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Midpoint conductivity dissipates the Yee discrete energy

Statement

For the midpoint-conductive update of Lossless or midpoint-conductive Yee leapfrog update, define the same staggered cross-energy as in Strict spectral CFL positivity and Yee modified energy. Then for every step $E^{n + 1} - E^{n} = - Δ t {(\frac{e^{n + 1} + e^{n}}{2})}^{T} j^{n + 1 / 2} - Δ t {(\frac{e^{n + 1} + e^{n}}{2})}^{T} S (\frac{e^{n + 1} + e^{n}}{2}) .$ With zero impressed current, this algebraic energy is nonincreasing because $S ⪰ 0$ . Under the strict spectral CFL hypothesis of Strict spectral CFL positivity and Yee modified energy it is also positive on admissible update states. Dissipation is asserted for this specified midpoint treatment, not for an arbitrary conductivity discretization.

Facts & Assumptions

Given: $Δ t > 0$ , symmetric positive definite $M_{e}, M_{b}$ , symmetric positive semidefinite $S$ , the midpoint conductivity row, and the stated cross-energy.

[F1]

The electric row contains $S (e^{n + 1} + e^{n}) / 2$ and the magnetic row is Faraday leapfrog. (Lossless or midpoint-conductive Yee leapfrog update)

[F2]

$S$ is positive semidefinite. (Oriented Maxwell cochains, masses, and the Yee grid)

[F3]

Under the strict spectral CFL bound the cross-energy is positive on admissible states. (Strict spectral CFL positivity and Yee modified energy)

Proof

1.1F1algebra

Put ${\overset{ˉ}{e}}^{n} = (e^{n + 1} + e^{n}) / 2$ . The electric quadratic-energy increment is $\frac{1}{2} ∥ e^{n + 1} ∥_{M_{e}}^{2} - \frac{1}{2} ∥ e^{n} ∥_{M_{e}}^{2} = ({\overset{ˉ}{e}}^{n})^{T} M_{e} (e^{n + 1} - e^{n})$ . By [F1], this is $Δ t ({\overset{ˉ}{e}}^{n})^{T} C^{T} M_{b} b^{n + 1 / 2} - Δ t ({\overset{ˉ}{e}}^{n})^{T} j^{n + 1 / 2} - Δ t ({\overset{ˉ}{e}}^{n})^{T} S {\overset{ˉ}{e}}^{n}$ .

2.1F1step 1.1algebra

The magnetic cross-energy increment is $\frac{1}{2} (b^{n + 1 / 2})^{T} M_{b} (b^{n + 3 / 2} - b^{n - 1 / 2})$ . Applying the magnetic row in [F1] at $n$ and $n + 1$ gives $b^{n + 3 / 2} - b^{n - 1 / 2} = - Δ t C (e^{n + 1} + e^{n}) = - 2 Δ t C {\overset{ˉ}{e}}^{n}$ . Therefore the magnetic increment is $- Δ t ({\overset{ˉ}{e}}^{n})^{T} C^{T} M_{b} b^{n + 1 / 2}$ and cancels the electric curl work in step 1.1. The remaining terms are exactly the claimed balance.

3.1F2F3step 2.1algebra∎

If $j^{n + 1 / 2} = 0$ , the balance and [F2] imply $E^{n + 1} \leq E^{n}$ . When the strict CFL condition also holds, [F3] shows these nonnegative values control the admissible electric and centered magnetic state.

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Sources

John B. Schneider, Understanding the Finite-Difference Time-Domain Method (standard reference, not scraped)