Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 2 / a

Let $V$ be a bounded region with piecewise smooth closed boundary $\partial V$, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus the volume integral of the divergence equals the outward boundary flux integral. For a region containing a hole, the outward unit normal vector on its inner boundary points into that hole.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Applying the product rule to the components of the vector field yieldsThe calculation is pointwise away from the origin; any extension of the vector field to the origin must be considered separately.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.