Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 14B / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

Fourier transformation in

x

changes Laplace's equation into

u_{yy} - k^{2} u = 0.

Imposing the two boundary values gives

u (k, y) = \frac{f ( k ) sinh ( ∣ k ∣ ( a - y )) + g ( k ) sinh ( ∣ k ∣ y )}{sinh ( ∣ k ∣ a )} .

The value at

k = 0

is understood by continuity.

ii

Solution

Part (b) gives

g (k) = - iπ sgn (k) (e^{- a ∣ k ∣} - e^{- 3 a ∣ k ∣}) .

Since

(e^{- a q} - e^{- 3 a q}) / sinh (a q) = 2 e^{- 2 a q}

,

u = - iπ sgn (k) (e^{- (2 a - y) ∣ k ∣} - e^{- (2 a + y) ∣ k ∣}) .

Inverting with part (b) yields

u (x, y) = \frac{x}{x ^{2} + ( 2 a - y ) ^{2}} - \frac{x}{x ^{2} + ( 2 a + y ) ^{2}} .

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