Attractivity and Lyapunov Stability I'm having some trouble grasping the difference between attractivity and Lyapunov stability.
Is it always the case that an equilibrium of a function which is attractive is also Lyapunov stable, and vice versa? If not, could you give an example of a function which is attractive, but not Lyapunov stable, and/or vice versa?
 A: There is an easy example for "stability $\not\Rightarrow$ attractivity":
$$
x'' + \omega^2x = 0, x(0) = x_0, x'(0) = 0
$$
for some $\omega \neq 0$. The solution is $x(t, x_0) = x_0\operatorname{cos}(\omega t), t \in \mathbb{R}$ and an equilibrium is given by $(0,0)$. This equilibrium is certainly stable since ($\epsilon > 0, \delta := \epsilon$ in the definition of stability) $$
|x(t,x_0)| = |x_0| |\operatorname{cos}(\omega t)| \leq |x_0| \leq \delta := \epsilon.$$
However, it is not attractive as for $x_0 \neq 0: \lim_{t \rightarrow \infty} x_0\operatorname{cos}(\omega t)$ does not even exist. 
Examples for "attractivity $\not\Rightarrow$ stability" are more difficult. I will only post one equation system and a link to a phase portrait as it requires some enhanced techniques:
\begin{cases} 
x' = x + xy - (x+y)(x^2+y^2)^\frac{1}{2} \\
y' = y - x^2 + (x-y)(x^2+y^2)^\frac{1}{2}
\end{cases}
which is in polar coordinates $$
r' = r(1-r), \varphi' = r(1-\operatorname{cos}(\varphi))$$
(1,0) is a (global) attractor which is not stable (phaseportrait).
You might also want to search for Vinograd (1957) who provided the first counterexample (according to one of my professors).
