# Rotationally Symmetric Harmonic Diffeomorphisms between Surfaces

Authors: Li Chen; Shi-Zhong Du; Xu-Qian Fan
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We show the nonexistence of rotationally symmetric harmonic diffeomorphism between the unit disk without the origin and a punctured disc with hyperbolic metric on the target.


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## Body

## 1. Introduction

The existence of harmonic diffeomorphisms between complete Riemannian manifolds has been extensively studied, please see, for example, [1–34]. In particular, Heinz [17] proved that there is no harmonic diffeomorphism from the unit disc onto ℂ with its flat metric. On the other hand, Schoen [25] mentioned a question about the existence, or nonexistence, of a harmonic diffeomorphism from the complex plane onto the hyperbolic 2-space. At the present time, many beautiful results about the asymptotic behavior of harmonic embedding from ℂ into the hyperbolic plane have been obtained, please see, for example, [4, 5, 14, 32] or the review [33] by Wan and the references therein. In 2010, Collin and Rosenberg [10] constructed harmonic diffeomorphisms from ℂ onto the hyperbolic plane. In [7, 24, 28, 29], the authors therein studied the rotational symmetry case. One of their results is the nonexistence of rotationally symmetric harmonic diffeomorphism from ℂ onto the hyperbolic plane.In this paper, we will study the existence, or nonexistence, of rotationally symmetric harmonic diffeomorphisms from the unit disk without the origin onto a punctured disc. For simplicity, let us denote(1)𝔻*=𝔻∖{0},P(a)=𝔻∖{|z|≤e-a}fora>0,
where 𝔻 is the unit disc and z is the complex coordinate of ℂ. We will prove the following results.Theorem 1.
For anya>0, there is no rotationally symmetric harmonic diffeomorphism from 𝔻* onto P(a) with its hyperbolic metric.And vice versa as shown below.Theorem 2.
For anya>0, there is no rotationally symmetric harmonic diffeomorphism from P(a) onto 𝔻* with its hyperbolic metric.We will also consider the Euclidean case and will prove the following theorem.Theorem 3.
For anya>0, there is no rotationally symmetric harmonic diffeomorphism from 𝔻* onto P(a) with its Euclidean metric; but on the other hand, there are rotationally symmetric harmonic diffeomorphisms from P(a) onto 𝔻* with its Euclidean metric.This paper is organized as follows. In Section2, we will prove Theorems 1 and 2. Theorem 3 will be proved in Section 3. At the last section, we will give another proof for the nonexistence of rotationally symmetric harmonic diffeomorphism from ℂ onto the hyperbolic disc.


## 2. Harmonic Maps from𝔻* to P(a) with Its Hyperbolic Metric and Vice Versa

For convenience, let us recall the definition about the harmonic maps between surfaces. LetM and N be two oriented surfaces with metrics τ2|dz|2 and σ2|du|2, respectively, where z and u are local complex coordinates of M and N, respectively. A C2 map u from M to N is harmonic if and only if u satisfies
(2)uzz-+2σuσuzuz-=0.Now let us prove Theorem1.Proof of Theorem1.
First of all, let us denote(r,θ) as the polar coordinates of 𝔻* and u as the complex coordinates of P(a) in ℂ; then the hyperbolic metric σ1d|u| on P(a) can be written as
(3)-π|du|a|u|sin((π/a)ln|u|).
Here |u| is the norm of u with respect to the Euclidean metric.
We will prove this theorem by contradiction. Supposeu is a rotationally symmetric harmonic diffeomorphism from 𝔻* onto P(a), with the metric σ1d|u|. Because 𝔻*, P(a) and the metric σ1d|u| are rotationally symmetric, we can assume that such a map u has the form u=f(r)eiθ. Substituting u,σ1 to (2), we can get
(4)f′′+f′r-fr2-sin((π/a)lnf)+(π/a)cos((π/a)lnf)fsin((π/a)lnf)×((f′)2-f2r2)=0
for 1>r>0. Since u is a harmonic diffeomorphism from 𝔻* onto P(a), we have
(5)f(0)=e-a,f(1)=1,f′(r)>0for1>r>0,
or
(6)f(0)=1,f(1)=e-a,(7)f′(r)<0for1>r>0.
We will just deal with the case that (5) is satisfied; the rest case is similar. Let F=lnf∈(-a,0), then we have
(8)f′=f′f>0,F′′=f′′f-(f′f)2.
Using this fact, we can get from (4) the following equation:
(9)F′′+1rF′-πactg(πaF)(F′)2+1r2πactg(πaF)=0for1>r>0,
with F(0)=-a, F(1)=0, and F′(r)>0 for 1>r>0.
Regardingr as a function of F, we have the following relations:
(10)Fr=rF-1,Frr=-rF-3rFF.
Using these facts, we can get from (9) the following equation:
(11)r′′r-(r′r)2+πactg(πaF)r′r-(r′r)3πactg(πaF)=0
for 0>F>-a. Let x=(lnr)′(F); from (11) we can get the following equation:
(12)x′+πactg(πaF)·x-πactg(πaF)·x3=0.
One can solve this Bernoulli equation to obtain
(13)x-2=1+c0(sin(πaF))2.
Here c0 is a constant depending on the choice of the function f. So
(14)x=11+c0(sin((π/a)F))2.
Since x=(lnr)′(F), we can get
(15)(lnr)(F)=∫0F‍x(t)dt=∫0F‍11+c0(sin((π/a)t))2dt.
Noting that x(F) is continuous in (-a,0) and is equal to 1 as F=-a, or 0, one can get x is uniformly bounded for F∈[-a,0]. So the right-hand side of (15) is uniformly bounded, but the left-hand side will tend to -∞ as F→-a. Hence, we get a contradiction. Therefore, such f does not exist, Theorem 1 has been proved.We are going to prove Theorem2.Proof of Theorem2.
First of all, let us denote(r,θ) as the polar coordinates of P(a) and u as the complex coordinates of 𝔻* in ℂ; then the hyperbolic metric σ2d|u| on 𝔻* can be written as
(16)|du||u|ln(1/|u|).
Here |u| is the norm with respect to the Euclidean metric.
We will prove this theorem by contradiction. The idea is similar to the proof of Theorem1. Suppose ψ is a rotationally symmetric harmonic diffeomorphism from P(a) onto 𝔻* with the metric σ2d|u|, with the form ψ=g(r)eiθ, then substituting ψ,σ2 to u,σ in (2), respectively, we can get
(17)g′′+g′r-gr2-1+lngglng((g′)2-g2r2)=0
for 1>r>e-a. Since v is a harmonic diffeomorphism from P(a) onto 𝔻*, we have
(18)g(e-a)=0,g(1)=1,g′(r)>0for1>r>e-a,
or
(19)g(e-a)=1,g(1)=0,(20)g′(r)<0for1>r>e-a.
We will only deal with the case that (18) is satisfied; the rest case is similar. Let G=lng, then (17) can be rewritten as
(21)G′′+1rG′-1G(G′)2+1r2G=0
for 1>r>e-a, with G(1)=0 and limr→e-aG(r)=-∞.
Regardingr as a function of G, using a similar formula of (10), from (21) we can get
(22)r′′r-(r′r)2+r′rG-1G(r′r)3=0,G∈(-∞,0).
Similar to solving (11), we can get the solution to (22) as follows:
(23)(lnr)′(G)=11+c1G2,G∈(-∞,0)
for some nonnegative constant c1 depending on the choice of g.
Ifc1 is equal to 0, then g=r; this is in contradiction to (18).
Ifc1 is positive, then taking integration on both sides of (23), we can get
(24)(lnr)(G)=∫0G‍11+c1t2dt=1c1ln(c1G+1+c1G2).
So
(25)r=(c1lng+1+c1ln2g)1/c1,
with limg→0+r(g)=0. On the other hand, from (18), we have r(0)=e-a. Hence, we get a contradiction. Therefore, such g does not exist, Theorem 2 has been proved.


## 3. Harmonic Maps from𝔻* to P(a) with Its Euclidean Metric and Vice Versa

Now let us consider the case of that the target has the Euclidean metric.Proof of Theorem3.
Let us prove the first part of this theorem, that is, show the nonexistence of rotationally symmetric harmonic diffeomorphism from𝔻* onto P(a) with its Euclidean metric. The idea is similar to the proof of Theorem 1, so we just sketch the proof here. Suppose there is such a harmonic diffeomorphism φ from 𝔻* onto P(a) with its Euclidean metric with the form φ=h(r)eiθ, and then we can get
(26)h′′+1rh′-1r2h=0for1>r>0,
with
(27)h(0)=e-a,h(1)=1,h′(r)>0for1>r>0,
or
(28)h(0)=1,h(1)=e-a,h′(r)<0for1>r>0.
We will just deal with the case that (27) is satisfied; the rest case is similar. Let H=(lnh)′; then we can get
(29)H′+H2+1rH=0for1>r>0.
Solving this equation, we can get
(30)H=1r+1c3r3-r2=1r+c32c3r-1-c3r-1r2.
Here c3 is a constant depending on the choice of h. So
(31)lnh=lnr+c3ln(c3r-1)-c3lnr+1r+c4.
Here c4 is a constant depending on the choice of h. Hence
(32)h=r(c3r-1)c3r-c3e1/rec4.
From (32), we can get limr→0h(r)=∞. On the other hand, from (27), h(0)=e-a. We get a contradiction. Hence such a function h does not exist; the first part of Theorem 3 holds.
Now let us prove the second part of this theorem, that is, show the existence of rotationally symmetric harmonic diffeomorphisms fromP(a) onto 𝔻* with its Euclidean metric. It suffices to find a map from P(a) onto 𝔻* with the form q(r)eiθ such that
(33)q′′+1rq′-1r2q=0for1>r>e-a
with q(e-a)=0,q(1)=1, and q′>0 for 1>r>e-a. Using the boundary condition and (32), we can get that
(34)q=e-1(ea-1)-ear(ear-1)ear-eae1/r
is a solution to (33).
Therefore, we finished the proof of Theorem3.


## 4. Harmonic Maps fromℂ to the Hyperbolic Disc

In this section, we will give another proof of the following result.Proposition 4.
There is no rotationally symmetric harmonic diffeomorphism fromℂ onto the hyperbolic disc.Proof.
It is well-known that the hyperbolic metric on the unit disc is(2/(1-z|2))|dz|. We will also use the idea of the proof of Theorem 1. Suppose there is such a harmonic diffeomorphism ϕ from ℂ onto 𝔻 with its hyperbolic metric with the form ϕ=k(r)eiθ, and then we can get
(35)k′′+1rk′-1r2k+2k1-k2[(k′)2-k2r2]=0forr>0
with
(36)k(0)=0,k′(r)>0forr>0.
Regarding r as a function of k, setting v=(lnr)′(k), (35) can be rewritten as
(37)(1-k2)v′-2kv+v3(k+k3)=0.
That is,
(38)(v-2)′+4k1-k2v-2=2(k+k3)1-k2.
One can solve this equation to obtain
(39)v-2=k2+c5(1-k2)2
for some nonnegative constant c5 depending on the choice of the function k.
Ifc5=0, then we can get r=c6k for some constant c6. On the other hand, ϕ is a diffeomorphism, so k→1 as r→∞. This is a contradiction.
Ifc5>0, then b1≥k2+c5(1-k2)2≥b2 for some positive constants b1 and b2. So
(40)|(lnr)′(k)|≤1b2.
This is in contradiction to the assumption that r→∞ as k→1.
Therefore, Proposition4 holds.



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*Source: 512383-2013-05-13.xml*