Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 / 1 / b

Table of contents
- Solution b

Solution

Reflection symmetry makes the two outer free tails identical. Their combined energy is

E_{o} = \frac{k _{c} δ _{o}^{2}}{ξ R ^{2}} .

Between the cylinders the free profile is even. Up to an irrelevant additive height it has the form

h_{i} (x) = C cosh (x / ξ), ∣ x ∣ \leq L - δ_{i} .

Slope matching at

x = L - δ_{i}

gives

C = \frac{ξ δ _{i}}{R s i n h [( L - δ _{i} ) / ξ ]} .

Direct integration, or the boundary expression from part (a), yields

E_{i} = \frac{k _{c} δ _{i}^{2}}{ξ R ^{2}} coth \frac{L - δ _{i}}{ξ} .

At the retained order

δ_{i} / ξ ≪ 1

, replace the argument by

L / ξ

. The four contact halves contribute bending plus adhesion energy

E_{c} = \frac{k _{c}}{R ^{2}} (1 - 2 U) (δ_{o} + δ_{i}) .

Thus

E_{2} = \frac{k _{c}}{ξ R ^{2}} [δ_{o}^{2} + δ_{i}^{2} coth (L / ξ) - 2 ξ (U - \frac{1}{2}) (δ_{o} + δ_{i})] .

Independent minimization gives

δ_{o} = ξ (U - \frac{1}{2}), δ_{i} = ξ (U - \frac{1}{2}) tanh (L / ξ),

and hence

E_{2} (L) = - \frac{k _{c} ξ}{R ^{2}} (U - \frac{1}{2})^{2} [1 + tanh (L / ξ)] .

At infinite separation the energy is twice the one-cylinder minimum. The membrane-mediated interaction potential is therefore

V (L) = E_{2} (L) - E_{2} (\infty) = \frac{k _{c} ξ}{R ^{2}} (U - \frac{1}{2})^{2} [1 - tanh (L / ξ)] .

It is positive and decreases monotonically to zero, so the two cylinders repel. The physical cause is the overlap of their exponentially relaxing membrane deformations.

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