Cahn-Hilliard equation

This example demonstrates the solution of the Cahn-Hilliard equation, a nonlinear, time-dependent fourth-order PDE.

  • A mixed finite element method

  • The \(\theta\)-method for time-dependent equations

  • Automatic linearisation

  • Use of the class NonlinearProblem

  • The built-in Newton solver (NewtonSolver)

  • Form compiler options

  • Interpolation of functions

  • Visualisation of a running simulation with pyvista

This demo is implemented in

Equation and problem definition

The Cahn-Hilliard equation is a parabolic equation and is typically used to model phase separation in binary mixtures. It involves first-order time derivatives, and second- and fourth-order spatial derivatives. The equation reads:

\[\begin{split} \begin{align} \frac{\partial c}{\partial t} - \nabla \cdot M \left(\nabla\left(\frac{d f}{dc} - \lambda \nabla^{2}c\right)\right) &= 0 \quad {\rm in} \ \Omega, \\ M\left(\nabla\left(\frac{d f}{d c} - \lambda \nabla^{2}c\right)\right) \cdot n &= 0 \quad {\rm on} \ \partial\Omega, \\ M \lambda \nabla c \cdot n &= 0 \quad {\rm on} \ \partial\Omega. \end{align} \end{split}\]

where \(c\) is the unknown field, the function \(f\) is usually non-convex in \(c\) (a fourth-order polynomial is commonly used), \(n\) is the outward directed boundary normal, and \(M\) is a scalar parameter.

Operator split form

The Cahn-Hilliard equation is a fourth-order equation, so casting it in a weak form would result in the presence of second-order spatial derivatives, and the problem could not be solved using a standard Lagrange finite element basis. A solution is to rephrase the problem as two coupled second-order equations:

\[\begin{split} \begin{align} \frac{\partial c}{\partial t} - \nabla \cdot M \nabla\mu &= 0 \quad {\rm in} \ \Omega, \\ \mu - \frac{d f}{d c} + \lambda \nabla^{2}c &= 0 \quad {\rm in} \ \Omega. \end{align} \end{split}\]

The unknown fields are now \(c\) and \(\mu\). The weak (variational) form of the problem reads: find \((c, \mu) \in V \times V\) such that

\[\begin{split} \begin{align} \int_{\Omega} \frac{\partial c}{\partial t} q \, {\rm d} x + \int_{\Omega} M \nabla\mu \cdot \nabla q \, {\rm d} x &= 0 \quad \forall \ q \in V, \\ \int_{\Omega} \mu v \, {\rm d} x - \int_{\Omega} \frac{d f}{d c} v \, {\rm d} x - \int_{\Omega} \lambda \nabla c \cdot \nabla v \, {\rm d} x &= 0 \quad \forall \ v \in V. \end{align} \end{split}\]

Time discretisation

Before being able to solve this problem, the time derivative must be dealt with. Apply the \(\theta\)-method to the mixed weak form of the equation:

\[\begin{split} \begin{align} \int_{\Omega} \frac{c_{n+1} - c_{n}}{dt} q \, {\rm d} x + \int_{\Omega} M \nabla \mu_{n+\theta} \cdot \nabla q \, {\rm d} x &= 0 \quad \forall \ q \in V \\ \int_{\Omega} \mu_{n+1} v \, {\rm d} x - \int_{\Omega} \frac{d f_{n+1}}{d c} v \, {\rm d} x - \int_{\Omega} \lambda \nabla c_{n+1} \cdot \nabla v \, {\rm d} x &= 0 \quad \forall \ v \in V \end{align} \end{split}\]

where \(dt = t_{n+1} - t_{n}\) and $\mu_{n+\theta} = (1-\theta) \mu_{n}

  • \theta \mu_{n+1}\(. The task is: given \)c_{n}\( and \)\mu_{n}\(, solve the above equation to find \)c_{n+1}\( and \)\mu_{n+1}$.

Demo parameters

The following domains, functions and time stepping parameters are used in this demo:

  • \(\Omega = (0, 1) \times (0, 1)\) (unit square)

  • \(f = 100 c^{2} (1-c)^{2}\)

  • \(\lambda = 1 \times 10^{-2}\)

  • \(M = 1\)

  • \(dt = 5 \times 10^{-6}\)

  • \(\theta = 0.5\)


This demo is implemented in the file.

import os

import numpy as np

import ufl
from basix.ufl import mixed_element, element
from dolfinx import log, plot
from dolfinx.fem import Function, FunctionSpace
from dolfinx.fem.petsc import NonlinearProblem
from import XDMFFile
from dolfinx.mesh import CellType, create_unit_square
from dolfinx.nls.petsc import NewtonSolver
from ufl import dx, grad, inner

from mpi4py import MPI
from petsc4py import PETSc

    import pyvista as pv
    import pyvistaqt as pvqt
    have_pyvista = True
    if pv.OFF_SCREEN:
except ModuleNotFoundError:
    print("pyvista and pyvistaqt are required to visualise the solution")
    have_pyvista = False

# Save all logging to file

Next, various model parameters are defined:

lmbda = 1.0e-02  # surface parameter
dt = 5.0e-06  # time step
theta = 0.5  # time stepping family, e.g. theta=1 -> backward Euler, theta=0.5 -> Crank-Nicholson

A unit square mesh with 96 cells edges in each direction is created, and on this mesh a FunctionSpace ME is built using a pair of linear Lagrange elements.

msh = create_unit_square(MPI.COMM_WORLD, 96, 96, CellType.triangle)
P1 = element("Lagrange", msh.basix_cell(), 1)
ME = FunctionSpace(msh, mixed_element([P1, P1]))

Trial and test functions of the space ME are now defined:

q, v = ufl.TestFunctions(ME)

For the test functions, TestFunctions (note the ‘s’ at the end) is used to define the scalar test functions q and v. Some mixed objects of the Function class on ME are defined to represent \(u = (c_{n+1}, \mu_{n+1})\) and \(u0 = (c_{n}, \mu_{n})\), and these are then split into sub-functions:

u = Function(ME)  # current solution
u0 = Function(ME)  # solution from previous converged step

# Split mixed functions
c, mu = ufl.split(u)
c0, mu0 = ufl.split(u0)

The line c, mu = ufl.split(u) permits direct access to the components of a mixed function. Note that c and mu are references for components of u, and not copies.

The initial conditions are interpolated into a finite element space:

# Zero u
u.x.array[:] = 0.0

# Interpolate initial condition
u.sub(0).interpolate(lambda x: 0.63 + 0.02 * (0.5 - np.random.rand(x.shape[1])))

The first line creates an object of type InitialConditions. The following two lines make u and u0 interpolants of u_init (since u and u0 are finite element functions, they may not be able to represent a given function exactly, but the function can be approximated by interpolating it in a finite element space).

The chemical potential \(df/dc\) is computed using UFL automatic differentiation:

# Compute the chemical potential df/dc
c = ufl.variable(c)
f = 100 * c**2 * (1 - c)**2
dfdc = ufl.diff(f, c)

The first line declares that c is a variable that some function can be differentiated with respect to. The next line is the function \(f\) defined in the problem statement, and the third line performs the differentiation of f with respect to the variable c.

It is convenient to introduce an expression for \(\mu_{n+\theta}\):

# mu_(n+theta)
mu_mid = (1.0 - theta) * mu0 + theta * mu

which is then used in the definition of the variational forms:

# Weak statement of the equations
F0 = inner(c, q) * dx - inner(c0, q) * dx + dt * inner(grad(mu_mid), grad(q)) * dx
F1 = inner(mu, v) * dx - inner(dfdc, v) * dx - lmbda * inner(grad(c), grad(v)) * dx
F = F0 + F1

This is a statement of the time-discrete equations presented as part of the problem statement, using UFL syntax.

The DOLFINx Newton solver requires a NonlinearProblem object to solve a system of nonlinear equations

# Create nonlinear problem and Newton solver
problem = NonlinearProblem(F, u)
solver = NewtonSolver(MPI.COMM_WORLD, problem)
solver.convergence_criterion = "incremental"
solver.rtol = 1e-6

# We can customize the linear solver used inside the NewtonSolver by
# modifying the PETSc options
ksp = solver.krylov_solver
opts = PETSc.Options()
option_prefix = ksp.getOptionsPrefix()
opts[f"{option_prefix}ksp_type"] = "preonly"
opts[f"{option_prefix}pc_type"] = "lu"

The setting of convergence_criterion to "incremental" specifies that the Newton solver should compute a norm of the solution increment to check for convergence (the other possibility is to use "residual", or to provide a user-defined check). The tolerance for convergence is specified by rtol.

To run the solver and save the output to a VTK file for later visualization, the solver is advanced in time from \(t_{n}\) to \(t_{n+1}\) until a terminal time \(T\) is reached:

# Output file
file = XDMFFile(MPI.COMM_WORLD, "demo_ch/output.xdmf", "w")

# Step in time
t = 0.0

#  Reduce run time if on test (CI) server
if "CI" in os.environ.keys() or "GITHUB_ACTIONS" in os.environ.keys():
    T = 3 * dt
    T = 50 * dt

# Get the sub-space for c and the corresponding dofs in the mixed space
# vector
V0, dofs = ME.sub(0).collapse()

# Prepare viewer for plotting the solution during the computation
if have_pyvista:
    # Create a VTK 'mesh' with 'nodes' at the function dofs
    topology, cell_types, x = plot.create_vtk_mesh(V0)
    grid = pv.UnstructuredGrid(topology, cell_types, x)

    # Set output data
    grid.point_data["c"] = u.x.array[dofs].real

    p = pvqt.BackgroundPlotter(title="concentration", auto_update=True)
    p.add_mesh(grid, clim=[0, 1])
    p.add_text(f"time: {t}", font_size=12, name="timelabel")

c = u.sub(0)
u0.x.array[:] = u.x.array
while (t < T):
    t += dt
    r = solver.solve(u)
    print(f"Step {int(t/dt)}: num iterations: {r[0]}")
    u0.x.array[:] = u.x.array
    file.write_function(c, t)

    # Update the plot window
    if have_pyvista:
        p.add_text(f"time: {t:.2e}", font_size=12, name="timelabel")
        grid.point_data["c"] = u.x.array[dofs].real


# Update ghost entries and plot
if have_pyvista:
    grid.point_data["c"] = u.x.array[dofs].real
    screenshot = None
    if pv.OFF_SCREEN:
        screenshot = "c.png"
    pv.plot(grid, show_edges=True, screenshot=screenshot)