Show that the function

maps the (open) unit disc onto the interior of a square.

This problem is difficult. Good luck! (Wink)

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- Feb 5th 2010, 10:38 AMBruno J.Complex integral (2)
Show that the function

maps the (open) unit disc onto the interior of a square.

This problem is difficult. Good luck! (Wink)

- Feb 5th 2010, 01:37 PMLaurent
Not a full proof (a few topological details missing), but convincing arguments...

__Spoiler__: - Feb 5th 2010, 03:05 PMBruno J.
Very good! You are very good, Laurent.

I had this problem to do on an assignment last week, and I knew that it followed from the following more general theorem. This one was also on the assignment and it was by far the hardest problem I've ever had on an assignment so far. I got together with two Ph.D students and 3 hours of discussion got us nowhere. I ended up solving it on my own! (Rock)

So here it is!

*Show that in the formula of the Schwarz-Christoffel transformation, the points on the real line which are to be mapped to the vertices of the polygon can be replaced by points on the unit circle, and the resulting function maps the unit disc to a polygon. In other words, if**are points on the unit circle, and**are the corresponding interior angles of the polygon*,*then*

*maps the unit disc to a polygon having the prescribed angles.*

The case above obviously follows right away from this one. Would you know how to show it?

- Feb 5th 2010, 03:48 PMLaurent
Do you mean show the formula for the Schwarz-Christoffel mapping? I wouldn't, but there seems to be a proof here for instance (Itwasntme). (for the half-plane, so one'd need a composition by something like to start from the disc...)

- Feb 5th 2010, 04:32 PMBruno J.
Well, yes, that was the problem. The hint is to use the transformation you suggest, but I was never able to bring it to look like the original formula again... I ended up finding another proof.

- Feb 6th 2010, 02:05 AMLaurent
I thought you asked for the usual formula for Schwarz-Christoffel mappings; if all you need is switch to the disc, you can do as follows: the function (where ) sends the upper half-plane to a polygon P with the given angles. Then the function (the second expression is for derivating) maps the unit disc to the upper half-plane. So the function you want is just .

In fact I think the other way around will be more natural... The function maps the upper half-plane to and we have to check that is a usual Schwarz-Christoffel mapping with the correct angles. We have . Note that (elementary geometry) hence if we reduce to the same denominator in the parentheses, the denominator simplifies with the first factor, and we are left with . Write , hence for some complex constant , . Since maps the unit circle to the real axis, we have a "usual" Schwarz-Christoffel mapping. This concludes. (Side note: so that : there is an additional translation (+ an affinitiy due to K). ) - Feb 6th 2010, 04:31 PMBruno J.
Nice! Thank you Laurent. Perhaps if I had followed through with the calculations I would have gotten it in this way. Here is an outline of the solution I found :

Consider where are points on the circle and . First, we want to show that the arcs of the circle corresponding to are mapped to line segments by . This is equivalent to showing that the tangent vectors to the circle along that arc are mapped to tangent vectors on the boundary of the image domain which do not change direction are stays within the interval . Let be a vector tangent to the unit circle at the point . Then is a vector tangent to the curve at the point ; moreover . So we want to show is constant when . To do this we write

and we see that for some constant . Moreover and so is constant.

To see that suddenly changes when goes over the point , notice that the factor changes sign, causing a change of in , and we are left with the same situation as with the original Schwarz-Christoffel mapping.