Given the following system:
$$ \left\{ \begin{array}{c} \dot x=y-x^2-x \\ \dot y=3x-x^2-y \end{array} \right. $$
I need to find the equilibrium points, and if stable, to find a Lyapunov function.
I have found two equilibrium points: $(0,0), (1,2)$. By linearization I'v found $(0,0)$ to be unstable, and $(1,2)$ to be stable. I tried to find a Lyapunov function for $(1,2)$ with no success. How can I find such a function?
Considering the surroundings for $(1,2)$ we have
$$ \left\{ \begin{array}{c} \dot x=y-x^2-x \\ \dot y=3x-x^2-y \end{array} \right. \approx \left\{ \begin{array}{c} \dot x=y-3x+1 \\ \dot y=x-y+1 \end{array} \right. $$
because linearizing about $(1,2)$ we have
$$ \left(\begin{array}{c} y-x^2-x \\ 3x-x^2-y \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right)+\left( \begin{array}{cc} -3 & 1 \\ 1 & -1 \\ \end{array} \right)\left(\begin{array}{c} x-1 \\ y-2 \end{array} \right)+O(x,y) $$
now considering a local Lyapunov function such as
$$ V = \alpha (x-1)^2+\beta(y-2)^2 $$
we have
$$ \dot V =-2 (\alpha (x-1) (3 x-y-1)-\beta (y-2) (x-y+1)) $$
or
$$ \dot V = (x-1,y-2)^{\dagger}\cdot H \cdot(x-1,y-2) $$
with
$$ H = \left( \begin{array}{cc} -6 \alpha & \alpha +\beta \\ \alpha +\beta & -2 \beta \\ \end{array} \right) $$
with eigenvalues
$$ \lambda = \pm\sqrt{2} \sqrt{5 \alpha ^2-2 \alpha \beta +\beta ^2}-3 \alpha -\beta $$
now it is easy to show that there exists $\alpha > 0, \beta > 0$ such that
$$ \lambda < 0 $$
hence this affirm the existence of a local Lyapunov function to assure the stability of $(1,2)$