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?

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.