Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

数学物理学报, 2021, 41(1): 69-80 doi:

论文

一类非线性复微分差分方程解的不存在性

林书情,, 陈俊凡,

The Non-Existence of Solutions of a Certain type of Nonlinear Complex Differential-Difference Equations

Lin Shuqing,, Chen Junfan,

通讯作者: 陈俊凡, E-mail: junfanchen@163.com

收稿日期: 2020-02-14  

基金资助: 国家自然科学基金.  11301076
福建省自然科学基金.  2018J01658
福建省自然科学基金.  2019J01672

Received: 2020-02-14  

Fund supported: Supported by the NSFC.  11301076
the NSF of Fujian Province.  2018J01658
the NSF of Fujian Province.  2019J01672

作者简介 About authors

林书情,E-mail:shuqinglin1996@163.com , E-mail:shuqinglin1996@163.com

Abstract

In this paper, we study transcendental entire solutions of a certain type of complex differential-difference equations

[f(z)f(z)]n+fm(z+η)=1,
[f(z)f(z)]n+[f(z+η)f(z)]m=1,
[f(z)f(z)]2+P2(z)f2(z+η)=Q(z)eα(z),
where P(z) and Q(z) are non-zero polynomials, α(z) is polynomial, m and n are positive integers, ηC{0}. Several sufficient conditions on the non-existence of transcendental entire solutions of such equations are supplied.

Keywords: Complex differential difference equation ; Transcendental entire solution ; Nevanlinna theory ; Finite order

PDF (334KB) 元数据 多维度评价 相关文章 导出 EndNote| Ris| Bibtex  收藏本文

本文引用格式

林书情, 陈俊凡. 一类非线性复微分差分方程解的不存在性. 数学物理学报[J], 2021, 41(1): 69-80 doi:

Lin Shuqing, Chen Junfan. The Non-Existence of Solutions of a Certain type of Nonlinear Complex Differential-Difference Equations. Acta Mathematica Scientia[J], 2021, 41(1): 69-80 doi:

1 引言与主要结果

f(z)是复平面C上的亚纯函数, 假定读者熟悉Nevanlinna值分布理论的通用记号(参见文献[1-2]), 如T(r,f), N(r,f), m(r,f),¯N(r,f),S(r,f), 其中, 当亚纯函数a(z)()满足

T(r,a)=S(r,f)=o(T(r,f)),r,

可能需除去r的一个线性测度为有限的集合, 则称a(z)为函数f(z)的一个小函数, 简称小函数.下面引入亚纯函数f(z)的级, 即ρ(f(z))的定义

ρ(f(z))=lim sup

众所周知, Nevanlinna值分布理论是处理复差分(或微分或微分差分)方程解的性质的有效工具(参见文献[3-14]).近年来, 关于复微分差分方程亚纯解的存在性问题受到很多学者们的关注, 并获得了一些重要结果(参见文献[15-19]).本文致力于研究一类复微分差分方程解的不存在性问题.

首先, 我们回顾Fermat型复微分差分方程

\begin{equation} f^{n}(z)+g^{n}(z) = 1 \end{equation}
(1.1)

解的结果:当 n\geq3 时, Montel[11]证明了方程(1.1)没有超越整函数解.但当 n = 2 时, Gross[5]证明了方程 f^{2}(z)+g^{2}(z) = 1 有整函数解 f(z) = \sin(u(z)), g(z) = \cos(u(z)) , 其中 u(z) 为任意的整函数. 1970年, Yang[13]研究了更一般的Fermat型复微分差分方程

\begin{equation} f^{n}(z)+g^{m}(z) = 1, \end{equation}
(1.2)

并获得以下结果.

定理A[13]  设 n, m 为正整数并满足 \frac{1}{m}+\frac{1}{n}<1, 则没有非常数整函数解 f(z), g(z) 满足方程 (1.2) .

在方程(1.2)中, 如果 f(z) g(z) 存在某种函数关系, 例如, g(z) f(z) 的差分算子: g(z) = f(z+\eta) , 那么将方程(1.2)改写为如下形式

\begin{equation} f^{n}(z)+f^{m}(z+\eta) = 1. \end{equation}
(1.3)

我们假设 n\geq m. n>m>1 n = m>2 时, 由定理A可知(1.3)式没有非常数整函数解; 当 n>m = 1 时, Shimomura[12]证明了(1.3)式存在一个整函数解; 但当 m\neq n 时, Liu等[10]证明了(1.3)式没有有穷级超越整函数解; 此外, Liu等[18]证明了差分方程 f(z)+f(z+\eta) = 1 存在周期为 2c 的函数解, 同时他们证明了差分方程 f^{2}(z)+f^{2}(z+\eta) = 1 的超越整函数解满足 f(z) = \sin (Az+B) , 其中 B 为常数, A = \frac{(4k+1)\pi}{2\eta} , k 为整数; 随后, Lü等[20]证明了差分方程 f^{3}(z)+f^{3}(z+\eta) = 1 没有有穷级亚纯解.

2010年, Yang等[14]研究了方程 f^{n}(z)+L(z, f) = h 的亚纯解, 其中 L(z, f) 是关于 f 的线性微分差分多项式, h\not\equiv0 为有穷级亚纯函数. 2012年, Liu等[18]研究了一类复微分差分方程超越整函数解的问题, 证明了如下定理.

定理B[18]  设 n\neq m, 则复微分差分方程

\begin{equation} {f'}^{n}(z)+f^{m}(z+\eta) = 1 \end{equation}
(1.4)

没有有穷级超越整函数解, 其中 m, n 为正整数, \eta\in{\Bbb C}\setminus\{0\} .

定理C[18]  设 m\neq n>1, 则复微分差分方程

\begin{equation} {f'}^{n}(z)+[f(z+\eta)-f(z)]^{m} = 1 \end{equation}
(1.5)

没有有穷级超越整函数解, 其中 m, n 为正整数, \eta\in{\Bbb C}\setminus\{0\} .

m>2, n>2 时, 由定理A可知方程(1.4)没有非常数整函数解.当 m = n = 2 时, Liu等[18]进一步研究方程(1.4)超越整函数解的存在性及解的形式, 主要结果如下.

定理D[18]  复微分差分方程

{f'}^{2}(z)+f^{2}(z+\eta) = 1

的有穷级超越整函数解必满足 f(z) = \sin(z\pm B{\rm i}), 其中 B 为常数, \eta = 2k\pi \eta = 2k\pi+\pi, k 为整数.

2017年, Chen等[16]推广了定理D中Fermat型复微分差分方程

\begin{equation} (f'(z))^{2}+P^{2}(z)f^{2}(z+\eta) = Q(z)e^{\alpha(z)}, \end{equation}
(1.6)

其中 P(z), Q(z) 为非零多项式, \alpha(z) 为多项式且 \eta\in{\Bbb C}\setminus\{0\} , 并获得以下结果.

定理E[16]  设 f(z) 为方程 (1.6) 的有穷级超越整函数解, 则 f(z) 必满足下列三种情形之一

(ⅰ)

f(z) = \frac{c_{1}e^{p_{1}z+q_{1}}}{2p_{1}}+\frac{c_{2}e^{p_{2} z+q_{2}}}{2p_{2}},

其中 P(z), Q(z) 退化为常数, p_{1} = {\rm i}e^{p_{1}\eta}A, p_{2} = -{\rm i}e^{p_{2}\eta}A, q_{1}, q_{2} 为常数 , p_{1}, p_{2}, c_{1}, c_{2}, A, \eta 为非零常数;

(ⅱ) P(z) 为常数, Q(z) 是次数为 1 的多项式, 则

f(z) = \frac{(a_{1}z+a_{0}-\frac{a_{1}}{p_{1}})e^{p_{1}z+q_{1} }}{2p_{1}}+\frac{c_{2}e^{p_{2}z+q_{2}}}{2p_{2}}, \quad \frac{1}{p_{1}} = \eta, \quad \frac{1}{p_{2}}\neq \eta

f(z) = \frac{c_{1}e^{p_{1}z+q_{1}}}{2p_{1}}+\frac{(b_{1}z+b_{0}-\frac{b_{1}}{p_{2}})e^{p_{2}z+q_{2}}}{2p_{2}}, \quad \frac{1}{p_{1}}\neq \eta, \quad \frac{1}{p_{2}} = \eta,

其中 p_{1} = {\rm i}e^{p_{1}\eta}A, p_{2} = -{\rm i}e^{p_{2}\eta}A, q_{1}, q_{2}, a_{0}, b_{0} 为常数, p_{1}, p_{2}, c_{1}, c_{2}, a_{1}, b_{1}, A, \eta 为非零常数;

(ⅲ)

f(z) = B(z)e^{pz}, \quad \alpha(z) = 2pz+D,

其中 p, \eta 为非零常数, D 为常数, B(z) 为多项式满足 [B'(z)+pB(z)]^{2}+P^{2}(z)B^{2}(z+\eta)e^{2p\eta} = Q(z)e^{D}.

在本文中, 我们考虑将方程(1.4), (1.5)和(1.6)中的项 f'(z) f(z)f'(z) 替换, 即

\begin{equation} [f(z)f'(z)]^{n}+f^{m}(z+\eta) = 1, \end{equation}
(1.7)

\begin{equation} [f(z)f'(z)]^{n}+[f(z+\eta)-f(z)]^{m} = 1 , \end{equation}
(1.8)

\begin{equation} \left[f(z)f'(z)\right]^{2}+P^{2}(z)f^{2}(z+\eta) = Q(z)e^{\alpha(z )}, \end{equation}
(1.9)

并证明了如下定理.

定理1.1  设 n = m, 则方程(1.7)没有有穷级超越整函数解, 其中 m, n 为正整数 , \eta\in{\Bbb C}\setminus\{0\} .

例1.1  在方程(1.7)中, 取 n = 1, m = 2, 则复微分差分方程 f(z)f'(z)+f^{2}(z+\eta) = 1 有一个超越整函数解 f(z) = ae^{{\rm i}(z+b)}+\frac{1}{2a}e^{-{\rm i} (z+b)} = \sqrt{2}\cos z, 其中 e^{{\rm i}b} = \frac{\sqrt{2}}{2a}, \eta = \frac{3\pi }{4}+k\pi, k\in{\Bbb Z}, a\in {\Bbb C}\setminus\{0\}, b\in {\Bbb C} .

注1.1  例 1.1 表明, 当 n = 1, m = 2 时, 方程(1.7)存在有穷级超越整函数解.

定理1.2  设 m\neq n, n>2, 则方程(1.8)没有有穷级超越整函数解, 其中 m, n 为正整数 , \eta\in{\Bbb C}\setminus\{0\} .

例1.2  在方程 (1.8) 中, 取 n = 1, m = 2, 则复微分差分方程 f(z)f'(z)+[ f(z+\eta)-f(z)]^{2} = 1 有一个超越整函数解 f(z) = \sin2z, 其中 \eta = \frac{\pi}{4}+k\pi, k\in{\Bbb Z} .

注1.2  例 1.2 表明, 当 n = 1, m = 2 时, 方程 (1.8) 存在有穷级超越整函数解.此外, 在方程(1.8)中, 当 n = 1, m = 1 时, 由引理2.1可知复微分差分方程 f(z)f'(z)+[f(z+\eta)-f(z)] = 1 没有超越整函数解.

定理1.3  设 P(z), Q(z) 为非零多项式, \alpha(z) 为多项式且 \eta\in{\Bbb C}\setminus\{0\}, 则方程(1.9)没有有穷级超越整函数解.

例1.3  在方程 (1.9) 中, 取 P(z)\equiv0, 则复微分差分方程 \left[f(z)f'(z)\right]^{2} = Q(z)e^{\alpha(z)} 有一个超越整函数解 f(z) = q(z)e^{\frac{\alpha(z)}{4}}, 其中 q(z) 为多项式满足 \left(q(z)q'(z)+q^{2}(z)\frac{\alpha'(z)}{4}\right)^{2} = Q(z) .

例1.4  复微分差分方程 f(z)f'(z)+f^{2}(z+\eta) = -e^{2z} 有一个超越整函数解 f(z) = e^{z}, 其中, e^{2\eta} = -2.

例1.5  复微分差分方程 \left[f(z)f'(z)+2f'(z)\right]+2f^{2}(z+\eta) = e^{\frac{z}{2}} 有一个超越整函数解 f(z) = e^{\frac{z}{2}}, 其中, e^{\eta} = -\frac{1}{4}.

注1.3  例1.3表明, 当 P(z)\equiv0 时, 方程 (1.9) 存在有穷级超越整函数解.例1.4和例1.5表明, 在方程 (1.9) 中, 项 \left[f(z)f'(z)\right]^{2} 不能用任意一个关于 f(z) 的次数为2的微分单项或微分多项式替换.此外, 在方程(1.9)中, 取 P(z)\equiv\pm1, Q(z)\equiv1 \alpha(z)\equiv0, 由定理1.1可知复微分差分方程 \left[f(z)f'(z)\right]^{2}+f^{2}(z+\eta) = 1 没有有穷级超越整函数解.

2 一些引理

引理2.1[14]  设 f(z) 为差分方程 U(z, f)P(z, f) = Q(z, f) 的有穷级 \rho 超越亚纯解, 其中 U(z, f), P(z, f), Q(z, f) 是关于 f 的差分多项式, 满足关于 f 和它的位移算子的 U(z, f) 的总次数是 n, Q(z, f) 的总次数 \leq n . U(z, f) 仅有一个最高次项, 则对任意的 \varepsilon>0,

m(r, P(z, f)) = O(r^{\rho-1+\varepsilon})+S(r, f),

除去一个有限对数测度的例外值集.

注2.1  证明这一引理的关键工具是值分布理论的核心部分, 即:对数导数引理.

下面的结果是亚纯函数的微分差分多项式的Clunie型引理[2, 21].它可以使用引理2.1中类似的方法证明, 具体如下.

命题2.1[2, 21]  若在引理2.1中, U(z, f) = f^{n}, P(z, f), Q(z, f) 是关于 f 的两个微分差分多项式, 则

m(r, P(z, f)) = S(r, f).

引理2.2[6]  设 f(z) 为有穷级 \rho 的超越亚纯函数, 则对任意 \varepsilon>0,

m\left (\frac{f(z+\eta)}{f(z)}\right) = O(r^{\rho-1+\varepsilon}) = S(r, f),
(2.1)

其中 \eta\in{\Bbb C}\setminus\{0\} .

引理2.3[4]  设 f(z) 为有穷级 \rho 的亚纯函数, 则对任意 \varepsilon>0,

T(r, f(z+\eta)) = T(r, f)+O(r^{\rho-1+\varepsilon})+O(\log r).
(2.2)

于是, 若 f(z) 为有穷级 \rho 的超越亚纯函数, 则

T(r, f(z+\eta)) = T(r, f)+S(r, f).
(2.3)

引理2.4[22]  设亚纯函数 f_{j}(z) (j = 1, 2, \cdots, n) (n\geq2) , 整函数 g_{j}(z) (j = 1, 2, \cdots, n) (n\geq2) 满足下列各条件

(ⅰ) \sum\limits_{j = 1}^{n}f_{j}e^{g_{j}}\equiv0;

(ⅱ)当 1\leq j<l\leq n 时, g_{j}-g_{l} 为非常数;

(ⅲ)当 1\leq j\leq n, 1\leq h<l\leq n 时, T(r, f_{j}) = o(T(r, e^{g_{h}-g_{l}})) (r\rightarrow\infty, r\not\in E) , 则 f_{j}\equiv0 (j = 1, 2, \cdots, n).

引理2.5[22]  设 f_{1}(z) , f_{2}(z), \cdots, f_{n}(z) (n\geq 3) 是不为常数的亚纯函数, 除了 f_{n}(z) , 且满足 \sum\limits_{j = 1}^{n}f_{j}\equiv1 . f_{n}(z)\not\equiv0 , 并有

\sum\limits_{j = 1}^{n}N(r, \frac{1}{f_{j}(z)})+(n-1)\sum\limits_{j = 1}^{n}\overline{N}(r, f_{j}(z))<(\lambda+o(1))T(r, f_{k}(z)),

其中 r\in I, k = 1, 2, \cdots, n-1 并且 \lambda<1 , 则 f_{n}(z)\equiv1 .

3 定理1.1的证明

假设 f(z) 为方程 (1.7) 的有穷级超越整函数解.接下来, 我们分三种情形讨论.

情形1  若 n = m = 1 , 则将方程(1.7)改写为如下形式

\begin{equation} f(z)f'(z)+f(z+\eta) = 1. \end{equation}
(3.1)

因此, (3.1)式可化为

f(z+\eta)f'(z)\frac{f(z)}{f(z+\eta)} = 1-f(z+\eta).

结合命题2.1及(2.1)式, 可得

m\left(r, f'(z)\frac{f(z)}{f(z+\eta)}\right) = S(r, f(z+\eta)),

从而

\begin{eqnarray*} T(r, f'(z))& = &m(r, f'(z))\leq m\left(r, f'(z)\frac{f(z)}{f(z+\eta)}\right)+m\left(r, \frac{f(z+\eta)}{f(z)}\right)\\ & = &S(r, f(z+\eta))+S(r, f (z)). \end{eqnarray*}

结合上述讨论及(2.3)式, 可得 T(r, f'(z)) = m(r, f'(z))\leq S(r, f). 这与 f(z) 是超越整函数相矛盾.

情形2  若 n = m = 2 , 则将方程(1.7)改写为如下形式

\begin{equation} [f(z)f'(z)]^{2}+[f(z+\eta)]^{2} = 1, \end{equation}
(3.2)

\begin{equation} [f(z)f'(z)+{\rm i}f(z+\eta)][f(z)f'(z)-{\rm i}f(z+\eta)] = 1. \end{equation}
(3.3)

由(3.3)式, 可知 f(z)f'(z)+{\rm i}f(z+\eta)\neq0 f(z)f'(z)-{\rm i}f(z+\eta)\neq0 .结合(3.3)式及Hadamard分解定理, 假设

f(z)f'(z)+{\rm i}f(z+\eta) = e^{p(z)}

f(z)f'(z)-{\rm i}f(z+\eta) = e ^{-p(z)},

其中 p(z) 为非常数多项式.由上述两个方程可得

\begin{equation} f(z)f'(z) = \frac{e^{p(z)}+e^{-p(z)}}{2} \end{equation}
(3.4)

\begin{equation} f(z+\eta) = \frac{e^{p(z)}-e^{-p(z)}}{2{\rm i}}. \end{equation}
(3.5)

由(3.4)式及(3.5)式, 可得

\frac{\left(p'(z)e^{2p(z)}-p'(z)e^{-2p(z)}\right)}{-4} = \frac{e^{p(z+\eta)}+e^{-p(z+\eta)}}{2},

\begin{equation} p'(z)e^{-2p(z)}-p'(z)e^{2p(z)}-2e^{p(z+\eta)}-2e^{-p(z+\eta)} = 0. \end{equation}
(3.6)

因为 \deg(-2p(z)) = \deg(2p(z)) = \deg(p(z+\eta)) = \deg(-p(z+\eta))\geq1; \deg[-2p(z)-2p(z)] = \deg(-4p(z))\geq1; \deg[-2p(z)-p(z+\eta)]\geq1, \deg[- 2p(z)+p(z+\eta)]\geq1; \deg[2p(z)-p(z+\eta)]\geq1; \deg[2p(z)-(-p(z+\eta))]\geq1; \deg[p(z+\eta)-(-p(z+\eta))] = \deg[2p( z+\eta)]\geq1 .由引理2.4, 可得 p'(z)\equiv-p'(z)\equiv0 .这与 p(z) 为非常数多项式相矛盾.

情形3  若 n = m>2 , 由方程(1.7)及(2.3)式, 并运用Nevanlinna第二基本定理得

\begin{eqnarray} (n-1)T(r, f(z)f'(z))&\leq&\overline{N}(r, f(z)f'(z))+\overline{N}\left(r , \frac{1}{(f(z)f'(z))^{n}-1}\right)+S(r, f){}\\ &\leq&\overline{N}\left(r, \frac{1}{f(z+\eta)}\right)+S(r, f){}\\ &\leq& T(r, f(z+\eta))+S(r, f){}\\ &\leq &T(r, f(z))+S(r, f). \end{eqnarray}
(3.7)

结合上面讨论及方程(1.7), 可得

\begin{eqnarray*} mT(r, f(z+\eta))& = &nT(r, f(z)f'(z))+S(r, f)\leq\frac{n}{n-1}T(r, f) +S(r, f)\\ &\leq&\frac{3}{2}T(r, f)+S(r, f). \end{eqnarray*}

由(2.3)式, 可知 T(r, f) = S(r, f). 这与 f(z) 是超越整函数相矛盾.

这就完成了定理1.1的证明.

4 定理1.2的证明

根据定理A, 我们只需证明下列复微分差分方程

\begin{equation} [f(z)f'(z)]^{n}+[f(z+\eta)-f(z)] = 1 \end{equation}
(4.1)

没有超越整函数解, 其中 n>2 .假设 f(z) 为方程 (4.1) 的有穷级超越整函数解.

对方程 (4.1) 两边同时求导得

\begin{equation} n{f'}^{n-1}(z)[f^{n-1}(z){f'}^{2}(z)+f^{n}(z)f''(z)]+f'(z+\eta)-f'(z) = 0. \end{equation}
(4.2)

\varphi = f^{n-1}(z){f'}^{2}(z)+f^{n}(z)f''(z) , g(z) = f'(z) , 则将 (4.2) 式改写为如下形式

\begin{equation} ng^{n-1}(z)\varphi(z) = -[g(z+\eta)-g(z)]. \end{equation}
(4.3)

n-1\geq2 , 并结合命题2.1, 可得

m(r, \varphi(z)) = S(r, g(z)), \quad m(r, g(z)\varphi(z)) = S(r, g(z)).

显然, \varphi(z)\not\equiv0 , 否则 f^{2}(z)\equiv c_{1}z+c_{2}, 其中 c_{1} 为非零常数, c_{2} 为常数, 矛盾.注意到 f(z) 为超越整函数, 可知 N(r, \varphi(z)) = S(r, g(z)). 因此

\begin{eqnarray*} T(r, g(z)) = m(r, g(z))&\leq & m(r, g(z)\varphi(z))+m\left(r, \frac{1}{\varphi(z)}\right)\\ &\leq & m(r, \varphi(z))+N(r, \varphi(z))+S(r, g(z)) = S(r, g(z)), \end{eqnarray*}

T(r, f'(z)) = T(r, g(z))\leq S(r, g(z)) = S(r, f'(z)) , 矛盾.

这就完成了定理1.2的证明.

5 定理1.3的证明

首先, 假设 f(z) 为方程 (1.9) 的有穷级超越整函数解, 则

\begin{equation} [f(z)f'(z)+{\rm i}P(z)f(z+\eta)][f(z)f'(z)-{\rm i}P(z)f(z+\eta)] = Q(z )e^{\alpha(z)}. \end{equation}
(5.1)

因此, f(z)f'(z)+{\rm i}P(z)f(z+\eta) f(z) f'(z)-{\rm i}P(z)f(z+\eta) 均有有限多个零点.结合(5.1)式及Hadamard分解定理, 假设

\begin{equation} f(z)f'(z)+{\rm i}P(z)f(z+\eta) = Q_{1}(z)e^{\alpha_{1}(z)} \end{equation}
(5.2)

\begin{equation} f(z)f'(z)-{\rm i}P(z)f(z+\eta) = Q_{2}(z)e^{\alpha_{2}(z)}. \end{equation}
(5.3)

由(5.2)式及(5.3)式, 可得

\begin{equation} f(z)f'(z) = \frac{Q_{1}(z)e^{\alpha_{1}(z)}+Q_{2}(z)e^{\alpha_{2} (z)}}{2} \end{equation}
(5.4)

\begin{equation} f(z+\eta) = \frac{Q_{1}(z)e^{\alpha_{1}(z)}-Q_{2}(z)e^{\alpha_{2}(z) }}{2{\rm i}P(z)}, \end{equation}
(5.5)

其中 \alpha_{1}(z) , \alpha_{2 }(z) 为多项式, Q_{1}(z) , Q_{2}(z) 为非零多项式且满足 \alpha(z) = \alpha_{1}(z)+ \alpha_{2}(z) , Q(z) = Q_{1}(z)Q_{2}(z) .由(5.5)式, 可知 \alpha_{1}(z) , \alpha_{2}(z) 不能同时为常数.否则, f(z) 为有理函数, 这显然与 f(z) 是超越整函数相矛盾.

由(5.4)式及(5.5)式, 可得

\begin{eqnarray} &&\frac{h_{1}(z)}{-4P^{3}(z)Q_{1 }(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)} +\frac{h_{2}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{1}(z)- \alpha_{1}(z+\eta)}\\ &&+\frac{h_{3}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{2}(z)- \alpha_{1}(z+\eta)} -\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+ \eta)}\equiv1, \end{eqnarray}
(5.6)

其中, h_{j}(z) (j = 1, 2, 3) 为多项式, 且

\left\{\begin{array}{ll} h_{1}(z) = -2P(z)[Q_{1}(z)Q_{2}(z)]'-2P(z)Q_{1}(z)Q_{2}(z) [\alpha_{1}'(z)+\alpha_{2}'(z)]+4P'(z)Q_{1}(z)Q_{2}(z), \\ h_{2}(z) = 2P(z)Q_{1}(z)Q_{1}'(z)+2P(z)Q_{1}^{2}(z)\alpha_{1}'(z)-2P'( z)Q_{1}^{2}(z), \\ h_{3}(z) = 2P(z)Q_{2}(z)Q_{2}'(z)+2P(z)Q_{2}^{2}(z)\alpha_{2}'(z)-2P'( z)Q_{2}^{2}(z). \end{array}\right.

\phi = \frac{h_{1}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)}.

接下来, 分两种情形讨论:   \phi\equiv0 \phi\not\equiv0 .

情形1  若 \phi\equiv0 , 则

h_{1}(z) = -2P(z)[Q_{1}(z)Q_{2}(z)]'-2P(z)Q_{1}(z)Q_{2}(z)[\alpha_{ 1}'(z)+\alpha_{2}'(z)]+ \\ 4P'(z)Q_{1}(z)Q_{2}(z)\equiv0,

P(z)[Q_{1}(z)Q_{2}(z)]'+P(z)[\alpha_{1}'(z)+\alpha_{2}'(z)]Q_ {1}(z)Q_{2}(z)\equiv2P'(z)Q_{1}(z)Q_{2}(z).

将上述方程改写为如下形式

P(z)\left[\frac{(Q_{1}(z)Q_{2}(z))'}{Q_{1}(z)Q_{2}(z)}+\left(\alpha_{1 }'(z)+\alpha_{2}'(z)\right)\right]\equiv2P'(z).

通过简单计算, 可得 P^{2}(z)\equiv C_{1}Q_{1}(z)Q_{2}(z)e^{\alpha_{1}(z)+\alpha_{2}(z) }, 其中 C_{1} 为非零常数.从而 \alpha_{1}(z)+\alpha_{2}(z) 为常数.否则, 上述恒等式的左边是多项式而右边是超越的, 矛盾.另一方面, 由于 \alpha_{1}(z) \alpha_{2}(z) 不能同时为常数, 可知 \alpha_{1}(z), \alpha_{2}(z) 均为非常数多项式.下面, 我们断言

h_{2}(z) = 2P(z)Q_{1}(z)Q_{1}'(z)+2P(z)Q_{1}^{2}(z)\alpha_{1}'(z)-2P'( z)Q_{1}^{2}(z)\not\equiv0

h_{3}(z) = 2P(z)Q_{2}(z)Q_{2}'(z)+2P(z)Q_{2}^{2}(z)\alpha_{2}'(z)-2P'( z)Q_{2}^{2}(z)\not\equiv0.

2P(z)Q_{1}(z)Q_{1}'(z)+2P(z)Q_{1}^{2}(z)\alpha_{1}'(z)-2P' (z)Q_{1}^{2}(z)\equiv0, 那么 P(z)\equiv C_{2} Q_{1}(z)e^{\alpha_{1}(z)}, 其中 C_{2} 为非零常数.由 P(z), Q_{1}(z) 为非零多项式, 则 \alpha_{1}(z) 必为常数, 矛盾.类似地, 有 2P(z)Q_{2}(z)Q_{2}'(z)+2P(z)Q_{2}^{2}(z)\alpha_{2}'(z)-2P'( z)Q_{2}^{2}(z)\not\equiv0 .

(5.6) 式改写为如下形式

\begin{eqnarray} &&\frac{h_{2}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)}\\ &&+\frac{h_{3}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{2}(z)- \alpha_{1}(z+\eta)} -\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)}\equiv1. \end{eqnarray}
(5.7)

e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} , e^{2\alpha_{2}(z)-\alpha_{1}(z+\eta)} e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)} 中的任意两个都不为常数, 由(5.7)式及引理2.5, 可知第三个必为常数.若它们中的任意两个都为常数, 则第三个必为常数.另一方面, 由 \alpha_{1}(z) 为非常数多项式, 可知 \deg(2\alpha_{1}(z)-\alpha_{1}(z+\eta))\geq1. 因此 e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} 不为常数.下面分两种子情形讨论:子情形1.1, e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} e^{2\alpha_{2}(z)-\alpha_{1}(z+\eta)} 不为常数; 子情形1.2, e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} e^{\alpha_{ 2}(z+\eta)-\alpha_{1}(z+\eta)} 不为常数.

子情形1.1  若 e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} e^{2\alpha_{2}(z)-\alpha_ {1}(z+\eta)} 不为常数, 结合(5.7)式及引理2.5, 可得

\begin{equation} -\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)}\equiv1, \end{equation}
(5.8)

这表明 \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta) 必为常数.由(5.7)式及(5.8)式, 可得

\begin{equation} \frac{h_{2}(z)}{h_{3}(z)}e^{2\alpha_{1}(z)-2\alpha_{2}( z)}\equiv-1, \end{equation}
(5.9)

这表明 2\alpha_{1}(z)-2\alpha_{2}(z) 为常数, 即 \alpha_{2}(z)-\alpha_{1}(z) 为常数.

不妨记 e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)} = e^{\alpha_{2}(z)-\alpha_{1}(z) } = k (k\neq0) .由(5.8)式, 可知 Q_{1}(z) = -kQ_{2}(z) .将其代入(5.9)式得

4\left[P'(z)Q_{2}^{2}(z)-P(z)Q_{2}(z)Q_{2}'(z)\right]\equiv2P(z)Q_{2 }^{2}(z)\left[\alpha_{1}'(z)+\alpha_{2}'(z)\right].

比较上述恒等式两边多项式的次数, 得 \alpha_{1}' (z)+\alpha_{2}'(z)\equiv0 , 从而 \alpha_{1}(z)+\alpha_{2}(z) 为常数.注意到 \alpha_ {2}(z)-\alpha_{1}(z) 为常数, 可知 \alpha_{1}(z) \alpha_{2}(z) 同时为常数, 矛盾.

子情形1.2  若 e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)} 不为常数, 可得 2\alpha_{1}(z)-\alpha_{1}(z+\eta), \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta) 不为常数.因此 \alpha_{2}(z)-\alpha_{1}(z) 不为常数.结合(5.7)式及引理2.5, 可得

\begin{equation} \frac{h_{3}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{2}(z)- \alpha_{1}(z+\eta)}\equiv1, \end{equation}
(5.10)

这表明 2\alpha_{2}(z)-\alpha_{1}(z+\eta) 为常数.由(5.10)式及(5.7)式, 可得

\begin{equation} \frac{h_{2}(z)}{-4P^{3}(z)Q_{2}(z+\eta)}e^{2\alpha_{1}(z)- \alpha_{2}(z+\eta)}\equiv1, \end{equation}
(5.11)

这表明 2\alpha_{1}(z)-\alpha_{2}(z+\eta) 为常数.由 2(\alpha_{2}(z)-\alpha_{1}(z))+(\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)) = (2\alpha_{2}(z)-\alpha_{1}(z+\eta))-(2\alpha_{1}(z)-\alpha_{2}(z+\eta)) 为常数, 则 \alpha_{2}(z)-\alpha_{1}(z) 为常数, 矛盾.

情形2  若 \phi\not\equiv0 , 则

h_{1}(z) = -2P(z)[Q_{1}(z)Q_{2}(z)]'-2P(z)Q_{1}(z)Q_{2}(z)[\alpha_{ 1}'(z)+\alpha_{2}'(z)]+ \\ 4P'(z)Q_{1}(z)Q_{2}(z)\not\equiv0.

接下来, 我们又分三种子情形讨论:子情形2.1, \alpha_{1}(z) 为常数, \alpha_{2}(z) 不为常数; 子情形2.2, \alpha_{1}(z) 不为常数, \alpha_{2}(z) 为常数; 子情形2.3, \alpha_{1}(z) \alpha_{2}(z) 不为常数.

子情形2.1  若 \alpha_{1}(z) 为常数, \alpha_{2}(z) 不为常数, 可得 \alpha_{2}(z)-\alpha_{1}(z) \alpha_{1}(z+\eta)-\alpha_{1}(z)-\alpha_{2}(z) 不为常数.另一方面, 由 \alpha_{1}(z) 为常数, \alpha_{2}(z) 不为常数, 可知 h_{2}(z)\equiv0, h_{3}(z)\not\equiv0. 将(5.6)式改写为如下形式

\begin{eqnarray} &&\frac{-4P^{3}(z)Q_{1}(z+\eta)}{h_{1}(z)}e^{\alpha_{1}(z+\eta)-\alpha_{1}(z)-\alpha_{2}(z)}\\ &&+\frac{-4P^{3}(z)Q_{2}(z+\eta)}{h_{1}(z)}e^{\alpha_{2}(z+\eta)-\alpha_{2}(z)-\alpha_{1}(z)} -\frac{h_{3}(z)}{h_{1}(z) }e^{\alpha_{2}(z)-\alpha_{1}(z)}\equiv1. \end{eqnarray}
(5.12)

注意到 \alpha_{2}(z)-\alpha_{1}(z) \alpha_{1}(z+\eta)-\alpha_{1}(z)-\alpha_{2}(z) 不为常数, 结合(5.12)式及引理2.5, 可得

\begin{equation} \frac{-4P^{3}(z)Q_{2}(z+\eta)}{h_{1}(z)}e^{\alpha_{2}(z+\eta)-\alpha_{2}(z)-\alpha_{1}(z)} \equiv1 . \end{equation}
(5.13)

由(5.13)式, 可知 \alpha_{2}(z+\eta)-\alpha_{2}(z)-\alpha_{1}(z) 为常数.将(5.13)式代入(5.12)式得

\frac{h_{3}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{2}(z)- \alpha_{1}(z+\eta)}\equiv1,

这表明 2\alpha_{2}(z)-\alpha_{1}(z+\eta) 是常数.由 [2\alpha_{2}(z)-\alpha_{1} (z+\eta)]+[\alpha_{2}(z+\eta)-\alpha_{2}(z)-\alpha_{1}(z)] = [\alpha_{2}(z)-\alpha_ {1}(z)]+[\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)] 为常数, 可知 \alpha_{2}(z)-\alpha_{1}(z) 为常数, 但这与假设的 \alpha_{1}(z) 为常数, \alpha_{2}(z) 为非常数相矛盾.

子情形2.2  若 \alpha_{1}(z) 不为常数, \alpha_{2}(z) 为常数, 可得 2\alpha_{1}(z)-\alpha_{1}(z+\eta) \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta) 不为常数.另一方面, 由 \alpha_{1}(z) 不为常数, \alpha_{2}(z) 为常数, 可知 h_{2}(z)\not\equiv0, h_{3}(z)\equiv0. 将(5.6)式改写为如下形式

\begin{eqnarray} & &\frac{h_{1}(z)}{-4P^{3}(z)Q_{1 }(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)}\\ &&+\frac{h_{2}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} -\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)}\equiv1. \end{eqnarray}
(5.14)

注意到 2\alpha_{1}(z)-\alpha_{1}(z+\eta) \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta) 不为常数, 结合(5.14)式及引理2.5, 可得

\begin{equation} \frac{h_{1}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1} (z+\eta)} \equiv1, \end{equation}
(5.15)

这表明 \alpha_{ 1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta) 为常数.将(5.15)式代入(5.14)式得

\frac{h_{2}(z)}{ -4P^{3}(z)Q_{2}(z+\eta)}e^{2\alpha_{1}(z)-\alpha_{2}(z+\eta)}\equiv1,

这表明 2\alpha_{1}(z)-\alpha_{2}(z+\eta) 为常数, 矛盾.

子情形2.3  若 \alpha_{1}(z) \alpha_{2}(z) 不为常数, 可得 2\alpha_{1}(z)-\alpha_{1}(z+\eta) 不为常数, 且 h_{2}(z)\not\equiv0, h_{3}(z)\not\equiv0. 接下来我们分三种子情形进行讨论.

子情形2.3.1  若 e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)} e^{2\alpha_ {2}(z)-\alpha_{1}(z+\eta)} 不为常数, 结合(5.6)式及引理2.5, 可得

\begin{equation} -\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta )}\equiv1, \end{equation}
(5.16)

这表明 \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta) , \alpha_{2}(z)-\alpha_{1}(z) 为常数.将(5.16)式代入(5.6)式得

\begin{equation} h_{1}(z)e^{\alpha_{1}(z)-\alpha_{ 2}(z)}+h_{2}(z)e^{2\alpha_{1}(z)-2\alpha_{2}(z)}+h_{3}(z)\equiv0. \end{equation}
(5.17)

不妨记 e^{\alpha_{2}(z)-\alpha_{1}(z)} = k (k\neq0) .由(5.16)式, 可知 Q_{1}(z) = -kQ_{2}(z) , 将其代入(5.17)式得

8P(z)Q'_{2}(z)Q_{2}(z)-8P'(z)Q_{2}^{2}(z)\equiv-4P(z)Q_{2} ^{2}(z)[\alpha_{1}'(z)+\alpha_{2}'(z)].

比较上述恒等式两边多项式的次数, 得 \alpha_{1}'(z)+\alpha_{2}' (z)\equiv0 , 从而 \alpha_{1}(z)+\alpha_{2}(z) 为常数.注意到 \alpha_{2}(z)-\alpha_{ 1}(z) 为常数, 可知 \alpha_{1}(z) \alpha_{2}(z) 均为常数, 矛盾.

子情形2.3.2  若 e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)} e^{ \alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)} 不为常数, 结合(5.6)式及引理2.5, 可得

\begin{equation} \frac{h_{3}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{2}(z)-\alpha_ {1}(z+\eta)}\equiv1, \end{equation}
(5.18)

这表明 2\alpha_{2}(z)-\alpha_{1}(z+\eta) 为常数.由 2\alpha_{2}(z)-2\alpha_{1}(z) = [2\alpha_{2}(z)-\alpha_{1}(z+\eta)]-[2\alpha_{ 1}(z)-\alpha_{1}(z+\eta)] 不为常数, 可知 \alpha_{2}(z)-\alpha_{1}(z) 不为常数.将(5.18)式代入(5.6)式得

\begin{equation} \frac{h_{2}(z)}{-4P^{3}(z)Q_{1}(z+\eta)}e^{2\alpha_{1}(z)}+\frac{h_{1}(z)}{-4P^{3}(z)Q_{1 }(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)}-\frac{Q_{2}(z+\eta)}{Q_{1}(z+\eta)}e^{\alpha_{2}(z+\eta)}\equiv0. \end{equation}
(5.19)

现只需证明 e^{2\alpha_{1}(z)}, e^{\alpha_{1}(z)+\alpha_{2}(z)}, e^{\alpha_{2 }(z+\eta)} 线性无关.相反地, 假设存在常数 a_{j} (j = 1, 2, 3) 使得

a_{1}e^{2\alpha_{1}(z)}+a_{2}e^{\alpha_{1}(z)+\alpha_{2}(z)}+a_{3}e^{\alpha_{2}(z+\eta)} = 0.

注意到 a_{j} (j = 1, 2, 3) 不能同时为零, 可知 a_{1}, a_{2}, a_{3} 至少有两个不为零.否则 a_{j} (j = 1, 2, 3) 全为零.

a_{1} = 0, 那么 a_{2}e^{\alpha_{1}(z)+\alpha_{2}(z)}+a_{3}e^{\alpha_ {2}(z+\eta)} = 0, 也就是 a_{2}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{2}(z+\eta)} = -a_{3} , 可得 \alpha_{1}(z)+\alpha_{2}(z)-\alpha_{2}(z+\eta) 必为常数.由 [\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{2}(z+\eta)]-[2\alpha_{2}(z)-\alpha_{1} (z+\eta)] = [\alpha_{1}(z)-\alpha_{2}(z)]+[\alpha_{1}(z+\eta)-\alpha_{2}(z+\eta)] 为常数, 可知 \alpha_{1}(z)-\alpha_{2}(z) 必为常数, 矛盾.

a_{2} = 0 , 那么 a_{1}e^{2\alpha_{1}(z)}+a_{3}e^{\alpha_{2}(z+\eta) } = 0, 也就是 a_{1}e^{2\alpha_{1}(z)-\alpha_{2}(z+\eta)} = -a_{3} , 可得 2\alpha_{1} (z)-\alpha_{2}(z+\eta) 必为常数.由 [2\alpha_{1}(z)-\alpha_{2}(z+\eta)]-[2\alpha_{2}(z)-\alpha_{1}(z+\eta)] = [2(\alpha_{1}(z)-\alpha_{2}(z))]+[\alpha_{1}( z+\eta)-\alpha_{2}(z+\eta)] 为常数, 可知 \alpha_{1}(z)-\alpha_{2}(z) 必为常数, 矛盾.

a_{3} = 0 , 那么 a_{1}e^{2\alpha_{1}(z)}+a_{2}e^{\alpha_{1}(z)+\alpha_{2}(z)} = 0, 也就是 a_{1}e^{\alpha_{1}(z)-\alpha_{2}(z)} = -a_{2}. \alpha_{1}(z)-\alpha_{2}(z) 不为常数, 可知上述方程的左边是超越的而右边是非零常数, 矛盾.

因此, e^{2\alpha_{1 }(z)}, e^{\alpha_{1}(z)+\alpha_{2}(z)}, e^{\alpha_{2}(z+\eta)} 线性无关, 但这与(5.19)式相矛盾.

子情形2.3.3  若 e^{2\alpha_{2}(z)-\alpha_{1}(z+\eta)} e^{\alpha_{2}(z+\eta)-\alpha_{1}(z+\eta)} 不为常数, 可得 2\alpha_{2}(z)-\alpha_{1}(z+\eta), \alpha_{2}(z+ \eta)-\alpha_{1}(z+\eta) 不为常数, 从而 \alpha_{2}(z)-\alpha_{1}(z) 不为常数.结合(5.6)式及引理2.5, 可得

\begin{equation} \frac{h_{1}(z)}{-4P^{3}(z)Q_{ 1}(z+\eta)}e^{\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)}\equiv1, \end{equation}
(5.20)

这表明 \alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta) 为常数.将(5.20)式代入(5.6)式得

\begin{equation} h_{2}(z)e^{2\alpha_{1}(z)}+h_{3}(z)e^{2\alpha_{2}(z)}+4P^{3}(z)Q_{2}(z+\eta)e^{\alpha_{2}(z+\eta)}\equiv0. \end{equation}
(5.21)

\deg(2\alpha_{2}(z)-\alpha_{2}(z+\eta))\geq 1 , 可知 2\alpha_{2}(z)-\alpha_{2}(z+\eta) 不为常数.另一方面, 易知 2\alpha_{1}(z)-\alpha_{2} (z+\eta) 为常数.否则, 由引理2.4, 可知 h_{2}(z) = h_{3}(z) = 4P^{3}(z)Q_{2}(z+\eta)\equiv 0, 矛盾.因此 2\alpha_{1}(z)-\alpha_{2}(z+\eta)-(\alpha_{1}(z)+\alpha_{2}(z)-\alpha_{1}(z+\eta)) = [\alpha_{1}(z)-\alpha_{2}(z)]+[\alpha_{1}(z+\eta)-\alpha_{2}(z+\eta)] 为常数, 那么 \alpha_{1}(z)-\alpha_{2}(z) 必为常数, 矛盾.

这就完成了定理1.3的证明.

参考文献

Hayman W K .

Meromorphic Functions

Oxford: Clarendon Press, 1964

[本文引用: 1]

Laine I .

Nevanlinna Theory and Complex Differential Equations

Berlin: Walter de Gruyter, 1993

[本文引用: 3]

Chen Z X .

Growth and zeros of meromorphic solution of some linear difference equations

J Math Anal Appl, 2011, 373: 235- 241

DOI:10.1016/j.jmaa.2010.06.049      [本文引用: 1]

Chiang Y M , Feng S J .

On the Nevanlinna characteristic of f(z+η) and difference equations in the complex plane

Ramanujan J, 2008, 16: 105- 129

DOI:10.1007/s11139-007-9101-1      [本文引用: 1]

Gross F .

On the equation fn+gn=hn

Amer Math Monthy, 1966, 73: 1093- 1096

DOI:10.2307/2314644      [本文引用: 1]

Halburd R G , Korhonen R J .

Meromorphic solutions of difference equations, integrability and the discrete Painlevé equations

J Phys A, 2007, 40: 1- 38

DOI:10.1088/1751-8113/40/1/001      [本文引用: 1]

Han Q , F .

On the equation fn(z)+gn(z)=eαz+β

J Contem Math Anal, 2019, 54 (2): 98- 102

Li S , Gao Z S .

Finite order meromorphic solutions of linear difference equations

Proc Japan Acad Ser A, 2011, 87: 73- 76

DOI:10.3792/pjaa.87.73     

Liu K , Yang L Z .

On entire solutions of some differential-difference equations

Comput Methods Funct Theory, 2013, 13: 433- 447

DOI:10.1007/s40315-013-0030-2     

Liu K , Yang L Z , Liu X L .

Existence of entire solutions of nonlinear difference equations

Czech Math J, 2011, 61: 565- 576

DOI:10.1007/s10587-011-0075-1      [本文引用: 1]

Montel P .

Lecons sur les Familles Normales de Fonctions Analytiques et leurs Applications

Paris: Gauthier-Villars, 1927: 135- 136

[本文引用: 1]

Shimomura S .

Entire solutions of a polynomial difference equation

J Fac Sci Univ Tokyo Sect IA Math, 1981, 28: 253- 266

[本文引用: 1]

Yang C C .

A generalization of a theorem of P

Montel on entire functions. Proc Amer Math Soc, 1970, 26: 332- 334

DOI:10.1090/S0002-9939-1970-0264080-X      [本文引用: 2]

Yang C C , Laine I .

On analogies between nonlinear difference and differential equations

Proc Japan Acad Ser A, 2010, 86: 10- 14

DOI:10.3792/pjaa.86.10      [本文引用: 3]

陈敏风, 高宗升.

某类非线性微差分方程的整函数解

数学物理学报, 2016, 36A (2): 297- 306

DOI:10.3969/j.issn.1003-3998.2016.02.008      [本文引用: 1]

Chen M F , Gao Z S .

Entire solutions of a certain type of nonlinear differential-difference equation

Acta Math Sci, 2016, 36A (2): 297- 306

DOI:10.3969/j.issn.1003-3998.2016.02.008      [本文引用: 1]

Chen M F , Gao Z S , Du Y F .

Existence of entire solutions of some non-linear differential-difference equations

J Inequal Appl, 2017, 90: 2- 17

[本文引用: 2]

Hu P C , Wang Q .

On meromorphic solutions of functional equations of Fermat type

Bull Malays Math Sci Soc, 2019, 42: 2497- 2515

DOI:10.1007/s40840-018-0613-1     

Liu K , Cao T B , Cao H Z .

Entire solutions of Fermat type differential-difference equations

Arch Math, 2012, 99: 147- 155

[本文引用: 6]

曾翠萍, 邓炳茂, 方明亮.

复微分-差分方程组的整函数解

数学学报, 2019, 62 (1): 123- 136

DOI:10.3969/j.issn.0583-1431.2019.01.010      [本文引用: 1]

Zeng C P , Deng B M , Fang M L .

Entire solutions of systems of complex differential-difference equations

Acta Math Sinica, 2019, 62 (1): 123- 136

DOI:10.3969/j.issn.0583-1431.2019.01.010      [本文引用: 1]

F , Han Q .

On the Fermat-type equation f3(z)+f3(z+c)=1

Aequat Math, 2017, 91: 129- 136

[本文引用: 1]

Clunie J .

On integral and meromorphic functions

J Lond Math Soc, 1962, 37: 17- 27

[本文引用: 2]

Yang C C , Yi H X .

Uniqueness Theory of Meromorphic Functions

Dordrecht: Kluwer, 2003

[本文引用: 2]

/