Logistic Equation

We can use the ideas introduced in fixed-point iteration to investigate the logistic equation. The logistic equation is written as

$$x_{n+1} = x_{n} + r x_{n}(1 - x_{n}),$$ (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,

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

Now we have

$$G^{\prime}(0) = 1 + r,$$

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

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

$$G^{\prime}(1) = 1 + r - 2r = 1 - r.$$

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

$$-1 < G^{\prime}(1) < 1,$$

and so

$$-1 < 1 - r < 1, \qquad \Rightarrow \qquad 0 < r < 2.$$ (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

$$x_{n+2} = x_{n} \hbox{ and } x_{n+3} = x_{n+1},$$

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

$$x_{n+2} = x_{n} = x,$$

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

$$r(x_{n} + rx_{n}(1-x_{n}))[1 - x_{n} - rx_{n}(1-x_{n})] = rx_{n}(1-x_{n})[1 + r(1-x_{n})](1 - rx_{n}).$$

Hence, we have

$$x_{n+2} = x_{n} + rx_{n}(1-x_{n})[1 + [1 + r(1-x_{n})](1 - rx_{n})],$$

Expanding the brackets and collecting powers of $x$ together we obtain

$$x_{n+2} = x_{n} + rx_{n}(1 - x_{n})[2 + r - (2r + r^{2})x_{n} + r^{2}x_{n}^{2}].$$ (1.19)

This is of the form

$$x_{n+2} = G(x_{n}).$$

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

$$0 = rx(1 - x)[2 + r - (2r + r^{2})x + r^{2}x^{2}].$$

The first two factors give the period one solutions and the square brackets factor gives the new period two solutions. Thus, they satisfy

$$r^{2}x^{2} - (2 + r)rx + 2 + r = 0.$$ (1.20)

Solving the quadratic equation gives

$$rx = {(2 + r) \pm \sqrt{(2 + r)^{2} - 4(2+r)}\over 2}$$

$$rx = 1 +{r\over 2} \pm {1\over 2}\sqrt{r^{2} - 4}.$$ (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

$$G^{\prime}(x) = -1.$$

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,

$$r^{2} = 6, \qquad \Rightarrow \qquad r = \sqrt{6}.$$

So period two solutions are stable for

$$2 < r < \sqrt{6}.$$

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

$$r_{0}, r_{1}, r_{2}, \cdots$$

until chaotic solutions (infinte period) arise.