FAME-matlab Package: Fast Algorithm for Maxwell Equations Tsung-Ming Huang

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1 FAME-matlab Package: Fast Algorithm for Maxwell Equations Tsung-Ming Huang Modelling, Simulation and Analysis of Nonlinear Optics, NUK, September, 4-8,

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3 FAME group Wen-Wei Lin Department of Applied Mathematics National Chiao-Tung University Weichung Wang Department of Mathematics National Taiwan University Chien-Chih Huang( 黃建智 ) Department of Mathematics National Taiwan Normal University Han-En Hsieh( 謝函恩 ) Department of Mathematics National Taiwan University 32

4 Fast Algorithm for Maxwell Eq FAME FAME.GPU FAME.m FAME.mpi 4

5 FAME Eigen-solvers (JD, SIRA) Photonic Crystals Dispersive Metallic materials Complex materials E(r) = λε(r)e(r) E(r) = ω 2 ε(r,ω )E(r) 0 0 E H = iω ζ µ ε ξ E H Simple Cubic Face-Centered Cubic 5

6 Generalized eigenvalue problems for 3D photonic crystal 6

7 E(r) = ω 2 ε(r)e(r) Curl operator Central edge points E = 0 z z y y 0 x 0 x E 1 E 2 E 3 Central face points where H (r) = ω 2 ε(r)e(r) C h = ω 2 Be E(r) = H (r) Ce = h C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 Resulting generalized eigenvalue problem C C ω 2 B with diagonal B! 3n 3n C 1 = I n2 n 3 K 1! n n, C 2 = I n3 K 2! n n, C 3 = K 3! n n ( )x ( A λ B)x = 0 7

8 Finite Diff. Assoc. with Quasi-Periodic Cond. K 1 = 1 δ x K 2 = 1 δ y K 3 = 1 δ z 1 1!! 1 1 e ı2πk a 1 I n1 e ı2πk a 2 J 2 I n1 1!! I n1! n 1 n 1, I n1 I n1 I n1 n 2 I n1 n 2!! I n1 n 2 I n1 n 2 e ı2πk a 3 J 3 I n1 n 2! (n 1n 2 ) (n 1 n ) 2,! n n 8

9 Finite Diff. Assoc. with Quasi-Periodic Cond. E(r + a l ) = e i2πk a l E(r) K 1 = 1 δ x K 2 = 1 δ y K 3 = 1 δ z 1 1!! 1 1 e ı2πk a 1 I n1 e ı2πk a 2 J 2 I n1 1!! I n1! n 1 n 1, I n1 I n1 I n1 n 2 I n1 n 2!! I n1 n 2 I n1 n 2 e ı2πk a 3 J 3 I n1 n 2! (n 1n 2 ) (n 1 n ) 2,! n n 8

10 Finite Diff. Assoc. with Quasi-Periodic Cond. K 1 = 1 δ x K 2 = 1 δ y K 3 = 1 δ z 1 1!! 1 1 e ı2πk a 1 I n1 e ı2πk a 2 J 2 I n1 1!! I n1! n 1 n 1, I n1 I n1 I n1 n 2 I n1 n 2!! I n1 n 2 I n1 n 2 e ı2πk a 3 J 3 I n1 n 2! (n 1n 2 ) (n 1 n ) 2,! n n 8

11 Finite Diff. Assoc. with Quasi-Periodic Cond. K 1 = 1 δ x K 2 = 1 δ y K 3 = 1 δ z 1 1!! 1 1 e ı2πk a 1 I n1 e ı2πk a 2 J 2 I n1 1!! I n1! n 1 n 1, I n1 I n1 I n1 n 2 I n1 n 2!! I n1 n 2 I n1 n 2 e ı2πk a 3 J 3 I n1 n 2! (n 1n 2 ) (n 1 n ) 2,! n n For SC lattice J 2 = I n1, J 3 = I n1 n 2 8

12 Finite Diff. Assoc. with Quasi-Periodic Cond. K 1 = 1 δ x K 2 = 1 δ y K 3 = 1 δ z 1 1!! 1 1 e ı2πk a 1 I n1 e ı2πk a 2 J 2 I n1 1!! I n1! n 1 n 1, I n1 I n1 I n1 n 2 I n1 n 2!! I n1 n 2 I n1 n 2 e ı2πk a 3 J 3 I n1 n 2! (n 1n 2 ) (n 1 n ) 2,! n n J 2 = J 3 = For SC lattice J 2 = I n1, J 3 = I n1 n 2 For FCC lattice 0 e ı2πk a 1 I n 1 /2 I n1 /2 0 0 e ı2πk a 2 I 1 3 n 2 I 2 J n 2!n 1 n 1, I n1! (n 1n 2 ) (n 1 n 2 ) 8

13 Power method Let x,, x 1 n be the eigenpairs of A where is linearly independent (λ i, x i ) for i = 1,,n For any nonzero vector u u = α 1 x 1 +!+ α n x n Since A k x i = λ i k x i, we have If for i >1 and, then Given shift value A k u = α 1 λ 1 k x 1 +!+ α n λ n k x n λ 1 > λ i α λ 1 k Ak u = α 1 x 1 + ( λ 2 λ 1 ) k α 2 x 2 +!+ α n ( λ n λ 1 ) k x n α 1 x 1 as k {(A σ I) 1 } k u = α { 1 (λ 1 σ ) 1 } k x 1 +!+ α { n (λ n σ ) 1 } k x n 9

14 Solving ( A λ B)x = 0 Use shift-and-invert Lanczos method In each iteration of shift-and-invert Lanczos method, we need to solve (A σ B)y = b How to efficiently solve this linear system? 10

15 Solving linear system (A σ B)y = b 11

16 Solve (A σ B)y = b Direct method (Gaussian elimination) Iterative method y = (A σ B) \ b Matrix vector multiplication with A σ B Preconditioner M sol = bicgstabl(coef_mtx, rhs, tol, diag_coef_mtx, lower_l)); 12

17 Solve (A σ B)y = b Direct method (Gaussian elimination) Iterative method y = (A σ B) \ b Matrix vector multiplication with A σ B Preconditioner M sol = bicgstabl(coef_mtx, rhs, tol, diag_coef_mtx, lower_l)); Demo performance 12

18 Eigen-decomp. of C 1, C 2, C 3 for SC lattice Define Define unitary matrix T as Then it holds that ( ), D a,m = diag( 1,e θ a,m,!,e (m 1)θ a,m ), Λ a,m = diag e θ m,1+θ a,m 1! e θ m,m +θ a,m 1 U m = 1 1! 1 e θ m,1 e θ m,2! 1!!! e (m 1)θ m,1 e (m 1)θ m,2! 1! m m, θ a,m = ı2πk a m, θ m,i = ı2πi m T = 1 ( n D a 3,n 3 D a2,n 2 D )( a1,n 1 U n3 U n2 U ) n1 ( ) T Λ 1, C 1 T = δ x 1 T I n3 I n2 Λ a1,n 1 C 2 T = δ 1 y T ( I n3 Λ a2,n 2 I ) n1 T Λ 2, C 3 T = δ 1 z T ( Λ a3,n 3 I n2 I ) n1 T Λ 3 13

19 Eigen-decomp. of C 1, C 2, C 3 for FCC lattice Define Define unitary matrix T as T = 1 n Then it holds that ψ x = ı2πk a 1, D x = diag 1,e ψ x ( 1 1)ψ n,!,e(n x ), 1 ψ y,i = ı2π k a 2 a 1 n 2 2 x i = D x U n1 (:,i), y i, j = D y,i U n2 (:, j) T 1 T 2! T n1 T i, j = ( D z,i+ j U n3 ) ( y i, j x ) i i 2 ψ z,i+ j = ı2π k a 3 a + a 1 2 n 3 3, i + j 3,!n n, T i = T i,1 T i,2! T i,n2 C 1 T = T ( Λ n1 I ) n2 n 3 T Λ 1, ( ) T Λ 2, C 2 T = T ( n 1 i=1 Λ i,n2 ) I n3 D = diag y,i 1,eψ y,i,!,e (n2 1)ψ y,i ( ), D z,i+ j = diag 1,e ψ y,i+ j,!,e (n3 1)ψ y,i+ j ( )!n (n 2n 3 ), C 3 T = T i=1 ( n 1 n 2 j=1 Λ ) i, j,n3 T Λ 3 14

20 Eigen-decomp. of C 1, C 2, C 3 for FCC lattice Define Define unitary matrix T as T = 1 n Then it holds that ψ x = ı2πk a 1, D x = diag 1,e ψ x ( 1 1)ψ n,!,e(n x ), 1 ψ y,i = ı2π k a 2 a 1 n 2 2 x i = D x U n1 (:,i), y i, j = D y,i U n2 (:, j) T 1 T 2! T n1 T i, j = ( D z,i+ j U n3 ) ( y i, j x ) i i 2 ψ z,i+ j = ı2π k a 3 a + a 1 2 n 3 3, i + j 3,!n n, T i = T i,1 T i,2! T i,n2 C 1 T = T ( Λ n1 I ) n2 n 3 T Λ 1, ( ) T Λ 2, C 2 T = T ( n 1 i=1 Λ i,n2 ) I n3 D = diag y,i 1,eψ y,i,!,e (n2 1)ψ y,i ( ), D z,i+ j = diag 1,e ψ y,i+ j,!,e (n3 1)ψ y,i+ j ( )!n (n 2n 3 ), Demo performance C 3 T = T i=1 ( n 1 n 2 j=1 Λ ) i, j,n3 T Λ 3 14

21 CPU Times for T*p and Tq with FCC MATLAB 10 9 Tq T * p T*p : fft CPU times (sec.) n x 10 7 Tq : ifft 15

22 Solving preconditioning linear system (C C τ I)y = d 16

23 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C

24 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG { I 3 (G G) τ I}y = d + GG y C = 0 C 3 C 2 C 3 0 C 1 C 2 C

25 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 CG = 0 GG y = τ 1 GG d { I 3 (G G) τ I}y = d + GG y 16

26 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 CG = 0 GG y = τ 1 GG d { I 3 (G G) τ I}y = d + GG y { I 3 (G G) τ I}y = d τ 1 GG d 16

27 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 CG = 0 GG y = τ 1 GG d { I 3 (G G) τ I}y = d + GG y { I 3 (G G) τ I}y = d τ 1 GG d Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 C 1 T = T Λ 1, C 2 T = T Λ 2, C 3 T = T Λ 3 16

28 Solving preconditioning linear system (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 CG = 0 GG y = τ 1 GG d { I 3 (G G) τ I}y = d + GG y { I 3 (G G) τ I}y = d τ 1 GG d Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 C 1 T = T Λ 1, C 2 T = T Λ 2, C 3 T = T Λ 3 Λ 1 ( I 3 Λ q τ I )y! = I τ 1 Λ 2 Λ 3 Λ 1 * Λ 2 * Λ 3 * ( I 3 T ) * d, y = ( I 3 T )y! 16

29 Solving preconditioning linear system Demo performance (C C τ I)y = d G = [C 1,C 2,C 3 ] C C = I 3 ( G G) GG C = 0 C 3 C 2 C 3 0 C 1 C 2 C 1 0 CG = 0 GG y = τ 1 GG d { I 3 (G G) τ I}y = d + GG y { I 3 (G G) τ I}y = d τ 1 GG d Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 C 1 T = T Λ 1, C 2 T = T Λ 2, C 3 T = T Λ 3 Λ 1 ( I 3 Λ q τ I )y! = I τ 1 Λ 2 Λ 3 Λ 1 * Λ 2 * Λ 3 * ( I 3 T ) * d, y = ( I 3 T )y! 16

30 Preconditioner M = C C τ I Iterative solver with preconditioner M: sol = bicgstabl(coef_mtx, rhs, tol, Lambda, tau, EigDecompDoubCurl_cell, fun_mtx_th_prod_vec, fun_mtx_t_prod_vec)); 17

31 Preconditioner M = C C τ I Iterative solver with preconditioner M: sol = bicgstabl(coef_mtx, rhs, tol, Lambda, tau, EigDecompDoubCurl_cell, fun_mtx_th_prod_vec, fun_mtx_t_prod_vec)); Since we have M 1 (A σ B) = M 1 (A- τ I + τ I σ B) = I + M 1 (τ I σ B) { I + M 1 (τ I σ B) }y = M 1 b No need to compute the matrix-vector multiplication involving A: sol = bicgstabl(@(vec)mtx_prod_vec_shift_invert_ls(vec, tau, Lambda_new, EigDecompDoubCurl_cell, mtx_b_sigma, fun_mtx_th_prod_vec, fun_mtx_t_prod_vec), rhs, tol, maxit); 17

32 Challenge in Solving Linear System SC lattice (dim = 46875) FCC lattice 18

33 Null-space free eigenvalue problem 19

34 Huge zero eigenvalues Eigen-decomposition where Q 0 Q 0 Q is unitary and Q A Q0 Q ( ) Π Π 0 1 := I 3 T ( ) I 3 T Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 = diag ( 0,Λ q,λ ) q diag( 0,Λ) Π 0,1 Π 1,1 Π 1,2 Π 0,2 Π 1,3 Π 1,4 Π 0,3 Π 1,5 Π 1,6 Q AQ = Λ 20

35 Huge zero eigenvalues Eigen-decomposition where Q 0 Q 0 Q is unitary and Q A Q0 Q ( ) Π Π 0 1 := I 3 T ( ) I 3 T Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 = diag ( 0,Λ q,λ ) q diag( 0,Λ) Π 0,1 Π 1,1 Π 1,2 Π 0,2 Π 1,3 Π 1,4 Π 0,3 Π 1,5 Π 1,6 Q AQ = Λ Ax = λ Bx 0 n zero eigenvalues k wanted interior eigenvalues other 20

36 Huge zero eigenvalues Eigen-decomposition where Q 0 Q 0 Q is unitary and Q A Q0 Q ( ) Π Π 0 1 := I 3 T ( ) I 3 T Λ q = Λ 1 Λ 1 + Λ 2 Λ 2 + Λ 3 Λ 3 = diag ( 0,Λ q,λ ) q diag( 0,Λ) Π 0,1 Π 1,1 Π 1,2 Π 0,2 Π 1,3 Π 1,4 Π 0,3 Π 1,5 Π 1,6 Q AQ = Λ Ax = λ Bx 0 n zero eigenvalues k wanted interior eigenvalues other Ritz values are dragged toward zero during the iteration 20

37 Null-space free method Theorem and span( B 1 QΛ 1/2 ) = span{ x Ax = λbx, λ 0} { λ 0 Ax = λbx} = { λ Λ 1/2 Q B 1 QΛ 1/2 u = λu} Q AQ = Λ 21

38 Null-space free method Theorem and span( B 1 QΛ 1/2 ) = span{ x Ax = λbx, λ 0} { λ 0 Ax = λbx} = { λ Λ 1/2 Q B 1 QΛ 1/2 u = λu} Q AQ = Λ Null-space free SEP Ax = λbx Ku ( Λ 1/2 Q B 1 QΛ 1/2 )u = λu Dim. of GEP and SEP are 3n and 2n, respectively GEP and SEP have same 2n nonzero eigenvalues. SEP has no zero eigenvalues 21

39 Null-space free method Theorem and span( B 1 QΛ 1/2 ) = span{ x Ax = λbx, λ 0} { λ 0 Ax = λbx} = { λ Λ 1/2 Q B 1 QΛ 1/2 u = λu} Q AQ = Λ Null-space free SEP Ax = λbx Ku ( Λ 1/2 Q B 1 QΛ 1/2 )u = λu Dim. of GEP and SEP are 3n and 2n, respectively GEP and SEP have same 2n nonzero eigenvalues. SEP has no zero eigenvalues Ku = λ u 0 n zero eigenvalues k wanted interior eigenvalues other 21

40 Solving Λ 1/2 Q B 1 QΛ 1/2 u = λu Invert Lanczos method In each step, we need to solve a linear system Λ 1/2 Q B 1 QΛ 1/2 v = b 22

41 Solving Λ 1/2 Q B 1 QΛ 1/2 u = λu Invert Lanczos method In each step, we need to solve a linear system Λ 1/2 Q B 1 QΛ 1/2 v = b Solve LS by CG method sol = pcg(@(vec)nfsep_mtx_prod_vec_lambda(vec, EigDecompDoubCurl_cell, FFT_parameter)), rhs, tol, maxit); 22

42 Solving Λ 1/2 Q B 1 QΛ 1/2 u = λu Invert Lanczos method In each step, we need to solve a linear system Λ 1/2 Q B 1 QΛ 1/2 v = b Demo performance Solve LS by CG method sol = pcg(@(vec)nfsep_mtx_prod_vec_lambda(vec, EigDecompDoubCurl_cell, FFT_parameter)), rhs, tol, maxit); 22

43 Solving Λ 1/2 Q B 1 QΛ 1/2 u = λu Invert Lanczos method In each step, we need to solve a linear system Solve LS by CG method sol = pcg(@(vec)nfsep_mtx_prod_vec_lambda(vec, EigDecompDoubCurl_cell, FFT_parameter)), rhs, tol, maxit); Rewrite linear system as Well condition number Solve it by CG method Λ 1/2 Q B 1 QΛ 1/2 v = b Q B 1 Qv! = Λ 1/2 b, v = Λ 1/2 v! κ (Q B 1 Q) κ (B 1 ) Demo performance 22

CPU Time Comparison R. L. CHERN et al. Ax = λbx Λ 1/2 Q B 1 QΛ 1/2 u = λu Fig. 2. Tetragonal square spiral structure comprising circular cylinders. 27,28) 11000 SILM IPLM 10000 9000 (b) n log(n) Fig.

44 CPU Time Comparison R. L. CHERN et al. Ax = λbx Λ 1/2 Q B 1 QΛ 1/2 u = λu Fig. 2. Tetragonal square spiral structure comprising circular cylinders. 27,28) SILM IPLM (b) n log(n) Fig. 3. diamond structure with sp 3 -like configuration comprising dielectric spheres and connecting spheroids CPU time for computing T p and T q with various n. cation (photopolymerization). 32) The order 0.9 of presentation of the paper is organized as follows. In 2, we show how to correctly formulate the finite 0.8 difference method for the double curl operator of the 0.7 photonic eigenvalue problem. In 3, we develop the 0.6 numerical method (inverse iteration with the full multigrid acceleration) 0.5 and present the fast algorithm, in which two alternative 0.4 methods of projection are proposed to avoid the frequency necessity of deflating zeros). In 4, we first present 0.3 numerical results that illustrate the efficiency of the Elapsed times (sec.) presently developed method. Then, the band structures are 0.1 computed for the modified simple cubic lattice, the tetragonal square 0 spiral structure (direct and inverse structure) and X U L G X W K the diamond structure with sp 3 -like configuration. Finally, concluding remarks with a summary of results are drawn in ructure 5. of the 3D photonic crystals with p FCC lattice. The vectors k s along X U L G X W K 23

45 Shift-Invert Residual Arnoldi method 24

46 Shift-Invert Residual Arnoldi method (SIRA) For a given search subspace V, let (θ,z! ) be an eigenpair of V (Λ 1/2 Q B 1 QΛ 1/2 λi)vz = 0 and let x! = V z! be the associated Ritz vector The new search direction v is chosen as ( ) 1 (Λ 1/2 Q B 1 QΛ 1/2 θi)x! v = Λ 1/2 Q B 1 QΛ 1/2 σ I where σ is a given shift value ( Λ1/2 Q B 1 QΛ 1/2 σ I ) 1 r After re-orthogonalizing v against V, the vector is appended to V and one repeats this process until converges to the desired eigenpair. (θ,x! ) 25

CPU Time Comparison Λ 1/2 Q B 1 QΛ 1/2 u = λu 0.9 0.

47 CPU Time Comparison Λ 1/2 Q B 1 QΛ 1/2 u = λu frequency X U L G X W K 26

Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems

Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems Tsung-Ming Huang Department of Mathematics National Taiwan Normal University, Taiwan April 28, 2011 T.M. Huang (Taiwan Normal Univ.)