I have a question about a differential equation I tried to analyse:
$$ \begin{align} \dfrac {dx}{dt} &= v \\ \dfrac {dv}{dt} &= -x+x^3-v^3 \\ \end{align} $$
I plotted this differential equation, and it looks like it has a spiral going inward (see picture). Now I tried to prove this: Therfore I searched a liapunov function and I think I found a local one:
$$ L = \dfrac 14(2v^2+2x^2-x^4)$$ for $v^2+x^2 > \dfrac { x^4}2 $ and $\dfrac {dL}{dt} = -v^4$
Now if I understand it right, the fixed point in the middle should attract all trajectories in the region where the lyapunov function is defined: $v^2+x^2 > x^4/2$ But if I compare the ‚region of attraction‘ of the lyapunov function with the computer simulation, it does not fit together.
Can someone help me and tell me what I am doing wrong? I appreciate every help!
You could write this as second order equation $$ \ddot x+\dot x^3+x-x^3=0 $$ which can be interpreted as a one-dimensional kinetic system with potential force $x^3-x$ and a friction term $c(x,\dot x)\dot x$ with friction coefficient $c=\dot x^2$. Consequently it appears as reasonable to take the energy $$ E(x,\dot x)=\frac12\dot x^2+\frac12x^2-\frac14x^4 $$ of the system as Lyapunov function, as you did.
The reasoning about the solutions crossing the level surfaces downwards is only correct as long as the level curves are concentric around the minimum. This region ends at the saddle points $(x,\dot x)=(\pm 1,0)$. This gives the region $$ E< \frac14\iff 2\dot x^2< (1-x^2)^2\iff \sqrt2|\dot x|< |1-x^2| $$ The component containing the origin is $$ \sqrt2|\dot x|+x^2< 1 $$ which looks rather compatible with your phase portrait.