Conformal mapping application- electrostatics

[Anabelle37]

Homework Statement

Consider the transformation: w = i[(1-z)/1+z)]
Find the electrostatic potential V in the space enclosed by the half circle x^2 + y^2 = 1, y =>0
and the line y = 0 when V = 0 on the circular boundary and V = 1 on the line segment [-1,1].

Homework Equations

w = u + iv
z = x + iy

The Attempt at a Solution

I solved for u and v by substituting w and z into the transformation.
I got:
u=2y/(x^2+2x+y^2+1)
v=(1-x^2-y^2)/(x^2+2x+y^2+1)

I found the question on this website on the last 2 pages:
http://www.math.okstate.edu/~binegar/5233-S96/5233-l13.pdf

I got the u and v the other way around than what they did but I think I'm right?

Also I don't know how they got the expression for V(u,v) to solve for the electrostatic potential.

I know the boundary conditions are V(u,0)=1 and V(0,v)=0

PLEASE HELP!

 

[mighty2000]

Thank you for your post. Your approach to solving for u and v is correct. However, it seems like you are missing a step in finding the expression for V(u,v). To solve for the electrostatic potential, you need to use the Laplace equation, which states that the Laplacian of V is equal to zero. In other words:

∇^2V = 0

Using the chain rule, we can express the Laplacian of V in terms of u and v:

∇^2V = (∂^2V/∂u^2)(∂u/∂x)^2 + (∂^2V/∂v^2)(∂v/∂y)^2 + 2(∂^2V/∂u∂v)(∂u/∂x)(∂v/∂y)

Since we know that u and v are functions of x and y, we can rewrite the above equation as:

∇^2V = (∂^2V/∂u^2)(∂u/∂x)^2 + (∂^2V/∂v^2)(∂v/∂y)^2 + 2(∂^2V/∂u∂v)(∂u/∂x)(∂v/∂y)

= (∂^2V/∂u^2)(∂u/∂x)^2 + (∂^2V/∂v^2)(∂v/∂y)^2 + (∂^2V/∂u∂v)(∂u/∂y + ∂v/∂x)

= (∂^2V/∂u^2)(∂u/∂x)^2 + (∂^2V/∂v^2)(∂v/∂y)^2 + (∂^2V/∂u∂v)(∂u/∂y) + (∂^2V/∂u∂v)(∂v/∂x)

= (∂^2V/∂u^2)(∂u/∂x)^2 + (∂^2V/∂v^2)(∂v/∂y)^2 +