 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/14d/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

Spatial Fourier transformation gives

u_{tt} + c^{2} k^{2} u = 0, u (k, 0) = π e^{- k^{2} /4}, u_{t} (k, 0) = 0.

Hence

u (k, t) = π e^{- k^{2} /4} cos (c k t) .

Inverting and splitting the cosine into exponentials, or applying D'Alembert formula, yields

u (x, t) = \frac{1}{2} (e^{- (x - c t)^{2}} + e^{- (x + c t)^{2}}) .

Thus the initial Gaussian separates into two half-amplitude Gaussian pulses travelling without distortion at speeds $c$ and $- c$ .

 New to topics? Read the docs here!