同济高数 第11章 第11-7-*6题

[AI解答]

[AI解答]

要证明旋度的线性性质，即
$$
\operatorname{rot}(a+b)=\operatorname{rot} a+\operatorname{rot} b,
$$
其中 $a, b$ 是三维空间中的向量场。

设
$$
a = (P_1, Q_1, R_1),\quad b = (P_2, Q_2, R_2),
$$
则
$$
a+b = (P_1+P_2,\; Q_1+Q_2,\; R_1+R_2).
$$

旋度定义为
$$
\operatorname{rot} F = \nabla \times F =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\displaystyle{\frac{\partial}{\partial x}} & \displaystyle{\frac{\partial}{\partial y}} & \displaystyle{\frac{\partial}{\partial z}} \\
F_x & F_y & F_z
\end{vmatrix}.
$$

因此，对于 $a+b$，有
$$
\operatorname{rot}(a+b) =
\left(
\frac{\partial (R_1+R_2)}{\partial y} - \frac{\partial (Q_1+Q_2)}{\partial z},\;
\frac{\partial (P_1+P_2)}{\partial z} - \frac{\partial (R_1+R_2)}{\partial x},\;
\frac{\partial (Q_1+Q_2)}{\partial x} - \frac{\partial (P_1+P_2)}{\partial y}
\right).
$$

利用偏导数的线性性质，将每一项拆开：
$$
\frac{\partial (R_1+R_2)}{\partial y} = \frac{\partial R_1}{\partial y} + \frac{\partial R_2}{\partial y},
$$
其他项同理。于是得到：
$$
\operatorname{rot}(a+b) =
\left(
\frac{\partial R_1}{\partial y} - \frac{\partial Q_1}{\partial z} + \frac{\partial R_2}{\partial y} - \frac{\partial Q_2}{\partial z},\;
\frac{\partial P_1}{\partial z} - \frac{\partial R_1}{\partial x} + \frac{\partial P_2}{\partial z} - \frac{\partial R_2}{\partial x},\;
\frac{\partial Q_1}{\partial x} - \frac{\partial P_1}{\partial y} + \frac{\partial Q_2}{\partial x} - \frac{\partial P_2}{\partial y}
\right).
$$

这恰好是：
$$
\operatorname{rot} a + \operatorname{rot} b =
\left(
\frac{\partial R_1}{\partial y} - \frac{\partial Q_1}{\partial z},\;
\frac{\partial P_1}{\partial z} - \frac{\partial R_1}{\partial x},\;
\frac{\partial Q_1}{\partial x} - \frac{\partial P_1}{\partial y}
\right)
+
\left(
\frac{\partial R_2}{\partial y} - \frac{\partial Q_2}{\partial z},\;
\frac{\partial P_2}{\partial z} - \frac{\partial R_2}{\partial x},\;
\frac{\partial Q_2}{\partial x} - \frac{\partial P_2}{\partial y}
\right).
$$

因此
$$
\operatorname{rot}(a+b) = \operatorname{rot} a + \operatorname{rot} b,
$$
得证。

难度：★☆☆☆☆