假定读者熟悉Nevanlinna值分布理论的标准记号和基本结果(参见文献[5, 10, 12]).设$f$为非常数亚纯函数,$S(r,f)$表示任一满足$S(r,f)=o(T(r,f))(r\longrightarrow \infty)$的量,可能须除去一个线测度有限的集合.

设$f$和$g$为非常数亚纯函数,$a$为任意有限复数,$k$是正整数或$+\infty$,定义$\overline{E}_{k)}(a,f)$为$f-a$的重数$\leq k$的零点集合,且不计重数; 若考虑重数,则记为$E_{k)}(a,f)$. 如果$\overline{E}_{\infty)}(a,f)=\overline{E}_{\infty)}(a,g)$ (或$E_{\infty)}(a,f)=E_{\infty)}(a,g)$),则称$f$和$g$IM(或CM)分担$a$. 定义$\overline{N}_{k)}\left(r,\frac{1}{f-a}\right)$为$f-a$的重数$\leq k$的零点的精简计数函数,$\overline{N}_{(k}\left(r,\frac{1}{f-a}\right)$为$f-a$的重数$> k$的零点的精简计数函数.记 $$N_{k}\left(r,\frac{1}{f-a}\right)=\overline{N}\left(r,\frac{1}{f-a}\right)+\overline{N}_{(2}\left(r,\frac{1}{f-a}\right)+\cdots+\overline{N}_{(k}\left(r,\frac{1}{f-a}\right), $$ $$\Theta (a,f)=1-\mathop{\overline{\lim}}\limits_{r\rightarrow\infty}\frac{\overline{N}\left(r,\frac{1}{f-a}\right)}{T(r,f)},\Theta_{k)}(a,f)=1-\mathop{\overline{\lim}}\limits_{r\rightarrow\infty}\frac{\overline{N}_{k)}\left(r,\frac{1}{f-a}\right)}{T(r,f)} $$(参见文献 ^{[1, 7, 8, 12]}等).

Hayman^[6]和 Clunie^[2]证明了如下结论.

定理 A 设$f(z)$为超越整函数,$n\geq 1$为整数,则$f^{n}f'=1$有无穷多个根.

相应于定理A,Fang和Hua^[4],Yang和Hua^[13]分别得到下述定理.

定理 B 设$f(z)$和$g(z)$为非常数整函数,$n\geq6$为整数. 如果$f^nf'$与$g^ng'$CM分担1,则$f(z)=c_1{\rm e}^{cz},g(z)=c_2{\rm e}^{-cz}$,其中$c_1$,$c_2$,$c$为满足$(c_1c_2)^{n+1}c^2=-1$的常数,或者$f(z)=tg(z)$,其中$t^{n+1}=1$.

2007年,Bhoosnurmath 和 Dyavanal推广定理A得到如下定理.

定理 C^[1] 设$f$为超越亚纯函数,$n,k$ 为正整数满足$n\geq k+3$,则 $\{f^{n}(f-1)\}^{^{(k)}}=1$ 有无穷多个根.

相应于定理C,他们也得到了如下的唯一性定理.

定理 D^[1] 设$f(z)$和$g(z)$为非常数亚纯函数满足$\Theta(\infty,f)>\frac{3}{n+1}$. $n,k$是正整数且$n\geq3k+13$.如果$[f^{n}(f-1)]^{(k)}$与 $[g^{n}(g-1)]^{(k)}$ CM分担1,则$f(z)\equiv g(z)$.

2008年,Chen 和Zhang等^[3]考虑将"CM分担1"条件改为"IM分担1",在 $f$ 和 $g$ 为非常数整函数的情况下,得到下述结果.

定理 E设$f(z)$和$g(z)$为非常数整函数,$n,k$为正整数且$n>5k+13$.如果 $[f^{n}(f-1)]^{(k)}$与$[g^{n}(g-1)]^{(k)}$ IM分担1,则$f(z)\equiv g(z)$.

近来,Li和Yi^[7]研究了当亚纯函数分担1个公共值改为分担1个多项式的唯一性问题,得到下述结果.

定理 F 设$f$和$g$为超越亚纯函数,$n,k$为正整数满足$n>9k+20$ 和 $\max\{\chi_1,$ $\chi_2\}<0$,其中

$ \begin{equation}\label{eq:a1}\chi_1=\frac{2}{n-2k+1}+\frac{2}{n+2k+1}+\frac{2k+1}{n+k+1}+1-\Theta_{k)}(1,f)-\Theta_{k-1)}(1,f),\end{equation}$

(1.1)

$\begin{equation}\label{eq:a2}\chi_2=\frac{2}{n-2k+1}+\frac{2}{n+2k+1}+\frac{2k+1}{n+k+1}+1-\Theta_{k)}(1,g)-\Theta_{k-1)}(1,g).\end{equation}$

(1.2)

如果$\Theta(\infty,f)>\frac{2}{n}$ 且$\{f^{n}(f-1)\}^{(k)}-P$和$\{g^{n}(g-1)\}^{(k)}-P$ IM分担$0$,其中$P$是非零多项式,则$f\equiv g$.

本文中,我们继续考虑减弱亚纯函数分担小函数的条件,得到如下结果,推广了定理E和定理F.

定理 1.1设$f$和$g$为超越亚纯函数,$n,k$为正整数满足$n>9k+20$ 和 $\max\{\chi_1,$ $\chi_2\}<0$,其中$\chi_1$ 和 $\chi_2$ 如(1.1)和(1.2)式所示. 如果$\Theta(\infty,f)>\frac{2}{n},$且 $$\overline{E}_{l)}\left(0,\{f^{n}(f-1)\}^{(k)}-P\right)=\overline{E}_{l)}\left(0,\{g^{n}(g-1)\}^{(k)}-P\right), $$其中$P$是非零多项式,$l$是正整数或$\infty$,则$f\equiv g$.

定理 1.2设$f$和$g$为超越整函数,$n,k$为正整数满足$n>5k+13$,如果 $$\overline{E}_{l)}\left(0,\{f^{n}(f-1)\}^{(k)}-P\right)=\overline{E}_{l)}\left(0,\{g^{n}(g-1)\}^{(k)}-P\right), $$其中$P$是非零多项式,$l$是正整数或$\infty$,则$f\equiv g$.

注在定理1.1和定理1.2中,当 $l=\infty$ 时,我们得到 $\{f^{n}(f-1)\}^{(k)}-P$ 和$\{g^{n}(g-1)\}^{(k)}-P$ IM分担$0$,因此定理1.1和定理1.2是定理E和定理F的推广.

设$f$和$g$为非常数亚纯函数,$a$为任意有限复数,$z_0$为$f-a$和$g-a$的公共零点,其重数分别为$r$和$q$. 我们用$\overline{N}_{L}(r,\frac{1}{f-a})$表示$f-a$和$g-a$的公共零点中满足$r>q\geq1$的那些零点的精简计数函数,用$\overline{N}^{1)}_{E}(r,\frac{1}{f-a})$表示$f-a$和$g-a$的公共零点中满足$r=q=1$的那些零点的精简计数函数,用$\overline{N}^{(2}_{E}(r,\frac{1}{f-a})$表示$f-a$和$g-a$的公共零点中满足$r=q\geq2$的那些零点的精简计数函数.类似地,可定义$\overline{N}_{L}(r,\frac{1}{g-a}),$$\overline{N}^{1)}_{E}(r,\frac{1}{g-a})$和$\overline{N}^{(2}_{E}(r,\frac{1}{g-a})$.同时我们用$\overline{N}_{\ast}(r,\frac{1}{f-a})$表示是$f-a$的零点但不是$g-a$的零点的精简计数函数. 类似地,可定义$\overline{N}_{\ast}(r,\frac{1}{g-a})$.

引理 2.1^[11] 设$f$为非常数亚纯函数,$k(\geq1)$为整数,$\varphi(\not\equiv0,\infty)$ 为$f$的小函数. 则 $$ T(r,f) \leq \overline{N}(r,f)+N\left(r,\frac{1}{f}\right)+N\left(r,\frac{1}{f^{(k)}-\varphi}\right)-N\Bigg(r,\frac{1}{\Big(\frac{f^{(k)}}{\varphi}\Big)'}\Bigg)+S(r,f). $$

引理 2.2^{[9, 14]} 设$f$为非常数亚纯函数,$k,p$为正整数. 则 $$ N_{p}\Big(r,\frac{1}{f^{(k)}}\Big)\leq N_{p+k}\Big(r,\frac{1}{f}\Big)+k\overline{N}(r,f)+S(r,f). $$

引理 2.3设$f$为超越亚纯函数,$P$为非零多项式,$k$为正整数. 令$F=\frac{f^{(k)}}{P}$,则 $$ \overline{N}_{(2}\Big(r,\frac{1}{F-1}\Big)\leq(k+1)\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+S(r,f). $$

证 由引理2.2 可得\begin{eqnarray*}\overline{N}_{(2}\Big(r,\frac{1}{F-1}\Big)&\leq&\overline{N}\Big(r,\frac{F}{F'}\Big)\\&\leq&T\Big(r,\frac{F'}{F}\Big)+O(1)\\&\leq&\overline{N}(r,F)+\overline{N}\Big(r,\frac{1}{F}\Big)+S(r,f)\\&\leq&\overline{N}(r,f)+N_1\Big(r,\frac{1}{f^{(k)}}\Big)+S(r,f)\\&\leq &(k+1)\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+S(r,f).\end{eqnarray*}证毕.

引理 2.4设$f$和$g$为超越亚纯函数,$P$是为非零多项式,令$F=\frac{f^{(k)}}{P},G=\frac{g^{(k)}}{P}$.如果$\overline{E}_{l)}\left(0,f^{(k)}-P\right) $$=\overline{E}_{l)}\left(0,g^{(k)}-P\right)$,其中$k$为正整数,$l$为正整数或$\infty$,则 $$ \overline{N}_{L}\Big(r,\frac{1}{F-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{F-1}\Big)\leq(k+1)\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+S(r,f), $$ $$ \overline{N}_{L}\Big(r,\frac{1}{G-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{G-1}\Big)\leq(k+1)\overline{N}(r,g)+N_{k+1}\Big(r,\frac{1}{g}\Big)+S(r,g). $$

证 由$F$的表达式可知$F-1$的零点可能来自$f^{(k)}-P$的零点和$P$的零点.若$z_0\not\in\{z:P(z)=0\}$是$F-1$的单零点,则$z_0$是$f^{(k)}-P$的单零点. 再结合条件$\overline{E}_{l)}(0,f^{(k)}-P)=\overline{E}_{l)}(0,g^{(k)}-P)$,我们得到$z_0$是$g^{(k)}-P$的零点. 因此$z_0$是$G-1$的零点. 由此可得 $$\begin{eqnarray*}\overline{N}_{L}\Big(r,\frac{1}{F-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{F-1}\Big)&\leq&\overline{N}_{(2}\Big(r,\frac{1}{F-1}\Big)+\overline{N}\Big(r,\frac{1}{P}\Big)\\&\leq& \overline{N}_{(2}\Big(r,\frac{1}{F-1}\Big)+O(\log r).\end{eqnarray*}$$由上述不等式和引理2.3,我们得到 $$\overline{N}_{L}\Big(r,\frac{1}{F-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{F-1}\Big)\leq(k+1)\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+S(r,f). $$类似地,我们有 $$\overline{N}_{L}\Big(r,\frac{1}{G-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{G-1}\Big)\leq(k+1)\overline{N}(r,g)+N_{k+1}\Big(r,\frac{1}{g}\Big)+S(r,g). $$证毕.

引理 2.5设$f$和$g$为超越亚纯函数,$P$为非零多项式,令

$F=\frac{{{f}^{(k)}}}{P},G=\frac{{{g}^{(k)}}}{P},$

(2.1)

$H=\left( \frac{F\text{''}}{{{F}'}}-\frac{2{F}'}{F-1} \right)-\left( \frac{G\text{''}}{{{G}'}}-\frac{2{G}'}{G-1} \right).$

(2.2)

如果$\overline{E}_{l)}\left(0,f^{(k)}-P\right)=\overline{E}_{l)}\left(0,g^{(k)}-P\right)$且$H\not\equiv 0$,其中$k$为正整数,$l$为正整数或$\infty$,则$$\begin{eqnarray*}T(r,f)&\leq&(2k+4)\overline{N}(r,f)+(2k+3)\overline{N}(r,g)+3N_{k+1}\left(r,\frac{1}{f}\right)+2N_{k+1}\left(r,\frac{1}{g}\right)\\& &+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)+S(r,f)+S(r,g).\end{eqnarray*}$$

证 由(2.1)和(2.2)式可得

$\begin{equation}m(r,H)=S(r,f)+S(r,g).\end{equation}$

(2.3)

由(2.2)式可知$H$的极点可能来自于$F$和$G$的极点,$F-1$和$G-1$的零点,$F'$和$G'$的零点,且$H$的每个极点都为单极点.

设$z_1$为$F$的单极点,通过简单的计算可知$F"/F'-2F'/(F-1)$在$z_1$处解析.类似地,如果$z_1$是$G$的单极点,则$G"/G'-2G'/(G-1)$在$z_1$处解析.

设$z_2$是为$F-1$和$G-1$具有相同重数的公共零点,则由(2.2)式可得$H$在$z_2$处解析.并且当$z_2$是$F-1$和$G-1$的公共单零点时,我们可得$H(z_2)=0$. 故由(2.3)式可得

$\begin{eqnarray}\overline{N}^{1)}_{E}\bigg(r,\frac{1}{F-1}\bigg)&=&\overline{N}^{1)}_{E}\bigg(r,\frac{1}{G-1}\bigg)\leq N\bigg(r,\frac{1}{H}\bigg)\\ &\leq& T(r,H)+O(1)\\ &\leq& N(r,H)+S(r,f)+S(r,g).\end{eqnarray}$

(2.4)

记$\overline{N_0}\Big(r,\frac{1}{F'}\Big)$为是$F'$零点,但不是$f(F-1)$零点的精简计数函数,由上述分析可得

$\begin{eqnarray}N(r,H)&\leq &\overline{N}_{(2}(r,F)+\overline{N}_{(2}(r,G)+\overline{N}_{L}\left(r,\frac{1}{F-1}\right)+\overline{N}_{L}\left(r,\frac{1}{G-1}\right)+\overline{N}_{*}\left(r,\frac{1}{F-1}\right)\\&&+\overline{N}_{*}\left(r,\frac{1}{G-1}\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)+\overline{N}_{0}\left(r,\frac{1}{F'}\right)+\overline{N}_{0}\left(r,\frac{1}{G'}\right)\\&\leq &\overline{N}(r,f)+\overline{N}(r,g)+\overline{N}_{L}\left(r,\frac{1}{F-1}\right)+\overline{N}_{L}\left(r,\frac{1}{G-1}\right)+\overline{N}_{*}\left(r,\frac{1}{F-1}\right)\\&&+\overline{N}_{*}\left(r,\frac{1}{G-1}\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)+\overline{N}_{0}\left(r,\frac{1}{F'}\right)+\overline{N}_{0}\left(r,\frac{1}{G'}\right)\\&&+O(\log r).\end{eqnarray}$

(2.5)

设$z_3\not\in \{z:P(z)=0\}$ 为$f$的$l(\geq k+2)$重零点,则由(2.1)式可知$z_3$为$F'$的$l-k-1(\geq 1)$重零点.于是由引理 2.1可得

$\begin{eqnarray}T(r,f) &\leq&\overline{N}(r,f)+N\Big(r,\frac{1}{f}\Big)+N\Big(r,\frac{1}{P(F-1)}\Big)-N\Big(r,\frac{1}{F'}\Big)+S(r,f)\\&\leq&\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+\overline{N}\Big(r,\frac{1}{F-1}\Big)-N_{0}\Big(r,\frac{1}{F'}\Big)+O(\log r)+S(r,f)\\&\leq&\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+\overline{N}\Big(r,\frac{1}{F-1}\Big)-N_{0}\Big(r,\frac{1}{F'}\Big)+S(r,f). \end{eqnarray}$

(2.6)

类似地

$ \begin{equation}T(r,g)\leq\overline{N}(r,g)+N_{k+1}\Big(r,\frac{1}{g}\Big)+\overline{N}\Big(r,\frac{1}{G-1}\Big)-N_{0}\Big(r,\frac{1}{G'}\Big)+S(r,g). \end{equation}$

(2.7)

由于$$\begin{eqnarray*} \overline{N}\Big(r,\frac{1}{F-1}\Big)&=&\overline{N}^{1)}_{E}\Big(r,\frac{1}{F-1}\Big)+\overline{N}^{(2}_{E}\Big(r,\frac{1}{F-1}\Big)+\overline{N}_{L}\Big(r,\frac{1}{F-1}\Big)\\&&+\overline{N}_{L}\Big(r,\frac{1}{G-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{F-1}\Big),\end{eqnarray*} $$结合(2.4)和(2.5)式得

$\begin{eqnarray} \overline{N}(r,\frac{1}{F-1})&\leq &\overline{N}(r,f)+\overline{N}(r,g)+2\overline{N}_{L}\left(r,\frac{1}{F-1}\right)+2\overline{N}_{L}\left(r,\frac{1}{G-1}\right)\\&&+2\overline{N}_{*}\left(r,\frac{1}{F-1}\right)+\overline{N}_{*}\left(r,\frac{1}{G-1}\right) +\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)\\&&+\overline{N}_{0}\left(r,\frac{1}{F'}\right)+ \overline{N}_{0}\left(r,\frac{1}{G'}\right)+\overline{N}^{(2}_{E}\Big(r,\frac{1}{F-1}\Big)+S(r,f)+S(r,g).\end{eqnarray} $

(2.8)

由引理 2.4 可得

$\begin{equation}\overline{N}_{L}\Big(r,\frac{1}{F-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{F-1}\Big)\leq(k+1)\overline{N}(r,f)+N_{k+1}\Big(r,\frac{1}{f}\Big)+S(r,f) \ \end{equation}$

(2.9)

及

$\begin{equation}\overline{N}_{L}\Big(r,\frac{1}{G-1}\Big)+\overline{N}_{*}\Big(r,\frac{1}{G-1}\Big)\leq(k+1)\overline{N}(r,g)+N_{k+1}\Big(r,\frac{1}{g}\Big)+S(r,g).\end{equation}$

(2.10)

结合(2.1)式有

$ \begin{eqnarray}\overline{N}_{L}\Big(r,\frac{1}{G-1}\Big)+\overline{N}^{(2}_{E}\Big(r,\frac{1}{G-1}\Big)+\overline{N}\Big(r,\frac{1}{G-1}\Big)&\leq& N\Big(r,\frac{1}{G-1}\Big)\\& \leq& T(r,G)+O(1)\\&\leq& T\left(r,g^{(k)}\right)+O(\log r) \\&\leq& T(r,g)+k\overline{N}(r,g)+S(r,g).\end{eqnarray}$

(2.11)

因此由(2.6)-(2.11)式得

$\begin{eqnarray}T(r,f)+T(r,g)&\leq&\overline{N}(r,f)+\overline{N}(r,g)+N_{k+1}\left(r,\frac{1}{f}\right)+N_{k+1}\left(r,\frac{1}{g}\right)+\overline{N}\left(r,\frac{1}{F-1}\right)\\&&+\overline{N}\left(r,\frac{1}{G-1}\right)-N_{0}\left(r,\frac{1}{F'}\right)-N_{0}\left(r,\frac{1}{G'}\right)+S(r,f)+S(r,g)\\&\leq& (2k+4)\overline{N}(r,f)+(2k+3)\overline{N}(r,g)+3N_{k+1}\left(r,\frac{1}{f}\right)+2N_{k+1}\left(r,\frac{1}{g}\right)\\&&+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)+T(r,g)+S(r,f)+S(r,g).\end{eqnarray}$

(2.12)

再由(2.12)式得$$\begin{eqnarray*}T(r,f)&\leq&(2k+4)\overline{N}(r,f)+(2k+3)\overline{N}(r,g)+3N_{k+1}\left(r,\frac{1}{f}\right)+2N_{k+1}\left(r,\frac{1}{g}\right)\\& &+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{g}\right)+S(r,f)+S(r,g).\end{eqnarray*}$$引理 2.5得证.

引理 2.6^{[5, 12]} 设$f(z)$为非常数亚纯函数, $$P(f)=a_{n}f^{n}+\cdots+a_{1}f+a_{0}, $$其中$a_{0}$,$a_{1},\cdots,a_{n}(\not\equiv 0)$是$f$的小函数,则$T(r,P(f))=n T(r,f)+S(r,f)$.

引理 2.7^[5] 设$f$为非常数亚纯函数. 则对任意正整数$k$,有$$\begin{eqnarray*} \frac{f^{(k)}}{f}&=&\left(\frac{f'}{f}\right)^{k}+\frac{k(k-1)}{2} \left(\frac{f'}{f}\right)^{k-2}\left(\frac{f'}{f}\right)'\\ &&+ \frac{k(k-1)(k-2)}{6}\left(\frac{f'}{f}\right)^{k-3}\left(\frac{f'}{f}\right)"+P_{k-2}\left(\frac{f'}{f}\right),\end{eqnarray*}$$其中$P_{k-2}\left(\frac{f'}{f}\right)$是关于$\frac{f'}{f}$的常系数微分多项式且次数$\leq k-2$.

引理 2.8^{[5, 12]} 设$f$为非常数亚纯函数,$\alpha_{1}(z)$,$\alpha_{2}(z)$为$f$的小亚纯函数,则 $$ T(r,f)\leq \overline{N}(r,f)+\overline{N}\Big(r,\frac{1}{f-\alpha_{1}}\Big)+\overline{N}\Big(r,\frac{1}{f-\alpha_{2}}\Big)+S(r,f). $$

引理 2.9设$f$和$g$为超越整函数,$a,b$为非零常数,$n,k$为正整数满足$n>k$. 令 $$ F=[f^{n}(af+b)]^{(k)},G=[g^{n}(ag+b)]^{(k)}, $$则$FG\not\equiv P^{2}$,其中$P$是非零多项式.

证 设$FG\equiv P^{2}$,则

$\begin{equation}[f^{n}(af+b)]^{(k)}[g^{n}(ag+b)]^{(k)}\equiv P^{2}.\end{equation}$

(2.13)

设$z_0\not\in \{z:P(z)=0\}$,由于$f$和$g$为超越整函数且$n>k$,故由(2.13)式可得$f(z_0)\neq 0$,$g(z_0)\neq 0$. 于是我们有

$\begin{equation}\overline{N}\Big(r,\frac{1}{f}\Big)\leq \overline{N}\Big(r,\frac{1}{P}\Big)=O(log r),\overline{N}\Big(r,\frac{1}{g}\Big)\leq\overline{N}\Big(r,\frac{1}{P}\Big)=O(\log r). \end{equation}$

(2.14)

由引理2.7可得

$\begin{equation}[f^{n}(af+b)]^{(k)}=a\left(f^{n+1}\right)^{(k)}+b\left(f^{n}\right)^{(k)}=aDf^{n+1}+bQf^{n}=aDf^{n}\left(f+\frac{b}{a}\cdot\frac{Q}{D}\right),\end{equation}$

(2.15)

其中

$\begin{eqnarray}D&=&(n+1)^k\left(\frac{f'}{f}\right)^k+\frac{k(k-1)}{2}(n+1)^{k-1}\left(\frac{f'}{f}\right)^{k-2}\left(\frac{f'}{f}\right)'\\&&+ \frac{k(k-1)(k-2)}{6}(n+1)^{k-2} \left(\frac{f'}{f}\right)^{k-3} \left(\frac{f'}{f}\right)" +A_{k-2}\left(\frac{f'}{f}\right),\end{eqnarray}$

(2.16)

$\begin{eqnarray}Q&=&n^k\left(\frac{f'}{f}\right)^k+\frac{k(k-1)}{2}n^{k-1}\left(\frac{f'}{f}\right)^{k-2}\left(\frac{f'}{f}\right)'\\&&+ \frac{k(k-1)(k-2)}{6}n^{k-2}\left(\frac{f'}{f}\right)^{k-3}\left(\frac{f'}{f}\right)" +B_{k-2}\left(\frac{f'}{f}\right),\end{eqnarray}$

(2.17)

这里$A_{k-2}\left(\frac{f'}{f}\right),B_{k-2}\left(\frac{f'}{f}\right)$是关于$\frac{f'}{f}$的常系数微分多项式且次数$\leq k-2$. 由(2.14)和(2.16)式可得

$\begin{equation}T(r,D)=m(r,D)+N(r,D)=S(r,f)+O\left(\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}(r,f)\right)=S(r,f).\end{equation}$

(2.18)

类似地,由(2.14)和(2.17)式可得

$\begin{equation}T(r,Q)=S(r,f).\end{equation}$

(2.19)

于是由(2.18)和(2.19)式得

$\begin{equation}T\left(r,\frac{b}{a}\cdot\frac{Q}{D}\right)=S(r,f).\end{equation}$

(2.20)

另一方面,由(2.13)-(2.15)式和(2.18)式得

$\begin{equation}\overline{ N}\Bigg(r,\frac{1}{f+\frac{b}{a}\frac{Q}{R}}\Bigg)=S(r,f).\end{equation}$

(2.21)

故由(2.14),(2.20),(2.21)式和引理 2.8,有 $$T(r,f)\leq \overline{N}(r,f)+\overline{N}\bigg(r,\frac{1}{f}\bigg)+\overline{N}\Bigg(r,\frac{1}{f+\frac{b}{a}\frac{Q}{R}}\Bigg)+S(r,f) \leq S(r,f). $$矛盾. 引理 2.9得证.

定理1.1的证明 令$F_{1}=f^{n}(f-1)$,$G_{1}=g^{n}(g-1)$,则由引理 2.6,可得

$\begin{equation}T(r,F_1)=(n+1)T(r,f)+S(r,f),T(r,G_1)=(n+1)T(r,g)+S(r,g).\end{equation}$

(3.1)

同时令

$\begin{equation}F=\frac{F_1^{(k)}}{P},G=\frac{G_1^{(k)}}{P},\end{equation}$

(3.2)

$ \begin{equation}H=\left(\frac{F"}{F'}-\frac{2F'}{F-1}\right)-\left(\frac{G"}{G'}-\frac{2G'}{G-1}\right).\end{equation}$

(3.3)

由于 $$\overline{E}_{l)}\Big(0,\{f^n(f-1)\}^{(k)}-P\Big)=\overline{E}_{l)}\Big(0,\{g^n(g-1)\}^{(k)}-P\Big), $$我们有

$\begin{equation}\overline{E}_{l)}\Big(0,F_1^{(k)}-P\Big)=\overline{E}_{l)}\Big(0,G_1^{(k)}-P\Big).\end{equation}$

(3.4)

假设$H\not\equiv 0$,则由(3.1),(3.4)式和引理 2.5,可得

$\begin{eqnarray}T(r,F_1)&\leq&(2k+4)\overline{N}(r,F_1)+(2k+3)\overline{N}(r,G_1)+3N_{k+1}\left(r,\frac{1}{F_1}\right)+2N_{k+1}\left(r,\frac{1}{G_1}\right)\\&&+\overline{N}\left(r,\frac{1}{F_1}\right)+\overline{N}\left(r,\frac{1}{G_1}\right)+S(r,f)+S(r,g)\end{eqnarray}$

(3.5)

及

$\begin{eqnarray}T(r,G_1)&\leq&(2k+4)\overline{N}(r,G_1)+(2k+3)\overline{N}(r,F_1)+3N_{k+1}\left(r,\frac{1}{G_1}\right)+2N_{k+1}\left(r,\frac{1}{F_1}\right)\\&&+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{F_1}\right)+S(r,f)+S(r,g).\end{eqnarray}$

(3.6)

故由(3.5)和(3.6)式,可得

$\begin{eqnarray}T(r,F_1)+T(r,G_1)&\leq&(4k+7)\Big[\overline{N}(r,F_1)+\overline{N}(r,G_1)\Big]\\&&+5\bigg[N_{k+1}\left(r,\frac{1}{F_1}\right)+N_{k+1}\left(r,\frac{1}{G_1}\right)\bigg]\\&&+2\bigg[\overline{N}\left(r,\frac{1}{F_1}\right)+\overline{N}\left(r,\frac{1}{G_1}\right)\bigg]+S(r,f)+S(r,g).\end{eqnarray}$

(3.7)

由$F_1$的表达式及$n>9k+20$,可得

$\begin{eqnarray}N_{k+1}\bigg(r,\frac{1}{F_{1}}\bigg)&=&N_{k+1}\bigg(r,\frac{1}{f-1}\bigg)+N_{k+1}\bigg(r,\frac{1}{f^{n}}\bigg)\\&\leq& N\bigg(r,\frac{1}{f-1}\bigg)+(k+1)N\bigg(r,\frac{1}{f}\bigg)\\&\leq&(k+2)T(r,f)+O(1)\end{eqnarray}$

(3.8)

及

$\begin{equation}\overline{N}\left(r,\frac{1}{F_1}\right)=\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{f-1}\right)\leq 2T(r,f)+O(1),\overline{N}(r,F_1)=\overline{N}(r,f).\end{equation}$

(3.9)

类似地,我们有

$\begin{equation}N_{k+1}\Big(r,\frac{1}{G_{1}}\Big)\leq(k+2)T(r,g)+O(1),~\overline{N}\left(r,\frac{1}{G_1}\right)\leq 2T(r,g)+O(1),~\overline{N}(r,G_1)=\overline{N}(r,g).\end{equation}$

(3.10)

故由(3.1),(3.7)-(3.10)式,可得 $$n\Big[T(r,f)+T(r,g)\Big]\leq (9k+20)\Big[T(r,f)+T(r,g)\Big]+S(r,f)+S(r,g). $$这与$n>9k+20$矛盾. 故有$H\equiv 0$.再结合(3.3)式可得

$\begin{equation}\frac{F"}{F'}-\frac{2F'}{F-1}=\frac{G"}{G'}-\frac{2G'}{G-1}.\end{equation}$

(3.11)

由(3.11)式可得 $$\frac{1}{F-1}=\frac{A}{G-1}+B, $$其中$A(\neq0),B$是两个复常数. 再结合(3.2)式可得

$\begin{equation}\frac{P}{F_1^{(k)}-P}=\frac{AP}{G_1^{(k)}-P}+B.\end{equation}$

(3.12)

下面我们分三种情形证明$F_1^{(k)}G_1^{(k)}\equiv P^2$或$F_1\equiv G_1$.

情形 1. $B\neq 0$且$A=B$. 若$B=-1$,则由(3.12)式得$F_1^{(k)}\cdot G_1^{(k)}\equiv P^2$.若$B\neq-1$,则由(3.12)式得

$\begin{equation}F_1^{(k)}=P\left\{\frac{1+B}{B}-\frac{P}{BG_1^{(k)}}\right\}=\frac{(1+B)P}{B}\cdot\frac{G_1^{(k)}-\frac{P}{1+B}}{G_1^{(k)}}.\end{equation}$

(3.13)

故由(3.13)式有

$ \begin{equation}\overline{N}\left(r,\frac{1}{F_1^{(k)}-(1+B)P/B}\right)=\overline{N}(r,G_1)+O(\log r)\end{equation}$

(3.14)

及

$\begin{equation}\overline{N}\left(r,\frac{1}{G_1^{(k)}-P/(1+B)}\right)=\overline{N}\Big(r,\frac{1}{F_1^{(k)}}\Big)+O(\log r).\end{equation}$

(3.15)

设$z_0\not\in \{z:P(z)=0\}$为$F_1$的$m(\geq k+2)$重零点,则$z_0$为$\Big(\frac{F_1^{(k)}}{P}\Big)'$的$m-k-1$重零点.类似地,设$z_0\not\in \{z:P(z)=0\}$ 是$F_1^{(k)}-\frac{1+B}{B}P$的$q(\geq 2)$重零点,则 $z_0$是$\Big(\frac{F_1^{(k)}}{P}\Big)'$的$q-1$重零点. 由以上分析及(3.1),(3.14)式和引理 2.1,可得

$\begin{align} & T(r,{{F}_{1}})\le \bar{N}(r,{{F}_{1}})+N\left( r,\frac{1}{{{F}_{1}}} \right)+ \\ & N\left( r,\frac{1}{F_{1}^{(k)}-P(1+B)/B} \right)- \\ & N(r,\frac{1}{(\frac{F_{1}^{(k)}}{P}{)}'})+S(r,f)\le \bar{N}(r,{{F}_{1}})+ \\ & {{N}_{k+1}}\left( r,\frac{1}{{{F}_{1}}} \right)+\bar{N}\left( r,\frac{1}{F_{1}^{(k)}-P(1+B)/B} \right)+ \\ & O(\log r)+S(r,f)\le \bar{N}(r,{{F}_{1}})+{{N}_{k+1}} \\ & (r,\frac{1}{{{F}_{1}}})+\bar{N}(r,{{G}_{1}})+S(r,f). \\ \end{align}$

(3.16)

类似地,我们有 $$T(r,G_1)\leq\overline{N}(r,G_1)+N_{k+1}\Big(r,\frac{1}{G_1}\Big)+\overline{N}\Big(r,\frac{1}{F_1^{(k)}}\Big)+S(r,g). $$由上式和引理 2.2可得

$\begin{equation}T(r,G_1)\leq\overline{N}(r,G_1)+N_{k+1}\Big(r,\frac{1}{G_1}\Big)+N_{k+1}\Big(r,\frac{1}{F_1}\Big)+k\overline{N}(r,F_1)+S(r,g)+S(r,f).\end{equation}$

(3.17)

再结合(3.1),(3.8)-(3.10),(3.16)和(3.17)式可得 $$n\Big[T(r,f)+T(r,g)\Big]\leq (3k+4)\Big[T(r,f)+T(r,g)\Big]+S(r,f)+S(r,g). $$这与$n>9k+20$矛盾.

情形 2. $B\neq 0$且$A\neq B$. 若$B=-1$,则由(3.12)式可得

$\begin{equation}F_1^{(k)}=\frac{-AP^2}{G_1^{(k)}-(A+1)P},G_1^{(k)}=\frac{(A+1)P\left(F_1^{(k)}-\frac{A}{A+1}P\right)}{F_1^{(k)}}.\end{equation}$

(3.18)

因此结合(3.18)式,我们有

$\begin{equation}\overline{N}\left(r,\frac{1}{F_1^{(k)}-\frac{A}{A+1}P}\right)=\overline{N}\left(r,\frac{1}{G_1^{(k)}}\right)+O(\log r)\end{equation}$

(3.19)

及

$\begin{equation}\overline{N}\left(r,\frac{1}{G_1^{(k)}-(A+1)P}\right)=\overline{N}\left(r,F_1\right)+O(\log r).\end{equation}$

(3.20)

使用类似于情形 1的证明方法,由(3.1),(3.19) (或(3.20))式,引理2.1 和引理2.2可得

$\begin{equation}T(r,F_1)\leq\overline{N}(r,F_1)+N_{k+1}\Big(r,\frac{1}{F_1}\Big)+N_{k+1}\Big(r,\frac{1}{G_1}\Big)+k\overline{N}(r,G_1)+S(r,f)+S(r,g)\end{equation}$

(3.21)

及

$\begin{equation}T(r,G_1)\leq\overline{N}(r,G_1)+N_{k+1}\Big(r,\frac{1}{G_1}\Big)+\overline{N}(r,F_1)+S(r,g).\end{equation}$

(3.22)

再由(3.1),(3.8)-(3.10),(3.21)和(3.22)式可得 $$n\Big[T(r,f)+T(r,g)\Big]\leq (3k+4)\Big[T(r,f)+T(r,g)\Big]+S(r,f)+S(r,g). $$这与$n>9k+20$矛盾.若$B\neq-1$,由(3.12)式可得

$\begin{equation}F_1^{(k)}-\frac{B+1}{B}P=\frac{-AP^{2}}{B^2\left(G_1^{(k)}-\frac{B-A}{B}P\right)},G_1^{(k)}-\frac{B-A}{B}P=\frac{-AP^{2}}{B^2\left(F_1^{(k)}-\frac{B+1}{B}P\right)}.\end{equation}$

(3.23)

故由(3.23)式,可得(3.14)式和

$\begin{equation}\overline{N}\left(r,\frac{1}{G_1^{(k)}-(B-A)P/B}\right)=\overline{N}(r,F_1)+O(\log r).\end{equation}$

(3.24)

使用类似于情形 1的证明过程,结合(3.1),(3.14) (或(3.24))式,引理2.1和引理2.2,可得到(3.16)式和

$\begin{equation}T(r,G_1)\leq\overline{N}(r,G_1)+N_{k+1}\Big(r,\frac{1}{G_1}\Big)+\overline{N}(r,F_1)+S(r,g).\end{equation}$

(3.25)

再由(3.1),(3.8)-(3.10),(3.16)和(3.25)式得到 $$n\Big[T(r,f)+T(r,g)\Big]\leq (k+3)\Big[T(r,f)+T(r,g)\Big]+S(r,f)+S(r,g). $$这与$n>9k+20$矛盾.

情形 3. $B=0$. 则由(3.12)式可得

$\begin{equation}G_1^{(k)}=AF_1^{(k)}+(1-A)P.\end{equation}$

(3.26)

若$A\neq 1$,则由(3.26)式有

$\begin{equation}G_1=AF_1+(1-A)P_1,\end{equation}$

(3.27)

其中$P_1$是次数为 $\deg P+k$的多项式.故由(3.27)式得到

$\begin{equation}\overline{N}\left(r,\frac{1}{G_1-(1-A)P_1}\right)=\overline{N}\left(r,\frac{1}{F_1}\right).\end{equation}$

(3.28)

另一方面,由(3.1)和(3.27)式可得

$\begin{equation}T(r,f)=T(r,g)+S(r,g).\end{equation}$

(3.29)

故结合(3.1),(3.9),(3.10),(3.28)和(3.29)式可得 $$\begin{eqnarray*}(n+1)T(r,g)&=&T(r,G_1)+S(r,g)\\&\leq&\overline{N}(r,G_1)+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{G_1-(1-A)P_1}\right)+S(r,g)\\&=&\overline{N}(r,G_1)+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{F_1}\right)+S(r,g)\\&\leq& 3T(r,g)+2T(r,f)+S(r,g)\\&\leq& 5T(r,g)+S(r,g).\end{eqnarray*}$$这与$n>9k+20$矛盾.若$A=1$,则由(3.26)式可得

$\begin{equation}G_1=F_1+P_2,\end{equation}$

(3.30)

其中$P_2$是次数为$\deg P_2< k$的多项式.下面我们证明$P_2\equiv 0$. 假设$P_2\not\equiv 0$,则由(3.30)式可得

$\begin{equation}\overline{N}\left(r,\frac{1}{G_1-P_2}\right)=\overline{N}\left(r,\frac{1}{F_1}\right),~~T(r,G_1)=T(r,F_1)+O(\log r).\end{equation}$

(3.31)

再结合(3.1),(3.9),(3.10)和(3.31)式可得$$\begin{eqnarray*}(n+1)T(r,g)&=&T(r,G_1)+S(r,g)\\&\leq&\overline{N}(r,G_1)+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{G_1-P_2}\right)+S(r,g)\\&=&\overline{N}(r,G_1)+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{F_1}\right)+S(r,g)\\&\leq& 5T(r,g)+S(r,g).\end{eqnarray*}$$这与$n>9k+20$矛盾.故$P_2\equiv 0$.再结合(3.30)式得$F_1\equiv G_1$.

由以上讨论,我们得到$F_1^{(k)}G_1^{(k)}\equiv P^2$ 或$F_1\equiv G_1$.结合(3.1)式即有$\{f^{n}(f-1)\}^{(k)}\{g^{n}(g-1)\}^{(k)}\equiv P^2$ 或$f^{n}(f-1)\equiv g^{n}(g-1)$. 再使用与定理E (见文献[7,定理1.1])一样的讨论,可得 $f\equiv g$. 定理 1.1得证.

定理1.2的证明设$F_1,G_1,F,G$和$H$分别如(3.1)-(3.3)式所定义. 设$H\not\equiv 0$,则由(3.1)式和引理2.5,可得

$\begin{eqnarray}T(r,F_1)&\leq&3N_{k+1}\left(r,\frac{1}{F_1}\right)+2N_{k+1}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{F_1}\right)+\overline{N}\left(r,\frac{1}{G_1}\right)\\& &+S(r,f)+S(r,g)\end{eqnarray}$

(3.32)

及

$\begin{eqnarray}T(r,G_1)&\leq&3N_{k+1}\left(r,\frac{1}{G_1}\right)+2N_{k+1}\left(r,\frac{1}{F_1}\right)+\overline{N}\left(r,\frac{1}{G_1}\right)+\overline{N}\left(r,\frac{1}{F_1}\right)\\& &+S(r,f)+S(r,g).\end{eqnarray}$

(3.33)

故由(3.32)和(3.33)式,可得

$\begin{eqnarray}T(r,F_1)+T(r,G_1)&\leq&5\Big[N_{k+1}\left(r,\frac{1}{F_1}\right)+N_{k+1}\left(r,\frac{1}{G_1}\right)\Big]\\& &+2\Big[\overline{N}\left(r,\frac{1}{F_1}\right)+\overline{N}\left(r,\frac{1}{G_1}\right)\Big]+S(r,f)+S(r,g).\end{eqnarray}$

(3.34)

由定理 1.2的条件可知(3.8),(3.9)和(3.10)式仍然成立.故由(3.1),(3.8)-(3.10)和(3.34)式,可得 $$n\Big[T(r,f)+T(r,g)\Big]\leq (5k+13)\Big[T(r,f)+T(r,g)\Big]+S(r,f)+S(r,g). $$这与$n>5k+13$矛盾. 故$H\equiv 0$. 使用类似于定理1.1中情形 1-3的讨论,可得$F_1^{(k)}G_1^{(k)}\equiv P^2$ 或 $F_1\equiv G_1$.

由于$f$和$g$是超越整函数,结合(3.1)式和引理 2.9,可得$F_1^{(k)}G_1^{(k)}\not\equiv P^2$.故$F_1\equiv G_1$. 再结合(3.1)式即得

$\begin{equation}f^{n}(f-1)=g^{n}(g-1).\end{equation}$

(3.35)

令$h=\frac{f}{g}$,并将$f=gh$代入(3.35)式,可得

$\begin{equation}g\left(h^{n+1}-1\right)=h^{n}-1.\end{equation}$

(3.36)

假设$h$不恒为常数,由(3.36)式可得

$\begin{equation}g=\frac{h^n-1}{h^{n+1}-1}=\frac{h^{n-1}+\cdots+h+1}{h^n+h^{n-1}+\cdots+h+1}.\end{equation}$

(3.37)

由于$g$是超越整函数,结合(3.37)式我们知道$h^n+h^{n-1}+\cdots+h+1$ 无零点. 故$h^{n+1}-1$的零点必是 $h-1$的零点.由第二基本定理可得$$\begin{eqnarray*} (n-1)T(r,h)&\leq& \sum\limits^{n+1}_{i=1}\overline{N}\left(r,\frac{1}{h(z)-w_i}\right)+S(r,h) \\&\leq &\overline{N}\left(r,\frac{1}{h^{n+1}-1}\right)+S(r,h)\\&\leq& \overline{N}\left(r,\frac{1}{h-1}\right)+S(r,h)\\&\leq& T(r,h)+S(r,h),\end{eqnarray*}$$其中$w_{i}(i=1,\cdots,n+1)$是复常数满足$w^{n+1}=1$. 这与$n>5k+13$矛盾.因此$h$恒为常数.若$h\not\equiv 1$,则由(3.36)式可得$g$恒为常数,矛盾.故$h\equiv 1$,即$f\equiv g$. 定理1.2得证.