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I'm studying lyapunov functions and i came across the following problem

Find a strict Lyapunov function for the equilibrium $(0,0)$ of

\begin{align*} x' &= -2x - y^2\\ y' &= -y -x^2 \end{align*}

Find $\delta > 0$ as large as you can such that the open disk of radius $\delta$ and center $(0,0)$ is contained in the basin of $(0,0)$.

So this is what i thought, seems like the Lyapunov function that i find, will only be defined for a ball with radius $\delta$ and center $(0,0)$, but i have problems finding the Lyapunon function since it won't be defined globally and i really don't know how to find such $\delta$.

Could u give me more ideas of what is happening and how do i find such $\delta$.

Thank you in advance

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Try the Lyapunov candidate function

$$V(x,y)=x^2+y^2$$

the time derivative is given by

$$\dot{V}=2x\dot{x}+2y\dot{y}=2x\left[-2x-y^2 \right]+2y\left[-y-x^2 \right]=-4x^2-2xy^2-2y^2-2yx^2$$ $$=-2(2+y)x^2-2(1+x)y^2$$

So for $x>-1$ and $y>-2$ this function is negative definite. As we have to consider the level sets of $V(x,y)\leq c$ for the basin of attraction (these are circles), we see that $c=1-\varepsilon$, in which $\varepsilon$ an arbitrarily small positive number. Hence, we conclude that $\delta = (1-\varepsilon)$.

I guess that it is possible to find an even larger radius for the basin of attraction but that would require a different Lyapunov function.

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