Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 1 / i

Table of contents
- Solution i

Solution

The steady radial equation first gives the base-pressure gradient

\frac{d p _{0}}{d r} = r Ω^{2} .

For the stated normal mode, put

D (r) = σ + im Ω (r) .

Retaining terms linear in the disturbance in the Euler equations for an inviscid fluid gives

D u^{'} - 2Ω v^{'} + \frac{d p ^{'}}{d r} = 0,

D v^{'} + (2Ω + r Ω^{'}) u^{'} + \frac{im}{r} p^{'} = 0,

D w^{'} + ik p^{'} = 0,

together with incompressible flow

\frac{1}{r} \frac{d}{d r} (r u^{'}) + \frac{im}{r} v^{'} + ik w^{'} = 0 .

The

- 2Ω v^{'}

and

(2Ω + r Ω^{'}) u^{'}

terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.

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