CHAPTER TEN

Preliminaries to theSolutionof Eigenproblems

0. INTRODUCTION

In various sections of the preceding chapters we encountered eigenproblems and the state ment of their solutions. We did not at that time discuss how to obtain the required eigenval ues and eigenvectors. It is the purpose of this and the next chapter to describe the actual solution procedures used to solve the eigenproblems of interest. Befare presenting the algorithms, we discuss in this chapter sorne important basic considerations for the solution of eigenproblems.First, let us briefty summarize the eigenproblems that we want to solve. The simplestproblem encountered is the standard eigenproblem,

K = A(10.1)

where K is the stiffness matrix of a single finite element or of an element assemblage. We recall that K has order n, and for an element assemblage the half-bandwidth mK (i.e., thetotal bandwidth is 2mK + 1), and that K is positive semidefinite or positive definite. Thereare n eigenvalues and corresponding eigenvectors satisfying (10.1). The ith eigenpair is denoted as (A, cfJ, where the eigenvalues are ordered according to their magnitudes:

(10.2)

The solution for p eigenpairs can be writtenKc, = c,A(10.3)

where fP is an n X p matrix with its columns equal to the p eigenvectors and A is a p X p diagonal matrix listing the corresponding eigenvalues. As an example, ( 10.3) may represent the solution to the lowest p eigenvalues and corresponding eigenvectors of K, in which casefP = [cf 1 , , cfp] andA = diag(A;), i = 1, ...,p. We recall that if K is positive definite,

838

Sec. 10.1lntroduction 839

A; > O, i = 1, ... , n, and if K is positive semidefinite, A; > O, i = 1, ... , n, where the number of zero eigenvalues is equal to the number of rigid body modes in the system.The solution of the eigenvalue problem in (10.1) is, for example, sought in theevaluation of an element stiffness matrix or in the calculation of the condition number of a structure stiffness matrix. We discussed in Section 4.3.2 that the representation of the element stiffness matrix in its canonical form (i.e., in the eigenvector basis) is used to evaluate the effectiveness of the element. In this case all eigenvalues and vectors of K must be calculated. On the other hand, to evaluate the condition number of a stiffness matrix, only the smallest and largest eigenvalues are required (see Section 8.2.6).Before proceeding to the generalized eigenproblems, we should mention that other standard eigenproblems may also need to be solved. For example, we may require the eigenvalues of the mass matrix M, in which case M replaces K in ( 10.1). Similarly, we may want to solve for the eigenvalues of a conductivity or heat capacity matrix in heat ftow analysis (see Section 7.2).A very frequently considered eigenproblem is the one to be solved in vibration mode superposition analysis (see Section 9.3). In this case we consider the generalized eigen problem,K

It is believed that the two eigenpairs of the problem K]wrMw = [ -2](JUJ =o

w2rMw2 =

[25-2J [o!

][_= 1

02wfMw2 = wfMw1 = O

Hence, the relations in (10.5) and (10.10) are satisfied and we have

EXAMPLE 10.2:Consider the eigenproblemK

11--v'2v'2

-

and in heat transfer analysis, using the notation in ( 10.7), we have

and of lts Associated Constraint Problems

An important property of the eigenvalues of the problem Kp(A) = det(K - AM)(10.16)

We can show that this property derives from the basic relation in (10.8). Rewriting (10.8) in the form(K - A;M)Gauss elimination, we have Snn = O. However, since

p(A;) = det LS = fi Si=l

(10.18)

it follows that p(A;) = O. Furthermore, if ,\; has multiplicity m, we also have Sn-!,n- 1 = = Sn-m+!,n-m+l = O. We should note that in the factorization of K - A;M, interchanges may be needed, in which case the factorization of K - ,\M with its rows and possibly its columns interchanged is obtained (each row and each column interchange then introduces a sign change in the determinant which must be taken into account; see Section 2.2). If no interchanges are carried out, or row and corresponding column interchanges are performed, which in practice is nearly always possible (but see Example 10.4 for a case where it is not possible), the coefficient matrix remains symmetric. In this case we can write for (10.18)

np(A;) = det LDLT = TI dui=l

(10.19)

where LDLr is the factorization ofK - ,\Mor ofthe matrix derived from it by interchang ing rows and corresponding columns, i.e., using a different ordering for the system degrees of freedom (see Section 8.2.5). The condition Snn = O is now dnn = O, and when ,\ has multiplicity m, the last m elements in D are zero.In Section 8.2.5 we discussed the Sturm sequence property of the characteristic polynomials of the constraint problems associated with the problem K. Theproof follows from the fact that the generalized eigenproblem Kformed to a standard eigenproblem for which the Sturm sequence property of the characteristic polynomials holds. Referring the proof to Section 10.2.5, Example 10.11, let us summarize the important result.The eigenproblem of the rth associated constraint problem corresponding toKK(r)The coefficient matrix K - A 1M in (a) can be factorized into LDLr without interchanges. Using the procedure described in Section 8.2.2, we obtain

-1[ o-1J J-!}=o

-11

(b)

We note that d33 = 0.0. To eva1uate

_J7

Il7

f.]['

-l3-n

Since all elements in D are larger than zero, we have A1 > l.1. We now try .t = 8, whereandSince all three diagonal elements are smaller than zero, it follows that ,\3 < 8.1. The next estimate .t should logically lie between l and 8; we choose .t = 5, for which

and

2-n

1-1

Since two negative e1ements are in D, we have A2 < 5.4. The next estmate must lie between 1 and 5. Let us use JL = 3, in which case

-1K-[-! -1-n

and

LDL'[-1,r -1

J -2n

o1

Hence A2 > 3, because there is on1y one negative e1ement in D.

The pattern of the so1ution procedure has now been established. So far we know that 3 < A2 < 5. In order to obtain a closer estmate on A2 we wou1d continue choosing a shift JL in the interva1 3 tr> 5 and investigate whether the new shift is smaller or 1arger than A2 By a1ways choosing an appropriate new shift, the required eigenva1ue can be determined very accurate1y (see Section 11.4.3). It shou1d be noted that we did not need to use interchanges in the factorizations of K - JLM carried out above.

0. Shifting

An important procedure that is used extensively in the solution of eigenvalues and eigenvec tors is shifting. The purpose of shifting is to accelerate the calculations of the requiredeigensystem. In the solution Kc, = AMe,, we perform a shift p on K by calculating

K= K- pM(10.24)

and we then consider the eigenproblem

K\fl = JLM\fl(10.25)

To identify how the eigenvalues and eigenvectors of Kc, = AMe, are related to those of the problem K\jl = JLM\jl we rewrite ( 10.25) in the form

(10.26)

where y = p + JL However, ( 10.26) is, in fact, the eigenproblem Kc, = AMe,, and since the solution of this problem is unique, we have

; = p + .L;; (10.27)

In other words, the eigenvectors oj:K\jl = JLM\jl are the same as the eigenvectors ofKc, = AMe,, but the eigenvalues have been decreased by p. A frequent application ofshifting occurs in the calculation of rigid body modes when an algorithm is to be used that is not designed explicitly to calculate zero eigenvalues. We illustrate such an application in the following example.

EXAMPLE 10.6: Consider the eigenproblem[- - J = A[J(a)Calculate the eigenvalues and eigenvectors. Then impose a shift p = -2 and solve again for the eigenvalues and corresponding eigenvectors.To calculate the eigenvalues we use the characteristic polynornialp(A) = det(K - AM) = 3A 2 - 18A

and thus obtain A 1 = O, A2 = 6. To calculate1 and 2 we use the relation in (10.17) and the mass orthonormality condition ;M = l. We have

[_-33].'!."..1 =o'

hence,=

1v'6Iv'6

-9].. ., =O1

(b)

and

-9[-9

v21-9 '!"2'hence, 2 =

v2

(e)

Imposing a shift of p = -2, we obtain the problem[- - J =A[J(d)

Proceeding as before, we have

p(A) = A 2 - lOA + 16

and obtain as the roots A1 = 2, A2 = 8. Hence the eigenvalues have increased by 2; i.e., they have decreased by p.The eigenvectors would be calculated using (10.17). However, we note that this relationagain yields the equations in (b) and (e), and therefore the eigenvectors ofthe problem in (d) arethose of the problem in (a).

An important observation resulting from the above discussion is that, in principie, we need only solution algorithms to calculate the eigenvalues and corresponding eigenvectorsof the problem Kc, = AMe, when all eigenvalues are larger than zero. This follows,because if rigid body modes are present, we may always operate on a shifted stiffness matrix that renders all eigenvalues positive.Extensive applications of shifting are given in Chapter 11, where the various eigensystem solution algorithms are discussed.

0. Effect of Zero Mass

When using a lumped mass matrix, M is diagonal with positive and possibly sorne zero diagonal elements. If all elements mu are larger than zero, the eigenvalues can usually not be obtained without the use of an eigenvalue solution algorithm as described in Chapter 11.

However, if M has sorne zero diagonal elements, say r diagonal elements in M are zero, we can immediately say that the problem Kc, = AMe, has the eigenvalues An = n-I =1. = n-r+l = oc and can also construct the corresponding eigenvectors by inspection.To obtain the above result, let us recall the fundamental objective in an eigensolution. It is important to remember that all we require is a vector e, and scalar A that satisfy the equation

Kcl> = AMe!>(10.4)

where e, is nontrivial; i.e., e, is a vector with at least one element in it nonzero. In other words, ifwe have a vector e, and scalar A that satisfy (10.4), then A ande, are an eigenvalue A; and eigenvector e,;, respectively, where it should be noted that it does not matter how e, and A have been obtained. If, for example, we can guess e, and A, we should certainly takeadvantage of it. This may :,e the case when rigid body modes are present in the structural element assemblage. Thus, if we know that the element assemblage can undergo a rigidbody mode, we have = O and need to seek c,I to satisfy the equation Kc,1 = O. In general, the solution of e, must be obtained using an equation solver, but in a simple finiteelement assemblage we may be able to identify c,I by inspection.In the case of r zero diagonal elements in a diagonal mass matrix M, we can alwaysimmediately establish r eigenvalues and corresponding eigenvectors. Rewriting the eigen problem in (10.4) in the form

(10.28)

where J.L = A-l, we find that if mkk = O, we have an eigenpair (J.L;, e,;) = (O, ek); i.e.,

c!>i = [OO. . .O1O. . .O];tkth element

J.L = o(10.29)

That e,; and J.L; in (10.29) are indeed an eigenvector and eigenvalue of ( 10.28) is verified by simply substituting into (10.28) and noting that (J.L;, e,;) is a nontrivial solution. Since J.L = A- 1, we therefore found that an eigenpair of Kc, = AMe, is given by (A;, e,;) = (oc, ek).Considering the case of r zero diagonal elements in M, it follows that there are r infinite eigenvalues, and the corresponding eigenvectors can be taken to be unit vectors with each unit vector having the 1 in a location corresponding to a zero mass element in M. Since An is then an eigenvalue of multiplicity r, the corresponding eigenvectors are not unique (see Section 10.2.1). In addition, we note that the length of an eigenvector cannot be fixed using the condition of M-orthonormality. We demonstrate how we establish the eigenvalues and eigenvectors by means of a brief example.

EXAMPLE 10.7: Consider the eigenproblem

-12-1-12-1

There are two zero diagonal elements in M; hence, A3 = oo, ,\, = oo. As corresponding eigenvectors we can use

(a)

Altematively, any linear combination of cl>3 and c1>4 given in (a) would representan eigenvector. We should note that ci>;Mcl>; = O for i = 3, 4, and therefore the magrritude of the elements in el>; cannot be fixed using the M-orthonormality condition.

0. Transformation of the Generalizad Eigenprob!em Kcf- = AM = ASSTel>(10.31)Premultiplying both sides of (10.31) by s-1 and defining a vector,, = STel>(10.32)

we obtain the standard eigenproblem,

where:K= s-IKS-r

(10.33)

(10.34)

One of two decompositions of M is used in general: the Cholesky factorization or the spectral decomposition of M. TJte fholesky factorization of M is obtained as described in Section 8.2.4 and yields M = LML&t. In (10.30) to (10.34) we therefore have

(l0.35)

The spectral decomposition of M requires solution of the complete eigensystem of M. Denoting the matrix of orthonormal eigenvectors by R and the diagonal matrix of eigenval ues by D2, we have

(10.36)

and we use in (10.30) to (10.34),

S= RD(10.37)

lt should be noted that when Mis diagonal, the matrices S in (10.35) and (10.37) are the same, but when M is banded, they are different.Considering the effectiveness ofthe sol tion ofthe required eigenvalues and eigenvec tors of (10.33), it is most important tJtat K has the same bandwidth as K when M is diagonal. However, when Mis banded, K in (10.33) is in general a full matrix, which makes the transformation ineffective in almost alllarge-order finite element analyses. This will become more apparent in Chapter 11 when various eigensystem solution algorithms are discussed.Comparing the Cholesky factorization and the spectral decomposition of M, it may be noted that the use of the Cholesky factors is in general computationally more efficient than t_he use of the spectral decomposition because fewer operations are involved in calculating LM than R and D. However, the spectral decomposition of M may yield a more accuratesolution of Kc, = AMe,. Assume that M is ill-conditioned with respect to inversion; thenthe transformation process to the standard eigenproblem is also ill-conditioned. In that caseit is important to employ the more stable transformation procedure. Using the Cholesky factorization of M without pivoting, we find that LM1 has large elements in many locations

because of the coupling in M and LM1

Consequently, K is calculated with little precision,

and the lowest eigenvalues and corresponding eigenvectors are determined inaccurately.

On the other hand, using the spectral decomposition of M, good accuracy may be obtained in the elements of R and nz, although sorne elements in nz are small in relationto the other elements. The ill-cof2ditioning of M is now concentrated in only the small elements of D2, and considering K, only those rows and columns that correspond to the small elements in D have large elements, and the eigenvalues of normal size are more likely to be preserved accurately.Consider the following examples of transforming the generalized eigenvalue problem Kc, = AMe, to a standard form.

EXAMPLE 10.8: Consider the prob1em Kcl> = AMe!>, where

o]3-1o]1K=-12-1 ;31[ o-1112Use the Cho1esky factorization of M to calcu1ate the matrix K of a corresponding standard eigenprob1em.We first ca1cu1ate the LDLT decomposition of M,

Hence, the Cho1esky factor of Mis (see Section 8.2.4)v2

'J31_2LM = v2

o

1v21

and(-M1-

---v'1o1

--1-

v4ov'1o

The matrix of the standard eigenprob1em K = L i KL l is in this case,

224v'52-3254-3v'5v'5

K=

v'55-3442

EXAMPLE 10.9: Consider the generalized eigenprob1em in Examp1e 10.8. Use the spectral decomposition of M to ca1cu1ate the matrix K of a corresponding standard eigenprob1em.The eigenva1ues and corresponding eigenvectors of the prob1em Me!> = Ael> can be ca1cu- 1ated as shown in Examp1e 10.4. We obtain A1 = 1, A2 = 2, A3 = 4, and

111v'3v2v'6

el>=

1--

cl>2 =oc!>3=2

v'3'v'6111v'3v2v'6

Sec. 10.2Fundamental Facts Used in the Solution of Eigensystems 857Hence, the decomposition M = RD2RT is111111

-v'3v2

v'6

-v'3v'3v'3

M=- 1o2

[' 2

.] 1o1

v'3

v'6

v2v2

211111-v'3v2v'6v'6v'6v'6Noting that S = RD and s-1 = n- 1RT because RRT = 1, we obtain

111---v'3v'3v'3

os-1 =11

2---21112v'6v'62v'6

101---13v'33V1v'3112v'6The matrix of the standard eigenprob1em is K = s-1KS-T; i.e.,

-

K=

--1-1-13V2v'66We shou1d note that the matrix K obtained here is different from the matrix K derived inExamp1e 10.8.

In the above discussion we considered only the factorization of M into M = SST and then the transformation of Ke, = AMe, into the form given in ( 10.33). We pointed out that this transformation can yield inaccurate results if M is ill-conditioned. In such a case it seems natural to avoid the decomposition of M and instead use a factorization of K. Rewriting Kc, = AMe, in the form Me, = (1/A)Kc,, we can use an analogous procedure to obtain the eigenproblem

- -1 -Me!> = -el>(10.38)A

where:M= s-1Ms-T(10.39)

K= SST(10.40)cT> = STel>(10.41)

and S is obtained using the Cholesky factor or spectral decomposition of K. If K is well-conditioned, the transformation is also well-conditioned. However, since K is alwaysbanded, M is always a full matrix, and the transformation is usually inefficient for thesolution of Kc, = AMe,.

As we pointed out earlier, the possibility of actually solving a generalized eigenprob lem by first transforming it into a standard form is only one reason why we considered the above transformations. The second reason is that the properties of the eigensolution of the generalized eigenproblem can be deduced from the properties of the solution of the corre sponding standard eigenproblem. Specifically, we can derive the orthogonality properties of the eigenvectors as given in ( 1O.1O) and ( 1O.11), and the Sturm sequence property of thecharacteristic polynomials of the eigenproblem K = AM and of its associated constraint problems as given in (10.22). In both cases no fundamentally new concepts need be proved; instead, the corresponding properties of the standard eigenproblem, which is ob tained from the generalized eigenproblem, are used. We give the proofs in the followingexamples asan application of the transformation of a generalized eigenproblem toa stan dard form.EXAMPLE 10.10: Show that the eigenvectors of the problem K = A.M are M- and K orthogonal and discuss the orthogonality of the eigenvectors of the problems 'K = A'-A'Kcfl given in (10.6) and K = A.C given in (10.7).The eigenvector orthogonality is proved by transforming the generalized eigenproblem to a standard form and using the fact that the eigenvectors of a standard eigenproblem with a symmetric matrix are orthogonal. Consider first the problem K = A.M and assume that Mis positive definite. Then we can use the transformation in (10.30) to (10.34) to obtain, asan equivalent eigenproblem,whereKJ = 5;J;iK1 = A;5;1(a)If M is not positive definite, we consider the eigenproblem M = (1/A.)K (with K positive definite ora shift must be imposed; see Section 10.2.3). We now use the transformationM- =(A1)whereand the propertiesSubstituting for was considered. To obtain the same vectors as in the problem K = A.M, we need to multiply the eigenvectors ; of the problem M = (1/A.)K by the factors y; i = 1, ... , n.Considering these proofs, we note that we arranged the eigenproblems in such a way thatthe matrix associated with the eigenvalue, i.e., the matrix on the right-hand side ofthe eigenvalueSec. 10.2Fundamental Facts Used in the Solution of Eigensystems 859problem, is positive definite. This is necessary to be able to crry out the transformation of the generalized eigenproblem to a standard form and thus derive the eigenvector orthogonality properties in (a) and (b). However, considering the problems 'K = A'- 'K and K = AC given in (10.6) and (10.7), we can proceed in a similar manner. The results would be the eigenvector orthogonality properties given in (10.14) and (10.15).EXAMPLE 10.11: Prove the Sturm sequence property of the characteristic polynomials of the problem K = A.M and the associated constraint problems. Demonstrate the proof for the following matrices: -12(a)-1The proof that we consider here is based on the transformation of the eigenproblems K = AM and Kpolynomials are known to form a Sturm sequence (see Sections 2.6 and 8.2.5).As in Example 10.10, we assume first that Mis positive definite. In this case we can transform the problem K = AM into the formKHence, the Sturm sequence property also holds for the characteristic polynomials ofK = ..\M and the associated constraint problerns.For the example to be considered, we have[2 o o]Hence,oLM=22oo22H2-1o][' -! -!J-1(e)K=H1JH2-!=7- ]122(d)2-!-1o-27 'Using K in (d) to obtain ((ll and K = A'- 'K and K = AC given in (10.6) and (10.7) and of their associated constraint problems also form a Sturm sequence.0. Exercises0. Consider the generalized eigenproblem6-1-14o][2 o o]-1 =A O21 [ o-12o110. Calculate the eigenvalues and eigenvectors and show explicitly that the eigenvectors are M-orthogonal.0. Find two vectors that are M-orthogonal but are not eigenvectors.0. Calculate the eigenvalues of the eigenproblem in Exercise 10.1 and of its associated constraint problems. Show that the eigenvalues satisfy the separation property (10.22).0. Consider the eigenproblemH -} { 2}2. Calculate the eigenvalues and eigenvectors of the problem. Also, calculate the eigenvalues of the associated constraint problems [see (10.20)].2. Establish two vectors that are M-orthogonal but are not eigenvectors.0. Calculate the eigenvalues and eigenvectors of the problem[-16-14] =A [2Oo]O Then apply a shift p = 3 on K and calculate the eigenvalues and eigenvectors of the new problem [see (10.25)].0. Transform the generalized eigenproblem in Exercise 10.1 into a standard form.0. (a) The eigenvalues and eigenvectors of the problemK = AareA = 1;Calculate K.1 = _1 [1]v212 = _1[ 1]v2-1(b) The eigenvalues and eigenvectors of the problemK = A.M1 = _1 [1]areA = 1;v31Calculate K and M. Are the K and M matrices in (a) and (b) unique?10.3 APPROXIMATE SOLUTION TECHNIQUESIt is apparent from the nature of a dynamic problem that a dynamic response calculation must be substantially more costly than a static analysis. Whereas in a static analysis the solution is obtained in one step, in dynamics the solution is required at a number of discrete time points over the time interval considered. Indeed, we found that in a direct step-by-step integration solution, an equation of statics, which includes the effects of inertia and damping forces, is considered at the end of each discrete time step (see Section 9.2). Considering a mode superposition analysis, the main computational effort is spent in the calculation ofthe required frequencies and mode shapes, which also requires considerably more effort than a static analysis. It is therefore natural that much attention has been directed toward effective algorithms for the calculation of the required eigensystem in the problemK = AM. In fact, because the "exact" solution of the required eigenvalues and corresponding eigenvectors can be prohibitively expensive when the order of the system is largeand a "conventional" technique is used, approximate techniques of solution have been developed. The purpose of this section is to present the major approximate methods that have been designed and are currently still in use.The approximate solution techniques have primarily been developed to calculate the lowest eigenvalues and corresponding eigenvectors in the problem K = AM when the order of the system is large. Most programs use exact solution techniques in the analysis of small-order systems. However, the problem of calculating the few lowest eigenpairs of relatively large-order systems is very important and is encountered in all branches of structural engineering and in particular in earthquake response analysis. In the followingsections we present three major techniques. The aim in the presentation is not to advocate the implementation of any one of these methods but rather to describe their practica/ use, their limitations, and the assumptions employed. Moreover, the relationships between the approximate techniques are described, and in Section 11.6 we will, in fact, find that the approximate techniques considered here can be understood to be a first iteration (and may be used as such) in the subspace iteration algorithm.10.3.1 Static CondensationWe have already encountered the procedure of static condensation in the solution of static equilibrium equations, where we showed that static condensation is, in fact, an application of Gauss elimination (see Section 8.2.4). In static condensation we eliminated those degreesof freedom that are not required to appear in the global finite element assemblage. Por example, the displacement degrees of freedom at the interna! nodes of a finite element can be condensed out because they do not take part in imposing interelement continuity. We mentioned in Section 8.2.4 that the term "static condensation" was actually coined in dynarnic analysis.The basic assumption of static condensation in the calculation of frequencies and mode shapes is that the mass of the structure can be lumped at only sorne specific degrees of freedom without much effect on the accuracy of the frequencies and mode shapes of interest. In the case of a lumped mass matrix with sorne zero diagonal elements, sorne ofthe mass lumping has already been carried out. However, additional mass lumping is ingeneral required. Typically, the ratio of mass degrees of freedom to the total number of degrees of freedom may be somewhere between 1 and 1 The more mass lumping isperformed, the less computer effort is required in the solution; however, the more probable it is also that the required frequencies and mode shapes are not predicted accurately. We shall have more to say about this later.Assume that the mass lumping has been carried out. By partitioning the matrices, wecan then write the eigenproblem in the form[KaaKac] [a] = A [MaO] [a]KcaKcccOOc(10.42)where c:l>a and c:l>c are the displacements at the mass and the massless degrees of freedom, respectively, and M, is a diagonal mass matrix. The relation in ( 10.42) gives the conditionKca a + Kcc e = Owhich can be used to elirninate c:l>c From ( 10.43) we obtainc = -K1Kcaaand substituting into ( 10.42), we obtain the reduced eigenproblemKa a = AMa awhere(10.43)(10.44)(10.45)(10.46)The solution of the generalized eigenproblem in ( 10.45) is in most cases obtained by transforrning the problem first into a standard form as described in Section 10.2.5. Since M,. is a diagonal mass matrix with all its diagonal elements positive and probably not small, the transformation is in general well-conditioned.The analogy to the use of static condensation in static analysis should be noted.Realizing that the right-hand side of ( 10.42) may be understood to be a load vector R, where (10.47)we can use Gauss elirnination on the massless degrees of freedom in the same way as we do on the degrees of freedom associated with the interior nodes of an element or a substructure (see Section 8.2.4).One important aspect should be observed when comparing the static condensationprocedure on the massless degrees of freedom in ( 10.42) to ( 10.46) on the one side, with Gauss elirnination or static condensation in static analysis on the other side. Considering ( 10.47), we find that the loads at the c:l>a degrees of freedom depend on the eigenvalue(free-vibration frequency squared) and eigenvector (mode shape displacements). This means that in ( 10.45) a further reduction of the number of degrees of freedom to be considered is not possible. This is a basic difference to static condensation as applied in static analysis, where the loads are given explicitly and their effect can be carried over to the remaining degrees of freedom.EXAMPLE 10.12: Use static condensation to calculate the eigenvalues and eigenvectors ofthe problem K = AM, whereo .2-1o01r=-12-1KO-12-1 'oo-11Firstwe rearrange columns and rows to obtain the form given in (10.42), which isHence, Ka given in (10.46) is in this case,Ka=[]- [-= J[J[=- J = [_The eigenproblem Kaa = AMaa is, therefore,1[-11-1]a = A [2Jaand we have Hence,det(Ka- AMa) = 2A 2 - 2A +1v'21v'2A = 2 - 4;A2 = 2 + 4The corresponding eigenvectors are calculated usingrHenceUsing (10.44), we obtainc, = -o,, -r:864Preliminaries to the Solution of EigenproblemsChap. 10Therefore, the solution to the eigenproblem K = AM is1-4121v'2-,\, = Z- 4;=1 + v'24v'22--141--22=-1 + v'24v'22In the above discusssion we gave the formal matrix equations for carrying out static condensation. The main computational effort is in calculating Ka given in ( 10.46), where it should be noted that in practice a formal inversion of Kcc is not performed. Instead, Kacan be obtained conveniently using the Cholesky factor Lec of Kcc. If we factorize KceoKcc = Lir(10.48)we can calculate Ka in the following way:Ka= Kaa- yry(10.49)where Y is solved from(10.50)As pointed out earlier, this procedure is, in fact, Gauss elimination of the massless degrees offreedom, i.e., elimination ofthose degrees offreedom at which no external forces (mass effects) are acting. Therefore, an alternative procedure to the one given in ( 10.42) to ( 10.50) is to directly use Gauss elirnination on the c:l>c degrees of freedom without partition ing K into the submatrices Kaa, Kceo Kaeo and Kca because Gauss elirnination can be performed in any order (see Example 8.1, Section 8.2.1). However, the bandwidth ofthe stiffness matrix will then, in general, increase during the reduction process, and problems of storage must be considered.Por the solution of the eigenproblem Kaa = AMaa, it is important to note that Ka is, in general, a full matrix, and the solution is relatively expensive unless the order of the matrices is small.lnstead of calculating the matrix Ka, it may be preferable to evaluate the ftexibility,matrix Fa = K;;- 1which is obtained using[KaaKac] [Fa][']KcaKccFe=O(10.51)where 1is a unit matrix of the same order as Kaa Therefore, in (10.51), we solve for the displacements of the structure when unit loads are applied in turn at the mass degrees of freedom. Although the degrees of freedom have been partitioned in ( 10.51), there is no need for it in this analysis (see Example 10.13). Having solved for Fa, we now consider instead of ( 10.45) the eigenproblem (10.52)Although this eigenproblem is of a slightly different form than the generalized problem K = AM, the transformation to a standard problem proceeds in much the same way (see Section 10.2.5). Por the transformation we define (10.53)where M!12 is a diagonal matrix with its ith diagonal element equal to the root of the ith diagonal element of Ma. Premultiplying both sides of ( 10.52) by M!12 and substituting the relation in ( 10.53), we obtain(10.54)(10.55)Once the displacements a have been calculated, we obtain the complete displace ment vector using (10.56)where Fe was calculated in (10.51). The relation in (10.56) is arrived at by realizing that the forces applied at the mass degrees of freedom to impose a are Kaa. Using (10.51), the corresponding displacements at all degrees of freedom are given in ( 10.56).EXAMPLE 10.13: Use the procedure given in (10.51) to (10.56) to calculate the eigenvalues and eigenvectors of the problem K = A.M considered in Example 10.12.The first step is to so1ve the equations-12-12-1ooro10o1(a)or 0o-12-1 [VVz] = 0 0o-111where we did not interchange rows and columns in K in order to obtain the form in (10.51).For the solution of the equations in (a), we use the LDLT decomposition of K, whereHence, we obtainand hence,vf = [12Fa= [ !l22];vr = [1234]Fe= [ !]F = [v'2 0][22][v'2o][ 42v'2]aO124O1 =2v'24The solution of the eigenproblemgivesJL = 4 - 2v'2;1v'2al=1v'21v'2(b)#1-2 = 4 + 2v'2;Since a = M;112 a. we havea2 =1v'2.. [0 :J-111-v'22211'a2 =1(e)-v'2v'2v'2The vectors c,, and c2 are calculated using ( 10.56); hence,(d)Since JL = 1/A. we realize that in (b) to (d) we have the same solution as obtained in Exam ple 10.12.Considering the different procedures of eliminating the massless degrees of freedom, the results of the eigensystem analysis are the same irrespective ofthe procedure followed, i.e., whether Ka or Fa is established and whether the eigenproblem in (10.45) or in (10.52) is solved. The basic assumption in the analysis is that resulting from mass lumping. As we discussed in Section 10.2.4, each zero mass corresponds to an infinite frequency in thesystem. Therefore, in approximating the original system equation K[ o-12oUsing the procedure given in (10.51) to (10.56), we obtain=F[.!]a3 'Hence, A1 = , a1 = [1/v'2], and12v'212v'2The solution of the eigenproblem in (a) for the smallest eigenvalue and corresponding eigenvectoris hence2'\ - 3./\ -12v'21v'212v'2whereas the solution of the original problem (see Example 10.4) is1V2 = 2;cf>,1= V21V2lt should be noted that using the mass lumping procedure, the eigenvalues can be smaller-as in this example-or larger than the eigenvalues of the original system.0. Rayleigh-Ritz AnalysisA most general technique for finding approximations to the lowest eigenvalues and corre sponding eigenvectors of the problem K = AM is the Rayleigh-Ritz analysis. The static condensation procedure in Section 10.3.1, the component mode synthesis described in the next section, and various other methods can be understood to be Ritz analyses. As we will see, the techniques differ only in the choice of the Ritz basis vectors assumed in the analysis. In the following we first present the Rayleigh-Ritz analysis procedure in general and then show how other techniques relate to it.The eigenproblem that we consider isKcf> = Mcf>(10.4)where we now first assume for clarity of presentation that K and M are both positive definite, which ensures that the eigenvalues are all positive; i.e., A, > O. As we pointed out in Section 10.2.3, K can be assumed positive definite because a shift can always be intro duced to obtain a shifted stiffness matrix that satisfies this condition. As for the mass matrix, we now assume that M is a consistent mass matrix or a lumped mass matrix with no zero diagonal elements, which is a condition that we shall later relax.Consider first the Rayleigh minimum principie, which states that = min p(cf>)(10.57)where the minimum is taken over all possible vectors . and p( ) is the Rayleigh quotientcf>TKcf>p( cf>) = cf>TMcf>(10.58)This Rayleigh quotient is obtained from the Rayleigh quotient of the standard eigenvalue problem :icj, = Acj, (see Sections 2.6 and 10.2.5). Since both K and M are positive definite, p( ) has finite values for all Referring to Section 2.6, the bounds on the Rayleighquotient areO < :::::; p(cf>) :::::; n < 00(10.59)In the Ritz analysis we consider a set of vectors. which are linear combinations of the Ritz basis vectors t!J;, i = 1, ... , q; i.e., a typical vector is given byqel> = L X;tll;i=l(10.60)where the X are the Ritz coordinates. Since e, is a linear combination of the Ritz basis vectors, e, cannot be any arbitrary vector but instead lies in the subspace spanned by theRitz basis vectors, which we call Vq (see Sections 2.3 and 11.6). lt should be noted that the vectors \fl, i = 1, ... , q, must be linearly independent; therefore, the subspace Vq has dimension q. Also, denoting the n-dimensional space in which the matrices K and M are defined by Vn, we have that Vq is contained in Vn.In the Rayleigh-Ritz analysis we aim to determine the specific vectors c,j, i = 1, ... , q, which, with the constraint of lying in the subspace spanned by the Ritz basis vectors, "best" approximate the required eigenvectors. For this purpose we invoke the Rayleigh minimum principie. The use of this principie determines in what sense the solution "best"approximates the eigenvectors sought, an aspect that we shall point out during the presentation of the solution procedure.To invoke the Rayleigh minimum principie on e,, we first evaluate the Rayleighquotient,whereqqL L XXi(,p(cf>) = :...,i;,..:.l-=-:i;,..:-1--L L XXffflj=! i=lkij = tlliKtll(10.61)(10.62)mij = tlliMtll(10.63)The necessary condition for a minimum of p(c,) given in (10.61) is ap(c,)jaxj = O,i = 1, ... , q, because the X are the only variables. However,q-- q""p(..!..)2t .L Xk - 2k L Xffl_u_'!"_ ==!J=I(10.64)and using p = k/r, the condition for a minimum of p(c,) isqL (k1 - pmij)x1 = ofor i = 1, ... , q(10.65)j=!In actual analysis we write the q equations in ( 10.65) in matrix form, thus obtaining the eigenproblemKx = pMx(10.66)where K and M are q X q matrices with typical elements defined in ( 10.62) and ( 10.63), respectively, and x is a vector of the Ritz coordinates sought: (10.67)The solution to ( 10.66) yields q eigenvalues p, ... , pq, which are approximations toA,, ... , q, and q eigenvectors,xf = [xlxix ]xr = [xrx = [x1XX...x ]x:J(10.68)The eigenvectors X; are used to evaluate the vectors, ... , q, which are approximations to the eigenvectors... , q Using (10.68) and (10.60), we haveqcf>; = L x; tllj;j=li = 1, ... , q(10.69)An important feature of the eigenvalue approximations calculated in the analysis is that they are upper bound approximations to the eigenvalues of interest; i.e., (10.70)meaning that since K and M are assumed to be positive definite, K and M are also positive definite matrices.The proof of the inequality in (10.70) shows the actual mechanism that is used to obtain the eigenvalue approximations p;. To calculate p we search for the rninimum of p( ) that can be reached by linearly combining all available Ritz basis vectors. The inequality A1 < p follows from the Rayleigh minimum principie in (10.57) and because Vq is con tained in the n-dimensional space Vn, in which K and M are defined.The condition that is employed to obtain pz is typical of the mechanism used to calculate the approximations to the higher eigenvalues. First, we observe that for the eigenvalue problem K = AM , we haveA2 = min p(cf>)(10.71)where the rninimum is now taken over all possible vectors in Vn that satisfy the orthog onality condition (see Section 2.6) (10.72)Considering the approximate eigenvectors; obtained in the Rayleigh-Ritz analysis, we observe that (10.73)where 5;j is the Kronecker delta, and that, therefore, in the above Rayleigh-Ritz analysis we obtained pz by evaluatingP2 = min p(cf>)(10.74)where the rninimum was taken over all possible vectorsin Vq that satisfy the orthogonality condition (10.75)To show that Az < pz, we consideran auxiliary problem; i.e., assume that we evaluate[h = min p((ji)where the rninimum is taken over all vectorsthat satisfy the conditioncf>TMcf> = 0(10.76)(10.77)The problem defined in (10.76) and (10.77) is the same as the problem in (10.71) and (10.72), except that in the latter case the minimum is taken over all . whereas in the problem in (10.76) and (10.77) we consider all vectors in Vq. Then since Vq is contained in Vn, we have Az < j)z. On the other hand, j)z < pz because the most severe constraint one, in (10.77) is e,. Therefore, we have(10.78)The basis for the calculation of c,z, and hence pz, is that the rninirnurn of p(c,) is sought with the orthogonality condition in (1O.75) on e,. Sirnilarly, to obtain p; ande, , we in fact rninirnize p( e,) with the orthogonality conditions c,rMc,j = O for j = 1, ... , i - l. Accordingly, the inequality on p in (10.70) can be proved in an analogous rnanner to the procedure used above for pz, but all i - 1 constraint equations need to be satisfied.The observation that i - 1 constraint equations need to be fulfilled in the evaluation of p also indicates that we can expect less accuracy in the approxirnation of the higher eigenvalues than in the approxirnation of the lower eigenvalues, for which fewer constraints are irnposed. This is generally also observed in actual analysis.Considering the procedure in practica! dynarnic analysis, the Ritz basis functions rnay be calculated frorn a static solution in which q load patterns are specified in R; i.e., we considerK\fl = R(10.79)where '\}f is an n X q rnatrix storing the Ritz basis vectors; i.e., '\}f = [\fl, ... , \flq]. The analysis is continued by evaluating the projections of K and M onto the subspace Vq spanned by the vectors \fl;, i = 1, ... , q; i.e., we calculateK = 'fiTK\fl(10.80)and where because of (10.79) we haveNext we solve the eigenproblern K-x = pM-x, the solution of which can be written(10.81)(10.82)K-X = M-Xp(10.83)where pis a diagonal J!!atrix listing the eigenvalue approxirnations p;, p = diag(p;), and X is a rnatrix storing the M- orthonorrnal eigenvectors x 1, ,Xq. The approxirnations to the eigenvectors of the problern Kc, = AMe, are thencll = \f!X(10.84)So far we have assurned that the rnass rnatrix of the finite elernent systern is positive definite; i.e., Mis nota diagonal rnass rnatrix with sorne zero diagonal elernents. The reason for this assurnption was to avoid the case c,rMc, equal to zero in the calculation of the Rayleigh quotient, in which case p( e,) gives an infinite eigenvalue. However, the Rayleigh Ritz analysis can be carried out as described above when M is a diagonal rnatrix with sorne zero diagonal elernents, provided the Ritz basis vectors are selected to lie in the subspace that corresponds to the finite eigenvalues. In addition, the Ritz basis vectors rnust be linearly independent wh n considering only the rnass degrees of freedorn in order to obtain a positive definite rnatrix M. One way of achieving this in practice is to excite different rnass degrees of freedorn in each of the load vectors in R in (10.79) (see Section 11.6.3 and Exam ple 10.16).Of particular interest are the errors that we rnay expect in the solution. Although we have shown that an eigenvalue calculated from the Ritz analysis is an upper bound on thecorresponding exact eigenvalue of the system, we did not establish anything about the actual error in the eigenvalue. This error depends on the Ritz basis vectors used because thevectors e, are linear combinations of the Ritz basis vectors \fl;, i = 1, ... , q. We can obtaingood results only if the vectors \fl; span a subspace Vq that is close to the least dominant subspace of K and M spanned by e,,, ... , c,q. It should be noted that this does not meanthat the Ritz basis vectors should each be close to an eigenvector sought but rather that linear combinations of the Ritz basis vectors can establish good approximations of the requiredeigenvectors of Kc, = AMe,. We further discuss the selection of good Ritz basis vectors andthe approximations involved in the analysis in Section 11.6 when we present the subspaceiteration method, because this method uses the Ritz analysis technique.To demonstrate the Rayleigh-Ritz analysis procedure, consider ,the following examples.EXAMPLE 10.15: Obtain approximate solutions to the eigenproblem Kcf> = AMcf> consid ered in Example 10.4, where2-1o]K=-14-1 ;[ o-12The exact eigenvalues are A, = 2, A2 = 4, A3 = 6.l. Use the following load vectors to generate the Ritz basis vectors2. Then use a different set of load vectors to generate the Ritz basis vectorsHIn the Ritz analysis we employ the relations in (10.79) to (10.84) and obtain, in case 1,-!J [i J\fl=fi fi][Hence,6617TI TIandThe solution of the eigenproblem Kx = pMx isM: = _1_[2911J144 11 29(p, X) = (2.4004,1.3418]1.3({12. x2) = (4.0032, [_ 2.008] )[);4182 0008Hence, we have as eigenvalue approximations p 1 = 2.40, P2 = 4.00, and evaluating4470 895- = [ ][1.34182.0008] = [ cll1.3418-2.00080.895TiTiwe have(j)1 =0.895] ;(j)2 = [ 1.00] [0.4470.000.895-1.00H-n [: !JNext we assume the load vectors in case 2 and salveHence,and[! !J[NK=21]'3 31 ]M= J..[4136 13The solution of the eigenprob1emKx = pMx gives0.70711]({12, X2) = (6.0000,-2.1213])(p1, X1) = (2.000, [0.70711 );[.6 3640[i !]Hence, we have as eigenvalue approximations P1 = 2.00, P2 = 6.00, and evaluatingcll =[0.70711510.707116 6-2.1213][0.707116.3640=0.707110.70711-0.70708]0.70713-0.70708we have0.70711]4>1 =0.70711 ;[0.70711-0.70708]4>2 =o.70713[-0.70708Comparing the resu1ts with the exact solution, it is interesting to note that in case 1, P1 > A1 and P2 = A2, whereas in case 2, P1 = A1 and P2 = 3. In both cases we did not obtain good approximations to the lowest two eigenvalues, and it is clearly demonstrated that the resultsdepend completely on the initial Ritz basis vectors chosen.EXAMPLE 10.16: Use the Ray1eigh-Ritz analysis to calculate an approximation to A1 and cf>1of the eigenproblem considered in Example 10.12.We note that in this case M is positive sernidefinite. Therefore, to carry out the Ritz analysis we need to choose a load vector in R that excites at least one mass. Assume that we useRT = [O1 oO]Then the solution of (10.79) yields (see Example 10.13)wr = [1222]Hence,K= [2];p = 61;M= [12]X= [2 ]andHence we have, as expected, p, > A, .The Ritz analysis procedure presented above is a very general tool, and, as pointed out earlier, various analysis methods known under different names can actually be shown to be Ritz analyses. In Section 10.3.3 we present the component mode synthesis as a Ritz analysis. In the following we briefly want to show that the technique of static condensation as described in Section 10.3.1 is, in fact, also a Ritz analysis.In the static condensation analysis we assumed that all mass can be lumped at q degrees of freedom. Therefore, as an approximation to the eigenproblem K = AM . we obtained the following problem:[KaaKae] [e!>] = "-[MO] [e!>]KeaKeecf>eOOcf>e(10.42)with q finite and n - q infinite eigenvalues, which correspond to the massless degrees of freedom (see Section 10.2.4). To calculate the finite eigenvalues, we used static condensa tion on the massless degrees of freedom and arrived at the eigenproblem (10.45)where K. is defined in (10.46). However, this solution is actually a Ritz analysis of the lumped mass model considered in ( 10.42). The Ritz basis vectors are the displacement patterns associated with the . degrees of freedom when the e degrees of freedom are released. Solving the equations[KaaKae] [F] = [1]KeaKeeFeO(10.51)in which F. = K;;', we find that the Ritz basis vectors to be used in (10.80), (10.81), and (10.84) are (10.85)To verify that a Ritz analysis with the base vectors in (10.85) yields in fact (10.45), we evaluate (10.80) and (10.81). Substituting for '\}f and K in (10.80), we obtainK= [1 (FeKaYJ[ ::J[Fe J(10.86)which, using ( 10.51), reduces toK= K.Sirnilarly, substituting for '\}f and M in (10.81), we haveM= [1 (FeK.Y][ ][Fe J(10.87)(10.88)or M=Ma (10.89)Hence, in the static condensation we actually perform a Ritz analysis of the lumped mass model. lt should be noted that in the analysis we calculate the q finite eigenvaluesexactly (i.e., p; = A; for i = 1, ... , q) because the Ritz basis vectors span the qdimensional subspace corresponding to the finite eigenvalues. In practice, the evaluation ofthe vectors '\}f in (10.85) is not necessary (and would be costly), and instead the Ritz analysis is better carried out using\fl = [ :](10.90)Since the vectors in (10.90) span the same subspace as the vectors in (10.85), the same eigenvalues and eigenvectors are calculated employing either set of base vectors.Specifically, using (10.90), we obtain in the Ritz analysis the reduced eigenproblem (10.91)To show that this eigenproblem is indeed equivalent to the problem in ( 10.45), we premul tiply both sides in ( 10.91)by K. and use the transformation x = Kai, giving Kai = AMai, i.e., the problem in (10.45).EXAMPLE 10.17: Use the Ritz ana1ysis procedure to perform static condensation of the mass1ess degrees of freedom in the problem Kcf> = AMcf> considered in Example 10.12.We first need to evaluate the Ritz basis vectors given in ( 10.90). This was done inExample 10.13, where we found thatThe Ritz reduction given in (10.91) thus yields the eigenprob1em[22]- ,\[1216]24 X-1624 XFinally, we should note that the use ofthe Ritz basis vectors in ( 10.85) [(or in (10.90)] is also known as the Guyan reduction (see R. J. Guyan [A]). In the Guyan scheme the Ritz vectors are used to operate on a lumped mass matrix with zero elements on the diagonal as in (10.88) or on general fulllumped or consistent mass matrices. In this reduction the a degrees of freedom are frequently referred to as dynamic degrees of freedom.0. Component Mode SynthesisAs for the static condensation procedure, the component mode synthesis is, in fact, a Ritz analysis, and the method might have been presented in the previous section as a specific application. However, as was repeatedly pointed out, the most important aspect in a Ritz analysis is the selection of appropriate Ritz basis vectors because the results can be only as good as the Ritz basis vectors allow them to be. The specific scheme used in the component mode synthesis is of particular interest, which is the reason we want to devote a separate section to the discussion of the method.The component mode synthesis has been developed to a large extent as a natural consequence of the analysis procedure followed in practice when large and complex struc tures are analyzed. The general practica! procedure is that different groups perform the analyses of different components of the structure under consideration. For example, in a plant analysis, one group may analyze a main pipe and another group a piping system attached to it. In a first preliminary analysis, both groups work separately and model the effects of the other components on the specific component that they consider in an approx imate manner. For example, in the analysis of the two piping systems referred to above, the group analyzing the side branch may assume full fixity at the point of intersection with the main pipe, and the group analyzing the main pipe may introduce a concentrated spring and mass to allow for the side branch. The advantage of considering the components of the structure separately is primarily one of time scheduling; i.e., the separate groups can work on the analyses and designs of the components at the same time. lt is primarily for this reason that the component mode synthesis is very appealing in the analysis and design of large structural systems.Assume that the preliminary analyses of the components have been carried out and that the complete structure shall now be analyzed. lt is at this stage that the component mode synthesis is a natural procedure to use. Namely, with the mode shape characteristics of each component known, it appears natural to use this information in estimating the frequencies and mode shapes of the complete structure. The specific procedure may vary (see R. R. Craig, Jr. [A]), but, in essence, the mode shapes of the components are used in a Rayleigh-Ritz analysis to calculate approximate mode shapes and frequencies of the complete structure.Consider for illustration that each component structure was obtained by fixing a11 its boundary degrees of freedom and denote the stiffness matrices of the component structures by K1, Kn, ... , KM (see Example 10.18). Assume that only component structures L - 1and L connect, L = 2, ... , M; then we can write for the stiffness matrix of the completestructure,KnK=(10.92)Using an analogous notation for the mass matrices, we also haveM.Mn M=(10.93)Assume that the lowest eigenvalues and corresponding eigenvectors of each componentstructure have been calculated; i.e., we have for each component structure,K, el, = M, 11A,Kn t + 21>2"'2 = 3f>t - 1>2where f>t and !>2 are the eigenvectors corresponding to At and A2 , then the Rayleigh-Ritz analysis will give the exact eigenvalues At and A2 and the corresponding eigenvectors f> 1 and 1>2. Show explicitly that this result is indeed obtained.0. Consider the following spring system.3. Evaluate the exact smallest frequency of the system.(b) Evaluate an approximation of the smallest frequency by using the component mode synthe sis technique in Section 10.3.3. Use only the eigenvector of the smallest frequency of each component in the system.Component 1-------------,/km*\\'l 1.... _ ------------""K= 10k= 1m=2 m*= 10. SOLUTION ERRORSAn important part of an eigenvalue and vector solution is to estimate the accuracy with which the required eigensystem has been calculated. Since an eigensystem solution is necessarily iterative, the solution should be terminated once convergence within the pre scribed tolerances giving the actual accuracy has been obtained. When one of the approx imate solution techniques outlined in Section 10.3 is used, an estimate ofthe actual solution accuracy obtained is of course also important.0. Error BoundsIn order to identify the accuracy that has been obtained in an eigensolution, we recall that the equation to be solved isK = AM(10.96)Let us first assume that using any one solution procedure we obtained an approximation X and toan eigenpair. Then without regard to how the values have been obtained, we can evaluate a residual vector that gives important information about the accuracy with whichX and approximate the eigenpair. The results are given in (10.101) to ( 10.104). We thenpresent also error bound calculations useful in solutions based on inverse iterations and asimple error measure.Standard EigenproblemConsider first that M = l. In that case we can writer =K - Xand using the relations in (10.12) and (10.13), we haver = ci(A - XI)clr(i)or because X is not equal but only close to an eigenvalue, we haveCi) = ci(A - Xlt1clrrHence, because 11 llz = 1, taking norms we obtain1 :s; II = [1]0Evaluate r and thus establish the relations given in ( 10.101) and ( 10.102).First, we calculate r as given in ( 10.97),Hence, llrll2 = 1 and (10.101) yieldsminiA- XI :511(a)Therefore, we can conclude that an eigenvalue has been approximated with about 1 percent or less error. Since we know At and A2, we can compare X with At or A2 and find that (a) does indeedbold.Considering now the eigenvector approximation f>, we note that f> does not approximate either f>t or !>2 This is also reflected by evaluating the relation ( 10.102). Assuming that f> is anapproximation to !>t. which gives s = 1, we have11 (j) - a1!>t lb :5 1Similarly, assuming thatis an approximation to 2. we obtain11 - a2 2ll2 S: 1and in both cases the bound obtained is very large (note that ll 1ll2 = 1 and ll 2lb = 1),indicating thatdoes not approximate an eigenvector.Generalizad EigenproblemConsider now that we wish to estimate the accuracy obtained in the solution of a generalized eigenproblem K = A.M. Assume that we have calculated as an approximation to ; and; the values X and . Then, in analogy to the calculations performed above, we cancalculate an error vector rM, whererM = K - XM(i)(10.103)In order to relate the error vector in ( 10.103) to the error vector that corresponds to the standard eigenproblem, we use M = ssr, and thenr = icf, - Xcf,(10.104)where r = s-rM, rMcl> /rM consists of two numbers that are easily calculated in the iterations.While the above error bounds are very effective, it is finally also of interest to consider the following simple error measure:IIK(i)- XM(i)ll2E =IIK 112(10.lOS)1 To avoid the factorization of M we may instead consider the problem Me, = A-1Kc, if the factorization of K is already available, and then establish bounds on A-1Since, physically, K represents the elastic nodal point forces and AM represents the inertia nodal point forces when the finite element assemblage is vibrating in the mode , we evaluate in ( 10.108) the norm of out-of-balance nodal point forces divided by the normof elastic nodal point forces. This quantity should be small if X and are an accuratesolution of an eigenpair.If M = 1, it should be noted that we can writeand hence,XE = /lrll2E 2:: mm. 1 A;- XI(10.109); '---'-A=----'(10.110)EXAMPLE 10.23: Consider the eigenproblem K = AM , whereK = [ 10-lO]M = [21]-10100 '1 4The exact eigenvalues and eigenvectors to 12-digit precision areA1 = 3.863385512876;0.640776011246]1 = [0.105070337503-0.401041986380]A2 = 33.279471629982;2 = [0.524093989558Assume that (i) = ( 1 + s 2)c, where e is such that (i)TM = 1 and S= 10-1, 10-3, and 10-6 For each value of S evaluate X as the Ray1eigh quotient of (i) and calcu1ate the error bounds based on (10.104), (10.106), and the error measure E given in (10.108).The following tab1e summarizes the resu1ts obtained. The equations used to evaluate the quantities are given in ( 10.103) to ( 10.108). The results in the tab1e show that for each value of S the error bounds are satisfied and that E is also small for an accurate solution.S10-110-310-60.5976907926560.6403746490730.6407756102040.1566981944810.1055943786950.1050708615974lK4.1546338902753.8634149289323.863385512905X4.1546338902753.8634149289323.863385512905-1.207470493734-0.008218153965-0.000008177422rM4.6056305811240.0498388032260.000049870085r1.6344194662420.0211067436170.0000211523641.4116792956810.0150425453270.000015049775JA1- XJ0.2912483773990.000029416056 0.000000000029Bound2.1596678970360.0259185801320.000025959936 (10.101/10.104)Bound2.9124837739830.0294160567440.000029433139(10.106)Measure0.4471132358130.0074582086600.000007491764(10.108)886Preliminaries to the Solutions of EigenproblemsChap. 100. Exercises0. The following error bound is discussed by J. Stoer and R. Bulirsch [A]. Let A be a symmetric matrix and A; be an eigenvalue of A; thenmin lA;- xrAx 1 :5iXTXfor any vector x * O.Show that (10.105) to (10.107) follow from this formula.0. Consider the eigenproblem in Exercise 10.1. LetCalculateusing (10.105) and p( ). These values, p(Ci)) and Ci), are now the best approxima tions to an eigenvalue and eigenvector.Establish the error bounds (10.101) [with (10.103)] and (10.106). Also evaluate the errormeasure (10.108).