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We have the Duffing equation, $\ddot{x}+ λ\dot{x}=x-x^3$, which can also be written as

$\dot{x}=y$

$\dot{y}=-U'(x)- λ y=x-x^3-\lambda y $

Show that the transformation $(t,x) \rightarrow (-t,-x)$ of time and place the cases $\lambda>0$ and $\lambda<0$ intertwines.

I've made a phase portrait where I saw it happen, but I don't know how I can show it. I don't know where to start, or how to use that $t \rightarrow -t$. I'd appreciate some hints on where to start.

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  • $\begingroup$ More precisely, the equation is invariant under $(t,x)\to(t,-x)$ and the time reversal $(t,x)\to(-t,x)$ by itself changes the sign of $λ$ resp. the friction term. $\endgroup$ – Dr. Lutz Lehmann Mar 11 at 9:19
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These things can be confusing, but this specific case is quite simple. We impose the substitutions $$ x=-X, \qquad t=-s, $$ which imply $$ \frac{d}{dt}=-\frac{d}{ds},\qquad \frac{d^2}{dt^2}=\frac{d^2}{ds^2}, $$ so that $$ \frac{d^2x}{dt^2}+\lambda \frac{dx}{dt}=-\frac{d^2X}{ds^2}+\lambda\frac{dX}{ds}, $$ and $$ x-x^3=-(X-X^3).$$ Thus, $\frac{d^2x}{dt^2}+\lambda \frac{dx}{dt}=x-x^3$ if and only if $$\frac{d^2X}{ds^2}-\lambda \frac{dX}{ds}=X-X^3.$$

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