Q:

Hamiltonian system with velocity singularity

I am studying a classical mechanical system for which Hamilton's equations are the following:
$$
\psi' = \frac{p}{\sqrt{1-p^2}},~~~~~p' = -\phi \sin \psi,
$$
where $\phi$ is a parameter. If $\phi>1/2$, then there are certain initial values of $p$ and $\psi$ that inevitably end up at one of the lines $p=1$ or $p=-1$, implying that the velocity $\psi'$ diverges. Specifically, if the initial conditions satisfy
$$
\psi \ge \cos^{-1}\left(1-\frac{1}{\phi}\right),~~~~~p = 0,
$$
we reach the $p=1$ line. Is there any way to understand these dynamics physically? 
If I attempt to numerically integrate the equations of motion, the solver gives a solution for $\psi$ with a discontinuous jump at the singularity, with $\psi$ going from some value $-\psi_0$ to $\psi_0$. Is there any reason to believe this, i.e. to trust any result past the singularity? How can this system be understood when $\phi>1/2$?
Any help or suggested reading material is appreciated.
EDIT:
The full Lagrangian is given by
$$L = \sqrt{1+\psi'^2} + \phi \cos\psi.$$
This does indeed have a rigorous analogy to a relativistic pendulum. 

A:

It is easy to understand the phenomenon just using energy conservation.
N.B. I henceforth assume $\phi \geq 0$.
Since the Lagrangian does not depend on time explicitly, the Hamiltonian is conserved along the solutions:
$$\sqrt{1-p(t)^2} + \phi \cos \psi(t) = \sqrt{1-p(0)^2} + \phi \cos \psi(0)\:.$$
The evolution of the system necessarily stays along a connected component of the curve above in the plane $\psi, p$, determined by the initial conditions.
To understand the dynamics  would be very helpful plotting the curves 
$$\sqrt{1-p^2} + \phi \cos \psi =E$$
for various values of $E$,  however I go on exploiting an analytic approach.
Suppose that a certain time the velocity of $\psi$ diverges, i.e.,  $p(t)=\pm 1$, thus 
$$ \phi \cos \psi(t) = \sqrt{1-p(0)^2} + \phi \cos \psi(0)\:.$$
If you also assume $p(0)=0$ as you wrote, you have
$$ \phi \cos \psi(t) = 1+\phi \cos \psi(0)\:.$$
Namely
$$\cos \psi(t) = \frac{1}{\phi} + \cos \psi(0)\:.$$
Since $|\cos \psi(t)|\leq 1$ and $|\cos \psi(0)|\leq 1$, this identity is possible only if $1/\phi \leq 2$ that is
$$\phi \leq \frac{1}{2}\:.\tag{1}$$
Assuming it, since $|\cos \psi(t)|\leq 1$ we have 
$$-1 -\frac{1}{\phi} \leq \cos \psi(0) \leq 1-\frac{1}{\phi}  \leq 1$$
The left hand side condition is always satisfied since $-1 -\frac{1}{\phi} < -1 \leq  \cos \psi(0) $. The other condition leads to 
$$ \cos \psi(0) \leq  1-\frac{1}{\phi} $$ which implies, dealing with $\psi(0) \in [0,2\pi]$ 
$$\pi -\cos^{-1}\left(1- \frac{1}{\phi}\right) \geq \psi(0) \geq \cos^{-1}\left(1- \frac{1}{\phi}\right)\:.\tag{2}$$
(1) and (2) are necessary conditions for  the existence of solutions with $p(0)=0$ such that $\psi'(t)$ diverges for some time $t$.
The fact that they also are sufficient  can only be proved through a closer scrutiny discussing when the point $(\psi(t),p(t))$ is allowed to stop along a connected component of the curve $$\sqrt{1-p(t)^2} + \phi \cos \psi(t) = \sqrt{1-p(0)^2} + \phi \cos \psi(0)\:.$$
Otherwise the point must moves along the complete connected component reaching 
the anomalous points with $p=\pm 1$ if they are allowed on the curve.
Stops along the curve are nothing but the zeros of the gradient of the Hamiltonian.
If $$\phi > \frac{1}{2}\:,$$
energy conservation prevents the system from admitting  pathological solutions if these solutions satisfy $p(t_0)=0$ for some $t_0$ in their domain (since the system is autonomous we can always shift $t_0$ to $0$).

If I attempt to numerically integrate the equations of motion, the solver gives a solution for ψ with a discontinuous jump at the singularity, with ψ going from some value $−ψ_0$ to $ψ_0$.

I think it is just a problem of the solver. These configurations $(\pm 1, \psi_0)$ in the configuration space $(p,\psi)$ have nothing pathological from the geometric point of view. Just the curve $\psi=\psi(p)$ passes through $(\pm 1, \psi_0)$ and  has vertical tangent  there (parallel to $p = \pm 1$) since  $$\frac{d\psi}{dp} = \frac{\psi'}{p'} = - \frac{p}{\phi \sqrt{1-p^2} \sin \psi}\:,$$
unless the disaster happens exactly  at $(\pm 1, k\pi)$ requiring a more sophisticated analysis.

