第十四章 球函数

Legendre 方程和连带 Legendre 方程

Helmholtz 方程在球坐标系下分离变量, 即可得连带 Legendre 方程

$$\begin{array}{}& \frac{1}{\sin \theta }\text{dv}\theta (\sin \theta \text{dv}\mathrm{\Theta }\theta )+\text{qty}(\lambda -\frac{\mu }{\sin [2]\theta })\mathrm{\Theta }=0\end{array}$$

以及它的特殊情形, Legendre 方程

$$\begin{array}{}& \frac{1}{\sin \theta }\text{dv}\theta (\sin \theta \text{dv}\mathrm{\Theta }\theta )+\lambda \mathrm{\Theta }=0,\end{array}$$

做变换 $x=\cos \theta$, $y\text{qty}(x)=\mathrm{\Theta }\text{qty}(\theta )$, 则又可将它们改写成

$$\begin{array}{}& \text{dv}x\text{qty}[\text{qty}(1-x^2)\text{dv}yx]+\text{qty}(\lambda -\frac{\mu }{1-x^2})y=0\end{array}$$

和

$$\begin{array}{}& \text{dv}x\text{qty}[\text{qty}(1-x^2)\text{dv}yx]+\lambda y=0.\end{array}$$

Legendre 多项式

本征值问题

$$\begin{array}{}\text{(1a)}& \text{dv}x\text{qty}[\text{qty}(1-x^2)\text{dv}yx]+\lambda y=0,\text{(1b)}& y\text{qty}(\pm 1)\text{有界}\end{array}$$

的解是

$$\begin{array}{}\text{(2a)}& & \text{本征值}& & {\lambda }_l=l\text{qty}(l+1),& & l=0,1,2,\dots ,\text{(2b)}& & \text{本征函数}& & y_l\text{qty}(x)={\mathrm{P}}_l\text{qty}(x).\end{array}$$

${\mathrm{P}}_l\text{qty}(x)$ 称为 $l$ 次 Legendre 多项式,

$$\begin{array}{}\text{(3)}& {\mathrm{P}}_l\text{qty}(x)& =\sum _{n=0}^l\frac{1}{\text{qty}(n!{)}^2}\frac{\text{qty}(l+n)!}{\text{qty}(l-n)!}\text{qty}(\frac{x-1}{2}{)}^n\text{(4)}& & =\sum _{r=0}^{\text{qty}[l/2]}\text{qty}(-1{)}^r\frac{l!}{r!\text{qty}(l-r)!}\frac{\text{qty}(2l-2r)!}{\text{qty}(l-2r)!}{x}^{l-2r}.\end{array}$$

Legendre 多项式的主要性质

微分表示 (Rodrigues 公式)

$$\begin{array}{}\text{(5)}& {\mathrm{P}}_l\text{qty}(x)=\frac{1}{2^{l}l!}\text{dv}[l]x(x^2-1{)}^l.\end{array}$$

生成函数

Legendre 多项式的生成函数是 $1/\sqrt{1-2xt+t^2}$,

$$\begin{array}{}\text{(6)}& \frac{1}{\sqrt{1-2xt+t^2}}=\sum _{t=0}^{\mathrm{\infty }}{\mathrm{P}}_l\text{pqty}x{t}^l\text{qc}\text{vqty}v<min\text{vqty}x\pm \sqrt{x^2-1},\end{array}$$

规定多值函数 $1/\sqrt{-2xt+t^2}$ 的单值分枝为 $\text{eval}{\frac{1}{\sqrt{1-2xt+t^2}}}_{t=0}=1$.

递推关系

Legendre 多项式的主要递推关系有

$$\begin{array}{}\text{(7)}& & \text{qty}(2l+1)x{\mathrm{P}}_l\text{qty}(x)=\text{qty}(l+1){\mathrm{P}}_{l+1}\text{qty}(x)+l{\mathrm{P}}_{l-1}\text{qty}(x),\text{(8)}& & {\mathrm{P}}_l\text{qty}(x)={\mathrm{P}}_{l+1}^{\prime }\text{qty}(x)-2x{\mathrm{P}}_l^{\prime }\text{qty}(x)+{\mathrm{P}}_{l-1}^{\prime }\text{qty}(x),\text{(9)}& & {\mathrm{P}}_{l+1}^{\prime }\text{qty}(x)=x{\mathrm{P}}_l^{\prime }\text{qty}(x)+\text{qty}(l+1){\mathrm{P}}_l\text{qty}(x),\text{(10)}& & {\mathrm{P}}_{l-1}^{\prime }\text{qty}(x)=x{\mathrm{P}}_l^{\prime }\text{qty}(x)-l{\mathrm{P}}_l\text{qty}(x),\text{(11)}& & {\mathrm{P}}_{l+1}^{\prime }\text{qty}(x)-{\mathrm{P}}_{l-1}^{\prime }\text{qty}(x)=\text{qty}(2l+1){\mathrm{P}}_l\text{qty}(x).\end{array}$$

把这些递推关系重新组合, 还能给出其他形式的递推关系.

正交完备性

正交性

不同次数的 Legendre 多项式在区间 $\text{comm}-11$ 上正交,

$$\begin{array}{}\text{(12)}& {\int }_{-1}^1{\mathrm{P}}_l\text{qty}(x){\mathrm{P}}_k\text{qty}(x)\text{dd}x=0\text{qc}k\ne l.\end{array}$$

Legendre 多项式的模方

$$\begin{array}{}\text{(13)}& {\int }_{-1}^1{\mathrm{P}}_l\text{qty}(x){\mathrm{P}}_l\text{qty}(x)\text{dd}x=\frac{2}{2l+1}.\end{array}$$

把 $\text{(12)}$ 和 $\text{(13)}$ 合并起来, 还可以写成

$$\begin{array}{}\text{(14)}& {\int }_{-1}^1{\mathrm{P}}_k\text{qty}(x){\mathrm{P}}_l\text{qty}(x)\text{dd}x=\frac{2}{2l+1}{\delta }_{kl},\end{array}$$

其中 ${\delta }_{ij}$ 是 Kronecker 的 $\delta$ 符号.

通过变换 $x=\cos \theta$ 变回到以 $\theta$ 为自变量, $\text{(14)}$ 就变为

$$\begin{array}{}\text{(14’)}& {\int }_0^{\pi }{\mathrm{P}}_k\text{qty}(\cos \theta ){\mathrm{P}}_l\text{qty}(\cos \theta )\sin \theta \text{dd}\theta =\frac{2}{2l+1}{\delta }_{kl},\end{array}$$

即 ${\mathrm{P}}_k\text{qty}(\cos \theta )$ 和 ${\mathrm{P}}_l\text{qty}(\cos \theta )$ 在区间 $\text{comm}0\pi$ 上以权函数 $\sin \theta$ 正交. 权函数 $\sin \theta$ 正好就是微分方程

$$\begin{array}{}& \text{dv}\theta (\sin \theta \text{dv}\mathrm{\Theta }\theta )+\lambda \sin \theta \mathrm{\Theta }=0\end{array}$$

中本征值 $\lambda$ 后的函数 $\sin \theta$.

Legendre 多项式的完备性

任意一个在区间 $\text{comm}-11$ 中分段连续的函数 $f\text{qty}(x)$, (在平均收敛意义下) 可以展开为级数

$$\begin{array}{}\text{(15)}& f\text{qty}(x)=\sum _{l=0}^{\mathrm{\infty }}{c}_l{\mathrm{P}}_l\text{qty}(x),\end{array}$$

其中的展开系数 $c_l$ 可以根据 Legendre 多项式的正交性求得

$$\begin{array}{}\text{(16)}& c_l=\frac{2l+1}{2}{\int }_{-1}^{1}f\text{qty}(x){\mathrm{P}}_l\text{qty}(x)\text{dd}x.\end{array}$$

连带 Legendre 函数

本征值问题

$$\begin{array}{}\text{(17a)}& & \text{dv}x\text{qty}[\text{qty}(1-x^2)\text{dv}yx]+\text{qty}(\lambda -\frac{m^2}{1-x^2})y=0,& & m=0,1,2,\dots ,\text{(17b)}& & y\text{qty}(\pm 1)\text{有界}\end{array}$$

的解是

$$\begin{array}{}\text{(18a)}& & \text{本征值}& & {\lambda }_l=l\text{qty}(l+1),& & l=m,m+1,m+2,\dots ,\text{(18b)}& & \text{本征函数}& & y_l\text{qty}(x)={\mathrm{P}}_l^m\text{qty}(x).\end{array}$$

${\mathrm{P}}_l^m\text{qty}(x)$ 称为 $m$ 阶 $l$ 次连带 Legendre 函数,

$$\begin{array}{}\text{(19)}& {\mathrm{P}}_l^m\text{qty}(x)\equiv \text{qty}(-1{)}^m\text{qty}(1-x^2{)}^{m/2}\text{dv}[m]{\mathrm{P}}_l\text{qty}(x)x.\end{array}$$

球面调和函数

定义

球面调和函数来自本征值问题

$$\begin{array}{}\text{(20a)}& \frac{1}{\sin \theta }\text{pdv}\theta \text{qty}[\sin \theta \text{pdv}S\text{qty}(\theta ,\varphi )\theta ]+\frac{1}{\sin [2]\theta }\text{pdv}[2]S\text{qty}(\theta ,\varphi )\varphi +\lambda S\text{qty}(\theta ,\varphi )=0,\end{array}$$$$\begin{array}{}\text{(20b)}& & \text{eval}{S}_{\theta =0}\text{有界},& & \text{eval}{S}_{\theta =\pi }\text{有界},\text{(20c)}& & \text{eval}{S}_{\varphi =0}=\text{eval}{S}_{\varphi =2\pi },& & \text{eval}{\text{pdv}S\varphi }_{\varphi =0}=\text{eval}{\text{pdv}S\varphi }_{\varphi =2\pi }.\end{array}$$

此本征值问题的本征值为

$$\begin{array}{}\text{(21)}& {\lambda }_l=l\text{qty}(l+1)\text{qc}l=0,1,2,3,\dots ;\end{array}$$

而对应于一个本征值 ${\lambda }_l$, 有 $2l+1$ 个本征函数 ($2l+1$ 度简并)

$$\begin{array}{}\text{(22a)}& & S_{lm1}\text{qty}(\theta ,\varphi )={\mathrm{P}}_l^m\text{qty}(\cos \theta )\cos m\varphi ,& & m=0,1,2,\dots ,l\text{(22b)}& & S_{lm2}\text{qty}(\theta ,\varphi )={\mathrm{P}}_l^m\text{qty}(\cos \theta )\sin m\varphi ,& & m=1,2,\dots ,l.\end{array}$$

或者组合成

$$\begin{array}{}\text{(22c)}& S_{lm}\text{qty}(\theta ,\varphi )={\mathrm{P}}_l^m\text{qty}(\cos \theta )e^{im\varphi }\text{qc}m=0,\pm 1,\pm 2,\dots ,\pm l,\end{array}$$

其中

$$\begin{array}{}& {\mathrm{P}}_l^{-m}\text{qty}(\cos \theta )=\text{qty}(-1{)}^m\frac{\text{qty}(l-m)!}{\text{qty}(l+m)!}{\mathrm{P}}_l^m\text{qty}(\cos \theta ).\end{array}$$

这些本征函数, 统称为球面调和函数, 或球谐函数.

在此基础上, 就可以写出球内 Laplace 方程边值问题

$$\begin{array}{}\text{(23a)}& \frac{1}{r^2}\text{pdv}r(r^2\text{pdv}ur)+\frac{1}{r^2\sin \theta }\text{pdv}\theta (\sin \theta \text{pdv}u\theta )+\frac{1}{r^2\sin [2]\theta }\text{pdv}[2]u\varphi =0,\end{array}$$$$\begin{array}{}\text{(23b)}& & \text{eval}{u}_{\theta =0}\text{有界},& & \text{eval}{u}_{\theta =\pi }\text{有界},\text{(23c)}& & \text{eval}{u}_{\varphi =0}=\text{eval}{u}_{\varphi =2\pi },& & \text{eval}{\text{pdv}u\varphi }_{\varphi =0}=\text{eval}{\text{pdv}u\varphi }_{\varphi =2\pi },\text{(23d)}& & \text{eval}{u}_{r=0}\text{有界},& & \text{eval}{u}_{r=a}=f\text{qty}(\theta ,\varphi )\end{array}$$

的一般解

$$\begin{array}{}& u\text{qty}(r,\theta ,\varphi )=\sum _{l=0}^{\mathrm{\infty }}\sum _{m=0}^l{r}^l{\mathrm{P}}_l\text{qty}(\cos \theta )\text{qty}(A_{lm}\cos m\varphi +B_{lm}\sin m\varphi ).\end{array}$$

正交性

$l$ 或 $m$ 不同的球面调和函数在整个 $4\pi$ 立体角上彼此正交, 即当 $\text{qty}(l,m)\ne \text{qty}(k,n)$ 时, 有

$$\begin{array}{}\text{(24a)}& {\int }_0^{\pi }{\mathrm{P}}_l^m\text{qty}(\cos \theta ){\mathrm{P}}_k^n\text{qty}(\cos \theta )\sin \theta \text{dd}\theta {\int }_0^{2\pi }\cos m\varphi \cos n\varphi \text{dd}\varphi =0,\text{(24b)}& {\int }_0^{\pi }{\mathrm{P}}_l^m\text{qty}(\cos \theta ){\mathrm{P}}_k^n\text{qty}(\cos \theta )\sin \theta \text{dd}\theta {\int }_0^{2\pi }\sin m\varphi \sin n\varphi \text{dd}\varphi =0,\text{(24c)}& {\int }_0^{\pi }{\mathrm{P}}_l^m\text{qty}(\cos \theta ){\mathrm{P}}_k^n\text{qty}(\cos \theta )\sin \theta \text{dd}\theta {\int }_0^{2\pi }\cos m\varphi \sin n\varphi \text{dd}\varphi =0.\end{array}$$

球面调和函数的模方

$$\begin{array}{}\text{(25a)}& & {\int }_0^{\pi }\text{qty}[{\mathrm{P}}_l^m\text{qty}(\cos \theta ){]}^2\sin \theta \text{dd}\theta {\int }_0^{2\pi }\cos [2]m\varphi \text{dd}\varphi =\frac{\text{qty}(l+m)!}{\text{qty}(l-m)!}\frac{2\pi }{2l+1}\text{qty}(1+{\delta }_{m0}),\text{(25b)}& & {\int }_0^{\pi }\text{qty}[{\mathrm{P}}_l^m\text{qty}(\cos \theta ){]}^2\sin \theta \text{dd}\theta {\int }_0^{2\pi }\sin [2]m\varphi \text{dd}\varphi =\frac{\text{qty}(l+m)!}{\text{qty}(l-m)!}\frac{2\pi }{2l+1}.\end{array}$$

(归一化的) 球面调和函数

$$\begin{array}{}\text{(26)}& {\mathrm{Y}}_l^m\text{pqty}\theta ,\varphi =\sqrt{\frac{\text{pqty}l-\text{vqty}m!}{\text{pqty}l+\text{vqty}m!}\frac{2l+1}{4\pi }}{\mathrm{P}}_l^{\text{vqty}m}\text{pqty}\cos \theta e^{im\varphi }\text{qc}m=0,\pm 1,\pm 2,\dots ,\pm l.\end{array}$$

相应的正交归一关系为

$$\begin{array}{}\text{(27)}& {\int }_0^{\pi }{\int }_0^{2\pi }{\mathrm{Y}}_l^m\text{qty}(\theta ,\varphi ){\mathrm{Y}}_k^{n\ast }\text{qty}(\theta ,\varphi )\sin \theta \text{dd}\theta \text{dd}\varphi ={\delta }_{lk}{\delta }_{mn}.\end{array}$$