Determine the stability of an equilibrium point for a given Lagrangian

Question

I have to determine the equilibria and the stability for a Lagrangian given by

\begin{align} \frac{\dot{x}^2+\dot{y}^2}{2}-V(x,y) \end{align}

where $V(x,y)=-(x^2+y^2)^4+(2x^2+y^2)^2-1$.

The origin $(0,0)$ is an equilibria, but since the Hessian of the potential energy is indefinite when evaluated at this point, I need an alternative way to move.

I've tried so far to see if $V(x,y)-V(0,0)>0$ in a neighbourhood of the origin, but I can't prove it. The previous point asked me to see if there's any first integral, and I know that $E(x,y,\dot x, \dot y)$ (the energy) is preserved by the "Jacobi function theorem", so maybe I have to use it but I don't know how.

Answer 1

We have that $$(2x^2+y^2)^2=x^4+2x^2(x^2+y^2)+(x^2+y^2)^2$$ And if $0<q<1$, we have that $q>q^2$, so we have that (for $0<x^2+y^2<1$) $$(x^2+y^2)^2 > (x^2+y^2)^4$$ Which means that $$x^4+2x^2(x^2+y^2)+(x^2+y^2)^2>x^4+2x^2(x^2+y^2)+(x^2+y^2)^2 +(x^2+y^2)^4$$ So $$(2x^2+y^2)^2-(x^2+y^2)^4 > x^4+2x^2(x^2+y^2)$$

Answer 2

Making $x^2+y^2 = r^2$ we have

$$ V(r,\theta) = -r^8+\frac 12(3+\cos(2\theta))r^4-1 $$

so for small $r^8 < r^4$ and $1 \le \frac 12(3+\cos(2\theta))\le 2$ so as we can observe $V(r,\theta)$ has a minimum at $r = 0$