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So far in my education career I have only met differential equations as small parts of courses on other stuff. Solving special cases as part of calculus, solving simple systems as a part of linear algebra. This coming semester I'm going to have two courses devoted entirely to differential equations, so I thought I would try to gain some understanding that isn't purely mechanical.

Here's an example I have some questions about.

This is example 10.2.1 from 'KALKULUS' (3rd edition) by Lindstrøm. The translation is mine. The part where the differential equation is solved has been removed.

An animal population consists today of $P$ animals and has a growth rate $r$. How big is the population in $t$ years?

Let $y(t)$ be the population size after $t$ years. In the time between $t$ and $t+\Delta t$ the population increases from $y(t)$ to $y(t+\Delta t)$, i.e. an increase of $y(t+\Delta t)-y(t)$. We can also derive this increase in another way: The growth rate is $r$, which means that the population increase per time unit is $ry(t)$. During a small time interval from $t$ to $t+\Delta t$ the population increase is approximately $ry(t)\Delta t$, an approximation that gets better with smaller $\Delta t$'s. If we equate these expressions, we get

$$y(t+\Delta t)-y(t)\approx ry(t)\Delta t$$

Dividing by $\Delta t$, we get

$$\frac{y(t+\Delta t)-y(t)}{\Delta t}\approx ry(t)$$

Letting $\Delta t$ go towards zero, this gives

$$y'(t)=ry(t)$$

Thus we have a differential equation that $y$ has to satisfy:

$$y'(t)-ry(t)=0$$

NOTE: We could have gotten this differential equation faster by using the fact that the growth rate $r$ by definition means that $y'(t)=ry(t).$ We have chosen the more elaborate approach because it shows a general thought process which can be used in several situations.

  1. Why is the population increase approximately $ry(t)Δt$ during a small time interval from $t$ to $t+Δt$?

  2. How is this thought process different than using the definition? In the real world it has been observed that the growth of a population tend to depend on the size of the population, so you want the change in population to depend on its size. In the language of calculus one way to write this out is $y'(t)=ry(t)$. What's the advantage of going through the more elaborate reasoning?

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  • $\begingroup$ The elaborate reasoning above is just the reasoning that gives your $y'(t)=ry(t)$. Once you know this, you generally don't need to elaborate reasoning. $\endgroup$ – Thomas Andrews Dec 25 '13 at 15:13
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This is not a question about differential equations, but a question about mathematical modeling. What you call "elaborate reasoning" is the thought process that leads from an observed phenomenon to the differential equation that describes it "in the limit". Note that the number $y$ of individuals in your population in reality is an integer at all times, and that you are counting them just once a year. But at the end of this thought process you have a differential equation that involves a real-valued function $t\mapsto y(t)$ of the continuous (i.e. real) variable $t$, and you are convinced that you have derived a "law of nature".

You are given a certain system, say a Petri dish containing an amount $y$ of bacteria. This amount will increase with time, so we have a certain function $t\mapsto y(t)$ that we would like to understand. Of course we can make hourly or even more frequent measurements and plot the results on graph paper. Connecting the data dots we obtain a nice curve. Can we somehow explain the shape of this curve?

Naturally we have some ideas about the reproduction process of these bacteria. The idea is the following: A bacterion living at time $t$ will, independently of its present age, in the next second split into two copies of itself with a certain very small probability $p\ll 1$. If there are $y(t)$ bacterias at time $t$ we therefore can expect $(1+p)y(t)$ of them one second later, and as $p^2$ can be neglected with respect to $p$ we shall see $(1+2p)y(t)$ of them two seconds later, $(1+3p)y(t)$ of them three seconds later, and so on. It follows that $$y(t+\Delta t)-y(t)\doteq p\>y(t)\>\Delta t$$ for reasonably small $\Delta t$, or that $${y(t+\Delta t)-y(t)\over\Delta t}\doteq p\>y(t)\qquad(|\Delta t|\ll1)\ .$$ (The numerical value of $p$ depends on the chosen time unit, "seconds" in our example.)

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  • $\begingroup$ Thank you for your reply. I'm having trouble with the last step though, I don't understand where the $\Delta t$ in $py(t)\Delta t$ comes from. What does multiplying with $\Delta t$ actually mean in this context? $\endgroup$ – john.abraham Dec 26 '13 at 12:36
  • $\begingroup$ @john.abraham: $\Delta t$ was $1$, $2$, and $3$ seconds in the immediately foregoing text. $\endgroup$ – Christian Blatter Dec 26 '13 at 12:55

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