= Single-mode smectic mean-field free energy
{title2=$f(A)=\frac14(a-a_c)A^2+\frac18Bq_0^2A^4$}
For $\phi=A\cos(q_0z)$, the <gradient-quartic Brazovskii model> has volume-averaged <free-energy density>
$$
f(A)=\frac{a-a_c}{4}A^2+\frac{Bq_0^2}{8}A^4,
\quad q_0^2=-\frac\kappa{2\gamma},
\quad a_c=\frac{\kappa^2}{4\gamma}.
$$
The averages are $\langle\cos^2\rangle=\langle\sin^2\rangle=1/2$ and $\langle\cos^2\sin^2\rangle=1/8$. For $B>0$, the stable amplitude is zero for $a\geq a_c$ and satisfies $A^2=(a_c-a)/(Bq_0^2)$ below it. The <mean-field approximation> therefore predicts a <continuous phase transition>.
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