Application of Finite Element Exterior Calculus to Elasticityarnold/dgcd-talks/Falk.pdf ·...

Application of Finite Element Exterior Calculusto Elasticity

Richard S. Falk

Department of MathematicsRutgers University

July 19, 2007

Joint work with:Douglas Arnold, IMA, University of Minnesota

Ragnar Winther, Centre of Mathematics for Applications,University of Oslo, Norway

Richard S. Falk Application of Finite Element Exterior Calculus to Elasticity

Outline Of Talk

I Variational formulations of the equations of linear elasticity

I Stability of discretizations of saddle-point problems

I Connections to exact sequences – continuous and discrete

I Exact sequences for elasticity

I From de Rham to elasticity

I Stability of continuous formulation of elasticity with weaklyimposed symmetry

I New finite element methods for the equations of elasticityfrom connections to de Rham


Equations of linear elasticity

For N = 2, 3, equations of linear elasticity can be written as systemof equations of form

Aσ = εu, div σ = f in Ω ⊂ RN .

Here• stressfield σ(x) ∈ S (symmetric matrices).• displacement field u(x) ∈ RN .• f = f (x) is given body force.• ε(u) is symmetric gradient of u.• div of matrix field taken row-wise.• If body clamped on boundary ∂Ω of Ω, proper boundarycondition is u = 0 on ∂Ω.• A = A(x) : S 7→ S is given, uniformly positive definite,compliance tensor (material dependent).


Weak formulations

Strongly imposed symmetry:

Find (σ, u) ∈ H(div,Ω, S)× L2(Ω, RN) such that:

(Aσ, τ) + (div τ, u) = 0, τ ∈ H(div,Ω, S),

(div σ, v) = (f , v) v ∈ L2(Ω, RN).

Weakly imposed symmetry:

Find (σ, u, p) ∈ H(div,Ω, M)× L2(Ω, RN)× L2(Ω, K) such that:

(Aσ, τ) + (div τ, u) + (τ, p) = 0, τ ∈ H(div,Ω, M),

(div σ, v) = (f , v), v ∈ L2(Ω, RN),

(σ, q) = 0, q ∈ L2(Ω, K).

M = N × N matrices, K = skew symmetric matrices.


Stable discretization of saddle point problems

Consider mixed formulation of Poisson’s equation ∆p = f .

(u, v) +(p,div v) = 0 ∀v ∈ H(div,Ω, RN),(div u, q) = (f , q) ∀q ∈ L2(Ω, R).

Sufficient conditions for stability of discretization:

Vh × Qh ⊂ H(div,Ω, RN)× L2(Ω, R)

Condition(A1) : div Vh ⊂ Qh

Condition (A2): There exist projections Πdh and Π0

h , boundeduniformly with respect to h (in suitable norms), and satisfyingcommuting diagram:

C∞(RN)div−−→ C∞(R)yΠd

h

yΠ0h

Vhdiv−−→ Qh


Commuting diagrams from de Rham

Starting from de Rham sequence, can construct finite elementspaces and interpolation operators with full commuting diagrams.

R ⊂−−→ C∞(R)grad−−−→ C∞(R3)

curl−−→ C∞(R3)div−−→ C∞(R) −−→ 0yid

yΠ1h

yΠch

yΠdh

yΠ0h

R ⊂−−→ Shgrad−−−→ Zh

curl−−→ Vhdiv−−→ Qh −−→ 0

R ⊂−−→ C∞(R)curl−−→ C∞(R2)

div−−→ C∞(R) −−→ 0yid

yΠ1h

yΠdh

yΠ0h

R ⊂−−→ Shcurl−−→ Vh

div−−→ Qh −−→ 0

If domain contractible, all sequences exact.

We will use these spaces to approximate elasticity problem.


Commuting diagrams for elasticity with strong symmetry

Find (σ, u) ∈ H(div,Ω, S)× L2(Ω, RN) such that

(Aσ, τ) +(u,div τ) = 0 ∀τ ∈ H(div,Ω, S),(div σ, v) = (f , v) ∀v ∈ L2(Ω, RN).

Corresponding exact sequences in this case:

RM⊂−→ C∞(R3)

ε−→ C∞(S)J−→ C∞(S)

div−−→ C∞(R3) → 0,

in 3-D, where Jσ = curl(curlσ)T .

P1⊂−→ C∞(R)

J−→ C∞(S)div−−→ C∞(R2) → 0

in 2-D, where

Jq =

(∂2q/∂y2 −∂2q/∂x∂y

−∂2q/∂x∂y ∂2q/∂x2

).






RM⊂−→ C∞(R3)

ε−→ C∞(S)J−→ C∞(S)

div−−→ C∞(R3) → 0,


P1⊂−→ C∞(R)

J−→ C∞(S)div−−→ C∞(R2) → 0

in 2-D, where

Jq =

(∂2q/∂y2 −∂2q/∂x∂y

−∂2q/∂x∂y ∂2q/∂x2

).






RM⊂−→ C∞(R3)

ε−→ C∞(S)J−→ C∞(S)

div−−→ C∞(R3) → 0,


P1⊂−→ C∞(R)

J−→ C∞(S)div−−→ C∞(R2) → 0

in 2-D, where

Jq =

(∂2q/∂y2 −∂2q/∂x∂y

−∂2q/∂x∂y ∂2q/∂x2

).


Approximation in 2D

In 2002, Arnold-Winther constructed commuting diagrams of form:

P1⊂−−→ C∞(R)

J−−→ C∞(S)div−−→ C∞(R2) −−→ 0yid

yΠ2h

yΠdh

yΠ0h

P1⊂−−→ Qh

J−−→ Σhdiv−−→ Vh −−→ 0

Stress space Σh = p. cubic functions with p. linear divergence.Displacement space Vh = piecewise linear functionsQh = Argyris space of C 1 quintics.

Proved stability for corresponding mixed finite element method forelasticity. Elements are complicated.

Also new elements in 3-D, but elements even more complicated(stress field in lowest order case – 162 dof).


Elasticity with weakly imposed symmetry

In 2-D, setting W = K× R2, relevant exact sequence:

P1⊂−→ C∞(R)

J−→ C∞(M)(skwdiv)−−−→ C∞(W) → 0.

In 3-D, with W = K× R3, relevant exact sequence:

· · · −→ C∞(M)J−→ C∞(M)

(skwdiv)−−−→ C∞(W) → 0.

Here J : C∞(M) 7→ C∞(M) denotes extension of previousoperator.

Jτ = curlS−1 curl τ, S algebraic


Elasticity with weakly imposed symmetry

In 2-D, setting W = K× R2, relevant exact sequence:

P1⊂−→ C∞(R)

J−→ C∞(M)(skwdiv)−−−→ C∞(W) → 0.

In 3-D, with W = K× R3, relevant exact sequence:

· · · −→ C∞(M)J−→ C∞(M)

(skwdiv)−−−→ C∞(W) → 0.

Here J : C∞(M) 7→ C∞(M) denotes extension of previousoperator.

Jτ = curlS−1 curl τ, S algebraic


New approach to discretization of elasticity sequences:

I Use procedure on continuous level to derive elasticitysequence from multiple copies of de Rham sequence.

I Use this connection to establish stability for continuousformulation of elasticity

I To discretize, start from known good discretizations of deRham sequence.

I Determine conditions so that an analogue of stability proof forcontinuous problem will give stability of discrete problem.

To see structure more clearly, adopt notation of differential forms.

For simplicity, mostly consider 2D examples.


de Rham sequences with values in a vector space

Write 2-D de Rham sequence in form:

R ⊂−→ Λ0 d0−→ Λ1 d1−→ Λ2 → 0.

Also consider sequences whose values lie in either V = Rn or K,space of skew-symmetric matrices. Both corresponding de Rhamsequences also exact, e.g.,

V ⊂−→ Λ0(V)d0−→ Λ1(V)

d1−→ Λ2(V) → 0.

Here Λk(V) consists of elements of form:

ω(x) =∑

I

fI (x)dxI

with coefficients fI ∈ C∞(Ω, V).


de Rham sequences with values in a vector space

Write 2-D de Rham sequence in form:

R ⊂−→ Λ0 d0−→ Λ1 d1−→ Λ2 → 0.

Also consider sequences whose values lie in either V = Rn or K,space of skew-symmetric matrices. Both corresponding de Rhamsequences also exact, e.g.,

V ⊂−→ Λ0(V)d0−→ Λ1(V)

d1−→ Λ2(V) → 0.

Here Λk(V) consists of elements of form:

ω(x) =∑

I

fI (x)dxI

with coefficients fI ∈ C∞(Ω, V).


Elasticity sequence from de Rham sequence

Following ideas of Eastwood: Start from two de Rham sequences:

· · · −→ Λn−2(K)dn−2−−−→ Λn−1(K)

dn−1−−−→ Λn(K) → 0,

· · · −→ Λn−2(V)dn−2−−−→ Λn−1(V)

dn−1−−−→ Λn(V) → 0.

Let X = (x1, . . . , xn)T and define Kk : Λk(Ω; V) → Λk(Ω; K) by

Kkω = XωT − ωXT .

Then define

Sk = dkKk − Kk+1dk : Λk(Ω; V) → Λk+1(Ω; K).



Following ideas of Eastwood: Start from two de Rham sequences:

· · · −→ Λn−2(K)dn−2−−−→ Λn−1(K)

dn−1−−−→ Λn(K) → 0,

· · · −→ Λn−2(V)dn−2−−−→ Λn−1(V)

dn−1−−−→ Λn(V) → 0.

Let X = (x1, . . . , xn)T and define Kk : Λk(Ω; V) → Λk(Ω; K) by

Kkω = XωT − ωXT .

Then define

Sk = dkKk − Kk+1dk : Λk(Ω; V) → Λk+1(Ω; K).


The operator Sk

Can show: Sk is an algebraic operator and:

Key property used to establish stability: dk+1Sk = −Sk+1dk .Two important operators: Sn−2 and Sn−1.

In both 2 and 3 dimensions, spaces Λn−1(Ω; V) are spaces ofstresses and can be identified with n × n matrices.

Operator Sn−1 can be identified with skw, i.e., taking skew part ofmatrix (i.e., (W −W T )/2).


The operator Sn−2

For n = 2, if ω = (ω1, ω2)T , then letting χ =

(0 −11 0

),

S0ω =

(0 ω2

−ω2 0

)dx1 +

(0 −ω1

ω1 0

)dx2 = −ω2χdx1 + ω1χdx2.

Note that S0 is invertible, i.e.,

S−10 [µ1χdx1 + µ2χdx2] = (µ2,−µ1)

T .

For n = 3, S1 more complicated, but still algebraic and invertible.



Picture is:

· · · −→ Λn−2(K)dn−2−−−→ Λn−1(K)

dn−1−−−→ Λn(K) → 0

Sn−2 Sn−1

· · · −→ Λn−2(V)dn−2−−−→ Λn−1(V)

dn−1−−−→ Λn(V) → 0.

Since Sn−2 invertible, combine to one sequence: Let W = K× V.

· · · −→ Λn−2(K)dn−2S−1

n−2dn−2−−−−−−−−−−→ Λn−1(V)

(Sn−1dn−1

)−−−−→ Λn(W) → 0

After proper identifications, (n = 2), this is elasticity sequence

C∞(R)J−→ C∞(M)

(skwdiv)−−−→ C∞(W) → 0.



Picture is:

· · · −→ Λn−2(K)dn−2−−−→ Λn−1(K)

dn−1−−−→ Λn(K) → 0

Sn−2 Sn−1

· · · −→ Λn−2(V)dn−2−−−→ Λn−1(V)

dn−1−−−→ Λn(V) → 0.


· · · −→ Λn−2(K)dn−2S−1

n−2dn−2−−−−−−−−−−→ Λn−1(V)

(Sn−1dn−1

)−−−−→ Λn(W) → 0


C∞(R)J−→ C∞(M)

(skwdiv)−−−→ C∞(W) → 0.



Picture is:

· · · −→ Λn−2(K)dn−2−−−→ Λn−1(K)

dn−1−−−→ Λn(K) → 0

Sn−2 Sn−1

· · · −→ Λn−2(V)dn−2−−−→ Λn−1(V)

dn−1−−−→ Λn(V) → 0.


· · · −→ Λn−2(K)dn−2S−1

n−2dn−2−−−−−−−−−−→ Λn−1(V)

(Sn−1dn−1

)−−−−→ Λn(W) → 0


C∞(R)J−→ C∞(M)

(skwdiv)−−−→ C∞(W) → 0.


Stability of continuous problem

To establish stability for continuous problem, use two results.

Lemma: (from PDE) Let Ω be a bounded domain in Rn with aLipschitz boundary. Then, for all µ ∈ L2Λn(Ω), there existsη ∈ H1Λn−1(Ω) satisfying dn−1η = µ.

Theorem: Given (ω, µ) ∈ L2Λn(Ω; K)× L2Λn(Ω; V), there existsσ ∈ HΛn−1(Ω; V) such that dn−1σ = µ, −Sn−1σ = ω. Moreover,we may choose σ so that

‖σ‖HΛ ≤ c(‖ω‖+ ‖µ‖),

for a fixed constant c .


Outline of key part of proof

By Lemma:

Can find η ∈ H1Λn−1(Ω; V) satisfying dn−1η = µ,

and then τ ∈ H1Λn−1(Ω; K) satisfying dn−1τ = ω + Sn−1η.

Since Sn−2 isomorphism from H1Λn−2(Ω; V) onto H1Λn−1(Ω; K),have % ∈ H1Λn−2(Ω; V) with Sn−2% = τ .

Define σ = dn−2% + η ∈ HΛn−1(Ω; V).

Then dn−1σ = dn−1dn−2ρ + dn−1η = µ and

− Sn−1σ = −Sn−1dn−2%− Sn−1η = dn−1Sn−2%− Sn−1η

= dn−1τ − Sn−1η = ω.


Remarks about proof

(i) Although elasticity problem only involves 3 spacesHΛn−1(Ω; V), L2Λn(Ω; V), and L2Λn(Ω; K), proof brings in 2additional spaces: HΛn−2(Ω; V) and HΛn−1(Ω; K).

(ii) Although Sn−1 is only S operator arising in formulation, Sn−2

plays key role in proof.

(iii) Do not fully use fact that Sn−2 is an isomorphism fromΛn−2(Ω; V) to Λn−1(Ω; K), only that it is a surjection.

(iv) Other slightly weaker conditions can be used in some places inthe proof: (needed for some choices of stable finite elementspaces).


Try to do analogous proof in discrete case

Let Λkh be discrete k–forms: ω(x) =

∑I fI (x)dxI , where fI are

piecewise polynomial functions with values in R.

Assume following discrete sequence is exact

R ⊂−→ Λ0h

d0−→ Λ1h

d1−→ Λ2h → 0

and that interpolation operators Πh onto Λkh are such that

following diagram commutes

R ⊂−−→ Λ0 d−−→ Λ1 d−−→ Λ2 −−→ 0yid

yΠh

yΠh

yΠh

R ⊂−−→ Λ0h

d−−→ Λ1h

d−−→ Λ2h −−→ 0


What is needed?

Define discrete version of operators Kk and Sk :

Kk,h = ΠhKk , Sk,h = ΠhSk .

Problem in using same stability proof on discrete level is step (3),i.e., can’t expect Sn−2,h to be an isomorphism from Λn−2

h (V) toΛn−1

h (K).

However, looking at proof, only need:

(A) The operator Sn−2,h : Λn−2h (V) 7→ Λn−1

h (K) is onto.


Choosing finite element spaces to get stable discretization

Need to find combinations of discrete de Rham sequences for whichAssumption (A) (Sn−2,h : Λn−2

h (V) 7→ Λn−1h (K) onto) is satisfied.

Let n = 2. Given ω1 = f1χdx1 + f2χdx2 ∈ Λ1h(K), find

ω0 = (g1, g2)T ∈ Λ0

h(V) such that

S0,hω0 = πhS0ω0 = πh(g2χdx1 − g1χdx2) = f1χdx1 + f2χdx2.

Can be reduced to checking degrees of freedom of discrete spaces.


Choosing finite element spaces to get stable discretization

Need to find combinations of discrete de Rham sequences for whichAssumption (A) (Sn−2,h : Λn−2

h (V) 7→ Λn−1h (K) onto) is satisfied.

Let n = 2. Given ω1 = f1χdx1 + f2χdx2 ∈ Λ1h(K), find

ω0 = (g1, g2)T ∈ Λ0

h(V) such that

S0,hω0 = πhS0ω0 = πh(g2χdx1 − g1χdx2) = f1χdx1 + f2χdx2.

Can be reduced to checking degrees of freedom of discrete spaces.


Simple stable choice

P1Λ0h(K)

d0−→ P−1 Λ1h(K)

d1−→ P0Λ2h(K) → 0

P2Λ0h(V)

d0−→ P1Λ1h(V)

d1−→ P0Λ2h(V) → 0.

A

AAAA• •

•

A

AAAA

HHY *

?A

AAAA

•d0-

d1-

A

AAAA• •

•

•

• •

A

AAAA

HHYHHY

**

??A

AAAA

•d0-

d1-

Top sequence: continuous P1, RT0, piecewise constants.Bottom sequence: continuous P2, BDM1, piecewise constants.


Family of stable finite elements

Mixed elasticity with weakly imposed symmetry: Find(σ, u, p) ∈ Σh × Vh × Qh ⊂ H(div,Ω, M)× L2(Ω, V)× L2(Ω, K)such that

(Aσ, τ) + (div τ, u) + (τ, p) = 0, τ ∈ Σh,

(div σ, v) = (f , v), v ∈ Vh,

(σ, q) = 0, q ∈ Qh.

A family of elements: r ≥ 0, for n = 2 and n = 3:• Σh

∼= Pr+1Λn−1h (V)

• Vh∼= PrΛ

nh(V)

• Qh∼= PrΛ

nh(K)


Error estimates

Theorem: Suppose (σ, u, p) solves elasticity system and(σh, uh, ph) solves discrete elasticity system. Under reasonablehypotheses, and for appropriate projection operators Πh,

‖σ − σh‖+ ‖p − ph‖ ≤ C (‖σ −Πn−1h σ‖+ ‖p −Πn

hp‖),‖u − uh‖ ≤ C (‖σ −Πn−1

h σ‖+ ‖p −Πnhp‖+ ‖u −Πn

hu‖),‖dn−1(σ − σh)‖ = ‖dn−1σ −Πn

hdn−1σ‖.

For family just discussed, and 1 ≤ k ≤ r + 1,

‖σ − σh‖+ ‖p − ph‖+ ‖u − uh‖ ≤ Chk(‖σ‖k + ‖p‖k + ‖u‖k).


Connections to Previous Work

Most methods previously developed (e.g., PEERS, Stenberg,Morley) fit this framework with one basic modification, i.e., needto insert L2 projection (Πn

h ) into top discrete exact sequence.

Λ0h(K)

d0−→ Λ1h(K)

Πnhd1−−−→ Λ2

h(K) → 0

Λ0h(V)

d0−→ Λ1h(V)

d1−→ Λ2h(V) → 0.


PEERS (Arnold-Brezzi-Douglas)

For n = 2, letting B3 denote cubic bubble function, choose

Λ1h(V) = P−1 Λ1(Th; V) + dB3Λ

0(Th; V), Λ2h(V) = P0Λ

2(Th; V),

Λ2h(K) = P1Λ

2(Th; K) ∩ H1Λ2(K),

Choose two remaining spaces:

Λ0h(V) = (P1 + B3)Λ

0(Th; V), Λ1h(K) = S0Λ

0h(V).

A

AAAA• •

•

•A

AAAA• •

•Πhd1-

A

AAAA• •

•

•A

AAAA

HHY *

?

•-

d0

A

AAAA

•d1-

Known error estimate:

‖σ − σh‖0 + ‖p − ph‖0 + ‖u − uh‖0 ≤ Ch(‖σ‖1 + ‖p‖1 + ‖u‖1).


A PEERS-like method with improved stress approximation

Choose

Λ1h(V) = P1Λ

1(Th; V), Λ2h(V) = P0Λ

2(Th; V),

Λ2h(K) = P1Λ

2(Th; K) ∩ H1Λ2(K),

and two remaining spaces as

Λ0h(V) = P2Λ

0(Th; V), Λ1h(K) = S0Λ

0h(V) ≡ P2Λ

1(Th; K)∩H1.

A

AAAA• •

•

•

• •

A

AAAA• •

•-

Πhd1

A

AAAA• •

•

•

• •

A

AAAA

HHYHHY

**

??

-d0

A

AAAA

•-

d1


Summary

1. Using variational formulation enforcing symmetry of stressweakly, can use standard mixed elements for second orderproblems.

2. Using relation between de Rham sequence and elasticitysequence, obtained new (and simpler) finite element methods forelasticity.

3. By slightly modifying approach, old methods fit new frameworkand improved version of PEERS method developed.


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