Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 3 / c

Solution

The spherically symmetric Laplace equation with

ϕ (R) = δ (R)

and

ϕ (\infty) = ε

has solution

ϕ (r) = ε + [δ (R) - ε] \frac{R}{r} .

Thus

J_{r} (R^{+}) = - M \partial_{r} μ ∣_{R^{+}} = - \frac{α M}{R} [ε - δ (R)] .

Conservation at the moving interface, whose composition jump is

2 ϕ_{B}

, gives

2 ϕ_{B} \dot{R} = - J_{r} (R^{+})

and hence

\dot{R} = \frac{α M}{2 ϕ _{B} R} [ε - δ (R)] .

Since

δ (R) = C / R

, the right side is proportional to

ε / R - C / R^{2}

. It is negative for

R < R^{*}

, zero at

R^{*} = C / ε

, positive for

R > R^{*}

, and approaches zero from above for large

R

. Thus

R^{*}

is the unstable critical nucleus: smaller droplets dissolve, while larger droplets grow.

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