Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 3 / 12C / ii

For a spherically symmetric function in three dimensions, the radial Laplacian is ${\nabla }^{2}u=r^{-2}(r^2{u}^{\prime }{)}^{\prime }$. On $0<r\le a$, integrate $(r^2{u}^{\prime }{)}^{\prime }=r^2$ twice to obtainThe region includes the origin. A classical solution must be regular there, so $A=0$: the $1/r$ term is singular and is not a solution of the stated full-ball problem. Hence the complete family on the ball isThe boundary cases below use this regular family. Allowing punctured-ball solutions would change the problem and the compatibility calculation.

The Dirichlet boundary condition gives $B=-a^2/6$, so the unique radial solution isIts radial Laplacian is one, it is smooth at the origin, and its boundary value is zero. The uniqueness result in part (i) shows that this is also the only solution among all classical functions on the ball, not merely among spherically symmetric ones.

Every regular radial solution has outward normal derivative $u^{\prime }(a)=a/3$, which is nonzero since $a>0$. Therefore there is no solution with the stated homogeneous Neumann boundary condition. This also follows without imposing spherical symmetry: the compatibility condition from part (i) would requirewhich is impossible. Thus the full-ball Poisson equation has no classical solution for these boundary data, radial or otherwise. A singular $1/r$ term could formally cancel the derivative at $r=a$, but would introduce a source at the origin and violate the problem's equation there.