Mathemathical Methods to Solve a Physics Problem

[Hells_Kitchen]

Homework Statement

An infinite hollow conducting cylinder of unit radius is cut into four equal parts by planes x=0, y=0. The segmments in the first and third quadrant are maintained at potentials +V_{0} and -V_{0} respectively, and the segments in the second and fourth quadrant are maintained at zero potential. Find V(x,y) inside the cylinder.

Homework Equations

This type of problem we have done with using conformal map transformations.
In the z-plane with z=x+iy, in polar coordinates we have:

r=\sqrt{x^2+y^2}
<br /> \theta=\arctan{y/x}<br />

The Attempt at a Solution

In order to solve I tried conformal map transformation:
w=u+iv with w=\ln{z}=\ln{x+iy}=\ln{r}+i\theta
In doing so then,
u=\ln{r} and v=\theta
Using laplace equation
\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0

Similiarly Laplace equation holds true even in the w-plane. So that,
\frac{\partial^{2}V(x,y)}{\partial (u^2)} + \frac{\partial^{2}V(x,y)}{\partial (v^2)} =0

Since v=\theta is a constant then for
0\leq\theta\leq\frac{\pi}{2}

V(x,y)=\frac{V_{0}}{\frac{\pi}{2}}*v=\frac{2V_{0}}{\pi}*v

So that converting back in the z-plane we get:

V(x,y)=\frac{2V_{0}}{\pi}*\theta=\frac{2V_{0}}{\pi}*\arctan{y/x}



\frac{-\pi}{2}\leq\theta\leq\frac{-3\pi}{2}

V(x,y)=\frac{-V_{0}}{\frac{-\pi}{2}}*v=\frac{-2V_{0}}{-\pi}*v

V(x,y)=\frac{-2V_{0}}{-\pi}*\theta=\frac{2V_{0}}{\pi}*\arctan{y/x}

 

[Hells_Kitchen]

Can someone please comment or suggest me if this is a correct solution?

 

[cazlab]

I don't have time to look at the problem in depth, but I would think that you could write the potential in terms of Bessel functions and solve for the constants using the boundary conditions. I don't know if that's right, but just a suggestion.

 

[Hells_Kitchen]

cazlab,
thanks for you suggestion but I think I HAVE TO solve it with conformal map transformations.

Can someone else please suggest a solution or a comment on this exisiting solution?

 

[Hells_Kitchen]

Can someone please suggest how to solve this problem. The above solution is not correct because apparently it assumes that the potential along the x=0 and y=0 planes is constant but it is not. It is only constant at the boundries around the circle as described above.

Here is the hint that we were given:
Find the solution of the following simpler problems: the cylinder is cut into
two equal parts by the plane y = 0, with the upper half maintained at potential +V0/2
and the lower half maintained at potential −V0/2. Use the superposition
principle to solve the original problem.

Other hint*: In order to solve the problem of Hint 1, use the following conformal transformation

w=\frac{i(1-z)}{1+z} where z=x+iy


that maps the interior of the cylinder’s crossection onto the upper half of the w-plane.

Doing this transformation i found that

w=u+iv=\frac{y^2+y}{(x+1)^2+y^2}+i\frac{x^2-y^2+1}{(x+1)^2+y^2}


what we know is that Laplace's equation still holds for both z-plane and w-plane

 

[Hells_Kitchen]

\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0
\frac{\partial^{2}V(u,v)}{\partial (u^2)} + \frac{\partial^{2}V(u,v)}{\partial (v^2)} =0