I’m learning geometric thinking of nonlinear dynamics. Example is: if $dx/dt=\sin x$, draw a diagram of $x$ and $dx/dt$ as two axes. If you draw vector fields, you can find fixed points. For ex, here, if you start at $\pi/2$, you will be attracted to $x=\pi$.
My question is.. how do you “assume” or how do you know that if you start at $\pi/2$, after moving incrementally with $dx/dt=\sin x$, you will move through continuous values of $x$ and get to $x=\pi$?
In other words, if $dx/dt$ is $\sin x$, then how do I know that $x$ would not jump “over” $x=\pi$?
Assuming the equation $dx/dt=f(x)$ satisfies conditions that guarantee existence and uniqueness of solutions, the integral curves cannot intersect. Of course, they are also continuous, because they are differentiable.
In your case, there are integral curves at $x=\pm k\pi$, where $k$ is an integer. So the integral curve with $x(0)=1/2\pi$ flows up continuously to the equilibrium solution $x(t)=\pi$. (It flows up rather than down because it is easy to see that $dx/dt$ is positive on the interval $(1/2\pi, \pi)$.) It cannot intersect $x(t)=\pi$ (by the non-intersection condition), nor can it jump over it (by the continuity condition).