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 Incoming links: Outgoing-energy identity for a radial defocusing wave

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 6 / Solution Created 2026-09-28 Updated 2026-10-03

Write

q = w_{t} + w_{r}

. The equation in part 5 says

q_{t} - q_{r} = - \frac{∣ w ∣ ^{p - 1} w}{r ^{p - 1}} .

Differentiate

J_{ψ}

, replace

q_{t}

by

q_{r} - ∣ w ∣^{p - 1} w / r^{p - 1}

, and integrate the

q q_{r}

and

∣ w ∣^{p - 1} w w_{r}

terms by parts. This gives the outgoing-energy identity for a radial defocusing wave

\frac{d J _{ψ}}{d t} = - \frac{1}{2} \int_{0}^{\infty} ψ^{'} q^{2} d r + \frac{1}{p + 1} \int_{0}^{\infty} \frac{∣ w ∣ ^{p + 1}}{r ^{p - 1}} (ψ^{'} - (p - 1) \frac{ψ}{r}) d r .

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