Write the free energy as
$$
f(\phi)=a\phi^3+bB_2\phi^2,
\qquad
a=\frac{c(k_BT)^2}{k_c\delta^3}>0,
\qquad
b=\frac{k_BT}{\delta^3}>0.
$$
When $A_H<A_H^*$, $B_2>0$ and the global minimum on $\phi\geq0$ is $\phi_*=0$, corresponding to an unbound stack. When $A_H>A_H^*$, $B_2<0$, and minimization gives
$$
f'(\phi)=\phi(3a\phi+2bB_2)=0,
\qquad
\boxed{\phi_*=-\frac{2bB_2}{3a}}.
$$
Using $B_2\sim-r(A_H-A_H^*)$ yields
$$
\phi_*\sim A_H-A_H^*.
$$
Since $d_*=\delta/\phi_*$,
$$
\boxed{
d_*\sim(A_H-A_H^*)^{-1}},
$$
so the continuous <membrane unbinding transition> has exponent
$$
\boxed{\nu=1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 355 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-355/3/d/solution