Fluid Dynamics 3 - 2011/2012

Jens Eggers

Preliminaries

Course information

• Lecturer: Prof. Jens Eggers, Room SM2.3

• Timetable: Weeks 1-12, Tuesday 2.00 (SM2), Thursday 4.10 (SM1) and Friday 11.10(SM1). Course made up of 32 lectures.

• Office hours: my office hours for this course are Tuesday 3.00 in my room, SM2.3

• Prerequisites: Mechanics 1, APDE2 and Calc2. Need ideas from vector calculus,complex function theory, separation solutions and PDE’s.

• Homework: Questions from 10 worksheets set and marked during the course. Willbe handed out each Friday, starting in the first week. Solutions to be returned thefollowing Friday in the box marked “Fluids3”.

• Assessment: 2.5 hours exam in April, best 4 out of 5 questions. No calculatorsallowed.

• Credit points: will be awarded for serious attempts at 40% of homework. Onlycomes into play for exam marks in the range 30-40.

• Web: Standard unit description includes detailed course information. Also a fluids3 web page (http://www.maths.bris.ac.uk/~majge/fluids3.html), which con-tains: Homework and solution sheets as well as lecture notes (will be put up on theweb as soon as possible after each lecture).

• Lectures: There is no need to take notes during the lecture, as all material relevantfor the exam will be put on the web. The main purpose of the lecture is the livedevelopment of the material, and a chance for you to ask questions!

Recommended texts

1. A.R. Paterson, A First Course in Fluid Dynamics, Cambridge University Press.(The recommended text to complement this course - costs ≈ £50 from Amazon;there are 6 copies in Queen’s building Library and 3 copies in the Physics Library)

2. D.J. Acheson, Elementary Fluid Dynamics. Oxford University Press

3. L.D. Landau and E.M. Lifshitz, Fluid Mechanics. Butterworth Heinemann

Films

There is a very good series of educational films on Fluid Mechanics available on YouTube,produced by the National Committee for Fluid Mechanics Films in the US in the 1960’s.Each film is also accompanied by a set of notes. I recommend them highly, and will pointout the appropriate ones throughout this course.

The following 3 sections are useful for the course. For the purposes of an examination, Iwould expect you to know the definition of div, grad, curl and the Laplacian in Cartesiansand grad and the Laplacian in plane polars. Other definitions would be provided.

Revision of vector operations

Let u = (u1, u2, u3), v = (v1, v2, v3) be Cartesian vectors. Let φ(x) be a scalar functionand f(x) = (f1(x), f2(x), f3(x)) a vector field of position x = (x, y, z) ≡ (x1, x2, x3). Then

• The dot product is u.v = u1v1 + u2v2 + u3v3

• The cross (or vector) product is u×v = (u2v3−u3v2)x+(u3v1−u1v3)y+(u1v2−u2v1)z

• The gradient is ∇φ =

(

∂φ

∂x1,∂φ

∂x2,∂φ

∂x3

)

• The divergence is ∇ · f = ∂f1∂x1

+∂f2∂x2

+∂f3∂x3

• The curl is ∇× f =

(

∂f3∂x2

− ∂f2∂x3

)

x+

(

∂f1∂x3

− ∂f3∂x1

)

y+

(

∂f2∂x1

− ∂f1∂x2

)

z

• The Laplacian is ∇2φ = ∇ · ∇φ =∂2φ

∂x21+∂2φ

∂x22+∂2φ

∂x23

Formulae in cylindrical polar coordinates

Coordinate system is x = (r, θ, z) where the relationship to Cartesians is x = r cos θ,y = r sin θ. The unit vectors are r = x cos θ+ y sin θ, θ = −x sin θ+ y cos θ and z. In thefollowing, f = (fr, fθ, fz) ≡ frr+ fθθ + fzz.

• The gradient is ∇φ = r∂φ

∂r+ θ

1

r

∂φ

∂θ+ z

∂φ

∂z

• The divergence is ∇ · f = 1

r

∂(rfr)

∂r+

1

r

∂fθ∂θ

+∂fz∂z

• The curl is ∇× f =1

r

∣

∣

∣

∣

∣

∣

r rθ z∂/∂r ∂/∂θ ∂/∂zfr rfθ fz

∣

∣

∣

∣

∣

∣

.

• The Laplacian is ∇2φ =∂2φ

∂r2+

1

r

∂φ

∂r+

1

r2∂2φ

∂θ2+∂2φ

∂z2

Formulae in spherical polar coordinates

Coordinate system is x = (r, θ, ϕ) where the relationship to Cartesians is x = r sin θ cosϕ,y = r sin θ sinϕ, z = r cos θ.

The unit vectors are r = x sin θ cosϕ + y sin θ sinϕ + z cos θ, θ = x cos θ cosϕ +y cos θ sinϕ− z sin θ and ϕ = −x sinϕ+ y cosϕ.

In the following, f = (fr, fθ, fϕ) ≡ frr+ fθθ + fϕϕ.

• The gradient is ∇φ = r∂φ

∂r+ θ

1

r

∂φ

∂θ+ ϕ

1

r sin θ

∂φ

∂ϕ

• The divergence is ∇ · f = 1

r2∂(r2fr)

∂r+

1

r sin θ

∂(sin θfθ)

∂θ+

1

r sin θ

∂fϕ∂ϕ

• The curl is ∇× f =1

r2 sin θ

∣

∣

∣

∣

∣

∣

r rθ r sin θϕ∂/∂r ∂/∂θ ∂/∂ϕfr rfθ r sin θfϕ

∣

∣

∣

∣

∣

∣

.

• The Laplacian is ∇2φ =1

r2∂

∂r

(

r2∂φ

∂r

)

+1

r2 sin θ

∂

∂θ

(

sin θ∂φ

∂θ

)

+1

r2 sin2 θ

∂2φ

∂ϕ2

1 Introduction & Basic ideas

1.1 What is a fluid ?

In this course we will treat the laws governing the motion of fluids and gases. A fluidor gas is characterised by the fact that there is no preferred rest state for the parts itis composed of. If a hole is made in a water bottle, the water will flow out. If a dropis placed on a solid surface, it will spread. By contrast, a solid retains a memory of itsoriginal state. If one deforms a piece of metal, it will relax back to its original state oncethe force is no longer applied.

Another important property of liquids and gases is that they are “featureless”. Viewedfrom a particular point in space, all directions are equivalent. Solids, on the other hand,often have an internal (lattice) structure. As a result, it makes a difference in whichdirection they are deformed relative to their internal structure.

Fluid dynamics is an example of ‘continuum’ mechanics:

Defn. A continuum is any medium whose state at a given instant can be describedin terms of a set of continuous functions of position x = (x, y, z). E.g. Density, ρ(x, t),velocity, u(x, t), temperature T (x, t). (These functions usually also depend upon time,t).

We know that this description fails if we observe matter on small enough length scales:e.g. typical molecule size (∼ 10−9m) or typical mean free path in a gas (∼ 10−7m). Themiracle is that on a scale only slightly larger than that, all microscopic features can beignored, and we end up with a “universal” description of all things fluid. We will alsoignore all effects of incompressibility, so for the purposes of this course, fluids and gasesare the same thing. We will almost always speak of a fluid, but which can mean either afluid or a gas for the purposes of this course.

The motion of fluids and gases is governed by the same underlying princi-ples

1.2 Viscosity

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x

xδ

tangential force F

stationary plate

plate moving at speed U

plate area, Au(y) = γy

A very important approximation we will be making throughout this course is the neglectof internal friction (or viscosity) of the fluid. For example, consider two plates of areaA being moved at speed V relative to each other (see figure). It is common sense that

a certain force is necessary to maintain this motion, just like it takes a certain force torub one hand against the other, and there is friction between the two. In the case of afluid, the origin of this force is that molecules “rub” against each other. This way a forceis transmitted from one plate to the other. One expects that the necessary shear force isproportional to

1. the speed U

2. the area A

and inversely proportional to

3. the distance δ between the plates.

Thus the force is given by

F = ηUA

δ, (1)

where the constant of proportionality η is called the viscosity η of the fluid. In thiscourse we will neglect all viscous effects, we will pretend that the viscosity is small. Notethat what constitutes a small viscosity depends on the dimensions of the system: if δis sufficiently small, the force can be made to be significant, no matter how small theviscosity! There is a wonderful way to avoid further discussion of the matter: if we canneglect all viscous effects, we speak of Ideal Fluids.

1.3 What we would like to do

In this lecture course, we will first develop an equation of motion for the velocity fieldu(x, t), which gives the fluid velocity at any instant in time t, everywhere in space.This equation (or set of equations) will necessarily have the form of a partial differential

equation. It will be based on Newton’s equations of motion, but for a continuum ofparticles, distributed over space.

With the equations in hand, it is down to our ability to deal with the mathematicalcomplexities of solving a partial differential equation (PDE) to solve physical problems.Some examples of problems dealt with rather successfully using the concept of ideal fluidsare the following:

There are many wonderful phe-nomena described quite well within the “ideal” framework of this course. One of them

are impact problems. The Jesus Christ lizard (see Figure) can walk on water by hitting itssurface with its feet. At each impact, it receives an upward impulse, which compensatesits weight.

Another spec-tacular success is the theory of flight. The ideal flow of air around a wing is able to describethe lift necessary for flight, and much more. What is much more difficult is the theory ofdrag. Inviscid theory suggests that there should be no energy cost to flight at all!

Another huge (and hugely impor-tant) area is the theory of water waves. This includes waves from the scale of millimetresup to huge tsunami waves. The absence of any solid boundary results in very little friction,so the ideal theory works very well.

1.4 Outside the realm of ideal fluids

Of course in reality there is always atleast a little bit of viscosity, so a viscous description will always be more accurate than aninviscid one. If the viscosity is large (think for example of Lyle’s golden syrup), a theorybased on neglecting viscosity is of course spectacularly wrong. We will not be able todescribe the pouring of syrup shown in the picture (note the interesting shape of bubblestrapped inside the syrup)! The same is true if the geometry under consideration is smalland confined. Then there is a lot of shearing going on, and frictional forces are large.

Unfortunately, thereare also many cases where inviscid theory fails even if the viscosity is small, as illustratedin the figure showing the flow around an obstacle. The fluid flow does not remain attached

to the body, but rather separates from it at the rear. We will see that ideal theory usu-ally predicts the flow lines to hug the body, even in the back. The reason for the failureof ideal fluid theory is very subtle. There is a very small region near the body (calledthe boundary layer), where viscosity becomes important. It is in this small region, notdescribed properly by inviscid theory, where separation occurs. More on this in Fluids4...

1.5 Lagrangian and Eulerian descriptions of the flow

We now begin to develop a dynamical description of fluid flow, which will lead us toformulate a PDE for fluid motion, known as the Euler equation. Before we can dothat, we must understand the motion of fluids a little better. The description of motionis called kinematics. In this chapter, we will deal with kinematics. I encourage you tolook at the film “Fluid Mechanics (Eulerian and Lagrangian description) parts 1-3” onYouTube.

There are two very different ways of describing fluid motion, known as the Eulerianand Lagrangian description. Ultimately, they are equivalent, as they describe the samething. However, they serve different purposes, so we need them both.

Eulerian description of the flow.This is what the stationary observer sees. Choose a fixed point, x to measure, for e.g. thevelocity u(x, t). This provides a spatial distribution of the flow at each instant in time.This is the way continuum equations are usually formulated, and our equation of motionwill indeed be an equation for the Eulerian field u(x, t). If the flow is steady, then u doesnot depend on time, t: u = u(x).

Why do we need anything else? The reason is that to make contact with Newton’sequations, we need to describe the flow as a moving particle would see it. This is theLagrangian description of the flow.The observer moves with the fluid. Choose a fluid particle (for example, we can place asmall drop of ink in the fluid), and follow it through the fluid. Measuring its velocity ata given time, t gives its ‘Lagrangian velocity’. Now we describe the whole velocity fieldthis way, by labelling all material points. A convenient way of doing so is to choose aninitial time t0, and to label all fluid particle by their position x = a at that time. Thenat time t > t0, the particle is at

x = x(a, t), where x(a, t0) = a.

Of course, x(a, t) is very interesting in its own right. For example, it describes the courseof a balloon, launched at time t0 and at position a into the atmosphere.

The Lagrangian velocity is defined as

v(a, t) =∂x

∂t

∣

∣

∣

∣

afixed

.

By definition, it is the velocity of a particle going with the flow. This is precisely howvelocity is defined in Newtonian Mechanics. As we said earlier, the Eulerian and theLagrangian velocity fields contain the same information; the relation between the two is:

v(a, t) = u(x(a, t), t). (2)

An example: Consider logs flowing along a narrowing section of river. A fixed observermeasures the velocity by observing the velocity of logs at a given point in space. Byobserving many logs at different positions, he will be able to obtain the entire Eulerianvelocity field u(x, t). Unless the flow conditions are changing, this field will be time-independent, u(x, t) = u(x). Now imagine each log being ridden by a moving observer,each of whom reports his velocity as time goes by. If all observers are labelled by theirposition a at some reference time t0, this will produce the Lagrangian velocity field v(a, t).

The logs travel with the fluid and will see the flow accelerating, as the river bed becomesnarrower. Thus v(a, t) is manifestly time-dependent although the Eulerian field is not!

Stationary observer sees logs passing at constant speed.

Observer moving with the flow sees the logsaccelerate

Defn: A flow is two-dimensional if it is independent of one of its components (in somefixed frame of reference). E.g. u = (u, v, 0).

Example 1: Let us consider a very simple model flow for the river shown above.Consider the two-dimensional, stationary Eulerian velocity field u = (u, v, 0), defined by

u = kx, v = −ky.

As can be seen in the Figure, the flow speeds up as the rivercontracts, which we can think of as being confined by the river banks, shown as the redlines.

1.6 Particle paths and streamlines

Defn. The particle paths (pathlines) are the paths followed by individual particles. Theyare determined by the solution of the differential equation:

dx

dt= u(x, t), with initial condition x(t0) = a = (a1, a2, a3) (3)

where u is assumed given.This system of equations specifies a unique curve. For some (simple) u, can be inte-

grated using elementary methods, but in general not. Let u = (u(x, t), v(x, t), w(x, t)),then in components, (3) is

dx

dt= u(x, y, z, t)

dy

dt= v(x, y, z, t)

dz

dt= w(x, y, z, t)

with x(t0) = a1, y(t0) = a2, z(t0) = a3.

Example 1: In the example above, the equations become

x = kx, y = −ky.

The solutions, with initial positions a = (a1, a2) at time t = 0, are

x(t) = a1ekt, y(t) = a2e

−kt,

which describes the path of particles in the flow. Using the relation (2) between Eulerianand Lagrangian fields, we find

v(a, t) =(

ka1ekt,−ka2e−kt

)

.

In particular, the Lagrangian velocity field is indeed time-dependent, while the Eulerianfield was steady. Now let us consider a case where the Eulerian field is also time dependent.

Example 2: Two-dimensional flow,

u = (y,−x) cos t,

with x(t0) = (1, 0). As seen in the Figure, this is a vortex whose strength (and sense ofrotation) oscillates.

Thendx

dt= y cos t, (∗) dy

dt= −x cos t,

which we can combine as.

0 = xdx

dt+ y

dy

dt=

d

dt(12x2 + 1

2y2),

so that the solution isx2 + y2 = 1.

Now we know that the particle paths lie on a circle, which is not a surprise when lookingat the velocity field. We still need to find the time dependence x(t) and y(t).

We can describe the curve by

x(t) = cos θ(t), y(t) = sin θ(t),

and so dx/dt = − sin θθ = −yθ and from (∗) must have θ = − cos t or θ = − sin t+ const.The constant is sin t0, since θ = 0 at t = t0. Thus the particle paths oscillate around acircle, with angle

θ(t) = sin t0 − sin t.

As shown in the Figure below, the particle position oscillates back and forth in a section

of the circle. θ=−1,

θ=1, t= 3π/2

t= π/2

0,π,2πt=x

y

0e.g. t =0

Defn: A streamline of a flow u(x, t) at a given instant in time t0, is a curve which iseverywhere parallel to u(x, t0).

Thus, along a streamline,dx

dt= u(x, t0). (4)

Eliminating t, we finddx

u=dy

v=dz

w(= dt), t = t0. (5)

Notice that time doesn’t go into the definition of a streamline, and we could have takenany quantity to parameterise the curve.

In general, a streamline varies with time, but are time-independent for a steady flow.Streamlines can be visualised by taking a short-time exposure of illuminated particles ina flow. Taking an arbitrary starting point, one always continues in the local flow direction.

For steady flows, streamlines and particle paths coincide.Evidently, the defining equations (4) and (3) become the same, if u does not depend ontime. Example 1: u = (kx,−ky).

Since this is a steady flow, we could eliminate t from the pathline to obtain the streamline.Alternatively, we have

dx

kx= −dy

kv

according to (5). Integrating, we find ln |x|+ ln |y| = C, or

|x||y| = eC ,

which is the equation for a hyperbola. Any such hyperbola is a possible equation for theriver bank, which must be a streamline of the flow.

Example 2: u = (y,−x, 0) cos t.At t = t0,

dx

y cos t0= − dy

x cos t0so that

dx

dy= −y

xand then integrating, x2 + y2 = const.

Thus streamlines are circles, and are different from pathlines, which just cover a sectionof a the circle. It is also important to draw arrows into the streamlines, to indicate thelocal direction of the flow. This has to be done only once for each streamline, since thedirection cannot reverse; it is enough to compute the velocity at one point along thestreamline. As seen in the Figure below, the direction of flow reverses periodically in our

example.

t <0y y

x x

π/2 3π/2<0t<π/2

Before we go on, we will revise some material from vector calculus.Appendix A goes in here

1.7 The Lagrangian derivative

(a.k.a. the convective derivative, or the material derivative).

We know how to measure the time derivative of a physical quantity associated with thefluid, for example that of the temperature T (x, t), at a fixed point in space (the Eulerianderivative). It’s just

∂T

∂t.

This quantity will describe the change of temperature at a fixed location, for example airtemperature in Bristol. However, this quantity will not be a measure of how a mass ofair heats up or becomes colder. The reason is that air is swept away by the prevailingflow field u(x, t). In other words, to describe the change of temperature of a piece of air,we need to consider the rate of change of T (x, t), following a fluid particle. This is calledthe Lagrangian derivative.

δx+ x

t+ δtx=uδ δt

O

x

t

Suppose a particle at time t is positioned at x, then at time t + δt it has moved tox+ uδt. The change in T in the instant δt is

T (x+ uδt, t + δt)− T (x, t) (6)

Taylor expanding about x, t we get

T (x, t) + uδtTx(x, t) + vδtTy(x, t) + wδtTz(x, t) + δtTt(x, t) +O((δt)2)− T (x, t).

The Lagrangian derivative is (6) divided by δt as δt→ 0, so that we get

∂T

∂t+ (u · ∇)T ≡ DT

Dt. (7)

A particularly important example is the change in velocity (the acceleration) of a fluidparticle, which we need to apply Newton’s equations to fluid motion. Of course, thevelocity is a vector quantity, which means we have to apply (7) to each component:

Du

Dt=

(

Du1Dt

,Du2Dt

,Du3Dt

)

.

The the acceleration of a fluid particle becomes

Du

Dt=∂u

∂t+ (u · ∇)u. (8)

Note: The Lagrangian derivative (i.e. following fluid particles) is given in terms ofEulerian (i.e. fixed point) measurements. It is vital to understand exactly how to com-pute expressions like (u · ∇)u for a given velocity.

Example: Consider an accelerating fluid flow, such as the log flowing through anarrowing channel. Suppose

u = (U + kx, U − ky, 0).

Then

• The flow is steady since∂u

∂t= 0.

• The advective term is

(u · ∇)u =

(

(U + kx)∂

∂x+ (U − ky)

∂

∂y

)

(U + kx, U − ky, 0).

Hence the acceleration of the log is

Du

Dt= (k(U + kx),−k(U − ky), 0) .

1.8 Mass conservation

One of the fundamental laws of continuum mechanics is the law of mass conservation.That is, fluid is neither created or destroyed.

Consider an arbitrary finite volume, V , which is fixed in a fixed frame of reference. Vis bounded by the surface S and n represents a unit normal on S outward from V .

A fluid occupies the space of which V is a subset. The fluid has velocity u(x, t) anddensity ρ(x, t). Fluid can flow in and out of V , and its density can change (within V ).

ds

n

V S

The mass contained in V is

∫

V

ρ(x, t)dV .

The flux of mass into V (n points outward) is −∫

S

ρu · ndS. The quantity j = ρu is

called the mass flux density.

n

u

dS

In time dt the fluid on the surface dS is transporteddistance u · n in a direction perpendicular to dS.Thus, the mass transported is ρ(u · n)dtdS and the

rate of mass transport is this quantity divided by dt.

So the rate of change of mass in V must equal the rate of change of mass in/out of Vthrough S. So

d

dt

∫

V

ρ(x, t)dV = −∫

S

ρu · ndS

Since V does not change with t and using the divergence theorem for the RHS we get∫

V

∂ρ

∂tdV = −

∫

V

∇ · (ρu)dV

or∫

V

(

∂ρ

∂t+∇ · (ρu)

)

dV = 0

This is true for any fixed V , so must have

∂ρ

∂t+∇ · (ρu) = 0 (9)

at every point in the fluid. This is called the mass conservation equation or the continuityequation.

Remark: The structure of this equation is very general, and applies to any conserva-tion law, which can be written

∂ρ

∂t+∇ · j = 0. (10)

Here j is the flux of the quantity in question.

1.9 Incompressibility

Defn A fluid is said to be incompressible if the density of each fluid ‘particle’ is constant,

(i.e.Dρ

Dt= 0).

From (9) we have

0 =∂ρ

∂t+∇ · (ρu) = ∂ρ

∂t+ u · ∇ρ+ ρ∇ · u =

Dρ

Dt+ ρ∇ · u

So (ρ > 0) an incompressible fluid satisfies

∇ · u = 0. (11)

No fluid is completely incompressible, but even gases are often sufficiently incompress-ible for (11) to apply to their motion. Incompressibility is a valid approximation if

• The flow speed is much less than the velocity of sound;

• Timescales are much larger than (sound frequency)−1.

Of course this leaves out everything to do with sound waves, which are due entirely tocompressible effects.

Before we go on, we will revise more material from vector calculus.Appendix B goes in here

1.10 Streamfunctions (for incompressible flows)

If ∇ · u = 0, then it follows that there exists a vector field A(x, t) s.t.

u = ∇×A

The so-called vector vector potential A is complicated unless the flow is two-dimensionalin which case it is very simple.

1.10.1 Two-dimensional flows

(i) Cartesians: Here A = ψ(x, y, t)z. In other words, the flow is determined by a singlefunction ψ. This corresponds exactly to the expected degrees of freedom of the flow: thereare two components of the velocity, and one constraint ∇ · u = 0. Then

u = ∇×A =

(

∂ψ

∂y,−∂ψ

∂x, 0

)

(and clearly ∇ · u = 0).

Reminder: streamlines are given bydx

u=dy

v, where u = (u, v, 0). So

vdx− udy = 0, ⇒ ∂ψ

∂xdx+

∂ψ

∂ydy = dψ = 0

(by Chain rule). Therefore ψ(x, y, t) = const on a streamline of the flow.

Defn: We call the function ψ(x, y, t) the streamfunction of the flow.

ψ= constant

Note: For steady flows, the streamlines do not cross each other and fluid does notcross the streamlines.

(ii) Cylindrical polar coordinates: Now let A = ψ(r, θ, t)z in cylindrical polars(r, θ, z). Then (confirm ...)

u = ∇×A =

(

1

r

∂ψ

∂θ,−∂ψ

∂r, 0

)

where u = (ur, uθ, 0) (check (i) ∇ · u = 0 satisfied again and (ii) streamlines given byψ = const)

Example 1: A simple source: the flow is purely radial. So uθ = 0 and ur is indepen-dent of θ.

ur = f(r) =1

r

∂ψ

∂θ

uθ = 0 = −∂ψ∂r

Must have ψ = ψ(θ) but ∂ψ/∂θ independent of θ. So ψ = Aθ where A constant. Thenf(r) = A/r.

Defn: The source strength is the flux of fluid from the source point. The flow isincompressible, so the flux at the origin equals the flux through any closed boundarysurrounding the origin. Found, most conveniently, by measuring the flux through a circleradius r centred at the origin. The source strength is

m =

∮

circleu · ds =

∫ 2π

0

u.nrdθ =

∫ 2π

0

A

rrdθ = 2πA.

where n = r = (1, 0, 0) in polars. This means the mass flow through a circle surroundingthe origin is independent of the radius of the circle, and equals the source of mass fluxat the origin. In fact, one can show that the mass flux through any closed surface is m.(how?)

Note: In Cartesians, ψ = A tan−1(y/x).

Example 2: The same can be done for a 3D point sourceThe flow is once more radially symmetric, that is we expect u = f(r)r with the source

at the origin. Now the total flux through a sphere of radius r (which should give thesource strength) is

m =

∫

sphereu · ndS = 4πr2f(r).

Therefore the radial dependence must be f(r) = m/4πr2, and thus

u =m

4πr2r, (12)

the flow of a three-dimensional point source of strength m.

2 Flow dynamics for an incompressible inviscid flow

2.1 Forces on a fluid

Fluids move in response to the forces that on each fluid particle. These forces are of twotypes:

• Force on parcel of fluid, volume δV may be proportional to δV . These are knownas body forces (e.g. gravitational force, ρδV g).

• Force is transmitted across the surface element δS of a fluid parcel. I.e. force exertedby the fluid on the exterior of δV or vice versa. These are surface forces.

When the fluid is a rest, the surface force must be in the direction of the normal, n.(The definition of a fluid is one that remains at rest unless a tangential force is applied).

When a fluid is in motion, tangential components of the force on δS can occur. Theseare associated with viscosity which describes the effect that one layer of fluid in motionhas on an adjacent layer.

Defn An fluid is said to be inviscid when viscous effects are small enough to be ide-alised as zero (e.g. ‘slidy’ fluids but depends on lengthscales).

For an inviscid fluid, the ‘surface stress’ (i.e. Force per unit area) is in the direction nnormal to the surface δS even when the fluid is in motion.

That is, the surface force Fs = −pnδS = force exerted by exterior fluid on fluid insideδV .

The magnitude of this force is −pδS where p(x, t) is the pressure and it is directedinwards because fluids are usually in a state of compression.

2.2 Equation of motion

We apply Newton’s law to a fixed (arbitrary) volume V with surface S. The momentumwithin V may change as a result of (i) body forces (ii) surface forces, and (iii) fluidmomentum flow out of V .

u.n( )z

y

x0

nS

VdS

nuρ

dSδ t

Total momentum in V =

∫

V

ρudV

In time δt, the fluid momentum leaving a small section of surface δS is ρu(u · n)δtδS.

So rate of momentum transport (i.e. the momentum flux) is ρu(u · n)δS.

Thusd

dt

∫

V

ρudV = −∫

S

ρu(u · n)dS −∫

S

pndS +

∫

V

FdV

In each i = 1, 2, 3 components,

∫

V

∂

∂t(ρui)dV = −

∫

S

ρui(ujnj)dS −∫

S

pnidS +

∫

V

FidV.

since V is fixed. Now apply the divergence theorem to get

∫

V

∂

∂t(ρui)dV = −

∫

V

∂

∂xj(ρuiuj)dV −

∫

V

∂p

∂xidV +

∫

V

FidV.

Now since V is an arbitrary volume within the fluid, we must have

∂

∂t(ρui) +

∂

∂xj(ρuiuj) = − ∂p

∂xi+ Fi. (13)

This is (one form of) the equation of motion, but it can be simplified by expanding as

ui∂ρ

∂t+ ρ

∂ui∂t

+ ρuj∂ui∂xj

+ ui∂

∂xj(ρuj) = − ∂p

∂xi+ Fi

and using the continuity equation (9) in component form:

∂ρ

∂t+

∂

∂xj(ρuj) = 0

so that we have

ρ∂ui∂t

+ ρuj∂ui∂xj

= − ∂p

∂xi+ Fi.

Finally, we can identify this in vector notation as

ρ∂u

∂t+ ρu · ∇u = −∇p+ F (14)

or, using the convective derivative, as

ρDu

Dt= −∇p + F. (15)

This is the momentum equation or Euler’s equation (due to Euler 1756).

2.3 Hydrostatics

If the body forces are conservative we can write F = −ρ∇Φ where Φ is a potential. Then,if ρ = const, (15) becomes

ρDu

Dt= −∇P, where P = p+ ρΦ is a modified pressure (16)

For a fluid at rest, u = 0 and so it follows

∇P = 0, or, in general, ∇p = F

I.e. P = const.

In the case of a fluid at rest under gravity, the gravitational force on a fluid parcel ofvolume δV is ρδV g and so the force per unit volume is −ρgk, where k is the unit vectorvertically upwards. Then Φ = gz so that F = −ρ∇(gz). So, for ρ = const,

P = p+ ρgz = const, in the fluid

And for ρ = ρ(z) (as in the oceans, atmospheres),

∇p = −ρ(z)gk, ⇒ p(z) = p(z0)− g

∫ z

z0

ρ(z′)dz′

2.3.1 Example: The milk bottle.

patm

R

r

h

Hρ

ρ

m

c

z

Suppose a milk bottle consists of a smaller cylinder on top of a larger cylinder. Creamof density ρc occupies the smaller cylinder and floats on milk of density ρm > ρc in thelarger cylinder below.

We need to solve∂p

∂z= −ρg where ρ =

ρc, H < z < H + hρm, 0 < z < H

and p = patm (at-

mospheric pressure) on z = H + h and p must be continuous at z = H . This givesfirst

p(z) = patm + ρcg(H + h− z), H < z < H + h

thenp(z) = patm + ρcgh+ ρmg(H − z), 0 < z < H

The pressure at the base is p1(0) = patm + ρcgh+ ρmgH .

Compare with homogenised (shaked-up) mixture: here ρh =πr2hρc + πR2HρmπR2H + πr2h

, for

0 < z < H + h. So

ρh = ρm − r2h

R2H + r2h(ρm − ρc)

andp(z) = patm + ρhg(H + h− z), 0 < z < H + h.

The pressure on the base is now

p2(0) = patm + ρhg(H + h) = patm + ρmg(H + h)− r2h(H + h)

R2H + r2hg(ρm − ρc)

The difference between the base pressures in problems 1 and 2 is

p1(0)− p2(0) =

(

r2h(H + h)

R2H + r2h− h

)

g(ρm − ρc) =hH(r2 − R2)

R2H + r2hg(ρm − ρc) < 0

since R > r. So pressure is greater for mixture. Why ?

2.3.2 Archimedes principle (∼ 250BC)

V

S n

z

Stationaryfluid, density ρ

Force on submerged body = −∫

S

pndS

= −∫

V

∇pdV, Divergence Theorem

=

∫

V

ρgkdV, k unit normal in z-direction

= ρgV k

Therefore force on body = weight of fluid displaced is ρgV .

What about a body intersecting two fluids of different densities (air/water) ?

2.3.3 Example: Lock gates

zH

H

p

pp

atm

atmp

2

1

21( )

( )zz

In x < 0, solve∂p1∂z

= −ρg with p = patm on z = H1:

p1(z) = patm + ρg(H1 − z), 0 < z < H1

and similarly, in x > 0,

p2(z) =

patm, z > H2,patm + ρg(H2 − z), 0 < z < H2,

Force exerted by fluid on gate from x < 0 is

F1 =

∫ H1

0

p1dz = patmH1 +12ρgH2

1

Force exerted by fluid on gate from x > 0 is

F2 =

∫ H1

0

p2dz = patmH1 +12ρgH2

2

Net force is 12ρg(H2

1 −H22 ). What about turning moment ?

2.4 The momentum integral

From §2.2, equation (13),∂

∂t(ρui) +

∂Mij

∂xj= 0 (17)

where Mij = ρuiuj + Pδij is called the momentum flux (recall P = p + ρΦ) and ρui iscalled the momentum density. Note that (17) has the same fundamental structure than(9), as is typical for conservation laws. For example, if I define the flux of x-momentumpx by

f(px)j =M1,j,

then it follows from (17) that∂px∂t

+∇ · f (px) = 0,

which describes the conservation of x-momentum. The equations for the other compo-nents are analogous.

For steady flows, LHS of (17) is zero. So

0 =

∫

V

∂Mij

∂xjdV =

∫

S

MijnjdS

by the divergence theorem. So, back in vector form,∫

S

ρu(u · n) + PndS = 0

for any closed surface S in/bounding the fluid.

This is the momentum integral theorem

2.5 Example: Axisymmetric jet impinging on a wall

. What is the force exterted by the flow on the wall ? Ignore the effects of gravity.

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z

r

film

wall

jet

p = patm

p = patm

S

u = −U z

n = z

u = Urrn = r

n = −z

Take a control surface S of cylindrical shape that is composed of the free surface of thejet on one side, and the wall on the other. The surface is closed by a circular surfacethrough the jet far from impact (marked “jet”), and a cylindrical surface where the fluidflows parallel to the wall (marked “film”). Assume uniform flow parallel to wall in thefilm region far from impact and across the cross section of the jet u = −U z. Also assumepressure on all boundaries apart from the wall is patm. Then P = p (no body forces) and

∫

S

ρu(u · n) + pndS = 0.

Now∫

S

pndS =

∫

S

(p− patm)ndS +

∫

S

patmndS = −∫

wall

(p− patm)zdS

since

∫

S

patmndS =

∫

V

∇patmdV = 0 and p = patm everywhere else. Also,

∫

S

ρu(u · n)dS =

∫

jet

ρU2zdS +

∫

film

ρU2r rdS

since u · n = 0 on the wall. Bringing it all together and equating terms in direction zgives

∫

wall

(p− patm)dS = ρU2A,

where A is the cross section of the jet. So there is an extra pressure, and the force extertedby the fluid on the wall is ρU2A. [What about Ur and U2 ? Later!]

2.6 Dynamic boundary condition

Consider a surface of discontinuity, C (e.g. interface between oil/water).

Vε

water

oil

ε

n

C

Integrate momentum equation throughout a small disc Vǫ of thickness ǫ and let ǫ→ 0:

∫

Vǫ

∂

∂t(ρui)dV → 0, as ǫ→ 0.

Therefore

0 = limǫ→0

∫

Vǫ

∂

∂xjMijdV = lim

ǫ→0

∫

Sǫ

MijnjdS = [Mijnj ]+−

So[ρu(u · n) + pn]+− = 0 (18)

Provided ρ = const in each layer, the kinematic boundary condition (see §1.9) implies[u · n]+− = 0.

So component of (18) in direction n is

[p]+− = 0

Which just says pressure is continuous across an interface.

2.7 Bernoulli’s equation for steady flows

We start with the vector identity (see Appendix A)

u · ∇u = ∇(12u · u)− u× ω

where ω = ∇× u is the vorticity. Write u2 = u · u and use above in (16)

∂u

∂t− u× ω = −∇(P/ρ+ 1

2u2) (19)

The flow is steady, so∂

∂t= 0 and taking the dot-product with u on both sides gives

u · (u× ω) = 0 = u · ∇(P/ρ+ 12u2)

The definition of a streamline isdx

ds= λu (s measured along a streamline), so

d

ds(P/ρ+ 1

2u2) =

dxids

∂

∂xi(P/ρ+ 1

2u2) = λu · ∇(P/ρ+ 1

2u2) = 0

Consequently,

P/ρ+ 12u2 = p/ρ+ Φ+ 1

2u2 = const, along any streamline in the flow (20)

This equation can be seen as a statement of energy conservation, where u2/2 is the kineticenergy, and Φ the potential energy.

2.7.1 Example: Flow out of a tank

A tank of uniform cross section A0 with a small hole, area Ae at a height h above thebase. The height of the fluid above the hole is H

H

h

U

U

Aatm

e

Ae

o

oarea

area

streamlinesz

p = patm

Q: What is the flow speed out of the drain ?First, conservation of mass gives

ρA0U0 = ρAeUe (21)

(mass fluxes across two boundaries equal). Assume steady flow (at any instant, flow isapproximately steady). Here, Φ = gz. Streamlines connect the surface to the exit. Alongany one of these streamlines apply Bernoulli’s equation.

• At surface: patm + ρg(h+H) + 12ρU2

0 = C

• At exit: patm + ρgh+ 12ρU2

e = C

where C = const. Eliminating C and patm and use (21) to give

U2e [1− (Ae/A0)

2] = 2gH

Reasonable to assume that Ae ≪ A0 so Ue ≈√2gH.

1) Draining time found by integrating up

dH

dt= −U0 ≈ −Ae

A0

√

2gH.

If the draining time is td and the initial height above the drain is H0 then we get

td =A0

Ae

√

2H0/g

2) Distance travelled by the jet. The flow speed is a maximum at t = 0 when H = H0

and Ue =√2H0g. Assume projectile motion of the jet of water. Then time taken to hit

the floor found from h = 12gt2 and distance travelled is x = Uet – simplifies to x = 2

√H0h.

To find the value of h for which x is a maximum, for a given fill height Hf = H0+h weneed to maximise x = 2

√

(Hf − h)h w.r.t. h and this gives h = 12Hf . Then xmax = Hf .

2.7.2 Example: Flow through a slowly diverging channel

streamline

areaarea

U1

U2

A

A2

1

Mass conservation: A1U1 = A2U2.

Bernoulli’s equation (ignoring gravity): 12ρU2

1 + p1 = const = 12ρU2

2 + p2.

[Note: fast flow implies low pressure and vice versa]

Hence, ∆p = p1 − p2 =12ρ(U2

2 − U21 ) =

12ρU2

2 (1− A22/A

21).

2.8 The vorticity equation

We have seen that the velocity field can be decomposed into a straining part and arotational part, which is given by the vorticity ω. We now try to find out what governsthe vorticity? We start with Euler’s equation written as in (19), namely

∂u

∂t− u× ω = −∇(P/ρ+ 1

2u2)

Taking the curl of this equation gives

∂ω

∂t−∇× (u× ω) = 0

We can use a vector identity:

∇× (u× ω) = u(∇ · ω)− ω(∇ · u) + (ω · ∇)u− (u · ∇)ω = (ω · ∇)u− (u · ∇)ω

since the flow is incompressible and ∇ · ω = ∇ · (∇× u) = 0. Hence

∂ω

∂t+ u · ∇ω = ω · ∇u

orDω

Dt= ω · ∇u

This is called the vorticity equation.

In 2D flows, u = (u(x, y), v(x, y), 0) and so, by definition ω = (0, 0, ω(x, y)) = ω(x, y)z.It is then clear that the term

ω · ∇u = ω(x, y)∂u

∂z= 0

and soDω

Dt= 0

That is, vorticity is conserved as it moves with the flow.

Note: If ω = 0 at time t = 0, then ω = 0 for all time. Vorticity cannot be generatedin a 2D flow.

2.9 Kelvin’s circulation theorem

Defn: Circulation of a velocity field is defined to be

Γ =

∮

C(t)

u · dl

where C(t) is a closed loop which moves with the fluid and dl is an infinitesimal linesegment along C(t).

Note: By Stokes’ theorem

Γ =

∮

C(t)

u · dl =∫

S(t)

(∇× u) · ndS =

∫

S(t)

ω · ndS

where S(t) is a surface whose edges connect with C(t).

Kelvin’s theorem states thatDΓ

Dt= 0.

Proof: Consider two points on C(t), x1 and x2 s.t. dl = x2 − x1. In a short time δt,

x1 → x1 + u(x1)δt

x2 → x2 + u(x2)δt

⇒ dl → dl+ (u(x2)− u(x1))δt

Now, by Taylor expanding,

u(x2) = u(x1 + dl) = u(x1) + (dl · ∇)u,

which is the derivative in the dl-direction, so that dl → dl+ (dl · ∇)uδt. That is

D

Dtdl = (dl · ∇)u (22)

NowD

Dt(u · dl) = dl · Du

Dt+ u · D

Dtdl = dl · Du

Dt+ u · (dl · ∇)u.

The last term can be written as dl · ∇(12u · u) since, using (22) it is

uiD(dli)

Dt= ui(dlj)

∂ui∂xj

= dlj∂ 1

2uiui

∂xj.

So now,D

Dt(u · dl) = dl ·

(

Du

Dt+∇(1

2u2)

)

Finally, using Euler’s equation (16) (for ρ const) we have

D

Dt

∮

C(t)

u · dl = −∮

C(t)

∇(P/ρ− 12u2) · dl = 0

because

∮

∇v · dl =∫

S(t)

(∇×∇v) · ndS = 0. Hence result.

Note: Irrotational flow can have circulation if it has point sources of vorticity. E.g inpolars, suppose u = (0, A/r, 0). Then ω = ∇× u = 0 apart from at r = 0. But

Γ =

∮

C

u · dl =∫ 2π

0

A

rrdθ = 2πA.

(steady flow, C is a circle radius r, so dl = (0, rdθ, 0)).

3 Irrotational flows: potential theory

Kelvin’s circulation theorem states that the circulation around any loop convected bythe flow cannot change, if the fluid is inviscid. In particular, if the circulation is zero, itwill remain zero. Now if there is no vorticity present in the flow at some intitial time,the circulation around any loop in the flow is zero. Hence Kelvin’s circulation theoremguarantees that there will never be any circulation in the flow, and thus ω = ∇× u = 0throughout the domain. No vorticity can be created in a flow domain if the viscosity issmall.

However, we will see in a little while that problems arise with this argument whenapplied near solid boundaries. The reason is that the circulation theorem applies toloops. However, it is impossible to surround a point on a solid boundary by a loop thatstays in the flow. Thus no conclusions can be drawn on parts of the flow that originatefrom a solid wall. For the time being we simply concentrate on those parts of the flowwhich are reasonably free of vorticity. In that case, very powerful tools from potentialtheory can be brought to bear on fluid problems, as we will see now.

3.1 The velocity potential

If ω = ∇ × u = 0 throughout the domain (apart from at isolated singularities) - i.e.irrotational - it follows that there exists a φ(x, t) s.t.

u = ∇φ.Then φ is called the velocity potential or simply the potential. If the velocity field u isgiven, the potential is calculated from the path integral over u,

φ =

∫

x

x0

u · dl,

provided of course the flow is indeed potential, i.e. there is path-independence.If the fluid is also incompressible then, ∇ · u = 0 and

(∇ · ∇)φ = ∇2φ ≡ φ = 0

This is Laplace’s equation. When the fluid domain communicates with a (moving) solidboundary, we impose the Kinematic boundary condition, §1.9, u.n = f(x) or n.∇φ = f(x)for some given f .

The field equation (Laplace’s) and the boundary conditions are an example of aNeumann boundary-value problem. An important feature of the equations are that theyare linear and so we can use superposition of solutions. This may seem odd, since theEuler equation is a nonlinear equation, and does not obey the superposition principle.However, as we will see very soon, the pressure depends on φ in a non-linear way (later),which reflects the nonlinear character of the Euler equation.

3.1.1 Uniqueness of the velocity potential

Suppose an incompressible irrotational fluid occupies a simply connected domain D, sou = ∇φ and that on the boundaries S of D, u · n = f(x).

Suppose that φ is non-unique. I.e. let φ1 and φ2 satisfy

∇2φi = 0, and n · ∇φi = f(x), for i = 1, 2

s.t. φ1 6= φ2. Let ψ = φ1 − φ2

∫

V

|∇ψ|2dV ≡∫

V

∇ψ · ∇ψdV =

∫

V

(

∇ · (ψ∇ψ)− ψ∇2ψ)

dV, and ∇2ψ = 0

=

∫

S

ψ∇ψ · ndS = 0, using divergence theorem

since ∇ψ · n = ∇φ1 · n − ∇φ1 · n = 0 on S. So must have |∇ψ| = 0 throughout thedomain, so ψ = const. So φ1 and φ2 only differ by a constant, which is irrelevant, as onlyterms in ∇φ appear.

3.2 Some simple flows and their potentials

3.2.1 Uniform flow

If the velocity field is constant, u = U = const, a simple integration yields

φ = U · x.

If U is chosen to point in the x-direction, U = U x, then φ = Ux, and the streamfunctionis ψ = Uy.

3.2.2 A three-dimensional source

The flow field has been given in section 1.10.1 cf. (12). Integrating radially from infinity,we find

φ = − m

4πr.

It is shown readily that Laplace’s equation is indeed satisfied:

∂2i1

r= −∂i

xir3

= − 3

r3+ 3

x2ir5

= 0.

The Stokes stream function (which has to be used because the flow is axisymmetric) forthis flow becomes simplest in spherical polars, and is defined by

A =Ψ(r, θ)

r sin θϕ,

so

ur =1

r2 sin θ

∂Ψ

∂θ, uθ = − 1

r sin θ

∂Ψ

∂r.

Since uθ = 0 it follows that Ψ = Ψ(θ), and then

m

4πr2=

1

r2 sin θ

∂Ψ

∂θ.

If we define Ψ = 0 for θ = 0 it follows that

Ψ(θ) =m

4π(1− cos θ).

Using simple geometry, Ψ can also be expressed in cylindrical polars as

Ψ =m

4π

(

1− z√z2 + r2

)

.

3.2.3 Stagnation point flow

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y

x

stagnation point

We begin with the two-dimensional case of a flow onto a flat plate, see Figure. Weconsider the simplest case in which the flow is two-dimensional, u = (u, v, 0). There mustbe a stagnation point u = 0 somewhere along the plate, which we place at the origin,x = y = 0. The x-component will increase linearly as one moves away from the line ofsymmetry: u = Ax, and correspondingly v = −Ay if one moves away from the plateagainst the flow. Integrating, one finds the potential:

φ =A

2(x2 − y2)

It is easy to confirm that this is a solution of Laplace’s equation, and thus satisfies the fluidequations even away from the stagnation point. More importantly, it will automaticallygive the structure of the fluid flow close to a (two-dimensional) stagnation point. Thestream function is

ψ = Axy.

The axisymmetric version of stagnation point flow is relevant if any axisymmetric bodyis placed in an oncoming stream. The potential is in that case (exercise)

φ =A

4(r2 − 2z2)

3.3 Bernoulli’s theorem for unsteady, irrotational flows

Within potential flow theory, we have the slightly confusing situation that we are seem-ingly solving flow problems only by solving Laplace’s equation, which in turn comes fromthe incompressibility condition ∇ · u = 0. We are not using Euler’s equation of motion(14) to solve flow problems (at least this is almost true; we have been using (14) indi-rectly inasmuch we used it to derive Kelvin’s theorem, which motivated the potential flowassumption).

However, the Euler equation does come in, albeit in a sneaky way, as we will nowshow. From (19), the modified version of Eulers equation is

∂u

∂t− u× ω = −∇(P/ρ+ 1

2u2)

For irrotational flows, u = ∇φ, and ω = 0 so

∇(

∂φ

∂t+P

ρ+ 1

2u2

)

= 0

It follows that∂φ

∂t+P

ρ+ 1

2u2 = C(t) (23)

where C(t) is an arbitrary function of time.Remember P = p+ ρΦ where F = −∇Φ is the body forces in the fluid (e.g. P = p+ ρgzfor gravity forces).

Note: (23) applies throughout the fluid, not just on a streamline.

Note: By defining a new velocity potential

φ = φ−∫ t

C(t′)dt′,

the function C(t) can always be eliminated from the problem. The potential φ leads tothe same flow field! (why?)The Bernoulli equation (23) is best viewed as an equation for the pressure. Solving apotential flow problem then becomes a two-step process: First, solve Laplance’s equationfor φ. Second, given φ, use (23) to compute the pressure. The pressure depends on u ina non-linear way, a reflection of the nonlinearity of the Euler equation.

3.4 The collapse of a spherical bubble

Suppose that at t = 0 we have a spherical bubble, initially at rest, radius R0 in an infinitefluid. Approximations:

(i) neglect gravity;

(ii) neglect pressure in the bubble (p = 0);

(iii) pressure at infinity is constant: p0 > 0.

The pressure difference causes the bubble to collapse, so R(t) < R0, where R(t) isradius at time t.

Problem: find R(t) and time of collapse.

Introduce velocity potential, u = ∇φ. The flow is spherically symmetric, so

φ(r, t) =A

r

satisfies ∇2φ = 0, and A = A(t). Now

ur =∂φ

∂r= −A

r2

Boundary condition is

ur = R ≡ dR

dt, on r = R, ⇒ A = −R2R

and hence

φ(r, t) = −R2R

r

Still don’t know R(t) of course. Use unsteady Bernoulli’s eqn so that

ρ∂φ

∂t+ p+ 1

2ρu2 = C(t) = p0

since φ→ 0 and u → 0 as r → ∞. Now, on surface of bubble p = 0 (by assumption) and

|u| = R, and∂φ

∂t= −R

2R + 2RR2

R.

Substituting in gives−RR− 2R2 + 1

2R2 = p0/ρ

which is an ODE for R(t). Trick: multiply by 2R2R, so that

d

dt

(

R3R2)

= 3R2R3 + 2RRR3 = −p0ρ2R2R

So

R3R2 = −2

3

p0ρR3 + C

where C is a constant, determined by the initial condition R = 0 when R = R0. Hence

R2 =2

3

p0ρ

(

R30

R3− 1

)

(24)

ordR

dt= −

√

2

3

p0ρ

(

R30 − R3

R3

)1/2

where the negative sign must be chosen because R < 0 by assumption. Integrating upgives

∫ R(t)

R0

R3/2 dR

(R30 −R3)1/2

= −∫ t

0

(

2

3

p0ρ

)1/2

dt = t

(

2

3

p0ρ

)1/2

.

If the bubble collapses at time t = tc then R(tc) = 0 implies (R = R0u)

∫ 1

0

u3/2 du

(1− u3)1/2=

tcR0

(

2

3

p0ρ

)1/2

and the integral can be evaluated numerically to a value of 0.747. Hence

tc = 0.915

(

ρ

p0

)1/2

R0

We can also analyze the collapse during its final moments, when the radius goes tozero. Such singular events are usually described by power laws, so we try

R = A(tc − t)α.

Now R2 ∝ (tc − t)2α−2, and the right hand side of (24) scales as R−3 ∝ (tc − t)−3α. Thus2α− 2 = −3α, and the power law exponent is found to be

α =2

5.

Plugging this back into (24), we find

R =

(

2p0R30

3ρ

)1/5

(tc − t)2/5.

What is remarkable about the collapse is that the inertia of the surrounding liquid isfocused into a point by the converging motion. If one calculates the speed of the collapse,one finds

R ∝ (tc − t)−3/5,

which is diverging as the radius goes to zero.If the cavity is filled with gas, the gas can be compressed to very high pressures during

the last moments, reaching around 10,000 kelvins in the interior of the bubble. Thismay cause ionization of the gas, which may start to glow. This phenomenon is knownas sonoluminescence, since the collapse of the bubble is started off by an external soundfield.

3.5 Flow past a sphere

z

θ

n

R

One of the most fundamnetal problems of fluid mechanics is to understand the flowaround obstacles, which stand in the way of a flow. Equivalently (think of Galileaninvariance) one can think of a body moving inside a fluid which is at rest (for examplean aeroplane). Let us begin studying this problem by considering the perhaps simplestpossible body, a sphere of radius R. The sphere is placed inside a uniform stream, whichwe choose in the direction of the z or symmetry axis: u = U z.

Far from the body, the flow will be uniform. The sphere introduces a perturbation tothe flow which decays as r → ∞, i.e. as one moves away from the body. According tothe superposition principle (the Laplace equation is a linear equation), this perturbationis to be added to the potential φ = Uz describing the uniform flow. The simplest suchflow we know is a source, with potential φ ∝ 1/r. However, such a flow cannot describethe physical situation at hand. Far from the sphere, the total mass flow through a closedcontrol surface should be zero, not finite as for a source. Another way of putting it is thatthe flow field of a source decays too slowly (like 1/r2) away from the body.

This can be helped by taking the derivative of the source solution. Taking the deriva-tive in the µ-direction, one arrives at the potential

φ = µ · ∇1

r= −µ · x

r3,

which is called the dipole flow (why?). The dipole is oriented in the µ-direction and hasstrength |µ|. This flow is also a solution of Laplace’s equation, since

(

µ · ∇1

r

)

= µ · ∇(

1

r

)

= 0.

In summary, our ansatz for the potential is

φ = Uz +µ · xr3

.

The velocity field is

ui = ∂iφ = Uδi3 + ∂iµjxjr3

= Uδi3 +µi

r3− 3µjxj

xir5

or

u = U z+1

r3

(

µ− 3µ · x x

r2

)

.

Now we have to satisfy the boundary condition on the surface of the sphere, which isu · n = 0. The normal vector to the sphere is n = x/r. Now

u · n = Uz

r− 2

µ · xr4

,

and the choice µ =UR3

2z will guarantee that this is zero for r = R. In summary, the

solution we seek, which satisfies all the boundary conditions is

φ = Uz +UR3

2

z

r3, (25)

and the velocity field is

u = U z+U

2

(

R

r

)3(

z− 3z

rn)

. (26)

Now we want to calculate the pressure field, and in particular the pressure on thesurface of the sphere. This will permit us to calculate the total force on the sphere. Onthe surface we have

u =3U

2

(

z− z

rn)

,

so

u2 =9U2

4

(

1− z2

r2

)

.

According to Bernoulli’s equation

p = patm +ρ(U2 − u2)

2= patm +

ρU2

8

(

9 cos2 θ − 5)

,

using thatz

r= cos θ.

This result has the remarkable property that the pressure distribution is symmetric

around the equator of the sphere θ = π/2. Thus while the sphere is pushed by the flow inthe downstream direction, there is an equally high overpressure in the back. As a result,without further calculation, we can conclude that the total drag force on the sphere iszero. By definition, the drag on a body in a uniform stream is the force in the directionof the stream. The force perpendicular to the stream is called the lift, which is of coursezero if the flow is axisymmetric, as is the case here.

3.6 d’Alambert’s paradox

At first sight, one might think that the fact that there is no force acting on a sphere in astream at steady state is related to the high symmetry of the sphere. In fact, this is notthe case: we now show that the force F acting on a body of arbitrary shape vanishes atsteady state. At least as far as the component of F in the direction of U is concerned,this is a statement about energy conservation, since F · U is the power. Since there isno friction, this energy input cannot be dissipated; the only other possible option wouldbe for the energy to be transported off to infinity. The fact that this cannot occur is astatement about how fast the perturbation introduced by the body into the stream decaysat infinity.

S∞

S

V

n

FU

Let us write the total velocity field in the form

u = U+ v,

where v = ∇φ is the perturbation introduced by the body. As argued before for the caseof the sphere, we will assume that

φ = O(

r−2)

for r → ∞.

Now the force on the body is calculated from integrating the pressure force over thebody:

F = −∫

S

pndS;

the minus sign comes from the normal pointing outward, opposite the direction of thepressure force. To find p, we use the Bernoulli equation (23) with φ = 0, since the flow issteady:

p+ ρU2/2 + ρv2/2 + ρU · v = patm + ρU2/2.

In other words,

F = −∫

S

patmndS +ρ

2

∫

S

v2ndS + ρ

∫

S

(U · v)ndS.

The first integral on the right is zero, as we have seen before. As a mathematical corrolary,we will show below that

∫

S

v2ndS = 2

∫

S

v(v · n)dS. (27)

But on the surface of the body the normal component of the total velocity vanishes:n · (U+ v) = 0. Armed with this insight, we have thus shown that

Fi = −ρUj

∫

S

(vinj − vjni) dS. (28)

From the symmetry of this expression it is clear immedeately that the drag of thebody, which is defined as

D = F ·Uis zero. However, the components of F normal to U are also zero. Let V be the volumeexterior to the body but bounded by a closed surface S∞ very far from the body (seeFigure). Then it follows from the divergence theorem that

∫

V

(

∂vi∂xj

− ∂vj∂xi

)

dV = −∫

S

(vinj − vjni) dS +

∫

S∞

(vinj − vjni) dS;

note that n points into the volume V . If v decays faster than 1/r2 at infinity, the integralover S∞ goes to zero as the radius of S∞ goes to infinity. The volume integral has acomponent of ∇ × v as its integrand, which is zero. It follows that the surface integralappearing on the left hand side of (28) is zero, and thus the force is altogether zero.The corollaryWe now demonstrate (27). From the divergence theorem we conclude that

∫

V

∇ (∇φ)2 dV = −∫

S

(∇φ)2 ndS +

∫

S∞

(∇φ)2 ndS.

The integrand of second integral on the right behaves like 1/r6, so its contribution cearlyvanishes. We have shown in Appendix A that ∇v2 = 2(v · ∇)v for an irrotational flow,and ((v · ∇)v)i = vj∂jvi = ∂j(vivj) for an incompressible fluid. Thus

∫

S

(∇φ)2 nidS = −∫

V

∂i (∇φ)2 dV = −2

∫

V

∂j (vivj) dV = 2

∫

S

vivjnjdS

by the divergence theorem and using the boundary condition, which proves the corollary.This is a truly remarkable result of potential flow theory, but it also presents a series

problem, as absence of drag is clearly not in accord with observation. If our argument werecompletely airtight, airline companies would surely be doing some serious overcharging!(maybe they do anyway).

3.7 Drag

The experimental picture of a sphere in a stream reveals what the problem is: the flowaround the sphere is far from rear-aft symmetric. While the flow in front of the sphereis nicely attached to the body and well described by (26), the back is very different. Theso-called Reynolds number of the flow is

Re =URρ

η≈ 1.5× 104 (29)

which is a dimensionless measure of the size of the viscosity η. Since the Reynolds numberis very large, one might believe that viscous effects are small and the potential flow as-sumption represents a faithful represenation. However, this is not the case. In particular,the flow does not remain attached to the body, and rather there are fluid particles whichoriginate from the surface of the sphere and the enter the liquid. This is also the placewhere vorticity enters the flow. Clearly, vortices have formed in so-called wake of the flow.

Apart from the vortices, the wake appears to be a reatively stagnant region of the flow,which is thus at constant pressure p = pcav, provided it remains closed (which it seems tobe in the eperimental Figure). What is the value of pcav ? Let us consider the streamlinethat separates from the body and bounds the cavity. Since the pressure is constant inthe cavity, the pressure on this streamline is also constant, by continuity of the pressure.Such a streamline is called a free streamline. According to the steady Bernoulli equation(20) the velocity along this streamline is also constant. Therefore, there is a jump ofthe velocity at the free streamline to the almost vanishing velocity in the wake. (Nothingprevents the velocity to be discontinuous if there is no viscosity). Such a surface overwhich the velocity changes abruptly is called a shear layer or a vortex sheet. Now pressurein the cavity will be the pressure on the separating streamline, as it leaves the body.

To do any calculation of the effect of the wake, one needs an estimate of the pointwhere the flow leaves the body, to produce the wake. A very simple criterion (whichcan be justified in more detail using full viscous theory), is the concept of the “adversepressure gradient”. At the front of the body the pressure maximum, and then decreasesto a minimum value at the equator, where the speed is maximum; from there the pressuredecreases again. It is intuitive and can be motivated in greater detail from the viscousflow theory, that the flow along the body is stable as long as the pressure decreases alongthe path of a fluid particle. If however a fluid article has to push against an increasingpressure (“adverse pressure gradient”), it prefers to leave the surface of the body.

Thus on the basis of the pressure distribution which we have calculated, one can givean estimate of the place where eparation is going to occur, which is at θ = π/2 in the caseof a sphere. Now we can attempt to recalculate the drag force on the sphere, assumingseparation at θs = π/2. The pressure in the cavity will be

pcav = patm − 5

8ρU2 :

in our calculation, there is an underpressure in the cavity which pulls the sphere along.We are now in a position to calculate the drag on the sphere, doing the front and the

back separately. The result cannot depend on patm which exters equal forces on front andback; therefore, we can put it to zero. Since n · z = cosθ, the force from the front in theflow direction is

F (front)z = −

∫

front

p(θ) cos θdθ = −π4ρU2R2

∫ π

π/2

sin θ cos θ(9 cos2 θ − 5)dθ =

π

4ρU2R2

∫ 1

0

x(9x2 − 5)dx = − π

16ρU2R2,

and from the back

F (back)z = −2πR2

∫ π/2

0

sin θ cos θpcavdθ =5π

4ρU2R2

∫ 1

0

xdx =5π

8ρU2R2.

Thus the total force or drag is

D = Fz =9π

16ρU2R2 =

9

16ρAU2,

where A = πR2 is the projected area of the sphere.For a general body, the drag coefficient Cd is defined by

D =ρ

2U2ACd,

and so Cd = 9/8 for our calculation of the sphere. We hasten to add that our calculationis not particularly good, see the Figure. In the relevant range of Reynolds numbers,

CD ≈ 0.5. Our intention is to give a general idea of where the drag is coming from, and

how d’Alambert’s paradox arizes.

3.8 Impact theory

If a body impacts on a fluid surface, an impulse upward force results. The surroundingsof the projectile show a characteristic splash profile, over which the the parts close to

the projectile show the greatest upward velocity.We now calculate both the impulse as well as the velocity profile right after impact.

Our assumption is that the speed of impact is so high that the acceleration of the fluid

∂u

∂tis large, but there is no significant transport yet, so (u · ∇)u is small in comparison.

Thus the Euler equation simplifies to

∂u

∂t= −1

ρ∇p.

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p

ttb ta

Now the acceleration and thus the pressure will be very great at the moment of impact,and small before and after: the pressure as function of time has a very sharp peak. Inview of Newton’s equation p = F we have to calculate the time integral over the peak(the area of the shaded region) to calculate the total momentum transferred to the body.This object is known as the impulse. We thus have

u(a) − u(b) = ∇(

φ(a) − φ(b))

= −1

ρ∇∫ ta

tb

pdt′ ≡ −∇Π

ρ,

where Π is the impulse per unit area, the superscripts refer to a time tb just before impact,and a time ta just after impact. Now φ(b) will be constant (describing a fluid at rest beforeimpact), and integrating we find

φ(a) = −Π

ρ+ const.

free surface

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body

U

V n

φ = 0

n · ∇φ = n ·UClearly, the impulse is zero along the unperturbed free surface of the liquid (see Fig-

ure). Thus the potential is constant along the free surface; we can choose this constantto be zero. On the lower side of the body, the fluid has to move with the impact velocityU of the body:

U · n = u · n = n · ∇φ.This defines the boundary conditions for φ on the boundaries of the fluid domaine V ,bounded by the free surface on one hand, and the underside of the body on the other.

In general this problem is quite difficult to solve. However in the special case that thefree surface lies in a symmetry plane of the body, the calculation becomes equivalent tothe problem of the body moving with speed U in an infinite expanse of fluid. As anexample, let us consider a sphere impacting on a bath of fluid, such that its lower halfis wet by the fluid. Equation (25) is the potential for a stationary sphere in a stream ofspeed U . If one subtracts the velocity u = U z, this is equivalent to the sphere moving inthe negative z-direction in a stationary fluid. Thus the solution we seek (sphere movingin the z-direction) is

φ = −UR3

2

z

r3.

It is checked directly that φ = 0 in the plane of symmetry z = 0, and by construction

U · n = n · ∇φ. Thus this is the solution we seek. To calculate the total impulse

∫

Fzdt

on the sphere (there only is a z-component), we have to integrate −Πnz = −Πcos θ overthe front Sf of the sphere:

∫

Fzdt = −∫

Sf

ΠnzdS.

On the surface of the sphere, φ(a) = −UR cos θ

2, and so

∫

Fzdt = ρ

∫

Sf

φ(a)nzdS = −ρUR2

∫ π/2

0

2πR2 sin θ cos θ cos θdθ = −πρUR3

∫ 1

0

x2dx = −π3ρUR3

(pointing opposite the impact direction).

The velocity in the z = 0 plane is

uz =∂φ

∂z= −U

2

(

R

r

)3

,

and ur = 0. After the impact, fluid particles will continue to travel along straight linesin the direction of this velocity field, so the free surface will have a shape given by theabove velocity profile! This is confirmed by comparison to the experimental picture.

There is a reptile, called the Jesus Christ lizard or Basiliscus basiliscus, who supportsits weight by a rapid succesion of impacts with their feet. (There are some nice movies ofit on youtube). To do that, they have to satisfy the equation

∫

Fzdt =MgTstep

For more information, see J. W. Glasheen and T. A. McMahon, Nature 380, 340 (1996).As a model for the feet of the lizard, they choose a disc of radius R. Using ellipticalcoordinates, it can be shown that the potential of a disc moving at speed U along its axisz is

φ = −2UR

πµ(1− ζ arctan ζ)

in front of the disc, where

z = Rµζ, r = R√

(1− µ2)(1 + ζ2)

are cylindrical coordinates. In the plane z = 0 we have ζ = 0, and thus

φ = −2UR

π

√

1−( r

R

)2

.

Thus the impulse on the plate is

∫

Fzdt = ρ

∫

Sf

φ(a)nzdS = −4ρUR3

∫ 1

0

ξ√

1− ξ2dξ = −4ρUR3

3

(not too far from the result for a sphere). Glasheen and McMahon’s measurements inPhys. Fluids 8, 2078 (1996) agree very well with this answer.

4 Two-dimensional flows

If the flow is in the plane, and there are only two independent variables x, y, the flow prob-lem is much simplified. Moreover, we will see that some of the physics is fundamentallydifferent from three dimensions. One fact we know of already is that no vorticity is createdin two dimensions, ω is simply convected with the flow. Of course, real flows are nevertruly two-dimensional; however if the geometry extends very far in one direction (thinkof the wing of an aeroplane), a two-dimensional description is a good approximation.

4.1 Flow past a cylinder

Find the flow around a stationary cylinder in a steady stream U = U x. This is veryclosely analogous to the flow around a sphere; the effect of the sphere is modelled by a(two-dimensional) dipole, which is the derivative of a source, whose velocity field behaveslike ur ∝ 1/r (section 1.9.1). Thus the potential is φ = ln r, and a dipole in the µ-directionhas potential

φ = (µ ·∇) ln r =µ · xr2

,

The ansatz for the velocity potential is thus

φ = Ux+µ · xr2

,

and so

u = U x +µ

r2− 2(µ · x)x

r4.

Now the boundary condition for r = R is u · n = u · x/r = 0, and from the velocity fieldwe find

u · n = Ux

r− µ · x

r3.

Thus the boundary condition is satisfied if we choose

µ = R2U .

Thus the final answer for the velocity field is

u =R2

r2[U − 2n(U · n)] +U . (30)

Now we plug this into Bernoulli’s equation to compute the pressure:

u2

2+p

ρ=U2

2+patmρ.

On the surface

u2∣

∣

r=R= (2U − 2n(U · n))2 = 4U 2 − 4(U · n)2 = 4U2(1− cos2 θ) = 4U2 sin2 θ,

so in other words

p = patm +ρU2

2(1− 4 sin2 θ).

The pressure is once more symmetric about θ = π/2, i.e. about the midsection of thecylinder. It follows that the total force on the cylinder is again zero.

4.2 Non-uniqueness of the potential

One particularly important aspect of two-dimensional flow is that any solid body placedin the flow domain will create a domain that is no longer simply connected.

Defn: A closed curve C is reducible in a domain D if it can be shrunk to a pointwithout ever leaving D. If every closed curve is reducible then D is simply connected.

E.g. (i) if D is the interior of a circle, then it is simply connected. (ii) if D is theexterior of a circle then it is not simply connected.

D

i ii

D

Example: Let D be the domain a < r < b, 0 < θ < 2π in cylindrical polars. D is notsimply connected.

Define a flow in this domain,

u = (0,Γ/2πr, 0),

which is called a point vortex of strength Γ. At r = 0 the flow is singular, but outsideit is irrotational [Check] and has circulation Γ. Hence there is a velocity potential s.t.

ur =∂φ

∂r= 0, uθ =

1

r

∂φ

∂θ= Γ/(2πr). The solution is

φ(r, θ) =Γ

2π(θ + A),

for any constant A. There’s a problem here: for example, φ(r, 0) 6= φ(r, 2π) so thepotential is discontinuous along θ = 0 or, alternatively, φ is multivalued. Evidently,whenever there is circulation

Γ =

∮

C

dl · ∇φ,

a finite amount of potential is picked up as one goes around the curve. Such a flow is thusnecessarily multivalued.

In practice, one places a branch-cut in D to make the domain simply connected. Insimply-connected domains, φ is unique:

Needs a branch cut to make domain simply connectedand the potential single−valued

b

a

4.3 Steady flow past a circular cylinder with circulation

In the previous sections we have shown that the potential for the steady flow past acylinder of radius R is

φ = U

(

r +R2

r

)

cos θ

(cylinder fixed at origin, flow speed U from left to right).Suppose now that there is circulation round the cylinder, by adding a vortex rotating

in the clockwise direction:

φ = U

(

r +R2

r

)

cos θ − Γ

2πθ

This is not a single-valued function. The streamfunction is

ψ = U

(

r − R2

r

)

sin θ +Γ

2πlog(r/a)

satisfying ψ = 0 on r = R. Other streamlines given by ψ = const. Find the stagnationpoints (u = (0, 0)) with, on r = R

ur =∂φ

∂r= 0

uθ =1

r

∂φ

∂θ= −U

(

1 +R2

R2

)

sin θ − Γ

2πR

So uθ = 0 implies

sin θ = − Γ

4πUR

which has two roots in 0 < θ < 2π (and hence two stagnation points) if Γ/UR < 4π.

If Γ/UR = 4π then two stagnation points coincide at θ = 32π. If Γ/UR > 4π then there

are no stagnation points: To calculate the force on the cylinder, use unsteady Bernoulli,

p+ 12ρu2 = C

noting that the flow is steady (C(t) = C, ∂φ/∂t = 0). With p → patm, u2 → U2 as

r → ∞,p = patm + 1

2ρU2 − 1

2ρu2.

On cylinder surface,

u = (ur, uθ) = (0, φθ/r)|r=R = (0, 2U sin θ + Γ/2πR)

and with n = (cos θ, sin θ),

F = −∫ 2π

0

pnR dθ

So F · x = Fx = 0 (no drag force, as before). But the transverse force is

Fy = F · y = −∫ 2π

0

[

p0 +12ρU2 − 1

2ρ(2U sin θ + Γ/2πR)2

]

sin θR dθ

=

∫ 2π

0

12ρ4U sin θ(Γ/2πR) sin θR dθ

= ρUΓ

so there is an upward lift force on the cylinder (Magnus Effect - 1853). This can explainthe forces acting on spinning objects (for example ping-pong balls). The effect is alsoclosely related to the lift forces on airplane wings, to which we will come back in moredetail. The big question in that case of course is what sets the value of Γ!

4.4 The complex potential

We now introduce the formulation of potential flow using a complex formulation, whereeach point in the xy plane is represented by a complex number z. Only this formulationreveals the full power of two-dimensional methods. The first remarkable result is thatthe potential and the streamfunction are simply the real and imaginary part of the samecomplex potential.

Let u = (u, v, 0). The flow is irrotational and incompressible as in chapter 3, sou = ∇φ and ψ exists such that (see §3.2.1)

u =∂φ

∂x=∂ψ

∂y

v =∂φ

∂y= −∂ψ

∂x

(31)

Now let z = x+ iy and define the complex potential w(z) = φ(x, y) + iψ(x, y).

If w(z) = w(x, y) is analytic then

dw

dz=∂φ

∂x+ i

∂ψ

∂x= −i∂φ

∂y+∂ψ

∂y

(w′(z) is independent of the direction of the derivative). Eqns (31) are the Cauchy-Riemannequations and are satisfied by any analytic functions.

Properties:

(i) ∇2φ = ∇2ψ = 0 follow directly from (31)

(ii) ∇φ · ∇ψ =∂φ

∂x

∂ψ

∂x+∂φ

∂y

∂ψ

∂y= 0 meaning equipotential curves (where φ = const)

are perpendicular to streamlines, (where ψ = const)

(iii)dw

dz= u− iv = qe−iχ where q = (u2 + v2)1/2 is the speed of the flow, χ is the angle

the flow makes to the x-axis;dw

dzis known as the complex velocity (note the minus

sign)!

4.4.1 Example: straining flow

Straining flow, ψ = 2kxy, k a constant. Then

u =∂ψ

∂y= 2kx =

∂φ

∂x

v = −∂ψ∂x

= −2ky =∂φ

∂y

, ⇒ φ = k(x2 − y2)

Hence w(z) = φ+ iψ = k(x2 − y2) + i2kxy = kz2 is the complex potential.

4.4.2 Example: Line source

In three dimensions, a two-dimensional source is a line of sources pushing fluid radiallyoutward. The flow field (see 1.9.1 (ii)) is

u =m

2πrr,

while the potential and stream functions are

φ =m

2πlog r, ψ =

m

2πθ.

It is recognised immediately that these are the real and imaginary parts of the complexpotential

w =m

2πlog z,

since in polars z = reiθ where r = |z| and θ = arg(z) and so log(z) = log(r) + log(eiθ) =log(r) + iθ.

4.4.3 Example: Line vortex

A vortex has the flow field

u =Γ

2πrθ,

and the potential is φ =Γ

2πθ, which is the same as the streamfunction of a source! Thus

we know directly that the complex potential of a vortex must be the same as that of asource, up to a complex rotation:

w = − iΓ

2πlog z.

In particular, we can conclude without further calculation that the stream function is

ψ = Imw = − Γ

2πln r.

This is a simple example for the application of a powerful theorem from complex analysis:once one knows either the real or the imaginary part of an analytic function, the wholecomplex function is determined completely. In three dimensions, this will be a line vortexwhich is completely straight and perpendicular to the xy plane. In general, line vorticesare curved, but we disregard this effect here.

4.5 Interaction of vortices

A line vortex intersects a plane in one point, in two dimensions one can thus describethe vortex by a point (the position z0 = x0 + iy0 of the vortex, and its strength Γ.According to Kelvin’s theorem, the circulation is conserved, and thus the strength of thevortex is constant in time. The only thing we need to keep track of is the position of thevortex.

In complex notation, the velocity field generated by a vortex located at z0 is

u− iv = − iΓ

2π

d log(z − z0)

dz= − iΓ

2π

1

z − z0= − iΓ

2π

z − z0|z − z0|

Now each vortex moves in the velocity field generated by all the other vortices. In thesimplest case of two vortices of strength Γ1 and Γ2, located at (x1, y1) and (x2, y2), re-spectively, the equations of motion are

dx1dt

= −Γ2

2π

y1 − y2(x1 − x2)2 + (y1 − y2)2

,dy1dt

=Γ2

2π

x1 − x2(x1 − x2)2 + (y1 − y2)2

,

and correspondingly for the other vortex. The generalisation for N vortices is obviousand involves the sum over all the vortices except for the one that is being convected.

This system of equations is beautiful and simple, but things are even better: thesystem has a Hamiltonian structure! The Hamiltonian is

H =Γ1Γ2

4πln(

(x1 − x2)2 + (y1 − y2)

2)

.

If now one defines variables (xi, yi) =√

|Γi|(xi, sign(Γi)yi), then the equations become

dxidt

=∂H

∂yi,

dyidt

= −∂H∂xi

,

which are the usual Hamilton’s equations. One particular of this is that since H isconserved, the distance between the two vortices must stay the same.

How does the motion of two vortices look like ? Let us look at the two possible cases:

4.5.1 Two corotating vortices

1 2

h

R1

R2

G

Γ1

2πh

Γ2

2πh

The vortex strengths are Γ1 and Γ2, both being positive (the two vortices rotatecounterclockwise). Let the distance between the two vortices be h. The speeds of the twovortices are then

U1 =Γ2

2πh, U2 =

Γ1

2πh,

respectively. From the construction it is clear that at each instant, the vortices movein a direction perpendicular to the line connecting them. This means they have to goaround in circles, whose centre G is along the line connecting the vortices. The point Gis stationary in space. The some of the radii R1, R2 must satisfy R1+R2 = h. Since bothvortices make a full turn in the same time, we must have

R1

R2=

Γ2

Γ1.

For two equal vortices of strength Γ1 = Γ2 = Γ, it follows that R1 = R2 = R. Thismeans the vortices are always at the opposing sides of a circle of radius R. The wholeconfiguration rotates at an angular velocity of

Ω =Γ

4πR2.

4.5.2 Two counterrotating vortices

Now consider the case that the two vortices have opposite sign. This situation isproduced easily by the stroke of a paddle. In fact since the total vorticity of the systemshould remain zero according to Kelvin’s theorem, typically Γ1 = −Γ2. After the paddleis withdrawn, the vortices continue to move according to the other vortex’ velocity field.Let the distance between the two vortices again be h.

1 2h

R1

R2Γ1

2πh

Γ2

2πh

The vortices now move in the same direction, and rotate about the same centre. Theradii of this circular motion obeys the same relationship has before. However, we now have

R2−R1 = h, and thus R1 =Γ2h

Γ1 − Γ2. Thus when the vortices are equal and opposite, the

radii of both circles tend to infinity. In this case the vortices move in a straight line, and

translate uniformly at speed U =Γ

2πh. This is what’s seen in the experimental pictures

of two vortices being created behind a paddle.

4.6 The method of images

4.6.1 A vortex next to a wall

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1

2

b

Γ

2π

ψ = 0

Consider the motion of a vortex located at some position z0 = a+ ib in the upper halfof the complex plane. The x axis is the location of a solid wall. Thus we have to satisfythe boundary condition of no flow through the wall: v = 0 for y = 0. Simply taking theflow field of the vortex,

u− iv = − iΓ

2π

x− a− i(y − b)

(x− a)2 + (y − b)2, (32)

does not satisfy this condition: the flow field must be modified by the presence of thewall.

The method of images is a technique that permits to find solutions to Laplace’s equa-tion which satisfy the right boundary condition on the solid wall. Near the vortex atz0, the solution should look like (32). The method of solution is indicated in the figure:imagine replacing the wall by another image vortex that is placed at an equal distanceon the other side of the wall. Its position is where an image in a mirror would appear.And equally like a mirror image, the sense of rotation of the image vortex is reversed.By symmetry it is clear that the v component of the image vortex has opposite sign, soon the line of symmetry (the locus of the wall), the two contributions cancel and v = 0.The boundary condition on the wall is satisfied and we have solved our problem! In otherwords, the flow problem we want to solve (one vortex in the presence of a wall) is exactlythe same as the flow problem of two vortices (the original vortex and its image) without

the wall.Let’s check if our reasoning was correct. It is easiest to do the calculation using

complex notation. If the vortex is at z0, it’s image is at z0. Since the image also has theopposite sense of rotation, there is a minus sign in front of it. The total potential (vortex+ image) becomes

w(z) = − iΓ

2πlog(z − z0) +

iΓ

2πlog(z − z0).

We have to verify that Imw = ψ = const for z = x real! Now since ln(z) = ln(z)(why?) we have

w(z) =iΓ

2πlog(z − z0)−

iΓ

2πlog(z − z0).

But if z is real, z = z, and thus w(x)− w(x) = 0, and so ψ = 0 along the wall.In fact, we have shown that the problem of a vortex near a wall is mathematically

equivalent to the problem of two counterrotating vortices of equal strength, which in theprevious section we showed to move at constant speed in the direction perpendicular tothe line connecting them, in other words, parallel to the wall. Let us check this fact aswell, using complex notation. The vortex in question moves in the flow field of its image,i.e.

u− iv =iΓ

2π

x− a− i(y + b)

(x− a)2 + (y + b)2,

evaluated at z = a+ ib. Thus

u− iv =iΓ

2π

(−2ib

4b2

)

=Γ

4πb,

and the vortex moves with a horizontal speed ofΓ

4πb, which corresponds to the earlier

result with h = 2b.

4.6.2 A recipe

This leads us to the following result, which permits to find the flow generated by any

combination of singularities next to a wall. Let the wall be the real axis of the complexplain, and let f(z) describe all singularities (vortices, sources, etc.) present in the flow

domain y > 0. For example, f(z) =iΓ

2πln(z − z0) for a vortex and f(z) = − µ

2π(z − z0)for a dipole.

Thus w(z) = f(z) would be the complex potential of the flow without the wall. Weclaim that

w(z) = f(z) + f(z)

is the potential in the presence of the wall. In the second member, we take the complexconjugate of everything except of z itself. We have to prove that ψ = Imw(z) = 0 ifz = x is real. But if this is the case then z = z, and so

w(x)− w(x) = f(x) + f(x)− (f(x) + f(x)) = 0,

which proves the theorem.

To apply this, consider a source of strength m placed at z = ib above a wall. Then

f(z) =m

2πln(z − ib) and f(z) =

m

2πln(z + ib). Thus the solution we are seeking is

w(z) =m

2π(ln(z − ib) + ln(z + ib)) =

m

2πln(z2 + b2).

4.7 von Karman vortex street

The Figure shows a cylinder in a uniform stream at a Reynolds number Re = 105. Theflow is not steady, but the cylinder sheds a periodic sequence of counterrotating vortices,which form a characteristic staggered pattern.

Von Karman became interested in the problem problem when he was appointed as agraduate assistant in Goettingen in Prandtl’s laboratory in 1911. Prandtl had a doctoralcandidate, K. Hiemenz, to whom he had given the task of constructing a water channelin order to observe the separation of the flow behind a cylinder. Much to his surprise,Hiemenz found that the flow in his channel oscillated violently, and he failed to achievesymmetrical flow about a circular cylinder. Von Karman writes: When Hiemenz reported

this to Prandtl, the latter told him: “Obviously your cylinder is not circular.” However,

even after very careful machining of the cylinder, the ow continued to oscillate. Then

Hiemenz was told that possibly the channel was not symmetric, and he started to adjust it.

I was not concerned with this problem, but every morning when I came into the laboratory

I asked him, “Herr Hiemenz, is the flow steady now?” He answered very sadly, “It always

oscillates.”

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a

b

Von Karman modelled the flow by an infinite array of point vortices as pictured.Sufficiently far behind the obstacle that produces the street, the effect of the obstacle canbe neglected. The characteristic feature of the street is that the pattern is stationary inthe frame of reference of the obstacle. To achieve this, fix a particular vortex and considerthe velocity field it experiences. The velocity generated by all the other vortices has tocompensate exactly the velocity u = U of the stream.

We take the street as infinitely extended in both directions. Thus there is one set

of vortices of strength Γ at z = na, and another row of vortices at z = (n+1

2)a+ ib,

where n = 0,±1,±2, . . .. Just a little bit of thought shows that all the contributions tothe velocity of one vortex coming from the same row cancel exactly, so we only have toconsider the other row.

Now fix a vortex, at z = a/2+ ib, say. The contribution of each of the members of the

other row is − iΓ

2πln(z − na), which is the same (up to a constant), as

− iΓ

2πln(

1− z

na

)

.

Thus the total contribution is

w = − iΓ

2π

∞∑

−∞, n 6=0

ln(

1− z

na

)

− iΓ

2πln z =

− iΓ

2πln

[

z∞∏

n=1

(

1− z2

n2a2

)

]

= − iΓ

2πln(a

πsin

πz

a

)

,

using a well-known relation from complex functions. Now the complex velocity is

u− iv =dw

dz= − iΓ

2acot(πz

a

)

,

and at the position of the vortex

dw

dz

∣

∣

∣

∣

z=a/2+ib

=iΓ

2atan(πz

a

)

= − Γ

2atanh

(

πb

a

)

.

We thus have the condition that

U =Γ

2atanh

(

πb

a

)

for the vortex street to stay in place.

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b

aWhy have we been assuming a staggered configuration rather then a symmetric one,

as shown above? The reason is that this configuration is unstable. Imagine shifting one ofthe vortices to the right. The main contribution to its motion comes from the vortex justbelow. As a result, the vortex will move farther away from it, and the horizontal velocityit receives becomes smaller. As a result, it will be convected even further downstream.Clearly, this is an unstable configuration.

In the staggered configuration, this argument does not apply: while the vortex movesfarther away from its upstream neighbour, it comes closer to its downstream neighbour.However, there is only one particular ratio of a/b for which the staggered configuration iscompletely stable. A more detailed calculation shows that

cosh

(

πb

a

)

=√2,

orb

a= 0.283 if the configuration is to be stable.

4.8 Mappings and transformations

The most powerful tool of complex analysis consists in mapping the flow domain intoanother. As soon as I have found a mapping that is conformal, I have transformed theproblem from one geometry to another.

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ζ = f(z)

z0

ζ0

DH

Imagine we want to solve a flow problem in a given domain D. This means we arelooking for a function w(z) (the complex potential) with the following properties:

• w(z) is analytic in the domain D, except perhaps for one or more singularities at z0(and other places).

• Imw(z) = const on the boundaries of the domain.

Now imagine that ζ = f(z) is a mapping that is analytic everywhere in D, and whichmaps onto a new domain D1. Then the singularity at z0 is mapped to a new placeζ0 = f(z0) and the boundary of D is mapped to the boundary of D1. For D1 one willchoose a very simple domain such as the upper half plane H, for which I can solve theflow problem; let this solution be w1(ζ) (which has a singularity at ζ0).

Then w(z) = w1(f(z)) is a solution to the original flow problem.

To prove this statement, it is enough to observe that since w1(ζ) is analytic in D1,w(z) is analytic in D. This is the first requirement. In addition, Imw(ζ) is constanton the boundary of D1. But this means that Imw(z) = Imw1(f(z)) is also constant,which meets the second requirement.

4.8.1 Example: A source in a corner

The function ζ = f(z) ≡ z2, z = ζ1/2 maps the first quadrant onto the half-plane.Namely, if z = reiθ, then ζ = r2e2iθ = r2eiφ. Thus the domain 0 ≤ θ < π/2 is mappedonto 0 ≤ φ < π. The source at z0 maps to a source at ζ0 = z20 . Now the solution in theζ-domain is

w1(ζ) =m

2πlog(ζ − ζ0) +

m

2πlog(ζ − ζ0)

as we have seen before. So

w(z) = w1(z2) =

m

2π

(

log(z2 − z20) + log(z2 − z20))

=m

2π(log(z − z0) + log(z + z0) + log(z − z0) + log(z + z0))

The plot is for the case z0 = 2 + i.

4.8.2 Example: rotation

Consider any flow problem, for example the flow around a cylinder:

w1(ζ) = U

(

ζ +R2

ζ

)

.

Let ζ = ze−iα ≡ f(z), which is a rotation of the axes by α in the anticlockwise direction.Namely, for z = reiθ, ζ = rei(θ−α).

Then in the z-plane,

w(z) = w1(ze−iα) = U

(

ze−iα +R2

zeiα)

.

Note: Velocities are most easily found from

dw

dz= u− iv =

dw1

dζ

dζ

dz

E.g. in above u− iv = (u1 − iv1)e−iα.

4.8.3 Example: a channel

Another useful mapping is the exponential ζ = f(z) ≡ eKz, (z = ln ζ/K), which maps theupper half plane to a channel of width π/K. Namely, if ζ = |ζ |eiϕ then z = ln |ζ |/K +iϕ/K. Since 0 ≤ ϕ < π, the imaginary part of z varies between 0 and π/K. Here thepositive x-axis is mapped onto the lower boundary of the channel, the negative x-axis onthe upper boundary. Now imagine a vortex situated in the channel of width π at z0 = i.Then ζ0 = ei = cos(1) + i sin(1). The potential in the half space is

w1(ζ) = − iΓ

2π

(

ln(ζ − ζ0)− ln(ζ − ζ0))

,

and thus

w(z) = − iΓ

2πln

[

ez − ei

ez − e−i

]

is the solution.

4.9 The Joukowski mapping: circles to ellipses

A particularly useful application of the mapping idea concerns the flow around bodies. Wehave solved the problem of the flow around a cylinder. Thus if we can find a conformalmapping between the unit circle and any given shape, we have solve the flow problemaround this shape.

z

ζ

R

2c

2d

z = ζ + b2

ζ

Consider the (inverse) mapping

z = ζ +b2

ζ, (33)

called the Joukowski mapping. Consider a circle of radius R, whose surface in the ζ-planeis described by the polar representation ζ = Reiθ. Under the Joukowski mapping,

z = Reiθ +b2

Re−iθ =

(

R +b2

R

)

cos θ + i

(

R− b2

R

)

sin θ = c cos θ + id sin θ

is an ellipse with axes length 2c and 2d.The semiaxes come out to be

c = R +b2

R, d = R− b2

R.

Now consider the uniform flow past an ellipse. To model an arbitrary angle α betweenthe direction of the flow and the semi-major axis, we consider the flow around a cylinderthat approaches the x-axis under an angle α:

w1(ζ) = U

(

ζe−iα +R2

ζeiα)

.

In principle, one can find w(z) using the inverse of the Joukowski mapping

ζ = f(z) =1

2(z +

√z2 − 4b2),

so that w(z) = w1(f(z)).However, the resulting expressions are often not so useful. For example, to find the

streamlines, it is much easier to find the streamlines of the w1(ζ) in the ζ-plane, and thento transform them using (33). This is how the pictures were produced. If one wants tocalculate the velocity, one uses

u− iv =dw

dz=

dw1

dζ

1

dz/ dζ=U(e−iα − R2eiα/ζ2)

(1− b2/ζ2).

On the cylinder, ζ = Reiθ, so

u− iv =U(e−iα − eiαe−2iθ)

(1− (b2/R2)e−2iθ)=

2iU sin(θ − α)

(eiθ − (b2/R2)e−iθ).

This means there are stagnation points at θ = α and θ = −π + α. This point is wherea streamline leaves the surface. In other words, this streamline (plotted in red) has thesame value of the streamfunction ψ then the surface of the ellipse.

4.10 Lift

Now we want to fly! In principle, we know how to construct the flow around a wing ofarbitrary shape, we only have to find the transformation, starting from a circle. We haveseen already that the key ingredient is to have circulation around the wing. We havecalculated the lift in the case of a cylindrical cross-section, but what is it for an arbitraryshape?

4.10.1 Blasius’ theorem

Ct = (cosχ, sinχ)

n = (sinχ,− cosχ)

χ

x

y

We now calculate the force on a fixed body in a stream, using complex notation. Thisis a close relative of the calculation done in section (3.6), leading to (28). Suppose a fixedrigid body, boundary C is in a steady flow, generating a potential w(z). We know

dw

dz= u− iv = qe−iχ

so |u| = q (reminder: χ is the angle the flow direction makes to the horizontal). So,according to Bernoulli (no gravity):

p = p0 − 12ρq2,

where p0 = patm + ρU2/2 is a constant.Let s be the arclength along C, which we use to integrate over the surface of the body.

Now the line element along C can be written in complex notation as

dz ≡ dx+ idy =

(

dx

ds+ i

dy

ds

)

ds.

Since the surface of the body is a streamline, the velocity vector (u, v) is parallel tothe tangent on the surface. Thus using the angle χ we have dz = dx + idy = eiχ ds =(cosχ + i sinχ) ds. Multiplying by −i we achieve a rotation by ninety degrees in theclockwise direction, which gives the direction of the outward normal: n = (sinχ,− cosχ).Now we are in a position to write the total force F = (Fx, Fy)

F = −∮

C

pn ds

in complex form, defining a complex force F = Fx − iFy. Then

F = Fx − iFy = −∮

C

p(sinχ+ i cosχ) ds = −i∮

C

pe−iχ ds

= −i∮

C

p0(dx− idy) +iρ

2

∮

C

q2e−2iχeiχ ds

=iρ

2

∮

C

(qe−iχ)2 dz =iρ

2ρ

∮

C

(

dw

dz

)2

dz

(the term proportional to p0 vanishes because the integral of a total differential over aclosed loop is zero). This is Blasius’ theorem.

Example A cylinder in a stream with circulation:

w(z) = Uz + UR2

z− iΓ

2πlog z

sodw

dz= U − U

R2

z2− iΓ

2πz

So Blasius around the circle C says

Fx − iFy =iρ

2

∮

C

(

U − UR2

z2− iΓ

2πz

)2

dz

=iρ

2

∮

C

(

U − iUΓ

πz+ terms z−2, z−3, z−4

)

dz

Now use Cauchy’s residue theorem

∮

C

dz

z= 2πi,

∮

C

dz

zn= 0, n 6= 1 and

Fx − iFy = iρUΓ,

which is the same as in 4.3.

4.10.2 The Kutta-Koukowski lift theorem

From the previous example one appreciates that Blasius’ theorem yields a much moregeneral result, since only the residue of the integral comes into play. Consider a body ofarbitrary cross-section C, in a uniform stream, U , which generates circulation of strengthΓ. Far away from the body,

w(z) → Uz − iΓ

2πlog z

(origin inside C) sodw

dz≈ U − iΓ

2πz, z → ∞

and it must be analytic outside C and so

∮

C

(

dw

dz

)2

dz = limR→∞

∮

|z|=R

(

dw

dz

)2

dz

since there are no singularities in between C and |z| = R, a circle of large radius (basictheorem from C.F.T). So

∮

C

(

dw

dz

)2

dz =

∮

|z|=R

(

U − iΓ

2πz

)2

dz = 2UΓ,

using the same reasoning as for the cylinder. Hence we have shown that

Fx − iFy = iUρΓ.

So the drag on an arbitrary body is zero and the lift force is Flift = −UρΓ. This is theKutta-Joukowski lift theorem. Note that the lift force is by definition the force in thedirection normal to the flow.

4.11 Oblique flow past plates

.

We have developed a wonderful theory for lift. The only problem is that we don’tknow how to determine the value of Γ which determines the lift. Indeed, the wings of anairplane do not have circular cross section: if cylinder is placed in a uniform stream (leftside of the Figure), there is no reason why the flow should turn either way, and thus thereis no circulation. The secret is to fashion the shape so as to induce lift. Namely, the tailof wing is shaped to have a sharp corner, so as to force the flow to separate at that point.

As the simplest possible model for such a situation consider the flow around a flatplate, which is placed in a uniform stream at an angle α. A flat plate is obtained from anellipse by letting d → 0, that is putting b = R in the Joukowski transformation (33):

z = ζ +R2

ζ.

The the width of the plate is 2c = 4R. As is seen from the Figure, the flow separates fromthe plate beyond the trailing edge, so that fluid particles very close to the plate have togo around the sharp corner before leaving the plate.

This situation is reflected by the velocity on the surface of the plate, which is

u− iv =U sin(θ − α)

sin θ,

and so u→ ∞ as θ → 0, π (at the sharp edges of the plate). This is a physically untenablesituation; fluid particles are not likely to make such a sharp turn without leaving the platealtogether, especially not at infinite speed! The same conclusion can be drawn from theconcept of “adverse pressure gradient”, introduced earlier. The speed is very great at thetrailing edge and decreases as one moves up the plate. According to Bernoulli, this meansthat the pressure increases, in other words fluid particles experience an adverse pressuregradient. Instead of following a path of adverse pressure, fluid particles prefer to leavethe body, producing a point of separation at the trailing edge. This requirement is knownas the Kutta condition.

As we have seen in our calculation of the flow around a cylinder with circulation, thepoint of separation can be made to move by adding circulation:

w1(ζ) = U

(

ζe−iα +R2

ζeiα)

− iΓ

2πlog ζ.

Then the velocity is

u− iv =U sin(θ − α)

sin θ− Γ

4πR sin θLet us focus on the trailing edge, corresponding to θ → 0. (assume for example that thefront of the plate is slightly ‘rounded’, so that separation is not as important there). Thuswe want u− iv to be finite for θ → 0, a requirement which is another incarnation of theKutta condition. This can be achieved by choosing

Γ = −4πRU sinα,

which is the case shown in the above Figure. Now the singularity cancels and the velocitybecomes

u− iv =U

sin θ(sin(θ − α) + sinα) → U cosα, as θ → 0.

This means that the flow leaves the plate smoothly in the direction of the orientation ofthe plate, as seen in the above Figure.

The wonderful thing is that we have now determined the circulation uniquely, so wecan calculate the lift using Blasius’ theorem:

Flift = −UρΓ = 4πRρU2 sinα,

where 4R is the width of the plate. Once more, note that this is the force acting normalto the flow, so it is at an angle α relative to the orientation of the plate. It is customaryto define a lift coefficient cL by

Flift = cLAwρU2

2,

where Aw = 4R is the area of the wing (per unit length), also called the chord. Thus wehave found that for the plate

C(plate)L = 2π sinα ≈ 2πα

In the Figure, this result is compared to experiment, and it works really well, as long asthe angle of attack α is small. If however α becomes too large the theory fails abruptly.The reason is clear from the two photographs below. As long as α is small, the flowremains laminar and attached to the wing. As α is too great, the flow separates and acompletely different type of description must be sought.

Finally, we comment on the presence of circulation around the wing, which is thecrucial ingredient needed for flying. In the first instance, there is nothing wrong with thatfrom the point of view of a potential flow description. However, where is this circulationcoming from when one imagines starting up the flow, with zero circulation. Since thenet circulation in a large circle around the wing must vanish initially, and the Kuttacondition requires a circulation Γ < 0 around the wing, this means that in the processof the point of separation moving to the trailing edge, vortices of positive circulation(rotating counterclockwise) are shed from the wing. Eventually the vortices are convecteddownstream, and no longer matter for the problem.

4.12 Joukowski wings

wing

ζ

R

R−R − 2λ

λ

Now we try to model a wing in a slightly more realistic fashion, as proposed byJoukowski. In particular, we account for the fact that a real wing will be rounded at theleading edge (for the oncoming flow to go around it smoothly) and to be very sharp atthe trailing edge (for the flow to separate).

Mapping an ellipse, we have seen that if b < R, the Joukowski transformation mapsa circle to an ellipse; If b = R, the ellipse degenerates to a flat plate. Now let us take theJoukowski transformation

z = ζ +R2

ζ,

but shift the circle by λ to the left, and with radius R + λ. The equation of the circle inthe ζ-plane is

ζ = −λ + (R + λ)eiθ, θ = 0..2π.

Thus it touches the circle of radius R at the right, but lies inside it everywhere else. Inother words, the aerofoil is described by the equation

z = −λ + (R + λ)eiθ +R2

−λ + (R + λ)eiθ.

It is clear that this shape is rounded everywhere, but sharp at the trailing edge to theright, where it has a cusp (it looks locally line the sharp end of a plate), as seen in theFigure. The length of the wing in the horizontal is called the chord. From the above

mapping, back and front ends are at ζ = R and ζ = −R − 2λ, respectively. Thus thechord is

A = 2R +R + 2λ+R2

R + 2λ=

4(R + λ)2

R + 2λ.

Now the complex potential around the shifted circle is evidently

w1(ζ) = U

(

(ζ + λ)e−iα +(R + λ)2

(ζ + λ)eiα)

− iΓ

2πlog(ζ + λ),

and the complex velocity around the aerofoil is

dw

dz=

U

(

e−iα −(

R + λ

(ζ + λ)

)2

eiα

)

− iΓ

2π(ζ + λ)

(

1− R2

ζ2

)−1

Once more, there are potential singularities at ζ = ±R. However, the singularityζ = −R lies inside the wing, and is therefore harmless. On the other hand, the pointζ = R lies on the surface, and will lead to a singularity, unless Γ is chosen appropriately.At ζ = R, the term in braces becomes

U(

e−iα − eiα)

− iΓ

2π(R + λ)= −2iU sinα− iΓ

2π(R+ λ),

which vanishes forΓ = −4πU(R + λ) sinα.

In other words, using Blasius’ theorem, the lift coefficient for the Joukowski wing is

F(wing)lift = 4πρU2(R + λ) sinα.

and the lift coefficient is

c(wing)L =

2π(R + 2λ) sinα

R + λ.

wingζ

R

Rλ

µ

β

It is seen from the above formula that there is no lift if the angle of attack vanishes.In reality, wings are cambered: the top is rounded, and the bottom is hollowed out. Thiscan be achieved by shifting the centre of the circle to the position −λ+ Iµ in the ζ-plane.To produce a sharp trailing edge, the circle still has to touch the point ζ = R, and thusits equation is

ζ = −λ + iµ+√

(R + λ)2 + µ2eiθ.

For future convenience, we define the angle

β = arctan

(

µ

R + λ

)

(see Figure). The chord is the same as for the profile without camber.

The resulting cambered profile is shown as the green outline above. To compute theflow, we note the complex potential in the ζ-plane:

w1(ζ) = U

(

(ζ + λ− iµ)e−iα +(R + λ)2 + µ2

(ζ + λ− iµ)eiα)

− iΓ

2πlog(ζ + λ− iµ),

and the velocity is now

u− iv =

U

(

e−iα − (R + λ)2 + µ2

(ζ + λ− iµ)2eiα)

− iΓ

2π(ζ + λ− iµ)

(

1− R2

ζ2

)−1

.

As before, the term in braces has to vanish for ζ = R for the velocity to remain finite, sothis condition gives

U

(

e−iα − (R + λ)2 + µ2

(R + λ− iµ)2eiα)

− iΓ

2π(R + λ− iµ)= 0.

Noting that R + λ+ iµ =√

(R + λ)2 + µ2eiβ , we find

Γ = −4π√

(R + λ)2 + µ2 sin(α + β)

as the Kutta condition. The lift coefficient becomes

c(wing)L =

2π√

(R + λ)2 + µ2(R + 2λ) sin(α + β)

(R + λ)2,

and so indeed there is lift for α = 0. As a result, the angle of attack can be kept small,

and stall is less likely.The theoretical result for such a cambered profile is compared to experiment in the

Figure. The pressure distribution over the wing as well as the total lift predicted byJoukowski theory is in good agreement with experiment.

5 Waves and free surface flows

Free surface flows are in some ways very different from what we have done before. Inall problems considered so far, the domain D in which to solve the problem is given (forexample some box or the exterior of an aeroplane wing). A free surface, on the otherhand, moves, so the domain varies in time. The key feature, however, is that the domaindoes not vary according to some predetermined program. Instead, it moves in responseto the flow itself, that is in response to the flow solution (which of course itself dependson the domain).

This leads to solutions which are quite different in character to the fixed-boundary solu-tions we were considering so far (and generally speaking much more interesting solutions)!Another nice feature of free surface flows is that they are easy to observe experimentally,one just needs to track the motion of the free surface.

5.1 Kinematic boundary condition

If a fluid particle is adjacent to a boundary then we must impose a condition which linksthe velocity of the boundary to that of the particle. This is known as the kinematicboundary condition.

Let S(x, t) = 0 describe the equation of a surface in (or on the boundary of) the fluid.As the flow evolves, particles remain on the surface S if

DS

Dt=∂S

∂t+ u.∇S = 0

Proof: Taylor’s theorem:

0 = S(x, t)− S(x+ uδt, t+ δt) = δt

(

u.∇S +∂S

∂t

)

Water

Oil

S=0

Example: Consider two fluids bounded by an interface S(x, t) = 0 (e.g. water/oil).Then

DS

Dt=∂S

∂t+ u(oil).∇S = 0, on S = 0 from above

DS

Dt=∂S

∂t+ u(water).∇S = 0, on S = 0 from below

Now ∇S is normal to the surface S = const, so n = ∇S/|∇S| is the unit normal to thesurface S = 0. It follows that

u(oil).n = u(water).n, on S = 0 (34)

I.e. the normal component of the velocity on either side of the interface must be equal(intuitive).

At a fluid/solid boundary, (34) holds, with u(oil) replaced by u(solid), the velocity vectorof the boundary. If fixed, u(water) · n = 0 on fixed surface S(x) = 0 with normal n.

5.2 Linear gravity waves

The simplest approximation (but still v. good in many cases): waves of small amplitude.Flow is irrotational and fluid incompressible. Assume 2D flow (for now).

z=−h

z

xζ(x,t) = z

z=0

Density = ρ

Pressure = patm

So u = ∇φ, ∇2φ = 0, where u = (u, 0, w) and φ = φ(x, z, t) now.

Choose z = 0 to coincide with the undisturbed free surface Bottom on the fluid is atz = −h.Let the surface of the water in motion be given by z = ζ(x, t).

Boundary conditions:

(i) On z = −h, 0 = n.∇φ =∂φ

∂z= w (Kinematic B.C.)

(ii) On z = ζ(x, t), Kinematic B.C. on a moving surface isDS

Dt= 0 (see section (1.10)),

where S = z−ζ(x, t) = 0 defines the surface of the fluid. This implies that particleson the surface, remain there ∀ time. I.e.

(

∂

∂t+ u

∂

∂x+ w

∂

∂z

)

(z − ζ(x, t)) = −∂ζ∂t

− ∂φ

∂x

∂ζ

∂x+∂φ

∂z= 0 on z = ζ

(iii) Dynamic boundary condition: p = patm (const) on z = ζ . So unsteady Bernoulli onsurface gives

patm/ρ+12|∇φ|2 + ∂φ

∂t+ gζ = C(t), on z = ζ

By the small amplitude assumption, |ζ | ≪ h and so

∣

∣

∣

∣

∂ζ

∂x

∣

∣

∣

∣

≪ 1. If we assume that

the flow is driven by the motion of the free surface (no additional driving from below thesurface), it must be that the flow is also week: |φ| ≪ 1. Thus we will throw away allterms quadratic in the two variables ζ and φ, i.e. containing ζ2, φ2, or products ζφ.

Note that (iii) contains the velocity at quadratic (nonlinear) order, which is a remnantof the nonlinear character of the Euler equation, as it remains in Bernoulli’s equation. The

kinematic boundary condition contains∂φ

∂x

∂ζ

∂x, which is quadratic as well. Both terms

will be neglected in our linearised treatment.However, the most significant problem is that in the formulation of the boundary

conditions, the position of the free surface is part of the solution, so don’t know where to

apply B.C’s ! We can deal with this difficulty by linearising about z = 0 using Taylor’sexpansion about z = 0 for all z-dependent terms. E.g.

∂φ

∂t

∣

∣

∣

∣

z=ζ

=∂φ

∂t

∣

∣

∣

∣

z=0

+ ζ∂

∂z

(

∂φ

∂t

)

z=0

+ . . .

and throw away any products involving ζ and/or φ, because quadratic terms are negligible.So now

(i) On z = −h, ∂φ∂z

= 0

(ii) On z = 0,∂ζ

∂t=∂φ

∂z

(iii) On z = 0,∂φ

∂t+ gζ = C(t)− patm/ρ.

Last condition is ugly, so redefine φ = φ′ − patmt/ρ+

∫ t

C(t′)dt′. The extra terms

are time dependent only, so they don’t affect the flow velocities, Laplace’s equationor the B.C’s. So now (dropping primes) (iii) is replaced with

(iii) On z = 0,∂φ

∂t+ gζ = 0.

Linearised pressure in the fluid is

p(x, z, t)

ρ+∂φ

∂t+ gz = C(t)

and because of φ → φ′ this transforms to

p(x, z, t)− patmρ

= −∂φ∂t

− gz (35)

Now look for particular solutions to ∇2φ = 0 with (i), (ii), (iii) motivated by observations.Assume surface of the form

ζ(x, t) = H sin(kx− ωt+ δ) (36)

• max/min of ζ is ±H .

• δ is just a phase shift (in space or time).

• periodic with period τ = 2π/ω (ω is angular frequency)

• Repeats in space every λ = 2π/k (λ is wavelength)

• Moves to the right with speed c = ω/k (c is phase speed). Why ? Let ξ = x − ct,then ζ = H sin(kξ + δ) is unchanged for ξ constant so 0 = dξ/dt = dx/dt − c anddx/dt = c = speed.

Now need φ. Since∂φ

∂t= −gζ reasonable to write

φ(x, z, t) = cos(kx− ωt+ δ)Z(z)

for some Z(z) – this is a separation solution φ = X(x)Z(z). Then from ∇2φ = 0,

−k2 cos(kx− ωt+ δ)Z(z) + cos(kx− ωt+ δ)Z ′′(z) = 0

and so Z ′′(z)− k2Z(z) = 0. Then

Z(z) = A cosh k(z + h) +B sinh k(z + h) (= A′e−kz +B′ekz)

for constants A,B,A′, B′. Because of (i), need B = 0, so

Z(z) = A cosh k(z + h)

Still have (ii) and (iii) to apply (but only A to find !) From (iii) first, gζ = −∂φ/∂t|z=0

gH sin(kx− ωt+ δ) = −Aω sin(kx− ωt+ δ) cosh kh

and then A = −gH/(ω cosh kh). Then from (ii), ∂φ/∂z|z=0 = ∂ζ/∂t

−ωH cos(kx− ωt+ δ) = kA cos(kx− ωt+ δ) sinh kh

implies

−ωH = k

( −gHω cosh kh

)

sinh kh

orω2

g= k tanh kh (37)

Notes:

• Changing k → −k, (37) still holds. So then

ζ(x, t) = H sin(kx+ ωt− δ)

is a wave travelling to the left, speed c = −ω/k.

• k = 2π/λ is called the wavenumber (the number of wavelengths that can be fit into2π).

The principal result of this calculation is the so called dispersion relation (37) of thewave problem. It establishes how the frequency of a monochromatic wave is related toits wavelength. Very roughly, large λ ⇒ small k ⇒ small ω ⇒ large τ = 2π/ω. So longwavelengths imply long periods, and vice versa.

If the dispersion relation between ω and k is non-linear, one speaks of a dispersiveproblem: parts of a wave containing different frequencies travel at different speeds, andthus disperse. Two important special cases of (37) exist:

5.2.1 Shallow depth

This is the case kh≪ 1 (or λ/h≫ 1), which means that the wave length is long comparedto the depth. Then tanh kh ≈ kh or

ω =√

ghk,

so the dispersion relation is a linear function. The constant of proportionality is the(constant) wave speed c = ω/k =

√gh. Evidently, the shallow water problem has no

dispersion.

5.2.2 Infinite depth

If on the other hand kh≫ 1 (or λ/h≪ 1), the water is deep relative to the length of thewave. Then tanh kh ≈ 1 and the dispersion relation is

ω =√

gk.

Thus the wave speed c = ω/k =√

g/k =√

gλ/2π does depend on the wavelength, andthere is dispersion. The velocity potential corresponding to the wave form (36) is

φ = −ωHk

ekz cos(kx− ωt+ δ).

Neglecting the irrelevant phase factor δ, the velocity components are

u = ωHekz sin(kx− ωt), w = −ωHekz cos(kx− ωt).

Assuming that a particle with position (x′, z′) in the flow only departs by a smallamount from its mean position (x, z), one can write

dx′

dt= ωHekz sin(kx− ωt),

dz′

dt= −ωHekz cos(kx− ωt).

This is integrated easily to yield the particle trajectory

x′ = Hekz sin(kx− ωt), z′ = −Hekz cos(kx− ωt).

Thus a particle moves along a circular path, whose radius decreases exponentially withdepth. Most of the motion is concentrated near the surface, and the wave energy decreasesexponentially as one descends into the water.

5.3 Group velocity

Now we investigate the propagation of so-called wave-packet, that is a burst of finitespatial extend, as it might be produced by throwing a stone into the water. We write thissituation as a superposition of infinitely extended plane waves:

ζ =

∫ ∞

−∞

a(k)ei(kx−ωt)dk.

The amplitude of each part of the wave is given by a(k). It is easy to show (a versionof the uncertainty relation), that the more peaked the distribution a(k) is around somemean value k0, the more oscillations the wave packet contains, and thus the broader itbecomes.

Let us assume that a(k) is sufficiently narrow relative to k0. In other words, it is safeto write

ω(k) = ω(k0) +dω

dk

∣

∣

∣

∣

k=k0

(k − k0).

But this means the wave packet can be written

ζ(x, t) = ei(k0x−ω(k0)t)

∫ ∞

−∞

a(k)ei(k−k0)(x−cgt)dk,

where

cg ≡dω

dk

∣

∣

∣

∣

k=k0

(38)

is called the group velocity. The first factor in the representation of the wave packetrepresents a harmonic modulation with a high-frequency wave. The second factor isenvelope, giving the shape of the wave packet. The key observation is that x only appearsin the combination x−cgt, and so the entire wave packet moves with the group velocity cg.This velocity is usually more significant than the phase velocity c = ω/k. For example,the energy of a wave is concentrated where the wave packet is localised. Therefore, energytransported by a wave moves at speed cg.

Taking waves of infinite depth as an example, the phase velocity is c =√

g/k, while

cg =√

g/k/2, or half the value!

5.3.1 Example: Tsunamis

The ocean is deep ∼ 4km, but in this case the waves are very long (∼ 200km). Asa result, λ/h ≫ 1 and the shallow water approximation applies. There is no dispersion,but the wave speed c = ω/k =

√gh (which is the same as the group velocity) depends on

the depth. Thus as the ocean depth changes, the Tsunami wave moves in a medium ofvariable refractive index, and does not move along a straight path. Instead, it is refractedin unexpected directions, and can “turn corners”. Here we only want to discuss theamplification of the wave amplitude near the shore.

The Tsunami moves very fast, the wave speed is

c = ω/k =√

gh =√10.4000 = 200ms−1 = 400mph.

Owing to energy conservation, the wave energy must be conserved as h changes, and thusthe energy flux of the wave must be constant. The wave contains potential and kineticenergy to equal proportion. From elementary mechanics, the potential energy of a liquidcolumn (per unit area) of height ζ is

∫ ζ

0

ρgzdz = ρgζ2

2.

Now for a sinusoidal wave, the average of the evaluation is

〈ζ2〉 = H2

2π

∫ 2π

0

sin(x)dx =H2

2,

so the potential energy becomes ρgH2/4. Since the same energy is contained in the kineticpart, the total energy (per unit area) is

e =ρ

2gH2.

Since the energy moves with the group velocity, the energy flux becomes

fe = cg1

2ρgH2 =

1

2

√

ghρgH2.

If this is to be constant for all x, this implies that H4h = const, and therefore H ∼1/h1/4.

If the depth goes from 4000m to 4m then factor of 1000 and the wave height increasesby a factor of 10001/4 ≈ 6. So a 0.5m wave in the ocean is a 3m wave at the coast.

5.4 Oscillations in a container

Liquids readily slosh back and forth in a closed container (e.g. tea in a tea-cup). Thereare many associated important practical problems: sloshing in road tankers, water ondecks of ships, resonance in harbours, etc...

z=0

z=−h

x=Lx=0

x

zz= (x,t)ζ

Example: consider a two-dimensional rectangular box with rigid walls at x = 0, L anda bottom on z = −h, filled with fluid to z = 0.

Use small-amplitude theory from before, so that linearised equations are ∇2φ = 0 for(x, y, z) in box,

(i)∂φ

∂z= 0 on z = −h.

(ii) Kinematic:∂ζ

∂t=∂φ

∂zon z = 0.

(iii) Dynamic:∂φ

∂t= −gζ on z = 0.

Also need no-flow conditions (kinematic) on walls:

(iv)∂φ

∂x= 0 on x = 0, L;

Different to before - no travelling waves - but still periodic in time, so write

ζ(x, y, t) = ζ(x, y) sinωt

and assume we can separate variables in φ:

φ(x, y, z, t) = X(x)Z(z) cosωt.

Then Laplace’s equation gives

X ′′(x)Z(z) cosωt+ Z ′′(z)X(x) cosωt = 0

which impliesX ′′(x)Z(z) + Z ′′(z)X(x) = 0

This separates:X ′′(x)

X(x)= −Z

′′(z)

Z(z)= −k2

where −k2 is the separation constant.Solving for Z(z), with (i) gives Z(z) = A cosh k(z + h) as before. Combining (ii) and

(iii) to eliminate ζ (that’s g(ii) +∂/∂t(iii) gives

∂2φ

∂t2= −g∂φ

∂z, on z = 0

which implies−ω2X(x)Z(0) cosωt = −gX(x)Z ′(0) cosωt

and it follows that ω2/g = k tanh kh as before. (We expect this, as the vertical structureof the fluid is independent of the vertical lateral walls)

Now solving for X(x) gives

X(x) = B cos(kx) + C sin(kx)

subject to (from (iv)), X ′(0) = X ′(L) = 0. Easy to show that must have C = 0 andk = nπ/L and so

X(x) = B cos(nπx/L)

So pulling everything together we have

φ(x, z, t) = A′ cos(nπx/L) cosh k(z + h) cosωt

for some constant A′ = AB while from (iii) the free surface is given by

ζ(x, t) =ωA′ cosh kh

gcos(nπx/L) sinωt

and here, k = nπ/L, so that (37) reads

ω2/g = (nπ/L) tanh(nπh/L) ≡ ω2n/g, n = 0, 1, 2, . . .

and therefore defines a set of discrete wave frequencies at which these oscillations mayoccur. Here, n is a mode number and tells you how many oscillations are occurring acrossthe box.

We cannot have n = 0 as then ω = 0 and φ = 1, and ζ = 0. So there is no motionin the fluid. (In fact, you can discount a sloshing mode with no x-dependence – a flatsurface oscillating up and down – as it would violate mass conservation in the tank).

The Fundamental frequency is the lowest frequency, given here by n = 1.E.g. Let L = 1m, h = 20cm. Then the fundamental frequency (n = 1) is

ω ≡ ω1 =√

9.81× π × tanh(π ∗ 0.2) = 4.14s−1

given a period of τ = 2π/4.14 = 1.51 seconds. The shape of the surface is given by thevariation of x in ζ(x, t), which is cos(πx/L) and it is modulated by the signal sinωt.

n=1 n=2 large n

ce

The n = 2 mode gives a period of 0.86 seconds and a shape function of cos(2πx/L)etc...

As n→ ∞, tanh(nπh/L) → 1 and so ωn →√

gnπ/L.

5.5 Liquid sheets

BA’ B’ C’C A

BB’

AA’

CC’

ζz

We now move on some of examples of truly nonlinear solutions of free surface flowproblems. The first example is that of a sheet coming out of a slit at the bottom of acontainer. One might think that the sheet has the same thickness as the width of theslit. However, the flow converges as it enters the slit, so the sheet is actually thinner; itcontracts as it flows out of the container. We want to find out what the flow is like exactlynear the entrance, and in particular what the shape of the free surface is like.

This is a two-dimensional problem, so we will use complex mapping techniques. Theypermit to deal with the difficulties associated with the non-linearity of free surface flow.We begin with the trivial problem of an infinite sheet flowing in the negative y-direction.With view to what is to come, we want to solve the problem by mapping to solve theproblem by mapping to the upper half plane, see the Figure above. We have already seenin section 4.8.3 that the map which accomplishes this is the logarithm, which producesa channel oriented in the x-direction. Thus we introduce a ninety degree rotation, andconsider the map

z = i ln ζ,

which maps points A,B,C,C’,B’,A’ as indicated. For simplicity, we consider a sheet ofwidth π; it is easy to change the length scales later on.

In particular, the lower limit of the sheet is near the origin of the ζ-plane. Therefore,one might guess that the representation of the sheet is a sink located at the origin.According to section 4.4.2 the potential is

w(ζ) = − ln ζ. (39)

We will henceforth adopt the somewhat more lax, but useful notation w1 ≡ w if w isrepresented as function of ζ . Now in the z-plane this becomes w = −z/i = iz, and thusu− iv = i. Thus in the sheet there is a flow with unit speed in the negative y-direction,exactly as anticipated.

AB

C,C’

A’B’

AA’B’ B

CC’

container

sheet

Ω

z

Of course the real problem at hand is much more difficult, as shown on the right ofthe above Figure. Between A and B as well as between B’ and A’, the flow is borderedby hard walls. At least in these places we know where the boundary is. However, atthe free surface of the sheet, which is between B and C as well as B’ and C’, the shapeof the domain becomes part of the problem. This is also the reason why the problembecomes non-linear. Complex mapping techniques are extremely powerful at approachingthis kind of problem, since the whole flow domain (bordered along A,B,C,C’,B’,A’) canstill be mapped onto the upper half plane. The name of the game is to find a complexmapping z(ζ) which maps the upper half plane conformally onto the flow domain. Oncerepresented in the ζ-plane, the flow will still be described by the potential (39). Thereason is that the only singularity of the problem is the sink located at ζ = 0, whichdescribes the sheet flowing off to infinity.

How do we find the mapping z(ζ)? The main idea is due to Helmholtz and Kirchhoff(1869), with additional simplifications suggested by Planck and Love. The trick is to mapthe flow domain onto a domain of known shape, for which also some of the boundaryconditions are known. According to complex mapping theory, this then determines themap uniquely. What are the boundary conditions? At a solid border, the flow must beparallel to the wall. Thus representing the velocity in complex notation,

dw

dz= u− iv = qe−iθ,

the flow direction θ is θ = 0 along B’,A’, and θ = −π along A,B.At the free surface, the pressure is constant (we are not considering gravity). This is

why the surface is called a free streamline. According to Bernoulli, the flow speed q isalso constant along the free streamline. We will normalise such that q = 1 all along thesheet. Now the ingenious idea is to introduce the function

Ω = lndz

dw= − ln q + iθ.

This maps the flow onto a rectangular domain, see above. Namely, B′, A′ is mapped ontothe straight line ℑΩ = 0, and A,B onto the straight line ℑΩ = −π. The free surface,on the other hand, is mapped onto a straight line parallel to the imaginary axis: sinceq = 1, we have ℜΩ = 0. Thus all the boundaries are as drawn in the Figure. As a checkof consistency, let us calculate Ω at the points C,C’ infinitely far down the sheet. The flowwill be that of an unperturbed sheet: w = iz, and thus Ω = ln(−i) = ln e−iπ/2 = −iπ/2.Indeed, this is midway between the upper and lower boundary of the domain.

Now we construct the mapping between the ζ and Ω planes. This mapping is guar-anteed by the so-called Schwarz-Christoffel mapping theorem, but we will take a more

constructive approach: we consider the derivative ∂Ω/∂ζ . On the boundary, this expres-sion has singularities at ζ = 1 and ζ = −1, corresponding to points B and B′, where theboundary of Ω makes 90 turns. We have considered this situation before in problem 1of worksheet 9. There we have shown that a turn by π/2 corresponds to a singularityζα/π−1 = ζ−1/2. Two such turns are to be executed at ζ = ±1, so we have

dΩ

dζ=

1√

ζ2 − 1.

This function does exactly what we need: For ζ > 1 the argument of the square root ispositive, and thus the expression on the right is real and positive. Thus as one integratesfrom A to B along the ζ-axis, one moves to the left parallel to the real z-axis. At ζ = 1,the right hand side becomes becomes purely imaginary, with a nagative coefficient. Thusthe path of integration moves up parallel to the imaginary axis as one goes from B to B’.Finally, at ζ = −1, the right hand side is once more real, and one moves back parallel tothe real z-axis.

Integrating, we have Ω = Arcoshζ + Ω0. Now for ζ > 1 we must have Ω = −iπ, andso since Arcoshζ is real, it follows that ℑΩ0 = −π. In fact Ω0 = −iπ; for −1 < ζ < 1,Ω must be purely imaginary: Ω = iΩ2. Thus

cosh iΩ2 = cosΩ2 = cosh(Arcoshζ − iπ) = −ζ,

and Ω2 runs from −π at ζ = 1 to 0 at ζ = −1, exactly as required. In summary,

Ω = Arcoshζ − iπ. (40)

This is the mapping we are looking for, which will essentially be the solution to theproblem. Streamlines, represented in the ζ-plane, are straight lines approaching the originζ = 0, thus with the mapping (40) streamlines can be represented in the Ω-plane, as shown.

To find z(ζ), we write

dz

dζ=dz

dw

dw

dζ= eΩ

dw

dζ= −eΩ

ζ=ζ +

√

ζ2 − 1

ζ,

where we have used Arcoshζ = ln(

ζ +√

ζ2 − 1)

and e−iπ = −1. Another integration

will yield the desired mapping. Since we are mainly interested in the shape of the sheetfor 0 < ζ < 1, we write

√

ζ2 − 1 = i√

1− ζ2, and the integral gives

z = ζ + i

(

√

1− ζ2 − ln1 +

√

1− ζ2

ζ

)

+ z0.

The constant of integration z0 is merely a shift, but we want to place the sheet sym-metrically about the y-axis. In the limit of ζ → 0, we have z ≈ i ln ζ as before for thefree sheet, which lies between −π and 0. Thus we have z0 = π/2 and finally

z = ζ + i

(

√

1− ζ2 − ln1 +

√

1− ζ2

ζ

)

+π

2. (41)

In particular, for 0 < ζ < 1, x = ℜz = ζ + π/2, and so

y = ℑz =√

1− (x− π/2)2 − ln1 +

√

1− (x− π/2)2

x− π/2

is the shape of the sheet.Of particular interest is the contraction of the sheet. The width x of the sheet decreases

from 2(1+π/2) at the opening to π far downstream. Thus the sheet suffers a contractionby the ratio

Cc =π

2 + π≈ 0.611 < 1.

The experimental value, found by J.S. McNown and E.-Y. Hsu, is C(exp)c = 0.632: pretty

good!

Appendix A: Vector calculus

We shall revise some vectors operations that you should have already met before thiscourse. These may be presented slightly differently to the way you have previously seenthem.

A.1 Suffix notation and summation convention

Suppose that we have two vectors u = (u1, u2, u3) and v = (v1, v2, v3). Then thedot product is defined to be

u · v =

3∑

i=1

uivi or, more simply, write u · v = uivi

(drop the summation symbol on the understanding that repeated suffices imply summa-tion.

Defn A.1.2: The Kronecker delta is defined by

δij =

1, i = j0, i 6= j

So in summation convention, δijaj = ai since

δijaj ≡3∑

j=1

δijaj = ai

Examples:

1. δii = 3

2. δijuivj = ujvj ≡ u · v.

Defn A.1.3: The antisymmetric symbol ǫijk is defined by

• ǫ123 = 1

• ǫijk is zero if there are any repeated suffices. E.g. ǫ113 = 0.

• Interchanging any two suffices reverses the sign: e.g. ǫijk = −ǫjik = −ǫkji

• Above implies invariant under cyclic rotation of suffices: ǫijk = ǫjki = ǫkij

With this definition all 27 permutations are defined. There are only 6 non-zero compo-nents,

ǫ123 = ǫ231 = ǫ312 = 1, ǫ213 = ǫ132 = ǫ321 = −1,

Defn A.1.4: The cross product is defined by

w = u× v = (u2v3 − u3v2)x+ (u3v1 − u1v3)y + (u1v2 − u2v1)z

But can now be written in component form as

wi = ǫijkujvk

where summation over j and k occurs. [Check].

Example: Consider the triple product,

w · (u× v) = wiǫijkujvk

where summation is over i, j, k so result is scalar. It follows that

w · (u× v) = ǫjkiwiujvk = u · (v ×w)

= ǫkijwiujvk = v · (w × u)

= −ǫjikwiujvk = −u · (w× v)

= −ǫikjwiujvk = −w · (v × u)

Defn A.1.5: The double product of ǫijk is

ǫijkǫilm = δjlδkm − δjmδkl

[Check]

Defn A.1.6: The vector triple product is defined by the result

w × (u× v) = (w · v)u− (w · u)v

Proof:

[w × (u× v)]i = ǫijkwj[u× v]k

= ǫijkwjǫklmulvm

= ǫkijǫklmwjulvm

= (δilδjm − δjlδim)wjulvm

= wjuivj − wjujvi = (w · v)ui − (w · u)vi

True for i = 1, 2, 3, hence result.

A.2 Differential operations

Here we consider operations on a function φ(x) and a vector field f(x) where x =

(x1, x2, x3). One can regard ∇ as the vector operator

(

∂

∂x1,∂

∂x2,∂

∂x3

)

. Without us-

ing any information (but always remembering the true meaning of the symbol!), one canalso use the much sleaker notation ∇ ≡ (∂1, ∂2, ∂3). Then

• The gradient is ∇φ. So [∇φ]i =∂φ

∂xi≡ ∂iφ

• The divergence is ∇ · f = ∂fi∂xi

≡ ∂ifi (in summation convention)

• The curl is ∇× f where [∇× f ]i = ǫijk∂fk∂xj

≡ ǫijk∂jfk.

Examples:

1. [∇(xj)]i =∂xj∂xi

= δij

2. ∇ · x =∂xi∂xi

= δii = 3

3. [∇× x]i = ǫijk∂xk∂xj

= ǫijkδkj = ǫijj = 0

4. ∇r = x

r, where r2 ≡ x · x.

Proof: [∇r]i = ∂i√xjxj =

∂ix2j

2√

x2j

=xj∂ixjr

=xjδijr

=xir.

A.2.1 Useful vector identities

1. ∇ · (φf) = ∂(φfi)

∂xi= φ

∂fi∂xi

+ fi∂φ

∂xi= φ∇ · f + f · ∇φ

2. ∇× (φf) = φ(∇× f) + (∇φ× f)

Proof: [∇× (φf)]i = ǫijk∂(φfk)

∂xj= φǫijk

∂fk∂xj

+ fkǫijk∂φ

∂xj= φ[∇× f ]i + [∇φ× f ]i.

3. f × (∇× f) = ∇(12f · f)− (f · ∇)f

Proof:

[f × (∇× f)]i = ǫijkfj [∇× f ]k = ǫijkfjǫklm∂fm∂xl

= ǫkijǫklmfj∂fm∂xl

= (δilδjm − δimδjl)fj∂fm∂xl

= fj∂fj∂xi

− fj∂fi∂xj

=∂ 1

2fjfj

∂xi− (f · ∇)fi

Hence result.

A.3 Integral results

A.3.1 The divergence theorem

d

n

V S

S

Very important. Consider a volume V bounded by a closed surface S with outward unitnormal n. Then for a vector field f(x)

∫

V

∇ · fdV =

∫

S

f · ndS

Corollary: Let f(x) = aφ(x) where a is an arbitrary constant vector, and φ(x) a scalarfunction. Since ∇ · (aφ(x)) = a · ∇φ(x), so the divergence theorem reduces to

a ·∫

V

∇φdV = a ·∫

S

φndS

True for any a, so∫

V

∇φdV =

∫

S

φndS

A.3.2 Stokes’ theorem

dl

ds

n

C

S

Let C be a closed curve bounding a surface S with unit normal n.Then for a vector field f(x),

∮

C

f · dl =∫

S

(∇× f) · ndS

where dl is a line element on C.

Corollary: Let f(x) = aφ(x) where a is an arbitrary constant vector. Then

∮

C

aφ · dl =∫

S

(∇× aφ) · ndS

and from §A.2.1(2), (∇×aφ) = φ(∇×a)+ (∇φ×a) = (∇φ×a). Using the vector tripleproduct result, (∇φ× a) · n = −a · (∇φ× n) and then

a ·∮

C

φdl = −a ·∫

S

∇φ× ndS

Therefore∮

C

φdl = −∫

S

∇φ× ndS

Appendix B: Curvilinear coordinate systems

r(q1, q2)

e1e2

q1 increasing

q2 increasing

Many problems can be approached more simply by choosing a coordinate system thatfits a given geometry. Instead of writing the position vector r as a function of Cartesiancoordinates (x, y, z), r is now written as function of three new coordinates: r(q1, q2, q3).The coordinate lines are swept out by varying one of the coordinates, keeping the other twoconstant. We will deal only with the by far most important case of orthogonal coordinatesystems, in which the coordinate lines always intersect one another at right angles.

Evidently,∂r

∂qiis a vector which points in the direction of the i-th coordinate line,

see the figure. If each of these vectors are normalized to unity, we obtain the local basissystem:

qi =1

hi

∂r

∂qi, hi ≡

∣

∣

∣

∣

∂r

∂qi

∣

∣

∣

∣

. (42)

The quantities hi(q1, q2, q3) are called scale factors or metric coefficients. The fact thatthe coordinate system is orthogonal means that all qi, computed at a point (q1, q2, q3),are orthogonal. However, the direction of qi of course changes as one goes along thecoordinate lines.

Example: The cylindrical polar coordinate system.The position vector is

r = (r cos θ, r sin θ, z) ,

and the coordinates are (r, θ, z). Thus the scale factors become h1 = 1, h2 = r, andh3 = 1, and the local basis is

r =∂r

∂r= (cos θ, sin θ, 0) ,

θ =1

r

∂r

∂θ= (− sin θ, cos θ, 0) ,

z =∂r

∂z= (0, 0, 1).

It is clear that r, θ, z are indeed mutially orthogonal.

Remark: As the coordinates are varied by δq1, δq2, and δq3, respectively, r describes avolume whose sides are orthogonal. The length of each side is hiδqi, and thus the volume

of the cuboid is d3x = h1h2h3dq1dq2dq3. Thus integration in a curvilinear coordinatesystem is achieved by the formula

∫

V

f(x)d3x =

∫

V

f(q1, q2, q3)h1h2h3dq1dq2dq3. (43)

In cylindrical polars, the volume element is rdrdθdz.

To do vector calculus in the curvilinear coordinate system, we have to work out whatthe ∇-operator is. For any scalar function f(r),

∂f

∂qi=∂rj∂qi

∂f

∂rj=

(

∂r

∂qi· ∇)

f = hi (qi · ∇) f

by the chain rule. This menas that in the local basis, ∇ has the representation

∇ =q1

h1

∂

∂q1+

q2

h2

∂

∂q2+

q3

h3

∂

∂q3. (44)

Thus in cylindrical polars, the gradient of a scalar function φ is

∇φ =∂φ

∂rr+

1

r

∂φ

∂θθ +

∂φ

∂zz,

in agreement with chapter 0.Now we are almost there. The only but crucial thing that remains to be considered

is the fact that both scale factors and unit vectors depend on the coordinates qi, so they

need to be differentiated. In particular, we want to represent∂qi

∂qjin terms of the basis

vectors. It is not difficult to write down general expressions for the derivatives of the basisvectors if only the hi are known. However, for a given coordinate system, it is generallymuch easier to simply calculate the derivatives directly. For example, in cylindrical polars

∂r

∂θ= θ,

∂θ

∂θ= −r,

and all the other derivatives are zero.Now it is easy to derive the other formulae in chapter 0. In cylindrical polars, the

Laplacian becomes

φ = ∇ · ∇φ =

(

r∂

∂r+

θ

r

∂

∂θ+ z

∂

∂z

)(

r∂φ

∂r+

θ

r

∂φ

∂θ+ z

∂φ

∂z

)

=

(r · r)∂2φ

∂r2+

1

r

(

θ · ∂r∂θ

)

∂φ

∂r+

1

r2

(

θ · ∂θ∂θ

)

∂φ

∂θ+(

θ · θ) 1

r2∂2φ

∂θ2+ (z · z) ∂

2φ

∂z2=

∂2φ

∂r2+

1

r

∂φ

∂r+

1

r2∂2φ

∂θ2+∂2φ

∂z2.

In local coordinates, a vector field is written as u = urr + uθθ + uzz. Then thedivergence is

∇ · u =

(

r∂

∂r+

θ

r

∂

∂θ+ z

∂

∂z

)

·(

urr+ uθθ + uzz)

=∂ur∂r

+urr

+1

r

∂uθ∂θ

+∂uz∂z

,

again in agreement with chpater 0.

Appendix C: Streamfunctions

If ∇ · u = 0, then it follows that there exists a vector field A(x, t) s.t.

u = ∇×A.

Conversely, it is clear from this representation that the flow is incompressible. The rep-resentation is particularly useful if there are two independent coordinates, in which caseA can be written in terms of a single scalar function, the streamfunction. By defini-tion, the streamfunction is constant along streamlines. There are two important cases:two-dimensional flows and three-dimensional, axisymmetric flows. In the latter case thestreamfunction is known as the Stokes streamfunction.

5.5.1 Two-dimensional flows

(i) Cartesians: Consider Cartesian coordinates (x, y, z), and take the flow in the (x, y)-plane. Then A = ψ(x, y, t)z, and

u =∂ψ

∂y, v = −∂ψ

∂x.

Now confirm that ψ is indeed constant along streamlines, which are defined by

dx

ds= λu,

and thus dx = λuds and dy = λvds. It follows that

dψ =∂ψ

∂xdx+

∂ψ

∂ydy = vdx− udy = 0,

if indeed ψ is varied along streamlines. Thus ψ is constant along a streamlines, as adver-tised.

Defn: We call the function ψ(x, y, t) the streamfunction of the flow.

ψ= constant

(ii) Cylindrical polar coordinates: Now consider flow in the same two dimensionalplane z = 0, which is represented in cylindrical polars (r, θ, z). The relation to Cartesiancoordinates is x = r cos θ, y = r sin θ. Since the stream function is defined completely bythe geometry of the streamlines, it must be the same in any coordinate system. In otherwords if ψc(x, y, t) is the Cartesian version, and ψp(x, y, t) the streamfunction in polarcoordinates, then ψp(r, θ, t) = ψc(r cos θ, r sin θ, t).

As before, the vector potential is A = ψ(r, θ, t)z, and thus from the curl in polarcoordinates

ur =1

r

∂ψ

∂θ, uθ = −∂ψ

∂r,

where u = urr+ uθθ.

Now verify that ψ(r, θ) is indeed constant along streamlines. In polar coordinates,x = rr, and thus

dx

ds=dr

dsr+ r

dθ

ds

dr

dθ=dr

dsr+ r

dθ

dsθ,

where we have used thatdr

dθ= θ. In other words, we find that dr = λurds and rdθ =

λuθds. Now as before,

dψ =∂ψ

∂rdr +

∂ψ

∂θdθ = −uθdr + rurdθ = 0,

if ψ is indeed varied along streamlines.

5.5.2 Three-dimensional axisymmetric flows

This is a different physical situation, as the flow is now three-dimensional. However, sinceit is axisymmetric, it can be written in terms of just two coordinates. Once more, wechoose two different coordinate systems.

(i) Cylindrical polars: Coordinates (r, θ, z) and and the velocity field is independentof θ and so u = ur(r, z)r+uz(r, z)z. For this to be true, A must be in the third coordinatedirection:

A =Ψ

rθ.

The definition is such that Ψ(r, z) is indeed a streamfunction, as we will now confirm; itis called the Stokes’ streamfunction.

First, calculating ∇×A gives

ur = −1

r

∂Ψ

∂z, uz =

1

r

∂Ψ

∂r.

Streamlines given by dr = λurds and dz = λuzds, so

dΨ =∂Ψ

∂rdr +

∂Ψ

∂zdz = ruzdr − rurdz = 0.

Therefore Ψ = const on streamlines (actually stream surfaces as 3D flow).

θ r

z

x−axis

z

x

y

φ

θr

Cylindrical polars Spherical Polars

(ii) Spherical polars: The coordinates are (r, θ, ϕ) and the velocity field is indepen-dent of ϕ. Thus u = ur(r, θ)r+ uθ(r, θ)θ, and A must be in the ϕ-direction:

A =Ψ(r, θ)

r sin θϕ.

We confirm that Ψ(r, θ) is the streamfunction, and thus it is the same as defined previouslyin cylindrical polar coordinates.

First, computing ∇×A we find

ur =1

r2 sin θ

∂Ψ

∂θ, uθ = − 1

r sin θ

∂Ψ

∂r.

As in polar coordinates, streamlines are given by dr = λurds and rdθ = λuθds. Thus

dψ =∂ψ

∂rdr +

∂ψ

∂θdθ = −r sin θuθdr + r2 sin θurdθ = 0

from the definition of the stream function. Thus Ψ(rθ) is indeed constant along stream-lines. If (ρ, ϕ, z) are cylindrical polars, we have z = r cos θ and ρ = r sin θ, and thusΨsp(r, θ) = Ψcp(r sin θ, r cos θ), defined in spherical polars and cylindrical polars, respec-tively.