Monday, January 12, 2015

Vector Calculus... with Poles?!

A couple days ago, my partner and I were about to go to sleep, when I wondered out loud whether Stokes' Theorem and the Divergence Theorem would hold for functions that were analytic except at finitely many poles (it's what engaged couples do in bed!).

Since my fiancé is a physicist, he already knew the answer for the Divergence Theorem, and was happy to clue me in: for $f(r, \theta,\phi)$ in polar coordinates,
$$\iiint_\Omega \nabla \cdot f(r,\theta,\phi) \; d\Omega=\oint_{\partial \Omega} f(r,\theta,\phi)\;d\partial\Omega.$$
appears to break down when a pole occurs in the interior of $\Omega$. In order to demonstrate this, we consider the function $f(r,\theta,\phi)=\frac{1}{r^2}$, which has the classical divergence (in polar coordinates)
$$\nabla \cdot \frac{1}{r^2} = \frac{\partial }{\partial r}\frac{r^2}{r^2}=0,$$
forcing the left-hand side to be zero. Yet, because $\frac{1}{r^2}$ is constant on the surface of the sphere, the right-hand side evaluates to
$$\oint_{\partial \Omega}\frac{1}{r^2}\hat{r}\cdot\hat{r}\;dA=\frac{1}{r^2}\oint_{\partial \Omega}dA=4\pi,$$
which is nonzero! Hence the Divergence Theorem does not hold in this case... for classical forms of divergence.

Knowing my interest in analysis (see blog title, above), my fiancé clarified that when a function has poles, we redefine divergence in the sense of distributions so that Green's Theorem does hold. Apparently then
$$\nabla \cdot \frac{1}{r^2} = 4\pi\delta(r)$$
which I plan to check rigorously in the future (no money for books). It hadn't occurred to me yet that that distributions could also be used to generalize multivariate forms of the derivative, so this was an interesting way for the conversation to go.

We quickly noted that for Stoke's Theorem in 2D, with $f(z)=p(z)+iq(z)$ and $z=x+iy$,
$$\int\int_\Omega \frac{\partial p}{\partial x} - \frac{\partial q}{\partial y} \;d\Omega=\int_{\partial \Omega} p\; dx - q\;dy$$
for the real part and
$$\int\int_\Omega \frac{\partial p}{\partial y} + \frac{\partial q}{\partial x} \;d\Omega=\int_{\partial \Omega} p\; dy + q\;dx$$
for the imaginary part. Assuming that $f(z)$ is analytic leads to one of the proofs of Cauchy's Integral Theorem, but, as differentiability is a condition for Stokes' Theorem, it is expected to break down under classical conditions for a complex function with finite poles. However, given my partner's earlier insight with the Divergence Theorem, I wouldn't be surprised if a distributional equivalent existed for Stokes' Theorem as well.

I'm certain there are a few texts that could clear up exactly how this happens rigorously... sounds like something fun to do in the future!

(Please excuse my slightly incorrect use of notation. Some symbols are not supported in MathJax.)