Is there a compact formula for this cross product involving $\nabla$?

Question

Let $\textbf{u}=(u_1(x,y,z),u_2(x,y,z),u_3(x,y,z))$ and $\textbf{v}=(v_1(x,y,z),v_2(x,y,z),v_3(x,y,z))$ be two smooth vector fields in $\mathbb{R}^3$, is there a simplified form of the following difference $$\nabla\times((\textbf{u}\cdot\nabla)\textbf{v})-\nabla\times((\textbf{v}\cdot\nabla)\textbf{u})~~?$$ where $\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$ is the del operator and $(\textbf{u}\cdot\nabla)\textbf{v}=\left(u_1\frac{\partial}{\partial x}+u_2\frac{\partial}{\partial y}+u_3\frac{\partial}{\partial z}\right)\textbf{v}$.

My goal is to simplify above difference in the special case where $\textbf{v} = \nabla\times\textbf{u}$ and $\nabla\cdot\textbf{u}=0$, but it will be nice if there is a universal simplification to the difference when $\textbf{u}$ and $\textbf{v}$ are unrelated, then one could apply that to the special case.

Answer 1

Using the sum convention $$\nabla \times \vec{a} = \epsilon_{ijk} \frac{\partial a_j}{\partial x_i} \vec{e_k} = \epsilon_{ijk} \partial_i a_j \vec{e_k};$$

$$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \nabla \times [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]$$ $$ = \epsilon_{ijk} \partial_i [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]_j \vec{e_k} $$ $$= \epsilon_{ijk} \partial_i [u_m \partial_m v_j - v_m \partial_m u_j]\vec{e_k} $$ It doesn't look like there is much more you can do.

If we assume as well that $\vec{v} = \nabla \times \vec{u}$ then $v_j = \epsilon_{lnj} \partial_{l} u_{n}.$ $$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \epsilon_{ijk} \partial_i [u_m \partial_m \epsilon_{lnj} \partial_{l} u_{n} - \epsilon_{lnm} \partial_{l} u_{n} \partial_m u_j]\vec{e_k} $$ $$=\left(\epsilon_{ijk} \epsilon_{lnj} \partial_i [u_m \partial_m \partial_l u_n] - \epsilon_{ijk} \epsilon_{lnm}\partial_i \partial_l[u_n \partial_m u_j]\right)\vec{e_k}$$ $$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right) - \epsilon_{ijk} \epsilon_{lnm}\partial_i \left(\frac{\partial u_n}{\partial x_l}\frac{\partial u_j}{\partial x_m} + u_n \frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$ $$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right) - \epsilon_{ijk} \epsilon_{lnm}\left(\frac{\partial^2 u_n}{\partial x_i \partial x_l}\frac{\partial u_j}{\partial x_m} + \frac{\partial u_n}{\partial x_l}\frac{\partial^2 u_j}{\partial x_i \partial x_m} + u_n \frac{\partial^3 u_j}{\partial x_i \partial x_l \partial x_m} + \frac{\partial u_n}{\partial x_i}\frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$

Which can possibly be simplified. Kronecker delta and Levi-Civita symbol