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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 183, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/183\hfil Growth of solutions]
{Growth of solutions to linear complex differential equations
in an angular region}

\author[N. Wu\hfil EJDE-2013/183\hfilneg]
{Nan Wu}  % in alphabetical order

\address{Nan Wu \newline
Department of Mathematics, School of Science,
China University of Mining and Technology (Beijing), Beijing 100083, China}
\email{wunan2007@163.com}

\thanks{Submitted April 26, 2013. Published August 10, 2013.}
\subjclass[2000]{30D10, 30D20, 30B10, 34M05}
\keywords{Meromorphic solutions; order of a function; angular region}

\begin{abstract}
 In this article, we consider the growth of solutions of higher-order
 linear differential equations in an angular region instead of the
 complex plane.
\end{abstract}

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

\section{Introduction and statement of results}

We assume that the reader is familiar with the fundamental results
and standard notations of the Nevanlinna theory in the unit disk
$\Delta=\{z:|z|<1\}$ and in the complex plane $\mathbb{C}$ (see
\cite{Hayman,Tsuji,Yang}), such as $T(r,f), N(r,f), m(r,f),\delta(a,f)$. 
The order and lower order of $f$ in $\mathbb{C}$ or in
$\Delta$ are defined as follows:
\begin{gather*}
\rho_\mathbb{C}(f)=\limsup_{r\to\infty}\frac{\log T(r,f)}{\log r}, \quad
\rho_\Delta(f)=\limsup_{r\to1-}\frac{\log T(r,f)}{-\log(1-r)},\\
\mu_\mathbb{C}(f)=\liminf_{r\to\infty}\frac{\log T(r,f)}{\log r}, \quad
\mu_\Delta(f)=\liminf_{r\to1-}\frac{\log T(r,f)}{-\log(1-r)}.
\end{gather*}

The meromorphic functions in the unit disk can be divided into the
following three classes:
\begin{itemize}
\item[(1)]  bounded type: $T(r,f)=O(1)$ as $r\to 1-$;

\item[(2)]  rational type: 
$T(r,f)=O(\log (1-r)^{-1})$ and $f(z)$ does not belong to (1);

\item[(3)] admissible in $\Delta$:
$$
\limsup_{r\to 1-}\frac{T(r,f)}{-\log(1-r)}=\infty.
$$
\end{itemize} 
Meromorphic functions in the complex plane can also be divided into the
following three classes:
\begin{itemize}
\item[(1)] bounded type: $T(r,f)=O(1)$ as $r\to \infty$;

\item[(2)] rational type: $T(r,f)=O(\log r)$ and $f(z)$ does not belong to (1);

\item[(3)] admissible in $\mathbb{C}$:
$$
\limsup_{r\to \infty}\frac{T(r,f)}{\log r}=\infty.
$$
\end{itemize}

The growth of solutions to higher-order linear differential
equations in $\mathbb{C}$ and in $\Delta$ has been investigated by
many authors. Gundersen \cite{G} and Heittokangas
\cite{Heitiokangas} considered the growth of solutions of the 
second-order linear differential equations and obtained a theorem in
$\mathbb{C}$ and in $\Delta$ respectively as follows.

\begin{theorem}[\cite{G,Heitiokangas}] \label{thm1.1} 
Let $B(z)$ and $C(z)$ be the analytic coefficients of
the equation
\begin{equation}\label{1.1}
g''+B(z)g'+C(z)g=0
\end{equation}
in $\mathbb{C}$ (or in $\Delta$). If either (i) $\rho(B)<\rho(C)$ or
(ii) $B(z)$ is non-admissible while $C(z)$ is admissible, then all
solutions $g\not\equiv0$ of \eqref{1.1} are of infinite order of
growth.
\end{theorem}

Chen \cite{Chen} generalized Theorem \ref{thm1.1} as follows.

\begin{theorem}[\cite{Chen}] \label{thm1.4} 
Let $A_0(z),\dots, A_k(z)$ be the analytic coefficients of the
equation
\begin{equation}\label{1.3}
A_k(z)f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_0(z)f=0
\end{equation}
 in $\mathbb{C}$ (or in $\Delta$). If either (i) 
$\max_{1\leq j\leq k}\rho(A_j)<\rho(A_0)$,
 or (ii) $A_j(z)$ $(j=1,2,\dots,k)$ are
non-admissible while $A_0(z)$ is admissible, then all solutions
$f\not\equiv0$ of \eqref{1.3} are of infinite order of growth.
\end{theorem}

In 1994, Wu \cite{Wu,Wu1} used the Nevanlinna theory in an angle
to study the growth of solutions of the second-order linear
differential equation in an angular region and obtained the
following two theorems.

\begin{theorem}[\cite{Wu}] \label{thm1.6} 
Let $A(z)$ and $B(z)$ be meromorphic in $\mathbb{C}$ with
$\rho(A)<\rho(B)$ and $\delta(\infty,B)>0$. Then every
nontrivial meromorphic solution $f$ of the equation
\begin{equation}\label{1.4}
f''+A(z)f'+B(z)f=0
\end{equation}  
has infinite order. Furthermore, if $\rho(B)\leq1/2$ and
$\delta(\infty,B)=1$, then $\rho_{\alpha,\beta}(f)=+\infty$ for
every angular region $\Omega(\alpha,\beta)$.
\end{theorem}

\begin{theorem}[\cite{Wu1}]\label{thm1.5} 
Let $A(z)$ and $B(z)$ be analytic on
$\overline{\Omega}(\alpha,\beta)$. If for any $K>0$, the measure of
$$ 
\Big\{ \theta:  \alpha<\theta<\beta,
\liminf_{r\to\infty}\frac{(|A(re^{i\theta})|+1)r^K}{|B(re^{i\theta})|}=0\Big\}
$$
is larger than zero, then any solution $f\not\equiv0$ of \eqref{1.4}
has $\varrho_{\alpha,\beta}(f)=+\infty$.
\end{theorem}

In 2009, Xu and Yi \cite{Xu} generalized Theorem \ref{thm1.5} to the
case of linear higher order differential equation and obtained the
following theorem.

\begin{theorem}\cite{Xu} \label{thm1.7}
Let $A_j(z)(j=0,1,\dots,k-1)$ be analytic on
$\Omega(\alpha,\beta)(0<\beta-\alpha\leq2\pi)$, if for any $K>0$ the
$\theta$'s which satisfy $\alpha\leq\theta\leq\beta$ and
\begin{equation}\label{1.6}
\liminf_{r\to\infty}\frac{(|A_1(re^{i\theta})|+\dots
+|A_{k-1}(re^{i\theta})|+1)r^K}{|A_0(re^{i\theta})|}=0
\end{equation}
form a set of positive measure. Then for every solution
$f\not\equiv0$ of \eqref{1.6} we have
$\varrho_{\alpha,\beta}(f)=+\infty$.
\end{theorem}

\begin{remark} \rm
The order $\varrho_{\alpha,\beta}(f)$ in Theorems
\ref{thm1.5} and \ref{thm1.7}  is defined by
$$
\varrho_{\alpha,\beta}(f)=\limsup_{r\to\infty}
\frac{\log\log M(r,\overline{\Omega},f)}{\log r},
$$
where
$M(r,\overline{\Omega},f)=\max_{\alpha\leq\theta\leq\beta}|f(re^{i\theta})|$
and $f\not\equiv0$ is a function analytic on the set
$\overline{\Omega}(\alpha,\beta)=\{z:\alpha\leq\arg
z\leq\beta\}(0<\beta-\alpha\leq2\pi)$. The order
$\rho_{\alpha,\beta}(f)$ in this paper is different from
$\varrho_{\alpha,\beta}(f)$.
\end{remark}

It is natural to pose the following question:
\begin{quote}
 How does the solutions of linear differential
equations with analytic or meromorphic coefficients grow in an
angular region?
\end{quote}

Before stating our results, we give some notation and definitions
of a meromorphic function in an angular region
$\Omega(\alpha,\beta)=\{z:\alpha<\arg z<\beta\}$. 
In this article, $\Omega$ usually denotes the angular region 
$\Omega(\alpha,\beta)$ and 
$ \Omega_\varepsilon=\{z:\alpha+\varepsilon<\arg
z<\beta-\varepsilon\}$, where 
$0<\varepsilon<(\beta-\alpha)/2$. Let
$f(z)$ be a meromorphic function on
$\overline{\Omega}(\alpha,\beta)=\{z:\alpha\leq\arg z\leq\beta\}$.
Recall the definition of Ahlfors-Shimizu characteristic in an
angular region (see \cite{Tsuji}). Set
$\Omega(r)=\Omega(\alpha,\beta)\cap\{z:0<|z|<r\}=\{z:\alpha<\arg
z<\beta,0<|z|<r \}$. Define
\begin{gather*}
\mathcal{S}(r,\Omega,f)=\frac{1}{\pi}\iint_{\Omega(r)} 
{\Big(|f'(z)|\over 1+|f(z)|^2\Big)^2 }d\sigma,
\\
\mathcal{T}(r,\Omega,f)=\int_0^r
{\mathcal{S}(t,\Omega,f)\over t}dt.
\end{gather*}
The order and lower order of $f$ on $\Omega$ are defined as
follows (see pp.93 in \cite{Zheng}):
$$
\rho_\Omega(f)=\limsup_{r\to \infty}\frac{\log \mathcal {T}(r,\Omega,f)}{\log
r}, \quad 
\mu_\Omega(f)=\liminf_{r\to \infty}\frac{\log \mathcal
{T}(r,\Omega,f)}{\log r}.
$$ 
We remark that the order
$\rho_\Omega(f)$ of a meormorphic function $f$ in an angular region
given here is reasonable, because $\mathcal {T}(r, \mathbb{C},
f)=T(r, f)+O(1)$ (see pp.20 in \cite{Goldberg}).

From \cite[Theorem 2.7.6]{Zheng}, we know that if
$\varrho_{\alpha,\beta}(f)=+\infty$, then
$\rho_{\alpha,\beta}(f)=+\infty$; if
$\rho_{\alpha+\varepsilon,\beta-\varepsilon}(f)=+\infty$ for some
$0<\varepsilon<\frac{\beta-\alpha}{2}$, then
$\varrho_{\alpha,\beta}(f)=+\infty$.

Now, we state our results in the following theorems.

\begin{theorem}\label{thm1.3}
Let $A(z)$ be analytic in an angular region $\Omega=\{z: \alpha<\arg
z<\beta\}(0<\beta-\alpha<2\pi)$ satisfying
\begin{equation}\label{1.5}
\limsup_{r\to\infty}\frac{\mathcal{T}(r,\Omega_\varepsilon,A)}{r^\omega\log
r}=\infty,
\end{equation}
where $\Omega_\varepsilon=\{z:\alpha+\varepsilon<\arg z<\beta-\varepsilon\}$, 
$0<\varepsilon<\frac{\beta-\alpha}{2}$, 
$\omega=\pi/(\beta-\alpha)$.
Then, all solutions $f\not\equiv0$ of the equation $f^{(k)}+A(z)f=0$
have the order $\rho_\Omega(f)=+\infty$.
\end{theorem}

\begin{theorem}\label{thm1.2}
Let $A_0, \dots, A_k$ be analytic in an angular region 
$\Omega=\{z: \alpha<\arg z<\beta\}(0<\beta-\alpha<2\pi)$. 
If either (i) for any small $0<\varepsilon<\frac{\beta-\alpha}{2}$, 
we have $ \rho_{\Omega}(A_j)<\rho_{\Omega_\varepsilon}(A_0)-\omega$
$(j=1,2,\dots,k)$,
or $(ii) A_j(z)$ $(j=1,2,\dots,k)$ satisfy 
$\mathcal{T}(r,\Omega, A_j)=O(\log r)$ while $A_0(z)$ satisfies
\begin{equation}\label{1.2}
\limsup_{r\to\infty}\frac{\mathcal{T}(r,\Omega_\varepsilon, A_0)}{r^\omega\log r}
=\infty,
\end{equation}
where $\Omega_\varepsilon=\{z:\alpha+\varepsilon<\arg
z<\beta-\varepsilon\}$, $\omega=\pi/(\beta-\alpha)$, then all
solutions $f\not\equiv0$ of \eqref{1.3} have the order
$\rho_\Omega(f)=+\infty$.
\end{theorem}

\section{Preliminaries}

In this section, we give some auxiliary results for the proof of the
theorems. The following result was proved in \cite{ZhangGH}.

\begin{lemma}[\cite{ZhangGH}] \label{lem2.1}
 The transformation
\begin{equation}\label{2.1}
\zeta(z)=\frac{(ze^{-i\theta_0})^{\pi/(\beta-\alpha)}-1}
{(ze^{-i\theta_0})^{\pi/(\beta-\alpha)}+1}
\quad \Big(\theta_0=\frac{\alpha+\beta}{2}\Big)
\end{equation} 
maps the angular region $X=\{z: \alpha<\arg z<\beta\}(0<\beta-\alpha<2\pi)$ 
conformally onto the unit disk $\{\zeta: |\zeta|<1\}$ in the $\zeta-$plane,
and maps $z=e^{i\theta_0}$ to $\zeta=0$. The image of
$X_\varepsilon(r)=\{z: 1\leq |z|\leq r, \alpha+\varepsilon\leq \arg
z\leq\beta-\varepsilon\}(0<\varepsilon<\frac{\beta-\alpha}{2})$ in
the $\zeta-$plane is contained in the disk $\{\zeta:|\zeta|\leq
h\}$, 
where
$$
h=1-\frac{\varepsilon}{\beta-\alpha}r^{-\frac{\pi}{\beta-\alpha}}.
$$
On the other hand, the inverse image of the disk 
$\{\zeta:|\zeta|\leq h\}(h<1)$ in the $z-$plane is contained in 
$X\cap\{z:|z|\leq r\}$,
where
$$
r=\Big(\frac{2}{1-h}\Big)^{(\beta-\alpha)/\pi}.
$$
The inverse transformation of \eqref{2.1} is
\begin{equation}\label{2.6}
z=e^{i\theta_0}\Big(\frac{1+\zeta}{1-\zeta}\Big)^{(\beta-\alpha)/\pi}.
\end{equation}
\end{lemma}

Using Lemma \ref{lem2.1}, we will prove the following lemma,
which to the best of our knowledge has not been published before.

\begin{lemma}\label{lem2.2}
Let $f(z)$ be meromorphic in an angular region
$\Omega=\{z:\alpha<\arg z<\beta\}(0<\beta-\alpha<2\pi)$. For any
small $\varepsilon>0$, write $\omega=\frac{\pi}{\beta-\alpha},
\eta=\frac{\varepsilon}{\beta-\alpha}$. Then the following
inequalities hold: 
\begin{gather}\label{2.2}
\mathcal {T}(r, \mathbb{C}, f(z(\zeta)))\leq
2\mathcal{T}\Big(\big(\frac{2}{1-r}\big)^{1/\omega},\Omega,f(z)\Big)+O(1),
\\
\label{2.3}
\mathcal {T}(r,\Omega_\varepsilon,f(z))
\leq \frac{r^\omega}{\omega\eta}\mathcal{T}(1-\eta
r^{-\omega},\mathbb{C}, f(z(\zeta)))+O(1),
\end{gather} 
where $z=z(\zeta)$ is the inverse transformation of \eqref{2.1}. 
Consequently,
\begin{equation}\label{2.4}
\rho_\Delta(f(z(\zeta)))\leq\frac{1}{\omega}\rho_\Omega(f(z)),\quad
\rho_{\Omega_\varepsilon}(f(z))\leq(\rho_\Delta(f(z(\zeta)))+1)\omega. 
\end{equation}
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.1}, for the inverse of the transformation
\eqref{2.1} it follows that
$$
z(\Delta_h)\subset \Omega\cap\Big\{z: |z|\leq \big(\frac{2}{1-h}\big)
^{1/\omega}\Big\}, \quad \text{where } 
\Delta_h=\{z:|z|<h\}.
$$
Since the term $\mathcal{S}$ is a conformal invariant, we derive
$$
\mathcal {S}(t, \mathbb{C}, f(z(\zeta)))\leq \mathcal
{S}\Big(\big(\frac{2}{1-t}\big)^{1/\omega}, \Omega, f(z)\Big).
$$
Dividing the above by $t$ and integrating from $0$ to $r$ gives
\begin{equation}
\begin{split}
\mathcal {T}(r, \mathbb{C}, f(z(\zeta)))
&=\int_0^r\frac{\mathcal {S}(t, \mathbb{C},
f(z(\zeta)))}{t}dt=\int_{1/2}^r\frac{\mathcal {S}(t, \mathbb{C},
f(z(\zeta)))}{t}dt+O(1)\\&\leq2\int_{1/2}^r\mathcal{S}(t,\mathbb{C},f(z(\zeta)))dt+O(1)\\
&\leq 2\int_{1/2}^r\mathcal
{S}\Big(\big(\frac{2}{1-t}\big)^{1/\omega}, \Omega,
f(z)\Big)dt+O(1)\\
&\leq2\int_1^{\left(\frac{2}{1-r}\right)^{1/\omega}}\frac{\mathcal
{S}(t,\Omega,f(z))}{t^{\omega+1}}dt+O(1)\\
&=2\mathcal{T}\Big(\big(\frac{2}{1-r}\big)^{1/\omega},\Omega,f(z)\Big)+O(1).
\end{split}
\end{equation}
Secondly, for the transformation \eqref{2.1}, we have
$$
\zeta(\{z: 1\leq |z|\leq r,
\alpha+\varepsilon\leq \arg z\leq\beta-\varepsilon\})\subset
\Delta_{(1-\eta r^{-\omega})}.
$$ 
Then,
$$
\mathcal {S}(r,\Omega_\varepsilon,f(z))\leq\mathcal{S}
\left(1-\eta r^{-\omega},\mathbb{C},f(z(\zeta))\right).
$$
Divide the above by $r$ and integrate from $1$ to $r$:
\begin{equation}
\begin{split}
\mathcal {T}(r,\Omega_\varepsilon,f(z))
&=\int_1^r\frac{\mathcal
{S}(t,\Omega_\varepsilon, f(z))}{t}dt+O(1)\\
&\leq\int_1^r\frac{\mathcal
{S}(1-\eta t^{-\omega}, \mathbb{C},
f(z(\zeta)))}{t}dt+O(1)\\&=\frac{1}{\omega}\int_{1-\eta}^{1-\eta
r^{-\omega}}\frac{\mathcal
{S}(x, \mathbb{C}, f(z(\zeta)))}{1-x}dx+O(1)\\
&\leq\frac{r^\omega}{\omega\eta}\int_{1-\eta}^{1-\eta
r^{-\omega}}\mathcal
{S}(x, \mathbb{C}, f(z(\zeta)))dx+O(1)\\
&\leq\frac{r^\omega}{\omega\eta}\int_{1-\eta}^{1-\eta
r^{-\omega}}\frac{\mathcal {S}(x, \mathbb{C},
f(z(\zeta)))}{x}dx+O(1)\\&=\frac{r^\omega}{\omega\eta}\mathcal{T}(1-\eta
r^{-\omega},\mathbb{C},f(z(\zeta)))+O(1).
\end{split}
\end{equation}
Using the definition of order, we  obtain \eqref{2.4}. 
The proof is complete.
\end{proof}

\begin{lemma}[\cite{Edrei1}]\label{lem2.3}
Let $f(z)$ be meromorphic in $\Omega=\{z:\alpha<\arg
z<\beta\}(0<\beta-\alpha<2\pi)$ and $z=z(\zeta)$ be the inverse
transformation of \eqref{2.1}. Write
$F(\zeta)=f(z(\zeta)),\psi(\zeta)=f^{(l)}(z(\zeta))$. Then,
\begin{equation}\label{2.5}
\psi(\zeta)=\sum_{j=1}^l\alpha_jF^{(j)}(\zeta)
\end{equation}
where the coefficients $\alpha_j$ are the polynomials (with
numerical coefficients) in the variables
$V(\zeta)(=\frac{1}{z'(\zeta)}),V'(\zeta),V''(\zeta),\dots$. Moreover, we have
$T(r,\alpha_j)=O(\log(1-r)^{-1}), j=1,2,\dots,l$.
\end{lemma}

Lemma \ref{lem2.3} can be proved by the same method of 
\cite[Lemma 1]{Edrei1}, where the lemma was stated for a different 
transformation:
\begin{equation}\label{2.7}
\zeta=\frac{(ze^{-i\theta_0})^\omega
-(ze^{-i\theta_0})^{-\omega}-\kappa}{(ze^{-i\theta_0})^\omega
 -(ze^{-i\theta_0})^{-\omega}+\kappa},\quad 
 \omega=\frac{\pi}{\beta-\alpha},
\end{equation}
where $\kappa$ is a positive parameter. However, the conformal
transformation \eqref{2.7} maps the sector
$\{z:|z|>1,\alpha<\arg z  theta<\beta\}(0<\beta-\alpha<2\pi)$ onto the unit
disk $\{\zeta:|\zeta|<1\}$ while the transformation \eqref{2.1} maps
the angular region $\{z:\alpha<\arg z <\beta\}(0<\beta-\alpha<2\pi)$
onto the unit disk $\{\zeta:|\zeta|<1\}$. For completeness, we give
the proof of Lemma \ref{lem2.3} using the method of 
\cite[Lemma 1]{Edrei1}.

\begin{proof}
Put 
$$
V(\zeta)=\frac{1}{z'(\zeta)}.
$$
By a simple calculation, we have
$$
f'(z(\zeta))=V(\zeta)F'(\zeta).
$$
An obvious induction shows that
$$
\psi(\zeta)=f^{(l)}(z(\zeta))=\sum_{j=1}^l\alpha_jF^{(j)}(\zeta)
$$
where the coefficients $\alpha_j$ are polynomials (with numerical coefficients) 
in the variables
$V, V', V'', \dots$.
Taking the derivative on both side of \eqref{2.6},
we obtain that
$$
\frac{dz}{d\zeta}=\frac{e^{i\theta_0}}{\omega}
\Big(\frac{1+\zeta}{1-\zeta}\Big)^{\frac{1}{\omega}-1}
\frac{2}{(1-\zeta)^2},\quad 
\omega=\frac{\pi}{\beta-\alpha},\theta_0=\frac{\alpha+\beta}{2}.
$$
Then
$$
\big|\frac{dz}{d\zeta}\big|\leq\frac{1}{\omega}\frac{2^{1/\omega}}{(1-|\zeta|)
^{\frac{1}{\omega}+1}}.
$$
Therefore,
$$
T(r,z'(\zeta))\leq\log M(r,z'(\zeta))\leq\Big(\frac{1}{\omega}+1\Big)
\log\frac{2}{1-r}+\log\frac{1}{\omega},
$$
By the first fundamental theorem,
$$
T\Big(r,\frac{1}{z'(\zeta)}\Big)
=T(r,z'(\zeta))+\log\frac{1}{|z'(0)|}
\leq\Big(\frac{1}{\omega}+1\Big)\log\frac{2}{1-r}+\log\frac{1}{\omega}
+\log2\omega.
$$
Thus,
$$
T(r,V(\zeta))=T\Big(r,\frac{1}{z'(\zeta)}\Big)
\leq\Big(\frac{1}{\omega}+1\Big)\log\frac{2}{1-r}+\log\frac{1}{\omega}
+\log2\omega,
$$
\begin{align*}
T(r,V^{(k)})&=m(r,V^{(k)})+N(r,V^{(k)})
\leq m\Big(r,\frac{V^{(k) }}{V}\Big)+m(r,V)+kN(r,V)\\
&\leq  m\Big(r,\frac{V^{(k)}}{V}\Big)+(k+1)T(r,V)
\leq O\Big(\log\frac{2}{1-r}\Big), \quad
k=1,2,\dots.
\end{align*}
In view of the coefficients $\alpha_j$ are polynomials (with
numerical coefficients) in the variables $V, V', V'', \dots$, we
have 
$$
T(r,\alpha_j)\leq O\Big(\log\frac{2}{1-r}\Big), \quad
j=1,2,\dots, l.
$$ 
The proof is complete.
\end{proof}

\section{Proof of Theorem \ref{thm1.3}}

Suppose that $f\not\equiv0$ is a solution of $f^{(k)}+A(z)f=0$ in
$\Omega$. Then $F(\zeta)=f(z(\zeta))$ is a solution of the
differential equation
\begin{equation}
\alpha_kF^{(k)}(\zeta)+\alpha_{k-1}F^{(k-1)}(\zeta)+\dots
+\alpha_1F'(\zeta)+B(\zeta)F(\zeta)=0
\end{equation}
in $\Delta$,
where $B(\zeta)=A(z(\zeta)), \alpha_j(j=1,2,\dots,k)$ are described
in Lemma \ref{lem2.3}. From the condition \eqref{1.5} and the
inequality \eqref{2.3}, we obtain that $B$ is admissible in
$\Delta$. By Lemma \ref{lem2.3}, we get that
$T(r,\alpha_j)=O(\log(1-r)^{-1})(j=1,2,\dots,k)$, so
$\alpha_j(j=1,2,\dots,k)$ are non-admissible in $\Delta$. By
Theorem \ref{thm1.4}, we have $\rho_\Delta(F)=\infty$. Combining
this with \eqref{2.4} leads to $\rho_\Omega(f)=\infty$. Theorem
\ref{thm1.3} follows.

\section{Proof of Theorem \ref{thm1.2}}

Suppose that $f\not\equiv0$ is a solution of \eqref{1.3} in
$\Omega$. In view of \eqref{2.5}, we have
\begin{align*}
&\sum_{i=1}^kA_i(z(\zeta))f^{(i)}(z(\zeta))+A_0(z(\zeta))f(z(\zeta))\\
&=\sum_{i=1}^kA_i(z(\zeta))\sum_{j=1}^i\alpha_jF^{(j)}(\zeta)
 +A_0(z(\zeta))f(z(\zeta))\\
&=\sum_{j=1}^k\alpha_j\sum_{i=j}^kA_i(z(\zeta))F^{(j)}(\zeta)
 +A_0(z(\zeta))f(z(\zeta)).
\end{align*} Then $F(\zeta)=f(z(\zeta))$ is a solution of the
differential equation
\begin{equation}
B_k(\zeta)F^{(k)}(\zeta)+B_{k-1}(\zeta)F^{(k-1)}(\zeta)+\dots
+B_1(\zeta)F'(\zeta)+B_0(\zeta)F(\zeta)=0
\end{equation}
in $\Delta$,
where $B_0(\zeta)=A_0(z(\zeta)),
B_j(\zeta)=\alpha_j\sum_{i=j}^kA_i(z(\zeta))(j=1,2,\dots,k)$. Since
$T(r,\alpha_j)=O(\log(1-r)^{-1})(j=1,2,\dots,k)$, it follows that
\begin{align*}
T(r, B_j)
&\leq T(r,\alpha_j)+\sum_{i=j}^kT(r,A_i(z(\zeta)))+O(1)\\
&=\sum_{i=j}^kT(r,A_i(z(\zeta)))+O(\log(1-r)^{-1}),\quad
j=1,2,\dots,k.
\end{align*} 
By Lemma \ref{lem2.2}, we have
\begin{gather*}
\frac{1}{\omega}\rho_{\Omega_\varepsilon}(A_0)\leq\rho_\Delta(B_0)+1,\\
\rho_\Delta(B_j)\leq \frac{1}{\omega}\max_{1\leq j\leq k}\rho_\Omega(A_j),\quad
\text{for } j=1,2,\dots,k.
\end{gather*}
Combining the above with the condition (i) gives
$$
\rho_\Delta(B_j)<\rho_\Delta(B_0), \quad\text{for } j=1,2,\dots,k.
$$
By Theorem \ref{thm1.4}, we have $\rho_\Delta(F)=\infty$. Combining
this with \eqref{2.4} leads to $\rho_\Omega(f)=\infty$.

In view of $\mathcal{T}(r,\Omega, A_j)=O(\log r)$ it follows that
$T(r,B_j)=O(\log(1-r)^{-1})$. From the condition \eqref{1.2} and the
inequality \eqref{2.3}, we obtain that $B_0$ is admissible in
$\Delta$. By Theorem \ref{thm1.4}, we have that
$\rho_\Delta(F)=\infty$. This leads to $\rho_\Omega(f)=\infty$.
Then Theorem \ref{thm1.2} follows.


\subsection*{Acknowledgements} 

The author would like to express his gratitude to the anonymous referee for
the valuable comments and suggestions in improving this paper.
The author is supported in part by the grant 11231009 from the  NSF of
China.

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