Serendipity Virtual Element Methods L. Beir˜ ao da Veiga, F. Brezzi, L.D. Marini, A. Russo IMATI-C.N.R., Pavia, Italy Polytopal Element Methods in Mathematics and Engineering Atlanta, October 26-th, 2015 Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 1 / 31
Outline Origins 1 2 Basic idea of VEMs General VEMs in 2 and 3 dimensions 3 4 VEMs and FEMs Serendipity VEM 5 The reduced degrees of freedom The operator D S The operator R S The Serendipity VEM spaces S-VEM, FEM, and S-FEM 6 Numerical results Conclusions 7 Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 2 / 31
In the beginning...MFD Assume that: E = pentagon and we want P 2 -accuracy. ~ E ~ point value average We consider 11 degrees of freedom (values at vertexes and midpoints, plus the average on E ). For every E ) we define D v ∈ R 11 to be the nodal values v ∈ C 0 ( ¯ � � and the average of v : D v = ( Dv ) 1 , ..., ( Dv ) 10 , ( Dv ) 11 Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 3 / 31
MFD-2 Our (local) discrete space will be R 11 For P and Q in R 11 we want [ P , Q ] E . to mimic, say � ~ ∇ p · ∇ q d E E If p is ( any ) function p.w. P 2 on ∂ E , with D p = P and q 2 a polynomial of degree ≤ 2 such that D q 2 = Q then � � � p ∂ q 2 � ∇ p ·∇ q 2 = − p ∆ q 2 + ∂ n ≃− P 11 ∆ q 2 + Simpson E E ∂ E E can be computed without knowing p (but only P = D p ). Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 4 / 31
MFD-3 According to the previous slide, we can compute � [ D p , D q ] E ≃ ∇ p · ∇ q d E E using only the values of D p and D q in R 11 whenever either p or q is a polynomial of degree ≤ 2. This obviously defines an 6 × 11 matrix R , from R 11 to R 6 that gives the values of [ P , D ( q 2 )] E for P ∈ R 11 and q 2 ∈ P 2 ∼ R 6 . Then we can define a sort of scalar product in R 11 , for all P and all Q in R 11 , as [ P , Q ] E :=( R P ) T · ( R Q )+ Stabilization (on the kernel of R ) Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 5 / 31
On the other side...FdV-based methods “Darcy problem”: u = −∇ p , div u = f + B.C. From now on: p = 0 on the boundary. Following B. M. Fraeijs de Veubeke ( 1965 ) we observe that: given an approximation λ h of p at the inter-element boundaries, and another approximation p h of p inside each element we can deduce an approximation u h of u by requiring, in each element E � � � u h · v = p h div v + λ h ( v · n ) for all v . E E ∂ E This defines an approximate gradient G h : ( λ h , p h ) → u h = G h ( λ h , p h ) Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 6 / 31
Use of FdV Darcy problem: u = −∇ p , div u = f , p = 0 at ∂ Ω � � � u h · v = p h div v + λ h ( v · n ) for all v . E E ∂ E You can then add a discretized conservation equation � � div u h q = f q for all q . E E Then you must close the system , with as many equations as there are λ h ’s , requiring “continuity” (for u h · n or for p h , or for a combination of the two) at the interelement boundaries. You can get zillions of methods. Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 7 / 31
Approximate Gradient methods Remember that given a λ h at the inter-element edges (or faces), and a p h inside each element, we can define an approximate gradient G h : ( λ h , p h ) → u h = G h ( λ h , p h ) by requiring, in each element E � � � G h · v = p h div v + λ h ( v · n ) for all v E E ∂ E where v ranges over the space where you look for G h . Then you look for a pair ( λ h , p h ) such that � � � G h ( λ h , p h ) · G h ( µ h , q h ) = f q h for all ( λ h , q h ) . E Ω E Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 8 / 31
Approximate Gradient methods-2 Actually, you also need to add a stabilizing term like � � G h ( λ h , p h ) · G h ( µ h , q h ) E E � � + C h − 1 � ( λ h − p h )( µ h − q h ) = f q h , ∀ µ h ∀ q h . Ω e e Various methods distinguish themselves for the space where you look for the approximate gradient and for the type of stabilization used . You can get zillions of methods Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 9 / 31
Basic idea of VEMs ~ E point value ~ average We take again 11 degrees of freedom (values at vertexes and midpoints, + the average on E ). We define the space V E := { v | such that v | e ∈ P 2 ( e ) ∀ edge e , and ∆ v ∈ P 0 ( E ) } It is easy to see that our 11 d.o.f.s are V E - unisolvent . Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 10 / 31
Basic idea of VEMs-2 To every v ∈ V E we associate Π E 2 v ∈ P 2 ( E ) defined by � ∇ (Π E 2 v − v ) · ∇ q 2 = 0 for all q 2 ∈ P 2 ( E ) E � (Π E 2 v − v ) d E = 0 . E Note that the quantity (= right-hand side) v ∂ q 2 � � � ∇ v · ∇ q 2 = − ∆ q 2 v + ∂ n E E ∂ E is computable (from the dofs) ∀ v ∈ V E and ∀ q 2 ∈ P 2 . Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 11 / 31
Basic idea of VEMs-3 Discretized problem: find p h ∈ V such that � � � � ∇ Π E 2 p h ·∇ Π E f Π E 2 q h + S ( p h , q h ) = 2 q h ∀ q h ∈ V E E E E where the stabilizing term S ( p h , q h ) can be taken as � T � � � � D ( q h − Π E D ( p h − Π E S ( p h , q h ) := 2 q h ) · 2 p h ) E where D is the degrees of freedom vector defined before. Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 12 / 31
Panorama In summary: AD-HOC functions (e.g. Rational, Baricentric,...); one field and numerical integration DG: One field, discontinuous MFD: Only dofs, no functions HDG WG HHO: Two/three polynomial fields ( λ, p , ( u )) VEM: One field, solution of a PDE. Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 13 / 31
Original VEM-General order For k integer ≥ 1 we define V k ( E )= { ϕ ∈ C 0 ( ¯ E ): ϕ | e ∈ P k ( e ) ∀ edge e , ∆ ϕ ∈ P k − 2 ( E ) } . The degrees of freedom in V k ( E ) are taken as • the values of ϕ at the vertices , � • ϕ q d s for all q ∈ P k − 2 ( e ) ∀ edge e , e � • ϕ q d E for all q ∈ P k − 2 ( E ) . E It is immediate to verify that the degrees of freedom are unisolvent Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 14 / 31
Classical VEM-General case For k integer ≥ 1 and k ∆ integer ≥ − 1 we define V k , k ∆ ( E )= { ϕ ∈ C 0 ( ¯ E ): ϕ | e ∈ P k ( e ) ∀ edge e , ∆ ϕ ∈ P k ∆ ( E ) } . The degrees of freedom in V k ( E ) are taken as • the values of ϕ at the vertices , � • ϕ q d s for all q ∈ P k − 2 ( e ) ∀ edge e , e � • ϕ q d E for all q ∈ P k ∆ ( E ) . E It is immediate to verify that the degrees of freedom are unisolvent. In general it is better to take k ∆ ≥ k − 2. Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 15 / 31
Classical VEM-General case -3D For k integer ≥ 1,and k ∆ , k f integers ≥ − 1 we define V k , k f , k ∆ ( E )= { ϕ ∈ C 0 ( ¯ E ): ϕ | f ∈ V k , k f ∀ face f , ∆ ϕ ∈ P k ∆ ( E ) } The following degrees of freedom are unisolvent • the values of ϕ at the vertices , � • ϕ q d s for all q ∈ P k − 2 ( e ) ∀ edge e , e � • ϕ q d s for all q ∈ P k f ( f ) ∀ face f , f � • ϕ q d E for all q ∈ P k ∆ ( E ) . E Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 16 / 31
VEM and FEM on triangles: degrees of freedom FEM k=1 FEM k=3 FEM k=2 VEM k=1 VEM k=2 VEM k=3 Figure: Triangles: Classical FEM and Original VEM Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 17 / 31
VEM and FEM on quads: degrees of freedom FEM k=1 FEM k=3 FEM k=2 VEM k=1 VEM k=2 VEM k=3 Figure: Quads: Classical FEM and Original VEM Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 18 / 31
The reduced dofs We choose an S with ( k + 1)( k + 2) / 2 ≤ S ≤ N E ( N E =dimensiion ov V k , k ), and assume that the degrees of freedom of V k , k are ordered in such a way that the first S δ 1 , δ 2 , ... δ S have the following properties: • ( B ) They include all the boundary dofs • ( S ) ∀ p k ∈ P k ( E ) we have: { δ 1 ( p k ) = δ 2 ( p k ) = ... = δ S ( p k ) = 0 } ⇒ { p k ≡ 0 } Property ( B ) is there to ensure conformity , and is easy. Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 19 / 31
The reduced dofs - Examples Once ( B ) is satisfied, you know that a polynomial p k that satisfies { δ 1 ( p k ) = δ 2 ( p k ) = ... = δ S ( p k ) = 0 } must be identically zero on the boundary. Let’s deal with ( S ) . To get Property ( S ), on top of the bounday dofs: on a triangle , you must include as many internal dofs as the dimension of P k − 3 , on a square , you must include as many internal dofs as the dimension of P k − 4 , on a regular n-gon you must include as many internal dofs as the dimension of P k − n . In general (even on very distorted polygons): you must have as many internal dofs as there are P k -bubbles . Beir˜ ao da Veiga, Brezzi, Marini, Russo (vv) VEM October 26-th, 2015 20 / 31
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