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2 Mathematical Models and Numerical Methods

2.1 Population Models

In Section 1.4 we introduced the exponential differential equation dP/dt=kP, $d P / d t = k P,$ with solution P(t)=P0ekt, $P (t) = P_{0} e^{k t},$ as a mathematical model for natural population growth that occurs as a result of constant birth and death rates. Here we present a more general population model that accommodates birth and death rates that are not necessarily constant. As before, however, our population function P(t) will be a continuous approximation to the actual population, which of course changes only by integral increments—that is, by one birth or death at a time.

Suppose that the population changes only by the occurrence of births and deaths—there is no immigration or emigration from outside the country or environment under consideration. It is customary to track the growth or decline of a population in terms of its birth rate and death rate functions defined as follows:

β(t) $β (t)$ is the number of births per unit of population per unit of time at time t;
δ(t) $δ (t)$ is the number of deaths per unit of population per unit of time at time t.

Then the numbers of births and deaths that occur during the time interval [t,t+Δt] $[t, t + Δ t]$ is given (approximately) by

births: β (t) \cdot P (t) \cdot Δ t, deaths: δ (t) \cdot P (t) \cdot Δ t .

$births: β (t) \cdot P (t) \cdot Δ t, deaths: δ (t) \cdot P (t) \cdot Δ t .$

Hence the change ΔP $Δ P$ in the population during the time interval [t,t+Δt] $[t, t + Δ t]$ of length Δt $Δ t$ is

Δ P = {births} - {deaths} \approx β (t) \cdot P (t) \cdot Δ t - δ (t) \cdot P (t) \cdot Δ t,

$Δ P = {births} - {deaths} \approx β (t) \cdot P (t) \cdot Δ t - δ (t) \cdot P (t) \cdot Δ t,$

Δ P Δ t \approx [β (t) - δ (t)] P (t) .

$\frac{Δ P}{Δ t} \approx [β (t) - δ (t)] P (t) .$

The error in this approximation should approach zero as Δt→0, $Δ t \to 0,$ so—taking the limit—we get the differential equation

d P d t = (β - δ) P,

$\frac{d P}{d t} = (β - δ) P,$ (1)

in which we write β=β(t), δ=δ(t), $β = β (t), δ = δ (t),$ and P=P(t) $P = P (t)$ for brevity. Equation (1) is the general population equation. If β $β$ and δ $δ$ are constant, Eq. (1) reduces to the natural growth equation with k=β−δ. $k = β - δ .$ But it also includes the possibility that β $β$ and δ $δ$ are variable functions of t. The birth and death rates need not be known in advance; they may well depend on the unknown function P(t).

Example 1

Population explosion Suppose that an alligator population numbers 100 initially, and that its death rate is δ=0 $δ = 0$ (so none of the alligators is dying). If the birth rate is β=(0.0005)P $β = (0.0005) P$ —and thus increases as the population does—then Eq. (1) gives the initial value problem

d P d t = (0.0005) P 2, P (0) = 100

$\frac{d P}{d t} = (0.0005) P^{2}, P (0) = 100$

(with t in years). Then upon separating the variables we get

\int 1 P 2 d P - 1 P = = \int (0.0005) d t; (0.0005) t + C .

$\begin{aligned} \int \frac{1}{P^{2}} d P & = & \int (0.0005) d t; \\ - \frac{1}{P} & = & (0.0005) t + C . \end{aligned}$

Substitution of t=0, P=100 $t = 0, P = 100$ gives C=−1/100, $C = - 1/100,$ and then we readily solve for

P (t) = 2000 20 - t .

$P (t) = \frac{2000}{20 - t} .$

For instance, P(10)=2000/10=200, $P (10) = 2000/10 = 200,$ so after 10 years the alligator population has doubled. But we see that P→+∞ $P \to + \infty$ as t→20, $t \to 20,$ so a real “population explosion” occurs in 20 years. Indeed, the direction field and solution curves shown in Fig. 2.1.1 indicate that a population explosion always occurs, whatever the size of the (positive) initial population P(0)=P0. $P (0) = P_{0} .$ In particular, it appears that the population always becomes unbounded in a finite period of time.

FIGURE 2.1.1.

Slope field and solution curves for the equation dP/dt=(0.0005)P2 $d P / d t = (0.0005) P^{2}$ in Example 1.

Bounded Populations and the Logistic Equation

In situations as diverse as the human population of a nation and a fruit fly population in a closed container, it is often observed that the birth rate decreases as the population itself increases. The reasons may range from increased scientific or cultural sophistication to a limited food supply. Suppose, for example, that the birth rate β $β$ is a linear decreasing function of the population size P, so that β=β0−β1P, $β = β_{0} - β_{1} P,$ where β0 $β_{0}$ and β1 $β_{1}$ are positive constants. If the death rate δ=δ0 $δ = δ_{0}$ remains constant, then Eq. (1) takes the form

d P d t = (β 0 - β 1 P - δ 0) P;

$\frac{d P}{d t} = (β_{0} - β_{1} P - δ_{0}) P;$

that is,

d P d t = a P - b P 2,

$\frac{d P}{d t} = a P - b P^{2},$ (2)

where a=β0−δ0 $a = β_{0} - δ_{0}$ and b=β1 $b = β_{1}$ .

If the coefficients a and b are both positive, then Eq. (2) is called the logistic equation. For the purpose of relating the behavior of the population P(t) to the values of the parameters in the equation, it is useful to rewrite the logistic equation in the form

d P d t = k P (M - P),

$\frac{d P}{d t} = k P (M - P),$ (3)

where k=b $k = b$ and M=a/b $M = a / b$ are constants.

Example 2

Logistic model In Example 4 of Section 1.3 we explored graphically a population that is modeled by the logistic equation

d P d t = 0.0004 P (150 - P) = 0.06 P - 0.0004 P 2 .

$\frac{d P}{d t} = 0.0004 P (150 - P) = 0.06 P - 0.0004 P^{2} .$ (4)

To solve this differential equation symbolically, we separate the variables and integrate. We get

∫dPP(150−P)1150∫(1P+1150−P) dPln |P|−ln |150 − P|P150−P====∫0.0004 dt,∫0.0004 dt0.06t+C,±eCe0.06t=Be0.06t[partial fractions],[where B=±eC].

$\begin{aligned} \int \frac{d P}{P (150 - P)} & = & \int 0.0004 d t, \\ \frac{1}{150} \int (\frac{1}{P} + \frac{1}{150 - P}) d P & = & \int 0.0004 d t & [partial fractions], \\ \ln | P | - \ln | 150 - P | & = & 0.06 t + C, \\ \frac{P}{150 - P} & = & \pm e^{C} e^{0.06 t} = B e^{0.06 t} & [where B = \pm e^{C}] . \end{aligned}$

If we substitute t=0 $t = 0$ and P=P0≠150 $P = P_{0} \neq 150$ into this last equation, we find that B=P0/(150−P0). $B = P_{0} / (150 - P_{0}) .$ Hence

P 150 - P = P 0 e 0.06 t 150 - P 0 .

$\frac{P}{150 - P} = \frac{P_{0} e^{0.06 t}}{150 - P_{0}} .$

Finally, this equation is easy to solve for the population

P (t) = 150 P 0 P 0 + ( 150 - P 0 ) e - 0.06 t

$P (t) = \frac{150 P_{0}}{P_{0} + (150 - P_{0}) e^{- 0.06 t}}$ (5)

at time t in terms of the initial population P0=P(0). $P_{0} = P (0) .$ Figure 2.1.2 shows a number of solution curves corresponding to different values of the initial population ranging from P0=20 $P_{0} = 20$ to P0=300. $P_{0} = 300 .$ Note that all these solution curves appear to approach the horizontal line P=150 $P = 150$ as an asymptote. Indeed, you should be able to see directly from Eq. (5) that limt→∞P(t)=150, $lim_{t \to \infty} P (t) = 150,$ whatever the initial value P0>0 $P_{0} > 0$ .

FIGURE 2.1.2.

Typical solution curves for the logistic equation P'=0.06P−0.0004P2 $P ′ = 0.06 P - 0.0004 P^{2}$ .

Limiting Populations and Carrying Capacity

The finite limiting population noted in Example 2 is characteristic of logistic populations. In Problem 32 we ask you to use the method of solution of Example 2 to show that the solution of the logistic initial value problem

d P d t = k P (M - P), P (0) = P 0

$\frac{d P}{d t} = k P (M - P), P (0) = P_{0}$ (6)

P (t) = M P 0 P 0 + ( M - P 0 ) e - k M t .

$P (t) = \frac{M P_{0}}{P_{0} + (M - P_{0}) e^{- k M t}} .$ (7)

Actual animal populations are positive valued. If P0=M, $P_{0} = M,$ then (7) reduces to the unchanging (constant-valued) “equilibrium population” P(t)≡M. $P (t) \equiv M .$ Otherwise, the behavior of a logistic population depends on whether 0<P0<M $0 < P_{0} < M$ or P0>M. $P_{0} > M .$ If 0<P0<M, $0 < P_{0} < M,$ then we see from (6) and (7) that P'>0 $P ′ > 0$ and

P (t) = M P 0 P 0 + ( M - P 0 ) e - k M t = M P 0 P 0 + { pos . number } < M P 0 P 0 = M .

$P (t) = \frac{M P_{0}}{P_{0} + (M - P_{0}) e^{- k M t}} = \frac{M P_{0}}{P_{0} + {pos . number}} < \frac{M P_{0}}{P_{0}} = M .$

However, if P0>M, $P_{0} > M,$ then we see from (6) and (7) that P'<0 $P ′ < 0$ and

P (t) = M P 0 P 0 + ( M - P 0 ) e - kMt = M P 0 P 0 + { neg. number } > M P 0 P 0 = M .

$P (t) = \frac{{M P}_{0}}{P_{0} + (M - P_{0}) e^{- kMt}} = \frac{{M P}_{0}}{P_{0} + {neg. number}} > \frac{{M P}_{0}}{P_{0}} = M .$

In either case, the “positive number” or “negative number” in the denominator has absolute value less than P0 $P_{0}$ and—because of the exponential factor—approaches 0 as t→+∞. $t \to + \infty .$ It follows that

lim t \to + \infty P (t) = M P 0 P 0 + 0 = M .

$\lim_{t \to + \infty} P (t) = \frac{M P_{0}}{P_{0} + 0} = M .$ (8)

Thus a population that satisfies the logistic equation does not grow without bound like a naturally growing population modeled by the exponential equation P'=kP. $P ′ = k P .$ Instead, it approaches the finite limiting population M as t→+∞. $t \to + \infty .$ As illustrated by the typical logistic solution curves in Fig. 2.1.3, the population P(t) steadily increases and approaches M from below if 0<P0<M, $0 < P_{0} < M,$ but steadily decreases and approaches M from above if P0>M. $P_{0} > M .$ Sometimes M is called the carrying capacity of the environment, considering it to be the maximum population that the environment can support on a long-term basis.

FIGURE 2.1.3.

Typical solution curves for the logistic equation P'=kP(M−P). $P ′ = k P (M - P) .$ Each solution curve that starts below the line P=M/2 $P = M / 2$ has an inflection point on this line. (See Problem 34.)

Example 3

Limiting population Suppose that in 1885 the population of a certain country was 50 million and was growing at the rate of 750,000 people per year at that time. Suppose also that in 1940 its population was 100 million and was then growing at the rate of 1 million per year. Assume that this population satisfies the logistic equation. Determine both the limiting population M and the predicted population for the year 2000.

Solution

We substitute the two given pairs of data in Eq. (3) and find that

0.75 = 50 k (M - 50), 1.00 = 100 k (M - 100) .

$0.75 = 50 k (M - 50), 1.00 = 100 k (M - 100) .$

We solve simultaneously for M=200 $M = 200$ and k=0.0001. $k = 0.0001 .$ Thus the limiting population of the country in question is 200 million. With these values of M and k, and with t=0 $t = 0$ corresponding to the year 1940 (in which P0=100 $P_{0} = 100$ ), we find that—according to Eq. (7)—the population in the year 2000 will be

P (60) = 100 \cdot 200 100 + ( 200 - 100 ) e - ( 0.0001 ) ( 200 ) ( 60 ),

$P (60) = \frac{100 \cdot 200}{100 + (200 - 100) e^{- (0.0001) (200) (60)}},$

about 153.7 $153.7$ million people.

Historical Note

The logistic equation was introduced (around 1840) by the Belgian mathematician and demographer P. F. Verhulst as a possible model for human population growth. In the next two examples we compare natural growth and logistic model fits to the 19th-century U.S. population census data, then compare projections for the 20th century.

Example 4

Percentage growth rate The U.S. population in 1800 was 5.308 million and in 1900 was 76.212 million. If we take P0=5.308 $P_{0} = 5.308$ (with t=0 $t = 0$ in 1800) in the natural growth model P(t)=P0ert $P (t) = P_{0} e^{r t}$ and substitute t=100, P=76.212, $t = 100, P = 76.212,$ we find that

76.212 = 5.308 e 100 r, so r = 1 100 ln 76.212 5.308 \approx 0.026643.

$76.212 = 5.308 e^{100 r}, so r = \frac{1}{100} \ln \frac{76.212}{5.308} \approx 0.026643.$

Thus our natural growth model for the U.S. population during the 19th century is

P (t) = (5.308) e (0.026643) t

$P (t) = (5.308) e^{(0.026643) t}$ (9)

(with t in years and P in millions). Because e0.026643≈1.02700, $e^{0.026643} \approx 1.02700,$ the average population growth between 1800 and 1900 was about 2.7% per year.

Example 5

Logistic modeling The U.S. population in 1850 was 23.192 million. If we take P0=5.308 $P_{0} = 5.308$ and substitute the data pairs t=50, P=23.192 $t = 50, P = 23.192$ (for 1850) and t=100, P=76.212 $t = 100, P = 76.212$ (for 1900) in the logistic model formula in Eq. (7), we get the two equations

( 5.308 ) M 5.308 + ( M - 5.308 ) e - 50 k M ( 5.308 ) M 5.308 + ( M - 5.308 ) e - 100 k M = = 23.192, 76.212

$\begin{aligned} \frac{(5.308) M}{5.308 + (M - 5.308) e^{- 50 k M}} & = & 23.192, \\ \frac{(5.308) M}{5.308 + (M - 5.308) e^{- 100 k M}} & = & 76.212 \end{aligned}$ (10)

in the two unknowns k and M. Nonlinear systems like this ordinarily are solved numerically using an appropriate computer system. But with the right algebraic trick (Problem 36 in this section) the equations in (10) can be solved manually for k=0.000167716, M=188.121. $k = 0.000167716, M = 188.121 .$ Substitution of these values in Eq. (7) yields the logistic model

P (t) = 998.546 5.308 + ( 182.813 ) e - ( 0.031551 ) t .

$P (t) = \frac{998.546}{5.308 + (182.813) e^{- (0.031551) t}} .$ (11)

The table in Fig. 2.1.4 compares the actual 1800–2010 U.S. census population figures with those predicted by the exponential growth model in (9) and by the logistic model in (11). Both models agree well with the 19th-century figures. But the exponential model diverges appreciably from the census data in the early decades of the 20th century, whereas the logistic model remains accurate until 1940. By the end of the 20th century the exponential model vastly overestimates the actual U.S. population; indeed it predicts a U.S. population of nearly 1.5 billion in the year 2010, over 3.6 times the actual value. The logistic model, on the other hand, underestimates the U.S. population, but with a percentage error of less than 43%.

FIGURE 2.1.4.

Comparison of exponential growth and logistic models with U.S. census populations (in millions).

Year	Actual U.S. Pop.	Exponential Model	Exponential Error	Logistic Model	Logistic Error
1800	5.308	5.308	0.000	5.308	0.000
1810	7.240	6.929	0.311	7.202	0.038
1820	9.638	9.044	0.594	9.735	−0.097 $- 0.097$
1830	12.861	11.805	1.056	13.095	−0.234 $- 0.234$
1840	17.064	15.409	1.655	17.501	−0.437 $- 0.437$
1850	23.192	20.113	3.079	23.192	0.000
1860	31.443	26.253	5.190	30.405	1.038
1870	38.558	34.268	4.290	39.326	−0.768 $- 0.768$
1880	50.189	44.730	5.459	50.034	0.155
1890	62.980	58.387	4.593	62.435	0.545
1900	76.212	76.212	0.000	76.213	−0.001 $- 0.001$
1910	92.228	99.479	−7.251 $- 7.251$	90.834	1.394
1920	106.022	129.849	−23.827 $- 23.827$	105.612	0.410
1930	123.203	169.492	−46.289 $- 46.289$	119.834	3.369
1940	132.165	221.237	−89.072 $- 89.072$	132.886	−0.721 $- 0.721$
1950	151.326	288.780	−137.454 $- 137.454$	144.354	6.972
1960	179.323	376.943	−197.620 $- 197.620$	154.052	25.271
1970	203.302	492.023	−288.721 $- 288.721$	161.990	41.312
1980	226.542	642.236	−415.694 $- 415.694$	168.316	58.226
1990	248.710	838.308	−589.598 $- 589.598$	173.252	75.458
2000	281.422	1094.240	−812.818 $- 812.818$	177.038	104.384
2010	308.745	1428.307	−1119.562 $- 1119.562$	179.905	128.839

The two models are compared in Fig. 2.1.5, where plots of their respective errors—as a percentage of the actual population—are shown for the 1800–1950 period. We see that the logistic model tracks the actual population reasonably well throughout this 150-year period. However, the exponential error is considerably larger during the 19th century and literally goes off the chart during the first half of the 20th century.

FIGURE 2.1.5.

Percentage errors in the exponential and logistic population models for 1800–1950.

In order to measure the extent to which a given model fits actual data, it is customary to define the average error (in the model) as the square root of the average of the squares of the individual errors (the latter appearing in the fourth and sixth columns of the table in Fig. 2.1.4). Using only the 1800–1900 data, this definition gives 3.162 for the average error in the exponential model, while the average error in the logistic model is only 0.452. Consequently, even in 1900 we might well have anticipated that the logistic model would predict the U.S. population growth during the 20th century more accurately than the exponential model.

The moral of Examples 4 and 5 is simply that one should not expect too much of models that are based on severely limited information (such as just a pair of data points). Much of the science of statistics is devoted to the analysis of large “data sets” to formulate useful (and perhaps reliable) mathematical models.

More Applications of the Logistic Equation

We next describe some situations that illustrate the varied circumstances in which the logistic equation is a satisfactory mathematical model.

Limited environment situation. A certain environment can support a population of at most M individuals. It is then reasonable to expect the growth rate β−δ $β - δ$ (the combined birth and death rates) to be proportional to M−P, $M - P,$ because we may think of M−P $M - P$ as the potential for further expansion. Then β−δ=k(M−P), $β - δ = k (M - P),$ so that

$d P d t = (β - δ) P = k P (M - P) .$ $\frac{d P}{d t} = (β - δ) P = k P (M - P) .$

The classic example of a limited environment situation is a fruit fly population in a closed container.
Competition situation. If the birth rate β $β$ is constant but the death rate δ $δ$ is proportional to P, so that δ=αP, $δ = α P,$ then

$d P d t = (β - α P) P = k P (M - P) .$ $\frac{d P}{d t} = (β - α P) P = k P (M - P) .$

This might be a reasonable working hypothesis in a study of a cannibalistic population, in which all deaths result from chance encounters between individuals. Of course, competition between individuals is not usually so deadly, nor its effects so immediate and decisive.
Joint proportion situation. Let P(t) denote the number of individuals in a constant-size susceptible population M who are infected with a certain contagious and incurable disease. The disease is spread by chance encounters. Then P'(t) $P ′ (t)$ should be proportional to the product of the number P of individuals having the disease and the number M−P $M - P$ of those not having it, and therefore dP/dt=kP(M−P). $d P / d t = k P (M - P) .$ Again we discover that the mathematical model is the logistic equation. The mathematical description of the spread of a rumor in a population of M individuals is identical.

Example 6

Spread of rumor Suppose that at time t=0, 10 $t = 0, 10$ thousand people in a city with population M=100 $M = 100$ thousand people have heard a certain rumor. After 1 week the number P(t) of those who have heard it has increased to P(1)=20 $P (1) = 20$ thousand. Assuming that P(t) satisfies a logistic equation, when will 80% of the city’s population have heard the rumor?

Solution

Substituting P0=10 $P_{0} = 10$ and M=100 $M = 100$ (thousand) in Eq. (7), we get

P (t) = 1000 10 + 90 e - 100 k t .

$P (t) = \frac{1000}{10 + 90 e^{- 100 k t}} .$ (12)

Then substitution of t=1, P=20 $t = 1, P = 20$ gives the equation

20 = 1000 10 + 90 e - 100 k

$20 = \frac{1000}{10 + 90 e^{- 100 k}}$

that is readily solved for

e - 100 k = 4 9, so k = 1 100 ln 9 4 \approx 0.008109.

$e^{- 100 k} = \frac{4}{9}, so k = \frac{1}{100} \ln \frac{9}{4} \approx 0.008109.$

With P(t)=80, $P (t) = 80,$ Eq. (12) takes the form

80 = 1000 10 + 90 e - 100 k t,

$80 = \frac{1000}{10 + 90 e^{- 100 k t}},$

which we solve for e−100kt=136. $e^{- 100 k t} = \frac{1}{36} .$ It follows that 80% of the population has heard the rumor when

t = ln 36 100 k = ln 36 ln 9 4 \approx 4.42,

$t = \frac{\ln 36}{100 k} = \frac{\ln 36}{\ln \frac{9}{4}} \approx 4.42,$

thus after about 4 weeks and 3 days.

Doomsday versus Extinction

Consider a population P(t) of unsophisticated animals in which females rely solely on chance encounters to meet males for reproductive purposes. It is reasonable to expect such encounters to occur at a rate that is proportional to the product of the number P/2 of males and the number P/2 of females, hence at a rate proportional to P2. $P^{2} .$ We therefore assume that births occur at the rate kP2 $k P^{2}$ (per unit time, with k constant). The birth rate (births/time/population) is then given by β=kP. $β = k P .$ If the death rate δ $δ$ is constant, then the general population equation in (1) yields the differential equation

d P d t = k P 2 - δ P = k P (P - M)

$\frac{d P}{d t} = k P^{2} - δ P = k P (P - M)$ (13)

(where M=δ/k>0 $M = δ / k > 0$ ) as a mathematical model of the population.

Note that the right-hand side in Eq. (13) is the negative of the right-hand side in the logistic equation in (3). We will see that the constant M is now a threshold population, with the way the population behaves in the future depending critically on whether the initial population P0 $P_{0}$ is less than or greater than M.

Example 7

Doomsday vs. extinction Consider an animal population P(t) that is modeled by the equation

d P d t = 0.0004 P (P - 150) = 0.0004 P 2 - 0.06 P .

$\frac{d P}{d t} = 0.0004 P (P - 150) = 0.0004 P^{2} - 0.06 P .$ (14)

We want to find P(t) if (a) P(0)=200; $P (0) = 200;$ (b) P(0)=100 $P (0) = 100$ .

Solution

To solve the equation in (14), we separate the variables and integrate. We get

∫dPP(P−150)−1150∫(1P−1P−150)dPln |P|−ln |P − 150|PP−150=±eCe−0.06t====∫0.0004 dt,∫0.0004 dt−0.06t+C,Be−0.06t[partial fractions],[where B=±eC].

$\begin{aligned} \int \frac{d P}{P (P - 150)} & = & \int 0.0004 d t, \\ - \frac{1}{150} \int (\frac{1}{P} - \frac{1}{P - 150}) d P & = & \int 0.0004 d t & [partial fractions], \\ \ln | P | - \ln | P - 150 | & = & - 0.06 t + C, \\ \frac{P}{P - 150} = \pm e^{C} e^{- 0.06 t} & = & B e^{- 0.06 t} & [where B = \pm e^{C}] . \end{aligned}$ (15)

Substitution of t=0 $t = 0$ and P=200 $P = 200$ into (15) gives B=4. $B = 4 .$ With this value of B we solve Eq. (15) for

$P (t) = 600 e - 0.06 t 4 e - 0.06 t - 1 .$ $P (t) = \frac{600 e^{- 0.06 t}}{4 e^{- 0.06 t} - 1} .$ (16)

Note that, as t increases and approaches T=ln (4)/0.06≈23.105, $T = \ln (4) / 0.06 \approx 23.105,$ the positive denominator on the right in (16) decreases and approaches 0. Consequently P(t)→+∞ $P (t) \to + \infty$ as t→T−. $t \to T^{-} .$ This is a doomsday situation—a real population explosion.
Substitution of t=0 $t = 0$ and P=100 $P = 100$ into (15) gives B=−2. $B = - 2 .$ With this value of B we solve Eq. (15) for

$P (t) = 300 e - 0.06 t 2 e - 0.06 t + 1 = 300 2 + e 0.06 t .$ $P (t) = \frac{300 e^{- 0.06 t}}{2 e^{- 0.06 t} + 1} = \frac{300}{2 + e^{0.06 t}} .$ (17)

Note that, as t increases without bound, the positive denominator on the right in (16) approaches +∞. $+ \infty .$ Consequently, P(t)→0 $P (t) \to 0$ as t→+∞. $t \to + \infty .$ This is an (eventual) extinction situation.

Thus the population in Example 7 either explodes or is an endangered species threatened with extinction, depending on whether or not its initial size exceeds the threshold population M=150. $M = 150 .$ An approximation to this phenomenon is sometimes observed with animal populations, such as the alligator population in certain areas of the southern United States.

Figure 2.1.6 shows typical solution curves that illustrate the two possibilities for a population P(t) satisfying Eq. (13). If P0=M $P_{0} = M$ (exactly!), then the population remains constant. However, this equilibrium situation is very unstable. If P0 $P_{0}$ exceeds M (even slightly), then P(t) rapidly increases without bound, whereas if the initial (positive) population is less than M (however slightly), then it decreases (more gradually) toward zero as t→+∞. $t \to + \infty .$ See Problem 33.

FIGURE 2.1.6.

Typical solution curves for the explosion/extinction equation P'=kP(P−M) $P ′ = k P (P - M)$ .

2.1 Problems

Separate variables and use partial fractions to solve the initial value problems in Problems 1–8. Use either the exact solution or a computer-generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution.

dxdt=x−x2, x(0)=2 $\frac{d x}{d t} = x - x^{2}, x (0) = 2$
dxdt=10x−x2, x(0)=1 $\frac{d x}{d t} = 10 x - x^{2}, x (0) = 1$
dxdt=1−x2, x(0)=3 $\frac{d x}{d t} = 1 - x^{2}, x (0) = 3$
dxdt=9−4x2, x(0)=0 $\frac{d x}{d t} = 9 - 4 x^{2}, x (0) = 0$
dxdt=3x(5−x), x(0)=8 $\frac{d x}{d t} = 3 x (5 - x), x (0) = 8$
dxdt=3x(x−5), $\frac{d x}{d t} = 3 x (x - 5),$ x(0)=2 $x (0) = 2$
dxdt=4x(7−x), x(0)=11 $\frac{d x}{d t} = 4 x (7 - x), x (0) = 11$
dxdt=7x(x−13), $\frac{d x}{d t} = 7 x (x - 13),$ x(0)=17 $x (0) = 17$
Population growth The time rate of change of a rabbit population P is proportional to the square root of P. At time t=0 $t = 0$ (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many rabbits will there be one year later?
Extinction by disease Suppose that the fish population P(t) in a lake is attacked by a disease at time t=0, $t = 0,$ with the result that the fish cease to reproduce (so that the birth rate is β=0 $β = 0$ ) and the death rate δ $δ$ (deaths per week per fish) is thereafter proportional to 1/P−−√. $1 / \sqrt{P} .$ If there were initially 900 fish in the lake and 441 were left after 6 weeks, how long did it take all the fish in the lake to die?
Fish population Suppose that when a certain lake is stocked with fish, the birth and death rates β $β$ and δ $δ$ are both inversely proportional to P−−√. $\sqrt{P} .$ (a) Show that

$P (t) = (1 2 k t + P 0 - - \sqrt) 2,$ $P (t) = {(\frac{1}{2} k t + \sqrt{P_{0}})}^{2},$

where k is a constant. (b) If P0=100 $P_{0} = 100$ and after 6 months there are 169 fish in the lake, how many will there be after 1 year?
Population growth The time rate of change of an alligator population P in a swamp is proportional to the square of P. The swamp contained a dozen alligators in 1988, two dozen in 1998. When will there be four dozen alligators in the swamp? What happens thereafter?
Birth rate exceeds death rate Consider a prolific breed of rabbits whose birth and death rates, β $β$ and δ, $δ,$ are each proportional to the rabbit population P=P(t), $P = P (t),$ with β>δ. $β > δ .$ (a) Show that

$P (t) = P 0 1 - k P 0 t, k constant .$ $P (t) = \frac{P_{0}}{1 - k P_{0} t}, k constant .$

Note that P(t)→+∞ $P (t) \to + \infty$ as t→1/(kP0). $t \to 1 / (k P_{0}) .$ This is doomsday. (b) Suppose that P0=6 $P_{0} = 6$ and that there are nine rabbits after ten months. When does doomsday occur?
Death rate exceeds birth rate Repeat part (a) of Problem 13 in the case β<δ. $β < δ .$ What now happens to the rabbit population in the long run?
Limiting population Consider a population P(t) satisfying the logistic equation dP/dt=aP−bP2, $d P / d t = a P - b P^{2},$ where B=aP $B = a P$ is the time rate at which births occur and D=bP2 $D = b P^{2}$ is the rate at which deaths occur. If the initial population is P(0)=P0, $P (0) = P_{0},$ and B0 $B_{0}$ births per month and D0 $D_{0}$ deaths per month are occurring at time t=0, $t = 0,$ show that the limiting population is M=B0P0/D0 $M = B_{0} P_{0} / D_{0}$ .
Limiting population Consider a rabbit population P(t) satisfying the logistic equation as in Problem 15. If the initial population is 120 rabbits and there are 8 births per month and 6 deaths per month occurring at time t=0, $t = 0,$ how many months does it take for P(t) to reach 95% of the limiting population M?
Limiting population Consider a rabbit population P(t) satisfying the logistic equation as in Problem 15. If the initial population is 240 rabbits and there are 9 births per month and 12 deaths per month occurring at time t=0, $t = 0,$ how many months does it take for P(t) to reach 105% of the limiting population M?
Threshold population Consider a population P(t) satisfying the extinction-explosion equation dP/dt=aP2−bP, $d P / d t = a P^{2} - b P,$ where B=aP2 $B = {a P}^{2}$ is the time rate at which births occur and D=bP $D = b P$ is the rate at which deaths occur. If the initial population is P(0)=P0 $P (0) = P_{0}$ and B0 $B_{0}$ births per month and D0 $D_{0}$ deaths per month are occurring at time t=0, $t = 0,$ show that the threshold population is M=D0P0/B0 $M = D_{0} P_{0} / B_{0}$ .
Threshold population Consider an alligator population P(t) satisfying the extinction-explosion equation as in Problem 18. If the initial population is 100 alligators and there are 10 births per month and 9 deaths per month occurring at time t=0, $t = 0,$ how many months does it take for P(t) to reach 10 times the threshold population M?
Threshold population Consider an alligator population P(t) satisfying the extinction-explosion equation as in Problem 18. If the initial population is 110 alligators and there are 11 births per month and 12 deaths per month occurring at time t=0, $t = 0,$ how many months does it take for P(t) to reach 10% of the threshold population M?
Logistic model Suppose that the population P(t) of a country satisfies the differential equation dP/dt=kP(200−P) $d P / d t = k P (200 - P)$ with k constant. Its population in 1960 was 100 million and was then growing at the rate of 1 million per year. Predict this country’s population for the year 2020.
Logistic model Suppose that at time t=0, $t = 0,$ half of a “logistic” population of 100,000 persons have heard a certain rumor, and that the number of those who have heard it is then increasing at the rate of 1000 persons per day. How long will it take for this rumor to spread to 80% of the population? (Suggestion: Find the value of k by substituting P(0) and P'(0) $P ′ (0)$ in the logistic equation, Eq. (3).)
Solution rate As the salt KNO3 ${KNO}_{3}$ dissolves in methanol, the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation dx/dt=0.8x−0.004x2 $d x / d t = 0.8 x - 0.004 x^{2}$ .
1. What is the maximum amount of the salt that will ever dissolve in the methanol?
2. If x=50 $x = 50$ when t=0, $t = 0,$ how long will it take for an additional 50 g of salt to dissolve?
Spread of disease Suppose that a community contains 15,000 people who are susceptible to Michaud’s syndrome, a contagious disease. At time t=0 $t = 0$ the number N(t) of people who have developed Michaud’s syndrome is 5000 and is increasing at the rate of 500 per day. Assume that N'(t) $N ′ (t)$ is proportional to the product of the numbers of those who have caught the disease and of those who have not. How long will it take for another 5000 people to develop Michaud’s syndrome?

Logistic model The data in the table in Fig. 2.1.7 are given for a certain population P(t) that satisfies the logistic equation in (3). (a) What is the limiting population M? (Suggestion: Use the approximation

P' (t) \approx P ( t + h ) - P ( t - h ) 2 h

$P ′ (t) \approx \frac{P (t + h) - P (t - h)}{2 h}$

with h=1 $h = 1$ to estimate the values of P'(t) $P ′ (t)$ when P=25.00 $P = 25.00$ and when P=47.54. $P = 47.54 .$ Then substitute these values in the logistic equation and solve for k and M.) (b) Use the values of k and M found in part (a) to determine when P=75. $P = 75 .$ (Suggestion: Take t=0 $t = 0$ to correspond to the year 1965.)

FIGURE 2.1.7.

Population data for Problem 25.

Year	`P` (millions)
1964	24.63
1965	25.00
1966	25.38
⋮ $⋮$	⋮ $⋮$
2014	47.04
2015	47.54
2016	48.04

Constant death rate A population P(t) of small rodents has birth rate β=(0.001)P $β = (0.001) P$ (births per month per rodent) and constant death rate δ. $δ .$ If P(0)=100 $P (0) = 100$ and P'(0)=8, $P ′ (0) = 8,$ how long (in months) will it take this population to double to 200 rodents? (Suggestion: First find the value of δ $δ$ .)
Constant death rate Consider an animal population P(t) with constant death rate δ=0.01 $δ = 0.01$ (deaths per animal per month) and with birth rate β $β$ proportional to P. Suppose that P(0)=200 $P (0) = 200$ and P'(0)=2. $P ′ (0) = 2 .$ (a) When is P=1000? $P = 1000 ?$ (b) When does doomsday occur?
Population growth Suppose that the number x(t) (with t in months) of alligators in a swamp satisfies the differential equation dx/dt=0.0001x2−0.01x $d x / d t = 0.0001 x^{2} - 0.01 x$ .
1. If initially there are 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.
2. Repeat part (a), except with 150 alligators initially.
Logistic model During the period from 1790 to 1930, the U.S. population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this period, P(t) remained close to the solution of the initial value problem

$d P d t = 0.03135 P - 0.0001489 P 2, P (0) = 3.9.$ $\frac{d P}{d t} = 0.03135 P - 0.0001489 P^{2}, P (0) = 3.9.$
1. What 1930 population does this logistic equation predict?
2. What limiting population does it predict?
3. Has this logistic equation continued since 1930 to accurately model the U.S. population?
[This problem is based on a computation by Verhulst, who in 1845 used the 1790–1840 U.S. population data to predict accurately the U.S. population through the year 1930 (long after his own death, of course).]
Tumor growth A tumor may be regarded as a population of multiplying cells. It is found empirically that the “birth rate” of the cells in a tumor decreases exponentially with time, so that β(t)=β0e−αt $β (t) = β_{0} e^{- α t}$ (where α $α$ and β0 $β_{0}$ are positive constants), and hence

$d P d t = β 0 e - α t P, P (0) = P 0 .$ $\frac{d P}{d t} = β_{0} e^{- α t} P, P (0) = P_{0} .$

Solve this initial value problem for

$P (t) = P 0 exp (β 0 α (1 - e - α t)) .$ $P (t) = P_{0} \exp (\frac{β_{0}}{α} (1 - e^{- α t})) .$

Observe that P(t) approaches the finite limiting population P0exp(β0/α) $P_{0} \exp (β_{0} / α)$ as t→+∞ $t \to + \infty$ .
Tumor growth For the tumor of Problem 30, suppose that at time t=0 $t = 0$ there are P0=106 $P_{0} = 10^{6}$ cells and that P(t) is then increasing at the rate of 3×105 $3 \times 10^{5}$ cells per month. After 6 months the tumor has doubled (in size and in number of cells). Solve numerically for α, $α,$ and then find the limiting population of the tumor.
Derive the solution

$P (t) = M P 0 P 0 + ( M - P 0 ) e - k M t$ $P (t) = \frac{M P_{0}}{P_{0} + (M - P_{0}) e^{- k M t}}$

of the logistic initial value problem P'=kP(M−P), P(0)=P0. $P ′ = k P (M - P), P (0) = P_{0} .$ Make it clear how your derivation depends on whether 0<P0<M $0 < P_{0} < M$ or P0>M $P_{0} > M$ .
1. Derive the solution
  
  $P (t) = M P 0 P 0 + ( M - P 0 ) e k M t$ $P (t) = \frac{M P_{0}}{P_{0} + (M - P_{0}) e^{k M t}}$
  
  of the extinction-explosion initial value problem P'=kP(P−M), P(0)=P0 $P ′ = k P (P - M), P (0) = P_{0}$ .
2. How does the behavior of P(t) as t increases depend on whether 0<P0<M $0 < P_{0} < M$ or P0>M $P_{0} > M$ ?
If P(t) satisfies the logistic equation in (3), use the chain rule to show that

$P'' (t) = 2 k 2 P (P - 1 2 M) (P - M) .$ $P ″ (t) = 2 k^{2} P (P - \frac{1}{2} M) (P - M) .$

Conclude that P''>0 $P ″ > 0$ if 0<P<12M; $0 < P < \frac{1}{2} M;$ P''=0 $P ″ = 0$ if P=12M; $P = \frac{1}{2} M;$ P''<0 $P ″ < 0$ if 12M<P<M; $\frac{1}{2} M < P < M;$ and P''>0 $P ″ > 0$ if P>M. $P > M .$ In particular, it follows that any solution curve that crosses the line P=12M $P = \frac{1}{2} M$ has an inflection point where it crosses that line, and therefore resembles one of the lower S-shaped curves in Fig. 2.1.3.
Approach to limiting population Consider two population functions P1(t) $P_{1} (t)$ and P2(t), $P_{2} (t),$ both of which satisfy the logistic equation with the same limiting population M but with different values k1 $k_{1}$ and k2 $k_{2}$ of the constant k in Eq. (3). Assume that k1<k2. $k_{1} < k_{2} .$ Which population approaches M the most rapidly? You can reason geometrically by examining slope fields (especially if appropriate software is available), symbolically by analyzing the solution given in Eq. (7), or numerically by substituting successive values of t.
Logistic modeling To solve the two equations in (10) for the values of k and M, begin by solving the first equation for the quantity x=e−50kM $x = e^{- 50 k M}$ and the second equation for x2=e−100kM. $x^{2} = e^{- 100 k M} .$ Upon equating the two resulting expressions for x2 $x^{2}$ in terms of M, you get an equation that is readily solved for M. With M now known, either of the original equations is readily solved for k. This technique can be used to “fit” the logistic equation to any three population values P0, P1, $P_{0}, P_{1},$ and P2 $P_{2}$ corresponding to equally spaced times t0=0, t1, $t_{0} = 0, t_{1},$ and t2=2t1 $t_{2} = 2 t_{1}$ .
Logistic modeling Use the method of Problem 36 to fit the logistic equation to the actual U.S. population data (Fig. 2.1.4) for the years 1850, 1900, and 1950. Solve the resulting logistic equation and compare the predicted and actual populations for the years 1990 and 2000.
Logistic modeling Fit the logistic equation to the actual U.S. population data (Fig. 2.1.4) for the years 1900, 1930, and 1960. Solve the resulting logistic equation, then compare the predicted and actual populations for the years 1980, 1990, and 2000.
Periodic growth rate Birth and death rates of animal populations typically are not constant; instead, they vary periodically with the passage of seasons. Find P(t) if the population P satisfies the differential equation

$d P d t = (k + b cos 2 π t) P,$ $\frac{d P}{d t} = (k + b \cos 2 π t) P,$

where t is in years and k and b are positive constants. Thus the growth-rate function r(t)=k+bcos 2πt $r (t) = k + b \cos 2 π t$ varies periodically about its mean value k. Construct a graph that contrasts the growth of this population with one that has the same initial value P0 $P_{0}$ but satisfies the natural growth equation P'=kP $P ′ = k P$ (same constant k). How would the two populations compare after the passage of many years?

2.1 Application Logistic Modeling of Population Data

These investigations deal with the problem of fitting a logistic model to given population data. Thus we want to determine the numerical constants a and b so that the solution P(t) of the initial value problem

d P d t = a P + b P 2, P (0) = P 0

$\frac{d P}{d t} = a P + b P^{2}, P (0) = P_{0}$ (1)

approximates the given values P0, P1, …, Pn $P_{0}, P_{1}, \dots, P_{n}$ of the population at the times t0=0, t1, …, tn. $t_{0} = 0, t_{1}, \dots, t_{n} .$ If we rewrite Eq. (1) (the logistic equation with kM=a $k M = a$ and k=−b $k = - b$ ) in the form

1 P d P d t = a + b P,

$\frac{1}{P} \frac{d P}{d t} = a + b P,$ (2)

then we see that the points

(P (t i), P ' ( t i ) P ( t i )), i = 0, 1, 2, \dots, n,

$(P (t_{i}), \frac{P ′ (t_{i})}{P (t_{i})}), i = 0, 1, 2, \dots, n,$

should all lie on the straight line with y-intercept a and slope b (as determined by the function of P on the right-hand side in Eq. (2)).

This observation provides a way to find a and b. If we can determine the approximate values of the derivatives P1',P2',… $P_{1} ′, P_{2} ′, \dots$ corresponding to the given population data, then we can proceed with the following agenda:

First plot the points (P1,P1'/P1), (P2,P2'/P2), … $(P_{1}, P_{1} ′ / P_{1}), (P_{2}, P_{2} ′ / P_{2}), \dots$ on a sheet of graph paper with horizontal P-axis.
Then use a ruler to draw a straight line that appears to approximate these points well.
Finally, measure this straight line’s y-intercept a and slope b.

But where are we to find the needed values of the derivative P'(t) $P ′ (t)$ of the (as yet) unknown function P? It is easiest to use the approximation

P i' = P i + 1 - P i - 1 t i + 1 - t i - 1

$P_{i} ′ = \frac{P_{i + 1} - P_{i - 1}}{t_{i + 1} - t_{i - 1}}$ (3)

suggested by Fig. 2.1.8. For instance, if we take i=0 $i = 0$ corresponding to the year 1790, then the U.S. population data in Fig. 2.1.9 give

P 1' = P 2 - P 0 t 2 - t 0 = 7.240 - 3.929 20 \approx 0.166

$P_{1} ′ = \frac{P_{2} - P_{0}}{t_{2} - t_{0}} = \frac{7.240 - 3.929}{20} \approx 0.166$

for the slope at (t1,P1) $(t_{1}, P_{1})$ corresponding to the year 1800.

FIGURE 2.1.8.

The symmetric difference approximation Pi + 1−Pi − 1ti + 1−ti − 1 $\frac{P_{i + 1} - P_{i - 1}}{t_{i + 1} - t_{i - 1}}$ to the derivative P'(ti). $P ′ (t_{i}) .$

Investigation A: Use Eq. (3) to verify the slope figures shown in the final column of the table in Fig. 2.1.9, then plot the points (P1,P1'/P1), …, (P11,P11'/P11) $(P_{1}, P_{1} ′ / P_{1}), \dots, (P_{11}, P_{11} ′ / P_{11})$ indicated by the dots in Fig. 2.1.10. If an appropriate graphing calculator, spreadsheet, or computer program is available, use it to find the straight line y=a+bP $y = a + b P$ as in (2) that best fits these points. If not, draw your own straight line approximating these points, and then measure its intercept a and slope b as accurately as you can. Next, solve the logistic equation in (1) with these numerical parameters, taking t=0 $t = 0$ corresponding to the year 1800. Finally, compare the predicted 20th- and 21st-century U.S. population figures with the actual data listed in Fig. 2.1.4.

FIGURE 2.1.9.

U.S. population data (in millions) and approximate slopes.

Year	`i`	$t_{i}$	Population $P_{i}$	Slope $P_{i}'$
1790	0	$- 10$	3.929
1800	1	0	5.308	0.166
1810	2	10	7.240	0.217
1820	3	20	9.638	0.281
1830	4	30	12.861	0.371
1840	5	40	17.064	0.517
1850	6	50	23.192	0.719
1860	7	60	31.443	0.768
1870	8	70	38.558	0.937
1880	9	80	50.189	1.221
1890	10	90	62.980	1.301
1900	11	100	76.212	1.462
1910	12	110	92.228

FIGURE 2.1.10.

Points and approximating straight line for U.S. population data from 1800 to 1900.

FIGURE 2.1.11.

World population data.

Year	World Population (billions)
1975	4.062
1980	4.440
1985	4.853
1990	5.310
1995	5.735
2000	6.127
2005	6.520
2010	6.930
2015	7.349

Investigation B: Repeat Investigation A, but take $t = 0$ in 1900 and use only the 20th-century population data listed in Fig. 2.1.4 to create a logistic model. How well does your model predict the U.S. population in the years 1990–2010?

Investigation C: Model similarly the world population data shown in Fig. 2.1.11. The Population Division of the United Nations predicts a world population of 9.157 billion in the year 2040. What do you predict?

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Table of Contents for 2 Mathematical Models and Numerical Methods

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Table of Contents for
2 Mathematical Models and Numerical Methods