Far from a planar interface, the equilibrium profile equals a bulk minimum and contributes the uniform density $f(\phi_B)$. 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,
$$
\boxed{\sigma A=F[\phi_E]-Vf(\phi_B).}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-344/1/e/solution