INTRODUCTION

In Section 20.1 we saw that a nonconstant linear mapping f(z) = az + b, a and b complex numbers, acts by rotating, magnifying, and translating points in the complex plane. As a result, it is easily shown that the angle between any two intersecting curves in the z-plane is equal to the angle between the images of the arcs in the w-plane under a linear mapping. Other complex mappings that have this angle-preserving property are the subject of our study in this section.

Angle-Preserving Mappings

A complex mapping w = f(z) defined on a domain D is called conformal at z = z₀ in D when f preserves the angles between any two curves in D that intersect at z₀. More precisely, if C₁ and C₂ intersect in D at z₀, and and are the corresponding images in the w-plane, we require that the angle θ between C₁ and C₂ equal the angle ϕ between and . See FIGURE 20.2.1.

These angles can be computed in terms of tangent vectors to the curves. If and denote tangent vectors to curves C₁ and C₂, respectively, then, applying the law of cosines to the triangle determined by and , we have

or (1)

Likewise, if and denote tangent vectors to curves and , respectively, then

(2)

The next theorem gives a simple condition that guarantees that θ = ϕ.

PROOF:

If a curve C in D is parameterized by z = z(t), then w = f(z(t)) describes the image curve in the w-plane. Applying the Chain Rule to w = f(z(t)) gives w′ = f′(z(t))z′(t). If curves C₁ and C₂ intersect in D at z₀, then = f′(z₀) and = f′(z₀). Since f′(z₀) ≠ 0, we can use (2) to obtain

We can apply the laws of absolute value to factor out f′(z₀)² in the numerator and denominator and obtain

Therefore, from (1), ϕ = θ. ≡

EXAMPLE 1 Conformal Mappings

If f′(z₀) = 0 but f″(z₀) ≠ 0, it is possible to show that f doubles the angle between any two curves in D that intersect at z = z₀. The next two examples will introduce two important complex mappings that are conformal at all but a finite number of points in their domains.

EXAMPLE 2 f(z) = sin z as a Conformal Mapping

The vertical strip −π/2 ≤ x ≤ π/2 is called the fundamental region of the trigonometric function w = sin z. A vertical line x = a in the interior of this region can be described by z(t) = a + it, −∞ < t < ∞. From (6) in Section 17.7, we have

sin z = sin x cosh y + i cos x sinh y

and so u + iv = sin(a + it) = sin a cosh t + i cos a sinh t.

From the identity cosh²t − sinh²t = 1, it follows that

The image of the vertical line x = a is therefore a hyperbola with ±sin a as u-intercepts, and since −π/2 < a < π/2, the hyperbola crosses the u-axis between u = −1 and u = 1. Note that if a = −π/2, then w = −cosh t, and so the line x = −π/2 is mapped onto the interval (−∞, −1] on the negative u-axis. Likewise, the line x = π/2 is mapped onto the interval [1, ∞) on the positive u-axis.

A similar argument establishes that the horizontal line segment described by z(t) = t + ib, −π/2 < t < π/2, is mapped onto either the upper portion or the lower portion of the ellipse

according to whether b > 0 or b < 0. These results are summarized in FIGURE 20.2.2(b), which shows the mapping by f(z) = sin z. Note that we have carefully used capital letters to indicate where portions of the boundary are mapped. Thus, for example, boundary segment AB is transformed to A′B′.

Since f′(z) = cos z, f is conformal at all points in the region except z = ±π/2. The hyperbolas and ellipses are therefore orthogonal since they are images of the orthogonal families of horizontal segments and vertical lines. Note that the 180° angle at z = −π/2 formed by segments AB and AC is doubled to form a single line segment at w = −1. ≡

EXAMPLE 3 f(z) = z + 1/z as a Conformal Mapping

The complex mapping f(z) = z + 1/z is conformal at all values of z except z = ±1 and z = 0. In particular, the function is conformal for all values of z in the upper half-plane that satisfy z > 1. If z = re^iθ, then w = re^iθ + (1/r)e^−iθ, and so

(3)

Note that if r = 1, then v = 0 and u = 2 cos θ. Therefore, the semicircle z = e^it, 0 ≤ t ≤ π, is mapped to the segment [−2, 2] on the u-axis. It follows from (3) that if r > 1, then the semicircle z = re^it, 0 ≤ t ≤ π, is mapped onto the upper half of the ellipse u²/a² + v²/b² = 1, where a = r + 1/r and b = r − 1/r. See FIGURE 20.2.3 for the mapping by f(z) = z + 1/z.

For a fixed value of θ, the ray z = te^iθ, for t ≥ 1, is mapped to the portion of the hyperbola u²/cos²θ − v²/sin²θ = 4 in the upper half-plane v ≥ 0. This follows from (3), since

Since f is conformal for z > 1 and a ray θ = θ₀ intersects a circle z = r at a right angle, the hyperbolas and ellipses in the w-plane are orthogonal. ≡

Conformal Mappings Using Tables

Conformal mappings are given in Appendix D. The mappings have been categorized as elementary mappings (E-1 to E-9), mappings to half-planes (H-1 to H-6), mappings to circular regions (C-1 to C-5), and miscellaneous mappings (M-1 to M-10). Some of these complex mappings will be derived in Sections 20.3 and 20.4.

The entries indicate not only the images of the region R but also the images of various portions of the boundary of R. This will be especially useful when we attempt to solve boundary-value problems using conformal maps. You should use the appendix much like you use integral tables to find antiderivatives. In some cases, a single entry can be used to find a conformal mapping between two given regions R and R′. In other cases, successive transformations may be required to map R to R′.

EXAMPLE 4 Using a Table of Conformal Mappings

EXAMPLE 5 Using a Table of Conformal Mappings

Harmonic Functions and the Dirichlet Problem

A bounded harmonic function u = u(x, y) that takes on prescribed values on the entire boundary of a region R is called a solution to a Dirichlet problem on R. In Chapters 13–15 we introduced a number of techniques for solving Laplace’s equation in the plane, and we interpreted the solution to a Dirichlet problem as the steady-state temperature distribution in the interior of R that results from the fixed temperatures on the boundary.

There are at least two disadvantages to the Fourier series and integral transform methods presented in Chapters 13–15. The methods work only for simple regions in the plane and the solutions typically take the form of either infinite series or improper integrals. As such, they are difficult to evaluate. In Section 17.5 we saw that the real and imaginary parts of an analytic function are both harmonic. Since we have a large stockpile of analytic functions, we can find closed-form solutions to many Dirichlet problems and use these solutions to sketch the isotherms and lines of flow of the temperature distribution.

We will next show how conformal mappings can be used to solve a Dirichlet problem in a region R once the solution to the corresponding Dirichlet problem in the image region R′ is known. The method depends on the following theorem.

PROOF:

We will give a proof for the special case in which D′ is simply connected. If U has a harmonic conjugate V in D′, then H = U + iV is analytic in D′, and so the composite function H(f(z)) = U(f(z)) + iV(f(z)) is analytic in D. By Theorem 17.5.3, it follows that the real part U(f(z)) is harmonic in D, and the proof is complete.

To establish that U has a harmonic conjugate, let g(z) = ∂U/∂x − i ∂U/∂y. The first Cauchy–Riemann equation (∂/∂x)(∂U/∂x) = (∂/∂y)(−∂U/∂y) is equivalent to Laplace’s equation ∂²U/∂x² + ∂²U/∂y² = 0, which is satisfied because U is harmonic in D′. The second Cauchy–Riemann equation (∂/∂y)(∂U/∂x) = −(∂/∂x)(−∂U/∂y) is equivalent to the equality of the second- order mixed partial derivatives. Therefore, g(z) is analytic in the simply connected domain D′ and so, by Theorem 18.3.3, has an antiderivative G(z). If G(z) = U₁ + iV₁, then g(z) = G′(z) = ∂U₁/∂x − i ∂U₁/∂y. Since g(z) = ∂U/∂x − i ∂U/∂y, it follows that U and U₁ have equal first partial derivatives. Therefore, H = U + iV₁ is analytic in D′, and so U has a harmonic conjugate in D′. ≡

Theorem 20.2.2 can be used to solve a Dirichlet problem in a region R by transforming the problem to a region R′ in which the solution U either is apparent or has been found by prior methods (including the Fourier series and integral transform methods of Chapters 13–15). The key steps are summarized next.

Solving Dirichlet Problems Using Conformal Mapping

1. Find a conformal mapping w = f(z) that transforms the original region R onto the image region R′. The region R′ may be a region for which many explicit solutions to Dirichlet problems are known.
2. Transfer the boundary conditions from the boundary of R to the boundary of R′. The value of u at a boundary point ξ of R is assigned as the value of U at the corresponding boundary point f(ξ). See FIGURE 20.2.4 for an illustration of transferring boundary conditions.

   

3. Solve the corresponding Dirichlet problem in R′. The solution U may be apparent from the simplicity of the problem in R′ or may be found using Fourier or integral transform methods. (Additional methods will be presented in Sections 20.3 and 20.5.)
4. The solution to the original Dirichlet problem is u(x, y) = U(f(z)).

EXAMPLE 6 Solving a Dirichlet Problem

The function U(u, v) = (1/π)Arg w is harmonic in the upper half-plane v > 0 since it is the imaginary part of the analytic function g(w) = (1/π)Ln w. Use this function to solve the Dirichlet problem in FIGURE 20.2.5(a).

SOLUTION

The analytic function f(z) = sin z maps the original region to the upper half-plane v ≥ 0 and maps the boundary segments to the segments shown in Figure 20.2.5(b). The harmonic function U(u, v) = (1/π) Arg w satisfies the transferred boundary conditions U(u, 0) = 0 for u > 0 and U(u, 0) = 1 for u < 0. Therefore, u(x, y) = U(sin z) = (1/π)Arg(sin z) is the solution to the original problem. If tan⁻¹(v/u) is chosen to lie between 0 and π, the solution can also be written as

≡

EXAMPLE 7 Solving a Dirichlet Problem

From C-1 in Appendix D of conformal mappings, the analytic function f(z) = (z − a)/(az − 1), where a = (7 + 2)/5, maps the region outside the two open disks z < 1 and z − < onto the annular region r₀ ≤ w ≤ 1, where r₀ = 5 − 2. FIGURE 20.2.6(a) shows the original Dirichlet problem, and Figure 20.2.6(b) shows the transferred boundary conditions.

In Problem 12 in Exercises 14.1, we discovered that U(r, θ) = (log_er)/(log_er₀) is the solution to the new Dirichlet problem. From Theorem 20.2.2 we can conclude that the solution to the original boundary-value problem is

≡

A favorite image region R′ for a simply connected region R is the upper half-plane y ≥ 0. For any real number a, the complex function Ln(z − a) = log_ez − a + i Arg(z − a) is analytic in R′. Therefore, Arg(z − a) is harmonic in R′ and is a solution to the Dirichlet problem shown in FIGURE 20.2.7.

It follows that the solution in R′ to the Dirichlet problem with

is the harmonic function U(x, y) = (c₀/π)(Arg(z − b) − Arg(z − a)). A large number of Dirichlet problems in the upper half-plane y ≥ 0 can be solved by adding together harmonic functions of this form.

20.2 Exercises Answers to selected odd-numbered problems begin on page ANS-48.

In Problems 1–6, determine where the given complex mapping is conformal.

1. f(z) = z³ − 3z + 1
2. f(z) = cos z
3. f(z) = z + e^z + 1
4. f(z) = z + Ln z + 1
5. f(z) = (z² − 1)^1/2
6. f(z) = πi − [Ln(z + 1) + Ln(z − 1)]

In Problems 7–10, use the results in Examples 2 and 3.

7. Use the identity cos z = sin(π/2 − z) to find the image of the strip 0 ≤ x ≤ π under the complex mapping w = cos z. What is the image of a horizontal line in the strip?
8. Use the identity sinh z = −i sin(iz) to find the image of the strip −π/2 ≤ y ≤ π/2, x ≥ 0, under the complex mapping w = sinh z. What is the image of a vertical line segment in the strip?
9. Find the image of the region defined by −π/2 ≤ x ≤ π/2, y ≥ 0, under the complex mapping w = (sin z)^1/4. What is the image of the line segment [−π/2, π/2] on the x-axis?
10. Find the image of the region z ≤ 1 in the upper half-plane under the complex mapping w = z + 1/z. What is the image of the line segment [−1, 1] on the x-axis?

In Problems 11–18, use the conformal mappings in Appendix D to find a conformal mapping from the given region R in the z-plane onto the target region R′ in the w-plane, and find the image of the given boundary curve.

In Problems 19–22, use an appropriate conformal mapping and the harmonic function U = (1/π)Arg w to solve the given Dirichlet problem.

In Problems 23–26, use an appropriate conformal mapping and the harmonic function U = (c₀/π)[Arg(w − 1) − Arg(w + 1)] to solve the given Dirichlet problem.

23. 

24. 

25. 

26. 

27. A real-valued function ϕ(x, y) is called biharmonic in a domain D when the fourth-order differential equation

    

    at all points in D. Examples of biharmonic functions are the Airy stress function in the mechanics of solids and velocity potentials in the analysis of viscous fluid flow.

    1. Show that if ϕ is biharmonic in D, then u = ∂²ϕ/∂x² + ∂²ϕ/∂y² is harmonic in D.
    2. If g(z) is analytic in D and ϕ(x, y) = Re( g(z)), show that ϕ is biharmonic in D.

*It is also possible to prove that f preserves the sense of direction between the tangent vectors.