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Example 4: steady-state solution of the Cahn–Hilliard equations

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Statement of the problem

This test consists of finding the steady-state solution of the Cahn–Hilliard equations. A transient simulation is performed to reach this steady-state, and a convergence analysis is carried out to ensure the consistency of the results.

The domain Ω\Omega is a square [0,1]×[0,1][0,1]\times[0,1]

ϕt=MΔμ in Ωμ=F(ϕ)λΔϕ in Ω \begin{align} \frac{\partial \phi}{\partial t}&=M \Delta \mu \text{ in }\Omega \\[6pt] \mu &= F'(\phi) - \lambda \Delta \phi \text{ in }\Omega \end{align}

where ϕ\phi is the phase indicator, μ\mu the generalized chemical potential and FF' the derivative against ϕ\phi of the potential FF defined by:

F(ϕ)=ωϕ2(1ϕ)2 \begin{align} F(\phi)&=\omega\phi^2(1-\phi)^2 \end{align}

Initial condition

The initial condition is given by:

ϕ=12+12tanh(x0.55103). \phi = \dfrac{1}{2} + \dfrac{1}{2} \tanh\left(\dfrac{x-0.5}{5\cdot 10^{-3}}\right).

Parameters used for the test

For this test, all parameters are equal to one.

Name Description Symbol Value
mob mobility coefficient MϕM_\phi 1.01.0
lambda energy gradient coefficient λ\lambda 32σϵ\dfrac{3}{2}{\sigma}{\epsilon}
omega depth of the double-well potential ω\omega 12σϵ12\dfrac{\sigma}{\epsilon}
sigma surface tension σ\sigma 1.01.0
epsilon thickness of interface ϵ\epsilon 0.10.1

Boundary conditions

Neumann boundary conditions are prescribed on boundary of the domain:

nλϕ=0 on Ωnλμ=0 on Ω \begin{align} {\bf{n}} \cdot{} \lambda \nabla \phi&=0 \text{ on }\partial\Omega \\[6pt] {\bf{n}} \cdot{} \lambda \nabla \mu&=0 \text{ on }\partial\Omega \end{align}

Numerical scheme

  • Time integration: Euler Implicit over the interval t[0,1]t\in[0,1] with a time-step δt=103\delta t=10^{-3}. The calculation stops when convergence criteria are reached.

ϕ(t+δt)ϕ(t)<ϵϕμ(t+δt)μ(t)<ϵμ \|{\phi(t+\delta t)-\phi(t)}\|<\epsilon_\phi \\[6pt] \|{\mu(t+\delta t)-\mu(t)}\|<\epsilon_\mu

where ϵϕ=1012\epsilon_\phi=10^{-12} and ϵμ=107\epsilon_\mu=10^{-7}.

  • Spatial discretization for convergence analysis: uniform grid with N=30,60,90,120N={30, 60, 90, 120} nodes in each spatial direction, with Q1\mathcal{Q}_1 and Q2\mathcal{Q}_2 finite elements
  • Newton solver: relative tolerance 101210^{-12}, absolute tolerance 101210^{-12}
  • Iterative solver: HYPRE_GMRES
  • Preconditioner: HYPRE_ILU

Results

The steady state solution is given by:

ϕ=12+12tanh(x0.5ϵ). \phi = \dfrac{1}{2} + \dfrac{1}{2} \tanh\left(\dfrac{x-0.5}{\epsilon}\right).

Figures 1 shows the results of convergence analysis with Q1\mathcal{Q}_1 and Q2\mathcal{Q}_2.

MMS
Figure 1: convergence analysis with Q1\mathcal{Q}_1 and Q2\mathcal{Q}_2 finite elements