1
$\begingroup$

As from the derivation we are able to see that the implicit density based solver (such as BTCS) is unconditionally stable for any courant number. So what is it, which limits the upper boundary of CFL while using it in practical applications?

I need this for simulations in ANSYS Fluent. I have been told that non-linear effects start coming in, but I don't know what exactly are the non-linearities?

$\endgroup$

1 Answer 1

1
$\begingroup$

Analytical stability results for backward-time, centered space (BTCS) discretizations only apply to the first order (linear) hyperbolic partial differential equation. The only reason why they are applicable to the navier stokes equation is because navier stokes tends to exhibit behaviors similar to this first order linear wave equation in some cases (e.g. slow moving laminar flow). The stability analysis is much easier to perform on linear equations such as the 1st order wave equation and provides somewhat of an estimate of timestepping stability in the non-linear hyperbolic PDE case.

The navier stokes equations have nonlinearities in the momentum equation (i.e. the nonlinear advection term $\nabla\cdot(\vec{u}\otimes\vec{u})$). As this non-linear term becomes stronger, the stability estimate based on BTCS discretization (for the 1st order linear wave equation) becomes less and less precise. Unfortunately, non-linear equations are extremely difficult (impossible?) to analytically derive stability conditions for different timestepping schemes. In practice, you will observe some deviations from the stability conditions for BTCS for the first order wave equation. Explicit timestepping schemes often need to be much smaller than 1 in order to observe convergence, while backward timestepping will have some limit to how large the time stepsize can be. Usually, this limit is larger than that which an explicit timestepping scheme would allow.

Heuristics on stability criterion for the navier stokes equations are generally obtained empirically via trial-and-error and are highly problem dependent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.