Lecture 15 – the Riemann Mapping Theorem 1 Normal Families

Home , Conformal map, Region (mathematics), Unit disk

Lecture 15 – The Riemann mapping theorem MATH-GA 2451.001 Complex Variables

The point of this lecture is to prove that the unit disk can be mapped conformally onto any simply connected open set in the plane, other than the plane itself. We will then know that any two such simply connected open set can be mapped conformally onto each other, since we can map either of these sets to the unit disk.

The result above is known as the Riemann mapping theorem. We will prove it using basic theory of normal families. We start this lecture with that.

1 Normal families 1.1 Deﬁnition

Let Ω be an open connected set in C and F a family of analytic functions on Ω. F is said to be (i) normal or relatively compact if every sequence from F has a subsequence that converges uniformly on every compact subset of Ω (ii) locally uniformly bounded if for any compact subset K of Ω, there exists M > 0 such that |f(z)| ≤ M ∀f ∈ F , ∀z ∈ K. (iii) equicontinuous on a compact set K of Ω if ∀ > 0 , ∃δ > 0 such that

∀(z, w) ∈ K2 , |z − w| < δ ⇒ |f(z) − f(w)| < ∀f ∈ F

Observe that if there exists M > 0 such that ∀ z ∈ Ω , ∀ f ∈ F, |f 0(z)| ≤ M, then F is equicontinuous.

1.2 Equivalence of the deﬁnitions Theorem (Montel’s theorem): Suppose F is a locally uniformly bounded family of analytic functions in an open connected set Ω. Then F is normal.

Proof : The proof of Montel’s theorem can be decomposed into the following two steps:

(1) Lemma: Suppose F is a family of analytic functions uniformly bounded on compact subsets of Ω. Then F is equicontinuous on every compact subset of Ω.

To show this, let us consider a compact subset K of Ω. There exists r > 0 such that the set K˜ deﬁned by K˜ := {z ∈ C : d(z, K) ≤ 2r} is contained in Ω. Let M = sup |f(z)| f∈F,z∈K˜ and (z, w) ∈ K2 such that |z − w| < r.

The closed disk D2r(z) is contained in Ω. If we call γ the boundary of this disk, we may write

1 Z f(ζ) f(ζ) 1 Z f(ζ)(z − w) f(z) − f(w) = − dζ = dζ 2πi γ ζ − z ζ − w 2πi γ (ζ − w)(ζ − z)

∀ > 0, let δ = min (r/M, r). 1 4πr |z − w| < δ ⇒ |f(z) − f(w)| ≤ M|z − w| ≤ 2π 2r2 Hence F is equicontinuous on K. (2) Now that we know that F in Montel’s theorem is equicontinuous on every compact subset of Ω, we can prove the theorem.

1 Let (fn)n∈N be a sequence from F. Consider a sequence (wj)j∈N which is everywhere dense in Ω. The sequence (fn(w1))n∈N is bounded, so we can extract a convergent subsequence, and write the corresponding functions (fn,1)n∈N.(fn,1(w2))n∈N is bounded, so we can extract another convergent subsequence ∗ (fn,2)n∈N. Repeating the process, we obtain the nested sequence (fn,k)k∈N ,n∈N such that (fn,k(wk))n∈N converges for every wj with j ∈ 〚1, k〛.

Let gn = fn,n be the diagonal sequence of functions. gn converges at all the points (wn)n∈N. We now show that (gn)n ∈ N converges uniformly on every compact subset K of Ω. Let K be such a compact subset. Since F is equicontinuous on K, ∀ > 0 ∃δ > 0 such that ∀(z, w) ∈ K2 , |z − w| < δ ⇒ |f(z) − f(w)| < 3

N Since K is compact, we can select N points from (wj)j∈N such that K ⊂ ∪p=1Dδ(wp). By construction of the functions gn, for that choice of , there exists N ∈ N such that m, n ≥ N ⇒ ∀p ∈ 〚1,N〛, |gn(wp) − gm(wp)| < 3

from which we conclude that gn converges uniformly on K, as it is uniformly Cauchy on K.

2 The Riemann mapping theorem 2.1 Statement of the theorem Theorem (Riemann mapping theorem): Given any simply connected open set Ω which is not the whole plane, and a point z0 in Ω, there exists a unique analytic function f in Ω, normalized by the conditions 0 f(z0) = 0, f (z0) > 0, such that f deﬁnes a one-to-one mapping of Ω onto the disk |w| < 1.

2.2 Proving the theorem −1 Uniqueness: Suppose there are two such functions f1 and f2. Then f1 ◦ f2 is a one-to-one mapping of the −1 unit disk to itself. You will prove in Homework 9, using the lemma of Schwarz, that this implies that f1 ◦ f2 is a M¨obiustransformation of the form z − a S(z) = λ 1 − az with |λ| = 1 and a ∈ C such that |a| < 1. Then the normalization conditions S(0) = 0 ,S0(0) > 0 imply that S(z) = z, so that f1 ≡ f2.

Existence:

• Lemma: If Ω is a simply connected open set which is not C, there exists a one-to-one conformal map h :Ω 7→ h(Ω) such that h(Ω) does not intersect a disk Dδ(w0) for some w0 ∈ C and δ > 0.

To prove the lemma, observe that by hypothesis, there exists a ∈ C \ Ω. Hence ϕ(z) = z − a is a nonvanishing analytic function on Ω. From Lecture 9, we then know that we can deﬁne an analytic function h on Ω such that h2(z) := z − a. 2 Observe that h does not take the same value twice, nor opposite values. Indeed if (z1, z2) ∈ Ω are such that h(z1) = ±h(z2), then z1 − a = z2 − a ⇒ z1 = z2.

Now, by the open mapping theorem, ∀z0 ∈ Ω, h(Ω) contains a disk Dδ(h(z0)). This means that h(Ω) does not contain the disk Dδ(−h(z0)), which proves the lemma, with w0 = −h(z0).

• We now show that there are one-to-one, analytic functions g from Ω to D1(0) such that g(z0) = 0 and 0 g (z0) > 0.

2 To see that there is at least one such function, consider the function G0 on Ω deﬁned by

0 δ |h (z0)| h(z0) h(z) − h(z0) G0(z) = 2 0 4 |h(z0)| h (z0) h(z) + h(z0)

with δ and h as deﬁned in the previous lemma. G0 is one-to-one since h is and G0 is obtained from h by a linear fractional transformation.

0 0 δ |h (z0)| G0(z0) = 0 ,G0(z0) = 2 > 0 8 |h(z0)| Finally, observe that

δ 1 h(z) − h(z0) δ 1 2 ∀z ∈ Ω , |G0(z)| = = − < 1 4 |h(z0)| h(z) + h(z0) 4 h(z0) h(z) + h(z0) This proves our point.

• The last step of the proof is to show that within the family F of functions g discussed previously, there exists an f with maximal derivative, and that this f has all the properties required for the mapping theorem. The key is to show that f is surjective.

Observe that by Cauchy’s estimate on Dr(z0) ⊂ Ω, ∀ g ∈ F, 1 |g0(z )| ≤ 0 r

0 0 The set {|g (z0)| , g ∈ F} is therefore bounded, and has a supremum B = supg∈F |g (z0)|. There exists 0 a sequence (gn)n∈N ∈ F such that gn(z0) → B as n → +∞.

Now, since |gn| < 1 on Ω, by Montel’s theorem we know that there exists a subsequence (gnk ) of gn converging to a function f uniformly on every compact subset of Ω. By Weierstrass’ theorem, f is analytic. Furthermore,

f(z ) = lim g (z ) = 0 , |f 0(z )| = lim |g0 (z )| = B > 0 0 nk 0 0 nk 0 nk→∞ nk→∞ From the latter result, we can say that f is not a constant function.

Now, we take some z1 in Ω, and defineg ñk := gnk (z) − gnk (z1).g ñk is nonzero on Ω \{z1}. Hence, by ˜ Hurwitz’ theorem, f := f(z) − f(z1) is nonzero on Ω \{z1}. We have just proved that f is one-to-one on Ω. The remaining question is: is f onto?

Suppose it is not: ∃w0 ∈ D1(0) such that w0 ∈/ f(Ω). Then, as before, we can consider the single-valued function F on Ω such that f(z) − w F 2(z) := 0 (1) 1 − w0f(z) F is one-to-one and satisﬁes |F (z)| < 1 for z ∈ Ω. F can be normalized as follows:

0 |F (z0)| F (z) − F (z0) G(z) := 0 F (z0) 1 − F (z0)F (z)

G is one-to-one, satisﬁes |G(z)| < 1 ∀z ∈ Ω, and G(z0) = 0. Furthermore,

|F 0(z )| |f 0(z )| 1 1 + |w | G0(z ) = 0 = 0 (1 − |w |2) = |f 0(z )| 0 > |f 0(z )| 0 2 p 0 0 p 0 1 − |F (z0)| 2 |w0| 1 − |w0| 2 |w0|

This is a contradiction, so we may say that f is onto, which concludes the proof of the theorem

3 2.3 Returning to our definition of simple connectedness You may recall that in Lecture 9, we introduced an unusual definition of simple connectedness, only valid in R2, but very convenient for our purposes at the time. I told you then that we would eventually be able to prove the equivalence of the more conventional definition – that any simple closed curve can be shrunk to a point continuously in the set. We are now ready for this.

We have already shown in Lecture 9 that if every closed curve in Ω is contractible to a point, then Ω is simply connected in the sense of Ahlfors. It remains to prove that if Ω is simply connected in the sense of Ahlfors, every closed curve in Ω is contractible to a point. This is easy to see: based on the deﬁnition of simple connectedness of Ahlfors, we were able to construct the square root h of ϕ and the function F in (1), and thus prove the Riemann mapping theorem. Therefore, if Ω is simply connected in the sense of Ahlfors, it is homeomorphic to the unit disk D1(0), which is simply connected in the usual sense

3 Visualizing maps

A streamlined way to visualize the transformation of a set by a map is to consider the eﬀect of the map on a mesh of the set. This is what we do below.

• f(z) = z2 We know that f is conformal on C \{0}. We consider the mesh ( zx0 = x0 + iy

zy0 = x + iy0

for a countable set of x0 and y0. Any vertical line zx0 = x0 + iy is mapped to

w = z2 = u2 + iv2 = x2 − y2 + 2ix y x0 x0 x0 0 0 We see that the real and imaginary parts satisfy

v2 = 4x2y2 = 4x2(x2 − u) x0 0 0 0 This is the equation of a parabola in the w-plane, with focus (0, 0) and pointed in the negative direction.

2 2 Any horizontal line zy0 = x + iy0 is mapped to w = x − y0 + 2ixy0. Hence

v2 = 4y2(u + y2) y0 0 0 This is the equation of a parabola in the w-plane, with focus (0, 0) and pointed in the positive direction. A mesh in the z-plane and its maps in the w-plane by the function f(z) = z2 is shown in Figure 1.

• The Cayley transform Consider the M¨obiustransformation z − i S(z) = z + i sometimes called the Cayley transform. By direct computation, it is easy to see that S maps the real line to the unit circle. Another way to see this is that the points {∞, 1, −1} are mapped to {1, −i, i}. Since S maps circlelines to circlelines, we have the desired result. Using the same theorem, it is easy to see that every horizontal line z = x + iy0, with y0 > 0 is mapped to a circle inside the unit circle. By direct computation, one ﬁnds that this is a circle of center (y0/(y0 + 1), 0) and radius 1/(y0 + 1). Likewise, every vertical line z = x0 + iy is mapped to a circle of center (1, −1/x0) and radius 1/x0. The arcs of these circles corresponding to y > 0 are inside the unit disk. The Cayley transform maps the upper half plane onto the unit disk. A few mesh lines in the upper half of the z-plane and their maps in the w-plane by the Cayley transform are shown in Figure 2.

4 Figure 1: Mesh in the z-plane and its image by the map z 7→ z2

Figure 2: Mesh lines in the upper half of the z-plane and their images by the Cayley transform

5 iθ Figure 3: (Left) Circles z = r0e , θ ∈ [0, 2π] in the z-plane, with r0 ∈ {1, 1.5, 2, 2.5} and rays x = ±y, and (Right) their images in the w-plane under the Joukowski map J(z) = 1/2(z + 1/z)

• The Joukowski map

A mapping which was historically important in ﬂuid dynamics is the Joukowski map, deﬁned by 1 1 w = z + 2 z

This map is conformal everywhere on C except at z = 0 where is it not deﬁned, and at z = ±1, where dw/dz = 0.

2 Observe that since f(z) = f(1/z), two points (z1, z2) ∈ C such that z1z2 = 1 are mapped onto the same image by f. f is two-to-one, and only one-to-one in a domain Ω if there are no two points z1 and z2 such 2 that (z1, z2) ∈ C and z1z2 = 1. f is for example one-to-one in D1(0), and in the exterior of D1(0).

iθ To further visualize the map, let us consider families of circles z = r0e , with r0 > 0. 1 iθ 1 −iθ 1 1 1 1 w = u + iv = r0e + e ⇒ u = r0 + cos θ , v = r0 − sin θ 2 r0 2 r0 2 r0

Hence, the circle Cr0 (0), r0 < 1 is mapped onto an ellipse, and this ellipse degenerates into the line segment [−1, 1] as r0 → 1. Observe that D1(0) is mapped to C \{[−1, 1]}, and since f(z) = f(1/z), the exterior of D1(0) is also mapped to C \{[−1, 1]}. The images of circles with radius r0 ≥ 1 and the images of the rays x = ±y perpendicular to the circles under the Joukowski map are shown in Figure 3 to illustrate this. This mapping has relevance in fluid dynamics because the image of certain circles not centered at the origin under the map resembles the cross section of an airplane wing. You can for instance see the image of the circle z = 0.1 + 0.2i + 0.85eiθ, θ ∈ [0, 2π] under the Joukowski map in Figure 4. The idea, then, at a time when numerical solvers for fluid dynamics were not available or very slow, was to compute the tractable problem of fluid potential flow around a cylinder with circular cross section and apply the Joukowski mapping to the solution to obtain the fluid flow around the airplane wing. We will look at this in the next lecture.

4 Conformal mapping of an annulus

The Riemann mapping theorem applies to simply connected open sets. In the proof of the theorem, that was the condition under which we could construct the analytic function h. Topologically, it is intuitive that the theorem cannot apply to non-simply connected domains: a hole cannot be made to disappear under a one-to-one continuous mapping.

6 Figure 4: (Left) Circle z = 0.1 + 0.2i + 0.85eiθ, θ ∈ [0, 2π] in the z-plane (its center is shown with the red square marker), and (Right) its image in the w-plane under the Joukowski map J(z) = 1/2(z + 1/z)

One may however ask if one can generalize the Riemann mapping theorem to non-simply connected domains by looking into mappings of non-simply connected open sets to the simplest non-simply connected set one can think of, i.e. the annulus. Even more simply, one can ask whether any two concentric annuli are conformally equivalent. The answer to this question is, perhaps surprisingly, negative.

Theorem: A(r1,R1) := {z : r1 < |z| < R1} and A(r2,R2) := {z : r2 < |z| < R2} are conformally equivalent iﬀ R1 = R2 . r1 r2

Proof : Without loss of generality, let us rescale A(r1,R1) = A1 and A(r2,R2) = A2 such that A1 = A(1,R1) and A2 = A(1,R2). Let us assume there is an analytic conformal map f such that f(A1) = A2. Then, since f is a homeomorphism between A1 and A2, either |f(z)| → 1 as |z| → 1 or |f(z)| → R2 as |z| → 1. If the second situation holds, we can always deﬁne g := R2/f which is such that |g(z)| → 1 as |z| → 1. Hence, without loss of generality, we assume that lim |f(z)| = 1 and thus lim |f(z)| = R2. Now, since f(z) 6= 0 |z|→1 |z|→R1 ∀z ∈ A1, ln |f| is harmonic in A1. Let

ln R2 u(z) := ln |f(z)| − α ln |z| , α = , ∀z ∈ A1 ln R1 Observe that lim u(z) = 0 = lim u(z) |z|→1 |z|→R1

We can extend u to a continuous function on A1, with u = 0 on ∂A1. Since u is harmonic in A1, u ≡ 0 on A1. Therefore, α ∀z ∈ A1 , |f(z)| = |z| (2)

Now, take z0 ∈ A1, and r > 0 such that Dr(z0) ⊂ A1. As we have shown in Homework 1 (or by the Maximum Modulus Principle), Eq.(2) implies that

iθ0 α ∀z ∈ Dr(z0) , f(z) = e z (3) for some θ0 ∈ R. Taking the logarithmic derivative of Eq.(3), we ﬁnd f 0(z) α ∀z ∈ D (z ) , = r 0 f(z) z

And since this is true for any z0 ∈ A1 (provided r is chosen small enough that Dr(z0) ⊂ A1) 1 f 0(z) 1 α ∀z ∈ A , = 1 2πi f(z) 2πi z

7 1/2 We integrate this equality over the circle of center 0 and radius R1 . The right-hand side gives α. The left-hand side is 1 Z dw = ±1 2πi f(γ) w and so it must be that α = 1, i.e. R1 = R2.

R1 R2 Conversely, consider A1(r1,R1) and A2(r2,R2) such that = = β. Let γ = r2/r1. The mapping z 7→ γz r1 r2 transforms A1(r1,R1) = A1(r1, βr1) into A(γr1, γβr1) = A2(r2,R2). This concludes our proof.

Note ﬁnally that it can be shown that any doubly connected region of C can be conformally mapped to an annulus. The proof of this result is however beyond the scope of this class. From what we have just seen, the annulus to which the doubly connected region is mapped has a ratio r/R which is uniquely speciﬁed.

5 Conformal mapping of the unit disk to polygons 5.1 Mapping the unit disk to a polygon The Riemann mapping theorem tells us that one can map any polygon to the unit disk. What is interesting in that particular case, is that the inverse map, from the unit disk to the polygon, has an explicit formula, called the Schwarz-Christoﬀel formula, which we give below without proof.

Theorem (Schwarz-Christoﬀel Formula): The functions z = F (w) which map D1(0) conformally onto polygons with angles παk with k ∈〚1, n〛 in counterclockwise order are of the form

Z w n Y −βk F (w) = A (ζ − wk) dζ + B 0 k=1 Pn where A and B are complex constants, βk = 1 − αk such that k=1 βk = 2, and (wk)k=1,...,n n points on the unit circle.

The (wk)k=1,...,n are called the prevertices: they are the points on the unit circle such that F (wk) = zk are the vertices of the polygon. 0 0 0 Now, observe that given two triplets of points on the unit circle {w1, w2, w3} and {w1, w2, w3}, there exists 0 0 0 a Möbiustransformation which maps the unit disk to the unit disk, and {w1, w2, w3} to {w1, w2, w3}. That means that we are always free to choose three of the wk as we like, provided we are consistent with the ordering of the points. Hence, for n ≤ 3, the Schwarz-Christoffel formula can be viewed as an explicit formula for the mapping. A is easily determined by scaling and rotating the polygon appropriately, and B by mapping the origin of the disk to the desired point. For n ≥ 4, different choices of prevertices lead to different polygons, all consistent with the angles αk. This is shown in Figure 5. In general, computing the non-free prevertices for a desired polygon must be done numerically. You may want to have a look at the great book Schwarz-Christoffel Mapping by T. Driscoll and L.N. Trefethen, Cambridge University Press, for a description of numerical algorithms and nice examples.

5.2 Generalizing the result: mapping the upper half plane to a polygon

We have seen that the Cayley transform z − i S(z) = z + i maps the upper half plane to the unit disk. Its inverse is 1 + w z = S−1(w) = i 1 − w

8 Figure 5: Two possible images of the unit disk for diﬀerent choices of the prevertices, and the same choice of angles

Composing F with S, we could have a mapping from the upper half plane to polygons. Deﬁning 1 + ζ u = S−1(ζ) = i 1 − ζ

u − i −βk 1 − w −βk 1 + w −βk −βk k k (ζ − wk) = − wk = u − i u + i u + i 1 − wk

We let vk := i(1 + wk)/(1 − wk), and observe that 2i dζ = du (u + i)2 to obtain the general form of the mapping from the upper half plane to polygons:

Z z n −βk Y 1 − wk du G(z) := F ◦ S(z) = 2iA (u − v )−βk + B u + i k (u + i)2 0 k=1 Pn Using the fact that k=1 βk = 2, this can be written in the more concise form Z z n Y −βk G(z) = α (u − vk) du + β 0 k=1 with α and β constants, and βk as before.

5.3 Examples •Mapping the upper half plane to the semi-inﬁnite strip −π/2 < <(w) < π/2 , =(w) > 0 We have w1 = −π/2, w2 = π/2. Let us choose v1 = −1, v2 = 1. The mapping is given by Z z Z z 1 G(z) = α (u + 1)−1/2(u − 1)−1/2du + β = α √ du + β = α0 arcsin z + β 2 0 0 u − 1 Setting G(−1) = −π/2 and G(1) = π/2 leads to the system

( π 0 π − 2 α + β = − 2 π 0 π 2 α + β = 2 from which we conclude that β = 0 and α0 = 1: f(z) = arcsin z is a map from the upper half plane to the desired semi-inﬁnite strip. • Mapping the upper half plane to a rectangle

9 Let us assume the rectangle is rotated and translated so that its vertices are w1 = −K1 + iK2, w2 = −K1, −1/2 w3 = K1, and w4 = K1 + iK2. By symmetry, we choose the prevertices such that v1 = −k , v2 = −1, −1/2 v3 = 1, and v4 = k , where k represents the fact that only 3 points can be freely speciﬁed. We thus have Z z 4 Z z Y −1/2 du w = G(z) = β + α (u − vk) = α q √ 0 0 2 1 2 k=1 u − k u − 1 where we have set β = 0 so that F (0) = 0. This is an elliptic integral of the ﬁrst kind, which can be written in the more concise form Z arcsin z dθ w = G(z) = α p 2 0 1 − k sin θ The constant α represents the freedom in rotating and scaling the rectangle, and the constant k must be varied to obtain the desired aspect ratio K2/K1 for the rectangle.