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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-344/3/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 / 3 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

The far-field composition

- ϕ_{B}

has chemical potential zero. Local curved-interface coexistence imposes

μ (R) = μ_{s} = γ / (ϕ_{B} R)

. The decaying spherically symmetric solution of the exterior Laplace equation is

μ (r) = μ_{s} \frac{R}{r} = \frac{γ}{ϕ _{B} r}, r \geq R .

(1)

With constant mobility, its outward current is

J_{r} (r) = - M \frac{d μ}{d r} = \frac{M γ}{ϕ _{B} r ^{2}}, J_{r} (R) = \frac{M γ}{ϕ _{B} R ^{2}} .

(2)

The linearized exterior composition is

ϕ (r) = - ϕ_{B} + μ (r) / α

, consistent with the surface shift in the previous part.

For diffusion-controlled droplet growth, the interior is uniform to leading order and has no diffusive current. Replacing exterior material by interior material requires the composition jump

2 ϕ_{B}

, so conservation gives

2 ϕ_{B} \dot{R} = - J_{r} (R)

. Hence

\dot{R} = - \frac{M γ}{2 ϕ _{B}^{2} R ^{2}} .

(3)

The negative sign represents evaporation. The total outward current is

4 π R^{2} J_{r} (R) = 4 π M γ / ϕ_{B}

, independent of

R

in this leading quasi-static capillary approximation.

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