 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/3/b/solution

Fix

x^{'}

inside

V

and put

ψ (x) = G (x, x^{'})

\int_{V} [ϕ (x) (- δ^{(3)} (x - x^{'})) - G (x, x^{'}) \cdot 0] d V = - ϕ (x^{'}) .

Here Laplace's equation eliminates the second term, and the Dirac delta evaluates the first. Reversing the signs in the boundary integral gives

ϕ (x^{'}) = \int_{S} [G (x, x^{'}) n \cdot \nabla_{x} ϕ (x) - ϕ (x) n \cdot \nabla_{x} G (x, x^{'})] d S .

All boundary derivatives act on the integration variable

x

, not on the fixed evaluation point

x^{'}

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