Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / 2

For $s=|\mathbf{x}-{\mathbf{x}}_0|>0$, the radial Laplacian gives ${\nabla }^2(1/s)=s^{-2}{\partial }_s(s^2{\partial }_s(1/s))=0$. The outward flux of $\nabla [-1/(4\pi s)]$ through a small sphere centred at the source is $[1/(4\pi s^2)]4\pi s^2=1$. Applying Green's second identity to a smooth test function outside that small sphere and shrinking its radius therefore gives the distributional identityThe kernel tends to zero at infinity. Thus it is the fundamental solution with the sign convention $\Delta G=+\delta$. The sign will reverse when converting it into the electrostatic potential of a positive point charge through Poisson's equation.

Let ${\mathbf{x}}_0^{-}=(x_0,y_0,-z_0)$ be the reflection of the source across the plane. The method of images gives the Dirichlet Green functionThe two source distances agree on $z=0$, so the boundary value vanishes. The reflected source lies outside the upper half-space, leaving only the required Dirac delta distribution inside. Both terms vanish at infinity.

Put $r_0=|{\mathbf{x}}_0|$ and ${\mathbf{x}}_0^{\ast }=a^2{\mathbf{x}}_0/r_0^2$ for $0<r_0<a$. The Dirichlet Green function inside a sphere isOn $|\mathbf{x}|=a$, the distance identity $|\mathbf{x}-{\mathbf{x}}_0^{\ast }|=(a/r_0)|\mathbf{x}-{\mathbf{x}}_0|$ cancels the two terms. The inverse point lies outside the sphere, so it introduces no interior source. This weighted inversion is the Kelvin transform. For a central source, take the limit to get $G_B(\mathbf{x};0)=-(4\pi {)}^{-1}(1/|\mathbf{x}|-1/a)$.

Reflect the source across the equatorial plane and subtract the two ball kernels. This gives the Dirichlet Green function inside a hemisphere $G_{\mathrm{hemi}}(\mathbf{x};{\mathbf{x}}_0)=G_B(\mathbf{x};{\mathbf{x}}_0)-G_B(\mathbf{x};{\mathbf{x}}_0^{-})$. It is zero on the spherical boundary because each ball kernel is zero there, and on the plane because reflection makes their values equal.

For a point charge $e$, Poisson's equation in SI units is $\Delta V=-e\delta (\mathbf{x}-{\mathbf{x}}_0)/{\epsilon }_0$, so $V=-e{G}_{\mathrm{hemi}}/{\epsilon }_0$. Writing ${\mathbf{x}}_0^{\ast -}$ for the reflected inverse point givesThe construction uses the real charge, its negative reflected image, a negative inversion image of magnitude $ea/r_0$, and its positive reflected image. Reflection and inversion commute. All three images lie outside the open upper half-ball, and the prescribed zero boundary values determine the electrostatic potential uniquely.