[PPT] - Matrix-Eigenvalue Problems in Stochastic Structural Dynamics S PowerPoint Presentation

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April 2004

Matrix-Eigenvalue Problems in Stochastic Structural Dynamics

S Adhikari

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

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April 2004

Outline of the Presentation

Random eigenvalue problem Perturbation Methods Asymptotic analysis of multidimensional integrals Moments and pdf of the eigenvalues Numerical Example & results Conclusions & open problems

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Random Eigenvalue Problem

The random eigenvalue problem of undamped or proportionally damped linear systems: K(x)φj = λjM(x)φj 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 px(x) = e−L(x) −L(x) is the log-likelihood function.

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April 2004

The Broad Issues

To obtain the joint probability density function of the eigenvalues and the eigenvectors If a matrix A = M−1K has pdf f(A) then the joint pdf of the eigenvalues (R. J. Muirhead, Theorem 3.2.17, pp 104) πN 2/2 ΓN((N/2))ΠN

i≤j(λi − λj)

O(N)

f(HΛHT)dH It is hard to get marginal distribution of the eigenvalues - too much information!!

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April 2004

Perturbation Method

Taylor series expansion of λj(x) about x = α λj(x) ≈ λj(α) + dT

λj(α) (x − α)

+ 1 2 (x − α)T Dλj(α) (x − α) In mean-centered approach α is the mean of x Alternatively, α can be obtained such that the any moment of each eigenvalue is calculated most accurately

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April 2004

α-centered perturbation

The rth moment of λj(x): λ(r)

j

=

R

m λr

j(x)e−L(x) dx = (2π)−m/2

R

m e−hj(x) dx

(1) where hj(x) = L(x) − r ln λj(x) (2) Expand the function h(x) in a Taylor series about a point where hj(x) attends its global minimum.

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April 2004

α-centered perturbation

Therefore, the optimal point can be obtained as ∂hj(x) ∂xk = 0, ∀k (3) Combining for all k we have dλj(α) = λj(α)dL(α)/r (4)

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April 2004

Multidimensional Integrals

We want to evaluate an m-dimensional integral over the unbounded domain Rm: J =

R

m e−f(x) dx

Assume f(x) is smooth and at least twice differentiable The maximum contribution to this integral comes from the neighborhood where f(x) reaches its global minimum, say θ ∈ Rm

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April 2004

Multidimensional Integrals

Therefore, at x = θ ∂f(x) ∂xk = 0, ∀k

r

df (θ) = 0 Expand f(x) in a Taylor series about θ: J =

R

m e

−

f(θ)+ 1

2(x−θ) TDf(θ)(x−θ)+ε(x,θ)

dx

= e−f(θ)

R

m e− 1 2(x−θ) TDf(θ)(x−θ)−ε(x,θ) dx

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Multidimensional Integrals

Use the coordinate transformation: ξ = (x − θ) D−1/2

f

(θ) The Jacobian: J = Df (θ)−1/2 The integral becomes: J ≈ e−f(θ)

R

m Df (θ)−1/2 e

− 1

2

ξ

Tξ

dξ
r

J ≈ (2π)m/2e−f(θ) Df (θ)−1/2

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April 2004

Moments of Single Eigenvalues

An arbitrary rth order moment of the eigenvalues can be obtained from µ(r)

j

= E

λr

j(x)

=
R

m λr

j(x)px(x) dx

=

R

m e−(L(x)−r ln λj(x)) dx,

r = 1, 2, 3 · · · Previous result can be used by choosing f(x) = L(x) − r ln λj(x)

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April 2004

Moments of Single Eigenvalues

After some simplifications µ(r)

j

≈ (2π)m/2λr

j(θ)e−L(θ)

DL(θ) + 1

rdL(θ)dL(θ)T − r λj(θ)Dλj(θ)

−1/2

r = 1, 2, 3, · · · θ is obtained from: dλj(θ)r = λj(θ)dL(θ)

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April 2004

Maximum Entropy pdf

Constraints for u ∈ [0, ∞]: ∞ pλj(u)du = 1 ∞ urpλj(u)du = µ(r)

j ,

r = 1, 2, 3, · · · , n Maximizing Shannon’s measure of entropy S = − ∞

0 pλj(u) ln pλj(u)du, the pdf of λj is

pλj(u) = e−{ρ0+n

i=1 ρiui} = e−ρ0e− n i=1 ρiui,

u ≥ 0

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Maximum Entropy pdf

Taking first two moments, the resulting pdf is a truncated Gaussian density function pλj(u) = 1 √ 2πσj Φ

λj/σj

exp      −

u −

λj 2 2σ2

j

     where σ2

j = µ(2) j

− λ2

j

Ensures that the probability of any eigenvalues becoming negative is zero

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April 2004

Central χ2 Approximation

Pdf of jth eigenvalue pλj(u) ≈ 1 γj pχ2

νj

u − ηj γj

= (u − ηj)νj/2−1e−(u−ηj)/2γj

(2γj)νj/2Γ(νj/2) The constants ηj, γj, and νj are such that the first three moments of λj are the same.

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April 2004

Joint Moments of Two Eigenvalues

Arbitrary r − s-th order joint moment of two eigenvalues µ(rs)

jl

= E

λr

j(x)λs l (x)

=
R

m exp {− (L(x) − r ln λj(x) − s ln λl(x))} dx,

Choose f(x) = L(x) − r ln λj(x) − s ln λl(x)

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April 2004

Joint Moments of Two Eigenvalues

After some simplifications µ(rs)

jl

≈ (2π)m/2λr

j(θ)λs l (θ) exp {−L (θ)} Df (θ)−1/2

where θ is obtained from: dL(θ) = r λj(θ)dλj(θ) + s λl(θ)dλl(θ) Df (θ) = DL(θ) +

r λ2

j(θ)dλj(θ)dλj(θ)T −

r λj(θ)Dλj(θ) + s λ2

l(θ)dλl(θ)dλl(θ)T −

s λl(θ)Dλl(θ)

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April 2004

Joint Moments of Multiple Eigenvalues

We want to obtain µ(r1r2···rn)

j1j2···jn

=

R

m

λr1

j1(x)λr2 j2(x) · · · λrn jn(x)

px(x) dx

It can be shown that µ(r1r2···rn)

j1j2···jn

≈ (2π)m/2 λr1

j1 (θ) λr2 j2 (θ) · · · λrn jn (θ)

exp {−L (θ)} Df (θ)−1/2

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April 2004

Joint Moments of Multiple Eigenvalues

Here θ is obtained from dL(θ) = r1 λj1(θ)dλj1(θ)+ r2 λj1(θ)dλj2(θ)+· · · rn λjn(θ)dλjn(θ) and the Hessian matrix is given by Df (θ) = DL(θ)+

jn,rn

j = j1, j2, · · ·

r = r1, r2, · · · r λ2

j (θ)dλj(θ)dλj(θ)T

− r λj (θ)Dλj(θ)

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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 m 2 k k2 3 k 1 1 1

2

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

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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.

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April 2004

Numerical example

The percentage error: Errorith method = {µ′

k}ith method − {µ′ k}MCS

{µ′

k}MCS

× 100 i = 1, · · · 5.

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April 2004

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

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April 2004

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

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April 2004

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

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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

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Conclusions

The statistics of the eigenvalues of linear stochastic dynamic systems has been considered A closed form expression is obtained for general order joint moments of the eigenvalues Pdf of the eigenvalues are obtained: using maximum entropy method in terms of central χ2 density

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Open Problems

Joint statistics of the eigenvectors Joint statistics of the eigenvalues and eigenvectors Systems with non-proportional damping

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