Posted on Categories:Ordinary Differential Equations, 常微分方程, 数学代写

# 数学代写|常微分方程代考Ordinary Differential Equations代写|MA26600 DEFINITION AND GENERAL PROPERTIES

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## 数学代写|常微分方程代考Ordinary Differential Equations代写|DEFINITION AND GENERAL PROPERTIES

Definition 1.1 The Bessel equation is defined to be (1.1)
$$z^2 \frac{d^2 w}{d z^2}+z \frac{d w}{d z}+\left(z^2-n^2\right) w=0$$
where $n$, any complex number, is the only parameter involved.
The point $z=0$ is the only singularity of the equation in the finite part of the $z$-plane which is regular, and on setting $z=1 / \zeta(1.1)$ transforms to
$$\zeta^4 \frac{d^2 w}{d \zeta^2}+\zeta^3 \frac{d w}{d \zeta}+\left(1-n^2 \zeta^2\right) w=0$$
which shows that $\zeta=0$ is not a regular singularity of this equation, i.e. $z=\infty$ is not a regular singularity of the Bessel equation. Hence the Bessel equation is not one of the Fuchsian type.
It also follows that the domain of the singularity $z=0$ is the entire $z$ plane except the point $z=0$, and any solution obtained will be valid everywhere except possibly $z=0$.
Note that the co-efficients of $(1.1)$ are respectively a quadratic function, a linear function and again a quadratic function so that the Bessel equation is not expected to have contour integral solutions of Jordan-Pochhammer type. In fact, special kernels and contours have to be tried for the Bessel equation.
For the Bessel equation it will be worthwhile to calculate the Wronskian of different pairs of solutions, and a general formula for this is given by Theorem $1.1$ If $P(z)$ and $Q(z)$ are any two solutions of the Bessel equation, then their Wronskian $\Delta(P, Q)=P Q^{\prime}-P^{\prime} Q$ is given by (1.2)
$$\Delta(P, Q)=C / z \quad(C=\text { constant })$$
Proof. Writing (1.1) in the form
$$\frac{d}{d z}\left(z \frac{d w}{d z}\right)+\left(z-n^2 z^{-1}\right) w=0$$
$P(z)$ and $Q(z)$ both satisfy this equation giving two relations, from which we get
$$\frac{d}{d z}\left[z\left(P Q^{\prime}-P^{\prime} Q\right)\right]=0$$
Hence the result.

## 数学代写|常微分方程代考Ordinary Differential Equations代写|SERIES SOLUTIONS NEAR z=0

Define the operator $L$ by
(2.1)
$$L=z^2 \frac{d^2}{d z^2}+z \frac{d}{d z}+\left(z^2-n^2\right)$$
so that (1.1) is $L(w)=0$. We have
(2.2)
\begin{aligned} L\left(z^\lambda\right) &=\lambda(\lambda-1) z^\lambda+\lambda z^\lambda+\left(z^2-n^2\right) z^\lambda \ &=\left(\lambda^2-n^2\right) z^\lambda+z^{\lambda+2} \end{aligned}
Since $z=0$ is a regular singularity of the equation having domain $\mathrm{C}-{0}$, we assume
$$w=z^\rho \sum_0^{\infty} a_k z^k=\sum_0^{\infty} a_k z^{\rho+k} \quad\left(a_0 \neq 0\right)$$
to be a solution of $L(w)=0$ in $\mathbf{C}-{0}$. Substituting the solution (2.3 ) in the equation $L(w)=0$, we get by virtue of (2.2)
(2.4) $z^\rho \sum_0^{\infty} a_k\left[\left{(\rho+k)^2-n^2\right} z^k+z^{k+2}\right] \equiv 0$
in $\mathbf{C}-{0}$. Equating to 0 the co-efficient of $z^\rho$ and noting that $q_0 \neq 0$, we get the indicial equation $\rho^2-n^2=0$ which gives
(2.5) $\rho=n,-n$
Equating to 0 the co-efficients of $z^{\rho+1}$ and $z^{\rho+k+2}$, we get respectively
$$\begin{array}{lc} (2.6) & a_1(\rho-n+1)(\rho+n+1)=0 \ (2.7) & a_{k+2}(\rho-n+k+2)(\rho+n+k+2)+a_k=0 \ \text { or } & \end{array}$$

(2.8) $\quad a_{k+2}=-\frac{1}{(\rho-n+k+2)(\rho+n+k+2)} a_k \quad(k=0,1,2, .$. From (2.6) we have $a_1=0$, by choice when necessary.
Case I. $2 n$, the exponent difference is not an integer or 0 . For $\rho=n,(2.8)$ reduces to
$$a_{k+2}=-\frac{1}{(k+2)(2 n+k+2)} a_k$$
As $a_1=0, a_3=a_5=\ldots=0$, and
$$a_2=-\frac{1}{2(2 n+2)} a_0, \quad a_4=\frac{1}{2.4(2 n+2)(2 n+4)} a_0, \ldots$$
giving
$$w_1=a_0 z^n\left[1-\frac{z^2}{2(2 n+2)}+\frac{z^4}{2.4(2 n+2)(2 n+4)}-\ldots\right]$$
which is valid everywhere except possibly $z=0$.

# 常微分方程代写

## 数学代写常微分方程代考Ordinary Differential Equations代写|DEFINITION AND GENERAL PROPERTIES

$$z^2 \frac{d^2 w}{d z^2}+z \frac{d w}{d z}+\left(z^2-n^2\right) w=0$$

$$\zeta^4 \frac{d^2 w}{d \zeta^2}+\zeta^3 \frac{d w}{d \zeta}+\left(1-n^2 \zeta^2\right) w=0$$

$$\Delta(P, Q)=C / z \quad(C=\text { constant })$$

$$\frac{d}{d z}\left(z \frac{d w}{d z}\right)+\left(z-n^2 z^{-1}\right) w=0$$
$P(z)$ 和 $Q(z)$ 两者都满足这个方程，给出两个关系，从中我们得到
$$\frac{d}{d z}\left[z\left(P Q^{\prime}-P^{\prime} Q\right)\right]=0$$

## 数学代写|常微分方程代考Ordinary Differential Equations代写|SERIES SOLUTIONS NEAR $z=0$

$(2.1)$
$$L=z^2 \frac{d^2}{d z^2}+z \frac{d}{d z}+\left(z^2-n^2\right)$$

$$L\left(z^\lambda\right)=\lambda(\lambda-1) z^\lambda+\lambda z^\lambda+\left(z^2-n^2\right) z^\lambda \quad=\left(\lambda^2-n^2\right) z^\lambda+z^{\lambda+2}$$

$$w=z^\rho \sum_0^{\infty} a_k z^k=\sum_0^{\infty} a_k z^{\rho+k} \quad\left(a_0 \neq 0\right)$$
(2.4)〈left 的分隔符缺失或无法识别

$$\text { (2.5) } \rho=n,-n$$

(2.6) $a_1(\rho-n+1)(\rho+n+1)=0(2.7) \quad a_{k+2}(\rho-n+k+2)(\rho+n+k+2)+a_k=0$ or
(2.8) $a_{k+2}=-\frac{1}{(\rho-n+k+2)(\rho+n+k+2)} a_k \quad\left(k=0,1,2, \ldots\right.$ 从 $(2.6)$ 我们有 $a_1=0$, 必要时选铎。

$$a_{k+2}=-\frac{1}{(k+2)(2 n+k+2)} a_k$$

$$a_2=-\frac{1}{2(2 n+2)} a_0, \quad a_4=\frac{1}{2.4(2 n+2)(2 n+4)} a_0, \ldots$$

$$w_1=a_0 z^n\left[1-\frac{z^2}{2(2 n+2)}+\frac{z^4}{2.4(2 n+2)(2 n+4)}-\ldots\right]$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。
The provided text is a mathematical passage discussing the Bessel equation and its properties. It does not contain any specific mathematical questions or problems to solve, nor does it provide answers to any such questions.

**Content Structure:**
The passage defines the Bessel equation, discusses its singularities, and provides a theorem and proof about the Wronskian of solutions to the Bessel equation. It also outlines the method for finding series solutions near \( z = 0 \).

**Analysis:**
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Posted on Categories:Ordinary Differential Equations, 常微分方程, 数学代写

# 数学代写|常微分方程代考Ordinary Differential Equations代写|MA26600 DEFINITION AND GENERAL PROPERTIES

## avatest™帮您通过考试

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## 数学代写|常微分方程代考Ordinary Differential Equations代写|DEFINITION AND GENERAL PROPERTIES

Definition 1.1 The Bessel equation is defined to be (1.1)
$$z^2 \frac{d^2 w}{d z^2}+z \frac{d w}{d z}+\left(z^2-n^2\right) w=0$$
where $n$, any complex number, is the only parameter involved.
The point $z=0$ is the only singularity of the equation in the finite part of the $z$-plane which is regular, and on setting $z=1 / \zeta(1.1)$ transforms to
$$\zeta^4 \frac{d^2 w}{d \zeta^2}+\zeta^3 \frac{d w}{d \zeta}+\left(1-n^2 \zeta^2\right) w=0$$
which shows that $\zeta=0$ is not a regular singularity of this equation, i.e. $z=\infty$ is not a regular singularity of the Bessel equation. Hence the Bessel equation is not one of the Fuchsian type.
It also follows that the domain of the singularity $z=0$ is the entire $z$ plane except the point $z=0$, and any solution obtained will be valid everywhere except possibly $z=0$.
Note that the co-efficients of $(1.1)$ are respectively a quadratic function, a linear function and again a quadratic function so that the Bessel equation is not expected to have contour integral solutions of Jordan-Pochhammer type. In fact, special kernels and contours have to be tried for the Bessel equation.
For the Bessel equation it will be worthwhile to calculate the Wronskian of different pairs of solutions, and a general formula for this is given by Theorem $1.1$ If $P(z)$ and $Q(z)$ are any two solutions of the Bessel equation, then their Wronskian $\Delta(P, Q)=P Q^{\prime}-P^{\prime} Q$ is given by (1.2)
$$\Delta(P, Q)=C / z \quad(C=\text { constant })$$
Proof. Writing (1.1) in the form
$$\frac{d}{d z}\left(z \frac{d w}{d z}\right)+\left(z-n^2 z^{-1}\right) w=0$$
$P(z)$ and $Q(z)$ both satisfy this equation giving two relations, from which we get
$$\frac{d}{d z}\left[z\left(P Q^{\prime}-P^{\prime} Q\right)\right]=0$$
Hence the result.

## 数学代写|常微分方程代考Ordinary Differential Equations代写|SERIES SOLUTIONS NEAR z=0

Define the operator $L$ by
(2.1)
$$L=z^2 \frac{d^2}{d z^2}+z \frac{d}{d z}+\left(z^2-n^2\right)$$
so that (1.1) is $L(w)=0$. We have
(2.2)
\begin{aligned} L\left(z^\lambda\right) &=\lambda(\lambda-1) z^\lambda+\lambda z^\lambda+\left(z^2-n^2\right) z^\lambda \ &=\left(\lambda^2-n^2\right) z^\lambda+z^{\lambda+2} \end{aligned}
Since $z=0$ is a regular singularity of the equation having domain $\mathrm{C}-{0}$, we assume
$$w=z^\rho \sum_0^{\infty} a_k z^k=\sum_0^{\infty} a_k z^{\rho+k} \quad\left(a_0 \neq 0\right)$$
to be a solution of $L(w)=0$ in $\mathbf{C}-{0}$. Substituting the solution (2.3 ) in the equation $L(w)=0$, we get by virtue of (2.2)
(2.4) $z^\rho \sum_0^{\infty} a_k\left[\left{(\rho+k)^2-n^2\right} z^k+z^{k+2}\right] \equiv 0$
in $\mathbf{C}-{0}$. Equating to 0 the co-efficient of $z^\rho$ and noting that $q_0 \neq 0$, we get the indicial equation $\rho^2-n^2=0$ which gives
(2.5) $\rho=n,-n$
Equating to 0 the co-efficients of $z^{\rho+1}$ and $z^{\rho+k+2}$, we get respectively
$$\begin{array}{lc} (2.6) & a_1(\rho-n+1)(\rho+n+1)=0 \ (2.7) & a_{k+2}(\rho-n+k+2)(\rho+n+k+2)+a_k=0 \ \text { or } & \end{array}$$

(2.8) $\quad a_{k+2}=-\frac{1}{(\rho-n+k+2)(\rho+n+k+2)} a_k \quad(k=0,1,2, .$. From (2.6) we have $a_1=0$, by choice when necessary.
Case I. $2 n$, the exponent difference is not an integer or 0 . For $\rho=n,(2.8)$ reduces to
$$a_{k+2}=-\frac{1}{(k+2)(2 n+k+2)} a_k$$
As $a_1=0, a_3=a_5=\ldots=0$, and
$$a_2=-\frac{1}{2(2 n+2)} a_0, \quad a_4=\frac{1}{2.4(2 n+2)(2 n+4)} a_0, \ldots$$
giving
$$w_1=a_0 z^n\left[1-\frac{z^2}{2(2 n+2)}+\frac{z^4}{2.4(2 n+2)(2 n+4)}-\ldots\right]$$
which is valid everywhere except possibly $z=0$.

# 常微分方程代写

## 数学代写常微分方程代考Ordinary Differential Equations代写|DEFINITION AND GENERAL PROPERTIES

$$z^2 \frac{d^2 w}{d z^2}+z \frac{d w}{d z}+\left(z^2-n^2\right) w=0$$

$$\zeta^4 \frac{d^2 w}{d \zeta^2}+\zeta^3 \frac{d w}{d \zeta}+\left(1-n^2 \zeta^2\right) w=0$$

$$\Delta(P, Q)=C / z \quad(C=\text { constant })$$

$$\frac{d}{d z}\left(z \frac{d w}{d z}\right)+\left(z-n^2 z^{-1}\right) w=0$$
$P(z)$ 和 $Q(z)$ 两者都满足这个方程，给出两个关系，从中我们得到
$$\frac{d}{d z}\left[z\left(P Q^{\prime}-P^{\prime} Q\right)\right]=0$$

## 数学代写|常微分方程代考Ordinary Differential Equations代写|SERIES SOLUTIONS NEAR $z=0$

$(2.1)$
$$L=z^2 \frac{d^2}{d z^2}+z \frac{d}{d z}+\left(z^2-n^2\right)$$

$$L\left(z^\lambda\right)=\lambda(\lambda-1) z^\lambda+\lambda z^\lambda+\left(z^2-n^2\right) z^\lambda \quad=\left(\lambda^2-n^2\right) z^\lambda+z^{\lambda+2}$$

$$w=z^\rho \sum_0^{\infty} a_k z^k=\sum_0^{\infty} a_k z^{\rho+k} \quad\left(a_0 \neq 0\right)$$
(2.4)〈left 的分隔符缺失或无法识别

$$\text { (2.5) } \rho=n,-n$$

(2.6) $a_1(\rho-n+1)(\rho+n+1)=0(2.7) \quad a_{k+2}(\rho-n+k+2)(\rho+n+k+2)+a_k=0$ or
(2.8) $a_{k+2}=-\frac{1}{(\rho-n+k+2)(\rho+n+k+2)} a_k \quad\left(k=0,1,2, \ldots\right.$ 从 $(2.6)$ 我们有 $a_1=0$, 必要时选铎。

$$a_{k+2}=-\frac{1}{(k+2)(2 n+k+2)} a_k$$

$$a_2=-\frac{1}{2(2 n+2)} a_0, \quad a_4=\frac{1}{2.4(2 n+2)(2 n+4)} a_0, \ldots$$

$$w_1=a_0 z^n\left[1-\frac{z^2}{2(2 n+2)}+\frac{z^4}{2.4(2 n+2)(2 n+4)}-\ldots\right]$$

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。
user
The provided text is a mathematical passage discussing the Bessel equation and its properties. It does not contain any specific mathematical questions or problems to solve, nor does it provide answers to any such questions.

**Content Structure:**
The passage defines the Bessel equation, discusses its singularities, and provides a theorem and proof about the Wronskian of solutions to the Bessel equation. It also outlines the method for finding series solutions near \( z = 0 \).

**Analysis:**
TYPE: NO_QUESTION