748 Practical Aspects of Optimization
14.5.2 Derivatives ofYi
The differentiation of Eqs. (14.39) and (14.45) with respect toxjresults in[Pi]∂Yi
∂xj= −
∂[Pi]
∂xjYi (14.47)2 YTi[M]∂Yi
∂xj= −YTi∂[M]
∂xjYi (14.48)where∂[Pi]/ x∂jis given by Eq. (14.44). Equations (14.47) and (14.48) can be shown
to be linearly independent and can be written together as[
[Pi]
2 YTi[M]]
(m+ 1 )×m∂Yi
∂xjm× 1= −
∂[Pi]
∂xjYTi∂[M]
∂xj
(m+ 1 )×mYim× 1(14.49)
By premultiplying Eq. (14.49) by
[
[Pi]
YTi[M]]T
=
[
[Pi] [M]Yi]
we obtain[[Pi][Pi] + 2 [M]YiYTi[ ]M]
m×m∂Yi
∂xj
m× 1= −
[
[Pi]∂[Pi]
∂xj+[M]YiYTi∂[M]
∂xj]
m×mYi
m× 1
(14.50)
The solution of Eq. (14.50) gives the desired expression for the derivative of the
eigenvector,∂Yi/∂xj, as
∂Yi
∂xj= − [[Pi][Pi] + 2 [M]YiYTi[ ]M]−^1×
[
[Pi]∂[Pi]
∂xj+[M]YiYTi∂[M]
∂xj]
Yi (14.51)Again it can be seen that only the eigenvalue and eigenvector under consideration are
involved in the evaluation of the derivatives of eigenvectors.Y 1 Y 3Y 2 Y 4 Y 6Y 51 in. 1 in. 1 in.x 1 • x 2 • x 3xi = 0.25′′ (i = 1, 2, 3), r = 0.283 lb/in^3 ,
E = 30 × 106 psi
Figure 14.4 Cylindrical cantilever beam.