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[Garth Wells <question133595@xxxxxxxxxxxxxxxxxxxxx>]

Question #133595 on DOLFIN changed:
https://answers.launchpad.net/dolfin/+question/133595

Garth Wells proposed the following answer:

On 11/11/10 12:45, Thomas Fraunholz wrote:
> New question #133595 on DOLFIN:
> https://answers.launchpad.net/dolfin/+question/133595
>
> I have a time-dependent Cahn-Hilliard-Type problem and I need to
> evaluate the angle from the x-axis of a gradient in order to define
> my variational problem. The angle of the gradient can be calculated
> by using phi = arctan( dc/dy / dc/dx ). This is needed to model an
> anisotropic N-fold surface tension with the help of cos(N*phi).
>
> A very native approach was to use the following lines of code. It
> doesn't work of course. But I think it helps to understand my
> problem.
>
> -------------------- import numpy as np from dolfin import *
>
> ...
>
> V = FunctionSpace(mesh, "CG", 2) w  = TestFunction(V)
>
> c = Function (V)  # current solution c0  = Function(V) # old
> solution
>
> ...
>
> # Anisotropic N-fold surface tension: def anisotropic_tension(c0):
> return np.cos(N*np.arctan2(c0.dx(1),c0.dx(0)))
>
> L = L - dt*anisotropic_tension(c0)*(inner(div(grad(w)),
> div(grad(c)))*dx)
>
> ...
>
> --------------------
>
> I think one basic problem is the interaction with numpy and the
> encapsuled evaluation of the derivatives. I'm glad for any hint how
> to proper evaluate the angle and how to use it in order to define L.

Try using 'Atan' (a UFL function) directly, e.g.

M = dt*N*Atan(....)* . . . . .

Garth

>

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