(1) 设 $(r,\theta)$ 是 $\bbR^2$ 的极坐标, 即 $$\bex x=r\cos\theta,\quad y=r\sin \theta. \eex$$ 证明 Laplace 算子 $\dps{\lap=\frac{\p^2}{\p x^2}+ \frac{\p^2}{\p y^2}}$ 可以表示为 $$\bex \lap u=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}. \eex$$ (2) 设 $(r,\theta,\phi)$ 是 $\bbR^3$ 的极坐标, 即 $$\bex x=r\sin \theta\cos \phi,\quad y=r\sin \theta\sin \phi,\quad z=r\cos \theta. \eex$$ 证明 Laplace 算子 $\dps{\lap=\frac{\p^2}{\p x^2}+ \frac{\p^2}{\p y^2}+\frac{\p^2}{\p z^2}}$ 可以表示为 $$\bex \lap u=\frac{1}{r^2}\frac{\p}{\p r}\sex{r^2\frac{\p u}{\p r}} +\frac{1}{r^2\sin \theta} \frac{\p}{\p \theta}\sex{\sin \theta\frac{\p u}{\p \theta}} +\frac{1}{r^2\sin^2\theta}\frac{\p^2u}{\p \phi^2}. \eex$$
证明: (1) 由 $$\bex x_r=\cos \theta,\quad x_\theta=-r\sin \theta=-y,\quad y_r=\sin \theta,\quad y_\theta=r\cos \theta=x \eex$$ 知 $$\beex \bea u_r&=u_x\cos \theta+u_y\sin \theta,\\ u_{rr}&=u_{xx}\cos^2\theta +2u_{xy}\sin \theta\cos \theta +u_{yy}\sin^2\theta,\\ u_\theta&=-yu_x+xu_y,\\ u_{\theta\theta}&= -xu_x-y(-yu_{xx}+xu_{xy})\\ &\quad-yu_y+x(-yu_{xy}+xu_{yy})\\ &=y^2u_{xx}-2xyu_{xy} +x^2u_{yy}-(xu_x+yu_y). \eea \eeex$$ 而 $$\bex u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta} =u_{xx}+u_{yy}=\lap u. \eex$$ (2) 设 $\rho=r\sin \theta$, 则 $$\bex \ba{ll} x=\rho \cos \phi,&y=\rho\sin \phi,\\ z=r\cos \theta,&\rho=r\sin \theta. \ea \eex$$ 而由 (1), $$\beex \bea u_{xx}+u_{yy}&=u_{\rho\rho} +\frac{1}{\rho}u_\rho+\frac{1}{\rho^2}u_{\phi\phi},\\ u_{zz}+u_{\rho\rho} &=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}. \eea \eeex$$ 两式相加得 $$\bee\label{3.2:1} \lap u=u_{rr}+\frac{1}{r}u_r+\frac{1}{\rho}u_\rho +\frac{1}{r^2}u_{\theta\theta}+\frac{1}{\rho^2}u_{\phi\phi}. \eee$$ 我们再计算 $u_\rho$ 如下: $$\bee\label{3.2:2} \bea u_\rho&=u_rr_\rho+u_\theta\theta_\rho\quad\sex{z=r\cos\theta,\ \rho=r\sin \theta,\quad u(z,\rho)=u(r,\theta)}\\ &=\frac{\rho}{r}u_r+\frac{z}{r^2}u_\theta\\ &=u_r\sin \theta+u_\theta\frac{\cos\theta}{r}. \eea \eee$$ 把 \eqref{3.2:2} 代入 \eqref{3.2:1}, 得 $$\beex \bea \lap u&=u_{rr}+\frac{1}{r}u_r+\frac{1}{r\sin\theta} \sex{u_r\sin \theta+u_\theta\frac{\cos \theta}{r}} +\frac{1}{r^2}u_{\theta\theta} +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=u_{rr}+\frac{2}{r}u_r+ \frac{1}{r^2\sin\theta}\cdot u_\theta\cos \theta +\frac{1}{r^2}u_{\theta\theta} +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=\frac{1}{r^2}(r^2u_{rr}+2ru_r) +\frac{1}{r^2\sin\theta} (u_\theta\cos \theta+u_{\theta\theta}\sin \theta) +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=\frac{1}{r^2}(r^2u_r)_r +\frac{1}{r^2\sin\theta}(u_\theta\sin \theta)_\theta +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}. \eea \eeex$$
