[petsc-users] How do I know it is steady state?

Tue Nov 3 12:55:20 CST 2015

> On Nov 3, 2015, at 9:38 AM, Zou (Non-US), Ling <ling.zou at inl.gov> wrote:
> 
> 
> 
> On Tue, Nov 3, 2015 at 8:24 AM, Matthew Knepley <knepley at gmail.com> wrote:
> On Tue, Nov 3, 2015 at 9:12 AM, Zou (Non-US), Ling <ling.zou at inl.gov> wrote:
> Matt, thanks for the reply.
> The simulation is a transient simulation, which eventually converges to a steady-state solution, given enough simulation time.
> My code runs fine and I could tell the simulation reaches steady state by looking at the residual monitored by SNES monitor function.
> 
> See an example screen output
> 
> Solving time step 90, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 8.85. 
>         NL step =  0, SNES Function norm =  1.47538E-02
>         NL step =  1, SNES Function norm =  8.06971E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 91, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 8.95. 
>         NL step =  0, SNES Function norm =  1.10861E-02
>         NL step =  1, SNES Function norm =  6.26584E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 92, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.05. 
>         NL step =  0, SNES Function norm =  7.21253E-03
>         NL step =  1, SNES Function norm =  9.93402E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 93, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.15. 
>         NL step =  0, SNES Function norm =  5.40260E-03
>         NL step =  1, SNES Function norm =  6.21162E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 94, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.25. 
>         NL step =  0, SNES Function norm =  3.40214E-03
>         NL step =  1, SNES Function norm =  6.16805E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 95, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.35. 
>         NL step =  0, SNES Function norm =  2.29656E-03
>         NL step =  1, SNES Function norm =  6.19337E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 96, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.45. 
>         NL step =  0, SNES Function norm =  1.53218E-03
>         NL step =  1, SNES Function norm =  5.94845E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 97, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.55. 
>         NL step =  0, SNES Function norm =  1.32136E-03
>         NL step =  1, SNES Function norm =  6.19933E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 98, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.65. 
>         NL step =  0, SNES Function norm =  7.09342E-04
>         NL step =  1, SNES Function norm =  6.18694E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 99, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.75. 
>         NL step =  0, SNES Function norm =  5.49192E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 100, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.85. 
>         NL step =  0, SNES Function norm =  5.49192E-04
>         total_FunctionCall_number: 0 
> converged, time step increased = 0.1
> Solving time step 101, using BDF1, dt = 0.1. 
> Current time (the starting time of this time step) = 9.95. 
>         NL step =  0, SNES Function norm =  5.49192E-04
>         total_FunctionCall_number: 0 
> 
> I observed that after time step 99, the residual never changed, so I believe the transient simulation converges at time step 99.
> I wonder can I use the criterion "SNES converges and it takes 0 iteration" to say the simulation reaches a steady state. Such that I don't have to look at the screen and the code knows it converges and should stop.
> 
> Put it another way, what's the common way people would implement a scheme to detect a transient simulation reaches steady state.
> 
> I don't think so. The above makes no sense to me. You are signaling SNES convergence with a relative
> residual norm of 5e-4? That does not sound precise enough to me.
> 
> I would argue that number (5.e-4) depends on the problem you are solving (actually I am solving).
> The initial residual of the problem starts at ~1e8.
> But you might be right, and I have to think about this issue more carefully.
>  
> As I said, I think the believable way to find steady states is to look for solutions to the algebraic equations,
> perhaps by using timestepping as a preconditioner.
> 
> You still need a numerical criterion to let the code understand it converges, right? For example, "a set of solutions have already been found to satisfy the algebraic equations because ___residuals drops below (a number here)__". 

  After each SNESSolve you could call SNESGetConvergedReason() and if the number of iterations was 0 and the reason was snorm then declare it steady state.

   Barry

> 
> Thanks,
> 
> Ling
> 
>  
>   Thanks,
> 
>      Matt
>  
> Thanks,
> 
> Ling
> 
> 
> On Tue, Nov 3, 2015 at 5:25 AM, Matthew Knepley <knepley at gmail.com> wrote:
> On Mon, Nov 2, 2015 at 7:29 PM, Barry Smith <bsmith at mcs.anl.gov> wrote:
> 
> > On Oct 30, 2015, at 12:23 PM, Zou (Non-US), Ling <ling.zou at inl.gov> wrote:
> >
> > Hi All,
> >
> > From physics point of view, I know my simulation converges if nothing changes any more.
> >
> > I wonder how normally you do to detect if your simulation reaches steady state from numerical point of view.
> > Is it a good practice to use SNES convergence as a criterion, i.e.,
> > SNES converges and it takes 0 iteration(s)
> 
>    Depends on the time integrator and SNES tolerance you are using. If you use a -snes_rtol 1.e-5 it will always try to squeeze 5 MORE digits out of the residual so won't take 0 iterations even if there is only a small change in the solution.
> 
> There are two different situations here:
> 
>   1) Solving for a mathematical steady state. You remove the time derivative and solve the algebraic system with SNES. Then
>       the SNES tolerance is a good measure.
> 
>   2) Use timestepping to advance until nothing looks like it is changing. This is a "physical" steady state.
> 
> You can use 1) with a timestepping preconditioner TSPSEUDO, which is what I would recommend if you
> want a true steady state.
> 
>   Thanks,
> 
>      Matt
>  
> >
> > Thanks,
> >
> > Ling
> 
> 
> 
> 
> -- 
> What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead.
> -- Norbert Wiener
> 
> 
> 
> 
> -- 
> What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead.
> -- Norbert Wiener
> 