Age Class | Age interval |

1 | [0,L/n) |

2 | [L/n, 2L/n) |

3 | [2L/n, 3L/n) |

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

Next, we use the vector to represent the population at some particular point in time. That is, the component of is the number of females in the 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.

- The populations are measured at equally spaced time
intervals . We will use to
represent the population
at the end of the time interval, when
*t*=*t*_{k}. The vector will be the initial population vector. - The time between measurements will be exactly the same as the
length of the age intervals used to classify the population. For
example, if
*L*=20 and the number of age classes is five, the time between measurements would be four years. - The only factors that will be used to determine how the populations change are birth, death, and aging. That is, we will not take into account factors like immigration or emigration.

To make our model quantitative we define the following parameters.
The birth coefficients *a*_{i}, denote the average
number of females born to a single female while she is a member of the
age class. For example, if *L*=20 and *n*=5 as above,
there would be five birth coefficients and *a _{3}*, 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

- 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.

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

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 is an
*eigenvalue* for an matrix *A* if we can find a vector
satisfying the equation

If we rewrite the equation above as the homogeneous system

then it should be clear that is an eigenvalue if and only if the following condition is satisfied. It can be shown that is a polynomial of degree
For example, suppose *A* is given by

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 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 . This eigenvalue is always a positive number and essentially determines the long-term behavior of the model as follows.

- If the dominant eigenvalue satisfies , then the population eventually decreases and decays to zero.
- If the dominant eigenvalue is exactly equal to 1, the population approaches a stable steady state.
- If the dominant eigenvalue is larger than 1, the population eventually grows without bound.

To get an idea of how this happens, suppose that is the
dominant eigenvalue and is the corresponding
eigenvector. If for some value of *k* we have , then we would have

This shows how the dominant eigenvalue controls the behavior if , 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.

- 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 age class at
*t*=*t*_{k+1}were in the age class at*t*=*t*_{k}. - 2.
- Explain how the structure of the matrix
*A*is consistent with the definitions of*a*_{i}and*b*_{i}. - 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 *a*_{i}*b*_{i}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 . What happens to the populations? - 5.
- Repeat the previous exercise with the following data.
Age Class *a*_{i}*b*_{i}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 *a*_{i}*b*_{i}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 and approaches a constant, equal to the dominant eigenvalue.

This document was generated using the
**LaTeX**2`HTML` translator Version 97.1 (release) (July 13th, 1997)

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

The command line arguments were:

**latex2html** `-split +0 project_template.tex`.

The translation was initiated by William W. Farr on 9/28/1999

9/28/1999