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 polygonwith $n$ 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 sidesof the polygon.

The fractional powers ${(\zeta -z_k)}^{{\alpha }_k-1}$serve to clamp up an arc of thecircle 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 spacelooks like a half-disc.For the same reason, the fractional power used in the formulais to be a single-valued branch continuous on the half-disc.Finally, the extra $-1$ exponents that appear in the integralrepresentation for $f$come from the power rule for differentiation.

0.1 Example: $n=3$

Figure 2 illustrates a mappingfrom the disc to a triangle ($n=3$).The contours are the approximate images, under $f$, of circlesof 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 parameterizedas $z=r{e}^{i\theta }$,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 arerays 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 valueand a new integral to be computed. And the new integralis computed using 32-point Gauss quadratureafter 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, thoughwe must be careful to respect the branch cuts prescribed above.The $\log$ function in most computer languagestakes 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 acurved Bézier splinethrough 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 obtainthat triangle. In the examples here, we simply avoid this difficultyby 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 anglesof 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 5shows an example with $n=10$ points.The strategy for computing this figure is similar to thatof 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