Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 3 / 11A (Vector Calculus)

For a real-valued function on a bounded regular region, apply the divergence theorem to $\varphi \nabla \varphi$. The product rule yields $\nabla \cdot (\varphi \nabla \varphi )=|\nabla \varphi {|}^2+\varphi {\nabla }^2\varphi$. The prescribed zero boundary values and Laplace equation therefore implyThe continuous nonnegative integrand must vanish everywhere. Hence $\varphi$ is constant on each connected component, and its boundary value makes each constant zero. This proves zero-boundary harmonic uniqueness. If complex-valued functions are allowed, apply the same argument separately to their real and imaginary parts. If ${\psi }_1,{\psi }_2$ solve the same Poisson equation and Dirichlet boundary data, their difference is harmonic with zero boundary value, so is zero. Thus there is at most one solution. An exterior unbounded region would require an additional condition at infinity; the preceding bounded-domain proof does not assert uniqueness without one.

For the radial calculation use Cartesian coordinates with $r=(x_i{x}_i{)}^{1/2}>0$. The chain rule gives ${\partial }_{i}r=x_i/r$ andDifferentiate again and sum the three coordinates:This derives the radial Laplacian directly in Cartesian coordinates. For ${\nabla }^2\psi =c$, set $u=r\psi$, so $u^{\prime \prime }=cr$. Integrating twice gives $u=c{r}^3/6+ar+b$, and consequentlyIf the solution must extend boundedly through the origin, $b=0$. The remaining polynomial is smooth there.

The radial Laplace equation has solution $a+b/r$. Boundedness at the origin requires $b=0$, and the value at $r=1$ fixes $a=1$. Thus

The radial Poisson equation with constant source one has $\psi =r^2/6+a+b/r$. The boundary conditions giveSubtracting gives $b=1$, then $a=-1/6$. HenceIts radial Laplacian is one and both endpoint values are one. The prescribed values ensure continuity with the adjacent region solutions, but do not impose derivative continuity.

The exterior radial Laplace equation has $\psi =a+b/r$. Decay at infinity forces $a=0$ and the value at $r=2$ forces $b=2$. ThereforeTogether the three solutions satisfy their separate regional equations and specified boundary values. Their radial derivatives jump at the interfaces, so they should not be asserted to solve one globally smooth Poisson equation with only the listed volume sources and no interface terms.