I come up a problem when solving the following question:
Let $\boldsymbol{u} = (u_1(x, y), u_2(x, y))$ be a smooth planar vector field and $P = P(x, y)$ be a scalar function in $\Omega$ satisfying $(u · \nabla)u = \nabla P, \operatorname{div} u = 0$. Here $(u · \nabla)u$ is understood as $\sum_{k=1}^2 u_k\partial_k \boldsymbol{u}$. Derive the partial differential equation of $\omega = \partial_2u_1 − \partial_1u_2$ which only involves $\boldsymbol{u}$ and $\nabla ω$ but not involve $P$.
I first write $\omega$ as $\dfrac{\partial u_1}{\partial y}-\dfrac{\partial u_2}{\partial x}$ and then compute the derivative \begin{align*} \mathrm{d}w &= \frac{\partial \omega}{\partial x}\mathrm{d}x+ \frac{\partial \omega}{\partial y}\mathrm{d}y\\ &= \left(\dfrac{\partial^2 u_1}{\partial y \partial x} - \dfrac{\partial^2 u_2}{\partial x^2}\right) \mathrm{d}x +\left( \dfrac{\partial^2 u_1}{\partial y^2} -\dfrac{\partial^2 u_2}{\partial x \partial y}\right) \mathrm{d}y. \end{align*} Then I can not see any relation between $\omega$ and $\boldsymbol{u}$. I do not know how to make use of the information $(u · \nabla)u = \nabla P, \operatorname{div} u = 0$. Any help will be appreciated.
The vector field $\mathbf{u} = (u_1(x,y),u_2(x,y))$ satifies $(\mathbf{u} \cdot \nabla)\mathbf{u} = \nabla P$ which reduces to two component equations
$$\tag{1a}u_1 \frac{\partial u_1}{\partial x}+u_2 \frac{\partial u_1}{\partial y}= \frac{\partial P}{\partial x}$$ $$\tag{1b}u_1 \frac{\partial u_2}{\partial x}+u_2 \frac{\partial u_2}{\partial y}= \frac{\partial P}{\partial y}$$
Take the partial derivative of both sides of (1a) with respect to $y$ and the partial derivative of both sides of (1b) with respect to $x$ to obtain
$$\tag{2a}\frac{\partial u_1}{\partial y}\frac{\partial u_1}{\partial x}+ u_1 \frac{\partial^2 u_1}{\partial y\partial x}+\frac{\partial u_2}{\partial y}\frac{\partial u_1}{\partial y}+u_2 \frac{\partial^2 u_1}{\partial y^2}= \frac{\partial^2 P}{\partial y\partial x}$$ $$\tag{2b}\frac{\partial u_1}{\partial x}\frac{\partial u_2}{\partial x}+ u_1 \frac{\partial^2 u_2}{\partial x^2}+\frac{\partial u_2}{\partial x}\frac{\partial u_2}{\partial y}+u_2 \frac{\partial^2 u_2}{\partial x\partial y}= \frac{\partial^2 P}{\partial x\partial y}$$
Subtracting (2b) from (2a) we get
$$\tag{3}\frac{\partial u_1}{\partial x}\left( \frac{\partial u_1}{\partial y}-\frac{\partial u_2}{\partial x}\right)+ \frac{\partial u_2}{\partial y}\left( \frac{\partial u_1}{\partial y}-\frac{\partial u_2}{\partial x}\right)\\+u_1 \left(\frac{\partial^2 u_1}{\partial y\partial x}- \frac{\partial^2 u_2}{\partial x^2}\right)+ u_2 \left(\frac{\partial^2 u_1}{\partial y^2}- \frac{\partial^2 u_2}{\partial x\partial y}\right)\\ = \frac{\partial^2 P}{\partial y\partial x}-\frac{\partial^2 P}{\partial x\partial y}$$
Assuming $P$, $u_1$, and $u_2$ are twice continuously differentiable, we have equality of mixed second-order partial derivatives by Clairaut's theorem. Consequently, the RHS of (3) is $0$ and we can rearrange to obtain
$$\tag{4}\left(\frac{\partial u_1}{\partial x}+ \frac{\partial u_2}{\partial y}\right)\left( \frac{\partial u_1}{\partial y}-\frac{\partial u_2}{\partial x}\right)+u_1\frac{\partial}{\partial x} \left(\frac{\partial u_1}{\partial y}- \frac{\partial u_2}{\partial x}\right)+ u_2 \frac{\partial}{\partial y}\left(\frac{\partial u_1}{\partial y}- \frac{\partial u_2}{\partial x}\right) = 0$$
Substituting into (4) with $\displaystyle \omega = \frac{\partial u_1}{\partial y}-\frac{\partial u_2}{\partial x}$ and $\displaystyle\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}= \nabla \cdot \mathbf{u} = 0$ , we get the result
$$\boxed{u_1 \frac{\partial \omega}{\partial x} + u_2 \frac{\partial\omega}{\partial y} = 0}$$