Lyapunov Functions

How do we prove stability of a critical point?
Idea: If we can find a function non-negative function ${V}$ which doesn’t increase with time and vanishes only at the critical point, then we see that the trajectories sufficiently close to the critical point will stay close to the critical point.

For ${\varepsilon>0,}$ consider the system
$\displaystyle \begin{aligned} &\dot{x}=y\\ &\dot{y}=-x+\varepsilon x^{2} y \end{aligned}$

The linearized approximation at ${(0,0)}$ is given by
$\displaystyle \begin{aligned} &\dot{x}=y\\ &\dot{y}=-x\end{aligned}$ and has eigenvalues ${\pm i.}$

Take $V(x, y)=\left(x^{2}+y^{2}\right) / 2.$
$\displaystyle \begin{aligned}\frac{dV}{dt} = &=\nabla V(x, y) \cdot(\dot{x}, \dot{y}) \\ &=(x, y) \cdot\left(y, -x+\varepsilon x^{2} y\right) \\ &=\varepsilon x^{2} y^{2}\end{aligned}$
Thus we see that ${(0,0)}$ is globally stable. Infact, it can be shown that it is asymptotically stable.

$\displaystyle \dot{x}=-x$
Using $\displaystyle V(x)=x^{2}$
we see that
$\displaystyle \dot{V}(x)=V^{\prime}(x) \dot{x}=2 x \cdot(-x)=-2 x^{2} \leq 0$
and hence ${0}$ is globally asymptotically stable.

$\displaystyle \frac{d x}{d t}=-x y^{2}, \frac{d y}{d t}=3 y x^{2}$
$\displaystyle V(x, y)=3 x^{2}+y^{2}$
$\displaystyle \frac{d V}{d t} =6 x\left(-x y^{2}\right)+2 y\left(3 y x^{2}\right)=0$
Hence, the trajectories are ellipses- you could obtain this by integrating by noting that it’s a separable equation.

$\displaystyle \frac{d x}{d t}=y-2 x, \frac{d y}{d t}=2 x-y-x^{3}$
$\displaystyle V(x, y)=(x+y)^{2}+\frac{x^{4}}{2}$
$\displaystyle \frac{d V}{d t}=-6 x^{4} \leq 0$
This shows stability of the system.