I am working on being able to recognize appropriate Lyapunov functions to show the stability (or instability) of equilibrium points. I have the following system:

$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} -x^3 \\ 2x^3 \end{pmatrix}$

with the equilibrium point $\bar{x}=(0,0)$. I wish to prove that $\bar{x}$ is asymptotically stable via an appropriate Lyapunov function. I started with

$V(x,y) = x^4 + y^4$

but I was not sure how to show that $\dot{V} < 0$. Does anyone have any ideas? Should I use a different Lyapunov function?

  • $\begingroup$ I was wondering the same thing! $\endgroup$ – MathIsHard Oct 25 '17 at 17:04

Consider $V(x,y) = x^4 + x^2 + y^2$. $V$ is positive definite, as $V(0,0) = 0$ and $V(x,y) > 0 \; \forall (x,y) \neq (0,0)$. Now let's compute $\dot V$:

$\dot V(x,y) = (4 x^3 + 2x)(y-x^3) + 2y(-x-2y+2x^3) = - 4 (x^3-y)^2 - 2x^4$

So, clearly we have $\dot V(x,y) \leq 0$. Besides, for $\dot V(x,y)$ to be $0$, we must have $x^4=0$ and $y=x^3$, i.e., $x=y=0$.

Therefore, we've proved local asymptotic stability of the origin. Besides, given that $V$ is positive definite over all $\mathbb{R}^2$ and $\dot V$ is negative definite over all $\mathbb{R}^2$, and also that $\lim_{||\mathbf{x}|| \to \infty} V(x,y) = \infty$, the origin is also globally asymptotically stable.

  • $\begingroup$ Thank you! I had a misconception that $V(x,y)$ had to contain both $x^4$ and $y^4$, but now I see it is more of an art form. Thank you for the help! $\endgroup$ – Rebecca Hardenbrook Oct 26 '17 at 19:08
  • 1
    $\begingroup$ You are welcome! Indeed, finding a Lyapunov function can be more of an art than a science. Please don't forget to mark this as the accepted answer if it fulfills your question. $\endgroup$ – LGenzelis Oct 26 '17 at 19:27

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