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March 27, 2004, 16:02 
steady unsteady

#1 
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Hi, I am a graduate student..Can somebody help me find an answer to following two doubts I have:
1) Do Problems that give rise to physical instabilities will always give rise to numerical instability? 2) What can be the expected consequence of trying to solve numerically the following laminar problem with a steadystate solver? Flow in a 2D channel with a cylinder installed at its center(Re=200, say) An actual experiment with shows a periodic shedding of vortices, the solution never gets timeindependent.. 

March 27, 2004, 18:25 
Re: steady unsteady

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1. No. Damping (numerical viscosity) can overpower physical instability.
2. Depends on the amount of numerical viscosity. The examples I've seen (hardly all of course) show a steadystate wake (separation, recirculation) behind the obstruction. 

March 27, 2004, 19:22 
Re: steady unsteady

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Thank you very much. Should I conclude :
(1) With high numerical viscosity what I get from the crazy idea of steadystatesolving is soln. corresponding to much lower reynold's no. than 200 (say 20). (2) With a low numerical viscosity (say by maintaining second or higher order everywhere ), no steady numerical solution should be possible. (divergence!) 

March 28, 2004, 16:56 
Re: steady unsteady

#4 
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1. Yes, that's what I've seen. A high level of damping will force a 'steady state' solution that would have at least some of the features of a steady flow at a much lower Reynolds number.
2. I don't know what would happen in this case. And I sure would be skeptical of any solution obtained. 

March 29, 2004, 04:32 
Re: steady unsteady

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1) What do you mean with physical instability? A blowup? The classical Runge Kutta method can solve the equation x'=x perfectly well. What makes numerical computations very difficult is, if changes in the initial data have a huge impact on the solution. Then you have to be very careful, but it is still possible to compute good solutions.
2) I don't know 

March 29, 2004, 04:47 
Re: steady unsteady

#6 
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1. Such a large drop in the Reynolds number would imply that your solution is not converged. In a viscous flow simulation the numerical viscosity should be smaller than the real viscosity.
2. There are two possible outcomes depending on your steady state solver: (i) you reach a steady state  which is a solution to the problem but may not be physically realizable or (ii) the iteration scheme oscillates and so does not converge. 

March 29, 2004, 05:58 
Re: steady unsteady

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What you get from steadystate solving of such problem  depends on the solver. I've made such an experiment myself once. The result was pretty weird  I've got a "frozen" Karman vortex street which was "switching" half a period per iteration


March 29, 2004, 13:12 
Re: steady unsteady

#8 
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The 'switching' really interesting and makes some sense too...


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