Numerical Integration exact integration is not needed to achieve the optimal convergence rate of nite element solutions ([, 9, 11], and Chapter 7). In

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Chapter 6 Numerical Integration 6.1 Introduction After transformation to a canonical element,typical integrals in the element stiness or mass matrices (cf. (5.5.8)) have the forms Q = T ( )N s Nt det(j e )dd (6.1.1a) where ( ) depends on the coecients of the partial dierential equation and the transformation to (cf. Section 5.4). The subscripts s and t are either nil,, or implying no dierentiation, dierentiation with respect to, or dierentiation with respect to, respectively. Assuming that N has the form N T =[N 1 N ::: N np ] then (6.1.1a) may be written in the more explicit form Q = ( ) 6 4 (N 1 ) s (N 1 ) t (N 1 ) s (N ) t (N 1 ) s (N np ) t (N ) s (N 1 ) t (N ) s (N ) t (N ) s (N np ) t... (N np ) s (N 1 ) t (N np ) s (N ) t (N np ) s (N np ) t 3 7 5 det(j e)dd: (6.1.1b) (6.1.1c) Integrals of the form (6.1.1b) may be evaluated exactly when the coordinate transformation is linear (J e is constant) and the coecients of the dierential equation are constant (cf. Problem 1 at the end of this section). With certain coecient functions and transformations it may be possible to evaluate (6.1.1b) exactly by symbolic integration however, we'll concentrate on numerical integration because: it can provide exact results in simple situations (e.g., when and J e are constants) and 1

Numerical Integration exact integration is not needed to achieve the optimal convergence rate of nite element solutions ([, 9, 11], and Chapter 7). Integration is often called quadrature in one dimension and cubature in higher dimensions however, we'll refer to all numerical approximations as quadrature rules. We'll consider integrals and quadrature rules of the form I = f( )dd W i f( i i ): (6.1.a) where W i, are the quadrature rule's weights and ( i i ) are the evaluation points, i = 1 ::: n. Of course, we'll want to appraise the accuracy of the approximate integration and this is typically done by indicating those polynomials that are integrated exactly. Denition 6.1.1. The integration rule (6.1.a) is exact to order q if it is exact when f( ) isany polynomial of degree q or less. When the integration rule is exact to order q and f( ) H q+1 ( ), the error satises an estimate of the form E = I ; W i f( i i ) (6.1.b) E Cjjf( )jj q+1 : (6.1.c) Example 6.1.1. Applying (6.1.) to (6.1.1a) yields Q W i ( i i )N( i i )N T ( i i )det(j e ( i i )): Thus, the integrand at the evaluation points is summed relative to the weights to approximate the given integral. Problems 1. A typical term of an element stiness or mass matrix has the form i j dd i j : Evaluate this integral when is the canonical square [ 1] [ 1] and the canonical right 45 unit triangle.

6.. One-Dimensional Quadrature 3 6. One-Dimensional Gaussian Quadrature Although we are primarily interested in two- and three-dimensional quadrature rules, we'll set the stage by studying one-dimensional integration. Thus, consider the onedimensional equivalent of(6.1.) on the canonical [ 1] element I = f()d = W i f( i )+E: (6..1) Most classical quadrature rules have this form. For example,the trapezoidal rule I f() + f(1) has the form (6..1) with n =,W 1 = W =1,; 1 = =1,and Similarly, Simpson's rule E = f () ; ( 1): 3 I 1 [f() + 4f() + f(1)] 3 has the form (6..1) with n =3,W 1 = W =4=W 3 =1=3, ; 1 = 3 =1, =,and E = f (iv) () ; ( 1): 9 Gaussian quadrature is preferred to these Newton-Cotes formulas for nite element applications because they have fewer function evaluations for a given order. With Gaussian quadrature, the weights and evaluation points are determined so that the integration rule is exact (E = ) to as high an order as possible. Since there are n unknown weights and evaluation points, we expect to be able to make (6..1) exact to order n ; 1. This problem has been solved [3, 6] and the evaluation points i, i =1 ::: n, are the roots of the Legendre polynomial of degree n (cf. Section.5). The weights W i, i =1 ::: n, called Christoel weights, are also known and are tabulated with the evaluation points in Table 6..1 for n ranging from 1 to 6. A more complete set of values appear in Abromowitz and Stegun [1]. Example 6..1. The derivation of the two-point (n = ) Gauss quadrature rule is given as Problem 1 at the end of this section. From Table 6..1 we see that W 1 = W =1 and ; 1 = =1= p 3. Thus, the quadrature rule is f()d f(= p 3) + f(1= p 3): This formula is exact to order three thus the error is proportional to the fourth derivative of f (cf. Theorem 6..1, Example 6..4, and Problem at the end of this section).

4 Numerical Integration n i W i 1...57735 691 8966 1. 3..88888 88888 88889.77459 6669 41483.55555 55555 55556 4.33998 1435 84856.6514 51548 6546.86113 63115 9453.34785 48451 37454 5..56888 88888 88889.53846 9311 5683.4786 8674 99366.9617 98459 38664.369 6885 56189 6.3861 9186 83197.46791 39345 7691.661 93864 6665.3676 1573 48139.9346 9514 315.1713 4493 7917 Table 6..1: Christoel weights W i and roots i, i =1 ::: n, for Legendre polynomials of degrees 1to6[1]. Example 6... Consider evaluating the integral I = e ;x dx = p erf(1) = :74684138143 (6..) by Gauss quadrature. Let us transform the integral to [ 1] using the mapping to get I = 1 The two-point Gaussian approximation is =x ; 1 I ~I = 1 p 1= 3 [e;( 1+ ;( e ) d: p ) + e ;( 1+1= 3 ) ]: Other approximations follow in similar order. Errors I ; ~I when I is approximated by Gaussian quadrature to obtain ~I appear in Table 6.. for n ranging from 1 to 6. Results using the trapezoidal and Simpson's rules are also presented. The two- and three-point Gaussian rules have higher orders than the corresponding Newton-Cotes formulas and this leads to smaller errors for this example.

6.. One-Dimensional Quadrature 5 n Gauss Rules Newton Rules Error Error 1 3.198(- ) -.94(- 4) -6.88(- ) 3-9.549(- 6) 3.563(- 4) 4 3.353(- 7) 5-6.46(- 9) 6 7.77(-11) Table 6..: Errors in approximating the integral of Example 6.. by Gauss quadrature, the trapezoidal rule (n =, right) and Simpson's rule (n = 3, right). Numbers in parentheses indicate a power of ten. Example 6..3. Composite integration formulas, where the domain of integration [a b] is divided into N subintervals of width x j = x j ; x j j =1 ::: N are not needed in nite element applications, except, perhaps, for postprocessing. However, let us do an example to illustrate the convergence of a Gaussian quadrature formula. Thus, consider where I = Z b a I j = f(x)dx = Z xj x j The linear mapping 1 ; x = x j transforms [x j x j ] to [ 1] and j=1 f(x)dx: I j + x j 1+ I j = x j f(x j 1 ; + x j 1+ )d: Approximating I j by Gauss quadrature gives I j x j W i f(x j 1 ; i 1+ i + x j ): We'll approximate (6..) using composite two-point Gauss quadrature thus, I j = x j [e;(x j=;x j =( p 3)) + e ;(x j=+x j =( p 3)) ]

6 Numerical Integration where x j= = (x j + x j )=. Assuming a uniform partition with x j = 1=N, j = 1 ::: N, the composite two-point Gauss rule becomes I 1 N j=1 The composite Simpson's rule, [e ;(x j==(n p 3)) + e ;(x j=+1=(n p 3)) ]: I 1 N X 3N [1 + 4 3 e ;x j + N; X i= 4 e ;x j + e ] on N= subintervals of width x has an advantage relative to the composite Gauss rule since the function evaluations at the even-indexed points combine. The number of function evaluations and errors when (6..) is solved by the composite two-point Gauss and Simpson's rules are recorded in Table 6..3. We can see that both quadrature rules are converging as O(1=N 4 ) ([6], Chapter 7). The computations were done in single precision arithmetic as opposed to those appearing in Table 6.., which were done in double precision. With single precision, round-o error dominates the computation as N increases beyond 16 and further reductions of the error are impossible. With function evaluations dened as the number of times that the exponential is evaluated, errors for the same number of function evaluations are comparable for Gauss and Simpson's rule quadrature. As noted earlier, this is due to the combination of function evaluations at the ends of even subintervals. Discontinuous solution derivatives at interelement boundaries would prevent such a combination with nite element applications. N Gauss Rules Simpson's Rule Fn. Eval. Abs. Error Fn. Eval. Abs. Error 4.8(- 4) 3.356(- 3) 4 8.161(- 5) 5.39(- 4) 8 16.358(- 6) 9.137(- 5) 16 3.364(- 5) 17.44( -5) Table 6..3: Comparison of composite two-point Gauss and Simpson's rule approximations for Example 6..3. The absolute error is the magnitude of the dierence between the exact and computational result. The number of times that the exponential function is evaluated is used as a measure of computational eort. As we may guess, estimates of errors for Gauss quadrature use the properties of Legendre polynomials (cf. Section.5). Here is a typical result.

6.. One-Dimensional Quadrature 7 Theorem 6..1. Let f() C n [ 1], then the quadrature rule (6..1) is exact to order n ; 1 if i, i = 1 ::: n, are the roots of P n (), the nth-degree Legendre polynomial, and the corresponding Christoel weights satisfy W i = 1 P n( i ) Additionally, there exists a point ( 1) such that Proof. cf. [6], Sections 7.3, 4. E = f (n) () n! P n () ; i d i =1 ::: n: (6..3a) ny ( ; i ) d: (6..3b) Example 6..4. Using the entries in Table 6..1 and (6..3b), the discretization error of the two-point (n = )Gauss quadrature rule is E = f iv () 4! ( + p 1 ) ( ; p 1 ) d = f iv () ( 1): 3 3 135 Problems 1. Calculate the weights W 1 and W and the evaluation points 1 and so that the two-point Gauss quadrature rule f(x) W 1 f( 1 )+W f( ) is exact to as high an order as possible. This should be done by a direct calculation without using the properties of Legendre polynomials.. Lacking the precise information of Theorem 6..1, we may infer that the error in the two-point Gauss quadrature rule is proportional to the fourth derivative of f() since cubic polynomials are integrated exactly. Thus, E = Cf iv () ( 1): We can determine the error coecient C by evaluating the formula for any function f(x) whose fourth derivative does not depend on the location of the unknown point. In particular, any quartic polynomial has a constant fourth derivative hence, the value of is irrelevant. Select an appropriate quartic polynomial and show that C =1=135 as in Example 6..4.

8 Numerical Integration 6.3 Multi-Dimensional Quadrature Integration on square elements usually relies on tensor products of the one-dimensional formulas illustrated in Section 6.. Thus, the application of (6..1) to a two-dimensional integral on a canonical [ 1] [ 1] square element yields the approximation I = f( )dd W i f( i )d = W i f( i )d and I = f( )dd j=1 W i W j f( i j ): (6.3.1) Error estimates follow the one-dimensional analysis. Tensor-product formulas are not optimal in the sense of using the fewest function evaluations for a given order. Exact integration of a quintic polynomial by (6.3.1) would require n = 3 or a total of 9 points. A complete quintic polynomial in two dimensions has 1 monomial terms thus, a direct (non-tensor-product) formula of the form I = f( )dd W i f( i i ) could be made exact with only 7 points. The 1 coecients W i, i, i, i = 1 ::: 7, could potentially be determined to exactly integrate all of the monomial terms. Non-tensor-product formulas are complicated to derive and are not known to very high orders. Orthogonal polynomials, as described in Section 6., are unknown in two and three dimensions. Quadrature rules are generally derived by a method of undetermined coecients. We'll illustrate this approach by considering an integral on a canonical right 45 triangle I = f( )dd = Example 6.3.1. Consider the one-point quadrature rule W i f( i i )+E: (6.3.) f( )dd = W 1 f( 1 1 )+E: (6.3.3) Since there are three unknowns W 1, 1,and 1,we expect (6.3.3) to be exact for any linear polynomial. Integration is a linear operator hence, it suces to ensure that (6.3.3) is exact for the monomials 1,, and. Thus,

6.3. Multi-Dimensional Quadrature 9 If f( ) =1: ; (1)dd = 1 = W 1: If f( ) =: ; ()dd = 1 6 = W 1 1 : If f( ) =: ; ()dd = 1 6 = W 1 1 : The solution of this system is W 1 =1=and 1 = 1 =1=3 thus, the one-point quadrature rule is f( )dd = 1 f(1 3 1 )+E: (6.3.4) 3 As expected, the optimal evaluation point is the centroid of the triangle. A bound on the error E may be obtained by expanding f( ) in a Taylor's series about some convenient point ( ) to obtain f( ) =p 1 ( )+R 1 ( ) (6.3.5a) where and p 1 ( ) =f( )+[( ; ) @ @ +( ; ) @ @ ]f( ) (6.3.5b) R 1 ( ) = 1 [( ; ) @ @ +( ; ) @ @ ] f(!) (!) : (6.3.5c) Integrating (6.3.5a) using (6.3.4) E = [p 1 ( )+R 1 ( )]dd ; Since (6.3.4) is exact for linear polynomials E = R 1 ( )dd ; 1 [p 1( 1 3 1 3 )+R 1( 1 3 1 3 )]: 1 R 1( 1 3 1 3 ): Not being too precise, we take an absolute value of the above expression to obtain jej jr 1 ( )jdd + 1 jr 1( 1 3 1 3 )j:

1 Numerical Integration For the canonical element, j ; j1andj ; j1 hence, jr 1 ( )j max jjd f jj 1 jj= where Since the area of is 1=, jjf jj 1 = max jf( )j: ( ) jej max jjd f jj 1 : (6.3.6) jj= Errors for other quadrature formulas follow the same derivation ([6], Section 7.7). Two-dimensional integrals on triangles are conveniently expressed in terms of triangular coordinates as e f(x y)dxdy = A e W i f( i 1 i i 3)+E (6.3.7) where ( i 1 i i 3) are the triangular coordinates of evaluation point i and A e is the area of triangle e. Symmetric quadrature formulas for triangles have appeared in several places. Hammer et al. [5] developed formulas on triangles, tetrahedra, and cones. Dunavant [4] presents formulas on triangles which are exact to order however, some formulas have evaluation points that are outside of the triangle. Sylvester [1] developed tensor-product formulas for triangles. We have listed some quadrature rules in Table 6.3.1 that also appear in Dunavant [4], Strang and Fix [9], and Zienkiewicz [1]. A multiplication factor M indicates the number of permutations associated with an evaluation point having a weight W i. The factor M = 1 is associated with an evaluation point at the triangle's centroid (1=3 1=3 1=3), M = 3 indicates a point onamedian line, and M =6indicates an arbitrary point intheinterior. The factor p indicates the order of the quadrature rule thus, E = O(h p+1 ) where h is the maximum edge length of the triangle. Example 6.3.. Using the data in Table 6.3.1 with (6.3.7), the three-point quadrature rule on the canonical triangle is f( )dd = 1 [f(=3 1=6 1=6) + f(1=6 1=6 =3) + f(1=6 =3 1=6)] + E: 6 The multiplicative factor of 1/6 arises because the area of the canonical element is 1/ and all of the weights are 1/3. The quadrature rule can be written in terms of the canonical variables by setting = and 3 = (cf. (4..6) and (4..7)). The discretization error associated with this quadrature rule is O(h 3 ).

6.3. Multi-Dimensional Quadrature 11 n W i i 1 i i 3 M p 1 1..333333333333333.333333333333333 1.333333333333333 1 3.333333333333333.666666666666667.166666666666667.166666666666667 3 4 -.565.333333333333333.333333333333333 3.333333333333333 1.5833333333333.6.. 3 6.199517436553.8168475798459.915761359771 4.915761359771 3.338158967811.1813181687.44594849915965.44594849915965 3 7.5.333333333333333.333333333333333 5.333333333333333 1.159391854487.7974698535387.11865733456.11865733456 3.133941578856.5971587178977.47146415115.47146415115 3 1.584496377.8738197116996.6389144915 6.6389144915 3.1167867576379.514659658179.49867451791.49867451791 3.885175618374.636549911399.313545133785.5314549844816 6 13 -.149574446767.333333333333333.333333333333333 7.333333333333333 1.175615574334.479386784193.63459667938.63459667938 3.533473568839.869739794195568.65131916.65131916 3.77113768957.63844418856989.318654964875.486931545316 6 Table 6.3.1: Weights and evaluation points for integration on triangles [4].

1 Numerical Integration Quadrature rules on tetrahedra have the form Z e f(x y z)dxdydz = V e W i f( i 1 i i 3 i 4)+E (6.3.8) where V e is the volume of Element e and ( i 1 i i 3 i 4) are the tetrahedral coordinates of evaluation point i. Quadrature rules are presented by Jinyun [7] for methods to order six and by Keast [8] for methods to order eight. Multiplicative factors are such that M =1for an evaluation point at the centroid (1=4 1=4 1=4 1=4), M =4for points on the median line through the centroid and one vertex, M = 6 for points on a line between opposite midsides, M = 1 for points in the plane containing an edge an and opposite midside, and M = 4 for points in the interior (Figure 6.3.1). n W i 1 i 3 4 M p 1 1..5.5 1.5.5 1 4.5.5854119664969.138196611511.138196611511.138196611511 4 5 -.8.5.5 3.5.5 1.45.5.166666666666667.166666666666667.166666666666667 4 11 -.13155555555556.5.5 4.5.5 1.76.7857148571486.714857148571.714857148571.714857148571 4.4888888888889.39943576166799.39943576166799.1596438331.1596438331 6 15.3836789789.5.5 5.5.5 1.6678571486..333333333333333.333333333333333.333333333333333 4.1164549869.77777777.99999991.99999991.99999991 4.1949141561386.6655153573664.6655153573664.43344984646336.43344984646336 6 Table 6.3.: Weights and evaluation points for integration on tetrahedra [7, 8].

6.3. Multi-Dimensional Quadrature 13 1 Q 1 4 1 11 11 111 1111 11111 11111 111111Q 34 1111111 1111111 111111111 11111111 11111111 P C 1111111111 14 1111111111 1111111111 1 1111111111 111111111 1111111 11111 111 Figure 6.3.1: Some symmetries associated with the tetrahedral quadrature rules of Table 6.3.. An evaluation point with M = 1 is at the centroid (C), one with M = 4 is on a line through a vertex and the centroid (e.g., line 3 ; P 134 ), one with M =6is on a line between two midsides (e.g., lineq 1 ; Q 34 ), and one with M =1isina plane through two vertices and an opposite midside (e.g., plane 3 ; 4 ; Q 1 ) 3 Problems 1. Derive a three-point Gauss quadrature rule on the canonical right 45 triangle that is accurate to order two. In order to simplify the derivation, use symmetry arguments to conclude that the three points have the same weight and that they are symmetrically disposed on the medians of the triangle. Show that there are two possible formulas: the one given in Table 6.3.1 and another one. Find both formulas.. Show that the mapping = 1+u (1 ; u)(1 + v) = 4 transforms the integral (6.3.) from the triangle to one on the square u v 1. Find the resulting integral and show how to approximate it using a tensor-product formula.

14 Numerical Integration

Bibliography [1] M. Abromowitz and I.A. Stegun. Handbook of Mathematical Functions, volume 55 of Applied Mathematics Series. National Bureau of Standards, Gathersburg, 1964. [] S.C. Brenner and L.R. Scott. The Mathematical Theory of Finite Element Methods. Springer-Verlag, New York, 1994. [3] R.L. Burden and J.D. Faires. Numerical Analysis. PWS-Kent, Boston, fth edition, 1993. [4] D.A. Dunavant. High degree ecient symmetrical Gaussian quadrature rules for the triangle. International Journal of Numerical Methods in Engineering, 1:119{1148, 1985. [5] P.C. Hammer, O.P. Marlowe, and A.H. Stroud. Numerical integration over simplexes and cones. Mathematical Tables and other Aids to Computation, 1:13{137, 1956. [6] E. Isaacson and H.B. Keller. Analysis of Numerical Methods. John Wiley and Sons, New York, 1966. [7] Y. Jinyun. Symmetric Gaussian quadrature formulae for tetrahedronal regions. Computer Methods in Applied Mechanics and Engineering, 43:349{353, 1984. [8] P. Keast. Moderate-degree tetrahedral quadrature formulas. Computer Methods in Applied Mechanics and Engineering, 55:339{348, 1986. [9] G. Strang and G. Fix. Analysis of the Finite Element Method. Prentice-Hall, Englewood Clis, 1973. [1] P. Sylvester. Symmetric quadrature formulae for simplexes. Mathematics of Computation, 4:95{1, 197. [11] R. Wait and A.R. Mitchell. The Finite Element Analysis and Applications. John Wiley and Sons, Chichester, 1985. 15

16 Numerical Integration [1] O.C. Zienkiewicz. The Finite Element Method. McGraw-Hill, New York, third edition, 1977.