To be concrete, for now, let's set the (net) growth rate $r$ to $r=0.2$ so that the dynamic system is
\begin{align*}
\diff{p}{t} &= 0.2 p\\
p(0) &= p_0.
\end{align*}
Notice that for this differential equation, to know the rate of change $\diff{p}{t}$ at any moment $t$, we don't actually care about what the value of $t$ is (there is no $t$ directly in the right hand side of the differential equation). All we need to know is the value of the state variable $p$, and we can directly compute the rate of change $\diff{p}{t}$ (by multiplying by 0.2).
If at $t=17$, the population size is $p(17)=1234$, what is $\diff{p}{t}$ at time $t=17$?
$p'(17) = $
If at the moment $t=31.987$, we knew that the population size was exactly $1234$ elk, what is rate of change in the population size at that moment?
$p'(31.987) = $
Typically, we are primarily interested in understanding the qualitative behavior of system rather than obtaining some formula to give us exactly what the population size will be at each time. In this case, we care if the population is growing or shrinking. If at some time $t$ (and again, it doesn't matter what that time is), the population contains $1234$ elk, the sign of $\diff{p}{t}$ is
. Therefore, at that moment the population size is
.
To represent where $p$ is increasing and decreasing, we'll use a “phase line” diagram, where the phase line is just a representation of the different values that $p$ can take. Since the rate of change $\diff{p}{t}$ doesn't depend directly on time, we don't need to represent time at all on the phase diagram. The below applet shows a phase line for our system. Our goal is to show on this phase line where $p(t)$ is increasing, decreasing, or staying constant. (Although negative values of $p$ don't make biological sense, we include some negative values as it helps get a fuller picture of the behavior of the mathematical equations.)
Feedback from applet
direction vectors: correct directions:
direction vectors: no stray vectors:
equilibria: locations:
equilibria: number of equilibria:
equilibria: stability:
solutions: final values:
solutions: initial conditions:
solutions: number of solutions:
If $p$ is zero, then $\diff{p}{t}$ is
so that the population size is
. This means, if for some reason, the population size is zero, it will remain zero for all time (which makes sense if we aren't allowing elk from somewhere else to move in an colonize our empty forest). The value of the state variable $p=0$ is called an equilibrium.
On the above phase line applet, increase $n_e$ to 1, as we have one equilibrium and move the resulting red point that appears to the location of the equilibrium ($p=0$).
For the case that $r=0.2$, if $p$ is positive, then the sign $\diff{p}{t}$ is
so that the population size is
. We use vectors (arrows) to show where $p$ is increasing (i.e., $\diff{p}{t}$ is positive) and where it is decreasing. Since $p$ is increasing for all positive values, draw some rightward pointing arrows on the positive part of the phase line. On the applet, increase $n_v$ to reveal arrows that you can move to phase line. (The applet only requires one vector, but it looks nicer if you draw more short vectors spread over the phase line.) Make sure all arrows are completely on the phase line, as it doesn't make sense to have anything off the line.
Although it doesn't make sense to have negative numbers of elk, if $p(t)$ were negative, the differential equation (with $r=0.2$) shows that in this case, the sign of $\diff{p}{t}$ is
so that the population size would be
to even more negative numbers of elk (whatever that means). Draw a leftward point arrow on the negative portion of the phase line to show this fact.
Notice how the arrows are pointing away from the equilibrium at $p=0$. Since solutions move away from the equilibrium, we say it is unstable. To show it is unstable, click the point so that it is an unfilled circle.
The last step is to use this information from the phase line to determine what a solution must look like. Let's start with the initial condition $p(0)=100$. If you look at the above phase line, when $p=100$, the arrows point to the
, meaning the solution $p(t)$ must be
. Since the arrows never change direction, it means $p(t)$ must keep increasing
.
On the above phase line, increase $n_s$ to one in order to plot a solution curve on the phase line. (Actually, the curve will have to be a straight line, since it going to be stuck on the phase line.) Drag the base of the vector representing the solution curve to the initial condition point, i.e., to $p=$
. Although $p(t)$ should be increasing forever, you'll have to make it stop when you reach the end $p=1000$, so end the arrow there.
If, on the other hand, the initial condition was $p_0=300$, will the solution look much different? Increase $n_s$ to two and plot what the solution starting at $p_0=300$ should look like with the resulting arrow.
It's hard to visualize the solutions $p(t)$ on the phase line, since there is no representation of time $t$. It is just an arrow pointing right. Let's also plot the solutions $p(t)$ like we would normally plot functions, on a graph versus time, using the below applet.
Feedback from applet
Equilibrium stability:
Final values:
Found equilibria:
Initial conditions:
Number of curves:
Valid shapes:
Start by plotting the equilibrium $p(t)=0$ for all time $t$. Increase $n_c$ to one to show one curve. Since in this case, the $p$ axis is the vertical axis, the equilibrium corresponds to a horizontal line at the height $p=0$. Move all three points along the line $p=0$. The curve should turn red to show it is an equilibrium (a constant solution). To show that the equilibrium is unstable, click the curve so that it is a dashed curve.
Next, plot the solution with initial condition $p(0)=100$. Increase $n_c$ to two and move the first point to the height of $p=100$. Since $p(t)$ must be
, move the remaining two points to give the appropriate shape for the curve. Remember, $p(t)$ must increase without bound. For the applet to recognize that you mean it should keep increasing, make sure it curves upward and hits the upper edge of the graph.
For the initial condition $p(0)=300$, the curve looks pretty the same. Increase $n_c$ to three to sketch that solution. (In reality, this solution curve should not cross the one you drew for $p(0)=100$, though the applet does not require the curves to avoid any crossing.)