stochastic processes in mechanical engineeringspeci ed by probability distributions. the theory of...

Stochastic Processesin Mechanical Engineering

J.J.H. Brouwers

Eindhoven University of Technology

[email protected]/woc/ptc

2

Copyright c©2006 by J.J.H.BrouwersAll rights reserved.No part of the material protected by this copyright notice may be reproducedin any form whatsoever or by any means, electronic or mechanical, includedphotocopying, recording or by any information storage and retrieval system,without permission from the author.

A catalogue record is available from the Library Eindhoven University of Tech-nology:ISBN-10: 90-386-2938-9ISBN-13: 978-90-386-2938-4

Preface

Stochastic or random vibrations occur in a variety of applications of mechani-cal engineering. Examples are: the dynamics of a vehicle on an irregular roadsurface; the variation in time of thermodynamic variables in municipal wasteincinerators due to fluctuations in heating value of the waste; the vibrationsof an airplane flying through turbulence; the fluctuating wind loads acting oncivil structures; the response of off-shore structures to random wave loading(figure 1).

Suppose we know the value of a random variable x at a certain time t0.For future times t > t0 the value of x will become more and more uncertain.Its value becomes less and less correlated to the starting value. It can only bespecified by probability distributions. The theory of stochastic processes aimsto describe this behavior. Although x is stochastic, by describing it in termsof probability distributions and other statistical characteristics (correlationfunctions, peak distributions, etc.) calculation procedures and codes can beestablished by which decisions on design issues can be made.

Attention will be focussed on problems of external noise. That is, weshall consider models of mechanical engineering structures where the sourceof random behavior comes from outside: e.g. a prescribed random force ora prescribed random displacement. The statistical behavior of the source issupposed to be known: for example, a Gaussian random process with knownpower density spectrum. No attention will be given to questions on the originof the noise: how deterministic laws of physics can produce random fluctu-ations. A classical example is the chaotic motion of colliding molecules inotherwise empty space. Another example, closer to the applications consid-ered here: how winds blowing over the sea surface can lead to the occurrenceof random waves. Answering these questions is not an easy task; they requirestudies of their own and we do not address them here. We focus attentionto the dynamical behavior of mechanical engineering structures subject to ex-ternal noise of specified form. The structures inhibit inertia, damping andrestoring, linear and non-linear. The main questions we intend to answer are:

4 Preface

how do these structures respond to random excitation, and how can we quan-tify the random behavior of response variables in a manner that an engineeris able to make rational design decisions.

In chapter 1, some basic notions on probability theory are recapitulated,followed in chapter 2 by an introduction to the description of a variable whichvaries randomly in time. In chapter 3, methods of Fourier transform are ex-tended to problems of random vibration. These methods are used in chapter 4,to analyze a linear spring-mass system subject to Gaussian random excitationin the frequency-domain. The description of a random signal in the time-domain is given in chapter 5; it forms the starting point for analysis in thetime-domain of the spring-mass system; this is presented in chapter 6.

The tensioned beam subject to random excitation provides an excursion topartial differential equations with random right-hand sides. Formulation andanalytical methods of solution are presented in chapters 7-9. The statistics ofpeak values of randomly and Gaussianly varying signals, and the statistics ofthe extreme value in a signal of a certain duration, are the subject of analysisin chapter 10.

The analysis of non-linear systems is presented in chapters 11-14. In chap-ter 11 we summarize numerical time-domain simulation techniques appliedto non-linear systems. Quasi-static non-linear response to random excitationis treated in chapter 12, non-linearly damped resonance in chapter 13. Sta-bility analysis of systems with parametric random excitation is presented inchapter 14.

Literature: a number of topics presented in subsequent chapters can befound in the introductory textbooks of Robson [1] and Crandall & Mark [2].More detailed treatises of stochastic theory are presented in the books ofStratonovich [3] and Van Kampen [4]. Other references are given where rele-vant.

The author wishes to express his thanks to Ir. R.J.E. Walpot for histremendous support in the completion of this manuscript. Dr. J.G.M. Kuertenis acknowledged for critical remarks.

5

Figure 1. Illustration of an off-shore structure; by courtesy of Royal Dutch ShellExploration & Production.

6 Preface

Contents

Preface 3

1 Some basic properties of probability 111.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Expectation or mean value . . . . . . . . . . . . . . . . . . . . 131.3 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Gaussian distribution . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Higher order moments . . . . . . . . . . . . . . . . . . . . . . . 17

2 Random Vibrations: Introductory remarks 192.1 Random events . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Ensemble and time averaging . . . . . . . . . . . . . . . . . . . 202.3 Correlations in time . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Fourier transform in case of random vibration 273.1 Fourier integrals and transforms . . . . . . . . . . . . . . . . . 273.2 Spectral density . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 The Fourier transform of the autocorrelation function . . . . . 30

4 Single degree of freedom system: analysis in the frequencydomain 334.1 Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Types of system response . . . . . . . . . . . . . . . . . . . . . 354.3 Quasi-static response (Ω∗ 1) . . . . . . . . . . . . . . . . . . 364.4 Dynamical response (Ω∗ 1) . . . . . . . . . . . . . . . . . . . 374.5 Resonant response (Ω∗ = O(1), δ 1) . . . . . . . . . . . . . . 38

5 The description of a stationary Gaussian signal in the time-domain 43

8 Contents

5.1 Random signal described by sinusoidal waves with random phaseangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.2 Gaussianity of the time-series . . . . . . . . . . . . . . . . . . . 455.3 Power density spectrum of the time-series . . . . . . . . . . . . 48

6 Single degree of freedom system: analysis in the time-domain 516.1 Time-domain description of the excitation . . . . . . . . . . . . 516.2 Solution in the time-domain . . . . . . . . . . . . . . . . . . . . 556.3 Resonant response . . . . . . . . . . . . . . . . . . . . . . . . . 566.4 Transient response . . . . . . . . . . . . . . . . . . . . . . . . . 61

7 Multi degree of freedom system: the tensioned beam 677.1 Basis equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 687.2 Tensioned string versus beam . . . . . . . . . . . . . . . . . . . 71

8 Random Vibrations of a tensioned string 758.1 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . 758.2 Expansion in eigenfunctions . . . . . . . . . . . . . . . . . . . . 77

9 Random vibrations of a beam 799.1 Quasi-static response . . . . . . . . . . . . . . . . . . . . . . . . 799.2 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . 819.3 Eigenfunction expansion . . . . . . . . . . . . . . . . . . . . . . 829.4 The general problem of the tensioned beam . . . . . . . . . . . 83

10 Peak and extreme statistics 8710.1 Peak statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8710.2 Fatigue damage . . . . . . . . . . . . . . . . . . . . . . . . . . . 9210.3 Extreme statistics . . . . . . . . . . . . . . . . . . . . . . . . . 94

11 Non-linear analysis: Some general observations and numericaltime-domain simulation 97

12 Non-linear quasi-static response 99

13 Non-linearly damped resonance 10313.1 Two-scale expansions . . . . . . . . . . . . . . . . . . . . . . . . 10313.2 Derivation of the Fokker-Planck equation . . . . . . . . . . . . 10713.3 Solutions for stationary response . . . . . . . . . . . . . . . . . 11013.4 Transient response behavior . . . . . . . . . . . . . . . . . . . . 114

9

14 Stability in case of randomly fluctuating restoring 12114.1 Basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 12514.2 Fokker-Planck equation . . . . . . . . . . . . . . . . . . . . . . 12614.3 Time-domain behavior . . . . . . . . . . . . . . . . . . . . . . . 13114.4 Solutions for transient probability density and statistical moments13414.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14314.6 Appendix: Derivation of asymptotic expressions for probability

density and statistical moments . . . . . . . . . . . . . . . . . . 146

Abstract 151

Bibliography 153

Index 155

10 Contents

Chapter 1

Some basic properties ofprobability

To describe random processes, probabilistic concepts are used. In this chapter,some basic notions of probability are explained.

1.1 Probability

We define the concept of probability here as the probability Pr of the occur-rence of a certain event n. This probability is equal to Pr[n] = 0 in case of anevent which can not possibly occur and Pr[n] = 1 if the event is certain to oc-cur. If we consider for example the case of a six-sided die, we can expect thatthe result N of any given throw is equally likely to be any number between 1and 6. We can therefore write: Pr[N = n] = p(n) = 1/6 for where n can beany number between 1 and 6.

Probability theory is the mathematical study of probability and plays animportant role in many aspects of life. As an example, imagine that a certaininvestment costs 1.000.000 EUR. The chance of earning 10.000.000 EUR withthe investment is 50%, but it’s equally probable that the investment resultsin no profit at all. If faced with the choice whether to invest or not, anindividual with a total fortune of just over 1.000.000 EUR won’t choose to goahead with the investment. Because when the gamble has a bad outcome forhim, he will lose all his money and can’t play the game a second time. A bigmultinational, on the other hand, has the capacity of playing this game manytimes and the theory of probability ensures the company that when playingthe game many times, ’on average’ the investment will result in a large profit.It illustrates the power of capital. Another example of how to deal with

12 Some basic properties of probability

uncertainty, in a rational manner, is insurance. The risk an individual isexposed to is covered by mutual insurance via an insurance company. In thisway, the risk is spread over many people and over a long period of time. Thepremium every individual will have to pay to the company will correspond tothe statistical expectation (expected or mean value) of the financial risk he isexposed to (supplemented with the profit of the insurance company!).

The above examples illustrate that uncertainty becomes a certainty oncethe game is played many times. It is this principle which makes probabil-ity theory a useful tool in practice. Scientific concepts based on probabilisticmodels are verifiable (as should be!) because we can repeat measurementsand check the outcome. Equally, designers can make rational decisions in en-gineering practice, for example by meeting the standards required by insurancecompanies (such as Lloyd’s insuring ships and offshore structures).

The quantities on which we will focus in the remainder of this text have acontinuous range of possible values as opposed to the discrete value which therandom variable N can have in the case of a die. It is convenient to define aquantity called the cumulative distribution function (CDF) as the probabilitythat a random variable has a value X less than a certain specified value x:

P (x) = Pr[X ≤ x] =∫ x

−∞p(x)dx. (1.1)

The previously defined quantity p(x), the probability density function (PDF),on its turn can be defined as:

p(x) =dP (x)dx

. (1.2)

Figure 1.1 shows the typical shapes of both quantities in the special case ofa zero-mean Gaussian process (which will be discussed in section 1.4). Theshapes of these graphs give a nice qualitative indication of the nature of therandom variable. For example, a variable whose values are closely clusteredaround a mean value would give a tall narrow p(x) curve and a P (x) curvesteeply rising near the mean value. The sum of all probabilities, and thereforethe area under the p(x) curve is equal to one:

∫∞−∞ p(x)dx = 1.

Suppose that we measured the room temperature x of a classroom infinitelymany times and that the CDF on the lower side of figure 1.1 characterizes thetemperature distribution P (x). The probability that the measured tempera-ture is within the range x1 < x < x2 is equal to P (x2) − P (x1), as illustratedin the figure.

1.2 Expectation or mean value 13

1.2 Expectation or mean value

An important quantity in probability theory is the expected value E[x] or meanvalue µ of the variable x. It is also called the first moment of the distributionp(x) and is given by

µ = 〈x〉 =∫ ∞

−∞xp(x)dx. (1.3)

The mean value operator 〈.〉 is a linear operator, which means that it satisfiesthe following conditions:

〈f + g〉 = 〈f〉 + 〈g〉〈αf〉 = α〈f〉 (1.4)

〈〈f〉g〉 = 〈f〉〈g〉

where x is random variable and α a constant.

1.3 Variance

To describe the nature of the spread of x completely would require the con-struction of either the P (x) or p(x) functions. A measure for the spread ofthe signal around its mean value is given by the variance of x and is com-monly denoted by σ2. The variance is second moment about the mean of thedistribution p(x) and is defined by:

σ2 =⟨[x− µ]2

⟩. (1.5)

In other words, the variance is the average of the square of the distance ofeach data point from the mean. In terms of the probability density function,the variance can be written as:

σ2 =∫ ∞

−∞[x− µ]2 p(x)dx. (1.6)

In the remainder of this text, all random processes will be considered to bezero-mean processes if not specified otherwise. In this case, the above equationobviously simplifies to σ2 =

∫∞−∞ x2p(x)dx: the variance is then equal to the

mean square value (MSV). The square root σ of the variance is known as theroot-mean-square value (RMS) and is also called the standard deviation of x.It is a direct measure for the amount of variation around the mean.


1.4 Gaussian distribution

Many real-life random processes can often be considered Gaussian meaningthat their probability density function, see figure 1.1, can be described by thefollowing relation:

p(x) =1

σ√

2πe−(x−µ)2/2σ2

, (1.7)

with µ the mean value. A Gaussian distribution is also known as a normaldistribution. Equation (1.7) shows that for Gaussian processes, the wholedistribution is defined by its MSV and mean value. For the CDF we canwrite:

P (x) =1

σ√

2π

∫ x

−∞e−(x−µ)2/2σ2

dx. (1.8)

With the definition of the error function as:

erf(ξ) =2√π

∫ ξ

0e−t

2dt, (1.9)

this can be written as:

P (x) =12

(1 + erf

(x− µ

σ√

2

)). (1.10)

The Gaussian distribution has several important properties:

• The probability density function is symmetric about its mean value;

• 68.27% of the area under the curve is within one standard deviation ofthe mean;

• 95.45% of the area is within two standard deviations;

• 99.73% of the area is within three standard deviations.

A distribution of two zero-mean variables x and y is said to be Gaussianif the joint probability density function (see figure 1.2) is given by:

p(x, y) =1

2π√σ2xσ

2y − σ4

xy

e−σ2

yx2−2σ2xyxy+σ2

xy2

2(σ2xσ2

y−σ4xy) , (1.11)

where σ2x and σ2

y are the variances of x and y respectively and σ2xy = 〈xy〉 is

called the covariance.

1.4 Gaussian distribution 15

Figure 1.1. The PDF (upper graph) and the CDF (lower graph) of a zero-meanGaussian process


p(x,y)

x y

Figure 1.2. Example of a joint Gaussian PDF of two random variables x and y.

1.5 Higher order moments 17

1.5 Higher order moments

In sections 1.2 and 1.3, we defined the first and second moment about themean value. In general, the n-th moment about the mean, or n-th centralmoment is defined as follows:

µn = 〈(x− µ)n〉. (1.12)

The third central moment normalized with the second central moment,µ3/µ

3/22 , is known as skewness. Skewness is a measure of the degree of asym-

metry of a distribution. If the left tail (tail at small end of the distribution)is more pronounced than the right tail (tail at the large end of the distribu-tion), the function is skewed to the left and is said to have negative skewness.If the reverse is true, it has positive skewness. If the two are equal, it haszero skewness as is the case for a Gaussian distribution. Examples of skeweddistributions are shown in figure 1.3.

The fourth central moment normalized with the second central moment,µ4/µ

22, is known as the kurtosis or flatness of a distribution. It is a measure

of the peakedness of a distribution and is equal to 3 for a Gaussian distri-bution. A kurtosis larger than 3 indicates a ”peaked” distribution and whenµ4/µ

22 < 3, the distribution is described as ”flat”. Examples of peaked and

flat distributions are shown in figure 1.4.Instead of moments, one can express statistical properties in terms of cu-

mulants. The n-th order cumulant , xn or κn, is defined as the momentof n-th order minus all sub-order-moments:

κ1 = 〈x〉κ2 = 〈x2〉 − 〈x〉〈x〉 = µ2 = σ2

κ3 = 〈x3〉 − 3〈x〉〈x2〉 + 2〈x〉3 = µ3

κ4 = 〈x4〉 − 4〈x〉〈x3〉 − 3〈x2〉2 + 12〈x〉2〈x2〉 − 6〈x〉4= µ4 − 3µ2

2 (1.13)

For a Gaussian distribution, all cumulants of order > 2 are equal to zero. Adistribution is almost Gaussian when the cumulants of order n (with n > 2),normalized with the standard deviation to the power n, are small comparedto unity: κn/σn 1 for n ≥ 3.


−3 −2 −1 0 1 2 30

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

x / σ

σ p(

x)

skewness = 0 (Gaussian)skewness = 0.45skewness = −0.45

Figure 1.3. Examples of skewed distributions.

−3 −2 −1 0 1 2 30

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

x / σ

σ p(

x)

kurtosis = 3 (Gaussian)kurtosis = 2.4kurtosis = 3.8

Figure 1.4. Examples of peaked and flat distributions.

Chapter 2

Random Vibrations:Introductory remarks

In the previous chapter we were concerned with the statistical descriptionof some variable. Variation of this variable with time was not taken intoconsideration. This will be subject of study in this chapter.

2.1 Random events

Let us consider a quantity x(t) randomly varying with an independent vari-able t. This independent variable will be considered to represent time. Thequantity x may be any physical quantity such as velocity, pressure or posi-tion. If we carefully design an experimental setup, it is possible to record arandomly varying signal (for example: the velocity of a turbulent flow) overa time interval T . Figure 2.1 shows a typical result for such an experiment.Although it is possible to plot the measured signal x(t), its random behav-ior makes it impossible to predict its value at a certain time t1 outside themeasured interval T . In order to develop a method to statistically describesuch random signals, it is helpful to first consider some elementary statisticaltheories.

Let x(t) represent the force acting on the wheel of a car during a trip.The forces working on the same wheel during different trips under identicalconditions are denoted by x1(t), x2(t), x3(t), ..., etc. The ensemble of all theserandom signals is said to constitute a random process x(t). If the conditionsof all the trips are identical, and if the conditions remain steady during eachtrip, we can in general assume that the process is both stationary and ergodic.

By stationary we mean that the probability distribution of the quantities

20 Random Vibrations: Introductory remarks

x(t)

T

< x(t) >

0t

Figure 2.1. Example of a random signal x(t) as measured over a time interval T .

x1(t), x2(t), x3(t), ...,etc. at any instant of time t1 is independent on the choiceof t1. In other words: all statistical averages are constant in time.

By ergodic we mean that the probability distribution of the process x(t)is equal to the distributions of all the member functions that constitute thewhole process.

So if we restrict our attention to stationary and ergodic random processes,by determining the statistical properties of a single member function x(t), wecan define the statistical properties of the whole random process x(t) ofwhich x(t) is a member.

2.2 Ensemble and time averaging

The mean, or average, value of a random signal has been discussed before.A distinction can be made between the ensemble average 〈x(t)〉 and the timeaverage value x(t) of a random signal. Imagine that we can perfectly isolate arandom process in a lab experiment. The process is turned on at time t = 0.We then wait until a certain time t has passed and at this time we measurequantity x. This is repeated N times, see figure 2.2. The ensemble averageor statistical moment is now calculated by taking the sum of all recordedvalues divided by N . We can thus write down the following definition for the

2.2 Ensemble and time averaging 21

ensemble average:

〈x(t)〉 = limN→∞

∑Ni=1 xi(t)N

. (2.1)

The time average is calculated by averaging the value of the signal over a longperiod of time:

x(t) = limT→∞

1T

∫ T

0xi(t)dt. (2.2)

In the special case of a stationary, ergodic process, time averages and ensembleaverages are equal 〈x(t)〉 = x(t).The above can be repeated for the k-th moment:

〈xk(t)〉 = limN→∞

N∑i=1

xki (t)N

, (2.3)

xk(t) = limT→∞

1T

∫ T

0xki (t)dt. (2.4)

For stationary ergodic processes:

〈xk(t)〉 = xk(t). (2.5)

The probability P (X) is defined as the number of realizations for which attime t the value of x(t) is smaller than a prescribed value X divided by thetotal number of realizations N and letting N → ∞. So, at some time t, onehas simply to count the number of realizations for which x(t) < X and dividethis number by N ; see figure 2.2. This can be repeated for any other value ofX and then results in a complete distribution of P (X) at time t. If the processis non-stationary, the distribution of P (X) will vary with time t! However, ifthe process is stationary, P (X) will be the same at any time. Furthermore, ifthe process is also ergodic, one can construct the probability distribution of asingle time record. In this case, P (X) is the amount of time that X < x(t)divided by the total time T and letting T → ∞: see figure 2.3. In general, forstationary ergodic process the statistics of x(t) can be derived from a singletime record of the process.


realization 1

0 t

realization 2

realization 3

realization N

Figure 2.2. Illustration of the principle of ensemble averaging over N realizations.

2.2 Ensemble and time averaging 23

x(t)

T

x

t

0 t2 t3 t4 t5t6 t7 t8 t9 t10t1

Figure 2.3. Determination of probability distribution from a stationary time recordaccording to P (x) = Pr(X < x) = 1

T

∑i ∆ti.


2.3 Correlations in time

The extent to which two random variables x(t) and y(t) are correlated can bequantitatively expressed in the magnitude of the covariance σ2

xy = 〈x(t)y(t)〉,which we already encountered in section 1.4. In the case that x(t) and y(t)are completely independent of each other σ2

xy = 0. The same concept can alsobe used to assess the correlation of signals measured at different times withthe definition of the cross-correlation

Rxy(τ) = 〈x(t)y(t+ τ)〉. (2.6)

τ is called the time separation ∗. For a stationary process, the cross-correlationonly depends on τ because statistical properties of such a process do not changewith time. In the same way, the correlation of a variable x(t) at time t withitself at time t+ τ is defined as

Rxx(τ) = 〈x(t)x(t+ τ)〉 (2.7)

and is called the auto-correlation function. The calculation process is as fol-lows: we multiply the value of a (zero-mean) quantity x(t) at time t withthe value of the same quantity when a τ amount of time has passed, wedo this for all realizations and then take the ensemble average of all suchproducts. If we assume to be dealing with a stationary ergodic process,Rxx(τ) will be the same for all member functions of the process. In thatcase, 〈x(t)x(t + τ)〉 = x(t)x(t+ τ); the latter can be calculated from a singlerealization by time averaging according to equation (2.2). That is, we multiplyx(t) with its value x(t + τ) when a time τ has passed and treat the productas xi in equation (2.2). Definition (2.7) shows that the autocorrelation forzero time-separation is equal to the MSV of x(t), Rxx(0) = 〈x2(t)〉. The auto-correlation function, and correlation functions in general, is therefore oftennormalized as follows:

Rxx(τ) =〈x(t)x(t+ τ)〉

σ2xx

(2.8)

An example shape of the auto-correlation function is shown in figure 2.4. Forvery large time separation τ the value of Rxx(τ) must be zero as the followingconsideration will show. For very large time separation τ , the two values x(t)and x(t + τ) will be quite uncorrelated to each other, so that some products

∗Definition (2.6) applies to zero-mean random variables. In case of non-zero-mean x(t)and y(t), we have to subtract their mean values when calculating their correlations. Cor-relations are then indicated by double brackets and are defined as: 〈〈x(t)y(t + τ)〉〉 =〈[x(t)−〈x(t)〉][y(t+τ)−〈y(t+τ)〉]〉 = 〈x(t)y(t+τ)〉−〈x(t)〉〈y(t+τ)〉. The general definitionof correlation is thus similar to that of cumulant: cf. eq. (1.13)

2.3 Correlations in time 25

x(t)x(t+τ) will be positive and some negative, the values being symmetricallyscattered on each side of the value zero. The ensemble average of these valueswill therefore be zero, resulting in Rxx(∞)=0.

0

R( )

Time separation ( )

1/e

c

1

Figure 2.4. Example of a correlation function

The shape of the autocorrelation function supplies us with informationabout a random signal. The value of τ at which R(τ) effectively falls to zerogives an indication for the suddenness of the fluctuations in the signal. Thelower this value of τ , the more sudden the fluctuations in the signal, in otherwords: the memory of the system gets shorter. This idea can be quantifiedwith the use of the typical correlation time, τc, of the process. This can bedefined as the time at which an autocorrelation function normalized with theMSV drops to a value of 1/e as illustrated in the figure.

Another property of the autocorrelation function, for a stationary randomprocess, is symmetry around τ = 0:

Rxx(τ) = Rxx(−τ), (2.9)

which logically results in:∂Rxx(0)∂τ

= 0. (2.10)


Property (2.9) can easily be demonstrated as follows:

Rxx(τ) = 〈x(t)x(t+ τ)〉 = 〈x(t− τ)x(t)〉 = 〈x(t)x(t− τ)〉 = Rxx(−τ). (2.11)

where, in step 2, we used the property that statistical averages of a stationaryprocess do not change with a shift in time.

Chapter 3

Fourier transform in case ofrandom vibration

A powerful method for analyzing the time-dependent behavior of linear sys-tems is the Fourier transform. With some amendments, this method can alsobe applied to random vibrations.

3.1 Fourier integrals and transforms

Any periodic signal can be expressed as a series of harmonically varying quan-tities, called a Fourier series. Non-periodic functions, such as a transientloading, can only be expressed as a Fourier series if we consider it to be pe-riodic, with infinite period. This gives rise to the concept of the Fourierintegral . The following set of equations give the Fourier integral expressionfor the non-periodic function x(t):

x(t) =12π

∫ ∞

−∞A(ω)eiωtdω (3.1)

A(ω) =∫ ∞

−∞x(t)e−iωtdt (3.2)

The quantity A(ω) is called the Fourier transform of x(t). It is in generalcomplex and shows how x(t) can be considered to be distributed over thefrequency range. On its turn, x(t) is said to be the inverse transform of A(ω).The two quantities together form a Fourier transform pair.

28 Fourier transform in case of random vibration

3.2 Spectral density

The theories of Fourier series and integrals can not be applied directly torandom signals. This is because the periodic requirement for the Fourierseries is not met which rules out the Fourier series. Moreover, if we considera random signal to continue over infinite time, neither the real or imaginarypart of the Fourier transform converges to a steady value which is why it is notpossible to use the concept of Fourier integrals: see also the text subsequentto eq. (5.28). Instead, we will introduce a new quantity, the spectral density ,which has no convergence problems.

Consider a stationary ergodic random process x(t) which is assumed tohave started at t = −∞ and continue until t = ∞. Such a signal is notperiodic and it is thus impossible to define its Fourier transform. It is possible,however, to determine the Fourier transform AT (iω) of a signal xT (t) whichis equal to x(t) over the interval −T

2 < t < T2 and zero at all other times. In

line with Robson [1]:

1T

∫ T/2

−T/2x2T (t)dt =

1T

∫ ∞

−∞xT (t)xT (t)dt

=1T

∫ ∞

−∞xT (t)

[12π

∫ ∞

−∞AT (ω)eiωtdω

]dt

=1

2πT

∫ ∞

−∞AT (ω)

[∫ ∞

−∞xT (t)eiωtdt

]dω (3.3)

If we define the complex conjugate of A(ω) as A∗(ω) =∫∞−∞ x(t)eiωtdt, equa-

tion (3.3) can be written as:

1T

∫ T/2

−T/2x2T (t)dt =

12πT

∫ ∞

−∞AT (ω)A∗

T (ω)dω

=1

2πT

∫ ∞

−∞|AT (ω)|2 dω

=1πT

∫ ∞

0|AT (ω)|2 dω (3.4)

where use is made of the fact that |AT (ω)|2 is an even function of ω. If wenow let T → ∞, we obtain an expression for the MSV of the original signalx(t):

⟨x2(t)

⟩= x2(t) = lim

T→∞1T

∫ T/2

−T/2x2(t)dt =

∫ ∞

0limT→∞

(1πT

|AT (ω)|2)dω

(3.5)

3.2 Spectral density 29

The spectral density or power density of x(t) is now defined as:

S(ω) = limT→∞

(1πT

|AT (ω)|2)

(3.6)

so that

σ2 =⟨x2(t)

⟩=∫ ∞

0S(ω)dω. (3.7)

The power density indicates how the harmonic content of x(t) is spread overthe frequency domain. The amount of

⟨x2(t)

⟩associated with a narrow fre-

quency band ∆ω is equal to S(ω)∆ω. Different realizations of a stationaryergodic random process have a common S(ω). The power density S(ω) is thusa constant, non-random, statistical parameter of the random process x(t).This contrasts with the Fourier transform A(ω) of x(t) which varies randomlyitself: see also section 5.3. If a signal has a spectrum that is uniform (constant)over the whole frequency domain, the spectrum is said to be ’white’ and thesignal is referred to as white noise. Clearly, this is a theoretical abstraction;real power densities are never constant. But the abstraction can be usefulto handle certain problems such as lightly damped resonance: see, amongstothers, chapters 13 and 14. A more realistic power density spectrum is thatof the algebraic function:

S(ω) =2σ2Ω−1

π(1 + (ω/Ω)2). (3.8)

Here, Ω is a parameter which scales the frequency of the spectrum, while σ isstandard deviation:

∫ ∞

0S(ω)dω =

2π

∫ ∞

0

σ2dη

1 + η2=

2σ2

πarctan(η)

∣∣∣∞0

= σ2. (3.9)

Another, more peaked, spectrum is:

S(ω) =2σ2Ωω2

e−2Ω/ω. (3.10)

In this expression, Ω is a parameter which determines at which frequency thepower density has its largest value. By differentiation with respect to ω, onecan show that S(ω) reaches its maximum at ω = Ω. The standard deviationis represented by σ. One can easily verify that the integral of S(ω) accordingto equation (3.10) equals σ2 as should be (see equation (3.7)). The abovementioned spectra are shown in figure 3.1.


0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

ω/Ω

Ωσ−

2 S

eq. (3.8)

eq. (3.10)

Figure 3.1. The power density spectra according to eqs. (3.8) and (3.10).

3.3 The Fourier transform of the autocorrelationfunction

There is a direct analytical relationship between the autocorrelation functionand the spectral density of a stationary random signal. To show this, weexpress the autocorrelation as:

R(τ) = 〈x(t)x(t+ τ)〉 = x(t)x(t+ τ) = limT→∞

1T

∫ T/2

−T/2x(t)x(t+ τ)dt. (3.11)

For its Fourier transform we have

∫ ∞

−∞R(τ)e−iωτdτ = lim

T→∞

[1T

∫ ∞

−∞dτ

∫ T/2

−T/2x(t)x(t+ τ)e−iωτdt

]. (3.12)

To evaluate the term between square brackets, we take the same approach asin the previous section: we consider a signal xT (t) which is equal to x(t) over

3.3 The Fourier transform of the autocorrelation function 31

the interval −T/2 ≤ t ≤ T/2 and zero at all other times. We then have

1T

∫ ∞

−∞dτ

∫ T/2

−T/2x(t)x(t+ τ)e−iωτdt

=1T

∫ ∞

−∞

[∫ ∞

−∞xT (t)xT (t+ τ)e−iωτdt

]dτ

=1T

∫ ∞

−∞

[∫ ∞

−∞xT (t)xT (t+ τ)eiωte−iω(t+τ)dt

]dτ

=1T

∫ ∞

−∞

[∫ ∞

−∞xT (t)xT (s)eiωte−iωsdt

]ds

=1T

∫ ∞

−∞xT (t)eiωtdt

∫ ∞

−∞xT (s)e−iωsds

=1TA∗T (ω)AT (ω)

=1T|AT (ω)|2 (3.13)

Implementing (3.13) in the right-hand side of (3.12), dividing by π and usingthe definition of the power spectrum density (cf. eq. (3.6)) we find:

S(ω) =1π

∫ ∞

−∞R(τ)e−iωτdτ

=1π

∫ ∞

−∞R(τ) cos(ωτ)dτ

=2π

∫ ∞

0R(τ) cos(ωτ)dτ. (3.14)

There is thus a Fourier transform relationship between spectral density andthe autocorrelation function, S(ω) being the Fourier transform of π−1R(τ).From the definition of the Fourier transform of section 3.1 it follows that:

R(τ) =12

∫ ∞

−∞S(ω)eiωτdω =

∫ ∞

0S(ω)eiωτdω =

∫ ∞

0S(ω) cos(ωτ)dω. (3.15)

As for a stationary random signal R(τ) is a symmetrical function with respectto τ (see eq. (2.9)), S(ω) is symmetric as well: S(ω) = S(−ω). These symme-tries were used in eqs. (3.14) and (3.15). In case of white noise, i.e. S(ω) is con-stant with respect to ω, R(τ) will behave like Dirac’s δ-function: R(τ) = δ(τ).White noise processes are therefore also called δ-correlated processes. Thecorrelation functions associated with the spectra of eqs. (3.8) and (3.10) areshown in figure 3.2. The correlation function associated with spectrum (3.8) is


a simple exponential function: R(τ) = σ2e−τ/Ω−1: see Stratonovich [3], Vol.I,

Table 1 of §2.1, where a number of analytical relationships for power densityspectra and correlation functions are given. Note that Stratonovich applieda different normalization to S(ω); his S(ω) equals the present one multipliedwith the factor 2π. The present normalization corresponds to that of VanKampen [4], §III.3.

The correlation function associated with eq. (3.10) has been calculated nu-merically. It is seen that also this correlation function decays with time. Whiledecaying, it becomes negative and then approaches zero. A behavior of de-caying and oscillating around the neutral axis is not uncommon to correlationfunctions.

0 1 2 3 4 5 6 7 8 9−0.2

0

0.2

0.4

0.6

0.8

1

τΩ

⟨ x(t

)x(t

+τ)

⟩ σ−

2

eq. (3.10)

eq. (3.8)

Figure 3.2. The auto-correlation functions associated with the spectra (3.8)and (3.10).

Chapter 4

Single degree of freedomsystem: analysis in thefrequency domain

Many systems encountered in mechanical engineering practice can be modeledby a single degree of freedom system. This holds for simple spring-mass sys-tems, but also for multi-degree of freedom systems, which are decomposed intheir natural modes: see chapters 7-9. In the present chapter, we shall payattention to an ordinary spring-mass system with linear damping and subjectto random forcing: see figure 4.1. The equation of motion of such systems

Figure 4.1. A spring-mass system.

34

Single degree of freedom system: analysis in the frequency

domain

follows directly from Newton’s second law (the sum of all forces equals masstimes acceleration):

mx+ dx+ cx = P (t) (4.1)

where x(t) is displacement response, m mass, c linear spring stiffness, d lineardamping constant and P (t) excitation force.

4.1 Fourier transform

Using the concepts of Fourier transform introduced in section 3.1, we can writethe excitation force as a Fourier integral:

P (t) =12π

∫ ∞

−∞AP (ω)eiωtdω. (4.2)

The response of the system, x(t), can be written in a similar form with adifferent, frequency dependent, amplitude Ax:

x(t) =12π

∫ ∞

−∞Ax(ω)eiωtdω. (4.3)

If we now differentiate equation (4.3) with respect to time we get the followingequations for the time derivatives of the response:

x(t) =12π

∫ ∞

−∞Ax(ω)iωeiωtdω (4.4)

x(t) = − 12π

∫ ∞

−∞Ax(ω)ω2eiωtdω. (4.5)

Substituting equations (4.2)-(4.5) into the governing equation for this sys-tem (4.1):

−m∫ ∞

−∞Ax(ω)ω2eiωtdω + d

∫ ∞

−∞Ax(ω)iωeiωtdω + c

∫ ∞

−∞Ax(ω)eiωtdω

=∫ ∞

−∞AP (ω)eiωtdω (4.6)

This can of course be simplified to:∫ ∞

−∞

[(−mω2 + diω + c)Ax(ω) −Ap(ω)

]eiωtdω = 0 (4.7)

⇒ [(−mω2 + diω + c)Ax(ω) −Ap(ω)

]eiωt = 0

4.2 Types of system response 35

⇒ Ax(ω) =AP (ω)

−mω2 + diω + c. (4.8)

which, in combination with equation (4.3), gives a complete description of thesystem response x(t).

For further analysis of the system, it is convenient to introduce the dimen-sionless frequency ω∗ = ω

ω0and the natural frequency or eigenfrequency of the

system ω0 =√

cm . With the introduction of these quantities, the expression

for the Fourier amplitude of the system response can be rewritten as:

Ax(ω) = AP (ω)(−mω2

ω20

ω20 + id

ω

ω0ω0 + c

)−1

= AP (ω)(−mω2

0ω∗2 + idω0ω

∗ +mω20

)−1

= AP (ω)((

−ω∗2 +idω∗

mω0+ 1

)mω2

0

)−1

. (4.9)

An even further simplification is possible after normalizing AP with mω20,

AnP (ω∗) = AP (ω)mω2

0, and introducing a dimensionless damping coefficient δ =

dmω0

:

Ax(ω∗) =AnP (ω∗)

−ω∗2 + iδω∗ + 1. (4.10)

If we multiply this equation with the complex conjugate of Ax(ω), A∗x(ω),

we can derive an expression for the spectral density of the response of thesystem:

Sx(ω∗) = limT→∞

1πT

|Ax(ω)|2

= limT→∞

1πT

∣∣∣∣ AnP (ω∗)−ω∗2 + iδω∗ + 1

∣∣∣∣2

=SnP

(1 − ω∗2)2 + δ2ω∗2 ,

(4.11)

becauseSnP = lim

T→∞1πT

|AnP (ω∗)|2 . (4.12)

4.2 Types of system response

Objective is to analyse the types of response we can expect from a spring-mass system under conditions of random excitation using solution (4.11). For

36


domain

this purpose, we shall use the spectrum of equation (3.10) to represent thepower density of excitation. Our spring-mass system is linear: the magnitudeof response will be linearly related to the magnitude of excitation. Therefore,the standard deviation of response is directly and linearly proportional to thestandard deviation of excitation. It is of no use to study this relationshipany further: hence, we put σ = 1 in eq. (3.10) (the same would be achievedby dividing the left- and right-hand side of solution (4.11) by σ2 and re-normalizing Sx by σ−2). The example excitation spectrum we henceforth useis:

SnP (ω∗) =2Ω∗

ω∗2 e−2Ω∗/ω∗

, Ω∗ = Ω/ω0 ,

∫ ∞

0SnP (ω∗)dω∗ = 1.

(4.13)The power spectrum of response is given by (solution (4.11)):

Sx(ω∗) =SnP (ω∗)

(1 − ω∗2)2 + δ2ω∗2 . (4.14)

System response is now governed by two parameters: δ and Ω∗. The pa-rameter δ determines the magnitude of damping; its effect will be analyzedbelow. The parameter Ω∗ determines the frequency where most excitationenergy occurs in comparison with the natural frequency ω0 of the spring-masssystem. The shape of SnP (ω∗) versus ω∗/Ω∗ = ω/Ω was shown in figure 3.1(curve corresponding to eq. (3.10)). Most excitation energy is located aroundω∗/Ω∗ = O(1). So if Ω∗ 1, i.e. Ω ω0, most energy is at low values ofthe dimensionless frequency ω∗; so at low frequencies compared to the naturalfrequency of the system. For Ω∗ 1, i.e. Ω ω0, the opposite is true: exci-tation occurs mainly at frequencies that are large in comparison to the naturalfrequency. Depending on the value of Ω∗, three types of system response cannow be distinguished. These are discussed in the subsequent sections.

4.3 Quasi-static response (Ω∗ 1)

The first type of response is known as quasi-static or sub-critical and is illus-trated in figure 4.2 for Ω∗ = 0.25 and δ = 1. The spectrum SnP can only reachhigh values in case ω∗/Ω∗ ∼ 1, and thus ω∗ ∼ 0.25. The major part of thesystem response is therefore found at small dimensionless frequencies wherethe denominator in solution (4.14) is practically equal to 1. When ω∗ is verysmall, we can approximate Sx by

Sx ∼ SnP =SPc2. (4.15)

4.4 Dynamical response (Ω∗ 1) 37

This corresponds to a spring-mass system with negligible mass and damping.Energy of excitation is concentrated at low frequencies where the response ofthe system is governed by stiffness only! This is illustrated in figure 4.2: SnPand Sx according to (4.14) are almost equal.

Figure 4.2. Quasi-static or sub-critical response.

4.4 Dynamical response (Ω∗ 1)

The second type of response is called dynamical or super-critical responseand is illustrated in figure 4.3 for Ω∗ = 4 and δ = 1. For this situation,phenomena occur at dimensionless frequencies ω∗ 1. Energy of excitationoccurs predominantly at frequencies that are large in comparison with thenatural (eigen)frequency of the spring-mass system. The term involving ω∗4

in the denominator of solution (4.14) will dominate. Therefore, response isgoverned by inertia forces only: i.e., for large values of ω∗ the system response

38


domain

can be approximated by

Sx ∼ SnPω∗4 , (4.16)

which is depicted in the figure with the dashed line. Note the scale on thehorizontal axes in comparison with those of figures 4.2 and 4.4.

Figure 4.3. Dynamical or super-critical response.

4.5 Resonant response (Ω∗ = O(1), δ 1)

The third response type is known as resonant response. The spectra areplotted in figure 4.4 for Ω∗ = 1 and δ = 1 and δ = 0.25. Note the differentscales on the vertical axes.

To analyze the behavior of resonant response, consider the response spec-trum Sx(ω∗) of equation (4.14) when δ 1. Disregarding terms containingδ one has: Sx → Sn

P(1−ω∗2)2 . Clearly, this expression explodes as ω∗ → 1, that

4.5 Resonant response (Ω∗ = O(1), δ 1) 39

=0.25

=1

Figure 4.4. Resonant response.

is: for frequencies close to the natural frequency of the system. The explo-sion is a consequence of dropping the damping term in the denominator ofthe response spectrum. For ω∗ → 1, the damping term cannot be neglectedanymore. To show this in mathematical terms, we introduce a ’new’ frequencyυ∗ such that: ω∗ = 1 + δυ∗. We consider now the situation of ω∗ close to thenatural frequency, such that υ∗ = O(1). Then we have for the denominator ofeq. (4.14):

(1 − ω∗2)2 = 4δ2υ∗2 + O(δ3), (4.17)

δ2ω∗2 = δ2 + O(δ3) (4.18)

andSnP (ω∗) = SnP (ω∗ = 1) + O(δ) (4.19)

The result is that, with a relative error of O(δ):

Sx(ω∗) = δ−2SnP (ω∗ = 1)4υ∗2 + 1

. (4.20)

40


domain

−5 50

1

ν*

Sx(ω*)

ω*=1+δν*

δ−2SPn(ω*=1)

Figure 4.5. The response spectrum as a function of frequencies υ∗ and ω∗.

This result is shown in figure 4.5. The response peaks at υ∗ = 0, i.e.ω∗ = 1. The height of the peak is δ−2SnP (ω∗ = 1); the width is δ, so the meansquare response related to resonance scales as δ−1SnP (ω∗ = 1). The exact valueof the mean square response can be calculated using relation (3.7):

σ2 =∫ ∞

0Sx(ω∗)dω∗ =

∫ ∞

0

SnP(1 − ω∗2)2 + δ2ω∗2 dω

∗ (4.21)

As we have seen, in case of resonant response, most of the response energyis concentrated around the natural frequency (ω∗ = 1). Therefore, we canapproximate SnP by SnP (ω∗ = 1), so that

σ2res =

∫ ∞

0

SnP (ω∗ = 1)(1 − ω∗2)2 + δ2ω∗2 dω

∗ = SnP (ω∗ = 1)∫ ∞

0

dω∗

(1 − ω∗2)2 + δ2ω∗2(4.22)

An exact solution can be found for this integral when using the following trans-formation: ω∗2 = x and thus ω∗ = x1/2 and dω∗ = 1/2x−1/2dx. Using integral

4.5 Resonant response (Ω∗ = O(1), δ 1) 41

rule 3.252.12 in Gradshteyn & Ryzhik (which gives a thorough overview ofexact integral solutions) [5], we can write:

∫ ∞

0

dω∗

(1 − ω∗2)2 + δ2ω∗2 =12

∫ ∞

0

x−1/2dx

(1 − x)2 + δ2x

=12

∫ ∞

0

x−1/2dx

1 + x2 − (2 − δ2)x

=π

2δ(4.23)

The expression for the resonant part of the variance of the system responsetherefore simplifies to:

σ2res =

π

2δSnP (ω∗ = 1) (4.24)

To check whether the system response is dominated by resonance we haveto compare the value of σres with the total value of the variance, σ2

tot. Even forsystems with an eigenfrequency that differs a lot from the typical excitationfrequency Ω, the response can still be dominated by σ2

res if only δ is smallenough. This is shown in figure 4.6 where the response of a system with highnatural frequency (Ω∗ 1) and light damping (δ 1) is shown (in this caseΩ∗ = 0.25 and δ = 0.25). The response spectrum reveals two peaks: one closeto ω∗ = 0.1 where the response is quasi-static (Sx = SnP ) and one aroundω∗ = 1 which is due to resonance. The total variance of the system is:

σ2tot = σ2

q−s + σ2res (4.25)

where the variance due to the quasi-static response is:

σ2q−s =

∫ ∞

0SnP (ω∗)dω∗ (4.26)

Although SnP (ω∗ = 1) is small as Ω∗ 1 (see equation (4.13)), the con-tribution of the resonance peak can be large if δ 1! With the help ofequation (4.24) one can calculate the minimum required damping necessaryto suppress resonance under random excitation.

42


domain

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

ω*

Sx(ω

* )

quasi−static response

resonant response

Figure 4.6. Quasi-static system with little damping applied.

Chapter 5

The description of astationary Gaussian signal inthe time-domain

An alternative to the frequency domain analysis of the previous chapter isanalysis in the time-domain. To be able to perform such an approach, it isnecessary to have explicit descriptions of the external noise in the time-domain.These descriptions are presented in this chapter.

5.1 Random signal described by sinusoidal waveswith random phase angles

A conventional way to describe a stationary Gaussian random time signal isby addition of large numbers of sinusoidal waves with random phase angles:

x(t) =N∑n=1

(2S(ωn)∆ω)1/2 cos(ωnt+ ϕn),

ωn = n∆ω , 0 ≤ t ≤ T , T = 2π/∆ω (5.1)

The amplitude of each wave is defined in accordance with (2S(ωn)∆ω)1/2

where S(ωn) is the value of the power density taken at the frequency ωn of thecorresponding sinusoidal wave, and ∆ωn is the width of the parts in which thepower spectrum is broken up: see figure 5.1. The phase angle ϕn is random.For each n its value is chosen at-random according to a probability which isconstant and equal to 1/(2π) in the range 0 ≤ ϕn ≤ 2π. The randomly selectedvalue of ϕn is constant in time. The duration of the time-series is limited

44

The description of a stationary Gaussian signal in the

time-domain

Figure 5.1. Break-up of the power density spectrum in discrete parts to generate arandom time-series.

to 2π/∆ω in order to avoid repetition. This implies: to achieve long timeduration, the parts in which the spectrum is divided, have to be sufficientlysmall. The argument of the sinusoidal wave can apart from ϕn be expressedas

ωnt = n∆ωt = 2πnt/T = 2πnz , 0 ≤ z ≤ 1. (5.2)

It shows that the series representation of eq. (5.1) corresponds to a Fourierseries expansion of the signal over the time interval for which it is valid. Such arepresentation of a random signal can also be found in Van Kampen [4], §III.3.In the subsequent sections we shall show that the signal generated accordingto equation (5.1) approaches for N → ∞ a stationary Gaussian process withpower density equal to S(ω).

5.2 Gaussianity of the time-series 45

5.2 Gaussianity of the time-series

The series of sinusoidal waves can also be written as

x(t) =N∑n=1

anxn (5.3)

wherean = (2S(ωn)∆ω)1/2 (5.4)

andxn = cos (ωnt+ ϕn) . (5.5)

Consider now a fixed moment in time. The variable xn is a random variableof which the probability distribution can be derived from the probability dis-tribution of ϕn using transformation of variables invoking relation (5.5). Thevariable ϕn is two-valued for −1 ≤ xn ≤ +1: for every xn there exist twovalues of ϕn in the interval 0 ≤ ϕn ≤ 2π. Let us first restrict attention to theinterval 0 ≤ ϕn ≤ π where the connection between xn and ϕn is single-valued.In this case, the probability that Xn ≤ xn is equal to the probability thatΦn ≤ ϕn:

P (xn) = P (ϕn). (5.6)

Using the connection between cumulative probability and probability density(cf. equation (1.2)), we have

p(xn)dxn = p(ϕn)dϕn, (5.7)

or

p(xn) = p(ϕn)∣∣∣∣dϕndxn

∣∣∣∣ . (5.8)

Now p(ϕn) is constant and equal to 1/(2π), while from equation (5.5) it followsthat

dxndϕn

= sin(ωnt+ ϕn) =[1 − cos2(ωnt+ ϕ)

]1/2 = (1 − x2n)

1/2. (5.9)

so that (5.8) becomes

p(xn) =1

2π(1 − x2n)1/2

, − 1 ≤ xn ≤ +1. (5.10)

We only considered the interval 0 ≤ ϕn ≤ π. Because the connection betweenxn and ϕn in the interval 0 ≤ ϕn ≤ 2π is the same, we just have to multiply the

46


time-domain

right hand side of (5.10) by 2 to obtain the complete probability distributionof p(xn):

p(xn) =1

π(1 − x2n)1/2

, − 1 ≤ xn ≤ +1. (5.11)

The surface underneath this probability density distribution is 1 as should be.Because p(xn) is a symmetrical function of xn, uneven moments of xn are zero.The second moment can be calculated as follows:∫ +1

−1x2np(xn)dxn =

1π

∫ π

0cos2 φdφ

=1π

∫ π

0

(12

+12

cos 2φ)dφ

=12. (5.12)

For the fourth moment, we can write∫ +1

−1x4np(xn)dxn =

1π

∫ π

0cos4 φdφ

=1π

∫ π

0

(38

+12

cos 2φ+18

cos 4φ)dφ

=38. (5.13)

We can now conclude from representation (5.3) that at any moment in timex equals the sum of a large number of independent stochastic variables xn,each having the same probability distribution. According to the central limittheorem (Van Kampen [4], §1.7), the probability distribution of x approachesthe Gaussian distribution. This is even true if xn, n fixed, is non-Gaussian,as is the case above. This feature is responsible for the dominant role of theGaussian distribution in all fields of statistics.

To make the Gaussianity of x plausible, consider the so-called characteristicfunction of p(xn), defined as:

Gxn(k) =∫ +1

−1eikxnpx(xn)dxn. (5.14)

Expanding the exponential term, noting that uneven moments of xn are zeroand using eqs. (5.12) and (5.13), we can write

Gxn(k) =∫ +1

−1

(1 + ikxn − 1

2k2x2

n −i

3!k3x3

n +14!k4x4

n + ...

)px(xn)dxn

= 1 − 14k2 +

164k4 + O(k6). (5.15)

5.2 Gaussianity of the time-series 47

The characteristic function associated with anxn then reads

Ganxn(k) = 1 − 14k2a2

n +164k4a4

n + O(k6). (5.16)

The characteristic function of the sum of random variables equals the productof the characteristic function of the individual terms. Therefore,

Gx(k) = Ga1x1(k)Ga2x2(k)...Ganxn(k)

=N∏n=1

Ganxn(k)

=N∏n=1

(1 − 1

4k2a2

n +164k4a4

n + O(k6))

=N∏n=1

exp(−k

2

4a2n −

k4

64a4n + O(k6)

)

= exp

(−k

2

4

N∑n=1

a2n −

k4

64

N∑n=1

a4n + O(k6)

). (5.17)

Here,N∑n=1

a2n = 2

N∑n=1

S(ωn)∆ω = 2σ2, (5.18)

if N → ∞, whileN∑n=1

a4n ∼ σ4

N, (5.19)

where N is the number of parts of width ∆ω into which the power densityspectrum is subdivided. We now have,

Gx(k) = exp(−k

2

2σ2 + O

(k4σ4

N

)), (5.20)

so that

Gx(k) ∼ exp(−k

2

2σ2

)as N → ∞. (5.21)

This is the characteristic function of the Gaussian distribution with zero-meanand standard deviation σ. In conclusion, the distribution of x approaches thatof a stationary Gaussian distribution with the desired properties (zero-mean,standard deviation equal to σ). The larger the number of terms in the series

48


time-domain

representation, i.e. the smaller ∆ω in the break-down of the power spectrum,the smaller the higher order terms will be in comparison to the term retained:the series representation will represent more and more a Gaussian distributionwith increasing N or decreasing ∆ω.

5.3 Power density spectrum of the time-series

The Fourier transform of x(t) can be expressed as:

A(ωm) =∫ T/2

−T/2x(t)e−iωmtdt, (5.22)

where, in line with (5.1), we have discretised the frequencies: i.e. ωm =m∆ω = m2π/T . Substituting (5.1), we can write:

A(ωm) =N∑n=1

(2S(ωn)∆ω)1/2∫ +T/2

−T/2e−iωmt cos(ωnt+ ϕn)dt

=N∑n=1

(2S(ωn)∆ω)1/2∫ +T/2

−T/2(cosωmt− i sinωmt)

(cosϕn cosωnt− sinϕn sinωnt)dt

=N∑n=1

(2S(ωn)∆ω)1/2∫ +T/2

−T/2(cosϕn cosωmt cosωnt

+i sinϕn sinωmt sinωnt) dt

−∫ +T/2

−T/2(sinϕn cosωmt sinωnt

+i cosϕn sinωmt cosωnt) dt

. (5.23)

5.3 Power density spectrum of the time-series 49

The term in the first integral is an even function with respect to t, the termin the second is uneven. Hence,

A(ωm) =N∑n=1

(2S(ωn)∆ω)1/2∫ +T/2

−T/2(cosϕn cosωmt cosωnt

+i sinϕn sinωmt sinωnt) dt

=N∑n=1

(2S(ωn)∆ω)1/2[

12

cosϕn∫ +T/2

−T/2cos(ωm − ωn)t

+ cos(ωm + ωn)t dt

+12i sinϕn

∫ +T/2

−T/2cos(ωm − ωn)t− cos(ωm + ωn)t dt

]

=N∑n=1

(2S(ωn)∆ω)1/2[

12(cosϕn + i sinϕn)

sin(ωm − ωn)t(ωm − ωn)

∣∣∣∣+T/2

−T/2

+12(cosϕn − i sinϕn)

sin(ωm + ωn)t(ωm + ωn)

∣∣∣∣+T/2

−T/2

]

=12

(2S(ωm)∆ω)1/2 T (cosϕm + i sinϕm), (5.24)

where we used the properties

(ωm − ωn)T/2 = (m− n)π, (5.25)

(ωm + ωn)T/2 = (m+ n)π, (5.26)

sin((ωm − ωn)T/2)ωm − ωn

= T/2 for n = m

sin((ωm − ωn)T/2)ωm − ωn

= 0 for n = m (5.27)

sin((ωm + ωn)T/2)ωm + ωn

= 0. (5.28)

From (5.24) it can be concluded that the Fourier transform of a random func-tion varies itself randomly through its dependency on ϕm. Furthermore, theFourier transform grows unboundedly with time T . Quite different is thesituation if we consider the power density of x(t) which is related to A(ω) as:

S(ω) = limT→∞

1πT

|A(ω)|2 . (5.29)

50


time-domain

Substituting (5.24), we obtain

S(ω) = limT→∞

[T

2πS(ωm)∆ω(cos2 ϕm + sin2 ϕm)

]= S(ωm). (5.30)

because T∆ω/(2π) = 1. So, we have shown that S(ωn) in series expres-sion (5.1) corresponds to the discretised value of the power density S(ω) ofx(t) at ω = ωn.

Chapter 6

Single degree of freedomsystem: analysis in thetime-domain

Methods are presented to analyze a single degree of freedom system in thetime-domain. The approach is complementary to the frequency-domain anal-ysis presented in chapter 4. Methods and results given can be used to back-updirect numerical simulations of systems in the time-domain.

6.1 Time-domain description of the excitation

In the previous chapter, a description in the time-domain was presented for astationary Gaussian process. This representation is used to generate a timesignal for the excitation of the spring-mass system we intend to analyze:

P ∗(t∗) =N∑n=1

(2SnP (ω∗n)∆ω

∗)1/2 cos(ω∗nt

∗ + ϕn),

ω∗n = n∆ω∗ , 0 ≤ t∗ ≤ T , T = 2π/∆ω∗ (6.1)

The break-up of the power density spectrum in discrete parts to generate theabove time-series was shown in figure 5.1. An illustration of a time-domainrealization of the excitation according to equations (6.1)is shown in figure 6.1.The power spectrum was described according to eq. (4.13). Matlab was used.We took Ω∗ = 1, ∆ω∗ = 3 · 10−2, T = 200, N = 103. Random phase angleswere generated by a random generator, normalized to values between 0 and2π. From the generated signal, various interesting statistical properties can

52 Single degree of freedom system: analysis in the time-domain

be calculated. For this purpose, one can adopt the time averaging methodsof chapter 2. In this way, results for standard deviation, auto-correlation andprobability distribution can be calculated. Note that the standard deviationshould be equal to 1, the auto-correlation should be equal to the correlationshown in figure 3.2 and the probability distribution should be close to theGaussian distribution. The accuracy of the results can be assessed as a func-tion of the value of ∆ω∗. Also a range of different realizations can be generatedfrom which ensemble averages can be determined. Stationarity and ergodicitycan thus be verified.

An illustration of the Gaussianity of series representation (6.1) with thepower density spectrum according to eq. (4.13) has been given in figure 6.2.The probability distribution was calculated according to the procedure out-lined in figure 2.3. While Ω∗ = 1, the spectrum was broken down taking∆ω = 3 and ∆ω = 3 · 10−3 with N = 10 and N = 103, respectively. It is seenthat Gaussianity improves with increasingN as forecasted by eqs. (5.17)-(5.21)in the previous chapter.

6.1 Time-domain description of the excitation 53

Figure 6.1. Illustration of a time-domain realization of the excitation for Ω∗ = 1.


Figure 6.2. Probability distribution of excitation according to representation (6.1)and spectrum (4.13) with Ω∗ = 1.

6.2 Solution in the time-domain 55

6.2 Solution in the time-domain

We shall analyze the response behavior of the spring-mass system formulatedin dimensionless form:

x+ δx+ x = P ∗(t∗), (6.2)

whereP ∗(t∗) = P (t)/c, (6.3)

t∗ = ω0t. (6.4)

The dots denote differentiation with respect to dimensionless time t∗, c isthe stiffness or spring constant and ω0 is the eigenfrequency of the system.Eq. (6.2) follows from eq. (4.1) using (6.3), (6.4) and

ω0 =√c/m δ = d/(mω0). (6.5)

Given P ∗(t∗) as described by equation (6.1) and disregarding any transientresponse, the solution of equation (6.2) can be found by describing x as:

x =N∑i=n

αn cos(ω∗nt

∗ + ϕn) +N∑n=1

βn sin(ω∗nt

∗ + ϕn). (6.6)

Substitute (6.6) in (6.2) and equate coefficients of terms like cos(ω∗nt

∗ + ϕn)and sin(ω∗

nt∗ + ϕn):

−αnω∗2n + δβnω

∗n + αn = (2SnP (ω∗

n)∆ω∗)1/2 , (6.7)

−βnω∗2n + δαnω

∗n + βn = 0, (6.8)

from which it follows that:

αn = βn1 − ω∗2

n

δω∗n

, (6.9)

βn = (2SnP (ω∗n)∆ω

∗)1/2δω∗

n

(1 − ω∗2n )2 + δ2ω∗2

n

. (6.10)

Using the property (we sometimes repeat basic algebra to brush-up the ana-lytical mind; for an overview on algebraic rules and properties of known math-ematical functions, see Abramowitz & Stegun [6] which is now also availablein digital form; property (6.11) can be found under 4.3.17 of [6]):

cos(z1 + z2) = cos z1 cos z2 − sin z1 sin z2. (6.11)


equation (6.6) can be written as:

x =N∑n=1

(α2n + β2

n

)1/2 cos (ω∗nt

∗ + φn − arctan(βn/αn)) , (6.12)

where

α2n + β2

n =2SnP∆ω∗

(1 − ω∗2n )2 + δ2ω∗2

n

, (6.13)

and

βn/αn =δω∗

n

1 − ω∗2n

. (6.14)

From (6.12) and (6.13) it can be seen that the spectral density of the responseequals solution (4.11) (as should be!).

The above results can be used to back-up direct numerical simulation re-sults of the spring-mass system. The equation, i.e. eq. (6.2), can be split-upinto two first order equations:

x = y

y = −δy − x− P ∗(t∗). (6.15)

Implementing the excitation according to eq. (6.1), the above two equationscan be integrated numerically by standard routines. As initial conditions, onecan take: x = y = 0 at t = 0. The numerically calculated result will exhibita transient response which decays as e−δt∗ : for further details on transientresponse, see section 6.4. Once the transient response has vanished, the resultshould, in statistical sense, be the same as the analytical solution (6.12) whichdoes not exhibit transient behavior.

6.3 Resonant response

In section 4.5 we have seen that resonant response (Ω∗ = O(1), δ 1) isgoverned by a small band of frequencies around ω∗ = 1. To describe thisbehavior, we write

ω∗n = 1 + δυ∗m , n = m+N0 , N0 = 1/∆ω. (6.16)

Oscillations at natural frequency ω∗ = 1 thus correspond to n = N0 or m = 0in the series representations of the solution. Substituting this in solution (6.12)

6.3 Resonant response 57

and dropping terms of magnitude δ directly compared with order unity we have

x =(πSnP (ω∗ = 1)

2δ

)1/2 N−N0∑m=−N0

(4∆υ∗

π(1 + 4υ∗2m )

)1/2

cost∗ + υ∗mT

∗ + ϕm + arctan(

12υ∗m

), (6.17)

whereT ∗ = δt∗. (6.18)

Note that υ∗m = m∆υ∗ and that υ∗m extends from −N0∆υ∗ to +(N −N0)∆υ∗.Solution (6.17) can also be written as:

x =(πSnP (ω∗ = 1)

2δ

)1/2

f1(T ∗) cos t∗ − f2(T ∗) sin t∗

=(πSnP (ω∗ = 1)

2δ

)1/2

a(T ∗) cos t∗ + ψ(T ∗) , (6.19)

where:a(T ∗) =

f21 (T ∗) + f2

2 (T ∗)1/2

, (6.20)

ψ(T ∗) = arctan f2(T ∗)/f1(T ∗) . (6.21)

Here, f1(T ∗) and f2(T ∗) are Gaussian random signals:

f1(T ∗) =N−N0∑m=−N0

(4∆υ∗

π(1 + 4υ∗2m )

)1/2

cos(υ∗mT

∗ + ϕm + arctan(

12υ∗m

)),

(6.22)

f2(T ∗) =N−N0∑m=−N0

(4∆υ∗

π(1 + 4υ∗2m )

)1/2

sin(υ∗mT

∗ + ϕm + arctan(

12υ∗m

)),

(6.23)which are stochastically equivalent to:

f1(T ∗) =N∑n=1

(8∆υ∗

π(1 + 4υ∗2n )

)1/2

cos (υ∗nT∗ + ϕn) , (6.24)

f2(T ∗) =N−N0∑n=1

(8∆υ∗

π(1 + 4υ∗2n )

)1/2

sin (υ∗nT∗ + ϕn) . (6.25)

These signals hold over the time interval

0 ≤ T ∗ ≤ T ∗0 , T ∗

0 = 2π/∆υ∗. (6.26)


The power densities of the signals equal 4/(π(1 + 4υ∗2)) implying that theirstandard deviations are equal to unity:

〈f21 (T ∗)〉 = 〈f2

2 (T ∗)〉 =∫ ∞

0

4dυ∗

π(1 + 4υ∗2)=

2π

arctan(2υ∗)

∣∣∣∣∣∞

0

= 1. (6.27)

Furthermore, f1(T ∗) and f2(T ∗) are uncorrelated. This can be shown as fol-lows:

〈f1(T ∗)f2(T ∗)〉 = limT ∗0 →∞

1T ∗

0

∫ T ∗0

0f1(T ∗)f2(T ∗)dT ∗

= limT ∗0 →∞

N∑n=0

N∑m=0

8∆υ∗

π(1 + 4υ∗2n )(1 + 4υ∗2m )T ∗0

×∫ T ∗

0

0cos

(2πn

T ∗

T ∗0

+ ϕn

)sin

(2πm

T ∗

T ∗0

+ ϕm

)dT ∗

(6.28)

Here,

1T ∗

0

∫ T ∗0

0cos

(2πn

T ∗

T0+ ϕn

)sin

(2πm

T ∗

T0+ ϕm

)dT ∗

=∫ 1

0cos(2πnz + ϕn) sin(2πmz + ϕm)dz

=12

∫ 1

0[sin(2π(m− n)z + ϕm − ϕn) + sin(2π(m+ n)z + ϕm + ϕn)] dz

=12

∫ 1

0[sin(ϕm − ϕn) cos(2π(m− n)z) + cos(ϕm − ϕn) sin(2π(m− n)z)

+ sin(ϕm + ϕn) cos(2π(m+ n)z) + cos(ϕm + ϕn) sin(2π(m+ n)z)] dz= 0. (6.29)

because the integrals of the oscillatory terms are zero over the interval 0 ≤z ≤ 1. Hence,

〈f1(T ∗)f2(T ∗)〉 = 0. (6.30)

6.3 Resonant response 59

The mean-square value of response can now be calculated as follows (usingeq. (6.19), (6.27) and (6.30)):

σ20 = 〈x2(t)〉 =

πSnP (ω∗ = 1)2δ

〈(f1(T∗) cos t∗ − f2(T ∗) sin t∗)2〉

=πSnP (ω∗ = 1)

2δ[〈f2

1 (T ∗)〉2 cos2 t∗ + 〈f22 (T ∗)〉 sin2 t∗

−2〈f1(T ∗)f2(T ∗)〉 cos t∗ sin t∗]

=πSnP (ω∗ = 1)

2δ(6.31)

which is the same as eq. (4.24)!From solution (6.19) it is seen that resonant response to random exci-

tation consists of quasi-deterministic sinusoidal behavior around the naturalfrequency with an amplitude and phase which vary randomly and slowly withtime (T ∗ = δt∗). An illustration of this behavior is given in figure 6.3, takingδ = 0.02 and T ∗

0 = 600.Calculating resonant response by direct numerical integration of the dif-

ferential equation of the spring-mass system can be quite cumbersome. Thedamping term in the equation is very small and at the same time, the determin-ing factor for the magnitude of the resonant response: a recipe for problems!While performing numerical integration, round-off errors are made. These cancreate an artificial form of damping, which is larger than the real damping. Toavoid this, very small time steps are required which leads to long calculationtimes.


x(t

*)/

0x(t

*)/

0

Figure 6.3. Resonant response (Ω∗ = 1, δ = 0.02)

6.4 Transient response 61

6.4 Transient response

The solutions presented so far did not take transient behavior into account.Both the solutions for response in the frequency-domain presented in chapter 4and the solutions in the time-domain given in the previous sections describe re-sponse that is statistically stationary ∗. In the present section, we shall presentdescriptions for transient behavior due to some initial condition imposed attime zero. More specifically, we shall consider the case where displacementand velocity are zero at time zero.

A description in the time-domain for transient response is simply obtainedby using the linearity of the governing differential equation: cf. eq. (6.2). Asolution can be written as

x(t∗) = xH(t∗) + xS(t∗), (6.32)

where xS(t∗) is the stationary solution given by eqs. (6.12)-(6.14). BecausexS(t∗) satisfies eq. (6.2) with random right hand side P ∗(t∗), xH(t∗) will bethe solution of the homogeneous problem

xH + δxH + xH = 0. (6.33)

The two basic solutions of this problem are:

xH = Ae−δt∗/2 cos

((1 − δ2/4)1/2t∗

)+Be−δt

∗/2 sin((1 − δ2/4)1/2t∗

). (6.34)

At time zero, we require that displacement and velocity are zero:

x(t∗) = x(t∗) = 0 at t∗ = 0 (6.35)

From eq. (6.32) it then follows that

xH(t∗) = −xS(0) , xH(t∗) = −xS(0) at t∗ = 0, (6.36)

where xS(0) and xS(0) are specified by eqs. (6.12)-(6.14) substituting t∗ = 0.Given conditions (6.36), the constants A and B of solution (6.33) can beevaluated to:

A = −xS(0) (6.37)

B =(1 − δ2

)−1/2(−xS(0) − δ

2xS(0)

)(6.38)

∗With stationary response we mean response of which the statistics (mean value, standarddeviation, etc.) do not vary with time. The response signal itself obviously varies with time(in a random way)


The statistics of response can be determined as follows. The excitation P ∗(t∗)is a Gaussian process. In that case, any variable that is linearly related toP ∗(t∗) will be Gaussian as well. Also variables that suffer from a time-lagwith respect to P ∗(t∗) because of inertia and/or damping. Therefore, theprobability distribution of x(t∗) will be Gaussian. All that remains to bespecified are the parameters that determine this distribution, i.e. mean andstandard deviation. The mean of x(t∗) equals the mean of xH(t∗) plus themean of xS(t∗), which are both equal to zero (because 〈P ∗(t∗)〉 equals zero).Hence,

〈x(t∗)〉 = 0. (6.39)

The standard deviation follows from the mean-square response as:

σ2 = 〈x2(t∗)〉 = 〈(xH(t∗) + xS(t∗))2〉= 〈x2

H(t∗) + x2S(t∗) + 2xH(t∗)xS(t∗)〉

= 〈x2H(t∗)〉 + 〈x2

S(t∗)〉 + 2〈xH(t∗)xS(t∗)〉 (6.40)

Substituting solutions (6.34) and (6.37)-(6.38) we end up with moments ofproducts of xS(t) with xS(0) and xS(0). The determination of these correla-tions, though not impossible, can be quite laborious, depending on the shapeof the power density spectrum of P ∗(t∗). A more simple situation arises incase of resonant response. Such response occurs when damping is light, δ 1,and the total response is governed by the resonance peak in the power densityspectrum. When δ 1, the time domain description for transient responsegiven by eqs. (6.34) and (6.37)-(6.38) can be simplified to:

xH(t∗) = −xS(0)e−T∗/2 cos t∗ − xS(0)e−T

∗/2 sin t∗, (6.41)

where, as before, T ∗ = δt∗. For resonant response, xS(t∗) can be described bythe right hand side of eq. (6.19):

xS(t∗) = σ0 f1(T ∗) cos t∗ − f2(T ∗) sin t∗ , (6.42)

where f1(T ∗) and f2(T ∗) are uncorrelated Gaussian signals described by eqs. (6.24)-(6.25). The total response (transient and stationary) is then given by:

x(t∗) = σ0

(f1(T ∗) − f1(0)e−T

∗/2)

cos t∗ − σ0

(f2(T ∗) − f2(0)e−T

∗/2)

sin t∗.(6.43)


The standard deviation can be assessed as:

σ2 = 〈x2(t∗)〉= σ2

0

⟨(f1(T ∗) − f1(0)e−T

∗/2)2⟩

cos2 t∗

+σ20

⟨(f2(T ∗) − f2(0)e−T

∗/2)2⟩

sin2 t∗

−2σ20

⟨(f2(T ∗) − f2(0)e−T

∗/2)(

f1(T ∗) − f1(0)e−T∗/2

)⟩

× sin t∗ cos t∗. (6.44)

Now ⟨(f1(T ∗) − f1(0)e−T

∗/2)2⟩

= 〈f21 (T ∗)〉 + 〈f2

1 (0)〉e−T ∗

−2〈f1(T ∗)f1(0)〉e−T ∗/2. (6.45)

Here,〈f2

1 (T ∗)〉 = 〈f21 (0)〉 = 1 : (6.46)

see eq. (6.27). The auto-correlation of f1(T ∗) follows from its power densityspectrum which is 4/π(1 + 4υ∗2. The Fourier transform of this spectrum(cf. eq. (3.15)) yields the autocorrelation function being e−T ∗/2: this resultis easily obtained from eq. (3.8) setting Ω = 1/2, σ = 1 and noting that theauto-correlation associated with eq. (3.8) is σ2e−ΩT ∗

(see text subsequent toeq. (3.15)). The net result for eq. (6.45) is now:

⟨(f1(T ∗) − f1(0)e−T

∗/2)2⟩

= 1 − e−T∗. (6.47)

Clearly, the same result is obtained for the second term on the right hand sideof eq. (6.44): ⟨(

f2(T ∗) − f2(0)e−T∗/2

)2⟩

= 1 − e−T∗. (6.48)

For the third term we can write:⟨(f1(T ∗) − f1(0)e−T

∗/2)(

f2(T ∗) − f2(0)e−T∗/2

)⟩

= 〈f1(T ∗)f2(T ∗)〉 + 〈f1(0)f2(0)〉e−T ∗

− (〈f1(0)f2(T ∗)〉 + 〈f2(0)f1(T ∗)〉) e−T ∗/2 (6.49)

In view of result (6.30),

〈f1(T ∗)f2(T ∗)〉 = 〈f1(0)f2(0)〉 = 0. (6.50)


Furthermore, using a procedure similar to that of eqs. (6.28)-(6.29) one canshow that:

〈f1(0)f2(T ∗)〉 + 〈f2(0)f1(T ∗)〉 = 0 (6.51)

Here it is noted that the separate correlations are not equal to zero! Thesevalues can be calculated in the following way:

〈f1(0)f2(T ∗)〉 = 〈f1(T )f2(T +T ∗)〉 = limT ∗0 →∞

∫ T ∗0

0f1(T )(f2(T +T ∗)dT. (6.52)

The right hand side can be evaluated upon substituting series representa-tions (6.24)-(6.25), using addition formulae for trigonometric functions to bringthe terms back to products of terms like sin(2mπT/T ∗

0 ) and cos(2mπT/T ∗0 )

and evaluating the integrals; all this similar to the procedure of eq. (6.29).The result is

〈f1(0)f2(T ∗)〉 =N∑n=1

2∆υ sin(υnT ∗)π(1 + 4υ2

n)=∫ ∞

0

2 sin(υT ∗)dυπ(1 + 4υ2)

(6.53)

Similarly it can be shown that

〈f2(0)f1(T ∗)〉 = −∫ ∞

0

2 sin(υT ∗)dυπ(1 + 4υ2)

(6.54)

The above cross-correlation functions are zero if T ∗ = 0 but have nonzero butopposite values for T ∗ > 0.

The standard deviation now follows from eq. (6.44) as

σ2(T ∗) = σ20(1 − e−T

∗). (6.55)

The Gaussian probability density distribution for transient response is thengiven by

p =1

σ(T ∗)√

2πe−x

2/(2σ2(T ∗)) (6.56)

The distribution approaches a Dirac δ-function around x = 0 as t → ∞. ForT ∗ → ∞ the distribution evolves to a stationary Gaussian distribution withstandard deviation σ0. An illustration of the distribution is given is figure 6.4.

The cumulative Gaussian distribution for transient response reads as

P (x) =12

(1 + erf

(x

σ(T ∗)√

2

)), (6.57)

where erf is the error function: cf. eq. (1.9). An illustration of the cumulativedistribution is given in figure 6.5. It is seen that the cumulative distribu-tion evolves from a Heaviside function at T ∗ = 0 to the stationary Gaussiandistribution as T ∗ → ∞.


−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

ρ/σ0

σ 0 p

T*=0.05

T*=1

T*=5

Figure 6.4. Transient cumulative Gaussian response behavior in case of zero initialresponse.


2 3-2-3

Figure 6.5. Transient cumulative Gaussian response behavior in case of zero initialresponse.

Chapter 7

Multi degree of freedomsystem: the tensioned beam

The beam is a basic element of many constructions of mechanical engineering.Its stiffness against lateral deflection can be increased by applying longitudinaltension at both ends. A guitar string is a nice example of such application.In off-shore structures we find beams and tensioned beams at various places.For example, as elements of steel jacket structures; as risers in which oil or gasflows from the well at the sea floor to the production platform (Brouwers [8]);or as anchoring device keeping a floating platform at its position, see figure 7.1for examples of such structures. Direct wave loading and wave-induced lat-eral displacement via a floating structure are the cause of random responsebehavior in time. Analyzing the random behavior of the tensioned beam isthe objective of subsequent chapters. In the present chapter, we shall focus onthe formulation of the governing partial differential equations. Furthermore,we shall identify the conditions under which bending stiffness and stiffnessby pre-tensioning prevail. In chapter 8, analysis is focussed to the case of atensioned string, in chapter 9 to the beam.

68 Multi degree of freedom system: the tensioned beam

Figure 7.1. Illustration of various types of off-shore structures; by courtesy of RoyalDutch Shell Exploration & Production

7.1 Basis equations

Attention is focussed on the so-called marine riser which connects a floatingoff-shore structure to the oil or gas well at the sea bottom: Brouwers [8]. Theriser can be modeled as a tensioned beam which undergoes lateral deflectionas a result of direct wave forcing and lateral motion of the floating structure,see the sketch in figure 7.2.

Basic equations can be formulated when applying Newton’s second law inhorizontal direction to an infinitesimal element of the beam, see figure 7.3:

−dD + d(Trθ) + Fdz =(m∂2x

∂t2+ d

∂x

∂t

)dz. (7.1)

Equilibrium of moments yields:

dM = Ddz. (7.2)

The bending moment is related to deformation according to the Bernoulli-Euler relation

M = EI∂2x

∂z2, (7.3)

7.1 Basis equations 69

Figure 7.2. Riser configuration.

while for small angles, angle and displacement are related to each other by thegeometrical relation

θ =∂x

∂z. (7.4)

Substituting equations (7.2)-(7.4) into equation (7.1) yields the partial differ-


Figure 7.3. Infinitesimal element of a tensioned beam.

ential equation of the tensioned beam:

EI∂4x

∂z4− ∂

∂z

(Tr(z)

∂x

∂z

)+m

∂2x

∂t2+ d

∂x

∂t= F (z, t), (7.5)

where:

x = horizontal deflectionz = vertical distance from riser baset = timeE = modulus of elasticityI = moment of stiffness against bendingTr = riser tension (Tr > 0)m = mass per unit lengthd = damping constant per unit lengthF = external force per unit length

7.2 Tensioned string versus beam 71

The bottom end of the riser is assumed to be fixed to the base:

x = 0 at z = 0 (7.6)

and subject to a rotational constraint

EI∂2x

∂z2= Cb

∂x

∂zat z = 0, (7.7)

where Cb is the rotational stiffness. At the top, the riser is connected to thefloating structure by a hinge:

x = V (t) at z = L, (7.8)

EI∂2x

∂z2= 0 at z = L, (7.9)

where V (t) is the time-dependent horizontal displacement of the floating struc-ture and L is the length of the riser.

7.2 Tensioned string versus beam

Two kinds of stiffness can be distinguished in the tensioned beam: bendingstiffness and stiffness due to tensioning. The question now is: which is theimportant one? To answer this question, consider a beam of length L subjectto a typical lateral deflection of magnitude a. The magnitudes of the bendingand tension terms in equation (7.5) are then:

EI∂4x

∂z4≈ EIa

L4, (7.10)

T∂2x

∂z2≈ Ta

L2. (7.11)

Hence, bending and tension forces compare to each other as:

D1 =EI

TL2. (7.12)

In case of low pre-tension, D1 will be large and bending stiffness dominates.The tension forces can be disregarded and we end up with the problem of thebeam: see chapter 9.

On the other hand, when D1 1, the stiffness of the beam against lat-eral deflection is mainly due to pre-tension (the guitar string!) and bendingforces can then be disregarded. However, when neglecting bending forces, thegoverning partial differential equation reduces from fourth- to second-order in


z, so that 2 out of 4 boundary conditions have to be dropped. A singularperturbation problem results (just as the resonance problem in case of lightdamping in the spring-mass system; see section 4.5). To solve this problem,boundary layers have to be introduced at both ends of the tensioned string(so boundary layers not only occur in fluid mechanics!). At the bottom end,the boundary layer is described by a local coordinate:

η = ε−1 z

L; η = O(1), ε 1. (7.13)

In terms of this coordinate, we have:

EI∂4x

∂z4≈ EI

L4ε4∂4x

∂η4, (7.14)

T∂2x

∂z2≈ T

L2ε2∂2x

∂η2. (7.15)

Bending and tension forces become of the same magnitude as:

EI

L4ε4=

T

L2ε2⇒ ε = D

1/21 . (7.16)

For large pre-tension, such that ε 1 or D1 1, bending forces will only beimportant in small areas of length Lε = (EI/T )1/2 at both ends of the beam:(EI/T )1/2 L when D1 1. Outside these areas, i.e. the main section ofthe beam, lateral deflection, angles and curvatures are governed by pre-tensionforces rather than bending forces. The boundary layers at the top and bottomserve to adjust deflections, angles and curvatures to the values prescribed bythe boundary conditions. Because bending and tension forces are 4th and 2ndorder derivatives with respect to vertical distance, these terms can be assumedto dominate in comparison to all other terms in the boundary layers. To showthis in explicit terms, introduce the boundary layer coordinate η defined byequation (7.13) into basic equation (7.5). This gives, after some re-ordering ofterms:

∂4x

∂η4− ∂2x

∂η2= ε4

L4

EI

(F −m

∂2x

∂t2− d

∂x

∂t

). (7.17)

For small ε, the right hand side becomes small and can be disregarded. Be-cause the boundary layer is very short in length, inertia, damping and externalforces are relatively small. The boundary layers are areas where response isquasi-static in nature. Solutions describe exponential decaying behavior overthe distance (EI/T )1/2. In case of clamped ends (Cb → ∞ in equation (7.7))they reveal local bending stress increase due to increasing curvature. To avoidany local stress increase, hinges or ball-joints should be incorporated at the

7.2 Tensioned string versus beam 73

ends of the tensioned string (like the guitar string). In the subsequent anal-ysis, we shall pay no further attention to the boundary layers at the top andbottom: for further details, see Brouwers [8]. We shall consider the case of thebeam without pre-tension (chapter 9) and the string without bending stiffness(chapter 8) only.

Chapter 8

Random Vibrations of atensioned string

As we have shown in the previous chapter, for cases where the dimensionlessratio D1 given by eq. (7.12) is small, D1 1, forces due to bending stiffnesscan be disregarded. Furthermore, the pre-tension is assumed to be constantwith respect to z: Tr(z) = Tr (which is justified if the total weight of the riseris much less than the pre-tension applied at the top). The governing equationof the tensioned beam, i.e. eq. (7.5) can then be simplified to that of thetensioned string:

−Tr ∂2x

∂z2+m

∂2x

∂t2+ d

∂x

∂t= F (z, t) (8.1)

We disregard lateral displacement imposed at the top. The boundary condi-tions are:

x(z, t) = 0 at z = 0 and z = L. (8.2)

The effects of boundary conditions imposed on angle and curvature are ac-commodated by boundary layer type solutions at the top and bottom, seesection 7.2 . These are not considered here.

8.1 Separation of variables

Drop the excitation and damping forces in equation (8.1):

Tr∂2x(z, t)∂z2

= m∂2x(z, t)∂t2

(8.3)

To solve this equation we will apply separation of variables [7]. This involvesexpressing the solution x(z, t) as a product of two functions which only depend

76 Random Vibrations of a tensioned string

on z and t, respectively:x(z, t) = g(z)h(t) (8.4)

Substitution in (8.3) and dividing through g(z)h(t) gives:

Tr1g(z)

∂2g(z)∂z2

= m1h(t)

∂2h(t)∂t2

(8.5)

The above equation can only hold if the terms on the left- and right-hand sideare equal to a constant −λ (the minus sign is chosen for convenience). Thisresults in the following ordinary differential equations:

d2g(z)dz2

= − λ

Trg(z) (8.6)

d2h(t)dt2

= − λ

mh(t) (8.7)

We will start with finding a solution for the first differential equation (8.6),using x(z, t) = 0 at z = 0, L as boundary conditions. These boundary condi-tions can now be expressed as g(z) = 0 at z = 0, L because h(t) = 0 leads to atrivial solution x = 0 for all times t. The general solution for such an equationis:

g(z) = A sin

(√λ

Trz

)+B cos

(√λ

Trz

)(8.8)

The constant A and B can be determined using the boundary conditions.The first boundary condition (g(0) = 0 leads to B = 0). The second boundary

condition only leads to non-trivial solutions in case√

λTrL = nπ with n an

integer. This leads to the following eigenvalues:

λn = Tr

(nπL

)2, n = 1, 2, 3, ...,∞ (8.9)

with the corresponding eigenfunctions or natural modes:

gn(z) = sin(nπzL

), n = 1, 2, 3, ...,∞ (8.10)

Following the same procedure for the second differential equation (8.7) onlyleads to trivial solutions using the initial conditions h(0) = 0 and dh(0)/dt = 0:h(t) = 0. This means that the total solution x(z, t) = g(z)h(t) equals zero inthis case. A non-zero solution is only obtained by introducing a forcing. Thiswill be considered in the next section.

8.2 Expansion in eigenfunctions 77

8.2 Expansion in eigenfunctions

The solution of equation (8.1) is written as an infinite series of eigenfunctionswith time-dependent amplitudes:

x(z, t) =∞∑n=1

xn(t)gn(z) (8.11)

with gn(z) = sin(nπz/L). The equation for the amplitudes xn can be foundby substituting the above equation into the PDE:

∞∑n=1

[Trn2π2

L2xn(t) sin

(nπzL

)+(md2xn(t)dt2

+ ddxndt

)sin

(nπzL

)]= F (z, t)

(8.12)This equation can be solved for xn(t) using well-known properties of orthogo-nality of eigenfunctions gk and gn:∫ L

0gn(z)gk(z)dz = 0 if k = n (8.13)

and ∫ L

0gn(z)gn(z)dz =

L

2. (8.14)

In order to use this property we multiply (8.12) with gk and integrate fromz = 0 to z = L:∫ L

0

∞∑n=1

[Trn2π2

L2xn(t) sin

(nπzL

)sin

(kπz

L

)+

(md2xn(t)dt2

+ ddxndt

)sin

(nπzL

)sin

(kπz

L

)]dz

=∫ L

0F (z, t) sin

(kπz

L

)dz. (8.15)

Using the orthogonality properties this can be simplified to:(Trn2π2

L2xn(t) +m

d2xn(t)dt2

+ ddxndt

)L

2

=∫ L

0F (z, t) sin

(nπzL

)dz. (8.16)

Equation (8.16) can also be written in a form which is equal to that of aspring-mass system (equation (4.1)):

mxn + dxn + cnxn = Pn(t), (8.17)

78 Random Vibrations of a tensioned string

where:cn = n2π2 Tr

L2, (8.18)

Pn(t) =2L

∫ L

0F (z, t) sin

(nπzL

)dz. (8.19)

We see that by expanding the solution in terms of eigenfunctions of the ten-sioned string, the amplitude of each eigenfunction can be described by theequation of a single degree of freedom spring-mass system. In chapters 4and 6, we have presented methods for analyzing the behavior of xn(t) in caseof random excitation.

An interesting question is how the overall response is dominated by one ofthe eigenfunctions or eigenmodes. One important factor is the axial shape ofthe excitation. If the axial shape of F (z, t) equals one of the eigenfunctions,say F (z, t) = gk(z)f(t), then Pn(t) will only have a value different from zeroif n = k. This follows directly from property (8.13)-(8.14). Furthermore, ifF (z, t) is constant with respect to z, F (z, t) = F (t), then:

Pn(t) =4nπ

F (t) for n odd, (8.20)

Pn(t) = 0 for n even. (8.21)

Another factor which can lead to domination of one eigenmode are the valuesof the natural frequency of the string:

ωn0 =(cnm

)1/2=nπ

L

(Trm

)1/2

, n = 1, 2, 3, ...,∞. (8.22)

If one particular natural frequency, say ωk, has a value which is in the centerof the power density spectrum of excitation and the others (ωk−1, ωk+1, etc.)have values where energies of excitation are much less, the standard deviationof xk(t) will dominate over the others. This follows directly from the resultspresented in chapters 4 and 6. In other words: as designers we can influenceresponse with the value of the eigenfrequencies which depend on, amongstothers, the amount of pre-tension which is applied.

Chapter 9

Random vibrations of a beam

As was shown in chapter 7, whenever D1 1, bending forces dominate overtension forces in differential equation (7.5). Disregarding the tension andexternal forces, we have:

EI∂4x

∂z4+m

∂2x

∂t2+ d

∂x

∂t= 0 (9.1)

The boundary conditions at z = 0 are:

x(0, t) = 0 and EI∂2x

∂z2= 0, (9.2)

and at z = L:

x(L, t) = V (t) and EI∂2x

∂z2= 0. (9.3)

The second boundary conditions in eqs. (9.2) and (9.3) describe the situationof zero bending stress apparent in case of frictionless ball-joints or hinges.Condition (9.3a) describes prescribed random motion at the top (for examplethe motion of a floating structure to which the beam is connected via a hinge).

9.1 Quasi-static response

Assume for the time-being that response is quasi-static, denoted by xp. Inthat case, we can disregard all dynamical terms in equation (9.1), so that:

EI∂4xp∂z4

= 0. (9.4)

Integrating this expression twice gives:

∂2xp∂z2

= c1z + c2. (9.5)

80 Random vibrations of a beam

The constants c1 and c2 are determined using the following boundary condi-tions:

EI∂2xp∂z2

= 0 at z = 0, (9.6)

EI∂2xp∂z2

= 0 at z = L, (9.7)

which yields c1 = c2 = 0. Hence:

∂2xp∂z2

= 0 (9.8)

Again, we integrate this equation twice:

xp(z) = c3z + c4 (9.9)

Apply the two remaining boundary conditions:

xp = 0 at z = 0 (9.10)

xp = V at z = L (9.11)

which yields c3 = V/L and c4 = 0, so that:

xp(z) = V · zL. (9.12)

The total solution of equation (9.1) is written as:

x = xp + xh. (9.13)

Substituting this in equation (9.1) and using equation (9.12) yields for xh:

EI∂4xh∂z4

+m∂2xh∂t2

+ d∂xh∂t

= − z

L

(mV + dV

). (9.14)

The boundary conditions for this problem at z = 0 as well as z = L are:

xh =∂2xh∂z2

= 0. (9.15)

In conclusion, by adding the solution for quasi-static response we have trans-formed our problem from a homogeneous PDE with inhomogeneous bound-ary conditions (equations (9.2)-(9.3)) to a PDE with prescribed excitation atthe right-hand side and homogeneous boundary conditions: equations (9.14)-(9.15). This problem is similar to that treated in the previous chapter, exceptthat we now have a fourth-order spatial derivative representing bending stiff-ness instead of a second order spatial derivative representing stiffness due totensioning.

9.2 Separation of variables 81

9.2 Separation of variables

To apply separation of variables, drop the damping and excitation terms:

EI∂4xh∂z4

+m∂2xh∂t2

= 0 (9.16)

Now write x(z, t) as a product of functions g(z) and h(t):

x(z, t) = g(z)h(t) (9.17)

Differential equation (9.16) then becomes:

EI1g(z)

∂4g(z)∂z4

= −m 1h(t)

∂2h(t)∂t2

(9.18)

This equation can only be valid if both functions are equal to a constant +λ.This results in the following ordinary differential equations

d4g(z)dz4

=λ

EIg(z) (9.19)

d2h(t)dt2

=−λmh(t) (9.20)

In order to obtain the eigenfunctions we solve equation (9.19). Therefore, wechoose eαz as basic solution for the differential equation. It then follows thatα2 = ±√λ/EI and we end up with the following four values for α:

α1 =(λ

EI

) 14

α2 = −(λ

EI

) 14

α3 = i

(λ

EI

) 14

α4 = −i(λ

EI

) 14

(9.21)The solution of g(z) can then be written as:

g(z) = A1eα1z +A2e

α2z +A3 cos

[(λ

EI

) 14

z

]+A4 sin

[(λ

EI

) 14

z

](9.22)

Applying 3 out of the 4 boundary conditions, we find

g(z) = A4 sin((λ

EI

) 14

z) (9.23)

Implementing the boundary condition g = 0 at z = L, one gets(λEI

)1/4L = nπ

with n an integer. The eigenvalues thus are:

λn = EI

(n4π4

L4

), n = 1, 2, 3, ...,∞. (9.24)


The associated eigenfunctions are:

gn(z) = sin(nπzL

)(9.25)

9.3 Eigenfunction expansion

If we now add the excitation and damping force, we can solve eq. (9.14) byexpansion in eigenfunctions:

EI∂4x

∂z4+m

∂2x

∂t2+ d

∂x

∂t= − z

L

(mV + dV

). (9.26)

Substitute x(z, t) =∑∞

n=1 xn(t)gn(z) into equation (9.26) to obtain:

∞∑n=1

[EI

n4π4

L4xn sin

(nπzL

)+(md2xndt2

+ ddxndt

)sin

(nπzL

)]=

− z

L

(mV + dV

). (9.27)

As in the previous chapter, multiply the left- and right-hand side of this equa-tion with gk(z) and integrate over z from z = 0 to z = L. Using the orthogo-nality property

∫ L

0gn(z)gk(z)dz = 0 for k = n (9.28)

∫ L

0gn(z)gk(z)dz =

L

2for k = n (9.29)

(9.30)

we have:

EIn4π4

L4xn +m

d2xndt2

+ ddxndt

= − 2L

(mV + dV

)∫ L

0

z

Lsin

(nπzL

)dz. (9.31)

9.4 The general problem of the tensioned beam 83

The integral can be evaluated as:

1L

∫ L

0

z

Lsin

(nπzL

)dz =

∫ 1

0x sin(nπx)dx

= − 1nπ

∫ 1

0xd[cos(nπx)]

= −x cos(nπx)nπ

∣∣∣∣1

0

+1nπ

∫ 1

0cos(nπx)dx

= −cos(nπ)nπ

+1

(nπ)2sin(nπx)

∣∣∣∣1

0

= −cos(nπ)nπ

=(−1)n+1

nπ(9.32)

so that equation (9.31) becomes

EIn4π4

L4xn +m

d2xndt2

+ ddxndt

=2(−1)n

nπ

(mV + dV

). (9.33)

Each amplitude of an eigenfunction is thus described according to an ordinarymass-spring system with random excitation. The eigenfrequencies are:

ωn0 =n2π2

L2

√EI

m. (9.34)

Note the difference with the eigenfrequencies of the tensioned string that aregiven in equation (8.22). Equation (9.33) can be solved further according tothe methods presented in chapters 4 and 6.

9.4 The general problem of the tensioned beam

The general problem of the tensioned beam as described by equation (7.5)and boundary conditions (7.3)-(7.4) can be solved by combining the methodspresented in chapter 6 and the previous sections of the present chapter:

1. Derive a quasi-static solution associated with the prescribed motion atthe top:

EI∂4xp∂z4

− ∂

∂z

(Tr(z)

∂xp∂z

)= 0 (9.35)

subject to the boundary conditions as given in equations (9.2)-(9.3). Thesolution will be of the form:

xp = f(z)V (t). (9.36)


2. The total solution is:x = xp + xh (9.37)

where xh is a solution of the problem

EI∂4xh∂z4

− ∂

∂z

(Tr(z)

∂xh∂z

)+m

∂2xh∂t2

+d∂xh∂t

= F (z, t)−f(z)(mV + dV

)

(9.38)subject to the homogeneous boundary conditions:

xh =∂2xh∂z2

= 0 at z = 0 and z = L. (9.39)

3. Expand xh in eigenfunctions En(z) with eigenvalues λn which are thesolutions of

EId4Endz4

− d

dz

(Tr(z)

dEndz

)− λnEn = 0 (9.40)

subject to the four boundary conditions:

En =d2Endz2

= 0 at z = 0 and z = L (9.41)

4. The solutions of equation (9.38) are an infinite series of eigenfunctions,n = 1, 2, 3, ...,∞ with prescribed values for λn, n = 1, 2, 3, ...,∞; theeigenvalues. The solution for xh can be written as:

xh =∞∑n=1

xn(t)En(z). (9.42)

The amplitude of each eigenfunction is described by an equation which issimilar to that of the single degree spring-mass system. It is obtained bysubstituting equation (9.42) into (9.38), multiplying the left- and right-hand side by the adjoined eigenfunction Ek(z), integrating with respectto z from z = 0 to z = L and using the orthogonality property:

∫ L

0En(z)Ek(z)dz = 0 for k = n (9.43)

∫ L

0En(z)Ek(z)dz = αn for k = n (9.44)

Note that Ek(z) is not necessarily the same as En(z). Methods forderiving Ek or its governing adjoined differential equation can be foundin Courant & Hilbert [7].

9.4 The general problem of the tensioned beam 85

5. The equation for each amplitude of the eigenfunction is now obtainedby substituting equation (9.42) in PDE (9.38), multiplying left and righthand with Ek(z) and integrating with respect to z from z = 0 to z = L:

∫ L

0

∞∑n=1

xn(t)

(EI

d4En(z)dz4

− d

dzTr(z)

dEn(z)dz

)Ek(z)

+(md2xndt2

+ ddxndt

)En(z)Ek(z)

dz

=∫ L

0F (z, t)Ek(z)dz −

(mV + dV

)∫ L

0f(z)Ek(z)dz (9.45)

Substituting equation (9.40) into the first term, using the orthogonalityproperty (9.43)-(9.44) and dividing all terms by αn, we have for eachamplitude of the eigenfunctions the ordinary spring-mass type problem.

λnxn +md2xndt2

+ ddxndt

= Pn(t), (9.46)

where

Pn(t) = α−1n

∫ L

0F (z, t)En(z)dz − α−1

n

(mV + dV

)∫ L

0f(z)En(z)dz.

(9.47)The natural frequencies are ωn0 = (λn/m)1/2, n = 1, 2, 3, ...,∞. Thisproblem can be further handled in the usual manner.

Chapter 10

Peak and extreme statistics

Given a Gaussian random signal of the variable x(t), the probability densityof all possible values of x is given by a Gaussian distribution. The Gaussiandistribution is determined by the mean value and standard deviations, seeequation (1.7). The previously presented methods enable specification of thestandard deviation of x(t); its mean value (if relevant) can be calculated bystandard methods of static structural analysis. So the probability distributionof instantaneous values of x, that is the percentage of time that x(t) lies be-tween certain values, is fully specified. Given a record of random fluctuationsx(t) in time (see figure 10.1), however, the statistics of the peak values of x(t)are different from those of the instantaneous values. Furthermore, the extremevalue of x(t), i.e. the value of the largest peak in a given record of certainduration is subject to uncertainty; it will be different for different realizationsof the process. Also, the statistics of extreme values will be different fromthose of Gaussianly distributed instantaneous values. Peak values are impor-tant for assessing fatigue damage in mechanical structures; extreme values areimportant for analyzing the possibility of yield. This means that there is agreat need in deriving expressions for these statistics in mechanical engineeringpractice. Deriving these expressions is the objective of this chapter.

10.1 Peak statistics

The simplest case for determining the statistics of peaks is that of narrow-bandresponse. That is, response which is governed by a narrow-band of frequen-cies as, for example, is the case for resonant response: sections 4.5 and 6.3.Response can then be described by a sinusoidal wave of which the amplitudevaries slowly and randomly with time. Clearly, the statistics of peaks in sucha signal are those of the amplitude: see figure 6.3. To derive its probability

88 Peak and extreme statistics

peakshighest peak = extreme

time

x(t)

Figure 10.1. Peaks and extreme of a random record.

distribution, we write (see also equation (6.19)):

x = ρ(T ∗) sin t∗ + ψ(T ∗) , (10.1)

where T ∗ is ’slow’ time defined as:

T ∗ = δt∗, (10.2)

with δ 1. For the time derivative, we have:

x = ρ(T ∗) cos t∗ + ψ(T ∗) + δdρ(T ∗)dT ∗ sin t∗ + ψ(T ∗)

+ δdψ(T ∗)dT ∗ ρ(T ∗) cos t∗ + ψ(T ∗) . (10.3)

Because δ 1 we can approximate x with a relative error of O(δ) by:

x = ρ(T ∗) cos t∗ + ψ(T ∗) . (10.4)

10.1 Peak statistics 89

Response x and its time derivative are uncorrelated, i.e. their covariance iszero, because for stationary processes we can write:

〈xx〉 = xx = limT→∞

1T

∫ T

0x(t∗)x(t∗)dt∗

= limT→∞

1T

∫ T

0xdx

= limT→∞

1T

(12x2(t∗)

)T0

= 0. (10.5)

The standard deviation of x is equal to that of x. This can be seen as fol-lows. The standard deviation of x is related to the power density of x by (seeequation (3.7)):

σ2x =

∫ ∞

0Sxdω. (10.6)

According to the theory of chapters 3 and 4, we have for the power density ofthe time derivative of x:

Sx = ω2Sx, (10.7)

so that the standard deviation of x is given by

σ2x =

∫ ∞

0ω2Sxdω. (10.8)

In case of lightly damped (δ 1) resonant response, the power density of xpeaks sharply in a narrow band of frequencies near the natural frequency. Asω is made dimensionless with the natural frequency, Sx will peak at ω = 1.So the value of the integral in equation (10.8) is mainly determined by valuesof Sx at and very close to ω = 1. Therefore, for δ 1,

σ2x =

∫ ∞

0Sxdω = σ2

x. (10.9)

The joint Gaussian distribution of x and x can now be expressed as (cf. equa-tion (1.11)):

p(x, x) =1

2πσ2x

e−x2+x2

2σ2x . (10.10)

The value of ρ and ψ, 0 ≤ ρ ≤ ∞, 0 ≤ ψ ≤ 2π is uniquely related to that ofx and x, −∞ ≤ x ≤ ∞, −∞ ≤ x ≤ ∞. That is, for any particular value ofx and x there is only one corresponding value for ρ and ψ, and vice versa. In


this case, the joint probability that X ≤ x and X ≤ x is equal to that of ρ ≤ρ

and Ψ ≤ ψ:

P (ρ, ψ) = P (x, x)∣∣∣∣x=x(ρ,ψ),x=x(ρ,ψ)

(10.11)

Cumulative probability is related to probability density as:

p(ρ, ψ) =∂2P (ρ, ψ)∂ρ∂ψ

, p(x, x) =∂2P (x, x)∂x∂x

. (10.12)

so that we can derive from (10.11) the following relation for probability den-sities:

p(ρ, ψ) =p(x, x)

∣∣∣∣dxdxdρdψ

∣∣∣∣x=x(ρ,ψ),x=x(ρ,ψ)

. (10.13)

The Jacobian can be evaluated as:∣∣∣∣dxdxdρdψ

∣∣∣∣ =

∣∣∣∣∣∂x∂ρ

∂x∂ψ

∂x∂ρ

∂x∂ψ

∣∣∣∣∣ =∣∣∣∣∂x∂ρ

∂x

∂ψ− ∂x

∂ψ

∂x

∂ρ

∣∣∣∣= ρ

[cos2(t∗ + ψ) + sin2(t∗ + ψ)

]= ρ. (10.14)

Noting that x2 + x2 = ρ2, the joint distribution of ρ and ψ now becomes

p(ρ, ψ) =ρ

2πσ2x

e−ρ2/(2σ2

x). (10.15)

This can also be written as the product of two independent probability densi-ties of amplitude and phase:

p(ρ, ψ) = p(ρ)p(ψ). (10.16)

wherep(ρ) =

ρ

σ2x

e−ρ2/(2σ2

x) 0 ≤ ρ ≤ ∞. (10.17)

andp(ψ) =

12π

0 ≤ ψ ≤ 2π. (10.18)

Note that ∫ ∞

0p(ρ)dρ = 1 and

∫ 2π

0p(ψ)dψ = 1 : (10.19)

the total surface underneath the probability density must be equal to one asshould be (because the cumulative probability must become equal to one atthe upper boundary of the domain, see equation (1.1)).

10.1 Peak statistics 91

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4σ

xp

ρ/σx

Figure 10.2. The Rayleigh distribution: equation (10.17).

The probability density according to equation (10.17) is known as theRayleigh distribution. It is shown in figure 10.2.

The above shown distribution for peaks applies to a narrow-band processas is the case in, for example, resonant response. For more general non-narrow-band processes, the peak distribution is more complicated. In fact, itcan then be described by Rice’s distribution: see Robson [1]. The high-tailsend of this distribution, that is the probability density of the high values ofρ in Rice’s distribution, can be conservatively approximated by the Rayleighdistribution [1]. Therefore, the Rayleigh distribution represents a conservativeestimate to the real peak distribution. In this sense it can be used to estimatefatigue damage due to a large number of stress peaks apparent in a mechanicalstructure.


10.2 Fatigue damage

The cumulative damage which occurs at some place in a mechanical engineer-ing structure due to cyclic stresses can be assessed using Miner’s hypothesis. Itimplies that the total damage equals the sum of the damage due to each indi-vidual stress cycle. The damage di due to a single cycle with stress amplitudeρi is given by

di = βρλi , (10.20)

where β and λ are material constants. Their values are assessed from cyclicfatigue tests. The total damage is:

d =n∑i=1

βρλi . (10.21)

To avoid failure during lifetime,

d < 1. (10.22)

The coefficient λ, λ > 1, in equation (10.21) reflects the fact that larger stresspeaks contribute more than smaller ones. For welds, λ can have values ofabout 4: American Welding Society. In that case, a stress amplitude whichis two times larger leads to an accumulated damage which is 16 times larger,implying that the total number of cycles before failure will be 16 times smaller.

In our problem, the stress cycle is random in nature. We can write forlarge numbers of cycles n:

d = βn∑i=1

ρλi = βn

(1n

n∑i=1

ρλi

)

= βn〈ρλ〉= βn

∫ ∞

0ρλp(ρ)dρ, (10.23)

where p(ρ) is probability density of peaks. The number of cycles can becalculated from the average frequency of the random stress record:

ω2 =

∫∞0 ω2Sdω∫∞

0 Sdω, (10.24)

where S is the power density of the stress record. The number of cycles nowis:

n =T

2π/ω, (10.25)

10.2 Fatigue damage 93

where 2π/ω is average time between peaks and T is total time.The damage can now be calculated once we know the probability density

of peaks. Because λ is relatively large, the contribution to the integral inequation (10.23) will mainly come from large values of ρ, i.e. the large tails endof p(ρ). For a Gaussian random process, this part of the peak distribution canbe conservatively described by Rayleigh’s distribution: cf. equation (10.17).In this case, ∫ ∞

0ρλp(ρ)dρ =

∫ ∞

0

ρλ+1

σ2e−ρ

2/(2σ2)dρ. (10.26)

Taking η = ρ/σ

∫ ∞

0

ρλ+1

σ2e−ρ

2/(2σ2)dρ = σλ∫ ∞

0ηλ+1eη

2/2dη. (10.27)

Taking t = (1/2)η2:

σλ∫ ∞

0ηλ+1e1/2η

2dη = σλ2λ/2

∫ ∞

0tλ/2e−tdt

= σλ2λ/2Γ(1 + λ/2) (10.28)

where Γ is the Gamma function which can be found in Abramowitz & Ste-gun [6], chapter 6. The fatigue damage then becomes:

d = βnσλ2λ/2Γ(1 + λ/2). (10.29)

For λ = 4, Γ(1 + λ/2) = 2 in which case d = 8βnσ4.The above relations can also be used to calculate the amplitude of an

equivalent deterministic sinusoidal stress cycle: that is, a sinusoidal stresscycle which leads to the same fatigue damage as that caused by the randomcycle. The equivalent amplitude ρeq follows from the equality

βnρλeq = βn

∫ ∞

0ρλp(ρ)dρ, (10.30)

which can be evaluated to

ρeq =(∫ ∞

0ρλp(ρ)dρ

)1/λ

. (10.31)

Substituting the expression for the integral given by eqs. (10.26)-(10.28), onehas

ρeq = σ√

2 (Γ(1 + λ/2))1/λ . (10.32)

For λ = 4, this becomes ρeq = 1.68σ.


10.3 Extreme statistics

Consider the probability P (ρ) that any of the peaks out of a time record havingn peaks will be less than a particular value ρ. The probability that all of then peaks are less than ρ is then (see also Longuet-Higgins [9]):

Pρ(ρ) = Pn(ρ)∣∣∣∣ρ=ρ

(10.33)

where P (ρ) is the CDF of peaks:

P (ρ) =∫ ρ

0p(ρ)dρ. (10.34)

As shown in section 10.1, for narrow-band processes peaks are described ac-cording to the Rayleigh distribution (which represents, as mentioned before,also a conservative estimate to the distribution of the larger peaks in case ofmore general non-narrow-band Gaussian processes):

P (ρ) = 1 − e−ρ2/(2σ2), (10.35)

so thatPρ(ρ) =

(1 − e−ρ

2/(2σ2))n

(10.36)

The PDF of extremes is then

pρ(ρ) =dPρdρ

= n

(1 − e−

ρ2

2σ2

)n−1 ρ

σ2e−ρ

2/(2σ2) (10.37)

In figure 10.3 we have shown the density of extremes for number of peaks n =103 and n = 104. The density shifts to higher values when n increases, as tobe expected! Knowing the duration of a stationary state of random excitation,i.e. the number of peaks n, we can quantify the possible values of the largestpeak which are some multiple of the standard deviation. This enables us toestimate the likelihood of overstressing leading to yield at sensitive places ina mechanical engineering structure. In off-shore structures, one considers ingeneral the so-called 100-years storm: a stationary sea-state lasting about 3hours and having a probability of occurring once in 100 years. The numberof peaks in such a period is typically 103. Methods of random analysis aspresented in previous sections are used to calculate the standard deviationof stress during such a sea-state. The statistics of extremes are subsequentlyused to analyze the possibilities of overstressing.

10.3 Extreme statistics 95

2 2.5 3 3.5 4 4.5 5 5.5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

ρ/σx

pρ(ρ)

n=104n=103

^

^

^

Figure 10.3. Probability density of the extreme peak normalized with standarddeviation for different values of n (number of peaks).

Chapter 11

Non-linear analysis: Somegeneral observations andnumerical time-domainsimulation

The methods of analysis presented in the previous chapters were concernedwith linear systems subject to Gaussian random excitation. In case of linearsystems subject to Gaussian excitation, response is also Gaussian. Governingprobability distributions of response variables are thus known; all that is nec-essary is specification of the parameters of these distributions, e.g. standarddeviations, using the methods presented in the previous chapters. If the sys-tem is non-linear, however, the situation is rather different. The probabilitydistributions of instantaneous response become non-Gaussian, peak statisticsbecome non-Rayleigh or non-Rice, etc. For each particular non-linear system,the probability distributions as well as their parameters have to be determined.

A crude method to handle non-linear systems is by numerical time-domainsimulation. It is in fact straightforward application of the results presented inchapter 5. Gaussian excitation is generated by a series of sinusoidal waves withrandomly chosen phase angle: equation (5.1). The differential equation whichdescribes the system to which the excitation is applied is solved numericallyin the time-domain (and space-domain if necessary). Statistical parameters ofresponse can be derived from the generated random record using previouslypresented methods. Ensuring that instationary transients are removed fromthe time-record, the probability distribution of instantaneous response can beassessed according to the procedure presented in figure 2.3. Moments of re-

98

Non-linear analysis: Some general observations and numerical

time-domain simulation

sponse can be calculated according to equation (2.4). By generating a largeseries of different realizations, statistical parameters can also be gathered fromensemble averaging: see figure 2.2. But in all these cases, the numerical ap-proach is lengthy and time-consuming; in particular, if one wants to achieveaccuracy in the assessment of the tails of the distributions, such tails being ofprime importance for the evaluation of fatigue damage and extreme response.Moreover, numerical results yield only data; obtaining insight into the effectof design variables on design criteria from these data is rather cumbersome.Analytical methods for handling random processes are therefore the preferredway for the designer. But such methods only exist for specific non-linear prob-lems. Some of these that are particularly important for mechanical engineeringpractice, will be considered in the next chapters.

Chapter 12

Non-linear quasi-staticresponse

Whenever there is a direct non-linear relation between two stochastic variables,the statistical relationships between these two variables can be derived usingwell-known methods of probability theory. With direct relationship we meana relation without time-delay. In dynamical systems such as the spring-masssystem of chapters 4 and 6 and the tensioned beam of chapters 7, 8 and 9,a direct relationship will occur when the inertia and damping forces play aminor role. This will happen if the center of gravity of the power densityspectrum of excitation is at frequencies which are much lower than the naturalfrequency of the system. Response is then primarily governed by excitationforces and static restoring forces. The restoring forces can be non-linear as aresult of non-linear, non-elastic behavior of the structure. But non-linearitycan also occur through the excitation force. An example of the latter is thenon-linear drag force acting on members of off-shore structures due to watervelocities associated with random waves on the sea surface. These drag forcesare particularly important in case of large waves acting on relatively smallmembers of off-shore structures. We shall treat this problem in some detail asan example of how to proceed in case of quasi-static non-linear behavior.

The drag force acting on a bluff body in fluid is proportional to the squareof the velocity v of the fluid perpendicular to the body. In case of alternatingfluid velocity, the force will be proportional to v|v|. For a linearly quasi-statically responding structure, the response variable will also be proportionalto v|v|. Therefore,

x = αv|v|, (12.1)

where x is response variance and α constant of proportionality; the value of α

100 Non-linear quasi-static response

can be calculated by applying static analysis under unit force. The velocity ofthe fluid due to wind-generated waves can, in general, be well-presented by aGaussian distribution. To derive the distribution of x given relation (12.1) wenote that there exists a one-to-one relationship between x and v: for each valueof v there is only one value of x and vise versa. In this case, the probabilitythat X < x is equal to the probability that V < v(x):

P (x) = Pv(v)

∣∣∣∣∣v=v(x)

, (12.2)

where v = v(x) follows from the inverse of relationship (12.1).

v = (α|x|)−1/2 x, (12.3)

(to derive this relation consider first v > 0, x > 0 and subsequently v < 0, x <0 and combine the results). Cumulative probabilities are related to probabilitydensities as

pv(v) =dPv(v)dv

, p(x) =dP (x)dx

, (12.4)

so that we can derive from equation (12.2):

p(x) =pv(v)

∣∣∣∣dvdx∣∣∣∣v=v(x)

. (12.5)

For pv(v) we take the Gaussian distribution

pv(v) =1

σv√

2πe−v

2/(2σ2v). (12.6)

Furthermore, ∣∣∣∣dvdx∣∣∣∣ =

12

(α|x|)−1/2 . (12.7)

Relation (12.4) can now be evaluated to the following probability density ofx:

p(x) =1

2σv(2πα|x|)1/2e−|x|/(2ασ2

v). (12.8)

The standard deviation σx of x follows from:

σ2x =

∫ +∞

−∞x2p(x)dx =

(2α)2σ4v

π1/2

∫ ∞

0t3/2e−tdt =

(2α)2σ4v

π1/2Γ(

52

)= 3α2σ4

v .

(12.9)

101

where we took x = 2ασ2vt. In (12.9) Γ is the Gamma function: Abramowitz

& Stegun [6], chapter 6. In terms of σx, equation (12.8) becomes:

p(x) =(

8πσx|x|√3

)−1/2

e−|x|√3/(2σx). (12.10)

For the cumulative probability density we can write:

P (x) =∫ x

−∞p(x)dx =

12√π

∫ t=√

3x/(2σx)

−∞|t|−1/2e−|t|dt, (12.11)

with x = 2σxt/√

3. The integral on the right-hand side can also be expressed interms of the incomplete Gamma function γ(x1, x2): Abramowitz & Stegun [6],chapter 6:

P (x) =12

+1

2√π

|x|xγ

(12,12

√3|x|σx

). (12.12)

In fact, this result could have been directly obtained by substituting eq. (12.3)in eq. (12.2) using eq. (1.10) and noting that the incomplete Gamma functioncan be expressed in the error function: Abramowitz & Stegun [6], property6.5.16. In figure 12.1, the above distribution has been shown and comparedwith the Gaussian distribution. It is seen that for large values of x/σx, theprobability of not being exceeded is less in case of the above distribution.It reflects a larger probability of obtaining large values at equal standarddeviation due to quadratic loading.

Given a non-linear direct relationship between wave loading and response,it is also possible to derive descriptions for the probability distributions ofpeaks invoking the narrow-band model. In this way it is possible to deriveanalytical expressions for fatigue damage and extreme statistics in case ofnon-linear quasi-static behavior: Brouwers and Verbeek [11]. This approachcan be applied to many other non-linear problems as long as the non-linearbehavior is quasi-static.

102 Non-linear quasi-static response

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Gaussian CDFDrag loading CDF

Figure 12.1. CDF for Gaussian process and for drag loading.

Chapter 13

Non-linearly dampedresonance

As mentioned before, analytical methods for handling non-linear problems ofrandom vibration are only available for specific cases. One of these casesis non-linearly damped resonance of a mass-spring system. In case of lightdamping, the governing equation can be reduced using two-scale expansions.The resulting stochastic problem can be treated by a Fokker-Planck equation.

13.1 Two-scale expansions

Attention is focussed to a mass-spring system with non-linear power-law damp-ing: i.e. the damping force equals dx|x|α. In dimensionless form applying thenon-dimensioning of eqs. (6.2)-(6.5), the equation of motion is:

x+ δx|x|α + x = P ∗(t∗), (13.1)

which becomes equal to eq. (6.2) for α = 0 (linear damping). The excitationis a stationary Gaussian process described according to equation (6.1). When-ever δ 1, damping will be unimportant in solutions of eq. (13.1) except forfrequencies close to the natural frequency ω∗ = 1. This becomes apparentwhen disregarding damping and analyzing the behavior of the solution of theresulting equation in the frequency domain: see section 4.5. Therefore, whenfocussing on the solution appropriate for resonance, we can limit attention tothe behavior of the excitation in a narrow band of frequencies around ω∗ = 1.This can be modeled as:

ω∗ = 1 + δβν∗ , ν∗ = O(1). (13.2)

104 Non-linearly damped resonance

The exponent β, β > 0, is 1 in case of linear damping; its value appropriatefor non-linear damping will be determined below. The excitation describedby eq. (6.1) can now be approximated by (analogous to the procedure ofeqs. (6.16)-(6.25)):

P (t∗) = 2δβ/2S1/21

+N−N0∑m=−N0

(∆ν∗

2

)1/2

cos(t∗ + ν∗mT∗ + ϕm), (13.3)

where

T ∗ = δβt∗ , ν∗m = m∆ν∗ , S1 = Snp (ω∗ = 1). (13.4)

Eq. (13.3) can also be written as (see again eqs. (6.16)-(6.25))

P (t∗) = 2δβ/2S1/21 (f1(T ∗) cos t∗ + f2(T ∗) sin t∗) , (13.5)

where

f1(T ∗) =N∑n=1

(∆ν∗)1/2 cos(ν∗nT∗ + ϕn), (13.6)

f2(T ∗) = −N∑n=1

(∆ν∗)1/2 sin(ν∗nT∗ + ϕn). (13.7)

Here, f1(T ∗) and f2(T ∗) are Gaussian white-noise processes of power density1/2 (compare the above equations with eq. (5.1)). Furthermore, f1(T ∗) andf2(T ∗) are uncorrelated:

〈f1f2〉 = f1f2 = limT→∞

1T

∫ +T/2

−T/2f1f2dT

∗ = 0 : (13.8)

the prove of this is similar to the procedure of eqs. (6.28)-(6.30).The excitation consists of sinusoidal waves with randomly and slowly vary-

ing amplitudes: cf. eqs. (13.5)-(13.7). In view of the two times t∗ and T ∗ ap-parent in the excitation, we also assume that the solution of eq. (13.1) exhibitsthe two times t∗ and T ∗:

x = x(t∗, T ∗), (13.9)

so thatx =

∂x

∂t∗+ δβ

∂x

∂T ∗ , (13.10)

x =∂2x

∂t∗2+ 2δβ

∂2x

∂t∗∂T ∗ + O(δ2β). (13.11)

13.1 Two-scale expansions 105

Substituting eqs. (13.9)-(13.11) and (13.5) into (13.1) yields:

∂2x

∂t∗2+ x+ 2δβ

∂2x

∂t∗∂T ∗ + δ∂x

∂t∗

∣∣∣∣ ∂x∂t∗∣∣∣∣α

= 2δβ/2S1/21 (f1(T ∗) cos t∗ + f2(T ∗) sin t∗)

+O(δ2β) + O(δ1+β) + O(δ1+αβ). (13.12)

For small δ we can disregard all terms except the first two:

∂2x

∂t∗2+ x = 0 if δ → 0, (13.13)

the solution of which is:

x = δγa(T ∗) cos(t∗ + φ(T ∗)), (13.14)

where amplitude and phase can vary with the second and slow time T ∗. Thevalue of the exponent γ has yet to be determined.

A problem like eq. (13.12) gives rise to supposing a solution for x in termsof a perturbation expansion in powers of δ as:

x = δγ(x1 + δµx2 + ...) , µ > 0 (13.15)

Equations for x1, x2, etc. are obtained by substituting expansion (13.15) ineq. (13.12) and equating terms of equal order in δ. The largest or leadingorder are the terms formed by the first two terms in eq. (13.12) expressed inx1:

∂2x1

∂t∗2+ x1 = 0, (13.16)

which is of course the problem formed by eq. (13.13) with solution: x1 =a(T ∗) cos(t∗ + φ(T ∗)). The next order of terms lead to the problem:

∂2x2

∂t∗2+ x2 = −2δβ−µ

∂2x1

∂t∗∂T ∗ − δ1+αγ−µ∂x1

∂t∗

∣∣∣∣∂x1

∂t∗

∣∣∣∣α

+2δβ/2−γ−µS1/21 (f1(T ∗) cos t∗ + f2(T ∗) sin t∗) (13.17)

Now, meaningful solutions are only obtained if the terms on the right handside are all of order δ0. Hence: β−µ = 1 +αγ−µ = β/2− γ−µ = 0, so that

µ =2

2 + α, β =

22 + α

, γ = − 12 + α

. (13.18)


In terms of these values and substituting the solution for x1, eq. (13.17) be-comes:

∂2x2

∂t∗2+ x2 = 2

∂

∂T ∗ (a sin(t∗ + φ)) + a1+α sin(t∗ + φ)| sin(t∗ + φ)|α

+2S1/21 (f1 cos t∗ + f2 sin t∗). (13.19)

Here, the term that is due to damping can be expanded in a Fourier series:

sin(t∗ + φ)| sin(t∗ + φ)|α = α1 sin(t∗ + φ) + α2 sin 2(t∗ + φ) + ... (13.20)

where

α1 =2π

∫ π

0sin2 η| sin η|αdη

=4π

∫ π/2

0sinα+2 ηdη

=2πB

(α+ 3

2,12

)

=2Γ((α+ 3)/2)

π1/2Γ((α+ 4)/2)(13.21)

where B is Beta function and Γ is Gamma function: Abramowitz & Stegun [6],chapter 6.

Implementing eq. (13.20) and using addition formulae for trigonometricfunctions, eq. (13.19) becomes

∂2x2

∂t∗2+ x2 =

2d

dT ∗ (a cosφ) + α1aα+1 cosφ+ 2S1/2

1 f2(T ∗)

sin t∗

+

2d

dT ∗ (a sinφ) + α1aα+1 sinφ+ 2S1/2

1 f1(T ∗)

cos t∗

+α2 cos 2φ sin 2t∗ + α2 sin 2φ cos 2t∗ + ... (13.22)

The basic solutions of the homogeneous part of the above differential equa-tion are sin t∗ and cos t∗. Terms like sin t∗ and cos t∗ on the right hand sideof eq. (13.22) then give rise to particular solutions of the type t∗ sin t∗ andt∗ cos t∗. These grow unboundedly with increasing t∗. To arrive at meaningfulsolutions, we have to avoid this behavior; that is, we have to set the coefficientsof the terms like sin t∗ and cos t∗ on the right-hand side of eq. (13.22) equal tozero. This criterion for avoiding secular behavior is at the heart of the two-scale method . It leads to specification of the equations governing amplitudeand phase:

2d

dT ∗ (a cosφ) + α1a1+α cosφ = −2S1/2

1 f2 (13.23)

13.2 Derivation of the Fokker-Planck equation 107

2d

dT ∗ (a sinφ) + α1a1+α sinφ = −2S1/2

1 f1 (13.24)

The above equations determine the amplitude and phase of solution (13.14).

13.2 Derivation of the Fokker-Planck equation

Eqs. (13.23) and (13.24) are two-coupled first order equations for amplitudeand phase with white-noise excitation. These equations can be transformedinto a so-called Fokker-Planck equation for the transient joint probability den-sity distribution of amplitude and phase. Advantage of this transformationis that the resulting Fokker-Planck equation is a linear partial differentialequation, in contrast to eqs. (13.23) and (13.24), which are non-linear forgeneral values of α. As we shall see later, the Fokker-Planck equation asso-ciated with eqs. (13.23)-(13.24) can be solved yielding explicit descriptions ofthe stationary probability distribution of amplitude and phase. The Fokker-Planck equation was initially derived in connection with models of randommotion of molecules. In later years, the approach has been extended to non-linear problems: e.g. Stratonovich [3] and Van Kampen [4]. Below, we shallpresent a heuristic derivation of the Fokker-Planck equation associated witheqs. (13.23)-(13.24). These two equations can be formulated as:

dudT ∗ = −g(u) + S

1/21 f(T ∗), (13.25)

whereu = (u1, u2) = (a cosφ, a sinφ), (13.26)

g(u) =12α1u(u2

1 + u22)α/2, (13.27)

f(T ∗) = −(f2(T ∗), f1(T ∗)), (13.28)

where f1 and f2 are white noise or δ-correlated processes.Objective is to derive an equation that describes the probability density

pu(u, T ∗) given equation (13.25). From (13.25) it follows that

u(T ∗ +∆T ∗) = u(T ∗)−g(u(T ∗ +∆T ∗))∆t+S1/21 F(∆T ∗)+o(∆T ∗), (13.29)

where o(∆T ∗) is used to indicate a rest term that goes to zero more rapidlythan ∆T ∗ if ∆T ∗ → 0. The function F(∆T ∗) is defined as

F(∆T ∗) =∫ T ∗+∆T ∗

T ∗f(τ)dτ. (13.30)


The white-noise is zero-mean. Hence,

〈F(∆T ∗)〉 = 0. (13.31)

Furthermore,

〈FmFn〉 =∫ T ∗+∆T ∗

T ∗dτ1

∫ T ∗+∆T ∗

T ∗dτ2〈fm(τ1)fn(τ2)〉

=π

2δmn

∫ T ∗+∆T ∗

T ∗dτ1

∫ T ∗+∆T ∗

T ∗dτ2δ(τ2 − τ1)

=π

2δmn

∫ ∆T ∗

0dτ1

∫ ∆T ∗

0dτ2δ(τ2 − τ1)

=π

2δmn

∫ ∆T ∗

0dτ1

∫ −τ1+∆T ∗

−τ1δ(τ)dτ

=π

2δmn∆T ∗, (13.32)

where we used the property that f1 and f2 are mutually uncorrelated whitenoise processes of power density 1/2, in which case the autocorrelation func-tions of f1 and f2 are Dirac δ-functions with intensity π

2 : see eq. (3.14). Fur-thermore, we note that higher order moments of Fm are o(∆T ∗).

Equation (13.29) provides a relation between the statistical values of u(T ∗+∆T ∗) denoted by u′ and of u(T ∗) and F(∆T ∗) denoted by u′′ and F :

u′′ = u′ + g(u′)∆T ∗ − S1/21 F (13.33)

Using methods of transformations of random variables (Van Kampen, §I5 [4]),the following relation between the probability densities of u′ , u′′ and F de-noted by pu(u′, T ∗ + ∆T ∗), pu(u′′, T ∗) and pF (F,∆T ∗) can be given:

pu(u′, T ∗ + ∆T ∗)dVu′ =∫pF (F,∆T ∗)pu(u′′, T ∗)dVFdVu′′ (13.34)

where use was made of the property that F is statistically independent of u′′.Invoking (13.33) we can expand as

pu(u′′, T ∗) = pu(u′ + g(u′)∆T ∗ − S1/21 F, T ∗)

= pu(u′, T ∗) +∑ ∂pu(u′, T ∗)

∂u′k(gk(u′)∆T ∗ − S

1/21 Fk)

+12S1

∑ ∂2pu(u′)∂u′j∂u

′k

FjFk + o(∆T ∗). (13.35)

13.2 Derivation of the Fokker-Planck equation 109

Expanding the left-hand side of equation (13.34) for small ∆T ∗ and imple-menting (13.35) in the right-hand side, we have

pu(u′, T ∗) +∂pu(u′, T ∗)

∂T ∗ ∆T ∗ =dVu′′

dVu′

∫pF (F,∆T ∗)

pu(u′, T ∗)

+∑ ∂pu(u′, T ∗)

∂u′k(gk(u)∆T ∗ − S

1/21 Fk)

+12S1

∑ ∂2pu(u′, T ∗)∂u′j∂u

′k

FjFk

dVF + o(∆T ∗)

=∣∣∣∣du

′′

du′

∣∣∣∣pu(u′, T ∗) +

∑ ∂pu(u′, T ∗)∂u′k

gk(u)∆T ∗

+14πS1

∑ ∂2pu(u′, T ∗)∂u′2k

∆T ∗

+ o(∆T ∗). (13.36)

where we used equations (13.31)-(13.32) to evaluate moments of F. The Ja-cobian can be assessed as follows:∣∣∣∣du

′′

du′

∣∣∣∣ =

∣∣∣∣∣1 + ∂g1

∂u′1∆T ∗ ∂g1

∂u′2∆T ∗

∂g2∂u′1

∆T ∗ 1 + ∂g2∂u′2

∆T ∗

∣∣∣∣∣ = 1 +∑ ∂gk

∂u′k∆T ∗ + o(∆T ∗). (13.37)

Leaving out the primes and collecting terms like ∆T ∗, one finds from equa-tions (13.36)and (13.37):

∂pu(u, T ∗)∂T ∗ =

∑ ∂

∂uk(gk(u)pu(u, T ∗)) +

π

4S1

∑ ∂2pu(u, T ∗)∂u2

k

, (13.38)

which, in vector notation, reads as

∂pu∂T ∗ = · (gpu) +

π

4S1∆pu. (13.39)

The above equation is the Fokker-Planck equation associated with the fluc-tuation equation (13.25). It describes the evolution of the joint probabilitydensity of u1 and u2 as a function of time given some initial distribution atT ∗ = 0. For example, if u1 and u2 have fixed deterministic values u10 and u20

at time zero, the initial conditions for equation (13.39) are:

pu(u, T ∗) = δ(u1 − u10)δ(u2 − u20) at T ∗ = 0, (13.40)

where δ is Dirac delta function. The solution of Fokker-Planck equation (13.39)then describes the evolution of a fixed initial value to an uncertain future valuedescribed by a probability distribution.


13.3 Solutions for stationary response

If the solution of eq. (13.39) evolves to a stationary probability distribution,this probability distribution is described by:

· (gpu) +π

4S1∆pu = 0. (13.41)

In terms of the cylindrical variables a and φ related to u1 and u2 by eq. (13.27),this becomes

∂

∂a(gap) +

π

4S1

(∂

∂aa∂

∂a

(pa

)+

1a2

∂2p

∂φ2

)= 0 (13.42)

where

p = p(a, φ) = pu(u1, u2)∣∣∣∣du1du2

dadφ

∣∣∣∣ = apu(u1, u2) (13.43)

andga =

12α1a

1+α (13.44)

The solution of (13.42) can be written as the product of two independentprobability distributions of a and φ:

p = p(a)p(φ), (13.45)

wherep(φ) =

12π

, 0 ≤ φ ≤ 2π (13.46)

while p(a) satisfies

∂

∂a

(gap+

π

4S1a

∂

∂a

(pa

))= 0. (13.47)

Integrating once and requiring that p tends to zero exponentially fast as a→∞, we have

gap+π

4S1a

∂

∂a

(pa

)= 0. (13.48)

The solution of this equation is:

p = C0a exp(− 4πS1

∫ a

gada

)= C0a exp

(− 2α1a

α+2

π(α+ 2)S1

)(13.49)

where the integration constant follows from the requirement that∫∞0 pda = 1:

C−10 =

∫ ∞

0a exp

(− 2α1a

α+2

π(α+ 2)S1

)da. (13.50)

13.3 Solutions for stationary response 111

Substituting2α1a

α+2

π(α+ 2)S1= t (13.51)

eq. (13.50) can be evaluated to:

C−10 =

(π(α+ 2)S1

2α1

) 2α+2 1

α+ 2

∫ ∞

0t

2α+2

−1e−tdt

=(π(α+ 2)S1

2α1

) 2α+2 Γ(2/(α+ 2))

α+ 2(13.52)

With this result, we have obtained an expression in closed-form for the prob-ability distribution of amplitudes for non-linearly damped resonant response.

The mean-square response can be assessed as follows:

σ2x = lim

T→∞1T

∫ +T/2

−T/2x2(t)dt

= limT→∞

δ−2

2+α

T

∫ +T/2

−T/2a2(T ∗) sin2(t∗ + φ(T ∗))dt∗

= limT→∞

δ−2

2+α

(1π

∫ +π/2

−π/2sin2 ηdη

)1T

∫ +T/2

−T/2a2(T ∗)dT ∗

=12δ−

22+α

∫ ∞

0a2p(a)da, (13.53)

where we substituted eqs. (13.14) and (13.18) and made use of the propertythat the sinusoidal wave sin(t + φ(T ∗)) varies much more rapidly (if δ 1)than the random variables a and φ. Upon substituting the distribution foundfor p(a) (cf. eqs. (13.49) and (13.52)), we have

σ2x =

12δ−

22+αC0

∫ ∞

0a3 exp

(− 2α1a

α+2

π(α+ 2)S1

)da

=δ−

22+α

2(α+ 2)C0

(π(α+ 2)S1

2α1

) 4α+2

∫ ∞

0t

4α+2

−1e−tdt

=Γ(4/(α+ 2))2Γ(2/(α+ 2))

(π3/2(α+ 2)Γ((α+ 4)/2)SnP (ω∗ = 1)

4δΓ((α+ 3)/2)

) 2α+2

,(13.54)

where we also used the latter of eqs. (13.4). Eq. (13.54) describes the standarddeviation of response in case of non-linearly damped resonance. For the specialcase of linear damping, α = 0, the right-hand side of eq (13.54) becomes equal


to the previous result obtained by Fourier transform: cf. eq. (4.24) (as shouldbe).

The probability distribution of peak response, ρ = aδ−1/(2+α) (cf. eqs. (13.15)and (13.18)), normalized with standard deviation of instantaneous response,σx, now becomes

p(ρ) =(α+ 2)

2Γ[4/(α+ 2)]Γ2[2/(α+ 2)]

ρ

σ2x

exp−[

Γ[4/(α+ 2)]2Γ[2/(α+ 2)]

1/2 ρ

σx

]α+2

,

(13.55)In figure 13.1 we have shown plots of the probability density of amplitude orpeak value considering the cases α = −1 (Coulomb or dry friction), α = 0(linear damping) and α = 1 (quadratic damping). For α = 0, the solutionbecomes equal to the Rayleigh distribution (as should be). With increasingvalue of α, the probability density of larger values of ρ/σx becomes less inmagnitude: higher peak values are better damped with higher values of α!

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

ρ/σx

σ x p(ρ

)

α=0: linear dampingα=−1: dry frictionα=+1: quadratic damping

Figure 13.1. Probability density of peaks or amplitudes for a lightly damped reso-nant system with power law damping.

With the distribution of amplitudes, it becomes possible to derive ana-

13.3 Solutions for stationary response 113

lytical expressions for statistical parameters that govern non-linearly dampedresonant response of mechanical structures subject to random excitation. Anexample is lightly damped response of a mooring riser of a floating oil pro-duction system: Brouwers [12]. Here, response is governed by resonance inthe first natural mode. The amplitude of this mode can be described by anequation equivalent to a non-linearly damped mass-spring system. Using themethods of the previous sections, it is possible to assess parameters as expectedfatigue damage and expected extreme response under conditions of non-lineardamping.

To illustrate the possibilities to assess fatigue damage consider the ampli-tude of the equivalent deterministic stress cycle discussed in section 10.2: i.e.the sinusoidal cycle with deterministic amplitude ρeq described by eq. (10.31)which leads to the same fatigue damage as that caused by the random cyclewith probability according to eq. (13.55) above. Substituting eq. (13.55) intoeq. (10.31) one has

ρeq = σx

(α+ 2)

2Γ(4/(α+ 2))

[Γ(2/(α+ 2))]2

∫ ∞

0x1+λ

exp

(−[Γ(4/(α+ 2))x2

2Γ(2/(α+ 2))

]α+22

)dx

1/λ

. (13.56)

Implementing [Γ(4/(α+ 2))x2

2Γ(2/(α+ 2))

]α+22

= t (13.57)

we obtain

ρeq = σx

√2[Γ(2/(α+ 2))]1/2−1/λ

[Γ(4/(α+ 2))]1/2

∫ ∞

0t

λ+2α+2

−1e−tdt1/λ

= σx

[2Γ(2/(α+ 2))Γ(4/(α+ 2))

]1/2 [Γ[(λ+ 2)/(α+ 2)]Γ(2/(α+ 2))

]1/λ

, (13.58)

The right hand side of this equation has been calculated taking a power λ = 4in Miner’s law for fatigue damage: cf. eq. (10.20). Furthermore, dampingpowers were taken as α = −1, α = 0 and α = 1. Results have been plottedin table 13.1. It is seen that with increasing α, ρeq decreases. Because withincreasing power of damping, the probability density of higher peaks becomesless (figure 13.1), fatigue damage normalized with standard deviation decreasesas well.


Table 13.1. Amplitude of deterministic stress cycle leading to equivalent fatiguedamage for the case that λ = 4 in Miner’s law. Considered are Coulomb frictionα = −1, linear damping α = 0 and quadratic damping α = 1.

α ρeq/σx-1 1.910 1.68+1 1.57

The probability density distribution of the extreme peak in a record ofn peaks can be calculated according to eqs. (10.33)-(10.34) substituting forp(ρ) the distribution according to eq. (13.55). The result has been shown infigure 13.2 taking a record of 1000 peaks and considering α = −1, α = 0 andα = 1. The extreme value normalized with standard deviation shifts to lowervalues with increasing α.

13.4 Transient response behavior

So far, emphasis was laid on solutions of the Fokker-Planck equation for sta-tionary response. In this section, attention is focussed on non-stationary solu-tions. We shall restrict ourselves to linear damping. It will be shown that thetime-dependent solutions of the Fokker-Planck equation for linear dampingare consistent with the previous results for non-stationary resonant responseof section 6.4. Time-dependent solutions of more complex versions of theFokker-Planck equation will be dealt with in the next chapter.

We shall return to the formulation of the Fokker-Planck equation in termsof the variables u1 and u2: cf. eq. (13.39). In case of linear damping, α = 0and α1 = 1 so that g = 1/2u: see eqs (13.1), (13.21) and (13.27). Eq. (13.39)now becomes:

∂pu∂T ∗ =

12∂

∂u1(u1pu) +

12∂

∂u2(u2pu) +

πS1

4

(∂2pu∂u2

1

+∂2pu∂u2

2

)(13.59)

In line with the situation considered in section 6.4, we assume zero dis-placement and velocity at time zero. Noting that (cf. eqs. (13.14), (13.18)and (13.26))

x = δ1/2(u1 sin t+ u2 cos t), (13.60)

x = δ1/2(u1 cos t− u2 sin t), (13.61)

the initial conditions taken for eq. (13.59) are:

u1 = u2 = 0 at T ∗ = 0. (13.62)

13.4 Transient response behavior 115

2 3 4 5 6 7 8 90

0.5

1

1.5

2

ρ/σx

p(ρ)

α=−1α=0α=+1

Figure 13.2. Probability densities of extremes for Coulomb friction α = −1, lineardamping α = 0 and quadratic damping α = 1.

Eq. (13.59) is the Fokker-Planck equation appropriate for a Gaussian process:Van Kampen [4], §VIII6. The solution of eq. (13.59) is thus given by thetwo-dimensional Gaussian distribution (eq. (1.11)):

pu =1

2π√σ2u1σ2u2

− σ4u1u2

exp(−σ

2u2u1

2 − 2σ2u1u2

u1u2 + σ2u1u2

2

2(σ2u1σ2u2

− σ4u1u2

)

), (13.63)

Here, σu1 , σu2 and σu1u2 are time-dependent standard deviations and covari-ance. Descriptions for σu1 , σu2 and σu1u2 can be obtained by substitutingsolution (13.63) into differential eq. (13.59) and equating coefficients of liketerms. An alternative way is presented here. Therefore, multiply each term ofeq. (13.59) with u2

1 and integrate with respect to u1 and u2 from −∞ to +∞.Because

σ2u1

=∫ +∞

−∞

∫ +∞

−∞u2

1pudu1du2, (13.64)


the first term yields,

∫ +∞

−∞

∫ +∞

−∞u2

1

∂pu∂T ∗ du1du2 =

dσ2u1

dT ∗ (13.65)

Adopting partial integration, the integral of the first term on the right handside yields:

12

∫ +∞

−∞du2

∫ +∞

−∞u2

1

∂

∂u1(u1pu)du1

=12

∫ +∞

−∞du2

∫ +∞

−∞u2

1d(u1pu)

=12

∫ +∞

−∞du2(u3

1pu)

∣∣∣∣∣u1=+∞

u1=−∞− 1

2

∫ +∞

−∞du2

∫ +∞

−∞u1pudu

21

= −∫ +∞

−∞du2

∫ +∞

−∞u2

1pudu1

= −σ2u1

(13.66)

where we assumed that u31pu tends to zero as u1 → ±∞, which is likely,

because probability densities generally tend exponentially fast to zero at theextreme ends of the distribution.

For the integral of the second term on the right hand side of eq. (13.59)one can write:

12

∫ +∞

−∞

∫ +∞

−∞u2

1

∂

∂u2(u2pu)du1du2 =

12

∫ +∞

−∞u2

1du1(u2pu)

∣∣∣∣∣+∞

−∞= 0 (13.67)

Similarly, the fourth term is also found to be equal to zero. For the third term,


we need to apply partial integration twice:

πS1

4

∫ +∞

−∞du2

∫ +∞

−∞u2

1

∂2pu∂u2

1

du1 =πS1

4

∫ +∞

−∞du2

∫ +∞

−∞u2

1d

(∂pu∂u1

)

=πS1

4

∫ +∞

−∞du2

(u2

1

∂pu∂u1

)+∞

−∞− πS1

4

∫ +∞

−∞du2

∫ +∞

−∞∂pu∂u1

du21

= −πS1

2

∫ +∞

−∞du2

∫ +∞

−∞u1∂pu∂u1

du1

= −πS1

2

∫ +∞

−∞du2

∫ +∞

−∞u1dpu

= −πS1

2

∫ +∞

−∞du2(u1pu)

∣∣∣+∞−∞

+πS1

2

∫ +∞

−∞du2

∫ +∞

−∞pudu1

=πS1

2, (13.68)

because ∫ +∞

−∞

∫ +∞

−∞pudu1du2 = 1. (13.69)

Collecting the above results, we obtain from eq. (13.59) an equation for σu1

which reads asdσ2

u1

dT ∗ = −σ2u1

+πS1

2. (13.70)

In view of condition (13.62), σu1 = 0 at T ∗ = 0. The solution to eq. (13.70)then is:

σ2u1

=πS1

2

(1 − e−T

∗)(13.71)

In a similar way, one can show that

σ2u2

=πS1

2

(1 − e−T

∗). (13.72)

Multiplying eq. (13.59) with u1u2, integrating with respect to u1 and u2 overthe range −∞ ≤ u1 ≤ +∞ and −∞ ≤ u2 ≤ +∞, and evaluating terms a.o.by partial integration as above, one arrives at:

dσu1u2

dT ∗ = −σu1u2 . (13.73)

Because of condition (13.62), σu1u2 = 0 at T ∗ = 0, so that eq. (13.73) yields

σu1u2 = 0. (13.74)


Similarly, it can be shown that the mean values of u1 and u2 are zero, a featurealready adopted in solution (13.63). The transient joint probability density ofu1 and u2 now becomes:

pu =1

2πσ2u(T ∗)

exp(− u2

1 + u22

2σ2u(T ∗)

)(13.75)

whereσ2u(T

∗) = σ∗20 (1 − e−T∗) , σ∗20 =

πS1

2, (13.76)

where σ∗0 is standard deviation of stationary u valid as T ∗ → ∞. Fromeqs. (13.60)-(13.61) one can verify that

u21 + u2

2 = δ(x21 + x2

1). (13.77)

Moreover,

px = pu

∣∣∣∣dudx∣∣∣∣ = δpu, (13.78)

so that

px =1

2πσ2(T ∗)exp

(− x2 + x2

2σ2(T ∗)

)(13.79)

where σ(T ∗) is the standard deviation of transient resonant response givenbefore (cf. eqs. (6.31) and (6.55)):

σ2(T ∗) =σu(T ∗)δ

= σ20(1 − e−T

∗) , σ2

0 =σ∗20

δ, (13.80)

σ0 being standard deviation of stationary resonant response. Because x andx are uncorrelated, we can express the right hand side of eq. (13.79) as theproduct of two one-dimensional Gaussian distributions for x and x with stan-dard deviations σ(T ∗). The result for x thus obtained is entirely consistentwith the results presented in section 6.4: see eqs. (6.55)-(6.56).

Applying eqs. (13.26) and (13.43), one can derive an expression for thejoint probability density of amplitude and phase from eqs. (13.75) and (13.76)as:

p(a, φ) =a

2πσ2u(T ∗)

exp(− a2

2σ2u(T ∗)

). (13.81)

Integrating over all possible phase angles in the range 0 ≤ φ ≤ 2π, we havefor the one-dimensional probability density of amplitude the description:

p(a) =a

σ2u(T ∗)

exp(− a2

2σ2u(T ∗)

). (13.82)

The evolution of the distribution with time has been shown in figure 13.3. Itillustrates the development of a Dirac type distribution near a = 0 at T ∗ = 0towards the Rayleigh distribution as T ∗ → ∞.


0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

a / σ*0

σ*0 p

T*=0.05

T*=1

T*=5

Figure 13.3. Transient probability density distributions of amplitude in case oflinear damping.

Chapter 14

Stability in case of randomlyfluctuating restoring

It is known that a sinusoidally varying component in the stiffness term of amass spring system can give rise to instability, even if the magnitude of theoscillation is small in comparison to the total stiffness: e.g. Verhulst [13]. Thephenomenon is also known as Mathieu instability . It occurs for oscillationswith a frequency which is twice the natural frequency of the system. Depend-ing on the damping, the system can exhibit an oscillation solely due to thesmall oscillating component in the stiffness terms and without any externalexcitation applied.

A nice example is the pendulum mounted in a frame that oscillates verti-cally, see figure 14.1. For small swing angles of the pendulum, the equation ofmotion becomes that of a linear spring-mass system with a fluctuating compo-nent in the restoring force. For frequencies of oscillation of the frame that aretwice the natural frequency, Ω = 2ω0, a solution exists other than that of thependulum at rest; i.e., a solution for which the pendulum exhibits a swingingmotion, even if the amplitude of vibration of the frame is very small (ε 1;see figure 14.1). The amplitude of swing-oscillations grows with time unlesssufficient damping is applied: Verhulst [13].

There are many examples of mechanical systems exhibiting an oscillatingparametric component: Notably floating structures in water where restoringoccurs according to Archimedes’ law. When the water surface oscillates inheight and as a result, the degree of submergence of the structure as well,restoring exhibits some degree of fluctuation. The system is prone to Mathieuinstability. Examples are among others roll motion of ships and horizontaldisplacement of tensioned leg platforms for oil production: see figure 14.2. Inthese applications, the fluctuating component in the restoring forces is caused

122 Stability in case of randomly fluctuating restoring

by waves that vary randomly in time. Question now is: does the phenomenonof instability also occur in case of random fluctuation of the restoring force?The answer will be given below. It is an illustration of how to apply stochastictheory in case of questions on stability.

123

s(t)

l

mg

Figure 14.1. Pendulum with oscillating support.


Figure 14.2. Horizontal motion of a tensioned leg platform.

14.1 Basic equations 125

14.1 Basic equations

The governing equation of motion can be written as:

x+ δx|x|α + (1 + δ1/2f(t∗))x = 0. (14.1)

The equation is in dimensionless form as before: cf. eqs. (6.3) and (13.1).Damping is light, δ 1, as well as parametric excitation. Now rewriteeq. (14.1) as:

x+ x = −δx|x|α − δ1/2xf(t∗). (14.2)

Setting δ = 0 gives rise to the solution

x = a cos(t∗ + φ) (14.3)

Amplitude and phase are chosen such that

x = −a sin(t∗ + φ), (14.4)

so thata = (x2 + x2)1/2 (14.5)

φ = −t∗ − arctan(x/x) (14.6)

Differentiating (14.5) and (14.6) with respect to t∗ and implementing (14.2),one obtains:

ada

dt∗= x(x+ x) = −δ|x|α+2 − δ1/2xxf(t∗) (14.7)

a2 dφ

dt∗= −x(x+ x) = δxx|x|α + δ1/2x2f(t∗) (14.8)

Upon substituting (14.3)-(14.4) and dividing by a and a2, respectively, wehave the following equations for amplitude and phase

da

dt∗= −δaα+1| sin(t∗ + φ)|α+2 + δ1/2a sin(t∗ + φ) cos(t∗ + φ)f(t∗) (14.9)

dφ

dt∗= −δaα cos(t∗+φ) sin(t∗+φ)| sin(t∗+φ)|α+δ1/2 cos2(t∗+φ)f(t∗) (14.10)

Question now is: how do amplitude and phase behave with increasing time t∗

given some initial value at t∗ = 0:

a = a0 , φ = φ0 at t∗ = 0. (14.11)

The answer to this question will give us a clue to whether the system is stableor not.


14.2 Fokker-Planck equation

Because f(t∗) is random, a and φ will be random as well. Their behaviorcan be studied from the transient probability density of a and φ which isgoverned by a Fokker-Planck equation associated with eqs. (14.9)-(14.10). Insection 13.2, we derived a Fokker-Planck equation for a fluctuating equationof the type of eq. (13.25). The present fluctuation equation is rather different:the derivation has to be revisited. For this purpose, rewrite eqs. (14.9)-(14.10)in the form:

dqndt∗

= −δgn(q, t∗) + δ1/2hn(q, t∗)f(t∗) (14.12)

where q = (q1, q2) = (a, φ), while

g1 = qα+11 | sin(q2 + t∗)|α+2 (14.13)

g2 = qα1 cos(q2 + t∗) sin(q2 + t∗)| sin(q2 + t∗)|α (14.14)

h1 = q1 sin(q2 + t∗) cos(q2 + t∗) (14.15)

h2 = cos2(q2 + t∗) (14.16)

Comparing eq. (14.12) with eq. (13.25) the following differences are observed.First of all, the functions gn and hn on the right hand side of eq. (14.12) varywith time. In fact, they oscillate with t∗ which is fast compared to the changeof q which is slow because the time derivative of q is small when δ 1. Sec-ondly, the second term on the right hand side of eq. (14.12) contains productsof functions of q and the random excitation f(t∗). It reflects parametric ex-citation in contrast to direct excitation in eq. (13.25). Thirdly, the randomexcitation is a regular finite band-width Gaussian process. In eq. (13.25),the excitation is white noise being the outcome of applying a limit processinvolving δ → 0. Such a limit process still needs to be applied to eq. (14.12).

The derivation of the Fokker-Planck equation associated with a fluctuatingequation like eq. (14.12) applying the limit process δ → 0 has been given byStratonovich: Vol. I, §4.8-4.9 [3]. The derivation is not repeated here. Interms of the present variables, the result of the Stratonovich derivation, i.e.eq. (4.194) of Stratonovich; Vol. I, §4.9 [3], is:

∂p

∂T ∗ =∑ ∂(〈gl〉p)

∂ql−∑ ∂

∂ql

∑(∫ ∞

0

⟨∂hl∂qm

hmτffτ

⟩dτp

)

+∑ ∂2

∂ql∂qm

(∫ ∞

0〈hlhmτffτ 〉 dτp

), (14.17)

where p = p(q, T ∗) is transient probability density of q, T ∗ = δt∗ and gl =gl(q, t∗), hl = hl(q, t∗), hmτ = hm(q, t∗ + τ), f = f(t∗), fτ = f(t∗ + τ).

14.2 Fokker-Planck equation 127

Angular brackets denote statistical averaging over fluctuations of f taking qfixed.

Compared to Fokker-Planck equation (13.38) derived in section 13.2 it isnoted that coefficients in eq. (14.17) have yet to be averaged. This will bedone in the paragraph below. Furthermore, it is seen that eq. (14.17) containsan extra term, viz. the second term on the right hand side, in comparison witheq. (13.38). This term is due to correlation between q and f when the termhn(q, t) preceding the random excitation in eq. (14.12) varies with q. In fact,the presence of this extra term has provoked debate amongst mathematicians.It is known as the Ito -Stratonovich dilemma: Van Kampen [4], IX5. It is nowagreed that the derivation of the Fokker-Planck equation by Stratonovich iscorrect. It leads to consistent formulations which are invariant to the choiceof the Markov variables q.

To evaluate the statistical averages in the coefficients of the Fokker-Planckequation, we note that the net contribution of rapidly oscillatory terms canbe calculated by averaging over one period of oscillation. Thereby, one canassume that q is constant because q varies over the much larger time scaleT ∗. Accordingly,

〈g1〉 =⟨qα+11 | sin(q2 + t∗)|α+2

⟩

= qα+11

12π

∫ 2π

0| sin η|α+2dη

=12α1a

α+1 (14.18)

〈g2〉 = 〈qα1 cos(q2 + t∗) sin(q2 + t∗)| sin(q2 + t∗)|α〉= qα1

12π

∫ 2π

0cos η sin η| sin η|αdη = 0, (14.19)

where α1 is given by eq. (13.21).

The diffusion terms in Fokker-Planck equation (14.17) contain integralsof correlations of the second term on the right hand side of the fluctuation


equation. To illustrate how to proceed in this case, consider:

∫ ∞

0〈h1h1τffτ 〉dτ =

∫ ∞

0dτq21〈sin(q2 + t∗) cos(q2 + t∗) sin(q2 + t∗ + τ)

cos(q2 + t∗ + τ)f(t∗)f(t∗ + τ)〉=

14q21

∫ ∞

0dτ〈sin(2t∗ + 2q2) sin(2t∗ + 2q2 + 2τ)f(t∗)f(t∗ + τ)〉

=18q21

∫ ∞

0dτ〈[cos(2τ) − cos(4t∗ + 4q2 + 2τ)] f(t∗)f(t∗ + τ)〉

=18q21

∫ ∞

0dτ cos(2τ)〈f(t∗)f(t∗ + τ)〉

=a2

16

∫ ∞

−∞〈f(t∗)f(t∗ + τ)〉 cos(2τ)dτ

=π

16a2S2 (14.20)

where we used multiplication and multiple-angle formulae of trigonometricfunctions. Furthermore, we implemented eq. (3.14) relating auto-correlationto power density: S2 is the value of the power density of f(t∗) at ω∗ = 2, thefrequency that corresponds to twice the undamped natural frequency of thesystem.

In a similar manner we can evaluate the other coefficients in the diffusionterms of eq. (14.17). We shall illustrate this for two terms:

∫ ∞

0〈h2h2τffτ 〉dτ =

∫ ∞

0dτ

⟨cos2(q2 + t∗) cos2(q2 + t∗ + τ)f(t∗)f(t∗ + τ)

⟩

=14

∫ ∞

0dτ〈[1 + cos(2q2 + 2t∗)][1 + cos(2q2 + 2t∗ + 2τ)]f(t∗)f(t∗ + τ)〉

=14

∫ ∞

0dτ〈[1 + cos(2q2 + 2t∗) + cos(2q2 + 2t∗ + 2τ) +

12

cos(2τ)

+12

cos(4q2 + 4t∗ + 2τ)]f(t∗)f(t∗ + τ)〉

=14

∫ ∞

0dτ [1 +

12

cos(2τ)]〈f(t∗)f(t∗ + τ)〉

=π

8S0 +

π

16S2, (14.21)

14.2 Fokker-Planck equation 129

where S0 is the value of the power density of f(t∗) at ω∗ = 0.∫ ∞

0

⟨∂h2

∂q2h2τffτ

⟩dτ =

∫ ∞

0dτ〈2 cos(q2 + t∗) sin(q2 + t∗)

cos2(q2 + t∗ + τ)f(t∗)f(t∗ + τ)〉=

12

∫ ∞

0dτ〈sin(2q2 + 2t∗)[1 + cos(2q2 + 2t∗ + 2τ)]f(t∗)f(t∗ + τ)〉

=12

∫ ∞

0dτ

⟨[sin(2q2 + 2t∗) − 1

2sin(2τ) +

12

sin(4q2 + 4t∗ + 2τ)]

f(t∗)f(t∗ + τ)⟩

= −14

∫ ∞

0sin(2τ)〈f(t∗)f(t∗ + τ)〉dτ

= −π8S2 (14.22)

where

S(ω) =2π

∫ +∞

0R(τ) sin(ωτ)dτ, (14.23)

andS2 = S(ω = 2). (14.24)

The procedures for eliminating oscillatory terms can also be found in Vol. IIof Stratonovich [3], chapter 10.

Eliminating all oscillatory terms in Fokker-Planck equation (14.17) ac-cording to the previously presented procedures and returning to the originalvariables a and φ, we obtain:

∂p

∂T ∗ =12α1

∂

∂a(aα+1p) +

πS2

16∂

∂a

(a3 ∂

∂a

(pa

))+π

16(S2 + 2S0)

∂2p

∂φ2+π

8S2∂p

∂φ,

(14.25)where p = p(a, φ, T ∗) is the joint probability of amplitude and phase as afunction of time and T ∗ = δt∗. One-dimensional probability densities foramplitude and phase follow from p as:

pa =∫ 2π

0pdφ , pφ =

∫ ∞

0pda. (14.26)

Integrating (14.25) with respect to φ and a respectively, and applying ap-propriate boundary conditions, i.e.: p → 0 and (∂/∂a)p → 0 as a → ∞;p(φ = 0) = p(φ = 2π), (∂/∂φ)p(φ = 0) = (∂/∂φ)p(φ = 2π),

∂pa∂T ∗ =

12α1

∂

∂a

(aα+1pa

)+πS2

16∂

∂a

(a3 ∂

∂a

(paa

)), (14.27)


∂pφ∂T ∗ =

π

8S2∂pφ∂φ

+π

16(S2 + 2S0)

∂2p

∂φ2, (14.28)

subject to initial conditions

pa = δ(a− a0) , pφ = δ(φ− φ0) at T ∗ = 0. (14.29)

It is also possible to formulate fluctuation equations for amplitude and phasethat correspond to eqs. (14.27)-(14.28): i.e.,

da

dT ∗ = −12α1a

1+α +πS2

8a+

√2S2

4af1(T ∗), (14.30)

dφ

dT ∗ = −πS2

8+

14(2S2 + 4S0)1/2f2(T ∗). (14.31)

where f1(T ∗) and f2(T ∗) are uncorrelated Gaussian white noise processesof power density unity. By applying transformation rules (14.17), one caneasily verify that the Fokker-Planck equations associated with eqs. (14.30)and (14.31) correspond to eqs. (14.27)-(14.28). In this way, it is possible toreconstruct fluctuation equations from a Fokker-Planck equation. The thusobtained equations, cf. eqs. (14.30) and (14.31) are stochastically equivalentto the original equations, viz. eqs. (14.9)-(14.10) taking the limit δ → 0. Thatis, they lead to the same statistics for transient behavior when δ → 0.

Unbounded growth of response will depend on the behavior of a(T ∗) as T ∗

increases. To study this behavior, we can focus attention to the descriptionof a(T ∗) in the time domain by eq. (14.30) or on the probability distributiongiven by eq. (14.27). This will be done in the subsequent sections. Thereby,attention will be focussed on a combination of linear and non-linear damping:i.e. in eq. (14.1) we replace the damping force by:

x|x|α → β0x+ βαx|x|α, (14.32)

in which case,α1a

1+α → β0a+ βαa1+α (14.33)

withβα = βαα1, (14.34)

in eqs. (14.27) and (14.30). So we shall consider the problems

da

dT= −1

2β0a− 1

2βαa

1+α +πS2

8a+

√2S2

4af1(T )

a = a0 at T = 0, (14.35)

14.3 Time-domain behavior 131

and∂pa∂T

=12β0∂apa∂a

+12βα∂a1+αpa∂a

+πS2

16∂

∂a

(a3 ∂

∂a

(paa

))

pa = δ(a− a0) at T = 0; (14.36)

for convenience, we left out the asterisk of T ∗ (hence T = δt∗). As we shall seein the next sections, the combination of linear and non-linear damping leadsto a range of interesting phenomenae as T → ∞.

14.3 Time-domain behavior

It appears possible to solve eq. (14.35) in closed form. The solution is givenby (Brouwers [14]):

a = y exp(γπS2

8T +

√2S2

4w(T )

), (14.37)

wherey = a0 if βα = 0, (14.38)

i.e. for linear damping, and more general

y =[a−α0 +

12αβα

∫ T

0exp

(αγ

πS2

8τ + α

√2S2

4w(τ)

)dτ

]−1/α

, (14.39)

i.e. for non-linear damping. In the above solutions,

γ = 1 − 4β0

πS2, (14.40)

while

w(T ) =∫ T

0f1(τ)dτ (14.41)

is a Wiener or Brownian motion process of unit intensity. The correctnessof the above solution can easily be checked by substitution in the originaldifferential equation and boundary conditions.

Statements about the behavior of a(T ) as T → ∞ can be made because it isknown that for Brownian motion with probability 1, |w(T )| ≤ (2T ln lnT )1/2:Ariaratnam and Tam [15]. We shall first illustrate this for linear damping(βα = 0). The first term in the exponent of solution (14.37), if γ = 0, growslinearly with T which is faster than w(T ). Therefore, the value of γ determinesthe behavior as T → ∞. If γ < 0, a decays exponentially, if γ > 0, a growsexponentially. Hence, stability is ensured if γ < 0, i.e. if β0 > (πS2)/4. If


the magnitude of linear damping is sufficiently large in comparison with thevalue of the power density at twice the natural frequency, the system is stable:any initial disturbance decays with time. Interestingly, also for deterministicsinusoidal parametric excitation, a threshold for the amount of linear dampingnecessary to ensure stability has been established: Verhulst [13].

We shall now investigate the case of non-linear damping. Distinction willbe made between (A) α > 0 and (B) α < 0, as well as between (I) γ > 0 and(II) γ < 0.

(AI) α > 0 and γ > 0Inspection of the right hand side of eq. (14.39) shows that for T → ∞, the sec-ond term in square brackets grows exponentially in magnitude in comparisonwith the first. The first term, which represents the effect of the initial condi-tion, can thus be disregarded when T → ∞. From eqs. (14.37) and (14.39) itthen follows that

a ∼[12αβα

∫ T

0exp

(αγ

πS2

8(τ − T ) + α

√2S2

4(w(τ) − w(T ))

)dτ

]−1/α

as T → ∞. (14.42)

From this expression, it can be concluded that the amplitude of response re-mains finite with probability 1 as T → ∞. Furthermore, the amplitude doesnot tend to zero with probability 1 as T → ∞.

(AII) α > 0 and γ < 0In view of the negative value of γ, rather than growing exponentially, theintegral on the right hand side of eq. (14.39) now approaches a constant asT → ∞, with the result that

a ∼(a−α0 +

12αβαC0

)−1/α

exp(γπS2

8T +

√2S2

4w(T )

)

as T → ∞ (14.43)

where

C0 =∫ ∞

0exp

(αγ

πS2

8+ α

√2S2

4w(τ)

)dτ (14.44)

Bearing in mind that |w(T )| ≤ (2T ln lnT )1/2 as T → ∞ with probability 1, itfollows from eq. (14.43) that the amplitude of response decays exponentiallyto zero with probability 1 as T → ∞. Furthermore, it is noted that, apartfrom the constant C0, the asymptotic result of eq. (14.43) corresponds to thesolution for linear damping given by eqs. (14.37) and (14.38). In fact, the right

14.3 Time-domain behavior 133

hand side of eq. (14.43) can be made a solution of the linear problem by adapt-ing the initial value for amplitude imposed at time zero. As the value of C0

varies with realizations of w(T ), the adapted initial value will also vary withrealizations of w(T ). The asymptotic result of eq. (14.43) thus corresponds tothe solution of the linear problem subject to variable initial condition at timezero.

(BI) α < 0 and γ < 0The integral on the right hand side of eq. (14.39) increases exponentially withtime. Because of the negative value of α, for each realization there is a momentin time, T0, defined by

a−α0 +12αβα

∫ T0

0exp

(αγ

πS2

8+ α

√2S2

4w(τ)

)dτ = 0, (14.45)

for which y and hence the amplitude of response becomes zero. The solutiongiven by eqs. (14.37) and (14.39) then only holds for 0 ≤ T ≤ T0.

The contribution of a(T ) as T > T0 is as follows. For −1 < α < 0 it canbe shown that a(T ) ≡ 0 as T ≥ 0. This continuation satisfies eq. (14.35) andjoins smoothly the solution appropriate for T ≤ T0. For α ≤ −1, however, wearrive at an irregularity. As can be verified from eq. (14.35) for α < 0,

da

dT∼ −1

2βαa

α+1 as a→ 0 (14.46)

from which it follows that da/dT = 0 as a → 0 when α = −1 and thatda/dT → ∞ as a → 0 when α < −1. Furthermore, for any positive value ofa, the slope da/dT is negative. Hence, when α < 0 and γ < 0, differentialequation (14.35) tends to reduce the amplitude to zero but the value zero itselfis not allowed when α ≤ −1.

The above mentioned irregularity suggests improper formulation. A de-scription of the damping force with α < −1 is not conceived as being realisticand will be excluded from the present analysis. For α = −1, which representsCoulomb friction, the irregularity may be ascribed to reducing the originalequation of motion (14.1) to approximate eq. (14.35). Such a reduction is jus-tified as long as the damping force in equation (14.1) is small in comparisonto the acceleration force; that is if:

δaα 1. (14.47)

For α < 0, this condition is no longer satisfied for small values of a such thata ∼ δ−1/α. Solutions derived from eq. (14.35) can thus be expected to becomeinaccurate or even invalid if a→ 0 and α < 0.


It can be concluded that for −1 ≤ α < 0 and γ < 0, the amplitude of re-sponse approaches in a finite time (T0) some finite value close to zero (∼ δ1/α).

(BII) α < 0 and γ > 0Just as for case (AII), α > 0 and γ < 0, the integral on the right hand sideof eq. (14.39) approaches a constant as T → ∞. The solution then reduces tothat given by eq. (14.43), according to which response grows exponentially asT → ∞. The asymptotic result corresponds to the solution for linear dampingsubject to variable initial condition at time zero.

It is noted, however, that the above result is only valid as long as thecoefficient preceding the exponential term in solution (14.43) is positive ofmagnitude, i.e. if

−12αβαC0 < a−α0 , (14.48)

a condition which, in view of the negative value of α, is not satisfied whenconsidering all possible realizations of the random process w(T ). For thoserealizations for which the above-mentioned condition is not satisfied, there isa moment in time, T0, defined by eq. (14.45), for which the amplitude becomeszero. Solution (14.43) is then only valid if 0 ≤ T ≤ T0. The continuation ofa(T ) as T > T0 is equal to that discussed under the previous case (BI).

It can be concluded that for realizations of the excitation to the extentthat condition (14.48) is satisfied, the amplitude of response increases expo-nentially as T → ∞. On the other hand, for those realizations for whichcondition (14.48) is not satisfied the amplitude of response approaches in afinite time, T0, some finite value close to zero (∼ δ1/α).

14.4 Solutions for transient probability density andstatistical moments

A general solution can be derived from partial differential equation (14.36):Brouwers [14]. To derive this solution, use is made of the Laplace transforma-tion (Abramowitz & Stegun [6], chapter 29):

pa(a, T1) =1

2πi

∫ e+i∞

e−i∞pa(a, s)esT1ds, T1 =

πS2

16T, (14.49)

For pa(a, s) we then have the ordinary differential equation:

spa − δ(a− a0) =8βαπS2

d

da(a1+αpa) +

d

da

(−2γapa + a

d(apa)da

)(14.50)

14.4 Solutions for transient probability density and statistical

moments 135

where γ is given by eq. (14.40) and δ is Dirac’s delta function.Solutions for pa(a, s) can be constructed analogous to the derivation of

Green’s functions (Morse and Feshbach [16]):

pa(a, s) =p1a(a, s) for 0 ≤ a ≤ a0

p2a(a, s) for a0 ≤ a ≤ ∞ (14.51)

where p1,2a (a, s) are solutions of the homogeneous equation

sp1,2a =

8βαπS2

d

da(a1+αp1,2

a ) +d

da

(−2γap1,2

a + ad(ap1,2

a )da

)(14.52)

As boundary conditions we require that ap1a and ap2

a tend to zero as a→ 0 anda → ∞, respectively. Furthermore, integrating eq. (14.50) over the intervala0 − ε ≤ a ≤ a0 + ε and letting ε→ 0, we have the additional conditions

p2a = p1

a at a = a0

dp2a

da=dp1

a

da− a−2

0 at a = a0 (14.53)

For further analysis, it is useful to make distinction between the cases (A)α > 0 and (B) α < 0.

(A) α > 0Upon substitution of the variable η, defined by

η =8βαπS2α

aα (14.54)

andp1,2a = aγ−1η−1/2e−η/2f(η), (14.55)

eq. (14.52) reduces to

d2f

dη2+(−1

4+κ

η+

1/4 − µ2

η2

)f = 0, (14.56)

where

µ =(s+ γ2)1/2

α; κ =

12

+γ

α(14.57)

Eq. (14.56) is known as confluent hypergeometric equation, the two basic so-lutions of which are given by Whittaker’s functions Wκ,µ(η) and Mκ,µ(η):


Abramowitz & Stegun [6], chapter 13; see also [17]. Application of the bound-ary conditions then yields the solution,

p1a(a, s) =

Γ(µ− γ/α)αaη0Γ(1 + 2µ)

(η

η0

)γ/α−1/2

e−η/2+η0/2Wκ,µ(η0)Mκ,µ(η) (14.58)

p2a(a, s) =

Γ(µ− γ/α)αaη0Γ(1 + 2µ)

(η

η0

)γ/α−1/2

e−η/2+η0/2Wκ,µ(η)Mκ,µ(η0) (14.59)

where Γ is the gamma function [6] and η0 follows from eq. (14.54) takinga = a0.

The problem that now remains is to re-transform the description givenabove from the s-domain to the T -domain using eq. (14.49). In this connectionit should be noted that the function pa(a, s) as defined by eqs. (14.51), (14.58)and (14.59) is a many-valued function of s with its branch line along −∞ ≤s ≤ −γ2: see figure 14.3. Furthermore, when γ > 0, pa(a, s) has simple polesat s = nα(nα− 2γ), where n = 0, 1, 2, ..., j with j < γ/α.

Laplace integral (14.49) is now evaluated using Cauchy’s theorem

12πi

∫pa(a, s)esT1ds

=∑

residues of pa(a, s)esT1 at the poles within the contour of integration.

As contour of integration, we have chosen the path ABCDEFA shown infigure 14.3 when γ > 0, and the path AGOHFA when γ < 0. Here, thepaths BCDE and GOH are defined by the equations (s + γ2)1/2 = ir and(s + γ2)1/2 = ir + |γ|, respectively, where r is real and −∞ ≤ r ≤ ∞. Theresulting expressions for the probability density of amplitude are given below.

(AI) α > 0 and γ > 0In this case

pa(a(η), T1) = 2a−1η2γ/αe−η<γ/α∑n=0

(γ − nα)n!(ηη0)−n

Γ(2γ/α+ 1 − n)

L(2γ/α−2n)n (η)L(2γ/α−2n)

n (η0)e−nα(2γ−nα)T1

+αη

2πaηκeη0/2−η/2

∫ ∞

0

|Γ(ir − γ/α)|2|Γ(2ir)|2

Wκ,ir(η)Wκ,ir(η0)e−γ2T1−α2r2T1dr, (14.60)

where i =√−1. The summation in the series term is up to γ/α (but excluding

n = γ/α). Furthermore, L(2γ/α−2)n (η) is Laguerre’s polynomial and Wκ,ir(η)

is Whittaker’s function [6].


moments 137

Figure 14.3. Branch line, poles and contours of integration in the complex plane.

According to eq. (14.60), the solution for the probability density consistsof a finite series and an integral. The finite series corresponds to a discrete andfinite number of eigenvalues associated with Fokker-Planck equation (14.36),whereas the integral corresponds to a continuous spectrum of eigenvalues.Such a hybrid situation is not special; it has also been encountered in a number


of solutions of the Schrodinger equation in quantum mechanics [16].From the series term in solution (14.60), it is noted that the first term

(n = 0) describes a stationary probability density distribution. The otherterms of the series, if present, decay exponentially with T1 as T1 → ∞. Fur-thermore, using expansions of Whittaker’s functions for r → 0 [6, 17], the in-tegral in solution (14.60) can be shown to decay as T 1/2

1 e−γ2T1 when T1 → ∞.Solution (14.60) thus describes a transient probability density which decaysexponentially with time to a stationary form given by

pa(a, T1) ∼ α

Γ(2γ/α)

(8βαπS2α

)2γ/α

a2γ−1 exp(−8βαaα

πS2α

)as T1 → ∞ (14.61)

Statistical moments of amplitude, defined by

E[ak] =∫ ∞

0akpa(a, T1)da, k ≥ 0, (14.62)

can be shown to approach a constant as T1 → ∞. The steady-state value isequal to the value obtained when substituting the result of eq. (14.61) into theright hand side of eq. (14.62).

(AII) α > 0 and γ < 0Using the variables and parameters η, T1 and κ, the probability density ofamplitude can be described as

pa(a(η), T1) =(η/η0)κ

2πaηeη0/2−η/2

∫ +∞

−∞Γ(−2γ/α+ ir/α)Γ(−2γ/α+ 2ir/α)

H(η, r)e−r2T1−2γirT1dr,

(14.63)where

H(η, r) =Mκ,µ(η)Wκ,µ(η0) for η ≤ η0

Mκ,µ(η0)Wκ,µ(η) for η ≥ η0(14.64)

withµ = ir/α− γ/α. (14.65)

Applying method of steepest descent to the right hand side of eq. (14.63), it ispossible to derive an asymptotic expression for the probability density whichis valid for T1 → ∞: see 14.6 Appendix. The asymptotic expression can bewritten as

pa(a, T1) ∼ 12a(πT1)1/2

∫ ∞

0G(r, a0)e−(2γT1−ln(a/r))2/(4T1)dr as T1 → ∞,

(14.66)


moments 139

where

G(r, a0) =iαη0

4πr

(a0

r

)αe(1/2)η0

∫C

(2t1)2+2γ/α(1 + 4t21(a0/r)α)−1/2

[−1 + (1 + 4t21(a0/r)α)1/2]1+2γ/α

e−η0t1−(1/2)η0(1+4t21(a0/r)α)1/2dt1. (14.67)

The path of integration C in eq. (14.67) starts at +∞ on the real axis, circlesthe origin in the counterclockwise direction and returns to the starting point.

The asymptotic result given by eq. (14.66) is equal to the solution forlinear damping, i.e. the solution of eq. (14.36) with βα = 0 subject to theinitial condition that pa(a, T1) = G(a0, a) at time zero. It represents theprobabilistic description of an exponentially decaying amplitude. The resultis in line with the asymptotic expression for time-domain solutions given byeqs. (14.43) and (14.44).

For the kth moment of the amplitude, defined by eq. (14.62), one canderive the asymptotic expressions

E[ak] ∼ ak0Γ2(−γ/α)eη0/2

2π1/2α3ηκ+k/α0

Wκ,0(η0)e−γ2T1

T3/21

∫ ∞

0η−3/2+κ/α+γ/αe−η/2Wκ,0(η)dη

as T1 → ∞ if k > −γ (14.68)

and

E[ak] ∼ ak0Γ(−2γ/α− k/α)eη0/2

Γ(−2k/α− 2γ/α)ηκ+k/α0

Wκ,−γ/α−k/α(η0)ek(k+2γ)T1

as T1 → ∞ if k < −γ. (14.69)

This shows that any statistical moment, except for the zero’s, decays expo-nentially to zero as time increases. The result presented in eq. (14.69) is inline with that for pure linear damping where the kth moment is found to beproportional to exp(k(k+2γ)T1) as T1 → ∞ when k > 0 [15]. The result givenin eq. (14.68) indicates that the presence of a non-linear term in the dampingforce which increases more than linearly with velocity causes statistical mo-ments with k > −2γ also to decay to zero (as opposed to the result for purelinear damping).

(B) α < 0A convenient transformation of variables is one involving q, defined by

q = − 8βαπS2α

a2 (14.70)


andp1,2a = aγ−1q−1/2eq/2f(q). (14.71)

Eq. (14.52) then reduces to the confluent hypergeometric equation

d2f

dq2+(−1

4+κ1

q+

1/4 − µ21

q2

)f = 0, (14.72)

where

κ1 = γ/α1 − 1/2 , µ1 = (s+ γ2)1/2/α1 , α1 = −α (14.73)

Applying conditions (14.53) and the boundary conditions imposed at a = 0and a = ∞, we obtain the solutions

p1a(a, s) =

Γ(1 − γ/α1 + µ1)α1aq0Γ(1 + 2µ1)

(q/q0)−γ/α1−1/2eq/2−q0/2Wκ1,µ1(q0)Mκ1,µ1(q),

(14.74)

p2a(a, s) =

Γ(1 − γ/α1 + µ1)α1aq0Γ(1 + 2µ1)

(q/q0)−γ/α1−1/2eq/2−q0/2Wκ1,µ1(q)Mκ1,µ1(q0).

(14.75)It is noted that p1,2

a (a, s) is a many-valued function of s with its branch linealong −∞ ≤ s ≤ −γ2: see figure 14.3. Furthermore, when γ > α1, p

1,2a (a, s)

has simple poles at s = nα1(nα1 − 2γ), where n = 1, 2, 3, ..., j with j < γ/α1.Note that there is no pole at s = 0!

To re-transform the descriptions given above into the time-domain us-ing Cauchy’s theorem, the contours discussed under the case (A) above andshown in figure 14.3 have been used again: contour ABCDEFA for γ < 0 andcontour AGOHFA for γ > 0. The resulting expressions for the probabilitydensity of amplitude are given below.

(BI) α > 0 and γ > 0In terms of the variables q and T1, the probability density of amplitude canbe expressed as

pa(a(q), T1) =α1(q0/q)κ1

2πaqeq/2−q0/2

∫ ∞

0

|Γ(1 − γ/α1 + ir)|2|Γ(2ir)|2

Wκ1,ir(q)Wκ1,ir(q0)e−(r2α2+γ2)T1dr. (14.76)

From solution (14.76) it can be verified that

pa(a(q), T1) ∼ (q0/q)κ

2π1/2α21aq

eq/2−q0/2Γ2(1 − γ/α1)Wκ1,0(η)Wκ1,0(η0)T−3/21 e−γ

2T1

as T1 → ∞. (14.77)


moments 141

According to this result, the probability density decreases to zero as T1 → ∞ sothat

∫∞0 p1(a, T1)da → 0. Apparently, probability ’escapes’ as time increases.

This may be explained by the time-domain solutions given in the previoussection. According to these solutions, for α < 0 and γ > 0, any realization ofa(T ) approaches zero (or almost zero) in a finite time. The above-mentionedprobability distribution now describes this process except for the behavior ata = 0 itself. The phenomenon of ‘escape of probability’ is not unfamiliar forstochastic processes where variables reach, in a finite time, some deterministicvalue [18].

The kth order moment of response can be shown to be

E[ak] ∼ ak0Γ2(1 − γ/α1)

2π1/2q20α31

Wκ1,0(q0)e−q0/2 e

−γ2T1

T3/21∫ ∞

0(q/q0)−γ/α1−k/α1−3/2Wκ1,0(q)e

q/2dq

as T1 → ∞ if k < −γ (14.78)

and

E[ak] ∼ ak0Γ(1 + k/α1)e−q0/2

Γ(2k/α1 + 2γ/α1)Wκ1,k/α1+γ/α1

(q0)ek(k+2γ)T1

as T1 → ∞ if k > −γ (14.79)

From these expressions, it is seen that the kth moment decreases exponentiallyto zero whenever k < −2γ. The kth moment, however, grows exponentially ifk > −2γ. A similar conclusion can be drawn for pure linear damping where thekth moment is found to be proportional to exp(k(k + 2γ)T1) as T1 → ∞ [15].In other words, provided that γ < 0, the presence of a non-linear term in thedamping force which increases less than linearly with velocity (α < 0) doesnot alter, in a qualitative sense, the conclusions with regard to the behaviorof E[ak] as T1 → ∞ obtained for pure linear damping.

(BII) α < 0 and γ > 0Using the variables and parameters q, T1, κ1 and α1, the probability densityof amplitude can be described as

pa(a(q), T1) =(q0/q)κ1

2πaqeq/2−q0/2

∫ +∞

−∞Γ(1 + ir/α1)

Γ(2γ/α1 + 2ir/α1)H(q, r)e−r

2T1−2γirT1dr,

(14.80)where

H(q, r) =Mκ1,µ1(q)Wκ1,µ1(q0) for q ≤ q0Mκ1,µ1(q0)Wκ1,µ1(q) for q ≥ q0

(14.81)


withµ1 = ir/α1 + γ/α1. (14.82)

Applying methods of steepest descent, it can be verified from solution (14.80)that the asymptotic expression for pa(a, T1) valid as T1 → ∞ is equal to thatgiven in eq. (14.66) but with G(r, a0) defined as

G(a0, r) =iαq04πr

(a0

r

)αe(1/2)q0

∫C

(2t1)2+2γ/α(1 + 4t21(a0/r)α)−1/2

[−1 + (1 + 4t21(a0/r)α)1/2]1+2γ/α

e−q0t1−(1/2)q0(1+4t21(a0/r)α)1/2dt1. (14.83)

The path of integration C is equal to that described for the integral in eq. (14.67).As discussed before, the asymptotic result given by (14.66) corresponds to

the solution for linear damping, subject to the initial condition that pa(a, T1) =G(a0, a) at time zero, but with G(a0, a) defined by eq. (14.83) above. In viewof the positive value of γ in the exponent of eq. (14.66), the result representsthe probabilistic description of an exponentially increasing amplitude. Theresult is in line with the asymptotic expressions for time-domain behavior ofamplitude as given by eqs. (14.43) and (14.44) in the previous section.

As we have seen in the previous section, however, apart from growingexponentially with time when α < 0 and γ > 0, there is also a chance of theamplitude approaching zero or some value close to zero as T1 → ∞. Indeed,if we integrate the right hand side of eq. (14.66) over the entire a range andtake account of the result given in eq. (14.83), we find∫ ∞

0pa(a, T1)da ∼ 1 − P (2γ/α1, q0) as T1 → ∞ (14.84)

where P is an incomplete Gamma function [5]. Noting that 0 < P < 1, itfollows from eq. (14.84) that the integral is less than unity, which implies thatthe solution given by eqs. (14.66) and (14.83) does not describe the entireprobability density of amplitude. As found under case (BI) above, the givenprobability density distributions do not describe the behavior of amplitude ata = 0.

The kth moment, k ≥ 0, associated with the above-mentioned probabilitydistribution can be shown to be

E[ak] ∼ ak0Γ(1 + k/α1)Γ(2k/α1 + 2γ/α1)

e−q0/2qκ1+k/α1

0 Wκ1,k/α1+γ/α1(q0)ek(k+2γ)T1

(14.85)as T1 → ∞

According to this description, any statistical moment, except from the zero’s,grows exponentially as T1 → ∞.

14.5 Conclusions 143

14.5 Conclusions

Analytical descriptions have been derived for the non-stationary response ofa randomly parametrically excited second-order system. Damping consistedof a combination of linear and non-linear power-law damping. Response wasinitiated by some disturbance at time zero.

The description of the response consists of a sinusoidal wave with fre-quency equal to the undamped natural frequency of the system. Amplitudeand phase of the sinusoidal wave vary slowly, instationary and randomly withtime. Amplitude and phase are described by fluctuation equations with thetime as independent variable and by Fokker-Planck equations for the proba-bility density of amplitude and phase. These descriptions hold asymptoticallyin the limit of small damping and small parametric excitation. Fluctuationequation and Fokker-Planck equation for amplitude have subsequently beensolved in closed-form using methods of mathematical physics involving highertranscendental functions. Although lengthy and complicated, the procedurewas worthwhile performing: it led to concrete results enabling clear and un-equivocal conclusions to be drawn on stability of the system. The designercan use this information to design safe structures. Results derived for the be-havior of time-domain realizations of amplitude, of probability distributionsof amplitude and of statistical moments of amplitude, as time increases, havebeen summarized in table 14.1.

The asymptotic results for time domain realizations and for probabilitydistributions as time tends to infinity are complementary to each other. Fromthese results, three types of response can be identified.

(a) When the linear part of the total damping force is sufficiently large incomparison to the parametric excitation force, i.e. when γ < 0 or β0 >πS2/4, response decays to zero as time increases. In other words, thereis no apparent effect due to parametric excitation when γ < 0 or β0 >πS2/4.

(b) For small linear part of damping force or large parametric excitation tothe extent that γ > 0 or β0 < πS2/4, on the other hand, every time-domain realization of response remains finite only if the non-linear partof the damping force increases more than linearly with velocity (α > 0).Response then tends with time to a stationary state of random finite-amplitude response. The magnitude of this ‘limit-cycle’ response typeis independent of the initial disturbances imposed at time zero but isstrongly determined by the magnitude of the parametric excitation forceand the non-linear part of the damping force.


(c) For small linear part of damping force in combination with a non-linearpart of the damping force which increases less than linearly with ve-locity, i.e. for γ > 0 and α < 0, however, time-domain realizations ofamplitudes may either tend in a finite time to zero (or almost zero) orgrow exponentially. Either possibility has a finite chance of occurrence.

Parametric excitation can be minimized by appropriately designing thenatural frequency such that S2 is small and γ < 0, i.e. S2 < 4β0/π, whereβ0 is the coefficient of linear damping. In many applications of mechanicalengineering, linear damping is provided by a fluid surrounding the vibratingstructure. But the Reynolds numbers of the fluid flow are generally large, sothat linear damping must come from boundary layer behavior. The magnitudeof this damping is generally small [19]. So it may well be that S2 > 4β0/π. Inthat case, a type of damping that is stronger than linear damping is necessaryto ensure that response does not grow unboundedly. Such damping is providedby fluid drag occurring on bluff bodies moving over distances comparable totheir own cross-sectional dimension. It leads to quadratic damping and theresulting response is of a stationary type as described under case (b) above.The magnitude of the quadratic damping can be enhanced by mounting platesto the vibrating structure in a direction perpendicular to its motion. Examplesare bilge keels on ships and small diameter braces on floating marine structures.They can generate the necessary quadratic damping even at relatively smallexcursions of the structure. The solutions given under case γ > 0 and α > 0can be used to quantify this response. Parameters as fatigue damage andextreme response can be calculated as a function of design variables.

14.5 Conclusions 145

Table 14.1. Behavior of time-domain realizations of amplitude a(T ), of probabilitydistributions of amplitude pa(a, T ) and of statistical moments of amplitude E[ak] astime increases.

a(T ) as T → ∞ pa(a, T ) as T →∞

E[ak] as T → ∞with k > 0

α > 0 Decays exponen-tially to zero

Describes anamplitude whichtends to zero

Decays exponen-tially to zero

γ < 0 βα = 0 Decays exponen-tially to zero

Describes anamplitude whichtends to zero

Decays to zero ifk > −2γ; growsexponentially ifk < −2γ

α < 0 Decays in a finitetime to zero (oralmost zero)

Vanishes anddoes not describethe behavior ata = 0

Decays to zero ifk < −2γ; growsexponentially ifk > −2γ

α > 0 Neither growsunboundedly nordecays to zero

Becomes a sta-tionary distribu-tion

Becomes a con-stant

γ > 0 βα = 0 Grows exponen-tially

Describes anamplitude whichtends to ∞

Grows exponen-tially

α < 0 For certain real-izations exponen-tial increase; forothers decay in afinite time to zeroor almost zero

Describes anamplitude whichtends to ∞; doesnot describe thebehavior at a = 0

Grows exponen-tially


14.6 Appendix: Derivation of asymptotic expres-sions for probability density and statistical mo-ments

Procedures are described for deriving expressions for probability density andstatistical moments of amplitude which are asymptotically valid as T1 → ∞.Use will be made of techniques to compute the asymptotic behavior of inte-grals such as the method of steepest descent or saddle-point method : e.g. seeMorse and Feshbach [16], §4.6 and De Bruin [20]. Results presented applyto the case where α > 0 and γ < 0, as presented in section 14.4. Similarprocedures can be adopted to derive asymptotic results valid for the otherparameter ranges of α and γ.

Probability density as T1 → ∞From the time-domain solutions, it is known that for α > 0 and γ < 0 theamplitude of response decays exponentially to zero as T1 → ∞. Accordingly,when studying the behavior as T1 → ∞, our main interest concerns the be-havior of pa(a(η), T1) given by eqs. (14.63) and (14.64) at η 1. Using theasymptotic expression [5]:

Mκ,µ ∼ η1/2+ir/α−γ/α(

1 +ir − 2γ

α− 2γ + 2irη + ...

)as η → 0, (14.86)

solution (14.63) reduces to

pa(a(η), T1) ∼ eη0/2

2πaη1/2+γ/α0

∫ +∞

−∞Γ(−2γ/α+ ir/α)Γ(−2γ/α+ 2ir/α)

ηir/αWκ,µ(η0)

e−r2T1−2γirT1dr as η → 0 (14.87)

Furthermore, use will be made of Hankel’s Contour Integral [5]

Γ−1(−2γ/α+ 2ir/α) =i

2π

∫C(−t1)2γ/α−2ir/αe−t1dt1, (14.88)

where the contour C is described in the discussion of eq. (14.67), and theintegral representation for Whittaker’s function [5]

Wκ,µ(η0) =e−η0/2η1/2+γ/α

0

Γ(ir/α− 2γ/α)

∫ ∞

0e−t2(1 + t2/η0)ir/αt

ir/α−2γ/α−12 dt2. (14.89)

Upon substituting the above relations into the right hand side of eq. (14.87)and introducing the coordinate transformation

r = −i[γ − 1

2αT1ln(η (1 + t2/η0) t2/t21

)]+ T

−1/21 ν, (14.90)

14.6 Appendix: Derivation of asymptotic expressions for

probability density and statistical moments 147

we obtain, after some algebraic manipulation, the expression

pa(a(η), T1) ∼ i

(2π)2aT 1/21

∫C

∫ ∞

t2=0(−t1)2γ/αt−1−2γ/α

2 e−t1−t2−F2(t1,t2)dt2t1

∫ +∞

ν=−∞e−v

2dv as η → 0, (14.91)

where

F (t1, t2) = γT1/21 − 1

2αT 1/21

ln(η (1 + t2/η0) t2/t21

). (14.92)

It is noted that the right hand side of eq. (14.91) will reach a maximumvalue for values of η where |F (t1, t2)| is of unit order of magnitude. Thesevalues of η define the location and extent of the main part of the probabilitydensity function. From eq. (14.92), it can then be verified that the main partof the probability density function shifts exponentially to zero as T1 → ∞.The same result would have been obtained if we had considered the completedescription of the probability density, as done in the next section, rather thanthe simplified expression valid for small values of η only, as given by eq. (14.87).

As the main part of the probability density shifts exponentially to zero asT1 → ∞, we can replace the limit η → 0 on the right hand side of eq. (14.91)by T1 → ∞. Furthermore,

∫ +∞

ν=−∞e−ν

2dν = π1/2. (14.93)

Applying the coordinate transformations

t1 = η0t∗1 , t2 = −1

2η0 +

12η0(1 + 4t∗21 (a0/r)α)1/2, (14.94)

and dropping the asterisk of t∗1 then enables the asymptotic result presentedby eqs. (14.66) and (14.67) to be derived from eqs. (14.91) and (14.92).

Statistical moments as T1 → ∞Making use of the integral representations for the Whittaker functions givenby eq. (14.89) and [5]

Mκ(η) =eη/2η1/2+ir/α−γ/αΓ(1 + 2ir/α− 2γ/α)

Γ(ir/α− 2γ/α)Γ(1 + ir/α)∫ 1

t1=0e−ηt1tir/α1 (1 − t1)ir/α−2γ/α−1dt1 (14.95)


substitution of the expression for the probability density of amplitude givenby eqs. (14.63) and (14.64) into eq. (14.62) yields

E[ak] =ak0e

η0/2

2παηk/α+κ0

(I1 + I2), (14.96)

where

I1 =∫ +∞

r=−∞

∫ η0

η=0

∫ 1

t1=0

2(ir − γ)Wκ,µ

αΓ(1 + ir/α)ηk/α−1+ir/αt

ir/α1

(1 − t1)−1−2γ/α+ir/αe−ηt1−r2T1−2γirT1dt1dηdt, (14.97)

and

I2 =∫ +∞

r=−∞

∫ ∞

η=η0

∫ ∞

t2=0

Mκ,µ(η0)ηγ/α+k/α−1

T (−2γ/α+ 2ir/α)t−2γ/α−1+ir/α2

(1 + t2/η)ir/αe−t2−η−r2T1−2γirT1dt2dηdr. (14.98)

Upon applying the coordinate transformation

r = i

(−γ +

12αT1

ln (ηt1(1 − t1)))

+ν

T1/21

(14.99)

eq. (14.97) becomes

I1 =2ek(k+2γ)T1

αT1/21

∫ η0

η=0

∫ +∞

ν=−∞

∫ 1

t1=0

(ir − γ)Wκ,µ(η0)Γ(1 + ir/α)

(1 − t1)−1−2γ/α−k/αt−k/α1 e−ηt1−F21 −ν2

dt1dνdη

η(14.100)

where

F1 = (γ + k)T 1/21 − 1

2αT 1/21

ln(ηt1(1 − t1)). (14.101)

Furthermore, introducing the transformations

r = i

(−γ +

12αT1

ln (t2(1 + t2/η)))

+ν

T1/21

(14.102)

andt2 = −η + ηt1 (14.103)

14.6 Appendix: Derivation of asymptotic expressions for

probability density and statistical moments 149

the integral I2 becomes

I2 =e−γ2T1

T1/21

∫ ∞

η=η0

∫ +∞

ν=−∞

∫ ∞

t1=1

Mκ,µ(η0)tγ/α1

Γ(−2γ/α+ 2ir/α)(t1 − 1)−γ/α−1ηk/α−1

exp(−ν2 − ηt1 − 1

4α2T1ln2(t1(t1 − 1)η)

)dt1dνdη. (14.104)

To derive asymptotic expressions valid as T1 → ∞ from the descriptionsof I1 and I2 given by eqs. (14.100) and (14.104), distinction is made betweenthe cases k < −γ and k > −γ.

k < −γThe contribution to integration with respect to r on the right hand side ofeq. (14.100) will be largest for values of η where |F1| is of unit order of mag-nitude. From eq. (14.101) it then follows that, provided that k < −γ, asT1 → ∞, the contribution to the integral stems predominantly from values ofη close to zero. Furthermore, taking |F1| = O(1), r as defined by eq. (14.99),reduces to ik as T1 → ∞. Using these properties and eq. (14.93) we can write

I1 ∼ −2(k + γ)π1/2ek(k+2γ)T1

αT1/21

Wκ,−k/α−γ/α(η0)

∫ ∞

η=0

∫ 1

t1=0(1 − t1)−1−2γ/α−k/αt−k/α1 e−F

21 dt1

dη

η

as T1 → ∞. (14.105)

The integrals on the right hand side of this equation can be evaluated bychanging from the coordinate η to the coordinate F1 defined by eq. (14.101).Integration with respect to t1 subsequently reduces to a Beta Function [5].Noting that I2 decays to zero as T−1/2

1 e−γ2T1 , it then follows from eqs. (14.96)and (14.105) that E[ak] approaches zero as T1 → ∞ in the manner describedby eq. (14.69). This result would also have been obtained when the asymptoticresult for the probability density given by eqs. (14.66) and (14.67) would havebeen substituted into the right hand side of eq. (14.62).

k > −γInspection of the integrals on the right hand sides of eqs. (14.100) and (14.104)shows that for k > −γ, there is no local area of η-values where the contributionto integration with respect to r is dominant. Accordingly, all values of ηbetween 0 and ∞ have to be considered. An asymptotic approximation can


then be derived by applying the transformation

r = −iγ +ν

T1/21

(14.106)

to the integral on the right hand side of eq. (14.63). Making use of the relations

µ =iν

αT1/21

(14.107)

andMκ,µ(η) ∼ −µWκ,0(η)Γ(−γ/α) as T1 → ∞ (14.108)

Γ(2µ) ∼ (2µ)−1 as T1 → ∞ (14.109)

one finds from eqs. (14.63) and (14.64) that

p(a(η), T1) ∼ e−γ2T1(η/η0)κeη0/2−η/2

πα2T3/21 aη

Γ(−γ/α)Wκ,0(η)Wκ,0(η0)

×∫ +∞

−∞ν2e−ν

2dν as T1 → ∞. (14.110)

Noting that ∫ +∞

−∞ν2e−ν

2dν =

π1/2

2, (14.111)

substitution of eq. (14.110) into eq. (14.62) yields the asymptotic result givenby eq. (14.68). Note that this result would not have been obtained when theasymptotic expression for pa(a(η), T1) given by eqs. (14.66) and (14.67) wouldhave directly been substituted into the right hand side of eq. (14.62). Hence,while evaluating the asymptotic value of the right hand side of eq. (14.62), it isnot allowed to interchange integration with respect to a and the limit processinvolving T1 → ∞!

Abstract

Stochastic or random vibrations occur in a variety of applications of mechani-cal engineering. Examples are: the dynamics of a vehicle on an irregular roadsurface; the variation in time of thermodynamic variables in municipal wasteincinerators due to fluctuations in heating value of the waste; the vibrationsof an airplane flying through turbulence; the fluctuating wind loads acting oncivil structures; the response of off-shore structures to random wave loading.

Attention is focussed on problems of external noise. That is, models ofmechanical engineering structures are considered where the source of randombehavior comes from outside: e.g. a prescribed random force or a prescribedrandom displacement. The structures inhibit inertia, damping and restor-ing, linear and non-linear. Questions addressed are: how do these structuresrespond to random excitation, and how to quantify the random behavior ofresponse variables in a manner that an engineer is able to make rational designdecisions.

152 Abstract

Bibliography

[1] Robson, J.D., An Introduction to Random Vibration, Elsevier PublishingCompany, 1964.

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[3] Stratonovich, R.L., Topics in the Theory of Random Noise, Vol. I&II,Gordon & Breach, 1967.

[4] Kampen, van N.G., Stochastic Processes in Physics and Chemistry, re-vised and enlarged edition, Elsevier Publishing Company, 1992.

[5] Gradshteyn, I.S. and Ryzhik, I.M., Table of Integrals, Series, and Prod-ucts. Academic Press, 1980.

[6] Abramowitz, M. and Stegun, I.A, Handbook of Mathematical Func-tions, Dover Publications, 1970. Also available in electronic form at:http://www.math.sfu.ca/ cbm/aands/

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[11] Brouwers, J.J.H. and Verbeek, P.H.J., Expected fatigue damage and ex-pected extreme response for Morison-type wave loading, Appl. Ocean Res.,5, 129, 1983.

154 BIBLIOGRAPHY

[12] Brouwers, J.J.H., Response near resonance of non-linearly damped sys-tems subject to random excitation with application to marine risers,Ocean Eng., 9, 235, 1982.

[13] Verhulst, F., Non-linear differential equations and dynamical systems,Springer, 1996.

[14] Brouwers, J.J.H., Stability of a non-linearly damped second-order systemwith randomly fluctuating restoring coefficient, Int. J. Non-linear Mechan-ics, 21, 1, 1–13, 1986.

[15] Ariatnam, S.T. and Tam, D.S.F., Random vibration and stability of alinear parametrically excited oscillator, Z. Angew. Math. Mech., 59, 79–84, 1979.

[16] Morse, P.M. and Feshbach, H., Methods of theoretical physics, McGraw-Hill, 1953.

[17] Erdelyi, A., Magnus, W., Oberhettinger, F. and Tricomi, F.G., Highertranscendal functions, Vol.I, McGraw-Hill, 1953.

[18] Feller, W., An introduction to probability theory and its applications,Vol.I, John Wiley, 1957.

[19] Brouwers, J.J.H. and Meijssen T.E.M., Viscous damping forces on oscil-lating cylinders, Appl. Ocean Res., 7, 1985.

[20] Bruin, de, N. G., Asymptotic methods in analysis, John Wiley, 1961

stochastic processes in mechanical engineeringspeci ed by probability distributions. the theory of...

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