Prove that the zero solution of the equation $$\ddot{x} + (1-x^2)\dot{x} + x=0$$ is stable by using Lyapunov's Stability Theorem.
An indeed straight forward question but in general hard to tackle!
Let $\dot{x}=y$. The we have the system $$\dot{x}=y$$ $$\dot{y}=(x^2-1)y-x$$ It remains to find a positive definite function $V$ to complete the proof. However, I fail to do so. The book suggests $V=x^2 + y^2$ but it leads me to nowhere. Can anyone give me some tips? Thanks in advance.
If you have the the equation $$\ddot{x}+ (1-x^2)\dot{x} + x=0$$ is stable by using Lyapunov's Stability Theorem
Let $\dot{x}=y$. The we have the system $$\dot{x}=y$$ $$\dot{y}=-(1-x^2)y-x$$
Let $V=x^2 + y^2$.
Then $$\dot{V}=2x\dot{x} + 2y\dot{y}=2xy+2y[-(1-x^2)y-x]=-2(1-x^2)y^2$$
In the neighborhood of the equilibrium $(0,0)$ we have $-2(1-x^2)y^2<0$, therefore the equilibrium $(0, 0)$ is asymptotically stable.