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Harmonic transmissibility across a one-dimensional material interface

Example

Let Ω=(0,1) with cells K=(0,1/2) and L=(1/2,1), chosen centers xK=1/4 and xL=3/4, cell coefficients λK=1 and λL=9, and boundary values u(0)=0, u(1)=1. The stationary continuous solution has constant flux across the material interface. Compare the TPFA transmissibility using the two half-cell resistances with the transmissibility obtained by substituting the arithmetic coefficient mean.

Facts & Assumptions

Given: The two interval cells, piecewise-constant coefficient, Dirichlet boundary data, and the face transmissibility formula from Orthogonal TPFA diffusion with harmonic face transmissibility.

[F1]

The centers xK=1/4 and xL=3/4 are the choices specified in this example, permitted by Finite-volume cell-face meshes and geometric regularity; the shared interface is x=1/2.

[F2]

Orthogonal TPFA diffusion with harmonic face transmissibility: the interface transmissibility is TF=∣F∣/(dK,F/λK+dL,F/λL), with one shared flux TF(uK−uL) outward from K.

Verification

1.1givenF1algebra

The exact solution is piecewise affine. The constant c in λu′=c satisfies 1=c(12/1+12/9), hence c=9/5. Thus u(1/2)=9/10, uK=u(1/4)=9/20, and uL=u(3/4)=19/20. The outward diffusive flux from K is −λKuK′=−9/5 and equals the outward flux from L with the opposite orientation.

2.1F1F2step 1.1algebra

The half-distances are both 1/4, so [F2] gives TF=1/(1/4+1/36)=18/5. The TPFA flux is (18/5)(9/20−19/20)=−9/5, exactly the continuous interface flux.

3.1F1step 1.1algebra∎

The arithmetic mean coefficient is (1+9)/2=5 and the center distance is 1/2, which would give transmissibility 5/(1/2)=10. Its flux is 10(9/20−19/20)=−5, not −9/5. This finite interface calculation verifies that the resistance-weighted harmonic transmissibility is essential for this discontinuous-coefficient example.

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