Proof of $-\nabla\times\omega = \nabla^2 U$

Question

What is a proof for $$ -\nabla\times\omega = \nabla^2 U $$ in the scope of fluid mechanics? I'm learning vector calculus for my project and stuck on this seemingly simple proof problem. Detailed answer with underlying concepts would be the most useful, as I want to have a clear understanding of what is happening and what concepts do I need to use to prove the expression. I suppose that I need to use the Levi Civita and Kronecker delta to get a proof.

Answer 1

This only holds in an incompressible flow.

Vorticity is defined by

$$\mathbb{\omega} = \nabla \times \mathbb{u}.$$

We have the vector identity

$$\nabla \times\mathbb{\omega} = \nabla \times \nabla \times \mathbb{u} = \nabla(\nabla \cdot \mathbb{u})- \nabla^2 \mathbb{u}.$$

This is derived as follows using Cartesian coordinates:

$$\begin{align}(\nabla \times \mathbb{\omega})_i &= \epsilon_{ijk} \partial_j [\epsilon_{klm}\partial_lu_m] \\ &= \epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lu_m \\ &= \epsilon_{kij}\epsilon_{klm}\partial_j\partial_lu_m \\ &= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\partial_j\partial_lu_m \\ &= \partial_i\partial_ju_j - \partial_j\partial_ju_i \end{align} $$

In an incompressible flow the velocity field is solenoidal to ensure conservation of mass,

$$\nabla \cdot \mathbb{u} = 0,$$

and your desired result follows.