Phase portrait of the simple pendulum. How much time does it take to get from one unstable equilibrium to the other? Consider a simple pendulum. I am trying to understand the phase portrait of its dynamics.

I read that the "time" in which the dynamical system goes from the the unstable equilibrium $\theta = - \pi$ to $\theta = \pi$, is infinite. I do not understand how to interpret, let alone formally prove this fact; I read that it should follow from Cauchy's existence and uniqueness theorem.
Can you explain to me both at the intuitive and formal level this fact?

 A: You can't literally start at the first equilibrium point and expect to go somewhere else. If you start at an equilibrium point, then you are going to stay there forever, precisely because it's an equilibrium.
If you start somewhere on the trajectory connecting the two equilibria in the phase portrait, then as $t \to +\infty$ you are going to approach the second equilibrium, and if you trace the dynamics backwards in time, then as $t \to -\infty$ you approach the first equilibrium. So in that sense the trajectory in question “starts” at the first equilibrium, but there is no time $t_0$ such that at time you are precisely located there.
To prove that you can't reach the second equilibrium point $P$ in finite time (say at $t=t_0$), consider the solution which stays at $P$ for all $t$, and the solution which arrives at $P$ from somewhere else at time $t_0$ and then stays at $P$ for all $t \ge t_0$. Can you see how that would contradict uniqueness?
A: If the pendulum equation (rescaled) is $\ddot\theta=-\sin(\theta)$, then around the upper equilibrium you can set $\theta = -\pi+u$ and get
$$
\ddot u=\sin(u)\approx u.
$$
The approximation has well-known solutions $u(t)=C\cosh(t)$ with the condition of starting at rest, $\dot u(0)=0$.
Pushing the pendulum (gently) out of its equilibrium means to position it some distance $\varepsilon$ away from it, $C=ε$. Let's say that the linear approximation of the sine is good for some distance $\Delta$, then the pendulum will leave that interval around the equilibrium at around
$$
t=\cosh^{-1}(Δ/ε)\approx \ln(2Δ/ε).
$$
So the smaller $ε$ the larger the time to move away from the equilibrium.
