Theory of Iterative Methods

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1 Based on Strang s Introduction to Applied Mathematics Theory of Iterative Methods The Iterative Idea To solve Ax = b, write Mx (k+1) = (M A)x (k) + b, k = 0, 1,,... Then the error e (k) x (k) x satisfies Me (k+1) = (M A)e (k), e (k+1) = Be (k) where the iteration matrix B = M 1 (M A). Now e (k) B = k e (0) ρ k 0 iff ρ < 1 where ρ = maximum of eigenvalues of B is the spectral radius of B. Convergence of Iterative Methods Convergence. Does x (k) x? If x (k) x, then Mx = (M A)x + b and Ax = b, which implies x = x. The rate of convergence is governed by powers of B = M 1 (M A). Since e (k) = B k e (0), e (k) 0 and x (k) x (convergence) iff B k 0 (stability). Theorem. B k 0 iff every eigenvalue of B satisfies λ i < 1. The rate of convergence is governed by the spectral radius ρ of B: ρ = max i λ i. Proof using eigenvectors. Expand the initial error in terms of the eigenvectors of B (assuming a complete set of eigenvectors) e (0) = c 1 v 1 + c v + + c n v n

2 where Bv i = λ i v i. Then as k, the error at iteration k is e (k) = B k e (0) = c 1 λ k 1v 1 + c λ k v + + c n λ k nv n 0 iff λ i < 1, i = 1,..., n and e (k) cj λ k j v j ρ k. Proof in matrix form. Assuming a complete set of eigenvectors, write B = S 1 ΛS, where S = [v 1 v v n ] is the eigenvector matrix of B and Λ = diag{λ 1, λ,..., λ n } is the eigenvalue matrix of B. Then B k = ( S 1 ΛS ) ( S 1 ΛS ) (S 1 ΛS ) = S 1 Λ k S 0 iff λ i < 1, i = 1,..., n. If B does not have a complete set of eigenvectors, the matrix form of the proof simply involves the Jordan form J of B instead of Λ. Choice of M Mx (k+1) = (M A)x (k) + b should be easy to solve. Diagonal or lower triangular M s are good choices. M should be close to A, so that the eigenvalues of B = M 1 (M A) = I M 1 A are as small in magnitude as possible (must be inside the unit circle in the complex plane). For A = L+D+U, Jacobi takes M = D while Gauss-Seidel takes M = D+L. SOR (successive over-relaxation) introduces a relaxation factor 1 < < in Gauss-Seidel which is adjusted to make the spectral radius ρ as small as possible. For a wide class of finite-difference matrices, Young s formula (1950) relates the eigenvalues µ of Jacobi and the eigenvalues λ of SOR: (λ + 1) = λ µ. Minimizing ρ SOR = max{ λ } = 1 (using the quadratic formula) gives ( ) 1 1 ρ J opt =, ρ SOR = opt 1. ρ J For the model Laplace problem on an N N grid with h = 1/N, ρ J = cos(πh), ρ GS = ρ J, ρ SOR = 1 sin πh 1 + sin πh.

3 Model Laplace Problem Spectral Radii To show that the Jacobi spectral radius ρ J = cos(πh) for Laplace s equation on the unit square with second-order accurate central differences, first consider a 1D problem. In 1D, set A = tridiag[ 1 1]. Then the iteration matrix B = 1 tridiag[1 0 1]. Then show that Bv = cos(πh) v where the 1D eigenvector v = [sin(πh), sin(πh),, sin(nπh)], n = N 1, h = 1 N = 1 n + 1. The other eigenvectors of B replace π with π, 3π,..., nπ in v, with eigenvalues cos(πh), cos(3πh),..., cos(nπh). In D, the eigenvector = [sin(jπh) sin(kπh)] = [sin(πh) sin(πh), sin(πh) sin(πh),, sin(πh) sin(nπh), sin(πh) sin(πh),, sin(πh) sin(nπh),, sin(nπh) sin(nπh)] with the same eigenvalue, so ρ J = cos(πh). Using ρ J in Young s formula yields the SOR spectral radius Iteration Matrices ρ SOR = 1 sin πh 1 + sin πh. Solve Ax = b iteratively. Decompose A = L + D + U and define r (k) = Ax (k) b. Jacobi Iteration M = D Dx (k+1) = (L + U)x (k) + b = Dx (k) r (k) x (k+1) = x (k) D 1 r (k) B J = D 1 (L + U) 3

4 Gauss-Seidel Iteration M = L + D (D + L)x (k+1) = Ux (k) + b Dx (k+1) = Dx (k) (Lx (k+1) + Dx (k) + Ux (k) b) Dx (k) r (k) SOR/SUR Iteration M = D + L x (k+1) = x (k) (k) D 1 r B GS = (L + D) 1 U x (k+1) = x (k) (k) D 1 r D x(k+1) = D x(k) (Lx (k+1) + Dx (k) + Ux (k) b) ( ) ( ) D D + L x (k+1) = D U x (k) + b ( ) D 1 ( ) D B SOR = + L D U = (D + L) 1 ((1 )D U) Note that for SOR/SUR, det{b} = (1 ) n, 0 < <, and ρ SOR = opt 1, ρ SUR = 1 opt. Example M GS = det{a} = λ 1 λ λ n, Tr{A} = λ 1 + λ + + λ n [ ] 1 A = 1 [ ] [ ] M J =, B 0 J = 1, λ 0 ± = ± 1, ρ J = 1 [ ] [ ] 0 0 1, B 1 GS = M SOR = [ 0 1 ], B SOR =, λ 1 = 0, λ = 1 4, ρ GS = (1 ) ( ) 1 4

5 3 3 Examples opt = 4( 3), λ 1 = λ = opt 1 = ρ SOR (i) Example where Jacobi converges but Gauss-Seidel diverges A = 1 1 1, M J = 0 1 0, B J = λ 1 = λ = λ 3 = 0, ρ J = 0, Jacobi converges. Note that here B 3 = 0 and e 3 = 0, which is an unusual situation! See the Cayley-Hamilton Theorem in the web notes Conjugate Gradients. M GS = , B GS = λ 1 = 0, λ = λ 3 =, ρ GS =, Gauss-Seidel diverges (ii) Example where Jacobi diverges but Gauss-Seidel converges A = 1 1, M J = 0 0, B J = λ 1 = 1, λ = λ 3 = 0.5, ρ J = 1, Jacobi diverges. 0 0 M GS = 1 0, B GS = λ 1 = 0, λ ± = ± i, ρ GS = , Gauss-Seidel converges. Chebyshev SOR By dynamically adjusting, the Chebyshev SOR method insures that the norm of the error always decreases. Make two half sweeps on even/odd (black/white) meshes. Odd (even) points depend only on even (odd) mesh values. Define (µ is the Jacobi spectral radius): (0) = 1 5

6 (k+ 1 ) = ( 1 ) = 1 1 µ 1 1 µ (k) 4 Then ( ) = opt and e (k) always decreases., k = 1, 1, 3,... 6