 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/14b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

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g

g (y) = \sum_{n = 1}^{\infty} g_{n} sin (k_{n} y), k_{n} = \frac{nπ}{a}, g_{n} = \frac{2}{a} \int_{0}^{a} g (y) sin (k_{n} y) d y .

By separation of variables, the decaying solution of Laplace equation in each half-strip has

n

th mode proportional to

e^{- k_{n} ∣ x ∣} sin (k_{n} y)

. Continuity at

x = 0

and integration of the equation across the Dirac delta function require the normal-derivative jump

ϕ_{x} (0^{+}, y) - ϕ_{x} (0^{-}, y) = g (y) .

The jump of

C_{n} e^{- k_{n} ∣ x ∣}

- 2 k_{n} C_{n}

, so

ϕ (x, y) = - n = 1 \sum \infty \frac{g _{n}}{2 k _{n}} e^{- k_{n} ∣ x ∣} sin (k_{n} y) .

For

g (y) = δ (y - c)

g_{n} = \frac{2}{a} sin \frac{nπ c}{a} .

Thus

ϕ (x, y) = - \frac{1}{π} \sum_{n = 1}^{\infty} \frac{1}{n} e^{- nπ ∣ x ∣/ a} sin \frac{nπ y}{a} sin \frac{nπ c}{a} .

x

(b, c)

G (x, y; b, c) = - \frac{1}{π} n = 1 \sum \infty \frac{1}{n} e^{- nπ ∣ x - b ∣/ a} sin \frac{nπ y}{a} sin \frac{nπ c}{a} .

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