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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 135, pp. 1--9.\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-2013/135\hfil Properties of meromorphic solutions]
{Properties of meromorphic solutions to certain 
differential-difference equations}

\author[X. Qi, L. Yang\hfil EJDE-2013/135\hfilneg]
{Xiaoguang Qi, Lianzhong Yang}  % in alphabetical order

\address{Xiaoguang Qi \newline
University of Jinan, School of Mathematics,
Jinan, Shandong 250022, China}
 \email{xiaogqi@gmail.com, xiaogqi@mail.sdu.edu.cn}

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

\thanks{Submitted March 15, 2013. Published June 20, 2013.}
\subjclass[2000]{34M05, 39B32, 30D35}
\keywords{Differential-difference equation; meromorphic solution;
\hfill\break\indent entire solution; finite order}

\begin{abstract}
 We consider the properties of meromorphic solutions to certain type of
 non-linear difference equations. Also we show the existence of meromorphic
 solutions with finite order for differential-difference equations
 related to the Fermat type functional equation.
 This article extends earlier results by Liu et al \cite{Liu}.
\end{abstract}

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

\section{Introduction}

In this article, we assume that the reader is familiar with standard
symbols and fundamental results of Nevanlinna Theory \cite{Ha, yyi}.
A meromorphic function will mean meromorphic in the whole complex
plane. In particular, we denote the order of growth of a meromorphic
function  $f(z)$ by $\sigma(f)$.
The values $m(r,f)$, $N(r,f)$, $\overline{N}(r,f)$ and $T(r,f)$ 
denote the proximity function, the counting function, the reduced 
counting function and the characteristic function of $f(z)$, respectively:
\begin{gather*}
m(r,f):=\frac{1}{2\pi}\int_{0}^{2\pi}\log^+|f(re^{i\theta})|\, d\theta,
\\
N(r,f):=\int_{0}^{r}\frac{n(t,f) - n(0,f)}{t}\,dt + n(0,f)\log r,
\\
\overline{N}(r,f):=\int_{0}^{r}\frac{\overline{n}(t,f) 
-\overline{n}(0,f)}{t}\,dt + \overline{n}(0,f)\log r,
\\
T(r,f):= m(r,f) + N(r,f),
\end{gather*}
where $\log^+x=\max(\log x, 0)$ for all $x\geq 0$,  $n(t,f)$ 
denotes the number of poles of $f(z)$ in the disc $|z|\leq t$, 
counting multiplicities; and $\overline{n}(t,f)$ denotes the number 
of poles of $f(z)$ in the disc $|z|\leq t$, ignoring multiplicities.


Nevanlinna's value distribution theory of meromorphic functions has
been used to study the growth, oscillation and existence of entire
or meromorphic solutions of differential equations. In 2001, Yang
\cite{yangc} started to study the existence and uniqueness of finite
order entire solutions of the following type of non-linear
differential equation
\begin{equation}\label{yang}
L(f)-p(z)f(z)^n=H(z).
\end{equation}
Subsequently, several papers have appeared in which the solutions of
equation \eqref{yang} are studied. The reader is referred to 
\cite{janne3, Lip1, Lip2}.


Recently, many articles focused on complex difference equations. The
background for these studies lies in the recent difference
counterparts of Nevanlinna theory. The key result here is the
difference analogue of the lemma on the logarithmic derivative
obtained by Halburd-Korhonen \cite{halburd6, halburd5} and
Chiang-Feng \cite{cfeng1}, independently.

Yang and Laine \cite{yL} gave difference, resp.
differential-difference, analogues of previous results concerning the
equation \eqref{yang}. In fact, they proved the following theorem.


\begin{theorem}[{\cite[Theorem 2.6]{yL}}] \label{thmA} 
 Let $n\geq4$ be an integer, $L(z, f)$ be a linear differential-difference
polynomial of $f(z)$, with small meromorphic coefficients, and
$H(z)$ be a meromorphic function of finite order. Then the
differential-difference equation
$$
f(z)^n+ L(z, f)=H(z)
$$
possesses at most one transcendental entire solution of finite
order, unless $L(z, f)$ vanishes identically. If such a solution
$f(z)$ exists, then $f(z)$ is of the same order as $H(z)$.
\end{theorem}


Using Theorem \ref{thmA}, the authors investigate the existence and 
the growth of meromorphic solutions with a few poles of the difference equation
\begin{equation}\label{1'}
f(z)^n+L(z, f)=H(z),
\end{equation}
where $L(z, f)=a_0f+a_1f(z+c_1)+\dots a_kf(z+c_k)$ is a linear
difference polynomials in $f(z)$ with small meromorphic functions as
the coefficients, and $c_i$ are constants, $i=1,2,\dots k$. Here,
$H(z)$ is meromorphic of finite order, and $n$ is an integer such
that $n\geq2$. In fact, if $n=0$ or $n=1$, then 
\eqref{1'} reduces to a linear difference equation, which has been
considered in \cite{chen, cfeng1, LY, zheng}.

AUTHORS: Please define $N(r, f)$ and $S(r, f)$  $S(r, f)$

\begin{theorem}\label{thm1}
Given $L(z, f)$ and $H(z)$ as above. If $f(z)$ is a finite order
meromorphic solution of  \eqref{1'} satisfying 
$N(r, f)=S(r, f)$ and $n\geq 4$, then one of the following statements
hold:

(1) Equation \eqref{1'} has $f(z)$ as its unique
transcendental meromorphic solution with finite order such that
$N(r, f)=S(r, f)$.

(2) Equation \eqref{1'} has exactly $n$ transcendental
meromorphic solutions,  $f_j$
$(j=1, 2, 3\dots n)$, with finite order
such that $N(r, f_j)=S(r, f_j)$.
\end{theorem}

Next, we consider the growth of meromorphic solutions of
 \eqref{1'}. In fact, using the same method as Theorem
\ref{thm1}, we prove the following result.

\begin{theorem}\label{thm3}
Given $L(z, f)$ and $H(z)$ as  above. Let $f_1(z)$ and $f_2(z)$ be
two distinct arbitrary solutions such that $N(r, f_i)=S(r, f_i)$
$(i= 1,2)$. Then
$$
T(r, f_1)=T(r, f_2)+S(r, f_1).
$$
\end{theorem}

\begin{theorem}\label{thm2}
Given $L(z, f)$ and $H(z)$ as above, assume that $f(z)$ is a
meromorphic solution of for \eqref{1'} with finite order. 
Then $\sigma(f)\leq\sigma(H)$. Furthermore, if $f(z)$ satisfies
any one of the following two conditions
\begin{itemize}
\item[(1)] $n\geq k+2$, or
\item[(2)] $N(r, f)=S(r, f)$,
\end{itemize}
then $\sigma(f)=\sigma(H)$.
\end{theorem}

\noindent\textbf{Remark.} It seems that replacing $L(z, f)$
with differential-difference polynomial in $f(z)$, the same
conclusions of Theorem \ref{thm1}--Theorem \ref{thm2} can be proved.


In a recent publication, Liu et al \cite{Liu,Liu1} discussed the existence
of entire solutions with finite order of the  Fermat type
differential-difference equation
\begin{equation}\label{00'}
(f'(z))^n+f(z+c)^m=1.
\end{equation}
They showed that the above equation has no transcendental entire
solutions with finite order, provided that $m\neq n$, where $n, m$
are positive integers. Here and in the following, $c$ is a non-zero
constant, unless otherwise specified. It is natural to ask what
happens if the right side of  \eqref{00'} is a
meromorphic function $H(z)$. Corresponding to this question, we give
the following results:

\begin{theorem}\label{thm0}
Let $f(z)$ be a transcendental meromorphic function with finite
order, $m$ and $n$ be two positive integers such that $m\geq 2n+4$
and $H(z)$ be a meromorphic function satisfying 
$\overline{N}(r,1/H)=S(r, f)$. Then $f(z)$ is not a solution 
of the equation
\begin{equation}\label{0}
(f'(z))^n+f(z+c)^m=H(z).
\end{equation}
\end{theorem}

Using a similar reasoning as in Theorem \ref{thm0}, we conclude have the following
result.

\begin{corollary}\label{coro1}
Let $f(z)$ be a transcendental entire function with finite order,
$m$ and $n$ be two positive integers such that $m\geq n+2$ and
$H(z)$ be a meromorphic function satisfying 
$\overline{N}(r, 1/H)=S(r,f)$. Then $f(z)$ is not a solution of 
\eqref{0}.
\end{corollary}


\noindent\textbf{Remarks} 
(1) Corollary \ref{coro1} does not hold when $n=m$.
In particular,
$$
f'(z)+f(z+2\pi i)=2e^z
$$
admits a transcendental entire solution, $e^z$. 
This implies that the restriction that $m\geq n+2$ is necessary. 
Meanwhile, we considered Corollary \ref{coro1} for $m=n+1$.
Unfortunately, we have not succeed.

(2) Let $f(z)=\cos z$, then  $f(z)$ is a transcendental
entire solution of the  equation
$$
f'(z)+f(z-\frac{\pi}{2})^3=\sin z(\sin^2z-1).
$$
Indeed, this example shows Corollary \ref{coro1} cannot hold when
$\overline{N}(r, 1/H)\neq S(r, f)$, 
(If $\overline{N}(r, 1/H)=\overline{N}(r,\frac{1}{\sin z(\sin^2z-1)})=S(r, f)$,
then we can have a contradiction by the second main theorem.)
which means the assumption $\overline{N}(r, 1/H)=S(r, f)$ in
Corollary \ref{coro1} is sharp.

(3) If we omit the restriction of the order of the solutions, then
 \eqref{0} may have an infinite order entire solution.
Indeed, $f(z)=e^{e^{z}}$ is an entire function with infinite order
and solves the equation
$$
f'(z)+f(z+\ln\frac{1}{3})^3=(e^z+1)e^{e^{z}}.
$$

(4) Some ideas in this paper are based on \cite{G, janne3}.


\begin{theorem}\label{thm7}
Let $f(z)$ be a transcendental entire function with finite order,
$m$ and $n$ be two positive integers such that $m\neq n$ and $H(z)$
be a small function of $f(z)$. Then $f(z)$ is not a solution of
equation \eqref{0}.
\end{theorem}

The proof of Theorem \ref{thm7} is similar to the proof of 
\cite[Theorem 1.2]{Liu}. One can apply the the second 
main theorem for small target functions, instead of the classical second main
theorem, and use an elementary computation. Therefore, we omit the
proof here.

\section{Preliminaries}

\begin{lemma}[{\cite[Theorem 2.1]{cfeng1}}] \label{lem2.5}
Let $f(z)$ be a finite order meromorphic function, then for each
$\varepsilon>0$,
$$ 
T(r, f(z + c)) = T(r,f(z))+ O(r^{\sigma(f)-1+\varepsilon}) + O(\log r)
$$ 
and
$$
\sigma(f(z+c))=\sigma(f(z)).
$$
Thus, if $f(z)$ is a transcendental meromorphic function with finite
order, then 
$$
T(r, f(z+c))=T(r, f)+S(r, f).
$$
\end{lemma}

 \begin{lemma}[{\cite[Theorem 2.1]{halburd5}}] \label{lem2.3}
Let $f(z)$ be a meromorphic function with finite order, and let
$c\in\mathbb{C}$ and $\delta\in(0, 1)$. Then
\begin{equation*}
m\Big(r,\frac{f(z+c)}{f(z)}\Big)+ m\Big(r,\frac{f(z)}{f(z+c)}\Big)
=o\Big(\frac{T(r, f)}{r^\delta}\Big)=S(r,f),
\end{equation*}
outside of a possible set $E$ with finite logarithmic
measure.
\end{lemma}

\begin{lemma}[{\cite[Lemma 5]{G1}}]\label{lem3}
Let $F$ and $G$ be non-decreasing functions on $(0, +\infty)$.
 If $F(r)\leq G(r)$ for $r\not\in E\cup[0, 1]$, where the set 
$E\subset(1, +\infty)$ has  finite logarithmic measure, then, 
for any constant $\alpha >1$, there exists a value
 $r_0>0$, such that $F(r)\leq G(\alpha r)$ for $r>r_0$.
\end{lemma}

\begin{lemma} \label{lem1}
Let $f(z)$ be a meromorphic solution of \eqref{0}, and
\begin{equation}\label{0'}
G(z)=\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}.
\end{equation} 
Then
$$
N(r,G)\leq \overline{N}(r, \frac{1}{H})+\overline{N}(r,
f)+\overline{N}(r, \frac{1}{f(z+c)})+S(r, f).
$$
\end{lemma}

\noindent\textbf{Remark}.In the following proof, first impression of
the reader is that the poles of $G(z)$ are at the poles of $H(z)$ as
well. But looking at the equation \eqref{0}, one realizes that the
poles of $H(z)$ should be at the poles of $f(z)$ and $f(z + c)$.
Hence, it is sufficient to discuss the poles of $f(z)$ and $f(z + c)$
here.


\begin{proof} 
Observe that, the poles of $G(z)$ are at the
zeros of $H(z)$ and $f(z+c)$, and at the poles of $f(z)$, $f(z+c)$
from \eqref{0} and \eqref{0'}. If $z_0$ is a zero of $H(z)$, zero of
$f(z+c)$ , or pole of $f(z)$, then $z_0$ is at most a simple pole of
$G(z)$ by \eqref{0} and \eqref{0'}. If $z_0$ is a pole of $f(z+c)$
but not a pole of $f(z)$, then by the Laurent expansion of $G(z)$ at
$z_0$, we obtain that $G(z)$ is analytic at $z_0$. Hence, from the
discussions above, we can conclude that
$$
N(r,G)\leq \overline{N}(r, \frac{1}{H})+\overline{N}(r,
f)+\overline{N}(r, \frac{1}{f(z+c)})+S(r, f).
$$
\end{proof}

\section{Proof of main resutls}

\begin{proof}[Proof of Theorem \ref{thm1}]
Suppose $f_1(z)$, $f_2(z)$ are two distinct finite order meromorphic
solutions of \eqref{1'} such that $N(r, f_i)=S(r, f_i)$ $(i=1, 2)$.
From  \eqref{1'}, we know that
  \begin{equation}\label{2013}
       \frac{L(z, f_1)-L(z,
       f_2)}{f_2-f_1}=\frac{f_{1}^{n}-f_{2}^{n}}{f_1-f_2}=F(z)
       =\frac{L(z, f_1-f_2)}{f_2-f_1},
  \end{equation}
where $F(z)=(f_1-t_1f_2)(f_1-t_2f_2) \dots (f_1-t_{n-1}f_2)$. Here
$t_j\neq 1$ $(j=1, 2, \dots n-1)$ are the distinct $n$-th roots of
the unity. From Lemma \ref{lem2.3} to \eqref{2013}, we obtain
$$
m(r, F)=S(r, f_1)+S(r, f_2). 
$$ 
Since $N(r, f_i)=S(r, f_i)$, it
follows that $N(r, F)=S(r, f_1)+S(r, f_2)$. 
Hence
 \begin{equation}\label{2012}
T(r, F)=S(r, f_1)+S(r, f_2).
  \end{equation}
We will discuss the following two cases.

Case 1. If $F(z)\equiv 0$, then we conclude that
$f_{1}^{n}=f_{2}^{n}$, that is, $f_2=t_jf_1$. Substituting
$f_2=t_jf_1$ into \eqref{2013}, we have $L(z, f_1)-L(z,
t_jf_1)=(1-t_j)L(z, f_1)=0$. Hence, $L(z, f_1)=0$ and $L(z,
t_jf_1)=0$. This means $f_1$ and $t_jf_1$ $(j=1, 2, \dots n-1)$ are
the solutions of \eqref{1'}, as asserted in part (2).

Case 2. If $F(z)\not\equiv 0$, then
 \begin{equation}\label{2'}
       F(z)=f_{2}^{n-1}\big(\frac{f_1}{f_2}-t_1\big)
\big(\frac{f_1}{f_2}-t_2\big)\dots \big(\frac{f_1}{f_2}-t_{n-1}\big).
      \end{equation}
Then equation \eqref{2'} gives
$$
\frac{F(z)}{f_{2}^{n-1}}=P(\frac{f_1}{f_2}),
$$
where $P$ is a polynomial in $f_1/f_2$ of degree $n-1$ with
constant coefficients. Applying Valiron-Mohon'ko theorem and
\eqref{2012} to the above equation, we have
\begin{equation}\label{3'}
(n-1)T(r, \frac{f_1}{f_2})=T(r, \frac{F(z)}{f_{2}^{n-1}})=(n-1)T(r,
f_2)+S(r, f_1)+S(r, f_2).
      \end{equation}
Using the same way, we obtain
\begin{equation}\label{33'}
(n-1)T(r, \frac{f_2}{f_1})=T(r, \frac{F(z)}{f_{1}^{n-1}})=(n-1)T(r,
f_1)+S(r, f_1)+S(r, f_2)
      \end{equation}
as well. Combining \eqref{3'} and \eqref{33'}, we have
$$
T(r, f_1)+S(r, f_1)=T(r, f_2)+S(r, f_2).
$$
Thus, $S(r, f_1)=S(r, f_2)$. Moreover, substituting 
$S(r, f_1)=S(r, f_2)$ into \eqref{3'}, we see
$$
T(r, \frac{f_1}{f_2})=T(r, f_2)+S(r, f_2),
$$
hence $S(r, \frac{f_1}{f_2})=S(r, f_2)$. Assume now that $z_0$ such
that $\frac{f_1(z_0)}{f_2(z_0)}=t_j$, then either $F(z_0)=0$ or
$f_2(z_0)=\infty$ by \eqref{2013}. That means that
$$
N(r, \frac{1}{\frac{f_1}{f_2}-t_j})=S(r, f_2)
$$ 
by the assumption and equation \eqref{2012}. From the arguments above 
and the second main theorem, we obtain
$$
(n-3)T(r, \frac{f_1}{f_2})\leq \sum_{j=1}^{n-1}N\big(r,
\frac{1}{\frac{f_1}{f_2}-t_j}\big)=S(r, f_2)=S\big(r, \frac{f_1}{f_2}\big),
$$ 
which contradicts the assumption that $n\geq 4$. Completing the
proof of the part (1).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 If $L(z, f)\equiv 0$, then the
conclusion follows. In the following, we suppose 
$L(z, f)\not\equiv 0$. Since $f(z)$ is a meromorphic solution of \eqref{1'},
with finite order, it follows from Lemma \ref{lem2.5} that
\begin{equation}\label{4'}
T(r, L(z, f))\leq (k+1)T(r, f)+S(r, f).
\end{equation}
From  \eqref{1'}, we  obtain
\begin{equation}\label{5'}
T(r, H)\leq T(r, f^n)+T(r, L(z, f))+S(r, f).
\end{equation}
Combining \eqref{4'} and \eqref{5'}, and applying Lemma \ref{lem3}, we
know that, for $\alpha>1$, there exists a value $r_0>0$, such that
$$
T(r, H)\leq (n+k+1)T(\alpha r, f)+S(r, f)
$$
for $r>r_0$. By the definition of $\sigma(f)$, we conclude that
$\sigma(H)\leq \sigma(f)$. Next, we investigate the special cases.
By the conclusion above, it suffices to show that 
$\sigma(H)\geq \sigma(f)$.


Case 1. If $n\geq k+2$, then  \eqref{1'} gives
\begin{equation}\label{6'}
T(r, f^n)\leq T(r, H)+T(r, L(z, f))+S(r, f).
\end{equation}
Substituting \eqref{4'} into \eqref{6'}, and from Lemma \ref{lem3}, 
we obtain that for $\alpha>1$ there exists a value $r_0>0$, such that
$$
(n-k-1)T(r, f)\leq T(\alpha r, H)+S(r, f)
$$
for $r>r_0$. By the assumption that $n\geq k+2$ and the definition
of $\sigma(f)$, it follows that $\sigma(H)\geq \sigma(f)$.


Case 2. If $N(r, f)=S(r, f)$, then by Lemma \ref{lem2.3}, we 
obtain 
\begin{equation}\label{7'}
        \begin{split}
T(r, L(z, f))
&=m(r,  L(z, f))\leq m(r, \frac{ L(z, f)}{f})+m(r, f)+S(r, f)\\
&\leq T(r, f)+S(r, f).
\end{split}
      \end{equation}
Substituting \eqref{7'} into \eqref{6'}, and using the same way as
in Case 1, we have
$$
(n-1)T(r, f)\leq T(\alpha r, H)+S(r, f)
$$ 
for $r>r_0$. The conclusion follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm0}]
If $H(z)$ is infinite order, then  \eqref{0} has no
meromorphic solution with finite order, by comparing the growth of
both sides of the equation. It remains to consider that $H(z)$ is
finite order. Suppose, contrary to the assertion, that $f(z)$ is a
transcendental meromorphic function with finite order satisfying
 \eqref{0}. Then we will distinguish two cases:


Case 1. If $T(r, H)\neq S(r, f)$. Then from  \eqref{0}, we
obtain
\begin{equation}\label{2}
f^m(z+c)=\frac{\frac{H'}{H}(f'(z))^n-((f'(z))^n)'}
{\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}}.
\end{equation}
First of all, we affirm that
$\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}$ cannot vanish
identically. Indeed, if
$\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}\equiv 0$, then we see
$$
H(z)=Af^m(z+c),
$$
 where $A$ is a non-zero constant.
Combining the above equality and equation \eqref{0},
$$
(f'(z))^n=(A-1)f^m(z+c)
$$
follows. By Lemma \ref{lem2.5} and the above equation, we obtain
$$
mT(r, f)\leq 2nT(r, f)+S(r, f),
$$ 
or $f'(z)\equiv 0$, which contradicts the assumptions.


From equation \eqref{2}, we obtain that
   \begin{equation}\label{3}
        \begin{split}
       T(r, f^m(z+c))
&=mT(r, f)+S(r, f)\leq m(r, (f'(z))^n)
 +m\Big(r,\frac{H'}{H}-\frac{((f'(z))^n)'}{(f'(z))^n}\Big)\\
&\quad +N\Big(r, \frac{H'}{H}(f'(z))^n-((f'(z))^n)'\Big)
 +m\Big(r, \frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}\Big)\\
&\quad +N\Big(r, \frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}\Big)+S(r, f).
\end{split}
\end{equation}
Then, Lemma \ref{lem2.5} together with equation \eqref{0}, implies
that
$$
T(r, H)\leq (m+2n)T(r, f)+S(r, f),
$$
which means all meromorphic functions $a(z)$ that satisfy
 $T(r, a)=S(r, H)$ must be $S(r, f)$. To apply Lemma \ref{lem2.5}, Lemma
\ref{lem2.3} and the Lemma on logarithmic derivative to equation
\eqref{3}, we obtain that
\begin{equation}\label{5}
 \begin{split}
mT(r, f)
&\leq n m(r, f)+N\Big(r, \frac{H'}{H}(f'(z))^n-((f'(z))^n)'\Big)\\
&\quad +N\Big(r, \frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}\Big)+S(r, f).
\end{split}
\end{equation}
We will estimate $N\big(r,\frac{H'}{H}(f'(z))^n-((f'(z))^n)'\big)$ and 
$N\big(r,\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}\big)$ next.
Set
\begin{gather}\label{6}
M(z)=\frac{H'}{H}(f'(z))^n-((f'(z))^n)',\\
\label{7}
G(z)=\frac{(f^m(z+c))'}{f^m(z+c)}-\frac{H'}{H}.
\end{gather}
From \eqref{0} and \eqref{6}, we know the poles of $M(z)$ are at the
zeros of $H(z)$, and at the poles of $f(z)$, $f(z + c)$. If $z_0$ is
a zero of $H(z)$ or $z_0$ is a pole of $f(z+c)$ but not a pole of
$f(z)$, then $z_0$ is at most a simple pole of $M(z)$ by \eqref{6}.
If $z_0$ is a pole of $f(z)$ but not a pole of $f(z+c)$, then $z_0$
is at most a simple pole of $M(z)$ by \eqref{2}. If $z_0$ is a pole
of $f(z)$ with multiplicity $p$ and a pole of $f(z+c)$ with
multiplicity $q$, then $z_0$ is a pole of $M(z)$ with the
multiplicity no more than $n(p+1)+1$ by \eqref{6}. From above
arguments and our assumption, we conclude that
\begin{equation}\label{8}
        \begin{split}
N(r,M)
&\leq \overline{N}(r, \frac{1}{H})+N(r,
(f'(z))^n)+\overline{N}(r, f(z+c))+S(r, f)\\
&\leq nN(r, f'(z))+\overline{N}(r, f(z+c))+S(r, f).
\end{split}
\end{equation}
On the other hand, by Lemma \ref{lem1} and our assumption, it follows
that
\begin{equation}\label{9}
\begin{split}
N(r,G)&\leq \overline{N}(r, \frac{1}{H})+\overline{N}(r,
f)+\overline{N}(r, \frac{1}{f(z+c)})+S(r, f)\\
&\leq \overline{N}(r, f)+\overline{N}(r, \frac{1}{f(z+c)})+S(r, f).
\end{split}
\end{equation}
From equations \eqref{5}, \eqref{8} and \eqref{9}, we have
\begin{equation*}
        \begin{split}
mT(r, f)&\leq n m(r, f)+n(N(r, f)+\overline{N}(r, f))+\overline{N}(r,f)\\
&\quad +\overline{N}(r, f(z+c))+\overline{N}(r, \frac{1}{f(z+c)})+S(r, f)\\
&\leq(2n+3)T(r, f)+S(r, f),
       \end{split}
      \end{equation*}
which contradicts the assumption that $m\geq 2n+4$.


Case 2. If $T(r, H)=S(r, f)$, then applying Lemma \ref{lem2.5} to
equation \eqref{0}, we have
$$
mT(r, f)\leq 2nT(r, f)+S(r, f),
$$
which contradicts the assumption that $m\geq 2n+4$. We get a
conclusion as well, completing the proof of Theorem \ref{thm0}.
\end{proof}

\subsection*{Acknowledgments}
This work was supported by grant 11226094 from the
NSFC Tianyuan Mathematics Youth Fund, grants
ZR2012AQ020 and ZR2010AM030 from the NSF of
Shandong Province, China,  and grant XBS1211 from
the Fund of Doctoral Program Research of University of Jinan.


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\end{document}