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.
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 $$\frac{d^2X}{ds^2}-\lambda \frac{dX}{ds}=X-X^3.$$