I need to find a strict Liapunov function for this system at the equilibrium point $(0,0)$ $$x'= -2x-y^{2}$$ $$y'=-y-x^{2}$$ Also need to determine $\delta > 0$ as large as possible so that the open disk of radius $\delta$ and center $(0,0)$ is contained in the basin of attraction of $(0,0)$.
(by definition, the basin of attraction is the set of all initial conditions whose solutions tend to the equilibrium.)
I've tried the famed function $$L(x,y)=x^{2} + y^{2}$$ as a Liapunov function, but the problem is that all I get as the basin of attraction is only the first quadrant, for which of course there's no open disk of radius $\delta$ at the center $(0,0)$.
Any help would be appreciated.
Hint: You can try the Lyapunov Function candidate $V =\mathbf{x}^T \mathbf{Px}$, for any positive-definite, symmetric $\mathbf P$.
The derivative along the lines of the system are of the form
$\dot V = -[\mathbf{x^T Q x + 2x^TPg}(x,y)]$,
where$\mathbf g(x,y) := (x^2 \quad y^2)^T$. Moreover, $\mathbf Q= \mathbf Q ^T >0$ satisfies the Algebraic Lyapunov Equation
$-\mathbf Q = \mathbf{PA + A^TP}$.
And
$\mathbf A := \begin{pmatrix} -2&\phantom{-}0\\\phantom{-}0&-1 \end{pmatrix}$.
Now, you can assume that $\sqrt{x^2+y^2}\leq\delta$ so as to find a bound on the term $\mathbf {x^TPg}(x,y)$ and ensure that $V>0$ and $\dot V<0~\forall \sqrt{x^2+y^2}\leq\delta$.
Hope this helps.
Cheers!