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My last question was about the condition number of the stiffness matrix. This question is about discretization error (which is probably more important to most people). The mesh on the right side of the figure below (source) produced better discretization error than the "perfect" quality mesh on the left side even though the mesh on the right side has quads with angles close to 180 degrees.

Burgers' Equation on regular and irregular grids

I assume the creator of the figure (professor Christopher Roy) used a finite volume or finite difference method since he is a fluid dynamics specialist. Based on your personal experience using FEM, could FEM get a good discretization error using a mesh like the one on the right side of the figure?

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  • $\begingroup$ Am I interpreting the figure labels accurately in thinking that this is solving moderate Reynolds number 2d Burgers' equation? If so, a lot of analysis based on elliptic equations can fail to translate. $\endgroup$ – origimbo Aug 15 '16 at 10:20
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    $\begingroup$ @origimbo, that being said, if the 180 degree elements have a zero Jacobian, the FEM is going to fail. I don't have my memory or books handy, but I'm going to suggest that these elements do have a zero Jacobain and should be modified so that they aren't so triangular. I'll try to make an answer later. Also, not all fluids people demand to use FVM or FDM. I do CFD and use the FEM. $\endgroup$ – Bill Barth Aug 15 '16 at 13:10
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I think, but I don't have the proof in front of me, that if you use a bilinear, Cartesian product element and the isoparametric map from the master element to the real element, then the Jacobian of the map will be zero at the flat corner on these elements. That might actually be OK if you never evaluate anything there. If you use Gaussian quadrature, then you may get away with only evaluating things in the interior at the element's Gauss points. If you use a Newton-Cotes rule that wants to do integration at the nodes of the element, then your ability to integrate at all is shot. So, the answer is probably "it depends". In the Gaussian case, you may also lose some accuracy even if you don't evaluate anything.

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    $\begingroup$ For the bilinear mapping from (0,0), (1,0), (0,1) to themselves and (1,1) to (1/2,1/2), the Jacobian is indeed zero at (1,1). I checked this yesterday and this made me curious how FEM would work on the mesh in the figure. $\endgroup$ – J. Heller Aug 16 '16 at 2:05
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There are multiple problems with the mesh on the right, if it comes to finite elements.

The elements are quite distorted, which will cause problems. Reason for that is that the derivatives calculated with the help of isoparametric elements will lose accuracy the more the elements are distorted.

Like it was mentioned before the quadrilateral elements which now have a flat corner will cause the determinant of the Jacobian to be zero in the element. The isoparametric transformation which requires the inverse of the Jacobian would fail.

It's highly unlikely that a finite element calculation with this mesh would be successful.

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