3.1.2 Numerical Integration

The advantage of the transformation of any element onto a standard element offers the possibility to use standardized integration methods. Thus, a set of sampling points has to be chosen with the aim of best accuracy. To integrate (3.21) a discrete formulation like

$\displaystyle \int_{0}^{1}$ $\displaystyle \int_{0}^{1}$ $\displaystyle \int_{0}^{1}$ $\displaystyle \bf F$ ( $\displaystyle \xi$ , $\displaystyle \eta$ , $\displaystyle \zeta$ ) d $\displaystyle \xi$ d $\displaystyle \eta$ d $\displaystyle \zeta$ = $\displaystyle \sum_{i=1}^{k}$ $\displaystyle \bf F$ ( $\displaystyle \xi_{i}^{}$ , $\displaystyle \eta_{i}^{}$ , $\displaystyle \zeta_{i}^{}$ )^.w_i

(3.22)

can be defined where k is the order of integration, $\xi_{i}^{}$ , $\eta_{i}^{}$ and $\zeta_{i}^{}$ are sampling points or so called Gaussian points within the standard element and w_i are belonging weighting functions.

In case of the previously discussed example the integration of the Laplace operator can be calculated using (3.22)

$\displaystyle \int_{\Omega}^{}$ $\displaystyle \nabla$ $\displaystyle \bf N^{T}_{}$ ( $\displaystyle \xi$ , $\displaystyle \eta$ , $\displaystyle \zeta$ )^. $\displaystyle \nabla$ $\displaystyle \bf N$ ( $\displaystyle \xi$ , $\displaystyle \eta$ , $\displaystyle \zeta$ ) d $\displaystyle \Omega$ = $\displaystyle \sum_{i=1}^{k}$ w_i $\displaystyle \nabla$ $\displaystyle \bf N^{T}_{}$ ( $\displaystyle \xi_{i}^{}$ , $\displaystyle \eta_{i}^{}$ , $\displaystyle \zeta_{i}^{}$ )^. $\displaystyle \nabla$ $\displaystyle \bf N$ ( $\displaystyle \xi_{i}^{}$ , $\displaystyle \eta_{i}^{}$ , $\displaystyle \zeta_{i}^{}$ )

(3.23)

with the values for sampling points and weighting functions taken from Table 3.1/3.2.

Table 3.1: Numerical integration formula for triangles

Order	Error	$\xi$	$\eta$	w
linear	O(h²)	${\frac{1}{3}}$	${\frac{1}{3}}$	${\frac{1}{2}}$
		${\frac{1}{2}}$	0	${\frac{1}{6}}$
quadratic	O(h³)	${\frac{1}{2}}$	${\frac{1}{2}}$	${\frac{1}{6}}$
		0	${\frac{1}{2}}$	${\frac{1}{6}}$
		${\frac{1}{3}}$	${\frac{1}{3}}$	- ${\frac{27}{96}}$
cubic	O(h⁴)	${\frac{1}{5}}$	${\frac{1}{5}}$	${\frac{25}{96}}$
		${\frac{3}{5}}$	${\frac{1}{5}}$	${\frac{25}{96}}$
		${\frac{1}{5}}$	${\frac{3}{5}}$	${\frac{25}{96}}$

Table 3.2: Numerical integration formula for tetrahedrons with $\alpha$ = 0.58541020 and $\beta$ = 0.13819660 [Zie77]

Order	Error	$\xi$	$\eta$	$\zeta$	w
linear	O(h²)	${\frac{1}{4}}$	${\frac{1}{4}}$	${\frac{1}{4}}$	${\frac{1}{2}}$
		$\alpha$	$\alpha$	$\alpha$	${\frac{1}{8}}$
quadratic	O(h³)	$\beta$	$\alpha$	$\alpha$	${\frac{1}{8}}$
		$\alpha$	$\beta$	$\alpha$	${\frac{1}{8}}$
		$\alpha$	$\alpha$	$\beta$	${\frac{1}{8}}$
		${\frac{1}{4}}$	${\frac{1}{4}}$	${\frac{1}{4}}$	- ${\frac{2}{5}}$
		${\frac{1}{6}}$	${\frac{1}{6}}$	${\frac{1}{6}}$	${\frac{9}{40}}$
cubic	O(h⁴)	${\frac{5}{6}}$	${\frac{1}{6}}$	${\frac{1}{6}}$	${\frac{9}{40}}$
		${\frac{1}{6}}$	${\frac{5}{6}}$	${\frac{1}{6}}$	${\frac{9}{40}}$
		${\frac{1}{6}}$	${\frac{1}{6}}$	${\frac{5}{6}}$	${\frac{9}{40}}$

With the numerical integration used to substitute the exact integration, an additional error is introduced and the first impression is, that it should be reduced as much as possible. Therefore it is of interest to determine

It turned out, that the minimum order for the numerical integration must be the same or higher than the order of the interpolation chosen by the shape functions. To preserve the accuracy of the results it is normally enough to increase the integration order by one in comparison to the interpolation order of the shape functions [Sch80][Sch97b].