Solving Polynomial Eigenproblems by Linearization

Transcription

1 Solving Polynomial Eigenproblems by Linearization Nick Higham School of Mathematics University of Manchester Joint work with D. Steven Mackey and Françoise Tisseur 13th ACME Annual Conference, Sheffield, March 2005 PEP Linearizations p. 1/21

2 P(λ) = Polynomial Eigenproblem m λ i A i, A i C n n, A m 0. i=0 P assumed regular (det P(λ) 0). Find scalars λ and nonzero vectors x and y satisfying P(λ)x = 0 and y P(λ) = 0. Standard eigenvalue problem (SEP): Ax = λx. Generalized eigenvalue problem (GEP): Ax = λbx. Quadratic eigenvalue problem (QEP): (λ 2 M + λc + K)x = 0. PEP Linearizations p. 2/21

3 P(λ) = Polynomial Eigenproblem m λ i A i, A i C n n, A m 0. i=0 P assumed regular (det P(λ) 0). Find scalars λ and nonzero vectors x and y satisfying P(λ)x = 0 and y P(λ) = 0. Standard eigenvalue problem (SEP): Ax = λx. Generalized eigenvalue problem (GEP): Ax = λbx. Quadratic eigenvalue problem (QEP): (λ 2 M + λc + K)x = 0. Have very good ways of solving SEP, GEP but not QEP. PEP Linearizations p. 2/21

4 Applications of QEP From, e.g., vibration analysis of structural systems, M q(t) + C q(t) + Kq(t) = f(t). Leads to QEP (λ 2 M + λc + K)x = 0. Damping term C is arbitrary. More applications Acoustic structural coupled systems Fluid mechanics MIMO systems in control theory Signal processing (time series forecasting) Constrained least squares problems PEP Linearizations p. 3/21

5 Spectrum of P m P(λ) = λ i A i, A i C n n. i=0 Recall p assumed regular. Note that det(p(λ)) = λ mn det(a m ) + α mn 1 λ mn det(a 0 ) = α r λ r + + α 1 λ + α 0, α r 0. P has r finite eigenvalues: the roots of det(p(λ)) = 0. P has mn r infinite eigenvalues. PEP Linearizations p. 4/21

6 Spectrum of P m P(λ) = λ i A i, A i C n n. i=0 Recall p assumed regular. Note that det(p(λ)) = λ mn det(a m ) + α mn 1 λ mn det(a 0 ) = α r λ r + + α 1 λ + α 0, α r 0. P has r finite eigenvalues: the roots of det(p(λ)) = 0. P has mn r infinite eigenvalues. A m singular implies λ = is an eigenvalue. A 0 singular implies λ = 0 is an eigenvalue. λ = for P(λ) corr. to λ = 0 for λ mn P(1/λ). PEP Linearizations p. 4/21

7 Polynomial Zeros The roots of p(λ) = α m λ m + a m 1 λ m a 0, a m 0, are the eigenvalues of the companion matrix a m 1 /a m a m 2 /a m a 0 /a m C = MATLAB sroots computes polynomial roots by applying eig to C. Nonlinear polynomial linear eigenproblem. PEP Linearizations p. 5/21

8 Linearization Reduction of 2nd order DE to 1st order: M q(t) + C q(t) + Kq(t) = f(t), q,f C n. Define p 1 = q, p 2 = p 1. Then p 1 = p 2, Mp 2 + Cp 2 + Kp 1 = f. PEP Linearizations p. 6/21

9 Linearization Reduction of 2nd order DE to 1st order: M q(t) + C q(t) + Kq(t) = f(t), q,f C n. Define p 1 = q, p 2 = p 1. Then p 1 = p 2, Mp 2 + Cp 2 + Kp 1 = f. Reduction of QEP to GEP: (λ 2 M + λc + K)x = 0. Define y 1 = x, y 2 = λx. Then ( [ ] [ I 0 0 I λ + 0 M C K ]) y = 0. PEP Linearizations p. 6/21

10 Linearization Definition The pencil L(λ) = λx + Y, X,Y C mn mn is a linearization of P(λ) = m i=0 λi A i if [ ] P(λ) 0 E(λ)L(λ)F(λ) = 0 I (m 1)n for some unimodular E(λ) and F(λ). Example: Companion form linearization ( [ ] [ ]) [ M 0 C K λ E(λ) λ + F(λ) = 2 M + λc + K 0 0 I I 0 0 I ]. PEP Linearizations p. 7/21

11 Some Linearizations of λ 2 M + λc + K L1 : L2 : L3 : L4 : [ ] M 0 λ 0 I [ ] M 0 λ 0 I [ M 0 λ 0 K [ ] 0 M λ M C [ ] C K + I 0 [ ] C I + K 0 ] [ ] C K + K 0 [ ] M K first companion second companion DL(Q) DL(Q) PEP Linearizations p. 8/21

12 Choice of Linearization? Which best preserves structure (symmetry, Hamiltonian,... ) in Q? Which gives the most accurate computed eigenvalues? Which is the best conditioned for a particular eigenvalue? How does conditioning of L compare with that of Q? PEP Linearizations p. 9/21

13 DL(P) Linearizations Mackey, Mackey, Mehl & Mehrmann (2005) identify an m-dimensional vector space of linearizations for P(λ) s.t. right (left) e vecs of P can be recovered from right (left) e vecs of L. For Q(λ) = λ 2 A + λb + C, DL(Q) is the pencils [ ] [ v1 A v L(λ) = λ 2 A v1 B v + 2 A v 1 C v 2 A v 2 B v 1 C v 1 C v 2 C ], v C 2. Right e vecs of L are [ λx x ], where x = right e vec of Q. (Technicality: L is a linearization if no e val of Q equals v 2 /v 1.) PEP Linearizations p. 10/21

14 Condition Number for P(λ) Simple λ, P(λ)x = 0, y P(λ) = 0. { λ κ P (λ) = lim sup ǫ 0 ǫ λ : ( P(λ + λ) + P(λ + λ) ) (x + x) = 0, } A i 2 ǫω i, i = 0:m. Tisseur (2000) showed κ P (λ) = ( m ) i=0 λ i ω i y 2 x 2 λ y P. (λ)x PEP Linearizations p. 11/21

15 Minimizing the Condition Number κ L Theorem 1 Consider pencils L = λx + Y DL(P). Set relative weights ω i = A i 2. Then κ L (λ;e 1 ) ρm 3/2 inf v κ L (λ,v) if A 0 nonsing, λ 1, κ L (λ;e m ) ρm 3/2 inf v κ L (λ,v) if A m nonsing, λ 1, where ρ = max i A i 2 min( A 0 2, A m 2 ). For ρ = O(1), one of v = e 1 and v = e m gives near optimal κ L for λ. Wrong choice of v = e 1 or v = e m can be disastrous: κ L (0,e 1 ) =, κ L (,e m ) =. PEP Linearizations p. 12/21

16 QEP Case For Q(λ) = λ 2 A + λb + C, v = e 1 L 1 (λ) = λ [ A 0 0 C ] + [ B C C 0 ], v = e m L 2 (λ) = λ [ 0 A A B ] + [ A 0 0 C ]. PEP Linearizations p. 13/21

17 Optimal κ L Versus κ P Theorem 2 Let λ be a simple eigenvalue of P. Then inf v κ L (λ;v) κ P (λ) m 2 ρ. If ρ = O(1): Best conditioned L DL(P) for a given λ is about as well conditioned as P itself for λ. Despite perturbations to L not respecting the block structure of X and Y! Combined with Thm. 1: one of pencils with v = e 1 and v = e m is about as well conditioned as P itself for λ. PEP Linearizations p. 14/21

18 Quadratic Case: Scaling Q(λ) = λ 2 A + λb + C, a = A 2, b = B 2, c = C 2. For optimality of κ L (λ) for v = e 1, v = e 2 need ρ = O(1), i.e. b < max(a,c) and a c. With the scaling : λ 2 A + λb + C µ 2 (γ 2 A) + µ(γb) + C (λ = µγ), γ = c/a [Fan, Lin & Van Dooren, 2004], it suffices that b < ac. ( ) ( ) is true if problem is not too heavily damped, for elliptic QEPs. PEP Linearizations p. 15/21

19 Companion Linearizations L1 : L2 : [ A 0 λ 0 I [ A 0 λ 0 I ] ] + + [ B C I 0 [ B I C 0 ] ] first companion second companion Further analysis shows these linearizations are as well conditioned as Q if a b c 1, potentially worse conditioned than Q if left e vec of L is much larger than left e vec of Q. PEP Linearizations p. 16/21

20 Example 1: Nuclear Power Plant (1) n = 8; A 2 = , B 2 = , C 2 = λ (17, 362); Unscaled; v = e 1, ρ = Condition number Companion DL(Q) Q Angular error 10 5 Companion DL(Q) PEP Linearizations p. 17/21

21 Example 1: Nuclear Power Plant (2) Scaled; v = e 2, ρ = Condition number Companion DL(Q) Q Angular error 10 5 Companion DL(Q) PEP Linearizations p. 18/21

22 Example 2: Damped Mass-Spring (1) n = 50; 50 λ ( 320, 6.4), 50 λ Unscaled; v = e 1, ρ = 320. Condition number 10 4 Companion DL(Q) 10 2 Q Angular error Companion DL(Q) PEP Linearizations p. 19/21

23 Example 2: Damped Mass-Spring (2) n = 50; 50 λ ( 320, 6.4), 50 λ Unscaled; v = e 2, ρ = Condition number Companion DL(Q) Q Angular error Companion DL(Q) PEP Linearizations p. 20/21

24 Conclusions Solve ei problem for P via linearized L: P(λ) = m i=0 λi A i, L(λ) = λx + Y. If max i A i 2 min( A 0 2, A m 2 ) then L for v = e 1 and v = e m optimally conditioned within DL(P) for λ 1 and λ 1, resp. and as well conditioned as P. For quadratics, assumption on coeffs weakened to B 2 < A 2 C 2 by use of scaling. Companion linearizations can be poorly conditioned. PEP Linearizations p. 21/21

25 Conclusions Solve ei problem for P via linearized L: P(λ) = m i=0 λi A i, L(λ) = λx + Y. If max i A i 2 min( A 0 2, A m 2 ) then L for v = e 1 and v = e m optimally conditioned within DL(P) for λ 1 and λ 1, resp. and as well conditioned as P. For quadratics, assumption on coeffs weakened to B 2 < A 2 C 2 by use of scaling. Companion linearizations can be poorly conditioned. Justifies solving P(λ)x = 0 by linearization. Guides choice of linearization. PEP Linearizations p. 21/21