> On Nov 3, 2015, at 9:38 AM, Zou (Non-US), Ling <[email protected]> wrote: > > > > On Tue, Nov 3, 2015 at 8:24 AM, Matthew Knepley <[email protected]> wrote: > On Tue, Nov 3, 2015 at 9:12 AM, Zou (Non-US), Ling <[email protected]> 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 <[email protected]> wrote: > On Mon, Nov 2, 2015 at 7:29 PM, Barry Smith <[email protected]> wrote: > > > On Oct 30, 2015, at 12:23 PM, Zou (Non-US), Ling <[email protected]> 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 >
