Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 1 / b

Solution

Since

D f / D t = 0

\frac{1}{ρ} \frac{D ρ}{D t} = \frac{ρ ˙ _{0}}{ρ _{0}} .

\nabla \cdot u = A + B + C = \frac{d}{d t} lo g (ab c) .

ρ_{0} \propto (ab c)^{- 1} .

Similarly,

\frac{1}{p} \frac{D p}{D t} = \frac{ρ ˙ _{0}}{ρ _{0}} + \frac{T ˙}{T} = - γ \nabla \cdot u,

and hence

T \propto (ab c)^{- (γ - 1)} .

The

x

component of the material acceleration is

\frac{D u _{x}}{D t} = (\dot{A} + A^{2}) x = \frac{a ¨}{a} x .

Because

d p / df = ρ

- \frac{1}{ρ} \partial_{x} p = - \frac{1}{ρ _{0} ρ} ρ_{0} T ρ (- \frac{2 x}{a ^{2}}) = \frac{2 T x}{a ^{2}} .

Combining this with

- \partial_{x} Φ = - Ω^{2} x

gives

\overset{a}{¨} + Ω^{2} a = \frac{2 T}{a} .

The

y

and

z

components give the analogous equations for

b

and

c

. Finally,

p (0) = 0

makes

p = 0

on the material free surface, so both dynamic and kinematic boundary condition requirements are satisfied.

Solved by gpt-5.6-sol high.

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