Ordinary damped SHM for x = x(t) with constant forcing is described by
\ddot{x} + \epsilon \dot{x} + x = 1
This has the property that the equilibrium solution x = 1 is approached as t \rightarrow \infty via damped oscillations which have period close to 2 \pi if \epsilon is small, as every second year student
of Physics or Mathematics knows. The term responsible for damping is "\epsilon \dot{x}", and it
is instructive to write the ODE as
\ddot{x} + x = 1 + D
where the RHS is a net motion-inducing force including a damping force D = - \epsilon \dot{x} which opposes motion when the velocity \dot{x} is positive.
If all we require is that x = 1 be approached, other forms of damping force D may do the job, so long
as they have the above motion-opposing property. For example, consider
D = - \epsilon \text{sgn}(\dot{x})
where
\text{sgn}(u) = \left\aligned -1 &, u < 0;\\
0&, u = 0\\
+1 &, u > 0 aligned \right.
This provides more damping than the usual case for small velocities, but limits the damping for large velocities.
Anyway, the ODE is trivially solvable. Here is the solution starting from rest with zero displacement. It is
convenient to write a separate expression for x in each half-period, namely
x = 1 + (-1)^n \epsilon - [1 - (2n - 1)\epsilon] \cos(t)
for (n - 1)\pi < t < n \pi, n = 1,2,\cdots,n_0. This solution holds only for the first n_0 half-periods,
where n_0 is about .5/\epsilon; more exactly, n_0 is the unique integer satisfying
\frac{1}{\epsilon} - 1 < 2n_0 < \frac{1}{\epsilon} + 1.
There is no solution for n > n_0, i.e. no solution for any time t > n_0 \pi! (Well at least
I can't find one.)
So long as t \leq n_0 \pi, the behaviour is reasonable. In each half-period \pi, there is a perfect (undamped) cosine wave, but the amplitude of that wave decreases by 2 \epsilon from half-period to half-period. The solution and its first derivative are continuous for all time, but the second derivative has a step jump at the end of each half-period, as the velocity changes sign.
What happens for t > n_0 \pi? Well, the first thing we must note is that during the last half-period n = n_0, we are "close" to equilibrium, in the sense that \mid x - 1 \mid < \epsilon and \mid \dot{x} \mid < 2 \epsilon. But that doesn't help us to continue the solution beyond t = n_0 \pi, if the bang-bang damping law is retained exactly. For example, if we assume that \dot{x} > 0 and hence D < 0 when n = n_0 + 1, the result is a "solution" with the incorrect property that \dot{x} < 0, and vice-versa. Identically zero velocity also does not satisfy the ODE unless we were exactly at the equilibrium point x = 1 when t = n_0 \pi, and in general we are not. So it seems there is nowhere to go.
I can only assume to resolve this paradox in a spineless way that the bang-bang model breaks down
for sufficiently small velocity \dot{x}. For
example, a more continuous damping force is
D = \left \aligned - (\epsilon/\delta) \dot{x}, &|\dot{x}| < \delta,\\
- \epsilon \text{sgn} (\dot{x}), &|\dot{x}| \geq \delta aligned \right.
For small \delta, this is the same as the bang-bang law, and the solution for small \delta is almost the same as that above until t = n_0 \pi. Then for t > n_o \pi, since the solution is already close to equilibrium, with the velocity \dot{x} small, the result is a (heavily) damped SHM, approaching the equilibrium state monotonically, with |\dot{x}| < \delta satisfied for all t > n_0 \pi.
Still, I don't entirely understand the need for this modification to the bang-bang law, which ought to be sufficient by itself, but actually fails to yield a solution at all for t > n_o \pi.
Department of Applied Mathematics
University of Adelaide