A posteriori error estimates Part II Complementary estimates Tom - PowerPoint PPT Presentation

A posteriori error estimates Part II – Complementary estimates Tom´ aˇ s Vejchodsk´ y vejchod@math.cas.cz Institute of Mathematics, Academy of Sciences ˇ Zitn´ a 25, 115 67 Praha 1 Czech Republic E M H A A T T I C M S f o E T U s e T c I n c T e i l b S i c u N S p f e I o R y h m c e e z d C a c A May 29, 2012, Technical University of Ostrava

T H E M A T M A I C S f o E T U s e T c Outline I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A ◮ Toy problem ◮ Derivation of complementary estimates ◮ Two options ◮ (A) Error majorant ◮ (B) Dual finite elements ◮ Energy minimization ◮ Method of hypercircle ◮ Numerical examples ◮ Conclusions

T H E M A T M A I C S f o E T U s e T c Toy problem I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A Classical formulation: − ∆ u = f in Ω , u = 0 on ∂ Ω Weak formulation: V = H 1 0 (Ω) u ∈ V : a ( u , v ) = F ( v ) ∀ v ∈ V Notation: ◮ a ( u , v ) = ( ∇ u , ∇ v ) ◮ F ( v ) = ( f , v ) � ◮ ( ϕ, ψ ) = ϕψ d x Ω ◮ Error: e = u − u h | 2 = a ( e , e ) = ( ∇ e , ∇ e ) = �∇ e � 2 ◮ Energy norm: | | | e | | 0

T H E M A T M A I C S f o E T U s e T c Derivation I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A Divergence thm.: (div y , v ) + ( y , ∇ v ) = 0 ∀ y ∈ H (div , Ω) , v ∈ V Friedrichs’ inequality: � v � 0 ≤ C F | | | v | | | ∀ v ∈ V Theorem: Let u h ∈ V be arbitrary then | | | u − u h | | | ≤ η ( u h , y ) η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) Proof: v ∈ V a ( u − u h , v ) = ( f , v ) − ( ∇ u h , ∇ v ) = ( f + div y , v ) + ( y − ∇ u h , ∇ v ) ≤ � f + div y � 0 � v � 0 + � y − ∇ u h � 0 �∇ v � 0 ≤ ( C F � f + div y � 0 + � y − ∇ u h � 0 ) | | | v | | | Set v = u − u h . �

T H E M A T M A I C S f o E T U s e T c Derivation I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A Divergence thm.: (div y , v ) + ( y , ∇ v ) = 0 ∀ y ∈ H (div , Ω) , v ∈ V Friedrichs’ inequality: � v � 0 ≤ C F | | | v | | | ∀ v ∈ V Theorem: Let u h ∈ V be arbitrary then | | | u − u h | | | ≤ η ( u h , y ) η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) Lemma: Let u ∈ V be the exact solution. Then | | | u − u h | | | = η ( u h , ∇ u ).

T H E M A T M A I C S f o E T U s e T c Two options I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A (A) Error majorant | | | u − u h | | | ≤ η ( u h , y ) η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) [S. Repin et al., 2000–] (B) Dual finite elements | | | u − u h | | | ≤ � η ( u h , y ) η ( u h , y ) = � y − ∇ u h � 0 � ∀ y ∈ Q ( f ) Q ( f ) = { y ∈ H (div , Ω) : f + div y = 0 } [J. Haslinger, I. Hlav´ aˇ cek, M. Kˇ r´ ıˇ zek, 1970s–80s]

T H E M A T M A I C S f o E T U s e T c (A) Error majorant I n c T e i l S i b c u N S p f e I o R y m h c e e d z C a c A | | | u − u h | | | ≤ η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) � � 1 + 1 | 2 ≤ � η 2 ( u h , y , β ) = C 2 F � f + div y � 2 0 +(1 + β ) � y − ∇ u h � 2 | | | u − u h | | 0 β ∀ β > 0 � � 1 + 1 Proof: ( A + B ) 2 ≤ A 2 + (1 + β ) B 2 ∀ β > 0 β Equality for β = A / B . �

T H E M A T M A I C S f o E T U s e T c (A) Error majorant I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A | | | u − u h | | | ≤ η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) � � 1 + 1 | 2 ≤ � η 2 ( u h , y , β ) = C 2 F � f + div y � 2 0 +(1 + β ) � y − ∇ u h � 2 | | | u − u h | | 0 β ∀ β > 0 Notation: W = H (div , Ω) Complementary problem (equivalent formulations): (i) Find y ∈ W : η ( u h , y ) ≤ η ( u h , w ) ∀ w ∈ W η 2 ( u h , w , � η 2 ( u h , y , β ) ≤ � (ii) Find y ∈ W and β > 0 : � β ) ∀ w ∈ W , � β > 0 If β > 0 fixed: η 2 ( u h , y , β ) ≤ � η 2 ( u h , w , β ) (iii) Find y ∈ W : � ∀ w ∈ W (iv) Find y ∈ W : (div y , div w ) + β ( y , w ) = β ( ∇ u h , w ) − ( f , div w ) C 2 C 2 F F ∀ w ∈ W

T H E M A T M A I C S f o E T U s e T c (A) Error majorant I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A | | | u − u h | | | ≤ η ( u h , y ) = C F � f + div y � 0 + � y − ∇ u h � 0 ∀ y ∈ H (div , Ω) � � 1 + 1 | 2 ≤ � η 2 ( u h , y , β ) = C 2 F � f + div y � 2 0 +(1 + β ) � y − ∇ u h � 2 | | | u − u h | | 0 β ∀ β > 0 Notation: W = H (div , Ω) Practical implementation: ◮ W h ⊂ W , dim W h < ∞ e.g. Raviart-Thomas elements of degree p : W p h = { w h ∈ H (div , Ω) : w h | K ∈ P p ( K ) ∀ K ∈ T h } ◮ Set values for β and C F ◮ Find y h ∈ W h : (div y h , div w h ) + β ( y h , w h ) = β ( ∇ u h , w h ) − ( f , div w h ) C 2 C 2 F F ∀ w h ∈ W h ◮ Compute η ( u h , y h )

T H E M A T M A I C S f o E T U s e T c Friedrichs’ constant C F I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A � v � 0 ≤ C F | | | v | | | ∀ v ∈ V (a) Analytical estimate (Mikhlin, 1986): V = H 1 0 (Ω) � 1 � − 1 / 2 C F ≤ 1 1 | a 1 | + · · · + , Ω ⊂ a 1 × · · · × a d , π | a d |

T H E M A T M A I C S f o E T U e s T c Friedrichs’ constant C F I n c T e l i S i b c u N S p f e I o R y m h c e e d z C a c A � v � 0 ≤ C F | | | v | | | ∀ v ∈ V (b) Numerical upper bound: | 2 � v � 0 | | | v | | C 2 C F = sup ⇔ λ 1 = inf , F = 1 /λ 1 � v � 2 | | | v | | | v ∈ V v ∈ V 0 Eigenvalue problem: u i ∈ V : a ( u i , v ) = λ i ( u i , v ) ∀ v ∈ V ∀ v h ∈ V h Galerkin approxim.: u h i ∈ V h : a ( u h i , v h ) = λ h i ( u h i , v h ) V h ⊂ V | 2 | | | v h | | ⇒ λ h λ 1 ≤ λ h 1 /λ h 1 ≤ C 2 1 = inf ⇒ ⇒ 1 � v h � 2 F v h ∈ V h 0 h h C 2 Sigillito, Kuttler (1970s): λ 1 ≤ λ 1 ⇒ F ≤ 1 /λ 1

T H E M A T M A I C S f o E T U s e T c (B) Dual finite elements I n c T e i l S i b c u N S p f e I o R y m h c e e d C z a c A | | | u − u h | | | ≤ � η ( u h , y ) = � y − ∇ u h � 0 ∀ y ∈ Q ( f ) Q ( f ) = { y ∈ H (div , Ω) : f + div y = 0 } Complementary problem: (i) Find y ∈ Q ( f ) : � η ( u h , y ) ≤ � η ( u h , w ) ∀ w ∈ Q ( f ) 1 2 � y � 2 0 ≤ 1 2 � w � 2 (ii) Find y ∈ Q ( f ) : ∀ w ∈ Q ( f ) 0 ∀ w 0 ∈ Q (0) (iii) Find y ∈ Q ( f ) : ( y , w 0 ) = 0

T H E M A T M A I C S f o E T U s e T c (B) Dual finite elements I n c T e i l S i b c u N S p f e I o R y m h c e e d z C a c A | | | u − u h | | | ≤ � η ( u h , y ) = � y − ∇ u h � 0 ∀ y ∈ Q ( f ) Q ( f ) = { y ∈ H (div , Ω) : f + div y = 0 } Complementary problem: (i) Find y ∈ Q ( f ) : � η ( u h , y ) ≤ � η ( u h , w ) ∀ w ∈ Q ( f ) 1 2 � y � 2 0 ≤ 1 2 � w � 2 (ii) Find y ∈ Q ( f ) : ∀ w ∈ Q ( f ) 0 ∀ w 0 ∈ Q (0) (iii) Find y ∈ Q ( f ) : ( y , w 0 ) = 0 Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) � y − ∇ u h � 2 0 ≤ � w − ∇ u h � 2 0 � y � 2 0 − 2( y , ∇ u h ) + �∇ u h � 2 0 ≤ � w � 2 0 − 2( w , ∇ u h ) + �∇ u h � 2 0

T H E M A T M A I C S f o E T U s e T c (B) Dual finite elements I n c T e i l S i b c u N S p f e I o R y m h c e e d z C a c A | | | u − u h | | | ≤ � η ( u h , y ) = � y − ∇ u h � 0 ∀ y ∈ Q ( f ) Q ( f ) = { y ∈ H (div , Ω) : f + div y = 0 } Complementary problem: (i) Find y ∈ Q ( f ) : � η ( u h , y ) ≤ � η ( u h , w ) ∀ w ∈ Q ( f ) 1 2 � y � 2 0 ≤ 1 2 � w � 2 (ii) Find y ∈ Q ( f ) : ∀ w ∈ Q ( f ) 0 ∀ w 0 ∈ Q (0) (iii) Find y ∈ Q ( f ) : ( y , w 0 ) = 0 Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) � y − ∇ u h � 2 0 ≤ � w − ∇ u h � 2 0 � y � 2 ≤ � w � 2 0 − 2( y , ∇ u h ) 0 − 2( w , ∇ u h )

T H E M A T M A I C S f o E T U s e T c (B) Dual finite elements I n c T e i l S i b c u N S p f e I o R y m h c e e d z C a c A | | | u − u h | | | ≤ � η ( u h , y ) = � y − ∇ u h � 0 ∀ y ∈ Q ( f ) Q ( f ) = { y ∈ H (div , Ω) : f + div y = 0 } Complementary problem: (i) Find y ∈ Q ( f ) : � η ( u h , y ) ≤ � η ( u h , w ) ∀ w ∈ Q ( f ) 1 2 � y � 2 0 ≤ 1 2 � w � 2 (ii) Find y ∈ Q ( f ) : ∀ w ∈ Q ( f ) 0 ∀ w 0 ∈ Q (0) (iii) Find y ∈ Q ( f ) : ( y , w 0 ) = 0 Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) � y − ∇ u h � 2 0 ≤ � w − ∇ u h � 2 0 � y � 2 ≤ � w � 2 0 − 2( f , u h ) 0 − 2( f , u h )