Question 7.3.4: Using the Schwarz-Christoffel Formula. Use the Schwarz-Chris......
Using the Schwarz-Christoffel Formula
Use the Schwarz-Christoffel formula (6) to construct a conformal mapping
from the upper half-plane onto the nonpolygonal region defined by v ≥ 0,
with the horizontal half-line v = π, −\infty < u ≤ 0, deleted. See Figure 7.26(c).
f^{\prime}(z)=A\left(z-x_{1}\right)^{(\alpha_{1}/\pi)-1}(z-x_{2})^{(\alpha_{2}/\pi)-1}\cdot\cdot\cdot(z-x_{n})^{(\alpha_{n}/\pi)-1}, (6)
Let u_{0} be a point on the nonpositive u-axis in the w-plane. We can approximate the non-polygonal region defined by v ≥ 0, with the half-line v = π, −\infty < u ≤ 0, deleted by the polygonal region whose boundary consists of the horizontal half-line v = π, −\infty < u ≤ 0, the line segment from πi to u_{0}, and the horizontal half-line v = 0, u_{0} ≤ u ≤ \infty. The vertices of this polygonal region are w_{1} = πi and w_{2} = u_{0}, with corresponding interior angles α_{1} \text{and} α_{2}. See Figure 7.26(b). If we choose the points z_{1} = −1 \text{and} z_{2} = 0 to map onto the vertices w_{1} = πi \text{and} w_{2} = u_{0}, respectively, then (6) gives the derivative
A\left(z+1\right)^{(\alpha_{1}/\pi)-1}z^{(\alpha_{2}/\pi)-1}. (14)
Observe in Figure 7.26(b) that as u_{0} approaches −\infty along the u-axis, the interior angle α1 approaches 2π and the interior angle α_{2} approaches 0. With these limiting values, (14) suggests that our desired mapping f has derivative
f^{\prime}(z)=A\left(z+1\right)^{1}z^{-1}=A\left(1+\frac{1}{z}\right). (15)
An antiderivative of the function in (15) is given by
f(z)=A\left(z+\mathrm{Ln}\,z\right)+B, (16)
where A and B are complex constants.
In order to determine the appropriate values of the constants A and B, we first consider the mapping g(z) = z +Lnz on the upper half-plane y ≥ 0. The function g has a point of discontinuityat z = 0; thus, we will consider separatelythe boundaryhalf-lines y = 0, −\infty < x < 0, and y = 0, 0 < x < \infty, of the half-plane y ≥ 0. If z = x + 0i is on the half-line y = 0, −\infty < x < 0, then Arg(z) = π, and so g(z) = x + log_{e}|x| + iπ.
When x < 0, x + log_{e}|x| takes on all values from −\infty to −1. Thus, the image of the negative x-axis under g is the horizontal half-line v = π, −\infty < u < −1. On the other hand, if z = x + 0i is on the half-line y = 0, 0 < x < \infty, then Arg(z) = 0, and so g(z) = x + log_{e}|x|. When x > 0, x + log_{e} |x| takes on all values from −\infty to \infty . Therefore, the image of the positive x-axis under g is the u-axis. It follows that the image of the half-plane y ≥ 0 under g(z) = z+ Lnz is the region defined by v ≥ 0, with the horizontal half-line v = π, −∞ < u < −1 deleted. In order to obtain the region shown in Figure 7.26(c), we should compose g with a translation by 1. Therefore, the desired mapping is given by
f(z)=z+\mathrm{Ln}\,(z)+1.