The best constant in Poincare-liked inequality in BVBV and BDBD space

This question has been posted on Math Stack exchange for a while and received no response. So I decide to move it here to get more attention.

Let $\Omega\subset \mathbb R^N$ be open, bounded and with smooth boundary. Then we could prove that for any $u\in BV(\Omega)$ and $\omega\in \operatorname{ker}\mathcal E$, where $\mathcal E$ denotes the distributional symmetric derivative $\mathcal E\omega = \frac{1}{2}(\nabla \omega+\nabla \omega^T)$ from $\mathbb R^N$ to $\mathbb R^N$, there exists a constant $C>0$ independent of $u$ and $\omega$ such that
$$\|Du\|_{\mathcal M(\Omega)}\leq C(\|Du-\omega\|_{\mathcal M(\Omega)}+\|u\|_{L^1})$$
The proof is not long and can be found here, Theorem 3.3, equation $(4)$.

It can also be shown that $\omega\in \operatorname{ker}\mathcal E$ iff $\omega = Ax+b$ where $A=-A^T$, $A\in\mathbb R^{N\times N}$ and $b\in \mathbb R^N$

However, the proof is done by using contradiction, which is short and simple but can not give any information of this constant $C$.

I am interested in finding the best constant $C>0$. I am sure this constant should only depend on $\Omega$ but I really want to know how it depends on $\Omega$. Can we make the best constant larger or smaller by changing $\Omega$? Also, if $\Omega:=[0,1]\times[0,1]$ in $\mathbb R^2$, can we explicitly compute this constant?

Thank you!

Answer

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Source : Link , Question Author : JumpJump , Answer Author : Community