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

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