Numerical Zoom and Domain Decomposition - PowerPoint PPT Presentation

Numerical Zoom and Domain Decomposition http://www.ann.jussieu.fr/pironneau Olivier Pironneau 1 1 University of Paris VI, Laboratoire J.-L. Lions, Olivier.Pironneau@upmc.fr with J.-B. Apoung Kamga, H. Hecht, A. Lozinski Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 1 / 36

The Site of Bure Figure: Schematic view of the Bure project (East of France) Nuclear waste is cooled, processed, then buried safely for 1M years Simulation requires a super computer, or does it really? Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 2 / 36

The COUPLEX I Test Case Figure: A 2D multilayered geometry 20km long, 500m high with permeability variations K + K − = O ( 10 9 ) . Hydrostatic pressure by a FEM. H or ∂ H ∇ · ( K ∇ H ) = 0 , ∂ n given on Γ Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 3 / 36

COUPLEX I : Concentration of Radio-Nucleides Figure: Concentration at 4 times with Discontinuous Galerkin FEM (Apoung-Despré). r ∂ t c + λ c + u ∇ c − ∇ · ( K ∇ c ) = q ( t ) δ ( x − x R ) Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 4 / 36

Couplex II: Geological figures Layer Permeability 3 . 10 − 5 Tithonien 3 . 10 − 4 Kimmeridgien I 10 − 12 Kimmeridgien II 2 . 10 − 7 Oxfordien I 8 . 10 − 9 Oxfordien II 4 . 10 − 12 Oxfordien III 10 − 13 Callovo-Oxfordien 2 . 510 − 6 . Dogger + = K − ∂ H − implies that ∂ H + = O ( K − Layer decomposition: K + ∂ H K + ) . ∂ n ∂ n ∂ n So ∂ H ∂ n | KI − KII ≈ 0 is a B.C. that decouples the top from the bottom. Later H − | KII = H + is used as B.C for the bottom. Note that the Callovo-Oxfordian+Oxfordian III have H | Γ given from top and bottom separate calculations. Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 5 / 36

COUPLEX II Hydrostatic Pressure Figure: Final result and comparison with a global solution on a supercomputer (Apoung) Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 6 / 36

The Clay Layer with the repository Figure: A computation within the clay layer only with Dirichlet B.C. from the surrounding layers (Apoung-Delpino). Left: a geometrical zoom Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 7 / 36

First Numerical Zoom Figure: Mesh and Sol of Darcy’s in a portion of the entire site. Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 8 / 36

Second Zoom Figure: Mesh and Sol around a single gallery capable of evaluating the impact of a lining around the gallery. Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 9 / 36

Last Zoom and upscale comp. of the concentration What are the errors in the end? Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 10 / 36

Other Examples: What are the errors in the end? Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 11 / 36

Why Numerical Zoom The dream is to combine graphical zoom and numerical zoom. Numerical zoom are needed when it is very expensive or impossible to solve the full problem For instance if the problem has multiple scales Improved precision may be found necessary a posteriori Numerical zoom methods exist: Steger’s Chimera method, J.L. Lions’s Hilbert space decomposition (HSD), Glowinski-He-Rappaz-Wagner’s Subspace correction methods (SCM), etc. The 3 methods are really the same: Schwarz-Hilbert Enrichment (SHE). We need error estimates . Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 12 / 36

The Schwarz-Zoom Method Find u m + 1 ∈ V H , u m + 1 ∈ V h , such that ∀ w H ∈ V 0 H , ∀ w h ∈ V 0 h H h a H ( u m + 1 , w H ) = ( f , w H ) , u m + 1 | S H = γ H u m h , u m + 1 | Γ H = g H , H H H a h ( u m + 1 , w h ) = ( f , w h ) , u m + 1 | S h = γ h u m H , u m + 1 | Γ h = g h h h h where γ H (resp γ h ) is the interpolation operator on V H (resp V h ), where S H and Γ H are the polygonal approximation of S 1 and Γ 1 and similarly for S h , Γ h with S 2 , Γ 2 . Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 13 / 36

Convergence of Discrete Schwarz-Zoom Method Hypothesis 1 Assume that the maximum principle holds for each system independently and that the solution ν H ∈ V H of a H ( ν H , w H ) = 0 , ∀ w H ∈ V 0 H , ν H | S H = 1 , ν H | Γ H = 0 satisfies | ν H | ∞ , S h := λ < 1. Theorem Then the discrete Schwarz algorithm converges to: a H ( u ∗ H , w H ) = ( f , w H ) , ∀ w H ∈ V 0 H , u ∗ H | S H = γ H u ∗ h , u ∗ H | Γ H = g H a h ( u ∗ h , w h ) = ( f , w h ) , ∀ w h ∈ V 0 h , u ∗ h | S h = γ h u ∗ H and max ( || u ∗ H − u || ∞ , Ω H , || u ∗ h − u || ∞ , Ω h ) ≤ C ( H 2 log 1 H || u || H 2 , ∞ (Ω H ) + h 2 log 1 h || u || H 2 , ∞ (Ω h ) ) (1) see also X.C. Cai and M. Dryja and M. Sarkis (SIAM 99) Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 14 / 36

Proof of Convergence By the maximum principle and the fact that γ H and γ h decrease the L ∞ norms, problems of the type: find v H ∈ V H , v h ∈ V h a H ( v H , w H ) = 0 , ∀ w H ∈ V 0 H , v H | S H = γ H u h , v m + 1 | Γ H = 0 H a h ( v h , w h ) = 0 , ∀ w h ∈ V 0 h , v m + 1 | S h = γ h v H h satisfy � v H � ∞ ≤ � u h � ∞ , S H , � v h � ∞ ≤ � v H � ∞ , S h . Combining this with the estimate on the solution of (1) we obtain � v h � ∞ ≤ � v H � ∞ , S h ≤ λ � v H � ∞ ≤ λ � u h � ∞ . Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 15 / 36

Proof of Error estimate (I of II) The solution u to the continuous problem satisfies u | Γ = g and a H ( u , w ) = ( f , w ) ∀ w ∈ H 1 0 (Ω H ) , u = γ H u + ( u − γ H u ) on S H , a h ( u , w ) = ( f , w ) ∀ w ∈ H 1 0 (Ω h ) , u = γ h u + ( u − γ h u ) on S h Let e = u ∗ H − u and ε = u ∗ h − u . Setting w = w H in the first equation and w = w h in the second, we have a H ( e , w H ) = 0 ∀ w H ∈ V 0 H , e = γ H ε − ( u − γ H u ) on S H , e | Γ = g H − g a h ( ε, w h ) = 0 ∀ w h ∈ V 0 h , ε = γ h e − ( u − γ h u ) on S h Let Π H u ∈ V H and Π h u ∈ V h be the solutions of a H (Π H u , w H ) = a H ( u , w H ) ∀ w H ∈ V 0 H , Π H u = γ H u on S H , Π H u | Γ = g H a h (Π h u , w h ) = a h ( u , w h ) ∀ w h ∈ V 0 h , Π h u = γ h u on S h By Schatz& Wahlbin, we have || Π H u − u || ∞ , Ω H ≤ H 2 log 1 H || u || H 2 , ∞ (Ω H ) , || Π h − u || ∞ , Ω h ≤ h 2 log 1 h || u || H 2 , ∞ (Ω h ) . Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 16 / 36

Proof of Error estimate (II) Finally let ε H = u H − Π H u = e + u − Π H u , ε h = u h − Π h u = ε + u − Π h u Then ε H ∈ V H , ε h ∈ V h and a H ( ε H , w H ) = 0 ∀ w H ∈ V 0 H , ε H = γ H ( ε h + Π h u − u ) on S H , ε H | Γ = 0 a h ( ε h , w h ) = 0 ∀ w h ∈ V 0 h , ε h = γ h ( ε H + Π H u − u ) on S h The maximum principle (like in (2) and (2)) again yields � ε H � ∞ ≤ � Π h u − u � ∞ , S H + � ε h � ∞ , S H , � ε h � ∞ ≤ � Π H u − u � ∞ , S h + � ε H � ∞ , S h , � ε H � ∞ , S h ≤ λ � ε H � ∞ Therefore 1 max ( � ε h � ∞ , � ε H � ∞ ) ≤ 1 − λ ( � Π H u − u � ∞ , Ω H + � Π h u − u � ∞ , Ω h ) Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 17 / 36

Hilbert Space Decomposition (JL. Lions) All would be well if Schwarz didn’t require to dig a hole in the zoom. u ∈ V : a ( u , v ) = < f | v > ∀ v ∈ V If V H is not rich enough, use V H + V h and solve u H ∈ V H , u h ∈ V h : a ( u H + u h , v H + v h ) = < f | v H + v h > ∀ v H ∈ V H , v h ∈ V h 0 "usmall.gnu" using 1:3 "ularge.gnu" using 1:3 "u.gnu" using 1:3 "log.gnu" using 1:3 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 f = 1 + δ 0 -0.8 -1 -0.5 0 0.5 1 If solved iteratively, it is similar to Schwarz’DDM or Steger’s Chimera at the continuous level: when Ω 1 ∪ Ω 2 = Ω , Ω 1 ∩ Ω 2 � = ∅ . Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 18 / 36

Discretization and Proof of Uniqueness (Brezzi) Find U H ∈ V 0 H ≈ H 1 0 (Ω) , u h ∈ V 0 h ≈ H 1 0 (Λ) a ( U H + u h , W H + w h ) = < f | W H + w h > ∀ W H ∈ V 0 H ∀ w h ∈ V 0 h Theorem The solution is unique if no vertex belong to both triangulations. Proof If u h = U H on Λ then they are linear on Λ because ∆ u h = ∆ U H and each is a distribution on the edges. The only singularity, if any, are at the intersection of both set of edges (which are points), but being in H − 1 it cannot be singular at isolated points. So ∆ u h = ∆ U H | Λ = 0 Olivier Pironneau (LJLL) Numerical Zoom and Domain Decomposition Bourgeat65 19 / 36