非一致格子上的超几何型差分方程:第二类解的Rodrigues型表示公式

非一致格子上的超几何型差分方程:第二类解的Rodrigues型表示公式

程金发^,, 贾鲁昆

Hypergeometric Type Difference Equations on Nonuniform Lattices: Rodrigues Type Representation for the Second Kind Solution

Cheng Jinfa^,, Jia Lukun

通讯作者: 程金发, E-mail:jfcehng@xmu.edu.cn

收稿日期: 2017-03-14

基金资助:

福建省自然科学基金. 2016J01032
the Natural Science Foundation of Fujian Province. 2016J01032

Received: 2017-03-14

Fund supported:

福建省自然科学基金. 2016J01032
the Natural Science Foundation of Fujian Province. 2016J01032

摘要

通过建立非一致格子上的二阶伴随方程，得到了非一致格子上超几何型差分方程第二类解的Rodrigues型表示公式，它们推广了经典Rodrigues公式.由此得到由经典Rodrigues公式和广义罗德里格斯公式线性组合而成的通解.

关键词： 特殊函数 ; 正交多项式 ; 伴随差分方程 ; 超几何型差分方程 ; 非一致格子

Abstract

By building a second order adjoint equation, the Rodrigues type representation for the second kind solution of a second order difference equation of hypergeometric type on nonuniform lattices is given. The general solution of the equation in the form of a combination of a standard Rodrigues formula and a "generalized" Rodrigues formula is also established.

Keywords： Special function ; Orthogonal polynomial ; Adjoint difference equation ; Hypergeometric difference equations ; Nonuniform lattice

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本文引用格式

程金发, 贾鲁昆. 非一致格子上的超几何型差分方程:第二类解的Rodrigues型表示公式. 数学物理学报[J], 2019, 39(4): 875-893 doi:

Cheng Jinfa, Jia Lukun. Hypergeometric Type Difference Equations on Nonuniform Lattices: Rodrigues Type Representation for the Second Kind Solution. Acta Mathematica Scientia[J], 2019, 39(4): 875-893 doi:

1 引言

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

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

(1.1)

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)

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

(1.2)

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

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

(1.3)

(这里 $\alpha , \beta$ 是常数)且

$\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}$

(1.4)

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)

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

(1.5)

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

(1.6)

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.

2 非一致格子上的差分及和分

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

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

(2.1)

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

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

(2.2)

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

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

3 Rodrigues公式

运用第二节中的记号,超几何型(1.2)的差分方程可以写成

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

(3.1)

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]

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

(3.2)

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

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

(3.3)

这里

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

(3.4)

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

(3.5)

我们发现

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

(3.6)

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

(3.7)

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

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

(3.8)

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]

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

4 $\tau _k (s), \mu _k$ 和 $\lambda_n$ 的显示表示

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

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

(4.1)

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

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

(4.2)

进一步,有

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

(4.3)

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

(4.4)

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

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

(4.5)

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

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

(4.6)

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

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

(4.7)

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

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

(4.8)

5 非一致格子上超几何差分方程的伴随方程

让

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

(5.1)

那么方程(5.1)有自相伴形式

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

(5.2)

这里 $\rho(s)$ 满足Pearson型方程

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

(5.3)

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

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

(5.4)

将(5.4)式代入(5.2)式,我们得

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

(5.5)

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

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

(5.6)

将(5.6)式代入(5.5)式,可得

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

(5.7)

这里

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

(5.8)

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

(5.9)

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

(5.10)

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

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

(5.11)

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 $\tau _{\nu }(s)$

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

(5.12)

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

(5.13)

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

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

(5.14)

且

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

(5.15)

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

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

(5.16)

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

(5.17)

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

(5.18)

证从(5.9)式我们有

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

(5.19)

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

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

(5.20)

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

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

(5.21)

且

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

(5.22)

命题 5.3 伴随方程(5.7)可以改写为

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

(5.23)

证由于

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

我们有

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

(5.24)

将(5.24)式代入(5.7)式,我们有

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

(5.25)

由(5.16)式,可得

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

(5.26)

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

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

(5.27)

也是非一致格子上的超几何型差分方程.

6 Rodrigues公式的一个推广

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

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

(6.1)

命题 6.1 如果 $u_1(s)$ 是以下差分方程的一个非平凡解

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

(6.2)

这里 $p_1(s)\neq0$ ,那么它满足差分方程

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

(6.3)

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

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

(6.4)

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

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

(6.5)

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

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

(6.6)

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

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

(6.7)

这里

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

(6.8)

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

(6.9)

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

(6.10)

由与第五节类似的方法,方程(6.7)可被改写成

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

(6.11)

记方程(6.7)另外的解为 $\widehat{Y}_n(s)$ .那么

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

(6.12)

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

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

(6.13)

由递推关系式(3.6)和(3.7),对任何非负整数 $k$ ,我们有

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

(6.14)

且

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

(6.15)

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

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

(6.16)

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

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

(6.17)

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

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

(6.18)

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

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

7 更一般的Rodrigues公式

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

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

(7.1)

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

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

(7.2)

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

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

(7.3)

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

(7.4)

由(7.3)式,我们得

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

(7.5)

进一步,从(7.2)式和命题2.1,我们得

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

(7.6)

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

(7.7)

从(7.5)和(7.6)式,我们得

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

(7.8)

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

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

(7.9)

现在,我们解方程(7.9).首先,我们考虑下面方程的解

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

(7.10)

方程(7.10)可被改写为

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

(7.11)

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

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

(7.12)

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.

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1944

... 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] ...

... 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 ...

1997

Sur la formule de rodrigues

1943

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

1953

A Note on the derivation of Rodrigues' formulae

1963

Generalizations of the formulas of Rodrigues and Schlafli

1963

2010

... 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]. ...

... 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]. ...

... 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) ...

... 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] ...

1988

1991

... 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) ...

... 那么对非负整数

$k$

$z_{k}(s)$

满足与方程(3.1)同型的方程^[10] ...

... 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] ...

... 命题 7.1^{[10, p62]} 设格子函数

$x(s)$

具有形式 ...

1989

... 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] ...

1999

A set of orthogonal polynomials that generalize the Racah coefficients or 6j-symbols

1979

Contiguous relations, basic hypergeometric functions, and orthogonal polynomials (I)

1989

On the theory of difference analogues of special functions of hypergeometric type

1989

... 引理 4.1^[18] 对

$\alpha (\mu ), \nu (\mu )$

,我们有 ...

... 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

$\tau _{\nu }(s)$

...

On classical orthogonal polynomials

1995

2004

Hypergeometric-type differential equations:second kind solutions and related integrals

2003

On the Rodrigues formula solution of the hypergeometric-type differential equation

2013

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

On the q-polynomials in the exponential lattice x(s)=c₁q^s + c₃

1999

Recurrence relations for discrete hypergeometric functions

2005

On the properties of special functions on the linear-type lattices

2011

Hypergeometric type q-difference equations:Rodrigues type representation for the second kind solution

2005

... 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. ...

... ]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. ...

Variational calculus on q-nonuniform lattices

2005

... 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 ...

q-deformation of meromorphic solutions of linear differential equations

2015

2002

A q-analogue of Kummer's equation

2017

On difference equations for orthogonal polynomials on nonuniform lattices

2008

On solutions of holonomic divideddifference equations on nonuniform lattices

2013

Semi-classical orthogonal polynomial systems on non-uniform lattices, deformations of the Askey table and analogs of isomonodromy

2015

1 引言

2 非一致格子上的差分及和分

3 Rodrigues公式

$\tau _k (s), \mu _k$ 和

$\lambda_n$ 的显示表示

〈

〉