New answers tagged

1

As stated there's no re-rejection mechanism, i.e. ability to decrease the stepsize after a step has potentially failed. This is required for implicit methods which have Newton steps since there's a chance the $\Delta t$ is large enough that the (quasi-)Newton is unstable, in which case it needs to pullback on time. This instability can sometimes be seen via ...


1

I do not think this is a bad approach, but it is not a very precise way to select timesteps either. Admittedly, I have not come across this sort of timestep heuristic before, but looking at a linear test problem provides some insight as to why this is reasonable. For $y' = \lambda y$, the conditions becomes $\Delta t = \frac{\alpha}{\lambda}$. This looks ...


Top 50 recent answers are included