Distribution of Eigenvalues of Linear Stochastic Systems S A DHIKARI - - PowerPoint PPT Presentation

distribution of eigenvalues of linear stochastic systems
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Distribution of Eigenvalues of Linear Stochastic Systems S A DHIKARI - - PowerPoint PPT Presentation

Distribution of Eigenvalues of Linear Stochastic Systems S A DHIKARI and R. S. L ANGLEY Department of Aerospace Engineering, University of Bristol, Bristol, U.K. Email: S.Adhikari@bristol.ac.uk URL:


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SLIDE 1

Distribution of Eigenvalues of Linear Stochastic Systems

S ADHIKARI† and R. S. LANGLEY‡

†Department of Aerospace Engineering, University of Bristol, Bristol, U.K.

Email: S.Adhikari@bristol.ac.uk URL: http://www.aer.bris.ac.uk/contact/academic/adhikari/home.html

‡Department of Engineering, University of Cambridge, Cambridge, UK

Random Eigenvalue Problems – p.1/28

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SLIDE 2

Outline of the talk

  • Random eigenvalue problem
  • Perturbation Methods
  • Mean-centered perturbation method
  • α-centered perturbation method
  • Asymptotic analysis
  • PDF of the eigenvalues
  • Numerical Example
  • Conclusions & Open Problems

Random Eigenvalue Problems – p.2/28

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SLIDE 3

Random eigenvalue problem

The random eigenvalue problem of undamped or proportionally damped linear systems:

K(x)φj = λjM(x)φj (1)

λj eigenvalues; φj eigenvectors; M(x) ∈ RN×N mass matrix and K(x) ∈ RN×N stiffness matrix.

x ∈ Rm is random parameter vector with pdf

p(x) = (2π)−m/2e−xTx/2

(2)

Random Eigenvalue Problems – p.3/28

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SLIDE 4

The fundamental aim

  • To obtain the joint probability density function
  • f the eigenvalues and the eigenvectors.
  • If the matrix M−1K is GUE (Gaussian unitary

ensemble) or GOE (Gaussian orthogonal ensemble) an exact closed-form expression can be obtained for the joint pdf of the eigenvalues.

  • In general the system matrices for real

structures are not GUE or GOE.

Random Eigenvalue Problems – p.4/28

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SLIDE 5

Mean-centered perturbation

Assume that M(0) = M0 and K(0) = K0 are ‘deterministic parts’. Deterministic eigenvalue problem: K0φj0 = λj0M0φj0. The eigenvalues λj(x) : Rm → R are non-linear functions of x. Expanding λj(x) by Taylor series about x = 0: λj(x) ≈ λj(0) + dT

λj(0)x + 1

2

xT Dλj(0)x (3) dλj(0) ∈ Rm: gradient vector, Dλj(0) ∈ Rm×m the

Hessian matrix of λj(x) evaluated at x = 0.

Random Eigenvalue Problems – p.5/28

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SLIDE 6

α-centered perturbation

We are looking for a point x = α in the x-space such that the Taylor series expansion of λj(x) about this point λj(x) ≈ λj(α) + dT

λj(α) (x − α)

+ 1 2 (x − α)T Dλj(α) (x − α)

(4)

is optimal in some sense. The optimal point α is selected such that the mean or the first moment

  • f each eigenvalue is calculated most accurately.

Random Eigenvalue Problems – p.6/28

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SLIDE 7

α-centered perturbation

The mean of λj(x) can be obtained as ¯ λj =

  • R

m λj(x)p(x) dx = (2π)−m/2

  • R

m e−h(x) dx

(5)

where h(x) = xT x/2 − ln λj(x)

(6)

Expand the function h(x) in a Taylor series about a point where h(x) attends its global minimum. By doing so the error in evaluating the integral (5) would be minimized.

Random Eigenvalue Problems – p.7/28

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SLIDE 8

α-centered perturbation

Therefore, the optimal point can be obtained as ∂h(x) ∂xk = 0

  • r

xk = 1 λj(x) ∂λj(x) ∂xk , ∀k

(7)

Combining for all k we have dλj(α) = λj(α)α. Rearranging α = dλj(α)/λj(α)

(8)

This equation immediately gives a recipe for an iterative algorithm to obtain α.

Random Eigenvalue Problems – p.8/28

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SLIDE 9

α-centered perturbation

Substituting dλj(α) in Eq. (4) λj(x) ≈ λj(α)

  • 1 − |α|2

+ 1 2αT Dλj(α)α + αT λj(α)I − Dλj(α)

  • x + 1

2

xT Dλj(α)x (9)

This, like the mean-centered approach, also re- sults in a quadratic form in the random variable x.

Random Eigenvalue Problems – p.9/28

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SLIDE 10

Eigenvalue statistics

Both approximations yield a quadratic form in Gaussian random variable of the form λj(x) ≈ cj + aT

j x + 1

2

xT Ajx (10)

The moment generating function: Mλj(s) = E

  • esλj(x)

≈ escj+ s2

2 aT j [I−sAj] −1aj

  • I − sAj

(11)

Random Eigenvalue Problems – p.10/28

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SLIDE 11

Eigenvalue statistics

Cumulants: κr =

  • cj + 1

2Trace (Aj)

if r = 1,

r! 2 aT j Ar−2 j

aj + (r−1)!

2

Trace

  • Ar

j

  • if

r ≥ 2

(12)

Thus ¯ λj = κ1 = cj + 1 2Trace (Aj)

(13)

Var [λj] = κ2 = aT

j aj + 1

2Trace

  • A2

j

  • (14)

Random Eigenvalue Problems – p.11/28

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SLIDE 12

Asymptotic analysis

We want to evaluate an integral of the following form: J =

  • R

m f(x)p(x) dx = (2π)−m/2

  • R

m e

  • h(x) dx

(15)

where

  • h(x) = ln f(x) − xT x/2

(16)

Assume f(x) : Rm → R is smooth and at least twice differentiable and h(x) reaches its global maximum at an unique point θ ∈ Rm.

Random Eigenvalue Problems – p.12/28

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SLIDE 13

Asymptotic analysis

Therefore, at x = θ ∂ h(x) ∂xk = 0 or xk = ∂ ∂xk ln f(x), ∀k, or θ = ∂ ∂x ln f(θ).

(17)

Further assume that h(θ) is so large that

  • 1
  • h(θ)

Dj( h(θ))

  • → 0

for j > 2

(18)

Dj( h(θ)): jth order derivative of h(x) at x = θ.

Random Eigenvalue Problems – p.13/28

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SLIDE 14

Asymptotic analysis

Under previous assumptions, using second-order Taylor series of h(x) the integral (12) can be evaluated asymptotically as J ≈ e

h(θ)

  • H(θ)

= f(θ)e

− θ

Tθ/2

  • H(θ)−1/2

(19)

  • H(θ) is the Hessian matrix of

h(x) at x = θ.

Random Eigenvalue Problems – p.14/28

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SLIDE 15

Asymptotic analysis

An arbitrary rth order moment of the eigenvalues µ′

r =

  • R

m λr

j(x)p(x) dx,

r = 1, 2, 3 · · ·

(20)

Comparing this with Eq. (12) it is clear that f(x) = λr

j(x)

and

  • h(x) = r ln λj(x)− xT x/2 (21)

The optimal point θ can be obtained from (14) as θ = r dλj(θ)/λj(θ)

(22)

Random Eigenvalue Problems – p.15/28

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SLIDE 16

Asymptotic analysis

The rth moment: µ′

r = λr j(θ)e− |θ|2

2

  • I + 1

rθθT − r λj(θ)

Dλj(θ)

  • −1/2

(23)

The mean of the eigenvalues (substitute r = 1): ¯ λj = λj(θ)e− |θ|2

2

I + θθT − Dλj(θ)/λj(θ)

  • −1/2

(24)

Central moments: E

  • (λj − ¯

λj)r = r

k=0

r

k

  • (−1)r−kµ′

λr−k

j

.

Random Eigenvalue Problems – p.16/28

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SLIDE 17

Pdf of the eigenvalues

Theorem 1 λj(x) is distributed as a non-central χ2 random variable with noncentrality parameter δ2 and degrees-of-freedom m′ if and only if (a)

A2

j = Aj, (b) Trace (Aj) = m′ and (c)

aj = Ajaj, δ2 = cj = aT

j aj/4.

This implies that the the Hessian matrix Aj should be an idempotent matrix. In general this require- ment is not expected to be satisfied for eigenval- ues of real structural systems.

Random Eigenvalue Problems – p.17/28

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SLIDE 18

Central χ2 approximation (Pearson’s)

Pdf of the jth eigenvalue pλj(u) ≈ 1

  • γpχ2

ν

u − η

  • γ
  • = (u −

η)ν/2−1e−(u−

η)/2 γ

(2 γ)ν/2Γ(ν/2)

(25)

where

  • η = −2κ22 + κ1κ3

κ3 , γ = κ3 4κ2 , and ν = 8κ23 κ32

(26)

Random Eigenvalue Problems – p.18/28

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SLIDE 19

Non-central χ2 approximation

Pdf of the jth eigenvalue pλj(u) ≈ 1 γj pQj u − ηj γj

  • (27)

where pQj(u) =

e−(δj+u/2)um/2−1 2m/2

r=0 (δu)r r! 2rΓ(m/2+r),

ηj = cj − 1

2aT j A−1 j aj, γj = Trace(Aj) 2m

, δ2

j = ρT j ρj and

ρj = A−1

j aj.

Random Eigenvalue Problems – p.19/28

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SLIDE 20

Numerical example

Undamped two degree-of-system system: m1 = 1 Kg, m2 = 1.5 Kg, ¯ k1 = 1000 N/m, ¯ k2 = 1100 N/m and k3 = 100 N/m.

m

1

m

2 1 2

k

1

k

2

k

3

Only the stiffness parameters k1 and k2 are uncer- tain: ki = ¯ ki(1 + ǫixi), i = 1, 2. x = {x1, x2}T ∈ R2 and the ‘strength parameters’ ǫ1 = ǫ2 = 0.25.

Random Eigenvalue Problems – p.20/28

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SLIDE 21

Numerical example

Following six methods are compared

  • 1. Mean-centered first-order perturbation
  • 2. Mean-centered second-order perturbation
  • 3. α-centered first-order perturbation
  • 4. α-centered second-order perturbation
  • 5. Asymptotic method
  • 6. Monte Carlo Simulation (10K samples) - can

be considered as benchmark.

Random Eigenvalue Problems – p.21/28

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SLIDE 22

Numerical example

The percentage error: Errorith method = {µ′

k}ith method − {µ′ k}MCS

{µ′

k}MCS

× 100 i = 1, · · · 5.

Random Eigenvalue Problems – p.22/28

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SLIDE 23

Numerical example

1 2 3 4 2 4 6 8 10 12 14 16 18 20 k−th order moment: E [λk

1]

Percentage error wrt MCS Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Percentage error for the first four raw moments of the first eigenvalue

Random Eigenvalue Problems – p.23/28

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SLIDE 24

Numerical example

1 2 3 4 −20 −18 −16 −14 −12 −10 −8 −6 −4 −2 k−th order moment: E [λk

2]

Percentage error wrt MCS Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Percentage error for the first four raw moments of the second eigenvalue

Random Eigenvalue Problems – p.24/28

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SLIDE 25

Numerical example

500 1000 1500 0.5 1 1.5 2 2.5 3 x 10−3 u pλ

1

(u) Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Probability density function of the first eigenvalue

Random Eigenvalue Problems – p.25/28

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SLIDE 26

Numerical example

400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10−3 u pλ

2

(u) Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Probability density function of the second eigenvalue

Random Eigenvalue Problems – p.26/28

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SLIDE 27

Conclusions

  • Two methods, namely (a) optimal point

expansion method, and (b) asymptotic moment method, are proposed.

  • The optimal point is obtained so that the

mean of the eigenvalues are estimated most accurately.

  • The asymptotic method assumes that the

eigenvalues are large compared to their 3rd

  • rder or higher derivatives.
  • Pdf of the eigenvalues are obtained in terms
  • f central and non-central χ2 densities.

Random Eigenvalue Problems – p.27/28

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SLIDE 28

Open problems

  • Joint statistics (moments/pdf/cumulants) of

the eigenvalues with non-Gaussian system parameters.

  • Statistics of the difference and ratio of the

eigenvalues.

  • Statistics of a single eigenvector (for

GUE/GOE and general matrices).

  • Joint statistics of the eigenvectors.
  • Joint statistics of the eigenvalues and

eigenvectors.

Random Eigenvalue Problems – p.28/28