I am trying to derive the vorticity equation and I got stuck when trying to prove the following relation using index notation:
$$
{\rm curl}((\textbf{u}\cdot\nabla)\mathbf{u}) = (\mathbf{u}\cdot\nabla)\pmb\omega – ( \pmb\omega \cdot\nabla)\mathbf{u}
$$
considering that the fluid is incompressible $\nabla\cdot\mathbf{u} = 0 $, $\pmb \omega = {\rm curl}(\mathbf{u})$ and that $\nabla \cdot \pmb \omega = 0.$
Here follows what I've done so far:
$$
(\textbf{u}\cdot\nabla) \mathbf{u} = u_m\frac{\partial u_i}{\partial x_m} \mathbf{e}_i = a_i \mathbf{e}_i \\
{\rm curl}(\mathbf{a}) = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \mathbf{e}_i = \epsilon_{ijk} \frac{\partial}{\partial x_j}\left( u_m\frac{\partial u_k}{\partial x_m} \right) \mathbf{e}_i = \\
= \epsilon_{ijk}\frac{\partial u_m}{\partial x_j}\frac{\partial u_k}{\partial x_m} \mathbf{e}_i + \epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i \\
$$
the second term $\epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i$ seems to be the first term "$(\mathbf{u}\cdot\nabla)\pmb\omega$" from the forementioned identity. Does anyone have an idea how to get the second term?
Best Answer
The trick is the following:
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_m}{\partial x_k} = 0 $$
by antisymmetry.
So you can rewrite
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_k}{\partial x_m} = \epsilon_{ijk} \frac{\partial u_m}{\partial x_j}\left( \frac{\partial u_k}{\partial x_m} - \frac{\partial u_m}{\partial x_k} \right) $$
Note that the term in the parentheses is something like $\pm\epsilon_{kml} \omega_l$
Lastly use the product property for Levi-Civita symbols
$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$