Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 3

Apply the product rule to the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two mixed dot products cancel. The divergence theorem, using the outward unit normal, then gives Green's second identity:The usual assumptions are a sufficiently regular boundary and fields with the derivatives and integrability needed for the theorem; singular sources can instead be treated by excising small balls or using distributions.

Define the candidate potential by convolving the decaying Green function with the source. Acting with the Laplacian in the observation variable gives, in the sense of distributions,The final equality is the sifting property of the Dirac delta. The kernel found in part (ii) is locally integrable, and its convolution with a sufficiently regular rapidly decaying source also vanishes at spatial infinity. Thus it satisfies the Poisson equation and the required condition at infinity.

The difference from any other regular decaying solution is a harmonic function on all of space vanishing at infinity, hence is zero by the maximum principle. ThereforeOne can also obtain this expression from Green's second identity on larger and larger balls, with the singularity of the kernel treated distributionally. The surface term at infinity vanishes for the usual $1/r$ potential decay and $1/r^2$ gradient decay.

Translation and rotation invariance allow a radial Green function $G(R)$, with $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Outside the source, $G^{\prime \prime }+2G^{\prime }/R=0$, so $G=A/R+B$. Decay at infinity gives $B=0$.

Integrate the defining equation over a small ball centred on the source. The divergence theorem fixes its flux:ConsequentlyThis is the Green function of the Laplacian with the stated negative-delta sign. A homogeneous $1/R$ addition would change the delta normalization, so the singular coefficient is fixed, not an arbitrary integration constant.

Write $r=|\mathbf{r}|$, $s=|{\mathbf{r}}^{\prime }|$. For a spherical source, the angular integral needed for the Green function of the Laplacian isThis is the angular form of the spherical shell theorem. Substituting the source into the Green function integral therefore yieldsThe elementary integrals areThe terms proportional to $e^{-\beta r}/\beta$ cancel, leavingThe value at the origin is the continuous limit. At large radius $\varphi \sim -1/({\beta }^{2}r)$, consistent with the integrated source $\int \rho dV=-4\pi /{\beta }^2$ and the factor $1/(4\pi r)$ in the far-field potential.

For a direct check, set $u=r\varphi$. Then $u^{\prime \prime }=e^{-\beta r}$, so ${\nabla }^2\varphi =u^{\prime \prime }/r=e^{-\beta r}/r=-\rho$ for $r>0$. Boundedness of $\varphi$ at zero cancels the singular $1/r$ terms and excludes an extra point source. The original density is itself singular but locally integrable; the equation at the origin is understood in the sense of distributions. This is the Poisson potential of an exponential-over-radius source with amplitude $A=-1$.