问题及解答

假设 $r=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$, 求 $\Delta r$, $\Delta r^{-1}$, $\Delta\ln r$. 这里 $\Delta$ 是 Laplace 算子,   $\Delta=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}$. 

一般的, 若 $r=\sqrt{\sum_{i=1}^{n}(x_i-x_i^0)^2}$, 且 $f(r)$ 二阶可导, 证明: $\Delta f(r)=f''(r)+\dfrac{n-1}{r}f'(r)$.

\[
r'_x=\frac{\partial r}{\partial x}=\frac{2(x-x_0)}{2\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}=\frac{x-x_0}{r},
\]

类似的, $r'_y=\dfrac{y-y_0}{r}$, $r'_z=\dfrac{z-z_0}{r}$.

(1)

\[
r''_{xx}=\frac{\partial^2 r}{\partial x^2}=(\frac{x-x_0}{r})'_x=\frac{1\cdot r-(x-x_0)r'_x}{r^2}=\frac{r-(x-x_0)\cdot\frac{x-x_0}{r}}{r^2}=\frac{r^2-(x-x_0)^2}{r^3}.
\]

类似地, 

\[
\begin{aligned}
r''_{yy}&=\frac{\partial^2 r}{\partial y^2}=\frac{r^2-(y-y_0)^2}{r^3},\\
r''_{zz}&=\frac{\partial^2 r}{\partial z^2}=\frac{r^2-(z-z_0)^2}{r^3},
\end{aligned}
\]

因此

\[
\begin{split}
\Delta r&=\frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 r}{\partial y^2}+\frac{\partial^2 r}{\partial z^2}\\
&=\frac{r^2-(x-x_0)^2}{r^3}+\frac{r^2-(y-y_0)^2}{r^3}+\frac{r^2-(z-z_0)^2}{r^3}\\
&=\frac{3r^2-[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]}{r^3}\\
&=\frac{3r^2-r^2}{r^3}=\frac{2}{r}.
\end{split}
\]

(2)

\[
\frac{\partial}{\partial x}r^{-1}=-\frac{1}{r^2}\cdot r'_x=-\frac{1}{r^2}\cdot\frac{x-x_0}{r}=-\frac{x-x_0}{r^3}.
\]

类似地,

\[
\frac{\partial}{\partial y}r^{-1}=-\frac{y-y_0}{r^3},\quad\frac{\partial}{\partial z}r^{-1}=-\frac{z-z_0}{r^3}.
\]

求二阶导数,

\[
\frac{\partial^2}{\partial x^2}r^{-1}=\frac{\partial}{\partial x}(-\frac{x-x_0}{r^3})=-\frac{1\cdot r^3-(x-x_0)3r^2\cdot r'_x}{r^6}=-\frac{r^3-(x-x_0)3r^2\cdot\frac{x-x_0}{r}}{r^6}=-\frac{r^2-3(x-x_0)^2}{r^5},
\]

类似地, 

\[
\begin{aligned}
\frac{\partial^2}{\partial y^2}r^{-1}&=-\frac{r^2-3(y-y_0)^2}{r^5},\\
\frac{\partial^2}{\partial z^2}r^{-1}&=-\frac{r^2-3(z-z_0)^2}{r^5},\\
\end{aligned}
\]

因此,

\[
\begin{split}
\Delta r^{-1}&=\frac{\partial^2}{\partial x^2}r^{-1}+\frac{\partial^2}{\partial y^2}r^{-1}+\frac{\partial^2}{\partial z^2}r^{-1}\\
&=-\frac{r^2-3(x-x_0)^2}{r^5}-\frac{r^2-3(y-y_0)^2}{r^5}-\frac{r^2-3(z-z_0)^2}{r^5}\\
&=-\frac{3r^2-3r^2}{r^5}=0.
\end{split}
\]

(3)

\[
\frac{\partial}{\partial x}\ln r=\frac{1}{r}\cdot r'_x=\frac{x-x_0}{r^2},
\]

类似地,

\[
\frac{\partial}{\partial y}\ln r=\frac{y-y_0}{r^2},\quad\frac{\partial}{\partial z}\ln r=\frac{z-z_0}{r^2}.
\]

求二阶导数

\[
\frac{\partial^2}{\partial x^2}\ln r=\frac{\partial}{\partial x}(\frac{x-x_0}{r^2})=\frac{1\cdot r^2-(x-x_0)2r\cdot r'_x}{r^4}=\frac{r^2-(x-x_0)2r\cdot\frac{x-x_0}{r}}{r^4}=\frac{r^2-2(x-x_0)^2}{r^4},
\]

于是

\[
\begin{aligned}
\frac{\partial^2}{\partial y^2}\ln r&=\frac{r^2-2(y-y_0)^2}{r^4},\\
\frac{\partial^2}{\partial z^2}\ln r&=\frac{r^2-2(z-z_0)^2}{r^4},\\
\end{aligned}
\]

因此

\[
\begin{split}
\Delta\ln r&=\frac{\partial^2}{\partial x^2}\ln r+\frac{\partial^2}{\partial y^2}\ln r+\frac{\partial^2}{\partial z^2}\ln r\\
&=\frac{r^2-2(x-x_0)^2}{r^4}+\frac{r^2-2(y-y_0)^2}{r^4}+\frac{r^2-2(z-z_0)^2}{r^4}\\
&=\frac{3r^2-2r^2}{r^4}\\
&=\frac{1}{r^2}.
\end{split}
\]

Pf. $\dfrac{\partial}{\partial x_i}f(r)=f'(r)\cdot\dfrac{\partial r}{\partial x_i}=f'(r)\cdot\dfrac{x_i-x_i^0}{r}$, 于是

\[
\begin{split}
\frac{\partial^2}{\partial x_i^2}f(r)&=\biggl(f'(r)\cdot\frac{x_i-x_0}{r}\biggr)'_{x_i}\\
&=f''(r)\cdot r'_{x_i}\cdot\frac{x_i-x_i^0}{r}+f'(r)\cdot\frac{1\cdot r-(x_i-x_i^0)\cdot r'_{x_i}}{r^2}\\
&=f''(r)\cdot\frac{x_i-x_i^0}{r}\cdot\frac{x_i-x_i^0}{r}+f'(r)\cdot\frac{r-(x_i-x_i^0)\cdot\frac{x_i-x_i^0}{r}}{r^2}\\
&=f''(r)\cdot\frac{(x_i-x_i^0)^2}{r^2}+f'(r)\cdot\frac{r^2-(x_i-x_i^0)^2}{r^3}.
\end{split}
\]

于是

\[
\begin{split}
\Delta f(r)=\sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}f(r)&=f''(r)\cdot\frac{1}{r^2}\sum_{i=1}^{n}(x_i-x_i^0)^2+\frac{f'(r)}{r^3}\sum_{i=1}^{n}\bigl[r^2-(x_i-x_i^0)^2\bigr]\\
&=f''(r)\cdot\frac{1}{r^2}\cdot r^2+\frac{f'(r)}{r^3}\cdot(n-1)r^2\\
&=f''(r)+\frac{n-1}{r}f'(r).
\end{split}
\]