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✓ 11 results · all verified · 9 also independently AI-judged
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ALE, interfaces and partitioned multiphysics coupling

1 · Prerequisites

2 · Summary

ALE, interfaces and partitioned multiphysics coupling

This page has two independent branches: a selected one-dimensional moving-mesh finite-volume scheme, and a finite-dimensional linear diffusion-transmission problem. It also proves the continuous ALE Jacobian and Reynolds identities. These results do not claim a high-dimensional discrete GCL estimate, a curved-mesh transfer rate, or nonlinear moving-domain fluid-structure well-posedness.

Continuous ALE map and transport

ALE moving-domain map and mesh velocity fixes the reference and moving domains, the positive Jacobian, and mesh-velocity convention. Jacobi identity for the ALE mesh Jacobian proves ∂tJ=Jdiv⁡wm. The Reynolds identity has separate dimension branches: Choice-free one-dimensional ALE Reynolds transport identity is choice-free and handles finite unions of intervals; Reynolds transport identity on an ALE moving domain covers finite piecewise-C1 domains in dimensions at least two and states its Countable Choice use through the divergence theorem and the explicit Riemann/Lebesgue volume-integral comparison.

Selected one-dimensional moving-grid branch

Periodic one-dimensional moving grid and Moving faces satisfy the one-dimensional discrete GCL fix the periodic face and Jacobian conventions. Conservative upwind ALE finite-volume step returns mesh_crossing before updating the state if any new cell Jacobian is nonpositive. On success, The upwind ALE step conserves mass and preserves constants proves exact periodic mass conservation and constant-state preservation. Under the stated CFL condition, CFL monotonicity and weighted L1 contraction for the ALE step proves positivity, monotonicity, and weighted L1 contraction. The CFL condition is needed for monotonicity, not for mass or constant-state preservation.

Linear diffusion transmission and partitioned solve

Finite-dimensional backward-Euler diffusion transmission model uses finite-dimensional backward-Euler subdomain blocks and the outward flux convention qi=Siλi−fi. Both subdomains use one trace space and the monolithic interface equations are trace continuity and q1+q2=0. The diagram fixes these signs and does not enter any proof.

\begin{tikzpicture}[scale=1.0, every node/.style={font=\small}]
  \draw[thick] (0,0) rectangle (2,1.4);
  \draw[thick] (2,0) rectangle (4,1.4);
  \draw[very thick] (2,0) -- (2,1.4);
  \node at (1,0.72) {$\Omega_1$};
  \node at (3,0.72) {$\Omega_2$};
  \node[above] at (2,1.4) {$\Gamma$};
  \node[below] at (2,0) {$\lambda_1=\lambda_2=\lambda$};
  \draw[->,blue,thick] (1.55,0.95) -- (2.35,0.95);
  \node[blue,above] at (1.95,0.95) {$n_1,\ q_1$};
  \draw[->,red,thick] (2.45,0.42) -- (1.65,0.42);
  \node[red,below] at (2.05,0.42) {$n_2,\ q_2$};
\end{tikzpicture}

The diffusion Schur complement is positive and minimizes energy proves positivity and the energy-minimizing property of each condensed Schur operator. Monolithic backward Euler transmission energy identity proves the exact monolithic energy identity after paired outward interface work cancels. Finite-budget Robin-Robin transmission iteration specifies a finite-budget sequence of two Robin solves. Robin-Robin iteration contracts to the monolithic trace identifies its unique fixed point with the monolithic trace, proves contraction, and gives the exact stop certificate. Schur residual controls trace and reconstruction energy error transfers a condensed residual bound to the reconstructed subdomain energy norm.

Added-mass warning and optional time splitting

Scalar added-mass staggered recurrence defines the specific scalar staggered recurrence used by the companion example; no instability claim is made for other fluid-structure algorithms. The optional finite-dimensional Lie/Strang route is limited to Conditional finite-dimensional Lie and Strang splitting residual orders and inherits its exact-subflow, common-tube, commutator-regularity, residual-budget, and stability hypotheses from the cited supplier. It supplies no unbounded-operator or general PDE splitting estimate.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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Conditional finite-dimensional Lie and Strang splitting residual orders

Statement

For the optional splitting branch of the finite-dimensional coupled model, suppose a,b∈C3(U,Rd) define the additive autonomous ODE of Additive autonomous ODE partitions, their exact subflows use the Lie and Strang compositions of Lie and Strang compositions of exact split flows, and all exact and numerical paths lie with positive margin in a common compact convex tube on [0,T]. Initialize both schemes at the exact initial value y0=y(0). Under the remaining Lipschitz, regularity, and step-size hypotheses of Lie and Strang splitting have global orders one and two, with separate subflow errors, the exact Lie and Strang schemes have global errors O(h) and O(h2) respectively. Approximate subflows retain these orders only under the separate per-subflow error budgets and composition stability condition stated in that theorem. This corollary makes no claim for unbounded operators or general moving-domain PDEs.

Facts & Assumptions

Given: The finite-dimensional autonomous split ODE, exact subflows, exact initial data, and common-tube hypotheses in the Statement.

[F1]

Additive autonomous ODE partitions: the total field is the autonomous sum a+b on an open subset of Rd.

[F2]

Lie and Strang compositions of exact split flows: the Lie map is Φbh∘Φah and the Strang map is Φah/2∘Φbh∘Φah/2.

[F3]

Lie and Strang splitting have global orders one and two, with separate subflow errors: on a common compact convex tube with the theorem's regularity and step hypotheses, the exact maps have global orders one and two; approximate subflows preserve these orders only under its separate local error budgets and composed-map stability bound.

Proof

1.1givenF1F2F3

By [F1] and [F2], the optional split system and its stage maps are exactly the finite-dimensional additive ODE, Lie map, and Strang map covered by [F3]. The common-tube, regularity, and exact-initial-data assumptions of [F3] are explicit hypotheses here.

2.1F3step 1.1∎

Apply [F3] to these maps. Its exact-flow conclusion gives global errors O(h) for Lie and O(h2) for Strang. Its approximate-subflow conclusion requires the additional individual error budgets and one-step stability bound and keeps these contributions separate from the splitting defects. No unbounded-operator conclusion is imported.

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Finite-dimensional backward-Euler diffusion transmission model

Definition

Let I1,I2 and the common trace space Λ be finite-dimensional real inner-product spaces, with dim⁡Λ≥1, and set Xi:=Ii⊕Λ. The trace map is γi(xi,λi)=λi. Let Mi and Ai be self-adjoint positive-definite operators on Xi, let Δt>0, and define the local backward-Euler block Bi:=Δt−1Mi+Ai. In coordinates Ii⊕Λ, write

Bi=(Bi,00Bi,0ΓBi,Γ0Bi,ΓΓ),gi=(gi,0,gi,Γ).

For a prescribed local trace λi, solve the interior equation Bi,00xi+Bi,0Γλi=gi,0 and define the outward algebraic interface flux qi:=Bi,Γ0xi+Bi,ΓΓλi−gi,Γ. Eliminating xi gives qi=Siλi−fi, where Si:=Bi,ΓΓ−Bi,Γ0Bi,00−1Bi,0Γ and fi:=gi,Γ−Bi,Γ0Bi,00−1gi,0. The sign convention is outward flux on each subdomain.

The monolithic interface equations are trace continuity λ1=λ2=λ and flux balance q1+q2=0, equivalently (S1+S2)λ=f1+f2. These finite-dimensional matrices represent a linear diffusion-transmission step; no nonlinear fluid-structure well-posedness is asserted.

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ALE moving-domain map and mesh velocity

Definition

Fix an integer d≥1, a time interval [0,T] with T>0, and a nonempty bounded open reference domain Ω^⊂Rd with compact Jordan closure; Jordan measurability means equality of finite-rectangle inner and outer contents (Jordan inner and outer content and Jordan measurable bounded sets in Rm). In dimension d=1, require Ω^ to be a finite union of bounded open intervals with pairwise disjoint closures. In dimensions d≥2, require a finite piecewise-C1 boundary presentation: a finite family of compact boundary faces S1,…,Sq and a compact edge set E such that the faces cover ∂Ω^, each face lies in a regular C1 hypersurface patch, Si∩E is surface-null, all face overlaps lie in E, and off E the boundary is locally one C1 graph with Ω^ on one side. The faces carry that side's outward unit normal.

An ALE moving-domain map is a map A:[0,T]×Ω^‾→Rd whose space-time restriction extends to a C2 map on a neighborhood of this set, and for each t the map At(ξ):=A(t,ξ) is an orientation-preserving diffeomorphism on neighborhoods of the closures. Write Ω(t):=At(Ω^) and J(ξ,t):=det⁡DξAt(ξ)>0. Define the mesh velocity in physical coordinates by wm(x,t):=∂tAt(At−1(x)). In d=1 the moving boundary consists of the transported interval endpoints. In d≥2, require the transported faces under At to satisfy the same finite piecewise-C1 boundary-presentation conditions; the diffeomorphism carries the reference faces and edge set to that presentation, with the transported outward normal.

This definition fixes the regularity and sign conventions for the continuous ALE identities. It does not assert a discrete approximation property, a nonintersection theorem for a computed mesh, or well-posedness of a coupled PDE.

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Periodic one-dimensional moving grid

Definition

Fix N≥1, a reference cell width Δξ>0, a time step Δt>0, and a physical period length L>0. Face positions xi+1/2n and face velocities wi+1/2n are indexed periodically by i∈Z, with xi+N+1/2n=xi+1/2n+L and wi+N+1/2n=wi+1/2n. The physical cell i is [xi−1/2n,xi+1/2n] and its dimensionless Jacobian is Jin=(xi+1/2n−xi−1/2n)/Δξ; require Jin>0.

For a constant physical advection speed a∈R, the relative face speed is ci+1/2n:=a−wi+1/2n. Write c+:=max⁡(c,0) and c−:=min⁡(c,0). The prescribed face move is xi+1/2n+1=xi+1/2n+Δt wi+1/2n. The cell average uˉin is the average over the corresponding physical cell.

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Scalar added-mass staggered recurrence

Definition

For positive scalars ms and mf, call ms the structure mass and mf the fluid added mass in the scalar toy added-mass recurrence. Given a scalar initial velocity increment δv0, define successive increments by

ms δvn+1=−mf δvn(n≥0).

This is a stipulated one-mode sequential update used only to expose its amplification factor. It is not derived from, and makes no stability claim about, a general fluid-structure discretization.

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Conservative upwind ALE finite-volume step

Definition

Inputs: The periodic grid and relative speeds of Periodic one-dimensional moving grid, cell averages uˉin∈R, a constant physical speed a, face velocities wi+1/2n, Δt>0, and Δξ>0.

Instructions:

  1. Without modifying the input mesh or cell averages, compute trial face positions xi+1/2n+1=xi+1/2n+Δt wi+1/2n and trial Jacobians Jin+1=(xi+1/2n+1−xi−1/2n+1)/Δξ.

  2. If any Jin+1≤0, discard the trial data and return the status mesh_crossing and no updated state, leaving the input mesh and cell averages unchanged.

  3. Otherwise, for every face set ci+1/2n=a−wi+1/2n and F^i+1/2n=(ci+1/2n)+uˉin+(ci+1/2n)−uˉi+1n, with periodic indices and c+=max⁡(c,0), c−=min⁡(c,0).

  4. Accept the trial mesh and return the status success and the unique cell averages defined by Jin+1uˉin+1=Jinuˉin−(Δt/Δξ)(F^i+1/2n−F^i−1/2n).

All state updates occur only after every new cell Jacobian has passed the positivity check.

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Jacobi identity for the ALE mesh Jacobian

Statement

For the ALE moving-domain map At of ALE moving-domain map and mesh velocity, let F(ξ,t):=DξAt(ξ) and J(ξ,t):=det⁡F(ξ,t). Then

∂tJ(ξ,t)=J(ξ,t) div⁡xwm(At(ξ),t).

This is the continuous geometric conservation identity. It makes no discrete or time-integration claim.

Facts & Assumptions

Given: The C2 orientation-preserving ALE map At, its Jacobian J>0, and mesh velocity wm as defined in ALE moving-domain map and mesh velocity.

[F1]

ALE moving-domain map and mesh velocity: A is C2 in space and time, F=DξAt is invertible, J=det⁡F, and wm(At(ξ),t)=∂tAt(ξ).

Proof

1.1givenF1algebra

Fix (ξ,t) and write H=∂tF(ξ,t). Differentiability gives F(ξ,t+h)=F+hH+o(h). Since F is invertible, factor this as F[I+hF−1H+o(h)]. The permutation expansion of the determinant gives det⁡(I+hK+o(h))=1+htr⁡K+o(h): the identity permutation contributes 1+h∑jKjj+o(h), while every nonidentity permutation contains at least two off-diagonal factors of order h. Hence ∂tJ=Jtr⁡(F−1H).

2.1F1step 1.1algebra∎

Differentiate wm(At(ξ),t)=∂tAt(ξ) in ξ. The chain rule and equality of the mixed derivatives of the C2 map give H=(Dxwm)(At(ξ),t)F. Therefore tr⁡(F−1H)=tr⁡(F−1(Dxwm)F)=tr⁡(Dxwm)=div⁡xwm, using cyclic invariance of trace. Substitution into step 1.1 proves the identity.

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Moving faces satisfy the one-dimensional discrete GCL

Statement

For the face motion of Periodic one-dimensional moving grid, every cell satisfies

Jin+1=Jin+ΔtΔξ(wi+1/2n−wi−1/2n).

Facts & Assumptions

Given: Periodic faces with Δξ>0 and the update xi+1/2n+1=xi+1/2n+Δt wi+1/2n from Periodic one-dimensional moving grid.

[F1]

Periodic one-dimensional moving grid: Jin=(xi+1/2n−xi−1/2n)/Δξ and each face moves by Δt times its prescribed velocity.

Proof

1.1givenF1algebra

Subtract the left-face update from the right-face update for cell i. This gives xi+1/2n+1−xi−1/2n+1=xi+1/2n−xi−1/2n+Δt(wi+1/2n−wi−1/2n).

2.1F1step 1.1algebra∎

Divide by the positive reference width Δξ and use the definition of Jin+1 and Jin in [F1]. The claimed identity follows.

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The diffusion Schur complement is positive and minimizes energy

Statement

Let B=(KCC∗D) be a self-adjoint positive-definite operator on a finite-dimensional real inner-product direct sum I⊕Λ. Then K is positive definite, S:=D−C∗K−1C is positive definite on Λ, and for every z∈Λ,

⟨Sz,z⟩=min⁡x∈I⟨B(x,z),(x,z)⟩.

In particular, the local Schur operators Si in Finite-dimensional backward-Euler diffusion transmission model are positive definite.

Facts & Assumptions

Given: The block matrix B is self-adjoint positive definite on I⊕Λ, as are the local backward-Euler blocks Bi in Finite-dimensional backward-Euler diffusion transmission model.

[F1]

Finite-dimensional backward-Euler diffusion transmission model: each Bi=Δt−1Mi+Ai is self-adjoint positive definite and its interface Schur complement is Si=Bi,ΓΓ−Bi,Γ0Bi,00−1Bi,0Γ.

Proof

1.1givenF1algebra

For x≠0, ⟨Kx,x⟩=⟨B(x,0),(x,0)⟩>0. Thus K has zero kernel and is invertible because it is an endomorphism of a finite-dimensional space.

2.1step 1.1algebra

For fixed z, put x∗=−K−1Cz. Expanding the quadratic form and completing the square gives ⟨B(x,z),(x,z)⟩=⟨K(x−x∗),x−x∗⟩+⟨(D−C∗K−1C)z,z⟩. The first term is nonnegative and vanishes exactly at x=x∗, proving the minimum formula.

3.1F1step 2.1givenalgebra∎

If z≠0, then (x∗,z)≠0, so positive definiteness of B gives ⟨Sz,z⟩=⟨B(x∗,z),(x∗,z)⟩>0. Hence S is positive definite; applying this to each Bi in [F1] proves the final claim.

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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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Finite-budget Robin-Robin transmission iteration

Definition

Inputs: Positive-definite real Schur operators S1,S2 on the nonzero finite-dimensional trace space Λ, loads f1,f2∈Λ, a parameter α>0, initial incoming datum r10∈Λ, tolerance ε>0, and integer pass budget M≥1.

All vector norms on Λ and operator norms on its endomorphisms are induced by its specified real inner product. Set Ri:=(αI−Si)(αI+Si)−1, qi:=∥Ri∥, and q:=q1q2. For each pass m=0,…,M−1, perform in order

(S1+αI)λ1m+1=r1m+f1,r2m+1=(αI−S1)λ1m+1+f1;(S2+αI)λ2m+1=r2m+1+f2,r1m+1=(αI−S2)λ2m+1+f2.

If ∥r1m+1−r1m∥≤(1−q)ε/∥(αI+S1)−1∥, return converged with trace λ1m+1 and pass count m+1. If all M passes are used without this test succeeding, return budget_exhausted with the last trace λ1M and pass count M. This procedure states the finite stop predicate but makes no claim about its error certificate.

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Schur residual controls trace and reconstruction energy error

Statement

Let S:=S1+S2, let λ∗ solve Sλ∗=f1+f2, and let λ~∈Λ have residual ρ:=Sλ~−(f1+f2). All trace-vector and operator norms below are induced by the given real inner product on Λ. Then

∥λ~−λ∗∥≤∥S−1∥ ∥ρ∥.

If each subdomain state is reconstructed by solving its interior equations at the supplied trace, define the product backward-Euler energy norm by ∥(z1,z2)∥B1⊕B22:=⟨B1z1,z1⟩+⟨B2z2,z2⟩ for zi∈Xi. Then

∥u~−u∗∥B1⊕B2≤∥S−1∥ ∥ρ∥.

Facts & Assumptions

Given: The positive-definite subdomain blocks and Schur complements of Finite-dimensional backward-Euler diffusion transmission model, their energy-minimization property, and the unique monolithic trace λ∗.

[F1]

Finite-dimensional backward-Euler diffusion transmission model: Si is the Schur complement of the positive-definite local block Bi, and the monolithic trace solves (S1+S2)λ∗=f1+f2.

[F2]

The diffusion Schur complement is positive and minimizes energy: for a fixed trace error z, its minimum local Bi energy is ⟨Siz,z⟩.

Proof

1.1givenF1algebra

Set e:=λ~−λ∗. By [F1], ρ=Se. Since S is positive definite, e=S−1ρ; the operator-norm inequality gives ∥e∥≤∥S−1∥ ∥ρ∥.

2.1F1F2step 1.1algebra

The reconstructed state difference on subdomain i has trace e. By [F2], its squared minimum Bi energy is ⟨Sie,e⟩. Adding the two subdomains gives ∥u~−u∗∥B1⊕B22=⟨Se,e⟩=⟨ρ,S−1ρ⟩.

3.1step 2.1algebra∎

Cauchy–Schwarz and the operator norm yield ⟨ρ,S−1ρ⟩≤∥S−1∥ ∥ρ∥2. Take nonnegative square roots to obtain the stated energy bound.

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Choice-free one-dimensional ALE Reynolds transport identity

Statement

Let d=1 and let At be an ALE moving-domain map of ALE moving-domain map and mesh velocity, so Ω^ is a finite union of bounded intervals with pairwise disjoint closures and ∂ξAt>0. For every C1 scalar field g on the moving closure,

ddt∫Ω(t)g(x,t) dx=∫Ω(t)∂tg(x,t) dx+∑x∈∂Ω(t)g(x,t)wm(x,t)n(x),

where n=−1 at each left endpoint and n=+1 at each right endpoint. This dimension-one branch is choice-free.

Facts & Assumptions

Given: A one-dimensional ALE map, a finite union of bounded reference intervals, and a C1 scalar field g on the moving closure.

[F1]

ALE moving-domain map and mesh velocity: in dimension one the reference domain is a finite union of intervals with compact closure, J=∂ξAt>0, and wm(At(ξ),t)=∂tAt(ξ).

[F2]

Jacobi identity for the ALE mesh Jacobian: ∂tJ=J ∂xwm.

[F3]

Change of variables for an injective C1 map on a compact Jordan set: an injective C1 map with invertible derivative sends a compact Jordan set to a compact Jordan set; for a bounded integrand, the image and pullback are Riemann integrable together, and when they are the formula uses the absolute Jacobian determinant.

[F4]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): if H is differentiable on a compact interval and H′ is integrable, then ∫abH′(x) dx=H(b)−H(a).

[F5]

A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: every continuous real function on a compact Jordan set is Riemann integrable there.

[F6]

The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a): a scalar function continuous on a closed interval and differentiable in its interior has a point where its derivative equals the interval's average rate of change.

[F7]

The Riemann integral of a bounded function over a bounded Jordan measurable set and Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm: Jordan integrals are integrals of zero extensions on bounding rectangles, where linearity, monotonicity, and the absolute-value estimate hold.

[F8]

A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content: the integral of the indicator of a compact Jordan set is its finite Jordan content.

[F9]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on a compact metric space is uniformly continuous.

[F10]

A bounded set in Rm is Jordan measurable iff its boundary is null, equivalently of content zero: a bounded set is Jordan measurable exactly when its boundary is null, equivalently has Jordan content zero.

[F11]

Lebesgue's criterion in Rm: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null: a bounded function on a closed bounding rectangle whose discontinuities lie in a null set is Riemann integrable.

Proof

1.1F1F10construct

Put K^=Ω^‾, Kt=At(K^)=Ω(t)‾, and let Bt=∂Ω(t) and B^=∂Ω^. These boundaries are finite endpoint sets, hence compact. Covering each point by intervals of arbitrarily small total length shows that both have Jordan content zero and are compact Jordan sets. By [F10] the boundaries are null and the two open domains are Jordan measurable.

2.1F3F5F7F8F11step 1.1algebra

For any continuous h on Kt, its zero extensions from Ω(t) and Kt to a nondegenerate closed interval containing Kt in its interior are bounded and can be discontinuous only at Bt. By [F11] they are integrable. Their difference is supported on Bt and bounded in absolute value by a constant times 1Bt; [F8] gives ∫1Bt=0, so [F7] shows their integrals agree. The same argument applies on K^ and Ω^. The map image in [F3] is compact Jordan, and [F5] makes g(⋅,t) and G(⋅,t) integrable on the compact closures. Change of variables on those closures, followed by the open/closed equalities just proved, gives I(t)=∫K^G(ξ,t) dξ=∫Ω^G(ξ,t) dξ, where G(ξ,t):=g(At(ξ),t)J(ξ,t).

3.1F5F6F7F8F9step 2.1algebra

For small nonzero h, apply [F6] to s↦G(ξ,s) between t and t+h. The difference quotient equals ∂tG(ξ,θξ) at an intermediate time. By [F9], ∂tG is uniformly continuous on K^ times a compact time neighborhood of t, so these quotients converge uniformly in ξ to ∂tG(ξ,t). The latter and each quotient are integrable on K^ by [F5]; [F7] and [F8] bound the integral of their difference by its uniform norm times cont⁡(K^). Since I(t)=∫K^G(ξ,t) dξ by step 2.1, it follows that I′(t)=∫K^∂tG(ξ,t) dξ.

4.1F1F2F3F5step 2.1step 3.1algebra

The chain and product rules together with [F2] give I′(t)=∫K^J(∂tg+∂xg wm+g ∂xwm)(At(ξ),t) dξ. This integrand is continuous on the compact reference closure; [F3] changes variables on the compact closures, and step 2.1 converts the resulting physical-domain integrals to open-domain integrals. Thus I′(t)=∫Ω(t)[∂tg+∂x(gwm)](x,t) dx.

5.1F4F5step 4.1algebra∎

The function x↦g(x,t)wm(x,t) is differentiable on each compact image interval and its derivative is continuous, hence integrable there by [F5]. Apply [F4] to obtain ∫αjβj∂x(gwm) dx=g(βj,t)wm(βj,t)−g(αj,t)wm(αj,t). Summing right endpoints with normal +1 and left endpoints with normal −1 proves the boundary term. No choice axiom is used.

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Reynolds transport identity on an ALE moving domain

Statement

Assume Countable Choice and d≥2. Let At be the ALE moving-domain map of ALE moving-domain map and mesh velocity, and let g(x,t) and its first derivatives be continuous on the moving closure. Then, for each t∈(0,T),

ddt∫Ω(t)g(x,t) dx=∫Ω(t)∂tg(x,t) dx+∫∂Ω(t)g(x,t)wm(x,t)⋅n(x,t) dS.

Here n is the outward normal from the transported finite piecewise-C1 boundary presentation. This is a continuous transport identity; no discrete high-dimensional geometric conservation estimate follows from it.

Facts & Assumptions

Given: Assume Countable Choice, an ALE moving-domain map At, its Jacobian J, its mesh velocity wm, and a scalar field g with the stated space-time regularity.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice, written ACω, is the axiom that every countable family of nonempty sets has a choice function.

[F1]

ALE moving-domain map and mesh velocity: At is a C2 orientation-preserving diffeomorphism, J=det⁡DξAt>0, and wm(At(ξ),t)=∂tAt(ξ).

[F2]

Jacobi identity for the ALE mesh Jacobian: ∂tJ=J div⁡xwm.

[F3]

Specified finite piecewise C1 boundary presentations: the boundary has a specified finite set of compact C1 faces and a compact edge set, with overlaps and edges surface-null and an outward normal on each face off the edges.

[F4]

Change of variables for an injective C1 map on a compact Jordan set: an injective C1 map with invertible derivative sends a compact Jordan set to a compact Jordan set; for a bounded integrand, the change-of-variables formula applies when the image integrand (equivalently its pullback) is Riemann integrable.

[F5]

Divergence for finite piecewise C1 presentations: under ACω, a C1 vector field on a specified finite piecewise-C1 domain satisfies ∫Ωdiv⁡F=∫∂ΩF⋅n dS.

[F6]

The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a): a scalar function continuous on a closed interval and differentiable in its interior has a point where its derivative equals the interval's average rate of change.

[F7]

The Riemann integral of a bounded function over a bounded Jordan measurable set and Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm: Jordan integrals are integrals of zero extensions on bounding rectangles, where linearity, monotonicity, and the absolute-value estimate hold.

[F8]

A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content: the integral of the indicator of a compact Jordan set is its finite Jordan content.

[F9]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on a compact metric space is uniformly continuous.

[F10]

A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: every continuous real function on a compact Jordan set is Riemann integrable there.

[F11]

A bounded set in Rm is Jordan measurable iff its boundary is null, equivalently of content zero: a bounded set is Jordan measurable exactly when its boundary is null, equivalently has Jordan content zero.

[F12]

Lebesgue's criterion in Rm: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null: a bounded function on a closed bounding rectangle whose discontinuities lie in a null set is Riemann integrable.

[F13]

Borel Darboux integrands in finite dimension: under ACω, a bounded Borel Riemann-integrable function on a nondegenerate closed box is Lebesgue integrable with the same integral. The volume integral in [F5] uses Lebesgue measure.

Proof

1.1F1F4F11construct

Put K^=Ω^‾, Kt=At(K^)=Ω(t)‾, and let Bt=∂Ω(t) and B^=∂Ω^. Cover each compact face in either finite boundary presentation by finitely many smaller regular C1 graph charts with bounded base and bounded graph derivative. For a base grid of side δ, each chart needs at most Cδ−(d−1) base cubes; the Lipschitz bound places each graph portion in an axis-parallel d-cube of side C′δ. The total covering volume is at most C′′δ, which tends to zero. Since there are finitely many faces, both boundaries are compact Jordan sets of content zero. By [F11] they are null and the open domains are Jordan measurable; by the compact-image conclusion in [F4], Kt is compact Jordan.

2.1F4F7F8F10F12step 1.1algebra

For any continuous h on Kt, its zero extensions from Ω(t) and Kt to a nondegenerate closed rectangle containing Kt in its interior are bounded and have discontinuities only on Bt, a null set, so [F12] makes them integrable. Their difference is supported on Bt and bounded in absolute value by a constant times 1Bt. By [F8], ∫1Bt=0, and [F7] shows the two integrals agree. The same argument applies to K^ and Ω^. By [F10], continuous integrands on the compact closures are integrable, so [F4] applies to the compact closures. Consequently, for G(ξ,t):=g(At(ξ),t)J(ξ,t), I(t):=∫Ω(t)g dx=∫K^G(ξ,t) dξ=∫Ω^G(ξ,t) dξ.

3.1F6F7F8F9F10step 2.1algebra

For small nonzero h, apply [F6] to s↦G(ξ,s) between t and t+h. The difference quotient equals ∂tG(ξ,θξ) at an intermediate time. By [F9], ∂tG is uniformly continuous on K^ times a compact time neighborhood of t, so these quotients converge uniformly in ξ to ∂tG(ξ,t). The derivative and difference quotients are continuous on the compact Jordan set K^, hence integrable by [F10]. [F7] and [F8] bound the integral of their difference by its uniform norm times cont⁡(K^). Since I(t)=∫K^G(ξ,t) dξ by step 2.1, I′(t)=∫K^∂tG(ξ,t) dξ. The chain and product rules then give I′(t)=∫K^J(∂tg+∇xg⋅wm)+g∂tJ dξ.

4.1F2F4F10step 2.1step 3.1algebra

Substitute [F2] into step 3.1 and use ∇xg⋅wm+gdiv⁡xwm=div⁡x(gwm). The resulting integrand is continuous on the moving closure. Apply [F4] to the compact closures and use the open/closed integral equality from step 2.1 to obtain I′(t)=∫Ω(t)(∂tg+div⁡x(gwm)) dx.

5.1A1F3F5F13step 2.1step 4.1∎

Set f=div⁡x(gwm) and choose a nondegenerate closed box Q containing Kt in its interior. The field gwm is C1 on the moving closure, so f is continuous and bounded there. Its zero extension f~ from the open set Ω(t) to Q is Borel: each strict superlevel is a relatively open subset of Ω(t), together with Q∖Ω(t) when the level is negative. Step 2.1 gives its Riemann integrability. By [A1] and [F13], its Riemann integral over Q equals its Lebesgue integral, which is the Lebesgue integral of f over Ω(t). Thus the Jordan volume integral in step 4.1 is exactly the volume integral used by [F5]. Apply [F5] to gwm on the finite piecewise-C1 presentation [F3] to obtain ∫Ω(t)div⁡x(gwm) dx=∫∂Ω(t)gwm⋅n dS. Substitution into step 4.1 proves the asserted identity. Countable Choice is used through [F5] and [F13].

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The upwind ALE step conserves mass and preserves constants

Statement

For every input satisfying Conservative upwind ALE finite-volume step, the procedure completes after finitely many operations and returns exactly one of mesh_crossing (with no updated state, if some Jin+1≤0) or success (with a unique updated array, if all Jin+1>0). Whenever it returns success, it preserves total periodic mass, ∑iJin+1uˉin+1=∑iJinuˉin. It also preserves every constant state: if uˉin=U for every cell, then uˉin+1=U for every cell. The mass and constant-state properties require no CFL condition.

Facts & Assumptions

Given: An admissible periodic moving grid, a constant physical speed, cell averages, and the inputs of Conservative upwind ALE finite-volume step.

[F1]

Periodic one-dimensional moving grid: indices are periodic, Jin=(xi+1/2n−xi−1/2n)/Δξ, and ci+1/2=a−wi+1/2.

[F2]

Conservative upwind ALE finite-volume step: after computing the finite trial face table it returns mesh_crossing with no state update if any new Jin+1≤0; otherwise it accepts the trial mesh and returns success with the cell averages defined by Jin+1uˉin+1=Jinuˉin−(Δt/Δξ)(F^i+1/2−F^i−1/2), where F^i+1/2=ci+1/2+uˉin+ci+1/2−uˉi+1n.

[F3]

Moving faces satisfy the one-dimensional discrete GCL: Jin+1=Jin+(Δt/Δξ)(wi+1/2−wi−1/2).

Proof

1.1givenF1F2algebra

By [F1], the input has a finite number N of cells and faces. The algorithm performs a fixed finite number of arithmetic operations and comparisons per cell or face, so its trial face calculation and Jacobian check terminate. If some Jin+1≤0, its specified branch returns mesh_crossing before any state update.

1.2givenF1F2algebra

Sum the update in [F2] over one period. Each face flux occurs once with a plus sign and once with a minus sign, so the periodic flux sum is zero. The remaining identity is ∑iJin+1uˉin+1=∑iJinuˉin.

1.3F1F2F3algebra

If every input average equals U, then c++c−=c gives F^i+1/2=(a−wi+1/2)U. Thus F^i+1/2−F^i−1/2=−U(wi+1/2−wi−1/2). Substituting this in [F2] and using [F3] gives Jin+1uˉin+1=Jin+1U.

2.1F1F2step 1.1algebra

If all checked Jacobians are positive, each update equation in [F2] has the unique solution obtained by dividing its specified right-hand side by Jin+1>0. There are finitely many such formulas, and no further failure branch, so the algorithm returns success with exactly this array.

3.1step 2.1step 1.3algebra∎

In the success branch of step 2.1, divide the equality in step 1.3 by Jin+1>0 to obtain uˉin+1=U.

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Robin-Robin iteration contracts to the monolithic trace

Statement

Assume the model, positive Schur operators, and algorithm of Finite-dimensional backward-Euler diffusion transmission model, The diffusion Schur complement is positive and minimizes energy, and Finite-budget Robin-Robin transmission iteration, with α>0. Every Robin solve in the algorithm exists uniquely, and for every admitted input it returns converged with the current first-subdomain trace on the first passing test, or returns budget_exhausted with λ1M after at most M passes. Let λ∗ be the unique monolithic trace solving (S1+S2)λ∗=f1+f2, and put Ri=(αI−Si)(αI+Si)−1 and qi=∥Ri∥, q=q1q2. Then q<1, the unique fixed point of the iteration corresponds exactly to λ∗, and the incoming-data error satisfies

r1m+1−r1∗=R2R1(r1m−r1∗),∥r1m+1−r1∗∥≤q∥r1m−r1∗∥.

If the algorithm's stop test succeeds at pass m+1, its returned trace satisfies ∥λ1m+1−λ∗∥≤ε. A budget_exhausted return carries no such certificate.

Facts & Assumptions

Given: The finite-dimensional diffusion transmission model, Si positive definite, α>0, loads fi, and the finite-pass algorithm.

[F1]

Finite-dimensional backward-Euler diffusion transmission model: the outward flux is pi:=Siλi−fi, and the monolithic trace equation is (S1+S2)λ=f1+f2.

[F2]

The diffusion Schur complement is positive and minimizes energy: each Si is self-adjoint positive definite.

[F3]

Finite-budget Robin-Robin transmission iteration: each pass uses (Si+αI)λi=ri+fi and exchanges ri+1=(αI−Si)λi+fi.

[F4]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: each finite-dimensional real self-adjoint endomorphism has an orthonormal eigenbasis.

Proof

1.1givenF2F3algebra

For each i, Si+αI is self-adjoint positive definite because ⟨(Si+αI)z,z⟩=⟨Siz,z⟩+α∥z∥2>0 for every z≠0. It is therefore invertible on the finite-dimensional space Λ; its inverse and the operators Ri have finite norms, and ∥(αI+S1)−1∥>0 because Λ≠{0}. Each first solve uniquely determines λ1m+1, the first exchange uniquely determines r2m+1, the second solve uniquely determines λ2m+1, and the last exchange uniquely determines r1m+1. Since M is finite, these operations define all passes up to M; the tests either return converged at the first pass that satisfies them or, if none does, return budget_exhausted with the last trace. Thus the algorithm is well-defined and terminates within its stated budget.

1.2givenF1F2algebra

By [F1] and [F2], S:=S1+S2 is positive definite, so the monolithic trace λ∗=S−1(f1+f2) exists uniquely. Set pi∗:=Siλ∗−fi; then p1∗+p2∗=0. Define r1∗:=αλ∗+p1∗ and r2∗:=αλ∗+p2∗.

2.1F1F3step 1.2algebra

The first Robin equation in [F3] is equivalent to r1=αλ1+p1, and its exchange is r2=αλ1−p1; the second equation and exchange have the same forms with indices 1 and 2 reversed. Since p2∗=−p1∗, the pair (r1∗,r2∗) is unchanged by both exchanges, so λ∗ gives a fixed point.

2.2F1F3step 1.2algebra

Conversely, at any fixed point the two exchange equalities give r2=αλ1−p1=αλ2+p2 and r1=αλ1+p1=αλ2−p2. Adding the equalities for r2 and r1 yields 2α(λ1−λ2)=0, hence λ1=λ2; subtracting yields p1+p2=0. Thus the common trace solves the unique monolithic equation from step 1.2, proving the fixed point is unique.

3.1F2F3F4step 2.1algebra

By [F2] and [F4], in an orthonormal eigenbasis of Si, Ri has eigenvalues (α−σ)/(α+σ) with σ>0. Each has absolute value strictly below 1, so qi<1. Subtracting the fixed-point equations from the iteration equations in [F3] gives r2m+1−r2∗=R1(r1m−r1∗) and then r1m+1−r1∗=R2(r2m+1−r2∗). Therefore the displayed error map holds and q=q1q2<1.

4.1F3step 3.1algebra∎

Put em=r1m−r1∗ and dm=r1m+1−r1m. The error map in step 3.1 gives dm=(R2R1−I)em. Since ∥R2R1∥≤q<1, the reverse triangle inequality yields ∥dm∥≥(1−q)∥em∥. The first solve in [F3] gives λ1m+1−λ∗=(αI+S1)−1em. Therefore the stop test in [F3] implies ∥λ1m+1−λ∗∥≤∥(αI+S1)−1∥ ∥dm∥/(1−q)≤ε. If the finite pass budget is exhausted without this test, no trace certificate has been issued.

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CFL monotonicity and weighted L1 contraction for the ALE step

Statement

Suppose the step of Conservative upwind ALE finite-volume step succeeds and, for every cell,

ΔtΔξ((ci+1/2n)+−(ci−1/2n)−)≤Jin.

Then the update is monotone, maps nonnegative cell averages to nonnegative cell averages, and for any two input arrays U,V with outputs U′,V′ it satisfies

∑iJin+1∣Ui′−Vi′∣≤∑iJin∣Ui−Vi∣.

Facts & Assumptions

Given: A periodic moving grid, the successful upwind ALE step, and the stated cellwise CFL inequalities.

[F1]

Periodic one-dimensional moving grid: ci+1/2=a−wi+1/2, Jin>0, and indices are periodic.

[F2]

Conservative upwind ALE finite-volume step: the update is linear in the input averages and uses the relative-speed upwind flux, provided every Jin+1>0.

[F3]

The upwind ALE step conserves mass and preserves constants: every successful step conserves weighted mass for arbitrary inputs and preserves constant input arrays.

Proof

1.1givenF1F2algebra

Put μ=Δt/Δξ. Expanding the flux difference in [F2] gives Jin+1Ui′=μ(ci−1/2n)+Ui−1+[Jin−μ((ci+1/2n)+−(ci−1/2n)−)]Ui−μ(ci+1/2n)−Ui+1. The first and third coefficients are nonnegative by the sign definitions of positive and negative parts; the middle coefficient is nonnegative by the CFL condition.

2.1F2F3step 1.1algebra

Divide by the positive Jin+1 from success. The resulting matrix P has nonnegative entries, so U≤V componentwise implies PU≤PV, and U≥0 implies PU≥0. By [F3], the update preserves constants, hence every row of P sums to 1.

3.1F3step 2.1algebra∎

For each coordinate vector ej, mass conservation in [F3] gives ∑iJin+1Pij=Jjn. Linearity and nonnegativity now imply ∑iJin+1∣Ui′−Vi′∣=∑iJin+1∣(P(U−V))i∣≤∑j(∑iJin+1Pij)∣Uj−Vj∣=∑jJjn∣Uj−Vj∣. This is the asserted weighted L1 contraction.

5 · Examples, counterexamples and false statements

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