Rotationally Symmetric Harmonic Diffeomorphisms between Surfaces

Theorem 1. For any a > 0, there is no rotationally symmetric harmonic diffeomorphism from 𝔻^* onto P(a) with its hyperbolic metric.

Theorem 2. For any a > 0, there is no rotationally symmetric harmonic diffeomorphism from P(a) onto 𝔻^* with its hyperbolic metric.

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.

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

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

for 1 > r > 0. Since u is a harmonic diffeomorphism from 𝔻^* onto P(a), we have or

We will just deal with the case that (5) is satisfied; the rest case is similar. Let F = lnf ∈ (−a, 0), then we have

Using this fact, we can get from (4) the following equation: 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:

Using these facts, we can get from (9) the following equation: for 0 > F > −a. Let x = (lnr)^′(F); from (11) we can get the following equation: One can solve this Bernoulli equation to obtain Here c₀ is a constant depending on the choice of the function f. So Since x = (lnr)^′(F), we can get 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.

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

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

for 1 > r > e^−a. Since v is a harmonic diffeomorphism from P(a) onto 𝔻^*, we have or

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

for 1 > r > e^−a, with G(1) = 0 and $$.

Regarding r as a function of G, 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 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

So 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.

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

with or

We will just deal with the case that (27) is satisfied; the rest case is similar. Let H = (lnh)^′; then we can get

Solving this equation, we can get Here c₃ is a constant depending on the choice of h. So Here c₄ is a constant depending on the choice of h. Hence 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

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 is a solution to (33).

Therefore, we finished the proof of Theorem 3.

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

with Regarding r as a function of k, setting v = (lnr)^′(k), (35) can be rewritten as That is, One can solve this equation to obtain 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

This is in contradiction to the assumption that r → ∞ as k → 1.

Therefore, Proposition 4 holds.