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

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## 数学代写|常微分方程代考Ordinary Differential Equations代写|FUNDAMENTAL SETS OF் SOLUTIONS

Let $z_{0}$ be an ordinary point of the differential equation $(1.2)$ and $D$ be its domain. Now corresponding to different sets of initial values at $z_{0}$, we get different analytic solutions of the equation in $D$. But these solutions are not all essentially different for we can find $m$ different solutions in $D$ such that any solution in $D$ (i.e. the solution corresponding to arbitrarily prescribed initial values at $z_{0}$ ) is a linear combination of these solutions.
Theorem $3.1$ Let $w_{1}, w_{2}, \ldots, w_{m}$ be manalytic solutions of the differential equation (1.2) in the domain $D$ of an ordinary point $z_{0}$, and let $\Delta(z)=$ $\Delta\left(w_{1}, w_{2}, \ldots, w_{m}\right)$ denote the Wronskian of $w_{1}, w_{2}, \ldots, w_{m}$. Then
$$\Delta(z)=\Delta\left(z_{0}\right) e^{\int_{z_{0}}^{z} p_{1}(\zeta) d \zeta} \quad \text { in } D$$
where the integral is taken along any Jordan arc joining $z_{0}$ to $z$ and lying in $D$
(ii) the set of analytic functions $w_{1}, w_{2}, \ldots, w_{m}$ is linearly independent iff $\Delta\left(z_{0}\right) \neq 0$, and

(iii) if $w_{1}, w_{2}, \ldots, w_{m}$ form a linearly independent set and $w$ is any analytic solution in $D$, then constants $c_{1}, c_{2}, \ldots, c_{m}$ can be found such that
$$w=c_{1} w_{1}+c_{2} w_{2}+\ldots+c_{m} w_{m} \text { in } D$$
Proof. (i) Since $w_{i}(i=1,2, \ldots, m)$ is a solution of $(1.2)$ in $D$, we have in $D$ (3.1)
$$w_{i}^{(m)}=p_{1} w_{i}^{(m-1)}+p_{2} w_{i}^{(m-2)}+\ldots+p_{m} w_{i} \quad(i=1, \ldots, m)$$
Now by differentiating the determinant
$$\Delta(z)=\left|\begin{array}{cccc} w_{1} & w_{2} & \ldots & w_{m} \ w_{1}^{\prime} & w_{2}^{\prime} & \ldots & w_{m}^{\prime} \ \cdots & \ldots & \ldots & \ldots \ w_{1}^{(m-1)} & w_{2}^{(m-1)} & \ldots & w_{m}^{(m-1)} \end{array}\right|$$

## 数学代写|常微分方程代考Ordinary Differential Equations代写|Special fundamental sets

1 The set of $m$ analytic solutions $w_{1}, w_{2}, \ldots, w_{m}$ in the domain $D$ of an ordinary point $z_{0}$ of the equation (1.2) subject to the following sets of initial values at $z_{0}$ :
$w_{1}\left(z_{0}\right)=1, w_{2}\left(z_{0}\right)=\ldots=w_{m}\left(z_{0}\right)=0$
$w_{1}^{\prime}\left(z_{0}\right)=0, w_{2}^{\prime}\left(z_{0}\right)=1, w_{3}^{\prime}\left(z_{0}\right)=\ldots=w_{m}^{\prime}\left(z_{0}\right)=0$
$w_{1}^{(m-1)}\left(z_{0}\right)=\ldots=w_{m-1}^{(m-1)}\left(z_{0}\right)=0, w_{m}^{(m-1)}\left(z_{0}\right)=1$
is a fundamental set of solutions since at $z_{0}, \Delta\left(w_{1}, w_{2}, \ldots, w_{m}\right)=1 \neq 0$, and any analytic solution $w$ in $D$ is given by
$$w=w\left(z_{0}\right) w_{1}+w^{\prime}\left(z_{0}\right) w_{2}+\ldots .+w^{(m-1)}\left(z_{0}\right) w_{m}$$
where $w\left(z_{0}\right), w^{\prime}\left(z_{0}\right), \ldots, w^{(m-1)}\left(z_{0}\right)$ are prescribed constants for the solution $w$.
2 Consider, for simplicity, the case $m=2$, i.e. the second-order differential equation
$$\frac{d^{2} w}{d z^{2}}=p_{1}(z) \frac{d w}{d z}+p_{2}(z) w$$
and suppose one non-zero solution $w_{1}$ in the domain $D$ of an ordinary point $z_{0}$ is known. Then we proceed to find another solution $w_{2}$ in $D$ such . that $w_{1}, w_{2}$ form a fundamental set as follows.
$$w=w_{1} \int v d z$$

# 常微分方程代写

## 数学代写|常微分方程代考Ordinary Differential Equations代写|FUNDAMENTAL SETS OF SOLUTIONS

$$\Delta(z)=\Delta\left(z_{0}\right) e^{\int_{z 0}^{z} p_{1}(\zeta) d \zeta} \quad \text { in } D$$

(ii) 分析函数集 $w_{1}, w_{2}, \ldots, w_{m}$ 是线性独立的 iff $\Delta\left(z_{0}\right) \neq 0$ ，和
(iii) 如果 $w_{1}, w_{2}, \ldots, w_{m}$ 形成一个线性独立的集合和 $w$ 是任何解析解 $D$ ，然后是常数 $c_{1}, c_{2}, \ldots, c_{m}$ 可以发现这样
$$w=c_{1} w_{1}+c_{2} w_{2}+\ldots+c_{m} w_{m} \text { in } D$$

$$w_{i}^{(m)}=p_{1} w_{i}^{(m-1)}+p_{2} w_{i}^{(m-2)}+\ldots+p_{m} w_{i} \quad(i=1, \ldots, m)$$

## 数学代写|常微分方程代考Ordinary Differential Equations代写|Special fundamental sets

1 集合 $m$ 分析解决方安 $w_{1}, w_{2}, \ldots, w_{m}$ 在域中 $D$ 个个普通的点 $z_{0}$ 等式 (1.2) 営以下初始直集的影响 $z_{0}$ :
\begin{aligned} &w_{1}\left(z_{0}\right)=1, w_{2}\left(z_{0}\right)=\ldots=w_{m}\left(z_{0}\right)=0 \ &w_{1}^{\prime}\left(z_{0}\right)=0, w_{2}^{\prime}\left(z_{0}\right)=1, w_{3}^{\prime}\left(z_{0}\right)=\ldots=w_{m}^{\prime}\left(z_{0}\right)=0 \ &w_{1}^{(m-1)}\left(z_{0}\right)=\ldots=w_{m-1}^{(m-1)}\left(z_{0}\right)=0, w_{m}^{(m-1)}\left(z_{0}\right)=1 \end{aligned}

$$w=w\left(z_{0}\right) w_{1}+w^{\prime}\left(z_{0}\right) w_{2}+\ldots+w^{(m-1)}\left(z_{0}\right) w_{m}$$

2 为简单起见，考虑以下情况 $m=2$ ，即二阶微分方程
$$\frac{d^{2} w}{d z^{2}}=p_{1}(z) \frac{d w}{d z}+p_{2}(z) w$$

$$w=w_{1} \int v d z$$

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## MATLAB代写

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