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kalish1
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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?
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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 $\frac{b}{c+d}>1.$
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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
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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