Show that $x' = y(1-x^2), y ' = 1 - y^2$ is reversible.

Question

Show that $x' = y(1-x^2), y ' = 1 - y^2$ is reversible.

The book defines a second order system to be reversible system iff "it is invariant under $t \rightarrow -t$ and $y \rightarrow -y$", but I have no idea how I would show this for this equation.

$\frac{dx}{d(-t)} = -y(-t)(1-[x(-t)]^2)$ and $\frac{dy}{d(-t)} = 1-[y(-t)]^2$

From here I have no idea how to show that these are equal.

Anyone have any ideas?

I think you have a typo. Do you mean $y\rightarrow -y$ or perhaps $y''\rightarrow -y''$? — Nathaniel Bubis, Jul 04 '17 at 17:11

score 1 · Answer 1 · answered Jul 04 '17 at 17:23

1

It may be less confusing if you use different names for the transformed variables. If $s = -t$ and $Y = -y$, $\dfrac{dx}{dt} = - \dfrac{dx}{ds}$ and $\dfrac{dy}{dt} = - \dfrac{dy}{ds} = \dfrac{dY}{ds}$. Thus the system becomes

$$ \eqalign{\dfrac{dx}{ds} &= -y (1-x^2) = Y (1-x^2)\cr \dfrac{dY}{ds} &= 1 - y^2 = 1 - Y^2 \cr}$$

which is the same as the original system for the new variables.

answered Jul 04 '17 at 17:23

Robert Israel

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How did you get $\frac{dx}{dt} = - \frac{dx}{ds}$ from $s = -t$? – Oliver G Jul 04 '17 at 17:38
The chain rule. – Robert Israel Jul 04 '17 at 20:06

Show that $x' = y(1-x^2), y ' = 1 - y^2$ is reversible.

1 Answers1