 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-1/27j/iv/solution

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

Assume

ψ_{+} (c) = e^{- A c}

with

A \geq 0

. Part (ii) gives

ψ_{-} (c) = \frac{q}{a + A} e^{- A c}, a = q + λ .

Insert this into (iii). If

A \neq = a

, the convolution becomes

e^{- A c} = e^{- a c} + \frac{q ^{2}}{( a + A ) ( a - A )} (e^{- A c} - e^{- a c}) .

Equality for all

c > 0

q^{2} = (a + A) (a - A)

, hence

A^{2} = a^{2} - q^{2} = λ (2 q + λ)

. The case

A = a

would instead give

e^{- a c}

times a nonzero term linear in

c

, so is impossible. Thus

A = λ (2 q + λ), ψ_{+} (c) = e^{- c λ (2 q + λ)}, ψ_{-} (c) = \frac{q}{q + λ + A} e^{- A c} .

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