Since it is the dynamic nature of a system that we want to model and
understand, the simplest form will be considered. This will involve onestate variable and will give rise to so-called "exponential" growth.
Two examples
First consider a mathematical model of a bank savings account. Assume
that there is an initial deposit but after that, no deposits orwithdrawals. The bank has an interest rate
${r}_{i}$ and service charge
rate
${r}_{s}$ that are used to calculate the interest and service charge
once each time period. If the net income (interest less service charge)is re-invested each time, and each time period is denoted by the integer
$n\phantom{\rule{0.166667em}{0ex}}$ , the future amount of money could be calculated from
$$M(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{2.em}{0ex}}$$
A net growth rate
$r$ is defined as the difference
$$r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{2.em}{0ex}}$$
and this is further combined to define
$R$ by
$$R\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}(r\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
The basic model in
[link] simplifies to give
$$M(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}(1\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s})\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{2.em}{0ex}}$$
$$=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}(1\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}r)\phantom{\rule{4pt}{0ex}}M\left(n\right)$$
$$M(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}.$$
This equation is called a
first-order difference equation , and the solution
$M\left(n\right)$ is found in a
fairly straightforward way. Consider the equation for the first fewvalues of
$n\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0,\phantom{\rule{0.166667em}{0ex}}1,\phantom{\rule{0.166667em}{0ex}}2,\phantom{\rule{0.166667em}{0ex}}...$
$$M\left(1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(0\right)$$
$$M\left(2\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{R}^{2}\phantom{\rule{4pt}{0ex}}M\left(0\right)$$
$$M\left(3\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(2\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{R}^{3}\phantom{\rule{4pt}{0ex}}M\left(0\right)$$
$$\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\xb7\phantom{\rule{4pt}{0ex}}\xb7\phantom{\rule{4pt}{0ex}}\xb7$$
$$M\left(n\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}M\left(0\right)\phantom{\rule{4pt}{0ex}}{R}^{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
The solution to
[link] is a geometric sequence that has an initial value of
$M\left(0\right)$ and increases as a function of
$n$ if
$R$ is greater than 1
$\left(r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\right)$ , and decreases toward zero as a function of
$n$ if
$R$ is
less than 1
$\left(r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\right)\phantom{\rule{0.166667em}{0ex}}$ . This makes intuitive sense. One's account
grows rapidly with a high interest rate and low service charge rate, andwould decrease toward zero if the service charges exceeded the interest.
A second example involves the growth of a population that has no
constraints. If we assume that the population is a continuous function of
time
$p\left(t\right)\phantom{\rule{0.166667em}{0ex}}$ , and that the birth rate
${r}_{b}$ and death rate
${r}_{d}$ are constants (
$\phantom{\rule{4pt}{0ex}}$ not functions of the population
$p\left(t\right)$ or time
$t\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$ ), then the rate of increase in population can be written
$$\frac{dp}{dt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}({r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d})\phantom{\rule{4pt}{0ex}}p\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
There are a number of assumptions behind this simple model, but we delay
those considerations until later and examine the nature of the solution ofthis simple model. First, we define a net rate of growth
$$r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d}\phantom{\rule{2.em}{0ex}}$$
which gives
$$\frac{dp}{dt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}rp\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
which is a first-order linear differential equation. If the value of the
population at time equals zero is
${p}_{o}\phantom{\rule{4pt}{0ex}}$ , then the solution of
[link] is
given by
$$p\left(t\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{e}^{rt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(0\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
The population grows exponentially if
$r$ is positive (if
${r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d\phantom{\rule{0.166667em}{0ex}}}$ ) and decays exponentially if
$r$ is negative (
${r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d\phantom{\rule{0.166667em}{0ex}}}$ ). The fact that
[link] is a solution of
[link] is easily verified by
substitution. Note that in order to calculate future values ofpopulation, the result of the past as given by
$p\left(0\right)$ must be known.
(
$\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(t\right)$ is a state variable and only one is necessary.)
Exponential and geometric growth
It is worth spending a bit of time considering the nature of the solution
of the difference
[link] and the differential
[link] . First,
note that the solutions of both increase at the same "rate". If we samplethe population function
$p\left(t\right)$ at intervals of
$T$ time units, a
geometric number sequence results. Let
${p}_{n}$ be the samples of
$p\left(t\right)$ given by
$${p}_{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(nT\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}n\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0,\phantom{\rule{4pt}{0ex}}1,\phantom{\rule{4pt}{0ex}}2,\phantom{\rule{4pt}{0ex}}...\phantom{\rule{2.em}{0ex}}$$
This give for
[link]
$${p}_{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(nT\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o\phantom{\rule{0.166667em}{0ex}}e}^{rnT}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}{\left({e}^{rT}\right)}^{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
which is the same as
[link] if
$$R\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{e}^{rT}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$$
This means that one can calculate samples of the exponential solution of
differential equations exactly by solving the difference
[link] if
$R$ is chosen by
[link] . Since difference equations are easily implemented
on a digital computer, this is an important result; unfortunately,however, it is exact only if the equations are linear. Note that if the
time interval
$T$ is small, then the first two terms of the Taylor's
series give