### Exponential growth and decay modeled by discrete dynamical systems

The solution to a linear discrete dynamical system is an exponential. If we choose conditions so that the state variable stays positive, this exponential solution will exhibit either exponential growth or exponential decay. We can determine whether we get growth or decay from the parameters of the system.

#### Growth versus decay

##### Video

*Exponential growth and decay in discrete dynamical systems.*

##### Video review questions

##### Video summary

If we have a linear discrete dynamical system in function iteration form, \begin{align} x_{t+1}&=R x_t\label{function_iteration_form}\tag{1}\\ x_{0} &= \text{a known positive value},\notag \end{align} then the solution is the exponential \begin{align*} x_t = R^tx_0. \end{align*} Given that $x_0 > 0$, the solution will be positive as long as $R > 0$. The value of $R$ determines whether we get exponential growth or decay.

- If $R > 1$, then at each time step, the value of the state variable increases. The solution exhibits exponential growth. The growth makes sense because in each time step, we are multiplying by a number greater than 1.
- If $0 < R <1$, then at each time step, the value of the state variable decreases. The solution exhibits exponential decay. The decay makes sense because in each time step, we are multiplying by a positive number less than 1.
- If $R=1$, then the solution is a constant. The constant solution makes sense because in each time step, we are multiplying by 1.

On the other hand, if we have a linear dynamical system in difference form, \begin{align} x_{t+1} - x_t &=r x_t\label{difference_form}\tag{2}\\ x_{0} &= \text{a known positive value},\notag \end{align} then, to solve the system, we first must add $x_t$ to both sides of the evolution rule, converting it into the equivalent system in function iteration form: \begin{align*} x_{t+1} &=(r +1)x_t\\ x_{0} &= \text{a known positive value}. \end{align*} Its solution is the exponential \begin{align*} x_t = (r+1)^tx_0. \end{align*} Given that $x_0 > 0$, the solution will be positive as long as $r > -1$. The value of $r$ determines whether we get exponential growth or decay.

- If $r > 0$, then at each time step, the value of the state variable increases. The solution exhibits exponential growth. The growth makes sense because in each time step, we are adding a positive quantity.
- If $-1 < r < 0$, then at each time step, the value of the state variable decreases. The solution exhibits exponential decay. The decay makes sense because in each time step, we are subtracting a positive quantity, but we aren't subtracting off any more than the value of $x_t$ in the previous time step.
- If $r = 0$, then the solution is a constant. The constant solution makes sense because in each time step, we are adding zero.

#### Examples

##### Human population growth

Suppose a human population is growing at 1% per year and initially has 1,000,000 individuals. Let $P_t$ denote the populations size $t$ years after the initial population of $P_0 = 1,000,000$ individuals. If one asks what the population will be in 50 years there are two options.

**Option 1.** At 1% per year growth, the dynamic equation would
be
\begin{gather*}
P_{t+1} - P_t = 0.01 \, P_t
\end{gather*}
and the corresponding iteration
equation is
\begin{gather*}
P_{t+1} = 1.01 \, P_t
\end{gather*}
Starting with the initial condition, $P_0 = 1,000,000$, we could calculate $P_1 = 1.01 \times 1,000,000 = 1,010,000$,
$P_2 = 1.01 \times 1,010,000 = 1,020,100$ and so on for 50 iterations.

**Option 2.** Alternatively, one may write the solution
\begin{gather*}
P_t = 1.01^t \left( 1,000,000 \right)
\end{gather*}
so that we immediately have a formula for the population after 50 years:
\begin{gather*}
P_{50} = 1.01^{50} \left( 1,000,000 \right) = 1,644,631.
\end{gather*}

##### Bacteria growth

When nutrients are plentiful and the population size isn't too large, a bacteria can exhibit exponential growth. See this page for details on developing an exponential growth model from data and solving it.

##### Penicillin clearance

If penicillin is given as a bolus injection (a one-time injection) and the kidneys remove a fixed fraction of the penicillin every 5 minutes, the amount of penicillin in the blood exhibits exponential decay. See this page for the steps to develop such an exponential decay model from data.

##### Chocolate decay

I start with 2 chocolate bars. Every day, I eat 10% of the chocolate that I started the day with. Write down a dynamical system describing the chocolate I have left after each day and determine how much chocolate I have left after two weeks.

**Solution**: Let $c_t$ be the amount of chocolate (in units of a chocolate bar) that I have after $t$ days. The dynamical system that $c_t$ follows is
\begin{align*}
c_{t+1} - c_t &= -0.1 c_t\\
c_0 &= 2.
\end{align*}
In function iteration form, the system is $c_{t+1}= 0.9c_t$, and its solution is
\begin{gather*}
c_t = 0.9^t c_0 = 0.9^t 2.
\end{gather*}

After two weeks, the chocolate I have left is \begin{gather*} c_{14} = 0.9^{14}2 \approx 0.46 \text{ chocolate bars}. \end{gather*}

#### Thread navigation

##### Elementary dynamical systems

- Previous: Penicillin clearance model exercises
- Next: Discrete exponential growth and decay exercises

##### Math 1241, Fall 2020

- Previous: Problem set: Solving linear discrete dynamical systems
- Next: Discrete exponential growth and decay exercises

##### Math 201, Spring 22

#### Similar pages

- Doubling time and half-life of exponential growth and decay
- More details on solving linear discrete dynamical systems
- Discrete exponential growth and decay exercises
- Discrete exponential growth and decay exercise answers
- A model of chemical pollution in a lake
- Chemical pollution model exercises
- Chemical pollution model exercises answers
- Limits of exponential functions at infinity
- Exponential growth and decay: a differential equation
- The idea of a dynamical system
- More similar pages