Consider the system
$$
x'=y(1-x)=f(x,y)
$$
$$
y'=x(1-y)=g(x,y)
$$
Does it have a periodic orbit?
I consider several methods:
I first try to apply Dulac's criteria:
$$
div(\mathbb{f})=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial x}=-x-y
$$
I cannot say if this one change the sign…
Also, if I consider the function $V(x,y)=\frac{1}{2}x^2+\frac{1}{2}y^2$ and compute
$$
\dot V(x,y)=xx'+yy'=xy(2-x-y)
$$
I still cannot find the annulus so that it change the sign…
Best Answer
As stated in Cesareo's comment, $(x-y)' = -(x-y)$, which implies that the quantity $x-y$ is exponentially decreasing and the points in the $\omega$-limit set of any point must have $x-y=0$, which means that they lie on the line $y=x$. However, there can be no periodic orbits on a line for an autonomous system other than a trivial fixed point. The existence of a nontrivial periodic orbit on the line would imply the orbit must pass through the same points on the line, but traveling in a different direction, which is impossible due to uniqueness. This can also be seen by explicitly solving the system along the line, which is the logistic equation.