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Fixed Points

It has already been shown that (6) was constructed so that it retains the fixed (period 1) points of the original system (1). Examining that equation, setting $x_{n+1} = x_n$ yields an equation for the fixed points


$\displaystyle x = \frac{-1 \pm \sqrt{1+4a}}{2}$     (7)

The stability of these fixed points can be determined in general by calculating the Jacobian of (6) and evaluating it at the fixed points1.

The Jacobian of (6) is


$\displaystyle A=
\left(
\begin{array}{cc}
-2x + b & -b \\
1 & 0
\end{array}\right)$     (8)

The eigenvalues of (8) are given by


$\displaystyle \lambda = \frac{ (2x-b) \pm \sqrt{(-2x+b)^2 - 4b}}{2}$     (9)

In the original system (1) there exists one eigenvalue $\lambda_o = -2x$. Substituting the values for the fixed points (7) it can be seen that one fixed point will always be unstable, while the other will change stability from stable to unstable when $a = \frac{3}{4}$. It will be shown later that a bifurcation occurs at this point as well.

One question to explore is whether the addition of feedback will affect where this change occurs, and if so whether it serves to "control" the system and inhibit the onset of this change in stability and its corresponding bifurcation.

In this analysis we will assume that $b<1$. By examing (6) it can be seen that this means that the feedback is not "amplified", that is, what is fed into the system from the previous time step will be reduced in magnitude.

The determinant of the Jacobian (8) is $b$. The product of the eigenvalues will thus also equal $b$. The change in stability discussed above occurs when $\left\vert \lambda \right\vert = 1$.



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Next: Eigenvalues Up: Analysis Previous: Analysis
URA Program Website 2004-12-02