# Lyapunov function for specific $2$D autonomous system with no potential

Consider the following two-dimensional autonomous system $$\left\{ \begin{array}{cc} \dot{u} = -2v+v^2 \\ \dot{v} = -3u^2 +6u \end{array} \right.$$

Determine whether the critical point $$(0,0)$$ is stable or unstable.

Looking at the phase portrait, I expect that $$(0,0)$$ is stable, but not asymptotically stable.

If we try to investigate the stability using the linearization of the system around the point $$(0,0)$$, then we cannot conclude anything, as the Jacobi matrix in $$(0,0)$$ has purely imaginary eigenvalues.

Hence, we have to use a Lyapunov function. However, I have not been able to determine it. I tried to use a quadractic form $$V(u,v) = au^2 + buv + cv^2,$$ with $$a > 0$$ and $$b^2 - 4ac < 0$$, but to no avail.

the system is Hamiltonian, meaning there is a function $$H$$ that stays constant. $$\dot{x} = -2y + y^2$$ $$\dot{y} = 6x - 3 x^2$$ with Hamiltonian $$H = x^3 - 3 x^2 + \frac{y^3}{3} - y^2$$ so trajectories near the origin are closed (periodic) and sort of egg shaped.
• I see, thank you! However, how did you determine the Hamiltonian? I know that $\dot{p_x} = - \frac{\partial H}{\partial x}$ and $\dot{x} = \frac{\partial H}{\partial p_x}$ and the same for $y$, where $p_x$ is the momentum in $x-$coordinate, but we need the Lagrangian to find the momentum. However, in this case, we have Cartesian coordinates, so we may not need the Lagrangian?
• Oh wait, $y$ is the momentum in the $x-$coordinate. Now it makes sense, thank you!