Consider a dynamical system of the form $$ \dot{x}=f(x), \quad x\in X, $$ and assume that the system possesses a set of non-isolated fixed points. Suppose moreover that there exists a Lyapunov $V(x)$ function whose derivative is non-decreasing along the trajectories of the system. By virtue of the LaSalle invariance principle we know that the system will converge to the largest invariance set of fixed points $\{x\in X\,:\, \dot{V}(x)=0\}$. Now, I know that there exist counter-examples showing that the system can converge to some trajectories in the invariant set without necessarily approaching to a point (see for instance this post). My question is whether there exist additional conditions (different from the fact that the set of fixed points is isolated) on the system and/or Lyapunov function $V(x)$ which guarantee asymptotic converge to fixed points and not to "moving" trajectories.

Thank you in advance.


Possibly, invariant set localization tecnique can be useful to you.

Each periodic trajectory in a compact invariant set contains at least two points of the set $$ S_h= \{ x\in X : \dot h(x)=0 \}, $$ where $h(x)$ is any $C^{\infty}$ function on $X$. This fact can be used to prove non-existence of non-zero trajectories, completely contained in $\{ x\in X: \dot V(x)=0 \}$.


Yes, there do exist sufficient conditions for asymptotic stability when the Lyapunov function is negative semi-definite, which I describe below.

Krasovsky's Theorem

Given an autonomous ODE $\dot x = f(x)$ with fixed point at the origin. Let $K$ be a manifold that does not contain entire trajectories of the ODE. If there exists a Lyapunov function $V$ such that the orbital derivative $\dot V$ satisfies:

  1. $\dot V<0$ outside of $K$
  2. $\dot V=0$ on $K$

then the dynamics is asymptotically stable.

If $K = \{ x ~:~ F(x)=0\}$, here is a sufficient condition for $K$ to not contain entire trajectories of the ODE: $$ (f^T \nabla F)(x) \ne 0 \quad \text{on $K \setminus \{ \mathbf{0} \}$} $$ where $f$ is the vector field of the ODE. (This just ensures that the vector field $f$ is never orthogonal to the normal to the surface.)


Consider the ODE: $ \dot x_1 = -x_1 + 3 x_2^2 \;, \dot x_2 = - x_1 x_2 - x_2^3 \;. $ In this case, a Lyapunov function is given by $V(x_1,x_2)=(x_1^2+x_2^2)/2$, whose orbital derivative is negative semi-definite. The set where $\dot V=0$ is given by the zero level set of the function $F(x_1,x_2)=x_1 - x_2^2$. Note that $(f^T \nabla F)(x_1,x_2) = 2 x_2^2 + 4 x_2^4 \ne 0$ on $K \setminus \{ \mathbf{0} \}$. Thus, by Krasovsky's Theorem the origin is asymptotically stable as illustrated in the graphic below.

graphical illustration

In this graphic, two of the axes correspond to state variables $x_1$ and $x_2$, and the other axis is the time variable $t$. The red line marks the state $x_1=0$ and $x_2=0$ for the time interval shown. Different grey shading is used for trajectories with different initial conditions.


Here is a cartoon from the book referenced below, which illustrates the idea behind Krasovsky's Theorem. The dark line labelled $\gamma$ represents a solution of the ODE, the lighter lines are contour lines of the semi-definite Lyapunov function $V$, and the dashed region is $K$ where $\dot V=0$.
cartoon This cartoon nicely illustrates how the dynamic avoids getting stuck inside $K$, and instead, asymptotically reaches a fixed point. Note that if the ODE solution enters $K$ then the value of the Lyapunov function does not change. This is illustrated by the curve remaining on a level curve of $V$ in the dashed region labelled $K$. However, eventually the ODE solution must exit $K$ (by hypothesis of Krasovky's Theorem), after which the Lyapunov function again decreases.

This picture suggests that a set of non-isolated fixed points can be reached in this fashion.


David R. Merkin [1997]. Introduction to the Theory of Stability. Texts in Applied Mathematics. Springer. Translated from Russian by Andrei L. Smirnov and Fred Afagh.

  • 1
    $\begingroup$ Thanks for the answer. In my question, I assume that the dynamical system possesses a set of non-isolated fixed points, and I ask whether there exist conditions which implies convergence to a point in the set of fixed points (and not to the set itself). It's not clear to me if your answer applies to this framework. $\endgroup$
    – Ludwig
    Sep 22 '16 at 19:46
  • $\begingroup$ As in Lasalle's principle, the set K (for Krasovsky) is allowed to be non-isolated. Krasovsky's theorem is a general purpose result for autonomous ODEs that guarantees asymptotic convergence to fixed points and not relative equilibria, as requested. $\endgroup$ Sep 22 '16 at 21:19
  • 1
    $\begingroup$ I'm still a bit confused. I'm not talking about the set $K$ but the set of fixed points of $f$, namely $\mathrm{Fix}(f):=\{x\in X : f(x)=0\}$. In particular, in your example the fixed point at the origin is isolated. $\endgroup$
    – Ludwig
    Sep 22 '16 at 23:06
  • $\begingroup$ I added the idea behind the proof. Since the proof relies on a lyapunov function and the fact that K does not contain an entire trajectory of the ODE, I think it can be adapted to your setting where the set of fixed points is non-isolated. $\endgroup$ Sep 23 '16 at 11:20
  • $\begingroup$ Unfortunately, the assumption that $K$ does not contain entire trajectories (other than the trajectory in $0$ I guess) rules out the case that the OP is interested in. If you have a set of nonisolated fixed points, then every such fixed point gives rise to an entire trajectory. So if $K$ is this set then it completely consists of entire trajectories. $\endgroup$ Oct 7 at 18:05

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