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\markboth{N.\,K.\,Cheriet and K.\,Hamani}{On the hyper-order of solutions of nonhomogeneous linear
differential equations}
\title[On the hyper-order of solutions of nonhomogeneous linear
differential equations]{On the hyper-order of solutions of nonhomogeneous linear
differential equations
}




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\author[N.\,K.\,Cheriet, K.\,Hamani]{Nour El Imane Khadidja Cheriet and Karima Hamani{\corrauth}}
\address{Department of Mathematics, Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB), B.\,P.\,227 Moslabelanem, Algeria
}
\emails{{\tt cheriet.nour@univ-mosta.dz}\,\,(N.\,K.\,Cheriet), {\tt  karima.hamani@univ-mosta.dz} (K.\,Hamani)}




%%%%% Begin Abstract %%%%%%%%%%%
\begin{abstract}
In this paper, we study the hyper-order of solutions of   higher order linear differential equation

\begin{equation*}
f^{(k)}+A_{k-1}(z)f^{(k-1)}+\ldots A_{1}(z)f^{\prime }+A_{0}(z)f=H(z),
\end{equation*}%
where $k\geq 2$ is an integer, $A_{j}\left( z\right) $ $(j=0,1,\ldots,k-1)$ and
$H\left( z\right) $ $\left( \not\equiv 0\right) $ are entire functions or
polynomials. We improve previous results given by Xu and Cao.


\end{abstract}
%%%%% end %%%%%%%%%%%

%%%%% Keywords %%%%%%%%%%%
\keywords{Linear differential equation, entire function,
hyper-order}

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\ams{34M10, 30D35}

%%%% maketitle %%%%%
\maketitle

%%%% Start %%%%%%
\section{Introduction and main results}
We assume that the reader is familiar with   usual notations and basic
results of Nevanlinna theory (see \cite{HAY2,YAN}). We also use basic results of Wiman-Valiron theory, (see \cite{JAN}). In addition, we use the notation $\sigma (f)$ to denote
the order of growth of a meromorphic function $f$, $\lambda
\left( f\right)$ and $\overline{\lambda }\left( f\right)$ to denote
 the exponent of convergence of a sequence of zeros and a
sequence of distinct zeros of $f$, respectively. We also by $\sigma_{2}(f)$ denote  the
hyper-order of $f$   defined by (see \cite{YAN})

\begin{equation*}
\sigma _{2}(f)=\underset{r\rightarrow +\infty }{\lim \sup }\frac{\log \log
T(r,f)}{\log r},
\end{equation*}%
where $T(r,f)$ is the Nevanlinna characteristic function of $f$.

\noindent The hyper-exponent of convergence of a sequence of zeros and
distinct zeros of $f$ are  defined by (see \cite{CHE1})

\begin{equation*}
\lambda _{2}(f)=\underset{r\rightarrow +\infty }{\lim \sup }\frac{\log \log
N(r,\frac{1}{f})}{\log r}
\end{equation*}%
and

\begin{equation*}
\overline{\lambda }_{2}(f)=\underset{r\rightarrow +\infty }{\lim \sup }\frac{%
\log \log \overline{N}(r,\frac{1}{f})}{\log r},
\end{equation*}%
respectively, where $N(r,\frac{1}{f})$ and $\overline{N}(r,\frac{1}{f})$ are
the counting functions of zeros and distinct zeros of $f$, respectively.


\smallskip
\noindent For a set $E\subset [1,+\infty)$, let $m(E)$ and $m_l(E)$ denote the linear measure
and the logarithmic
measure of $E$, respectively. Moreover, the upper logarithmic density and lower
logarithmic density of $E$ are defined by
\begin{equation*}
\overline{\log dens}(E)=\limsup_{r\rightarrow \infty }\frac{m_l(E\cap
\lbrack 1,r])}{\log r},\underline{\text{ }\log dens}(E)=\liminf_{r\rightarrow \infty }\frac{m_l(E\cap \lbrack 1,r])}{\log r}.
\end{equation*}%

%\quad

\noindent For the second order linear differential equation

\begin{equation}\label{1}
f^{^{\prime \prime }}+e^{-z}f^{^{\prime }}+Q(z)f=0,
\end{equation}%
where $Q\left( z\right) $ is an entire function of finite order, it is well known that every solution of equation (1) is an entire function and most solutions of (1) have an infinite order. But equation (1) with $Q(z)=-(4+2e^{-z})$ possesses a solution $f(z)=e^{2z}$ of finite order.

\smallskip
\noindent
Thus a natural question is: what condition on $Q(z)$ will guarantee that every solution $f(\not\equiv 0)$ of equation (1) has an infinite order?

\smallskip\noindent
In $[2]$, Chen studied the problem, where $Q(z)=h(z) e^{bz}$, $h(z)$ is a nonzero polynomial and $b$ is a complex number. He proved that if $b\neq -1$, then
every solution $f(\not\equiv 0)$ of equation (1) is of infinite order and $\sigma_2(f)=1$.

\smallskip
\noindent
In the same paper, he also considered  more general equations of second order and proved the following two results:

\begin{theorem}[see \cite{CHE2}]
Let $A_{j}(z)(\not\equiv 0)$ $(j=0,1)$ be entire functions with $\sigma(A_j)<1$, $a$, $b$   complex constants such that $ab\neq 0$ and $a=cb$ $(c>1)$. Then every solution $f(\not\equiv 0)$ of equation
\begin{equation}
f^{^{\prime \prime }}+A_1(z)e^{az}f^{^{\prime }}+A_0(z)e^{bz}f=0,
\label{2}
\end{equation}%
has an infinite order.
\end{theorem}
\begin{theorem}[see \cite{CHE2}]
Let $A_{j}(z)(\not\equiv 0)$, $D_j(z)$ $(j=0,1)$ be entire functions with $\sigma(A_j)<1$, $\sigma(D_j)<1$ $a$, $b$ be complex constants such that $ab\neq 0$ and $\arg a\neq \arg b$ or $a=cb$ $(0<c<1)$. Then every solution $f(\not\equiv 0)$ of equation
\begin{equation}
f^{^{\prime \prime }}+(A_1(z)e^{az}+D_1(z))f^{^{\prime }}+(A_0(z)e^{bz}+D_0(z))f=0,
\label{3}
\end{equation}%
has an infinite order.
\end{theorem}
\noindent In 2008, Wang and Laine have investigated   nonhomogeneous equations related to (2) and (3) and obtained the following two results:
\begin{theorem}[see \cite{WAN}]Suppose that $A_{0}(z)\not\equiv 0$\textit{, }$A_{1}(z)\not\equiv 0$, and $H$\textit{\ are entire functions of order less than one, and the
complex constants }$a$\textit{, }$b$\textit{\ satisfy }$ab\neq 0$\textit{\
and }$b\neq a$.\textit{\ Then every nontrivial solution }$f$\textit{\ of
equation }$\left( 2\right) $\textit{\ is of infinite order.}
\end{theorem}
\begin{theorem}[see \cite{WAN}]Suppose that $A_{0}(z)\not\equiv 0$\textit{, }$A_{1}(z)\not\equiv 0$%
\textit{, }$D_{0}(z)$\textit{, }$D_{1}(z)$, and $H$\textit{\ are entire
functions of order less than one, and the complex numbers }$a$\textit{, }$b$%
\textit{\ satisfy }$ab\neq 0$\textit{\ and }$b/a<0$\textit{. Then every
nontrivial solution }$f$\textit{\ of equation }$\left( 3\right) $\textit{\
is of infinite order.}
\end{theorem}
\noindent In \cite{XU}, Xu and Cao have studied the above problem for
higher order linear differential equations and proved the following two
results:

\begin{theorem}[see \cite{XU}]Let $k\geq 2$\textit{\ be an integer, }$%
P(z)=a_{n}z^{n}+\ldots+a_{1}z+a_{0}$,\textit{\ and }$%
Q(z)=b_{n}z^{n}+\ldots+b_{1}z+b_{0}$\textit{\ be nonconstant polynomials, where
}$a_{i},b_{i}$\textit{\ }$(i=0,1,\ldots,n)$\textit{\ are complex numbers with }$%
a_{n}b_{n}\neq 0$\textit{\ and }$a_{n}\neq b_{n}$\textit{. Suppose that }$%
h_{i}(z)$\textit{\ }$(2\leq i\leq k-1)$\textit{\ are polynomials with degree
no more than }$n-1$\textit{\ in }$z$,\textit{\ }$A_{j}(z)$\textit{\ }$%
\not\equiv 0$\textit{\ }$(j=0,1)$\textit{\ and }$H$\textit{\
are entire functions satisfying }$\sigma \left( A_{j}\right) <n$, $(j=0,1)$\textit{\
and }$\sigma \left( H\right) <n$,\textit{\ and }$\varphi $\textit{\ is an
entire function of finite order. Then every nontrivial solution $f$ of equation }

\begin{equation}
f^{(k)}+h_{k-1}f^{(k-1)}+\ldots+h_{2}f^{^{\prime \prime
}}+A_{1}e^{P(z)}f^{^{\prime }}+A_{0}e^{Q(z)}f=H \label{4}
\end{equation}%
\textit{satisfies }$\sigma \left( f\right) =+\infty $\textit{, }$\sigma
\left( f\right) =\lambda (f)=\overline{\lambda}(f)=\overline{\lambda }%
(f-\varphi )=+\infty $\textit{\ and }$\sigma _{2}(f)=\lambda _{2}(f)=%
\overline{\lambda }_{2}(f)=\overline{\lambda }_{2}(f-\varphi )\leq n.$
\end{theorem}
\begin{theorem}[see \cite{XU}]Let $k\geq 2$\textit{\ be an integer. Suppose that }$%
A_{j}(z)\not\equiv 0$\textit{, }$D_{j}(z)$\textit{\ }$(j=0,1)$\textit{, and }%
$H$\textit{\ are entire functions satisfying }$\sigma (A_{j})<n$\textit{,
}$\sigma (D_{j})<n$\textit{, }$(j=0,1)$\textit{, }$\sigma (H)<n$\textit{, and }%
$P(z)$\textit{, }$Q(z)$\textit{, }$h_{i}$\textit{\ }$(2\leq i\leq k-1)$%
\textit{\ are as in Theorem 5 satisfying }$a_{n}b_{n}\neq 0$\textit{\ and }$%
a_{n}/b_{n}<0$\textit{. Then every nontrivial solution }$f$\textit{\ of
equation}

\begin{equation}
f^{(k)}+h_{k-1}f^{(k-1)}+\ldots+h_{2}f^{^{\prime \prime }}+\left(
A_{1}e^{P(z)}+D_{1}\right) f^{^{\prime }}+\left( A_{0}e^{Q(z)}+D_{0}\right)
f=H \label{5}
\end{equation}%
\textit{is of infinite order.}
\end{theorem}
\noindent In this paper, we investigate the hyper-order of   nontrivial solutions
of equations $\left( 4\right) $ and $\left( 5\right)$. We obtain the
following two results:
\quad
\begin{theorem}Let $k\geq 2$\textit{\ be an integer}, $P(z)$,\textit{\ }$Q(z)$,\textit{\ }$a_{n}$,
\textit{\ }$b_{n}$\textit{, }$%
h_{i}(z)$\textit{\ }$(2\leq i\leq k-1)$,\textit{\ }$A_{j}(z)$\textit{\ }$%
\not\equiv 0$\textit{\ }$(j=0,1)$\textit{, }$H (\equiv 0)$\textit{\ and
}$\varphi $\textit{\ satisfy  additional hypotheses of Theorem 5. Then
every nontrivial solution }$f$\textit{\ of equation }$\left( 4\right) $%
\textit{\ satisfies }$\sigma _{2}(f)=\lambda _{2}(f)=%
\overline{\lambda }_{2}(f)=\overline{\lambda }_{2}(f-\varphi )=n.$
\end{theorem}
\begin{theorem}Let $k\geq 2$\textit{\ be an integer}, $P(z)$,\textit{\ }$Q(z)$,\textit{\ }$a_{n}$,\textit{\ }$b_{n}$\textit{, }$%
h_{i}(z)$\textit{\ }$(2\leq i\leq k-1)$,\textit{\ }$A_{j}(z)$\textit{\ }$%
\not\equiv 0$,\textit{\ }$D_{j}(z)$ $(j=0,1)$\textit{\ and }$H (\equiv 0)$\textit{\ satisfy  additional hypotheses of Theorem 6. Then
every nontrivial solution }$f$\textit{\ of equation }$\left( 5\right) $%
\textit{\ satisfies }$\sigma _{2}(f)=n.$
\end{theorem}



\section{Preliminary lemmas}
\begin{lemma} [see \cite{XU}]Suppose that $k\geq 2$\textit{\ is an integer}$,$\textit{\ }$A_{0},$%
\textit{\ }$A_{1},\ldots,A_{k-1}$\textit{\ and }$F$\textit{\ }$\left(
\not\equiv 0\right) $\textit{\ are entire functions of finite order. Then
every solution }$f$\textit{\ of infinite order of equation}

\begin{equation*}
f^{\left( k\right) }+A_{k-1}f^{\left( k-1\right) }+\ldots+A_{1}f^{\prime
}+A_{0}f=F
\end{equation*}%
s\textit{atisfies }$\sigma _{2}(f)\leq \max \left\{ \sigma \left(
A_{j}\right) ,\sigma \left( F\right) :j=0,1,\ldots,k-1\right\} $\textit{.}
\end{lemma}
\quad

\begin{lemma}[see \cite{GUN}]Let $f(z)$\textit{\ be a transcendental meromorphic function and
let }$\alpha >1$\textit{\ and }$\varepsilon >0$\textit{\ be given constants. Then there exist a set }$E_{1}\subset \lbrack 1,+\infty )$\ \textit{having
finite logarithmic measure and a constant }$B>0$\textit{\ that depends only
on }$\alpha $\textit{\ and} $\left( i,j\right) $ $\left( i,j\text{ \textit{%
positive integers with} }i>j\right) $\textit{\ such that for all }$z$\textit{%
\ satisfying }$|z|=r\notin \left[ 0,1\right] \cup E_{1},$\textit{\ we have }%
\begin{equation*}
\left\vert \frac{f^{\left( i\right) }(z)}{f^{\left( j\right) }(z)}%
\right\vert \leq B\left[ \frac{T(\alpha r,f)}{r}\left( \log ^{\alpha
}r\right) \log T(\alpha r,f)\right] ^{i-j}\text{.}
\end{equation*}
\end{lemma}
\begin{lemma}[see \cite{HAY1}, p. 344]Let $f\left( z\right) =\underset{n=0}{\overset{+\infty }{\sum }}
a_{n}z^{n}$ be an entire function,  $\nu _{f}\left( r\right)
$\textit{\ denote the central index of }$f$, and $\mu \left( r\right) $%
\textit{\ denote the maximum term,   }$\mu \left( r\right) =\left\vert a_{\nu
\left( r\right) }\right\vert r^{\nu \left( r\right) }.$\textit{\ Then}

\begin{equation*}
\nu _{f}\left( r\right) =r\frac{d}{dr}\log \mu \left( r\right) <\left[ \log
\mu \left( r\right) \right] ^{2}\leq \left[ \log M\left( r,f\right) \right]
^{2}
\end{equation*}%
\textit{holds outside a set }$E_{2}\subset \left( 1,+\infty \right) $\textit{%
\ that has finite logarithmic measure.}
\end{lemma}


\begin{lemma}[see \cite{HAM}]Let $P\left( z\right) =\left( \alpha +i\beta \right) z^{n}+\ldots$%
\textit{(}$\alpha $\textit{, }$\beta $\textit{\ are real numbers, }$%
\left\vert \alpha \right\vert +\left\vert \beta \right\vert \neq 0$\textit{%
)\ be a polynomial with degree }$n\geq 1$\textit{\ and }$A\left( z\right) $%
\textit{\   a meromorphic function with }$\sigma \left( A\right) <n.$%
\textit{\ Set }$f\left( z\right) =A\left( z\right) e^{P\left( z\right) },$%
\textit{\ }$z=re^{i\theta },$\textit{\ }$\delta \left( P,\theta \right)
=\alpha \cos (n\theta) -\beta \sin (n\theta) .$\textit{\ Then for any given }$%
\varepsilon >0$\textit{, there exists a set }$E_{3}\subset \lbrack 1,+\infty
)$\textit{\ having finite logarithmic measure such that for any }$\theta \in %
\left[ 0,2\pi \right) \setminus H_1$\textit{\ and for }$|z|=r\notin \left[ 0,1%
\right] \cup E_{3},$\textit{\ }$r\rightarrow +\infty $,\textit{\ we have}
\begin{itemize}
\item[\textbf{(i)}] if $\delta \left( P,\theta \right) >0$\textit{, then}

\begin{equation*}
\exp \left\{ \left( 1-\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} \leq \left\vert f\left( re^{i\theta }\right) \right\vert \leq
\exp \left\{ \left( 1+\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} ,
\end{equation*}

\item[\textbf{(ii)}] if $\delta \left( P,\theta \right) <0$\textit{, then}

\begin{equation*}
\exp \left\{ \left( 1+\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} \leq \left\vert f\left( re^{i\theta }\right) \right\vert \leq
\exp \left\{ \left( 1-\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} ,
\end{equation*}
\end{itemize}
\noindent where $H_1=\left\{ \theta \in \left[ 0,2\pi \right) :\delta \left( P,\theta
\right) =0\right\} .$
\end{lemma}


\begin{lemma}[see \cite{LAI}]
\textit{Let }$f\left( z\right) $\textit{\ be a transcendental entire
function, and let }$\nu _{f}(r)$\textit{\ be the central index of }$f$%
\textit{\ and }$\delta $\textit{\   a constant such that }$0<\delta <\frac{1%
}{4}$. Then there exists a set $E_{4}$\textit{\ of finite logarithmic
measure such that for all }$z$\textit{\ satisfying }$\left\vert z\right\vert
=r\notin E_{4}$\textit{\ and }$\left\vert f\left( z\right) \right\vert \geq
M\left( r,f\right) \nu _{f}(r)^{-\frac{1}{4}+\delta }$\textit{, we have}

\begin{equation*}
\frac{f^{\left( n\right) }(z)}{f(z)}=\left( \frac{\nu _{f}\left( r\right) }{z%
}\right) ^{n}\left( 1+o\left( 1\right) \right) \text{ }\left( n\geq 1\text{
\textit{is an integer}}\right).
\end{equation*}
\end{lemma}
\begin{lemma}[see \cite{ZHA}]Let $f\left( z\right) $\textit{\ be an entire function and }$%
M\left( r,f\right) =\left\vert f\left( re^{i\theta _{r}}\right) \right\vert $%
\textit{\ for every }$r.$\textit{\ Set }$\theta _{r}\rightarrow \theta
_{0}\in \left[ 0,2\pi \right) $\textit{\ as }$r\rightarrow +\infty .$\textit{%
\ Then there exist a constant }$l_{0}>0$\textit{\ and a set }$E$\textit{\ of
positive lower logarithmic density such that }

\begin{equation}
M\left( r,f\right) ^{1/5}\leq \left\vert f\left( re^{i\theta }\right)
\right\vert  \label{6}
\end{equation}%
f\textit{or all }$r\in E$\textit{\ large enough and all }$\theta $\textit{\
such that }$\left\vert \theta -\theta _{0}\right\vert <l_{0}.$
\end{lemma}


\begin{lemma}[see \cite{CHE2}]Let $f\left( z\right) $\textit{\ be an entire function of infinite
order and }$\sigma _{2}\left( f\right) =\alpha <+\infty ,$\textit{\ and let
a set }$E_{5}\subset \left( 1,+\infty \right) $\textit{\ that has finite
logarithmic measure. Then there exists a sequence of points }$\left\{
z_{m}=r_{m}e^{i\theta _{m}}\right\} $\textit{\ such that }$\left\vert
f\left( z_{m}\right) \right\vert =M\left( r_{m},f\right) ,$\textit{\ }$%
\theta _{m}\in \left[ 0,2\pi \right) ,\underset{m\rightarrow +\infty }{\lim }%
\theta _{m}=\theta _{0}\in \left[ 0,2\pi \right) ,$\textit{\ }$r_{m}\notin
E_{5},$\textit{\ }$r_{m}\rightarrow +\infty $\textit{, }
\begin{equation*}
\lim_{r_{m}\rightarrow +\infty } \frac{\log \nu _{f}\left(
r_{m}\right) }{\log r_{m}}=+\infty % \nolabel
\end{equation*}
\textit{and for any given }$\varepsilon >0$\textit{, we have for a
sufficiently large }$r_{m}$

\begin{equation*}
\exp \left\{ r_{m}^{\alpha -\varepsilon }\right\} <\nu _{f}\left(
r_{m}\right) <\exp \left\{ r_{m}^{\alpha +\varepsilon }\right\},
\end{equation*}%
where $\nu _{f}\left( r\right) $\textit{\ is the central index of }$f.$
\end{lemma}

\section{Proof of Theorem 7}
\begin{proof}
Assume that $f$ is a nontrivial solution of equation $\left( 4\right) $.
 Then by Theorem 5, we have $\sigma(f)=+\infty$ and $\sigma _{2}(f)=\lambda_2(f)=\overline{\lambda }_{2}(f)=\overline{\lambda }_{2}(f-\varphi )\leq n$. We assert that $\sigma _{2}(f)=n$. Now we assume that $\sigma _{2}(f)=\alpha <n$ and we prove that $\sigma
_{2}(f)=\alpha $ fails. By Lemma 2, there exist a constant $B>0$ and a set $%
E_{1}\subset \lbrack 1,+\infty )$ having finite logarithmic measure such
that for all $z$ satisfying $|z|=r\notin \lbrack 0,1]\cup E_{1}$, we have%
\begin{equation}
\left\vert \frac{f^{(j)}(z)}{f(z)}\right\vert \leq Br\left[ T(2r,f)\right]
^{j+1}(j=1,\ldots,k)  \label{7}
\end{equation}%
Let $\nu _{f}\left( r\right) $ denote  the central index of $f.$ By Lemma 3, there is a set $E_{2}\subset \left( 1,+\infty \right) $\ that has finite
logarithmic measure such that for $\left\vert z\right\vert =r\notin \left[
0,1\right] \cup E_{2},$ we have

\begin{equation}
\nu _{f}\left( r\right) <\left[ \log M\left( r,f\right) \right] ^{2}.
\label{8}
\end{equation}%
By Lemma 4,\ for any given $\varepsilon >0$, there exists a set $%
E_{3}\subset \lbrack 1,+\infty )$\ having finite logarithmic measure such
that for any $\theta \in \left[ 0,2\pi \right) \setminus H_2,$ where
\[
H_2=\left\{ \theta \in \left[ 0,2\pi \right) :\delta \left( P,\theta \right) =0%
\text{ or }\delta \left( Q-P,\theta \right) =0\text{ or }\delta \left(
Q,\theta \right) =0\right\} \] and for $|z|=r\notin \left[ 0,1\right] \cup
E_{3},$\ $r\rightarrow +\infty $,\ we have
\begin{itemize}
\item[-] if $\delta \left( P,\theta \right) >0$, then
\begin{equation}
\exp \left\{ \left( 1-\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} \leq \left\vert A_{1}\left( z\right) e^{P\left( z\right)
}\right\vert \leq \exp \left\{ \left( 1+\varepsilon \right) \delta \left(
P,\theta \right) r^{n}\right\} ,  \label{9}
\end{equation}

\item[-] if $\delta \left( P,\theta \right) <0$, then
\begin{equation}
\exp \left\{ \left( 1+\varepsilon \right) \delta \left( P,\theta \right)
r^{n}\right\} \leq \left\vert A_{1}\left( z\right) e^{P\left( z\right)
}\right\vert \leq \exp \left\{ \left( 1-\varepsilon \right) \delta \left(
P,\theta \right) r^{n}\right\} ,  \label{10}
\end{equation}

\item[-] if $\delta \left( Q-P,\theta \right) >0$, then
\begin{align}
\begin{split}
\exp \left\{ \left( 1-\varepsilon \right) \delta \left( Q-P,\theta \right)
r^{n}\right\} &\leq \left\vert \frac{A_{0}\left( z\right) }{A_{1}\left(
z\right) }e^{Q\left( z\right) -P\left( z\right) }\right\vert \\
&\leq \exp
\left\{ \left( 1+\varepsilon \right) \delta \left( Q-P,\theta \right)
r^{n}\right\} , % \label{11}
\end{split}\end{align}

\item[-] if $\delta \left( Q-P,\theta \right) <0$, then
\begin{align}\begin{split}
\exp \left\{ \left( 1+\varepsilon \right) \delta \left( Q-P,\theta \right)
r^{n}\right\}& \leq \left\vert \frac{A_{0}\left( z\right) }{A_{1}\left(
z\right) }e^{Q\left( z\right) -P\left( z\right) }\right\vert\\
& \leq \exp
\left\{ \left( 1-\varepsilon \right) \delta \left( Q-P,\theta \right)
r^{n}\right\} ,  %\label{12}
\end{split}
\end{align}
\item[-] if $\delta \left( Q,\theta \right) >0$, then

\begin{equation}
\exp \left\{ \left( 1-\varepsilon \right) \delta \left( Q,\theta \right)
r^{n}\right\} \leq \left\vert A_{0}\left( z\right) e^{Q\left( z\right)
}\right\vert \leq \exp \left\{ \left( 1+\varepsilon \right) \delta \left(
Q,\theta \right) r^{n}\right\} ,  \label{13}
\end{equation}

\item[-] if $\delta \left( Q,\theta \right) <0$, then

\begin{equation}
\exp \left\{ \left( 1+\varepsilon \right) \delta \left( Q,\theta \right)
r^{n}\right\} \leq \left\vert A_{0}\left( z\right) e^{Q\left( z\right)
}\right\vert \leq \exp \left\{ \left( 1-\varepsilon \right) \delta \left(
Q,\theta \right) r^{n}\right\} .  \label{14}
\end{equation}%
\end{itemize}
By Lemma 5, for any given constant $0<\delta <\frac{1}{4}$\textit{,%
} there\ exists a set $E_{4}$\ of finite logarithmic measure such that for
all $z$\ satisfying $\left\vert z\right\vert =r\notin E_{4}$\ and $%
\left\vert f\left( z\right) \right\vert \geq M\left( r,f\right) \nu
_{f}(r)^{-\frac{1}{4}+\delta }$, we have

\begin{equation}
\frac{f^{(j)}(z)}{f(z)}=\left( \frac{\nu _{f}(r)}{z}\right) ^{j}(1+o(1))%
\text{ }\left( j=1,\ldots,k\right).  \label{15}
\end{equation}%
\noindent Since $m_l(E_{1}\cup E_{2}\cup E_{3}\cup E_{4})<+\infty $, then $m_l(E\setminus \left( \left[ 0,1\right] \cup
E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right) )$ is infinite, where $E$ is the
set defined in Lemma 6. Thus by Lemma 7, there exists a sequence of\textit{\
}points\textit{\ }$\left\{ z_{m}=r_{m}e^{i\theta _{m}}\right\} $\textit{\ }%
such that\textit{\ }$\left\vert f\left( z_{m}\right) \right\vert =M\left(
r_{m},f\right) ,$\textit{\ }$\theta _{m}\in \left[ 0,2\pi \right) ,\underset{%
m\rightarrow +\infty }{\lim }\theta _{m}=\theta _{0}\in \left[ 0,2\pi
\right) ,$\textit{\ }$r_{m}\in E\setminus \left( \left[ 0,1\right] \cup
E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right) ,$\textit{\ }$r_{m}\rightarrow
+\infty $\textit{, }%
\begin{equation}
\underset{r_{m}\rightarrow +\infty }{\lim }\frac{\log \nu _{f}\left(
r_{m}\right) }{\log r_{m}}=+\infty  \label{16}
\end{equation}%
and for any given $\varepsilon >0$, we have for a sufficiently large $r_{m}$

\begin{equation}
\exp \left\{ r_{m}^{\alpha -\varepsilon }\right\} <\nu _{f}\left(
r_{m}\right) <\exp \left\{ r_{m}^{\alpha +\varepsilon }\right\} .  \label{17}
\end{equation}%
By $\left( 16\right)$, for any sufficiently large $A>2\sigma \left(
H\right)$ and $m$ sufficiently large, we have

\begin{equation}
\nu _{f}\left( r_{m}\right) >r_{m}^{A}.  \label{18}
\end{equation}%
By $\left( 8\right) $ and $\left( 18\right),$  for $m$ sufficiently large we obtain

\begin{equation}
M\left( r_{m},f\right) >\exp \left\{ r_{m}^{A/2}\right\}.  \label{19}
\end{equation}%
On the other hand, for any given $\varepsilon $ $\left( 0<2\varepsilon
<A-2\sigma \left( H\right) \right)$ and $m$ sufficiently large, we have

\begin{equation}
\left\vert H\left( z_{m}\right) \right\vert \leq \exp \left\{ r_{m}^{\sigma
\left( H\right) +\varepsilon }\right\} .  \label{20}
\end{equation}%
From $\left( 19\right) $ and $\left( 20\right),$ it follows that

\begin{equation}
\frac{\left\vert H\left( z_{m}\right) \right\vert }{M\left( r_{m},f\right) }%
\rightarrow 0  \label{21}
\end{equation}%
as $r_{m}\rightarrow +\infty $.

\smallskip
\noindent For the above $\theta _{0}$, there are three cases: $\delta (P,\theta
_{0})>0$, $\delta (P,\theta _{0})<0$ and $\delta (P,\theta _{0})=0$.

\noindent \textbf{Case 1.} $\delta (P,\theta _{0})>0.$ From the continuity of
$\delta \left( P,\theta \right) $, we have

\begin{equation}
\frac{1}{2}\delta (P,\theta _{0})<\delta \left( P,\theta _{m}\right) <\frac{3%
}{2}\delta (P,\theta _{0})  \label{22}
\end{equation}%
for $m$ sufficiently large. For any given $\varepsilon $ $\left(
0<2\varepsilon <\min \left\{ 1,n-\alpha ,\text{ }A-2\sigma \left( H\right)
\right\} \right),$ from $\left( 9\right) $ and $\left( 22\right),$ we have
\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){
r_{m}^{n}}\right\}\leq \left\vert A_{1}\left( z_{m}\right) e^{P\left( z_{m}\right) }\right\vert\leq
\exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\delta (P,\theta _{0}){
r_{m}^{n}}\right\}  \label{23}
\end{equation}
for $m$ sufficiently large.

\smallskip

\noindent \textbf{Subcase 1.1. }We first assume that $\theta_0$ satisfies $\eta:=\delta (Q-P,\theta _{0})>0$. From
the continuity of $\delta \left( Q-P,\theta \right) $, we have

\begin{equation}
\frac{1}{2}\delta \left( Q-P,\theta _{0}\right) <\delta \left( Q-P,\theta
_{m}\right) <\frac{3}{2}\delta \left( Q-P,\theta _{0}\right) \text{.}
\label{24}
\end{equation}%
Hence by $\left( 11\right) $ and $\left( 24\right),$ for the above $%
\varepsilon $, we have

\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\eta r_{m}^{n}\right\}\leq \left\vert\frac{A_{0}\left( z_{m}\right) }{A_{1}\left(
z_{m}\right) }e^{Q\left( z_{m}\right) -P\left( z_{m}\right) }\right\vert\leq \exp \left\{
\frac{3\left( 1+\varepsilon \right) }{2}\eta r_{m}^{n}\right\}
\label{25}
\end{equation}%
for $m$ sufficiently large.

\noindent From $\left( 4\right)$ we obtain
\begin{align}
\begin{split}
\left\vert \frac{A_{0}(z)}{A_{1}(z)}e^{Q\left( z\right) -P\left(
z\right) }\right\vert\leq & \left\vert \frac{f^{\prime }(z)}{f(z)}\right\vert+\frac{1}{
\left\vert A_{1}(z)e^{P\left( z\right) }\right\vert } \\
&\times\left(
\left\vert \frac{f^{\left( k\right) }(z)}{f(z)}\right\vert +\underset
{j=2}{\overset{k-1}{\sum }}\left\vert h_{j}\left( z\right)
 \frac{%
f^{\left( j\right) }(z)}{f(z)}\right\vert +\left\vert\frac{ H\left(
z\right) }{ f(z) }\right\vert\right) .  \label{26}
\end{split}\end{align}%
Substituting $\left( 15\right) $ into $\left( 26\right) $ and from $\left(
17\right) ,$ $\left( 21\right) ,$ $\left( 23\right) $ and $\left( 25\right) $,  for $m$ sufficiently large we have

\begin{align}
%\begin{split}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\eta r_{m}^{n}\right\}
\leq& \exp \left\{ r_{m}^{\alpha +\varepsilon }\right\} r_{m}^{-1}\left\vert
1+o(1)\right\vert +
\nonumber\\
&+\exp \left\{-\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){%
r_{m}^{n}}\right\}\left[ \exp \left\{ kr_{m}^{\alpha +\varepsilon }\right\}
r_{m}^{-k}\left\vert 1+o\left( 1\right) \right\vert \right]
\nonumber\\
&+\exp \left\{-\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){%
r_{m}^{n}}\right\}\nonumber\\&\times
\left[ M_{1}r_{m}^{d_{1}}\exp \left\{ \left( k-1\right)\right.
r_{m}^{\alpha +\varepsilon }\right\}\left. \left\vert 1+o(1)\right\vert +o\left(
1\right) \right] ,  \label{27}
%\end{split}
\end{align}%
where $M_{1}$ $\left( >0\right) $ is a constant and $d_{1}$ is an
entire number. This is a contradiction.

\smallskip

\noindent \textbf{Subcase 1.2. }$\eta:=\delta (Q-P,\theta _{0})<0$. From the continuity of $\delta (Q-P,\theta _{0})$ and $\left(
12\right),$  for $m$ sufficiently large we have
\begin{equation}
\exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\eta r_{m}^{n}\right\}\leq \left\vert\frac{A_{0}\left( z_{m}\right) }{A_{1}\left(
z_{m}\right) }e^{Q\left( z_{m}\right) -P\left( z_{m}\right) }\right\vert\leq \exp \left\{
\frac{\left( 1-\varepsilon \right) }{2}\eta r_{m}^{n}\right\}
\label{28}
\end{equation}%
From $\left( 4\right) $ we obtain

\begin{align}
\begin{split}
\left\vert \frac{f^{^{\prime }}(z)}{f(z)}\right\vert \leq&\left\vert
\frac{A_{0}(z)}{A_{1}(z)}e^{Q\left( z\right) -P\left(
z\right) }\right\vert
+\frac{1}{\left\vert A_{1}(z)e^{P\left( z\right) }\right\vert }%
\\
&\times\left( \left\vert \frac{f^{\left( k\right) }(z)}{f(z)}\right\vert +%
\underset{j=2}{\overset{k-1}{\sum }}\left\vert h_{j}\left( z\right)
\frac{f^{\left( j\right) }(z)}{f(z)}\right\vert +\left\vert\frac{ H\left(
z\right)}{ f(z)}\right\vert\right) .
\label{29}
\end{split}
\end{align}%
Substituting $\left( 15\right) $ into $\left( 29\right) $ and from $\left(
17\right) ,$ $\left( 21\right),$ $\left( 23\right) $ and $\left( 28\right)$,  for $m$ sufficiently large we have

\begin{align}
\left(\frac{\nu _{f}(r_{m})}{r_{m}}\right)\left\vert 1+o(1)\right\vert \leq& \exp \left\{
\frac{\left( 1-\varepsilon \right) }{2}\eta r_{m}^{n}\right\}
\nonumber\\
&+\exp \left\{-\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){%
r_{m}^{n}}\right\}\left[ \exp \left\{ kr_{m}^{\alpha +\varepsilon }\right\}
r_{m}^{-k}\left\vert 1+o\left( 1\right) \right\vert \right]
\nonumber\\
&+\exp \left\{-\frac{\left( 1-\varepsilon \right)}{2} \delta (P,\theta _{0})r_{m}^{n}\right\}\nonumber\\
&\times\left[
M_{2}r_{m}^{d_{2}}\exp \left\{ \left( k-1\right) r_{m}^{\alpha +\varepsilon
}\right\} \left\vert 1+o(1)\right\vert +o\left( 1\right) \right] ,  \label{30}
\end{align}%
where $M_{2}$ $\left( >0\right) $ is a constant and $d_{2}$ is an entire
number. This implies that $\nu _{f}(r_{m})\rightarrow 0$ as $m\rightarrow
+\infty $, which is impossible.

\smallskip

\noindent \textbf{Subcase 1.3. }$\eta:=\delta (Q-P,\theta _{0})=0$. Here $%
\left( 6\right) $ may be used to construct another sequence of points $%
\left\{ {z_{m}^{\ast }=r}_{m}e^{i\theta _{m}^{\ast }}\right\} $ with $%
\underset{m\rightarrow +\infty }{\lim }\theta _{m}^{\ast }=\theta _{0}^{\ast
}$\textit{\ }such that\textit{\ }$\eta_1:=\delta (Q-P,\theta _{0}^{\ast })>0.$
Indeed,  without loss of generality, we may suppose  that

\begin{align}
\delta (Q-P,\theta) &>0,\text{ }\theta \in \left( \frac{\theta _{0}+2k\pi}{n}
,\frac{\theta _{0}+\left( 2k+1\right) \pi}{n} \right) ,  %\nolabel
\nonumber\\
\delta (Q-P,\theta) &<0,\text{ }\theta \in \left( \frac{\theta _{0}+\left(
2k-1\right) \pi}{n} ,\frac{\theta _{0}+2k\pi}{n} \right) \label{31}
\end{align}%
with $k\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ When $m$ is large enough, we have $\left\vert \theta _{m}-\theta
_{0}\right\vert \leq l_{0},$ where $l_{0}$ is a small constant. Choose now $%
\theta _{m}^{\ast }$ such that $\frac{l_{0}}{2}\leq \theta _{m}^{\ast }-\theta
_{m}\leq l_{0}.$ Then $\theta _{0}+\frac{l_{0}}{2}\leq \theta _{0}^{\ast }\leq
\theta _{0}+l_{0}.$

\noindent For  $m$ sufficiently large, we have $\left( 6\right) $ for ${%
z_{m}^{\ast }}$ and $\delta (Q-P,\theta _{0}^{\ast })>0.$ Therefore

\begin{equation}
\left\vert \frac{H\left( {z_{m}^{\ast }}\right) }{f\left( {z_{m}^{\ast }}%
\right) }\right\vert \leq \frac{\exp \left\{ r_{m}^{\sigma \left( H\right)
+\varepsilon }\right\} }{\left( M\left( r_{m},f\right) \right) ^{1/5}}%
\rightarrow 0,\text{ }m\rightarrow +\infty  \label{32}
\end{equation}%
and

\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\eta_1 r_{m}^{n}\right\}\leq \left\vert\frac{A_{0}\left( z_{m}^{\ast }\right) }{A_{1}\left(
z_{m}^{\ast }\right) }e^{Q\left( z_{m}^{\ast }\right) -P\left( z_{m}^{\ast
}\right) }\right\vert\leq \exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\eta_1 r_{m}^{n}\right\}  \label{33}
\end{equation}%
for  $m$  sufficiently large. Taking now $l_{0}$ small enough, we have $\delta
(P,\theta _{0}^{\ast })>0$ by the continuity of $\delta (P,\theta )$. Hence

\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}^{\ast }){%
r_{m}^{n}}\right\}\leq \left\vert A_{1}\left( z_{m}^{\ast }\right) e^{P\left( z_{m}^{\ast
}\right) }\right\vert\leq \exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\delta
(P,\theta _{0}^{\ast }){r_{m}^{n}}\right\}  \label{34}
\end{equation}%
for $m$ sufficiently large. By $\left( 7\right) $, $\left( 26\right) $, $\left( 32\right) $-$\left( 34\right) $, we obtain
\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\eta_1 r_{m}^{n}\right\}\leq M_3 r_m^{d_3}[T(2r,f)]^{k+1}, \label{35}
\end{equation}
where $M_{3}$ $\left( >0\right) $ is a constant and $d_{3}$ is an entire
number. Thus $\sigma_2(f)\geq n$ and  this contradicts $\sigma_2(f)<n$.

\smallskip
\noindent \textbf{Case 2.} $\delta (P,\theta _{0})<0.$ From the continuity
of $\delta \left( P,\theta \right) $ and $\left( 10\right)$. For any given $%
\varepsilon $ $\left( 0<2\varepsilon <\min \left\{ 1,\text{ }n-\alpha,\text{
}A-2\sigma \left( H\right) \right\} \right),$ we have

\begin{equation}
\exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\delta (P,\theta _{0}){%
r_{m}^{n}}\right\}\leq \left\vert A_{1}\left( z_{m}\right) e^{P\left( z_{m}\right) }\right\vert\leq
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){%
r_{m}^{n}}\right\}  \label{36}
\end{equation}%
for $m$ sufficiently large.

\smallskip
\noindent \textbf{Subcase 2.1. }$\delta (Q,\theta _{0})>0$. From
the continuity of $\delta \left( Q,\theta \right) $ and $\left( 13\right),$ for the above $\varepsilon $ and $m$ sufficiently large, we have

\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (Q,\theta _{0}){%
r_{m}^{n}}\right\}\leq \left\vert A_{0}\left( z_{m}\right) e^{Q\left( z_{m}\right) }\right\vert\leq
\exp \left\{\frac{3\left( 1+\varepsilon \right) }{2}\delta (Q,\theta _{0}){%
r_{m}^{n}}\right\}.  \label{37}
\end{equation}%
From $\left( 4\right) $ we obtain

\begin{align}\begin{split}
|A_{0}\left( z\right) e^{Q\left( z\right) }|\leq& \left\vert \frac{%
f^{\left( k\right) }(z)}{f(z)}\right\vert +\underset{j=2}{\overset{%
k-1}{\sum }}\left\vert h_{j}\left( z\right) \frac{f^{\left( j\right)
}(z)}{f(z)}\right\vert
\\
\hspace{3cm}&+\left\vert A_{1}(z)e^{P\left( z\right) }\right\vert \left\vert
\frac{f^{^{\prime }}(z)}{f(z)}\right\vert +\left\vert\frac{ H\left(
z\right) }{ f(z) }\right\vert.  \label{38}
\end{split}
\end{align}%
Substituting $\left( 15\right) $ into $\left( 38\right) $ and from $\left(
17\right),$$\left(21\right),$$\left(36\right) $ and $\left(37\right)$, for $m$ sufficiently large we have

\begin{align}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (Q,\theta _{0}){
r_{m}^{n}}\right\}\leq&\exp \left\{ kr_{m}^{\alpha +\varepsilon }\right\}
r_{m}^{-k}\left\vert 1+o\left( 1\right) \right\vert
\nonumber\\
&+M_{4}r_{m}^{d_{4}}\exp \left\{ \left( k-1\right) r_{m}^{\alpha +\varepsilon
}\right\} \left\vert 1+o(1)\right\vert
\nonumber\\
&+\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta _{0}){
r_{m}^{n}}\right\}r_{m}^{-1}\exp \left\{ r_{m}^{\alpha +\varepsilon }\right\}\nonumber\\
&\times\left\vert 1+o(1)\right\vert +o(1),  \label{39}
\end{align}%
where $M_{4}$ $\left( >0\right) $ is a constant and $d_{4}$ is an entire
number. This is a contradiction.

\smallskip

\noindent \textbf{Subcase 2.2. }$\delta (Q,\theta _{0})<0$. From
the continuity of $\delta \left( Q,\theta \right) $ and $\left( 14\right)$ for the above $\varepsilon $ and $m$ sufficiently large, we have
\begin{equation}
\exp \left\{ \frac{3\left( 1+\varepsilon \right) }{2}\delta \left( Q,\theta_0
\right) r_{m}^{n}\right\} \leq \left\vert A_{0}\left( z_m\right) e^{Q\left(
z_m\right) }\right\vert \leq \exp \left\{ \frac{\left( 1-\varepsilon \right) }{%
2}\delta \left( Q,\theta_0 \right) r_{m}^{n}\right\} .  \label{40}
\end{equation}%
From $\left( 4\right)$ we obtain%
\begin{align}
\begin{split}
\left\vert \frac{f^{\left( k\right) }(z)}{f(z)}\right\vert \leq&
\underset{j=2}{\overset{k-1}{\sum }}\left\vert h_{j}\left( z\right)
\frac{f^{\left( j\right) }(z)}{f(z)}\right\vert +\left\vert
A_{1}(z)e^{P\left( z\right) }\right\vert \left\vert \frac{f^{\prime
}(z)}{f(z)}\right\vert \\
&+\left\vert A_{0}(z)e^{Q\left( z\right)
}\right\vert +\left\vert\frac{ H\left(
z\right) }{ f(z) }\right\vert.  \label{41}
\end{split}
\end{align}%
Substituting $\left( 15\right) $ into $\left( 41\right) $ and from $\left(
17\right),$$\left( 21\right),$$\left( 36\right) $ and $\left( 40\right) $, for $m$ sufficiently large  we have
\begin{align}\nonumber
%\begin{split}
\left( \nu _{f}(r_{m})\right) ^{k}r_{m}^{-k}\left\vert 1+o(1)\right\vert
\leq& M_{5}r_{m}^{d_{5}}\left( \nu _{f}(r_{m})\right) ^{k-1}\left\vert
1+o(1)\right\vert
\\\nonumber
&+\exp \left\{ \frac{\left( 1-\varepsilon \right) }{2}\delta (P,\theta
_{0})r_{m}^{n}\right\} \exp \left\{ r_{m}^{\alpha +\varepsilon }\right\}
r_{m}^{-1}\left\vert 1+o(1)\right\vert \\
&+
\exp \left\{ \frac{\left(
1-\varepsilon \right) }{2}\delta (Q,\theta _{0})r_{m}^{n}\right\} +o\left(
1\right) ,  \label{42}
%\end{split}
\end{align}%
where $M_{5}$ $\left( >0\right) $ is a constant and $d_{5}$ is an entire
number. This is a contradiction.

\smallskip

\noindent \textbf{Subcase 2.3. }$\delta (Q,\theta _{0})=0$. By
using a similar argument as in Subcase 1.3, we may again construct another
sequence of points $\left\{ {z_{m}^{\ast }=r}_{m}e^{i\theta _{m}^{\ast
}}\right\} $ satisfying $\frac{l_{0}}{2}\leq \theta _{m}^{\ast }-\theta _{m}\leq
l_{0}$ with $\underset{m\rightarrow +\infty }{\lim }\theta _{m}^{\ast
}=\theta _{0}^{\ast }$\textit{\ }such that\textit{\ }$\delta (P,\theta
_{0}^{\ast })<0<\delta (Q,\theta _{0}^{\ast }).$ Replace $\delta (P,\theta
_{0})$ with $\delta (P,\theta _{0}^{\ast })$ in $\left( 36\right) $ and $
\delta (Q,\theta _{0})$ with $\delta (Q,\theta _{0}^{\ast })$ in $\left(
37\right) $. Using the same argument as in Subcase 1.3, we also have $\left(32\right)$ for
the sequence of points $z_{m}^{\ast }$. From $\left(38\right)$ and  $m$ sufficiently large, we have
\begin{equation}
\exp \left\{\frac{\left( 1-\varepsilon \right) }{2}\delta(Q,\theta_{0}^{\ast }) r_{m}^{n}\right\}\leq M_6 r_m^{d_6}[T(2r,f)]^{k+1}, \label{43}
\end{equation}
where $M_{6}$ $\left( >0\right) $ is a constant and $d_{6}$ is an entire
number. Thus $\sigma_2(f)\geq n$ and  this contradicts $\sigma_2(f)<n$.\\

\smallskip
\noindent \textbf{Case 3.} $\delta (P,\theta _{0})=0$. We discuss three
cases according to $\delta (Q,\theta _{0})$ as follows:

\smallskip

\noindent \textbf{Subcase 3.1. }$\delta (Q,\theta _{0})>0.$ By an argument
similar to   Subcase 1.3, we can choose another sequence of points $%
\left\{ {z_{m}^{\ast }=r}_{m}e^{i\theta _{m}^{\ast }}\right\} $ satisfying $\frac{l_{0}}{2}\leq \theta _{m}^{\ast }-\theta _{m}\leq l_{0}$ with $\underset{%
m\rightarrow +\infty }{\lim }\theta _{m}^{\ast }=\theta _{0}^{\ast },$%
\textit{\ }such that $z_{m}^{\ast }$ satisfies $\left( 32\right)$ and $\delta (P,\theta _{0}^{\ast })<0<\delta
(Q,\theta _{0}^{\ast }).$ Similarly to   Subcase 2.3, a contradiction
follows as $m$ is large enough.

\smallskip

\noindent \textbf{Subcase 3.2. }$\delta (Q,\theta _{0})<0.$ By the
definition of $\delta (P,\theta )$ in Lemma 4, we may define

\begin{equation*}
\delta ^{\prime }(P,\theta)=-n\alpha \sin \left( n\theta \right)
-n\beta \cos \left( n\theta \right),
\end{equation*}%
where $a_{n}=\alpha +i\beta.$ Since $a_{n}\neq 0,$ we have $\delta ^{\prime
}(P,\theta _{0})\neq 0$. Take ${z_{m}^{^{\prime }}=r}_{m}e^{i\theta
_{m}^{^{\prime }}}$ satisfying $0<\left\vert \theta _{m}^{^{\prime }}-\theta
_{0}\right\vert \leq l_{0}$, we know that ${z_{m}^{^{\prime }}}$ satisfies $%
\left( 32\right) $ and $\delta (P,\theta _{m}^{\prime })\neq 0.$ By the
continuity of $\delta (Q,\theta )$, we may assume that $\delta (Q,\theta
_{m}^{\prime })<0<\delta (P,\theta _{m}^{\prime })$ for a suitable $l_{0},$ $%
0<\theta _{m}^{^{\prime }}-\theta _{0}\leq l_{0}$. Then $\delta ^{\prime
}(P,\theta _{0})>0,$ which means that for a suitable $l_{0},$

\begin{equation}
\frac{1}{2}\delta ^{\prime }(P,\theta _{0})<\delta ^{\prime }\left( P,\theta
\right) <\frac{3}{2}\delta ^{\prime }(P,\theta _{0}),\text{ }\theta \in
\left( \theta _{0},\theta _{0}+l_{0}\right).  \label{44}
\end{equation}%
Since we have chosen $z_{m}$ such that $|f(z_{m})|=M(r_{m},f)$ and $\theta
_{m}\rightarrow \theta _{0}$ as $m\rightarrow \infty $, we have $%
|f(r_{m}e^{i\theta _{0}})|\geq M(r_{m},f)v_{f}(r_{m})^{-\frac{1}{4}+\delta }$
for $m$ sufficiently large. From $\left( 4\right) $, we have
\begin{align}
\begin{split}
\left\vert\frac{f^{^{\prime }}(z)}{f(z)}\right\vert\leq &\left\vert\frac{1}{
A_{1}(z)e^{P(z)}}\right\vert\left( |\frac{f^{\left(
k\right) }(z)}{f(z)}|+
\sum_{j=2}^{k-1}|h_{j}(z)||\frac{f^{\left( j\right)
}(z)}{f(z)}|\right)
\\
&+\left\vert\frac{1}{A_{1}(z)e^{P(z)}}\right\vert\left(|A_{0}(z)e^{Q(z)}|+\left\vert\frac{H(z)}{f(z)}
\right\vert\right).  \label{45}
\end{split}\end{align}%
By $\left( 9\right) $ and $\left( 14\right),$ for the above $%
\varepsilon $ and for $m$ sufficiently large  we have

\begin{equation}
\exp \{\left( 1+\varepsilon \right) \delta (Q,\theta _{m}^{\prime
})r_{m}^{n}\}\leq |A_{0}\left( z_{m}^{\prime }\right) e^{Q(z_{m}^{\prime
})}|\leq \exp \{\left( 1-\varepsilon \right) \delta (Q,\theta
_{m}^{\prime})r_{m}^{n}\}  \label{46}
\end{equation}%
and
\begin{equation}
\exp \{-\left( 1+\varepsilon \right) \delta (P,\theta _{m}^{\prime
})r_{m}^{n}\}\leq \left\vert\frac{e^{-P(z_{m}^{\prime })}}{A_{1}\left( z_{m}^{\prime
}\right) }\right\vert\leq \exp \{-\left( 1-\varepsilon \right) \delta (P,\theta
_{m}^{\prime })r_{m}^{n}\}  \label{47}
\end{equation}%
for  $m$ sufficiently large. From the definition of the hyper-order, it
follows that
\begin{equation}
T(2r_{m},f)\leq \exp \{\left( 2r_{m}\right) ^{\alpha +\varepsilon }\} \label{48}
\end{equation}%
for  $m$ sufficiently large. By $\left( 7\right) $, $\left( 32\right),$ $%
\left( 45\right) -\left( 48\right) $,  for $m$ sufficiently large we can get

\begin{equation}
\left\vert\frac{f^{^{\prime }}(z'_{m})}{f(z'_{m})}\right\vert\leq \exp \{-(1-2\varepsilon )\delta
(P,\theta _{m}^{\prime }){r_{m}}^{n}\}  \label{49}
\end{equation}%
Since $\theta _{m}^{\prime }$ is arbitrary in $(\theta _{0},\theta
_{0}+l_{0})$, for  $m$ sufficiently large, we can obtain
\begin{equation}
\left\vert\frac{f^{^{\prime }}(r_{m}e^{i\theta })}{f(r_{m}e^{i\theta })}\right\vert\leq \exp
\{-(1-2\varepsilon )\delta (P,\theta ){r_{m}^{n}}\},\text{ }\theta \in
(\theta _{0},\theta _{0}+l_{0})  \label{50}
\end{equation}%
Therefore, for $\theta \in (\theta _{0},\theta _{0}+l_{0})$, we have
\begin{align}
\xi (r_{m},\theta )=&r_{m}\int_{\theta _{0}}^{\theta }\left\vert\frac{f^{^{\prime
}}(r_{m}e^{i\theta })}{f(r_{m}e^{i\theta })}\right\vert d\theta \leq r_{m}\int_{\theta
_{0}}^{\theta }e^{-\eta _{2}(\theta ){r_{m}^{n}}}d\theta\nonumber
\\
&=
\int_{\theta_{0}}^{\theta }\frac{1}{\eta _{1}(\theta ){r_{m}}^{n-1}}e^{-\eta
_{2}(\theta ){r_{m}^{n}}}d(\eta _{2}(\theta ){r_{m}^{n}}),  \label{51}
\end{align}%
where $\eta _{1}(\theta )=(1-2\varepsilon )\delta ^{\prime }(P,\theta )$ and
$\eta _{2}(\theta )=(1-2\varepsilon )\delta (P,\theta )$.

\noindent Since $\delta (P,\theta )>0$ for all $\theta \in (\theta
_{0},\theta _{0}+l_{0})$, we can get
\begin{equation*}
0\leq \xi (r_{m},\theta )\leq \frac{2}{(1-2\varepsilon )\delta ^{\prime
}(P,\theta _{0})r_{m}^{n-1}}(e^{-\eta _{2}(\theta _{0}){r_{m}^{n}}}-e^{-\eta
_{2}(\theta ){r_{m}^{n}}}).
\end{equation*}%
Thus for  $m$ sufficiently large, we can get
\begin{equation}
0\leq \xi (r_{m},\theta )\leq \frac{2}{\eta _{1}(\theta _{0})}.  \label{52}
\end{equation}%
By the proof of Lemma 2.4 in \cite{WAN}, we have

\begin{equation*}
\log |f(r_{m}e^{i\theta _{0}})|-\xi (r_{m},\theta )\leq \log
|f(r_{m}e^{i\theta })|+2\pi .
\end{equation*}%
From this and $\left( 52\right) $ it follows  that
\begin{equation}
\nu _{f}(r_{m})^{-\frac{1}{4}+\delta ^{\prime }}M(r_{m},f)=\exp \{-2\pi
-2/\eta _{1}(\theta _{0})\}\nu _{f}(r_{m})^{-\frac{1}{4}+\delta
}M(r_{m},f)\leq |f(r_{m}e^{i\theta })|  \label{53}
\end{equation}%
for $\theta \in (\theta _{0},\theta _{0}+l_{0})$, where $0<\delta ^{\prime
}<\delta <\frac{1}{4}$. Therefore, we choose another sequence of points $%
z_{m}^{\ast }=r_{m}e^{i\theta _{m}^{\ast }}$ satisfying $\theta _{m}^{\ast }=%
\frac{l_{0}}{2}+\theta _{0}$ and $\left( 32\right) $ for $z_{m}^{\ast }$. Furthermore, from $\left( 53\right)$, we have $\left( 15\right)$ for $%
z_{m}^{\ast }$ when $m$ is sufficiently large. Thus, from $\left( 15\right) $
and $\left( 50\right) $, we can deduce that $\nu _{f}(r_{m})\rightarrow0$ as $m\rightarrow \infty $, which is impossible.

\noindent When $\delta (Q,\theta _{m}^{\prime })<0<\delta (P,\theta
_{m}^{\prime })$ for $-l_{0}<\theta _{m}^{\prime }-\theta _{0}<0$. Clearly $%
\xi (r_{m},\theta )\leq 0$ for all $\theta \in (\theta _{0}-l_{0},\theta
_{0})$. Similarly, we can get
\begin{equation}
\nu _{f}(r_{m})^{-\frac{1}{4}+\delta ^{\prime }}M(r_{m},f)=\exp \{-2\pi
\}\nu _{f}(r_{m})^{-\frac{1}{4}+\delta }M(r_{m},f)\leq |f(r_{m}e^{i\theta })|
\label{54}
\end{equation}%
for $\theta \in (\theta _{0}-l_{0},\theta _{0})$, where $0<\delta ^{\prime
}<\delta <\frac{1}{4}$. Thus we can also get a contradiction.

\smallskip
\noindent \textbf{Subcase 3.3.} Finally, suppose that $\delta (Q,\theta
_{0})=0$. We now have $a_{n}=cb_{n}$, $c\in \mathbb{R}\setminus \{0,1\}$. Then we have $P(z)=cb_{n}z^{n}+a_{n-1}z^{n-1}+\ldots+a_{0}$, $%
Q(z)-P(z)=(1-c)b_{n}z^{n}+R_{n-1}(z)$, where $R_{n-1}(z)$ is a polynomial of
degree at most $n-1$.

\smallskip \noindent
If $c<0$, we may take $l_0$ small enough such that $\delta (Q,\theta
)<0<\delta (P,\theta )$, provided that either $\theta \in (\theta
_{0},\theta _{0}+l_{0})$ or $(\theta _{0}-l_{0},\theta _{0})$. By an argument similar to that in Subcase 3.2, we can get a
contradiction.

\smallskip \noindent
If $0<c<1$, we similarly obtain $\delta (Q-P,\theta )>0$ and $\delta
(P,\theta )>0$, provided that either $\theta \in (\theta _{0},\theta
_{0}+l_{0})$ or $(\theta _{0}-l_{0},\theta _{0})$ for $l_{0}$ small enough.
By an argument similar to that in Subcase 1.3, a contradiction follows.

\smallskip \noindent
Finally, if $c>1$, we obtain $\delta (Q-P,\theta )<0$ and $\delta (P,\theta
)>0$ for either $\theta \in (\theta _{0},\theta _{0}+l_{0})$ or $(\theta
_{0}-l_{0},\theta _{0})$. Furthermore, $z_{m}^{\prime }=r_{m}e^{i\theta _{m}^{\prime }}$ satisfies $\left( 32\right)
$ provided that either $\theta _{m}^{\prime }\in (\theta _{0},\theta
_{0}+l_{0})$ or $(\theta _{0}-l_{0},\theta _{0})$. Similarly to Subcase 3.2, we get $\left( 50\right) $ and $\left(
53\right) $. By a standard Wiman-Valiron argument, a contradiction also
follows. Therefore  from these three cases it results that $\sigma _{2}(f)=n$.
\end{proof}
\section{Proof of Theorem 8}
\begin{proof}
Assume that $f(z)$ is a non-trivial solution of $\left( 5\right) $. We know
that $\sigma (f)=+\infty $. By Lemma 1, it follows that $\sigma _{2}(f)\leq
n $. Set $\sigma _{2}(f)=\alpha $ and we assert that $\alpha =n$. Now we
assume that $\alpha <n.$
Since $\rho =\max \{\rho (D_{j}):j=0,1\}<n$, then for any $\varepsilon $ $%
\left( 0<2\varepsilon <n-\rho \right) $, we have
\begin{equation}
|D_{j}(z)|\leq \exp \{r^{\rho +\varepsilon }\}\text{ }\left( j=0,1\right).
\label{55}
\end{equation}%
Similarly to the proof of Theorem 7, we can take a sequence of points $%
z_{m}=r_{m}e^{i\theta _{m}}$, $r_{m}\rightarrow \infty $, such that $%
\underset{m\rightarrow +\infty }{\lim }\theta _{m}=\theta _{0}$ and $%
|f(z_{m})|=M(r_{m},f),$ $r_{m}\in E\setminus (\left[ 0,1\right] \cup
E_{1}\cup E_{2}\cup E_{3}\cup E_{4})$ and the sequence of points satisfies $%
\left( 15\right) -\left( 17\right) $ and $\left( 21\right).$

\smallskip \noindent
Since $a_{n}/b_{n}=c<0$, there are three cases to be discussed, according to
the signs of $\delta (P,\theta _{0})$ and $\delta (Q,\theta _{0})$.

\smallskip \noindent
\noindent \textbf{Case 1.} Suppose that $\delta (Q,\theta _{0})<0<\delta
(P,\theta _{0})$. By $\left( 9\right),$ $\left( 14\right) $ and the
continuity of $\delta (Q,\theta )$ and $\delta (P,\theta )$, for any given $%
\varepsilon $ $\left( 0<2\varepsilon <\min \left\{ 1,n-\alpha ,A-2\sigma \left( H\right),n-\rho \right\} \right)$ we have $\left(
23\right) $ and $\left( 40\right) $ for  $m$ sufficiently large. From $\left(
5\right) $, we have
\begin{align}
\begin{split}
\left\vert A_{1}(z)e^{P(z)}+D_{1}(z)\right\vert\left\vert\frac{f^{\prime }(z)}{f(z)}\right\vert\leq &\left\vert\frac{f^{(k)}(z)}{f(z)}\right\vert
+\sum_{j=2}^{k-1}|h_{j}(z)|\left\vert\frac{f^{(j)}(z)}{f(z)}\right\vert\\&
+\left\vert(A_{0}e^{Q(z)}+D_{0}(z))\right\vert
+\left\vert\frac{H(z)}{f(z)}\right\vert.  \label{56}
\end{split}\end{align}%
Combining $\left( 55\right) $ with $\left( 23\right) $ and $\left( 40\right)
$, we conclude
\begin{equation}
|A_{0}\left( z_{m}\right) e^{Q(z_{m})}+D_{0}(z_{m})|\leq \exp \{r^{\rho
+2\varepsilon }\}  \label{57}
\end{equation}%
and%
\begin{equation}
|A_{1}(z_{m})e^{P(z_{m})}+D_{1}(z_{m})|\geq \exp \left\{\frac{(1-2\varepsilon )}{2
}\delta (P,\theta _{0}){r_{m}^{n}}\right\}  \label{58}
\end{equation}%
for  $m$ large enough. Substituting $\left( 15\right) $, $\left( 21\right) $, $\left( 57\right) $ and
$\left( 58\right) $ into $\left( 56\right) $,
for $m$ sufficiently large we get
\begin{align}\nonumber
\left(\frac{\nu _{f}(r_{m})}{r_{m}}\right)\left\vert 1+o\left( 1\right) \right\vert \leq &
\exp \left\{\frac{-(1-2\varepsilon )}{2}\delta (P,\theta _{0}){r_{m}^{n}}\right\}\left[
\exp \{kr_{m}^{\alpha +\varepsilon }\}r_{m}^{-k}\left\vert 1+o\left(
1\right) \right\vert \right.
\\
&+\left. M_{7}r_{m}^{d_{7}}\exp \{\left( k-1\right) r_{m}^{\alpha
+\varepsilon }\}\left\vert 1+o\left( 1\right) \right\vert\nonumber\right.\\
& +\exp
\{r_{m}^{\rho +2\varepsilon }\}\left.+o\left( 1\right) \right] ,  \label{59}
\end{align}%
where $M_{7}$ $\left( >0\right) $ is a constant and $d_{7}$ is an entire
number. This implies that $\nu _{f}(r_{m})\rightarrow 0,$ $m\rightarrow
+\infty ,$ which is impossible.

\smallskip
\noindent \textbf{Case 2.} Suppose that $\delta (P,\theta _{0})<0<\delta
(Q,\theta _{0})$. By $\left( 10\right) ,$ $\left( 13\right) $ and the
continuity of $\delta (Q,\theta )$ and $\delta (P,\theta )$, for any given $%
\varepsilon $ $\left( 0<2\varepsilon <\min \left\{ 1,n-\alpha ,\text{ }%
A-2\sigma \left( H\right) ,n-\rho \right\} \right) ,$ we have $\left(
36\right) $ and $\left( 37\right) $ for  $m$ sufficiently large. From $\left(
5\right) $, we have
\begin{align}
\nonumber\left\vert A_{0}(z)e^{Q(z)}+D_{0}(z)\right\vert\leq &\left\vert\frac{f^{(k)}(z)}{f(z)}\right\vert+\sum_{j=2}^{k-1}|h_{j}(z)|\left\vert\frac{f^{(j)}(z)}{f(z)}\right\vert
\\&+
|(A_{1}(z)e^{P(z)}+D_{1}(z))|\left\vert\frac{f^{\prime }(z)}{f(z)
}\right\vert+\left\vert\frac{H(z)}{f(z)}\right\vert.  \label{60}
\end{align}%
Combining $\left( 55\right) $ with $\left( 36\right) $ and $\left( 37\right)
$, we conclude
\begin{equation}
|A_{1}\left( z_{m}\right) e^{P(z_{m})}+D_{1}(z_{m})|\leq \exp \{r^{\rho
+2\varepsilon }\}  \label{61}
\end{equation}%
and%
\begin{equation}
|A_{0}(z_{m})e^{Q(z_{m})}+D_{0}(z_{m})|\geq \exp \left\{\frac{(1-2\varepsilon )}{2%
}\delta (Q,\theta _{0}){r_{m}^{n}}\right\}  \label{62}
\end{equation}%
for  $m$ large enough. Substituting $\left( 15\right) $, $\left( 21\right) $,$\left( 61\right) $ and
$\left( 62\right) $ into $\left( 60\right) $,
for $m$ sufficiently large we get%
\begin{equation}
\exp \left\{\frac{(1-2\varepsilon )}{2}\delta (Q,\theta _{0}){r}_{m}^{n}\right\}\leq
M_{8}r_{m}^{d_{8}}\exp \{kr_{m}^{\alpha +\varepsilon }\}\exp \{r_{m}^{\rho
+2\varepsilon }\},  \label{63}
\end{equation}%
where $M_{8}$ $\left( >0\right) $ is a constant and $d_{8}$ is an entire
number. This is a contradiction.

\smallskip

\noindent \textbf{Case 3.} Suppose that $\delta (Q,\theta _{0})=0=\delta
(P,\theta _{0})$. Similarly to  Subcase 1.3 of the proof of Theorem 7, we may
again construct another sequence of points $z_{m}^{\ast }=r_{m}e^{i\theta
_{m}^{\ast }}$ with $\underset{m\rightarrow +\infty }{\lim }\theta
_{m}^{\ast }=\theta _{0}^{\ast }$, such that $\delta (P,\theta _{0}^{\ast
})<0$ and $\left( 32\right) $ holds for $z_{m}^{\ast }.$

Without loss of generality, we can assume that

\begin{equation*}
\delta (P,\theta )>0,\text{ }\theta \in \left( \frac{\theta _{0}+2q\pi}{n} ,\frac{\theta
_{0}+\left( 2q+1\right) \pi}{n} \right)
\end{equation*}%
and

\begin{equation*}
\delta (P,\theta )<0,\text{ }\theta \in \left( \frac{\theta _{0}+\left(
2q-1\right) \pi}{n} ,\frac{\theta _{0}+2q\pi}{n} \right)
\end{equation*}%
for all $q\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ Provided $m$ is large enough, we have $\left\vert \theta -\theta
_{m}\right\vert \leq l_{0}.$ Choosing now $\theta _{0}^{\ast }$ such that $%
\frac{l_{0}}{2}\leq \theta _{m}-\theta _{m}^{\ast }\leq l_{0},$ then $\theta
_{0}-l_{0}\leq \theta _{0}^{\ast }\leq \theta _{0}-\frac{l_{0}}{2}$ and $%
\delta (P,\theta _{0}^{\ast })<0$. Since $\delta (Q,\theta _{0}^{\ast
})>0$, a contradiction follows as in case 2 above.
\end{proof}
\section*{Acknowledgement}

The authors would like to thank the referees for their helpful suggestions.

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%\end{linenumbers}
\end{document}
