Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 14B (Methods)

Table of contents
- a 14B
  - Solution a
- b 14B
  - Solution b
- c 14B
  - i c
    - Solution i
  - ii c
    - Solution ii

a

Solution

Integration by parts, using decay of

m

, gives the Fourier transform of a derivative

h (k) = \int m^{'} (x) e^{- ik x} d x = ik m (k) .

b

Solution

The standard contour transform

\int_{- \infty}^{\infty} \frac{x}{x ^{2} + α ^{2}} e^{- ik x} d x = - iπ sgn (k) e^{- α ∣ k ∣}

holds for

Re α > 0

, with the appropriate upper or lower pole chosen according to the sign of

k

. Therefore

m (x) = \frac{x}{x ^{2} + α ^{2}} .

c

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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