# Shape function in Finite Element Method

Why is it that the choice of polynomial for 6-nodes rectangular element(linear in sides 1 and 3, quadratic in sides 2 and 4) in FEM does not follow normal pascal triangle regular arrangement? i.e $u=c_1+c_2x+c_3y+c_4xy+c_5xy^2+c_6x^2y$ .

## 1 Answer

See above picture: there are only (linear) line segments in vertical direction. And because of the three nodal points, there are two quadratic elements in horizontal direction. A reference for the latter:

Therefore the interpolation must be linear in $\,y\,$ and quadratic in $\,x\,$. According to a Cartesian product: $$(1,y) \times (1,x,x^2) = (1,x,x^2,y,yx,yx^2)$$ Mind that your term with $y^2$ is absent now and two terms with $x^2$ are present. So the correct interpolation may be formulated as: $$u=c_1+c_2x+c_3y+c_4xy+\color{red}{c_5x^2}+c_6x^2y$$