Wikipedia:Reference desk/Archives/Mathematics/2016 February 3

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February 3[edit]

Conformal mapping[edit]

As I understand it, in two dimensions conformal maps can be almost uniquely defined by specifying a simple closed curve in the domain and another such curve in the codomain, such that everything within the curve in the domain is mapped bijectively to the area inside the codomain. I have two questions relating to this:

  1. Using complex-valued functions to represent the conformal map, is there a general procedure for finding the holomorphic function that maps the boundary of an arbitrary simple closed curve to the unit circle?
  2. Is the function thus found guaranteed to be representable by a single power series throughout the simple closed curve in the domain? I believe that the inverse map always has this property (the inverse function must be holomorphic throughout the unit disk so is representable by a single power series about the origin), but I'm not so sure about the original function.--Leon (talk) 08:47, 3 February 2016 (UTC)[reply]
1. There is the Schwarz-Christoffel mapping, which only applies to certain types of regions (but can be adapted as a numerical scheme for more general regions). There is a book by that title by Driscoll and Trefethen with lots of details and examples. 2. In general, the map will not have a single power series expansion throughout the domain. Most likely, this is never true except when the domain and codomain are both discs (or halfplanes). Sławomir
Biały
12:27, 3 February 2016 (UTC)[reply]
Thanks, but I didn't mean a power series that converges throughout the entire domain, I meant a power series with a radius of convergence that covers the simple closed curve.--Leon (talk) 12:36, 3 February 2016 (UTC)[reply]
If a disc contains a Jordan curve, then the disc must contain a component of the complement of the curve. So if the power series converges on the whole curve, then it must converge on a connected component of the complement. Sławomir
Biały
12:39, 3 February 2016 (UTC)[reply]