| Author | Message | |||
     JM (neo) Junior Member Username: neo Post Number: 3 Registered: 04-2006 |
Please find attached the script of a nonlinear parabolic PDE. This script is giving me an error 'time step fallen below 1e-9' when I reduce the errlim below 1e-8. I do not understand the reason. Please advice. PS: Is there a way of writing the J (which is a fucntion of x)so as to make it smooth. JM
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 592 Registered: 06-2003 |
Your initial condition is by implication the solution of a PDE operating at times prior to the start time of the problem. At the start time, the conditions and equation terms change to the values you specify in your script. If the initial value is not a solution of the new equation, there will be a strong transient while the solution adapts to the new conditions. A non-zero source term turned on instantaneously is not a physically realizable event. All natural processes turn on over some finite time. FlexPDE adaptive solution processes are based on the assumption that at some timestep and mesh size, the solution will look polynomial in time and space. Instantaneous switches are not polynomial at any time scale. Therefore, the timestep controller continues to cut the timestep until the deviation from a polynomial behavior is less than ERRLIM. If it has cut until the timestep falls below the cutoff timestep (defaulted to 1e-9 times the trial initial timestep), the problem will halt. By specifying a wildly optimistic ERRLIM of 1e-8, you force the timestep down to a value that can resolve the initial transient to eight decimal places (an accuracy greater than the known values of fundamental physical constants). You have several choices to resolve this issue: 1) Specify a more realistic ERRLIM. The default value of 0.002 is usually adequate. 1e-4 or 1e-5 is demanding. 2) Ramp the turn-on of the source over a meaningful time interval in the context of your problem. 3) Specify a very small initial timestep. You have given 1e-12, for which the cutoff is 1e-9*1e-12. Make it smaller until the solution succeeds. 4) Specify a very small HALT timestep (ie, replace the default cutoff of 1e-9 times the specified or default trial initial timestep). (See Halt in the Help Index). | |||
| JM (neo) Member Username: neo Post Number: 4 Registered: 04-2006 |
Thanks RG for the helpful tips. I will do as you suggested. JM |
Nonlinear parabolic PDE