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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-80/3/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 80 / 3 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use positive carrier wavenumber

k

and retain waves traveling mainly in the positive

x

direction. Factor out their rapid wave phase by writing

ψ = e^{ik x} E (x, z)

. Substitution into the Helmholtz equation is exact and gives

E_{xx} + 2 ik E_{x} + E_{zz} = 0.

For a transverse Fourier mode

e^{i ν z}

at a small angle,

∣ ν ∣/ k ≪ 1

and its forward longitudinal wavenumber is

k_{x} = k^{2} - ν^{2} = k - \frac{ν ^{2}}{2 k} + O (ν^{4} / k^{3}) .

Thus the envelope varies along

x

on scale

k / ν^{2}

, much more slowly than the carrier wavelength, and

∣ E_{xx} ∣/∣ k E_{x} ∣ = O (ν^{2} / k^{2})

. Dropping this smaller term derives the parabolic wave equation,

2 ik E_{x} + E_{zz} = 0, E_{x} = \frac{i}{2 k} E_{zz} .

With transform

E (x, ν) = \int E (x, z) e^{- i ν z} d z

, its free propagation is

E (x, ν) = E (0, ν) e^{- i ν^{2} x / (2 k)}, E (x, z) = \frac{1}{2 π} \int E (0, ν) e^{i ν z - i ν^{2} x / (2 k)} d ν .

This is a one-way paraxial approximation, not a replacement for backward waves or large-angle Helmholtz components. Its Fourier multiplier has unit modulus and therefore does not attenuate any admitted component.

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