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Consider an ordinary differential equation (ODE) of a 1D damped oscillator of the form

$$\ddot{x}+\gamma\dot{x}+\omega^2x=0.~~(\omega^2,\gamma>0)$$

I want to know if this ODE is reversible i.e., given the information at present i.e. $x(0)=x_0$ and $\dot{x}(0)=v_0$, I want to know whether we can reconstruct the past uniquely. If yes, I would like to see how. I think one needs to make the change $t\to -t$ in the original ODE and then solve for $x(t)$ to predict the past.

If not, I would like to understand why. The answer here by Vincent Fraticelli suggests to me that the time-reversed solution uniquely predicts the past. But I may be wrong.

Qmechanic
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Solidification
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2 Answers2

2

I )

$$ {\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +{\gamma}\,{\frac {d}{dt}} x \left( t \right) +{\omega}^{2}x \left( t \right)=0\tag 1 $$

you can solve this differential equation and obtain the solution $x(t)$ and $x(t\mapsto -t)$

II )

for $~\tau=-t~$ from equation (1) you obtain:

$${\frac {d^{2}}{d{\tau}^{2}}}x \left( \tau \right) -{\gamma}\,{\frac {d }{d\tau}}x \left( \tau \right) +{\omega}^{2}x \left( \tau \right) $$

if the ODE is reversible the solution $~x(\tau)~$ must be equal to $x(-t)$ , this is the case if the initial velocity $\dot x(0)=0~$ equal zero, in this case is the ODE reversible!!

Eli
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One can solve this equation exactly and obtain the values of $x,v$ at any past moment... unless we are at the special point $x=v=0$. But this is not what we call reversibility of fundamental laws, since damping here is a phenomenological force.

Roger V.
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