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Subsection 3.3.3 Population Growth

Suppose that we wish to predict the size \(P(t)\) of a population as a function of the time \(t\text{.}\) In the most naive model of population growth, each couple produces \(\beta\) offspring (for some constant \(\beta\)) and then dies. Thus over the course of one generation \(\beta\tfrac{P(t)}{2}\) children are produced and \(P(t)\) parents die so that the size of the population grows from \(P(t)\) to

\begin{gather*} P(t+t_g)= \underbrace{P(t)+\beta\frac{P(t)}{2}}_{\text{parents+offspring}} -\underbrace{P(t)}_{\text{parents die}}=\frac{\beta}{ 2 } P(t) \end{gather*}

where \(t_g\) denotes the lifespan of one generation. The rate of change of the size of the population per unit time is

\begin{gather*} \frac{P(t+t_g)-P(t)}{t_g} =\frac{1}{t_g}\Big[\frac{\beta}{2}P(t) -P(t)\Big] = b P(t) \end{gather*}

where \(b=\tfrac{\beta-2}{2t_g}\) is the net birthrate per member of the population per unit time. If we approximate

\begin{gather*} \tfrac{P(t+t_g)-P(t)}{t_g}\approx\diff{P}{t}(t) \end{gather*}

we get the differential equation

By Corollary 3.3.4, with \(-k\) replaced by \(b\text{,}\)

\begin{align*} P(t) &= P(0)\cdot e^{bt} \end{align*}

This is called the Malthusian  10 This is named after Rev. Thomas Robert Malthus. He described this model in a 1798 paper called “An essay on the principle of population”. growth model. It is, of course, very simplistic. One of its main characteristics is that, since \(P(t+T) = P(0)\cdot e^{b(t+T)} = P(t)\cdot e^{bT}\text{,}\) every time you add \(T\) to the time, the population size is multiplied by \(e^{bT}\text{.}\) In particular, the population size doubles every \(\frac{\log 2}{b}\) units of time. The Malthusian growth model can be a reasonably good model only when the population size is very small compared to its environment  11 That is, the population has plenty of food and space to grow.. A more sophisticated model of population growth, that takes into account the “carrying capacity of the environment” is considered in the optional subsection below.

In 1927 the population of the world was about 2 billion. In 1974 it was about 4 billion. Estimate when it reached 6 billion. What will the population of the world be in 2100, assuming the Malthusian growth model?

Solution We follow our usual pattern for dealing with such problems.

  • Let \(P(t)\) be the world's population \(t\) years after 1927. Note that 1974 corresponds to \(t=1974-1927 = 47\text{.}\)
  • We are assuming that \(P(t)\) obeys equation 3.3.12. So, by Corollary 3.3.4 with \(-k\) replaced by \(b\text{,}\)
    \begin{gather*} P(t)=P(0)\cdot e^{bt} \end{gather*}
    Notice that there are 2 unknowns here — \(b\) and \(P(0)\) — so we need two pieces of information to find them.
  • We are told \(P(0)=2\text{,}\) so
    \begin{gather*} P(t)=2\cdot e^{bt} \end{gather*}
  • We are also told \(P(47)=4\text{,}\) which gives
    \begin{align*} 4 &=2\cdot e^{47b} & \text{clean up}\\ e^{47b}&=2 & \text{take the log and clean up}\\ b&=\frac{\log 2}{47} = 0.0147 & \text{to 3 significant digits} \end{align*}
  • We now know \(P(t)\) completely, so we can easily determine the predicted population  12 The 2015 Revision of World Population, a publication of the United Nations, predicts that the world's population in 2100 will be about 11 billion. They are predicting a reduction in the world population growth rate due to lower fertility rates, which the Malthusian growth model does not take into account. in 2100, i.e. at \(t=2100-1927 = 173\text{.}\)
    \begin{gather*} P(173) = 2 e^{173 b} = 2 e^{173\times 0.0147} = 25.4\text{ billion} \end{gather*}
  • Finally, our crude model predicts that the population is 6 billion at the time \(t\) that obeys
    \begin{align*} P(t) &= 2 e^{b t} = 6 & \text{clean up}\\ e^{b t}&=3 & \text{take the log and clean up}\\ t&=\frac{\log 3}{b} = 47\frac{\log 3}{\log 2} = 74.5 \end{align*}
    which corresponds  13 The world population really reached 6 billion in about 1999. to the middle of 2001.

Subsubsection 3.3.3.1 (Optional) — Logistic Population Growth

Logistic growth adds one more wrinkle to the simple population model. It assumes that the population only has access to limited resources. As the size of the population grows the amount of food available to each member decreases. This in turn causes the net birth rate \(b\) to decrease. In the logistic growth model \(b=b_0\left(1-\tfrac{P}{K}\right)\text{,}\) where \(K\) is called the carrying capacity of the environment, so that

\begin{gather*} P'(t) =b_0\left(1-\frac{P(t)}{K}\right)P(t) \end{gather*}

We can learn quite a bit about the behaviour of solutions to differential equations like this, without ever finding formulae for the solutions, just by watching the sign of \(P'(t)\text{.}\) For concreteness, we'll look at solutions of the differential equation

\begin{gather*} \diff{P}{t}(t)=\big(\,6000-3P(t)\,\big)\,P(t) \end{gather*}

We'll sketch the graphs of four functions \(P(t)\) that obey this equation.

  • For the first function, \(P(0)=0\text{.}\)
  • For the second function, \(P(0)=1000\text{.}\)
  • For the third function, \(P(0)=2000\text{.}\)
  • For the fourth function, \(P(0)=3000\text{.}\)

The sketches will be based on the observation that \((6000-3P)\,P=3(2000-P)\,P\)

  • is zero for \(P=0,\ 2000\text{,}\)
  • is strictly positive for \(0 \lt P \lt 2000\) and
  • is strictly negative for \(P \gt 2000\text{.}\)

Consequently

\begin{align*} \diff{P}{t}(t)\ \begin{cases} =0 & \text{if }P(t)=0\\ \gt 0 & \text{if }0 \lt P(t) \lt 2000\\ =0 & \text{if }P(t)=2000 \\ \lt 0 & \text{if }P(t) \gt 2000 \end{cases} \end{align*}

Thus if \(P(t)\) is some function that obeys \(\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)\text{,}\) then as the graph of \(P(t)\) passes through \(\big(t,P(t)\big)\)

\begin{align*} \text{the graph has } \begin{cases} \text{slope zero,}& \text{i.e. is horizontal, if }P(t)=0 \\ \text{positive slope,}& \text{i.e. is increasing, if } 0 \lt P(t) \lt 2000 \\ \text{slope zero,}& \text{i.e. is horizontal, if }P(t)=2000 \\ \text{negative slope,}& \text{i.e. is decreasing, if }P(t) \gt 2000 \end{cases} \end{align*}

as illustrated in the figure

As a result,

  • if \(P(0)=0\text{,}\) the graph starts out horizontally. In other words, as \(t\) starts to increase, \(P(t)\) remains at zero, so the slope of the graph remains at zero. The population size remains zero for all time. As a check, observe that the function \(P(t)=0\) obeys \(\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)\) for all \(t\text{.}\)
  • Similarly, if \(P(0)=2000\text{,}\) the graph again starts out horizontally. So \(P(t)\) remains at \(2000\) and the slope remains at zero. The population size remains 2000 for all time. Again, the function \(P(t)=2000\) obeys \(\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)\) for all \(t\text{.}\)
  • If \(P(0)=1000\text{,}\) the graph starts out with positive slope. So \(P(t)\) increases with \(t\text{.}\) As \(P(t)\) increases towards 2000, the slope \((6000-3P(t)\big)P(t)\text{,}\) while remaining positive, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from below 2000 to above 2000, because to do so, it would have to have strictly positive slope for some value of \(P\) above 2000, which is not allowed.
  • If \(P(0)=3000\text{,}\) the graph starts out with negative slope. So \(P(t)\) decreases with \(t\text{.}\) As \(P(t)\) decreases towards 2000, the slope \((6000-3P(t)\big)P(t)\text{,}\) while remaining negative, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from above 2000 to below 2000, because to do so, it would have to have negative slope for some value of \(P\) below 2000. which is not allowed.

These curves are sketched in the figure below. We conclude that for any initial population size \(P(0)\text{,}\) except \(P(0)=0\text{,}\) the population size approaches \(2000\) as \(t\rightarrow\infty\text{.}\)

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