20.4 Schwarz–Christoffel Transformations

INTRODUCTION

If D′ is a simply connected domain with at least one boundary point, then the famous Riemann mapping theorem asserts the existence of an analytic function g that conformally maps the unit open disk z < 1 onto D′. The Riemann mapping theorem is a pure existence theorem that does not specify a formula for the conformal mapping. Since the upper half-plane y > 0 can be conformally mapped onto this disk using a linear fractional transformation, it follows that there exists a conformal mapping f between the upper half-plane and D′. In particular, there are analytic functions that map the upper half-plane onto polygonal regions of the types shown in FIGURE 20.4.1. Unlike the Riemann mapping theorem, the Schwarz–Christoffel specifies a form for the derivative f′(z) of a conformal mapping from the upper half-plane to a bounded or unbounded polygonal region. This formula is named for the Polish mathematician Karl Hermann Amandus Schwarz (1843–1921) and the German mathematician/physicist Elwin Bruno Christoffel (1829–1900), who discovered it independently in the late 1860s.

Two images. The first image shows a polygon with five vertices labeled z subscript 1, z subscript 2, z subscript 3, z subscript 4, and z subscript 5. The first four angles are labeled alpha subscript 1, alpha subscript 2, alpha subscript 3, and alpha subscript 4, respectively. The region bound by the polygon is shaded. The second image shows an unbound polygon with one side open. Its four vertices are labeled z subscript 1, z subscript 2, z subscript 3, and z subscript 4. Its four angles labeled alpha subscript 1, alpha subscript 2, alpha subscript 3, and alpha subscript 4, respectively. The unbound region is shaded.

FIGURE 20.4.1 Polygonal regions

Special Cases

To motivate the general Schwarz–Christoffel formula, we first examine the effect of the mapping f(z) = (z − x1)α/π, 0 < α < 2π, on the upper half-plane y ≥ 0 shown in FIGURE 20.4.2(a). This mapping is the composite of the translation ζ = z − x1 and the real power function w = ζα/π. Since w = ζα/π changes the angle in a wedge by a factor of α/π, the interior angle in the image region is (α/π)π = α. See Figure 20.4.2(b).

Two images. The first image shows a line segment labeled A B with center labeled x subscript 1. The graph also shows a 180-degree angle labeled pi. The upper plane of the line is shaded. The second image shows an obtuse angle at 0 labeled alpha. The sides are labeled A prime 0 and 0 B. The region bound by the angle is shaded.

FIGURE 20.4.2 Image of upper half-plane

Note that f′(z) = A(z − x1)(α/π)−1 for A = α/π. Next assume that f(z) is a function that is analytic in the upper half-plane and that has the derivative

(1)

where x1 < x2. In determining the images of line segments on the x-axis, we will use the fact that a curve w = w(t) in the w-plane is a line segment when the argument of its tangent vector w′(t) is constant. From (1), an argument of f′(t) is given by

(2)

Since Arg(t − x) = π for t < x, we can find the variation of arg f′(t) along the x-axis. The results are shown in the following table.

A table has three columns and three rows. The columns are as follows: Column 1, Interval. Column 2, arg F prime (t). Column 3, Change in Argument. The row entries are as follows: Row 1. Interval, (minus infinite, x subscript 1). arg F prime (t). Arg A plus (alpha subscript 1 minus pi) plus (alpha subscript 2 minus pi). Change in Argument, 0. Row 2. Interval, (x subscript 1, x subscript 2). arg F prime (t), Arg A plus (alpha subscript 2 minus pi). Change in Argument, pi minus alpha subscript 1. Row 3. Interval, (x subscript 2, infinite). arg F prime (t), Arg A. Change in Argument, pi minus alpha subscript 2.

Since arg f′(t) is constant on the intervals in the table, the images are line segments, and FIGURE 20.4.3 shows the image of the upper half-plane. Note that the interior angles of the polygonal image region are α1 and α2. This discussion generalizes to produce the Schwarz–Christoffel formula.

An image of an unbound polygon with one side open. The three sides are labeled w = f (t), t is greater than x subscript 2, w = f(t), x subscript 1 is less than t is less than x subscript 2, and w = f(t), t is less than x subscript 1. The two angles are labeled alpha subscript 1 and alpha subscript 2. The adjacent exterior angle of alpha subscript 1 is pi minus alpha subscript 1 and alpha subscript 2 is alpha subscript 2 is pi minus alpha subscript 2. The interior of the polygon is shaded.

FIGURE 20.4.3 Image of upper half-plane

THEOREM 20.4.1 Schwarz–Christoffel Formula

Let f(z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative

(3)

where x1 < x2 < ... < xn and each αi satisfies 0 < αi < 2π. Then f(z) maps the upper half-plane y ≥ 0 to a polygonal region with interior angles α1, α2, ..., αn.

In applying formula (3) to a particular polygonal target region, the reader should carefully note the following comments:

  1. One can select the location of three of the points xk on the x-axis. A judicious choice can simplify the computation of f(z). The selection of the remaining points depends on the shape of the target polygon.
  2. A general formula for f(z) is

    and therefore f(z) may be considered as the composite of the conformal mapping

    and the linear function w = Az + B. The linear function w = Az + B allows us to magnify, rotate, and translate the image polygon produced by g(z). (See Section 20.1.)

  3. If the polygonal region is bounded, only n − 1 of the n interior angles should be included in the Schwarz–Christoffel formula. As an illustration, the interior angles α1, α2, α3, and α4 are sufficient to determine the Schwarz–Christoffel formula for the pentagon shown in Figure 20.4.1(a).

EXAMPLE 1 Constructing a Conformal Mapping

Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane to the strip v ≤ 1, u ≥ 0.

SOLUTION

We may select x1 = −1 and x2 = 1 on the x-axis, and we will construct a conformal mapping f with f(−1) = −i and f(1) = i. See FIGURE 20.4.4. Since α1 = α2 = π/2, the Schwarz–Christoffel formula (3) gives

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis and a point labeled minus 1 on the negative horizontal axis. A horizontal line passes through the horizontal axis labeled A on the extreme left of the negative horizontal axis and B on the extreme right of the positive horizontal axis. The first and second quadrants are shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line labeled B prime intersecting the positive vertical axis at i and a horizontal line labeled A prime intersecting the negative vertical axis at minus 1. The angles forming at i and minus i is pi over 2. The region bound by the two horizontal lines is shaded.

FIGURE 20.4.4 Image of upper half-plane in Example 1

Therefore, f(z) = −Ai sin−1z + B. Since f(−1) = −i and f(1) = i, we obtain, respectively,

and conclude that B = 0 and A = −2/π. Thus, . ≡

EXAMPLE 2 Constructing a Conformal Mapping

Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane to the region shown in FIGURE 20.4.5(b).

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis and a point labeled minus 1 on the negative horizontal axis. A horizontal line passes through the horizontal axis labeled A on the extreme left of the negative horizontal axis and B on the extreme right of the positive horizontal axis. The first and second quadrants are shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line labeled B prime passes through the positive horizontal axis, a vertical line passes through the positive vertical axis from the origin at a i, and a horizontal line labeled A prime intersecting the positive vertical axis at a i in the second quadrant. The upper part of the horizontal line is shaded.

FIGURE 20.4.5 Image of upper half-plane in Example 2

SOLUTION

We again select x1 = −1 and x2 = 1, and we will require that f(−1) = ai and f(1) = 0. Since α1 = 3π/2 and α2 = π/2, the Schwarz–Christoffel formula (3) gives

f′(z) = A(z + 1)1/2(z − 1)−1/2.

If we write f′(z) as A(z/(z2 − 1)1/2 + 1/(z2 − 1)1/2), it follows that

f(z) = A[(z2 − 1)1/2 + cosh−1z] + B.

Note that cosh−1(−1) = πi and cosh−1 1 = 0, and so ai = f(−1) = A(πi) + B and 0 = f(1) = B. Therefore, A = a/π and . ≡

The next example will show that it may not always be possible to find f(z) in terms of elementary functions.

EXAMPLE 3 Constructing a Conformal Mapping

Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane to the interior of the equilateral triangle shown in FIGURE 20.4.6(b).

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis and a point labeled 0 at the origin. A horizontal line passes through the horizontal axis labeled A on the extreme left of the negative horizontal axis and B on the extreme right of the positive horizontal axis. The first and second quadrants are shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows an increasing line labeled A prime from the origin labeled 0 to the center, a line with negative slope labeled B’ from the horizontal axis at a point labeled 1 merging with the line A prime at the center, and a horizontal line passes through the horizontal axis from 0 to 1 forming a triangle. The region bound by the triangle is shaded.

FIGURE 20.4.6 Image of upper half-plane in Example 3

SOLUTION

Since the polygonal region is bounded, only two of the three 60° interior angles should be included in the Schwarz–Christoffel formula. If x1 = 0 and x2 = 1, we obtain f′(z) = Az−2/3(z − 1)−2/3. It is not possible to evaluate f(z) in terms of elementary functions; however, we can use Theorem 18.3.3 to construct the antiderivative

If we require that f(0) = 0 and f(1) = 1, it follows that B = 0 and

It can be shown that this last integral is , where Γ denotes the gamma function. Therefore, the required conformal mapping is

≡

The Schwarz–Christoffel formula can sometimes be used to suggest a possible conformal mapping from the upper half-plane onto a nonpolygonal region R′. A key first step is to approximate R′ by polygonal regions. This will be illustrated in the final example.

EXAMPLE 4 Constructing a Conformal Mapping

Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane to the upper half-plane with the horizontal line v = π, u ≤ 0, deleted.

SOLUTION

The nonpolygonal target region can be approximated by a polygonal region by adjoining a line segment from w = πi to a point u0 on the negative u-axis. See FIGURE 20.4.7(b). If we require that f(−1) = πi and f(0) = u0, the Schwarz–Christoffel transformation satisfies

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled minus 1 on the negative horizontal axis and a point labeled 0 at the origin. A horizontal line passes through the horizontal axis labeled A on the extreme left of the negative horizontal axis and B on the extreme right of the positive horizontal axis. The first and second quadrants are shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line passes through the horizontal axis from the second quadrant to the first quadrant and the line lies in the first quadrant is labeled B prime. The graph shows an increasing line at a point labeled u subscript 0 at the extreme left of the negative horizontal axis intersecting at the center of the positive vertical axis and a horizontal line labeled A prime in the second quadrant intersecting the positive vertical axis at the same point. The horizontal line is also labeled with v = pi. The angle forming at u subscript 0 is labeled alpha subscript 2 and the exterior angle forming at the positive vertical axis is labeled alpha subscript 1. The two sections are shaded differently.

FIGURE 20.4.7 Image of upper half-plane in Example 4

Note that as u0 approaches −∞, the interior angles α1 and α2 approach 2π and 0, respectively. This suggests we examine conformal mappings that satisfy w′ = A(z + 1)1z−1 = A(1 + 1/z) or w = A(z + Ln z) + B.

We will first determine the image of the upper half-plane under g(z) = z + Ln z and then translate the image region if needed. For t real,

g(t) = t + loget + i Arg t.

If t < 0, Arg t = π and u(t) = t + loget varies from −∞ to −1. It follows that w = g(t) moves along the line v = π from −∞ to −1. When t > 0, Arg t = 0 and u(t) varies from −∞ to ∞. Therefore, g maps the positive x-axis onto the u-axis. We can conclude that g(z) = z + Ln z maps the upper half-plane onto the upper half-plane with the horizontal line v = π, u ≤ −1, deleted. Therefore, w = z + Ln z + 1 maps the upper half-plane onto the original target region. ≡

Many of the conformal mappings in Appendix D can be derived using the Schwarz–Christoffel formula, and we will show in Section 20.6 that these mappings are especially useful in analyzing two-dimensional fluid flows.

20.4 Exercises Answers to selected odd-numbered problems begin on page ANS-49.

In Problems 1–4, use (2) to describe the image of the upper half-plane y ≥ 0 under the conformal mapping w = f(z) that satisfies the given conditions. Do not attempt to find f(z).

  1. f′(z) = (z − 1)−1/2, f(1) = 0
  2. f′(z) = (z + 1)−1/3, f(−1) = 0
  3. f′(z) = (z + 1)−1/2(z − 1)1/2, f(−1) = 0
  4. f′(z) = (z + 1)−1/2(z − 1)−3/4, f(−1) = 0

In Problems 5–8, find f′(z) for the given polygonal region using x1 = −1, x2 = 0, x3 = 1, x4 = 2, and so on. Do not attempt to find f(z).

  1. f(−1) = 0, f(0) = 1
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled 1 at the extreme right of the horizontal axis and a point labeled i at the top of the vertical axis. The graph shows a horizontal line passes through the horizontal axis from origin to 1, a vertical line intersecting the horizontal axis at 1, and a horizontal line from i intersecting the vertical line forming a square. The region bound by the square is shaded.

    FIGURE 20.4.8 Polygonal region for Problem 5

  2. f(−1) = −1, f(0) = 0
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled minus 1 on the negative horizontal axis. The graph shows a horizontal line passes through the negative horizontal axis from origin to minus 1, an increasing line from minus 1 intersecting at the top of the positive vertical axis, and a vertical line passes through the positive vertical axis from origin to the same point forming a right triangle right angled at the origin. The adjacent angle is labeled pi over 3. The region bound by the triangle is shaded.

    FIGURE 20.4.9 Polygonal region for Problem 6

  3. f(−1) = −1, f(0) = 1
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled minus 1 on the negative horizontal axis and a point labeled 1 on the positive horizontal axis. The graph shows a horizontal line passes through horizontal axis from minus 1 to 1, an increasing line from 1 to top right, and a line with negative slope from minus 1 to top left. The angles forming at 1 and minus 1 are labeled 2 pi over 3. The region bound by the lines is shaded.

    FIGURE 20.4.10 Polygonal region for Problem 7

  4. f(−1) = i, f(0) = 0
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line from the second quadrant intersecting the positive vertical axis at point labeled i, a vertical line passes through the positive vertical axis from origin to i, and an increasing line from the origin to the top right in the first quadrant. The angle adjacent to the horizontal axis in the first quadrant is labeled pi over 4. The upper section of the lines is shaded.

    FIGURE 20.4.11 Polygonal region for Problem 8

  5. Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane y ≥ 0 to the region in FIGURE 20.4.12. Require that f(−1) = πi and f(1) = 0.
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled pi i at the top of the vertical axis. The graph shows a horizontal line passes through the horizontal axis from the origin to the extreme right, a vertical line passes through the vertical axis from the origin to pi i, and a horizontal line intersecting the vertical axis at pi i. The region bound by the lines is shaded.

    FIGURE 20.4.12 Image of upper half-plane in Problem 9

  6. Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane y ≥ 0 to the region in FIGURE 20.4.13. Require that f(−1) = −ai and f(1) = ai.
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled a i at the center of the positive vertical axis and a point labeled minus a i at the center of the negative vertical axis. The graph also shows a vertical line passes through the vertical axis from a i to minus a i, a horizontal line intersecting at a i in the first quadrant, and another horizontal line intersecting at minus a i in the fourth quadrant. The exterior of the lines covering all the quadrants is shaded.

    FIGURE 20.4.13 Image of upper half-plane in Problem 10

  7. Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane y ≥ 0 to the horizontal strip 0 ≤ v ≤ π by first approximating the strip by the polygonal region shown in FIGURE 20.4.14. Require that f(−1) = πi, f(0) = w2 = −, and f(1) = 0, and let w1 → ∞ in the horizontal direction.
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line passes through the horizontal axis from the second quadrant to the first quadrant, a horizontal line intersecting the positive vertical axis at point labeled pi i parallel to the line on the horizontal axis. The graph also shows an increasing line from the origin to the center right and a decreasing line from the positive vertical axis merging the increasing line at w subscript 1. It also shows a line with negative slope from the origin to the center left and a line with positive slope from top of the positive vertical axis merging the line with negative slope at w subscript 2 forming a rhombus. The interior of the rhombus and the exterior of rhombus bound by the horizontal lines are shaded differently.

    FIGURE 20.4.14 Image of upper half-plane in Problem 11

  8. Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper half-plane y ≥ 0 to the wedge 0 ≤ Arg w ≤ π/4 by first approximating the wedge by the region shown in FIGURE 20.4.15. Require that f(0) = 0 and f(1) = 1 and let θ → 0.
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled 1 at the center of the horizontal axis. The graph shows an increasing line from the origin to top right and an increasing line from 1 to top right intersecting the other line. The exterior angle formed at 1 is labeled theta. The region bound by the lines and the region on the right of the line is shaded differently.

    FIGURE 20.4.15 Image of upper half-plane in Problem 12

  9. Verify M-4 in Appendix D by first approximating the region R′ by the polygonal region shown in FIGURE 20.4.16. Require that f(−1) = −u1, f(0) = ai, and f(1) = u1 and let u1 → 0 along the u-axis.
    A graph. The horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled u subscript 1 on the positive horizontal axis and a point labeled minus u subscript 1 on the negative horizontal axis close to the origin. A horizontal line passes through the horizontal axis from the first quadrant to the second quadrant. A decreasing line from the positive vertical axis at point labeled a i to u subscript 1, an increasing line from minus u subscript 1 intersecting at a i, and a vertical line passes through the vertical axis from the origin to a i forming a triangle. The interior and exterior of the triangle are shaded differently and labeled R prime.

    FIGURE 20.4.16 Image of upper half-plane in Problem 13

  10. Show that if a curve in the w-plane is parameterized by w = w(t), a ≤ t ≤ b, and arg w′(t) is constant, then the curve is a line segment. [Hint: If w(t) = u(t) + iv(t), then tan(arg w′(t)) = dv/du.]