Open sets and Poincaré's inequality

In many references, Poincaré inequality is presented in the following way :

Let $\Omega\subset \mathbb R^d$ an open bounded set. We can find a constant $C$ which depend of $\Omega$ such that for all $u\in H^1_0(\Omega)$, we have $$\lVert u\rVert_{L^2}\leq C\lVert \nabla u\rVert_{(L^2(\Omega))^d}.$$

In fact it works if $\Omega$ is bounded in one direction. An other sufficient condition is that we can find $v\neq 0$ such that Lebesgue measure of $\{\lambda\in\mathbb R,\lambda v\in \Omega\}$ is finite).

My question, maybe a little vague, is the following: is there a "nice" necessary and sufficient condition on $\Omega$ to have Poincaré's inequality?

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You know Ziemers Book 'Weakly Differentiable Functions'? Chapter 4 is dedicated to Poincaré type inequalities. –  user20266 Jan 7 '12 at 11:19
Yes, but when I looked at it I didn't think about this question. And some pages are missing in Gooble book (which is normal). Anyway, this book is at the library of my university, so I will have a look at it Monday. –  Davide Giraudo Jan 7 '12 at 11:29
One generalization I know from one of my teachers can be found here: mathproblems123.wordpress.com/2011/10/05/… This needs $\Omega$ to have Lipschitz boundary, and increases the space of admissible functions $H_0^1(\Omega)$ to a closed subspace of $H^1(\Omega)$ which does not contain the non-zero constant functions. –  Beni Bogosel Jan 7 '12 at 17:38
Beni Bogosel: Thanks, I didn't know this result. @Thomas I look at this book, but I didn't find the answer. Maybe should I ask it at MathOverfow. –  Davide Giraudo Jan 11 '12 at 10:40
@DavideGiraudo: I suppose such condition can be that $\Omega$ is regular enough such that the Rellich Kondrachov theorem holds. en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem –  Beni Bogosel Apr 30 '12 at 8:02
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