Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)

It seems to me that Hamiltonian formalism does not suit well for problems involving instantaneous change of momentum, like particle collisions with hard wall or hard sphere gas model. At least I could not apply it straightforwardly to the simplest possible problem of 1D particle hitting a wall:

            │/ wall
            │/

  particle  │/
───o────────┼────────────> x
            │/
            │/
            │/

My attempt was quite direct. I took the Hamiltonian to be

$$H = \frac{p^2}{2m} + U(x)$$

with potential $U$ defined as

$$U(x) = \cases{ 0, \; x < 0, \\[.5em] K, \; x > 0, \\[.5em] E, \; x= 0.}$$

where $E$ is the particle energy and $K > E$. Hamiltonian equations should read as

$$\cases{\dot x = \frac{p}{m}, \\[.6em] \dot p = - \frac{d U}{d x}.}$$

It is not hard to integrate the first equation, but my attempts to integrate the second one did not lead to any meaningful result (that's why I do not share them here, it was a complete failure).

So I ask whether it is possible to obtain the solution to the problem by directly integrating Hamiltonian equations in the form above, without relying on general mechanical theorems/principles like energy conservation? Or is such an approach completely unsuitable for the task?

If so, what is the general (and elegant) approach to such systems?

There exist a related question on PSE "Hamiltonian function for classical hard-sphere elastic collision", but the setting is more cumbersome.

Answer

I think I got how to deal with the problem in a straightforward way, without the passage to a limit.

Let the phase space of the initial problem be the half plane

$$\{ \,(q,p) \; | \; q > 0\},$$

the wall is at $q=0$. In this phase space when the particle reaches the point $(0, p)$ it instantaneously teleports to the point $(0, -p)$. Particle trajectories are thus discontinuous.

The trick now is to glue the half plane into a cone, so that particle trajectories will become continuous. At the same time the particles are considered free on the entire trajectory, yielding the Hamiltonian being just a kinetic energy. Sorry for the lack of appropriate drawings, but I hope it is not so hard to imagine. The operation can be formally achieved with a non-canonical change of coordinates to $(r, \varphi)$:

$$\begin{cases} p = 2 r \sin \dfrac{\varphi}{2}, \\[.5em] q = 2 r \cos \dfrac{\varphi}{2}. \end{cases}$$

These are actually polar coordinates in a plane perpendicular to the cone axis. The symplectic form $dp \land dq$ transforms to $2r \, d \varphi \land dr$, or, in other words, the Poisson matrix becomes

$$ \begin{bmatrix} 0 & \dfrac{1}{2r} \\ -\dfrac{1}{2r} & 0 \end{bmatrix} $$

The Hamiltonian (kinetic energy) is given by

$$H = \frac{2}{m} r^2 \sin^2 \frac{\varphi}{2}.$$

All of the above leads to the Hamiltonian flow of

$$X_H = \frac{1}{2m} r \sin \varphi \frac{\partial}{\partial r} - \frac{1 - \cos \varphi}{m} \frac{\partial}{\partial \varphi}$$

and the equations of motion of

$$\begin{cases} \dfrac{d r}{dt} = \dfrac{1}{2m} r \sin \varphi, \\[.5em] \dfrac{d \varphi}{dt} = -\dfrac{1}{m} (1 - \cos \varphi). \end{cases}$$

These are amenable to a relatively simple integration, free of the subtleties of special functions. As result one gets as a solution

$$\begin{gather} \cot \dfrac{\varphi(t)}{2} = C_1+ \dfrac{t}{m}, \\[.5em] r(t) = C_2 \sqrt{1 + \cot^2 \dfrac{\varphi(t)}{2}}, \end{gather}$$

which also can be obtained directly from the known solution in $(q,p)$ half-plane and coordinate transformation rules.

To sum up, the effect of the wall is accounted for via the change of the phase space topology. Thus, particles are considered free, with a Hamiltonian being a kinetic energy which is preserved. As far as I reckon the phase space is not a cotangent bundle of a configuration space anymore. If this is true along with all the above derivation, this represents probably the most simple case of a phase space that is not a cotangent bundle.

---

Although at first I was guided by geometrical reasoning about the phase space, now I have been thinking about the coordinate transformation itself. I came up with another transform, which is much closer to the original coordinates:

$$\begin{cases} p = \operatorname{sgn} \! \left(\, \widetilde q \,\right) \, \widetilde p, \\[.5em] q = \left|\, \widetilde q \, \right| . \end{cases}$$

Here it is assumed that $\frac{d}{dx} \operatorname{sgn} x = 0$. Then $dp \land dq = d \, \widetilde p \land d \, \widetilde q$ and the Hamiltonian is the one of a free particle:

$$H = \frac{\; \widetilde p^{\, 2} \!}{2m}.$$

Like with trigonometric coordinate transformation there seems to be a need to choose the right branch for the transform $q \mapsto \widetilde q$, but since $q \geqslant 0$ there is no ambiguity.