Suppose I have a reference quadrilateral on $[-1, -1] \times [-1, 1]$ with reference coordinates $\xi, \eta$ and a mapping to an isoparametric quadrilateral in 'physical space' described by coordinates $x, y$.

For a triangular element, the isoparametric mapping is handled the following way for the global nodal basis functions $\phi_i$ in 'physical space': \begin{align} \begin{bmatrix} \frac{\partial \phi_i}{\partial x} \\ \frac{\partial \phi_i}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial \xi}{\partial x} & \frac{\partial \eta}{\partial x} \\ \frac{\partial \xi}{\partial y} & \frac{\partial \eta}{\partial y} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial \phi_i}{\partial \xi} \\ \frac{\partial \phi_i}{\partial \eta} \end{bmatrix} = J^{-1} \begin{bmatrix} \frac{\partial \phi_i}{\partial \xi} \\ \frac{\partial \phi_i}{\partial \eta} \end{bmatrix} \end{align}

Via the chain rule, so that I can calculate \begin{align} \frac{\partial \phi_i}{\partial x} &= \frac{\partial \phi_i}{\partial \xi}J^{-1}_{11} + \frac{\partial \phi_i}{\partial \eta}J^{-1}_{12} \\ \frac{\partial \phi_i}{\partial y} &= \frac{\partial \phi_i}{\partial \xi}J^{-1}_{21} + \frac{\partial \phi_i}{\partial \eta}J^{-1}_{22} \end{align}

I would then build $J$ by first constructing

\begin{align} J = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \\ \end{bmatrix} = \begin{bmatrix} \sum x_j^K\frac{\partial \psi_j}{\partial \xi} & \sum y_j^K\frac{\partial \psi_j}{\partial \xi} \\ \sum x_j^K\frac{\partial \psi_j}{\partial \eta} & \sum y_j^K\frac{\partial \psi_j}{\partial \eta} \end{bmatrix} \end{align}

where $x^K$ are the interpolation points on the 'physical space' element $K$, and $\psi_i$ are the nodal shape functions on the master element. I then invert $J$ to find $J^{-1}$.

Does this procedure also work for calculating $J^{-1}$ for isoparametric quads? Or does the chain rule expression become more complicated / nonlinear?

EDIT: The idea is to be able to calculate integrals like $$ \int_K \nabla \phi_j \cdot \nabla \phi_i \, dx = \int_K \left( \frac{\partial \phi_i}{\partial x} \frac{\partial \phi_j}{\partial x} + \frac{\partial \phi_i}{\partial y} \frac{\partial \phi_j}{\partial y}\right)\, dx $$


1 Answer 1


The rule is exactly the same. The only difference is that the matrix $J$ still depends on the coordinates $\xi,\eta$ because the functions $\psi_j$ are not linear but bilinear. Other than that, everything remains the same.

  • 1
    $\begingroup$ Of course the functions on "triangles" are also generically nonlinear for orders greater than one, it's just that people tend to use low order triangular meshes. $\endgroup$
    – origimbo
    Apr 13, 2018 at 16:54
  • $\begingroup$ Yes, that is correct. $\endgroup$ Apr 13, 2018 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.