# Symmetry in P1 basis elements on a reference triangle in 2D-FEM

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.

• 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. Oct 29, 2020 at 22:32
• 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. Oct 30, 2020 at 9:42
• Also, with linear basis functions $\phi1(x, y) = 1 - x - y$, $\phi_2(x, y) = x$ and $\phi_3(x, y) = y$ I have $M(1, 2) = M(2, 3)$. Oct 30, 2020 at 9:48

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.