 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-321/3/b/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

The pressureless dust fluid in a shearing sheet has no pressure restoration and is driven by drag force toward the prescribed gas velocity; gas backreaction is omitted in the given model. After linearizing about the common background shear, axisymmetry gives

(\partial_{t} + ϵ Ω) v_{x}^{'} - 2Ω v_{y}^{'} = ϵ Ω u_{x}^{'}, (\partial_{t} + ϵ Ω) v_{y}^{'} + \frac{Ω}{2} v_{x}^{'} = ϵ Ω u_{y}^{'} .

Apply

D_{ϵ} = \partial_{t} + ϵ Ω

to the first equation and use the second to eliminate

v_{y}^{'}

. The exact linear scalar equation is

[\partial_{t}^{2} + 2 ϵ Ω \partial_{t} + (1 + ϵ^{2}) Ω^{2}] v_{x}^{'} = ϵ Ω \partial_{t} u_{x}^{'} + ϵ^{2} Ω^{2} u_{x}^{'} + 2 ϵ Ω^{2} u_{y}^{'} .

For the supplied leading-order epicycle, set

θ = k x - Ω t

. Since

\partial_{t} u_{x}^{'} = Ω u sin θ

and

u_{y}^{'} = u sin θ /2

, this becomes

\partial_{t}^{2} v_{x}^{'} + 2 ϵ Ω \partial_{t} v_{x}^{'} + (1 + ϵ^{2}) Ω^{2} v_{x}^{'} = 2 ϵ Ω^{2} u sin θ + ϵ^{2} Ω^{2} u cos θ .

The final cosine term is missing from the printed equation. Dropping it is justified at leading order in

0 < ϵ ≪ 1

, recovering the intended forcing, but the printed equation retains an order-

ϵ^{2}

frequency term on the left and is not an exact derivation at that order. A direct countercheck is

v^{'} = u^{'}

: the drag vanishes and the pair solves both dust equations exactly, while

v_{x}^{'} = u cos θ

leaves residual

ϵ^{2} Ω^{2} u cos θ

(ϵ Ω)^{- 1}

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