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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 114, pp. 1--17.\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-2014/114\hfil Growth of solutions]
{Growth of solutions to higher-order linear
differential equations with entire coefficients}

\author[H. Habib, B. Bela\"idi \hfil EJDE-2014/114\hfilneg]
{Habib Habib, Benharrat Bela\"idi}  % in alphabetical order

\address{Habib Habib \newline
Department of Mathematics,
Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB),
B. P. 227 Mostaganem, Algeria}
\email{habibhabib2927@yahoo.fr}

\address{Benharrat Bela\"idi  \newline
Department of Mathematics,
Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB),
B. P. 227 Mostaganem, Algeria}
\email{belaidi@univ-mosta.dz}

\thanks{Submitted November 22, 2013. Published April 21, 2014.}
\subjclass[2000]{34M10, 30D35}
\keywords{Linear differential equation; entire solution; order of growth;
 \hfill\break\indent hyper-order of growth; fixed point}

\begin{abstract}
 In this article, we discuss the order and
 hyper-order of the linear differential equation
 \[
 f^{(k) }+\sum_{j=1}^{k-1} (B_je^{b_jz}+D_je^{d_jz}) f^{(j) }+(
 A_1e^{a_1z}+A_2e^{a_2z}) f=0,
 \]
 where $A_j(z), B_j(z), D_j(z)$  are entire functions  $(\not\equiv 0)$
 and $a_1,a_2, d_j$  are complex numbers $(\neq 0)$, and $b_j$  are real numbers.
 Under certain conditions, we prove that every solution $f\not\equiv 0$
 of the above equation is of infinite order. Then, we obtain an estimate
 of the hyper-order. Finally, we give an estimate of the exponent of convergence
 for distinct zeros of the functions $f^{(j) }-\varphi $ $(j=0,1,2) $,
 where $\varphi$ is an entire function $(\not\equiv 0) $ and
 of order $\sigma (\varphi ) <1$, while the solution $f$ of the
 differential equation is of infinite order. Our results extend the previous
 results due to Chen, Peng and Chen and others.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\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's value
distribution theory (see \cite{h2,y1}).
 Let $\sigma (f) $ denote the order of growth of an entire function $f$
and the hyper-order $\sigma _2(f) $ of $f$ is defined by (see \cite{y1})
\[
\sigma _2(f) =\limsup_{r\to +\infty} \frac{\log \log T(r,f) }{\log r}
=\limsup_{r\to +\infty} \frac{\log \log \log M(r,f) }{\log r},
\]
where $T(r,f) $ is the Nevanlinna characteristic function of $f$
and
\[
M(r,f) =\max_{| z| =r}| f(z) |.
\]
 To give some estimates of fixed points, we recall the
following definition.

 \begin{definition}[\cite{c2,l2}] \label{def1.1} \rm
 Let $f$ be a meromorphic function. Then the exponent of convergence of the
sequence of distinct fixed points of $f(z) $ is defined by
\[
\overline{\tau }(f) =\overline{\lambda }(f-z) =
\limsup_{r\to +\infty} \frac{\log \overline{N}(r,
\frac{1}{f-z}) }{\log r},
\]
where $\overline{N}(r,1/f) $ is the counting function of
distinct zeros of $f(z) $ in $\{z:| z|\leq r\} $. We also define
\[
\overline{\lambda }(f-\varphi ) =\limsup_{r\to +\infty }
\frac{\log \overline{N}(r,\frac{1}{f-\varphi })}{\log r}
\]
for any meromorphic function $\varphi (z) $.
\end{definition}

 For the second-order linear differential equation
\begin{equation}
f''+e^{-z}f'+B(z) f=0,  \label{e1.1}
\end{equation}
where $B(z) $ is an entire function, it is well-known that each
solution $f$ of equation \eqref{e1.1} is an entire function, and that if $f_1$
and $f_2$ are two linearly independent solutions of \eqref{e1.1},
then by \cite{f1}, there is at least one of $f_1$,  $f_2$ of
infinite order. Hence, ``most'' solutions
of \eqref{e1.1} will have infinite order. But equation \eqref{e1.1} with
$B(z)=-(1+e^{-z})$ possesses a solution $f(z) =e^{z}$ of finite order.

  A natural question arises: What conditions on $B(z)$ will
guarantee that every solution $f\not\equiv 0$ of \eqref{e1.1}
has infinite order? Many authors, Frei \cite{f2}, Ozawa \cite{o1},
Amemiya-Ozawa \cite{a1} and Gundersen \cite{g2}, Langley \cite{l1} have studied
this problem. They proved that when $B(z)$
is a nonconstant polynomial or $B(z)$ is a transcendental entire function
with order $\sigma (B)\neq 1$, then every solution $f\not\equiv 0$ of
\eqref{e1.1} has infinite order.

 In 2002, Chen \cite{c3} considered the question: What conditions on
$B(z)$ when $\sigma (B)$ $=1$ will guarantee that every nontrivial solution
of \eqref{e1.1} has infinite order? He proved the following result, which
improved results of Frei, Amemiya-Ozawa, Ozawa, Langley and Gundersen.


 \begin{theorem}[\cite{c3}] \label{thmA}
Let $A_j(z)$ $(\not\equiv 0) $ $(j=0,1) $ be
entire functions with $\max \{\sigma (A_j) $ $(j=0,1) \}<1$.
and let $a,b$ be complex constants that satisfy
$ab\neq 0$ and $a\neq b$.  Then every solution
$f\not\equiv 0$  of the differential equation
\[
f''+A_1(z) e^{az}f'+A_0(z) e^{bz}f=0
\]
is of infinite order.
\end{theorem}


 In \cite{p1}, Peng and Chen  investigated the
order and hyper-order of solutions of some second order linear differential
equations and have proved the following result.


 \begin{theorem}[\cite{p1}] \label{thmB}
Let $A_j(z)$ $(\not\equiv 0)$ $(j=1,2) $  be entire functions with
$\sigma (A_j) <1$, $a_1$, $a_2$
be complex numbers such that $a_1a_2\neq 0$,
$a_1\neq a_2$  (suppose that $| a_1| \leq |a_2| $).
If $\arg a_1\neq \pi $ or $a_1<-1 $, then every solution
$f(\not\equiv 0) $  of the differential equation
\[
f''+e^{-z}f'+(A_1e^{a_1z}+A_2e^{a_2z}) f=0
\]
has infinite order and $\sigma _2(f) =1$.
\end{theorem}

  Recently in \cite{h1}, the authors extend and
improve the results of Theorem \ref{thmB} to some higher order linear differential
equations as follows.

 \begin{theorem}[\cite{h1}] \label{thmC}
Let $A_j(z) $ $(\not\equiv 0) $ $(j=1,2)$,
$B_{l}(z) $ $(\not\equiv 0) $ $(l=1,\dots ,k-1) $,
$D_{m}$ $(m=0,\dots ,k-1) $  be entire
functions with $\max \{\sigma (A_j) ,\sigma (B_{l}) ,\sigma (D_{m}) \} <1$,
$b_{l}$ $(l=1,\dots ,k-1) $ be complex constants such that
\begin{itemize}
\item[(i)] $\arg b_{l}=\arg a_1$  and $b_{l}=c_{l}a_1$
$(0<c_{l}<1) $  $(l\in I_1) $ and

\item[(ii)]  $b_{l}$ is a real constant such that
$b_{l}\leq 0$  $(l\in I_2) $, where
$I_1\neq \emptyset $, $I_2\neq \emptyset $,
$I_1\cap I_2=\emptyset $, $I_1\cup I_2=\{1,2,\dots ,k-1\} $, and
$a_1$, $a_2$ are complex numbers such that $a_1a_2\neq 0$,
$a_1\neq a_2$ (suppose that $| a_1| \leq | a_2|$).
\end{itemize}
 If $\arg a_1\neq \pi $ or $a_1$ is a
real number such that $a_1<\frac{b}{1-c}$, where
$c=\max \{c_{l}:l\in I_1\} $ and
$b=\min \{b_{l}:l\in I_2\} $, then every solution
$f\not\equiv 0$ of the differential equation
\[
f^{(k) }+(D_{k-1}+B_{k-1}e^{b_{k-1}z}) f^{(k-1) }+\dots +(D_1+B_1e^{b_1z}) f'
+(D_0+A_1e^{a_1z}+A_2e^{a_2z}) f=0
\]
satisfies $\sigma (f) =+\infty $ and $\sigma_2(f) =1$.
\end{theorem}

In this paper, we continue the research in this type of
problems, the main purpose of this paper is to extend and improve the
results of Theorems \ref{thmA}--\ref{thmC} to some higher order linear differential equations.
In fact we will prove the following results.

\begin{theorem} \label{thm1.1} Let
$k\geq 2$ be an integer, $A_j(z) $ $(\not\equiv 0) $ $(j=1,2) $  and
$B_j(z) $ $(\not\equiv 0) $, $D_j(z) $ $(\not\equiv 0) $
$(j=1,\dots ,k-1) $  be entire functions with
\[
\max \{\sigma (A_j) (j=1,2),\sigma (B_j)
(j=1,\dots ,k-1),\sigma (D_j) (j=1,\dots ,k-1)\}<1,
\]
$a_1,a_2$ be complex numbers such that
$a_1a_2\neq 0,a_1\neq a_2,d_j\neq 0$ $(j=1,\dots ,k-1)$
 be complex numbers and $b_j$ $(j=1,\dots ,k-1) $ be
real numbers such that $b_j<0$. Suppose that there exists
$ \alpha _j$, $\beta _j$ $(j=1,\dots ,k-1)$ where
$0<\alpha _j<1$, $0<\beta _j<1$  and
$d_j=\alpha _ja_1+\beta _ja_2$. Set
$\alpha =\max \{\alpha _j:j=1,\dots ,k-1\} $,
$\beta =\max \{\beta _j:j=1,\dots ,k-1\} $ and
$b=\min \{b_j:j=1,\dots ,k-1\} $.
If
\begin{itemize}
\item[(1)]  $\arg a_1\neq \pi $ and $\arg a_1\neq \arg a_2$; or

\item[(2)]  $\arg a_1\neq \pi $, $\arg a_1=\arg a_2$  and
(i)  $| a_2| >\frac{| a_1| }{1-\beta }$ or (ii)
 $| a_2| <(1-\alpha ) |a_1| $; or

\item[(3)]  $a_1<0$ and $\arg a_1\neq \arg a_2$; or

\item[(4)]  (i)  $(1-\beta ) a_2-b<a_1<0$, $a_2<\frac{b}{1-\beta }$
or (ii) $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$,
\end{itemize}
then every solution $f(\not\equiv 0)$ of the differential equation
\begin{equation}
f^{(k) }+\sum_{j=1}^{k-1} (B_je^{b_jz}+D_je^{d_jz}) f^{(j) }
+(A_1e^{a_1z}+A_2e^{a_2z}) f=0  \label{e1.2}
\end{equation}
satisfies $\sigma (f) =+\infty $ and $\sigma_2(f) =1$.
\end{theorem}

 Set
\begin{gather*}
I_1=\{2a_1, 2a_2, a_1+a_2, a_1, a_2,
a_1+b_i, a_2+b_i, a_1+d_i, a_2+d_i\; (i=1,\dots ,k-1)\},\\
I_2=\{2a_1, 2a_2, a_1+a_2, a_1+b_1, a_2+b_1, a_1+d_1, a_2+d_1\},\\
\begin{aligned}
I_3=\Big\{&3a_1, 3a_2, 2a_1+a_2, a_1+2a_2, 2a_1, 2a_2, a_1+a_2, a_1+b_1,
a_2+b_1, a_1+d_1,\\
&a_2+d_1, 2a_1+b_i, 2a_2+b_i, 2a_1+d_i, 2a_2+d_i,
a_1+a_2+b_i, a_1+a_2+d_i,\\
& a_1+b_1+b_i, a_2+b_1+b_i, a_1+d_1+d_i, a_2+d_1+d_i,
 a_1+b_1+d_i,\\
& a_2+b_1+d_i (i=1,\dots ,k-1), a_1+d_1+b_i,
a_2+d_1+b_i\; (i=2,\dots ,k-1)\Big\}.
\end{aligned}
\end{gather*}

 \begin{theorem} \label{thm1.2}
Let $A_j(z) $ $(j=1,2) $, $B_j(z) $, $D_j(z) $ $(j=1,\dots ,k-1)$,
$a_1$, $a_2$, $b_j$, $d_j$, $\alpha _j$, $\beta _j$
$(j=1,\dots ,k-1)$, $\alpha $, $\beta $ and $b$
satisfy the additional hypotheses of Theorem \ref{thm1.1}.
If $\varphi (\not\equiv 0) $ is an entire function of order
$\sigma (\varphi) <1$, then every solution $f(\not\equiv 0) $
 of equation \eqref{e1.2}  satisfies
\[
\overline{\lambda }(f-\varphi ) =+\infty .
\]
Furthermore, we have
\begin{itemize}
\item[(1)] If $(2a_1) \notin I_1\setminus \{2a_1\} $ or
 $(2a_2) \notin I_1\setminus\{2a_2\} $, then
\[
\overline{\lambda }(f'-\varphi ) =+\infty .
\]
\item[(2)] If (i) $(2a_1) \notin I_2\setminus \{2a_1\} $ or
$(2a_2)\notin I_2\setminus \{2a_2\} $ and (ii)
$(3a_1) \notin I_3\setminus \{3a_1\}$  or
$(3a_2) \notin I_3\setminus \{3a_2\} $,  then
\[
\overline{\lambda }(f''-\varphi ) =+\infty .
\]
\end{itemize}
\end{theorem}
Now set
\begin{gather*}
J_1=\Big\{2a_1, 2a_2, a_1+a_2, a_1+b_i, a_2+b_i, a_1+d_i, a_2+d_i\; (i=1,2)\Big\}, \\
\begin{aligned}
J_2=\Big\{&3a_1, 3a_2, 2a_1+a_2, a_1+2a_2, 2a_1+b_i, 2a_2+b_i, 2a_1+d_i,\\
&2a_2+d_i, a_1+a_2+b_i, a_1+a_2+d_i, a_1+b_1+b_i, a_2+b_1+b_i, a_1\\
&+d_1+d_i,a_2+d_1+d_i, a_1+b_1+d_i, a_2+b_1+d_i \; (i=1,2,3), \\
&a_1+d_1+b_i, a_2+d_1+b_i\; (i=2,3)\Big\}.
\end{aligned}
\end{gather*}

 From Theorem \ref{thm1.2}, we obtain the following corollary.


 \begin{corollary} \label{coro1.1}
Let $A_j(z) $ $(j=1,2) $, $B_j(z) $, $D_j(z) $ $(j=1,\dots ,k-1) $, $a_1$,
$a_2$, $b_j$, $d_j$, $\alpha _j$, $\beta _j$ $(j=1,\dots ,k-1)$, $\alpha $,
$\beta $ and $b$ satisfy the additional hypotheses of Theorem \ref{thm1.1}.
If $f(\not\equiv 0) $ is any solution of  \eqref{e1.2}, then
$f$ has infinitely many fixed points and satisfies
\[
\overline{\tau }(f) =\infty .
\]
Furthermore, we have
\begin{itemize}
\item[(1)]  If $(2a_1) \notin J_1\setminus\{2a_1\} $ or
$(2a_2) \notin J_1\setminus \{2a_2\} $, then
$f'$ has infinitely many fixed points and satisfies
\[
\overline{\tau }(f') =\infty .
\]
\item[(2)] If (i)  $(2a_1) \notin I_2\setminus \{2a_1\} $ or
 $(2a_2) \notin I_2\setminus \{2a_2\} $ and (ii)
 $(3a_1) \notin J_2\setminus \{3a_1\}$ or $(3a_2) \notin J_2\setminus \{
3a_2\} $, then $f''$ has infinitely many fixed points and satisfies
\[
\overline{\tau }(f'') =\infty .
\]
\end{itemize}
\end{corollary}

\section{Preliminary lemmas}

 We define the linear measure of a set $E\subset [ 0,+\infty) $ by
$m(E)=\int_0^{+\infty }\chi _{E}(t)dt$ and the logarithmic measure
of a set $F\subset (1,+\infty ) $ by
$lm(F)=\int_1^{+\infty } \frac{\chi _{F}(t)}{t}dt$, where
$\chi _{H}$ is the characteristic function of a set $H$.



 \begin{lemma}[\cite{g3}] \label{lem2.1}
Let $f$ be a transcendental meromorphic function with
$\sigma (f) =\sigma <+\infty $. Let $\varepsilon >0$\ be a
given constant, and let $k$,  $j$  be integers satisfying
$k>j\geq 0$. Then, there exists a set
$E_1\subset [-\frac{\pi }{2},\frac{3\pi }{2}) $ with linear measure zero, such
that, if $\psi \in [-\frac{\pi }{2},\frac{3\pi }{2}) \setminus
E_1$, then there is a constant $R_0=R_0(\psi ) >1$, such that for all
$z$ satisfying $\arg z=\psi $ and $| z| \geq R_0$, we have
\begin{equation}
| \frac{f^{(k) }(z) }{f^{(j)}(z) }| \leq | z| ^{(k-j) (\sigma -1+\varepsilon ) }.  \label{e2.1}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{c3}] \label{lem2.2}  Suppose that
$P(z) =(\alpha +i\beta ) z^n+\dots $ ($\alpha ,\beta $
 are real numbers, $| \alpha |+| \beta | \neq 0$) is a polynomial with degree
$n\geq 1$, that $A(z) $ $(\not\equiv 0)$ is an entire function with
$\sigma (A) <n$. Set $g(z) =A(z) e^{P(z) }$, $z=re^{i\theta }$,
$\delta (P,\theta ) =\alpha \cos n\theta -\beta \sin n\theta $.
Then for any given $\varepsilon >0$, there is a set
$E_2\subset [0,2\pi )$ that has linear measure zero, such that for any
$\theta \in [0,2\pi ) \setminus (E_2\cup E_3) $,
there is $R>0$, such that for $| z| =r>R$, we have
\begin{itemize}
\item[(i)]  If $\delta (P,\theta) >0$, then
\begin{equation}
\exp \{(1-\varepsilon ) \delta (P,\theta )
r^n\} \leq | g(re^{i\theta }) | \leq
\exp \{(1+\varepsilon ) \delta (P,\theta )r^n\} .  \label{e2.2}
\end{equation}

\item[(ii)]  If $\delta (P,\theta ) <0$, then
\begin{equation}
\exp \{(1+\varepsilon ) \delta (P,\theta )
r^n\} \leq | g(re^{i\theta }) | \leq
\exp \{(1-\varepsilon ) \delta (P,\theta )r^n\} ,  \label{e2.3}
\end{equation}
where $E_3=\{\theta \in [0,2\pi ) :\delta (P,\theta ) =0\} $  is a finite set.
\end{itemize}
\end{lemma}


 \begin{lemma}[\cite{p1}]  \label{lem2.3}
Suppose that $n\geq 1$ is a natural number. Let
$P_j( z) =a_{jn}z^n+\dots $ $(j=1,2) $ be
nonconstant polynomials, where
$a_{jq}$ ($q=1,\dots ,n$)
are complex numbers and $a_{1n}a_{2n}\neq 0$. Set
$z=re^{i\theta }$, $a_{jn}=| a_{jn}| e^{i\theta _j}$,
$\theta _j\in [-\frac{\pi }{2},\frac{3\pi }{2}) $,
$\delta (P_j,\theta ) =| a_{jn}| \cos (\theta _j+n\theta ) $,
then there is a set $E_4\subset[-\frac{\pi }{2n},\frac{3\pi }{2n}) $
that has linear measure zero such that if $\theta _1\neq \theta _2$,
then there exists a ray $\arg z=\theta $,
$\theta \in (-\frac{\pi }{2n}, \frac{\pi }{2n}) \setminus (E_4\cup E_5) $,
satisfying
\begin{equation}
\delta (P_1,\theta ) >0,\quad \delta (P_2,\theta) <0  \label{e2.4}
\end{equation}
or
\begin{equation}
\delta (P_1,\theta ) <0,\quad \delta (P_2,\theta) >0,  \label{e2.5}
\end{equation}
where $E_5=\{\theta \in [-\frac{\pi }{2n},\frac{3\pi
}{2n}) :\delta (P_j,\theta ) =0\} $ is a
finite set, which has linear measure zero.
\end{lemma}


 \begin{remark}[\cite{p1}] \label{rmk2.1} \rm
In Lemma \ref{lem2.3}, if $\theta \in (-\frac{\pi }{2n},\frac{\pi }{2n})
\setminus (E_4\cup E_5) $ is replaced by
 $\theta \in (\frac{\pi }{2n},\frac{3\pi }{2n}) \setminus (E_4\cup E_5) $,
then we obtain the same result.
\end{remark}

 \begin{lemma}[\cite{c4}] \label{lem2.4} Suppose that
$k\geq 2$  and $B_0,B_1,\dots ,B_{k-1}$ are entire
functions of finite order and let $\sigma =\max \{\sigma (
B_j) :j=0,\dots ,k-1\} $. Then every solution $f$
 of the differential equation
\begin{equation}
f^{(k) }+B_{k-1}f^{(k-1) }+\dots +B_1f'+B_0f=0  \label{e2.6}
\end{equation}
satisfies $\sigma _2(f) \leq \sigma $.
\end{lemma}


 \begin{lemma}[\cite{g3}] \label{lem2.5}
 Let $f(z)$ be a transcendental meromorphic function, and let
$\alpha >1$ be a given constant. Then there exist a set
$E_6\subset (1,\infty ) $ with finite logarithmic measure and a
constant $B>0$ that depends only on $\alpha $ and
$i,j$  $(0\leq i<j\leq k)$, such that for all
$z$ satisfying $| z| =r\notin [ 0,1]\cup E_6$, we have
\begin{equation}
| \frac{f^{(j) }(z)}{f^{(i) }(z)}|
\leq B\{\frac{T(\alpha r,f)}{r}(\log ^{\alpha
}r) \log T(\alpha r,f)\} ^{j-i}.  \label{e2.7}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{g4}] \label{lem2.6}
Let $\varphi:[0,+\infty ) \to \mathbb{R}$
 and $\psi :[0,+\infty ) \to  \mathbb{R} $ be monotone non-decreasing functions
such that $\varphi (r) \leq \psi (r) $ for all
$r\notin E_7\cup [0,1] $, where $E_7\subset (1,+\infty ) $
 is a set of finite logarithmic measure. Let
$\gamma >1$ be a given constant. Then there exists an
$r_1=r_1(\gamma ) >0$ such that $\varphi (r) \leq \psi (\gamma r) $
 for all $r>r_1$.
\end{lemma}

 \begin{lemma}[\cite{c1}] \label{lem2.7} Let
$A_0,A_1,\dots ,A_{k-1}$, $F\not\equiv 0$ be finite order
meromorphic functions. If $f(z) $ is an infinite order
meromorphic solution of the equation
\begin{equation}
f^{(k) }+A_{k-1}f^{(k-1) }+\dots +A_1f'+A_0f=F,  \label{e2.8}
\end{equation}
then $f$ satisfies $\overline{\lambda }(f)=\lambda (f) =\sigma (f) =\infty $.
\end{lemma}


 The following lemma, due to Gross \cite{g1}, is
important in the factorization and uniqueness theory of meromorphic
functions, playing an important role in this paper as well.


 \begin{lemma}[\cite{g1,y1}] \label{lem2.8}
Suppose that $f_1(z) ,f_2(z),\dots ,f_{n}(z) (n\geq 2) $ are
meromorphic functions and $g_1(z) ,g_2(z),\dots ,g_{n}(z) $
 are entire functions satisfying the following conditions:
\begin{itemize}
\item[(i)]  $\sum_{j=1}^nf_j(z) e^{g_j(z) }\equiv 0$;

\item[(ii)]  $g_j(z) -g_{k}(z) $ are not constants for
 $1\leq j<k\leq n$;

\item[(iii)]  For $1\leq j\leq n$, $1\leq h<k\leq n$,
$T(r,f_j) =o\{T(r,e^{g_{h}(z)-g_{k}(z) }) \} $ ($r\to \infty $, $r\notin
E_8$), where $E_8$  is a set with finite linear measure.
\end{itemize}
Then $f_j(z) \equiv 0$ ($j=1,\dots ,n$).
\end{lemma}

 \begin{lemma}[\cite{x1}] \label{lem2.9}
Suppose that $f_1(z) ,f_2(z) ,\dots,f_{n}(z) (n\geq 2) $ are meromorphic
functions and $g_1(z) ,g_2(z) ,\dots,g_{n}(z) $
 are entire functions satisfying the following conditions:
\begin{itemize}
\item[(i)]  $\sum_{j=1}^nf_j(z) e^{g_j(z) }\equiv f_{n+1}$;

\item[(ii)] If $1\leq j\leq n+1,1\leq k\leq n$, the order of
$f_j$  is less than the order of $e^{g_{k}(z) }$. If
$n\geq 2$, $1\leq j\leq n+1$, $1\leq h<k\leq n$, and the order of
$f_j$  is less than the order of $e^{g_{h}-g_{k}}$.
\end{itemize}
Then $f_j(z) \equiv 0$ $(j=1,2,\dots ,n+1) $.
\end{lemma}

\section{Proof of Theorem \ref{thm1.1}}

\noindent \textbf{First step.}
 Assume that $f(\not\equiv 0) $ is a solution of equation \eqref{e1.2}.
We prove that $\sigma (f) =+\infty $. Suppose that
$\sigma (f) =\sigma <+\infty $. We rewrite \eqref{e1.2} as
\begin{equation}
\frac{f^{(k) }}{f}+\sum_{j=1}^{k-1} (B_je^{b_jz}+D_je^{(\alpha _ja_1+\beta _ja_2)
z}) \frac{f^{(j) }}{f}+A_1e^{a_1z}+A_2e^{a_2z}=0.
\label{e3.1}
\end{equation}
Set
\[
\gamma =\max \{\sigma (B_j) \; (j=1,\dots,k-1) \} <1.
\]
Then, for any given $\varepsilon $ $(0<\varepsilon <1-\gamma ) $
and for sufficiently large $r$, we have
\begin{equation}
| B_j(z) | \leq \exp \{r^{\gamma
+\varepsilon }\} \quad (j=1,\dots ,k-1) .  \label{e3.2}
\end{equation}
By Lemma \ref{lem2.1}, for any given $\varepsilon $
$(0<\varepsilon <1-\gamma) $, there exists a set
$E_1\subset [-\frac{\pi }{2},\frac{3\pi }{2}) $ of linear measure zero,
such that if $\theta \in [-\frac{\pi }{2},\frac{3\pi }{2}) \setminus E_1$,
then there is a constant $R_0=R_0(\theta ) >1$, such that for all $z$
satisfying $\arg z=\theta $ and $| z| =r\geq R_0$, we have
\begin{equation}
| \frac{f^{(j) }(z) }{f(z) }
| \leq r^{j(\sigma -1+\varepsilon ) }\quad (j=1,\dots ,k) .  \label{e3.3}
\end{equation}
Let $z=re^{i\theta }$, $a_1=| a_1| e^{i\theta _1}$,
$a_2=| a_2| e^{i\theta _2}$,
$\theta _1,\theta_2\in [-\frac{\pi }{2},\frac{3\pi }{2}) $. We know that
$\delta (\alpha _ja_1z,\theta ) =\alpha _j\delta (a_1z,\theta ) $,
$\delta (\beta _ja_2z,\theta ) =\beta _j\delta (a_2z,\theta ) $
$(j=1,\dots ,k-1) $ and $\alpha <1$, $\beta <1$.
\smallskip


\noindent \textbf{Case 1.} Assume that $\arg a_1\neq \pi $ and
$\arg a_1\neq \arg a_2$, which is $\theta _1\neq \pi $ and $\theta _1\neq
\theta _2$. By Lemma \ref{lem2.2} and Lemma \ref{lem2.3}, for any given $\varepsilon $,
\[
0<\varepsilon <\min \{1-\gamma ,\frac{1-\alpha }{2(1+\alpha
) },\frac{1-\beta }{2(1+\beta ) }\} ,
\]
there is a ray $\arg z=\theta $ such that $\theta \in (-\frac{\pi }{2},
\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $
(where $E_4$ and $E_5$ are defined as in Lemma \ref{lem2.3},
$E_1\cup E_4\cup E_5$ is of the linear measure zero), and satisfying
\[
\delta (a_1z,\theta ) >0\text{, }\delta (a_2z,\theta) <0
\]
or
\[
\delta (a_1z,\theta ) <0\text{, }\delta (a_2z,\theta) >0.
\]
(a) When $\delta (a_1z,\theta ) >0$, $\delta (a_2z,\theta ) <0$,
for sufficiently large $r$, we obtain by Lemma \ref{lem2.2},
\begin{gather}
| A_1e^{a_1z}| \geq \exp \{(1-\varepsilon) \delta (a_1z,\theta ) r\} ,  \label{e3.4}
\\
| A_2e^{a_2z}| \leq \exp \{(1-\varepsilon) \delta (a_2z,\theta ) r\} <1,  \label{e3.5}
\\
\begin{aligned}
| D_je^{\alpha _ja_1z}|
&\leq \exp \{(1+\varepsilon ) \alpha _j\delta (a_1z,\theta )r\} \\
&\leq \exp \{(1+\varepsilon ) \alpha \delta (
a_1z,\theta ) r\} \quad (j=1,\dots ,k-1) ,
\end{aligned} \label{e3.6}
\\
| e^{\beta _ja_2z}| \leq \exp \{(
1-\varepsilon ) \beta _j\delta (a_2z,\theta )
r\} <1\ (j=1,\dots ,k-1) .  \label{e3.7}
\end{gather}
By \eqref{e3.6}  and \eqref{e3.7}, we obtain
\begin{equation}
| D_je^{(\alpha _ja_1+\beta _ja_2)
z}| =| D_je^{\alpha _ja_1z}| |
e^{\beta _ja_2z}| \leq \exp \{(1+\varepsilon
) \alpha \delta (a_1z,\theta ) r\} ,  \label{e3.8}
\end{equation}
where $j=1,\dots ,k-1$. For
$\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) $, by \eqref{e3.2}, we have
\begin{equation}
| B_je^{b_jz}| =| B_j|\,| e^{b_jz}| \leq \exp \{r^{\gamma +\varepsilon
}\} e^{b_jr\cos \theta }\leq \exp \{r^{\gamma +\varepsilon
}\}  \label{e3.9}
\end{equation}
because $b_j<0$ and $\cos \theta >0$ $(j=1,\dots ,k-1) $. By
\eqref{e3.1}, we obtain
\begin{equation}
| A_1e^{a_1z}| \leq | \frac{f^{(k) }}{f}|
+\sum_{j=1}^{k-1} \Big(| B_je^{b_jz}| +| D_je^{(\alpha
_ja_1+\beta _ja_2) z}| \Big) | \frac{f^{(j) }}{f}| +| A_2e^{a_2z}| .
\label{e3.10}
\end{equation}
Substituting \eqref{e3.3} -\eqref{e3.5}, \eqref{e3.8} and
\eqref{e3.9} in \eqref{e3.10}, we have
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\}
& \leq | A_1e^{a_1z}| \\
& \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp
\{(1+\varepsilon ) \alpha \delta (a_1z,\theta) r\} ,
\end{aligned} \label{e3.11}
\end{equation}
where $M_1>0$ and $M_2>0$ are some constants.
By $0<\varepsilon <\frac{1-\alpha }{2(1+\alpha ) }$ and \eqref{e3.11},
we obtain
\begin{equation}
\exp \{\frac{1-\alpha }{2}\delta (a_1z,\theta )r\}
\leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} .
\label{e3.12}
\end{equation}
By $\delta (a_1z,\theta ) >0$ and $\gamma +\varepsilon <1$ we
know that \eqref{e3.12} is a contradiction.
\smallskip


\noindent(b) When $\delta (a_1z,\theta ) <0$, $\delta (a_2z,\theta ) >0$,
for sufficiently large $r$, we obtain
\begin{gather}
| A_2e^{a_2z}| \geq \exp \{(1-\varepsilon) \delta (a_2z,\theta ) r\},\label{e3.13}
\\
| A_1e^{a_1z}| \leq \exp \{(1-\varepsilon) \delta (a_1z,\theta ) r\} <1,\label{e3.14}
\\
| D_je^{\alpha _ja_1z}| \leq \exp \{(1-\varepsilon ) \alpha _j\delta (a_1z,\theta )
r\} <1\quad (j=1,\dots ,k-1) ,  \label{e3.15}
\\
\begin{aligned}
| e^{\beta _ja_2z}|
&\leq \exp \{(1+\varepsilon ) \beta _j\delta (a_2z,\theta ) r\}\\
&\leq \exp \{(1+\varepsilon ) \beta \delta (
a_2z,\theta ) r\} \quad  (j=1,\dots ,k-1) .
\end{aligned} \label{e3.16}
\end{gather}
By \eqref{e3.15} and \eqref{e3.16}, we have
\begin{equation}
| D_je^{(\alpha _ja_1+\beta _ja_2)z}|
=| D_je^{\alpha _ja_1z}|\, |e^{\beta _ja_2z}|
\leq \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} ,  \label{e3.17}
\end{equation}
where $j=1,\dots ,k-1$. By \eqref{e3.1}, we obtain
\begin{equation}
| A_2e^{a_2z}| \leq | \frac{f^{(k) }}{f}| 
+\sum_{j=1}^{k-1} \Big(| B_je^{b_jz}| +| D_je^{(\alpha _ja_1+\beta _ja_2) z}| 
\Big) | \frac{ f^{(j) }}{f}| +| A_1e^{a_1z}| .
\label{e3.18}
\end{equation}
Substituting \eqref{e3.3}, \eqref{e3.9}, \eqref{e3.13},
\eqref{e3.14} and \eqref{e3.17} in \eqref{e3.18},
we have
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\}
&\leq | A_2e^{a_2z}| \\
& \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp
\{(1+\varepsilon ) \beta \delta (a_2z,\theta) r\} .
\end{aligned}\label{e3.19}
\end{equation}
By $0<\varepsilon <\frac{1-\beta }{2(1+\beta ) }$ and
\eqref{e3.19}, we obtain
\begin{equation}
\exp \big\{\frac{1-\beta }{2}\delta (a_2z,\theta ) r\big\}
\leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} .
\label{e3.20}
\end{equation}
By $\delta (a_2z,\theta ) >0$ and $\gamma +\varepsilon <1$ we
know that \eqref{e3.20} is a contradiction.
\smallskip

\noindent\textbf{Case 2.}
Assume that $\arg a_1\neq \pi $, $\arg a_1=\arg a_2$, which is
 $\theta _1\neq \pi $, $\theta _1=\theta _2$. By Lemma \ref{lem2.3}, for any given
$\varepsilon $
\[
0<\varepsilon <\min \big\{1-\gamma ,\frac{(1-\alpha )
| a_1| -| a_2| }{2[(1+\alpha ) | a_1| +| a_2|] },
\frac{(1-\beta ) | a_2| -| a_1| }{2[(1+\beta ) |a_2| +| a_1| ] }\big\} ,
\]
there is a ray $\arg z=\theta $ such that
$\theta \in (-\frac{\pi }{2}, \frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $
and $\delta (a_1z,\theta ) >0$. Since $\theta _1=\theta _2$, then
$\delta (a_2z,\theta ) >0$.
\smallskip

\noindent (i)  $| a_2| >\frac{| a_1| }{1-\beta }$. For sufficiently large $r$, we
have \eqref{e3.6}, \eqref{e3.13} , \eqref{e3.16}  hold and
\begin{equation}
| A_1e^{a_1z}| \leq \exp \{(1+\varepsilon
) \delta (a_1z,\theta ) r\} .  \label{e3.21}
\end{equation}
By \eqref{e3.6} and \eqref{e3.16}, we obtain
\begin{equation}
| D_je^{(\alpha _ja_1+\beta _ja_2)
z}| \leq \exp \{(1+\varepsilon ) \alpha \delta
(a_1z,\theta ) r\} \exp \{(1+\varepsilon
) \beta \delta (a_2z,\theta ) r\} ,  \label{e3.22}
\end{equation}
where $j=1,\dots ,k-1$. Substituting \eqref{e3.3}, \eqref{e3.9},
\eqref{e3.13}, \eqref{e3.21} and \eqref{e3.22} in \eqref{e3.18}, we have
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} \\
&\leq | A_2e^{a_2z}| \\
& \leq k\exp \{r^{\gamma +\varepsilon }\} \exp \{(
1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\}
\exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta
) r\} r^{k(\sigma -1+\varepsilon ) } \\
&\quad +\exp \{(1+\varepsilon ) \delta (a_1z,\theta) r\}
\\
&\leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp
\{(1+\varepsilon ) \delta (a_1z,\theta )
r\} \exp \{(1+\varepsilon ) \beta \delta (
a_2z,\theta ) r\} .
\end{aligned}  \label{e3.23}
\end{equation}
From \eqref{e3.23}, we obtain
\begin{equation}
\exp \{\eta _1r\} \leq M_1r^{M_2}\exp \{r^{\gamma+\varepsilon }\} ,  \label{e3.24}
\end{equation}
where
\[
\eta _1=(1-\varepsilon ) \delta (a_2z,\theta )
-(1+\varepsilon ) \delta (a_1z,\theta )
-(1+\varepsilon ) \beta \delta (a_2z,\theta ) .
\]
Since
\[
0<\varepsilon <\frac{(1-\beta ) |a_2| -| a_1| }{2[(1+\beta) | a_2| +| a_1| ] },
\]
$\theta _1=\theta _2$ and $\cos (\theta _1+\theta ) >0$,
we have
\begin{align*}
\eta _1&=[1-\beta -\varepsilon (1+\beta ) ] \delta
(a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) \\
&=[1-\beta -\varepsilon (1+\beta ) ] |a_2| \cos (\theta _1+\theta )
-(1+\varepsilon ) | a_1| \cos (\theta_1+\theta )\\
&=\{[1-\beta -\varepsilon (1+\beta ) ]
| a_2| -(1+\varepsilon ) |a_1| \} \cos (\theta _1+\theta )\\
&=\{(1-\beta ) | a_2| -|a_1| -\varepsilon [(1+\beta ) |a_2|
+| a_1| ] \} \cos (\theta _1+\theta )\\
& >\frac{(1-\beta ) | a_2| -|a_1| }{2}\cos (\theta _1+\theta ) >0.
\end{align*}
Since $\eta _1>0$ and $\gamma +\varepsilon <1$, we know that \eqref{e3.24}
 is a contradiction.
\smallskip

\noindent (ii) $| a_2| <(1-\alpha ) | a_1| $. For sufficiently large $r$,
we have \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16} and
\eqref{e3.22} hold; then we obtain
\begin{equation}
| A_2e^{a_2z}| \leq \exp \{(1+\varepsilon
) \delta (a_2z,\theta ) r\} .  \label{e3.25}
\end{equation}
Substituting \eqref{e3.3}, \eqref{e3.4}, \eqref{e3.9}, \eqref{e3.22}
 and \eqref{e3.25} in \eqref{e3.10}, we have
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon ) \delta (a_1z,\theta ) r\} \\
&\leq | A_1e^{a_1z}| \\
&\leq k\exp \{r^{\gamma +\varepsilon }\} \exp \{(
1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\}
\exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta
) r\} r^{k(\sigma -1+\varepsilon ) }\\
&\quad +\exp \{(1+\varepsilon ) \delta (a_2z,\theta) r\} \\
& \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp
\{(1+\varepsilon ) \alpha \delta (a_1z,\theta
) r\} \exp \{(1+\varepsilon ) \delta (
a_2z,\theta ) r\} .
\end{aligned}  \label{e3.26}
\end{equation}
From the above inequality we obtain
\begin{equation}
\exp \{\eta _2r\} \leq M_1r^{M_2}\exp \{r^{\gamma+\varepsilon }\} ,  \label{e3.27}
\end{equation}
where
\[
\eta _2=(1-\varepsilon ) \delta (a_1z,\theta )
-(1+\varepsilon ) \alpha \delta (a_1z,\theta )
-(1+\varepsilon ) \delta (a_2z,\theta ) .
\]
Since $0<\varepsilon <\frac{(1-\alpha ) |a_1| -| a_2| }{2[(1+\alpha)
| a_1| +| a_2| ] }$,
$\theta _1=\theta _2$ and $\cos (\theta _1+\theta ) >0$,
then we obtain
\begin{align*}
\eta _2&=\{(1-\alpha ) | a_1|-| a_2| -\varepsilon [(1+\alpha )
| a_1| +| a_2| ] \}\cos (\theta _1+\theta ) \\
&>\frac{(1-\alpha ) | a_1| -| a_2| }{2}\cos (\theta _1+\theta ) >0.
\end{align*}
By $\eta _2>0$ and $\gamma +\varepsilon <1$ we know that
\eqref{e3.27} is a contradiction.
\smallskip

\noindent\textbf{Case 3.} Assume that $a_1<0$ and
$\arg a_1\neq \arg a_2$, which is $\theta _1=\pi $ and
 $\theta _2\neq \pi $. By Lemma \ref{lem2.2}, for the above $\varepsilon $,
there is a ray $\arg z=\theta $ such that
$\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $ and
$\delta (a_2z,\theta ) >0$. Because $\cos \theta >0$,
$\delta (a_1z,\theta )=| a_1| \cos (\theta _1+\theta )=-| a_1| \cos \theta <0$.
Using the same reasoning as in Case 1 (b), we can get a contradiction.
\smallskip

\noindent \textbf{Case 4.} Assume that
(i)  $(1-\beta ) a_2-b<a_1<0$ and $a_2<\frac{b}{1-\beta }$ or
(ii)  $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$, which is
$\theta _1=\theta _2=\pi $. By Lemma \ref{lem2.2}, for any given $\varepsilon $ satisfying
\[
0<\varepsilon <\min \big\{1-\gamma ,\frac{(1-\alpha )
| a_1| -| a_2| +b}{2[(1+\alpha ) | a_1| +| a_2|] },\frac{(1-\beta ) | a_2|
-| a_1| +b}{2[(1+\beta ) |a_2| +| a_1| ] }\big\} ,
\]
there is a ray $\arg z=\theta $ such that
$\theta \in (\frac{\pi }{2}, \frac{3\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $,
then $\cos \theta <0$, $\delta (a_1z,\theta ) =|a_1| \cos (\theta _1+\theta ) =-|
a_1| \cos \theta >0$ and
\[
\delta (a_2z,\theta )
=| a_2| \cos (\theta _2+\theta ) =-| a_2| \cos \theta >0.
\]
(i)  $(1-\beta ) a_2-b<a_1<0$ and $a_2<\frac{b}{1-\beta }$.
For sufficiently large $r$, we obtain \eqref{e3.6}, \eqref{e3.13}, \eqref{e3.16},
\eqref{e3.21} and \eqref{e3.22} hold.
For $\theta \in (\frac{\pi }{2},\frac{3\pi}{2}) $, by \eqref{e3.2} we have
\begin{equation}
| B_je^{b_jz}| =| B_j|| e^{b_jz}| \leq \exp \{r^{\gamma +\varepsilon
}\} e^{b_jr\cos \theta }\leq \exp \{r^{\gamma +\varepsilon
}\} e^{br\cos \theta }  \label{e3.28}
\end{equation}
because $b\leq b_j<0$ and $\cos \theta <0$ $(j=1,\dots ,k-1) $.
Substituting \eqref{e3.3}, \eqref{e3.13}, \eqref{e3.21}, \eqref{e3.22}
 and \eqref{e3.28} in \eqref{e3.18}, we obtain
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} \\
&\leq | A_2e^{a_2z}| \\
& \leq M_1r^{M_2}e^{br\cos \theta }\exp \{r^{\gamma +\varepsilon
}\} \exp \{(1+\varepsilon ) \delta (
a_1z,\theta ) r\} \exp \{(1+\varepsilon )
\beta \delta (a_2z,\theta ) r\} .
\end{aligned}  \label{e3.29}
\end{equation}
From \eqref{e3.29} we have
\begin{equation}
\exp \{\eta _3r\} \leq M_1r^{M_2}\exp \{r^{\gamma
+\varepsilon }\} ,  \label{e3.30}
\end{equation}
where
\[
\eta _3=(1-\varepsilon ) \delta (a_2z,\theta )
-(1+\varepsilon ) \delta (a_1z,\theta ) -(
1+\varepsilon ) \beta \delta (a_2z,\theta ) -b\cos
\theta .
\]
Since $(1-\beta ) a_2-b<a_1$, $a_2=-|a_2| $ and $a_1=-| a_1| $, then we obtain
$(1-\beta ) | a_2| -|a_1| +b>0$. We can see that
$0<(1-\beta ) |a_2| -| a_1| +b<(1-\beta )| a_2| -| a_1|
<2[(1+\beta ) | a_2| +| a_1|] $. Therefore,
\[
0<\frac{(1-\beta ) | a_2| -|a_1| +b}{2[(1+\beta ) |a_2| +| a_1| ] }<1.
\]
From $0<\varepsilon <\frac{(1-\beta ) | a_2|-| a_1| +b}{2[(1+\beta ) |
a_2| +| a_1| ] }$, $\theta _1=\theta _2=\pi $ and $\cos \theta <0$, we obtain
\begin{align*}
\eta _3
&=[1-\beta -\varepsilon (1+\beta ) ] \delta
(a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) -b\cos \theta\\
&=-[1-\beta -\varepsilon (1+\beta ) ] |a_2| \cos \theta +(1+\varepsilon ) |
a_1| \cos \theta -b\cos \theta \\
&=(-\cos \theta ) \{[1-\beta -\varepsilon (
1+\beta ) ] | a_2| -(1+\varepsilon) | a_1| +b\} \\
&=(-\cos \theta ) \{(1-\beta ) |a_2| -| a_1| +b-\varepsilon [(1+\beta ) | a_2|
 +| a_1| ] \} \\
&>\frac{-1}{2}[(1-\beta ) | a_2|-| a_1| +b] \cos \theta >0.
\end{align*}
From $\eta _3>0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.30}
 is a contradiction.
\smallskip

\noindent (ii) $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$. For sufficiently large
$r$, we obtain \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16}, \eqref{e3.22}, and
\eqref{e3.25} hold. Substituting \eqref{e3.3}, \eqref{e3.4},
\eqref{e3.22}, \eqref{e3.25} and \eqref{e3.28} in
\eqref{e3.10}, we obtain
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\} 
&\leq | A_1e^{a_1z}|\\
&\leq M_1r^{M_2}e^{br\cos \theta }
 \exp \{r^{\gamma +\varepsilon}\} 
 \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \\
&\quad\times \exp \{(1+\varepsilon ) \delta (a_2z,\theta ) r\} . 
\end{aligned} \label{e3.31}
\end{equation}
From this inequality we have
\begin{equation}
\exp \{\eta _4r\} \leq M_1r^{M_2}\exp \{r^{\gamma
+\varepsilon }\} ,  \label{e3.32}
\end{equation}
where
\[
\eta _4=(1-\varepsilon ) \delta (a_1z,\theta )
-(1+\varepsilon ) \alpha \delta (a_1z,\theta )
-(1+\varepsilon ) \delta (a_2z,\theta ) -b\cos \theta .
\]
Since $a_1<\frac{a_2+b}{1-\alpha },a_2=-| a_2| $
and $a_1=-| a_1| $, then we obtain $(1-\alpha) | a_1| -| a_2| +b>0$. We
can see that $0<(1-\alpha ) | a_1|
-| a_2| +b<(1-\alpha ) |a_1| -| a_2| <2[(1+\alpha) | a_1| +| a_2| ] $.
Therefore,
\[
0<\frac{(1-\alpha ) | a_1| -|a_2| +b}{2[(1+\alpha ) |a_1| +| a_2| ] }<1.
\]
From
\[
0<\varepsilon <\frac{(1-\alpha ) | a_1|
-| a_2| +b}{2[(1+\alpha ) |a_1| +| a_2| ] },
\]
$\theta _1=\theta_2=\pi $ and $\cos \theta <0$, we obtain
\begin{align*}
\eta _4
&=(-\cos \theta ) \{(1-\alpha )| a_1| -| a_2|
 +b-\varepsilon [(1+\alpha ) | a_1| +|a_2| ] \}\\
&>\frac{-1}{2}[(1-\alpha ) | a_1| -| a_2| +b] \cos \theta >0.
\end{align*}
By $\eta _4>0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.32}
is a contradiction. Concluding the above proof, we obtain
$\sigma (f) =+\infty $.
\smallskip


\noindent\textbf{Second step.}
 We prove that $\sigma _2(f) =1$. By
\[
\max \{\sigma (B_je^{b_jz}+D_je^{d_jz})\; (j=1,\dots ,k-1) ,
\sigma (A_1e^{a_1z}+A_2e^{a_2z}) \} =1
\]
and Lemma \ref{lem2.4}, we obtain $\sigma _2(f) \leq 1$. By Lemma \ref{lem2.5},
we know that there exists a set $E_6\subset (1,+\infty ) $
with finite logarithmic measure and a constant $C>0$, such that for all $z$
satisfying $| z| =r\notin [0,1] \cup E_6$,
we obtain
\begin{equation}
| \frac{f^{(j) }(z)}{f(z)}| \leq C[
T(2r,f)] ^{j+1}\quad (j=1,\dots ,k) .  \label{e3.33}
\end{equation}

\noindent\textbf{Case 1.}
 $\arg a_1\neq \pi $ and $\arg a_1\neq \arg a_2$. In
first step, we have proved that there is a ray $\arg z=\theta $ where
 $ \theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $,
 satisfying
\[
\delta (a_1z,\theta ) >0,\quad
\text{ }\delta (a_2z,\theta ) <0\quad\text{or}\quad
\delta (a_1z,\theta ) <0,\quad \delta (a_2z,\theta ) >0.
\]
(a) When $\delta (a_1z,\theta ) >0$,
 $\delta ( a_2z,\theta ) <0$, for sufficiently large $r$, we obtain
\eqref{e3.4}--\eqref{e3.8} hold. Substituting \eqref{e3.4}, \eqref{e3.5},
\eqref{e3.8}, \eqref{e3.9} and \eqref{e3.33} in \eqref{e3.10}, we obtain
that for all $z=re^{i\theta }$
satisfying $| z| =r\notin [0,1] \cup E_6$,
$\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (
E_1\cup E_4\cup E_5) $,
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\}
&\leq | A_1e^{a_1z}| \\
&\leq M\exp \{r^{\gamma +\varepsilon }\} \exp \{(
1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\}[T(2r,f) ] ^{k+1},
\end{aligned}  \label{e3.34}
\end{equation}
where $M>0$ is a  constant. From \eqref{e3.34} and
$0<\varepsilon <\frac{1-\alpha }{2(1+\alpha ) }$, we obtain
\begin{equation}
\exp \big\{\frac{1-\alpha }{2}\delta (a_1z,\theta )r\big\}
\leq M\exp \{r^{\gamma +\varepsilon }\} [T(2r,f) ] ^{k+1}.  \label{e3.35}
\end{equation}
Since $\delta (a_1z,\theta ) >0$ and $\gamma +\varepsilon <1$,
then by using Lemma \ref{lem2.6} and \eqref{e3.35}, we obtain
$\sigma_2(f) \geq 1$. Hence $\sigma _2(f) =1$.
\smallskip

(b) When $\delta (a_1z,\theta ) <0,\delta (a_2z,\theta ) >0$,
for sufficiently large $r$, we obtain \eqref{e3.13}--\eqref{e3.17}
 hold. By using the a same reasoning as
above, we can get $\sigma _2(f) =1$.
\smallskip


\noindent\textbf{Case 2.}
$\arg a_1\neq \pi $, $\arg a_1=\arg a_2$.
In the first step, we have proved that there is a ray $\arg z=\theta $ where
$\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $,
satisfying $\delta (a_1z,\theta) >0$ and $\delta (a_2z,\theta ) >0$.
\smallskip


\noindent (i)  $| a_2| >\frac{| a_1| }{1-\beta }$. For sufficiently large $r$, we
have \eqref{e3.6}, \eqref{e3.13}, \eqref{e3.16}, \eqref{e3.21} and \eqref{e3.22} hold.
Substituting \eqref{e3.9}, \eqref{e3.13}, \eqref{e3.21}, \eqref{e3.22} and
\eqref{e3.33} in \eqref{e3.18}, we obtain that for all $z=re^{i\theta }$ satisfying
$| z| =r\notin [0,1] \cup E_6$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}
) \setminus (E_1\cup E_4\cup E_5) $,
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} 
&\leq | A_2e^{a_2z}| \\
&\leq M\exp \{r^{\gamma +\varepsilon }\} 
\exp \{(1+\varepsilon ) \delta (a_1z,\theta ) r\} \\
&\quad \times \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\}
 [T(2r,f) ] ^{k+1}.
\end{aligned}  \label{e3.36}
\end{equation}
From this inequality, we obtain
\begin{equation}
\exp \{\eta _1r\} \leq M\exp \{r^{\gamma +\varepsilon}\} [T(2r,f) ] ^{k+1},
  \label{e3.37}
\end{equation}
where
\[
\eta _1=(1-\varepsilon ) \delta (a_2z,\theta )
-(1+\varepsilon ) \delta (a_1z,\theta )
-(1+\varepsilon ) \beta \delta (a_2z,\theta ) .
\]
Since $\eta _1>0$ and $\gamma +\varepsilon <1$, then by using Lemma \ref{lem2.6}
and \eqref{e3.37}, we obtain $\sigma _2(f) \geq 1$.
Hence $\sigma _2(f) =1$.
\smallskip


\noindent (ii) $| a_2| <(1-\alpha ) | a_1| $. For sufficiently large $r$,
we have \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16},
\eqref{e3.22} and \eqref{e3.25} hold. By using the same
reasoning as above, we can get $\sigma _2(f) =1$.
\smallskip


\noindent\textbf{Case 3.} $a_1<0$ and $\arg a_1\neq \arg a_2$. In the
first step, we have proved that there is a ray $\arg z=\theta $ where
$\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $,
satisfying $\delta (a_2z,\theta) >0$ and $\delta (a_1z,\theta ) <0$. Using the same
reasoning as in second step (Case 1 (b)), we
can get $\sigma _2(f) =1$.
\smallskip

\noindent \textbf{Case 4.}
(i)  $(1-\beta )a_2-b<a_1<0$ and $a_2<\frac{b}{1-\beta }$ or
(ii) $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$.
In the first step, we have proved that there is a ray $\arg z=\theta $,
where $\theta \in (\frac{\pi }{2},\frac{3\pi }{2}) \setminus (E_1\cup E_4\cup
E_5) $, satisfying $\delta (a_2z,\theta ) >0$ and $
\delta (a_1z,\theta ) >0$.
\smallskip


\noindent(i)  $(1-\beta ) a_2-b<a_1<0$
and $a_2<\frac{b}{1-\beta }$. For sufficiently large $r$, we obtain
\eqref{e3.6}, \eqref{e3.13}, \eqref{e3.16}, \eqref{e3.21} and
\eqref{e3.22} hold. Substituting \eqref{e3.13}, \eqref{e3.21}, \eqref{e3.22},
\eqref{e3.28} and \eqref{e3.33} in \eqref{e3.18}, we
obtain that for all $z=re^{i\theta }$ satisfying
$| z| =r\notin [0,1] \cup E_6$, $\theta \in (\frac{\pi }{2},
\frac{3\pi }{2}) \setminus (E_1\cup E_4\cup E_5) $,
\begin{equation}
\begin{aligned}
\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\}
&\leq | A_2e^{a_2z}| \\
&\leq Me^{br\cos \theta }\exp \{r^{\gamma +\varepsilon }\} \exp
\{(1+\varepsilon ) \delta (a_1z,\theta )r\} \\
&\quad\times \exp \{(1+\varepsilon ) \beta \delta ( a_2z,\theta ) r\}
 [T(2r,f) ] ^{k+1}.
\end{aligned} \label{e3.38}
\end{equation}
From this inequality we obtain
\begin{equation}
\exp \{\eta _3r\} \leq M\exp \{r^{\gamma +\varepsilon
}\} [T(2r,f) ] ^{k+1},  \label{e3.39}
\end{equation}
where
\[
\eta _3=(1-\varepsilon ) \delta (a_2z,\theta )
-(1+\varepsilon ) \delta (a_1z,\theta ) -(
1+\varepsilon ) \beta \delta (a_2z,\theta ) -b\cos
\theta .
\]
Since $\eta _3>0$ and $\gamma +\varepsilon <1$, then by using Lemma \ref{lem2.6}
and \eqref{e3.39}, we obtain $\sigma _2(f) \geq 1$.
Hence $\sigma _2(f) =1$.
\smallskip

\noindent(ii) $a_1<\frac{a_2+b}{1-\alpha }$ and
$a_2<0$. For sufficiently large $r$, we obtain \eqref{e3.4},
\eqref{e3.6}, \eqref{e3.16}, \eqref{e3.22} and \eqref{e3.25} hold.
 By using the same reasoning as above, we can get $\sigma
_2(f) =1$.  Concluding the above proof, we obtain that every
solution $f(\not\equiv 0) $ of \eqref{e1.2} satisfies $\sigma _2(f) =1$.
The proof of Theorem \ref{thm1.1} is complete.


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

 Set $R_0(z) =A_1e^{a_1z}+A_2e^{a_2z}$ and
$R_i(z) =B_ie^{b_iz}+D_ie^{d_iz}$ $(i=1,\dots,k-1) $. Assume
$f(\not\equiv 0) $ is a solution of \eqref{e1.2}.
Then $\sigma (f) =+\infty $ by Theorem \ref{thm1.1}. Set
$g_0(z) =f(z) -\varphi (z) $. Then we have
$\sigma (g_0) =\sigma (f) =\infty $. Substituting $f=g_0+\varphi $ into
\eqref{e1.2}, we obtain
\begin{equation}
\begin{aligned}
&g_0^{(k) }+R_{k-1}g_0^{(k-1) }+\dots +R_2g_0''+R_1g_0'+R_0g_0\\
&=-[\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1)
}+\dots +R_2\varphi ''+R_1\varphi '+R_0\varphi ] .
\end{aligned} \label{e4.1}
\end{equation}
We can rewrite \eqref{e4.1} in the form
\begin{equation}
g_0^{(k) }+h_{0,k-1}g_0^{(k-1) }+\dots
+h_{0,2}g_0''+h_{0,1}g_0'+h_{0,0}g_0=h_0,
\label{e4.2}
\end{equation}
where
\[
h_0=-[\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1)
}+\dots +R_2\varphi ''+R_1\varphi '+R_0\varphi ] .
\]
We prove that $h_0\not\equiv 0$. In fact, if $h_0\equiv 0$, then
\[
\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1) }+\dots
+R_2\varphi ''+R_1\varphi '+R_0\varphi =0.
\]
Hence, $\varphi \not\equiv 0$ is a solution of \eqref{e1.2}
with $\sigma (\varphi ) =+\infty $ by Theorem \ref{thm1.1}, which is a
contradiction. Hence, $h_0\not\equiv 0$ is proved. By Lemma \ref{lem2.7} and
\eqref{e4.2} we know that $\overline{\lambda }(g_0) =
\overline{\lambda }(f-\varphi ) =\sigma (g_0)
=\sigma (f) =\infty $.

 Now we prove that $\overline{\lambda }(f'-\varphi) =\infty $.
Set $g_1(z) =f'(z) -\varphi (z) $. Then we have
$\sigma (g_1) =\sigma (f') =\sigma (f) =\infty $. Differentiating
both sides of equation \eqref{e1.2}, we obtain
\begin{equation}
\begin{aligned}
f^{(k+1) }+R_{k-1}f^{(k) }+(R_{k-1}'+R_{k-2}) f^{(k-1) }+(R_{k-2}'+R_{k-3})
f^{(k-2) } \\
+\dots +(R_3'+R_2) f'''+(R_2'+R_1) f''+(R_1'+R_0) f'+R_0'f=0.
\end{aligned}  \label{e4.3}
\end{equation}
By \eqref{e1.2}, we have
\begin{equation}
f=-\frac{1}{R_0}[f^{(k) }+R_{k-1}f^{(k-1)
}+\dots +R_2f''+R_1f'] .  \label{e4.4}
\end{equation}
Substituting \eqref{e4.4} into \eqref{e4.3}, we have
\begin{equation}
\begin{aligned}
&f^{(k+1) }+\Big(R_{k-1}-\frac{R_0'}{R_0}\Big)
f^{(k) }+\Big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\Big) f^{(k-1) }\\
&+\Big(R_{k-2}'+R_{k-3}-R_{k-2}\frac{R_0'}{R_0}
\Big) f^{(k-2) }+\dots +\Big(R_3'+R_2-R_3
\frac{R_0'}{R_0}\Big) f'''\\
&+\Big(R_2'+R_1-R_2\frac{R_0'}{R_0}\Big)
f''+\Big(R_1'+R_0-R_1\frac{R_0'}{
R_0}\Big) f'=0.
\end{aligned}  \label{e4.5}
\end{equation}
We can write equation \eqref{e4.5} in the form
\begin{equation}
f^{(k+1) }+h_{1,k-1}f^{(k) }+h_{1,k-2}f^{(
k-1) }+\dots +h_{1,2}f'''+h_{1,1}f''+h_{1,0}f'=0,  \label{e4.6}
\end{equation}
where
\begin{gather*}
h_{1,i}=R_{i+1}'+R_i-R_{i+1}\frac{R_0'}{R_0}\quad (i=0,1,\dots ,k-2) , \\
h_{1,k-1}=R_{k-1}-\frac{R_0'}{R_0}.
\end{gather*}
Substituting
$f^{(j+1) }=g_1^{(j) }+\varphi ^{(j) }$ $(j=0,\dots ,k) $ into \eqref{e4.6},
we obtain
\begin{equation}
g_1^{(k) }+h_{1,k-1}g_1^{(k-1)}+h_{1,k-2}g_1^{(k-2) }+\dots
+h_{1,2}g_1''+h_{1,1}g_1'+h_{1,0}g_1=h_1,  \label{e4.7}
\end{equation}
where
\[
h_1=-[\varphi ^{(k) }+h_{1,k-1}\varphi ^{(
k-1) }+h_{1,k-2}\varphi ^{(k-2) }+\dots +h_{1,2}\varphi
''+h_{1,1}\varphi '+h_{1,0}\varphi ] .
\]
We can get
\begin{equation}
h_{1,i}(z) =\frac{N_i(z) }{R_0(z) }\quad (i=0,1,\dots ,k-1) ,  \label{e4.8}
\end{equation}
where
\begin{gather}
N_0=R_1'R_0+R_0^{2}-R_1R_0',  \label{e4.9} \\
N_i=R_{i+1}'R_0+R_iR_0-R_{i+1}R_0'\quad  (i=1,2,\dots ,k-2) ,  \label{e4.10}\\
N_{k-1}=R_{k-1}R_0-R_0'.  \label{e4.11}
\end{gather}
Now we prove that $h_1\not\equiv 0$. In fact, if $h_1\equiv 0$, then
$\frac{h_1}{\varphi }\equiv 0$. Hence, by \eqref{e4.8} we obtain
\begin{equation}
\frac{\varphi ^{(k) }}{\varphi }R_0+\frac{\varphi ^{(k-1) }}{\varphi }N_{k-1}
+\frac{\varphi ^{(k-2) }}{\varphi }
N_{k-2}+\dots +\frac{\varphi ''}{\varphi }N_2+\frac{
\varphi '}{\varphi }N_1+N_0=0.  \label{e4.12}
\end{equation}
Obviously, $\frac{\varphi ^{(j) }}{\varphi }$
 $(j=1,\dots ,k) $ are meromorphic functions with
$\sigma (\frac{\varphi ^{(j) }}{\varphi }) <1$. By
\eqref{e4.9}--\eqref{e4.11} we can rewrite \eqref{e4.12} in the form
\begin{equation}
A_1^{2}e^{2a_1z}+A_2^{2}e^{2a_2z}+\underset{\lambda \in
I_1'}{\sum }f_{\lambda }e^{\lambda z}=0,  \label{e4.13}
\end{equation}
where $I_1'=I_1\setminus \{2a_1,2a_2\} $ and $f_{\lambda }$
 $(\lambda \in I_1') $ are meromorphic
functions with order less than $1$.
\smallskip


\noindent(1) If $(2a_1) \notin I_1\setminus \{2a_1\} $, then we write
\eqref{e4.13} in the form
\[
A_1^{2}e^{2a_1z}+\underset{\lambda \in \Gamma _1}{\sum }g_{1,\lambda
}e^{\lambda z}=0,
\]
where $\Gamma _1\subseteq I_1\setminus \{2a_1\} $,
$g_{1,\lambda }$ $(\lambda \in \Gamma _1) $ are meromorphic
functions with order less than $1$ and $2a_1$, $\lambda $
$(\lambda \in \Gamma _1) $ are distinct numbers. By Lemmas \ref{lem2.8}
and \ref{lem2.9},
we obtain $A_1\equiv 0$, which is a contradiction.
\smallskip

\noindent (2) If $(2a_2) \notin I_1\setminus \{2a_2\} $, then we write \eqref{e4.13}
 in the form
\[
A_2^{2}e^{2a_2z}+\underset{\lambda \in \Gamma _2}{\sum }g_{2,\lambda}e^{\lambda z}=0,
\]
where $\Gamma _2\subseteq I_1\setminus \{2a_2\} $, $g_{2,\lambda }$
$(\lambda \in \Gamma _2) $ are meromorphic
functions with order less than $1$ and $2a_2$, $\lambda $
$(\lambda \in \Gamma _2) $ are distinct numbers. By Lemmas \ref{lem2.8} and
\ref{lem2.9},
we obtain $A_2\equiv 0$, which is a contradiction. Hence, $h_1\not\equiv 0$
is proved. By Lemma \ref{lem2.7} and \eqref{e4.7} we know that
$\overline{ \lambda }(g_1) =\overline{\lambda }(f'-\varphi )
=\sigma (g_1) =\sigma (f) =\infty $.


 Now we prove that $\overline{\lambda }(f''-\varphi ) =\infty $.
 Set $g_2(z) =f''(z) -\varphi (z) $. Then we have
$\sigma (g_2) =\sigma (f'') =\sigma (f) =\infty $. Differentiating both sides
of equation \eqref{e4.3}, we have
\begin{equation}
\begin{aligned}
&f^{(k+2) }+R_{k-1}f^{(k+1) }+(2R_{k-1}'+R_{k-2}) f^{(k) }
+(R_{k-1}''+2R_{k-2}'+R_{k-3}) f^{(k-1) } \\
& +(R_{k-2}''+2R_{k-3}'+R_{k-4}) f^{(k-2) }+\dots +(R_3''+2R_2'+R_1) f'''\\
&+(R_2''+2R_1'+R_0) f''+(R_1''+2R_0') f'+R_0''f=0.
\end{aligned}  \label{e4.14}
\end{equation}
By \eqref{e4.4} and \eqref{e4.14}, we have
\begin{equation}
\begin{aligned}
&f^{(k+2) }+R_{k-1}f^{(k+1) }+\Big(2R_{k-1}'+R_{k-2}-\frac{R_0''}{R_0}\Big) f^{(k) }\\
&+\Big(R_{k-1}''+2R_{k-2}'+R_{k-3}-R_{k-1}\frac{R_0''}{R_0}\Big) f^{(k-1) } +\dots \\
&+\Big(R_4''+2R_3'+R_2-R_4\frac{
R_0''}{R_0}\Big) f^{(4) }+\Big(R_3''+2R_2'+R_1-R_3\frac{R_0''}{R_0}\Big) f'''\\
&+\Big(R_2''+2R_1'+R_0-R_2\frac{R_0''}{R_0}\Big) f''+\Big(
R_1''+2R_0'-R_1\frac{R_0''}{R_0}\Big) f'=0.
\end{aligned}  \label{e4.15}
\end{equation}
Now we prove that $R_1'+R_0-R_1\frac{R_0'}{R_0}\not\equiv 0$. Suppose
that $R_1'+R_0-R_1\frac{R_0'}{R_0}\equiv 0$, then we have
\begin{equation}
A_1^{2}e^{2a_1z}+A_2^{2}e^{2a_2z}+\underset{\lambda \in
I_2'}{\sum }f_{\lambda }e^{\lambda z}=0,  \label{e4.16}
\end{equation}
where $I_2'=I_2\setminus \{2a_1,2a_2\} $ and $f_{\lambda }$
$(\lambda \in I_2') $ are meromorphic functions with order less than $1$.
 By using the same reasoning as above, we
can get a contradiction. Hence, $R_1'+R_0-R_1\frac{R_0'}{R_0}\not\equiv 0$
 is proved. Set
\begin{equation}
\psi (z) =R_1'R_0+R_0^{2}-R_1R_0'
\text{ and }\phi (z) =R_1''R_0+2R_0'R_0-R_1R_0''.  \label{e4.17}
\end{equation}
By \eqref{e4.5} and \eqref{e4.17}, we obtain
\begin{equation}
\begin{aligned}
f'&=\frac{-R_0}{\psi (z) }\Big\{f^{(k+1)
}+\Big(R_{k-1}-\frac{R_0'}{R_0}\Big) f^{(k)
}+\Big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\Big) f^{(k-1) }\\
&\quad +\Big(R_{k-2}'+R_{k-3}-R_{k-2}\frac{R_0'}{R_0}
\Big) f^{(k-2) }+\dots +\Big(R_2'+R_1-R_2
\frac{R_0'}{R_0}\Big) f''\Big\} .
\end{aligned} \label{e4.18}
\end{equation}
Substituting \eqref{e4.17} and \eqref{e4.18} into \eqref{e4.15}, we obtain
\begin{equation}
\begin{aligned}
&f^{(k+2) }+[R_{k-1}-\frac{\phi }{\psi }] f^{(k+1) }
+\Big[2R_{k-1}'+R_{k-2}-\frac{R_0''}{
R_0}-\frac{\phi }{\psi }\big(R_{k-1}-\frac{R_0'}{R_0}
\big) \Big] f^{(k) }\\
&+\Big[R_{k-1}''+2R_{k-2}'+R_{k-3}-R_{k-1}\frac{
R_0''}{R_0}-\frac{\phi }{\psi }\big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\big)
\Big] f^{(k-1) } \\
&+\dots +\Big[R_3''+2R_2'+R_1-R_3\frac{
R_0''}{R_0}-\frac{\phi }{\psi }\big(R_3'+R_2-R_3\frac{R_0'}{R_0}\big) \Big] f'''
\\
&+\Big[R_2''+2R_1'+R_0-R_2\frac{
R_0''}{R_0}-\frac{\phi }{\psi }\big(R_2'+R_1-R_2\frac{R_0'}{R_0}\big) \Big] f''=0.
\end{aligned}\label{e4.19}
\end{equation}
We can write \eqref{e4.19}  in the form
\begin{equation}
f^{(k+2) }+h_{2,k-1}f^{(k+1) }+h_{2,k-2}f^{(
k) }+\dots +h_{2,2}f^{(4) }+h_{2,1}f'''+h_{2,0}f''=0,  \label{e4.20}
\end{equation}
where
\begin{gather*}
\begin{aligned}
h_{2,i}&=R_{i+2}''+2R_{i+1}'+R_i-R_{i+2}\frac{R_0''}{R_0} \\
&\quad -\frac{\phi (z) }{\psi (z) }\Big(R_{i+2}'+R_{i+1}-R_{i+2}\frac{R_0'}{R_0}\Big)
\quad (i=0,1,\dots ,k-3),
\end{aligned}
\\
h_{2,k-2}=2R_{k-1}'+R_{k-2}-\frac{R_0''}{R_0}-
\frac{\phi (z) }{\psi (z) }\Big(R_{k-1}-\frac{R_0'}{R_0}\Big) , \\
h_{2,k-1}=R_{k-1}-\frac{\phi (z) }{\psi (z) }.
\end{gather*}
Substituting $f^{(j+2) }=g_2^{(j) }+\varphi^{(j) }$ $(j=0,\dots ,k) $ in
 \eqref{e4.20}  we have
\begin{equation}
g_2^{(k) }+h_{2,k-1}g_2^{(k-1)
}+h_{2,k-2}g_2^{(k-2) }+\dots +h_{2,1}g_2'+h_{2,0}g_2=h_2,  \label{e4.21}
\end{equation}
where
\[
h_2=-[\varphi ^{(k) }+h_{2,k-1}\varphi ^{(
k-1) }+h_{2,k-2}\varphi ^{(k-2) }+\dots +h_{2,2}\varphi
''+h_{2,1}\varphi '+h_{2,0}\varphi ] .
\]
We obtain
\begin{equation}
h_{2,i}=\frac{L_i(z) }{\psi (z) }\quad (i=0,1,\dots ,k-1) ,  \label{e4.22}
\end{equation}
where
\begin{gather}
\begin{aligned}
L_0(z) &=R_2''R_1'R_0+R_2''R_0^{2}-R_2''R_1R_0'+2{R_1'}^2 R_0+3R_1'R_0^{2}
-2R_1'R_1R_0'+R_0^{3} \\
&\quad -3R_1R_0'R_0-R_2R_1'R_0''-R_2R_0''R_0-R_2'R_1''R_0-2R_2'R_0'R_0+R_2'R_1R_0''
\\
&\quad -R_1''R_1R_0+R_1^{2}R_0''+R_2R_1''R_0'+2R_2{R_0'}^2,
\end{aligned} \label{e4.23}
\\
\begin{aligned}
L_i&=R_{i+2}''R_1'R_0+R_{i+2}''R_0^{2}-R_{i+2}''R_1R_0'+2R_{i+1}'R_1'R_0
+2R_{i+1}'R_0^{2}-2R_{i+1}'R_1R_0' \\
&\quad +R_iR_1'R_0+R_iR_0^{2}-R_iR_1R_0'-R_{i+2}R_1'R_0''-R_{i+2}R_0''R_0
-R_{i+2}'R_1''R_0 \\
&\quad -2R_{i+2}'R_0'R_0+R_{i+2}'R_1R_0''-R_{i+1}R_1''R_0-2R_{i+1}R_0'R_0+R_{i+1}R_1R_0''\\
&\quad +R_{i+2}R_1''R_0'+2R_{i+2}{R_0'}^2\quad (i=1,2,\dots ,k-3) ,
\end{aligned}  \label{e4.24}
\\
\begin{aligned}
L_{k-2}&=2R_{k-1}'R_1'R_0+2R_{k-1}'R_0^{2}-2R_{k-1}'R_1R_0'+R_{k-2}R_1'R_0
 +R_{k-2}R_0^{2} \\
&\quad -R_{k-2}R_1R_0'-R_1'R_0''-R_0''R_0-R_{k-1}R_1''R_0-2R_{k-1}R_0'R_0 \\
&\quad +R_{k-1}R_1R_0''+R_1''R_0'+2{R_0'}^2,
\end{aligned} \label{e4.25}
\\
L_{k-1}=R_{k-1}R_1'R_0+R_{k-1}R_0^{2}-R_{k-1}R_1R_0'-R_1''R_0-2R_0'R_0+R_1R_0''.  \label{e4.26}
\end{gather}
Therefore,
\begin{equation}
\frac{-h_2}{\varphi }=\frac{1}{\psi }[\frac{\varphi ^{(
k) }}{\varphi }\psi +\frac{\varphi ^{(k-1) }}{\varphi }
L_{k-1}+\dots +\frac{\varphi ''}{\varphi }L_2+\frac{
\varphi '}{\varphi }L_1+L_0] .  \label{e4.27}
\end{equation}
Now we prove that $h_2\not\equiv 0$. In fact, if $h_2\equiv 0$, then
$\frac{-h_2}{\varphi }\equiv 0$. Hence, by \eqref{e4.27} we obtain
\begin{equation}
\frac{\varphi ^{(k) }}{\varphi }\psi +\frac{\varphi ^{(
k-1) }}{\varphi }L_{k-1}+\dots +\frac{\varphi ''}{
\varphi }L_2+\frac{\varphi '}{\varphi }L_1+L_0=0.  \label{e4.28}
\end{equation}
Obviously, $\frac{\varphi ^{(j) }}{\varphi }$
$(j=1,\dots,k) $ are meromorphic functions with
$\sigma (\frac{\varphi^{(j) }}{\varphi }) <1$. By \eqref{e4.17} and
\eqref{e4.23}--\eqref{e4.26}, we can rewrite \eqref{e4.28} in the form
\begin{equation}
A_1^{3}e^{3a_1z}+A_2^{3}e^{3a_2z}+\sum_{\lambda \in I_3'}
f_{\lambda }e^{\lambda z}=0,  \label{e4.29}
\end{equation}
where $I_3'=I_3\setminus \{3a_1,3a_2\} $ and
$f_{\lambda }$ $(\lambda \in I_3') $ are meromorphic
functions with order less than $1$.
\smallskip

\noindent(1) If $(3a_1) \notin I_3\setminus \{3a_1\} $,
then we write \eqref{e4.29}  in the form
\[
A_1^{3}e^{3a_1z}+\underset{\lambda \in \Gamma _1}{\sum }g_{1,\lambda
}e^{\lambda z}=0,
\]
where $\Gamma _1\subseteq I_3\setminus \{3a_1\} $,
$g_{1,\lambda }$ $(\lambda \in \Gamma _1) $ are meromorphic
functions with order less than $1$ and $3a_1$, $\lambda $
$(\lambda \in \Gamma _1) $ are distinct numbers. By Lemmas \ref{lem2.8}
and \ref{lem2.9},
we obtain $A_1\equiv 0$, which is a contradiction.
\smallskip

\noindent(2) If $(3a_2) \notin I_3\setminus \{3a_2\} $, then we write \eqref{e4.29}
 in the form
\[
A_2^{3}e^{3a_2z}+\underset{\lambda \in \Gamma _2}{\sum }g_{2,\lambda
}e^{\lambda z}=0,
\]
where $\Gamma _2\subseteq I_3\setminus \{3a_2\} $,
$g_{2,\lambda }$ $(\lambda \in \Gamma _2) $ are meromorphic
functions with order less than $1$ and $3a_2$, $\lambda $
$(\lambda \in \Gamma _2) $ are distinct numbers.
By Lemmas \ref{lem2.8} and \ref{lem2.9}, we obtain $A_2\equiv 0$, which is a contradiction.
Hence, $h_2\not\equiv 0$ is proved. By Lemma \ref{lem2.7} and \eqref{e4.21}, we have
$\overline{ \lambda }(g_2) =\overline{\lambda }(f''-\varphi ) =\sigma (g_2)
=\sigma (f) =\infty $. The proof of Theorem \ref{thm1.2} is complete.


\begin{proof}[Proof of Corollary \ref{coro1.1}]
Setting $\varphi (z) =z$ and using the same reasoning
as in the proof of Theorem \ref{thm1.2}, we obtain Corollary \ref{coro1.1}.
\end{proof}

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\end{document}