新讲 第12章 多元微分学 第3题
📝 题目
例 3 二维拉普拉斯 (Laplace) 算子 $\Delta$ 定义如下:
$$ {\Delta u} = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}. $$
试对 $u = \ln \frac{1}{r}\left( {r > 0}\right)$ 计算 ${\Delta u}$ ,这里 $r = \sqrt{{x}^{2} + {y}^{2}}$ .
💡 答案解析
解 我们有
$$ u = \ln \frac{1}{r} = - \frac{1}{2}\ln \left( {{x}^{2} + {y}^{2}}\right) , $$
$$ \frac{\partial u}{\partial x} = - \frac{x}{{x}^{2} + {y}^{2}},\;\frac{\partial u}{\partial y} = - \frac{y}{{x}^{2} + {y}^{2}}, $$
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} = - \frac{\left( {{x}^{2} + {y}^{2}}\right) - 2{x}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}} = \frac{{x}^{2} - {y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}, $$
$$ \frac{{\partial }^{2}u}{\partial {y}^{2}} = \frac{{y}^{2} - {x}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}, $$
因而
$$ {\Delta u} = \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} = 0. $$