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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.

copper.hat
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Li Xun
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2 Answers2

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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.

enter image description here

copper.hat
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  • Thanks! Also, are these the only equilibria? – Li Xun Oct 24 '15 at 00:44
  • There is one equilibrium (if $a \neq 0$). The equations are not difficult to solve. Remember that if $a = c \cdot d$ and $a \neq 0$ then neither $c$ nor $d$ can be zero. – copper.hat Oct 24 '15 at 00:45
  • It is also entirely feasible that I made a mistake. – copper.hat Oct 24 '15 at 00:47
  • And how would we use the Jabobian matrix to linearize the system and determine the type of the equilibrium? – Li Xun Oct 24 '15 at 00:49
  • Evaluate the Jacobian at the above point. – copper.hat Oct 24 '15 at 00:50
  • Can you show me that as well? I do not think I am doing it correctly – Li Xun Oct 24 '15 at 00:55
  • Are you sure you have the correct equation? – copper.hat Oct 24 '15 at 01:01
  • Thank you again. How would this point be classified (seeing as it has the constants present)? – Li Xun Oct 24 '15 at 01:02
  • It is not immediately obvious to me, you need to compute the eigenvalues, but that is beyond me right now :-). It ain't pretty (that's why I was asking you to check the equations). – copper.hat Oct 24 '15 at 01:03
  • Alright, I will look at this. So f((x,y))=(a−x(1+4y1+x2),bx((1−y1+x2))Tf((x,y))=(a−x(1+4y1+x2),bx((1−y1+x2))T is the equilbrium point? – Li Xun Oct 24 '15 at 01:07
  • No. That is the right hand side of the differential equation. The equilibrium points are given by solving $f((x,y)) = 0$. The equilibrium point is $({a \over 5}, 1+({a \over 5})^2)$. – copper.hat Oct 24 '15 at 01:08
  • Oh sorry, I mean the equilibrium point would just be f((x,y))=((a/5),(1+(a/5)^2)) – Li Xun Oct 24 '15 at 01:12
  • Actually, can I please have help with the equilibrium points classification as well? The constants a and b are really throwing me off. – Li Xun Oct 24 '15 at 01:52
  • I'm not sure what you are asking. The equilibrium points are solutions to $f((x,y)) = 0$. As far as I can determine, if $a \neq 0$, there is exactly one equilibrium point, the formula is given above by $(x,y) = ({a \over 5}, 1+({a \over 5})^2)$. If the Jacobian of $f$, evaluated at an equilibrium, has eigenvalues with all strictly negative real parts, then the equilibrium is stable. If there is an eigenvalue with a positive real part, the equilibrium is unstable, otherwise it is indeterminate. However, I have been unable to come up with a nice formula for the real parts of the eigenvalues. – copper.hat Oct 24 '15 at 05:20
  • Sorry, I meant to say I have not been able to successfully find the eigenvalues and use them to determine the type that the equilibrium point is. – Li Xun Oct 24 '15 at 16:55
  • @JohnWatson see the edit to my answer for a discussion of this. Some of the comments there also spell things out in more detail. – Chris Kerridge Oct 24 '15 at 17:20
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$\DeclareMathOperator{\Tr}{Tr}$ 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).

Graph of discriminant to classify fixed points

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.

Chris Kerridge
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  • Can you please show me how to implement this into our nonlinear system? – Li Xun Oct 24 '15 at 02:38
  • As far as I can see, everything you have above is correct, but the implied clarity of the solution is not obvious to me. For example, the discriminant is zero iff $160a^3b^2−25a^2b^2−30a^3b+1250ab−9a^4+750a^2−15625=0$. – copper.hat Oct 24 '15 at 16:41
  • $\DeclareMathOperator{\Tr}{Tr}$ It depends how much you want to know about the solution. If $\det(J) > 0$ (which it is) it is enough to show $\Tr(J) < 0$ to show the fixed point is stable. If you want to know how it is attracted to the fixed point (i.e. spiral or not) then you need to look closer at the discriminant. – Chris Kerridge Oct 24 '15 at 16:52
  • If $\det(J) > 0$ then the sign of the real parts of eigenvalues are decided by the sign of the trace. – Chris Kerridge Oct 24 '15 at 17:13
  • Oops, I am embarrassed. I should know better. – copper.hat Oct 24 '15 at 17:15
  • $$\DeclareMathOperator{\Tr}{Tr}\lambda = \frac{-\Tr(J) \pm \sqrt{\Tr(J)^2-4\det(J)}}{2}$$. If $\det(J) > 0$ then $\left|\Tr(J)\right| > \left|\sqrt{\Tr(J)^2 -4 \det(J)}\right|$ – Chris Kerridge Oct 24 '15 at 17:16