\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 22, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/22\hfil Entire solutions] {Entire solutions for nonlinear differential-difference equations} \author[N. Xu, T.-B. Cao, K. Liu \hfil EJDE-2015/22\hfilneg] {Na Xu, Ting-Bin Cao, Kai Liu} \address{Na Xu \newline School of Mathematical Sciences, Xiamen University, Xiamen 361005, China} \email{xuna406@163.com} \address{Ting-Bin Cao (corresponding author)\newline Department of Mathematics, Nanchang University, Nanchang, Jiangxi 330031, China} \email{tbcao@ncu.edu.cn} \address{Kai Liu \newline Department of Mathematics, Nanchang University, Nanchang, Jiangxi 330031, China} \email{liukai418@126.com} \thanks{Submitted July 15, 2014. Published January 27, 2015.} \subjclass[2000]{30D35, 39A05} \keywords{Nevanlinna theory; differential-difference equation; entire function} \begin{abstract} In this article, we study entire solutions of the nonlinear differential-difference equation $$ q(z)f^{n}(z)+a(z)f^{(k)}(z+1)=p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)} $$ where $p_1(z)$, $p_2(z)$ are nonzero polynomials, $q_1(z)$, $q_2(z)$ are nonconstant polynomials, $q(z)$, $a(z)$ are nonzero entire functions of finite order, $n\geq2$ is an integer. We obtain additional results for case: $n=3$, $q_1(z)=-q_2(z)$, and $p_1(z)$, $p_2(z)$ nonzero constants. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction and main results} In this article, we assume that the reader is familiar with standard symbols and fundamental results of Nevanlinna Theory. We denote by $S(r,f)$ any quantity satisfying $S(r,f)=o(T(r,f))$, as $r \to \infty$, possibly outside of a set $E$ with finite linear measure. We use $\lambda(\frac{1}{f})$ and $\lambda(f)$ to denote the exponents of convergence of poles and zeros of $f(z)$ respectively, $\sigma(f)$ to denote the order of $f(z)$. The hyper-order of $f(z)$ is defined as $$ \sigma_2(f)=\limsup_{r\to\infty}\frac{\log{\log{T(r,f)}}}{\log r}, $$ the lower hyper-order of $f(z)$ is defined as $$ \mu_2(f)=\liminf_{r\to\infty}\frac{\log{\log{T(r,f)}}}{\log r}, $$ the hyper exponent of convergence of zeros of $f(z)$ is defined by $$ \lambda_2(f)=\limsup_{r\to\infty}\frac{\log{\log N (r,\frac{1}{f})}}{\log r}, $$ and the deficiency of $a$ with respect to $f(z)$ is defined by $$ \delta(a,f)=1-\limsup_{r\to\infty}\frac{N(r,\frac{1}{f-a})}{T(r,f)}. $$ A differential polynomial of $f(z)$ means that it is a polynomial in $f(z)$ and its derivatives with small functions of $f(z)$ as coefficients. A differential-difference polynomial of $f(z)$ means that it is a polynomial in $f(z)$, its derivatives and its shifts $f(z+c)$ with small functions of $f(z)$ as coefficients. We shall use $P_d(f)$ to denote a differential polynomial or a differential-difference polynomial of $f(z)$ with degree $d$. In previous two decades, the existence and growth of meromorphic solutions of difference equations have been investigated in many papers [1-7, 9-12, 15]. Recently, there has been a renewed interest in studying meromorphic solutions of differential-difference equations, see \cite{Peng, Yang1, Zhang}. For instance, many authors have considered the equation $f^{n}(z)+P_d(f)=p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}$. when $P_d(f)$ is a differential polynomial, Li and Yang \cite{Li, Yang2} investigated the properties of solutions of the above equation. When $P_d(f)$ is a differential-difference polynomial, Zhang and Liao \cite{Zhang} proved that if the above equation satisfies some conditions, it doesn't have any transcendental entire solution of finite order. \begin{theorem}[{\cite[Theorem 3]{Zhang}}] \label{thm1.1} Let $n\geq4$ be an integer and $P_d(f)$ denote an algebraic differential-difference polynomial in $f(z)$ of degree $d\leq n-3$. If $p_1(z)$, $p_2(z)$ are nonzero polynomials, $\alpha_1$, $\alpha_2$ are nonzero constants with $\frac{\alpha_1}{\alpha_2}\neq (\frac{d}{n})^{\pm1},1$. Then the equation $$ f^{n}(z)+P_d(f)=p_1(z)e^{\alpha_1z}+p_2(z)e^{\alpha_2z}, $$ does not have any transcendental entire solution of finite order. \end{theorem} Peng and Chen \cite{Peng} considered the special case for difference equations and obtained some results. \begin{theorem}[{\cite[Theorem 2.1]{Peng}}] Consider the nonlinear difference equation $$ f^{n}(z)+a(z)f(z+1)=c\sin bz, $$ where $a(z)$ is a nonconstant polynomial, $b$, $c$ are nonzero constants and $n\geq2$ is an integer. Suppose that an entire function $f(z)$ satisfies any one of the following three conditions: \begin{itemize} \item[(1)] $\lambda(f)<\sigma(f)=\infty$; \item[(2)] $\lambda_2(f)<\sigma_2(f)$; \item[(3)] $\mu_2(f)<1$. \end{itemize} Then $f(z)$ can not be an entire solution of this equation. \end{theorem} In this paper, we consider a general differential-difference equation and obtain the following theorem. \begin{theorem}\label{T-1} Consider the nonlinear differential-difference equation \begin{equation}\label{E-1.1} q(z)f^{n}(z)+a(z)f^{(k)}(z+1)=p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}, \end{equation} where $p_1(z)$, $p_2(z)$ are two nonzero polynomials, $q(z)$, $a(z)$ are two nonzero entire functions of finite order, $q_1(z)$, $q_2(z)$ are two nonconstant polynomials, $n\geq2$ is an integer. Suppose that an entire function $f(z)$ satisfies any one of the following two conditions: \begin{itemize} \item[(1)] $\lambda(f)<\sigma(f)=\infty$, $\sigma_2(f)<\infty;$\par \item[(2)] $\lambda_2(f)<\sigma_2(f)<\infty$. \end{itemize} Then $f(z)$ can not be an entire solution of \eqref{E-1.1}. \end{theorem} Zhang and Liao \cite{Zhang} also considered the existence of transcendental entire solutions of finite order to $$ f^{3}(z)+a(z)f(z+1)=p_1e^{\lambda z}+p_2e^{-\lambda z} $$ and obtained the following theorem. \begin{theorem}[{\cite[Theorem 4]{Zhang}}] \label{thm1.4} Let $p_1$, $p_2$ and $\lambda$ be nonzero constants, for the difference equation $$ f^{3}(z)+a(z)f(z+1)=p_1e^{\lambda z}+p_2e^{-\lambda z}, $$ where $a(z)$ is a polynomial, we have \begin{itemize} \item[(1)] if $a(z)$ is not a constant, then the equation does not have any transcendental entire solution of finite order; \item[(2)] if $a(z)$ is a nonzero constant, then the equation admits transcendental entire solutions of finite order if and only if the condition $e^{\lambda/3}=\mp1$ and $p_1p_2=\pm(a/3)^{3}$ holds. Furthermore if the condition above holds, then the transcendental entire solution of finite order of the equation has the form $$ f(z)=\sigma_je^{2k\pi iz}-\frac{a}{3\sigma_j}e^{-2k\pi iz} $$ or $$ f(z)=\sigma_je^{2k\pi iz+\pi iz}+\frac{a}{3\sigma_j}e^{-(2k\pi iz+\pi iz)}. $$ \end{itemize} \end{theorem} In this article, we consider the more general case for differential-difference equations and obtain the following theorem. \begin{theorem} \label{T-2} Let $p_1, p_2$ and $\lambda$ be nonzero constants, $a(z)$ be an entire function with zero order, $q(z)$ be a nonconstant polynomial. Then any transcendental entire solution $f(z)$ of finite order of the equation \begin{equation}\label{E-1.2} f^{3}(z)+a(z)f^{(k)}(z+1)=p_1e^{\lambda q(z)}+p_2e^{-\lambda q(z)}, \end{equation} satisfies $\delta(0,f)=0$. \end{theorem} For the special case of $q(z)\equiv z$, we have the following result. \begin{theorem}\label{T-3} Consider the differential-difference equation \begin{equation}\label{E-1.3} f^{3}(z)+a(z)f^{(k)}(z+1)= p_1e^{\lambda z}+p_2e^{-\lambda z}, \end{equation} where $p_1$, $p_2$ and $\lambda$ are nonzero constants, $a(z)$ is an entire function with zero order. We have \begin{itemize} \item[(1)] if $a(z)$ is not a constant, then the equation does not have any transcendental entire solution of finite order; \item[(2)] if $a(z)$ is a nonzero constant, $k$ is an even number, then the equation admits transcendental entire solutions of finite order if and only if the condition $e^{\lambda/3}=\mp1$ and $p_1p_2=\pm(a/3)^{3}$ holds. Furthermore if the condition above holds, then the transcendental entire solution of finite order of the equation has the form $$ f(z)=\sigma_je^{2k\pi iz}-\frac{a}{3\sigma_j}e^{-2k\pi iz} $$ or $$ f(z)=\sigma_je^{2k\pi iz+\pi iz}+\frac{a}{3\sigma_j}e^{-(2k\pi iz+\pi iz)}; $$ \item[(3)] if $a(z)$ is a nonzero constant, $k$ is an odd number, then the equation admits transcendental entire solutions of finite order if and only if the condition $e^{\frac{1}{3}\lambda}=\mp i$ and $p_1p_2=\pm(\frac{ai}{3})^{3}$ holds. \end{itemize} Furthermore if the condition above holds, then the transcendental entire solution of finite order of the equation has the form $$ f(z)=\sigma_je^{2k\pi iz+\frac{\pi}{2}iz} -\frac{ai}{3\sigma_j}e^{-(2k\pi iz+\frac{\pi}{2}iz)} $$ or $$ f(z)=\sigma_je^{2k\pi iz-\frac{\pi}{2}iz} +\frac{ai}{3\sigma_j}e^{-(2k\pi iz-\frac{\pi}{2}iz)}. $$ \end{theorem} \section{Lemmas} To prove our results, we need some lemmas. \begin{lemma}[\cite{yi-yang}] \label{L-1} Suppose that $ f_1(z), f_2(z), \ldots, f_n(z), (n\geq2) $ are meromorphic functions and $ g_1(z), g_2(z), \ldots, g_n(z)$ are entire functions satisfying the following conditions: \begin{itemize} \item[(1)] $\sum_{j=1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv 0$; \item[(2)] $ g_j(z)-g_k(z)$ are not constants for $1\leq j0$, $$ m(r,P(z,f))=O(r^{\sigma-1+\varepsilon})+S(r,f) $$ holds possibly outside of an exceptional set of finite logarithmic measure. \end{lemma} \begin{lemma}[\cite{Yang2}] \label{L-4} Suppose that $c$ is a nonzero constant and $\alpha$ is a nonconstant meromorphic function. Then the equation $$ f^2(z)+(cf^{(n)}(z))^2=\alpha $$ has no transcendental meromorphic solution $f(z)$ satisfying $T(r,\alpha)=S(r,f)$. \end{lemma} \section{Proofs main results} \begin{proof}[Proof of Theorem \ref{T-1}] (1) Let $f$ be an entire solution of equation \eqref{E-1.1} and satisfy $\lambda(f)<\sigma(f)=\infty, \sigma_2(f)<\infty$. By Lemma \ref{L-2}, $f(z)$ can be rewritten as $f(z)=Q(z)e^{g(z)}$, where $Q$ is an entire function, $g$ is a transcendental entire function such that $\lambda(Q)=\sigma(Q)=\lambda(f)$, $\lambda_2(Q)=\sigma_2(Q)=\lambda_2(f)$, $\sigma_2(f)=\max\{\sigma_2(Q),\sigma_2(e^{g})\}$. From condition $\sigma_2(f)<\infty$, so $\sigma(g)=\sigma_2(e^{g})<\infty$. Substituting $f(z)=Q(z)e^{g(z)}$ into \eqref{E-1.1} we obtain that \begin{equation}\label{E-3.1} q(z)Q^{n}(z)e^{ng(z)}+a(z)H(z)e^{g(z+1)}=p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)} , \end{equation} where $H(z)$ is a differential polynomial in $Q(z+1)$ and $g(z+1)$, $\sigma(H)<\infty.$ Set $G(z)=g(z+1)-ng(z)$, then \eqref{E-3.1} becomes \begin{equation}\label{E-3.2} q(z)Q^{n}(z) +a(z)H(z)e^{G(z)}=e^{-ng(z)} \Big(p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}\Big). \end{equation} If $G(z)$ is a polynomial, then $$ \sigma\Big(q(z)Q^{n}(z)+a(z)H(z)e^{G(z)}\Big)<\infty, $$ but $$ \sigma\Big(e^{-ng(z)}\Big(p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}\Big)\Big) =\infty. $$ Then by \eqref{E-3.2}, we obtain a contradiction. If $G(z)$ is a transcendental entire function, then \eqref{E-3.1} can be rewritten as \begin{equation}\label{E-3.3} q(z)Q^{n}(z)e^{ng(z)}+a(z)H(z)e^{g(z+1)}-e^{h(z)} \Big(p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}\Big)=0, \end{equation} where $h(z)\equiv0.$ By Lemma \ref{L-1}, we deduce $$ q(z)Q^{n}(z)\equiv0, a(z)H(z)\equiv0, -p_1(z)e^{q_1(z)}-p_2(z)e^{q_2(z)}\equiv0, $$ for $Q^{n}(z)\equiv0$, so $f(z)\equiv0$, but $\sigma(f)=\infty$, this is a contradiction. (2) Suppose that $f$ is an entire solution of equation \eqref{E-1.1} and satisfies $\lambda_2(f)<\sigma_2(f)<\infty$. By Lemma \ref{L-2}, $f(z)$ can be rewritten as $f(z)=Q(z)e^{g(z)}$, where $Q$ is an entire function, $g$ is a transcendental entire function such that $$ \lambda(Q)=\sigma(Q)=\lambda(f), \lambda_2(Q)=\sigma_2(Q) =\lambda_2(f), \sigma_2(f)=\max\{\sigma_2(Q), \sigma_2(e^{g})\}. $$ From condition, we obtain $\sigma_2(f)=\sigma_2(e^{g})<\infty$, so $\sigma_2(Q)<\sigma_2(e^{g})=\sigma(g)<\infty$. Substituting $f(z)=Q(z)e^{g(z)}$ into \eqref{E-1.1}, we obtain \eqref{E-3.2}. Since $\sigma(q(z))<\infty$, so $\sigma_2(q(z))=0$. If $\sigma(G)<\sigma(g)$, then \begin{align*} \sigma_2\Big(q(z)Q^{n}(z)+a(z)H(z)e^{G(z)}\Big) &\leq \max \{\sigma_2(Q),\sigma(G)\}<\sigma(g)\\ &= \sigma_2\Big(e^{-ng(z)}\Big(p_1(z)e^{q_1(z)}+p_2(z)e^{q_2(z)}\Big)\Big), \end{align*} which is a contradiction. If $\sigma(G)=\sigma(g)$, then we can get \eqref{E-3.3}. Using the same method as in the proof of $(1)$, by Lemma \ref{L-1}, we also get a contradiction. \end{proof} \begin{proof}[Proof of Theorem \ref{T-2}] Let $f(z)$ be a transcendental entire solution of finite order of \eqref{E-1.2} with $\delta(0,f)>0$. By differentiating both sides of \eqref{E-1.2}, we obtain \begin{equation}\label{E-3.4} 3f^2(z)f'(z)+a'(z)f^{(k)}(z+1)+a(z)f^{(k+1)}(z+1) =\lambda q'(z)\Big(p_1 e^{\lambda q(z)}-p_2 e^{-\lambda q(z)}\Big). \end{equation} By taking both squares of \eqref{E-1.2} and \eqref{E-3.4}, and eliminating $e^{\pm\lambda q(z)}$, we deduce \begin{equation}\label{E-3.5} \begin{aligned} &(\lambda q'(z))^2\Big(f^{3}(z)+a(z)f^{(k)}(z+1)\Big)^2\\ &-\Big(3f^2(z)f'(z)+a'(z)f^{(k)}(z+1)+a(z)f^{(k+1)}(z+1)\Big)^2\\ &=4p_1p_2\lambda^2(q'(z))^2, \end{aligned} \end{equation} Set $\alpha(z)=\lambda^2(q'(z))^2f^2(z)-9(f'(z))^2$, thus $\alpha(z)$ is an entire function. Then we rewrite \eqref{E-3.5} in the form $f^{4}\alpha=Q(f)$, where $Q(f)$ is a differential-difference polynomial in $f(z)$ with total degree $4$. By Lemma \ref{L-3}, we obtain $$ T(r,\alpha)=m(r,\alpha)=O(r^{\sigma-1+\varepsilon})+S(r,f). $$ Thus $\alpha$ is a small function of $f(z)$. Next, we consider two cases. \smallskip \noindent\textbf{Case 1.} $\alpha\equiv 0.$ Then $f(z)=ce^{\pm\frac{1}{3}\lambda q(z)}$. By substituting this into \eqref{E-1.2}, we obtain $$ (c^{3}-p_1)e^{\lambda q(z)}+\frac{1}{3}\lambda a(z)q'(z+1) e^{\frac{1}{3}\lambda q(z+1)}-p_2 e^{-\lambda q(z)}=0, $$ or $$ (c^{3}-p_2)e^{-\lambda q(z)}-\frac{1}{3}\lambda a(z)q'(z+1) e^{-\frac{1}{3}\lambda q(z+1)}-p_1 e^{\lambda q(z)}=0. $$ Since $q(z)$ is a nonconstant polynomial, by Lemma \ref{L-1}, we obtain $p_1=0$ or $p_2=0$. This is a contradiction. \noindent\textbf{Case 2.} $\alpha \not\equiv 0.$ We rewrite $\alpha$ as $$ \alpha=f^2A(z), $$ where $A(z)=\lambda^2q'-9(\frac{f'}{f})^2$, by the Lemma of Logarithmic Derivative of meromorphic function, then $m(r,A)=S(r,f)$. Since $\alpha \not\equiv 0$, then $A \not\equiv 0$. For any Small $\varepsilon>0$, we have \begin{align*} O(1)+2T(r,f) &=T(r, f^2)=m(r,f^2)=m(r,\frac{\alpha}{A})\\ &\leq m(r,\alpha)+m(r,\frac{1}{A})\\ &\leq S(r,f)+T(r,A)\\ &\leq S(r,f)+N(r,A)\\ &\leq S(r,f)+2N(r,\frac{1}{f}) \\ &\leq 2(1-\delta(0,f)+\varepsilon)T(r,f). \end{align*} This is impossible for $0<\varepsilon<\delta(0,f)$. The proof of Theorem \ref{T-2} is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{T-3}] Suppose that $f(z)$ is a transcendental entire solution of \eqref{E-1.3} with finite order. By differentiating both sides of \eqref{E-1.3}, we obtain \begin{equation}\label{E-3.6} 3f^2(z)f'(z)+a'(z)f^{(k)}(z+1)+a(z)f^{(k+1)}(z+1) =\lambda p_1 e^{\lambda z}-\lambda p_2 e^{-\lambda z}. \end{equation} By taking both squares of \eqref{E-1.3} and \eqref{E-3.6}, and eliminating $e^{\pm\lambda z}$, we deduce \begin{align*} 4\lambda^2p_1p_2 &=\lambda^2\left(f^{3}(z)+a(z)f^{(k)}(z+1)\right)^2\\ &\quad -\left(3f^2(z)f'(z)+a'(z)f^{(k)}(z+1)+a(z)f^{(k+1)}(z+1)\right)^2, \end{align*} set $\alpha(z)=\lambda^2f^2(z)-9(f'(z))^2$, thus $\alpha(z)$ is an entire function. Then we rewrite\eqref{E-3.6} in the form $f^{4}\alpha=Q(f)$, where $Q(f)$ is a differential-difference polynomial in $f(z)$ with total degree $4$. By Lemma \ref{L-3}, we obtain $$ T(r,\alpha)=m(r,\alpha)=O(r^{\sigma-1+\varepsilon})+S(r,f). $$ Thus $\alpha$ is a small function of $f(z)$. Next, we consider two cases. \smallskip \noindent\textbf{Case 1.} $\alpha \equiv0.$ Then $f(z)=ce^{\pm\frac{1}{3}\lambda z}$. By substituting this into \eqref{E-1.3}, we obtain $$ (c^{3}-p_1)e^{\lambda z}+(\frac{1}{3}\lambda)^{k}a(z) e^{\frac{1}{3}\lambda(z+1)}-p_2 e^{-\lambda z}=0, $$ or $$ (c^{3}-p_2)e^{-\lambda z}+(-\frac{1}{3}\lambda)^{k}a(z) e^{-\frac{1}{3}\lambda(z+1)}-p_1 e^{\lambda z}=0. $$ By Lemma \ref{L-1}, we obtain $p_1=0$ or $p_2=0$. This is a contradiction. \smallskip \noindent\textbf{Case 2.} $\alpha \not\equiv 0$. By Lemma \ref{L-4}, we obtain $\alpha$ is a nonzero constant. Thus $$ \alpha'=2f'(\lambda^2f-9f^{''})=0. $$ Since $f(z)$ is transcendental, then $$ \lambda^2f-9f^{''}=0. $$ By a simple calculation, $$ f(z)=c_1 e^{\frac{1}{3}\lambda z}+c_2 e^{-\frac{1}{3}\lambda z}, $$ where $c_1, c_2$ are nonzero constants. By substituting this into \eqref{E-1.3} and simple calculation, get \begin{align*} &(c_1^{3}-p_1)e^{\lambda z}+(c_2^{3}-p_2)e^{-\lambda z} +\Big(3c_1^2c_2+c_1a(z)(\frac{1}{3}\lambda)^{k}e^{\frac{1}{3}\lambda}\Big) e^{\frac{1}{3}\lambda z}\\ &+\Big(3c_1c_2^2+c_2a(z)(-\frac{1}{3}\lambda)^{k}e^{-\frac{1}{3}\lambda}\Big) e^{-\frac{1}{3}\lambda z}=0, \end{align*} by Lemma \ref{L-1}, we deduce $$ c_1^{3}=p_1, c_2^{3}=p_2, 3c_1c_2+a(z)(\frac{1}{3}\lambda)^{k} e^{\frac{1}{3}\lambda}\equiv 0, 3c_1c_2 +a(z)(-\frac{1}{3}\lambda)^{k}e^{-\frac{1}{3}\lambda}\equiv 0. $$ If $a(z)$ is not a nonzero constant, we can get a contradiction. Then equation \eqref{E-1.3} does not admit any transcendental entire solution of finite order. If $a(z)$ is a nonzero constant, $k$ is an even number, then $$ a(\frac{1}{3})^{k}\lambda^{k} \Big(e^{\frac{1}{3}\lambda}-e^{-\frac{1}{3}\lambda}\Big)=0, $$ so $$ e^{\frac{1}{3}\lambda}=\mp 1,p_1p_2=\pm(\frac{a}{3})^{3},c_1c_2=\pm\frac{a}{3}. $$ Thus $c_1$ can assume $\sigma_j(j=1, 2, 3)$, where $\sigma_j$ satisfies $\sigma_j^{3}=p_1(j=1, 2, 3)$ and $c_2=\pm\frac{a}{3c_1}$. Hence $f(z)$ is of the following forms $f(z)=\sigma_j e^{2k\pi iz}-\frac{a}{3\sigma_j}e^{-2k\pi iz}$ or $f(z)=\sigma_j e^{2k\pi iz+\pi iz}+\frac{a}{3\sigma_j}e^{-{(2k\pi iz+\pi iz)}}$. If $a(z)$ is a nonzero constant, $k$ is an odd number, then $$ a(\frac{1}{3})^{k}\lambda^{k}\Big(e^{\frac{1}{3}\lambda}+e^{-\frac{1}{3}\lambda} \Big)=0, $$ so $$ e^{\frac{1}{3}\lambda}=\mp i, p_1p_2=\pm (\frac{ai}{3})^{3}, c_1c_2 =\pm \frac{ai}{3}. $$ Thus $c_1$ can assume $\sigma_j$ $(j=1, 2, 3)$, where $\sigma_j$ satisfies $\sigma_j^{3}=p_1(j=1, 2, 3)$ and $c_2=\pm\frac{ai}{3c_1}$. Hence $f(z)$ is of the following forms $f(z)=\sigma_j e^{2k\pi iz+\frac{\pi}{2}iz} -\frac{a i}{3\sigma_j}e^{-(2k\pi iz+\frac{\pi}{2}iz)}$ or $f(z)=\sigma_j e^{2k\pi iz-\frac{\pi}{2}iz}+\frac{a i}{3\sigma_j} e^{-(2k\pi iz-\frac{\pi}{2}iz)}$. Therefore, the proof of Theorem \ref{T-3} is complete. \end{proof} \subsection*{Acknowledgements} This research was partly supported by NSFC (nos. 11101201, 11461042, 11301260), CPSF (no.2014M551865), PSF of Jiangxi (no.2013KY10)， NSF of Jiangxi (20132BAB211003). \begin{thebibliography}{99} \bibitem{Bergweiler} W. Bergweiler, J. K. Langley; \emph{Zeros of differences of meromorphic functions,} Math. Proc. Cambridge Philos. Soc. \textbf{142} (2007), 133--147. \bibitem{Chen1} Z. X. Chen; \emph{Growth and zeros of meromorphic solution of some linear difference equations,} J. Math. Anal. Appl. \textbf{373} (2011), 235--241. \bibitem{Chen2} Z. X. Chen; \emph{On the entire function sharing one value CM with k-th derivatives,} J. Korean. Math. Soc. \textbf{42} (2005), 85--99. \bibitem{chiang} Y. M. Chiang and S. J. Feng; \emph{On the Nevanlinna characteristic $f(z+\eta)$ and difference equations in the complex plane,} The Ramanujan J. \textbf{16} (2008), 105--129. \bibitem{Halburd1} R. G. Halburd, R. J. Korhonen; \emph{Difference analogue of the lemma on the logarithmic derivative with applications to difference equations,} J. Math. Anal. Appl. \textbf{314} (2006), 477--487. \bibitem{Halburd2}R. G. Halburd, R. J. Korhonen; \emph{Nevanlinna theory for the difference operator,} Ann. Acad. Sci. Fenn. Math. \textbf{31} (2006), 463--478. \bibitem{huang} Z. B. Huang, Z. X. Chen; \emph{A Clunie lemma for difference and $q$-difference polynomials,} Bull. Aust. Math. Soc. \textbf{81} (2010), 23--32. \bibitem{Laine1} I. Laine; \emph{Nevanlinna Theory and Complex Differential Equations,} Studies in Mathematics 15, Walter de Gruyter, Berlin-New York (1993). \bibitem{Laine2} I. Laine, C. C. Yang; \emph{Clunie theorems for difference and q-difference polynomials,} J. London. Math. Soc. \textbf{76}(2007), 556--566. \bibitem{Laine3} I. Laine, C. C. Yang; \emph{Value distribution of difference polynomials,} Proc. Japan Acad. Ser. A \textbf{83} (2007), 148--151. \bibitem{Li} P. Li, C. C. Yang; \emph{On the nonexistence of entire solutions of certain type of nonlinear differential equations,} J. Math. Anal. Appl. \textbf{320} (2006), 827--835. \bibitem{Liu} K. Liu, L. Z. Yang; \emph{Value distribution of the difference operator}, Arch. Math. \textbf{92} (2009), 270--278. \bibitem{Peng} C. W. Peng, Z. X. Chen; \emph{On a conjecture concerning some nonlinear difference equations,} Bull. Malays. Math. Sci. Soc. (2) \textbf{36}(1) (2013), 221šC227. \bibitem{Yang1}C. C. Yang, I. Laine; \emph{On analogies between nonlinear difference and differential equations,} Proc. Japan Acad. Ser. A \textbf{86} (2010), 10--14. \bibitem{Yang2} C. C. Yang, P. Li; \emph{On the transcendental solutions of a certain type of nonlinear differential equations,} Arch. Math. \textbf{82} (2004), 442--448. \bibitem{yi-yang} H. X. Yi, C. C. Yang; \emph{Uniqueness Theory of Meromorphic Functions}, Science Press 1995/Kluwer 2003. \bibitem{Zhang} J. Zhang, L. W. Liao; \emph{On entire solutions of a certain type of nonlinear differential and difference equations,} Taiwan. J. Math. \textbf{15} (2011), 2145--2157. \end{thebibliography} \end{document}