How Do Conformal Mappings Aid in Solving Laplace's Equation?

[aaaa202]

Exam tomorrow and I am lacking understanding of conformal transformations and their applications. Can someone therefore point the main properties of conformal mappings that are used to make the conclusions in the following type of exercises:

the mapping f(z) = 1 + 1/z maps the unit circle to y=0 and -2≤x≤2. Use that to find a solution to the laplacian equation ∇²ψ = 0 with ψ=0 for y=0 and -2≤x≤2. It is to be assumed that ψ diverges logarithmically for lzl->∞.

My teacher did this by inverting the above mapping, which gives z=½(w±√(w²-4)). He was then out of nowhere able to write down the solution as:

ψ(w) = Alog( mod(w/2±√((w/2)²-4)))

Question one: Where did this come from? I know that any analytic function of z transforms to an analytic function of w, and thus they both satisfy laplaces equation in their respective domain. But how is this used specifically here? And where does that modulus and log come from?

He then proceeds to find the complex potential which is the same as the above without the modulus.

Question two: I know that the potential satisfies laplaces equations for regions without charge. But how is it specifically used here - and what is meant by the complex potential?

He then finds the E-field by taking the derivative with respect to w of the complex potential but conjugates it all. Again I want some basic explanation.

I do realize that my questions require perhaps a bit too long of an answer but then just please try to focus on the main points needed to understand what is done. My book's chapter on this is extremely poor, so if you have some notes on conformal mappings and their applications which are quick to read and give a good overall understanding, then I would be pleased to have them.

Cheers :-)

 

[lippyka]

Conformal mappings are used to transform a region from one complex plane to another. They preserve angles and can be used to map one region to another. This is useful for solving Laplace's equation as it preserves the form of the equation, thus allowing us to solve it in one region and then use the mapping to find the solution in another region. In the example given, your teacher used the inverse of the conformal mapping f(z) = 1 + 1/z to map the unit circle to the line y=0 and the interval -2≤x≤2. The inverse of this mapping is z=½(w±√(w2-4)). Your teacher then used this inverse to write down the solution to Laplace's equation as: ψ(w) = Alog( mod(w/2±√((w/2)2-4))). This is because the inverse of a conformal mapping maps analytic functions to analytic functions, thus preserving the form of the equation. The modulus and logarithm come from the fact that this solution must logarithmically diverge as lwl->∞. The complex potential is the same as the above solution, but without the modulus. The complex potential is useful for finding the electric field as it takes into account the charge distribution in a region. By taking the derivative of the complex potential with respect to w and conjugating it, we can find the electric field in the region.