# Stability of singular points in a discrete control system

## Homework Statement

Give an example of a non-linear discrete-time system of the form

x1(k + 1) = f1(x1(k), x2(k))
x2(k + 1) = f2(x1(k), x2(k))

With precisely four singular points, two of which are unstable, and two other singular points which are asymptotically stable.

## Homework Equations

$$J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \\[1ex] \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \end{bmatrix}$$

## The Attempt at a Solution

I know that the singular is asymptotically stable if the eigenvalues of the Jacobian are under 1, and unstable if an eigenvalue is greater than 1. I've found a solution in MATLAB through brute force, but this is a practice exam question; how would I be able to do this by hand just by looking at it?

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jambaugh
Gold Member
By "...to do this by hand.." I presume you mean find the eigen-values? This is a classic linear algebra problem and you can read up on it on wikipedia, or any of thousands of good and even bad linear algebra textbooks.

But the basic method is to solve the characteristic equation:
$$det(J - \lambda \boldsymbol{1})=0$$
where $\boldsymbol{1}$ is the identity matrix $\boldsymbol{1}=\left( \begin{array}{cc}1 & 0\\0& 1\end{array}\right)$.

Let's see... if you want a simple formula for the 2x2 case, for matrix:
$$M = \left( \begin{array}{cc}a & b\\c& d\end{array}\right)$$
the eigen-values will be solutions to:
$$(a-\lambda)(d-\lambda) -bc =0, \quad \lambda^2 -(a+d)\lambda +ad-bc = 0$$

By "...to do this by hand.." I presume you mean find the eigen-values? This is a classic linear algebra problem and you can read up on it on wikipedia, or any of thousands of good and even bad linear algebra textbooks.

But the basic method is to solve the characteristic equation:
$$det(J - \lambda \boldsymbol{1})=0$$
where $\boldsymbol{1}$ is the identity matrix $\boldsymbol{1}=\left( \begin{array}{cc}1 & 0\\0& 1\end{array}\right)$.

Let's see... if you want a simple formula for the 2x2 case, for matrix:
$$M = \left( \begin{array}{cc}a & b\\c& d\end{array}\right)$$
the eigen-values will be solutions to:
$$(a-\lambda)(d-\lambda) -bc =0, \quad \lambda^2 -(a+d)\lambda +ad-bc = 0$$
I understand how to find the eigenvalues, the problem is that I need to find 2 functions such that when solved for its singular points (by setting xn(k+1) = xn(k)), the Jacobian matrix has four pairs of eigenvalues (to do this the functions will have to be at least quadratic, and solving them means I end up with a quartic which I don't think I'll be expected to solve by hand I think I'm expected just look at it and find a solution without calculations via some property?), two of which will have an eigenvalue greater than 1 (i.e. unstable), and two of which have all eigenvalues under 1 (i.e. asymptotically stable).

Last edited:
FactChecker
Gold Member
You say that you found a solution using MATLAB. So you have a nonlinear example with 4 eigenvalues. What is the example? @jambaugh has given the method for finding the eigenvalues of the linearization Jacobian. It has only 2 eigenvalues. I am interested in how your nonlinear example can be so different from its linearization.

You say that you found a solution using MATLAB. So you have a nonlinear example with 4 eigenvalues. What is the example? @jambaugh has given the method for finding the eigenvalues of the linearization Jacobian. It has only 2 eigenvalues. I am interested in how your nonlinear example can be so different from its linearization.
My example found was of the form $$x_1(k+1) = G_1x_1(k)^2+G_2x_1(k)+G_3x_2(k) \\ x_2(k+1) = G_4x_2(k)^2+G_5x_2(k)+G_6x_1(k)$$
To find the singular points, I set x1(k+1) = x1(k) and x2(k+1) = x2(k) to give a set of two simultaneous quadratic equations. Solving these said equations simultaneously in MATLAB gave a solution with four pairs of roots (p1, p2)
Putting each these pairs of roots into the Jacobian matrix and finding the eigenvalues shows the stability of each of these roots.

I did this by trial and error after some fiddling and I don't think I could possibly be expected to solve a quartic equation by hand, so I feel like there's something I'm missing and I should be able to find a solution that fills the criteria without going through the calculations.

EDIT: I don't really want someone to come across this post and write down the exact same solution as me if a similar question comes up during the exam so I've removed the coefficients/roots, PM me if you need them.

Last edited:
jambaugh