Joint Distribution of Eigenvalues of Linear Stochastic Systems S Adhikari Department of Aerospace Engineering, University of Bristol, Bristol, U.K. URL: http://www.aer.bris.ac.uk/contact/academic/adhikari/home.html 19 April 2005 Joint Distribution of Eigenvalues – p.1/36
Outline of the Presentation Random eigenvalue problem Existing methods Exact methods Perturbation methods Asymptotic analysis of multidimensional integrals Joint moments and pdf of the natural frequencies Numerical examples & results Conclusions 19 April 2005 Joint Distribution of Eigenvalues – p.2/36
Random Eigenvalue Problem The random eigenvalue problem of undamped or proportionally damped linear systems: K ( x ) φ j = ω 2 j M ( x ) φ j (1) ω j natural frequencies; 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. 19 April 2005 Joint Distribution of Eigenvalues – p.3/36
The Objectives The aim is to obtain the joint probability density function of the natural frequencies and the eigenvectors in this work we look at the joint statistics of the eigenvalues while several papers are available on the distribution of individual eigenvalues, only first-order perturbation results are available for the joint pdf of the eigenvalues 19 April 2005 Joint Distribution of Eigenvalues – p.4/36
Exact Joint pdf Without any loss of generality the original eigenvalue problem can be expressed by H ( x ) ψ j = ω 2 (2) j ψ j where H ( x ) = M − 1 / 2 ( x ) K ( x ) M − 1 / 2 ( x ) ∈ R N × N ψ j = M 1 / 2 φ j and 19 April 2005 Joint Distribution of Eigenvalues – p.5/36
Exact Joint pdf The joint probability (following Muirhead, 1982) density function of the natural frequencies of an N -dimensional linear positive definite dynamic system is given by π N 2 / 2 � � � ω 2 j − ω 2 p Ω ( ω 1 , ω 2 , · · · , ω N ) = i Γ( N/ 2) i<j ≤ N � � ΨΩ 2 Ψ T � p H ( d Ψ ) (3) O ( N ) where H = M − 1 / 2 KM − 1 / 2 & p H ( H ) is the pdf of H . 19 April 2005 Joint Distribution of Eigenvalues – p.6/36
Limitations of the Exact Method the multidimensional integral over the orthogonal group O ( N ) is difficult to carry out in practice and exact closed-form results can be derived only for few special cases the derivation of an expression of the joint pdf of the system matrix p H ( H ) is non-trivial even if the joint pdf of the random system parameters x is known 19 April 2005 Joint Distribution of Eigenvalues – p.7/36
Limitations of the Exact Method even one can overcome the previous two problems, the joint pdf of the natural frequencies given by Eq. (3) is ‘too much information’ to be useful for practical problems because it is not easy to ‘visualize’ the joint pdf in the space of N natural frequencies, and the derivation of the marginal density functions of the natural frequencies from Eq. (3) is not straightforward, especially when N is large. 19 April 2005 Joint Distribution of Eigenvalues – p.8/36
Eigenvalues of GOE Matrices Suppose the system matrix H is from a Gaussian orthogonal ensemble (GOE). The pdf of H : � � H 2 � � p H ( H ) = exp − θ 2 Trace + θ 1 Trace ( H ) + θ 0 The joint pdf of the natural frequencies: � � N �� � θ 2 ω 4 j − θ 1 ω 2 p Ω ( ω 1 , ω 2 , · · · , ω N ) = exp − j − θ 0 j =1 � � � � ω 2 j − ω 2 � i i<j 19 April 2005 Joint Distribution of Eigenvalues – p.9/36
Perturbation Method Taylor series expansion of ω j ( x ) about the mean x = µ ω j ( x ) ≈ ω j ( µ ) + d T ω j ( µ ) ( x − µ ) + 1 2 ( x − µ ) T D ω j ( µ ) ( x − µ ) Here d ω j ( µ ) ∈ R m and D ω j ( µ ) ∈ R m × m are respec- tively the gradient vector and the Hessian matrix of ω j ( x ) evaluated at x = µ . 19 April 2005 Joint Distribution of Eigenvalues – p.10/36
Joint Statistics Joint statistics of the natural frequencies can be obtained provided it is assumed that the x is Gaussian. Assuming x ∼ N ( µ , Σ ) , first few cumulants can be obtained as � � = E [ ω j ] = ω j + 1 κ (1 , 0) 2Trace , D ω j Σ jk = E [ ω k ] = ω k + 1 κ (0 , 1) 2Trace ( D ω k Σ ) , jk �� � � = Cov ( ω j , ω k ) = 1 κ (1 , 1) + d T 2Trace ( D ω k Σ ) D ω j Σ ω j Σd ω k jk 19 April 2005 Joint Distribution of Eigenvalues – p.11/36
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 19 April 2005 Joint Distribution of Eigenvalues – p.12/36
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 19 April 2005 Joint Distribution of Eigenvalues – p.13/36
Multidimensional Integrals The error ε ( x , θ ) depends on higher derivatives of f ( x ) at x = θ . If they are small compared to f ( θ ) their contribution will negligible to the value of the integral. So we assume that f ( θ ) is large so that � � � � 1 � f ( θ ) D ( j ) ( f ( θ )) � � → 0 for j > 2 � where D ( j ) ( f ( θ )) is j th order derivative of f ( x ) eval- uated at x = θ . Under such assumptions ε ( x , θ ) → 0 . 19 April 2005 Joint Distribution of Eigenvalues – p.14/36
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 19 April 2005 Joint Distribution of Eigenvalues – p.15/36
Moments of Single Eigenvalues An arbitrary r th order moment of the natural frequencies can be obtained from � � � µ ( r ) ω r m ω r = E j ( x ) = j ( x ) p x ( x ) d x j R � m e − ( L ( x ) − r ln ω j ( x )) d x , = r = 1 , 2 , 3 · · · R Previous result can be used by choosing f ( x ) = L ( x ) − r ln ω j ( x ) 19 April 2005 Joint Distribution of Eigenvalues – p.16/36
Moments of Single Eigenvalues After some simplifications j ( θ ) e − L ( θ ) µ ( r ) ≈ (2 π ) m/ 2 ω r j � � − 1 / 2 � � � D L ( θ ) + 1 r r d L ( θ ) d L ( θ ) T − � � ω j ( θ ) D ω j ( θ ) � r = 1 , 2 , 3 , · · · θ is obtained from: d ω j ( θ ) r = ω j ( θ ) d L ( θ ) 19 April 2005 Joint Distribution of Eigenvalues – p.17/36
Maximum Entropy pdf Constraints for u ∈ [0 , ∞ ] : � ∞ p ω j ( u ) du = 1 0 � ∞ u r p ω j ( u ) du = µ ( r ) j , r = 1 , 2 , 3 , · · · , n 0 Maximizing Shannon’s measure of entropy � ∞ S = − 0 p ω j ( u ) ln p ω j ( u ) du , the pdf of ω j is p ω 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 19 April 2005 Joint Distribution of Eigenvalues – p.18/36
Maximum Entropy pdf Taking first two moments, the resulting pdf is a truncated Gaussian density function � � ω j ) 2 1 − ( u − � √ p ω j ( u ) = ω j /σ j ) exp 2 σ 2 2 πσ j Φ ( � j j = µ (2) where σ 2 ω 2 − � j j Ensures that the probability of any natural frequencies becoming negative is zero 19 April 2005 Joint Distribution of Eigenvalues – p.19/36
Joint Moments of Two Eigenvalues Arbitrary r − s -th order joint moment of two natural frequencies � � µ ( rs ) ω r j ( x ) ω s = E l ( x ) jl � = m exp {− ( L ( x ) − r ln ω j ( x ) − s ln ω l ( x )) } d x , R r = 1 , 2 , 3 · · · Choose f ( x ) = L ( x ) − r ln ω j ( x ) − s ln ω l ( x ) 19 April 2005 Joint Distribution of Eigenvalues – p.20/36
Joint Moments of Two Eigenvalues After some simplifications µ ( rs ) l ( θ ) exp {− L ( θ ) } � D f ( θ ) � − 1 / 2 ≈ (2 π ) m/ 2 ω r j ( θ ) ω s jl where θ is obtained from: r s d L ( θ ) = ω j ( θ ) d ω j ( θ ) + ω l ( θ ) d ω l ( θ ) j ( θ ) d ω j ( θ ) d ω j ( θ ) T − r D f ( θ ) = D L ( θ ) + and ω 2 l ( θ ) d ω l ( θ ) d ω l ( θ ) T − r s s ω j ( θ ) D ω j ( θ ) + ω l ( θ ) D ω l ( θ ) ω 2 19 April 2005 Joint Distribution of Eigenvalues – p.21/36
Joint Moments of Multiple Eigenvalues We want to obtain � � � µ ( r 1 r 2 ··· r n ) ω r 1 j 1 ( x ) ω r 2 j 2 ( x ) · · · ω 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 ) ω r 1 j 1 ( θ ) ω r 2 j 2 ( θ ) · · · ω r n j n ( θ ) j 1 j 2 ··· j n exp {− L ( θ ) } � D f ( θ ) � − 1 / 2 19 April 2005 Joint Distribution of Eigenvalues – p.22/36
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