= Solution
Since $q=-\kappa\nabla T$ and $\nabla\cdot q=H$, the divergence theorem gives
$$
\oint_Sq\cdot dS=\int_VH\,dV.
$$
Spherical symmetry gives $(r^2\kappa T')'=-r^2H$. Regularity at zero and decay at infinity yield
$$
T(r)=\begin{cases}
\displaystyle \frac{H_0}{3(\alpha+1)}+\frac{H_0}{3(2-\alpha)}(1-r^{2-\alpha}),&r\leq1,\\[6pt]
\displaystyle \frac{H_0}{3(\alpha+1)}r^{-\alpha-1},&r>1.
\end{cases}
$$
For $\alpha\leq-1$, the exterior solution cannot tend to zero: it is logarithmic at $-1$ and grows below it.
On the exterior domain, the energy-minimizing solution with boundary value one is $\phi=r^{-\alpha-1}$. Applying the <Dirichlet principle> to any admissible $w$ gives
$$
\boxed{\int_1^\infty r^{\alpha+2}(w')^2dr
\geq\int_1^\infty r^{\alpha+2}(\phi')^2dr
=\alpha+1.}
$$
Back to article page
