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The volume far-field map is bounded and is the uniform far-field limit

Statement

Assume Countable Choice. Let D,q,k be as in The three-dimensional volume integral scattering model, with 0<∣D∣<∞, and put R0>sup⁡y∈D∣y∣. Then Fk:L2(D;C)→C(S2;C) is linear and ∥Fkv∥∞≤∥q∥∞∣D∣4π∥v∥2. It is Lipschitz in direction with constant at most ∣k∣R0∥q∥∞∣D∣∥v∥2/(4π). For every r≥2R0, the exterior volume potential of The three-dimensional volume integral scattering model satisfies sup⁡θ∈S2∣re−ikrVkv(rθ)−Fkv(θ)∣≤2R0+∣k∣R024πr∥q∥∞∣D∣∥v∥2. Thus Fkv is the uniform far-field amplitude under the outgoing eikr/r convention.

Facts & Assumptions

Given: Countable Choice, D,q,k as in the Statement, v∈L2(D;C), and R0>sup⁡y∈D∣y∣.

[F1]

Cauchy–Schwarz gives ∥qv∥L1(D)≤∥q∥∞∣D∣∥v∥2 (The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz).

[F2]

For real s,t, let γ(r)=exp⁡(i(t+r(s−t))) on [0,1]. If s=t, it is constant. Otherwise, with z=i(t+r(s−t)), the complex derivative theorem The complex exponential is entire and its complex derivative is itself gives γ(r+h)−γ(r)h=i(s−t)exp⁡(z+ih(s−t))−exp⁡zih(s−t)⟶i(s−t)exp⁡z. Thus ∣γ′(r)∣=∣s−t∣ by exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0, and identifying C with R2, If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz gives ∣eis−eit∣≤∣s−t∣.

[F3]

The integral convention for the exterior potential and the far-field map is fixed in The three-dimensional volume integral scattering model.

Proof

Given: The data in the Statement.

1.1F1F3algebra

By [F1], the defining integral of Fkv is absolutely convergent and ∣Fkv(θ)∣≤(4π)−1∥qv∥1. Taking the supremum proves the norm bound; linearity follows from linearity of the integral.

1.2F1F2F3

For θ,θ′∈S2, [F2] gives ∣e−ikθ⋅y−e−ikθ′⋅y∣≤∣k∣ ∣y∣ ∣θ−θ′∣. Integrating against ∣qv∣/(4π) and using [F1] proves the stated Lipschitz bound and hence continuity in direction.

1.3givenF2algebra

Fix r≥2R0, θ∈S2, and y∈D, and set s:=∣rθ−y∣. Then s≥r−R0≥r/2, ∣r−s∣≤R0, and ∣s−(r−θ⋅y)∣=∣y∣2−(θ⋅y)2s+r−θ⋅y≤R02r, because the denominator is at least (r−R0)+(r−R0)≥r. Therefore ∣r/s−1∣≤2R0/r and, by [F2], ∣eik(s−r)−e−ikθ⋅y∣≤∣k∣R02/r.

2.1F1F3step 1.3algebra∎

The integrand identity re−ikrGk(rθ−y)=(4π)−1(r/s)eik(s−r) and step 1.3 give a uniform pointwise error at most (2R0+∣k∣R02)/(4πr). Integrating against ∣qv∣, applying [F1], and taking the supremum over θ proves the far-field estimate.

Depends on

Used by

Cited to discharge well-definedness by The three-dimensional volume integral scattering model.

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