Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 2 / c / Solution

For candidate (i), $\mathbf{p}=p_0\hat{\mathbf{p}}\cos q_{0}x$. Since $G(q_0)=\hat{a}:=a-a_c$, $⟨{\cos }^2{q}_{0}x⟩=1/2$, and $⟨{\cos }^4{q}_{0}x⟩=3/8$, its mean-field free-energy density isFor $\hat{a}<0$, stationarity givesand substitution yieldsFor $\hat{a}\ge 0$, the minimum is $p_0=0$. The amplitude therefore vanishes continuously as $p_0\propto (a_c-a{)}^{1/2}$ on approaching $a_c$ from below, which is a continuous mean-field transition.