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Outline Finite element differential forms Background on the continuous problem Douglas N. Arnold Requirements for discretization joint with R. Falk and R. Winther Subspaces of finite element differential forms School of Mathematics University of Minnesota Application to elasticity April 2009 2 / 41 References Finite element exterior calculus, homological techniques, and applications, Acta Numerica 2006, p. 1–155 Background on the continuous problem “Any young (or not so young) mathematician who spends the time to master this paper will have tools that will be useful for his or her entire career.” — Math Reviews Exterior calculus The de Rham complex Hodge theory and the Hodge Laplacian Mixed finite element methods for linear elasticity with weakly imposed symmetry, Math. Comp. 2007 Geometric decompositions and local bases for spaces finite element differential forms, to appear in CMAME everything is at http://umn.edu/∼arnold 3 / 41 Differential forms Exterior calculus Exterior derivative. A k-form ω can be differentiated to get a (k + 1)-form d k ω : take the directional derivative of ωx (v1 , . . . , vk ) in the direction vk+1 and skew-symmetrize. n Let Ω be a domain in R , f : Ω → R smooth. ∀x ∈ Ω, dfx : Rn → R is a linear map: dfx (v ) = v · ∇f (x). d k+1 ◦ d k = 0, i.e., Bk := range(d k−1 ) ⊂ Zk := ker(d k ) (More generally, on any smooth manifold, dfx : Tx Ω → R is linear) f is a 0-form, df is a 1-form, Λ0 (Ω), If F : Ω → Ω0 , the pullback F ∗ maps Λk (Ω0 ) → Λk (Ω): (F ∗ ω)x (v1 , . . . , vk ) = ωF (x) (dFx v1 , . . . , dFx vk ) Λ1 (Ω) ω is a k-form ⇐⇒ ∀x, ωx is an alternating k-linear form on Rn (Tx Ω) dx 1 , . . . , dx n : standard dual basis of Rn =⇒ dx 1 ∧ · · · ∧dx k , i1 < · · · < ik , basis for alternating k-linear forms. ω ∈ Λk (Ω) ⇐⇒ ω = A k-form has n k X If S ⊂ Ω is a submanifold, the pullback of the inclusion is the trace, tr : Λk (Ω) → Λk (S). This is not just the restriction. There is Ra natural integral of a k-form ω over a k-dimensional surface: S ω ∈ R fi1 ···ik dx 1 ∧ · · · ∧dx k . coefficients (so Λk (Ω) = 0 for k > n). Z Z Stokes theorem: dω = Ω tr ω, ω ∈ Λn−1 (Ω) ∂Ω 6 / 41 5 / 41 The case of Ω ⊂ R3 The de Rham complex For Ω ⊂ R3 , the de Rham complex boils down to We work in a Hilbert space setting. d = d k is a closed densely-defined unbounded operator L2 Λk (Ω) → L2 Λk+1 (Ω) with closed range. grad scalar fn Its domain is HΛk (Ω) = { ω ∈ L2 Λk (Ω) | dω ∈ L2 Λk+1 (Ω) }, a Hilbert space with the graph norm kωk2HΛ = kωk2 + kdωk2 . d1 vector field div vector field scalar fn Physical vector quantities may be divided into two classes, in one of which the quantity is dened with reference to a line, while in the other They connect to form the de Rham complex: d0 curl 0 → H 1 (Ω) −−→ H(curl, Ω) −−→ H(div, Ω) −−→ L2 (Ω) → 0 the quantity is dened with reference to an area. d n−1 James Clerk Maxwell, 0 → HΛ0 (Ω) −→ HΛ1 (Ω) −→ · · · −−→ HΛn (Ω) → 0 The kth de Rham cohomology group is Zk /Bk . Treatise on Electricity & Magnetism, 1891 # components of Ω, # tunnels thru Ω, dim(Hk ) = # voids in Ω, 0, Since we are in a Hilbert setting, the cohomology space is isomorphic to the space of harmonic k-forms: Hk := { ω ∈ Zk | ω ⊥ Bk } 7 / 41 k k k k =0 =1 =2 =3 For Ω contractible (e.g., convex) H0 = R, Hk = 0, k > 0 8 / 41 Hodge theory Naive weak formulation of the Hodge Laplacian An obvious weak formulation (in the case of no harmonic forms): Find u ∈ HΛk ∩ H ∗ Λk such that Using the inner product, we have: Hodge decomposition: L2 Λk (Ω) = Bk ⊕ Poincaré’s inequality: kωkL2 ≤ c kdωkL2 , Hk ⊕ Zk⊥ hdu, dv i + hd ∗ u, d ∗ v i = hf , v i, adjoint: d ∗ : L2 Λk+1 (Ω) → L2 Λk (Ω), closed densely-defined This formulation is problematic for discretization. Lagrange elements, or any conforming finite elements, converge. . . but not to the solution! H = ker(d) ∩ ker(d ∗ ) Hodge Laplace problem: Given f ∈ Λk (0 ≤ k ≤ n), find u ∈ Λk such that (d ∗ d + d d ∗ )u = f v ∈ HΛk ∩ H ∗ Λk Equivalently, 12 kduk2 + 12 kd ∗ uk2 − hf , ui → minimum ω ∈ HΛk , ω ⊥ Zk curl curl u − grad div u = f (with natural BC) plus BC The harmonic functions determine well-posedness: (1) ∃u ⇐⇒ f ⊥ Hk , (2) u is determined only mod Hk 9 / 41 Mixed formulation of the Hodge Laplacian The mixed formulation is well-posed Given f ∈ L2 Λk (Ω), find σ ∈ HΛk−1 , u ∈ HΛk , p ∈ Hk : hσ, τ i − hdτ, ui = 0 hdσ, v i + hdu, dv i+hp, v i = hf , v i hu, qi = 0 10 / 41 Well-posedness of the mixed formulation follows directly from the Hodge decomposition and the Poincaré inequality. We must prove the inf-sup condition for the bilinear form ∀τ ∈ HΛk−1 ∀v ∈ HΛk ∀q ∈ Hk B(σ, u, p; τ, v , q) = hσ, τ i−hdτ, ui−hdσ, v i−hdu, dv i−hv , pi−hu, qi, Equivalently 12 hσ, σi − 12 hdu, dui − hdσ, ui − hu, pi + hf , ui → saddle point Special cases: k = 0: ordinary Laplacian k = n: mixed Laplacian k = 1, n = 3: σ = − div u, grad σ + curl curl u = f k = 2, n = 3: σ = curl u, curl σ − grad div u = f i.e., given (σ, u, p) ∈ HΛk−1 × HΛk × Hk , we need to control kσkHΛ + kukHΛ + kpk by a bounded choice of τ , v , and q. τ =σ v =u controls kσk, v = dσ controls kdσk, v = p controls kduk. How to control kuk? Hodge decomp.: u = dη + s + z, controls kpk η ∈ HΛk−1 , s ∈ Hk , z ∈ Zk⊥ Includes the problems τ = η controls kdηk and q = s controls ksk. How to control kzk? curl curl u = f , div u = 0 and div u = f , curl u = 0 Poincaré’s inequality: kzk ≤ ckdzk = ckduk 11 / 41 12 / 41 Discretization Choose finite dimensional subspaces Λk−1 ⊂ HΛk−1 , Λkh ⊂ HΛk . h Assume that dΛk−1 ⊂ Λkh so we get a subcomplex. h d k−1 · · · −−→ HΛk−1 −−−→ HΛk −−→ · · · x x ∪ ∪ Requirements for discretization d k−1 · · · −−→ Λk−1 −−−→ Λkh −−→ · · · h Subcomplexes Bounded cochain projections We may then define Bkh , Zkh , Hkh = (Bkh )⊥ ∩ Zkh (and obtain the discrete Hodge decomp: Λkh = Bkh ⊕ Hkh ⊕ (Zkh )⊥ ) Galerkin’s method: discretize the mixed formulation with k k Λk−1 , Λk , Hk −→ Λk−1 h , Λ h , Hh When is it stable, consistent, and convergent? 14 / 41 A simple case Bounded cochain projections Key property: Suppose that there exists a bounded cochain projection. Stable discretization is not obvious, even in simple cases. σ ∈ H(div ), u ∈ L2 : hσ, τ i − hdiv τ, ui = 0 hdiv σ, v i = hf , v i d k−1 · · · −−→ HΛk−1 −−−→ HΛk −−→ · · · πk yπhk−1 y h ∀τ ∈ H(div) ∀v ∈ L2 d k−1 · · · −−→ Λk−1 −−−→ Λkh −−→ · · · h πhk bounded πhk a projection πhk d k−1 = d k−1 πhk−1 Theorem If kv − πhk v k < kv k ∀v ∈ Hk , then the induced map on cohomology is an isomorphism. gap Hk , Hkh ≤ sup kv − πhk v k v ∈Hk kv k=1 The discrete Poincaré inequality holds uniformly in h. P1 -P0 Galerkin’s method is stable and convergent. Raviart–Thomas - P0 15 / 41 16 / 41 Proof of discrete Poincaré inequality Subspaces of finite element differential forms Thm. There is a positive constant c, independent of h, such that kωk ≤ ckdωk, ω ∈ Zk⊥ h . Shape functions Degrees of freedom Dual bases and explicit bases Finite element de Rham subcomplexes Bounded cochain projections Applications k⊥ ⊂ HΛk (Ω) by dη = dω. By Proof. Given ω ∈ Zk⊥ h , define η ∈ Z the Poincaré inequality, kηk ≤ ckdωk, so it is enough to show that k kωk ≤ ckηk. Now, ω − πh η ∈ Λh and d(ω − πh η) = 0, so ω − πh η ∈ Zkh . Therefore kωk2 = hω, πh ηi + hω, ω − πh ηi = hω, πh ηi ≤ kωkkπh ηk, whence kωk ≤ kπh ηk, and the result follows. 17 / 41 Finite element differential forms Construction of FE differential forms Key to the construction is the Koszul differential κ : Λk → Λk−1 : Let T = Th be a triangulation of Ω ⊂ Rn . We wish to construct finite element spaces Λkh ⊂ HΛk (Ω) which form a finite dimensional subcomplex with bounded cochain projections. (κω)x (v 1 , . . . , v k−1 ) = ωx (x, v 1 , . . . , v k−1 ) κ κ Koszul complex •X ×X X 0 ← Pr Λ0 ←−− Pr −1 Λ1 (R3 ) ←−− Pr −2 Λ2 (R3 ) ←−− Pr −3 Λ3 (R3 ) ← 0 a finite dimensional space of polynomial forms on the simplex, and a decomposition of its dual space into subspaces associated to the subsimplices (degrees of freedom) C.f., the polynomial de Rham complex d d d 0 −−→ Pr Λ0 −−→ Pr −1 Λ1 −−→ · · · −−→ Pr −n Λn −−→ 0 Special cases: ∀ω ∈ Hr Λk (homogeneous polys) κ is a contracting chain homotopy Key relation: (dκ + κd)ω = (r + k)ω Lagrange finite element spaces Pr Λ0 (T ) Discontinuous piecewise polynomials of degree r : Pr κ 0 ←−− Pr Λ0 ←−− Pr −1 Λ1 ←−− · · · ←−− Pr −n Λn ←−− 0 The Λkh are finite element spaces in the sense that they can be assembled from the following data on each simplex: Λn (T ). Ex: grad(u · x) + (curl u)×x = 3u, u ∈ H2 (R3 , R3 ) Whitney k-forms (one DOF per k-face): P1− Λk (T ) Hr Λk = dHr +1 Λk−1 ⊕ κHr −1 Λk+1 19 / 41 20 / 41 Pr Λk and Pr− Λk Degrees of freedom Using the Koszul differential, we define a special space of polynomial differential k-forms between Pr Λk and Pr −1 Λk : To define the finite element spaces, we must specify degrees of freedom, i.e., a decomposition of the dual spaces Pr Λk (T )∗ and Pr− Λk (T )∗ , into subspaces associated to subsimplices f of T . X Pr−Λk := Pr −1 Λk + κHr −1 Λk+1 + dHr +1 Λk−1 Pr− Λk (Th ) dim Pr Λk = dim Pr− Λk = 8 k > <Pr Λ (Th ), = Pr −1 Λk (Th ), > : strictly between, DOF for Pr− Λk (T ): to a subsimplex f of dim. d ≥ k we associate Z ω 7→ Trf ω ∧ η, η ∈ Pr +k−d−1 Λd−k (f ) Hiptmair f Z − k DOF for Pr Λ (T ): ω 7→ Trf ω ∧ η, η ∈ Pr +k−d Λd−k (f ) k = 0, k = n, 0<k<n n+r n n+r r +k = n k n−k k n+r r +k −1 n−k k f The resulting FE spaces have precisely the continuity required by HΛk : Theorem: Pr Λk (T ) = { ω ∈ HΛk (Ω) : ω|T ∈ Pr Λk (T ) ∀T ∈ T } and similarly for Pr− . k−1 d For each form degree k and polynomial degree r , Pr Λk and Pr− Λk are the two natural finite element subspace of HΛk (Ω). The projections associated to these DOF commutes with d (Stokes theorem). For example, they are invariant under pull-pack by any affine map Rn → Rn , and they are almost the only such spaces of polynomial k-forms. Λk−1 −−−→ Λk k yPhk−1 yPh d k−1 Λk−1 −−−→ Λkh h 21 / 41 Ex: DOFs for P3 Λ1 on a tetrahedron Finite element differential forms/Mixed FEM Pr− Λ0 (T ) = Pr Λ0 (T ) ⊂ H 1 22 / 41 Lagrange elts Cubic Nedelec 2nd kind edge elements, dim = 60 Pr− Λn (T ) = Pr −1 Λn (T ) ⊂ L2 discontinuous elts n = 2: Pr− Λ1 (T ) ⊂ H(curl) Raviart–Thomas elts n = 2: Pr Λ1 (T ) ⊂ H(curl) Brezzi–Douglas–Marini elts n = 3: Pr− Λ1 (T ) ⊂ H(curl) Nedelec 1st kind edge elts n = 3: Pr Λ1 (T ) ⊂ H(curl) Nedelec 2nd kind edge elts n = 3: Pr− Λ2 (T ) ⊂ H(div) Nedelec 1st kind face elts n = 3: Pr Λ2 (T ) ⊂ H(div) Nedelec 2nd kind face elts DOFs are: For each edge, moments versus P3− Λ0 (e) (4 DOF/edge × 6 edges) For each face, moments versus P2− Λ1 (f ) (8 DOF/face × 4 faces) On the tet, moments versus P1− Λ2 (f ) (4 DOF) 23 / 41 24 / 41 Dual bases Explicit geometric bases As a computational basis for Pr Λk (T ) and Pr− Λk (T ) one choice is the dual basis to the degrees of freedom. The Bernstein basis is an explicit alternative to the Lagrange basis for the Lagrange finite elts. αn 0 Pr = span{ λα := λα 0 · · · λn | |α| = r } Pr (T , f ) := span{ λα | spt α = {i0 , . . . , ik }, |α| = r } For k = 0 this is the standard Lagrange basis. For P1− Λk (T ) there is one basis element for each k-simplex f = [xi0 , . . . , xik ], namely the Whitney form given in barycentric coordinates by φi0 ···ik := k X Pr (T ) = M Pr (T , f ) f ∼ = Pr (T , f ) − → P̊r (f ) ∼ = Pr −dim f −1 (f ) | z c i ∧ · · · ∧dλi (−1)p λip dλi0 ∧ · · · ∧dλ p k {z }| tr } { There are similar geometric bases for all k: p=0 Pr Λk (T ) = λi P1− Λ0 = P1 Λ0 : λi dλj − λj dλi P1− Λ1 : λi dλj ∧dλk − λj dλi ∧dλk + λk dλi ∧dλj P1− Λ2 : etc. M ∼ = Pr Λk (T , f ), Pr Λk (T , f ) − → P̊r Λk (f ) ∼ = Pr−+k−dim f Λdim f −k (f ) tr dim f ≥k Pr− Λk (T ) = M ∼ = Pr− Λk (T , f ), Pr− Λk (T , f ) − → P̊r− Λk (f ) ∼ = Pr +k−dim f −1 Λdim f −k (f ) tr dim f ≥k 26 / 41 25 / 41 Example: explicit bases for Pr− Λ2 and Pr Λ2 on a tet Finite element de Rham subcomplexes We don’t only want spaces, we also want them to fit together into discrete de Rham complexes. Pr− Λ2 One such FEdR subcomplex uses Pr− Λk spaces of constant degree r : r Face [xi , xj , xk ] 1 φijk 2 Tet [xi , xj , xk , xl ] λi φijk , λj φijk , λk φijk d d λl φijk , λk φijl , λj φikl grad −−→ curl d Face [xi , xj , xk ] 1 λk dλi ∧dλj , λj dλi ∧dλk , λi dλj ∧dλk 2 λ2k dλi ∧dλj , λj λk dλi ∧d(λk − λj ) λ2j dλi ∧dλk , λi λj d(λj − λi )∧dλk λ2i dλj ∧dλk , λi λk dλj ∧d(λk − λi ) div −−→ −−→ →0 Whitney 1957, Bossavit 1988 Another uses Pr Λk spaces with decreasing degree: Pr Λ2 r d 0 → Pr− Λ0 (T ) −−→ Pr− Λ1 (T ) −−→ · · · −−→ Pr− Λn (T ) → 0 0→ d d 0 → Pr Λ0 (T ) −−→ Pr −1 Λ1 (T ) −−→ · · · −−→ Pr −n Λn (T ) → 0 Tet [xi , xj , xk , xl ] 0→ λk λl dλi ∧dλj , λj λl dλi ∧dλk λj λk dλi ∧dλl , λi λl dλj ∧dλk λi λk dλj ∧dλl , λi λj dλk ∧dλl grad −−→ curl −−→ div −−→ →0 Demkowicz–Monk–Vardapetyan–Rachowicz 2000 These are extreme cases. For every r ∃ 2n−1 such FEdR subcomplexes. 27 / 41 28 / 41 Constructing bounded cochain projections Stable finite elements for the Hodge Laplacian We also need projections πhk : HΛk → Λkh which are uniformly bounded in h and commute with d. Putting this altogether we obtain four different stable families of mixed finite elements for the Hodge Laplacian. (reduces to 2 for k = n and 1 for k = 0) The projections Phk defined via the DOFs, commute with d. Unfortunately they are not bounded on HΛk . So first we regularize, and then project: Qhk u := Phk (ρ ∗ u) This map is bounded, uniformly in h if = δh, and still commutes with d. Unfortunately it is not a projection. Pr− Λk−1 (T ) × Pr− Λk (T ) Pr Λk−1 (T ) × Pr− Λk (T ) If we choose the δ sufficiently small, we can prove that Qhk |Λk : Λkh → Λkh is close to the identity, uniformly in h, hence h invertible. Then the composition πhk = (Qhk |Λk )−1 h ◦ Pr−+1 Λk−1 (T ) × Pr Λk (T ) Pr +1 Λk−1 (T ) × Pr Λk (T ) Qhk For each we obtain optimal order estimates for σ and u. is the desired bounded cochain projection. Christiansen, Schöberl, AFW 29 / 41 30 / 41 Other applications Maxwell’s equations and related EM problems Mixed eigenvalue problems (bounded cochain projections =⇒ discrete compactness) Application to elasticity Preconditioning and multigrid A posteriori error estimation Stable mixed FEM for elasticity 31 / 41 Stress–displacement mixed finite elements for elasticity Recent progress coming from the FEEC perspective 3 Find stress σ : Ω → R3×3 sym , displacement u : Ω → R such that Aσ = (u), Z Ω 1 A σ : σ + div σ · u + f · u 2 div σ = f First stable elements based on polynomials, 2D (Arnold–Winther 2002), all degrees r ≥ 1: σ,u dx −−−−−−−−−−→ stationary point H(div;S)×L2 (Rn ) Search for stable finite elements dates back to the ’60s, very limited success. It is, of course, possible to derive elements that exhibit complete continuity of the appropriate components along interfaces and indeed this was achieved by Raviart and Thomas in the case of the heat conduction problem discussed previously. Extension to the full stress problem is dicult and as yet such elements have not been successfully noted. 3D stable elements, all degrees r ≥ 1 (Arnold–Awanou–Winther 2007): for r = 1 stress space has 162 degrees of freedom (27 per component on average) Zienkiewicz, Taylor, Zhu The Finite Element Method: Its Basis & Fundamentals, 6th ed., 2005 Thanks to FEEC, we can retire that statement! 33 / 41 A computation using the new elements 34 / 41 Mixed formulation with weak symmetry Idea goes back to Fraeijs de Veubeke 1975, Amara–Thomas 1979 In the classical Hellinger–Reissner principle, symmetry of the stress tensor (balance of angular momentum) is assumed to hold exactly. Instead we impose it weakly with a Lagrange multiplier (the rotation). From Eberhard, Hueber, Jiang, Wohlmuth 2006 Z Ω Z Ω 1 A σ : σ + div σ · u + f · u 2 σ,u dx −−−−−−−−−−→ stationary point H(div;S)×L2 (Rn ) 1 A σ : σ + div σ · u + σ : p + f · u 2 σ,u,p dx −−−−−−−−−−−−−−−→ S.P. H(div;M)×L2 (Rn )×L2 (K) FEEC has led to very simple stable elements σ 35 / 41 u p 36 / 41 The elasticity complex New mixed finite elements for elasticity There is a complex for elasticity analogous to the de Rham complex. It has versions both for strong symmetry and weak symmetry. displacement rotation y y 1 3 strain y (grad,−I ) 2 The elasticity complex can be derived from the de Rham complex by an intricate construction. Mimicking this construction on the discrete level we have derived stable mixed finite elements for elasticity. (Arnold-Falk-Winther 2006, 2007). Main result Choose two discretizations of the de Rham complex: J 0 → H (Ω; R )×L (Ω, K) −−−−−−→ H(J, Ω; M) −−→ 0 @ J div skw A load curl div grad curl div 0 −−→ Λ̃0h −−→ Λ̃1h −−→ Λ̃2h −−→ Λ̃3h −−→ 0 −−→ H(div, Ω; M) −−−−−→ L2 (Ω; R3 )×L2 (Ω; K) → 0 x x x stress grad 0 −−→ Λ0h −−→ Λ1h −−→ Λ2h −−→ Λ3h −−→ 0 1 Surjectivity Hypothesis: (roughly) for each DOF of Λ2h there is a corresponding DOF of Λ̃1h . Λ̃2h (R3 ) stress: Then is a stable element choice. displacement: Λ̃3h (R3 ) rotation: Λ3h (K) couple J is second order! 38 / 41 37 / 41 The simplest choice Features of the new mixed elements Based on HR formulation with weak symmetry; very natural 0→ 0→ grad −−→ −−→ curl −−→ →0 grad curl div →0 −−→ −−→ div −−→ Lowest degree element is very simple: full P1 for stress, P0 for displacement and rotation Works for every polynomial degree Works the same in 2 and 3 (or more) dimensions Robust to material constraints like incompressibility Provably stable and convergent σ u p 39 / 41 40 / 41 Conclusions Capturing the right geometric structure on the discrete level can be essential to get stable methods. For PDE related to the de Rham complex, this is achieved by using subspaces which form a subcomplex with bounded cochain projections. There are two natural families of finite element subspaces of HΛk , the spaces Pr Λk and Pr− Λk . They can be assembled into complexes with bounded cochain projections (in numerous ways). They are inter-related and must be studied together. These spaces unify and clarify many known mixed finite element methods that were derived independently. This approach allows for a systematic way of choosing degrees of freedom, dual bases, and explicit geometric bases. Through FEEC we believe we have completed the long search for “just the right” mixed finite elements for elasticity. 41 / 41