Robust Preconditioning in Elasticity Joachim Sch oberl Center for - PowerPoint PPT Presentation

Robust Preconditioning in Elasticity Joachim Sch¨ oberl Center for Computational Engineering Sciences (CCES) RWTH Aachen University Germany DD17, Strobl, 2006, July 3-7 Joachim Sch¨ oberl Page 1

System of PDEs Linear elasticity: � A ( u, v ) = µ ε ( u ) : ε ( v ) + λ div u div v dx displacement u ∈ [ H 1 0 ,D ] d , strain operator ε ( u ) := 0 . 5( ∇ u + ( ∇ u ) T ) Lam´ e parameters µ, λ . Timoshenko beam model: � 1 � 1 β ′ δ ′ dx + t − 2 ( w ′ − β )( v ′ − δ ) dx A ( w, β ; v, δ ) = 0 0 t vertical displacement w , rotation β , thickness t , w β In principle the same as a scalar PDE Joachim Sch¨ oberl Parameter Dependent Problems Page 2

System of PDEs Linear elasticity: � A ( u, v ) = µ µ ε ( u ) : ε ( v ) + λ λ div u div v dx Nearly incompressible materials: λ ≫ µ Timoshenko beam model: � 1 � 1 β ′ δ ′ dx + t − 2 ( w ′ − β )( v ′ − δ ) dx A ( w, β ; v, δ ) = 0 0 Thin beam: t ≪ 1 In principle the same as a scalar PDE but dependency on parameters Joachim Sch¨ oberl Parameter Dependent Problems Page 2

Parameter Dependent Problems [Arnold 81] Find u ∈ V : A ε ( u, v ) = f ( v ) ∀ v ∈ V with A ε ( u, v ) = a ( u, v ) + 1 ε c (Λ u, Λ v ) small parameter: ε ∈ (0 , 1] symmetric bilinear form: a ( u, u ) ≥ 0 ∀ u ∈ V Hilbert space: ( Q, c ( ., . )) operator: Λ : V → Q with kernel: V 0 := kern Λ A 1 ( u, u ) ≃ � u � 2 Well posed for ε = 1 : V Joachim Sch¨ oberl Parameter Dependent Problems Page 3

A priori estimates Univorm V -coercivity:: A ε ( u, u ) ≥ A 1 ( u, u ) � � u � 2 V Non-uniform V -continuity: A ε ( u, u ) ≤ ε − 1 A 1 ( u, u ) � ε − 1 � u � 2 V Non-robust a priori error estimate: � u − u h � V ≤ ε − 1 / 2 inf � u − v h � V v h ∈ V h 0.12 t=1e-1 t=1e-2 t=1e-3 0.1 0.08 Numerical example: Timoshenko beam w(1) 0.06 Vertical load f = 1 , compute w (1) : 0.04 0.02 0 1 10 100 1000 10000 Elements Joachim Sch¨ oberl Parameter Dependent Problems Page 4

Primal FEM with Reduction Operators The primal FEM a ( u h , v h ) + 1 Find u h ∈ V h s.t.: εc (Λ u h , Λ v h ) = f ( v h ) ∀ v h ∈ V h often leads to bad results, knwon as locking phenomena. (One) explanation: This is a penalty approximation to Λ u = 0 , but no FE functions fulfill Λ u h = 0 , i.e. V 0 ∩ V h too small. Joachim Sch¨ oberl Parameter Dependent Problems Page 5

Primal FEM with Reduction Operators The primal FEM a ( u h , v h ) + 1 Find u h ∈ V h s.t.: εc (Λ u h , Λ v h ) = f ( v h ) ∀ v h ∈ V h often leads to bad results, knwon as locking phenomena. (One) explanation: This is a penalty approximation to Λ u = 0 , but no FE functions fulfill Λ u h = 0 , i.e. V 0 ∩ V h too small. Weaken the high energy term by reduction operator R h (reduced integration, B-bar method, mixed method , EAS, ...) a ( u h , v h ) + 1 Find u h ∈ V h s.t.: εc ( R h Λ u h , R h Λ v h ) = f ( v h ) ∀ v h ∈ V h Large enough kernel V h, 0 = kern R h Λ ∩ V h Joachim Sch¨ oberl Parameter Dependent Problems Page 5

Numerical example: Timoshenko beam Vertical load f = 1 , compute w (1) : Conforming FEM: With reduction operator: 0.12 0.12 t=1e-1 t=1e-1 t=1e-2 t=1e-2 t=1e-3 t=1e-3 0.1 0.1 0.08 0.08 w(1) w(1) 0.06 0.06 0.04 0.04 0.02 0.02 0 0 1 10 100 1000 10000 1 10 100 1000 Elements Elements Joachim Sch¨ oberl Parameter Dependent Problems Page 6

Analysis by mixed formulation Primal method: a ( u, v ) + ε − 1 c (Λ u, Λ v ) = f ( v ) Find u ∈ V : ∀ v ∈ V Introduce new variable p = ε − 1 Λ u ∈ Q . a ( u, v ) + c (Λ v, p ) = f ( v ) ∀ v ∈ V c (Λ u, q ) − εc ( p, q ) = 0 ∀ q ∈ Q Joachim Sch¨ oberl Parameter Dependent Problems Page 7

Analysis by mixed formulation Primal method: a ( u, v ) + ε − 1 c (Λ u, Λ v ) = f ( v ) Find u ∈ V : ∀ v ∈ V Introduce new variable p = ε − 1 Λ u ∈ Q . a ( u, v ) + c (Λ v, p ) = f ( v ) ∀ v ∈ V c (Λ u, q ) − εc ( p, q ) = 0 ∀ q ∈ Q Mixed bilinear-from B ( · , · ) : ( V × Q ) × ( V × Q ) → R B (( u, p ) , ( v, q )) = a ( u, v ) + c (Λ u, q ) + c (Λ v, p ) − εc ( p, q ) Mixed problem: Find ( u, p ) ∈ V × Q : B (( u, p ) , ( v, q )) = f ( v ) ∀ ( v, q ) ∈ V × Q Joachim Sch¨ oberl Parameter Dependent Problems Page 7

Well-posed mixed formulation Define norm � . � Q, 0 such that the LBB condition is fulfilled by definition: c (Λ v, p ) � q � Q, 0 := sup � v � V v ∈ V Product space norm � ( v, q ) � 2 V × Q = � v � 2 V + � q � 2 Q, 0 + ε � q � 2 c Then B ( ., . ) is uniformely continuous: B (( u, p ) , ( v, q )) sup sup � 1 � ( u, p ) � V × Q � ( v, q ) � V × Q ( u,p ) ( v,q ) and uniformely inf − sup stable: B (( u, p ) , ( v, q )) ( u,p ) sup inf � 1 � ( u, p ) � V × Q � ( v, q ) � V × Q ( v,q ) Joachim Sch¨ oberl Parameter Dependent Problems Page 8

Example: Nearly incompressible elasticity 0 ,D ] 2 and p ∈ Q = L 2 such that Find u ∈ V = [ H 1 � � � µ ε ( u ) : ε ( v ) dx + div v p dx = f · v dx ∀ v ∈ V λ − 1 � � div u q dx − p q dx = 0 ∀ q ∈ Q The limit problem for λ → ∞ is a Stokes-like problem. Mixed finite element discretization by Stokes-stable (discrete LBB !) element pairs, e.g., V h = { v ∈ V : v | T ∈ P 2 } Q h = { q ∈ Q : q | T ∈ P 0 } . Joachim Sch¨ oberl Parameter Dependent Problems Page 9

Example: Nearly incompressible elasticity 0 ,D ] 2 and p ∈ Q = L 2 such that Find u ∈ V = [ H 1 � � � µ ε ( u ) : ε ( v ) dx + div v p dx = f · v dx ∀ v ∈ V λ − 1 � � div u q dx − p q dx = 0 ∀ q ∈ Q The limit problem for λ → ∞ is a Stokes-like problem. Mixed finite element discretization by Stokes-stable (discrete LBB !) element pairs, e.g., V h = { v ∈ V : v | T ∈ P 2 } Q h = { q ∈ Q : q | T ∈ P 0 } . A priori estimates by stability and approximation: � ( u − v h , p − p h ) � V × Q � h α ( � u � H 1+ α + � p � H α ) � ( u − u h , p − p h ) � V × Q � inf v h ∈ V h ,q h ∈ Q h Joachim Sch¨ oberl Parameter Dependent Problems Page 9

Solvers for linear system Indefinite matrix equation B T � � � � � � A u f = B − εC p 0 • Block Transformation: Inexact Uzawa, SIMPLE, GMRES Axelsson-Vassilevski, Bramble-Pasciak, Langer-Queck, Rusten-Winther, Bank-Welfert-Yserentant, Klawonn, Bramble-Pasciak-Vassilev, Zulehner, Benzi-Golub-Liesen, ... Use (standard) preconditioners for A and for Schur-complement B T A − 1 B + εC . • Multigrid for indefinite problem: Braess-Bl¨ omer, Brenner, Huang, Wittum, Braess-Sarazin, Zulehner, Sch¨ oberl-Zulehner Use special smoothers (squared system, Vanka, SIMPLE) Joachim Sch¨ oberl Parameter Dependent Problems Page 10

Schur complement system Indefinite matrix equation B T � � � � � � A u f = B − εC p 0 Elimination of p from second line leads to the Schur complement system � A + 1 � εB T C − 1 B u = f Cheap if C is (block-)diagonal. Positive definite matrix of smaller dimension, but very ill conditioned for ε → 0 Goal: Design of ε -robust solver Joachim Sch¨ oberl Parameter Dependent Problems Page 11

Elimination of dual variable on the finite element level Finite element system: Find u h ∈ V h and p h ∈ Q h such that a ( u h , v h ) + c (Λ u h , p h ) = f ( v h ) ∀ v h ∈ V h c (Λ u h , q h ) − εc ( p h , q h ) = 0 ∀ q h ∈ Q h Second line defines p h : p h = ε − 1 P c Q h Λ u h Use in first line: a ( u h , v h ) + ε − 1 c ( P c Q h Λ u h , P c Q h Λ p h ) = f ( v h ) ∀ v h ∈ V h Joachim Sch¨ oberl Parameter Dependent Problems Page 12

Elimination of dual variable on the finite element level Finite element system: Find u h ∈ V h and p h ∈ Q h such that a ( u h , v h ) + c (Λ u h , p h ) = f ( v h ) ∀ v h ∈ V h c (Λ u h , q h ) − εc ( p h , q h ) = 0 ∀ q h ∈ Q h Second line defines p h : p h = ε − 1 P c Q h Λ u h Use in first line: a ( u h , v h ) + ε − 1 c ( P c Q h Λ u h , P c Q h Λ p h ) = f ( v h ) ∀ v h ∈ V h Elasticity with reduction operators: � h div v h dx A ε h ( u, v ) = µε ( u ) : ε ( v ) + λ div u Discrete kernel: � V h 0 = { v h ∈ V h : div v h dx = 0 ∀ T ∈ T } T Joachim Sch¨ oberl Parameter Dependent Problems Page 12

Timoshenko beam Conforming bilinear form: � � β ′ δ ′ dx + t − 2 ( w ′ − β )( v ′ − δ ) dx A (( w, β ) , ( v, δ )) = has the kernel V 0 = { ( v, δ ) : δ = v ′ } t → 0 is a penalty approximation to the 4 th -order Bernoulli model A ( w, v ) = f ( v ) with � w ′′ v ′′ dx A ( w, v ) = Reduction of a (stable !) mixed system with w ∈ P 1 , β ∈ P 1 , q ∈ P 0 leads to h ( v ′ h dx � � h dx + t − 2 β ′ h δ ′ A h (( w h , β h ) , ( v h , δ h )) = ( w ′ h − β h ) h − δ h ) Joachim Sch¨ oberl Parameter Dependent Problems Page 13