Introduction to finite element exterior calculus

Transcription

1 Introduction to finite element exterior calculus Ragnar Winther CMA, University of Oslo Norway

2 Why finite element exterior calculus? Recall the de Rham complex on the form: R H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0

3 Why finite element exterior calculus? Recall the de Rham complex on the form: R H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 and the corresponding elasticity complex T H 1 (Ω; R 3 ) ɛ H(J, Ω; S) J H(div, Ω; S) div L 2 (Ω; R 3 ) 0. Here J is the second order operator Jτ = curl(curl τ) T mapping symmetric matrix fields into symmetric matrix fields, and T is the six dimensional space of rigid motions.

4 Why finite element exterior calculus? Recall the de Rham complex on the form: R H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 and the corresponding elasticity complex T H 1 (Ω; R 3 ) ɛ H(J, Ω; S) J H(div, Ω; S) div L 2 (Ω; R 3 ) 0. Here J is the second order operator Jτ = curl(curl τ) T mapping symmetric matrix fields into symmetric matrix fields, and T is the six dimensional space of rigid motions. Reference: Arnold, Falk, Winther, Mixed finite elements for linear elasticity with weakly imposed symmetry, Math.Comp

5 Mapping of the domain Consider a map φ : R 3 R 3 mapping a domain Ω to Ω. Ω ϕ Ω

6 Mapping of the domain Consider a map φ : R 3 R 3 mapping a domain Ω to Ω. Ω ϕ Ω Desired commuting diagram: grad curl R H 1 (Ω ) H(curl, Ω ) H(div, Ω ) F 1 φ R H 1 (Ω) F c φ grad H(curl, Ω) F d φ curl H(div, Ω) div L 2 (Ω ) 0 F 0 φ div L 2 (Ω) 0

7 Piola transforms The proper Piola transforms are given by (Fφ 1 u)(x) = u(φ(x)) (F c φ u)(x) = (Dφ(x))T u(φ(x) (F d φ u)(x) = (det Dφ(x))(Dφ(x)) 1 u(φ(x)) (Fφ 0 u)(x) = (det Dφ(x))u(φ(x))

8 Alternative de Rham complex We replace the de Rham complex R H 1 (Ω) grad H(curl, Ω) curl H(div, Ω) div L 2 (Ω) 0 by an equivalent complex R HΛ 0 (Ω) d HΛ 1 (Ω) d HΛ 2 (Ω) d L 2 Λ 3 (Ω) 0 In fact, the second complex can be defined for any dimension n.

9 Exterior calculus and differential forms Instead of consider scalar fields and vector fields, i.e., functions of the form u : Ω R or u : Ω R 3 where Ω R 3, we consider instead functions of the form u : Ω Alt k R n Here Ω is allowed to be a subset of R n.

10 Exterior calculus and differential forms Instead of consider scalar fields and vector fields, i.e., functions of the form u : Ω R or u : Ω R 3 where Ω R 3, we consider instead functions of the form u : Ω Alt k R n Here Ω is allowed to be a subset of R n. For any vector space V, the space Alt k V is the space of alternating k-linear maps on V

11 The space Alt k V Assume dim V = n and that v 1, v 2,... v n is a basis. If ω Alt k V and j 1, j 2,..., j k are integers in {1, 2,..., n} then ω(v j1, v j2,..., v jk ) R.

12 The space Alt k V Assume dim V = n and that v 1, v 2,... v n is a basis. If ω Alt k V and j 1, j 2,..., j k are integers in {1, 2,..., n} then ω(v j1, v j2,..., v jk ) R. Furthermore, ω is linear in each argument, and if two neighboring vectors change position then the value of ω change sign. Therefore, it is enough to specify ω(v j1, v j2,..., v jk ) for an increasing sequence {j i }.

13 The space Alt k V Assume dim V = n and that v 1, v 2,... v n is a basis. If ω Alt k V and j 1, j 2,..., j k are integers in {1, 2,..., n} then ω(v j1, v j2,..., v jk ) R. Furthermore, ω is linear in each argument, and if two neighboring vectors change position then the value of ω change sign. Therefore, it is enough to specify ω(v j1, v j2,..., v jk ) for an increasing sequence {j i }. As a consequence Alt k V is a vector space with ( ) n dim Alt k V =. k

14 The space Alt k V Assume dim V = n and that v 1, v 2,... v n is a basis. If ω Alt k V and j 1, j 2,..., j k are integers in {1, 2,..., n} then ω(v j1, v j2,..., v jk ) R. Furthermore, ω is linear in each argument, and if two neighboring vectors change position then the value of ω change sign. Therefore, it is enough to specify ω(v j1, v j2,..., v jk ) for an increasing sequence {j i }. As a consequence Alt k V is a vector space with ( ) n dim Alt k V =. k Note that dim Alt k V = dim Alt n k V. The canonical map between these spaces is the Hodge star operator.

15 Interior and exterior products If ω Alt k V and v V then the interior product, or the contraction of ω by v, ω v Alt k 1 V is given by ω v(v 1, v 2,... v k 1 ) = ω(v, v 1, v 2,... v k 1 )

16 Interior and exterior products If ω Alt k V and v V then the interior product, or the contraction of ω by v, ω v Alt k 1 V is given by ω v(v 1, v 2,... v k 1 ) = ω(v, v 1, v 2,... v k 1 ) If ω Alt j V and η Alt k V then the exterior product, or wedge product, ω η Alt j+k is given by (ω η)(v 1,..., v j+k ) = σ (sign σ)ω(v σ(1),..., v σ(j) )η(v σ(j+1),..., v σ(j+k) ), v i V, where the sum is over all permutations σ of {1, 2,..., j + k} such that σ(1) <... < σ(j) and σ(j + 1) <... < σ(j + k).

17 Basis for Alt k R n We let {dx i } be the natural dual basis in Alt 1 R n given by dx i (e j ) = δ i,j

18 Basis for Alt k R n We let {dx i } be the natural dual basis in Alt 1 R n given by dx i (e j ) = δ i,j A basis for Alt 2 R n is given by dx i dx j, i < j,

19 Basis for Alt k R n We let {dx i } be the natural dual basis in Alt 1 R n given by dx i (e j ) = δ i,j A basis for Alt 2 R n is given by dx i dx j, i < j, while dx j1 dx j2 dx jk, where {j i } is increasing is a basis for Alt k R n.

20 Any ω Alt k R n can be represented uniquely on the form u σ dx σ(1) dx σ(2)... dx σ(k) σ where the sum is over all incresing maps σ mapping {1, 2,..., k} into {1, 2,..., n}, and u σ R.

21 Any ω Alt k R n can be represented uniquely on the form u σ dx σ(1) dx σ(2)... dx σ(k) σ where the sum is over all incresing maps σ mapping {1, 2,..., k} into {1, 2,..., n}, and u σ R. Furthermore, dx σ(1) dx σ(2)... dx σ(k) (v 1,... v k ) = det(dx σ(i) v j )

22 Correspondences Alt k and vector fields: Alt 0 R 3 = R Alt 1 R 3 = R 3 Alt 2 R 3 = R 3 c c u 1 dx 1 + u 2 dx 2 + u 3 dx 3 u Alt 3 R 3 = R c c dx1 dx 2 dx 3 u 3 dx 1 dx 2 u 2 dx 1 dx 3 + u 1 dx 2 dx 3 u

23 Correspondences Alt k and vector fields: Alt 0 R 3 = R Alt 1 R 3 = R 3 Alt 2 R 3 = R 3 c c u 1 dx 1 + u 2 dx 2 + u 3 dx 3 u Alt 3 R 3 = R c c dx1 dx 2 dx 3 Exterior product: u 3 dx 1 dx 2 u 2 dx 1 dx 3 + u 1 dx 2 dx 3 u : 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

24 Differential forms Differential forms are functions of the form ω : Ω Alt k Alt k R n.

25 Differential forms Differential forms are functions of the form ω : Ω Alt k Alt k R n. We let Λ k = Λ k (Ω) = C (Ω; Alt k ). For such smooth differentiable forms the exterior derivative, d : Λ k Λ k+1 is defined by k+1 dω x (v 1, v 2,..., v k+1 ) = ( 1) j+1 vj ω x (v 1,..., ˆv j,..., v k+1 ), j=1 ω Λ k, v 1,..., v k+1 R n

26 Differential forms Differential forms are functions of the form ω : Ω Alt k Alt k R n. We let Λ k = Λ k (Ω) = C (Ω; Alt k ). For such smooth differentiable forms the exterior derivative, d : Λ k Λ k+1 is defined by k+1 dω x (v 1, v 2,..., v k+1 ) = ( 1) j+1 vj ω x (v 1,..., ˆv j,..., v k+1 ), j=1 ω Λ k, v 1,..., v k+1 R n Using the fact that vi vj = vj vi it follows that d d = 0.

27 Check of d d = 0 k+1 dω x (v 1, v 2,..., v k+1 ) = ( 1) j+1 vj ω x (v 1,..., ˆv j,..., v k+1 ), j=1 ω Λ k, v 1,..., v k+1 R n Start with a zero form ω.

28 Check of d d = 0 k+1 dω x (v 1, v 2,..., v k+1 ) = ( 1) j+1 vj ω x (v 1,..., ˆv j,..., v k+1 ), j=1 ω Λ k, v 1,..., v k+1 R n Start with a zero form ω. dω x (v) = v ω x

29 Check of d d = 0 k+1 dω x (v 1, v 2,..., v k+1 ) = ( 1) j+1 vj ω x (v 1,..., ˆv j,..., v k+1 ), j=1 ω Λ k, v 1,..., v k+1 R n Start with a zero form ω. dω x (v) = v ω x (d d)ω(v 1, v 2 ) = v1 dω(v 2 ) v2 dω(v 1 ) = 0.

30 Representation by a basis If ω is given on the form ω x = σ u σ dx σ(1) dx σ(2)... dx σ(k), where each u σ is a scalar field on Ω then dω x = σ n j=1 u σ x j dx j dx σ(1) dx σ(2)... dx σ(k)

31 The de Rham complex and differential forms Using differential forms the de Rham complex (smooth version) can be written as R Λ 0 d Λ 1 d d Λ n 0.

32 Mapping of the domain Ω ϕ Ω

33 Mapping of the domain Ω ϕ Ω The corresponding pull backs, φ, which maps Λ k (Ω ) to Λ k (Ω), is given by (φ ω) x (v 1, v 2,..., v k ) = ω φ(x) (Dφ x (v 1 ), Dφ x (v 2 ),..., Dφ x (v k )), where Dφ x is the derivative of φ at x mapping T x Ω into T φ(x) Ω.

34 Mapping of the domain Ω ϕ Ω The corresponding pull backs, φ, which maps Λ k (Ω ) to Λ k (Ω), is given by (φ ω) x (v 1, v 2,..., v k ) = ω φ(x) (Dφ x (v 1 ), Dφ x (v 2 ),..., Dφ x (v k )), where Dφ x is the derivative of φ at x mapping T x Ω into T φ(x) Ω. This expression should be compared with the expressions we had for the Piola maps before.

35 Commuting diagram R Λ 0 (Ω ) R Λ 0 (Ω) d Λ 1 (Ω ) d d Λ n (Ω ) 0. φ φ φ d Λ 1 (Ω) d d Λ n (Ω) 0.

36 Commuting diagram R Λ 0 (Ω ) R Λ 0 (Ω) d Λ 1 (Ω ) d d Λ n (Ω ) 0. φ φ φ d Λ 1 (Ω) d d Λ n (Ω) 0. So the pullback commutes with the exterior derivative, i.e., φ (dω) = d(φ ω), ω Λ k (Ω ), and distributes with respect to the wedge product: φ (ω η) = φ ω φ η.

37 Commuting diagram R Λ 0 (Ω ) R Λ 0 (Ω) d Λ 1 (Ω ) d d Λ n (Ω ) 0. φ φ φ d Λ 1 (Ω) d d Λ n (Ω) 0. So the pullback commutes with the exterior derivative, i.e., φ (dω) = d(φ ω), ω Λ k (Ω ), and distributes with respect to the wedge product: φ (ω η) = φ ω φ η. Note this also gives us a way to define the exterior derivative on a manifold.

38 Commuting diagram R Λ 0 (Ω ) R Λ 0 (Ω) d Λ 1 (Ω ) d d Λ n (Ω ) 0. φ φ φ d Λ 1 (Ω) d d Λ n (Ω) 0. So the pullback commutes with the exterior derivative, i.e., φ (dω) = d(φ ω), ω Λ k (Ω ), and distributes with respect to the wedge product: φ (ω η) = φ ω φ η. Note this also gives us a way to define the exterior derivative on a manifold. Key identity: (ψ φ) = φ ψ

39 Exterior derivative and wedge product By combining the wedge product with the exterior derivative we obtain a Leibniz rule of the form d(ω η) = dω η + ( 1) j ω dη, ω Λ j, η Λ k.

40 Exterior derivative and wedge product By combining the wedge product with the exterior derivative we obtain a Leibniz rule of the form d(ω η) = dω η + ( 1) j ω dη, ω Λ j, η Λ k. The pullback of the inclusion Ω Ω is the trace map, Tr. The vanishing of Tr ω for ω Λ k (Ω) does not imply that ω x vanishes for x Ω, only that it vanishes when applied to k-tuples of tangent vectors to Ω.

41 Exterior derivative and wedge product By combining the wedge product with the exterior derivative we obtain a Leibniz rule of the form d(ω η) = dω η + ( 1) j ω dη, ω Λ j, η Λ k. The pullback of the inclusion Ω Ω is the trace map, Tr. The vanishing of Tr ω for ω Λ k (Ω) does not imply that ω x vanishes for x Ω, only that it vanishes when applied to k-tuples of tangent vectors to Ω. Since Tr is a pull back we will have Tr d = d Ω Tr

42 Integration of differential forms If ω Λ n (Ω), Ω R n, with the representation ω x = u x dx 1 dx 2... dx n, then we can define ω = u dx Ω Ω

43 Integration of differential forms If ω Λ n (Ω), Ω R n, with the representation ω x = u x dx 1 dx 2... dx n, then we can define ω = If φ : Ω Ω then we have φ ω = Ω Ω Ω u dx Ω ω

44 Integration of differential forms If ω Λ n (Ω), Ω R n, with the representation then we can define If φ : Ω Ω then we have ω x = u x dx 1 dx 2... dx n, Ω Ω ω = Ω φ ω = u dx Ω ω Note that this gives us a way to define the integral of an n-form on a manifold.

45 Stokes theorem With the three concepts exterior derivative, trace operator, and integral we can formulate Stokes theorem as dω = Tr ω, ω Λ n 1 (Ω) Ω Ω

46 Stokes theorem With the three concepts exterior derivative, trace operator, and integral we can formulate Stokes theorem as dω = Tr ω, ω Λ n 1 (Ω) Ω Ω Furthermore, by combining this with the Leibniz rule we obtain the following integration by parts identity dω η = ( 1) k 1 ω dη + Tr ω Tr η Ω where ω Λ k, η Λ n k 1. Ω Ω

47 L 2 differential forms If ω is a k form given by ω = σ u σ dx σ(1) dx σ(2)... dx σ(k) then ω is defined to be in L 2 Λ k (Ω) if each of the coefficients u σ are in L 2.

48 L 2 differential forms If ω is a k form given by ω = σ u σ dx σ(1) dx σ(2)... dx σ(k) then ω is defined to be in L 2 Λ k (Ω) if each of the coefficients u σ are in L 2. Furthernmore, we let HΛ k (Ω) = { ω L 2 Λ k (Ω) dω L 2 Λ k+1 (Ω) }

49 L 2 differential forms If ω is a k form given by ω = σ u σ dx σ(1) dx σ(2)... dx σ(k) then ω is defined to be in L 2 Λ k (Ω) if each of the coefficients u σ are in L 2. Furthernmore, we let HΛ k (Ω) = { ω L 2 Λ k (Ω) dω L 2 Λ k+1 (Ω) } We then ontain the L 2 de Rham complex: 0 HΛ 0 (Ω) d HΛ 1 (Ω) d d HΛ n (Ω) 0

50 Polynomial de Rham complex We define P r Λ k = {ω Λ k ω(v 1,... v k ) P r, v 1,... v k }

51 Polynomial de Rham complex We define P r Λ k = {ω Λ k ω(v 1,... v k ) P r, v 1,... v k } Alternatively, P r Λ k consists of all u σ dx σ(1) dx σ(2)... dx σ(k) σ where the coefficients u σ are in P r.

52 Polynomial de Rham complex We define P r Λ k = {ω Λ k ω(v 1,... v k ) P r, v 1,... v k } Alternatively, P r Λ k consists of all u σ dx σ(1) dx σ(2)... dx σ(k) σ where the coefficients u σ are in P r. The polynomial de Rham complex is exact. 0 P r Λ 0 d Pr 1 Λ 1 d d Pr n Λ n 0

53 The Kozul complex The Koszul differential κ of a k-form ω is the (k 1)-form given by (κω) x (v 1,..., v k 1 ) = ω x x(v 1,..., v k 1 ) = ω x ( x, v1,..., v k 1 ),

54 The Kozul complex The Koszul differential κ of a k-form ω is the (k 1)-form given by (κω) x (v 1,..., v k 1 ) = ω x x(v 1,..., v k 1 ) = ω x ( x, v1,..., v k 1 ), For each r, κ maps P r 1 Λ k to P r Λ k 1,

55 The Kozul complex The Koszul differential κ of a k-form ω is the (k 1)-form given by (κω) x (v 1,..., v k 1 ) = ω x x(v 1,..., v k 1 ) = ω x ( x, v1,..., v k 1 ), For each r, κ maps P r 1 Λ k to P r Λ k 1, and the Koszul complex 0 P r n Λ n κ Pr n+1 Λ n 1 κ κ Pr Λ 0 R 0, is exact.

56 The spaces P r Λ k The operators d and κ are related by the homotopy relation (dκ + κd)ω = (r + k)ω, ω H r Λ k, where H r denotes the homogeneous polynomials of degree r.

57 The spaces P r Λ k The operators d and κ are related by the homotopy relation (dκ + κd)ω = (r + k)ω, ω H r Λ k, where H r denotes the homogeneous polynomials of degree r. As a consequence we obtain the identity P r Λ k = P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1

58 The spaces P r Λ k The operators d and κ are related by the homotopy relation (dκ + κd)ω = (r + k)ω, ω H r Λ k, where H r denotes the homogeneous polynomials of degree r. As a consequence we obtain the identity P r Λ k = P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1 We then define the space P r Λ k by P r Λ k = P r 1 Λ k + κp r 1 Λ k+1.

59 The spaces P r Λ k The operators d and κ are related by the homotopy relation (dκ + κd)ω = (r + k)ω, ω H r Λ k, where H r denotes the homogeneous polynomials of degree r. As a consequence we obtain the identity P r Λ k = P r 1 Λ k + κh r 1 Λ k+1 + dh r+1 Λ k 1 We then define the space P r Λ k by P r Λ k = P r 1 Λ k + κp r 1 Λ k+1. We note that P r Λ 0 = P r Λ 0 and P r Λ n = P r 1 Λ n. Furthermore, 0 P r Λ 0 d P r Λ 1 d d P r Λ n 0 is an exact complex, and the space P r Λ k is affine invariant.

60 Characterization of affine invariant polynomial spaces Theorem Assume that the polynomial space QΛ k, satisfies P r 1 Λ k QΛ k P r Λ k. Then QΛ k = P r Λ k or QΛ k = P r 1 Λ k + dp r+1 Λ k 1

61 The four exact sequences ending with P r Λ 3 (T ) in 3D 0 P r+1 Λ 0 d P r+1 Λ 1 d P r+1 Λ 2 d Pr Λ P r+2 Λ 0 d Pr+1 Λ 1 d P r+1 Λ 2 d Pr Λ P r+2 Λ 0 d P r+2 Λ 1 d Pr+1 Λ 2 d Pr Λ P r+3 Λ 0 d Pr+2 Λ 1 d Pr+1 Λ 2 d Pr Λ 3 0

62 The four exact sequences ending with P r Λ 3 (T ) in 3D 0 P r+1 Λ 0 d P r+1 Λ 1 d P r+1 Λ 2 d Pr Λ P r+2 Λ 0 d Pr+1 Λ 1 d P r+1 Λ 2 d Pr Λ P r+2 Λ 0 d P r+2 Λ 1 d Pr+1 Λ 2 d Pr Λ P r+3 Λ 0 d Pr+2 Λ 1 d Pr+1 Λ 2 d Pr Λ 3 0 In general there are 2 n 1 exact sequences composed of the families of P and P spaces.

63 The four sequences ending with P 0 Λ 3 (T ) in 3D 0 grad curl div 0 0 grad curl div 0 0 grad curl div 0 0 grad curl div 0

64 Piecewise smooth differential forms It is a consequence of Stokes theorem that a piecewise smooth k form ω, with respect to a simplicial mesh T h of Ω, is in HΛ k (Ω) if and only if the trace of ω, Tr ω, is continuous on the interfaces. Here Tr ω is defined by restricting the spatial variable x to the interface, and by applying ω only to tangent vectors of the interface.

65 Degrees of freedom To obtain finite element differential forms not just pw polynomials we need degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (P r Λ k (T )) (with T a simplex), into subspaces associated to subsimplices f of T.

66 Degrees of freedom To obtain finite element differential forms not just pw polynomials we need degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (Pr Λ k (T )) (with T a simplex), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dimension d we associate ω Tr f ω η, η P r+k d Λd k (f ) f

67 Degrees of freedom To obtain finite element differential forms not just pw polynomials we need degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (Pr Λ k (T )) (with T a simplex), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dimension d 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 ) f

68 Degrees of freedom To obtain finite element differential forms not just pw polynomials we need degrees of freedom, i.e., a decomposition of the dual spaces (P r Λ k (T )) and (Pr Λ k (T )) (with T a simplex), into subspaces associated to subsimplices f of T. DOF for P r Λ k (T ): to a subsimplex f of dimension d 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 ) f Given a triangulation T, we can then define P r Λ k (T ), P r Λ k (T ). They are subspaces of HΛ k (Ω).

69 De Rham cohomology 0 Λ 0 d (Ω) Λ 1 d (Ω) Λ 2 (Ω) 0 0 C (Ω) grad C (Ω; R 2 ) rot C (Ω) 0

70 Cohomology The de Rham complex HΛ k 1 (Ω) is called exact if for all k, d k 1 HΛ k (Ω) d k HΛ k+1 (Ω) Z k := ker d k = range d k 1 =: B k.

71 Cohomology The de Rham complex HΛ k 1 (Ω) is called exact if for all k, d k 1 HΛ k (Ω) d k HΛ k+1 (Ω) Z k := ker d k = range d k 1 =: B k. In general, B k Z k and we assume throughout that the kth cohomology group Z k /B k is finite dimensional.

72 Cohomology The de Rham complex HΛ k 1 (Ω) is called exact if for all k, d k 1 HΛ k (Ω) d k HΛ k+1 (Ω) Z k := ker d k = range d k 1 =: B k. In general, B k Z k and we assume throughout that the kth cohomology group Z k /B k is finite dimensional. The space of harmonic k-forms, H k, consists of all q Z k such that q, µ = 0 µ B k.

73 Cohomology The de Rham complex HΛ k 1 (Ω) is called exact if for all k, d k 1 HΛ k (Ω) d k HΛ k+1 (Ω) Z k := ker d k = range d k 1 =: B k. In general, B k Z k and we assume throughout that the kth cohomology group Z k /B k is finite dimensional. The space of harmonic k-forms, H k, consists of all q Z k such that q, µ = 0 µ B k. This leads to the Hodge decomposition HΛ k (Ω) = Z k Z k = B k H k Z k. Note that H k = Z k /B k.

74 Hodge Laplace problem Λ k 1 (Ω) d k 1 Λ k (Ω) d k Λ k+1 (Ω)

75 Hodge Laplace problem Λ k 1 (Ω) d k 1 Λ k (Ω) d k Λ k+1 (Ω) Formally: Given f Λ k, find u Λ k such that (d k 1 δ k 1 + δ k d k )u = f. Here δ k is a formal adjoint of d k.

76 Hodge Laplace problem Λ k 1 (Ω) d k 1 Λ k (Ω) d k Λ k+1 (Ω) Formally: Given f Λ k, find u Λ k such that (d k 1 δ k 1 + δ k d k )u = f. Here δ k is a formal adjoint of d k. The following mixed formulation is always well-posed: Given f L 2 Λ k (Ω), find σ HΛ k 1, u HΛ k and p H k such that σ, τ dτ, u = 0 τ HΛ k 1 dσ, v + du, dv + p, v = f, v v HΛ k u, q = 0 q H k

77 Hodge Laplacian Well-posedness of the Hodge Laplace problem follows from the Hodge decomposition and Poincaré s inequality: ω L 2 c dω L 2, ω (Z k ). Special cases: Recall that if dim H k = 0: Find σ HΛ k 1, and u HΛ k σ, τ dτ, u = 0 τ HΛ k 1 dσ, v + du, dv = f, v v HΛ k

78 Hodge Laplacian Well-posedness of the Hodge Laplace problem follows from the Hodge decomposition and Poincaré s inequality: ω L 2 c dω L 2, ω (Z k ). Special cases: Recall that if dim H k = 0: Find σ HΛ k 1, and u HΛ k σ, τ dτ, u = 0 τ HΛ k 1 dσ, v + du, dv = f, v v HΛ k 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,

79 Discretizations of the de Rham complex in three dimensions R H 1 (Ω) I 1 h grad H(curl, Ω) I c h curl H(div, Ω) I d h div L 2 (Ω) 0 I 0 h R H 1 h grad H h (curl) curl H h (div) div L 2 h 0.

80 Discretizations of the de Rham complex in three dimensions R H 1 (Ω) I 1 h grad H(curl, Ω) I c h curl H(div, Ω) I d h div L 2 (Ω) 0 I 0 h R H 1 h grad H h (curl) curl H h (div) div L 2 h 0. The operators I h are the canonical interpolation operators defined from the degrees of freedom. They commute with grad curl and div, but the are not bounded on the spaces H 1, H(curl) or H(div).

81 Discretizations of the de Rham complex in n dimensions R HΛ 0 (Ω) R d HΛ 1 (Ω) d d HΛ n (Ω) 0. I h I h I h Λ 0 h d Λ 1 h d d Λ n h 0.

82 Discretizations of the de Rham complex in n dimensions R HΛ 0 (Ω) R d HΛ 1 (Ω) d d HΛ n (Ω) 0. I h I h I h Λ 0 h d Λ 1 h d d Λ n h 0. The operators I h are the canonical interpolation operators defined from the degrees of freedom. They commute with d, but the are not bounded on the HΛ.

83 Discretization, Abstract setting Λ k 1 d k 1 Λ k Complex of Hilbert spaces with d k bounded and closed range.

84 Discretization, Abstract setting Λ k 1 d k 1 Λ k Complex of Hilbert spaces with d k bounded and closed range. Hodge decomposition and Poincaré inequality follow.

85 Discretization, Abstract setting Λ k 1 d k 1 Λ k Λ k 1 h d k 1 Λ k h Complex of Hilbert spaces with d k bounded and closed range. Hodge decomposition and Poincaré inequality follow. For discretization, construct a finite dimensional subcomplex.

86 Discretization, Abstract setting Λ k 1 d k 1 Λ k Λ k 1 h d k 1 Λ k h Complex of Hilbert spaces with d k bounded and closed range. Hodge decomposition and Poincaré inequality follow. For discretization, construct a finite dimensional subcomplex. Define H k h = (Bk h ) Z k h.

87 Discretization, Abstract setting Λ k 1 d k 1 Λ k Λ k 1 h d k 1 Λ k h Complex of Hilbert spaces with d k bounded and closed range. Hodge decomposition and Poincaré inequality follow. For discretization, construct a finite dimensional subcomplex. Define H k h = (Bk h ) Z k h. Discrete Hodge decomposition follows: Λ k h = Bk h Hk h (Zk h )

88 Discretization, Abstract setting Λ k 1 d k 1 Λ k Λ k 1 h d k 1 Λ k h Complex of Hilbert spaces with d k bounded and closed range. Hodge decomposition and Poincaré inequality follow. For discretization, construct a finite dimensional subcomplex. Define H k h = (Bk h ) Z k h. Discrete Hodge decomposition follows: Λ k h = Bk h Hk h (Zk h ) Galerkin s method: Λ k 1, Λ k, H k Λ k 1 h, Λ k h, Hk h When is it stable?

89 Bounded cochain projections Key property: Suppose that there exists a bounded cochain projection. 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 k h uniformly bounded πh k ω ω 0. π k h a projection π k h d k 1 = d k 1 π k 1 h

90 Bounded cochain projections Key property: Suppose that there exists a bounded cochain projection. Λ k 1 d k 1 Λ k π k 1 h Λ k 1 h Theorem π k h d k 1 Λ k h π k h uniformly bounded πh k ω ω 0. π k h a projection π k h d k 1 = d k 1 π k 1 h If v π k h v < v v Hk, then the induced map on cohomology is an isomorphism. gap ( H k, H k h) sup v πh k v v H k, v =1 The discrete Poincaré inequality holds uniformly in h. Galerkin s method is stable and convergent.

91 Proof of discrete Poincaré inequality Theorem: There is a positive constant c, independent of h, such that ω c dω, ω Z k h. In fact, c c P c π, where c P is continuous Poincaré constant, and c π bounds π h.

92 Proof of discrete Poincaré inequality Theorem: There is a positive constant c, independent of h, such that ω c dω, ω Z k h. In fact, c c P c π, where c P is continuous Poincaré constant, and c π bounds π h. Proof: Given ω Z k h, define η Zk HΛ k (Ω) by dη = dω. By the Poincaré inequality, η c dω, so it is enough to show that ω c η. Now, ω π h η Λ k h and d(ω π hη) = 0, so ω π h η Z k h. Therefore ω 2 = ω, π h η + ω, ω π h η = ω, π h η ω π h η, whence ω π h η. The result follows from the uniform boundedness of π h.

93 Construction of bounded cochain projections The canonical projections, I h, determined by the degrees of freedom, commute with d. But they are not bounded on HΛ k.

94 Construction of bounded cochain projections The canonical projections, I h, determined by the degrees of freedom, commute with d. But they are not bounded on HΛ k. If we apply the three operations: extend (E) regularize (R) canonical projection (I h ) we get a map Qh k : HΛk (Ω) Λ k h which is bounded and commutes with d. But it is not a projection.

95 Construction of bounded cochain projections The canonical projections, I h, determined by the degrees of freedom, commute with d. But they are not bounded on HΛ k. If we apply the three operations: extend (E) regularize (R) canonical projection (I h ) we get a map Qh k : HΛk (Ω) Λ k h which is bounded and commutes with d. But it is not a projection. However the composition π k h = (Qk h Λ k h) 1 Q k h can be shown to be a bounded cochain projection.

96 Bounded projections, definitions We combine extension and smoothing into an operator R ɛ h : L2 Λ k (Ω) CΛ k (Ω) given by (Rh ɛ ω) x = ρ(y)(eω) x+ɛhy dy. B 1

97 Bounded projections, definitions We combine extension and smoothing into an operator R ɛ h : L2 Λ k (Ω) CΛ k (Ω) given by (Rh ɛ ω) x = ρ(y)(eω) x+ɛhy dy. B 1 Assume that the mesh is quasi uniform: Lemma For each ɛ > 0 sufficiently small there is a constant c(ɛ) such that for all h I h R ɛ h L(L 2 Λ k (Ω),L 2 Λ k (Ω)) c(ɛ). There is constant c, independent of ɛ and h such that I I h R ɛ h L(L 2 Λ k h,l2 Λ k h ) cɛ.

98 Scaling and compactness

99 Bounded cochain projections The operator π h = π k h defined by π h = (I h R ɛ h Λ k h) 1 I h R ɛ h, where ɛ is taken small, but not too small, has all the desired properties, i.e. π k h is bounded in L(L2 Λ k (Ω), L 2 Λ k (Ω)) and L(HΛ k (Ω), HΛ k (Ω)) π k h π k h commutes with the exterior derivative is a projection

100 Removing the assumption of a quasi uniform mesh If the family of meshes is shape regular, but not necessarily quasi uniform, we introduce a piecewise linear and uniformly Lipschitz continuous function g h (x) to represent the local mesh size.

101 Removing the assumption of a quasi uniform mesh If the family of meshes is shape regular, but not necessarily quasi uniform, we introduce a piecewise linear and uniformly Lipschitz continuous function g h (x) to represent the local mesh size. For each vertex x we let g h (x) be the average of the values of diam T for the simplices meeting x, and we define Φ ɛy h (x) = x + ɛg h(x)y.

102 Removing the assumption of a quasi uniform mesh If the family of meshes is shape regular, but not necessarily quasi uniform, we introduce a piecewise linear and uniformly Lipschitz continuous function g h (x) to represent the local mesh size. For each vertex x we let g h (x) be the average of the values of diam T for the simplices meeting x, and we define Φ ɛy h (x) = x + ɛg h(x)y. Then we replace the operator (Rh ɛ ω) x = ρ(y)(eω) x+ɛhy dy B 1 by (Rh ɛ ω) x = ρ(y)((φ ɛy h ) Eω) x dy. B 1

103 References The development of FEEC leans heavily on earlier results taken from Whitney, Bossavit, Raviart and Thomas, Nedelec, Hiptmair,... as well as on the theory of finite elements in general.

104 References The development of FEEC leans heavily on earlier results taken from Whitney, Bossavit, Raviart and Thomas, Nedelec, Hiptmair,... as well as on the theory of finite elements in general. The presentation here is mostly based on D.N. Arnold, R.S. Falk, R. Winther, Finite element exterior calculus, homological techniques, and applications, Acta Numerica D.N. Arnold, R.S. Falk and R. Winther Finite element exterior calculus: From Hodge theory to numerical stability Bulletin of the Amer. Math. Soc. 47 (2010),