 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/14a/c/solution

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

At the first loss of stability, the minimum of

Δ (q)

just touches zero. The threshold condition from part (b) is

β = 1 + α D_{c},

D_{c} = \frac{( β - 1 ) ^{2}}{α ^{2}} .

The minimizing squared wavenumber is

q_{*} = - \frac{D _{c} α ^{2} - ( β - 1 )}{2 D _{c}} = \frac{β - 1 - D _{c} α ^{2}}{2 D _{c}} .

At threshold,

β - 1 - D_{c} α^{2} = 2 α D_{c}

, and therefore

k_{*}^{2} = q_{*} = \frac{α}{D _{c}} = \frac{α ^{2}}{β - 1} .

k_{*} = \frac{α}{β - 1} .

 New to topics? Read the docs here!