1. Finding Rightmost Eigenvalues of Large, Sparse, Nonsymmetric Parameterized Eigenvalue Problems Minghao Wu AMSC Program mwu@math.umd.edu Advisor: Dr. Howard Elman Department of Computer Science elman@cs.umd.edu

2. 2 Motivation To determine the stability of the linearized system of the form: The steady state solution x* is - stable, if all the eigenvalues of Ax = λBx have negative real parts; - unstable, otherwise.

3. 3 Problem Statement To find the rightmost eigenvalues of: where matrices A and B are Real N-by-N Large Sparse Nonsymmetric Depend on one or several parameters

4. 4 Method Eigensolver: Arnoldi Algorithm - Iterative method - Based on Krylov subspace: - Computes Arnoldi decomposition: where: [U k u k+1 ]: an orthonormal basis of H k : k-by-k upper Hessenberg matrix, k « N β k: scalar, e k : k-by-1 vector [0 0 … 0 1]’

5. 5 Arnoldi Algorithm (continue) Eigenvalues of H k approximate eigenvalues of A Premultiply previous equation by transpose(Uk): Let (λ,z) be an eigenpair of H k, then: As k increases, ||Residual|| will decrease. When k = N, Residual = 0. Residual

6. 6 Matrix Transformation Motivation - Arnoldi Algorithm cannot solve generalized eigenvalue problem - It converges to well – separated extremal eigenvalues, not rightmost eigenvalues Shift – Invert Transformation

7. 7 Matlab Code of Arnoldi Method Arnoldi Algorithm (with Shift – Invert matrix transformation) routine: [v,X,U,H]=SI_Arnoldi(A,B,k,sigma) Input: A, B: matrix A and B in Ax = λBx k: number of eigenpairs wanted sigma: the shift σ in shift – invert matrix transformation Output: v: a vector of k computed eigenvalues X: k eigenvectors associated with the eigenvalues U: the Krylov basis U k+1 H: the upper Hessenberg matrix H k

8. 8 Test Problem Olmstead Model (see Olmstead et al (1986)): with boundary conditions: This model represents the flow of a layer of viscoelastic fluid heated from below. u: the speed of the fluid S: related to viscoelastic forces b,c: scalars, R: scalar, Rayleigh number

9. 9 Test Problem (continue) Discretize the model with finite differences - grid – size: h = 1/(N/2) - (4) can be written as dy/dt = f(y) with Evaluate the Jacobian matrix A = df/dy at steady state solution y* - N = 1000, b = 2, c = 0.1, R = 0.6 - y* = 0 - A = df/dy(y*) is a nonsymmetric sparse matrix with bandwidth 6

10. 10 Test Problem (continue) Computational Result: Rightmost eigenvalues: λ 1,2 = 0 ± 0.4472i Residual: ||Ax i - λ i x i || = 8.4504e-012, i=1,2 The result agrees with the literature.

11. 11 Implicitly Restarted Arnoldi (IRA) Motivation - Large k is not practical Example: Size of A:10,000 Value of k:100 Memory required to store U 100 in double precision: 10 MB - When B is singular, Arnoldi algorithm may give rise to spurious eigenvalues

12. 12 IRA (continue) Basic idea of Implicitly Restarted Arnoldi Filter out the unwanted eigendirections from the starting vector by using the most recent spectrum information and a clever filtering technique IRA steps 1. Compute m eigenpairs (k<m«N) by Arnoldi method with starting vector u 1 2. Filter out the m-k unwanted eigendirections from u 1 (Key Technique: shifted QR algorithm) 3. Restart the process with filtered starting vector till the k eigenvalues of interest converge

13. 13 Test Problem K: 200-by-200 matrix, full rank; C: 200-by-100 matrix, full rank; M: 200-by-200 matrix, full rank. Eigenvalue problem with this kind of block structure appears in the stability analysis of steady state solution of Navier – Stokes equations for incompressible flow.

14. 14 Test Problem (continue) Use Matlab function “rand” to generate K, C, M -2.7377 ≤ Re(λ) ≤ 49.9129 Find out 10 rightmost eigenvalues Use the IRA code written by Fei Xue

15. 15 Test Problem (continue) Exact Eigenavalues 49.9129 + 0i 2.9112 + 1.1256i 2.9112 - 1.1256i 2.5036 + 0.0624i 2.5036 - 0.0624i 2.3792 + 0i 2.1318 + 0.9356i 2.1318 - 0.9356i 2.1081 + 1.3539i 2.1081 - 1.3539i Computed Eigenavalues (Arnoldi) 49.9129 + 0i 193.8412 - 7113830.9524i 193.8412 + 7113830.9524i 3.0891 + 0i 2.6112 + 0i -0.5752 - 2.7079i -0.5752 + 2.7079i -1.0901 - 0.7532i -1.0901 + 0.7532i -47.3106 + 0i Computed Eigenavalues (IRA) 49.9129 + 0i 2.9112 - 1.1256i 2.9112 + 1.1256i 2.5036 - 0.0624i 2.5036 + 0.0624i 2.3792 + 0i 2.1318 - 0.9356i 2.1318 + 0.9356i 2.1081 - 1.3539i 2.1081 + 1.3539i Computational Result: (shift σ = 60)

16. 16 Future Work (AMSC 664) Solve the third test problem Implement iterative solvers for linear systems