I have an isoparametric mapping $F_{E}: \hat{E} \to E$ where $\hat{E}$ is a reference quadrilateral (square) and $E$ is some quadrilateral in the domain $\Omega$. If $E$ is a parallelogram, then the mapping is affine and computing $F_{E}^{-1}$ is straightforward. But for the general case, we have: $$ F_{E}(\hat{x},\hat{y}) = \left(\begin{array}{c}\sum_{i=1}^{4}x_{i}\hat{\psi}_{i}(\hat{x},\hat{y})\\ \sum_{i=1}^{4}y_{i}\hat{\psi}_{i}(\hat{x},\hat{y})\end{array}\right) = \left(\begin{array}{c}x\\ y\end{array}\right) $$
where $\hat{\psi}_{i}$ are the basis functions on $\hat{E}$.
For the discontinuous Galerkin method, I need to compute some integrals over the interfaces (2D) of the elements. A simple example would be the following: $$ \int_{e}\langle\langle\nabla\psi_{i}\cdot\mathbf{n}\rangle\rangle[[\psi_{j}]] $$
where $\langle\langle\cdot\rangle\rangle$ and $[[\cdot]]$ denote the average and jump operators, respectively.
We have that $\psi_{j}(x,y) = \hat{\psi}_{j}(F_{E}^{-1}(x,y)) = \hat{\psi}_{j}(\hat{x},\hat{y})$. Furthermore, on a given edge $e$, the point $(x,y) \in e$ maps to two values due to the discontinuities of the basis functions across the interfaces, i.e. we have $\psi_{j}^{-}$ and $\psi_{j}^{+}$ along $e$.
To compute the above integral, I use a quadrature rule on the interval $\hat{e} = [-1,1]$. For a quadrature point $\hat{\zeta} \in \hat{e}$, I map this 1D point to $e$ and call it $\zeta$.
My goal is to compute $\hat{\psi}_{j}^{-}(F_{E-}^{-1}(\zeta))$ and $\hat{\psi}_{j}^{+}(F_{E+}^{-1}(\zeta))$, where $F_{E-}$ and $F_{E+}$ are the mappings associated with the neighboring elements to $e$. For triangular elements and parallelograms, this is really straightforward since the mapping is affine and thus, computing $F_{E}^{-1}$ is easy.
But for general quadrilaterals, is this even possible? I found this paper, but it's quite old and the process described seemed a bit complicated. Is there any other literature anyone can recommend on this? Or some insight into a simpler way to do this?
SymPy
and got a solution here. If you do it by hand it is prettier (of course), but I don't have much time now. $\endgroup$