 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/14a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Extend the boundary data oddly to the whole real axis. The extended function is precisely the function

f

from part (a). Taking the Fourier transform in

x

\partial_{y}^{2} u - k^{2} u = 0.

Decay as

y \to \infty

selects

u (k, y) = f (k) e^{- ∣ k ∣ y} .

Part (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernel

P_{y} (s) = \frac{y}{π ( s ^{2} + y ^{2} )} .

For

x, y > 0

this gives

u (x, y) = \frac{y}{π} \int_{0}^{1} [\frac{1}{( x - v ) ^{2} + y ^{2}} - \frac{1}{( x + v ) ^{2} + y ^{2}}] d v = \frac{4 x y}{π} \int_{0}^{1} \frac{v d v}{[( x - v ) ^{2} + y ^{2} ] [( x + v ) ^{2} + y ^{2} ]} .

The odd extension enforces

u (0, y) = 0

, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

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