The spherically symmetric <Laplace equation> with $\widetilde\phi(R)=\delta(R)$ and $\widetilde\phi(\infty)=\varepsilon$ has solution
$$
\widetilde\phi(r)=\varepsilon+[\delta(R)-\varepsilon]\frac Rr.
$$
Thus
$$
J_r(R^+)=-M\partial_r\mu|_{R^+}
=-\frac{\alpha M}{R}[\varepsilon-\delta(R)].
$$
Conservation at the moving interface, whose composition jump is $2\phi_B$, gives $2\phi_B\dot R=-J_r(R^+)$ and hence
$$
\boxed{\dot R=\frac{\alpha M}{2\phi_BR}[\varepsilon-\delta(R)].}
$$

Since $\delta(R)=C/R$, the right side is proportional to $\varepsilon/R-C/R^2$. It is negative for $R<R^*$, zero at $R^*=C/\varepsilon$, 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-344/3/c/solution