Section 2.6
Population Dynamics

We are going to look at some equations of the form

$$\frac{dy}{dt}= f(y)$$

These equations are called autonomous because the independent variable, t, does not appear in the equation.

Let $y = \phi(t)$ measure a population at time $t$

The simplest model for population is exponential growth.

$$\frac{dy}{dt}= ry$$

With the initial condition that $y(0) = y_0$, we obtain the familiar solution

$$y(t) = y_0e^{rt}$$

Logistic Growth

To take into account that populations cannot grow in an uninhibited fashion indefinitely, we can replace the proportionality constant, $r$, with a function $h(y)$:

$$\frac{dy}{dt}= h(y)y$$

If we assume that the growth rate decreases in proportion to the population, we can choose $h(y) = r - ay \quad (a>0)$ to obtain

$$\frac{dy}{dt}= (r - ay)y$$

This is the Verhulst Equation, or the logistic equation.

Rewrite this equation as

$$\frac{dy}{dt}= r(1-\frac{y}{K})y$$

where $K=\frac{r}{a}$.

We refer to $r$ as the intrinsic growth rate.

Logistic Growth

In solving the Verhulst equation,

$$\frac{dy}{dt}= r(1-\frac{y}{K})y$$

we first search for the simplest type of solutions: constant solutions.

If $y = \phi(t)$ is some constant solution, then $\frac{dy}{dt}= 0$, so we just solve the algebraic equation

$$r(1-\frac{y}{K})y = 0$$

and determine that either $y=0$ or $y=K$.

These solutions are called equilibrium solutions.

Given the general autonomous equation

$$\frac{dy}{dt}= f(y),$$

the roots of f(y) are called critical points of the equation.

At this point, we spent time looking at Figures 2.6.2 and 2.6.3 from section 2.6.

Logistic Growth

We can determine the concavity of the solutions and find their inflection points using the second derivative test:

$$\frac{d^2y}{dt^2} = f'(y)\frac{dy}{dt}= f'(y)f(y)$$

The graph of $y$ is concave up when $y'' > 0$ ($f$ and $f'$ have the same sign), and it is concave down when $y'' < 0$ ($f$ and $f'$ have different signs)

It is also important to note that the value $K$ represents the saturation level, or environmental carrying capacity for the model

Logistic Growth

$$\frac{dy}{dt}= r(1-\frac{y}{K})y$$

We've learned a lot about the solutions of this equation without even explicitly solving it, but just for fun, let's try and solve it.

$$\frac{dy}{(1-\frac{y}{K})y} = r \: dt$$

Obtaining a partial fraction decomposition for the left side yields

$$\biggl( \frac{1}{y} + \frac{\frac{y}{K}}{1 - \frac{y}{K}} \biggr) dy = r \: dt$$

Integration gives

$$\ln{\vert y\vert} - \ln{\biggl\vert 1 - \frac{y}{K} \biggr\vert } = rt + c$$

Next, exponentiate

$$\frac{y}{1 - \frac{y}{K}} = Ce^{rt}$$

Logistic Growth

Finally, given an initial condition $y=y_0$, we see that $C=y(1-\frac{y_0}{K})$. Plugging in we get

$$y = \frac{y_0K}{y_0 + (K-y_0)e^{-rt}}$$

Taking the limit as $t \rightarrow \infty$, we see that $y \rightarrow K$

We refer to the constant solution $y=K$ as an
asymptotically stable solution, whereas
the solution $y=0$ is an unstable equilibrium solution.