STRUCTURAL DYNAMICS, VOL. 9 Computational Dynamics · 1.1 Fundamentals of Linear Structural...

STRUCTURAL DYNAMICS, VOL. 9

Computational Dynamics

Søren R. K. Nielsen

λλ1 λ2 λ3

µkµk−1

µk+1

P (λ)

y(λ) = P (µk) +(P (µk) − P (µk−1)

) λ − µk

µk − µk−1

Aalborg tekniske UniversitetsforlagJune 2005

Contents

1 INTRODUCTION 71.1 Fundamentals of Linear Structural Dynamics . . . . . . . . . . . . . . . . . . 71.2 Solution of Initial Value Problem by Modal Decomposition Techniques . . . . 251.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2 NUMERICAL INTEGRATION OF EQUATIONS OF MOTION 312.1 Newmark Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.1.1 Numerical accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.1.2 Numerical stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.1.3 Period Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.1.4 Numerical Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.2 Generalized Alpha Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3 LINEAR EIGENVALUE PROBLEMS 533.1 Gauss Factorization of Characteristic Polynomials . . . . . . . . . . . . . . . . 533.2 Eigenvalue Separation Principle . . . . . . . . . . . . . . . . . . . . . . . . . 593.3 Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.4 Transformation of GEVP to SEVP . . . . . . . . . . . . . . . . . . . . . . . . 653.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4 APPROXIMATE SOLUTION METHODS 694.1 Static Condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.2 Rayleigh-Ritz Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3 Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5 VECTOR ITERATION METHODS 895.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.2 Inverse and Forward Vector Iteration . . . . . . . . . . . . . . . . . . . . . . . 905.3 Shift in Vector Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.4 Inverse Vector Iteration with Rayleigh Quotient Shift . . . . . . . . . . . . . . 1065.5 Vector Iteration with Gram-Schmidt Orthogonalization . . . . . . . . . . . . . 1095.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

— 3 —

4 Contents

6 SIMILARITY TRANSFORMATION METHODS 1156.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.2 Special Jacobi Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.3 General Jacobi Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.4 Householder Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1286.5 QR Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

7 SOLUTION OF LARGE EIGENVALUE PROBLEMS 1477.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1477.2 Simultaneous Inverse Vector Iteration . . . . . . . . . . . . . . . . . . . . . . 1497.3 Subspace Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1567.4 Characteristic Polynomial Iteration . . . . . . . . . . . . . . . . . . . . . . . . 1647.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

8 INDEX 172

A Solutions to Exercises 175A.1 Exercise 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175A.2 Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177A.3 Exercise 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179A.4 Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180A.5 Exercise 3.4: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182A.6 Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185A.7 Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188A.8 Exercise 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189A.9 Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191A.10 Exercise 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193A.11 Exercise 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.12 Exercise 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200A.13 Exercise 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204A.14 Exercise 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206A.15 Exercise 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

Preface

This book has been prepared for the course on Computational Mechanics given at the 8thsemester at the structural engineering program in civil engineering at Aalborg University. Thecourse presumes undergraduate knowledge of linear algebra and ordinary differential equations,as well as a basic graduate course in structural dynamics. Some of these prerequisites have beenreviewed in an introductory chapter. The author wants to thank Jesper W. Larsen, Ph.D., andPh.D. student Kristian Holm-Jørgensen for help with the preparation of figures and illustrationsthroughout the text.

Answers to all exercises given at the end of each chapter can be downloaded from the homepage of the course at the address: www.civil.auc.dk/i5/engelsk/dyn/index/htm

Aalborg University, June 2005Søren R.K. Nielsen

— 5 —

6 Contents

CHAPTER 1INTRODUCTION

In this chapter the basic results in structural dynamics and linear algebra have been reviewed.

In Section 2.1 the relevant initial and eigenvalue problems in structural dynamics are formu-lated. The initial value problems form the background for the numerical integration algorithmsdescribed in Chapter 2, whereas the related undamped generalized eigenvalue problem consti-tute the generic problem for the numerical eigenvalue solvers described in Chapters 3-7. Formalsolutions to various formulations of the initial value problem are indicated, and their shortcom-ings in practical applications are emphasized.

In Section 2.2 the semi-analytical solution approaches to the basic initial value problem of amulti degrees-of-freedom system in terms of expansion in various modal bases are presented.The application of these methods in relation to various reduction schemes, where typicallymerely the low-frequency modes are required, has been outlined.

1.1 Fundamentals of Linear Structural Dynamics

The basic equation of motion for forced vibrations of a linear viscous damped n degree-of-freedom system reads1

Mx(t) + Cx(t) + Kx(t) = f(t) , t > t0

x(t0) = x0 , x(t0) = x0

}(1–1)

x(t) is the vector of displacements from the static equilibrium state, x(t) is the velocity vector,x(t) is the acceleration vector, and f(t) is the dynamic load vector. x0 and x0 denote the initialvalue vectors for the displacement and velocity, respectively. K, M and C indicate the stiffnessmatrix, mass matrix and damping matrices, all of the dimension n × n. For any vector a �= 0these fulfill the following positive definite and symmetry properties

aTKa > 0 , K = KT

aTMa > 0 , M = MT

aTCa > 0

⎫⎪⎪⎬⎪⎪⎭ (1–2)

1S.R.K. Nielsen: Structural Dynamics, Vol. 1. Linear Structural Dynamics, 4th Ed.. Aalborg tekniske Univer-sitetsforlag, 2004.

— 7 —

8 Chapter 1 – INTRODUCTION

If the structural system is not supported against stiff-body motions, the stiffness matrix is merelypositive semi-definite, so aTKa ≥ 0. Correspondingly, if some degrees of freedom are notcarrying kinetic energy (pseudo degrees of freedom with zero mass or zero mass moment ofinertia), the mass matrix is merely positive semi-definite, so aTMa ≥ 0. The positive definiteproperty of the damping matrix is a formal statement of the physical property that any non-zerovelocity of the system should be related with energy dissipation. However, C needs not fulfillany symmetry properties, although energy dissipation is confined to the symmetric part of thematrix. So-called aerodynamic damping loads are external dynamic loads proportional to thestructural velocity, i.e. f(t) = −Cax(t). If the aerodynamic damping matrix Ca is absorbed inthe total damping matrix C, no definite property can be stated.

The solution of the initial value problem (1-1) can be written in the following way1

x(t) =

∫ t

t0

h(t − τ)f(τ)dτ + a0(t − t0)x0 + a1(t − t0)x0

a0(t) = h(t)C + h(t)M

a1(t) = h(t)M

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭

(1–3)

h(t) is the impulse response matrix . Formally, this matrix is obtained as a solution to the initialvalue problem

Mh(t) + Ch(t) + Kh(t) = I δ(t)

h(0−) = 0 , h(0−) = 0

}(1–4)

I is the unit matrix of the dimension n × n, and δ(t) is Dirac’s delta function.

The frequency response matrix H(iω) related to the system (1-1) is given as

H(iω) =((

iω)2M + (iω)C + K)−1

(1–5)

where i =√−1 is the complex unit. The impulse response matrix is related to the frequency

response matrix in terms of the Fourier transform

h(t) =1

2π

∫ ∞

−∞H(iω)eiωtdt (1–6)

The convolution quadrature in (1-3) is relative easily evaluated numerically. Hence, the solutionof (1-1) is available, if the impulse response matrix h(t) is known. In turn, the n×n componentsof this matrix can be calculated by the Fourier transforms (1-6). Although these transforms maybe evaluated numerically, the necessary calculation efforts become excessive even for a moder-ate number of degrees of freedom n. Hence, more direct analytical or numerical approaches are

1.1 Fundamentals of Linear Structural Dynamics 9

mandatory.

Undamped eigenvibrations(C = 0, f(t) ≡ 0

)are obtained as linear independent solutions to

the homogeneous matrix differential equation

Mx(t) + Kx(t) = 0 (1–7)

Solutions are searched for on the form

x(t) = Φ(j)eiωjt (1–8)

Insertion of (1-8) into (1-7) provides the following homogeneous system of linear equations forthe determination of the amplitude Φ(j) and the unknown constant ωj

(K− λjM

)Φ(j) = 0 , λj = ω2

j (1–9)

(1-9) is a so-called generalized eigenvalue problem (GEVP). If M = I, the eigenvalue problemis referred to as a special eigenvalue problem (SEVP).

The necessary condition for non-trivial solutions (i.e. Φ(j) �= 0) is that the determinant of thecoefficient matrix is different from zero. This lead to the characteristic equation

P (λ) = det(K − λM

)= 0 (1–10)

P (λ) indicates the characteristic polynomial. This may be expanded as

P (λ) = a0λn + a1λ

n−1 + · · ·+ an−1λ + an (1–11)

The constants a0, a1, . . . , an are known as the invariants of the GEVP. This designation stemsfrom the fact that the characteristic polynomial (1-11) is invariant under any rotation of the co-ordinate system. Obviously, a0 = (−1)n det(M), and an = det(K). The nth order equation(1-10) determines n solutions, λ1, λ2, . . . , λn.

Assume that either M or K are positive definite. Then, all eigenvalues λj are non-negative real,which may be ordered in ascending magnitude as follows1

0 ≤ λ1 ≤ λ2 ≤ · · · ≤ λn−1 ≤ λn ≤ ∞ (1–12)

λn = ∞, if det(M) = 0. Similarly, λ1 = 0, if det(K) = 0. The eigenvalues are denotes assimple, if λ1 < λ2 < · · · < λn−1 < λn. The undamped circular eigenfrequencies are related tothe eigenvalues as follows


ωj =√

λj (1–13)

The corresponding solutions for the amplitude functions, Φ(1), . . . ,Φ(n), are denoted the un-damped eigenmodes of the system, which are real as well.

The eigenvalue problems (1-9) can be assembled into following matrix formulation

K[Φ(1) Φ(2) · · ·Φ(n)

]= M

[Φ(1) Φ(2) · · ·Φ(n)

]⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn

⎤⎥⎥⎥⎦ ⇒

KΦ = MΦΛ (1–14)

where

Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn

⎤⎥⎥⎥⎦ (1–15)

and Φ is the so-called modal matrix of dimension n × n, defined as

Φ =[Φ(1) Φ(2) · · ·Φ(n)

](1–16)

If the eigenvalues are simple, the eigenmodes fulfill the following orthogonality properties1

Φ(i) TMΦ(j) =

{0 , i �= j

Mi , i = j(1–17)

Φ(i) TKΦ(j) =

{0 , i �= j

ω2i Mi , i = j

(1–18)

where Mi denotes the modal mass.

The orthogonality properties (1-17) can be assembled in the following matrix equation

[Φ(1) Φ(2) · · ·Φ(n)

]TM[Φ(1) Φ(2) · · ·Φ(n)

]=

⎡⎢⎢⎢⎣

M1 0 · · · 0

0 M2 · · · 0...

.... . .

...0 0 · · · Mn

⎤⎥⎥⎥⎦ ⇒

ΦTMΦ = m (1–19)


where

m =

⎡⎢⎢⎢⎣M1 0 · · · 0

0 M2 · · · 0...

.... . .

...0 0 · · · Mn

⎤⎥⎥⎥⎦ (1–20)

The corresponding grouping of the orthogonality properties (1-18) reads

ΦTKΦ = k (1–21)

where

k =

⎡⎢⎢⎢⎣

ω21M1 0 · · · 0

0 ω22M2 · · · 0

......

. . ....

0 0 · · · ω2nMn

⎤⎥⎥⎥⎦ (1–22)

If the eigenvalues are all simple, the eigenmodes become linear independent, which means thatthe inverse Φ−1 exists.

In the following it is generally assumed that the eigenmodes are normalized to unit modal mass,so m = I. For the special eigenvalue problem, where M = I, it then follows from (1-19) that

Φ−1 = ΦT (1–23)

A matrix fulfilling (1-23) is known as orthonormal or unitary, and specifies a rotation of thecoordinate system. All column and row vectors have the length 1, and are mutually orthogonal.It follows from (1-19) and (1-21) that in case of simple eigenvalues a so-called similarity trans-formation exists, defined by the modal matrix Φ, that reduce the mass and stiffness matrices toa diagonal form. In case of multiple eigenvalues the problem becomes considerable more com-plicated. For the standard eigenvalue problem with multiple eigenvalues it can be shown thatthe stiffness matrix merely reduces to the so-called Jordan normal form under the consideredsimilarity transformation, given as follows

k =

⎡⎢⎢⎢⎣k1 0 · · · 0

0 k2 · · · 0...

.... . .

...0 0 · · · km

⎤⎥⎥⎥⎦ (1–24)

where m ≤ n denotes the number of different eigenvalues, and ki signifies the so-called Jordanboxes, which are block matrices of the form


ω2i ,

[ω2

i 1

0 ω2i

],

⎡⎢⎣ω2

i 1 0

0 ω2i 1

0 0 ω2i

⎤⎥⎦ ,

⎡⎢⎢⎢⎣ω2

i 1 0 0

0 ω2i 1 0

0 0 ω2i 1

0 0 0 ω2i

⎤⎥⎥⎥⎦ , . . . (1–25)

Assume that the mass matrix is non-singular so M−1 exists. Then, the equations of motion(1-1) may be reformulated in the following state vector form of coupled 1st order differentialequations2

z(t) = Az(t) + F(t) , t > t0

z(t0) = z0

}(1–26)

z(t) =

[x(t)

x(t)

], z0 =

[x0

x0

], A =

[0 I

−M−1K −M−1C

], F(t) =

[0

M−1f(t)

](1–27)

z(t) denotes the state vector. The corresponding homogeneous differential system reads

z(t) = Az(t) (1–28)

The solution of (1-26) becomes2

z(t) = eAt

(e−At0 z0 +

∫ t

t0

e−AτF(τ)dτ

)(1–29)

The n × n matrix eAt is denoted the matrix exponential function. This forms a fundamentalmatrix to (1-28), i.e. the columns of eAt form 2n linearly independent solutions to (1-28).Actually, eAt is the fundamental matrix fulfilling the matrix initial value problem

d

dteAt = AeAt , t > 0

eA·0 = I

⎫⎬⎭ (1–30)

where I denotes a 2n × 2n unit matrix. Now,(eAt)−1

= e−At as shown in Box 1.1. Usingthis relation for t = t0, (1-29) is seen to fulfil the initial value of (1-26). Since conventionaldifferentiation rules also applies to matrix products, the fulfilment of the differential equation

2D.G. Zill and M.R. Cullen: Differential Equations with Boundary-Value Problems, 5th Ed. Brooks/Cole,2001.


in (1-26) follows from differentiation of the right hand side of (1-29), and application of (1-30),i.e.

d

dtz(t) =

d

dteAt

(e−At0 z0 +

∫ t

t0

e−AτF(τ)dτ

)+ eAt

(0 + e−AtF(t)

)=

AeAt

(e−At0 z0 +

∫ t

t0

e−AτF(τ)dτ

)+ I F(t) = Az(t) + F(t) (1–31)

The solution to (1-30) can be represented by the following infinite series of matrix products

eAt = I + tA +t2

2!A2 +

t3

3!A3 + · · · (1–32)

where A2 = AA, A3 = AAA etc. (1-32) is seen to fulfil the initial value eA·0 = I. Thefulfilment of the matrix differential equation (1-30) follows from termwise differentiation ofthe right-hand side of (1-32)

d

dteAt = 0 + A +

t

1!A2 +

t2

2!A3 + · · · = 0 + A

(I + tA +

t2

2!A2 + · · ·

)= AeAt (1–33)

The right-hand side of (1-32) converges for arbitrary values of t as the number of terms in-creases beyond limits on the right-hand side. Hence, eAt can in principle be calculated usingthis representation. However, for large values of t the convergence is very slow. In (1-29) thefundamental matrix eAt is needed for arbitrary positive and negative values of t. Hence, theuse of (1-32) as an algorithm for eAt in the solution (1-29) becomes increasingly computationalexpensive as the integration time interval is increased. In Box 1.1 an analytical solution for eAt

has been indicated, which to some extent circumvents this problem. However, this approachrequires that all eigenvectors and eigenvalues of A are available.

Damped eigenvibrations are obtained as linear independent solutions to the homogeneous dif-ferential equation (1-28). Analog to (1-8) solutions are searched for on the form

z(t) = Ψ(j)eλjt (1–34)

Insertion of (1-34) into (1-28) provides the following special eigenvalue problem of the dimen-sion 2n for the determination of the damped eigenmodes Ψ(j) and the damped eigenvalues λj(

A− λjI)Ψ(j) = 0 (1–35)

Since A is not symmetric, λj and Ψ(j) are generally complex. Upon complex conjugation of(1-35), it is seen that if (λ,Ψ) denotes an eigen-pair (solution) to (1-35), then (λ∗,Ψ∗) is also aneigen-pair, where * denotes complex conjugation. For lightly damped structures all eigenvaluesare complex. In this case only n eigen-pairs (λj,Ψ

(j)), j = 1, 2, . . . , n need to be considered,where no eigen-pair is a complex conjugate of another in the set.


Let the first n components of Ψ(j) be assembled in the n-dimensional sub-vector Φ(j). Then,from (1-27) and (1-34) it follows that

x(t) = Φ(j)eλjt ⇒ x(t) = λjΦ(j)eλjt (1–36)

Consequently, the damped eigenmodes must have the structure

Ψ(j) =

[Φ(j)

λjΦ(j)

](1–37)

Hence, merely the first n components of Ψ(j) need to be determined.

The eigenvalue problems (1-35) can be assembled into the following matrix formulation, cf.(1-14)-(1-16)

A[Ψ(1) Ψ(2) · · ·Ψ(2n)

]=[Ψ(1) Ψ(2) · · ·Ψ(2n)

]⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λ2n

⎤⎥⎥⎥⎦ ⇒

AΨ = ΨΛA (1–38)

where

ΛA =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λ2n

⎤⎥⎥⎥⎦ (1–39)

Ψ =[Ψ(1) Ψ(2) · · ·Ψ(2n)

](1–40)

The following representation of A in terms of the damped eigenmodes and eigenvalues followsfrom (1-38)

A = ΨΛAΨ−1 (1–41)

Assume that another 2n × 2n matrix B has the same eigenvectors Ψ(j) as A, whereas theeigenvalues as stored in the diagonal matrix ΛB are different. Then, similar to (1-41), B has therepresentation

B = ΨΛBΨ−1 (1–42)

The matrix product of A and B becomes

AB = ΨΛAΨ−1 ΨΛBΨ−1 = ΨΛAΛBΨ−1 (1–43)


Since ΛA and ΛB are diagonal matrices, matrix multiplication of these is commutative, i.e.ΛAΛB = ΛBΛA. Then, (1-43) may be written

AB = ΨΛBΛAΨ−1 = ΨΛBΨ−1 ΨΛAΨ−1 = BA (1–44)

Consequently, if two matrices have the same eigenvectors, matrix multiplication of two matri-ces is commutative. Identical eigenvectors of two matrices can also be shown to constitute thenecessary condition (the ”only if” requirement) for commutative matrix multiplication.

The so-called adjoint eigenvalue problem to (1-35) reads(AT − νiI

)Ψ(i)

a = 0 (1–45)

Hence, (νi,Ψ(i)a ) denotes the eigenvalue and eigenvector to the transposed matrix AT . In Box

1.2 it is shown that the eigenvalues of the basic eigenvalue problem and the adjoint eigenvalueproblem are identical, i.e. νj = λj . Further, it is shown that the eigenvectors Ψ(j) and Ψ

(j)a

fulfill the orthogonality properties

Ψ(i) Ta Ψ(j) =

{0 , i �= j

mi , i = j(1–46)

Ψ(i) Ta AΨ(j) =

{0 , i �= j

λimi , i = j(1–47)

where mi is denoted the complex modal mass. Without any restriction this may be chosenas mj = 1. Then, the orthogonality conditions (1-46) and (1-47) may be assembled into thefollowing matrix relation

ΨTa Ψ = I (1–48)

ΨTa AΨ = ΛA (1–49)

where

Ψa =[Ψ(1)

a Ψ(2)a · · ·Ψ(2n)

a

](1–50)

From (1-48) follows that

Ψa =(Ψ−1

)T(1–51)

Hence, the eigenvectors Ψ(i)a of the adjoint eigenvalue problem (the column vectors in Ψa)

normalized to unit modal mass are determined as the row vectors of Ψ−1. The eigenvectorsΨ(i) of the direct eigenvalue problem may be arbitrarily normalized. Of course, if (λi,Ψ

(i)a ) is

an eigen-solution to the adjoint eigenvalue problem, so is the complex conjugate (λ∗i ,Ψ

(i)∗a ).


Box 1.1: Matrix exponential function

Multiple application of (1-41) provides for k = 1, 2, . . .

A2 = AA = ΨΛAΨ−1 ΨΛAΨ−1 = ΨΛ2AΨ−1

A3 = AA2 = ΨΛAΨ−1 ΨΛ2AΨ−1 = ΨΛ3

AΨ−1

...

Aj+1 = AAj = ΨΛAΨ−1 ΨΛjAΨ−1 = ΨΛj+1

A Ψ−1

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

(1–52)

Λj+1A is a product of diagonal matrices, and then becomes a diagonal matrix itself. The

diagonal elements become λj+1k , where λk is the corresponding diagonal element in ΛA.

Consider the matrix exponential function, cf. (1-32)

eΛAt = I + tΛA +t2

2!Λ2

A +t3

3!Λ3

A + · · · (1–53)

Since all addends on the right-hand side of (1-53) are diagonal matrices, it follows thatalso eΛAt becomes diagonal with the diagonal elements

1 + tλk +t2

2!λ2

k +t3

3!λ3

k + · · · = eλk t (1–54)

where the Maclaurin series for the exponential function has been used in the last state-ment. Then, from (1-32), (1-52) and (1-53) follows

eAt = Ψ(I + tΛA +

t2

2!Λ2

A +t3

3!Λ3

A + · · ·)Ψ−1 = ΨeΛAtΨ−1 (1–55)

For arbitrary positive or negative t1 and t2 it then follows that

eAt1eAt2 = ΨeΛAt1Ψ−1 ΨeΛAt2Ψ−1 = ΨeΛAt1eΛAt2Ψ−1 = ΨeΛA(t1+t2)Ψ−1 =

eA(t1+t2) (1–56)

(1-56) represents the fundamental multiplication rule of matrix exponential functions.Especially for t1 = t and t2 = −t we have

eAte−At = eA·0 = I ⇒ e−At =(eAt)−1

(1–57)

Further,

A−n = A−1 · · ·A−1 = ΨΛ−nA Ψ−1 , n = 1, 2, . . . (1–58)

(1-58) is proved by insertion of (1-52) and (1-58) into the identity AnA−n = I. As seen,eAt and A−n have identical eigenvectors. Then, from (1-44) it follows that

A−neAt = eAtA−n , n = 1, 2, . . . (1–59)


Box 1.2: Proof of orthogonality properties of eigenvectors and adjoint eigenvectors

(1-35) is pre-multiplied with Ψ(i) Ta , and (1-45) is pre-multiplied with Ψ(j) T , lead-

ing to the identities

Ψ(i) Ta AΨ(j) = λjΨ

(i) Ta Ψ(j) (1–60)

Ψ(j) TATΨ(i)a = νiΨ

(j) TΨ(i)a ⇒

Ψ(i) Ta AΨ(j) = νiΨ

(i) Ta Ψ(j) (1–61)

The last statement follows from transposing the previous one. Withdrawal of (1-61) from(1-60) provides(

λj − νi

)Ψ(i) T

a Ψ(j) = 0 (1–62)

For i = j, (1-62) can only be fulfilled for νi = λi, since Ψ(i) Ta Ψ(i) �= 0.

Next, presume simple eigenvalues, so λi �= λj. Then, for i �= j, (1-62) can only befulfilled, if Ψ

(i) Ta Ψ(j) = 0, corresponding to (1-46).

Since the right-hand side of (1-60) is zero for i �= j, this must also hold true for the left-hand side, i.e. Ψ

(i) Ta AΨ(j) = 0 for i �= j. Then for i = j, (1-60) provides the result

Ψ(i) Ta AΨ(i) = λimi, which completes the proof of (1-47).

Example 1.1: Equations of motion of linear viscous damped 2DOF system

m1

m1

m2

m2

f1

f1

f2

f2

x1 x2

k1 k2 k3

c1 c2 c3

c1x1

k1x1

c2(x2 − x1)

k2(x2 − x1)

c3x2

k3x2

Fig. 1–1 Equation of motion of linear viscous damped 2DOF system.

The two-degree-of-freedom system shown on Fig. 1-1 consists of the masses m 1 and m2 connected with linearelastic springs with the spring constants k1, k2, k3, and linear viscous damper elements with the damper constants


c1, c2, c3. The displacement of the masses from the static equilibrium state are denoted as x 1(t) and x2(t). Thevelocities xi(t) and accelerations xi(t) are considered positive in the same direction as the displacements x i(t)and the external forces fi(t). The masses are cut free from the springs and dampers in the deformed state, andthe damper- and spring forces are applied as equivalent external forces. Next, Newton’s 2nd law of motion isformulated for each of the masses leading to

m1x1 = −k1x1 + k2(x2 − x1) − c1x1 + c2(x2 − x1) + f1(t)

m2x2 = −k3x2 − k2(x2 − x1) − c3x2 − c2(x2 − x1) + f2(t)

}(1–63)

(1-63) may be formulated as the following matrix differential equations

Mx(t) + Cx(t) + Kx(t) = f(t) , t > t0

x(t) =

[x1(t)x2(t)

], f(t) =

[f1(t)f2(t)

]

M =

[m1 00 m2

], C =

[c1 + c2 −c2

−c2 c2 + c3

], K =

[k1 + k2 −k2

−k2 k2 + k3

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(1–64)

For each of the masses an initial displacement xi(t0) = xi,0 from the static equilibrium state and an initial velocityxi(t0) = xi,0 are specified. These are assembled into the following initial value vectors

x0 = x(t0) =

[x1,0

x2,0

], x0 = x(t0) =

[x1,0

x2,0

](1–65)

The presented system will be further analyzed in various numerical examples throughout the book.

Example 1.2: Discretized equations of motion of a vibrating string

FF

x

j

∆l

l

1 2 n − 1 n

u1 u2 u(x, t) uj uj+1 un−2 un−1

Fig. 1–2 Discretization of vibrating string.

Fig. 1-2 shows a vibrating string with the pre-stress force F , and the mass per unit length µ. The string has beendivided into n identical elements, each of the length ∆l. Hence, the total length of the string is l = n∆l. Thedisplacement u(x, t) of the string at the position x and time t in the transverse direction is given by the waveequation with homogeneous boundary conditions 1

µ∂2u

∂t2− F

∂2u

∂x2= 0 , x ∈]0, l[

u(0, t) = u(l, t) = 0

⎫⎪⎬⎪⎭ (1–66)


where x is measured from the left support point. The spatial differential operator in (1-66) is discretized by meansof a central difference operator,2 i.e.

F∂2u(xi, t)

∂x2 F

∆l2(ui+1 − 2ui + ui−1

), i = 1, . . . , n − 1 (1–67)

where ui(t) = u(xi, t), xi = i∆l. Further, let ui(t) = ∂∂t2 u(xi, t). The boundary conditions imply that u0(t) =

un(t) = 0. Then, the discretized wave equation may be represented by the matrix differential equation

Mx(t) + Kx(t) = 0 (1–68)

x(t) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1(t)u2(t)u3(t)

...

un−2(t)un−1(t)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

, M = µ

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 · · · 0 00 1 0 · · · 0 00 0 1 · · · 0 0...

......

. . ....

...

0 0 0 · · · 1 00 0 0 · · · 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

, K =F

∆l2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 −1 0 · · · 0 0−1 2 −1 · · · 0 0

0 −1 2 · · · 0 0...

......

. . ....

...

0 0 0 · · · 2 −10 0 0 · · · −1 2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1–69)

Alternatively, the wave equation may be discretized by means of a finite element approach. Assuming linear inter-polation between the nodal values stored in the vector x(t), and using the same interpolation for the displacementfield and the variational field (Galerkin variation), the following mass- and stiffness matrices are obtained

M =µ∆l

6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 1 0 · · · 0 01 4 1 · · · 0 00 1 4 · · · 0 0...

......

. . ....

...

0 0 0 · · · 4 10 0 0 · · · 1 4

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

, K =F

∆l

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 −1 0 · · · 0 0−1 2 −1 · · · 0 0

0 −1 2 · · · 0 0...

......

. . ....

...

0 0 0 · · · 2 −10 0 0 · · · −1 2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1–70)

(1-70) represents the so-called consistent mass matrix, for which the same interpolation algorithm is used for dis-cretizing the kinetic and the potential energy.1 By contrast the diagonal mass matrix in (1-69) is referred to as alumped mass matrix. As seen the central difference operator and Galerkin variation with piecewise linear inter-polation leads to the same stiffness matrix. The presented system will be further analyzed in various numericalexamples in what follows.

The calculated eigenvalues based on the system matrices (1-69) and (1-70) are shown in Fig. 1-3 as a function ofthe number of elements n. The solutions based on the lumped mass matrix (1-69) and the consistent mass (1-70)are shown with dotted and dashed signature, respectively. The numerical solutions have been given relative to theanalytical solutions

ωj,a = jπ

√F

µl2, j = 1, . . . , 4 (1–71)

As seen, the consistent mass matrix provides upper-bounds in accordance with the Rayleigh-Ritz principle de-scribed in Section 4.2. By contrast the lumped mass matrix provides lower bounds, when used in combinationwith the consistent stiffness matrix. There is no formal proof of this property, which merely is an empirical obser-vation fulfilled in many dynamical problems. The indicated observation immediately suggest that an improvement


of the numerical solutions may be obtained by using a linear combination of the consistent and the lumped massmatrix. Typically, the mean value is used leading to the mass matrix

M =12

µ∆l

6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 1 0 · · · 0 01 4 1 · · · 0 00 1 4 · · · 0 0...

......

. . ....

...

0 0 0 · · · 4 10 0 0 · · · 1 4

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦+

12µ∆l

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 · · · 0 00 1 0 · · · 0 00 0 1 · · · 0 0...

......

. . ....

...

0 0 0 · · · 1 00 0 0 · · · 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

=µ∆l

12

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

10 1 0 · · · 0 01 10 1 · · · 0 00 1 10 · · · 0 0...

......

. . ....

...

0 0 0 · · · 10 10 0 0 · · · 1 10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1–72)

(1-72) is solved with the consistent stiffness matrix (1-70). The results are showed with a dashed-dotted signatureon Fig. 1-3. As expected the results show a significant improvement. A theoretical argument for using the meanvalue of the consistent and lumped mass matrices for the combined mass matrix has been given by Krenk. 3

5 10 15 20 25 300.98

0.99

1

1.01

1.02

5 10 15 20 25 30

0.94

0.96

0.98

1

1.02

1.04

1.06

5 10 15 20 25 300.85

0.9

0.95

1

1.05

1.1

5 10 15 20 25 30

0.8

0.9

1

1.1

1.2

nn

nn

ω1/ω

1,a

ω2/ω

2,a

ω3/ω

3,a

ω4/ω

4,a

Fig. 1–3 Undamped eigenvibrations of string. —: Analytical solution. - - - : Consistent mass matrix. .... :Lumped mass matrix. -.-. : Combined mass matrix.

3S. Krenk: Dispersion-corrected explicit integration of the wave equation. Computer Methods in AppliedMechanics and Engineering, 191, pp. 975-987, 2001.


Example 1.3: Verification of eigensolutions

Given the following mass- and stiffness matrices

M =

[54 00 1

5

], K =

[5 −2

−2 2

](1–73)

Verify that the eigensolutions with modal masses normalized to 1 are given by

Λ =

[ω2

1 00 ω2

2

]=

[2 00 12

], Φ =

[Φ(1) Φ(2)

]=

[45

25

1 −2

](1–74)

Based on the proposed eigensolutions the following calculations are performed, cf. (1-14)

KΦ =

[5 −2

−2 2

][45

25

1 −2

]=

[2 625 − 24

5

]

MΦΛ =

[54 00 1

5

][45

25

1 −2

] [2 00 12

]=

[2 625 − 24

5

]

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭

(1–75)

This proofs the validity of the proposed eigensolutions. The orthonormality follows from the following calcula-tions, cf. (1-19) and (1-21)

ΦT MΦ =

[45

25

1 −2

]T [54 00 1

5

][45

25

1 −2

]=

[1 00 1

]

ΦT KΦ =

[45

25

1 −2

]T [5 −2

−2 2

] [45

25

1 −2

]=

[2 00 12

]

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭

(1–76)

Example 1.4: M- and K-orthogonal vectors


M =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ (1–77)

Additionally, the following vectors are considered

v1 =

⎡⎢⎣ 1

√2

2

0

⎤⎥⎦ , v2 =

⎡⎢⎣ 1

−√

22

0

⎤⎥⎦ (1–78)

From (1-78) the following matrix is formed

V = [v1 v2] =

⎡⎢⎣ 1 1

√2

2 −√

22

0 0

⎤⎥⎦ (1–79)


We may then perform the following calculations, cf. (1-19) and (1-21)

VT MV =

⎡⎢⎣ 1 1

√2

2 −√

22

0 0

⎤⎥⎦

T ⎡⎢⎣

12 0 0

0 1 0

0 0 12

⎤⎥⎦⎡⎢⎣ 1 1

√2

2 −√

22

0 0

⎤⎥⎦ =

[1 00 1

]

VT KV =

⎡⎢⎣ 1 1

√2

2 −√

22

0 0

⎤⎥⎦

T ⎡⎢⎣ 2 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣ 1 1

√2

2 −√

22

0 0

⎤⎥⎦ =

[2.5858 0

0 5.4142

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(1–80)

(1-80) shows that the vectors v1 and v2 are mutual orthogonal with weights M and K, and that both have beennormalized to unit modal mass. As will be shown in Example 1.5 neither v 1 nor v2 are eigenmodes, and the eigen-values are different from 2.5858 and 5.4142. However, if three linear independent vectors are mutual orthogonalweighted with the three-dimensional matrices M and K, they will be eigenmodes to the system.

Example 1.5: Analytical calculation of eigensolutions

The mass- and stiffness matrices defined in Example 1.4 are considered again. Now, an analytical solution of theeigenmodes and eigenvalues is wanted.

The generalized eigenvalue problem (1-9) becomes

⎡⎢⎣2 − 1

2λj −1 0

−1 4 − λj −1

0 −1 2 − 12λj

⎤⎥⎦⎡⎢⎣Φ(j)

1

Φ(j)2

Φ(j)3

⎤⎥⎦ =

⎡⎢⎣000

⎤⎥⎦ (1–81)

The characteristic equation (1-10) becomes

P (λ) = det

⎛⎜⎝⎡⎢⎣2 − 1

2λj −1 0−1 4 − λj −10 −1 2 − 1

2λj

⎤⎥⎦⎞⎟⎠ =

(2 − 1

2λj

)((4 − λj

)(2 − 1

2λj

)− 1)

+ 1 ·(−(2 − λj

12

))=(2 − 1

2λj

)(6 − 4λj +

12λ2

j

)= 0 ⇒

λj =

⎧⎪⎨⎪⎩

2 , j = 14 , j = 26 , j = 3

(1–82)

Initially, the eigenmodes are normalized by setting an arbitrary component to 1. Here we shall choose Φ (j)3 = 1.

The remaining components Φ(j)1 and Φ(j)

2 are then determined from any two of the three equations (1-81). Thefirst and the second equations are chosen, corresponding to

[2 − 1

2λj −1

−1 4 − λj

] [Φ(j)

1

Φ(j)2

]=

[01

]⇒

⎡⎢⎣Φ(j)

1

Φ(j)2

Φ(j)3

⎤⎥⎦ =

⎡⎢⎢⎣

214−8λj+λ2

j

4−λj

14−8λj+λ2j

1

⎤⎥⎥⎦ (1–83)

The modal matrix with eigenmodes normalized as indicated in (1-83) is denoted as Φ. This becomes


Φ =

⎡⎢⎣1 −1 1

1 0 −1

1 1 1

⎤⎥⎦ (1–84)

The modal masses become, cf. (1-19)

m = ΦT MΦ =

⎡⎢⎣2 0 0

0 1 0

0 0 2

⎤⎥⎦ (1–85)

Φ(1) denotes the 1st eigenmode normalized to unit modal mass. This is related to Φ(1) in the following way

Φ(1) =1√M1

Φ(1) =1√2

⎡⎢⎣1

1

1

⎤⎥⎦ (1–86)

The other modes are treated in the same manner, which results in the following eigensolutions

Λ =

⎡⎢⎣ω2

1 0 0

0 ω22 0

0 0 ω23

⎤⎥⎦ =

⎡⎢⎣2 0 0

0 4 0

0 0 6

⎤⎥⎦ , Φ =

[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣

√2

2 −1√

22√

22 0 −

√2

2√2

2 1√

22

⎤⎥⎦ (1–87)

Example 1.6: Undamped and damped eigenvibrations of 2DOF system

1 kg 2 kg

x1 x2

100 N/m 200 N/m 300 N/m

3 kg/s 2 kg/s 1 kg/s

Fig. 1–4 Eigenvibrations of 2DOF system.

The system in Example 1.1 is considered again with the structural parameters defined in Fig. 1-3. The mass-damping and stiffness matrices become, cf. (1-64)

M =

[1 00 2

]kg , C =

[5 −2

−2 3

]kgs

, K =

[300 −200

−200 500

]Nm

(1–88)

The eigensolutions with modal masses normalized to 1 become

Λ =

[ω2

1 00 ω2

2

]=

[131.39 0

0 418.61

]s−2 , Φ =

[Φ(1) Φ(2)

]=

[0.64262 0.766180.54177 −0.45440

](1–89)


The matrix A defined by (1-27) becomes

A =

[0 I

−M−1K −M−1C

]=

⎡⎢⎢⎢⎢⎣

0 0 1 00 0 0 1

−300 200 −5.0 2.0100 −250 1.0 −1.5

⎤⎥⎥⎥⎥⎦ (1–90)

The eigenvalues and eigenfunctions become

ΛA =

⎡⎢⎢⎢⎣λ1 0 0 00 λ2 0 00 0 λ3 00 0 0 λ4

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣−0.7763 + 11.480i 0 0 0

0 −2.4737 + 20.231i 0 00 0 −0.7763− 11.480i 00 0 0 −2.4737− 20.231i

⎤⎥⎥⎥⎦

(1–91)

Ψ =[Ψ(1) Ψ(2) Ψ(3) Ψ(4)

]=

⎡⎣ Φ(1) Φ(2) Φ(1)∗ Φ(2)∗

λ1Φ(1) λ2Φ(2) λ∗1Φ

(1)∗ λ∗2Φ

(2)∗

⎤⎦ =

⎡⎢⎢⎢⎢⎣

1.1693− 0.1414i −1.6846− 0.3657i 1.1693 + 0.1414i −1.6846 + 0.3657i

1 1 1 1

0.7153 + 13.534i 11.565− 33.177i 0.7153− 13.534i 11.565− 33.177i

−0.7763 + 11.480i −2.4737 + 20.231i −0.7763− 11.480i −2.4737− 20.231i

⎤⎥⎥⎥⎥⎦

(1–92)

As seen from (1-92) the second component of the sub-vectors Φ (1) and Φ(2) has been normalized to one. Hence,the entire modal matrix with 16 components is defined by merely 4 entities, namely the the first component of thesub-vectors Φ(1) and Φ(2) and the eigenvalues λ1 and λ2.

The eigenvectors of the adjoint eigenvalue problem follows from (1-51) and (1-92)

Ψa =(Ψ−1

)T =[Ψ(1)

a Ψ(2)a Ψ(3)

a Ψ(4)a

]=[Ψ(1)

a Ψ(2)a Ψ(1)∗

a Ψ(2)∗a

]=

⎡⎢⎢⎢⎣

0.1723− 0.0430i −0.1723 + 0.0388i 0.1723 + 0.0430i −0.1723− 0.0388i

0.3007 + 0.0411i 0.1993− 0.0592i 0.3007− 0.0411i 0.1993 + 0.0592i

−0.0004− 0.0154i 0.0004 + 0.0087i −0.0004 + 0.0154i 0.0004− 0.0087i

0.0025− 0.0260i −0.0025− 0.0098i 0.0025 + 0.0260i −0.0025 + 0.0098i

⎤⎥⎥⎥⎦

(1–93)

As seen Ψ(3)a and Ψ(3)

a become the complex conjugates of Ψ(1)a and Ψ(2)

a , cf. remarks subsequent to (1-51).

1.2 Solution of Initial Value Problem by Modal Decomposition Techniques 25

1.2 Solution of Initial Value Problem by Modal Decom-position Techniques

Assume that undamped eigenmodes Φ(i) in addition to the orthogonality properties (1-17) and(1-18) also are orthogonal weighted with the damping matrix, i.e.

Φ(i) TCΦ(j) =

{0 , i �= j

2ζiωiMi , i = j(1–94)

ζi denotes the modal damping ratio. In practice (1-94) is fulfilled, if the structure is lightlydamped and the eigenfrequencies are well separated.1 The orthogonality properties may beassembled into the following matrix relation similar to (1-19) and (1-21)

ΦTCΦ = c (1–95)

where

c =

⎡⎢⎢⎢⎣

2ω1ζ1M1 0 · · · 0

0 2ω2ζ2M2 · · · 0...

.... . .

...0 0 · · · 2ωnζnMn

⎤⎥⎥⎥⎦ (1–96)

The undamped eigenmodes are linear independent and may be used as a basis in the n-dimensionalvector space. Hence, the displacement vector x(t) may be written as

x(t) =

n∑j=1

Φ(j)qj(t) = Φq(t) , q(t) =

⎡⎢⎢⎢⎣

q1(t)

q2(t)...

qn(t)

⎤⎥⎥⎥⎦ (1–97)

where q1(t), . . . , qn(t) represent the undamped modal coordinates, i.e. the coordinates in thevector basis formed by the undamped eigenmodes Φ(1), . . . , Φ(n). Insertion of (1-95) into (1-1), followed by a pre-multiplication with ΦT and use of (1-19), (1-21), (1-94), provides thefollowing matrix differential equation for the modal coordinates

mq(t) + cq(t) + kq(t) = F(t) , t > t0

q(t0) = Φ−1x0 , q(t0) = Φ−1x0

}(1–98)

where


F(t) = ΦT f(t) =

⎡⎢⎢⎢⎣

F1(t)

F2(t)...

Fn(t)

⎤⎥⎥⎥⎦ (1–99)

F1(t), . . . , Fn(t) are denoted the modal load. Since m, c and k are diagonal matrices thecomponent differential equations related to (1-98) decouple completely. This is caused by theorthogonality condition (1-94) for which reason this relation is referred to as the decouplingcondition. The differential equation for the kth modal coordinate reads

Mk

(qk(t) + 2ζkωkqk(t) + ω2

kqk(t))

= Fk(t) , k = 1, . . . , n (1–100)

Hence, the decoupling condition reduces the integration of a linear n degrees-of-freedom sys-tem to the integration of n single-degree-of-freedom oscillators.

Typically, the dynamic response is carried by lowest modes in the expansion (1-97). Assumethat the modal response above the first n1 n may be disregarded. Then (1-97) reduces to

x(t) n1∑j=1

Φ(j)qj(t) = Φ1q1(t) (1–101)

where Φ1 is a reduced modal matrix of dimension n × n1, and q1(t) is a sub-vector of modalcoordinates defined as

Φ1 =[Φ(1) Φ(2) · · ·Φ(n1)

], q1(t) =

⎡⎢⎢⎢⎣

q1(t)

q2(t)...

qn1(t)

⎤⎥⎥⎥⎦ (1–102)

(1-101) completely ignores the influence of the higher modes. Although the dynamic responseof these modes are ignorable, they may still influence the low-frequency components via aquasi-static response component. A consistent correction taken this effect into considerationreads1

x(t) n1∑j=1

Φ(j)qj(t) +

(K−1 −

n1∑j=1

1

ω2j Mj

Φ(j)Φ(j) T

)f(t) (1–103)

(1-103) may be represented in terms of the following equivalent matrix formulation

x(t) Φ1q1(t) +(K−1 − Φ1 k−1

1 ΦT1

)f(t) (1–104)

1.2 Solution of Initial Value Problem by Modal Decomposition Techniques 27

where

k1 =

⎡⎢⎢⎢⎣ω2

1M1 0 · · · 0

0 ω22M2 · · · 0

......

. . ....

0 0 · · · ω2n1

Mn1

⎤⎥⎥⎥⎦ (1–105)

Both (1-101) and (1-103) requires knowledge of the first n1 eigen-pairs (ω2j ,Φ

(j)). The corre-sponding modal coordinates are determined from the first n1 equations in (1-100).Correspondingly, the 2n eigenvectors Ψ(j), j = 1, . . . , 2n of the matrix A form a vector basisin the 2n-dimensional vector space. Then, the state vector z(t) admits the representation

z(t) =

2n∑j=1

Ψ(j)qj(t) = Ψq(t) (1–106)

where

q(t) =

⎡⎢⎢⎢⎣

q1(t)

q2(t)...

q2n(t)

⎤⎥⎥⎥⎦ (1–107)

q1(t), . . . , q2n(t) represent the damped modal coordinates, i.e. the coordinates in the vectorbasis made up of the damped eigenmodes Ψ(1), . . . Ψ(2n). Insertion of (1-106) into (1-26),followed by pre-multiplication of ΨT

a and use of (1-48), (1-49), the following matrix differentialdifferential equations for the damped modal coordinates is obtained

q(t) = ΛAq(t) + G(t) , t > t0

q(t0) = Ψ−1z0 = ΨTa z0

}(1–108)

where

G(t) = ΨTa F(t) =

⎡⎢⎢⎢⎣

G1(t)

G2(t)...

G2n(t)

⎤⎥⎥⎥⎦ (1–109)

In the initial value statement of (1-106) the relation (1-51) between the adjoint and direct modalmatrices has been used. Gj(t) = Ψ

(j)Ta F(t) denotes the jth damped modal load.


(1-108) indicates 2n decoupled complex 1st order differential equations. The differential equa-tion for the jth modal coordinate reads

qj(t) = λjqj(t) + Gj(t) , j = 1, . . . , 2n (1–110)

Since,(λj+n,Ψ

(j+n)a

)=(λ∗

j ,Ψ(j)∗a

)for n = 1, . . . , n, it follows that Gj+n(t) = G∗

j (t), andin turn that qj+n(t) = q∗j (t). Hence, merely the first n differential equations (1-110) need to beintegrated. Then, (1-104) may be written

z(t) = 2Re

(n∑

j=1

Ψ(j)qj(t)

)(1–111)

As is the case for expansion in undamped modal coordinates the response is primarily carriedby the lowest n1 modes leading to the following reduced form of (1-111)

z(t) 2Re

(n1∑j=1

Ψ1q1(t)

)(1–112)

where

Ψ1 =[Ψ(1) Ψ(2) · · ·Ψ(n1)

], q1(t) =

⎡⎢⎢⎢⎣

q1(t)

q2(t)...

qn1(t)

⎤⎥⎥⎥⎦ (1–113)

(1-101), (1-103) and (1-112) describes the dynamic system with less coordinates than the orig-inal formulation (1-1). For this reason such formulations are referred to as system reductionschemes. A system reduction scheme with due consideration to the quasi-static response mayalso be formulated as a correction to (1-112).1

1.3 Conclusions

On condition that the convolution integral is evaluated numerically an analytical solution tothe initial value problem (1-1) is provided by the result (1-3). Since this solution relies on theFourier transform of the frequency matrix (1-6) for the impulse response matrix, the approachbecomes computational prohibitive for a large number of degrees of freedom. Alternatively,if the initial value problem is reformulated in the state vector form (1-26) the analytical solu-tion (1-29) is obtained. This solution relies on the fundamental matrix in terms of the matrixexponential function for the corresponding homogeneous differential system (1-28). The ma-trix exponential function may be calculated analytically as indicated by (1-55), but the solutionrequires all eigen-solutions to the system matrix A. Again, the calculation of these becomes

1.3 Conclusions 29

prohibited for large systems. Hence, both analytical or semi-analytical solution approaches areout of the question for large degree-of-freedom systems.

The state vector formulation (1-26) directly admits the application of vectorial generaliza-tions of standard ordinary differential equation solvers such as the Euler method, the extendedEuler method, the various Runge-Kutta algorithms or the Adams-Bashforth/Adams-Moultonalgorithm.2 As is the case for all conditional stable algorithms the numerical stability of theseschemes is determined by the length of the time step in proportion to the eigenperiod of highestmode of the system. Hence, in order to insure stability for large scale systems excessive smalltime steps becomes necessary, which means that the high accuracy of some of these algorithmscannot be utilized. Consequently, there is a need for numerical matrix differential solvers forwhich the length of the time step is determined from accuracy rather than stability. These al-gorithms predict stable although inaccurate responses for the highest modes. Instead, the timestep is adjusted to predict accurate results for the lowest modes, which carry the global responseof the structure. The devise of such algorithms will be the subject of Chapter 2.

System reduction schemes such as (1-101), (1-103) and (1-112) require a limited number oflow-frequency eigen-pairs to be know. Since, the high frequency components have been filteredout the numerical integration of the modal coordinate differential equations (1-100) and (1-110)may be performed by standard ordinary differential solvers or by modification of the methodsdevised in Chapter 2. Hence, the primary obstacle in using these methods is the determinationof the low frequency eigen-pairs. This problem will be the subject of the Chapters 3-7 of thebook. Moreover, only solutions to the GEVP (1-9) will be considered, i.e. the involved systemmatrices are assumed to be symmetric and non-negative definite.


1.4 Exercises

1.1 Given the following mass- and stiffness matrices

M =

⎡⎢⎣1 0 0

0 2 0

0 0 12

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0

−1 2 0

0 0 3

⎤⎥⎦

(a.) Calculate the eigenvalues and eigenmodes normalized to unit modal mass.(b.) Determine two vectors that are M-orthonormal without being eigenmodes.

1.2 The eigensolutions with eigenmodes normalized to unit modal mass of a 2-dimensionalgeneralized eigenvalue probem are given as

Λ =

[λ1 0

0 λ2

]=

[1 0

0 4

], Φ =

[Φ(1) Φ(2)

]=

[√2

2

√2

2√2

2−

√2

2

]

(a.) Calculate M and K.

1.3 Write a MATLAB program, which solves the undamped generalized eigenvalue problemfor the vibrating string problem considered in Example 1.2 for both the finite difference andthe finite element discretized equations of motion. The circular eigenfrequencies shouldbe presented in ascending order of magnitude, and the related eigenmodes should be nor-malized to unit modal mass.

(a.) Use the program to evaluate the 4 lowest circular eigenfrequencies of the string as afunction of the number of elements n for both discretization methods, and comparethe numerical results with the analytical solution (1-71).

(b.) Based on the obtained results suggest a mass matrix, which will do better.

1.4 Write a MATLAB program, which solves the undamped and damped generalized eigen-value problems considered in Example 1.6.

CHAPTER 2NUMERICAL INTEGRATION OF

EQUATIONS OF MOTION

This chapter deals with the numerical time integration in the finite interval [t0, t0 + T ] of theinitial value problem (1-1). The solution is searched for . The idea of the numerical integrationscheme is to determine the solution of (1-1) approximately at the discrete instants of time tj =t0 + j∆t, j = 1, 2, . . . , n, where ∆t = T/n. To facilitate notations the following symbols areintroduced

xj = x(tj) , xj = x(tj) , xj = x(tj) , fj = f(tj) , j = 0, 1, . . . , n (2–1)

Single step algorithms in numerical time integration in structural dynamics determines the dis-placement vector xj+1, the velocity vector xj+1 and the acceleration vector xj+1 at the newtime tj+1, on condition of knowledge of xj , xj , xj at the previous instant of time, as well asthe load vectors fj and fj+1 at the ends of the considered sub-interval [tj , tj+1]. In multi stepalgorithms the solution at the time tj+1 also depends on one or more solutions prior to the timetj . Additionally, distinction will be made between single value algorithms, which solves solelyfor the displacement vector xj , and multi value algorithms, where the solution is obtained fora state vector encompassing the displacement vector xj, the velocity vector xj , and in somecases even the acceleration vector xj . Classical algorithms in numerical analysis such as thevector generalization of the Runge-Kutta methods1 may be used for the solution of (1-1). How-ever, given that large scale structural models contain very high frequency components, theseschemes become numerical unstable unless extremely small time steps are used. For this reasonthe devise of useful algorithms in structural dynamics is governed by different objectives thanin numerical analysis, as will be further explained below.

Newmark algorithms2 treated in Section 2.1 are probably the most widely used algorithms instructural dynamics for solving (1-1). The derived single step multi value formulation of themethods serves as a generic example for specification of accuracy, stability, and numericaldamping.

1D.G. Zill and M.R. Cullen: Differential Equations with Boundary-Value Problems, 5th Ed. Brooks/Cole,2001.

2N.M. Newmark: A Method of Computation for Structural Dynamics. J.Eng.Mech., ASCE, 85(EM3), 1959,67-94.

— 31 —

32 Chapter 2 – NUMERICAL INTEGRATION OF EQUATIONS OF MOTION

High frequencies and mode shapes of the spatially discretized equations (1-1) does not rep-resent the behavior of the underlying physical problem very well. The corresponding modalcomponents merely behave as numerical noise at the top of the response carried by the lowerfrequency modes. For this reason it is desirable to filter these components out of the response.In numerical time integrators in structural dynamics this is achieved by the introduction of nu-merical (artificial) damping, which are affecting merely the high frequency modes. However,it turns out that numerical damping cannot be introduced in the Newmark algorithms withoutcompromising the accuracy of the response of the lower modes. Several suggestions to rem-edy this problem have been suggested. Here, we shall consider the so-called generalized alphaalgorithm suggested by Chung and Hulbert,3 which seems to be the most favorable single stepsingle valued algorithm for this purpose. The outline of the text relies primarily on the mono-graphs of Hughes 4, 5 and Krenk.6

2.1 Newmark Algorithm

The Newmark family consists of the following equations

Mxj+1 + Cxj+1 + Kxj+1 = fj+1 , j = 1 , . . . , n (2–2)

xj+1 = xj + xj ∆t +

((1

2− β)xj + β xj+1

)∆t2 (2–3)

xj+1 = xj +((

1 − γ)xj + γ xj+1

)∆t (2–4)

(2-2) indicates the differential equation at the time tj+1, which is required to be fulfilled for thenew solution for xj+1, xj+1, xj+1. (2-3) and (2-4) are approximate Taylor expansions, whichhave been derived in Box 2.2. The parameters β and γ determines the numerical stability andaccuracy of the algorithms. The Newmark family contains several wellknown numerical algo-rithms as special cases. Examples are the central difference algorithm treated in Example 2.2,which corresponds to (β, γ) = (0, 1

2), and the Crank-Nicolson algorithm treated in Example

2.3, which corresponds to (β, γ) = ( 14, 1

2).

There are several implementations of the methods. The most useful is the following single stepsingle value implementation. At first, define the following predictors

3J. Chung and G.M. Hulbert: A time Integration Algorithm for Structural Dynamics with Improved NumericalDissipation: The Generalized α Method. Journal of Applied Mechanics, 60, 1993, 371-375.

4T.J.R. Hughes: The Finite Element Method. Linear Static and Dynamic Finite Element Analysis. Printice-Hall,Inc., 1987.

5T.J.R. Hughes: Analysis of Transient Algorithms with Particular Reference to Stability Behavior. Chapter2 in Computational Methods for Transient Analysis. Vol. 1 in Computational Methods in Mechanics, Eds. T.Belytschko and T.J.R. Hughes, North-Holland, 1983.

6S. Krenk: Dynamic Analysis of Structures. Numerical Time Integration. Lecture Notes, Department of Me-chanical Engineering, Technical University of Denmark, 2005.

2.1 Newmark Algorithm 33

xj+1 = xj + xj ∆t +(1

2− β)∆t2 xj (2–5)

˙xj+1 = xj +(1 − γ

)∆t xj (2–6)

(2-5) and (2-6) specify predictions (preliminary solutions) for xj+1 and xj+1 based on the infor-mation available at the time tj . The idea of the algorithm is to insert (2-3) and (2-4) into (2-2).Given that the solution is required to fulfill the equations of motion at the time t j+1, and using(2-5) and (2-6), the following equations are obtained for the new acceleration vector in terms ofknown solution quantities from the previous time and the load vector fj+1

(M + γ∆tC + β∆t2K

)xj+1 = fj+1 −C ˙xj+1 − Kxj+1 (2–7)

Next, based on the solution for xj+1 obtained from (2 − 7) corrected (new) solutions for xj+1

and xj+1 may be obtained from (2-3) and (2-4). These may be written as

xj+1 = xj+1 + β∆t2 xj+1 (2–8)

xj+1 = ˙xj+1 + γ∆t xj+1 (2–9)

To start the algorithm the acceleration x0 at the time t0 is needed. This is obtained from theequation of motion

Mx0 = f0 − Cx0 − Kx0 (2–10)

The algorithm has been summarized in Box 2.1. In stability and accuracy analysis a single stepmulti value formulation for the state vector made up of the displacement and velocity vectorsis preferred. In order to derived this, eqs. (2-3) and (2-4) are multiplied with M. Next, theaccelerations are eliminated by means of the differential equations at the times tj and tj+1,leading to

Mxj+1 = Mxj + M∆txj +((1

2− β)(

fj −Cxj − Kxj

)+ β

(fj+1 − Cxj+1 −Kxj+1

))∆t2

Mxj+1 = Mxj +((1 − γ

)(fj −Cxj − Kxj

)+ γ(fj+1 − Cxj+1 −Kxj+1

))∆t

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

[M + β∆t2K β∆t2C

γ∆tK M + γ∆tC

][xj+1

xj+1

]=[

M − (12− β)∆t2K ∆tM − (1

2− β)∆t2C

−(1 − γ)∆tK M− (1 − γ)∆tC

][xj

xj

]+

[(12− β)∆t2 β∆t2

(1 − γ)∆t γ∆t

][fj

fj+1

]⇒


zj+1 = Dzj + Ej (2–11)

where

zj =

[xj

xj

]

D =


γ∆tK M + γ∆tC

]−1 [M− (1

2− β)∆t2K ∆tM − (1

2− β)∆t2C

−(1 − γ)∆tK M − (1 − γ)∆tC

]

Ej =


γ∆tK M + γ∆tC

]−1 [(12− β)∆t2 β∆t2

(1 − γ)∆t γ∆t

][fj

fj+1

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2–12)

D denotes the so-called amplification matrix. The bar indicates that this is an approximation tothe exact amplification matrix, which has has been derived in Example 2.1.

Box 2.1: Newmark algorithm

Given the initial displacement vector x0 and the initial velocity vector x0 at the time t0.Calculate the initial acceleration vector x0 from

Mx0 = f0 − Cx0 − Kx0

Repeat the following items for j = 0, 1, . . . , n

1. Calculate predictors for the new displacement and velocity vectors

xj+1 = xj + xj ∆t +(1

2− β)∆t2 xj

˙xj+1 = xj +(1 − γ

)∆t xj

2. Calculate new acceleration vector from(M + γ∆tC + β∆t2K

)xj+1 = fj+1 −C ˙xj+1 − Kxj+1

3. Calculate new displacement and velocity vectors

xj+1 = xj+1 + β∆t2 xj+1

xj+1 = ˙xj+1 + γ∆t xj+1


Box 2.2: Derivation of (2-3) and (2-4)

Based on conventional integration theory the following identities may be formulated

x(tj+1) = x(tj) +

∫ tj+1

tj

x(τ)dτ

x(tj+1) = x(tj) +

∫ tj+1

tj

x(τ)dτ

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (2–13)

Integration by parts of the first relation provides

x(tj+1) = x(tj) −[(

tj+1 − τ)x(τ)

]tj+1

tj+

∫ tj+1

tj

(tj+1 − τ

)x(τ)dτ ⇒

xj+1 = xj + ∆t xtj +

∫ tj+1

tj

(tj+1 − τ

)x(τ)dτ (2–14)

The indicated derivation is due to Krenk.6 (2-14) may interpreted as a truncated Taylor ex-pansion, where the integrals represent the remainder. Correspondingly, the 2nd equationin (2-13) is written as

xj+1 = xj +

∫ tj+1

tj

x(τ)dτ (2–15)

Next, the integrals in (2-14) and (2-15) are represented by the following linear combina-tions of the value of the acceleration vector at the end of the integration interval

∫ tj+1

tj

(tj+1 − τ

)x(τ)dτ

(1

2− β

)∆t2 xj + β∆t2 xj+1∫ tj+1

tj

x(τ)dτ (1 − γ)∆t xj + γ∆t xj+1

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (2–16)

It is seen that the result in (2-16) becomes correct in case of constant acceleration, wherex(τ) ≡ xj = xj+1. In any case the values of β and γ reflect the actual variation of theacceleration during the interval. If x(τ) is assumed to be constant and equal to the meanof the end-point values, one obtains (β, γ) = ( 1

4, 1

2), whereas a linear variation between

the end-point values provides (β, γ) = ( 16, 1

2).

The modal expansion (1-97) defines a one-to-one transformation from the physical to the modalcoordinates. Hence, the time integration may equally well be performed on the differential equa-tions for the modal coordinate equations. It follows that the synthesized motion (1-97) becomes


numerical unstable, if the integration of just one of the modal coordinates render into instability.Similarly, the accuracy of the synthesized motion is determined by the accuracy of those modalcoordinates, which are retained in the truncated modal expansion (1-101). On condition of themodal decoupling condition (1-94) the time integration of the modal coordinates is reduced tothe integration of n decoupled single-degrees-of-freedom systems. Since the stability and accu-racy analysis of a SDOF system can be performed analytically, the important role of the modaldecomposition assumption in the stability and accuracy analysis of numerical time integratorsbecomes clear. In this respect let q(t) denote an arbitrary of the n modal coordinates, and ζ ,ω0 and F (t) the corresponding modal damping ratio, undamped circular eigenfrequency andmodal load. On condition that the eigenmodes have been normalized to unit modal mass, thedifferential equation of the said modal coordinate reads

q(t) + 2ζω0q(t) + ω20q(t) = F (t) (2–17)

The corresponding Newmark integration of (2-17) is given by (2-15), using M = [1], C =[2ζω0], K = [ω2

0] and f(t) = [F (t)] in (2-16), resulting in the system matrices

zj =

[qj

qj

]

D =

[1 + β∆t2ω2

0 2ζβω0∆t2

γω20∆t 1 + 2ζγω0∆t

]−1 [1 − (1

2− β)ω2

0∆t2 ∆t − (12− β)2ζω0∆t2

−(1 − γ)ω20∆t 1 − (1 − γ)2ζω0∆t

]

=

[D11 D12

D21 D22

]

Ej =

[1 + β∆t2ω2

0 2ζβω0∆t2

γω20∆t 1 + 2ζγω0∆t

]−1 [(12− β)∆t2 β∆t2

(1 − γ)∆t γ∆t

][Fj

Fj+1

]

=

[E11 E12

E21 E22

][Fj

Fj+1

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2–18)

where

D11 =1 + 2γζκ + (β − 1

2)κ2 + (2β − γ)ζκ3

1 + 2γζκ + βκ2

D12 =1 + (2γ − 1)ζκ + 2(2β − γ)ζ2κ2

1 + 2γζκ + βκ2∆t

D21 = −1 + 12(2β − γ)κ2

1 + 2γζκ + βκ2

κ2

∆t

D22 =1 + 2(γ − 1)ζκ + (β − γ)κ2 − (2β − γ)ζκ3

1 + 2γζκ + βκ2

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2–19)


E11 = −β − 12

+ (2β − γ)ζκ

1 + 2γζκ + βκ2∆t2

E12 =β

1 + 2γζκ + βκ2∆t2

E21 =1 − γ + 1

2(2β − γ)κ2

1 + 2γζκ + βκ2∆t

E22 =γ

1 + 2γζκ + βκ2∆t

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2–20)

κ = ω0∆t (2–21)

Example 2.1: Exact single step multi value method

Assume that the initial time in the analytical solution (1-29) is chosen as the time t j = j∆t. Then, the initial valueis changed into z(tj) = zj , and the solution is modified to

z(t) = eA(t−tj)

(zj +

∫ t

tj

e−A(τ−tj)F(τ)dτ

), t > tj (2–22)

Next, (2-22) is considered at the following time t j+1 = tj + ∆t, which leads to following integration algorithm

zj+1 = eA∆t

(zj +

∫ tj+1

tj

e−A(τ−tj)F(τ)dτ

)= Dzj + Ej (2–23)

where

zj =

[xj

xj

](2–24)

D = eA∆t (2–25)

Ej = eA∆t

∫ tj+1

tj

e−A(τ−tj)F(τ)dτ eA∆t((

1 − α)e−A·0Fj + αe−A∆tFj+1

)∆t =((

1 − α)e−A∆tFj + αFj+1

)∆t (2–26)

In (2-26) the integral has been evaluated by a generalized trapezoidal rule defined by the parameter α ∈ [0, 1] interms of a weighted average of the values e−A·0Fj and e−A∆tFj+1 at the ends of the integration interval [tj , tn+1].D denotes the exact amplification matrix. Correspondingly, the evolutionary equation (2-11) determines the exactsolution for the state vector to the accuracy of the approximation (2-26) for the load vector.

The matrix exponential functions entering (2-25) and (2-26) are given by, cf. (1-55)

eA∆t = ΨeΛA∆tΨ−1

e−A∆t = ΨeΛA∆tΨ−1

}(2–27)


The modal matrix Ψ contains the eigenmodes of the matrix D, which are identical to the eigenmodes of the matrixA. The diagonal matrix eΛA∆t stores the eigenvalues λD,j = eλA,j∆t of D in the main diagonal. λA,j denotes thecorresponding eigenvalue of A, which may be written in the form 7

λA,j = ωj

(− ζj + i

√1 − ζ2

j

)(2–28)

ωj and ζj defines the equivalent undamped circular eigenfrequency and damping ratio in case of damped of dampedeigenvibration in the jth mode. These definitions correspond to the conventional definition of these quantities incase of the modal decoupling condition (1-94). 7 From (2-28) follows, that the modal damping ratio is related tothe magnitude of the eigenvalues of D as

|λD,j | = e−ζjωj∆t ⇒

ζj = − ln |λD,j |ωj∆t

(2–29)

The damped circular eigenfrequency ωd,j is related to the eigenvalues of the amplification matrix as follows

ωd,j = ωj

√1 − ζ2

j = Im[λA,j

]=

1∆t

Im[ln λD,j

]=

1∆t

Im[ln(|λD,j |eiθD,j

)]=

θD,j

∆t(2–30)

where θD,j denotes the argument of λD,j .

Example 2.2: Displacement difference equation form of Newmark algorithm

In this example an implementation of the Newmark algorithms will be derived, where the solution for the unknowndisplacement vector xj+1 at the time tj+1 is determined as a function of the previous known displacement vectorsxj and xj−1, as well as the load vectors fj+1, fj and fj−1. At first, (2-2) is formulated at the times tj+1, tj andtj−1 as follows

Mxj+1 + Cxj+1 + Kxj+1 = fj+1

Mxj + Cxj + Kxj = fj

Mxj−1 + Cxj−1 + Kxj−1 = fj−1

⎫⎪⎬⎪⎭ (2–31)

The first equation in (2-29) is multiplied with β∆t2, the 2nd equation is multiplied with(

12 − 2β + γ

)∆t2, and the

third equation is multiplied with(

12 + β − γ

)∆t2. Finally, the resulting equations are added, leading to

∆t2M[βxj+1 − 2βxj + βxj−1 +

(12

+ γ

)xj +

(12− γ

)xj−1

]+

∆t2C[βxj+1 − 2βxj + βxj−1 +

(12

+ γ

)xj +

(12− γ

)xj−1

]+

∆t2K[βxj+1 − 2βxj + βxj−1 +

(12

+ γ

)xj +

(12− γ

)xj−1

]=

β∆t2 fj+1 +(

12− 2β + γ

)∆t2 fj +

(12

+ β − γ

)∆t2 fj−1 (2–32)

Next, (2-4) and (2-5) are formulated at the time t j+1 and tj as follows

7S.R.K. Nielsen: Structural Dynamics, Vol. 1. Linear Structural Dynamics, 4th Ed. Aalborg tekniske Univer-sitetsforlag, 2004.


xj+1 = xj + ∆t xj +(

12− β

)∆t2 xj + β∆t2 xj+1

xj = xj−1 + ∆t xj−1 +(

12− β

)∆t2 xj−1 + β∆t2 xj

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (2–33)

xj+1 = xj +(1 − γ

)∆t xj + γ∆t xj+1

xj = xj−1 +(1 − γ

)∆t xj−1 + γ∆t xj

}(2–34)

Withdrawal of the last equations in (2-33) and (2-34) from the the first equations provides the identities

β∆t2(xj+1 − 2xj + xj−1

)= xj+1 − 2xj + xj−1 − ∆t

(xj − xj−1

)− 12∆t2(xj − xj−1

)(2–35)

γ∆t(xj+1 − 2xj + xj−1

)= xj+1 − 2xj + xj−1 − ∆t

(xj − xj−1

)(2–36)

Next, ( 12 + γ)∆t2xj + 1

2 − γ)∆t2xj−1 is added on both sides of (2-35), and (2-3) is used on the resulting righthand side, which provides

∆t2(βxj+1 − 2βxj + βxj−1

)+(

12

+ γ

)∆t2 xj +

(12− γ

)∆t2 xj−1 =

xj+1 − 2xj + xj−1 − ∆t(xj − xj−1

)− 12∆t2(xj − xj−1

)+(

12

+ γ

)∆t2 xj +

(12− γ

)∆t2 xj−1 =

xj+1 − 2xj + xj−1 (2–37)

(2-36) is solved for the velocity terms on the right hand side, and the resulting equation is multiplied with β∆t 2.Next, ( 1

2 + γ)∆t2xj + 12 − γ)∆t2xj−1 is added on both sides of the equation, resulting in

∆t2(βxj+1 − 2βxj + βxj−1

)+(

12

+ γ

)∆t2 xj +

(12− γ

)∆t2 xj−1 =

βγ∆t3(xj+1 − 2xj + xj−1

)+ β∆t3

(xj − xj−1

)+(

12

+ γ

)∆t2 xj +

(12− γ

)∆t2 xj−1 =

γ∆t(xj+1 − 2xj + xj−1

)− γ∆t2

(xj − xj−1

)+(β − 1

2γ)∆t3(xj − xj−1

)+(

12

+ γ

)∆t2 xj +

(12− γ

)∆t2 xj−1 = γ∆t

(xj+1 − 2xj + xj−1

)+ ∆t

(xj − xj−1

)(2–38)

The 3rd line in (2-38) follows from the 2nd line by eliminating the term βγ∆t 3(xj+1 − 2xj + xj−1) by means of(2-35). The final result is based on the following identity, which is obtained by a multiplication of the last equationin (2-33) with ∆t, and the last equation in (2-34) with 1

2∆t2, following by a withdrawal of the resulting equations

(β − 1

2γ)∆t3(xj − xj−1

)= ∆t

(xj − xj−1

)− 12∆t2(xj + xj−1

)(2–39)

(2-37) and (2-38) are inserted into (2-32). After grouping terms with common multipliers x j+1, xj , xj−1 thefollowing final result is obtained

[M + γ∆tC + β∆t2 K

]xj+1 −

[2M− (1 − 2γ

)∆tC−

(12− 2β + γ

)∆t2 K

]xj +[

M − (1 − γ)∆tC +

(12

+ β − γ

)∆t2 K

]xj−1 =

β∆t2 fj+1 +(

12− 2β + γ

)∆t2 fj +

(12

+ β − γ

)∆t2 fj−1 (2–40)


(2-40) represents the so-called displacement difference equation form of the Newmark algorithm, which constitutesa multi step single value formulation of the method. At the calculation of x 1 the previous solution x0 is given asone of the initial value vectors, whereas x−1 is unknown. Hence, the algorithm has a starting problem. Instead, x 1

is calculated by the standard implementation given in Box 2.1, before the algorithm (2-40) is used for j ≥ 1.

Next, consider the special case of β = 0 and γ = 12 . Then, (2-40) reduces to

[M +

12∆tC

]xj+1 −

[2M − ∆t2 K

]xj +

[M − 1

2∆tC

]xj−1 = ∆t2 fj (2–41)

Consider the central difference1 approximations to the acceleration and the velocity vectors

xj xj+1 − 2xj + xj−1

∆t2, xj xj+1 − xj−1

2∆t(2–42)

(2-41) is obtained, if the finite difference approximations (2-42) are inserted into the middlemost equation of (2-31). Hence, the central difference solution to (1-1) constitutes a special case of the Newmark family correspondingto the parameters (β, γ) = (0, 1

2 ).The central difference algorithm is only conditional stable. However, if M and C are diagonal it provides anexplicit solution for xj+1, which makes it highly economical. In cases where the time step is controlled byaccuracy rather than stability, which is often the case in many wave propagation problems, the central differencealgorithm is widely used.

Example 2.3: Crank-Nicolson algorithm

For (β, γ) = ( 14 , 1

2 ), eqs. (2-11), (2-12) may be written

[M + 1

4∆t2K 14∆t2C

12∆tK M + 1

2∆tC

][xj+1

xj+1

]=

[M − 1

4∆t2K ∆tM − 14∆t2C

− 12∆tK M − 1

2∆tC

] [xj

xj

]+

[14∆t2 1

4∆t2

12∆t 1

2∆t

][fj

fj+1

](2–43)

The 2nd equation in (2-43) is multiplied with 12∆t, and is withdraw from the 1st equation, resulting in

[M − 1

2∆tM12∆tK M + 1

2∆tC

] [xj+1

xj+1

]=

[M 1

2∆tM− 1

2∆tK M − 12∆tC

][xj

xj

]+

[0 0

12∆t 1

2∆t

] [fj

fj+1

](2–44)

The Crank-Nicolson algorithm is single step multi value method, where the amplification matrix in (2-11) is givenas

D = D−11 D2 (2–45)

where, cf. (1-27)

D1 = I − 12∆tA =

[I − 1

2∆t I12∆tM−1K I + 1

2∆tM−1C

]=

[M−1 0

0 M−1

][M − 1

2∆tM12∆tK M + 1

2∆tC

](2–46)

D2 = I +12∆tA =

[I 1

2∆t I− 1

2∆tM−1K I − 12∆tM−1C

]=

[M−1 0

0 M−1

][M 1

2∆tM− 1

2∆tK M − 12∆tC

](2–47)


A systematic derivation of the Crank-Nicolson algorithm along with other high-accuracy methods will be given inExample 2.x.

Insertion of (2-46) and (2-47) into (2-45) provides

D =

[M − 1

2∆tM12∆tK M + 1

2∆tC

]−1 [M−1 0

0 M−1

]−1 [M−1 0

0 M−1

][M 1

2∆tM− 1

2∆tK M − 12∆tC

]

=

[M − 1

2∆tM12∆tK M + 1

2∆tC

]−1 [M 1

2∆tM− 1

2∆tK M − 12∆tC

](2–48)

The amplification (2-48) is identical to the one obtained from (2-44). From this is concluded that the Crank-Nicolson algorithm constitutes a special case of the Newmark family corresponding to the parameters (β, γ) =(14 , 1

2 ). As mentioned in Box 2.2 this parameter combination is obtained in case of a constant variation of theacceleration during the interval [tj , tj+1] given by x(t) ≡ 1

2 (xj + xj+1).

2.1.1 Numerical accuracy

The approximate Newmark algorithm develops after (2-11), whereas the exact developmentis given by (2-23). Assume that the load vectors Ej in (2-11) and (2-23) are identical. Oncondition of an identical state vector zj at the time tj , the deviation between the Newmarksolution zj+1 and the exact solution zj+1 at the succeeding time tj+1 is given by

zj+1 − zj+1 =(D − D

)zj = ε (2–49)

The error vector ε depends on the difference D − D, which in term depends on the magnitudeof the time step ∆t. Assume that the error vector has the form

τl =∣∣ε(∆tk+1

)∣∣ (2–50)

τ =∥∥O(∆tk+1

)∥∥ (2–51)

D = ΨΛDΨ−1 (2–52)

D = ΨΛDΨ−1 (2–53)

ΛD = eΛA∆t (2–54)


ΛD = eΛA∆t (2–55)

ω =θD

∆t(2–56)

ζ0 = − ln |λD|ω∆t

(2–57)

2.1.2 Numerical stability

Given that xj−1 and xj are known exactly, (2-5) determines xj+1 with an error of the magnitudeO(∆t2). Hence, at each calculation step new errors are introduced in the solution. As is the

case for the initial conditions or any other external disturbance, errors introduced at a certaininstant of time will be dissipated by the numerical scheme. However, if the errors are introducedat a rate and magnitude overruling the dissipating capability of algorithm, the obtained resultswill eventually oscillate with larger and larger amplitudes. We say that the algorithm becomesnumerical unstable. Typically, stability problems may be cured by selecting a smaller time step.In order to identify the mechanism trigging the numerical instability, the generic form (2-5) isiterated to obtain the following sequence of results

z1 = Dz0 + E0

z2 = Dz1 + E1 = D2z0 + DE0 + E1

z3 = Dz2 + E2 = D3z0 + D2E0 + DE1 + E2

...

zj+1 = Dzj + Ej = Dj+1z0 + DjE0 + Dj−1E1 + · · ·+ DEj−1 + Ej = Dj+1z0 +

j∑m=0

Dj−mEm

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2–58)

Additionally, the eigenvalue problem related to the transfer matrix D is considered. In the fol-lowing the kth eigen-pair of the transfer matrix D and the system matrix A given by (1-27) aredenoted as (λk,D,Ψ

(k)D ) and (λk,A,Ψ

(k)A ), respectively. Although in some cases the eigenvectors

of D and A are identical, as for the Crank-Nicolson algorithm described in Example 2.2, theeigenvectors of the two matrices will in general be different and may even be of different di-mension. Nevertheless, the kth eigenvectors Ψ

(k)D and Ψ

(k)A describes the same eigenmode of the

system. Similarly, λk,D is an approximation to the kth eigenvalue λk,A. The eigenvalues λk,D

are stored in the diagonal matrix ΛD, whereas the eigenvectors Ψ(k) are stored column-wise inthe modal matrix ΨD. Similar to (1-38) the eigenvalue problems may represented in the matrixform


DΨD = ΨDΛD (2–59)

From (2-39) follows, cf. (1-41), (1-52)

D = ΨDΛDΨ−1D ⇒ Dj = ΨDΛj

DΨ−1D (2–60)

ΛjD signifies a diagonal matrix with the quantities λj

k,D in the main diagonal. Insertion of (2-40)into (2-38) provides the relation

qj+1 = Λj+1D q0 +

j∑m=0

Λj−mD em (2–61)

where

qj = Ψ−1D zj , em = Ψ−1

D Em (2–62)

All matrices entering (2-41) are diagonal. Hence, the relation among the kth component of the2n-dimensional vectors qj+1, q0 and em become

qk,j+1 = λj+1k,Dqk,0 +

j∑m=0

λj−mk,D ek,m , k = 1, . . . , N (2–63)

where N is the dimension of the state vector. Next, using the triangle inequality the followingupper bound for the right hand side of (2-43) may be formulated

∣∣qk,j+1

∣∣ =∣∣∣∣∣λj+1

k,Dqk,0 +

j∑m=0

λj−mk,D ek,m

∣∣∣∣∣ ≤ ∣∣λk,D

∣∣j+1∣∣qk,0

∣∣ + j∑m=0

∣∣λk,D

∣∣j−m∣∣ek,m

∣∣ (2–64)

Let,

ek,max = maxm=1, 2, ...

∣∣ek,m

∣∣ (2–65)

Then, the upper bound in (2-44) may be further elaborated as

∣∣qk,j+1

∣∣ ≤ ∣∣λk,D

∣∣j+1∣∣qk,0

∣∣+ ek,max

j∑m=0

∣∣λk,D

∣∣j−m=

∣∣λk,D

∣∣j+1∣∣qk,0

∣∣ + ek,max

1 − ∣∣λk,D

∣∣j+1

1 − ∣∣λk,D

∣∣ (2–66)


In the last statement the following well-known sum formula has been used

1 + x + x2 + · · ·+ xj =1 − xj+1

1 − x(2–67)

The right hand side of (2-46), and hence∣∣yk,j+1

∣∣, remains finite as j → ∞, if

∣∣λk,D

∣∣ < 1 , k = 1, . . . , N (2–68)

(2-48) is a sufficient condition for stability of the numerical integration algorithm. From (2-41)follows that the stability of the algorithm requires that all the diagonal matrices Λj+1

D and Λj−mD

remain finite as as j → ∞. This is the case if

∣∣λk,D

∣∣ ≤ 1 , k = 1, . . . , N (2–69)

(2-49) is a necessary condition for stability of the algorithm. The sufficiency of the algorithm,if one or more eigenvalues fulfill

∣∣λk,D

∣∣ = 1 is not guaranteed.

In conclusion it is shown that the stability analysis of the numerical integration algorithm can bereduced to an eigenvalue analysis of the transfer matrix of the related state vector formulation.

Next, consider the expansion in modal coordinates (1-95). Due to the one-to-one correspon-dence the solution for the displacement xj = x(j∆t) will be of finite length as j → ∞, if andonly if the modal coordinate vector qj = q(j∆t) also remains of finite length. Assume thatthe modal decoupling condition (1-92) is valid. Then, the stability of the numerical integrationalgorithm can be investigated by checking the stability of the same numerical integration algo-rithm applied to the sequence of SDOF differential equations (1-98). In this way the stabilityof the central difference method has been investigated in Example 2.5. From this observation isconcluded that the stability is determined by the highest frequency mode in the expansion. Asa consequence truncated modal expansions such as (1-99) will have better numerical stabilityproperties than the full expansion (1-95). This observation is often formulated by the phrasethat filtration of the high-frequency modes from the response improves the numerical stability.

A numerical integration algorithm, which is stable for arbitrary length of the time step ∆t isdenoted unconditional stable. The Crank-Nicolson algorithm is unconditional stable as shownin Example 2.4. If stability is only achieved for time steps below a certain critical length thealgorithm is referred to as conditional stable. The central difference algorithm is conditionalstable as shown in Example 2.5 below. Generally, the length of the time step should be deter-mined by accuracy requirements. Hence, conditional stable algorithms should be avoided incases where the time step is determined by stability requirements.


2.1.3 Period Distortion

Consider damped eigenvibrations of the system (1-1), i.e. the dynamic load vector f(t) ≡ 0. Inthis case the iterated solution in modal coordinates (2-43) reduces to

qk,j+1 = λj+1k,Dqk,0 (2–70)

Let the eigenvalues be written in the polar notation

λk,D =∣∣λk,D

∣∣ei θk,D ⇒λj

k,D =∣∣λk,D

∣∣jei jθk,D =∣∣λk,D

∣∣j( cos jθk,D + i sin jθk,D

)(2–71)

where θk,D signifies the argument of the kth eigenvalue. The last multiplier on the right-handside specifies the harmonic varying part of the kth damped eigenvibration, whereas the factor∣∣λk,D

∣∣j determines the diminish of the amplitude due to energy dissipation.

Assume that the harmonic varying part of the numerical solution repeat itself after J time steps.Then, Jθk,D = 2π. Hence, the damped eigenperiod in the kth mode predicted by the numericalalgorithm becomes equal to

Tk,d,D = J∆t = 2π∆t

θk,D(2–72)

In order to calculate the corresponding damped eigenperiod of the system (1-26) the eigensolu-tion (1-34) is written on the form

z(t) = Ψ(k)A eλk,A t = Ψ

(k)A e−µk,A t

(cos νk,A t + i sin νk,A t

)(2–73)

where the representation (2-51) for the eigenvalues λj,A has been used. From (2-61) followsthat the analytical solution for the damped eigenperiod in the kth mode becomes

Tk,d,A =2π

νk,A(2–74)

Quite often it is seen that the numerical and analytical damped eigenperiod differ. This phe-nomenon is denoted period distortion. As a measure of the period distortion the followingnon-dimensional ratio is introduced

τk =Tk,d,D − Tk,d,A

Tk,d,A(2–75)

τk indicates the period distortion per period. If τk > 0 we talk of period elongation, whereasτk < 0 signifies period shortening. If the algorithm is related with period distortion the nu-merical and analytical eigensolutions become more and more in disagreement, and may at a


sufficient large elapsed time even become in counter phase. Similar to numerical stability prob-lems the period distortion may be cured by reducing the time step. Since the structural responseis determined by the low frequency modes only these modes need to be adjusted for perioddistortion. Hence, the requirement to the time step for elimination of period distortion is not asrestrictive as for the fulfillment of the stability requirement for conditional stable numerical al-gorithms. For undamped or lightly damped systems shown in Example 2.5 the Crank-Nicolsonalgorithm has period elongation, whereas the central difference algorithm has period shorteningas demonstrated in Examples 2.7 and 2.8 below.

A numerical example demonstrating the period shortening of the central difference method hasbeen given in Example 2.9 below.

2.1.4 Numerical Damping

Damped eigenvibrations in the kth mode is given by (2-61). At the time t = j∆t the analyticalsolution for the kth modal coordinate may be written

qj,k,A = q0,k,Aeλk,A j∆t = q0,k,Ae−µk,A j∆t(cos νk,A j∆t + i sin νk,A j∆t

)(2–76)

The corresponding numerical solution is given by (2-41)

qj,k,D = λjk,Dq0,k,D (2–77)

As a measure of the dissipation of the numerical solution relative to the analytical solution thefollowing fraction is considered

∣∣qj,k,D

∣∣∣∣qj,k,A

∣∣ = aj

∣∣q0,k,D

∣∣∣∣q0,k,A

∣∣ (2–78)

where

a =

∣∣λk,D

∣∣e−µk,A ∆t

(2–79)

Quite often the decrease of the numerical solution for the damped eigenvibrations in the kthmode relative to the analytical solution is determined by the fraction a. If a < 1 the numericalsolution will decrease faster then the numerical solution. We say that the numerical algorithmis related with positive numerical damping in the kth mode. On the other hand, if a > 1, thenumerical solution will decrease at a slower rate than the analytical solutions, for which reasonthe numerical algorithm is said to be related with negative numerical damping.


Occationally, numerical damping is investigated by checking whether the algorithm predictsdissipation in the undamped case, where C = 0. Then, µk,A = 0, and the criteria a < 1 fornumerical damping reduces to

∣∣λk,D

∣∣ < 1 (2–80)

where the eigenvalue λk,D is calculated for a transfer matrix D of an undamped structural sys-tem.

As we shall see the two criteria are not unique. As shown in Examples 2.8 and 2.9 both theCrank-Nicolson and the central difference algorithms are free of numerical damping accordingto the criteria (2-76). Nevertheless it turns out that a > 1 for the Crank-Nicolson algorithm cor-responding to negative numerical damping, whereas a < 1 for the central difference algorithmcorresponding to positive numerical damping. As demonstrated in Examples 2.8 and 2.9, thedeviation of the two criteria is of the magnitude O(∆t3). a depends linearly on the dampingratio for the Crank-Nicolson algorithm, whereas the dependence is cubic for the central differ-ence algorithm. For this reason the discrepancy between the two criteria for numerical dampingis most pronounced for the Crank-Nicolson algorithm as will be demonstrated in the numericalexamples.

Positive numerical damping is a useful property, since it tends to stabilize the numerical integra-tion of the high frequency modes. Actually, the numerically damping tends to reduce or filterthese modes out of the response. Of course, the time step should be adjusted so the dampingeffect is not visible in the low frequency modes determining the global response.

Example 2.4: Numerical damping of the central difference algorithm and Crank-Nicolson algorithms for a


0 1 2 3 4 5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t/Tk

q k(t

)

Fig. 2–1 Crank-Nicolson algorithm. Damped eigen-vibrations of SDOF oscillator. ωk = 1, ζk = 0.01,(qk,0, qk,0) = (1, 0). —-: Analytical solution, - - -:∆tTk

=0.1, -.-.-: ∆tTk

= 0.2, .....: ∆tTk

= 1.05π .

0 1 2 3 4 5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

t/Tk

q k(t

)/e−

ζ kω

kt

Fig. 2–2 Negative numerical damping of Crank-Nicolson algorithm. Normalized damped eigenvi-brations of SDOF oscillator. ωk = 1, ζk = 0.1,(qk,0, qk,0) = (1, 0), ∆t

Tk= 0.2.

Fig. 2-3 shows the numerical results for eigenvibrations of a SDOF oscillator with ω k = 1 and ζk = 0.1 obtainedby the Crank-Nicolson algorithm with the time steps ∆t

Tk= 0.1, 0.2, 1.05

π . The solutions remain numerical stableeven at very large time steps as predicted by (2-53). Further, all numerical solutions are related with period elon-gation, which increases proportional with ∆t2 as indicated by (2-68).

No numerical damping is visible in the results shown on Fig. 2-3. In order to investigate this problem further,the numerical results are normalized with the amplitude e−ζkωkt of the analytical solution. The results have beenshown on Fig. 2-4 for the relatively large damping ratio ζ k = 0.1 and the time step ∆t

Tk= 0.2. As seen the

numerical solution is dissipated at a slower rate than the analytical solution, verifying the negative numericaldamping effect predicted by (2-84).

2.2 Generalized Alpha Algorithm 49

0 1 2 3 4 5−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t/Tk

q k(t

)

Fig. 2–3 Central difference algorithm. Dampedeigenvibrations of SDOF oscillator. ωk = 1, ζk =0.01, (qk,0, qk,0) = (1, 0). —-: Analytical solution, -- -: ∆t

Tk=0.1, -.-.-: ∆t

Tk= 0.2, .....: ∆t

Tk= 1.05

π .

0 10 20 30 40 50−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t/Tk

q k(t

)/e−

ζ kω

kt

Fig. 2–4 Positive numerical damping of central dif-ference algorithm. Normalized damped eigenvibra-tions of SDOF oscillator. ωk = 1, ζk = 0.1,(qk,0, qk,0) = (1, 0), ∆t

Tk= 0.2.

Fig. 2-5 shows the obtained numerical results. The solutions are numerical stable at the time steps ∆tTk

=0.1, 0.2,

and unstable at the time step ∆tTk

= 1.05π as predicted by (2-57), demonstrating the conditional stability of the

central difference algorithm. As shown on Fig. 2-3 the Crank-Nicolson algorithm remain stable at the time step∆tTk

= 1.05π . All numerical solutions are related with period shortening, which increases proportional with ∆t 2 as

predicted by (2-71).

No numerical damping is visible in the results shown on Fig. 2-5. In order to highlight this problem, the numericalsolution is normalized with respect to the envelope function of the analytical solution, e −ζkωkt. The results havebeen shown on Fig. 2-6 for the damping ratio ζk = 0.1 and time step ∆t

Tk= 0.2. As seen the numerical solution

is dissipated at a slower rate than the analytical solution, verifying the positive numerical damping effect predictedby (2-84). However, the numerical damping effect is much weaker than on Fig. 2-4 due to the dependence on ζ 3

k

in (2-88), versus the linear dependence on the damping ration in (2-84).

2.2 Generalized Alpha Algorithm

Box 2.3: Derivation of alpha algorithm

Historical review:Newmark algorithms,8

8N.M. Newmark: A Method of Computation for Structural Dynamics. J.Eng.Mech., ASCE, 85(EM3), 1959,67-94.


Numerical damping is introduced in the Newmark method by chosen the parameter γ > 12.

However, this will introduce numerical damping also in the lower modes,and hence compromisethe accuracy. Methods to improve upon this situation are the socalled collocation methods, firstintroduced by Wilson9, and improved by Hilber and Hughes10

α method of Hilber, Hughes and Taylor11 α-modification of damping, stiffness and externalforces.α method of Wood, Bossak and Zienkiewicz12 α-modification of inertial forces alone.Generalized α-method of Chung and Hulbert13 α-modification of inertial terms and of damping,stiffness and external forces with different α parameters.

Example 2.5: Numerical integration of SDOF oscillator by generalized alpha algorithms

Example 2.6: Unconditional stable high accuracy algorithms

The considered algorithms are based on the following factorization of the exact amplification matrix (2-25)

D = eA∆t = eA(1+α)∆t2 eA(1−α)∆t

2 =(e−A(1+α)∆t

2)−1eA(1−α)∆t

2 = D−11 D2 (2–81)

where α is a number in the interval ]0, 1[. Additionally, (1-57) has been used with t = −(1 + α) ∆t2 . From (1-53)

follows

D1 = e−A(1+α)∆t2 = I − 1 + α

2∆tA +

(1 + α)2

8∆t2A2 − (1 + α)3

48∆t3A3 +

(1 + α)4

384∆t4A4 + · · ·

D2 = eA(1−α)∆t2 = I +

1 − α

2∆tA +

(1 − α)2

8∆t2A2 +

(1 − α)3

48∆t3A3 +

(1 − α)4

384∆t4A4 + · · ·

⎫⎪⎪⎬⎪⎪⎭

(2–82)

From (1-55) follows, that the matrices D1 and D2 alternatively may be written as

D1 = ΨΛD1Ψ−1

D2 = ΨΛD2Ψ−1

}(2–83)

ΛD1 and ΛD2 are diagonal matrices, storing the eigenvalues of D1 and D2 in the main diagonal. These are givenas

ΛD1 = I − 1 + α

2∆tΛA +

(1 + α)2

8∆t2Λ2

A − (1 + α)3

48∆t3Λ3

A +(1 + α)4

384∆t4Λ4

A + · · ·

ΛD2 = I +1 − α

2∆tΛA +

(1 − α)2

8∆t2Λ2

A +(1 − α)3

48∆t3Λ3

A +(1 − α)4

384∆t4Λ4

A + · · ·

⎫⎪⎪⎬⎪⎪⎭ (2–84)

9E.L. Wilson: A Computer Program for the Dynamic Stress Analysis of Underground Structures. SESM ReportNo- 68-1, Division of Structural Engineering and Structural Mechanics, University of Califirnia, Berkeley, 1968.

10H.H. Hilber and T.J.R. Hughes and R.L. Taylor: Collocation, Dissipation and ”Overshoot” for Time Integra-tion Schemes in Structural Dynamics. Earthquake Engineering and Structural Dynamics, 6, 1978, 99-118.

11H.H. Hilber, T.J.R. Hughes and R.L. Taylor: Improved Numerical Dissipation for Time Integration in Struc-tural Dynamics. Earthquake Engineering and Structural Dynamics, 5, 1977, 283-292.

12W.L. Wood, M. Bossak and O.C. Zienkiewicz: An Alpha Modification of Newmarks Method. InternationalJournal for Numerical Methods in Engineering, 15, 1981, 1562-1566.

13J. Chung and G.M. Hulbert: A time Integration Algorithm for Structural Dynamics with Improved NumericalDissipation: The Generalized α Method. Journal of Applied Mechanics, 60, 1993, 371-375.

2.2 Generalized Alpha Algorithm 51

Ψ is a modal matrix containing the eigenvectors of D1 and D2 stored column-wise. It is remarkable that theseeigenvectors without any approximation are equal to the eigenvectors of the matrix of A, independently of thenumber of terms retained in the expansion (2-94). Hence, the local truncation error is entirely related to thetruncation errors of these expansion.


2.3 Exercises

2.1 Consider the damped eigenvibrations of the two-degrees-of-freedom system defined in Ex-ample 1.2 subjected to the initial values

x1(0) = 0.01 m , x2(0) = x1(0) = x2(0) = 0

(a.) Write a MATLAB program, which perform Newmark integration.

(b.) Perform and compare the calculation for (β, γ) = (0.25, 0.50), (β, γ) = (0.25, 0.25)with the time-step h = T1/10 and h = T1/100, where T1 denotes the fundamentalundamped eigenperiod.

2.2 Consider the same problem as in Exercise 2.1.

(a.) Write a MATLAB program, which perform generalized alpha integration.

(b.) Perform and compare the calculation for |λ∞| = 0.8, |λ∞| = 1.0 with the time-steph = T1/10 and h = T1/100, where T1 denotes the fundamental undamped eigenpe-riod.

CHAPTER 3LINEAR EIGENVALUE PROBLEMS

In this chapter the generalized eigenvalue problem (1-9), and the related characteristic polyno-mial will be further analyzed. At first in Section 3.1, the Gauss factorization of the coefficientmatrix of the generalized eigenvalue problem is treated. This factorization plays an importantrole in several iterative numerical eigenvalue solvers based on the characteristic polynomial.Furthermore, the factorization provides a simple way for calculating a sequence of characteris-tic polynomials known as the Sturm sequence, which makes it possible to formulate upper andlower bounds of the eigenvalues of the problem. These bounds are contained in the so-calledeigenvalue separation principle treated in Section 3.2. Various iterative schemes for solving theindicated generalized eigenvalue problem require that the stiffness matrix is non-singular. Forstructures, which admit stiff-body motions, it then becomes necessary to perform a so-calledshift, where an artificial non-singular stiffness matrix is introduced. The eigenvectors of theshifted system are identical to those of the original problem, whereas the eigenvalues of thetwo systems deviate with the specified shift parameter. Shifting of a generalized eigenvalueproblem is treated in Section 3.3. Some iterative eigensolvers presume a special eigenvalueproblem corresponding to M = I in (1-9). In this case an introductory transformation fromthe anticipated generalized eigenvalue problem into an equivalent special eigenvalue problembecomes necessary. This can be achieved in several ways. In Section 3.4 a so-called similaritytransformation has been used, which preserves the symmetry of the transformed stiffness ma-trix. A similarity transformation leave the eigenvalues unaffected, whereas the eigenvectors arechanged in a known manner.

3.1 Gauss Factorization of Characteristic Polynomials

Since the coefficient matrix of the generalized eigenvalue problem K − λM is symmetric, itmay be Gauss factorized on the form

K − λM = LDLT (3–1)

where L is a lower triangular matrix with units in the main diagonal, and D is a diagonal matrix,given as

— 53 —

54 Chapter 3 – LINEAR EIGENVALUE PROBLEMS

L =

⎡⎢⎢⎢⎢⎢⎢⎣

1

l21 1

l31 l32 1...

......

. . .

ln1 ln2 ln3 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎦ (3–2)

D =

⎡⎢⎢⎢⎣

d11 0 · · · 0

0 d22 · · · 0...

.... . .

...0 0 · · · dnn

⎤⎥⎥⎥⎦ (3–3)

The details of the Gauss factorization of an symmetric matrix has been given in Box. 3.1. Since,det(L) = det(LT ) = 1, the following representation of the characteristic polynomial (1-10) isobtained

P (λ) = det(LDLT

)= det

(L)det(D)det(LT)

= det(D)

= d11d22 · · · dnn (3–4)

At the same time (1-11) can be written on the form

P (λ) = a0

(λ − λ1

)(λ − λ2

) · · · (λ − λn

)(3–5)

Despite the striking similarity between (3-4) and (3-5), dii is very different from the correspond-ing factor

(λ − λi

)in (3-5), as demonstrated in Example 3.2 below.

Let λ be monotonously increased in the interval [0,∞[. From (1-10) follows that P (0) =an = det(K) ≥ 0. Since all factors in (3-5) have negative sign for λ ∈]0, λ1[, it follows thatP (λ) > 0 throughout this interval. As λ passes λ1 from below, the factor

(λ − λ1

)changes its

sign from negative to positive, while the other factors remain negative. This means that the signof P (λ) changes from positive to negative at the passage of λ1. Then, a similar sign changemust occur in (3-4). For λ < λ1 all the diagonal elements d11, d22, . . . , dnn are positive. Asλ passes λ1, exactly one of these factors (not necessarily d11) changes its sign from positiveto negative, while the other factors remain positive. The sign of the characteristic polynomialremain negative for λ ∈]λ1, λ2[. As λ passes λ2 from below, the factor

(λ − λ2

)in the same

way changes its sign from negative to positive, making the characteristic polynomial positive inthe interval λ ∈]λ2, λ3[. A similar sign change occurs in (3-4), meaning that an additional diag-onal element has changed its sign from positive to negative. Hence, for λ ∈]λ2, λ3[ exactly twoof the diagonal elements d11, d22, . . . , dnn are negative. Proceeding in this manner, it is seenthat if λ is placed somewhere in the interval ]λm, λm+1[, exactly m diagonal elements amongd11, d22, . . . , dnn are negative. This observation is contained in the following theorem.

3.1 Gauss Factorization of Characteristic Polynomials 55

Theorem 3.1: Let D be the diagonal matrix in the Gauss factorization of the coefficient matrixK − λM of a generalized eigenvalue problem, and λ is an arbitrary parameter. Then, the num-ber of negative components in the main diagonal of D is equal to the number of eigenvalues ofthe generalized eigenvalue problem, which are smaller than the parameter λ entering the factor-ization.

The theorem can be used to formulate bounds for any of the eigenvalues as demonstrated belowin Example 3.3. Actually, one can calculate say the jth eigenvalue λj with arbitrary accuracy.The method is simply to make an initial sequence of calculations of the characteristic polyno-mial P (λ) as a function of λ by (3-4), until j components in the main diagonal of D are negative.Next, one can perform additional calculations to reduce the interval, where the jth sign changetakes place. This procedure for calculation of eigenvalues is known as the telescope method.

Box 3.1: Gauss factorization of symmetric matrix

Gauss factorization reduces a symmetric matrix K of dimension n × n to an upper trian-gular matrix S in a sequence of n− 1 matrix multiplications. After the first (i− 1) matrixmultiplications the following matrix is considered

K(i) = L−1i−1L

−1i−2 · · ·L−1

1 K , i = 2, . . . , n (3–6)

where K(1) = K. Sequentially, the indicated matrix multiplications produce zeros belowthe main diagonal of the columns j = 1, . . . , i − 1. Then, pre-multiplication of K(i) withL−1

i will produce zeroes below the main diagonal of the ith column without affecting thezeroes in the previous columns. L−1

i is a lower triangular matrix with units in the principaldiagonal, and where only the ith column is non-zero, given as

L−1i =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

0 1...

.... . .

0 0 · · · 1

0 0 · · · 0 1

0 0 · · · 0 −li+1,i 1...

......

......

. . .

0 0 · · · 0 −ln,i 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(3–7)


The components lj,i entering the ith column are given as

lj,i =K

(i)j,i

K(i)i,i

, j = i + 1, . . . , n (3–8)

where K(i)j,i denotes the component in the jth row and ith column of K(i). By insertion it

is proved that the inverse of (3-7) is given as

Li =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

0 1...

.... . .

0 0 · · · 1

0 0 · · · 0 1

0 0 · · · 0 li+1,i 1...

......

......

. . .

0 0 · · · 0 ln,i 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(3–9)

Then, K(n) obtained after the (n−1)th multiplication with L−1n−1, has zeroes in all the first

(n− 1) columns below the main diagonal, corresponding to an upper triangular matrix S.Hence

L−1n−1L

−1n−2 · · ·L−1

1 K = S ⇒K = LS , L = L1L2 · · ·Ln−1 (3–10)

Since, L defined by (3-10) is the product of lower triangular matrices with 1 in the maindiagonal, it becomes a matrix with the same structure as indicated by (3-2).

Because K is symmetric, S must have the structure

S = DLT (3–11)

where D is a diagonal matrix, given by (3-3). This proofs the validity of the factorization(3-1).

3.1 Gauss Factorization of Characteristic Polynomials 57

Example 3.1: Gauss factorization of a three-dimensional matrix

Given the symmetric matrix

K = K(1) =

⎡⎢⎣ 5 −4 1−4 6 −4

1 −4 6

⎤⎥⎦ (3–12)

L−11 =

⎡⎢⎣ 1 0 0

45 1 0

− 15 0 1

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

K(2) = L−11 K(1) =

⎡⎢⎣5 −4 1

0 2.8 −3.20 −3.2 5.8

⎤⎥⎦

L(1) = L1 =

⎡⎢⎣ 1 0 0− 4

5 1 015 0 1

⎤⎥⎦

(3–13)

L−12 =

⎡⎢⎣1 0 0

0 1 00 3.2

2.8 1

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

K(3) = L−12 K(2) = S =

⎡⎢⎣5 −4 1

0 2.8 −3.20 0 2.1429

⎤⎥⎦

L(2) = L1L2 = L =

⎡⎢⎣ 1 0 0−0.8 1 0

0.2 −1.1429 1

⎤⎥⎦

(3–14)

From this follows that

L =

⎡⎢⎣ 1 0 0−0.8 1 0

0.2 −1.1429 1

⎤⎥⎦ , D =

⎡⎢⎣5 0 00 2.8 00 0 2.1429

⎤⎥⎦ (3–15)

Example 3.2: Gauss factorization of a three-dimensional generalized eigenvalue problem

Of course for a given value of λ the matrix K − λM may be factorized according to the method explained inBox 3.1. However, for smaller problems explicit expressions may be derived, as demonstrated in the following.Given the mass- and stiffness matrices defined in Example 1.4, the components of L and D are calculated from thefollowing identities, cf. (1-79), (3-1)

⎡⎢⎣ 1 0 0l21 l 0l31 l32 1

⎤⎥⎦⎡⎢⎣d11 0 0

0 d22 00 0 d33

⎤⎥⎦⎡⎢⎣1 l21 l31

0 1 l32

0 0 1

⎤⎥⎦ =

⎡⎢⎣ d11 d11l21 d11l31

d11l21 d22 + d11l221 d11l21l31 + d22l32

d11l31 d11l21l31 + d22l32 d33 + d11l231 + d22l

232

⎤⎥⎦ =

⎡⎢⎣2 − 1

2λ −1 0

−1 4 − λ −1

0 −1 2 − 12λ

⎤⎥⎦ (3–16)


Equating the corresponding components on the left and right hand sides, provides the following equations for thedetermination of the unknown quantities

d11 = 2 − 12λ = −1

2(λ − 4

)d11l21 = −1 ⇒ l21 =

2λ − 4

d22 + d11l221 = 4 − λ ⇒ d22 = 4 − λ +

12(λ − 4

) 4(λ − 4

)2 = −λ2 − 8λ + 14λ − 4

d11l21l31 + d22l32 = d22l32 = −1 ⇒ l32 =λ − 4

λ2 − 8λ + 14

d33 + d11l231 + d22l

232 = d33 − l32 = 2 − 1

2λ ⇒

d33 = 2 − 12

+λ − 4

λ2 − 8λ + 14= −1

2λ3 − 12λ2 + 44λ− 48

λ2 − 8λ + 14= −1

2

(λ − 2

)(λ − 4

)(λ − 6

)λ2 − 8λ + 14

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(3–17)

Then, the following expression for the characteristic equation is obtained, in agreement with (1-80)

P (λ) = d11d22d33 = −12(λ−4

)(−λ2 − 8λ + 14λ − 4

)(−1

2

(λ − 2

)(λ − 4

)(λ − 6

)λ2 − 8λ + 14

)= −1

4(λ−2

)(λ−4

)(λ−6

)(3–18)

Example 3.3: Bounds on eigenvalues

In this example bounds on the eigenvalues of the GEVP in Example 1.4 is constructed from the number of negativecomponents in the diagonal of the matrix D, using Theorem 3.1.

For λ = 1 one gets:

K− λM =

⎡⎢⎣

32 −1 0

−1 3 −10 −1 3

2

⎤⎥⎦ ⇒ LDLT =

⎡⎢⎣ 1− 2

3 10 − 3

7 1

⎤⎥⎦⎡⎢⎣

32 0 00 7

3 00 0 15

14

⎤⎥⎦⎡⎢⎣1 − 2

3 01 − 3

7

1

⎤⎥⎦ (3–19)

The components of the matrices L and D may be calculated by the formulas indicated in (3-17). As seend11 = 3

2 > 0, d22 = 73 > 0, d33 = 15

14 > 0. Hence, all three diagonal components are positive, from whichit is concluded that λ1 > λ = 1.


K− λM =

⎡⎢⎣−2 −1 0−1 −4 −1

0 −1 −2

⎤⎥⎦ ⇒ LDLT =

⎡⎢⎣1

12 10 2

7 1

⎤⎥⎦⎡⎢⎣−2 0 0

0 − 72 0

0 0 − 127

⎤⎥⎦⎡⎢⎣1 1

2 01 2

7

1

⎤⎥⎦ (3–20)

As seen d11 = −2 < 0, d22 = − 72 < 0, d33 = − 12

7 < 0. Hence, all three diagonal components are negative, fromwhich it is concluded that λ3 < λ = 8.


3.2 Eigenvalue Separation Principle 59

K− λM =

⎡⎢⎣−

12 −1 0

−1 −1 −10 −1 − 1

2

⎤⎥⎦ ⇒ LDLT =

⎡⎢⎣1

2 10 −1 1

⎤⎥⎦⎡⎢⎣−

12 0 00 1 00 0 − 3

2

⎤⎥⎦⎡⎢⎣1 2 0

1 −11

⎤⎥⎦ (3–21)

As seen d11 = − 12 < 0, d22 = 1 > 0, d33 = − 3

2 < 0. Hence, two diagonal components are negative and one ispositive, from which it is concluded that λ2 < λ = 5 < λ3.


K− λM =

⎡⎢⎣

12 −1 0

−1 1 −10 −1 1

2

⎤⎥⎦ ⇒ LDLT =

⎡⎢⎣ 1−2 1

0 1 1

⎤⎥⎦⎡⎢⎣

12 0 00 −1 00 0 3

2

⎤⎥⎦⎡⎢⎣1 −2 0

1 11

⎤⎥⎦ (3–22)

As seen d11 = 12 > 0, d22 = −1 < 0, d33 = 3

2 < 0. Hence, two diagonal components are positive and one isnegative, from which it is concluded that λ1 < λ = 3 < λ2.

In conclusion the following bounds prevail

1 < λ1 < 33 < λ2 < 55 < λ3 < 8

⎫⎪⎬⎪⎭ (3–23)

3.2 Eigenvalue Separation Principle

The matrices M(m) and K(m) of dimension (n − m) × (n − m) are obtained from M and K,if the last m rows and columns are omitted in these matrices. Then, consider the sequence ofrelated characteristic polynomials of the order (n − m)

P (m)(λ(m)

)= det

(K(m) − λ(m)M(m)

), m = 0, 1, . . . , n − 1 (3–24)

where M(0) = M, K(0) = K, λ(0) = λ and P (0)(λ) = P (λ). The eigenvalues corresponding toM(m) and K(m) are denoted as λ

(m)1 , λ

(m)2 , . . . , λ

(m)n−m.

Now, for any m = 0, 1, . . . , n − 1 it can be proved that the roots of P (m+1)(λ(m+1)

)= 0 are

separating the roots of P (m)(λ(m)

)= 0, i.e.

0 ≤ λ(m)1 ≤ λ

(m+1)1 ≤ λ

(m)2 ≤ λ

(m+1)2 ≤ · · · ≤ λ

(m)n−m−1 ≤ λ

(m+1)n−m−1 ≤ λ

(m)n−m ≤ ∞ (3–25)

A formal proof of (3-25) has been given by Bathe.1 The sequence of polynomials P (m)(λ) withroots fulfilling the property (3-25), is denoted a Sturm sequence. (3-25) is illustrated in Exam-ple 3.4.

1K.-J. Bathe: Finite Element Procedures. Printice Hall, Inc., 1996.


Next, consider the Gauss factorization (3-1). Omitting the last m rows and columns in M andK is tantamount to omitting the last m rows and columns in L and D. Then

P (m)(λ(m)

)= det

(K(m) − λ(m)M(m)

)= det

(L(m)D(m)L(m)T

)= det

(D(m)

)=

d11d22 · · ·dn−m,n−m (3–26)

where

L(m) =

⎡⎢⎢⎢⎢⎢⎢⎣

1

l21 1

l31 l32 1...

......

. . .

ln−m,1 ln−m,2 ln−m,3 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎦ (3–27)

D(m) =

⎡⎢⎢⎢⎣d11 0 · · · 0

0 d22 · · · 0...

.... . .

...0 0 · · · dn−m,n−m

⎤⎥⎥⎥⎦ (3–28)

The bounding property explained in Theorem 3.1 for the case m = 0 can then easily be gener-alized. Let λ(m) = µ, and perform a Gauss factorization on the matrix K(m) − µM(m). Thenthe number of eigenvalues, λ(m)

j < µ, will be equal to number of negative diagonal componentsd11, . . . , dn−m,n−m in the matrix D.

The number of negative elements in main diagonal of the matrix D in the Gauss factorization ofK − λM = LDLT , and hence the number of eigenvalues smaller than λ, can then be retrievedfrom the signs of the sequence P (0)(λ), P (1)(λ), . . . , P (n−1)(λ) as seen in the following way.

Introduce P (n)(λ) as an arbitrary positive quantity. Since P (n−1)(λ) = d11, it follows that thesequence P (n)(λ), P (n−1)(λ) has the sign sequence sign(P (n)(λ)), sign(P (n−1)(λ)) = +,−, ifd11 < 0, and the sign sequence +, +, if d11 > 0. d11 < 0 indicates that at least one eigenvalueis smaller than λ, in which case one sign change, namely from + to −, has occurred in theindicated sign sequence.

Next, P (n−2)(λ) = d11d22 is considered. d11 < 0 ∧ d22 < 0 indicates that two eigenvaluesare smaller than λ. This in turns implies that P (n−1)(λ) has a negative sign, and P (n−2)(λ)has a positive sign. Then, one additional sign change has occurred in the sequence of signof the characteristic polynomials sign(P (n)(λ)), sign(P (n−1)(λ)), sign(P (n−2)(λ))=+,−, +. Ifd22 > 0, then P (n−1)(λ) and P (n−2)(λ) have the same sign, and no additional sign change isrecorded in the sequence of signs of the characteristic polynomials.

3.2 Eigenvalue Separation Principle 61

Proceeding in this way it is seen that the number of sign changes in the sequence of signssign(P (n)(λ)), sign(P (n−1)(λ)), . . . , sign(P (0)(λ)) determines the total number of eigenvaluessmaller than λ. This property of the sequence of characteristic polynomials is known as a Sturmsequence check. In Example 3.5 it is illustrated, how the sign of the components d11, d22, d33

for the case n = 3 can be retrieved from the sequence of signs of the Sturm sequence.

Example 3.4: Bounds on eigenvalues by eigenvalue separation principle

For the mass- and stiffness matrices defined in Example 1.4, the matrices M (1) and K(1) become

M(1) =

[12 00 1

], K(1) =

[2 −1

−1 4

](3–29)


det

([2 − 1

2λ(1)j −1

−1 4 − λ(1)j

])= 0 ⇒

{λ

(1)1 = 4 −√

2 = 2.59λ

(1)2 = 4 +

√2 = 5.41

(3–30)

The matrices M(2) and K(2) become

M(2) =[12

], K(2) =

[2]

⇒ λ(2)1 = 4 (3–31)

The relation (3-25) becomes

λ1 ≤ λ(1)1 ≤ λ2

λ(1)1 ≤ λ

(2)1 ≤ λ

(1)2

λ2 ≤ λ(1)2 ≤ λ3

⎫⎪⎪⎬⎪⎪⎭ ⇒

⎧⎪⎪⎨⎪⎪⎩

0 ≤ λ1 ≤ λ(1)1

λ(1)1 ≤ λ2 ≤ λ

(1)2

λ(1)2 ≤ λ3 ≤ ∞

⇒

⎧⎪⎨⎪⎩

0 ≤ λ1 ≤ 2.592.59 ≤ λ2 ≤ 5.415.41 ≤ λ3 ≤ ∞

(3–32)

The exact solutions are λ1 = 2, λ2 = 4, and λ3 = 6, cf. Example 1.4.

Example 3.5: Sturm sequences and correspondence to sign of components in D-matrix

Consider a generalized eigenvalue of order n = 3. For a given value of λ, the Sturm sequence P (3)(λ), P (2)(λ),


P (1)(λ), P (0)(λ) is calculated. Below are shown the 8 possible sign sequences of the Sturm sequence.

+

P (3)(λ) > 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

++

P (2)(λ) > 0 ⇒ d11 > 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+++

P (1)(λ) > 0 ⇒ d22 > 0

⎧⎪⎪⎨⎪⎪⎩

++++

P (0)(λ) > 0 ⇒ d33 > 0

+++−P (0)(λ) < 0 ⇒ d33 < 0

++−P (1)(λ) < 0 ⇒ d22 < 0

⎧⎪⎪⎨⎪⎪⎩

++−+

P (0)(λ) > 0 ⇒ d33 < 0

++−−P (0)(λ) < 0 ⇒ d33 > 0

+−P (2)(λ) < 0 ⇒ d11 < 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+−+

P (1)(λ) > 0 ⇒ d22 < 0

⎧⎪⎪⎨⎪⎪⎩

+−++

P (0)(λ) > 0 ⇒ d33 > 0

+−+−P (0)(λ) < 0 ⇒ d33 < 0

+−−P (1)(λ) < 0 ⇒ d22 > 0

⎧⎪⎪⎨⎪⎪⎩

+−−+

P (0)(λ) > 0 ⇒ d33 < 0

+−−−P (0)(λ) < 0 ⇒ d33 > 0

(3–33)

With an arbitrary positive value for P (3)(λ) the first curly bracket indicate how the sign of d11 is retrieved fromthe possible signs of P (2)(λ). The sign sequences ++ and +− have been indicated atop of P (2)(λ). At thenext level the sign of P (1)(λ) in combination to the previous sign sequence makes it possible to retrieve thesign of d22. Finally, at the 3rd level the sign of the characteristic polynomial P (0)(λ) in combination to theprevious sign sequence makes it possible to retrieve the sign of d33. As an example, the sequence of signssign(P (3)(λ)), sign(P (2)(λ)), sign(P (1)(λ)), sign(P (0)(λ))=+ − +− are obtained for the specific sign combi-nation d11 < 0, d22 < 0 and d33 < 0. Moreover, there are three sign changes in the indicated sign sequence+ − +−, and correspondingly all three components d11, d22 and d33 are negative. The reader is encouraged toverify, that the number of sign changes in the sequence of signs at the lowest level in (3-33) always is equal to thenumber of negative components in the specific combination of d 11, d22 and d33 producing this sequence of signs.

Example 3.6: Physical interpretation of the eigenvalue separation principle

FF

x

j

∆l

1 2µ

n − 1 n

u1 u2 u(x, t) uj uj+1 un−2 un−1

ω(0)1

ω(1)1

ω(1)2

ω(0)2

a)

b)

Fig. 3–1 Vibrating string. a) Definition of elements and degrees of freedom. b) Undamped eigenmodes.

3.3 Shift 63

The vibrating string problem in Example 1.2 is considered again. The eigenvibrations of the discretized string isgiven by (1-68) with M(0) = M and K(0) = K given by (1-69) or (1-70).

Next, consider the system defined by the matrices M(1) and K(1) of dimension (n − 2) × (n − 2), where thelast row and column are omitted in M(0) and K(0). Physically, this corresponds to constraining the displacementun−1(t) = 0, as indicated by the additional support in Fig. 1.3b. The corresponding eigenmodes of the continuoussystem have been shown with a dashed signature. As seen in Fig. 1.3b the wave-lengths related to the circulareigenfrequencies ω

(0)1 , ω

(1)1 , ω

(0)2 and ω

(1)2 decreases in the indicated order. Hence, the following ordering of these

eigenfrequencies prevails

ω(0)1 < ω

(1)1 < ω

(0)2 < ω

(1)2 (3–34)

Since λ(m)j =

(ω

(m)j

)2, the corresponding ordering of the eigenvalues become

λ(0)1 < λ

(1)1 < λ

(0)2 < λ

(1)2 (3–35)

which corresponds to (3-25).

3.3 Shift

Occasionally, a shift on the stiffness matrix may be used to enhance the speed of calculationof the considered GEVP. In order to explain this the eigenvalue problem (1-9) is written in thefollowing way

(K− ρM + ρM − λjM

)Φ(j) = 0 (3–36)

Obviously, we have withdrawn and added the quantity ρM inside the bracket, where ρ is asuitable real number, which will not affect neither the eigenvalues λj , nor the eigenvectors Φ(j).(3-36) is rearranged on the form

(K− µjM

)Φ(j) = 0 (3–37)

where

K = K − ρM , µj = λj − ρ (3–38)

Hence, instead of the original generalized eigenvalue problem defined by the matrices(K,M

),

the system with the matrices(K,M

)is considered in the shifted system, where K is calculated

as indicated in (3-38). The two systems have identical eigenvectors. However, the eigenvaluesof the shifted system become (λ1 − ρ), (λ2 − ρ), . . . , (λn − ρ), where λ1, λ2, . . . , λn denote theeigenvalues of the original system.


For non-supported systems (e.g. ships and aeroplanes) a stiff-body motion Φ �= 0 exists, whichfulfills

KΦ = 0 (3–39)

(3-39) shows that λ = 0 is an eigenvalue for such systems. Correspondingly, det(K) = 0 forsystems, which possesses a stiff-body motion. However, some numerical algorithms presumethat det(K) �= 0. In such cases a preliminary shift on the stiffness matrix must be performed,because det(K − ρM) �= 0, if det(K) = 0.

Example 3.7: Shift on stiffness matrix

Given the mass- and stiffness matrices

M =

[2 11 2

], K =

[3 −3

−3 3

](3–40)


det

([3 − 2λ −3 − λ

−3 − λ 3 − 2λ

])= 0 ⇒

{λ1 = 0λ2 = 6

(3–41)

λ1 = 0, since det(K) = 0.

Next, a shift on the stiffness matrix with ρ = −2 is performed, which provides

K =

[3 −3

−3 3

]+ 2

[2 11 2

]=

[7 −1

−1 7

](3–42)

Now, the characteristic equation becomes

det

([7 − 2µ −1 − µ

−1 − µ 7 − 2µ

])= 0 ⇒

{µ1 = 2µ2 = 8

(3–43)

3.4 Transformation of GEVP to SEVP 65

3.4 Transformation of GEVP to SEVP

Some eigenvalue solvers are written for the special eigenvalue problem. Hence, their use pre-sumes an initial transformation of the generalized eigenvalue problem (1-9). Of course, thismay be performed, simply by a pre-multiplication of (1-9) with M−1. However, then the result-ing system matrix M−1K is no longer symmetric. In this section a similarity transformation isindicated, which preserves the symmetry of the system matrix.

Since, M = MT it can be factorized on the form

M = SST (3–44)

The generalized eigenvalue problem (1-9) may then be written in the form

K(ST)−1

STΦ(j) = λjSST Φ(j) ⇒

S−1K(S−1)T

STΦ(j) = λj STΦ(j) (3–45)

where the identity(ST)−1

=(S−1)T

has been used. (1-9) can then be formulated in terms ofthe following standard EVP

KΦ(j) = λjΦ(j) (3–46)

where

K = S−1K(S−1)T

(3–47)

Φ(j) = ST Φ(j)

Φ(j) =(S−1)T

Φ(j)

⎫⎬⎭ (3–48)

(3-47) defines a similarity transformation with the transformation matrix S−1, which diagonal-ize the mass matrix. Similarity transformations is further explained in Chapter 6. Obviously,K = KT . As seen from (3-46) the eigenvalues λ1, . . . , λn are identical for the original andthe transformed eigenvalue problem, whereas the eigenvectors Φ(j) and Φ(j) are related by thetransformation (3-48).

The determination of a matrix S fulfilling (3-44) is not unique. Actually, infinite many solutionsto this problem exist. Below, two approaches have been given. In both cases it is assumed thatM = MT is positive definite.


Generally, Choleski decomposition is considered the most effective way of solving the problem.In this case a lower triangular matrix S is determined, so (3-44) is fulfilled. Obviously, S isrelated to the Gauss factorization as follows

S = LD12 , D

12 =

⎡⎢⎢⎢⎣√

d11 0 · · · 0

0√

d22 · · · 0...

.... . .

...0 0 · · · √

dnn

⎤⎥⎥⎥⎦ (3–49)

The diagonal matrix D12 does only exist, if the components dii of the matrix D are all positive.

This is indeed the case, if M is positive definite. Although, S may be calculated from (3-49),there exists a faster and more direct algorithm for the determination of this quantity.

Alternatively, a so-called spectral decomposition of M may be used. The basis of this methodis the following SEVP for M

Mv(j) = ρjv(j) (3–50)

ρj and v(j) denotes the jth eigenvalue and eigenvector of M. Both are real, since M is symmet-ric. The eigenvalue problems (3-50) can be assembled into the matrix formulation, cf. (1-14)

Mµ = VR (3–51)

µ =

⎡⎢⎢⎢⎣µ1 0 · · · 0

0 µ2 · · · 0...

.... . .

...0 0 · · · µn

⎤⎥⎥⎥⎦ , V = [v(1) v(2) · · ·v(n)] (3–52)

The eigenvectors are normalized to magnitude 1, i.e. v(i) Tv(j) = δij . Then, the modal matrixV fulfills, cf. (1-23)

V−1 = VT (3–53)

From (3-51) and (3-53) the following representation of M is obtained

M = VµVT (3–54)

Finally, from (3-44) and (3-54) the following solution for S is obtained

3.4 Transformation of GEVP to SEVP 67

S = Vµ12 , µ

12 =

⎡⎢⎢⎢⎣√

µ1 0 · · · 0

0√

µ2 · · · 0...

.... . .

...0 0 · · · √

µn

⎤⎥⎥⎥⎦ (3–55)

The drawback of the spectral approach is that an initial SEVP must be solved, before the trans-formed eigenvalue problem (3-46) can be analyzed. Hence, the method requires the solution oftwo SEVP of the same dimension.

Box 3.2: Choleski decomposition of symmetric positive definite matrix

Choleski decomposition factorizes a symmetric positive definite matrix M into the matrixproduct of a lower triangular matrix S and its transpose, as follows

M = SST ⇒⎡⎢⎢⎢⎣

m11 m21 · · · mn1

m21 m22 · · · mn2...

.... . .

...mn1 mn2 · · · mnn

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

s11 0 · · · 0

s21 s22 · · · 0...

.... . .

...sn1 sn2 · · · snn

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

s11 s21 · · · sn1

0 s22 · · · sn2...

.... . .

...0 0 · · · snn

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

s211

s21s11 s222 + s2

21 symmetric...

.... . .

sn1s11 sn2s22 + sn1s21 · · · s2nn + s2

n,n−1 + · · ·+ s2n2 + s2

n1

⎤⎥⎥⎥⎦ (3–56)

Equating the components of the final matrix product with the component on and belowthe main diagonal of M, equations can be formulated for the determinations of sij , whichare solved sequentially. First s11 =

√m11 is calculated. Next, si1, i = 2, . . . , n are

determined from si1 = mi1/s11. Next, s22 =√

m22 − s221 is calculated, and si2, i =

3, . . . , n can be determined from si2 = (m2i − si1s21)/s22. Next, the 3th column can becalculated and so forth. The general algorithm for calculating the components sij in thejth column reads

sjj =√

mjj − s2j,j−1 − · · · − s2

j1 , j = 1, . . . , n

sij = (mij − si,j−1sj,j−1 − · · · − si1sj1)/sjj , i = j + 1, . . . , n(3–57)


3.5 Exercises

3.1 Given the same mass- and stiffness matrices as in Exercise 1.1.

(a.) Show that the eigenvalue separation principle is valid for the considered example.


M =

[2 0

0 0

], K =

[6 −1

−1 4

]

(a.) Calculate the eigenvalues and eigenmodes normalized to unit modal mass.(b.) Perform a shift ρ = 3 on K and calculate the eigenvalues and eigenmodes of the new

problem.

3.3 Given a symmetric matrix K.

(a.) Write a MATLAB program, which determines the matrices L and D of a Gauss fac-torization as well as the matrix (S−1)T , where S is a lower triangular matrix fulfillingSST = K.

3.4 Given a symmetric positive definite matrix K.

(a.) Write a MATLAB program, which performs Choleski decomposition.

CHAPTER 4APPROXIMATE SOLUTION METHODS

This chapter deals with various approximate solution methods for solving the generalized eigen-value problem.

Section 4.1 consider the application of static condensation or Guyan reduction.1 The idea ofthe method is to reduce the magnitude of the generalized eigenvalue problem from n to n1 ndegrees of freedom. Next, the reduced system is solved exact. In principle no approximation isrelated to the procedure.

Section 4.2 deals with the application of Rayleigh-Ritz analysis. Similar to static condensationthis is a kind of system reduction procedure. As shown the method can be given a formulationidentical to static condensation. However, exact results are no longer obtained.

Section 4.3 deals with the bounding of the error related to a certain approximate eigenvalue.

4.1 Static Condensation

The basic assumption of static condensation is that inertia is confined to the first n1 degrees offreedom, whereas inertia effects are ignored for the remaining n2 = n−n1 degrees of freedom.The approximation of the method stems from the ignorance of these inertial couplings. Thiscorresponds to the following partitioning of the mass and stiffness matrices

M =

[M11 0

0 0

], K =

[K11 K12

K21 K22

](4–1)

M11 and K11 are sub-matrices of dimension n1 × n1, K12 = KT21 is a sub-matrix of dimension

n1 × n2, and K22 is of the dimension n2 × n2. The eigenvalue problems for the first n1 andthe last n2 eigenvectors can be assembled in the following partitioned matrix formulations, cf.(1-14)

1S.R.K. Nielsen: Structural Dynamics, Vol. 1. Linear Structural Dynamics, 4th Ed. Aalborg tekniske Univer-sitetsforlag, 2004.

— 69 —

70 Chapter 4 – APPROXIMATE SOLUTION METHODS

[K11 K12

K21 K22

][Φ11

Φ21

]=

[M11 0

0 0

][Φ11

Φ21

]Λ1

[K11 K12

K21 K22

][Φ12

Φ22

]=

[M11 0

0 0

][Φ12

Φ22

]Λ2

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭

(4–2)

where Λ1 and Λ2 are diagonal matrices of the dimension n1 × n1 and n2 × n2

Λ1 =

⎡⎢⎢⎢⎣

λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ , Λ2 =

⎡⎢⎢⎢⎣λn1+1 0 · · · 0

0 λn1+2 · · · 0...

.... . .

...0 0 · · · λn

⎤⎥⎥⎥⎦ (4–3)

Φ(j)1 and Φ

(j)2 denote sub-vectors encompassing the first n1 and the last n2 components of the

jth eigenmode Φ(j). Then, the matrices Φ11, Φ12, Φ21 and Φ22 entering (4-2) are defined as

Φ11 =[Φ

(1)1 Φ

(2)1 · · ·Φ(n1)

1

], Φ12 =

[Φ

(n1+1)1 Φ

(n1+2)1 · · ·Φ(n)

1

]Φ21 =

[Φ

(1)2 Φ

(2)2 · · ·Φ(n1)

2

], Φ22 =

[Φ

(n1+1)2 Φ

(n1+2)2 · · ·Φ(n)

2

]⎫⎬⎭ (4–4)

At first the solution for the first n1 eigenmodes is considered. From the lower lower half of thefirst matrix equation in (4-2) follows

K21Φ11 + K22Φ21 = 0 ⇒Φ21 = −K−1

22 K21Φ11 (4–5)

From the corresponding upper half of the said matrix equation, and (4-5), follows

K11Φ11 − K12K−122 K21Φ11 = M11Φ11Λ1 ⇒

K11Φ11 = M11Φ11Λ1 (4–6)

where

K11 = K11 − K12K−122 K21 (4–7)

4.1 Static Condensation 71

(4-6) is a generalized eigenvalue problem of reduced dimension n1, which is solved for(Λ1,Φ11

).

Next, the remaining components of the first n1 eigenmodes are calculated from (4-5). Themodal masses become

m1 =

[Φ11

Φ21

]T [M11 0

0 0

][Φ11

Φ21

]= ΦT

11M11Φ11 (4–8)

Hence, the total eigenmodes will be normalized to unit modal mass with respect to M, if thesub-vectors Φ11 are normalized to unit modal mass with respect to M11.

Next, the solution for the last n2 eigenmodes are considered. From the last matrix equation in(4-2) follows

[K11 K12

K21 K22

][Φ12

Φ22

]Λ−1

2 =

[M11Φ12

0

](4–9)

Obviously, (4-9) is fulfilled for Λ−12 = 0 ∧ Φ12 = 0. Λ−1

2 = 0 implies that all n2 eigenvaluesare equal to infinity. Hence, the following eigensolutions are obtained

Λ2 =

⎡⎢⎢⎢⎣∞ 0 · · · 0

0 ∞ · · · 0...

.... . .

...0 0 · · · ∞

⎤⎥⎥⎥⎦ ,

[Φ12

Φ22

]=

[0

Φ22

](4–10)

The matrix Φ22 is undetermined. Any matrix with linear independent column vectors will do.Then, this quadratic matrix may simply be chosen as an n2 × n2 unit matrix

Φ22 = I (4–11)

The modal masses become

m2 =

[0

Φ22

]T [M11 0

0 0

][0

Φ22

]= 0 (4–12)

Generally, if the ith row and the ith column in M are equal to zero, then ΦT = [0, . . . , 0, 1, 0, . . . , 0]is an eigenvector with the eigenvalue λ = ∞. The modal mass is 0.

In praxis the calculation of K11 is solved by means of an initial Choleski decomposition of K22,cf. Box 3.2


K22 = SST ⇔ K−122 =

(S−1)T

S−1 (4–13)

where both S and S−1 are lower triangular matrices. Then, K11 is determined from

K11 = K11 − RTR (4–14)

where the n2 × n2 matrix R is obtained as solution to the matrix equation

SR = K21 (4–15)

In principle (4-15) represent n2 linear equations with n2 right-hand sides. Given that S is alower triangular matrix, this is relatively easily solved.

Finally, it should be noticed that the static condensation approach is only of value if n1 n.

Example 4.1: Static condensation


M =

⎡⎢⎢⎢⎣

0 0 0 00 2 0 00 0 0 00 0 0 1

⎤⎥⎥⎥⎦ , K =

⎡⎢⎢⎢⎣

2 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 1

⎤⎥⎥⎥⎦ (4–16)

The rows and columns are interchanged the mass and stiffness matrices, so the following eigenvalue problems areobtained

⎡⎢⎢⎢⎢⎣

2 0 −1 −10 1 −1 0

−1 −1 2 0−1 0 0 2

⎤⎥⎥⎥⎥⎦⎡⎢⎣Φ11

Φ21

⎤⎥⎦ =

⎡⎢⎢⎢⎢⎣

2 0 0 00 1 0 0

0 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎦⎡⎢⎣Φ11

Φ21

⎤⎥⎦⎡⎢⎣λ1 0

0 λ2

⎤⎥⎦

⎡⎢⎢⎢⎢⎣

2 0 −1 −10 1 −1 0

−1 −1 2 0−1 0 0 2

⎤⎥⎥⎥⎥⎦⎡⎢⎣Φ12

Φ22

⎤⎥⎦ =

⎡⎢⎢⎢⎢⎣

2 0 0 00 1 0 0

0 0 0 00 0 0 0

⎤⎥⎥⎥⎥⎦⎡⎢⎣Φ12

Φ22

⎤⎥⎦⎡⎢⎣λ3 0

0 λ4

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4–17)

A formal procedure for obtaining the mass and stiffness matrices in (4-17) by means of a similarity transformationhas been demonstrated in Example 6.1. Then,

K11 =

[2 00 1

], K12 = K21 =

[−1 −1−1 0

], K22 =

[2 00 2

], M11 =

[2 00 1

](4–18)

4.1 Static Condensation 73

From (4-7) follows

K11 = K11 − K12K−122 K21 =

[2 00 1

]−[−1 −1−1 0

][2 00 2

]−1 [−1 −1−1 0

]=

[1 − 1

2

− 12

12

](4–19)

The reduced eigenvalue problem (4-6) becomes

[1 − 1

2

− 12

12

]Φ11 =

[2 00 1

]Φ11Λ1 (4–20)

The eigensolutions with eigenmodes normalized to modal mass 1 become

Λ1 =

[λ1 00 λ2

]=

[12 −

√2

4 00 1

2 +√

24

], Φ11 =

[12 − 1

2√2

2

√2

2

](4–21)

From (4-5) follows

Φ21 = −[2 00 2

]−1 [−1 −1−1 0

][12 − 1

2√2

2

√2

2

]=

[14 +

√2

4 − 14 +

√2

414 − 1

4

](4–22)

From (4-10) and (4-11) follows

Λ2 =

[λ3 00 λ4

]=

[∞ 00 ∞

], Φ12 =

[0 00 0

], Φ22 =

[1 00 1

](4–23)

After interchanging the degrees of freedom back to the original order (the 1st and 2nd components of Φ 11 and Φ12

are placed as the 2nd and 4th component of Φ (j), the 1st and 2nd components of Φ21 and Φ22 are placed as the3rd and 1st component Φ(j), the following eigensolution is obtained

Λ =

⎡⎢⎢⎢⎣

λ1 0 0 00 λ2 0 00 0 λ3 00 0 0 λ4

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

12 −

√2

4 0 0 00 1

2 +√

24 0 0

0 0 ∞ 00 0 0 ∞

⎤⎥⎥⎥⎦

Φ =[Φ(1) Φ(2) Φ(3) Φ(4)

]=

⎡⎢⎢⎢⎢⎣

14 − 1

4 0 112 − 1

2 0 014 +

√2

4 − 14 +

√2

4 1 0√

22

√2

2 0 0

⎤⎥⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4–24)


4.2 Rayleigh-Ritz Analysis

Consider the generalized eigenvalue problem (6-5). If M is positive definite, so vTMv > 0 forany v �= 0, the so-called Rayleigh quotient may be defined as

ρ(v) =vTKv

vTMv(4–25)

It can be proved that ρ(v) fulfills the bounding, see Box 4.1

λ1 ≤ ρ(v) ≤ λn (4–26)

where λ1 and λn denote the smallest and the largest eigenvalues of the generalized eigenvalueproblem.

Especially, if v = Φ(1), where Φ(1) has been normalized to unit modal mass, it follows thatΦ(1) TMΦ(1) = 1 and Φ(1) TKΦ(1) = λ1, see (4-34) and (4-35) below. Then, ρ(v) = λ1. Thisproperty is contained in the so-called Rayleigh’s principle

λ1 = minv∈Rn

ρ(v) (4–27)

Next, assume that v is M-orthogonal to Φ(1), so Φ(1) TMv = 0. Then, the following boundingof the Rayleigh quotient may be proved, see Box 4.1

λ2 ≤ ρ(v) ≤ λn (4–28)

Correspondingly, λ2 may be evaluated by the following extension of the Rayleigh principle,where the M-orthogonality of the test vector v to the first eigenmode Φ(1) has been included asa restriction

λ2 =

⎧⎨⎩

minv∈Rn

ρ(v)

Φ(1) TMv = 0

(4–29)

The corresponding optimal vector will be v = Φ(2).

Generally, if v is M-orthogonal to the first m − 1 eigenmodes Φ(1),Φ(2), . . . ,Φ(m−1), soΦ(j) TMv = 0 , j = 1, . . . , m − 1, the following bounding of the Rayleigh quotient maybe proved, see Box 4.1

λm ≤ ρ(v) ≤ λn , m < n (4–30)

4.2 Rayleigh-Ritz Analysis 75

Correspondingly, λm may be evaluated by the following extension of the Rayleigh variationalprinciple, where restriction of M-orthogonal of the test vector v to the eigenmodes Φ(j) =0 , j = 1, . . . , m − 1 are included

λm =

⎧⎨⎩

minv∈Rn

ρ(v)

Φ(j) TMv = 0 , j = 1, . . . , m − 1

(4–31)

The corresponding optimal vector will be v = Φ(m).

The Rayleigh quotient may be used to calculate an upper bound for the lowest eigenvalue λ1.The quality of the estimate depends on the choice of v. The better the qualitative and quantita-tive resemblance of v to the shape of the lowest eigenmode, the sharper will be the calculatedupper bound.

Box 4.1: Proof of boundings of the Rayleigh quotient

Given the linear independent eigenmodes, normalized to unit modal massΦ(1),Φ(2), . . . ,Φ(n). Using the eigenmodes as a vector basis, any n-dimensionalvector may be written as

v = q1Φ(1) + q2Φ

(2) + · · · + qnΦ(n) (4–32)

Insertion of (4-32) into (4-25) provides

ρ(v) =

n∑i=1

n∑j=1

qiqjΦ(i) TKΦ(j)

n∑i=1

n∑j=1

qiqjΦ(i) TMΦ(j)

=q21λ1 + q2

2λ2 + · · · + q2nλn

q21 + q2

2 + · · ·+ q2n

(4–33)

where the orthonormality conditions of the eigenmodes have been used in the last state-ment, i.e.

Φ(i) TMΦ(j) =

{0 , i �= j

1 , i = j(4–34)

Φ(i) TKΦ(j) =

{0 , i �= j

λi , i = j(4–35)


Given the following ordering of the eigenvalues

0 ≤ λ1 ≤ λ2 ≤ · · · ≤ λn−1 ≤ λn ≤ ∞ (4–36)

it follows directly from (4-33) that

ρ(v) ≥ q21λ1 + q2

2λ1 + · · ·+ q2nλ1

q21 + q2

2 + · · ·+ q2n

= λ1

ρ(v) ≤ q21λn + q2

2λn + · · ·+ q2nλn

q21 + q2

2 + · · ·+ q2n

= λn

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (4–37)

which proves the bounding (4-26).

(4-32) is pre-multiplied with Φ(j) TM. Then, use of (4-34) provides the following expres-sion for the jth modal coordinate

qj = Φ(j) TMv (4–38)

Hence, if v is M-orthogonal to Φ(j), j = 1, . . . , m − 1 it follows that q1 = q2 = · · · =qm−1 = 0. In this case (4-33) attains the form

ρ(v) =q2mλm + q2

m+1λm+1 + · · · + q2nλn

q2m + q2

m+1 + · · ·+ q2n

(4–39)

Proceeding as in (4-37) it then follows that

ρ(v) ≥ q2mλm + q2

m+1λm + · · · + q2nλm

q2m + q2

m+1 + · · ·+ q2n

= λm

ρ(v) ≤ q2mλn + q2

m+1λn + · · · + q2nλn

q2m + q2

m+1 + · · · + q2n

= λn

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

(4–40)

which proves the bounding (4-30).

The so-called Ritz analysis m linearly independent base vectors, Ψ(1), . . . ,Ψ(m), are defined,which span an m-dimensional subspace Vm ⊆ Vn. Often the base vectors are determined asthe static deflections from m linearly independent load vectors f1, . . . , fm. This is preferred, be-cause it often is simpler to specify static load, which will produce displacements qualitatively inagreement with the eigenmodes to be determined by the analysis. The Ritz-basis is determinedfrom the equilibrium equation


KΨ = f ⇒ Ψ = K−1f (4–41)

Ψ =[Ψ(1) Ψ(2) · · ·Ψ(m)

], f = [f1 f2 · · · fm] (4–42)

Then, any vector v ∈ Vm can be written on the form

v = q1Ψ(1) + q2Ψ

(2) + · · ·+ qmΨ(m) =[Ψ(1) Ψ(2) · · ·Ψ(m)

]⎡⎢⎢⎢⎣

q1

q2...

qm

⎤⎥⎥⎥⎦ = Ψq , q =

⎡⎢⎢⎢⎣

q1

q2...

qm

⎤⎥⎥⎥⎦

(4–43)

The idea in Ritz analysis is to insert (4-43) into the Rayleigh quotient (4-25), and determinethe modal coordinates q1, q2, . . . , qm, which minimizes this quantity. Hence, the following re-formulation of the Rayleigh quotient is considered

ρ(q) =

(Ψq)T

KΨq(Ψq)T

MΨq=

qT Kq

qTMq(4–44)

where

M = ΨTMΨ = [Mij ] , Mij = Ψ(i) TMΨ(j)

K = ΨTKΨ = [Kij ] , Kij = Ψ(i) TKΨ(j)

}(4–45)

M and K are denoted as the projected mass- and stiffness matrices on the subspace spanned bythe Ritz basis Ψ.

The approximation to λ1 then follows from (4-27)

λ1 ≤ ρ1 = minq∈Vm

ρ(q) = minq1,...,qm

m∑i=1

m∑j=1

qiKijqj

m∑i=1

m∑j=1

qiMijqj

(4–46)

Generally, ρ1 is larger than λ1 in agreement with (4-26). Only for Φ(1) ∈ Vm will modal coor-dinates q, . . . , qm exist, so Φ(1) = q1Ψ

(1) + · · ·+ qmΨ(m), with the implication that ρ1 = λ1.

The necessary condition for a minimum is that


∂

∂qi

(qT Kq

qTMq

)= 0 , i = 1, . . . , m ⇒

qTMq · ∂∂qi

(qT Kq

)− qT Kq · ∂∂qi

(qT Mq

)(qT Mq

)2 = 0 ⇒

∂

∂qi

(qT Kq

)− ρ∂

∂qi

(qTMq

)= 0 (4–47)

Now, ∂∂qi

(qT Kq

)= ∂

∂qi

∑mj=1

∑mk=1 qjKjkqk = 2

∑mk=1 Kikqk, where the symmetry property,

Kjk = Kkj (K = KT ), has been applied. Similarly, ∂∂qi

(qTMq

)= 2

∑mk=1 Mikqk. Then, the

minimum condition (4-47) reduces to

m∑j=1

Kijqj − ρm∑

j=1

Mijqj = 0 (4–48)

From (4-48) follows that ρ1 is determined as the lowest eigenvalue to the following generalizedeigenvalue problem of dimension m

Kq − ρMq = 0 (4–49)

(4-49) has m eigensolutions (ρi,q(i)), i = 1, . . . , m. ρi becomes an approximation to the ith

eigenvalue λi. The corresponding approximation to the ith eigenmode is calculated from

Φ(i) = q1,iΨ(1) + · · · + qm,iΨ

(m) = Ψq(i) , i = 1, . . . , m (4–50)

where q1,i, . . . , q1m,i denote the components of q(i).

The relations (4-50) can be assembled into the matrix equation

Φ = ΨQ (4–51)

Φ =[Φ(1) Φ(2) · · · Φ(m)

], Q =

[q(1) q(2) · · ·q(m)

](4–52)

We shall assume that the eigenvectors q(i) are normalized to unit modal mass with respect tothe projected mass matrix, i.e. the following orthonormality properties are fulfilled

q(i) TMq(j) =

{0 , i �= j

1 , i = j(4–53)


q(i) T Kq(j) =

{0 , i �= j

ρi , i = j(4–54)

Then, the modal mass of the eigenmodes Φ become

ΦTMΦ = (ΨQ)TMΨQ = QTMQ = I (4–55)

Hence, the approximate eigenmodes Φ(i) will be normalized to unit modal mass, if this is thecase for the eigenvectors q(i) with respect to the projected mass matrix. Φ forms an alternativeRitz-basis in V m, which in addition is M-orthonormal. Similarly, the approximate eigenmodesare K-orthogonal as follows

ΦTKΦ = (ΨQ)TKΨQ = QT KQ = R (4–56)

where R is m-dimensional diagonal matrix with the eigenvalues ρ1, . . . , ρm in the main diago-nal.

Obviously, the Rayleigh quotient approach corresponds to m = 1. Hence, Ritz analysis ismerely a multi-dimensional generalization, for which reason the name Rayleigh-Ritz analysishas been coined for the method.

As a generalization to (4-26) the following boundings can be proved2

λ1 ≤ ρ1 , λ2 ≤ ρ2 , . . . , λm ≤ ρm ≤ λn (4–57)

2K.-J. Bathe: Finite Element Procedures. Printice Hall, Inc., 1996.


Box 4.2: Rayleigh-Ritz algorithm

1. Estimate m linearly independent static load vectors f1, . . . , fm, assembled column-wise in the n × m matrix f = [f1 f2 · · · fm].

2. Calculate the Ritz basis from Ψ = K−1f , Ψ =[Ψ(1) Ψ(1) · · ·Ψ(m)

].

3. Calculate projected mass and stiffness matrices in the m-dimensional subspacespanned by the Ritz basis: M = ΨTMΨ , K = ΨTKΨ.

4. Solve the generalized eigenvalue problem of dimension m: KQ = MQR.

5. Determine approximations to the lowest m eigenvector from the transformation Φ =ΨQ , Φ =

[Φ(1) Φ(2) · · · Φ(m)

]. The corresponding approximate eigenvalues are

contained in the main diagonal of R.

Returning to the static condensation problem in Section 4.1, let us define a Ritz basis of thedimension m = n1 as

Ψ1 =

⎡⎣ I

−K−122 K21

⎤⎦ (4–58)

where I is a unit matrix of dimension n1 × n1. Given the structure of the mass and stiffnessmatrices in (4-1), we may then evaluate the following projected matrices

M = ΨT1 MΨ1 =

⎡⎣ I

−K−122 K21

⎤⎦T ⎡⎣M11 0

0 0

⎤⎦⎡⎣ I

−K−122 K21

⎤⎦ = M11 (4–59)

K = ΨT1 KΨ1 =

⎡⎣ I

−K−122 K21

⎤⎦T ⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣ I

−K−122 K21

⎤⎦ = K11 − K12K

−122 K21 = K11

(4–60)

Hence, (4-49) reduce to the generalized eigenvalue problem (4-6), with Q = Φ11, and R = Λ1.Consequently, static condensation may be interpreted as merely a Rayleigh-Ritz analysis withthe Ritz basis (4-58).

The following identity may be proved by insertion

⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣ I

−K−122 K21

⎤⎦(K11 − K12K

−122 K21

)−1

=

⎡⎣I

0

⎤⎦ (4–61)


Then, we may construct an alternative Ritz basis from (4-41) with the static load given as theright hand side of (4-61), i.e.

Ψ2 = K−1f =

⎡⎣K11 K12

K21 K22

⎤⎦−1 ⎡⎣I

0

⎤⎦ =

⎡⎣ I

−K−122 K21

⎤⎦(K11 − K12K

−122 K21

)−1

= Ψ1K−111

(4–62)

Hence, the base vectors in Ψ2 is a linear combination of the base vectors in Ψ1. Then, Ψ1 andΨ2 span the same subspace Vn1, for which reason both bases will determine the same eigenval-ues and eigenvectors.

The projected mass and stiffness matrices become

M = ΨT2 MΨ2 = K−1

11 ΨT1 MΨ1K

−111 = K−1

11 M11K−111 (4–63)

K = ΨT2 KΨ2 = K−1

11 ΨT1 KΨ1K

−111 = K−1

11 (4–64)

Then, the modal matrices Q1 and Q2, obtained as solutions to (4-49) for the respective Ritzbases, are seen to be related as

Q1 = K−111 Q2 (4–65)

(4-65) follows from Φ = Ψ1Q1 = Ψ2Q2.

Example 4.2: Rayleigh-Ritz analysis


M =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ (4–66)

which have the exact eigensolutions, cf. Example 1.5

Λ =

⎡⎢⎣λ1 0 0

0 λ2 0

0 0 λ3

⎤⎥⎦ =

⎡⎢⎣2 0 0

0 4 0

0 0 6

⎤⎥⎦ , Φ =

[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣

√2

2 −1√

22√

22 0 −

√2

2√2

2 1√

22

⎤⎥⎦ (4–67)

A two dimensional Rayleigh-Ritz analysis is performed, where the static load vectors are estimated as

f =

⎡⎢⎣1 00 00 1

⎤⎥⎦ (4–68)


The Ritz basis becomes

Ψ =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣1 0

0 00 1

⎤⎥⎦ =

112

⎡⎢⎣7 1

2 21 7

⎤⎥⎦ (4–69)


M =1

144

[29 1111 29

], K =

112

[7 11 7

](4–70)


R =

[ρ1 00 ρ2

]=

[2.4 00 4

], Q =

[q(1) q(2)

]=

[3√5

23√5

−2

](4–71)

The solutions for the eigenvectors become

Φ =[Φ(1) Φ(2)

]=

112

⎡⎢⎣7 1

2 2

1 7

⎤⎥⎦[

3√5

23√5

−2

]=

⎡⎢⎢⎣

2√5

11√5

02√5

−1

⎤⎥⎥⎦ (4–72)

As seen from (4-71) and (4-72) ρ2 = 4 and Φ(2) are calculated exact, cf. (1-86). This is so, because Φ (2) is placedin the subspace spanned by the selected Ritz basis as seen from the expansion

Φ(2) = 2Ψ(1) − 2Ψ(2) =212

⎡⎢⎣721

⎤⎥⎦− 2

12

⎡⎢⎣127

⎤⎥⎦ =

⎡⎢⎣ 1

0−1

⎤⎥⎦ (4–73)

Next, a new analysis is performed, where the static load vectors are estimated as

f =

⎡⎢⎣1 01 11 0

⎤⎥⎦ (4–74)

The Ritz basis becomes

Ψ =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣1 0

1 11 0

⎤⎥⎦ =

16

⎡⎢⎣5 14 25 1

⎤⎥⎦ (4–75)


4.3 Error Analysis 83

M =136

[41 1313 5

], K =

13

[7 22 1

](4–76)


R =

[ρ1 00 ρ2

]=

[2 00 6

], Q =

[q(1) q(2)

]=

[√2

2 − 3√

22√

22

9√

22

](4–77)

The solutions for the eigenvectors become

Φ =[Φ(1) Φ(2)

]=

16

⎡⎢⎣5 1

4 2

5 1

⎤⎥⎦[√

22 − 3

√2

2√2

29√

22

]=

⎡⎢⎣

√2

2 −√

22√

22

√2

2√2

2 −√

22

⎤⎥⎦ (4–78)

In this case(ρ1, Φ(1)

)=(λ1,Φ(1)

)and(ρ2, Φ(2)

)=(λ3,Φ(3)

). This is so, because Φ(1) and Φ(3) are placed

in the sub-space spanned by the selected Ritz basis.

4.3 Error Analysis

Given a certain approximation to the jth eigen-pair (λj, Φ(j)), the error vector is defined as

εj =(K − λjM

)Φ(j) (4–79)

Presuming that the eigenvectors have been normalized to unit modal mass, it follows from (1-19) and (1-21) that

M =(Φ−1

)TIΦ−1 , K =

(Φ−1

)TΛΦ−1 (4–80)

Insertion of (4-80) into (4-79) provides

εj =(Φ−1

)T(Λ− λjI

)Φ−1Φ(j) ⇒

Φ(j) = Φ(Λ − λjI

)−1

ΦT εj (4–81)

We shall use the Euclidean vector norm ‖·‖E and the Hilbert matrix norm ‖·‖H in the following.For a definition of these quantities, see Box 4.3. The mentioned norms are compatible, so

∥∥Φ(j)∥∥

E≤∥∥∥Φ(Λ−λjI

)−1

ΦT∥∥∥

H‖εj‖E ≤ ∥∥Φ∥∥

H

∥∥∥(Λ−λjI)−1∥∥∥

H

∥∥ΦT∥∥

H‖εj‖E (4–82)


The last statement of (4-82) follows from the defining properties of matrix norms, see Box 4.3.

(Λ−λjI) is a diagonal matrix. Then, (Λ− λjI)−1 is also a diagonal matrix with the components

(λk − λj)−1, k = 1, . . . , n in the main diagonal. The eigenvalues of a diagonal matrix is equal

to the components in the main diagonal. Since, the Hilbert norm of a symmetric matrix is equalto the numerical largest eigenvalue, it follows that

∥∥∥(Λ − λjI)−1∥∥∥

H= max

k=1,...,n

(1

|λk − λj |)

=1

mink=1,...,n

|λk − λj|(4–83)

The Hilbert norms of Φ and ΦT are identical as stated in Box 4.3. Further, it can be shown that,see Box 4.4

‖Φ‖2H =

1

µ1(4–84)

where µ1 is the lowest eigenvalue of M.

Then, (4-82), (4-83) and (4-84) provides the following bounding of the calculated eigenvalueλj

mink=1,...,n

|λk − λj | ≤ 1

µ1

‖εj‖E

‖Φ(j)‖E

=1

µ1

|εj||Φ(j)| (4–85)

(4-85) is only of value, if µ1 can be calculated relatively easily. This is the case for the specialeigenvalue problem, where M = I, which means that µ1 = · · · = µn = 1, so (4-85) reduces to

mink=1,...,n

|λk − λj | ≤ ‖εj‖E

‖Φ(j)‖E

=|εj ||Φ(j)| (4–86)


Box 4.3: Vector and matrix norms

A vector norm is a real number ‖v‖ associated to any n-dimensional vector v, whichfulfills the following conditions

1. ‖v‖ > 0 for v �= 0 and ‖0‖ = 0.

2. ‖cv‖ = |c| · ‖v‖ for any complex or real number c.

3. ‖u + v‖ ≤ ‖u‖ + ‖v‖ (triangle inequality).

The most common vector norms are

1. p-norm (p ∈]0,∞[): ‖v‖p =( n∑

i=1

|vi|p)1/p

.

2. One norm (p = 1): ‖v‖1 =n∑

i=1

|vi|.

3. Two norm (p = 2, Euclidean norm): ‖v‖2 = |v| =( n∑

i=1

|vi|2)1/2

.

4. Infinity norm (p = ∞): ‖v‖∞ = maxi=1,...,n

|vi|.

where vi denotes the components of v. Given

v =

⎡⎢⎣ 1

−3

2

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

‖v‖1/2 =(√

1 +√

3 +√

2)2

= 17.19

‖v‖1 =(1 + 3 + 2

)= 6

‖v‖2 =(12 + 32 + 22

)1/2= 3.74

‖v‖∞ = max(1, 3, 2) = 3

(4–87)

A matrix norm is a real number ‖A‖ associated to any n × n matrix A, which fulfills thefollowing conditions

1. ‖A‖ > 0 for A �= 0 and ‖0‖ = 0.

2. ‖cA‖ = |c| · ‖A‖ for any complex or real number c.

3. ‖A + B‖ ≤ ‖A‖ + ‖B‖ (triangle inequality).

4. ‖AB‖ ≤ ‖A‖‖B‖.


The most common matrix norms are

1. One norm: ‖A‖1 = maxj=1,...,n

n∑i=1

|aij|.

2. Infinity norm: ‖A‖∞ = maxi=1,...,n

n∑j=1

|aij |.

3. Euclidean norm: ‖A‖E =( n∑

i=1

n∑j=1

a2ij

)1/2

.

4. Hilbert norm (spectral norm): ‖A‖H =(

maxi=1,...,n

λi

)1/2

, where λi is the ith eigen-

value of AAT identical to the eigenvalues of ATA, so ‖A‖H = ‖AT‖H .

aij denotes the components of A. Notice, if A = AT the eigenvalues of AAT = A2

becomes equal to the square of the eigenvalues of A. Given

A =

[2 −5

3 −1

]⇒ AAT =

[29 11

11 10

]⇒

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

‖A‖1 = max(2 + 3, 5 + 1) = 6

‖A‖∞ = max(2 + 5, 3 + 1) = 7

‖A‖E =(4 + 25 + 9 + 1

)1/2= 6.24

‖A‖H =

√13

2

(3 +

√5)

= 5.83

(4–88)

A matrix norm ‖ · ‖m is said to be compatible to a given vector norm ‖ · ‖v , if

‖Av‖v ≤ ‖A‖m · ‖v‖v (4–89)

It can be shown that the Hilbert matrix norm is compatible to the Euclidean vector norm,that the one matrix norm is compatible to the one vector norm, and that the infinity matrixnorm is compatible to the infinity vector norm. However, the Euclidean matrix norm isnot compatible to the Euclidean vector norm.


Box 4.4: Hilbert norm of modal matrix

Presuming that the columns of the modal matrix have been normalized to unit modalmass, so m = I, it follows from (1-19) that

M =(ΦT)−1

Φ−1 ⇒ M−1 = ΦΦT (4–90)

From the definition of the Hilbert norm in Box 4.3 and (4-90) follows that ‖Φ‖2H becomes

equal to the maximum eigenvalue of M−1. If µ1, µ2, . . . , µn denote the eigenvaluesof M in ascending order, then the eigenvalues in ascending order of M−1 become1

µn, . . . , 1

µ2, 1

µ1, so the maximum eigenvalue of M−1 is equal to 1

µ1. This proves (4-84).

Example 4.3: Bound on calculated eigenvalue

Given the mass- and stiffness matrices for the following special eigenvalue problem

M =

⎡⎢⎣1 0 0

0 1 00 0 1

⎤⎥⎦ , K =

⎡⎢⎣ 3 −1 0−1 2 −1

0 −1 3

⎤⎥⎦ (4–91)

The eigensolutions with modal masses normalized to 1 are given as

Λ =

⎡⎢⎣λ1 0 0

0 λ2 0

0 0 λ3

⎤⎥⎦ =

⎡⎢⎣1 0 0

0 3 0

0 0 4

⎤⎥⎦ , Φ =

[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣

√6

6

√2

2

√3

32√

66 0 −

√3

3√6

6 −√

22

√3

3

⎤⎥⎦ (4–92)

Assume that the following approximate solution , ( λ2, Φ(2)), has been calculated to the 2nd eigen-pair (λ2,Φ(2))

λ2 = 3.1 , Φ(2) =

⎡⎢⎣ 1.0

0.2−1.0

⎤⎥⎦ ⇒ ∣∣Φ(2)

∣∣ = 1.4283 (4–93)

Then, the error vector becomes, cf. (4-79)

ε2 =

⎛⎜⎝⎡⎢⎣ 3 −1 0−1 2 −1

0 −1 3

⎤⎥⎦− 3.1

⎡⎢⎣1 0 00 1 00 0 1

⎤⎥⎦⎞⎟⎠⎡⎢⎣ 1.0

0.2−1.0

⎤⎥⎦ =

⎡⎢⎣−0.30−0.22−0.10

⎤⎥⎦ ⇒ ∣∣ε2

∣∣ = 0.3852 (4–94)

Since, M = I we may use the simplified result (4-86), which provides

|λ2 − λ2| ≤ 0.38521.4283

= 0.26971 (4–95)

Actually, |λ2 − λ2| = 0.1.


4.4 Exercises


M =

⎡⎢⎣0 0 0

0 2 1

0 1 1

⎤⎥⎦ , K =

⎡⎢⎣ 6 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦

(a.) Perform a static condensation by the conventional procedure based on (4-5), (4-6),and next by a Rayleigh-Ritz analysis with the Ritz basis given by (4-62).


M =

⎡⎢⎣2 0 0

0 2 1

0 1 1

⎤⎥⎦ , K =

⎡⎢⎣ 6 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦

(a.) Calculate approximate eigenvalues and eigenmodes by a Rayleigh-Ritz analysis us-ing the following Ritz basis

Ψ = [Ψ(1) Ψ(2)] =

⎡⎢⎣1 1

1 −1

1 1

⎤⎥⎦

4.3 Consider the mass- and stiffness matrices in Exercise 4.2, and let

v =

⎡⎢⎣1

1

1

⎤⎥⎦

(a.) Calculate the vector Φ(1) = K−1Mv, and next λ1 = ρ(Φ(1)

), as approximate solu-

tions to the lowest eigenmode and eigenvalue.(b.) Establish the error bound for the obtained approximation to the lowest eigenvalue.

CHAPTER 5VECTOR ITERATION METHODS

5.1 Introduction

In structural dynamics only a small number n1 of the lowest eigen-pairs,(λ1,Φ

(1)),(λ2,Φ

(2)),

. . . ,(λn1 ,Φ

(n1)), where n1 n, are of structural significance. Hence, there is a need for meth-

ods, which concentrate on the determination of the low-order modes. This is the underlyingmotivation for most of the methods described in the following chapters.

It should be noticed that if λj is known, then Φ(j) can be determined from the linear, homoge-neous equations, cf. (1-9)

(K − λjM

)Φ(j) = 0 (5–1)

If λj is an eigenvalue, the coefficient matrix K−λjM is singular. Then, Φ(j) can be determinedwithin a common factor by solving a linear system of n− 1 equations as illustrated in Example1.5.

On the other hand, if Φ(j)is known, the eigenvalue λj can be determined from the Rayleighquotient, cf. (4-25)

λj =Φ(j) TKΦ(j)

Φ(j) TMΦ(j)(5–2)

Since, the eigenvalues are determined as solutions to the characteristic equation (1-10), whichcan only be solved analytically for n ≤ 4, all solution methods for practical problems reliesimplicitly or explicitly on iterative numerical schemes.

Iterative numerical solution methods may be classified in the following categories

Vector iteration methods operate directly on the generalized eigenvalue problem (5-1), so thata certain eigenvalue and associated eigenmode are determined iteratively with increasing accu-racy. Vector iteration methods are considered both in Chapters 5 and 7.

— 89 —

90 Chapter 5 – VECTOR ITERATION METHODS

Similarity transformation methods transform the generalized eigenvalue problem via a sequenceof similarity transformations, so the transformed mass and stiffness matrices eventually attain adiagonal form. These methods are considered in Chapter 6.

Characteristic polynomial iteration methods operates directly or indirectly on the characteristicequation (1-10). These methods are dealt with in Section 7.4.

5.2 Inverse and Forward Vector Iteration

The principle in inverse vector iteration may be explained in the following way. Given a startvector, Φ0. Based on the generalized eigenvalue problem (5-1), one may then calculate a newvector Φ1 as follows

KΦ1 = MΦ0 ⇒ Φ1 = K−1MΦ0 = AΦ0 (5–3)

where

A = K−1M (5–4)

Clearly, if Φ0 = Φ(j) is an eigenmode, then Φ1 = 1λj

Φ0. If not so, we may consider Φ1

as another, and hopefully better approximation to the eigenmode. Next, based on Φ1 we mayproceed to calculate a still better approximation Φ2 from

KΦ2 = MΦ1 ⇒ Φ2 = AΦ1 (5–5)

This proceed may be continued until the convergence criteria Φk+1 = 1λj

Φk is fulfilled withsufficient accuracy.

The inverse vector iteration algorithm may then be summarized as follows

Box 5.1: Inverse vector iteration algorithm

Given start vector Φ0, which needs not be normalized to unit modal mass. Repeat thefollowing items for k = 0, 1, . . .

1. Calculate Φk+1 = AΦk.

2. Normalize solution vector to unit modal mass, so ΦTk+1MΦk+1 = 1:

Φk+1 =Φk+1√

ΦTk+1MΦk+1

5.2 Inverse and Forward Vector Iteration 91

Obviously, the algorithm requires that the stiffness matrix is non-singular, so the inverse K−1

exists. By contrast the mass matrix needs not be non-singular as is the case in Example 5.1 be-low. After convergence the lowest eigenvalue is most accurately calculated from the Rayleighquotient (4-25).

In case the lowest eigenvalue is simple, i.e. that λ1 < λ2, the inverse iteration algorithm con-verges towards the lowest eigenpair (λ1,Φ

(1)). The solution vector obtained after the kth itera-tion step, Φk, is an n-dimensional vector, which may be expanded in the basis formed by the nundamped eigenmodes as follows

Φk = q1,kΦ(1) + q2,kΦ

(2) + · · ·+ qn,kΦ(n) = Φqk

Φ = [Φ(1) Φ(2) · · ·Φ(n)] , qk =

⎡⎢⎢⎢⎣

q1,k

q2,k...

qn,k

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭

(5–6)

The components of the vector qk denote the modal coordinates of the vector Φk. The expansion(5-6) should be considered as formal, since the base vectors Φ(1),Φ(2), . . . ,Φ(n) are unknown.Actually, the whole analysis deals with the determination of these quantities. Similarly, theexpansion for Φk+1 reads

Φk+1 = Φqk+1 (5–7)

where qk+1 denotes a vector of modal coordinates of Φk+1. Insertion of (5-6) and (5-7) into theiteration algorithm provides

KΦqk+1 = MΦqk ⇒

ΦTKΦqk+1 = ΦTMΦqk ⇒Λqk+1 = qk (5–8)

where the orthogonality properties (1-19) and (1-21) have been used, and the eigenmodes areassumed to be normalized to unit modal mass. The diagonal matrix Λ is given by (1-15). Ask → ∞ convergence implies that λjqk+1 = qk = Ψ(j), where Ψ(j) signifies the eigenmode inthe modal space. This means that


ΛΨ(j) = λjΨ(j) ⇒⎡

⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

Ψ1

Ψ2...

Ψn

⎤⎥⎥⎥⎦ = λj

⎡⎢⎢⎢⎣

Ψ1

Ψ2...

Ψn

⎤⎥⎥⎥⎦ ⇒

Ψ(j) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Ψ1...

Ψj−1

Ψj

Ψj+1...

Ψn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0...0

1

0...0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(5–9)

The jth component of Ψ(j) is equal to 1, and the remaining components are zero.

Let the start vector be given as q0 = [1, . . . , 1]T . Then, the following sequence of results maybe calculated from (5-8)

q1 = Λ−1q0 =

⎡⎢⎢⎢⎢⎣

1λ1

0 · · · 0

0 1λ2

· · · 0...

.... . .

...

0 0 · · · 1λn

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎣

1

1...

1

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎣

1λ1

1λ2

...1

λn

⎤⎥⎥⎥⎥⎦ ⇒

q2 = Λ−1q1 =

⎡⎢⎢⎢⎢⎣

1λ1

0 · · · 0

0 1λ2

· · · 0...

.... . .

...

0 0 · · · 1λn

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎣

1λ1

1λ2

...1

λn

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎣

1λ211λ22...1

λ2n

⎤⎥⎥⎥⎥⎥⎦ ⇒ · · · ⇒

qk = Λ−1qk−1 =

⎡⎢⎢⎢⎢⎣

1λ1

0 · · · 0

0 1λ2

· · · 0...

.... . .

...

0 0 · · · 1λn

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎣

1

λk−111

λk−12...1

λk−1n

⎤⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎣

1λk11λk2...1

λkn

⎤⎥⎥⎥⎥⎥⎦ =

1

λk1

⎡⎢⎢⎢⎢⎣

1(λ1

λ2

)k...(λ1

λn

)k

⎤⎥⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–10)

If λ1 < λ2 ≤ · · · ≤ λn it follows from (5-10) that


limk→∞

λk1qk =

⎡⎢⎢⎢⎣

1

0...0

⎤⎥⎥⎥⎦ = Ψ(1) (5–11)

Hence, the algorithm converge to Ψ(1) in the modal space. The corresponding convergence toΦ(1) then takes place in the physical space.

As seen from (5-11), |qk| → 0 if λ1 > 1, and |qk| → ∞ if λ1 < 1. This is the rationale behindthe normalization to unit modal mass of the iteration vector, performed at each iteration step inthe algorithm in Box 5.1.

The relative error of the iteration vector after the kth iteration step is defined from

ε1,k =

∣∣λk1qk − Ψ(1)

∣∣∣∣Ψ(1)∣∣ =

∣∣λk1qk − Ψ(1)

∣∣∣ =√(

λ1

λ2

)2k

+

(λ1

λ3

)2k

+ · · ·+(

λ1

λn

)2k

=

(λ1

λ2

)k√

1 +

(λ2

λ3

)2k

+ · · · +(

λ2

λn

)2k

(5–12)

From (5-12) follows, that the relative error at large values of k has the magnitude ε1,k (λ1

λ2

)k.

Based on the asymptotic behavior of the relative error, the convergence rate is defined from

r1 = limk→∞

ε1,k+1

ε1,k= lim

k→∞

∣∣λk+11 qk+1 −Ψ(1)

∣∣∣∣λk1qk −Ψ(1)

∣∣ =

limk→∞

λ1

λ2

√1 +(

λ2

λ3

)2k+2+ · · ·+ ( λ2

λn

)2k+2√1 +(

λ2

λ3

)2k+ · · ·+ ( λ2

λn

)2k=

λ1

λ2

(5–13)

The last statement of (5-13) presumes that the eigenvalue λ2 is simple, i.e. that λ2 < λ3. Itfollows from (5-12) that the smaller is the fraction λ1

λ2the faster will the convergence to the first

eigenmode be. Hence, the convergence rate as defined by (5-13) should be small (despite lin-guistic logics suggests the opposite). An vector iteration scheme, where the convergence rate isproportional to λ1

λ2is said to have linear convergence. Hence, inverse vector iteration has linear

convergence.

The Rayleigh quotient based on Φk = Φqk becomes

ρ(qk) =ΦT

k KΦk

ΦTk MΦk

=qT

k ΦTKΦqk

qTk ΦTMΦqk

=qT

k Λqk

qTk qk

(5–14)


The relative error of the Rayleigh quotient after the kth iteration step is defined from

ε2,k =ρ(qk) − λ1

λ1

(5–15)

From (5-10) follows that

qTk Λqk =

⎡⎢⎢⎢⎢⎢⎣

1λk11λk2...1

λkn

⎤⎥⎥⎥⎥⎥⎦

T ⎡⎢⎢⎢⎢⎣

λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...

0 0 · · · λn

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎣

1λk11λk2...1

λkn

⎤⎥⎥⎥⎥⎥⎦ =

1

λ2k−11

+1

λ2k−12

+1

λ2k−13

+ · · ·+ 1

λ2k−1n

qTk qk =

⎡⎢⎢⎢⎢⎢⎣

1λk11λk2...1

λkn

⎤⎥⎥⎥⎥⎥⎦

T ⎡⎢⎢⎢⎢⎢⎣

1λk11λk2...1

λkn

⎤⎥⎥⎥⎥⎥⎦ =

1

λ2k1

+1

λ2k2

+1

λ2k3

+ · · · + 1

λ2kn

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–16)

Then, (5-15) may be written as

ε2,k =1

λ1

1

λ2k−11

+ 1

λ2k−12

+ 1

λ2k−13

+ · · · + 1

λ2k−1n

1λ2k1

+ 1λ2k2

+ 1λ2k3

+ · · ·+ 1λ2k

n

− 1 =

1 +(

λ1

λ2

)2k−1+(

λ1

λ3

)2k−1+ · · ·+ ( λ1

λn

)2k−1

1 +(

λ1

λ2

)2k+(

λ1

λ3

)2k+ · · · + ( λ1

λn

)2k− 1 =

(λ1

λ2

)2k−11 − λ1

λ2+(

λ2

λ3

)2k−1(1 − λ1

λ3

)+ · · · + ( λ2

λn

)2k−1(1 − λ1

λn

)1 +(

λ1

λ2

)2k+(

λ1

λ3

)2k+ · · ·+ ( λ1

λn

)2k=

(λ1

λ2

)2k−1(1 − λ1

λ2+ · · ·

)(5–17)

where the dots denote terms, which converge to zero as k → ∞. (5-17) shows that the relativeerror of the Rayleigh quotient at large values of k has the magnitude ε2,k (λ1

λ2

)2k−1. Hence,

the relative error on the components of the eigenmode at a certain iteration step, as measuredby ε1,k, is significantly larger than the relative error on the eigenvalue estimate, as determinedby the Rayleigh quotient.

The convergence rate of the Rayleigh quotient is defined from


r2 = limk→∞

ε2,k+1

ε2,k= lim

k→∞

(λ1

λ2

)2k+1(1 − λ1

λ2+ · · · )(

λ1

λ2

)2k−1(1 − λ1

λ2+ · · · ) =

(λ1

λ2

)2

(5–18)

Hence, the Rayleigh quotient has quadratic convergence in inverse vector iteration.

Example 5.1: Inverse vector iteration

Consider the generalized eigenvalue problem defined by the mass and stiffness matrices in Example 4.1. Calculatethe lowest eigenvalue and eigenvector by inverse vector iteration using the inverse iteration algorithm described inBox 5.1 with the start vector

Φ0 =

⎡⎢⎢⎢⎣1111

⎤⎥⎥⎥⎦ (5–19)

The matrix A becomes, cf. (5-5), (4-16)

A =

⎡⎢⎢⎢⎣

2 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 1

⎤⎥⎥⎥⎦−1 ⎡⎢⎢⎢⎣0 0 0 00 2 0 00 0 0 00 0 0 1

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣0 2 0 10 4 0 20 4 0 30 4 0 4

⎤⎥⎥⎥⎦ (5–20)

At the 1st and 2nd iteration step the following calculations are performed

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎢⎢⎣0 2 0 10 4 0 20 4 0 30 4 0 4

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

1111

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

3678

⎤⎥⎥⎥⎦ ⇒ ΦT

1 MΦ1 = 136

Φ1 =1√136

⎡⎢⎢⎢⎣3678

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣0.257250.514500.600250.68599

⎤⎥⎥⎥⎦

(5–21)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ2 =

⎡⎢⎢⎢⎣0 2 0 10 4 0 20 4 0 30 4 0 4

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

0.257250.514500.600250.68599

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣1.71503.43004.11604.8020

⎤⎥⎥⎥⎦ ⇒ ΦT

2 MΦ2 = 46.588

Φ2 =1√

46.588

⎡⎢⎢⎢⎣

1.71503.43004.11604.8020

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0.251260.502520.603020.70353

⎤⎥⎥⎥⎦

(5–22)


The Rayleigh quotient based on Φ2 provides the following estimate for λ1, cf. (4-25)

ρ(Φ2) =

⎡⎢⎢⎢⎣

0.251260.502520.603020.70353

⎤⎥⎥⎥⎦

T ⎡⎢⎢⎢⎣

2 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

0.251260.502520.603020.70353

⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣0.251260.502520.603020.70353

⎤⎥⎥⎥⎦

T ⎡⎢⎢⎢⎣0 0 0 00 2 0 00 0 0 00 0 0 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

0.251260.502520.603020.70353

⎤⎥⎥⎥⎦

= 0.1464646 (5–23)

The exact solutions are given as, cf. (4-24)

λ1 =12−

√2

4= 0.1464466 , Φ(1) =

⎡⎢⎢⎢⎢⎣

1412

14 +

√2

4√2

2

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎣0.25000

0.50000

0.60355

0.70711

⎤⎥⎥⎥⎥⎦ (5–24)

The relative errors, ε1 and ε2, on the calculation of the eigenvalue and the 1st component of Φ (1) becomes

ε1,2 =|Φ2 − Φ(1)|

|Φ(1)| =0.004581.0848

= 4.22 · 10−3

ε2,2 =ρ(Φ2) − λ1

λ1=

0.1464646− 0.14644660.1464466

= 1.23 · 10−4

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

(5–25)

As seen the relative error on the components of the eigenmode is significantly larger than the error on the Rayleighquotient.

The generalized eigenvalue problem (1-9) may be reformulated on the form

MΦ(1) = λ1MK−1MΦ(1) ⇒

Ψ(1) = λ1MK−1Ψ(1) ⇒

K−1Ψ(1) = λ1K−1MK−1Ψ(1) , Ψ(1) = MΦ(1) (5–26)

From (5-26) the following Rayleigh quotient may be defined

ρ(v) =vT K−1 v

vT K−1MK−1 v(5–27)

If v = Ψ(1) = MΦ(1) then (5-4) provides the limit λ1. An inverse vector iteration procedurebased on the formulation (5-26), (5-27) has been indicated in Box 5.2. The lowest eigenmode


Φ(1) can only be retrieved after convergence, if M−1 exists.

Box 5.2: Alternative inverse vector iteration algorithm

Given start vector Ψ0. Repeat the following items for k = 0, 1, . . .

1. Calculate vk+1 = K−1Ψk.

2. Calculate Ψk+1 = Mvk+1.

3. Calculate the Rayleigh quotient (5-29) for the test vector Ψk by

ρ(Ψk

)=

vTk+1Ψk

vTk+1Ψk+1

(=

ΨTk K−1 Ψk

ΨTk K−1MK−1 Ψk

)

4. Normalize the new solution vector, so ΨTk+1 K−1MK−1 Ψk+1 = 1

Ψk+1 =Ψk+1√

vTk+1Ψk+1

(=

Ψk+1√ΨT

k K−1MK−1 Ψk

)

5. After convergence the lowest eigenmode at the same iteration step is calculated fromΦk+1 = M−1Ψk+1.

Example 5.2: Alternative inverse vector iteration

Consider the generalized eigenvalue problem defined by the mass and stiffness matrices in Example 1.5. Calculatethe lowest eigenvalue and eigenvector by inverse vector iteration using the alternative inverse vector iterationalgorithm in Box 5.2 with the start vector

Φ0 =

⎡⎢⎣111

⎤⎥⎦ (5–28)

The inverse stiffness matrix becomes, cf. (1-77)

K−1 =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1

=112

⎡⎢⎣7 2 1

2 4 21 2 7

⎤⎥⎦ (5–29)

At the 1st and 2nd iteration steps the following calculations are performed


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v1 =112

⎡⎢⎣7 2 12 4 21 2 7

⎤⎥⎦⎡⎢⎣111

⎤⎥⎦ =

16

⎡⎢⎣545

⎤⎥⎦

Ψ1 =16

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣5

45

⎤⎥⎦ =

112

⎡⎢⎣585

⎤⎥⎦ , vT

1 Ψ1 =1

6 · 12

⎡⎢⎣5

45

⎤⎥⎦

T ⎡⎢⎣585

⎤⎥⎦ =

4136

ρ(Ψ0

)=

vT1 Ψ0

vT1 Ψ1

=36

6 · 41

⎡⎢⎣5

45

⎤⎥⎦

T ⎡⎢⎣111

⎤⎥⎦ =

8441

= 2.0488

Ψ1 =Ψ1√vT

1 Ψ1

=1

12 ·√

4136

⎡⎢⎣585

⎤⎥⎦ =

⎡⎢⎣0.39040.62470.3904

⎤⎥⎦

(5–30)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v2 =112

⎡⎢⎣7 2 12 4 21 2 7

⎤⎥⎦⎡⎢⎣0.39040.62470.3904

⎤⎥⎦ =

⎡⎢⎣0.36440.33840.3644

⎤⎥⎦

Ψ2 =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣0.36440.33840.3644

⎤⎥⎦ =

⎡⎢⎣0.18220.33840.1822

⎤⎥⎦ , vT

2 Ψ2 =

⎡⎢⎣0.3644

0.33840.3644

⎤⎥⎦

T ⎡⎢⎣0.18220.33840.1822

⎤⎥⎦ = 0.2473

ρ(Ψ1

)=

vT2 Ψ1

vT2 Ψ2

=1

0.2473

⎡⎢⎣0.3644

0.33840.3644

⎤⎥⎦

T ⎡⎢⎣0.39040.62470.3904

⎤⎥⎦ = 2.0055

Ψ2 =Ψ2√vT

2 Ψ2

=1√

0.2473

⎡⎢⎣0.18220.33840.1822

⎤⎥⎦ =

⎡⎢⎣0.36640.68050.3664

⎤⎥⎦

(5–31)

The lowest eigenvector at the end of the 2nd iteration step becomes

Φ2 = M−1Ψ2 =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦−1 ⎡⎢⎣0.3664

0.68050.3664

⎤⎥⎦ =

⎡⎢⎣0.7328

0.68050.7328

⎤⎥⎦ (5–32)

The exact solutions are given as, cf. (1-87)

Φ(1) =

⎡⎢⎣

√2

2√2

2√2

2

⎤⎥⎦ =

⎡⎢⎣0.7071

0.7071

0.7071

⎤⎥⎦ , λ1 = 2 (5–33)

As for the simple formulation of the inverse vector iteration algorithm the convergence towards the exact eigen-value takes place as a monotonously decreasing sequence of upper values, ρ(Ψ 0), ρ(Ψ1), . . ..


The principle in forward vector iteration may also be explained based on the eigenvalue problem(5-1). Given a start vector, Φ0, a new vector Φ1 may be calculated as follows

KΦ0 = MΦ1 ⇒ Φ1 = M−1KΦ0 = BΦ0 (5–34)

where

B = M−1K (5–35)

Clearly, if Φ0 = Φ(j) is an eigenmode, then Φ1 = λjΦ0. If not so, a new and better approxi-mation Φ2 may be calculated based on Φ1 as follows

Φ2 = BΦ1 (5–36)

The process may be continued until converge is obtained. The forward vector iteration algo-rithm may be summarized as follows

Box 5.3: Forward vector iteration algorithm


1. Calculate Φk+1 = BΦk.

2. Normalize solution vector to unit modal mass, so ΦTk+1MΦk+1 = 1

Φk+1 =Φk+1√

ΦTk+1MΦk+1

Obviously, the algorithm requires that the mass matrix is non-singular, so the inverse M−1

exists. By contrast the stiffness matrix needs not be non-singular. After convergence the eigen-value is calculated from the Rayleigh quotient.

In case the largest eigenvalue is simple, i.e. that λn−1 < λn, the forward iteration algorithmconverges towards the largest eigenpair

(λn,Φ(n)

). The convergence rate of the eigenmode

estimate is linear, and the convergence rate of the Rayleigh quotient is quadratic in the fractionλn−1

λn. A proof of this has been given in Section 5.3.


Example 5.3: Forward vector iteration

Consider the generalized eigenvalue problem defined by the mass and stiffness matrices in Example 1.4. Calculatethe largest eigenvalue and eigenvector by forward vector iteration using the forward vector iteration algorithm inBox 5.3 with the start vector

Φ0 =

⎡⎢⎣100

⎤⎥⎦ (5–37)

The matrix B becomes, cf. (5-35), (1-77)

B =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦−1 ⎡⎢⎣ 2 −1 0

−1 4 −10 −1 2

⎤⎥⎦ =

⎡⎢⎣ 4 −2 0−1 4 −1

0 −2 4

⎤⎥⎦ (5–38)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎣ 4 −2 0−1 4 −1

0 −2 4

⎤⎥⎦⎡⎢⎣1

00

⎤⎥⎦ =

⎡⎢⎣ 4−1

0

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 9

Φ1 =1√9

⎡⎢⎣ 4−1

0

⎤⎥⎦ =

⎡⎢⎣ 1.3333−0.3333

0

⎤⎥⎦

(5–39)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ2 =

⎡⎢⎣ 4 −2 0−1 4 −1

0 −2 4

⎤⎥⎦⎡⎢⎣ 1.3333−0.3333

0

⎤⎥⎦ =

⎡⎢⎣ 6.0000−2.6667

0.6667

⎤⎥⎦ ⇒ ΦT

2 MΦ2 = 25.333

Φ2 =1√

25.333

⎡⎢⎣ 6.0000−2.6667

0.6667

⎤⎥⎦ =

⎡⎢⎣ 1.1921−0.5298

0.1325

⎤⎥⎦

(5–40)

The Rayleigh quotient based on Φ2 becomes

ρ(Φ2) =

⎡⎢⎣ 1.1921−0.5298

0.1325

⎤⎥⎦

T ⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣ 1.1921−0.5298

0.1325

⎤⎥⎦

⎡⎢⎣ 1.1921−0.5298

0.1325

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣ 1.1921−0.5298

0.1325

⎤⎥⎦

= 5.404 (5–41)

The results for the iteration vector and the Rayleigh quotient in the succeeding iteration steps become

5.3 Shift in Vector Iteration 101

Φ3 =

⎡⎢⎣ 1.0622−0.6276

0.2897

⎤⎥⎦ , ρ(Φ3) = 5.697

Φ4 =

⎡⎢⎣ 0.9584−0.6726

0.4204

⎤⎥⎦ , ρ(Φ4) = 5.855

Φ5 =

⎡⎢⎣ 0.8811−0.6923

0.5149

⎤⎥⎦ , ρ(Φ5) = 5.933

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–42)

The exact solutions becomes, cf. (6-54)

Φ(3) =

⎡⎢⎣

√2

2

−√

22√2

2

⎤⎥⎦ =

⎡⎢⎣ 0.7071

−0.7071

0.7071

⎤⎥⎦ , λ3 = 6 (5–43)

The relative slow convergence of the algorithm to the exact solution is because the fraction λ2λ3

= 46 is relatively

high. Theoretically the relatively errors of the Rayleigh quotient after 5 iterations should be of the magnitude, cf.(5-17)

ε2,5 (4

6

)2·5−1(1 − 4

6

)= 0.0087 (5–44)

Actually, the error is slightly larger, namely

ε2,5 =6 − 5.933

6= 0.0112 (5–45)

5.3 Shift in Vector Iteration

Shift on the stiffness matrix in the eigenvalue problem (5-1) as indicated by (3-36)-(3-38) maybe appropriate both in relation to inverse and forward vector iteration, either in order to obtainconvergence to other eigen-pairs than (λ1,Φ

(1)) or (λn,Φ(n)), or to improve the convergencerate of the iteration process.

Let K = K − ρM. denote a shift on the stiffness matrix as indicated by (3-38). The vectoriteration is next performed on the shifted eigenvalue problem (3-37). The algorithms in Box 5.1and 5.3 remain unchanged, if the matrices A and B in (5-6) and (5-35) are redefined as follows

A = K−1M (5–46)


B = M−1K (5–47)

The Rayleigh quotient estimate of the eigenvalue λj after the kth iteration step becomes

λj = ρ(Φk) + ρ =ΦT

k KΦk

ΦTk MΦk

+ ρ (5–48)

In the modal space the inverse vector iteration with shift on the stiffness matrix can be writtenas, cf. (5-8)

KΦqk+1 = MΦqk ⇒

ΦT(K− ρM

)Φ qk+1 = ΦTMΦqk ⇒

(Λ− ρI

)qk+1 = qk (5–49)

(5-49) is identical to (5-8), if λj is replaced with λj − ρ. With the same start vector q0 =[1, . . . , 1]T as in (5-10), the solution vector after the kth iteration step becomes, cf. (5-10)

qk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1(λ1−ρ)k

...1

(λj−1−ρ)k

1(λj−ρ)k

1(λj+1−ρ)k

...1

(λn−ρ)k

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=1

(λj − ρ)k

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(λj−ρ

λ1−ρ

)k...( λj−ρ

λj−1−ρ

)k1( λj−ρ

λj+1−ρ

)k...( λj−ρ

λn−ρ

)k

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(5–50)

where the jth eigenvalue fulfills

|λj − ρ| = mini=1,...,n

|λi − ρ| (5–51)

It then follows from (5-50) that

limk→∞

(λj − ρ

)kqk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0...0

1

0...0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= Ψ(j) (5–52)


Hence, for a value of ρ fulfilling (5-51) the algorithm converge to Ψ(j) in the modal space. Inphysical space the algorithm then converge to Φ(j). The convergence rate of the eigenmodebecomes, cf. (5-13)

r1 = max

(∣∣∣∣ λj − ρ

λj−1 − ρ

∣∣∣∣ ,∣∣∣∣ λj − ρ

λj+1 − ρ

∣∣∣∣)

(5–53)

Then, the corresponding convergence rate of the Rayleigh quotient is given as r2 = r21.

0

0

0

λ1

λ1

λ1

λj−1

λj−1

λj−1

λj

λj

λj

λj+1

λj+1

λj+1

λn−1

λn−1

λn−1

λn

λn

λn

a)

b)

c) λ

λ

λ

λ

ρ

ρ

ρ

Fig. 5–1 Optimal position of shift parameter at inverse vector iteration. a) Convergence towards λ j . b) Conver-gence towards λ1. c) Convergence towards λn.

In case inverse vector iteration towards the jth eigenmode is attempted, the shift parametershould be place in the vicinity of λj as shown on Fig. 5.1a in order to obtain a small con-vergence rate. It should be emphasized that any inverse vector iteration with shift should beaccompanied with a Sturm sequence check to insure that the calculated eigenvalue is indeed theλj.

At inverse vector iteration towards the lowest eigenmode the convergence rate r1 = |λ1 −ρ|/|λ2 − ρ| should be minimized. Hence, ρ should be placed close to but below λ1, as shownon Fig. 5.1b.

At inverse vector iteration towards the highest eigenmode the convergence rate r1 = |λn−1 −ρ|/|λn − ρ| should be minimized. Hence, ρ should be placed close to but above λn, as shownon Fig. 5.1c.

In case of forward iteration with shift, (5-49) provides the solution after k iterations


qk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(λ1 − ρ)k

...(λj−1 − ρ)k

(λj − ρ)k

(λj+1 − ρ)k

...(λn − ρ)k

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= (λj − ρ)k

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(λ1−ρλj−ρ

)k...(λj−1−ρ

λj−ρ

)k1(λj+1−ρ

λj−ρ

)k...(

λn−ρλj−ρ

)k

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(5–54)

where the jth eigenvalue fulfills

|λj − ρ| = maxi=1,...,n

|λi − ρ| (5–55)

Clearly, (5-55) has the solutions λj = λ1 or λj = λn. The former occurs, if ρ is closest to λn,and the latter if ρ is closest to λ1. Then, it follows from (5-54) that

limk→∞

1(λj − ρ

)k qk = Ψ(j) , j = 1, n (5–56)

For a value of ρ fulfilling (5-55) the algorithm converge to Ψ(j) in the modal space, or to Φ(j)

in the physical space. Forward iteration with shift always converge to either the lowest or thehighest eigenmode depending on the magnitude of the shift parameter. The convergence rate ofthe iteration vector becomes

r1 = max

(∣∣∣∣λ1 − ρ

λj − ρ

∣∣∣∣ , . . . ,∣∣∣∣λj−1 − ρ

λj − ρ

∣∣∣∣ ,∣∣∣∣λj+1 − ρ

λj − ρ

∣∣∣∣ , . . . ,∣∣∣∣λn − ρ

λj − ρ

∣∣∣∣)

(5–57)

Shift in forward vector iteration is not as useful as in inverse vector iteration, because the optimalchoice of the shift parameter is more difficult to specify. At forward vector iteration towards thehighest eigenmode the optimal shift parameter is typically placed somewhere in the middle ofthe eigenvalue spectrum. Especially for ρ = 0, (5-57) becomes

r1 =λn−1

λn

(5–58)

as stated in Section 5.2 on forward iteration without shift.

Example 5.4: Forward vector iteration with shift


The problem in Example 5.3 is considered again. However, now a shift with ρ = 3 is performed on the stiffnessmatrix.

The matrix K becomes, cf. (3-38), (1-77)

K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦− 3

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ =

⎡⎢⎣

12 −1 0

−1 1 −10 −1 1

2

⎤⎥⎦ (5–59)

The matrix B becomes, cf. (5-47), (1-77)

B =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦−1 ⎡⎢⎣

12 −1 0

−1 1 −10 −1 1

2

⎤⎥⎦ =

⎡⎢⎣ 1 −2 0−1 1 −1

0 −2 1

⎤⎥⎦ (5–60)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎣ 1 −2 0−1 1 −1

0 −2 1

⎤⎥⎦⎡⎢⎣1

00

⎤⎥⎦ =

⎡⎢⎣ 1−1

0

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 1.5

Φ1 =1√1.5

⎡⎢⎣ 1−1

0

⎤⎥⎦ =

⎡⎢⎣ 0.8165−0.8165

0

⎤⎥⎦

(5–61)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ2 =

⎡⎢⎣ 1 −2 0−1 1 −1

0 −2 1

⎤⎥⎦⎡⎢⎣ 0.8165−0.8165

0

⎤⎥⎦ =

⎡⎢⎣ 2.4495−1.6330

1.6330

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 7

Φ2 =1√7

⎡⎢⎣ 2.4495−1.6330

1.6330

⎤⎥⎦ =

⎡⎢⎣ 0.9258−0.6172

0.6172

⎤⎥⎦

(5–62)

The Rayleigh quotient estimate of λ3 based on Φ2 becomes, cf. (5-48)

λ3 = ρ(Φ2) + 3 =

⎡⎢⎣ 0.9258−0.6172

0.6172

⎤⎥⎦

T ⎡⎢⎣12 −1 0

−1 1 −10 −1 1

2

⎤⎥⎦⎡⎢⎣ 0.9258−0.6172

0.6172

⎤⎥⎦

⎡⎢⎣ 0.9258−0.6172

0.6172

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣ 0.9258−0.6172

0.6172

⎤⎥⎦

+ 3 = 2.9048 + 3 = 5.9048 (5–63)

The results for the iteration vector and the eigenvalue estimate in the succeeding iteration steps become


Φ3 =

⎡⎢⎣ 0.7318−0.7318

0.6273

⎤⎥⎦ , λ3 = 5.9891

Φ4 =

⎡⎢⎣ 0.7331−0.6982

0.6982

⎤⎥⎦ , λ3 = 5.9988

Φ5 =

⎡⎢⎣ 0.7100−0.7100

0.6983

⎤⎥⎦ , λ3 = 5.9999

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–64)

The results in (5-64) should be compared to those in (5-42). As seen the convergence of the shifted problem ismuch faster.

5.4 Inverse Vector Iteration with Rayleigh Quotient Shift

As demonstrated in Section 5.3 the convergence properties of inverse vector towards the lowestmode are improved if a shift on the stiffness matrix is performed with a shift parameter fulfillingρ λ1. The idea in the present section is to update the shift parameter at each iteration step withthe most recent estimate of the lowest eigenvalue. Assume, that an estimate of the eigenvalueλ1 is known after the kth iteration step. Then, a shift with the parameter ρk = λ1 is performed,so a new un-normalized eigenmode estimate is calculated at the (k + 1)th iteration step from

Φk+1 =(K − ρkM

)−1

MΦk (5–65)

where Φk is the normalized estimate of the eigenmode after the kth iteration step.

A new estimate of the eigenvalue, and hence the shift parameter, then follows from (5-48)

ρk+1 =ΦT

k+1

(K − ρkM

)Φk+1

ΦTk+1 MΦk+1

+ ρk (5–66)

The convergence towards(λ1,Φ

(1))

is not safe, since the first shift determined by ρ1 may causeconvergence towards other eigen-pairs, especially if the first and second eigenvalue are close.For this reason the first couples of iteration steps are often performed without shift. When theconvergence towards the first eigen-pair takes place, the convergence rate of the Rayleigh quo-tient estimate of the eigenvalue will be cubic, i.e. r2 = (λ1

λ2)3. Additionally, the length of the

converge process is very much dependent on the start vector, as demonstrated in the succeedingExample 5.5. Even though the convergence may be fast it should be realized that the processrequires inversion of the matrix K − ρkM at each iteration step, which may be expensive for

5.4 Inverse Vector Iteration with Rayleigh Quotient Shift 107

large systems.

Box 5.4: Algorithm for inverse vector iteration with Rayleigh quotient shift

Given start vector Φ0, which needs not be normalized to unit modal mass, and set theinitial shift to ρ0 = 0. Repeat the following items for k = 0, 1, . . .

1. Calculate Φk+1 =(K − ρkM

)−1

MΦk.

2. Calculate new shift parameter (new estimate on the eigenvalue) from the Rayleighquotient estimate based on Φk+1 by

ρk+1 =ΦT

k+1

(K − ρkM

)Φk+1

ΦTk+1 MΦk+1

+ ρk

(estimate on λ1

)3. Normalize the new solution vector to unit modal mass

Φk+1 =Φk+1√

ΦTk+1 MΦk+1

Example 5.5: Inverse vector iteration with Rayleigh quotient shift

Consider the generalized eigenvalue problem defined by the mass and stiffness matrices in Example 1.4. Calculatethe lowest eigenvalue and eigenvector by inverse vector iteration with Rayleigh quotient shift with the start vector

Φ0 =

⎡⎢⎣100

⎤⎥⎦ (5–67)



⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦− 0 ·

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦

Φ1 =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣100

⎤⎥⎦ =

⎡⎢⎣0.29170.08330.0417

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 0.05035

ρ1 =1

0.05035

⎡⎢⎣0.2917

0.08330.0417

⎤⎥⎦

T ⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.29170.08330.0417

⎤⎥⎦+ 0 = 2.8966

Φ1 =1√

0.05035

⎡⎢⎣0.29170.08330.0417

⎤⎥⎦ =

⎡⎢⎣1.29990.37140.1857

⎤⎥⎦

(5–68)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦− 2.8966 ·

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ =

⎡⎢⎣ 0.5517 −1.0000 0.0000−1.0000 1.1034 −1.0000

0.0000 −1.0000 0.5517

⎤⎥⎦

Φ2 =

⎡⎢⎣ 0.5517 −1.0000 0.0000−1.0000 1.1034 −1.0000

0.0000 −1.0000 0.5517

⎤⎥⎦−1 ⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣1.29990.37140.1857

⎤⎥⎦ =

⎡⎢⎣−0.0567−0.6812−1.0664

⎤⎥⎦ ⇒ ΦT

2 MΦ2 = 1.0342

ρ2 =1

1.0342

⎡⎢⎣−0.0567−0.6812−1.0664

⎤⎥⎦

T ⎡⎢⎣ 0.5517 −1.0000 0.0000−1.0000 1.1034 −1.0000

0.0000 −1.0000 0.5517

⎤⎥⎦⎡⎢⎣−0.0567−0.6812−1.0664

⎤⎥⎦+ 2.8966 = 2.5206

Φ2 =1√

1.0342

⎡⎢⎣−0.0567−0.6812−1.0664

⎤⎥⎦ =

⎡⎢⎣−0.0557−0.6698−1.0486

⎤⎥⎦

(5–69)

The results for the iteration vector and the eigenvalue estimate in the succeeding iteration steps become

Φ3 =

⎡⎢⎣0.90110.68300.5049

⎤⎥⎦ , ρ3 = 2.0793

Φ4 =

⎡⎢⎣−0.6985−0.7073−0.7152

⎤⎥⎦ , ρ4 = 2.0001

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–70)

Despite the shifts the convergence is very slow during the 1st and 2nd iteration step. Not until the 3rd and 4th stepa fast speed-up of the convergence takes place. This is due to the poor guess of the start vector.

5.5 Vector Iteration with Gram-Schmidt Orthogonalization 109

5.5 Vector Iteration with Gram-Schmidt Orthogonaliza-tion

Inverse vector iteration or forward vector iteration with Gram-Schmidt orthogonalization isused, when other eigen-pairs than

(λ1,Φ

(1))

or(λn,Φ(n)

)are wanted.

Assume, that the eigenmodes Φ(1),Φ(2), . . . ,Φ(m), m < n, have been determined. Next, theeigenmode Φ(m+1) is wanted using inverse vector iteration by means of the algorithm in Box5.1. In order to prevent the algorithm to converge toward Φ(1) a cleansing of the vector Φk+1

for information about the first m eigenmodes is performed by a so-called Gram-Schmidt or-thogonalization . In this respect the following modified iteration vector iteration algorithm isconsidered

Φk+1 = Φk+1 −m∑

j=1

cjΦ(j) (5–71)

Inspired by the variational problem (4-31), where the test vector v is chosen to be M-orthogonalto the previous determined eigenmodes, the modified iteration vector Φk+1 is chosen to be M-orthogonal on Φ(1),Φ(2), . . . ,Φ(m), i.e.

Φ(i) TMΦk+1 = 0 , i = 1, . . . , m (5–72)

(5-71) is premultiplied with Φ(i) TM. Assuming that the calculated eigenmodes have beennormalized to unit modal mass, it follows from (1-17), (5-71) and (5-72) that the expansioncoefficients c1, c2, . . . , cm are determined from

0 = Φ(i) TMΦk+1 −m∑

j=1

cjΦ(i) TMΦ(j) = Φ(i) TMΦk+1 − ci ⇒

ci = Φ(i) TMΦk+1 (5–73)

After insertion of the calculated expansion coefficients into (5-71), Φk+1 is considered as theestimate to Φ(m+1) at the (k + 1)th iteration step. The convergence takes place with the linearconvergence rate r1 = λm+1

λm+2.

In principle the orthogonalization process need only to be performed after the first iterationstep, since all succeeding iteration vectors then will be orthogonal to the subspace spanned byΦ(1),Φ(2), . . . ,Φ(m). However, round-off errors inevitable introduce information about the firsteigenmode. Obviously, the use of this so-called vector deflation method becomes increasinglycumbersome as m increases.

A similar orthogonalization process can be performed in relation to forward vector iteration toensure convergence to eigenmodes somewhat lower than the highest.


Box 5.5: Algorithm for inverse vector iteration with Gram-Schmidt orthogonalization


1. Calculate Φk+1 = K−1MΦk.

2. Orthogonalize iteration vector to previous calculated eigenmodes Φ(j), j = 1, . . . , m

Φk+1 = Φk+1 −m∑

j=1

cjΦ(j) , cj = Φ(j) TMΦk+1

3. Normalize the orthogonalized iteration vector to unit modal mass

Φk+1 =Φk+1√

ΦTk+1 MΦk+1

Example 5.6: Inverse and forward vector iteration with Gram-Schmidt orthogonalization


M =

⎡⎢⎢⎢⎣

2 0 0 00 2 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦ , K =

⎡⎢⎢⎢⎣

5 −4 1 0−4 6 −4 1

1 −4 6 −40 1 −4 5

⎤⎥⎥⎥⎦ (5–74)

Further, assume that the lowest and highest eigenmodes have been determined by inverse and forward vectoriteration

Φ(1) =

⎡⎢⎢⎢⎣0.312630.495480.479120.28979

⎤⎥⎥⎥⎦ , Φ(4) =

⎡⎢⎢⎢⎣−0.10756

0.25563−0.72825

0.56197

⎤⎥⎥⎥⎦ (5–75)

Calculate Φ(2) by inverse vector iteration with deflation, and Φ (3) by forward vector iteration with deflation. Inboth cases the following start vector is used

Φ0 =

⎡⎢⎢⎢⎣1111

⎤⎥⎥⎥⎦ (5–76)

The matrices A and B become

5.5 Vector Iteration with Gram-Schmidt Orthogonalization 111

A = K−1M =

⎡⎢⎢⎢⎣2.4 3.2 1.4 0.83.2 5.2 2.4 1.42.8 4.8 2.6 1.61.6 2.8 1.6 1.2

⎤⎥⎥⎥⎦

B = M−1K =

⎡⎢⎢⎢⎣

2.5 −2.0 0.5 0.0−2.0 3.0 −2.0 0.5

1.0 −4.0 6.0 −4.00.0 1.0 −4.0 5.0

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–77)

At the 1st iteration step in the inverse iteration process towards Φ (2) the following calculations are performed

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎢⎢⎣2.4 3.2 1.4 0.83.2 5.2 2.4 1.42.8 4.8 2.6 1.61.6 2.8 1.6 1.2

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣1111

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

7.811.211.87.2

⎤⎥⎥⎥⎦ ⇒ c1 = Φ(1) T MΦ1 = 24.7067

Φ1 =

⎡⎢⎢⎢⎣

7.811.211.87.2

⎤⎥⎥⎥⎦− 24.7067

⎡⎢⎢⎢⎣0.312630.495480.479120.28979

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0.07595−0.04158−0.03740

0.04016

⎤⎥⎥⎥⎦ ⇒ ΦT

1 MΦ1 = 0.01801

Φ1 =1√

0.01801

⎡⎢⎢⎢⎣

0.07595−0.04158−0.03740

0.04016

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0.56599−0.30989−0.27871

0.29927

⎤⎥⎥⎥⎦

(5–78)

The results for the iteration vector in the succeeding iteration steps become

Φ2 =

⎡⎢⎢⎢⎣

0.61639−0.14318−0.42383−0.13960

⎤⎥⎥⎥⎦

Φ3 =

⎡⎢⎢⎢⎣

0.534120.02582

−0.48439−0.43985

⎤⎥⎥⎥⎦

...

Φ13 =

⎡⎢⎢⎢⎣

0.445270.12443

−0.48944−0.57702

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–79)

The process converged with the indicated digit after 13 iterations.


At the 1st iteration step in the forward iteration process towards Φ (3) the following calculations are performed

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎢⎢⎣

2.5 −2.0 0.5 0.0−2.0 3.0 −2.0 0.5

1.0 −4.0 6.0 −4.00.0 1.0 −4.0 5.0

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣1111

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

1.0−0.5−1.0

2.0

⎤⎥⎥⎥⎦ ⇒ c4 = Φ(4) T MΦ1 = 1.38144

Φ1 = Φ1 − c4Φ(4) =

⎡⎢⎢⎢⎣

1.0−0.5−1.0

2.0

⎤⎥⎥⎥⎦− 1.38144

⎡⎢⎢⎢⎣−0.10756

0.25563−0.72825

0.56197

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

1.14859−0.85314

0.006041.22367

⎤⎥⎥⎥⎦ ⇒ ΦT

1 MΦ1 = 5.59161

Φ1 =1√

5.59161

⎡⎢⎢⎢⎣

1.14859−0.85314

0.006041.22367

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0.48573−0.36079

0.002560.51748

⎤⎥⎥⎥⎦

(5–80)

The results for the iteration vector in the succeeding iteration steps become

Φ2 =

⎡⎢⎢⎢⎣

0.44542−0.41392−0.02891

0.50962

⎤⎥⎥⎥⎦

Φ3 =

⎡⎢⎢⎢⎣

0.44063−0.41617−0.02534

0.51445

⎤⎥⎥⎥⎦

...

Φ9 =

⎡⎢⎢⎢⎣

0.43867−0.41674−0.02322

0.51696

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5–81)

The process converged with the indicated digit after 9 iterations.

Based on the Rayleigh quotient estimates of the obtained eigenmodes the following eigenvalues may be calculated,cf. (5-2)

Λ =

⎡⎢⎢⎢⎢⎣λ1 0 0 0

0 λ2 0 0

0 0 λ3 0

= 0 0 λ4

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎣0.09654 0 0 0

0 1.39147 0 0

0 0 4.37355 0

0 0 0 10.6384

⎤⎥⎥⎥⎥⎦ (5–82)

5.6 Exercises 113

5.6 Exercises


M =

⎡⎢⎣2 0 0

0 2 1

0 1 1

⎤⎥⎦ , K =

⎡⎢⎣ 6 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦

(a.) Perform two inverse iterations, and then calculate an approximation to λ1.(b.) Perform two forward iterations, and then calculate an approximation to λ3.


M =

⎡⎢⎣

12

0 0

0 1 0

0 0 12

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦

The eigenmodes Φ(1) are Φ(3) are known to be, cf. (1-87)

Φ(1) =

⎡⎢⎣

√2

2√2

2√2

2

⎤⎥⎦ , Φ(3) =

⎡⎢⎣

√2

2

−√

22√2

2

⎤⎥⎦

(a.) Calculate Φ(2) by means of Gram-Schmidt orthogonalization, and calculate all eigen-values.


M =

⎡⎢⎢⎢⎢⎢⎣

4 1 0 0 0

1 4 1 0 0

0 1 4 1 0

0 0 1 4 1

0 0 0 1 4

⎤⎥⎥⎥⎥⎥⎦ , K =

⎡⎢⎢⎢⎢⎢⎣

2 −1 0 0 0

−1 2 −1 0 0

0 −1 2 −1 0

0 0 −1 2 −1

0 0 0 −1 2

⎤⎥⎥⎥⎥⎥⎦

(a.) Write a MATLAB program, which calculates the lowest three eigenvalues and eigen-modes of the related generalized eigenvalue problem by means of inverse vector iter-ation with Gram-Schmidt ortogonalization.

CHAPTER 6SIMILARITY TRANSFORMATION

METHODS

6.1 Introduction

Iterative similarity transformation methods are based on a sequence of similarity transforma-tions of the original generalized eigenvalue problem in order to reduce this to a simpler form.The general form of a similarity transformation is defined by the following coordinate transfor-mation of the eigenmodes

Φ(j) = PΨ(j) (6–1)

where P is the transformation matrix, and Φ(j) and Ψ(j) signify the old and the new coordinatesof the eigenmode. Then, the eigenvalue problem (1-9) may be written

KΦ(j) = λjMΦ(j) ⇒KPΨ(j) = λjMPΨ(j) ⇒KΨ(j) = λjMΨ(j)

K = PTKP , M = PTMP

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭

(6–2)

The eigenvalues λj are unchanged under a similarity transformation, whereas the eigenmodesare related by (6-1). In the iteration process the transformation matrix P is determined, so thismatrix converge toward the modal matrix Φ = [Φ(1) Φ(2) · · ·Φ(n)]. Hence, after convergence ofthe iteration process the eigenmodes are stored column-wise in P = Φ. On condition that theeigenmodes have been normalized to unit modal mass it then follows from (1-19) and (1-21)that K = PTKP = Λ, and M = PTMP = I, so the transformed stiffness and mass matricesbecome diagonal at convergence, and the eigenvalues are stored in the main diagonal of K. Bycontrast to vector iteration methods similarity transformation methods determine all eigen-pairs(λj,Φ

(j)), j = 1, . . . , n.

The general format of the similarity iteration algorithm has been summarized in Box 6.1.

— 115 —

116 Chapter 6 – SIMILARITY TRANSFORMATION METHODS

Box 6.1: Iterative similarity transformation algorithm

Let M0 = M, K0 = K and Φ0 = I. Repeat the following items for k = 0, 1, . . .

1. Calculate appropriate transformation matrix Pk at the kth iteration step.

2. Calculate updated transformation matrix and transformed mass and stiffness matrices

Φk+1 = ΦkPk , Mk+1 = PTk MkPk , Kk+1 = PT

k KkPk

After convergence:

k = K∞ , m = M∞

Λ =

⎡⎢⎢⎢⎣

λj1 0 · · · 0

0 λj2 · · · 0...

.... . .

...0 0 · · · λjn

⎤⎥⎥⎥⎦ = m−1k , Φ =

[Φ(j1) Φ(j2) · · ·Φ(jn)

]= Φ∞m− 1

2

Orthonormal transformation matrices fulfill, cf. (1-23)

P−1k = PT

k (6–3)

For transformation methods operating on the generalized eigenvalue problem, such as the gen-eral Jacobi iteration method considered in Section 6.2, the transformation matrices Pk are notorthonormal, in which case Mk and Kk converge towards the diagonal matrices m and k asgiven by (1-20) and (1-22). Then, PT

k Pk �= I, and an original SEVP will change into a GEVPduring the iteration process. The eigenvalue matrix Λ and the normalized modal matrix Φ areretrieved as indicated in Box 6.1, where m− 1

2 denotes a diagonal matrix with the components1/√

Mj in the main diagonal.

Some similarity transformation algorithms are devised for the special eigenvalue problem, asis the case for the special Jacobi iteration method in Section 6.1, and the Householder-QRiteration method in Section 6.3. Hence, application of these methods require an initial similaritytransformation from a GEVP to a SEVP as explained in Section 3.4. This may be achieved byspecifying the transformation matrix of the transformation k = 0 in Box 6.1 as, cf. (3-48)

P0 =(S−1)T

(6–4)

where S fulfills (3-44). Then, M1 = I. If the succeeding similarity transformation matrices areorthonormal, then all transformed mass matrices become identity matrices as seen by inductionfrom Mk+1 = PT

k MkPk = PTk IPk = I. Moreover, Φk+1 is orthonormal at each iteration step,

6.1 Introduction 117

as seen by induction from ΦTk+1Φk+1 = PT

k ΦTk ΦkPk = PT

k IPk = I.

Finally, it should be noticed that after convergence the sequence of eigenvalues in the main diag-onal of Λ and the eigenmodes in Φ is not ordered in ascending magnitude of the correspondingeigenvalues as indicated in Box 6.1, where the set of indices (j1, j2, . . . , jn) denotes an arbitrarypermutation of the numbers (1, 2, . . . , n).

Example 6.1: Interchange of rows and columns in GEVP by means of a similarity transformation

Interchange of rows and columns in a matrix may be performed by a similarity transformation. Assume, that theif the ith row and column are to be interchanged with the jth row and colums. Then the similarity transformationmatrix is given as

i j

P =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 · · · 0 · · · 0 · · · 00 1 · · · 0 · · · 0 · · · 0...

.... . .

... · · · ... · · · ...

0 0 · · · 0 · · · 1 · · · 0...

... · · · .... . .

... · · · ...

0 0 · · · 1 · · · 0 · · · 0...

... · · · ... · · · .... . .

...

0 0 · · · 0 · · · 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

i

j

(6–5)

The rule is that ones are placed at the positions (i, j) and (j, i) of P. Consider the generalized eigenvalue problemdefined by (4-16). It is easily verified that P−1 = PT , so the transformation matrix is orthonormal, cf. (3-4). Theintention is to interchange the 1st row and column with the 2nd row and column, and next the new 2nd row andcolumn with the 4th row and column. This is achieved by two similarity transformations with the transformationmatrices P1 and P2, given the following combined transformation matrix obtained as a product of tweo matricesof the type (6-5)

P = P1P2 =

⎡⎢⎢⎢⎣0 1 0 01 0 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

1 0 0 00 0 0 10 0 1 00 1 0 0

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦ (6–6)

The transformed stiffness and mass matrices become

M =

⎡⎢⎢⎢⎣

0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦

T ⎡⎢⎢⎢⎣

0 0 0 00 2 0 00 0 0 00 0 0 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣2 0 0 00 1 0 00 0 0 00 0 0 0

⎤⎥⎥⎥⎦

K =

⎡⎢⎢⎢⎣

0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦

T ⎡⎢⎢⎢⎣

2 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

2 0 −1 −10 1 −1 0

−1 −1 2 0−1 0 0 2

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–7)

In the formulation (6-1 the solutions for the eigenmodes in the transformed system as given by (4-21) - (4-23) maybe written as


Ψ =

⎡⎢⎢⎢⎢⎢⎣

12 − 1

2 0 0√

22

√2

2 0 0

14 +

√2

4 − 14 +

√2

4 1 014 − 1

4 0 1

⎤⎥⎥⎥⎥⎥⎦ (6–8)

The corresponding eigenmodes of the original system is obtained from, cf. (6-1)

Φ = PΨ =

⎡⎢⎢⎢⎣0 0 0 11 0 0 00 0 1 00 1 0 0

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎢⎣

12 − 1

2 0 0√

22

√2

2 0 014 +

√2

4 − 14 +

√2

4 1 014 − 1

4 0 1

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎣

14 − 1

4 0 112 − 1

2 0 014 +

√2

4 − 14 +

√2

4 1 0√

22

√2

2 0 0

⎤⎥⎥⎥⎥⎦ (6–9)

6.2 Special Jacobi Iteration

The special Jacobi iteration algorithm operates on the special eigenvalue problem, so M = Iat the outset. The idea is to ensure during the kth transformation that the off-diagonal compo-nent Kij,k, entering the ith and jth row and column of Kk, becomes zero after the similaritytransformation. The transformation matrix is given as

i j

Pk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 · · · 0 0 · · · 0 0 · · · 0

0 1 · · · 0 0 · · · 0 0 · · · 0...

.... . .

...... · · · ...

... · · · ...0 0 · · · cos θ 0 · · · − sin θ 0 · · · 0

0 0 · · · 0 1 · · · 0 0 · · · 0...

... · · · ......

. . ....

... · · · ...0 0 · · · sin θ 0 · · · cos θ 0 · · · 0

0 0 · · · 0 0 · · · 0 1 · · · 0...

... · · · ...... · · · ...

.... . .

...0 0 · · · 0 0 · · · 0 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

i

j

(6–10)

Basically, (6-10) is a identity matrix, where only the components Pii, Pij, Pji and Pjj are differ-ing. Obviously, (6-10) is orthonormal. The components of the updated similarity transformationmatrix Φk+1 = ΦkPk and the transformed stiffness matrix Kk+1 = PT

k KkPk become

{Φli,k+1 = Φli,k cos θ + Φlj,k sin θ , l = 1, . . . , n

Φlj,k+1 = Φlj,k cos θ − Φli,k sin θ , l = 1, . . . , n(6–11)

6.2 Special Jacobi Iteration 119

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Kii,k+1 = Kii,k cos2 θ + Kjj,k sin2 θ + 2Kij,k cos θ sin θ

Kjj,k+1 = Kjj,k cos2 θ + Kii,k sin2 θ − 2Kij,k cos θ sin θ

Kij,k+1 =(Kjj,k − Kii,k

)cos θ sin θ + Kij,k

(cos2 θ − sin2 θ

)Kli,k+1 = Kil,k+1 = Kli,k cos θ + Klj,k sin θ , l �= i, j

Klj,k+1 = Kjl,k+1 = Klj,k cos θ − Kli,k sin θ , l �= i, j

(6–12)

The remaining components of Φk+1 and Kk+1 are identical to those of Φk and Kk. Hence, onlythe ith and jth row and column of Kk are affected by the transformation.

Box 6.2: Special Jacobi iteration algorithm

Let M0 = I, K0 = K and Φ0 = I. Repeat the following items for the sweeps m =1, 2, . . .

1. Specify omission criteria εm in the mth sweep.

2. Check, if the component Kij,k in the ith row and jth column of Kk fulfills the criteria√K2

ij,k

Kii,kKjj,k< εm

3. If the criteria is fulfilled, then skip to the next component in the sweep. Else performthe following calculations

(a.) Calculate the transformation angle θ from (6-13), and then the transformationmatrix Pk as given by (6-10).

(b.) Calculate the components of the updated similarity transformation matrixΦk+1 = ΦkPk, and the transformed stiffness matrix Kk+1 = PT

k KkPk from(6-11) and (6-12). Notice that k after the mth sweep is of the magnitude12(n − 1)n · m.

After convergence:

Φ∞ = Φ = [Φ(j1) Φ(j2) · · ·Φ(jn)] , K∞ = Λ =

⎡⎢⎢⎢⎣

λj1 0 · · · 0

0 λj2 · · · 0...

.... . .

...0 0 · · · λjn

⎤⎥⎥⎥⎦

Next, the angle θ is determined, so the off-diagonal component Kij,k+1 becomes equal to zero


Kij,k+1 = −1

2

(Kii,k − Kjj,k

)sin 2θ + Kij,k cos 2θ = 0 ⇒

⎧⎪⎨⎪⎩

θ =1

2arctan

(2Kij,k

Kii,k − Kjj,k

), Kii,k �= Kjj,k

θ =π

4, Kii,k = Kjj,k

(6–13)

Notice, that even though Kij,k+1 = 0 after the transformation, a subsequent transformationinvolving either the ith or jth row or column may reintroduce a non-zero value at this position.Optimally, Kij,k should be selected as the numerically largest off-diagonal component in Kk.However, in practice the iteration process is often performed in so-called sweeps, where all12(n − 1)n components above the main diagonal in turn are selected as the critical element to

become zero after the transformation. In this case the method is combined with a criteria foromission of the similarity transformation, in case the component is numerically small. Thetransformation is omitted, if

√K2

ij,k

Kii,kKjj,k

< εm (6–14)

where εm is the omission value in the mth sweep.

Finally, it should be noticed that if K0 has a banded structure, so non-zero components aregrouped in a band around the main diagonal, the banded structure is not preserved during thetransformation process as seen from Example 6.1, where the initial matrix K0 is on a three di-agonal form, whereas the transformed matrix K1 is full, see (6-16) below.

The special Jacobi iteration algorithm can be summarized as indicated in Box 6.2.

Example 6.2: Special Jacobi iteration

Given a special eigenvalue problem with the stiffness matrix

K = K0 =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ , Φ0 =

⎡⎢⎣1 0 00 1 00 0 1

⎤⎥⎦ (6–15)

6.2 Special Jacobi Iteration 121

In the 1st sweep the following calculations are performed for (i, j) = (1, 2) :

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · (−1)2 − 4

)= 0.3927 ⇒

{cos θ = 0.9239sin θ = 0.3827

P0 =

⎡⎢⎣0.9239 −0.3827 00.3827 0.9239 0

0 0 1

⎤⎥⎦

Φ1 = Φ0P0 =

⎡⎢⎣0.9239 −0.3827 00.3827 0.9239 0

0 0 1

⎤⎥⎦ , K1 = PT

0 K0P0 =

⎡⎢⎣ 1.5858 0 −0.3827

0 4.4142 −0.9239−0.3827 −0.9239 2

⎤⎥⎦(6–16)

Next, the calculations are performed for (i, j) = (1, 3) :

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · (−0.3827)1.5858− 2

)= 0.5374 ⇒

{cos θ = 0.8591sin θ = 0.5119

P1 =

⎡⎢⎣0.8591 0 −0.5119

0 1 00.5119 0 0.8591

⎤⎥⎦

Φ2 = Φ1P1 =

⎡⎢⎣0.7937 −0.3827 −0.47290.3287 0.9238 −0.19590.5119 0 0.8591

⎤⎥⎦ , K2 = PT

1 K1P1 =

⎡⎢⎣ 1.3578 −0.4729 0−0.4729 4.4142 −0.7937

0 −0.7937 2.2280

⎤⎥⎦

(6–17)

Finally, to end the 1st sweep the calculations are performed for (i, j) = (2, 3) :

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · (−0.7937)4.4142− 2.2280

)= −0.3140 ⇒

{cos θ = 0.9511sin θ = −0.3089

P2 =

⎡⎢⎣1 0 0

0 0.9511 0.30890 −0.3089 0.9511

⎤⎥⎦

Φ3 = Φ2P2 =

⎡⎢⎣0.7937 −0.2179 −0.56800.3287 0.9392 0.09910.5119 −0.2653 0.8171

⎤⎥⎦ , K3 = PT

2 K2P2 =

⎡⎢⎣ 1.3578 −0.4498 −0.1461−0.4498 4.6720 0−0.1461 0 1.9703

⎤⎥⎦

(6–18)

Φ3 and K3 represents the estimates of the modal matrix Φ and Λ after the 1st sweep. As seen the K 12,1 = 0,whereas K12,2 = −0.4729. This is in agreement with the statement above, that off-diagonal components set tozero in one iteration, may attain non-zero values in a later iteration. Comparison of K 0 to K3 shows that thenumerical maximum off-diagonal component has decreased from | − 1| = 1 to | − 0.4498| after the 1st sweep.Hence, the algorithm is converging.

At the end of the 2nd and 3rd sweep the following estimates are obtained for the modal matrix and the eigenvalues


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ6 =

⎡⎢⎣0.6276 −0.3258 −0.70710.4607 0.8876 0.00000.6276 −0.3258 0.7071

⎤⎥⎦ , K6 =

⎡⎢⎣ 1.2680 0.0039 −0.0000

0.0039 4.7320 0−0.0000 0 2.0000

⎤⎥⎦

Φ9 =

⎡⎢⎣0.6280 −0.3251 −0.70710.4597 0.8881 0.00000.6280 −0.3251 0.7071

⎤⎥⎦ , K9 =

⎡⎢⎣ 1.2679 −0.0000 −0.0000−0.0000 4.7321 0−0.0000 0 2.0000

⎤⎥⎦

(6–19)

As seen the eigenmodes are stored column-wise in Φ according to the permutation (j1, j2, j3) =(1, 3, 2).

6.3 General Jacobi Iteration

The general Jacobi iteration method operates on the generalized eigenvalue problem, i.e. M �=I. The idea of the transformation is to ensure that during the kth transformation the off-diagonalcomponent Mij,k and Kij,k, entering the ith and jth row and column of Mk and Kk, simultane-ous become zero after the similarity transformation.

xi1

1

xj[− sin θ

cos θ

] [cos θ

sin θ

][β

1

]

[1

α

]

Fig. 6–1 Projection of ith and jth column vectors of similarity transformation matrix in the (x i, xj)-plane.

The transformation matrix is given as

6.3 General Jacobi Iteration 123

i j

Pk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 · · · 0 0 · · · 0 0 · · · 0

0 1 · · · 0 0 · · · 0 0 · · · 0...

.... . .

...... · · · ...

... · · · ...0 0 · · · 1 0 · · · β 0 · · · 0

0 0 · · · 0 1 · · · 0 0 · · · 0...

... · · · ......

. . ....

... · · · ...0 0 · · · α 0 · · · 1 0 · · · 0

0 0 · · · 0 0 · · · 0 1 · · · 0...

... · · · ...... · · · ...

.... . .

...0 0 · · · 0 0 · · · 0 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

i

j

(6–20)

Because we have to specify requirements for both Mij,k+1 and Kij,k+1, we need two free pa-rameters α and β in the transformation matrix, where only the angle θ appears in (6-10). As aconsequence (6-20) is not orthonormal. Actually, the ith and jth column vectors neither havethe length 1 nor are mutual orthogonal, by contrast to the corresponding vectors in (6-10), seeFig. 6-1. The components of the updated similarity transformation matrix Φk+1 = ΦkPk

and the transformed mass and stiffness matrices, Mk+1 = PTk MkPk and Kk+1 = PT

k KkPk,become

{Φli,k+1 = Φli,k + α Φlj,k , l = 1, . . . , n

Φlj,k+1 = Φlj,k + β Φli,k , l = 1, . . . , n(6–21)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Mii,k+1 = Mii,k + α2Mjj,k + 2αMij,k

Mjj,k+1 = Mjj,k + β2Mii,k + 2βMij,k

Mij,k+1 = βMii,k + αMjj,k + Mij,k

(1 + αβ

)Mli,k+1 = Mil,k+1 = Mli,k + αMlj,k , l �= i, j

Mlj,k+1 = Mjl,k+1 = Mlj,k + βMli,k , l �= i, j

(6–22)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Kii,k+1 = Kii,k + α2Kjj,k + 2αKij,k

Kjj,k+1 = Kjj,k + β2Kii,k + 2βKij,k

Kij,k+1 = βKii,k + αKjj,k + Kij,k

(1 + αβ

)Kli,k+1 = Kil,k+1 = Kli,k + αKlj,k , l �= i, j

Klj,k+1 = Kjl,k+1 = Klj,k + βKli,k , l �= i, j

(6–23)

The remaining components of Φk+1, Mk+1 and Kk+1 are identical to those of Φk, Mk and Kk.Hence, only the ith and the jth row and columns of Kk and Mk are affected by the transforma-tion.


Next, the parameters α and β are determined, so the off-diagonal components Mij,k+1 andKij,k+1 become equal to zero

Mij,k+1 = βMii,k + αMjj,k + Mij,k

(1 + αβ

)= 0

Kij,k+1 = βKii,k + αKjj,k + Kij,k

(1 + αβ

)= 0

⎫⎬⎭ (6–24)

The solution of (6-24) becomes, see Box 6.3

α =1

a

(1

2−√

1

4+ ab

), β = −a

bα

a =Kjj,kMij,k − Mjj,kKij,k

Kii,kMjj,k − Mii,kKjj,k

b =Kii,kMij,k − Mii,kKij,k

Kii,kMjj,k − Mii,kKjj,k

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

, if Kii,kMjj,k �= Mii,kKjj,k

α =

√Kii,kMij,k − Mii,kKij,k

Kjj,kMij,k − Mjj,kKij,k

, β = − 1

α, if Kii,kMjj,k = Mii,kKjj,k

(6–25)


Box 6.3: Proof of equation (6-25)

From (6-19) follows

βKii,k + αKjj,k

βMii,k + αMjj,k

=Kij,k

Mij,k

⇒ β = −Kjj,kMij,k − Mjj,kKij,k

Kii,kMij,k − Mii,kKij,k

α (6–26)

Elimination of β in the 1st equation in (6-24) by means of (6-26) provides the followingquadratic equation in α

Mij,k

(Kjj,kMij,k − Mjj,kKij,k

)α2 − Mij,k

(Kii,kMjj,k − Mii,kKjj,k

)α−

Mij,k

(Kii,kMij,k − Mii,kKij,k

)= 0 (6–27)

If Kii,kMjj,k = Mii,kKjj,k the coefficient in front of α cancels. Then, in combination to(6-26) the following solutions are obtained for α and β

α = ±√

Kii,kMij,k − Mii,kKij,k

Kjj,kMij,k − Mjj,kKij,k

, β = − 1

α(6–28)

If Kii,kMjj,k �= Mii,kKjj,k solutions of the quadratic equation for α in combination to(6-26) provides

α =1

a

(1

2±√

1

4+ ab

), β = −a

bα (6–29)

where a and b are as given in (6-25). Both sign combinations in (6-28) and (6-29) will do.

The transformations are performed in sweeps as for the special Jacobi method. In this case thecriteria for omitting a transformation during the mth sweep may be formulated as

√K2

ij,k

Kii,kKjj,k

+M2

ij,k

Mii,kMjj,k

< εm (6–30)

where εm is the omission value in the mth sweep.

The general Jacobi iteration algorithm can be summarized as indicated in Box 6.4.


Box 6.4: General Jacobi iteration algorithm

Let M0 = M, K0 = K and Φ0 = I. Repeat the following items for the sweeps m =1, 2, . . .

1. Specify omission criteria εm in the mth sweep.

2. Check, if the components Mij,k and Kij,k in the ith row and jth column of Mk Kk

fulfill the criteria√K2

ij,k

Kii,kKjj,k+

M2ij,k

Mii,kMjj,k< εm

3. If the criteria is fulfilled, then skip to the next component in the sweep. Else performthe following calculations

(a.) Calculate the parameters α and β as given by (6-29), and then the transforma-tion matrix Pk as given by (6-20).

(b.) Calculate the components of the updated similarity transformation matrixΦk+1 = ΦkPk, and the transformed mass and stiffness matrices Mk+1 =PT

k MkPk and Kk+1 = PTk KkPk from (6-21), (6-22) and (6-23). Notice that k

after the mth sweep is of the magnitude 12(n − 1)n · m.

After convergence:

k = K∞ , m = M∞

Λ =

⎡⎢⎢⎢⎣

λj1 0 · · · 0

0 λj2 · · · 0...

.... . .

...0 0 · · · λjn

⎤⎥⎥⎥⎦ = m−1k , Φ = [Φ(j1) Φ(j2) · · ·Φ(jn)] = Φ∞m− 1

2

Example 6.3: General Jacobi iteration

Given a generalized eigenvalue problem with the mass and stiffness matrices

M = M0 =

⎡⎢⎣0.5 0.5 00.5 1 0.50 0.5 1

⎤⎥⎦ , K = K0 =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ , Φ0 =

⎡⎢⎣1 0 00 1 00 0 1

⎤⎥⎦ (6–31)


In the 1st sweep the following calculations are performed for (i, j) = (1, 2) :

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

⎧⎪⎪⎪⎨⎪⎪⎪⎩

α =

√2 · 0.5 − 0.5 · (−1)4 · 0.5 − 1 · (−1)

= 0.7071

β = − 10.7071

= −1.4142

(NB : K11,0M22,0 = K22,0M11,0

)

P0 =

⎡⎢⎣ 1 −1.4142 00.7071 1 0

0 0 1

⎤⎥⎦ , Φ1 = Φ0P0 =

⎡⎢⎣ 1 −1.4142 00.7071 1 0

0 0 1

⎤⎥⎦

M1 = PT0 M0P0 =

⎡⎢⎣1.7071 0 0.3536

0 0.5858 0.50.3536 0.5 1

⎤⎥⎦ , K1 = PT

0 K0P0 =

⎡⎢⎣ 2.5858 0 −0.7071

0 10.8284 −1−0.7071 −1 2

⎤⎥⎦

(6–32)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

a =2 · 0.3536− 1 · (−0.7071)

2.5858 · 1 − 1.7071 · 2 = −1.7071

b =2.5858 · 0.3536− 1.7071 · (−0.7071)

2.5858 · 1 − 1.7071 · 2 = −2.5607

⎫⎪⎪⎬⎪⎪⎭ ⇒

{α = 0.9664β = −0.6443

P1 =

⎡⎢⎣ 1 0 −0.6443

0 1 00.9664 0 1

⎤⎥⎦ , Φ2 = Φ1P1 =

⎡⎢⎣ 1 −1.4142 −0.64430.7071 1 −0.45560.9664 0 1

⎤⎥⎦

M2 = PT1 M1P1 =

⎡⎢⎣3.3243 0.4832 00.4832 0.5858 0.5

0 0.5 1.2530

⎤⎥⎦ , K2 = PT

1 K1P1 =

⎡⎢⎣ 3.0869 −0.9664 0−0.9664 10.8284 −1

0 −1 3.9844

⎤⎥⎦

(6–33)

Finally, to end the 1st sweep the calculations are performed for (i, j) = (2, 3) :⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

a =3.9844 · 0.5 − 1.2530 · (−1)

10.8284 · 1.2530− 0.5858 · 3.9844= 0.2889

b =10.8284 · 0.5 − 0.5858 · (−1)

10.8284 · 1.2530− 0.5858 · 3.9844= 0.5341

⎫⎪⎪⎬⎪⎪⎭ ⇒

{α = −0.4702β = 0.2543

P2 =

⎡⎢⎣1 0 00 1 0.25430 −0.4702 1

⎤⎥⎦ , Φ3 = Φ2P2 =

⎡⎢⎣ 1 −1.1113 −1.00390.7071 1.2142 −0.20120.9664 −0.4702 1

⎤⎥⎦

M3 = PT2 M2P2 =

⎡⎢⎣3.3243 0.4832 0.12290.4832 0.3926 00.1229 0 1.5452

⎤⎥⎦ , K3 = PT

2 K2P2 =

⎡⎢⎣ 3.0869 −0.9664 −0.2458−0.9664 12.6498 0−0.2458 0 4.1761

⎤⎥⎦

(6–34)

At the end of the 2nd and 3rd sweep the following estimates are obtained for the modal matrix and the transformedmass and stiffness matrices


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ6 =

⎡⎢⎣0.7494 −1.2825 −1.07420.8195 1.0999 −0.28651.0376 −0.6084 0.9213

⎤⎥⎦

M6 =

⎡⎢⎣ 3.4931 −0.0024 0.0000−0.0024 0.3225 0

0.0000 0 1.5517

⎤⎥⎦ , K6 =

⎡⎢⎣ 3.0336 0.0048 −0.0000

0.0048 13.029 0−0.0000 0 4.2464

⎤⎥⎦

Φ9 =

⎡⎢⎣0.7501 −1.2820 −1.07420.8189 1.1005 −0.28651.0379 −0.6076 0.9213

⎤⎥⎦

M9 =

⎡⎢⎣ 3.4932 −0.0000 0.0000−0.0000 0.3225 0

0.0000 0 1.5517

⎤⎥⎦ , K9 =

⎡⎢⎣ 3.0336 0.0000 −0.0000

0.0000 13.029 0−0.0000 0 4.2464

⎤⎥⎦

(6–35)

Presuming that the process has converged after the 3rd sweep the eigenvalues and normalized eigenmodes are nextretrieved by the following calculations, cf. Box. 6.4

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

m = M9 =

⎡⎢⎣ 3.4932 −0.0000 0.0000−0.0000 0.3225 0

0.0000 0 1.5517

⎤⎥⎦ , m− 1

2 =

⎡⎢⎣0.5350 0 0

0 1.7608 00 0 0.8028

⎤⎥⎦ ⇒

Λ =

⎡⎢⎣λ1 0 0

0 λ3 00 0 λ2

⎤⎥⎦ = M−1

9 K9 =

⎡⎢⎣ 0.8684 0.0000 −0.0000

0.0000 40.395 −0.0000−0.0000 −0.0000 2.7365

⎤⎥⎦

Φ =[Φ(1) Φ(3) Φ(2)

]= Φ9m− 1

2 =

⎡⎢⎣0.4013 −2.2573 −0.86230.4381 1.9378 −0.23000.5553 −1.0698 0.7396

⎤⎥⎦

(6–36)

The reader should verify that the solution matrices within the indicated accuracy fulfill ΦTMΦ =I and ΦTKΦ = Λ.

6.4 Householder Reduction

The Householder reduction method operates on the special eigenvalue problem . Hence, a pre-liminary similarity transformation of a GEVP to a SEVP must be performed as explained inSection 3.4.

The Householder method reduces a symmetric matrix K1 to three diagonal form by totally n−2consecutive similarity transformations. After the (n − 2)th transformation the stiffness matrixhas the form

6.4 Householder Reduction 129

Kn−1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 β1 0 · · · 0 0

β1 α2 β2 · · · 0 0

0 β2 α3 · · · 0 0...

......

. . ....

...0 0 0 · · · αn−1 βn−1

0 0 0 · · · βn−1 αn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6–37)

During the reduction process the numbers α1, . . . , αn and β1, . . . , βn−1, as well as the sequenceof transformation matrices P1, . . . ,Pn−2 are determined. Since all transformation matrices be-come orthonormal all transformed mass matrices remain unit matrices.

After completing the Householder reduction process the special eigenvalue problem with thethree diagonal matrix Kn−1 must be solved by some kind of iteration method, which preservesthe three diagonal structure of the reduced system matrix, and benefits from this reduced struc-ture in order to improve the calculation time. As mentioned in Section 6.2 this requirementrules out the special Jacobi iteration method. Since, the inverse of a three diagonal matrix isfull, inverse vector iteration with Gram-Schmidt orthogonalization must also be avoided. Of themethods discussed hitherto only forward vector iteration with Gram-Schmidt orthogonalizationmeets the requirement. As wee shall see the requirements are also met by the QR iterationmethod to be discussed in Section 6.5. Finally, an initial Householder reduction is favorable inrelation to characteristic polynomial iteration methods discussed in Section 7.4.

The transformation matrix during the kth similarity transformation is given as follows

Pk = I− 2wkwTk , |wk| = 1 (6–38)

wk denotes a unit column vector to be determined below. Hence, wTk wk = 1.

Obviously, Pk is symmetric, i.e. Pk = PTk . Moreover, Pk is orthonormal as seen from the

following derivation

PkPTk =

(I − 2wkw

Tk

)(I − 2wkw

Tk

)=

I − 2wkwTk − 2wkw

Tk + 4

(wT

k wk

)wkw

Tk = I ⇒

PTk = P−1

k (6–39)

As mentioned, this means that the mass matrix remains an identity matrix during the House-holder similarity transformations, because this is ensured in the initial transformation from aGEVP to a SEVP, as explained in the remarks subsequent to (6-4).


l

x

Pkx

wTk x

wTk x

−2wk(wTk x)

wk

Fig. 6–2 Geometrical interpretation of the Householder transformation.

Consider a given column vector x. Then,

Pkx =(I − 2wkw

Tk

)x = x − 2

(wT

k x)wk (6–40)

Notice that wTk x is a scalar. The transformed vector, Pkx, may be interpreted as a reflection of

x in the line l, which is orthogonal to the vector wk and placed in the plane spanned by x andwk as illustrated in Fig. 6-2.

At the kth transformation the applied unit vector wk is taken on the following form

wk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0...0

wk+1...

wn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎣ 0

wk

⎤⎦}

k rows}n − k rows

(6–41)

where

wTk wk = wT

k wk = w2k+1 + · · ·+ w2

n = 1 (6–42)

Then, the transformation matrix may be written on the following matrix form

kcolumns︷︸︸︷

n − kcolumns︷︸︸︷

Pk =

⎡⎣In−k 0

0 Pk

⎤⎦}

k rows}n − k rows

, Pk = Ik − 2wkwTk

(6–43)

where Ik denotes a unit matrix of dimension (n − k) × (n − k).


In order to determine the sub-vector wk defining the transformation matrix, the stiffness matrixbefore the kth similarity transformation is considered, at which stage the stiffness matrix hasbeen reduced to three diagonal form down to and including the (k − 1)th row and column.Hence, the stiffness matrix has the structure

n − kcolumns

k ︷︸︸︷

Kk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 β1 0 · · · 0 0 0

β1 α2 β2 · · · 0 0 0

0 β2 α3 · · · 0 0 0...

......

. . ....

......

0 0 0 · · · αk−1 βk−1 0

0 0 0 · · · βk−1 Kkk kk

0 0 0 · · · 0 kTk Kk

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ k}

n − k rows

(6–44)

kk is a row vector of the dimension (n − k), and Kk is a symmetric matrix of the dimension(n − k) × (n − k) defined as

kk = [Kk k+1 Kk k+2 · · ·Kkn] (6–45)

Kk =

⎡⎢⎢⎢⎣Kk+1 k+1 Kk+1 k+2 · · · Kk+1n

Kk+2 k+1 Kk+2 k+2 · · · Kk+2n...

.... . .

...Kn k+1 Kn k+2 · · · Knn

⎤⎥⎥⎥⎦ (6–46)

Then, with the transformation matrix given by (6-43) the stiffness matrix after the kth transfor-mation becomes

n − kcolumns

k ︷︸︸︷

Kk+1 = PTk KkPk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 β1 0 · · · 0 0 0

β1 α2 β2 · · · 0 0 0

0 β2 α3 · · · 0 0 0...

......

. . ....

......

0 0 0 · · · αk−1 βk−1 0

0 0 0 · · · βk−1 Kkk kkPk

0 0 0 · · · 0 PTk kT

k PTk KkPk

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ k}

n − k rows

(6–47)


where

αk = Kkk (6–48)

kkPk = kk

(Ik − 2wkw

Tk

)= kk − 2

(kkwk

)wT

k (6–49)

PTk KkPk = Kk − 2wkw

Tk Kk − 2Kkwkw

Tk + 4

(wT

k Kkwk

)wkw

Tk (6–50)

Since, kk is a row vector and wk is a column vector, kkwk is a scalar. Similarly, wTk Kkwk

becomes a scalar.

If the kth row and column in (6-45) should be on a three-diagonal form, it is required that

PTk kT

k = βkek , ek =

⎡⎢⎢⎢⎣

1

0...0

⎤⎥⎥⎥⎦ (6–51)

where ek is a unit column vector of dimension (n−k). The transformation matrix is symmetric,so PT

k kTk = Pkk

Tk . Moreover, Pkk

Tk is a reflection of the vector kT

k in the line l as depicted inFig. 6-2, and hence has the length |kT

k |. Hence, it follows that βk should be selected as

βk = ±|kk| (6–52)

Then it follows from (6-49) that

kTk − 2

(kkwk

)wk = ±|kk|ek ⇒

wk = a(kT

k ∓ |kk|ek

)(6–53)

where it is noticed that 2(kT

k wk

)is a scalar, which may be absorbed in the coefficient a. a is

determined so the vector wk is of unit length.


Box 6.5: Householder reduction algorithm

Transform the GEVP to a SEVP by the similarity transformation matrix P =(S−1)T

,where S is a solution to M = SST , and define the initial transformation and stiffnessmatrices as

K1 = S−1K(S−1)T

, Φ1 =(S−1)T

Next, repeat the following items for k = 1, . . . , n − 2

1. Calculate the similarity transformation matrix Pk at the kth similarity transformationby (6-43), (6-55).

2. Calculate updated transformation and stiffness matrices from (6-47), (6-57)

Φk+1 = ΦkPk , Kk+1 = PTk KkPk

After completion of the reduction process the following standard eigenvalue problem issolved by some iteration method

Kn−1V = VΛ

Λ is the diagonal eigenvalue matrix of the original GEVP, and V is the orthonormaleigenvector matrix of the three diagonal matrix Kn−1. Then, the eigenmodes normalizedto unit modal mass of the original GEVP are retrieved from the matrix product

Φ = Φn−1V

Both sign combinations in (6-52) and (6-53) will do. However, in order to prevent numericalproblems of the algorithm in the case, where kk Kk k+1ek the following choice of sign in thesolutions for βk and wk should be preferred

βk = −sign(Kk k+1

)|kk| (6–54)

wk =kT

k + sign(Kk k+1)|kk|ek∣∣kTk + sign(Kk k+1)|kk|ek

∣∣ (6–55)

The updated transformation matrix before the kth transformation is partitioned as follows



Φk =

⎡⎣Φ11 Φ12

Φ21 Φ22

⎤⎦}

k rows}n − k rows

(6–56)


With the transformation matrix as given by (6-43) the transformation matrix after the kth trans-formation becomes



Φk+1 =

⎡⎣Φ11 Φ12Pk

Φ21 Φ22Pk

⎤⎦}

k rows}n − k rows

(6–57)

Finally, it should be noticed that alternative algorithms for reduction to three diagonal form havebeen indicated by Givens1 and Lanczos.2

Example 6.4: Householder reduction

Given a generalized eigenvalue problem with the mass and stiffness matrices given by (5-74). The similaritytransformation matrix transforming from a GEVP to a SEVP becomes

S = M12 =

⎡⎢⎢⎢⎣√

2 0 0 00

√2 0 0

0 0 1 00 0 0 1

⎤⎥⎥⎥⎦ ⇒ (

S−1)T =

⎡⎢⎢⎢⎣

√2

2 0 0 00

√2

2 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦ (6–58)

Then, the stiffness matrix and updated transformation matrix before the 1st Householder similarity transformationbecomes, cf. (3-47), (3-48)

K1 = S−1K(S−1)T =

⎡⎢⎢⎢⎣

√2

2 0 0 00

√2

2 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

5 −4 1 0−4 6 −4 1

1 −4 6 −40 1 −4 5

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

√2

2 0 0 00

√2

2 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦ ⇒

K1 =

⎡⎢⎢⎢⎣

52 −2

√2

2 0−2 3 −2

√2

√2

2√2

2 −2√

2 6 −40

√2

2 −4 5

⎤⎥⎥⎥⎦

Φ1 =

⎡⎢⎢⎢⎣

√2

2 0 0 00

√2

2 0 00 0 1 00 0 0 1

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–59)

At the Householder transformation k = 1 one has

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

α1 =52

k1 =

[−2

√2

20

], |k1| =

3√

22

(6–60)

1W. Givens: Numerical Computation of the Characteristic Values of a Real Symmetric Matrix. Oak RidgeNational Laboratory, ORNL - 1574, 1954.

2C. Lanczos: An Iterative Method for the Solution of the Eigenvalue Problem of Linear Differential and IntegralOperators. Journal of Research of the National Bureau of Standards, 45(4), 1950, 255-282.


Then, cf. (6-43), (6-54) and (6-55)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

β1 = −sign(−2)3√

22

=3√

22

= 2.1213

w1 = a

⎛⎜⎝⎡⎢⎣−2√

22

0

⎤⎥⎦+ sign(−2)

3√

22

⎡⎢⎣100

⎤⎥⎦⎞⎟⎠ = a

⎡⎢⎣−2 − 3

√2

2√2

2

0

⎤⎥⎦ ⇒ w1 =

⎡⎢⎣−0.9856

0.16910

⎤⎥⎦

P1 = I1 − 2w1wT1 =

⎡⎢⎣−0.9828 0.3333 0

0.3333 0.9428 00 0 1

⎤⎥⎦

(6–61)

The stiffness matrix and updated transformation matrix after the Householder transmission k = 1 becomes

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

K2 = PT1 K1P1 =

⎡⎢⎢⎢⎣

2.5000 2.1213 0 02.1213 5.1111 3.1427 −2.0000

0 3.1427 3.8889 −3.53550 −2.0000 −3.5355 5.000

⎤⎥⎥⎥⎦

Φ2 = Φ1P1 =

⎡⎢⎢⎢⎣0.7071 0 0 0

0 −0.6667 0.2357 00 0.3333 0.9428 00 0 0 1

⎤⎥⎥⎥⎦

(6–62)

where the transformed matrices are calculated by means of (6-47) and (6-57), respectively.

At the Householder transformation k = 2 the following calculations are performed

⎧⎨⎩

α2 = 5.1111

k2 = [3.1427 − 2.0000] , |k2| = 3.7251(6–63)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

β2 = −sign(3.1427) · 3.7251 = −3.7251

w2 = a

([3.1427

−2.0000

]+ sign(3.1427) · 3.7251

[10

])= a

[6.8678

−2.0000

]⇒ w2 =

[0.9601

−0.2796

]

P2 = I2 − 2w2wT2 =

[−0.8436 0.5369

0.5369 0.8436

] (6–64)

The stiffness matrix and updated transformation matrix after the Householder transmission k = 2 becomes


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

K3 = PT2 K2P2 =

⎡⎢⎢⎢⎣

2.5000 2.1213 0 02.1213 5.1111 −3.7251 −0.0000

0 −3.7251 7.4120 2.00050 −0.0000 2.0005 1.4769

⎤⎥⎥⎥⎦

Φ3 = Φ2P2 =

⎡⎢⎢⎢⎣0.7071 0 0 0

0 −0.6667 −0.1988 0.12650 0.3333 −0.7954 0.50620 0 0.5369 0.8436

⎤⎥⎥⎥⎦

(6–65)

The reader should verify that the solution matrices within the indicated accuracy fulfill Φ T3 MΦ3 = I and

ΦT3 KΦ3 = K3.

6.5 QR Iteration

As is the case for the Householder reduction method QR-iteration operates on the standardeigenvalue problem, so an initial similarity transformation of the GEVP to a SEVP is presumed.Let K1 = S−1K

(S−1)T

denote the stiffness matrix after the initial similarity transformation,where S is a solution to M = SST , cf. (3-44), (3-47).

QR iteration is based on the following property that any non-singular matrix K can be factorizedon the following form

K = QR (6–66)

where Q is an orthonormal matrix, and R is an upper triangular matrix. Hence, Q and R havethe form

Q =[q1 q2 · · ·qn

], qT

k qj = δkj (6–67)

R =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

r11 r12 r13 r14 · · · r1n

0 r22 r23 r24 · · · r2n

0 0 r33 r34 · · · r3n

0 0 0 r44 · · · r4n...

......

.... . .

...0 0 0 0 · · · rnn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6–68)

where δij denotes Kronecker’s delta. It should be noticed that the factorization (6-66) holdseven for non-symmetric matrices. The orthonormality of Q, which implies that Q−1 = QT , is

6.5 QR Iteration 137

essential to the method.

Based on K1 a sequence of transformed stiffness matrices Kk are next constructed with the QRfactors Qk and Rk according to the algorithm

Kk = QkRk

Kk+1 = QTk KkQk = QT

k QkRkQk = RkQk

⎫⎬⎭ (6–69)

Hence, Kk+1 is obtained by a similarity transformation with the transformation matrix Qk. Thetransformation is reduced to an evaluation of RkQk due to the orthonormality property of Qk.For the same reason all transformed mass matrices remain unit matrices.


Box 6.6: Proof of equation (6-61)

Let k1 k2, . . . ,kn denote the column vectors of the matrix K, i.e.

K =[k1 k2 · · ·kn

](6–70)

Since K is non-singular, k1 k2, . . . ,kn are linearly independent, and hence form a vectorbasis. A new orthonormal vector basis q1 q2 · · ·qn linearly dependent on k1 k2, . . . ,kn

may then be constructed by a process, which resembles the Gram-Schmidt orthogonaliza-tion described in Section 5.5. (6-66) is identical to the following relations

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k1 = r11q1

k2 = r12q1 + r22q2...

kj = r1jq1 + r2jq2 + · · · + rjjqj =

j∑k=1

rkjqk

...

kn =

n∑k=1

rknqk

(6–71)

(6-71) is solved sequentially downwards using the properties of orthonormality of qj.From the 1st equation follows by scalar multiplication with q1

r11 = |k1| ⇒ q1 =1

r11

k1 (6–72)

Now, q1 and r11 are known. Scalar multiplication of the 2nd equation with q1, and use ofthe orthogonality property qT

1 q2 = 0, provides

r12 = qT1 k2 ⇒ r22 = |k2 − r12q1| ⇒ q2 =

1

r22

(k2 − r12q1

)(6–73)

At the determination of qj , 1 < j ≤ n, the mutually ortonormal basis vectorsq1,q2, . . . ,qj−1 have already been determined. Scalar multiplication of the jth equationwith qk, k = 1, 2, . . . , j − 1, and use of the orthogonality property qT

k qj = 0, provides

rkj = qTk kj ⇒ rjj =

∣∣∣∣∣kj −j−1∑k=1

rkjqk

∣∣∣∣∣ ⇒ qj =1

rjj

(kj −

j−1∑k=1

rkjqk

)(6–74)

Hence a solution fulfilling all requirements has been obtained for the components rkj ofR and the column vectors qj of Q, which proves the validity of the factorization (6-66).


Box 6.7: QR iteration algorithm

Transform the GEVP to a SEVP by the similarity transformation matrix P =(S−1)T

,where S is a solution to M = SST , and define the initial transformation and stiffnessmatrices as

K1 = S−1K(S−1)T

, Φ1 =(S−1)T

Repeat the following items for k = 1, 2, . . .

1. Perform a QR factorization of the stiffness matrix before the kth similarity transfor-mation

Kk = QkRk

2. Calculate updated transformation and stiffness matrices by a similarity transforma-tion with the orthonormal transformation matrix Qk

Φk+1 = ΦkQk , Kk+1 = QTk KkQk = RkQk

After convergence:

Λ =

⎡⎢⎢⎢⎣λn 0 · · · 0

0 λn−1 · · · 0...

.... . .

...0 0 · · · λ1

⎤⎥⎥⎥⎦ = K∞ = R∞ , Φ =

[Φ(n) Φ(n−1) · · ·Φ(1)

]= Φ∞

Now, it can be proved that

K∞ = R∞ = Λ =

⎡⎢⎢⎢⎣λn 0 · · · 0

0 λn−1 · · · 0...

.... . .

...0 0 · · · λ1

⎤⎥⎥⎥⎦ , Φ∞ = Φ =

[Φ(n) Φ(n−1) · · ·Φ(1)

](6–75)

Qk converge to a unit matrix, as a consequence of K∞ = R∞.

As seen, at convergence the eigen-pairs are ordered in descending order of the eigenvalues.Moreover, the algorithm converges faster to the lowest eigenmode than to the largest, as is thecase for subspace iteration as describes in Section 7.3, a method which has some resemblanceto QR iteration. The rate of convergence seems to be rather comparable to that of subspace iter-ation. These properties have been illustrated in Example 6.4 below. The proof of convergenceand the associated determination of the convergence rate is rather tedious and involved, and willbe omitted here.


The general QR iteration algorithm can be summarized as indicated in Box 6.7.

Usually, the QR algorithm becomes computational expensive when applied to large full ma-trices, due to the time consuming orthogonalization process involved in the QR factorization.However, if Kk is on the three diagonal form (6-37), it can be shown that matrices Rk and Qk

have the form

Rk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

r11 r12 r13 0 0 · · · 0

0 r22 r23 r24 0 · · · 0

0 0 r33 r34 r35 · · · 0

0 0 0 r44 r45 · · · 0

0 0 0 0 r55 · · · 0...

......

......

. . ....

0 0 0 0 0 · · · rnn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6–76)

Qk =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

q11 q12 q13 q14 q15 · · · q1n

q21 q22 q23 q24 q25 · · · q2n

0 q32 q33 q34 q35 · · · q3n

0 0 q42 q44 q45 · · · q4n

0 0 0 q54 q55 · · · q5n...

......

......

. . ....

0 0 0 0 0 · · · qnn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6–77)

Hence, Rk becomes an upper three diagonal matrix with only 3n− 3 nontrivial coefficients rjk

versus 12n(n + 1) for a full matrix Kk. Similarly, Qk contains zeros below the first lower

diagonal. As a consequence of the indicated structure of Rk and Qk, the matrix productKk+1 = RkQk will again be a symmetric three diagonal matrix. Hence, this property is pre-served for the transformed stiffness matrices during the iteration process. This motivates theapplication of QR iteration in combination to an initial Householder reduction of the initialgeneralized eigenvalue problem to three diagonal form, which is known as the HOQR method.

Example 6.5: HOQR iteration

QR iteration is performed on the stiffness matrix of Example 6.3, which has been reduced to three diagonal formby Householder reduction. Hence, the initial stiffness matrix and updated transformation matrix reads, cf. (6-65)

K1 =

⎡⎢⎢⎢⎣2.5000 2.1213 0 02.1213 5.1111 −3.7251 −0.0000

0 −3.7251 7.4120 2.00050 −0.0000 2.0005 1.4769

⎤⎥⎥⎥⎦ , Φ1 =

⎡⎢⎢⎢⎣0.7071 0 0 0

0 −0.6667 −0.1988 0.12650 0.3333 −0.7954 0.50620 0 0.5369 0.8436

⎤⎥⎥⎥⎦

(6–78)

At the determination of q1 and r11 in the 1st QR iteration the following calculations are performed, cf. (6-72)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k1 =

⎡⎢⎢⎢⎣2.50002.1312

00

⎤⎥⎥⎥⎦ , r11 =

∣∣∣∣∣∣∣∣∣

⎡⎢⎢⎢⎣

2.50002.1312

00

⎤⎥⎥⎥⎦∣∣∣∣∣∣∣∣∣= 3.2787

q1 =1

3.2787

⎡⎢⎢⎢⎣

2.50002.1312

00

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

0.76250.6470

00

⎤⎥⎥⎥⎦

(6–79)

q2 and r12, r22 are determined from the following calculations, cf. (6-73)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k2 =

⎡⎢⎢⎢⎣

2.12135.1111

−3.72510

⎤⎥⎥⎥⎦ , r12 =

⎡⎢⎢⎢⎣

0.76250.6470

00

⎤⎥⎥⎥⎦

T ⎡⎢⎢⎢⎣

2.12135.1111

−3.72510

⎤⎥⎥⎥⎦ = 4.9244

r22 =

∣∣∣∣∣∣∣∣∣

⎡⎢⎢⎢⎣

2.12135.1111

−3.72510

⎤⎥⎥⎥⎦− 4.9244 ·

⎡⎢⎢⎢⎣0.76250.6470

00

⎤⎥⎥⎥⎦∣∣∣∣∣∣∣∣∣= 4.5001

q2 =1

4.5001

⎛⎜⎜⎜⎝⎡⎢⎢⎢⎣

2.12135.1111

−3.72510

⎤⎥⎥⎥⎦− 4.9244 ·

⎡⎢⎢⎢⎣0.76250.6470

00

⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠ =

⎡⎢⎢⎢⎣−0.3630

0.4278−0.8278

0

⎤⎥⎥⎥⎦

(6–80)

q3 and r13, r23, r33 are determined from the following calculations, cf. (6-74)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k3 =

⎡⎢⎢⎢⎣

0−3.7251

7.41202.0005

⎤⎥⎥⎥⎦ , r13 = qT

1 k3 = −2.4101 , r23 = qT2 k3 = −7.7292

r33 =∣∣∣k3 + 2.4101q1 + 7.7292q2

∣∣∣ = 2.6959

q3 =1

2.6959

(k3 + 2.4101q1 + 7.7292q2

)=

⎡⎢⎢⎢⎣−0.3590

0.42310.37610.7421

⎤⎥⎥⎥⎦

(6–81)


Finally, q4 and r14, r24, r34, r44 are determined from the following calculations, cf. (6-74)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k4 =

⎡⎢⎢⎢⎣

00

2.00051.4769

⎤⎥⎥⎥⎦ , r14 = qT

1 k4 = 0 , r24 = qT2 k4 = −1.6560

r34 = qT3 k4 = 1.8483 , r44 =

∣∣∣k4 − 0q1 + 1.6560q2 − 1.8483q3

∣∣∣ = 0.1571

q4 =1

0.1571

(k4 − 0q1 + 1.6560q2 − 1.8483q3

)=

⎡⎢⎢⎢⎣

0.3974−0.4684−0.4163

0.6703

⎤⎥⎥⎥⎦

(6–82)

Then, at the end of the 1st iteration the following matrices are obtained

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q1 =

⎡⎢⎢⎢⎣0.7625 −0.3630 −0.3590 0.39740.6470 0.4278 0.4231 −0.4684

0 −0.8278 0.3761 −0.41630 0 0.7421 0.6703

⎤⎥⎥⎥⎦

R1 =

⎡⎢⎢⎢⎣3.2787 4.9244 −2.4101 0

0 4.5001 −7.7292 −1.65600 0 2.6959 1.84830 0 0 0.1571

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ2 = Φ1Q1 =

⎡⎢⎢⎢⎣

0.5392 −0.2567 −0.2539 0.2810−0.4313 −0.1206 −0.2629 0.4799

0.2157 0.8010 0.2175 0.51430 −0.4444 0.8280 0.3420

⎤⎥⎥⎥⎦

K2 = R1Q1 =

⎡⎢⎢⎢⎣5.6860 2.9115 0 02.9115 8.3232 −2.2317 0

0 −2.2317 2.3854 0.11660 0 0.1166 0.1053

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–83)

As seen the matrices R1 and Q1 have the structure (6-76) and (6-77). Additionally, K 2 has the same three diagonalstructure as K1. The corresponding matrices after the 2nd and 3rd iteration become


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q2 =

⎡⎢⎢⎢⎣0.8901 −0.4279 −0.1566 0.01170.4558 0.8356 0.3058 −0.0229

0 −0.3445 0.9362 −0.07020 0 0.0748 0.9972

⎤⎥⎥⎥⎦

R2 =

⎡⎢⎢⎢⎣6.3881 6.3850 −1.0171 0

0 6.4780 −2.6866 −0.04020 0 1.5595 0.11700 0 0 0.0968

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ3 = Φ2Q2 =

⎡⎢⎢⎢⎣

0.3629 −0.3577 −0.3795 0.3103−0.4389 0.1744 −0.1796 0.4947

0.5570 0.5021 0.4533 0.4818−0.2026 −0.6566 0.6648 0.2931

⎤⎥⎥⎥⎦

K3 = R2Q2 =

⎡⎢⎢⎢⎣8.5962 2.9525 0 02.9525 6.3386 −0.5372 0

0 −0.5372 1.4687 0.00720 0 0.0072 0.0966

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–84)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q3 =

⎡⎢⎢⎢⎣0.9458 −0.3230 −0.0345 0.00020.3248 0.9404 0.1003 −0.0005

0 −0.1061 0.9943 −0.00510 0 0.0051 1.0000

⎤⎥⎥⎥⎦

R3 =

⎡⎢⎢⎢⎣9.0891 4.8514 −0.1745 0

0 5.0643 −0.6610 −0.00080 0 1.4065 0.00770 0 0 0.0965

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ4 = Φ3Q3 =

⎡⎢⎢⎢⎣

0.2270 −0.4134 −0.4242 0.3125−0.3584 0.3248 −0.1434 0.4954

0.6899 0.2442 0.4844 0.4793−0.4049 −0.6226 0.6036 0.2900

⎤⎥⎥⎥⎦

K4 = R3Q3 =

⎡⎢⎢⎢⎣10.172 1.6451 0 01.6451 4.8328 −0.1492 0

0 −0.1492 1.3986 0.00050 0 0.0005 0.0965

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–85)

As seen from R3 and K4 the terms in the main diagonal have already after the 3rd iteration grouped in descendingmagnitude, corresponding to the ordering of the eigenvalues at convergence indicated in Box 6.7. Moreover, forboth matrices convergence to the lowest eigenvalue λ1 = 0.0965 has occurred, illustrating the fact that the QRalgorithm converge faster to the lowest eigenmode than to the highest.


The matrices after the 14th iteration become

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q14 =

⎡⎢⎢⎢⎣1.0000 −0.0000 −0.0000 0.00000.0000 1.0000 0.0000 −0.0000

0 −0.0000 1.0000 −0.00000 0 0.0051 1.0000

⎤⎥⎥⎥⎦

R14 =

⎡⎢⎢⎢⎣10.638 0.0003 −0.0000 00.0000 4.3735 −0.0000 −0.0008

0 0 1.3915 0.00770 0 0 0.0965

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ15 = Φ14Q14 =

⎡⎢⎢⎢⎣

0.1076 −0.4387 −0.4453 0.3126−0.2556 0.4167 −0.1244 0.4955

0.7283 0.0232 0.4894 0.4791−0.5620 −0.5170 0.5770 0.2898

⎤⎥⎥⎥⎦

K15 = R14Q14 =

⎡⎢⎢⎢⎣10.638 0.0001 0 00.0001 4.3735 −0.0000 0

0 −0.0000 1.3915 0.00000 0 0.0000 0.0965

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–86)

Presuming that convergence has occurred after the 14th iteration the following solutions are obtained for the eigen-values and eigenmodes of the original general eigenvalue problem

Λ =

⎡⎢⎢⎢⎣

λ4 0 0 00 λ3 0 00 0 λ2 00 0 0 λ1

⎤⎥⎥⎥⎦ = K15 =

⎡⎢⎢⎢⎣10.638 0 0 0

0 4.3735 0 00 0 1.3915 00 0 0 0.0965

⎤⎥⎥⎥⎦

Φ =[Φ(4) Φ(3) Φ(2) Φ(1)

]= Φ15 =

⎡⎢⎢⎢⎣

0.1076 −0.4387 −0.4453 0.3126−0.2556 0.4167 −0.1244 0.4955

0.7283 0.0232 0.4894 0.4791−0.5620 −0.5170 0.5770 0.2898

⎤⎥⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6–87)

The reader should verify that the solution matrices within the indicated accuracy fulfill Φ T MΦ = I and ΦT KΦ =Λ, where M and K are the mass and stiffness matrices given by (5-74). (6-87) agrees with the results (5-75), (5-

79), (5-81) and (5-82) in Example 5.6.

6.6 Exercises 145

6.6 Exercises

6.1 Given a symmetric matrix K in a special eigenvalue problem.

(a.) Write a MATLAB program, which performs special Jacobi iteration.

6.2 Given the symmetric matrices M and K.

(a.) Write a MATLAB program, which performs general Jacobi iteration.

6.3 Given the following mass- and stiffness matrices defined in Exercise 4.2.

(a.) Perform an initial transformation to a special eigenvalue problem, and calculate theeigenvalues and eigenvectors by means of standard Jacobi iteration.

(b.) Calculate the eigenvalues and normalized eigenvectors by means of general Jacobiiteration operating on the original general eigenvalue problem.

6.4 Given the symmetric matrices M and K of dimension n ≥ 3.

(a.) Write a MATLAB program, which performs a Householder reduction to three diago-nal form.

6.5 Given the symmetric matrices M and K.

(a.) Write a MATLAB program, which performs QR iteration.

6.6 Consider the mass- and stiffness matrices defined in Exercise 4.2 after the transformationto the special eigenvalue problem as performed in Exercise 6.3.

(a.) Calculate the eigenvalues and normalized eigenvectors by means of QR iteration.

CHAPTER 7SOLUTION OF LARGE EIGENVALUE

PROBLEMS

7.1 Introduction

In civil engineering large numerical models with n = 104 − 106 degrees of freedom have be-come common practise along with the development of computer technology. However, mostnatural and man made loads such as wind, waves, earthquakes and traffic have spectral contentsin the low frequency range. As a consequence only a relatively small number n1 n of thelowest structural modes will contribute to the global structural dynamic response. In this chap-ter methods will be discussed, which have been devised with this specific fact in mind.

Sections 7.2 and 7.3 deals with simultaneous inverse vector iteration and socalled subspaceiteration, respectively. In both cases a sequence of subspaces are defined, each of which arespanned by a specific system of basis vectors. The idea is that these subspaces at the end of theiteration process contains the n1 lowest eigenmodes Φ(1),Φ(2), . . . ,Φ(n1) of the general eigen-value problem (1-9). These eigenvalue problems may be assembled on the following matrixform, cf. (1-14), (1-15), (1-16)

K[Φ(1) Φ(2) · · ·Φ(n1)] = M[Φ(1) Φ(2) · · ·Φ(n1)]

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ ⇒

KΦ = MΦΛ (7–1)

Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ (7–2)

By contrast to the formulation in Chapter 6 the modal matrix Φ is no longer quadratic, but hasthe dimension n × n1, defined as

— 147 —

148 Chapter 7 – SOLUTION OF LARGE EIGENVALUE PROBLEMS

Φ =[Φ(1) Φ(2) · · ·Φ(n1)

](7–3)

V0

V∞

Φ(1)0

Φ(2)0

Φ(1)∞ = Φ(1)

Φ(2)∞ = Φ(2)

Fig. 7–1 Principle of subspace iteration.

The principle of iterating through a sequence of subspaces has been illustrated in Fig. 7-1. V0

denotes a start subspace, which is spanned by the start basis Φ0 =[Φ

(1)0 Φ

(2)0

]. The iteration

process passes through a sequence of subspaces V1, V2, . . ., where Vk is spanned by the basis

Φk =[Φ

(1)k Φ

(2)k

]. At convergence, Φ∞ =

[Φ

(1)∞ Φ

(2)∞]

=[Φ(1) Φ(2)

]is spanning the limiting

subspace V∞ containing the eigenmodes searched for.

Simultaneous inverse inverse vector iteration is a generalization of the inverse vector iterationand inverse vector iteration with deflation described in Sections 5.2 and 5.5. The start vectorbasis converges towards a basis made up of the wanted eigenmodes as shown in Fig. 7-1.

The subspace iteration method and socalled subspace iteration described in Section 7.2 is inprinciple a sequence of Rayleigh-Ritz analyses, where the Ritz base vectors are forced to con-verge to each of the eigenmodes. As a consequence, if the start basis contains the n1 eigenmodesthe subspace iteration converge in a single step as described in Section 7.2, which is generallynot the case for simultaneous inverse vector iteration. Being based on a convergence of a se-quence of vector bases both methods are in fact subspace iteration methods, although this namehas been coined solely for the latter method. A more informative name for this method wouldprobably be Rayleigh-Ritz iteration.

Section 7.4 deals with characteristic polynomial iteration methods, which operates on the char-acteristic equation (1-10). These methods form an alternative to inverse or forward vector it-eration with deflation in case some specific eigenmode different from the smallest or largest issearched for. To be numerical effective these methods require that the generalized eigenvalue

7.2 Simultaneous Inverse Vector Iteration 149

problem has been reduced to a standard eigenvalue problem on three diagonal form, such as theHouseholder reduction described in Section 6.4. Polynomial methods may be based either onthe numerical iteration of the characteristic polynomial directly, or based on a Sturm sequenceiteration. Even in the first mentioned case a Sturm sequence check should be performed afterthe calculation to verify that the calculated n1 eigenmodes are indeed the lowest.

It should be noticed that some problems in structural dynamics, such as acoustic transmissionand noise emission, are governed by high frequency structural response. Additional to the nu-merical problems in calculating these modes, lack of accuracy of the underlying mechanicalmodels in the high-frequency range adds to the problems in using modal analysis in such highfrequency cases.

7.2 Simultaneous Inverse Vector Iteration

Let Φ0 =[Φ

(1)0 Φ

(2)0 · · ·Φ(n1)

0

]denote n1 arbitrary linearly independent vectors, which span

an n1 dimensional start subspace. Next, the algorithm for simultaneous inverse vector iterationtakes place according to the algorithm

Φk+1 = AΦk , k = 0, 1, . . . (7–4)

where A = K−1M, cf. (5-4). (7-4) is identical to the inverse vector iteration algorithm de-scribed by (5-4). The only difference is that now n1 vectors are simultaneous iterated.

At convergence the iterated base vectors obtained from (7-4) will span an n1-dimensional sub-space containing the n1 lowest eigenmodes. However, due to the inherent properties of theinverse vector iteration algorithm all the iterated base vectors tend to become mutually parallel,and parallel to the lowest eigenmode Φ(1). Hence, the vector basis becomes more and more illconditioned. For the case shown on Fig. 7-1 this means that the subspace Vk will converge tothe limit plane V∞, but the iterated base vectors Φ

(1)k and Φ

(2)k become more and more parallel.

In order to prevent this the method is combined with a Gram-Schmidt orthogonalization pro-cedure. Similar to the QR factorization procedure described in Box 6.6 the iterated basis Φk+1

can be written on the following factorized form

Φk+1 = Φk+1Rk+1 (7–5)

where Φk+1 is an M-orthonormal basis in the iterated subspace, and Rk+1 is an upper triangularmatrix. Hence, Φk+1 and Rk+1 have the properties

Φk+1 =[Φ

(1)k+1 Φ

(2)k+1 · · ·Φ(n1)

k+1

], Φ

(i) Tk+1 MΦ

(j)k+1 = δij (7–6)


Rk+1 =

⎡⎢⎢⎢⎢⎢⎢⎣

r11 r12 r13 · · · r1n1

0 r22 r23 · · · r2n1

0 0 r33 · · · r3n1

......

.... . .

...0 0 0 · · · rn1n1

⎤⎥⎥⎥⎥⎥⎥⎦ (7–7)

The M-orthonormal base vectors Φk+1 =[Φ

(1)k+1 Φ

(2)k+1 · · ·Φ(n1)

k+1

]spanning the iterated subspace

Vk+1, as well as the components of the triangular matrix Rk+1, are determined sequentiallyin much the same way as the determination of the matrices Q and R in the QR factorizationdescribed by (6-66) - (6-69). At first, it is noticed that (7-5) is identical to the following relations⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(1)k+1 = r11Φ

(1)k+1

Φ(2)k+1 = r12Φ

(1)k+1 + r22Φ

(2)k+1

...

Φ(j)k+1 = r1jΦ

(1)k+1 + r2jΦ

(2)k+1 + · · ·+ rjjΦ

(j)k+1 =

j∑i=1

rijΦ(i)k+1

...

Φ(n1)k+1 =

n1∑i=1

rin1Φ(i)k+1

(7–8)

(7-8) is solved sequentially downwards using the M-orthonormality of the already determinedbase vectors Φ

(j)k+1. The details of the derivation has been given in Box 7.1.

After convergence the eigenvalues are obtained from the Rayleigh quotients evaluated with thecalculated eigenvectors, cf. (4-25). Since each of the n1 eigenmodes have been normalized tounit modal mass the quotients become

λj = Φ(j) TKΦ(j) , j = 1, . . . , n1 (7–9)

The Rayleigh quotients in (7-9) may be assembled in the following matrix equation

Λ = ΦTKΦ (7–10)

where

Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ , Φ =

[Φ(1) Φ(2) · · ·Φ(n1)

]= Φ∞ (7–11)


It can be proved that the upper triangular matrix Rk+1 converges towards the diagonal matrixΛ−1. Although the Rayleigh quotients (7-10) provides more accurate estimates, the eigenvaluesmay then as an alternative be retrieved from

Λ = R−1∞ (7–12)


Box 7.1: M-orthonormalization of iterated basis

Evaluating the modal mass on both sides of the 1st equation of (10-8) provides

r11 =∥∥Φ(1)

k+1

∥∥ ⇒ Φ(1)k+1 =

1

r11Φ

(1)k+1 (7–13)

where the norm∥∥Φ(1)

k+1

∥∥ represents the square root of the modal mass of Φ(1)k+1 defined as

∥∥Φ(1)k+1

∥∥ =(Φ

(1) Tk+1 MΦ

(1)k+1

) 12

(7–14)

Now, Φ(1)k+1 and r11 are known. Scalar pre-multiplication of the 2nd equation

with Φ(1) Tk+1 M, and use of the orthonormality properties Φ

(1) Tk+1 MΦ

(2)k+1 = 0 and

Φ(1) Tk+1 MΦ

(1)k+1 = 1, provides

r12 = Φ(1) Tk+1 MΦ

(2)k+1 ⇒ r22 =

∥∥Φ(2)k+1 − r12Φ

(1)k+1

∥∥ ⇒

Φ(2)k+1 =

1

r22

(Φ

(2)k+1 − r12Φ

(1)k+1

)(7–15)

At the determination of Φ(j)k+1, 1 < j ≤ n1, the mutually ortonormal basis vectors

Φ(1)k+1,Φ

(2)k+1, . . . ,Φ

(j−1)k+1 have already been determined. Scalar pre-multiplication of the

jth equation with Φ(i) Tk+1 M, i = 1, 2, . . . , j − 1, and use of the orthogonality property

Φ(i) Tk+1 MΦ

(j)k+1 = 0 provides

rij = Φ(i) Tk+1 MΦ

(j)k+1 ⇒ rjj =

∥∥∥∥∥Φ(j)k+1 −

j−1∑i=1

rijΦ(i)k+1

∥∥∥∥∥ ⇒

Φ(j)k+1 =

1

rjj

(Φ

(j)k+1 −

j−1∑i=1

rijΦ(i)k+1

)(7–16)

It is characteristic for simultaneous inverse vector method in contrast to the subspace iterationmethod described in Section 7.3, that eigenmodes which at one level of the iteration processis contained in the iterated subspace, may move out of the iterated subspace at later levels asillustrated in Example 7.1.


Box 7.2: Simultaneous inverse vector iteration algorithm

Given the n1-dimensional start vector basis Φ0 =[Φ

(1)0 Φ

(2)0 · · ·Φ(n1)

0

]. The base vectors

must be linearly independent, but need not be normalized to unit modal mass. Repeat thefollowing items for k = 0, 1, . . .

1. Perform simultaneous inverse vector iteration:

Φk+1 = AΦk , A = K−1M

2. Perform Gram-Schmidt orthogonalization Gram-Schmidt orthogonalization to ob-tain a new M-orthonormal iterated vector basis Φk+1 as explained by (7-12) - (7-16)corresponding to the factorization:

Φk+1 = Φk+1Rk+1

After convergence has been achieved the eigenvalues and eigenmodes normalized to unitmodal mass are obtained from:

Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ = ΦT

∞KΦ∞ = R−1∞ , Φ =

[Φ(1) Φ(2) · · ·Φ(n1)

]= Φ∞

As for all kind of inverse vector iteration methods the convergence rate of the iteration vector islinear in the quantity

r1 = max(λ1

λ2

,λ2

λ3

, . . . ,λn1

λn1+1

)(7–17)

Correspondingly, the Rayleigh quotients (7-9) have quadratic convergence rate r2 = r21.

The simultaneous inverse vector iteration algorithm always converge towards the lowest n1

eigenmodes. Hence, no Sturm sequence check is needed to ensure that these modes have in-deed been calculated. Further, the rate of convergence seems to be comparable for all modescontained in the subspace, as demonstrated in Example 7.1 below.

The simultaneous inverse vector iteration algorithm may be summarized as indicated in Box7.2.


Example 7.1: Simultaneous inverse vector iteration

Consider the generalized eigenvalue problem defined in Example 1.4. Calculate the two lowest eigenmodes andcorresponding eigenvalues by simultaneous inverse vector iteration with the start vector basis

Φ0 =[Φ(1)

0 Φ(2)0

]=

⎡⎢⎣0 21 12 0

⎤⎥⎦ (7–18)

The matrix A becomes, cf. (6-44)

A = K−1M =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ =

⎡⎢⎣0.2917 0.1667 0.0417

0.0833 0.3333 0.08330.0417 0.1667 0.2917

⎤⎥⎦ (7–19)

Then, the 1st iterated vector basis becomes, cf. (7-4)

Φ1 =[Φ(1)

1 Φ(2)1

]= AΦ0 =

⎡⎢⎣0.2917 0.1667 0.0417

0.0833 0.3333 0.08330.0417 0.1667 0.2917

⎤⎥⎦⎡⎢⎣0 21 12 0

⎤⎥⎦ =

⎡⎢⎣0.2500 0.75000.5000 0.50000.7500 0.2500

⎤⎥⎦ (7–20)

At the determination of Φ(1)1 and r11 in the 1st vector iteration the following calculations are performed, cf. (7-13)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(1)1 =

⎡⎢⎣0.25000.50000.7500

⎤⎥⎦ , r11 =

∥∥∥Φ(1)1

∥∥∥ =

⎛⎜⎜⎝⎡⎢⎣0.25000.50000.7500

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣0.2500

0.50000.7500

⎤⎥⎦⎞⎟⎟⎠

12

= 0.7500

Φ(1)1 =

10.7500

⎡⎢⎣0.25000.50000.7500

⎤⎥⎦ =

⎡⎢⎣0.33330.66671.0000

⎤⎥⎦

(7–21)

Φ(2)1 and r12, r22 are determined from the following calculations, cf. (7-15)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(2)1 =

⎡⎢⎣0.75000.50000.2500

⎤⎥⎦ , r12 =

⎡⎢⎣0.33330.66671.0000

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣0.75000.50000.2500

⎤⎥⎦ = 0.5833

r22 =

∥∥∥∥∥∥∥⎡⎢⎣0.7500

0.50000.2500

⎤⎥⎦− 0.5833 ·

⎡⎢⎣0.33330.66671.0000

⎤⎥⎦∥∥∥∥∥∥∥ = 0.4714

Φ(2)1 =

10.4714

⎛⎜⎝⎡⎢⎣0.75000.50000.2500

⎤⎥⎦− 0.5833 ·

⎡⎢⎣0.33330.66671.0000

⎤⎥⎦⎞⎟⎠ =

⎡⎢⎣ 1.1785

0.2357−0.7071

⎤⎥⎦

(7–22)



⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R1 =

[0.7500 0.5833

0 0.4714

]

Φ1 =

⎡⎢⎣0.3333 1.17850.6667 0.23571.0000 −0.7071

⎤⎥⎦

(7–23)

The reader should verify that Φ1R1 = Φ1. The corresponding matrices after the 2nd and 3rd iteration become

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R2 =

[0.4787 0.1231

0 0.2611

]

Φ2 =

⎡⎢⎣0.5222 1.10780.6963 0.12310.8704 −0.8616

⎤⎥⎦

(7–24)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R3 =

[0.4943 0.0650

0 0.2529

]

Φ3 =

⎡⎢⎣0.6163 1.05830.7043 0.06230.7924 −0.9339

⎤⎥⎦

(7–25)

Convergence of the eigenmodes with the indicated number of digits were achieved after 14 iterations, where

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R14 =

[0.5000 0.0000

0 0.2500

]

Φ14 =

⎡⎢⎣0.7071 1.00000.7071 0.00000.7071 −1.0000

⎤⎥⎦

(7–26)

Presuming that convergence has occurred after the 14th iteration the following eigenvalues are obtained from(7-10) and (7-12)

Λ =

[λ1 00 λ2

]= ΦT

14KΦ14 = R−1∞ =

[2.0000 −0.0000

−0.0000 4.0000

]

Φ =[Φ(1) Φ(2)

]= Φ14 =

⎡⎢⎣0.7071 1.00000.7071 0.00000.7071 −1.0000

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–27)


λ3 = 6, see (1-87). Then, the convergence rate of the iteration vectors becomes r 1 = max(

λ1λ2

, λ2λ3

)= max

(24 , 4

6

)=

23 , cf. (7-17). This is a relatively large number, which is displayed in the rather slow convergence of the iterativeprocess. The convergence towards Φ(1) and Φ(2) occurred within the same iteration step. This suggests that theconvergence rate is uniform to all considered modes in the subspace.

Further it is noted that

Φ(1) =√

22

⎡⎢⎣111

⎤⎥⎦ =

√2

4·

⎡⎢⎣012

⎤⎥⎦+

√2

4·

⎡⎢⎣210

⎤⎥⎦ =

√2

4·Φ(1)

0 +√

24

· Φ(2)0

Φ(2) =

⎡⎢⎣ 1

0−1

⎤⎥⎦ = −1

2·

⎡⎢⎣012

⎤⎥⎦+

12·

⎡⎢⎣2

10

⎤⎥⎦ = −1

2·Φ(1)

0 +12·Φ(2)

0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–28)

Hence, the 1st and 2nd eigenmode are originally in the subspace spanned by the basis Φ 0. As seen during theiteration process these eigenmodes are moving out of the iterated subspace.

7.3 Subspace Iteration

As is the case for the simultaneous inverse vector iteration algorithm the subspace iteration algo-

rithm presumes that a start subspace V0, spanned by the vector basis Φ0 =[Φ

(1)0 Φ

(2)0 · · ·Φ(n1)

0

],

has been defined.

At the kth iteration step of the iteration process a vector basis Φk =[Φ

(1)k Φ

(2)k · · ·Φ(n1)

k

],

which spans the iterated subspace Vk, has been obtained. Based on this a simultaneous inversevector iteration is performed

Φk+1 = AΦk , k = 0, 1, . . . (7–29)

where A = K−1M, cf. (5-4). Next, a Rayleigh-Ritz analysis is performed using Φk+1 as aRitz basis, in order to obtain approximate solutions to the lowest n1 eigenmodes and eigenval-ues. This requires the solution of the following reduced generalized eigenvalue problem of thedimension n1, cf. (1-14), (4-49)

Kk+1Qk+1 = Mk+1Qk+1Rk+1 , k = 0, 1, . . . (7–30)

Mk+1 and Kk+1 denote the mass and stiffness matrices projected on the subspace Vk+1, cf.(4-45)

7.3 Subspace Iteration 157

Mk+1 = ΦTk+1MΦk+1

Kk+1 = ΦTk+1KΦk+1

⎫⎬⎭ (7–31)

Qk+1 =[q

(1)k+1q

(2)k+1 · · ·q(n1)

k+1

]of the dimension n1 × n1 contains the eigenvectors of the eigen-

value problem (7-30). In what follows the eigenvectors q(i)k+1 are assumed to normalized to unit

modal mass with respect to the projected mass matrix, i.e.

q(i) Tk+1 Mk+1q

(j)k+1 =

⎧⎨⎩

0 , i �= j

1 , i = j(7–32)

Rk+1 is a diagonal matrix containing the corresponding eigenvalues of (7-29) in the main diag-onal

Rk+1 =

⎡⎢⎢⎢⎣ρ1,k+1 0 · · · 0

0 ρ2,k+1 · · · 0...

...... 0

0 0 · · · ρn1,k+1

⎤⎥⎥⎥⎦ (7–33)

The eigenvalues ρj,k+1, j = 1, . . . , n1 indicates the estimate of the eigenvalues after the kthiteration. These are all upperbounds to the corresponding eigenvalues of the full problem, cf.(4-57).

At the end of the kth iteration step a new estimate of the lowest n1 eigenvectors are determinedfrom, cf. (4-51)

Φk+1 = Φk+1Qk+1 (7–34)

If the column vectors in Qk+1 have been normalized to unit modal mass with respect to Mk+1,the M-orthogonal column vectors of Φk+1 will automatically be normalized to unit modal masswith respect to M, cf. (4-55).

Next, the calculations in (7-30), (7-31), (7-34) are repeated with the new estimate of the nor-malized eigenmodes Φk+1.

At convergence of the subspace iteration algorithm the lowest n1 eigenvectors and eigenvaluesare retrieved from

Φ =[Φ(1) Φ(2) · · ·Φ(n1)

]= Φ∞ , Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

...... 0

0 0 · · · λn1

⎤⎥⎥⎥⎦ = R∞ = ±Q∞ (7–35)


At convergence, Q∞ can be shown to be a diagonal matrix, where the numerical value of thecomponents are equal to the eigenvalue of the original problem as indicated in (7-35).

It should be realized that subspace iteration involves iteration at two levels. Primary, a globalsimultaneous inverse vector iteration loop as defined by the index k is performed. Inside thisloop a secondary iteration process is performed at the solution of the eigenvalue problem (7-30).Usually, the latter problem is solved iteratively by means of a general Jacobi iteration algorithmas described in Section 6.3. Because the applied similarity transformations in the general Jacobialgorithm are not orthonormal, the eigenvectors q

(j)k are not normalized to unit modal mass at

convergence. Hence, in order to fulfill the requirements (7-32) this normalization should beperformed after convergence. Further, the eigenvalues will not be ordered in ascending order ofmagnitude as presumed in (7-35), cf. Box 6.4.

The convergence rate for the components in the kth eigenmode and the kth eigenvalue, r1,k andr2,k, are defined as

r1,k =λk

λn1+1

r2,k =λ2

k

λ2n1+1

= r21,k

⎫⎪⎪⎪⎬⎪⎪⎪⎭ , k = 1, . . . , n1 (7–36)

Hence, convergence is achieved at first for the lowest mode and latest for mode k = n1, ashas been demonstrated in Example 7.2 below. This represents a marked difference from simul-taneous inverse vector iteration, where as mentioned the convergence rate seems to be almostidentical for all modes contained in the subspace. A rule of thumb says that approximately 10subspace iterations are needed to obtain a solution for the components of Φ(1) with 6 correctdigits.


Box 7.3: Subspace iteration algorithm

Given the n1-dimensional start vector basis Φ0 =[Φ

(1)0 Φ

(2)0 · · ·Φ(n1)

0

]. The base vectors

must be linearly independent, but the base vectors need not be normalized to unit modalmass. Repeat the following items for k = 0, 1, . . .

1. Perform simultaneous inverse vector iteration:

Φk+1 = AΦk , A = K−1M

2. Calculate projected mass and stiffness matrices:

Mk+1 = ΦTk+1MΦk+1 , Kk+1 = ΦT

k+1KΦk+1

3. Solve the generalized eigenvalue problem of dimension n1 by means of a generalJacobi iteration algorithm with the eigenvectors Qk+1 normalized to unit modal massat exit:Kk+1Qk+1 = Mk+1Qk+1Rk+1

4. Calculate new solution to eigenvectors:

Φk+1 = Φk+1Qk+1

After convergence has been achieved the eigenvalues and eigenmodes normalized to unitmodal mass are obtained from:

Λ =

⎡⎢⎢⎢⎣λ1 0 · · · 0

0 λ2 · · · 0...

.... . .

...0 0 · · · λn1

⎤⎥⎥⎥⎦ = R∞ = ±Q∞ , Φ =

[Φ(1) Φ(2) · · ·Φ(n1)

]= Φ∞

Finally, a Sturm sequence check should be performed to ensure that the lowest n1 eigen-pairs have been calculated.

In order to speed up the iteration process towards the n1 modes actually wanted, the dimensionof the iterated subspace is sometimes increased to n2 > n1. Then, the convergence rate of theiteration vector the highest mode of interest decreases to

r1,n1 =λn1

λn2+1(7–37)

In case of an adverse choice of the start basis vector Φ0 it may happen that one of the eigen-


modes searched for, Φ(j), j = 1, . . . , n1, is M-orthogonal to start subspace, i.e.

Φ(j) TMΦ(k)0 = 0 , k = 1, 2, . . . , n1 (7–38)

In this case the subspace iteration algorithm converges towards the eigenmodes Φ(1), . . . ,Φ(j−1),Φ(j+1), . . . ,Φ(n1),Φ(n1+1). In principle a similar problem occurs in simultaneous inverse vectoriteration, although round-off errors normally eliminates this possibility.

Singular to subspace iteration is that eigenmodes contained in the initial basis Φ0 remain inlater iterated bases. Hence, if Φ(j), j = n1 + 1, . . . , n is contained in Φ0, this mode will beamong the calculated modes.

In both cases we are left with the problem to decide whether the calculated n1 eigenmodesare the lowest n1 modes of the full system. For this reason a subspace iteration should alwaysbe followed by a Sturm sequence check. This is performed in the following way. Let µ be anumber slightly larger than the largest calculated eigenvalue ρn1,∞, and perform the followingGauss factorization of the matrix K − µM

K − µM = LDLT (7–39)

where L and D are given by (3-2), (3-3). The number of eigenvalue less than µ is equalto the number of negative elements in the diagonal of the diagonal matrix D, cf. Section3.1. Hence, the analysis should show exactly n1 negative elements in D. Alternatively, thesame information may be withdrawn from the number of sign changes in the sign sequencesign(P (n)(µ)

), sign

(P (n−1)(µ)

), . . . , sign

(P (0)(µ)

), where P (n−1)(µ), . . . , P (0)(µ) denotes the

Sturm sequence of characteristic polynomials, and P (n)(µ) is a dummy positive component inthe sequence, cf. Section 3.2.

The marked difference between the subspace iteration algorithm and and the simultaneous in-verse vector iteration algorithm is that the orthonormalization process to prevent ill-conditioningof the iterated vector base in the former case is performed by an eigenvector approach relatedto the Rayleigh-Ritz analysis, whereas a Gram-Schmidt orthogonalization procedure is used inthe latter case. There are no marked difference in the rate of convergence of the two algorithms.

Example 7.2: Subspace iteration

The generalized eigenvalue problem defined in Example 6.2 is considered again. Using the same initial start basis(7-18) as in Example 7.1, the problem is solved in this example by means of subspace iteration.

At the 1st iteration step (k = 0) the simultaneous inverse vector iteration produces the vector basis Φ1, which isunchanged given by (7-20).

Based on Φ1 the following projected mass and stiffness matrices are calculated, cf. (4-45), (7-20), (7-31)


M1 = ΦT1 MΦ1 =

⎡⎢⎣0.2500 0.7500

0.5000 0.50000.7500 0.2500

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣0.2500 0.7500

0.5000 0.50000.7500 0.2500

⎤⎥⎦ =

[0.5625 0.43750.4375 0.5625

]

K1 = ΦT1 KΦ1 =

⎡⎢⎣0.2500 0.75000.5000 0.50000.7500 0.2500

⎤⎥⎦

T ⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.2500 0.75000.5000 0.50000.7500 0.2500

⎤⎥⎦ =

[1.2500 0.75000.7500 1.2500

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–40)

The corresponding eigenvalue problem (7-30) becomes

K1Q1 = M1Q1R1 ⇒[1.2500 0.75000.7500 1.2500

] [q(1)

1 q(2)1

]=

[0.5625 0.43750.4375 0.5625

] [q(1)

1 q(2)1

] [ρ1,1 00 ρ2,1

]⇒

R1 =

[2 00 4

]=

[λ1 00 λ2

], Q1 =

[√2

2 −2√

22 2

](7–41)

The estimate of the lowest eigenvectors after the 1st iteration becomes, cf. (7-34)

Φ1 = Φ1Q1 =

⎡⎢⎣0.2500 0.75000.5000 0.50000.7500 0.2500

⎤⎥⎦[√

22 −2√

22 2

]=

⎡⎢⎣

√2

2 −1√

22 0√

22 1

⎤⎥⎦ =

[Φ(1) Φ(2)

](7–42)

(7-41) and (7-42) indicate the exact eigenvalues and eigenmodes, cf. (1-87). Hence, convergence is obtained in justa single iteration. This is so because the start subspace V0, spanned by the vector basis Φ0 contains the eigenmodesΦ(1) and Φ(2) as shown by (7-28). This property is singular to the subspace iteration algorithm compared to thesimultaneous inverse vector iteration technique.

Next, let us perform the same calculations using the start basis

Φ0 =[Φ(1)

0 Φ(2)0

]=

⎡⎢⎣1 −12 23 −3

⎤⎥⎦ (7–43)

The simultaneous inverse vector iteration (7-29) provides, cf. (7-19)

Φ1 = AΦ0 =

⎡⎢⎣0.2917 0.1667 0.04170.0833 0.3333 0.08330.0417 0.1667 0.2917

⎤⎥⎦⎡⎢⎣1 −12 23 −3

⎤⎥⎦ =

⎡⎢⎣0.7500 −0.08331.0000 0.33331.2500 −0.5833

⎤⎥⎦ (7–44)



M1 =

⎡⎢⎣0.7500 −0.0833

1.0000 0.33331.2500 −0.5833

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣0.7500 −0.0833

1.0000 0.33331.2500 −0.5833

⎤⎥⎦ =

[2.0625 −0.0625

−0.0625 0.2847

]

K1 =

⎡⎢⎣0.7500 −0.08331.0000 0.33331.2500 −0.5833

⎤⎥⎦

T ⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.7500 −0.0833

1.0000 0.33331.2500 −0.5833

⎤⎥⎦ =

[4.2500 −0.2500

−0.2500 1.5833

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–45)

The solution of the corresponding generalized eigenvalue problem (7-30) becomes

R1 =

[2.0534 0

0 5.5656

], Q1 =

[−0.6982 −0.0254−0.0851 −1.8784

](7–46)

The estimate of the lowest eigenmode after the 1st iteration becomes, cf. (7-34)

Φ1 = Φ1Q1 =

⎡⎢⎣0.7500 −0.08331.0000 0.33331.2500 −0.5833

⎤⎥⎦[−0.6982 −0.0254−0.0851 −1.8784

]=

⎡⎢⎣−0.5165 0.1375

−0.7265 −0.6516

−0.8231 1.0640

⎤⎥⎦ (7–47)

Correspondingly, after the 2nd, 7th and 14th iteration steps the following matrices are calculated

R2 =

[2.0118 0

0 5.2263

], Q2 =

[−2.0171 0.1513−0.0887 −5.3145

]

Φ2 =

⎡⎢⎣0.6195 0.08210.7241 0.56860.7535 −1.1604

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–48)

R7 =

[2.0000 0

0 4.0533

], Q7 =

[−2.0000 0.0011−0.0007 −4.0661

]

Φ7 =

⎡⎢⎣−0.7067 −0.8711−0.7074 −0.1155−0.7069 1.1020

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–49)

R14 =

[2.0000 0

0 4.0002

], Q14 =

[−2.0000 0.0000−0.0000 −4.0002

]

Φ14 =

⎡⎢⎣0.7071 0.99310.7071 0.00680.7071 −1.0068

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–50)


As seen the subspace iteration process determines the 1st eigenvalue and eigenvector after 7 iteration, whereas the2nd eigenvector has not yet been calculated with the sufficiently accuracy even after 14 iterations. By contrastthe simultaneous inverse vector iteration managed to achieve convergence for this quantity after 14 iterations, see(7-26).

The 2nd calculated eigenvalue becomes ρ2,14 = 4.0002. Then, let µ = 4.05 and perform a Gauss factorization ofthe matrix K− 4.05M, i.e.

K− 4.05M =

⎡⎢⎣−0.0250 −1.0000 0.0000−1.0000 −0.0500 −1.0000

0.0000 −1.0000 −0.0250

⎤⎥⎦ =

LDLT =

⎡⎢⎣ 1 0 0

40 1 00 −0.0250 1

⎤⎥⎦⎡⎢⎣−0.0250 0 0

0 39.950 00 0 −0.0500

⎤⎥⎦⎡⎢⎣1 40 00 1 −0.02500 0 1

⎤⎥⎦ (7–51)

It follows that two components in the main diagonal of D are negative, from which is concluded that two eigen-values are smaller than µ = 4.05. In turn this means that the two eigensolutions obtained by (7-46) are indeed thelowest two eigensolutions of the original system.

Finally, consider the start vector basis

Φ0 =[Φ(1)

0 Φ(2)0

]=

⎡⎢⎣ 0 2−1 −1

2 0

⎤⎥⎦ (7–52)

Now,

Φ(1) T MΦ(1)0 =

√2

2

⎡⎢⎣1

11

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣ 0−1

2

⎤⎥⎦ = 0

Φ(1) T MΦ(2)0 =

√2

2

⎡⎢⎣1

11

⎤⎥⎦

T ⎡⎢⎣12 0 00 1 00 0 1

2

⎤⎥⎦⎡⎢⎣ 2−1

0

⎤⎥⎦ = 0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–53)

It follows that the lowest eigenmode Φ(1) is M-orthogonal to the selected start vector basis. Hence, it should beexpected that the algorithm converges towards Φ (2) and Φ(3). Moreover, in the present three dimensional case astart subspace, which is M-orthogonal to Φ(1), must contain Φ(2) and Φ(3). Actually, cf. (1-87)

Φ(2) =

⎡⎢⎣−1

01

⎤⎥⎦ = −1

2·

⎡⎢⎣ 0−1

2

⎤⎥⎦+

12·

⎡⎢⎣ 2−1

0

⎤⎥⎦ = −1

2·Φ(1)

0 +12·Φ(2)

0

Φ(3) =√

22

⎡⎢⎣ 1−1

1

⎤⎥⎦ =

√2

4·

⎡⎢⎣ 0−1

2

⎤⎥⎦+

√2

4·

⎡⎢⎣ 2−1

0

⎤⎥⎦ =

√2

4·Φ(1)

0 +√

24

·Φ(2)0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–54)


Hence, convergence towards Φ(2) and Φ(3) should take place in a single iteration step. Actually, after the 1stsubspace iteration the following matrices are calculated

R1 =

[4 00 6

], Q1 =

[−2.0000 −2.1213−2.0000 −2.1213

]

Φ2 =

⎡⎢⎣ 1.0000 −0.7071

0.0000 0.7071−1.0000 −0.7071

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–55)

The 2nd calculated eigenvalue becomes ρ2,1 = 6. In order to check whether ρ2,1 = λ2 or ρ2,1 = λ3 we chooseµ = 6.05, and perform a Gauss factorization of the matrix K− 6.05M, i.e.

K− 6.05M =

⎡⎢⎣−1.0250 −1.0000 0.0000−1.0000 −2.0500 −1.0000

0.0000 −1.0000 −1.0250

⎤⎥⎦ =

LDLT =

⎡⎢⎣ 1 0 0

0.9756 1 00 0.9308 1

⎤⎥⎦⎡⎢⎣−1.0250 0 0

0 −1.0744 00 0 −0.0942

⎤⎥⎦⎡⎢⎣1 0.9756 00 1 0.93080 0 1

⎤⎥⎦ (7–56)

It follows that three components in the main diagonal of D are negative, from which is concluded that the largest of

the two calculated eigenvalues must be equal to the largest eigenvalue of the original system, i.e. ρ 2,1 = λ3. Still,

we do not know whether ρ1,1 = λ1 or ρ1,1 = λ2. In order to investigate this another calculation is performed with

µ = 4.05. The Gauss factorization of the matrix K − 4.05M has already been performed as indicated by (7-51).

Since this result shows that two eigenvalues exist, which are smaller than µ = 4.05, ρ1,1 = 4 must be the largest

of these, and hence the 2nd eigenvalue of the original system.

7.4 Characteristic Polynomial Iteration

In this section it is assumed that the stiffness and mass matrices have been reduced to a threediagonal form through a series of similarity transformations as explained in Section 6.4, corre-sponding to, cf. (6-32)

K =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 β1 0 · · · 0 0

β1 α2 β2 · · · 0 0

0 β2 α3 · · · 0 0...

......

. . ....

...0 0 0 · · · αn−1 βn−1

0 0 0 · · · βn−1 αn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7–57)

7.4 Characteristic Polynomial Iteration 165

M =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

γ1 δ1 0 · · · 0 0

δ1 γ2 δ2 · · · 0 0

0 δ2 γ3 · · · 0 0...

......

. . ....

...0 0 0 · · · γn−1 δn−1

0 0 0 · · · δn−1 γn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7–58)

The Helmholz reduction in Section 6.4 results in M = I. The slightly more general case,where M is three diagonal has been assumed in what follows. In principle polynomial iterationmethods works equally well on fully populated stiffness and mass matrices. However, the com-putational efforts become too extensive to make them competitive in this case.

Now, the characteristic equation of the generalized eigenvalue problem can be written in thefollowing form, cf. (1-10)

P (λ) = P (0)(λ) = det(K − λM

)=

det

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 − λγ1 β1 − λδ1 0 · · · 0 0

β1 − λδ1 α2 − λγ2 β2 − λδ2 · · · 0 0

0 β2 − λδ2 α3 − λγ3 · · · 0 0...

......

. . ....

...0 0 0 · · · αn−1 − λγn−1 βn−1 − λδn−1

0 0 0 · · · βn−1 − λδn−1 αn − λγn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

(αn − λγn

) · P (1)(λ) − (βn−1 − λδn−1

)2 · P (2)(λ) (7–59)

The last statement in (7-59) is obtained by expanding the determinant after the components inthe last row. P (1)(λ) and P (2)(λ) denote the characteristic polynomials obtained by omittingthe last row and column, and the last two rows and columns in the matrix K − λM, respec-tively, cf. (3-26). The validity of the result (7-59) has been demonstrated for a 4-dimensionalcase in Example 7.3. In turn, P (1)(λ) may be expressed in terms of P (2)(λ) and P (3)(λ) by asimilar expression. Actually, the complete Sturm sequence of characteristic polynomials maybe calculated recursively from the algorithm


P (n−1)(λ) =(α1 − λγ1

)P (n−2)(λ) =

(α1 − λγ1

)(α2 − λγ2

)− (β1 − λδ1

)2P (n−m)(λ) =

(αm − λγm

) · P (n−m+1)(λ) − (βm−1 − λδm−1

)2 · P (n−m+2)(λ) , m = 3, 4, . . . , n

⎫⎪⎪⎬⎪⎪⎭

(7–60)

The effectiveness of characteristic polynomial iteration methods for matrices on three diagonalform relies on the result (7-60).

Assume, that the jth eigensolution(λj,Φ

(j))

is wanted. At first one needs to determine twofigures µ0 and µ1 fulfilling λj−1 < µ0 < λj < µ1 < λj+1. This is done based on the sequence ofsigns sign(P (n)(µ)), sign(P (n−1)(µ)), . . . , sign(P (1)(µ)), sign(P (0)(µ)), in which the numberof sign changes indicates the total number of eigenvalues smaller than µ, and where P (n)(µ) isa dummy positive figure, cf. Section 6.3.Below, on Fig. 7-2 are marked two points µk−1 and µk on the λ-axis in the vicinity of theeigenvalue searched for, which is λ1 in the illustrated case. The values of the characteristicpolynomial in these points, P (µk−1) and P (µk), may easily be calculated by means of (7-60)(notice that P (µ) = P (0)(µ)). The line through the points

(µk−1, P (µk−1)

)and

(µk, P (µk)

)has the equation

y(λ) = P (µk) +(P (µk) − P (µk−1)

) λ − µk

µk − µk−1(7–61)

λλ1 λ2 λ3

µkµk−1

µk+1

P (λ)

y(λ) = P (µk) +(P (µk) − P (µk−1)

) λ − µk

µk − µk−1

Fig. 7–2 Secant iteration of characteristic equation towards λ1.

The line defined by (7-61) intersects the λ-axis at the point µk+1. It is clear that this point will becloser to λj than both µk−1 and µk. The intersection point of the line with the λ-axis is obtainedas the solution to the equation y(λ) = y(µk+1) = 0, which is given as

µk+1 = µk − P (µk)

P (µk) − P (µk−1)

(µk − µk−1

)(7–62)

Next, the iteration index is raised to k + 1, and a new intersection point µk+2 is obtained. Thesequence µ0, µ1, µ2, . . . converges relatively fast to the eigenvalue λj as demonstrated below in


Example 7.4.

Box 7.4: Characteristic polynomial iteration algorithm

In order to calculate the jth eigenvalue λj and the jth eigenvector Φ(j) the following itemsare performed

1. Based on the sequence of signs sign(P (n)(µ)), sign(P (n−1)(µ)), . . . , sign(P (1)(µ)),

sign(P (0)(µ)) of the Sturm sequence of characteristic polynomials determine two

figures µ0 and µ1 fulfilling the inequalities:

λj−1 < µ0 < λj < µ1 < λj+1

2. Perform secant iteration in search for λj = µ∞ according to the algorithm:

µk+1 = µk − P (µk)P (µk)−P (µk−1)

(µk − µk−1

)3. Determine the unnormalized eigenmode Φ(j) from the algorithm (7-64).

4. Normalize the eigenmode to unit modal mass:

Φ(j) = Φ(j)√Φ(j) T MΦ(j)

Alternatively, the eigenvalue λj may be determined by means of Sturm sequence check, wherethe interval ]µ0, µ1[ is increasingly narrowed around the eigenvalue λj by bisection of the pre-vious interval. This algorithm, which is merely telescope method described in Section 6.2, willgenerally converge much slower than the secant iteration algorithm.

Finally, the components[Φ

(j)1 , Φ

(j)2 , . . . , Φ

(j)n

]of the eigenmode Φ(j) are determined as non-

trivial solutions to the linear equations

(K− λjM

)Φ(j) = 0 ⇒

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

α1 − λjγ1 β1 − λjδ1 0 · · · 0 0

β1 − λjδ1 α2 − λjγ2 β2 − λjδ2 · · · 0 0

0 β2 − λjδ2 α3 − λjγ3 · · · 0 0...

......

. . ....

...0 0 0 · · · αn−1 − λjγn−1 βn−1 − λjδn−1

0 0 0 · · · βn−1 − λjδn−1 αn − λjγn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

Φ(j)1

Φ(j)2

Φ(j)3...

Φ(j)n−1

Φ(j)n

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0

0

0...0

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7–63)


Let Φ(j) = [Φ(j)1 , Φ

(j)2 , . . . , Φ

(j)n ] denote the eigenmode with components arbitrarily normalized.

Setting Φ(j)1 = 1 the equations (7-63) may be solved recursively from above by the following

algorithm

Φ(j)2 = −α1 − λjγ1

β1 − λjδ1· 1

Φ(j)3 = −β1 − λjδ1

β2 − λjδ2· 1 − α2 − λjγ2

β2 − λjδ2· Φ(j)

2

Φ(j)m = −βm−2 − λjδm−2

βm−1 − λjδm−1· Φ(j)

m−2 −αm−1 − λjγm−1

βm−1 − λjδm−1Φ

(j)m−1 , m = 4, . . . , n

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭

(7–64)

Hence, the determination of the components of the vector Φ(j) is almost free. Obvious, theindicated algorithm breaks down, if any of the denominators βm−1 − λjδm−1 = 0. This meansthat the algorithm should be extended with alternatives to deal with such exceptions.

Finally, the eigenmode Φ(j) should be normalized to unit modal mass as follows

Φ(j) =Φ(j)

√Φ(j) TMΦ(j)

(7–65)

Example 7.3: Evaluation of determinant

The determinant of the following matrix on a three diagonal form of the dimension 4 × 4 is wanted

K =

⎡⎢⎢⎢⎣α1 β1 0 0β1 α2 β2 00 β2 α3 β3

0 0 β3 α4

⎤⎥⎥⎥⎦ (7–66)

Expansion of the determinant after the components in the 4th row provides

det(K)

= P (0) = α4 · det

⎛⎜⎝⎡⎢⎣α1 β1 0

β1 α2 β2

0 β2 α3

⎤⎥⎦⎞⎟⎠− β3 · det

⎛⎜⎝⎡⎢⎣α1 β1 0

β1 α2 00 β2 β3

⎤⎥⎦⎞⎟⎠ =

α4 · det

⎛⎜⎝⎡⎢⎣α1 β1 0

β1 α2 β2

0 β2 α3

⎤⎥⎦⎞⎟⎠− β2

3 · det

([α1 β1

β1 α2

])= α4 · P (1) − β2

3 · P (2) (7–67)

(7-67) has the same recursive structure as described by (7-59).


Example 7.4: Characteristic polynomial iteration

The generalized eigenvalue problem defined in Example 1.4 is considered again. Calculate the 3rd eigenvalue bysecant iteration on the characteristic polynomial, and next determine the corresponding eigenvector.

At first a calculation with µ = 2.5 is performed, which produces the following results

K− 2.5M =

⎡⎢⎣ 0.7500 −1.0000 0.0000−1.0000 1.5000 −1.0000

0.0000 −1.0000 0.7500

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

P (3)(2.5) = 1 , sign(P (3)(2.5)) = +

P (2)(2.5) = 0.7500 , sign(P (2)(2.5)) = +

P (1)(2.5) = 0.7500 · 1.5000− (−1)2 = 0.1250 , sign(P (1)(2.5)) = +

P (0)(2.5) = 0.7500 · 0.1250− (−1)2 · 0.7500 = −0.6563 , sign(P (0)(2.5)) = −

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–68)

Hence, the sign sequence of the Sturm sequence becomes + + +−. One sign change occurs in this sequence fromwhich is concluded that the lowest eigenvalue λ1 is smaller than µ = 2.5.

Next, a calculation with µ = 5.5 is performed, which provided the results

K− 5.5M =

⎡⎢⎣−0.7500 −1.0000 0.0000−1.0000 −1.5000 −1.0000

0.0000 −1.0000 −0.7500

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

P (3)(5.5) = 1 , sign(P (3)(5.5)) = +

P (2)(5.5) = −0.7500 , sign(P (2)(5.5)) = −P (1)(5.5) = (−0.7500) · (− 1.5000

)− (−1)2 = 0.1250 , sign(P (1)(5.5)) = +

P (0)(5.5) = (−0.7500) · 0.1250− (−1)2 · (−0.7500) = 0.6563 , sign(P (0)(5.5)) = +

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–69)

Now, the sign sequence of the Sturm sequence becomes +−++, in which two sign changes occur, from which isconcluded that the lowest two eigenvalues λ1 and λ2 are both smaller than µ = 5.5.

Finally a calculation with µ = 6.5 is performed, which provided the results

K− 6.5M =

⎡⎢⎣−1.2500 −1.0000 0.0000−1.0000 −2.5000 −1.0000

0.0000 −1.0000 −1.2500

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

P (3)(6.5) = 1 , sign(P (3)(6.5)) = +

P (2)(6.5) = −1.2500 , sign(P (2)(6.5)) = −P (1)(6.5) = (−1.2500) · (− 2.5000

)− (−1)2 = 2.1250 , sign(P (1)(6.5)) = +

P (0)(6.5) = (−1.2500) · 2.1250− (−1)2 · (−1.2500) = −1.4063 , sign(P (0)(6.5)) = −

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7–70)

In this cse the sign sequence of the Sturm sequence becomes +−+−, corresponding to three sign changes. Hence,it is concluded that all three eigenvalues λ1, λ2 and λ3 are smaller than µ = 6.5.


From the Sturm sequence checks it is concluded that 5.5 < λ3 < 6.5. Then, we may use the following startvalues, µ0 = 5.5 and µ1 = 6.5, in the secant iteration algorithm. Moreover P (5.5) = P (0)(5.5) = 0.6563 andP (6.5) = P (0)(6.5) = −1.4063, cf. (7-69) and (7-70). Then, from (7-61) it follows for k = 1

µ2 = 6.5 − (−1.4063)(−1.4063)− 0.6563

(6.5 − 5.5) = 5.8182 (7–71)

Next, P (µ2) = P (5.8182) = 0.3156 is calculated by means of the algorithm (7-60), and a new value µ 3 can beobtained from

µ3 = 5.8182− 0.31560.3156− (−1.4063)

(5.8182− 6.5) = 5.9431 (7–72)

During the next 5 iterations the following results were obtained

µ4 = 6.00900500472288µ5 = 5.99960498912941µ6 = 5.99999734553262µ7 = 6.00000000078659µ8 = 6.00000000000000

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭

(7–73)

As seen the convergence of the secant iteration algorithm is very fast.

The linear equation (7-63) attains the form

(K− 6.0000M

)Φ(3) = 0 ⇒

⎡⎢⎣−1 −1 0−1 −2 −1

0 −1 −1

⎤⎥⎦⎡⎢⎣Φ(3)

1

Φ(3)2

Φ(3)3

⎤⎥⎦ =

⎡⎢⎣0

00

⎤⎥⎦ (7–74)

Setting Φ(3)1 = 1 the algorithm (7-64) now provides

Φ(3)2 = − (−1)

(−1)· 1 = −1

Φ(3)3 = − (−1)

(−1)· 1 − (−2)

(−1)· (−1) = 1

⎫⎪⎪⎪⎬⎪⎪⎪⎭ ⇒ Φ(3) =

⎡⎢⎣ 1−1

1

⎤⎥⎦ (7–75)

Normalization to unit modal mass provides, cf. (1-87)

Φ(3) =√

22

⎡⎢⎣ 1−1

1

⎤⎥⎦ (7–76)

7.5 Exercises 171

7.5 Exercises

7.1 Given the following mass- and stiffness matrices defined in Exercise 4.2.

(a.) Calculate the two lowest eigenmodes and corresponding eigenvalues by simultaneousinverse vector iteration with the start vector basis

Φ0 =[Φ

(1)0 Φ

(2)0

]=

⎡⎢⎣1 1

1 0

1 −1

⎤⎥⎦

7.2 Given the symmetric matrices M and K of dimension n.

(a.) Write a MATLAB program, which for given start basis performs simultaneous inversevector iteration for the determination of the lowest n1 eigenmodes and eigenvalues.

7.3 Consider the general eigenvalue problem in Exercise 4.2.

(a.) Calculate the two lowest eigenmodes and corresponding eigenvalues by subspace it-eration using the same start basis as in Exercise 7.1.

7.4 Given the symmetric matrices M and K of dimension n.

(a.) Write a MATLAB program, which for given start basis performs subspace iterationfor the determination of the lowest n1 eigenmodes and eigenvalues.

7.5 Consider the general eigenvalue problem in Exercise 4.2.

(a.) Calculate the 3rd eigenmode and eigenvalue by Sturm sequence iteration (telescopemethod).

7.6 Given the symmetric matrices M and K of dimension n on three diagonal form.

(a.) Write a MATLAB program, which performs Sturm sequence check and secant iter-ation iteration for the determination of the jth eigenvalue, and next determines thecorresponding eigenvector.

Index

acceleration, 18acceleration vector, 7adjoint eigenvalue problem, 15, 24aerodynamic damping load, 8aerodynamic damping matrix, 8alternative inverse vector iteration, 96, 97amplification matrix, 34, 37, 50argument of complex eigenvalue, 45

central difference algorithm, 32, 40, 44, 46, 47,49

central difference operator, 19, 40characteristic equation, 9, 22, 61, 64, 89, 148,

165characteristic polynomial, 9, 53, 59, 165characteristic polynomial iteration, 90, 129, 148,

164, 166, 167, 169Choleski decomposition, 66, 67, 71commutative matrix multiplication, 15compatible matrix and vector norms, 83, 86complex conjugation, 13complex modal mass, 15complex unit, 8conditional stable algorithm, 44, 49consistent mass matrix, 19convergence rate of iteration vector, 93, 103, 104,

109, 153, 156convergence rate of Rayleigh quotient, 94, 103,

106, 153convergence rate of the iteration vector, 159Crank-Nicolson algorithm, 32, 40, 42, 44, 46–

48cubic convergence, 106

damped eigenmode, 13damped eigenperiod, 45damped eigenvalue, 13damped modal coordinate, 27

damping matrix, 7decoupling condition, 26, 36degree of freedom, 69, 147diagonal matrix, 91, 151, 157, 158Dirac’s delta function, 8displacement difference equation form of New-

mark algorithm, 40displacement difference form of Newmark algo-

rithm, 38displacement vector, 7dynamic load vector, 7, 45

eigenmode, 10, 115eigenvalue, 9eigenvalue separation principle, 53, 59, 62error analysis of calculated eigenvalues, 69, 83,

87error vector, 41, 83Euclidean matrix norm, 86Euclidean vector norm, 83, 85

forward vector iteration, 90, 99–101forward vector iteration with Gram-Schmidt or-

thogonalization, 109, 110, 129forward vector iteration with shift, 104Fourier transform, 8frequency response matrix, 8fundamental matrix, 12

Galerkin variation, 19Gauss factorization, 53, 55, 60, 66, 160, 163,

164general Jacobi iteration method, 116, 122, 126,

158, 159generalized eigenvalue problem, 159generalized alpha algorithm, 32, 49generalized eigenvalue problem, 9, 22, 53, 57,

58, 63, 65, 71, 78, 90, 95–97, 100, 107,

172

INDEX 173

115, 116, 122, 126, 128, 133, 134, 139,147, 149, 154, 156, 162, 165

Gram-Schmidt orthogonalization, 109, 138, 149,160

Guyan reduction, 69

Hilbert matrix norm, 83, 86, 87HOQR iteration method, 116, 140Householder reduction method, 128, 133, 134,

140, 149

impulse response matrix, 8infinity matrix norm, 86infinity vector norm, 85initial value vector, 7, 18inverse vector iteration, 90, 95, 101, 149inverse vector iteration with Gram-Schmidt or-

thogonalization, 109, 110, 129inverse vector iteration with Rayleigh quotient

shift, 106, 107inverse vector iteration with shift, 102, 103, 106iterative similarity transformation method, 115,

116

Jordan boxes, 11Jordan normal form, 11

Kronecker’s delta, 136

linear convergence, 93, 109linear viscous damping, 7lower triangular matrix, 53, 56, 66, 67, 72lumped mass matrix, 19

M-orthonormalization of vector basis, 152mass matrix, 7, 19, 129, 164matrix exponential function, 12, 37matrix norm, 85modal coordinate, 25, 26, 76, 77, 91modal damping ratio, 36modal damping ration, 25modal load, 36modal loads, 26modal mass, 10, 23, 36, 71, 75, 79, 83, 87, 90,

99, 107, 109, 110, 115, 150, 152, 153,159, 167

modal matrix, 10, 22, 87, 115, 147modal space, 102multi step algorithm, 31multi value algorithm, 31

negative numerical damping, 46Newmark algorithm, 31, 34numerical accuracy, 31, 41numerical damping, 31, 46numerical stability, 31, 42

omission criteria for similarity transformation,119, 120, 125, 126

one matrix norm, 86one vector norm, 85orthogonality property, 10, 15, 25, 91orthonormal matrix, 11, 116, 118, 129, 136

p vector norm, 85partitioned matrix, 69period distortion, 45period distortion per period, 45period elongation, 45period shortening, 45permutation of numbers, 117positive definite matrix, 7, 66positive numerical damping, 46positive semi-definite matrix, 8projected mass matrix, 77, 81, 82, 156, 159–161projected stiffness matrix, 77, 81, 82, 156, 159–

161

QR iteration method, 129, 136, 139, 140quadratic convergence, 95quasi-static response, 26

Rayleigh quotient, 74, 75, 89, 91, 93, 96, 97, 99,100, 102, 106, 107, 112, 150

Rayleigh’s principle, 74Rayleigh-Ritz analysis, 69, 74, 76, 79, 80, 148,

156, 160relative error of iteration vector, 93, 96relative error of Rayleigh quotient, 94, 96relatively errors of Rayleigh quotient, 101Ritz basis, 76, 79, 82, 148, 156

secant iteration, 167, 169, 170shift, 53shift on stiffness matrix, 63, 64, 101similarity transformation, 11, 53, 65, 72, 90, 115,

117, 122, 128, 137, 139, 164similarity transformation matrix, 65, 115, 117,

122, 129, 133, 134, 137, 139simple eigenvalues, 9

174 INDEX

simultaneous inverse vector iteration, 147–149,152–154, 156, 158–161

single step algorithm, 31single step multi value algorithm, 31, 37single value algorithm, 31special eigenvalue problem, 9, 13, 65, 66, 84,

87, 116, 118, 120, 128, 129, 133, 134,136, 139

special Jacobi iteration method, 116, 118–120,129

spectral decomposition, 66spectral matrix norm, 86standard eigenvalue problem, 149state vector, 12state vector formulation, 12static condensation, 69, 72, 80stiff-body motion, 64stiffness matrix, 7, 19, 131, 136, 164Sturm sequence, 53, 59, 160, 165, 167Sturm sequence check, 61, 103, 149, 153, 159,

160, 167, 169, 170Sturm sequence iteration, 149subspace iteration, 139, 147, 148, 152, 156, 159–

161sweep, 120, 121, 125, 127symmetric matrix, 7, 128, 129system reduction, 28, 29

telescope method, 55, 167three diagonal matrix, 128, 133, 140, 149, 164,

168transfer matrix, 42transposed matrix, 15two vector norm, 85

unconditional stable algorithm, 44undamped circular eigenfrequency, 9, 36undamped eigenmode, 25undamped eigenvibration, 9unit matrix, 8, 12unitary matrix, 11upper three diagonal matrix, 140upper triangular matrix, 56, 136, 149–151

vector basis, 75, 76, 138, 148vector iteration method, 89vector iteration with deflation, 109vector iteration with Gram-Schmidt orthogonal-

ization, 109

vector iteration with shift, 101vector norm, 85velocity, 18velocity vector, 7vibrating string, 18

wave equation, 18

APPENDIX ASolutions to Exercises

A.1 Exercise 1.1Given the following mass- and stiffness matrices

M =

⎡⎢⎣1 0 00 2 00 0 1

2

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0−1 2 0

0 0 3

⎤⎥⎦ (1)

1. Calculate the eigenvalues and eigenmodes normalized to unit modal mass.

2. Determine two vectors that are M-orthonormal, but are not eigenmodes.

SOLUTIONS:

Question 1:

The generalized eigenvalue problem (1-9) becomes

⎡⎢⎣2 − λj −1 0

−1 2 − 2λj 0

0 0 3 − 12λj

⎤⎥⎦⎡⎢⎢⎣Φ(j)

1

Φ(j)2

Φ(j)3

⎤⎥⎥⎦ =

⎡⎢⎣0

00

⎤⎥⎦ (2)

Upon evaluating the determinant of the coefficient matrix after the 3rd row, the characteristic equation(1-10) becomes

P (λ) = P (0)(λ) = det

⎛⎜⎝⎡⎢⎣2 − λj −1 0

−1 2 − 2λj 0

0 0 3 − 12λj

⎤⎥⎦⎞⎟⎠ =

(3 − 1

2λj

)((2 − λj

)(2 − 2λj

)− (− 1

)2) =(3 − 1

2λj

)(3 − 6λj + 2λ2

j

)= 0 ⇒

λj =

⎧⎪⎪⎨⎪⎪⎩

12

(3 −√

3)

, j = 112

(3 +

√3)

, j = 2

6 , j = 3

(3)

— 175 —

176 Chapter A – Solutions to Exercises

The largest eigenvalue λ3 = 6 is obtained when the 1st factor in (3) is equal to 0, whereas the two lowestsolutions corresponds vanishing of the 2nd factor.

Because the 3rd eigenmode is decoupled from the 1st and 2nd the solution method is slightly different inthis case. As seen be inspection the solutions have the form

Φ(j) =

⎡⎢⎣Φ(j)

1

Φ(j)2

0

⎤⎥⎦ , j = 1, 2

Φ(3) =

⎡⎢⎣0

01

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4)

The 1st and 2nd components of the 1st and 2nd eigenmodes, Φ(j)1 and Φ(j)

2 are determined from the two

first equations in (2). We choose to set Φ(j)1 = 1, and determine Φ(j)

2 from the 1st equations. Notice that

we may as well have determined Φ(j)2 from the 2nd equation. Then

(2 − λj

) · 1 − Φ(j)2 = 0 ⇒ Φ(j)

2 =

{12

(1 +

√3)

, j = 112

(1 −√

3)

, j = 2(5)

The modal masses become

Mj = Φ(j) TMΦ(j) =

⎡⎢⎣ 1

Φ(j)2

0

⎤⎥⎦

T ⎡⎢⎣1 0 00 2 00 0 1

2

⎤⎥⎦⎡⎢⎣ 1

Φ(j)2

0

⎤⎥⎦ = 1 + 2

(Φ(j)

2

)2=

{3 +

√3 , j = 1

3 −√3 , j = 2

(6)

M3 = Φ(3) TMΦ(3) =

⎡⎢⎣001

⎤⎥⎦

T ⎡⎢⎣1 0 00 2 00 0 1

2

⎤⎥⎦⎡⎢⎣0

01

⎤⎥⎦ =

12

(7)

Φ(1) denotes the 1st eigenmode normalized to unit modal mass. This is related toΦ(1) in the followingway

Φ(1) =1√M1

Φ(1) =1√

3 +√

3

⎡⎢⎣ 1

12

(1 +

√3)

0

⎤⎥⎦ =

⎡⎢⎣0.4597

0.62800

⎤⎥⎦ (8)

The other modes are treated in the same manner, which results in the following eigensolutions

A.2 Exercise 1.2 177

Λ =

⎡⎢⎣λ1 0 0

0 λ2 0

0 0 λ3

⎤⎥⎦ =

⎡⎢⎣

12

(3 −√

3)

0 0

0 12

(3 +

√3)

0

0 0 6

⎤⎥⎦

Φ =[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣0.4597 0.8881 0

0.6280 −0.3251 0

0 0 1.4142

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(9)

Question 2:

Consider the vectors

v1 =

⎡⎢⎣1

00

⎤⎥⎦ , v2 =

⎡⎢⎣ 0√

22

0

⎤⎥⎦ (10)

Upon insertion the following relations are seen to be valid

vT1 Mv1 = 1 , vT

2 Mv2 = 1 , vT1 Mv2 = 0 (11)

Hence, v1 and v2 are mutually M-orthonormal. However,

Kv1 =

⎡⎢⎣ 2

−1

0

⎤⎥⎦ �= λ1Mv1 =

⎡⎢⎣

12

(3 −√

3)

0

0

⎤⎥⎦

Kv2 =

⎡⎢⎢⎣−

√2

2√2

0

⎤⎥⎥⎦ �= λ2Mv2 =

⎡⎢⎣ 0√

24

(3 +

√3)

0

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(12)

Hence, neither v1 nor v2 are eigenmodes.

A.2 Exercise 1.2The eigensolutions with eigenmodes normalized to unit modal mass of a 2-dimensional generalizedeigenvalue problem are given as


Λ =

[λ1 0

0 λ2

]=

[1 00 4

], Φ =

[Φ(1) Φ(2)

]=

[√2

2

√2

2√2

2 −√

22

](1)

1. Calculate M and K.

SOLUTIONS:

Question 1:


M =(Φ−1

)TmΦ−1 (2)

K =(Φ−1

)TkΦ−1 (3)

Since it is known that the eigenmodes have been normalized to unit modal mass it follows from (1-20)and (1-22) that

m = I , k = Λ (4)

The inverse of the modal matrix becomes

Φ−1 =

[√2

2

√2

2√2

2 −√

22

]−1

=

[√2

2

√2

2√2

2 −√

22

](5)

Of course, (5) can be obtained by direct calculation. Alternatively, the result may be obtained from thefollowing arguments. Notice that Φ is orthonormal, so Φ−1 = ΦT , cf. (1-23). Additionally, the modalmatrix is symmetric, i.e. Φ = ΦT , from which the indicated result follows.

Insertion of (4) and (5) into (1) and (2) provides

M =

[√2

2

√2

2√2

2 −√

22

]T [1 0

0 1

][√2

2

√2

2√2

2 −√

22

]=

[1 0

0 1

](6)

K =

[√2

2

√2

2√2

2 −√

22

]T [1 0

0 4

][√2

2

√2

2√2

2 −√

22

]=

[2.5 −1.5

−1.5 2.5

](7)

Actually, since M = I, the considered eigenvalue problem is of the special type, cf. the remarks subse-quent to (1-9).



M =

⎡⎢⎣1 0 00 2 00 0 1

2

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0−1 2 0

0 0 3

⎤⎥⎦ (1)

1. Show that the eigenvalue separation principle is valid for the considered example.

SOLUTIONS:

Question 1:

The eigenvalues λ(0)j , which have been calculated in Exercise 1.1, are

λj =

⎧⎪⎪⎨⎪⎪⎩

12

(3 −√

3)

, j = 112

(3 +

√3)

, j = 2

6 , j = 3

(2)

Correspondingly, the eigenvalues λ(1)j and λ

(1)j are calculated from

P (1)(λ(1)) = det

([2 − λ

(1)j −1

−1 2 − 2λ(1)j

])=

((2 − λ

(1)j

)(2 − 2λ(1)

j

)− (− 1

)2) =(3 − 6λ(1)

j ) + 2(λ

(1)j

)2) = 0 ⇒

λ(1)j =

{12

(3 −√

3)

, j = 112

(3 +

√3)

, j = 2(3)

P (2)(λ(2)) = det([

2 − λ(2)j

])⇒ λ

(2)j = 2 (4)

Then, (3-25) attains the following forms for m = 0 and m = 1

0 ≤ λ(0)1 ≤ λ

(1)1 ≤ λ

(0)2 ≤ λ

(1)2 ≤ λ

(0)3 ≤ ∞ ⇒

0 ≤ 12(3 −

√3) ≤ 1

2(3 −

√3) ≤ 1

2(3 +

√3) ≤ 1

2(1 +

√3) ≤ 6 ≤ ∞ (5)

0 ≤ λ(1)1 ≤ λ

(2)1 ≤ λ

(1)2 ≤ ∞ ⇒

0 ≤ 12(3 −

√3) ≤ 2 ≤ 1

2(3 +

√3) ≤ ∞ (6)

Hence, (3-25) holds for the considered example. λ(0)1 = λ

(1)1 and λ

(0)2 = λ

(1)2 , because of the decoupling

of the 3rd eigenmode from the 1st and 2nd eigenmode.



M =

[2 00 0

], K =

[6 −1

−1 4

](1)

1. Calculate the eigenvalues and eigenmodes normalized to unit modal mass.

2. Perform a shift ρ = 3 on K and calculate the eigenvalues and eigenmodes of the new problem.

SOLUTIONS:

Question 1:

The generalized eigenvalue problem (1-9) is written on the form

[2 00 0

][Φ(j)

1

Φ(j)2

]=

1λj

[6 −1

−1 4

][Φ(j)

1

Φ(j)2

], j = 1, 2 (2)

Obviously, (2) has the solution

λ2 = ∞ , Φ(2) =

[Φ(2)

1

Φ(2)2

]=

[01

](3)

Hence, λ2 = ∞ is an eigenvalue. This is so because the mass matrix is singular, and has zeroes in thelast row and column. Since, the modal mass M2 related to eigenmode Φ(2) is zero, this mode cannot benormalized in the usual manner. In Section 4.1 the problem of infinite eigenvalues will be thoroughlydealt with.

The other eigensolution may be obtained by the standard approach. Then, the eigenvalue problem (2) iswritten on the form

[6 − 2λ1 −1

−1 4

][Φ(1)

1

Φ(1)2

]=

[00

](4)


det

([6 − 2λ1 −1

−1 4

])= 4(6 − 2λ1

)− (− 1)2 = 23 − 8λ1 = 0 ⇒ λ1 =

238

(5)


We choose to set Φ(1)1 = 1, and determine Φ(1)

2 from the 1st equations. Then

(6 − 2λ1

) · 1 − Φ(1)2 = 0 ⇒ Φ(1)

2 =14

(6)

The modal mass becomes

M1 = Φ(1) TMΦ(1) =

[114

]T [2 00 0

][114

]= 2 (7)

Then, the eigenmode normalized to unit modal mass Φ(1) becomes

Φ(1) =1√M1

Φ(1) =1√2

[114

]=

[0.70710.1768

](8)

Hence, the following eigensolutions have been obtained

Λ =

[λ1 0

0 λ2

]=

[238 0

0 ∞

], Φ =

[Φ(1) Φ(2)

]=

[0.7071 0

0.1768 1

](9)

Question 2:

(3-38) attains the form

K = K − 3M =

[6 −1

−1 4

]− 3

[2 00 0

]=

[0 −1

−1 4

](10)

The eigenvalue problem (3-37) becomes

([0 −1

−1 4

]− λj

[2 0

0 0

])[Φ(1)

1

Φ(1)2

]=

[00

](11)

For the same reason as in question 1, λ2 = ∞ is still an eigenvalue with the eigenmode given by (3).The characteristic equation for the 1st eigenvalue becomes, cf. (5)

det

([−2λ1 −1

−1 4

])= 4(−2λ1

)−(−1)2 = −1−8λ1 = 0 ⇒ λ1 = −1

8

(=

238−3)

(12)


Let Φ(1)1 = 1, and determine Φ(1)

2 from the 1st equations of (11)

(− 2λ1

) · 1 − Φ(1)2 = 0 ⇒ Φ(1)

2 =14

(13)

which is identical to (6). Hence Φ(1) is unaffected by the shift as expected, cf. the comments following(3-38). The eigensolutions are unchanged as given by (9), save that λ1 = −1

8 .

A.5 Exercise 3.4: Theory

Gauss EliminationGiven a symmetric matrix K of the dimension n × n with the components Kij = Kji. Consider thestatic equilibrium equation

Kx = f ⇒⎡⎢⎢⎢⎢⎢⎢⎣

K11 K12 K13 · · · K1n

K21 K22 K23 · · · K2n

K31 K32 K33 · · · K3n...

......

. . ....

Kn1 Kn2 Kn3 · · · Knn

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

x1

x2

x3...

xn

⎤⎥⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎢⎣

f1

f2

f3...

fn

⎤⎥⎥⎥⎥⎥⎥⎦ (1)

In order to have a one at the 1st element of the main diagonal of the coefficient matrix the 1st equation isdivided with K11 resulting in

⎡⎢⎢⎢⎢⎢⎢⎣

1 K(1)12 K

(1)13 · · · K

(1)1n

K21 K22 K23 · · · K2n

K31 K32 K33 · · · K3n...

......

. . ....


⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

x1

x2

x3...

xn

⎤⎥⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎢⎣

f(1)1

f2

f3...

fn

⎤⎥⎥⎥⎥⎥⎥⎦ (2)

where

K(1)1j =

K1j

K11, j = 2, . . . , n

f(1)1 =

f1

K11

⎫⎪⎪⎪⎬⎪⎪⎪⎭ (3)

In turn, the 1st equation of (2) is multiplied with Ki1, i = 2, . . . , n, and the resulting equation iswithdrawn from the ith equation. This will produce a zero in the ith row of the 1st column, correspondingto the following system of equations

A.5 Exercise 3.4: Theory 183

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 K(1)12 K

(1)13 · · · K

(1)1n

0 K(1)22 K

(1)23 · · · K

(1)2n

0 K(1)32 K

(1)33 · · · K

(1)3n

......

.... . .

...

0 K(1)n2 K

(1)n3 · · · K

(1)nn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

x1

x2

x3...

xn

⎤⎥⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎢⎣

f(1)1

f(1)2

f(1)3...

f(1)n

⎤⎥⎥⎥⎥⎥⎥⎦ (4)

where

K(1)ij = Kij − Ki1K

(1)1j , i = 2, . . . , n , j = 2, . . . , n

f(1)i = fi − Ki1f

(1)1 , i = 2, . . . , n

⎫⎪⎬⎪⎭ (5)

Next, the 2nd equation is divided with K(1)22 , so the coefficient in the 2nd component in the main diagonal

becomes equal to 1. In turn, the resulting 2nd equation is multiplied with K(1)i2 , i = 3, . . . , n, and the

resulting equation is withdrawn from the ith equation. This will produce zeros in the ith row of the 2ndcolumn below the main diagonal, corresponding to the system of equations

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 K(1)12 K

(1)13 · · · K

(1)1n

0 1 K(2)23 · · · K

(2)2n

0 0 K(2)33 · · · K

(2)3n

......

.... . .

...

0 0 K(2)n3 · · · K

(2)nn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎣

x1

x2

x3

...

xn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

f(1)1

f(2)2

f(2)3...

f(2)n

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(6)

where

K(2)2j =

K(1)2j

K(1)22

, j = 3, . . . , n

f(2)2 =

f(1)2

K(1)22

K(2)ij = K

(1)ij − K

(1)i2 K

(2)2j , i = 3, . . . , n , j = 3, . . . , n

f(2)i = f

(1)i − K

(1)i2 f

(2)2 , i = 3, . . . , n

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7)

The process of producing ones in the main diagonal, and zeros below the main diagonal is continued forall n columns resulting in the following system of linear equations


⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 K(1)12 K

(1)13 · · · K

(1)1n

0 1 K(2)23 · · · K

(2)2n

0 0 1 · · · K(3)3n

......

.... . .

...

0 0 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎣

x1

x2

x3

...

xn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

f(1)1

f(2)2

f(3)3

...

f(n)n

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(8)

Next, (1) are solved simultaneous with n righthand sides, where the loads form the columns in a unitmatrix. The n solution vectors X = [x1 x2 x3 · · ·xn] are organized in the matrix equationi.e.

KX = I ⇒⎡⎢⎢⎢⎢⎢⎢⎣

K11 K12 K13 · · · K1n

K21 K22 K23 · · · K2n

K31 K32 K33 · · · K3n...

......

. . ....


⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

x11 x12 x13 · · · x1n

x21 x22 x23 · · · x2n

x31 x32 x33 · · · x3n...

......

. . ....

xn1 xn2 xn3 · · · xnn

⎤⎥⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎢⎣

1 0 0 · · · 00 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎦ (9)

Following the steps (2)-(8), simultaneous Gauss elimination of the coefficient matrix and the n righthandsides provides the following equivalent matrix equation

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 K(1)12 K

(1)13 · · · K

(1)1n

0 1 K(2)23 · · · K

(2)2n

0 0 1 · · · K(3)3n

......

.... . .

...

0 0 0 · · · 1

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎣

x11 x12 x13 · · · x1n

x21 x22 x23 · · · x2n

x31 x32 x33 · · · x3n

......

.... . .

...

xn1 xn2 xn3 · · · xnn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

f(1)11 0 0 · · · 0

f(2)21 f

(2)22 0 · · · 0

f(3)31 f

(3)32 f

(3)33 · · · 0

......

.... . .

...

f(n)n1 f

(n)n2 f

(n)n3 · · · f

(n)nn

⎤⎥⎥⎥⎥⎥⎥⎥⎦(10)

As indicated the identity matrix on the righthand side is transformed into a lower triangular matrix F.

In the program the triangulation of the matrix K and the calculation of the matrix F is performed in amatrix A of the dimension n × 2n, which at the entry of the triangulation loop has the form

A =[KI]

(11)

At exit from the triangulation loop the matrix A stores the triangulized stiffness at the position originallyoccupied by K, and the matrix F at the position occupied by the unit matrix.


Calculation of L, D and (S−1)T

Using the Gauss factorization of the stiffness matrix (9) may be written, cf. (3-1)

KX = LDLTX = I ⇒

LTX = ID−1L−1 = D−1L−1 = F(12)

Upon comparison of (10) and (12) it becomes clear that LT is stored as the coefficient matrix in (10),whereas the righthand sides store the matrix F = D−1L−1. Since, L−1 is a lower triangular matrix withones in the main diagonal, the main diagonal must contain the main diagonal of D−1. Hence,

D−1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

f(1)11 0 0 · · · 0

0 f(2)22 0 · · · 0

0 0 f(3)33 · · · 0

......

.... . .

...

0 0 0 · · · f(n)nn

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⇒ D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

f(1)11

0 0 · · · 0

0 1

f(2)22

0 · · · 0

0 0 1

f(3)33

· · · 0...

......

. . ....

0 0 0 · · · 1

f(n)nn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(13)

Finally, cf. (3-49)

S = LD12 ⇒ S−1 = D− 1

2L−1 = D12 D−1L−1 = D

12F (14)

The matrices D and (S−1)T are retrieved from the righthand sides of (10) as stored in the matrix Faccording to the indicated relations at the end of the program.


M =

⎡⎢⎣0 0 00 2 10 1 1

⎤⎥⎦ , K =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ (1)

1. Perform a static condensation by the conventional procedure based on (4-5), (4-6), and next byRayleigh-Ritz analysis with the Ritz basis given by (4-62).

SOLUTIONS:

Question 1:

The 1st and 3rd row, and next the 1st and 3rd column of the are interchanged, which brings the matricesof the general eigenvalue problem on the following form, cf. (4-1), (4-2)


M =

⎡⎣M11 M12

M21 M22

⎤⎦ =

⎡⎢⎢⎣

1 1 01 2 0

0 0 0

⎤⎥⎥⎦

K =

⎡⎣K11 K12

K21 K22

⎤⎦ =

⎡⎢⎢⎣

2 −1 0−1 4 −1

0 −1 6

⎤⎥⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2)

Notice that the interchange of two rows or two columns may change the sign, but not the numerical value,of the characteristic polynomial. However, since the characteristic polynomial is zero at the eigenvaluethe determination of the eigenvalues is unaffected by the sign change.

The reduced stiffness matrix (4-7) becomes

K11 = K11 − K12K−122 K21 =

[2 −1

−1 4

]−[

0−1

][6]−1

[0 −1

]=

[2 −1

−1 236

](3)

The reduced eigenvalue problem (4-6) is solved

[2 −1

−1 236

]Φ11 =

[1 11 2

]Φ11Λ1 (4)

The eigensolutions with eigenmodes normalized to modal mass 1 with respect to M11 becomes

Λ1 =

[λ1 00 λ2

]=

[0.7325 0

0 9.1008

], Φ11 =

[Φ(1)

1 Φ(2)1

]=

[0.5320 1.3103

0.3892 −0.9212

](5)

From (4-5) follows

Φ21 =[Φ(1)

2 Φ(2)2

]= − [6]−1

[0 −1

] [0.5320 1.3103

0.3892 −0.9212

]=[0.0649 −0.1535

](6)


Λ2 = [λ3] = [∞] , Φ12 =

[00

], Φ22 = [1] (7)


After interchanging the degrees of freedom back to the original order (the 1st components of Φ11 andΦ12 are placed as the 3rd component of Φ(j), and the components of Φ21 and Φ22 are placed as the 1stcomponent Φ(j), the following eigensolution is obtained

Λ =

⎡⎢⎣λ1 0 0

0 λ2 00 0 λ3

⎤⎥⎦ =

⎡⎢⎣0.7325 0 0

0 9.1008 00 0 ∞

⎤⎥⎦

Φ =[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣0.0649 −0.1535 10.3892 −0.9212 00.5320 1.3103 0

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(8)

Next, the same problem is solved by means of Rayleigh-Ritz analysis. The Ritz basis is constructed from(4-62)

Ψ2 =

⎡⎢⎢⎣

2 −1 0−1 4 −1

0 −1 6

⎤⎥⎥⎦−1 ⎡⎢⎢⎣

1 00 1

0 0

⎤⎥⎥⎦ =

⎡⎢⎢⎣0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦ (9)

The projected mass and stiffness matrices become, cf. (4-63), (4-64)

M =

⎡⎢⎢⎣0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦

T ⎡⎢⎣1 1 0

1 2 00 0 0

⎤⎥⎦⎡⎢⎢⎣

0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦ =

[0.548125 0.3712500.371250 0.292500

]

K =

⎡⎢⎢⎣

0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦

T ⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 6

⎤⎥⎦⎡⎢⎢⎣0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦ =

[0.5750 0.15000.1500 0.3000

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(10)

The eigensolutions to the eigenvalue problem defined byM and K with modal masses normalized to 1with respect to M become, cf. Box. 4.2

R =

[ρ1 00 ρ2

]=

[0.7325 0

0 9.1008

], Q =

[q(1) q(2)

]=

[0.6748 3.54180.9599 −4.8415

](11)


The solutions for the eigenvectors become, cf. (4-51)

Φ =[Φ(1) Φ(2)

]=

⎡⎢⎢⎣0.5750 0.15000.1500 0.3000

0.0250 0.0500

⎤⎥⎥⎦[0.6748 3.54180.9599 −4.8415

]=

⎡⎢⎣0.5320 1.3103

0.3892 −0.92120.0649 −0.1535

⎤⎥⎦ (12)

As seen the eigenvalues (11) are identical to the lowest two eigenvalues from the static condensationprocedure (8). The two lowest eigenmodes in (8) are retrieved from (12) upon interchanging the 1st and3rd components in the latter.


M =

⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦ , K =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ (1)

1. Calculate approximate eigenvalues and eigenmodes by Rayleigh-Ritz analysis using the followingRitz basis

Ψ = [Ψ(1) Ψ(2)] =

⎡⎢⎣1 11 −11 1

⎤⎥⎦

SOLUTIONS:

Question 1:

The projected mass and stiffness matrices become, cf. (4-45)

M =

⎡⎢⎣1 11 −11 1

⎤⎥⎦

T ⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦⎡⎢⎣1 1

1 −11 1

⎤⎥⎦ =

[7 11 3

]

K =

⎡⎢⎣1 1

1 −11 1

⎤⎥⎦

T ⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣1 1

1 −11 1

⎤⎥⎦ =

[8 44 16

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(2)

The eigensolutions to the eigenvalue problem defined byM and K with modal masses normalized to 1with respect to M become, cf. Box. 4.2


R =

[ρ1 00 ρ2

]=

[1.0459 0

0 5.3541

], Q =

[q(1) q(2)

]=

[−0.3864 0.0269

0.0887 −0.5849

](3)

The solutions for the eigenvectors become, cf. (4-51)

Φ =[Φ(1) Φ(2)

]=

⎡⎢⎣1 11 −11 1

⎤⎥⎦[−0.3864 0.0269

0.0887 −0.5849

]=

⎡⎢⎣−0.2976 −0.5580−0.4751 0.6118−0.2976 −0.5580

⎤⎥⎦ (4)

The exact eigensolutions can be shown to be

Λ =

⎡⎢⎣λ1 0 0

0 λ2 00 0 λ3

⎤⎥⎦ =

⎡⎢⎣0.7245 0 0

0 2.9652 00 0 9.3104

⎤⎥⎦

Φ =[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎣−0.0853 −0.6981 −0.0458−0.3884 −0.0486 0.5778−0.5251 0.1997 −0.8149

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5)

As seen ρ1 and ρ2 are upperbounds to the exact eigenvalues λ1 and λ2, and ρ2 is smaller than λ3, cf.(4-57). The estimates of the eigenmodes are not useful. Not even the signs of the components ofΦ(2)

are correctly represented. These poor results are obtained because the chosen Ritz basis is far away fromthe basis spanned by Φ(1) and Φ(2).

A.8 Exercise 4.3Consider the mass- and stiffness matrices in Exercise 4.2, and let

v =

⎡⎢⎣1

11

⎤⎥⎦ (1)

1. Calculate the vector Φ(1) = K−1Mv, and next λ1 = ρ(Φ(1)

), as approximate solutions to the

lowest eigenmode and eigenvalue.

2. Establish the error bound for the obtained approximation to the lowest eigenvalue.


SOLUTIONS:

Question 1:

From the given formula we calculate

Φ(1) =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦⎡⎢⎣1

11

⎤⎥⎦ =

⎡⎢⎣0.55

1.301.65

⎤⎥⎦ (2)

The Rayleigh quotient based on Φ(1) becomes, cf. (4-25)

λ1 = ρ(Φ(1)

)=

⎡⎢⎣0.55

1.301.65

⎤⎥⎦

T ⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.55

1.301.65

⎤⎥⎦

⎡⎢⎣0.55

1.301.65

⎤⎥⎦

T ⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦⎡⎢⎣0.55

1.301.65

⎤⎥⎦

= 0.7547 (3)

The obtained un-normalized eigenmode Φ(1) resembles Φ(1) much better than the corresponding ap-proximation for Φ(1) indicated in eq. (4) of Exercise 7.2. As a consequence the obtained eigenvalueλ1

is a much better approximation to the exact eigenvalue λ1 = 0.7245 given in eq. (5) of Exercise 4.2,than the approximation ρ1 = 1.0459 obtained by the Rayleigh-Ritz analysis. The indicated formula forobtaining Φ(1) represents the 1st iteration step in the socalled inverse vector iteration algorithm describedin Section 5.2

Question 2:

From (2) follows that

∣∣Φ(1)∣∣ = 2.1714 (4)

The error vector becomes, cf. (4-79)

ε1 =

⎛⎜⎝⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦− 0.7547 ·

⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦⎞⎟⎠⎡⎢⎣0.551.301.65

⎤⎥⎦ =

⎡⎢⎣ 1.1698−0.2075−0.2264

⎤⎥⎦ ⇒

∣∣ε1

∣∣ = 1.2095 (5)


The lowest eigenvalue of M can be shown to be

µ1 = 0.3820 (6)

Then, from (4-85) the following bound is obtained

|λ1 − λ1| ≤ 10.3820

· 1.20952.1714

= 2.1714 (7)

Actually, |λ1 − λ1| = |0.7245− 0.7547| = 0.0302. Hence, the bounding method provides a rather crudeupperbound in the present case.

A.9 Exercise 5.1Given the following mass- and stiffness matrices defined in Exercise 4.2.

1. Perform two inverse iterations, and then calculate an approximation to λ1.

2. Perform two forward iterations, and then calculate an approximation to λ3.

SOLUTIONS:

Question 1:

The calculations are performed with the start vector

Φ0 =

⎡⎢⎣1

11

⎤⎥⎦ (1)


A =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦ =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦ (2)

At the 1st and 2nd iteration steps the following calculations are performed, cf. Box 5.1


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦⎡⎢⎣111

⎤⎥⎦ =

⎡⎢⎣0.551.301.65

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 10.9975

Φ1 =1√

10.9975

⎡⎢⎣0.55

1.301.65

⎤⎥⎦ =

⎡⎢⎣0.16585

0.392010.49755

⎤⎥⎦

(3)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ2 =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦⎡⎢⎣0.165850.392010.49755

⎤⎥⎦ =

⎡⎢⎣0.14436

0.534490.71202

⎤⎥⎦ ⇒ ΦT

2 MΦ2 = 1.8812

Φ2 =1√

1.8812

⎡⎢⎣0.144360.534490.71202

⎤⎥⎦ =

⎡⎢⎣0.10526

0.389700.51914

⎤⎥⎦

(4)

Since, Φ2 has been normalized to unit modal mass, so ΦT2 MΦ2 = 1, an approximation is obtained from

the following Rayleigh fraction, cf. (4-25)

λ1 = ΦT2 KΦ2 =

⎡⎢⎣0.10526

0.389700.51914

⎤⎥⎦

T ⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.105260.389700.51914

⎤⎥⎦ = 0.72629 (5)

The exact solution is λ1 = 0.72446.

Question 2:

The calculations are performed with the start vector given by (1).

The matrix B becomes, cf. (5-35)

B =

⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦−1 ⎡⎢⎣ 6 −1 0

−1 4 −10 −1 2

⎤⎥⎦ =

⎡⎢⎣ 3.0 −0.5 0.0−1.0 5.0 −3.0

1.0 −6.0 5.0

⎤⎥⎦ (6)

At the 1st and 2nd iteration steps the following calculations are performed, cf. Box 5.3


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ1 =

⎡⎢⎣ 3.0 −0.5 0.0−1.0 5.0 −3.0

1.0 −6.0 5.0

⎤⎥⎦⎡⎢⎣1

11

⎤⎥⎦ =

⎡⎢⎣2.5

1.00.0

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 14.5

Φ1 =1√14.5

⎡⎢⎣2.51.00.0

⎤⎥⎦ =

⎡⎢⎣0.65653

0.262610.00000

⎤⎥⎦

(7)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ2 =

⎡⎢⎣ 3.0 −0.5 0.0−1.0 5.0 −3.0

1.0 −6.0 5.0

⎤⎥⎦⎡⎢⎣0.65653

0.262610.00000

⎤⎥⎦ =

⎡⎢⎣ 1.83829

0.65653−0.91915

⎤⎥⎦ ⇒ ΦT

1 MΦ1 = 7.25862

Φ1 =1√

7.25862

⎡⎢⎣ 1.83829

0.65653−0.91915

⎤⎥⎦ =

⎡⎢⎣ 0.68232

0.24369−0.34116

⎤⎥⎦

(8)

Again, Φ2 has been normalized to unit modal mass, so ΦT2 MΦ2 = 1, an approximation is obtained

from the following Rayleigh fraction

λ3 = ΦT2 KΦ2 =

⎡⎢⎣ 0.68232

0.24369−0.34116

⎤⎥⎦

T ⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣ 0.68232

0.24369−0.34116

⎤⎥⎦ = 3.09739 (9)

The exact solution is λ3 = 9.31036. The poor result is obtained because Φ2 is a rather bad approximationto Φ(3).


M =

⎡⎢⎣

12 0 00 1 00 0 1

2

⎤⎥⎦ , K =

⎡⎢⎣ 2 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ (1)

The eigenmodes Φ(1) are Φ(3) are known to be, cf. (1-87)

Φ(1) =

⎡⎢⎢⎣

√2

2√2

2√2

2

⎤⎥⎥⎦ , Φ(3) =

⎡⎢⎢⎣

√2

2

−√

22√2

2

⎤⎥⎥⎦ (2)


1. Calculate Φ(2) by means of Gram-Schmidt orthogonalization, and calculate all eigenvalues.

SOLUTIONS:

Question 1:

Consider an arbitrary vector

x =

⎡⎢⎣1

22

⎤⎥⎦ (3)

Since, Φ(1), Φ(2) and Φ(3) form a vector basis, we may write

x = c1Φ(1) + c2Φ(2) + c3Φ(3) (4)

In order to determine the expansion coefficient cj , (4) is premultiplied with Φ(j) TM, and the M-othonormality of the eigenmodes are used, i.e. that Φ(i) TMΦ(j) = δij . For j = 1, 3 the followingresults are obtained

cj = Φ(j) TMx =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

⎡⎢⎢⎣

√2

2√2

2√2

2

⎤⎥⎥⎦

T ⎡⎢⎣

12 0 0

0 1 0

0 0 12

⎤⎥⎦⎡⎢⎣1

2

2

⎤⎥⎦ = 7

√2

4 , j = 1

⎡⎢⎢⎣

√2

2

−√

22√2

2

⎤⎥⎥⎦

T ⎡⎢⎣

12 0 0

0 1 0

0 0 12

⎤⎥⎦⎡⎢⎣1

2

2

⎤⎥⎦ = −

√2

4 , j = 3

(5)

Then, from (3), (4) and (5) follows

c2Φ(2) =

⎡⎢⎣1

2

2

⎤⎥⎦− 7

√2

4·

⎡⎢⎢⎣

√2

2√2

2√2

2

⎤⎥⎥⎦+

√2

4·

⎡⎢⎢⎣

√2

2

−√

22√2

2

⎤⎥⎥⎦ =

⎡⎢⎣−0.5

0.0

0.5

⎤⎥⎦ ⇒

c22 =

⎡⎢⎣−0.5

0.0

0.5

⎤⎥⎦

T ⎡⎢⎣

12 0 0

0 1 0

0 0 12

⎤⎥⎦⎡⎢⎣−0.5

0.0

0.5

⎤⎥⎦ = 0.25 ⇒

Φ(2) =1

0.5·

⎡⎢⎣−0.5

0.0

0.5

⎤⎥⎦ =

⎡⎢⎣−1

0

1

⎤⎥⎦ (6)


Hence, the modal matrix becomes, cf. (1-87)

Φ =[Φ(1) Φ(2) Φ(3)

]=

⎡⎢⎢⎣

√2

2 −1√

22√

22 0 −

√2

2√2

2 1√

22

⎤⎥⎥⎦ (7)

Given that all eigenmodes have been normalized to unit modal mass the eigenvalues may be calculatedfrom the Rayleigh quotient, cf. (1-21), (4-25)

Λ = ΦTKΦ =

⎡⎢⎣2 0 00 4 00 0 6

⎤⎥⎦ (8)

Generally, if n − 1 eigenmodes to a general eigenvalue problem is known the remaining eigenmode canlways be determined solely from the M-orthonormality conditions.

A.11 Exercise 6.3Given the mass- and stiffness matrices defined in Exercise 4.2.

1. Perform an initial transformation to a special eigenvalue problem, and calculate the eigenvalues andeigenvectors by means of standard Jacobi iteration.

2. Calculate the eigenvalues and normalized eigenvectors by means of general Jacobi iteration oper-ating on the original general eigenvalue problem.

SOLUTIONS:

Question 1:

Initially, a Choleski decomposition of the mass matrix is performed, cf. (3-44). As indicated by thealgorithm in Box 3.2 the following calculations are performed

s11 =√

m11 =√

2

s21 =m21

s11=

0√2

= 0

s31 =m31

s11=

0√2

= 0

s22 =√

m22 − s221 =

√2 − 02 =

√2

s32 =1

s22(m32 − s31 · s21) =

1√2(1 − 0 · 0) =

√2

2

s33 =√

m33 − s232 − s2

31 =

√1 −(√2

2

)2− 02 =

√2

2

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(1)


Hence, the matrices S and S−1 become

S =

⎡⎢⎣√

2 0 00

√2 0

0√

22

√2

2

⎤⎥⎦ ⇒

S−1 =

⎡⎢⎢⎣

√2

2 0 0

0√

22 0

0 −√

22

√2

⎤⎥⎥⎦ (2)

The initial value of the updated similarity transformation matrix and the stiffness matrix becomes, cf.(3-48), (6-4)

Φ0 = (S−1)T =

⎡⎢⎢⎣

√2

2 0 0

0√

22 −

√2

2

0 0√

2

⎤⎥⎥⎦

K0 = K = S−1K(S−1)T =⎡⎢⎢⎣

√2

2 0 0

0√

22 0

0 −√

22

√2

⎤⎥⎥⎦⎡⎢⎣ 6 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎢⎣

√2

2 0 0

0√

22 −

√2

2

0 0√

2

⎤⎥⎥⎦ =

⎡⎢⎣ 3.0 −0.5 0.5−0.5 2.0 −3.0

0.5 −3.0 8.0

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(3)

In the 1st sweep the following calculations are performed for (i, j) = (1, 2) :⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · (−0.5)3.0 − 2.0

)= −0.3927 ⇒

{cos θ = 0.9239sin θ = −0.3827

P0 =

⎡⎢⎣ 0.9239 0.3827 0−0.3827 0.9239 0

0 0 1

⎤⎥⎦

Φ1 = Φ0P0 =

⎡⎢⎣ 0.6533 0.2706 0−0.2706 0.6533 −0.7071

0 0 1.4142

⎤⎥⎦

K1 = PT0 K0P0 =

⎡⎢⎣3.2071 0 1.6070

0 1.7929 −2.58031.6070 −2.5803 8.0000

⎤⎥⎦

(4)


Next, the calculations are performed for (i, j) = (1, 3) :⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · 1.60703.2071 − 8.0000

)= −0.2958 ⇒

{cos θ = 0.9566sin θ = −0.2915

P1 =

⎡⎢⎣ 0.9566 0 0.2915

0 1 0−0.2915 0 0.9566

⎤⎥⎦

Φ2 = Φ1P1 =

⎡⎢⎣ 0.6249 0.2706 0.1904−0.0527 0.6533 −0.7553−0.4122 0 1.3528

⎤⎥⎦

K2 = PT1 K1P1 =

⎡⎢⎣2.7165 0.7521 00.7521 1.7929 −2.4682

0 −2.4682 8.4906

⎤⎥⎦

(5)

Finally, to end the 1st sweep the calculations are performed for (i, j) = (2, 3) :⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

θ =12

arctan(

2 · (−2.4682)1.7929 − 8.4906

)= 0.3176 ⇒

{cos θ = 0.9500sin θ = 0.3123

P2 =

⎡⎢⎣1 0 0

0 0.9500 −0.31230 0.3123 0.9500

⎤⎥⎦

Φ3 = Φ2P2 =

⎡⎢⎣ 0.6249 0.3165 0.0964−0.0527 0.3848 −0.9215−0.4122 0.4224 1.2852

⎤⎥⎦

K3 = PT2 K2P2 =

⎡⎢⎣ 2.7165 0.7145 −0.2349

0.7145 0.9816 0−0.2349 0 9.3019

⎤⎥⎦

(6)

At the end of the 2nd and 3rd sweep the following estimates are obtained for the modal matrix and theeigenvalues

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ6 =

⎡⎢⎣ 0.6980 0.0862 0.0729

0.0481 0.3885 −0.9202−0.2004 0.5249 1.2978

⎤⎥⎦ , K6 =

⎡⎢⎣2.9652 0.0028 0.0000

0.0028 0.7245 00.0000 0 9.3104

⎤⎥⎦

Φ9 =

⎡⎢⎣ 0.6981 0.0853 0.0729

0.0486 0.3884 −0.9202−0.1997 0.5251 1.2978

⎤⎥⎦ , K9 =

⎡⎢⎣2.9652 0.0000 0.0000

0.0000 0.7245 00.0000 0 9.3104

⎤⎥⎦

(7)

As seen the eigenmodes are stored column-wise in Φ according to the permutation (j1, j2, j3) = (2, 1, 3),cf. Box 6.2.


Question 2:

The following initializations are introduced, cf. Box 6.4

M0 = M =

⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦ , K0 = K =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦ , Φ0 =

⎡⎢⎣1 0 0

0 1 00 0 1

⎤⎥⎦ (8)

In the 1st sweep the following calculations are performed for (i, j) = (1, 2) :⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

a =4 · 0 − 2 · (−1)

6 · 2 − 2 · 4 = 0.5

b =6 · 0 − 2 · (−1)

6 · 2 − 2 · 4 = 0.5

⎫⎪⎪⎬⎪⎪⎭ ⇒

{α = −0.4142β = 0.4142

P0 =

⎡⎢⎣ 1 0.4142 0−0.4142 1 0

0 0 1

⎤⎥⎦ , Φ1 = Φ0P0 =

⎡⎢⎣ 1 0.4142 0−0.4142 1 0

0 0 1

⎤⎥⎦

M1 = PT0 M0P0 =

⎡⎢⎣ 2.3431 0 −0.4142

0 2.3431 1−0.4142 1 1

⎤⎥⎦

K1 = PT0 K0P0 =

⎡⎢⎣7.5147 0 0.4142

0 4.2010 −10.4142 −1 2

⎤⎥⎦

(9)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

a =2 · (−0.4142) − 1 · 0.4142

7.5147 · 1 − 2.3431 · 2 = −0.4393

b =7.5147 · (−0.4142) − 2.3431 · 0.4142

7.5147 · 1 − 2.3431 · 2 = −1.4437

⎫⎪⎪⎬⎪⎪⎭ ⇒

{α = 1.0023β = −0.3050

P1 =

⎡⎢⎣ 1 0 −0.3050

0 1 01.0023 0 1

⎤⎥⎦ , Φ2 = Φ1P1 =

⎡⎢⎣ 1 0.4142 −0.3050−0.4142 1 0.1263

1.0023 0 1

⎤⎥⎦

M2 = PT1 M1P1 =

⎡⎢⎣2.5174 1.0023 0

1.0023 2.3431 10 1 1.4707

⎤⎥⎦

K2 = PT1 K1P1 =

⎡⎢⎣10.3542 −1.0023 0−1.0023 4.2010 −1

0 −1 2.4465

⎤⎥⎦

(10)


Finally, to end the 1st sweep the calculations are performed for (i, j) = (2, 3) :

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

a =2.4465 · 1 − 1.4707 · (−1)

4.2010 · 1.4707 − 2.3407 · 2.4465 = 8.7838

b =4.2010 · 1 − 2.3431 · (−1)

4.2010 · 1.4707 − 2.3431 · 2.4465 = 14.6745

⎫⎪⎪⎬⎪⎪⎭ ⇒

{α = −1.2369β = 0.7404

P2 =

⎡⎢⎣1 0 0

0 1 0.74040 −1.2369 1

⎤⎥⎦ , Φ3 = Φ2P2 =

⎡⎢⎣ 1 0.7915 0.0016−0.4142 0.8437 0.8667

1.0023 −1.2369 1

⎤⎥⎦

M3 = PT2 M2P2 =

⎡⎢⎣2.5174 1.0023 0.7421

1.0023 2.1193 00.7421 0 4.2357

⎤⎥⎦

K3 = PT2 K2P2 =

⎡⎢⎣10.3542 −1.0023 −0.7421−1.0023 10.4174 0−0.7421 0 3.2684

⎤⎥⎦

(11)

At the end of the 2nd and 3rd sweep the following estimates are obtained for the modal matrix and thetransformed mass and stiffness matrices

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ6 =

⎡⎢⎣ 1.6779 −0.0129 0.1846

0.0350 1.3400 0.8282−0.3741 −1.9075 1.1188

⎤⎥⎦

M6 =

⎡⎢⎣ 5.7469 0.1959 −0.0118

0.1959 2.1179 0−0.0118 0 4.5448

⎤⎥⎦ , K6 =

⎡⎢⎣ 17.0856 −0.1959 0.0118−0.1959 19.6067 0

0.0118 0 3.2925

⎤⎥⎦

Φ9 =

⎡⎢⎣ 1.6780 −0.1060 0.1819

0.1169 1.3379 0.8281−0.4801 −1.8869 1.1195

⎤⎥⎦

M9 =

⎡⎢⎣5.7769 0.0000 0.0000

0.0000 2.1139 00.0000 0 4.5448

⎤⎥⎦ , K9 =

⎡⎢⎣ 17.1296 −0.0000 −0.0000−0.0000 19.6810 0−0.0000 0 3.2925

⎤⎥⎦

(12)

Presuming that the process has converged after the 3rd sweep the eigenvalues and normalized eigenmodesare next retrieved by the following calculations, cf. Box. 6.4


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

m = M9 =

⎡⎢⎣5.7769 0.0000 0.0000

0.0000 2.1139 00.0000 0 4.5448

⎤⎥⎦ , m− 1

2 =

⎡⎢⎣0.4161 0 0

0 0.6878 00 0 0.4691

⎤⎥⎦ ⇒

Λ =

⎡⎢⎣λ2 0 0

0 λ3 00 0 λ1

⎤⎥⎦ = M−1

9 K9 =

⎡⎢⎣ 2.9652 −0.0000 −0.0000−0.0000 9.3104 0.0000−0.0000 0.0000 0.7245

⎤⎥⎦

Φ =[Φ(2) Φ(3) Φ(1)

]= Φ9m− 1

2 =

⎡⎢⎣ 0.6981 −0.0729 0.0853

0.0486 0.9202 0.3884−0.1997 −1.2978 0.5251

⎤⎥⎦

(13)

The solutions (13) are identical to those obtained in (8) for the special Jacobi iteration algorithm. In thepresent case the eigenmodes are stored column-wise in Φ according to the permutation (j1, j2, j3) =(2, 3, 1), cf. Box 6.4. The convergence rates of the special nd the general Jacobi iteration algorithmseems to be rather alike.


1. Calculate the eigenvalues and normalized eigenvectors by means of QR iteration.

SOLUTIONS:

Question 1:

At first, as indicated in Box 6.7 an initial similarity transformation of the indicated general eigenvalueproblem into a special eigenvalue problem is performed with the similarity transformation matrix P =(S−1)T

, where S is a solution to M = SST . In case S is determined from an Choleski decomposition ofthe mass matrix the initial updated transformation and stiffness matrices have been calculated in Exercise6.3, eq. (4). The result becomes

Φ1 = (S−1)T =

⎡⎢⎢⎣

√2

2 0 0

0√

22 −

√2

2

0 0√

2

⎤⎥⎥⎦

K1 = S−1K(S−1)T =⎡⎢⎢⎣

√2

2 0 0

0√

22 0

0 −√

22

√2

⎤⎥⎥⎦⎡⎢⎣ 6 −1 0

−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎢⎣

√2

2 0 0

0√

22 −

√2

2

0 0√

2

⎤⎥⎥⎦ =

⎡⎢⎣ 3.0 −0.5 0.5−0.5 2.0 −3.0

0.5 −3.0 8.0

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(1)


As seen the original three diagonal structure of K is destroyed by the similarity. This may be reestab-lished by means of a Householder transformation as described in Section 6.4. However, this will beomitted here, so the QR-iteration is performed on the full matrix K1.

At the determination of q1 and r11 in the 1st QR iteration the following calculations are performed, cf.(6-72)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k1 =

⎡⎢⎣ 3.0−0.5

0.5

⎤⎥⎦ , r11 =

∣∣∣∣∣∣∣⎡⎢⎣ 3.0−0.5

0.5

⎤⎥⎦∣∣∣∣∣∣∣ = 3.0822

q1 =1

3.0822

⎡⎢⎣ 3.0−0.5

0.5

⎤⎥⎦ =

⎡⎢⎣ 0.9733−0.1622

0.1622

⎤⎥⎦

(2)

q2 and r12, r22 are determined from the following calculations, cf. (6-73)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k2 =

⎡⎢⎣−0.5

2.0−3.0

⎤⎥⎦ , r12 =

⎡⎢⎣ 0.9733−0.1622

0.1622

⎤⎥⎦

T ⎡⎢⎣−0.52.0

−3.0

⎤⎥⎦ = −1.2978

r22 =

∣∣∣∣∣∣∣⎡⎢⎣−0.5

2.0−3.0

⎤⎥⎦+ 1.2978 ·

⎡⎢⎣ 0.9733−0.1622

0.1622

⎤⎥⎦∣∣∣∣∣∣∣ = 3.4009

q2 =1

3.4009

⎛⎜⎝⎡⎢⎣−0.5

2.0−3.0

⎤⎥⎦+ 1.2978 ·

⎡⎢⎣ 0.9733−0.1622

0.1622

⎤⎥⎦⎞⎟⎠ =

⎡⎢⎣ 0.2244

0.5262−0.8202

⎤⎥⎦

(3)

q3 and r13, r23, r33 are determined from the following calculations, cf. (6-74)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k3 =

⎡⎢⎣ 0.5−3.0

8.0

⎤⎥⎦ , r13 = qT

1 k3 = 2.2711 , r23 = qT2 k3 = −8.0282

r33 =∣∣∣k3 − 2.2711q1 + 8.0282q2

∣∣∣ = 1.9080

q3 =1

1.9080

(k3 − 2.2711q1 + 8.0282q2

)=

⎡⎢⎣0.0477

0.83480.5486

⎤⎥⎦

(4)



⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q1 =

⎡⎢⎣ 0.9733 0.2244 0.0477−0.1622 0.5662 0.8348

0.1622 −0.8202 0.5486

⎤⎥⎦

R1 =

⎡⎢⎣3.0822 −1.2978 2.2711

0 3.4009 −8.02820 0 1.9080

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ2 = Φ1Q1 =

⎡⎢⎣ 0.6882 0.1587 0.0337−0.2294 0.9521 0.2024

0.2294 −1.1600 0.7758

⎤⎥⎦

K2 = R1Q1 =

⎡⎢⎣ 3.5789 −1.8540 0.3095−1.8540 8.3744 −1.5650

0.3095 −1.5650 1.0466

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(5)

The corresponding matrices after the 2nd and 3rd iteration become

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q2 =

⎡⎢⎣ 0.8853 0.4648 0.0115−0.4586 0.8689 0.1861

0.0766 −0.1700 0.9825

⎤⎥⎦

R2 =

⎡⎢⎣4.0425 −5.6020 1.0719

0 6.6809 −1.39400 0 0.7405

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ3 = Φ2Q2 =

⎡⎢⎣ 0.5391 0.4521 0.0706−0.6243 0.6862 0.3734

0.7945 −1.0332 0.5489

⎤⎥⎦

K3 = R2Q2 =

⎡⎢⎣ 6.2303 −3.1708 0.0567−3.1708 6.0422 −0.1259

0.0567 −0.1259 0.7275

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q3 =

⎡⎢⎣ 0.8912 0.4536 0.0021−0.4536 0.8610 0.0219

0.0081 −0.0205 0.9998

⎤⎥⎦

R3 =

⎡⎢⎣6.9910 −5.5673 0.1135

0 3.9475 −0.10140 0 0.7274

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ4 = Φ3Q3 =

⎡⎢⎣ 0.2760 0.6459 0.0816−0.8645 0.3206 0.3871

1.1811 −0.5714 0.5277

⎤⎥⎦

K4 = R3Q3 =

⎡⎢⎣ 8.5763 −1.7913 0.0059−1.7913 3.5192 −0.0148

0.0059 −0.0148 0.7245

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7)

As seen from R3 and K4 the terms in the main diagonal have already after the 3rd iteration groupedin descending magnitude, corresponding to the ordering of the eigenvalues at convergence indicated inBox 6.7. Moreover, for both matrices convergence to the lowest eigenvalue λ1 = 0.7245 has occurred,illustrating the fact that the QR algorithm converge faster to the lowest eigenmode than to the highest.

The matrices after the 14th iteration become

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Q14 =

⎡⎢⎣ 1.0000 0.0000 0.0000−0.0000 1.0000 0.0000

0.0000 −0.0000 1.0000

⎤⎥⎦

R14 =

⎡⎢⎣9.3104 −0.0000 0.0000

0 2.9652 −0.00000 0 0.7245

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

⇒

Φ15 = Φ14Q14 =

⎡⎢⎣ 0.0729 0.6981 0.0853−0.9202 0.0486 0.3884

1.2978 −0.1997 0.5251

⎤⎥⎦

K15 = R14Q14 =

⎡⎢⎣ 9.3104 −0.0000 0.0000−0.0000 2.9652 −0.0000

0.0000 −0.0000 0.7245

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(8)


Presuming that convergence has occurred after the 14th iteration the following solutions are obtained forthe eigenvalues and eigenmodes of the original general eigenvalue problem

Λ =

⎡⎢⎣λ3 0 0

0 λ2 00 0 λ1

⎤⎥⎦ = K15 =

⎡⎢⎣ 9.3104 −0.0000 0.0000−0.0000 2.9652 −0.0000

0.0000 −0.0000 0.7245

⎤⎥⎦

Φ =[Φ(3) Φ(2) Φ(1)

]= Φ15 =

⎡⎢⎣ 0.0729 0.6981 0.0853−0.9202 0.0486 0.3884

1.2978 −0.1997 0.5251

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(9)

The solution (9) agrees with the corresponding solutions for the special and general Jacobi iterationalgorithms obtained in Exercise 6.3, eq. (8) and (14), respectively.


1. Calculate the two lowest eigenmodes and corresponding eigenvalues by simultaneous inverse vec-tor iteration with the start vector basis

Φ0 =[Φ(1)

0 Φ(2)0

]=

⎡⎢⎣1 11 01 −1

⎤⎥⎦

SOLUTIONS:

Question 1:


A = K−1M =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦ =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦ (1)


Φ1 =[Φ(1)

1 Φ(2)1

]= AΦ0 =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦⎡⎢⎣1 11 01 −1

⎤⎥⎦ =

⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦ (2)


At the determination of Φ(1)1 and r11 in the 1st vector iteration the following calculations are performed,

cf. (7-13)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(1)1 =

⎡⎢⎣0.550

1.3001.650

⎤⎥⎦ , r11 =

∥∥∥Φ(1)1

∥∥∥ =

⎛⎜⎜⎝⎡⎢⎣0.550

1.3001.650

⎤⎥⎦

T ⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦⎡⎢⎣0.550

1.3001.650

⎤⎥⎦⎞⎟⎟⎠

12

= 3.3162

Φ(1)1 =

13.3162

⎡⎢⎣0.5501.3001.650

⎤⎥⎦ =

⎡⎢⎣0.1659

0.39200.4976

⎤⎥⎦

(3)

Φ(2)1 and r12, r22 are determined from the following calculations, cf. (7-15)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(2)1 =

⎡⎢⎣ 0.275−0.350−0.675

⎤⎥⎦ , r12 =

⎡⎢⎣0.1659

0.39200.4976

⎤⎥⎦

T ⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦⎡⎢⎣ 0.275−0.350−0.675

⎤⎥⎦ = −0.9578

r22 =

∥∥∥∥∥∥∥⎡⎢⎣ 0.275−0.350−0.675

⎤⎥⎦+ 0.9578 ·

⎡⎢⎣0.16590.39200.4976

⎤⎥⎦∥∥∥∥∥∥∥ = 0.6380

Φ(2)1 =

10.6380

⎛⎜⎝⎡⎢⎣ 0.275−0.350−0.675

⎤⎥⎦+ 0.9578 ·

⎡⎢⎣0.16590.39200.4976

⎤⎥⎦⎞⎟⎠ =

⎡⎢⎣ 0.6800

0.0399−0.3111

⎤⎥⎦

(4)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R1 =

[3.3162 −0.9578

0 0.6380

]

Φ1 =

⎡⎢⎣0.1659 0.6800

0.3920 0.03990.4976 −0.3111

⎤⎥⎦

(5)

The reader should verify that Φ1R1 = Φ1. The corresponding matrices after the 2nd and 3rd iterationbecome

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R2 =

[1.3716 −0.1507

0 0.3392

]

Φ2 =

⎡⎢⎣0.1053 0.6944

0.3897 0.04920.5191 −0.2311

⎤⎥⎦

(6)


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R3 =

[1.3798 −0.0371

0 0.3374

]

Φ3 =

⎡⎢⎣0.0902 0.6972

0.3888 0.04960.5237 −0.2086

⎤⎥⎦

(7)

Convergence of the eigenmodes with the indicated number of digits were achieved after 9 iterations,where

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

R14 =

[1.3803 −0.0000

0 0.3372

]

Φ9 =

⎡⎢⎣0.0853 0.6981

0.3884 0.04860.5251 −0.1997

⎤⎥⎦

(8)

Presuming that convergence has occurred after the 9th iteration the following eigenvalues are obtainedfrom (10-10) and (10-12)

Λ =

[λ1 00 λ2

]= ΦT

9 KΦ9 = R−1∞ =

[0.7245 0.00000.0000 2.9652

]

Φ =[Φ(1) Φ(2)

]= Φ9 =

⎡⎢⎣0.0853 0.69810.3884 0.04860.5251 −0.1997

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(9)

The solution (9) agrees with the corresponding solutions obtained in Exercises 6.2 and 6.6.


1. Calculate the two lowest eigenmodes and corresponding eigenvalues by subspace iteration with thestart vector basis

Φ0 =[Φ(1)

0 Φ(2)0

]=

⎡⎢⎣1 11 01 −1

⎤⎥⎦

SOLUTION:


Question 1:


A = K−1M =

⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦−1 ⎡⎢⎣2 0 0

0 2 10 1 1

⎤⎥⎦ =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦ (1)


Φ1 =[Φ(1)

1 Φ(2)1

]= AΦ0 =

⎡⎢⎣0.350 0.125 0.075

0.100 0.750 0.4500.050 0.875 0.725

⎤⎥⎦⎡⎢⎣1 11 01 −1

⎤⎥⎦ =

⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦ (2)

In order to perform the Rayleigh-Ritz analysis in the 1st subspace iteration the following projected massand stiffness matrices are calculated based onΦ1 , cf. (4-59), (4-60), (7-31)

M1 = ΦT1 MΦ1 =

⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦

T ⎡⎢⎣2 0 00 2 10 1 1

⎤⎥⎦⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦ =

[10.998 −3.1763

−3.1763 1.3244

]

K1 = ΦT1 KΦ1 =

⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦

T ⎡⎢⎣ 6 −1 0−1 4 −1

0 −1 2

⎤⎥⎦⎡⎢⎣0.550 0.2751.300 −0.3501.650 −0.675

⎤⎥⎦ =

[8.300 −1.850

−1.850 1.575

]

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(3)

The corresponding eigenvalue problem (7-30) becomes

K1Q1 = M1Q1R1 ⇒[8.300 −1.850

−1.850 1.575

] [q(1)

1 q(2)1

]=

[10.998 −3.1763

−3.1763 1.3244

] [q(1)

1 q(2)1

] [ρ1,1 00 ρ2,1

]⇒

R1 =

[0.7246 0

0 2.9752

], Q1 =

[−0.2471 −0.4845

0.1813 −1.5569

](4)

The estimate of the lowest eigenvectors after the 1st iteration becomes, cf. (7-34)

Φ1 = Φ1Q1 =

⎡⎢⎣0.550 0.275

1.300 −0.3501.650 −0.675

⎤⎥⎦[−0.2471 −0.4845

0.1813 −1.5569

]=

⎡⎢⎣−0.0861 −0.6947

−0.3848 −0.0850

−0.5302 0.2514

⎤⎥⎦ (5)


Correspondingly, after the 2nd and 9th iteration steps the following matrices are calculated

R2 =

[0.7245 0

0 2.9662

], Q2 =

[−0.7245 −0.0013

0.0004 −2.9673

]

Φ2 =

⎡⎢⎣0.0854 0.6972

0.3881 0.06030.5255 −0.2162

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(6)

R9 =

[0.7245 0

0 2.9652

], Q9 =

[−0.7245 −0.0000

0.0000 −2.9652

]

Φ9 =

⎡⎢⎣−0.0853 −0.6981−0.3884 −0.0486−0.5251 0.1997

⎤⎥⎦

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

(7)

The subspace iteration process converged with the indicated accuracy after 8 iterations.

Finally, it should be checked that th calculated eigenvalues are inded the lowest two by a Sturm sequenceor Gauss factorization check. The 2nd calculated eigenvalue becomes ρ2,9 = 2.9652. Then, let µ = 3.1and perform a Gauss factorization of the matrix K − 3.1M, i.e.

K − 3.1M =

⎡⎢⎣−0.2 −1.0 0−1.0 −2.2 −4.1

0 −4.1 −1.1

⎤⎥⎦ =

LDLT =

⎡⎢⎣1 0 0

5 1 00 −1.4643 1

⎤⎥⎦⎡⎢⎣−0.2 0 0

0 2.8 00 0 −7.1036

⎤⎥⎦⎡⎢⎣1 5 00 1 −1.46430 0 1

⎤⎥⎦ (8)

It follows that two components in the main diagonal of D are negative, from which is concluded that twoeigenvalues are smaller than µ = 3.1. In turn this means that the two eigensolutions obtained by (8) areindeed the lowest two eigensolutions of the original system.

The solution (8) agrees with the corresponding solutions obtained in Exercises 6.3, 6.6 and 7.1.


1. Calculate the 3rd eigenmode and eigenvalue by Sturm sequence iteration (telescope method).


SOLUTION:

Question 1:

At first a calculation with µ = 2.5 is performed, which produces the following results

K − 2.5M =

⎡⎢⎣ 1.0 −1.0 0.0−1.0 −1.0 −3.5

0.0 −3.5 −0.5

⎤⎥⎦ ⇒

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

P (3)(2.5) = 1 , sign(P (3)(2.5)) = +

P (2)(2.5) = 1.0 , sign(P (2)(2.5)) = +

P (1)(2.5) = 1.0 · (−1.0) − (−1.0)2 = −2.0 , sign(P (1)(2.5)) = −P (0)(2.5) = −0.5 · (−2.0) − (−3.5)2 · (1.0) = −11.25 , sign(P (0)(2.5)) = −

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(1)

Hence, the sign sequence of the Sturm sequence becomes + +−−. corresponding to the number of signchanges nsign = 1 in the sequence. One eigenvalue is smaller than µ = 2.5.

Similar calculations are performed for µ = 3.5, 4.5, . . . , 9.5

µ = 3.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 4.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 5.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 6.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 7.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 8.5 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.5 : Sign sequence = + − + − ⇒ nsign = 3

(2)

From this is concluded that the 3rd eigenvalue is placed somewhere in the interval 8.5 < λ3 < 9.5.

Next, similar calculations are performed for µ = 8.6, 8.7, . . . , 9.4

µ = 8.6 : Sign sequence = + − + + ⇒ nsign = 2µ = 8.7 : Sign sequence = + − + + ⇒ nsign = 2µ = 8.8 : Sign sequence = + − + + ⇒ nsign = 2µ = 8.9 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.0 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.1 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.2 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.3 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.4 : Sign sequence = + − + − ⇒ nsign = 3

(3)

From this is concluded that the 3rd eigenvalue is confined to the interval 9.3 < λ3 < 9.4.


Next, calculations are performed for µ = 9.31, 9.32, . . . , 9.39

µ = 9.31 : Sign sequence = + − + + ⇒ nsign = 2µ = 9.32 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.33 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.34 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.35 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.36 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.37 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.38 : Sign sequence = + − + − ⇒ nsign = 3µ = 9.39 : Sign sequence = + − + − ⇒ nsign = 3

(4)

From this is concluded that the 3rd eigenvalue is confined to the interval 9.31 < λ3 < 9.32.

Proceeding in this manner, it may be shown after totally 52 Sturm sequence calculations that the 3rdeigenvalue is confined to the interval 9.31036 < λ3 < 9.31037. Each extra digit requires 9 calculations.

Setting λ3 9.310365, the linear equation (7-63) attains the form

(K − 9.310365M

)Φ(3) = 0 ⇒

⎡⎢⎣−12.6207 −1 0

−1 −14.6207 −10.31040 −10.3104 −7.3104

⎤⎥⎦⎡⎢⎣Φ(3)

1

Φ(3)2

Φ(3)3

⎤⎥⎦ =

⎡⎢⎣0

00

⎤⎥⎦ (5)

Setting Φ(3)1 = 1 the algorithm (7-64) now provides

Φ(3)2 = −(−12.6207)

(−1)· 1 = −12.6207

Φ(3)3 = − (−1)

(−10.3104)· 1 − (−14.6207)

(−10.3104)· (−12.6207) = 1

⎫⎪⎪⎪⎬⎪⎪⎪⎭ ⇒ Φ(3) =

⎡⎢⎣ 1−12.6207

17.7800

⎤⎥⎦ (6)

Normalization to unit modal mass provides

Φ(3) =

⎡⎢⎣ 0.0729−0.9202

1.2978

⎤⎥⎦ (7)

The eigenvalue λ3 9.310365 and the corresponding eigenmode Φ(3) as given by (7) agree with thecorresponding results obtained in Excercises 6.3 and 6.6.

STRUCTURAL DYNAMICS, VOL. 9 Computational Dynamics · 1.1 Fundamentals of Linear Structural...

Documents

Transcript of STRUCTURAL DYNAMICS, VOL. 9 Computational Dynamics · 1.1 Fundamentals of Linear Structural...