Population Growth Models.
In this discussion,
\begin{align*}
P &= \text{ size of population}\\
t &= \text{ time}\\
\beta &= \text{ birth rate}\\
\delta &= \text{ death rate}
\end{align*}
Exponential Growth Model (Review).
I use the word βgrowthβ here to mean growth or decay. Some books separate exponential growth and exponential decay as though they are two totally different topics. You treat them in exactly the same way, and there is no need to study them separately.
An exponential growth model occurs when \(\beta\) and \(\delta\) are constants and the population is modeled by the differential equation
\begin{equation*}
\frac{dP}{dt}=(\beta-\delta)P=kP
\end{equation*}
This differential equation is and has solution
\begin{equation*}
P(t)=P_0e^{kt}\text{.}
\end{equation*}
If \(k>0\text{,}\) then the population is and
\begin{equation*}
\lim_{t \rightarrow \infty}P(t) =
\fillinmath{XXXXXXXXXXXXXXX}\text{.}
\end{equation*}
If \(k<0\text{,}\) then the population is and
\begin{equation*}
\lim_{t \rightarrow \infty}P(t) =
\fillinmath{XXXXXXXXXXXXXXX}\text{.}
\end{equation*}
More General Population Growth Models.
The birth rate \(\beta\) and the death rate \(\delta\) are not always constants as in the exponential growth model. This leads to a more general population growth model:
\begin{equation*}
\frac{dP}{dt}=(\beta(t,P)-\delta(t,P))P\text{.}
\end{equation*}
Example 56. A population model with a variable rate.
(Exercise 1.4.13 from Bazettβs website)
A population has a constant birth rate
\(\beta(t)=2\) and a variable death rate of
\(\delta(t)=0.1t\) If
\(P_0=1000\text{,}\) find the size of the population at time
\(t=5\text{.}\)
In some situations, the birth rate
as the
increases. For example, there may be limited
If the birth rate decreases linearly \(\beta(t,P)=\mu_0-\mu_1P\) and the death rate is constant \(\delta(t,P)=\gamma\) where \(\mu_0\text{,}\) \(\mu_1\text{,}\) and \(\gamma\) are constants, then this yields the population growth model
\begin{align*}
\frac{dP}{dt} & = & (\mu_0-\mu_1P -\gamma)P \\
& = & aP-bP^2
\end{align*}
where \(a=
\fillinmath{XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX}\) and \(b=\fillinmath{XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX}\text{.}\)
Definition 57.
A logistic equation is a differential equation of the form
\begin{equation*}
\frac{dP}{dt}=aP-bP^2
\end{equation*}
where \(a\) and \(b\) are positive constants. Equivalently, a logistic equation is a differential equation of the form
\begin{equation*}
\frac{dP}{dt}=kP(M-P)
\end{equation*}
where \(k\) and \(M\) are positive constants.
Logistic Equations can be solved using
and
.
Theorem 59.
The solution of the IVP with logistic equation
\begin{equation*}
\frac{dP}{dt}=kP(M-P), \qquad P(0)=P_0
\end{equation*}
is
\begin{gather}
P(t) = \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}\tag{βΆ}
\end{gather}
Notice that the
equilibrium solutions (i.e., constant function solutions)
\(P \equiv 0\) and
\(P \equiv M\) can be obtained from the general form of the solution given in
(βΆ) by selecting the necessary
\(P_0\text{.}\)
Some comments about logistic growth equations:
\begin{equation*}
\frac{dP}{dt}=kP(M-P)
\end{equation*}
-
\(P\equiv 0\) and
\(P\equiv M\) are constant solutions of the logistic growth equation
\(\frac{dP}{dt}=kP(M-P)\) that follow the format of the general solution.
-
The limiting population is given by
\begin{equation*}
\lim_{t \rightarrow \infty} P(t)
\end{equation*}
In a logistic model, \(\frac{dP}{dt}=kP(M-P)\text{,}\) \(k\) and \(M\) are positive constants, so
\begin{align*}
\lim_{t \rightarrow \infty} P(t) & = &
\lim_{t \rightarrow \infty} \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}\\
& = & \fillinmath{XXXXXXXXXXXX}\\
& = & \fillinmath{XXXXXXXXXXXX}
\end{align*}
Definition 60.
In a logistic growth model
\begin{equation*}
\frac{dP}{dt}=kP(M-P)\text{,}
\end{equation*}
the constant \(M\) is called the carrying capacity of the model.
Intuitively, the carrying capacity is the maximum population that the environment can handle.