For a profile depending only on $x$, equation (2) gives $\Sigma_{yy}=-\Pi$ and $\Sigma_{xx}=-\Pi-\kappa(\phi')^2$. Thus
$$
\Sigma_{yy}-\Sigma_{xx}=\kappa(\phi')^2.
$$
For $\phi_E=\phi_B\tanh[(x-x_0)/\xi_0]$,
$$
\phi_E'=\frac{\phi_B}{\xi_0}\operatorname{sech}^2\left(\frac{x-x_0}{\xi_0}\right).
$$
Changing variable to $u=(x-x_0)/\xi_0$ yields
$$
\sigma=\frac{\kappa\phi_B^2}{\xi_0}
\int_{-\infty}^{\infty}\operatorname{sech}^4u\,du
=\frac{4\kappa\phi_B^2}{3\xi_0}.
$$
Using $\phi_B^2=-a/b$ and $\xi_0^2=-2\kappa/a$ gives the positive <interfacial tension of a phi-four diffuse interface>
$$
\boxed{\sigma=\sqrt{\frac{-8a^3\kappa}{9b^2}}.}
$$

Solution

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

{title2=$\sigma$}

The excess free energy per area of the phi-four diffuse interface is
$$
\sigma=\int_{-\infty}^{\infty}\kappa(\phi')^2dx
=\sqrt{\frac{-8a^3\kappa}{9b^2}}.
$$


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