 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-2/13e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

u = 1 + u (t) cos (s / L)

and

v = 1 + v (t) cos (s / L)

and retain linear terms. With

k = 1/ L

- k^{2} cos (k s)

. Hence

\frac{d}{d t} (u v) = (1 - k^{2} 2 Q - 1 - Q (1 + P k^{2})) (u v) .

The trace and determinant of this linearization are

T (k) = 1 - Q - (1 + QP) k^{2}, D (k) = Q [1 + (1 - P) k^{2} + P k^{4}] .

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