Dynamics that often are calculated in terms of differential equations are growth processes.

Linear growth

Think of a population of size \(N\) which grows with a steady inflow, for example an inflow of \(I\) immigrants. \(N\) in this case is a constant amount, added to the population (or a stock of any kind) in each moment of time. As such \(I\) is the difference between the size of \(N\) in the moment \(t\) and the size of \(N\) in the moment \(t+1\).


If we know the size of \(N\) at time zero (\(N_0\)) and the size of \(I\), we can calculate the size of \(N\) at time \(t\) simply as \(N_t=I*t+N_0\). If we would want to express this in terms of a difference (or recursive) equation we can write: \(N_{t+1}=N_t+I\)

Exponential growth

More often of course one finds populations growing in terms of a reproduction of their elements. In this case the rate of growth is proportional to the size of the population. The larger the size, the larger the amount that is added to the population in each moment of time. We can denote the proportion with \(\gamma\) (for growth rate) and, simply define the gain per time step as \(\gamma*N_t * \Delta t\). Exponential growth would proceed according to

\[N_{t+1}=N_t + (\gamma*N_t\ * \Delta t)\]

with \(\Delta t\) indicating the difference of population sizes between two consecutive time steps.

The growth rate

Not knowing the growth rate of a development, but having consecutive data for \(N_t\) and \(N_{t+1}\) in a time series allows calculating the growthrate:

\[\gamma = \frac{N_{t+1} - N_t}{N_t}\]

Assume that data points are \(N_1 = 12\) and \(N_2 = 14\). The growth rate hence will be \(\gamma = \frac{14 - 12}{12} = 0.1667\)

If we consider the length of a time step \(\Delta t\) to approach zero (lim → 0), we can denote the above difference equation in terms of a differential equation (in Leibniz’ notation*)

\[\frac{dN}{dt}=\gamma * N\]

which, using the Euler-number \(e\), has the solution

\[N_t=N_0*e^{\gamma * t}\]

or alternatively written

\[N_t=N_0*exp(\gamma * t)\]

*Alternative notations: Lagrange’s: N'=\(\gamma*N\), Newton’s: Ń=\(\gamma*N\), Euler’s: DN=\(\gamma*N\)

Example – Population growth of the United States

In 1970 the size of the population of the United States was \(203.3\) million and in 1980 the population size was \(226.5\) million. We let \(N_t\) represent the population \(t\) years after 1970. Assuming that the development follows exponential growth, we have \(N_t=203.3e^{\gamma t}\) for some growth rate \(\gamma\). Since \(N_{10}=226.5\), we can find \(\gamma\) by solving \(226.5 = 203.3e^{\gamma 10}\). We need to find a value for \(\gamma\) such that \(e^{\gamma 10}= \frac{226.5}{203.3} = 1.114\).

Referring to the natural logarithm \(ln\), with \(ln b = a\) when \(e^a=b\), we see that \(\gamma= 0.0108\) satisfies the equation. Thus, assuming exponential growth, this model would predict the population of the United States \(t\) years after 1970 to be \(N_t= 203.3e^{0.0108 t}\)

In 1990 for instance, the population size should be \(N_{20} = 203.3e^{0.0108∗20} = 252.3\) million. In the year 2000 it should be \(281.1\) million and in the year 2120, 150 years from 1970, according to this model of exponential growth, it should be \(1027.29\) million.

The predictions for 1990 and 2000 seem fairly accurate. The actual population was about \(249\) and \(281\) million respectively. The prediction for 2120 however seems rather unrealistic. Experts doubt that human population growth can proceed uninhibited in exponential way.

The US-population example is taken from Sloughter, Dan (2000). Difference to Differential Equations. Section 6.1., S.9