Alphabeta Math
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Harmonic TPFA flux on a rectangular Voronoï mesh

Example

Take Ω=(0,2)×(0,1) and the two rectangular cells K=(0,1)×(0,1) and L=(1,2)×(0,1). Their centers are (1/2,1/2) and (3/2,1/2), so the shared vertical face is orthogonal to the center segment. Set λK=λL=1, use u(x,y)=x, and impose its exact face-center values on the six cell faces on the outer boundary. The exact source is f=0.

Facts & Assumptions

Given: The two-cell mesh and data above, and the TPFA transmissibilities and signed fluxes from Orthogonal TPFA diffusion with harmonic face transmissibility.

[F1]

Finite-volume cell-face meshes and geometric regularity: ∣K∣=∣L∣=1; the shared face has length 1, and each outer face has length 1.

[F2]

Orthogonal TPFA diffusion with harmonic face transmissibility: TF=∣F∣/(dK,F/λK+dL,F/λL) on an interior face, and TK,F=∣F∣λK/dK,F on a boundary face.

Verification

1.1F1F2givenalgebra

The center-to-center distance across the shared face is 1, with half-distances dK,F=dL,F=1/2. Thus TF=1. The left and right boundary half-distances are 1/2, so their transmissibilities are 2; top and bottom center-to-face displacements are vertical and hence orthogonal to those faces. The mesh satisfies the TPFA orthogonality requirements.

2.1F1F2step 1.1algebra∎

The cell values are uK=1/2 and uL=3/2. At the shared face, QK,F=1(1/2−3/2)=−1 and QL,F=1. On the left boundary the outward flux is 2(1/2−0)=1, and on the right boundary it is 2(3/2−2)=−1. The top and bottom boundary fluxes vanish because their face-center values equal the adjacent cell-center values. Each cell balance therefore sums to zero, and the shared flux matches the exact outward diffusive flux −∇u⋅(1,0)=−1 from K.

Depends on

Used by

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