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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 199, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/199\hfil Schwarzian difference equations]
{Properties of Schwarzian difference equations}

\author[S.-T. Lan, Z.-X. Chen \hfil EJDE-2015/199\hfilneg]
{Shuang-Ting Lan, Zong-Xuan Chen}

\address{Shuang-Ting Lan \newline
Department of Mathematics,
 Guangzhou Civil Aviation College, Guangzhou 510403, China}
\email{wqh200811@126.com}

\address{Zong-Xuan Chen (corresponding author) \newline
School of Mathematical Sciences, South China Normal University,
Guangzhou 510631, China}
\email{chzx@vip.sina.com}

\thanks{Submitted April 22, 2015. Published August 4, 2015.}
\subjclass[2010]{30D35, 34A20}
\keywords{Schwarzian difference equation; rational solution; value distribution}

\begin{abstract}
 We consider the Schwarzian type difference equation
 $$
 \Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
 \Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}=R(z),
 $$
 where $R(z)$ is a nonconstant rational function. We study the existence of
 rational solutions and value distribution of transcendental meromorphic
 solutions with finite order of the above equation.
\end{abstract}

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

\section{Introduction and statement of main results}

In this article, we use the basic notions of Nevanlinna's theory
\cite{ha2, ya}. In addition, $\sigma(f)$
denotes the order of growth of the meromorphic function $f(z)$;
$\lambda(f)$ and $\lambda\left(\frac{1}{f}\right)$ denote the exponents
of convergence of zeros and poles of $f(z)$.
 Let $S(r,w)$ denote any quantity satisfying $S(r,w) = o\big(T(r,w)\big)$
for all $r$ outside of a set with finite logarithmic measure.
A meromorphic solution $w$ of a difference (or differential) equation is
called \textit{admissible} if the characteristic function of all coefficients
of the equation are $S(r,w)$. For every $n\in\mathbb{N}^{+}$, the forward
differences $\Delta^{n}f(z)$ are defined in the standard way \cite{wh} by
\[
\Delta f(z)=f(z+1)-f(z),\quad \Delta^{n+1} f(z)=\Delta^{n}f(z+1)-\Delta^{n}f(z).
\]

The Schwarzian differential equation
 \begin{equation}\label{eq0}
 \big[\frac{f'''}{f'}-\frac{3}{2}\big(\frac{f''}{f'}\big)^{2}\big]^{k}
=R(z,f)=\frac{P(z,f)}{Q(z,f)}
 \end{equation}
was studied by Ishizaki \cite{is}, and obtained some important results.
 Chen and Li \cite{ch2} investigated Schwarzian difference equation,
and obtained the following theorem.

\begin{theorem} \label{thmA}
Let $f(z)$ be an admissible solution of difference equation
 \[%\label{eq4}
 \Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}
=R(z,f)
=\frac{P(z,f)}{Q(z,f)}
 \]
such that $\sigma_{2}(f)<1$, where $k(\geq1)$ is an integer, $P(z,f)$ and
$Q(z,f)$ are polynomials with $\deg_{f}P(z,f)=p,~ \deg_{f}Q(z,f)=q$,
$d=\max\{p,q\}$. Let $\alpha_{1}, \ldots, \alpha_{s}$ be $s(\geq2)$ distinct
complex constants. Then
 $$
 \sum_{j=1}^{s}\delta(\alpha_{j}, f)\leq4-\frac{q}{2k}.
 $$
In particular, if $N(r,f)=S(r,f)$, then
 $$
 \sum_{j=1}^{s}\delta(\alpha_{j}, f)\leq2-\frac{d}{2k}.
 $$
\end{theorem}

Set $\deg_{f}P(z,f)=\deg_{f}Q(z,f)=0$ in equation \eqref{eq0}, then
$R(z,f)\equiv R(z)$ is a small function with respect to $f(z)$.
Liao and Ye \cite{li} studied this type of Schwarzian differential equation,
and obtained the following result.

\begin{theorem} \label{thmB}
Let $P$ and $Q$ be polynomials with $\deg P=p, ~\deg Q=q$, and let
$R(z)=\frac{P(z)}{Q(z)}$ and $k$ a positive integer. If $f(z)$ is a
transcendental meromorphic solution of equation
$$
\Big[\frac{f'''}{f'}-\frac{3}{2}\Big(\frac{f''}{f'}\Big)^{2}\Big]^{k}=R(z),
$$
then $p-q+2k>0$ and the order $\sigma(f)=\frac{p-q+2k}{2k}$.
\end{theorem}

In this article, we study a Schwarzian difference equation, and obtain the
following result.

\begin{theorem}\label{thm1}
Let $R(z)=\frac{P(z)}{Q(z)}$ be an irreducible rational function with
$\deg P(z)=p$, $\deg Q(z)=q$. Consider the difference equation
 \begin{equation}\label{eq1}
 \Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}
 -\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}=R(z),
 \end{equation}
where $k$ is a positive integer. Then
 \begin{itemize}
 \item[(i)] every transcendental meromorphic solution $f(z)$ of \eqref{eq1}
satisfies $\sigma(f)\geq 1$; if $p-q+2k>0$, then \eqref{eq1} has no rational
solutions;

\item[(ii)] if $f(z)$ is a mereomorphic solution of \eqref{eq1} with finite order,
 terms $\frac{\Delta^{2}f(z)}{\Delta f(z)}$ and
 $\frac{\Delta^{3}f(z)}{\Delta f(z)}$ in \eqref{eq1} are nonconstant rational
functions;

\item[(iii)] every transcendental meromorphic solution $f(z)$ with finite order
has at most one Borel exceptional value unless
 \begin{equation}\label{eq00}
 f(z)=b+R_0(z)e^{az},
 \end{equation}
 where $b\in\mathbb{C}, ~a\in\mathbb{C}\setminus\{0\}$ and $R_0(z)$
is a nonzero rational function.

\item[(iv)] if $p-q+2k>0, \sigma(f)<\infty$, then $\Delta f(z)$ has at most
one Borel exceptional value unless
 \begin{equation}\label{eq01}
 \Delta f(z)=R_{1}(z)e^{az},
 \end{equation}
 where $a\in\mathbb{C},~ a\neq i2k_{1}\pi$ for any $k_{1}\in\mathbb{Z}$, and $R_{1}(z)$
is a nonzero rational function.
 \end{itemize}
\end{theorem}

\begin{corollary}\label{cor1}
Let $f(z)$ be a finite order meromorphic solution of \eqref{eq1}, if $p-q+2k>0$,
then $f(z)$, $\Delta f(z)$, $\Delta^{2}f(z)$ and $\Delta^{3}f(z)$ cannot be
 rational functions, and $\frac{\Delta^{2}f(z)}{\Delta f(z)}$ and
$\frac{\Delta^{3}f(z)}{\Delta f(z)}$ are nonconstant rational functions.
\end{corollary}


\begin{remark}\label{re0} \rm
Let $f(z)$ be the function in the form \eqref{eq00}, then the Schwarzian
difference is an irreducible rational function $R(z)=\frac{P(z)}{Q(z)}$ with
$\deg P\leq\deg Q$.
\end{remark}

\begin{proof}
 Suppose that $f(z)$ has the form \eqref{eq00}. Since $R_0(z)$ is a rational
function, we see $R_0(z)$ satisfies
 \begin{equation}\label{043}
 \frac{R_0(z+j)}{R_0(z)}\to 1,\quad z\to \infty,\; j=1,2,3.
 \end{equation}
By \eqref{eq00}, we have
 \begin{gather*}
\Delta f(z)=e^{az}(e^{a}R_0(z+1)-R_0(z));\\
\Delta^{2}f(z)=e^{az}(e^{2a}R_0(z+2)-2e^{a}R_0(z+1)+R_0(z));\\
\Delta^{3}f(z)=e^{az}(e^{3a}R_0(z+3)-3e^{2a}R_0(z+2)+3e^{a}R_0(z+1)-R_0(z)).
\end{gather*}
Combining these with \eqref{043}, we have
\begin{equation} \label{001}
\begin{aligned}
\frac{\Delta^{3}f(z)}{\Delta f(z)}
&=\frac{e^{3a}R_0(z+3)-3e^{2a}R_0(z+2)+3e^{a}R_0(z+1)-R_0(z)}
 {e^{a}R_0(z+1)-R_0(z)}\\
&=\frac{e^{3a}\frac{R_0(z+3)}{R_0(z)}-3e^{2a}\frac{R_0(z+2)}{R_0(z)}
 +3e^{a}\frac{R_0(z+1)}{R_0(z)}-1}{e^{a}\frac{R_0(z+1)}{R_0(z)}-1} \\
&\to \frac{e^{3a}-3e^{2a}+3e^{a}-1}{e^{a}-1}=(e^{a}-1)^{2},\quad z\to \infty,
 \end{aligned}
\end{equation}
and
\begin{equation} \label{003}
\begin{aligned}
\frac{\Delta^{2}f(z)}{\Delta f(z)}
&=\frac{e^{2a}R_0(z+2)-2e^{a}R_0(z+1)+R_0(z)}{e^{a}R_0(z+1)-R_0(z)}\\
&=\frac{e^{2a}\frac{R_0(z+2)}{R_0(z)}-2e^{a}\frac{R_0(z+1)}{R_0(z)}+1}
{e^{a}\frac{R_0(z+1)}{R_0(z)}-1} \\
&\to \frac{e^{2a}-2e^{a}+1}{e^{a}-1}=e^{a}-1,\quad z\to \infty.
 \end{aligned}
\end{equation}
Thus,
 \begin{equation}\label{004}
 \frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
\to (e^{a}-1)^{2}-\frac{3}{2}(e^{a}-1)^{2}
=-\frac{1}{2}(e^{a}-1)^{2},\quad z\to \infty.
\end{equation}

By \eqref{001}, \eqref{003} and $R_0(z)$ begin a rational function, we see that
$$
R(z)=\Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}
$$
is a rational function. Denote $R(z)=\frac{P(z)}{Q(z)}$, where $P(z)$ and
$Q(z)$ are prime polynomials. By \eqref{004}, we see
$$
R(z)=\frac{P(z)}{Q(z)}
=\Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}\to
\frac{(-1)^{k}}{2^{k}}(e^{a}-1)^{2k}, \quad z\to \infty.
$$
If $e^{a}\neq1$, then $\deg P=\deg Q$; if $e^{a}=1$, then $\deg P<\deg Q$.
So, $\deg P\leq \deg Q$.
\end{proof}

\begin{remark}\label{re1}\rm
Checking the proof of Theorem \ref{thm1} (iv), we see that
for $f(z)$ a function such that $\Delta f(z)$ in the form \eqref{eq01},
then the Schwarzian difference satisfies
$$
\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
=e^{2a}\frac{R_{1}(z+2)}{R_{1}(z)}-\frac{3}{2}e^{2a}
\Big(\frac{R_{1}(z+1)}{R_{1}(z)}\Big)^{2}
 +e^{a}\frac{R_{1}(z+1)}{R_{1}(z)}-\frac{1}{2}.
$$
\end{remark}

Examples \ref{ex1} and \ref{ex2} below show that the condition ``$p-q+2k>0$''
in Theorem \ref{thm1} (i) cannot be omitted.

\begin{example}\label{ex1} \rm
Consider the Schwarzian type difference equation
$$
\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
=-\frac{6}{(2z+1)^{2}},
$$
where $k=1$, $p=0$, $q=2$, and $p-q+2k=0$. This equation has a rational
solution $f_{1}(z)=z^{2}$, and a transcendental meromorphic solution
$f_{2}(z)=e^{i2\pi z}+z^{2}$.
\end{example}

\begin{example}\label{ex2} \rm
Consider the Schwarzian type difference equation
$$
\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
=\frac{-6}{(z+3)(z+2)^{2}},
$$
where $k=1$, $p=0$, $q=3$, and $p-q+2k=-1<0$. This equation has a rational
solution $f_{1}(z)=\frac{1}{z}$, and a transcendental meromorphic solution
$f_{2}(z)=e^{i2\pi z}+\frac{1}{z}$.
\end{example}


\begin{example}\label{ex3} \rm
The function $f(z)=ze^{(\log3)z}$ satisfies Schwarzian type difference equation
$$
\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
=\frac{-8z^{2}-48z-108}{(2z+3)^{2}}.
$$
We see $\sigma(f)=1$ and $f(z)$ has finitely many zeros and poles.
It shows the result of Theorem \ref{thm1} (iii) is precise.
\end{example}



\section{Preliminaries}

\begin{lemma}[\cite{ch1}] \label{le11}
Let $f(z)$ be a meromorphic function of finite order $\sigma$ and let $\eta$
 be a nonzero complex constant. Then for each $\varepsilon(0<\varepsilon<1)$,
we have
\[
 m\Big(r, \frac{f(z+\eta)}{f(z)}\Big)+m\Big(r, \frac{f(z)}{f(z+\eta)}\Big)
=O(r^{\sigma-1+\varepsilon}).
\]
\end{lemma}

\begin{lemma}[\cite{ch1}] \label{le1}
Let $f(z)$ be a meromorphic function with order $\sigma=\sigma(f), \sigma<\infty$,
and let $\eta$ be a fixed nonzero complex number, then for each $\varepsilon>0$,
\[
 T(r, f(z+\eta))=T(r, f(z))+O\big(r^{\sigma-1+\varepsilon}\big)+O(\log r).
\]
\end{lemma}

\begin{lemma}[{\cite[Theorem 1.8.1]{ch3}, \cite{la2}}] \label{lecha}
Let $c\in\mathbb{C}\setminus\{0\}$ and $f(z)$ be a finite order meromorphic function
with two finite Borel exceptional values $a$ and $b$. Then for every $n\in\mathbb{N}^{+}$,
$$
T(r,\Delta^{n}f)=(n+1)T(r,f)+S(r,f)
$$
unless $f(z)$ and $c$ satisfy
\begin{gather*}
f(z)=b+\frac{b-a}{pe^{dz}-1}, \quad p, d\in\mathbb{C}\setminus\{0\},\\
mdc=i2k_{1}\pi\quad \text{for some $k_{1}\in\mathbb{Z}$ and } m\in\{1, 2, \ldots, n\}.
\end{gather*}
\end{lemma}

\begin{remark}\label{re2} \rm
Checking the proof of Lemma \ref{lecha}, we point out that when
$c\in\mathbb{C}\setminus\{0\}$ and $f(z)$ is a finite order meromorphic function
with two finite Borel exceptional values, for every $n\in\mathbb{N}^{+}$,
if $c,2c,\ldots, nc$ are not periods of $f(z)$, then
$$
T(r,\Delta^{n}f)=(n+1)T(r,f)+S(r,f).
$$
\end{remark}

\begin{lemma}[\cite{be}] \label{le7}
Let $f(z)$ be a function transcendental and meromorphic in the plane
which satisfies
$$
\liminf_{r\to \infty}\frac{T(r,f)}{r}=0.
$$
Then $\Delta f$ and $\Delta f/f$ are both transcendental.
\end{lemma}


\begin{lemma}\label{le4}
Suppose that $f(z)=H(z)e^{az}$, where $a\neq0$ is a constant, $H(z)$
is a transcendental meromorphic function with $\sigma(H)<1$.
Then $\frac{\Delta f(z)}{f(z)}$ is transcendental.
\end{lemma}

\begin{proof}
Substituting $f(z)=H(z)e^{az}$ into $\frac{\Delta f(z)}{f(z)}$, we see that
\begin{equation} \label{030}
\begin{aligned}
\frac{\Delta f(z)}{f(z)}
&=\frac{f(z+1)-f(z)}{f(z)}
 =\frac{H(z+1)e^{a(z+1)}-H(z)e^{az}}{H(z)e^{az}} \\
&=e^{a}\frac{H(z+1)}{H(z)}-1
 =e^{a}\Big(\frac{H(z+1)}{H(z)}-1\Big)+e^{a}-1\\
&=e^{a}\frac{\Delta H(z)}{H(z)}+e^{a}-1.
 \end{aligned}
\end{equation}
From the fact $\sigma(H)<1$, we see that
$$
\limsup_{r\to \infty}\frac{\log T(r,H)}{\log r}=\sigma(H)<1.
$$
Then for large enough $r$, choose $\varepsilon=\frac{1-\sigma(H)}{2}>0$, we have
$$
\log T(r,H)<(\sigma(H)+\varepsilon)\log r;
$$
that is,
$$
T(r,H)<r^{\sigma(H)+\varepsilon}.
$$
Thus,
 \begin{equation}\label{031}
 \liminf_{r\to \infty}\frac{T(r,H)}{r}
 \leq\liminf_{r\to \infty}\frac{r^{\sigma(H)+\varepsilon}}{r}
 =\liminf_{r\to \infty}r^{\sigma(H)+\varepsilon-1}
 =\liminf_{r\to \infty}r^{-\varepsilon}=0.
 \end{equation}
So, $H(z)$ is a transcendental meromorphic function which satisfies \eqref{031}.
From Lemma \ref{le7}, we see $\frac{\Delta H(z)}{H(z)}$ is transcendental.
By \eqref{030}, $\frac{\Delta f(z)}{f(z)}$ is transcendental too.
\end{proof}

\begin{lemma}[{\cite[Lemma 5.2.2]{ch3}}] \label{le10}
Let $f(z)$ be a transcendental meromorphic function with $\sigma(f)<1$,
 and let $g_{1}(z)$ and $g_{2}(z)(\not\equiv0)$ be polynomials, 
$c_{1}$, $c_{2}$ ($c_1\neq c_2$) be constants. Then
$$
h(z)=g_{2}(z)f(z+c_{2})+g_{1}(z)f(z+c_{1})
$$
is transcendental.
\end{lemma}

\begin{lemma}[\cite{ha1,la1}] \label{le3}
Let $w$ be a transcendental meromorphic solution with finite order of 
difference equation
 \[
 P(z,w) = 0,
 \]
where $P(z,w)$ is a difference polynomial in $w(z)$. 
If $P(z,a)\not\equiv0$ for a meromorphic function
$a$, where $a$ is a small function with respect to $w$, then
 \[
 m\big(r, \frac{1}{w-a}\big) = S(r,w).
 \]
\end{lemma}

\begin{remark}\label{re3}\rm
Ishizaki \cite[Remark 1]{is} pointed out that
if $P(z,w)$ and $Q(z,w)$ are mutually prime, there exist polynomials of
 $w, U(z,w)$ and $V(z,w)$ such that
$$
U(z,w)P(z,w)+V(z,w)Q(z,w)=s(z),
$$
where $s(z)$ and coefficients of $U(z,w)$ and $V(z,w)$ are small functions
 with respect to $w(z)$.
\end{remark}

\begin{lemma}\label{le6}
Let $R(z)$ be a nonconstant rational function. Suppose that $f(z)$ is a 
transcendental meromorphic solution of equation \eqref{eq1} with finite order, 
then in \eqref{eq1}, terms $\frac{\Delta^{2}f(z)}{\Delta f(z)}$ and 
$\frac{\Delta^{3}f(z)}{\Delta f(z)}$ are both nonconstant rational functions.
\end{lemma}

\begin{proof} 
Set $G(z)=\frac{\Delta f(z+1)}{\Delta f(z)}$. Then $G(z)$ is a meromorphic 
function with finite order, and
\begin{gather*}
\Delta f(z+1)=G(z)\Delta f(z),\\
 \Delta f(z+2)=G(z+1)\Delta f(z+1) =G(z+1)G(z)\Delta f(z).
\end{gather*}
Hence,
 \begin{equation}\label{029}
 \Delta^{2}f(z)=\Delta f(z+1)-\Delta f(z)=(G(z)-1)\Delta f(z),
 \end{equation}
and
\begin{equation} \label{027}
\begin{aligned}
 \Delta^{3}f(z)
&=\Delta^{2}(\Delta f(z))=\Delta f(z+2)-2\Delta f(z+1)+\Delta f(z) \\
 &=(G(z+1)G(z)-2G(z)+1)\Delta f(z).
 \end{aligned}
\end{equation}
From \eqref{eq1},
$$
R(z)= \Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}
$$
is a nonconstant rational function, then
$$
\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}
$$
is also a nonconstant rational function. Denote
 \begin{equation}\label{028}
 \frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}=R_{2}(z),
 \end{equation}
where $R_{2}(z)$ is a nonconstant rational function.

It follows from \eqref{029}--\eqref{028} that
 \begin{equation}\label{z18}
 G(z+1)G(z)-2G(z)+1-\frac{3}{2}(G(z)-1)^{2}=R_{2}(z);
 \end{equation}
that is,
 \begin{equation}\label{018}
 G(z+1)=\frac{\frac{3}{2}G^{2}(z)-G(z)+R_{2}(z)+\frac{1}{2}}{G(z)}.
 \end{equation}
Since $R_{2}(z)$ is a nonconstant rational function, by \eqref{z18}, $G(z)$ 
cannot be a constant. Suppose that $G(z)$ is transcendental. We see that
$$
\frac{3}{2}G^{2}(z)-G(z)+R_{2}(z)+\frac{1}{2}+(-\frac{3}{2}G(z)+1)G(z)
=R_{2}(z)+\frac{1}{2}.
$$
Together with Remark \ref{re3}, $\frac{3}{2}G^{2}(z)-G(z)+R_{2}(z)+\frac{1}{2}$ 
and $G(z)$ are irreducible. Applying Valiron-Mohon'ko Theorem to \eqref{018}, 
we have
$$
T(r,G(z+1))=2T(r,G(z))+S(r,G),
$$
which contradicts Lemma \ref{le1}. So, $G(z)$ is a nonconstant rational function. 
By \eqref{029} and \eqref{027}, we see that 
$\frac{\Delta^{2}f(z)}{\Delta f(z)}$ and $\frac{\Delta^{3}f(z)}{\Delta f(z)}$ 
are nonconstant rational functions.
\end{proof}


\section{Proofs of theorems}

\begin{proof}[Proof of Theorem \ref{thm1}]

(i) Suppose that $f(z)$ is a transcendental meromorphic solution of equation 
\eqref{eq1} with $\sigma(f)<1$. Lemma \ref{le7} shows 
$g(z)=\Delta f(z)$ is transcendental with $\sigma(g)<1$. 
Again by Lemma \ref{le7}, we see 
$\frac{\Delta^{2}f(z)}{\Delta f(z)}=\frac{\Delta g(z)}{g(z)}$ 
is also transcendental, which contradicts with Lemma \ref{le6}. 
Thus, $\sigma(f)\geq1$.

Next, we prove that if $f(z)$ is a rational solution of equation \eqref{eq1}, 
then $p-q+2k\leq0$. Set $g(z)=\Delta f(z)$. By \eqref{eq1}, we see
 \begin{equation}\label{eq3}
 \Big[\frac{\Delta^{2}g(z)}{g(z)}
-\frac{3}{2}\Big(\frac{\Delta g(z)}{g(z)}\Big)^{2}\Big]^{k}=R(z).
 \end{equation}
Thus, $g(z)$ is a rational solution of equation
 \[
 \frac{\Delta^{2}g(z)}{g(z)}
-\frac{3}{2}\Big(\frac{\Delta g(z)}{g(z)}\Big)^{2}=R_{2}(z),
 \]
 or
 \begin{equation}\label{014}
 g(z)\Delta^{2}g(z)-\frac{3}{2}(\Delta g(z))^{2}=R_{2}(z)g^{2}(z),
 \end{equation}
where $R_{2}(z)$ is some rational function such that $R_{2}^{k}(z)=R(z)$. 
Since $R(z)=Az^{p-q}(1+o(1))$, where $A$ is some nonzero constant, then
 \begin{equation}\label{014z}
 R_{2}(z)=Bz^{\frac{p-q}{k}}(1+o(1)),
 \end{equation}
where $B$ is some nonzero constant.

Suppose that
 \begin{equation}\label{010}
 g(z)=h(z)+\frac{m(z)}{n(z)},
 \end{equation}
where $h(z), m(z)$ and $n(z)$ are polynomials with 
$\deg h(z)=l(\geq 0)$, $\deg m(z)=m$, $\deg n(z)=n$ with $m<n$. Denote
 \begin{equation}\label{011}
 h(z)=c_0z^{l}+\cdots+c_{l},\quad
 m(z)=a_0z^{m}+\cdots+a_{m},\quad
 n(z)=b_0z^{n}+\cdots+b_{n},
 \end{equation}
where $c_0, \ldots, c_{l}$, $a_0, \ldots, a_{m}$,
$b_0, \ldots,  b_{n}$ are constants, with $a_0\neq 0$ and $b_0\neq 0$.

We divide this proof into the following three cases.
\smallskip

\noindent\textbf{Case 1.} $l>0$. 
By \eqref{010} and \eqref{011}, when $z$ is large enough, $g(z)$ can 
be written as
 \begin{equation}\label{012}
 g(z)=c_0z^{l}(1+o(1)).
 \end{equation}
Hence,
 \begin{equation}\label{013}
 \Delta g(z)=lc_0z^{l-1}(1+o(1)), \quad  \Delta^{2}g(z)=l(l-1)c_0z^{l-2}(1+o(1)).
 \end{equation}
Substituting \eqref{014z}, \eqref{012}, \eqref{013} in \eqref{014}, we obtain
$$
c_0z^{l}l(l-1)c_0z^{l-2}(1+o(1))-\frac{3}{2}(lc_0z^{l-1})^{2}(1+o(1))
=Bz^{\frac{p-q}{k}}c_0^{2}z^{2l}(1+o(1));
$$
that is,
$$
-\big(\frac{l}{2}+1\big)lc_0^{2}z^{2l-2}(1+o(1))
=Bz^{\frac{p-q}{k}}c_0^{2}z^{2l}(1+o(1)),
$$
from which it follows
$$
2l-2=\frac{p-q}{k}+2l.
$$
So, $p-q+2k=0$.
\smallskip

\noindent\textbf{Case 2.}
 $l=0$, $c_0\neq0$. By \eqref{010} and \eqref{011}, when $z$ is large enough, 
$g(z)$ can be written as
 \begin{equation}\label{015}
 g(z)=c_0+\frac{m(z)}{n(z)}=c_0+o(1).
 \end{equation}
By calculation and $m<n$, we see that
\begin{gather*}
n(z)n(z+1)=b_0^{2}z^{2n}(1+o(1)), \\
 m(z+1)n(z)-m(z)n(z+1)=(m-n)a_0b_0z^{m+n-1}(1+o(1)).
\end{gather*}
Thus,
 \begin{equation}\label{016}
 \Delta g(z)=\frac{m(z+1)n(z)-m(z)n(z+1)}{n(z)n(z+1)}
=(m-n)\frac{a_0}{b_0}z^{m-n-1}(1+o(1)).
 \end{equation}
Again by calculations, we have
 \begin{equation}\label{017}
 \Delta^{2}g(z)=(m-n)(m-n-1)\frac{a_0}{b_0}z^{m-n-2}(1+o(1)).
 \end{equation}
Submitting \eqref{014z}, \eqref{015}--\eqref{017} in \eqref{014}, 
since $2(m-n-1)<m-n-2<0$, we have
 \begin{align*}
 Bz^{\frac{p-q}{k}}(c^{2}_0+o(1)) 
&=c_0(m-n)(m-n-1)\frac{a_0}{b_0}z^{m-n-2}(1+o(1))\\
&\quad -\frac{3}{2}\Big((m-n)\frac{a_0}{b_0}z^{m-n-1}\Big)^{2}(1+o(1))\\
&= c_0(m-n)(m-n-1)\frac{a_0}{b_0}z^{m-n-2}(1+o(1)).
 \end{align*}
Hence, $p-q=k(m-n-2)=k(m-n)-2k<-2k$. That is, $p-q+2k<0$.
\smallskip

\noindent\textbf{Case 3.}$l=0$, $c_0=0$. Because $m<n$, we see that
 \begin{equation}\label{017z}
 g(z)=\frac{m(z)}{n(z)}=\frac{a_0}{b_0}z^{m-n}(1+o(1)).
 \end{equation}
We also obtain \eqref{016} and \eqref{017}. Substituting \eqref{014z}, 
\eqref{016}--\eqref{017z} into \eqref{014}, we have
 \begin{equation}\label{018z}
 \frac{n-m-2}{2}(m-n)\frac{a_0^{2}}{b_0^{2}}z^{2m-2n-2}(1+o(1))
=Bz^{\frac{p-q}{k}}\frac{a_0^{2}}{b_0^{2}}z^{2m-2n}(1+o(1)).
 \end{equation}
If $n\neq m+2$, by \eqref{018z},
$$
2m-2n-2=\frac{p-q}{k}+(2m-2n);
$$
thus, $p-q+2k=0$.

If $n=m+2$, by \eqref{018z},
$$
2m-2n-2>\frac{p-q}{k}+(2m-2n),
$$
thus, $p-q+2k<0$.

By the above Cases 1--3, we see if \eqref{eq1} has a rational solution 
$f(z)$, then $p-q+2k\leq0$.

(ii) By Lemma \ref{le6}, we see that Theorem \ref{thm1} (ii) holds.


(iii) Set $G(z)=\frac{\Delta^{2}f(z)}{\Delta f(z)}$. 
Lemma \ref{le6} shows $G(z)$ is a nonconstant rational function. Then
 \begin{equation}\label{019}
 \Delta^{2}f(z)=G(z)\Delta f(z),
 \end{equation}
By \eqref{eq1}, we easily see $\Delta f(z)\not\equiv0$, that is 
$f(z+1)\not\equiv f(z)$. Assert that $f(z+2)\not\equiv f(z)$. Otherwise,
$$
\Delta^{2}f(z)=f(z+2)-2f(z+1)+f(z)=2f(z)-2f(z+1)=-2\Delta f(z).
$$
Together with \eqref{019},
$$
G(z)=\frac{\Delta^{2}f(z)}{\Delta f(z)}\equiv-2,
$$
which contradicts with the fact $G(z)$ is a nonconstant rational function.

If $f(z)$ has two finite Borel exceptional values, by
 $f(z+2)\not\equiv f(z)$, $f(z+1)\not\equiv f(z)$ and Remark \ref{re2}, we have
$$
T(r,\Delta^{2}f)=3T(r, f)+S(r,f),\quad 
T(r,\Delta f)=2T(r, f)+S(r,f).
$$
On the other hand, \eqref{019} shows that
$$
T(r,\Delta^{2}f)=T(r,\Delta f)+O(\log r).
$$
The last two equalities follows $T(r,f)=S(r,f).$ It is a contradiction. 
So, $f(z)$ cannot have two finite Borel exceptional values.

Suppose that $f(z)$ has two Borel exceptional values $b\in\mathbb{C}$ and $\infty$.
By Hadamard's factorization theory, $f(z)$ takes the form
 \begin{equation}\label{040}
 f(z)=b+R_0(z)e^{h(z)},
 \end{equation}
where $R_0(z)$ is a meromorphic function, and $h(z)$ is a polynomial such that
$$
\sigma(R_0)=\max\big\{\lambda(f-b),
\lambda\big(\frac{1}{f}\big)\big\}<\deg h.
$$
Thus,
 \begin{equation}\label{041}
 \Delta f(z)=\left(R_0(z+1)e^{h(z+1)-h(z)}-R_0(z)\right)e^{h(z)}=R_{1}(z)e^{h(z)},
 \end{equation}
where $R_{1}(z)=R_0(z+1)e^{h(z+1)-h(z)}-R_0(z)$. Obviously,
 \begin{equation}\label{042}
 \sigma(R_{1})=\sigma\Big(R_0(z+1)e^{h(z+1)-h(z)}-R_0(z)\Big)
\leq\max\{\sigma(R_0), \deg h-1\}<\deg h.
 \end{equation}

From \eqref{041} and \eqref{042}, we see that $\sigma(\Delta f)=\sigma(f)$, 
and $\Delta f(z)$ has two Borel exceptional values $0$ and $\infty$. 
Substituting $\Delta f(z)=R_{1}(z)e^{h(z)}$ into \eqref{019}, we have
 \begin{equation}\label{042z}
 R_{1}(z+1)e^{h(z+1)-h(z)}=R_{1}(z)(G(z)+1).
 \end{equation}
If $\deg h\geq2$, then $\sigma(e^{h(z+1)-h(z)})=\deg h-1\geq1$. 
By \eqref{042z} and Lemma \ref{le11}, for any given $\varepsilon>0$, we have
\begin{align*}
m(r,e^{h(z+1)-h(z)})
&\leq m\Big(r,\frac{R_{1}(z)}{R_{1}(z+1)}\Big)+m(r, G(z)+1)\\
&=O(r^{\sigma(R_{1})-1+\varepsilon})+O(\log r),
\end{align*}
which yields $\deg h-1\leq \sigma(R_{1})-1+\varepsilon$. 
Letting $\varepsilon\to 0$, we have $\deg h\leq\sigma(R_{1})$, 
which contradicts with \eqref{042}. Hence, if $\deg h\geq2$, 
then $f(z)$ has at most one Borel exceptional value.

If $\deg h=1$, then $F(z)=\Delta f(z)=R_{1}(z)e^{az}$, 
where $a\in\mathbb{C}\setminus\{0\}$. If $R_{1}(z)$ is transcendental with
$\sigma(R_{1})<1$, by Lemma \ref{le4}, we see 
$G(z)=\frac{\Delta^{2}f(z)}{\Delta f(z)}=\frac{\Delta F(z)}{F(z)}$ 
is also transcendental. This contradicts with the fact $G(z)$ is a rational function. 
Therefore, $R_{1}(z)$ is a rational function. Combining this with \eqref{040} 
and \eqref{041}, we have
 \begin{equation}\label{043z}
 f(z)=b+R_0(z)e^{az}
 \end{equation}
and
$$
R_{1}(z)=e^{a}R_0(z+1)-R_0(z),
$$
where $\sigma(R_0)<1$. If $R_0(z)$ is transcendental, by Lemma \ref{le10}, 
we see $e^{a}R_0(z+1)-R_0(z)$ is transcendental, which contradicts 
with $R_{1}(z)=e^{a}R_0(z+1)-R_0(z)$ is a rational function. 
Hence, $R_0(z)$ is a rational function.

(iv) Suppose that $f(z)$ is a meromorphic solution of equation \eqref{eq1}, 
then $g(z)=\Delta f(z)$ is a meromorphic solution of equation \eqref{eq3}. 
Checking the proof of (i), we see if $g(z)$ is a rational solution of \eqref{eq3}, 
then $p-q+2k\leq0$. Since $p-q+2k>0$, we know $\Delta f(z)$ is transcendental. 
\eqref{019} still hold. By \eqref{019}, set
$$
P(z,\Delta f):=\Delta^{2}f(z)-G(z)\Delta f(z)=0.
$$
Since $G(z)$ is a nonconstant rational function, then for any given 
$a\in\mathbb{C}\setminus\{0\}$, we have $P(z,a)=-aG(z)\not\equiv0$.
Together with Lemma \ref{le3} , we have
$m\big(r,\frac{1}{\Delta f-a}\big)=S(r,\Delta f)$.
Thus, $\delta(a, \Delta f)=0$. By this and the proof of (iii), we see 
taht $\Delta f(z)$ has at most one Borel exceptional value $0$ or $\infty$ unless
 \begin{equation}\label{047}
 \Delta f(z)=R_{1}(z)e^{az}
 \end{equation}
where $a\in\mathbb{C}\setminus\{0\}, R_{1}(z)$ is a nonzero rational function.
Now we prove that $a\neq i2k_{1}\pi$ for any $k_{1}\in\mathbb{Z}$.
We see $R_{1}(z)$ satisfies
 \begin{equation}\label{044}
 \frac{R_{1}(z+2)}{R_{1}(z)}\to 1, \quad
 \frac{R_{1}(z+1)}{R_{1}(z)}\to 1, \quad z\to \infty.
 \end{equation}
By \eqref{047}, we have
 \begin{equation}\label{048}
\begin{gathered}
 \Delta^{2}f(z)=\Delta(\Delta f(z))=e^{az}(e^{a}R_{1}(z+1)-R_{1}(z)),\\
 \Delta^{3}f(z)=\Delta^{2}(\Delta f(z))
  =e^{az}(e^{2a}R_{1}(z+2)-2e^{a}R_{1}(z+1)+R_{1}(z)).
\end{gathered}
 \end{equation}
From \eqref{047}--\eqref{048}, we deduce that
 \begin{align*}% \label{045}
&\frac{\Delta^{3}f(z)}{\Delta f(z)}
 -\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\\
&=e^{2a}\frac{R_{1}(z+2)}{R_{1}(z)}
 -\frac{3}{2}e^{2a}\Big(\frac{R_{1}(z+1)}{R_{1}(z)}\Big)^{2}
 +e^{a}\frac{R_{1}(z+1)}{R_{1}(z)}-\frac{1}{2} \\
&\to e^{2a}-\frac{3}{2}e^{2a}+e^{a}-\frac{1}{2}
 =-\frac{1}{2}(e^{a}-1)^{2},\quad z\to \infty.
 \end{align*}
Combining this with \eqref{eq1}, we have
$$
\Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2}
\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}
=R(z)\to \frac{(-1)^{k}}{2^{k}}(e^{a}-1)^{2k}, \quad z\to \infty.
$$
If $e^{a}=1$, by \eqref{048}, we have
$$
\Delta^{2}f(z)=e^{az}\Delta R_{1}(z),\quad
\Delta^{3}f(z)=e^{az}\Delta^{2}R_{1}(z).
$$
Combining this with \eqref{eq1} and \eqref{047}, we obtain
 \[
 \Big[\frac{\Delta^{3}f(z)}{\Delta f(z)}
-\frac{3}{2}\Big(\frac{\Delta^{2}f(z)}{\Delta f(z)}\Big)^{2}\Big]^{k}
= \Big[\frac{\Delta^{2}R_{1}(z)}{R_{1}(z)}
 -\frac{3}{2}\Big(\frac{\Delta R_{1}(z)}{R_{1}(z)}\Big)^{2}\Big]^{k}=R(z).
 \]
Hence, $R_{1}(z)$ is a rational solution of the equation
 \begin{equation}\label{046}
 \Big[\frac{\Delta^{2}g(z)}{g(z)}-\frac{3}{2}\Big(\frac{\Delta g(z)}{g(z)}\Big)^{2}
\Big]^{k}=R(z).
 \end{equation}
By the conclusion of (i), we see if $p-q+2k>0$, equation \eqref{046} has no 
rational solutions. It is a contradiction. Thus, $e^{a}\neq1$.
 So, $a\neq i2k_{1}\pi$ for any $k_{1}\in\mathbb{Z}$.
\end{proof}


\subsection*{Acknowledgements}
The project was supported by the National Natural Science Foundation of 
China (11171119).

\begin{thebibliography}{00}

\bibitem{be} W. Bergweiler, J. K. Langley;
 Zeros of differences of meromorphic functions. \emph{Math. Proc. Cambridge Philos.
Soc.} {\bf 142} (2007), 133-147.

\bibitem{ch1} Y. M. Chiang, S. J. Feng;
 On the Nevanlinna characteristic of $f(z+\eta)$ and difference equations 
in the complex plane, \emph{Ramanujan J.}, {\bf 16} (2008), 105--129.

\bibitem{ch2} B. Q. Chen, S. Li;
 Admissible solutions of the Schwarzian type difference equation, 
\emph{Abstr. Appl. Anal.}, {\bf 2014} (2014), Article ID 306360.

\bibitem{ch3} Z. X. Chen;
 \emph{Complex Differences and Difference Equations}, 
Mathematics Monograph Series 29, Science Press, Beijing, 2014.

\bibitem{ha1} R. G. Halburd, R. J. Korhonen;
 Difference analogue of the Lemma on the Logarithmic Derivative with applications 
to difference equations, \emph{J. Math. Anal. Appl.}, {\bf 314} (2006), 477--487.


\bibitem{ha2} W. K. Hayman;
 \emph{ Meromorphic Functions.} Clarendon Press, Oxford, 1964.

\bibitem{is} K. Ishizaki;
 Admissible solutions of the Schwarzian differential equation, 
AustralianMathematical SocietyA: \emph{Pure Mathematics
and Statistics,} {\bf 50}, (2) (1991), pp. 258-278.

\bibitem{la1} I. Laine, C. C. Yang;
 Clunie theorems for difference and q-difference polynomials,
 \emph{J. Lond. Math. Soc.} {\bf 76} (2007), 556--566.

\bibitem{la2} S. T. Lan, Z. X. Chen;
 Relationships between characteristic functions of $\Delta^{n}f$ and $f$, 
\emph{J. Math. Anal. Appl.,} {\bf 412}(2014), 922--942.

\bibitem{li} L. W. Liao, Z. Ye;
 On the growth of meromorphic solutions of the Schwarzian differential equations,
 \emph{J. Math. Anal. Appl.,} {\bf 309} (2005), 91--102.

\bibitem{wh} J. M. Whittaker;
 \emph{Interpolatory Function Theory}, Cambridge Tracts in Math. and 
Math. Phys., vol. 33, Cambridge University Press, 1935.

\bibitem{ya} L. Yang;
 \emph{Value distribution theory and its new research} (in Chinese). 
Beijing: Science Press. 1982.


\end{thebibliography}

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