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A two-cell Cahn–Hilliard mass and energy ledger

Example

Take two unit cells on (0,2), one interior face with transmissibility T=1, no-flux boundary faces, ε=κ=M=τ=1, and u0=(1,−1). The discrete mean is zero, and the one-step convex-splitting state is u1=(a,−a) where a is the unique root in (0,1) of 2a3+7a−5=0. The initial energy is 2 and the new energy is 2a2+12(1−a2)2<2.

Facts & Assumptions

Given: The two-cell finite-volume realization and the convex-splitting Cahn–Hilliard step.

[F1]

No-flux finite-volume and conforming realizations for phase fields: for this two-cell mesh, mh(u,v)=u1v1+u2v2, Ah(u1,u2)=(u1−u2,u2−u1), and Eh(u)=12⟨Ahu,u⟩h+F(u1)+F(u2).

[F2]

Existence, uniqueness, mass conservation, and energy decay of a convex-splitting step: the Cahn–Hilliard step is unique, preserves the discrete mean, and does not increase Eh.

Verification

1.1F1F2givenalgebra

The initial mean is zero, so mass conservation [F2] forces u1=(a,−a). Since Ah(a,−a)=(2a,−2a) and Fc′(s)=s3+s, Fe′(s)=2s, the first chemical-potential component is μ1=2a+(a3+a)−2=a3+3a−2. The first state equation gives a−1=−2μ1, hence 2a3+7a−5=0. The polynomial is strictly increasing because its derivative is 6a2+7>0, and its values at 0 and 1 are −5 and 4; therefore it has exactly one root a∈(0,1).

2.1F1step 1.1algebra∎

The initial gradient energy is 12⟨Ahu0,u0⟩h=2 and both potential terms vanish, so Eh(u0)=2. For u1=(a,−a) the gradient energy is 2a2 and the potential energy is 12(1−a2)2. Since 0<a<1, (1−a2)2<1−a2, hence Eh(u1)<2=Eh(u0). This is the exact finite ledger for the step.

Depends on

Used by

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