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} for a>0,
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 <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M21"><mml:mrow><mml:msup><mml:mrow><mml:mi>𝔻</mml:mi></mml:mrow><mml:mrow><mml:mi>*</mml:mi></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula> to <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M22"><mml:mi>P</mml:mi><mml:mo mathvariant="bold">(</mml:mo><mml:mi>a</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> 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
(2)uzz-+2σuσuzuz-=0.

Now let us prove Theorem 1.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.
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. 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)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)>0 for 1>r>0,
or
(6)f(0)=1, f(1)=e-a,(7)f′(r)<0 for 1>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) =0 for 1>r>0,
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:
(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 Theorem 2.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.
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 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)eiθ, then substituting ψ, σ2 to u, σ in (2), respectively, we can get
(17)g′′+g′r-gr2-1+lngg lng((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)>0 for 1>r>e-a,
or
(19)g(e-a)=1, g(1)=0,(20)g′(r)<0 for 1>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)=-∞.

Regarding r 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.

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
(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 from <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M137"><mml:mrow><mml:msup><mml:mrow><mml:mi>𝔻</mml:mi></mml:mrow><mml:mrow><mml:mi>*</mml:mi></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula> to <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M138"><mml:mi>P</mml:mi><mml:mo mathvariant="bold">(</mml:mo><mml:mi>a</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> with Its Euclidean Metric and Vice Versa
Now let us consider the case of that the target has the Euclidean metric.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.
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)eiθ, and then we can get
(26)h′′+1rh′-1r2h=0 for 1>r>0,
with
(27)h(0)=e-a, h(1)=1,h′(r)>0 for 1>r>0,
or
(28)h(0)=1, h(1)=e-a,h′(r)<0 for 1>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=0 for 1>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 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)eiθ such that
(33)q′′+1rq′-1r2q=0 for 1>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 Theorem 3.

4. Harmonic Maps from <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M170"><mml:mrow><mml:mi>ℂ</mml:mi></mml:mrow></mml:math></inline-formula> 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)eiθ, and then we can get
(35)k′′+1rk′-1r2k+2k1-k2[(k′)2-k2r2] =0 for r>0
with
(36)k(0)=0, k′(r)>0 for r>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.

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
(40)|(lnr)′(k)|≤1b2.
This is in contradiction to the assumption that r→∞ as k→1.

Therefore, Proposition 4 holds.