Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 1

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.