conditions for unsteady problems
Hi,
I am solving unsteady state convective diffusion scalar equation using finite volume method(explicitly). My problem is: For steady state problem the solution converges at some time step and we get results. That means steady state problem is independent of initial conditions. For unsteady state problem I am printing the values of scalar at a particular time step. So My question is, for unsteady problem what initial conditions I have to take..? means if I run unsteady problem for long time as in case of steady problem I am getting the same result of Steady state problem. So, how I have to solve unsteady problem. What initial conditions I should take. What is the difference between unsteady state and steady state solutions 
If the problem admits a steady state solution, the unsteady solution procedure should lead to the same solution, if it is run long enough.

Thank You................

Quote:
Best, 
solution profile
Actually I want to know whether there is any difference in solution procedure of unsteady state and steady state scalar convective diffusion equation.
Also I want to know if we solve scalar equation with diffusion term only(without convection term) we will get one solution. Now, if we solve the same including convection term, we get another solution. So, adding convection term changes the solution profile( with only diffusion term) or it shifts the solution profile..? 
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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