Introduction to finite element exterior calculus
Transcript of Introduction to finite element exterior calculus
Introduction to finite element exterior calculus
Ragnar Winther
CMA, University of OsloNorway
Why finite element exterior calculus?
Recall the de Rham complex on the form:
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
and the corresponding elasticity complex
T →H1(Ω; R3)ε−→ H(J,Ω; S)
J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.
Here J is the second order operator
Jτ = curl(curl τ)T
mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.
Why finite element exterior calculus?
Recall the de Rham complex on the form:
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
and the corresponding elasticity complex
T →H1(Ω; R3)ε−→ H(J,Ω; S)
J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.
Here J is the second order operator
Jτ = curl(curl τ)T
mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.
Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.
Why finite element exterior calculus?
Recall the de Rham complex on the form:
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
and the corresponding elasticity complex
T →H1(Ω; R3)ε−→ H(J,Ω; S)
J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.
Here J is the second order operator
Jτ = curl(curl τ)T
mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.
Mapping of the domain
Consider a map φ : R3 → R3 mapping a domain Ω to Ω′.
Ω’Ωϕ
Desired commuting diagram:
R →H1(Ω′)grad−−→ H(curl,Ω′)
curl−−→ H(div,Ω′)div−−→ L2(Ω′) −→ 0yF 1
φ
yF cφ
yF dφ
yF 0φ
R → H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
Mapping of the domain
Consider a map φ : R3 → R3 mapping a domain Ω to Ω′.
Ω’Ωϕ
Desired commuting diagram:
R →H1(Ω′)grad−−→ H(curl,Ω′)
curl−−→ H(div,Ω′)div−−→ L2(Ω′) −→ 0yF 1
φ
yF cφ
yF dφ
yF 0φ
R → H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
Piola transforms
The proper “Piola transforms” are given by
(F 1φu)(x) = u(φ(x))
(F cφu)(x) = (Dφ(x))T u(φ(x)
(F dφ u)(x) = (det Dφ(x))(Dφ(x))−1u(φ(x))
(F 0φu)(x) = (det Dφ(x))u(φ(x))
Alternative de Rham complex
We replace the de Rham complex
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0
by an equivalent complex
R →HΛ0(Ω)d−→ HΛ1(Ω)
d−→ HΛ2(Ω)d−→ L2Λ3(Ω) −→ 0
In fact, the second complex can be defined for any dimension n.
Exterior calculus and differential forms
Instead of consider scalar fields and vector fields, i.e., functions ofthe form
u : Ω→ R or u : Ω→ R3
where Ω ⊂ R3, we consider instead functions of the form
u : Ω→ AltkRn
Here Ω is allowed to be a subset of Rn.
For any vector space V , the space AltkV is the space ofalternating k-linear maps on V
Exterior calculus and differential forms
Instead of consider scalar fields and vector fields, i.e., functions ofthe form
u : Ω→ R or u : Ω→ R3
where Ω ⊂ R3, we consider instead functions of the form
u : Ω→ AltkRn
Here Ω is allowed to be a subset of Rn.
For any vector space V , the space AltkV is the space ofalternating k-linear maps on V
The space AltkV
Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.
Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with
dim AltkV =
(n
k
).
Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.
The space AltkV
Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.
Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.
As a consequence AltkV is a vector space with
dim AltkV =
(n
k
).
Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.
The space AltkV
Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.
Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with
dim AltkV =
(n
k
).
Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.
The space AltkV
Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.
Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with
dim AltkV =
(n
k
).
Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.
Interior and exterior products
If ω ∈ AltkV and v ∈ V then the interior product, or thecontraction of ω by v , ωyv ∈ Altk−1V is given by
ωyv(v1, v2, . . . vk−1) = ω(v , v1, v2, . . . vk−1)
If ω ∈ AltjV and η ∈ AltkV then the exterior product, or wedge
product, ω ∧ η ∈ Altj+k is given by
(ω ∧ η)(v1, . . . , vj+k)
=∑σ
(signσ)ω(vσ(1), . . . , vσ(j))η(vσ(j+1), . . . , vσ(j+k)), vi ∈ V ,
where the sum is over all permutations σ of 1, 2, . . . , j + k suchthat σ(1) < ... < σ(j) and σ(j + 1) < ... < σ(j + k).
Interior and exterior products
If ω ∈ AltkV and v ∈ V then the interior product, or thecontraction of ω by v , ωyv ∈ Altk−1V is given by
ωyv(v1, v2, . . . vk−1) = ω(v , v1, v2, . . . vk−1)
If ω ∈ AltjV and η ∈ AltkV then the exterior product, or wedge
product, ω ∧ η ∈ Altj+k is given by
(ω ∧ η)(v1, . . . , vj+k)
=∑σ
(signσ)ω(vσ(1), . . . , vσ(j))η(vσ(j+1), . . . , vσ(j+k)), vi ∈ V ,
where the sum is over all permutations σ of 1, 2, . . . , j + k suchthat σ(1) < ... < σ(j) and σ(j + 1) < ... < σ(j + k).
Basis for AltkRn
We let dxi be the natural dual basis in Alt1Rn given by
dxi (ej) = δi ,j
A basis for Alt2Rn is given by
dxi ∧ dxj , i < j ,
while
dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing
is a basis for AltkRn.
Basis for AltkRn
We let dxi be the natural dual basis in Alt1Rn given by
dxi (ej) = δi ,j
A basis for Alt2Rn is given by
dxi ∧ dxj , i < j ,
while
dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing
is a basis for AltkRn.
Basis for AltkRn
We let dxi be the natural dual basis in Alt1Rn given by
dxi (ej) = δi ,j
A basis for Alt2Rn is given by
dxi ∧ dxj , i < j ,
while
dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing
is a basis for AltkRn.
Any ω ∈ AltkRn can be represented uniquely on the form∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
where the sum is over all incresing maps σ mapping 1, 2, . . . , kinto 1, 2, . . . , n, and uσ ∈ R.
Furthermore,
dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)(v1, . . . vk) = det(dxσ(i)vj)
Any ω ∈ AltkRn can be represented uniquely on the form∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
where the sum is over all incresing maps σ mapping 1, 2, . . . , kinto 1, 2, . . . , n, and uσ ∈ R.
Furthermore,
dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)(v1, . . . vk) = det(dxσ(i)vj)
Correspondences
Altk and vector fields:
Alt0 R3 = R c ↔ c
Alt1 R3 ∼=−→ R3 u1 dx1 + u2 dx2 + u3 dx3 ↔ u
Alt2 R3 ∼=−→ R3 u3 dx1 ∧ dx2 − u2 dx1 ∧ dx3 + u1 dx2 ∧ dx3 ↔ u
Alt3 R3 ∼=−→ R c ↔ c dx1 ∧ dx2 ∧ dx3
Exterior product:
∧ : Alt1 R3 × Alt1 R3 → Alt2 R3 × : R3 × R3 → R3
∧ : Alt1 R3 × Alt2 R3 → Alt3 R3 · : R3 × R3 → R
Correspondences
Altk and vector fields:
Alt0 R3 = R c ↔ c
Alt1 R3 ∼=−→ R3 u1 dx1 + u2 dx2 + u3 dx3 ↔ u
Alt2 R3 ∼=−→ R3 u3 dx1 ∧ dx2 − u2 dx1 ∧ dx3 + u1 dx2 ∧ dx3 ↔ u
Alt3 R3 ∼=−→ R c ↔ c dx1 ∧ dx2 ∧ dx3
Exterior product:
∧ : Alt1 R3 × Alt1 R3 → Alt2 R3 × : R3 × R3 → R3
∧ : Alt1 R3 × Alt2 R3 → Alt3 R3 · : R3 × R3 → R
Differential forms
Differential forms are functions of the form
ω : Ω→ Altk ≡ Altk Rn.
We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.
Differential forms
Differential forms are functions of the form
ω : Ω→ Altk ≡ Altk Rn.
We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.
Differential forms
Differential forms are functions of the form
ω : Ω→ Altk ≡ Altk Rn.
We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.
Check of d d = 0
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Start with a zero form ω.
dωx(v) = ∂vωx
(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.
Check of d d = 0
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Start with a zero form ω.
dωx(v) = ∂vωx
(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.
Check of d d = 0
dωx(v1, v2, . . . , vk+1) =k+1∑j=1
(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),
ω ∈ Λk , v1, . . . , vk+1 ∈ Rn
Start with a zero form ω.
dωx(v) = ∂vωx
(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.
Representation by a basis
If ω is given on the form
ωx =∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k),
where each uσ is a scalar field on Ω then
dωx =∑σ
n∑j=1
∂uσ∂xj
dxj ∧ dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
The de Rham complex and differential forms
Using differential forms the de Rham complex (smooth version)can be written as
R →Λ0 d−→ Λ1 d−→ · · · d−→ Λn −→ 0.
Mapping of the domain
Ω’Ωϕ
The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by
(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),
where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.This expression should be compared with the expressions we hadfor the Piola maps before.
Mapping of the domain
Ω’Ωϕ
The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by
(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),
where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.
This expression should be compared with the expressions we hadfor the Piola maps before.
Mapping of the domain
Ω’Ωϕ
The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by
(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),
where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.This expression should be compared with the expressions we hadfor the Piola maps before.
Commuting diagram
R →Λ0(Ω′)d−→ Λ1(Ω′)
d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)
d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.
So the pullback commutes with the exterior derivative, i.e.,
φ∗(dω) = d(φ∗ω), ω ∈ Λk(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 amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗
Commuting diagram
R →Λ0(Ω′)d−→ Λ1(Ω′)
d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)
d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.
So the pullback commutes with the exterior derivative, i.e.,
φ∗(dω) = d(φ∗ω), ω ∈ Λk(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 amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗
Commuting diagram
R →Λ0(Ω′)d−→ Λ1(Ω′)
d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)
d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.
So the pullback commutes with the exterior derivative, i.e.,
φ∗(dω) = d(φ∗ω), ω ∈ Λk(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 amanifold.
Key identity: (ψ φ)∗ = φ∗ ψ∗
Commuting diagram
R →Λ0(Ω′)d−→ Λ1(Ω′)
d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)
d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.
So the pullback commutes with the exterior derivative, i.e.,
φ∗(dω) = d(φ∗ω), ω ∈ Λk(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 amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗
Exterior derivative and wedge product
By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form
d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .
The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.
Since Tr is a pull back we will have
Tr d = d∂Ω Tr
Exterior derivative and wedge product
By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form
d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .
The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.
Since Tr is a pull back we will have
Tr d = d∂Ω Tr
Exterior derivative and wedge product
By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form
d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .
The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.
Since Tr is a pull back we will have
Tr d = d∂Ω Tr
Integration of differential forms
If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation
ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,
then we can define ∫Ωω =
∫Ω
u dx
If φ : Ω→ Ω′ then we have∫Ωφ∗ω =
∫Ω′ω
Note that this gives us a way to define the integral of an n-form ona manifold.
Integration of differential forms
If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation
ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,
then we can define ∫Ωω =
∫Ω
u dx
If φ : Ω→ Ω′ then we have∫Ωφ∗ω =
∫Ω′ω
Note that this gives us a way to define the integral of an n-form ona manifold.
Integration of differential forms
If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation
ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,
then we can define ∫Ωω =
∫Ω
u dx
If φ : Ω→ Ω′ then we have∫Ωφ∗ω =
∫Ω′ω
Note that this gives us a way to define the integral of an n-form ona manifold.
Stokes theorem
With the three concepts exterior derivative, trace operator, andintegral we can formulate Stokes theorem as∫
Ωdω =
∫∂Ω
Trω, ω ∈ Λn−1(Ω)
Furthermore, by combining this with the Leibniz rule we obtain thefollowing integration by parts identity∫
Ωdω ∧ η = (−1)k−1
∫Ωω ∧ dη +
∫∂Ω
Trω ∧ Tr η
where ω ∈ Λk , η ∈ Λn−k−1.
Stokes theorem
With the three concepts exterior derivative, trace operator, andintegral we can formulate Stokes theorem as∫
Ωdω =
∫∂Ω
Trω, ω ∈ Λn−1(Ω)
Furthermore, by combining this with the Leibniz rule we obtain thefollowing integration by parts identity∫
Ωdω ∧ η = (−1)k−1
∫Ωω ∧ dη +
∫∂Ω
Trω ∧ Tr η
where ω ∈ Λk , η ∈ Λn−k−1.
L2 differential forms
If ω is a k form given by
ω =∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.
Furthernmore, we let
HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)
We then ontain the L2 de Rham complex:
0→ HΛ0(Ω)d−→ HΛ1(Ω)
d−→ · · · d−→ HΛn(Ω)→ 0
L2 differential forms
If ω is a k form given by
ω =∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.
Furthernmore, we let
HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)
We then ontain the L2 de Rham complex:
0→ HΛ0(Ω)d−→ HΛ1(Ω)
d−→ · · · d−→ HΛn(Ω)→ 0
L2 differential forms
If ω is a k form given by
ω =∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.
Furthernmore, we let
HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)
We then ontain the L2 de Rham complex:
0→ HΛ0(Ω)d−→ HΛ1(Ω)
d−→ · · · d−→ HΛn(Ω)→ 0
Polynomial de Rham complex
We define
Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk
Alternatively, Pr Λk consists of all∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
where the coefficients uσ are in Pr .The polynomial de Rham complex
0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0
is exact.
Polynomial de Rham complex
We define
Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk
Alternatively, Pr Λk consists of all∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
where the coefficients uσ are in Pr .
The polynomial de Rham complex
0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0
is exact.
Polynomial de Rham complex
We define
Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk
Alternatively, Pr Λk consists of all∑σ
uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)
where the coefficients uσ are in Pr .The polynomial de Rham complex
0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0
is exact.
The Kozul complex
The Koszul differential κ of a k-form ω is the (k − 1)-form given by
(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)
= ωx
(x , v1, . . . , vk−1
),
For each r , κ maps Pr−1Λk to Pr Λk−1, and the Koszul complex
0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,
is exact.
The Kozul complex
The Koszul differential κ of a k-form ω is the (k − 1)-form given by
(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)
= ωx
(x , v1, . . . , vk−1
),
For each r , κ maps Pr−1Λk to Pr Λk−1,
and the Koszul complex
0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,
is exact.
The Kozul complex
The Koszul differential κ of a k-form ω is the (k − 1)-form given by
(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)
= ωx
(x , v1, . . . , vk−1
),
For each r , κ maps Pr−1Λk to Pr Λk−1, and the Koszul complex
0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,
is exact.
The spaces P−r Λk
The operators d and κ are related by the homotopy relation
(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,
where Hr denotes the homogeneous polynomials of degree r .
As a consequence we obtain the identity
Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1
We then define the space P−r Λk by
P−r Λk = Pr−1Λk + κPr−1Λk+1.
We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,
0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0
is an exact complex, and the space P−r Λk is affine invariant.
The spaces P−r Λk
The operators d and κ are related by the homotopy relation
(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,
where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity
Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1
We then define the space P−r Λk by
P−r Λk = Pr−1Λk + κPr−1Λk+1.
We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,
0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0
is an exact complex, and the space P−r Λk is affine invariant.
The spaces P−r Λk
The operators d and κ are related by the homotopy relation
(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,
where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity
Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1
We then define the space P−r Λk by
P−r Λk = Pr−1Λk + κPr−1Λk+1.
We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,
0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0
is an exact complex, and the space P−r Λk is affine invariant.
The spaces P−r Λk
The operators d and κ are related by the homotopy relation
(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,
where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity
Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1
We then define the space P−r Λk by
P−r Λk = Pr−1Λk + κPr−1Λk+1.
We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,
0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0
is an exact complex, and the space P−r Λk is affine invariant.
Characterization of affine invariant polynomial spaces
TheoremAssume that the polynomial space QΛk , satisfiesPr−1Λk ( QΛk ( Pr Λk . Then
QΛk = P−r Λk or QΛk = Pr−1Λk + dPr+1Λk−1
The four exact sequences ending with Pr Λ3(T ) in 3D
0→ Pr+1Λ0 d−→ P−r+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+2Λ0 d−→ Pr+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+2Λ0 d−→ P−r+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+3Λ0 d−→ Pr+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0
In general there are 2n−1 exact sequences composed of the familiesof P and P− spaces.
The four exact sequences ending with Pr Λ3(T ) in 3D
0→ Pr+1Λ0 d−→ P−r+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+2Λ0 d−→ Pr+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+2Λ0 d−→ P−r+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0
0→ Pr+3Λ0 d−→ Pr+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0
In general there are 2n−1 exact sequences composed of the familiesof P and P− spaces.
The four sequences ending with P0Λ3(T ) in 3D
0→ grad−−→ curl−−→ div−−→ → 0
0→ grad−−→ curl−−→ div−−→ → 0
0→ grad−−→ curl−−→ div−−→ → 0
0→ grad−−→ curl−−→ div−−→ → 0
Piecewise smooth differential formsIt is a consequence of Stokes theorem that a piecewise smoothk–form ω, with respect to a simplicial mesh Th of Ω, is in HΛk(Ω)if and only if the trace of ω, Trω, is continuous on the interfaces.
Here Trω is defined by restricting thespatial variable x to the interface, andby applying ω only to tangent vectors ofthe interface.
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
(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces
associated to subsimplices f of T .
DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate
ω 7→∫
fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )
DOF for P−r Λk(T ):
ω 7→∫
fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )
Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They
are subspaces of HΛk(Ω).
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
(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces
associated to subsimplices f of T .
DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate
ω 7→∫
fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )
DOF for P−r Λk(T ):
ω 7→∫
fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )
Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They
are subspaces of HΛk(Ω).
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
(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces
associated to subsimplices f of T .
DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate
ω 7→∫
fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )
DOF for P−r Λk(T ):
ω 7→∫
fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )
Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They
are subspaces of HΛk(Ω).
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
(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces
associated to subsimplices f of T .
DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate
ω 7→∫
fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )
DOF for P−r Λk(T ):
ω 7→∫
fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )
Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They
are subspaces of HΛk(Ω).
De Rham cohomology
0 → Λ0(Ω)d−→ Λ1(Ω)
d−→ Λ2(Ω) → 0
0 → C∞(Ω)grad−−→ C∞(Ω; R2)
rot−→ C∞(Ω) → 0
Cohomology
The de Rham complex
HΛk−1(Ω)dk−1
−−−→ HΛk(Ω)dk
−→ HΛk+1(Ω)
is called exact if for all k,
Zk := ker dk = range dk−1 =: Bk .
In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat
〈q, µ〉 = 0 µ ∈ Bk .
This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .
Cohomology
The de Rham complex
HΛk−1(Ω)dk−1
−−−→ HΛk(Ω)dk
−→ HΛk+1(Ω)
is called exact if for all k,
Zk := ker dk = range dk−1 =: Bk .
In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.
The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat
〈q, µ〉 = 0 µ ∈ Bk .
This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .
Cohomology
The de Rham complex
HΛk−1(Ω)dk−1
−−−→ HΛk(Ω)dk
−→ HΛk+1(Ω)
is called exact if for all k,
Zk := ker dk = range dk−1 =: Bk .
In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat
〈q, µ〉 = 0 µ ∈ Bk .
This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .
Cohomology
The de Rham complex
HΛk−1(Ω)dk−1
−−−→ HΛk(Ω)dk
−→ HΛk+1(Ω)
is called exact if for all k,
Zk := ker dk = range dk−1 =: Bk .
In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat
〈q, µ〉 = 0 µ ∈ Bk .
This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .
Hodge Laplace problem
Λk−1(Ω)dk−1
−−−→ Λk(Ω)dk
−→ Λk+1(Ω)
Formally: Given f ∈ Λk , find u ∈ Λk such that
(dk−1δk−1 + δkdk)u = f .
Here δk is a formal adjoint of dk .
The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that
〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1
〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk
〈u, q〉 = 0 ∀q ∈ Hk
Hodge Laplace problem
Λk−1(Ω)dk−1
−−−→ Λk(Ω)dk
−→ Λk+1(Ω)
Formally: Given f ∈ Λk , find u ∈ Λk such that
(dk−1δk−1 + δkdk)u = f .
Here δk is a formal adjoint of dk .
The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that
〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1
〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk
〈u, q〉 = 0 ∀q ∈ Hk
Hodge Laplace problem
Λk−1(Ω)dk−1
−−−→ Λk(Ω)dk
−→ Λk+1(Ω)
Formally: Given f ∈ Λk , find u ∈ Λk such that
(dk−1δk−1 + δkdk)u = f .
Here δk is a formal adjoint of dk .
The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that
〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1
〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk
〈u, q〉 = 0 ∀q ∈ Hk
Hodge LaplacianWell-posedness of the Hodge Laplace problem follows from theHodge decomposition and Poincare’s inequality:
‖ω‖L2 ≤ c ‖dω‖L2 , ω ∈ (Zk)⊥.
Special cases:Recall that if dim Hk = 0: Find σ ∈ HΛk−1, and u ∈ HΛk
〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1
〈dσ, v〉+ 〈du, dv〉 = 〈f , v〉 ∀v ∈ HΛk
I k = 0: ordinary LaplacianI k = n: mixed LaplacianI k = 1, n = 3: σ = − div u, gradσ + curl curl u = fI k = 2, n = 3: σ = curl u, curl σ − grad div u = f ,
Hodge LaplacianWell-posedness of the Hodge Laplace problem follows from theHodge decomposition and Poincare’s inequality:
‖ω‖L2 ≤ c ‖dω‖L2 , ω ∈ (Zk)⊥.
Special cases:Recall that if dim Hk = 0: Find σ ∈ HΛk−1, and u ∈ HΛk
〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1
〈dσ, v〉+ 〈du, dv〉 = 〈f , v〉 ∀v ∈ HΛk
I k = 0: ordinary LaplacianI k = n: mixed LaplacianI k = 1, n = 3: σ = − div u, gradσ + curl curl u = fI k = 2, n = 3: σ = curl u, curl σ − grad div u = f ,
Discretizations of the de Rham complex in threedimensions
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0yI1
h
yIch
yIdh
yI0h
R → H1h
grad−−→ Hh(curl)curl−−→ Hh(div)
div−−→ L2h −→ 0.
The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with grad curl anddiv, but the are not bounded on the spaces H1, H(curl) or H(div).
Discretizations of the de Rham complex in threedimensions
R →H1(Ω)grad−−→ H(curl,Ω)
curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0yI1
h
yIch
yIdh
yI0h
R → H1h
grad−−→ Hh(curl)curl−−→ Hh(div)
div−−→ L2h −→ 0.
The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with grad curl anddiv, but the are not bounded on the spaces H1, H(curl) or H(div).
Discretizations of the de Rham complex in n dimensions
R →HΛ0(Ω)d−→ HΛ1(Ω)
d−→ · · · d−→ HΛn(Ω) −→ 0.yIh yIh yIhR → Λ0
hd−→ Λ1
hd−→ · · · d−→ Λn
h −→ 0.
The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with d , but the arenot bounded on the HΛ.
Discretizations of the de Rham complex in n dimensions
R →HΛ0(Ω)d−→ HΛ1(Ω)
d−→ · · · d−→ HΛn(Ω) −→ 0.yIh yIh yIhR → Λ0
hd−→ Λ1
hd−→ · · · d−→ Λn
h −→ 0.
The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with d , but the arenot bounded on the HΛ.
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·
x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·
x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Discretization, Abstract setting
· · · −→ Λk−1 dk−1
−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1
hdk−1
−−→ Λkh −→ · · ·
Complex of Hilbert spaces with dk bounded and closed range.
Hodge decomposition and Poincare inequality follow.
For discretization, construct a finite dimensional subcomplex.
Define Hkh = (Bk
h)⊥ ∩ Zkh .
Discrete Hodge decomposition follows: Λkh = Bk
h ⊕ Hkh ⊕ (Zk
h)⊥
Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk
h , Hkh
When is it stable?
Bounded cochain projections
Key property: Suppose that there exists a bounded cochainprojection.
· · · −→ Λk−1 dk−1
−−−→ Λk −→ · · ·yπk−1h
yπkh
· · · −→ Λk−1h
dk−1
−−−→ Λkh −→ · · ·
I πkh uniformly bounded
I ‖πkhω − ω‖ → 0.
I πkh a projection
I πkh dk−1 = dk−1πk−1
h
Theorem
I If ‖v − πkhv‖ < ‖v‖ ∀v ∈ Hk , then the induced map on
cohomology is an isomorphism.
I gap(Hk ,Hk
h
)≤ sup
v∈Hk ,‖v‖=1
‖v − πkhv‖
I The discrete Poincare inequality holds uniformly in h.
I Galerkin’s method is stable and convergent.
Bounded cochain projections
Key property: Suppose that there exists a bounded cochainprojection.
· · · −→ Λk−1 dk−1
−−−→ Λk −→ · · ·yπk−1h
yπkh
· · · −→ Λk−1h
dk−1
−−−→ Λkh −→ · · ·
I πkh uniformly bounded
I ‖πkhω − ω‖ → 0.
I πkh a projection
I πkh dk−1 = dk−1πk−1
h
Theorem
I If ‖v − πkhv‖ < ‖v‖ ∀v ∈ Hk , then the induced map on
cohomology is an isomorphism.
I gap(Hk ,Hk
h
)≤ sup
v∈Hk ,‖v‖=1
‖v − πkhv‖
I The discrete Poincare inequality holds uniformly in h.
I Galerkin’s method is stable and convergent.
Proof of discrete Poincare inequality
Theorem: There is a positive constant c , independent of h, suchthat
‖ω‖ ≤ c‖dω‖, ω ∈ Zk⊥h .
In fact, c ≤ cPcπ, where cP is continuous Poincare constant, andcπ bounds ‖πh‖.
Proof: Given ω ∈ Zk⊥h , define η ∈ Zk⊥ ⊂ HΛk(Ω) by dη = dω. By
the Poincare inequality, ‖η‖ ≤ c‖dω‖, so it is enough to show that‖ω‖ ≤ c‖η‖. Now, ω − πhη ∈ Λk
h and d(ω − πhη) = 0, soω − πhη ∈ Zk
h . Therefore
‖ω‖2 = 〈ω, πhη〉+ 〈ω, ω − πhη〉 = 〈ω, πhη〉 ≤ ‖ω‖‖πhη‖,
whence ‖ω‖ ≤ ‖πhη‖. The result follows from the uniformboundedness of πh.
Proof of discrete Poincare inequality
Theorem: There is a positive constant c , independent of h, suchthat
‖ω‖ ≤ c‖dω‖, ω ∈ Zk⊥h .
In fact, c ≤ cPcπ, where cP is continuous Poincare constant, andcπ bounds ‖πh‖.Proof: Given ω ∈ Zk⊥
h , define η ∈ Zk⊥ ⊂ HΛk(Ω) by dη = dω. Bythe Poincare inequality, ‖η‖ ≤ c‖dω‖, so it is enough to show that‖ω‖ ≤ c‖η‖. Now, ω − πhη ∈ Λk
h and d(ω − πhη) = 0, soω − πhη ∈ Zk
h . Therefore
‖ω‖2 = 〈ω, πhη〉+ 〈ω, ω − πhη〉 = 〈ω, πhη〉 ≤ ‖ω‖‖πhη‖,
whence ‖ω‖ ≤ ‖πhη‖. The result follows from the uniformboundedness of πh.
Construction of bounded cochain projections
The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .
If we apply the three operations:
I extend (E )
I regularize (R)
I canonical projection (Ih)
we get a map Qkh : HΛk(Ω)→ Λk
h which is bounded and commuteswith d . But it is not a projection.
However the composition
πkh = (Qk
h |Λkh)−1 Qk
h
can be shown to be a bounded cochain projection.
Construction of bounded cochain projections
The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .
If we apply the three operations:
I extend (E )
I regularize (R)
I canonical projection (Ih)
we get a map Qkh : HΛk(Ω)→ Λk
h which is bounded and commuteswith d . But it is not a projection.
However the composition
πkh = (Qk
h |Λkh)−1 Qk
h
can be shown to be a bounded cochain projection.
Construction of bounded cochain projections
The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .
If we apply the three operations:
I extend (E )
I regularize (R)
I canonical projection (Ih)
we get a map Qkh : HΛk(Ω)→ Λk
h which is bounded and commuteswith d . But it is not a projection.
However the composition
πkh = (Qk
h |Λkh)−1 Qk
h
can be shown to be a bounded cochain projection.
Bounded projections, definitionsWe combine extension and smoothing into an operatorRε
h : L2Λk(Ω)→ C Λk(Ω) given by
(Rεhω)x =
∫B1
ρ(y)(Eω)x+εhy dy .
Assume that the mesh is quasi–uniform:
Lemma
I For each ε > 0 sufficiently small there is a constant c(ε) suchthat for all h
‖IhRεh‖L(L2Λk (Ω),L2Λk (Ω)) ≤ c(ε).
I There is constant c, independent of ε and h such that
‖I − IhRεh‖L(L2Λk
h ,L2Λk
h) ≤ cε.
Bounded projections, definitionsWe combine extension and smoothing into an operatorRε
h : L2Λk(Ω)→ C Λk(Ω) given by
(Rεhω)x =
∫B1
ρ(y)(Eω)x+εhy dy .
Assume that the mesh is quasi–uniform:
Lemma
I For each ε > 0 sufficiently small there is a constant c(ε) suchthat for all h
‖IhRεh‖L(L2Λk (Ω),L2Λk (Ω)) ≤ c(ε).
I There is constant c, independent of ε and h such that
‖I − IhRεh‖L(L2Λk
h ,L2Λk
h) ≤ cε.
Scaling and compactness
Bounded cochain projections
The operator πh = πkh defined by
πh = (IhRεh|Λk
h)−1 IhRε
h,
where ε is taken small, but not too small, has all the desiredproperties, i.e.
I πkh is bounded in L(L2Λk(Ω), L2Λk(Ω)) andL(HΛk(Ω),HΛk(Ω))
I πkh commutes with the exterior derivative
I πkh is a projection
Removing the assumption of a quasi–uniform mesh
If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.
For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy
h (x) = x + εgh(x)y .
Then we replace the operator
(Rεhω)x =
∫B1
ρ(y)(Eω)x+εhy dy
by
(Rεhω)x =
∫B1
ρ(y)((Φεyh )∗Eω)x dy .
Removing the assumption of a quasi–uniform mesh
If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.
For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy
h (x) = x + εgh(x)y .
Then we replace the operator
(Rεhω)x =
∫B1
ρ(y)(Eω)x+εhy dy
by
(Rεhω)x =
∫B1
ρ(y)((Φεyh )∗Eω)x dy .
Removing the assumption of a quasi–uniform mesh
If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.
For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy
h (x) = x + εgh(x)y .
Then we replace the operator
(Rεhω)x =
∫B1
ρ(y)(Eω)x+εhy dy
by
(Rεhω)x =
∫B1
ρ(y)((Φεyh )∗Eω)x dy .
References
The development of FEEC leans heavily on earlier results takenfrom
I 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
I D.N. Arnold, R.S. Falk, R. Winther, Finite element exteriorcalculus, homological techniques, and applications, ActaNumerica 2006.
I D.N. Arnold, R.S. Falk and R. Winther Finite element exteriorcalculus: From Hodge theory to numerical stability Bulletin ofthe Amer. Math. Soc. 47 (2010), 281-354.
References
The development of FEEC leans heavily on earlier results takenfrom
I 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
I D.N. Arnold, R.S. Falk, R. Winther, Finite element exteriorcalculus, homological techniques, and applications, ActaNumerica 2006.
I D.N. Arnold, R.S. Falk and R. Winther Finite element exteriorcalculus: From Hodge theory to numerical stability Bulletin ofthe Amer. Math. Soc. 47 (2010), 281-354.