CHAPTER 6 COMPLEX VARIABLE METHOD AND …huhui/teaching/2012FX/pdf-files/chapter6.pdfComplex Variable Method and Applications in Potential Flows For a certain type of flow, it is possible

CHAPTER 6 COMPLEX VARIABLE METHOD AND APPLICATIONS IN POTENTIAL FLOWS

- 126 -

Chapter 6

Complex Variable Method and Applications

in Potential Flows

For a certain type of flow, it is possible to introduce complex variables to aid in the solution of

the flow problem.

Complex variable can only used if the irrotational flow that is two-dimensional and expressible

in Cartesian coordinate system (x,y) or polar coordinate system ( θ,r ). i,e., the flow must satisfy

both 02 =∇ φ and 02 =∇ ψ . For no other coordinate system can complex variable be used.

6.1 Nomenclature and Algebra of Complex Variables

Extend the real number system by including 1−=i , then, we consider numbers of the form

yixZ += represent points on the x-y plane.

)sin(cos θθ ⋅+=+= iryixZ

22yxZ +=

)/(tan)( 1 xyZArg −==θ

)Re(Zx = Real part

)Im(Zy = Imaginary part

The complex number yixZ += can be also

represented as a vector in a plane by an ordered pair

of real numbers (x, y).

i.e., ),( yxZ =

θ

x 0

θ

x

iy

r

Real axis

Imaginary axis

Z=x+ y i


- 127 -

Complex conjugate:

Conjugate if Z=x+iy is defined as yixZ −=

ZZ −=

)()( ZArgZArg −=−

Z is the reflection of the point Z in the real axis.

2)Re(

ZZZ

+=

2)Im(

ZZZ

−=

The complex variable Z can also be expressed in polar coordinates in ( θ,r ).

Since θcosrx = , θsinry = , then

)sin(cos θθ ⋅+=+= iryixZ

Consider:

θθθθθ diid )cossin()sin(cos ⋅+−=⋅+

Divided by θθ sincos ⋅+ i at two side

Then:

θθ

θθθ

θθ

θθ

sincos

)cossin(

sincos

)sin(cos

⋅+

⋅+−=

⋅+

⋅+

i

di

i

id

⇒ θθ

θθθ

θθ

θθθ

θθ

θθ

sincos

)cossin(

sincos

)cossin)1((

sincos

)sin(cos 2

⋅+

⋅+⋅=

⋅+

⋅+⋅−=

⋅+

⋅+

i

dii

i

di

i

id

⇒ θθθ

θθθ

θθ

θθdi

i

dii

i

id=

⋅+

+⋅=

⋅+

⋅+

sincos

)cossin(

sincos

)sin(cos

Integrating both sides leads to: ∫∫ =⋅+

⋅+θ

θθ

θθdi

i

id

sincos

)sin(cos

⇒ θθθ ii =⋅+ ]sincos[ ln ⇒ θθθ iei =⋅+ sincos

Therefore,

θθθ ireiryixZ =⋅+=+= )sin(cos

θ

y

0

θ

x

y

Z

θ

(x,y)

-Z (x,-y)


- 128 -

Using the exponential form, we see that the product of two complex numbers 1

11

θierZ = and

2

22

θierZ = is given by

)(

212121 θθ +⋅=⋅ i

errZZ

212121 ZZrrZZ ⋅=⋅=⋅

2121 )( θθ +=⋅ ZZArg

If ZZZ == 12 , then ierZZZ θ222

21 ==⋅

If 1=r , ieZ

θ22 = . i.e., θθθθ 2sin2cos)sin(cos 2 ⋅+=⋅+ ii .

In a general form known as De Moivre’s theorem:

θθθθ nini n sincos)sin(cos ⋅+=⋅+

Similar relations can be written for any complex plane.

Example: for the ς - plane, φφφηξζ ieRiRi ⋅=⋅+=+= )sin(cos

)/tan(;22 ξηφηξζ =+== R


- 129 -

6.2 Analytic Complex Functions

If )(Zf=ξ is a complex function in the Z-plane ( yixZ += ), we define a derivative of a

complex function by

Z

ZfZZfZf

z ∆

−∆+=

→∆

)()(lim)('

0 for all possible Z∆ .

That is, since yixZ ∆+∆=∆ for the derivative should be unique, and the limit must exist

independent of how 0→∆Z .

A complex functions that are differentiable are called analytic functions.

For the points in the complex domain of Z-plane, where the )(Zf=ξ is analytical, are called

regular points of the function.

For the points in the complex domain Z, where the function )(Zf=ξ is not analytical, are

called singular points of the function.

The rules for real function of one variable calculus can be generally applied to analytic functions.

)()()()(

)]()([ ''ZgZf

dZ

Zdg

dZ

ZdfZgZf

dZ

d+=+=+

)()()()()]()([ ''ZgZfZgZfZgZf

dZ

d⋅+⋅=⋅

2

''

)(

)()()()(]

)(

)([

Zg

ZgZfZgZf

Zg

Zf

dZ

d ⋅−⋅=

For )(()( ZWZg ξ=

dZ

d

d

dW

dZ

Zdg ξ

ξ⋅=

)(

i.e., an analytical function of an analytic function is analytic.


- 130 -

6.3 Cauchy- Riemann Conditions

6.3.1 Cauchy- Riemann Conditions in Cartesian

Coordinate System

Consider:

),(),()( yxviyxuZf ⋅+=

If )(' Zf exists, then there are some conditions that u

and v must satisfy:

Z

ZfZZfZf

z ∆

−∆+=

→∆

)()(lim)('

0

Consider the path along x∆ , xZ ∆=∆ 1 and along y∆

given by yiZ ∆⋅=∆ 2 .

For the path 1

x

ZfxZf

Z

ZfZZf

Z

ZfZZfZf

xzz ∆

−∆+=

∆

−∆+=

∆

−∆+=

→∆→∆→∆

)()(lim

)()(lim

)()(lim)('

01

1

00 1

Therefore,

x

vi

x

u

x

yxviyxxvi

x

yxuyxxu

x

yxviyxuyxxviyxxuZf

xx

x

∂

∂⋅+

∂

∂=

∆

⋅−∆+⋅+

∆

−∆+=

∆

⋅−−∆+⋅+∆+=

→∆→∆

→∆

),(),(lim

),(),(lim

),(),(),(),(lim)('

00

0

For the path 2,

yi

ZfyiZf

Z

ZfZZf

Z

ZfZZfZf

yzz s ∆

−∆+=

∆

−∆+=

∆

−∆+=

→∆→∆→∆

)()(lim

)()(lim

)()(lim)('

02

2

00

θ

x

y

r

∆x

i∆y

f(Z)


- 131 -

y

v

y

ui

y

vi

y

u

i

yi

yxviyyxvi

yi

yxuyyxu

yi

yxviyxuyyxviyyxuZf

xx

x

∂

∂+

∂

∂−=

∂

∂⋅+

∂

∂=

∆

⋅−∆+⋅+

∆

−∆+=

∆

⋅−−∆+⋅+∆+=

→∆→∆

→∆

)(1

),(),(lim

),(),(lim

),(),(),(),(lim)('

00

0

Since )(' Zf should be the same in both paths, therefore,

x

vi

x

u

y

v

y

uiZf

∂

∂+

∂

∂=

∂

∂+

∂

∂−=)('

⇒

∂

∂−=

∂

∂

∂

∂=

∂

∂

y

u

x

v

y

v

x

u

Such conditions for an analytic function are called Cauchy- Riemann Conditions.


- 132 -

6.3.2 Cauchy- Riemann Conditions in Polar Coordinate

We consider a complex analytic function in polar coordinate system as :

),(),()( θθ rviruZf ⋅+=

If )(' Zf exists, then there are some conditions that u

and v must satisfy:

Z

ZfZZfZf

z ∆

−∆+=

→∆

)()(lim)('

0

Consider the path along r , rZ ∆=∆ 1 and along θ

given by θ∆=∆ riZ 2 .

For the path 1

r

ZfrZf

Z

ZfZZf

Z

ZfZZfZf

rzz ∆

−∆+=

∆

−∆+=

∆

−∆+=

→∆→∆→∆

)()(lim

)()(lim

)()(lim)('

01

1

00 1

Therefore,

r

vi

r

u

r

rvirrvi

r

ruyrru

r

rvirurrvirruZf

rx

r

∂

∂⋅+

∂

∂=

∆

⋅−∆+⋅+

∆

−∆+=

∆

⋅−−∆+⋅+∆+=

→∆→∆

→∆

),(),(lim

),(),(lim

),(),(),(),(lim)('

00

0

θθθ

θθθθ

For the path 2,

θ

θ

∆

−∆+=

∆

−∆+=

∆

−∆+=

→∆→∆→∆ ri

ZfriZf

Z

ZfZZf

Z

ZfZZfZf

rzz

)()(lim

)()(lim

)()(lim)('

01

1

00 1

Therefore,

θ

x

iy er

∆r

f(Z)

Z’ eθ

dθ


- 133 -

θθθθ

θ

θθθ

θ

θθθθ

θθθθθθ

∂

∂+

∂

∂−=

∂

∂⋅+

∂

∂=

∆⋅

⋅−∆+⋅+

∆⋅

−∆+=

∆⋅

⋅−−∆+⋅+∆+=

→∆→∆

→∆

r

vi

r

u

r

vi

r

u

i

ri

rvirvi

ri

ruru

ri

rvirurviruZf

rx

r

)(1

),(),(lim

),(),(lim

),(),(),(),(lim)('

00

0

)(' Zf should be the same in both paths, therefore,

θθ ∂

∂+

∂

∂−=

∂

∂+

∂

∂=

r

v

r

ui

r

vi

r

uZf )('

⇒

∂

∂−=

∂

∂

∂

∂=

∂

∂

θ

θ

r

u

r

v

r

v

r

u

Such conditions for an analytic function are called Cauchy- Riemann Conditions in polar

coordinate system.


- 134 -

6.3.3 Sufficient Conditions for a Complex Function

to be Analytic

Consider:

),(),()( yxiyxiZFF ηξηξ ⋅+=⋅+== and yixZ ⋅+=

dZ

dF

x

Z

Z

F

xi

xx

F=

∂

∂

∂

∂=

∂

∂+

∂

∂=

∂

∂ ηξ (1)

idZ

dF

y

Z

Z

F

yi

yy

F=

∂

∂

∂

∂=

∂

∂+

∂

∂=

∂

∂ ηξ ⇒

dy

di

dy

d

dZ

dF ξη−= (2)

From (1) and (2), we can get ⇒

∂

∂−=

∂

∂

∂

∂=

∂

∂

xy

yx

ηξ

ηξ

ξ and η are called conjugated function.

Cauchy –Riemann conditions provide the necessary condition for ηξ ⋅+== iZFW )( to be

analytic.

Continuity of the first order derivative terms such as yxyx ∂

∂

∂

∂

∂

∂

∂

∂ ηηξξ,,, provides the sufficient

conditions for ηξ ⋅+== iZFF )( to be analytic.

Thus, if F satisfies Cauchy –Riemann conditions and have continuous first order derivatives,

then, F is an analytic function and can be expressed as )(ZF .


- 135 -

6.3.4 Some Properties of Analytic Functions

For a complex function

viuZf ⋅+=)( where yixZ ⋅+= , ),( yxuu = , ),( yxvv =

According to Cauchy –Riemann conditions, if )(Zf is analytic then,

x

v

y

uand

y

v

x

u

∂

∂−=

∂

∂

∂

∂=

∂

∂

Conversely, if a function viuZf ⋅+=)( satisfies Cauchy –Riemann conditions, then, f can be

written as a function of Z only.

Some properties of analytic functions:

Let Z be a complex position variable yixZ ⋅+= , and ),(),(),( yxiyxyx ηξς ⋅+= a complex

function, if ),( yxς is analytic function,

Then:

(1). ),( yxς is a function of Z, i. e., )(Zςς =

(2). ς has a definite value for every value of Z. i.e., ς→Z is unique.

(3). ∞≠dz

dς

(4). dz

dς at every point in the prescribed domain is independent of the path used to reach the

point where the derivative is being evaluated.

(5). )(),(),(),( zyxiyxyx ςηξς =⋅+= , if it satisfies Cauchy –Riemann conditions:

xyyx ∂

∂−=

∂

∂

∂

∂=

∂

∂ ηξηξ;

Then

yxyxxxx ∂∂

∂=

∂

∂

∂

∂=

∂

∂

∂

∂=

∂

∂ ηηξξ 2

2

2

)()(


- 136 -

yxxyyyy ∂∂

∂−=

∂

∂−

∂

∂=

∂

∂

∂

∂=

∂

∂ ηηξξ 2

2

2

)()(

Therefore 022

2

2

2

2

=∂∂

∂−

∂∂

∂=

∂

∂+

∂

∂

yxyxyx

ηηξξ

Similarly 02

2

2

2

=∂

∂+

∂

∂

yx

ηη

i.e., both the real and imaginary parties of an analytic function satisfy Laplace equation. They

both are conjugate harmonic function.

(6). Consider two families of curves given by constyx =),(ξ and constyx =),(η , where

),( yxξ and ),( yxη are the real and imaginary parts of an analytic function. If we consider two

direction, one tangential to curve (alongξ ) and the other normal to it in some unknown direction,

then,

nnt en

en

et

ˆˆˆ∂

∂=

∂

∂+

∂

∂=∇

ξξξξ ⇐ Gradient of constyx =),(ξ curve is normal to the curve!

Since:

jy

ix

ˆˆ∂

∂+

∂

∂=∇

ξξξ

jy

ix

ˆˆ∂

∂+

∂

∂=∇

ηηη

0)ˆˆ()ˆˆ( =∂

∂

∂

∂−

∂

∂

∂

∂=

∂

∂

∂

∂+

∂

∂

∂

∂=

∂

∂+

∂

∂•

∂

∂+

∂

∂=∇•∇

xyyxyyxxj

yi

xj

yi

x

ξξξξηξηξηηξξηξ

Therefore constyx =),(ξ and constyx =),(η are orthogonal


- 137 -

6.4 Complex Potential and Complex Velocity

Complex potential

According to the continuity equation of ideal flow 0=•∇ Vr

For irrotational flows 0=×∇ Vr

For a 2-D incompressible flow, the stream function ψ determined by

xvand

yu

∂

∂−=

∂

∂=

ψψ;

If the 2-D incompressible flow is irrotational, potential function φ will exit:

∂

∂=

∂

∂=

⇒∇=

yv

xu

Vφ

φ

φr

Therefore:

∂

∂−=

∂

∂=

∂

∂=

∂

∂=

xyv

yxu

ψφ

ψφ

Whenever ψ and φ are related by the above relations, it is possible to form complex function as

a linear combination of ψ and φ , and call it, W, the complex potential:

ψφ iF += or ),(),(),( yxiyxyxF ψφ +=

However, since ψ and φ satisfy the Cauchy –Riemann conditions, therefore, we can represent

),( yxF as )(ZF , where yixZ ⋅+=

The above concept is valid for any two real variables ξ and η similar to ψ and φ .

Assume ),( yxξξ = , ),( yxηη = and ),(),(),( yxiyxyx ηξς +=

And if ξ and η satisfy Cauchy –Riemann conditions, then ),( yxς can be represented as )(Zς ,

where yixZ ⋅+= .


- 138 -

Complex velocity

If ψφ iF +=

1). For the path1, find dZ

dF along x∆ therefore,

dx

dF

dZ

dF=

Along path 1 ⇒ ivux

ixdx

dF

dZ

dF⋅−=

∂

∂+

∂

∂==

ψφ

2). Path2, find dZ

dF along yi∆ therefore,

idy

dF

dZ

dF=

Along path 2 ⇒ uiviuvi

iiuv

iyii

yiidy

dF

dZ

dF+−=⋅+

−=⋅+=

∂

∂+

∂

∂== )()(

1 2ψφ

VvudZ

dF=+= 2/122 ][

i.e., if )(ZW is complex potential, ivudZ

dF−= is called complex velocity.

θθ

θ

θ

θ

θ

θ

θθ

θθ

θθ

θθ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

i

r

r

r

r

r

r

rr

rr

rr

eViV

iViV

iViiV

iiViiV

iViiV

iViV

ViVViV

ViVViV

VViVV

ivudZ

dF

−⋅−=

⋅−⋅−=

⋅−⋅−⋅−=

⋅+⋅⋅+⋅−=

⋅++⋅−=

⋅+−⋅−=

⋅+−⋅−=

⋅−−⋅−=

+⋅−−=

−=

)(

]sin)[cos(

]sin[cos]sin[cos

]cossin[]sin[cos

]cos[sin)(]sin[cos

]cos[sin]sin[cos

]cossin[]sincos[

cossinsincos

)cossin(sincos

2

2

i.e. complex velocity θθ

i

r eViVdZ

dF −⋅−= )( in polar coordinate system.

θ

x

iy

r u

Vr Vθ

v


- 139 -

6.5 Uniform Flows

6.5.1 Uniform Flow to the Right

For a 2-D incompressible flow with the velocity

distribution is given by:

0; == ∞ vVu

Is the flow physically possible?

The continuity of incompressible flow is 0=•∇ Vr

000 =+=∂

∂+

∂

∂=•∇

y

v

x

uVr

Therefore, the flow is physically possible.

The stream function ψ can be determined by

xv

yu

∂

∂−=

∂

∂=

ψψ;

Constfdx

df

xxv

xfyVy

Vy

u

=⇒+=⇒∂

∂=⇒

∂

∂−=

+=⇒∂

∂=⇒

∂

∂= ∞∞

000

)(

ψψ

ψψψ

Therefore yV∞=ψ

Is the flow is irrotational?

0

00

ˆˆˆ

=∂

∂

∂

∂

∂

∂=×∇

∞V

zyx

kji

Vr

Therefore the flow is irrotational.

Since the flow is irrotational, potential function φ will exit:

Cosntfdy

ydf

yyv

yfxVx

Vx

u

=⇒+=⇒∂

∂=⇒

∂

∂=

+=⇒∂

∂=⇒

∂

∂= ∞∞

)(000

)(

φφ

φφφ

Therefore, xV∞=φ

ZVyixVyVixViW ∞∞∞∞ =⋅+=⋅+=+= )(ψφ

In Cartesian system ZVW ∞=

In polar system θierVW ∞=

x

V∞ y


- 140 -

6.5.2 Uniform flow at an Angle of Attack of α

Since yixZ ⋅+=

We define '' yixZ ⋅+=

The complex function will be

Since

+−=

+=⇒

+=

−=

αα

αα

αα

αα

cossin'

sincos'

cos'sin'

sin'cos'

yxy

yxx

yxy

yxx

Then the potential complex will be

α

αα

αααα

αααα

αααα

αααα

iZeV

iyixV

iyiixV

iiyixV

iyixV

yxiyxV

yixV

ZVF

−∞

∞

∞

∞

∞

∞

∞

∞

=

−+=

+−⋅+−=

+−+−=

++−=

+−++=

⋅+=

=

)sin)(cos(

)]cossin()sin(cos[

)]cossin()sin(cos[

)]cos(sin)sin(cos[

)]cossin()sincos[(

)''(

'

2

Therefore

In Cartesian coordinate system αiZeVF −∞=

In polar coordinate system )( αθ −∞= ierVF

X’

V∞

Y’

X

Y

α θ


- 141 -

6.6 2-D Source and Sink

6.6.1 2-D Source

Definition: A 2-D source is defined as infinity line from

which flow issues along radial lines.

To calculate the volume flow rate from the source

r

qVrVq rr

ππ

22 =⇒=

Then the velocity field will be:

0

2

=

=

θ

π

V

r

qVr

Is the flow physically possible?

The continuity of ideal flow is 0=•∇ Vr

0000])()(

[1

=++=∂

∂+

∂

∂+

∂

∂=•∇

Z

rVV

r

rV

rV zr

θθ

r


To determine stream functionψ .

constrfdr

rdf

rV

rfq

r

q

rVr

=⇒+=⇒=∂

∂−=

+=⇒=∂

∂=

)()(

000

)(22

ψ

θπ

ψπθ

ψ

θ

Thus constq

+= θπ

ψ2

Let us assume 0=ψ when 0=θ ⇒ 0=const

Therefore θπ

ψ2

q=

For a sink flow θπ

ψ2

q−=

Y

θ X

Z


- 142 -

Is the flow irrotational?

0

002

ˆˆˆ

1=

∂

∂

∂

∂

∂

∂=×∇

r

qzr

eere

rV

zr

π

θ

θr

Therefore the flow is irrotational.


To determine potential functionφ

Cosntfd

df

rV

frq

r

q

rVr

=⇒+=⇒=∂

∂=

+=⇒=∂

∂=

)()(

0001

)(ln22

θθ

θ

θ

φ

θπ

φπ

φ

θ

Thus, constrq

+= ln2π

φ

Let us assume 0=φ when ar = , ⇒ aq

const ln2π

=

Therefore, a

rqln

2πφ =

For a sink flow a

rqln

2πφ −=

Therefore, for a source flow, the complex potential will be:

a

Zqa

qir

qqi

a

rqiF ln

2ln

2)(ln

22ln

2 ππθ

πθ

ππψφ =−+=⋅+=+=

i.e., a

ZqF ln

2π=

For a sink flow: a

ZqF ln

2π−=

To determine the velocity field from the potential complex

For a source flow


- 143 -

Z

q

a

a

Z

q

dZ

a

Zd

q

dZ

dF

πππ 2

11

2

)(ln

2===

To determine the potential complex function by integrating the velocity field

In polar coordinate system:

Z

qe

r

qei

r

qeiVV

dZ

dF iii

rπππ

θθθθ

22)0.

2()( ==−=−= −−−

Integrating the above equation, we can have

constZq

F += ln2π

We assume constrq

F += ln2

)Re(π

when ar = , therefore,

aq

constconstaq

ln2

0ln2 ππ

−=⇒=+

Thus a

ZqF ln

2π=


- 144 -

6.6.2 A Source at (-b,0) Plus a Sink at (b,0)

Without constant term, the expression of a source flow will be

Zq

F ln2π

=

For a source situated at a position

0ZZ = in the complex plane, the

expression will be

)ln(2

0ZZq

F −=π

For a source at (-b,0) and a sink at

(b,0)

)ln(2

bZq

Fsource +=π

)ln(2

sin bZq

F k −−=π

)]ln()[ln(2

ln2

)ln(2

)ln(2

21

21

sin

θθ

π

π

ππ

ii

ksourcecombined

ererq

bZ

bZq

bZq

bZq

FFF

−=

−

+=

−−+=

+=

To determine the geometry of the streamlines

For a source +sink combined flow, the ψ and φ are given by the expression of

112

θπ

ψq

= and 222

θπ

ψq

−=

)(2

2121 θθπ

ψψψ −=+=q

combined

][tan 1

1bx

y

+= −θ or

bx

y

+=1tanθ

θ

P

θ1 θ2

X

y


- 145 -

][tan 1

2bx

y

−= −θ or

bx

y

−=2tanθ

Then

222

22

222

22

21

2121

2

)(

1tantan1

tantan)tan(

ybx

yb

bx

ybx

bx

bxbxy

bx

y

bx

ybx

y

bx

y

+−

−=

−

+−−

−−−

=

+⋅

−+

−−

+=+

−=−

θθ

θθθθ

Therefore, ]2

[tan222

1

21ybx

yb

+−

−=− −θθ

Then ]2

[tan2

)(2 222

1

21ybx

ybqqcombined

+−

−=−= −

πθθ

πψ

If

)1

1(][

)()(2

02

02

2tan

]2

[tan2

]2

[tan2

2

222

22222

22

222

222

222

1

222

1

cb

c

byx

c

bb

c

by

c

byx

c

ybbx

c

ybybx

ybx

ybkc

ybx

yb

qk

kybx

ybq

+=++⇒

+=+++⇒

=+−⇒

=++−⇒

+−

−==⇒

+−

−==⇒

=+−

−=

−

−

πψ

πψ

There are circles with center at ),0(c

b− and radius of )

11(

2c

bR += , i.e., the circles on Y – axis.

To determine the geometry of the iso-potential lines

Equipotent lines:

Since 2

12121 ln

2ln

2ln

2 r

rqr

qr

q

πππφφφ =−=+=

Since 22

1 )( ybxr ++= and 22

2 )( ybxr +−=

Therefore

22

222/1

22

22

2

1

)(

)(ln

4]

)(

)(ln[

2ln

2 ybx

ybxq

ybx

ybxq

r

rq

+−

++=

+−

++==

πππφ


- 146 -

For const=φ lines

2

2

2

22222

2

2

2222

22222

222

222

2222

22

224

)1(

4]

)1(

)1()1([]

)1(

)1([

]1)1(

)1([]

)1(

)1([

)1(

)1(2

])1(

)1([]

)1(

)1([

)1(

)1(2

0)1()1(2)1()1(

0)1()()(

])[()(

)(

)(

k

kb

k

kkby

k

kbx

k

kb

k

kby

k

kxbx

bk

kb

k

kby

k

kxbx

ykkxbbkkx

ykbxkbx

ybxkybx

kybx

ybxe

q

−=

−

−−+=+

−

++⇒

−−

+=

−

+++

−

++⇒

−−

+=

−

+++

−

++⇒

=−+++−+−⇒

=−+−−+⇒

+−=++⇒

=+−

++=⇒

πφ

2

222

)1(

4]

)1(

)1([

k

kby

k

kbx

−=+

−

++

Therefore, the const=φ lines are circles with center at )0,)1(

)1((

k

kb

−

+− and radius of

2

2

)1(

4

k

kbR

−= , i.e., the circles on X – axis [except k=1].


- 147 -

6.7 2-D Doublet

Definition: A double is obtained when a source and sink of equal strength approach each other so

that the produce of their strength and the distance apart remains constant.

i.e., µ==⋅ constbq 2

b

bZ

bZ

b

bZ

bZqb

bZq

bZq

FFF ksourcecombined

−

+

=

−

+⋅

=

−−+=+=

ln

4

ln

4

2

)ln(2

)ln(2

sin

π

µ

π

ππ

Using L’Hopital’s roles

When

0→b

ZbZ

Z

bZ

Z

bZ

bZ

db

dbdb

bZ

bZd

b

bZ

bZ

Fbbbb π

µ

π

µ

π

µ

π

µ

π

µ

2

2

4lim]

)(

2

4lim

)(ln

4lim

ln

4lim

2202000=

−=

−+

−=

−

+

=−

+

=→→→→

Therefore, for a 2-D doublet

)sin(cos222

θθπ

µ

π

µ

π

µθ i

re

rZF i

combined −=== −

i.e.,

θπ

µψ

θπ

µφ

sin2

cos2

r

r

−=

=


- 148 -

6.8 2-D Vortex

Definition: A 2-D vortex is a mathematical concept that induced a velocity field given by:

0;;0 ≠== rr

CVVr θ

Is the flow field physical possible?

The continuity of ideal flow is 0=•∇ Vr

0000])()(

[1

=++=∂

∂+

∂

∂+

∂

∂=•∇

Z

rVV

r

rV

rV zr

θθ

r


To determine the stream functionψ .

Constfd

df

rV

frCr

C

rV

r =⇒+=⇒=∂

∂=

+−=⇒=∂

∂−=

)()(

000

)(ln

θθ

θ

θ

ψ

θψψ

θ

Thus constrC +−= lnψ

When assume 0=ψ when ar = , then constaC +−= ln0

Therefore, the stream function will be a

rCoraCrC lnlnln −=+−= ψψ

Is the flow irrotational?

0

00

ˆˆˆ

1=

∂

∂

∂

∂

∂

∂=×∇

r

Cr

zr

eere

rV

zr

θ

θr

Therefore, the flow is irrotational.



- 149 -

Cosntrfdr

rdf

rV

rfCr

C

rV

r =⇒+=⇒=∂

∂=

+=⇒=∂

∂=

)()(

000

)(1

φ

θφθ

φθ

Thus, constC += θφ

Let us assume 0=φ when 0=θ , ⇒ 0=const

Then θφ C=

If Γ is defined as the circulation of a circle

(counter-clockwise) surrounding the 2-D vortex

Then

Crr

CrdeeVldV

cc

ππθθθθ 22ˆˆ ==•=•=Γ ∫∫rr

⇒ π2

Γ=C

[Is the Stokes theorem still applicable? Think about it!]

Therefore: a

rln

2πψ

Γ−= ; θ

πφ

2

Γ=

For a 2-D vortex flow:

a

Ziaireiaiiri

airiiairia

riiF

i ln2

ln2

)ln(2

ln2

)(ln2

}ln]ln){[(2

]ln)ln[(2

ln22

2

ππππθ

π

θπ

θππ

θπ

ψφ

θ Γ−=

Γ+

Γ−=

Γ++

Γ−=

+−−Γ

=+−Γ

=Γ

⋅−Γ

=+=

i.e., a

ZiF ln

2π

Γ−=

The singularity of this expression is located at Z=0.

The potential complex for a positive vortex located at Z=Z0 will be

a

ZZiF

)(ln

2

0−Γ−=

π

If the vortex is rotate in clockwise, then the potential complex will be:

a

ZZiF

)(ln

2

0−Γ=

π

X

Y


- 150 -

6.9 2-D Flow around a Circular Cylinder

Flow around a circular cylinder can be decomposed as a uniform flow plus a doublet flow.

For a uniform flow to the right ZVF ∞=1

For a double flow sit at Z=0 Z

Fπ

µ

22 =

The potential complex of the combined flow will be

)2

1(sin)2

1(cos

)2

(sin)2

(cos

)sin(cos2

)sin(cos

2)sin(cos

2

22

22

21

rVriV

rVrV

rVri

rVr

ir

irV

er

iVZ

ZVFFF i

∞

∞

∞

∞

∞∞

∞

−∞∞

−++=

−⋅++=

−++=

++=+=+=

π

µθ

π

µθ

π

µθ

π

µθ

θθπ

µθθ

π

µθθ

π

µθ

Therefore:

)2

1(sin

)2

1(cos

rVrV

rVrV

∞

∞

∞

∞

−=

+=

π

µθψ

π

µθφ

Then velocity field will be

)2

1(sin

)2

1(cos)2(2

cos)2

1(cos232

rVV

rV

rVV

rVrV

rVV

rVr

∞

∞

∞

∞

∞

∞

∞

∞

+−=∂

∂=

−=−++=∂

∂=

π

µθ

θ

φ

π

µθ

π

µθ

π

µθ

φ

θ

To determine the stagnation points:

At stagnation point 00 == rr VandV

πθθθπ

µθθ ==⇒=⇒=+−=

∞

∞ orrV

VV 00sin0)2

1(sin

When 0=θ

∞∞

∞

∞

∞ =⇒=−=−=V

rrV

VrV

VVrπ

µ

π

µ

π

µθ

20)

21()

21(cos

22

When πθ =


- 151 -

∞∞

∞

∞

∞ =⇒=−−=−=V

rrV

VrV

VVrπ

µ

π

µ

π

µθ

20)

21()

21(cos

22

If we define ∞

=V

aπ

µ

2

There are two stagnation points with coordinates of ),( θr being )0,(a and ),( πa

To determine the geometry of the streamline passing the stagnation points:

The streamline passing the stagnation point will be

0)1(sin2

2

=−= ∞r

arVstagn θψ

Then the equation for the solid body will be

=⇒=

−⇒=⇒=−∞

aR radius with cirleA

0sin0)1(sin

2

2

ar

axisX

r

arV

θθ

Therefore, the flow field is to the flow around a cylinder.

To summarize the flow around a circular cylinder:

)(2

Z

aZVW += ∞

)1(sin

)1(cos

2

2

2

2

r

arV

r

arV

−=

+=

∞

∞

θψ

θφ

)1(sin

)1(cos

2

2

2

2

r

aVV

r

aVVr

+−=

−=

∞

∞

θ

θ

θ

To evaluation of the flow velocity field from the potential complex

Since )(2

Z

aZVF += ∞

θθ

θ

θ

θθθθ

θ

θθ

θθθθ

i

r

i

i

iiii

i

eViV

er

ai

r

aV

eir

aiV

eer

aeVe

r

aV

re

aV

Z

aV

dZ

dF

−

−∞

−∞

−−∞

−∞∞∞

⋅−=

++−=

−−+=

−=−=−=−=

)(

)]1(sincos)1[(

)]sin(cossin[cos

][])(1[))(

1()1(

2

2

2

2

2

2

2

22

2

2

2

2

2

2


- 152 -

Therefore

+−=

−=

∞

∞

)1(sin

)1(cos

2

2

2

2

r

aVV

r

aVVr

θ

θ

θ

Pressure coefficient distribution on the surface of the circular cylinder

Since 2

2

1∞

∞−=

V

PPC p

ρ

At the surface of the cylinder, ar = ⇒

−=

=

∞ θθ sin2

0

VV

Vr

According to Bernoulli’s equation

2

2

2

22 1

2

12

1

2

1

∞∞

∞∞∞ −=

−⇒+=+

V

V

V

PPVPVP

ρρρ

Pressure coefficient distribution on the surface of the circular cylinder will be:

θθ

ρ

2

2

2

2

2

2

sin41)sin2(

11

2

1−=

−−=−=

−=

∞

∞

∞∞

∞

V

V

V

V

V

PPC p

-3

-2

-1

0

1

0 100 200 300 400

Angle, Deg.

Cp


- 153 -

6.10 Flow around a Spinning Cylinder

Flow around a spinning cylinder can be decomposed as the combination of a uniform flow, a 2-D

doublet and a vortex flow.

The potential complex of a uniform flow ZVF ∞=1

A double flow sit at Z=0 Z

Fπ

µ

22 =

A 2-D vortex in clockwise a

ZiF ln

23

π

Γ=

If we make ∞

=V

aπ

µ

2

Then, the potential complex of the combined flow will be:

a

Zi

Z

aZV

a

Zi

ZZVFFFF

ln2

)(

ln22

2

321

π

ππ

µ

Γ++=

Γ++=++=

∞

∞

When we set 0=Γ , then the potential complex will be for a non-spinning cylinder.

θθ

θ

θθθθ

θθ

θ

θθ

θθ

πθθ

πθθθθ

π

ππ

ππ

i

r

i

iiii

ii

i

ii

ii

eiVV

err

aVi

r

aV

er

iir

aiVe

rie

r

aeV

er

ieer

aeeV

rei

re

aV

aZ

ai

Z

aV

a

Zi

Z

aZV

dZ

dF

−

−∞∞

−∞

−−∞

−−−∞∞

∞∞

−=

Γ+++−=

Γ+−−+=

Γ+−=

Γ+−⋅=

Γ+−=

Γ+−=

Γ++=

}(

]}2

)1(sin[)1(cos{

}2

)]sin(cossin[cos{]2

)([

2)(

1

2)

)(1(

1

2)1(ln

2)(

2

2

2

2

2

2

2

2

2

2

2

2

2

22

Then:

)1(cos2

2

r

aVVr −= ∞ θ

rr

aVV

πθθ

2)1(sin

2

2 Γ−+−= ∞


- 154 -

To determine the stagnation points:

Since 0=θV and 0=rV at the stagnation point

−=

=

⇒=−= ∞

22

0)1(cos2

2

ππθ

θ

or

or

ar

r

aVVr

CASE 1. If ar =

aV

rr

aVV

∞

∞

Γ−=⇒

=Γ

−+−=

πθ

πθθ

4sin

02

)1(sin2

2

If Γ =0, we get 0=θ or π . This is the case for

non-spinning cylinder we studied before.

Since 04

>Γ

∞ aVπ, then, sθ has to be in the 3

rd

and 4th

quadrant.

Since 1sin ≤θ

Then aVaV

∞

∞

≤Γ⇒≤Γ

ππ

414

i.e., the spinning velocity of the cylinder can

not be too high in order to find stagnation point

in the surface of the cylinder.

When the stagnation points are on the surface of the cylinder ar =

∞

Γ−==

VaYs

πθ

4sin

22

ss YaX −±=

The solution is valid only when aVoraV

∞

∞

≤Γ≤Γ

ππ

414

. When aV∞>Γ π4 , the

stagnation point will not be on the surface of the spining cylinder any more.

X

Y

r=a


- 155 -

CASE 2 If 2

πθ =

02

)1(2

2

<Γ

−+−= ∞rr

aVV

πθ Therefore 0=rV is impossible

CASE 3 If 2

πθ = ,

22

22

22

2

2

2

2

)4

(4

02

02

2)1(

02

)1(

aVV

r

rV

ar

rV

ar

Vr

ar

rr

aVV

−Γ

±Γ

=⇒

=Γ

−+⇒

=Γ

−+⇒

Γ=+⇒

=Γ

−++=

∞∞

∞

∞

∞

∞

ππ

π

π

π

πθ

In order to make the solution physically possible, it has

to be aV

>Γ

∞π4 (i.e. ∞>Γ Va π4 ).

X

Y


- 156 -

6.11 Cauchy Integral Theorem

Consider the complex function ),(),(),( yxiyxyx ηξς += not necessarily analytic defined in a

certain region in the Z-plane.

The line integral of ),( yxς from Z0 to Z1 along line C is

∫1

0

),(

Z

Z

dZyxς .

The value of this integral in general dependent on the

path C and the end points Z0 and Z1.

Since idydxdZ +=

Then

∫∫

∫∫

++−=

++=

),(

),(

),(

),(

),(

),(

11

00

11

00

11

00

1

0

}),(),({}),(),({

))}(,(),({),(

yx

yx

yx

yx

yx

yx

Z

Z

dxyxdyyxidyyxdxyx

idydxyxiyxdZyx

ηξηξ

ηξς

We are interested only in complex integrals whose values depend only on the end points and not

on the path that joints them.

For this to be true, the integrand of the integral on the right hand side of the above equation must

be an exact differential.

∫∫ ++−),(

),(

),(

),(

11

00

11

00

}),(),({}),(),({

yx

yx

yx

yx

dxyxdyyxidyyxdxyx ηξηξ

Suppose the above expression can be written as

∫∫ +),(

),(

),(

),(

11

00

11

00

),(),(

yx

yx

yx

yx

yxdiyxd λβ

By comparison, we can write this as:

dyy

dxx

dyyxdxyxyxd∂

∂+

∂

∂=−=

ββηξβ ),(),(),(

dyy

dxx

dyyxdxyxyxd∂

∂+

∂

∂=+=

λλξηλ ),(),(),(

Z0

Z1

X

Y


- 157 -

i.e.,

yxyx

∂

∂=

∂

∂=

λβξ ),(

yxyx

∂

∂−=

∂

∂=

βλη ),(

By differentiating and equating, we get:

yx ∂

∂=

∂

∂ ηξ and

xy ∂

∂−=

∂

∂ ηξ which are the Cauchy- Riemann conditions.

In other words, the complex function ),( yxς should be analytical for the complex integral to be

independent of the path.

Cauchy integral theorem

If ),( yxς is analytic in a simply connects region, then ),( yxς will be the function of Z only (i.e.,

)(Zςς = ), and the integral ∫1

0

),(

Z

Z

dZyxς is a function of the end points only.

i.e., )()()(),( 01

1

0

1

0

ZFZFZdFdZyx

Z

Z

Z

Z

−== ∫∫ς

Then for a closed curve C, 0),( =∫ dZyxς .

The statement is true when there are no points or enclosed by curve C where ∞=dZ

dς

To consider the case when a singular point is enclosed in the enclosed curve C

The positive sense of integration is that which the region enclosed by curve C is to the left.

Consider a region where )(Zς is analytic at every point except Z0 where ∞=0zdZ

dς, i.e., Z0 is

the singular point.

For any closed curve C not enclosing Z0, 0),( =∫ dZyxς .

If the closed curve C encloses the singular point Z0, we enclosed this point in a closed curve C1

and focus our attention on the analytic region between C and C1.


- 158 -

We render this region into a closed region bounded by a continuous curve by introducing the cut

ab, then we apply the Cauchy integral theorem by performing the line integral along the

boundary in a positively sense always keeping the enclose region to the left.

0)()()()()()(

'

'

'

''

''

''

''

''

=++++= ∫∫∫∫∫∫a

b

b

b

b

a

a

M

M

a

dZZdZZdZZdZZdZZdZZ ςςςςςς

In a limiting sense: ''' aa → and ''' bb → then

0)()(

'

'

''

''

=+ ∫∫a

b

b

a

dZZdZZ ςς

Also

∫∫∫ =+C

a

M

M

a

dZZdZZdZZ )()()(

''

''

ςςς and

∫∫ −=1

)()(

''

'' C

b

a

dZZdZZ ςς

Therefore ∫∫ =1

)()(CC

dZZdZZ ςς

i.e., all the line integrals over all closed circuits

enclosing the singular point are equal.

If curve C encloses n singular points,

Then

∫∫∫∫ +++=nCCCC

dZZdZZdZZdZZ )()()()(

21

ςςςς K

Singular point at Z0

a’

a’’

b’

b’’

M

Curve C

Curve C 111

Singular point at Z3

a’

a’’

b’

b’’

M

Curve C

Curve C3

Z1

C1

Z2

C2


- 159 -

Integral of a complex function of n

ZZZ )()( 0−=ς

Consider a complex function

L2,1,0,)()( 0 ±±=−= nZZZnς

1)()( −−= n

ZZndZ

ZDς

for n<0

n

n

ZZnZZn

dZ

ZD−

−

−=−=

1

1

)(

1)(

)(ς

Therefore, 0ZZ = is a singular point.

When n>0 , 1)()( −−= n

ZZndZ

ZDς, 0ZZ = is a regular point.

If we consider a closed curve in the form of a circle, i.e., θieRZZ =− 0

1]1[1

)1(

1])1([

1

)()()(

2)1(1

2

0

1)1(

1)1(1

00

−≠−+

=

+

+=+

+==

=−=−

++

++

+++

∫∫

∫∫∫

nforen

R

ni

n

Rnide

n

RdieR

dieReRdZZZdZZZ

nin

n

RCircle

nin

RCircle

nin

RCircle

ini

RCircle

n

C

n

e

π

πθθ

θθ

θθθ

θ

For 1−≠n , 0]1]2)1sin[(]2)1[cos[(1

]1[1

)(1

2)1(1

0 =−++++

=−+

=−+

++

∫ πππnin

n

Re

n

RdZZZ

nni

n

C

n

When 1−=n iideReRdZZZ

dZZZC

ii

CC

n πθθθ 2)(1

)( 1

0

0 ==−

=− ∫∫∫−

As described before, for any closed curve C enclosing the same singular point, the line integral

will be the same. Therefore, for any closed curve C idZZZC

π2)( 1

0 =−∫−

In summary:

Given L2,1,0,)()( 0 ±±=−= nZZZnς

−=

−≠=−∫

0

0

0 Zenclosing is C Curve and12

Zenclosingnot is C Curveor 10)(

nwheni

nwhendZZZ

C

n

π

θ

x

y

Z0


- 160 -

6.12 Blasius Integral Law

We can calculate force acting on a solid body laced in fluid flow in by evaluating velocity

component on the surface of the body, and then integrating pressure by using the Bernoulli

equation. Therefore, the difficulty to calculate aerodynamic forces is to conduct the body shape

integration.

Blasius integral law states that if the complex potential is known for a flow around a body, then

it is possible to evaluate the forces and the turning moment acting on the body by means of

simple contour integrals.

Force on a body:

Let )(ZF be the complex potential describing the flow

about a 2-D body whose boundary in the Z-plane is a

closed curve C. Then, the component of the net force on

the body are obtained from

∫=−C

YX dZdZ

dFiFiF

2)(2

''ρrr

Where )'( denotes per unit normal to 2-D plane.

β

β

cos'

sin'

'

⋅=

⋅−=

=

dsPFd

dsPFd

dsPFd

y

xr

r

r

β

ββ

ββ

ββ

i

yx

eiPds

iiPds

dsiPdsPi

dsiPdsP

FiFFd

=

+=

+=

+−=

+=

)cossin(

cossin)(

cossin

'''

2

rrr

Since βββ iedsidsdsidydxdZiyxZ =⋅+⋅=+=⇒+= sincos

Therefore dZiPdF ='

From Bernoulli’s equation, constVP =+ 2

2

1ρ ⇒ 2

2

1VconstP ρ−=

Then dZVconstiFd )2

1(' 2ρ−=

r

Z1

X

Y

β

ds


- 161 -

Integrating 'Fdr

around the body the closed curve C

∫∫∫∫∫ ⋅⋅−=⋅⋅−⋅⋅=−==CCCCC

dZVidZVidZconstidZVconstidFF222

22)

2

1(''

ρρρ

r

Therefore

])sincos([2

])sin(cos[222

'''

2

222

∫

∫∫∫

−⋅⋅−=

+⋅⋅−=⋅⋅−=⋅⋅−=+=

C

CC

i

C

yx

dsiV

dsiVidseVidZViFiFF

ββρ

ββρρρ β

rrr

Then the conjugate of 'Fr

is given by

])([2

][2

])sin(cos[2

])sin)(cos([2

])sincos([2

'''

2222

222

dseeVidseVidsiVi

dsiiVdsiVFiFF

i

C

i

C

i

C

CC

yx

βββ ρρββ

ρ

ββρ

ββρ

∫∫∫

∫∫

−− ⋅=⋅=−⋅=

−⋅=−−⋅⋅−=−=rrr

On the body- a streamline – the velocity is given by

ivudZ

dFZW −==)( where βcosVu = ; βsinVv =

Then βββ ieViVV

dZ

dF −=−= sincos

Therefore

∫∫∫∫ ⋅=⋅=⋅⋅=⋅⋅=−= −−

CC

i

C

ii

C

ii

yx dZdZ

dFidse

dZ

dFidseeVidseeViFiFF

22222 )(2

)(2

)(2

)(2

'''ρρρρ βββββ

rrr

Moment: Clockwise moments are positive (stalling)

])(2

Re[ 2

0 ∫ ⋅=C

dZZdZ

dWM

ρ

Assume positive forces, then

β

β

cos'

sin'

''0

⋅=

⋅−=

−=

dsPdF

dsPdF

xdFydFdM

y

x

yx


- 162 -

)()2

1(

)(cossin''

2

0

dxxdyyVconst

dxxdyyPxdsPydsPxdFydFdM yx

+−−=

+−=⋅−⋅−=−=

ρ

ββ

])(2

Re[

])(2

1Re[]

2

1Re[]

2

1Re[

])sin)(cos(2

1Re[

)cossin(2

10

)(2

1)((

2

)()2

1)()()

2

1(

2

2222

2

2

222

22

00

ZdZdz

dF

dsZeeVdsZeeVdsZeV

dsiiyxV

dsxdsyV

dxxdyyVyxdconst

dxxdyyVdxxdyyconstdxxdyyVconstdMM

C

i

C

ii

C

i

C

i

C

C

CC

CCCC

∫

∫∫∫

∫

∫

∫∫

∫∫∫∫

=

===

−+=

++=

+++−=

−++−=+−−==

−−−

ρ

ρρρ

ββρ

ββρ

ρ

ρρ

βββββ

Force and Moment on a spinning circular cylinder

For a spinning cylinder, the potential complex function is a

Zi

Z

aZVF ln

2)(

2

π

Γ++= ∞

Therefore,

Zi

Z

aV

dZ

dF 1

2)1(

2

2

π

Γ+−= ∞

Since ∫⋅=−=C

yx dZdZ

dFiiFFF

2)(2

'''ρ

ZZ

aiV

ZZ

aV

Zi

Z

aV

dZ

dF 1

2)1(2)

1

2()1(]

1

2)1([)(

2

222

2

222

2

22

πππ

Γ−+

Γ−−=

Γ+−= ∞∞∞

∞∞

∞∞∞∞

∞∞

∞∞

∞∞

Γ−=Γ

−=

Γ+

Γ−=

Γ+

Γ−=

Γ−

Γ⋅=

Γ−⋅++=

Γ−⋅+

Γ−⋅+−⋅=

Γ−+

Γ−−⋅=−=

∫∫∫∫

∫∫

∫∫∫

∫

ViiV

dZZ

Va

idrere

VdZZ

Va

dZZ

V

dZZ

ia

Z

iVidZ

ZZ

aiVi

dZZZ

aiVidZ

ZidZ

Z

aVi

dZZZ

aiV

ZZ

aViFiFF

CC

i

i

CC

CC

CCC

C

yx

ρππ

ρ

π

ρθ

π

ρ

π

ρ

π

ρ

ππ

ρ

π

ρ

π

ρ

π

ρρ

ππ

ρ

θ

θ

2*2

1

2

1

2

1

2

1

2

]2

2

2

2[

2]

1

2)1(2

200

]1

2)1(2

2)

1

2(

2)1(

2

]1

2)1(2)

1

2()1([

2'''

3

2

3

2

3

2

2

2

2

222

2

22

2

222

2

22

rrr

Therefore:


- 163 -

∞Γ=

=

VF

F

y

x

ρ'

0'r

r

0}2)4

2(2

1Re{})]

1

42[

2

1Re{

}2

)1(22

1Re{})

41([

2

1Re{)]

1

42[

2

1Re{

}2

)1(22

1Re{}

1)

2(

2

1Re{})

421([

2

1Re{

}1

2)1(2

2

1Re{})

1

2(

2

1Re{}])1([

2

1Re{

}]1

2)1(2)

1

2()1([

2

1Re{)(

2

1Re[

])(2

1Re[

2

2

22

2

222

2

2

3

2

2

222

2

22

3

22

2

222

2

22

2

222

2

222

2

0

=Γ

−−=Γ

−−=

Γ−+++

Γ−−=

Γ−+

Γ−++−=

Γ−+

Γ−+−=

Γ−+

Γ−−==

=

∞∞

∞∞∞

∞∞

∞∞

∞∞

∫

∫∫∫

∫∫∫

∫∫∫

∫ ∫

∫

iVadZZZ

Va

dZZ

aiVdZ

Z

aVdZ

ZZ

Va

dZZ

aiVdZ

ZdZ

Z

a

Z

aV

ZdZZZ

aiVZdZ

ZZdZ

Z

aV

ZdZZZ

aiV

ZZ

aV

dZ

dF

ZdZdz

dFM

C

CCC

CCC

CCC

C C

C

ππ

ρπ

ρ

πρρ

πρ

πρ

πρρ

πρ

πρρ

ππρρ

ρ

Therefore: 00 =M


- 164 -

6.13 Conformal Transformation

Fore the curves 1Z and

2Z in the Z-plane transform to curve 1ς and

2ς in the ς -plane,

conformal transformation is defined as such that: Two line elements ab and cd in the Z-plane

intersecting at angle β provides 2 line segments AB and CD in the ς -plane intersecting at the

same angle β with the same sense and having the same ratio of length as the original elements:

(a). angle β in Z-plane = β

in ς -plane

(b). CD

AB

cd

ab=

For certain point of the

transformation:

Given the analytic

mapping )(Zς , points in the original complex Z-plane where 0≠dZ

dς or ∞≠

dZ

dς are regular

points on the Z-plane.

Points in the original complex Z-plane where 0=dZ

dς are critical points on the Z-plane.

The points in the original complex Z-plane where ∞=dZ

dς are singular points on the Z-plane.

)(Zς

dZ

dς

Nature of points in Z-

plane

transformation

Analytic function

∞≠≠ and0

Regular points

Conformal

transformation

Analytic function

0=

Critical points

Not conformal

transformation

Analytic function

∞=

Singular points

Not conformal

transformation

At a critical point, the transformation has a property of multiplying angle by n where n is the

other of the derivative n

n

dZ

d ςwhich first become non-zero at the critical point.

Z-plane ζ-plane

a

b

c

d

A

B C

D

β β


- 165 -

If a transformation is a conformal transformation, it has following properties:

(1).Laplace equation is invariant.

i.e., Lapalace equation in the Z-plane will be transformed into Laplace’s equation in ς -plane,

provided these two planes are related by a conformal transformation.

For example: Let )(Zςς = be a conformal transformation. Both φ and ψ are solution of Laplace

equation in Z-plane.

Therefore: 02

2

2

2

=∂

∂+

∂

∂

yx

ψψ and 0

2

2

2

2

=∂

∂+

∂

∂

yx

φφ

Since 2∇ is invariant under conformal transformation,

Therefore: 02

2

2

2

=∂

∂+

∂

∂

η

ψ

ξ

ψ and 0

2

2

2

2

=∂

∂+

∂

∂

η

φ

ξ

φ

In other words, a complex potential in the Z-plane is also a valid complex in the ς -plane and

vice versa.

(2). Velocity field

Consider the complex potential

ψφς iFZFF +=== )()(

dZ

dW

dZ

d

d

dF

dZ

dFZW

ςς

ς

ς)(ˆ)( ===

(3). The strengths of the circulation and source are invariant for a conformal

transformation,

Γ is the net strength of all vortices inside the

contours.

q is the net strength of all source and sinks inside the

contours.

∫ •= dseVq nˆ

r

jviuV ˆˆ +=r

vdxudyjdxidyjviudseV n −=−+=• )ˆˆ)(ˆˆ(ˆr

∫∫ −=•= vdxudydseVq nˆ

r

Z-plane

dx

dy

X

Y

ds

ds

ne

dy

dx


- 166 -

∫∫ +=•=ΓCC

t vdyudxdseV ˆr

qivdxudyivdyudxidydxivudZZWCCCC

+Γ=−++=+−= ∫∫∫∫ )()())(()(

Now consider )(ZC transforming into )(ςC and dZ

dW

dZ

d

d

dW

dZ

dWZW

ςς

ς

ς)(ˆ)( ===

Then ∫∫∫ +Γ===CCC

qidWdZdZ

dWdZZW ςς

ςς )(ˆ)(ˆ)(

How to use conformal transformation to solve problems of aerodynamics

1. Pose a problem in Z-plane with complex body geometry such as airfoil.

2. Map body to a simple body in ς -plane such as circular cylinder

3. Find a solution to the Laplace equation in ς -plane. i.e., find )(ςF in the ς -plane such

that )(ςF satisfy both boundary conditions at infinity (transformed from Z-plane). Let

Im )(ςF is a constant along the surface

4. Transform the solution back to Z-plane. i.e., Knowing the transformation )(Zςς = and

),(),()( ηξψηξφς iF += . Transform )()( ZFF =ς by satisfy )(Zςς = in )(ςF .

Example: ς -plane as the flow around a circular cylinder of radius a spinning with

circulation Γ could become flow around an airfoil through conformal transformation.

a

Zi

Z

aZVF ln

2)(

2

π

Γ++= ∞


- 167 -

6.14 Joukowski Transformation and Joukowski Airfoil

The transformation function of Joukowski transformation is

Z

aZZ

2

)( +== ςς where 2a is real.

For large values of Z , we will have Z→ς when ∞→Z . In other words, far from the origin,

the mapping is identical and the complex velocity is the same in both the planes far from the

origin.

Therefore, if a uniform flow of a certain magnitude is approaching a body in the Z-plane at same

angle of attack, a uniform flow of the same magnitude and angle of attack will approaching the

corresponding body in ς - plane.

Find the single and critic points of the Joukowski transformation:

2

2

1Z

a

dZ

d−=

ς ⇒ Single point at Z=0 (Z=0 is usually within a body)

2

2

1Z

a

dZ

d−=

ς ⇒ Critical point at aZ ±= ,

Since areZi ±== θ

⇒ ar = and πθ or0=

If we use Cartesian in ς -plane and polar in Z-plane,

ire

are

Z

aZ

i

i ηξςθ

θ +=+=+=22

Then:

θθ

θθθθηξθ

θ

sin)(cos)(

)sin(cos)sin(cos

22

22

r

ari

r

ar

ir

air

re

arei

i

i

−++=

−++=+=+

−=

+=

⇒

θη

θξ

sin)(

cos)(

2

2

r

ar

r

ar


- 168 -

Case 1. For a circle located in the origin with radius ar = in Z-plane

What it will be in ς -plane through the Joukowski transformation?

When θξη cos2,0 aar ==⇒=

When } aar

2,00

==⇒=

=ξη

θ

When } aar

2,0 −==⇒=

=ξη

πθ

a x

y

2a ξ

η

-2a


- 169 -

Case 2 For a circle located in the origin with radius br = ( b>a) in Z-plane

What it will be in ς -plane through the Joukowski transformation?

θθ ii beerZ == where b>a.

Since ib

abi

b

ab

Z

aZZ ηξθθςς +=−++=+== sin)(cos)()(

222

⇒

θη

θξ

sin)(

cos)(

2

2

b

ab

b

ab

−=

+= ⇒

)(

sin

cos

2

2

b

ab

b

ab

−

=

+

=

ηθ

ξθ

Since 1sincos 22 =+ θθ ⇒ 1]

)(

[][ 2

2

2

2=

−

+

+b

ab

b

ab

ηξ

It is the equation of an ellipse with major axis b

ab

2

+ and semi-minor axis of b

ab

2

− .

a x

y

2a

ξ

η

-2a

b

Z-plane

ζ-plane


- 170 -

Case 3. Joukowski Airfoils

From and aerodynamics point of view, the most interesting application of the Joukowski

transform is to an offset circle. If we consider a circle slightly offset from the origin along the

negative real axis, one obtains a symmetric Joukowski airfoil.

The equation of the offset circle is z = be

iθ-ea where the constant e is a small number. If the

cylinder is displaced slightly along the complex axis as well, one obtains a cambered airfoil

shape.

Here, the points A and B are the intercepts of the displaced circle on the real axis and their

corresponding points in the transformed plane. The angle β is the angle formed by the line

joining the point A (or B) and the origin with the real axis. If lifting flow about the original circle

had been imposed, the Joukowski transformation would have generated a lifting flow about the

Joukowski airfoil;

ς- plane

x

y

ξ

η z plane

z

bz

2

+=ς

a

b

e.a -2a 2a x

y

ξ

η z plane ζ-plane

z

az

2

+=ς

a

b

ea -2a 2a x

y

ξ

η z plane ζ-plane

A

β B B A

z

az

2

+=ς


- 171 -

Although such a flow is mathematically possible, in reality it may not be realistic. The stagnation

points on the cylinder map to stagnation points that are not always realistic. For instance the

stagnation point on the top surface of the airfoil cannot exist is steady flight since the velocity

would tend to infinity as one moves very close to the trialing edge. The only means of making a

realistic flow is to impose the Kutta condition where the stagnation point is forced to exist at the

trailing edge thus making the streamlines flow smoothly from this point. This is done by

adjusting the value of vorticity strength Γ, such that the stagnation points on the cylinder reside

at the cylinder’s intercepts of the real axis. In this case, when the cylinder is transformed, one

stagnation point will be forced to the trailing edge.

The lift force generated by the lifting flow over the cylinder is proportional to the circulation

about the cylinder imposed by the added vortex flow according to the Kutta-Joukowski relation,

L’ = ρV∞ Γ. The lifting force on the resulting Joukowski airfoil is not clear. To evaluate the lift,

the circulation is needed and therefore the velocity field. The velocity fields in each plane can be

related to each other through the chain rule of differentiation. If the lifting flow about the

cylinder is defined as function F where F = F(z) in the z- plane and F = F(ζ) in the ζ -plane, the

velocities in each plane are;

ς

ςς

∂

∂=

∂

∂=

)()(ˆ)(

FW

z

FZW

By chain rule:

z

F

Z

F

∂

∂

∂

∂=

∂

∂ ς

ς

ZWZW

∂

∂=

ςς )()(

Using the Joukowski transformation;

2

22

Z

aZ

Z

−=

∂

∂ς

Clearly, the velocity field very close to the cylinder and its transformed counterpart are

dissimilar as one would expect. However, farther away from these objects the velocity fields

become identical as the magnitude of z becomes larger than the constant value of a. Since the

ζ - plane

x

y

ξ

η z plane

z

az

2

+=ς


- 172 -

circulation can be calculated about any closed path, including paths very far from the object

surface, the circulations must be the same in both planes.

Joukowskicylinder VV Γ=Γ ∞∞ ρρ

Vortex strength

The appropriate vortex strength to impose the Kutta condition must be determined. Consider the

lifting flow about a cylinder. The velocity in the θ direction is,

Γ+−= ∞

RVV

πθθ

2)sin(2

Here, R is the radius of the cylinder surface.

This velocity is zero on the surface of the cylinder at the

stagnation points. At the points of θ =-β.

RV

πβ

2)sin(20

Γ−= ∞

)sin(4 βπ RV∞=Γ

If the field is rotated by α to simulate an angle of attack,

)sin(4 αβπ +=Γ ∞RV

Since the cord length of the Joukowski airfoil is 4a, the lift coefficient can be written,

aV

RV

aVaV

V

cV

LCL 2

2

222 2

)sin(4

24

2

1

2

1∞

∞

∞∞∞

+=

Γ=

Γ=

′=

βαπ

ρ

ρ

ρ

Making the assumption that a ≈ R,

)(2)sin(2 βαπβαπ +≈+=LC

x

y z plane

-β


- 173 -

Example

A Joukowski airfoil is formed by displacing a circle of radius 1 by ∆x = -0.08 (real axis) and ∆y

= 0.05 (imaginary axis). Find,

a) Vortex strength Γ if α = 0o, and V∞ = 10 m/s

b) CL at α = 0o and α = 10

o

O87.21

05.0sin 1 =

= −β

b

O

+=

08.0

05.0)87.2tan(

a = 0.9187

a) Γ = 4πV∞R⋅sin(α+β) =

4π(10)(1)sin(2.87) = 6.2831

b) CL = 2πsin(2.87) = 0.31415

CL = 2πsin(10 + 2.87) = 1.40

x

y

β 0.05

-

cylinder

stagnation

point

a

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CHAPTER 6 COMPLEX VARIABLE METHOD AND …huhui/teaching/2012FX/pdf-files/chapter6.pdfComplex Variable Method and Applications in Potential Flows For a certain type of flow, it is possible