Saturated damped harmonic oscillator

Question

Consider a critically damped harmonic oscillator:

$$x' = v$$ $$v' = -2v - x$$

Such a system has the property that if $x(0) < 0$ and $x(0)+v(0)\leq 0$, then $x(t) < 0$ for all $t$. Now suppose the system is modified so that the acceleration is subject to some minimum value, i.e. $v' \geq a_{min}$. Is there some way of maintaining the above property, i.e. $x(t) < 0$ for all $t$? More precisely, is it possible to do one or both of the following:

1. Add additional constraints on the initial state ($x(0)$, $v(0))$ so that $v'(t) = -2v(t) - x(t) \geq a_{min}$ for all $t$.
2. Add a term to the differential equations, i.e. $v' = -2v-x + g(x,v)$ so that $v'(t) = -2v-x + g(x,v) \geq a_{min}$, perhaps also under additional constraints on the initial state.

In other words, I want to make a small tweak to the critically damped harmonic oscillator to take into account saturation of acceleration.

Answer 1

Previous related work at Mathematics Stack Exchange:

• General ODE question
• Inhomogeneous 2nd-order linear differential equation

First eliminate $\,v$ , resulting in: $$ x' = v \quad ; \quad v' = -2v - x \quad \Longrightarrow \quad x'' + 2x' + x = 0 $$ My knowledge about e.g. critical damping has become rusty; so please allow me to do it my way and step by step. With Operator Calculus: $$ \left[\frac{d^2}{dt^2} + 2\frac{d}{dt} + 1\right] x(t) = \left[\frac{d}{dt} + 1\right]^2 x(t) = \left[e^{-t}\,\frac{d}{dt}e^{+t}\right]^2x(t) =\\ e^{-t}\,\frac{d}{dt}e^{+t}e^{-t}\,\frac{d}{dt}e^{+t}x(t) = e^{-t}\,\left(\frac{d}{dt}\right)^2e^{+t}x(t) = 0 \quad \Longrightarrow \\ \left(\frac{d}{dt}\right)^2e^{+t}x(t) = 0 \quad \Longrightarrow \quad \frac{d}{dt}e^{+t}x(t) = A \quad \Longrightarrow \quad e^{+t}x(t) = At+B \quad \Longrightarrow \\ x(t) = (At+B)e^{-t} $$ Boundary conditions: $$ \begin{cases} x(t) = (At+B)e^{-t} \quad \Longrightarrow \quad x(0) = B \\ x'(t) = Ae^{-t}-(At+B)e^{-t} \quad \Longrightarrow \quad x'(0) = A-B \end{cases} \\ \Longrightarrow \quad A = x(0)+x'(0) \quad ; \quad B = x(0) \quad \Longrightarrow \\ x(t) = \left[\left\{x(0)+x'(0)\right\}t+x(0)\right]e^{-t} $$ In concordance with the updated question, it will be assumed in the sequel that $\,x(0)+x'(0) < 0\,$ i.e. $\,x'(0) < -x(0)\,$ as well as $\,x(0) < 0$ . Derivatives: $$ x'(t) = Ae^{-t}-(At+B)e^{-t} = -(At+B-A)e^{-t}\\ x''(t) = -Ae^{-t} + (At+B-A)e^{-t} = (At+B-2A)e^{-t}\\ x'''(t) = Ae^{-t} - (At+B-2A)e^{-t} = -(At+B-3A)e^{-t}\\ x'''(t) = 0 \quad \Longrightarrow \quad At+B-3A = 0\quad \Longrightarrow \\ t_{min} = -\frac{B-3A}{A} = 3 - \frac{x(0)}{x(0)+x'(0)} \\ x''''(t) = Ae^{-t} - (At+B-3A)e^{-t} = (At+B-4A)e^{-t}\\ At_{min}+B-3A = 0 \quad \Longrightarrow \\ x''''(t_{min}) = -Ae^{-t_{min}} = -\left[x(0)+x'(0)\right]e^{-t_{min}} > 0 \quad \Longrightarrow \quad \mbox{minimum}\\ x''(t_{min})=(At_{min}+B-2A)e^{-t_{min}} = (At_{min}+B-3A+A)e^{-t_{min}} = Ae^{-t_{min}} < 0 $$ Conclusions. Assuming that $\,x(0)+x'(0) < 0\,$ i.e. $\,x'(0) < -x(0)$ and $\,x(0) < 0$ :

• The acceleration has a minimum for $\,t_{min} = 3 - x(0)/\left[x(0)+x'(0)\right]$
• And it's minimum value is $\,a_{min} = \left[x(0)+x'(0)\right]e^{-t_{min}} < 0\;$ (i.e. negative)

EDIT. The derivatives that have been used to determine the minimum "have no knowledge" about the fact that physical time is positive or zero, especially $\,t_{min} \ge 0$ . This means that, for a minimal acceleration to occur in the realm $\,t \ge 0$ : $$ 3-\frac{B}{A} \ge 0 \quad \Longrightarrow \quad x(0) \ge 3\left[x(0)+x'(0)\right] \quad \Longrightarrow \quad x'(0) \le -\frac{2}{3}x(0) $$ Otherwise, if $\,t_{min} \le 0$ , then the acceleration minimum is to be expected simply at $\,t = 0$ : $$\,a_{min} = x''(0) = B-2A = x(0)-2\left[x'(0)+x(0)\right]$$