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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 51, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/51\hfil Growth of solutions]
{Growth of solutions to second-order complex differential equations}

\author[N. Li, L. Yang \hfil EJDE-2014/51\hfilneg]
{Nan Li, Lianzhong Yang}  % in alphabetical order

\address{ Nan Li \newline
School of Mathematics, Shandong University\\
Jinan, Shandong Province, 250100, China. \newline
Department of Physics and Mathematics, Joensuu Campus, 
University of Eastern Finland, P.O. Box 111, 80101 Joensuu, Finland}
\email{nanli32787310@gmail.com}

\address{Lianzhong Yang (corresponding author)\newline
School of Mathematics, Shandong University,
Jinan, Shandong Province, 250100, China}
\email{lzyang@sdu.edu.cn}

\thanks{Submitted September 25, 2013. Published February 19, 2014.}
\subjclass[2000]{30D35, 34M10}
\keywords{Differential equation; subnormal solution; hyper order}

\begin{abstract}
 In this article, we study the  existence of non-trivial subnormal
 solutions for second-order linear differential equations.
 We show that under certain conditions some differential equations
 do not have subnormal solutions, also that the hyper-order of
 every solution equals one.
\end{abstract}

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

\section{Introduction}

In this article, we use standard notation from the value
distribution theory of meromorphic functions (see \cite{Laine,Yi1}).
In addition, we denote the order of growth of $f(z)$ by $\sigma(f)$.
The hyper-order of $f(z)$ is defined by
\begin{equation*}
  \sigma_2(f)=\limsup_{r\to\infty}\frac{\log\log T(r,f)}{\log r}.
\end{equation*}
Consider the second order homogeneous linear periodic differential equation
\begin{equation}\label{iaaaa}
  f''+P(e^{z})f'+Q(e^{z})f=0,
\end{equation}
where $P(z)$ and $Q(z)$ are polynomials in $z$ and not both constants.
It is well known that every solution $f$ of \eqref{iaaaa} is entire.

For be a meromorphic function  $f$, define
\begin{equation}\label{iaaab}
  \sigma_{e}(f)=\limsup_{r\to\infty}\frac{\log T(r,f)}{r}
\end{equation}
to be the e-type order of $f$.
If $f\not\equiv0$ is a solution of  \eqref{iaaaa} satisfying $\sigma_{e}(f)=0$, then we say that $f$ is a nontrivial subnormal solution
of \eqref{iaaaa}.

Wittich \cite{Wittich}, Gundersen and Steinbart \cite{Gundersen1994},  Xiao \cite{Lpxiao} etc. have investigated the
subnormal solution of \eqref{iaaaa}, and obtained good results.
In 2007, Chen and Shon \cite{chen2007} studied the existence of subnormal solutions of the general equation
\begin{equation}\label{xaac}
f''+\big(P_1(e^{z})+P_2(e^{-z})\big)f'+\big(Q_1(e^{z})+Q_2(e^{-z})\big)f
=0,
\end{equation}
and obtained the following results.

\begin{theorem} \label{thmA}
 Let $P_j(z)$, $Q_j(z)$  $(j=1,2)$ be the polynomials in $z$. If
\begin{equation} \label{yaaaab}
  \deg Q_1> \deg P_1 \quad \text{or} \quad  \deg Q_2 > \deg P_2
\end{equation}
then \eqref{xaac} has no nontrivial subnormal solution,
and every solution of \eqref{xaac} satisfies $\sigma_2(f)=1$.
\end{theorem}

\begin{theorem} \label{thmB}
 Let $P_j(z)$, $Q_j(z)$ $(j=1,2)$ be the polynomials in $z$. If
\begin{equation} \label{yaaaaa}
  \deg Q_1< \deg P_1 \quad \text{and}  \quad  \deg Q_2 < \deg P_2
\end{equation}
and $Q_1+Q_2\not\equiv 0$, then \eqref{xaac}
has no nontrivial subnormal solution, and every solution of \eqref{xaac}
satisfies $\sigma_2(f)=1$.
\end{theorem}

\subsection*{Question}
 What can we said when $\deg P_1= \deg Q_1$ and $\deg P_2= \deg Q_2$ for \eqref{xaac}? 
We will prove the following theorem.

\begin{theorem} \label{thm1}.
 Let
\begin{gather*}
P_1(z)=a_nz^{n}+\dots+a_1z+a_0, \\
Q_1(z)=b_nz^{n}+\dots+b_1z+b_0, \\
P_2(z)=c_{m}z^{m}+\dots+c_1z+c_0, \\
Q_2(z)=d_{m}z^{m}+\dots+d_1z+d_0,
\end{gather*}
where $a_i, b_i$ $(i=0,\dots, n)$, $c_j, d_j$ $(j=0,\dots, m)$ are constants, 
$a_nb_nc_{m}d_{m}\neq 0$.
Suppose that $a_nd_{m}=c_{m}b_n$ and any one of the following three hypotheses holds:
\begin{itemize}
\item[(i)] there exists $i$ satisfying
$\bigl(-\frac{b_n}{a_n}\bigr)a_i+b_i\neq 0$, $0 < i< n$;
(ii) there exists $j$ satisfying
$\bigl(-\frac{b_n}{a_n}\bigr)c_j+d_j\neq 0$, $0 < j < m$;
\item[(iii)]
$$
\bigl(-\frac{b_n}{a_n}\bigr)^{2} +\bigl(-\frac{b_n}{a_n}\bigr)
(a_0+c_0)+b_0+d_0\neq 0.
$$
\end{itemize}
Then \eqref{xaac} has no non-trivial subnormal solution, and every non-trivial 
solution $f$
satisfies $\sigma_2(f)=1$.
\end{theorem}

We remark that the equation
\[
f''+(e^{2z}+e^{-z}+1)f'+(2e^{2z}+2e^{-z}-2)f=0
\]
has a subnormal solution $f_0=e^{-2z}$.  Here $n=2$, $m=1$,
$a_2=1$, $b_2=2$, $a_1=b_1=0$, $c_1=1$, $d_1=2$, $a_0+c_0=1$, $b_0+d_0=-2$,
$(-\frac{b_2}{a_2})\cdot a_1+b_1=0$, and
$(-\frac{b_2}{a_2})^{2} +(-\frac{b_2}{a_2})(a_0+c_0)+b_0+d_0= 0$.
This shows that the restrictions (i)--(iii) in
Theorem \ref{thm1} are sharp.

Another problem we want to consider in this paper is what condition
will guarantee the more general form
\begin{equation} \label{qaaaaa}
f''+\big(P_1(e^{\alpha z})+P_2(e^{-\alpha z})\big)f'
+\big(Q_1(e^{\beta z}) +Q_2(e^{-\beta z})\big)f=0,
\end{equation}
where $P(z), Q(z)$ are polynomials in $z, \alpha, \beta$ are complex constants, does not have a non-trivial subnormal
solution? We will prove the following theorems.

\begin{theorem} \label{thm2} Let
\begin{gather*}
 P_1(z)=a_{1m_1}z^{m_1}+\dots+a_{11}z+a_{10}, \\
 P_2(z)=a_{2m_2}z^{m_2}+\dots+a_{21}z+a_{20}, \\
 Q_1(z)=b_{1n_1}z^{n_1}+\dots+b_{11}z+b_{10}, \\
 Q_2(z)=b_{2n_2}z^{n_2}+\dots+b_{21}z+b_{20},
\end{gather*}
where $m_k\geq 1,\,  n_k \geq 1\, (k=1,2)$ are integers,
$a_{1i_1}$ $(i_1=0,1, \dots, m_1)$, $a_{2i_2}$
$(i_2=0,1, \dots, m_2)$,
$b_{1j_1}$ $(j_1=0,1, \dots, n_1)$, $b_{2j_2}$  $(j_2=0,1, \dots, n_2)$,
$\alpha$ and $\beta$ are complex constants, 
$a_{1m_1}a_{2m_2}b_{1n_1}b_{2n_2}\neq 0$, $\alpha \beta\neq 0$. Suppose
$m_1\alpha=c_1n_1\beta$ $(0<c_1<1)$
or $m_2\alpha=c_2n_2\beta$   $(0<c_2<1)$.
Then \eqref{qaaaaa} has no non-trivial subnormal solution and every 
non-trivial solution $f$ satisfies $\sigma_2(f)=1$.
\end{theorem}

\begin{theorem} \label{thm3} Let
\begin{gather*}
P_1(z)=a_{1m_1}z^{m_1}+\dots+a_{11}z+a_{10}, \\
P_2(z)=a_{2m_2}z^{m_2}+\dots+a_{21}z+a_{20}, \\
Q_1(z)=b_{1n_1}z^{n_1}+\dots+b_{11}z+b_{10}, \\
Q_2(z)=b_{2n_2}z^{n_2}+\dots+b_{21}z+b_{20},
\end{gather*}
where $m_k \geq 1$, $n_k \geq 1$ $(k=1,2)$ are integers,
$a_{1i_1}$  $(i_1=0,1, \dots, m_1)$, $a_{2i_2}$  $(i_2=0,1, \dots, m_2)$,
$b_{1j_1}$  $(j_1=0,1, \dots, n_1)$, $b_{2j_2}$  $(j_2=0,1, \dots, n_2)$,
$\alpha$ and $\beta$ are complex constants, 
$a_{1m_1}a_{2m_2}b_{1n_1}b_{2n_2}\neq 0$,
 $\alpha \beta\neq 0$. Suppose
$m_1\alpha=c_1n_1\beta$  $(c_1>1)$
and $m_2\alpha=c_2n_2\beta$ $(c_2>1)$.
Then \eqref{qaaaaa} has no non-trivial subnormal solution and every 
non-trivial solution $f$ satisfies $\sigma_2(f)=1$.
\end{theorem}

Note that a subnormal solution $f_0=e^{-z}+1$ satisfies the equation
\[
  f''-[e^{3z}+e^{2z}+e^{-z}]f'-[e^{2z}+e^{-z}]f=0.
\]
Here $\alpha=\frac{1}{2}$, $\beta=1/3$, $m_1=6$, $m_2=2$, $n_1=6$, $n_2=3$, 
$m_1\alpha=\frac{3}{2}n_1\beta$ and
$m_2\alpha=n_2\beta $. This shows that the restrictions that 
$m_1\alpha=c_1n_1\beta$ $(c_1>1)$
and $m_2\alpha=c_2n_2\beta\, (c_2>1)$ can not be omitted.

\section{Some lemmas}

 
Let $P(z)=(a+ib)z^{n}+\dots$ be a polynomial with
degree $n\geq 1$. and $z=re^{i\theta}$.
We will we denote $\delta(P,\theta)=a\cos (n\theta)-b\sin (n\theta)$.

\begin{lemma}[\cite{Laine}] \label{lem1}
  Let $P(z)=a_nz^{n}+a_{n-1}z^{n-1}+\dots +a_0$ be a polynomial with $a_n\neq 0$.
Then, for every $\varepsilon>0$, there exists $r_0>0$ such that for 
all $r=|z|>r_0$ we have the inequalities
\[
  (1-\varepsilon) |a_n|r^{n}\leq |P(z)| \leq (1+\varepsilon) |a_n|r^{n}\,.
\]
\end{lemma}

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


\begin{lemma}\cite{Gundersen1} \label{lem3}
 Let $f(z)$ be a transcendental meromorphic function with $\sigma(f)=\sigma<\infty$.
Let $H=\{(k_1,j_1),(k_2,j_2), \dots, (k_{q},j_{q})\}$ be a finite set of
 distinct pairs of
integers that satisfy $k_i>j_i \geq 0$, for $i=1,2, \dots, q$.
 And let $\varepsilon>0$ be a given constant. 
Then there exists a set $E\subset [0, 2\pi)$ that has linear
measure zero, such that if $\psi \in [0,2\pi) \setminus E$, 
then there is a constant $R_0=R_0(\psi)>1$
such that for all $z$ satisfying $\arg z =\psi$ and $|z|\geq R_0$ 
and for all $(k,j)\in H$, we have
\begin{equation}\label{baa}
\big|\frac{f^{(k)}(z)}{f^{(j)}(z)}\big| \leq |z|^{(k-j)(\sigma-1+\varepsilon)}.
\end{equation}
\end{lemma}

\begin{lemma}[\cite{Gundersen2,Laine-Yang}] \label{lem4}
 Let $f(z)$ be an entire function and suppose that $|f^{(k)}(z)|$ is unbounded 
on some ray $\arg z=\theta$. Then, there exists an infinite sequence of
 points $z_n=r_ne^{i\theta}$ $(n=1,2,\dots)$,
where $r_n\to \infty$, such that $f^{(k)}(z_n)\to \infty$ and
\begin{equation}\label{bab}
\big|\frac{f^{(j)}(z_n)}{f^{(k)}(z_n)}\big|
\leq  \frac{1}{(k-j)!} |z_n|^{(k-j)}(1+o(1))  \quad (j=0, \dots , k-1).
\end{equation}
\end{lemma}

\begin{lemma}[\cite{chen2003}] \label{lem5}
 Let $f(z)$ be an entire function with $\sigma(f)=\sigma<\infty$. 
Let there exists a set $E\subset[0,2\pi)$ with linear measure zero, 
such that for any $\arg z= \theta_0$ $\in [0,2\pi) \setminus E $,
$|f(re^{i\theta_0})|\leq Mr^{k}$ ($M=M(\theta_0)>0$ is a constant, $k(>0)$ 
is constant independent of $\theta_0$). 
Then $f(z)$ is a polynomial of $\deg f \leq k$.
\end{lemma}

\begin{lemma}[\cite{chenshon2011}] \label{lem6}
 Let $A$ and $B$ be entire functions of finite order.
If $f(z)$ is a solution of the equation
$$
f''+Af'+Bf=0,
$$
then $\sigma_2(f)\leq \max \{\sigma(A), \sigma(B)\}$.
\end{lemma}

\begin{lemma}[\cite{chenshon}] \label{lem7}
 Let $f(z)$ be an entire function of infinite order with
 $\sigma_2=\alpha\, (0\leq \alpha <\infty)$, and a set $E\subset[1,\infty)$ 
have a finite logarithmic measure. Then, there exists $\{z_k=r_ke^{i \theta_k}\}$ 
such that
$|f(z_k)|=M(r_k,f)$, $\theta_k\in[0, 2\pi)$,
$\lim_{k\to\infty}\theta_k=\theta_0\in[0, 2\pi)$,
$r_k\not\in E$,  $r_k\to\infty$, and such that
\begin{itemize}
\item[(1)] if $\sigma_2(f)=\alpha$ $(0< \alpha <\infty)$, then for any given
$\varepsilon_1$  $(0<\varepsilon_1<\alpha$),
\begin{equation}\label{bac}
  \exp\{r_k^{\alpha-\varepsilon_1}\}<\nu(r_k)<\exp \{r_k^{\alpha+\varepsilon_1}\},
\end{equation}

\item[(2)] if $\sigma (f)=\infty$ and $\sigma_2(f)=0$, then for any given
$\varepsilon_2$ $(0<\varepsilon_2<1/2$), and any large $M\,(>0)$, we have,
 for $r_k$ sufficiently large,
\begin{equation}\label{bad}
  r_k^{M}<\nu(r_k)<\exp \{r_k^{\varepsilon_2}\}.
\end{equation}
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{Gundersen1}] \label{lem8}
 Let $f$ be a transcendental meromorphic function, and $\alpha>1$ be a
given constant. Then there exists a set $E\subset(1,\infty)$ with finite 
logarithmic measure and a constant $B>0$ that depends only on $\alpha$ and 
$i, j$ $(0\leq i< j \leq 2)$,
such that for all $z$ satisfying $|z|=r \not\in E \cup [0,1]$,
\begin{equation}\label{bae}
  \big|\frac{f^{(j)}(z)}{f^{(i)}(z)}\big|
\leq B\Big(\frac{T(\alpha r,f)}{r}(\log^{\alpha}r)\log T(\alpha r, f)\Big)^{j-i}.
\end{equation}
\end{lemma}

\begin{remark}[\cite{chen2007}] \label{rmk3} \rm
From the proof of Lemma \ref{lem8}, we can see that the exceptional set $E$
satisfies that if $a_n$ and $b_{m}$ ($n,m=1,2,\dots$) denote all zeros and poles
of $f$, respectively, $O(a_n)$ and $O(b_{m})$ denote sufficiently small
neighborhoods of $a_n$ and $b_{m}$, respectively, then
$$ 
E=\{|z|:z\in(\cup_{n=1}^{+\infty}O(a_n))\cup (\cup_{m=1}^{+\infty}O(b_{m}))\}.
$$
Hence, if $f(z)$ is a transcendental entire function, and $z$ is a point that
satisfies $|f(z)|$ to be sufficiently large, then \eqref{bae} holds.
\end{remark} 

\section{Proof of Theorem \ref{thm1}}

Suppose that $f(z)$ is a non-trivial subnormal solution of \eqref{xaac}. Let
\[
  h(z)=e^{(b_n/a_n)z}f(z),
\]
then $h(z)$ is a non-trivial subnormal solution of 
\begin{align*}
 &h''+\Big(2(-\frac{b_n}{a_n})+P_1(e^{z})+P_2(e^{-z})\Big)h'\\
+&\Big((-\frac{b_n}{a_n})^{2} + (-\frac{b_n}{a_n})
\big(P_1(e^{z})+P_2(e^{-z})\big)+Q_1(e^{z})+Q_2(e^{-z})\Big)h=0.
\end{align*}
Since any one of the following three hypotheses holds:
\begin{itemize}
\item[(i)] there exists $i$ satisfying
$(-\frac{b_n}{a_n})a_i+b_i\neq 0$, $0 < i< n$;

\item[(ii)] there exists $j$ satisfying
$(-\frac{b_n}{a_n})c_j+d_j\neq 0$, $0 < j< m$;

\item[(iii)]
$$
\Big((-\frac{b_n}{a_n})^{2} +(-\frac{b_n}{a_n})(a_0+c_0)+b_0+d_0\Big)\neq 0,
$$
\end{itemize}
we obtain
\begin{equation} \label{zxbbbb}
(-\frac{b_n}{a_n})^{2} + (-\frac{b_n}{a_n})
(P_1(e^{z})+P_2(e^{-z}))+Q_1(e^{z})+Q_2(e^{-z})\not\equiv 0.
\end{equation}
From $a_nd_{m}=c_{m}b_n$, we obtain
\begin{equation} \label{zxbbbc}
\deg P_2(z) >m-1 \geq \deg [(-\frac{b_n}{a_n})P_2(z)+Q_2(z)].
\end{equation}
Combining \eqref{zxbbbb} and \eqref{zxbbbc} with
\begin{equation}
\deg P_1(z) > n-1 \geq \deg [ (-\frac{b_n}{a_n})P_1(z)+Q_1(z)],
\end{equation}
we obtain the conclusion by using Theorem \ref{thmB}.

\section{Proof of Theorem \ref{thm2}}

Suppose $f (\not\equiv 0)$ is a solution of \eqref{qaaaaa}, then $f$ 
is an entire function.
Next we will prove that $f$ is transcendental. 
Since $Q_1(e^{\beta z}) +Q_2(e^{-\beta z})\not\equiv 0$, we see that 
any nonzero constant can not be a solution of the \eqref{qaaaaa}.
Now suppose that $f_0=b_nz^{n}+\dots+b_1z+b_0$, 
($n\geq 1, b_n, \dots, b_0$ are constants, $b_n\neq 0$) is
a polynomial solution of \eqref{qaaaaa}.
\smallskip

\noindent (1) $m_1\alpha=c_1n_1\beta\, (0<c_1<1)$. Take $z=re^{i\theta}$, 
such that $\delta(\beta z, \theta)=|\beta|\cos(\arg \beta+\theta)>0$, then
$\delta(\alpha z, \theta)=\frac{n_1c_1}{m_1}\delta(\beta z, \theta)>0$.
 From \eqref{qaaaaa} and Lemma \ref{lem1}, that
for a sufficiently large $r$ and $\varepsilon>0$, we have
\begin{equation} \label{wwaaaaa}
\begin{aligned}
&(1-\varepsilon)|b_n|r^{n}| b_{1n_1}|e^{n_1\delta(\beta z, \theta)r}(1-o(1))
&\leq |Q_1(e^{\beta z})+Q_2(e^{-\beta z})|\cdot |f_0|  \\
&\leq  |f_0''|+  |P_1(e^{\alpha z})+P_2(e^{-\alpha z})| \cdot |f_0'| \\
&\leq  |a_{1m_1}|e^{m_1\delta(\alpha z,\theta)r} n(n-1)(1+\varepsilon)|b_n|r^{n-1}(1+o(1))\\
&\leq  M_1e^{m_1\cdot \frac{n_1c_1}{m_1}\delta(\beta z, \theta)r}r^{n-1}(1+o(1))\\
&\leq  M_1e^{n_1c_1\delta(\beta z, \theta)r}r^{n-1}(1+o(1)),
\end{aligned}
\end{equation}
where $M_1>0$ is some constant. Since $0<c_1<1$, we see that \eqref{wwaaaaa}
is a contradiction.
\smallskip

\noindent (2) $m_2\alpha=c_2n_2\beta\, (0<c_2<1)$. Take $z=re^{i\theta}$, 
such that $\delta(\beta z, \theta)=|\beta|\cos(\arg \beta+\theta)<0$, then
$\delta(\alpha z, \theta)=\frac{n_2c_2}{m_2}\delta(\beta z, \theta)<0$.
 From \eqref{qaaaaa} and Lemma \ref{lem1}, that
for a sufficiently large $r$ and $\varepsilon>0$, we have
\begin{equation}\label{waaaab}
\begin{aligned}
&(1-\varepsilon)|b_n|r^{n}| b_{2n_2}|e^{-n_2\delta(\beta z, \theta)r}(1-o(1))\\
&\leq |Q_1(e^{\beta z})+Q_2(e^{-\beta z})|\cdot |f_0|  \\
&\leq  |f_0''|+  |P_1(e^{\alpha z})+P_2(e^{-\alpha z})| \cdot |f_0'| \\
&\leq  |a_{2m_2}|e^{-m_2\delta(\alpha z,\theta)r} n(n-1)(1+\varepsilon)|b_n|r^{n-1}
 (1+o(1))\\
&\leq  M_2e^{-m_2\cdot \frac{n_2c_2}{m_2}\delta(\beta z, \theta)r}r^{n-1}(1+o(1))\\
&\leq  M_2e^{-n_2c_2\delta(\beta z, \theta)r}r^{n-1}(1+o(1)),
\end{aligned}
\end{equation}
where $M_2>0$ is some constant. Since $0<c_2<1$, we see that \eqref{waaaab} 
is also a contradiction.
Thus we obtain that $f$ is transcendental.

By Lemma \ref{lem6} and $\max \{\sigma(P_1(e^{\alpha z})), \sigma(P_2(e^{-\alpha z})),
 \sigma(Q_1(e^{\beta z})), \sigma(Q_2(e^{-\beta z})) \}=1$, 
we see that $\sigma_2(f)\leq 1$. 
By Lemma \ref{lem8}, we can see that there exists a subset $E\subset (1, \infty)$ 
having a logarithmic measure
$m_{l}E<\infty$ and a constant $B >0$ such that for all $z$ satisfying 
$|z|=r \not\in [0,1]\cup E$, we have
\begin{equation} \label{dddaaa}
 | \frac{f^{(j)}(z)}{f(z)} | \leq B[T(2r,f)]^{j+1}, \quad  j=1,2.
\end{equation}
(1) Suppose $m_1\alpha=c_1n_1\beta\, (0<c_1<1)$. Take $z=re^{i\theta}$,
such that $\delta(\beta z, \theta)>0$, then
$\delta(\alpha z, \theta)=\frac{n_1c_1}{m_1}\delta(\beta z, \theta)>0$.
From \eqref{qaaaaa}, \eqref{dddaaa}, that
for a sufficiently large $r$ and $r\not\in [0,1]\cup E$, we have
\begin{equation} \label{dddaaaa}
\begin{aligned}
&(1-\varepsilon)| b_{1n_1}|e^{n_1\delta(\beta z, \theta)r}(1-o(1))\\
& \leq |Q_1(e^{\beta z})+Q_2(e^{-\beta z})|   \\
  &\leq  | \frac{f''(z)}{f(z)}|+|P_1(e^{\alpha z})+P_2(e^{-\alpha z})|
 | \frac{f'(z)}{f(z)} |  \\
  &\leq  B[T(2r,f)]^3+(1+\varepsilon)|a_{1m_1}|e^{m_1\delta(\alpha z,\theta)r}
 B[T(2r,f)]^{2}(1+o(1)) \\
  &\leq  C[T(2r,f)]^3e^{m_1\cdot \frac{n_1c_1}{m_1}\delta(\beta z, \theta)r}
 (1+o(1)) \\
  &\leq  C[T(2r,f)]^3e^{n_1c_1\delta(\beta z, \theta)r}(1+o(1)) .
\end{aligned}
\end{equation}
Since $0<c_1<1$, by lemma \ref{lem2}, \eqref{dddaaaa}, we obtain $\sigma_2(f)\geq 1$.
So $\sigma_2(f)= 1$.

Next we prove that any $f (\not\equiv 0)$ is not subnormal. 
If $f$ is subnormal, then for any $\varepsilon > 0$,
\begin{equation}\label{dddaab}
 T(r,f)\leq e^{\varepsilon r}.
\end{equation}
When taking $z=re^{i\theta}$, such that $\delta(\beta z, \theta)>0$, 
by \eqref{dddaaaa} and \eqref{dddaab}, we deduce that
\begin{equation} \label{dddaac}
\begin{aligned}
  (1-\varepsilon)| b_{1n_1}|e^{n_1\delta(\beta z, \theta)r}(1-o(1))
&\leq  C[T(2r,f)]^3e^{n_1c_1\delta(\beta z, \theta)r}(1+o(1))  \\
&\leq  C e^{6\varepsilon r} \cdot e^{n_1c_1\delta(\beta z, \theta)r}(1+o(1)).
\end{aligned}
\end{equation}
We see that \eqref{dddaac} is a contradiction when
$0<\varepsilon<\frac{1}{6}n_1\delta(\beta z,\theta)(1-c_1)$.
Hence \eqref{qaaaaa} has no non-trivial subnormal solution and every
solution $f$ satisfies $\sigma_2(f)=1$.
\smallskip

\noindent (2) Suppose $m_2\alpha=c_2n_2\beta\, (0<c_2<1)$. 
Take $z=re^{i\theta}$, such that $\delta(\beta z, \theta)<0$, then
$\delta(\alpha z, \theta)=\frac{n_2c_2}{m_2}\delta(\beta z, \theta)<0$. 
Using the similar method as
in the proof of  (1), we obtain the conclusion.

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

Suppose that $f(\not\equiv 0)$ is a solution of \eqref{qaaaaa}, then $f$ 
is an entire function. 
Next we will prove that $f$ is transcendental. 
Since $Q_1(e^{\beta z}) +Q_2(e^{-\beta z})\not\equiv 0$, we see that any 
nonzero constant can not be a solution of the Eq.\eqref{qaaaaa}.
Now suppose that $f_0=b_nz^{n}+\dots+b_1z+b_0$, ($n\geq 1, b_n, \dots, b_0$ 
are constants, $b_n\neq 0$) is a polynomial solution of \eqref{qaaaaa}.

 Take $z=re^{i\theta}$, such that 
$\delta(\alpha z, \theta)=|\alpha|\cos(\arg \alpha + \theta) >0$, then
 $\delta(\beta z,\theta)=\frac{m_1}{c_1n_1}\delta(\alpha z,\theta)>0$. 
From \eqref{qaaaaa} and Lemma \ref{lem1}, that
for a sufficiently large $r$ and $\varepsilon>0$, we have
\begin{equation} \label{waaaaa}
\begin{aligned}
&(1-\varepsilon)|b_n|n r^{n-1} |a_{1m_1}|e^{m_1\delta(\alpha z, \theta)r}(1-o(1))
&\leq | P_1(e^{\alpha z})+P_2(e^{-\alpha z}) |\cdot |f_0'|  \\
&\leq  |f_0''|+  |Q_1(e^{\beta z})+Q_2(e^{-\beta z})| \cdot |f_0| \\
&\leq  |b_{1n_1}|e^{n_1\delta(\beta z, \theta)r} n(n-1)(1+\varepsilon)|b_n|r^{n}(1+o(1))\\
&\leq  Me^{n_1\cdot \frac{m_1}{c_1n_1}\delta(\alpha z,\theta)r}r^{n}(1+o(1))\\
&\leq  Me^{\frac{m_1}{c_1}\delta(\alpha z, \theta)r}r^{n}(1+o(1)),
\end{aligned}
\end{equation}
where $M>0$ is some constant. Since $c_1>1$, we see that \eqref{waaaaa}
is a contradiction.
Thus we obtain that $f$ is transcendental.

First step. We prove that $\sigma(f)=\infty$. We assume that 
$\sigma(f)=\sigma<\infty$. By Lemma \ref{lem3}, we
know that for any given $\varepsilon>0$, there exists a set 
$E\subset [0, 2\pi)$ which has linear measure zero, such that
if $\psi\in [0, 2\pi)\setminus E$, then there is a constant $R_0=R_0(\psi)>1$, 
such that for all $z$ satisfying $\arg z=\psi$ and $|z|=r \geq R_0$, we have
\begin{equation} \label{dddaad}
 |\frac{f''(z)}{f'(z)}|\leq r^{\sigma-1+\varepsilon}.
\end{equation}
Let $H=\{\theta\in [0,2\pi): \delta(\alpha z, \theta)=0\}$; then $H$ is a
finite set. Now we take a ray $\arg z=\theta \in [0,2\pi)\setminus(E\cup H)$, then
$\delta(\alpha z, \theta)>0$ or $\delta(\alpha z, \theta)<0$. We divide the proof
into the following two cases.
\smallskip

\noindent\textbf{Case 1.}
 If $\delta(\alpha z, \theta)>0$, then 
$\delta(\beta z,\theta)=\frac{m_1}{c_1n_1}\delta(\alpha z,\theta)>0$,
$\delta(-\alpha z, \theta)<0$ and $\delta(-\beta z,\theta)<0$.
We assert that $|f'(re^{i\theta})|$ is bounded on the ray $\arg z=\theta$. 
If $|f'(re^{i\theta})|$ is unbounded on the ray $\arg z=\theta$, 
then by Lemma \ref{lem4}, 
there exists a sequence of points
$z_{t}=r_{t}e^{i\theta} (t=1, 2, \dots)$ such that as 
$r_{t}\to \infty$, $f'(z_{t})\to \infty$
and
\begin{equation}\label{dddaae}
 | \frac{f(z_{t})}{f'(z_{t})} | \leq r_{t}(1+o(1)).
\end{equation}
By \eqref{qaaaaa}, we obtain that
\begin{equation}\label{dddaaf}
  -[ P_1(e^{\alpha z_{t}})+P_2(e^{-\alpha z_{t}}) ]
=\frac{f''(z_{t})}{f'(z_{t})}+
  [ Q_1(e^{\beta z_{t}})+Q_2(e^{-\beta z_{t}})] \cdot \frac{f(z_{t})}{f'(z_{t})}.
\end{equation}
From $\delta(\alpha z, \theta)>0$, we have
\begin{gather}\label{dddaag}
 | P_1(e^{\alpha z_{t}})+P_2(e^{-\alpha z_{t}}) |\geq
  (1-\varepsilon)|a_{1m_1}|e^{m_1\delta(\alpha z_{t}, \theta)r_{t}}(1-o(1)),\\
\label{dddaah}
 | Q_1(e^{\beta z_{t}})+Q_2(e^{-\beta z_{t}}) | \leq
  Me^{n_1\delta(\beta z_{t}, \theta)r_{t}}(1+o(1)).
\end{gather}
Substituting \eqref{dddaad}, \eqref{dddaae}, \eqref{dddaag} and \eqref{dddaah}
in \eqref{dddaaf}, we obtain 
\begin{equation}\label{dddaai}
\begin{aligned}
&(1-\varepsilon)|a_{1m_1}|e^{m_1\delta(\alpha z_{t}, \theta)r_{t}}(1-o(1))\\
&\leq r_{t}^{\sigma-1+\varepsilon} 
  +  Me^{n_1\delta(\beta z_{t}, \theta)r_{t}}(1+o(1))r_{t} (1+o(1))  \\
&\leq  Mr_{t}^{\sigma+\varepsilon}e^{\frac{m_1}{c_1}\delta(\alpha z_{t}, 
\theta)r_{t}} (1+o(1)).
\end{aligned}
\end{equation}
Since $c_1>1$, $\delta(\alpha z_{t}, \theta)>0$, when $r_{t}\to \infty$, 
\eqref{dddaai} is a contradiction. Hence $|f'(re^{i\theta})|\leq C$. So
\begin{equation}\label{dddaaj}
|f(re^{i\theta})|\leq Cr.
\end{equation}
\smallskip

\noindent\textbf{Case 2.}
 If $\delta(\alpha z, \theta)<0$, then 
$\delta(\beta z,\theta)=\frac{m_2}{c_2n_2}\delta(\alpha z,\theta)<0$,
$\delta(-\alpha z, \theta)>0$ and $\delta(-\beta z,\theta)>0$. 
Using the similar method as above, we can obtain that
\begin{equation}\label{dddaak}
 |f(re^{i\theta})|\leq Cr.
\end{equation}
Since the linear measure of $E\cup H$ is zero, by \eqref{dddaaj}, \eqref{dddaak} 
and Lemma \ref{lem5}, we know that $f(z)$
is a polynomial, which contradicts the assumption that $f(z)$ is transcendental. 
Therefore $\sigma(f)=\infty$.

Second step. We prove that \eqref{qaaaaa} has no non-trivial subnormal solution. 
Now suppose that \eqref{qaaaaa} has a non-trivial subnormal solution $f_0$.
 By the conclusion in the first step, $\sigma(f_0)=\infty$.
By Lemma \ref{lem6}, we see that $\sigma_2(f_0)\leq 1$. Set $\sigma_2(f_0)=\omega \leq 1$.
 By Lemma \ref{lem8}, we see that there exists a subset $E_1\subset(1, \infty)$ having 
finite logarithmic measure and a constant $B>0$ such that
for all $z$ satisfying $|z|=r\not\in [0,1]\cup E_1$, we have
\begin{equation}\label{dddaal}
 | \frac{f_0^{(j)}(z)}{f_0(z)} | \leq B [T(2r,f_0)]^3, \quad (j=1,2).
\end{equation}
From the Wiman-Valiron theory, there is a set $E_2\subset (1, \infty)$ 
having finite logarithmic measure,
so we can choose $z$ satisfying $|z|=r \not\in E_2$ and $|f_0(z)|=M(r,f_0)$.
 Thus, we have
\begin{equation}\label{dddaam}
   \frac{f_0^{(j)}(z)}{f_0(z)}  = \big( \frac{\upsilon(r)}{z} \big)^{j}(1+o(1)),
 \quad j=1,2,
\end{equation}
where $\upsilon(r)$ is the central index of $f_0(z)$.

By Lemma \ref{lem7}, we see that there exists a sequence $\{z_n=r_ne^{i\theta_n}\}$ such that
$|f_0(z_n)|=M(r_n,f_0)$, $\theta_n\in [0,2\pi)$, 
$\lim_{n\to \infty} \theta_n=\theta_0\in[0,2\pi)$,
$r_n\not\in [0,1]\cup E_1\cup E_2$, $r_n\to \infty$, and if $\omega>0$, 
we see that for any given $\varepsilon_1$  $(0<\varepsilon_1<\omega)$, 
and for sufficiently large $r_n$,
\begin{equation}\label{dddaan}
 \exp\{r_n^{\omega-\varepsilon_1}\}< \upsilon(r_n)
<\exp\{r_n^{\omega+\varepsilon_1}\},
\end{equation}
and if $\omega=0$, then by $\sigma(f_0)=\infty$ and Lemma \ref{lem7},
 we see that for any given $\varepsilon_2$ $(0<\varepsilon_2<1/2)$,
and for any sufficiently large $M$, as $r_n$ is sufficiently large,
\begin{equation}\label{dddaao}
 r_n^{M}< \upsilon(r_n)<\exp\{r_n^{\varepsilon_2}\}.
\end{equation}
From \eqref{dddaan} and \eqref{dddaao}, we obtain that
\begin{equation}\label{jjjjjj}
\upsilon(r_n)>r_n, \quad  r_n\to \infty.
\end{equation}
For $\theta_0$, let 
$\delta=\delta(\alpha z, \theta_0)=|\alpha|\cos(\arg \alpha+\theta_0)$, then 
$\delta <0$, or $\delta>0$, or $\delta=0$.
We divide this proof into three cases.
\smallskip

\noindent\textbf{Case 1.}
 $\delta >0$. By $\theta_n\to\theta_0$, we see that
there is a constant $N(>0)$ such that, as $n>N$, $\delta(\alpha z_n, \theta_n)>0$. 
Since $f_0$ is a subnormal solution, for any  given 
$\varepsilon (0< \varepsilon <\frac{1}{12}(1-\frac{1}{c_1})
\delta(\alpha z_n, \theta_n))$, we have
\begin{equation}\label{dddaap}
[T(2r_n,f_0)]^3\leq e^{6\varepsilon r_n}
\leq e^{\frac{1}{2}(1-\frac{1}{c_1})\delta(\alpha z_n, \theta_n)r_n}.
\end{equation}
By \eqref{dddaal}, \eqref{dddaam}, \eqref{dddaap},
 we have
\begin{equation}\label{dddaaq}
\begin{aligned}
  \big(\frac{\upsilon(r_n)}{r_n} \big)^{j}(1+o(1))
&= |\frac{f_0^{(j)}(z_n)}{f_0(z_n)}| \\
&\leq  B[T(2r_n, f_0)]^3  \\
&\leq  Be^{\frac{1}{2}(1-\frac{1}{c_1})\delta (\alpha z_n, \theta_n)r_n},  \quad 
 j=1,2.
\end{aligned}
\end{equation}
Since $\delta(\alpha z_n, \theta_n)>0$, from \eqref{qaaaaa}, \eqref{dddaam}, 
we obtain that
\begin{equation}\label{dddaar}
\begin{aligned}
 & (1-\varepsilon)\frac{\upsilon(r_n)}{r_n}|a_{1m_1}|
 e^{m_1\delta(\alpha z_n, \theta_n)r_n}(1-o(1))\\
&\leq|\frac{f_0'(z_n)}{f_0(z_n)}
 \big(P_1(e^{\alpha z_n})+P_2(e^{-\alpha z_n}) \big)|  \\
  & = | \frac{f''_0(z_n)}{f_0(z_n)}+[Q_1(e^{\beta z_n})+Q_2(e^{-\beta z_n}) ] | \\
  & \leq \big(\frac{\upsilon(r_n)}{r_n}\big) ^{2}(1+o(1))
 +(1+\varepsilon)|b_{1n_1}|e^{n_1\delta(\beta z_n, \theta_n)r_n}(1+o(1))\\
  &\leq M_1\big(\frac{\upsilon(r_n)}{r_n}\big) ^{2}
e^{\frac{m_1}{c_1}\delta(\alpha z_n, \theta_n)r_n}(1+o(1)).
 \end{aligned}
 \end{equation}
From \eqref{dddaaq} and \eqref{dddaar}, we can obtain
\begin{equation}\label{dddaas}
\begin{aligned}
&(1-\varepsilon)|a_{1m_1}|e^{m_1(1-\frac{1}{c_1})
 \delta(\alpha z_n, \theta_n)r_n}(1-o(1))   \\
&\leq  M_1Be^{\frac{1}{2}(1-\frac{1}{c_1})
 \delta(\alpha z_n,\theta_n)r_n}(1+o(1)).
\end{aligned}
\end{equation}
Since $c_1>1$ and $m_1\geq 1$, we see that \eqref{dddaas} is a contradiction.
\smallskip

\noindent\textbf{Case 2.}
 $\delta <0$. By $\theta_n\to\theta_0$, we see that
there is a constant $N(>0)$ such that, as $n>N$, $\delta(\alpha z_n, \theta_n)<0$. 
Since $f_0$ is a subnormal solution, for any  given $\varepsilon$
$ (0< \varepsilon <-\frac{1}{12}(1-\frac{1}{c_2})\delta(\alpha z_n, \theta_n))$, we
have
\begin{equation}\label{dddaat}
[T(2r_n,f_0)]^3\leq e^{6\varepsilon r_n} 
\leq e^{-\frac{1}{2}(1-\frac{1}{c_2})\delta(\alpha z_n, \theta_n)r_n}.
\end{equation}
By \eqref{dddaal}, \eqref{dddaam}, \eqref{dddaat} we have
\begin{equation}\label{dddaav}
\begin{aligned}
\big(\frac{\upsilon(r_n)}{r_n} \big)^{j}(1+o(1))
&= |\frac{f_0^{(j)}(z_n)}{f_0(z_n)}|\leq B[T(2r_n, f_0)]^3 \\
&\leq  Be^{-\frac{1}{2}(1-\frac{1}{c_2})\delta (\alpha z_n, \theta_n)r_n},  \quad  
j=1,2.
\end{aligned}
\end{equation}
By  \eqref{dddaam} and \eqref{qaaaaa}, we obtain
\begin{equation}\label{dddaau}
\begin{aligned}
 & (1-\varepsilon)\frac{\upsilon(r_n)}{r_n}|a_{2m_2}|e^{-m_2
 \delta(\alpha z_n, \theta_n)r_n}(1-o(1))\\
&\leq|\frac{f_0'(z_n)}{f_0(z_n)}\left(P_1(e^{\alpha z_n})+P_2(e^{-\alpha z_n}) \right)|  \\
& = | \frac{f''_0(z_n)}{f_0(z_n)}+[Q_1(e^{\beta z_n})+Q_2(e^{-\beta z_n}) ] | \\
& \leq \big(\frac{\upsilon(r_n)}{r_n}\big) ^{2}(1+o(1))+(1+\varepsilon)|b_{2n_2}|e^{-n_2\delta(\beta z_n, \theta_n)r_n}(1+o(1)) \\
& \leq M_2\big(\frac{\upsilon(r_n)}{r_n}\big) ^{2}e^{-\frac{m_2}{c_2}\delta(\alpha z_n, \theta_n)r_n}(1+o(1)).
 \end{aligned}
 \end{equation}
From \eqref{dddaav} and \eqref{dddaau}, we can deduce that
\begin{equation}\label{dddaaw}
\begin{aligned}
&(1-\varepsilon)|a_{2m_2}|e^{-m_2(1-\frac{1}{c_2})
 \delta(\alpha z_n, \theta_n)r_n}(1-o(1))  \\
&\leq  M_2Be^{-\frac{1}{2}(1-\frac{1}{c_2})\delta(\alpha z_n,\theta_n)r_n}(1+o(1)).
\end{aligned}
\end{equation}
Since $c_2>1$ and $m_2\geq 1$, we see that \eqref{dddaaw} is a contradiction.
\smallskip

\noindent\textbf{Case 3.}
 $\delta=0$. Then 
$\theta_0\in H=\{\theta|\theta\in[0,2\pi), \delta(\alpha z, \theta)=0\}$.
Since $\theta_n\to \theta_0$, for any given $\varepsilon>0$, we see that there
is an integer $N\, (>0)$, as $n>N$, 
$\theta_n\in [\theta_0-\varepsilon, \theta_0+\varepsilon]$ and
$z_n=r_ne^{i\theta_n}\in \overline{\Omega}=\{z:\theta_0-\varepsilon 
\leq \arg z \leq \theta_0+\varepsilon\}$.
By Lemma \ref{lem8}, there exists a subset $E_{3}\subset (1,\infty)$ having finite 
logarithmic measure and a constant $B>0$, such that for all $z$ satisfying
 $|z|=r\not\in[0,1]\cup E_{3}$, we have
\begin{equation}\label{dddaax}
  |\frac{f_0''(z)}{f_0'(z)} |\leq B[T(2r,f_0')]^{2}.
\end{equation}
Now we consider the growth of $f_0(re^{i\theta})$ on a ray
$\arg z=\theta \in \overline{\Omega}\setminus \{\theta_0\}$.
Denote $\Omega_1=[\theta_0-\varepsilon, \theta_0)$,  
$\Omega_2=(\theta_0, \theta_0+\varepsilon]$.
We can easily see that when $\theta_1\in \Omega_1, \theta_2\in \Omega_2$, then
$\delta(\alpha z, \theta_1)\cdot\delta(\alpha z, \theta_2)<0$.
 Without loss of generality, we suppose that
$\delta(\alpha z, \theta)>0 \, (\theta\in\Omega_1)$ and 
$\delta(\alpha z, \theta)<0 \, (\theta\in\Omega_2)$.

Since when $\theta\in \Omega_1$, $\delta(\alpha z, \theta)>0$.
 Recall $f_0$ is subnormal, then
for any given $\varepsilon$ 
$(0<\varepsilon<\frac{1}{8}(1-\frac{1}{c_1})\delta(\alpha z, \theta))$,
\begin{equation}\label{dddaay}
[T(2r,f_0')]^{2}\leq  e^{4\varepsilon r}
\leq e^{\frac{1}{2}(1-\frac{1}{c_1})\delta(\alpha z, \theta)r}.
\end{equation}
We assert that $|f_0'(re^{i\theta})|$ is bounded on the ray $\arg z=\theta$. 
If $|f_0'(re^{i\theta})|$ is unbounded on the ray $\arg z=\theta$, 
then by Lemma \ref{lem4}, there exists a sequence $\{y_j=R_je^{i\theta}\}$
such that $R_j\to \infty$, $f_0'(y_j)\to \infty$ and
\begin{equation}\label{dddaaz}
| \frac{f_0(y_j)}{f_0'(y_j)} | \leq R_j(1+o(1)).
\end{equation}
By \eqref{dddaax}, \eqref{dddaay}, we see that for sufficiently large $j$,
\begin{equation}\label{dddbaa}
 |\frac{f_0''(y_j)}{f_0'(y_j)} |\leq B[T(2R_j,f_0')]^{2} 
\leq Be^{\frac{1}{2}(1-\frac{1}{c_1})\delta(\alpha y_j,\theta)R_j}.
\end{equation}
By  \eqref{qaaaaa}, we deduce that
\begin{equation}\label{dddbab}
\begin{aligned}
&(1-\varepsilon)|a_{1m_1}|e^{m_1\delta(\alpha y_j, \theta)R_j}(1-o(1))\\
&\leq|-\left(P_1(e^{\alpha y_j})+P_2(e^{-\alpha y_j})\right)  |  \\
&\leq  | \frac{f_0''(y_j)}{f_0'(y_j)}  |+ | Q_1(e^{\beta y_j})+
  Q_2(e^{-\beta y_j}) |\cdot |\frac{f_0(y_j)}{f_0'(y_j)}|  \\
&\leq  C_1e^{\frac{1}{2}(1-\frac{1}{c_1})\delta(\alpha y_j, \theta)R_j}
 e^{n_1\delta(\beta y_j,\theta)R_j}R_j(1+o(1))\\
&\leq  C_1e^{[\frac{1}{2}(1-\frac{1}{c_1})+\frac{m_1}{c_1}]\delta(\alpha y_j, 
\theta)R_j}R_j(1+o(1)).
\end{aligned}
\end{equation}
Since $\delta(\alpha y_j, \theta)>0$, $c_1>1$, we know that when 
$R_j\to \infty$, \eqref{dddbab} is a contradiction.
Hence
\begin{equation}\label{dddbac}
|f_0(re^{i\theta})|\leq Cr,
\end{equation}
on the ray $\arg z=\theta \in \Omega_1$.

When
 $\theta\in \Omega_2$, $\delta(\alpha z, \theta)<0$. Recall $f_0$ is subnormal, then
for any given $\varepsilon$ 
$(0<\varepsilon<-\frac{1}{8}(1-\frac{1}{c_2})\delta(\alpha z, \theta))$,
\begin{equation}\label{dddbad}
 [T(2r,f_0')]^{2}\leq  e^{4\varepsilon r}
\leq e^{-\frac{1}{2}(1-\frac{1}{c_2})\delta(\alpha z, \theta)r}.
\end{equation}

We assert that $|f_0'(re^{i\theta})|$ is bounded on the ray $\arg z=\theta$. 
If $|f_0'(re^{i\theta})|$
is unbounded on the ray $\arg z=\theta$, using the similar proof as above,
 we can obtain that
\begin{equation}\label{dddbae}
\begin{aligned}
 & (1-\varepsilon) |a_{2m_2}|e^{-m_2(1-\frac{1}{c_2})
\delta(\alpha y_j, \theta)R_j}(1-o(1))\\
  &\leq   C_2e^{-\frac{1}{2}(1-\frac{1}{c_2})
\delta(\alpha y_j,\theta)R_j}R_j(1+o(1))
 \end{aligned}
\end{equation}
Since $\delta(\alpha y_j, \theta)<0$ and $c_2>1$, we know that when
 $R_j\to \infty$, \eqref{dddbae} is a contradiction. Hence
\begin{equation}\label{dddbaf}
|f_0(re^{i\theta})|\leq Cr,
\end{equation}
on the ray $\arg z=\theta \in \Omega_2$.
By \eqref{dddbac}, \eqref{dddbaf}, we see that $|f_0(re^{i\theta})|$ satisfies
\begin{equation}\label{dddbag}
|f_0(re^{i\theta})|\leq Cr,
\end{equation}
on the ray $\arg z=\theta \in \overline{\Omega}\setminus \{\theta_0\}$.
However, since $f_0$ is transcendental and $\{z_n=r_ne^{i\theta_n}\}$ 
satisfies $|f_0(z_n)|=M(r_n,f_0)$,
we see that for any large $N(>2)$, as $n$ is sufficiently large,
\begin{equation}\label{dddbah}
|f_0(z_n)|=|f_0(r_ne^{i\theta_n})|\geq r_n^{N}.
\end{equation}
Since $z_n\in \overline{\Omega}$, by \eqref{dddbag}, \eqref{dddbah},
 we see that for sufficiently,large $n$,
$$\theta_n=\theta_0.$$
Thus for sufficiently large $n$, $\delta(\alpha z_n, \theta_n)=0$ and
\begin{equation}\label{dddbai}
 |P_1(e^{\alpha z_n})+P_2(e^{-\alpha z_n}) |\leq C, \quad  
 |Q_1(e^{\beta z_n})+Q_2(e^{-\beta z_n}) |\leq C.
\end{equation}
By \eqref{qaaaaa}, \eqref{dddaam}, we obtain that
\begin{equation}\label{dddbaj}
\begin{aligned}
&-\big(\frac{\upsilon(r_n)}{z_n}\big)^{2}(1+o(1))\\
&= \left(P_1(e^{\alpha z_n})+P_2(e^{-\alpha z_n})\right)
  \big(\frac{\upsilon(r_n)}{z_n} \big)(1+o(1)) 
+[Q_1(e^{\beta z_n})+Q_2(e^{-\beta z_n}) ].
\end{aligned}
\end{equation}
By \eqref{dddbai}, \eqref{dddbaj} and \eqref{jjjjjj} we obtain that
\begin{equation}\label{dddbak}
 \upsilon(r_n)\leq 2Cr_n,
\end{equation}
by \eqref{dddaan} (or \eqref{dddaao}), we see that \eqref{dddbak} is a contradiction.
Hence \eqref{qaaaaa} has no non-trivial subnormal solution.

Third step. We prove that all solutions of \eqref{qaaaaa} satisfies 
$\sigma_2(f)=1$. If
there is a solution $f_1$ satisfying $\sigma_2(f_1)<1$, then $\sigma_{e}(f_1)=0$, 
that is to say $f_1$ is subnormal, but this contradicts the conclusion in step 2.
 Hence $\sigma_2(f)=1$. This completes the proof of Theorem \ref{thm3}.


\subsection*{Acknowledgements}
The authors would like to thank the editor and the referee for their valuable 
suggestions. This work was  supported by the NNSF of China 
(No. 11171013 and No. 11371225).

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