Nonlinear Piola Transform

 

This is an extension for Linear Piola Transform, where we discuss the cases where the transformation from the reference element to the physical element is nonlinear such as a parametric element. The main tool here is actually from the exterior calculus framework, and will just be from the FEEC book by Arnold.

Let $\Omega, \hat \Omega$ by the physical element and reference element with an orientation-preserving, differentiable map $\phi: \hat \Omega \to \Omega$. Suppose $p: \Omega \to \mathbb{R}, v: \Omega \to \mathbb{R}^3$ are the functions on the physical element. The quantity $(p, \nabla \cdot v) = \int_\Omega p \nabla \cdot v \, dx$ is of importance.

The key observation here is that by letting $\nu = v^1 dx^2 \wedge dx^3 – v^2 dx^1 \wedge dx^3 + v^3 dx^1 \wedge dx^2$, we have $d\nu = v$ where $d$ is the exterior derivative. A key fact here is that with integral of differential forms is it preserves pullbacks (see second answer) hence
\begin{align*}
(p, \nabla \cdot v) = \int_\Omega p \nabla \cdot v \, dx = \int_\Omega p \wedge d\nu = \int_{\hat \Omega} \phi^*(p \wedge d\nu)
\end{align*}
where $\phi^*$ is the pullback of differential form $\phi$.

Now, it’s just a matter of algebra
\begin{align*}
\int_{\hat \Omega} \phi^*(p \wedge d\nu) = \int_{\hat \Omega} \phi^*p \wedge \phi^* d\nu = \int_{\hat \Omega} \phi^*p \wedge d\phi^*\nu.
\end{align*}
Now, we need to look up what the pullback does to a 0-form $p$ and on the 2-form $\nu$. Well, we can look this up or just algebraically do it (which I’ll do it sooner or later because I couldn’t find a good source online), we have $\phi^* p = p \cdot \phi$ and
\begin{align*}
\phi^* \nu = (\det \phi’ )(\phi’)^{-1} (v \cdot \phi): \hat \Omega \to \mathbb{R}^3
\end{align*}
which is exactly the Piola transform.

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