negative time step, possible to set?
hi there,
Do you know if it possible to introduce in OF, icoFoam algorithm in particular, a time step (adjusted or not) which is negative? I mean, can I run a simulation in OF that start from t=0 and it goes till t=tstar with tstar<0 with a deltaT<0? Thank you in advance Regards |
I don't believe you can do that with the default time classes, as there is only has increment operator:
http://foam.sourceforge.net/docs/cpp/a02579.html I'm not sure in what situation these would be useful. If you really wanted it you would have to have your own time counter and decrement it, although this would only mean that your simulation was still evolving forward in time and you've flipped your time axis. That, or you could find a flux capacitor and drive 88 mph... |
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Anyway my question was bad posed. Indeed the equation I want to solve are not NS classic ones but a motification of them with the temporal term which is going to be -ddt(U)...OF icofoam solver comes to be just the skeleton of the algorithm... How can I create "my own time counter"? Thank you in advance to support my idea |
I don't think you would need a negative time-step to do what you want. Is ddt(U) negative because it absolutely needs to be, or just because its on the RHS of the equation you are solving (a reference to the equations you are solving would be helpful)?
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I mean leaving the time flows as positive but introducing at the place of Dt the term -1*Dt in the CrankNicholson schemes... My question at this point could be. If it is possible, where is that OF does compute the CrankNicholson scheme, where is the exact file where it does the discretization? |
Dear Tom, Dear Foamers,
In my point of view, the problem here is to set the right boundary conditions in time for the adjoint NSE. To make sure, that the problem is well posed, one definitely is forced to compute the solution of ANSE backward in time. Tom, did you find a way to realize a computation in openFOAM with negative timeSteps? I tried a lot so far, but haven't find a satisfying answer yet, so even short hints might be very useful ( I hope you're still familiar with the subject posted above...). Kind regards Tom |
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