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\noindent
{\bf Math 1120}
\hfill {\Large \bf Logistic Growth} \hfill 
{\bf Calculus 2}

\vspace{0.2in}

When populations are small relative to their environment and resources, population growth tends to be proportional to the size of the population. In other words, if $P(t)$ is a population and $t$ is time, we expect to see $P'(t) \approx k \cdot P(t)$ for some growth constant $k$ when $P(t)$ is relatively small. 

However, when the population starts getting big relative to its environment and resources become scarce (e.g., food or land), we expect to see this growth to come to a near stop. Imagine we are situated in a region that can only reasonably sustain a population of $M$ (i.e., the location's {\it carrying capacity}). Then as $P(t)$ approaches $M$, we should see $P'(t) \approx 0$. 

A simple model that incorporates both of these features is {\it logistic growth}. Here we have $P'(t) = kP\left(1-\frac{P}{M}\right)$ and let $P(0)=P_0$ denote some initial population.  Notice that if $P$ is much smaller than $M$, $1-\frac{P}{M} \approx 1-0 =1$ so $P' \approx kP$ and if $P$ is close to $M$, then $1-\frac{P}{M} \approx 1-1=0$ so $P' \approx 0$. 

\vspace{0.15in}

We now solve the logistic equation using separation of variables and a partial fraction decomposition. We have $\frac{dP}{dt} = kP(1-P/M)$ so that $\frac{dP}{P(1-P/M)} = k\,dt$. We then seek to integrate both sides. The right hand side is easy: $\int k\,dt = kt+C_1$ for some constant $C_1$. The left hand side requires a partial fraction decomposition.

We have $\dfrac{1}{P(1-P/M)} = \dfrac{M}{M} \cdot \dfrac{1}{P(1-P/M)} = \dfrac{M}{P(M-P)} = \dfrac{-M}{P}{P-M} = \dfrac{A}{P}+\dfrac{B}{P-M}$ for some $A$ and $B$.
Clearing the denominators (i.e., multiplying both sides by $P(P-M)$), we get $-M = A(P-M)+B\cdot P$. Now plug in our roots. At $P=0$, we get $-M=A(-M)+0$ so $A=1$ and at $P=M$, we get $-M=0+B \cdot M$ so $B=-1$. 

Therefore, we integrate $\displaystyle \int \left( \dfrac{1}{P} + \dfrac{-1}{P-M} \right) \,dP = \ln|P|-\ln|P-M|$ (plus some constant). Putting this together with our other (easy) integral, we get $\ln|P|-\ln|P-M| = kt+C_1$. Next, use a law of logs: $\ln\left|\dfrac{P}{P-M}\right|=kt+C_1$. Then exponentiate: $\left|\dfrac{P}{P-M}\right|=e^{kt+C_1}=e^{kt}e^{C_1}$ and drop absolute values: $\dfrac{P}{P-M} = \pm e^{C_1} e^{kt}$. Acknowledge that we lost a zero solution when dividing during the separation of variables and so $\pm e^{C_1}$ (plus that zero solution) account for all possible constants, so we rename this as $C$. Therefore, $\dfrac{P}{P-M} = Ce^{kt}$.

Next, we solve for $P$. We have $P=Ce^{kt}(P-M) = PCe^{kt}-MCe^{kt}$ so $-MCe^{kt} = P-PCe^{kt}= P(1-Ce^{kt})$. Thus $P = \dfrac{-MCe^{kt}}{1-Ce^{kt}} = \dfrac{MCe^{kt}}{Ce^{kt}-1} = \dfrac{Ce^{kt}}{Ce^{kt}} \cdot \dfrac{M}{1-Ce^{-kt}}$. Therefore, our general solution is $P(t) = \dfrac{M}{1-Ce^{-kt}}$ (as well as $P(t)=0$  which was another solution that we lost due to an earlier division). 

Finally, assuming that $P(0)=P_0$, we have $P_0 = \dfrac{M}{1-Ce^0} = \dfrac{M}{1-C}$ so that $M = P_0(1-C) = P_0 - P_0\cdot C$. Thus $P_0 \cdot C = P_0-M$ and so $C=\dfrac{P_0-M}{P_0}$. Therefore, $P(t) = \dfrac{M}{1-(P_0-M)/P_0 e^{-kt}} = \dfrac{M}{1-(P_0-M)/P_0 e^{-kt}} \cdot \dfrac{P_0}{P_0}$. We then arrive at our solution's final form: \fbox{$P(t) = \dfrac{M \cdot P_0}{P_0-(P_0-M)e^{-kt}}$}.

\vspace{0.15in}

Note that if $P_0=0$, then $P(t)=0$ (no population stays that way). Also, if $P_0=M$, then $P(t)=M$ (if we are already at our carrying capacity, we stay there). Also, notice that, assuming $k>0$  (i.e., we have a {\it growth} constant), then $e^{-kt} \to 0$ as $t \to \infty$. Thus $P(t) \to \dfrac{M\cdot P_0}{P_0-0} = M$ as $t \to \infty$. In other words, we should expect our population to tend toward our carrying capacity regardless of where it started. So a small population over a long time will grow to approximately $M$ and a very large population over a long time will shrink to $M$.  


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