The textbook claims that this system of nonlinear equations is reversible. However I can't seem to work it out.

\begin{align} \dot{x} &= -2cos(x) - cos(y) \\ \dot{y} &= -2cos(y) - cos(x) \end{align}

By definition of reversible systems,

\begin{align} -\dot{x} &= -2cos(x) - cos(-y) = -2cos(x) - cos(y) \\ \dot{y} &= -2cos(-y) - cos(x) = -2cos(y) - cos(x) \end{align}

It seems like the reversibility holds for the 2nd equation but not for the first. Isn't the system known to be reversible if the system is invariant when t-> -t and y -> -y?


The system is reversible with respect to the transformation

$$t\rightarrow -t\\x\rightarrow -x\\y\rightarrow -y$$

as you can easily confirm. A transformation that changes only $y$ is not general. In the general sense, a set of ODE is called reversible if it is invariant under the transformation

$$t\rightarrow -t\\\boldsymbol{x}\rightarrow R\left(\boldsymbol{x}\right)$$

where $R^{2}\left(\boldsymbol{x}\right)=\boldsymbol{x}$.

You can check Strogatz's "Nonlinear Dynamics and Chaos" section $6.6$ on reversible systems. He gives there an example that is almost exactly identical to your ODE set (just replace $-2$ with $2$).

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  • $\begingroup$ What does x→R(x) line mean? I'm not 100% sure on what this indicates. I always thought that the reversible equation satisfied t→−t and y→−y. $\endgroup$ – Betelgeuse Dec 10 '17 at 0:33
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    $\begingroup$ Take for example $$R=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$ Then $$\boldsymbol{x}\rightarrow R\left(\boldsymbol{x}\right)$$ means $$\begin{pmatrix}x\\y\end{pmatrix}\rightarrow\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-x\\-y\end{pmatrix}$$ or equivalently $$x\rightarrow -x\\y\rightarrow -y$$ The transformation $$t\rightarrow -t\\y\rightarrow -y$$ is just a special case where $$R=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ $\endgroup$ – eranreches Dec 10 '17 at 0:38
  • $\begingroup$ Thanks so much! Really really appreciate the explanation $\endgroup$ – Betelgeuse Dec 10 '17 at 2:10

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