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12.3. FLUID FLOW AND ANALYTIC FUNCTIONS 159

Problem c: Consider the velocity eld (4.9) due to a point source at r = 0. Show that

this ow eld follows from the potential

V

f(x y) = 2 ln r (12.20)

p

where r = x2 + y2 .

Problem d: Verify explicitly that this potential satis es (12.19) except for r = 0. What

is the physical reason that (12.19) is not satis ed at r = 0?

We now want to identify the potential f(x y) with the real part of an analytic function

h(x+iy). We know already that the ow follows from the requirement that the velocity is

the gradient of the potential, hence it follows by taking the gradient of the real part of h.

The curves f = const: are perpendicular to the ow because v = is perpendicular to

rf

these curves. However, it was shown in section (12.1) that the curves g = = const:

=(h)

are perpendicular to the curves f = = const. This means that the ow lines are

<(h)

given by the curves g = = const. In order to use this we rst have to nd an analytic

=(h)

function with a real part given by (12.20). The simplest guess is to replace r in (12.20)

by the complex variable z:

V

h(z) = 2 ln z : (12.21)

Problem e: Verify that the real part of this function indeed satis es (12.20) and that

the imaginary part g of this function is given by:

V

g=2 ' (12.22)

where ' = arctan (y=x) is the argument of the complex number. Hint: use the

representation z = r exp i' for the complex numbers.

Problem f: Sketch the lines g(x y) = const: and verify that these lines indeed represent

the ow lines in this ow.

Now we will consider the more complicated problem of section (4.2) where the ow

has a source at r+ = (L 0) and a sink at r; = (;L 0). The velocity eld is given by the

equations (4.10) and (4.11) and our goal is to determine the slow lines without solving

the di erential equation dr=dt = v(r). The points r+ and r; can be represented in the

complex plane by the complex numbers z+ = L + i0 and z; = + i0 respectively. For

;L

the source at r+ ow is represented by the potential 2V ln z+ this follows from the

jz ; j,

solution (12.21) for a single source by moving this source to z = z+

Problem g: Using a similar reasoning determine the contribution to the potential f by

the sink at r; . Show that the potential of the total ow is given by:

V z+

f(z) = 2 ln : (12.23)

jz ; j

z;

jz ; j

CHAPTER 12. ANALYTIC FUNCTIONS

160

z

r+

r-

Ï•- Ï•+

z- z+

Figure 12.3: De nition of the geometric variables for the uid ow with a source and a

sink.

Problem h: Express this potential in x and y, compute the gradient and verify that this

potential indeed gives the ow of the equations (4.10)-(4.11). You may nd gure

(12.3) helpful.

We have found that the potential f is the real part of the complex function

h(z) = 2 ln z z+ :

V z (12.24)

;

z; ;

Problem i: Write z z = r exp i' (with r and ' de ned in gure (12.3)), take

;

the imaginary part of (12.24), and show that g = is given by:

=(h)

g = V (' ' ) : ; (12.25)

+;

2

Problem j: Show that the streamlines of the ow are given by the relation

arctan x y L y

arctan x + L = const: (12.26)

;

;

A gure of the streamlines can thus be found by making a contour plot of the function in

the left hand side of (12.26). This treatment is de nitely much simpler than solving the

di erential equation dr=dt = v(r).

Chapter 13

Complex integration

In chapter (12) the properties of analytic functions in the complex plane were treated.

One of the key-results is that the contour integral of a complex function is equal to zero

when the function is analytic everywhere in the area of the complex plane enclosed by

that contour, see expression (12.10). From this it follows that the integral of a complex

function along a closed contour is only nonzero when the function is not analytic in the

area enclosed by the contour. Functions that are not analytic come in di erent types. In

this section complex functions are considered that are not analytic only at isolated points.

These points where the function is not analytic are called the poles of the function.

13.1 Non-analytic functions

When a complex function is analytic at a point z0 , it can be expanded in a Taylor series

around that point. This implies that within a certain region around z0 the function can

be written as: 1

X

h(z) = an(z z0 )n : (13.1)

;

n=0

Note that in this sum only positive powers of (z z0 ) appear.

;

Problem a: Show that the function h(z) = sin z can be written as a Taylor series around

z

the point z0 = 0 of the form (13.1) and determine the coe cients an .

Not all functions can be represented in a series of the form (13.1). As an example consider

the function h(z) = exp(1=z). The function is not analytic at the point z = 0 (why?).

Expanding the exponential leads to the expansion

1

1+ 1 1 + =X 1 1 :

h(z) = exp(1=z) = 1 + z 2! z2 (13.2)

n! zn

n=0

In this case an expansion in negative powers of z is needed to represent this function. Of

course, each of the terms 1=z n is for n 1 singular at z = 0, this re ects the fact that the

function exp(1=z) has a pole at z = 0. In this section we consider complex functions that

can be expanded around a point z0 as an in nite sum of integer powers of (z z0 ):

;

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