Checking Lyapunov stability of non linear system

Question

I need to check the stability of the equilibrium point of the following system, $n \in \Bbb N$: $$ \left\{ \begin{array} \dot \dot x_1=x_2 \\ \dot x_2=-x_1^n \end{array} \right. $$

I tried using linearization, but the eigenvalues are zero, which means it's not the way to go. I also searched for a Lyapunov function, but couldn't find one. Any ideas?

Answer 1

Hints:

If $n$ is even it is not hard to directly show that $(0,0)$ is unstable. Consider a starting point $(-\epsilon, -\epsilon)$ for small $\epsilon >0$. Note that $x_2$ is non increasing, hence $x_2(t) \le -\epsilon$ for all $t$. What does that say about $x_1$?

If $n$ is odd, look at the function $V(x) = {1 \over n+1} x_1^{n+1} + {1 \over 2} x_2^2$.

Answer 2

Choose a Lyapunov function candidate:

$V(x) = a*x1^2 + b*x2^2$

Calculate its time derivative along the trajectories:

$\dot V(x) = 2ax_1x_2 - 2bx_2x_1^n$

To ensure stability, we choose a and b as positive constants to make V(x) positive for all x except (0, 0). Without loss of generality, we set a = b = 1 to simplify the analysis.

Now, $\dot V(x) = 2x_1x_2(1 - x_1^{(n-1)})$

This derivative is negative for all x except at (0, 0) when n is even.

When n is odd, the stability of the equilibrium point depends on the specific values of a and b.

Thus, the equilibrium point (0, 0) is stable for even values of n, and for odd values of n, further analysis is needed to determine stability. This derivative is negative for all x except at (0, 0).