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Monolithic backward Euler transmission energy identity

Statement

Assume the finite-dimensional diffusion-transmission model of Finite-dimensional backward-Euler diffusion transmission model. Let V:={(v1,v2)∈X1⊕X2:γ1v1=γ2v2} be the common-trace space, take un∈V, and let Fin+1∈Xi. In the block notation of the definition, set the local right-hand side to gi:=Δt−1Miuin+Fin+1; assemble the local equations with common trace and the outward algebraic fluxes qi from that definition. There is a unique un+1∈V satisfying

∑i=12(⟨Miuin+1−uinΔt,vi⟩+⟨Aiuin+1,vi⟩)=∑i=12⟨Fin+1,vi⟩(v∈V).

It obeys the exact identity

12(∥un+1∥M2−∥un∥M2+∥un+1−un∥M2)+Δt∥un+1∥A2=Δt∑i=12⟨Fin+1,uin+1⟩,

where ∥v∥M2:=∑i⟨Mivi,vi⟩ and ∥v∥A2:=∑i⟨Aivi,vi⟩. In the unforced case the monolithic backward-Euler energy is nonincreasing.

Facts & Assumptions

Given: The finite-dimensional spaces, positive-definite mass and diffusion operators, common trace map, outward flux convention, and monolithic flux balance of Finite-dimensional backward-Euler diffusion transmission model.

[F1]

Finite-dimensional backward-Euler diffusion transmission model: Mi and Ai are self-adjoint positive definite, local fluxes use the outward sign, and monolithic traces and fluxes satisfy λ1=λ2 and q1+q2=0.

Proof

1.1givenF1algebra

The interior rows of the two local blocks give their backward-Euler subdomain equations; after eliminating those rows, the common trace row is q1+q2=0 by the definition in [F1]. For a common trace λ, the paired interface work is therefore ⟨q1,λ⟩+⟨q2,λ⟩=⟨q1+q2,λ⟩=0; this cancellation uses the shared trace and outward-flux sign convention.

1.2givenF1algebra

On V, the assembled bilinear form associated with Δt−1Mi+Ai is positive definite: for v∈V∖{0}, at least one vi≠0, and every corresponding summand Δt−1⟨Mivi,vi⟩+⟨Aivi,vi⟩ is positive. Hence the finite-dimensional restricted operator has zero kernel and is invertible, so the variational step has a unique solution. Test the monolithic equation with v=un+1 and multiply by Δt. For each i, self-adjointness gives 2⟨Mi(uin+1−uin),uin+1⟩=∥uin+1∥Mi2−∥uin∥Mi2+∥uin+1−uin∥Mi2. Summing this identity and retaining Δt∑i⟨Aiuin+1,uin+1⟩ yields the stated equality.

2.1step 1.2algebra∎

If Fin+1=0 for both subdomains, the right-hand side is zero and both the time-jump term and diffusion term on the left are nonnegative. Therefore ∥un+1∥M2≤∥un∥M2.

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