 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/20/c/i/solution

For the two travelling profiles, the initial conditions imply

f (x) + g (x) = cos^{2} (x /2), - v f^{'} (x) + v g^{'} (x) = 0.

Since

v \neq = 0

g - f

is constant. One may therefore choose

f (x) = \frac{1}{2} cos^{2} (x /2) + C, g (x) = \frac{1}{2} cos^{2} (x /2) - C,

where the constants cancel in the sum. The D'Alembert formula then gives

ψ (x, t) = \frac{1}{2} cos^{2} (\frac{x - v t}{2}) + \frac{1}{2} cos^{2} (\frac{x + v t}{2}) = \frac{1}{2} + \frac{1}{2} cos x cos (v t) .

This agrees exactly with the restricted-domain solution from part (b). The whole-line cosine profile already has zero spatial derivative at both

x = 0

and

x = π

for every time, so its restriction satisfies the Neumann boundary conditions without further modification.

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