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Modeling Population Growth

Introduction

This project deals with a linear model of growth of a population. Developed in the 1940's, this model has been used by demographers to model human population growth as well as the growth of animal populations. The mathematical background for this project is in this document, but you might want to look at section 5.1 on eigenvalues. (This project is based on an example taken from Applications of Linear Algebra by Rorres and Anton.)

Background

Defining the Model

The model described here is commonly referred to as the ``Leslie model'', after one of the original developers of the model. It models only the female portion of a population and starts by assuming that the maximum age that can be attained by any female of the species is L. By dividing the interval [0,L] into n equal subintervals, the population is divided up into n classes by age, as shown in the following table.

Age Class Age interval
1 [0,L/n)
2 [L/n, 2L/n)
3 [2L/n, 3L/n)
$\vdots$ $\vdots$
n-1 [(n-2)L/n, (n-1)L/n)
n [(n-1)L/n, L)

Note that the intervals do not include the right-hand endpoint. For example, the age a of a female in class 3 must satisfy $2L/n \leq a
< 3L/n$.

Next, we use the vector $\mathbf{x}$ to represent the population at some particular point in time. That is, the $k\mbox{th}$ component of $\mathbf{x}$ is the number of females in the $k\mbox{th}$ age class. As we have done before, we want to create a model that lets us relate the present population distribution to the population distribution in the past. To do so, we will use the following assumptions.

To make our model quantitative we define the following parameters. The birth coefficients ai, $i=1, \ldots, n$ denote the average number of females born to a single female while she is a member of the $i\mbox{th}$ age class. For example, if L=20 and n=5 as above, there would be five birth coefficients and a3, for example, would be the average number of female babies born to a female between the ages of 8 and 12. We also define the parameters bi, $i=1, \ldots, n$by saying that bi is the fraction of the females in class i that can be expected to survive and pass into the $(i+1)\mbox{st}$class. For example, suppose that at the last measurement period there were 100 females in age class 4 and that b4 = 3/4. Then we would predict that there would be 75 females in class 5 the next time the population is measured. Note that our assumption that no member of the population lives longer than L years means that bn = 0. Given these parameters, $\mathbf{x}^{(k+1)}$ can be calculated from $\mathbf{x}^{(k)}$ with the equation

\begin{displaymath}
\mathbf{x}^{(k+1)} = A \mathbf{x}^{(k)} \end{displaymath}

where the matrix A looks like the following.

\begin{displaymath}
A = \left[ \begin{array}
{cccccc}
a_1 &a_2 &a_3 &\cdots &a_{...
 ...vdots &\vdots \\ 0 &0 &0 &\cdots &b_{n-1} &0\end{array} \right]\end{displaymath}

A matrix of this form is called a Leslie matrix and turns out to have some very special properties.

Population Stability

There are really three things that can happen to a population as the number of time intervals gets large.
1.
The population can eventually be increasing as the number of time intervals increases.
2.
The population can eventually be decreasing as the number of time intervals increases.
3.
The population can approach a steady state. That is, the population in each of the classes approaches a fixed value.
In the simulations you are asked to do for the exercises, you will encounter examples of each of these behaviors. In this section, we come up with a criterion for deciding which kind of behavior a particular model will exhibit without having to do a lot of simulation.

The case of a stable steady state is the easiest one to begin with. We already know that for there to be a steady state, there has to be a solution to the equation

\begin{displaymath}
A \mathbf{x} = \mathbf{x} \end{displaymath}

So being able to find such a vector is necessary for there to be a stable population. However, even if we can find such a steady state, there is no guarantee that our simulations will approach the steady state.

It turns out that the key to stability, as well as the key to understanding why a population eventually grows or eventually decreases, is the notion of eigenvalue. We say that $\lambda$ is an eigenvalue for an $n \times n$ matrix A if we can find a vector $\mathbf{u}$ satisfying the equation

\begin{displaymath}
A \mathbf{u} = \lambda \mathbf{u} \end{displaymath}

The vector $\mathbf{u}$ is called an eigenvector of the matrix A corresponding to the eigenvalue $\lambda$.

If we rewrite the equation above as the homogeneous system

\begin{displaymath}
(A - \lambda I)\mathbf{u} = \mathbf{0} \end{displaymath}

then it should be clear that $\lambda$ is an eigenvalue if and only if the following condition is satisfied.

\begin{displaymath}
\det(A - \lambda I) = 0 \end{displaymath}

It can be shown that $\det(A - \lambda I)$ is a polynomial of degree n in $\lambda$. This polynomial turns out to be so important in understanding the properties of the matrix A that it is given a special name. The polynomial $\det(A - \lambda I)$ is called the characteristic polynomial of the matrix A.

For example, suppose A is given by

\begin{displaymath}
A = \left[ \begin{array}
{cc}
 -2 & 1 \\  1 & -2
 \end{array} \right]\end{displaymath}

then the characteristic polynomial is

\begin{displaymath}
\left\vert\begin{array}
{cc}
 -2-\lambda & 1 \\  1 & -2-\lam...
 ...\lambda) -1
= \lambda^2 +4 \lambda + 3 = (\lambda+1)(\lambda+3)\end{displaymath}

The roots of the characteristic polynomial are the eigenvalues of the matrix A. In our simple example above, the eigenvalues are easily found to be -1 and -3. It isn't always so easy. In the exercises you will be working with $5 \times 5$ matrices so the characteristic polynomials are fifth order, which usually can't be solved analytically. This means that you usually have to resort to numerical techniques, which can make the problem very difficult.

Now, what does this have to do with our population growth model? The answer is that a Leslie matrix, under certain conditions satisfied by all of the examples you will see in this project, has a single dominant eigenvalue $\lambda$. This eigenvalue is always a positive number and essentially determines the long-term behavior of the model as follows.

To get an idea of how this happens, suppose that $\lambda$ is the dominant eigenvalue and $\mathbf{u}$ is the corresponding eigenvector. If for some value of k we have $\mathbf{x}^{(k)} =
\mathbf{u}$, then we would have

\begin{displaymath}
\mathbf{x}^{(k+1)} = A \mathbf{x}^{(k)} = A \mathbf{u} = \lambda
\mathbf{u} \end{displaymath}

On the next iteration, you would get $\mathbf{x}^{(k+2)} = \lambda^2
\mathbf{u}$, and so on, with the next population just being the previous population multiplied by $\lambda$. So, if $\lambda \gt 1$, the population keeps getting larger and larger. If $\lambda=1$, the population stays the same and if $0 < \lambda < 1$, the population keeps decreasing.

This shows how the dominant eigenvalue controls the behavior if $\mathbf{x}^{(k)} =
\mathbf{u}$, but it doesn't tell the whole story. It is a little beyond the scope of this project, but it can be shown that no matter what the initial population vector is, it eventually approaches a multiple of the dominant eigenvector. When this happens, you get the behavior described above.

Exercises

1.
Explain why making the time between population measurements equal to the length of the age intervals guarantees that all of the females in the $(i+1)\mbox{st}$ age class at t=tk+1 were in the $i\mbox{th}$ age class at t=tk.
2.
Explain how the structure of the matrix A is consistent with the definitions of ai and bi.

3.
Explain why it makes sense to only model the female members of a population.

4.
Consider the following parameter values, for L=20 and n=5.

Age Class ai bi
1   1/2
2 1 2/3
3 2 3/4
4 2 1/2
5 1 -

Model this population for several different choices of initial population vector $\mathbf{x}^{(0)}$. What happens to the populations?

5.
Repeat the previous exercise with the following data.

Age Class ai bi
1   1/4
2 1 1/3
3 1 1/2
4 2 1/4
5 1 -

6.
Repeat the previous exercise with the following data.

Age Class ai bi
1   1/4
2 1 3/4
3 2 3/4
4 2 2/3
5 1 -

7.
For each of the three cases above, compare the values of the coefficients and try to explain why each behaves as it does.
8.
For each of the three cases above, find the dominant eigenvalue and its corresponding eigenvector. Show that your simulation results above are consistent with the theory. That is, show that in each case that the ratio between components of $\mathbf{x}^{(k+1)}$ and $\mathbf{x}^{(k)}$ approaches a constant, equal to the dominant eigenvalue.

About this document ...

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Copyright © 1993, 1994, 1995, 1996, 1997, Nikos Drakos, Computer Based Learning Unit, University of Leeds.

The command line arguments were:
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The translation was initiated by William W. Farr on 9/28/1999


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William W. Farr
9/28/1999