Constant in the Poincare inequality for curl square integrable vector fields

$\newcommand{\v}[1]{\boldsymbol{#1}}$For an $u\in H^1(\Omega) = W^{1,2}(\Omega)$ where $\Omega$ is Lipschitz, we have

$${\|u - \frac{1}{\Omega} \int_{\Omega} u\|}_{L^2}\leq C {\|\nabla u \|}$$

and the Poincare constant is basically a multiple of diameter of the domain.

However in $\mathbb{R}^3$, the only similar result for $\mathbf{curl}$-square integrable vector fields $\v{u}$ would be:

$${\|\v{u} - \frac{1}{\Omega} \int_{\Omega} \v{u}\|}_{L^2} \leq {\|\mathbf{curl} \ \v{u}\|}$$

if $\v{u}$ is divergence free.

If not, suppose $\v{u}$'s divergence is not well-defined, then we could have:

$${\|\v{u} - \frac{1}{\Omega} \int_{\Omega} \v{u}\|}_{L^2(\Omega)} \leq C_1 \left(\|\v{u}\|_{L^2(\Omega)}^2+{\|\mathbf{curl} \ \v{u}\|}^2 \right)^{1/2}$$

Since we could still do Helmholtz decomposition $\v{u} = \v{w}+\nabla p$, qualitatively speaking, locally in the a compact subdomain $K\subset \Omega$, if the $\v{u}$ is more irrotational(that gradient field $\nabla p$ is dominant), then $C_1$ is closed to $1$, if the $\v{u}$ is more weakly-solenoidal($\v{w}$ is dominant), then the $C_1$ is more closed to $\sqrt{1+d^2}$.

I am curious if we could decompose the domain $\Omega$ into different parts, such that we estimate the Poincare constant locally and then put together to get a sharper bound?

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I don't quite grok your paragraph starting "Since we could..." Isn't $w$ the divergence free part? Then when $w$ is dominant, shouldn't you get closer to the divergence free case, and the reverse when $\nabla p$ is dominant? – Willie Wong Sep 11 '12 at 11:09
BTW, have you seen arxiv.org/abs/1208.6045 ? – Willie Wong Sep 11 '12 at 11:12
@Willie Wong, thanks for the comment, the main reason I am doing this is because I want to circumvent the divergence part, since for general $H(\mathbf{curl})$ vector fields, the divergence is not defined, and $w$ is in the weakly divergence free, such that I could get a Poincare inequality solely for $H(\mathbf{curl})$ vector fields not having the divergence free constraint. Thanks for the heads up of the uniformity paper as well, I will look into it. – Shuhao Cao Sep 11 '12 at 16:02