5.1.1球坐标系下的拉普拉斯方程分离变数法求解

合集 - 数学物理方程 姜颖版 笔记(11)

1.3.4施图姆-刘维尔本征值问题2024-11-142.4.1 线性齐次常微分方程解的相关性2024-11-133.4.2二阶线性齐次常微分方程在正常点附近邻域内的级数解法2024-11-144.4.3勒让德方程正常点附近邻域上的求解2024-11-145.4.4二阶齐次常微分方程在正则奇点附近邻域的级数解法2024-11-146.4.5贝塞尔方程求解2024-11-147.4.6二阶非齐次常微分方程的求解2024-11-14

8.5.1.1球坐标系下的拉普拉斯方程分离变数法求解2024-11-14

9.5.1.2勒让德多项式2024-11-1410.5.1.3 勒让德多项式的正交性及相应的广义傅里叶级数2024-11-1411.5.1.4具有极轴转动对称性的拉普拉斯问题求解2024-11-14

收起

拉普拉斯方程的球坐标系解法

$$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial u}{\partial \theta })+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}u}{\partial {\varphi }^2}=0$$

分离变量：$u(r,\theta ,\varphi )=R(r)\mathrm{径}\mathrm{向}\mathrm{函}\mathrm{数}Y(\theta ,\varphi )\mathrm{球}\mathrm{谐}\mathrm{函}\mathrm{数}$

$$\frac{Y(\theta ,\varphi )}{r^2}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)+\frac{R(r)}{r^2\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial Y(\theta ,\varphi )}{\partial \theta })+\frac{R(r)}{r^2{\sin }^2\theta }\frac{{\partial }^{2}Y(\theta ,\varphi )}{\partial {\varphi }^2}=0$$

上式两边同乘以$\times r^2/(YR)$

$$\frac{1}{R(r)}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=-\frac{1}{Y(\theta ,\varphi )\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial Y(\theta ,\varphi )}{\partial \theta })-\frac{1}{Y(\theta ,\varphi ){\sin }^2\theta }\frac{{\partial }^{2}Y(\theta ,\varphi )}{\partial {\varphi }^2}$$

$$\{\begin{array}{ll}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0& \text{径向方程}\\ \frac{1}{\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial Y(\theta ,\varphi )}{\partial \theta })+\frac{1}{{\sin }^2\theta }\frac{{\partial }^{2}Y(\theta ,\varphi )}{\partial {\varphi }^2}+l(l+1)Y(\theta ,\varphi )=0& \text{球谐方程}\end{array}$$

继续分离变量：$Y(\theta ,\varphi )=\mathrm{\Theta }(\theta )\mathrm{\Phi }(\varphi )$

$$\frac{\mathrm{\Phi }(\varphi )}{\sin \theta }\frac{d}{d\theta }(\sin \theta \frac{d\mathrm{\Theta }(\theta )}{d\theta })+\frac{\mathrm{\Theta }(\theta )}{{\sin }^2\theta }\frac{d^2\mathrm{\Phi }(\varphi )}{d{\varphi }^2}+l(l+1)\mathrm{\Theta }(\theta )\mathrm{\Phi }(\varphi )=0$$

上式两边同乘以$\times {\sin }^2\theta /(\mathrm{\Theta }\mathrm{\Phi })$

$$\frac{\sin \theta }{\mathrm{\Theta }(\theta )}\frac{d}{d\theta }(\sin \theta \frac{d\mathrm{\Theta }(\theta )}{d\theta })+l(l+1){\sin }^2\theta =-\frac{1}{\mathrm{\Phi }(\varphi )}\frac{d^2\mathrm{\Phi }(\varphi )}{d{\varphi }^2}$$

$$\{\begin{array}{l}\frac{d^2\mathrm{\Phi }(\varphi )}{d{\varphi }^2}+\lambda \mathrm{\Phi }(\varphi )=0\\ \frac{1}{\sin \theta }\frac{d}{d\theta }(\sin \theta \frac{d\mathrm{\Theta }(\theta )}{d\theta })+[l(l+1)-\frac{\lambda }{{\sin }^2\theta }]\mathrm{\Theta }(\theta )=0\\ \frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0\end{array}$$

径向方程求解

$$\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0⟹r^2\frac{d^{2}R(r)}{d{r}^2}+2r\frac{dR(r)}{dr}-l(l+1)R(r)=0$$

令 $R(r)\sim r^n$：$n(n-1)r^n+2n{r}^n-l(l+1)r^n=0⟹n(n+1)-l(l+1)=0$

径向方程通解：$n_1=l,n_2=-(l+1)\to R_l(r)=A_l{r}^l+B_l\frac{1}{r^{l+1}}$

$\mathrm{\Phi }(\varphi )$ 的方程的求解

$$\frac{d^2\mathrm{\Phi }(\varphi )}{d{\varphi }^2}+\lambda \mathrm{\Phi }(\varphi )=0$$

函数单值性：$\mathrm{\Phi }(\varphi )=\mathrm{\Phi }(\varphi +2\pi )$ (自然边条件)

刘维尔本征值问题

$${\lambda }_m=m^2,m=0,\pm 1,\pm 2,\cdots ,{\mathrm{\Phi }}_m(\varphi )=e^{im\varphi }$$

$${\lambda }_m=m^2,m=0,1,2,\cdots ,{\mathrm{\Phi }}_m(\varphi )=\{\begin{array}{l}\sin m\varphi \\ \cos m\varphi \end{array}$$

$\mathrm{\Theta }(\theta )$ 满足的方程

$$\frac{1}{\sin \theta }\frac{d}{d\theta }(\sin \theta \frac{d\mathrm{\Theta }(\theta )}{d\theta })+[l(l+1)-\frac{m^2}{{\sin }^2\theta }]\mathrm{\Theta }(\theta )=0,0\le \theta \le \pi$$

变量代换：$x\equiv \cos \theta ,y(x)\equiv \mathrm{\Theta }(\theta )$

$$-\frac{1}{\sin \theta }\frac{d}{d\theta }=\frac{d}{dx},{\sin }^2\theta =1-x^2$$

$$\frac{d}{dx}[(1-x^2)\frac{dy(x)}{dx}]+[l(l+1)-\frac{m^2}{1-x^2}]y(x)=0$$

$$\downarrow$$

$$(1-x^2)\frac{d^{2}y(x)}{d{x}^2}-2x\frac{dy(x)}{dx}+[l(l+1)-\frac{m^2}{1-x^2}]y(x)=0-1\le x\le 1\text{缔合勒让德方程}$$

系统具有极轴转动对称性

$\mathrm{\Phi }(\varphi )=\text{Const.}m=0$

$$(1-x^2)\frac{d^{2}y(x)}{d{x}^2}-2x\frac{dy(x)}{dx}+l(l+1)y(x)=0\text{勒让德方程}$$