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The following is from Introduction to Dynamics, by Percival and Richards:

At each zero $x_{k}$ of the velocity field $v\left[x\right],$

$$v\left[x_{k}\right]=0,$$

so that a system initially at $x_{k}$ remains there for all time. The points $x_{k}$ represent states of equilibrium: they are named fixed points. At all other points the state of the system changes. A system in an open interval between two fixed points cannot pass either of them. Such open intervals, together with those that extend from a fixed point to infinity, are invariant, as are the fixed points. Such fixed points and intervals represent invariant sets of states which are defined by the property that if any system is in such a set at some time, then it remains in that set for all times.

The discussion pertains to first-order autonomous systems in 1-dimensional phase space.

The mathematically correct reading of their statement appears to be that a system in some invariant set of states which is not a fixed point will not transition to a fixed point. That is, the idealized system will never come completely to rest in a finite amount of time. Is that correct?

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  • $\begingroup$ Is dissipation allowed? Is the system Hamiltonian? $\endgroup$ – Qmechanic Feb 10 '18 at 21:46
  • $\begingroup$ There has been no discussion of Hamiltonian systems to this point in the book. Such systems as exponential decay have been introduced as examples, so that implies that dissipation is allowed. That's why I find their statement problematic. Perhaps I am overthinking its meaning. The example of a bouncing ball with a coefficient of elasticity seems to contradict a pedantic interpretation of their words. In that the ball will come to rest in a finite period. $\endgroup$ – Steven Thomas Hatton Feb 10 '18 at 22:15
  • $\begingroup$ In 1d phase space, once you reach a point with phase velocity $0$ you cannot move beyond it since, by definition the system has stopped there. $\endgroup$ – ZeroTheHero Feb 10 '18 at 22:19
  • $\begingroup$ @ZeroTheHero That part I understand. The part I question is the definition of the invariant sets of states being open intervals. That implies that a system in motion will never reach a fixed point. $\endgroup$ – Steven Thomas Hatton Feb 10 '18 at 22:24
  • $\begingroup$ A bouncing ball with a coefficient of elasticity does not represent an autonomous system. The phase space changes after every bounce, and therefore depends on time. $\endgroup$ – probably_someone Feb 11 '18 at 0:22
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the idealized system will never come completely to rest in a finite amount of time. Is that correct?

Yes, a fixed point $x^*$ in such a system can only be reached asymptotically.

The simplest way to see why it's so is probably through the uniqueness of a $C^1$ differential equation: $dx/dt\equiv \dot x(x) = 0$ admits as solution the constant function $x(t)= x^*$, thus, by uniqueness, there cannot be a different solution; in particular, there can not be a solution that has $x(t_0) \ne x^*$ and $x(t^*)=x^*$ for some finite $t^*>t_0$. If an equilibrium point can be reached in finite time, then the system is not a smooth first order autonomous ODE.

Another way of proving this results is via the Hartman–Grobman theorem, which tells us that, for a smooth system, we can use its linearization to assess the system's behavior in the vicinity of a fixed point $x^*$ that's hyperbolic (i.e., for which $\dot v(x^*)\ne 0$). And, for a autonomous $1$-D system $$ \dot x(x) = v(x) $$ the linearization $y\equiv x-x^*$ about the fixed point $x^*$ is $$ \dot y = A y,$$ where $A\equiv \dot x(x^*)$, and whose solution is $$ y(t) = y(0) e^{At}.$$

Which means that, for a stable $x^*$ (i.e., $A<0$), $y$ approaches $0$ (i.e., $x$ approaches $x^*$) only asymptotically.

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  • $\begingroup$ And conversely, a system that reaches a fixed point in finite time must not be first-order autonomous. $\endgroup$ – probably_someone Feb 11 '18 at 1:03
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    $\begingroup$ @probably_someone Interesting point. And showing that that's true also provides a simpler way to answer the OP's question. I've added these as the 1st paragraph of my (v2) answer. $\endgroup$ – stafusa Feb 11 '18 at 3:43

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