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conditions for unsteady problems

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Old   November 9, 2009, 02:15
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Hello .... dvdromnu (actually it would be nice to have a name here),
If the equations you're solving admit a solution, and if that solution is unique, then it doesn't matter what initial conditions you start with (as long as they are physical or representative for your equations).
If you're interested in the steady state solution, then again it doesn't matter what approach you use (steady/unsteady).
If the convection/diffusion equation has the same steady state solution as the diffusion only equation, then you will not see any "profile shift".
The only difference would be in the choice of the solver, and the amount of computational time requested (shorter for steady state and longer for unsteady).

I hope this was useful,
Dragos
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