Consider the motion of a simple pendulum as a 2nd-order ODE:
$\ddot{\theta} + \frac{g}{l} \sin \theta =0 $.
I can write this as a coupled system of first-order equations as:
$\dot{\theta} = u$,
$\dot{u} = -\frac{g}{l}\sin \theta$.
I have seen several definitions/derivations online and in textbooks where they authors say that the phase space of this system is given by a manifold which is a cylinder. I don't understand. Isn't the phase space of this system just $\mathbb{R}^{2}$ where the state vector is given by $\left[\theta, u\right] \in \mathbb{R}^{2}$?
Thanks. Thomas
Note that vector field you have constructed is $2 \pi$ periodic in $\theta$ so it makes sense to consider this variable as being on the circle $S^1$, this has the benefit of making these trajectories closed.
So for example here is the phase space with level curves of the Hamiltonian in your original coordinates along with several highlighted trajectories, note in particular how the red curve seems to run off to infinity:
Now lets take the position variable $\theta$ and map it onto the circle and so the image of the trajectories look like this
Which seems to fit better with the fact that the angle variable is $2\pi$-periodic, all of the trajectories are now closed curves on the cylinder.