Does Gauss' theorem and Stokes' theorem still hold for distributions?

Question

We can define the derivative of distributions and then we can define their grad, div and curl, so do the Stokes' theorem and Gauss' theorem still hold for distributions? if not, under which condition can these two theorems hold?

Answer 1

The divergence theorem is in general inapplicable but if $\Omega \subset \mathbb{R}^3$ such that $\operatorname{sing supp}\vec{F} \cap \partial\Omega = \emptyset$ then $\iint \vec{F}\cdot\vec{n}\,dS$ is defined and one can even define $\vec{F} \chi_\Omega$ as a distribution. Take $\rho \in C^\infty_c(\mathbb{R}^n)$ with $\rho\equiv 1$ on a neighborhood of $\Omega$. Then $\langle (\nabla\cdot\vec{F}) \chi_\Omega, \rho \rangle = \iint \vec{F}\cdot\vec{n}\,dS$. Abusing integral notation we can write $\langle (\nabla \cdot \vec{F}) \chi_\Omega, \rho\rangle$ as $\iiint_\Omega \nabla\cdot\vec{F}\,dV$ so that Gauss' theorem holds.

Computation: $$ 0 = -\langle \vec{F}\chi_\Omega \stackrel{\bullet}{,} \nabla\rho \rangle = \langle \nabla\cdot(\vec{F}\chi_\Omega), \rho \rangle = \langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle + \langle \vec{F}\cdot\nabla\chi_\Omega, \rho \rangle \\ = \langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle - \iint \vec{F}\cdot\vec{n}\,dS . $$