Mapping problems and harmonic univalent mappings

Mapping problems and harmonic univalent mappings Antti Rasila Helsinki University of Technology antti.rasila@tkk.fi (Mainly based on P. Duren s book Harmonic mappings in the plane) Helsinki Analysis Seminar, 2009-11-16 FILE: plharm3-1.tex, 2009-11-04, printed: 2009-11-16, 17.52 Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 1 / 36

1 Introduction 2 Preliminaries Differential operators / z and / z Canonical representation Jacobian and local univalence Connections to quasiconformal mappings Complex dilatation Second complex dilatation Harmonic univalent mappings 3 Mapping problems Generalizing the Riemann mapping theorem Collapsing Boundary behavior Existence theorems Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 2 / 36

Introduction A real-valued function u(x, y) is harmonic if it satisfies the Laplace equation u = 2 x 2 u + 2 y 2 u = 0. A complex-valued function f (x + iy) = u(x, y) + iv(x, y) from a domain f : D C is harmonic if the two coordinate functions are harmonic. A complex-valued harmonic function is a harmonic mapping if it is univalent (one-to-one). Harmonic mappings in the plane are univalent complex-valued mappings whose real and imaginary parts are not necessarily conjugate i.e. do not need to satisfy the Cauchy Riemann equations. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 3 / 36

Compositions of complex-valued harmonic functions If f : D C, g : f (D) C are harmonic functions, g f is not necessarily harmonic. If f : D C is analytic and g : f (D) C is harmonic, then g f is harmonic. If f : D C is harmonic and g : f (D) C is analytic, then g f is not necessarily harmonic. In fact, even a square or the reciprocal of a harmonic function need not be harmonic. Example: f (x, y) = x 2 y 2 + ix. Then Im(f 2 ) = 16x 0. Inverse of a harmonic mapping need not be harmonic. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 4 / 36

Example 1 The simplest example of a harmonic mapping that is not necessarily conformal is an affine mapping f (z) = αz + γ + β z, α β. Every composition of a harmonic mapping with an affine mapping is harmonic, i.e. if f is harmonic, then so is αf + γ + β f. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 5 / 36

Example 2 Consider the function f (z) = z + 1 2 z2, which maps the open unit disk D onto the region outside a hypocycloid of three cusps in the circle w = 3/2. To verify its univalence, suppose f (z 1 ) = f (z 2 ) for z 1, z 2 D. Then z 2 1 z 2 2 = 2(z 2 z 1 ) or ( z 1 + z 2 )( z 1 z 2 ) = 2(z 2 z 1 ). By taking absolute values on both sides, we see that this is impossible unless z 1 = z 2 because z 1 z 2 < 2. A similar argument shows that f (z) = z 1 n zn is univalent for each n 2. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 6 / 36

1 0.5 0 0.5 1 2 1.5 1 0.5 0 0.5 1 Figure 1: Image of the unit disk in the mapping f (z) = z + 1 2 z2. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 7 / 36

1 0.5 0 0.5 1 1.5 1 0.5 0 0.5 1 Figure 2: Image of the unit disk in the mapping f (z) = z + 1 4 z4. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 8 / 36

Differential operators / z and / z Define z = 1 ( 2 x i ) and y z = 1 ( 2 x + i ), y where z = x + iy. The equation f / z = 0 is just another way of writing the Cauchy Riemann equations. The Laplacian of f is f = 4 2 f z z. Function f with continuous second partial derivatives is harmonic if and only if f / z is analytic. If f is analytic, then f / z = f (z), i.e. the ordinary derivative. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 9 / 36

Properties of / z and / z The operators / z and / z are linear and have the usual properties of differential operators. For example, the product and quotient rules hold: g (fg) = f z z + g f z and ( f ) ( = g f z g z f g ) 1 z g 2, and similarly for / z. The two derivatives are connected by the property ( f ) = f z z. The notation f z = f / z and f z = f / z is often more convenient. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 10 / 36

Canonical representation Lemma In a simply connected domain D C a complex valued harmonic function f has the representation f = h + ḡ, where h and g are analytic in D. The representation is unique up to an additive constant. Proof. Since f is harmonic, f z is analytic. Let h = f z, where h is analytic in D. Let g = f h. By definition of h and thus g is analytic in D. g z = f z h z = 0 in D, The uniqueness follows from the fact that a function both analytic and anti-analytic is constant. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 11 / 36

Canonical representation (continued) For a harmonic mapping f of the unit disk D, it is convenient to choose the additive constant so that g(0) = 0. The representation f = h + ḡ is then unique and is called the canonical representation of f. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 12 / 36

Jacobian and local univalence The Jacobian of a function f = u + iv is J f (z) = u x v x = u xv y u y v x = f z 2 f z 2. If f is analytic, then u y v y J f (z) = (u x ) 2 + (v x ) 2 = f (z) 2. For analytic functions, it is a classical result that J f (z) 0 if and only if f is locally univalent at z. Hans Lewy showed in 1936 that this remains true for harmonic mappings. Consequently f is locally univalent and sense-preserving if f z (z) > f z (z), and sense reversing if f z (z) < f z (z). Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 13 / 36

Connections to quasiconformal mappings For sense-preserving mappings one sees that ( f z f z ) dz dw ( f z + f z ) dz. These sharp inequalities have the geometric interpretation that f maps an infinitesimal circle onto an infinitesimal ellipse with D f = f z + f z [1, ) f z f z as the ratio of the major and minor axes. The quantity D f (z) is called the dilatation of f at the point z. A sense-preserving homeomorphism f is said to be K-quasiconformal, D f (z) K throughout the given region, where K [1, ) is a constant. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 14 / 36

Complex dilatation The ratio µ f = f z /f z is called the complex dilatation of f. If f is sense-preserving, then µ f (z) [0, 1). It may be observed that D f (z) K if and only if µ f (z) K 1 K + 1. It follows that a sense-preserving homeomorphism f is quasiconformal if and only if µ f (z) k < 1. The mapping f is conformal if and only if µ f = 0. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 15 / 36

Second complex dilatation In the theory of harmonic mappings, the quantity ν f = f z /f z, known as the second complex dilatation, is often more natural than µ f. Since ν f = µ f, it is clear that f is quasiconformal if and only if ν f (z) k < 1. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 16 / 36

Lemma Let f : D C be a function with continuous second order partial derivatives with J f (z) > 0. Then f is harmonic if and only if ω = ν f is analytic. Proof. Let ω = ν f = f z /f z. Then ω(z) < 1 in D. Differentiating the equation f z = ωf z with respect to z, one finds f zz = f z z ω + f z ω z. Now if f is harmonic in D, then f z z = 1 4 f = 0, and hence ω z = 0 in D, so ω is analytic. Conversely, if ω is analytic, then f zz = f z z ω. Since ω(z) < 1, we have f z z = 0, and f is harmonic. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 17 / 36

Theorem The only harmonic mappings of C onto C are the affine mappings f (z) = αz + γ + β z, where α, β and γ are complex constants and α β. Proof. Let f = h + ḡ map C harmonically onto C. We may assume that f is sense preserving. Now g (z) < h (z), or g (z)/h (z) < 1 for all z C. By Liouville s theorem g (z)/h (z) b for some complex constant b with b < 1. Integration gives g(z) = bh(z) + c. Thus f has the form f = h + c + bh = F h, where F is an (invertible) affine mapping. It follows that h = F 1 f maps C univalently onto C. But h is analytic, so h(z) = αz + β for some complex constants α, β. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 18 / 36

Harmonic univalent mappings Let f = h + ḡ with g(0) = 0, where h(z) = a n z n and g(z) = n=0 b n z n. The class of all sense-preserving harmonic mappings of the unit disk with a 0 = b 0 = 0 and a 1 = 1 is denoted by S H. n=0 S H is a normal family; every sequence of functions in S H has a subsequence that converges locally uniformly in D. S H is not a compact family; it is not preserved under passage to locally uniform limits. The limit function is necessarily harmonic in D, but it need not be univalent. The class of functions f S H with g (0) = 0 will be denoted by S 0 H. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 19 / 36

Generalizing the Riemann mapping theorem The classical Riemann mapping theorem states that any simply connected plane domain D, which is not the whole plane, can be conformally mapped onto the unit disk. Next we consider generalizations of this result to the harmonic setting. Obviously, existence of harmonic mappings is not in question as every conformal mapping is harmonic. Rather, the problem is that there are far too many such mappings. Hence, the problem is to develop a meaningful way to classify such mappings, hoping that some additional conditions will uniquely determine the mapping. We will start by taking a look to the theory of quasiconformal mappings in the plane. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 20 / 36

The measurable Riemann mapping theorem Theorem (e.g. [Iwaniec-Martin, Thm 11.1.2]) Suppose that µ: D C is a measurable function with µ < 1. Then there exists a quasiconformal mapping g : D C whose complex dilatation (the Beltrami coefficient) µ g is equal to µ a.e. on D. Moreover, every W 1,2 loc is of the form solution f to the Beltrami equation f z (z) = µ(z)f z (z) f (z) = F (g(z)) where F : g(d) C is an analytic function. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 21 / 36

The harmonic case (idea) In the light of this result, it seems plausible to assume that the second complex dilatation ω f = f z /f z would play the similar role in classifying the harmonic mappings f of the unit disk onto a given domain D. Suppose that ω : D D is an analytic function. We try to find a sense-preserving harmonic mapping of D onto a simply connected domain with dilatation ω = f z /f z, unique upto to the normalization f (0) = w 0 D and f z (0) > 0. First we note that if ω f 0, then we have the Riemann mapping theorem. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 22 / 36

Constant dilatation Suppose that f is a harmonic function with dilatation ω f, and F = αf + β f, β < α, is an affine transformation of f. Then F has the dilatation F z /F z = αω f + β α + βω f. If ω f c, c < 1, then the choice β = α c will make F analytic. It follows that harmonic mappings with constant dilatation will take the form f = h + c h, where h is a conformal mapping. By taking the inverse image of D in the affine mapping z z + c z, and using the Riemann mapping theorem, we obtain a harmonic mapping with the desired properties. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 23 / 36

Collapsing 1/4 However, it turns out that given a simply connected domain D, D C, and an analytic function ω : D D there does not necessarily exist a harmonic mapping of D onto D with dilatation ω. The counterexample is the following. Hengartner and Schober have shown that there is no harmonic mapping of the disk onto itself with ω(z) = z. The idea of the proof is show that if f maps D harmonically into itself, and has dilatation ω = z, then the image must have area less than π. Hence it cannot fill the disk. Suppose that f = h + ḡ is an arbitrary sense-preserving harmonic mapping of D, and write h(z) = a n z n and g(z) = n=0 b n z n. n=1 Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 24 / 36

Collapsing 2/4 Then the area of the image set f (D) is ( fz A = J f (z) dx dy = 2 f z 2) dx dy because = D D D ( h (z) 2 g (z) 2) dx dy = π n 2 a n 2 z 2(n 1) dx dy = n 2 a n 2 1 n ( a n 2 b n 2), n=1 r=0 2π θ=0 r 2(n 1) r dθ dr 1 = 2πn 2 a n 2 r 2(n 1)+1 dr = 2πn2 a n 2 0 2n = πn a n 2. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 25 / 36

Collapsing 3/4 Because f has the dilatation ω f (z) = z, from the differential equation f z = ωf z we have f (z) = zh (z), and hence n=1 (n + 1)b n+1 = na n, for all n = 0, 1, 2,.... (3.1) In particular, b 1 = 0. It follows that ( A = π n an 2 (n + 1) b n+1 2) n = π n + 1 a n 2. (3.2) Next we estimate the integral mean We have M 2 (r, f ) 2 = 1 2π 2π 0 M 2 (r, f ) 2 = 1 2π 2π 0 n=1 f (re iθ ) 2 dθ. ( h(re iθ ) 2 + g(re iθ ) 2) dθ = a n 2 r 2n + b n 2 r 2n. n=0 n=1 Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 26 / 36

Collapsing 4/4 If f (D) D, then M 2 (r, f ) 1 and thus lim M 2(r, f ) 2 = M 2 (1, f ) 2 = r 1 ( an 2 + b n+1 2) 1. n=0 Because ω f (z) = z, we obtain from (3.1) M 2 (1, f ) 2 = ( 1 + n 2 /(n + 1) 2) a n 2 1. This, together with the estimate 2n/(n + 1) < 1 + n 2 /(n + 1) 2 and (3.2) gives A < π 2 M 2(1, f ) 2 π 2, which is a contradiction because the area of the unit disk is equal to π. Hence, the image set cannot be a the whole disk. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 27 / 36

Boundary behavior Hopf s lemma Suppose that D is a Jordan domain with smooth boundary, and u : D R is a non-constant harmonic function with a smooth extension to the boundary. If u has a local minimum at some point w D, then its inner normal derivative is strictly positive at this point: ( u/ n)(w) > 0. The proof of Hopf s lemma is based on Harnack s inequality: R r R + r R + r u(0) u(z) u(0), r = z (3.3) R r for a positive harmonic function u : D(0, R) R +. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 28 / 36

Proof We may assume that u(w) = 0 and u(z) > 0 for points in D near w. Then u(z) > 0 for some disk D 0 = D(z 0, r) D whose boundary is tangent to D at w. By taking a similarity transformation if necessary, we may assume that z 0 = 0 and w = r. Then, by Harnack s inequality (3.3), It follows that u(x) u(0) r x r + x, 0 < x < r. u(x) u(r) r x By letting x r, we obtain u(0) r + x, 0 < x < r. u u(0) (r) = u (r) > 0. n x 2r Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 29 / 36

Boundary dilatation of unit modulus We obtain the following result: Theorem Suppose that f : D D = f (D) is a sense-preserving harmonic mapping with C 1 extension to some open arc I D that maps univalently onto a convex arc γ D. Denote by s(θ) the arc length along γ for e iθ I. If ds/dθ 0 at some point z 0 = e iα I, then the dilatation of f has a continuous extension to a sub-arc of I containing z 0 and ω f (z 0 ) < 1. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 30 / 36

Proof 1/2 By taking a Möbius transformation if necessary, we may assume that f is a mapping of the upper half-plane onto D and I R with 0 I. After a similarity transformation, we may assume that D is in the upper half-plane, tangent to the real axis and f (0) = 0 = z 0. Then, f has a smooth homeomorphic boundary extension to I and f (I ) is a convex boundary arc of D. Write w = f (z), where z = x + iy and w = u + iv. It is clear that v x (0) = 0, since v has a local minimum at 0. Because ds/dθ(0) 0, it follows that u x (0) > 0, and the Jacobian of f at the origin is J f (0) = u x (0)v y (0) u y (0)v x (0) = u x (0)v y (0). Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 31 / 36

Proof 2/2 Because v is a positive harmonic function in the upper half-plane that attains its minimum value 0 at the origin, it follows by Hopf s lemma that v y (0) > 0 as it is the inner normal derivative of v at the origin. It follows that J f (0) > 0, or equivalently, h (0) > g (0). This shows that the dilatation ω f = g /h is continuous up to a sub-arc of I containing the origin, and ω f (0) < 1, proving the claim. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 32 / 36

A connection to harmonic quasiconformal mappings In particular, we obtain the following generalization of a result of O. Martio [Martio]. Corollary Suppose that f : D D = f (D) is a sense-preserving harmonic mapping onto a convex domain D with a homeomorphic C 1 extension to the boundary such that ds/dθ > 0 for all e iθ D. Then f is quasiconformal. Proof. The dilatation ω g has a continuous boundary extension satisfying ω f (z) < 1 for all z D. Hence ω(z) k for some constant k < 1, and f is quasiconformal in D. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 33 / 36

Existence theorems 1/2 Theorem Suppose that D is a bounded simply connected domain, and D is an analytic Jordan curve, and let w 0 D. Let ω : D C be analytic with ω(z) k for some constant k < 1. Then there exists a harmonic mapping f : D D = f (D) with ω f = ω, f (0) = w 0 and f z (0) > 0. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 34 / 36

Existence theorems 2/2 Theorem Suppose that D is a bounded simply connected domain, and D is an analytic Jordan curve Γ, and let w 0 D. Let ω : D C be an analytic function. Then there exists a harmonic mapping f : D D with ω f = ω, f (0) = w 0 and f z (0) > 0 such that radial limits belong to Γ for almost every angle θ. ˆf (e iθ ) = lim r 1 f (re iθ ) Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 35 / 36

References James Clunie and Terry Sheil-Small, Harmonic univalent functions, Ann. Acad. Sci. Fenn. Ser. A.I. 9 (1984), 3 25. Peter Duren, Harmonic mappings in the plane, Cambridge University Press, 2004. Tadeusz Iwaniec and Gaven Martin, Geometric Function Theory and Non-linear Analysis, Claredon Press, Oxford, 2001. Olli Martio, On harmonic quasiconformal mappings, Ann. Acad. Sci. Fenn. Ser A.I. (1968), 3 10. Antti Rasila () Mapping problems and harmonic mappings 2009-11-16 36 / 36