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


Try the Lyapunov candidate function


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.