Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $U\subset \mathbb{R}^{n}$ be an open set and $f:U \to \mathbb{R}$ a continuous function which is piecewise $C^{1}$. This is: there is a partition of $U$ by (say, a finite number of) open sets $U_{\alpha}$, each with piecewise $C^{1}$ boundary, such that $f$ restricted to each of these is $C^{1}$.

My question is the following: (when) are the weak derivatives of $f$, $D_{i} f$, given by the piecewise defined functions $g_{i}|_{U_{\alpha}}=D_{x_{i}}(f|_{U_{\alpha}})$? Where can I find a proof if the case?

share|improve this question
1  
I am sorry that I couldn't understand your notation: What is $g_i$ here? –  Paul Dec 3 '11 at 7:11
    
Are the Lakes of Wada a counterexample? –  user7530 Dec 3 '11 at 8:59
    
@Paul: maybe the notation is a bit dumb. I have add a parenthesis that hopefully makes it clear (remember that weak derivatives are well defined if specified up to a set of measure zero, so this assumption is implicit w.r.t the union of the $\partial U_{\alpha}$). –  David Dec 3 '11 at 17:55
    
@ user7530: I don't have time to study that example right now (I have just look for "Lakes of Wada" at Wikipedia), but it probably goes beyond the assumptions of my problem. For example: is the boundary of these "lakes" piecewise $C^{1}$? Also: what function are you expecting to define over them? –  David Dec 3 '11 at 17:57

1 Answer 1

With the exception of some pathological cases, this will usually be true. Just verify the definition of weak derivative:

$$ \int_U f D_i \varphi dx = \sum_{\alpha} \left( \int_{\partial U_\alpha} f \varphi \nu_i dS - \int_{U_\alpha} \varphi D_i f dx\right)$$

where $\nu_i$ is the $i^{th}$ component of the normal vector to $\partial U_\alpha$. Then we have

$$ \int_U f D_i \varphi dx = -\int_{U} \varphi D_i f dx + \sum_{\alpha} \int_{\partial U_\alpha} f \varphi \nu_i dS.$$

I would expect the term on the right side to be zero, besides maybe in pathological cases, since each boundary should be shared by two regions and the normal vector $\nu_i$ will be of opposite sign in each region.

share|improve this answer
    
Yes, this is an application of integrarion by parts. Still there is a technicality that I'd like to discuss: the integration by parts formula works when one has $C^{1}$ boundaries and functions. What happens if $D_{i}f$ blows up near some point of $\partial U_{\alpha}$? And, when can we allow the int. by parts formula over open sets with piecewise smooth boundaries? (this last case is not so "pathological": try to give a partition of the ball by (closures of) open sets with nondisjoint, $C^{1}$, boundaries). –  David Dec 3 '11 at 19:53
    
Is it obvious that almost all boundary points appear on the boundary of only two open sets, and with opposite orientation? –  user7530 Dec 3 '11 at 22:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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