 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/14a/d/solution

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

x

\partial_{t}^{2} u (k, t) + k^{2} u (k, t) = 0,

with initial data

u (k, 0) = f (k)

and

\partial_{t} u (k, 0) = g (k)

. Therefore

u (k, t) = f (k) cos (k t) + g (k) \frac{s i n ( k t )}{k} .

\frac{1}{2} {f (x + t) + f (x - t)} .

For the second term choose an antiderivative

p

g

. By part (b),

g = ik p

, and hence

g (k) \frac{s i n ( k t )}{k} = \frac{1}{2} p (k) {e^{ik t} - e^{- ik t}} .

Its inverse transform is

\frac{1}{2} {p (x + t) - p (x - t)} = \frac{1}{2} \int_{x - t}^{x + t} g (ξ) d ξ .

Combining the terms gives the D'Alembert formula

u (x, t) = \frac{1}{2} {f (x + t) + f (x - t)} + \frac{1}{2} \int_{x - t}^{x + t} g (ξ) d ξ .

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