Finding Rightmost Eigenvalues of Large, Sparse, Nonsymmetric

Parameterized Eigenvalue Problems

Minghao WuAMSC Program

mwu@math.umd.eduAdvisor:

Dr. Howard ElmanDepartment of Computer Science

elman@cs.umd.edu

2

Motivation

• To determine the stability of the linearized system of the form:

AxxB

• The steady state solution x* is

- stable, if all the eigenvalues of Ax = λBx have

negative real parts;

- unstable, otherwise.

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Problem Statement

To find the rightmost eigenvalues of:

BxAx

where matrices A and B are

• Real N-by-N

• Large

• Sparse

• Nonsymmetric

• Depend on one or several parameters

4

Method• Eigensolver: Arnoldi Algorithm - Iterative method

- Based on Krylov subspace:

111

2111, uAuAAuuuAK k

k

- Computes Arnoldi decomposition:Tkkkkkk euHUAU 1

where:

[Uk uk+1]: an orthonormal basis of

Hk: k-by-k upper Hessenberg matrix, k « N

βk: scalar, ek: k-by-1 vector [0 0 … 0 1]’

111 uAAuu k

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Arnoldi Algorithm (continue)

• Eigenvalues of Hk approximate eigenvalues of A Premultiply previous equation by transpose(Uk):

.1 kTkk

Tkkkk

Tkk

Tk HeuUHUUAUU

Let (λ,z) be an eigenpair of Hk, then:

.zUλzUA kkTkk

Tkk

Tk UzUUzzHU

As k increases, ||Residual|| will decrease.

When k = N, Residual = 0.

0 zUzUAU kkTk

Residual

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Matrix Transformation

• Motivation - Arnoldi Algorithm cannot solve generalized eigenvalue problem

- It converges to well – separated extremal eigenvalues, not

rightmost eigenvalues

• Shift – Invert Transformation

1

;, 1

xxTBxAx

BBABAT

SI

SI

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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 Uk+1

H: the upper Hessenberg matrix Hk

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Test ProblemOlmstead Model (see Olmstead et al (1986)):

SucbS

uRucuSu

t

xxxxt

1

3

with boundary conditions:

,0,0 xSu

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

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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

].[ 2/2/2211 NNT SuSuSuy

• 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

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Test Problem (continue)

Computational Result:

• Rightmost eigenvalues: λ1,2 = 0 ± 0.4472i

• Residual: ||Axi - λi xi|| = 8.4504e-012, i=1,2

• The result agrees with the literature.

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Implicitly Restarted Arnoldi (IRA)

• Motivation - Large k is not practical Example:

Size of A: 10,000

Value of k: 100

Memory required to

store U100 in double precision:

10 MB

- When B is singular, Arnoldi algorithm may give rise to spurious

eigenvalues

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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 u1

2. Filter out the m-k unwanted eigendirections from u1

(Key Technique: shifted QR algorithm) 3. Restart the process with filtered starting vector till the k eigenvalues of interest converge

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Test Problem

xM

xC

CKT

00

0

0

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.

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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

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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)

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Future Work (AMSC 664)

• Solve the third test problem

• Implement iterative solvers for linear systems