Population growth with a time varying growth rate

Sun, 17 Oct 2021 00:00:00 +0000

This article considers the exponential population growth model where the population growth rate is not constant over time.

I have just started working my way through Andrei Rogers’ Applied Multiregional Demography Through Problems: A Programmed Learning Workbook with Exercises and Solutions (Rogers 2020). There is an exercise that has the population growth rate decline linearly over time to zero growth. In the preceeding text, the explanation only looks at a case where the growth rate is constant. I was unable to generalise this method to deal with a non-constant case.

I pulled out my copy of Demographic Methods (Hinde 2014) to see how it presents population growth. It derives the case for constant growth using the same approach as Rogers (2020). So I was no further ahead.

In this post I:

outline the approach used in these books; and then
describe a generalised method to derive a solution for the growth rate to vary with regard to time.

Constant growth

The approach used in the two books moves from an annual constant rate of growth to the continuous case. Their derivation uses the following approach.

Say we had a population growing at constant rate of \(r\) per annum. If the population size at the start was \(P_0\) then after a year the population size would be \(P_1 = P_0(1+r)\). Since this can compound over time, then after year \(n\) the population size would be \(P_n = P_0(1+r)(1+r)...(1+r) = P_0(1+r)^n\). The compounding in this formulation only happens once a year.

In order to move to the continous case, the compunding can be spit into finer time periods. So for example, growth compounding six monthly would be \(P_1 = P_0(1+\frac{r}{2})^{2}\) or for n years \(P_n = P_0(1+\frac{r}{2})^{2n}\). If the compounding period is reduced to shorter time periods then the equation can be generalised to \(P_n = P_0(1+\frac{r}{k})^{kn}\), where \(k\) is the number of compounding time periods within a year.

At this point they move to focussing on increasing \(k\) toward infinity and using a limit. Taking the compounding part of the formula they derive the limit as \(e^r\):\[\lim_{k \to \infty}(1+\frac{r}{k})^{k} = e^r\]

This leads to the growth formula as: \[P_n = P_0e^{rn}\]

This is a rather beautiful mathematical result. To explain how \(e\) pops out of the above limit Hinde (2014) takes a further step to show the binomial expansion of the \((1+\frac{r}{k})^{k}\) for \(k=3\), to show how this looks like the power series expansion of the \(e^x\).

Constant growth, now with calculus

So while above is very interesting, I was not able to generalise it to \(r\) changing with time. As a result, I was struggling to solve Rogers’ exercise. Stepping back, I thought the answer might involve calculus.

My calculus is very rusty. So I turned to my old friend Thomas and Finney (1988). With population growth we are seeking to change the population size over time. To express this idea as a differential equation for the constant growth rate expression we would write this as: \[ \frac{dP(t)}{dt} = rP(t)\].

In words, the change in the size of the population is proportional to the size of the population. With the proportion in our case being \(r\). To solve this we can rearrange and then integrate. \[ \frac{1}{P(t)}dP(t) = r dt\]

This integrates to: \[ln(P(t)) + C = rt\]

Since we want \(P(t)\) we can raise both sides to the power of \(e\) to get rid of the log: \[P(t)e^C = e^{rt}\]

We can then move the constant \(e^C\) to the otherside and since at time zero the equation has to equal the starting population size we can set this to \(P(0)\). This leaves us with: \[P(t) = P(0)e^{rt}\]

So we are now back at where we started, but now we have a nice approach for moving to a non-constant growth rate.

Non-constant growth

Going back to our original calculus based formula and substuting the constant \(r\) with a time based function \(r(t)\) we get: \[ \frac{dP(t)}{dt} = r(t)P(t)\].

So now we can solve this as:\[P(t) = P(0)e^{\int{r(t)dt}}\]

Admittedly, the \(\int{r(t)dt}\) looks daunting. But in solving this, all we are looking for is the area under the \(r(t)\) curve. In Rogers’ example, he was using a linearly declining \(r(t)\) and solved it using geometry — essentially he recognises that the area under of the \(r\) curve is half of the \(r \times n\) rectangle.

A worked example

Say we have a city population that starts with 200,000 people in 2020. Let’s say the growth rate starts at 2.5% but is declining linearly to zero over 25 years - ie it is decling by 0.1 percentage points per year. What is the population size expected to be in 2045?

Let’s solve \(\int{r(t)dt}\).

From the description \[r(t) = 0.025 - 0.001t\].

So \(\int_{t=0}^{25} (0.025 - 0.001t) dt\) equals \[0.025t - \dfrac{0.001}{2}t^2 \Big|_{t=0}^{25} = 0.3125\]

So, the population size in 2045 would be \(P_{2045} = 200000 e^{0.3125}\) or 273,368 people.

We can check this geometrically. If we had constant growth of 2.5% we would have had the growth component as \(e^{0.025 \times 25}\), but since the area under the curve is \(r(t)\) is only half of the \(r \times n\) rectangle we could write \(e^{\frac{0.025}{2} \times 25}\) which equals \(e^{0.3125}\) giving us the same result.

Conclusion

I found it quite fun to revisit calculus. I realise that introductory demography textbooks want to step away from this complexity, but I think that in this case avoiding calculus limited the class of problems that could be solved without feeling like I just had to trust the author.

References

Hinde, A. (2014). Demographic Methods. Routledge.

Rogers, A. (2020). Applied Multiregional Demography Through Problems: A Programmed Learning Workbook with Exercises and Solutions. Springer Nature.

Thomas, G.B. and Finney, R.L. (1988). Calculus and Analytic Geometry. 7th ed. Reading, Mass: Addison-Wesley Pub. Co.