Proof of vector identity from vorticity transport equation

Question

In fluid dynamics, the vorticity transport equation can be derived by taking the curl of the Navier-Stokes equations. In 2D [$\boldsymbol \omega = (0,0,\omega)$], the vorticity transport equation can be written as \begin{align} \frac{\partial \boldsymbol\omega}{\partial t}+(\mathbf{u}\cdot\nabla)\boldsymbol\omega=\frac{1}{Re}\nabla^2\boldsymbol\omega \end{align} where the vorticity $\boldsymbol\omega$ is the curl of the velocity $\mathbf{u}$ \begin{align} \boldsymbol\omega=\nabla\times\mathbf{u} \end{align} The curl of the curl vector identity \begin{align} \nabla \times \left( \nabla \times \mathbf{A} \right) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^{2}\mathbf{A}\\\\ \nabla \times \left( \nabla \times \mathbf{u} \right) = \nabla(\nabla \cdot \mathbf{u}) - \nabla^{2}\mathbf{u} \end{align} is used together with the incompressible flow condition $\nabla\cdot\mathbf{u}=0$ to get a Poisson equation for the velocity field \begin{align} \nabla \times \boldsymbol \omega = - \nabla^{2}\mathbf{u} \end{align} This closes the system of equations for 2D incompressible flow using the vorticity-velocity formulation.

My question is how to proof, or get, the above Poisson equation from the vorticity transport equation. What I have tried is to apply the divergence operator to the whole vorticity transport equation which leads to \begin{align} \frac{\partial \nabla\cdot\boldsymbol\omega}{\partial t}+\nabla\cdot\left[(\mathbf{u}\cdot\nabla)\boldsymbol\omega\right]=\frac{1}{Re}\nabla\cdot\left(\nabla^2\boldsymbol\omega\right) \end{align} since $\nabla\cdot\boldsymbol\omega=0$ \begin{align} \nabla\cdot\left[(\mathbf{u}\cdot\nabla)\boldsymbol\omega\right]&=\frac{1}{Re}\nabla^2\left(\nabla\cdot\boldsymbol\omega\right)\\\\ \nabla\cdot\left[(\mathbf{u}\cdot\nabla)\boldsymbol\omega\right]&=0 \end{align} I guess that from the equation above somehow one can get the velocity Poisson equation, but I have not been able to solve this yet.

There is a vector identity which can be written as \begin{align} \nabla \times (\mathbf{A} \times \mathbf{B}) &\ =\ \mathbf{A}\ (\nabla \cdot \mathbf{B}) - \mathbf{B}\ (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B}\\\\ \nabla \times \left(\mathbf u \times \boldsymbol \omega \right) &= \mathbf u \left(\nabla \cdot \boldsymbol \omega\right) -\boldsymbol \omega \left(\nabla \cdot \mathbf u\right) + \left(\boldsymbol \omega \cdot \nabla \right) \mathbf u - \left(\mathbf u \cdot \nabla\right) \boldsymbol \omega \end{align} Which for 2D incompressible flow is \begin{align} \nabla \times \left(\mathbf u \times \boldsymbol \omega \right) &=- \left(\mathbf u \cdot \nabla\right) \boldsymbol \omega \end{align} So probably with the relation above the Poisson equation for the velocity field can be recovered, but I got stuck here. Any help would be very much appreciated.

Answer 1

The vorticity transport equation is obtained from the (dimensionless) Navier-Stokes momentum equation

$$\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla\mathbf{u}= -\nabla p \, + \, \frac{1}{Re} \nabla^2 \mathbf{u}.$$

Taking the curl of both sides and applying some vector identities we get

$$\frac{\partial \boldsymbol\omega}{\partial t} + \mathbf{u} \cdot \nabla\boldsymbol\omega= \boldsymbol\omega \cdot \nabla \mathbf{u} - \boldsymbol\omega \, (\nabla \cdot \mathbf{u}) \, + \, \frac{1}{Re} \nabla^2 \boldsymbol\omega.$$

The first term on the RHS (vortex-line stretching) vanishes in 2D flow and the second on the RHS vanishes for incompressible flow since $\nabla \cdot \mathbf{u} = 0,$ resulting in

$$\frac{\partial \boldsymbol\omega}{\partial t} + \mathbf{u} \cdot \nabla\boldsymbol\omega= \frac{1}{Re} \nabla^2 \boldsymbol\omega.$$

You have already derived the equation

$$\tag{*}-\nabla^2 \mathbf{u} = \nabla \times \boldsymbol\omega,$$

using the definition of vorticity, $ \boldsymbol\omega = \nabla \times \mathbf{u},$ and the incompressibility condition, $ \nabla \cdot \mathbf{u} = 0.$ These are independent of the momentum equation and vorticity transport equations.

You are not going to derive (*) from the vorticity transport equation. It is just a kinematic relationship that incorporates the incompressibility condition. The incompressibility condition expresses conservation of mass. The vorticity transport equation expresses conservation of angular momentum.

Answer 2

There is no need to invoke any hydro equation, apart from the incompressible flow condition $\nabla \cdot {\bf{u}}=0$. Your Poisson equation is simply the by-product of the vector identity

$$ \nabla \times(\nabla\times {\bf{v}}) = -\nabla^2 {\bf{v}} + \nabla (\nabla \cdot {\bf{v}}) $$

Using ${\bf{u}}$ in place of the generic vector field ${\bf{v}}$ gives

$$ \nabla \times(\nabla\times {\bf{u}}) = \nabla \times {\boldsymbol{\omega}} =-\nabla^2 {\bf{u}} $$

In other words, this Poisson-like equation is not a particular feature of the system (i.e. it does not arise necessarily from the hydro equations), but it is a generic identity valid for any solenoidal field. In other words: there is no need to consider the Navier-Stokes, you just need to know that $\mathbf{u}$ is solenoidal, plus give a certain name to $\nabla \times \mathbf{u}$.