Matrix square root and interpolation spaces
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1 Matrix square root and interpolation spaces Mario Arioli and Daniel Loghin STFC-Rutherford Appleton Laboratory, and University of Birmingham Sparse Days,CERFACS, Toulouse, 2008 p.1/30
2 Outline Norms and duality in finite dimensional Hilbert spaces Discrete Interpolation Norms The continuous case and finite-element approximation An example: Domain Decomposition Summary and open problems Sparse Days,CERFACS, Toulouse, 2008 p.2/30
3 TRUTH may be misleading In a finite dimensional space all norms are equivalent i.e. Sparse Days,CERFACS, Toulouse, 2008 p.3/30
4 TRUTH may be misleading In a finite dimensional space all norms are equivalent i.e. c(n) v 1 v 2 C(N) v 1 Sparse Days,CERFACS, Toulouse, 2008 p.3/30
5 TRUTH may be misleading In a finite dimensional space all norms are equivalent i.e. c(n) v 1 v 2 C(N) v 1 Identify the norms for which we have c v 1 v 2 C v 1 Sparse Days,CERFACS, Toulouse, 2008 p.3/30
6 (, ) : H H IR scalar product and u H = (u,u) u H norm. Finite dimensional Hilbert spaces and IR N Sparse Days,CERFACS, Toulouse, 2008 p.4/30
7 (, ) : H H IR scalar product and u H = (u,u) u H norm. Finite dimensional Hilbert spaces and IR N {ψ i } i=1,...,n a basis for H u H u = N i=1 u iψ i u i IR i = 1,...,N Sparse Days,CERFACS, Toulouse, 2008 p.4/30
8 (, ) : H H IR scalar product and u H = (u,u) u H norm. Finite dimensional Hilbert spaces and IR N {ψ i } i=1,...,n a basis for H u H u = N i=1 u iψ i u i IR i = 1,...,N Representation of scalar product in IR N. Let u = N i=1 u iψ i and v = N i=1 v iψ i. Then N N (u,v) = u i v j (ψ i,ψ j ) = v T Hu i=1 j=1 where H ij = H ji = (ψ i,ψ j ) and u,v IR N. Moreover, u T Hu > 0 iff u 0 and, thus H SPD. Sparse Days,CERFACS, Toulouse, 2008 p.4/30
9 Dual space H f H : H IR (functional); f(αu + βv) = αf(u) + βf(v) u,v H H is the space of the linear functionals on H f H = sup u 0 f(u) u H Sparse Days,CERFACS, Toulouse, 2008 p.5/30
10 Dual space H f H : H IR (functional); f(αu + βv) = αf(u) + βf(v) u,v H H is the space of the linear functionals on H f H = sup u 0 f(u) u H If H finite dimensional and u = N i=1 u iψ i, then f(u) = N i=1 u if(ψ i ) = f T u Sparse Days,CERFACS, Toulouse, 2008 p.5/30
11 Dual space H f H : H IR (functional); f(αu + βv) = αf(u) + βf(v) u,v H H is the space of the linear functionals on H f H = sup u 0 f(u) u H If H finite dimensional and u = N i=1 u iψ i, then f(u) = N i=1 u if(ψ i ) = f T u Dual vector Let u H, u 0, then f u H such that (Hahn-Banach). f u (u) = u H Sparse Days,CERFACS, Toulouse, 2008 p.5/30
12 Dual space H Let H be a Hilbert finite dimensional space and H the real N N matrix identifying the scalar product. Sparse Days,CERFACS, Toulouse, 2008 p.6/30
13 Dual space H Let H be a Hilbert finite dimensional space and H the real N N matrix identifying the scalar product. f u (u) = f T u = (u T Hu) 1/2 The dual vector of u has the following representation: Sparse Days,CERFACS, Toulouse, 2008 p.6/30
14 Dual space H Let H be a Hilbert finite dimensional space and H the real N N matrix identifying the scalar product. f u (u) = f T u = (u T Hu) 1/2 The dual vector of u has the following representation: and f u 2 H f = Hu u H = ut Hu = f T H 1 f Sparse Days,CERFACS, Toulouse, 2008 p.6/30
15 Linear operator A : H V H and V finite dimensional Hilbert spaces. Sparse Days,CERFACS, Toulouse, 2008 p.7/30
16 Linear operator A : H V H and V finite dimensional Hilbert spaces. A H,V = max u 0 Au V u H = V 1/2 AH 1/2 2 Sparse Days,CERFACS, Toulouse, 2008 p.7/30
17 Linear operator A : H V H and V finite dimensional Hilbert spaces. A H,V = max u 0 Au V u H = V 1/2 AH 1/2 2 The result follows from the generalized eigenvalue problem in IR N A T VAu = λhu Sparse Days,CERFACS, Toulouse, 2008 p.7/30
18 Linear operator A : H V H and V finite dimensional Hilbert spaces. A H,V = max u 0 Au V u H = V 1/2 AH 1/2 2 The result follows from the generalized eigenvalue problem in IR N A T VAu = λhu κ H (M) = M H,H 1 M 1 H 1,H. Sparse Days,CERFACS, Toulouse, 2008 p.7/30
19 Linear operator A : H V H and V finite dimensional Hilbert spaces. A H,V = max u 0 Au V u H = V 1/2 AH 1/2 2 The result follows from the generalized eigenvalue problem in IR N A T VAu = λhu κ H (M) = M H,H 1 M 1 H 1,H. The interesting case is κ H (M) independent of N Sparse Days,CERFACS, Toulouse, 2008 p.7/30
20 Interpolation spaces H M = (IR N,(u,v) H = u T Hv) = (IR N,(u,v) M = u T Mv) (u,v) H = (u,sv) M = (Su,v) M S = M 1 H S self-adjoint in the good scalar product! { } Sx = µx Hx = µmx µ = δ 2 > 0 W s.t. M = W T W, H = W T 2 W, diagonal Sparse Days,CERFACS, Toulouse, 2008 p.8/30
21 Interpolation spaces H M = (IR N,(u,v) H = u T Hv) = (IR N,(u,v) M = u T Mv) (u,v) H = (u,sv) M = (Su,v) M S = M 1 H S self-adjoint in the good scalar product! { } Sx = µx Hx = µmx µ = δ 2 > 0 W s.t. M = W T W, H = W T 2 W, diagonal 0 Sparse Days,CERFACS, Toulouse, 2008 p.8/30
22 Interpolation spaces H M = (IR N,(u,v) H = u T Hv) = (IR N,(u,v) M = u T Mv) (u,v) H = (u,sv) M = (Su,v) M S = M 1 H S self-adjoint in the good scalar product! { } Sx = µx Hx = µmx µ = δ 2 > 0 W s.t. M = W T W, H = W T 2 W, diagonal 0 Sparse Days,CERFACS, Toulouse, 2008 p.8/30
23 Interpolation spaces Λ = W 1 W Λ 1/2 = W 1 1/2 W S = M 1 H = W 1 W T W T 2 W = W 1 WW 1 W = Λ 2 MΛ = W T WW 1 W T W T W = Λ T M (u,λv) M = (Λu,v) M (Λ 1/2 u,λ 1/2 u) M = (u,λu) M Sparse Days,CERFACS, Toulouse, 2008 p.9/30
24 Interpolation spaces [ { ) 1/2 } H, M ]ϑ = u IR N ; ((u,u) M + (u,s 1 ϑ u) M Sparse Days,CERFACS, Toulouse, 2008 p.10/30
25 Interpolation spaces [ { ) 1/2 } H, M ]ϑ = u IR N ; ((u,u) M + (u,s 1 ϑ u) M [ { ) 1/2 } H, M ]1/2 = u IR N ; ((u,u) M + (u,λu) M Sparse Days,CERFACS, Toulouse, 2008 p.10/30
26 Interpolation spaces [ { ) 1/2 } H, M ]ϑ = u IR N ; ((u,u) M + (u,s 1 ϑ u) M [ { ) 1/2 } H, M ]1/2 = u IR N ; ((u,u) M + (u,λu) M v 2 ϑ = v 2 H ϑ = v T( M + MS 1 ϑ) v ( H ϑ = M I + S 1 ϑ) = W T( I + 2(1 ϑ)) W Sparse Days,CERFACS, Toulouse, 2008 p.10/30
27 Interpolation spaces (duality) M and H dual spaces of M and H [ ] [ H, M ϑ = M, H ] 1 ϑ S = MH 1 = W T 2 W T Sparse Days,CERFACS, Toulouse, 2008 p.11/30
28 Interpolation spaces (duality) M and H dual spaces of M and H [ ] [ H, M ϑ = M, H ] 1 ϑ S = MH 1 = W T 2 W T H 1 ϑ = H 1 1 ϑ = W 1 2ϑ W T Sparse Days,CERFACS, Toulouse, 2008 p.11/30
29 Interpolation spaces ( dimension case) X, Y two Hilbert spaces with X Y, X dense and continuously embedded in Y., X,, Y and X, Y the respective norms. (Riesz representation theory) J : X Y positive and self-adjoint with respect to, Y such that u,v X = u, J v Y. E = J 1/2 : X Y, X = D(E) with u X u E := ( u 2 Y + Eu 2 Y ) 1/2. u θ := ( u 2 Y + E1 θ u 2 Y ) 1/2. The interpolation space of index θ [X,Y ] θ := D(E 1 θ ), 0 θ 1, with the inner-product u,v θ = u,v Y + u, E 1 θ v is a Hilbert Y space (Lions Magenes 1968). [X,Y ] 0 = X and [X,Y ] 1 = Y. If 0 < θ 1 < θ 2 < 1 then X [X,Y ] θ1 [X,Y ] θ2 Y. Sparse Days,CERFACS, Toulouse, 2008 p.12/30
30 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). Sparse Days,CERFACS, Toulouse, 2008 p.13/30
31 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). Let X X h and Y Y h finite-dimensional spaces i h : L(X h ;X) L(Y h ;Y ) the continuous injection operator i h L([X h,y h ] θ ;[X,Y ] θ ). Sparse Days,CERFACS, Toulouse, 2008 p.13/30
32 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). u h [X h,y h ] θ, i h u h θ = u h θ C 1 u h θ,h. Sparse Days,CERFACS, Toulouse, 2008 p.13/30
33 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). u h [X h,y h ] θ, i h u h θ = u h θ C 1 u h θ,h. Assume now that there exists an interpolation operator I h such that I h : L(X;X h ) L(Y ;Y h ) and I h u = u h for all u h X h. I h L([X,Y ] θ ;[X h,y h ] θ ) Sparse Days,CERFACS, Toulouse, 2008 p.13/30
34 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). u h [X h,y h ] θ, i h u h θ = u h θ C 1 u h θ,h. u [X,Y ] θ, I h u θ,h C 2 u θ. Sparse Days,CERFACS, Toulouse, 2008 p.13/30
35 Interpolation Theorem Let X,Y Hilbert spaces X Y with X dense in Y, and with inclusion compact and continuous. Let X, Y satisfy similar properties. Let π L(X; X) L(Y ; Y). Then for all θ (0,1), π L([X,Y ] θ ;[X, Y] θ ). u h [X h,y h ] θ, i h u h θ = u h θ C 1 u h θ,h. u [X,Y ] θ, I h u θ,h C 2 u θ. Since [X h,y h ] θ [X,Y ] θ 1 C 1 u h θ u h θ,h C 2 u h θ. Sparse Days,CERFACS, Toulouse, 2008 p.13/30
36 Interpolation spaces ( dimension case) Ω R n open bounded with smooth boundary Γ and let α denote a multi-index of order m where m is a positive integer H m (Ω) = { u : D α u L 2 (Ω), α m } (H 0 (Ω) = L 2 (Ω)) H s (Ω) := [H m (Ω),H 0 (Ω)] 1 s/m H0 s(ω) completion of C 0 (Ω) in Hm (Ω), where s > 0. For 0 s 2 < s 1, { [H s 1 0 (Ω),Hs 2 0 (Ω)] θ = H (1 θ)s 1+θs 2 0 (Ω) if (1 θ)s 1 + θs 2 k + 1/2 ( [H s 1 0 (Ω),Hs 2 0 (Ω)] θ = Hk+1/2 00 (Ω) H k+1/2 0 if (1 θ)s 1 + θs 2 = k + 1/2 ( If (1 θ)s 1 + θs 2 = 1/2 H s (Ω) = (H s 0(Ω)) s > 0 [H s 1 (Ω),H s 2 (Ω)] θ = ( H 1/2 00 (Ω) ). Sparse Days,CERFACS, Toulouse, 2008 p.14/30
37 Finite-element example H 1/2 00 (Ω) = [H1 0(Ω),L 2 (Ω)] 1/2. Let X h H 1 0 (Ω),Y h L 2 (Ω). Let {φ i } 1 i n X h be a spanning set for Y h and let L k R n n denote the Grammian matrices corresponding to the, H k 0 (Ω) -inner product (H0 (Ω) = L 2 (Ω)): H = L 1, M = L 0 and (L k ) ij = φ i,φ j H k 0 (Ω). H 1/2,h = L 0 ( I + (L 1 0 L 1) 1/2) Moreover, we have ( ) 1/2 H 1/2,h H 1/2 = L 0 L 1 0 L 1 Sparse Days,CERFACS, Toulouse, 2008 p.15/30
38 Finite-element example ( H 1/2 00 (Ω) ) = [H (1) (Ω),H 0 (Ω)] 1/2 = ( [H 1 0(Ω),H 0 0(Ω)] 1/2 ). Let X h,y h be defined as above. Let Y h X h = span {ψ i} 1 i n where ψ i are basis functions dual to φ i, i.e., ψ i,φ j H 1 (Ω) H 1 0(Ω) = δ ij. n If Y h z = z i ψ i = i=1 n w i φ i, φ l = i=1 n K li ψ i i=1 δ ij = ψ i,φ j H 1 (Ω) H 1 0(Ω) = n i=1 K 1 il φ l,φ j H 1 0 (Ω) = n i=1 K 1 il (L 1 ) lj so that z = L 0 w and z H Y = w L 1, z 0 H X = w L0 L 1 1 L 0 and the matrix representation of H Y,H X are respectively L 0 and L 0 L 1 1 L 0. H 1/2 = L 0(L 1 0 L 1) 1/2 = H 1/2. Sparse Days,CERFACS, Toulouse, 2008 p.16/30
39 Evaluation of H θ z Generalised Lanczos H X V k = H Y V k T k + β k+1 H y v k+1 e T k, V k TH Y V k = I k (T k tridiagonal). Sparse Days,CERFACS, Toulouse, 2008 p.17/30
40 Evaluation of H θ z Generalised Lanczos H X V k = H Y V k T k + β k+1 H y v k+1 e T k, V k TH Y V k = I k (T k tridiagonal). v 0 = z H θ z H Y V k T 1 θ k e 1 z HY and H θ,h z H Y V k (I k + T 1 θ k )e 1 z HY. Sparse Days,CERFACS, Toulouse, 2008 p.17/30
41 Evaluation of H θ z Generalised Lanczos H X V k = H Y V k T k + β k+1 H y v k+1 e T k, V k TH Y V k = I k (T k tridiagonal). v 0 = z H θ z H Y V k T 1 θ k e 1 z HY and H θ,h z H Y V k (I k + T 1 θ k )e 1 z HY. v 0 = H 1 H 1 θ H 1 Y z z V k T θ 1 k θ,h z V k(i k + T 1 θ k e 1 z H 1 Y and ) 1 e 1 z H 1 Y. Sparse Days,CERFACS, Toulouse, 2008 p.17/30
42 Evaluation of H θ z Generalised Lanczos H X V k = H Y V k T k + β k+1 H y v k+1 e T k, V k TH Y V k = I k (T k tridiagonal). v 0 = z H θ z H Y V k T 1 θ k e 1 z HY and H θ,h z H Y V k (I k + T 1 θ k )e 1 z HY. v 0 = H 1 H 1 θ H 1 Y z z V k T θ 1 k θ,h z V k(i k + T 1 θ k e 1 z H 1 Y and ) 1 e 1 z H 1 Y Alternative: N. Hale, and N. J. Higham and L. N. Trefethen, SIAM J. Numer. Anal.. Sparse Days,CERFACS, Toulouse, 2008 p.17/30
43 Preconditioners for the Steklov-Poincaré operator Let Ω be an open subset of R d with boundary Ω and consider the model problem { u = f in Ω, u = 0 on Ω. Given a partition of Ω into two subdomains Ω Ω 1 Ω 2 with common boundary Γ this problem can be equivalently written as { { u 1 = f in Ω 1, u 2 = f in Ω 2, u 1 = 0 on Ω 1 \ Γ, u 2 = 0 on Ω 2 \ Γ, with the interface conditions { u 1 = u 2 u 1 n 1 = u 2 n 2 on Γ Sparse Days,CERFACS, Toulouse, 2008 p.18/30
44 Preconditioners for the Steklov-Poincaré operator Given λ 1,λ 2 H 1/2 00 (Γ), ψ 1,ψ 2 denote the harmonic extensions of λ 1,λ 2 respectively into Ω 1,Ω 2, i.e., for i = 1,2, ψ i satisfy ψ i = 0 in Ω i, ψ i = λ i on Γ, ψ i = 0 on Ω i \ Γ. The Steklov-Poincaré operator S : H 1/2 00 (Γ) H 1/2 (Γ) Sλ 1,λ 2 H 1/2 (Γ) = ψ 1, ψ 2 L2 (Ω) =: s(λ 1,λ 2 ). c 1 λ 2 H 1/2 (Γ) s(λ,λ) c 2 λ 2 H 1/2 (Γ). Sparse Days,CERFACS, Toulouse, 2008 p.19/30
45 Preconditioners for the Steklov-Poincaré operator (i) (ii) (iii) { { { The resulting solution is u {1} i = f in Ω i, u {1} i = 0 on Ω i, Sλ = u{1} 1 u{1} 2 on Γ, n 1 n 2 u {2} i = 0 in Ω i, u {2} i = λ on Ω i. u Ωi = u {1} i + u {2} i. Sparse Days,CERFACS, Toulouse, 2008 p.20/30
46 An other problem { ν u + b u = f in Ω, u = 0 on Ω. Sparse Days,CERFACS, Toulouse, 2008 p.21/30
47 Discrete Formulation V h = V h,r := { w C 0 (Ω) : w t P k t T h } H 1 (Ω) be a finite-dimensional space of piecewise polynomial functions defined on some subdivision T h of Ω into simplices t of maximum diameter h. Let further VI h,v B h V h satisfy VI h VB H V h where VI h = { w V h : w Ω = 0 }. Let X h H0 1 (Γ) denote the space spanned by the restriction of the basis functions of VI h to the internal boundary Γ. Sparse Days,CERFACS, Toulouse, 2008 p.22/30
48 Discrete Formulation (i) A II,i u {1} i = f I,i, (ii) Su B = f B A T IB,1 u{1} 1 A T IB,2 u{2} 2, (iii) A II,i u {2} i = A T IB,1 u B A T IB,2 u B, where S is the Schur complement corresponding to the boundary nodes S = S 1 + S 2, S i = A BB,i A T IB,iA 1 II,i A IB,i. The resulting solution is (u I,1, u I,2, u B ) where u I,i = u {1} i + u {2} i. Sparse Days,CERFACS, Toulouse, 2008 p.23/30
49 H 1/2 00 -preconditioners Let X h = span {φ i,1 i m} be defined as above and let (L k ) ij = φ i,φ j H k 0 (Γ) for k = 0,1. Let Then for all λ R m \ {0} H 1/2 := L 0 (L 1 0 L 1) 1/2. κ 1 λt Sλ λ T H 1/2 λ κ 2 with κ 1,κ 2 independent of h. Sparse Days,CERFACS, Toulouse, 2008 p.24/30
50 Discrete DD P = ( A II A IB 0 P S with A II = νl II + N II where L II is the direct sum of Laplacians assembled on each subdomain and N II is the direct sum of the convection operator b assembled also on each subdomain. ) Sparse Days,CERFACS, Toulouse, 2008 p.25/30
51 Numerical results: Poisson equatione Linear Quadratic #dom n m H 1/2,h H 1/2 H1/2 b H 1/2,h H 1/2 H1/2 b 45, , , , , , , , ,050,625 14, FGMRES iterations for model problem. Sparse Days,CERFACS, Toulouse, 2008 p.26/30
52 An other problem Linear Quadratic #dom n m ν = 1 ν = 0.1 ν = 0.01 ν = 1 ν = 0.1 ν = , , , , , , , , ,050,625 14, FGMRES iterations for 2nd model problem Sparse Days,CERFACS, Toulouse, 2008 p.27/30
53 Exotic domains QUANTUM graphs (metric graph) Sparse Days,CERFACS, Toulouse, 2008 p.28/30
54 Exotic domains QUANTUM graphs (metric graph) 2-D simplex Sparse Days,CERFACS, Toulouse, 2008 p.28/30
55 Exotic domains QUANTUM graphs (metric graph) 2-D simplex Same theory with Laplace-Beltrami operator Sparse Days,CERFACS, Toulouse, 2008 p.28/30
56 Green functions on wirebasket Steklov-Poincaré Green function Sparse Days,CERFACS, Toulouse, 2008 p.29/30
57 Green functions on wirebasket Steklov-Poincaré Green function Neumann-Neumann Green function Sparse Days,CERFACS, Toulouse, 2008 p.29/30
58 Green functions on wirebasket Steklov-Poincaré Green function H 1/2 Green function Sparse Days,CERFACS, Toulouse, 2008 p.29/30
59 Open problems Domain decomposition: preliminary results show total independence from mesh size Sparse Days,CERFACS, Toulouse, 2008 p.30/30
60 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH Sparse Days,CERFACS, Toulouse, 2008 p.30/30
61 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH 3D PDEs Sparse Days,CERFACS, Toulouse, 2008 p.30/30
62 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH 3D PDEs Utilization in Image Denoising Sparse Days,CERFACS, Toulouse, 2008 p.30/30
63 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH 3D PDEs Utilization in Image Denoising H s s > 1 Sparse Days,CERFACS, Toulouse, 2008 p.30/30
64 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH 3D PDEs Utilization in Image Denoising H s s > 1 Generalization to Banach spaces (K-method Peetre) Sparse Days,CERFACS, Toulouse, 2008 p.30/30
65 Open problems Domain decomposition: preliminary results show total independence from mesh size Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH 3D PDEs Utilization in Image Denoising H s s > 1 Generalization to Banach spaces (K-method Peetre) For which class of matrices the Schur complement is spectrally equivalent to an algebraic interpolation space matrix Sparse Days,CERFACS, Toulouse, 2008 p.30/30
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