Advanced Modeling in Science

8/13/2019 Advanced Modeling in Science

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Chapter 1

of the draft version of the SIAM-book

Advanced Modeling in Science

by E.W.C. van Groesen and J.Molenaar


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Contents

1 Dimensional Analysis and Scaling 51.1 Mathematical models . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . 91.4 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5 Challenging problems . . . . . . . . . . . . . . . . . . . . . . 23

1.5.1 The Prandtl-Blasius problem for a flat plane . . . . . 231.5.2 Heat conduction in a bar . . . . . . . . . . . . . . . . 261.5.3 Surface waves . . . . . . . . . . . . . . . . . . . . . . . 27

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4 CONTENTS


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Chapter 1

Dimensional Analysis and

Scaling

1.1 Mathematical models

A mathematical model describes the behaviour of a real-life system in termsof mathematical equations. These equations represent the relations betweenthe relevant properties of the system under consideration. In these modelswe meet with variables and parameters. As for the variables we discern

betweendependent and independentvariables. E.g., in mechanical systemsone is usually interested in the positions of the different parts as functionsof time, so in these systems the positions act as the dependent variables andtime as the independent one. Parameters are properties like masses, pre-scribed temperatures, currents, voltages, and friction coefficients. Parame-ters that can be influenced by the observer are referred to as adjustable. Theother parameters act as constants in the model. For example, in atmospher-ical models used in weather forecasting one is interested in properties liketemperature and humidity - the dependent variables - as functions of positionand time - the independent variables. Important parameters are then thegravity field and the rotational speed of the earth, and these clearly belong

to the class of non-adjustableparameters. The solutionof a mathematicalmodel is known if we can determine the relations between dependent andindependent variables. Since the solution depends on the values of the ad-justable parameters, mathematical models are a powerful tool to determinewhich values of the adjustable parameters yield specific required behaviour.

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1.2 Dimensions

If the variables or parameters in a model correspond to physical propertiesthey have physical dimensions. The fundamental dimensions used in thisbook are given in the following table:

dimension symbol MKS-unit

length L m (meter)mass M kg (kilogram)time T s (second)temperature C (degree Celsius)current I A(Ampere)

The dimension of any physical quantity can be expressed in terms of the fun-damental dimensions. For most quantities this is clear from the definition.For example we have:

quantity dimension

area L2

volume L3

velocity L/Tacceleration L/T2

mass density M/L3

mechanical energy M L

2

/T

2

In other cases the dimensionality of a quantity is deduced from the rule thatall terms in a particular equation must have the same dimensionality. Thisrule is a consequence of the condition that the form of any equation in amathematical model may not depend on the units used. E.g., the dimensionof force directly follows from the second law of Newton, which states thatfor a single mass the mass times the acceleration equals the total force ex-erted on the mass. In standard notation: F = ma. So, the dimensionalityof a force F, denoted as [F], equals the dimensionality [ma] of the prod-uct of mass m and acceleration a. Since [ma] = [m][a], we conclude that[F] =ML/T2.

For coefficients, the dimensionality may vary with specific choices made bythe modeller. E.g., if a frictional force is introduced with strength propor-tional to the velocity of the object, the constant of proportionality will havethe dimension of the quotient of force and velocity. However, if the frictionis assumed to be proportional to the velocity squared, the proportionality


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1.2. DIMENSIONS 7

Figure I.1: A vibrating mass attached to a spring is the prototype of har-

monic motion if the spring response is linear, i.e. if the spring forceFs islinearly proportional to the deviation u (see b.) measured with respect tothe equilibrium position (see a.). Apart fromFs, often an external drivingforceFd and a friction forceFf apply. The latter is usually taken linearlyproportional to the velocity of the mass but in reverse direction.

constant will have the dimension of force divided by velocity squared. Seealso Example 1.3c. For a dimensionless quantity, say q, we have [q] = 1.Examples are angles and universal constants like and e. In dimensionalanalysis, to be treated in1.3, dimensionless quantities play a central role.There it is shown how they can be constructed. The existence of these so-

called dimensionless numbers allows us to draw important conclusions aboutthe system without solving the governing mathematical model.

Example 1.2a. Driven, damped, harmonic oscillator

Consider a bead-spring system in one dimension under influence of frictionand a driving force. The position of the bead with massm is denoted byits displacement u measured with respect to its equilibrium position. SeeFig. I.1. We are interested inuas a function of timet. So,uis the dependentand tthe independent variable of the system. As for the notation, we shalluse the convention u du/dt, u du2/dt2, etc. The second law of Newtonstates that the inertia force, i.e. massm times acceleration u, equals the

sum of the forces exerted on the bead. These forces are the driving forceFd, which is taken harmonic with angular frequency and amplitude F0,the spring force Fs, which is linearly proportional to the displacement andreversely directed to it, and the frictional force Ff, which is assumed to belinearly proportional to the velocity and reversely directed to it. This leadsto the balance of forces

mu= Ff+ Fs+ Fd= cu ku + F0sin t .

The conventional form to write this equation of motion is

mu + cu + ku= F0sin t . (1.1)

Since m, c, k, and F0 can all be influenced, they are adjustable parameters.Every term in this equation has the dimension of force, so ML/T2. Fromthis it follows that


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[c] =ML/T2

L/T

=M

T

, [k] =ML/T2

L

= M

T2

, [F0] =ML

T2

.

The argumenttof a function must be dimensionless, so [t] = 1. We thushave

[] = 1

T .

A mathematical model in terms of an ordinary differential equation is notyet complete if the initial values are left unspecified. This introduces twoextra parameters into the system: initial position u0 u(t0) and initialvelocityv0 u(t0). The solution u(t) thus depends on 7 parameters and we

could write it as

u= u(t; m,c,k,F0, , u0, v0) .

For such a simple system this is a huge number to handle, since in anexperiment all these parameters could in principle be varied. However, in thefollowing we show that such a system can essentially be described with muchless parameters, since it does not make sense to vary them all independently.

The fact that the variables and parameters have physical dimensions canbe fruitfully exploited. The techniques of non-dimensionalising and scal-

ingare extremely powerful tools in analyzing the models. Their importanceis only fully appreciated through examples, which account for the largestpart of this chapter. The basic idea is to apply a transformation to thevariables and parameters such that simplified equations result. It is oftenamazing how much structure is revealed simply by non-dimensionalising,without solving the model explicitly. Thanks to these techniques it is oftenon beforehand known that the system does not depend on all parametersseparately, but only on certain combinations. In an experimental situationit is of great importance to know how the system depends on the parame-ters, so this insight may save much time, costs, and energy.

In practice two methods are applied, dimensional analysisand scaling, whicheach have their own merits. They are dealt with in the subsections belowrespectively. Dimensional analysis fully exploits the information containedin the physical dimensions of the variables and parameters. Scaling has amore restricted scope and aims at reduction of the number of parameters.


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1.3. DIMENSIONAL ANALYSIS 9

1.3 Dimensional Analysis

Non-dimensionalising a mathematical model is a constructive way to formu-late the model in terms of dimensionless quantities only. A big achievementis that dimensional analysis yields insight in the scaling relations of the sys-tem without using knowledge of any governing equation. An advantageouscorollary is that the total number of variables and/or parameters is minimal.Reduction of the number of parameters is also the purpose ofscaling, a tech-nique to be dealt with in the next section. However, dimensional analysisis more general than scaling in that it is based on a transformation of bothvariables and parameters on the same footing, whereas in scaling only thevariables are transformed. Another difference is that scaling starts from thegoverning equations, whereas dimensional analysis starts much more basi-

cally, namely from the dimensions involved in the system, and may evenpredict from them some quantitative features of the model without knowl-edge of the model equations. The basic idea of dimensional analysis is easilyexplained. Consider a system with scalar variables x1,...,xk and scalar pa-rametersp1,...,p. So, the total number of quantities involved is N=k + .Note that in the model vectors, matrices, etc. may figure, but for this anal-ysis all their components have to be treated separately. We now form theproducts

xr11 . . . xrkk p

rk+11 . . . p

rN ,

with ri

Q| , i= 1,...,N, and ask for which choices of the ri these products

are dimensionless. The answer follows from replacing eachxi and pi by itsfundamental dimensions. If, say, m dimensions d1, . . . , dm are involved, thereplacement gives rise to another type of products:

ds11 . . . dsmm ,

with the rational numbers si, i = 1,...,m, being linear functions of theri. The procedure is illustrated several times in the examples below. Byrequiring

si= 0, i= 1, ..., m ,

we obtain a set ofm linear equations for the N unknowns r1,...,rN. From

linear algebra it then follows that there are (at most) N m linearly in-dependent solutions, corresponding to N m dimensionless quantities qi,i= 1,..., (N m). Buckingham formalized this in the following theorem.

Theorem(Buckingham).


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Figure I.2: The main scaling characteristics of a mass m, launched with

initial speedv0, are easily predicted by dimensional analysis

Consider a system with variablesx1,...,xk and parametersp1,...,p, inwhichm fundamental dimensions are involved. Then, k + m dimen-sionless quantitiesqi can be defined, which are products and quotients ofthe original. Each (scalar) relation between thexi andpi

f(x1,...,xk, p1,...,p) = 0

of a mathematical model for the system can be replaced by a correspond-ing relation between theqi:

f(q1,...,qk+m) = 0 .

Since Buckingham denoted the dimensionless quantities by i, this lemmais often referred to as the -theorem of Buckingham. We shall not fol-low his notation since it is not common in the literature. As follows fromthe construction of the qi as solutions of an underdetermined set of linearequations, they are not uniquely defined by the procedure. If the procedureyields a set ofqi, we can apply a transformation, e.g. by taking algebraic oreven functional combinations of them, obtaining another set of dimension-less quantities of the system. It is a matter of expertise and partly of taste,

to determine a convenient set ofqi for the system under consideration. Es-pecially, if the number of variables and parameters is not small, the freedomof choice has to be exploited with care.We shall work out the non-dimensionalising procedure for a considerablenumber of examples pointing out both the practical aspects of the tech-nique and the insight it may yield about the behavior of the system withoutsolving the equations explicitly.

Example 1.3a. CatapultingLet us start with an example in which the mathematics is very basic, butthe ideas behind dimensional analysis are clearly illustrated. A projectile

with mass m is launched vertically. See Fig. I.2. At launching it has veloc-ity v0. Its trajectory, i.e. its vertical position z as a function of time t, isassumed to be completely determined by the influence of gravity. The effectof friction due to the air is ignored here (but dealt with in Example 1.3e).The projectile will decelerate because of gravity until it reaches its highest


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Figure I.3: The heightz of a mass, launched at speed v0, as a function of

time t. It will reach a maximum height zmax at time tmax and reach theearths surface again at timetfinal.

position zmax at time tmax. After that it falls back with increasing velocityand arrives at earth at time tfinal. Since we take v0 such that zmaxremainssmall compared to the earths radius, we may take the gravity field uniformwith gravity constant g .In this system the variables are z and t, and the parameters m, v0, and g.The relevant physical dimensions are M, L, and T. So, k = 2, = 3, andm= 3, and the theorem of Buckingham states that the system has 2 dimen-sionless quantities. All properties of the system can be expressed in onlythese 2 quantities. In this simple case the dimensionless quantities can beeasily found from inspection of the dimensions: [z] =L, [t] =T, [m] =M,[v0] =L/T, and [g] =L/T

2. An evident choice is

t = gt

v0, z =

gz

v20.

Remark that the mass m is not present in t and z, since the physicaldimension M is not present in one of the other variables and parameters.This immediately leads to the conclusion that the motion of the projectileis independent of its mass. The Buckingham theorem yields that its motionis described by a relation between z and t. From experimental evidence

we know that the relation between z and t is more or less as sketched inFig. I.3. The function z(t) reaches a maximumzmaxattmaxand vanishes attfinal. Since z

and t are just scaled versions ofz and t, z can apparentlybe written as an explicit function oft:

z =f(t) . (1.2)

The theorem does not specify any information about f, but only assuresits existence and the insight that the form off does not depend on any ofthe parametersm,v0, and g separately. The latter property thus also holdsfor the dimensionless quantities zmax, t

max, and t

final. Using the relationsbetween dimensional and dimensionless quantities we directly conclude that

zmax=v20

gzmax , tmax=

v0g

tmax , tfinal=v0

gtfinal .

This yields the insight that zmaxscales withv20 and bothtmaxandtfinal with

v0, for fixed value ofg. We denote this as


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Figure I.4: Explicit form of the dimensionless functionf in (1.2). Note that

this function is independent of the parametersm, g, andv0 of the system.The dimensionless heightz reaches a maximum valuezmax = 1/2 at timetmax = 1 and hits the earths surface again at time t

final= 2.

Figure I.5: Swinging pendulum of mass m and length l. The motion isconfined to a vertical plane and the position of the pendulum can be indicatedby the angle with the vertical.

zmax v20 , tmax v0 , tfinal v0 .So, launching with a twice as large velocity leads to a four times larger

maximal height of the projectile. In the same way we conclude that

zmax 1g

, tmax 1g

, tfinal 1g

for fixed value ofv0. So, catapulting on the moon where g is (approximately6 times) smaller than on earth, enhances zmax, tmax, and tfinal all by thesame factor.

Exercise 1.3a.

Check these conclusions on catapulting by explicitly solving the equationof motion

m d2z

dt2 = mg .

Show that f in (1.2) has the explicit form as given in Fig. I.4. Notethat this function cannot be found from dimensional analysis only.

Example 1.3b. Swinging pendulum.

Consider the motion of a mathematical swing: this pendulum has mass mconcentrated in a point at the end of a rigid rod of length . The motionis restricted to a vertical plane. See Fig. I.5. The position of the swingingpendulum is completely specified by the anglewith the vertical. This is theindependent variable, and timet the dependent one. Parameters are massm,

rod length, gravitational accelerationg, and the initial position0= (0).For convenience we take the initial velocity vanishing. So, k+ = 6, andsince the 3 fundamental dimensions M, L, and Tare involved, the systemhas 3 dimensionless quantities. Since and 0 are already dimensionless,they form an obvious choice. To find the third one we form the products


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tr1r2mr3gr4 .

The condition that this product must be dimensionless leads to the linearequations

r1 2r4= 0r2+ r4= 0

r3= 0

The choice (r1, r2) = (1, 0) then yields

t

=tg

.

Note that the mass m is not present in any of the dimensionless quantities, 0, and t

. This implies that pendulum motion is independent from m.The movement of the pendulum is given by some relation between , 0,and t. With 0 constant and t

monotonously increasing, we may write as an explicit function oft:

= f(t, 0) .

This allows for a conclusion about the period of the system. One shouldrealize that dimensional analysis as such does not reveal us that is a

periodic function of time. However, if we take this for granted in view of theobservations, we have that f(t +) = f(t), with the dimensionlessperiod. Since =

g/and does not depend on any of the parameters,

we find that scales with

/g, so

g .

Exercise 1.3b.

a. Give the dimensionless form of the exact pendulum equation

m + mg sin = 0 .

b. If || 1, the reduced pendulum equation


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Figure I.6: Sketch of a shock wave propagating from a explosion with energy

E. Dimensional analysis shows that the energyEcan be estimated from thepropagation velocity of the front.

m + mg = 0

is a good approximation. Give its dimensionless form.

c. Write down the solution of the equation under b) and check that theperiod indeed scales with

/g as derived in Example 1.3b. Determine

how the period is influenced if the length is doubled and also when thependulum is placed on the moon.

Example 1.3c. Estimating the power of explosions.Details of the strength of the first atomic bomb in 1945 were classified untilthe sixties. However, the British physicist G.I. Taylor (see ?? for a de-scription) was able to give a very accurate estimate of the strength fromdimensional analysis using a movie picture that was available of the expan-sion of the mushroom-shape of the explosion. His arguments proceed asfollows.The basic appearance of the explosion is an expanding spherical fireballwhose edge corresponds to a powerful shock wave as sketched in Fig. I.6.LetR be the radius of the shock wave. It will depend on E, the energy re-

leased by the explosion, t, the time elapsed since the explosion, the initialand ambient air density; and p the initial and ambient air pressure. In totalwe recognize 5 variables and parameters. There are 3 dimensions involved.Hence 2 dimensionless quantities can be found.

Exercise 1.3c.

Show that from requiring the products

Rr1tr2Er3r4pr5

to be dimensionless, the fol lowing dimensionless quantities can be de-rived:

q1= p t6

E23

1/5, q2 =

R5

Et2 .


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Figure I.7: The path of a water droplet on a train window yields enough in-

formation to estimate the train speed, as follows from dimensional analysis.

The theorem of Buckingham assures that the motion of the shock front isgoverned by some relation between q1 and q2. Since q1, which is essentiallya scaled time, is monotonously increasing, we may write q2 as an explicitfunction ofq1: q2= f

(q1). Thus R can be expressed as

R=Et2f(q1)

1/5. (1.3)

From this relation we conclude in the first instance that Rdepends ontbothvia the prefactor off and via q1. This complicates the analysis. Taylorfound a way out by first plotting measured values ofR as a function oft ina double logarithmic plot. Taking logarithms of both sides of (1.3) we have

log R= 25

log t + 15

logEf(q1)

.

The data turned out to nearly lie on a straight line. This suggested thatf(q1) hardly depends on time, so that it can be replaced by its initial valuef(0). Then, still two unknown parameters figure in the model: E andf(0). Taylor estimated the value off(0) from performing an experimentunder well defined conditions, for which Ewas known and R(t) measured.This led to the conclusion that f(0)

1. Every shock wave is thus fairly

well described by the model equation

log R= 25

log t + 15

logE

.

Since is usually known, it is easy to estimate Efrom fitting this equationto measuredR(t) data.

Example 1.3d. Estimating train speed from drop tracks.

Let us imagine we are traveling by train on a rainy day. Looking through thewindow we see rain drops attached to the glass following straight trajectoriesalong the window downwards. The angle of inclination of the drop paths

appears to depend on the train speed. We wonder whether the speed canbe estimated from this angle. To answer this question, the system of a dropsliding along a moving glass plate has to be modeled. See Fig. I.7. Threeforces act on the droplet: the gravitational force with strengthFg, the dragforce with strength Fd due to friction between the drop and surrounding


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air, and the adhesion force with strength Fa between drop and glass. Fg is

directed vertically and its strength is equal to mg with m the drop mass.Fd is directed horizontally but its strength is not known beforehand. Fc isa friction force and directed in the reverse direction of the drop speed. Thisforce influences the speed of the drop, but not its direction. This impliesthat the angle of inclination of the drop trajectory is determined by Fgand Fd. From Fig. I.7 we conclude that

tan =FgFd

.

To estimate the train speed vtrain from this relation, we must know how Fddepends on vtrain. We investigate how dimensional analysis can help us to

discover this relation. The friction between the drop and the passing airwill depend on drop diameterD with [D] = L, air density with [] =M/L3, and air speed given by vtrain with [vtrain] = L/T. Fd is a force, so[Fd] = ML/T

2. The friction force exerted by a flow on an object movingthrough the flow is measured by the viscosity with [] = M/LT. Itmeasures the internal friction. Daily experience tells us that syrup has alarger viscosity than water. The value of for a gas or fluid can be measuredby dropping an object in the medium. Under influence of gravity it willinitially accelerate. After some transient time its speed will become constant,since then the friction force comes into equilibrium with the gravitationalforce. This phenomenon is experienced by, e.g., parachutists. The viscosity

can directly be deduced from the equilibrium speed of a standardized object.In determining an expression for Fd the five quantitiesFd,vtrain,D , L, and are involved. Since three physical dimensions play a role, the system hastwo dimensionless quantities.

Exercise 1.3d.

Check that we can choose for these

Fd = Fd

D2v2train, Re=

Dvtrain

.

The dimensionless Reynolds number Replays an important role in all flow

problems. It measures the ratio of the convective and viscous forces and iscalled after Osborne Reynolds, a researcher in fluid mechanics. We expectthat for a still unknown function f it will hold that

Fd =f(Re) .


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Figure I.8: The drag force felt by an object in a flow as function of the

Reynolds numberRe.

For the drag force we thus have

Fd= D2v2trainf(Re) .From this we cannot deduce howFdscales withvtrainsinceRe also containsvtrain. To answer this intricate question one has to determine the form off

from measurements. These data are given in Fig. I.8. Note that this formis universal and holds for all flows, thanks to the dimensionless formulation.The conclusion from these data is that

f(Re)

1Re

if Re 1000

The range 100 < Re < 1000 is a transition region. To find the order ofmagnitude ofRe for a moving train, we substitute some data. For air wehave 1.3 kg/m3 and 1.5 105 kg/(m.s). The size of a droplet isD 5.103 m. The velocity of the train varies from 0 to, say, 50 m/s.Substituting these numbers we find that Re >1000 ifvtrain> 2.5 m/s (i.e.10 km/h). This leads to the conclusion that for all relevant train speeds wehave

Fd= c1D2v2trainfor some constantc1 which does not depend on , D, andvtrain. Eventually,we arrive at

vtrain= ( mg

c1D2 tan )1/2 .

This expression can be reduced a bit by noting that the drop is approxi-mately a half sphere, so m = 1

12D3 if we take the density of water equal to

1. Since g and hardly vary, we may write

vtrain= c2

Dtan The value ofc2can be determined from one observation. This is the relationwe were aiming at. In Fig. I.9 we plot the dimensionless velocity vtrainvtrain/(c2

D) as a function of.


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Figure I.9: The dimensionless train velocity as a function of the angle

indicated in Fig. I.7.

Figure I.10: Study of scaled ship models is only useful if one realizes theconsequences of dimensional analysis.

Note that the accuracy with which must be measured becomes moreand more important if becomes smaller and smaller and thus the trainspeed higher and higher.

Example 1.3e. Ship modeling.

Let us model a ship of length sailing at constant speed v as sketched inFig. I.10. The motion of the ship transfers energy from the ship to the wateras a result of viscous friction. This energy is used partly to induce surfacewaves and partly to overcome the internal friction of the turbulent motion ofthe water. In view of these effects the acceleration of gravity g , the densityof water , and the viscosity will play a role, with dimensions [g] =L/T2,[] = M/L3, and [] = M/LT. If we assume that the ship is streamlinedsuch that its height and width are not of importance, the system has 5 vari-ables and parameters. Because 3 dimensions are involved, the number ofdimensionless quantities is 2. We can choose for these

F r= v

g , Re=

v

.

F ris called the Froude numberafter William Froude, a famous ship builder.Re is the Reynolds number we already met in Example 1.3.e. Because real-life experiments are hard for these systems, it is very attractive to performexperiments on (physical) models in which all sizes are scaled down by acertain factor. The conclusions from these experiments are valid for theoriginal system only if both systems are described by the same dimension-less (mathematical) model. So, F r and Re have to remain constant uponscaling. In practice the values ofg, , and can hardly be adjusted. To keepF r constant, v/

may not change, and to keep Re constant, v must be

preserved. The experimenter must be aware that this can never be managed

in the same experiment. This implies that in one experiment with scaledship models only a restricted aspect of the real situation can be studied.

Exercise 1.3e.

a. Check that we can indeed choose


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F r= v

g

, Re=v

as dimensionless numbers.

b. Why is it not possible to design the test facilities in such a way that boththe Froude and the Reynolds number are preserved? Can you give anargument why the scaling of a ship is usually determined by the Froudenumber and not the Reynolds number? Will this be the same when testingan airplane in a (low speed) wind tunnel?

c. To do experiments on a ship of 100m length that sails with a maximumspeed of 35 km/hr, one uses in a laboratory a model of the ship. Thetowing in the laboratory is restricted to velocities of at most 7 km/hr.

What is the smallest scale of the model that can be used?

Example 1.3f. Harmonic oscillatorHere, we revisit the harmonic oscillator introduced in Example 1.2a. Settingthe initial values at zero for convenience, the model equation

mu + cu + ku = F0sin t

has the two variables u and t, and the five parameters m,c,k,F0, and .So, N= 7 in this case. The fundamental dimensions involved are mass M,length L, and time T. Forming the products

ur1tr2mr3cr4kr5Fr60

r7

and substituting the dimensions, we arrive at the products

Lr1Tr2Mr3M

T

r4MT2

r5M LT2

r6 1T

r7Collecting powers ofM, L, andT, we obtain the following three linear equa-tions for the ri:

r1+ r6= 0

r2 r4 2r5 2r6 r7= 0r3+ r4+ r5+ r6= 0

Here, we meet with 3 equations for 7 unknowns, so 4 unknowns can betreated as free parameters. E.g., we could take r1,...,r4. The choices(r1, r2, r3, r4) = (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1), respectivelyyield the dimensionless quantities


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u =uk

F0

, t =t, m =m2

k

, c = c

k

.

The dimensionless spring equation then reads as

mu + cu + u = sin t ,

where the time derivative is taken with respect to t.

Exercise 1.3f.

The approach used above for the driven spring system is based on theassumptionF0= 0. Apply dimensional analysis to the caseF0 = 0, butwith the initial positionu0 and initial velocityv0 both non-vanishing.

1.4 Scaling

The aim ofscalingis to reduce the number of parameters in a given model.So, a prerequisite of the technique of scaling is knowledge of the equationsgoverning the system. Scaling does not necessarily yield dimensionless quan-tities. We show the technique via examples.

Example 1.4a. Harmonic oscillator scaled

We apply the scaling technique to the harmonic oscillator introduced inExample 1.2a. As pointed out there and in Example 1.3f, this system has

the 2 variables u and t and the 7 parameters m, c, k,F0, , u0, v0. Considerthe scaling of variables

=u

a, =

t

b

with the scaling factors a and b still unspecified. Substitution of this trans-formation into equation of motion (1.1) leads to

ma

b2+

ca

b+ ka = F0sin( ) ,

where the time derivative is now with respect to. Dividing all terms by thefactor ma/b2 and choosingb=

m/kand a = F0/kwe obtain the reduced

equation

+ c+ = sin() ,


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1.4. SCALING 21

Figure I.11: Graphical representation of the chemical reaction in Example

1.4b.

with the new parameters c =c/

km and =

m/k. The initial valuesu0 and v0 have to be scaled accordingly to u

0 and v

0. Note, that thisreduction reveals that the harmonic oscillator depends on only 4 parameters(including the initial parameters). To study the behaviour of the solution asa function of the parameters, it suffices to vary only the friction coefficientc and the angular frequency (apart from the initial values). We remarkthat the scaling procedure does not lead to a unique choice for the scalingfactors a andb. An alternative choice is b= 1/ and a = F0/(

2m), whichyields the equation

+ c+ k= sin() ,

with c = c/(m) and k = k/(2m). So, for a complete analysis of thesystem it also suffices to vary only these c and k (apart from the initialvalues).

Exercise 1.4a.

Compare the methods and results of scaling and dimensional analysis asapplied to the driven, linear spring in Examples 1.3 and 1.3b.

Example 1.4b. Chemical reaction

Consider a hypothetical chemical reaction, the so-called Brusselator, withsubstancesA ,B,C ,D,X,Yinvolved. The situation is sketched in Fig. I.11.CandD are produced fromA and B , withXandYas intermediates. Thereaction has the following irreversible stages:

Ak1 X

B+ Xk2

Y + C

2X+ Yk3 3X

Xk4 D


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22

Figure I.12: Solution sets of equation (1.4) after scaling

The capital letters denote reagents, while the constants ki over de arrowsindicate the reaction rates. It is assumed that A and B are excessivelyavailable so that the concentrations ofA and B can be taken to be constant.We denote the concentrations of A ,B,X, and Y by a,b,x, and y. Thereaction equations for xand y are then

x= k1a k2bx k4x + k3x2y

y= k2bx k3x2y .

Exercise 1.4b.

Use scaling of (x,y ,t) to ( , , ) to show that these equations can bereduced to

= (+ 1)+ 2

= 2 .

Note that after this procedure the number of parameters is considerablyreduced, since instead of the original parameters a, b, k1,...,k4, the final set

of equations contains only the two parameters and .

Exercise 1.4c.

Use the method of scaling to show that the equation

x2 + axy+ b= 0 , (1.4)

with variablesx, y and parametersa, b, can be reduced to the equivalentequation

2 + + sign(b) = 0 ,

with variables , . This implies that the structure of the solutions ofequation (1.4) can be caught without varying the parameters. It sufficesto study the second equation for the two cases sign(b) = +1 and1.These curves are plotted in Fig. I.12.


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1.5. CHALLENGING PROBLEMS 23

Figure I.13: Time development of the velocity profile of air flow approaching

and behind a trailing edge.

1.5 Challenging problems

In this section we challenge the reader to apply the techniques of dimensionalanalysis and scaling to situations that are quite difficult to model. Forapplication of dimensional analysis we only need to know the dimensions ofthe variables and parameters involved in the system. For scaling one needsto start from the governing equations. In the present cases the reader hasto take these equations for granted. They will be derived in Chapter 2.

Here, the focus is on the reduction of the models via scaling and not on thederivation of the models themselves.

1.5.1 The Prandtl-Blasius problem for a flat plane

The following description of flow above a flat plate is a simple model tostudy for instance the following phenomena:- The disturbance of the air by a thin airfoil of an airplane that cruises atconstant altitude with constant speed U.- The disturbance of a strong, uniform wind by vegetation, buildings, etcetera.- The disturbance of water flowing over a rough surface.

In all these cases the flow (air, water) near the surface experiences resistancefrom the presence of the airfoil, obstacles, etcetera. In an idealised way, wemodel the airfoil or the ground by a horizontal flat plate (standing still) withthe air or water flowing over it, assuming that the flow is uniform before itreaches the plate. We choose coordinates such that the plate is in the halfplane (x 0, < y < , z= 0). See Fig. I.13. Since the y-coordinate isirrelevant, we shall omit it further. In front of the plate edge, so for x 0,the flow is uniform with velocity U in the positive x-direction. The fluidhas mass density with [] =M L3 and viscosity with [] =M L1T1.Close to the plate the friction between plate and flow decelerates the flow.This friction causes the flow velocity near the plate to go down. For increas-

ing x the boundary condition will approach the no-slip condition with thevelocity at the plate exactly vanishing. For large xvalues the velocity profilewill become independent ofxand approach a stationary profile, independentofx and thus t. We want to know how fast this convergence takes place asa function of time t, and thus of positionx. In Fig. I.13 the velocity profiles


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24

above the plate are sketched for differentx-positions. Let the velocity of the

flow be denoted by (u, w) with u the velocity in the x-direction and w thevelocity in the normal z-direction. The so-called shear rateis the variationof the horizontal velocity u in the normal direction. It is common use todenote it as :

(x, z) =u

z(x, z) .

Its value at the plate, thus for z = 0, is denoted as 0(x) = (x, 0). It onlydepends on the distance x from the edge of the plate, and it is this depen-dence that we want to investigate. In Fig. I.13 an angle(x) is indicated.It is related to 0(x) via 0(x) = tan (x). Far away from the plate, where

z is large, the flow is hardly influenced by the presence of the plate, so therewe may take (u, w) = (U, 0).

Except of the dependence on the distancexfrom the edge,will also dependon the viscosity, the velocity U, and the density . In the steady state wecan generally write:

0= 0(x; ,U,).

In the following steps we want to find this relationship as precisely as possi-ble. We first determine how far we can come with dimensional analysis andthen we will use more information about the governing equations.

a. Determine two dimensionless variables from the set0, x , , U , and.b. Show that for some functionf it holds that

0=U2

f(

U x

).

To apply scaling we need information about the governing equations. Theactual equations of motion are given by conservation equations ?? and ??in Chapter 2. For steady flow they read as

u

x+

w

z = 0 ,

(uux

+ wuz

) = 2u

z2 .

The first equation (vanishing of the divergence of the velocity field) expressesthe incompressibility of the flow. The second equation expresses the balance


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between the convection force (at the left hand side) and the viscous friction

force (at the right hand side).We apply the following scaling:

u= u

U, w=

w

W, x=

x

X, z=

z

Z

with Ugiven and X, Z, and Wto be chosen later on.

c. Show that in the equations above the number of parameters reduces if thefollowing two conditions are satisfied:

X W

Z U = 1,

X

Z2 U = 1.

Given Z, the second condition determines X, after which W follows fromthe first condition. Hence these two conditions do not uniquely define thescaling. Note that the scaled variables are not necessarily dimensionless.

d. Determine a scaling for which the scaled variables are dimensionless.

The non-uniqueness of the scaling can be exploited to find an explicit expres-sion for the stress. Note that this can be done without solving the equationsexplicitly. The strategy is as follows. KeepZ arbitrary. The shear rate atthe plate only depends on x, i.e. for some function h it holds that

0= h(x).

e. Determine the dimension of gamma0 andx expressed in the dimensionof Z. Then apply dimensional arguments to show that for some constantc >0

h(x) = c

x .

We remark that this argument can only be applied since Z was kept ar-bitrary. IfZ is chosen such that all quantities become dimensionless, thisrelation cannot be derived.

f. Now, translating to the original quantities, show that

0(x; ,U,) =c

U3

x

which is the relation we aimed at. Compare this with the result under b. anddetermine the explicit form of the functionfmentioned there.


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26

1.5.2 Heat conduction in a bar

We consider heat diffusion in a rod of length . The rod is assumed to bethermally isolated everywhere. Starting with a given temperature distri-bution u(x, 0) over the rod, we are interested in the time evolution of thetemperature profile u(x, t). As will be explained in Examples 2.4c and 2.4din Chapter 2, the speed of the heat diffusion along the rod is determinedby the so-called thermal conductivity. The dimension of is L2/T. Asinitial profile we take a distribution which is everywhere vanishing exceptfor a peak in the origin:

u(x, 0) =u0 (x) (1.5)

where(x) is Diracs delta-function.

a. Determine the dimension of the constantu0. To that end, integrate theinitial condition over some interval includingx= 0 and use the propertiesof the delta-function.

The variables are thus u, x, and t and the parameters ,, and u0.

b. Find three dimensionless quantities, choosing them such that they are justscalings of the variablesu, x, andt.

Now we assume that the rod is infinitely long. This implies that the param-eter is no longer relevant and that the number of dimensionless quantitiesreduces to two.

c. Show that these dimensionless quantities are appropriately chosen asq1=x

2/(t) andq2= u

t/u0.

In view of the expected behaviour of the system we may write

q2= f(q1) (1.6)

for some functionf(q1). The important conclusion from dimensional anal-

ysis is that in an infinitely long rod the temperature is not a function ofposition x separately, but of the quotient x2/t. Note that this does not holdfor a rod of finite length . To find the explicit form off one needs thegoverning equation. The equation that describes the time evolution of thetemperature in the bar follows from conservation of heat. It will be derived


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Figure I.14: Temperature profiles for timest1< t2< t2 in a long, thin rod.

in Example 2.4d of Chapter 2. The resulting partial differential equation,the so-called heat diffusion equation, reads as

u

t =

2u

x2. (1.7)

d. Show that the dimension of given above agrees with the dimension fol-lowing from this equation.

e. Use (1.6) to rewrite the heat diffusion equation (1.7) in terms of q1,q2, and f

, and show that this leads to the following ordinary differentialequation forf(q1):

4q12f

q21+ (q1+ 2)

f

q1+

1

2f = 0 , (1.8)

for0< q1< .

f. Determine the boundary conditions atq1 = 0 andq1 = . Note that theformer condition follows from the symmetry in the system.

g. Check that the dimensional solution reads as

u(x, t) = u0

4 tex

2 / 4 t.

So, at each time t the spatial temperature distribution is a Gaussian distri-bution in x, with mean 0 and standard deviation determined by t. Thetemperature decays with increasing time since the inital peak of heat dif-fuses away along the rod. This is sketched in Fig. I.14 for different values oft.

1.5.3 Surface waves

Consider a layer of fluid, for example water, above a horizontal bottom.See Fig. I.15. When the fluid is set into motion, by whatever cause, the


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28

Figure I.15: In the modeling of moving surface waves one wants to find the

elevationu(x, t) of the surface with respect to the completely flat profile.

fluid particles will start to move, interacting with each other and influencedby gravity. When the upper surface of the fluid is free, this surface willalso be deformed by the particle motions. Looking only at the surface, thedeformation is often of a characteristic type, like waves that are caused bythrowing a stone in a pond. In principle, the particles below the surfacedetermine the surface elevation. However, when the fluid is incompressible,and when so called irrotational flow is considered, it turns out to be possibleto describe approximately the elevation of the surface without reference tothe internal motion. We will not derivethe governing equations from firstprinciples, but investigate a postulated description of the phenomenon viadimensional analysis and scaling.

We make the following assumptions:

- The fluid is incompressible with uniform mass density, which we setequal to unity for convenience.

- No forces act from above on the free surface; for instance, effects ofwind are neglected.

- In the horizontal plane, the motion is uniform in one direction; thatis, we consider plane waves. This implies that, if we take the x axis,

say, as the direction of propagation, the motion does not depend on thetransverse horizontal direction.

We take thez axis in the vertical direction, opposite to the direction of grav-ity with its origin at the undisturbed water surface. The distance betweenbottom and undisturbed water surface is H; g denotes the acceleration ofgravity. The modelling focusses on the elevationu(x, t) of the surface withrespect to the undisturbed level z = 0.

a. With the (constant) mass density left outside consideration, the problem

has 3 variables (u,x,t), 2 parameters (g, H) and 2 dimensions (length andtime). Find 3 dimensionless quantities.

A rough idea about wave propagation can be obtained by studying harmonic


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profiles, say of the form

u(x, t) =a cos(2

(x + V t) ,

wherea is the wave amplitudeand is thewave length. This surface profilepropagates with wave velocityV .

b. Take these three quantities into account, together with the parametersg, H, and show that the problem is described by a relation between the nor-malized amplitude, the wave length and the dimensionless velocity:

f(a

H,

H,

V

gH) = 0.

c. From the above result, try to explain the observation - which can be madeat any coast when one looks at the waves running into the shore - that wavesapproach the coast perpendicular, even when the coastal boundary is ratherirregular.

In 1895 Korteweg and De Vries published an equation for u(x, t) that de-scribes the surface elevation in a certain order of approximation. In thederivation of the equation it was assumed that the waves were rather low(small amplitude) and rather long. This equation is known as theKorteweg-de Vries equation (KdV-equation). This equation became famous when itwas found that the equation has very special mathematical properties. Weshall not discuss them here, but focus on the original purpose of this model.It reads as

u

t = cu

xcH

2

6

3u

x3 3c

2Hu

u

x , (1.9)

where the parameter c =

gHhas been introduced since it plays an im-portant role in the physical phenomenon, as we shall see later.This equation

shows the time evolution of the free surface: at a fixed position, the timederivative of the elevation depends (in a complicated way) on the spatialderivatives of the elevation: temporal variations and spatial variations arecoupled, which is characteristic for a partial differential equation. Under-standing the coupling would mean that the meaning of each of the three


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30

terms in the right hand side should be clear, which at this moment is not

possible. The equation is rather difficult in the sense that it is not easy tofind explicit solutions. In the following we will try to interpret the various

terms in the right hand side, and particularly, try to understand how theunderlying modeling assumptions of long and low waves show itself in thisequation.

The starting point is rather characteristic: we perform a scaling of the vari-ables without specifying at this moment the scaling factors; these will bedetermined, or chosen, at a later instant.

x= x/L , t= t/ , u(x, t) =u(x, t)/a .

It should be remarked that at this moment we do not aim at making the newvariables u,x, tdimensionless. To simplify the notation and since confusionis hardly possible, we drop the overhead bar, keeping in mind that theequations below are in terms of scaled quantities.

d. Show that the scaling leads to the following form of the equation:

tu + xu + 3xu + u xu= 0, (1.10)

where

= cL, = c H26L3 , =

3 c a2H L .

Here and often in the following, the spatial derivatives are indicated by xufor ux , and tu for

ut , and similarly for higher order derivatives. Observe

that by the scaling we introduced 3 additional parameters (L, and a),which together with H and g (or equivalently H and c) brings the totalnumber of parameters to 5. However, the equation shows that only specificcombinations of these parameters (namely , and) play a role. We showhow scaling can be used for various purposes.

In order to study the original KdV-equation (1.9) it is sufficient to studyequation (1.10). Therefore we would like to reduce the numbers of parame-ters in this equation as much as possible. This can be done by choosing thescaling coefficients in an appropriate way:


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e. Determine scaling coefficients such that the KdV-equation (1.10) get be

put into the following, parameterless form:tu + xu +

1

63xu +

1

6u xu= 0. (1.11)

This shows that by means of scaling a lot of progress can be made. Oneonly needs to study equation (1.11), and the results are directly applicableto various different physical situations.

Scaling can also be used for another purpose as we shall show now. Inparticular we show how scaling arguments can give insight in the meaning

and relative importance of the three terms in the right hand side of (1.9) or,equivalently, of (1.10). As for comparisons in magnitude, it is necessary tomake explicit that, when dealing with the scaled variables, we assume theseto be of unit order. So, in the followingx, tanduandall of the derivatives ofuare considered to be of order unity. Small is then measured with respectto unity.

Below some limiting cases are considered.

- Small amplitude waves.Since u is of order one, the value ofa is a measure of the amplitude of thesurface elevation: for large a the physical wave heights are large, whereas

the limita 0 means that waves of infinitesimal amplitude are considered.Sincea only appears in the coefficient , we can simply take the limit a 0by taking 0. The resulting equation is

tu + xu + 3xu= 0. (1.12)

Note that this equation is linear. It can be solved by means of Fouriertransformation techniques as shown in Chapter 4. This observation explainsthe statement that the nonlinear term in the KdV equation describes effectsthat are due to the finiteness of the wave heights.

- Long and short waves.The value of the parameter L determines the length of the physical spa-

tial interval in which changes take place: for small L this interval is small,whereas for large L changes occur over a large physical interval. Small andlarge values ofL correspond to short and long waves, respectively.


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32

g. Investigate the relation between the functionsf andg that are related by

f(x) =g(X) =g(Lx) .

Relate the derivatives off to those ofg and see the effect of the value ofL.

All three parameters , and in (1.10) contain the parameter L, andtherefore it is not immediately clear what the effect is of taking a limitingvalue ofL. Therefore, we argue as follows. The appearance ofL in caneasily be scaled away by taking = Lc . Then

= 1, =1

6

H

L

2, =

3

2

a

H

. (1.13)

h. Show that ifL is given the dimension of length, will have the dimen-sion of time, and that the variablesx, tandu and the parameters, , aredimensionless.

i. Observe that nowL only appears in the coefficient. Keeping all otherparameters fixed, look at the limit for long waves, and explain the statementthat the third order spatial derivative in the KdV equation describes effectsthat are due to the length of the waves under consideration; the longer thewaves the less this term contributes. Find the equation obtained in the limitfor infinite long waves of finite amplitude.

- Long waves with small amplitudes.

j. Consider the limit of infinitesimally small, infinitely long, waves. Showthat by taking the limitsa 0, L in (1.10), the equation reduces to:

tu + xu= 0.

The solutions of this equation are easily found. Show that for arbitraryf itholds that

u(x, t) =f(x t).This solution represents that the profile f is translated in the direction ofthe positive x-axis, undisturbed in shape and at unit velocity. For this rea-son, this equation is called the translation equation.


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k. Show, by realizing thatx andt are scaled variables here, that infinitesimalwaves of long wave length propagate at speedc=

gH.

Now, let us return to (1.10) with the scaling given by (1.13):

tu + xu + H2

6L23xu +

3 a2Hu xu= 0. (1.14)

In the reasoning of Korteweg and de Vries, the last two terms are improve-ments of the translation equation in the sense that (some) effects of thenonlinearity and dependence on wave length are taken into account. Being

corrections, the coefficients should be small compared to unity, so of order

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