Differential Equations: Time-Reversal Symmetry In Strogatz's Nonlinear Dynamics and Chaos (page: 163-164), he claims that any mechanical system of the form mx" = F(x) is T-R symmetric. He considers a system that has T-R symmetry to be one that is invariant under the change of variable
$ t -> -t $
First, I see that if we reverse time along a trajectory, that the first derivative would inherit a negative sign. Second, I'm not positive I see why the second derivative, x", would remain unchanged. Is this due to the following?
$ x''= d^2x/dt^2 $  ... which since t is squared would yield $t^2$ = $(-t)^2$ under the mapping of t -> -t. 
However, it seems a bit odd that you can manipulate the $dx/dt$ notation like that. If someone could show why $dt^2$ is able to obey the above I would appreciate it. Namely, why is $d(t)^2$ considered a product or function where we can first compute $t^2$ then apply d? 
Thanks! 
 A: This is the chain rule. If you have a function 
$$
f(t)
$$
and you compose with the function $g(t)=-t$, then 
$$
\frac{d}{dt}f(-t)=\frac{d}{dt}f(g(t))=g'(t)f'(g(t))=-f'(-t)\\
\implies \frac{d^2}{dt^2}f(-t)=\frac{d}{dt}-f'(-t)=-\frac{d}{dt}f'(-t)=f''(-t)
$$
A: You correctly deduce that x' will change signs (if the velocity is positive, then under time reversal--the movie played backwards--the velocity will be negative). Very intuitive.
Now x" (the acceleration) is just the first derivative of the velocity. So you  already convinced yourself that taking a derivative (of say x) will change sign, then it must be that taking the derivative of the velocity will also change sign.
From the point of view of the position x, the velocity changes sign, but the acceleration will see two changes of sing, so x" doesn't change sign. That means the differential equation is invariant under time reversal, i.e., symmetric.
I hope you can see that any DE with only even order terms will obey time reversal invariance.  But the presence of an odd order term will mess it up. And it should. The moment you stick in something like y' in the DE, you are putting in a damping term. Now the DE won't be invariant anymore under t-> -t. You will be able to tell forward in time from backwards. Consider a damped pendulum. You will certainly know if the movie is being shown forward or backward (what pendulum with no forcing term increases exponentially in amplitude?!). That's because the odd order term does not conserve energy (energy is lost through friction). So we see some principles tied together here: parity properties of the derivatives, conservation of energy, and time-reversal symmetry.
And, for a bonus, when the process is not time-reversable, we say the process is not reversal. Remember what that means? Entropy must be increasing. So in addition to the above properties, we can add whether entropy is increasing or not. The whole thing ties it all together from different topics very nicely.
Hope that helped.
A: Due to the power rule, if the first derivative is odd such that $F(-x)=-F(x)$, than the second derivative is even. That is why it is invariant to the transformation $t\to-t$
