Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 12A (Vector Calculus)

In Cartesian coordinates, put $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$. Away from the origin the partial derivatives of $\psi =1/r$ areSumming the three second partial derivatives yieldsThus $1/r$ is a harmonic function on the punctured space and solves Laplace equation there. The origin is excluded: the singularity cannot be treated as an ordinary harmonic point.

Use the product rule for the divergence of the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two dot products cancel. Apply the divergence theorem with outward unit normal vector $n$ to obtain Green's second identity:Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.