I'm doing this exercise:

> 
> Consider the following differential equation system:
> 
> 
> $$\begin{cases}\dot{x}=x(1-x-ay) \\\dot{y}=y(1-y-bx)\end{cases}$$
> 
> 
> With $a,b>0$ and different to $1$. Sketch the phase portrait on the first quadrant and prove it has no periodic orbits.
> 
> 
> 

What I've done so far is to find the equilibrium points. We make 

$$\begin{cases}0=x(1-x-ay) \\0=y(1-y-bx)\end{cases}$$

and we find equilibrium points at $(0,0), (1,0), (0,1)$ and at the intersection point between $x=1-ay\ $ and $y=1-bx$, wich is

$$x=\frac{1-a}{1-ab},\ y=\frac{1-b}{1-ab}.$$

Once we have this, we have to check what's the local behaviour og the orbits near every equilibrium point:

We first find $$J=\begin{pmatrix}
-x+(-1+x+ay) & -xa\\ 
 -yb & -y+(-1+y+bx)
\end{pmatrix},$$

but checking every equilibrium point $(J\_{(0,0)},J\_{(1,0)},J\_{(0,1)},\dots )$ doesn't give me nothing.

Am I missing something? Once I have the local behavior on the equilibrium points, how can I prove it has no periodic orbits?

Thanks for your time.