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Finite-time Yee error for smooth homogeneous periodic Maxwell fields

Statement

Assume the hypotheses and residual bounds of Second-order consistency of the homogeneous periodic Yee scheme on [0,T]. Write δttun=(un+1−2un+un−1)/Δt2. Let ehn,bhn+1/2 be a lossless, source-free periodic Yee solution, with S=0 and jn+1/2=0 at every step, and let e∗n,b∗n+1/2 be the exact samples. Set wn=ehn−e∗n and zn+1/2=bhn+1/2−b∗n+1/2. Suppose the initial electric rows are compatible with the Yee equations and satisfy ∥w0∥Me+∥Cw0∥Mb+∥w1∥Me+∥Cw1∥Mb+∥w1−w0Δt∥Me+∥z1/2∥Mb≤C0(h2+Δt2). Let λmax⁡ be the largest generalized curl-curl eigenvalue in Strict spectral CFL positivity and Yee modified energy. If the strict spectral margin obeys γ:=1−Δt2λmax⁡/4≥γ0>0, then for every integer n≥0 with tn+1≤T, ∥ehn−e∗n∥Me+∥bhn+1/2−b∗n+1/2∥Mb+∥C(ehn−e∗n)∥Mb≤C(h2+Δt2), where C is independent of the grid counts and time step. The statement is for the smooth homogeneous periodic branch and compatible starts; it makes no Taylor error claim across material jumps or an unspecified PEC boundary stencil.

Facts & Assumptions

Given: A homogeneous periodic grid, exact smooth source-free Maxwell samples, the lossless source-free Yee rows, a strict CFL margin, and the stated initial-error bound.

[F1]

With S=0 and jn+1/2=0 at every step, eliminating the magnetic half-step from the Yee rows gives the electric recurrence Meδtten+CTMbCen=0 and the Faraday half-step row. (Lossless or midpoint-conductive Yee leapfrog update)

[F2]

The generalized Rayleigh maximum λmax⁡ gives ∥Cv∥Mb2≤λmax⁡∥v∥Me2 for every electric cochain v. The strict spectral margin is 1−Δt2λmax⁡/4. (Strict spectral CFL positivity and Yee modified energy)

[F3]

Exact samples satisfy ∥rhn∥Me−1≤CT(h2+Δt2) and the unscaled Faraday-row residual satisfies ∥shn∥Mb≤CTΔt3. (Second-order consistency of the homogeneous periodic Yee scheme)

[F4]

Me is positive definite, and ∥r∥Me−12=rTMe−1r is the dual norm to ∥⋅∥Me. (Oriented Maxwell cochains, masses, and the Yee grid)

Proof

1.1F1F3algebra

Subtract the exact-sample equations with residuals from the numerical rows. With K=CTMbC, this gives Meδttwn+Kwn=−rhn,zn+1/2−zn−1/2=−Δt Cwn−shn. The sign convention for shn is the one in the consistency statement.

1.2F2givenalgebra

Put Vn=(wn+1−wn)/Δt, wˉn=(wn+1+wn)/2, and En+1/2=12∥Vn∥Me2+12⟨Cwn+1,Cwn⟩Mb. Expanding about Cwˉn gives En+1/2=12∥Vn∥Me2+12∥Cwˉn∥Mb2−Δt28∥CVn∥Mb2. The generalized Rayleigh bound in [F2] yields En+1/2≥γ02∥Vn∥Me2+12∥Cwˉn∥Mb2. The initial assumptions therefore give E1/2≥0 and E1/2≤C(h2+Δt2).

2.1F4step 1.1algebra

The kinetic-energy increment is Δt⟨(Vn+Vn−1)/2,Meδttwn⟩, and the curl-potential increment is Δt⟨(Vn+Vn−1)/2,Kwn⟩. Substituting step 1.1 gives En+1/2−En−1/2=−Δt⟨Vn+Vn−12,rhn⟩. By [F4] and Cauchy–Schwarz, the absolute value is at most Δt∥rhn∥Me−1(∥Vn∥Me+∥Vn−1∥Me)/2.

3.1F2F3step 1.2step 2.1algebra

Coercivity in step 1.2 bounds both velocities by the square roots of their adjacent energies times 2/γ0. Factoring the difference of squares and summing (with the zero-energy case obtained by continuity, or by first replacing each square root by E+δ and letting δ↓0) gives max⁡n(∥Vn∥Me+∥Cwˉn∥Mb)≤C(E1/2+∑nΔt∥rhn∥Me−1)≤CT(h2+Δt2). The sum has at most T/Δt terms, so [F3] gives the final inequality.

4.1F2givenstep 1.2step 3.1algebra

Since wn=w0+Δt∑m<nVm, the initial bound and step 3.1 imply max⁡n∥wn∥Me≤CT(h2+Δt2). For n≥1, Cwn=Cwˉn−1+ΔtCVn−1/2 and Δt∥CVn−1∥Mb≤Δtλmax⁡∥Vn−1∥Me≤2∥Vn−1∥Me by the strict CFL condition. Thus the same order holds for ∥Cwn∥Mb for n≥1; the n=0 curl bound is part of the initial assumptions.

5.1F3givenstep 1.1step 4.1algebra∎

Summing the second equation in step 1.1 from the controlled start z1/2 gives ∥zn+1/2∥Mb≤∥z1/2∥Mb+Δt∑m=1n∥Cwm∥Mb+∑m=1n∥shm∥Mb. The curl sum is bounded by Tmax⁡m∥Cwm∥Mb, and the residual sum is at most (T/Δt)CTΔt3=CTTΔt2 by [F3]. The initial bound and step 4.1 therefore bound both terms by CT(h2+Δt2). Combining the electric, curl and magnetic bounds proves the claim.

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