 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/39a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

ρ_{t} + u ρ_{x} + ρ u_{x} = 0, u_{t} + u u_{x} + \frac{1}{ρ} p_{x} = 0.

c^{2} = \frac{d p}{d ρ} = \frac{γ p}{ρ}, \frac{c}{c _{0}} = (\frac{ρ}{ρ _{0}})^{(γ - 1) /2} .

The continuity equation therefore implies

c_{t} + u c_{x} + \frac{γ - 1}{2} c u_{x} = 0,

while the momentum equation becomes

u_{t} + u u_{x} + \frac{2 c}{γ - 1} c_{x} = 0.

Adding and subtracting these equations gives

[\partial_{t} + (u \pm c) \partial_{x}] [u \pm \frac{2 ( c - c _{0} )}{γ - 1}] = 0.

R_{\pm} = u \pm \frac{2 ( c - c _{0} )}{γ - 1}

are constant along the characteristic curves

C_{\pm} : \frac{d x}{d t} = u \pm c .

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