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Affine anisotropic flux on a skew parallelogram

Example

Take the single parallelogram cell with vertices (0,0),(2,0),(3,1),(1,1), tensor A=diag⁡(2,1), and affine state u(x,y)=x+2y. Prescribe on each boundary face its exact face mean. The cell center is (3/2,1/2) and the face centers, in bottom/right/top/left order, are (1,0),(5/2,1/2),(2,1),(1/2,1/2). This cell is skew: its right and left face normals are not parallel to the vectors from the cell center to those face centers.

Facts & Assumptions

Given: The skew parallelogram, A=diag⁡(2,1), the affine function u=x+2y, and the hybrid mimetic reconstruction and flux conventions.

[F1]

Hybrid mimetic finite-volume diffusion on general meshes: GKu=∣K∣−1∑F∣F∣uFnK,F, the residual is uF−uK−GKu⋅(xF−xK), and the outward mimetic flux is −AK(u,ϕK,F).

[F2]

Coercivity, monotonicity, conservation, and affine exactness of the mimetic schemes: for affine states the mimetic flux is exactly −∣F∣(A∇u)⋅nK,F.

Verification

1.1F1givenalgebra

The cell area is ∣K∣=2. The face data (∣F∣,xF,nK,F), in bottom/right/top/left order, are (2,(1,0),(0,−1)),(2,(5/2,1/2),(1,−1)/2),(2,(2,1),(0,1)),(2,(1/2,1/2),(−1,1)/2). Their weighted normals sum to (0,0) and ∑F∣F∣nK,F⊗xF=2I2, so the cell satisfies the face-moment closure in [F1].

2.1F1step 1.1algebra

The cell and face means of u are uK=5/2 and (uF)=(1,7/2,4,3/2). Substitution into [F1] gives ∑F∣F∣uFnK,F=(2,4) and hence GKu=(1,2)=∇u. Each residual is zero: for example, on the bottom face uF−uK=−3/2=GKu⋅((−1/2,−1/2)), and the same dot-product identity holds for the other three face centers.

3.1F2givenstep 2.1algebra∎

We have A∇u=(2,2). By [F2], the outward integrated fluxes in bottom/right/top/left order are 4,0,−4,0, respectively. Their sum is zero, which is the exact cell balance for the affine state since its flux is constant and has zero divergence. This verifies affine consistency and local conservation on this skew, anisotropic cell.

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