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I've been ready G.Teschl ODE book and I got really confused with how to apply this theorem:

(Straighten out of vector fields)

Suppose that $f(x_{0})\neq 0$. Then, there is a local coordinate transform $y=\phi(x)$ s.t $\dot x = f(x)$ is transformed to $\dot y = (1,0,...,0)$.

I think I can picture the theorem and I understand the proof, but my real question is how to apply it. For example, there is this problem in the book which seems to be a direct use for this theorem and I have no idea how to solve it.

REAL PROBLEM

Find a transformation which straighten out the flow $\dot x = x$, defined in $M=\mathbb{R}$

Any help would be really appreciated <3

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If $x\neq 0$ (this is equivalent to the condition $f(x_{0})\neq 0$ in the hypotheses for the "straightening" lemma), then we can write $$\dfrac{x'}{x}=1.$$ We know that the general solution of the ODE is $x(t)=Ce^t$ for some constant $C$. Thus, $x(t)$ and $x'(t)$ have the same sign as $C$ for all $t$, and consequently, $\frac{d}{dt}\log \left\lvert x(t) \right\rvert = \dfrac{x'(t)}{x(t)}$ by the chain rule. Then, we can choose $y=\phi(x)=\log\left\lvert x \right\rvert$ as our local coordinate transform. We conclude that for all non-zero $x$, the equation $x'=x$ can be written as $y'=1$ under this change of coordinates.

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    $\begingroup$ +1, but you need to be a bit more careful with the case $x<0$. $\endgroup$ – Hans Lundmark Nov 9 at 10:42
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    $\begingroup$ Good point. One could argue that $\phi(x) = \log \left\lvert x \right\rvert$ works, if we assume that the general solution of the ODE is $x(t)=Ce^t$ and thus $x(t)$ and $x'(t)$ always have the same sign as $C$. I'll edit my answer. $\endgroup$ – B. Núñez Nov 9 at 18:12

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