Question on Partial differential toolbox

I am using FEM to solve a pde and I get the following error message quite often. Each time I get this message, I have to reduce the mesh size and redo it. I am very frustrated because this message will not show up after the program is running for a very very very long time (1 hour sometimes). Is there any way I can increase the ode15s tolerance so that I can speed the process? My problem does not require a very accurate solution. I tried to change PDESolverOptions which did not do anything.
By the way, I have over 90000 triangles in my mesh (I hope this is not a problem for matlab).
Warning: Failure at t=2.174639e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (4.440892e-16) at time t. > In ode15s (line 668) In pde.EquationModel/solveTimeDependent (line 83) In pde.PDEModel/solvepde (line 54) In SI_Solve1 (line 103) Warning: Solution failed to reach the requested end time. > In pde.EquationModel/solveTimeDependent (line 87) In pde.PDEModel/solvepde (line 54) In SI_Solve1 (line 103) Error using pde.TimeDependentResults (line 74) Solution does not correspond to time-dependent PDE.
Error in pde.PDEModel/solvepde (line 59) R = pde.TimeDependentResults(self,u,tlist);
Error in SI_Solve1 (line 103) sol=solvepde(pdem,tlist);

2 个评论

It is difficult to identify the problem without knowing the full details of the setup. In generally, the behavior you are observing could be due to stiff-problem. Decreasing the mesh size might get you a little further in solution, or it might just take longer to reach the same breakdown point. If you provide further details or attach the reproduction code, then someone could identify the root cause for the error.
Thank you for your reply! I think you are right. The initial value of my pde is extremely non-smooth. That may cause the stiffness. I am able to solve the problem in one hour sometimes (for some parameters), but I also get the error message very often (for some other parameters). Unfortunately, my initial value is based on a 500 M data set, so I can not upload it here.

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提问:

woo
2017-10-18

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woo
2017-10-20

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