VEM for general second order elliptic problems · 2014. 11. 3. · VEM for general second order elliptic problems Donatella Marini Dipartimento di Matematica, Universit a di Pavia,

VEM for general second order elliptic problems

Donatella Marini

Dipartimento di Matematica, Universita di Pavia, Italy

IMATI- C.N.R., Pavia, Italy

joint work with L. Beirao da Veiga, F. Brezzi and A. Russo

Structure-Preserving Discretizations of Partial Differential Equationsin honour of Doug Arnold 60th birthday

Minneapolis, October 22-24, 2014

Donatella Marini (Pavia) VEM variable IMA 2014 1 / 36

Outline

1 The problem - Variational formulationVirtual Element approximation of the primal formulationVarious choices for the discrete bilinear formsError estimatesNumerical results

2 Mixed formulationVirtual Element approximation of the mixed formulationError estimates


The continuous problem

Ω ⊂ R2 (polygonal) computational domain, κ, γ,b smooth,Assume that the problem

L p := div(−κ(x)∇p + b(x)p) + γ(x) p = f (x) in Ω

p = 0 on Γ

is solvable for any f ∈ H−1(Ω), and

‖p‖1,Ω ≤ C‖f ‖−1,Ω, ‖p‖2,Ω ≤ C‖f ‖0,Ω

Existence and uniqueness as well for the adjoint operator

L∗p := div(−κ(x)∇p)− b(x) · ∇p + γ(x) p

In particular, ∀f ∈ L2(Ω) ∃ϕ ∈ H2(Ω) ∩ H10 (Ω):

L∗ϕ = f , ‖ϕ‖2,Ω ≤ C ∗‖f ‖0,Ω


Variational formulation

Set:

a(p, q) :=

∫Ωκ∇p · ∇q dx , b(p, q) := −

∫Ωp(b · ∇q) dx

c(p, q) :=

∫Ωγp q dx , B(p, q) := a(p, q) + b(p, q) + c(p, q).

Variational formulation:Find p ∈ Q := H1

0 (Ω) such that

B(p, q) = (f , q) ∀q ∈ Q

B(p, q) ≤ M‖p‖1‖q‖1, p, q ∈ H1, supq∈H1

0

B(p, q)

‖q‖1≥ CB‖p‖1 p ∈ H1

0


Virtual Element Approximation

We need to define:

• Qkh : a finite dimensional space (⊂ Q = H1

0 (Ω)) (Pk ⊂ Qkh⊂ Q, k ≥ 1)

• a bilinear form Bh(·, ·) : Qkh ×Qk

h → R• an element fh ∈ (Qk

h)′

in such a way that the problem

find ph ∈ Qkh such that Bh(ph, qh) = (fh, qh) ∀qh ∈ Qk

h

has a unique solution, and optimal error estimates hold.


Virtual Element Approximation

Th a decomposition of Ω into polygons E

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


The discrete spaces

On each element E ∈ Th we define (k ≥ 1)

Qkh(E ) := q ∈ H1(E ) : q|e ∈ Pk(e) ∀e ∈ ∂E , ∆q ∈ Pk−2(E ).

Degrees of freedom in Qkh(E ):

(D1) The values q(Vi ) at the vertices Vi of E ,

and for k ≥ 2

(D2) The moments∫e q pk−2 ds, pk−2 ∈ Pk−2(e), on each edge e of E ,

(D3) The moments∫E q pk−2 dx , pk−2 ∈ Pk−2(E ).

Easy to check that D1–D3 are unisolvent.


How to construct a globally computable Bh(·, ·)

First attempt: extend to div(κ(x)∇p) the techniques used for ∆p

Let Π∇k : H1(E ) → Pk(E ) be defined by:

Def: Π∇k p ∈ Pk(E )

∫E∇Π∇k p · ∇qdx =

∫E∇p · ∇qdx ∀q ∈ Pk(E )∫

∂EΠ∇k pds =

∫∂E

pds

Π∇k p easily computable using the d.o.f. of pChoose

aEh (ph, qh) := aE (Π∇k ph,Π∇k qh) + SE ((I − Π∇k )ph, (I − Π∇k )qh)

with SE (·, ·) any symmetric bilinear form that scales like aE (·, ·):

c0aE (qh, qh) ≤ SE (qh, qh) ≤ c1a

E (qh, qh) ∀qh with Π∇k qh = 0

(fh, qh)0,E = (Π0k−2f , qh)0,E


Case k=1, variable diffusion κ = x2 + y 3 + 1

u(x , y) = y−x+log(y3+x+1)−xy−xy2+x2y+x2+x3+sin(5x) sin(7y)−1

Exact solution

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−1

−0.5

0

0.5

1

1.5

2

x

−x+y+log(x+y3+1.0)−x y−x y2+x2 y+x2+...−1.0

yDonatella Marini (Pavia) VEM variable IMA 2014 9 / 36


−div (κ(x)∇u) = f (x) ∇vh ≈ ∇Π∇k vh

10−2

10−1

100

10−3

10−2

10−1

h=hmean slope: ||Π0uh−u

e||

0,Ω = 1.873, ||Π∇ uh−u

e||

0,Ω = 1.873

h1

h2

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−2

10−1

100

101

h=hmean slope: ||∇Π 0uh−∇ u

e||

0,Ω = 0.997, ||∇Π ∇ uh−∇ u

e||

0,Ω = 0.997

h1

h2

||∇Π 0uh−∇ u

e||

0,Ω

L2 error k = 1 H1 error

triangle0.0188 polygons, Ndof=178, h

max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02

Generated by ploterrors on 01-Oct-2014 00:01:17Donatella Marini (Pavia) VEM variable IMA 2014 10 / 36



10−2

10−1

100

10−5

10−4

10−3

10−2


e||

0,Ω = 3.011, ||Π∇ uh−u

e||

0,Ω = 3.011

h1

h2

h3

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−3

10−2

10−1

100


e||

0,Ω = 1.996, ||∇Π ∇ uh−∇ u

e||

0,Ω = 1.996

h1

h2

h3

||∇Π 0uh−∇ u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02




10−2

10−1

100

10−6

10−5

10−4

10−3

10−2


e||

0,Ω = 2.092, ||Π∇ uh−u

e||

0,Ω = 2.074

h1

h2

h3

h4

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−4

10−3

10−2

10−1


e||

0,Ω = 1.053, ||∇Π ∇ uh−∇ u

e||

0,Ω = 1.177

h1

h2

h3

h4

||∇Π 0uh−∇ u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02



−div (κ(x)∇u) = f (x) ∇vh ≈ Π0k−1∇vh

10−2

10−1

100

10−3

10−2

10−1


e||

0,Ω = 1.873, ||Π∇ uh−u

e||

0,Ω = 1.873

h1

h2

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−2

10−1

100

101


e||

0,Ω = 0.997, ||∇Π ∇ uh−∇ u

e||

0,Ω = 0.997

h1

h2

||∇Π 0uh−∇ u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02



−div (κ(x)∇u) = f (x) ∇vh ≈ Π0k−1∇vh

10−2

10−1

100

10−5

10−4

10−3

10−2


e||

0,Ω = 3.047, ||Π∇ uh−u

e||

0,Ω = 3.047

h1

h2

h3

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−3

10−2

10−1

100


e||

0,Ω = 2.017, ||∇Π ∇ uh−∇ u

e||

0,Ω = 2.017

h1

h2

h3

||∇Π 0uh−∇ u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02



−div (κ(x)∇u) = f (x) ∇vh ≈ Π0k−1∇vh

10−2

10−1

100

10−7

10−6

10−5

10−4

10−3


e||

0,Ω = 3.966, ||Π∇ uh−u

e||

0,Ω = 3.962

h1

h2

h3

h4

||Π0uh−u

e||

0,Ω

10−2

10−1

100

10−5

10−4

10−3

10−2

10−1


e||

0,Ω = 2.967, ||∇Π ∇ uh−∇ u

e||

0,Ω = 2.966

h1

h2

h3

h4

||∇Π 0uh−∇ u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02


How to compute the L2− projection?

No problem for Π0k−1∇vh in the diffusion term:∫

EΠ0k−1∇vh · pk−1dx =

∫E∇vh · pk−1dx

= −∫Evh div pk−1dx +

∫∂E

vhpk−1 · nds

For advection and reaction terms we need L2− projections of scalars:∫E

Π0k−1vh pk−1dx =

∫Evh pk−1dx =????? NO WAY


The new discrete spaces

On each element E ∈ Th we define (k ≥ 1)

Qkh(E ) := q ∈ H1(E ) : q|e ∈ Pk(e) ∀e ∈ ∂E , ∆q ∈ Pk(E ),

and

∫Eqprdx =

∫E

Π∇k qprdx , r = k − 1, k

pk = homogeneous polynomial of degree k

The degrees of freedom in Qkh(E ) are the same:

(D1) The values q(Vi ) at the vertices Vi of E ,

and for k ≥ 2

(D2) The moments∫e q pk−2 ds, pk−2 ∈ Pk−2(e), on each edge e of E ,

(D3) The moments∫E q pk−2 dx , pk−2 ∈ Pk−2(E ).

We can compute the L2− projection onto Pk(E ):∫E

(q − Π0kq)pkdx = 0 ∀pk ∈ Pk(E )


How to construct a globally computable Bh(·, ·)

For the diffusion term keep

aEh (ph, qh) :=∫E κΠ0

k−1∇ph · Π0k−1∇qhdx + SE ((I − Π∇k )ph, (I − Π∇k )qh)

with SE (·, ·) any symmetric bilinear form that scales like a(·, ·):

c0a(qh, qh) ≤ SE (qh, qh) ≤ c1a(qh, qh) ∀qh with Π∇k qh = 0

For the other terms choose:bEh (ph, qh) := −

∫E Π0

k−1ph b · Π0k−1∇qhdx

cEh (ph, qh) :=∫E γΠ0

k−1ph Π0k−1qhdx

(fh, qh)0,E =∫E Π0

k−1f qhdx

Bh(ph, qh) := ah(ph, qh) + bh(ph, qh) + ch(ph, qh)


Consistency error

With this choice “Galerkin orthogonality” (Patch test ) does not hold, noteven when the coefficients are constant. Indeed, when p ∈ Pk∫

E[Π0

k−1p] [b · Π0k−1∇q]dx 6=

∫Ep[b · Π0

k−1∇q]dx∫Eγ[Π0

k−1p] [Π0k−1q]dx 6=

∫Eγp [Π0

k−1q]dx


Consistency error

We have the following result

Lemma

For all p sufficiently regular and for all qh ∈ Qkh it holds

BE (Π0kp, qh)− BE

h (Π0kp, qh) ≤ Cκ,b,γh

kE‖p‖k+1,E‖qh‖1,E ∀E ∈ Th.


Continuity and Inf-Sup

Lemma

The bilinear form Bh(·, ·) is continuous in Qkh ×Qk

h

Bh(p, q) ≤ Cκ,b,γ‖p‖1‖q‖1 p, q ∈ Qkh ,

with Cκ,b,γdepending on κ,b, γ but independent of h.Moreover, the discrete Inf-Sup condition holds: there exists an h0 > 0 anda constant CB such that, for all h < h0:

supqh∈Qk

h

Bh(ph, qh)

‖qh‖1≥ CB‖ph‖1 ∀ ph ∈ Qk

h .


Convergence in H1

The proof follows the approach of Schatz (Math Comp 1974) fornoncoercive operators, using:

The Inf-Sup for the continuous problem;

The ellipticity of a(·, ·) and ah(·, ·);

The compactness of the lower-order terms bh(·, ·) and ch(·, ·)

Theorem

For h sufficiently small, the discrete problem has a unique solutionph ∈ Qk

h , and the following error estimate holds:

‖p − ph‖1 ≤ Chk (‖p‖k+1 + |f |k).

with C a constant depending on κ,b, and γ but independent of h.


Convergence in L2

Via classical duality arguments we have:

Theorem

For h sufficiently small, the following error estimate holds:

‖p − ph‖0 ≤ Chk+1 (‖p‖k+1 + |f |k),

where C is a constant depending on κ,b, and γ but independent of h.


General case - a numerical test

κ = x2 + y3 + 1, b = [x y ], γ = sin(xy) + 1

u(x , y) = y−x+log(y3+x+1)−xy−xy2+x2y+x2+x3+sin(5x) sin(7y)−1Exact solution

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−1

−0.5

0

0.5

1

1.5

2

x

−x+y+log(x+y3+1.0)−x y−x y2+x2 y+x2+...−1.0

y


Case k=3

div (−κ(x)∇u + b(x)u) + γ(x)u = f (x) ∇vh ≈ Π0k−1∇vh, vh ≈ Π0

k−1vh

10−2

10−1

100

10−7

10−6

10−5

10−4

10−3

h=hmean slope: ||Π0u

h−u

e||

0,Ω = 3.966, ||Π

∇u

h−u

e||

0,Ω = 3.962

h1

h2

h3

h4

||Π0u

h−u

e||

0,Ω

10−2

10−1

100

10−5

10−4

10−3

10−2

10−1

h=hmean slope: ||∇Π0u

h−∇u

e||

0,Ω = 2.967, ||∇Π

∇u

h−∇u

e||

0,Ω = 2.966

h1

h2

h3

h4

||∇Π0u

h−∇u

e||

0,Ω



max = 1.74e−01, h

mean = 1.49e−01


max = 1.34e−01, h

mean = 1.07e−01


max = 6.34e−02, h

mean = 4.91e−02


max = 4.49e−02, h

mean = 3.45e−02


Case k=3, squares

div (−κ(x)∇u + b(x)u) + γ(x)u = f (x) ∇vh ≈ Π0k−1∇vh, vh ≈ Π0

k−1vh

10−2

10−1

100

10−6

10−5

10−4

10−3

10−2

h=hmean slope: ||Π0u

h−u

e||

0,Ω = 3.998, ||Π

∇u

h−u

e||

0,Ω = 3.998

h1

h2

h3

h4

||Π0u

h−u

e||

0,Ω

10−2

10−1

100

10−5

10−4

10−3

10−2

10−1

100

h=hmean slope: ||∇Π0u

h−∇u

e||

0,Ω = 2.998, ||∇Π

∇u

h−∇u

e||

0,Ω = 2.998

h1

h2

h3

h4

||∇Π0u

h−∇u

e||

0,Ω


quadrati4x416 polygons, Ndof=153, h

max = 3.54e−01, h

mean = 3.54e−01


max = 1.77e−01, h

mean = 1.77e−01


max = 8.84e−02, h

mean = 8.84e−02


max = 4.42e−02, h

mean = 4.42e−02


Case k=3, comparison polygons/squares

10−2 10−1 10010−7

10−6

10−5

10−4

10−3

10−2 L2 error

polygonssquaresh4

10−2 10−1 10010−5

10−4

10−3

10−2

10−1 H1 error

polygonssquaresh3


Mixed formulation

Setting: ν = κ−1, β = κ−1b

u := ν−1(−∇p + βp), div u + γ p = f in Ω, p = 0 on Γ

V = H(div,Ω), Q = L2(Ω)Find (u, p) ∈ V × Q such that

(νu, v)− (p, div v)− (β · v, p) = 0 ∀v ∈ V

(div u, q) + (γp, q) = (f , q) ∀q ∈ Q


VEM approximation (RT-like)

The spaces: We define, for k integer ≥ 0,

V kh (E ) := v ∈ H(div;E ) ∩ H(rot;E ) : v · n|e ∈ Pk(e) ∀e ∈ ∂E ,

div v ∈ Pk(E ), and rot v ∈ Pk−1(E ).

V kh := v ∈ H(div; Ω) such that v|E ∈ V k

h (E ) ∀E ∈ ThQk

h := q ∈ L2(Ω) such that: q|E ∈ Pk(E ) ∀ element E inThd.o.f in V k

h (Brezzi-Falk-M, M2AN 2014):∫ev · n pk ds ∀e, ∀ pk ∈ Pk(e)∫

Ev · ∇pk−1dx ∀E ,∀pk−1 ∈ Pk−1(E )∫

Erot vpk−1dx ∀E ,∀pk−1 ∈ Pk−1


A linear “Fortin” operator ΠFh from W := (H1(Ω))2 → V k

h can be definedthrough the degrees of freedom, and

divW −→ Q

ΠFh ↓ ↓ Π0

k

V kh −→ Qk

h −→ 0div

so thatdiv ΠF

h v = Π0k div v

Moreover,

‖u− ΠFh u‖0 ≤ Chk+1‖u‖k+1, ‖p − Π0

kp‖0 ≤ Chk+1‖p‖k+1,

‖ div(u− ΠFh u)‖0 ≤ Chk+1‖ div u‖k+1.


The bilinear forms

aEh (v,w) := (νΠ0kv,Π

0kw)0,E + SE (v − Π0

kv,w − Π0kw)

α∗aE (v, v) ≤ SE (v, v) ≤ α∗aE (v, v) ∀v ∈ V k

h

Setah(v,w) :=

∑E

aEh (v,w).

Lemma

The bilinear form ah(·, ·) is continuous and elliptic in (L2)2:

∃M > 0 such that |ah(v,w)| ≤ M‖v‖0‖w‖0 ∀v,w ∈ V kh ,

∃α > 0 such that ah(v, v) ≥ α‖v‖20 ∀v ∈ V k

h ,

with M and α depending on ν but independent of h.


The discrete problem

(∗)

Find (uh, ph) ∈ V k

h × Qkh such that

ah(uh, vh)− (ph, div vh)− (β · Π0kvh, ph) = 0 ∀vh ∈ V k

h

(div uh, qh) + (γph, qh) = (f , qh) ∀qh ∈ Qh.

Theorem

Assume that the continuous problem has a unique solution p. Then, for hsufficiently small, problem (*) has a unique solution (uh, ph) ∈ V k

h × Qkh ,

and the following error estimates hold:

‖p − ph‖0 ≤ Chk+1(‖u‖k+1 + ‖p‖k+1

),

‖u− uh‖0 ≤ Chk+1(‖u‖k+1 + ‖p‖k+1

),

with C a constant depending on ν,β, and γ but independent of h.


Sketch of the proof

For the proof (very long) we follow the arguments of Douglas-Roberts,Math. Comp. 85.

We use directly duality arguments, i.e., the adjoint problem in mixedform, with right-hand-side (ν(u− uh), p − ph);

Prove error estimates for “any possible solution of the discreteproblem”;

Prove uniqueness of the solution of the discrete problem for h < h0;

Use finite dimension to have existence of the solution of the discreteproblem.


US mesh (from Google maps)

−120 −110 −100 −90 −80 −70

25

30

35

40

45

50

USA − 49 polygons, 5685 vertices, 5733 edges


New York state (zoom)

−80 −78 −76 −74 −72 −70

40

41

42

43

44

45

polygon New York243 vertices


US mesh: results

k = 3, Ndof = 17298L2 error: 7.717488e-03, H1 error: 3.542968e-02

−140−120

−100−80

−60

20

30

40

50−2

−1

0

1

2

VEM solution


Documents

VEM for general second order elliptic problems · 2014. 11. 3. · VEM for general second order elliptic problems Donatella Marini Dipartimento di Matematica, Universit a di Pavia,