Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 3 / a

Solution

For an axisymmetric surface

r = a (z, t)

, twice the mean curvature with the outward-normal convention is

κ = \frac{1}{a 1 + a _{z}^{2}} - \frac{a _{zz}}{( 1 + a _{z}^{2} ) ^{3/2}} .

a = a_{0} + η

and retaining linear terms gives

κ = \frac{1}{a _{0}} - \frac{η}{a _{0}^{2}} - η_{zz}, κ^{'} = - \frac{η}{a _{0}^{2}} - η_{zz} .

At the unperturbed boundary

r = a_{0}

u = η_{t}, σ_{rz} = μ (u_{z} + w_{r}) = 0, p^{'} - 2 μ u_{r} = γ κ^{'} .

e^{ik z + s t}

these become

u (a_{0}) = sη

σ_{rz} (a_{0}) = 0

, and

p^{'} (a_{0}) - 2 μ u_{r} (a_{0}) = γ (k^{2} - a_{0}^{- 2}) η .

2 μ u = \nabla (x \cdot Φ + χ) - 2 Φ, p = \nabla \cdot Φ,

where

Φ

and

χ

Let

x = k r

and

E = e^{ik z + s t}

. Substitution of the given radial vector potential and scalar potential gives

u = \frac{E}{2 μ} [P (x I_{1}^{'} (x) - I_{1} (x)) + Q I_{0}^{'} (x)],

w = \frac{i E}{2 μ} [P x I_{1} (x) + Q I_{0} (x)], p = P k I_{0} (x) E .

(x I_{1})^{'} = x I_{0}

, the tangential stress simplifies to

σ_{rz} = ik E [P x I_{1}^{'} (x) + Q I_{0}^{'} (x)] .

The stress-free condition at

r = a

therefore yields

P x I_{1}^{'} (x) + Q I_{0}^{'} (x) = 0 at x = ka .

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