# Population Growth Modelling

1. Apr 6, 2007

### Absolut

The Problem:
Consider a species in which no individuals live beyond 3 years. Divide the population into four age groups labelled by n = 0,1,2 and 3. Assume that only the second and third groups can reproduce. For each group n let bn denote the birthrate and $$d_n$$ the death rate. Find a matrix A such that P(t + 1) = AP(t), with:
$$\ P = \left(\begin{array}{cc}P_0\\P_1\\P_2\\P_3\end{array}\right)$$
Where $$]P_n$$(t) denotes the population size of age group n at time t.

It's a question on an age cohort model. I've began with using the following basic information/equations:

My initial work:
$$\begin{array}{l | c|c|c|c |}\&0&1&2&3\\\hline\b_i&0&b_1&b_2&0\\\hline d_i&d_0&d_1&d_2&1\\\hline\end{array}$$
(Where the first column is year zero, the second year one, the third year two, the fourth year three. $$b_i$$ indicates the birth rate, $$d_i$$ indicates the death rate, where the "1" in column four indicates that individuals die at age three).

I then created a 4x4 matrix as below:
A = $$\left(\begin{array}{cccc}o&b_1&b_2&0\\1-d_0&0&0&0\\0&1-d_1&0&0\\0&0&1-d_2&0\end{array}\right)$$

My attempt at a solution:
After creating that matrix, I move on to the next section:

Section 2: Show that the eigenvalues of A satisfy the following equation:
$$\lambda^4 - b_1(1 - d_0)\lambda^2 - b_2(1-d_0)(1-d_1)\lambda = 0$$

To try to find the determinant of my above matrix, I took $$\lambda$$I from the above matrix (A - $$\lambda$$I) and set it equal to 0 to find the eigenvalues. But when I try to solve for these I keep getting 0. Now I'm thinking that perhaps my intial matrix (A) is incorrect? I'm trying to find the determinants of the matrix by the following method:

Reducing the 4x4 to 3x3 and then to 2x2 and then finding the determinant of that. But as I said, it's coming out to be 0. (I haven't done any matrix algebra in a few years, so I'm a bit out of practice).

This question is part of my revision for a final exam, so I'd appreciate any help you can give me!

Last edited: Apr 6, 2007
2. Apr 6, 2007

### Dick

What did you get when you took the determinant?? If you got the fourth degree equation that you quote, then you've already shown the eigenvalues satisfy that equation. You don't have to explicitly find them. It is true that zero is an eigenvalue but there could be others depending on the values of the constants.

3. Apr 6, 2007

### Absolut

Hi,

Sorry if I wasn't clear, but the fourth degree equation is actually part of the question - I have to show that the equation can be gotten from the matrix I found in the first part of the question - but that's where I'm having the problem, since when I find the determinant of the 4x4 matrix I can't get any eigenvalues, since the determinant is zero (or at least the way I'm doing it results in 0).

4. Apr 6, 2007

### Dick

I don't get zero for the determinant. Are you sure you are putting -lambda's down the diagonal of the matrix before evaluating the determinant?

5. Apr 6, 2007

### Absolut

$$\left(\begin{array}{cccc}-\lambda&b_1&b_2&0\\1-d_0&-\lambda&0&0\\0&1-d_1&-\lambda&0\\0&0&1-d_2&-\lambda\end{array}\right)$$

That's the (A-$$\lambda$$I) matrix that I'm using

Then I'm trying to find the determinant by taking each element of the first row as follows:

$$-\lambda \left(\begin{array}{ccc}-\lambda&0&0\\1-d_1&-\lambda&0\\0&1-d_2&-\lambda\end{array}\right)$$

Then reducing this further to a 2x2 matrix (where most of the matrices cancel out, since there are so many 0 elements in the matrix):
$$-\lambda^2 \left(\begin{array}{ccc}-\lambda&0\\1-d_2&-\lambda\end{array}\right)$$
When I get the determinant of this matrix:
$$-\lambda^2\left(\begin{array}{ccc}-\lambda&0\\1-d_2&-\lambda\end{array}\right) = -\lambda*(-\lambda*-\lambda)*(1-d_2 * 0)$$ which just multiplies out to zero (or not.... a stupid slip!)

Last edited: Apr 6, 2007
6. Apr 6, 2007

### Absolut

Appologies for my poor latex!

7. Apr 6, 2007

### Absolut

Ok, so here's my full work on finding the determinant so far:

$$\left(\begin{array}{cccc}-\lambda&b_1&b_2&0\\1-d_0&-\lambda&0&0\\0&1-d_1&-\lambda&0\\0&0&1-d_2&-\lambda\end{array}\right)$$

$$-\lambda \left(\begin{array}{ccc}-\lambda&0&0\\1-d_1&-\lambda&0\\0&1-d_2&-\lambda\end{array}\right)$$ $$-b_1 \left(\begin{array}{ccc}\1-d_0&0&0\\0&-\lambda&0\\0&1-d_2&-\lambda\end{array}\right)$$ + $$b_2\left(\begin{array}{ccc}1-d_0&-\lambda&0\\0&1-d_1&0\\0&0&-\lambda\end{array}\right)$$ - $$0\left(\begin{array}{ccc}-1-d_0&-\lambda&0\\0&1-d_1&-\lambda\\0&0&1-d_2\end{array}\right)$$

Then I reduce these 3x3 matrices to 2x2 matrices:

$$-\lambda\left(\begin{array}{cc}-\lambda&0\\1-d_2&-\lambda\end{array}\right)$$ (the other two 2x2 matrices from this 3x3 matrix are excluded, since they're multiplied by 0) and I continue along...

8. Apr 6, 2007

### Dick

Ok, so the first term in your sum of 3x3 matrices gives you a lambda^4. The second gives -b1*(1-d0)*lambda^2 etc. This looks like your polynomial coming out - not zero.

9. Apr 6, 2007

### Dick

$$-\lambda^2\left(\begin{array}{ccc}-\lambda&0\\1-d_2&-\lambda\end{array}\right) = -\lambda*(-\lambda*0)*(1-d_2 * 0)$$

Where did you get the -lambda*0???????? Review 2x2 determinants!

10. Apr 6, 2007

### Absolut

I only spotted my mistake when I took the time to type it all out in latex code - and it was a small mistake with mixing up a zero in a place where it shouldn't have been! So I did manage to get out my eigenvalue equation, but now I am stuck on another section:

If $$d_0 = 0.15$$, $$d_1 = 0.2$$, $$d_2 = 0.5$$, $$b_1 = 0.5$$ and $$b_2 = 1.0$$, show that there is one real eigenvalue greater than one. What is the significance of this?

So my initial reaction is to sub these numbers into my eigenvalue equation, setting $$\lambda = 1$$and hope for a positive number. However, I seem to be getting:
$$f(1) = (1)^4 - (0.5)*(0.85)*(-\lambda^2) - (1)(0.85)(0.8)(\lambda)$$
Which yields -0.105, which goes against my intial theory that I would get a positive number, implying that one value of lambda is indeed greater than 1?

I would also have guessed that a small number would mean that one eigenvalue would be close to 1 (and hence there would be a slow growth pattern), and a large number would mean that one eigenvalue would imply large growth, while a negative number would indicate falling numbers leading to extinction? Since the question specifically states that the number should be greater than one, I'm a bit confused.

Last edited: Apr 6, 2007
11. Apr 6, 2007

### Dick

All you know so far is f(1)=(-0.105). The roots are where f(lambda)=0. You don't have to actually solve the equation. What's f(2)? Compare with f(1).

12. Apr 6, 2007

### Absolut

f(2) is 12.94...

13. Apr 6, 2007

### Dick

So??? f(1) is negative, f(2) is positive. Somewhere in between it must be what?

14. Apr 6, 2007

### Absolut

My head is fried.... sorry for all the obvious questions!

f(1) is negative, f(2) is positive... so somewhere in between there is an f(x) that is equal to zero. I was concentrating on the magnitude instead of the signs!

15. Apr 6, 2007

### Staff: Mentor

What can you say about P(t+1), P(t+2), ... if P(t) is an eigenvector of A with eigenvalue lambda?

16. Apr 6, 2007

### Dick

P(t+1)=A(P(t)). What happens when you apply a matrix to one of it's eigenvectors with eigenvalue lambda?

17. Apr 6, 2007

### Absolut

That's actually part of the next question, which I've only just seen:

Consider the evolution of the species over the next three years, where the initial population is $$P_0 = 3, P_1 = 2, P_2 = 2, P_3 = 0$$. Using the following birth and death rates:
$$d_0 = 0.15, b_1 = 0.5, d_1 = 0.2 b_2 = 1.0, d_2 = 0.5$$

Last edited: Apr 6, 2007
18. Apr 6, 2007

### Dick

Ok. So all clear?

19. Apr 6, 2007

### Absolut

I'd imagine I start off by subbing the values into the given matrix:

$$\ P = \left(\begin{array}{cc}P_0 \\P_1\\P_2\\P_3\end{array}\right) = \left(\begin{array}{cc}3\\2\\2\\0\end{arra y}\right)$$

I also have P(t + 1) = AP(t).

So maybe sub in values for t = 0, 1 and 2 (but sub them into what?).

So, no, not really too clear. I don't actually have any of the eigenvectors (or eigenvalues) though?

Last edited: Apr 6, 2007
20. Apr 6, 2007

### Dick

You haven't really answered this or the question about the significance of having an eigenvalue bigger than one. But you are just being asked to apply the matrix to the vector and then iterate the process.