 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/3/b/solution

Write

C = A π / (2 k)

, so the radial field is

Φ = C e^{- k r} / r

. Integrate its equation over a ball of radius

ϵ

centred at the source. The divergence theorem gives

4 π ϵ^{2} Φ^{'} (ϵ) - k^{2} \int_{r < ϵ} Φ d V = 1.

The second term is

4 π k^{2} C \int_{0}^{ϵ} r e^{- k r} d r = O (ϵ^{2})

and vanishes in the limit. Meanwhile

4 π ϵ^{2} Φ^{'} (ϵ) = - 4 π C (1 + k ϵ) e^{- k ϵ} ⟶ - 4 π C .

Thus

C = - \frac{1}{4 π}, A = - \frac{1}{4 π} \frac{2 k}{π}, Φ (r) = - \frac{e ^{- k r}}{4 π r} .

The minus sign follows from the convention

(Δ - k^{2}) Φ = δ^{(3)}

, opposite to the positive static Yukawa Green function for

- Δ + k^{2}

. For arbitrary real

k

, replace it in the exponential by

∣ k ∣

to select decay at infinity. The

k = 0

limit is

- 1/ (4 π r)

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