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

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- Solution b

Solution

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},

where

F_{c} = \frac{A π ^{2}}{L ^{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}} .

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}) .

On the buckled branch,

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

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