Krylov methods for fast frequency response computations Karl - - PowerPoint PPT Presentation

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Krylov methods for fast frequency response computations

Karl Meerbergen February 28, 2007

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Outline

1 Motivation 2 Rayleigh damping 3 Modal truncation 4 Lanczos’ method 5 Numerical example 6 Conclusions

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Vibration problems

Bridge vibrating under footsteps and Thames wind

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Vibration problems

A car window is subjected to vibrations from outside, including

• wind. Glass manufacturers want to compute the transmission of

noise through windscreens.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Finite element analysis

Numerical simulation of vibration problems. Spatial (finite element) discretization: M¨ x(t) + C ˙ x(t) + Kx(t) = f (t) with initial values x(0) and ˙ x(0) f and x : vectors of length n K, C and M : n × n sparse matrices. In real applications n varies from 103 to over 106.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Fourier analysis

If f (t) = ˜ f eiωt, then (under certain conditions) for t → ∞, x(t) = ˜ xeiωt where (K + iωC − ω2M)˜ x = ˜ f The engineer is usually interested in the periodic regime solution, i.e. after a long integration time. Material properties are often frequency dependent.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Fourier analysis

(K + iωC − ω2M)˜ x = ˜ f ˜ x is called the frequency response function.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Traditional frequency response computation

1. For ω = ω1, . . . , ωp 1.1. Solve the linear system (K + iωC − ω2M)x = f for x For each frequency, a large system of algebraic equations needs to be solved. This requires

a sparse matrix factorization LU = K − ω2M + iωC (expensive) and a backward solve LUx = f (relatively cheap).

The goal is to reduce the number of matrix factorizations. Important is speed, not good reduction.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Damping

We make the damping ω dependent: D(ω) (K − ω2M + D(ω))x = f Rayleigh damping : D = γK + δM f is independent of ω

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

No damping

D(ω) ≡ 0 Linear system: (K − ω2M)x = f Corresponding eigenvalue problem: Ku = λMu

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Structural Rayleigh damping

D(ω) = iγK Linear system: ((1 + iγ)K − ω2M)x = f Corresponding eigenvalue problem: (1 + iγ)Ku = λMu

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Fluid Rayleigh damping

D(ω) = iω(α0M + α1K) Linear system: (K + iω(α0M + α1K) − ω2M)x = f Corresponding eigenvalue problem: (K + iλ(α0M + α1K) − λ2M)u = 0

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Undamped problem

Consider the eigendecomposition Kuj = λjMuj The solution of (K − ω2M)x = f is x =

n

• j=1

uj u∗

j f

λj − ω2 Rational function with poles λj.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Rayleigh damping

Define U = [u1, . . . , un] and Λ = diag(λ1, . . . , λn) KU = MUΛ D = βK + γM DU = MU(βI + γΛ) U∗MU = I U∗KU = Λ U∗DU = βI + γΛ

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Rayleigh damping

Simultaneous diagionalization of K, M, and D: u∗

j Mui = u∗ j Kui = u∗ j Dui = 0 iff i = j.

u∗

j Muj = 1

u∗

j Kuj = λj

u∗

j D(ω)uj = ζj(ω)

The solution of (K − ω2M + D)x = f x =

n

• j=1

uj u∗

j f

λj − ω2 + ζj(ω)

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Truncation

x =

n

• j=1

uj u∗

j f

λj − ω2 + ζj ≈

k

• j=1

uj u∗

j f

λj − ω2 + ζj

0.01 0.1 1 10 2 4 6 8 10 12 "damped" "damped10" "damped7"

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Lanczos process

Krylov space: {K −1f , K −1MK −1f , . . . , (K −1M)jK −1f , . . .} Lanczos method: 1. Compute the initial vector v1 = K −1f . 2. For j = 1, . . . , k 2.1. Compute Krylov vector vj+1 = K −1Mvj. 2.2. Orthogonalize vj+1 against v1, . . . , vj so that v ∗

j+1MVj = 0.

Lanczos vectors Vk = [v1, . . . , vk].

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Eigenvalue solver

Lanczos vectors Vk = [v1, . . . , vk]. Projection Tk = V ∗

k MK −1MVk (k × k tridiagonal matrix)

Ritz values: Tkz = θz Ritz vectors: ˜ u = Vkz Form an approximate eigenpair of K −1M: K −1M˜ u − θ˜ uM is small for the large |θ|’s.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Convergence

Lanczos method computes the dominant eigenvalues of K −1M Ku = λMu λ−1u = K −1Mu The Lanczos method computes large λ−1 i.e. small λ’s.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Shifted linear systems

Analyzed in the context of model reduction methods Feldman, Freund, Bai, Grimme, Sorensen, Van Dooren, Ruhe, Skoogh, Olsson, Simoncini, M., ... Connection with eigendecomposition Connection with iterative linear solvers Connection with rational approximation (Pad´ e)

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Iterative solver connection

Linear system (K − ω2M)x = f Precondition: K −1(K − ω2M)x = K −1f Use Lanczos (Conjugate Gradients) :

fast convergence when ω small

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Apply Lanczos to K −1(K − ω2M) − → Vk(ω), Tk(ω) If we apply Lanczos to K −1M − → Vk, Tk:

Vk(ω) = Vk Tk(ω) = I + ω2Tk

So, there is no need to recompute the Krylov space for each ω separately.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Conjugate gradient connection

eigenvalues of K −1(K − ω2M) are λj − ω2 λj Eigenvalues are clustered around one.

spectrum of K − λM spectrum of K −1(K − ωM) ω2 1

When there are no eigenvalues λ between 0 and ω2, then we have a positive definite linear system

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Define the error e(ω) = y(ω) − ˜ y(ω) MINRES minimizes the residual, i.e.

n

• j=1

λj − ω2 λj 2 (u∗

j Me(ω2))2

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Define the error e(ω) = y(ω) − ˜ y(ω) MINRES minimizes the residual, i.e.

n

• j=1

λj − ω2 λj 2 (u∗

j Me(ω2))2

Lanczos (CG) minimizes

n

• j=1

λj − ω2 λj

• (u∗

j Me(ω2))2

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Define the error e(ω) = y(ω) − ˜ y(ω) MINRES minimizes the residual, i.e.

n

• j=1

λj − ω2 λj 2 (u∗

j Me(ω2))2

Lanczos (CG) minimizes

n

• j=1

λj − ω2 λj

• (u∗

j Me(ω2))2

MINRES is less sensitive to the eigenvalues near ω than

• Lanczos. Lanczos usually produces smaller errors.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Lanczos produces an approximation of the form ˜ x =

k

• j=1

˜ uj u∗

j f

˜ λj − ω2 where (˜ λj, ˜ uj) is a Ritz pair of K, M.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Lanczos produces an approximation of the form ˜ x =

k

• j=1

˜ uj u∗

j f

˜ λj − ω2 where (˜ λj, ˜ uj) is a Ritz pair of K, M. MINRES produces an approximation of the form ˆ x =

k

• j=1

ˆ uj(ω) uj(ω)∗f ˆ λj(ω) − ω2 where (ˆ λj, ˆ uj) is a harmonic Ritz value

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

MINRES versus Lanczos

Lanczos produces an approximation of the form ˜ x =

k

• j=1

˜ uj u∗

j f

˜ λj − ω2 where (˜ λj, ˜ uj) is a Ritz pair of K, M. MINRES produces an approximation of the form ˆ x =

k

• j=1

ˆ uj(ω) uj(ω)∗f ˆ λj(ω) − ω2 where (ˆ λj, ˆ uj) is a harmonic Ritz value For the MINRES method, there never is a ˆ λj(ω) = ω2.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Example

20 40 60 80 100 120 140 160 180 200 Lanczos MINRES

10−1 10−2 10−3 10−4 10−5 10−6 10−7 10−8

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Pad´ e connection

Suppose we solve (K − ω2M)x = f The solution can be written as a Taylor series: x = x0 + ω2x2 + ω4x4 + · · · The Lanczos method uses starting vector K −1f Let the solution computed by the projection on the Ritz vectors be ˜ x = ˜ x0 + ω2˜ x2 + ω4˜ x4 + · · · then ˜ x2j = x2j for j = 0, . . . , k − 1

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Implementation

For (K − ω2M)x = f : Apply Lanczos to K −1M: Vk, Tk For each ω, solve (I − ω2Tk)z = e1K −1f M and compute the solution x = Vkz.

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Windscreen

Glaverbel-BMW windscreen with 10% structural damping grid : 3 layers of 60 × 30 HEX08 elements (n = 22, 692) unit point force at one of the corners Direct method : 2653 seconds Lanczos method : 14 seconds wanted : displacement for ω = [0.5Hz, 200Hz].

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

1e-16 1e-14 1e-12 1e-10 1e-08 1e-06 1e-04 0.01 1 100 20 40 60 80 100 120 140 160 180 200 "direct.exact" "kry.err"

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Reorthogonalization

The Lanczos process builds an M orthogonal basis in exact arithmetic Influence of reorthogonalization:

1e-14 1e-12 1e-10 1e-08 1e-06 0.0001 0.01 1 20 40 60 80 100 120 140 160 180 200 "exact" "noreorth" "reorth"

With reorthogonalization, we can solve for more frequencies

Karl Meerbergen Krylov methods for fast frequency response computations

Outline Motivation Rayleigh damping Modal truncation Lanczos’ method Numerical example Conclusions

Conclusions

Large reduction in computation time possible with model reduction methods Rayleigh damping is special structure that can be exploited using the Lanczos method

Karl Meerbergen Krylov methods for fast frequency response computations