Solution

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem is

u_{tt} + k^{2} u = 0, u (k, 0) = f (k), u_{t} (k, 0) = g (k),

and hence

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

Part (c) and the convolution theorem turn the first term into

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

For the second term, choose

p

with

p^{'} = g

. Part (b) says

g = ik p

, so

g (k) \frac{s i n ( k t )}{k} = i p (k) sin (k t) .

Part (c) now makes its inverse transform

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

Combining the two 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 .

Solved by gpt-5.6-sol high.

Let

A = 2 V

and

B = W - V

. Part (c) makes

A

and

B

independent identically distributed exponential variables of rate

λ

. The change of variables

(A, B) = (rs, (1 - r) s)

has Jacobian

s

, and integrating

λ^{2} s e^{- λ s}

over

s > 0

gives density one for

0 < r < 1

\frac{2 V}{W + V} = \frac{A}{A + B} \sim Uniform (0, 1) .

Solved by gpt-5.6-sol high.

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