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Given the dynamical system

$$\frac{dx}{dt}(t) = sin(x(t))$$

let $\xi(t)$ denote the solution of this system with initial condition $\xi(0)=\pi/2$

What is the $\omega$-limit set $\mathcal{S}_\omega(\xi)$ of $\xi(t)$?

In my lecture notes, I saw that what can be done, is to plot the right-hand side of the ODE, in this case the $sin(x(t))$ and from there, it is easy to spot the equilibrium points of the system and then it is possible to tell what is the $\omega$-limit set.

I don't understand why, plotting the am I able to tell that $\mathcal{S}_\omega(\xi)=\{\pi\}$. Shouldn't I need to solve the ODE to be able to say this?

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One dimensional systems are often straightforward to analyse because solution are limited by equilibrium points (that is, there is no 'way around' an equilibrium point).

The system is globally Lipschitz hence a unique solution exists for all $\mathbb{R}$. It is also time invariant, so trajectories cannot cross.

Since $\sin$ is positive on $(0, \pi)$, we see that $\xi$ must be strictly increasing when $\xi(t) \in (0, \pi)$.

We have $\sin \pi = 0$, and $\zeta(t) = \pi$ is a solution to the system $\dot{x} = \sin x$ with $\zeta(0) = \pi$. Since the trajectories cannot cross, we must have $\xi(t) < \zeta(t) = \pi$ for all $ t$. Hence $\xi(t) \in (0,\pi)$ for all $t \ge 0$, and $\hat{\xi}=\lim_{t \to \infty} \xi(t) $ exists (and must be in $[\xi(0),\pi]$). We must have $\hat{\xi} = \pi$, otherwise we would have $\dot{\xi}(t) \ge \epsilon >0$ for some $\epsilon$, which would be a contradiction.

Hence $\hat{\xi}=\lim_{t \to \infty} \xi(t) = \pi$, and so the $\omega$-limit set is just $\{\pi\}$.

In general, you can see that if you start at $n \pi$, then the trajectory will remain there. If $\sin \xi(0) >0$, then the system will converge to $\pi \lceil {\xi(0) \over \pi } \rceil$, and similarly, if $\sin \xi(0) <0$, the system will converge to $\pi \lfloor {\xi(0) \over \pi } \rfloor$.

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    $\begingroup$ Thanks for the answer. I kind of followed your explanation but still not getting it deepl. how can you tell that $\xi$ is strictly increasing? Can you put it in a simple way, probably something that can help me picture it? $\endgroup$
    – BRabbit27
    Jan 5, 2014 at 10:35
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    $\begingroup$ If $\xi(t) \in (0,\pi)$, then $\dot{\xi(t)} >0$. $\endgroup$
    – copper.hat
    Jan 5, 2014 at 10:48

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