Dear Daniel, Thanks a lot for the help. I got it to work by adding a transientterm().
If I try to use your second trick though: eq = (DiffusionTerm(coeff=(1.0), var=phi) - numerix.sin(phi) + phi * numerix.cos(phi)-ImplicitSourceTerm(numerix.cos(phi)) == TransientTerm(var=phi)) I get the error: ExplicitVariableError: Terms with explicit Variables cannot mix with Terms with implicit Variables. Also, I dont fully understand the expression: I assume the implicit term is to neutralize the phi * numerix.cos(phi), so should it not be: phi*ImplicitSourceTerm(numerix.cos(phi))? Also: if I would want a second derivative in time in there to solve the time-dependent problem, would the best way to do that to define a second variable: dphidt=TransientTerm(var=phi) and solve the coupled equations or is there a different way? Best, Quintin On 21 Mar 2019, at 16:48, Daniel Wheeler <daniel.wheel...@gmail.com<mailto:daniel.wheel...@gmail.com>> wrote: Hi Qunitin, I think you need to relax the updates on this. Introducing a transient term does that. I think that the b vector causes and instability. The b vector is phi_old - delta_t * sin(phi_old) with a transient term in the equation. For small phi, this should never be negative, since phi should only grow, but for delta_t > 1 this can be negative. It's the same reason that explicit diffusion has a time step restriction. You can get around this restriction by linearizing the source term so that the source looks like - numerix.sin(phi) + phi * numerix.cos(phi) - ImplicitSourceTerm(numerix.cos(phi)) There is now no time step restriction since the b vector is just "phi_old" when phi is small. After 100 sweeps, the non-linearized source is at a residual of 0.0022 (with dt=1) and the linearized source is at 0.0009 (with dt=1000), which is somewhat better. Possibly using Newton iterations could improve the convergence rate further. Hope that helps. Cheers, Daniel ~~~~ import numpy as np from fipy import * nx = 10000 dx = 0.01 mesh = Grid1D(nx=nx, dx=dx) phi = CellVariable(mesh=mesh, name="phi", hasOld=True) phi.constrain(2*np.pi, where=mesh.facesLeft) phi.constrain(0., where=mesh.facesRight) # requires dt < 1 #eq = TransientTerm() == DiffusionTerm() - numerix.sin(phi) # no dt requirment eq = TransientTerm() == DiffusionTerm() - numerix.sin(phi) + phi * numerix.cos(phi) - ImplicitSourceTerm(numerix.cos(phi)) sweeps=1000 view = Viewer(phi) for i in range(sweeps): res = eq.sweep(var=phi, dt=1000) print(res) view.plot() phi.updateOld() raw_input('stopped') ~~~~ On Thu, Mar 21, 2019 at 7:26 AM Meier Quintin <quintin.me...@mat.ethz.ch<mailto:quintin.me...@mat.ethz.ch>> wrote: Dear fipy community, I’m new to FiPy, so forgive me if this is a trivial question. I’m trying to find the kink solution to the (static) sine-gordon equation using this simply piece of code. import numpy as np from fipy import * nx = 10000 dx = 0.01 mesh = Grid1D(nx=nx,dx=dx) phi = CellVariable(mesh=mesh, name=“Phi") phi.constrain(2*np.pi, where=mesh.facesLeft) phi.constrain(0., where=mesh.facesRight) eq = (DiffusionTerm(coeff=(1.0), var=phi) - numerix.sin(phi) == 0.0) sweeps=10 for i in range(sweeps): eq.sweep(var=phi) MatplotlibViewer(phi) Unfortunately, I do not manage to get a numerically stable solution, the first sweep gives me a straight line and additional sweeps lead to numerical instabilities without convergence. I would be very thankful if someone could hint me to what I’m doing wrong. Best wishes, Quintin Meier _______________________________________________ fipy mailing list fipy@nist.gov<mailto:fipy@nist.gov> http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] -- Daniel Wheeler _______________________________________________ fipy mailing list fipy@nist.gov<mailto:fipy@nist.gov> http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
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