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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 81, 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 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/81\hfil Oscillation of linear differential equations]
{Oscillation of higher-order linear differential
equations with entire coefficients}

\author[Z. G. Huang, G. R. Sun\hfil EJDE-2010/81\hfilneg]
{Zhi Gang Huang, Gui Rong Sun}  % in alphabetical order

\address{Zhi Gang Huang \newline
School of Mathematics and Physics,
Suzhou University of Science and Technology,
Suzhou, 215009, China}
\email{alexehuang@yahoo.com.cn}

\address{Gui Rong Sun \newline
School of Mathematics and Physics,
Suzhou University of Science and Technology,
Suzhou, 215009, China}
\email{sguirong@pub.sz.jsinfo.net}

\thanks{Submitted February 10, 2010. Published June 15, 2010.}
\thanks{Supported by grant 07KJD110189 from the Natural Science
Foundation of Education \hfill\break\indent
Commission of Jiangsu Province.}
\subjclass[2000]{30D35, 34M10}
\keywords{Linear differential equation; order; hyper order}

\begin{abstract}
 This article is devoted to studying the solutions to
 the differential equation
 $$
 f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_0(z)f=0,\quad k\ge 2,
 $$
 where coefficients $A_j(z)$ are entire functions of integer
 order.  We obtain estimates on the orders and the hyper orders
 of the solutions to the above equation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction and main results}

In this note, we  apply standard notation of the Nevanlinna
theory, see \cite{hay1}. Let $f(z)$ be a nonconstant meromorphic
function. As usual,  $\sigma (f)$ denote the order. In addition,
we use the notation $\sigma_2(f)$ to denote the hyper-order of
$f(z)$,
$$
\sigma_2(f)=\limsup_{r\to\infty}\frac{\log\log T(r,f)}{\log r}.
$$
Chen \cite{chen1} studied  this differential equation when all the
coefficients are of order 1.

\begin{theorem} \label{thmA}
Let $a$, $b$ be nonzero complex numbers and $a\ne b$, $Q(z)$
be a non-constant polynomial or $Q(z)=h(z)e^{bz}$ where
$h(z)$ is nonzero polynomial. Then every solution $f(\not\equiv
0)$ of the equation
\begin{equation} \label{e1}
f''+e^{az}f'+Q(z)f=0
\end{equation}
is of infinite order.
\end{theorem}

Later on, Li and Huang\cite{li1}, Chen and Shon\cite{chen2}
extended this result to the equation
\begin{equation} \label{e2}
f^{(k)}+A_{k-1}f^{(k-1)}+\dots +A_0f=0, \quad k\ge 2.
\end{equation}
 Chen and Shon\cite{chen2} obtained the following result.

\begin{theorem} \label{thmB}
 Let $A_j(z)=B_j(z)e^{P_j(z)}(0\le j\le k-1)$,
where $B_j(z)$ are entire functions with $\sigma(B_j)<1$ and
$P_j(z)=a_jz$ with $a_j$ are complex numbers. Suppose that there
exists $a_s$ such that $B_s\not\equiv 0$, and for $j\ne s$, if
$B_j\not\equiv 0$, $a_j=c_ja_s$, $0<c_j<1$; If $B_j\equiv 0$, we
define $c_j=0$. Then every transcendental solution $f$ of the
 \eqref{e2} satisfies $\sigma (f)=\infty$.
Furthermore, if $\max\{c_1,\dots,c_{s-1}\}<c_0$,
then every solution $f(\not\equiv 0)$ of (1) is of infinite order.
\end{theorem}

\begin{theorem} \label{thmC}
Let $P_j$ be polynomials, $s$ and $A_j$,$a_j,
 B_j$ satisfy the other additional hypotheses of Theorem \ref{thmB}. Then
 every transcendental solution $f$ of the \eqref{e2} satisfies
 $\sigma(f)=\infty$ and $\sigma_2(f)=1$. Furthermore, if
 $max\{c_1,\dots,c_{s-1}\}<c_0$, then every solution
 $f(\not\equiv 0)$ of \eqref{e2} is of infinite order
and $\sigma_2(f)=1$.
\end{theorem}

The aim of this paper is to improve Theorems \ref{thmB} and  \ref{thmC}.

\begin{theorem} \label{thm1}
Let $A_j(z)=B_j(z)e^{P_j(z)}$ $(j=0,1,\dots, k-1)$,
where $B_j(z)$ are entire functions, $P_j(z)$ are non-constant
polynomials with $\deg(P_j(z)-P_i(z))\ge 1$ and $\max
\{\sigma(B_j),\sigma(B_i)\}<\deg(P_i-P_j)(i\ne j)$. Then every
transcendental solution $f$ of \eqref{e2} satisfies
$\sigma(f)=\infty$.
\end{theorem}

\begin{theorem} \label{thm2}
 Let $P_j(z)=a_{j,n}z^n+a_{j,n-1}z^{n-1}+\dots
+a_{j,0}(0\le j\le k-1)$ be non-constant polynomials, where
$a_{j,n}\ne 0$ and $\deg(P_j(z)-P_i(z))=n$, and let $Q_j(z)$ and
$B_j(z)(0\le j\le k-1)$ be entire functions with $\max
\{\sigma(B_j),\sigma(Q_j),0\le j\le k-1\}<n$. Set
$A_j(z)=B_j(z)e^{P_j(z)}+Q_j(z)$. Suppose that one of the following
occurs:
\begin{itemize}
\item[(1)] There exist $t,s\in\{0,1,\dots,k-1\}$, such that
$\frac{a_{t,n}}{a_{s,n}}<0$;

\item[(2)] $\arg  a_{0,n}\ne \arg
a_{1,n}$ and  $a_{j,n}=c_ja_{1,n}(c_j>0$, $j=2,3,\dots,k-1)$.
\end{itemize}
Then every transcendental solution $f$ of \eqref{e2} satisfies
$\sigma(f)=\infty$.
\end{theorem}

\begin{theorem} \label{thm3}
Let $A_j=P_j(e^{R(z)})+Q_j(e^{-{R(z)}})$ for
$j=1,2,\dots,k-1$ where $P_j(z), Q_j(z)$ and
$R(z)=c_sz^s+\dots+c_1 z+c_0(s(\ge 1)$ is an integer) are
polynomials. Suppose that $P_0(z)+Q_0(z)\not\equiv 0$ and there
exists $d(0\le d\le k-1)$, such that for $j\ne d$, $\deg P_d>\deg
P_j$ and $\deg Q_d>\deg Q_j$. Then every solution $f(z)$ of
 \eqref{e2} is of infinite order and satisfies $\sigma_2(f)=s$.
\end{theorem}

We remark that many authors have studied the order and the
hyper order  of solutions of \eqref{e2}. But, they always
require that there exists some  coefficient
 $A_j$ ($j\in\{0,1,\dots,k-1\}$) such that the order of $A_j$
is greater than the order of other coefficients. We note that our
theorems do not need the hypothesis. Our hypothesis of Theorem
\ref{thm3} are partly motivated by \cite{chen3}.

\section{Preliminary lemmas}

Assume that $R(z)=c_sz^s+\dots+c_1 z+c_0(s(\ge 1)$ is a
polynomial. Below, for $\theta\in [0,2\pi)$, we denote
 $\delta_j(R,\theta)=\mathop{\rm Re}(c_j(e^{i\theta})^j)$
for $j\in\{1,2,\dots,s\}$. Especially,
 we write $\delta(R,\theta)=\delta_s(R,\theta)$.

For $j\in \{0,1,\dots,k-1\}$, let
$$
P_j(e^{R(z)})=a_{jm_j}e^{m_jR(z)}+a_{j(m_j-1)}e^{(m_j-1)R(z)}
+\dots+a_{j1}e^{R(z)}+a_{j0}
$$
and
$$
Q_j(e^{-R(z)})=b_{jt_j}e^{-t_j{R(z)}}+b_{j(t_j-1)}e^{-(t_j-1)R(z)}
+\dots+b_{j1}e^{-R(z)}+b_{j0},
$$
where $a_{jm_j},\dots,a_{j0}$ and $b_{jt_j},\dots,b_{j0}$ are
constants, $m_j\ge 0$ and $t_j\ge0$ are integers,  $a_{jm_j}\ne 0$,
$b_{jt_j}\ne 0$. So we have
\begin{equation} \label{e3}
\begin{aligned}
&|P_j(e^{R(z)})+Q_j(e^{-R(z)})|\\
&=\begin{cases}
 |a_{jm_j}|e^{m_jr^s\delta(R,\theta)}(1+o(1)),
 & \arg z=\theta , \delta(R,\theta)>0, r\to\infty, \\
 |b_{jt_j}|e^{-t_jr^s\delta(R,\theta)}(1+o(1)),
& \arg z=\theta , \delta(R,\theta)<0, r\to\infty;
 \end{cases}
\end{aligned}
\end{equation}
To prove our results, some lemmas are needed.

\begin{lemma} \label{lem1}
Let $f(z)$ be a transcendental meromorphic function
with $\sigma(f)=\sigma <\infty$. Let $\Gamma=\{(k_1,j_1),\dots
,\; (k_m,j_m)\}$ be a finite set of distinct pairs of integers
satisfying $k_i>j_i\ge 0$ for $i=1,2,\dots ,m$. Also let
$\epsilon >0$ be a given constant. Then, there exists a set
$E_1\subset [0,2\pi)$ that has linear measure zero, such
that if $\psi_0\in  [0,2\pi)\setminus E_1$, then there is a constant
$R_0=R_0(\psi_0)>1$ such that for all $z$ satisfying $\arg
z=\psi_0$ and $|z|\ge R_0$, and for all $(k,j)\in \Gamma$, we have
$$
\frac{|w^{(k)}(z)|}{|w^{(j)}(z)|}\le
|z|^{(k-j)(\sigma-1+\varepsilon)}.
$$
\end{lemma}

The above lemma is \cite[Corollary 1]{gun2}. We also need the
following lemma  given in Chen \cite{chen2}.

\begin{lemma} \label{lem2}
Suppose that $P(z)$ is a non-constant polynomial,
$w(z)$ is a meromorphic function with $\sigma(w)<\deg P(z)=n$. Let
$g(z)=w(z)e^{P(z)}$, then there exists a set $H_1\subset [0,2\pi)$
that has linear measure zero, such that for $\theta\in
[0,2\pi)\setminus (H_1\cup H_2)$ and arbitrary constant
$\epsilon(0<\epsilon<1$), when $r>r_0(\theta,\epsilon)$, we have
\begin{itemize}
\item[(1)] if $\delta(P,\theta)<0$, then $\exp((1+\epsilon)\delta
(P,\theta)r^n)\le |g(re^{i\theta})|\le
\exp((1-\epsilon)\delta(P,\theta)r^n)$,

\item[(2)] if $\delta(P,\theta)>0$, then $\exp((1-\epsilon)\delta
(P,\theta)r^n)\le |g(re^{i\theta})|\le
\exp((1+\epsilon)\delta(P,\theta)r^n)$, where $H_2=\{\theta
:\delta(P,\theta)=0,0\le\theta <2\pi\}$ is a finite set.
\end{itemize}
\end{lemma}

We shall use a special version of Phragm\'en-Lindel\"of-type
theorem to prove our results. We refer to
Titchmarsh \cite[p.177]{tit1}.

\begin{lemma} \label{lem3}
 Let $f(z)$ be an analytic function of
$z=r e^{i\theta}$, regular in the region $D$ between two straight
lines making an angle $\frac{\pi}{\beta-\alpha}$ at the origin and
on the lines themselves. Suppose that $|f(z)|\le M$ on the lines,
and for any given $\epsilon>0$, as $r\to\infty$,
$|f(z)|<O(e^{\epsilon r^{\frac{\pi}{\beta-\alpha}}})$, uniformly
in the angle. Then actually the inequality $|f(z)|\le M$ holds
throughout the region $D$.
\end{lemma}


\begin{lemma} \label{lem4}
Let $n\ge 2$ and $A_j(z)=B_j(z)e^{P_j(z)}(1\le j\le n)$
where each $B_j(z)$ is an entire function ,  and $P_j(z)$ is a
non-constant polynomial. Suppose that $\deg(P_j(z)-P_i(z))\ge 1$,
$\max \{\sigma(B_j),\sigma(B_i)\}<\deg(P_i-P_j)$ for $i\ne j$. Then
there exists a set $H_1\subset [0,2\pi)$ that has linear measure
zero such that for any given constant $M>0$ and $z=re^{i\theta}$,
$\theta\in [0,2\pi)\setminus (H_1\cup H_2)$, we have some
$s=s(\theta)\in\{1,\dots ,n\}$, for $j\ne s$,
$$
\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to
0,\quad \text{as }  r\to\infty,
$$
where
$H_2=\{\theta:\delta(P_j,\theta)=0 \text{ or }
\delta(P_i,\theta)=\delta(P_j,\theta),\;i,j\in\{1,2,\dots,n\},\;
i\ne j,\; 0\le\theta<2\pi\}$ is a finite set.
\end{lemma}

\begin{proof}
 We use mathematical induction.
For $n=2$, Lemma \ref{lem4} can be proved by applying
Lemma \ref{lem2} to
$\frac{A_1}{A_2}$ or $\frac{A_2}{A_1}$.

Assume that Lemma \ref{lem4} holds for $n\le k-1$. For the case
$n=k$. Take $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$, there
exists some $t=t(\theta)\in\{1,2,\dots ,k-1\}$, such that
$\frac{|A_j(re^{i\theta})||z|^M}{|A_t(re^{i\theta})|}\to 0$ for
$j\in\{1,\dots ,t-1,t+1,\dots k-1\}$. Now we compare
$A_t(re^{i\theta})$ with $A_k(re^{i\theta})$. If
$\delta(P_t-P_k,\theta,)<0$, from Lemma \ref{lem2},
$\deg(P_t-P_k)\ge 1$ and
$max\{\sigma(B_t),\sigma(B_k)\}<\deg(P_t-P_k)$, for any given
$1>\epsilon >0$, we have
$$
|\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\le
e^{(1-\epsilon)\delta (P_t-P_k,\theta)r^{\deg(P_t-P_k)}}\le
e^{(1-\epsilon)\delta (P_t-P_k,\theta)r},
$$
thus $|\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\,|z^M|\to 0$ as
$r\to\infty$. Therefore, for $j\ne k$,
$$
|\frac{A_j(re^{i\theta})}{A_k(re^{i\theta})}||z^M|
=|\frac{A_j(re^{i\theta})}{A_t(re^{i\theta})}||z^M|
|\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\to 0.
$$
If $\delta(P_t-P_k,\theta)>0$, then
$\delta(P_k-P_t,\theta)<0$, by the similar discussion as above, we
have $|\frac{A_k(re^{i\theta})}{A_t(re^{i\theta})}||z^M|\to 0$ as
$r\to\infty$. The proof  is now complete.
\end{proof}

Observe that (1) For $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$,
set $v=\deg P_s$ and $\delta=\delta(P_s,\theta)$ as in Lemma
\ref{lem4}. Since for $j\ne s$,
$\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to 0$, as
$r\to\infty$. Then if $\deg P_j>v, j\ne s$, we have
$\deg(P_t-P_s)=\deg P_t$, so $\delta(P_j,\theta)<0$. If $\deg P_j=v,
j\ne s$, then $\delta(P_j,\theta)<\delta$. If $\deg P_j<v, j\ne t$,
$\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to 0$ no
matter that $\delta(P_j,\theta)$ is positive or negative.

(2) From the proof of Lemma \ref{lem4}, if there exist a polynomial
$P_v(z)$  which is a constant $(v\in\{1,2,\dots ,n\})$, then the lemma is
also true. In fact, the hypothesis $\deg(P_j(z)-P_i(z))\ge 1(j\ne i)$
implies that there is at most one polynomial which can be a
constant.

From the proof of Lemma \ref{lem4}, we can easily obtain the
following lemma.

\begin{lemma} \label{lem5}
Let $P_j(z)(1\le j\le m)$ be non-constant polynomial
with degree $n$. Let $B_j(z)$ and $Q_j(z)(1\le j\le m)$ be entire
functions with $\max \{\sigma(B_j),\sigma(Q_j),1\le j\le m\}<n$.
Set $A_j(z)=B_j(z)e^{P_j(z)}+Q_j(z)$. For $\theta\in [0,2\pi)$,
suppose that not all $\delta(P_j,\theta)(1\le j\le n)$ are
negative. Then there exists some $s=s(\theta)\in\{1,\dots ,n\}$,
for $j\ne s$, as $r\to\infty$,
$$
\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to 0,
$$
where $M$ is a constant.
\end{lemma}

\section{Proof of main results}

\subsection*{Proof of Theorem \ref{thm1}}
Without loss of generality, we can
assume that each $A_j\not\equiv 0$, $j\in\{0,1,\dots,k-1\}$.

\noindent\textbf{Claim:}
Each transcendental solution $f$ of equation \eqref{e1} is
infinite order.

Suppose to the contrary, there exists a transcendental solution
$f(z)$ which has order $\sigma (f)=\sigma <\infty$. By Lemma \ref{lem1}, for
any given $\epsilon_0(0<\epsilon_0<1),$ there exists a set
$E_1\subset[0,2\pi)$ that has linear measure zero, such that if
$\psi_0\in[0,2\pi)\setminus E_1$, then
\begin{equation} \label{e4}
\frac{|f^{(j)}(z)|}{|f^{(i)}(z)|}\le|z|^{k\sigma},\quad
\ i=0,1,\dots ,k-1;\ j=i+1,\dots ,k
\end{equation}
as $z\to\infty$ along $\arg z=\psi_0$. Denote
$E_2=\{\theta\in [0,2\pi) :\delta (P_j,\theta)=0,
0\le j\le k\}\cup\{\theta\in [0,2\pi):\delta (P_j-P_i,\theta)=0,
0\le j\le k,0\le i\le k\}$, so $E_2$Î
is a finite set. Suppose that $H_j\subset [0,2\pi)$ is
 the exceptional set applying Lemma \ref{lem2} to $A_j(j=0,1,\dots ,k-1)$.
Then $E_3=\bigcup^{k-1}_{j=0}H_j$ has linear measure zero.
Take $\arg z=\psi_0\in
[0,2\pi)-(E_1\cup E_2\cup E_3)$ and write
$\delta_j=\delta(P_j,\psi_0)$. We need to treat two cases:


Case (i): Not all $\delta_0$, $\delta_1$,\dots ,
$\delta_{k-1}$ are negative. By Lemma \ref{lem4}, there exists some
$t\in\{0,1,2,\dots ,k-1\}$ such that for $j\ne t$, $M>0$,
\begin{equation}
\label{e5}
|\frac{A_j(re^{i\psi_0})}{A_t(re^{i\psi_0})}||z^M|\to
0,
\end{equation}
as $r\to\infty$. Let $v=\deg(P_t)$,
$\delta=\delta(P_t,\psi_0)$. From the observation, it is obvious
$\delta>0$. Now we prove $|f^{(t)}(z)|$ is bounded on the ray
$\arg z=\psi_0$. Suppose that it is not.
Let
$$
M(r,f^{(t)},\psi_0)=\max\{|f^{(t)}(z)|:0\le|z|\le r, \arg
z=\psi_0\}.
$$
There exists an infinite sequence of points
$z_n=r_ne^{i\psi_0}$ such that
$$
M(r_n,f^{(t)}, \psi_0)=|f^{(t)}(r_ne^{i\psi_0})|, r_n\to\infty.
$$
Take a curve
$C_n:z=re^{i\psi_0}$, $0\le r\le |z_n|$, for each $n$, we have
$$
f^{(t-1)}(z_n)=f^{(t-1)}(0)+\int_{C}f^{(t)}(u)du.
$$
And hence
$$
|f^{(t-1)}(z_n)|\le |f^{(t-1)}(0)|+|z_n|\cdot |f^{(t)}(z_n)|
$$
holds, which leads to
$$
\frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}\le
(1+\circ(1))|z_n|,\quad z_n\to\infty.
$$
Furthermore,
\begin{equation} \label{e6}
\frac{|f^{(t-j)}(z_n)|}{|f^{(t)}(z_n)|}\le
(1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,t.
\end{equation}
as $z_n\to\infty$. Since $f^{(t)}\not\equiv 0$, then by \eqref{e1},
\begin{equation} \label{e7}
\begin{aligned}
|A_t(z_n)|
&\le\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)|}+\dots
 +|A_{t+1}(z_n)|\cdot
\frac{|f^{(t+1)}(z_n)|}{|f^{(t)}(z_n)|}\\
&\quad +|A_{t-1}(z_n)|\cdot
\frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}+\dots
 +|A_{0}(z_n)| \cdot \frac{|f(z_n)|}{|f^{(t)}(z_n)|}
\end{aligned}
\end{equation}
holds as $z_n\to\infty$. So we obtain
\begin{equation} \label{e8}
\begin{aligned}
1&\le\frac{1}{|A_t(z_n)|}(\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)|}+\dots
+|A_{t+1}(z_n)|\cdot
\frac{|f^{(t+1)}(z_n)|}{|f^{(t)}(z_n)|}\\
&\quad +|A_{t-1}(z_n)|\cdot
\frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}+\dots +|A_{0}(z_n)| \cdot
\frac{|f(z_n)|}{|f^{(t)}(z_n)|}).
\end{aligned}
\end{equation}
Since $\delta>0$, by Lemma \ref{lem2} and \eqref{e5}, it is easy to deduce
$\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)||A_t(z_n)|}\to 0$.
Then from \eqref{e5}, \eqref{e6} and \eqref{e7},
the right hand of \eqref{e8} tends to $0$ as
$z_n\to\infty$, a contradiction. Thus, $|f^{(t)}|$ is bounded on
$\arg z=\psi_0\in [0,2\pi)\setminus(E_1\cup E_2\cup E_3)$. Assume
that $|f^{(t)}(re^{i\psi_0})|\le M_1 (M_1>0$ is a constant). Take
a curve $C'=\{z:\arg z=\psi_0,0\le|z|\le r\}$. Since
$$
f^{(t-1)}(z)=f^{(t-1)}(0)+\int_{C'}f^{(t)}(u)du,
$$
 for large $z=re^{i\psi_0}$, we have  $|f^{(t-1)}(z)|\le M_2|z|$
($M_2>0$ is a constant). By induction, we obtain
\begin{equation} \label{e9}
|f(z)|\le M_3|z|^t\le M_4|z|^k.
\end{equation}


(ii) Assume that for any $j:0\le j\le k-1$,
$\delta(P_j,\psi_0)<0$. By Lemma \ref{lem4}, there exists some
$s\in\{0,1,2,\dots ,k-1\}$, for $j\ne s$, we have
$$
|\frac{A_j(re^{i\psi_0})}{A_s(re^{i\psi_0})}|\to 0
$$
 as $r\to\infty$. Let $v=\deg(P_s)$,
$\delta=\delta(P_s,\psi_0)$, then $\delta<0$. From Lemma \ref{lem2},
 for any given $\epsilon(0<\epsilon<1/2)$,
\begin{equation} \label{e10}
|A_j(re^{i\psi_0})|\le |A_s(re^{i\psi_0})|
\le\exp{((1-\epsilon)\delta r^v)}.
\end{equation}
Suppose that
$|f^{(k)}(z)|$ is unbounded on the ray $\arg z=\psi_0$. Let
$$
M(r,f^{(k)}, \psi_0)=\max\{|f^{(k)}(z)|:0\le|z|\le r, \arg
z=\psi_0\}.
$$
There exists a infinite sequence of points
$z_n=r_ne^{i\psi_0}$ such that
$$
M(r_n,f^{(k)}, \psi_0)=|f^{(k)}(r_ne^{i\psi_0})|
$$
holds as $r_n\to\infty$. Take a
curve $C_n:z=re^{i\psi_0}$, $0\le r\le |z_n|$. Since
$f^{(k-1)}(z_n)=f^{(k-1)}(0)+\int_{C_n}f^{(k)}(u)du$, and on
$C_n$, $|f^{(k)}(z)|\le|f^{(k)}(z_n)|$, we have
$$
|f^{(k-1)}(z_n)|\le |f^{(k-1)}(0)|+|z_n|\cdot
|f^{(k)}(z_n)|.
$$
It follows that
$$
\frac{|f^{(k-1)}(z_n)|}{|f^{(k)}(z_n)|}\le (1+\circ(1))|z_n|.
$$
So we have
\begin{equation} \label{e11}
\frac{|f^{(k-j)}(z_n)|}{|f^{(k)}(z_n)|}\le
(1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,k.
\end{equation}
Since
$f^{(k)}\not\equiv 0$, by \eqref{e1}, \eqref{e10} and \eqref{e11}, for
sufficiently large $n$,
 we have
\begin{equation} \label{e12}
1\le|A_{k-1}(z_n)|\cdot
\frac{|f^{(k-1)}(z_n)|}{|f^{(k)}(z_n)|}+\dots +|A_{0}(z_n)|\cdot
\frac{|f(z_n)|}{|f^{(k)}(z_n)|}\le\exp{\{(1-\epsilon)\delta|z_n|^v\}}
\cdot |z_n|^{M_5},
\end{equation}
where $M_5$ is a positive constant. This
is impossible since $\delta<0$. Then $f^{(k)}(z)$ is bounded on
$\arg z=\psi_0$. Assume that $|f^{(k)}(re^{i\psi_0})|\le
M_6(M_6>0$). We take a curve $C'=\{z:\arg z=\psi_0,0\le|z|\le
r\}$. Since
$$
f^{(k-1)}(z)=f^{(k-1)}(0)+\int_{C'}f^{(k)}(u)du,
$$
for sufficiently large
$z=re^{i\psi_0}$, by induction, we have
\begin{equation} \label{e13}
|f(z)|\le
M_7|z|^k\quad\quad(M_7>0).
\end{equation}
Combine case (i) and case (ii),  for $\arg
z=\psi_0\in[0,2\pi)\setminus(E_1\cup E_2\cup E_3)$ and $|z|=r\ge
r_0(\psi_0)>0$, we obtain
\begin{equation} \label{e14}
|f(z)|\le M(\psi_0)|z|^k,
\end{equation}
where $M(\psi_0)>0$ is a constant
 dependent only on $\psi_0$.

  On the other hand, we can choose
$\theta_j\in [0,2\pi)\setminus(E_1\cup E_2\cup E_3)$
($j=1,2,\dots ,n,n+1$) such that
$$
0\le \theta_1 <\theta_2\dots <\theta_n <2\pi,\theta_{n+1}
=\theta_1 +2\pi
$$
and
$$
\max\{\theta_{j+1}-\theta_j|1\le j\le
n\}<\frac{\pi}{\sigma+1}.
$$
For any given positive number
$\epsilon$, we have
$$
\frac{|f(z)|}{|z^k|}\le |f(z)|\le \exp\{\epsilon
r^{\sigma+1}\}
$$
for sufficiently large $r=|z|$. From \eqref{e13} and
Lemma \ref{lem3},
 $\frac{|f(z)|}{|z^k|}\le M'$$(M'$ is a positive constant)
holds in the sectors $\{z:\theta_j\le arg z\le \theta_{j+1},|z|\ge r\}$
($j=1,2,\dots,n$) for sufficiently large $r$.
 Therefore,  $\frac{|f(z)|}{|z^k|}\le M''$ holds in the whole plane,
where $M''$ is a positive constant. Thus $f(z)$ is a polynomial.
It is a contradiction, and hence $\sigma(f)=\infty$.

\subsection*{Proof of Theorem \ref{thm2}}
Assume that $f(z)$ ia a transcendental
solution of \eqref{e2} with $\sigma(f)=\sigma<+\infty$. Set
$\omega=\max\{\sigma(B_j),\sigma(Q_j),0\le j\le k-1\}$.

(1) If there exist $t,s\in\{0,1,\dots,k-1\}$, such that
$\frac{a_{t,n}}{a_{s,n}}<0$. By the similar discussion to
Theorem \ref{thm1}, we take $\arg  z=\psi_0\in [0,2\pi)-(E_1\cup E_2\cup E_3)$. So
either $\delta(P_t,\psi_0)>0$ or $\delta(P_s,\psi_0)>0$. Therefore,
not all $\delta_0$, $\delta_1,\dots , \delta_{k-1}$ are
negative. By Lemma \ref{lem5}, we can obtain \eqref{e5}. Following the proof of
(i) of Theorem \ref{thm1}, we can get \eqref{e9} and \eqref{e13}. Then
$\sigma(f)=\infty$.

(2) By Lemma \ref{lem1}, for any given
$\epsilon_0$ with $0<\epsilon_0<\min\{\frac{1}{2},\frac{n-\omega}{2}\}$,
there exists a set $E_4\subset[0,2\pi)$ that has linear measure
zero, such that if $\theta\in[0,2\pi)\setminus E_4$, we have
\begin{equation} \label{e15}
\frac{|f^{(j)}(z)|}{|f^{(i)}(z)|}\le|z|^{k\sigma},\quad
i=0,1,\dots ,k-1;\; j=i+1,\dots ,k
\end{equation}
as $z\to\infty$ along $\arg z=\theta$. For $B_je^{P_j}$,
suppose that $H'_j\subset [0,2\pi)$ is  the exceptional set
applying Lemma \ref{lem2} to $B_je^{P_j}(j=0,1,\dots ,k-1)$.
Then $E_5=\bigcup^{k-1}_{j=0}H_j$ has linear measure zero.
Since $\arg  a_{0,n}\ne \arg  a_{1,n}$,
it is obvious that there exists a ray $\arg z=\phi_0\in
[0,2\pi)\setminus (E_4\cup E_5)$ such that $\delta(P_0,\phi_0)>0$
and $\delta(P_1,\phi_0)<0$. By Lemma \ref{lem2}, for sufficiently large $r$,
we have
\begin{equation} \label{e16}
|B_0(re^{i\phi_0})e^{P_0(re^{i\phi_0})}+Q_0(re^{i\phi_0})|\ge
\exp\{(1-\epsilon_0)\delta(P_0,\phi_0)r^n\}
\end{equation}
and
\begin{equation} \label{e17}
\begin{aligned}
&|B_1(re^{i\phi_0})e^{P_1(re^{i\phi_0})}+Q_1(re^{i\phi_0})|\\
&\le \exp\{(1-\epsilon_0)\delta(P_1,\phi_0)r^n\}
\exp\{r^{\omega+\epsilon_0}\}+exp\{r^{\omega+\epsilon_0}\}.
\end{aligned}
\end{equation}
So for $j=2,3,\dots,k-1$, we obtain
\begin{equation} \label{e18}
\begin{aligned}
&|B_j(re^{i\phi_0})e^{P_j(re^{i\phi_0})}+Q_j(re^{i\phi_0})|\\
&\le \exp\{(1-\epsilon_0)c_j\delta(P_1,\phi_0)r^n\}
\exp\{r^{\omega+\epsilon_0}\}+exp\{r^{\omega+\epsilon_0}\}.
\end{aligned}
\end{equation}
 From \eqref{e2}, we have
\begin{equation} \label{e19}
\begin{aligned}
&|A_0(re^{i\phi_0})|\\
&\le\frac{|f^{(k)}(re^{i\phi_0})|}{|f(re^{i\phi_0})|}
+|A_{k-1}(re^{i\phi_0})|\cdot
\frac{|f^{(k-1)}(re^{i\phi_0})|}{|f(re^{i\phi_0})|}+\dots+
|A_{1}(re^{i\phi_0})|
\frac{|f'(re^{i\phi_0})|}{|f(re^{i\phi_0})|}.
\end{aligned}
\end{equation}
Combine \eqref{e15}--\eqref{e19}, we have
\begin{align*}
&\exp\{(1-\epsilon_0)\delta(P_0,\phi_0)r^n\}\\
&\le  r^{k\sigma}+ r^{k\sigma}[(\exp\{(1-\epsilon_0)
 \delta(P_1,\phi_0)r^n\}\exp\{r^{\omega+\epsilon_0}\}
+\exp\{r^{\omega+\epsilon_0}\})\nonumber\\
&\quad +
\Sigma_{j=2}^{k-1}(\exp\{(1-\epsilon_0)c_j\delta(P_1,\phi_0)r^n\}
\exp\{r^{\omega+\epsilon_0}\}+\exp\{r^{\omega+\epsilon_0}\})].
\end{align*} %20
This is impossible, since $\omega+\epsilon_0<n$.

\section{Proof of Theorem \ref{thm3}}

\begin{lemma}[\cite{chen2}] \label{lem6}
Let $f(z)$ be an entire function with
$\sigma(f)=\infty$ and $\sigma_2(f)=\alpha<+\infty$, let a set
$E\subset [1,\infty)$ has finite logarithmic measure. Then there
exists a sequence $\{z_k=r_ke^{i\theta_k}\}$ satisfying
$|f(z_k)|=M(r_k,f)$, $\theta_k\in [0,2\pi)$ ,
$\lim_{k\to\infty}\theta_k=\theta_0\in[0,2\pi)$, $r_k\not\in E$,
and for any given $\epsilon_1>0$, as $r_k\to\infty$, we have the
following properties:

(i) If $\sigma_2(f)=\alpha$ $(0<\alpha<\infty)$, then
$$
\exp\{r_k^{\alpha-\epsilon_1}\}<v(r_k)<exp\{r_k^{\alpha+\epsilon_1}\},
$$
where $v(f)$ is the central index of $f$.

(ii) If $\sigma(f)=\infty$ and $\sigma_2(f)=0$, then for any given
constant $M(>0)$,
$$
r_k^M<v(r_k)<exp\{r_k^{\epsilon_1}\}.
$$
\end{lemma}

\begin{lemma}[\cite{chen2}] \label{lem7}
Let $A_j(0\le j\le k-1)$ be an entire
function with $\sigma(A_j)\le\sigma<\infty$. Then every non-trivial
solution $f$ of \eqref{e2} satisfies $\sigma_2(f)\le \sigma$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm3}]
Assume that $f(z)$ is a solution of \eqref{e2}. Clearly $f$ is entire.
Since $P_0+Q_0\not\equiv 0$, $f$ can not be a constant function.
Compare with two sides of \eqref{e2}, $f$ can not be a
polynomial whose degree is equal or greater than 1.

{\bf Step 1:} We prove that $\sigma(f)=\infty$. If it is not true.
Assume $\sigma(f)=\sigma<+\infty$. By Lemma \ref{lem1}, for any given
$\epsilon_0(0<\epsilon_0<1),$ there exists a subset
$E_1\subset[0,2\pi)$ that has linear measure zero such that if
$\psi_0\in[0,2\pi)\setminus E_1$, there is a constant $R_0>1$, such
that for $\arg z=\psi_0$ and $|z|>R_0$, we have
\begin{equation} \label{e21}
\frac{|f^{(j)}(z)|}{|f^{(i)}(z)|}\le|z|^{k\sigma},\quad
 i=0,1,\dots ,k-1;\; j=i+1,\dots ,k.
\end{equation}
Take a ray $\arg  z=\psi_0\in [0,2\pi)\setminus E_1$,
we consider the following two cases:

{\bf Case A1:}  $\delta(R,\psi_0)>0$. We claim that $|f^{(d)}(z)|$
is bounded on the ray $\arg z=\psi_0$. Suppose that it is not.
Following the proof of Theorem \ref{thm1}, we have
\begin{equation} \label{e22}
\frac{|f^{(d-j)}(z_n)|}{|f^{(d)}(z_n)|}\le
(1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,d,
\end{equation}
as $z_n\to\infty$. Since $f^{(d)}\not\equiv 0$, from \eqref{e2},
\begin{align*}
A_d(z)&=(-1)(\frac{f^{(k)}(z)}{f^{(d)}(z)}+\dots
+A_{d+1}(z)\cdot \frac{f^{(d+1)}(z)}{f^{(d)}(z)}+A_{d-1}(z)\cdot
\frac{f^{(d-1)}(z)}{f^{(d)}(z)}\\
&\quad +\dots +A_{0}(z) \cdot \frac{f(z)}{f^{(d)}(z)})
\end{align*}
holds, as $z\to\infty$. By \eqref{e21} and
\eqref{e22}, as  $z_n\to\infty$, we obtain
\begin{equation} \label{e23}
|P_d(e^{R(z_n)})+Q_d(e^{-R(z_n)})|\le
r^M\cdot\Sigma_{j\ne d}|P_j(e^{R(z_n)})+Q_j(e^{-R(z_n)})|,
\end{equation}
where $M$ is a constant. By (3), we obtain
 \begin{equation} \label{e24}
|P_d(e^{R(z_n)})+Q_d(e^{-R(z_n)})|
=|a_{dm_d}|e^{m_dr^s\delta(R,\theta)}(1+o(1))
\end{equation}
 and
\begin{equation} \label{e25}
|P_j(e^{R(z_n)})+Q_j(^{-R(z_n)})|
\le |a_{jm_j}|e^{m_jr^s\delta(R,\theta)}(1+o(1))+M_1,j\ne d,
\end{equation}
where $M_1$ is a positive constant.
Substituting \eqref{e24} and \eqref{e25} into \eqref{e23},
we obtain a contradiction since
$m_d>m_j(j\ne d)\ge 0$. Hence, $|f^{(d)}(re^{\psi_0})|$ is bounded
on the ray $\arg  z=\psi_0$. By the similar discussion to
Theorem \ref{thm1},
we can obtain \eqref{e9}.

{\bf Case A2:} $\delta(R,\psi_0)<0$. By a similar discussion to
subcase A1 and noting that \eqref{e24} and \eqref{e25}
 can be substituted by
\begin{equation} \label{e26}
|P_d(e^{R(z)})+Q_d(e^{-R(z)})|
=|b_{dt_d}|e^{t_dr^s\delta(R,\psi_0)}(1+o(1)),
\end{equation}
and
\begin{equation} \label{e27}
|P_j(e^{R(z)})+Q_j(e^{-R(z)})|\le
|b_{jt_j}|e^{t_jr^s\delta(R,\psi_0)}(1+o(1))+M_2.
\end{equation}
Thus, we can deduce \eqref{e13}.

Combine Case A1 and Case A2, we have \eqref{e14}. Following the proof of
Theorem \ref{thm1}, we can also obtain a contradiction.

{\bf Step 2:} In this step, we prove $\sigma_2(f)=s$.
By Lemma \ref{lem7}, we have
\begin{equation} \label{e28}
\sigma_2(f)\le s.
\end{equation}
Now we assume that $\sigma_2(f)=\alpha<s$, we will get a
contradiction.

 Recall the Wiman-Valiron
theory \cite{val1}, there exists a subset $E_3\subset(1,\infty)$ that
has finite logarithmic measure, such that for $|z|=r\not\in
E_3\cup[0,1]$ and $|f(z)|=M(r,f)$, we
have
\begin{equation} \label{e29}
\frac{f^{(j)}(z)}{f(z)}=(\frac{v(r)}{z})^j(1+o(1))(j=1,2,\dots,k),
\end{equation}
where $v(r)$ is central index of $f(z)$.

If $\sigma_2(f)=\alpha$ $(0<\alpha<s)$, from Lemma \ref{lem6},
we can take a sequence of points $\{z_n=r_ne^{i\theta_n}\}$
satisfying
$|f(z_n)|=M(r_n,f)$, $\theta_n\in [0,2\pi),$
$lim_{n\to\infty}\theta_n=\theta_0\in[0,2\pi)$, for any given
$\epsilon_1(0<\epsilon_1<min\{\alpha,s-\alpha\})$ and $r_n\not\in
E_2\cup E_3\cup [0,1]$, we obtain
\begin{equation} \label{e30}
\exp\{r_n^{\alpha-\epsilon_1}\}<v(r_k)<\exp\{r_n^{\alpha+\epsilon_1}\},
\end{equation}
as $r_n\to\infty$.
If $\sigma_2(f)=\alpha=0$, then for any positive constant $M$, we
have
\begin{equation} \label{e31}
r_n^M<v(r_n)<exp\{r_n^{\epsilon_1}\},
\end{equation}
as $r_n\to\infty$.

In the following, we consider three cases:

{\bf Case B1:} $\delta(R,\theta_0)>0$. From \eqref{e2},
we have
\begin{equation} \label{e32}
\begin{aligned}
A_d(z)(\frac{f^{(d)}(z)}{f(z)})
&=(-1)\Big\{\frac{f^{(k)}(z)}{f(z)}+\dots
+A_{d+1}(z)\cdot \frac{f^{(d+1)}(z)}{f(z)}\\
&\quad +A_{d-1}(z)\cdot
\frac{f^{(d-1)}(z)}{f(z)}+\dots +A_{0}(z) \Big\}.
\end{aligned}
\end{equation}
For sufficiently large $n$, $\delta(R,\theta_n)>0$ since
$\theta_n\to\theta_0$. For the point range $\{z_n=r_ne^{i\theta_n}\}$,
combine \eqref{e3}, \eqref{e29} and \eqref{e32}, we
obtain
\begin{align*}
&|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)|
(\frac{v(r_n)}{r_n})^d\\
&\le (\frac{v(r_n)}{r_n})^k+\dots
 +(\frac{v(r_n)}{r_n})^{d+1}(|a_{d+1m_{d+1}}|e^{m_{d+1}r_n^s
\delta(R,\theta_n)})|1+o(1)|\\
&\quad + (\frac{v(r_n)}{r_n})^{d-1}(|a_{d-1m_{d-1}}
|e^{m_{d-1}r_n^s\delta(R,\theta_n)}|1+o(1)|\\
&\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)|.
\end{align*}
By \eqref{e30} or \eqref{e31}, we obtain
 \begin{align*}
&|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)|
(\frac{\exp(dr_n^{\alpha-\epsilon_1})}{r_n^d})\\
&\le (\frac{\exp(kr_n^{\alpha+\epsilon_1})}{r_n^k})+\dots
+(\frac{\exp((d+1)r_n^{\alpha+\epsilon_1})}{r_n^{d+1}})
(|a_{d+1m_{d+1}}|e^{m_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +(\frac{\exp((d-1)r_n^{\alpha+\epsilon_1})}{r_n^{d-1}})(|a_{d-1m_{d-1}}|
e^{m_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)|
\end{align*}
or
 \begin{align*}
&|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)|
(\frac{r_n^M}{r_n^d})\\
&\le
(\frac{\exp(kr_n^{\epsilon_1})}{r_n^k})+\dots
+(\frac{\exp((d+1)r_n^{\epsilon_1})}{r_n^{d+1}})(|a_{d+1m_{d+1}}|
e^{m_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +
(\frac{\exp((d-1)r_n^{\epsilon_1})}{r_n^{d-1}})(|a_{d-1m_{d-1}}|
e^{m_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)|.
\end{align*}
Since $m_d>m_j(j\ne d)$ and $\alpha+\epsilon_1<s$, the above two
inequalities are impossible. This shows case B1 can not occur.

{\bf Case B2:} $\delta(R,\theta_0)<0$. For sufficiently large $n$,
$\delta(R,\theta_n)<0$ since $\theta_n\to\theta_0$. Following the
discussion of Subcase B1, we have
\begin{align*}
&|b_{dt_d}|e^{-t_dr_n^s\delta(R,\theta_n)}|1+o(1)|
(\frac{\exp(dr_n^{\alpha-\epsilon_1})}{r_n^d})\\
&\le (\frac{\exp(kr_n^{\alpha+\epsilon_1})}{r_n^k})+\dots
 +(\frac{\exp((d+1)r_n^{\alpha+\epsilon_1})}{r_n^{d+1}})(|b_{d+1t_{d+1}}|
 e^{-t_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad + (\frac{\exp((d-1)r_n^{\alpha+\epsilon_1})}{r_n^{d-1}})
 (|b_{d-1t_{d-1}}|e^{-t_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +\dots+|b_{0t_0}|e^{-t_0r_n^s\delta(R,\theta_n)}|1+o(1)|
\end{align*}
or
 \begin{align*}
&|b_{dt_d}|e^{-t_dr_n^s\delta(R,\theta_n)}|1+o(1))|
(\frac{r_n^M}{r_n^d})\\
&\le (\frac{\exp(kr_n^{\epsilon_1})}{r_n^k})+\dots
+ (\frac{\exp((d+1)r_n^{\epsilon_1})}{r_n^{d+1}})(|b_{d+1t_{d+1}}|
 e^{-t_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad + (\frac{\exp((d-1)r_n^{\epsilon_1})}{r_n^{d-1}})
 (|b_{d-1t_{d-1}}|e^{-t_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\
&\quad +\dots+|b_{0t_0}|e^{-t_0r_n^s\delta(R,\theta_n)}|1+o(1)|.
\end{align*}
Since $t_d>t_j(j\ne d)$ and $\alpha+\epsilon_1<s$, we also obtain
a contradiction.

{\bf Case B3:} $\delta(R,\theta_0)=0$. If there exists a
subsequence of $\{\theta_n\}$ such that $\delta(R,\theta_n)>0$ or
$\delta(R,\theta_n)<0$. Then by case B1 and case B2, we can get a
contradiction.

Now, suppose that for sufficiently large $n$,
$\delta(R,\theta_n)=0$. Then we consider three
subcases: $\delta_{s-1}(R,\theta)<0$;$\delta_{s-1}(R,\theta)>0$;
$\delta_{s-1}(R,\theta)=0$.
If $\delta_{s-1}(R,\theta)<0$ or $\delta_{s-1}(R,\theta)>0$. Then
replace $\delta(R,\theta)$ by $\delta_{s-1}(R,\theta)$ in the case
B1 and B2, we can obtain a contradiction. If
$\delta_{s-1}(R,\theta)=0$, from the previous discussion in case
B3, the remain case is $\delta_{s-1}(R,\theta)=0$ and
$\delta_{s-1}(R,\theta_n)=0$ for sufficiently large $n$. Then we
can consider $\delta_{s-2}(R,\theta)$, and we  can also obtain a
contradiction. On the analogy by this, the remain case is that
$\delta_j(R,\theta_n)=0$ for $j\in \{1,2,\dots,s$\} and for
sufficiently large $n$.

Rewriting \eqref{e2}, we have
\begin{equation} \label{e33}
\begin{aligned}
(-\frac{v(r_n)}{z_n})^k(1+o(1))
&= A_{k-1}(z_n)(\frac{v(r_n)}{z_n})^{k-1}(1+o(1))+\dots\\
&\quad +A_d(z_n)(\frac{v(r_n)}{z_n})^{d}(1+o(1))+\dots+A_0(z_n).
\end{aligned}
\end{equation}
For $z_n=r_ne^{\theta_n}$, since $\delta_j(R,\theta_n)=0$ for
$j\in \{1,2,\dots,s\}$, it leads to
\begin{equation} \label{e34}
|A_j(z_n)|=|P_j(e^{R(z_n)})+Q_j(e^{-R(z_n)})|
\le M,j\in \{1,2,\dots,k\},
\end{equation}
where $M$ is a constant. From  \eqref{e34}, we obtain
\begin{equation} \label{e35}
v(r_n)\le Br_n^k,
\end{equation}
where $B$ is a constant. However, this
contradicts \eqref{e30} and \eqref{e31}.
Therefore, case B3 can not occur.

Combining case B1, B2 and B3, we have $\sigma_2(f)=s$.
\end{proof}

\subsection*{Acknowledgements}
The authors want to thank the anonymous referees for their
 valuable suggestions.

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\end{document}