 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-52/2/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 52 / 2 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

Let

e = ρ u^{2} /2 + p / (γ - 1)

be the total gas energy density, with

γ > 1

. The one-dimensional continuity and momentum equations are

\partial_{t} ρ + \partial_{z} (ρ u) = 0, ρ (\partial_{t} u + u \partial_{z} u) = - \partial_{z} p .

Using continuity to put the momentum equation into conservative form gives

\partial_{t} (ρ u) + \partial_{z} (ρ u^{2} + p) = 0.

The internal energy density

e_{int} = p / (γ - 1)

satisfies

\partial_{t} e_{int} + \partial_{z} (u e_{int}) = - p \partial_{z} u,

by the adiabatic pressure equation. Multiplying the momentum equation by

u

and using continuity gives

\partial_{t} (\frac{ρ u ^{2}}{2}) + \partial_{z} (\frac{ρ u ^{3}}{2}) = - u \partial_{z} p .

Adding these equations combines the pressure work into

- \partial_{z} (p u)

. The conservation law variables and conservation law fluxes are therefore

U = ρ ρ u ρ u^{2} /2 + p / (γ - 1), F = ρ u ρ u^{2} + p u (ρ u^{2} /2 + γ p / (γ - 1)) .

The three rows express conservation of mass, momentum and total energy. Their integral conservation laws remain meaningful across a shock wave, where the differential pressure and velocity equations cannot be applied pointwise.

 Read the full article

 New to topics? Read the docs here!