The Taylor-Green vortex sheet is a solution to the 2D Navier-Stokes equations for an incompressible Newtonian fluid: $latex \frac{d \mathbf{u}}{d t}= \nu \nabla^2 \mathbf{u} – \nabla p/\rho$ where $latex \mathbf{u}$ is the velocity field, $latex p$ is the pressure,  $latex \nu=\mu/\rho$ is the kinematic viscosity, and $latex \rho$ is the fixed density of the fluid. The time derivative is a total derivative: $latex \frac{d \mathbf{u}}{d t} = \frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u}$.

error:  $latex d \mathbf{u} = mathbf{u} $,

$latex \frac{d \mathbf{u}}{d t}$

It is common to choose parameters that simplify the equations, but that can obscure the role of the different parameters. In the following, I provide expressions with all relevant parameters included, with their physical dimensions. I later pass to dimensionless, or reduced, units, in terms of the Reynolds and Courant numbers.  

The solution

The solution is a periodic array of vortices, that repeats itself in the $latex x$ and $latex y$ directions with a periodic length $latex L$: $latex u_x =: u = f(t) u_0  \sin k x \cos k y $ $latex v_y =: v = -f(t)  u_0 \cos k x  \sin k y $ here, $latex k=2\pi/L$, and the function $latex f(t)$ is $latex f(t)= \exp(-2\nu k^2 t) ,$ so that the decay time of the vortices due to viscosity is given by $latex \tau=1/(2\nu k^2)$. The maximum modulus of the velocity field at time zero is $latex u_0$. The pressure field is given by $latex p = \frac{\rho u_0^2 }{4}  f(t)^2 \left( \cos (2kx) + \cos (2ky) \right) .$ Hence the vortices go around zones of low pressure, either clockwise or counter-clockwise (pictures will come, eventually.) Plugging these two fields into the Navier-Stokes equation shows that indeed this is a solution. It is interesting that the pressure gradient term exactly cancels the convective one, while the viscosity term cancels the partial derivative. That means that in the inviscid limit the vortices will never decay. The vorticity field is given by $latex \omega = 2 f(t) \sin (kx) \sin (ky) $ and the stream function is just $latex \psi=\omega /2$. Notice the vorticity satisfies the convection-diffusion equation $latex \frac{d \omega}{d t}= \nu \nabla^2 \omega .$

Recuced units

Let us introduce the dimensionless time, built from time, maximum initial velocity, and typical length L (another choice would be L/2, which is the actual length of a vortex) $latex t^*= t u_0 / L .$ Notice that $latex t^*=1$ is the time a fluid particle would need to travel a distance L. Function f can be written as $latex f(t)= \exp(-2\nu (2\pi/L)^2 (L t^* / L) ) = \exp(-8 \pi^2 t^* / \mathrm{Re})  , $ where the Reynolds number appears naturally: $latex \mathrm{Re} :=  L u_0 / \nu . $ The decay time is then seen to be $latex \tau^*=\mathrm{Re}/(8\pi^2)$ in reduced units. In a simulation, there is an important dimensionless parameter, the Courant number: $latex \mathrm{Co} :=  u_0 (\Delta t)/ (\Delta x) , $ where $latex \Delta t $ is the simulation time-step, and $latex \Delta x $ the size of the simulation cells (or, the interparticle distance in a particle simulation, aka h ). If each Cartesian direction is discretized in n cells (for a total of n x n cells), then $latex \mathrm{Co} = n  u_0 (\Delta t)/ L = n (\Delta t)^*, $ where the dimensionless time-step arises naturally. For a series of simulations at fixed Co, the product of  n and $latex \Delta t^*$ must remain constant. E.g. if we had 16 x 16 bins at some time-step, we should use 32 x 32 at half that time-step. A simulation that must then consider twice the number of  time steps in order to reach the same final time, and with a system that has four times as many cells as the original one ! We can suppose this will scale badly, as the eighth power or so. It turns out that this is more like the worst-case scenario, though (I think…).

Some choices

The typical instance corresponds to $latex L=2\pi \quad u_0=1$, in which case $latex \mathrm{Re}= 2\pi/\nu$, and the function f takes the simple form $latex f=\exp(-2\nu t)$. Also, $latex t^*= t / (2 \pi) $. If the viscosity is also taken to be one, the expressions are even simpler, but then the Reynolds number is only $latex 2\pi$!   https://www.youtube.com/watch?v=hFTCcyVX6nM

SPH simulations of the Taylor Green vortex sheet by INSEAN. Recall the time quoted is dimensioned. Hence, the onset of instability at about 70 would actually be about 10 in reduced units. The decay time is $latex 2000/(8 pi^2)$, which is about 25. Hence the velocity field has decayed exp(-10/25) at that time, which is about two thirds. Color by vorticity.

In my simulations, I took  $latex L=2 \quad u_0=1$ (I should have chosen  $latex L=1$, but…), so then $latex \mathrm{Re}= 2/\nu$, and $latex t^*= t / 2 $, a factor of two that I missed in an early version of a manuscript. I also took $latex \nu=0.01$, which would give a Reynolds number of 200 (not, as I thought, 100). I have also taken $latex \nu=0.001$ for a number of 2000, as in the simulations by the INSEAN group in Rome. https://www.youtube.com/watch?v=sc0BIgaloWU

pFEM simulations by myself, at the same Re=2000 as those above. Instability sets in at a reduced time of about 15. Later times seem to indicate a vortex "hopping". However, I have more recent simulations that show no instability at all. Color is by pressure on these simulations (left), and by vorticity, more or less (right)

If we take one of these vortices as a good model for a storm, we may set a pressure difference of 100 HPa. This would give maximum wind speeds of $latex \sqrt{p/\rho} \approx 90$ m/s, or 330 km/h… quite the storm indeed. Anyway, with a storm size of 10 km, the Reynolds number would be a huge 1.6 10^10 (using a dynamic viscosity of 1.5 10^-5 m^2/s for air). That's about 24 years, so, no, it seems clear storms do not decay due to air viscosity.