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Math 337 - Differential Equations
Spring Semester, 2002
Linear Differential Equations
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© 2000, All Rights Reserved, SDSU
& Joseph M. Mahaffy
San Diego State University -- This page last updated 28-Jan-02
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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) - 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
- 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 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
Solution:
- 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
- This is just the radioactive decay problem that we just
solved
- The solution is
z(t) = 8 e-kt
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-k or ek
= 4/3
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
8 e-kt =
37- 22 = 15 or e-kt =
15/8 = 1.875
-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
Te
- 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
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)
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.