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Exact spherical-mode reflection under the local impedance truncation

Statement

Fix k,R>0 and ℓ≥0. On r=R, write one radial mode as u(r)=Ahℓ(1)(kr)+Bhℓ(2)(kr), where hℓ(2)=hℓ(1)‾. If u satisfies ∂ru−iku=0 at R, then B=−(hℓ(1))′(kR)−ihℓ(1)(kR)(hℓ(2))′(kR)−ihℓ(2)(kR)A. The denominator is nonzero. For A≠0, the exact reflected modal-amplitude ratio is therefore BA=−(hℓ(1))′(kR)−ihℓ(1)(kR)(hℓ(2))′(kR)−ihℓ(2)(kR). If A=0, the coefficient equality forces B=0 and no ratio is defined. No uniform-in-ℓ smallness is asserted.

Facts & Assumptions

Given: The outgoing/incoming radial pair at positive real argument.

[F1]

First-order local impedance condition at an artificial plane: the spherical truncation condition uses the outgoing convention ∂ru−iku=0.

[F2]

Outgoing spherical Hankel modes have positive flux and no positive-real zeros: Im⁡(hℓ(1)‾(hℓ(1))′)=z−2 and the incoming conjugate has the negative of this flux.

Proof

Given: k,R>0, ℓ≥0, and [F1]–[F2].

1.1F1F2algebra

Let z=kR. The boundary condition, divided by k, is A((hℓ(1))′(z)−ihℓ(1)(z))+B((hℓ(2))′(z)−ihℓ(2)(z))=0. If the incoming factor in parentheses were zero, then (hℓ(2))′=ihℓ(2), so Im⁡(hℓ(2)‾(hℓ(2))′)=∣hℓ(2)∣2>0 by nonvanishing, contradicting its negative flux from [F2]. Hence that factor is nonzero.

2.1step 1.1algebra∎

Divide the equation in step 1.1 by its nonzero incoming factor to obtain the displayed coefficient equality for every A,B. For A≠0, division by A gives the displayed ratio. If A=0, the coefficient equality forces B=0, so the boundary condition has no nonzero purely incoming mode and the zero mode carries no ratio.

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