# Duffing equation transformation $(t,x) \rightarrow (-t,-x)$

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.

• 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. – Dr. Lutz Lehmann Mar 11 at 9:19

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