How to show that every non-trivial orbit of ODE is periodic

Question

I'm currently looking at the system of first order ODEs

$$ \begin{cases} x' = -y-x^2y \\ y' = x+xy^2 \end{cases} $$

and I try to show that every non-trivial orbit of the system is periodic. I'm lost how to approach this. I tried converting the system into polar coordinates (but this got ugly and the radius didn't stay constant). I know how to show that no periodic orbits exists using Bendixson, but this is of no use here.

The system looks kind of symmetric, since the first line is equivalent to $-y(1-x^2)$ and the second line to $x(1+y^2)$, but I don't know where is could help me. I also looked the only fix point $(0, 0)$. It is non-hyperbolic (the Jacobian at $(0, 0)$ hat Eigenvalues $\lambda_{1/2} = \pm i$). Any hints on how to proceed are welcome!

Answer 1

This system admits a first integral: $$ \frac{dx}{-y-x^2y}=\frac{dy}{x+xy^2} $$ or $$ \frac{x\,dx}{1+x^2}=-\frac{y\,dy}{1+y^2} $$ gives us $$\tag{1} (1+x^2)(1+y^2)=C. $$ The level sets defined by (1) are closed curves, except for the case $C=1$, which corresponds to the equilibrium point. Hence, all solutions are periodic.

Answer 2

if you don't object to partial derivatives: given system $$ \dot{x} = X(x,y) \; , \; $$ $$ \dot{y} = Y(x,y) \; , \; $$ the system is called Hamiltonian when $$ \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} = 0. $$

When this happens, we can solve an easy PDE system, finding (guessing) a function $H$ such that $$ X = \frac{\partial H}{\partial y} \; \; , \; \; \; Y = - \frac{\partial H}{\partial x} . $$

We need $$ \frac{\partial H}{\partial x} = -x-xy^2 \; \; , \; \; \frac{\partial H}{\partial y} = -y - x^2 y $$ Just by inspection, we can take $$ H = \frac{1}{2} \left( - x^2 - x^2 y^2 - y^2 \right) $$ This is constant on an orbit if and only if $$ x^2 y^2 + x^2 + y^2 $$ is constant, if and only if $$ x^2 y^2 + x^2 + y^2 + 1 = \color{blue} {\left(x^2 + 1 \right)\left( y^2 + 1 \right)}$$ is constant. So, we agree with the first answer.

Here, $H$ is called the Hamiltonian function.
The orbits of the ODE lie within level curves of $H,$ as the chain rule says that $H$ stays constant on an orbit. When $H$ has a strict local minimum or strict local maximum at the equilibrium point, this means the orbits near the equilibrium are little simple closed curves, ellipses or whatever.

This is pages 78-79 in the Third edition of Jordan and Smith, Nonlinear Ordinary Differential Equations.