上海理工大学学报  2019, Vol. 41 Issue (5): 429-432   PDF    
代数体函数及其线性微分多项式的唯一性
黄尧, 刘晓俊     
上海理工大学 理学院,上海 200093
摘要: 利用现有的亚纯函数和代数体函数的相关结论,研究代数体函数的分担值与分支点之间的关系,得到一个代数体函数与其线性微分多项式具有公共值的唯一性定理,将关于亚纯函数与其线性微分多项式的一个唯一性定理推广到代数体函数。
关键词: 代数体函数     线性微分多项式     分担值     唯一性    
Uniqueness for Algebroid Function and its Linear Differential Polynomial
HUANG Yao, LIU Xiaojun     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: By using the existing meromorphic functions and the related conclusions about algebroid functions, the relationship between the shared value of algebroid functions and the branch points was studied. The uniqueness theorem as regards an algebroid function and its linear differential polynomial with common values was presented. So, the uniqueness theorem for meromorphic functions and its linear differential polynomial was extended to algebroid functions.
Key words: algebroid function     linear differential polynomial     shared value     uniqueness    
1 问题的提出

代数体函数是由不可约方程

$\psi \left( {z,W} \right) \equiv {A_v}\left( z \right){W^v} \!+\! {A_{v \!-\! 1}}\left( z \right){W^{v \!-\! 1}} \!+\! \cdots \!+\! {A_0}\left( z \right) \!=\! 0$ (1)

确定的$v$值解析函数,其中,${A_j}\left( z \right)( j = 0,$$1, \cdot \cdot \cdot ,v )$是关于$z$的全纯函数,并且没有公共零点。特别地,如果${A_j}\left( z \right)\left( {j = 0,1, \cdot \cdot \cdot ,v} \right)$都是多项式时,则$W\left( z \right)$$v$值代数函数。本文假设${A_j}\left( z \right)\left( {j = 0,1, \cdot \cdot \cdot ,v} \right)$中至少有一个是超越整函数。当$v = 1$时,$W\left( z \right)$为亚纯函数;当$v$≥2时,$W\left( z \right)$为多值函数。显然,$v$值代数体函数是亚纯函数的延伸。

本文中代数体函数$W(z)$的一阶导函数记为$W'(z)$$k$阶导函数记为${W^{(k)}}(z)$及其线性微分多项式函数记为$L\left( W \right)$$L\left( W \right) \!=\! {c_k}{W^{\left( k \right)}} \!+\! {c_{k - 1}}{W^{\left( {k \!-\! 1} \right)}} \!+\! \cdots+$$ {c_0}W $,其中,${c_0},{c_1}, \cdot \cdot \cdot ,{c_k}$为不全为零的有穷复数,且${c_k} \ne 0$。它们都是$v$值代数体函数,具体证明参见文献[1]。在本文中若不加其他说明,$W\left( z \right)$是指在整个复平面$C$上由不可约方程(1)所确定的$v$值代数体函数。本文所使用的记号和相关结论参见文献[2-5]。

定义1[6] 设$W\left( z \right)$$v$值代数体函数,${z_0} \in C$。定义:

${W^{ \!-\! 1}}\left( a \right) \!=\! \left\{ {\left( {{w_{{\lambda _j}}}\left( z \right),{z_0}} \right)\!:\!\left( {{w_{{\lambda _j}}}\left( z \right),B\left( {{z_0}} \right)} \right) \!\in\! F\!\left( W \right),{w_{{\lambda _j}}}\left( {{z_0}} \right) \!=\! a} \right\}$
${W^{ \!-\! 1}}\left( \infty \right) \!=\! \left\{ {\left( {{w_{{\lambda _j}}}\left( z \right),{z_0}} \right)\!:\!\left( {{w_{{\lambda _j}}}\!\left( z \right),B\left( {{z_0}} \right)} \right) \!\!\in\!\! F\!\left( W \right),{w_{{\lambda _j}}}\!\left( {{z_0}} \right) \!=\!\! \infty } \right\}$

其中

$F\left( W \right) = \left\{ {\left( {{w_{{\lambda _j}}}\left( z \right),B\left( {{z_0}} \right)} \right):j = 1,2, \cdot \cdot \cdot ,l,\sum\limits_{j = 1}^l {{\lambda _j} = v,{z_0} \in C} } \right\}$

${W^{ - 1}}\left( a \right)$被称为$W\left( z \right)$$a$值点,${W^{ - 1}}\left( \infty \right)$被称为$W\left( z \right)$的极点。

定义2[6] 设$W\left( z \right)$$v$值代数体函数,${z_0} \in C$。如果${W^{ - 1}}\left( {{z_0}} \right) = {\left( {L\left( W \right)} \right)^{ - 1}}\left( {{z_0}} \right)$,且对于每一$\left( {{w_{{\lambda _j}}},{z_0}} \right) \in $${W^{ - 1}}\left( a \right)$,有$\tau \left( {{z_0},{w_{{\lambda _j}}}\left( z \right) = a} \right) = \tau \left( {{z_0},L\left( {{w_{{\lambda _j}}}} \right) = a} \right)$成立,其中,$\tau \left( {{z_0},{w_{{\lambda _j}}}\left( z \right) = a} \right)$表示在$z = {z_0}$${w_{{\lambda _j}}}\left( z \right) - a$的零点重级,则称$W\left( z \right)$$L\left( W \right)$ ${\rm{CM}}$分担$a$

关于亚纯函数及其线性微分多项式的唯一性问题,文献[7]证明了定理1。

定理1 设$f$为非常数亚纯函数,$k$为一正整数,$L\left( f \right)$为关于$f$的线性微分多项式,$L\left( f \right) = {c_k}{f^{\left( k \right)}} + $${c_{k - 1}}{f^{\left( {k - 1} \right)}} + \cdot \cdot \cdot + {c_0}f$,其中,${c_0},{c_1}, \cdot \cdot \cdot ,{c_k}$$f$的不恒为$\infty $的小函数,且${c_k}\not \equiv 0$,再设${a_1},{a_2},{a_3}$$f$的不恒为$\infty $的小函数,如果$f$$L\left( f \right)$几乎${\rm{CM}}$分担${a_1},{a_2}$,几乎${\rm{IM}}$分担${a_3}$,则$f \equiv L\left( f \right)$

关于代数体函数的唯一性问题,文献[8-9]分别得到了定理2和定理3。

定理2 设$W\left( z \right)$$v$值代数体函数,设${a_1}, $${a_2}, \cdot \cdot \cdot ,{a_{2v}}$$2v$个判别的有穷复数,如果$W\left( z \right)$$W'\left( z \right)$ ${\rm{CM}}$分担${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v}}$${\rm{IM}}$分担$\infty $,则$W\left( z \right) \equiv $$W'\left( z \right)$

定理3 设$W\left( z \right)$$v$($v$≥2)值代数体函数,设${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v - 1}}$$2v - 1$个判别的有穷非零复数,如果$W\left( z \right)$$W'\left( z \right)$ ${\rm{CM}}$分担$0,{a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v - 1}}$,如果存在有穷实数$c,R > 0$,当$\left| z \right| = r > R$时,有$\left| {{W_j}\left( z \right)}\right|{ > c} $$( j = 1,2, \cdots, v)$$\left| {{{W'}_j}\left( z \right)} \right| > c\left( {j = 1,2, \cdot \cdot \cdot ,v} \right)$,则$W\left( z \right) \equiv $$W'\left( z \right) $

现结合定理2和定理3,将定理1推广到代数体函数,得到定理4。

定理4 设$W\left( z \right)$$v$值代数体函数,$k$为一正整数,$L\left( W \right)$$W$的线性微分多项式,$ L\left( W \right) = {c_k}{W^{\left( k \right)}} + $${c_{k - 1}}{W^{\left( {k - 1} \right)}} + \cdot \cdot \cdot + {c_0}W$,其中,${c_0},{c_1}, \cdot \cdot \cdot, {c_k}$为不全为零的有穷复数,且${c_k} \ne 0$,再设${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$$2v + 1$个互相判别的有穷复数,如果$W\left( z \right)$$L\left( W \right)$ ${\rm{CM}}$分担${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$,则有:

a.$v$≥3时,且${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$是有穷非零复数,则$W\left( z \right) \equiv L\left( W \right)$

b.$v$≥4时,且${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$是包括0的有穷复数,则$W\left( z \right) \equiv L\left( W \right)$

2 引 理

引理1 设$W\left( z \right)$$v$值代数体函数,设${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$$({2v + 1})$个判别的有穷非零复数,如果$W\left( z \right)$$L\left( W \right)$ ${\rm{CM}}$分担${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$,则

$ \sum\limits_{i = 1}^{2v + 1} {N\left( {r,\frac{1}{{W - {a_i}}}} \right)} = \sum\limits_{i = 1}^{2v + 1} {N\left( {r,\frac{1}{{L - {a_i}}}} \right)}≤ N\left( {r,\frac{1}{{L - W}}} \right) $

证明 对任意${a_i}\left( {i \!=\! 1,2, \cdot \cdot \cdot ,2v \!+\! 1} \right)$,如果$ \left( {{w_{^{{\lambda _j}}}}\!\left( z \right),{z_0}} \right) \in$$ {W^{ - 1}}\left( {{a_i}} \right)$,则有

$ {w_{{\lambda _j}}}\left( z \right) = {a_i} + \sum\limits_{p = u}^\infty {{b_p}{{\left( {z - {z_0}} \right)}^{\frac{p}{{{\lambda _j}}}}}},\;{b_u} \ne 0 $

其中,${b_p}\left( {p = u,u + 1,u + 2, \cdot \cdot \cdot } \right)$是有穷复数,求$k$阶导可得

$ {w_{{\lambda _j}}}\!\!^{\left( k \right)}\left( z \right) \!=\! \sum\limits_{p = u}^\infty {\!{b_p}\frac{p}{{{\lambda _j}}}\frac{{p - {\lambda _j}}}{{{\lambda _j}}} \!\cdots \!\frac{{p - \left( {k - 1} \right){\lambda _j}}}{{{\lambda _j}}}{{\left( {z - {z_0}} \right)\!\!}^{\frac{{p \!-\! k{\lambda _j}}}{{{\lambda _j}}}}}}\!\!,\;{b_u} \!\ne\! 0 $
$L\left( {{w_{{\lambda _j}}}} \right) = {c_k}{w_{{\lambda _j}}}^{\left( k \right)}\left( z \right) + {c_{k - 1}}{w_{{\lambda _j}}}^{\left( {k - 1} \right)}\left( z \right) + \cdot \cdot \cdot + {c_0}{w_{{\lambda _j}}}\left( z \right)$

$W\left( z \right)$$L\left( W \right)$ ${\rm{CM}}$分担${a_i}$可得

$ L\left( {{w_{{\lambda _j}}}} \right) = {a_i} + \sum\limits_{p = u}^\infty {{d_p}{{\left( {z - {z_0}} \right)}^{\frac{p}{{{\lambda _j}}}}}},\;{d_u} \ne 0 $

其中,${d_p}\left( {p = u,u + 1,u + 2, \cdot \cdot \cdot } \right)$是有穷复数。

由以上分析可得

$\begin{split} & \!\!\!\!\!\!\!\!\!\!\tau \left( {{z_0},L\left( {{w_{{\lambda _j}}}} \right) - {w_{{\lambda _j}}}\left( z \right) = 0} \right)\geqslant u =\qquad\qquad \\ &\qquad \!\!\!\!\!\!\!\! \tau \left( {{z_0},{w_{{\lambda _j}}}\left( z \right) = {a_i}} \right) =\tau \left( {{z_0},L\left( {{w_{{\lambda _j}}}} \right) = {a_i}} \right) \end{split}$

$\begin{split}&\quad\sum\limits_{i = 1}^{2v + 1} {n\left( {r,\frac{1}{{W\left( z \right) - {a_i}}}} \right)} = \sum\limits_{i = 1}^{2v + 1} {\sum\limits_{( {{w_{{\lambda _j}\left( z \right)}},{z_0}} ) \in {w^{ - 1}}\left( {{a_i}} \right),\left| {{z_0}} \right| < r} {\tau \left( {{z_0},{w_{{\lambda _j}}}\left( z \right) = {a_i}} \right)} }= \sum\limits_{( {{w_{{\lambda _j}\left( z \right)}},{z_0}} ) \in {w^{ - 1}}\left( {{a_i}} \right),\left| {{z_0}} \right| < r} {\sum\limits_{i = 1}^{2v + 1} {\tau \left( {{z_0},{w_{{\lambda _j}}}\left( z \right) = {a_i}} \right)} }=\qquad\qquad\qquad\qquad\\ & \qquad\qquad\qquad\qquad\qquad \sum\limits_{i = 1}^{2v + 1} {n\left( {r,\frac{1}{{L\left( W \right) - {a_i}}}} \right)} = \sum\limits_{i = 1}^{2v + 1} {\sum\limits_{( {L( {{w_{{\lambda _j}}}}),{z_0}}) \in {{( {L( {{w_{{\lambda _j}}}})})}^{ - 1}}\left( {{a_i}} \right),\left| {{z_0}} \right| < r} {\tau \left( {{z_0},L\left( {{w_{{\lambda _j}}}} \right) = {a_i}} \right)} }=\\ & \qquad\qquad\qquad\qquad\qquad \sum\limits_{( {L( {{w_{{\lambda _j}}}}),{z_0}} ) \in {{( {L( {{w_{{\lambda _j}}}})})}^{ - 1}}\left( {{a_i}} \right),\left| {{z_0}} \right| < r} {\sum\limits_{i = 1}^{2v + 1} {\tau \left( {{z_0},L\left( {{w_{{\lambda _j}}}} \right) = {a_i}} \right)} }\leqslant n\left( {r,\frac{1}{{L\left( W \right) - W\left( z \right)}}} \right)\end{split} $

所以,

$ \sum\limits_{i = 1}^{2v + 1} {N\left( {r,\frac{1}{{W - {a_i}}}} \right)} = \sum\limits_{i = 1}^{2v + 1} {N\left( {r,\frac{1}{{L - {a_i}}}} \right)} \leqslant N\left( {r,\frac{1}{{L - W}}} \right) $

证毕。

引理2[10] 设$W\left( z \right)$$v$值代数体函数,设${a_1},{a_2}, \cdot \cdot \cdot ,{a_q}$$q$个判别的有穷复数,则

$m\left( {r,\sum\limits_{i = 1}^q {\frac{1}{{W - {a_i}}}} } \right) = \sum\limits_{i = 1}^q {m\left( {r,\frac{1}{{W - {a_i}}}} \right)} + {\rm O}\left( 1 \right)$

引理3[10] 设$W\left( z \right)$$v$值代数体函数,则

$ {N_x}\left( {r,W} \right) \leqslant 2\left( {v - 1} \right)T\left( {r,W} \right) + {\rm O}\left( 1 \right) $

引理4[11-12] 设$W\left( z \right)$$v$值代数体函数,$k$为一正整数,$L\left( W \right)$$W$的线性微分多项式,$ L\left( W \right) =$$ {c_k}{W^{\left( k \right)}} + {c_{k - 1}}{W^{\left( {k - 1} \right)}} + \cdot \cdot \cdot + {c_0}W$,其中,${c_0},{c_1}, \cdot \cdot \cdot, {c_k}$为不全为零的有穷复数,且${c_k} \ne 0$,再设${a_1},{a_2}, \cdot \cdot \cdot ,{a_q}$$q$$q$≥3)个互相判别的复数,则有

$\left( {q - 2} \right)T\left( {r,L} \right) < \sum\limits_{i = 1}^q {\bar N\left( {r,\frac{1}{{L - {a_i}}}} \right)} + {N_x}\left( {r,W} \right) + S\left( {r,L} \right)$

其中,$r \to \infty $$r \notin {E}$,这里E是一个线性测度为有穷的集合。

3 定理4的证明

现证明定理4。

证明 假设$W\left( z \right)\not \equiv L\left( W \right)$,令$q = 2v + 1$

a.${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$为有穷非零复数时,由引理1可得

$\begin{split} &\quad \sum\limits_{i = 1}^{2v + 1} N \left( {r,\frac{1}{{W - {a_i}}}} \right) \leqslant N\left( {r,\frac{1}{{L - W}}} \right) \leqslant \\ &\qquad T\left( {r,L - W} \right) + {\rm O}\left( 1 \right) \leqslant T\left( {r,L} \right) + T\left( {r,W} \right) + {\rm O}\left( 1 \right)\qquad\qquad\qquad\qquad\\[-10pt]\end{split} $ (2)

由引理2可得

$\begin{split} &\quad \sum\limits_{i = 1}^{2v + 1} {m\left( {r,\frac{1}{{W - {a_i}}}} \right)} = m\left( {r,\sum\limits_{i = 1}^{2v + 1} {\frac{1}{{W - {a_i}}}} } \right) + {\rm O}\left( 1 \right)\leqslant m\left( {r,\frac{1}{{W'}}} \right) + m\left( {r,\sum\limits_{i = 1}^{2v + 1} {\frac{{W'}}{{W - {a_i}}}} } \right) + {\rm O}\left( 1 \right)\leqslant T\left( {r,W'} \right) + S\left( {r,W} \right)\leqslant\qquad\qquad\qquad\qquad\qquad\\ &\qquad\;\;\quad m\left( {r,W} \right) + m\left( {r,\frac{{W'}}{W}} \right) + N\left( {r,W'} \right) + S\left( {r,W} \right)\leqslant m\left( {r,W} \right) + N\left( {r,L} \right) + S\left( {r,W} \right)\leqslant T\left( {r,W} \right) + T\left( {r,L} \right) + S\left( {r,W} \right)\\[-18pt] \end{split}$ (3)

结合式(2)和式(3),可得

$ \left( {2v + 1} \right)T\left( {r,W} \right) \leqslant 2T\left( {r,L} \right) + 2T\left( {r,W} \right) + S\left( {r,W} \right) $

整理上式,可得

$ \qquad\qquad\quad \left( {2v - 1} \right)T\left( {r,W} \right) \leqslant 2T\left( {r,L} \right) + S\left( {r,W} \right) $ (4)

由引理3,引理4和式(2)可得

$\begin{split} &\quad\left( {2v \!-\! 1} \right)T\left( {r,L} \right)\! <\! \sum\limits_{i = 1}^{2v \!+ \!1} {\bar N\left( {r,\frac{1}{{L \!-\! {a_i}}}} \right)} \!+\! {N_x}\left( {r,W} \right) + \qquad\\ &\qquad S\left( {r,L} \right)\leqslant N\left( {r,\frac{1}{{L - W}}} \right) + {N_x}\left( {r,W} \right) + S\left( {r,L} \right)\leqslant \\ &\qquad T\left( {r,W} \right) + T\left( {r,L} \right) + 2\left( {v - 1} \right)T\left( {r,W} \right) + S\left( {r,L} \right)\end{split} $

整理上式,可得

$ \quad\quad \left( {2v - 2} \right)T\left( {r,L} \right) \leqslant \left( {2v - 1} \right)T\left( {r,W} \right) + S\left( {r,L} \right) $ (5)

由式(5)可得,$S\left( {r,L} \right) = S\left( {r,W} \right)$。因此,式(5)可变为

$ \qquad \left( {2v - 2} \right)T\left( {r,L} \right) \leqslant \left( {2v - 1} \right)T\left( {r,W} \right) + S\left( {r,W} \right) $ (6)

再结合式(4)和式(6),可得

$\begin{split} &\quad\left( {2v - 1} \right)T\left( {r,W} \right) + \left( {2v - 2} \right)T\left( {r,L} \right) \leqslant \\ & \;\qquad\qquad \left( {2v - 1} \right)T\left( {r,W} \right) + 2T\left( {r,L} \right) + S\left( {r,W} \right)\qquad\qquad\qquad\qquad\qquad\qquad\end{split} $

从而有

$ \left( {2v - 4} \right)T\left( {r,L} \right) \leqslant S\left( {r,W} \right),{v \geqslant 3} $

矛盾。

b.${a_1},{a_2}, \cdot \cdot \cdot ,{a_{2v + 1}}$中有一个为$0$,设${a_{2v \!+\! 1}} \!=\! 0$。由引理1可得

$ \begin{split} &\quad\displaystyle\sum\limits_{i = 1}^{2v + 1} N \left( {r,\dfrac{1}{{W - {a_i}}}} \right) = \displaystyle\sum\limits_{i = 1}^{2v} N \left( {r,\dfrac{1}{{W - {a_i}}}} \right) + N\left( {r,\dfrac{1}{W}} \right) \leqslant\qquad\qquad\\ &\qquad N\left( {r,\dfrac{1}{{L - W}}} \right) + N\left( {r,\dfrac{1}{L}} \right)\leqslant T\left( {r,L - W} \right) + N\left( {r,\frac{1}{L}} \right) + \\ &\qquad {\rm O}\left( 1 \right) \leqslant T\left( {r,L} \right) + T\left( {r,W} \right) + T\left( {r,\frac{1}{L}} \right) + {\rm O}\left( 1 \right) \leqslant\\ &\qquad 2T\left( {r,L} \right) + T\left( {r,W} \right) + {\rm O}\left( 1 \right)\\[-10pt] \end{split} $ (7)

结合式(3)和式(7),可得

$ \left( {2v + 1} \right)T\left( {r,W} \right) \leqslant 3T\left( {r,L} \right) + 2T\left( {r,W} \right) + S\left( {r,W} \right) $

整理上式,可得

$ \qquad\qquad \left( {2v - 1} \right)T\left( {r,W} \right) \leqslant 3T\left( {r,L} \right) + S\left( {r,W} \right) $ (8)

另一方面,类似于式(5)的证明,有

$ \begin{split} &\quad\left( {2v \!-\! 1} \right)T\left( {r,L} \right)\! <\! \sum\limits_{i \!= \!1}^{2v} {\bar N\left( {r,\frac{1}{{L \!-\! {a_i}}}} \right)} \!+\! N\left( {r,\frac{1}{L}} \right) + {N_x}\left( {r,W} \right) +\qquad\qquad\\ & \qquad S\left( {r,L} \right)\!\leqslant\! N\left( {r,\frac{1}{{L \!-\! W}}} \right) \!+\! N\left( {r,\frac{1}{L}} \right) \!+\! {N_x}\left( {r,W} \right) \!+\! S\left( {r,L} \right)\leqslant\\ &\qquad T\left( {r,W} \right) \!+\! 2T\left( {r,L} \right) \!+\! 2\left( {v\! -\! 1} \right)T\left( {r,W} \right) \!+\! S\left( {r,L} \right) \end{split} $

整理上式,可得

$ \qquad \left( {2v - 3} \right)T\left( {r,L} \right) \leqslant \left( {2v - 1} \right)T\left( {r,W} \right) + S\left( {r,L} \right) $ (9)

由式(9)可得,$S\left( {r,L} \right) = S\left( {r,W} \right)$。因此,式(9)可变为

$ \;\;\;\;\quad \left( {2v - 3} \right)T\left( {r,L} \right) \leqslant \left( {2v - 1} \right)T\left( {r,W} \right) + S\left( {r,W} \right) $ (10)

再结合式(8)和式(10),可得

$\begin{split} &\quad \left( {2v - 1} \right)T\left( {r,W} \right) + \left( {2v - 3} \right)T\left( {r,L} \right) \leqslant \\ & \;\;\qquad\qquad 3T\left( {r,L} \right) + \left( {2v - 1} \right)T\left( {r,W} \right) + S\left( {r,W} \right)\qquad\qquad\qquad\qquad\qquad\end{split} $

从而有

$ \left( {2v - 6} \right)T\left( {r,L} \right) \leqslant S\left( {r,W} \right), {v \geqslant 4} $

矛盾,定理4得证。

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