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A two-domain heat trace balances outward fluxes

Statement

Take a one-dimensional trace space Λ=R, two interior spaces Ii=R, Δt=2, M1=M2=I2, A1=diag⁡(1/2,3/2), and A2=diag⁡(1/2,1/2), with the second coordinate as trace. Let the condensed loads be f1=1, f2=5. The monolithic trace is 2 and the outward fluxes are 3 and −3. With α=1, r10=2, tolerance 0.1, and budget M=2, the Robin–Robin algorithm returns the certified trace 2 on its second pass.

Facts & Assumptions

Given: The finite-dimensional local blocks, trace coordinates, condensed loads, Robin parameter, incoming datum, tolerance, and pass budget in the Statement.

[F1]

Finite-dimensional backward-Euler diffusion transmission model: Bi=Δt−1Mi+Ai, qi=Siλi−fi, and the monolithic trace solves (S1+S2)λ=f1+f2.

[F2]

The diffusion Schur complement is positive and minimizes energy: each positive-definite local block has a positive-definite Schur complement.

[F3]

Monolithic backward Euler transmission energy identity: when traces match and outward fluxes balance, paired interface work cancels in the monolithic energy identity.

[F4]

Finite-budget Robin-Robin transmission iteration: each pass performs the two Robin solves and exchanges and stops only when the stated residual test passes.

[F5]

Robin-Robin iteration contracts to the monolithic trace: a passing stop test certifies trace error at most ε relative to the monolithic trace.

Verification

1.1givenF1F2algebra

The local backward-Euler blocks are B1=diag⁡(1,2) and B2=diag⁡(1,1), which are positive definite. With zero off-diagonal blocks between the interior and trace coordinates, their Schur operators are S1=2 and S2=1, so [F2] applies.

1.2F1F4algebra

On pass one, λ11=(2+1)/3=1, r21=(1−2)1+1=0, λ21=(0+5)/2=2.5, and r11=(1−1)2.5+5=5. Here q1=∣(1−2)/(1+2)∣=1/3, q2=0, and the stop threshold is 0.1/(1/3)=0.3; the incoming-data change is ∣5−2∣=3, so the algorithm does not stop.

2.1F1F3step 1.1algebra

The monolithic scalar equation is 3λ=6, hence λ∗=2. Its fluxes are q1=2(2)−1=3 and q2=1(2)−5=−3, so they balance and the interface work λ∗(q1+q2) vanishes as in [F3].

3.1F4F5step 1.2step 2.1algebra∎

On pass two, λ12=(5+1)/3=2, r22=(1−2)2+1=−1, λ22=(−1+5)/2=2, and r12=(1−1)2+5=5. The incoming-data change is 0≤0.3, so [F4] returns converged with trace 2; [F5] certifies its error is at most 0.1 relative to λ∗=2 from step 2.1.

Depends on

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