数学物理学报  2016, Vol. 36 Issue (5): 901-910   PDF    
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本文作者相关文章
王钥
高凌云
关于复差分方程组的允许解的形式
王钥1, 高凌云2     
1. 河北经贸大学数学与统计学学院 石家庄 050061 ;
2. 暨南大学数学系 广州 510632
摘要:利用亚纯函数的值分布理论,该文主要研究了复差分方程组的允许解的形式,得到一个结论,将复微分(差分)方程的相关结论推广到复差分方程组中,例子表明该文结论精确.
关键词值分布     允许解     复差分方程组    
On the Behavior of Systems of Complex Difference Equations with Admissible Solutions
Wang Yue1, Gao Lingyun2     
1. College of Mathematics and Statistics, Hebei University of Economics and Business, Shijiazhuang 050061 ;
2. Department of Mathematics, Jinan University, Guangzhou 510632
Abstract: Using Nevanlinna theory of the value distribution of meromorphic functions, we will mainly investigate the behavior of systems of complex difference equations when we add some conditions to the quality of the solutions, and obtain an interesting result. It extends some results concerning complex differential (difference) equations to the systems of difference equations. Examples show that our results are precise.
Key words: Value distribution     Admissible solution     Systems of complex difference equations    
1 引言与主要结果

全文采用亚纯函数值分布理论的基本概念和通常记号(参见文献[1]).

Toda N[2]研究了复微分方程(1.1)的亚纯解

$ \Omega_1(z,w)^m=\sum\limits_{j=0}^{p}a_jw^j,$ (1.1)

其中 $\Omega_1(z,w)= \sum\limits_{(i)}a_{(i)}(z)w^{i_0}(w^{\prime})^{i_1}\cdots (w^{(n)})^{i_n},$ $ 0\leq p\leq m \max\{i_0+2i_1+\cdots+(n+1)i_n\},$ $ a_p\neq 0$.

他得到如下定理.

定理 A[2] 当$0\leq p\leq m-1$时,复微分方程(1.1)除了如下形式

$\Omega_1(z,w)^m=a_p(w+b)^p $

外,没有允许解,其中 $b= \frac{a_{p-1}}{pa_p}$.

近来,复差分方程解的一些性质研究成为时下复分析的热点之一. 许多学者[3-8]讨论了复差分方程解的存在性和增长级问题,并得到了许多理想的结论.

2014年,我们研究了如下复差分方程

$ [\Omega(z,w)]^m =\displaystyle\sum\limits_{j=0}^{p}a_j(z)w^j,\ \ \ \ 0\leq p\leq m\lambda,a_p\neq 0,$ (1.2)

其中 $\Omega(z,w)= \sum\limits_{(i)}a_{(i)}(z)w^{i_{0}} (w(z+c_1))^{i_{1}}\cdots(w(z+c_n))^{i_{n}}$,$(i)$是一有限指标集,系数 $\{a_i(z)\}$,$\{a_{(i)}(z)\}$ 是亚纯函数,$c_i\in {\bf C}\setminus \{0\},$ $ T(r,a_{(i)})=o(T(r,w)),$ $ T(r,a_{i})=o(T(r,w))$. 并得到如下结论.

定理 B[9]当 $0\leq p\leq m-1$时,复差分方程(1.2)除了如下形式

$ [\Omega(z,w)]^m =a_p(z)[w+b(z)]^p $

外,没有有限级允许解,其中$b= \frac{a_{p-1}}{pa_p}$.

令$c_j\in {\bf C}$,$j=1,\cdots ,n.$ $I,J$分别是$(i_0,\cdots ,i_n)$,$(j_0,\cdots ,j_n)$的两有限指标集. 差分多项式$\Omega_1(z,w_1,w_2),\,\Omega_2(z,w_1,w_2)$可以表示为

$ \Omega_1(z,w_1,w_2)=\displaystyle\sum\limits_{(i)}a_{(i)}(z)\prod\limits_{k=1}^2 w_k^{i_{k0}}(w_k(z+c_1))^{i_{k1}}\cdots(w_k(z+c_n))^{i_{kn}},$
$ \Omega_2(z,w_1,w_2)=\displaystyle\sum\limits_{(j)}b_{(j)}(z)\prod\limits_{k=1}^2 w_k^{j_{k0}}(w_k(z+c_1))^{j_{k1}}\cdots(w_k(z+c_n))^{j_{kn}}. $

$ l_m=\max\limits_{0\leq m\leq n}\{i_{1m}\},\,\,\,\overline{l}_m=\max\limits_{0\leq m\leq n}\{i_{2m}\},\,\,\,q_r=\max\limits_{0\leq r\leq n}\{j_{1r}\},\,\,\,\overline{q}_r=\max\limits_{0\leq r\leq n}\{j_{2r}\},$
$ s_m=\min\limits_{0\leq m\leq n}\{i_{1m}\},\,\,\,\overline{s}_m=\min\limits_{0\leq m\leq n}\{i_{2m}\},\,\,\,t_r=\min\limits_{0\leq r\leq n}\{j_{1r}\},\,\,\,\overline{t}_r=\min\limits_{0\leq r\leq n}\{j_{2r}\},$
$\lambda_1=\max\limits_{(i)} \Big\{\sum^n_{m=0}l_{m}\Big\},\,\,\,\lambda_2=\max\limits_{(i)}\Big\{\sum^n_{m=0}\overline{l}_{m}\Big\},\,\,\,\overline{\lambda}_1=\max\limits_{(j)}\Big\{\sum^n_{r=0}q_{r}\Big\},\,\,\,\overline{\lambda}_2=\max\limits_{(j)}\Big\{\sum^n_{r=0}\overline{q}_{r}\Big\},$
$ \sigma_1=\min\limits_{(i)}\Big\{\sum^n_{m=0}s_m\Big\},\,\,\,\sigma_2=\min\limits_{(i)}\Big\{\sum^n_{m=0}\overline{s}_{m}\Big\},\,\,\,\overline{\sigma}_1=\min\limits_{(j)}\Big\{\sum^n_{m=0}t_{r}\Big\},\,\,\,\overline{\sigma}_2=\min\limits_{(j)}\Big\{\sum^n_{m=0}\overline{t}_{r}\Big\},$

其中 $\sigma_k\neq 0,\,\,\overline{\sigma}_k\neq 0,\,\,k=1,2$.

我们已经讨论了多类复差分方程组的解的存在性和增长级问题,并得到一些结论,见文献[10-13]. 我们自然想到如果用如下的复差分方程组(1.3)去替代复差分方程(1.2),定理B中的条件或结论是否有所改变?

$ \left\{ \begin{array}{ll} \displaystyle [\Omega_1 (z,w_1,w_2)]^{m_1} = \displaystyle\sum\limits_{i=0}^pa_i(z)w_1^i,\\ [4mm] \displaystyle [\Omega_2 (z,w_1,w_2)]^{m_2} =\displaystyle\sum\limits_{j=0}^qb_j(z)w_2^j,\end{array} \right.$ (1.3)

其中$p\leq m_1\sigma_1,q\leq m_2\overline{\sigma_2}$,系数$\{a_{(i)}(z)\},\{b_{(j)} (z)\}$,$\{a_i(z)\},\{b_j(z)\}$是亚纯函数,并且

$T(r,a_{(i)} (z))=o(T(r,w_i(z))),~~ T(r,a_{i}(z))=o(T(r,w_i(z))),$
$ T(r,b_{(j)}(z))=o(T(r,w_i(z))),~~ T(r,b_{j}(z))=o(T(r,w_i(z))),~~ i=1,2,$
$ a_p(z)\neq 0,~~ b_q(z)\neq 0. $

本文主要目的是研究复差分方程组(1.3)存在允许解的形式.

定义1.1 设$(w_1(z),w_2(z))$是方程组(1.3)的亚纯解,$S(r)$是方程组(1.3)的所有系数的特征函数之和. 若 $w_1(z),w_2(z)$满足 $S(r)=o(T(r,w_k)),k=1,2,$ 除去一线测度为有限的例外值集,称$(w_1(z),w_2(z))$是方程组(1.3)的允许解.

方程组 (1.3)的亚纯解$(w_1(z),w_2(z))$的增长级定义为 $\rho=\rho(w_1,w_2)=\max\{\rho(w_1),\rho(w_2)\},\,\,\rho(w_k)=\limsup\limits_{r\rightarrow\infty}\frac{\log T(r,w_k)}{\log r},k=1,2.$

定义1.2 若$\rho(w_1,w_2) <\infty$,称$(w_1(z),w_2(z))$为有限级亚纯解.

本文主要结果如下:

定理1.1 当$p <m_1,q<m_2$时,复差分方程组(1.3)除了如下形式

$ \displaystyle \left\{ \begin{array}{ll} \displaystyle [\Omega_1 (z,w_1,w_2)]^{m_1} = a_p[w_1+c(z)]^p ,\\ \displaystyle [\Omega_2 (z,w_1,w_2)]^{m_2} = b_q[w_2+d(z)]^q \end{array} \right. $

外,不存在有限级为$\rho(w_1,w_2),i=1,2$的允许解,其中 $c(z)= \frac{a_{p-1}}{pa_p},$ $d(z)= \frac{b_{q-1}}{qb_q}$.

注1.1 我们不能肯定定理$1.1$中排除的方程组形式里存在允许解.

例 1.1 $(w_1,w_2)=(z,z)$是如下复差分方程组的一非允许解

$ \displaystyle \left\{ \begin{array}{ll} \displaystyle [\Omega_1 (z,w_1,w_2)]^{5} = [w_1(z)-(z+1)]^2 ,\\ \displaystyle [\Omega_2 (z,w_1,w_2)]^{5} = [w_2(z)-(z+1)]^2 ,\end{array} \right. $

其中

$ \Omega_1 (z,w_1,w_2) = \displaystyle\frac{1}{z^2-1}w_1(z-1)w_2(z+1),$
$ \Omega_2 (z,w_1,w_2) = \displaystyle\frac{1}{z^2-1} w_1(z+1) w_2(z-1). $

此时

$ p=2,\,m_1=5,\,q=2,\,m_2=5. $
2 一些引理

引理2.1[2] 设$g_0(z)$和$g_1(z)$是$|z| <\infty$中的亚纯函数,且在${\bf C}$中是线性无关的,记

$g_0(z)+g_1(z)=\Phi,$

则有

$ T(r,g_0)\leq m(r,\Phi)+N(r,g_0)+\overline{N}(r,g_0)+\overline{N}(r,g_1) +\overline{N}(r,\frac{1}{g_0})+\overline{N}(r,\frac{1}{g_1})+S_1(r),$

其中当$ g_0,g_1 $是有理函数时,

$ S_1(r)=O(1); $

当$ g_0,g_1 $是有限级时,

$ S_1(r)=O(\log r); $

当$ g_0,g_1 $是其他情形时,

$ S_1(r)=O(\log T(r,g_0))+\log T(r,g_1))+O(\log r),r\to\infty,r\not\in I,$

其中$I$是一线测度为有限的例外值集.

引理2.2[1] 令$w,a_0,\cdots ,a_k$是亚纯函数,则有

${\rm (i)}m(r,\displaystyle\sum\limits_{j=0}^ka_jw^j)\leq km(r,w)+\displaystyle\sum\limits_{j=0}^km(r,a_j) +O(1),$
${\rm (ii)} T(r,\displaystyle\sum\limits_{j=0}^ka_jw^j)\leq kT(r,w)+\displaystyle\sum\limits_{j=0}^kT(r,a_j) +O(1). $

引理2.3[7] 设$T:[0,+\infty)\rightarrow [0,+\infty)$是非减连续函数,$\delta\in (0,1)$,$s\in (0,\infty)$. 若$T$是有限级,即

$\lim\limits_{r\rightarrow\infty}\frac{\log T(r)}{\log r} <\infty,$

则,除去一对数测度为有限的例外值集,有

$T(r+s)=T(r)+o\Big(\frac{T(r)}{r^{\delta}}\Big).$

引理2.4 设$(w_1(z),w_2(z))$是方程组(1.3)的有限级亚纯允许解. 若$p <m_1,q<m_2$,则复差分方程组(1.3)为下列三种情形之一:

$ \left\{ \begin{array}{ll} \displaystyle [\Omega_1 (z,w_1,w_2)]^{m_1} = a_p[w_1+c(z)]^p,\\ \displaystyle [\Omega_2 (z,w_1,w_2)]^{m_2} =b_q[w_2+d(z)]^q ; \end{array} \right.$ (2.1)
$\left\{ \begin{matrix} {{[{{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})]}^{{{m}_{1}}}}=\sum\limits_{i=0}^{p}{{{a}_{i}}}(z)w_{1}^{i}, \\ {{[{{\Omega }_{2}}(z,{{w}_{1}},{{w}_{2}})]}^{{{m}_{2}}}}={{b}_{q}}[{{w}_{2}} \\ \end{matrix} \right.$ (2.2)
$\left\{ \begin{matrix} {{[{{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})]}^{{{m}_{1}}}}={{a}_{p}}{{[{{w}_{1}}+c(z)]}^{p}}, \\ \left\{ \begin{array}{*{35}{l}} {{[{{\Omega }_{2}}(z,{{w}_{1}},{{w}_{2}})]}^{{{m}_{2}}}}=\sum\limits_{j=0}^{q}{{{b}_{j}}}(z)w_{2}^{j}. \\ \end{array} \right. \\ \end{matrix} \right.$ (2.3)

其中$c(z)= \frac{a_{p-1}}{pa_p},$ $d(z)= \frac{b_{q-1}}{qb_q}$.

设$(w_1(z),w_2(z))$为方程组(1.3)的有限级允许解. 反设方程组(1.3)不是(2.1),(2.2),(2.3)式的三种形式之一. 重新将方程组(1.3)改写如下

$ \left\{ \begin{array}{ll} \displaystyle [\Omega_1 (z,w_1,w_2)]^{m_1} =a_p[w_1+c(z)]^p +\displaystyle\sum\limits_{l=0}^ta_l(z)w_1^l ,\\ [4mm] \displaystyle [\Omega_2 (z,w_1,w_2)]^{m_2} = b_q[w_2+d(z)]^q+\displaystyle\sum\limits_{s_1=0}^sb_{s_1}(z)w_2^{s_1},\end{array} \right.$ (2.4)

其中 $t\leq p-2,\,s\leq q-2,\,c(z)= \frac{a_{p-1}}{pa_p},$ $d(z)= \frac{b_{q-1}}{qb_q}$.

$A_{0}=-a_p(z)[w_1+c(z)]^p,\,\,\Phi_0=\sum\limits_{l=0}^ta_l(z)w_1^l,\,B_{0}=[\Omega_1(z,w_1,w_2)]^{m_1},$
$A_{1}=-b_q(z)[w_2+d(z)]^q,\,\,\Phi_1=\sum\limits_{s_1=0}^sb_{s_1}(z)w_2^{s_1},\,B_{1}=[\Omega_2(z,w_1,w_2)]^{m_2},$

则有$A_0+B_0=\Phi_0,\,\,A_1+B_1=\Phi_1.$

类似于文献[12,定理1]的证明,易证$A_0$和$B_0$线性无关,$A_1$和$B_1$线性无关.

由引理2.1,引理2.2和 引理2.3得

$ \begin{array}[b]{rl} T(r,A_0)\leq & m(r,\Phi_0)+N(r,A_0)+\overline{N}(r,A_0)+\overline{N}(r,B_0)\\ [2mm] &+ \overline{N}(r,\displaystyle\frac{1}{A_0})+\overline{N}(r,\displaystyle\frac{1}{B_0})+S(r)+S(r,w_1)+S(r,w_2),\end{array}$ (2.5)
$T(r,A_0)=pT(r,w_1)+S(r),$
$ T(r,B_0) =m_1T(r,\Omega_1(z,w_1,w_2))\leq pT(r,w_1)+S(r),$
$T(r,\Omega_1(z,w_1,w_2))\leq \frac{p}{m_1}T(r,w_1)+S(r),$
$ N(r,A_0)\leq pN(r,w_1)+N(r,a_p)+N(r,c(z)),$
$ \overline{N}(r,A_0)\leq \overline{N}(r,w_1)+N(r,a_p)+N(r,c(z)),$
$\begin{align} & \bar{N}(r,{{B}_{0}})\le \sum\limits_{l=0}^{n}{\left[ {{\lambda }_{1}}N\left( r,{{w}_{1}}\left( z+{{c}_{l}} \right) \right)+{{\lambda }_{2}}N\left( r,{{w}_{1}}\left( z+{{c}_{l}} \right) \right) \right]+}\sum\limits_{\left( i \right)}{N}\left( r,{{a}_{\left( i \right)}}\left( z \right) \right) \\ & \le \sum\limits_{k=0}^{2}{{{\lambda }_{k}}}N\left( r,{{w}_{k}} \right)+\sum\limits_{\left( i \right)}{N\left( r,{{a}_{\left( i \right)}}\left( z \right) \right)}+S\left( r,{{w}_{1}} \right)+S\left( r,{{w}_{2}} \right) \\ \end{align}$

其中 $c_0=0$,

$ \displaystyle\overline{N}\Big(r,\frac{1}{A_0}\Big) = \overline{N}\Big(r,\displaystyle\frac{1}{w_1+c(z)}\Big)+S(r) \leq T(r,w_1)+S(r),$
$\begin{align} & \bar{N}(r,\frac{1}{{{B}_{0}}})=\bar{N}(r,\frac{1}{{{[{{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})]}^{{{m}_{1}}}}}) \\ & =\bar{N}(r,\frac{1}{{{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})}) \\ & \le T(r,{{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})) \\ & \le \frac{p}{{{m}_{1}}}T(r,{{w}_{1}})+S(r), \\ \end{align}$
$m(r,\Phi_0)\leq (p-2)m(r,w_1)+\sum\limits_{l=0}^tm(r,a_l).$

将上述不等式代入(2.5)式得

$\Big(1-\frac{p}{m_1}\Big)T(r,w_1)\leq \sum_{k=1}^{2}\lambda_kN(r,w_k)+S(r)+S(r,w_1)+S(r,w_2).$ (2.6)

同理可得

$\Big(1-\frac{q}{m_2}\Big)T(r,w_2)\leq \sum_{k=1}^{2}\overline{\lambda}_kN(r,w_k)+S(r)+S(r,w_1)+S(r,w_2).$ (2.7)

下面证明$w_1(z)$,$w_2(z)$的极点由系数 $a_{(i)}(z),\,a_i(z),\,b_{(j)}(z),\,b_j(z)$的极点或零点控制.

情形(i)若$z_0$是$w_1(z)$的$\tau$重极点,而不是$w_2(z)$的极点,也不是复差分方程组(1.3)中系数$a_{(i)}(z),\,a_i(z),\,b_{(j)}(z),\,b_j(z)$的极点或零点,则知$z_0$应该是 $(w_1(z))^{i_{10}}$的$i_{10}\tau$重极点,$z_0-c_k$应该是$(w_1(z+c_k))^{i_{1k}}$的$i_{1k}\tau$重极点. 所以,方程组(1.3)的第一个方程的右边极点重数为$p\tau$.

对于复差分方程组(1.3)的第一个方程,有

$ n(r,[\Omega_1(z,w_1,w_2)]^{m_1}) = m_1n(r,\Omega_1(z,w_1,w_2)) \geq m_1\sigma_1n(r,w_1)+O(1). $

另一方面

$ n(r,[\Omega_1(z,w_1,w_2)]^{m_1}) = n\bigg(r,\displaystyle\sum\limits_{i=0}^{p}a_iw_1^j\bigg) = pn(r,w_1)+\displaystyle\sum n(r,a_{j}). $

$ pn(r,w_1)\geq m_1\sigma_1 n(r,w_1)+\sum n(r,a_{i}),$

$ [p-m_1\sigma_1]n(r,w_1)\geq \sum n(r,a_{i}). $

则有

$ p> m_1\sigma_1>m_1. $

矛盾.

情形(ii)若$z_0$是$w_2(z)$的$\tau$重极点,而不是$w_1(z)$的极点,也不是复差分方程组(1.3)中系数$a_{(i)}(z),\,a_i(z),\,b_{(j)}(z),\,b_j(z)$的极点或零点,类似于情形(i)的证明,易得

$ q> m_2\overline{\sigma}_2>m_2. $

矛盾.

情形(iii)若$z_0$是$w_1(z)$的$\tau$重极点,且是$w_2(z)$的$k$ 重极点,但不是复差分方程组(1.3)中系数$a_{(i)}(z),\,a_i(z),\,b_{(j)}(z),\,b_j(z)$的极点或零点,则

$ pn(r,w_1)\geq m_1(\sigma_1 n(r,w_1)+\sigma_2 n(r,w_2))+\sum n(r,a_{i}),$
$ qn(r,w_2)\geq m_2(\overline{\sigma}_1 n(r,w_1)+\overline{\sigma}_2 n(r,w_2))+\sum n(r,b_{j}). $

又因为 $p <m_1,\,q<m_2$,$\sigma_k \neq 0,\overline{\sigma}_k\neq 0,k=1,2 $,故上式不成立.

即有

$\begin{align} & N(r,{{w}_{i}})\le \sum\limits_{(i)}{[}N(r,{{a}_{(i)}})+N(r,\frac{1}{{{a}_{(i)}}})]+\sum\limits_{i=1}^{p}{[}N(r,{{a}_{i}})+N(r,\frac{1}{{{a}_{i}}})] \\ & +\sum\limits_{(j)}{[}N(r,{{b}_{(j)}})+N(r,\frac{1}{{{b}_{(j)}}})]+\sum\limits_{j=1}^{q}{[}N(r,{{b}_{j}})+N(r,\frac{1}{{{b}_{j}}})], \\ \end{align}$

其中$i=1,2$.

故 (2.6),(2.7)式变形为

$ \Big(1-\displaystyle\frac{p}{m_1}+o(1)\Big)T(r,w_1)\leq S(r)+S(r,w_2).$ (2.8)
$\Big(1-\frac{q}{m_2}+o(1)\Big)T(r,w_2)\leq S(r)+S(r,w_1).$ (2.9)

由(2.8),(2.9)式得到

$ 1-\frac{p}{m_1}+o(1) \leq \frac{S(r)}{T(r,w_1)}+\frac{S(r,w_2)}{T(r,w_1)},$ (2.10)
$ 1-\frac{q}{m_2}+o(1) \leq \frac{S(r)}{T(r,w_2)}+\frac{S(r,w_1)}{T(r,w_2)}.$ (2.11)

由$p <m_1,q<m_2$,(2.10),(2.11)式得到

$\begin{align} & (1-\frac{p}{{{m}_{1}}}+o(1))(1-\frac{q}{{{m}_{2}}}+o(1))\le (\frac{S(r)}{T(r,{{w}_{1}})}+\frac{S(r,{{w}_{2}})}{T(r,{{w}_{1}})})(\frac{S(r)}{T(r,{{w}_{2}})}+\frac{S(r,{{w}_{1}})}{T(r,{{w}_{2}})})\le \\ & \frac{S(r)}{T(r,{{w}_{1}})}\frac{S(r)}{T(r,{{w}_{2}})}+\frac{S(r)}{T(r,{{w}_{1}})}\frac{S(r,{{w}_{1}})}{T(r,{{w}_{2}})}+ \\ & \frac{S(r,{{w}_{2}})}{T(r,{{w}_{1}})}\frac{S(r,{{w}_{1}})}{T(r,{{w}_{2}})}+\frac{S(r,{{w}_{2}})}{T(r,{{w}_{1}})}\frac{S(r)}{T(r,{{w}_{2}})}= \\ & \frac{S(r)}{T(r,{{w}_{1}})}\frac{S(r)}{T(r,{{w}_{2}})}+\frac{S(r)}{T(r,{{w}_{2}})}\frac{S(r,{{w}_{1}})}{T(r,{{w}_{1}})}+ \\ & \frac{S(r,{{w}_{1}})}{T(r,{{w}_{1}})}\frac{S(r,{{w}_{2}})}{T(r,{{w}_{2}})}+\frac{S(r,{{w}_{2}})}{T(r,{{w}_{2}})}\frac{S(r)}{T(r,{{w}_{1}})}. \\ \end{align}$

又因为$p <m_1,\,q<m_2$,除去一线测度有穷的例外值集,得到

$1\leq 0,$

矛盾.

引理2.4证毕.

3 定理1.1的证明

由引理2.4知,复差分方程组(1.3)除(2.1),(2.2),(2.3)式的形式外,不存在有限级允许解.

下面证明$(w_1(z),w_2(z))$是复差分方程组(1.3)的有限级允许解,但不具有(2.2),(2.3)式的形式. 不失一般性,证明不具有形式 (2.2).

将(2.2)式重新改写如下

$$\left\{ \begin{matrix} {{\left[ {{\Omega }_{1}}\left( z,{{w}_{1}},{{w}_{2}} \right) \right]}^{{{m}_{1}}}}={{a}_{p}}{{\left[ {{w}_{1}}+c\left( z \right) \right]}^{p}}+\sum\limits_{l=0}^{t}{{{a}_{l}}}\left( z \right)w_{1}^{l},t\le p-2, \\ \left\{ \begin{array}{*{35}{l}} {{\left[ {{\Omega }_{1}}\left( z,{{w}_{1}},{{w}_{2}} \right) \right]}^{{{m}_{2}}}}={{b}_{q}}{{\left[ {{w}_{2}}+d\left( z \right) \right]}^{q}}. \\ \end{array} \right. \\ \end{matrix} \right.$$ (3.1)

类似引理2.4的证明,易证

$\Big(1-\frac{p}{m_1}\Big)T(r,w_1)\leq S(r)+S(r,w_2).$ (3.2)

由(3.1)式的第二个方程得

$ qT(r,w_2)\geq m_2\overline{\sigma}_2T(r,w_2)+S(r,w_1)+S(r,w_2). $

$ (q-m_2\overline{\sigma}_2+o(1))T(r,w_2)\geq S(r,w_1).$ (3.3)

由(3.2),(3.3)式得

$ \Big(1-\frac{p}{m_1}\Big)\frac{T(r,w_1)}{T(r,w_2)}\leq \frac{S(r,w_2)}{T(r,w_2)}.$ (3.4)
$ (m_2\overline{\sigma}_2-q+o(1))\frac{T(r,w_2)}{T(r,w_1)} \leq \frac{S(r,w_1)}{T(r,w_1)}.$ (3.5)

(3.4),(3.5)式分别取上极限,又因为$(w_1,\,w_2)$是允许解,则有

$ \lim\sup\limits_{r\rightarrow \infty}\frac{T(r,w_1)}{T(r,w_2)}=0,$ (3.6)
$ \lim\sup\limits_{r\rightarrow \infty}\frac{T(r,w_2)}{T(r,w_1)}=0.$ (3.7)

由(3.6),(3.7)式得

$\begin{align} & 1=\lim \underset{r\to \infty }{\mathop{\sup }}\,\frac{T(r,{{w}_{1}})}{T(r,{{w}_{2}})}\frac{T(r,{{w}_{2}})}{T(r,{{w}_{1}})} \\ & \le \lim \underset{r\to \infty }{\mathop{\sup }}\,\frac{T(r,{{w}_{1}})}{T(r,{{w}_{2}})}\lim \underset{r\to \infty }{\mathop{\sup }}\,\frac{T(r,{{w}_{2}})}{T(r,{{w}_{1}})}=0. \\ \end{align}$

$ 1\leq 0,$

矛盾.

同理可证不具有形式(2.3).

定理1.1证毕.

4 几个例子

例4.1表明定理1.1成立.

例 4.1 $(w_1(z),w_2(z))=({\rm e}^{2z-2}+z-1,{\rm e}^{-3z-3}+z+1)$是如下复差分方程组的有限级允许解

$\left\{ \begin{matrix} {{\left[ {{\Omega }_{1}}\left( z,{{w}_{1}},{{w}_{2}} \right) \right]}^{2}}={{\text{e}}^{2}}\left[ {{w}_{1}}\left( z \right)-\left( z-1 \right) \right], \\ {{\left[ {{\Omega }_{2}}\left( z,{{w}_{1}},{{w}_{2}} \right) \right]}^{3}}={{\text{e}}^{-27}}\left[ {{w}_{2}}\left( z \right)-\left( z-1 \right) \right], \\ \end{matrix} \right.$

其中

$\begin{align} & {{\Omega }_{1}}\left( z,{{w}_{1}},{{w}_{2}} \right)=w_{1}^{2}\left( z+1 \right){{w}_{2}}\left( z-1 \right)-2z{{w}_{1}}\left( z+1 \right){{w}_{2}}\left( z-1 \right) \\ & +{{z}^{2}}{{w}_{2}}\left( z-1 \right)-zw_{1}^{2}\left( z+1 \right)+2{{z}^{2}}{{w}_{1}}\left( z+1 \right)-{{z}^{3}}, \\ & {{\Omega }_{2}}\left( z,{{w}_{1}},{{w}_{2}} \right)={{w}_{1}}\left( z-1 \right){{w}_{2}}\left( z+1 \right)-\left( z+2 \right){{w}_{1}}\left( z-1 \right) \\ & -\left( z-2 \right){{w}_{2}}\left( z+1 \right)+{{z}^{2}}-4. \\ \end{align}$

此时

$ p=1,\,m_1=2,\,q=1,\,m_2=3. $

例4.2表明定理1.1中“允许解”的条件不能去掉.

例 4.2 $(w_1(z),w_2(z))=({\rm e}^z+z,z^2)$是如下复差分方程组的有限级非允许解

$\left\{ \begin{matrix} {{\Omega }_{1}}(z,{{w}_{1}},{{w}_{2}})=\frac{{{z}^{2}}+2z+1}{\text{e}}[{{w}_{1}}(z)-z], \\ {{[{{\Omega }_{2}}(z,{{w}_{1}},{{w}_{2}})]}^{2}}={{P}_{3}}({{w}_{2}}), \\ \end{matrix} \right.$

其中

$ \Omega_1 (z,w_1,w_2)=w_1(z-1 )w_2(z+1 ) -(z-1 )w_2(z+1 ),$
$\begin{align} & {{\Omega }_{2}}(z,{{w}_{1}},{{w}_{2}})={{w}_{1}}(z-1){{w}_{2}}(z+1)-\frac{1}{{{\text{e}}^{2}}}{{w}_{1}}(z+1){{w}_{2}}(z+1) \\ & +\frac{z+1-{{\text{e}}^{2}}(z-1)}{{{\text{e}}^{2}}}{{w}_{2}}(z+1)+w_{2}^{2}(z+1) \\ & -{{w}_{1}}(z-1){{w}_{2}}(z-1)+\frac{1}{{{\text{e}}^{2}}}{{w}_{1}}(z+1){{w}_{2}}(z-1) \\ & -\frac{z+1-{{\text{e}}^{2}}(z-1)}{{{\text{e}}^{2}}}{{w}_{2}}(z-1)-{{w}_{2}}(z+1){{w}_{2}}(z-1), \\ \end{align}$

$P_3(w_2)$是$w_2$的多项式,且$\deg_{w_2} P_3(w_2) =3$.

此时

$ p=m_1=1,\,\,q=3>2=m_2. $
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