note-math

separable ODE in 1 dimension

\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 $𝐺 = \int \frac{1}{𝑔 (𝑥)} 𝑑 𝑥, 𝐹 = \int 𝑓 (𝑡) 𝑑 𝑡$ , initial value undetermined

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 $ℝ^{𝑛}$

vector-field is analytic function $𝑣 : 𝑈 \to ℝ^{𝑛}$

The exponential-of-vector-field $(exp 𝑣) (𝑥)$ generated by the vector field $𝑣$ should be invariant

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

Taylor series of vector field

\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)

make harmonic-oscillator $\ddot{𝑥} = \mp 𝜔^{2} 𝑥$ first order

(\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 the case of trigonometric

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})

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

or written in the form of complex number and exponential

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}

similar for the case of hyperbolic

The characteristic polynomial equation of harmonic-oscillator is $𝜉^{2} \pm 𝜔^{2} = 0$ or $𝜉^{2} = \pm 𝜔^{2}$ . We are interested in the trigonometric case $𝜉^{2} + 𝜔^{2} = 0$ or $𝜉 = 𝜔 i 𝜔$ , its prototype is $𝜉^{2} = \pm 1$ or $𝜉^{2} = \pm 𝑖$

In the case where the harmonic oscillator $𝑥$ is a real value, for the complex exponential form, in order to remain in $ℝ$ , when $𝑥_{0}, 𝑣_{0} \in ℝ$ , the coefficients in front of $𝑒^{\pm i 𝜔 𝑡}$ are complex conjugates of each other $𝑥_{0} \pm \frac{𝑣_{0}}{i 𝜔}$

$ℝ \to ℝ$ , $𝑣 (𝑥) = 𝑥^{2}$

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

The $𝑓 (𝑡, 𝑥)$ in $𝑓 (0, 𝑥) = 𝑥, 𝑓 (1, 𝑥) = 𝑓 (𝑥)$ should correspond to the case of a dilation vector field $𝑡 \cdot 𝑣 (𝑥)$

One-parameter homomorphism embedding $𝑓 (𝑡, 𝑥) : ℝ ↪ Diff$

$- 𝑣$ and initial value $𝑦 = 𝑓 (𝑥)$ gives $𝑓^{- 1}$ . $𝑓^{- 1} (𝑡, 𝑦) = 𝑓 (- 𝑡, 𝑦)$

$𝑓 (𝑡, 𝑥)$ is called flow. exp road emission-like coordinates

vector-field-as-δ-diffeomorphism_(tag) Near $𝟙$ , the vector field is the coordinate of the diffeomorphism group $𝑣 ⇝ exp 𝑣$ , similar to #link(<geodesic-coordinate>)[]

ODE

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

wiki:Cauchy-Kovalevskaya_theorem, the estimation of the radius of convergence uses a special upper bound control method, similar to what is done in #link(<inverse-analytic>)[]

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

integral-curve_(tag) Picard iteration of ODE solution (wiki) representation or integral curve 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)

If it is a linear ODE, then (alias Dyson series)

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)

Linear ODE. The solution of a constant coefficient ODE can be written by transforming it into a first-order differential equation system + Jordan normal form

Lie-bracket_(tag) Lie bracket

as generator of conjugate-action of $Diff$

conjugation action of the $Diff$ group $𝑔, 𝑓 ⇝ 𝑓 𝑔 𝑓^{- 1}$

Differential := $ad (𝑣) (𝑤) = [𝑣, 𝑤]$

note that $𝑔, 𝑓 ⇝ 𝑔 𝑓 𝑔^{- 1}$ is a different map, if we consider the order of $𝑔, 𝑓$

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

$[𝑣, 𝑤] = [𝑤, 𝑣]$

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

Lie-derivative_(tag) Lie derivative alias drag derivative

let $𝑣$ generate a one-parameter diffeomorphism $𝑡 ⇝ 𝑓_{𝑡} \in Diff$

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 $[𝑣_{1}, [𝑣_{2}, 𝑤_{3}]] + [𝑣_{3}, [𝑣_{1}, 𝑣_{2}]] + [𝑣_{2}, [𝑣_{3}, 𝑣_{1}]] = 0$

Lie derivative can also be defined for tensor fields …

linear-PDE-integrable-condition_(tag) related to the symmetry of the second-order derivative. This condition allows the solution of a first-order linear PDE to come from successive integration curve of ODEs, and the result does not depend on the choice of path