A two-level enriched finite element method for the Darcy equation - PowerPoint PPT Presentation

A two-level enriched finite element method for the Darcy equation Gabriel R. Barrenechea Department of Mathematics, University of Strathclyde, Scotland in collaboration with: Alejandro Allendes Erwin Hern´ andez Fr´ ed´ eric Valentin Valpara´ ıso, Chile Valpara´ ıso, Chile LNCC, Brazil Scaling up and modeling for transport and flow in porous media Dubrovnik, October 13-16, 2008

Plan of the talk ☞ The Darcy equation and the enrichment strategy. ☞ The semi-discrete method and error estimates. ☞ The two-level method and its analysis. ☞ Numerical results. ☞ Concluding remarks. Teo-level FEM for the Darcy equation Page 1

The Darcy equation and the enrichment strategy The problem statement : Find ( u , p ) such that u + ∇ p = f , ∇· u = g in Ω , u · n = 0 on ∂ Ω , � where Ω g = 0 . (Ω) × L 2 Weak problem : Find ( u , p ) ∈ H div 0 (Ω) such that 0 ∀ ( v , q ) ∈ H div (Ω) × L 2 A (( u , p ) , ( v , q )) = F ( v , q ) 0 (Ω) , 0 where A (( u , p ) , ( v , q )) :=( u , v ) Ω − ( p, ∇· v ) Ω − ( q, ∇· u ) Ω , F ( v , q ) :=( f , v ) Ω − ( g, q ) Ω . Teo-level FEM for the Darcy equation Page 2

The PGEM for the Darcy problem Derivation of the Method : Find u H := u 1 + u e ∈ P 1 (Ω) 2 + H div (Ω) and 0 p H := p 0 + p e ∈ P 0 (Ω) ⊕ L 2 0 ( T H ) such that A (( u 1 + u e , p 0 + p e ) , ( v H , q H )) = F ( v H , q H ) , for all v H := v 1 + v b ∈ P 1 (Ω) 2 ⊕ H div ( T H ), q H = q 0 + q e ∈ P 0 (Ω) ⊕ L 2 0 ( T H ), 0 where ( T H ) := { w ∈ L 2 (Ω) 2 : w | K ∈ H div H div ( K ) ∀ K ∈ T H } , 0 0 L 2 0 ( T H ) := { q ∈ L 2 (Ω) : q | K ∈ L 2 0 ( K ) , ∀ K ∈ T H } . Equivalent system : A (( u 1 + u e , p 0 + p e ) , ( v 1 , q 0 )) = L ( v 1 , q 0 ) ∀ ( v 1 , q 0 ) ∈ V H × Q H , ( u 1 + u e , v b ) K − ( p 0 + p e , ∇ · v b ) K − ( q e , ∇ · ( u 1 + u e )) K = ( f , v b ) K − ( g, q e ) K , for all ( v b , q e ) ∈ H 0 ( div, K ) × L 2 0 ( K ) and all K ∈ T H . Teo-level FEM for the Darcy equation Page 3

Derivation of the Method (continued) Strong problem for ( u e , p e ) : u e + ∇ p e = − u 1 , ∇· u e = C K in K, u e · n = α H F � p 0 � on each F ⊆ ∂K ∩ Ω . In order to make this problem compatible, we set 3 � 1 � C K = � p 0 � . α H F i | K | F i i =1 Teo-level FEM for the Darcy equation Page 4

Derivation of the Method (continued) • Splitting u e = u M e + u D e and p e = p M e + p D e • ( u M e , p M e ) solves u M e + ∇ p M ∇· u M e = − u 1 , e = 0 in K, u M e · n = 0 on ∂K • ( u D e , p D e ) solves u D e + ∇ p D ∇· u D e = 0 in K, e = C K in K, u D e · n = αH F � p 0 � on each F ⊆ ∂K ∩ Ω . Teo-level FEM for the Darcy equation Page 5

Derivation of the Method (continued) Remarks: • u D e is a Raviart-Thomas field. Indeed, there holds � u D e = αH F � p 0 � ϕ F , F ⊆ ∂K ∩ Ω where | K | ϕ F (x) = (x − x F ) . 2 H F Teo-level FEM for the Darcy equation Page 6

Derivation of the Method (continued) Returning to the first equation : For all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω): ( u 1 + u e , v 1 ) Ω − ( p 0 + p e , ∇ · v 1 ) Ω + ( q 0 , ∇ · ( u 1 + u e )) Ω = F ( v 1 , q 0 ) . Remark : • ( p e , ∇· v 1 ) K = 0 for all K ∈ T H , and hence the enrichment of the pressure has no effect on the formulation. , v 1 ) Ω + � • ( u 1 + u e , v 1 ) Ω = ( u 1 + u M K ∈T H ( u D e ( − u 1 ) e ( � p 0 � ) , v 1 ) K ; � �� � = −M K ( u 1 ) • ( q 0 , ∇· u e ) Ω = � e · n , q 0 ) ∂K = � K ∈T H ( u D F ∈E H ( αH F � p 0 � , � q 0 � ) F ; Teo-level FEM for the Darcy equation Page 7

Derivation of the Method (continued) Find ( u 1 , p 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω) such that � � ( u D (( I − M K )( u 1 ) , v 1 ) Ω + e ( � p 0 � ) , v 1 ) K − ( p 0 , ∇ · v 1 ) Ω K ∈T H K ∈T H � − ( q 0 , ∇ · u 1 ) Ω − αH F ( � p 0 � , � q 0 � ) F = F ( v 1 , q 0 ) , F ∈E H for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω). Lemma: The operator M K satisfies ∀ v , w ∈ L 2 ( K ) 2 . ( v − M K ( v ) , M K ( w )) K = 0 Furthermore � ( u D e ( � p 0 � ) , v 1 ) K ≈ O ( H 2 ) , K ∈T H and then this term may be neglected. Teo-level FEM for the Darcy equation Page 8

The semi-discrete problem Find ( u 1 , p 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω) such that B (( u 1 , p 0 ) , ( v 1 , q 0 )) = F ( v 1 , q 0 ) , for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω), where � B (( u 1 , p 0 ) , ( v 1 , q 0 )) := (( I − M K )( u 1 ) , ( I − M K )( v 1 )) K K ∈T H � − ( p 0 , ∇ · v 1 ) Ω − ( q 0 , ∇ · u 1 ) Ω − αH F ( � p 0 � , � q 0 � ) F . F ∈E H Remark : This method is symmetric. Teo-level FEM for the Darcy equation Page 9

The semi-discrete problem Remark : u H has discontinuous tangential component (unlike u 1 ) and it satisfies the following local mass conservation property: � [ ∇ · ( u 1 + u D e ) − g ] = 0 ∀ K ∈ T H . K The same argument may be applied to any jump-based stabilized method for the Darcy equation. Teo-level FEM for the Darcy equation Page 10

The semi-discrete problem Numerical analysis of the semi-discrete problem : Lemma: The bilinear forms B ( ., . ) satisfies � B (( v 1 , q 0 ) , ( v 1 , − q 0 )) = � ( I − M K )( v 1 ) � 2 τ F � � q 0 � � 2 0 , Ω + 0 ,F , F ∈E H for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω). Lemma: There exists C > 0 such that ∀ v 1 ∈ P 1 ( K ) 2 . � v 1 � 0 ,K ≤ C ( � ( I − M K )( v 1 ) � 0 ,K + �∇ · v 1 � 0 ,K ) Teo-level FEM for the Darcy equation Page 11

The semi-discrete problem Mesh-dependent norm : � � ( w , t ) � 2 H = � w � 2 div, Ω + α � t � 2 αH F � � t � � 2 0 , Ω + 0 ,F . F ∈E H Theorem: Let α small enough, then there exists β > 0, independent of H and α , such that B (( v 1 , q 0 ) , ( w 1 , t 0 )) sup ≥ β � ( v 1 , q 0 ) � H , � ( w 1 , t 0 ) � H ( w 1 ,t 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω) −{ 0 } for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω). Teo-level FEM for the Darcy equation Page 12

The semi-discrete problem Theorem: There exists C > 0 such that � ( u − u 1 , p − p 0 ) � H ≤ CH ( � u � 2 , Ω + | p | 1 , Ω ) , � � u − ( u 1 + u D e ) � div, Ω ≤ C H ( � u � 2 , Ω + | p | 1 , Ω . Teo-level FEM for the Darcy equation Page 13

The two-level FEM Remember: To implement the method, M K ( u 1 ) must be computed, i.e., we must solve the local problem u M e + ∇ p M ∇· u M e = u 1 , e = 0 in K, u M e · n = 0 on ∂K Teo-level FEM for the Darcy equation Page 14

The two-level FEM Starting remark : v 1 − M K ( v 1 ) = ∇ p e ( v 1 ) . Then our method may be rewritten in the following equivalent way � ( ∇ p e ( u 1 ) , ∇ p e ( v 1 )) K − ( p 0 , ∇ · v 1 ) Ω − ( q 0 , ∇ · u 1 ) Ω K ∈T H � − αH F ( � p 0 � , � q 0 � ) F = ( f , v 1 ) Ω − ( g, q 0 ) , F ∈E H for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω). Here, p e ( v 1 ) solves − ∆ p e ( v 1 )= −∇ · v 1 in K , ∂ n p e ( v 1 )= v 1 · n on ∂K . Teo-level FEM for the Darcy equation Page 15

The two-level FEM Discrete local problems : Find p h ( v 1 ) ∈ R K h such that � � ∀ ξ h ∈ R K ∇ p h ( v 1 ) · ∇ ξ h = v 1 · ∇ ξ h h , K K where R K h are Lagrangian finite elements of degree l ≥ 1. Two-level method : Find ( u 1 ,h , p 0 ,h ) ∈ P 1 (Ω) 2 × P 0 (Ω) such that: ∀ ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω) , B h (( u 1 ,h , p 0 ,h ) , ( v 1 , q 0 )) = F( v 1 , q 0 ) where � B h (( v 1 , q 0 ) , ( w 1 , t 0 )) := ( ∇ p h ( v 1 ) , ∇ p h ( w 1 )) K − ( q 0 , ∇ · w 1 ) Ω K ∈T H � − ( t 0 , ∇ · v 1 ) Ω − τ F ( � q 0 � , � t 0 � ) F . F ∈E H Teo-level FEM for the Darcy equation Page 16

The two-level FEM Lemma: Let � · � h be the mesh-dependent norm given by � � ( v 1 , q 0 ) � 2 �∇ p h ( v 1 ) � 2 0 ,K + �∇ · v 1 � 2 h := 0 , Ω + K ∈T H � α � q 0 � 2 τ F � � q 0 � � 2 0 , Ω + 0 ,F , F ∈E H and let us suppose that there exists C 0 > 0 such that h ≤ C 0 H K . Then � ( v 1 , q 0 ) � H ≤ C � ( v 1 , q 0 ) � h . Theorem: There exists β 2 > 0 independent of H, h and α such that B h (( v 1 , q 0 ) , ( w 1 , t 0 )) sup ≥ β 2 � ( v 1 , q 0 ) � H , � ( w 1 , t 0 ) � H ( w 1 ,t 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω) for all ( v 1 , q 0 ) ∈ P 1 (Ω) 2 × P 0 (Ω). Teo-level FEM for the Darcy equation Page 17

The two-level FEM Theorem: There exists C > 0 such that hH t | g | t, Ω + ( H + h ) � u � 2 , Ω + H | p | 1 , Ω � � � ( u − u 1 ,h , p − p 0 ,h ) � H ≤ C , for t = 0 , 1. Remark : The condition h ≤ C 0 H means that a fixed mesh may be used for all the elements and all the refinements, hence making the computation cheap. In fact, in all the numerical results, only one P 1 element is used in each element. Teo-level FEM for the Darcy equation Page 18