One dimension and tensor products. 2.1 Legendre and Jacobi polynomials

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1 Part I, Chapter 2 One dimension and tensor products This chapter presents important examples of finite elements, first in one space dimension, then in multiple space dimensions using tensor product techniques. Important computational issues related to the manipulation of high-order polynomial bases are addressed. (JLG) revised again Feb 9, 26, 22:33 2. Legendre and Jacobi polynomials This section reviews some basic properties of the Legendre and Jacobi polynomials which are often used in the design of high-order finite elements. Definition 2. (Legendre polynomials). The Legendre polynomials are univariate polynomial functions R R defined by L m(t) = ()m 2 m m! for all t R and all m 0. d m dt m(( t)m (+t) m ), (2.) Thesuperscript referstothefollowingimportantorthogonalityproperty. Proposition 2.2 (L 2 -orthogonality). The Legendre polynomials are L 2 - orthogonal over the interval (, ), and the following holds: L m(t)l n(t)dt = Proof. Use induction and integration by parts. 2 2m+ δ mn, m,n 0. (2.2) Legendre polynomials satisfy many other properties. We now list some of the most important ones. The following Bonnet s recursion formula holds: (m+)l m+(t) = (2m+)tL m(t) ml m(t), m.

2 6 Chapter 2. One dimension and tensor products L m is a polynomial of degree m and L m() = () m, L m() =. The m roots of L m are in (,) and are called Gauss Legendre nodes; they play an important role in the design of quadratures; see Chapter 56. The first four Legendre polynomials are L 0 (t) =, L (t) = t, L 2 (t) = 2 (3t2 ), and their graphs are shown in Figure 2.. L 3 (t) = 2 (5t3 3t), Fig. 2.. Legendre polynomials of degree at most 3 in the interval [,]. Definition 2.3 (Jacobi polynomials). Let α,β R be such that α > and β >. The Jacobi polynomials are univariate polynomial functions R R defined by Jm α,β ( (t) = ()m 2 m m! ( t) α β dm (+t) dt ( t) α+m (+t) β+m), (2.3) m for all t R and all m 0. The Jacobi polynomials satisfy the following L 2 -orthogonality property on the interval (, ): ( t) α (+t) β J α,β with c m,α,β = 2α+β+ 2m+α+β+ m (t)j α,β n (t)dt = c m,α,β δ mn, m,n 0, (2.4) Γ(m+α+)Γ(m+β+) m!γ(m+α+β+), where Γ is the Gamma function (recall that Γ(n+) = n! for any natural number n). The Jacobi polynomials satisfy the following recursion formula for all m : 2(m+)(m+α+β +)(2m+α+β)J α,β m+ (t) = (2m+α+β +)((2m+α+β +2)(2m+α+β)t+α 2 β 2 )J α,β m (t) 2(m+α)(m+β)(2m+α+β +2)J α,β m (t).

3 Part I. Getting Started 7 Jm α,β is a polynomial of degree m and Jm α,β () = () m( ) m+β m, J α,β m () = ( m+α ) m. The Legendre polynomials are Jacobi polynomials with parameters α = β = 0, i.e., L m(t) = Jm 0,0 (t) for all m 0. The first three Jacobi polynomials corresponding to the parameters α = β = are J, 0 (t) =, J, (t) = 2t, J, 2 (t) = 3 4 (5t2 ) The Jacobi polynomials J, m are related to the integrated Legendre polynomials as follows (see Exercise 2.4): t L m(s)ds = 2m ( t2 )J, m (t), m. (2.5) We refer to Abramowitz and Stegun [, Chap. 22] for further results. 2.2 One-dimensional finite elements In this section, we present important examples of one-dimensional finite elements. Recall that P k, k 0, is the real vector space composed of univariate polynomial functions of degree at most k. For convenience, degrees of freedom and shape functions of one-dimensional finite elements using the polynomial space P k are numbered from 0 to k Lagrange (nodal) finite elements Following Definition.8, the degrees of freedom for Lagrange finite elements are chosen as the values at some set of nodes. Proposition 2.4 (Lagrange finite element). Let k 0. Let K be a compact interval with non-empty interior, and let P = P k. Consider a set of n sh = k+ distinct nodes {a l } l {0:k} in K. Let Σ = {σ l } l {0:k} be the linear forms on P such that σ l (p) = p(a l ) for all l {0:k}. Then, (K,P,Σ) is a Lagrange finite element. Proof. We verify unisolvence. We observe that dimp = k+ = n sh = cardσ and that a polynomial p P such that σ l (p) = p(a l ) = 0 for all l {0:k} is of degree at most k and has (k +) distinct roots. Hence, p 0. The shape functions of a one-dimensional Lagrange finite element are the Lagrange interpolation polynomials {L [a] l } i {0:k} defined as in (.2). Following (.0), the Lagrange interpolation operator acts as follows: IK(v)(t) L = v(a l )L [a] l (t), t K, (2.6) l {0: k}

4 8 Chapter 2. One dimension and tensor products and possible choices for the domain of I L K are V(K) = C0 (K) or V(K) = W, (K), see Exercise.7. The Lagrange interpolation polynomials based on equidistant nodes (including both endpoints) in the interval K = [, ] are henceforth denoted {L k l } l {0:k}. The graphs of these polynomials are illustrated in Figure. for k {,2,3}, and explicit expressions are as follows: L 0(t) = 2 ( t), L2 0(t) = 2 t(t), L3 0(t) = 9 6 (t+ 3 )(t 3 )( t), L (t) = 2 (+t), L2 (t) = (t+)( t), L 3 (t) = 27 6 (t+)(t 3 )(t), L 2 2(t) = 2 (t+)t, L3 2(t) = 27 6 (t+)(t+ 3 )( t), L 3 3(t) = 9 6 (t+)(t+ 3 )(t 3 ). While the choice of equidistant nodes appears somewhat natural, it is appropriate only when working with low-degree polynomials, see Modal finite elements In this section, we illustrate the construction of.4.2 in the one-dimensional setting, say on the interval [, ]. Proposition 2.5 (Legendre finite element). Let k 0. Let K = [,] and let P = P k. Let Σ = {σ l } l {0:k} be the n sh = k + linear forms on P such that σ l (p) = 2l+ L l (t)p(t)dt, l {0:k}. (2.7) 2 Then, (K,P,Σ) is a finite element. The shape functions are the (k +) Legendre polynomials of degree k. Proof. Use (2.2) and Proposition.9. Following (.3), the Legendre interpolation operator acts as follows: ) L l (s)v(s)ds L l (t) for all t K and all IK m(v)(t) = l {0:k} v V(K) = L (K). ( 2l Hybrid finite elements ( ) Hybrid finite elements mix nodal and modal degrees of freedom. When constructing so-called H -conforming approximation spaces (see Chapter 3 for a full treatment in multiple space dimensions), it is important that all the basis functions but one vanish at each endpoint of the interval, say K = [,]. This calls for using the values at ± as nodal degrees of freedom on P k. For k 2, some or all of the remaining degrees of freedom can be taken to be of modal type. Taking all of them as moments against polynomials in P k 2 leads to the so-called canonical hybrid finite element; a multi-dimensional extension in a simplex is presented in 3.4.

5 Part I. Getting Started 9 Proposition 2.6 (Canonical hybrid finite element). Let k. Set K = [,] and P = P k. Define σ 0 (p) = p(), σ k (p) = p(), and, if k 2, let {µ l } l {:k} be a basis of P k 2 and define σ l (p) = K pµ lds for all l {:k}. Set Σ = {σ l } l {0:k}. Then, the triple (K,P,Σ) is a finite element. Proof. See Exercise 2.5. The corresponding interpolation operator is denoted I g K (the superscript is consistent with the notation introduced in 0.2 where the letter g refers to the gradient operator). Its action on functions v W, (K) is such that I g K (v)(±) = v(±) and (Ig K (v) v)qds = 0 for all q P k 2. Proposition 2.7 (Commuting with derivative). Let k 0. Let I g K be the interpolation operator built from the canonical hybrid finite element of order (k+). Let IK m be the interpolation operator built from the modal finite element of order k. Then, I g K (v) = IK m(v ) for all v W, (K). Proof. Integrating by parts and using the properties of the interpolation operators (i.e., (Ig K (v) v)rds = 0 for all r P k and I g K (v)(±) = v(±)), we infer that, for all q P k, I g K (v) qdt = I g K (v)q dt+[i g K (v)q] = vq dt+[vq] = v qdt = I m K(v )qdt. This proves that I g K (v) = I m K (v ) since both functions are in P k. The shape functions of the canonical hybrid finite element can be computed explicitly once a choice for the basis functions {µ l } l {:k} of P k 2 is made; an example is proposed in Exercise 2.5 using Jacobi polynomials Hierarchical bases ( ) The notion of hierarchical polynomial bases is important when working with high-order polynomials. It is particularly convenient in simulations where the degree k varies from one element to the other. Definition 2.8 (Hierarchical basis). A sequence of polynomials (P k ) k N is said to be a hierarchical polynomial basis if the set {P l } l {0:k} is a basis of P k for all k N. The monomial basis (i.e., P k (t) = t k ) is the simplest example of hierarchical polynomial basis. The Jacobi polynomials introduced in 2. form a hierarchical basis which is L 2 -orthogonal with respect to the weight ( t) α (+t) β. The Lagrange shape functions do not form a hierarchical basis, i.e., increasing k to (k+) requires to recompute the whole basis of shape functions.

6 20 Chapter 2. One dimension and tensor products Instead, the shape functions of modal elements form, by construction, a hierarchical basis. Finally, in the hybrid case, the shape functions do not form, in general, a hierarchical basis. One can obtain a hierarchical basis though by slightly modifying the degrees of freedom. For instance, the following shape functions form a hierarchical basis: θ 0 (t) = 2 ( t), θ l (t) = 4 ( t)(+t)j, l (t), l {:k }, (2.8b) θ k (t) = 2 (+t). (2.8a) (2.8c) Their degrees of freedom can be taken to be σ 0 (p) = p(), σ k (p) = p(), and σ l (p) = α l pj, l dt+β l p()+β+ l p() for all l {:k }, where α l = 4c l,,, β± l = 2c l,, (±t)j, l dt, and c l,, defined in (2.4) High-order Lagrange elements ( ) The main difficulty comes from the oscillatory nature of the Lagrange polynomials as the number of interpolation nodes grows. This phenomenon is often referred to as the Runge phenomenon [400] (see also Meray [349]). A classic example illustrating this phenomenon consists of considering the function f(x) = ( + x 2 ), x [ 5,5]; the Lagrange interpolation polynomial of f using n equidistant points over [ 5, 5] converges uniformly to f in the interval ( x c,x c ) with x c 3.63 and diverges outside this interval. The approximation quality in the maximum norm of the one-dimensional Lagrange interpolation operator using n sh = k + distinct nodes is quantified by the Lebesgue constant; see Theorem.. Since the Lebesgue constant is invariant by linear transformation of the interval K, we henceforth restrict the discussion to K = [,]. It can be shown that the Lebesgue constant for equidistant nodes grows exponentially with n sh ; more precisely, 2 n en(logn+γ) Λ nsh as n where γ =.5772 is the Euler constant; see Trefethen and Weideman [45]. A lower bound for the Lebesgue constant for any set of points is 2 π ln(n sh) C for some positive C; see Erdős [208]. If the Chebyshev nodes a l = cos( (2l)π 2n sh ), l N = {:n sh }, are used instead of the equidistant nodes, the Lebesgue constant behaves as 2 π ln(n sh)+c +α n, with C = 2 π (γ+log 8 π ) = and α n 0 as n, showing that this choice is asymptotically optimal; see Luttmann and Rivlin [340] and Rivlin [396, Chap. 4]. One drawback of Chebyshev nodes is that they do not contain the interval endpoints ±, which make them unsuitable for constructing H -conforming finite element spaces; see the discussion at the beginning of One set of interpolating nodes realizing this constraint are the Gauss Lobatto Legendre nodes, which are, for k, the (k + ) roots of the polynomial ( t)( + t)(l k ) (t); all these nodes are in [,] since all the roots of L k are in (,). The corresponding Lagrange interpolation polynomials for

7 Part I. Getting Started 2 k = 3 are shown in the left panel of Figure 2.2. The Gauss Lobatto Legendre nodes lead to an asymptotically optimal Lebesgue constant with upper bound 2 π log(n sh) + C, with C 0.685; see Hesthaven [280, Conj. 3.2] and Hesthaven et al. [28, p. 06]. Note that the Lagrange polynomial bases using the Gauss Lobatto Legendre nodes are not hierarchical since the set of n sh Gauss Lobatto Legendre nodes is not included in the set of (n sh +) nodes Fig Left: Lagrange polynomials associated with the Gauss Lobatto Legendre nodes for k = 3. Right: Hybrid nodal/modal shape functions for k = 4. Another important class of sets of nodes is that consisting of the so-called Fekete points. Fekete points are defined from a maximization problem. Let {a i } i N be a set of nodes in K = [,] and let {φ i } i N be a basis of P k (recall that N = {:n sh }). Consider the (generalized) n sh n sh Vandermonde matrix V with entries V ij = φ i (a j ), for all i,j N. Since the Lagrange polynomials can be expressed as L [a] i (t) = j N (V ) ij φ j (t) (see Exercise 2.), a reasonable criterion for selecting the interpolation nodes is to maximize the determinant of V with respect to {a i } i N (observe that the solution to this problem does not depend on the chosen basis, since a change of basis only multiplies the determinant by a factor equal to the determinant of the change of basis matrix). It is shown in Fejér [229] that the Fekete points and the Gauss Lobatto Legendre nodes coincide on any interval. The notion of Fekete points extends naturally to any space dimension, but the concept of Gauss Lobatto Legendre nodes can be extended in higher space dimensions only by invoking tensor product techniques; see 2.3.

8 22 Chapter 2. One dimension and tensor products 2.3 Tensor product elements We show in this section that the one-dimensional finite elements presented in 2.2 can be extended to higher dimensions by using tensor product techniques when K R d has the Cartesian product structure K = d i= [z i,z+ i ] with real numbers z ± i, i {:d}, such that z i < z + i. A set of this form in Rd is called a cuboid The polynomial space Q k,d A useful ingredient to build tensor-product finite elements in R d is the polynomial space Q k,d = P k... P }{{ k. (2.9) } d times Thisspaceiscomposedofd-variatepolynomialfunctionsq : R d Rofpartial degree at most k with respect to each variable. Thus, { } Q k,d = span x β...xβ d d, 0 β,...,β d k. (2.0) We omit the subscript d and simply write Q k when the context is unambiguous. Let β = (β,...,β d ) N d be a multi-index, define β l = max i {:d} β i, and consider the multi-index set B k,d = {β N d β l k}. Polynomial functions q Q k,d can be written in the following generic form: q(x) = β B k,d a β x β, x β = x β...xβ d d, (2.) with real numbers a β. Note that card(b k,d ) = dim(q k,d ) = (k +) d Tensor-product finite elements We begin with tensor-product Lagrange finite elements with nodes obtained by invoking the tensor product of nodes along each Cartesian direction. This leads to the following construction. Proposition 2.9 (Tensor-product Lagrange). Let K = d i= [z i,z+ i ] be a cuboid in R d. Let k and let P = Q k,d. For all i {:d}, consider (k+) distinct nodes {a i,l } l {0:k} in [z i,z+ i ]. For any multi-index β = (β,...,β d ) B k,d, define the node a β in K with Cartesian components (a i,βi ) i {:d}. Let Σ = {σ β } β Bk,d be the linear forms on P such that σ β (p) = p(a β ) for all β B k,d. Then, (K,P,Σ) is a Lagrange finite element. Proof. See Exercise 2.6.

9 Part I. Getting Started 23 The shape functions of a tensor-product Lagrange finite element are products of the one-dimensional Lagrange interpolation polynomials defined in (.2): θ β (x) = d i= L [ai] β i (x i ), x K, (2.2) for all β B k,d. The Lagrange interpolation operator acts as follows: I L K(v)(x) = β B k,d v(a β )θ β (x), x K, (2.3) and possible choices for its domain are V(K) = C 0 (K) or V(K) = W s,p (K) with real numbers p [, ] and s > d p. Note that, in general, IL K (v)(x) cannot be factorized as a product of univariate functions, except when the function v has the separated form v(x) = Π d i= v i(x i ) with v i C 0 ([x i,x+ i ]), in which case one can verify that I L K (v)(x) = d i= IL [x i,x+ i ](v i)(x i ). Q Q 2 Q 3 Table 2.. Two- and three-dimensional Lagrange finite elements Q, Q 2, and Q 3; only visible degrees of freedom are shown in three dimensions. Table 2. presents examples for k {, 2, 3} in dimensions d {2, 3} with equidistant nodes in each Cartesian direction. The bullets conventionally indicate the location of the nodes. It is useful to use the tensor product of Gauss Lobatto Legendre nodes when k is large, since it can be shown that the Fekete points in cuboids are the tensor products of the one-dimensional Gauss Lobatto Legendre nodes, see Bos et al. [69]. The tensor-product technique can also be used to build modal and hybrid nodal/modal finite elements in cuboids; see [69]. Following the early work of Patera [372], finite element methods based on nodal bases using tensor products are often referred to as spectral methods.

10 24 Chapter 2. One dimension and tensor products Serendipity finite elements ( ) It is possible to remove some nodes inside the cuboid while maintaining the approximation properties obtained when using full tensor products. This is the idea of the serendipity finite elements. The corresponding polynomial space, S k, is then a proper subspace of Q k. The main motivation is to reduce computational costs without sacrificing the possibility to build H -conforming finite element spaces and without sacrificing the accuracy of the interpolation operator. Classical two-dimensional examples consist of using the 8 boundary nodes for k = 2 and the 2 boundary nodes for k = 3 (see Table 2.2). For k = 4, one uses the 6 boundary nodes plus the barycenter of K. A systematic construction of the serendipity finite elements for any space dimension and polynomial degree is devised in Arnold and Awanou [6]. S 2 S 3 S 4 Table 2.2. Two-dimensional serendipity Lagrange finite elements S 2, S 3, and S 4. Exercises Exercise 2. (Vandermonde matrix). Consider the Lagrange finite element from Proposition 2.4 with nodes {a i } i N in K. Let {L [a] i } i N be the corresponding Lagrange polynomials, see (.2). Let {φ i } i N be a basis of P k. Show that the Lagrange polynomials can be written in the form L [a] i (t) = j N(V ) ij φ j (t), where the matrix V has entries V ij = φ i (a j ) for all i,j N. (Note: when the functions φ i are the monomial basis, the matrix V is called a Vandermonde matrix; otherwise, it is called a generalized Vandermonde matrix.) Exercise 2.2 (Cross approximation). Let f : X Y R be a bivariate function where X,Y are two nonempty subsets of R. Let n sh, set N = {:n sh }, and consider n sh points {x i } i N in X and n sh points {y j } j N in

11 Part I. Getting Started 25 Y. Assume that the matrix F R n sh n sh with entries F ij = f(x i,y j ) is invertible. Let I CA (f) : X Y R be defined by I CA (f)(x,y) = (F T ) ij f(x,y j )f(x i,y). i,j N Prove that I CA (f)(x,y k ) = f(x,y k ) for all x X and all k N, and that I CA (f)(x k,y) = f(x k,y) for all y Y and all k N. Exercise 2.3 (Mass and Vandermonde matrices). Let K = [,], let {a,...,a nsh } be nodes in K, and let {L [a] i } i nsh be the corresponding Lagrange polynomials. Let M R n sh n sh be the so-called mass matrix with entries M ij = L[a] i (t)l [a] j (t)dt for all i,j N. Prove that M = (VT V) where V R n sh n sh is the (generalized) Vandermonde matrix with entries V ij = ( 2 2i )/2 L i (a j) and the L i s are the Legendre polynomials (see Definition 2.). Exercise 2.4 (Integrated Legendre polynomials). Let k 2 and set P (0) k = {p P k p(±) = 0}. (i) Show that a basis for P (0) k are the integrated Legendre polynomials { t L l (s)ds} l {:k}. (ii) Prove (2.5). (Hint: consider moments against polynomials in P m 2 and the derivative at t =.) Exercise 2.5 (Canonical hybrid element). (i) Prove Proposition 2.6. (Hint: verify unisolvence.) (ii) Compute the shape functions when µ l = J, l for all l {:k}. (Hint: consider the polynomials J,0, l {:k }, and J0, k, J, l k.) Exercise 2.6 (Q k,d Lagrange). Prove Proposition 2.9. (Hint: observe that any polynomial p Q k,d can be written p(x) = k i d =0 q i d (x,...,x d )x i d d and use induction on d.) Exercise 2.7 (Bi-cubic Hermite rectangle). Let K be a (non-degenerate) rectangle with vertices {z,z 2,z 3,z 4 }. Set P = Q 3,2 and Σ = {p(z i ), x p(z i ), x2 p(z i ), 2 x x 2 p(z i )} i 4. Show that (K,P,Σ) is a finite element. (Hint: write p Q 3,2 in the form p(x) = i,j 4 γ ijθ i (x )θ j (x 2 ) where {θ,...,θ 4 } are the shape functions of the one-dimensional Hermite finite element; see Exercise.3.)