📝 题目
例 2 设 $\mathbf{F} = f\left( r\right) \mathbf{r}$ ( $r$ 与 $\mathbf{r}$ 意义同例 1).
(1)求证: $\operatorname{rot}\mathbf{F} \equiv 0$ ；
(2) $f\left( r\right)$ 是什么函数时, $\operatorname{div}\mathbf{F} \equiv 0$ .
💡 答案与解析
解 (1) 令 $P = f\left( r\right) x,Q = f\left( r\right) y,R = f\left( r\right) z$ ,则
$$ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = {f}^{\prime }\left( r\right) \frac{yz}{r} - {f}^{\prime }\left( r\right) \frac{yz}{r} \equiv 0, $$
同理 $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \equiv 0,\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \equiv 0$ . 所以 $\operatorname{rot}\mathbf{F} \equiv 0$ .
(2) $\frac{\partial P}{\partial x} = f\left( r\right) + {f}^{\prime }\left( r\right) \frac{{x}^{2}}{r},\frac{\partial Q}{\partial y} = f\left( r\right) + {f}^{\prime }\left( r\right) \frac{{y}^{2}}{r}$ ,
$$ \frac{\partial R}{\partial z} = f\left( r\right) + {f}^{\prime }\left( r\right) \frac{{z}^{2}}{r}. $$
要使
$$ \operatorname{div}\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = {3f}\left( r\right) + r{f}^{\prime }\left( r\right) = 0, $$
只要
$$ 3{r}^{2}f\left( r\right) + {r}^{3}{f}^{\prime }\left( r\right) = {\left( {r}^{3}f\left( r\right) \right) }^{\prime } = 0, $$
所以 $f\left( r\right) = \frac{c}{{r}^{3}}$ ( $c$ 为任意常数)时, $\operatorname{div}\mathbf{F} \equiv 0$ .