同济高数 第11章 第11-7-*7题

[AI解答]

[AI解答]

我们要求的是旋度作用于梯度，即 $$ \operatorname{rot}(\operatorname{grad} u) $$ 其中 $ u = u(x, y, z) $ 具有二阶连续偏导数。

首先，梯度的表达式为： $$ \operatorname{grad} u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right) $$

旋度的计算公式为： $$ \operatorname{rot} \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \displaystyle{\frac{\partial}{\partial x}} & \displaystyle{\frac{\partial}{\partial y}} & \displaystyle{\frac{\partial}{\partial z}} \\ F_x & F_y & F_z \end{vmatrix} $$

将 $\mathbf{F} = \operatorname{grad} u$ 代入，即： $$ F_x = \frac{\partial u}{\partial x}, \quad F_y = \frac{\partial u}{\partial y}, \quad F_z = \frac{\partial u}{\partial z} $$

于是： $$ \operatorname{rot}(\operatorname{grad} u) = \left( \frac{\partial}{\partial y}\left( \frac{\partial u}{\partial z} \right) - \frac{\partial}{\partial z}\left( \frac{\partial u}{\partial y} \right), \; \frac{\partial}{\partial z}\left( \frac{\partial u}{\partial x} \right) - \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial z} \right), \; \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial y} \right) - \frac{\partial}{\partial y}\left( \frac{\partial u}{\partial x} \right) \right) $$

由于 $u$ 具有二阶连续偏导数，混合偏导数与顺序无关，即： $$ \frac{\partial^2 u}{\partial y \partial z} = \frac{\partial^2 u}{\partial z \partial y}, \quad \frac{\partial^2 u}{\partial z \partial x} = \frac{\partial^2 u}{\partial x \partial z}, \quad \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x} $$

因此，每个分量均为零： $$ \operatorname{rot}(\operatorname{grad} u) = (0, 0, 0) = \mathbf{0} $$

结论： $$ \boxed{\operatorname{rot}(\operatorname{grad} u) = \mathbf{0}} $$

难度：★☆☆☆☆（这是向量分析中的基本恒等式，只需直接代入定义并利用混合偏导可交换性即可证明。）