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In an atlantic fishery, anchovies are extracted at a constant rate of 6000 anchovies per year. The per-capita death rate by other causes is 0.35 anchovies per year and the per-capita birth rate is 0.60 anchovies per year.

i) Derive a difference equation to describe the anchovie population.

ii) Solve this difference equation and determine the steady state. Is it stable or unstable.

Having problems with deriving an equation, I should be abe to solve it and determine it's steady state if I have an equation.

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We like to see some evidence of effort beyond copying a problem out of a textbook. What do you know about the problem? What progress can you make? Where do you get stuck? Oh, and if it's homework, please add the "homework" tag. –  Gerry Myerson Aug 12 '12 at 13:10

1 Answer 1

Let $a(t)$ be the anchovy population at time $t$. You’re told that between time $t$ and time $t+1$ you lose $6000$ anchovies by extraction and $0.35a(t)$ by death from other causes, and you gain $0.60a(t)$ by births. Thus,

$$\begin{align*} a(t+1)&=a(t)-6000-0.35a(t)+0.60a(t)\\ &=(1-0.35+0.60)a(t)-6000\\ &=1.25a(t)-6000\;. \end{align*}$$

The steady-state population is the one that doesn’t change from one year to the next, so to find it, you need to solve the equation $a(t+1)=a(t)$:


That should be very straightforward.

To see whether the system is stable, see what happens if the population is a little more or a little less than the solution to $(1)$. Let $a_0$ be that steady-state population. If you start with a population a little bigger or a little smaller than $a_0$ does the population tend back to $a_0$, or does it do something else, like increasing without bound?

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