Random Eigenvalue Problem for Linear Dynamic Systems S. A DHIKARI - PowerPoint PPT Presentation

Random Eigenvalue Problem for Linear Dynamic Systems S. A DHIKARI Cambridge University Engineering Department Cambridge, U.K. Random Eigenvalue Problems – p.1/23

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 Random Eigenvalue Problems – p.2/23

Random eigenvalue problem The random eigenvalue problem of undamped or proportionally damped linear systems: K ( x ) φ j = λ j M ( x ) φ j . (1) 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 ) = (2 π ) − m/ 2 e − x T x / 2 . (2) Random Eigenvalue Problems – p.3/23

The Fundamental aim To obtain the joint probability density function of the eigenvalues and the eigenvectors. If the matrix M − 1 K 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.4/23

Mean-centered perturbation method Assume that M ( 0 ) = M 0 and K ( 0 ) = K 0 are ‘deterministic parts’ (in general different from the mean matrices). The deterministic eigenvalue problem K 0 φ j 0 = λ j 0 M 0 φ j 0 . The eigenvalues λ j ( x ) : R m → R are non-linear functions of x . Here λ j ( x ) is replaced by its Taylor series about the point x = 0 λ j ( 0 ) x + 1 x T D λ j ( 0 ) x . λ j ( x ) ≈ λ j ( 0 ) + d T (3) 2 d λ j ( 0 ) ∈ R m and D λ j ( 0 ) ∈ R m × m are respectively the gra- dient vector and the Hessian matrix of λ j ( x ) evaluated at x = 0 . Random Eigenvalue Problems – p.5/23

α -centered perturbation method We are looking for a point x = α in the x -space such that the Taylor series expansion of λ j ( x ) about this point λ j ( α ) ( x − α ) + 1 2 ( x − α ) T D λ j ( α ) ( x − α ) λ j ( x ) ≈ λ j ( α ) + d T (4) is optimal in some sense. The optimal point α is selected such that the mean or the first moment of each eigenvalue is calculated most accurately. Random Eigenvalue Problems – p.6/23

α -centered perturbation method The mean of λ j ( x ) can be obtained as � � m e − h ( x ) d x ¯ m λ j ( x ) p ( x ) d x = (2 π ) − m/ 2 λ j = (5) R R h ( x ) = x T x / 2 − ln λ j ( x ) . where (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. Therefore, the optimal point can be obtained as ∂h ( x ) 1 ∂λ j ( x ) = 0 or x k = , ∀ k. (7) ∂x k λ j ( x ) ∂x k Random Eigenvalue Problems – p.7/23

α -centered perturbation method Combining for all k we have d λ j ( α ) = λ j ( α ) α . Rearranging α = d λ j ( α ) /λ j ( α ) . (8) This equation immediately gives a recipe for an iterative algorithm to obtain α . Substituting d λ j ( α ) in Eq. (4) � 1 − | α | 2 � + 1 2 α T D λ j ( α ) α λ j ( x ) ≈ λ j ( α ) + α T � � x + 1 x T D λ j ( α ) x . λ j ( α ) I − D λ j ( α ) (9) 2 Random Eigenvalue Problems – p.8/23

Eigenvalue statistics using theory of quadratic forms Both approximations yield a quadratic form in Gaussian 2 x T A j x . j x + 1 random variable λ j ( x ) ≈ c j + a T The moment generating function: ≈ e sc j + s 2 − 1 a j j [ I − s A j ] 2 a T � e sλ j ( x ) � M λ j ( s ) = E � (10) � I − s A j � Cumulants: � c j + 1 2 Trace ( A j ) if r = 1 , � � κ r = (11) a j + ( r − 1)! r ! j A r − 2 2 a T A r Trace if r ≥ 2 . j j 2 Random Eigenvalue Problems – p.9/23

Asymptotic analysis We want to evaluate an integral of the following form: � � h ( x ) d x � m f ( x ) p ( x ) d x = (2 π ) − m/ 2 J = m e (12) R R h ( x ) = ln f ( x ) − x T x / 2 . � where (13) Assume f ( x ) : R m → R is smooth and at least twice differentiable and � h ( x ) reaches its global maximum at an unique point θ ∈ R m . Therefore, at x = θ ∂ � h ( x ) ∂ ln f ( x ) , ∀ k, or θ = ∂ = 0 or x k = ∂ x ln f ( θ ) . (14) ∂x k ∂x k Random Eigenvalue Problems – p.10/23

Asymptotic analysis Further assume that � h ( θ ) is so large that � � � � 1 � � D j ( � h ( θ )) � → 0 for j > 2 � � (15) � � h ( θ ) where D j ( � h ( θ )) is j th order derivative of � h ( x ) evaluated at x = θ . Under such assumptions, using second-order Taylor series of � h ( x ) the integral (12) can be evaluated as � θ � h ( θ ) � e T θ / 2 − H ( θ ) � − 1 / 2 . � � J ≈ � = f ( θ ) e (16) � � H ( θ ) � Random Eigenvalue Problems – p.11/23

Asymptotic analysis An arbitrary r th order moment of the eigenvalues � m λ r µ ′ r = j ( x ) p ( x ) d x , r = 1 , 2 , 3 · · · (17) R Comparing this with Eq. (12) it is clear that h ( x ) = r ln λ j ( x ) − x T x / 2 . � f ( x ) = λ r j ( x ) and (18) The optimal point θ can be obtained from (14) as θ = r d λ j ( θ ) /λ j ( θ ) . (19) Random Eigenvalue Problems – p.12/23

Asymptotic analysis Using the asymptotic approximation, the r th moment: � � − 1 / 2 � � j ( θ ) e − | θ | 2 � I + 1 r r θθ T − � � r = λ r µ ′ D λ j ( θ ) . (20) 2 � λ j ( θ ) The mean of the eigenvalues (by substituting r = 1 ): � � λ j = λ j ( θ ) e − | θ | 2 � − 1 / 2 . � I + θθ T − D λ j ( θ ) /λ j ( θ ) ¯ (21) 2 � λ j ) r � � r � = � r ( λ j − ¯ k ¯ λ r − k ( − 1) r − k µ ′ Central moments: E . j k =0 k Random Eigenvalue Problems – p.13/23

Pdf of the eigenvalues Theorem 1 λ j ( x ) is distributed as a non-central χ 2 random variable with noncentrality parameter δ 2 and degrees-of-freedom m ′ if and only if (a) A 2 j = A j , (b) Trace ( A j ) = m ′ and (c) a j = A j a j , δ 2 = c j = a T j a j / 4 . This implies that the the Hessian matrix A j should be an idempotent matrix. In general this requirement is not ex- pected to be satisfied for eigenvalues of real structural sys- tems. Random Eigenvalue Problems – p.14/23

Pearson’s approximation (central χ 2 ) Pdf of the j th eigenvalue � u − � � η ) ν/ 2 − 1 e − ( u − � η ) / 2 � γ p λ j ( u ) ≈ 1 η = ( u − � γ p χ 2 . (22) γ ) ν/ 2 Γ( ν/ 2) γ (2 � � � ν where η = − 2 κ 22 + κ 1 κ 3 , and ν = 8 κ 23 γ = κ 3 � , � κ 32 . (23) κ 3 4 κ 2 Random Eigenvalue Problems – p.15/23

Non-central χ 2 approximation Pdf of the j th eigenvalue � u − η j � p λ j ( u ) ≈ 1 p Q j (24) γ j γ j where � ∞ p Q j ( u ) = e − ( δ j + u/ 2) u m/ 2 − 1 ( δu ) r r ! 2 r Γ( m/ 2 + r ) . (25) 2 m/ 2 r =0 Trace ( A j ) where η j = c j − 1 j A − 1 2 a T , δ 2 j = ρ T j a j , γ j = j ρ j and 2 m ρ j = A − 1 j a j . Random Eigenvalue Problems – p.16/23

✂ ✂ ✁ ✁ ✁ ✁ ✁ ✁ ✂ ✂ ✂ ✂ ✁ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ✁ ✁ ✂ ✂ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✂ ✁ � � � � � � � � � � ✂ Numerical Example Undamped two degree-of-system system: m 1 = 1 Kg, m 2 = 1 . 5 Kg, ¯ k 1 = 1000 N/m, ¯ k 2 = 1100 N/m and k 3 = 100 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 = k i (1 + ǫ i x i ) , i = 1 , 2 . x = { x 1 , x 2 } T ∈ R 2 and the ‘strength ¯ parameters’ ǫ 1 = ǫ 2 = 0 . 25 . Random Eigenvalue Problems – p.17/23

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. The percentage error: Error i th method = { µ ′ k } i th method − { µ ′ k } MCS × 100 { µ ′ k } MCS Random Eigenvalue Problems – p.18/23 .

Numerical Example 20 Mean−centered 1st−order Mean−centered 2nd−order 18 α −centered 1st−order α −centered 2nd−order 16 Asymptotic Method Percentage error wrt MCS 14 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 Random Eigenvalue Problems – p.19/23

Numerical Example 0 −2 −4 Percentage error wrt MCS −6 −8 −10 Mean−centered 1st−order −12 Mean−centered 2nd−order α −centered 1st−order −14 α −centered 2nd−order 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 Random Eigenvalue Problems – p.20/23

Numerical Example 3 x 10 −3 Mean−centered 1st−order Mean−centered 2nd−order α −centered 1st−order 2.5 α −centered 2nd−order 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 Random Eigenvalue Problems – p.21/23