# What is the Lyapunov function that I should consider?

Consider the system $$\begin{split} x'&=-x^3 \\ y'&=-y(x^2+z^2+1)\\ z'&=-\sin(z) \end{split}$$ Find all the equilibrium points and find stability at origin.

This is what I've done:

I found the equilibrium points and I compute the jacobian matrix but it turns out one eigenvalue has real part $0$, therefore the only way to know if the point is stable it's with a Lyapunov function.

Could anyone tell me what is the Lyapunov function that I should consider?

• "the only way to know if the point is stable it's with a Lyapunov function" Certainly not, actually the signs of $x'$, $y'$ and $z'$, quite direct in the present case, suffice to solve this completely.
– Did
Feb 1, 2018 at 23:32
• @Did so no need of Lyapunov function? Feb 2, 2018 at 0:12
• In this particular case you can work out where trajectories go with respect to equilibrium point using the signs of derivatives as @Did suggested. So, by doing this here you are working directly with a definition of stability of equilibrium. Feb 2, 2018 at 10:28

$V(x,y,z) = \frac{1}{2}x^2 + \frac{1}{2}y^2 + \frac{1}{2}z^2$.
\begin{align} \dot V & = \dot x x +\dot y y + \dot z z\\ &= -x^4 - y^2(x^2+y^2+1) - z\,\sin(z) \\ &< 0\quad \text{for $(x,\,y,\,z)\ne0$ and $z \in (-\pi,\pi)$}. \end{align}
However, as also discussed in the comments, the equilibrium of each first order ODE is stable and attractive since in each equation the sign of the variable is always opposed to the sign of its corresponding derivative, except at the origin. (E.g. in $\dot x = -x^3$, any $x>0$ forces $x(t)$ to decrease due to $\dot x<0$, and $x<0$ yields $\dot x>0$.) The Lyapunov function approach says the same since $\dot x x<0$, $\dot y y<0$, $\dot z z<0$, $(x,\,y,\,z)\ne0$, $z \in (-\pi,\pi)$.
• Why did you consider only $z\in(-\pi,\pi)$ ? Feb 2, 2018 at 17:07
• The set which contains the origin and fro which elements $z\,\sin(z)>0$ is $\{z \in\mathbb R :-\pi<z<\pi\}$. If a trajectory $z(t)$ starts outside this set it tends to another equilibrium: $0 = \sin(z)$ holds for $z = n\pi$, $n\in \mathbb Z$. But only those where the graph of $\sin(z)$ crosses the $\dot z=0$ axis from below are asymptotically stable. This becomes clear when you sketch $\dot z$ over $\sin(z)$. Feb 2, 2018 at 17:20