证明 rot(gradu)=∇×∇u=0 div(rotA→)=∇⋅∇×A

1. 证明 $\operatorname{rot}(\operatorname{grad}u)=\nabla\times\nabla u=0$

根据向量分析中的恒等式，有：

$$
\begin{aligned}
\operatorname{rot}(\operatorname{grad}u) &= \nabla\times(\nabla u)\\
&=\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}\\
\end{vmatrix}\\
&=\left(\frac{\partial^2 u}{\partial y \partial z}-\frac{\partial^2 u}{\partial z \partial y}\right)\mathbf{i}+\left(\frac{\partial^2 u}{\partial z \partial x}-\frac{\partial^2 u}{\partial x \partial z}\right)\mathbf{j}+\left(\frac{\partial^2 u}{\partial x \partial y}-\frac{\partial^2 u}{\partial y \partial x}\right)\mathbf{k}\\
&=\mathbf{0}.
\end{aligned}
$$

因此，$\operatorname{rot}(\operatorname{grad}u)=\nabla\times\nabla u=0$。

2. 证明 $\operatorname{div}(\operatorname{rot}\mathbf{A})=\nabla\cdot(\nabla\times\mathbf{A})$

根据向量分析中的恒等式，有：

$$
\begin{aligned}
\operatorname{div}(\operatorname{rot}\mathbf{A}) &= \nabla\cdot(\nabla\times\mathbf{A})\\
&=\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
\frac{\partial}{\partial y} & \frac{\partial}{\partial z} & \frac{\partial}{\partial x}\\
A_x & A_y & A_z\\
\end{vmatrix}\\
&=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\\
&=\mathbf{0}.
\end{aligned}
$$

因此，$\operatorname{div}(\operatorname{rot}\mathbf{A})=\nabla\cdot(\nabla\times\mathbf{A})=0$。