Bounding the determinant of the Jacobian between a set and its polyhedral approximation

Question

My question is, essentially, suppose I have two simply connected subset of $R^n$, if I know that the boundary's of both are very close, how can I bound the determinant of the Jacobian between them. I can probably assume anything reasonable on $\Omega$, and possibly $\delta \Omega$.

More precisely:

Suppose I have a bounded simply connected set $\Omega \subset R^2$, with a boundary $\delta \Omega$ that is a smooth as I like. Think of $\Omega$ as some kind of distorted circle. $\delta \Omega$ is parametrised by $X : [0, 2 \pi ] \rightarrow \delta \Omega$.

Suppose I also have an approximation to this, given by $\Omega_h$ where $\Omega_h$ is a polyhedral shape. It has a boundary $\delta \Omega_h$ that is parametrised by $X_h : [0, 2 \pi ] \rightarrow \delta \Omega_h$.

Since $\Omega_h$ approximates $\Omega$, we have that $|| X - X_h ||^2_{L^2 [0, 2 \pi] } \leq C_1 h^2$, where $C_1$ is a positive constant and h is a small positive number that I control.

If I have a mapping $ \omega : \Omega_h \rightarrow \Omega $. Let $J$ be the determinant of the Jacobian of $\omega$. Are there any theorems that state something like:

$ || X_h - X ||^2_{L^2 [0, 2 \pi]} \leq C_1 h^2 \implies |J - 1|^2 \leq C_2 h^2 $,

where $C_2$ is once again a positive constant. Anything similar would be greatly appreciated.

Edit: Perhaps my question is more about whether such an $\omega$ exists.

Answer 1

First, since you are mapping a piecewise linear curve ($\partial\Omega_h$) onto a smooth one ($\partial\Omega$), a diffeomorphism $\omega: \Omega_h \rightarrow \Omega$ does not exist in the strong sense. What you want is a piecewise differentiable homeomorphism.

With that being said, as shown in this paper, Proposition 4.7, there exists a piecewise differentiable homeomorphism $\omega:\Omega_h \rightarrow \Omega$ that fulfils a second-order pointwise estimate

\begin{equation} \tag{1} \|\omega - Id\|_{L^\infty(\Omega_h)} \leq Ch^2, \end{equation}

but its Jacobian fulfils only a first-order pointwise estimate

\begin{equation} \tag{2} \label{jacobian} \|J - 1\|_{L^\infty(\Omega_h)} \leq Ch, \end{equation}

where $J$ is the weak Jacobian, because $\omega$ is piecewise differentiable only. Estimate \eqref{jacobian} is optimal, so you cannot get a second-order pointwise estimate for the Jacobian. A quick way to see this is that you are mapping the piecewise linear curve $\partial\Omega_h$ onto the smooth one $\partial\Omega$, which is interpolation in disguise. Now, piecewise linear interpolation provides first-order accuracy for the gradients, only.

If you are happy with an $L^p$, $1\leq p <+\infty$ estimate instead of $L^\infty$ in \eqref{jacobian}, then you can even get arbitrarily good accuracy, i.e. for any $\varepsilon > 0$ you can choose $\omega$ sufficiently close to the identity such that

\begin{equation} \tag{3} \|J - 1\|_{L^p(\Omega_h)} \leq \varepsilon, \end{equation}

as you correctly said in your comment.