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Old   December 6, 2006, 07:27
Default Unsteady simulation convergence
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Tomislav
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When I perform an unsteady simulation, do I have to wait for acheaving convergence (satisfy the convergence limits) in each of the timesteps or this is not critical?
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Old   December 6, 2006, 08:53
Default Re: Unsteady simulation convergence
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ganesh
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Dear Tomislav,

If you are referring to the concept of Dual Time Stepping, wherein a psuedo-steady problem is solved at each physical timestep of the unsteady solver, then the answer is yes. It might seem that, and many users (including me) would sometimes not look for convergence (upto a prescribed tolerance) and get the code running. This is a good idea, to test the code for validation and to have a quick glimpse on whether the code is robust and moving in the right direction. The fact that you are solving a pseudo-steady problem clearly indicates that a convergence is necessary. I have had experiences where an insufficient convergence gives goog looking results, but the temporal order of accuracy degenerates from second order to first order. Though outwardly this may seem to be not much of a problem, impact of loss in temporal accuracy could surface in several simulations of practical interest such as in aeeroelasticity. It must however be noted that certain simulations, esp. turbulent flows on stretched meshes, in my experience do not have good convergence characteristics, which could also be due to the use of limiters in shocked and high gradient flows. In such cases, a compromise needs to be done, as the prescribed tolerance need not be the true indicator of convergence, and the code can still get through with a larger tolerance (not converged in sense of tolerance), but still be second order accurate and yield right results. On a whole, the short answer to your question is YES, but you must be judicial enough in the way you judge the convergence at each time step.

Hope this helps

Regards,

Ganesh
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