# Finding a Lyapunov function to show stability

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.

• Try higher powers. Commented Apr 6, 2023 at 4:54

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

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

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