 Solution

With $s=r\sin \theta$ and $z=r\cos \theta$, radial differential rotation gives ${\partial }_z\omega =\cos \theta {\omega }^{\prime }(r)$. Thus the source is $B_{0}r{\omega }^{\prime }(r)\sin \theta \cos \theta$. Put $B=f(r)g(\theta )$ with $g=\sin \theta \cos \theta$. The toroidal diffusion operator isDirect differentiation of $g$ makes the bracket $-6g$. Therefore the steady radial equation isDefine $I(r)={\int }_0^r{x}^4\omega (x)dx$ and $h=I/r^3$. Since $I^{\prime }=r^4\omega$,so $h^{\prime \prime }+2h^{\prime }/r-6h/r^2=r{\omega }^{\prime }$. Taking $f=-B_{0}h/\eta$ proves the steady toroidal induction by spherical differential rotation:Near zero, $I(r)=\omega (0)r^5/5+O(r^6)$, hence $B=O(r^2)\sin \theta \cos \theta$ and is regular. Its leading vector field is proportional to $z(-y,x,0)$, so the apparent azimuthal-basis singularity on the axis is harmless. The rapid decay of $\omega$ makes $I_{\infty }={\int }_0^{\infty }{x}^4\omega (x)dx$ finite, giving $B\sim -(B_0{I}_{\infty }/\eta )r^{-3}\sin \theta \cos \theta$ at large radius. The homogeneous radial solutions are $r^2$ and $r^{-3}$; regularity at zero excludes the latter, and decay at infinity excludes the former. Thus these boundary requirements also single out the displayed solution in this angular sector.