该文主要研究方程

(1.1) $ \begin{equation} f^{n}(z)+g^{m}(z) = e^{Az+B} \end{equation} $

亚纯解, 其中$ A, B\in{\mathbb C} $为复数, $ n, m\geq 1 $为正整数.

特别地, 当$ A = B = 0 $时, 方程(1.1) 退化为著名的费马型函数方程

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

1927年, Montal^[19]首次研究了费马丢番图方程$ x^{n}+y^{m} = 1 $的函数模拟方程(1.2). $ 1965 $年, Jategaonkar^[13]给出了完整的证明, 并得到了单位盘上全纯函数的一些类似结果. 当$ n = m\geq 2 $时, Gross^[7-8]和Baker^[1]研究了方程(1.2) 的非常数亚纯解. 一般来说, 根据Yang^{[21, p332, 定理1]}可得: 当$ \frac{1}{n}+\frac{1}{m}<1 $时, 方程(1.2) 没有非平凡整函数解. 对于亚纯解, 可参考文献[7-8, 15-17]. 为了方便读者, 我们列出方程$ (1.2) $的所有可能的亚纯解, 其中$ n, m\geq 2 $为正整数.

命题1.1  在复平面$ {\mathbb C} $上, 对于$ n, m\geq 2 $, 函数方程(1.2) 有非平凡亚纯解, 当且仅当

(ⅰ) 当$ n = m = 2 $时, $ f = \frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $且$ g = \frac{2\alpha(z)}{1+\alpha^{2}(z)} $, 其中$ \alpha(z) $为非常数亚纯函数;

(ⅱ) 当$ n = m = 3 $时, $ f = \frac{1}{2}\left\{1+\frac{\wp'(u)}{\sqrt{3}}\right\}/\wp(u) $且$ g = \frac{\eta}{2}\left\{1-\frac{\wp'(u)}{\sqrt{3}}\right\}/\wp(u) $, 其中$ u $为非常数整函数, $ \eta^{3} = 1 $以及$ \wp $表示魏尔斯特拉斯$ \wp $ -函数并满足$ (\wp')^{2}\equiv 4\wp^{3}-1 $;

(ⅲ) 当$ n = 2 $且$ m = 3 $时, $ f = {\rm i}\wp'(u) $, $ g = \eta\sqrt[3]{4}\wp(u) $为生成的一对解, 其中$ u $为非常数整函数, $ \eta^{3} = 1 $以及$ \wp $满足$ (\wp')^{2}\equiv 4\wp^{3}-1 $;

(ⅳ) 当$ n = 2 $且$ m = 4 $时, $ f_{1} = \frac{1-4\wp^{2}(u)}{1+4\wp^{2}(u)} $, $ g_{1} = 2\zeta\frac{\wp(u)}{\wp'(u)} $以及$ f_{2} = {\rm i}\frac{4\wp^{2}(u)-1}{4\wp(u)} $, $ g_{2} = \zeta\frac{{\rm i}\wp'(u)}{2\wp(u)} $, 为生成的两对解, 其中$ u $为非常数整函数, $ \zeta^{4} = 1 $以及$ \wp $满足$ (\wp')^{2}\equiv 4\wp^{3}+\wp $;

(ⅴ) 当$ n = 3 $且$ m = 2 $时, 根据对称性, 由(ⅲ) 可得结论;

(ⅵ) 当$ n = 4 $且$ m = 2 $时, 根据对称性, 由(ⅳ) 可得结论.

注1.1  根据文献[12]中的方程$ (4) $及(7)–(7.3), 令$ b_{0} = 1 $, $ b_{1} = b_{2} = b_{3} = 0 $以及$ b_{4} = -1 $可得结论(ⅳ) 和(ⅵ). 但是文献[17, p217]中的结论取$ f_{1} = \wp' $是不对的. 与此同时, 陈玮等^{[4, p426]}的结论取$ f = \frac{-4\wp^{3}(u)+\frac{1}{12}\wp(u)+\frac{1}{3}}{4\wp^{3}(u)+\frac{1}{12}\wp(u)+\frac{1}{6}} $或者$ f = {\rm i}\frac{4\wp^{3}(u)-\frac{1}{12}\wp(u)-\frac{1}{3}}{4\wp^{2}(u)} $也是不对的.

本文, 假设读者熟悉以下标准记号, 比如特征函数$ T(r, f) $, 均值函数$ m(r, f) $以及计数函数$ N(r, f) $ (计重数), 以及奈望林纳理论第一和第二基本定理^{[9, 14]}. 另外, 用$ \rho(f) $表示亚纯函数$ f(z) $的增长级. 用$ S(r, f) $表示任意满足$ S(r, f) = o(T(r, f)) $ ($ r\rightarrow \infty $) 的量, 可能需除去一个对数测度有限的例外集.

为了方便读者, 先回顾一下形如

(1.3) $ \begin{equation} f(z) = P_{1}(z)e^{Q_{1}(z)}+\cdots+P_{k}(z)e^{Q_{k}(z)} \end{equation} $

的指数多项式的定义, 其中$ P_{j}'s $和$ Q_{j}'s $为多项式. 设$ q = \max\{Q_{j}(z):Q_{j}(z)\not\equiv 0\} $, $ \omega_{1}, \omega_{2}, $$ \cdots , \omega_{m} $ ($ 1\leq k\leq m $) 互不相同, 分别为最高次数为$ q $的多项式$ Q_{j}(z) $的首项系数. 根据文献[20, p156], 多项式(1.3) 可化为规范型

(1.4) $ \begin{equation} f(z) = H_{0}(z)+H_{1}(z)e^{\omega_{1}z^{q}}+\cdots+H_{m}(z)e^{\omega_{m}z^{q}}, \end{equation} $

其中$ H_{j}(z)\not\equiv 0 $ ($ j = 1, 2, \cdots , m $), 要么为次数$ \leq q-1 $的指数多项式, 要么为一般多项式.

该文研究了在复平面$ {\mathbb C} $上方程(1.1) 及(1.2) 的亚纯解的性质, 其中$ g(z): = f(z+c) $, $ c\in{\mathbb C}\setminus\{0\} $, 得到如下结论.

定理1.1  考虑如下差分方程

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

其中$ n, m\geq 1 $为正整数, $ c\in{\mathbb C}\setminus\{0\} $为复数. 则以下结论成立.

(ⅰ) 当$ n = m = 1 $时, 方程(1.5) 有非平凡亚纯解$ f(z) = \frac{1}{2}+h(z)e^{\frac{{\rm i}\pi z}{c}} $, 其中$ h(z) $为亚纯函数并满足$ h(z+c) = h(z) $; 如果$ f(z) $为方程(1.5) 的一个指数多项式解, 则$ f(z) = \frac{1}{2}+\sum_{i = 1}^{m}\limits b_{i}e^{\omega_{i}z} $, 其中$ e^{\omega_{i}c}+1 = 0 $, $ b_{i}\in{\mathbb C} $$ (i = 1, 2, \cdots , m) $不全为零;

(ⅱ) 当$ n = m = 2 $时, 方程(1.5) 有非平凡亚纯解$ f(z) = \frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = \frac{2\alpha(z)}{1+\alpha^{2}(z)} $或者$ f(z) = \frac{2\alpha(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = \frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $, 其中$ \alpha(z) $为非常数亚纯函数;

(ⅲ) 当$ n = m\geq3 $或者$ n\neq m $时, 方程(1.5) 没有有限级非平凡亚纯解.

注1.2  在定理$ 1.1 $中, 如果$ \rho(f) = \infty $, 则结论(ⅲ) 不成立, 见如下例题.

例1.1^{[18, 例6]}  设$ f(z) = \frac{1}{2}\left\{1+\frac{\wp'(e^{z})}{\sqrt{3}}\right\}/\wp(e^{z}) $, 则$ \rho(f) = \infty $. 令$ c = {\rm i}\pi $, 则$ f(z+c) = \frac{1}{2}\left\{1-\frac{\wp'(e^{z})}{\sqrt{3}}\right\}/\wp(e^{z}) $, $ f^{3}(z)+f^{3}(z+c) = 1 $.

定理1.2  考虑如下差分方程

(1.6) $ \begin{equation} f^{n}(z)+f^{m}(z+c) = e^{Az+B}, \end{equation} $

其中$ n, m\geq 1 $为正整数, $ A, c\in{\mathbb C}\setminus\{0\} $, $ B\in{\mathbb C} $为复数. 则以下结论成立.

(ⅰ) 当$ n = m = 1 $时, 方程(1.6) 有非平凡亚纯解$ f(z) = e^{Az+B}(\frac{1}{2}+h(z)e^{\frac{{\rm i}\pi z}{c}}) $, 其中$ h(z) $为亚纯函数并满足$ h(z+c) = h(z) $, $ e^{Ac} = 1 $; 如果$ f(z) $为方程(1.6) 的一个指数多项式解, 则$ f(z) = e^{Az+B}\Big(\frac{1}{2}+\sum_{i = 1}^{m}\limits b_{i}e^{\omega_{i}z}\Big) $, 其中$ e^{\omega_{i}c}+1 = 0 $, $ e^{Ac} = 1 $, $ b_{i}\in{\mathbb C} $$ (i = 1, 2, \cdots , m) $不全为零;

(ⅱ) 当$ n = m = 2 $时, 方程(1.6) 有非平凡亚纯解$ f(z) = e^{\frac{Az+B}{2}}\frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = e^{\frac{Az+B}{2}}\frac{2\alpha(z)}{1+\alpha^{2}(z)} $或者$ f(z) = e^{\frac{Az+B}{2}}\frac{2\alpha(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = e^{\frac{Az+B}{2}}\frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $, 其中$ \alpha(z) $为非常数亚纯函数;

(ⅲ) 当$ n = m\geq3 $或者$ (n, m) = (2, 3), (2, 4), (3, 2) $, $ (4, 2) $时, 方程(1.6) 没有有限级非平凡亚纯解.

以下引理在定理$ 1.1 $及$ 1.2 $的证明中起着至关重要的作用.

引理2.1^{[22, 定理1.51]}  设$ f_{1}(z), f_{2}(z), \cdots , f_{n}(z)(n\geq 2) $为亚纯函数, $ g_{1}(z), g_{2}(z), \cdots , g_{n}(z) $为整函数并满足以下条件

$ (1) $$ \sum_{j = 1}^{n}\limits f_{j}(z)e^{g_{j}(z)}\equiv 0; $

$ (2) $当$ 1\leq j<k\leq n $时, $ g_{j}(z)-g_{k}(z) $为非常数;

$ (3) $当$ 1\leq j\leq n, 1\leq h<k\leq n $时, 有

$ T(r, f_{j}(z)) = o(T(r, e^{g_{h}(z)-g_{k}(z)})) (r\rightarrow \infty, r\not\in E). $

则$ f_{j}(z)\equiv 0\ (j = 1, \cdots , n) $.

下面为对数导数引理的差分模拟, 参见文献[10, 引理2.2]或[5, 引理2.6].

引理2.2  设$ f(z) $为增长级$ \rho(f) = \rho<+\infty $的亚纯函数, $ \eta $为固定的复常数, 则对任意的$ \varepsilon>0 $, 有

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

引理2.3^{[5, 定理2.1]}  设$ f(z) $为增长级$ \rho(f) = \rho<+\infty $的亚纯函数, $ \eta $为非零固定的复常数, 则对任意的$ \varepsilon>0 $, 有

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

下面这个大家熟知的结论, 关于复合亚纯函数增长级是由文献[6, 引理1.2]以及文献[3, 定理1]得到.

引理2.4  设$ f(z) $为非零级亚纯函数, $ g(z) $为整函数且不为多项式. 则$ \rho(f\circ g) = \infty $. 如果$ \rho(f\circ g)<\infty $且$ g(z) $超越, 则$ \rho(f) = 0 $.

证  (ⅰ) 当$ n = m = 1 $时, 方程(1.5) 退化为

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

记$ g(\frac{z}{c}) = f(z) $, $ g(\frac{z+c}{c}) = f(z+c) $, 则方程$ (3.1) $可以写为

(3.2) $ \begin{equation} g\left(\frac{z}{c}\right)+g\left(\frac{z}{c}+1\right) = 1. \end{equation} $

方程(3.2) 的通解为$ g\left(\frac{z}{c}\right) = \frac{1}{2}+\gamma\left(\frac{z}{c}\right)e^{\frac{{\rm i}\pi z}{c}} $, 其中$ \gamma(z) $为复平面$ {\mathbb C} $上的亚纯函数并满足$ \gamma(\frac{z}{c}) = \gamma(\frac{z}{c}+1) $. 则方程(3.1) 的通解为

$ \begin{eqnarray*} f(z) = g\left(\frac{z}{c}\right) = \frac{1}{2}+\gamma\left(\frac{z}{c}\right)e^{\frac{{\rm i}\pi z}{c}} = \frac{1}{2}+h(z)e^{\frac{{\rm i}\pi z}{c}}, \end{eqnarray*} $

其中$ h(z) $为亚纯函数并满足$ h(z) = h(z+c) $.

如果$ f(z) $为方程(3.1) 的指数多项式解, 把式(1.4) 代入方程(3.1) 得

(3.3) $ \begin{eqnarray} H_{0}(z)+H_{0}(z+c)+[H_{1}(z)+H_{1}(z+c)e^{\omega_{1}(z+c)^{q}-\omega_{1}z^{q}}]e^{\omega_{1}z^{q}}&&{}\\ +\cdots+[H_{m}(z)+H_{m}(z+c)e^{\omega_{m}(z+c)^{q}-\omega_{m}z^{q}}]e^{\omega_{m}z^{q}}& = &1. \end{eqnarray} $

由引理2.1及方程(3.3) 得

(3.4) $ \begin{equation} \left\{\begin{array}{ll} H_{0}(z)+H_{0}(z+c) = 1;\\ H_{1}(z)+H_{1}(z+c)e^{\omega_{1}(z+c)^{q}-\omega_{1}z^{q}}\equiv 0;\\ \cdots\cdots\cdots\cdots\\ H_{m}(z)+H_{m}(z+c)e^{\omega_{m}(z+c)^{q}-\omega_{m}z^{q}}\equiv 0. \end{array}\right. \end{equation} $

如果$ q\geq 2 $, 由方程组(3.4) 得

$ \begin{eqnarray*} H_{i}(z)+H_{i}(z+c)e^{\omega_{i}(z+c)^{q}-\omega_{i}z^{q}}\equiv 0, \quad i = 1, 2, \cdots , m, \end{eqnarray*} $

即

$ e^{\omega_{m}(z+c)^{q}-\omega_{m}z^{q}}\equiv -\frac{H_{i}(z)}{H_{i}(z+c)}. $

由引理2.2, 对任意$ \varepsilon>0 $, 有

$ m(r, e^{\omega_{m}(z+c)^{q}-\omega_{m}z^{q}}) = m\left(r, -\frac{H_{i}(z)}{H_{i}(z+c)}\right) = O(r^{q-2+\varepsilon}) = o(r^{q-1}), $

矛盾. 则$ q = 1 $,

$ f(z) = H_{0}(z)+H_{1}(z)e^{\omega_{1}z}+\cdots+H_{m}(z)e^{\omega_{m}z}, $

其中$ H_{0}(z), H_{1}(z), \cdots , H_{m}(z) $为多项式.

设

$ H_{0}(z) = a_{k}z^{k}+a_{k-1}z^{k-1}+\cdots+a_{0}, $

$ H_{0}(z+c) = a_{k}(z+c)^{k}+a_{k-1}(z+c)^{k-1}+\cdots+a_{0}. $

把上述两个等式代入$ H_{0}(z)+H_{0}(z+c) = 1 $中, 然后比较$ z^{k}, z^{k-1}, \cdots , z $的系数得

$ \begin{eqnarray*} \left\{\begin{array}{ll} 2a_{k} = 0;\\ a_{k-1}+k a_{k}c = 0;\\ a_{k-2}+(k-1)a_{k-1}c+ {2\choose k} a_{k}c^{2} = 0;\\ \cdots\cdots\cdots\cdots\\ 2a_{0}+a_{1}c+\cdots+a_{k-1}c^{k-1}+a_{k}c^{k} = 1, \end{array}\right. \end{eqnarray*} $

其中$ {j\choose k} = \frac{k!}{j!(k-j)!} $, 则$ a_{k} = a_{k-1} = \cdots = a_{1} = 0 $, $ a_{0} = \frac{1}{2} $. 故$ H_{0}(z) = \frac{1}{2} $. 记

$ \begin{eqnarray*} H_{i}(z) = a_{l_{i}i}z^{l_{i}}+a_{l_{i}-1i}z^{l_{i}-1}+\cdots+a_{1i}z+b_{i}, \end{eqnarray*} $

$ \begin{eqnarray*} H_{i}(z+c) = a_{l_{i}i}(z+c)^{l_{i}}+a_{l_{i}-1i}(z+c)^{l_{i}-1}+\cdots+a_{1i}(z+c)+b_{i}, i = 1, 2, \cdots , m. \end{eqnarray*} $

把上述两个等式代入$ H_{i}(z)+H_{i}(z+c)e^{\omega_{i}c}\equiv 0 $中, 然后比较$ z^{l_{i}}, z^{l_{i}-1}, \cdots , z $的系数得

$ \begin{eqnarray*} \left\{\begin{array}{ll} (1+e^{\omega_{i}c})a_{l_{i}i} = 0;\\ e^{\omega_{i}c}a_{l_{i}i} {{1}\choose {l_{i}}} c+(1+e^{\omega_{i}c})a_{l_{i}-1i} = 0;\\ \cdots\cdots\cdots\cdots\\ e^{\omega_{i}c}\left(a_{l_{i}i} {{j}\choose {l_{i}}} c^{j}+a_{l_{i}-1i} {{j-1}\choose {l_{i}-1}} c^{j-1}+\cdots+ a_{l_{i}-j+1i} {{1}\choose {l_{i}-j+1}} c\right)+(1+e^{\omega_{i}c})a_{l_{i}-ji} = 0;\\ \cdots\cdots\cdots\cdots\\ e^{\omega_{i}c}(a_{l_{i}i}+a_{l_{i}-1i}c^{l_{i}-1}+\cdots+a_{1i}c)+(1+e^{\omega_{i}c})b_{i} = 0, i = 1, 2, \cdots , m. \end{array}\right. \end{eqnarray*} $

如果$ 1+e^{\omega_{i}c}\neq 0 $, 则$ a_{l_{i}i} = a_{l_{i}-1i} = \cdots = a_{1i} = b_{i} = 0 $, 从而$ H_{i}(z)\equiv 0 $, 矛盾. 如果$ 1+e^{\omega_{i}c} = 0 $, 则$ a_{l_{i}i} = a_{l_{i}-1i} = \cdots = a_{1i} = 0 $, 故$ f(z) = \frac{1}{2}+\sum\limits_{i = 1}^{m}\limits b_{i}e^{\omega_{i}z} $, 其中$ e^{\omega_{i}c}+1 = 0 $, $ b_{i}\in{\mathbb C} $ ($ i = 1, 2, \cdots , m $) 不全为零.

(ⅱ) 当$ n = m = 2 $时, 由文献[8, p86, 定理1] 得$ f(z) = \frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = \frac{2\alpha(z)}{1+\alpha^{2}(z)} $, 或者$ f(z) = \frac{2\alpha(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = \frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $, 其中$ \alpha(z) $为非常数亚纯函数.

(ⅲ) 当$ n = m = 3 $时, 由文献[18, p130, 定理2]可知, 方程(1.5) 没有任何非常数有限级亚纯解. 当$ n = m\geq 4 $时, 由文献[8, p87, 定理3]可知, 方程(1.5) 也没有任何非常数有限级亚纯解. 当$ n\neq m $时, 由引理2.3及方程(1.5) 得

$ \begin{eqnarray*} T(r, 1-f^{n}(z))& = & nT(r, f(z))+O(1) = T(r, f^{m}(z+c))\\ & = & mT(r, f(z+c)) = mT(r, f(z))+S(r, f), \end{eqnarray*} $

从而$ T(r, f) = S(r, f) $, 矛盾. 因此, 在这种情况下, 方程(1.5) 没有任何非常数有限级亚纯解. 定理$ 1.1 $证毕.

证  (ⅰ) 当$ n = m = 1 $时, 方程(1.6) 退化为

(4.1) $ \begin{equation} f(z)+f(z+c) = e^{Az+B}. \end{equation} $

把方程$ (4.1) $改写为

(4.2) $ \begin{equation} e^{-(Az+B)}f(z)+e^{-(Az+B)}f(z+c) = 1. \end{equation} $

运用定理$ 1.1 $(ⅰ) 中同样的证明方法可得方程(4.2) 的通解为$ f(z) = e^{Az+B}(\frac{1}{2}+h(z)e^{\frac{{\rm i}\pi z}{c}}) $, 其中$ h(z) $为亚纯函数并满足$ h(z+c) = h(z) $, $ e^{Ac} = 1 $. 如果$ F(z) $为方程(4.2) 的指数多项式解, 把式(1.4) 代入方程(4.2), 并运用定理$ 1.1 $(i) 中同样的证明方法可得$ f(z) = e^{Az+B}\Big(\frac{1}{2}+\sum_{i = 1}^{m}\limits b_{i}e^{\omega_{i}z}\Big) $, 其中$ e^{\omega_{i}c}+1 = 0 $, $ e^{Ac} = 1 $, $ b_{i}\in{\mathbb C} $ ($ i = 1, 2, \cdots , m $) 不全为零.

(ⅱ) 当$ n = m = 2 $时, 把方程(1.6) 改写为

(4.3) $ \begin{equation} (e^{-\frac{Az+B}{2}}f(z))^{2}+(e^{-\frac{Az+B}{2}}f(z+c))^{2} = 1, \end{equation} $

由文献[8, p86, 定理1]得$ f(z) = e^{\frac{Az+B}{2}}\frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = e^{\frac{Az+B}{2}}\frac{2\alpha(z)}{1+\alpha^{2}(z)} $, 或者$ f(z) = e^{\frac{Az+B}{2}}\frac{2\alpha(z)}{1+\alpha^{2}(z)} $且$ f(z+c) = e^{\frac{Az+B}{2}}\frac{1-\alpha^{2}(z)}{1+\alpha^{2}(z)} $, 其中$ e^{Ac} = 1 $, $ \alpha(z) $为非常数亚纯函数.

(ⅲ) 当$ n = m = 3 $时, 由文献[11, p100, 定理2.2]可知, 方程(1.6) 没有任何非常数有限级亚纯解. 当$ n = m\geq 4 $时, 方程(1.6) 可写为

(4.4) $ \begin{equation} (e^{-\frac{Az+B}{4}}f(z))^{4}+(e^{-\frac{Az+B}{4}}f(z+c))^{4} = 1. \end{equation} $

由文献[8, p87, 定理3]可知, 在复平面$ {\mathbb C} $上, 方程(4.4) 没有任何非常数有限级亚纯解.

当$ n = 2, m = 3 $时, 方程(1.6) 可写为

(4.5) $ \begin{equation} (e^{-\frac{Az+B}{2}}f(z))^{2}+(e^{-\frac{Az+B}{3}}f(z+c))^{3} = 1. \end{equation} $

由命题$ 1.1 $(ⅲ) 可得

(4.6) $ \begin{equation} f(z) = {\rm i}e^{\frac{Az+B}{2}}\wp'(u(z))\quad \rm{且} \quad f(z+c) = e^{\frac{Az+B}{3}}\eta\sqrt[3]{4}\wp(u(z)). \end{equation} $

又因为$ (\wp')^{2}\equiv4\wp^{3}-1 $, 经计算可得

(4.7) $ \begin{equation} f^{2}(z) = -e^{Az+B}(\wp'(u(z)))^{2} = e^{Az+B}(1-4\wp^{3}(u(z))). \end{equation} $

因为$ \rho(f)<\infty $, 由式(4.7) 可得

$ \begin{eqnarray*} 3T(r, \wp(u(z))) = T(r, \wp^{3}(u(z)))\leq T(r, e^{-(Az+B)})+T(r, f^{2}(z))+O(1), \end{eqnarray*} $

则$ \rho(\wp(u))<\infty $. 利用文献[2, p7]的估计式$ (2.7) $可知

(4.8) $ \begin{equation} T(r, \wp) = \frac{\pi}{K}r^{2}(1+O(1))\quad \rm{且}\quad \rho(\wp) = 2, \end{equation} $

其中$ K $为顶点在$ 0, \omega_{1}, \omega_{2}, \omega_{1}+\omega_{2} $上的平行四边形的面积. 由引理2.4及式(4.8) 可知$ u(z) $为多项式. 由式(4.6) 及式(4.7) 可得

(4.9) $ \begin{equation} \eta^{2}4^{\frac{2}{3}}\wp^{2}(u(z)) = e^{\frac{Az+B}{3}+Ac}(1-4\wp^{3}(u(z+c))). \end{equation} $

此时, 对某一正整数$ N\geq 1 $, $ T(r, \wp(u(z))) = O(r^{2N}) $. 由引理2.3及式(4.9) 可得

$ \begin{eqnarray*} 2T(r, \wp(u(z)))+O(1)& = & T(r, \eta^{2}4^{\frac{2}{3}}\wp^{2}(u(z)))\\ & = & T(r, e^{\frac{Az+B}{3}+Ac}(1-4\wp^{3}(u(z+c))))\\ & = & 3T(r, \wp(u(z+c)))+S(r, \wp(u(z+c)))\\ & = & 3T(r, \wp(u(z)))+S(r, \wp(u(z))), \end{eqnarray*} $

从而$ T(r, \wp(u(z))) = S(r, \wp(u(z))) $, 矛盾. 当$ n = 3, m = 2 $时, 类似于$ n = 2, m = 3 $的情形.

当$ n = 2, m = 4 $时, 方程(1.6) 可写为

(4.10) $ \begin{equation} (e^{-\frac{Az+B}{2}}f(z))^{2}+(e^{-\frac{Az+B}{4}}f(z+c))^{4} = 1. \end{equation} $

根据命题$ 1.1 $(ⅳ) 得

(4.11) $ \begin{equation} f(z) = e^{\frac{Az+B}{2}}\frac{1-4\wp^{2}(u(z))}{1+4\wp^{2}(u(z))}\quad \rm{且} \quad f(z+c) = e^{\frac{Az+B}{4}}2\zeta\frac{\wp(u(z))}{\wp'(u(z))}. \end{equation} $

由式(4.11) 可得

$ \begin{eqnarray*} 2T(r, \wp(u(z)))& = & T\left(r, \frac{1-4\wp^{2}(u(z))}{1+4\wp^{2}(u(z))}\right)\\ & = & T(r, e^{-\frac{Az+B}{2}}f(z))\\ &\leq & T(r, e^{-\frac{Az+B}{2}})+T(r, f(z)). \end{eqnarray*} $

又因为$ \rho(f)<\infty $, 则$ \rho(u(z))<\infty $. 由引理2.4及式(4.8) 可知$ u(z) $为多项式. 根据式(4.11) 及$ (\wp')^{2}\equiv 4\wp^{3}+\wp $可得

(4.12) $ \begin{equation} 4\zeta^{2}\frac{\wp(u(z))}{1+4\wp^{2}(u(z))} = e^{\frac{Az+B}{2}+Ac}\left(\frac{1-4\wp^{2}(u(z+c))}{1+4\wp^{2}(u(z+c))}\right)^{2}. \end{equation} $

此时, 对某一正整数$ N\geq 1 $, $ T(r, \wp(u(z))) = O(r^{2N}) $, 则$ T(r, e^{\frac{Az+B}{3}+Ac}) = S(r, \wp(u(z))) $. 由引理2.3及式(4.12) 可得

$ \begin{eqnarray*} 2T(r, \wp(u(z)))+O(1)& = & T\left(r, 4\zeta^{2}\frac{\wp(u(z))}{1+4\wp^{2}(u(z))}\right)\\ & = & T\left(r, e^{\frac{Az+B}{2}+Ac}\left(\frac{1-4\wp^{2}(u(z+c))}{1+4\wp^{2}(u(z+c))}\right)^{2}\right)\\ & = & 4T(r, \wp(u(z+c)))+S(r, \wp(u(z+c)))\\ & = & 4T(r, \wp(u(z)))+S(r, \wp(u(z))), \end{eqnarray*} $

这表明$ T(r, \wp(u(z))) = S(r, \wp(u(z))) $, 矛盾. 如果$ f(z) = {\rm i}e^{\frac{Az+B}{2}}\frac{4\wp^{2}(u(z))-1}{4\wp(u(z))} $且$ f(z+c) = e^{\frac{Az+B}{4}}\zeta\frac{{\rm i}\wp'(u(z))}{2\wp(u(z))} $, 同理可得矛盾. 当$ n = 4, m = 2 $时, 类似于$ n = 2, m = 4 $的情形. 定理$ 1.2 $证毕.