Linear Differential Equations

Introduction

This section examines two examples of linear first order differential equations that we are easily solvable. The applications are to Newton's Law of cooling and Pollution in a lake.

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: 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.

More examples of Malthusian growth, radioactive decay, Newton's law of cooling, and pollution in a lake are found in the hyperlinked Worked Examples section.

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