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Periodic points

Now we will investigate the existance and behavior of period two points.

In order to investigate the behavior of period 2 points, it is neccessary to create a new map based on the original equation (5).

For a period two point,


$\displaystyle x_{n} = x_{n+2}$ $\textstyle =$ $\displaystyle a - x_{n+1}^2 + b( x_{n+1} - y_{n+1})$ (14)
$\displaystyle y_{n} = y_{n+2}$ $\textstyle =$ $\displaystyle x_{n+1}$ (15)

where $x_{n+1}$ and $y_{n+1}$ are given by (6).

Plugging these in and dropping subscripts yields the equations to be satisfied at a period 2 point:


$\displaystyle x$ $\textstyle =$ $\displaystyle a- (a-x^2 + b(x-y))^2 + b(a-x^2 + b(x-y) - x)$ (16)
$\displaystyle y$ $\textstyle =$ $\displaystyle a-x^2 +b(x-y)$ (17)

Plugging (17) into (16) yields an equation which is quartic in $x$. However, this expression can be simplified since it is known that a period 1 point will also solve this equation. In other words, this equation will be of the form


$\displaystyle f(x) = 0$     (18)

and $f$ can be separated into a factor which represents period 1 points and a factor which represents period 2 points. Doing so yields the following expression for the period 2 points:


$\displaystyle x^2 - (1+2b)x + (1+3b+3b^2-a) = 0$     (19)

which can be solved for x:


$\displaystyle x$ $\textstyle =$ $\displaystyle \frac{ (1+2b) \pm \sqrt{(1+2b)^2 - 4(1+3b+2b^2-a)} } {2}$ (20)

Examining the radical term reveals that there won't be a real solution to this expression unless


$\displaystyle a > b^2 + 2b + \frac{3}{4}$     (21)

In other words, the $b$ delay term requires a larger $a$ forcing term, thus pushing off the onset of the emergence of periodic points when viewed in terms of the the control parameter $a$. This is directly observable from the following figure which depicts the onset of the bifurcation point in terms of the control parameter $a$ and the $b$ delay term:

Figure 1: The dependence of $a$ upon $b$
Image ../plotab.png

In the figure above, the numerical values calculated are shown as + signs and (21) is shown as a solid line. Note as $b$ approaches 1.00, the predescribed condition from (21) causes non-real solutions.


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URA Program Website 2004-12-02