Verhulst Process

Discrete Dynamical System

A Dynamical System^[1], also known as a Map can be defined informally, and without the need to delve into more advanced regions of mathematics, as a way of describing the evolution of a state over fixed time intervals^[2]. Such systems are discrete when we consider changes over integer intervals. They can be expressed using the general iterative formula: \begin{equation} x_{n+1} = f(x_n), \end{equation} where $f$ is a known continuous function and $n \in\mathbb{Z^+}$. Equation $(1)$ is a first order formula as the next value, $x_{n+1}$, depends only on the previous value, $x_{n}$. Equations of this kind can be used to describe the fluctuations of a population over a number of generations. A prime example of the biological applications of discrete dynamical systems is the Verhulst process...

The Verhulst Process

The Verhulst Process^[3], developed by French mathematician Pierre-François Verhulst in the 19^th century is used model bounded population growth. The formula describing this process can be written as: \begin{equation} \ x_{n+1} = x_n + rx_n(1-x_n), \end{equation} where $x_{n+1}$ is the population of the next generation, $r\in\mathbb{R}_{>0}$ is a fixed parameter, which determines how the population changes and $x_n$ is the current population. This map is one-dimensional as we are only interested in the evolution of a single quantity (the population $x_n$) which in order for it to make physical sense we need to restrict to $x_n \geq 0\ \forall n\in\mathbb{N}$. The parameter $r$ plays a very important role in the evolution of the population or, in more abstract terms, in the evolution of the process and we will examine the impact of the value of $r$ in the iterations, i.e. as $n$ increases.

The graph above, which is an example of a Cobweb plot, shows how varying the value of $r$ affects the values of $x_n$ as $n$ becomes big.In the graphs we have set the maximum number of iterations to be 1000, in order to be computationally efficient and because a greater number of iterations has a minuscule impact on the graph.

As one can easily observe, in the range $0 < r < 2$ as $n$ increases the process converges to a single point but as $r$ becomes greater than 2, in fact as $r$ is in the range $2< r < 2.45$ we see that for big $n$ the process does not remain fixed at one point but it oscillates between two different points. This is an instance of a period-doubling which reoccurs when $r$ becomes greater than 2.45, after which the map oscillates between four points. As $r$ continues to grow, these period doublings get more and more frequent until we reach $r=2.57$, after which the number of points between which the graph oscillates becomes chaotic. The following graph shows a more complete numerical approach to the relation between the value of $r$ and not only the number of fixed points, but also their values.

For an analytical approach to the relation between the value of $r$ and the number of points between which the graph oscillates we need to introduce a few definitions.

This point is fixed since $(3)$ implies that after $\mathbb{X}$ we have: \begin{equation} \ x_{n+1} = x_n \end{equation} So no matter how many times we iterate after the point $\mathbb{X}$ has been reached the population will remain constant. In the Verhulst process a fixed point could be found by solving: \begin{equation} \ \mathbb{X} = \mathbb{X} + r\mathbb{X}(1-\mathbb{X}). \end{equation}

Solving equation $(5)$ we can easily determine that the fixed points for the Verhulst process are $\mathbb{X} = 0$ or $\mathbb{X} = 1$. We can also easily determine $\frac{d}{dx} f(x)$ for the Verhulst process: \begin{equation} \frac{d}{dx} f(x) = 1 + r(1-2x). \end{equation} Now substituting for the two fixed values of $\mathbb{X}$ that we found we get that for $\mathbb{X} = 0$ \begin{equation} | \frac{d}{dx} f(x) | = |1 + r| \end{equation} and for $\mathbb{X} = 1$: \begin{equation} | \frac{d}{dx} f(x) | = |1 - r|. \end{equation} It is clear that for $0 < r < 2$ we have $\mathbb{X} = 0$ is unstable, since $|1 + r|>1$ for $r>0$, however $\mathbb{X} = 1$ is stable since $|1 - r|< 1$ for $r< 2$. But at $r=2$ we can see that both points are unstable. But as we can see from the numerical approach the value of $x_n$ for big $n$ oscillates between two values, so we need to consider more than just the previous iteration. This means that we must have that: \begin{equation} x_{n+2} = f(x_{n+1}) = f(f(x_n)) \end{equation} i.e. \begin{equation} \begin{aligned} x_{n+2} & = [x_n + rx_n(1-x_n)] + r[x_n + rx_n(1-x_n)][1-x_n - rx_n(1-x_n)]\\ & = x_n+2rx_n+r^2x_n-2rx_n^2-3r^2x_n^2-r^3x_n^2+2r^2x_n^3+2r^3x_n^3-r^3x_n^4 \end{aligned} \end{equation} In order to find the fixed points for $(12)$ we would need to solve: \begin{equation} \mathbb{X} = \mathbb{X}+2r\mathbb{X}+r^2\mathbb{X}-2r\mathbb{X}^2-3r^2\mathbb{X}^2-r^3\mathbb{X}^2+2r^2\mathbb{X}^3+2r^3\mathbb{X}^3-r^3\mathbb{X}^4, \end{equation} this an equation of 4^th degree, which is normally fairly complicated to solve; however, since the fixed points we had already determined remain fixed points, we already know two of the solutions. This is easy to see by $(3)$ and $(11)$ as, for $\mathbb{X}$ fixed we have: \begin{equation} f(f(\mathbb{X})) = f(\mathbb{X}) =\mathbb{X} \end{equation} Now if we go back to $(13)$ we can factor out $\mathbb{X}(\mathbb{X}-1)$ to get: \begin{equation} \mathbb{X}(\mathbb{X}-1)(-r^3\mathbb{X}^2+\mathbb{X}(r^3+2r^2)-r^2-2r)=0 \end{equation} Solving the quadratic factor gives us: \begin{equation} \begin{aligned} \mathbb{X}_1 &= \frac{r +2 +\sqrt{r^2-4}}{2r} \text{, and} \\ \mathbb{X}_2 &= \frac{r +2 -\sqrt{r^2-4}}{2r} \end{aligned} \end{equation} If we look at the derivatives of the two new fixed values $\mathbb{X}_1$ and $\mathbb{X}_2$, keeping in mind that the fact that this is a 2-cycle, i.e. the function oscillates between the two values implies the symmetric relation between : \begin{equation} \begin{aligned} f(\mathbb{X}_1)&= \mathbb{X}_2 \text{, and}\\ f(\mathbb{X}_2)&= \mathbb{X}_1; \end{aligned} \end{equation} we get that: \begin{equation} \begin{aligned} | \frac{d}{dx}f(f(\mathbb{X}))|_{\mathbb{X}_1,\mathbb{X}_2} & = |f'(\mathbb{X})f'(f(\mathbb{X}))|_{\mathbb{X}_1,\mathbb{X}_2} \\ & = |f'(\mathbb{X}_1)f'(\mathbb{X}_2)| \\ & = |5 - r^2|. \end{aligned} \end{equation} We can then determine that within a specific range of $r$ they are both stable by simply solving: \begin{equation} \begin{aligned} |5 - r^2|< 1 . \end{aligned} \end{equation}

This gives us that the two points $\mathbb{X}_1 = \frac{r +2 +\sqrt{r^2-4}}{2r} $ and $\mathbb{X}_2 = \frac{r +2 -\sqrt{r^2-4}}{2r}$ are stable for $2< r <\sqrt{6}$, but at $r = \sqrt{6}$ these two points become unstable, so all four points are now unstable. This would indicate another period doubling which we would deal by using similar methods to solve: \begin{equation} x_{n+4} = f(x_{n+2}) = f(f(f(f(x_n))))=f^{(4)}(x_n) \end{equation} This time, when trying to find the solutions for the fixed values we would end up with an equation of 16^th degree of which we only know four solutions this level of mathematics is too intricate for the scope of this investigation.

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