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   Author  Topic: crescent-shaped region mapping  (Read 9544 times)
trusure
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crescent-shaped region mapping  
« on: Apr 21st, 2009, 7:29am »
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how I can construct a conformal map from the region between  
the circles D(1,1) and D(2,2) to D(0,1)Huh
 
 
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 ...
 
Thanks in advance
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Obob
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Re: crescent-shaped region mapping  
« Reply #1 on: Apr 21st, 2009, 7:52am »
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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 »
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Thank you...
 
but I didn't get the IDEA ..?!!!
 
Could anyone suggest something Huh
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Eigenray
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Re: crescent-shaped region mapping  
« Reply #3 on: Apr 21st, 2009, 8:37pm »
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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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