For many things that grow, the rate of growth depends upon the amount of growth made and the remainder left to be grown. That is sufficient description for you to write a differential equation as a model. If the total (call it adult) growth is one (unity) and the amount grown at the moment (now) is x, then the amount left to grow is 1-x and so

dx/dt = kx(1-x).

Looks easy, doesn’t it? Did you do that yourself, or just read on and say “Yeah”? That is so easy to do – and it is surprising how hard people find it is to turn words into equations. I suspect I will write more of just that.

Separation of variables lies within the ordinary A-level syllabus, so, starting from the position described above:

1. Separate the variables (put all the terms including x on the same side of the equation), find some partial fractions and do the integration.

Show your answer can be written f(x) = Ae^{-kt} and thence (from there) x = (1+Be^{-kt})^{-1}

2. A baby is 65cm at birth (t=0.75 years) and grows to full height of 180 cm at say t=20. Sketch the curve (possibly using a gadget of choice) and make comment based on what you have observed for yourself – is this a reasonable model?

3. For the case in Q1, assume x=0.5 at t=0. Decide whether this function is odd or even.

4. Assume that at t=18 you are close to your full height, say 98% of it (172cm of 175.5, for example). Assume that you were 60cm at birth and this time let us call that t=0. Show that B = 1.925 and find k. Be clear that x is not height, since the model requires it to be in the range 0≤x≤1

Consider a population of size N in an environment suitable for population K. The rate of growth depends upon (is proportional to) the proportion of population we have and the proportion we have remaining to capacity. Let x=N/K and see that this is the same equation as before. The value K is called the carrying capacity. Historically in this model we use r as the constant, dx/dt = rx(1-x) = r N/K (1-N/K). The argument says that as N/K approaches the limit, so resources such as food or living space become critical. At stability the population is called mature (it has finished growing) and other factors - hence other models - apply. Species are described as using strategies for r or for K (r-strategist, K-strategist). To a mathematician, it is obvious that lim(N) = K as t heads to infinity. If N0 is the initial population (meaning when you start the model, then N = K N_{0}e^{rt} / (K + N_{0}(e^{rt} - 1)) follows in a straightforward way you can do for yourself (not hard at all after the other work, you might even say “Obvious”).

There’s an extension of this model that says the carrying capacity, K, varies with time; an obvious example is the cycle of the seasons, making K periodic. If the period is T, then K(T+t) = K(t) describes the situation.

There may also (or separately) be a delay (d) reflecting how population reacts to environment, so that K(t) = f(N(t-d)). This rapidly gets complicated, what one source I was reading calls a ‘rich’ behaviour - by which is meant that there are many possibilities reflected in the solutions. This would be a good topic for university-level research, such as in ecology. It specifically lends itself to computer-based modelling.

The collection of curves called __ logistic functions__ are found (well, almost found) in the FM Stats course; even ordinary students of Stats recognise z=(x-mean) / std dev as the standardised mean (yes you do, this is how you handle the Normal Distribution). If I use s for std dev then the probability density function f(z) = e

^{-z }/ s(1+e

^{-z})². Yes, I know s is a dependent variable too, but I’m trying to make it readable and not use mu, x and s.

5. FM students should prove that f(z) above = 1/4s sech² (z/2).

6. FM students can explain (and A-level students can simply integrate, for practice) the exponential f(z) to see the Cumulative function F(z)=(1+e^{-z})-1. We have seen this already on this page.

7. FM students should integrate the sech form to 1/2 (1+tanh(z/2))

8. Students of other disciplines and those who see Maths as a tool for actually doing things should write themselves a list of things which might be productively modelled by logistic distributions. Think of the S-shaped curve of the cumulative function.

I’ll offer some less likely ones, leaving you a wealth of opportunity to write to me about;

(i) the money sent on a project, especially in construction; subsidiary parts of the same project follow the same curve;

(ii) as a substitute for the Normal distribution, since this curve can be solved analytically and is very similar (define similar?);

(iii) the Hubbert curve (see next page, and Dying Away).

Q3: B=1. The exponential term cannot be negative, even if t could be. When t <0 the resulting exponent is the same as the equal value of opposite sign, so f(t) = f(-t), which is even.

Q4 If the model x = (1+Be-kt)-1 is suitable, then at t=0, x=1/(1+B), so if x(0)= 60/175.5 =>B=1.925 => k=0.253, I think. Please confirm that you agree. (0.25256 to 5sf).