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In FEM, with a triangular mesh over $R^2$, could $\phi\left(x\right)=x_1\cdot\left[x\in T\right]$ be a basis function for the triangle $T$ with vertices in $\left(0,0\right), \left(0,1\right), \left(1,0\right)$? My doubts come from the fact it is not contiuous, $\phi\left(\left(1,0\right)\right)=1$ but $\phi\left(\left(1+\epsilon,0\right)\right)=0\ for\ \epsilon>0$.

Edit: basis of the trial space.

Edit: I forgot to add I want piecewise linear trial space, so the the question pretty much is about a convenient basis for it.

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What does the notation $x_1\cdot\left[x\in T\right]$ mean? What is $x_1$? What does the bracket mean? Are you taking a dot product? – user2468 Mar 4 '12 at 21:55
It means $x_1$ when $x\in T$ and $0$ otherwise. – Adrian Panasiuk Mar 4 '12 at 21:58
@J.D.: I suspect it's the Iverson bracket. – joriki Mar 4 '12 at 21:58
Wouldn't be a more suitable place for this question? – user2468 Mar 4 '12 at 22:02
Here is a nice pdf for FEM triangle basis functions. It might be helpful:… – chemeng Mar 4 '12 at 22:09

1 Answer 1

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Well l I guess not, $\phi$ may be like that on triangle T, but (with the mesh consisting of triangles with horizontal, vertical and slope=-1 edges) in the other five mesh triangles which have $\left(0,0\right)$ as a vertex, $\phi$ (I'm talking piecewise linear here) should be 1 at $\left(0,0\right)$ and disappear at the edge that is opposite to (0,0).

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