# Two-dimensional Shape Functions

(Difference between revisions)
 Revision as of 09:49, 4 November 2009 (view source)JMora (Talk | contribs)← Older edit Revision as of 10:17, 4 November 2009 (view source)JMora (Talk | contribs) Newer edit → Line 9: Line 9: with:   $\qquad p=\frac{(n+1)(n+2)}{2}$   the number of terms. 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: + + + :$f(x,y)=\alpha_1+\alpha_2 x + \alpha_3 y$

## Revision as of 10:17, 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:

number of terms p | $f(x,y) \,$
$1 \,$ | $\alpha \,$
$3 \,$ | $\alpha_1+\alpha_2 x + \alpha_3 y \,$
$6 \,$ | $\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,$

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