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!

up vote 2 down vote accepted

(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 $(1,0)$.

  • @user539887 Yes, you are right. I updated the answer – AVK Jul 10 at 15:01
  • how did you know to choose these Lyapunov functions? We didn't cover them much in my modeling class, so I have always been unclear on how to choose the correct ones. – obewanjacobi Jul 11 at 0:58
  • 1
    @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 at 5:34

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.