Schwarz-Christoffel transformation (circular version)

The complex-variables function

$$f(z)={\int }_0^z\prod _{k=1}^n{(\zeta -z_k)}^{{\alpha }_k-1}d\zeta ,$$

maps the closed unit disc $\overline{D}=\{|z|\le 1\}$ in the complex plane conformally onto a polygon with $n$ sides, interior angles , and vertices $f(z_k)$. (The polygon is assumed to be not self-intersecting.) The parameters $z_k$ lie on the unit circle, and depend, generally in a complicated way, on the length of the sides of the polygon.

The fractional powers ${(\zeta -z_k)}^{{\alpha }_k-1}$ serve to clamp up an arc of the circle into a pointy angle of measure ${\alpha }_k\pi$. Indeed, the proof of the Schwarz-Christoffel formula shows that the function $f$ can be decomposed near $z_k$ as

$$f(z)=f(z_k)+{(z-z_k)}^{{\alpha }_k}{g}_k(z),$$

where $g_k$ is an analytic function with $g_k(z_k)\ne 0$. See Figure 1.

Figure 1: Mapping in a neighborhood of a boundary point

Note that the exponent is ${\alpha }_k$ — not ${\alpha }_k/2$ — because the neigbourhood of a point $z_k$ in the domain space looks like a half-disc. For the same reason, the fractional power used in the formula is to be a single-valued branch continuous on the half-disc. Finally, the extra $-1$ exponents that appear in the integral representation for $f$ come from the power rule for differentiation.

0.1 Example: $n=3$

Figure 2 illustrates a mapping from the disc to a triangle ($n=3$). The contours are the approximate images, under $f$, of circles of radius .

Figure 2: Image of a Schwarz-Christoffel mapping for a triangle

Figure 3: Corresponding contours of the domain

We describe the method used to compute the figure. Points in the domain $\overline{D}$ are first parameterized as $z=r{e}^{i\theta }$, with $0\le r\le 1$ and ranging over a discrete grid, shown schematically in Figure 4. The integral defining the function $f$ is path-independent, and a natural choice for the paths are rays emanating from the origin. When computing the integrals along each ray, we exploit the additivity of the complex path integral:

$$f((r+\mathrm{\Delta }r)e^{i\theta })=f(r{e}^{i\theta })+{\int }_{r{e}^{i\theta }}^{(r+\mathrm{\Delta }r)e^{i\theta }}\prod _{k=1}^n{(\zeta -z_k)}^{{\alpha }_k-1}d\zeta ,$$

so that $f(z)$ is found by summing a previously-computed value and a new integral to be computed. And the new integral is computed using 32-point Gauss quadrature after reparameterizing the path with $d\zeta =e^{i\theta }dr$.

Figure 4: Parameterization of the domain for computing $f$

The computation of the integrand

$$\prod _{k=1}^n{(\zeta -z_k)}^{{\alpha }_k-1}=\exp \left(\sum _{k=1}^n({\alpha }_k-1)\log (\zeta -z_k)\right)$$

is straightforward, though we must be careful to respect the branch cuts prescribed above. The $\log$ function in most computer languages takes a branch cut on the negative axis. To get the single-valued branches we need in this situation, we must instead compute $\zeta \mapsto \log (\zeta -z_k)$ via the expression

$$\zeta \mapsto \log i{z}_k+\log \frac{\zeta -z_k}{i{z}_k},$$

where $i{z}_k$ is the direction of the tangent to the circle at the point $z_k$.

Finally, after having obtained a discrete set of image points $f(z)$ traced along each circle $z=r{e}^{i\theta }$, the contours in the figure are obtained by interpolating a curved Bézier spline through the image points.

If a triangle is prescribed with the vertex locations, it is not immediately obvious what the parameters $z_k$ should be to obtain that triangle. In the examples here, we simply avoid this difficulty by arbitrarily choosing the parameters $z_k=e^{2\pi i(k-1)/n}$ to be equally spaced on the unit circle, and hope that nice figures result.

The ${\alpha }_k$ parameters are easily determined from the angles of the desired figure; they are, in this example:

$${\alpha }_1=\frac{1}{4},{\alpha }_2=\frac{1}{2},{\alpha }_3=\frac{1}{4}.$$

0.2 Example: $n=10$

Figure 5 shows an example with $n=10$ points. The strategy for computing this figure is similar to that of the triangle.

The parameters for this star are (rounded to four decimal places):

${\alpha }_1=0.2422,{\alpha }_2={\alpha }_1=1.3263,{\alpha }_3={\alpha }_9=0.3026,$

${\alpha }_4={\alpha }_8=1.3026,{\alpha }_5={\alpha }_7=0.2754,{\alpha }_6=1.3440.$

Figure 5: Image of a Schwarz-Christoffel mapping for a star-shaped figure

0.3 Demonstration computer programs

• http://svn.gold-saucer.org/repos/PlanetMath/SchwarzChristoffelTransformationCircularVersion/schwarz-christoffel.pyPython source code for producing images of the Schwarz-Christoffel transformation

• http://svn.gold-saucer.org/repos/PlanetMath/SchwarzChristoffelTransformationCircularVersion/explanation.pyPython source code for the explanatory diagrams

Title	Schwarz-Christoffel transformation (circular version)
Canonical name	SchwarzChristoffelTransformationcircularVersion
Date of creation	2013-03-22 16:52:57
Last modified on	2013-03-22 16:52:57
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	8
Author	stevecheng (10074)
Entry type	Result
Classification	msc 31A99
Classification	msc 30C20