问题及解答

设 $r=\sqrt{x^2+y^2+z^2}$, 证明: $\frac{\partial^2(\ln r)}{\partial x^2}+\frac{\partial^2(\ln r)}{\partial y^2}+\frac{\partial^2(\ln r)}{\partial z^2}=\frac{1}{r^2}$. 即 $\Delta(\ln r)=\frac{1}{r^2}$.

\[
\frac{\partial}{\partial x}(\ln r)=\frac{1}{r}\cdot\frac{\partial r}{\partial x}=\frac{1}{r}\cdot\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac{1}{r}\cdot\frac{x}{r}=\frac{x}{r^2}
\]

类似的,

\[
\frac{\partial}{\partial y}(\ln r)=\frac{y}{r^2},\quad \frac{\partial}{\partial z}(\ln r)=\frac{z}{r^2}
\]

\[
\begin{split}
\frac{\partial^2}{\partial x^2}(\ln r)&=\frac{\partial}{\partial x}(\frac{x}{r^2})=\frac{1\cdot r^2-x\cdot 2r\cdot r'_x}{r^4}\\
&=\frac{r^2-2rx\cdot\frac{x}{r}}{r^4}=\frac{r^2-2x^2}{r^4}.
\end{split}
\]

类似的, 有

\[
\frac{\partial^2}{\partial y^2}(\ln r)=\frac{r^2-2y^2}{r^4},\quad \frac{\partial^2}{\partial z^2}(\ln r)=\frac{r^2-2z^2}{r^4}.
\]

于是

\[
\begin{split}
\Delta(\ln r)&=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(\ln r)\\
&=\frac{r^2-2x^2}{r^4}+\frac{r^2-2y^2}{r^4}+\frac{r^2-2z^2}{r^4}\\
&=\frac{3r^2-2r^2}{r^4}=\frac{1}{r^2}.
\end{split}
\]