设$\vec{v},\vec{\omega}$散度和旋度均相等，问：在什么条件下，$\vec{v}=\vec{\omega}$
问题等价于当
$$
\nabla\times\vec{v}=0,\nabla\cdot\vec{v}=0
$$
什么条件下$\vec{v}=0$?
设$\Omega\subset\mathbb{R}^3$区域。$\partial \Omega$为光滑曲面。$\vec{v}$是$\overline{\Omega}$上的向量场
$$
\nabla\times\vec{v}=0,\nabla\cdot\vec{v}=0
$$
设$\vec{v}=P\vec{i}+Q\vec{j}+R\vec{k}$
$$
\nabla\times\vec{v}=0\Rightarrow
\left\lbrace
\begin{aligned}
\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=0\\
\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=0\\
\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0
\end{aligned}
\right.
$$
$$
\nabla\cdot\vec{v}=0\Rightarrow
\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0
$$
$$
\frac{\partial \nabla\cdot\vec{v}}{\partial x}=\Delta P=0
$$
$$
\Delta P=\Delta Q=\Delta R=0
$$
命题：设$f,\overline{\Omega}\rightarrow\mathbb{R}$光滑函数$\Delta f=0$且$f|_{\partial\Omega}\equiv 0\rightarrow f\equiv 0$
证明
$$
\nabla\cdot(f\nabla f)=\nabla f\cdot\nabla f+f\nabla\cdot\nabla f=|\nabla f|^2+f\nabla f=|\nabla f|^2
$$
积分
$$\int_\Omega|\nabla f|^2\mathop{}\mathrm{d}V=\int_\Omega\nabla \cdot(f\nabla f)\mathop{}\mathrm{d}V=\int_{\partial\Omega}(f\nabla f)\mathop{}\mathrm{d}\vec{S}=0\rightarrow|\nabla f|\equiv 0$$
所以$f$是常值函数