Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 1

Impose the fixed conserved order parameterwith a Lagrange multiplier ${\mu }_0$. Stationarity ofunder every local variation givesThus the chemical potential is spatially constant at constrained equilibrium.

For constant $\kappa$ and a planar profile,Far inside either bulk phase, derivatives vanish and $\varphi =\pm {\varphi }_B$, where ${\varphi }_B^2=-a/b$. ThereforeSince part a makes $\mu$ constant, it vanishes throughout the interface.

The Euler–Lagrange equation is $\kappa {\varphi }^{\prime \prime }=a\varphi +b{\varphi }^3$. Multiplication by ${\varphi }^{\prime }$ and integration gives the first integralWith $\varphi ={\varphi }_{B}g$, $u=x/{\xi }_0$, and ${\xi }_0^2=-2\kappa /a$, this becomesThe monotone heteroclinic solutions areTranslation invariance leaves $x_0$ undetermined in an infinite system. A fixed global composition, boundary condition, or pinning field fixes this midpoint position.

Forthe functional derivative isThe constrained minimum again makes $\mu$ constant. Evaluating it in either uniform bulk phase gives $\mu =f^{\prime }(\pm {\varphi }_B)=0$, so $\mu =0$ everywhere.

For a planar equilibrium, multiply $\mu =0$ by ${\varphi }^{\prime }$ and use the chain rule. The result isMatching to either bulk phase fixes the constant:Consequentlyand the inverse profile is

The interfacial tension is the excess free energy per unit interfacial area:The first integral from part e equates the first two terms inside the brackets to $\kappa (\varphi )({\varphi }^{\prime }{)}^2/2$. Hence

For the quartic free energy,With constant $\kappa$, part e gives, between the two bulk values,Integration giveswhose inversion is exactly the hyperbolic-tangent profile in part c.