CIMPA Summer School on Current Research in Finite Element Methods

Transcription

1 CIMPA Summer School on Current Research in Finite Element Methods Local zooming techniques for the finite element method Alexei Lozinski Laboratoire de mathématiques de Besançon Université de Franche-Comté, France Based on a a joint work with Jacques Rappaz EPFL Roland Glowinski, Jiwen He University of Houston Olivier Pironneau LJLL, Paris 6 Patrick Laborde IMT, Toulouse 3 Mickaël Duval, Jean-Charles Passieux ICA, Toulouse Franz Chouly LMB, Besançon

2 Three types of multi-scale problems Multiple scales treated numerically Localized multi-scales The multi-scales everywhere Multiple scales treated on modeling level

3 Three types of multi-scale problems Multiple scales treated numerically Disclaimer : only this case will be covered Localized multi-scales The multi-scales everywhere Multiple scales treated on modeling level

4 An illustration of the numerical zoom The meshes The numerical solution The exact solution The numerical zoom is An alternative to the classical mesh adaptation Less intrusive approach Hopefully useful in the complexes industrial applications

5 Outline Three approaches to construct a numerical zoom 1. A superposition-based numerical zoom : The method of finite element patches aka Numerical Zoom aka Chimera aka Overset Grids F. Brezzi, J.L. Lions and O. Pironneau (CRAS 2001), JB Apoung Kamga and O. Pironneau (JCP 2006), R. Glowinski, J. He, AL, J. Rappaz and J. Wagner (Num.Math. 2005), J. He, AL, J. Rappaz, (C.R.A.S. 2007), F. Hecht, AL, J. Perronet, O. Pironneau (D.C.D.S. 2009) 2. A substitution-based numerical zoom : The multi-model numerical zoom AL and P. Laborde (2015) Non-intrusive coupling L. Gendre, O. Allix, P. Gosselet, (Comp. Mech. 2009), M. Duval, JC Passieux, M. Salaün, S. Guinard (2014) 3. Classical overlapping Schwarz AL and O. Pironneau (2011) A parareal extention to time dependant problems F. Chouly. AL, C.R.A.S. (2014)

6 Motivation example I : nuclear waste storage JB Apoung Kamga and O. Pironneau Schematic view of the Bure project and a numerical test inspired by it (Hecht et al 09) Hydrostatic pressure due to a well in the center of the main gallery Hydrostatic pressure in the zoom

7 Motivations example II : structural mechanics ANR ICARE project : non-intrusive model coupling for the industrial problems (EADS) Duval, Passieux, Salaün, Guinard (2014) Linear elasticity in the coarse global simulation Non linear contact in the local refined problem Linear elasticity in the coarse global simulation Taking into account the non linear plasticity inside each of 12 zooms

8 The toy problems Laplace Dirichlet u = f in Ω, u = 0 on Ω, with f highly oscillating in Λ. More generally d ( K ij (x) u ) (x) = f in Ω, u = 0 on Ω, x i x j i,j=1 The weak formulation Find u H0 1 (Ω) such that a(u, v) =< f, v > v H0 1(Ω) where a(u, v) = d Ω i,j=1 K ij (x) u v dx = K u v dx x i x j Ω

9 Two approaches to the global/local approximation 1. Superposition : u := u Ω + u Λ with u Ω : Ω R, u Ω = 0 on Ω, u Λ : Λ R, u Λ = 0 outside Λ, (u Ω H 1 0 (Ω)) (u Λ H 1 0 (Λ)) such that a(u Ω +u Λ, v Ω +v Λ ) =< f, v Ω +v Λ > v Ω H 1 0 (Ω) and v Λ H 1 0 (Λ) 2. Substitution : u := u Ω in Ω\ Λ and u := w Λ in Λ such that u Ω : Ω R, u Ω = 0 on Ω, a(u Ω, v Ω ) =< f, v Ω > a Λ (w Λ, v Λ ) =< f, v Λ > Λ w Λ = u Ω K w Λ n = K u Ω n and w Λ : Λ R v Ω H0 1(Ω) v Λ H0 1(Λ) on Λ on Λ

10 On the discrete level(finite elements ) Two meshes T H on Ω and T h on Λ. Two approaches to the multiscale approximation : 1. Superposition We search for the multiscale solution u Hh such that u Hh := u H + u h V Hh := V 0 H + V 0 h where V 0 H is P k finite element space on T H (vanishing on Ω) V 0 h is P k finite element space on T h (vanishing on Λ). a(u Hh, v Hh ) =< f, v Hh > v Hh V Hh

11 On the discrete level(finite elements ) Two meshes T H on Ω and T h on Λ. Two approaches to the multiscale approximation : 2. Substitution We search for the multiscale solution u Hh such that { wh, on Λ u Hh = u H, on Ω \ Λ with u H V 0 H and w h V h s.t. Λ (w h u H )µ h = 0 µ h M h

12 Examples of finite element spaces in 1D and in 2D The case of nested meshes : A priori V 0 H V 0 h {0} Coarse and fine meshes The non nested case : The bases of V H and V h form typically an ill conditioned basis for V Hh Conforming interface Non conforming interface

13 Superposition approach : an iterative method Chimera, the method of finite element patches,... Recall that we are looking for u Hh = u H + u h such that a(u H +u h, v H +v h ) = (f, v H +v h ) v H V H and v h V h The alternating iterative method : for n = 1, 2, 3,... Find u n h V 0 h such that a(u n h, v h ) = (f, v h ) a(u n 1 H, v h) v h V 0 h Find u n H V 0 H such that a(u n H, v H ) = (f, v H ) a(u n h, v H ) v h V 0 H Set u n Hh = u n H + un H

14 Superposition approach : an iterative method Chimera, the method of finite element patches,... Recall that we are looking for u Hh = u H + u h such that a(u H +u h, v H +v h ) = (f, v H +v h ) v H V H and v h V h Attention to the intrusive terms Find u n h V 0 h such that a(u n h, v h ) = (f, v h ) a(u n 1 H, v h) v h V 0 h Find u n H V 0 H such that a(u n H, v H ) = (f, v H ) a(u n h, v H ) v h V 0 H Set u n Hh = u n H + un H

15 Superposition approach : an iterative method Chimera, the method of finite element patches,... Recall that we are looking for u Hh = u H + u h such that a(u H +u h, v H +v h ) = (f, v H +v h ) v H V H and v h V h Attention to the intrusive terms Find u n h Vh 0 such that a(uh, n v h ) = (f, v h ) Λ K u n 1 H v h dx v h V 0 h Find u n H VH 0 such that a(uh, n v H ) = (f, v H ) K uh n v H dx Λ v h V 0 H Set u n Hh = u n H + un H

16 Superposition approach : corrections with relaxation General framework : Successive Subspace Correction (Xu, SIAM Review 1992) Choose the relaxation parameter ω ]0, 2[ Initialization : Find u 0 Hh V H s.t. a(uhh 0, v) = (f, v) v V H For n = 1, 2, 3,... Find w h V h such that a(w h, v) = (f, v) a(u n 1 Hh, v) v V h Set u n 1 2 Hh = u n 1 Hh + ωw h Find wh V H such that Set u n Hh = un 1 2 Hh a(w H, v) = (f, v) a(u n 1 2 Hh, v) v V H + ωw H

17 The discretization error Theorem Upon the convergence of the iterative method ( ) u u Hh H 1 (Ω) C H k u H k+1 (Ω\Λ) + h k u H k+1 (Λ). The main ingredients of the proof : a smooth extension ũ of u Ω\Λ ũ Ω\Λ = u Ω\Λ and ũ H k+1 (Ω) C u H k+1 (Ω\Λ).

18 Details of the proof There exists a bounded extension operator E : H q (Ω \ Λ) H q (Ω). Define ũ H q (Ω) by ũ = E(u Ω\Λ ). Observe and ũ Ω\Λ = u Ω\Λ ũ H q (Ω) C u H q (Ω\Λ). We can assume without loss of generality that u ũ H q (Λ) 2 u H q (Λ). By defining ũ H = r H ũ, ũ h = r h (u ũ) and ũ Hh = ũ H + ũ h, we see that u ũ Hh ũ u H + u ũ u h C ( H r u H q (Ω\Λ) + h s u H q (Λ)). The method is conforming (V Hh H 1 0 (Ω)) so that Céa s lemma gives the result.

19 Problems with singular solutions For given f L 2 (Ω), find u H 1 (Ω) such that Problem 1 u = f in Ω, u = 0 on Γ 1, u n = 0 on Γ 2, u = 0 on Γ 3. Problem 2 { u = f in Ω, u = 0 on Ω. Singular part of u : c 0 r sin θ 2 H 3 2 ε (Ω) Singular part of u : c 0 r 2/3 sin 2 ( ) 5 3 θ π 2 H 3 ε (Ω)

20 Improving the convergence order with patches

21 A sharp a priori error estimate (He Lozinski Rappaz, 2007) Assume : Ω is a polygon contained in the sector 0 r R, 0 θ 1 θ θ 2 2π The exact solution is of the form u = r α g(θ) + w with 0 < α < 1, g(θ) H 2 ([θ 1, θ 2 ]), w H 2 (Ω) C ɛ Λ where C ɛ is the intersection of Ω with the ball of radius a centered at 0 Proposition ( H u u Hh H 1 (Ω) C + hα ε1 α with a constant C independent of H, h and ε ) g H 2 + CH w H 2 (Ω)

22 Sketch of the proof It suffices to consider u = r α f (θ) Observe that u H 2 (Ω\C C ε) ε 1 α f H 2 We can explicitly construct an extension ũ a H 2 (Ω) of u Ω\C ε such that ũ Ω\C ε = u Ω\C ε, ũ H 2 (Ω) C ε 1 α f H 2 Take u H = r H ũ, u h = r h (u ũ) and u Hh = u H + u h, Then ũ u H H 1 (Ω) CH ũ H 2 (Ω) C H ε 1 α f H 2 u ũ u h H 1 (Λ ε) u r h u H 1 (Λ ε) + ũ r h ũ H 1 (Λ ( ε) Ch α f H 2 + Ch ũ H 2 (Ω) C h α + h ) ε 1 α

23 Numerical tests We pick Problem 1 (Neuman-Dirichlet boundary conditions) with f = 0, u = r sin(θ/2) The estimate with α = 1 2 gives ( H ε u u Hh H 1 (Ω) C + ) h Choose ε = ε 0 ( H H 0 ) β and h = h0 ( H H 0 ) 2 β, then the estimate becomes u u Hh H 1 (Ω) CH (1 β/2) DOF in V H = N DOF in V h = M N 2(1 β)

24 Order of convergence in H β ɛ h M TO EO H 2 N 2 / /4 H 1/4 H 7/4 N 3/2 / 2 7/8 = /2 H H 3/2 N 3/4 = /4 H 3/4 H 5/4 N 1/2 2 5/8 = H H const no patch H 1 (Ω)-norm convergence orders : TO = theoretical a priori order in H, EO = order obtained by numerical experience

25 Convergence with respect to the total number of nodes Best choice : ε H, h H 3/2, M N Note : ɛ H as H 0

26 Convergence of the iterative method Evolution on the iterations u n+1 Hh u Hh = P H P h (un Hh u Hh) where P H : V Hh VH 0 and P h : V Hh Vh 0 are the orthogonal projectors Thus u n Hh u Hh C γ 2n x 3 V 0 = V 0 H V 0 h with V 0 H V 0 h γ = sup v h V h 0 V 0,vh 0 v H V H 0 V 0,v H 0 a(v h, v H ) v h v H, α x 2 V 0 = V 0 H V 0 h x 1 γ = cos α

27 Iterations operator Let us introduce the operator B : V Hh V Hh that describes the error evolution from one iteration to another with u n+1 Hh u Hh = B(u n Hh u Hh) B = (I ωp H )(I ωp h ) where P H : V Hh V H et P h : V Hh V h are the orthogonal projectors on the sub-spaces wrt the scalar product a(, ). Theorem Assume ω ]0; 2[. The spectral radius of B is ρ(b) = ρ( γ, ω) where ρ( γ, ω) = { ω 2 γ 2 ω ω γ ω2 γ ω + 4, ω 1, if ω ω 0 ( γ), otherwise, ω 0 ( γ) = γ 2 γ 2 with γ (0; 1].

28 The spectral radius and the norm of B Remarques γ is always < 1. On a ρ(b) < 1 for all ω ]0, 2[ Si ω = 1, one has ρ(b) = γ 2. Theorem Provided ω ]0; 2[, the norm of B is B = N( γ, ω) where N( γ, ω) = 1 2 ω (2 ω) γ ω2 (2 ω) 2 γ 2 + (ω 1) 2. In particular, B < 1 for all γ [0, 1[ and ω ]0, 2[.

29 The main tool for the proof : a theorem of decomposition of a space into the sum of planes Let V be a vector space of finite dimension with the scalar product (, ) and V = V 1 + V 2. There exist 2p (p 0) vectors v (m) 1 V 1 et v (m) 2 V 2, m = 1,..., p, such that with v (m) 1 = v (m) 2 = 1, (v (m) 1, v (m) 2 ) = γ m, m = 1,..., p, 1 > γ 1 γ 2 γ p > 0, (1) V can be decomposed into the direct sum V = (V 1 V 2 ) (V 1 V 2 ) (V 1 V 2 ) L 1 L p, (2) where L m = Vec{v (m) 1, v (m) 2 }, m = 1,..., p. The terms in (2) are invariant wrt the orthogonal projectors P 1 : V V 1 et P 2 : V V 1.

30 Idea of the proof By induction on k = 0, 1,... : construct the decomposition V = V 0 W k L 1 L k (3) where V 0 = V 1 V 2, L m = Vec{v (m) 1, v (m) 2 }, v (m) 1 V 1, v (m) 2 V 2 and all the sub-spaces V 0, L 1,..., L k, W k V0 are invariant wrt the projectors P 1 and P 2. Induction step k : let V (k) 1 = V 1 W k 1, V (k) 2 = V 2 W k 1 and γ k = max v 1 V (k) 1, v 2 V (k) 2, v 1 = v 2 =1 Observe that the inequalities (v (k) 1, v (k) 2 ) = (v (k) 1, P 1 v (k) 2 ) P 1 v (k) 2 = (v 1, v 2 ) v (k) 1, v (k) 2 ( ) P 1 v (k) 2 P 1 v (k) 2, v (k) 2 become equalities, which means that P 1 v (k) 2 = αv (k) 1 and vice-versa. (v (k) 1, v (k 2

31 Numerical illustrations { u = f in Ω = ( 1; 1) 2 R 2, u = 0 on Ω. H = 0.1, H/h = 10 nested non-nested non-structured ω = ω = ω opt 5 6 6

32 General intuition about the convergence rate The method is efficient in two cases : 1. Triangulations T H and T h are nested, i.e. V H Λ V h 2. All the functions in V H are very different from the functions in V h Best case : convergence in one iteration (Laplace problem) Worst case : +

33 Possibly bad convergence (example in 1D) There are pairs of functions v h V h and v H V H, which are very close to one another. Hence, γ 1 as h/h 0 Remark. No problem in the case of nested triangulations

34 If we get rid of the coarse basis functions inside Λ L H L γ provided h << H or in the case of nested triangulations

35 A better choice γ = 1 (only for 1D Laplace problem) Restriction : the boundary of the patch should consist of 2 nodes from T H

36 Harmonic patches (He Lozinski Rappaz, 2007) Remark : We now consider the boundary conforming case, i.e. the boundary of Λ is a union of the edges of triangles from T H Introduce the subspace ṼH V H where Ṽ H = {v H V H : a(v H, φ H ) = 0 φ H V 0 H } V 0 H = {v H V H : v H Ω\Λ = 0} The problem like : find u H ṼH such that a(u H, v) = (f v) a(u h, v), v Ṽ H. can be rewritten as a sequence of two subproblems : 1. Find λ H VH 0 such that 2. Find u H V H such that a(λ H, µ) = (f µ) a(u h, µ), µ V 0 H. a(u H, v) = (f v) a(u h, v) a(λ H, v), v V H.

37 Iterative procedure with ω = 1 1. Set u 0 Hh = u H where u H Ṽ H is the solution to a(u H, v) = ( f v), v ṼH 2. For n = 1, 2, 3,... find (i) u n h V h such that f = { f in Ω \ Λ 0 in Λ a(u n h, v) = f v a(u n 1 H, v), v V h (ii) λ n H V 0 H such that a(λ n H, µ) = (f µ) a(u n h, µ), µ V 0 H (iii) u n H V H such that a(u n H, v) = f v a(u n 1 h, v) a(λ n H, v), v V H Set u n Hh = un H + un h

38 Convergence rate We have γ = sup v H Ṽ H,v H 0 v h V h,v h 0 a(v H, v h ) v H 1 v h C 2 v H ṼH, v H 0 consider the harmonic extension v of v H Λ inside Λ : v = 0 in Λ, v Λ = v H Λ Let u H be Scott-Zhang interpolant of v inside Λ : u H Λ = v H Λ and u H v 1,Λ C v 1,Λ. Since a(v H v, u H v H ) = 0 Since a(v H v, v) = 0 v H v 1,Λ u H v 1,Λ C v 1,Λ v H 2 1,Λ = v H v 2 1,Λ + v 2 1,Λ (1 + C 2 ) v H v 2 1,Λ

39 Numerical experiments with harmonic patches { u = f in Ω = ( 1; 1) 2 R 2, u = 0 on Ω. where we substitute the exact solution u = cos π 2 x cos π 2 y + 20χ(x 2 + y 2 < R 2 )e 1 R 2 1 R 2 x 2 y 2, R = 0.3 Patch Λ = ( 0.2; 0.2) 2

40 The geometry and meshes H = 0.1, h = , Patch ( 0.2; 0.2) 2 The meshes T H and T h Zoom ( O.4, 0.4) 2

41 Evolution of the H 1 norm of the error 100 iterations Zoom on the first 10 iterations

42 Evolution of the L 2 norm of the error 100 iterations Zoom on the first 10 iterations

43 Algorithm convergence and mesh refinement Algorithm 1 H 1/10 1/20 1/40 iter. H 1 L 2 H 1 L 2 H 1 L E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E 3 conv. 9.87E E E E E E 4 N iter γ Algorithm 2 H b 1/10 1/20 1/40 iter. H 1 L 2 H 1 L 2 H 1 L E E E E E E E E E E E E E E E E E E E E E E E E E E E E E 4 conv. 7.82E E E E E E 4 N iter γ Relative H 1 - and L 2 -error on iterations Final error with the stopping criterion uhh n un 1 Hh 1 < 10 4 uhh n 1 γ 2 asymptotic convergence rate obtained experimentally

44 A completely non conforming case What to do in the situation when the boundary of Λ is not a union of the edges of triangles from T H? We define Λ H to be the union of triangles of T H which lie completely inside Λ Introduce the subspaces of V H V 0 H = {v H V H : supp v H Λ H } and Ṽ H = {v H V H : a(v H, φ H ) = 0 φ H V 0 H } Proceed as before

45 Numerical results H = 0.1, h = 0.011, Patch ( 0.22; 0.22) 2 The meshes T H and T h Zoom ( O.4, 0.4) 2

46 Evolution of the H 1 norm of the error 100 iterations Zoom on the first 10 iterations

47 Algorithm convergence and mesh refinement Algorithm 1 H 1/10 1/20 1/40 iter. H 1 L 2 H 1 L 2 H 1 L E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E 3 conv. 1.53E E E E E E 4 iter γ Algorithm 1 H 1/10 1/20 1/40 iter. H 1 L 2 H 1 L 2 H 1 L E E E E E E E E E E E E E E E E E E E E E E E E E E E E E 4 conv. 8.72E E E E E E 4 iter γ Relative H 1 - and L 2 -error on iterations Final error with the stopping criterion uhh n un 1 Hh 1 < 10 4 uhh n 1 γ 2 asymptotic convergence rate obtained experimentally

48 Implementation details Find w h V h such that a(w h, v) = (f, v) a(u Hh, v) v V h with u Hh = u H + u h = i u H i ϕ H i + i u h i ϕ h i Therefore we have to evaluate a(ϕ H i, ϕ h j ) = ϕ H i Ω ϕ h j dx Remeshing problem numerical quadrature problem

49 Numerical quadrature When h << H : use the patch for numerical quadrature, for example for P 1 elements a(ϕ H i, ϕ h j ) = ϕ h j dx ϕ H i Ω K T h K ϕ H i (C K ) ϕ h j where C K is the barycenter of the triangle K Finite element assembly on the patch T h. Modified data structure : for each vertex P of the patch T h the triangle of the coarse mesh T H containing P K P

50 Numerical results with quadrature : original method Quadrature rules : qf1 : one point in the middle of triangle qf2 : three middle points of the edges qf5 : six points inside the triangle

51 Numerical results with quadrature : harmonic patches 100 iterations Zoom on the first 15 iterations

52 The worst case The meshes T H and T h Zoom on the mesh T h

53 Numerical results for both methods Evolution of the error in the case f = 0, u = 0, u 0 Hh 0