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# 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。

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

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

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

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