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 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-Nicolson

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 = ufl.FiniteElement("Lagrange", msh.ufl_cell(), 1)
ME = FunctionSpace(msh, 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"
opts[f"{option_prefix}pc_factor_mat_solver_type"] = "mumps"

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)