The optimality of IPG methods for odd degrees of polynomial - PowerPoint PPT Presentation

The optimality of IPG methods for odd degrees of polynomial approximation Oto Havle Vít Dolejší Charles University Prague Workshop Dresden-Prague on Numerical Analysis 2010 1 / 21

Contents Introduction 1 Main result and numerical evidence 2 3 Sketch of the proof Conclusion 4 2 / 21

Contents Introduction 1 Main result and numerical evidence 2 3 Sketch of the proof Conclusion 4 3 / 21

DGFEM for the elliptic model problem Interior Penalty Discontinuous Finite Element Galerkin discretization. u ∈ H 2 (Ω) : − ∆ u = f in Ω + boundary conditions ↓ ↓ u h ∈ S hp : B h ( u h , v h ) = L h ( v h ) for all v h ∈ S hp Test and trial space S hp of piecewise polynomial functions of degree at most p ≥ 1. 4 / 21

IIPG, NIPG, SIPG; Interior penalty � � B h ( u , v ) = ∇ u · ∇ v d x K K ∈ T h � � � � − � u � [ v ] + θ � v � [ u ] dS Γ Γ ∈ F h c W � � + [ u ] [ v ] dS H Γ Γ Γ ∈ F h where [ v ] denotes the jump and � v � the average value of v on the mesh face Γ , and  1 , SIPG (symmetric variant)   θ = 0 , IIPG (incomplete variant)  − 1 , NIPG (nonsymetric variant)  5 / 21

IIPG, NIPG, SIPG; Interior penalty � � B h ( u , v ) = ∇ u · ∇ v d x K K ∈ T h � � � � − � u � [ v ] + θ � v � [ u ] dS Γ Γ ∈ F h c W � � + [ u ] [ v ] dS H Γ Γ Γ ∈ F h where [ v ] denotes the jump and � v � the average value of v on the mesh face Γ , and H Γ = h L + h R , max ( h L , h R ) , diam (Γ) etc. 2 c W > 0 , large enough 5 / 21

What is the optimal convergence behavior? A priori error estimate in a given norm � · � � u − u h � = O ( h k ) Best possible approximation in the trial space S hp � u − v h � = O ( h ℓ ) inf v h ∈S hp The error estimate is optimal if k = ℓ . For example: � · � H 1 (Ω , T h ) � · � L 2 (Ω) DGFEM - SIPG p p + 1 DGFEM - NIPG, IIPG p p classical FEM p p + 1 L 2 -projection p p + 1 6 / 21

Facts about L 2 -convergence of nonsymetric DGFEM Numerical experiments Optimal O ( h p + 1 ) for odd p , suboptimal O ( h p ) for even p . Suboptimal behavior possible for odd p on special nonuniform meshes [Guzmán and Riviére, 2009]. General theoretical results Suboptimal estimate O ( h p ) follows from discrete Poincaré-Friedrichs inequality. Optimal estimates can be proven for overpenalized DGFEM. 7 / 21

Facts about L 2 -convergence of nonsymetric DGFEM (contd.) The optimal L 2 convergence has been proved for: One-dimensional NIPG/IIPG with odd p and uniform meshes. [Larson, Niklasson, 2004] 2D NIPG, linear triangles, special assumptions on the mesh. [Burmann, Stamm, 2008] 2D/3D IIPG/NIPG, bilinear/trilinear elements on uniform Cartesian meshes. [Wang, Wang, Sun, Wheeler 2009] Our results: Optimal error estimate for one-dimensional IIPG with odd p , nonuniform meshes, special assumption on the penalty parameter. 8 / 21

Contents Introduction 1 Main result and numerical evidence 2 3 Sketch of the proof Conclusion 4 9 / 21

1D model problem Continuous problem − u ′′ = f u ′ ( 0 ) = g N in ( 0 , 1 ) , u ( 1 ) = u D locally quasi-uniform mesh 1 h k h k = x k + 1 − x k , ≤ ≤ C H . C H h k + 1 IIPG discretization � x k + 1 N − 1 N � � u ′ h v ′ u ′ � � B h ( u h , v h ) = h d x − k [ v h ] k h x k k = 0 k = 1 N c W � + [ u h ] k [ v h ] k H k k = 1 � 1 fv h d x − g N v h ( 0 ) + c W L h ( v h ) = u D v h ( 1 ) H N 0 10 / 21

Assumptions The interior and boundary penalization term N c W � J h ( u h , v h ) = [ u h ] k [ v h ] k H k k = 1 depends on parameters { c W , H k : k = 1 , . . . , N } . H k is given by H k = H ( h k − 1 , h k ) for k < N , H N = H ( h N − 1 , h N − 1 ) H ( · , · ) is a continuous function and H ( a , b ) > 0 , H ( a , b ) = H ( b , a ) , H ( κ a , κ b ) = κ H ( a , b ) c W ≥ c ⋆ W 11 / 21

The main result Then two following assertions are equivalent. (A) There exists a constant C E > 0 such that � u h − u � L 2 ( 0 , 1 ) ≤ C E h p + 1 where u is the weak solution. (B) The degree of approximation p is an odd number and the function H is a multiple of a p + 1 − b p + 1  , a � = b   a p − b p   H p ( a , b ) =   p + 1 p a , a = b   12 / 21

Numerical experiments Continuous problem: − u ′′ ( x ) = x 10 for x ∈ ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0 Computational meshes: uniform, h k = h = 1 / N non-uniform, N is a multiple of 3 3 3 3 h 3 i = 16 N , h 3 i + 1 = 5 · 16 N , h 3 i + 2 = 10 · 16 N various choices of H ( · , · ) 13 / 21

Results on uniform meshes p= 1 NIPG IIPG SIPG N � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC 12288 1.087E-09 2.000 2.207E-10 2.000 1.319E-10 1.999 24576 2.717E-10 2.000 5.518E-11 2.000 3.297E-11 2.000 49152 6.792E-11 2.000 1.380E-11 2.000 8.244E-12 2.000 p= 2 12288 3.407E-10 2.000 1.278E-11 2.001 5.863E-15 3.000 24576 8.516E-11 2.000 3.194E-12 2.000 7.329E-16 3.000 49152 2.129E-11 2.000 7.985E-13 2.000 9.161E-17 3.000 p= 3 12288 1.554E-18 4.001 4.372E-19 4.000 3.178E-19 4.000 24576 9.711E-20 4.000 2.732E-20 4.000 1.987E-20 4.000 49152 6.069E-21 4.000 1.708E-21 4.000 1.242E-21 4.000 p= 4 12288 3.169E-19 4.000 1.288E-20 4.001 9.840E-24 5.000 24576 1.981E-20 4.000 8.048E-22 4.000 3.075E-25 5.000 49152 1.238E-21 4.000 5.030E-23 4.000 9.610E-27 5.000 14 / 21

Results on nonuniform meshes, H ( a , b ) = max ( a , b ) p= 1 NIPG IIPG SIPG N � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC 12288 2.082E-07 0.974 2.076E-08 1.020 3.360E-10 1.999 24576 1.050E-07 0.987 1.031E-08 1.010 8.403E-11 2.000 49152 5.275E-08 0.994 5.136E-09 1.005 2.101E-11 2.000 p= 2 12288 3.283E-10 2.003 1.680E-11 2.002 3.501E-14 3.000 24576 8.198E-11 2.002 4.196E-12 2.001 4.375E-15 3.000 49152 2.048E-11 2.001 1.049E-12 2.001 5.469E-16 3.000 p= 3 12288 6.677E-15 2.993 3.040E-17 3.088 3.023E-18 4.000 24576 8.366E-16 2.997 3.694E-18 3.040 1.889E-19 4.000 49152 1.047E-16 2.998 4.557E-19 3.019 1.181E-20 4.000 p= 4 12288 9.695E-19 4.004 5.214E-20 4.002 1.903E-22 5.000 24576 6.052E-20 4.002 3.256E-21 4.001 5.946E-24 5.000 49152 3.780E-21 4.001 2.034E-22 4.001 1.858E-25 5.000 15 / 21

Results on nonuniform meshes, H ( a , b ) = H p ( a , b ) p= 1 NIPG IIPG SIPG N � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC � e h � L 2 (Ω) EOC 12288 3.178E-07 0.977 9.167E-10 1.999 3.274E-10 1.999 24576 1.602E-07 0.989 2.292E-10 2.000 8.189E-11 1.999 49152 8.040E-08 0.994 5.732E-11 2.000 2.048E-11 2.000 p= 2 12288 3.380E-10 2.004 1.837E-11 2.004 3.479E-14 3.000 24576 8.440E-11 2.002 4.587E-12 2.002 4.349E-15 3.000 49152 2.109E-11 2.001 1.146E-12 2.001 5.437E-16 3.000 p= 3 12288 6.755E-15 2.993 3.993E-18 4.000 3.020E-18 4.000 24576 8.465E-16 2.997 2.496E-19 4.000 1.887E-19 4.000 49152 1.059E-16 2.998 1.560E-20 4.000 1.180E-20 4.000 p= 4 12288 9.714E-19 4.004 5.306E-20 4.003 1.902E-22 5.000 24576 6.063E-20 4.002 3.313E-21 4.001 5.944E-24 5.000 49152 3.787E-21 4.001 2.069E-22 4.001 1.857E-25 5.000 16 / 21

Contents Introduction 1 Main result and numerical evidence 2 3 Sketch of the proof Conclusion 4 17 / 21

The proof of B = ⇒ A ⇒ � u h − u � L 2 ≤ Ch p + 1 p odd , H = H p = Step 1 The Aubin-Nitsche duality trick. − ψ ′′ = u h − u , ψ ′ ( 0 ) = ψ ( 1 ) = 0 N � � u h − u � 2 ψ ′ ( x k ) [ u h − u ] k L 2 ( 0 , 1 ) = B h ( u h − u , ψ ) − k = 1 18 / 21

The proof of B = ⇒ A ⇒ � u h − u � L 2 ≤ Ch p + 1 p odd , H = H p = Step 1 The Aubin-Nitsche duality trick. Step 2 Identify the leading part of the jump [ u h ] k . c W [ u h ] k = B h ( u h , w ⋆ h , p , k ) = L h ( w ⋆ h , p , k ) H k � 1 f w ⋆ = h , p , k d x (interior nodes) 0 f ( p − 1 ) ( x k ) + O ( h p + 1 ( − 1 ) p − 1 h p k − 1 − h p � � = K p ) k k h p k − 1 − h p f ( p − 1 ) ( x k ) + O ( h p + 1 � � = K p ) k k 18 / 21

The proof of B = ⇒ A ⇒ � u h − u � L 2 ≤ Ch p + 1 p odd , H = H p = Step 1 The Aubin-Nitsche duality trick. Step 2 Identify the leading part of the jump [ u h ] k . Step 3 Summation by parts. N N − 1 � � h p + 1 k − 1 − h p + 1 � � f ( p − 1 ) ( x k ) ψ ′ ( x k ) [ u h − u ] k ≈ ψ ′ ( x k ) k k = 1 k = 1 N − 2 ψ ′ f ( p − 1 ) � x k + 1 � h p + 1 � ≈ k x k k = 1 � Ch p + 1 � f � H p ( 0 , 1 ) � ψ � H 2 ( 0 , 1 ) 18 / 21

The proof of A = ⇒ B � u h − u � L 2 ≤ Ch p + 1 = ⇒ p odd , H = H p Step 1 Perturbation of the penalization parameter c W . � u c W 1 − u c W 2 � L 2 ≤ � u c W 1 − u � L 2 + � u c W 2 − u � L 2 ≤ Ch p + 1 h h h h The difference u c W 1 − u c W 2 constant on each element h h N � 1 − c W 1 � � � u c W 1 − u c W 2 � u C W 1 � ( x ℓ , x ℓ + 1 ) = � h h h c W 2 k k = ℓ + 1 We already have an expression for [ u h ] k . We can estimate � u c W 1 − u c W 2 � L 2 for particular meshes and h h right hand side function f . 19 / 21