Mimetic scalar products of discrete differential forms Annalisa Buffa IMATI “E. Magenes” - Pavia Consiglio Nazionale delle Ricerche In collaboration with F. Brezzi and M. Manzini A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 1 / 25
What is this talk about 1 Differential forms and Finite Elements 2 Differential forms and Mimetic Finite Differences 3 Open problems - Conclusions 4 A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 2 / 25
The aim of the talk We are interested in providing numerical solutions to PDEs whose unknown can be sought as differential forms Examples are diffusion, fluid flows, elasticity, electromagnetics... There are many attempts in the literature ! A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 3 / 25
The aim of the talk We are interested in providing numerical solutions to PDEs whose unknown can be sought as differential forms Examples are diffusion, fluid flows, elasticity, electromagnetics... There are many attempts in the literature ! A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 3 / 25
The aim of the talk We are interested in providing numerical solutions to PDEs whose unknown can be sought as differential forms Examples are diffusion, fluid flows, elasticity, electromagnetics... There are many attempts in the literature ! A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 3 / 25
The aim of the talk We are interested in providing numerical solutions to PDEs whose unknown can be sought as differential forms Examples are diffusion, fluid flows, elasticity, electromagnetics... There are many attempts in the literature ! We consider here and all along the talk equations of electrostatics/magnetostatics as “toy” problems: Ω ⊂ R d , bounded, Lipschitz. n div a u = g Ω Ω curl u = f Ω a u · n = 0 ∂ Ω . In what follows: Ω is simply connected. A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 3 / 25
Our example n Ω ⊂ R 3 , bounded, Lipschitz. Ω div a u = g Ω curl u = f Ω + B.C. u ∈ Λ 1 , ⋆ a : Λ 1 → Λ 2 v ∈ Λ 2 , ⋆ a − 1 : Λ 2 → Λ 1 = g = g d ⋆ a u d v d u = f d ⋆ a − 1 v = f Tr ( ⋆ a u ) = 0 . Tr ( v ) = 0 . If f = 0, u = d χ , χ ∈ Λ 0 and If g = 0, v = d ξ , ξ ∈ Λ 1 and d ⋆ a d χ = g d ⋆ a − 1 d ξ = f A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 4 / 25
De-Rham diagram Solvability is due to the (trivial) De-Rham complex: when Ω is simply connected d d d → Λ 3 → 0 0 → Λ 0 → Λ 1 → Λ 2 − − − − − − − − − − − − On proxy fields this is 0 → H 1 (Ω) div → L 2 (Ω) → 0 ∇ curl − − − − → H ( curl , Ω) − − − − → H ( div , Ω) − − − − with H ( curl , Ω) = { u ∈ L 2 (Ω) 2 : curl u ∈ L 2 (Ω) 3 } H ( div , Ω) = { u ∈ L 2 (Ω) 2 : div u ∈ L 2 (Ω) } A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 5 / 25
De-Rham diagram Solvability is due to the (trivial) De-Rham complex: when Ω is simply connected d d d → Λ 3 → 0 0 → Λ 0 → Λ 1 → Λ 2 − − − − − − − − − − − − On proxy fields this is div ∇ curl 0 → H 10 (Ω) → L 20 (Ω) → 0 − − − − → H 0 ( curl , Ω) − − − − → H 0 ( div , Ω) − − − − with H 0 ( curl , Ω) = { u ∈ L 2 (Ω) 2 : curl u ∈ L 2 (Ω) 3 u × n = 0 on ∂ Ω } H 0 ( div , Ω) = { u ∈ L 2 (Ω) 2 : div u ∈ L 2 (Ω) u · n = 0 on ∂ Ω } � L 2 0 (Ω) = { u ∈ L 2 (Ω) : Ω u = 0 } A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 5 / 25
Variational formulations: proxy fields All those problems admits a variational formulations: back A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 6 / 25
Variational formulations: proxy fields All those problems admits a variational formulations: if f = 0, we just wrote the equation for the electric scalar potential χ ∈ Λ 0 : d ⋆ a d χ = g . On the proxy field: � � a ∇ χ · ∇ χ t = ∀ χ t ∈ H 1 (Ω) . χ ∈ H 1 (Ω) : g χ t Ω Ω Alternatively, setting D = a ∇ χ : Find D ∈ H 0 ( div , Ω), χ ∈ L 2 0 (Ω) � � � � a − 1 D · D t + χ div D t = 0 ∀ D t div D χ t = g χ t ∀ χ t . Ω Ω Ω Ω back A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 6 / 25
Variational formulations: proxy fields All those problems admits a variational formulations: if g = 0, we just wrote v = d ξ , ξ ∈ Λ 1 is the magnetic vector potential and verifies d ⋆ a − 1 d ξ = f which on the proxy field is: find ξ ∈ H ( curl , Ω), p ∈ H 1 (Ω) � � � a − 1 curl ξ · curl ξ t − ∇ p · ξ t = f · ξ t ∀ ξ t Ω Ω Ω � ξ · ∇ q = 0 ∀ q . Ω back A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 6 / 25
Compatible discretization Mimic the geometric structure at the discrete level The idea is to construct discretizations which mimic, or preserve the geometric structure... A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 7 / 25
Compatible discretization Mimic the geometric structure at the discrete level The idea is to construct discretizations which mimic, or preserve the geometric structure... Finite Elements Techniques Arnold et al, 2006 , Bossavit ’87-, Hiptmair ’99-, Dular 90s-, Demkowicz ’98- . . . , Christiansen 05-, B. 05- Mimetic Finite Differences Hyman-Shashkov ’90s, Brezzi et al. ’04-’05, B.’07 Finite Volumes, Finite Differences, Primal-Dual methods, Finite Integration Techniques Yee ’66, Bossavit-Kettunen ’99-, Tonti 95- Weiland ’77-, Rain ’02-’07, Bochev ’04- . . . , And for a pure “geometric” approach: Electromagnetic theory on lattices Teixeira-Chew, ’99 Discrete exterior calculus Leok et al. ’04- Applied differential geometry Bossavit, Kotiuga A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 7 / 25
From continuous to discrete The De Rham map Polyhedral subdivision of the domain: • ( C 0 , C 1 , C 2 , C 3 ), chain complex, denote the set of vertices , edges , faces and elements , all endowed with an orientation. • ( C 0 , C 1 , C 2 , C 3 ) cochain complex: functionals on C i , i = 0 , . . . 3 A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 8 / 25
From continuous to discrete The De Rham map Polyhedral subdivision of the domain: • ( C 0 , C 1 , C 2 , C 3 ), chain complex, denote the set of vertices , edges , faces and elements , all endowed with an orientation. • ( C 0 , C 1 , C 2 , C 3 ) cochain complex: functionals on C i , i = 0 , . . . 3 There is a standard way to go from Λ i to C i : the De Rham map Π i , i = 0 , . . . 3 �� � u i ∈ Λ i c i u i , c i ∈ C i Π i ( u i ) ∈ C i Π i ( u i ) = A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 8 / 25
From continuous to discrete The De Rham map Polyhedral subdivision of the domain: • ( C 0 , C 1 , C 2 , C 3 ), chain complex, denote the set of vertices , edges , faces and elements , all endowed with an orientation. • ( C 0 , C 1 , C 2 , C 3 ) cochain complex: functionals on C i , i = 0 , . . . 3 There is a standard way to go from Λ i to C i : the De Rham map Π i , i = 0 , . . . 3 �� � u 1 ∈ Λ 1 e u 1 , e ∈ C 1 Π 1 ( u 1 ) ∈ C 1 Π 1 ( u 1 ) = A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 8 / 25
From continuous to discrete The De Rham map Polyhedral subdivision of the domain: • ( C 0 , C 1 , C 2 , C 3 ), chain complex, denote the set of vertices , edges , faces and elements , all endowed with an orientation. • ( C 0 , C 1 , C 2 , C 3 ) cochain complex: functionals on C i , i = 0 , . . . 3 There is a standard way to go from Λ i to C i : the De Rham map Π i , i = 0 , . . . 3 �� � u 1 ∈ Λ 1 e u 1 , e ∈ C 1 Π 1 ( u 1 ) ∈ C 1 Π 1 ( u 1 ) = Approximation of Differential forms C i are chosen as approximations of differential forms Λ i . A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 8 / 25
How to use cochains as discretizations C 0 � nodal values of the unknown Λ 0 at vertices C 1 � circulations of the unknown Λ 1 along edges C 2 � fluxes of the unknown Λ 2 through faces C 3 � average of the unknown Λ 3 on elements Main requirements: Differential operators : we need to compute all first order differential operators; Scalar products : we need L 2 -like scalar products to implement our variational principles A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 9 / 25
Coboundary operators as differential operators Once the orientation of edges , faces and elements is fixed, we have: � � GRAD : C 0 → C 1 GRAD u e = u | V 1 − u | V 2 1 2 back A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 10 / 25
Coboundary operators as differential operators Once the orientation of edges , faces and elements is fixed, we have: � � GRAD : C 0 → C 1 GRAD u e = u | V 1 − u | V 2 � � � CURL : C 1 → C 2 CURL φ e ∈ f = δ ( e , f ) φ | e i =1 1 2 back A. Buffa (IMATI-CNR Italy) Discrete Differential Forms 10 / 25
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