新讲 第12章 多元微分学 第4题
📝 题目
例 4 三维拉普拉斯算子 $\Delta$ 定义如下:
$$ {\Delta u} = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} + \frac{{\partial }^{2}u}{\partial {z}^{2}}. $$
试对 $u = \frac{1}{r}\left( {r \neq 0}\right)$ 计算 ${\Delta u}$ ,这里 $r = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$ .
💡 答案解析
解 我们有
$$ u = {\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{-1/2}, $$
$$ \frac{\partial u}{\partial x} = - \frac{1}{2}{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{-3/2} \cdot {2x} $$
$$ = - x{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{-3/2}, $$
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} = - {\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{-3/2} + 3{x}^{2}{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{-5/2} $$
$$ = \frac{2{x}^{2} - {y}^{2} - {z}^{2}}{{r}^{5}}, $$
$$ \frac{{\partial }^{2}u}{\partial {y}^{2}} = \frac{2{y}^{2} - {z}^{2} - {x}^{2}}{{r}^{5}}, $$
$$ \frac{{\partial }^{2}u}{\partial {z}^{2}} = \frac{2{z}^{2} - {x}^{2} - {y}^{2}}{{r}^{5}}, $$
因而
$$ {\Delta u} = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} + \frac{{\partial }^{2}u}{\partial {z}^{2}} = 0. $$