Show that the system $\ddot x+x\dot x+x=0$ is reversible

My attempt was to use a property from the book that if the system

$\dot x = f(x,y)$

$\dot y = g(x,y)$

is reversible then $f(x,-y)=-f(x,y)$ and $g(x,-y)=g(x,y)$.

I tried to convert the system into a system like the one above by letting

$\dot x=y$

$\ddot x=\dot y=-x-xy$

However, this system does not satisfy the second criteria, namely that $g(x,y)=-x-xy \ne g(x,-y)=-x+xy$

Is there another way to show that this system is reversible?

  • $\begingroup$ As a side note: I am not familiar with this property, but is it an if and only if? Should it just be an implication, from being reversible to the properties on $f$ and $g$, then it isn't conclusive to show it "can factor out" the sign. $\endgroup$ – Cehhiro Nov 27 '15 at 6:24
  • $\begingroup$ I do not think that the property is an if and only if property. The more general definition given: a reversible system is any second-order system that is invariant under $t \to -t$ and $y \to -y$ $\endgroup$ – Chad Nov 27 '15 at 6:26
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    $\begingroup$ Then let us use the definition. Consider some solution $x:t\mapsto x(t)$ and define a new function $\xi:t\mapsto\xi(t)=-x(-t)$, then $\xi'(t)=x'(-t)$ and $\xi''(t)=-x''(-t)$ hence $\xi''(t)+\xi(t)\xi'(t)+\xi(t)=\ldots$. $\endgroup$ – Did Nov 27 '15 at 7:00
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    $\begingroup$ Strictly speaking, more general definition involves some mapping $R$ such that $R^2 = {\rm id}$ which commutes with the flow. $\endgroup$ – Evgeny Nov 27 '15 at 8:32

To combine the (very useful!) comments of @Did and @Evgeny in an answer:

The definition of reversibility I'm using is that of time reversal symmetry, i.e. the motion described by the system backward in time is equivalent to that forward in time (I'm following James D. Meiss, Differential Dynamical Systems, SIAM MM14, 2007, p. 212 here). That means that there exists a diffeomorphism $S$ that conjugates the flow $\phi_t$ of the dynamical system with its inverse, so that $(\phi_{-t} \circ S)(z) = (S \circ \phi_t)(z)$, for a dynamical system $\dot{z} = f(z)$. A bit more concretely in terms of the dynamical system itself: this implies that \begin{equation} -f(S(z)) = D S(z) f(z). \end{equation} So, what's this diffeomorphism $S$ in the case at hand? Well, the idea is that we're looking for a coordinate transformation on $(x,y)$ that will give us 'the same' dynamical system as we originally had, before we inverted the time direction.

Inverting time, i.e. mapping $t \mapsto -t$ leads to the system \begin{align} -\dot{x} &= y, \\ - \dot{y} &= -x(1+y). \end{align} We now see that there's a particularly simple coordinate transformation that will return the original system, namely $(x,y) \mapsto (-x,y)$. Using this coordinate transformation, we obtain \begin{align} \dot{x} &= y, \\ - \dot{y} &= x(1+y), \end{align} which is indeed equivalent to the original dynamical system. That means we've found our diffeomorphism $S$, which acts as $S(x,y) = (-x,y)$. Note that, as $S$ is linear, $D S = S$. Moreover, $S$ is an involution, i.e. $S \circ S = \text{Id}$.

Admittedly, this is somewhat different from the 'usual' notion of time reversibility. In that case, the system would be invariant under simultaneously changing the direction of time ($t \mapsto -t$) and the direction of the velocity ($y = \dot{x} \mapsto -y$). This is for example the case with the ODE $\ddot{x} + \dot{x}^2 + x = 0$. Also, this holds for every Hamiltonian system.

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  • $\begingroup$ There is a confusion here: the dynamical system at hand involves only one quantity, namely $x(t)$, and specifies that $x''(t)+x(t)x'(t)+x(t)=0$ at all times $t$, hence reversibility should be a property of the function $\bar x(t)=-x(-t)$ only, namely that $\bar x''(t)+\bar x(t)\bar x'(t)+\bar x(t)=0$. In other words the transformation involved is $x\mapsto\bar x$, and this implies that $x'$ is transformed into $\bar x'(t)=+x'(-t)$. Thus, in the $(x,y)$ coordinate, the fact that $y(t)\to y(-t)$ is not a choice, but a consequence of the fact that $x(t)\to-x(-t)$. $\endgroup$ – Did Nov 28 '15 at 9:32
  • $\begingroup$ No, not really. Using the usual definition of a dynamical system (continuous time, differentiable), you can infer that this is written in the form $\dot{z} = f(z)$, $z \in M$ for some $n$-dimensional manifold $M$. So the proper way to turn the second order ODE in a dynamical system is to introduce a second variable, thereby turning it into a two-dimensional system of first order ODE's. That being said, the rest of your argument holds without question: indeed, it is even easier to spot the 'right' mirror symmetry from the second order ODE than from the two-dimensional dynamical system. $\endgroup$ – Frits Veerman Nov 28 '15 at 16:36
  • $\begingroup$ It seems you agree with my point, notwithstanding the first sentence of your comment. So, everything is well. $\endgroup$ – Did Nov 28 '15 at 17:37
  • $\begingroup$ Everything is well indeed, I agree with your point. $\endgroup$ – Frits Veerman Nov 28 '15 at 20:27

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