Verification of argument for global asymptotic stability of equilibrium point

[kalish1]

I am working on proving that an equilibrium point of a two-dimensional dynamical system is globally asymptotically stable. The background and justifications are below. I have gotten to the final steps (in bold), but cannot justify it. It seems right intuitively. Can someone navigate the argument and help me out?----------Let a two-dimensional dynamical system be described as follows:
\begin{align*}
x'=f(x,y)&= a-b\frac{xy}{x+y}-c x \\
y'=g(x,y)&=b\frac{xy}{x+y}-(c+d)y,
\end{align*}

where $a,b,c,d,x,y > 0, \frac{b}{c+d}>1,$ and $N(t) = x(t)+y(t).$

Let $E_1$ denote the equilibrium point of the system when $y=0.$ Then $E_1 = (\frac{a}{c},0).$

Let $E_2$ denote the equilibrium point of the whole system. Then $E_2 = \left(\frac{a}{b-d},\frac{a}{b-d}\left(\frac{b}{c+d}-1\right)\right).$ Note that $E_2$ exists if $\frac{b}{c+d}>1.$

Since
\begin{align*}
\dfrac{dN}{dt} &= a-cx-(c+d)y=a-cN-dy \leq a-cN,
\end{align*}
if $N > a/c,$ then $dN/dt<0.$ Since $dN/dt$ is bounded by $a-cN,$ a standard comparison theorem can be used to show that $N(t) \leq N(0)e^{-ct} + \frac{a}{c}(1-e^{-ct}).$ In particular, $N(0) \leq \frac{a}{c} \Rightarrow N(t) \leq \frac{a}{c}.$ Also, every solution of the system with initial conditions in $\mathscr{D} = \{(x,y) \in \mathbb{R}^{2}_{+}: N \leq a/c\}$ remains there for $t>0$. The $\omega$-limit sets of the system are contained in $\mathscr{D}$. Thus, $\mathscr{D}$ is positively invariant and attracting. Hence the trajectories of the system are bounded.

Now, given the local asymptotic stability of $E_2$ (verified in Maple by establishing the negativity of the eigenvalues of the Jacobian linearization of the system at $E_2$) and the boundedness of trajectories of the system, we can apply the Poincare-Bendixson theorem.

To show that $E_2$ is stable, we must rule out the possibility of a
periodic trajectory or a trajectory whose $\omega$-limit set contains $E_1.$ Using the Bendixson-Dulac criterion, we can rule out a periodic orbit if we can find a function, $\alpha(x,y),$ such that $\frac{\partial}{\partial x}(\alpha f)+ \frac{\partial}{\partial y}(\alpha g)$ does not change sign and is not identically zero on any open set of the $x-y$ plane. If we choose $\alpha(x,y)=\frac{1}{xy},$ then $\frac{\partial}{\partial x}(\alpha f)+ \frac{\partial}{\partial y}(\alpha g) = \frac{-a}{x^2y},$ which does not
change sign on any interval since $a$ is constant. Therefore, we have ruled out
the possibility of a periodic trajectory. Note that $E_1$ is a saddle node because the condition $\frac{b}{c+d}>1$ implies that one of the eigenvalues of the Jacobian linearization of the system at $E_1$, obtained by Maple, is positive.

We can rule out the possibility that the $\omega$-limit set of some trajectory contains $E_1$, because...

Hence, $E_2$ is globally asymptotically stable if $\frac{b}{c+d}>1.$


----------Thanks in advance!

This question has been crossposted here: http://math.stackexchange.com/questions/1242755/verification-of-argument-for-global-asymptotic-stability-of-equilibrium-point

 

[jvicens]

-in-2d-dynamical-systemSure, I can help you out with the argument. Let's break it down and go through it step by step.

First, we have the system described by the equations $x'=f(x,y)$ and $y'=g(x,y)$. The equilibrium point $E_1$ is the point where $y=0$, which is given by $E_1=(\frac{a}{c},0)$. This is a valid equilibrium point as it satisfies both equations of the system.

Next, we have the equilibrium point $E_2$, which is the point where both equations of the system are equal to 0. This point exists if $\frac{b}{c+d}>1$ and is given by $E_2=(\frac{a}{b-d},\frac{a}{b-d}(\frac{b}{c+d}-1))$. This can be verified by plugging in the values of $E_2$ into the system equations and showing that they both equal 0.

Now, we want to show that $E_2$ is globally asymptotically stable. To do this, we need to show that the trajectories of the system are bounded and that there are no periodic trajectories or trajectories whose $\omega$-limit set contains $E_1$.

To show that the trajectories are bounded, we use the fact that $\frac{dN}{dt}=a-cN-dy$. Since $dN/dt$ is bounded by $a-cN$, we can use a standard comparison theorem to show that $N(t) \leq N(0)e^{-ct} + \frac{a}{c}(1-e^{-ct})$. This means that $N(t)$ is always less than or equal to a certain value, which we can call $M=\frac{a}{c}$. This shows that the trajectories are bounded.

Next, we need to rule out the possibility of a periodic trajectory or a trajectory whose $\omega$-limit set contains $E_1$. To do this, we use the Bendixson-Dulac criterion. This criterion states that if we can find a function $\alpha(x,y)$ such that $\frac{\partial}{\partial x}(\alpha f) + \frac{\partial}{\partial y}(\alpha g)$ does not change sign and is not identically zero on any open set, then