I am required to find the stability of this nonlinear system and for which values of $k$ is the system stable.
$\dot x=x.(x^2-1-k)$
I am trying to use quadratic Lyapunov function, and used the function $g(x)=x^2/\sqrt {k}$ to constraint $k$.
I am confused if I am doing it the right way, or if this is the wrong lyapunov function to use. Keep in mind I tried other functions.
It seems like for whatever Lyapunov candidate I use, $\dot V(x)\le0$ for all values of x does not seem logical.
Any help would be appreciated.
Best Answer
If you want to investigate the local stability of an equilibrium point of a nonlinear system it is usually easier to linearize. If all eigenvalues of linearized system have a negative real part then that equilibrium point is (asymptotically and exponentially) stable and if any of the eigenvalues has a positive real part it is unstable. Only in the edge case, when it is not unstable but there are eigenvalues with zero real part, would it be required to consider more of the nonlinear dynamics in order to identify whether or not the equilibrium point is locally stable.
It can be noted that if you are able to show that the system is stable near the equilibrium point $x_{eq}$ by using the linearization
$$ \dot{e} \approx A\,e $$
with $e = x - x_{eq}$ and $\|e\|$ close to zero, you can also always find a Lyapunov function by solving the (continuous) Lyapunov equation
$$ A^\top P + P\,A = -Q, $$
for any given positive definite $Q=Q^\top$ the solution for $P$ is also positive definite. Namely, the associated Lyapunov function is given by $V(e) = e^\top P\,e$, for which it holds that $\dot{V}(e) \approx -e^\top Q\,e$ when $\|e\|$ is close to zero. A key thing to note here is that in general it is not true that $\dot{V}(e) \nleq 0\ \forall\, e$. Namely, if it would hold for all $e$ then you would have shown global instead of local stability. One final note is that the set of all $e$ such that $\dot{V}(e) < 0$ is a lower bound of the basin of attraction of that equilibrium point.