Show that the $\omega$-limit set of any orbit is on the set $K = \{(x,y) \in \mathbb{D} : x+y = 1\}$. The full problem is as follows:
Consider the following model defined in $\mathbb{D} = \{(x,y) \in \mathbb{R}^2 : x \geq 0, y\geq 0\}$: $\dot{x} = 1-x-\frac{2xy}{2+x}$, $\dot{y} = y \left( \frac{2x}{2+x} -1\right)$ where $x(0) > 0$ and $y(0) > 0$. (a) Show that the $\omega$-limit set of any orbit is on the set $K = \{(x,y) \in \mathbb{D} : x+y = 1\}$. (b) Show that the system has a unique steady state in $\mathbb{D}$ and the unique equilibrium point is globally stable in $\mathbb{D}$.
I know that $(0,0)$ is an equilibrium point here, but I am out of practice and can't show the $\omega$-limit set, or the unique steady state. Any help and advice would be greatly appreciated!
 A: (a) Consider the function
$$
V=(x+y-1)^2.
$$
Its derivative along the trajectories of the system is
$$
\dot V= 2(x+y-1)(\dot x+\dot y)=
2(x+y-1)\left(1-x-\frac{2xy}{2+x}+ \frac{2xy}{2+x} -y\right)
$$
$$
=2(x+y-1)(1-x-y)=-2V^2\le 0,
$$
thus, according to the LaSalle's invariance principle, the $\omega$-limit set of the system is contained in $K$. 
(b) The right part of the system is equal to zero iff $x=1$, $y=0$, thus, $(1,0)$ is an unique steady state. $\dot y$ is negative on $K$ except for the point $(1,0)$ (because $\frac{2x}{2+x}<1$ for any $x<2$), thus, $y$ is decreasing along the trajectories on $K$, thus, any solution that contained in $K$ moves toward the point $(1,0)$. It means that the $\omega$-limit set of the system contains the point $(1,0)$ only,
i.e. for any initial point in $\mathbb D$ $\lim_{t\to\infty} (x(t),y(t))=(1,0)$. 
In order to prove the global asymptotic stability of $(1,0)$ we need also to prove its (local) Lyapunov stability. It is given by the Lyapunov function
$W(x,y)=V(x,y)+y^2$ since the derivative
$$
\dot W(x,y)= -2V^2+2y^2\left( \frac{2x}{2+x} -1 \right)
$$
is negative in some neighborhood of the equilibrium point $(1,0)$.
