Invariant set of Lotka-Volterra equation

Question

I have the Lotka-Volterra equation

$\dot{x}=x(1-y),$

$\dot{y}=y(x-1),$

where $x$ and $y$ are non-negative. It is easy to see that the $x$- and $y$-axis are invariant sets. I can see from plots that a periodic orbit exists. Now I want to prove this by hand calculations according to the Poincare-Bendixon criterion.

So, is there any invariant set that is a closed and bounded trajectory and contains no equilibrium point, which I can find by hand calculations?

Answer 1

To understand the dynamics of this LV equation you do not even need the existence theorem of ODE (Cauchy-Picard-Lindelöf-Lipschitz). Just apply the implicit function theorem to the function $V(x,y)=x-\log x + y - \log y$. It tells you that for any $c > \min V$ the level sets $\{V=c\}$ are closed simple curves that bounds nbds of the minimum point $(1,1)$ of $V$; these curves suitably parametrized are periodic solutions of the LV equations, turning around the equilibrium $(1,1)$. (Alternatively, the fact that the level set $\{V=c\}$ are invariant set for the ODE just comes computing $\frac{d}{dt}V\big(x(t),y(t)\big)= 0$ ).

Answer 2

If you substitute $x=e^u$, $y=e^v$, then your system becomes Hamiltonian.