examples.cahnHilliard.tanh1D

This example solves the Cahn-Hilliard equation given by,

\[\frac{\partial \phi}{\partial t} = \nabla \cdot D \nabla \left( \frac{\partial f}{\partial \phi} - \epsilon^2 \nabla^2 \phi \right)\]

where the free energy functional is given by,

\[f = \frac{a^2}{2} \phi^2 (1 - \phi)^2\]

The Cahn-Hilliard equation can be rewritten in the following form,

\[\frac{\partial \phi}{\partial t} = \nabla \cdot D \left( \frac{\partial^2 f}{\partial \phi^2} \nabla \phi - \epsilon^2 \nabla^3 \phi \right)\]

The above form of the equation makes the non-linearity part of the diffusion coefficient for the first term on the RHS. This is the correct way to express the equation to FiPy.

We solve the problem on a 1D mesh

>>> from fipy import CellVariable, Grid1D, NthOrderBoundaryCondition, DiffusionTerm, TransientTerm, LinearLUSolver, DefaultSolver, Viewer
>>> from fipy.tools import numerix

>>> L = 40.
>>> nx = 1000
>>> dx = L / nx
>>> mesh = Grid1D(dx=dx, nx=nx)

and create the solution variable

>>> var = CellVariable(
...     name="phase field",
...     mesh=mesh,
...     value=1.)

The boundary conditions for this problem are

\[\begin{split}\left. \begin{aligned} \phi &= \frac{1}{2} \\ \frac{\partial^2 \phi}{\partial x^2} &= 0 \end{aligned} \right\} \qquad \text{on $x = 0$}\end{split}\]

and

\[\begin{split}\left. \begin{aligned} \phi &= 1 \\ \frac{\partial^2 \phi}{\partial x^2} &= 0 \end{aligned} \right\} \qquad \text{on $x = L$}\end{split}\]

or

>>> BCs = (
...     NthOrderBoundaryCondition(faces=mesh.facesLeft, value=0, order=2),
...     NthOrderBoundaryCondition(faces=mesh.facesRight, value=0, order=2))

>>> var.constrain(1, mesh.facesRight)
>>> var.constrain(.5, mesh.facesLeft)

Using

>>> asq = 1.0
>>> epsilon = 1
>>> diffusionCoeff = 1

we create the Cahn-Hilliard equation:

>>> faceVar = var.arithmeticFaceValue
>>> freeEnergyDoubleDerivative = asq * ( 1 - 6 * faceVar * (1 - faceVar))

>>> diffTerm2 = DiffusionTerm(
...     coeff=diffusionCoeff * freeEnergyDoubleDerivative)
>>> diffTerm4 = DiffusionTerm(coeff=(diffusionCoeff, epsilon**2))
>>> eqch = TransientTerm() == diffTerm2 - diffTerm4

>>> import fipy.solvers.solver
>>> if fipy.solvers.solver_suite in ['pysparse', 'pyamgx']:
...     solver = LinearLUSolver(tolerance=1e-15, iterations=100)
... else:
...     solver = DefaultSolver()

The solution to this 1D problem over an infinite domain is given by,

\[\phi(x) = \frac{1}{1 + \exp{\left(-\frac{a}{\epsilon} x \right)}}\]

or

>>> a = numerix.sqrt(asq)
>>> answer = 1 / (1 + numerix.exp(-a * (mesh.cellCenters[0]) / epsilon))

If we are running interactively, we create a viewer to see the results

>>> if __name__ == '__main__':
...     viewer = Viewer(vars=var, datamin=0., datamax=1.0)
...     viewer.plot()

We iterate the solution to equilibrium and, if we are running interactively, we update the display and output data about the progression of the solution

>>> dexp=-5
>>> from builtins import range
>>> for step in range(100):
...     dt = numerix.exp(dexp)
...     dt = min(10, dt)
...     dexp += 0.5
...     eqch.solve(var=var, boundaryConditions=BCs, solver=solver, dt=dt)
...     if __name__ == '__main__':
...         diff = abs(answer - numerix.array(var))
...         maxarg = numerix.argmax(diff)
...         print('maximum error: {}'.format(diff[maxarg]))
...         print('element id:', maxarg)
...         print('value at element {} is {}'.format(maxarg, var[maxarg]))
...         print('solution value: {}'.format(answer[maxarg]))
...
...         viewer.plot()

We compare the analytical solution with the numerical result,

>>> print(var.allclose(answer, atol=1e-4))
1

Last updated on Nov 20, 2024. Created using Sphinx 7.1.2.