Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 3

Use the product rule for the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The mixed dot products cancel. Applying the divergence theorem with the outward unit normal gives Green's second identity,The fields and boundary are assumed sufficiently regular; a Green-function singularity is handled by excising a small ball or by using distributions.

Fix ${\mathbf{x}}^{\prime }$ inside $V$ and put $\psi (\mathbf{x})=G(\mathbf{x},{\mathbf{x}}^{\prime })$ in Green's second identity. The volume integral becomesHere Laplace's equation eliminates the second term, and the Dirac delta evaluates the first. Reversing the signs in the boundary integral givesAll boundary derivatives act on the integration variable $\mathbf{x}$, not on the fixed evaluation point ${\mathbf{x}}^{\prime }$.

Impose the homogeneous Dirichlet boundary condition on the boundary variable:This removes the term containing the unknown normal derivative of $\varphi$. The prescribed boundary values alone then determine the solution throughFor an unbounded domain one also imposes the appropriate decay condition at infinity. This defines the required Dirichlet Green function.

Let the evaluation point be ${\mathbf{r}}_0=(x,y,z)$, $z>0$, and reflect it to ${\mathbf{r}}_0^{\ast }=(x,y,-z)$. The method of images constructsThe two Euclidean distances coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-{\nabla }_{\mathbf{r}}^{2}G={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$. The difference also decays at infinity.

On the plane write $\mathbf{r}=(\xi ,\eta ,s)$ and set $s=0$. The outward unit normal of the upper half-space is $n=-e_z$. Direct differentiation givesSubstituting into the boundary representation from part (c) yields the Poisson kernel for the upper half-space:Its sign is positive, as expected from the maximum principle for nonnegative boundary data. Its normalization can be checked in polar coordinates:As $z\downarrow 0$ the kernel is an approximate identity, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.

On the axis, the Poisson kernel for the upper half-space depends only on the radial boundary coordinate. Integrating over the disk givesTherefore the axial Poisson extension of a circular boundary patch isIt tends to one as the axis approaches the boundary inside the patch, and tends to zero as $a^2/(2z^2)$ far above it.

The boundary values jump at the circular rim. They are understood almost everywhere, or as limits at continuity points; changing the assigned value on that measure-zero circle does not change the integral. At a rim point the vertical limit is one-half, rather than a continuous extension with the specified value on both sides of the jump.