Dear Dr. Guyer, Thank you for you precious advices:
Indeed, I had misunderstood how to do the meshing and that was the source of the apparent shift in the solver behaviour. Thank you also for your suggestions regarding the results viewing optimization: it's true indeed that this way is more elegant and would have exhibited where my problem came from. With my best regards; and a good week-end to all, Fanny -----Message d'origine----- De : [email protected] [mailto:[email protected]] De la part de Jonathan Guyer Envoyé : jeudi 11 août 2011 20:13 À : Multiple recipients of list Objet : Re: time issues with implicit method + full script On Aug 11, 2011, at 10:59 AM, JAMBON Fanny 213198 wrote: > Here is the full script (I do not use the fipy viewer but pylab, make sure > you have it installed!). A better way to do this is to use the MatplotlibViewer and then manipulate its axes as you desire: view = MatplotlibViewer(vars=cH) view.axes.plot(X_scale, Sol, marker='o') view.plot() This allows FiPy to properly position its solution values, which is a problem in your script. > ## Mesh > X_scale= n.linspace(0,2.0e+4,1000)# grid 0-> 20 um ; units[nm] > Nbx=len(X_scale) > deltax= 1.0/float(Nbx) You calculate the analytical solution over a domain of [0, 2.0e+4 nm], but your FiPy grid is only [0, 1.0 nm]. Try: deltax= 2.0e+4/float(Nbx) instead. > # Graph plotting > Y_scale=tuple(cH) > plt.plot(X_scale,Y_scale, marker='x') The values of cH are not evaluated at [0., 20.02002002, 40.04004004, ...] as given by linspace, but rather at [1.00000000e+01, 3.00000000e+01, 5.00000000e+01, ...]. Better to do plt.plot(mesh.getCellCenters()[0], cH.getValue(), marker='x') or better still to use the MatplotlibViewer as I show above.
