The curved interface requires exterior composition $-\phi_B+\delta$, but the far field supplies only $-\phi_B$. Material therefore diffuses away from the droplet, which evaporates. Write $\phi=-\phi_B+g$ outside. Linearization gives $\mu=\alpha g$, and the quasistatic condition $\dot\phi=M\nabla^2\mu\simeq0$ gives
$$
\nabla^2g=0,
\qquad
g(R)=\delta,
\qquad
g(\infty)=0.
$$
Spherical symmetry yields $g(r)=\delta R/r$. Taking the normal outward from the droplet,
$$
J_n=-M\partial_r\mu=-M\alpha\partial_rg,
\qquad
J_n(R)=\frac{M\alpha\delta}{R}>0.
$$
The interface converts positive-phase material into negative-phase material. Integrating the continuity equation through the moving interface gives the Stefan condition
$$
v_n[\phi]=-J_n,
$$
where $[\phi]=2\phi_B$, and hence
$$
\boxed{v_n=-\frac{J_n}{2\phi_B}<0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-344/2/c/solution