I'm starting Strogatz's Dynamical Systems and Chaos. At the beginning he wants to develop intuition towards the concepts in dynamical systems before jumping into rigorous theory. As such, he emphases graphical analysis as opposed to analysis.

One of the first examples he gives is the simple one dimensional system $$\dot{x} = \sin x$$

The following question is asked:

For an arbitrary initial condition $x_0$ what is the behavior of $x(t)$ as $t \to \infty$?

First, lets turn to the graph of $\dot{x}$. enter image description here

Strogatz's Answer

Strogatz answers that the particle will asymptotically approach the nearest stable point on the graph; here's an example trajectory where $x_0=\frac{\pi}{2}$

enter image description here

While I intuitively understand why this would be so, being newly introduced to rigorous mathematics, I'm intellectually skeptical of myself if I think I understand an answer without first seeing some rigorous reasoning as to why it is so.

I'm hoping someone can confirm my more rigorous solution or suggest what reasoning allows him to make his argument so confidently.

My answer

From the graph, we see that a particle dropped at any point $x_0$ will continue moving in a direction determined by $\text{sgn}(\dot{x}(x_0))$. By the graph, the magnitude of the particle's velocity decreases as it nears a stable point of $\dot{x}(x)$. Consider that after some time, the particle is sufficiently close to a point $\tilde{x}$ s.t. $\dot{x}(\tilde{x})=0$ and we can make a small angle approximation for velocity, $\dot{x}=\sin x\approx x'$, where $x'$ shifts the function such that the origin is at $\tilde{x}$. Now the derivative of velocity linearized at $\tilde{x}$ is $\dot{x}'=sgn{x'}x$ and solving this yields $x(t)=x'e^{-t}$. It can trivially be shown that $x(t)$ converges to zero as $t \to \infty$. $\square$

This answer is very heuristic – it only describes the magnitude of distance, not direction or sign and helps understand the qualitative nature of Fig 2.1.2, but not rigorously. I'm looking for guidance as to make my argument more precise. Assume my formal background ends at Calc I and that I won't appreciate any advanced real analysis concepts.

EDIT: I had originally said that the particle approaches the nearest zero of the graph, but I edited that to say stable point which is the correct limit of the system.


I hope Strogatz didn't actually write "the nearest zero". It's actually the nearest stable equilibrium (coloured black in the picture) assuming it doesn't start at an unstable equilibrium. And "nearest" is rather misleading, because that's only when the equilibria are regularly spaced, as in this example. Closeness is not important, rather what happens in general is that when you start in an interval between two consecutive equilibria, you go to an endpoint of that interval; which endpoint is determined by the sign of $\dot{x}$ in that interval.

EDIT: The point is basically this. Consider a differential equation $\dot{x} = f(x)$, where $f$ is continuously differentiable, and initial condition $x(0) = x_0$ where $x_0$ is in an interval $(a,b)$ where $f(x) > 0$, with $f(b) = 0$. As long as $x(t)$ stays in that interval, the differential equation says it must be increasing. But $x(t)$ can never get to $b$, because the constant $x(t)=b$ is a solution and the Uniqueness Theorem says two solutions can never meet or cross, so $x(t)$ must be in the interval for all $t > 0$.

The next fact is that an increasing function that is bounded above has a limit as $t \to \infty$. Let $\lim_{t \to \infty} f(t) = L$. Of course, $L \le b$. But it's impossible that $L < b$, because then $f(L) > 0$ and (by continuity) there would be some $\delta > 0$ and $\epsilon > 0$ such that $f(x) > \epsilon$ whenever $|x - L| < \delta$. But that means that when $x$ gets is withing distance $\delta$ of $L$, it is increasing at a speed of more than $\epsilon$, so within time $\delta/\epsilon$ it will have gone past $L$, and since it must keep increasing this contradicts the assumption that the limit is $L$. So the only possibility is that $L = b$.

  • $\begingroup$ Thank you so much for this very clear and insightful answer. The one thing I don't understand is how you get that after $\delta/\epsilon$ it will have gone past $L$?> $\endgroup$ – theideasmith Dec 2 '16 at 0:55
  • $\begingroup$ Mean Value Theorem. Or informally, at velocity $> \epsilon$, in time $\delta/\epsilon$ you travel a distance $> \delta$. $\endgroup$ – Robert Israel Dec 2 '16 at 2:11

First off, either his statement or your understanding of it contains a small but significant error: It will asymptotically approach the nearest zero of $\dot{x}$ that has a negative slope $\frac{d\dot{x}}{dx}$.

For example, starting at $x = 0.01$, you will aysmptotically approach $x=\pi$, not $x=0$.

A rigorous proof that $x$ asymptotically approaches $\frac{2n+1}{2}\pi$ for some integer $n$ is available because you can actually solve the differential equation: $$ x = 2 \tan^{-1}(\alpha e^t)$$

As $t \to \infty$ the arctan will go to some value of the form shown above.

If the function were less nice, however, showing the asymptotic behavior would require techniques which are a major point of this course's content.


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