The system $$\begin{align} x'&=y-x^3, \\ y'&=-x^3, \end{align}$$ has a fixed point in the origin. The function $$V(x,y):=x^4+2y^2 $$ is a Lyapunov function, since $$\dot{V}=\frac{\mathrm{d}}{\mathrm{d} t} V(x(t),y(t))=-4x^6(t) \leq 0 .$$ It follows that solutions of the system travel through contour lines of $V$ (which resemble ellipses) in such a way that the value of $V$ is non-increasing. This proves that the origin is stable in the sense of Lyapunov. My question is: is it also asymptotically stable? I can't deduce it immediately since $\dot{V}=0$ on the $y$-axis (that is $\dot V$ is not negative definite).
Thank you!
Yes, it is globally asymptotically stable. We can use the following theorem (Khalil, Corollary 4.2):
$\bullet$ Let $x=0$ be an equilibrium point for $\dot x=f(x)$. Let $V:\mathbb R^n\to\mathbb R$ be a continuously differetiable, radially unbounded, positive definite function such that $\dot V\le 0$ for all $x\in\mathbb R^n$. Let $S=\{x\in\mathbb R^n \;|\; \dot V(x)=0\}$ and suppose that no solution can stay identically in $S$, other than the trivial solution $x(t)\equiv 0$. Then, the origin is globally asymptotically stable. $\bullet$
For this system $$S=\{(x,y)\in\mathbb R^2: x=0\},$$ $$x=0\;\Rightarrow\; \dot x=y;$$ if $y\ne 0$, then the solution can't stay in $S$. Thus, the conditions of the theorem are satisfied.