Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 3 / 14A / Solution

Away from $x=y$ the Newton kernel is harmonic, while integrating its normal derivative over a small sphere gives one, proving ${\nabla }^2\hat{G}=\delta$. Green’s second identity is ${\int }_V(u{\nabla }^{2}v-v{\nabla }^{2}u)={\int }_{\partial V}(u{\partial }_{n}v-v{\partial }_{n}u)$. For the half-space, images give $G(x,y)=-(4\pi |x-y|{)}^{-1}+(4\pi |x-y^{\ast }|{)}^{-1}$. Substitution in the boundary formula yields the Poisson kernel $z/[2\pi ((x_1-y_1{)}^2+(x_2-y_2{)}^2+z^2{)}^{3/2}]$; polar integration against the Gaussian gives exactly the stated one-dimensional integral.

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