Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 11C (Vector Calculus)

The product rule givesApply the divergence theorem, with outward unit normal $\mathbf{n}$, and use the vanishing Laplacian of the harmonic function $g$:The boundary value of $f$ is one. Another application of the divergence theorem therefore givesThis is the gradient pairing with a boundary-constant function and a harmonic function, obtained directly from Green's first identity. The harmonic function must be regular throughout the volume; an interior singularity can supply an additional boundary flux.

Here $v=z$, so $\nabla v={\mathbf{e}}_z$, while $\nabla u=\hat{\mathbf{r}}$ away from the origin. Their inner product is $\cos \theta$. The spherical coordinates volume element givesThis agrees with the boundary-constant gradient pairing with a boundary-constant function and a harmonic function: $u$ has constant boundary value $a$ and $v=z$ is a regular harmonic function. For comparison with the boundary value one in part (a), use $u/a$ and multiply back by $a$. Strictly, $u=r$ is not differentiable at the origin. Excising a ball of radius $\epsilon$ resolves this: the inner boundary term $u{\partial }_{n}v=-\epsilon \cos \theta$ has zero angular integral, so it contributes nothing in the limit. Thus the improper volume integral has the same zero value.

Now $v=z^2$ and $\nabla v=2z{\mathbf{e}}_z$. Since $\nabla u=\hat{\mathbf{r}}/a$, their inner product is $2r{\cos }^2\theta /a$. ConsequentlyAlthough $u=1$ on the boundary, the Laplacian is ${\nabla }^{2}v=2$, so $v$ is not a harmonic function. There is no contradiction with part (a). The origin singularity in the derivative of $u$ is harmless for this integral: its inner boundary contribution is of order ${\epsilon }^4/a$ and tends to zero. More explicitly, Green's first identity on the punctured ball gives outer flux $8\pi a^3/3$ and volume term ${\int }_{V}2r/adV=2\pi a^3$, whose difference is the displayed answer.

Away from the origin the gradients are $\nabla u=\hat{\mathbf{r}}/a$ and $\nabla v=-\hat{\mathbf{r}}/r^2$. The improper volume integral is thereforeThe function $v=1/r$ is a harmonic function only off the origin. It is singular inside the integration volume, so it does not meet part (a)'s regularity hypothesis. On the punctured ball, Green's first identity has outer flux $-4\pi$ and inner flux $4\pi \epsilon /a$: the inner outward unit normal is $-\hat{\mathbf{r}}$, so $u{\partial }_{n}v=(\epsilon /a)/{\epsilon }^2$. Their sum is $-4\pi +4\pi \epsilon /a$, exactly the integral before taking the limit. Equivalently the distributional Laplacian of $1/r$ is $-4\pi {\delta }_0$, rather than zero throughout the ball. This is boundary flux from a singular harmonic potential; it explains why applying part (a) across the origin would be invalid.