I am trying to understand the finite element method and want to apply it to a 2D equation with a triangular mesh.

I have chosen the reference element to be the triangle with vertices $(0, 0), (0, 1)\text{ and }(1, 0)$. On this reference element, I define the three basis functions $$\phi_1(x, y) = (1 - x)(1 - y) \quad \phi_2(x, y) = x(1 - y) \quad \phi_3(x,y) = (1 - x)y$$ where each of them is $1$ at exactly one of the vertices and $0$ at the others. Now, when I try to find the elements of the mass matrix, I need to solve $$M(i, j) = \int \phi_i \phi_j \mathrm{d}x$$ for some $i, j \in 1..3$ on the reference element. The problem I have now is that $M(1, 2) \ne M(2, 3)$. Is that right? Why would you not choose the reference element such that the corners are interchangeable, for example an equilateral triangle centered on the origin?

To me it looks like this way some of the elements are just arbitrarily chosen to "interact" less with one another just based on the choice of how we identify the the corners of the domain triangle with the reference element.

  • 1
    $\begingroup$ Your last two corners would be interchangeable, but that's not the case with the one in the origin. Also, your basis functions are not linear, and that's intriguing. $\endgroup$ – nicoguaro Oct 29 '20 at 22:32
  • $\begingroup$ Thank you for noticing this. I extended a 1D basis constructed in the same way to 2D and thought that bilinear functions would be fine. But I will go with a linear basis then. $\endgroup$ – Marten Oct 30 '20 at 9:42
  • $\begingroup$ Also, with linear basis functions $\phi1(x, y) = 1 - x - y$, $\phi_2(x, y) = x$ and $\phi_3(x, y) = y$ I have $M(1, 2) = M(2, 3)$. $\endgroup$ – Marten Oct 30 '20 at 9:48

You could choose the equilateral triangle as the reference element. But the exact shape of the reference element doesn't actually matter: You have to integrate over the concrete elements of your triangulation. Between the reference element and the actual element is the mapping that does the transformation from one to the other. If you choose the reference element differently, the mapping would also be different, but the matrix elements you compute on actual cells would be the same.

So you "could". But you shouldn't: The reference element should be chosen in such a way that it is easy to write down the basis functions and the mapping, and that happens to be especially so if you choose as reference the one that goes from zero to one for both the $x$ and the $y$ coordinates.


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