Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 335 / 1 / ii

For large

r = ∣ r ∣

, the outgoing kernel has

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

ψ_{s} = e^{ik r} f_{\infty} / r + O (r^{- 2})

, gives

f_{\infty} (r, r_{0}, k) = - \frac{1}{4 π} \int_{\partial D} e^{- ik r \cdot r^{'}} [\partial_{n^{'}} ψ_{s} + ik (r \cdot n^{'}) ψ_{s}] d S^{'} .

Conjugating the Herglotz wave function gives

\overline{v_{g}} (r^{'}) = \int_{S^{2}} e^{- ik r \cdot r^{'}} \overline{g (r)} d S

\int_{S^{2}} f_{\infty} (r, r_{0}, k) \overline{g (r)} d S = \frac{1}{4 π} \int_{\partial D} [ψ_{s} \partial_{n^{'}} \overline{v_{g}} - \overline{v_{g}} \partial_{n^{'}} ψ_{s}] d S^{'} .

This contains only the scattered-field boundary traces and the Herglotz function, with the outward-obstacle normal convention fixing the sign.

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