Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 321 / 2 / c

Solution

Inside the ring,

ν Σ = A Σ^{3} = A σ^{3} (1 - x^{2} / w^{2})^{3/2}

. Part a therefore gives

u_{x} = \frac{9 A σ ^{2}}{w ^{2}} x .

The edge is material, so

\overset{w}{˙} = u_{x} (w)

and

\overset{w}{˙} = \frac{9 A σ ^{2}}{w} .

Direct substitution of the profile into the diffusion equation gives

\overset{σ}{˙} = - \frac{σ w ˙}{w},

hence

σ w = σ_{0} w_{0}

. This is exactly mass conservation, since

M_{ring} = \int_{- w}^{w} Σ d x = σ w \int_{- 1}^{1} 1 - y^{2} d y = \frac{π}{2} σ w .

Using

σ = σ_{0} w_{0} / w

in the width equation and integrating,

w^{4} = w_{0}^{4} + 36 A σ_{0}^{2} w_{0}^{2} t .

Therefore

w = w_{0} (1 + \frac{36 A σ _{0}^{2}}{w _{0}^{2}} t)^{1/4}, σ = σ_{0} (1 + \frac{36 A σ _{0}^{2}}{w _{0}^{2}} t)^{- 1/4} .

In particular,

w \propto t^{1/4}

at late times.

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