 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/12b/iii/solution

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

Let

U (s, x) = L_{t} {u (t, x)}

. The initial data turn the wave equation into

- \frac{\partial ^{2} U}{\partial x ^{2}} + s^{2} U = f (x) .

Using the Green function from part (ii) with

m = s

gives

U (s, x) = \frac{1}{2 s} \int_{- \infty}^{\infty} e^{- s ∣ x - y ∣} f (y) d y .

Since

L^{- 1} {\frac{e ^{- a s}}{s}} = H (t - a),

u (t, x) = \frac{1}{2} \int_{- \infty}^{\infty} H (t - ∣ x - y ∣) f (y) d y = \frac{1}{2} \int_{x - t}^{x + t} f (y) d y .

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