Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 3 / 12C (Vector Calculus)

For two solutions of the Poisson equation with the same Dirichlet boundary condition, let $w=u_1-u_2$. Then ${\nabla }^{2}w=0$ in $V$ and $w=0$ on its boundary. By Green's first identity,Thus $w$ is constant on each connected component, and the zero boundary values make each constant zero. The Dirichlet solution is unique whenever it exists. This argument requires the regularity needed for the displayed integration by parts; it does not assert existence for arbitrary rough data.

With a Neumann boundary condition, integrating the Poisson equation and using the divergence theorem instead gives the necessary compatibility conditionOn a disconnected region this condition must hold on each connected component separately. If two solutions have the same normal derivative, their difference is harmonic with ${\partial }_{n}w=0$. The same Green's first identity again forces $\nabla w=0$. Thus Neumann solutions are unique only up to an additive constant on each connected component. Adding a constant preserves both the equation and normal derivative, so uniqueness without a normalization cannot hold. On a connected region, prescribing the mean value removes this one-dimensional freedom.

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.