By showing that $V(x_1,x_2) = (x_1)^2 + (x_2)^2$ is a strong lyapunov function for the system:
$x_1’ = -x_2$
$x_2’ = x_1 + (x_2)^3 - x_2$
determine a region of ''attraction'' for the origin.
I have the conditions for a strong Lyapunov Function to be:
L1:V(x) has local positive definiteness
L2) ΔV(x).f(x) ≤ 0
L3) ΔV(x).f(x)=0 if and only if x=0
Where ΔV(x) = [dV/d$x_1$ dV/d$x_2$]
I found V($x_1,x_2$) to satisfy L1, that the equation is positive definite, however I computed ΔV($x_1,x_2$).f($x_1,x_2$) to be ($x_2)^4 - (x_2)^2$ which I thought is always more than or equal to 0, contrary to that it should be less than or equal to zero. Moreover, the ΔV($x_1,x_2$).f($x_1,x_2$) equation I have has no $x_1$, so it can equal zero for any $x_1$, which wouldn't satisfy the Strong Lyapunov function conditions. Please tell me how to do this, it is driving me insane.
$$V = x_1^2 + x_2^2 $$ is a weak Lyapunov function for this system. Nevertheless you can get some local results since $$ \dot V = 2x_2^4 + 2x_2^2 $$ $$ -2x_2^2(1-x_2^2), $$therefore $$\dot V \leq -2x_2^2, \quad |x_2| < 1.$$ The derivative w.r.t. time of $V$ is negative semidefinite. However since the unique equilibrium of your system is $(0,0)$ the smallest invariant set is also $\underline 0$. Due to Krasovskii-LaSalle theorem the equlibrium is asymptotically stable. Of course, it is merely locally asymp. stable since for $|x_2|\geq 0,$ $V$ is increasing. However, you may find another Lyapunov function which derivative is negative definite, and therefore, a strict Lyapunov function.