Recovery-Based A Posteriori Error Estimation

Transcription

1 Recovery-Based A Posteriori Error Estimation Zhiqiang Cai Purdue University Department of Mathematics, Purdue University Slide 1, March 2, 2011

2 Outline Introduction Diffusion Problems Higher Order Elements Convection-Diffusion-Reaction Problems Acknowledgement This work is supported in part by the National Science Foundation and by Lawrence Livermore National Laboratory. Department of Mathematics, Purdue University Slide 2, March 2, 2011

3 Recovery Estimators of ZZ Type recovery-based estimators (Zienkiewicz & Zhu 87) ξ = G( u h ) u h where G : L 2 (Ω) d U h C 0 (Ω) d is gradient recovery operator saturation assumption: u G( u h ) β u u h = 1 β efficiency and reliability bounds reliability bound efficiency bound there exists a constant β [0, 1) s.t. ξ G( u h ) u 1 + β (Carstensen, Zhou, etc.) u u h C r η η K C e u u h ωk K T Department of Mathematics, Purdue University Slide 3, March 2, 2011

4 Recovery Estimators of ZZ Type recovery-based estimators ξ = G( u h ) u h where G : L 2 (Ω) d U h C 0 (Ω) d is gradient recovery operator recovery operators Zienkiewicz & Zhu estimator (87) G(u h )(z) = 1 u h dx ω z ω z z N L 2 -projection find G(u h ) U h such that (G(u h ), τ ) = ( u, τ ) τ U h other recovery techniques survey article by Z. Zhang 07 Department of Mathematics, Purdue University Slide 4, March 2, 2011

5 Recovery Error Estimators of ZZ Type recovery-based estimators + simple, universal, asymptotically exact inefficiency for nonsmooth problems, unreliable on coarse meshes, higher-order finite elements, complex systems, etc. Department of Mathematics, Purdue University Slide 5, March 2, 2011

6 Diffusion Problems elliptic interface problems (A(x) u) = f in Ω R d u = g on Γ D and n (A(x) u) = h on Γ N where A(x) = a(x) I d d and a(x) is positive piecewise constant w.r.t Ω = n i=1 Ω i : a(x) = a i > 0 in Ω i smoothness u H 1+α (Ω) where α > 0 could be very small. Department of Mathematics, Purdue University Slide 6, March 2, 2011

7 A Test Problem with Intersecting Interfaces the test problem Ω = ( 1, 1) 2, Γ D = Ω, f = 0 and a(x) = in (0, 1) 2 ( 1, 0) 2 1 in Ω \ ([0, 1] 2 [ 1, 0] 2 ) exact solution with µ(θ) being smooth u(r, θ) = r 0.1 µ(θ) H 1.1 ɛ (Ω) Department of Mathematics, Purdue University Slide 7, March 2, 2011

8 3443 nodes mesh generated by η ZZ for 50% relative error Department of Mathematics, Purdue University Slide 8, March 2, 2011

9 Why Does It Fail? true gradient and flux for interface problems u / C 0 (Ω) d and σ = A u C 0 (Ω) d recovery spaces G( u h ) C 0 (Ω) d and G( A u h ) C 0 (Ω) d the reason of the failure approximating discontinuous functions by continuous functions Department of Mathematics, Purdue University Slide 9, March 2, 2011

10 How to Fix It? true gradient and flux u H 1 (Ω) = u H(curl, Ω) σ = A u H(div, Ω) conforming elements u h H 1 (Ω) = u h H(curl, Ω) σ h = A u h / H(div, Ω) = G(σ h) RT 0 or BDM 1 mixed elements u h L 2 (Ω) σ h H(div, Ω) = G( A 1 σ h ) D 1 or N 1 Department of Mathematics, Purdue University Slide 10, March 2, 2011

11 Efficient Recovery Estimator for Linear Elements flux σ = A u H(div; Ω) (Cai-Zhang 09, SINUM) L 2 flux recovery find σ h V = RT 0 or BDM 1 H(div; Ω) s.t. (A 1 σ h, τ ) = ( u h, τ ) τ V efficient recovery estimator ξ 1,K = A 1/2 ( σ h + A ũ h ) 0,K K T ξ 1 = A 1/2 ( σ h + A ũ h ) 0,Ω reliability bound u h u h C (ξ 1 + H f ) efficiency bound ξ 1 C u ũ h + C ( ) h 2 1/2 K f α f K 2 0,K K Department of Mathematics, Purdue University Slide 11, March 2, 2011

12 3557 nodes mesh generated by ξ RT for 10% relative error Department of Mathematics, Purdue University Slide 12, March 2, 2011

13 Explicit L 2 Flux Recovery RT 0 (Cai & Zhang SINUM 09) Let τ = a(x) u h, then σ T = e E Ω E D σ e φ e (x), where σ e = γ e (τ + e n e ) + (1 γ e ) (τ e n e ) for e E Ω, τ e n e for e E D for some constant γ e [0, 1], e.g., a e γ e = a + e + a e, γ e = a e φ e 2 K + e a e φ e 2 K + e + a + e φ e 2 K e Department of Mathematics, Purdue University Slide 13, March 2, 2011

14 Recovery Estimators using the Hypercircle Method (Lade veze & Leguillon 83, Vejchodsky 04, Braess & Schöberl 07) Prager-Synge identity A 1/2 (u u h ) 2 + A 1/2 (u + A 1 τ ) 2 = A 1/2 (u h + A 1 τ ) 2 where τ H(div; Ω) satisfying τ = f recovery estimators by the hypercircle method construct σ h = η K = A 1/2 (u h + A 1 σ h ) K Department of Mathematics, Purdue University Slide 14, March 2, 2011

15 Recovery Estimators for Higher-Order Elements existing recovery estimators: Naga and Zhang (05), Bank, Xu, and Zheng (07) Department of Mathematics, Purdue University Slide 15, March 2, 2011

16 Recovery Estimators for Higher-Order Elements (Cai-Zhang 10, SINUM) sources of errors for conforming elements: the element residual h K f + div (A u h ) 0,K the edge jump residual h 1/2 e [n e (A u h )] 0,e H(div) flux recovery find σ 2 h V = RT k 1 or BDM k s.t. (A 1 σ 2 h, τ ) + ( σ 2 h, τ ) = ( u h, τ ) + (f, τ ) τ V H(div) recovery-based estimator ξ 2,K = A 1/2 (σ 2 h + A u h ) 0,K K T ξ 2 = A 1/2 (σ 2 h + A u h ) 0,Ω Department of Mathematics, Purdue University Slide 16, March 2, 2011

17 Recovery Estimators for Higher-Order Elements Set e = u ũ h, E = σ σ 2 h, and (E, e) 1,Ω = ( A 1/2 E 2 0,K + A1/2 e 2 0,K ) reliability bound (E, e) 1,Ω C ( ξ 2 + h(f div σ 2 ) h Xũ h ) 0,Ω ) = C (ξ 2 + h(a 1/2 E + A 1/2 e) 0,Ω efficiency bound ξ 2,K A 1/2 E 0,K + A 1/2 e 0,K 2 (E, e) 1,K Department of Mathematics, Purdue University Slide 17, March 2, 2011

18 Convection-Diffusion-Reaction Problems convection-diffusion-reaction problems (A(x) u) + b u + a 0 (x)u = f in Ω R d u = g on Γ D and n (A(x) u) = h on Γ N where A d d (x) is uniformly elliptic in Ω possible computational difficulties corner and interface singularities discontinuities in the form of shock-like fronts, and of interior and boundary layers oscillations of various scales Department of Mathematics, Purdue University Slide 18, March 2, 2011

19 Efficient and Reliable Recovery Estimators (Cai-Zhang 10, SINUM) L 2 -based recovery estimator η 1,K = ( ξk 2 + βk 2 σ 1 h + Xũ h f 2 1/2 0, K) K T ( ) 1/2 η 1 = ξ 2 + K T β 2 K σ 1 h + Xũ h f 2 0, K (a similar but different idea used by Fierro & Veeser 2006) H(div)-based recovery estimator η 2,K = ( ξ 2 K + σ 2 h + Xũ h f 2 0, K) 1/2 K T ( ) 1/2 η 2 = ξ 2 + K T σ 2 h + Xũ h f 2 0, K Department of Mathematics, Purdue University Slide 19, March 2, 2011

20 Efficient and Reliable Recovery Estimators true errors e = u ũ h, E 1 = σ σ 1 h, and E 2 = σ σ 2 h reliability bound for ũ h = u h (E 1, e) 1,Ω C r η 1 efficiency bound exactness η 1,K C e (E 1, e) 1,K + β K f f T 0,K ( ) 1/2 η 1 Ĉe e Ω + βk f 2 f T 2 0,K K T η 2,K = (E 2, e) K and η 2 = (E 2, e) Ω where (E 2, e) Ω = ( A 1/2 (E 2 + A e) 2 0,Ω + E 2 + Xe 2 0,Ω ) 1 2 Department of Mathematics, Purdue University Slide 20, March 2, 2011

21 A Test Problem with Highly Oscillatory Solution Poisson equation u = µ sin(2 m πx) in I = (0, 1), and u(0) = u(1) = 0. where m is a fixed integer and µ is an arbitrary constant exact solution u = µ 4 m π 2 sin(2m πx) which is highly oscillatory for large m existing recovery-based estimators mesh x k = k/2 n for k = 0,..., 2 n, when m > n, estimators of recovery type equal to zero starting with uniform the true error is proportional to µ, arbitrarily large Department of Mathematics, Purdue University Slide 21, March 2, 2011

22 Department of Mathematics, Purdue University Slide 22, March 2, 2011

23 A Test Problem with Intersecting Interfaces the test problem Ω = ( 1, 1) 2, Γ D = Ω, f = 0, a 0 = 0, A(x) = a(x)i and a(x) = in (0, 1) 2 ( 1, 0) 2 1 in Ω \ ([0, 1] 2 [ 1, 0] 2 ) exact solution with µ(θ) being smooth u(r, θ) = r 0.25 µ(θ) H 1.25 ɛ (Ω) Department of Mathematics, Purdue University Slide 23, March 2, 2011

24 number of V-cycle iterations and number of unknowns (edges) (different from observation of Cascon-Nochetto-Siebert 07) Department of Mathematics, Purdue University Slide 24, March 2, 2011

25 Figure 1: mesh generated by ξ with H(div) MG solver Figure 2: ξ vs error Department of Mathematics, Purdue University Slide 25, March 2, 2011

26 Figure 3: mesh generated by ξ with one step V-cycle MG solver Figure 4: ξ with one step V- cycle MG solver and error Department of Mathematics, Purdue University Slide 26, March 2, 2011

27 A Test Problem with Interior Layer reaction-dominant-diffusion problem u + κ 2 u = f in Ω = ( 1, 1) 2 with κ = 100 u = 0 on Ω exact solution u = tanh(κ(x 2 + y 2 1/4)) 1 and u r ( ) 1 2 = 2κ which has an interior layer along the circle centered at the origin with radius 1/2 Department of Mathematics, Purdue University Slide 27, March 2, 2011

28 Figure 5: mesh generated by ξ Figure 6: ξ vs error Department of Mathematics, Purdue University Slide 28, March 2, 2011

29 Figure 7: mesh generated by η Figure 8: η vs error Department of Mathematics, Purdue University Slide 29, March 2, 2011

30 Concluding Remarks recovery-based estimator + detecting interface singularities accurate in the energy norm unreliable on coarse meshes reliable estimator + local and global exactness on any given mesh detecting interface singularities, interior/boundary layers, etc. working well for highly oscillatory solutions requiring a fast solver Department of Mathematics, Purdue University Slide 30, March 2, 2011