Apostol Problem on ODE applied to Population Growth

  • #1
zenterix
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83
Homework Statement
The following two problems are from chapter 8.7 "Introduction to Differential Equations" of Apostol's Calculus Volume I.

Problem 13. (Prelude to Problem 14, about which my question is)

Express ##x## as a function of ##t## for the growth law

$$\frac{dx}{dt}=kx(M-x)$$

with ##k## and ##M## both constant. Show that

$$x(t)=\frac{M}{1+e^{-kM(t-t_1)}}\tag{8.23}$$

where ##t_1## is the time at which ##x=M/2##.

Problem 14.

Assume the growth law in formula 8.23 of exercise 13, and suppose a census is taken at three equally spaced times ##t_1,t_2,t_3##, the resulting numbers being ##x_1,x_2,x_3##. Show that this suffices to determine ##M## and that, in fact, we have

$$M=x_2\frac{x_3(x_2-x_1)-x_1(x_3-x_2)}{x_2^2-x_1x_3}\tag{8.24}$$
Relevant Equations
My question is strictly about exercise 14.

Exercise 13 is a relatively simple matter (and I include at the end my solution to it).
First of all, a few observations

1) It is not clear if the ##t_1## used in problem 14 is the same ##t_1## from problem 13 where ##x(t_1)=\frac{M}{2}##.

However, if it were, then the problem seems like it wouldn't make too much sense because we'd have ##M=2x_1## and that'd be it (though this wouldn't match the expression the problem has for ##M##).

We would be able to solve for ##k## as well in this case with just one census reading.

2) From (8.23) it seems that the population never actually reaches ##M##. It just gets really close from below.

Now let me rewrite the equation for ##x## as follows

$$x(t)=\frac{M}{1+e^{-kM(t-t_h)}}$$

where now ##t_h## is such that ##x(t_h)=\frac{M}{2}##.

I've tried a few things but am a bit stuck.

Each attempt is a lot to type so I will have to screenshot them.

First I tried to just plug the values in

1697739984963.png


But this doesn't seem to tell me much.

Well, I will post now, as I continue to think about this problem.Since it will be too much work to write out all my work for problem 13 in equations, here is a screenshot of the work

1697736249797.png


Note that I used partial fractions to solve the integral in purple, and the calculations were as follows

1697738887613.png
 

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  • #2
zenterix said:
First of all, a few observations

1) It is not clear if the ##t_1## used in problem 14 is the same ##t_1## from problem 13 where ##x(t_1)=\frac{M}{2}##.

It isn't, for the following reason:

However, if it were, then the problem seems like it wouldn't make too much sense because we'd have ##M=2x_1## and that'd be it (though this wouldn't match the expression the problem has for ##M##).

We would be able to solve for ##k## as well in this case with just one census reading.

2) From (8.23) it seems that the population never actually reaches ##M##. It just gets really close from below.

Now let me rewrite the equation for ##x## as follows

$$x(t)=\frac{M}{1+e^{-kM(t-t_h)}}$$

where now ##t_h## is such that ##x(t_h)=\frac{M}{2}##.

I've tried a few things but am a bit stuck.

The key is that the times are equally spaced, so [itex]t_2 = t_1 + T[/itex], [itex]t_3 = t_1 + 2T[/itex]. Then you have [tex]\begin{split}
x_1 &= \frac{M}{1 + A} \\
x_2 &= \frac{M}{1 + e^{-kM(t_1 + T - t_h)}} = \frac{M}{1 + AB} \\
x_3 &= \frac{M}{1 + e^{-kM(t_1 + 2T - t_h)}} = \frac{M}{1 + AB^2} \end{split}[/tex] where [itex]
A = e^{-kM(t_1 - t_h)}[/itex] and [itex]B = e^{-kMT}[/itex]. Here I think the strategy is to find an expression for [itex]B[/itex] from the second equation, and set the square of that equal to the expression for [itex]B^2[/itex] obtained from the third equation; then the first equation can be used to eliminate [itex]A[/itex].
 

FAQ: Apostol Problem on ODE applied to Population Growth

What is the Apostol Problem on ODE applied to population growth?

The Apostol Problem on ODE (Ordinary Differential Equations) applied to population growth typically refers to a mathematical model that describes how a population changes over time. It involves setting up an ODE that incorporates factors like birth rates, death rates, and possibly other influences such as immigration or emigration, to predict the future size of a population.

How do you set up the differential equation for population growth?

To set up the differential equation for population growth, you start with the basic form: dP/dt = rP, where P is the population size, t is time, and r is the intrinsic growth rate. This is a simple exponential growth model. For more complex models, additional terms can be added to account for factors like carrying capacity (logistic growth) or other population dynamics.

What is the solution to the basic population growth ODE?

The solution to the basic population growth ODE, dP/dt = rP, is P(t) = P0 * e^(rt), where P0 is the initial population size at time t=0, r is the growth rate, and e is the base of the natural logarithm. This solution describes exponential growth or decay, depending on the sign of r.

What are the limitations of the simple exponential growth model?

The simple exponential growth model assumes that the growth rate r is constant and that there are no limits to growth. This is often unrealistic because it doesn't account for factors like resource limitations, environmental changes, or interactions with other species. For more realistic modeling, terms that represent these factors are added, such as in the logistic growth model: dP/dt = rP(1 - P/K), where K is the carrying capacity.

How can the Apostol Problem be extended to include more complex factors?

The Apostol Problem can be extended to include more complex factors by modifying the differential equation to incorporate additional variables and parameters. For example, one can include terms for immigration/emigration, age structure, spatial distribution, and interactions with other species. These extensions often lead to systems of differential equations or partial differential equations that can capture more realistic population dynamics.

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