Equilibrium minimizes $F$ at fixed total composition $\int\phi$, so a <Lagrange multiplier> gives $\delta F/\delta\phi=\mu_0$, independent of position. In the two homogeneous phases,
$$
f'(\pm\phi_B)=a(\pm\phi_B)+b(\pm\phi_B)^3=0,
\qquad \phi_B=\sqrt{-a/b}.
$$
The symmetric coexistence pair therefore has $\mu_0=0$. For a planar profile depending only on the normal coordinate $x$,
$$
0=\mu=a\phi+b\phi^3-\kappa\phi'',
$$
or
$$
\boxed{\kappa\phi''=a\phi+b\phi^3,
\qquad \phi(\pm\infty)=\pm\phi_B}
$$
up to reversal of the two phases.

Solution

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

 Solution

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