Let $y,z$ two functions defined on $\mathbb{R}$.
Show that the solution of the differential system : $$ y'=z^3 \qquad z'=-y^3 $$
are periodic.
My attempt : With some works I can show that the "path" is included in the curve of equation $y^4+z^4= K$
I can only say that traverses a connected part (image of $\mathbb{R}$ by a function of $\mathbb{R}$ in $\mathbb{R}^2$ continuous) of this curve but it does not prove the periodicity because if the image is not a closed, one could never go through the same point on the curve I tried to eliminate one of the unknown functions.
We have $y''=3z^2z'=-3y^3(K-y^4)^1/2$ but I cannot see how can I do next ? Perhaps one has a direct way to prove the claim ?
This is an autonomous differential system and the solutions stay on a bounded closed curve hence either they converge to a point of the curve when $t\to\infty$ without ever reaching it, or they cycle in the sense that there exists some finite $T$ such that $(y(t+T),z(t+T))=(y(t),z(t))$ for every $t$.
In the present case, the square of the velocity $(y')^2+(z')^2=y^6+z^6$ is uniformly bounded below by a positive constant on each curve $y^4+z^4=c$ with $c\gt0$, hence the solutions cycle.