Computation of time step for FR

Dear All,

My name is Antonio Garcia-Uceda, and I’m working on Flux Reconstruction. I’d like to know the way you compute the local time step in FR, for steady state solution. Originally I was using an approach inherited from FV (i.e. cellwise computation using cell-averaged solution values), and then apply this value to all solution points throughout the cell However I’m aware that this may over-restrict the real CFL limit at each point. Do you compute the time step rather in a pointwise manner? Thanks in advance.

Best regards,
Antonio

PyFR does not use local time stepping; the time step is the same for all
elements in the simulation. If you're interested in solving steady
state problems there are better tools for the job (fully implicit
schemes with HP adaptivity, for example).

Regards, Freddie.

Hi Antonio

I would add that in the next release of PyFR (v0.2.3) we will have support for adaptive step-size control based on error estimating RK schemes, however this will adapt the global dt, not local dt. As Freddie says, we do not have element local dt in PyFR at the moment.

Cheers

Peter

Dear All,

Thanks for the reply. Could you please indicate why exactly pyFR does not have local dt for steady state solutions? I mean, is there any constraint in FR different from the normal local dt used in FV?

Cheers,
Antonio

There is no technical reason why one could not go ahead and implement
local time stepping into PyFR. However, as I said in my previous
e-mail, if you're interested in steady state solutions to the
Navier-Stokes equation there are more suitable tools and algorithms.

Regards, Freddie.

Thanks Freddie, once more, :).

Btw, just curiosity, at the solution points, Do you store primitive variables, or conserved Variables x JacDet Value (i.e. u or bar_u as indicated in some papers)?

Cheers,
Antonio

In PyFR we work exclusively in terms of conservative variables. We only
convert back to primitive files in pyfr-postp when writing out a .vtu
file for visualisation.

Regards, Freddie.