14.5 Derivatives of Eigenvalues and Eigenvectors 747
14.5 Derivatives of Eigenvalues and Eigenvectors
Let the eigenvalue problem be given by [14.4, 14.6, 14.10]
[K]
m×m
Y
m× 1
= λ[M]
m×m
Y
m× 1
(14.38)
whereλis the eigenvalue,Ythe eigenvector, [K] the stiffness matrix, and [M] the
mass matrix corresponding to the design vectorX= {x 1 , x 2 , · ·· , xn}T. Let the solution
of Eq. (14.38) be given by the eigenvaluesλiand the eigenvectorsYi, i = 1 , 2 ,... , m:
[Pi]Yi= 0 (14.39)
where [Pi] is a symmetric matrix given by
[Pi]=[K]−λi[M] (14.40)
14.5.1 Derivatives ofλi
Premultiplication of Eq. (14.39) byYTi gives
YTi[Pi]Yi= 0 (14.41)
Differentiation of Eq. (14.41) with respect to the design variablexjgives
YTi,j[Pi]Yi+YTi
∂[Pi]
∂xj
Yi+YTi[Pi]Yi,j= 0 (14.42)
whereYi,j=∂Yi/∂xj. In view of Eq. (14.39), Eq. (14.42) reduces to
YTi
∂[Pi]
∂xj
Yi= 0 (14.43)
Differentiation of Eq. (14.40) gives
∂[Pi]
∂xj
=
∂[K]
∂xj
−λi
∂[M]
∂xj
−
∂λi
∂xj
[M] (14.44)
where∂[K]/∂xjand ∂[M]/∂xjdenote the matrices formed by differentiating the ele-
ments of [K] and [M] matrices, respectively, with respect toxj. If the eigenvalues are
normalized with respect to the mass matrix, we have [14.10]
YTi[M]Yi= 1 (14.45)
Substituting Eq. (14.44) into Eq. (14.43) and using Eq. (14.45) gives the derivative of
λiwith respect toxjas
∂λi
∂xj
=YTi
[
∂[K]
∂xj
−λi
∂[M]
∂xj
]
Yi (14.46)
It can be noted that Eq. (14.46) involves only the eigenvalue and eigenvector under
consideration and hence the complete solution of the eigenvalue problem is not required
to find the value of∂λi/∂xj.
