Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 3 / ii / Solution

Fix the evaluation point $r^{\prime }$ inside $V$ and put $\psi (r)=G(r,r^{\prime })$ in Green's second identity. The two $k^2$ terms cancel because ${\nabla }^2\varphi =-k^2\varphi$ and ${\nabla }^{2}G={\delta }^{(3)}-k^{2}G$. The remaining volume integral is $\int \varphi (r){\delta }^{(3)}(r-r^{\prime })dV=\varphi (r^{\prime })$. ThusThe Dirac delta has a positive sign in this problem's Green-function equation, fixing the order of the two terms. All boundary derivatives act on the integration variable $r$.