Conformal mappings on the complex plane

[Gray]

Homework Statement

a) Using the conformal mapping w=cosh(z), find a rectangle R in the z-plane which maps to the region in the w-plane with boundaries as follows:

- a plate of constant temperature on the line segment {w=u+iv : -1<u<1, v=0}
- an outer boundary of cooler constant temperature given by the ellipse u^2/cosh^2(1) + v^2/sinh^2(1) = 1

b) The elliptical surface is at temperature 0 and the line segment is at tempertaure 1. Choose a complex function g, defined on R, such that the real part of g(inv_cosh(w)_ defines a temperature with appropriate boundary conditions in the w-plane.

Homework Equations

z=x+iy
cosh(z) = (e^z+e^-z)/2 = ... = cosh(x)cos(y) + i*sinh(x)sin(y)
general equations for ellipse and hyperbola

The Attempt at a Solution

Firstly I drew the plate and outer boundary.
To map into a rectangle in the z-plane I first recognise that hyperbolae transform to horizontal lines through w=cosh(z) and similarly ellipses transform to vertical lines.
With this in mind I can recognise that the 'plate' is simply a hyperbolae with v=0.
So for the plate
v=sinh(x)sin(y)=0
u=cosh(x)cos(y)=-1...1

sin(y) is not = 0 because then cos(y)=1 and hence cosh(x)=-1...1
Therefore sinh(x)=0
x=0

u=cosh(0)cos(y)=-1...1
y=inv_cos(-1...1)
=0...Pi

So we have mapped the plate to z=y, 0<y<Pi
But my problem here is this is sinusoidal so really n*Pi, i.e. an infinite line?
Also to make my rectangle R of boundary conditions won't I need two horizontal lines? Are these y=0 and y=Pi?

Anyway my attempt on transforming the ellipse to two vertical lines is similar:
cosh^2(1)=2.381
sinh^2(1)=1.381

u^2/2.381 + v^2/1.381 = 1
cosh^2(x)cos^2(y)/2.381 + sinh^2(x)sin^2(y)/1.381 = 1

Presumably y=0 but only because I know a priori that it should map to vertical lines

Anyway... cosh^2(x) = 2.381
x=inv_cosh(+-1.543)
=~2.5
only 1 value :(

I think my methods are wrong but don't know what else to do. (I tried starting the map from the other direction, i.e. from lines to ellipses/hyperbolae but got confused with the multiple values).
Thanks.

 

[Dick]

You have that cosh(z) defines the conformal mapping u=cosh(x)cos(y), v=sinh(x)sin(y). Compare this to the parametric form for an ellipse centered at the origin with semimajor axes r_u and r_v. u=r_u*cos(t), v=r_v*sin(t) for t=0 to 2pi. There aren't any hyperbolae in the problem at all. All you have to do is figure out the correct x and y intervals defining the rectangle.

 

[Gray]

So I set the two forms equal and eliminate y to obtain
1 = (r_u*cos(t)/cosh(x))^2 + (r_v*sin(t)/sinh(x))^2

where I know r_u=cosh(1) and r_v=sinh(1) looking at my specific ellipse

and somehow solve this to get my two values for x?

 

[Dick]

I would say r_u=cosh(x) and r_v=sinh(x) and y=t. Can you figure out the x and y ranges of the rectangle now? You want r_u to go from 1 to cosh(1) and r_v to go from 0 to sinh(1). The suggestion w=cosh(z) basically gave you everything. You just have to extract the boundaries of the rectangle from it.

 

[Gray]

Why from 1 to cosh(1)?
Is it because we want the gap between the plate and boundary to form our rectangle boundaries?

 

[Dick]

Because the ellipse starts with a u semimajor axis of 1 and ends with a semimajor axis of cosh(1). Draw a picture! That's cosh(0) to cosh(1).

 

[Gray]

We have
u=cosh(x)cos(y)
v=sinh(x)sin(y)

and we want u to vary from 1 to cosh(1) and v to vary from o to sinh(1)

so I want to solve the four equations
cosh(x)cos(y)=1
sinh(x)sin(y)=cosh(1)
cosh(x)cos(y)=0
sinh(x)sin(y)=sinh(1)
to get my four boundary lines to the rectangle?

I think I am making this too complicated!

 

[Dick]

You ARE making this way too complicated! cosh(x) and sinh(x) are the semimajor and semiminor axes of the ellipse. You want the semimajor axis to go from cosh(0) to cosh(1) and the semiminor to go from sinh(0) to sinh(1). Why don't you just take x to go from 0 to 1? I've been trying to avoid actually telling you that but there seems to be no choice. The x range of your rectangle is [0,1]. Now you really have to do the y range on your own. And I'm serious this time. It's EASY.