﻿ 代数体函数及其线性微分多项式的唯一性
 上海理工大学学报  2019, Vol. 41 Issue (5): 429-432 PDF

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

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

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

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

${W^{ - 1}}\left( a \right)$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

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

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

$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

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

3 定理4的证明

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)

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

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

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