This question is inspired by Arnold's book (Mathematical Methods of Classical Mechanics). In chapter 2 he defines the diffeomorphism $g^t$ as the mapping $g^t: \Bbb{R}^2 \to \Bbb{R}^2$ that propagates a point $M = (x,y=\dot{x})$ forward in time

$$ M(t) = g^t M $$

It's intuitively clear that $g^t$ forms an Abelian group, indeed $g^0 = 1$ and $g^{t + s} = g^s g^t$, he calls this group the phase flow. I always believed that for any continuos potential $U(x) = -\int_{x_0}^x{\rm d}\zeta~f(\zeta)$ , the solution to the equations

$$ \dot{x} = y ~~~\mbox{and}~~~ \dot{y} = f(x) $$

was guaranteed and, therefore, so was the existence of the phase flow. But then, he asks to show that for the potential

$$ U(x) = -x^4 $$

it is not possible to define a phase flow.

I am completely lost here. Question Under which conditions it is not possible to define the group $g^t$? Thanks in advance

  • $\begingroup$ Interesting question, especially since he implies (in his example at the bottom of the same page) that the phase flow for the "inverted pendulum" with $U(x) = -x^2$ is well-defined. My only guess would be that it has something to do with the existence of a differentiable inverse near the origin, but I don't know for sure. $\endgroup$ May 8, 2017 at 2:20
  • $\begingroup$ @MichaelSeifert Just wanted to draw your attention to a reply to this question, it is very interesting indeed. Thanks $\endgroup$
    – caverac
    May 8, 2017 at 22:56

1 Answer 1


Yeah, it looks like the group property terminates at the end of all times, T, and goes no further, in sharp contrast to the inverted quadratic potential that @MichaelSeifert mentions, which also has monotonically increasing acceleration.

Let's start from the latter, to set a baseline for normality, $$ H=y^2/2 -x^2. $$ Start from the origin x=0, at t=0, and discuss systems with positive energy $E=y^2/2-x^2$; and, w.l.o.g., absorb E into the variables so, effectively, E=1, and $$ y=\sqrt{2 (1+x^2)}. $$

As a consequence, integrating the trivial Hamilton's equation $y=\dot{x}$ nets t as a function of x, $$ t\sqrt{2} =\int_0^x \frac{dz}{\sqrt{1+z^2}} = \ln |x+\sqrt{1+x^2} |~. $$

So time goes on to infinity as x gets there, and vice versa. The Lie advective flow is $$ g^t x= e^{t \sqrt{2 (1+x^2)} ~ \partial_x} x~, $$ and one need not explicitly invert the above penultimate equation for x(t) to verify this.

What is radically different for the inverted quartic potential, is that, in fact (cf elliptic integral of the first kind ), x reaches infinity (monotonically) at finite time T, $$ T \sqrt{2}= \int_0^\infty \frac{dz}{\sqrt{1+z^4}} =4\Gamma(5/4)^2/\sqrt\pi \approx 1.85 ~. $$ So the group stops at T, and there is no meaningful flow past that time.

Edit on asymptotics : For very large x, the integrand goes like $1/z^2$, hence $$ t\sqrt{2}=\int_0^x \frac{dz}{\sqrt{1+z^4}} \sim T\sqrt{2}-\int_x^\infty \frac{dz}{z^2}=T\sqrt 2 -1/x ~, $$ hence $\sqrt 2 (T-t)\sim 1/x $, that is the almost trajectory terminates/expires as $$ x(t) \sim \frac{1}{\sqrt 2 (T-t)}, \qquad y(t)\sim \frac{1}{\sqrt 2 (T-t)^2}. $$

(Technically, slipping the velocity in the would-be-group element $\exp ( t \sqrt{2 (1+x^4)} ~ \partial_x) $ to the new canonical coordinate supplanting x in the partial derivative collapses it to T, terminating the Lie advective flow, as you need an invertible canonical coordinate to run it, but never mind, that's a whole side story...)

  • $\begingroup$ Great answer, thanks for taking the time for writing it. Although your last remark remains a mystery, would you mind elaborating a bit on it? $\endgroup$
    – caverac
    May 8, 2017 at 22:53
  • 1
    $\begingroup$ Thanks, but...Ach, ach... you put me on the spot... It's a whole different question. Briefly or else, returning to the late 19th century and its masters recasting to a canonical coordinate h(x) reduces the group element to a dumb translation operator, as they all are. But here $T=h(\infty)$... This is the renormalization group integrated β-fctn eqn, but, really another question, indeed a chapter, in its own right.... $\endgroup$ May 8, 2017 at 23:10
  • $\begingroup$ Thanks once again, I wasn't aware of this argument, but sure it makes a lot of sense $\endgroup$
    – caverac
    May 8, 2017 at 23:42
  • $\begingroup$ Might read up on p 294-5, Ch 8-6, in Hamermesh's book: "A one-parameter continuous group is equivalent to a group of translations". $\endgroup$ May 9, 2017 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.