由球坐标, $r=\sqrt{x^2+y^2+z^2}$, $\theta=\arccos\frac{z}{r}$, $\phi=\arctan\frac{y}{x}$. 于是
\[
r_x=\frac{2x}{2\sqrt{x^2+y^2+z^2}}=\frac{x}{r},\quad r_y=\frac{y}{r},\quad r_z=\frac{z}{r}.
\]
\[
\theta_x=-\frac{(\frac{z}{r})'_x}{\sqrt{1-(\frac{z}{r})^2}}=-\frac{-\frac{z}{r^2}\cdot r'_x}{\frac{\sqrt{r^2-z^2}}{r}}=\frac{xz}{r^2\sqrt{r^2-z^2}},
\]
同理
\[
\theta_y=\frac{yz}{r^2\sqrt{r^2-z^2}},
\]
\[
\theta_z=-\frac{(\frac{z}{r})'_z}{\sqrt{1-(\frac{z}{r})^2}}=-\frac{1\cdot\frac{1}{r}+z\cdot\frac{-1}{r^2}\cdot r'_z}{\frac{\sqrt{r^2-z^2}}{r}}=-\frac{1-\frac{z^2}{r^2}}{\sqrt{r^2-z^2}}=-\frac{r^2-z^2}{r^2\sqrt{r^2-z^2}}=-\frac{\sqrt{r^2-z^2}}{r^2}.
\]
\[
\phi_x=\frac{1}{1+(\frac{y}{x})^2}\cdot(\frac{y}{x})'_x=\frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}}=-\frac{y}{x^2+y^2}
\]
\[
\phi_y=\frac{\frac{1}{x}}{1+(\frac{y}{x})^2}=\frac{x}{x^2+y^2},
\]
\[
\phi_z=0.
\]
下面分别计算 $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$, $\frac{\partial}{\partial z}$.
\[
\begin{split}
\partial_x &=r_x\partial_r+\theta_x\partial_{\theta}+\phi_x\partial_{\phi}\\
&=\frac{x}{r}\partial_r+\frac{xz}{r^2\sqrt{r^2-z^2}}\partial_{\theta}-\frac{y}{x^2+y^2}\partial_{\phi}
\end{split}
\]
注意到 $r^2-z^2=x^2+y^2=r^2\sin^2\theta$, 故
\[
\frac{xz}{r^2\sqrt{r^2-z^2}}=\frac{\sin\theta\cos\theta\cos\phi}{r\sin\theta}=\frac{\cos\theta\cos\phi}{r},
\]
\[
\frac{y}{x^2+y^2}=\frac{r\sin\theta\sin\phi}{r^2\sin^2\theta}=\frac{\sin\phi}{r\sin\theta},
\]
\[
\frac{x}{x^2+y^2}=\frac{r\sin\theta\cos\phi}{r^2\sin^2\theta}=\frac{\cos\phi}{r\sin\theta},
\]
代入 $\partial_x$ 的表达式, 得
\[
\partial_x=\sin\theta\cos\phi\partial_r+\frac{\cos\theta\cos\phi}{r}\partial_{\theta}-\frac{\sin\phi}{r\sin\theta}\partial_{\phi}.
\]
类似地, 可以计算得
\[
\begin{split}
\partial_y&=r_y\partial_r+\theta_y\partial_{\theta}+\phi_y\partial_{\phi}\\
&=\frac{y}{r}\partial_r+\frac{yz}{r^2\sqrt{r^2-z^2}}\partial_{\theta}+\frac{x}{x^2+y^2}\partial_{\phi}\\
&=\sin\theta\sin\phi\partial_r+\frac{\cos\theta\sin\phi}{r}\partial_{\theta}+\frac{\cos\phi}{r\sin\theta}\partial_{\phi}.
\end{split}
\]
\[
\begin{split}
\partial_z&=r_z\partial_r+\theta_z\partial_{\theta}+\phi_z\partial_{\phi}\\
&=\frac{z}{r}\partial_r-\frac{\sqrt{r^2-z^2}}{r^2}\partial_{\theta}\\
&=\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}.
\end{split}
\]
然后分别计算 $\frac{\partial^2}{\partial x^2}$, $\frac{\partial^2}{\partial y^2}$, $\frac{\partial^2}{\partial z^2}$.
\[
\begin{split}
\partial_{xx}&=\Bigl(\sin\theta\cos\phi\partial_r+\frac{\cos\theta\cos\phi}{r}\partial_{\theta}-\frac{\sin\phi}{r\sin\theta}\partial_{\phi}\Bigr)\Bigl(\sin\theta\cos\phi\partial_r+\frac{\cos\theta\cos\phi}{r}\partial_{\theta}-\frac{\sin\phi}{r\sin\theta}\partial_{\phi}\Bigr)\\
&=\sin^2\theta\cos^2\phi\partial_{rr}+\bigl[(\sin\theta\cos\phi\cdot\cos\theta\cos\phi)\frac{-1}{r^2}\partial_{\theta}+\frac{\sin\theta\cos\theta\cos^2\phi}{r}\partial_{\theta r}\bigr]+\bigl[\sin\theta\cos\phi\cdot\frac{\sin\phi}{\sin\theta}\frac{1}{r^2}\partial_{\phi}-\frac{\sin\phi\cos\phi}{r}\partial_{\phi r}\bigr]\\
&\quad+\bigl[\frac{\cos^2\theta\cos^2\phi}{r}\partial_r+\frac{\sin\theta\cos\theta\cos^2\phi}{r}\partial_{r\theta}\bigr]+\bigl[\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\partial_{\theta}+\frac{\cos^2\theta\cos^2\phi}{r^2}\partial_{\theta\theta}\bigr]\\
&\qquad\qquad+\bigl[\frac{\cos\theta\cos\phi}{r}\cdot(-\frac{\sin\phi}{r})\cdot\frac{-\cos\theta}{\sin^2\theta}\partial_{\phi}-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\phi}\partial_{\phi\theta}\bigr]\\
&\quad+\bigl[\frac{\sin^2\phi}{r}\partial_r-\frac{\sin\phi\cos\phi}{r}\partial_{r\phi}\bigr]+\bigl[\frac{\cos\theta\sin^2\phi}{r^2\sin\theta}\partial_{\theta}-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}\partial_{\theta\phi}\bigr]\\
&\qquad\qquad+\bigl[\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}\partial_{\phi}+\frac{\sin^2\phi}{r^2\sin^2\theta}\partial_{\phi\phi}\bigr].
\end{split}
\]
我们将其整理一下, 二阶导数项放在左侧, 一阶导数项放在右侧.
\[
\begin{split}
\partial_{xx}&=\sin^2\theta\cos^2\phi\partial_{rr}+\frac{\sin\theta\cos\theta\cos^2\phi}{r}\partial_{\theta r}-\frac{\sin\phi\cos\phi}{r}\partial_{\phi r}+0\partial_r-\frac{\sin\theta\cos\theta\cos^2\phi}{r^2}\partial_{\theta}+\frac{\sin\phi\cos\phi}{r^2}\partial_{\phi}\\
&\quad+\frac{\sin\theta\cos\theta\cos^2\phi}{r}\partial_{r\theta}+\frac{\cos^2\theta\cos^2\phi}{r^2}\partial_{\theta\theta}-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}\partial_{\phi\theta}+\frac{\cos^2\theta\cos^2\phi}{r}\partial_r-\frac{\sin\theta\cos\theta\cos^2\phi}{r^2}\partial_{\theta}+\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta}\partial_{\phi}\\
&\quad-\frac{\sin\phi\cos\phi}{r}\partial_{r\phi}-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}\partial_{\theta\phi}+\frac{\sin^2\phi}{r^2\sin^2\theta}\partial_{\phi\phi}+\frac{\sin^2\phi}{r}\partial_r+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta}\partial_{\theta}+\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}\partial_{\phi}.
\end{split}
\]
