For an inextensible planar filament,

r_{s} = (cos θ, sin θ)

. The bending energy and compressive-force potential are

E [θ] = \frac{A}{2} \int_{0}^{L} θ_{s}^{2} d s + F \int_{0}^{L} cos θ d s .

(1)

The straight state has

E = F L

. Free pivoting means zero end moment,

θ_{s} (0) = θ_{s} (L) = 0 .

(2)

The endpoints sharing the

x

axis impose

\int_{0}^{L} sin θ d s = 0

; near the straight state this removes the constant angular mode. The Neumann eigenfunctions therefore give

θ (s) = n = 1 \sum \infty a_{n} cos \frac{nπ s}{L} .

(3)

b

 0  0

Solution

 0  0

For

θ = a_{1} cos (π s / L)

, expansion through fourth order gives

E := E - F L = \frac{L}{4} (F_{c} - F) a_{1}^{2} + \frac{F L}{64} a_{1}^{4},

(1)

where

F_{c} = \frac{A π ^{2}}{L ^{2}} .

(2)

Stationarity gives

a_{1} = 0 or a_{1}^{2} = \frac{8 ( F - F _{c} )}{F} = \frac{8 ϵ}{1 + ϵ}, ϵ = \frac{F - F _{c}}{F _{c}} .

(3)

The two signs of

a_{1}

are the symmetry-related buckled states of a supercritical pitchfork bifurcation.

The projected length is

L_{∥} = \int_{0}^{L} cos θ d s

. Hence

C = 1 - \frac{L _{∥}}{L} = \frac{a _{1}^{2}}{4} - \frac{a _{1}^{4}}{64} + O (a_{1}^{6}) .

(4)

On the buckled branch,

C (F) = \frac{2 ϵ + ϵ ^{2}}{( 1 + ϵ ) ^{2}} = 2 ϵ + O (ϵ^{2}) .

(5)

c

 0  0

Solution

 0  0

The quadratic energy is

E_{2} = \sum_{n = 1}^{\infty} \frac{A π ^{2} n ^{2} - F L ^{2}}{4 L} a_{n}^{2} .

(1)

Each mode is a centered Gaussian, so

⟨ a_{n} ⟩ = 0, ⟨ a_{n}^{2} ⟩ = \frac{2 k _{B} T L}{A π ^{2} n ^{2} - F L ^{2}} .

(2)

The

n = 1

variance diverges as

F ↑ F_{c}

because the first bending mode becomes soft; Gaussian theory then fails and the quartic term controls the fluctuations.

At zero force,

C (0) ≃ \frac{1}{4} \sum_{n = 1}^{\infty} ⟨ a_{n}^{2} ⟩ = \frac{k _{B} T L}{2 A π ^{2}} \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{k _{B} T L}{12 A} .

(3)

With the common persistence length convention

L_{p} = A / (k_{B} T)

C (0) = \frac{L}{12 L _{p}} .

(4)

Thermal bending therefore shortens the mean projected length even without compression. Under the strictly planar tangent-correlation convention

L_{p}^{(2 D)} = 2 A / (k_{B} T)

, the same result is

L / (6 L_{p}^{(2 D)})

d

 0  0

Solution

 0  0

In the single-mode approximation write

E = L r a_{1}^{2} + Lu a_{1}^{4}, r = \frac{F _{c} - F}{4}, u = \frac{F}{64} .

(1)

For the reference energy

U_{0} = L g_{1} a_{1}^{2}

Z_{0} = \frac{π k _{B} T}{L g _{1}}, F_{0} = \frac{k _{B} T}{2} lo g \frac{L g _{1}}{π k _{B} T},

(2)

⟨ a_{1}^{2} ⟩_{0} = \frac{k _{B} T}{2 L g _{1}}, ⟨ a_{1}^{4} ⟩_{0} = \frac{3 ( k _{B} T ) ^{2}}{4 L ^{2} g _{1}^{2}} .

(3)

Up to an irrelevant constant from subtracting

⟨ U_{0} ⟩_{0}

, the variational free energy is

F (g_{1}) = \frac{k _{B} T}{2} lo g \frac{L g _{1}}{π k _{B} T} + \frac{r k _{B} T}{2 g _{1}} + \frac{3 u ( k _{B} T ) ^{2}}{4 L g _{1}^{2}} .

(4)

Minimization gives

g_{1}^{2} - r g_{1} - \frac{3 u k _{B} T}{L} = 0,

(5)

so the positive optimum is

g_{1}^{*} = \frac{1}{8} [F_{c} - F + (F_{c} - F)^{2} + \frac{3 F k _{B} T}{L}] .

(6)

The compression estimate through quartic order is

C (F) ≃ C_{2} - \frac{3}{4} C_{2}^{2}, C_{2} = \frac{k _{B} T}{8 L g _{1}^{*}} .

(7)

Using

f = F / F_{c}

and

L_{p} = A / (k_{B} T)

C_{2} = \frac{L / ( π ^{2} L _{p} )}{1 - f + ( 1 - f ) ^{2} + 3 f L / ( π ^{2} L _{p} )} .

(8)

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.

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

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