What are the continuity requirements on a vector field $\boldsymbol{A}$ such that Stokes' theorem,

$$ \iint_S\nabla\times\left[(\boldsymbol{\hat{x}}\cdot\nabla\phi)\boldsymbol{A}\right]\cdot d\boldsymbol{s} = \oint_C \left(\boldsymbol{\hat{x}}\cdot\nabla\phi\right)\boldsymbol{A}\cdot d\boldsymbol{l}, $$

holds? Here $\boldsymbol{\hat{x}}$ is a constant vector, $\phi$ is a smooth function (1) over all of $\mathbb{R}^3$, $S\subset\mathbb{R}^3$ is a bounded, connected and open set with boundary $C$.

More specifically, can it suffer a jump discontinuity on $C$?

To put this into context (2), there is an article that supposes that the vector field must be continuous on $S$ and across $C$. Since $\boldsymbol{A}$ suffers a jump discontinuity on $C$, Stokes' theorem fails. This is used to argue that some integral representation of electromagnetic fields fail. On the other hand, I have this article saying that even if $\boldsymbol{A}$ is discontinuous across $C$, Stokes' theorem still holds, citing the work of Whitney (3, p.100).

(1): It can be singular at isolated points on $S$, but for our purposes it can be considered smooth.
(2): I can provide PDFs via email, see my bio.
(3): H. Whitney, Geometric Integration Theory, Princeton University Press: Princeton, 1957.

Update: This first article seems to say that the divergence theorem (not Stokes' theorem, but I suppose the arguments would be the same as for the divergence theorem) holds if the vector field is discontinuous in a set $Z$, contained in $S+C$, that is of logarithmic capacity zero. The way I understand it, this seems to preclude a jump discontinuity on $C$. This second article, however, includes an additional term due to the discontinuity at $C$. Is that second article right? Would that carry over to Stokes' theorem?

  • $\begingroup$ I don't think the continuity conditions are well defined here if you just say the field is continuous on S and has a jump discontinuity on C. What can you say about the limit of the vector field as you approach a point on C from S? $\endgroup$ – Paul Oct 14 '15 at 18:59
  • $\begingroup$ You're right, I'll flesh this out in a little more detail tonight. I think I might have to enter the specifics a little more. Typically, the surface $S$ is divided in two subsurfaces, $S=S_1+S_2$ with the contour $C$ delimiting the two surfaces. On $S_2$, $\boldsymbol{A}$ vanishes while it has a nonzero value on $S_1$. The limit exists on both sides of $C$, but differ. There is thus a jump across $C$. In optics, this is the black screen problem. $\endgroup$ – Joey Dumont Oct 14 '15 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.