Local asymptotical stability for an ODE

Question

Consider a system: \begin{align*} \frac{d}{dt} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_2 \\ a x_1 + b x_1^2 \end{pmatrix} \end{align*} with $a < 0$ and $b\ne0$. My question is whether the equilibrium point $x=0$ is a locally asymptotically stable equilibrium point. I tried some Lyapunov functions, for example: $$ V = -\frac{a}{2} x_1^2 + \frac{1}{2} x_2^2 - \frac{b}{3}x_1^3 $$ give $$ L_f V = \frac{\partial V}{\partial x}f = 0,$$ hence the Lyapunov-stability, but not asymptotic stability.

Answer 1

The fact that $dV/dt = 0$ says that the trajectories in the $x_1,x_2$ plane lie on the curves $V = \text{constant}$. The origin is a strict local minimum of $V$, so near the origin these form closed curves around the origin.
A solution starting near the origin will not approach the origin, but will stay on such a curve. Therefore the equilibrium is a center: stable, but not asymptotically stable.

From another point of view, the lack of asymptotic stability is the result of the invariance of the system under the symmetry $x_1 \to x_1$, $x_2 \to -x_2$, $t \to -t$. If such a system is started at, say, $x_1 = p, x_2 = 0, t = 0$, and reaches $x_2 = 0$ again at $x_1 = q, t = T$, then by symmetry we will have $(x_1(T+t),x_2(T+t)) = (x_1(T-t),-x_2(T-t))$, so that at $t=2T$ it comes back to $x_1 = p, x_2 = 0$, and will continue in a closed orbit of period $2T$.