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Consider the dynamical system $$x'=y^3-x^5$$ $$y'=-x^3-y^5$$ This has only one equilibrium point, which is $(0,0)$. I would like to find out whether this is stable, asymptotically stable or unstable. By linearizing the equation at $(0,0)$, we obtain the zero matrix, which is not hyperbolic, so the linearization theorem cannot be used. The system is not Hamiltonian either. So I think I need to find a Lyapunov function, but I am not really sure how to proceed. Most of the examples I have seen use a function of the form $V=x^2+y^2$, but this does not work here.

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    $\begingroup$ Try higher powers. $\endgroup$ Commented Apr 6, 2023 at 4:54

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Consider the Lyapunov function candidate

$$ V = \frac{1}{4} x^{4} + \frac{1}{4} y^{4} .$$

Taking the time derivative of $V$ along the trajectories of the system

$$ \dot{V} = x^{3} \dot{x} + y^{3} \dot{y} .$$

Making substitutions

$$ \dot{V} = x^{3} \left(- x^{5} + y^{3}\right) + y^{3} \left(- x^{3} - y^{5}\right) $$

and cancelling like terms

$$ \dot{V} = - x^{8} + x^{3} y^{3} - x^{3} y^{3} - y^{8} $$

yields

$$ \dot{V} = - x^{8} - y^{8} < 0, \quad \; x \neq 0, \; y \neq 0 .$$

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    $\begingroup$ I guess a hint would have been enough for the OP. This looked like a homework exercise. $\endgroup$
    – KBS
    Commented Apr 6, 2023 at 17:22

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