Finding Lyapunov function for a stable equilibruim of a non linear system 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? 
 A: 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 $(1,2)$

