Abstract and Applied Analysis

Volume 2013, Article ID 512383, 5 pages

http://dx.doi.org/10.1155/2013/512383

## Rotationally Symmetric Harmonic Diffeomorphisms between Surfaces

^{1}Faculty of Mathematics & Computer Science, Hubei University, Wuhan 430062, China^{2}The School of Natural Sciences and Humanities, Shenzhen Graduate School, The Harbin Institute of Technology, Shenzhen 518055, China^{3}Department of Mathematics, Jinan University, Guangzhou 510632, China

Received 12 February 2013; Revised 22 April 2013; Accepted 22 April 2013

Academic Editor: Yuriy Rogovchenko

Copyright © 2013 Li Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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.

#### 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 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 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 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 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 where is the unit disc and is the complex coordinate of . We will prove the following results.

Theorem 1. *For any , there is no rotationally symmetric harmonic diffeomorphism from onto with its hyperbolic metric. *

And vice versa as shown below.

Theorem 2. *For any , there is no rotationally symmetric harmonic diffeomorphism from onto with its hyperbolic metric. *

We will also consider the Euclidean case and will prove the following theorem.

Theorem 3. *For any , there is no rotationally symmetric harmonic diffeomorphism from onto with its Euclidean metric; but on the other hand, there are rotationally symmetric harmonic diffeomorphisms from onto with its Euclidean metric. *

This paper is organized as follows. In Section 2, 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 onto the hyperbolic disc.

#### 2. Harmonic Maps from to with Its Hyperbolic Metric and Vice Versa

For convenience, let us recall the definition about the harmonic maps between surfaces. Let and be two oriented surfaces with metrics and , respectively, where and are local complex coordinates of and , respectively. A map from to is harmonic if and only if satisfies

Now let us prove Theorem 1.

*Proof of Theorem 1. *First of all, let us denote as the polar coordinates of and as the complex coordinates of in ; then the hyperbolic metric on can be written as
Here is the norm of with respect to the Euclidean metric.

We will prove this theorem by contradiction. Suppose is a rotationally symmetric harmonic diffeomorphism from onto , with the metric . Because , and the metric are rotationally symmetric, we can assume that such a map has the form . Substituting to (2), we can get
for . Since is a harmonic diffeomorphism from onto , we have
or

We will just deal with the case that (5) is satisfied; the rest case is similar. Let , then we have
Using this fact, we can get from (4) the following equation:
with , , and for .

Regarding as a function of , we have the following relations:
Using these facts, we can get from (9) the following equation:
for . Let ; from (11) we can get the following equation:
One can solve this Bernoulli equation to obtain
Here is a constant depending on the choice of the function . So
Since , we can get
Noting that is continuous in and is equal to as , or , one can get is uniformly bounded for . So the right-hand side of (15) is uniformly bounded, but the left-hand side will tend to as . Hence, we get a contradiction. Therefore, such does not exist, Theorem 1 has been proved.

We are going to prove Theorem 2.

*Proof of Theorem 2. *First of all, let us denote as the polar coordinates of and as the complex coordinates of in ; then the hyperbolic metric on can be written as
Here is the norm with respect to the Euclidean metric.

We will prove this theorem by contradiction. The idea is similar to the proof of Theorem 1. Suppose is a rotationally symmetric harmonic diffeomorphism from onto with the metric , with the form , then substituting to in (2), respectively, we can get
for . Since is a harmonic diffeomorphism from onto , we have
or

We will only deal with the case that (18) is satisfied; the rest case is similar. Let , then (17) can be rewritten as
for , with and .

Regarding as a function of , using a similar formula of (10), from (21) we can get
Similar to solving (11), we can get the solution to (22) as follows:
for some nonnegative constant depending on the choice of .

If is equal to , then ; this is in contradiction to (18).

If is positive, then taking integration on both sides of (23), we can get
So
with . On the other hand, from (18), we have . Hence, we get a contradiction. Therefore, such does not exist, Theorem 2 has been proved.

#### 3. Harmonic Maps from to with Its Euclidean Metric and Vice Versa

Now let us consider the case of that the target has the Euclidean metric.

*Proof of Theorem 3. *Let us prove the first part of this theorem, that is, show the nonexistence of rotationally symmetric harmonic diffeomorphism from onto 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 from onto with its Euclidean metric with the form , and then we can get
with
or

We will just deal with the case that (27) is satisfied; the rest case is similar. Let ; then we can get
Solving this equation, we can get
Here is a constant depending on the choice of . So
Here is a constant depending on the choice of . Hence
From (32), we can get . On the other hand, from (27), . We get a contradiction. Hence such a function 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 from onto with its Euclidean metric. It suffices to find a map from onto with the form such that
with , and for . Using the boundary condition and (32), we can get that
is a solution to (33).

Therefore, we finished the proof of Theorem 3.

#### 4. Harmonic Maps from 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 from onto the hyperbolic disc. *

*Proof. *It is well-known that the hyperbolic metric on the unit disc is . We will also use the idea of the proof of Theorem 1. Suppose there is such a harmonic diffeomorphism from onto with its hyperbolic metric with the form , and then we can get
with
Regarding as a function of , setting , (35) can be rewritten as
That is,
One can solve this equation to obtain
for some nonnegative constant depending on the choice of the function .

If , then we can get for some constant . On the other hand, is a diffeomorphism, so as . This is a contradiction.

If , then for some positive constants and . So
This is in contradiction to the assumption that as .

Therefore, Proposition 4 holds.

#### Acknowledgments

The author (Xu-Qian Fan) would like to thank Professor Luen-fai Tam for his very worthy advice. The first author is partially supported by the National Natural Science Foundation of China (11201131); the second author is partially supported by the National Natural Science Foundation of China (11101106).

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