Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 67 / 3 / b / Solution

Fix counterclockwise arclength on each side and take the outward normal derivative. Write

h = \frac{l}{2 3}, z^{(j)} (s) = c_{j} (h + i s), (c_{1}, c_{2}, c_{3}) = (1, \overset{a}{ˉ}, a), a = e^{2 πi /3} .

The unit tangent is

i c_{j}

and the outward unit normal is

c_{j}

. On side

j

this gives the chain-rule identity

i c_{j} q_{z} = \frac{1}{2} q_{s}^{(j)} + \frac{i}{2} q_{N}^{(j)} .

Put

w = c_{j} k

. The exponential in the closed differential one-form pulls back to

W (z^{(j)} (s), k) = E (- i w) e^{(w + λ / w) s}, E (k) = exp [\frac{l}{2 3} (k + \frac{λ}{k})] .

Also

- λ q d \overset{z}{ˉ} / (ik) = (λ / w) q^{(j)} d s

. Consequently the contribution of side

j

is

ρ_{j} (c_{j} k)

, where the required local side spectra are

Ψ_{j} (k) Φ_{j} (k) ρ_{j} (k) = \int_{- l /2}^{l /2} e^{(k + λ / k) s} q_{N}^{(j)} (s) d s, = \int_{- l /2}^{l /2} e^{(k + λ / k) s} [\frac{1}{2} q_{s}^{(j)} (s) + \frac{λ}{k} q^{(j)} (s)] d s, = E (- ik) [\frac{i}{2} Ψ_{j} (k) + Φ_{j} (k)] .

These are spectral integral transforms in the local coordinate of each side. Keeping the tangential-derivative term is important: premature integration by parts in a single side integral would leave corner values, which cannot simply be discarded. At the special samples used below, those endpoint terms do vanish for the rotationally invariant solution.

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