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I am interested under which regularity condition is Stokes' theorem is still valid.

For concreteness I am interested in the following problem

Let's consider a domain $\Omega$ in $\mathbb{R}^{3}$ given in cylindrical coordinates by $0\le z\le 1 $,$0\le \theta\le2\pi$,$0\le \rho\le b$.

Now let $f\in H^{2}(\Omega)$.

If we apply Stokes' theorem for the domain $\Omega_{\epsilon}$ in $\mathbb{R}^{3}$ given in cylindrical coordinates by $0\le z\le 1 $,$0\le \theta\le2\pi$,$\epsilon\le \rho\le b$ we have: \begin{equation} \int_{\Omega}\nabla\cdot \nabla fd\Omega=\int_{\partial\Omega_{\epsilon}}\frac{\partial f}{\partial x^{i}} n^{i} dS \end{equation} which is well-defined because the trace theorem warranty that $\frac{\partial f}{\partial x^{i}}\in H^{1/2}$.

Now in the case when $\epsilon\rightarrow0$ the boundary integral is: \begin{equation} \lim_{\epsilon\rightarrow 0}\int_{\partial\Omega_{\epsilon}}\frac{\partial f}{\partial x^{i}} n^{i} dS=\int\frac{\partial f}{\partial x^{i}}b d\theta dz-\lim_{\epsilon\rightarrow 0}\int\frac{\partial f}{\partial x^{i}}\epsilon d\theta dz \end{equation}

If $f$ is smooth then the limit vanishes. However in the low differentiability case I have the following questions:

In the limit $\frac{\partial f}{\partial x^{i}}$ has to be restricted to a codimension 2 domain. What can I say about the regularity of the trace in the codimension 2?

In case $f$ has even lower regularity such that the trace in the codimension 2 case is in a negative Sobolev space,should the limit be consider to be a functional?

What's is the lower differentiability required for Stokes' theorem to hold?

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  • $\begingroup$ You said that $f\in H^2(\Omega)$, but $\Omega$ depends on $\epsilon$. So, do you assume that $f\in H^2(\Omega(\epsilon))$ for all $\epsilon>0$? With a uniform bound on $H^2 $ norm? $\endgroup$ – user127096 Apr 21 '14 at 21:26
  • $\begingroup$ I have rewritten the question. $\endgroup$ – yess Apr 21 '14 at 21:41
  • $\begingroup$ @yess what are situations where Stokes theorem fails? I am curious $\endgroup$ – cactus314 Apr 27 '14 at 22:50
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To simplify the question, let's just get rid of a derivative by considering $V = \nabla f$. Now when does the divergence theorem hold:

$$ \int_{\partial \Omega} V \cdot n \, dS = \int_\Omega \nabla \cdot V dx $$

The conditions on the domain are as follows: Suppose $\Omega$ to be a bounded domain with a $C^1$ boundary: $\forall \xi \in \partial \Omega$ there is an open ball $B_r(\xi)$ and a $C^1$ diffeomorphism $$\psi: B_r(\xi) \to B_1(0)$$ such that $\psi( \partial \Omega \cap B_r(\xi) ) \subset \{ x \in \mathbb{R}^n : x_n =0 \}$ and $\psi ( \Omega \cap B_r(\xi) ) \subset \{ x \in \mathbb{R}^n : x_n >0\}$. Notice that $\partial \Omega$ is a codimension 1 hypersurface.

Next the conditions on the vector field $V$ on $\Omega$, we basically need $V_i \in C^1(\bar{ \Omega} )$ for the derivative to be well defined.

Slightly weaker, $\partial \Omega$ is only piecewise $C^1$(i.e. $\partial \Omega$ is a finite union of $C^1$ boundaries). Then we can approximate inside $\Omega$ by taking $V_i \in C^1( \Omega) \cap C ( \bar{\Omega} )$ and it'll still hold if the right hand side converges.

Now we have our conditions, I'll basically say you got it figured out. Either $f$ is good enough, or we can treat it as a functional through a limit. As an example, take one of the many approximations of the delta function, like $f = \ln(r)$ in $\mathbb{R}^2$

For a reference , have a look at Evan's or McOwen's PDE Book

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  • $\begingroup$ There are result for vector fields in at least $W^{1,1}_{loc}(\Omega)$ arxiv.org/abs/1112.5779 $\endgroup$ – yess Apr 27 '14 at 21:59

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