教材：DLI-BSc-Mathematics-Documents/Ordinary Differential Equatrions.pdf at main · JavaZeroo/DLI-BSc-Mathematics-Documents (github.com)

第一章

Explicit First Order Equations

这种形式的 \[ y' = f(x, y) \] 称为 'Explicit First Order Equations' 。

$y'=f(x)$

1. Equations with Separated Variables

\[ y' = f(x)g(y) \]

这种可以直接变成 $\frac{dy}{g(y)}=f(x)dx$

积分完后 \[ \int\frac{dy}{g(y)}=\int f(x)dx \] IVP: $y(\xi)=\eta$ \[ \int^y_{\eta}\frac{dy}{g(y)}=\int^x_{\xi} f(x)dx \] 这里需要注意的是如果$g(y(\xi))=g(\eta)=0$ 那么，直接就有 $y'=0$ 因此$y=\eta$ ;

2. 普通的替换

\[ y'=f(ax+by+c) \]

这种情况用$u(x) = ax+by+c$ 去替换掉变量$x$ 。原理是$u'=a+by'(x)=a+bf(u)$。很关键的一点是，$u$ 和$y$ 都是一次的。求导后刚好是线性关系。

因此最后得出的$u(x)$后可以直接利用$u(x)=ax+by+c$得到$y$。

3. 普通的Homogeneous Differential Equation

\[ y'=f\left(\frac{y}{x}\right) \]

同样的道理用$u(x) = \frac{y(x)}{x}, (x\neq0)$替换掉变量$y$。有$y'=u+xu'=f(\frac{y}{x})$ ，可得$u'=\frac{f(u)-u}{x}$

因此最后得出的$u(x)$后可以直接利用$u(x)=\frac{y(x)}{x}$得到$y$。

4. 高级的 Homogeneous Differential Equation

\[ y'=f\left(\frac{ax+by+c}{\alpha x+\beta y+ \gamma}\right) \]

这个的核心思想是转换成“普通的Homogeneous Differential Equation”。

首先分析行列式 \[ \left | \begin{matrix} a &b \\ \alpha &\beta \\ \end{matrix} \right | \]

行列式为零时，说明$a=\lambda_a \alpha, b=\lambda_b \beta$ 此时可以直接转换成“普通的Homogeneous Differential Equation”
行列式不为零的时候，说明方程组有唯一解。

首先解方程组 \[ \left\{\begin{align} ax+by+c&=0\\ \alpha x+\beta y+ \gamma&=0\\ \end{align}\right. \] 可以解出一组$(x_0, y_0)$ 利用这组解将“高级的 Homogeneous Differential Equation”转换成“普通的Homogeneous Differential Equation”。

原理是新建坐标系得$\bar{x}:=x-x_0, \bar{y}:=y-y_0$ 那么在这个坐标系下面原方程就变成了$\bar{y}(\bar{x}):=y(\bar{x}+x_0)-y_0$。对这个方程求导可以将$y$消掉，将原问题变$\bar{y}$和$x$的关系。这样做的目的就是将“高级的 Homogeneous Differential Equation”转换成“普通的Homogeneous Differential Equation”。

对$\bar{y}$求导后可以发现（利用方程$\bar{y}(\bar{x}):=y(\bar{x}+x_0)-y_0, ax_0+by_0+c=0, \alpha x_0+\beta y_0+ \gamma=0$） \[ \frac{\bar{y}(x)}{d\bar{x}} = y'(\bar{x}+x_0)=f\left(\frac{a\bar{x}+b\bar{y}(\bar{x})}{\alpha\bar{x}+\beta\bar{y}(\bar{x})}\right) \] 这里就可以像刚才的普通的Homogeneous Differential Equation一样去做了。值得注意的是，这样解出来的$\bar{y}$需要转换成原来的$y$。这里可以用 \[ y(x):=y_0+\bar{y}(x-x_0) \] 得到最后的$y$。

The Linear Differential Equation

这种形式的 \[ y' + g(x)y=h(x) \] 称为 'The Linear Differential Equation' 。

这时有两种情况：$h(x)=0$和$h(x)\neq0$，分别称为"homogeneous" 和"nonhomogeneous"

事实上当$h(x)=0$也就是"homogeneous"时，就是上面的"Explicit First Order Equations"，这里就不在赘述了。

对于$h(x)\neq0$的情况，也就是"nonhomogeneous"时，我们需要用到"Method of variation of constants"。

Method of variation of constants：这个方法首先计算出齐次的时候的通解。对于方程$y'+g(x)y=h(x)$他的齐次方程的通解是通过解$y'+g(x)y=0$，可得 \[ y=C\cdot e^{-\int g(x)dx} \] 此时我们将常数$C$作为一个与$x$的函数$C(x)$。这个时候，只需要解出一个$C(x)$就是非齐次情况下的答案。

那么现在的问题就变成了，如何得到一个$C(x)$使得$y' + g(x)y=h(x)$？

首先先计算$C'(x)$ \[ \begin{align} y &=C(x)\cdot e^{-\int g(x)dx}\\ y'&=\left( C(x) \right)'\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\right)' \\ y'&=C'(x)\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\cdot\left(-\int g(x)\right)'\right) \\ y'&=C'(x)\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\cdot -g(x)\right) \\ &\Downarrow \\ y'&=C'\cdot e^{-\int g(x)}-gC\cdot e^{-\int g(x)dx} \end{align} \] 然后代入$y' + g(x)y=h(x)$ \[ \begin{align} L_y \equiv y' + g(x)y&=h(x)\\ L_y&=C'\cdot e^{-\int g(x)dx}-gC\cdot e^{-\int g(x)dx}+gC\cdot e^{-\int g(x)dx}\\ L_y&=C'\cdot e^{-\int g(x)dx}=h(x) \\ &\Downarrow \\ C'(x)&=h(x)\cdot e^{\int g(x)dx} \\ &\Downarrow \\ C(x)&=\int h(x)\cdot e^{\int g(x)dx}dx + C_0 \end{align} \] 现在我们可以知道，当$y=C(x)\cdot e^{-\int g(x)dx}$且$C(x)=\int h(x)\cdot e^{\int g(x)dx} + C_0$时。，有$y' + g(x)y=h(x)$。这里有个二级结论

If $y, \bar{y}$, yare two solutions to the nonhomogeneous equation $L_y = h$, then $L(y - y) = L_y - L_{\bar{y}} = 0, i.e., z(x) = y - \bar{y}$ is a solution of the homogeneous equation $L_y = 0$. Thus all solutions $y(x)$ of the nonhomogeneous equation can be written in the form \[ y(x)=\bar{y}+z(x) \]

这里面$z(x)$就是刚刚的$y=C\cdot e^{-\int g(x)dx}$，$\bar{y}$就是$C(x)=\int h(x)\cdot e^{\int g(x)dx} + C_0$和$y=C\cdot e^{-\int g(x)dx}$的结合中的$y$，也就是$\bar{y}=\left(\int h(x)\cdot e^{\int g(x)dx}dx + C_0\right)\cdot e^{-\int g(x)dx}$

Bernoulli's Equation

这种形式的 \[ y'+g(x)y+h(x)y^{\alpha}=0.\alpha \neq 1 \] 非常的简单，只需要把$y^{\alpha}$解决了就可以了。等式去除$y^{\alpha}$有 \[ y'y^{-\alpha}+g(x)y^{(1-\alpha)}+h(x)=0 \] 利用$z=y^{(1-\alpha)} \implies z'=(1-\alpha)y^{-\alpha}\cdot y'$替换原式得 \[ \frac{1}{1-\alpha}z'+g(x)z+h(x)=0 \] 现在，就变成了nonhomogeneous的"The Linear Differential Equation"。最后解出$z$，别忘了替换回$y$。

Exact differential equations

这种形式的 \[ M(x,y)dx+N(x,y)dy=0,\\ \exists\ U(x, y)\ s.t.\ U_x(x,y)=M(x,y),U_y(x,y)=N(x,y) \]

$xdx+ydy=0$ is an exact equation, and $U(x,y)=1/2 (x^2+y^2 )$is a potential function.

Integrating Factors

Integrating Factors是用来让非 Exact 变成Exact differential equations。

E.g. $ydx + 2xdy = 0$ is not exact. However, it can easily be made an exact differential equation (in the domain $x > 0$) by multiplying the equation by $1/\sqrt{x}$. The resulting differential equation \[ \frac{y}{\sqrt{x}}dx+2\sqrt{x}dy=0 \] is exact, and a potential function is given by \[ F(x,y)=2y\sqrt{x}=0\ (x>0) \]

对于一个not excat differential equation我们需要找到一个Factor $U(x,y)$ 使得$U(x,y)\cdot M(x,y)dx+U(x,y)\cdot N(x,y)dy=0$变成一个Exact differential equations。

这里有一个"Theorem on potential functions"保证可以找到$U(x,y)$

现在的问题就是，如何去找？首先令$M' = U\cdot M,N'=U\cdot N$如果$F_x=M',F_y=N'$则有$M'_y=N'_x$。利用这个关系可以知道

事实上就是$F_{xy}=F_{yx}$ (Jacobian Matrix)

\[ \begin{align} &(U\cdot M)_y=(U\cdot N)_x\\ \implies &U_y\cdot M+U\cdot M_y=U_x\cdot N+U\cdot N_x \\ \end{align} \] 此时需要考虑，Integrating Factors是只与$x$有关还是只与$y$有关（只需要选一个）

假如只与$x$有关则$U_y=0$则有

\[ \begin{align} U_y\cdot M+U\cdot M_y&=U_x\cdot N+U\cdot N_x \\ U\cdot M_y &= U' \cdot N+U \cdot N_x \\ \frac{1}{U}U'&=\frac{M_y-N_x}{N} \\ (\ln U)'&=\frac{M_y-N_x}{N} \\ U&=e^{\int \frac{M_y-N_x}{N}dx} \end{align} \]

这里的答案不是$U=C\cdot e^{\int \frac{M_y-N_x}{N}dx}$的原因是，我们只需要找到一个$U$，因此你可以认为我们选择$C=1$作为答案。下面的情况同理。

假如只与$y$有关则$U_x=0$则有

\[ \begin{align} U_y\cdot M+U\cdot M_y&=U_x\cdot N+U\cdot N_x \\ U'\cdot M+U\cdot M_y &= U \cdot N_x \\ \frac{1}{U}U'&=\frac{N_x-M_y}{M} \\ (\ln U)'&=\frac{N_x-M_y}{M} \\ U &=e^{\int \frac{N_x-M_y}{M}} \end{align} \]

Implicit First Order Differential Equations

这种形式的 \[ F(x, y, y')=0 \] 一般来说有两种解决办法。要么通过一些方法获得explicit differential equation，要么就用参数化。

在这里我们只讨论两种情况：

$F(x, y')=0 $, $F(y, y')=0$
$y=f(x,y')$, $x=f(y,y')$

第一种情况

对于$F(x, y')=0 $我们使用参数化： \[ \left\{\begin{array}{l} x=\phi(t) \\ y'=\psi(t) \end{array}\right. \] 此时方程变为$F(\phi(t), \psi(t))=0$，同时我们有 $$ \[\begin{align} y'&=\frac{dy}{dx} \ \text{and} \ \phi'(t)=\frac{d\phi(t)}{dt} \\ dy&=y'dx \ \ \text{and} \ d\phi(t)=\phi'(t)dt\\ y&=\int y'dx + C\\ y&=\int \psi(t)d\phi(t) + C \\ y&=\int \psi(t)\phi'(t)dt +C \end{align}\] \[ 最后得到： \] { \[\begin{array}{l} x=\phi(t) \\ y=\int \psi(t)\phi'(t)dt +C \end{array}\]

. $$

第一种情况2

对于$F(y,y')=0$我们仍然参数化： \[ \left\{\begin{array}{l} y=\phi(t) \\ y'=\psi(t) \end{array}\right. \] 此时有$F(\phi(t),\psi(t)=0$，同时有： \[ \begin{align} y'&=\frac{dy}{dx} \ \text{and} \ \phi'(t)=\frac{dy}{dt} \\ dx&=\frac{dy}{\psi(t)} \ \text{and} \ dy=\phi'(t)dt \\ dx&=\frac{\phi'(t)dt}{\psi(t)} \\ \int dx&= \int \frac{\phi'(t)dt}{\psi(t)} \\ x&=\int \frac{\phi'(t)dt}{\psi(t)} \end{align} \]

An Existence and Uniqueness Theorem

中文名是存在唯一性定理。首先我们介绍Lipschitz condition:

We consider the following initial value problem \[ y' = f(x,y),\ \text{for}\ \xi \leq x\leq \xi+a,\ y(\xi)=\eta \] The main assumptions in the following theorem are that $f$ is continuous in the strip $S=J\times\mathbb{R}$ with $J=[\xi,\xi+a]$ and satisfies a Lipschitz condition with respect to $y$ in $S$ \[ |f(x,y)-f(x, \bar{y})|\leq L|y-\bar{y}| \] No restrictions are placed on the value of the Lipschitz constant $L\geq 0$

然后引出存在唯一性定理：

Let $f \in C(S)$ satisfy the Lipschitz condition. Then the IVP has exactly one solution $y(x)$. The solution exists in the interval $J: \xi\leq x \leq\xi +a$

The extension of solutions.

以下三个定理是用于 The extension of solutions.

1️⃣Local Lipschitz condition. The function $f(x,y)$ is said to satisfy a local Lipschitz condition with respect to $y$ in $D\subset R^2$ if for every $(x_0,y_0 )\in D$ there exists a neighborhood $U=U(x_0,y_0 )$ and an $L=L(x_0,y_0 )$ such that in $U\cup D$ the function $f$ satisfies the Lipschitz condition $|f(x,y)-f(x,\bar{y})|\leq L|y-\bar{y}|$.

注意！我们一般通过连续性来判断Local Lipschitz condition

If $D$ is open and if $f \in C(D)$ has a continuous derivative $f_y$ in $D$, then f satisfies a local Lipschitz condition in this set.

2️⃣Theorem on local solvability If $D$ is open and $f\in C(D)$ satisfies a local Lipschitz condition in $D$, then the IVP is locally uniquely solvable for$ (x_0,y_0 )∈D$; i.e., there is a neighborhood $I$ of$ x_0$ such that exactly one solution exists in $I$.

3️⃣Theorem on the extension of solutions Let $f \in C (D)$ satisfy a local Lipschitz condition with respect to $y$ in $D$. Then for every $(x_0, y_0)\in D$ the initial value problem $y' = f (x, y), y(x_0) = y_0$ has a solution that can be extended to the left and to the right comes arbitrarily close to the boundary of $D$.

最后我们有：

The Peano existence theorem. If $f(x,y)$ is continuous in a domain $D$ and $(\xi,\eta)$ is a point in $D$, then at least one solution of the differential equation $y′=f(x,y)$ goes through $(\xi,\eta)$. Every solution can be extended to the left and to the right up to the boundary of $D$.

Linear System

这里开始就是在讨论，常微分方程组。

Systems of n Linear Differential Equations

给出常微分方程组的形式： $$ \[\begin{align} y_1' &= a_{11}(t)y_1+\cdots+ a_{1n}(t)y_n+b_1(t) \\ &\ \ \vdots \\ y_1' &= a_{11}(t)y_1+\cdots+a_{1n}(t)y_n+b_1(t) \\ \end{align}\] \[ 或者： \] '=A(t)+(t)\ \ A(t)=(a_{ij}(t)), (t)=(b_1(t), , b_n(t))^ $$ 值得一提的是，存在唯一性定理在常微分方程组也同样适用。

Homogeneous Linear Systems

对于齐次的形式，常微分方程组就变成了： \[ \mathbf{y}'=A(x)\mathbf{y} \] 此时，根据存在唯一性定理有： \[ \exist \text{ exactly one solution } \mathbf{y}=\mathbf{y}(t;\tau,\boldsymbol{\eta})\ \forall \tau \in J, \boldsymbol{\eta}\in \R^n\text{ or }\C^n \] 当然，齐次常微分方程组有一些重要的性质：

$ $ in $ J$ is a solution of the homogeneous linear systems.
There exist $n$ linearly independent solutions $_1,\dots,\mathbf{y}_n$. Every such set of $n$ linearly independent solutions is called a fundamental system of solutions. If $\mathbf{y}_1,\dots,\mathbf{y}_n$ is a fundamental system, then every solution $\mathbf{y}$ can be written in a unique way as a linear combination $\mathbf{y}=C_1 \mathbf{y}_1+\dots+C_n \mathbf{y}_n$.
A system of $n$ solutions $\mathbf{y}_1,…,\mathbf{y}_n$ can be assembled into an $n\times n$ solution matrix $\Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n )$. If $n$ solutions $\mathbf{y}_1,\dots,\mathbf{y}_n$ are linearly independent, then $\Phi(x)$ is a system of $n$ solutions $\mathbf{y}_1,\dots,\mathbf{y}_n$ can be assembled into an $n\times n$ solution matrix $\Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n )$. If $n$ solutions $\mathbf{y}_1,\dots,\mathbf{y}_n$ are linearly independent, then $\Phi(x)$is a Fundamental Matrix.

The Wronskian

现在讨论一下，齐次常微分方程的解，是线性无关还是线性相关。

The Wronskian. If $\Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n )$ is a solution matrix of $\mathbf{y}^′=A(x)\mathbf{y}$, then its determinant $W(x)=|\Phi(x)|$is called the Wronskian determinant.
Theorem If $\mathbf{y}_1,\dots,\mathbf{y}_n$ are linearly dependent in $J$, then the Wronskian $W(x)\equiv0$.
Theorem If $\mathbf{y}_1,…,\mathbf{y}_n$ is a fundamental system of equation $\mathbf{y}'=A(x)\mathbf{y}$, then the Wronskian $W(x)\neq0$ in $J$.
Theorem. There exists a fundamental system of solutions for equation $\mathbf{y}'=A(x)\mathbf{y}$.

因此，我们先求出$n$个解，然后再去判断这$n$个解的Fundamental Matrix的行列式，也就是The Wronskian，是否为零。

Inhomogeneous Systems

对于非齐次的常微分方程组，就是最开始样子： \[ \mathbf{y}'=A(t)\mathbf{y}+\mathbf{b}(t) \] 下面这个定理类似线性代数，非齐次方程组的通解是，齐次方程组的通解+非齐次方程组的特解：

Theorem. Let $\tilde{\mathbf{y}}(x)$be a fixed solution of the inhomogeneous equation (1). If $\mathbf{y}_0 (x)$ is an arbitrary solution of the homogeneous equation, then $\mathbf{y}(x)=\tilde{\mathbf{y}}(x)+\mathbf{y}_0 (x)$is a solution of the inhomogeneous equation, and all solutions of the inhomogeneous equation are obtained in this way.

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第一章