A Finite Element Method for Nonvariational Elliptic Problems · 2010-05-21 · Discretisation Solution of the system Numerical experiments Further applications A Finite Element Method

Discretisation Solution of the system Numerical experiments Further applications

A Finite Element Method for NonvariationalElliptic Problems

Tristan Pryerjoint work with Omar Lakkis

May 20, 2010


Outline

1 Discretisation

2 Solution of the system

3 Numerical experiments

4 Further applications


Model problem in nonvariational form

Model problem

Given f ∈ L2(Ω), find u ∈ H2(Ω) ∩ H10(Ω) such that⟨

A:D2u |φ⟩

= 〈f , φ〉 ∀ φ ∈ H10(Ω),

X:Y := trace (XᵀY) is the Frobenius inner product of two matrices.



Model problem


A:D2u |φ⟩

= 〈f , φ〉 ∀ φ ∈ H10(Ω),


Don’t want to rewrite in divergence form!⟨A:D2u |φ

⟩= 〈div (A∇u) |φ〉 − 〈div (A)∇u, φ〉 .



Model problem


A:D2u |φ⟩

= 〈f , φ〉 ∀ φ ∈ H10(Ω),


Don’t want to rewrite in divergence form!⟨A:D2u |φ

⟩= 〈div (A∇u) |φ〉 − 〈div (A)∇u, φ〉 .


What should we do?

Main idea

Define appropriately the Hessian of a function who’s not twicedifferentiable, i.e., as a distribution.

Construct a finite element approximation of this distribution. Whatis meant by the Hessian of a finite element function?[Aguilera and Morin, 2008]

Discretise the strong form of the PDE directly.


What should we do?

Main idea





What should we do?

Main idea





Hessians as distributions

generalised Hessian

Given v ∈ H1(Ω) it’s generalised Hessian is given by⟨D2v |φ

⟩= −〈∇v ⊗∇φ〉+ 〈∇u ⊗ n φ〉∂Ω ∀ φ ∈ H1(Ω),

x⊗ y := xyᵀ is the tensor product between two vectors.

finite element space notation

V :=Φ ∈ H1(Ω) : Φ|K ∈ Pp ∀ K ∈ T

,

V := V ∩ H10(Ω),

finite element Hessian

For each V ∈ V there exists a unique H[V ] ∈ Vd×d such that

〈H[V ],Φ〉 =⟨D2V |Φ

⟩∀ Φ ∈ V.



generalised Hessian


⟩= −〈∇v ⊗∇φ〉+ 〈∇u ⊗ n φ〉∂Ω ∀ φ ∈ H1(Ω),



V :=Φ ∈ H1(Ω) : Φ|K ∈ Pp ∀ K ∈ T

,

V := V ∩ H10(Ω),



〈H[V ],Φ〉 =⟨D2V |Φ

⟩∀ Φ ∈ V.



generalised Hessian


⟩= −〈∇v ⊗∇φ〉+ 〈∇u ⊗ n φ〉∂Ω ∀ φ ∈ H1(Ω),



V :=Φ ∈ H1(Ω) : Φ|K ∈ Pp ∀ K ∈ T

,

V := V ∩ H10(Ω),



〈H[V ],Φ〉 =⟨D2V |Φ

⟩∀ Φ ∈ V.


Nonvariational finite element method

Substitute the finite element Hessian directly into the model problem.We seek U ∈ V such that⟨

A:H[U], Φ⟩

=⟨f , Φ

⟩∀ Φ ∈ V.

Discretisation

⟨f , Φ

⟩=

d∑α=1

d∑β=1

⟨Aα,βHα,β[U], Φ

⟩

=d∑

α=1

d∑β=1

⟨Φ,Aα,βΦᵀ

⟩hα,β .




A:H[U], Φ⟩

=⟨f , Φ

⟩∀ Φ ∈ V.

Discretisation

⟨f , Φ

⟩=

d∑α=1

d∑β=1


⟩

=d∑

α=1

d∑β=1

⟨Φ,Aα,βΦᵀ

⟩hα,β .




A:H[U], Φ⟩

=⟨f , Φ

⟩∀ Φ ∈ V.

Discretisation

⟨f , Φ

⟩=

d∑α=1

d∑β=1


⟩

=d∑

α=1

d∑β=1

⟨Φ,Aα,βΦᵀ

⟩hα,β .




A:H[U], Φ⟩

=⟨f , Φ

⟩∀ Φ ∈ V.

Discretisation

⟨f , Φ

⟩=

d∑α=1

d∑β=1


⟩

=d∑

α=1

d∑β=1

⟨Φ,Aα,βΦᵀ

⟩hα,β .

〈Φ,Φᵀ〉hα,β = 〈Φ,Hα,β[U]〉

=(−

⟨∂βΦ, ∂αΦ

ᵀ⟩

+⟨Φnβ , ∂αΦ

ᵀ⟩

∂Ω

)u.




A:H[U], Φ⟩

=⟨f , Φ

⟩∀ Φ ∈ V.

Discretisation

⟨f , Φ

⟩=

d∑α=1

d∑β=1


⟩

=d∑

α=1

d∑β=1

⟨Φ,Aα,βΦᵀ

⟩hα,β .

〈Φ,Φᵀ〉hα,β = 〈Φ,Hα,β[U]〉

=(−

⟨∂βΦ, ∂αΦ

ᵀ⟩

+⟨Φnβ , ∂αΦ

ᵀ⟩

∂Ω

)u.


Linear system

U = Φᵀu, where u ∈ RN is the solution to the following linear system

Du :=d∑

α=1

d∑β=1

Bα,βM−1Cα,βu = f.

Components of the linear system

Bα,β :=⟨Φ,Aα,βΦᵀ

⟩∈ RN×N ,

M := 〈Φ,Φᵀ〉 ∈ RN×N ,

Cα,β := −⟨∂βΦ, ∂αΦ

ᵀ⟩

+⟨Φnβ , ∂αΦ

ᵀ⟩

∂Ω∈ RN×N ,

f :=⟨f , Φ

⟩∈ RN .


Linear system


Du :=d∑

α=1

d∑β=1


Components of the linear system

Bα,β :=⟨Φ,Aα,βΦᵀ

⟩∈ RN×N ,

M := 〈Φ,Φᵀ〉 ∈ RN×N ,

Cα,β := −⟨∂βΦ, ∂αΦ

ᵀ⟩

+⟨Φnβ , ∂αΦ

ᵀ⟩

∂Ω∈ RN×N ,

f :=⟨f , Φ

⟩∈ RN .


The system is hard to solve

Linear system


Du :=d∑

α=1

d∑β=1


Remarks

The matrix D is not sparse

Mass lumping only works for P1 elements AND in this case onlygives a reasonable solution U for very simple A

Notice D is the sum of Schur complements

We can create a block matrix to exploit this



Linear system


Du :=d∑

α=1

d∑β=1


Remarks







Linear system


Du :=d∑

α=1

d∑β=1


Remarks







Linear system


Du :=d∑

α=1

d∑β=1


Remarks






Block system

E =

M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1 B1,2 . . . Bd,d−1 Bd,d 0

,

v = (h1,1,h1,2, . . . ,hd,d−1,hd,d ,u)ᵀ,

h = (0, 0 . . . , 0, 0, f)ᵀ.

Equivalence of systems

Then solving the system Du = f is equivalent to solving

Ev = h.

for u.


Block system

E =

M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1 B1,2 . . . Bd,d−1 Bd,d 0

,

v = (h1,1,h1,2, . . . ,hd,d−1,hd,d ,u)ᵀ,

h = (0, 0 . . . , 0, 0, f)ᵀ.

Equivalence of systems

Then solving the system Du = f is equivalent to solving

Ev = h.

for u.


structure of the block matrix

the discretisation presented nothing but:

Find U ∈ V such that

〈H[U],Φ〉 = −〈∇U ⊗∇Φ〉+ 〈∇U ⊗ n Φ〉∂Ω

∀ Φ ∈ V

⟨A:H[U], Φ

⟩=

⟨f , Φ

⟩∀ Φ ∈ V.

M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1 B1,2 . . . Bd,d−1 Bd,d 0

h1,1

h1,2

...hd,d−1

hd,d

u

=

00...00f

.


structure of the block matrix

the discretisation presented nothing but:

Find U ∈ V such that

〈H[U],Φ〉 = −〈∇U ⊗∇Φ〉+ 〈∇U ⊗ n Φ〉∂Ω

∀ Φ ∈ V

⟨A:H[U], Φ

⟩=

⟨f , Φ

⟩∀ Φ ∈ V.

M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1 B1,2 . . . Bd,d−1 Bd,d 0

h1,1

h1,2

...hd,d−1

hd,d

u

=

00...00f

.


A Linear PDE in NDform

Operator choice - heavily oscillating

A =

[1 00 a(x)

]

a(x) = sin

(1

|x1|+ |x2|+ 10−15

)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1


A Linear PDE in NDform

Operator choice - heavily oscillating

A =

[1 00 a(x)

]

a(x) = sin

(1

|x1|+ |x2|+ 10−15

)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1


Figure: Choosing f appropriately such that u(x) = exp (−10 |x|2).

102 103 10410−5

10−4

10−3

10−2

10−1

100

101

dim V

||u−u

h|| X

EOC = 1.6815

EOC = 0.94816

EOC = 2.0751

EOC = 0.96833

EOC = 2.0465

EOC = 1.0129

EOC = 2.0345

EOC = 1.0149

EOC = 2.02

EOC = 1.0094

EOC = 2.0107

EOC = 1.0052

X = L2

X = H1


Another Linear PDE in NDform

Operator choice

A =

[1 00 a(x)

]

a(x) :=(arctan

(5000(|x|2 − 1)

)+ 2

).


Another Linear PDE in NDform

Operator choice

A =

[1 00 a(x)

]a(x) :=

(arctan

(5000(|x|2 − 1)

)+ 2

).


Figure: Choosing f appropriately such that u(x) = sin (πx1) sin (πx2).

102 103 10410−4

10−3

10−2

10−1

100

101

dim V

||u−u

h|| X

EOC = 1.8547

EOC = 0.91495

EOC = 2.0811

EOC = 1.0384

EOC = 2.069

EOC = 1.0339

EOC = 2.0402

EOC = 1.02

EOC = 2.0214

EOC = 1.0107

EOC = 2.011

EOC = 1.0055

X = L2

X = H1


The same Linear PDE in NDform

Figure: On the left we present the maximum error of the standard FE-solution.Notice the oscillations apparant on the unit circle. On the right we show themaximum error of the NDFE-solution


Fully nonlinear PDEs

Model problem

Given f ∈ L2(Ω), find u ∈ H2(Ω) ∩ H10(Ω) such that

N [u] = F (D2u)− f = 0

Newton’s method

Given u0 for each n ∈ N find un+1 such that⟨N ′ [un] | un+1 − un

⟩= −N [un]

〈N ′ [u] | v〉 = limε→0

N [u + εv ]−N [u]

ε

= limε→0

F (D2u + εD2v)−F (D2u)

ε

= limε→0

F ′(D2u) : D2v + O(ε).



Model problem


N [u] = F (D2u)− f = 0

Newton’s method


⟩= −N [un]

〈N ′ [u] | v〉 = limε→0

N [u + εv ]−N [u]

ε

= limε→0


ε

= limε→0

F ′(D2u) : D2v + O(ε).



Model problem


N [u] = F (D2u)− f = 0

Newton’s method


⟩= −N [un]

〈N ′ [u] | v〉 = limε→0

N [u + εv ]−N [u]

ε

= limε→0


ε

= limε→0

F ′(D2u) : D2v + O(ε).



Model problem


N [u] = F (D2u)− f = 0

Newton’s method


⟩= −N [un]

〈N ′ [u] | v〉 = limε→0

N [u + εv ]−N [u]

ε

= limε→0


ε

= limε→0

F ′(D2u) : D2v + O(ε).


Discretisation

VERY roughly

Given U0 = Λu0 find Un+1 such that

F ′(H[Un]):H[Un+1 − Un] = f −F (H[Un])

H[Un] is given in the solution of the previous iterate!

M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1n−1 B1,2

n−1 . . . Bd,d−1n−1 Bd,d

n−1 0

hn1,1

hn1,2...

hnd,d−1

hnd,d

un

=

00...00f

.


Discretisation

VERY roughly


F ′(H[Un]):H[Un+1 − Un] = f −F (H[Un])


M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1n−1 B1,2

n−1 . . . Bd,d−1n−1 Bd,d

n−1 0

hn1,1

hn1,2...

hnd,d−1

hnd,d

un

=

00...00f

.


Discretisation

VERY roughly


F ′(H[Un]):H[Un+1 − Un] = f −F (H[Un])


M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1n−1 B1,2

n−1 . . . Bd,d−1n−1 Bd,d

n−1 0

hn1,1

hn1,2...

hnd,d−1

hnd,d

un

=

00...00f

.

Saves us postprocessing another one! [Vallet et al., 2007][Ovall, 2007]


Discretisation

VERY roughly


F ′(H[Un]):H[Un+1 − Un] = f −F (H[Un])


M 0 · · · 0 0 −C1,1

0 M · · · 0 0 −C1,2

......

. . ....

......

0 0 · · · M 0 −Cd,d−1

0 0 . . . 0 M −Cd,d

B1,1n−1 B1,2

n−1 . . . Bd,d−1n−1 Bd,d

n−1 0

hn1,1

hn1,2...

hnd,d−1

hnd,d

un

=

00...00f

.

Saves us postprocessing another one! [Vallet et al., 2007][Ovall, 2007]


Bibliography I

Aguilera, N. E. and Morin, P. (2008).On convex functions and the finite element method.online preprint arXiv:0804.1780v1, arXiv.org.

Ovall, J. (2007).Function, gradient and hessian recovery using quadratic edge-bumpfunctions.J. Sci. Comput., 45(3):1064–1080.

Vallet, M.-G., Manole, C.-M., Dompierre, J., Dufour, S., andGuibault, F. (2007).Numerical comparison of some Hessian recovery techniques.Internat. J. Numer. Methods Engrg., 72(8):987–1007.

Documents

A Finite Element Method for Nonvariational Elliptic Problems · 2010-05-21 · Discretisation Solution of the system Numerical experiments Further applications A Finite Element Method