Prove that the system with $\dot{x} = x+y-4x^3$ and $\dot{y} = y-x-4y^3$ has a limit cycle 
Prove that the system with $\dot{x} = x+y-4x^3$ and $\dot{y} = y-x-4y^3$ has a limit cycle.

How do I begin diving in on this problem? Should it be solved through the use of a Lyapunov function, or is this a Poincare-Bendixon problem? Also some guidance on where to go after that would be helpful as well.
 A: Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
\dot V = 2V-8(x^4+y^4)
$$
and using
$$
\frac12(x^2+y^2)^2+\frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2\le \dot V\le 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $V\le\frac14$ and inwards towards zero for $V>\frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as 
$$
4\le \frac{d}{dt}(V^{-1})+2(V^{-1})\le 8,\\~\\
2(e^{2t}-1)\le e^{2t}V(t)^{-1}-V_0^{-1}\le 4(e^{2t}-1),\\~\\
\frac{V_0}{e^{-2t}+4V_0(1-e^{-2t})}\le V(t) \le\frac{V_0}{e^{-2t}+2V_0(1-e^{-2t})},
$$
which tells us that the dynamic moves every solution towards the annulus $\frac14\le V(t)\le\frac12$,
$$
\frac12\le r\le \sqrt{\frac12}.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
