Constant in the Poincare inequality for curl square integrable vector fields

Question

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

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

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:

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

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

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

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

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?

Answer 1

I don't understand what you are trying to say in your third inequality, as $C_1$ is always $1$. Consider the minimization problem $$ I_u=\inf_{k\in\mathbb R} \left\| u -k \right\|_{L^2(\mathbb \Omega)}^2 $$ On one hand, it is clear that $I_u \leq \left\| u \right\|_{L^2(\mathbb \Omega)}^2$, since this upper bound corresponds to $k=0$. On the other hand, since $$ \left\| u -k \right\|_{L^2(\mathbb \Omega)}^2 = k^2 \left|\Omega\right| -2 k \int_\Omega u ~\text d x+ \int_\Omega u^2 \text d x $$ the infimum is a minimum corresponding to $k= \frac 1 {\left|\Omega\right|} \int_\Omega u \text d x$ thus $$ \left\| u -\frac 1 {\left|\Omega\right|} \int_\Omega u ~\text d x \right\|_{L^2(\mathbb \Omega)} \leq \left\| u \right\|_{L^2(\mathbb \Omega)}, $$ and equality is attained for any function with zero average. Surely you meant something else?