Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 335 / 2 / b

Solution

Write

r = R θ

∣ θ ∣ = 1

. Uniformly for

r^{'}

in the source ball,

∣ r - r^{'} ∣ = R - θ \cdot r^{'} + O (r_{0}^{2} / R), ∣ r - r^{'} ∣^{- 1} = R^{- 1} + O (r_{0} / R^{2}) .

For fixed

k

and

r_{0}

G_{k} (R θ, r^{'}) = \frac{e ^{ik R}}{4 π R} e^{- ik θ \cdot r^{'}} + O (R^{- 2}) .

Substitution into the source integral yields

ψ (R θ, k) = \frac{e ^{ik R}}{R} f_{\infty} (θ, k) + O (R^{- 2}), f_{\infty} (θ, k) = - \frac{1}{4 π} \int_{A} Q (r^{'}) e^{- ik θ \cdot r^{'}} d^{3} r^{'} .

T : L^{2} (A) ⟶ L^{2} (S_{1}), (TQ) (θ) = - \frac{1}{4 π} \int_{A} Q (r^{'}) e^{- ik θ \cdot r^{'}} d^{3} r^{'} .

It samples the Fourier transform of the source on the sphere of radius

k

∥ T ∥_{HS}^{2} = \frac{∣ A ∣∣ S _{1} ∣}{16 π ^{2}} = \frac{∣ A ∣}{4 π} = \frac{r _{0}^{3}}{3} .

Use complex inner products linear in the first argument:

(f, g) = \int f \overline{g}

. Interchanging the volume and surface integrals gives

(TQ, g)_{L^{2} (S_{1})} = \int_{A} Q (r^{'}) \overline{[- \frac{1}{4 π} \int_{S_{1}} g (θ) e^{ik θ \cdot r^{'}} d S (θ)]} d^{3} r^{'} .

Hence the adjoint source-to-far-field operator is

(T^{*} g) (r^{'}) = - \frac{1}{4 π} \int_{S_{1}} g (θ) e^{ik θ \cdot r^{'}} d S (θ) .

This is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

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