Constant in the Poincare inequality for curl square integrable vector fields

2012-09-11T03:38:17Z

$\newcommand{\v}[1]{\boldsymbol{#1}}$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?

Constant in the Poincare inequality for curl square integrable vector fields - MathOverflow

Constant in the Poincare inequality for curl square integrable vector fields