Gray
Aug22-08, 01:32 AM
1. The problem statement, all variables and given/known data
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.
2. Relevant 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
3. 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.
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.
2. Relevant 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
3. 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.