I'm wondering if there's a reasonable way to define or evaluate curl of a vector field that isn't differentiable everywhere. I'm interested in two specific vector fields. $$v_1=\frac{[-y,x]}{x^2+y^2}$$ $$v_2=\frac{[cos(tan^{-1}(x,y)/4),sin(tan^{-1}(x,y)/4)]}{(x^2+y^2)^{\frac{1}{8}}}$$
Both vector fields have 0 curl almost everywhere. But $v_1$ has a singularity at $x=y=0$. Using stokes thm I suspect that $\nabla \times v_1 = 2 \pi \delta$ but I'm not confident because I'm not sure it's allowed to use stokes thm on a vector field over a singular region. Is this ok?
$v_2$ is even more problematic. It has a discontinuity along the ray $x\leq 0,y=0$. I tried using stokes thm again but got a completely nonsensical result. Is there a reasonable definition of the curl of $v_2$ at $x=y=0$? (If it helps, the context in which $v_2$ appears is as one component of a cross field. Alternatively, $v_2$ is a vector field on one sheet of a quad cover. On the quad cover, $v_2$ is continuous, but still has a singularity.)
For the first vector field, we have that
$$\nabla \tan^{-1}\left(\frac{y}{x}\right) = v_1$$
mod piecewise discontinuities that can be somewhat fixed by choosing constants. As functions, you can't use Stokes' theorem directly because we cannot conclude that every infinite point is a delta or even constant times delta. What if it was delta composed with another function?. This is why we need to use distributions. In the language of distributions, we would want to prove that
$$\text{curl}\hspace{2 pt}\nabla u = c\delta_0$$
where $u$ is a modified version of the arctan above that is continuous on $\mathbb{R}^2$ without the negative $x$ axis, but is still bounded on $\mathbb{R}^2$. We can use the distributional derivative to calculate:
$$\text{curl}\nabla u(\phi) \equiv u(\text{curl}\nabla \phi) = \lim_{\epsilon\to 0}\iint_{\mathbb{R}^2\text{\\}B(0,\epsilon)} u \hspace{2 pt}\text{curl}\nabla \phi dx$$
$$ = \lim_{\epsilon\to 0}\int_{\partial B(0,\epsilon)}(\phi\nabla u - u\nabla \phi)\cdot \nu ds + \lim_{\epsilon\to 0}\iint_{\mathbb{R}^2\text{\\}B(0,\epsilon)} \phi \hspace{2 pt}\text{curl}\nabla u dx$$
by using "integration by parts" with Green's theorem. The last term is $0$ since $\text{curl}\hspace{1 pt}u=0$ everywhere on its integration domain and$\phi$ by assumption is continuous (which is why we couldn't say the first term initially was $0$ because even though the curl of gradient is $0$, $u$ had a series of jump discontinuities).
Now looking at the line integral, we have that the tangent vector to the circle is given by $\nu ds = \frac{(-y,x)}{\sqrt{x^2+y^2}} = \frac{(-y,x)}{\epsilon} \epsilon d\theta = (-y,x)d\theta$. Plugging in to the integral,
$$\lim_{\epsilon \to 0}\int_{\partial B(0,\epsilon)}\left(\phi\frac{y^2+x^2}{x^2+y^2} - \epsilon u\nabla \phi\cdot \nu\right) d\theta = \lim_{\epsilon \to 0}\int_{\partial B(0,\epsilon)}\left(\phi - \epsilon u\nabla \phi\cdot \nu\right) d\theta$$
Since by assumption $\phi$ is smooth and $u$ is bounded, we can use dominated convergence to prove that the second term goes to $0$. Then
$$\lim_{\epsilon \to 0}\int_{\partial B(0,\epsilon)}\phi d\theta = 2\pi\phi(0) \equiv 2\pi\delta_0(\phi)$$
Therefore we can say that $\text{curl}\hspace{2 pt}v_1 = 2\pi\delta_0$. A similar distribution argument can be made for $v_2$.