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

Matrix-Eigenvalue Problems in Stochastic Structural Dynamics S Adhikari Department of Aerospace Engineering, University of Bristol, Bristol, U.K. April 2004 Random Eigenvalue Problems – p.1/28

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 April 2004 Random Eigenvalue Problems – p.2/28

Random Eigenvalue Problem The random eigenvalue problem of undamped or proportionally damped linear systems: K ( x ) φ j = λ j M ( x ) φ j λ j eigenvalues; φ j eigenvectors; M ( x ) ∈ R N × N mass matrix and K ( x ) ∈ R N × N stiffness matrix. x ∈ R m is random parameter vector with pdf p x ( x ) = e − L ( x ) − L ( x ) is the log-likelihood function. April 2004 Random Eigenvalue Problems – p.3/28

The Broad Issues To obtain the joint probability density function of the eigenvalues and the eigenvectors If a matrix A = M − 1 K has pdf f ( A ) then the joint pdf of the eigenvalues (R. J. Muirhead, Theorem 3.2.17, pp 104) � π N 2 / 2 Γ N (( N/ 2))Π N f ( HΛH T ) d H i ≤ j ( λ i − λ j ) O ( N ) It is hard to get marginal distribution of the eigenvalues - too much information!! April 2004 Random Eigenvalue Problems – p.4/28

Perturbation Method Taylor series expansion of λ j ( x ) about x = α λ j ( x ) ≈ λ j ( α ) + d T λ j ( α ) ( x − α ) + 1 2 ( x − α ) T D λ j ( 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 April 2004 Random Eigenvalue Problems – p.5/28

α -centered perturbation The r th moment of λ j ( x ) : � � j ( x ) e − L ( x ) d x = (2 π ) − m/ 2 m e − h j ( x ) d x λ ( r ) m λ r = j R R (1) where h j ( x ) = L ( x ) − r ln λ j ( x ) (2) Expand the function h ( x ) in a Taylor series about a point where h j ( x ) attends its global minimum. April 2004 Random Eigenvalue Problems – p.6/28

α -centered perturbation Therefore, the optimal point can be obtained as ∂h j ( x ) = 0 , ∀ k (3) ∂x k Combining for all k we have d λ j ( α ) = λ j ( α ) d L ( α ) /r (4) April 2004 Random Eigenvalue Problems – p.7/28

Multidimensional Integrals We want to evaluate an m -dimensional integral over the unbounded domain R m : � m e − f ( x ) d x J = R 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 ∈ R m April 2004 Random Eigenvalue Problems – p.8/28

Multidimensional Integrals Therefore, at x = θ ∂f ( x ) = 0 , ∀ k or d f ( θ ) = 0 ∂x k Expand f ( x ) in a Taylor series about θ : � � � T D f ( θ )( x − θ ) + ε ( x , θ ) f ( θ ) + 1 2 ( x − θ ) − J = m e d x R � T D f ( θ )( x − θ ) − ε ( x , θ ) d x = e − f ( θ ) 2 ( x − θ ) m e − 1 R April 2004 Random Eigenvalue Problems – p.9/28

Multidimensional Integrals Use the coordinate transformation: ξ = ( x − θ ) D − 1 / 2 ( θ ) f The Jacobian: � J � = � D f ( θ ) � − 1 / 2 The integral becomes: � � ξ � T ξ m � D f ( θ ) � − 1 / 2 e − 1 J ≈ e − f ( θ ) d ξ 2 R J ≈ (2 π ) m/ 2 e − f ( θ ) � D f ( θ ) � − 1 / 2 or April 2004 Random Eigenvalue Problems – p.10/28

Moments of Single Eigenvalues An arbitrary r th order moment of the eigenvalues can be obtained from � � � µ ( r ) λ r m λ r = E j ( x ) = j ( x ) p x ( x ) d x j R � m e − ( L ( x ) − r ln λ j ( x )) d x , = r = 1 , 2 , 3 · · · R Previous result can be used by choosing f ( x ) = L ( x ) − r ln λ j ( x ) April 2004 Random Eigenvalue Problems – p.11/28

Moments of Single Eigenvalues After some simplifications j ( θ ) e − L ( θ ) µ ( r ) ≈ (2 π ) m/ 2 λ r j � � − 1 / 2 � � � D L ( θ ) + 1 r r d L ( θ ) d L ( θ ) T − � � λ j ( θ ) D λ j ( θ ) � r = 1 , 2 , 3 , · · · θ is obtained from: d λ j ( θ ) r = λ j ( θ ) d L ( θ ) April 2004 Random Eigenvalue Problems – p.12/28

Maximum Entropy pdf Constraints for u ∈ [0 , ∞ ] : � ∞ p λ j ( u ) du = 1 0 � ∞ u r p λ j ( u ) du = µ ( r ) j , r = 1 , 2 , 3 , · · · , n 0 Maximizing Shannon’s measure of entropy � ∞ S = − 0 p λ j ( u ) ln p λ j ( u ) du , the pdf of λ j is p λ j ( u ) = e − { ρ 0 + � n i =1 ρ i u i } = e − ρ 0 e − � n i =1 ρ i u i , u ≥ 0 April 2004 Random Eigenvalue Problems – p.13/28

Maximum Entropy pdf Taking first two moments, the resulting pdf is a truncated Gaussian density function   � � 2   u − �   λ j 1 � exp p λ j ( u ) = � − √ 2 σ 2   � 2 πσ j Φ λ j /σ j   j j = µ (2) − � where σ 2 λ 2 j j Ensures that the probability of any eigenvalues becoming negative is zero April 2004 Random Eigenvalue Problems – p.14/28

Central χ 2 Approximation Pdf of j th eigenvalue � u − η j � = ( u − η j ) ν j / 2 − 1 e − ( u − η j ) / 2 γ j p λ j ( u ) ≈ 1 p χ 2 (2 γ j ) ν j / 2 Γ( ν j / 2) γ j γ j νj The constants η j , γ j , and ν j are such that the first three moments of λ j are the same. April 2004 Random Eigenvalue Problems – p.15/28

Joint Moments of Two Eigenvalues Arbitrary r − s -th order joint moment of two eigenvalues � � µ ( rs ) λ r j ( x ) λ s = E l ( x ) jl � = m exp {− ( L ( x ) − r ln λ j ( x ) − s ln λ l ( x )) } d x , R Choose f ( x ) = L ( x ) − r ln λ j ( x ) − s ln λ l ( x ) April 2004 Random Eigenvalue Problems – p.16/28

Joint Moments of Two Eigenvalues After some simplifications µ ( rs ) l ( θ ) exp {− L ( θ ) } � D f ( θ ) � − 1 / 2 ≈ (2 π ) m/ 2 λ r j ( θ ) λ s jl where θ is obtained from: r s d L ( θ ) = λ j ( θ ) d λ j ( θ ) + λ l ( θ ) d λ l ( θ ) j ( θ ) d λ j ( θ ) d λ j ( θ ) T − r D f ( θ ) = D L ( θ ) + λ 2 l ( θ ) d λ l ( θ ) d λ l ( θ ) T − r s s λ j ( θ ) D λ j ( θ ) + λ l ( θ ) D λ l ( θ ) λ 2 April 2004 Random Eigenvalue Problems – p.17/28

Joint Moments of Multiple Eigenvalues We want to obtain � � � µ ( r 1 r 2 ··· r n ) λ r 1 j 1 ( x ) λ r 2 j 2 ( x ) · · · λ r n = j n ( x ) p x ( x ) d x j 1 j 2 ··· j n m R It can be shown that ≈ (2 π ) m/ 2 � � µ ( r 1 r 2 ··· r n ) λ r 1 j 1 ( θ ) λ r 2 j 2 ( θ ) · · · λ r n j n ( θ ) j 1 j 2 ··· j n exp {− L ( θ ) } � D f ( θ ) � − 1 / 2 April 2004 Random Eigenvalue Problems – p.18/28

Joint Moments of Multiple Eigenvalues Here θ is obtained from r 1 r 2 r n d L ( θ ) = λ j 1 ( θ ) d λ j 1 ( θ )+ λ j 1 ( θ ) d λ j 2 ( θ )+ · · · λ j n ( θ ) d λ jn ( θ ) and the Hessian matrix is given by j n ,r n � r j ( θ ) d λ j ( θ ) d λ j ( θ ) T D f ( θ ) = D L ( θ )+ λ 2 j = j 1 , j 2 , · · · r = r 1 , r 2 , · · · r − λ j ( θ ) D λ j ( θ ) April 2004 Random Eigenvalue Problems – p.19/28

Numerical example Undamped two degree-of-system system: ¯ m 1 = 1 Kg, m 2 = 1 . 5 Kg, k 1 = 1000 ¯ k 2 = 1100 N/m and k 3 = 100 N/m. N/m, 1 2 m m 2 1 �� �� � � �� �� � � �� �� � � �� �� � � �� �� � � �� �� � � �� �� k k2 � � k 3 �� �� Only � � 1 �� �� � � �� �� � � the stiffness parameters k 1 and k 2 are uncertain: k i (1 + ǫ i x i ) , i = 1 , 2 . x = { x 1 , x 2 } T ∈ R 2 and the k i = ¯ ‘strength parameters’ ǫ 1 = ǫ 2 = 0 . 25 . April 2004 Random Eigenvalue Problems – p.20/28

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. April 2004 Random Eigenvalue Problems – p.21/28

Numerical example The percentage error: Error i th method = { µ ′ k } i th method − { µ ′ k } MCS × 100 { µ ′ k } MCS i = 1 , · · · 5 . April 2004 Random Eigenvalue Problems – p.22/28

Numerical example 20 Mean−centered 1st−order Mean−centered 2nd−order 18 α −centered 1st−order α −centered 2nd−order 16 Asymptotic Method 14 Percentage error wrt MCS 12 10 8 6 4 2 0 1 2 3 4 k−th order moment: E [ λ k 1 ] Percentage error for the first four raw moments of the first eigenvalue April 2004 Random Eigenvalue Problems – p.23/28

Numerical example 0 −2 −4 −6 Percentage error wrt MCS −8 −10 Mean−centered 1st−order −12 Mean−centered 2nd−order α −centered 1st−order α −centered 2nd−order −14 Asymptotic Method −16 −18 −20 1 2 3 4 k−th order moment: E [ λ k 2 ] Percentage error for the first four raw moments of the second eigenvalue April 2004 Random Eigenvalue Problems – p.24/28

Numerical example 3 x 10 −3 Mean−centered 1st−order Mean−centered 2nd−order α −centered 1st−order α −centered 2nd−order 2.5 Asymptotic Method 2 (u) 1.5 1 p λ 1 0.5 0 0 500 1000 1500 u Probability density function of the first eigenvalue April 2004 Random Eigenvalue Problems – p.25/28