How to find a conserved quantity in this differential equation.

Question

Consider the system: $$\ddot x = x^3 -x$$ What is the method to follow to find a conserved quantity for this system?

So far what I have is:

$\dot x = y$ and $\dot y = x^3 - x$ and I can find the Jacobian of the system and find and classify fixed points.

How would I find a conserved quantity?

Answer 1

For a second order ODE the following should do the trick: $$ \ddot x = x^3 - x \\ \ddot x\dot x = (x^3 - x)\dot x = \frac d{dt} \left( \frac{x^4}4 - \frac{x^2}2 \right) \\ \frac{\dot x ^2}2 - \frac{x^4}4 + \frac{x^2}2= const. $$

Answer 2

Multiplying the original 3rd order ODE by $\dot x$ we obtain $$ \dot x\ddot x=(x^3-x)\dot x $$ or equivalently $$ \left(\frac{(\dot x)^2}{2}\right)^{\bf\dot{}}=\left( \frac{x^4}{4}\right)^{\bf\dot{}}-\left(\frac{x^2}{2}\right)^{\bf\dot{}} $$ which means that there exists a constant $c$ such that $$ \frac{(\dot x)^2}{2}=\frac{x^4}{4}-\frac{x^2}{2}+c. $$

And as $y=\dot x$, we obtain $$ 2x^2+2y^2-x^4=c_1. $$ Then conserved quantity is $$ F(x,y)=2x^2+2y^2-x^4. $$