Editing Lyapunov stability (section)

==Example==
This example shows a system where a Lyapunov function can be used to prove Lyapunov stability but cannot show asymptotic stability.
Consider the following equation, based on the [[Van der Pol oscillator]] equation with the friction term changed:

:<math> \ddot{y} + y -\varepsilon \left( \frac{\dot{y}^{3}}{3} - \dot{y}\right) = 0.</math>

Let

:<math> x_{1} = y , x_{2} = \dot{y} </math>

so that the corresponding system is

:<math> \begin{align}
&\dot{x}_{1} =  x_{2}, \\
&\dot{x}_{2} = -x_{1} + \varepsilon \left( \frac{{x_{2}^{3}}}{3} - {x_{2}}\right).
\end{align}
 </math>

The origin <math> x_1= 0,\ x_2=0</math> is the only equilibrium point.
Let us choose as a Lyapunov function

:<math> V = \frac {1}{2} \left(x_{1}^{2}+x_{2}^{2} \right) </math>

which is clearly [[positive-definite function|positive definite]]. Its derivative is

:<math>
\dot{V} = x_{1} \dot x_{1} + x_{2} \dot x_{2}
= x_{1} x_{2} - x_{1} x_{2}+\varepsilon
\frac{x_{2}^4}{3} - \varepsilon {x_{2}^2}
=   \varepsilon \frac{x_{2}^4}{3} -\varepsilon {x_{2}^2}.
</math>

It seems that if the parameter <math> \varepsilon </math> is positive, stability is asymptotic for <math> x_{2}^{2} < 3.</math> But this is wrong, since <math>  \dot{V} </math> does not depend on <math>x_1</math>, and will be 0 everywhere on the <math>x_1</math> axis. The equilibrium is Lyapunov stable but not asymptotically stable.