 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/40b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

ρ = ρ_{0} + ρ^{'}

and

p = p_{0} + p^{'}

, and let

u

ρ_{t}^{'} + ρ_{0} \nabla \cdot u = 0, ρ_{0} u_{t} = - \nabla p^{'}, p^{'} = c_{0}^{2} ρ^{'},

c_{0}^{2} = (\frac{\partial p}{\partial ρ})_{s} .

If the flow is irrotational, write

u = \nabla ϕ

. The momentum equation integrates spatially to

p^{'} = - ρ_{0} ϕ_{t},

a purely time-dependent integration term having been absorbed into

ϕ

. Substitution into continuity gives the wave equation

ϕ_{tt} - c_{0}^{2} \nabla^{2} ϕ = 0,

so the wave speed is

c_{0}

To obtain energy conservation, dot the momentum equation with

u

and use the product rule:

\frac{\partial}{\partial t} (\frac{1}{2} ρ_{0} ∣ u ∣^{2}) = - \nabla \cdot (p^{'} u) + p^{'} \nabla \cdot u .

The continuity and pressure-density relations give

p^{'} \nabla \cdot u = - \frac{p ^{'} p _{t}^{'}}{ρ _{0} c _{0}^{2}} = - \frac{\partial}{\partial t} (\frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}) .

\frac{\partial E}{\partial t} + \nabla \cdot I = 0,

E = \frac{1}{2} ρ_{0} ∣ u ∣^{2} + \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = p^{'} u .

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