Finite element exterior calculus

Transcription

1 Finite element exterior calculus Douglas N. Arnold School of Mathematics, University of Minnesota NSF/CBMS Conference, ICERM, June 11 15, 2012 collaborators: R. Falk, R. Winther, G. Awanou, F. Bonizzoni, D. Boffi, J. Gopalakrishnan, H. Chen c 2012 Douglas N. Arnold. All rights reserved.

2 Computational examples

3 Standard P 1 finite elements for 1D Laplacian u u 2 / 94

4 Mixed finite elements for Laplacian σ u u P 1 -P u RT-P / 94

5 Vector Laplacian, L-shaped domain curl curl u grad div u = f in Ω u n = 0, curl u n = 0 on Ω (curl u curl v + div u div v) = Ω f v Ω v Lagrange finite elements converge nicely but not to the solution! (same problem with any conforming FE) 4 / 94

6 Vector Poisson equation curl curl u grad div u = f in Ω u n = 0, curl u n = 0 on Ω f 0 does not imply u 0: dim H = b 1 harmonic forms 1st Betti number (solutions for f = 0) (number of holes) curl curl u grad div u = f (mod H), u H, b.c. 5 / 94

7 Maxwell eigenvalue problem Find 0 u H(curl) s.t curl u curl v = λ u v Ω Ω v H(curl) λ = m 2 + n 2 = 0, 1, 1, 2, 4, 4, 5, 5, 8, P 1 P 1 Λ1 6 / 94

8 Maxwell eigenvalue problem, crisscross mesh λ = m 2 + n 2 = 1, 1, 2, 4, 4, 5, 5, 8, Boffi-Brezzi-Gastaldi 99 7 / 94

9 EM calculations based on the generalized RT elements Schöberl, Zaglmayr 2006, NGSolve 2K tets, P 6 Λ1 8 / 94

10 Homology 101

11 Chain complexes A chain complex (V, ) is a seq. of vector spaces and linear maps k+1 V k+1 k Vk Vk 1 with k k+1 = 0. typically, non-negative and finite: V k = 0 for k < 0 for k large In other words, V = k V k is a graded vector space and : V V is a graded linear operator of degree 1 such that = 0 ( k = Vk : V k V k ) V k : Z k = N ( k ): B k = R( k+1 ): H k = Z k /B k : k-chains k-cycles k-boundaries k-th homology space Thus the elements of H k are equivalence classes of k-cycles 9 / 94

12 Simplices and simplicial complexes A k-simplex in R n is the convex hull f = [x 0,..., x k ] of k + 1 vertices in general position. A subset determines a face of f : [x i0,..., x id ]. Simplicial complex: A finite set S of simplices in R n, such that 1. Faces of simplices in S are in S. 2. If f g for f, g S, then it is a face of f and of g. If we order all vertices of S, then an ordering of the vertices of the simplex determines an orientation / 94

13 The boundary operator on chains k (S): the set of k-simplices in S C k (k-chains): formal linear combinations c = k : k C k 1 : f k (S) c f f [x 0, x 1,..., x k ] = k i=0 ( 1)i [..., ˆx i,...] k : C k C k 1 : c = c f f / 94

14 The simplicial chain complex 0 C n Cn 1 C0 0 β k := dim H k (C) is the kth Betti number 1, 1, 0, 0 1, 1, 0, 0 1, 2, 1, 0 1, 2, 0, 0 1, 0, 1, 0 12 / 94

15 Chain maps k+1 V k+1 V k f k+1 f k f k 1 k Vk 1 V k+1 k+1 V k k V k 1 f(z) Z, f(b) B, so f induces f : H(V) H(V ). If V is a subcomplex (V k V k and = V ), and fv = v for v V, we call f a chain projection. Proposition A chain projection induces a surjection on homology. 13 / 94

16 Cochain complexes A cochain complex is like a chain complex but with increasing indices. V k 1 d k 1 V k d k V k+1 cocycles Z k, coboundaries B k, cohomology H k,... The dual of a chain complex is a cochain complex: k+1 : V k+1 V k = k+1 : V k V k+1 d k V k 14 / 94

17 The de Rham complex for a domain in R n 1-D: 2-D: 3-D: 0 C (Ω) d/dx C (Ω) 0 0 C (Ω) grad C (Ω, R 2 ) rot C (Ω) 0 0 C (Ω) grad C (Ω, R 3 ) curl C (Ω, R 3 ) div C (Ω) 0 n-d: 0 Λ 0 (Ω) d Λ 1 (Ω) d Λ 2 (Ω) d d Λ n (Ω) 0 The space Λ k (Ω) = C (Ω, R n n skw ), the space of smooth differential k-forms on Ω. Exterior derivative: d k : Λ k (Ω) Λ k+1 (Ω) Integral of a k-form over an oriented k-simplex: f v R Stokes theorem: c du = c u, All this works on any smooth manifold u Λk, c C k 15 / 94

18 De Rham s Theorem De Rham map: Λ k (Ω) C k (S) := C k (S) u (c c u) Stokes theorem says it s a cochain map, so induces a map from de Rham to simplicial cohomology. d Λ k (Ω) C k d Λ k+1 (Ω) C k+1 d Theorem (De Rham s theorem) The induced map is an isomorphism on cohomology. 16 / 94

19 Nonzero cohomology classes u = grad θ, 0 ū H 1 on cylindrical shell u = grad 1 r, 0 ū H2 on spherical shell 17 / 94

20 Unbounded operators on Hilbert space

21 Unbounded operators X,Y H-spaces (extensions to Banach spaces, TVSs,... ) T : D(T ) Y linear, D(T ) X subspace (not necessarily closed), T not necessarily bounded Not-necessarily-everywhere-defined-and-not-necessarily-bounded linear operators Densely defined: D(T ) = X Ex: X = L 2 (Ω), Y = L 2 (Ω; R n ), D(T ) = H 1 (Ω), Tv = grad v (changing D(T ) to H1 (Ω) gives a different example) S, T unbdd ops X Y = D(S + T ) = D(S) D(T ) (may not be d.d.) X S Y, Y T Z unbdd ops = D(T S) = {v D(S) Sv D(T )} Graph norm (and inner product): v 2 D(T ) := v 2 X + Tv 2 Y, v D(T ) Null space, range, graph: N (T ), R(T ), Γ(T ) 18 / 94

22 Closed operators T is closed if Γ(T ) is closed in X Y. Equivalent definitions: 1. If v 1, v 2,... D(T ) satisfy v n X x and Tv n Y y for some x X and y Y, then x D(T ) and Tx = y. 2. D(T ) endowed with the graph norm is complete. If D(T ) = X, then T is closed T is bdd (Closed Graph Thm) Many properties of bounded operators extend to closed operators. E.g., Proposition Let T be a closed operator X to Y. 1. N (T ) is closed in X. 2. γ > 0 s.t. Tx Y γ x X N (T ) = 0, R(T ) closed 3. If dim Y / R(T ) <, then R(T ) is closed 19 / 94

23 Adjoint of a d.d.unbdd operator Let T be a d.d.unbdd operator X Y. Define D(T ) = {w Y the map v D(T ) w, Tv Y R is bdd in X-norm } For w D(T )! T w X s.t. T w, v X = w, Tv Y, v D(T ), w D(T ). T is a closed operator (even if T is not). Define the rotated graph Γ(T ) = { ( T w, w) w D(T ) } X Y, Then Γ(T ) = Γ(T ), Γ(T ) = Γ(T ). Proposition Let T be a closed d.d. operator X Y. Then 1. T is closed d.d. 2. T = T. 3. R(T ) = N (T ), N (T ) = R(T ), R(T ) = N (T ), N (T ) = R(T ). 20 / 94

24 Closed Range Theorem Theorem Let T be a closed d.d.operator X Y. If R(T ) is closed in Y, then R(T ) is closed in X. Proof. 1. Reduce to case T is surjective. 2. Restrict to orthog comp of N (T ) in D(T ) (w/ graph norm). Get bounded linear isomorphism. bounded inverse: y Y x X s.t. Tx = y, x X c y Y 3. This implies y Y c T y X, y D(T ). 21 / 94

25 Grad and div Assume Ω R 3 with Lipschitz boundary (so trace theorem holds). 1. Start with div : C c L 2 (Ω; R 3 ) L 2 (Ω) 2. Its adjoint is grad with domain H 1 (this proves H 1 is complete). 3. The adjoint of (grad, H 1 ) is div with domain H(div) = {w H(div) γ n w := w n Ω = 0 } 4. div H(div) is finite-codimensional so closed. So grad H 1 is closed. 22 / 94

26 Curl 1. Start with curl : C c L 2 (Ω; R 3 ) L 2 (Ω; R 3 ) 2. Its adjoint is curl with domain H(curl) (complete). 3. Adjoint of (curl, H(curl)) with domain H(curl) = {w H(curl) γ t w := w n Ω = 0 } 4. We shall see that curl H(curl) is closed. 23 / 94

27 Hilbert complexes

28 Hilbert complexes Definition A Hilbert complex is a sequence of Hilbert spaces W k and a sequence of closed d.d.linear operators d k from W k to W k+1 such that R(d k ) N (d k+1 ). V k = D(d k ) H-space with graph norm: v 2 = v 2 + d k v 2 V k W k W k+1 The domain complex 0 V 0 d V 1 d d V n 0 is a bounded Hilbert complex (with less information). It is a cochain complex, so it has (co)cycles, boundaries, and homology. An H-complex is closed if B k is closed in W k (or V k ). An H-complex is Fredholm if dim H k <. Fredholm = closed 24 / 94

29 The dual complex Define d k : V k W k W k 1 as the adjoint of d k 1 : V k W k 1 W k. It is closed d.d.and, since R(d k 1 ) N (d k ), R(d k+1) R(d k+1 ) = N (d k ) R(d k 1 ) = N (d k ), so we get a Hilbert chain complex with domain complex 0 V n d n V d n 1 n 1 d 1 V 0 0. If (W, d) is closed, then (W, d ) is as well, by the Closed Range Theorem. From now on we mainly deal with closed H-complexes / 94

30 Harmonic forms The Hilbert structure of a closed H-complex allows us to identify the homology space H k = Z k /B k with a subspace H k of W k : H k := Z k B k = Z k Z k = {u V k V k du = 0, d u = 0}. An H-complex has the compactness property if V k V k compact in W k. This implies dim H k <. is dense and compactness property = Fredholm = closed 26 / 94

31 Two key properties of closed H-complexes Theorem (Hodge decomposition) For any closed Hilbert complex: W k = B } k {{ H } k V k = Z k Bk }{{} Z k {}}{ B k H k Z k V Theorem (Poincaré inequality) For any closed Hilbert complex, a constant c P s.t. z V c P dz, z Z k V. 27 / 94

32 L 2 de Rham complex on Ω R 3 k W k d k V k d k V k dim H k 0 L 2 (Ω) grad H 1 0 L 2 β 0 1 L 2 (Ω; R 3 ) curl H(curl) div H(div) β1 2 L 2 (Ω; R 3 ) div H(div) curl H(curl) β2 3 L 2 (Ω) 0 L 2 grad H1 0 0 H 1 grad H(curl) curl H(div) div L L 2 div H(div) curl H(curl) grad H / 94

33 The abstract Hodge Laplacian W k 1 d W k d d W k+1 L := d d + dd W k L W k d D(L k ) = { u V k V k du V k+1, d u V k 1 } N (L k ) = H k, H k R(L k ) Strong formulation: Find u D(L k ) s.t. Lu = f P H f, u H. Primal weak formulation: Find u V k V k Hk s.t. du, dv + d u, d v = f P H f, v, v V k V k H k. Mixed weak formulation. Find σ V k 1, u V k, and p H k s.t. σ, τ u, dτ = 0, τ V k 1, dσ, v + du, dv + p, v = f, v, v V k, u, q = 0, q H k. 29 / 94

34 Equivalence and well-posedness Theorem Let f W k. Then u W k solves the strong formulation it solves the primal weak formulation. Moreover, in this case, if we set σ = d u and p = P H u, then the triple (σ, u, p) solves the mixed weak formulation. Finally, if some (σ, u, p) solves the mixed weak formulation, then σ = d u, p = P H u, and u solves the strong and primal formulations of the problem. Theorem For each f W k there exists a unique solution. Moreover u + du + d u + dd u + d du c f P H f. The constant depends only on the Poincaré inequality constant c P. 30 / 94

35 Proof of well-posedness We used the mixed formulation. Set B(σ, u, p; τ, v, q) = σ, τ u, dτ dσ, v du, dv p, v u, q We must prove the inf-sup condition: (σ, u, p) (τ, v, q) s.t. B(σ, u, p; τ, v, q) γ( σ V + u V + p )( τ V + v V + q ), with γ = γ(c P ) > 0. Via the Hodge decomposition, with ρ Z V. Then take u = u B + u H + u B = dρ + u H + u B τ = σ 1 (c P ) 2 ρ, v = u dσ p, q = p u H. 31 / 94

36 Hodge Laplacian and Hodge decomposition f = dd u + P H f + d du is the Hodge decomposition of f Define K : W k D(L k ) by Kf = u (bdd lin op). P B = dd K, P B = d dk If f V, then Kdf = dkf. If f B, then dkf = 0. Since Kf H, Kf B. B problem: If f B, then u = Kf solves dd u = f, du = 0, u H. B problem: If f B, then u = Kf solves d du = f, d u = 0, u H. 32 / 94

37 The Hodge Laplacian on a domain in 3D 0 H 1 grad H(curl) curl H(div) div L L 2 div H(div) curl H(curl) grad H1 0 k L k = d d + dd BCs imposed on... V k 1 V k 0 u/ n H 1 1 curl curl grad div u n curl u n H 1 H(curl) 2 grad div + curl curl u n div u H(curl) H(div) 3 u H(div) L 2 essential BC for primal form. natural BC for primal form. 33 / 94

38 Approximation of Hilbert complexes

39 Why mixed methods? Naively, we might try to discretize the primal formulation with finite elements. This works in some circumstances, but we have seen two ways in which it can fail. It is not easy to construct a dense family of subspaces of the primal energy space V k V k Hk. We therefore consider finite element discretizations of the mixed formulation: Given f W k, find σ V k 1, u V k, and p H k s.t. σ, τ u, dτ = 0, τ V k 1, dσ, v + du, dv + p, v = f, v, v V k, u, q = 0, q H k. 34 / 94

40 Galerkin method Choose f.d. subspaces V j h V j Z j h = {v V j h dv = 0} Zj B j h H j h = {v Zj h v Bj h } j 1 = {dv v Vh } B j Given f W k, find σ h V k 1 h, u h Vh k, and p h H k h s.t. σ h, τ u h, dτ = 0, τ V k 1 h, dσ h, v + du h, dv + p h, v = f, v, v V k h, u h, q = 0, q H k h. If H k h Hk this is a nonconforming method. For any choice of the V j h there exists a unique solution. However, the consistency, stability, and accuracy of the discrete solution depends vitally on the choice of subspaces. 35 / 94

41 Key assumptions We need the spaces V j h V j (at least for j = k 1, k, k + 1) to satisfy three properties: 1. Approximation property: Of course V j h must afford good approximation of elements of V j. This can be formalized with respect to a family of subspaces parametrized by h by requiring lim inf h 0 v V j h w v V = 0, w V j (or = O(h r ) for w in some dense subspace, or... ) 2. Subcomplex property: dv k 1 h Vh k and dv h k V k+1 h, so is a subcomplex. V k 1 h d V k h d V k+1 h 36 / 94

42 Bounded cochain projection 3. Bounded cochain projection: Most important, we assume that there exists a cochain map from the H-complex to the subcomplex which is a projection and is bounded. V k 1 d V k d V k+1 π k+1 π k 1 h V k 1 h π k h d V k h h d V k+1 h For now, boundedness is in V -norm: π h v V c v V. But later we will need W -boundedness, which is a stronger requirement. A bounded projection is quasioptimal: v π h v V c inf v w V, w V j h v V j 37 / 94

43 First consequences from the assumptions From the subcomplex property V k 1 h d V k h d V k+1 h is itself a closed H-complex. (We take W k h = V k h Therefore there is a discrete adjoint operator d h ), a discrete Hodge decomposition W k h but with the W -norm.) (its domain is all of and a discrete Poincaré inequality V k h = B k h H k h B kh. z V c P h dz, z Z k V h. 38 / 94

44 Preservation of cohomology Theorem Given: a closed H-complex, and a choice of f.d. subspaces satisfying the subcomplex property and admitting a V -bdd cochain projection π h. Assume also the (very weak) approximation property q π h q < q, 0 q H k. Then π h induces an isomorphism from H k onto H k h. Moreover, ( ) gap H, H h sup q H q =1 q π h q V. gap(h, H h ) := max ( ) sup inf u v V, sup inf u v V. u H v H h v H h u H u =1 v =1 39 / 94

45 Uniform Poincaré inequality and stability Theorem Given: a closed H-complex, and a choice of f.d. subspaces satisfying the subcomplex property and admitting a V -bdd cochain projection π h. Then v V c P π h dv V, v Z k h V k h. Corollary (Stability and quasioptimality of the mixed method) The mixed method is stable (uniform inf-sup condition) and satisfies σ σ h V + u u h V + p p h C( inf τ V k 1 h where µ = µ h = σ τ V + inf u v V + inf p q V v Vh k q Vh k sup (I π h )r. r H k, r =1 + µ inf P B u v V ), v Vh k 40 / 94

46 Improved error estimates In addition to µ = (I π h )P H, define δ, η = o(1) by δ = (I π h )K Lin(W,W), η = (I π h )d [ ] K Lin(W,W). { When Vh k P r, µ = O(h r+1 O(h 2 ), r > 0, ), η = O(h), δ = O(h), r = 0, Theorem Given: an H-complex satisfying the compactness property, and a choice of f.d. subspaces satisfying the subcomplex property and admitting a W -bdd cochain projection π h. Then d(σ σ h ) ce(dσ), σ σ h c[e(σ) + ηe(dσ)], d(u u h ) c{e(du) + η[e(dσ) + E(p)]}, u u h c{e(u) + η[e(du) + E(σ)] +(η 2 + δ)[e(dσ) + E(p)] + µe(p B u)}. 41 / 94

47 Numerical tests grad div u + curl rot u = f in Ω (unit square), u n = rot u = 0 on Ω (magnetic BC) 0 H 1 grad H(rot) rot L 2 0 σ h V 0 h H1, u h V 1 h H(rot) σ h, τ u h, grad τ = 0, τ V k 1 h, grad σ h, v + rot u h, rot v + p h, v = f, v, v V k h, u h, q = 0, q H k h. V 0 h Lagrange V 1 h R-T V 2 h DG All hypotheses are met / 94

48 Numerical solution of vector Laplacian, magnetic BC σ σ h rate (σ σ h) rate u u h rate rot(u u h) rate 2.16e e e e e e e e e e e e e e e e / 94

49 Numerical solution of vector Laplacian, Dirichlet BC For Dirichlet boundary conditions, σ = div u is sought in H 1, u is sought in H(rot) (the BC u t = 0 is essential, u n = 0 is natural). There is no complex, so our theory does not apply. σ σ h rate (σ σ h) rate u u h rate rot(u u h) rate 1.90e e e e e e e e e e e e e e e e DNA-Falk-Gopalakrishnan M3AS / 94

50 Eigenvalue problems Find λ R, 0 u D(L) s.t. Lu = λu, u H λ u 2 = du 2 + d u 2 > 0 so λ > 0 and Ku = λ 1 u. By the compactness property, K : W k W k is compact and self-adjoint, so 0 < λ 1 λ 2. Denote by v i corresponding orthonormal eigenvalues, E i = Rv i. Mixed discretization: Find λ h R, 0 (σ h, u h, p h ) V k 1 h V k h Hk h s.t. σ h, τ u h, dτ = 0, τ V k 1 h, dσ h, v + du h, dv + p h, v = λ h u h, v, v V k h, u h, q = 0, q H k h. 0 < λ 1h λ 2h... λ Nh h, v ih orthonormal, E ih = Rv ih 45 / 94

51 Convergence of eigenvalue problems Let m(j) i=1 E i be the span of the eigenspaces of the first j distinct eigenvalues. The method converges if j, ɛ > 0, h 0 > 0 s.t. max λ i λ ih ɛ and gap 1 i m(j) ( m(j) ) m(j) E i, E i,h ɛ if h h 0. A sufficient (and necessary) condition for eigenvalue convergence is operator norm convergence of the discrete solution operator K h P h to K (Kato, Babuska Osborn, Boffi Brezzi Gastaldi): i=1 i=1 W W h orthog. The mixed discretization of the eigenvalue problem converges if lim K hp h K L(W,W) = 0. h 0 46 / 94

52 Eigenvalue convergence follows from improved estimates u u h c{e(u)+η[e(du)+e(σ)]+(η 2 +δ)[e(dσ)+e(p)]+µe(p B u)} E(dσ) + E(p) + E(P B u) dσ + p + u f E(u) δ f, E(du) + E(σ) η f Therefore (K K h P h )f δ + η 2 + µ 0 Rates of convergence also follow, included doubled convergence rates for eigenvalues / 94

53 Exterior calculus

54 The de Rham complex for a domain in R n 1-D: 2-D: 3-D: 0 C (Ω) d/dx C (Ω) 0 0 C (Ω) grad C (Ω, R 2 ) rot C (Ω) 0 0 C (Ω) grad C (Ω, R 3 ) curl C (Ω, R 3 ) div C (Ω) 0 n-d: 0 Λ 0 (Ω) d Λ 1 (Ω) d Λ 2 (Ω) d d Λ n (Ω) 0 The space Λ k (Ω) = C (Ω, R n n skw ), the space of smooth differential k-forms on Ω. Exterior derivative: d k : Λ k (Ω) Λ k+1 (Ω) Integral of a k-form over an oriented k-simplex: f v R Stokes theorem: c du = c u, All this works on any smooth manifold u Λk, c C k 48 / 94

55 Exterior algebra Multilinear forms on an n-dimensional vector space V k times {}}{ Lin k V : k-linear maps ω : V V R Lin 0 V := R Lin 1 V = V covectors, 1-forms tensor product: (ω µ)(v 1,..., v j+k ) = ω(v 1,..., v j )µ(v j+1,..., v j+k ) dim Lin k V = n k dual basis for Lin 1 R n : dx 1,...,dx n with dx i (e j ) = δ ij basis for Lin k R n : dx σ1 dx σk, 1 σ 1,..., σ k n Alternating multilinear forms (algebraic k-forms) ω Alt k V if ω(..., v i,..., v j,...) = ω(..., v j,..., v i,...) dim Alt k V = ( ) n k skew part: (skw ω)(v 1,..., v k ) = 1 k! σ Σ k sign(σ)ω(v σ1,..., v σk ) exterior product: ω µ = ( ) j+k j skw(ω µ) basis for Alt k R n : dx σ1 dx σk, 1 σ 1 <... < σ k n 49 / 94

56 Exterior algebra continued If ω Alt k V, v V, the interior product ω v(v 1,..., v k 1 ) = ω(v, v 1,..., v k 1 ) ω v Alt k 1 V is (ω η) v = (ω v) η ± ω (η v) If V has an inner product, pick any orthonormal basis v 1,..., v n and define ω, η = σ ω(v σ(1),..., v σ(k) )η(v σ(1),..., v σ(k) ), (sum over σ : {1,..., k} {1,..., n} increasing). dim Alt n V = 1. Fix the volume form by vol(v 1,..., v n ) = ±1. An orientation for V fixes the sign. Hodge star: : Alt k V = Alt n k V defined by ω, η Alt k V ω µ = ω, µ vol, ω Alt k V, µ Alt n k V ω = ±ω On R n, vol = dx 1 dx n = det, dx σ1 dx σk = ±dx σ 1 dx σ n k Pullback: If L : V W linear, L : Alt k W Alt k V is defined L ω(w 1,..., w k ) = ω(lw 1,..., Lw n ) 50 / 94

57 Exterior algebra in R 3 vector proxy exterior prod. pullback interior prod. Alt 0 R 3 = R c c Alt 1 R 3 = R 3 u 1 dx 1 + u 2 dx 2 + u 3 dx 3 u Alt 2 R 3 = R 3 u 1 dx 2 dx 3 u 2 dx 1 dx 3 + u 3 dx 1 dx 2 u Alt 3 R 3 = R c c dx 1 dx 2 dx 3 : Alt 1 R 3 Alt 1 R 3 Alt 2 R 3 : R 3 R 3 R 3 : Alt 1 R 3 Alt 2 R 3 Alt 3 R 3 : R 3 R 3 R L : Alt 0 R 3 Alt 0 R 3 id : R R L : Alt 1 R 3 Alt 1 R 3 L T : R 3 R 3 L : Alt 2 R 3 Alt 2 R 3 adj L : R 3 R 3 L : Alt 3 R 3 Alt 3 R 3 (det L) : R R (c c det L) v : Alt 1 R 3 Alt 0 R 3 v : R 3 R v : Alt 2 R 3 Alt 1 R 3 v : R 3 R 3 v : Alt 3 R 3 Alt 2 R 3 v : R R 3 (c cv) inner prod. inner product on Alt k R 3 dot product on R and R 3 Hodge star : Alt 0 R 3 Alt 3 R 3 id : R R : Alt 1 R 3 Alt 2 R 3 id : R 3 R 3 51 / 94

58 Exterior calculus A differential k-form on a manifold M is a map x M ω x Alt k T x M. ω takes a point x M and k-tangent vectors and returns a number. 0-forms are functions, 1-forms are covector fields. We write ω Λ k (M) or CΛ k (M) if its continuous, C Λ k (M) if its smooth, etc. If M = Ω R n, ω : Ω Alt k R n. The general element of Λ k (Ω) is ω = σ a σ dx σ1 dx σk with a σ : Ω R. A smooth map φ : M M, induces a linear maps φ x : T x M T x M and so a pullback φ : Λ k (M ) Λ k (M) on differential forms: (φ ω) x = φ ( ) x ω φ(x) or (φ ω) x (v 1,..., v k ) = ω φ(x) φ x v 1,..., φ x v k φ (ω µ) = (φ ω) (φ µ) pullback of inclusion defines trace For ω = a dx σ1 dx σk Λ k (Ω), the exterior derivative dω = n a σ x dx k dx σ1 k dx σk Λ k+1 (Ω). It satisfies σ k=1 d d = 0, d(ω µ) = (dω) µ ± ω dµ, φ (dω) = d(φ ω) We may use any coordinate chart to define d : Λ k (M) Λ k+1 (M). 52 / 94

59 Exterior calculus continued A differential n-form on an oriented n-dim l manifold M may be integrated very geometrically (w/o requiring a metric or measure): ω R, ω Λ n (M) M φ ω = M ω, ω Λ n (M ) if φ preserves orientation. For ω = f(x) vol on Ω R n we get what notation suggests. Stokes theorem: dω = tr ω, ω Λ k 1 (Ω) Ω Ω Combining with Leibniz, we get the integration by parts formula dω η = ± ω dη + tr ω tr η, ω Λ k (Ω), η Λ n k 1 (Ω) Ω Ω Ω For M an oriented Riemannian manifold we have inner prod and on Alt k T x M and can define: ω, η L2 Λ = ω k x, η x vol = ω η Ω This allows us to rewrite the integration by parts formula dω, µ = ω, δµ + tr ω tr µ, ω Λ k 1, µ Λ k, Ω where δµ := ± d µ is the coderivative operator. M 53 / 94

60 Vector proxies in R n k 0 1 n 1 n Λ k (Ω) functions vector fields vector fields functions Λ k ( Ω) functions tang vctr flds functions 0 tr : Λ k (Ω) Λ k ( Ω) u Ω π t u Ω u Ω n 0 d : Λ k (Ω) Λ k+1 (Ω) grad curl div 0 δ : Λ k (Ω) Λ k 1 (Ω) 0 div curl grad : Λ k (Ω) R u(f) f f u t dh 1 f u n dh n 1 f u dh n φ : Λ k (Ω ) Λ k (Ω) u φ (φ x) T (u φ) (adj φ x)(u φ) (det φ x)(u φ) dim f = k φ : Ω Ω Piola transform 54 / 94

61 L 2 differential forms on a domain in R n Ω R n Lipschitz boundary HΛ k := {u L 2 Λ k du L 2 Λ k+1 } H Λ k = {u L 2 Λ k δu L 2 Λ k 1 } = HΛ n k u HΛ k (Ω) = tr u H 1/2 Λ k ( Ω) u H Λ k (Ω) = tr u H 1/2 Λ n k ( Ω) HΛ k = {u HΛ k tr u = 0}, H Λ k = {u H Λ k tr u = 0} Theorem If we view d as an unbdd operator L 2 Λ k L 2 Λ k+1 with domain HΛ k, then d = δ with domain H Λ k. Consequently H k = {ω L 2 Λ k dω = 0, δω = 0, tr ω = 0}. 55 / 94

62 L 2 de Rham complex on a domain in R n L 2 de Rham complex: Dual complex: 0 HΛ 0 d HΛ 1 d d HΛ n 0 0 H Λ 0 δ H Λ 1 Theorem (R. Picard 84) δ δ H Λ n 0 For a domain (or Riemannian manifold) w/ Lipschitz boundary the compactness property holds: HΛ k H Λ k is compact in L 2 Λ k. compactness property = Fredholm = closed 56 / 94

63 Finite element spaces of differential forms

64 Finite element spaces Goal: define finite element spaces Λ k h HΛk (Ω) satisfying the hypotheses of approximation, subcomplexes, and bounded cochain projections. A FE space is constructed by assembling three ingredients: Ciarlet 78 A triangulation T consisting of polyhedral elements T For each T, a space of shape functions V(T ), typically polynomial For each T, a set of DOFs: a set of functionals on V(T ), each associated to a face of T. These must be unisolvent, i.e., form a basis for V(T ). The FE space V h is defined as functions piecewise in V(T ) with DOFs single-valued on faces. The DOFs determine (1) the interelement continuity, and (2) a projection operator into V h. 57 / 94

65 Example: HΛ 0 = H 1 : the Lagrange finite element family Elements T T h are simplices in R n. Shape fns: V(T ) = P r (T ), some r 1. DOFs: u f (tr f u)q, q P r d 1 (f), f (T ), d = dim f v 0 (T ): u u(v) e 1 (T ): u e (tr e u)q, q P r 2 (e) f 2 (T ): u f (tr f u)q, q P r 3 (f) T : u uq, T q P r 4(T ) Theorem The number of DOFs = dim P r (T ) and they are unisolvent. The imposed continuity exactly forces inclusion in H / 94

66 Unisolvence for Lagrange elements in n dimensions Shape fns: V(T ) = P r (T ), DOFs: u f (tr f u)q, q P r d 1 (f), d = dim f DOF count: #DOF = n ( )( ) ( ) n + 1 r 1 r + n = = dim P r (T ). d + 1 d n d=0 # d (T ) dim P r d 1 (f d ) dim P r (T ) Unisolvence proved by induction on dimension (n = 1 is obvious). Suppose u P r (T ) and all DOFs vanish. Let f be a facet of T. Note tr f u P r (f) the DOFs associated to f and its subfaces applied to u coincide with the Lagrange DOFs in P r (f) applied to tr f u Therefore tr f u vanishes by the inductive hypothesis. Thus u = ( n i=0 λ i)p, p P r n 1 (T ). Choose q = p in the interior DOFs to see that p = / 94

67 Polynomial differential forms Polynomial diff. forms: P r Λ k (Ω) Homogeneous polynomial diff. forms: σ aσ dxσ 1 dx σ k, a σ P r (Ω) H r Λ k (Ω) ( )( ) ( )( ) r + n n r + n r + k dim P r Λ k = = r k r + k k ( )( ) r + n 1 n dim H r Λ k = = n ( )( ) r + n r + k r k n + r r + k k (Homogeneous) polynomial de Rham subcomplex: 0 P r Λ 0 d P r 1 Λ 1 d 0 H r Λ 0 d H r 1 Λ 1 d d P r n Λ n d H r n Λ n / 94

68 The Koszul complex For x Ω R n, T x Ω may be identified with R n, so the identity map can be viewed as a vector field. The Koszul differential κ : Λ k Λ k 1 is the contraction with the identity: κω = ω id. Applied to polynomials it increases degree. κ κ = 0 giving the Koszul complex: 0 P r Λ n κ Pr+1 Λ n 1 κ Pr+n Λ 0 0 κ dx i = x i, κ(ω µ) = (κω) µ ± ω (κµ) κ(f dx σ 1 dx σk ) = f k i=1 ( )i dx σ 1 dx σ i dx σk 3D Koszul complex: 0 P r Λ 3 x P r+1 Λ 2 x P r+2 Λ 1 Theorem (Homotopy formula) x P r+3 Λ 0 0 (dκ + κd)ω = (r + k)ω, ω H r Λ k. 61 / 94

69 Proof of the homotopy formula (dκ + κd)ω = (r + k)ω, ω H r Λ k, ω H r Λ k Proof by induction on k. k = 0 is Euler s identity. Assume true for ω H r Λ k 1, and verify it for ω dx i. dκ(ω dx i ) = d(κω dx i + ( 1) k 1 ω x i ) = d(κω) dx i + ( 1) k 1 (dω) x i + ω dx i. κd(ω dx i ) = κ(dω dx i ) = κ(dω) dx i + ( 1) k dω x i. (dκ + κd)(ω dx i ) = [(dκ + κd)ω] dx i + ω dx i = (r + k)(ω dx i ). 62 / 94

70 Consequences of the homotopy formula The polynomial de Rham complex is exact (except for constant 0-forms in the kernel). The Koszul complex is exact (except for constant 0-forms in the coimage). κdω = 0 = dω = 0, dκω = 0 = κω = 0 H r Λ k = κh r 1 Λ k+1 dh r+1 Λ k 1 Define P r Λ k = P r 1 Λ k + κh r 1 Λ k+1 P r Λ 0 = P r Λ 0, P r Λ n = P r 1 Λ n, else P r 1 Λ k P r Λ k P r Λ k ( )( ) dim P r + n r + k 1 r Λ k = = r r + k k r + k dim P rλ k R(d P r Λ k ) = R(d P r Λ k ), N (d P r Λ k ) = N (d P r 1 Λ k ) The complex (with constant r) 0 P r Λ 0 d P r Λ 1 d d P r Λ n 0 is exact (except for constant 0-forms). 63 / 94

71 Complexes mixing P r and P r On an n-d domain there are 2 n 1 complexes beginning with P r Λ 0 (or ending with P r Λ n ). At each step we have two choices: P r Λ k 1 P r Λ k P r 1 Λ k or P r Λ P k 1 r Λ k P r 1 Λ k In 3-D: 0 P r Λ 0 d P r Λ 1 d P r Λ 2 d P r 1 Λ P r Λ 0 d P r Λ 1 d P r 1 Λ 2 d P r 2 Λ 3 0, 0 P r Λ 0 d P r 1 Λ 1 d P r 1 Λ2 d P r 2 Λ 3 0, 0 P r Λ 0 d P r 1 Λ 1 d P r 2 Λ 2 d P r 3 Λ 3 0, 64 / 94

72 The P r Λ k family of simplicial FE differential forms Given: a mesh T h of simplices T, r 1, 0 k n, we define P r Λ k (T h ) via: Shape fns: P r Λ k (T ) DOFs: u (tr f u) q, q P r+k d 1 Λ d k (f), f (T ), d = dim f k f Theorem The number of DOFs = dim P r Λ k (T ) and they are unisolvent. The imposed continuity exactly enforces inclusion in HΛ k. 65 / 94

73 Dimension count #DOFs = # d (T ) dim P r+k d 1 Λ k (R d ) d k ( )( )( ) n + 1 r + k 1 d = d k d + 1 d k = j 0 ( n + 1 j + k + 1 )( r + k 1 j + k )( j + k ) j Simplify using the identities ( )( ) ( )( ) a b a a c = b c c a b ( )( ) ( ) a c a + c = b + j j a b j 0 to get ( )( ) r + n r + k 1 #DOFs = = dim P r Λ k r + k k 66 / 94

74 Proof of unisolvence for P r Λ k Lemma If u P r 1 Λ k (T ) and T u q = 0 q P r+k n 1Λ n k (T ), then u = 0. Proof: This can be proved by an explicit choice of test function. Proof of unisolvence: Suppose u P r Λ k (T ) and all the DOFS vanish: f (tr f u) q = 0, q P r+k d 1 Λ d k (f), f (T ). Then tr f u P r Λ k (f) and all its DOFs vanish. By induction on dimension, tr u vanishes on the boundary. So we need to show: u P r Λ k (T ), T u q = 0 q P r+k n 1Λ n k (T ) = u = 0 In view of lemma, we just need to show u P r 1 Λ k (T ). By the homotopy formula, u P r Λ k, du = 0 = u P r 1 Λ k. So it remains to show that du = 0. du P r 1 Λ k+1 (T ), T du p = ± T u dp = 0 p P r+k nλ n k 1 (T ). Therefore du = 0 by the lemma (with k k +1). 67 / 94

75 The P r Λ k family of simplicial FE differential forms Given: a mesh T h of simplices T, r 1, 0 k n, we define P r Λ k (T h ) via: Shape fns: P r Λ k (T ) DOFs: u (tr f u) q, f q P r+k d Λd k (f), f (T ), d = dim f k Theorem The number of DOFs = dim P r Λ k (T ) and they are unisolvent. The imposed continuity exactly enforces inclusion in HΛ k. 68 / 94

76 The P r family in 2D P r Λ 0 P r Λ 1 P r Λ 2 Lagrange Raviart-Thomas 76 DG r = 1 r = 2 r = 3 69 / 94

77 The P r Λ k family in 3D P r Λ 0 P r Λ 1 P r Λ 2 P r Λ 3 Lagrange Nédélec 80 Nédélec 80 DG r = 1 r = 2 r = 3 70 / 94

78 The P r Λ k family in 2D P r Λ 0 P r Λ 1 P r Λ 2 Lagrange Brezzi-Douglas-Marini 85 DG r = 1 r = 2 r = 3 71 / 94

79 The P r Λ k family in 3D P r Λ 0 P r Λ 1 P r Λ 2 P r Λ 3 Lagrange Nédélec 86 Nédélec 86 DG r = 1 r = 2 r = 3 72 / 94

80 Application of the P r and P r families to the Hodge Laplacian The shape function spaces P r Λ k (T ) and P r Λ k (T ) combine into de Rham subcomplexes. The DOFs connect these spaces across elements to create subspaces of HΛ k (Ω). Therefore the assembled finite element spaces P r Λ k (T h ) and P r Λ k (T h ) combine into de Rham subcomplexes (in 2 n 1 ways). The DOFs of freedom determine projections from Λ k (Ω) into the finite element spaces. From Stokes thm, these commute with d. Suitably modified, we obtain bounded cochain projections. Thus the abstract theory applies. We may use any two adjacent spaces in any of the complexes. P r Λ k 1 (T ) or P r Λ k 1 (T ) d P r Λ k (T ) or P r 1 Λ k (T ) 73 / 94

81 Rates of convergence Rates of convergence are determined by the improved error estimates from the abstract theory. They depend on The smoothness of the data f. The amount of elliptic regularity. The degree of of complete polynomials contained in the finite element spaces. The theory delivers the best possible results: with sufficiently smooth data and elliptic regularity, the rate of convergence for each of the quantities u, du, σ, dσ, and p in the L 2 norm is the best possible given the degree of polynomials used for that quantity. Eigenvalues converge as O(h 2r ). 74 / 94

82 Historical notes The P 1 Λk complex is in Whitney 57 (Bossavit 88). In 76, Dodziuk and Patodi defined a finite difference approximation based on the Whitney forms to compute the eigenvalues of the Hodge Laplacian, and proved convergence. In retrospect, that method can be better viewed as a mixed finite element method. This was a step on the way to proving the Ray-Singer conjecture, completed in 78 by W. Miller. The P r Λ k complex is in Sullivan 78. Hiptmair gave a uniform treatment of the P r Λ k spaces in 99. The unified treatment and use of the Koszul complex is in DNA-Falk-Winther / 94

83 Bounded cochain projections The DOFs defining P r Λ k (T h ) and P r Λ k (T h ) determine canonical projection operators Π h from piecewise smooth forms in HΛ k onto Λ k h. However, Π h is not bounded on HΛ k (much less uniformly bounded wrt h). Π h is bounded on CΛ k. If we have a smoothing operator R ɛ,h Lin(L 2 Λ k, CΛ k ) such that R ɛ,h commutes with d, we can define Q ɛ,h = Π h R ɛ,h and obtain a bounded operator L 2 Λ k Λ k h which commutes with d (as suggested by Christiansen). However Q h will not be a projection. We correct this by using Schöberl s trick: if the finite dimensional operator Q ɛ,h Λ k h : Λ k h Λ k h is invertible, then π h := (Q ɛ,h Λ k h ) 1 Q ɛ,h, is a bounded commuting projection. It remains to get uniform bds on π h. 76 / 94

84 The two key estimates For this we need two key estimates for Q ɛ,h := Π h R ɛ,h : For fixed ɛ, Q ɛ,h is uniformly bounded: ɛ > 0 suff. small c(ɛ) > 0 s.t. sup Q ɛ,h Lin(L2,L 2 ) c(ɛ) h lim I Q ɛ,h Lin(L2,L ɛ 0 2 ) = 0 uniformly in h Theorem Suppose that these two estimates hold and define π h := (Q ɛ,h Λ k h ) 1 Q ɛ,h, where Λ k h is either P rλ k (T h ) or P r+1 Λk (T h ). Then, for h sufficiently small, π h is a cochain projection onto Λ k h and ω π h ω ch s ω Hs Λ k, ω Hs Λ k, 0 s r / 94

85 The smoothing operator The simplest definition is to take R ɛ,h u to be an average over y B 1 of (F y ɛ,h ) u where F y ɛ,h (x) = x + ɛhy: R ɛ,h u(x) = ρ(y)[(f y eh ) u](x) dy B 1 Needs modification near the boundary and for non-quasiuniform meshes. The key estimates can be proven using macroelements and scaling. 78 / 94

86 Bases for the spaces

87 Bases Since the DOFs determine a basis for the dual space of a FE space, there is a corresponding basis for the FE space. An alternative is to use the Bernstein basis fns which are given explicitly in terms of the barycentric coordinates λ i : Basis of P 3 dual to the nodal DOFs Bernstein basis functions λ i 0λ j 1. For the P r Λ k and P r Λ k families in n-dimensions, there is of course again the basis determined by the DOFs. In addition, there is an explicit basis analogous to the Bernstein basis (DNA-Falk-Winter 2009). 79 / 94

88 Some computations with barycentric coordinates The barycentric coordinates λ 0,..., λ n form the dual basis for P 1 (T ) = P 1 Λ0 (T ) dλ 1 dλ n = c vol Alt n T with c = (dλ1 dλ n )(x 1 x 0,..., x n x 0 ) vol(x 1 x 0,..., x n x 0 ) = 1 n! T More generally, dλ 0... dλ i... dλ n = ( 1)i n! T vol. κdλ i = λ i λ i (0), so κ(dλ σ0 dλ σk ) = ψ P 0 Λ k. k ( 1) i λ σi dλ σ0... dλσi... dλ σk + ψ, i=0 80 / 94

89 The Whitney forms Define the Whitney form associated to the k-face f with vertices x σ0,..., x σk by φ f = k ( 1) i λ σi dλ σ0... dλσi... dλ σk P 1 Λk i=0 vertices: edges: triangles: etc. If f, g k (T ) then λ i λ i dλ j λ j dλ i λ i dλ j dλ k λ j dλ i dλ k + λ k dλ i dλ j g tr g φ f = { 0, g f, 1/k!, g = f after normalization, the Whitney forms are a basis for P 1 Λk dual to the DOFs. 81 / 94

90 Explicit geometric bases The Bernstein basis is an explicit alternative to the Lagrange basis for the Lagrange finite elts. P r = span{ λ α := λ α0 0 λαn n α = r } P r (T, f) :=span{ λ α supp α = {σ 0,..., σ k }, α = r } P r (T ) = f P r (T, f) }{{}{}}{ P r (T, f) = tr P r (f) = P r dim f 1 (f) P r There are similar geometric bases for all k: P r Λ k (T ) = P r Λ k (T, f), P r Λ k (T, f) = P r Λ k (f) = P r+k dim f Λdim f k (f) tr dim f k Λ k (T ) = P r Λ k (T, f), P r Λ k (T, f) = P r Λ k (f) = P r+k dim f 1 Λ dim f k (f) tr dim f k 82 / 94

91 Basis for P r Λ k To create a basis for P r Λ k (the easier case), we consider all products λ α 0 0 λαn n φ f, f k (T ), αi = r 1 This form is associated the the face whose vertices are in f or for which α i > 0. E.g., λ 2 3 φ [1,2] is associated with the face [1, 2, 3]. These span P r Λ k. However they are not linearly independent since k ( 1) i λ σi φ [σ0 σ i σ k ] = 0. i=0 To get a linearly independent spanning set, we impose the extra condition that if α i 0 then i σ 0 (the least vertex index of f ). E.g., λ 1 λ 2 φ [1,2] and λ 2 3 φ [1,2] are included in the basis for P 3 Λ2 but λ 0 λ 3 φ [1,2] is not. 83 / 94

92 Example: explicit bases for P r Λ 1 and P r Λ 2 on a tet P r Λ 1 (T 3 ) r Edge [x i, x j] Face [x i, x j, x k] Tet [x i, x j, x k, x l] 1 φ ij 2 λ iφ ij, λ jφ ij λ kφ ij, λ jφ ik 3 {λ 2 i, λ 2 j, λ iλ j}φ ij {λ i, λ j, λ k}λ kφ ij, {λ i, λ j, λ k}λ jφ ik λ kλ lφ ij, λ jλ lφ ik, λ jλ kφ il P r Λ 2 (T 3 ) r Edge [x i, x j] Face [x i, x j, x k] Tet [x i, x j, x k, x l] 1 φ ijk 2 λ iφ ijk, λ jφ ijk, λ kφ ijk λ lφ ijk, λ kφ ijl, λ jφ ikl 3 {λ 2 i, λ 2 j, λ 2 k}φ ijk {λ i, λ j, λ k, λ l}λ lφ ijk {λ iλ j, λ iλ k, λ jλ k}φ ijk {λ i, λ j, λ k, λ l}λ kφ ijl {λ i, λ j, λ k, λ l}λ jφ ikl 84 / 94

93 Finite element differential forms on cubical meshes

94 The tensor product construction Again there are two families (only?). One results from a tensor product construction. (DNA Boffi Bonizzoni) Suppose we have a finite element de Rham subcomplex V on an element S R m : V k d V k+1 V k Λ k (S) and another, W, on another element T R n : W k d W k+1 The tensor-product construction produces a new complex V W, a subcomplex of the de Rham complex on S T. 85 / 94

95 The tensor product construction Again there are two families (only?). One results from a tensor product construction. (DNA Boffi Bonizzoni) Suppose we have a finite element de Rham subcomplex V on an element S R m : V k d V k+1 V k Λ k (S) and another, W, on another element T R n : W k d W k+1 The tensor-product construction produces a new complex V W, a subcomplex of the de Rham complex on S T. Shape fns: (V W) k = i+j=k π S V i π T W j (π S : S T S) DOFs: (η ρ)(π S v π T w) := η(v)ρ(w) 85 / 94

96 Finite element differential forms on cubes: the Q r Λ k family Start with the simple 1-D degree r finite element de Rham complex, V r : 0 P r Λ 0 (I) d P r 1 Λ 1 (I) 0 u(x) u (x) dx Take tensor product n times: Q r Λ k (I n ) := (V r V r ) k Q r = P r P r, P r 1 P r dx 1 + P r P r 1 dx 2, P r 1 P r 1 dx 1 dx 2 Q 1 Λ0 Q 1 Λ1 Q 1 Λ2 Raviart-Thomas 76 Nedelec / 94

97 The 2nd family of finite element differential forms on cubes The S r Λ k (I n ) family of FEDFs: (DNA Awanou 12) Shape fns: For a form monomial m = x α i 1 xαn n dx σ1 dx σk, define deg m = α i, ldeg m = #{ i α i = 1, α i {σ 1,..., σ k } }. Ex: If m = x 1 x 2 x3 5 dx 1, deg m = 7, ldeg m = 1. H r,l Λ k (I n ) = span of monomials with deg = r, ldeg l, J r Λ k (I n ) = l 1 κh r+l 1,l Λ k+1 (I n ), S r Λ k (I n ) = P r Λ k (I n ) J r Λ k (I n ) dj r+1 Λ k 1 (I n ). DOFs: u f u q, q P r 2dΛ d k (f), f (I n ) 87 / 94

98 Key properties For any n 1, r 1, 0 k n: Degree property: P r Λ k (I n ) S r Λ k (I n ) P r+n k Λ k (I n ) Inclusion property: S r Λ k (I n ) S r+1 Λ k (I n ) Trace property: For each face f of I n, tr f S r Λ k (I n ) = S r Λ k (f). Subcomplex property: ds r Λ k (I n ) S r 1 Λ k+1 (I n ) Unisolvence: The indicated DOFs are correct in number and are unisolvent. Commuting projections: The DOFs determine commuting projections from the de Rham complex to the subcomplex S r Λ 0 (I n ) d S r 1 Λ 1 (I n ) d d S r n Λ n (I n ). 88 / 94

99 The case of 0-forms (H 1 elements) Define sdeg m of a monomial m to be the degree ignoring variables that enter linearly: sdeg x 3 yz 2 = 5. For a polynomial p, sdeg p is the maximum over its monomials. S r (I n ) = { p P(I n ) sdeg p r } DNA-Awanou 10 1D: S r (I) = P r (I), 2D: S r (I 2 ) = P r (I 2 ) + span[x r y, xy r ] serendipity S 1 (I 2 ) S 2 (I 2 ) S 3 (I 2 ) S 4 (I 2 ) S 5 (I 2 ) 89 / 94

100 Serendipity 0-forms in more dimensions P r (I n ) n Dimensions S r (I n ) Q r (I n ) / 94

101 The 2nd cubic family in 2-D S 2 Λ 0 S 1 Λ 1 S 0 Λ 2 S 3 Λ 0 S 2 Λ 1 S 1 Λ 2 S r Λ k (I 2 ) k / 94

102 The 2nd cubic family in 3-D 92 / 94

103 Dimensions and low order cases S r Λ k (I 3 ) k S 1 Λ 1 (I 3 ) new element S 1 Λ 2 (I 3 ) corrected element 93 / 94

104 The 3D shape functions in traditional FE language S r Λ 0 : polynomials u such that sdeg u r S r Λ 1 : (v 1, v 2, v 3 ) + (x 2 x 3 (w 2 w 3 ), x 3 x 1 (w 3 w 1 ), x 1 x 2 (w 1 w 2 )) + grad u, v i P r, w i P r 1 independent of x i, sdeg u r + 1 S r Λ 2 : (v 1, v 2, v 3 ) + curl(x 2 x 3 (w 2 w 3 ), x 3 x 1 (w 3 w 1 ), x 1 x 2 (w 1 w 2 )), v i, w i P r (I 3 ) with w i independent of x i S r Λ 3 : v P r 94 / 94