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I have to find Lyapunov functions to stablish the stability of equilibrium points for the equation $\begin{cases}x' = -xy^4 \\ y' = yx^4\end{cases}$ which coincide with the axis of the plane $\{(x,y) \in \mathbb{R}^2. xy = 0\}$. Note that this points can't be assymptotically stable and that the version of Chetaev theorem I have is not sufficient to stablish instability of any of them (since $\dot V = 0$).

Using a traditional strategy for $p = (0,0)$ I searched for a Lyapunov function of the form $V(x,y) = V_1(x) + V_2(y)$ and I found that $V(x,y) = x^4+y^4$ is a proper function.

What about the rest of the points how can I show their stability?

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    $\begingroup$ @Moo: No, all of the coordinate axes are critical points, you only need one of $x$ or $y$ be equal to zero to have zero velocity in both directions. $\endgroup$ – LutzL Jun 1 at 7:14
  • $\begingroup$ If $V(\mathbf{0})=0, V(\mathbf{x}) > 0 \forall x \in \ D \ - \{0\}$ and $\dot{V}(\mathbf{x}) \leq 0$ then the origin is stable in the sense of the Lyapunov. I think you met all the conditions. $\endgroup$ – CroCo Jul 17 at 18:19
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You established that any solution will be contained in the level surfaces of $V=x^4+y^4$. Next look at the movement along the curve, how it moves along the tangent direction $(-y^3,x^3)$ (oriented counterclockwise) giving $$ -y^3\dot x+x^3\dot y=xy(x^6+y^6). $$ By the sign of this scalar product, this gives a movement toward the $y$ axis inside all of the quadrants, making the stationary points on the $y$-axis stable (but not asymptotically stable) and on the $x$-axis outside the origin unstable.

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  • $\begingroup$ There are two things I do not get. How can I see that $(-y^3,x^3)$ goes counterclockwise and how does the scalar product of the field and the tangent reflect a "movement" $\endgroup$ – Javier Jun 1 at 9:19
  • $\begingroup$ It is the (scaled) projection of the vector field on the tangent. This is essentially a trivial operation, as the vector field is already tangent, but it gives the direction of motion. That it is counter-clockwise follows the same geometric idea as that the vector field $(-y,x)$ is counter-clockwise. $\endgroup$ – LutzL Jun 1 at 10:05
  • $\begingroup$ Ok, one last thing, there is no evident lyspunov function for the rest of the points? $\endgroup$ – Javier Jun 1 at 10:14

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