Equilibrium Points and Linearization

Question

Consider the planar nonlinear system,

$x' = a − x − \frac{4xy}{1 + x^2}$
$y' = bx(1− \frac{ y}{1 + x^2})$

where $x$ and $y$ represent the concentrations of $I-$ (iodine ions) and $ClO_2-$ (chlorine dioxide ions), respectively, and $a$ and $b$ are positive parameters.

Find all equilibrium points for this system and linearize the system at your equilibria and determine the type of each equilibrium.

Answer 1

Setting $y'=x' = 0$ gives $x \neq 0$ since $a>0$ and so $1+x^2 = y$. Substituting into the other equation gives $x = {a \over 5}$ and hence $y = 1+({a \over 5})^2$.

If we let $f((x,y)) = (a-x(1+{4y \over 1+x^2}), bx((1-{y \over 1+x^2}) )^T $, then $Df(({a \over 5}, 1+({a \over 5})^2)) = {1 \over a^2 + 25} \begin{bmatrix} 3 a^2 -125 & -20 a \\ 2 a^2 b & -5 ab \end{bmatrix}$.

Following Chris' remarks: $\det Df(({a \over 5}, 1+({a \over 5})^2)) = {25 ab \over a^2+25 }$, and since this is strictly positive, determination of stability reduces to examining the sign of $\operatorname{tr} Df(({a \over 5}, 1+({a \over 5})^2))$.

We have $\operatorname{tr} Df(({a \over 5}, 1+({a \over 5})^2)) = { 3 a^2 - 5 a b -125\over 1+({a \over 5})^2 } $.

Hence we see that if $b > {3 \over 5} a -{25 \over a}$, the system is stable, if $b < {3 \over 5} a -{25 \over a}$, the system has an unstable mode and if equal, the eigenvalues are inconclusive in determining the stability of the original system.

In the following graph, $b$ is on the vertical axis and the stable region (that is, pairs $(a,b)$ for which the above equilibrium is stable) is above the blue line.

Answer 2

$\DeclareMathOperator{\Tr}{Tr}$ I think you now have how to get the fixed points. To classify them you need to look at the eigenvalues of the Jacobian at those points. For a 2D system, with Jacobian $$J =\left(\begin{matrix}a & b \\ c & d\end{matrix}\right):$$

$$\det(\lambda I - J) = \left|\begin{matrix}\lambda - a & -b \\ -c & \lambda-d\end{matrix}\right|=0$$

$$\Leftrightarrow \lambda^2 - \Tr(J) \lambda + \det(J) = 0 \tag{1}\label{1}$$

Looking at the discriminant of the quadratic, $\Tr(J)^2 - 4\det(J)$, you can determine the nature of the eigenvalues (complex/real, their sign). Here's a picture of what you should conclude ($\tau=\Tr(J)$, $\Delta = \det(J)$, up and right are increasing values of axes).

Continued:

Evaluating the Jacobian at a point with $y=1+x^2$ gives

$$J = \frac{1}{1+x^2}\left(\begin{matrix}-5+3x^2 & -4x \\ 2bx^2 & -bx\end{matrix}\right)$$ As the eigenvalues of some matrix $kM$ are the same as the eigenvalues of the matrix $M$, we'll just study $$J = \left(\begin{matrix}-5+3x^2 & -4x \\ 2bx^2 & -bx\end{matrix}\right)$$ This has $\Tr(J) = 3x^2 - bx - 5$, $\det(J) = 5bx(1+x^2)$.

This is 99% of your work. If you are still confused I encourage you to look at the general solution to (\ref{1}) for different values of $\Tr(J)$, $\det(J)$ to understand the eigenvalues you get from it. For instance, if $\Tr(J)^2 - 4\det(J) > 0$ there are 2 distinct eigenvalues which are both real, if $\Tr(J) < 0$ all eigenvalues have negative real part, etc... You should be able to come up with the graph I have shown. Apply this to the values of $\Tr(J)$, $\det(J)$ from your problem.