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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 177, 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 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/177\hfil Differential equations in the unit disc]
{Properties of solutions to linear differential equations with analytic
coefficients in \\ the unit disc}

\author[S. Hamouda \hfil EJDE-2012/177\hfilneg]
{Saada Hamouda}

\address{Saada Hamouda \newline
Laboratory of Pure and Applied Mathematics\\
University of Mostaganem, B.P. 227 Mostaganem, Algeria}
\email{hamoudasaada@univ-mosta.dz}

\thanks{Submitted July 6, 2012. Published October 12, 2012.}
\subjclass[2000]{34M10, 30D35}
\keywords{Complex differential equations; growth of solutions; unit disc}

\begin{abstract}
 In this article we study the growth of solutions  of linear
 differential equations with analytic coefficients in the unit disc. 
 Our  investigation is based on the behavior of the coefficients 
 on a neighborhood of a point on the boundary of the unit disc.
\end{abstract}

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

\section{Introduction and statement of results}

Throughout this paper, we assume that the reader is familiar with the
fundamental results and the standard notations of the Nevanlinna value
distribution theory of meromorphic function on the complex plane
$\mathbb{C}$ and in the unit disc $D=\{ z\in\mathbb{C}:| z| <1\}$
 (see \cite{haym,yang}). In addition, the order of meromorphic function
$f(z) $ in $D$ is defined by
\begin{equation*}
\sigma (f) =\limsup_{r\to 1^{-}}
\frac{\log^{+}T(r,f) }{\log \frac{1}{1-r}},
\end{equation*}
where $T(r,f) $ is the Nevanlinna characteristic
function of $f$; and for an analytic function $f(z) $ in $D$, we
have also the definition
\begin{equation*}
\sigma _{M}(f) =\limsup_{r\to 1^{-}}
\frac{\log ^{+}\log ^{+}M(r,f) }{\log \frac{1}{1-r}},
\end{equation*}
where $M(r,f) =\max_{| z| =r}|f(z) |$.
Tsuji \cite[p. 205]{tsu} states that
\begin{equation*}
\sigma (f) \leq \sigma _{M}(f) \leq \sigma (f) +1.
\end{equation*}
For example, the function $g(z) =\exp \{ \frac{1}{(
1-z) ^{\mu }}\} $ satisfies $\sigma (g) =\mu -1$ and
$\sigma _{M}(g) =\mu$.
Obviously, we have
\begin{equation*}
\sigma (f) <\infty \text{ if and only if }\sigma _{M}(
f) <\infty .
\end{equation*}

\begin{definition}[\cite{heitto}] \label{def1} \rm
 A meromorphic function $f$ in $D$ is called admissible if
\begin{equation*}
\limsup_{r\to 1^{-}} \frac{\log T(r,f) }{\log \frac{1}{1-r}}=\infty ;
\end{equation*}
and $f$ is called nonadmissible if
\begin{equation*}
\limsup_{r\to 1^{-}} \frac{\log T(r,f) }{\log \frac{1}{1-r}}<\infty .
\end{equation*}
\end{definition}

There are some similarities between results of linear differential equations
in the complex plane and the unit disc. For example, Heittokangas  \cite{heitto}
obtained the following results.

\begin{theorem}[\cite{heitto}] \label{thm1}
Let $A(z) $ and $B(z) $ be analytic functions in the unit disc.
 If $\sigma (A) <\sigma (B) $ or $A(z) $ is nonadmissible while $B(z) $
is admissible, then all solutions $f(z) \not\equiv 0$ of the
linear differential equation
\begin{equation}
f''+A(z) f'+B(z) f=0, \label{eq1}
\end{equation}
are of infinite order of growth.
\end{theorem}

\begin{theorem}[\cite{heitto}]  \label{thm2}
Let $A_0(z) ,\dots,A_{k-1}(z) $ be
analytic coefficients in the unit disc of the linear differential equation
\begin{equation}
f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots+A_0(z)f=0.  \label{eq2}
\end{equation}
Let $A_{d}(z) $ be the last coefficient not being an
 $\mathcal{H}$-function while the coefficients
$A_{d+1}(z) ,\dots,A_{k-1}(z)$ are $\mathcal{H}$-functions.
Then  possesses at most $d$ linearly
independent analytic solutions of finite order of growth.
\end{theorem}

These theorems are analogous of the following results respectively.

\begin{theorem}[\cite{gund1}] \label{thm3}
 Let $A(z) $ and $B(z) $ be entire functions.
If $\sigma (A) <\sigma (B) $ or $A(z) $ is a polynomial and $B(z) $ is transcendental,
then every solution $f(z) \not\equiv 0$ of \eqref{eq1} has infinite
order.
\end{theorem}

\begin{theorem}[\cite{frei}] \label{thm4}
 Let $A_0(z) ,\dots,A_{k-1}(z) $ be
entire functions. Let $A_{d}(z) $ be the last transcendental
coefficient in \eqref{eq2} while $A_{d+1}(z) ,\dots,A_{k-1}(z)$
are polynomials. Then \eqref{eq2} possesses at most $d$ linearly independent
entire solutions of finite order of growth.
\end{theorem}

In general, the study of growth of solutions of linear differential
equations in the complex plane or in the unit disc is based on the dominant of
some coefficient by using the order, iterated order, type and the degree;
see for example \cite{chen3,gund1,heit,kin}. In this paper, we will get out
of these methods by using only the behavior of the coefficients near a point
on the boundary of the unit disc. By this concept, we can study certain
class of linear differential equations with analytic coefficients in the
unit disc having the same order and type. We are motivated by certain
results in the complex plane concerning the linear differential equation
\begin{equation}
f''+A(z) e^{az}f'+B(z)e^{bz}f=0,  \label{1}
\end{equation}
where $A(z) $ and $B(z) $ are entire functions, see
for example \cite{ozaw,chen1,chen2,gund2}.
Chen \cite{chen1} proved that if $ab\neq 0$ and $\arg a\neq \arg b$ or
$a=cb$ $(0<c<1) $, then  every solution $f(z) \not\equiv 0$ of $\eqref{1} $
is of infinite order.

We will see that there are similarities and differences between our results
and those of the complex plane. In fact, we will prove the following results.

\begin{theorem}\label{t1}
Let $A(z) $ and $B(z) \not\equiv 0$ be
analytic functions in the unit disc. Suppose that $\mu >1$ is a real
constant, $b$ and $z_0$ are complex numbers such that $b\neq 0$,
 $| z_0| =1$. If $A(z) $ and $B(z) $ are analytic on $z_0$ then every solution
 $f(z)\not\equiv 0$ of the differential equation
\begin{equation}
f''+A(z) f'+B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}f=0,  \label{2}
\end{equation}
is of infinite order.
\end{theorem}

\begin{example} \label{examp1} \rm
Every solution $f(z) \not\equiv 0$ of the differential equation
\begin{equation*}
f''+e^{\frac{1}{(1+z) ^{\alpha }}}f'+e^{
\frac{1}{(1-z) ^{\beta }}}f=0,
\end{equation*}
is of infinite order, where $\alpha >1$ and $\beta >1$ are real constants.
We see that, in the case $\alpha =\beta $, the coefficients have the same
order and type.
\end{example}

\begin{theorem}\label{t2}
Let $A(z) $ and $B(z) \not\equiv 0$ be
analytic functions in the unit disc. Suppose that $\mu >1$ is a real
constant, $a,b$ and $z_0$ are complex numbers such that
$ab\neq 0$, $\arg a\neq \arg b$, $| z_0| =1$. If $A(z) $ and
$B(z) $ are analytic on $z_0$ then every solution $f(z) \not\equiv 0$ of
the differential equation
\begin{equation}
f''+A(z) e^{\frac{a}{(z_0-z) ^{\mu}}}f'+B(z) e^{\frac{b}{(z_0-z) ^{\mu }}
}f=0,  \label{3}
\end{equation}
is of infinite order.
\end{theorem}

\begin{theorem}\label{t3}
Let $A(z) $ and $B(z) \not\equiv 0$  be
analytic functions in the unit disc. Suppose that $\mu >1$ is a real
constant, $a,b$ and $z_0$ are complex numbers such that $ab\neq 0$,
$a=cb$ $(0<c<1) $, $| z_0| =1$. If $A(z) $ and $B(z) $ are analytic
on $z_0$ then every solution $f(z) \not\equiv 0$ of the differential equation
\begin{equation}
f''+A(z) e^{\frac{a}{(z_0-z) ^{\mu}}}f'+B(z) e^{\frac{b}{(z_0-z) ^{\mu }}
}f=0,  \label{4}
\end{equation}
is of infinite order
\end{theorem}

\begin{remark} \label{rmk1} \rm
In all these our theorems, it may happen that the order of growth of
 $A(z) $ and $B(z) $ is greater than $\mu $ not like
of the complex plane case.
\end{remark}

We can generalize our previous Theorems to the higher differential equations
as follows.

\begin{theorem}\label{t4}
Consider the linear differential equation
\begin{equation}
f^{(k) }+A_{k-1}(z) f^{(k-1)
}+\dots+A_1(z) f'+B(z) e^{\frac{b}{(
z_0-z) ^{\mu }}}f=0,  \label{eq}
\end{equation}
where $\mu >1$ is a real constant, $b$ and $z_0$ are complex numbers such
that $b\neq 0$, $| z_0| =1$, $B(z) \not\equiv 0$, $A_0(z) ,\dots,A_{k-1}(z) $
are analytic functions in the unit disc such that either $A_{j}(z) $ is
analytic on $z_0$ or
$A_{j}(z) =B_{j}(z) e^{\frac{b_{j}}{(z_0-z) ^{\mu }}}$
with $B_{j}(z) $ is analytic on $z_0$ and $b_{j}=c_{j}b$ $(0<c_{j}<1) $ or
$\arg b_{j}\neq \arg b$ for at most one possibility. Then every solution
$f(z) \not\equiv 0$ of $\eqref{eq} $ is of infinite order.
\end{theorem}

\begin{remark} \label{rmk2} \rm
Throughout  this paper, we choose the principal branch of logarithm of
the function
$e^{\frac{\lambda }{(z_0-z) ^{\mu }}}\ (\lambda \in\mathbb{C}\backslash \{ 0\} ) $.
\end{remark}

\section{Preliminaries}

To prove our results we need the following lemmas.

\begin{lemma}[\cite{chyz-gund}] \label{lem1}
Let $f$ be a meromorphic function in the unit disc $D$ of
finite order $\sigma $. Let $\varepsilon >0$ be a constant; $k$ and $j$ be
integers satisfying $k>j\geq 0$. Assume that
 $f^{(j) }\not\equiv 0$. Then there exists a set $E\subset [ 0,1) $ which satisfies
 $
\int_{E}\frac{1}{1-r}dr<\infty $, such that for all $z\in D$
satisfying $| z| \notin E$, we have
\begin{equation*}
\big| \frac{f^{(k) }(z) }{f^{(j)
}(z) }\big| \leq \Big(\frac{1}{1-| z|
}\Big) ^{(k-j) (\sigma +2+\varepsilon ) }.
\end{equation*}
\end{lemma}

\begin{lemma} \label{lem2}
Let $A(z) $ be an analytic function on a point
$z_0\in\mathbb{C}$. Set
$g(z) =A(z) e^{\frac{a}{(z_0-z) ^{\mu }}}$,
($\mu >0\ $is a real constant), $a=\alpha +i\beta \neq 0$,
$z_0-z=Re^{i\varphi }$, $\delta _{a}(\varphi )
=\alpha \cos (\mu \varphi ) +\beta \sin (\mu \varphi) $, and
$H=\{ \varphi \in [ 0,2\pi ) :\delta_{a}(\varphi ) =0\}$,
(obviously, $H$ is of linear measure zero).
Then for any given $\varepsilon >0$ and  for any
$\varphi \in [ 0,2\pi ) \backslash H$, there exists
$R_0>0$ such that for $0<R<R_0$, we have:

(i) if $\delta _{a}(\varphi ) >0$, then
\begin{equation}
\exp \{ (1-\varepsilon ) \delta _{a}(\varphi )
\frac{1}{R^{\mu }}\} \leq | g(z) | \leq
\exp \{ (1+\varepsilon ) \delta _{a}(\varphi )
\frac{1}{R^{\mu }}\} ,  \label{e1}
\end{equation}

(ii) if $\delta _{a}(\varphi ) <0$, then
\begin{equation}
\exp \{ (1+\varepsilon ) \delta _{a}(\varphi )
\frac{1}{R^{\mu }}\} \leq | g(z) | \leq
\exp \{ (1-\varepsilon ) \delta _{a}(\varphi )
\frac{1}{R^{\mu }}\} .  \label{e2}
\end{equation}
\end{lemma}

\begin{proof}
We have
\begin{equation}
\big| e^{\frac{a}{(z_0-z) ^{\mu }}}\big| =\exp
\{ \delta _{a}(\varphi ) \frac{1}{R^{\mu }}\} .
\label{1l}
\end{equation}

If $z_0$ is a zero of order $m$ of $A(z) $, then there exist
$c_1>0,\ c_2>0$ such that
\begin{equation}
c_1R^{m}\leq | A(z) | \leq c_2R^{m},\ \
\text{for $z$ near enough  }z_0.  \label{2l}
\end{equation}
Using \eqref{1l}  and \eqref{2l} , we obtain
\begin{equation}
c_1R^{m}\exp \{ \delta _{a}(\varphi ) \frac{1}{R^{\mu }}\}
 \leq | g(z) | \leq c_2R^{m}\exp \{ \delta _{a}(\varphi ) \frac{1}{R^{\mu }}\} ,
\label{3l}
\end{equation}
for $z$ near enough $z_0$.

Now, if $z_0$ is not a zero of $A(z) $, then there exist $
c_1'>0,\ c_2'>0\ $such that
\begin{equation*}
c_1'\leq | A(z) | \leq
c_2',\ \ \text{for }z\text{ near enough }z_0,
\end{equation*}
and so
\begin{equation}
c_1'\exp \{ \delta _{a}(\varphi ) \frac{1}{
R^{\mu }}\} \leq | g(z) | \leq
c_2'\exp \{ \delta _{a}(\varphi ) \frac{1}{
R^{\mu }}\} ,  \label{4l}
\end{equation}
 for $z$ near enough $z_0$.
From $\eqref{3l} $ and $\eqref{4l} $, we can easily
obtain $\eqref{e1} $ and $\eqref{e2} $.
\end{proof}

\begin{remark} \label{rmk3} \rm
In general, we can write $\delta _{a}(\varphi ) =c\cos (
\mu \varphi +\varphi _0) $,
 where $c>0$, $\varphi _0\in [0,2\pi ) $. By this formula, it is easy
to prove that if $\mu >1$,
$\delta _{a}(\varphi ) $ changes its sign on each interval
$(\varphi _1,\varphi _2) $ of linear measure equal to $\pi $.
\end{remark}

\section{Proof of theorems}

\begin{proof}[Proof of Theorem \ref{t1}]
Suppose that $f\not\equiv 0$ is a solution of \eqref{2} of
finite order $\sigma (f) =\sigma <\infty $. Since $\mu >1$, By
Remark \ref{rmk3}, there exist $(\varphi _1,\varphi _2) \subset [0,2\pi ) $
such that for $z\in D$ and $\arg (z_0-z)=\varphi \in (\varphi _1,\varphi _2) $
we have $\delta_{b}(\varphi ) >0$.

From $\eqref{2} $, we obtain
\begin{equation}
\big| B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}\big|
 \leq | \frac{f''}{f}| +| A(z) | | \frac{f'}{f}| .  \label{p1}
\end{equation}
From Lemma \ref{lem1}, for a given $\varepsilon >0$ there exists a set
 $E\subset [ 0,1) $ which satisfies $\int_{E}\frac{1}{1-r}dr<\infty $,
such that for all $z\in D$ satisfying $| z| \notin E$, we
have
\begin{equation}
\big| \frac{f^{(k) }(z) }{f(z) }\big|
\leq \Big(\frac{1}{1-| z| }\Big) ^{k(\sigma +2+\varepsilon ) },\quad (k=1,2) .
\label{p2}
\end{equation}
From Lemma \ref{lem2}, for any given $0<\varepsilon <1$ and for for $z\in D$ and
$\arg (z_0-z) =\varphi \in (\varphi _1,\varphi_2) $ with $| z_0-z| =R$,
there exists $R_0>0 $ such that for $0<R<R_0$, we have
\begin{equation}
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} \leq | B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}| .  \label{p3}
\end{equation}
Since $A(z) $ is analytic on $z_0$,  for $z$ near enough $z_0$, we have
\begin{equation}
| A(z) | \leq M,\quad M>0.  \label{p4}
\end{equation}
Using \eqref{p2}, \eqref{p3} and \eqref{p4} in \eqref{p1}, we obtain
\begin{equation}
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )\frac{1}{R^{\mu }}\}
\leq (\frac{1}{1-| z| }) ^{2(\sigma +2+\varepsilon ) }
+M(\frac{1}{1-| z| }) ^{(\sigma +2+\varepsilon ) },
\label{p5}
\end{equation}
where $z\in D$, $| z| \notin E$, $\arg (z_0-z) =\varphi \in (\varphi _1,\varphi _2) $
 with $| z_0-z| =R$ and $0<R<R_0$. By the metric relations
in the triangle $(oz_0z) $, we have
$| z|^{2}=1+R^{2}-2R\cos \varphi ^{\ast }$ and then
\begin{equation}
1-| z| =R(\frac{2\cos \varphi ^{\ast }-R}{1+| z| }) .  \label{p6}
\end{equation}
For $z$ near enough  $z_0$ and by considering that $\varphi $ is
fixed, there exists certain $\varepsilon _0>0$ such that
\begin{equation}
\frac{2\cos \varphi ^{\ast }-R}{1+| z| }>\varepsilon_0;  \label{p7}
\end{equation}
we signal here that $0\leq \varphi ^{\ast }<\frac{\pi }{2}$. By combining
\eqref{p6} and \eqref{p7}, we obtain
\begin{equation}
\frac{1}{1-| z| }<\frac{1}{\varepsilon _0R}.  \label{p8}
\end{equation}
Now by \eqref{p5}  and \eqref{p8}, we obtain
\begin{equation*}
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} \leq M'(\frac{1}{\varepsilon _0R}
) ^{2(\sigma +2+\varepsilon ) },\quad M'>1,
\end{equation*}
which gives a contradiction as $R\to 0$.
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{t2}]
Suppose that $f\not\equiv 0$ is a solution of \eqref{3} of
finite order $\sigma (f) =\sigma <\infty $. Since
$\arg a\neq \arg b$ and $\mu >1$, then there exist
 $(\varphi _1,\varphi _2) \subset [ 0,2\pi ) $ such that for $z\in D$ and
 $\arg (z_0-z) =\varphi \in (\varphi _1,\varphi _2) $
we have $\delta _{b}(\varphi ) >0$ and $\delta _{a}(\varphi ) <0$.

From \eqref{3}, we obtain
\begin{equation}
\big| B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}\big|
\leq | \frac{f''}{f}| +| A(z) | | A(z) e^{\frac{a}{(z_0-z) ^{\mu }}}| | \frac{
f'}{f}| .  \label{p9}
\end{equation}
From Lemma \ref{lem1}, for a given $\varepsilon >0$ there exists a set
$E\subset[ 0,1) $ which satisfies $\int_{E}\frac{1}{1-r}dr<\infty $,
such that for all $z\in D$ satisfying $| z| \notin E$, we have
\begin{equation}
\big| \frac{f^{(k) }(z) }{f(z) }
\big| \leq \Big(\frac{1}{1-| z| }\Big)^{k(\sigma +2+\varepsilon ) },\quad
 (k=1,2) .\label{p10}
\end{equation}
From Lemma \ref{lem2}, for any given $0<\varepsilon <1$ and for for $z\in D$ and
$\arg (z_0-z) =\varphi \in (\varphi _1,\varphi_2) $ with $| z_0-z| =R$,
there exists $R_0>0 $ such that for $0<R<R_0$, we have
\begin{equation}
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} \leq \big| B(z) e^{\frac{b}{
(z_0-z) ^{\mu }}}\big| ,  \label{p11}
\end{equation}
and
\begin{equation}
\big| A(z) e^{\frac{a}{(z_0-z) ^{\mu }}}\big|
\leq \exp \{ (1-\varepsilon ) \delta_a (\varphi ) \frac{1}{R^{\mu }}\} .  \label{p12}
\end{equation}
Using \eqref{p9} -\eqref{p12}  and \eqref{p8}, we obtain
\[
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} 
\leq (\frac{1}{\varepsilon _0R}) ^{2(\sigma +2+\varepsilon ) }
+ (\frac{1}{\varepsilon _0R}) ^{(\sigma +2+\varepsilon) }
\exp \{ (1-\varepsilon ) \delta _{a}(\varphi) \frac{1}{R^{\mu }}\} ,
\]
where $z\in D$, $| z| \notin E$, $\arg (z_0-z) =\varphi \in (\varphi _1,\varphi _2) $
 with $| z_0-z| =R$ and $0<R<R_0$. A contradiction follows
as $R\to 0$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t3}]
Suppose that $f\not\equiv 0$ is a solution of \eqref{4} of
finite order $\sigma (f) =\sigma <\infty $. Since $\mu >1$, then
there exist $(\varphi _1,\varphi _2) \subset [ 0,2\pi) $ such that for
$z\in D$ and $\arg (z_0-z) =\varphi
\in (\varphi _1,\varphi _2) $ we have $\delta _{a}(\varphi ) >0$.

From \eqref{4}, we obtain
\begin{equation}
\big| B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}\big|
\leq | \frac{f''}{f}| +\big| A(z) | | A(z) e^{
\frac{a}{(z_0-z) ^{\mu }}}\big|
| \frac{f'}{f}| .  \label{p13}
\end{equation}
From Lemma \ref{lem1}, for a given $\varepsilon >0$ there exists a set
$E\subset[ 0,1) $ which satisfies $\int_{E}\frac{1}{1-r}dr<\infty $,
such that for all $z\in D$ satisfying $| z| \notin E$, we
have
\begin{equation}
| \frac{f^{(k) }(z) }{f(z) }
| \leq \Big(\frac{1}{1-| z| }\Big)^{k(\sigma +2+\varepsilon ) },\quad (k=1,2) .
\label{p14}
\end{equation}

From Lemma \ref{lem2}, for any $\varepsilon >0$ and for for $z\in D$ and $\arg
(z_0-z) =\varphi \in (\varphi _1,\varphi _2) $
 with $| z_0-z| =R$, there exists $R_0>0$ such
that for $0<R<R_0$, we have
\begin{equation}
\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} \leq \big| B(z) e^{\frac{b}{
(z_0-z) ^{\mu }}}\big| ,  \label{p15}
\end{equation}
and
\begin{equation}
\big| A(z) e^{\frac{a}{(z_0-z) ^{\mu }}}\big|
\leq \exp \{ (1+\varepsilon ) \delta_{a}(\varphi ) \frac{1}{R^{\mu }}\} .  \label{p16}
\end{equation}
By using $\eqref{p14} -\eqref{p16} $ in \eqref{p13}  and taking into account
 that $\delta _{a}(\varphi) =c\delta _{b}(\varphi ) $ $(0<c<1) $, we
obtain
\begin{align*} %\label{p17b}
&\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )
\frac{1}{R^{\mu }}\} 
&\leq \big(\frac{1}{\varepsilon _0R}\big)^{2(\sigma +2+\varepsilon ) }
+   \big(\frac{1}{\varepsilon _0R}\big) ^{(\sigma +2+\varepsilon
) }\exp \{ (1+\varepsilon ) c\delta _{b}(
\varphi ) \frac{1}{R^{\mu }}\} . 
\end{align*}
By taking $0<\varepsilon <\frac{1-c}{1+c}$,
we obtain a contradiction to $R\to 0$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t4}]
Let
\begin{equation*}
A_{j_1}(z) =B_{j_1}(z) e^{\frac{b_{j_1}}{(z_0-z) ^{\mu }}}
\end{equation*}
such that $B_{j_1}(z) $ is analytic on $z_0$ and $\arg b_{j_1}\neq \arg b$;
\begin{equation*}
A_{j_m}(z) =B_{j_m}(z) e^{\frac{b_{j_m}}{
(z_0-z) ^{\mu }}}\quad (m=2,\dots,s)
\end{equation*}
such that $B_{j_m}(z) $ are analytic on $z_0$ and
 $b_{j_m}=c_{j_m}b$ $(0<c_{j_m}<1) $, and the remaining
coefficients $A_{j_m}(z) $ are analytic on $z_0$ 
$(m=s+1,\dots,k-1) $. Suppose that $f\not\equiv 0$ is a solution of 
\eqref{eq} of finite order $\sigma (f) =\sigma<\infty $. 
Since $\arg b_{j_1}\neq \arg b$ then there exist 
$(\varphi _1,\varphi _2) \subset [ 0,2\pi ) $ such that
for $z\in D$ and $\arg (z_0-z) =\varphi \in (\varphi_1,\varphi _2) $
 we have $\delta _{b}(\varphi ) >0$
and $\delta _{b_{j_1}}(\varphi ) <0$. 
Set $c=\max \{c_{j_m}:m=2,\dots,s\} $. We have 
$\delta _{b_{j_m}}(\varphi) =c_{j_m}\delta _{b}(\varphi ) \leq c\delta _{b}(
\varphi ) $. From \eqref{eq}, we can write
\begin{equation} \label{p18}
\begin{aligned}
\big| B(z) e^{\frac{b}{(z_0-z) ^{\mu }}}\big|
&\leq | \frac{f^{(k) }}{f}|+\sum_{m=s+1}^{k-1}| A_{j_m}(z) |
| \frac{f^{(j_m) }}{f}|   \\
&\quad +\sum_{m=2}^{s}| A_{j_m}(z) | \big| \frac{f^{(j_m) }}{f}\big| +|
A_{j_1}(z) | \big| \frac{f^{(j_1)}}{f}\big| .
\end{aligned}
\end{equation}
Using the same reasoning as above, from \eqref{p18}, we obtain
\begin{align*} %  \label{p19b}
&\exp \{ (1-\varepsilon ) \delta _{b}(\varphi )\frac{1}{R^{\mu }}\} \\
&\leq \big(\frac{1}{\varepsilon _0R}\big)^{k(\sigma +2+\varepsilon ) }
\Big(M+  (s-1) \exp \{ (1+\varepsilon ) c\delta
_{b}(\varphi ) \frac{1}{R^{\mu }}\}
+\exp \{ (1-\varepsilon ) \delta _{b_{j_1}}(\varphi ) \frac{1}{R^{\mu }}\} \Big).
\end{align*}
By taking $0<\varepsilon <\frac{1-c}{1+c}$,
we obtain a contradiction to $R\to 0$.
\end{proof}

\begin{remark} \label{rmk4} \rm
Concerning the case when $0<\mu \leq 1$, our method is not valid in general.
For example, for the differential equation
\begin{equation*}
f''+A(z) f'+B(z) e^{\frac{-1}{(1-z) }}f=0,
\end{equation*}
where $A(z) $ and $B(z) $ are analytic on $z_0=1$.
However, we cannot apply our method because for all $z\in D$ we have 
$\delta _{-1}(\varphi ) <0$, where $\varphi =\arg (1-z) $.
Furthermore, for the differential equation
\begin{equation*}
f''+A(z) f'+B(z) e^{\frac{1}{(1-z) }}f=0,
\end{equation*}
our method is valid. So we can deduce that every solution $f\not\equiv 0$
of this differential equation is of infinite order.
In general, our method is valid for $0<\mu \leq 1$ except the case when we
have $\delta _{b}(\varphi ) <0$ for 
$\arg z_0-\frac{\pi }{2}<\varphi <\arg z_0+\frac{\pi }{2}$.
\end{remark}

\subsection*{Acknowledgments}
The author wants to thank the anonymous referee for his/her valuable
suggestions and helpful remarks. This research is supported by
ANDRU (Agence Nationale pour le Developpement de la Recherche Universitaire)
and University of Mostaganem (UMAB), (PNR Project Code 8/u27/3144).

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\end{document}