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I've been struggling with solving the following system:

$\dot x = -x+xy$

$\dot y = x-y -x^2-y^3$

I already know that the only fixed point is $(0,0)$ and I also know that the system is exponentially stable(calculating the Jacobian). But what I really want to know is: do all trajectories of this system converges to $0$?

What can I do to prove that? Is it enough to create a Lyapunov function? What exactly means that it is asymptotically stable in a Lyapunov sense?

Thanks so much. :)

Rushabh Mehta
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skt1o2
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1 Answers1

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Consider the Lyapunov function $V(x,y) = \frac{1}{2}(x^2+y^2)$. For a Lyapunov function $V$, $\dot{V}<0$ for $(x,y)\neq0$ implies that the dynamical system is asymptotically stable. In this case, we have $$ \dot{V} = x\dot{x}+y\dot{y} = -x^2+xy-y^2-y^4. $$ You can use basic calculus to show that $\dot{V}(x,y)$ has a global maximum at $(x,y)=0$, (take the gradient and set equal to 0). We also have $\dot{V}(0,0) = 0$, so $\dot{V}<0$ for $(x,y)\neq0$ and the system is asymptotically stable

whpowell96
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  • Thanks for your answer, but, the fact that we can find a lyapunov function for the origin (0,0), means that all trajectories converges to (0,0), right? If we had a liapunov function for (z,w), means that all trajectories converge to (z,w)? – skt1o2 Dec 07 '19 at 23:01
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    Yes. In the literature you will typically find Lyapunov functions defined with the origin in mind but this is because you can apply a change of variables to move any critical point to the origin – whpowell96 Dec 07 '19 at 23:03
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    To show that the system is globally asymptotically stable, it's important to also point out that your Lyapunov function tends to infinity at infinity (which it does in this case). See this question: https://math.stackexchange.com/questions/2472608/system-with-a-lyapunov-function-over-mathbbrn-but-not-globally-asymptotica – Hans Lundmark Dec 08 '19 at 08:14