# Condition For Existence Of Phase Flows

I am a bit confused about the existence of one-parameter groups of diffeomorphisms/phase flows for various types of ODE's. Specifically, there is a problem in V.I. Arnold's Classical Mechanics text that asks to prove that a positive potential energy guarantees a phase flow, and also one asking to prove that $U(x) = -x^4$ does not define a phase flow- and these having me thinking.

Consider the following two (systems of) differential equations:

$\dot x(t) = y(t)$, $\dot y(t) = 4x(t)^3$

and

$\dot a(t) = b(t)$, $\dot b(t) = -4a(t)^3$.

Both phase flows might, as far as I see it, have issues with the fact that the functions $\dot y(t)$ and $\dot b(t)$ have inverses which are not $C^\infty$ everywhere. However, the $(x,y)$ phase flow has an additional, apparently (according to Arnold's ODE text) more important issue- it approaches infinity in a finite time.

Why, though, do I care about the solutions "blowing up" more than I care about the vector fields' differentiability issues?

$\textbf{What is, actually, the criterion for the existence of a phase flow, given a (sufficiently differentiable) vector field?}$

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## 2 Answers

How does Arnold define "phase flow"? As far as I know, part of the definition of a flow requires the solutions to exist for all $t > 0$. If they go to $\infty$ in a finite time, they don't exist after that. On the other hand, I don't see why not having a $C^\infty$ inverse would be an issue.

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Good point- you're right, a flow has to be defined for all real $t$. However, I seem to have poorly worded the question. What I'm most concerned with is whether or not I can tease 'definedness for all $t$' out of the properties of $\dot y(t) = f(x)$ function- i.e. its positivity. – specterhunter Sep 28 '11 at 1:25
For the system $x' = y$, $y' = f(x)$ where $f = - V'$ is continuous, there is a solution that goes to $+\infty$ in finite time iff $V$ is bounded above on $[0, \infty)$ with $$\int_0^\infty \frac{dx}{\sqrt{K-V(x)}} < \infty$$ for some $K > \sup_{x \ge 0} V(x)$. – Robert Israel Sep 28 '11 at 2:06

Arnold's defines a phase flow as a one parameter group of diffeomorphisms from R^2 -> R^2 that map points on the phase space to the points they evolve to after time t. In the case of U = -x^4 ,the map g(t,M) is not differentiable on the separatrix, which is defined as E = 0, in the second and fourth quadrants of the phase plane. Physically, particles with negative energy do not cross the potential barrier. Thus, a particle having E<0 , x>0 and p<0 will be "reflected" off the point where U(x) = E and continue to have x > 0 for all time. Particles having positive energy do cross the barrier, which is to say that if a particle has E>0 , x>0 , p<0 , will cross over to negative x in finite time. However a particle with E = 0, x > 0 and p < 0 will only go upto the point x = 0 , and it will take infinitely long time to do so. So if you look at a point on the separatrix having x > 0 , p < 0 , or x < 0 and p > 0 , in any neighbourhood of that point will lie points which will evolve to be far away on the phase plane for a large enough time t.

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