IntroDiffEq

Math 337 - Differential Equations
Spring Semester, 2002
Linear Differential Equations

Linear Differential Equations

Introduction
Newton's Law of Cooling
Pollution in a Lake
Worked Examples

Newton's Law of Cooling

After a murder (or death by other causes), the forensic scientist takes 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 are used to extrapolate back to when the murder occurred
This property 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
If T(t) is the temperature of the body, then it satisfies the differential equation

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

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

Example:

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

Solution:

Normal temperature of a human body when it is alive is 37^oC
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

This is just the radioactive decay problem that we just solved
The solution is

z(t) = 8 e^-^kt

However,

z(t) = T(t) - 22, so T(t) = z(t) + 22

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^-^kor e^k= 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 37^oC
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)

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)

Differential Equation for Pollution in a Lake

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 look at the units associated the concentration of the pollutant, p, which are stated in mass/volume (For example, mg/m³)
The flow rate, f, has units of volume/time (For example, 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)

Differential Equation for the Concentration of the Pollutant in the Lake

Since the lake maintains a constant volume V, then 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 the lake is initially clean, the initial condition for this differential equation is

c(0) = 0

Solution of the Differential Equation

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

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

Make the substitution, z(t) = c(t) - p
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

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:

Suppose that you begin with a 10,000 m³ clean lake
Assume 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)
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, 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
Assume 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

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

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

Complicating Factors to Consider in a Model

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.