Finite element exterior calculus: A new approach to the stability of finite elements

Transcription

1 Finite element exterior calculus: A new approach to the stability of finite elements Douglas N. Arnold joint with R. Falk and R. Winther Institute for Mathematics and its Applications University of Minnesota July 25, / 39

2 Outline 1 Motivating the search for stable finite elements: finding the right element for the job 2 The mathematical framework: exterior calculus 3 The star of the show: finite element differential forms 4 Application to elasticity: the holy grail attained? 2 / 39

3 Steady heat conduction problem: finite elements in H 1 Ω Ω div C grad u = f C grad u grad v dx = ( ) 1 C grad u grad u f u 2 Ω f v dx dx v u minimum 3 / 39

4 Steady heat conduction problem: finite elements in H 1 Ω Ω div C grad u = f C grad u grad v dx = ( ) 1 C grad u grad u f u 2 Ω f v dx dx v u minimum Ω grad u 2 dx < u H 1 (Ω) H 1 : u L 2 (Ω), grad u L 2 (Ω; R n ) 3 / 39

5 Steady heat conduction problem: finite elements in H 1 Ω Ω div C grad u = f C grad u grad v dx = ( ) 1 C grad u grad u f u 2 Ω f v dx dx v u minimum Ω grad u 2 dx < u H 1 (Ω) H 1 : u L 2 (Ω), grad u L 2 (Ω; R n ) The right FE spaces: Lagrange elements: { v H 1 (Ω) v T P r (Ω) } 3 / 39

6 Elasticity in displacement formulation u : Ω R n displacement field ( ) 1 C ɛ(u) : ɛ(u) dx f u dx 2 Ω u minimum ɛ(u) 2 dx grad u 2 dx Korn s inequality u [H 1 (Ω)] n Again, Lagrange elements have the right stuff. 4 / 39

7 First order (mixed) formulations Thermal A σ = grad u, divσ = f ( 1 2 A σ σ + div σ u + f u) dx σ, u stationary point σ H(div, Ω), u L 2 (Ω) 5 / 39

8 First order (mixed) formulations Thermal A σ = grad u, divσ = f ( 1 2 A σ σ + div σ u + f u) dx σ, u stationary point σ H(div, Ω), u L 2 (Ω) Elasticity A σ = ɛ(u), divσ = f ( 1 2 A σ : σ + div σ u + f u) dx σ,u stationary point σ H(div, Ω; S), u L 2 (Ω; R n ) Lagrange elements? 5 / 39

9 First order (mixed) formulations Thermal A σ = grad u, divσ = f ( 1 2 A σ σ + div σ u + f u) dx σ, u stationary point σ H(div, Ω), u L 2 (Ω) Elasticity A σ = ɛ(u), divσ = f ( 1 2 A σ : σ + div σ u + f u) dx σ,u stationary point σ H(div, Ω; S), u L 2 (Ω; R n ) Lagrange elements? Unstable! 5 / 39

10 Thermal problem in 1D σ = u, σ = f on ( 1, 1) Babuška Narasimhan (σ 2 + σ u + f u) dx σ, u stationary point H 1 L 2 6 / 39

11 Thermal problem in 1D σ = u, σ = f on ( 1, 1) Babuška Narasimhan (σ 2 + σ u + f u) dx σ, u stationary point H 1 L 2 P 1 -P 1 (20 elts) 6 / 39

12 Thermal problem in 1D σ = u, σ = f on ( 1, 1) Babuška Narasimhan (σ 2 + σ u + f u) dx σ, u stationary point H 1 L 2 P 1 -P 1 (20 elts) P 1 -P 1 (40 elts) 6 / 39

13 Thermal problem in 1D σ = u, σ = f on ( 1, 1) Babuška Narasimhan (σ 2 + σ u + f u) dx σ, u stationary point H 1 L 2 P 1 -P 1 (20 elts) P 1 -P 1 (40 elts) P 1 -P 0 (40 elts) 6 / 39

14 Thermal problem in 2D Ω σ = grad u, ( 1 2 σ 2 + div σ u + f u) dx div σ = f σ, u stationary point H(div) L 2 7 / 39

15 Thermal problem in 2D Ω σ = grad u, ( 1 2 σ 2 + div σ u + f u) dx div σ = f σ, u stationary point H(div) L 2 P 1 -P 0 7 / 39

16 Thermal problem in 2D Ω σ = grad u, ( 1 2 σ 2 + div σ u + f u) dx div σ = f σ, u stationary point H(div) L 2 P 1 -P 0 Raviart Thomas - P 0 7 / 39

17 Raviart Thomas elements A mixed FEM for 2nd order elliptic problems, Proc. conf. Math l Aspects of the FEM, Rome Springer Lect. Notes in Math #606, Shape functions: ( ( 1 0), 0 ( 1), x ) y DOFs: Generalizes to all degrees, and all dimensions (n = 3: Nédélec 80) Math & CS SIAM J. Numerical Analysis Numerische Mathematik Mathematics of Computation RAIRO M 2 AN Num. Methods for PDEs Eng. & Apps CMAME Computational Geosciences J. Computational Physics IJNME COMPEL 8 / 39

18 Maxwell eigenvalue problem, unstructured mesh µ 1 curl E curl Ẽ = ω 2 ɛe Ẽ Ẽ Ω Ω Right space is H(curl) λ = m 2 + n 2 = 0, 1, 1, 2, 4, 4, 5, 5, 8,... 9 / 39

19 Maxwell eigenvalue problem, unstructured mesh µ 1 curl E curl Ẽ = ω 2 ɛe Ẽ Ẽ Ω Ω Right space is H(curl) λ = m 2 + n 2 = 0, 1, 1, 2, 4, 4, 5, 5, 8,... (Lag.P 1 ) 2 9 / 39

20 Maxwell eigenvalue problem, unstructured mesh µ 1 curl E curl Ẽ = ω 2 ɛe Ẽ Ẽ Ω Ω Right space is H(curl) λ = m 2 + n 2 = 0, 1, 1, 2, 4, 4, 5, 5, 8,... (Lag.P 1 ) 2 P 1 Λ1 9 / 39

21 Maxwell eigenvalue problem, regular mesh λ = m 2 + n 2 = 1, 1, 2, 4, 4, 5, 5, 8, / 39

22 Maxwell eigenvalue problem, regular mesh λ = m 2 + n 2 = 1, 1, 2, 4, 4, 5, 5, 8, Boffi-Brezzi-Gastaldi / 39

23 Vector Laplacian curl curl u grad div u = f in Ω u t = 0, div u = 0 on Ω Ω 11 / 39

24 Vector Laplacian curl curl u grad div u = f in Ω u t = 0, div u = 0 on Ω Ω Ω 1 2 ( curl u 2 + div u 2 ) f v Lagrange finite elements will converge u minimum 11 / 39

25 Vector Laplacian curl curl u grad div u = f in Ω u t = 0, div u = 0 on Ω Ω Ω 1 2 ( curl u 2 + div u 2 ) f v u minimum Lagrange finite elements will converge to the wrong solution! 11 / 39

26 Vector Laplacian curl curl u grad div u = f in Ω u t = 0, div u = 0 on Ω Ω Ω 1 2 ( curl u 2 + div u 2 ) f v u minimum Lagrange finite elements will converge to the wrong solution! In fact same holds for any conforming finite elements. 11 / 39

27 Vector Laplacian curl curl u grad div u = f in Ω u t = 0, div u = 0 on Ω Ω Ω 1 2 ( curl u 2 + div u 2 ) f v u minimum Lagrange finite elements will converge to the wrong solution! In fact same holds for any conforming finite elements. A mixed formulation based on appropriate finite elements works just fine ( 1 2 σ2 curl σ u 1 2 div u 2 σ,u f v) dx stationary point H(curl) H(div) Ω 11 / 39

28 EM calculations based on the generalized RT elements Schöberl, Zaglmayr 2006, NGSolve 2K tets, P 6 Λ1 Also: White EMSolve, Demkowicz 3Dhp90, Durufle Montjoie,... 26K tets, P 3 Λ1 12 / 39

29 The Mathematical Framework: Exterior Calculus 13 / 39

30 Differential forms An algebraic k-form F on R n is a skew-symmetric k-linear form: it takes k vectors and delivers a number. } R n {{ R n } R, (v 1,..., v k ) F (v 1,..., v k ) k 14 / 39

31 Differential forms An algebraic k-form F on R n is a skew-symmetric k-linear form: it takes k vectors and delivers a number. } R n {{ R n } R, (v 1,..., v k ) F (v 1,..., v k ) k For example if u = ( u x, u y, u z ) denotes a vector dx(u) := u x is a 1-form, dx dy (u, v) := u x v y u y v x is a 2-form 14 / 39

32 Differential forms An algebraic k-form F on R n is a skew-symmetric k-linear form: it takes k vectors and delivers a number. } R n {{ R n } R, (v 1,..., v k ) F (v 1,..., v k ) k For example if u = ( u x, u y, u z ) denotes a vector dx(u) := u x is a 1-form, dx dy (u, v) := u x v y u y v x is a 2-form A differential k-form on Ω R n is a field of algebraic k-forms: (v 1,..., v k ) ω x (v 1,..., v k ) R x Ω. 0-form = function, 1-form = covector field f (x, y)dx + g(x, y)dy 14 / 39

33 Differential forms An algebraic k-form F on R n is a skew-symmetric k-linear form: it takes k vectors and delivers a number. } R n {{ R n } R, (v 1,..., v k ) F (v 1,..., v k ) k For example if u = ( u x, u y, u z ) denotes a vector dx(u) := u x is a 1-form, dx dy (u, v) := u x v y u y v x is a 2-form A differential k-form on Ω R n is a field of algebraic k-forms: (v 1,..., v k ) ω x (v 1,..., v k ) R x Ω. 0-form = function, 1-form = covector field f (x, y)dx + g(x, y)dy 0-forms: 1-forms: 2-forms: 3-forms: temperature, electric field potential electric field, magnetic field electric flux, magnetic flux, heat flux charge density, heat density, mass density 14 / 39

34 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R 15 / 39

35 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. 15 / 39

36 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. The finite energy k-forms are HΛ k (Ω) = { ω L 2 Λ k (Ω) dω HΛ k (Ω) } 15 / 39

37 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. The finite energy k-forms are HΛ k (Ω) = { ω L 2 Λ k (Ω) dω HΛ k (Ω) } They connect via the exterior derivative to form the de Rham complex: 0 HΛ 0 (Ω) d 0 HΛ 1 (Ω) d 1 d n 1 HΛ n (Ω) 0 15 / 39

38 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. The finite energy k-forms are HΛ k (Ω) = { ω L 2 Λ k (Ω) dω HΛ k (Ω) } They connect via the exterior derivative to form the de Rham complex: 0 HΛ 0 (Ω) d 0 HΛ 1 (Ω) d 1 d n 1 HΛ n (Ω) 0 d k+1 d k = 0, 15 / 39

39 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. The finite energy k-forms are HΛ k (Ω) = { ω L 2 Λ k (Ω) dω HΛ k (Ω) } They connect via the exterior derivative to form the de Rham complex: 0 HΛ 0 (Ω) d 0 HΛ 1 (Ω) d 1 d n 1 HΛ n (Ω) 0 d k+1 d k = 0, range(d k ) ker(d k ), 15 / 39

40 Exterior calculus and the de Rham complex A k-form ω can be naturally integrated over a k-dimensional surface: S ω R A k-form can be differentiated to get a (k + 1)-form dω: take the directional derivative of ω x (v 1,..., v k ) in the direction v k+1 and skew-symmetrize. This is the exterior derivative. The finite energy k-forms are HΛ k (Ω) = { ω L 2 Λ k (Ω) dω HΛ k (Ω) } They connect via the exterior derivative to form the de Rham complex: 0 HΛ 0 (Ω) d 0 HΛ 1 (Ω) d 1 d n 1 HΛ n (Ω) 0 d k+1 d k = 0, range(d k ) ker(d k ), cohomology: ker/range 15 / 39

41 The case of Ω R 3 For Ω R 3, the de Rham complex boils down to 0 H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 16 / 39

42 The case of Ω R 3 For Ω R 3, the de Rham complex boils down to 0 H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 16 / 39

43 The case of Ω R 3 For Ω R 3, the de Rham complex boils down to 0 H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 16 / 39

44 The case of Ω R 3 For Ω R 3, the de Rham complex boils down to 0 H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 # components of Ω, k = 0 dim(ker/range) = # tunnels thru Ω, k = 1 # voids in Ω, k = 2 16 / 39

45 The case of Ω R 3 For Ω R 3, the de Rham complex boils down to 0 H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 # components of Ω, k = 0 dim(ker/range) = # tunnels thru Ω, k = 1 # voids in Ω, k = 2 16 / 39

46 PDEs closely connected to the de Rham sequence div grad u = f (curl curl grad div)u = f curl curl u = f, div u = 0 div u = f, curl u = 0 dynamic problems, eigenvalue problems, lower order-terms Maxwell s equations variable coefficients, nonlinearities / 39

47 PDEs closely connected to the de Rham sequence div grad u = f (curl curl grad div)u = f curl curl u = f, div u = 0 div u = f, curl u = 0 dynamic problems, eigenvalue problems, lower order-terms Maxwell s equations variable coefficients, nonlinearities... The well-posedness of these PDEs is intimately tied to the cohomology of the de Rham complex. 17 / 39

48 PDEs closely connected to the de Rham sequence div grad u = f (curl curl grad div)u = f curl curl u = f, div u = 0 div u = f, curl u = 0 dynamic problems, eigenvalue problems, lower order-terms Maxwell s equations variable coefficients, nonlinearities... The well-posedness of these PDEs is intimately tied to the cohomology of the de Rham complex. To get a stable numerical method, our discretization must capture the essential structure of the de Rham complex, in particular the cohomology. 17 / 39

49 Design principles for discretizing PDEs related to de Rham Treat things for what they are: treat 1-forms as 1-forms, 2-forms as 2-forms,... A finite element subspace Λ k h of some HΛk should fit together with finite element subspaces of all HΛ j dλ k 1 h should be contained in Λ k h so we get a discrete de Rham subcomplex Λ k 1 h d Λ k h d Λ k+1 h The subcomplex should relate to the full complex via commuting projections Λ k 1 d k 1 Λ k π k 1 h Λ k 1 h π k h d k 1 Λ k h π h d = dπ h 18 / 39

50 Design principles for discretizing PDEs related to de Rham Treat things for what they are: treat 1-forms as 1-forms, 2-forms as 2-forms,... A finite element subspace Λ k h of some HΛk should fit together with finite element subspaces of all HΛ j dλ k 1 h should be contained in Λ k h so we get a discrete de Rham subcomplex Λ k 1 h d Λ k h d Λ k+1 h The subcomplex should relate to the full complex via commuting projections Λ k 1 d k 1 Λ k π k 1 h Λ k 1 h π k h d k 1 Λ k h π h d = dπ h 18 / 39

51 Design principles for discretizing PDEs related to de Rham Treat things for what they are: treat 1-forms as 1-forms, 2-forms as 2-forms,... A finite element subspace Λ k h of some HΛk should fit together with finite element subspaces of all HΛ j dλ k 1 h should be contained in Λ k h so we get a discrete de Rham subcomplex Λ k 1 h d Λ k h d Λ k+1 h The subcomplex should relate to the full complex via commuting projections Λ k 1 d k 1 Λ k π k 1 h Λ k 1 h π k h d k 1 Λ k h π h d = dπ h 18 / 39

52 Design principles for discretizing PDEs related to de Rham Treat things for what they are: treat 1-forms as 1-forms, 2-forms as 2-forms,... A finite element subspace Λ k h of some HΛk should fit together with finite element subspaces of all HΛ j dλ k 1 h should be contained in Λ k h so we get a discrete de Rham subcomplex Λ k 1 h d Λ k h d Λ k+1 h The subcomplex should relate to the full complex via commuting projections Λ k 1 d k 1 Λ k π k 1 h Λ k 1 h π k h d k 1 Λ k h π h d = dπ h 18 / 39

53 The payoff A finite element method based on these principles generally captures all the essential structure: dimension of the cohomology spaces the cohomology classes Hodge decomposition (Helmoltz decomposition) Poincaré inequality If the continuous problem is well-posed, the discretization inherits this, i.e., is stable. 19 / 39

54 The Star of the Show: Finite Element Differential Forms 20 / 39

55 Constructing spaces of finite element differential forms To construct a finite element space of differential forms, we have to specify for a given simplex T R n : Shape functions: a finite dimensional space of polynomial forms on the simplex Degrees of freedom: grouped into subspaces associated to the subsimplices 21 / 39

56 Constructing spaces of finite element differential forms To construct a finite element space of differential forms, we have to specify for a given simplex T R n : Shape functions: a finite dimensional space of polynomial forms on the simplex Degrees of freedom: grouped into subspaces associated to the subsimplices Prototypical case: Lagrange finite elements Shape functions: V (T ) = P r (T ) DOFs associated to a subsimplex f : W (T, f ) = { u f tr T,f u v dx : v P r 1 dim f (f ) } The assembled space is then precisely { u H 1 (Ω) : u T V (T ) T } 21 / 39

57 The spaces P r Λ k and P r Λ k For general form degree k there are two families of spaces of polynomial differential forms, P r Λ k and P r Λ k, which, when assembled lead to the natural finite element subspaces of HΛ k (Ω). They assemble into complexes with commuting projections in numerous ways. The two families are inter-related and should be studied together. 22 / 39

58 The spaces P r Λ k and P r Λ k For general form degree k there are two families of spaces of polynomial differential forms, P r Λ k and P r Λ k, which, when assembled lead to the natural finite element subspaces of HΛ k (Ω). They assemble into complexes with commuting projections in numerous ways. The two families are inter-related and should be studied together. Special cases: P r Λ 0 = P r Λ 0, the Lagrange finite elements P r Λ n (T ) = P r+1 Λn, all piecewise polynomials of degree r P 1 Λk (T ) is the space of Whitney k-forms (1 DOF per k-face) Whitney, / 39

59 Finite element differential forms and classical mixed FEM P r Λ 0 (T ) = P r Λ 0 (T ) H 1 Lagrange elts P r Λ n (T ) = P r 1 Λ n (T ) L 2 discontinuous elts n = 2: Pr Λ 1 (T ) H(curl) P r Λ 1 (T ) H(curl) n = 3: Pr Λ 1 (T ) H(curl) P r Λ 1 (T ) H(curl) Pr Λ 2 (T ) H(div) P r Λ 2 (T ) H(div) Raviart Thomas elts Brezzi Douglas Marini elts Nedelec 1st kind edge elts Nedelec 2nd kind edge elts Nedelec 1st kind face elts Nedelec 2nd kind face elts 23 / 39

60 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 24 / 39

61 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 0 P r Λ 0 κ P r 1 Λ 1 κ κ P r n Λ n 0 24 / 39

62 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 0 P r Λ 0 κ P r 1 Λ 1 C.f., the polynomial de Rham complex 0 P r Λ 0 d P r 1 Λ 1 κ d κ P r n Λ n 0 d P r n Λ n 0 24 / 39

63 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 0 P r Λ 0 κ P r 1 Λ 1 C.f., the polynomial de Rham complex 0 P r Λ 0 d P r 1 Λ 1 For Ω R 3 0 P r (Ω) X P r 1 (Ω; R 3 ) κ d X P r 2 (Ω; R 3 ) κ P r n Λ n 0 d P r n Λ n 0 X P r 3 (Ω) 0 24 / 39

64 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 0 P r Λ 0 κ P r 1 Λ 1 C.f., the polynomial de Rham complex 0 P r Λ 0 d P r 1 Λ 1 For Ω R 3 0 P r (Ω) X P r 1 (Ω; R 3 ) κ d X P r 2 (Ω; R 3 ) κ P r n Λ n 0 d P r n Λ n 0 X P r 3 (Ω) 0 Key relation: (dκ + κd)ω = (r + k)ω ω H r Λ k (homogeneous polys) 24 / 39

65 The Koszul complex The key to the construction is the Koszul differential κ : Λ k Λ k 1 : (κω) x (v 1,..., v k 1 ) = ω x (X, v 1,..., v k 1 ), X = x x 0 0 P r Λ 0 κ P r 1 Λ 1 C.f., the polynomial de Rham complex 0 P r Λ 0 d P r 1 Λ 1 For Ω R 3 0 P r (Ω) X P r 1 (Ω; R 3 ) κ d X P r 2 (Ω; R 3 ) κ P r n Λ n 0 d P r n Λ n 0 X P r 3 (Ω) 0 Key relation: (dκ + κd)ω = (r + k)ω ω H r Λ k (homogeneous polys) H r Λ k = dh r+1 Λ k 1 κh r 1 Λ k+1 24 / 39

66 Definition of P r Λ k Using this decomposition, we define the space P r Λ k contained between P r Λ k and P r 1 Λ k : P r Λ k := P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1 25 / 39

67 Definition of P r Λ k Using this decomposition, we define the space P r Λ k contained between P r Λ k and P r 1 Λ k : Note X P r Λ k := P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1 P r Λ 0 = P r Λ 0 P r Λ n = P r 1 Λ n P r 1 Λ k P r Λ k P r Λ k otherwise 25 / 39

68 Definition of P r Λ k Using this decomposition, we define the space P r Λ k contained between P r Λ k and P r 1 Λ k : Note X P r Λ k := P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1 P r Λ 0 = P r Λ 0 P r Λ n = P r 1 Λ n P r 1 Λ k P r Λ k P r Λ k otherwise God made P r Λ k and P r Λ k, all the rest is the work of man. 25 / 39

69 Degrees of freedom The other ingredient of a finite element space are the degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (P r Λ k (T )), into subspaces associated to subsimplices f of T. 26 / 39

70 Degrees of freedom The other ingredient of a finite element space are the degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (P r Λ k (T )), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dim. d k we associate ω Tr f ω η, η P r+k d Λd k (f ) f 26 / 39

71 Degrees of freedom The other ingredient of a finite element space are the degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (P r Λ k (T )), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dim. d k we associate ω Tr f ω η, η P r+k d Λd k (f ) f DOF for Pr Λ k (T ): ω Tr f ω η, η P r+k d 1 Λ d k (f ) Hiptmair f 26 / 39

72 Degrees of freedom The other ingredient of a finite element space are the degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (P r Λ k (T )), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dim. d k we associate ω Tr f ω η, η P r+k d Λd k (f ) f DOF for Pr Λ k (T ): ω Tr f ω η, η P r+k d 1 Λ d k (f ) Hiptmair f The resulting FE spaces have exactly the continuity required by HΛ k : Theorem. P r Λ k (T ) = { ω HΛ k (Ω) : ω T P r Λ k (T ) T T }. Similarly for P r. 26 / 39

73 Bases for P r Λ k and P r Λ k As a basis for P r Λ k (T ) and P r Λ k (T ) we may take the dual basis to the degrees of freedom. For k = 0 this is the standard Lagrange basis. For P1 Λk (T ) there is one basis element for each k-simplex, the Whitney form k φ σ0 σ k := ( 1) i λ σi dλ σ0 dλ σi dλ σk i=0 27 / 39

74 Geometric bases A useful alternative to the Lagrange basis for the Lagrange finite elements is the Bernstein basis, given by monomials in the barycentric coords. P r (T ) = P r (T, f ) f subsimplex P r (T, f ) = trace P r (f ) = P r dim f 1 (f ) 28 / 39

75 Geometric bases A useful alternative to the Lagrange basis for the Lagrange finite elements is the Bernstein basis, given by monomials in the barycentric coords. P r (T ) = P r (T, f ) f subsimplex P r (T, f ) = trace P r (f ) = P r dim f 1 (f ) P r Λ k (T ) = dim f k P r Λ k (T, f ), P r Λ k (T, f ) = trace P r Λ k (f ) = P r+k dim f Λdim f k (f ) 28 / 39

76 Geometric bases A useful alternative to the Lagrange basis for the Lagrange finite elements is the Bernstein basis, given by monomials in the barycentric coords. P r (T ) = P r (T, f ) f subsimplex P r (T, f ) = trace P r (f ) = P r dim f 1 (f ) P r Λ k (T ) = dim f k P r Λ k (T ) = P r Λ k (T, f ), P r Λ k (T, f ) dim f k P r Λ k (T, f ) P r Λ k (T, f ), = trace P r Λ k (f ) = P r+k dim f Λdim f k (f ) = trace P r Λ k (f ) = P r+k dim f 1 Λ dim f k (f ) 28 / 39

77 Finite element de Rham subcomplexes We don t only want spaces, we also want them to fit together into discrete de Rham complexes. One such FEdR subcomplex uses the P r Λ k spaces of constant degree r: 0 P r Λ 0 (T ) d P r Λ 1 (T ) d d P r Λ n (T ) 0 29 / 39

78 Finite element de Rham subcomplexes We don t only want spaces, we also want them to fit together into discrete de Rham complexes. One such FEdR subcomplex uses the P r Λ k spaces of constant degree r: 0 P r Λ 0 (T ) d P r Λ 1 (T ) d d P r Λ n (T ) 0 Another uses the P r Λ k spaces with decreasing degree: 0 P r Λ 0 (T ) d P r 1 Λ 1 (T ) d d P r n Λ n (T ) 0 29 / 39

79 Finite element de Rham subcomplexes We don t only want spaces, we also want them to fit together into discrete de Rham complexes. One such FEdR subcomplex uses the P r Λ k spaces of constant degree r: 0 P r Λ 0 (T ) d P r Λ 1 (T ) d d P r Λ n (T ) 0 Another uses the P r Λ k spaces with decreasing degree: 0 P r Λ 0 (T ) d P r 1 Λ 1 (T ) d d P r n Λ n (T ) 0 These are extreme cases. For every r there are 2 n 1 such FEdR subcomplexes. 29 / 39

80 The 4 FEdR subcomplexes ending with P 0 Λ 3 in 3D 0 grad curl div 0 0 grad curl div 0 0 grad curl div 0 0 grad curl div 0 30 / 39

81 Application to Elasticity: The Holy Grail Attained? 31 / 39

82 Stress displacement mixed finite elements for elasticity The search for such elements dates back to Fraeijs de Veubeke, Pian, Watwood and Hartz, Zienkiewicz,... in the 1960 s. 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 difficult and as yet such elements have not been successfully noted. Zienkiewicz, Taylor, Zhu The Finite Element Method: Its Basis & Fundamentals, 6th ed., vol / 39

83 Stress displacement mixed finite elements for elasticity The search for such elements dates back to Fraeijs de Veubeke, Pian, Watwood and Hartz, Zienkiewicz,... in the 1960 s. 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 difficult and as yet such elements have not been successfully noted. Zienkiewicz, Taylor, Zhu The Finite Element Method: Its Basis & Fundamentals, 6th ed., vol. 1 Thanks to FEEC, it is time to retire that statement! 32 / 39

84 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). 33 / 39

85 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). Ω ( ) 1 A σ :σ + div σ u + f u dx 2 σ,u H(div;S) L 2 (R n ) stationary point 33 / 39

86 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). Ω ( ) 1 A σ :σ + div σ u + f u dx 2 σ,u H(div;S) L 2 (R n ) stationary point Ω ( ) 1 A σ :σ + div σ u + σ :p + f u dx 2 σ,u,p H(div;M) L 2 (R n ) L 2 (K) S.P. 33 / 39

87 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). Ω ( ) 1 A σ :σ + div σ u + f u dx 2 σ,u H(div;S) L 2 (R n ) stationary point Ω ( ) 1 A σ :σ + div σ u + σ :p + f u dx 2 Arnold Brezzi Douglas 84: PEERS element σ,u,p H(div;M) L 2 (R n ) L 2 (K) S.P / 39

88 The elasticity complex There is a complex for elasticity analogous to the de Rham complex. It has versions both for strong symmetry and weak symmetry. displacement rotation strain 0 H 1 (Ω; R 3 ) L 2 (Ω, K) J is second order! J (grad, I ) H(J, Ω; M) H(div, Ω; M) 0 div A skw J L 2 (Ω; R 3 ) L 2 (Ω; K) 0 stress load couple 34 / 39

89 New mixed finite elements for elasticity 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: 0 Λ 0 h 0 Λ 0 h grad Λ 1 h grad Λ 1 h curl Λ 2 h curl Λ 2 h div Λ 3 h 0 div Λ 3 h 0 Surjectivity Hypothesis: (roughly) for each DOF of Λ 2 h there is a corresponding DOF of Λ 1 h. stress: Λ2 h (R 3 ) Then displacement: Λ3 h (R 3 ) is a stable element choice. rotation: Λ 3 h (K) 35 / 39

90 The simplest choice 0 grad curl div 0 0 grad curl div 0 σ u p 36 / 39

91 Features of the new mixed elements Based on HR formulation with weak symmetry; very natural Lowest degree element is very simple: full P 1 for stress, P 0 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 37 / 39

92 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. 38 / 39

93 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. Capturing the right structure on the discrete level can be essential to get stable methods. 38 / 39

94 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. Capturing the right structure on the discrete level can be essential to get stable methods. FEEC provides a very natural framework for the design and understanding of subtle stability issues that arise in the discretization of a wide variety of PDE systems. It brings to bear tools from geometry, topology, and algebra to develop discretizations which are compatible with the geometric, topological, and algebraic structure of the PDE system, and so obtain stability. 38 / 39

95 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. Capturing the right structure on the discrete level can be essential to get stable methods. FEEC provides a very natural framework for the design and understanding of subtle stability issues that arise in the discretization of a wide variety of PDE systems. It brings to bear tools from geometry, topology, and algebra to develop discretizations which are compatible with the geometric, topological, and algebraic structure of the PDE system, and so obtain stability. FEEC has been used to unify, clarify, and refine many known finite element methods. It is a mathematically rigorous theory. 38 / 39

96 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. Capturing the right structure on the discrete level can be essential to get stable methods. FEEC provides a very natural framework for the design and understanding of subtle stability issues that arise in the discretization of a wide variety of PDE systems. It brings to bear tools from geometry, topology, and algebra to develop discretizations which are compatible with the geometric, topological, and algebraic structure of the PDE system, and so obtain stability. FEEC has been used to unify, clarify, and refine many known finite element methods. It is a mathematically rigorous theory. The P r Λ k and P r Λ k spaces are the natural finite element discretizations for differential forms and the de Rham complex. 38 / 39

97 Conclusions Exterior calculus clarifies the nature of physical quantities and the structure of the PDEs involving them. Capturing the right structure on the discrete level can be essential to get stable methods. FEEC provides a very natural framework for the design and understanding of subtle stability issues that arise in the discretization of a wide variety of PDE systems. It brings to bear tools from geometry, topology, and algebra to develop discretizations which are compatible with the geometric, topological, and algebraic structure of the PDE system, and so obtain stability. FEEC has been used to unify, clarify, and refine many known finite element methods. It is a mathematically rigorous theory. The P r Λ k and P r Λ k spaces are the natural finite element discretizations for differential forms and the de Rham complex. Through FEEC we believe we have completed the long search for just the right mixed finite elements for elasticity. 38 / 39

98 References everything is at arnold Finite element exterior calculus, homological techniques, and applications, Acta Numerica 2006 Mixed finite element methods for linear elasticity with weakly imposed symmetry, Math. Comp Differential complexes and stability of finite element methods. I. The de Rham complex II. The elasticity complex in: Compatible Spatial Discretizations, IMA Volumes in Mathematics and its Applications / 39