Conditions for convergence to non-isolated fixed points 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.
 A: 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 $\{ x\in X: \dot V(x)=0 \}$.
A: 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: 


*

*$\dot V<0$ outside of $K$

*$\dot V=0$ 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.)
Example
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.

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.
ADD
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 $\dot V=0$.

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. 
Reference
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.
