# 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) 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 $(1,0)$.
• @obewanjacobi In the first case the choice is obvious as we know that the function must be equal to zero on $K$ and positive on $\mathbb D\setminus K$. In the second case it was just a guess. The simpliest possible Lyapunov function is a quadratic form. We need a positive definite quadratic form; $V$ is not positive definite. What can we add to $V$ in order to obtain a positive definite quadratic form? $(x-1)^2$ and $y^2$, for instance. $y^2$ is suitable for us because its derivative is $\le 0$ and the overall derivative is negative definite – AVK Jul 11 '18 at 5:34