Posted on Categories:Riemann surface, 数学代写, 黎曼曲面

# 数学代写|黎曼曲面代写Riemann surface代考|Models on the unit disk

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|黎曼曲面代写Riemann surface代考|Models on the unit disk

The goal of this section is to prove that every $f \in \operatorname{Hol}(\mathbb{D}, \mathbb{D})$ which is not superattracting elliptic admits a model in the sense of Definition 3.5.2. When $f$ is attracting elliptic, we know this already (Corollary 4.1.3); so we shall focus on hyperbolic and parabolic maps.

The idea is to apply Theorem 3.5.10; therefore, we must build a simply connected $f$-absorbing domain where $f$ is injective. To do so, we need a couple of lemmas that might be interesting on their own.

Lemma 4.5.1. Let $\Omega \subset \mathbb{C}$ be a convex domain and take $f \in \operatorname{Hol}(\Omega, \mathbb{C})$ such that $\operatorname{Re} f^{\prime} \geq 0$ on $\Omega$. Then $f$ is either constant or injective.

Proof. If there is $z_0 \in \Omega$ so that $\operatorname{Re} f^{\prime}\left(z_0\right)=0$, then by the minimum principle for harmonic functions $f^{\prime}$ is constant, and hence $f$ is either constant or injective. We henceforth can suppose that $\operatorname{Re} f^{\prime}>0$ everywhere.

Assume, by contradiction, that $f\left(z_1\right)=f\left(z_2\right)$ for two distinct points $z_1, z_2 \in \Omega$. Integrating along the segment $\sigma$ from $z_1$ to $z_2$, we obtain
$$0=f\left(z_2\right)-f\left(z_1\right)=\int_\sigma f^{\prime}(\zeta) d \zeta=\left(z_2-z_1\right) \int_0^1 f^{\prime}\left(z_1+t\left(z_2-z_1\right)\right) d t$$

## 数学代写|黎曼曲面代写Riemann surface代考|The hyperbolic step

In the previous section, we saw that if $f \in \operatorname{Hol}(\mathbb{D}, \mathbb{D})$ is not superattracting elliptic then it admits a model. A natural question is whether we can understand the model just by looking at the function $f$. The answer is essentially affirmative; to show how, we need a couple of definitions and preliminary results.

Definition 4.6.1. Let $X$ be a hyperbolic Riemann surface and take $f \in \operatorname{Hol}(X, X)$ and $z \in X$. For any $\mu \geq 1$ the Schwarz-Pick lemma implies that the sequence $\left{\omega_X\left(f^v(z), f^{v+\mu}(z)\right)\right}_{v \in \mathbb{N}} \subset \mathbb{R}^{+}$is decreasing, and hence it has a limit $s_\mu^f(z) \in \mathbb{R}^{+}$, that we shall call the hyperbolic $\mu$-step of $f$ at $z$. When the function $f$ is clear by the context we shall write $s_\mu$ instead of $s_\mu^f$. If $\mu=1$, we shall write $s^f(z)$ instead of $s_1^f(z)$ and we shall call $s^f(z)$ the hyperbolic step of $f$ at $z$. We shall say that $f$ has positive hyperbolic step if there exists $z_0 \in \mathbb{D}$ such that $s^f\left(z_0\right)>0$; otherwise, we say that $f$ has a zero hyperbolic step. If $X$ is an elliptic or parabolic Riemann surface, we shall put $s_\mu^f \equiv 0$ for all $\mu \geq 1$ and $f \in \operatorname{Hol}(X, X)$.

Remark 4.6.2. In the literature, parabolic maps with a positive hyperbolic step are sometimes called of simply parabolic type or of parabolic II type or of automorphic type. Analogously, parabolic maps with a zero hyperbolic step are sometimes called of doubly parabolic type or of parabolic I type or of nonautomorphic type.

## 数学代写|黎曼曲面代写Riemann surface代考|Models on the unit disk

$$0=f\left(z_2\right)-f\left(z_1\right)=\int_\sigma f^{\prime}(\zeta) d \zeta=\left(z_2-z_1\right) \int_0^1 f^{\prime}\left(z_1+t\left(z_2-z_1\right)\right) d t$$

## 数学代写|黎曼曲面代写Riemann surface代考|The hyperbolic step

4.6.1.定义设$X$为双曲黎曼曲面取$f \in \operatorname{Hol}(X, X)$和$z \in X$。对于任何$\mu \geq 1$, Schwarz-Pick引理表明，序列$\left{\omega_X\left(f^v(z), f^{v+\mu}(z)\right)\right}{v \in \mathbb{N}} \subset \mathbb{R}^{+}$是递减的，因此它有一个极限$s\mu^f(z) \in \mathbb{R}^{+}$，我们称之为$f$在$z$的双曲$\mu$ -步。当上下文明确了$f$函数时，我们应该写$s_\mu$而不是$s_\mu^f$。如果是$\mu=1$，我们将写成$s^f(z)$而不是$s_1^f(z)$，我们将称$s^f(z)$为$z$的$f$的双曲步。我们说$f$有正双曲阶跃如果存在$z_0 \in \mathbb{D}$使得$s^f\left(z_0\right)>0$;否则，我们说$f$的双曲步长为零。如果$X$是椭圆型或抛物线型黎曼曲面，我们将把$s_\mu^f \equiv 0$代入所有的$\mu \geq 1$和$f \in \operatorname{Hol}(X, X)$。

4.6.2。在文献中，具有正双曲阶跃的抛物型映射有时被称为单纯抛物型或抛物II型或自同构型。类似地，具有零双曲阶跃的抛物线映射有时称为双抛物线型或抛物线I型或非自同构型。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。