Question 8.2: Phase diagram for equation d²x/dt² + Ω²x + Λx³ = 0 Sketch th...

This equation is equivalent to the pair of first order equations

$\begin{aligned}&\frac{d x}{d t}=v ,\\&\frac{d v}{d t}=-\Omega^{2} x-\Lambda x^{3},\end{aligned}$

which is an autonomous system. The phase paths satisfy the equation

$\frac{d v}{d x}=-\frac{\Omega^{2} x+\Lambda x^{3}}{v},$

which is a first order separable ODE whose general solution is

$v^{2}=C-\Omega^{2} x^{2}-\frac{1}{2} \Lambda x^{4},$

where C is a constant of integration. Each positive value of C corresponds to a phase path. The phase diagram for the case Λ < 0 is shown in Figure 8.4. There are three equilibrium points at (0, 0), $\left(\pm \Omega /|\Lambda|^{1 / 2}, 0\right)$. The closed loops around the origin correspond to periodic oscillations of the particle about x = 0. Such oscillations can therefore exist for any amplitude less than $\Omega /|\Lambda|^{1 / 2}$; this confirms the prediction of the energy argument used earlier. Outside this region of closed loops, the paths are unbounded and correspond to unbounded motions of the particle. These two regions of differing behaviour are separated by the dashed paths (known as separatrices) that ‘terminate’ at the equilibrium points $\left(\pm \Omega /|\Lambda|^{1 / 2}, 0\right)$.