 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/40a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

ρ_{0} ϕ_{t} / c_{0}^{2}

and use product rules. This gives

\frac{\partial}{\partial t} [\frac{ρ _{0}}{2} ∣\nabla ϕ ∣^{2} + \frac{ρ _{0}}{2 c _{0}^{2}} ϕ_{t}^{2}] + \nabla \cdot (- ρ_{0} ϕ_{t} \nabla ϕ) = 0.

Therefore

E = \frac{ρ _{0}}{2} ∣\nabla ϕ ∣^{2} + \frac{ρ _{0}}{2 c _{0}^{2}} ϕ_{t}^{2} = \frac{ρ _{0}}{2} ∣ u ∣^{2} + \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = - ρ_{0} ϕ_{t} \nabla ϕ = p^{'} u .

These are respectively the kinetic-plus-compressional acoustic energy density and its flux.

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