拉普拉斯方程的球坐标系解法
\[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2u}{\partial\phi^2}=0 \]分离变量：$u(r,\theta,\phi)=R(r) 径向函数 Y(\theta,\phi) 球谐函数 $
\[\frac{Y(\theta,\phi)}{r^2}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)+\frac{R(r)}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y(\theta,\phi)}{\partial\theta}\right)+\frac{R(r)}{r^2\sin^2\theta}\frac{\partial^2Y(\theta,\phi)}{\partial\phi^2}=0 \]上式两边同乘以\(\times r^2/(YR)\)
\[\frac{1}{R(r)}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=-\frac{1}{Y(\theta,\phi)\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y(\theta,\phi)}{\partial\theta}\right)-\frac{1}{Y(\theta,\phi)\sin^2\theta}\frac{\partial^2Y(\theta,\phi)}{\partial\phi^2} \]\[\begin{cases} \frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0 & \text{径向方程} \\ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y(\theta,\phi)}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2Y(\theta,\phi)}{\partial\phi^2}+l(l+1)Y(\theta,\phi)=0 & \text{球谐方程} \end{cases}\]继续分离变量：\(Y(\theta,\phi)=\Theta(\theta)\Phi(\phi)\)
\[\frac{\Phi(\phi)}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta(\theta)}{d\theta}\right)+\frac{\Theta(\theta)}{\sin^2\theta}\frac{d^2\Phi(\phi)}{d\phi^2}+l(l+1)\Theta(\theta)\Phi(\phi)=0 \]上式两边同乘以\(\times \sin^2\theta/(\Theta\Phi)\)
\[\frac{\sin\theta}{\Theta(\theta)}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta(\theta)}{d\theta}\right)+l(l+1)\sin^2\theta=-\frac{1}{\Phi(\phi)}\frac{d^2\Phi(\phi)}{d\phi^2} \]\[\begin{cases} \frac{d^2\Phi(\phi)}{d\phi^2}+\lambda\Phi(\phi)=0 \\ \frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta(\theta)}{d\theta}\right)+\left[l(l+1)-\frac{\lambda}{\sin^2\theta}\right]\Theta(\theta)=0 \\ \frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0 \end{cases}\]径向方程求解
\[\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-l(l+1)R(r)=0 \implies r^2\frac{d^2R(r)}{dr^2}+2r\frac{dR(r)}{dr}-l(l+1)R(r)=0 \]令 \(R(r)\sim r^n\)：\(n(n-1)r^n+2nr^n-l(l+1)r^n=0 \implies n(n+1)-l(l+1)=0\)
径向方程通解：\(n_1=l,\quad n_2=-(l+1) \rightarrow R_l(r)=A_lr^l+B_l\frac{1}{r^{l+1}}\)
\(\Phi(\phi)\) 的方程的求解
\[\frac{d^2\Phi(\phi)}{d\phi^2}+\lambda\Phi(\phi)=0 \]函数单值性：\(\Phi(\phi)=\Phi(\phi+2\pi)\) (自然边条件)
刘维尔本征值问题
\[\lambda_m=m^2,\quad m=0,\pm1,\pm2,\cdots,\quad \Phi_m(\phi)=e^{im\phi} \]\[\lambda_m=m^2,\quad m=0,1,2,\cdots,\quad \Phi_m(\phi)=\begin{cases}\sin m\phi \\ \cos m\phi\end{cases} \]\(\Theta(\theta)\) 满足的方程
\[\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta(\theta)}{d\theta}\right)+\left[l(l+1)-\frac{m^2}{\sin^2\theta}\right]\Theta(\theta)=0,\quad 0\leq\theta\leq\pi \]变量代换：\(x\equiv\cos\theta,\quad y(x)\equiv\Theta(\theta)\)
\[-\frac{1}{\sin\theta}\frac{d}{d\theta}=\frac{d}{dx},\quad \sin^2\theta=1-x^2 \]\[\frac{d}{dx}\left[(1-x^2)\frac{dy(x)}{dx}\right]+\left[l(l+1)-\frac{m^2}{1-x^2}\right]y(x)=0 \]\[\downarrow \]\[(1-x^2)\frac{d^2y(x)}{dx^2}-2x\frac{dy(x)}{dx}+\left[l(l+1)-\frac{m^2}{1-x^2}\right]y(x)=0 \quad -1\leq x\leq1 \quad \text{缔合勒让德方程} \]系统具有极轴转动对称性
\(\Phi(\phi)=\text{Const.}\quad m=0\)
\[(1-x^2)\frac{d^2y(x)}{dx^2}-2x\frac{dy(x)}{dx}+l(l+1)y(x)=0 \quad \text{勒让德方程} \] 标签：5.1,phi,right,frac,拉普拉斯,sin,theta,坐标系,partial From： https://www.cnblogs.com/RES-HON/p/18546765