When implementing FEM for solving two-dimensional partial differential equations, integrals of the form
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1 Quadrature Formulas in Two Dimensions Math 57 - Finite Element Method Section, Spring Shaozhong Deng, PhD (shaodeng@unccedu Dept of Mathematics and Statistics, UNC at Charlotte When implementing FEM for solving two-dimensional partial differential equations, integrals of the form I = F (x, y dxdy have to be evaluated to obtain local stiffness matrices and local load vectors, where usually is an either quadrilateral or triangular element Because the integrand generally depends on user-specified information, such as p(x, y, q(x, y and f(x, y in the second-order self-adjoint elliptic equation: (p(x, y u + q(x, yu = f(x, y, (x, y Ω, in computer programs the analogous integrals shall be evaluated numerically A Recall - Gaussian quadrature in one dimension ( Gaussian quadrature of order N for the standard interval I st = [, ]: g(ξ dξ N w i g(ξ i, ( where ξ i and w i are Gaussian quadrature points and weights Remember that a Gaussian quadrature using N points can provide the exact integral if g(ξ is a polynomial of degree N or less (i Gaussian quadrature of order (one point: g(ξ dξ g( (ii Gaussian quadrature of order (two points: ( g(ξ dξ g ( + g (iii Gaussian quadrature of order (three points: g(ξ dξ 5 ( 9 g 5 ( g ( g 5 ( Gaussian quadrature for general intervals I = [a, b]: Therefore, we have b a x = a + b a x a ( + ξ, or ξ = b a + x b b a F (x dx = b a b a F (x dx b a ( F N w i F a + b a ( + ξ dξ ( a + b a ( + ξ i
2 B Gaussian quadrature for quadrilateral elements ( Gaussian quadrature for the standard quadrilateral element R st = [, ] : (, η (, R st ξ (, (, Figure : The standard quadrilateral element R st I = g(ξ, η dξdη = R st g(ξ, η dξdη For any fixed η, we can integrate numerically with respect to ξ: ( M g(ξ, η dξdη w i g(ξ i, η dη, where ξ i and w i are Gaussian quadrature points and weights of order M in the ξ direction Next integrating numerically with respect to η we have g(ξ, η dξdη M j= N w i ŵ j g(ξ i, η j, where η j and ŵ j are Gaussian quadrature points and weights of order N in the η direction This integration rule is exact if the integrand g(ξ, η contains only the monomials ξ i η j (i =,,, M, j =,,, N Usually M = N (so η i = ξ i, ŵ i = w i, and we have Gaussian quadrature of order N for the standard quadrilateral element g(ξ, η dξdη j= N N w i w j g(ξ i, ξ j ( Example: Gaussian quadrature of order for the standard quadrilateral element R st = [, ] : ( g(ξ, η dξdη g g g, 5 5 (, 5 (, 5 ( g g (, ( g, (, g, 5 5 (, 5 9 g g (, 5 5
3 Figure : Gaussian quadrature points (N = for the standard quadrilateral element R st = [, ] ( Gaussian quadrature for general quadrilateral elements: Let be a quadrilateral element with straight boundary lines and vertices (x i, y i, i =,,, 4 arranged in the counter-clockwise order: y P =(x, y P 4 =(x 4, y 4 x P =(x, y P =(x, y Figure : A quadrilateral element with straight boundary lines We would like to evaluate I = F (x, y dxdy The idea is simple: first transform the quadrilateral element to the standard quadrilateral element R st and then apply the Gaussian quadrature ( How to transform: Using nodal shape functions! Nodal shape functions for quadrilaterals: N (ξ, η = ( ξ( η, 4 (, η (, N (ξ, η = ( + ξ( η, 4 R st N (ξ, η = ( + ξ( + η, 4 ξ N 4 (ξ, η = ( ξ( + η 4 Note N i (ξ, η = at Node i ; at other nodes (, (,
4 Construct a linear mapping to map the quadrilateral element with straight boundary lines to the standard quadrilateral element R st : (x 4, y 4 (x, y y (x, y x (x, y x=p(ξ,η y=q(ξ,η (, R st (, η (, ξ (, Figure 4: Linear mapping between and R st The mapping can be achieved conveniently by using the nodal shape functions as follows: Then we have x = P (ξ, η = y = Q(ξ, η = 4 x i N i (ξ, η = x N (ξ, η + x N (ξ, η + x N (ξ, η + x 4 N 4 (ξ, η, 4 y i N i (ξ, η = y N (ξ, η + y N (ξ, η + y N (ξ, η + y 4 N 4 (ξ, η F (x, y dxdy = F (P (ξ, η, Q(ξ, η J(ξ, η dξdη, R st where J(ξ, η is the Jacobian of the transformation defined by x J(ξ, η = (x, y (ξ, η = ξ x η y ξ y η Applying the Gaussian quadrature ( for the standard quadrilateral element yields Gaussian quadrature of order N for general quadrilateral elements N N F (x, y dxdy w i w j F (P (ξ i, ξ j, Q(ξ i, ξ j J(ξ i, ξ j j= 4
5 C Gaussian quadrature for triangular elements ( Gaussian quadrature for the standard triangular element = {(ξ, η : ξ, η, ξ + η }: (, η (, (, ξ Figure 5: The standard triangular element I = g(ξ, η dξdη = η g(ξ, η dξdη = ξ (a Tensor product-type Gaussian quadrature - Simple but less efficient: g(ξ, η dηdξ Idea: Transform the standard triangular element to the standard quadrilateral element R st, and then apply the Gaussian quadrature for R st Such transformation is defined by ( + a ( b ξ =, 4 η = + b or, a = ξ η, b = η ( b η C a T A B ξ Figure 6: Illustration of the mapping between the square R st and the triangle The transformations defined in ( basically collapse the top edge of the square R st into the top vertex (, of the triangle The Jacobians of the transformations are J(a, b = (ξ, η (a, b = b = η 8 4, J (ξ, η = (a, b (ξ, η = 4 η Therefore, we have I = g(ξ, η dξdη = g(ξ(a, b, η(a, b J(a, b dadb R st 5
6 Remark : Note that one direction of the transformation defined in ( is linear and the corresponding Jacobian is bounded However, the other one is non-linear, and its Jacobian has a singularity Fortunately, this Jacobian is NOT used! Remark : Tensor product-type quadrature have several advantages In particular, their derivation and application is straightforward They are versatile in that many one-dimensional rules are available for several different integrands Extremely high-order polynomials may be evaluated, although precision may be limited since most references provide points and weights to significant digits at most The primary disadvantage is inefficiency since for high N, a relatively large number of points is required, and other quadrature rules are available with many fewer points The secondary disadvantage is that the location of the points is unsymmetrical Except for rules of low order, a large number of points will be concentrated in a relatively small region near one vertex (the top (, vertex Such an arrangement, although correct, maybe considered aesthetically undesirable For details, please read Ref [] (b Symmetrical Gaussian quadrature on triangles: Goal: The goal is to develop quadrature rules of the form g(ξ, η dξdη N g w i g(ξ i, η i, (4 where is the number of quadrature points, (ξ i, η i are quadrature points located inside the standard triangle and w i are weights (normalized with respect to the triangle area In addition to the criteria that the resulting quadrature should use as less as possible number of quadrature points to achieve as high as possible accuracy, we also would like the quadrature points possess some kind of symmetry Gaussian quadrature of degree N for triangles: The basic idea is the same as that used in developing Gaussian quadrature for the standard interval I st = [, ] We want to choose points (ξ i, η i and weights w i in (4 so that the quadrature (4 is as accurate as possible in some sense Generally speaking, a Gaussian quadrature of degree N for triangles is defined as a quadrature of (4 that is exact for arbitrary polynomial of degree N, namely, g(ξ, η dξdη = N g w i g(ξ i, η i, g(ξ, η P N (ξ, η, where P N (ξ, η represents the complete polynomial space of degree N in two dimensions For examples, P N (ξ, η = span { ξ i η j, i, j, i + j N } P (ξ, η = span {, ξ, η}, P (ξ, η = span {, ξ, η, ξ, ξη, η } Note that dim(p N = (N + (N + A useful identity: Therefore, we have ξ i η j dξdη = {, ξ, η, ξ, ξη, η, ξ, ξ η, ξη, η } dξdη = i!j! (i + j +! {, 6, 6,, 4,,, 6, 6, } 6
7 Gaussian quadrature of degree for triangles: By definition, the quadrature should be accurate for g(ξ, η =, ξ, and η We get g(ξ, η = = w i g(ξ, η = ξ 6 = N g w i ξ i g(ξ, η = η 6 = w i η i It is easy to see that =, w = and ξ = η = is a solution Therefore, we have Gaussian quadrature of degree for the standard triangle g(ξ, η dξdη = g Gaussian quadrature of degree for triangles: (, By definition, the quadrature should be accurate for g(ξ, η =, ξ, η, ξ, ξη, and η We get g(ξ, η = = N g w i g(ξ, η = ξ 6 = w i ξ i g(ξ, η = η 6 = w i η i g(ξ, η = ξ = g(ξ, η = ξη 4 = g(ξ, η = η = w i ξi w i ξ i η i w i ηi It is obvious that = will NOT work since we have 6 equations In theory, = may work since in this case we are going to have 6 unknowns (ξ, η, ξ, η, w, w, but the resulting quadrature is NOT symmetric! So we choose = Using = (three points, we have 9 unknowns Because there are only 6 equations, generally speaking the solution is not unique We can verify that (ξ, η = ( 6, 6 is a solution Therefore, we have, (ξ, η = (, 6, (ξ, η = ( 6,, w = w = w = Gaussian quadrature of degree for the standard triangle g(ξ, η dξdη = 6 [ ( g 6, ( + g 6, ( + g 6 6, ] 7
8 Remember that the solution is NOT unique! For example, we can also verify that ( (ξ, η =, ( (, (ξ, η =,, (ξ, η =,, w = w = w = is also a solution Therefore, we have another Gaussian quadrature of degree for the standard triangle g(ξ, η dξdη = 6 Gaussian quadrature of degree for triangles: [ ( g, ( ( + g, + g, ] By definition, the quadrature shall be accurate for g(ξ, η =, ξ, η, ξ, ξη, η, ξ, ξ η, ξη, and η So g(ξ, η = = N g w i g(ξ, η = ξ 6 = w i ξ i g(ξ, η = η 6 = w i η i g(ξ, η = ξ = g(ξ, η = ξη 4 = g(ξ, η = η = g(ξ, η = ξ = g(ξ, η = ξ η 6 = g(ξ, η = ξη 6 = g(ξ, η = η = w i ξi w i ξ i η i w i ηi w i ξi w i ξi η i w i ξ i ηi w i ηi It is obvious that = will NOT work since we have equations So we choose = 4 Using = 4 (four points, we have unknowns Because there are only equations, again generally speaking the solution is not unique We can verify that (ξ, η = (, is a solution Therefore, we have, (ξ, η = ( 5, 5, (ξ, η = w = 7, w = w = w 4 = 5 ( 5, (, (ξ 4, η 4 = 5 5,, 5 8
9 Gaussian quadrature of degree for the standard triangle g(ξ, η dξdη = 7 96 g (, + 5 [ ( g 96 5, ( + g 5 5, ( + g 5 5, ] 5 Again recall that the solution is NOT unique! For example, we can verify that ( (ξ, η =, (, (ξ, η = 5, (, (ξ, η = 5 5, (, (ξ 4, η 4 = 5 5,, 5 w = 7, w = w = w 4 = 5 is also a solution Therefore, we have another Gaussian quadrature of degree for the standard triangle g(ξ, η dξdη = 7 96 g Graphical illustration of quadrature points: (, + 5 [ ( g 96 5, ( + g 5 5, ( + g 5 5, ] 5 * c * b a * a * a * b * c * d (a Linear (b Quadratic (c Cubic a = (, (, w = a = 6, 6, w = a = (,, w = 7 b = (, 6, w = b = ( 5, 5, w = 5 c = ( 6,, w = c = ( 5, 5, w = 5 d = ( 5, 5, w = 5 Figure 7: First set of quadrature rules for triangular elements b * a* c * a a * b * c * d * (a Linear (b Quadratic (c Cubic a = (, (, w = a =,, w = a = (,, w = 7 b = (,, w = b = ( 5, 5, w = 5 c = (,, w = c = ( 5, 5, w = 5 d = ( 5, 5, w = 5 Figure 8: Second set of quadrature rules for triangular elements 9
10 Number of quadrature points : The most commonly referenced Gauss-Legendre locations and weights for triangles are the symmetric quadrature rules of [] This reference provides tables varying from degrees to (79 quadrature points, which are reproduced (for degrees between and in DTriGaussPointsm The weights in these tables are normalized with respect to triangle area, ie, g(ξ, η dξdη N g w i g(ξ i, η i The following tables list the number of quadrature points for degrees to as given in Ref [] It should be mentioned that for some N, the corresponding is not necessarily unique If interested, please see [] and references therein N dim(p N N dim(p N ( Gaussian quadrature for general triangular elements : Let be a triangular element with straight boundary lines and vertices (x i, y i, i =,, arranged in the counter-clockwise order: P =(x, y y P =(x, y x P =(x, y Figure 9: A triangular element with straight boundary lines We would like to evaluate I = F (x, y dxdy, Again the idea is very simple: first transform the triangular element to the standard triangular element and then apply the Gaussian quadrature for as described above How to transform: Using nodal shape functions!
11 Nodal shape functions for triangles: η (, N (ξ, η = ξ η, N (ξ, η = ξ, N (ξ, η = η (, (, ξ Construct a linear mapping to map the general triangular element with straight boundary lines to the standard triangular element : (x, y (x, y y x (x, y x=p(ξ,η y=q(ξ,η η (, (, ξ (, Figure : Linear mapping between and The mapping can be achieved conveniently by using the nodal shape functions as follows: Then we have x = P (ξ, η = y = Q(ξ, η = x i N i (ξ, η = x N (ξ, η + x N (ξ, η + x N (ξ, η, y i N i (ξ, η = y N (ξ, η + y N (ξ, η + y N (ξ, η F (x, y dxdy = F (P (ξ, η, Q(ξ, η J(ξ, η dξdη, where J(ξ, η is the Jacobian of the transformation, namely, x J(ξ, η = (x, y (ξ, η = ξ x η y ξ y = A η Here A represents the area of the triangle, which can be evaluated by Therefore, we have A = x (y y + x (y y + x (y y F (x, y dxdy = A k F (P (ξ, η, Q(ξ, η dξdη Applying the Gaussian quadrature of degree N for the standard triangular element (4 yields
12 Gaussian quadrature of degree N for general triangular elements N g F (x, y dxdy A w i F (P (ξ i, η i, Q(ξ i, η i (5 Note that the inverse of the transformation shown in Fig is NOT needed, but it can be found that ξ = A [(y y (x x (x x (y y ], η = A [ (y y (x x + (x x (y y ] ( Implementation: The Matlab code using the Gaussian quadrature (5 to evaluate I = F (x, y dxdy is int fm It uses fm which defines the function F (x, y and TriGaussPointsm which provides Gaussian points and weights The calling sequence of int fm is int_f(quadrature_degree, x, x, x, y, y, y where (x, y, (x, y, and (x, y are the coordinates of the three vertices of the triangular element A simple test: Consider the following integral I = ( x y dxdy =, where is the triangle defined by the three vertices: (,, (, / and (, Since f(x, y = x y has degree of, Gaussian quadrature with N = should be able to provide the exact integral Actually if we run int_f(,,,,, 5, in Matlab, it returns, which is / D Contour integrals In many cases (such as when natural or mixed boundary conditions are involved we have to evaluate integrals along the boundary of a (boundary element which are of the form: I = Pj P i B(x, y ds, where P i and P j are consecutive nodal points in the counter-clockwise order and ds = dx + dy is the differential arc length along the boundary of the element The idea is: first transform the straight contour P i P j to an interval I = [a, b], and then apply the Gaussian quadrature for the interval
13 y P j =(x j, y j x P i =(x i, y i Figure : A (straight contour Suppose that the following transformation is used to transform a general quadrilateral/triangular element to the standard quadrilateral/triangular element R st or : x = P (ξ, η, y = Q(ξ, η Then, we have dx = x x dξ + ξ η dη = J dξ + J dη, dy = y y dξ + ξ η dη = J dξ + J dη, (6a (6b where J, J, etc, are elements of the Jacobian matrix J(ξ, η The contour P i P j is a boundary line of a quadrilateral element: (x 4, y 4 (x, y y (x, y x (x, y x=p(ξ,η y=q(ξ,η (, R st (, η (, ξ (, Along each side of a quadrilateral element, either ξ or η is fixed For example, along side (P P, η = so dη = In this case, we have ( x dx = dξ = J (ξ, dξ, ξ η= ( y dy = dξ = J (ξ, dξ ξ η= Therefore, ds = J (ξ, + J (ξ, dξ,
14 and accordingly, we have Similarly, we have P P P B(x, y ds = P B(x, y ds = P4 P P P 4 B(x, y ds = B(x, y ds = B(P (ξ,, Q(ξ, J (ξ, + J (ξ, dξ B(P (, η, Q(, η J (, η + J (, η dη, B(P (ξ,, Q(ξ, J (ξ, + J (ξ, dξ, B(P (, η, Q(, η J (, η + J (, η dη The contour P i P j is a boundary line of a triangular element: (x, y (x, y y x (x, y x=p(ξ,η y=q(ξ,η η (, (, ξ (, Along side (P P or (P P of a triangular element, either ξ or η is fixed Similarly, we have P P B(x, y ds = P P B(x, y ds = B(P (ξ,, Q(ξ, J (ξ, + J (ξ, dξ, B(P (, η, Q(, η J (, η + J (, η dη On the other hand, along side (P P, ξ + η = so dξ = dη In this case, we have dx = J dξ + J dη = [J ( η, η J ( η, η] dη, dy = J dξ + J dη = [J ( η, η J ( η, η] dη Therefore, ds = [J ( η, η J ( η, η] + [J ( η, η J ( η, η] dη = H(ηdη, and accordingly, we have P P B(x, y ds = B(P ( η, η, Q( η, ηh(η dη 4
15 Appendix A - Simplex coordinates The Gaussian quadrature for triangles described above can be applied to integrals over any triangular domain T in either two or three dimensions: I = F (r dr T In three dimensions, however, it is more convenient to consider these integrals in terms of simplex coordinates, also called area coordinates or barycentric coordinates To develop the simplex coordinate transformation, consider the triangle T defined by the vertices v = (x, y, z, v = (x, y, z, and v = (x, y, z and the edges e = v v, e = v v, and e = v v Any point r located on the triangle can be written as a weighted sum of these three nodes via where α, β, and γ are the simplex coordinates defined by r = γv + αv + βv, (7 α = A A, β = A A, γ = A A, and A is the area of T These coordinates are subject to the constraint Therefore, and α + β + γ = γ = α β r = ( α βv + αv + βv Note that (7 has the form of a linear interpolation Indeed, simplex coordinates will also allow us to perform a linear interpolation of a function F (x, y at points inside the triangle if its values are known at the vertices v (, β A v A A z y x v (, (, α (a The original triangle (b The transformed triangle Figure : Simplex coordinates for a triangle The resulting transformation of the integral to simplex coordinates is F (r dr = F (α, β J(α, β dαdβ T Again, it can be proved that the Jacobian is J(α, β = A, where A is the area of the triangle T, which can be evaluated by A = (v v (v v 5
16 Finally, we have T F (r dr = A F (α, β dαdβ = A α F (α, β dβdα From (x, y, z to simplex coordinates: Given a point r = (x, y, z inside a triangle, sometimes it is also desirable to obtain the simplex coordinates (α, β, γ We can write the barycentric expansion of r = (x, y, z in terms of the components of the triangle vertices as Substituting γ = α β into the above gives Rearranging, these are x = γx + αx + βx, y = γy + αy + βy, z = γz + αz + βz x = ( α βx + αx + βx, y = ( α βy + αy + βy, z = ( α βz + αz + βz α(x x + β(x x + x x =, α(y y + β(y y + y y =, α(z z + β(z z + z z = Solving for α and β gives us and where α = β = B(F + I C(E + H A(E + H B(D + G A(F + I C(D + G B(D + G A(E + H, A = x x, B = x x, C = x x, D = y y, E = y y, F = y y, G = z z, H = z z, I = z z 6
17 Appendix B - Matlab code of Gaussian quadrature for triangles ( Main routine: int fm function z = int_f(n,x,x,x,y,y,y % This function evaluates \iint_ f(x,y dxdy using % the Gaussian quadrature of order N where is a % triangle with vertices (x,y, (x,y and (x,y xw = TriGaussPoints(N; % get quadrature points and weights % calculate the area of the triangle A=abs(x*(y-y+x*(y-y+x*(y-y/ % find number of Gauss points NP=length(xw(:,; z = ; for j = :NP x = x*(-xw(j,-xw(j,+x*xw(j,+x*xw(j, y = y*(-xw(j,-xw(j,+y*xw(j,+y*xw(j, z = z + f(x,y*xw(j,; end z = A*z; return end ( Function to define F (x, y: fm function z=f(x,y z=-x-*y; return; end ( Function to provide Gaussian points (ξ i, η i and weights w i : TriGaussPointsm function xw = TriGaussPoints(n %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Function TriGaussPoints provides the Gaussian points and weights % % for the Gaussian quadrature of order n for the standard triangles % % % % Input: n - the order of the Gaussian quadrature (n<= % % % % Output: xw - a n by matrix: % % st column gives the x-coordinates of points % % nd column gives the y-coordinates of points % % rd column gives the weights % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 7
18 xw = zeros(n,; if (n == xw=[ ]; elseif (n == xw=[ ]; elseif (n == xw=[ ]; elseif (n == 4 xw=[ ]; elseif (n == 5 xw=[ ]; elseif (n == 6 xw=[ ]; elseif (n == 7 xw=[
19 ]; elseif (n == 8 xw=[ ]; (cases for n>=9 omitted in the print-out else error( Bad input n ; end return end References [] D A Dunavant, High degree efficient symmetrical Gaussian quadrature rules for the triangle, Int J Num Meth Engng, (985:9- [] B Szabó, I Babu ska, Finite Element Analysis, Wiley, New York, 99 9
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