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Example 5: MMS benchmark

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Statement of the problem

This test is taken from Zhang & al.1. It consists of spatial convergence analysis based on a manufactured solution benchmark.

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

ϕt=Δμ+S(x,y,t) in Ωμ=F(ϕ)λΔϕ in Ω \begin{align} \frac{\partial \phi}{\partial t}&= \Delta \mu +S(x,y,t)\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)=ϕ44ϕ22 \begin{align} F(\phi)&=\frac{\phi^4}{4} - \frac{\phi^2}{2} \end{align}

In equation (1), S(x,y,t)S(x,y,t) is the source term allowing the exact solution:

ϕ=(t+1)sin(απx) \phi = (t+1) \sin(\alpha \pi x)

Initial condition

The initial condition is given by:

ϕ=sin(απx) \phi = \sin(\alpha \pi x)

For this test, α\alpha is set to 11.

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 1.01.0
omega depth of the double-well potential ω\omega 1.01.0

Boundary conditions

Neumann boundary conditions are prescribed on the top and bottom of the domain:

nλϕ=0 on ΩtopΩbottomnλμ=0 on ΩtopΩbottom \begin{align} {\bf{n}} \cdot{} \lambda \nabla \phi&=0 \text{ on }\partial\Omega_{top}\cup\partial\Omega_{bottom} \\[6pt] {\bf{n}} \cdot{} \lambda \nabla \mu&=0 \text{ on }\partial\Omega_{top}\cup\partial\Omega_{bottom} \end{align}

Dirichlet boundary conditions are prescribed on the right and left of the domain:

ϕ=0 on ΩleftΩrightμ=0 on ΩleftΩright \begin{align} \phi&=0 \text{ on }\partial\Omega_{left}\cup\partial\Omega_{right} \\[6pt] \mu&=0 \text{ on }\partial\Omega_{left}\cup\partial\Omega_{right} \end{align}

Numerical scheme

  • Time integration: Euler Implicit over the interval t[0,1]t\in[0,1] with a time-step δt=1\delta t=1.
  • Spatial discretization for convergence analysis: uniform grid with N=160,80,40,20N={160, 80, 40, 20} nodes in each spatial direction, with Q1\mathcal{Q}_1 and Q2\mathcal{Q}_2 finite elements
  • Newton solver: relative tolerance 10810^{-8}, absolute tolerance 101210^{-12}
  • Iterative solver: HYPRE_GMRES
  • Preconditioner: HYPRE_ILU

Results

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

  1. Liangzhe Zhang, Michael R Tonks, Derek Gaston, John W Peterson, David Andrs, Paul C Millett, and Bulent S Biner. A quantitative comparison between c0 and c1 elements for solving the cahn–hilliard equation. Journal of Computational Physics, 236:74–80, 2013.