# 微积分网课代修|函数代写Function theory代考|MAEN5060 The Poincar´e Metric

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|The Poincar´e Metric

The Poincaré metric on the disk has occurred frequently in this book. This metric is the paradigm for much of what we want to do in the present chapter, and we shall treat it in some detail here. The Poincaré metric on the disk $D$ is given by
$$\rho(z)=\frac{1}{1-|z|^{2}} .$$
(For the record, we note that there is no agreement in the literature as to what constant goes in the numerator; many references use a factor of 2.)
In this and succeeding sections, we shall use the phrase “conformal map” to refer to a holomorphic mapping of one planar region to another that is both one-to-one and onto.

Proposition 2.3.12. Let $\rho$ be the Poincaré metric on the disk D. Let $h$ : $D \rightarrow D$ be a conformal self-map of the disk. Then $h$ is an isometry of the pair $(D, \rho)$ with the pair $(D, \rho)$.
Proof. We have that
$$h^{} \rho(z)=\rho(h(z)) \cdot\left|h^{\prime}(z)\right| .$$ We now have two cases: (i) If $h$ is a rotation, then $h(z)=\mu \cdot z$ for some unimodular constant $\mu \in \mathbb{C}$. So $\left|h^{\prime}(z)\right|=1$ and $$h^{} \rho(z)=\rho(h(z))=\rho(\mu z)=\frac{1}{1-|\mu z|^{2}}=\frac{1}{1-|z|^{2}}=\rho(z)$$
as desired.
(ii) If $h$ is a Möbius transformation, then
$$h(z)=\frac{z-a}{1-\bar{a} z}, \quad \text { some constant } a \in D$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|The Schwarz Lemma

One of the important facts about the Poincaré metric is that it can be used to study not just conformal maps but all holomorphic maps of the disk. The key to this assertion is the classical Schwarz lemma. We begin with an elegant geometric interpretation of the Schwarz-Pick lemma (see Section 2.1).

Proposition 2.3.22. Let $f: D \rightarrow D$ be holomorphic. Then $f$ is distancedecreasing in the Poincaré metric $\rho$. That is, for any $z \in D$,
$$f^{} \rho(z) \leq \rho(z)$$ The integrated form of this assertion is that if $\gamma:[0,1] \rightarrow D$ is a continuously differentiable curve, then $$\ell_{\rho}\left(f_{} \gamma\right) \leq \ell_{\rho}(\gamma) .$$
Therefore, if $P$ and $Q$ are elements of $D$, we may conclude that
$$d_{\rho}(f(P), f(Q)) \leq d_{\rho}(P, Q)$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|The Poincar’e Metric

$$\rho(z)=\frac{1}{1-|z|^{2}} .$$
（作为记录，我们注意到文献中对于分子中的常数没有达成一致；许多参考文献使用因 子 2 。)

$$h \rho(z)=\rho(h(z)) \cdot\left|h^{\prime}(z)\right| .$$

$$h \rho(z)=\rho(h(z))=\rho(\mu z)=\frac{1}{1-|\mu z|^{2}}=\frac{1}{1-|z|^{2}}=\rho(z)$$

(ii) 如果 $h$ 是一个莫比乌斯变换，那么
$$h(z)=\frac{z-a}{1-\bar{a} z}, \quad \text { some constant } a \in D$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|The Schwarz Lemma

$$f \rho(z) \leq \rho(z)$$

$$\ell_{\rho}(f \gamma) \leq \ell_{\rho}(\gamma) .$$

$$d_{\rho}(f(P), f(Q)) \leq d_{\rho}(P, Q)$$