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!

where dr = dx and with the vectors v and w de ned by:

dy

! !

f g

v= w= f : (12.9)

;g

Note that we now use x and y both as the real and imaginary part of a complex number,

but also as the Cartesian coordinates in a plane. In the following problem we will (perhaps

confusingly) use the notation z both for a complex number in the complex plane, as well

as for the familiar z-coordinate in a three-dimensional Cartesian coordinate system.

Problem j: Show that the Cauchy-Riemann relations (12.4)-(12.5) imply that the z-

component of the curl of v and w vanishes: (r v)z = (r w)z = 0, and use

(12.8) and the theorem of Stokes (7.2) to show that when h(z) is analytic everywhere

within the contour C that:

I

h(z)dz = 0 h(z) analytic within C : (12.10)

C

This means that the line integral of a complex functions along any contour that encloses

a region of the complex plane where that function is analytic is equal to zero. We will

make extensive use of this property in section (13) where we deal with integration in the

complex plane.

12.2 The electric potential

Analytic functions are often useful in the determination of the electric eld and the po-

tential for two-dimensional problems. The electric eld satis es the eld equation (6.12):

(r E) = (r)="0 . In free space the charge density vanishes, hence (r E) = 0. The

electric eld is related to the potential V through the relation

E= : (12.11)

;rV

12.2. THE ELECTRIC POTENTIAL 157

Problem a: Show that in free space the potential is a harmonic function:

2 V (x y) = 0 : (12.12)

r

We can exploit the theory of analytic functions by noting that the real and imaginary parts

of analytic functions both satisfy (12.12). This implies that if we take V (x y) to be the

real part of a complex analytic function h(x + iy), the condition (12.12) is automatically

satis ed.

Problem b: If follows from (12.11) that the electric eld is perpendicular to the lines

where V is constant. Show that this implies that the electric eld lines are also

perpendicular to the lines V = const. Use the theory of the previous section to

argue that the eld lines are the lines where the imaginary part of h(x + iy) is

constant.

This means that we receive a bonus by expressing the potential V as the real part of a

complex analytic function, because the eld lines simply follow from the requirement that

= const.

=(h)

Suppose we want to know the potential in the half space y 0 when we have speci ed

the potential on the x-axis. (Mathematically this means that we want to solve the equation

2 V = 0 for y 0 when V (x y = 0) is given.) If we can nd an analytic function h(x+iy)

r

such that on the x-axis (where y = 0) the real part of h is equal to the potential, we have

solved our problem because the real part of h satis es by de nition the required boundary

condition and it satis es the eld equation (12.12).

Problem c: Consider a potential that is given on the x-axis by

2 =a2

V (x y = 0) = V0 exp : (12.13)

;x

Show that on the x-axis this function can be written as V = with

<(h)

2 =a2

h(z) = V0 exp : (12.14)

;z

Problem d: This means that we can determine the potential and the eld lines through-

out the half-plane y 0. Use the theory of this section to show that the potential

is given by !

y2 x2 cos 2xy :

V (x y) = V0 exp a2 (12.15)

;

a2

Problem e: Verify explicitly that this solution satis es the boundary condition at the

x-axis and that is satis es the eld equation (12.12).

Problem f: Show that the eld lines are given by the relation

2!

2

exp y a2 x sin 2xy = const: (12.16)

;

a2

CHAPTER 12. ANALYTIC FUNCTIONS

158

Problem g: Sketch the eld lines and the lines where the potential is constant in the half

space y 0.

In this derivation we have extended the solution V (x y) into the upper half plane

by identifying it with an analytic function in the half plane that has on the x-axis a

real part that equals the potential on the x-axis. Note that we found the solution to this

problem without explicitly solving the partial di erential equation (12.12) that governs the

potential. The approach we have taken is called analytic continuation since we continue

an analytic function from one region (the x-axis) into the upper half plane. Analytic

continuation turns out to be a very unstable process. This can be veri ed explicitly for

this example.

Problem h: Sketch the potential V (x y) as a function of x for the values y = 0, y = a

and y = 10a. What is the wavelength of the oscillations in the x-direction of the

potential V (x y) for these values of y? Show that when we slightly perturb the

constant a that the perturbation of the potential increases for increasing values of y.

This implies that when we slightly perturb the boundary condition, that the solution

is more perturbed as we move away from that boundary!

12.3 Fluid ow and analytic functions

As a second application of the theory of analytic functions we consider uid ow. At the

end of section (4.2) we have seen that the computation of the streamlines by solving the

di erential equation dr=dt = v(r) for the velocity eld (4.10)-(4.11) is extremely complex.

Here the theory of analytic functions is used to solve this problem in a simple way. We

consider once again a uid that is incompressible ( (r v) = 0 ) and will specialize to the

special case that the vorticity of the ow vanishes:

v=0: (12.17)

r

Such a ow is called irrotational because it does not rotate (see the sections (5.3) and (5.4)).

The requirement (12.17) is automatically satis ed when the velocity is the gradient of a

scalar function f:

v= : (12.18)

rf

Problem a: Show this.

The function f plays for the velocity eld v the same role as the electric potential V for

the electric eld E. For this reason, ow with a vorticity equal to zero is called potential

ow.

Problem b: Show that the requirement that the ow is incompressible implies that

2f = 0 : (12.19)

r

We now specialize to the special case of incompressible and irrotational ow in two

dimensions. In that case we can use the theory of analytic functions again to describe

the ow eld. Once we can identify the potential f(x y) with the real part of an analytic

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