Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 1

For $r>0$, the partial derivatives $r_x=x/r$ and $r_y=y/r$ giveAdding these partial derivatives yields the divergence:The original vector field is undefined at the origin. Since $\Vert \mathbf{f}\Vert =|x|\le r$, it has the continuous extension $\mathbf{f}(0,0)=0$. This extension is not differentiable: on $y=0$ its first component is $|x|$, whose derivative at zero does not exist. Thus the displayed divergence has no classical value at the origin.

The triangle is $0\le y\le x\le 1$. In polar coordinates this becomesbecause the outer side is $x=r\cos \theta =1$. The vector field and the Jacobian determinant giveEvaluate the double integral component by component:Consequently,The origin causes no ambiguity in this double integral: the continuous extension from part (a) is bounded there, and changing a value at a single point does not change the integral.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.