设$ f(z) $是复平面$ {\Bbb 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), $$ \overline{N}(r, f), $$ S(r, f) $, 其中, 当亚纯函数$ a(z) (\not\equiv\infty) $满足

$ T(r, a) = S(r, f) = o\left(T(r, f)\right), \qquad r\rightarrow\infty, $

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) $的级, 即$ \rho(f(z)) $的定义

$ \rho(f(z)) = \limsup\limits_{r\to\infty}\frac{\log ^{+}T(r, f(z))}{\log r}. $

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

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

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

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

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) $和$ g(z) $存在某种函数关系, 例如, $ g(z) $是$ 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

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

我们假设$ n\geq m. $当$ n>m>1 $或$ 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

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

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

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

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\} $.

当$ 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 $或$ \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

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

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 $

或

$ 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) $用$ f(z)f'(z) $替换, 即

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

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

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

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 $且$ \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.

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 $.若$ U(z, f) $仅有一个最高次项, 则对任意的$ \varepsilon>0, $有

$ 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, $有

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

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, $有

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

于是, 若$ f(z) $为有穷级$ \rho $的超越亚纯函数, 则

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

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 $.若$ 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 $.

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

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

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

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

即

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

由(3.3)式, 可知$ f(z)f'(z)+{\rm i}f(z+\eta)\neq0 $和$ 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)} $

和

$ 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

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

和

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

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}, $

即

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

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

(3.7) $ \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} $

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.

根据定理A, 我们只需证明下列复微分差分方程

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

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

(4.2) $ \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} $

记$ \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

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

由$ 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*} $

即$ 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.

首先, 假设$ f(z) $为方程$ (1.9) $的有穷级超越整函数解, 则

(5.1) $ \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} $

因此, $ f(z)f'(z)+{\rm i}P(z)f(z+\eta) $和$ 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

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

和

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

由(5.2)式及(5.3)式, 可得

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

和

(5.5) $ \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} $

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

(5.6) $ \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} $

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. $

记

$ \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 $和$ \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, $

即

$ 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) $和$ \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 $

和

$ 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. $

若$ 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 $.

将$ (5.6) $式改写为如下形式

(5.7) $ \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} $

若$ e^{2\alpha_{1}(z)-\alpha_{1}(z+\eta)} $, $ e^{2\alpha_{2}(z)-\alpha_{1}(z+\eta)} $和$ 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)} $和$ 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)} $和$ 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)} $和$ 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

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

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

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

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) $和$ \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)} $和$ 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

(5.10) $ \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} $

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

(5.11) $ \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} $

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) $和$ \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) $和$ \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

(5.12) $ \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} $

注意到$ \alpha_{2}(z)-\alpha_{1}(z) $和$ \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

(5.13) $ \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} $

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) $和$ \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

(5.14) $ \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} $

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

(5.15) $ \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} $

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) $和$ \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)} $和$ 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

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

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

(5.17) $ \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} $

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) $和$ \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)} $和$ 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

(5.18) $ \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} $

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

(5.19) $ \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} $

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.

若$ 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.

若$ 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.

若$ 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}. $由$ \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)} $和$ 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

(5.20) $ \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} $

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

(5.21) $ \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} $

由$ \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.