# Talk:Schwarz–Christoffel mapping

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Field: Analysis

## Title

What is the correct spelling and why is this called a theorem? The interesting part is the formula for the mapping; once you have this, the proof is obvious. The PlanetMath article is Schwarz-Christoffel transformation, the MathWorld article is Schwarz-Christoffel mapping, and Google gives 32 results for "Schwarz-Christoffel theorem", about 1300 for "Schwarz-Christoffel mapping", 8 for "Schwarz-Christoffel function", 748 for "Schwarz-Christoffel transformation" and 6, 18, 0, 184. Furthermore, there is a book Schwarz-Christoffel mapping stating on page 17 that it is named after Elwin Bruno Christoffel and Hermann Amandus Schwarz. Hence, I'm inclined to move the article to Schwarz-Christoffel mapping. -- Jitse Niesen 02:03, 8 Jun 2005 (UTC)

I would agree with that. Oleg Alexandrov 03:32, 8 Jun 2005 (UTC)
seconded (although I must confess to being a little intimidated by folk to whom the proof is "obvious" ;-)

Robinh 07:10, 8 Jun 2005 (UTC)

You are right, I shouldn't have written that the proof is obvious, especially since I didn't actually try to prove it. I didn't mean that the proof is easy, only that it seems straightforward. -- Jitse Niesen 09:23, 8 Jun 2005 (UTC)

I now moved the page. -- Jitse Niesen 11:24, 8 Jun 2005 (UTC)

## Z vs zeta

I'm confused. Looks like the sense of z and zeta are inverted between the main text and the example. I don't have time to try to fix this right now. Help!? (I may have contributed to the confusion by trying to tweak the intro, which was not self-consistent; revert my last edit if it will help restore order. However, the values a,b,c are on the number line and not on the polygon, and they are traditionally ordered via right-hand rule to be same order as polygon). I'm confused, as also the formula in the defn. looks like its for the unit disk and not the half-plane. They're almost the same but not quite. linas 14:36, 15 Jun 2005 (UTC)

Yes, you're right. I don't have access to any books, but I tried to fix it. Any better now? -- Jitse Niesen 17:27, 15 Jun 2005 (UTC)
Yes I think so. I did not check my disk vs. half-plane concern yet, though. linas 23:54, 15 Jun 2005 (UTC)
The first example is quite easy to go through by hand. The result (the function arccosh) maps indeed the half-plane to the semi-infinite strip, since the inverse ζ = cosh z = cos iz maps the infinite strip to the whole complex plane and semi-infinite strip to the half-plane (I cannot see the last statement immediately myself, but I guess it is obvious once you think about it; besides, you seem to have more experience with special functions). -- Jitse Niesen 13:32, 16 Jun 2005 (UTC)

## J-invariant

Robinh, I know you know what the J-invariant is, since you've worked that article. There is some mapping that takes upper half-plane to the fundamental domain, its some hypergeometric series, its effectively the inverse of the J-invariant. Do you have a good reference on this? Note the fundamental domain is the like the semi-infinte strip given in the example, except that the bottom edge is a semi-circle. What I'm really trying to find an expression for the derivative jacobean thingy $(dj/d\tau)\circ j^{-1}$; haven't seen that anywhere? Errr, wait, I'm confused by something and it suddenly strikes me that maybe its real simple ... linas 14:49, 15 Jun 2005 (UTC)

See below, I may be confusing this with Schwarz map! linas 23:54, 15 Jun 2005 (UTC)
No, I'm not. And I found an OK ref. Ah, the magic of google. linas 00:15, 16 Jun 2005 (UTC)

## Schwarz map

It seems I have also confused this with another thing called the "Schwarz map", which is a ratio of hypergeometric functions. Its not clear to me if these things are somehow related, or just accidentally share a name. The shwarz map seems to be similar, except that the polygon seems to always (??) be a triangle (??) and the sides are semi-circles, (??) not straight lines (??). Any clue here? linas 23:54, 15 Jun 2005 (UTC)

The schwarz s-map does seem to be a hyperbolic-triangle version of the shwarz-christoffel map. I found a good reference at
which seems to cover part of what I wanted. linas 00:15, 16 Jun 2005 (UTC)

## The second example

Can somebody explain the second example to me? I cannot find the given result. In fact, when I try to do the integral, I substitute $z = x^2 - 1$, and I find

$\int_0^\zeta \frac {dz}{\sqrt{z(z^2-1)}} = \int_0^{\sqrt{\zeta+1}} \frac{2x\,dx}{\sqrt{(x^2-1)(x^4-2x^2)}} = \int_0^{\sqrt{\zeta+1}} \frac{2\,dx}{\sqrt{(1-x^2)(2-x^2)}} = \sqrt{2} \, F \left( \sqrt{\zeta+1}; \frac12\sqrt2 \right),$

where F is the incomplete elliptic integral of the first kind. Maple gives the same result. This is close to the result in the article, but not quite the same, as far as I can see. Jitse Niesen 14:33, 16 Jun 2005 (UTC)

Hi Jitse. You're absolutely correct. My algebra isn't what it used to be. Article now changed.

best wishes

Robinh 14:54, 16 Jun 2005 (UTC)

## New Developments

I found a page on StumbleUpon with a URL of http://www.eurekalert.org/pub_releases/2008-03/icl-1yo030308.php

Will these revisions be added into the formulas?

71.83.47.129 (talk) 04:11, 4 March 2008 (UTC)

I replaced the link with what I consider to be a better reference (I don't like press releases). In principle, there is nothing against adding these new results. However, the article has first to be expanded with other results that are not mentioned yet before we can talk about the new work on multiple connected domains. -- Jitse Niesen (talk) 11:48, 4 March 2008 (UTC)

## Error in Article

The Schwarz-Christoffel transformation is not conformal at locations corresponding to the vertices of the polygon. Commutator (talk) 04:16, 16 September 2009 (UTC)

I changed the first few lines, since they where mathematically wrong:

The SCT doesn't map the upper half-plane confromally to a polygon, since such a mapping cannot map real numbers conformal the the vertices of the polygon. Also, a polygon is defined as a set of lines and point, and not as a set. Better would be: the SCT maps the upper half-plane conformally to a open connected set which has a boundary in the shape of a polygon, in addition this SCT maps the real numbers continious onto this polygon....

It can be more readible...

• Also, a mapping which maps the open unit circle to a open set which is the inside of a polygon is also considered as a SCT. —Preceding unsigned comment added by Whendrik (talkcontribs) 12:26, 21 October 2009 (UTC)

## Example

The evaluation of the integral in the first example is false. The correct integral should be Ln[ζ+sqrt(ζ2-1)] — Preceding unsigned comment added by 198.208.251.21 (talk) 13:58, 29 July 2013 (UTC) I retract this correction. The above formula is equal to arccosh(ζ) — Preceding unsigned comment added by 198.208.251.21 (talk) 14:40, 30 July 2013 (UTC)