人邮高数 第6章 第6-3-26题

[AI解答]

[AI解答] 已知 $ u = f\left(xy, \frac{x}{y}\right) $，其中 $ f $ 具有二阶连续偏导数。 令 $$ \xi = xy,\quad \eta = \frac{x}{y} $$ 则 $ u = f(\xi, \eta) $。

首先求一阶偏导：

对 $ x $ 求偏导： $$ \frac{\partial u}{\partial x} = f_1 \cdot \frac{\partial \xi}{\partial x} + f_2 \cdot \frac{\partial \eta}{\partial x} = f_1 \cdot y + f_2 \cdot \frac{1}{y} $$ 其中 $ f_1 = \frac{\partial f}{\partial \xi} $，$ f_2 = \frac{\partial f}{\partial \eta} $。

再对 $ y $ 求偏导，即求 $\frac{\partial^2 u}{\partial y \partial x} = \frac{\partial}{\partial y} \left( y f_1 + \frac{1}{y} f_2 \right)$：

$$ \frac{\partial}{\partial y} (y f_1) = f_1 + y \cdot \frac{\partial f_1}{\partial y} $$ 而 $$ \frac{\partial f_1}{\partial y} = f_{11} \cdot \frac{\partial \xi}{\partial y} + f_{12} \cdot \frac{\partial \eta}{\partial y} = f_{11} \cdot x + f_{12} \cdot \left(-\frac{x}{y^2}\right) $$ 所以 $$ \frac{\partial}{\partial y}(y f_1) = f_1 + y\left( x f_{11} - \frac{x}{y^2} f_{12} \right) = f_1 + x y f_{11} - \frac{x}{y} f_{12} $$

再处理第二项： $$ \frac{\partial}{\partial y} \left( \frac{1}{y} f_2 \right) = -\frac{1}{y^2} f_2 + \frac{1}{y} \cdot \frac{\partial f_2}{\partial y} $$ 而 $$ \frac{\partial f_2}{\partial y} = f_{21} \cdot \frac{\partial \xi}{\partial y} + f_{22} \cdot \frac{\partial \eta}{\partial y} = f_{21} \cdot x + f_{22} \cdot \left( -\frac{x}{y^2} \right) $$ 由于 $ f $ 二阶连续偏导，所以 $ f_{12} = f_{21} $。于是 $$ \frac{\partial}{\partial y} \left( \frac{1}{y} f_2 \right) = -\frac{1}{y^2} f_2 + \frac{1}{y} \left( x f_{21} - \frac{x}{y^2} f_{22} \right) = -\frac{1}{y^2} f_2 + \frac{x}{y} f_{12} - \frac{x}{y^3} f_{22} $$

将两部分相加： $$ \frac{\partial^2 u}{\partial x \partial y} = \left( f_1 + x y f_{11} - \frac{x}{y} f_{12} \right) + \left( -\frac{1}{y^2} f_2 + \frac{x}{y} f_{12} - \frac{x}{y^3} f_{22} \right) $$ 其中 $-\frac{x}{y} f_{12} + \frac{x}{y} f_{12} $ 抵消，得到最终结果：

$$ \boxed{\frac{\partial^{2} u}{\partial x \partial y} = f_1 + x y f_{11} - \frac{1}{y^{2}} f_2 - \frac{x}{y^{3}} f_{22}} $$

难度：★★★☆☆