 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/3/b/solution

Φ (r)

and

G (r, r^{'})

, differentiating in

r

\int_{V} (ΦΔ G - G ΔΦ) d V = \int_{S} (Φ \frac{\partial G}{\partial n} - G \frac{\partial Φ}{\partial n}) d S .

The volume terms become

Φ (r^{'}) - \int_{V} Gρ d V

. On the boundary

G = 0

and

Φ = f

, so

Φ (r^{'}) = \int_{V} ρ (r) G (r, r^{'}) d V + \int_{S} f (r) \frac{\partial G}{\partial n} (r, r^{'}) d S .

Here

\partial / \partial n

is the outward normal derivative in the integration variable. One may justify the identity at the singularity by excising a small sphere around

r^{'}

; its unit flux supplies the evaluation

Φ (r^{'})

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