# An Inverse Quadratic Eigenvalue Problem for Damped Structural Systems

Authors: Yongxin Yuan
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

We first give the representation of the general solution of the following inverse quadratic
eigenvalue problem (IQEP): givenΛ=diag{λ1,…,λp}∈Cp×p
, X=[x1,…,xp]∈Cn×p, and both Λ and X are closed under complex conjugation in the sense that λ2j=λ¯2j−1∈C, x2j=x¯2j−1∈Cn for j=1,…,l, and λk∈R,xk∈Rn for k=2l+1,…,p, find real-valued symmetric (2r+1)-diagonal
matrices M, D and K such that MXΛ2+DXΛ+KX=0. We then consider an optimal approximation
problem: given real-valued symmetric (2r+1)-diagonal matrices Ma,Da,Ka∈Rn×n, find (M∧,D∧,K∧)∈SE such that ∥M∧−Ma∥2+∥D∧−Da∥2+∥K∧−Ka∥2=inf(M,D,K)∈SE(∥M−Ma∥2+∥D−Da∥2+∥K−Ka∥2), where SE is the solution set of IQEP. We show that the optimal approximation solution (M∧,D∧,K∧) is unique and derive an explicit formula for it.


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## Body

## 1. Introduction

Throughout this
paper, we will adopt the following notation.Cm×n and Rm×n denote the set
of all m×n complex and
real matrices, respectively. SRn×n denotes the set
of all symmetric matrices in Rn×n. AT and A+ stand for the
transpose and the Moore-Penrose generalized inverse of a real matrix A. In represents the
identity matrix of size n; α¯ denotes the
conjugate of the complex number α. For A,B∈Rm×n, an inner product in Rm×n is defined by (A,B)=trace(BTA), then Rm×n is a Hilbert
space. The matrix norm ∥⋅∥ induced by the
inner product is the Frobenius norm. Given two matrices A=[aij]∈Rm×n and B=[bij]∈Rp×q, the Kronecker product of A and B is defined by A⊗B=[aijB]∈Rmp×nq, and the
stretching function Vec(A) is defined by Vec(A)=[a1T,a2T,…,anT]T∈Rmn, where ai,i=1,…,n, is the ith column
vector of A. Furthermore,
for a matrix A∈Rm×n, let EA and FA stand for the
two orthogonal projectors EA=Im−AA+ and FA=In−A+A.Using finite
element techniques, vibrating structures such as beams, buildings, bridges,
highways, and large space structures can be discretized to matrix second-order
models (referred to as analytical models). A matrix second-order model of the
free motion of a vibrating system is a system of differential equations of the
form(1.1)Max¨(t)+Dax˙(t)+Kax(t)=0,where Ma,Da, and Ka are the n×n analytical
mass, damping, and stiffness matrices. The system represented by (1.1) is called
damped structural system. It is well known that all solutions of  (1.1) can be obtained via the algebraic equation(1.2)(λ2Ma+λDa+Ka)x=0.Complex numbers λ and nonzero
complex vectors x for which this
relation holds are, respectively, the eigenvalues and eigenvectors of the
system. It is known that (1.2) has 2n finite
eigenvalues over the complex field, provided that the leading matrix
coefficient Ma is nonsingular.Due to the
complexity of the structure, the finite element model is only an approximation
to the practical structure. On the other hand, a part of the natural
frequencies (eigenvalues) and corresponding mode shapes (eigenvectors) of the
structure can be obtained experimentally by performing vibration tests [1].
Generally speaking, very often natural frequencies and mode shapes of an
analytical model described by (1.2) do not match very well with experimentally
measured frequencies and mode shapes. Thus, engineers would like to improve the
analytical model of the structure such that the updated model predicts the
observed dynamic behavior. Then, the updated model may be considered to be a
better dynamic representation of the structure. This model can be used with
greater confidence for the analysis of the structure under different boundary
conditions or with physical structural changes.For undamped
systems (i.e.,Da=0), various
techniques for updating mass and stiffness matrices using measured response
data have been discussed by Baruch [2], Baruch and Bar-Itzhack [3], Berman [4],
Berman and Nagy [5], and Wei [6, 7]. For damped structural   systems, the
theory and computation were first proposed by Friswell et al. [8, 9]; they applied the ideas in [2, 3] to minimize changes between the analytical
and updated model subject to the spectral constraints. Kuo et al. [10]
have recently proposed a direct method to close the weaknesses in [8] which
seems more efficient and reliable. All these existing methods can reproduce the
given set of measured data while updated matrices symmetry, but the connectivity
of the original finite element model is not necessarily preserved, causing the
addition of unwanted load paths.The purpose of
the work presented in this paper is to develop a new method for finite element
model updating problems which preserves the connectivity of the original model.
Assume thatMa,Da, and Ka are real-valued
symmetric (2r+1)-diagonal
matrices. Thus, the problem of updating mass, damping, and stiffness matrices
simultaneously can be mathematically formulated as follows.Problem IQEP.
Given matricesΛ=diag{λ1,…,λp}∈Cp×p,X=[x1,…,xp]∈Cn×p, and both Λ and X are closed
under complex conjugation in the sense that λ2j=λ¯2j−1∈C, x2j=x¯2j−1∈Cn for j=1,…,l, and λk∈R, xk∈Rn for k=2l+1,…,p, find real-valued symmetric (2r+1)-diagonal
matrices M,D, and K such
that
(1.3)MXΛ2+DXΛ+KX=0.Problem II.
LetSE be the solution
set of IQEP. Find (M^,D^,K^)∈SE such
that
(1.4)∥M^−Ma∥2+∥D^−Da∥2+∥K^−Ka∥2=inf(M,D,K)∈SE(∥M−Ma∥2+∥D−Da∥2+∥K−Ka∥2).The paper is
organized as follows. In Section2, using the Kronecker product and stretching
function of matrices, we give an explicit representation of the solution set SE of Problem
IQEP. In Section 3, we show that there exists a unique solution in Problem II
and present the expression of the unique solution (M^,D^,K^). Finally, in
Section 4, a numerical algorithm to acquire the optimal approximation solution
under the Frobenius norm sense is described, and a numerical example is
provided.


## 2. The Solution of Problem IQEP

To begin with, we introduce a lemma [11].
Lemma 2.1.
IfL∈Rm×q,b∈Rm, then Ly=b has a solution y∈Rq if and only if LL+b=b. In this case, the general solution of the equation
can be described as y=L+b+(Iq−L+L)z, where z∈Rq is an arbitrary
vector.LetS0 be the set of
all n×n real-valued
symmetric (2r+1)-diagonal
matrices, then S0 is a linear
subspace of SRn×n, and the dimension of S0 is N=(1/2)(2n−r)(r+1).DefineYij as
(2.1)Yij={22(eiejT+ejeiT),i=1,…,n−1;j=i+1,…,ti,eieiT,i=j=1,…,n,where ti=min{i+r,n} and ei,i=1,…,n, is the ith column
vector of the identity matrix In. It is easy to
verify that {Yij} forms an
orthonormal basis of the subspace S0, that
is,
(2.2)(Yij,Ykl)={0,i≠korj≠l,1,i=kandj=l.Now, if M,D,K∈SRn×n are (2r+1)-diagonal
matrices, then M,D,K can be
expressed as
(2.3)M=∑i,jαijYij,D=∑i,jβijYij,K=∑i,jγijYij,where the real numbers αij,βij,γij,i=1,…,n;j=i,…,ti,ti=min{i+r,n}, are yet to be
determined.Define a matrixTp as
(2.4)Tp=diag{12[1−i1i],…,12[1−i1i],Ip−2l}∈Cp×p,where i=−1. It is easy to
verify that Tp is a unitary
matrix, that is, T¯pTTp=Ip. Using this
transformation matrix, we have
(2.5)Λ=T¯pTΛTp=diag{[ζ1η1−η1ζ1],…,[ζ2l−1η2l−1−η2l−1ζ2l−1],λ2l+1,…,λp}∈Rp×p,(2.6)X=XTp=[2y1,2z1,…,2y2l−1,2z2l−1,x2l+1,…,xp]∈Rn×p,where ζj and ηj are,
respectively, the real part and the imaginary part
of the complex number λj; yj and zj, are,
respectively, the real part and the imaginary part of the complex vector xj for j=1,3,…,2l−1.It follows from (2.5) and (2.6) that (1.3) can be
equivalently written as(2.7)MXΛ2+DXΛ+KX=0.Substituting (2.3) into (2.7), we
have(2.8)∑i,jαijYijXΛ2+∑i,jβijYijXΛ+∑i,jγijYijX=0.When setting
(2.9)α=[α11,…,α1,r+1,…,αn−r,n−r,…,αn−r,n,…,αn−1,n−1,αn−1,n,αn,n]T,β=[β11,…,β1,r+1,…,βn−r,n−r,…,βn−r,n,…,βn−1,n−1,βn−1,n,βn,n]T,γ=[γ11,…,γ1,r+1,…,γn−r,n−r,…,γn−r,n,…,γn−1,n−1,γn−1,n,γn,n]T,G=[Vec(Y11),…,Vec(Y1,r+1),…,Vec(Yn−r,n−r),…,Vec(Yn−r,n),…,Vec(Yn−1,n−1),Vec(Yn−1,n),Vec(Yn,n)]∈Rn2×N,A=((Λ2)TXT⊗In)G,B=(ΛTXT⊗In)G,C=(XT⊗In)G.We see that (2.8) is
equivalent to(2.10)Aα+Bβ+Cγ=0.It follows from Lemma 2.1 that (2.10) with unknown vector α has a solution
if and only if(2.11)EABβ=−EACγ.Using Lemma 2.1 again, we know
that (2.11) with respect to β has a solution
if and only if(2.12)EHEACγ=0,where H=EAB. It follows from
Lemma 2.1 that (2.12) with respect to γ is always
solvable, and the general solution to the equation is(2.13)γ=FWu,where W=EHEAC, and u∈RN is an arbitrary
vector. Substituting (2.13) into (2.11) and applying Lemma 2.1, we
obtain(2.14)β=−H+EACFWu+FHv,where v∈RN is an arbitrary
vector. Inserting (2.14) and (2.13) into (2.10) yields  (2.15)α=A+BH+EACFWu−A+BFHv−A+CFWu+FAq,where q∈RN is an arbitrary
vector.As a summary
of the above discussion, we have proved the following result.Theorem 2.2.
Suppose thatΛ=diag{λ1,…,λp}∈Cp×p,X=[x1,…,xp]∈Cn×p
, and both Λ and X are closed
under complex conjugation. The real matrices Λ and X are given by
(2.5) and (2.6). Let {Yij},G,A,B,C be given as in
(2.1), (2.9). Write N=(1/2)(2n−r)(r+1),H=EAB,W=EHEAC. Then the
solution set SE of problem IQEP
can be expressed as(2.16)SE={(M,D,K)∈SRn×n×SRn×n×SRn×n∣M=S(α⊗In),D=S(β⊗In),K=S(γ⊗In)},where
(2.17)S=[Y11,…,Y1,r+1,…,Yn−r,n−r,…,Yn−r,n,…,Yn−1,n−1,Yn−1,n,Yn,n]∈Rn×nN,α,β,γ are,
respectively, given by (2.15), (2.14), and (2.13) with u,v,q∈RN being arbitrary
vectors.


## 3. The Solution of Problem II

It follows from Theorem2.2 that the set SE is always nonempty.
It is easy to verify that SE is a closed
convex subset of SRn×n×SRn×n×SRn×n. From the best approximation theorem (see [12]), we
know there exists a unique solution (M^,D^,K^) in SE such that (1.4)
holds.We now focus
our attention on seeking the unique solution(M^,D^,K^) in SE. For the real-valued symmetric (2r+1)-diagonal
matrices Ma,Da, and Ka, it is easily seen that Ma,Da, and Ka can be
expressed as the linear combinations of the orthonormal basis {Yij}, that is,
(3.1)Ma=∑i,jδijYij,Da=∑i,jξijYij,Ka=∑i,jφijYij,where δij,ξij,φij,i=1,…,n;j=i,…,ti,ti=min{i+r,n}, are uniquely
determined by the elements of Ma,Da, and Ka. Let
(3.2)δ=[δ11,…,δ1,r+1,…,δn−r,n−r,…,δn−r,n,…,δn−1,n−1,δn−1,n,δn,n]T,ξ=[ξ11,…,ξ1,r+1,…,ξn−r,n−r,…,ξn−r,n,…,ξn−1,n−1,ξn−1,n,ξn,n]T,φ=[φ11,…,φ1,r+1,…,φn−r,n−r,…,φn−r,n,…,φn−1,n−1,φn−1,n,φn,n]T.
Then, for any triple of matrices (M,D,K)∈SE in (2.16), by the
relations of (2.2) and (3.1), we see that(3.3)f=∥M−Ma∥2+∥D−Da∥2+∥K−Ka∥2=∥∑i,j(αij−δij)Yij∥2+∥∑i,j(βij−ξij)Yij∥2+∥∑i,j(γij−φij)Yij∥2=(∑i,j(αij−δij)Yij,∑i,j(αij−δij)Yij)+(∑i,j(βij−ξij)Yij,∑i,j(βij−ξij)Yij)+(∑i,j(γij−φij)Yij,∑i,j(γij−φij)Yij)=∑i,j(αij−δij)(Yij,∑i,j(αij−δij)Yij)+∑i,j(βij−ξij)(Yij,∑i,j(βij−ξij)Yij)+∑i,j(γij−φij)(Yij,∑i,j(γij−φij)Yij)=∑i,j(αij−δij)2+∑i,j(βij−ξij)2+∑i,j(γij−φij)2=∥α−δ∥2+∥β−ξ∥2+∥γ−φ∥2.Substituting (2.13), (2.14), and
(2.15) into the relation of f, we have(3.4)f=∥Ju−A+BFHv+FAq−δ∥2+∥−Lu+FHv−ξ∥2+∥FWu−φ∥2=uTJTJu−2uTJTA+BFHv−2uTJTδ+vTFHBT(A+)TA+BFHv+2vTFHBT(A+)Tδ+qTFAq−2qTFAδ+δTδ+uTLTLu+2uTLTξ+vTFHv−2vTFHξ+ξTξ+uTFWu−2uTFWφ+φTφ,where L=H+EACFW,J=A+BL−A+CFW. Therefore,
(3.5)∂f∂u=2JTJu−2JTA+BFHv−2JTδ+2LTLu+2LTξ+2FWu−2FWφ,∂f∂v=−2FHBT(A+)TJu+2FHBT(A+)TA+BFHv+2FHBT(A+)Tδ+2FHv−2FHξ,∂f∂q=2FAq−2FAδ.Clearly, f=∥M−Ma∥2+∥D−Da∥2+∥K−Ka∥2=min if and only if(3.6)∂f∂u=0,∂f∂v=0,∂f∂q=0.Note that LFW=L,JFW=J. Therefore, ∂f/∂u=0 implies
that
(3.7)FWu=P−1JTA+BFHv+P−1(JTδ−LTξ+FWφ),where P=JTJ+LTL+IN. Substituting
(3.7) into ∂f/∂v=0 yields
(3.8)FHv=Q+FHBT(A+)TJP−1(JTδ−LTξ+FWφ)−Q+(FHBT(A+)Tδ−FHξ),where Q=IN+FHBT(A+)TA+BFH−FHBT(A+)TJP−1JTA+BFH.Clearly,∂f/∂q=0 is equivalent
to(3.9)FAq=FAδ.Upon substituting (3.7) and (3.9)
into (2.13), (2.14), and (2.15), we obtain(3.10)α^=(JP−1JTA+B−A+B)FHv+JP−1(JTδ−LTξ+FWφ)+FAδ,β^=(IN−LP−1JTA+B)FHv−LP−1(JTδ−LTξ+FWφ),γ^=P−1JTA+BFHv+P−1(JTδ−LTξ+FWφ),where FHv is given by
(3.8).By now, we have
proved the following result.Theorem 3.1.
Let the real-valued symmetric(2r+1)-diagonal
matrices Ma,Da, and Ka be given. Then,
Problem II has a unique solution, and the unique solution of Problem II can be
expressed as(3.11)M^=S(α^⊗In),D^=S(β^⊗In),K^=S(γ^⊗In),
where α^,β^, and γ^ are given by
(3.10).


## 4. A Numerical Example

Based on Theorems2.2 and 3.1, we can describe an
algorithm for solving problem IQEP and Problem II as follows.Example 4.1.
Consider a five-DOF system
modelled analytically with mass, damping, and stiffness matrices given by(4.1)Ma=[2100014100014100014100014],Da=[11.0−8.0000−8.014.0−3.5000−3.513.0−7.8000−7.813.5−9.0000−9.015.4],Ka=[100−20000−20120−35000−3580−12000−1295−40000−40124].
That is, Ma,Da,Ka are symmetric
3-diagonal matrices. The measured eigenvalue and eigenvector matrices Λ=diag{λ1,λ2,λ3,λ4} and X=[x1,x2,x3,x4] are given by
(4.2)Λ=diag{−3.9525+6.8578i,−3.9525−6.8578i,−0.1622+3.3465i,−0.1622−3.3465i},X=[0.1099+0.4083i0.1099−0.4083i−0.0283+0.1570i−0.0283−0.1570i0.0684−0.1754i0.0684+0.1754i0.0525+0.3742i0.0525−0.3742i−0.1272−0.2703i−0.1272+0.2703i0.0594+0.6146i0.0594−0.6146i0.3403+0.5185i0.3403−0.5185i−0.1132+0.5663i−0.1132−0.5663i−0.3783−0.4108i−0.3783+0.4108i−0.0862+0.3309i−0.0862−0.3309i].
Using Algorithm 1, we obtain the
unique solution of Problem II as follows:(4.3)M^=[1.60100.31520000.31522.79091.53180001.53182.93320.97710000.97713.77791.20940001.20943.6053],D^=[9.0659−6.4506000−6.4506−5.50017.36770007.36774.1831−6.4524000−6.452410.2736−5.8430000−5.843010.3568],K^=[98.7229−25.5781000−25.5781124.0044−32.9787000−32.978777.7282−12.0909000−12.090999.5825−32.9904000−32.9904120.7917].
We define the residual as(4.4)res(λi,xi)=∥(λi2M^+λiD^+K^)xi∥,and the numerical results shown in the following table.
Therefore, the
prescribed eigenvalues (the diagonal elements of the matrixΛ) and
eigenvectors (the column vectors of the matrix X) are embedded
in the new model (λ2M^+λD^+K^)x=0 and the updated
matrices M^,D^,K^ are also
symmetric 3-diagonal matrices, which implies that the structural connectivity
information of the analytical model is preserved.Algorithm 1
(1) InputMa,Da,Ka,Λ,X.(2) Compute
the unitary transformation matrixTp by (2.4).(3) Compute
real matricesΛ,X by (2.5) and
(2.6), respectively.(4) Form the
orthonormal basis{Yij} by (2.1).(5) ComputeG,A,B,C according to
(2.9).(6) Compute
the matrixH=EAB,W=EHEAC.(7) ComputeL=H+EACFW,J=A+BL−A+CFW,P=JTJ+LTL+IN.(8) Compute
the matrixQ=IN+FHBT(A+)TA+BFH−FHBT(A+)TJP−1JTA+BFH.(9) Form
vectorsδ,ξ,φ by (3.1) and (3.2).(10) ComputeFHv by (3.8).(11) ComputeS,α^,β^,γ^ by (2.17) and (3.10), respectively.(12) Compute
the unique solution(M^,D^,K^) of Problem II
by (3.11).Table 1(λi,xi)(λ1,x1)(λ2,x2)(λ3,x3)(λ4,x4)res(λi,xi)1.9584e-0131.9584e-0139.0344e-0149.0344e-014



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