I want to ask what advantage of using vorticity equations in fluid dynamics.
Does it help to find large curls? Does it have singularities connected to presence of curls?



Vorticity equation for incompressible 2D fluid dynamics
The vorticity equation for the Euler equation in 3D is, with $\omega=\text{curl } v$,
$$
\dot\omega + (v\cdot\nabla)\omega-(\omega\cdot\nabla)v=0,
$$
so that if $v$ is two-dimensional, i.e.
$
v=\begin{pmatrix}v\_1(x\_1, x\_2)\\
v\_2(x\_1, x\_2)\\
0\end{pmatrix}
$
you get that
$$
\text{curl } v=\begin{pmatrix}0\\
0\\
\partial\_1v\_2-\partial\_2 v\_1\end{pmatrix}
\quad \text{so that}\quad \omega\cdot \nabla= 
(\partial\_1v\_2-\partial\_2 v\_1)\partial \_3
\quad \text{and }\quad(\omega\cdot \nabla) v=0,
$$
yielding
$\dot\omega + (v\cdot\nabla)\omega=0$ which is simply a transport equation from which you get (under mild assumptions on $v$)
$$
\Vert{\omega (t, \cdot)}\Vert\_{L^\infty}=\Vert{\omega (t=0, \cdot)}\Vert\_{L^\infty},
$$
yielding existence and uniqueness results for the 2D Euler equation.
This argument breaks down in 3D, since, even for the Navier-Stokes equation, you get the vorticity equation
$$
\dot\omega + (v\cdot\nabla)\omega+\nu\ \text{curl}^2 \omega
=(\omega\cdot\nabla)v.
$$
On the other hand it remains an elegant way of getting rid of the pressure since you can easily verify that
$
\text{div}\bigl((v\cdot\nabla)\omega-(\omega\cdot\nabla)v\bigr)=0
$ (true since the commutator of vector fields with null divergence has also null divergence).
By the way, you also get rid of the Leray-Hopf projection by writing the vorticity equation.


