 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-335/1/i/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 335 / 1 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Use the outgoing Green function for the three-dimensional Helmholtz equation

G_{k} (r, r^{'}) = \frac{e ^{ik ∣ r - r^{'} ∣}}{4 π ∣ r - r ^{'} ∣}, (Δ_{r^{'}} + k^{2}) G_{k} = - δ_{r} .

(1)

Let

n^{'}

point out of the obstacle

D

into the exterior. Apply Green second identity to the field and

G_{k}

in a large exterior region. Their Sommerfeld radiation conditions make the outer-sphere contribution vanish. The normal on the exterior region's inner boundary is

- n^{'}

, so the Kirchhoff–Helmholtz representation is

ψ (r) = \int_{V} G_{k} (r, r^{'}) Q (r^{'}) d r^{'} + \int_{\partial D} [ψ (r^{'}) \partial_{n^{'}} G_{k} - G_{k} \partial_{n^{'}} ψ (r^{'})] d S^{'}, r \in / \overline{D} .

(2)

The first integral is the incident source field

ψ_{i}

. Therefore the scattered field is the surface term. Since

ψ_{i}

solves the homogeneous Helmholtz equation within the obstacle, its surface term is zero at exterior observation points; the same representation can be written using only scattered-field traces:

ψ_{s} (r) = \int_{\partial D} [ψ_{s} \partial_{n^{'}} G_{k} - G_{k} \partial_{n^{'}} ψ_{s}] d S^{'} .

(3)

The unknown boundary condition determines these traces, but the representation itself does not require choosing that condition.

 Read the full article

 New to topics? Read the docs here!