Where does a population end up?

1. 1. Read the rate

   A fish stock of $y$ thousand changes at $\frac{dy}{dt} = 0.8y\left(1 - \frac{y}{5}\right)$ a year. Above is that rate against $y$; below, the stock against time $t$ in years, from $y_0 = 0.5$. Where the rate is positive, the stock grows.

   Question: Starting at half a thousand fish, what does the stock do over the years?

   It levels off at $5$. The rate is positive for every $y$ between $0$ and $5$, so the stock rises the whole time, more slowly as the rate falls to $0$ at $y = 5$. It cannot pass $5$: there the rate is $0$, so $y = 5$ is a steady state. The fastest growth, $1$ thousand a year at $y = 2.5$, is only on the way.

   The figure: The rate dy/dt = 0.8y(1 − y/5) − h with a catch h = 0.00, against the stock y from 0 to 5.8 thousand. It is zero at y = 0.00 and y = 5.00. The stock y against time t over 15.0 years, from y₀ = 0.5 thousand: it settles at 5.00 thousand.

2. 2. Arrows on the axis

   Mark the sign of the rate on the $y$-axis: an arrow to the right where it is positive, to the left where it is negative. That is the state-space diagram, or phase line, and every solution follows its arrows. Filled points are stable steady states, hollow ones unstable. Drag the start.

   Drag the starting stock to a value from which the stock falls.

   Above $5$ the rate is negative: the arrow points left and the stock falls back to $5$. Both arrows point at $5$, so it is stable. Both point away from $0$, so it is unstable: any stock above $0$ leaves it.

   $$f(y^*) = 0, \qquad y^* \text{ stable when } f > 0 \text{ just below and } f < 0 \text{ just above}$$$f(y^{\ast })$ $=0,$$y^{\ast }\text{ stable when }f$ $>0\text{ just below and }f$ $<0\text{ just above}$

   The figure: The rate dy/dt = 0.8y(1 − y/5) − h with a catch h = 0.00, against the stock y from 0 to 5.8 thousand. It is zero at y = 0.00 and y = 5.00. Arrows on the y-axis point right where the rate is positive and left where it is negative. The stock y against time t over 15.0 years, from y₀ = 5.5 thousand: it settles at 5.00 thousand. Faint curves from other starts follow the arrows too.

3. 3. Fish it

   Now catch $h$ thousand fish a year: $\frac{dy}{dt} = 0.8y\left(1 - \frac{y}{5}\right) - h$. The catch lowers the whole graph of the rate by $h$, and the steady states are where it crosses the axis.

   Question: As the catch grows, what happens to the two steady states?

   They close in. At $h = 0.64$ the zeros are $1$ and $4$: the stock now settles at $4$ thousand, and $1$ is unstable. Below $1$ thousand the catch outpaces the growth, the arrow points left, and the stock dies out.

   The figure: The rate dy/dt = 0.8y(1 − y/5) − h with a catch h = 0.64, against the stock y from 0 to 5.8 thousand. It is zero at y = 1.00 and y = 4.00. Arrows on the y-axis point right where the rate is positive and left where it is negative. The stock y against time t over 15.0 years, from y₀ = 4.5 thousand: it settles at 4.00 thousand. Faint curves from other starts follow the arrows too.

4. 4. Not the nearest

   With $h = 0.64$, a stock of $1.5$ thousand is much nearer the steady state at $1$ than the one at $4$. A common slip sends a stock to whichever steady state is nearest.

   Question: Where does a stock of $1.5$ thousand end up?

   At $4$. The rate there is $0.8(1.5)(0.7) - 0.64 = 0.2 > 0$: the arrow points right, so the stock rises, all the way to $4$. The amber curve is the slip's guess. Distance decides nothing; the sign of the rate where the stock starts decides everything.

   The figure: The rate dy/dt = 0.8y(1 − y/5) − h with a catch h = 0.64, against the stock y from 0 to 5.8 thousand. It is zero at y = 1.00 and y = 4.00. Arrows on the y-axis point right where the rate is positive and left where it is negative. The stock y against time t over 15.0 years, from y₀ = 1.5 thousand: it settles at 4.00 thousand. A dashed amber curve drifts to the nearest steady state, 1.00, which the stock does not do.

5. 5. Too big a catch

   The stable steady state is the stock a fishery can keep; the unstable one is the edge below which it dies out. Raise the catch and the two close in further.

   Raise the catch until no steady state is left.

   At $h = 1$ the top of the graph just touches the axis at $y = 2.5$: the two steady states have met. Past it the rate is negative at every $y$, every arrow points left, and the stock dies out from any start. No catch above $1$ thousand a year, the peak of the growth term, can last.

   The figure: The rate dy/dt = 0.8y(1 − y/5) − h with a catch h = 1.20, against the stock y from 0 to 5.8 thousand. It is below zero for every y. Arrows on the y-axis point left everywhere. The stock y against time t over 15.0 years, from y₀ = 4.5 thousand: it dies out. Faint curves from other starts follow the arrows too.

The film, as said, 1:47

1. 0:00

   A fish stock of $y$ thousand changes at $\frac{dy}{dt} = 0.8y\left(1 - \frac{y}{5}\right)$ a year, drawn above against $y$.

2. 0:05

   From $0.5$ thousand the rate is positive, so the stock grows: fastest at $y = 2.5$, then more and more slowly.

3. 0:15

   It levels off at $5$, where the rate is $0$: a steady state, which it never passes.

4. 0:21

   Draw the sign of the rate on the axis as arrows: right where it is positive, left where it is negative.

5. 0:28

   Every start follows the arrows: from above $5$ the arrow points left, and the stock falls back to $5$.

6. 0:36

   Both arrows point at $5$: it is stable. Both point away from $0$: it is unstable.

7. 0:42

   Now catch $h$ thousand a year: the graph sinks by $h$, and its zeros close in, to $1$ and $4$ at $h = 0.64$.

8. 0:52

   The stock settles at $4$ now, and below $1$ the catch outpaces the growth: the stock dies out.

9. 0:58

   Start at $1.5$ thousand, much nearer the steady state at $1$ than the one at $4$.

10. 1:03

    A common slip sends it to the nearest steady state, in amber; the stock itself climbs to $4$.

11. 1:13

    The rate at $1.5$ is $0.2$, positive, so the arrow points right: the sign where a stock starts decides, not the distance.

12. 1:21

    Raise the catch to $h = 1$: the top of the graph just touches the axis at $y = 2.5$, and the two steady states meet.

13. 1:31

    Past it the rate is negative everywhere: every arrow points left, and the stock dies out from any start.

14. 1:40

    So no catch above $1$ thousand a year can last, and the arrows between the steady states say where every stock ends up.