方企勤 第五章 多元函数微分学 第8题
📝 题目
例 8 设 $u = f\left( r\right) ,r = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$ ,若 $u$ 满足调和方程
$$ {\nabla }^{2}u = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} + \frac{{\partial }^{2}u}{\partial {z}^{2}} = 0, $$
试求出函数 $u$ .
💡 答案解析
解 因
$$ \frac{\partial u}{\partial x} = {f}^{\prime }\left( r\right) \frac{x}{r},\;\frac{\partial u}{\partial y} = {f}^{\prime }\left( r\right) \frac{y}{r},\;\frac{\partial u}{\partial z} = {f}^{\prime }\left( r\right) \frac{z}{r}, $$
所以
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} = {f}^{\prime \prime }\left( r\right) \frac{{x}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{r - {x}^{2}/r}{{r}^{2}} $$
$$ = {f}^{\prime \prime }\left( r\right) \frac{{x}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{{y}^{2} + {z}^{2}}{{r}^{3}}. $$
同理可得
$$ \frac{{\partial }^{2}u}{\partial {y}^{2}} = {f}^{\prime \prime }\left( r\right) \frac{{y}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{{z}^{2} + {x}^{2}}{{r}^{3}}, $$
$$ \frac{{\partial }^{2}u}{\partial {z}^{2}} = {f}^{\prime \prime }\left( r\right) \frac{{z}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{{x}^{2} + {y}^{2}}{{r}^{3}}. $$
由条件得
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} + \frac{{\partial }^{2}u}{\partial {z}^{2}} = {f}^{\prime \prime }\left( r\right) + \frac{2}{r}{f}^{\prime }\left( r\right) = 0, $$
或
$$ {r}^{2}{f}^{\prime \prime }\left( r\right) + {2r}{f}^{\prime }\left( r\right) = 0, $$
于是有 ${\left\lbrack {r}^{2}{f}^{\prime }\left( r\right) \right\rbrack }^{\prime } = 0$ ,推得 ${f}^{\prime }\left( r\right) = \frac{C}{{r}^{2}}$ ,解出 $f\left( r\right) = - \frac{C}{r} + {C}_{1}$ ,其中 $C,{C}_{1}$ 为任意常数.