Rotationally Symmetric Harmonic Diffeomorphisms between Surfaces

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 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 𝔻^∗=𝔻⧵{0},𝑃(𝑎)=𝔻⧵|𝑧|≤𝑒^−𝑎for𝑎>0, (1) where 𝔻 is the unit disc and 𝑧 is the complex coordinate of ℂ. We will prove the following results.

Theorem 1. For any 𝑎>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 any 𝑎>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 any 𝑎>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 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 ℂ onto the hyperbolic disc.

2. Harmonic Maps from 𝔻^∗ to 𝑃(𝑎) with Its Hyperbolic Metric and Vice Versa

For convenience, let us recall the definition about the harmonic maps between surfaces. Let 𝑀 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 𝑢_𝑧𝑧+2𝜎_𝑢𝜎𝑢_𝑧𝑢_𝑧=0. (2)

Now let us prove Theorem 1.

Proof of Theorem 1. 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 −𝜋|𝑑𝑢|𝑎|𝑢|)|𝑢.sin((𝜋/𝑎ln|) (3) Here |𝑢| is the norm of 𝑢 with respect to the Euclidean metric.
We will prove this theorem by contradiction. Suppose 𝑢 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)𝑒^𝑖𝜃. Substituting 𝑢,𝜎₁ to (2), we can get 𝑓^′′+𝑓^′𝑟−𝑓𝑟²−sin(𝜋/𝑎)ln𝑓+(𝜋/𝑎)cos(𝜋/𝑎)ln𝑓()×𝑓𝑓sin𝜋/𝑎ln𝑓_′2−𝑓²𝑟²=0 (4) for 1>𝑟>0. Since 𝑢 is a harmonic diffeomorphism from 𝔻^∗ onto 𝑃(𝑎), we have 𝑓(0)=𝑒^−𝑎(1)𝑓,𝑓=1,_′(𝑟)>0for1>𝑟>0, (5) or 𝑓(0)=1,𝑓(1)=𝑒^−𝑎,𝑓^′(𝑟)<0for1>𝑟>0. (6)(7)
We will just deal with the case that (5) is satisfied; the rest case is similar. Let 𝐹=ln𝑓∈(−𝑎,0), then we have 𝑓^′=𝑓^′𝑓>0,𝐹^′′=𝑓^′′𝑓−𝑓^′𝑓². (8) Using this fact, we can get from (4) the following equation: 𝐹^′′+1𝑟𝐹^′−𝜋𝑎𝜋ctg𝑎𝐹𝐹^′2+1𝑟²𝜋𝑎𝜋ctg𝑎𝐹=0for1>𝑟>0, (9) with 𝐹(0)=−𝑎, 𝐹(1)=0, and 𝐹^′(𝑟)>0 for 1>𝑟>0.
Regarding 𝑟 as a function of 𝐹, we have the following relations: 𝐹_𝑟=𝑟_𝐹⁻¹,𝐹_𝑟𝑟=−𝑟_𝐹⁻³𝑟_𝐹𝐹. (10) Using these facts, we can get from (9) the following equation: 𝑟^′′𝑟−𝑟^′𝑟²+𝜋𝑎𝜋ctg𝑎𝐹𝑟^′𝑟−𝑟^′𝑟³𝜋𝑎𝜋ctg𝑎𝐹=0 (11) for 0>𝐹>−𝑎. Let 𝑥=(ln𝑟)^′(𝐹); from (11) we can get the following equation: 𝑥^′+𝜋𝑎𝜋ctg𝑎𝐹𝜋⋅𝑥−𝑎𝜋ctg𝑎𝐹⋅𝑥³=0. (12) One can solve this Bernoulli equation to obtain 𝑥⁻²=1+𝑐₀𝜋sin𝑎𝐹². (13) Here 𝑐₀ is a constant depending on the choice of the function 𝑓. So 1𝑥=1+𝑐₀(sin((𝜋/𝑎)𝐹))². (14) Since 𝑥=(ln𝑟)^′(𝐹), we can get (ln𝑟)(𝐹)=^𝐹₀𝑥(𝑡)𝑑𝑡=^𝐹₀11+𝑐₀(sin((𝜋/𝑎)𝑡))²𝑑𝑡. (15) 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 Theorem 2.

Proof of Theorem 2. 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 |𝑑𝑢||𝑢|(|𝑢.ln1/|) (16) 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 Theorem 1. Suppose 𝜓 is a rotationally symmetric harmonic diffeomorphism from 𝑃(𝑎) onto 𝔻^∗ with the metric 𝜎₂𝑑|𝑢|, with the form 𝜓=𝑔(𝑟)𝑒^𝑖𝜃, then substituting 𝜓,𝜎₂ to 𝑢,𝜎 in (2), respectively, we can get 𝑔^′′+𝑔^′𝑟−𝑔𝑟²−1+ln𝑔𝑔𝑔ln𝑔^′2−𝑔²𝑟²=0 (17) for 1>𝑟>𝑒^−𝑎. Since 𝑣 is a harmonic diffeomorphism from 𝑃(𝑎) onto 𝔻^∗, we have 𝑔𝑒^−𝑎(1)𝑔=0,𝑔=1,_′(𝑟)>0for1>𝑟>𝑒^−𝑎, (18) or 𝑔𝑒^−𝑎(1)𝑔=1,𝑔=0,_′(𝑟)<0for1>𝑟>𝑒^−𝑎. (19)(20)
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 𝐺^′′+1𝑟𝐺^′−1𝐺𝐺^′2+1𝑟²𝐺=0 (21) for 1>𝑟>𝑒^−𝑎, with 𝐺(1)=0 and lim_{𝑟→𝑒−𝑎}𝐺(𝑟)=−∞.
Regarding 𝑟 as a function of 𝐺, using a similar formula of (10), from (21) we can get 𝑟^′′𝑟−𝑟^′𝑟²+𝑟^′−1𝑟𝐺𝐺𝑟^′𝑟³=0,𝐺∈(−∞,0). (22) Similar to solving (11), we can get the solution to (22) as follows: (ln𝑟)^′(𝐺)=11+𝑐₁𝐺²,𝐺∈(−∞,0) (23) for some nonnegative constant 𝑐₁ depending on the choice of 𝑔.
If 𝑐₁ is equal to 0, then 𝑔=𝑟; this is in contradiction to (18).
If 𝑐₁ is positive, then taking integration on both sides of (23), we can get (ln𝑟)(𝐺)=^𝐺₀11+𝑐₁𝑡²=1𝑑𝑡√𝑐₁ln√𝑐₁𝐺+1+𝑐₁𝐺². (24) So 𝑟=√𝑐₁ln𝑔+1+𝑐₁ln²𝑔^√1/𝑐1, (25) with lim_𝑔→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 from 𝔻^∗ 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 Theorem 3. Let us prove the first part of this theorem, that is, show the nonexistence of rotationally symmetric harmonic diffeomorphism from 𝔻^∗ 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)𝑒^𝑖𝜃, and then we can get ℎ^′′+1𝑟ℎ^′−1𝑟²ℎ=0for1>𝑟>0, (26) with ℎ(0)=𝑒^−𝑎(1)ℎ,ℎ=1,_′(𝑟)>0for1>𝑟>0, (27) or ℎ(0)=1,ℎ(1)=𝑒^−𝑎,ℎ^′(𝑟)<0for1>𝑟>0. (28)
We will just deal with the case that (27) is satisfied; the rest case is similar. Let 𝐻=(lnℎ)^′; then we can get 𝐻^′+𝐻²+1𝑟𝐻=0for1>𝑟>0. (29) Solving this equation, we can get 1𝐻=𝑟+1𝑐₃𝑟³−𝑟²=1𝑟+𝑐²₃𝑐₃−𝑐𝑟−1𝑟³−1𝑟². (30) Here 𝑐₃ is a constant depending on the choice of ℎ. So lnℎ=ln𝑟+𝑐₃ln𝑐₃𝑟−1−𝑐₃1ln𝑟+𝑟+𝑐₄. (31) Here 𝑐₄ is a constant depending on the choice of ℎ. Hence ℎ=𝑟𝑐₃𝑟−1^𝑐3𝑟^−𝑐3𝑒^1/𝑟𝑒^𝑐4. (32) From (32), we can get lim_𝑟→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 from 𝑃(𝑎) onto 𝔻^∗ with its Euclidean metric. It suffices to find a map from 𝑃(𝑎) onto 𝔻^∗ with the form 𝑞(𝑟)𝑒^𝑖𝜃 such that 𝑞^′′+1𝑟𝑞^′−1𝑟²𝑞=0for1>𝑟>𝑒^−𝑎 (33) 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 𝑞=𝑒⁻¹𝑒^𝑎−1^−𝑒𝑎𝑟𝑒^𝑎𝑟−1^𝑒𝑎𝑟^−𝑒𝑎𝑒^1/𝑟 (34) is a solution to (33).
Therefore, we finished the proof of Theorem 3.

4. Harmonic Maps from ℂ 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 ℂ 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)𝑒^𝑖𝜃, and then we can get 𝑘^′′+1𝑟𝑘^′−1𝑟²𝑘+2𝑘1−𝑘²𝑘^′2−𝑘²𝑟²=0for𝑟>0 (35) with 𝑘(0)=0,𝑘^′(𝑟)>0for𝑟>0. (36) Regarding 𝑟 as a function of 𝑘, setting 𝑣=(ln𝑟)^′(𝑘), (35) can be rewritten as 1−𝑘²𝑣^′−2𝑘𝑣+𝑣³𝑘+𝑘³=0. (37) That is, 𝑣^−2′+4𝑘1−𝑘²𝑣⁻²=2𝑘+𝑘³1−𝑘². (38) One can solve this equation to obtain 𝑣⁻²=𝑘²+𝑐₅1−𝑘²² (39) for some nonnegative constant 𝑐₅ depending on the choice of the function 𝑘.
If 𝑐₅=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.
If 𝑐₅>0, then 𝑏₁≥𝑘²+𝑐₅(1−𝑘²)²≥𝑏₂ for some positive constants 𝑏₁ and 𝑏₂. So (ln𝑟)^′(𝑘≤1)𝑏₂. (40) This is in contradiction to the assumption that 𝑟→∞ as 𝑘→1.
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).