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I found two different formulations of the Schwarz-Christoffel formula (e.g. Link1, p.20 and Link2, p. 9). The first is

\begin{align*} z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(\zeta-z_k\right)^{\alpha_k-1}d\zeta \end{align*}

The second is

\begin{align*} z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(1-\frac{\zeta}{z_k}\right)^{\alpha_k-1}d\zeta \end{align*}

In both equations

  • $A$ and $C$ are complex constants,
  • $z_k$ are the coordinates of point $k$ on the unit circle corresponding to the vertex $k$ of the rectangle,
  • $\alpha_k$ are the interior angles of the vertices by means of multiples of $\pi$
  • and $\zeta$ is a point outside the unit circle such that $|\zeta| > 1$.

In the second link on page 20 it is said that: "Composing the first equation with standard conformal maps leads to variations of the Schwarz-Christoffel formula for mapping from other fundamental domains [...]. The simplest such modification has the unit disk as domain. The vertices then lie in counterclockwise order on the unit circle and equation one can be transformed to equation two."

Why is this transformation possible for the upper complex half-plane and how is it carried out?

40 votes
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Martin
  • 61
  • Is,by definition, the Schw. Christoffel map given by the integral of a function $f$ along any path in the disk/upper half plane? Then, you can consider the transformation formula of such integrals under re-parametrization of the paths . In our case the re parametrization is given by the composition w.r.t. any conformal mapping from the disk to the upper half plane (for example $\varphi(z)=\dfrac{z-i}{z+i}$) – Avitus Jul 17 '13 at 20:48

1 Answers1

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First of all, there is no real difference between the two formulas in your post: in the second, $C$ includes $\prod (-z^k)^{\alpha_k-1}$, which is a constant.

It is somewhat remarkable that the transformation between the disk and the plane preserves the general appearance of the S-C formula. (Not so for other domains, such as an infinite strip). The steps of this transformation can be found on pages 192-193 of Conformal mapping by Zeev Nehari (published by Dover, highly recommended). I paraphrase the text with occasional relabeling of indices, as I hate seeing $\nu$ used as an index... For the halfplane, Nehari has $$f(z)=\alpha\int_0^z \prod_{k=1}^n (z-a_k)^{-\mu_k}\,dz+\beta \tag{50}$$ By the linear transformation $$z=i\frac{1+\zeta}{1-\zeta},\quad \zeta=\frac{z-i}{z+i}$$ which maps $|\zeta|<1$ onto $\operatorname{Im}z>0$, we can also obtain from (50) a formula for the conformal map of the unit circle onto the polygon $D$. We have $$(z-a_k)^{\mu_k} = \left[ i\frac{1+\zeta}{1-\zeta} -a_k\right]^{\mu_k} =\left(\frac{a_k+i}{1-\zeta}\right)^{\mu_k} \left[ \zeta - \frac{a_k-i}{a_k+i} \right]^{\mu_k} \tag{a}$$ and $$dz=\frac{2i\,d\zeta}{(1-\zeta)^2}\tag{b}$$ Let $$b_k=\frac{a_k-i}{a_k+i} $$ which is the point on the unit circle corresponding to $a_k$. Since $\sum \mu_k=2$, the denominators $(1-\zeta)$ in (a) cancel the denominator in (b). We end up with
$$f(\zeta)=\alpha_2\int_0^\zeta \prod_{k=1}^n (\zeta-b_k)^{-\mu_k}\,d\zeta+\beta_2 $$ where $\alpha_2,\beta_2$ are constants.

40 votes
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  • nice answer (+1) – Avitus Jul 18 '13 at 08:22
  • Thanks for the great explanation 40 votes. Is it common to change the value of the constant $C$ without changing its name? So for clarification I would write \begin{align} z=f(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(\zeta-z_k\right)^{\alpha_k-1}d\zeta\ =&A+\tilde{C}\int\limits^{z}\prod\limits_{k=1}^n\left(1-\frac{\zeta}{z_k}\right)^{\alpha_k-1}d\zeta \end{align} –  Jul 18 '13 at 13:10
  • with $\tilde{C}=C\prod (-z^k)^{\alpha_k-1}$. I know that $C$ or $\tilde{C}$ only has an influence on the scale, but nonetheless, I think the difference should be seen in an equation. The exterior angle $\mu_k=1-\alpha_k$. So the exponent $-\mu_k=\alpha_k-1$.

    In Stress Distribution around Holes by G.N. Savin I found yet another formulation on page 14:

    \begin{align} z=f(\zeta)=&A+C\int\limits^{\xi} \prod\limits_{k=1}^n \left(1-\frac{z_k}{\zeta}\right)^{\gamma_k-1} d\zeta\tag{I.47} \end{align}

     –  Jul 18 '13 at 13:10
  • where, as shown in figure I.1 on page 15, $\gamma_k=2-\alpha_k$ such that $\gamma_k-1=\mu_k$. In my opinion, if (I.47) is the same as the last equation in 40 votes answer, the exponent has to be $-\mu_k$ and the fraction has to be turned around. What am I missing? $\zeta$ isn't constant ?! –  Jul 18 '13 at 13:10