Linear Applications

This section is devoted to applications of first-order linear differential equations in the biological sciences, the physical sciences, and the social sciences. Each of these applications is closely related to the notions of exponential growth and decay.

Growth and Decay

It is interesting to note that a wide range of problems such as the growth of bacteria, fungi, rabbits, cities, and continuous compound interest, and the decay of radioactive substances all can be modeled by a single kind of differential equation, namely,

dy

dt = ry. (4.1)

The loss of temperature in warm-body cooling satisfies a closely related equation dy dt = r(c- y),

where c is a constant.

If r > 0, equation 4.1 describes a growth phenomenon (See Figure 3.4), and if r < 0 a decay phenomenon (See Figure 3.5). Whatever the value of r, y = 0 is a constant solution and every solution has the form y(t) = Ce'1•

There are certain key phrases that identify a process as satisfying a differential equation of the form y' = ry. Such phrases as "rate of change is proportional to

the amount present," or "the relative rate of change (of a quantity) is constant" are key to the proper statement of the problem. The relative rate of change is (dy/ dt)ly.

Sometimes r is given in absolute terms and sometimes as a percent. Be careful to express r in absolute terms. The word "initially" usually indicates a condition at timet= 0.

Applications in the Biological Sciences

Example 4.4 Suppose that the rate of growth of a population of organisms is 5% of the number present, t being measured in days. If there are 10,000 individuals present initially, how many are present in 10 days? When will the initial population have doubled?

Solution. Let p(t) denote the number present at timet. Then dp/dt = 0.05p and p(O) = 10,000, and the solution has the form p(t) = Ce0051 . From the given initial condition we find that p(O) = C = 10,000. Therefore there are p(t) = 10,000e⁰⁰⁵¹ individuals present t days later. Hence there are lO,OOOe0·5 = 16, 487.2, which we round to 16,487, individuals present at the end of day 10.

In order to find when the population has doubled, we need to solve the equation p(T) = 10,000e005T = 2(10,000) = 20,000 forT. This means e005T = 2, which means that T = (In 2)/0.05 = 13.8629 days. (From the size of the population after 10 days, we knew that doubling had to occur shortly after 10 days.) 0

A common variation on this problem is the following:

Example 4.5 The population of a city was 20,000 according to the 1980 census and 25,000 according to the 1990 census. If the rate of growth of population is propor-tional to the population and is constant,

1. what was the population in 1960? and 2. when will the population be 40,000?

Fig. 4.5. Exponential decay.

Solution. The statement of the problem suggests that dP/dt = rP, where P(t) is the population at timet (in years), so P(t) = Cerr. We are given that

P(1980) = Ce1980' = 20,000 and

P(1990) = Ce1990' = 25,000.

Divide these to find that e10'

=

25,000/20,000

=

5/4, so that lOr

=

ln(5/4), and r = (1110) ln(5/4) ~ 0.0223144. From Ce<1980110lln(S/4l = 20,000, we find that

C

=

20,000e(-l980110)ln(S/4) ~ 1.29672(10-15).

This means that

P(t)

=

20,000e((t-1980YIO)ln(5/4).

To answer the question of part (a),

P(l960)

=

20,000e((1960-1980YIO)ln(5/4))

= 20,000e-²¹n(S/4)

= 12,800.

To answer the question of part (b), we need to determine t (remember that the ques-tion asked when) such that

20,000e((t-1980Y10)ln(5/4) = 40,000.

This means that

so

and

e((t-1980YIO)ln(5/4)

=

2, t-1980

10 ^{ln(5/4) = ln2}

1980 +

10 1 ^n~~~ 4 ⁾

=

2011.06 years,

or sometime in early 2011. The time of doubling of the 1980 population was

there-fore 31 years. 0

The Spread of an Epidemic

In 1959, H. Muench introduced an elementary model for the progress of an epidemic.

His model is

-dy = r(l-y)

dt '

where r is a constant rate. Here "1" denotes the entire population, and y denotes the portion of the population that has come down with the disease. Initially some small fraction y(O) has the disease. From then on, the course of the outbreak is described completely by the solution

y(t) = 1 - (1 - y(O))e-rr,

where 0 ::; y ::; 1, and tis in days. This model says that the disease remains present in the population forever. However, when everyone in the population has had the dis-ease, we would say that the differential equation no longer applies, and the epidemic has passed. It is perhaps clearer to state the problem with a base population of size n that is large (tens of thousands or millions). Then the equation reads

dy dt = r(n-y),

and y(O) is a small positive integer. The epidemic is over when t ~ T implies y(t) ~

n - 1. This makes the duration of the epidemic finite. The solution in this case is y(t) = n- (n-y(O))e-rr, 0::; t::; T.

Adding Tap Water to a Fish Bowl

Anyone who keeps or is interested in keeping freshwater fish in a bowl or aquarium should want to know how to renew the water that is lost naturally by evaporation. It is important to do more than just maintain the water level. If water is merely added, the concentration of sodium (as ordinary table salt, NaCl) that is present in almost every commercial or private water supply will increase until deadly levels can occur.

We compare this scenario with that of systematically removing an extra quantity of water whenever fresh water is added. (As it has been stated, the problem is natu-rally a difference equation, with water additions being discrete events. We consider the fresh water to be (slowly) added continuously. When water is removed, whether by evaporation or manually, we assume that it is continuously being removed at a constant rate. This allows us to study the problem as a differential equation.)

In the first scenario, fresh (tap) water exactly replaces the water lost through evaporation. Suppose that r is the rate of addition of water, that e is the rate of evapo-ration, x(t) is the amount of salt present at timet, with x₀ being present initially. Say that the concentration of NaCl in the fresh water is s. Then

dt

dV = r- e = 0, V(O) = V0.

This means that V(t) = V0 fort > 0. Also

dx - = rs - 0 = rs > 0

dt '

and

x(O) = x0 ~ 0.

Thus fort > 0, the amount of salt, x(t)

=

(rs)t + x_0. This function increases with-out bound, and hence will cause the concentration x(t)!V0 of salt in the fishbowl to exceed the survivable level for the fish living in the fishbowl. Of course the described process must effectively halt at some time because the amount of salt in the fishbowl cannot exceed the volume of the fishbowl (See Fig. 4.6).

In the second scenario, more water than that which evaporates is removed (and disposed of) at the rate

o

and fresh water exactly replaces the water being lost through evaporation and deliberate removal. Under these circumstances,

- =

dV dt r -

o -

^e

=

0 , since r

= o

+ e, so that again

V(t) = V0 fort > 0.

Under these circumstances, the amount of salt in the fishbowl has a different differ-ential equation,

dx

=

rs-

o(~).

dt V0

but it satisfies the same initial condition x(O) = x₀ ~ 0. We can already see potential benefits in that now it is possible for d.x:!dt < 0 and therefore the amount of salt in solution can decrease. The differential equation has a constant, or equilibrium, solution x(t) = rsVofo for which x'(t)

=

0 (that may or may not satisfy the initial condition), and the solution

rsV0 ( rsV0 ) _.2..r

x(t) = -_0- + x₀ -

T

e vo •

This solution says that the limiting concentration

rsV⁰ rs ( r ) xc

= ~o = If =

^{r - e s} > s.

However, xc can be large if r/

o

is large. (That is, if the amount of water disposed of is only a small fraction of the fresh water being added.) If, for instance, extra water is disposed of at the same rate that evaporation takes place, then

o =

e, r

=

2e and xc

=

(2e/e)s

=

2s, which should be an acceptable concentration, because presumably the available tap water has an acceptably low concentration of salt. For a saltwater aquarium this procedure is inappropriate because it can cause the concentration of sodium to decrease to levels that may be dangerous to saltwater organisms that expect high concentrations of sodium. See Fig. 4.7.

The limiting value of x(t) as t ~ oo is the equilibrium solution noted above. This means that no matter what the initial amount x0 of salt, the solution ultimately ap-proaches the equilibrium solution. There is more discussion of equilibrium solutions in Sect. 1 0.2.

It might interest some of you to ask a representative of your local water supplier about the concentration of sodium in your water supply. Then you could use real data

to model the situations described here. In our small town in Tennessee, the concen-tration of sodium is 0.9 mg per liter. Nationally, the concenconcen-tration ranges from 0.5 mg per liter to 25 mg per liter. When the concentration exceeds 20 mg per liter some patients with heart disease may have to take special precautions under the guidance of a physician.

Of course fishbowls are not the only places where sodium can concentrate if water is able to evaporate but sodium cannot be readily removed: just consider the Dead Sea and the Great Salt Lake! The Great Lakes have a substantial concentration of sodium that is not substantially increasing because the Great Lakes are constantly emptying to the sea by way of the St. Lawrence River. The oceans themselves are the ultimate example of extreme sodium concentration produced when only evaporation is able to take place.

Applications in the Physical Sciences Cooling

Newton's law of cooling states that the rate that a warm body cools is proportional to the difference between its temperature, T, and the temperature, T_{0 ,} of the surround-ings. As an equation, this says

dT

- =

k(T, - T).

dt ⁰ (4.2)

This is a linear differential equation whose solution is obvious when the equation is written as

NaCl 2L

d(T - T0 )

=

dT

=

-k(T _ T, ),

dt dt ⁰

time Fig. 4.6. Amount of NaCl with simple replacement of lost water.

NaCl 2L

xO

time

Fig. 4.7. Amount of NaCI with systematic water removal.

so that or

T = T₀ + ce-k1•

The constant c is the initial temperature differential T(O) - T_{0 .} So the solution is

which decays to T0 as t ~ oo. The parameter k is related to several factors such as the volume, surface area, mass, and the heat capacity of the body that is cooling. In many situations where this cooling law is used, values for k are tabulated.

Example 4.6 Suppose that a cake is removed from a 325 degree (Fahrenheit) oven into a 75 degree kitchen. If the cake cools to 180 degrees in 10 minutes, how long afterward will its temperature have fallen to 80 degrees?

Solution. From the discussion above, T(t) = 75 + (325 -75)e-k¹ = 75 + 250e-kt.

From T(IO)

=

^75+250e-10k

=

180, we find that e-IOk

=

(180-75)/250

=

^105/250

=

21150, so that -10k

=

ln(21!50), or

k = -(1/10) ln(21!50) ""0.086750.

This means that

T(t) = 75 + 250eto In( ~ )¹• We need to solve

75 + 250e to In(~)t = 80 for t. This requires that

1 ( 21)

10 ln 50 ^{t =} ln(S/250) = ln(l/50),

so t

=

lO(ln( l/50)/ ln(2l/50))

=

45.1 minutes. This is the time since removal from the oven, so the time since the temperature was 180 degrees is 45.1 - 10 = 35.1 minutes for the temperature to drop the remaining 100 degrees. It is of interest to note that the temperature at the end of one hour of cooling is 76.4 degrees. Most people would say that the cake had cooled. See Fig. 4.8. ⁰

Those of you who thought carefully about the problem will realize that the heat that leaves the cake has to go into the environment, thereby raising the temperature of the environment. So our assumption about the temperature of the kitchen "being"

75 degrees might have been slightly incorrect. There are two ways to think about this. First, that there is another differential equation that is associated with the one we defined, but which accounts for the heat that the cake loses. This is absolutely correct. On the other hand, the heat capacity of the kitchen environment is certainly much greater than the heat capacity of the cake, so the temperature in the kitchen will rise only negligibly, and can safely be called constant. This, too, is correct. There are delicate situations where it is vitally important to keep track of everything, and there are situations where a high degree of accuracy is wasteful. What is being suggested here is that there is a sort of continuity about differential equations: One can say when two differential equations are near one another. Under the appropriate assumptions, the solutions of two differential equations that are near one another are also near one another. This model of cooling does not apply to cooling near a phase transition point, such as where water becomes ice or where steam becomes water. Other laws apply in such situations. It does apply on intervals that are not too near such points.

Nor does this model apply when the temperature changes over too great a range.

There are nonlinear differential equations that apply in more extreme situations such

0 10 20

Fig. 4.8. Cooling of a cake, Example 4.6.

as these, but they all are approximated very closely by Newton's law of cooling when the temperature range is small.

A second application of Newton's law of cooling that you may have heard about for years on your local television news is the procedure used by coroners and med-ical examiners to determine the time of death of a body. The coroner measures the temperature of the body. Then, using this information the coroner can estimate the time of death. It is of some interest that coroners in Tennessee use the approximate rule H = (99 - T)/1.5 to estimate hours since death from measured (rectal) tem-perature. The "more precise" rule in use is that body temperature falls three degrees Fahrenheit per hour for the first three hours and one degree per hour thereafter until

"ambient temperature is reached." You will recognize that this is merely a working approximation to the shape of a decreasing exponential. It implicitly assumes that the body is indoors, because a body outdoors in cold weather would certainly lose heat faster than one inside a building that is at a normal residential temperature.

We all know that "cooling" can go the other way. The temperature of a cold glass of water rises to the temperature of its surroundings, and Newton's law of cooling can be used to estimate the times and temperatures involved.

Mixtures

There are many types of problems that can be classified as mixture problems. Typi-cally, a container holds a solution of a substance that is initially at some concentra-tion. A new source of the substance is coming in at a different concentraconcentra-tion. Op-tionally, the solution in the container is being drawn off. Assuming that the solution in the container is being kept well-stirred at all times, how does the concentration of the substance in the container vary over time? In later chapters we will consider a network of interconnected containers and ask about the concentrations in each of them. This will require several related differential equations, one for each container.

Two principles that must be understood in order to be able to set up problems of this type are: Rate of change of volume = Incoming change of volume - Outgoing rate of change of volume,

dV = dVin dVout . dt

dt-dt,

and Change of mass = Incoming mass - Outgoing mass,

so that where c is the mass of the subject in question. This relationship also holds for weight.

Example 4.7 A container initially contains 50 pounds of salt dissolved in I 000 gal-lons of water. A brine mixture of 1/4 pounds of salt per gallon is entering the con-tainer at 6 gallons per minute. The well-mixed contents of the concon-tainer are being discharged from the container at the rate of 6 gallons per minute. Express the amount of salt in the container as a function of time. What is the limiting concentration of salt in the container?

Solution. Let x(t) denote the amount of salt in the container at timet. First note that the volume of the mixture in the container is unchanging since dV I dt

=

6-6

=

0.

Salt is coming into the container at the rate of

dx

d;rr =

(6gaVmin)(l/4lblgal)

=

(3/2)lblmin.

Also, since the concentration of salt in the container is amount in the container x

- - - . - = - -

^lblgal,

volume of the contamer 1000 it follows that salt is leaving the container at the rate of

dx 6x

d~ut

=

(6gaVmin)(x/1000 lblgal)

=

1000 lblmin.

This means that the rate of change of salt in the container is dx

and the initial condition is x(O) = 50. This is a linear differential equation. You should verify that the solution is

x(t)

=

250 - 200e-311500 .

See Fig. 4.9. 0

This expresses the amount of salt in the container at time ^t as was requested, and it is easy to see that the limiting amount of salt in the container is 250 pounds, which results in a limiting concentration of 25011000

=

114 pound per gallon of salt in the container. Is it an accident that the limiting concentration is the same as the entering concentration? See the exercises.

Example 4.8 Suppose that the rate of discharge of the solution in the previous ex-ample is reduced to 5 gallons per minute, and the container can hold a maximum of 2000 gallons. Now express the amount of salt in the container as a function of time.

When does the container begin to overflow? What is the concentration of salt in the container when the container begins to overflow?

Solution. Here dV I dt

=

6 - 5

=

1 gallons per minute, so that V (t)

= (

1000 + t) gallons after t minutes. This means that the container will overflow when 1000 ^{+ t}

=

2000. or t = 1000. We need to account for the fact that the volume is not constant

when we consider the concentration of salt. The rate of change of salt in the container

is dx

dt

dxin _ dx0111 _ ~ ^{_ _ _} 5x __

dt dt 2 1000 + t

This is still a linear differential equation, but it no longer has constant coefficients.

Using the methods we studied before, the solution is found to be

t 2(1017 )

x(t) = 250 + -4 - (1000 + t) ^{5 .}

This has a much different character than the solution with constant volume. In fact it is equivalent to a rational function: the quotient of two polynomials.

Notice in Figure 4.10 that the amount of salt is increasing without apparent limit (until the container overflows). But the concentration of the salt at t

=

1000 when the container begins to overflow is x(lOOO)/V(lOOO))

=

493.75/2000

=

^0.246875,

which is near 0.250, the concentration of the incoming stream. It is also interesting to note that if the process is not stopped, but the container is permitted to continue overflowing, then the differential equation of the process becomes

dx dt

dxin _ dxout

dt dt

3 6x

2 -

^{2000' t ~ 1000,}

with initial condition x(1000) = 493.75, the amount of salt in the container when overflow first begins. This overflow equation has solution

25e3(1000-rYIOOO

x(t) = 500 ,4

-from which can be easily seen that the limiting concentration in the overflowing container is 1/4 pound/gallon, the salt concentration in the entering stream. 0

250

200 400 600 800 1000

Fig. 4.9. The amount of salt, constant volume.

500 400 300 200 100

200 400 600 800 1000

Fig. 4.10. The amount of salt when volume is increasing.

Example 4.8 M Solve and check example 4.8 by Mathematica.

Solution. We let Mathematica solve the system.

3 x[t]

In[l]:= DSolve[{x'[t] ==--5 ,x[O] ==50},x[t],t]

2 1000 + t Out [1]= {

{x

^{[t] __,} - - - c c 1

4(1000+t)⁵

(200000000000000000 + 6000000000000000 t +15000000000000t² + 20000000000 t³ +15000000 t⁴ + 6000 t⁵ + t6 )}}

Capture the solution. Write it expanded by partial fractions using Apart.

In [2] := Soln[t_] =Apart [x[t] I .First[%] ] t 200000000000000000 Out[2]= 250 + ,

-4 (1000+t)⁵ Have Mathematica check this answer.

( 3 So1n[t]) In [ 3]: = Simplify [ Soln' [t] -

2 - ^s-

1-0-0-0-+-t Out[3]= True

The initial condition checks easily.

In[4] := Soln[O] ==50 Out[4]= True

Applications in the Social Sciences Compound Interest

== 0]

0

An amount P0 of money earns interest at an annual rate of i percent compounded n times per year at times tk

=

kin, k

=

l, 2, .... Let Pk denote the amount present at time tk. Then

Po+ lOOn Po. i

p +--P i

1 lOOn 1•

and in general

This says that

Pk+l- Pk __ 1_· p 1 n - 100 k'

If n is permitted to increase without bound, we have the law of continuously compounded interest:

dP i

dt = lOOP, P(O) =Po.

This has as solution P(t)

=

P0ei11100• Note also that continuously compounded interest obeys the same law as that of unrestricted growth.

Radioactive Decay

Radioactive substances decay at a rate proportional to the amount present. That is, dyl dt

=

-ry. If an amount y0 is present initially, then for any timet > 0, y(t)

=

y0^e-rt.

A common term that occurs in discussions of radioactive phenomena is half-life. A substance has half-life t112 if y(t + t112 ) = (1!2)y(t). The value of t₁₁₂ is determined completely by rand is independent of y0 and t. We find t₁₁₂ from the calculation

y(t + t_{112 )}

=

y₀e-r(t+tuz)

=

(112)y₀e-rt

=

(1!2)y(t).

Simplify to get

or _e-rtl/2 = -1

2

This simplifies tot ₁₁₂ = (1/r) In 2. The smaller r is, the larger t112 is. Values of t112 for various substances range from small fractions of a second to many billions of years.

This simplifies tot ₁₁₂ = (1/r) In 2. The smaller r is, the larger t112 is. Values of t112 for various substances range from small fractions of a second to many billions of years.

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