Andrea Can Giani

8/10/2019 Andrea Can Giani

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VIRTUAL ELEMENT METHODS

Andrea Cangiani

UNIVERSITY OF LEICESTER

EFEF 2012

BCAM, Bilbao, 8-9 June 2012

A. Cangiani (University of Leicester) Virtual Element Method (VEM) EFEF 2012 1 / 21


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FEM on general meshesAugmented FE are mentioned as early as Strang & Fix (1973)!

GFEM [Babuska & Osborn (1983)]Partition of Unity Method [Babuska & Melenk (1997)]XFEM [eg. Fries & Belytschko (2010)]PolygonalFEM [Tabarraei & Sukumar (2004 & 2007)]BEM-based FEM [Copeland, Langer, & Pusch (2009)]Various dG approaches, eg.

Polymorphic Nodal dG [Gassner, Lorcher, Munz, & Hesthaven (2009)]Agglomeration dG [Bassi, Botti, Colombo, & Rebay (2011)]

...

Driving idea: generalize FE to polygons considering/adding particularshape functions that may help capturing the solution

The VEM approach: Generalize FE to polygons maintaining the ease of

implementation of (polynomial) FE.A. Cangiani (University of Leicester) Virtual Element Method (VEM) EFEF 2012 2 / 21


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The Virtual Element Methods (VEM) principlefor the generalization of FEM on polygons

The VEM trial space contains, on each element, a space of polynomials, plus other functions.

The degrees of freedom (dof) are carefully chosen in order to allow

accurate dof-based computation when the polynomials are involved.

It is shown that for the remaining part a result with the right order of magnitude and stability properties suffices.



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Model problem and mesh partitionModel problem: On R2, consider the symmetric variational problem

nd u V := H 10 () such that

a(u , v ) = ( f , v ) v V ,

with f L2() and

a(u , v ) M |u |1|v |1, a(v , v ) |v |21 u , v V .

Polygonal decompositions: {T h }h decompositions of intonon-overlapping polygons K

Broken H 1-seminorm: On H 1(T h ) :=K T h

H 1(K ) dene

|v |h ,1 :=K T h

| v |20,K 1/ 2

.



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Abtract discrete problemGiven:

A space V h V ;A symmetric bilinear form ah : V h V h R such that

ah (u h , v h ) =K T h

aK h (u h , v h ) u h , v h V h ,

where aK h (, ) is a bilinear form on V h |K V h |K ;An element f h V h .

We dene the discrete problem

nd u h V h such thatah (u h , v h ) = f h , v h v h V h



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Assumptions on the discrete bilinear form

There exists an integer k 1 such that for all K T h

P k (K ) V h .k -consistency: for all p Pk (K ) and for all v h V h |K ,

aK h (p , v h ) = aK (p , v h ).

Stability: there exist two positive constants and , independentof h and of K , such that

v h V h |K aK (v h , v h ) aK h (v h , v h ) aK (v h , v h ).



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Abstract convergence theorem

Theorem

Under all the above assumptions the discrete problem has a uniquesolution u h and

|u u h |1 inf u I V h

|u u I |1 + inf u K P k (K )

|u u |h ,1 + supv V h \{ 0}

|(f , v ) f h , v ||v |1

| h |21 f h , h K aK h (u I , h ) ( h := u h u I )

= f h , h K aK h (u I u , h ) + a

K h (u , h )

= f h , h K aK h (u I u , h ) + aK (u , h )= f h , h (f , h ) K a

K h (u I u , h ) + a

K (u u , h ) .



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A Virtual Element MethodConsider the case a(u , v ) = ( u , v ).

Fix K T h with n edges, x

K barycenter of K , hK diameter of K , k 1,Bk ( K ) := {v C 0( K ) : v |e Pk (e ) e K };

note that dim( Bk ( K )) = n + n(k 1) = nk . Then, dene

V K ,k = {v H 1(K ) : v | K Bk ( K ), v |K Pk 2(K )},

with P 1(K ) := {0}. Note that

P k (K ) V K ,k ;NK := dim V K ,k = nk + k (k 1)/ 2.

Then, the global conforming VEM space is given by

V h := {v V : v |K V K ,k K T h }



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VEM degrees of freedomFor instance, we may choose as dof for any v h V K ,k ,

V K ,k - The values of v h

at the vertices .

E K ,k - For k > 1, e the values of v h at k 1 distinct points

P K ,k - For k > 1, the moments1

|K | K

m(x )v h (x )d x m x x K

hK

s

, |s | k 2 .



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VEM vs FEM (recall that NK = nk + k (k 1)/ 2)VEM k = 1 k = 2 k = 3

FEM k = 1 k = 2 k = 3

Q.: What is VEM for k = 1?



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VEM vs ??? (recall that NK = nk + k (k 1)/ 2)

VEM k = 1 k = 2 k = 3



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Construction of admissible bilinear forms

Observe that, as for any v V K ,k and for any p Pk (K ),

aK (v , p ) = K v p dx = K v p dx + K v p n ds ,is exacly computable just using the local degrees of freedom.Example. Take k = 2. Then,

K v p dx = |K | p 1|K | K v dx .Now, it is convenient to dene the projection K

k : V K ,k P

k (K ) as

aK (K k v , q ) = aK (v , q ) q Pk (K )

K k v = v := 1n

n

i =1

v (V i ) V i = vertices of K



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Then,

aK h (u , v ) := aK (K k u ,

K k v ) + S

K (u K k u , v K k v ) u , v V

K ,k ,

satises k -consistency and stability for every S K (u , v ) such that

c 0aK (v , v ) S K (v , v ) c 1aK (v , v ) v V K ,k with K k v = 0 .

Computing the RHS

Similarly, dene f h := P K k 2f as the L2(K ) projection of f onto Pk 2(K )

Then, the associated right-hand side

f h , v h K = K f h v h dx = K (P K k 2f ) v h dx = K f (P K k 2v h ) dx is exactly computable using the internal moments.



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VEM on quadrilaterals: the case k = 1Recall that if v V K ,1 then

v = 0 in K , v P1(e ) e K , v | K C 0( K )

On parallelograms, V K ,1 = Q1(K ).

And

aK h (u , v ) := a

K

(K 1 u ,

K 1 v ) + S

K

(u K 1 u , v

K 1 v )

Thus, on parallelograms, VEM Bilinear FEM plus quadrature



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An alternative point of view

Actually, we may as well dene: V K ,1 := span { 1, 2, 3, 4} with

i |e P1(e ) e K ; i (x j ) = ij ;P 1(K ) span{ 1, 2, 3, 4}.



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An alternative point of view

Actually, we may as well dene: V K ,1 := span { 1, 2, 3, 4} with

i |e P1(e ) e K ; i (x j ) = ij ;P 1(K ) span{ 1, 2, 3, 4}.

And, with respect to this basis, aK h

is given by

aK h ( i , j ) = K i j dx = 1K d i d j + ( 1)i + j T i T j K 2 K where

K = K | ( 1 + 3)|2 dx A. Cangiani (University of Leicester) Virtual Element Method (VEM) EFEF 2012 17 / 21


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Method robustness

10-4

10-2

100

102

104

10-2

10 -1

100

L2 error

H1 error

The values = 2 / 3 (dashed line) and = 2 (dotted line) are shown.


l


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Examples

5-points 9-points

Laplace Laplace

Exact Q1 Hourglass

All these result from applying quadrature to the Q1FE and from the VEM

bilinear form using different spaces.A. Cangiani (University of Leicester) Virtual Element Method (VEM) EFEF 2012 20 / 21


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