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。

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