Finite Element Exterior Calculus. Douglas N. Arnold, University of Minnesota The 41st Woudschoten Conference 5 7 October PDF Free Download

Finite Element Exterior Calculus Douglas N. Arnold, University of Minnesota The 41st Woudschoten Conference 5 7 October 2016

The Hilbert complex framework Discretization of Hilbert complexes Finite element differential forms FEDF on cubical meshes New complexes from old

The Hilbert complex framework

Short closed Hilbert complex W 0 (d0,v 0 ) W 1 (d1,v 1 ) W 2 d i are closed, densely-defined w/ closed range, d 1 d 0 = 0 Null space & range: Z i := N (d i ), B i := R(d i 1 ), B 1 Z 1 Dual complex: W 0 (d 1,V 1 ) W 1 (d 2,V 2 ) W 2 Harmonic forms: Hodge decomposition: H = Z 1 Z 1 = (B1 ) Z 1 = B 1 /Z 1 (homology) Z {}}{ Z {}}{ W 1 = B 1 H B1 }{{}}{{} Z Z Poincaré inequality: u C P du, u V Z 3 / 40

Hodge Laplacian d 0 W 0 W 1 W 2 L = dd + d d : W 1 W 1, D(L) = {u V V du V, d u V} d 1 L = L, N (L) = R(L) = H Strong form: Given f W find u D(L) s.t. Lu = f (mod H), u H Primal weak form: Find u V V H s.t. d 1 d 2 du, dv + d u, d v = f, v, v V V H Mixed weak form: Find σ V 0, u V 1, p H s.t. σ, τ u, dτ = 0, τ V 0, dσ, v + du, dv + p, v = f, v, v V 1, u, q = 0, q H. 4 / 40

Well-posedness THEOREM Let W 0 d W 1 d W 2 be a closed H-complex. Then 1. f W 1! (σ, u, p) V 0 V 1 H solving the weak mixed form. 2. c depending only on the Poincaré constant for the complex s.t. σ V + u V + p c f 3. u D(L) and satisfies Lu = f (mod H), u H. (weak strong) To show: inf-sup cond n for σ, τ u, dτ + dσ, v + du, dv + p, v + u, q. So we need to bound σ V + u V + p by bounded choice of τ, v, q. Easy to bound σ, dσ, p, du, but... what about u? Hodge decomp of W 1 : u = u B + u H + u B. Easy to bound u H. To bound u B take τ V 0 Z with dτ = u B and use Poincaré s inequality in V 0. Finally u B C P du B = C P du by Poincaré s inequality in V 1. 5 / 40

Examples Ex 0. Ω R 3 0 L 2 (Ω) (grad,h1 ) L 2 (Ω) R 3 Mixed=Primal: Find u H 1, p R s.t. grad u, grad v = f p, v, v H 1, u = 0. std. Poisson Neumann bc 6 / 40

Examples Ex 0. Ω R 3 0 L 2 (Ω) (grad,h1 ) L 2 (Ω) R 3 Mixed=Primal: Find u H 1, p R s.t. grad u, grad v = f p, v, v H 1, Ex 0a. 0 L 2 (Ω) (grad, H 1 ) L 2 (Ω) R 3 u = 0. std. Poisson Neumann bc Find u H 1 s.t. grad u, grad v = f, v, v H 1. Ex 3. Mixed: L 2 (Ω) R 3 (div,h(div)) L 2 (Ω) 0 Find σ H(div), u L 2 s.t. σ, τ u, div τ = 0, τ H(div), div σ, v = f, v, v L 2. mixed Poisson 6 / 40

The de Rham complex in 3D 0 L 2 grad,h 1 L 2 R 3 curl,h(curl) L 2 R 3 div,h(div) L 2 0 7 / 40

The de Rham complex in 3D grad,h 1 0 L 2 L 2 R 3 L 2 R 3 L 2 0 div, H(div) curl,h(curl) curl, H(curl) div,h(div) grad, H 1 7 / 40

The de Rham complex in 3D grad,h 1 0 L 2 L 2 R 3 L 2 R 3 L 2 0 div, H(div) curl,h(curl) curl, H(curl) div,h(div) grad, H 1 Ex 1. dd + d d = grad div + curl curl Natural magnetic b.c.: u n = 0, curl u n = 0 σ = div u Ex 2. dd + d d = curl curl grad div Natural electric b.c.: u n = 0, div u = 0 σ = curl u The dimensions of the homology groups (harmonic forms) are the Betti numbers of Ω. 7 / 40

The Hodge wave equation Ü + (dd + d d)u = 0, U(0) = U 0, U(0) = U 1 Then σ := d U, ρ := du, u := U satisfy σ 0 d 0 σ u + d 0 d u = 0 ρ 0 d 0 ρ Find (σ, u, ρ) : [0, T] V 0 V 1 W 2 s.t. σ, τ u, dτ = 0, τ V 0, u, v + dσ, v + ρ, dv = 0, v V 1, ρ, η du, η = 0, η W 2. strong weak THEOREM Given initial data (σ 0, u 0, ρ 0 ) V 0 V 1 W 2,! solution (σ, u, ρ) C 0 ([0, T], V 0 V 1 W 2 ) C 1 ([0, T], W 0 W 1 W 2 ). Proof: Uniqueness: (τ, v, η) = (σ, u, ρ). Existence: Hille Yosida. 8 / 40

Example: Maxwell s equations Ḋ = curl H div D = 0 D = ɛe Ḃ = curl E div B = 0 B = µh W 0 = L 2 (Ω) W 1 = L 2 (Ω, R 3, ɛ dx) W 2 = L 2 (Ω, R 3, µ 1 dx) W 0 grad W 1 curl W 2 (σ, E, B) : [0, T] Ω R R 3 R 3 solves σ, τ ɛe, grad τ = 0 τ, ɛė, F + ɛ grad σ, F µ 1 B, curl F = 0 F, µ 1 Ḃ, C + µ 1 curl E, C = 0 C. THEOREM If σ, div ɛe, and div B vanish for t = 0, then they vanish for all t, and E, B, D = ɛe, and H = µ 1 B satisfy Maxwell s equations. 9 / 40

Simulation of radar scattering off building Stowell Fassenfass White, IEEE Trans. Ant. & Prop. 2008 Solved time-dependent Maxwell equations using Q 1 Λ1 for E and Q 1 Λ2 for B (Nédélec elements of the first kind on bricks) 10,114,695,855 brick elements ( 1 cm resolution) 60, 000, 000, 000 unknowns 12, 000 time steps of 14 picoseconds concrete walls cinder block/rebar walls 10 / 40

Some other complexes sym grad curl T curl div 0 L 2 R 3 L 2 S 3 L 2 S 3 L 2 R 3 0 displacement strain stress load 0 L 2 R3 sym grad L 2 S 3 L 2 S 3 div L 2 R 3 0 primal method for elasticity mixed method for elasticity R 3 3 trace-free 0 L 2 grad grad L 2 S 3 curl L 2 div T L 2 R 3 0 0 L2 grad grad L 2 S 3 primal method for plate equation L2 grad grad L 2 S 3 curl L 2 V Einstein Bianchi eqs (GR) 11 / 40

Discretization of Hilbert complexes

Discretization Choose finite dimensional spaces V 0 h V0, V 1 h V1. Of course they must have reasonable approximation: lim inf u v W = 0, u W. h 0 v V h Let d h = d Vh, Z h = N (d h ), B h = R(d h ), H h = Z h Bh Galerkin s method: Find (σ h, u h, p h ) Vh 0 V1 h Hh s.t. σ h, τ u h, dτ = 0, τ Vh 0, dσ h, v + du h, dv + p h, v = f, v, v Vh 1, u h, q = 0, q H h. When is this approximation stable, consistent, and convergent? 12 / 40

Assumptions on the discretization Besides approximation, there are two key requirements: Subcomplex assumption: d V k h Vk+1 h Bounded Cochain Projection assumption: V k π k h V k h d k V k+1 π k+1 h d k V k+1 h π k h : Vk V k h πh k is bounded, uniformly in h π k h preserves Vk h π k+1 h d k = d k πh k 13 / 40

Consequences of the assumptions The subcomplex property implies that Vh 0 V 1 h W 2 is itself an H-complex. So it has its own harmonic forms, Hodge decomposition, and Poincaré inequality. The Hodge Laplacian for the discrete complex is the Galerkin method for the original problem. THEOREM Given the approximation, subcomplex, and BCP assumptions: H = H h and gap ( H, H h ) 0. The Galerkin method is consistent. The discrete Poincaré inequality ω c dω, ω Zh k, holds with c independent of h. The Galerkin method is stable. The Galerkin method is convergent with the error estimate: σ σ h V + u u h V + p p h V c[ E(σ) + E(u) + E(p) + ɛ ] where ɛ E(P B u) sup E(r). r H k, r =1 d h d h 14 / 40

Finite element differential forms

Differential forms on a domain Ω R n Differential k-forms are functions Ω Alt k R n 0-forms: functions; 1-forms: covector fields; k-forms: ( n k ) components u = f σ dx σ := f σ1 σ k dx σ 1 dx σ k σ 1 σ 1 < <σ k n The wedge product of a k-form and an l-form is a (k + l)-form The exterior derivative du of a k-form is a (k + 1)-form A k-form can be integrated over a k-dimensional subset of Ω Given F : Ω Ω, a k-form on Ω can be pulled back to a k-form on Ω. The trace of a k-form on a submanifold is the pull back under inclusion. Stokes theorem: du = tr u, u Λ k 1 (Ω) Ω Ω The exterior derivative can be viewed as a closed, densely-defined op L 2 Λ k L 2 Λ k+1 with domain HΛ k (Ω) = { u L 2 Λ k du L 2 Λ k+1 }. If Ω has a Lipschitz boundary, it has closed range. 15 / 40

The L 2 de Rham complex and its discretization 0 L 2 Λ 0 d L 2 Λ 1 d d L 2 Λ n 0 Our goal is to define spaces V k h HΛk satisfying the approximation, subcomplex, and BCP assumptions. In the case k = 0, V k h H1 will just be the Lagrange elements. It turns out that for k > 0 there are two distinct generalizations. 16 / 40

The Lagrange finite element space P r Λ 0 (T h ) nothing new (but notation) Elements: A triangulation T h consisting of simplices T Shape functions: V(T) = P r Λ 0 (T) = P r (T), some r 1 Degrees of freedom (unisolvent): v 0 (T): u u(v) e 1 (T): u e (tr e u)q dt, q P r 2 (e) f 2 (T): u f (tr f u)q ds, q P r 3 (f ). u (tr f u) q, q P r d 1 Λ d (f ), f d (T), d 0 f The resulting finite element space belongs to HΛ 0 = H 1. In fact, it equals { u HΛ 0 (Ω) : u T V(T) T T h } 17 / 40

The P r Λ k (T h ) and P r Λ k (T h ) finite element spaces Elements: A triangulation T h consisting of simplices T to be defined Shape functions: V(T) = P r Λ k (T) or Pr Λ k (T), some r 1 Degrees of freedom for P r Λ k (unisolvent): u (tr f u) q, q P r d+k Λd k (f ), f d (T), d k f Degrees of freedom for Pr Λ k (unisolvent): u (tr f u) q, q P r d+k 1 Λ d k (f ), f d (T), d k f Hiptmair 99 The resulting finite element spaces belongs to HΛ k. In fact, they equal { u HΛ 0 (Ω) : u T V(T) T T h } 18 / 40

The Koszul differential κ is a simple algebraic operation take k-forms to k 1-forms. It takes polynomial forms to polynomial forms of 1 degree higher. κ(dx i ) = x i, κ(u v) = (κu) v + ( ) k u (κv) κ(f dx σ 1 dx σk ) = k i=1 ( )i f x σ i dx σ 1 dx σ i dx σk In R 3 : P r Λ 3 x P r+1 Λ 2 x P r+2 Λ 1 x P r+3 Λ 0 Two basic properties: κ is a differential: κ κ = 0 Homotopy property: (dκ + κd)u = (r + k)u, u H r Λ k e.g., curl( x v) + x (div v) = (deg v + 2) v Consequences: homog. deg. r The Koszul complex 0 H r Λ n κ H r+1 Λ n 1 κ is exact. H r Λ k = κ H r 1 Λ k+1 d H r+1 Λ k 1 κ H r+n Λ 0 0 Definition: P r Λ k = P r 1 Λ k + κh r 1 Λ k+1 19 / 40

Finite element discretizations of the de Rham complex P r 1 Λ k P r Λ k P r Λ k except P r Λ 0 = P r Λ 0, P r Λ n = P r 1 Λ n ) The DOFs given above for P r Λ k and P r Λ k are unisolvent. The DOFs define cochain projections for the complexes 0 P r Λ 0 (T h ) 0 P r Λ 0 (T h ) d P r 1 Λ 1 (T h ) d P r Λ 1 (T h ) d d d P r n Λ n (T h ) 0 d P r Λ n (T h ) 0 decreasing degree constant degree This leads to four stable discretizations of the mixed k-form Laplacian: P r Λ k 1 (T h ) P r 1 Λ k (T h ) P r Λ k 1 (T h ) P r Λ k (T h ) P r Λ k 1 (T h ) P r 1 Λ k (T h ) P r Λ k 1 (T h ) P r Λ k (T h ) 20 / 40

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange DG n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange Raviart- Thomas 75 DG n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange Raviart- Thomas 75 DG n = 3 r = 2 Nedelec face elts 80

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange Raviart- Thomas 75 DG n = 3 r = 2 Nedelec edge elts 80 Nedelec face elts 80

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 Whitney 57 n = 2 r = 2 Lagrange Raviart- Thomas 75 DG n = 3 r = 2 Nedelec edge elts 80 Nedelec face elts 80

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange DG n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange BDM 85 DG n = 3 r = 2

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange BDM 85 DG n = 3 r = 2 Nedelec face elts, 2nd kind 86

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 n = 2 r = 2 Lagrange BDM 85 DG n = 3 r = 2 Nedelec edge elts, 2nd kind 86 Nedelec face elts, 2nd kind 86

P r Λ k k = 0 k = 1 k = 2 k = 3 n = 1 r = 2 Sullivan 78 n = 2 r = 2 Lagrange BDM 85 DG n = 3 r = 2 Nedelec edge elts, 2nd kind 86 Nedelec face elts, 2nd kind 86

Finite element differential forms on cubical meshes

The tensor product construction DNA Boffi Bonizzoni 2012 Suppose we have a de Rham subcomplex V on an element S R m : V k d V k+1 V k HΛ 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. 23 / 40

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 Take tensor product n times: u(x) u (x) dx Q r Λ 0 = Q r, Q r Λ 1 = Q r 1,r,r,... dx 1 + Q r,r 1,r,... dx 2 +, Q r Λ 2 = Q r 1,r 1,r,... dx 1 dx 2 +,... Q r Λ k (I n ) := (V r V r ) k Q 2 Λ0 Q 2 Λ1 Q 2 Λ2 constant degree 24 / 40

Qr Λk n=1 k=0 r=1 r=2 r=3 r=1 n=2 r=2 r=3 r=1 n=3 r=2 r=3 k=1 k=2 k=3

The 2nd family on cubes: 0-forms DNA Awanou 2011 The Q r Λ k family reduces to Q r when k = 0. For the second family, we get the serendipy space S r. 26 / 40

The 2nd family on cubes: 0-forms DNA Awanou 2011 The Q r Λ k family reduces to Q r when k = 0. For the second family, we get the serendipy space S r. 2-D shape fns: S r (I 2 ) = P r (I 2 ) span[x r 1 x 2, x 1 x r 2 ] DOFs: u f tr f u q, q P r 2d (f ), f d (I n ) n-d shape fns: S r (I n ) = P r (I n ) l 1 H r+l,l (I n ) H r,l (I n ) = span of monomials of degree r, linear in l variables 26 / 40

The 2nd family of finite element differential forms on cubes DNA Awanou 2012 The S r Λ k (I n ) family of FEDFs, uses the serendipity spaces for 0-forms, and serendipity-like DOFs. DOFs: u f tr f u q, q P r 2(d k) Λ d k (f ), f d (I n ), d k Shape fns: S r Λ k (I n ) = P r Λ k (I n ) [κh r+l 1,l Λ k+1 (I n ) dκh r+l,l Λ k (I n )] }{{} l 1 deg=r+l H r,l Λ k (I n ) = span of monomials x α i 1 xα n n dx σ1 dx σk, α = r, linear in l variables not counting the x σi Unisolvence holds for all n 1, r 1, 0 k n. 27 / 40

The 2nd cubic family in 2-D S 2 Λ 0 S 1 Λ 1 S 0 Λ 2 decreasing degree S 3 Λ 0 S 2 Λ 1 S 1 Λ 2 S r Λ k (I 2 ) k 1 2 3 4 5 0 4 8 12 17 23 1 8 14 22 32 44 2 3 6 10 15 21 Q r Λ k (I 2 ) k 1 2 3 4 5 0 4 9 16 25 36 1 4 12 24 40 60 2 1 4 9 16 25 28 / 40

The 3D shape functions in traditional FE language S r Λ 0 : polynomials u such that deg u r + ldeg u 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, deg u r + ldeg u + 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 29 / 40

Dimensions and low order cases S r Λ k (I 3 ) k 1 2 3 4 5 0 8 20 32 50 74 1 24 48 84 135 204 2 18 39 72 120 186 3 4 10 20 35 56 Qr Λ k (I 3 ) k 1 2 3 4 5 0 8 27 64 125 216 1 12 54 96 200 540 2 6 36 108 240 450 3 1 8 27 64 125 S 1 Λ 1 (I 3 ) new element S 1 Λ 2 (I 3 ) corrected element 30 / 40

Approximation properties On cubes the Q r Λ k and S r Λ k spaces provide the expected order of approximation. Same is true on parallelotopes, but accuracy is lost by non-affine distortions, with greater loss, the greater the form degree k. The L 2 approximation rate of the space Q r = Q r Λ 0 is r + 1 on either affinely or multilinearly mapped elements. The rate for S r = S r Λ 0 is r + 1 on affinely mapped elements, but only max(2, r/n + 1) on multilinearly mapped elements. The rate for Q r Λ k, k > 0, is r on affinely mapped elements, r k + 1 on multilinearly mapped elements. The rate for P r Λ n = S r Λ n is r + 1 for affinely mapped elements, r/n n + 2 for multilinearly mapped. DNA-Boffi-Bonizzoni 2012 31 / 40

Sr Λk n=1 k=0 r=1 r=2 r=3 r=1 n=2 r=2 r=3 r=1 n=3 r=2 r=3 k=1 k=2 k=3

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New complexes from old

Elasticity with weak symmetry The mixed formulation of elasticity with weak symmetry is more amenable to discretization than the standard mixed formulation. Fraeijs de Veubeke 75 p = skw grad u, Aσ = grad u p Find σ L 2 (Ω) R n n, u L 2 (Ω) R n, p L 2 (Ω) R n n skw s.t. Aσ, τ + u, div τ + p, τ = 0, div σ, v = f, v, σ, q = 0, τ L 2 (Ω) R n n v L 2 (Ω) R n q L 2 (Ω) R n n skw This is exactly the mixed Hodge Laplacian for the complex: L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 34 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S L 2 (Ω) R n R n n skw skw div L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S L 2 (Ω) R n R n n skw skw div q L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 v Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S L 2 (Ω) R n R n n skw skw div q L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 ρ v Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S L 2 (Ω) R n R n n skw skw div q+ skw ρ L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 ρ v Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S ψ L 2 (Ω) R n R n n skw skw div q+ skw ρ L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 ρ v Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S ψ L 2 (Ω) R n R n n skw skw div q+ skw ρ L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 φ ρ v Sτ = τ T tr(τ)i (invertible) 35 / 40

Well-posedness L 2 ( div, skw) A (Ω) Rn n [L 2 (Ω) R n ] [L 2 (Ω) R n n skw ] 0 Well-posedness depends on the exactness of the complex. This can be shown by relating the complex to two de Rham complexes: S ψ L 2 (Ω) R n R n n skw skw div q+ skw ρ L 2 (Ω) R n n skw 0 L 2 (Ω) R n n curl L 2 (Ω) R n n div L 2 (Ω) R n 0 φ curl φ+ ρ v Sτ = τ T tr(τ)i (invertible) 35 / 40

Discretization To discretize we select discrete de Rham subcomplexes with commuting projs Ṽ 0 h curl Ṽh 1 div Ṽh 2 0, V1 h div V 2 h 0 to get the discrete complex Ṽ 1 h Rn ( div, skw) (Ṽ 2 h Rn ) (V 2 h Rn n skw ) 0 We get stability if we can carry out the diagram chase on: π 1 h S curl V 1 h Rn n skw V 2 h Rn n skw 0 Ṽ 0 h Rn Ṽ 1 h Rn Ṽ 2 h Rn 0 div π 2 h skw div This requires that π 1 h S : Ṽ0 h Rn V 1 h Rn n skw is surjective. 36 / 40

Stable elements The requirement that πh 1S : Ṽ0 h Rn Vh 1 Rn n skw can be checked by looking at DOFs. The simplest choice is is surjective P r+1 Λ n 2 curl P r Λ n 1 div P r 1 Λ n 0, P r Λ n 1 div P r Λ n 0 which gives the elements of DNA Falk Winther 07 σ u p Other elements: Cockburn Gopalakrishnan Guzmán, Gopalakrishnan Guzmán, Stenberg,... 37 / 40

More complexes from complexes 0 V 1 d W 2 S D 0 Γ W 2 0 Ṽ 1 d W 2 where Γ = {(v, τ) V 1 Ṽ 1 dv = Sτ }, D(v, τ) = dτ. 0 Vh 1 d Vh 2 πh 2S D 0 Γ h W 2 0 Ṽh 1 d W 2 where Γ h = {(v, τ) Vh 1 Ṽ1 h dv = π2 hsτ }. Find u h V 1 h, σ h Ṽ 1 h, λ h V 2 h s.t. dσ h, dτ + λ h, dv π h Sτ = f, v, v V 1 h, τ Ṽ1 h, du h π h Sσ h, µ = 0, µ V 2 h. 38 / 40

FEEC discretization of the biharmonic grad 0 H 1 (Ω) L 2 (Ω; R n ) I 0 H 1 (Ω; R n ) grad L 2 C (Ω; Rn n ) grad 0 P r Λ 0 P r Λ 1 π h 0 P r+1 Λ 0 R n grad L 2 C (Ω; Rn n ) Find u h P r Λ 0, σ h P r+1 Λ 0, λ h P r Λ 1 s.t. C grad σ h, grad τ + λ h, grad v π h τ = f, v, v P r Λ 0, τ P r+1 Λ 0, grad u h π h σ h, µ = 0, µ P r Λ 1. u σ λ 39 / 40

References DNA, Falk, Winther, Finite element exterior calculus, homological techniques, and applications. Acta Numerica 2006, p. 1 155 DNA, Falk, Winther, Finite element exterior calculus: from Hodge theory to numerical stability. AMS Bulletin 2010, p. 281-354 everything at http://umn.edu/ arnold/publications.html 40 / 40

Finite Element Exterior Calculus. Douglas N. Arnold, University of Minnesota The 41st Woudschoten Conference 5 7 October 2016