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Consider the system $$ \dot{x}=y,\qquad\dot{y}=-x+y^2. $$ Then, it is said that the system is reversible $(t\to -t, y\to -y)$. What does this mean?

If I put this into the equations, I get $$ \dot{x}(-t)=y(-t),\qquad \dot{y}(-t)=-x(-t)+y(-t)^2. $$

So does reversible mean here that when replacing $t$ by $-t$ and $y$ by $-y$, the differential equations still hold?

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That the system is reversible indicates there is inherent symmetry in the problem. To show reversibility, you are correct.

Notice that $$\dot{x} = \frac{dx}{dt},\dot{y} = \frac{dy}{dt}$$So$$t\rightarrow -t \Rightarrow (\dot{x} \rightarrow -\dot{x})$$

$$t\rightarrow -t \text{ and } y \rightarrow -y \Rightarrow \dot{y} \rightarrow \dot{y} $$ Getting back to the system, making the given substitution yields,

$$-\dot{x} = -y\Rightarrow \dot{x} = y$$ $$\dot{y} = -x+(-y)^2 \Rightarrow \dot{y} =-x+y^2$$

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  • $\begingroup$ Why irreversible in your first sentence? $\endgroup$ – Rhjg Feb 5 '16 at 14:21
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    $\begingroup$ It was a typo, I've edited it. $\endgroup$ – fosho Feb 5 '16 at 14:26
  • $\begingroup$ How do you know that $t \rightarrow -t \Rightarrow (\dot x \rightarrow - \dot x)$? $\endgroup$ – Oliver G Jun 27 '17 at 19:43
  • $\begingroup$ Well, dt becomes -dt $\endgroup$ – fosho Jun 27 '17 at 21:11

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