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Suppose we have the system $$ \left(\begin{array}{c} \dot{x} \\ \dot{y} \end{array}\right) = \left(\begin{array}{c} f(x) + y \\ g(x) \end{array}\right). $$ Here $f,g: \mathbb{R} \to \mathbb{R}$ are smooth analytic functions of $x$ such that $$ \lim_{x\to 0} \frac{f(x)}{x^k} $$ and $$ \lim_{x\to 0} \frac{g(x)}{x^l} $$ exists and are non-zero for some $k, l \geq 2$. What are necessary and sufficient conditions on $f$ and $g$ so that $(0, 0)$ is (asymptotically) stable?

I have no idea where to start. I thought it has to do something with the signs of $f$ and $g$ near $0$, but the $y$ term causes a disturbance, and I don't know how to handle that, so to say.

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The existence of a Lyapunov function is a necessary and sufficient condition for the stability of dynamical systems, in the sense of Lyapunov.

You may start with a candidate Lyapunov functon $V(x,y)=(x^2+y^2)/2$. Its Lie derivative along the system dynamics yields \begin{equation*} \dot{V}(x,y)=x f(x)+xy+yg(x)=xf(x)+y(g(x)+1), \end{equation*} where $\dot{V}$ stands for the inner product between the gradient of $V$ and the vector field. To ensure that $\dot{V}$ is a negative definite definite function, you may assume that $f$ is unbounded, and satisfies $xf(x)<0$, for every $x\in\mathbb{R}$ and $f(x)\Leftrightarrow x=0$. Concerning $g$, you may assume that there exists a function $\alpha:\mathbb{R}\to\mathbb{R}$ such that, for every $y\in\mathbb{R}$, $y\alpha(y)<0$ and $\alpha(y)=0\Leftrightarrow y=0$, and for every $x\in\mathbb{R}$, $g(x)+1\leq \alpha(y)$.

Note that the previous assumptions are dependent on the candidate Lyapunov function that you chose.

If you want to consider only the differential equation itself, you may start with a necessary condition regarding the stabilizability of the $x$-subsystem:

  • if $y=0$, then 0 is asymptotically stable for the system $\dot{x}=f(x)$. Otherwise, whenever $g(x)=0$ and $f(x)\neq0$, there's nothing you can do.

You might find more information on the book Khalil, Nonlinear Systems.

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    $\begingroup$ The existence of a Lyapunov function is a sufficient condition for stability. Only in some cases can one establish necessity. $\endgroup$ – copper.hat Oct 28 '15 at 0:16
  • $\begingroup$ @copper.hat Could you provide a reference with counter-example? Or at least mention what could be an obstruction for creating Lyapunov function using the fact that equilibrium is asymptotically Lyapunov stable? $\endgroup$ – Evgeny Nov 12 '15 at 11:03

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