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Logistic Equation

We can use the ideas introduced in fixed-point iteration to investigate the logistic equation. The logistic equation is written as
\begin{displaymath}
x_{n+1} = x_{n} + r x_{n}(1 - x_{n}),
\end{displaymath} (1.17)

with the starting value $x_{0}$ satisfying $0 < x_{0} < 1$ and $r$ is a parameter. Choosing a value for $r$, (1.17) is used repeatedly and the limit of $x_{n}$ as $n \rightarrow \infty$ is obtained. Changing the value of the parameter $r$, the procedure is repeated and the limiting value of $x_{n}$ obtained. Thus, the limiting value as a function of $r$ is built up.

The logistic equation can be used as a simple model of how a population changes in time. It illustrates some interesting ideas and produces features such as bifurcations and period doubling when the parameter $r$ is adjusted.

There are two fixed-points at $x=0$ and $x=1$ as can be verified by setting $x_{n} = x_{n+1} = 0$ and $x_{n} = x_{n+1} = 1$ in (1.17). To investigate whether these fixed-points are stable or not, we look at the convergence properties by calculating $G^{\prime}(x)$. Thus,

\begin{displaymath}
G^{\prime}(x) = 1 + r - 2rx.
\end{displaymath}

Now we have

\begin{displaymath}
G^{\prime}(0) = 1 + r,
\end{displaymath}

and so the fixed-point at $x=0$ is unstable for $r > 0$.

For the fixed-point at $x=1$ we have

\begin{displaymath}
G^{\prime}(1) = 1 + r - 2r = 1 - r.
\end{displaymath}

Thus, for (1.17) to converge to $x=1$ we need

\begin{displaymath}
-1 < G^{\prime}(1) < 1,
\end{displaymath}

and so
\begin{displaymath}
-1 < 1 - r < 1, \qquad \Rightarrow \qquad 0 < r < 2.
\end{displaymath} (1.18)

Thus, for $0 < r < 2$ the logistic equation converges to the root at $x=1$. That is, $x_{n} \rightarrow 1$ as $n \rightarrow \infty$. Note that for $r=2$, $G^{\prime}(1) = -1$ and the error oscillates between positive and negative values without growing or decaying appreciably.

What happens for $r > 2$? For $r$ just slightly larger than 2, numerical solutions to the logistic equation show that the scheme settles down and oscillates between two values. So instead of getting $x_{n+1} = x_{n}$ as $n \rightarrow \infty$ we obtain

\begin{displaymath}
x_{n+2} = x_{n} \hbox{ and } x_{n+3} = x_{n+1},
\end{displaymath}

where $x_{n}$ and $x_{n+1}$ are the two different solutions. This is called a period two solution to the logistic equation and they exist for the parameter $r > 2$. The single solution $x=1$ is said to bifurcate when the parameter $r=2$ and this process is called bifurcation.

To obtain the period two solutions we apply the iteration scheme twice and assume that

\begin{displaymath}
x_{n+2} = x_{n} = x,
\end{displaymath}

as $n \rightarrow \infty$. Thus, we have

\begin{eqnarray*}
x_{n+1} & = & x_{n} + rx_{n}(1 - x_{n}), \\
x_{n+2} & = & x...
...n}) + r(x_{n} + rx_{n}(1-x_{n}))[1 -
x_{n} - rx_{n}(1-x_{n})].
\end{eqnarray*}



This is quite complicated but we know that $x_{n}=0$ and $x_{n}=1$ must be solutions (since $x_{n+2} = x_{n+1} = x_{n}$ for the original solutions) but they are unstable solutions. To proceed we note that the last term in the above equation can be expressed as

\begin{displaymath}
r(x_{n} + rx_{n}(1-x_{n}))[1 - x_{n} - rx_{n}(1-x_{n})] =
rx_{n}(1-x_{n})[1 + r(1-x_{n})](1 - rx_{n}).
\end{displaymath}

Hence, we have

\begin{displaymath}
x_{n+2} = x_{n} + rx_{n}(1-x_{n})[1 + [1 + r(1-x_{n})](1 - rx_{n})],
\end{displaymath}

Expanding the brackets and collecting powers of $x$ together we obtain
\begin{displaymath}
x_{n+2} = x_{n} + rx_{n}(1 - x_{n})[2 + r - (2r + r^{2})x_{n} +
r^{2}x_{n}^{2}].
\end{displaymath} (1.19)

This is of the form

\begin{displaymath}
x_{n+2} = G(x_{n}).
\end{displaymath}

To obtain the fixed points we set $x_{n+2} = x_{n} = x$ and so

\begin{displaymath}
0 = rx(1 - x)[2 + r - (2r + r^{2})x + r^{2}x^{2}].
\end{displaymath}

The first two factors give the period one solutions and the square brackets factor gives the new period two solutions. Thus, they satisfy
\begin{displaymath}
r^{2}x^{2} - (2 + r)rx + 2 + r = 0.
\end{displaymath} (1.20)

Solving the quadratic equation gives
$\displaystyle rx$ $\textstyle =$ $\displaystyle {(2 + r) \pm \sqrt{(2 + r)^{2} - 4(2+r)}\over 2}$  
$\displaystyle rx$ $\textstyle =$ $\displaystyle 1 +{r\over 2} \pm {1\over 2}\sqrt{r^{2} - 4}.$ (1.21)

Notice that the square root only gives real solutions when $r > 2$ and for $r=2$ we have $x=1$. The period two solutions bifurcate from the $x=1$ solution.

As the parameter $r$ is increased we will eventaully reach a point at which the period two solutions will become unstable and the two solutions will each bifurcate giving a period four solution. The value of $r$ at which a bifurcation occurs is given by the condition

\begin{displaymath}
G^{\prime}(x) = -1.
\end{displaymath}

The calculation of $G^{\prime}(x)$ is complicated. However, setting $G^{\prime}(x) = -1$, where $x$ is given by (1.21), we obtain

\begin{eqnarray*}
2 & = & rx(1-x)((2+r)r - 2r^{2}x) \\
2 & = & r^{2} - 4.
\end{eqnarray*}



Hence,

\begin{displaymath}
r^{2} = 6, \qquad \Rightarrow \qquad r = \sqrt{6}.
\end{displaymath}

So period two solutions are stable for

\begin{displaymath}
2 < r < \sqrt{6}.
\end{displaymath}

For $r > \sqrt{6}$ they become unstable and we get period four solutions. This process of bifurcations and period doubling occurs at particular values of $r$, namely

\begin{displaymath}
r_{0}, r_{1}, r_{2}, \cdots
\end{displaymath}

until chaotic solutions (infinte period) arise.
next up previous
Next: Summary Up: Other Iterative Schemes Previous: Summary
Prof. Alan Hood
2000-02-01