1.2: Solution to the dynamic equation \(P_{t+1} − P_{t} = r P_{t}\).

The dynamic equation with initial condition, \(P_0\),

\[P_{t+1}-P_{t}=r P_{t}, \quad t=0,1, \cdots, \quad P_{0} \quad \text { a known value } \label{1.6}\]

arises in many models of elementary biological processes. A solution to the dynamic equation \ref{1.6} is a formula for computing \(P_t\) in terms of \(t\) and \(P_0\).

Assume that \(r \neq 0\) and \(P_{0} \neq 0\). The equation \(P_{t+1}-P_{t}=r P_{t}\) can be changed to iteration form by

\[\begin{align}
P_{t+1}-P_{t} &=r P_{t} \\
P_{t+1} &=(r+1) P_{t} \\
P_{t+1} &=R P_{t}
\end{align} \label{1.7}\]

where \(R = r + 1\). The equation \(P_{t+1} = R P_t\) is valid for \(t = 0, t = 1, ... \) and represents a large number of equations, as in

\[\left.\begin{array} \\
P_{1} =R P_{0} && t=0 \\
P_{2} =R P_{1} && t=1 \\
\vdots && \vdots \\
P_{n-1} =R P_{n-2} && t=n-2 \\
P_{n} =R P_{n-1} && t=n-1
\end{array}\right) \label{1.8}\]

where \(n\) can be any stopping value.

Cascading equations. These equations \ref{1.8} may be ‘cascaded’ as follows.

1. The product of all the numbers on the left sides of Equations 1.8 is equal to the product of all of the numbers on the right sides. Therefore \[P_{1} \times P_{2} \times \cdots \times P_{n-1} \times P_{n}=R P_{0} \times R P_{1} \times \cdots \times R P_{n-2} \times R P_{n-1}\]
2. The previous equation may be rearranged to \[P_{1} P_{2} \cdots P_{n-1} P_{n}=R^{n} P_{0} P_{1} \cdots P_{n-2} P_{n-1}\].
3. \(P_{1}, P_{2}, \cdots P_{n-1}\) are factors on both sides of the equation and (assuming no one of them is zero) may be divided from both sides of the equation, leaving \[P_{n}=R^{n} P_{0}\] Because n is arbitrary and the dynamic equation is written with \(t\), we write \[P_{t}=R^{t} P_{0}=P_{0}(1+r)^{t} \label{1.9}\] as the solution to the iteration \[P_{t+1}=R P_{t} \quad \text { with initial value, } P_{0}\] and the solution to \[P_{t+1}-P_{t}=r P_{t} \quad \text{ with initial value, } P_{0}\].

Explore 1.2.1

1. Suppose \(r = 0\) in \(P_{t+1} − P_{t} = r P_{t}\) so that \(P_{t+1} − P_{t} = 0\). What are \(P_{1}, P_{2}, \cdots\)?
2. Suppose \(P_{0} = 0\), and \(P_{t+1} − P_{t} = r P_{t}\) for \(t = 0, 1, 2, \cdots\). What are \(P_{1}, P_{2}, \cdots\)?

Example 1.2.1 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.

1. Option 1. At 1% per year growth, the dynamic equation would be \[P_{t+1}-P_{t}=0.01 P_{t}\] and the corresponding iteration equation is \[P_{t+1}=1.01 P_{t}\] With \(P_{0}=1,000,000, P_{1}=1.01 \times 1,000,000=1,010,000, P_{2}=1.01 \times 1,010,000=1,020,100\) and so on for 50 iterations.
2. Option 2. Alternatively, one may write the solution \[P_{t}=1.01^{t}(1,000,000)\] so that \[P_{50}=1.01^{50}(1,000,000)=1,644,631\] The algebraic form of the solution, \(P_{t}=R^{t} P_{0}\) with \(r>0\) and \(R>1\) is informative and gives rise to the common description of exponential growth attached to some populations. If \(r\) is negative and \(R=1+r<1\), the solution equation \(P_{t}=R^{t} P_{0}\) exhibits exponential decay.

Exercises for Section 1.2, Solution to the dynamic equation, \(P_{t+1}-P_{t}=r P_{t}\)

Exercise 1.2.1 Write a solution equation for the following initial conditions and difference equations or iteration equations. In each case, compute \(B_100\).

1. \(B_{0}=1,000 \quad B_{t+1}-B_{t}=0.2 B_{t}\)
2. \(B_{0}=138 \quad B_{t+1}-B_{t}=0.05 B_{t}\)
3. \(B_{0}=138 \quad B_{t+1}-B_{t}=0.5 B_{t}\)
4. \(B_{0}=1,000 \quad B_{t+1}-B_{t}=-0.2 B_{t}\)
5. \(B_{0}=1,000 \quad B_{t+1}-B_{t}=1.2 B_{t}\)
6. \(B_{0}=1,000 \quad B_{t+1}-B_{t}=-0.1 B_{t}\)
7. \(B_{0}=1,000 \quad B_{t+1}-B_{t}=0.9 B_{t}\)

Exercise 1.2.2 The equation, \(B_{t} − B_{t−1} = r B_{t−1}\), carries the same information as \(B_{t+1} − B_{t} = r B_{t}\).

1. Write the first four instances of \(B_{t}-B_{t-1}=r B_{t-1}\) using \(t=1, t=2, t=3,\) and \(t=4\).
2. Cascade these four equations to get an expression for \(B_{4}\) in terms of \(r\) and \(B_{0}\).
3. Write solutions to and compute \(B_{40}\) for
   1. \(B_{0}=50 \quad B_{t}-B_{t-1}=0.2 B_{t-1}\)
   2. \(B_{0}=50 \quad B_{t}-B_{t-1}=0.1 B_{t-1}\)
   3. \(B_{0}=50 \quad B_{t}-B_{t-1}=0.05 B_{t-1}\)
   4. \(B_{0}=50 \quad B_{t}-B_{t-1}=-0.1 B_{t-1}\)

Exercise 1.2.3 Suppose a population is initially of size 1,000,000 and grows at the rate of 2% per year. What will be the size of the population after 50 years?

Exercise 1.2.4 The polymerase chain reaction is a means of making multiple copies of a DNA segment from only a minute amounts of original DNA. The procedure consists of a sequence of, say, 30 cycles in which each segment present at the beginning of a cycle is duplicated once; at the end of the cycle that segment and one copy is present. Introduce notation and write a difference equation with initial condition from which the amount of DNA present at the end of each cycle can be computed. Suppose you begin with 1 picogram = 0.000000000001 g of DNA. How many grams of DNA would be present after 30 cycles.

Exercise 1.2.5 Write a solution to the dynamic equation you obtained for growth of V. natriegens in growth medium of pH 7.85 in Exercise 1.1.5. Use your solution to compute your estimate of \(B_{4}\).

Exercise 1.2.6 There is a suggestion that the world human population is growing exponentially. Shown below are the human population numbers in billions of people for the decades 1940 - 2010.

Year	1940	1950	1960	1970	1980	1990	2000	2010
Index, t	0	1	2	3	4	5	6	7
Human Population \(\times 10^6\)	2.30	2.52	3.02	3.70	4.45	5.30	6.06	6.80

1. Test the equation \[P_{t}=2.2 (1 .19)^{t}\] against the data where \(t\) is the time index in decades after 1940 and \(P_t\) is the human population in billions.
2. What percentage increase in human population each decade does the model for the equation assume?
3. What world human population does the equation predict for the year 2050?