note-math

一维可分离变量 ODE

\begin{matrix} \frac{𝑑 𝑥}{𝑑 𝑡} = 𝑓 (𝑡) 𝑔 (𝑥) & ⟶ & \frac{1}{𝑔 (𝑥)} 𝑑 𝑥 = 𝑓 (𝑡) 𝑑 𝑡 \\ ⟶ & 𝑥 = 𝐺^{- 1} (𝐹 (𝑡)) \end{matrix}

where $𝐺 = \int \frac{1}{𝑔 (𝑥)} 𝑑 𝑥, 𝐹 = \int 𝑓 (𝑡) 𝑑 𝑡$ , 初值未定

Example

$𝑓 (𝑡) = 𝑎, 𝑔 (𝑥) = 𝑥$ . $𝑥 (𝑡) = 𝑥 (0) exp (𝑎 𝑡)$
$𝑓 (𝑡) = 1, 𝑔 (𝑥) = 𝑥^{2}$ . $𝑥 (𝑡) = \frac{1}{\frac{1}{𝑥 (0)} - 𝑡}$

exponential-of-vector-field_(tag) Question

let $𝑈$ open in $ℝ^{𝑛}$

向量场是解析函数 $𝑣 : 𝑈 \to ℝ^{𝑛}$

由向量场 $𝑣$ 生成的 exponential-of-vector-field $(exp 𝑣) (𝑥)$ 应该是 invariant 的

\partial_{𝑣 (𝑥)} (exp 𝑣) (𝑥) = 𝑣 (𝑓 (𝑥))

向量场的 Taylor 级数

\begin{matrix} (exp 𝑣) (𝑥) & = & 𝑥 + 𝑣 (𝑥) + \frac{1}{2!} \partial_{𝑣 (𝑥)} 𝑣 (𝑥) + \frac{1}{3!} \partial_{𝑣 (𝑥)} (\partial_{𝑣 (𝑥)} 𝑣 (𝑥)) + \dots \\ = & 𝑥 + \sum_{𝑛 = 0}^{\infty} \frac{1}{(𝑛 + 1)!} {(\partial_{𝑣 (𝑥)})}^{𝑛} 𝑣 (𝑥) \end{matrix}

$𝑛 + 1$ polynomial like ${(\partial_{𝜆 𝑣 (𝑥)})}^{𝑛} 𝜆 𝑣 (𝑥) = 𝜆^{𝑛 + 1} {(\partial_{𝑣 (𝑥)})}^{𝑛} 𝑣 (𝑥)$

Example

compare to the result from separable ODE in 1 dimension

$𝑣 (𝑥) = 𝐴 (𝑥), 𝐴 \in gl (𝑑, 𝕂)$

compare $\dot{𝑥} = 𝐴 𝑥$ , expect $(\sum_{0 . . \infty} \frac{1}{𝑛!} {(𝐴 𝑡)}^{𝑛}) (𝑥)$ with $𝑡 = 1$

$𝑑 𝑣 (𝑥) = 𝐴$ , $𝑛 \geq 2 ⟹ 𝑑^{𝑘} 𝑣 (𝑥) = 0$

$\begin{matrix} \partial_{𝑣 (𝑥)} 𝑣 (𝑥) & = & 𝐴 (𝑣 (𝑥)) \\ = & 𝐴^{2} (𝑥) \end{matrix}$

$\begin{matrix} {(\partial_{𝑣 (𝑥)})}^{2} 𝑣 (𝑥) & = & 𝑑 (𝐴^{2} (𝑥)) (𝑣 (𝑥)) \\ = & 𝐴^{2} (𝑣 (𝑥)) \\ = & 𝐴^{3} (𝑥) \end{matrix}$

…

${(\partial_{𝑣 (𝑥)})}^{𝑛} 𝑣 (𝑥) = 𝐴^{𝑛 + 1} (𝑥)$

$𝑓 (𝑥) = (𝟙 + 𝐴 + \frac{1}{2!} 𝐴^{2} + \dots) (𝑥) = (\sum \frac{1}{𝑛!} 𝐴^{𝑛}) (𝑥) = (exp 𝐴) (𝑥)$

$\begin{matrix} 𝑑 𝑓 (𝑥) 𝑣 (𝑥) & = & (exp 𝐴) (𝐴 (𝑥)) \\ = & 𝐴 (exp (𝐴) (𝑥)) \\ = & 𝑣 (𝑓 (𝑥)) \end{matrix}$

Example harmonic-oscillator_(tag)

谐振子 $\ddot{𝑥} = \mp 𝜔^{2} 𝑥$ 一阶化

(\begin{matrix} \frac{𝑑}{𝑑 𝑡} \\ \frac{𝑑}{𝑑 𝑡} \end{matrix}) (\begin{matrix} 𝑥 \\ 𝑣 \end{matrix}) = (\begin{matrix} 1 \\ \mp 𝜔^{2} \end{matrix}) (\begin{matrix} 𝑥 \\ 𝑣 \end{matrix})

with 三角情况

exp 𝑡 (\begin{matrix} 1 \\ - 𝜔^{2} \end{matrix}) = (\begin{matrix} cos 𝜔 𝑡 & \frac{1}{𝜔} sin 𝜔 𝑡 \\ - 𝜔 sin 𝜔 𝑡 & cos 𝜔 𝑡 \end{matrix})

从而

𝑥 (𝑡) = 𝑥_{0} cos 𝜔 𝑡 + \frac{𝑣_{0}}{𝜔} sin 𝜔 𝑡

或者写为复指数的形式

𝑥 (𝑡) = \frac{1}{2} (𝑥_{0} - i \frac{𝑣_{0}}{𝜔}) 𝑒^{i 𝜔 𝑡} + \frac{1}{2} (𝑥_{0} + i \frac{𝑣_{0}}{𝜔}) 𝑒^{- i 𝜔 𝑡} ≕ 𝑎 (𝜔, i) 𝑒^{i 𝜔 𝑡} + 𝑎 (𝜔, - i) 𝑒^{- i 𝜔 𝑡}

双曲情况类似

谐振子方程的特征多项式方程是 $𝜉^{2} \pm 𝜔^{2} = 0$ or $𝜉^{2} = \pm 𝜔^{2}$ . 我们对三角情况 $𝜉^{2} + 𝜔^{2} = 0$ pr $𝜉 = \pm i 𝜔$ 感兴趣, 其原型是 $𝜉^{2} = - 1$ or $𝜉^{2} = \pm i$

在谐振子 $𝑥$ 是实数值的情况, 对于复指数形式, 为了维持在 $ℝ$ , 当 $𝑥_{0}, 𝑣_{0} \in ℝ$ 时, $𝑒^{\pm i 𝜔 𝑡}$ 前面的系数互为复共轭 $𝑥_{0} \pm \frac{𝑣_{0}}{i 𝜔}$

$ℝ \to ℝ$ , $𝑣 (𝑥) = 𝑥^{2}$

compare $\dot{𝑥} = 𝑥^{2}$ , expect $\frac{1}{\frac{1}{𝑥} - 𝑡}$ with $𝑡 = 1$

$\partial_{𝑣 (𝑥)} 𝑣 (𝑥) = 𝑣^{'} (𝑥) 𝑣 (𝑥) = 2 \cdot 𝑥^{3}$

$\partial_{𝑣 (𝑥)} (\partial_{𝑣 (𝑥)} 𝑣 (𝑥)) = {(2 𝑥^{3})}^{'} 𝑣 (𝑥) = (2 \cdot 3) \cdot 𝑥^{4}$

…

${(\partial_{𝑣 (𝑥)})}^{𝑛} 𝑣 (𝑥) = (𝑛 + 1)! \cdot 𝑥^{𝑛 + 2}$

$\begin{matrix} 𝑓 (𝑥) & = & 𝑥 + \sum_{𝑛 = 0}^{\infty} \frac{1}{(𝑛 + 1)!} {(\partial_{𝑣 (𝑥)})}^{𝑛} 𝑣 (𝑥) \\ = & \sum_{𝑚 = 1}^{\infty} 𝑥^{𝑚} \\ = & 1 - \frac{1}{1 - 𝑥} \end{matrix}$

$\begin{matrix} 𝑑 𝑓 (𝑥) 𝑣 (𝑥) & = & {(\frac{1}{1 - 𝑥})}^{2} 𝑥^{2} \\ = & {(\frac{𝑥}{1 - 𝑥})}^{2} \\ = & 𝑣 (𝑓 (𝑥)) \end{matrix}$

Question

$𝑓 (0, 𝑥) = 𝑥, 𝑓 (1, 𝑥) = 𝑓 (𝑥)$ 中间的 $𝑓 (𝑡, 𝑥)$ 应该对应伸缩的向量场 $𝑡 \cdot 𝑣 (𝑥)$ 的情况

单参数同态嵌入 $𝑓 (𝑡, 𝑥) : ℝ ↪ Diff$

$- 𝑣$ 和初值 $𝑦 = 𝑓 (𝑥)$ 给出 $𝑓^{- 1}$ . $𝑓^{- 1} (𝑡, 𝑦) = 𝑓 (- 𝑡, 𝑦)$

$𝑓 (𝑡, 𝑥)$ is called flow. exp 道路发射状坐标

vector-field-as-δ-diffeomorphism_(tag) 在 $𝟙$ 附近, 向量场是微分同胚群的坐标 $𝑣 ⇝ exp 𝑣$ , 类似于 #link(<geodesic-coordinate>)[]

ODE

\frac{𝑑}{𝑑 𝑡} (exp 𝑡 𝑣) (𝑥) = 𝑣 ((exp 𝑡 𝑣) (𝑥))

wiki:Cauchy-Kovalevskaya_theorem, 收敛半径估计使用了特殊上界控制方法, 类似 #link(<inverse-analytic>)[] 中所作的

$𝐹 (𝑥, 𝛾) = \frac{𝑐 𝑥}{𝑥 - 𝛾}$ , $\frac{𝑑}{𝑑 𝑡} 𝛾 = 𝐹 (𝑥, 𝛾)$ ==> $𝛾 (𝑡, 𝑥) = 𝑥 - {(𝑥^{2} - 2 𝑐 𝑡 𝑥)}^{\frac{1}{2}}$

integral-curve_(tag) ODE 解的 Picard 迭代 (wiki) 表示 or 积分曲线 e.g.

𝑥 (𝑡) = \sum_{𝑛 = 0 . . \infty} \int_{0}^{𝑡} 𝑑 𝑡_{𝑛} \int_{0}^{𝑡_{𝑛}} 𝑑 𝑡_{𝑛 - 1} \dots \int_{0}^{𝑡_{0}} 𝑑 𝑡_{1} 𝑓 (𝑡_{𝑛}, \dots 𝑓 (𝑡_{1}, 𝑥 (0)) \dots)

如果是线性 ODE 则 (alias Dyson 级数)

𝑥 (𝑡) = \sum_{𝑛 = 0 . . \infty} \int_{0}^{𝑡} 𝑑 𝑡_{𝑛} \int_{0}^{𝑡_{𝑛}} 𝑑 𝑡_{𝑛 - 1} \dots \int_{0}^{𝑡_{0}} 𝑑 𝑡_{1} 𝐴 (𝑡_{𝑛}) \dots 𝐴 (𝑡_{1}) 𝑥 (0)

线性 ODE. 常系数 ODE 的解可以写出 by 转为一阶微分方程组 + Jordan normal form

Lie-bracket_(tag) Lie bracket

作为 #link(<conjugate-action>)[] of $Diff$ 的生成元

$Diff$ 的共轭作用 $𝑔, 𝑓 ⇝ 𝑓 𝑔 𝑓^{- 1}$

微分 := $ad (𝑣) (𝑤) = [𝑣, 𝑤]$

注意 $𝑔, 𝑓 ⇝ 𝑔 𝑓 𝑔^{- 1}$ 是不同的映射, 如果考虑 $𝑔, 𝑓$ 的顺序的话

$[𝑣, 𝑤] (𝑥) = \partial_{𝑣 (𝑥)} 𝑤 (𝑥) - \partial_{𝑤 (𝑥)} 𝑣 (𝑥)$

$[𝑣, 𝑤] = [𝑤, 𝑣]$

for $GL, gl$ , $[𝐴, 𝐵] \sim 𝐴 𝐵 - 𝐵 𝐴$

Lie-derivative_(tag) Lie derivative alias drag derivative

let $𝑣$ 生成单参数微分同胚 $𝑡 ⇝ 𝑓_{𝑡} \in Diff$

let $𝑤_{𝑡} = 𝑑 𝑓_{𝑡}^{- 1} (𝑓_{𝑡} (𝑥) : base, 𝑤 (𝑓_{𝑡} (𝑥)) : vector)$

$𝐿_{𝑣} (𝑤) ≔ lim_{𝑡 \to 0} \frac{1}{𝑡} (𝑤_{𝑡} - 𝑤_{0})$

$𝐿_{𝑣} (𝑤) = [𝑣, 𝑤]$

Jacobi identity $𝐿_{[𝑣, 𝑤]} = [𝐿_{𝑣}, 𝐿_{𝑤}]$ or $[𝑣_{1}, [𝑣_{2}, 𝑤_{3}]] + [𝑣_{3}, [𝑣_{1}, 𝑣_{2}]] + [𝑣_{2}, [𝑣_{3}, 𝑣_{1}]] = 0$

可以对 tensor field 也定义 Lie derivative …

linear-PDE-integrable-condition_(tag) 相关于二阶导数的对称性. 这个条件使得一阶线性 PDE 的解可以来自接连的 ODE 积分曲线, 且结果不依赖于道路的选取