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。）