# Checking Lyapunov stability of non linear system

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?

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 $V(x) = {1 \over n+1} x_1^{n+1} + {1 \over 2} x_2^2$.

• Thank you. For even $n$, I'm not sure I understand how to show that the point is unstable. For odd $n$, how did you come up with that funcion? Commented Jul 3, 2018 at 17:05
• By analogy with the $n=1$ case. Take $ax_1^b+c x_2^d$ and pick $a,b,c,d$ accordingly. Commented Jul 3, 2018 at 17:09
• I added an additional hint. Commented Jul 3, 2018 at 17:13
• @user401516 Notice that $V(x)$ is related to the first integral of the system written as a single equation: differentiate the first equation and substitute the second to get $\ddot{x}_1 = -x_1^n$, multiply by $\dot{x}_1$ and integrate to get $\frac{1}{2}\dot{x}_1^2 = -\frac{1}{n+1}x_1^{n+1} + C$, then rearrange and substitute to get $C = \frac{1}{n+1}x_1^{n+1} + \frac{1}{2}x_2^2 = V(x)$, which is the first integral. Commented Jul 3, 2018 at 19:07

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).