# Two-dimensional Shape Functions

(Difference between revisions)
 Revision as of 10:19, 4 November 2009 (view source)JMora (Talk | contribs)← Older edit Revision as of 10:19, 4 November 2009 (view source)JMora (Talk | contribs) Newer edit → Line 15: Line 15: {| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;" {| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;" ! polynomial order ''n'' !!  number of terms ''p'' !! $f(x,y) \,$ ! polynomial order ''n'' !!  number of terms ''p'' !! $f(x,y) \,$ − |- + |- align="center" | Constant: $0 \,$ || $1 \,$ || $\alpha \,$ | Constant: $0 \,$ || $1 \,$ || $\alpha \,$ − |- + |- align="center" | Linear: $1 \,$ || $3 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 y \,$ | Linear: $1 \,$ || $3 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 y \,$ − |- + |- align="center" | Quadratic: $2 \,$ || $6 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,$ | Quadratic: $2 \,$ || $6 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,$ |} |}

## Revision as of 10:19, 4 November 2009

Shape functions are selected to fit as exact as possible the Finite Element Solution. If this solution is a combination of polynomial functions of nth order, these functions should include a complete polynomial of equal order.

That is, a complete polynomial of nth order can be written as:

$f(x,y)=\sum_{i=1}^p \alpha_i x^j y^k \qquad j+k \le n$

with:   $\qquad p=\frac{(n+1)(n+2)}{2}$   the number of terms.

More specifically:

polynomial order n number of terms p $f(x,y) \,$
Constant: $0 \,$ $1 \,$ $\alpha \,$
Linear: $1 \,$ $3 \,$ $\alpha_1+\alpha_2 x + \alpha_3 y \,$
Quadratic: $2 \,$ $6 \,$ $\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,$

For example, in the case of a lineal polynomial:

f(x,y) = α1 + α2x + α3y

can only fit polynomial functions of pth order if they content a polynomial function

for any polynomial function of pth order it is enough to use p-1 integration points.

## References

1. Pascal's triangle