At fixed uniform $\phi$, the nonconserved variable $p$ can relax freely, so <mean-field approximation> minimizes
$$
f(\phi,p)=\frac a2\phi^2+\frac b4\phi^4+\frac{A+c\phi}{2}p^2+\frac B4p^4
$$
with respect to $p$. The stationary solutions obey
$$
p\left(A+c\phi+Bp^2\right)=0.
$$
Thus $p=0$ when $A+c\phi\geq0$, while $p^2=-(A+c\phi)/B$ when $A+c\phi<0$. Substitution gives
$$
f(\phi)=\frac a2\phi^2+\frac b4\phi^4-\frac{(A+c\phi)^2}{4B}\,	heta(-A-c\phi).
$$
Writing this in the requested form yields
$$
\boxed{\phi_o=-\frac Ac},
\qquad
\boxed{C=\frac{c^2}{2B}},
$$
because $A+c\phi=c(\phi-\phi_o)$. The value $\phi_o$ is the composition at which the coefficient of $p^2$ changes sign, so it is the mean-field threshold for polar ordering.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-344/1/c/solution