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I am interested in a generalization of the divergence theorem: Given an open subseteq $U \subseteq \mathbb{R}^n$, a compact set $G \subseteq U$ with smooth boundary $\partial G$ and a $C^1$-vector field $\vec{F} : U \rightarrow \mathbb{R}^n$, then $$ \int_{\partial G} \langle \vec{F},d\vec S\rangle \quad = \quad \int_G \mathrm{div}(\vec F) $$
In physics one often encounters situations where $\vec{F}$ is not $C^1$ (or not even defined) everywhere, such as $\vec F = \frac1{\|\vec r\|}\vec r$. My question: Is there a generalization of this equation to the case that $\vec F$ is a vector-valued distribution on $U$ ? One problem might be that one cannot integrate a distribution over a measurable set in general, but at least in the case that the distribution is induced by a (signed) Radon measure on $U$ this might be reasonable.

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    $\begingroup$ My understanding is this is, more or less, the definition of $div(\vec{F})$ (not really, you need to throw in a generic $C^\infty$ function that multiplies, but close enough that to passing from there to the formula you want is more or less a formality). Yeah it is a cheat, but it makes the formulas work :) $\endgroup$ Commented Feb 21, 2016 at 15:32
  • $\begingroup$ You need to assume at least enough regularity so that both side of the equation make sense. I believe that having the coefficients of $\vec{F}$ be in the Sobolev space $W^{1,1}$ suffices. The right side is bounded by the definition of the norm for $W^{1,1}$.. The left side requires a well-defined way to restrict $\vec{F}$ to the boundary. The operator that maps $\vec{F}$ to its restriction along the boundary is called the trace operator. See: en.wikipedia.org/wiki/Trace_operator (as well as the cited book by Evans ). $\endgroup$
    – Deane Yang
    Commented Feb 21, 2016 at 20:24
  • $\begingroup$ @Deane Yang: Thank you for the reference, I will have a look at Evans' book. Hope that I can formulate my own statement more precisely afterwards. $\endgroup$ Commented Feb 23, 2016 at 14:28

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