SDSU

 

Math 336 - Mathematical Modeling
Spring Semester, 2000

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San Diego State University -- This page last updated 14-Feb-00


Linear Differential Equations

  1. Introduction
  2. Malthusian Growth
  3. Radioactive Decay
  4. Newton's Law of Cooling
  5. Pollution in a Lake
  6. Problems

Introduction

This section examines several examples of linear first order differential equations that we are able to solve. The applications are to Malthusian growth of a population, Radioactive decay, and Newton's Law of cooling.

Malthusian Growth

In our introduction to differential equations, we developed the continuous Malthusian growth model. If P(t) is the population at any time t and r is the rate of growth of the population per unit time per animal in the population, then the differential equation for this model is given by

P'(t) = rP(t).

If the initial population satisfies P(0) = P0, then the solution becomes

P(t) = P0 ert.

Example 1: Suppose that a culture of Escherichia coli is growing according to the Malthusian growth model

P'(t) = rP(t).

If the population doubles in 30 minutes, then find the value of r.

Solution: From above we see that the solution is given by

P(t) = P0 ert.

This population doubles when 2P0 = P0 e30r. Thus, e30r = 2 or 30r = ln(2) or r = ln(2)/30 = 0.0231 min-1.

Radioactive Decay

Radioactive elements play an important role in many biological applications. Many experiments use radioactive tracers to determine what sequence of molecular events occurred. For example, 3H (tritium) can be used to tag certain DNA base pairs, which can then be added to mutant strains of E. coli that are unable to manufacture one particular DNA base. By treating the culture with the appropriate antibiotics, one can use the radioactive signal to determine how much DNA was replicated under a particular set of experimental conditions. Radioactive iodine is often used to detect thyroid problems in humans. In most cases, the experiments are run in such a way that radioactive decay is not an issue. (3H has a half-life of 12.5 yrs or 125I has a half-life of 56 days.)

One important application of radioactive decay is the dating of biological specimens. While an organism is living, it is continually changing its carbon with the environment. (Plants directly absorb CO2 from the atmosphere, while animals get their carbon either directly or indirectly from plants.) Gamma radiation that bombards the Earth keeps the ratio of 14C to 12C fairly constant in the atmospheric CO2. The 14C that entered a living organism can be used to determine how much time has passed since it has died. It has been determined that radioactive carbon, 14C, decays with a half-life of 5730 years, i.e., in 5730 years the quantity of 14C decreases to half its original amount. Living tissue shows a radioactivity of about 15.3 disintegrations per minute (dpm) per gram of carbon, which means that 5730 years after the organism has died, it will exhibit only half or 7.65 dpm per gram of carbon.

Let R(t) represent the number of dpm per gram of carbon from an ancient object. The loss of 14C from a sample at any time t is proportional to the amount of 14C remaining. Thus, the amount of 14C is directly related to R(t) which satisfies the differential equation

R'(t) = -kR(t), with initial condition R(0) = 15.3.

From our solution for the Malthusian growth model, it is clear that this differential equation has the solution, R(t) = 15.3e-kt, where k = ln(2)/5730 = 0.000121.

Example 2: Suppose that an object is found to have a radioactive count of 5.2 dpm per g of carbon. Find the age of this object.

From the solution above

5.2 = 15.3e-kt or ekt = 15.3/5.2 = 2.94 or kt = ln(2.94).

Thus, t = ln(2.94)/k = 8915 yr., so the object is about 9000 yrs old.

Newton's Law of Cooling

After a murder (or death by other causes), one of the first things that the forensic scientist does is to take the temperature of the body. A little later the temperature of the body is taken again to give the investigator an idea of the rate at which the body is cooling. These two (or more) data points can be used to extrapolate back to when the murder occurred. The property that is being applied is known as Newton's Law of Cooling. Newton's Law of Cooling states that the rate of change in temperature of a cooling body is proportional to the difference between the temperature of the body and the surrounding environmental temperature. Thus, if T(t) is the temperature of the body, then it satisfies the differential equation

T '(t) = -k(T(t) - Te) , with initial condition T(0) = T0.

The parameter k is dependent on the specific properties of the particular object (body in this case), Te is the environmental temperature, and T0 is the initial temperature of the object.

Example 3: Let us suppose that a murder victim is found at 8:30 am and that the temperature of the body at that time is 30oC. Assume that the room in which the murder victim lay was a constant 22oC. Suppose that an hour later the temperature of the body is 28oC. Use this information to determine the approximate time that the murder occurred.

We know that the normal temperature of a human body when it is alive is 37oC. From the model for Newton's Law of Cooling and the information that is given, if we set t = 0 to be 8:30 am, then we need to solve the initial value problem

T '(t) = -k(T(t) - 22) , with T(0) = 30.

Suppose that we make a change of variables. Let z(t) = T(t) - 22, then z'(t) = T '(t), so the differential equation above becomes

z'(t) = -kz(t), with z(0) = T(0) - 22 = 8.

But this is just the radioactive decay problem that we just solved. Thus, the solution is given by

z(t) = 8 e-kt.

However, z(t) = T(t) - 22, so T(t) = z(t) + 22 or T(t) = 22 + 8 e-kt.

We use the second piece of information to determine the constant k. T(1) = 28 = 22 + 8 e-k, so 6 = 8 e-k or ek = 4/3. Thus, k = ln(4/3) = 0.2877.

It only remains to find out when the murder occurred. At the time of death, the body was 37oC. So we need to solve

T(t) = 22 + 8 e-kt = 37

for t. Equivalently, 8 e-kt = 37- 22 = 15 or e-kt = 15/8 = 1.875. This gives

-kt = ln(1.875) or t = -ln(1.875)/k = -2.19.

From this information, we can conclude that the murder occurred about 2 hours and 11 minutes before the body was found, which places the time of death at around 6:19 am.

Pollution in a Lake

One of the most urgent problems in modern society is how we reduce the pollution and toxicity of our water sources. These are very complex issues that require a multidisciplinary approach and are often politically very intractable because of the key role that water plays in human society and the many competing interests. In this section we will examine only a very simplistic model for pollution of a lake, but it should demonstrate some basic elements from which more complicated models could be built and analyzed.

Consider the scenario of a new pesticide that is applied to fields upstream from a clean lake with volume V. We assume that a river receives a constant amount of this new pesticide into its water, and that it flows into the lake at a constant rate, f. This assumption implies that the river has a constant concentration of the new pesticide, p. Assume that the lake is well-mixed and maintains a constant volume by having a river exiting the lake with the same flow rate, f, that the inflowing river has. We will use this information to set up and solve a linear first order differential equation for the concentration of the pesticide in the lake, c(t).

The key to studying pollution problems of this type is setting up a differential equation that describes the mass balance of the pollutant. Descriptively, per unit time we examine

The change in amount of pollutant = Amount entering - Amount leaving.

The amount entering is simply the concentration of the pollutant in the river times the flow rate of the river. To see this we look at the units associated the concentration of the pollutant, p, which are stated in mass/volume. (For example, we might have something like mg/m3. The flow rate, f, has units of volume/time. (For example, we might have m3/day.) Thus, fp gives us the amount entering the lake per unit time, which by assumption is simply a constant.

The amount leaving has the same flow rate, f. Since the lake is assumed to be well-mixed, the concentration in the outflowing river will be equal to the concentration of the pollutant in the lake, c(t). The product f c(t) gives the amount of pollutant leaving the lake per unit time. Thus, if a(t) is the amount of pollutant in the lake at any time t, then a differential equation describing a(t) is given by:

a'(t) = fp - f c(t).

However, we would like this differential equation to be simply described by the concentration of the pollutant in the lake. Since we are assuming that the lake maintains a constant volume V, we have the relationship that c(t) = a(t)/V, which also implies that c'(t) = a'(t)/V. Dividing the above differential equation by the volume V, we have

c'(t) = (f/V)(p - c(t)).

Since we are assuming that the lake is initially clean, then the initial condition for this differential equation is given by c(0) = 0.

One solves this differential equation in exactly the same manner as solving the problem we saw before for Newton's Law of cooling. Notice that the constant f/V acts like the constant k in Newton's law of cooling, while p acts like the constant Te. The equation above can be rewritten in the form of Newton's Law of cooling:

c'(t) = -(f/V)(c(t) - p).

Now we make the substitution, z(t) = c(t) - p in a manner similar to before. The derivatives z'(t) = c'(t). The initial condition becomes z(0) = c(0) - p = - p, the initial value problem in z(t) becomes,

z'(t) = -(f/V)z(t), with z(0) = - p.

The solution to this problem is again like the solution to the radioactive decay problem and can be seen as

z(t) = - p exp(-ft/V).

Since z(t) = c(t) - p, the solution is easily solved to produce

c(t) = p - p exp(-ft/V).

The exponential decay of the second term in this solution shows that as t becomes "large," the solution approaches p. This is exactly what you would expect, as the entering river has a concentration of p.

Example:

Let us work the above problem with some specific conditions. Suppose that you begin with a 10,000 m3 clean lake. If the river entering has a flow of 100 m3/day and the concentration of some pesticide in the river is measured to have a concentration of 5 ppm (parts per million), then we can form the differential equation describing the concentration of pollutant in the lake at any time t and solve it. Find out how long it takes for this lake to have a concentration of 2 ppm.

In a second part of the example, let us suppose that when the concentration reaches 4 ppm, the pesticide is banned. For simplicity, assume that the concentration of pesticide drops immediately to zero in the river. (We are also assuming that the pesticide is not degraded or lost by any means other than dilution.) Find how long until the concentration reaches 1 ppm.

Solution:

The differential equation for this problem is given by

c'(t) = -(100/10000)(c(t) - 5) = -0.01 (c(t) - 5) with c(0) = 0.

We let z(t) = c(t) - 5, then the differential equation becomes,

z'(t) = -0.01 z(t), with z(0) = - 5.

This has a solution

z(t) = - 5 exp(-0.01t) or equivalently, c(t) = 5 - 5 exp(-0.01t).

To find how long it takes for the concentration to reach 2 ppm, we solve the equation

2 = 5 - 5 exp(-0.01t).

This is equivalent to

exp(-0.01t) = 3/5 or exp(0.01t) = 5/3.

Solving this for t, we obtain

t = 100 ln(5/3) = 51.1 days.

For the second part of the problem, there is no pollutant in the river and the initial concentration in the lake is assumed to be 4 ppm. The new initial value problem becomes

c'(t) = -0.01 (c(t) - 0) = -0.01 c(t) with c(0) = 4.

In this case, we don't need a substitution as the differential equation above is already in the form of a radioactive decay problem. This has the solution

c(t) = 4 exp(-0.01t).

To find how long it takes for the concentration to return to 1 ppm, we solve the equation

1 = 4 exp(-0.01t) or exp(0.01t) = 4.

Solving this for t, we obtain

t = 100 ln(4) = 138.6 days.

The above discussion for pollution in a lake fails to account for many significant complications. There are considerations of degradation of the pesticide, stratefication in the lake, and uptake and reentering of the pesticide through interaction with the organisms living in the lake. The river will vary in its flow rate, and the leeching of the pesticide into river is highly dependent on rainfall, ground water movement, and rate of pesticide application. Obviously, there are many other complications that would increase the difficulty of analyzing this model. In the next section we will examine how to handle certain complications in this model. 

Problems:

1. a. The population of Canada 1 was 24,070,000 in 1980, while in 1990 it was 26,620,000.  Assuming the population is growing according to the principle of Malthusian growth (with no food or space limitations), find the population as a function of time, and determine its doubling time.

b. For the same years, the populations of Kenya were 16,681,000  and  24,229,000 respectively. Find the population as a function of time.

c. In what year do the populations of Canada and Kenya  become equal?

2.  A radioactive substance satisfies the differential equation

R'(t) = -kR(t)

for some constant k.

Suppose that initially there are 10gm of the substance. After 25 days there are 8gm remaining. Find k and determine the half-life of the substance(time when R(t) =5 )

3.  When Strontium-90 (90Sr) is ingested, it can displace calcium in the formation of bones. After a beta decay it becomes an isotope of krypton (an inert gas), and diffuses out of the bone, leaving the bones porous.

a. Suppose that a particular bone contains 20 grams of 90Sr, which has a half-life of 28 years. Write an equation describing the amount of 90Sr remaining at any time, and determine the amount after 10 years.

b. Find how long until only 7 grams of 90Sr remain.

4.  You have just boiled a new batch of broth for your important culture of E. coli, so it is at 100oC. You have it sitting in a room that is at 22oC, and you find 5 minutes later that itís cooled to 93oC. You want to innoculate the culture when it reaches 40oC. You are interested in knowing if you can safely go off to exercise while the broth cools.

a. Let T(t) be the temperature of the broth. Assume that the broth satisfies Newton's Law of Cooling, and set up the differential equation for the temperature of the broth, and solve it.

b. Find how long it will be until you need to innoculate the broth with your culture. Sketch a graph of the function of T(t) for the first hour showing its starting and ending temperatures for the hour.

5. a. Paramecia in a pond sample are growing according to the principle of Malthusian growth (with no food or space limitations).

Initially, there are 1500 Paramecia. Four hours later, the population has 2000 individuals. Find the population of Paramecia as a function of time, and determine its cell doubling time.

b. A large population of 5000 is transferred to where a limited diet affects their growth dynamics. The Paramecia now satisfy the population dynamics of the differential equation:

dP/dt = -0.1P + 100

Solve this differential equation and find what happens to the population as t®¥ (Hint: Recall your techniques for Newton's Law of Cooling.)

6.  A well mixed pond, V = 200,000 m3, is initially clean (c(0) = 0). A polluted stream with a concentration of dioxin, Q = 5 ppb enters the pond, flowing at a rate of f= 4000 m3/day. Another stream carries the water away at the same rate. The differential equation for this problem is given by:

dc/dt = {f/V}(Q-c)

a. Set up the initial value problem and solve it.

b. Find how long before the pond has a concentration of 4 ppb

c. Find the limiting concentration.