_0^i|\geq\exp_p\{(\tau-\varepsilon)(\log_{q-1}r)^{\sigma_2}\},
   \end{equation}
where  $i\leq n$ and $j=1,2,\dots,k-1$.
From $U_0^{n+1}={U_{1}^n}'+U_0^n-\frac{{U_0^n}'}{U_0^n}U_{1}^n$ and
$U_j^{n+1}={U_{j+1}^n}'+U_j^n-\frac{{U_0^n}'}{U_0^n}U_{j+1}^n$
$(j=0,1,\dots,k-1)$ and $U_k^n\equiv 1$, we have
\begin{equation}
|U_0^{n+1}|\geq
  |U_0^n|-|U_{1}^n|\Big(|\frac{{U_{1}^n}'}{U_{1}^n}|
  +|\frac{{U_{0}^n}'}{U_{0}^n}|\Big)
\end{equation}
and
\begin{equation} \label{e32}
  |U_j^{n+1}|\leq
  |U_j^n|+|U_{j+1}^n|\Big(|\frac{{U_{j+1}^n}'}{U_{j+1}^n}|
  +|\frac{{U_{0}^n}'}{U_{0}^n}|\Big), \quad
  j=1,2,\dots,k-1.
\end{equation}
Then, from Lemma \ref{lem2.8}, Lemma \ref{lem2.9} and \eqref{e30}--\eqref{e32}, for all 
$|z|=r\in E_5$, we have
\begin{equation} \label{e33}
   \begin{aligned}
    |U_j^{n+1}|
&\leq 2M\exp_p\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\}
    \big(\exp_p\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\}\big)^2\\
&\quad +\exp_p\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\}\\
&\leq  \exp_p\{(\tau_1+2\varepsilon)(\log_{q-1}r)^{\sigma_2}\},\quad j\neq0,
   \end{aligned}
\end{equation}
   and
\begin{equation} \label{e34}
   \begin{aligned}
|U_0^{n+1}|
&\geq-2M\exp\{(\tau_1+\varepsilon)(\log_{q-1}r)^{\sigma_2}\}
    (\exp_p\{(\sigma_2+\frac{\varepsilon}{8})(\log_{q}2r)\})^2
\\
&\quad + \exp_p\{(\tau-\varepsilon)(\log_{q-1}r)^{\sigma_2}\} \\
&\geq \exp_p\{(\tau-2\varepsilon)(\log_{q-1}r)^{\sigma_2}\}.
\end{aligned}
\end{equation}
Thus, the proof is complete.
\end{proof}

\begin{lemma} \label{lem2.11}
  Let $B_j(z)$  $(j=0,1,\dots,k-1)$ be meromorphic functions such that
  $\max\{\sigma_{[p,q]}(B_j):
  j=1,2,\dots,k-1\}=\sigma_4<\sigma_{[p,q]}(B_0)=\sigma_3$ and
$\delta:=\delta(\infty,B_0)=\lim_{\overline{r\to\infty}}\frac{m(r,B_0)}{T(r,B_0)}>0$.
  Then every meromorphic solution $f\not\equiv0$ of equation
  \begin{equation} \label{e35}
    f^{(k)}+B_{k-1}f^{(k-1)}+\dots+B_1f'+B_0f=0
  \end{equation}
  satisfies $\sigma_{[p+1,q]}(f)\geq \sigma_3$.
\end{lemma}

\begin{proof}
Let $f\not\equiv0$ be a meromorphic solution of equation \eqref{e35}. Then
from \eqref{e35}, we have
\begin{equation} \label{e36}
\begin{aligned}
  m(r,B_0)
&\leq  m(r,\frac{f^{(k)}}{f})
  +m(r,\frac{f^{(k-1)}}{f})+\dots+m(r,\frac{f'}{f})
  +\sum_{j=1}^{k-1}m(r,B_j)+O(1) \\
&\leq O\{\log rT(r,f)\}+\sum_{j=1}^{k-1}T(r,B_j), \quad r\not\in
  E_6,
\end{aligned}
\end{equation}
where $E_6\subset[1,+\infty)$ is a set with finite linear measure.
By Lemma \ref{lem2.5}, there exists a set $E$ with infinite logarithmic
measure such that for all $|z|=r\in E$, we have
\begin{equation} \label{e37}
  \lim_{r\to\infty}\frac{\log_p T(r,B_0)}{\log_q r}=\sigma_3.
\end{equation}
Since $\delta:=\delta(\infty,B_0)>0$, then for any given
$\varepsilon(0<2\varepsilon<\min\{\delta,\sigma_3-\sigma_4\})$ and
for all $r\in E$, by (37), we have
\begin{equation} \label{e38}
  m(r,B_0)\geq (\delta-\varepsilon)\exp_p\{(\sigma_3-\varepsilon)\log_qr\}.
\end{equation}
From \eqref{e36} and \eqref{e38}, we have
\begin{equation} \label{e39}
  (\delta-\varepsilon)\exp_p\{(\sigma_3-\varepsilon)\log_qr\}\leq O\{\log
  rT(r,f)\}+(k-1)\exp_p\{(\sigma_4+\varepsilon)\log_qr\},
\end{equation}
where $ r\in E-E_6.$ From \eqref{e39}, we obtain $\sigma_{[p+1,q]}(f)\geq
\sigma_3=\sigma_{[p,q]}(B_0)$.
\end{proof}

\begin{lemma} \label{lem2.12}
  Let $B_j(z)$, $j=0,1,\dots,k-1$ be meromorphic
functions of finite $[p,q]$ order. If there exist positive constants
$\sigma_5, \beta_{3}, \beta_{4} (0<\beta_{3}<\beta_{4})$ and a set
$E_{8}$ with infinite logarithmic measure such that
$$
\max\{|B_j(z)|:j=1,2,\dots,k-1\}\leq\exp_p\{\beta_{3}(\log_{q-1}r)^{\sigma_5}\},
$$
and 
$$ 
|B_0(z)|\geq\exp_p\{\beta_{4}(\log_{q-1}r)^{\sigma_5}\}
$$
hold for all $|z|=r\in E_{8}$, then every meromorphic solution
$f\not\equiv0$ of \eqref{e35} satisfies $\sigma_{[p+1,q]}(f)\geq\sigma_5$.
\end{lemma}

\begin{proof}
Suppose that $f\not\equiv0$ is a meromorphic function of \eqref{e35}. Then
it follows that
\begin{equation} \label{e40}
  |B_0(z)|\leq
  |\frac{f^{(k)}}{f}|+\sum_{j=1}^{k-1}|B_j(z)||\frac{f^{(j)}}{f}|.
\end{equation}
By Lemma \ref{lem2.8}, there exists a set $E_7$ with finite logarithmic
measure such that for all $|z|=r\not\in E_7$, we have
\begin{equation} \label{e41}
  |\frac{f^{(j)}}{f}|\leq M[T(2r,f)]^{2j}, \quad
  j=1,2,\dots,k.
\end{equation}
 By \eqref{e40}, \eqref{e41} and the assumptions of Lemma \ref{lem2.12}, 
for all $|z|=r\in E_8-E_7$, we have
\begin{equation} \label{e42}
  \exp_p\{\beta_4(\log_{q-1}r)^{\sigma_5}\}\leq
  Mk[T(2r,f)]^{2k}\exp_p\{\beta_3(\log_{q-1}r)^{\sigma_5}\}.
\end{equation}
Since $0<\beta_3<\beta_4$ and by \eqref{e42}, we have
$\sigma_{[p+1,q]}(f)\geq \sigma_5$.
\end{proof}

\begin{lemma}[{\cite[Lemma 3.12]{Liu-Tu}}] \label{lem2.13}
Let $A_{0}, A_{1},\dots, A_{k-1}$, $F\not\equiv0$ be meromorphic
functions, if $f$ is a meromorphic solution of the equation
$$
f^{(k)}+A_{k-1}f^{(k-1)}+\cdot\cdot\cdot+A_{0}f=F, 
$$ 
satisfying
$\max\{\sigma_{[p,q]}(F), \sigma_{[p,q]}(A_{j});
j=0,1,\dots,k-1\}<\sigma_{[p,q]}(f)$, then we have
$\sigma_{[p,q]}(f)=\lambda_{[p,q]}(f)=\overline{\lambda}_{[p,q]}(f)$.
\end{lemma}

\begin{lemma}[{\cite[Theorem 2.3]{Liu-Tu}}] \label{lem2.14}
 Let $A_j(z)$  $(j=0,1,\dots,k-1)$ be entire
functions satisfying $\max\{\sigma_{[p,q]}(A_{j}):
j=1,2,\dots,k-1\}\leq\sigma_{[p,q]}(A_0)<\infty$ and 
$$
\max\{\tau_{[p,q]}(A_j)|\sigma_{[p,q]}(A_{j})
 =\sigma_{[p,q]}(A_0)>0\}<\tau_{[p,q]}(A_0).
$$
Then every nontrivial solution $f$ of \eqref{e2} satisfies
$\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$.
\end{lemma}

\section{Proofs of Theorems}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
We will consider two cases as follows.

\textbf{Case 1}. Suppose that $\max\{\sigma_{[p,q]}(A_j):
  j=1,2,\dots,k-1\}<\sigma_{[p,q]}(A_0)<\infty$.

  (i) First, we prove that
  $\overline{\lambda}_{[p+1,q]}(f-\varphi)=\sigma_{[p+1,q]}(f)$. Assume that
  $f$ is a nontrivial solution of \eqref{e2}, from \cite[Theorem 2.2]{Liu-Tu}, 
we have
  $\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$. Set $g=f-\varphi$. Since
  $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$, then
  $\sigma_{[p+1,q]}(g)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$ and
  $\overline{\lambda}_{[p+1,q]}(g)=\overline{\lambda}_{[p+1,q]}(f-\varphi)$. By
  Lemma \ref{lem2.1}, we get that $g$ satisfies the equation \eqref{e3}. Set
  $F=\varphi^{(k)}+A_{k-1}\varphi^{(k-1)}+\dots+A_0\varphi$. If $F\equiv0$, 
then from  \cite{Liu-Tu}, we have $\sigma_{[p+1,q]}(\varphi)=\sigma_{[p,q]}(A_0)$, a
  contradiction. Then $F\not\equiv0$. From Lemma \ref{lem2.4} and assumption of Case 1, 
we have
$$
\sigma_{[p+1,q]}(F)\leq\max\{\sigma_{[p+1,q]}(\varphi),\sigma_{[p+1,q]}(A_0)\}
  =\max\{\sigma_{[p+1,q]}(\varphi),0\}. 
$$
Since $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$, we have
$$
\max\{\sigma_{[p+1,q]}(F),\sigma_{[p+1,q]}(A_j):
  j=0,1,2,\dots,k-1\}<\sigma_{[p+1,q]}(f).
$$
 By Lemma \ref{lem2.13}, we have
  $\overline{\lambda}_{[p+1,q]}(g)=\lambda_{[p+1,q]}(g)=\sigma_{[p+1,q]}(g)=\sigma_{[p,q]}(A_0)$.
  Thus, we have
  $$
\overline{\lambda}_{[p+1,q]}(f-\varphi)
  =\lambda_{[p+1,q]}(f-\varphi)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0).
$$

  (ii) Second, we prove that
  $\overline{\lambda}_{[p+1,q]}(f'-\varphi)=\sigma_{[p+1,q]}(f)$. Set
  $g_1=f'-\varphi$, then 
$\sigma_{[p+1,q]}(g_1)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$.
  From Lemma \ref{lem2.2}, we get that $g_1$ satisfies the equation \eqref{e4}. Set
  $F_1=\varphi^{(k)}+U^1_{k-1}\varphi^{(k-1)}+\dots+U^1_0\varphi$,
  where $U^1_j (j=0,1,\dots,k-1)$ are stated as in Lemma \ref{lem2.2}. If
  $F_1\equiv0$, from Lemma \ref{lem2.6} and Lemma \ref{lem2.7}, we have
  $\sigma_{[p+1,q]}(\varphi)\geq \sigma_{[p,q]}(A_0)$, a contradiction with
  $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$. 
Hence $F_1\not\equiv0$. From the definition of
  $U_j^1 (j=0,1,\dots,k-1)$, we have $\sigma_{[p+1,q]}(U_j^1)\leq
  \sigma_{[p+1,q]}(A_j)$ $j=0,1,\dots,k-1$. Thus, we can get
  $\sigma_{[p+1,q]}(F_1)\leq\max\{\sigma_{[p+1,q]}(\varphi),\sigma_{[p+1,q]}(U_j^1):
  j=0,1,\dots,k-1\}$. Since $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$, we have
  $\max\{\sigma_{[p+1,q]}(F_1),\sigma_{[p+1,q]}(U_j^1):
  j=0,1,\dots,k-1\}<\sigma_{[p,q]}(A_0)=\sigma_{[p+1,q]}(g_1)$.
  By Lemma  \ref{lem2.13}, we obtain
  $$
\overline{\lambda}_{[p+1,q]}(f'-\varphi)=\lambda_{[p+1,q]}(f'-\varphi)
=\sigma_{[p+1,q]}(f).
$$

  (iii) We will prove that
$\overline{\lambda}_{[p+1,q]}(f^{(i)}-\varphi)=\sigma_{[p+1,q]}(f),
(i>1, i\in \mathbb{N})$. Set $g_i=f^{(i)}-\varphi$, then
$\sigma_{[p+1,q]}(g_i)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$.
From Lemma \ref{lem2.3}, we have $g_i$ satisfies equation \eqref{e5}. Set
$F_i=\varphi^{(k)}+U^i_{k-1}\varphi^{(k-1)}+\dots+U^i_0\varphi$,
where $U^i_j (j=0,1,\dots,k-1; i\in \mathbb{N})$ are stated as in
Lemma \ref{lem2.3}. If $F_i\equiv0$, from Lemma \ref{lem2.6} and 
Lemma \ref{lem2.7}, we have
$\sigma_{[p+1,q]}(\varphi)\geq \sigma_{[p,q]}(A_0)$, a contradiction
with
  $\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$. Hence $F_i\not\equiv0$. By
  using the same argument as in Case 1(ii), we can get
  $$
\overline{\lambda}_{[p+1,q]}(f^{(i)}-\varphi)
=\lambda_{[p+1,q]}(f^{(i)}-\varphi)=\sigma_{[p+1,q]}(f).
$$


\textbf{Case 2}. Suppose that $\max\{\sigma_{[p,q}(A_{j}):
j=1,2,\dots,k-1\}\leq\sigma_{[p,q]}(A_0)<\infty$ and 
$\max\{\tau_{[p,q]}(A_j)|\sigma_{[p,q]}$
$(A_{j})=\sigma_{[p,q]}(A_0)>0\}<\tau_{[p,q]}(A_0)$.

(i) We first prove that
$\overline{\lambda}_{[p+1,q]}(f-\varphi)=\sigma_{[p+1,q]}(f)$. Since
$f$ is a nontrivial solution of \eqref{e2}, by Lemma \ref{lem2.14}, we have
$\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)>0$. Set $g=f-\varphi$.
Since $\varphi\not\equiv0$ is an entire function satisfying
$\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$, then we have
$\sigma_{[p+1,q]}(g)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$ and
  $\overline{\lambda}_{[p+1,q]}(g)=\overline{\lambda}_{[p+1,q]}(f-\varphi)$. From
  Lemma \ref{lem2.1}, we get that $g$ satisfies equation \eqref{e3}.  We will
  affirm $F\not\equiv0$. If $F\equiv0$, by Lemma \ref{lem2.14}, we get
  $\sigma_{[p+1,q]}(\varphi)=\sigma_{[p,q]}(A_0)$, a contradiction. Hence
  $F\not\equiv0$. From the assumptions of Case 2, we get
  $$
  \max\{\sigma_{[p+1,q]}(F), \sigma_{[p+1,q]}(A_j): j=0,1,\dots,k-1\}<
  \sigma_{[p+1,q]}(g)=\sigma_{[p,q]}(A_0).
$$
   From Lemma \ref{lem2.13}, we have
   $$
\overline{\lambda}_{[p+1,q]}(f-\varphi)
   =\lambda_{[p+1,q]}(f-\varphi)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0).
$$


(ii) Now we prove that
$\overline{\lambda}_{[p+1,q]}(f'-\varphi)=\sigma_{[p+1,q]}(f)$. Let
$g_1=f'-\varphi$. Since
$\sigma_{[p+1,q]}(\varphi)<\sigma_{[p,q]}(A_0)$, we have
$\sigma_{[p+1,q]}(g_1)=\sigma_{[p+1,q]}(f)=\sigma_{[p,q]}(A_0)$. By
Lemma \ref{lem2.2}, we get that $g_1$ satisfies equation \eqref{e4}. If
$F_1\equiv0$, from Lemma \ref{lem2.10} and Lemma \ref{lem2.12}, we have
$\sigma_{[p+1,q]}(\varphi)\geq \sigma_{[p,q]}(A_0)$. Then we can get
a contradiction with $\sigma_{[p+1,q]}(\varphi)<
\sigma_{[p,q]}(A_0)$. Therefore, we have $F_1\not\equiv0$.  By \eqref{e4}
and Lemma \ref{lem2.13}, we have
$$
\overline{\lambda}_{[p+1,q]}(f'-\varphi)=\lambda_{[p+1,q]}(f'-\varphi)
=\sigma_{[p+1,q]}(f)=\sigma _{[p,q]}(A_0).
$$ 
Similar to the arguments as in Case 1 (iii) and by
using Lemmas \ref{lem2.3}, \ref{lem2.10} and \ref{lem2.12}, we obtain
$$
\overline{\lambda}_{[p+1,q]}(f^{(i)}-\varphi)
=\lambda_{[p+1,q]}(f^{(i)}-\varphi)=\sigma_{[p+1,q]}(f)
=\sigma_{[p,q]}(A_0), \quad (i\in \mathbb{N}).
$$
Thus, the proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
  According to the conditions of Theorem 1.2, we can easily obtain the
  conclusions  by using the similar argument as in
  Theorem \ref{thm1.1} and Lemma \ref{lem2.11}.
\end{proof}

\subsection*{Acknowledgments}
This work was supported by the  NSF of China (11301233, 61202313)
and the Natural Science Foundation of Jiang-Xi Province in China
(20132BAB211001 and 20132BAB211002).


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\end{document}