For a <spherically symmetric function> $\Phi(r)$, the <Laplacian in spherical coordinates> is
$$
\nabla^2\Phi=\frac1{r^2}\frac{d}{dr}(r^2\Phi').
$$
Integrating from zero to $r$ and using regularity at the origin gives
$$
\boxed{4\pi r^2\Phi'(r)=4\pi\int_0^rs^2\rho(s)\,ds}.
$$

For the stated density, integration and the boundary condition $\Phi(a)=0$ give
$$
\boxed{
\Phi(r)=
\begin{cases}
\displaystyle \rho_0\left(\frac{r^2}{6}-\frac{b^2}{2}+\frac{b^3}{3a}\right),&0\leq r\leq b,\\[6pt]
\displaystyle \frac{\rho_0b^3}{3}\left(\frac1a-\frac1r\right),&b<r\leq a.
\end{cases}}
$$
The two pieces and their first derivatives agree at $r=b$. If another solution existed, its difference from $\Phi$ would be <harmonic function>[harmonic] with zero boundary data, so part (a) proves <Uniqueness of the Dirichlet problem>.

On the shell $U(b,a)$, normalize the outer radial part to obtain the harmonic function
$$
v(r)=\frac{b(a-r)}{(a-b)r},
\qquad v(b)=1,\quad v(a)=0.
$$
By the <Dirichlet principle>, this function minimizes the energy among all $w$ with the same boundary values. Since
$$
v'(r)=-\frac{ab}{(a-b)r^2},
$$
its energy is
$$
4\pi\int_b^a r^2|v'(r)|^2dr
=\frac{4\pi a^2b^2}{(a-b)^2}\left(\frac1b-\frac1a\right)
=\frac{4\pi ab}{a-b}.
$$
Therefore
$$
\boxed{\int_{U(b,a)}|\nabla w|^2dV\geq\frac{4\pi ab}{a-b}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 3 / 12B / b / Solution

A scalar function on Euclidean space is spherically symmetric when it depends only on the distance $r=|x|$ from a fixed center.


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