Prove that the following system of ODEs has a focus at the origin.
$$\begin{aligned} \dot{x} &= -x^3-y^3\\
\dot{y} &= x^3 \end{aligned}$$
Plotting the vector field confirms this fact. I've tried to convert the system to polar coordinates and I've shown that $\dot{\theta}\neq 0$ except at the origin. I believe it would suffice to show that
$$r(\theta+2\pi)-r(\theta) < 0$$
for al $\theta$ when close to the origin. However, this seems to involve expressing $r$ in terms of $\theta$, which I don't know how to do. Any help? Additionaly, is there an easier method? Thank you in advance.
I also had the following idea: according to the fundamental theorem of calculus, $$r(\theta+2\pi)-r(\theta)=\int_\theta^{\theta+2\pi}\frac{dr}{d\theta}d\theta$$ If only I could somehow show this is strictly less than $0$…
Best Answer
You could consider a Lyapunov function of the form $V(x,y)=F(x)+G(y)$ with $F,G$ positive, convex, minimum in $0$, so that $$ \dot V=F'(x)(-x^3-y^3)+G'(y)x^3 $$ The term that is not directly compatible with a center is the $y$ derivative. Thus to eliminate the last term one can try to get $$G'(y)x^3-F'(x)y^3=0.$$
To see what happens, left is a plot of several solution trajectories, while right the same solution are plotted using $(x|x|,y|y|)$ as coordinates, so that the level curves of $x^4+y^4=r^4$ in that picture would be circles.