数学物理的特殊函数,即经典正交多项式和超几何及圆柱函数,是超几何型差分方程的解.

设$ \sigma (x) $和$ \tau (x) $分别为至多二次和一次多项式,且$ \lambda$为常数.以下二阶微分方程

(1.1) $ \begin{equation} \sigma (x)y^{\prime \prime }(x)+\tau (x)y^{\prime }(x)+\lambda y(x) = 0, \end{equation} $

称为超几何型微分方程.如果对于正整数$ n $,有

$ \lambda = \lambda _{n}: = -\frac{n(n-1)\sigma ^{\prime \prime }}{2}-n\tau ^{\prime }, \; \mbox{且}\; \lambda _{m}\neq \lambda _{n}\; \mbox{对}\; m = 0, 1, \cdots , n-1, $

方程(1.1)有一个次数为$ n $阶的多项式解$ y_{n}(x) $,它用Rodrigues公式表示^[1-11]

$ y_{n}(x) = \frac{1}{\rho (x)}\frac{{\rm d}^{n}}{{\rm d}x^{n}}(\rho (x)\sigma ^{n}(x)), $

这里$ \rho (x) $满足Pearson方程

$ (\sigma (x)\rho (x))^{\prime } = \tau (x)\rho (x). $

这些解在量子力学,群表示理论和计算数学上是非常有用的.基于此,经典超几何型方程理论被著名数学家如Andrews, Askey^[12-13], Wilson, Ismail^[14-17]; Nikiforov, Suslov, Uvarov, Atakishiyev^{[8-10, 18-20]}; George, Rahman^[21]; Koornwinder^[22]等大大向前发展;以及其他许多研究者如Álvarez-Nodarse, Cardoso, Area, Godoy, Ronveaux, Zarzo, Robin, Dreyfus, Kac, Cheung, Jia, Cheng, Feng等^[23-32].

如果已知方程(1.1)的一个多项式解,就可以用许多方式建立该方程线性独立的解:例如常数变易法^[1],用Cauchy积分表示公式^{[4, 9]}.然而, Area等在文献[23]中首先给出了方程(1.1)第二类型解的推广的Rodrigues型表示公式,其表达式是

$ y_{n}(x) = \frac{C_{1}}{\rho (x)}\frac{{\rm d}^{n}}{{\rm d}x^{n}}\left( \rho (x)\sigma ^{n}(x)\right) +\frac{C_{2}}{\rho (x)}\frac{{\rm d}^{n}}{{\rm d}x^{n}}\left( \rho (x)\sigma ^{n}(x)\int \frac{{\rm d}x}{\rho (x)\sigma ^{n+1}(x)}\right) , $

这里$ C_{1}, C_{2} $是任意常数.

最近,受文献[3]启发, Robin^[24]给出了更一般的Rodrigues公式

$ y_{n}(x) = \frac{1}{\rho (x)}\frac{{\rm d}^{n}}{{\rm d}x^{n}}\left[ \rho (x)\sigma ^{n}(x)\left( \int \frac{P_{n}(x)+D_{n}}{\rho (x)\sigma ^{n+1}(x)}{\rm d}x+C_{n}\right) \right] , $

这里$ P_{n}(x) $是一个任意的$ n $阶多项式,且$ C_{n}, D_{n} $是一个任意常数.

1983年以来,超几何型微分方程经典理论已经被Nikiforov, Suslov和Uvarov^[8-10]等数学家极大地向前推进,他们是从以下推广开始的.他们用变步长的非一致格子$ \nabla x(s) = x(s)-x(s-1) $上的差分方程替换方程(1.1)

(1.2) $ \begin{equation} \widetilde{\sigma }[x(s)]\frac{\Delta }{\Delta x(s-1/2)}\left[ \frac{\nabla y(s)}{\nabla x(s)}\right] +\frac{1}{2}\widetilde{\tau }[x(s)]\left[ \frac{\Delta y(s)}{\Delta x(s)}+\frac{\nabla y(s)}{\nabla x(s)}\right] +\lambda y(s) = 0, \end{equation} $

这里$ \widetilde{\sigma }(x) $和$ \widetilde{\tau }(x) $分别是关于$ x(s) $的至多二阶和一阶多项式, $ \lambda$是一个常数,

$ \Delta y(s) = y(s+1)-y(s), \quad \nabla y(s) = y(s)-y(s-1), $

$ x(s) $满足

(1.3) $ \begin{equation} \frac{x(s+1)+x(s)}{2} = \alpha x(s+\frac{1}{2})+\beta\end{equation} $

(这里$ \alpha , \beta$是常数)且

(1.4) $ \begin{equation} \mbox{ $x^{2}(s+1)+x^{2}(s)$是关于$x(s+\frac{1}{2})$的至多二阶多项式.}\end{equation} $

在非一致格子上,通过逼近微分方程(1.1)得到的差分方程(1.2)具有独立的重要意义,并且引出其它许多重要的问题.它的解实质上拓广了原超几何微分方程的解,且其本身有重要独立意义.它的一些解在量子力学、群论和计算数学中已经被长期使用了.有关非一致格子上超几何型差分方程(1.2)的更多信息,读者可以查阅相关文献如Koekoek, Lesky, Swarttouw^[7], Nikiforov, Suslov, Uvarov, Atakishiyev, Rahman^{[8-10, 18-20]}, Magnus^[33], Foupouagnigni^[34-35], Witte^[36].

定义 1.1    两类格子函数$ x(s) $被称为非一致格子如果它们满足条件(1.3)和(1.4)

(1.5) $ \begin{equation}x(s) = c_1q^s+c_2q^{-s}+c_3, \end{equation} $

(1.6) $ \begin{equation}x(s) = \widetilde{c}_1s^2+\widetilde{c}_2s+\widetilde{c}_3, \end{equation} $

这里$ c_i, \widetilde{c}_i $是任意常数且$ c_1c_2\neq0 $, $ \widetilde{c}_1\widetilde{c}_2\neq0 $.

对于方程(1.2)中$ \lambda $的某些值,通过Rodrigues公式的差分模拟,也有一个关于$ x(s) $的多项式解.自然地,人们可以问,在非一致格子差分方程中是否有Rodrigues公式的推广.据我们所知,在一致格子如$ x(s) = s $和$ x(s) = q^s $的情形,文献[28]已经做出有效的推广.然而,在非一致格子(1.5)或(1.6)的情况下,这将是十分复杂和困难的,因此在这种情况下,包括文献[28]在内,自2005年以来没有出现过任何该方面的相关进展和结果.

本文的目的,将在非一致格子(1.5)和(1.6)情形下,给出Rodrigues公式的两个推广.本论安排组织如下.第2节中介绍了非一致格子上的差分与和分基本概念、性质.在第3节和第4节,回顾非一致格子上的经典Rodrigues公式,并给出了一些必需的记号和一些引理.第5节,我们在非一致格子上,建立并化简一个二阶超几何超几何差分方程的伴随差分方程,这个新结果也有独立的重要意义.利用伴随差分方程,第6节,我们在定理6.2中给出了罗德里格斯公式的一种拓展.在第7节中,用不同于第6节的方法,建立另一个更一般的罗德里格斯公式定理7.1.

设$ x(s) $是一个格子,这里$ s\in{\Bbb C} $.对任何整数$ k $, $x_k(s) = x(s+\frac{k}{2}) $也是一个格子.给定函数$ f(s) $,定义关于$ x_k(s) $的两种差分算子如下

(2.1) $ \begin{eqnarray}\Delta_k f(s) = \frac{\Delta f(s)}{\Delta x_k(s)}, \qquad \nabla_kf(s) = \frac{\nabla f(s)}{\nabla x_k(s)}. \end{eqnarray} $

进一步,对任何非负整数$ n $,让

$ \Delta_k^{(n)}f(s) = \left\{ \begin{array}{ll} f(s), & \mbox{ $n = 0$, } \\ \frac{\Delta}{\Delta x_{k+n-1}(s)}\cdots\frac{\Delta}{\Delta x_{k+1}(s)}\frac{\Delta}{\Delta x_{k}(s)}f(s), \quad &\mbox{ $n\geq1$}. \end{array} \right.$

$ \nabla_k^{(n)}f(s) = \left\{ \begin{array}{ll} f(s), & \mbox{ $n = 0$, } \\ \frac{\nabla}{\nabla x_{k-n+1}}\cdots\frac{\nabla}{\nabla x_{k-1}(s)}\frac{\nabla}{\nabla x_k(s)}f(s), \quad & \mbox{ $n\geq1$}. \end{array} \right. $

下面这些性质容易验证.

命题 2.1    给定具有复变量$ s $的两个函数$ f(s), g(s) $,则有

$ \Delta_k(f(s)g(s)) = f(s+1)\Delta_k g(s)+g(s)\Delta_k f(s) = g(s+1)\Delta_k f(s)+f(s)\Delta_k g(s), $

$ \Delta_k\left(\frac{f(s)}{g(s)}\right) = \frac{g(s+1)\Delta_k f(s)-f(s+1)\Delta_k g(s)}{g(s)g(s+1)} = \frac{g(s)\Delta_k f(s)-f(s)\Delta_k g(s)}{g(s)g(s+1)}, $

$ \nabla_k(f(s)g(s)) = f(s-1)\nabla_k g(s)+g(s)\nabla_k f(s) = g(s-1)\nabla_k f(s)+f(s)\nabla_k g(s), $

$ \nabla_k\left(\frac{f(s)}{g(s)}\right) = \frac{g(s-1)\nabla_k f(s)-f(s-1)\nabla_k g(s)}{g(s)g(s-1)} = \frac{g(s)\nabla_k f(s)-f(s)\nabla_k g(s)}{g(s)g(s-1)}. $

为了处理逆算子$ \nabla_k $,它是一种和分,延续Bangerezako在文献[29]的记号,令$ \nabla_kf(t) = g(t). $那么

$ f(t)-f(t-1) = g(t)\left[x_k(t)-x_k(t-1)\right]. $

选取$ N, s\in{\Bbb C} $.从$ t = N $到$ t = s $相加,有

$ f(s)-f(N-1) = \sum\limits_{t = N}^{t = s}g(t)\nabla x_k(t). $

因此,我们定义

(2.2) $ \begin{eqnarray} \int_{N}^{s}g(t){\rm d}_\nabla x_k(t) = \sum\limits_{t = N}^{t = s}g(t)\nabla x_k(t). \end{eqnarray} $

容易验证

命题 2.2    给定具有复变量$ N, s $的两个函数$f(s), g(s) $,则有

$ \begin{eqnarray*} (1)\quad&&\nabla_k\left[\int_{N}^{s}g(t){\rm d}_\nabla x_k(t)\right] = g(s), \\ (2)\quad&&\int_{N}^{s}\nabla_kf(t){\rm d}_\nabla x_k(t) = f(s)-f(N). \end{eqnarray*} $

运用第二节中的记号,超几何型(1.2)的差分方程可以写成

(3.1) $ \begin{equation}\widetilde{\sigma}[x(s)]\Delta_{-1}\nabla_0y(s)+\frac{\widetilde{\tau}[x(s)]}{2}\left[\Delta_0y(s)+\nabla_0y(s)\right]+\lambda y(s) = 0. \end{equation} $

在下文中,我们假设格子$ x(s) $有(1.5)式和(1.6)式两种形式.

让

$ z_{k}(s) = \Delta _{0}^{(k)}y(s) = \Delta _{k-1}\Delta _{k-2}\cdots \Delta _{0}y(s). $

那么对非负整数$ k $, $ z_{k}(s) $满足与方程(3.1)同型的方程^[10]

(3.2) $ \begin{equation} \widetilde{\sigma }_{k}[x_{k}(s)]\Delta _{k-1}\nabla _{k}z_{k}(s)+\frac{\widetilde{\tau }_{k}[x_{k}(s)]}{2}\left[ \Delta _{k}z_{k}(s)+\nabla _{k}z_{k}(s)\right] +\mu _{k}z_{k}(s) = 0, \end{equation} $

这里$ \widetilde{\sigma }_{k}(x_{k}) $和$ \widetilde{\tau }_{k}(x_{k}) $分别是关于$ x_{k} $的至多二阶和一阶多项式, $ \mu _{k} $是一常数,且

$ \begin{eqnarray*} \widetilde{\sigma }_{k}[x_{k}(s)] & = &\frac{\widetilde{\sigma }_{k-1}[x_{k-1}(s+1)]+\widetilde{\sigma }_{k-1}[x_{k-1}(s)]}{2} \\ &&+\frac{1}{4}\Delta _{k-1}\widetilde{\tau }_{k-1}(s)\frac{\Delta x_{k}(s)+\nabla x_{k}(s)}{2\Delta x_{k-1}(s)}[\Delta x_{k-1}(s)]^{2} \\ &&+\frac{\widetilde{\tau }_{k-1}[x_{k-1}(s+1)]+\widetilde{\tau }_{k-1}[x_{k-1}(s)]}{2}\frac{\Delta x_{k}(s)-\nabla x_{k}(s)}{4}, \\ \widetilde{\sigma }_{0}[x_{0}(s)] & = &\widetilde{\sigma }[x(s)]; \end{eqnarray*} $

$ \begin{eqnarray*} \widetilde{\tau }_{k}[x_{k}(s)] & = &\Delta _{k-1}\widetilde{\sigma }_{k-1}\left[ x_{k-1}(s)\right] +\Delta _{k-1}\widetilde{\tau }_{k-1}\left[ x_{k-1}(s)\right] \frac{\Delta x_{k}(s)-\nabla x_{k}(s)}{4} \\ &&+\frac{\widetilde{\tau }_{k-1}[x_{k-1}(s+1)]+\widetilde{\tau }_{k-1}[x_{k-1}(s)]}{2}\frac{\Delta x_{k}(s)+\nabla x_{k}(s)}{2\Delta x_{k-1}(s)}, \\ \widetilde{\tau }_{0}[x_{0}(s)] & = &\widetilde{\tau }[x(s)]; \end{eqnarray*} $

$ \mu _{k} = \mu _{k-1}+\Delta _{k-1}\widetilde{\tau }_{k-1}\left[ x_{k-1}(s)\right] , \; \mu _{0} = \lambda . $

为了研究方程(3.2)解的更多性质,下面的方程是有用的

$ \begin{eqnarray*} \frac{1}{2}\left[ \Delta _{k}z_{k}(s)+\nabla _{k}z_{k}(s)\right] = \Delta _{k}z_{k}(s)-\frac{1}{2}\Delta \left[ \nabla _{k}z_{k}(s)\right]. \end{eqnarray*} $

将方程(3.2)改写为等价形式

(3.3) $ \begin{equation} \sigma _{k}(s)\Delta _{k-1}\nabla _{k}z_{k}(s)+\tau _{k}(s)\Delta _{k}z_{k}(s)+\mu _{k}z_{k}(s) = 0, \end{equation} $

这里

(3.4) $ \begin{eqnarray} && \sigma _{k}(s) = \widetilde{\sigma }_{k}[x_{k}(s)]-\frac{1}{2}\widetilde{\tau }_{k}[x_{k}(s)]\nabla x_{k+1}(s), \end{eqnarray} $

(3.5) $ \begin{eqnarray} && \tau _{k}(s) = \widetilde{\tau }_{k}[x_{k}(s)].\end{eqnarray} $

我们发现

(3.6) $ \begin{eqnarray} & &\tau _{k}(s) = \frac{\sigma (s+k)-\sigma (s)+\tau (s+k)\nabla x_{1}(s+k)}{\nabla x_{k+1}(s)}, \end{eqnarray} $

(3.7) $ \begin{eqnarray} && \mu _{k} = \lambda +\sum\limits_{j = 0}^{k-1}\Delta _{j}\tau _{j}(s).\end{eqnarray} $

注 3.1    当$ k $是一负整数时,我们也记等式(3.6)右边为$ \tau_k $.

将方程(3.3)改写成自相伴形式

$ \Delta _{k-1}\left[ \sigma _{k}(s)\rho _{k}(s)\nabla _{k}z_{k}(s)\right] +\mu _{k}\rho _{k}(s)z_{k}(s) = 0, $

这里$ \rho _{k}(s) $满足Pearson型差分方程

$ \Delta _{k-1}\left[ \sigma _{k}(s)\rho _{k}(s)\right] = \tau _{k}(s)\rho _{k}(s). $

让$ \rho (s) = \rho _{0}(s) $,我们发现

$ \rho _{k}(s) = \rho (s+k)\prod\limits_{i = 1}^{k}\sigma (s+i). $

如果对正整数$ n $,有

(3.8) $ \begin{equation} \lambda = \lambda _{n}: = -\sum\limits_{j = 0}^{n-1}\Delta _{j}\tau _{j}(s), \; \mbox{且}\; \lambda _{m}\neq \lambda _{n}\; \mbox{对}\; m = 0, 1, \cdots , n-1, \end{equation} $

那么方程(3.1)有一个关于$ x(s) $的$ n $阶多项式解$ y_{n}[x(s)] $,它可以表示成Rodriguez型公式的差分模拟^[8-10]

$ y_{n}[x(s)] = \frac{1}{\rho (s)}\nabla _{n}^{(n)}\left[ \rho _{n}(s)\right] = \frac{1}{\rho (s)}\Delta _{-n}^{(n)}\left[ \rho _{n}(s-n)\right] . $

现在,我们在非一致格子(1.5)和(1.6)下,分别给出$ \tau _k (s), \mu _k $和$ \lambda_n $的显示表达式.

命题 4.1    给定任意整数$ k $,如果$ x(s) = c_{1}q^{s}+c_{2}q^{-s}+c_{3} $,那么

$ \begin{eqnarray*} \tau _{k}(s)& = &\left[ \frac{q^{k}-q^{-k}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}\frac{\widetilde{\sigma }^{\prime \prime }}{2}+\left( q^{k}+q^{-k}\right) \frac{\widetilde{\tau }^{\prime }}{2}\right] x_{k}(s)+c(k) \\ & = &[\nu (2k)\frac{{\tilde{\sigma}^{\prime \prime }}}{2}+\alpha (2k)\tilde{\tau}^{\prime }]x_{k}(s)+c(k) = \kappa _{2k+1}x_{k}(s)+c(k), \end{eqnarray*} $

这里

$ \nu (\mu ) = \left\{{ \begin{array}{ll}{\frac{{q^{\frac{\mu }{2}} - q^{ - \frac{\mu }{2}} }}{{q^{\frac{1}{2}} - q^{ - \frac{1}{2}} }}} \\\mu\end{array}}\right. , \qquad \alpha (\mu ) = \left\{ {\begin{array}{ll} {\frac{{q^{\frac{\mu }{2}} + q^{ - \frac{\mu }{2}} }}{2}} \\1\end{array}} \right. , $

且

$ \kappa _{\mu } = \alpha (\mu -1)\tilde{\tau}^{\prime }+\nu (\mu -1)\frac{{\tilde{\sigma}^{\prime \prime }}}{2}. $

如果$ x(s) = \widetilde{c}_{1}s^{2}+\widetilde{c}_{2}s+\widetilde{c}_{3} $,那么

$ \tau _{k}(s) = \left[ k\widetilde{\sigma }^{\prime \prime }+\widetilde{\tau }^{\prime }\right] x_{k}(s)+\widetilde{c}(k) = \kappa _{2k+1}x_{k}(s)+\widetilde{c}(k), $

这里$ c(k), \widetilde{c}(k) $是关于$ k $的函数

$ \begin{eqnarray*} c(k) & = &c_{3}(1-q^{\frac{k}{2}})(q^{\frac{k}{2}}-q^{-k})+c_{3}\frac{(2-q^{\frac{k}{2}}-q^{-\frac{k}{2}})(q^{\frac{k}{2}}-q^{-\frac{k}{2}})}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}} \\ &&+\widetilde{\tau }(0)(q^{\frac{k}{2}}+q^{-\frac{k}{2}})+\widetilde{\sigma }(0)\frac{q^{\frac{k}{2}}-q^{-\frac{k}{2}}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}, \end{eqnarray*} $

$ \widetilde{c}(k) = \frac{\widetilde{\sigma }\prime \prime }{4}\widetilde{c}_{1}k^{3}+\frac{3\widetilde{\tau }\prime }{4}\widetilde{c}_{1}k^{2}+\widetilde{\sigma }(0)k+2\widetilde{\tau }(0). $

证    我们仅仅证明$ x(s) = c_{1}q^{s}+c_{2}q^{-s}+c_{3} $的情形.由(3.4), (3.5)和(3.6)式,我们有

(4.1) $ \begin{equation} \tau _{k}(s) = \frac{\widetilde{\sigma }[x(s+k)]-\widetilde{\sigma }[x(s)]+\frac{1}{2}\widetilde{\tau }[x(s+k)]\Delta x(s+k-\frac{1}{2})+\frac{1}{2}\widetilde{\tau }[x(s)]\Delta x(s-\frac{1}{2})}{\Delta x_{k-1}(s)}. \end{equation} $

一些简单的计算后,我们得到

$x(s+k)-x(s) = (q^{\frac{k}{2}}-q^{-\frac{k}{2}})(c_{1}q^{s+\frac{k}{2}}-c_{2}q^{-s-\frac{k}{2}}), $

$x(s+k)+x(s) = (q^{\frac{k}{2}}+q^{-\frac{k}{2}})x_{k}(s)+c_{3}(2-q^{\frac{k}{2}}-q^{-\frac{k}{2}}), $

$\Delta x_{k-1}(s) = (q^{\frac{1}{2}}-q^{-\frac{1}{2}})(c_{1}q^{s+\frac{k}{2}}-c_{2}q^{-s-\frac{k}{2}}). $

进一步, $ \widetilde{\sigma }[x(s)] = \frac{\widetilde{\sigma }^{\prime \prime }}{2}x^{2}(s)+\widetilde{\sigma }^{\prime }(0)x(s)+\widetilde{\sigma }(0). $那么

(4.2) $ \begin{eqnarray} \frac{\widetilde{\sigma }[x(s+k)]-\widetilde{\sigma }[x(s)]}{\Delta x_{k-1}(s)}& = &\frac{\widetilde{\sigma }^{\prime \prime }}{2}\frac{x^{2}(s+k)-x^{2}(s)}{\Delta x_{k-1}(s)}+\widetilde{\sigma }^{\prime }(0)\frac{x(s+k)-x(s)}{\Delta x_{k-1}(s)} \\ & = &\frac{\widetilde{\sigma }^{\prime \prime }}{2}\frac{q^{k}-q^{-k}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}x_{k}(s)+\widetilde{\sigma }^{\prime }(0)\frac{q^{\frac{k}{2}}-q^{-\frac{k}{2}}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}\\ &&+c_{3} \frac{(2-q^{\frac{k}{2}}-q^{-\frac{k}{2}})(q^{\frac{k}{2}}-q^{-\frac{k}{2}})}{q^{\frac{k}{2}}-q^{-\frac{k}{2}}}.\end{eqnarray} $

进一步,有

(4.3) $ \begin{eqnarray} &&x(s+k)\Delta x(s+k-\frac{1}{2})+x(s)\Delta x(s-\frac{1}{2}) \\ & = &(q^{k}+q^{-k})(q^{\frac{1}{2}}-q^{-\frac{1}{2}})(c_{1}q^{s+\frac{k}{2}}-c_{2}q^{-s-\frac{k}{2}})x_{k}(s) \\ &&+c_{3}(q^{\frac{1}{2}}-q^{-\frac{1}{2}})(c_{1}q^{s+\frac{k}{2}}-c_{2}q^{-s-\frac{k}{2}})(1-q^{\frac{k}{2}})(q^{\frac{k}{2}}-q^{-k}), \end{eqnarray} $

(4.4) $ \begin{equation} \Delta x(s+k-\frac{1}{2})+\Delta x(s-\frac{1}{2}) = (q^{\frac{1}{2}}-q^{-\frac{1}{2}})(q^{\frac{k}{2}}+q^{-\frac{k}{2}})(c_{1}q^{s+\frac{k}{2}}-c_{2}q^{-s- \frac{k}{2}}), \end{equation} $

且$ \tau \lbrack x(s)] = \widetilde{\tau }^{\prime }x(s)+\widetilde{\tau } (0).$因此

(4.5) $ \begin{eqnarray} && \frac{1}{2}\frac{\widetilde{\tau }[x(s+k)]\Delta x(s+k-\frac{1}{2})+ \widetilde{\tau }[x(s)]\Delta x(s-\frac{1}{2})}{\Delta x_{k-1}(s)} \\ & = &\frac{\widetilde{\tau }^{\prime }}{2}\frac{x(s+k)\Delta x(s+k-\frac{1}{2} )+x(s)\Delta x(s-\frac{1}{2})}{\Delta x_{k-1}(s)}+\frac{\widetilde{\tau }(0) }{2}\frac{\Delta x(s+k-\frac{1}{2})+\Delta x(s-\frac{1}{2})}{\Delta x_{k-1}(s)} \\ & = &\frac{\widetilde{\tau }^{\prime }}{2}(q^{k}+q^{-k})x_{k}(s)+c_{3}(1-q^{ \frac{k}{2}})(q^{\frac{k}{2}}-q^{-k})+\widetilde{\tau }(0)(q^{\frac{k}{2} }+q^{-\frac{k}{2}}).\end{eqnarray} $

将(4.2)和(4.5)式代入(3.5)式,就可得结论.

引理 4.1^[18]    对$ \alpha (\mu ), \nu (\mu ) $,我们有

$ \sum\limits_{j = 0}^{k-1}{\alpha (2j)} = \alpha (k-1)\nu (k), \qquad \sum\limits_{j = 0}^{k-1}{\nu (2j)} = \nu (k-1)\nu (k). $

从(3.7)式, (3.8)式和引理4.1,我们有

命题 4.2    若$ x(s) = c_1q^s+c_2q^{-s}+c_3 $或$ x(s) = \widetilde{c}_1s^2+\widetilde{c}_2s+\widetilde{c}_3 $,那么

(4.6) $ \begin{eqnarray} \mu_k = \lambda+\kappa _k \nu (k), \end{eqnarray} $

这里$\kappa _k = \alpha (k - 1)\tilde \tau ^{\prime }+ \frac{1}{2}\nu (k - 1)\tilde \sigma ^{\prime \prime }.$

证    如果$ x(s) = c_1q^s+c_2q^{-s}+c_3 $,那么

$ \begin{eqnarray*} \mu_k& = &\lambda+\sum\limits_{j = 0}^{k-1}\left[\frac{q^{j}-q^{-j}}{q^\frac{1}{2}+q^{-\frac{1}{2}}}\frac{\widetilde{\sigma}^{\prime \prime }}{2}-(q^j+q^{-j})\frac{\widetilde{\tau}^{\prime }}{2}\right] \\ & = & \lambda+\sum\limits_{j = 0}^{k - 1} {\nu (2j)} \frac{{\tilde \sigma ^{\prime \prime }}}{2} + \sum\limits_{j = 0}^{k - 1} {\alpha (2j)} \tilde \tau ^{\prime } \\ & = & \lambda+\nu (k - 1)\nu (k)\frac{{\tilde \sigma ^{\prime \prime }}}{2} + \alpha (k - 1)\nu (k)\tilde \tau ^{\prime } \\ & = & \lambda+\kappa _k \nu (k), \end{eqnarray*} $

这里

(4.7) $ \begin{equation} \kappa _k = \alpha (k - 1)\tilde \tau ^{\prime }+ \frac{1}{2}\nu (k - 1)\tilde \sigma ^{\prime \prime }. \end{equation} $

如果$ x(s) = \widetilde{c}_1s^2+\widetilde{c}_2s+\widetilde{c}_3 $,那么

$ \mu_k = \lambda+\sum\limits_{j = 0}^{k-1}\left[j\widetilde{\sigma}^{\prime \prime }+ \widetilde{\tau}^{\prime }\right] = \lambda+\frac{{(k - 1)k}}{2}\tilde \sigma ^{\prime \prime }+ k\tilde \tau ^{\prime } = \lambda+\kappa _k \nu (k). $

证毕.

由命题4.2和(3.8)式,我们有

(4.8) $ \begin{equation} \lambda _n = - n\kappa _n . \end{equation} $

让

(5.1) $ \begin{equation}L[y] = \sigma(s)\Delta_{-1}\nabla_0 y(s)+\tau(s)\Delta_0y(s)+\lambda y(s) = 0. \end{equation} $

那么方程(5.1)有自相伴形式

(5.2) $ \begin{equation}\Delta_{-1}\left[\sigma(s)\rho(s)\nabla_0y(s)\right]+\lambda\rho(s)y(s) = 0, \end{equation} $

这里$ \rho(s) $满足Pearson型方程

(5.3) $ \begin{equation}\Delta_{-1}\left[\sigma(s)\rho(s)\right] = \tau(s)\rho(s). \end{equation} $

为了得到非一致格子上Rodriguez型公式的推广,如何定义和建立非一致格子上超几何差分方程的伴随方程至关重要.

让$ w(s) = \rho(s)y(s) $.那么

(5.4) $ \begin{equation}\nabla_0y(s) = \nabla_0\frac{w(s)}{\rho(s)} = \frac{\rho(s-1)\nabla_0w(s)-w(s-1)\nabla_0\rho(s)}{\rho(s)\rho(s-1)}. \end{equation} $

将(5.4)式代入(5.2)式,我们得

(5.5) $ \begin{equation}\Delta_{-1}\left[\sigma(s)(\nabla_0w(s)-w(s-1)\frac{\nabla_0\rho(s)}{\rho(s-1)})\right]+\lambda w(s) = 0. \end{equation} $

由Pearson型方程(5.3),有

$ \frac{\Delta[\sigma(s)\rho(s)]}{\Delta x_{-1}(s)} = \frac{\sigma(s+1)\Delta\rho(s)+\Delta\sigma(s)\rho(s)}{\Delta x_{-1}(s)} = \tau(s)\rho(s). $

那么

(5.6) $ \begin{equation}\frac{\nabla\rho(s)}{\rho(s-1)} = \frac{\tau(s-1)\nabla_{-1}(s)-\nabla\sigma(s)}{\sigma(s)}. \end{equation} $

将(5.6)式代入(5.5)式,可得

(5.7) $ \begin{equation}L^*[w]: = \sigma^*(s)\Delta_{-1}\nabla_0 w(s)+\tau^*(s)\Delta_0 w(s)+\lambda^*w(s) = 0, \end{equation} $

这里

(5.8) $ \begin{equation} \sigma^*(s) = \sigma(s-1)+\tau(s-1)\nabla x_{-1}(s), \end{equation} $

(5.9) $ \begin{equation} \tau^*(s) = \frac{\sigma(s+1)-\sigma(s-1)}{\Delta x_{-1}(s)}-\tau(s-1)\frac{\nabla x_{-1}(s)}{\Delta x_{-1}(s)}, \end{equation} $

(5.10) $ \begin{equation} \lambda^* = \lambda-\Delta_{-1}\left(\tau(s-1)\frac{\nabla x_{-1}(s)}{\nabla x(s)}-\frac{\nabla\sigma(s)}{\nabla x(s)}\right).\end{equation} $

定义 5.1    方程(5.7)被称为对应于方程(5.1)的相伴方程.

由定义5.1,容易得到

命题 5.1    对$ y(s) $,我们有

(5.11) $ \begin{equation}L^*[\rho y] = \rho L[y]. \end{equation} $

引理 5.1^[18]    让$ x = x(s) $是一个满足(1.5)和(1.6)式的非一致格子,那么由下面等式所定义的函数$ \tilde{\sigma} _{\nu }(s) $和$ \tau _{\nu }(s) $

(5.12) $ \begin{equation} \tilde{\sigma}_{\nu }(s) = \sigma (s)+\frac{1}{2}\tau _{\nu }(s)\nabla x_{\nu +1}(s), \end{equation} $

(5.13) $ \begin{equation} \tau _{\nu }(s)\nabla x_{\nu +1}(s) = \sigma (s+\nu )-\sigma (s)+\tau (s+\nu )\nabla x_{1}(s+\nu ), \end{equation} $

分别是关于变量$ x_{\nu } = x(s+\frac{\nu }{2}), \nu \in {\Bbb R} $的至多二阶或一阶多项式.

利用命题4.1和引理5.1,不难得到

推论 5.1    对(5.9)和(5.10)式,我们有

(5.14) $ \begin{equation}\tau ^* (s) = - \tau _{ - 2} (s + 1) = [\gamma (4)\frac{{\tilde \sigma ^{\prime \prime }}}{2} - \alpha (4)\tilde \tau ^{\prime }]x_0 (s) + c( - 2), \end{equation} $

且

(5.15) $ \begin{equation}\lambda ^* = \lambda - \Delta _{ - 1} \tau _{ - 1} (s) = \lambda - \gamma (2)\frac{{\tilde \sigma ^{\prime \prime }}}{2} - \alpha (2)\tilde \tau^{\prime } = \lambda - \kappa _{ - 1} . \end{equation} $

证    由于

$ \sigma (s - 1) - \sigma (s + 1) + \tau (s - 1)\nabla x_{ - 1} (s) = \sigma (s - 1) - \sigma (s + 1) + \tau (s - 1)\nabla x_1 (s - 1). $

置$ s+1 = z $,则由(5.13)式,我们有

$ \begin{eqnarray*} &&\sigma (s - 1) - \sigma (s + 1) + \tau (s - 1)\nabla x_1 (s - 1)\\ & = & \sigma (z - 2) - \sigma (z) + \tau (z - 2)\nabla x_1 (z - 2) \\ & = & \tau _{ - 2} (z)\nabla x_{ - 1} (z) = \tau _{ - 2} (s + 1)\nabla x_{ - 1} (s + 1)\\ & = & \tau _{ - 2} (s + 1)\Delta x_{ - 1} (s), \end{eqnarray*} $

现在由(5.9)式和命题4.1,可得

$ \begin{eqnarray*} \tau ^* (s) & = & -\tau _{ - 2} (s + 1) = -[\gamma ( - 4)\frac{{\tilde \sigma ^{\prime \prime }}}{2} + \alpha ( - 4)\tilde \tau ^{\prime }]x_{ - 2} (s + 1) + c( - 2) \\ & = & [\gamma (4)\frac{{\tilde \sigma ^{\prime \prime }}}{2} - \alpha (4)\tilde \tau ^{\prime }]x_0 (s) + c( - 2). \end{eqnarray*} $

同理,从(5.10)和(5.13)式,我们得到

$ \begin{eqnarray*} \lambda ^* & = & \lambda - \Delta _{ - 1} \tau _{ - 1} (s) = \lambda - \Delta _{ - 1} \{ [\gamma (-2)\frac{{\tilde \sigma ^{\prime \prime }}}{2} + \alpha (-2)\tilde \tau ]x_{ - 1} (s)\} \\ & = & \lambda + \gamma (2)\frac{{\tilde \sigma ^{\prime \prime }}}{2} - \alpha (2)\tilde \tau ^{\prime } = \lambda - \kappa _{ - 1} . \end{eqnarray*} $

证毕.

关于伴随方程(5.7),我们发现它具有下面有趣的对偶性质.

命题 5.2    对于伴随方程(5.7),我们有

(5.16) $ \begin{equation} \sigma(s) = \sigma^*(s-1)+\tau^*(s-1)\nabla x_{-1}(s), \end{equation} $

(5.17) $ \begin{equation} \tau(s) = \frac{\sigma^*(s+1)-\sigma^*(s-1)}{\Delta x_{-1}(s)}-\tau^*(s-1)\frac{\nabla x_{-1}(s)}{\Delta x_{-1}(s)}, \end{equation} $

(5.18) $ \begin{equation} \lambda = \lambda^*-\Delta_{-1}\left(\tau^*(s-1)\frac{\nabla x_{-1}(s)}{\nabla x(s)}-\frac{\nabla\sigma^*(s)}{\nabla x(s)}\right).\end{equation} $

证    从(5.9)式我们有

(5.19) $ \begin{eqnarray} \tau ^* (s)\Delta x_{ - 1} (s) = \sigma (s + 1) - \sigma (s - 1) - \tau (s - 1)\nabla x_{ - 1} (s), \end{eqnarray} $

从(5.8)和(5.19)式,我们有

$ \sigma (s + 1) = \sigma ^* (s) + \tau ^* (s)\Delta x_{ - 1} (s), $

因此

$ \sigma (s) = \sigma ^* (s - 1) + \tau ^* (s - 1)\nabla x_{ - 1} (s). $

由(5.8)式,我们得

$ \tau (s - 1) = \frac{{\sigma ^* (s) - \sigma (s - 1)}}{{\nabla x_{ - 1} (s)}} = \frac{{\sigma ^* (s) - \sigma ^* (s - 2) - \tau ^* (s - 2)\nabla x_{ - 1} (s - 1)}}{{\nabla x_{ - 1} (s)}}, $

因此

$ \tau (s) = \frac{{\sigma ^* (s\mathrm{\ + }1) - \sigma ^* (s - 1) - \tau ^* (s - 1)\nabla x_{ - 1} (s)}}{{\Delta x_{ - 1} (s)}}. $

进一步

(5.20) $ \begin{eqnarray} \tau (s - 1)\nabla x_{ - 1} (s) - \nabla \sigma (s) & = & \sigma ^* (s) - \sigma (s - 1) - \nabla \sigma (s)\\ & = & \sigma ^* (s) - \sigma (s) \\ & = & \sigma ^* (s) - [\sigma ^* (s - 1) + \tau ^* (s - 1)\nabla x_{ - 1} (s)] \\ & = &\nabla \sigma ^* (s) - \tau ^* (s - 1)\nabla x_{ - 1} (s), \end{eqnarray} $

因此,由(5.10)和(5.20)式,可得

$ \lambda = \lambda ^* + \Delta _{ - 1} \bigg[\frac{{\tau (s - 1)\nabla x_{ - 1} (s) - \nabla \sigma (s)}}{{\nabla x(s)}}\bigg] = \lambda ^* - \Delta _{ - 1}\bigg[ \frac{{\tau ^* (s - 1)\nabla x_{ - 1} (s) - \nabla \sigma ^* (s)}}{{\nabla x(s)}}\bigg]. $

证毕.

用与推论5.1相同的方法,我们得到

推论 5.2    对方程(5.17)和(5.18),我们有

(5.21) $ \begin{equation} \tau (s) = - \tau^* _{ - 2} (s + 1), \end{equation} $

且

(5.22) $ \begin{equation} \lambda = \lambda^* - \kappa^* _{ - 1} . \end{equation} $

命题 5.3    伴随方程(5.7)可以改写为

(5.23) $ \begin{equation} \sigma (s + 1)\Delta _{ - 1} \nabla _0 w(s) - \tau _{ - 2} (s + 1)\nabla _0 w(s) + (\lambda - \kappa _{ - 1} ) w(s) = 0.\end{equation} $

证    由于

$ \Delta _0 w(s) - \nabla _0 w(s) = \Delta (\frac{{\nabla w(s)}}{{\nabla x(s)}}), $

我们有

(5.24) $ \begin{eqnarray} \tau ^* (s)\Delta _0 w(s)& = & \tau ^* (s)\nabla _0 w(s) + \tau ^* (s)\Delta (\frac{{\nabla w(s)}}{{\nabla x(s)}})\\& = & \tau ^* (s)\nabla _0 w(s) + \tau ^* (s)\Delta x_{ - 1} (s)\frac{\Delta }{{\Delta x_{ - 1} (s)}}(\frac{{\nabla w(s)}}{{\nabla x(s)}}), \end{eqnarray} $

将(5.24)式代入(5.7)式,我们有

(5.25) $ \begin{eqnarray}[\sigma ^* (s) + \tau ^* (s)\Delta x_{ - 1} (s)]\frac{\Delta }{{\Delta x_{ - 1} (s)}}(\frac{{\nabla w(s)}}{{\nabla x(s)}}) + \tau ^* (s)\nabla _0 w(s) + \lambda ^* w(s) = 0. \end{eqnarray} $

由(5.16)式,可得

(5.26) $ \begin{equation}\sigma ^* (s) + \tau ^* (s)\Delta x_{ - 1} (s) = \sigma (s + 1). \end{equation} $

将(5.26)式代入(5.25)式,且由(5.14)式,我们得

$ \sigma (s + 1)\Delta _{ - 1} \nabla _0 w(s) - \tau _{ - 2} (s + 1)\nabla _0 w(s) + (\lambda - \kappa _{ - 1}) w(s) = 0. $

证毕.

接下来我们要证明伴随方程(5.7)或(5.23)也是非一致格子上的超几何型差分方程.这仅需证明

$ \tilde \sigma ^* (s) = \sigma ^* (s) + \frac{1}{2}\tau ^* (s)\Delta x_{ - 1} (s) = \sigma (s + 1) +\frac{1}{2}\tau _{ - 2} (s + 1)\Delta x_{ - 1} (s) $

是关于变量$ x_0(s) $的至多二阶多项式.

事实上,由引理5.1和(5.12)式,可得

$ \tilde \sigma ^* (s) = \sigma (s + 1) + \frac{1}{2}\tau _{ - 2} (s + 1)\nabla x_{ - 1} (s + 1) = \tilde \sigma _{ - 2} (s + 1) $

是关于变量$ x_{-2}(s+1) = x_0(s) $的至多二阶多项式.

因此,我们得到

定理 5.1    伴随方程(5.23)或

(5.27) $ \begin{eqnarray}\tilde \sigma _{ - 2} (s + 1)\Delta _{ - 1} \nabla _0 w(s) - \frac{1}{2}\tau _{ - 2} (s + 1)[\Delta _0 w(s) + \nabla _0 w(s)] + (\lambda - \kappa _{ - 1} ) w(s) = 0 \end{eqnarray} $

也是非一致格子上的超几何型差分方程.

让

$ Y_n(s) = \rho_n(s-n) = \rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-i). $

现在构造一个形如(3.3)式的差分方程,它有解$ Y_n(s) $.改写为$ Y_n(s) = \rho(s)\sigma(s)\prod\limits_{j = 1}^{n-1} \sigma(s-i) $.那么利用命题2.1和Pearson方程(5.3),我们有

$ \begin{eqnarray*} \nabla_{-n}Y_n(s)& = &\frac{\nabla Y_n(s)}{\nabla x_{-n}(s)} \\ & = &\frac{1}{\nabla x_{-n}(s)}\left[\rho(s)\sigma(s)\prod\limits_{j = 1}^{n-1}\sigma(s-i)-\rho(s-1)\sigma(s-1)\prod\limits_{i = 1}^{n-1}\sigma(s-1-i)\right] \\ & = &\frac{1}{\nabla x_{-n}(s)}\bigg\{\lbrack \tau(s-1)\rho(s-1)\nabla x_{-1}(s)+\sigma(s-1)\rho(s-1)]\prod\limits_{i = 1}^{n-1}\sigma(s-i) \\ &&-\rho(s-1)\sigma(s-1)\prod\limits_{i = 1}^{n-1}\sigma(s-1-i)\bigg\}. \end{eqnarray*} $

那么我们得到一个解$ Y_n(s) $的差分方程

(6.1) $ \begin{eqnarray}\sigma(s-n)\nabla_{-n}Y_n(s) = \left(\frac{\sigma(s-1)-\sigma(s-n)}{\nabla x_{-n}(s)}+\tau(s-1)\frac{\nabla x_{-1}(s)}{\nabla x_{-n}(s)}\right)Y_n(s-1). \end{eqnarray} $

命题 6.1    如果$ u_1(s) $是以下差分方程的一个非平凡解

(6.2) $ \begin{eqnarray}p_1(s)\nabla_k u(s) = p_0(s)u(s-1), \end{eqnarray} $

这里$ p_1(s)\neq0 $,那么它满足差分方程

(6.3) $ \begin{eqnarray}&&\left(p_1(s)-p_0(s)\nabla x_k(s)\right)\Delta_{k-1}\nabla_k u(s) \\ &&+\left(\Delta_{k-1}p_1(s)-p_0(s)\frac{\nabla x_{k}(s)}{\Delta x_{k-1}(s)}\right)\Delta_{k}u(s)-\Delta_{k-1}p_0(s)u(s) = 0. \end{eqnarray} $

进一步,差分方程(6.3)的其它解是

$ \begin{eqnarray*} u_2(s) = Cu_1(s)\int_N^s \frac{1}{p_1(t)u_1(t)}d_\nabla x_k(t), \end{eqnarray*} $

这里$ C $是常数.

证    将算子$ \Delta_{k-1} $作用到方程(6.2)两边,我们有

(6.4) $ \begin{eqnarray}p_1(s)\Delta_{k-1}\nabla_k u(s)+\Delta_{k-1}p_1(s)\Delta_k u(s) = u(s)\Delta_{k-1}p_0(s)+p_0(s)\Delta_{k-1}u(s-1). \end{eqnarray} $

由于

$ \Delta_{k-1}\nabla_k u(s) = \frac{1}{\Delta x_{k-1}(s)}\left(\Delta_ku(s)-\frac{\nabla u(s)}{\nabla x_k(s)}\right), $

则有

(6.5) $ \begin{equation}\Delta_{k-1}u(s-1) = \frac{\nabla x_k(s)}{\Delta x_{k-1}(s)}\left(\Delta_ku(s)-\Delta_{k-1}\nabla_k u(s)\Delta x_{k-1}(s)\right). \end{equation} $

将(6.5)式代入(6.4)式,我们得到关于$ u(s) $的方程(6.3).记方程(6.3)其它另一解为$ u_2(s) $,那么

(6.6) $ \begin{eqnarray}\nabla_k\left(\frac{u_2(s)}{u_1(s)}\right) = \frac{u_1(s-1)\nabla_k u_2(s)-u_2(s-1)\nabla_k u_1(s)}{u_1(s)u_1(s-1)}. \end{eqnarray} $

由以上推导, $ u_2(s) $满足

$ \Delta_{k-1}[p_1(s)\nabla_k u(s)-p_0(s)u(s-1)] = 0. $

因此对任意常数$ C $,有

$ p_1(s)\nabla_ku_2(s)-p_0(s)u_2(s-1) = C. $

那么,将(6.2)式代入(6.6)式,可得

$ \nabla_k\left(\frac{u_2(s)}{u_1(s)}\right) = \frac{C}{p_1(s)u_1(s)}. $

由命题2.2,有

$ u_2(s) = Cu_1(s)\int_N^s\frac{1}{p_1(t)u_1(t)}{\rm d}_\nabla x_k(t). $

证毕.

由上面命题,可得

(6.7) $ \begin{eqnarray}\widehat{\sigma}(s)\Delta_{-(n+1)}\nabla_{-n} Y_n(s)+\widehat{\tau}(s)\Delta_{-n}Y_n(s)+\widehat{\lambda}Y_n(s) = 0, \end{eqnarray} $

这里

(6.8) $ \begin{equation} \widehat{\sigma}(s) = \sigma(s-1)+\tau(s-1)\nabla x_{-1}(s) = \sigma^*(s), \end{equation} $

(6.9) $ \begin{equation} \widehat{\tau}(s) = \frac{\sigma(s-n+1)-\sigma(s-1)}{\Delta x_{-(n+1)}(s)}-\tau(s-1)\frac{\nabla x_{-1}(s)}{\Delta x_{-(n+1)}(s)} = - \tau _{ - (n + 2)} (s + 1), \end{equation} $

(6.10) $ \begin{equation} \widehat{\lambda} = -\Delta_{-(n+1)}\left(\frac{\sigma(s-1)-\sigma(s-n)}{\nabla x_{-n}(s)}+\tau(s-1)\frac{\nabla x_{-1}(s)}{\nabla x_{-n}(s)}\right) = - \Delta _{ - (n + 1)} \tau _{ - (n + 1)} (s).\end{equation} $

由与第五节类似的方法,方程(6.7)可被改写成

(6.11) $ \begin{eqnarray} \sigma (s + 1)\Delta _{ - (n + 1)} \nabla _{ - n} Y_n (s) - \tau _{ - (n + 2)} (s + 1)\nabla _{ - n} Y_n (s) +\widehat{\lambda} Y_n (s) = 0.\end{eqnarray} $

记方程(6.7)另外的解为$ \widehat{Y}_n(s) $.那么

(6.12) $ \begin{eqnarray}\widehat{Y}_n(s) = \rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-j)\int_N^s\frac{1}{\rho(t)\prod\limits_{j = 0}^{n}\sigma(t-j)}{\rm d}_\nabla x_{-n}(t). \end{eqnarray} $

现在,对方程(6.7),让$ Y_n^{(n)}(s) = \Delta_{-n}^{(n)}Y_n(s) $.那么$ Y_n^{(n)}(s) $满足

(6.13) $ \begin{eqnarray}\widehat{\sigma}(s)\Delta_{-1}\nabla_0Y_n^{(n)}(s)+\widehat{\tau}_n(s)\Delta_0Y_n^{(n)}(s)+\widehat{\mu}_nY_n^{(n)}(s) = 0. \end{eqnarray} $

由递推关系式(3.6)和(3.7),对任何非负整数$ k $,我们有

(6.14) $ \begin{eqnarray}\widehat{\tau}_k(s)& = &\frac{\widehat{\sigma}(s+k)-\widehat{\sigma}(s)+\widehat{\tau}(s+k)\Delta x_{-n}(s+k-\frac{1}{2})}{\Delta x_{-n}(s+\frac{k-1}{2})} \\ & = &\frac{\sigma(s-n+k+1)-\sigma(s-1)-\tau(s-1)\nabla x_{-1}(s)}{\Delta x_{-n}(s+\frac{k-1}{2})} \\ & = &- \tau _{ - (n - k + 2)} (s + 1), \end{eqnarray} $

且

(6.15) $ \begin{equation}\hat \mu _n = \hat \lambda + \sum\limits_{k = 0}^{n - 1} {\Delta _{k - n} \hat \tau _k (s)}. \end{equation} $

当$ k $是一个负整数,我们也将(6.14)式的右边记为$ \widehat{\tau}_k(s) $.当$ k = n $,我们得到

$ \widehat{\tau}_n(s) = \frac{\sigma(s+1)-\sigma(s-1)}{\Delta x(s-\frac{1}{2})}-\tau(s-1)\frac{\nabla x_{-1}(s)}{\Delta x_{-1}(s)} = - \tau _{ - 2} (s + 1) = \tau^*(s). $

由与第5节相同的方法,方程(6.13)可被改写为

(6.16) $ \begin{eqnarray}\sigma (s + 1)\Delta _{ - 1} \nabla _0 Y_n (s) - \tau _{ - 2} (s + 1)\nabla _0 Y_n (s) + \hat \mu _n Y_n (s) = 0. \end{eqnarray} $

为了计算$ \widehat{\mu}_n $,我们需要一个引理,它的证明类似于命题4.1.

引理 6.1    给定一个整数$ k $,如果$ x(s) = c_1q^s+c_2q^{-s}+c_3 $,那么

$ \begin{eqnarray*} \widehat{\tau}_k(s)& = &\left[\frac{q^{k-n+2}-q^{n-k-2}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}\frac{\widetilde{\sigma}^{\prime \prime }}{2}-\left(q^{k-n+2}+q^{n-k-2}\right)\frac{\widetilde{\tau}^{\prime }}{2}\right]x_{k-n}(s)+\widehat{c}_1(k) \\ & = & \{ \nu [ - 2(n - k - 2)]\frac{{\tilde \sigma ^{\prime \prime }}}{2} - \alpha [ - 2(n - k - 2)]\tau ^{\prime }\} x_{n - k} (s) + \hat c_1 (k) \\ & = &- \kappa _{2(n - k - 2) + 1}x_{k - n} + \hat c_1 (k); \end{eqnarray*} $

如果$ x(s) = \widetilde{c}_1s^2+\widetilde{c}_2s+\widetilde{c}_3 $,那么

$ \begin{eqnarray*} \widehat{\tau}_k(s) = \left[(k-n+2)\widetilde{\sigma}^{\prime \prime }-\widetilde{\tau}^{\prime }\right]x_{k-n}(s)+\widehat{c}_2(k) = - \kappa _{2(n - k - 2) + 1}x_{k - n} + \hat c_2 (k); \end{eqnarray*} $

这里$ \widehat{c}_1(k), \widehat{c}_2(k) $是关于$ k $的函数

$ \begin{eqnarray*} \widehat{c}_1(k)& = &\frac{q^\frac{k-n+2}{2}-q^\frac{n-k-2}{2}}{q^\frac{1}{2}-q^{-\frac{1}{2}}}[\widetilde{\sigma}^{\prime }(0)+c_3(2-q^\frac{k-n+2}{2}-q^\frac{n-k-2}{2})] \\ &&+\widetilde{\tau}(0)(q^\frac{k-n+2}{2}+q^\frac{n-k-2}{2})+c_3\widetilde{\tau}^{\prime \frac{k-n+2}{2}})(q^{\frac{k-n+2}{2}}-q^{n-k-2}), \\ \widehat{c}_2(k)& = &\widetilde{\sigma}'(0)(k-n+2)+\frac{\widetilde{\sigma}''}{4}\widetilde{c}_1(k-n+2)^3+\frac{3\widetilde{\tau}'}{4}\widetilde{c}_1(k-n+2)^2+2\widetilde{\tau}(0). \end{eqnarray*} $

推论 6.1    如果$ x(s) = c_1q^s+c_2q^{-s}+c_3 $或$ x(s) = \widetilde{c} _1s^2+\widetilde{c}_2s+\widetilde{c}_3 $,那么

$ \begin{eqnarray*} \widehat{\mu}_n = - \kappa _{ - 1} - \kappa _n \nu (n) = - \kappa _{n-1} \nu (n+1). \end{eqnarray*} $

证    由(6.10)式,我们看到$ \hat \lambda \mathrm{\ = }\Delta _{ - (n + 1)} \tau _{ - (n + 1)} (s) = \Delta _{ - (n + 1)} \hat \tau _{ - 1} (s) $.那么,我们将(6.15)式写成

(6.17) $ \begin{eqnarray}\widehat{\mu}_n = {\hat \lambda }+\sum\limits_{k = 0}^{n-1}\Delta_{k-n}\widehat{\tau}_k(s) = \sum\limits_{k = -1}^{n-1}\Delta_{k-n}\widehat{\tau}_k(s). \end{eqnarray} $

那么由引理6.1,如果$ x(s) = c_1q^s+c_2q^{-s}+c_3 $,则

$ \begin{eqnarray*} \widehat{\mu}_n & = &\sum\limits_{k = -1}^{n-1}\left[\frac{q^{k-n+2}-q^{n-k-2}}{q^{\frac{1}{2}}-q^{\frac{1}{2}}}\frac{\sigma^{\prime \prime }}{2}-\left(q^{k-n+2}+q^{n-k-2}\right)\frac{\tau^{\prime }}{2}\right] \\ & = &\sum\limits_{k = -1}^{n-1}\left[\frac{q^{-k}-q^{k}}{q^{\frac{1}{2}}-q^{\frac{1}{2}}}\frac{\sigma^{\prime \prime }}{2}-\left(q^{-k}+q^{k}\right)\frac{\tau^{\prime }}{2}\right] \\ & = &- \kappa _{ - 1}+ \sum\limits_{k = 0}^{n-1}\left[\frac{q^{-k}-q^{k}}{q^{\frac{1}{2}}-q^{\frac{1}{2}}}\frac{\sigma^{\prime \prime }}{2}-\left(q^{-k}+q^{k}\right)\frac{\tau^{\prime }}{2}\right] \\ & = & - \kappa _{ - 1} - \kappa _n \nu (n) = - \kappa _{n-1} \nu (n+1). \end{eqnarray*} $

如果$ x(s) = \widetilde{c}_1s^2+\widetilde{c}_2s+\widetilde{c}_3 $,则

$ \begin{eqnarray*} \widehat{\mu}_n & = &\sum\limits_{k = -1}^{n-1}\left[(k-n+2)\sigma^{\prime \prime }-\tau^{\prime }\right] = \sum\limits_{k = -1}^{n-1}\left[-k\sigma^{\prime \prime }-\tau^{\prime }\right] \\ & = &- \kappa _{ - 1}+\sum\limits_{k = 0}^{n-1}\left[-k\sigma^{\prime \prime }-\tau^{\prime }\right] = - \kappa _{ - 1} - n\kappa _n = - (n+1)\kappa _{n-1}. \end{eqnarray*} $

证毕.

定理 6.1    如果$ \lambda = \lambda _n = - \kappa _n \nu (n) $,那么方程(6.13)是

$ \begin{eqnarray*} L^*[Y_n^{(n)}(s)] = 0. \end{eqnarray*} $

证    我们仅证明当$ \lambda = \lambda_n $的情形,有$ \widehat{\mu}_n = \lambda^* $.由(5.10)式和命题4.1,有

(6.18) $ \begin{eqnarray}\lambda^*& = &\lambda_n-\Delta_{-1}\left(\tau(s-1)\frac{\nabla x_{-1}(s)}{\nabla x(s)}-\frac{\nabla \sigma(s)}{\nabla x(s)}\right) \\ & = &\lambda_n-\Delta_{-1}\tau_{-1}(s) = - \kappa _n \nu (n) - \kappa _{ - 1} = - \kappa _{n-1} \nu (n+1) . \end{eqnarray} $

因此,由推论6.1,我们得到$ \widehat{\mu}_n = \lambda^* $.

由等式(5.11),我们有$ L[\frac{1}{\rho}Y_n^{(n)}] = L^*[Y_n^{(n)}] = 0 $.那么我们得到Rodrigues型公式的一个推广:

定理 6.2    如果

$ \lambda = \lambda_n \; \mbox{且}\;\lambda_m\neq \lambda_n\; \mbox{对}\; m = 0, 1, \cdots, n-1, $

那么方程(3.1)的通解是

$ \begin{eqnarray*} y_n(s)& = &\frac{C_1}{\rho(x)}\Delta_{-n}^{(n)}[\rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-j)] \\ &&+\frac{C_2}{\rho(x)}\Delta_{-n}^{(n)}\left[\rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-j)\int_N^s\frac{1}{\rho(t)\prod\limits_{j = 0}^{n}\sigma(t-j)}d_\nabla x_{-n}(t)\right]. \end{eqnarray*} $

命题 7.1^{[10, p62]}     设格子函数$ x(s)$具有形式

$x(s) = c_{1}q^{s}+c_{2}q^{-s}+c_{3}\; \mbox{或}\; x(s) = c_{1}s^{2}+c_{2}s+c_{3}, $

这里$q, c_{1}, c_{2}, c_{3}$为常数.如果$P_{n}[x_{k}(s)]$是关于$x_{k}(s)$(k是一任意整数)的n阶多项式,那么$\Delta _{k}P_{n}[x_{k}(s)]$是一个关于$x_{k+1}(s)$的$ n-1 $阶多项式.

我们把(5.7)式与同类型的差分方程联系起来

(7.1) $ \begin{equation}\sigma^*(s)\Delta_{-(n+1)}\nabla_{-n} v(s)+\gamma(s, n)\Delta_{-n} v(s)+\eta(n) v(s) = P_{n-1}[x_{-n}(s)], \end{equation} $

这里$\gamma(s, n)$和$\eta(n)$待定,且$P_{n-1}[x_{-n}(s)]$是一个关于$x_{-n}(s)$的$n-1$阶任意多项式.现假定

$ \begin{eqnarray*} &&\Delta_{-n}^{(n)}\left[\sigma^*(s)\Delta_{-(n+1)}\nabla_{-n} v(s)+\gamma(s, n)\Delta_{-n} v(s)+\eta(n) v(s)\right] \\ & = &\sigma^*(s)\Delta_{-1}\nabla_0 w(s)+\tau^*(s)\Delta_0 w(s)+\lambda^*w(s) = 0 \end{eqnarray*} $

且

$ w(s) = \Delta_{-n}^{(n)}v(s). $

进一步,我们假设

(7.2) $ \begin{eqnarray}&&\sigma^*(s)\Delta_{-n-1}\nabla_{-n} v(s)+\gamma(s, n)\Delta_{-n} v(s)+\eta(n) v(s) \\ & = &\Delta_{-(n+1)}\left[\sigma^*(s)\nabla_{-n} v(s)+\ell(s, n)v(s)\right], \end{eqnarray} $

这里$\ell(s, n)$待定.现在,由递推关系式(3.6)和(3.7),对任何非负整数$k$,我们有

$ \gamma_k(s, n) = \frac{\sigma^*(s+k)-\sigma^*(s)+\gamma(s+k, n)\Delta x_{-n}(s+k-\frac{1}{2})}{\Delta x_{-n+k-1}(s)}, \; \gamma_0(s, n) = \gamma(s, n), $

$ \eta_k(n) = \eta(n)+\sum\limits_{j = 0}^{k-1}\Delta_{j-n}\gamma_j(s, n), \; \eta_0(n) = \eta(n). $

当$k=n$,则有

(7.3) $ \begin{eqnarray}\tau^*(s)& = &\gamma_n(s, n) = \frac{\sigma^*(s+n)-\sigma^*(s)+\gamma(s+n, n)\Delta x_{-n}(s+n-\frac{1}{2})}{\Delta x_{-n}(s+\frac{n-1}{2})} \\ & = &\frac{\sigma(s+1)-\sigma(s-1)}{\Delta x_{-1}(s)}-\tau(s-1)\frac{\nabla x_{-1}(s)}{\Delta x_{-1}(s)}, \end{eqnarray} $

(7.4) $ \begin{equation}\lambda^* = \eta(n)+\sum\limits_{j = 0}^{n-1}\Delta_{j-n}\gamma_j(s, n) = \lambda-\nabla_{-1}\left(\tau(s-1)\frac{\nabla x_{-1}(s)}{\nabla x(s)}-\frac{\nabla\sigma(s)}{\nabla x(s)}\right). \end{equation} $

由(7.3)式,我们得

(7.5) $ \begin{eqnarray}\gamma(s, n) = \frac{\sigma(s-n+1)-\sigma(s-1)-\tau(s-1)\nabla x_{-1}(s)}{\Delta x_{-n}(s-\frac{1}{2})}. \end{eqnarray} $

进一步,从(7.2)式和命题2.1,我们得

(7.6) $ \begin{equation} \ell(s+1, n)\frac{\Delta x_{-n}(s)}{\Delta x_{-(n+1)}(s)}+\Delta_{-(n+1)}\sigma^*(s) = \gamma(s, n), \end{equation} $

(7.7) $ \begin{equation} \Delta_{-(n+1)}\ell(s, n) = \eta(n). \end{equation} $

从(7.5)和(7.6)式,我们得

(7.8) $ \begin{eqnarray}\ell(s, n) = \frac{\sigma(s-n)-\sigma(s-1)-\tau(s-1)\nabla x_{-1}(s)}{\nabla x_{-n}(s)}. \end{eqnarray} $

那么, $ \eta(n)=\widehat{\lambda}$.现在,我们计算$\lambda$.注意到,对任何非负整数$k$,有

$ \gamma_k(s, n) = \frac{\sigma(s+k-n+1)-\sigma(s-1)-\tau(s-1)\nabla x_{-1}(s)}{\Delta x_{k-n-1}(s)} = \widehat{\tau}_k(s). $

由(6.17)和(6.18)式,可得

$ \lambda_n = \nabla_{-1}\tau_{-1}(s)+\widehat{\lambda}+\sum\limits_{j = 0}^{n-1}\Delta_{j-n}\widehat{\tau}_k(s). $

那么由(7.4),我们得$\lambda=\lambda_n$.

由命题7.1,可得

(7.9) $ \begin{equation}\sigma^*(s)\nabla_{-n} v(s)+\ell(s, n)v(s) = P_n[x_{-(n+1)}(s)]. \end{equation} $

现在,我们解方程(7.9).首先,我们考虑下面方程的解

(7.10) $ \begin{eqnarray}\sigma^*(s)\nabla_{-n} v(s)+\ell(s)v(s) = 0. \end{eqnarray} $

方程(7.10)可被改写为

(7.11) $ \begin{equation}\sigma(s-n)v(s) = \left(\sigma(s-1)+\tau(s-1)\nabla x_{-1}(s)\right)v(s-1). \end{equation} $

进一步,由Pearson方程(5.3),可得

$ \rho(s)\sigma(s) = \left[\sigma(s-1)+\tau(s-1)\nabla x_{-1}(s)\right]\rho(s-1). $

那么方程(7.11)变成

$ \frac{v(s)}{v(s-1)} = \frac{\rho(s)\sigma(s)}{\rho(s-1)\sigma(s-n)}. $

容易证明以上方程的解是

$ v(s) = C\rho(s)\prod\limits_{i = 0}^{n-1}\sigma(s-i), $

这里$C$是一个常数.

现在,让

(7.12) $ \begin{equation}v(s) = C(s)\rho(s)\prod\limits_{i = 0}^{n-1}\sigma(s-i). \end{equation} $

那么,将(7.12)式代入(7.9)式,则有

$ \nabla_{-n}C(s) = \frac{P_n[x_{-(n+1)}(s)]}{\rho(s)\prod\limits_{i = 0}^n\sigma(s-i)}. $

然后,应用命题2.2,可得

$ C(s) = \int_N^s\frac{P_n[x_{-(n+1)}(t)]}{\rho(t)\prod\limits_{i = 0}^n\sigma(t-i)}{\rm d}_\nabla x_{-n}(t)+\widetilde{C}, $

这里$\widetilde{C}$是一个常数.由于

$ \rho(s)y(s) = w(s) = \Delta_{-n}^{(n)}v(s), $

我们得如下定理.

定理 7.1    如果

$ \lambda = \lambda_n \; \mbox{和}\;\lambda_m\neq \lambda_n\; \mbox{对}\; m = 0, 1, \cdots, n-1, $

那么方程(3.1)的通解是

$ \begin{eqnarray*} y_n(s)& = &\frac{C}{\rho(x)}\Delta_{-n}^{(n)}[\rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-j)] \\ &&+\frac{1}{\rho(x)}\Delta_{-n}^{(n)}\left[\rho(s)\prod\limits_{j = 0}^{n-1}\sigma(s-j)\int_N^s\frac{P_n[x_{-(n+1)}(t)]}{\rho(s)\prod\limits_{j = 0}^{n}\sigma(t-j)}{\rm d}_\nabla x_{-n}(t)\right], \end{eqnarray*} $

这里$C$是任意常数且$P_n(\cdot)$是一个$n$-阶多项式.