In the single-mode approximation write
$$
\widetilde E=Lra_1^2+Lua_1^4,
\qquad
r=\frac{F_c-F}{4},
\qquad
u=\frac F{64}.
$$
For the reference energy $U_0=Lg_1a_1^2$,
$$
Z_0=\sqrt{\frac{\pi k_BT}{Lg_1}},
\quad
F_0=\frac{k_BT}{2}\log\frac{Lg_1}{\pi k_BT},
$$
$$
\langle a_1^2\rangle_0=\frac{k_BT}{2Lg_1},
\qquad
\langle a_1^4\rangle_0=\frac{3(k_BT)^2}{4L^2g_1^2}.
$$
Up to an irrelevant constant from subtracting $\langle U_0\rangle_0$, the variational free energy is
$$
\boxed{\mathcal F(g_1)=
\frac{k_BT}{2}\log\frac{Lg_1}{\pi k_BT}
+\frac{rk_BT}{2g_1}
+\frac{3u(k_BT)^2}{4Lg_1^2}}.
$$
Minimization gives
$$
g_1^2-rg_1-\frac{3u k_BT}{L}=0,
$$
so the positive optimum is
$$
\boxed{g_1^*=\frac18\left[
F_c-F+\sqrt{(F_c-F)^2+\frac{3Fk_BT}{L}}
\right]}.
$$

The compression estimate through quartic order is
$$
\boxed{C(F)\simeq C_2-\frac34C_2^2},
\qquad
C_2=\frac{k_BT}{8Lg_1^*}.
$$
Using $f=F/F_c$ and $L_p=A/(k_BT)$,
$$
\boxed{C_2=
\frac{L/(\pi^2L_p)}
{1-f+\sqrt{(1-f)^2+3fL/(\pi^2L_p)}}}.
$$
This is finite and smooth at $f=1$, unlike the zero-temperature branch point. The width of the <thermal rounding of a buckling transition> is controlled by the dimensionless flexibility $L/L_p$ and vanishes for an increasingly stiff or cold filament.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 355 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-355/3/d/solution