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

Use the stable parameters $b,\kappa >0$. The bulk stationary points satisfy $f^{\prime }(\varphi )=a\varphi +b{\varphi }^3=0$. For $a<0$, $\varphi =0$ has negative curvature, while the two minima areBoth have $f(\pm {\varphi }_B)=-a^2/(4b)$ and $f^{\prime \prime }(\pm {\varphi }_B)=-2a>0$. To establish minimization at fixed total composition, writeFor mean composition $\bar{\varphi }$ with $|\bar{\varphi }|<{\varphi }_B$, a mixture of the two minima with volume fraction $\lambda =(\bar{\varphi }+{\varphi }_B)/(2{\varphi }_B)$ in the positive phase attains this lower bound in the bulk. Macroscopic slabs have only a subextensive interfacial excess free energy, so the bound is attained per unit volume in the thermodynamic limit. Thus constrained minimization gives phase coexistence in the mean-field approximation; at the endpoints only one phase is needed.

The two equilibrium conditions also follow directly by constrained optimization. Introduce a Lagrange multiplier $\mu$ for $\lambda {\varphi }_{+}+(1-\lambda ){\varphi }_{-}=\bar{\varphi }$ in the mixture's bulk free-energy density. Varying ${\varphi }_{+}$ and ${\varphi }_{-}$ gives $f^{\prime }({\varphi }_{+})=f^{\prime }({\varphi }_{-})=\mu$; varying $\lambda$ gives $f({\varphi }_{+})-\mu {\varphi }_{+}=f({\varphi }_{-})-\mu {\varphi }_{-}$. These equate the chemical potentials and pressures, respectively.

Equivalently, their bulk chemical potentials are both zero and their pressures are equal:These are precisely the conditions of the common-tangent construction for phase coexistence: the horizontal supporting tangent touches both minima. Equal chemical potentials prevent net composition exchange, and equal pressure gives mechanical equilibrium at a flat interface.

Apply force balance to one hemisphere. The pressure force is $\pi R^2({\Pi }_{\mathrm{in}}-{\Pi }_{\mathrm{out}})$, while the interfacial tension around its equator gives $2\pi R\gamma$. The Young–Laplace equation therefore yieldsTo leading order in curvature, let the two bulk compositions be ${\varphi }_{+}={\varphi }_B+{\delta }_{+}$ and ${\varphi }_{-}=-{\varphi }_B+{\delta }_{-}$. Their curvatures agree, $\alpha =f^{\prime \prime }(\pm {\varphi }_B)=-2a$, so equal chemical potentials require $\alpha {\delta }_{+}=\alpha {\delta }_{-}$ and hence a common shift $\delta$.

Since $d\Pi /d\varphi =\varphi f^{\prime \prime }(\varphi )$, expansion about the flat coexistence compositions gives ${\Pi }_{+}-{\Pi }_{-}=2{\varphi }_B\alpha \delta +O({\delta }^2)$. Matching the capillary pressure proves the Gibbs--Thomson relation in these variables:The pressure jump balances interfacial forces; it does not permit unequal chemical potentials. A common shift in both compositions supplies a common interfacial chemical potential while their pressures differ. The expansion assumes $R$ large compared with the phi-four diffuse interface width.

The far-field composition $-{\varphi }_B$ has chemical potential zero. Local curved-interface coexistence imposes $\mu (R)={\mu }_s=\gamma /({\varphi }_{B}R)$. The decaying spherically symmetric solution of the exterior Laplace equation isWith constant mobility, its outward current isThe linearized exterior composition is $\varphi (r)=-{\varphi }_B+\mu (r)/\alpha$, 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 $2{\varphi }_B$, so conservation gives $2{\varphi }_B\dot{R}=-J_r(R)$. HenceThe negative sign represents evaporation. The total outward current is $4\pi R^2{J}_r(R)=4\pi M\gamma /{\varphi }_B$, independent of $R$ in this leading quasi-static capillary approximation.

Integrate the diffusion-controlled droplet growth equation: $d(R^3)/dt=-3M\gamma /(2{\varphi }_B^2)$. ThusSetting the radius to zero givesThis is the lifetime predicted by the sharp-interface quasi-static model. The final stage at a radius comparable to the interface width falls outside that approximation, while the leading lifetime for a large initial droplet is dominated by the larger radii.

The flat interface is an equilibrium reservoir with $\mu =0$, while the droplet has the constant surface value ${\mu }_s=\gamma /({\varphi }_{B}R)$. The exterior chemical potential satisfies the Laplace equation, tends to zero far away, and obeys these two surface values. These are exactly the electrostatic boundary conditions at a conductor for a sphere held at potential ${\mu }_s$ above a grounded infinite plane.

Introduce an arbitrary permittivity $ϵ$. In the electrostatic problem, $Q=-ϵ{\int }_{S_R}{\partial }_n\mu dS=C(R,h){\mu }_s$. Therefore the diffusion-capacitance analogy gives the outward currentFor an isolated sphere, its capacitance is $C(R,\infty )=4\pi ϵR$, so $I(R,\infty )=4\pi MR{\mu }_s=4\pi M\gamma /{\varphi }_B$. Dividing givesThus sphere-plane capacitance determines the evaporation enhancement. The assumed spherical shape gives the corresponding law $\dot{R}=-M\gamma c(R/h)/(2{\varphi }_B^2{R}^2)$ for droplet evaporation near a planar reservoir. The local surface current is generally nonuniform, even though its total is described by a single capacitance.

For a fixed center height $h$, the radius decreases according to the preceding current enhancement. ThusWithin the stipulated logarithmic approximation, $c(y)=1+y/2+O(y^2)$, so $1/c(y)=1-y/2+O(y^2)$. Expanding uniformly over $0\le R\le R_0\ll h$ and integrating givesTherefore the leading correction to the isolated-droplet lifetime isThe negative correction reflects faster evaporation into the nearby reservoir. This leading coefficient also agrees with the small-$R/h$ expansion of the exact sphere-plane capacitance; the stipulated logarithmic approximation need not match higher orders.