Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 1

Because $\varphi$ is a conserved composition density, an infinitesimal material displacement $\mathbf{r}\mapsto \mathbf{r}+\mathbf{u}$ changes it at fixed position byThe second term is essential when the deformation is compressible. By the definition of the chemical potential, and taking $\mathbf{u}$ to vanish on the boundary,The same free-energy change written in terms of the stress tensor isSince $\mathbf{u}$ is arbitrary,This is the Korteweg force density of a diffuse-interface mixture.

Functional differentiation givesFor $\Pi =\mu \varphi -\mathbb{F}$ and ${\Sigma }_{ij}=-\Pi {\delta }_{ij}-\kappa ({\nabla }_i\varphi )({\nabla }_j\varphi )$,The bracket vanishes by the expression for $\mu$, while the two mixed second-derivative terms cancel. Therefore

Equilibrium minimizes $F$ at fixed total composition $\int \varphi$, so a Lagrange multiplier gives $\delta F/\delta \varphi ={\mu }_0$, independent of position. In the two homogeneous phases,The symmetric coexistence pair therefore has ${\mu }_0=0$. For a planar profile depending only on the normal coordinate $x$,orup to reversal of the two phases.

Put $\varphi ={\varphi }_{B}g(u)$, $u=(x-x_0)/{\xi }_0$, and ${\xi }_0^2=-2\kappa /a$. Since $b{\varphi }_B^2=-a$, the interface equation reduces toMultiplication by $g^{\prime }$ and use of $g(\pm \infty )=\pm 1$, $g^{\prime }(\pm \infty )=0$ gives the first integralFor the increasing profile, $g^{\prime }=1-g^2$, so $\mathrm{artanh}g=u$ after shifting $x_0$. Hence the phi-four diffuse interface isThe sign chooses the orientation and the translation zero mode $x_0$ sets the interface position.

Far from a planar interface, the equilibrium profile equals a bulk minimum and contributes the uniform density $f({\varphi }_B)$. The localized transition layer contributes an additional free energy proportional to its area $A$, because translation invariance makes the excess per unit area independent of position. By the definition of surface tension as interfacial excess free energy per area,

For a profile depending only on $x$, equation (2) gives ${\Sigma }_{yy}=-\Pi$ and ${\Sigma }_{xx}=-\Pi -\kappa ({\varphi }^{\prime }{)}^2$. ThusFor ${\varphi }_E={\varphi }_B\tanh [(x-x_0)/{\xi }_0]$,Changing variable to $u=(x-x_0)/{\xi }_0$ yieldsUsing ${\varphi }_B^2=-a/b$ and ${\xi }_0^2=-2\kappa /a$ gives the positive interfacial tension of a phi-four diffuse interface