 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/3/a/solution

L = d^{2} / d t^{2} - 4

and choose

L_{t} G (t, t^{'}) = δ (t - t^{'})

, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are

e^{\pm 2 t}

. Decay at each end and continuity at

t = t^{'}

therefore give

G = C e^{- 2∣ t - t^{'} ∣}

. Integrating its equation across the source gives a unit jump in its first derivative:

\partial_{t} G (t^{' +}, t^{'}) - \partial_{t} G (t^{' -}, t^{'}) = - 2 C - 2 C = 1.

Thus

A = - 2, C = - \frac{1}{4}, G (t, t^{'}) = - \frac{1}{4} e^{- 2∣ t - t^{'} ∣} .

x (t) = - \frac{1}{4} \int_{- \infty}^{\infty} e^{- 2∣ t - τ ∣} f (τ) d τ .

There is no extra homogeneous term: no nonzero combination of

e^{2 t}, e^{- 2 t}

decays at both infinities. The forcing must of course permit the imposed decay; integrable

f

, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

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