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I am interested in ways that can help me find a diffeomorphism $\psi$ that "strightens" the vector field of an OED. That is:

Let $F:U\to \mathbb R^n$ be a continuously differentiable vector field where $U\subset \mathbb R^n$ is a bounded domain. $\dot x = F(x)$ is the differential equation ($x:I\to \mathbb R^n$).

There's a theorem that states that if $F(x_0) \neq 0$ then there exists a neighborhood $U'$ of $x_0$, and there is a diffeomorphism $\psi:U'\to V\subset \mathbb R^n$ such that for every $x\in U'$:

$D_\psi (x) \cdot F(x) = e_1$.

My question is: how exactly can I find (or wisely guess) this diffeomorphism?

For example, if I have $\dot x = y, \: \dot y = -x$ then $\psi (x,y) = (\theta, r)^T$ can be proven to be such diffeomorphism, where $\theta$ and $r$ are the angle and radius of the polar representation of x and y (this fits well since the equations describe a harmonic oscillator, in which the radius is constant).

Some examples of my attempts - I am afraid that the $\psi$ I find isn't a diffeomorphism:

  1. For $\dot x = 1, \: \dot y = \sin x$ I tried $\psi \left(x,\,y\right) = (x,\, \cos x+y)^T:=\left(a,\,b\right)^T$
  2. For $\dot x = x, \: \dot y = 2y, \: x>0$ I tried $\psi \left(x,\,y\right) = \left(\ln x,\, \dfrac{y}{x^2}\right)^T$

My attempt was to search for $a,b$ such that:

  • $\dot a = a_x \cdot \dot x + a_y \cdot \dot y = 1$

  • $\dot b = b_x \cdot \dot x + b_y \cdot \dot y = 0$

Usually I find $a$ quite easily (make $a_y=0$ and guess) but struggle to find b.

I would like to know if there is a general approach and if not, how can I make smarter guesses.

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    $\begingroup$ Have you seen a proof of this theorem? It usually constructs the sought diffeomorphism in an explicit way. $\endgroup$
    – Artem
    Nov 15, 2015 at 21:44
  • $\begingroup$ As of yet I haven't seen a proof, just tried to fiddle with the theorem. $\endgroup$
    – Trouble
    Nov 16, 2015 at 6:48
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    $\begingroup$ Then you will find an answer in your question in the proof of this theorem. $\endgroup$
    – Artem
    Nov 16, 2015 at 12:06

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