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trusure
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 crescent-shaped region mapping   « on: Apr 21st, 2009, 7:29am » Quote Modify

how I can construct a conformal map from the region between
the circles D(1,1) and D(2,2) to D(0,1)

If I just pick some points on the boundary of
D(2,2), I'll end up mapping D(2,2) onto D(0,1), which is not what I want.

I'm not sure how you do this. The Riemann mapping theorem doesn't apply because the region
I'm trying to map to D(0,1) isn't simply connected. I don't think an LFT can work either,
because an LFT will have to take the boundary of
D(0,1) to the boundary of the original region--
that's impossible because the boundary isn't connected.

So, this is my PROBLEM ...

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Obob
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 Re: crescent-shaped region mapping   « Reply #1 on: Apr 21st, 2009, 7:52am » Quote Modify

Actually that region is simply connected.  The point 0 does not lie in the region, so there is no loop in the region which goes around the hole.  The boundary is also connected; its two circles meeting at a point, and it can be parameterized by a single curve.

As far as actually constructing the mapping, I'm terrible at these things.  I have a table of important types of conformal maps that I use to try and transform one region into the other.

You won't find a single linear fractional transformation that does the trick; however, you might be able to first apply a LFT that simplifies the region, then compose it with other things to get to a disk.
 « Last Edit: Apr 21st, 2009, 7:54am by Obob » IP Logged
trusure
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 Re: crescent-shaped region mapping   « Reply #2 on: Apr 21st, 2009, 8:26am » Quote Modify

Thank you...

but I didn't get the IDEA ..?!!!

Could anyone suggest something
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Eigenray
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 Re: crescent-shaped region mapping   « Reply #3 on: Apr 21st, 2009, 8:37pm » Quote Modify

There is an FLT taking your region to { z :  |z| > 1, Re z < 1 }.  If the conformal map doesn't have to be injective, you can just square and invert.  But otherwise, I dunno.  There is a variation of Schwarz-Christoffel for circular-arc polygons (for example, in this book) but it looks like it's given as a differential equation.  I would expect there to be something more explicit in this case though.

On second thought, it's actually pretty simple : just invert and you get a strip!
 « Last Edit: Apr 21st, 2009, 9:36pm by Eigenray » IP Logged
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