Welcome to calculus. I'm Professor Ghrist, we're about to
begin lecture 18, on a simple ODE. >> In this lesson we're going to
do one thing and one thing only. We're going to look at a simple
single differential equation. Consider some of its many applications. This is the subject of today's lesson, dx/dt equals a times x
where a is a constant. This is a linear autonomous
ordinary differential equation. It is extremely important that
you know this and its solution, x equals the initial condition
x nought times e to the at. And what does this solution mean? What does it look like? Well, at time, t equals 0, it passes
through x nought, the initial condition. What it does from there depends on a. If a is positive then you get what
is called exponential growth. You get an exponential function in
t that is growing as t increases. On the other hand, if a is negative
then you have exponential decay. Where the values of x are getting
closer and closer to zero over time. Much of the importance of
this equation comes from its ubiquity in modeling processes. You may have seen this equation
used to model radioactive decay. If you look at, say,
the radioactive isotope of carbon, carbon-14, then the amount
i of that isotope changes to a degree
that is proportional to i where that constant of
proportionality is negative. Likewise if you look at the amount of
a drug or medicine that is in a body, this changes over time according
to the same differential equation. Though perhaps with
a different constant of decay. Both of these have
negative coefficients and thus connote an exponential decay. There are some situations where you
have exponential growth, however. If we look at the most simplistic
model for population growth, one in which the size of the population,
P changes at a rate proportional to P, that constant of
proportionality is a birth rate. And this doesn't work so well for
modeling human populations, but for certain settings, it's reasonable. On the other hand,
continuously compounded interest is a fairly realistic model,
at least in the short term. It says that the rate of
change of the amount of money you have invested is proportional
to how much is there. What is that constant of proportionality? Well, that is an investment rate. But what do we mean by
continuous compounding? Let's assume an annual
interest rate of r and an initial investment of money at time 0. If we were to compound
our interest yearly, then after one year, the amount of money
we would have would be our initial investment plus that initial investment
times the annual interest rate. We could write that as the initial
investment times quantity (1 + r). If however, we compounded that interest
monthly, then what would we get? After one year we would have
the initial investment times quantity (1 + r/12), which is how much
we would earn in January. Then we would multiple all of that by quantity (1 + r/12)
at the end of February. We would continue on until the end of the
year, giving another factor of (1 + r/12). That means at the end of the year
we would have the initial investment times quantity (1
+ r/12) to the 12th power. That's actually a little bit more than
we would get from an annual compounding. Now, of course≤ monthly
compounding is maybe not so accurate because they have different days
per month, so we can do it daily and get the initial investment times
quantity (1 + r/365) to the 365th power. I think you can see where this is going. If we take the limit, then after one
year we get the initial investment times the limit as n goes to infinity
of quantity (1 + r/n) to the nth power. Now I know you've seen that limit before. You may recall that it is,
in fact, e to the r. This is a hint that the standard
linear differential equation is really behind what's
happening in continuous compounding. Let's see how that falls out. If we look at the amount of money at
time t plus a small increment, 1/n. Then according to our previous
reasoning this is the money at time t times quantity (1 + r/n). What we want to do is compute
a limit as n goes to infinity. Now let's change variables a little bit,
and instead of looking at n going to infinity, let us replace 1/n by h, where h is a small amount going to zero. Then the amount of money at time t + h is,
from the above, the amount of money at time t
times quantity 1 + r times h. And now,
expanding out that multiplication, we see a familiar sight from
our definition of a derivative. The coefficient in front of the h term tells us the derivative of
the amount of money, function of t. And so
we can see the differential equation, d dt of the amount of money equals
r times the amount of money. That is our standard
differential equation. Now there are many applications of
this equation that go far beyond radioactivity and population and money. Let's look at an unusual one
coming from linguistics. If we read English poetry from antiquity,
we'll see some unfamiliar words. And if you look at Chaucer,
the Canterbury Tales, many of the words are unfamiliar to us. But if we move the clock
ahead to Shakespeare's time, then some of that poetry is familiar. Some is unfamiliar. Moving ahead a little
bit further to Milton, things are still a little obtuse but more, the words make sense, and
we could keep going with Blake and others. What we might guess is
the following linguistic model. Namely, that the number of words
in English remaining in common use decreases at a rate proportional
to the remaining words. That means that words fall out of use. But this model states that words fall
out of use according to a linear differential equation where W is
the amount of words remaining in usage and alpha is a decay constant. Assuming this model,
let's answer the following question. If we find 20% of the words in
Milton's poem to be unusual, then what fraction of Chaucer's poetry
did Shakespeare's audience recognize? Whew, that sounds difficult. What can we do with that? Well, let's say that t
corresponds to the year. We're going to be interested
in two functions of t. The first, W sub M, is the number of words in Milton's usage
currently used at time t. That is, according to the solution
to this differential equation, W sub M at 1667, the initial condition,
times e to the minus alpha times the number of years
that have elapsed since 1667. Likewise, we'll be interested in W c of t, the number of words from Chaucer's
time still in common usage. This has the same solution with
1667 being replaced by 1400. Most importantly, the alphas are assumed
same because this is all English. Now what is it that we are given? We are told that the number of words
from Milton's time in common use today, let's say in 2012,
is 80% of those originally available. Plugging in the solution to the
differential equation for W sub M when t equals 2012, and
then dividing by the initial condition gives us a single equation that
has only alpha as an unknown. We can therefore solve for alpha and we get a quantity that is
about 6.5 times 10 to the -4. Now, if we use that alpha
in our solution for W sub C of t, then what can we get? Well, our goal is to find
the fraction of words from Chaucer's time that people in
Shakespeare's time would have understood. That is,
W c of 1600 divided by W c of 1400. Plugging in our solution to
the differential equation for W sub C gives us an answer of e to
the negative 200 alpha. Since we know alpha we can compute it. It's about 0.88. Meaning that about 88% of
Chaucer's words would have been recognizable to Shakespeare's audience. Well, that's pretty cool. But is that actually true? Well, I don't think so. This simple model that we've written down
ignores so many features of linguistics, such as the creation of new words. Such as the evolution of words. Words are constantly
changing their meaning. So that even though a word does
not fall out of common usage, it's really a different word with
an altogether different meaning. Lastly, there is whole scale
evolution of language. I'm speaking American English which is
in some respects very different then British English. If you look at, say, computer languages,
you can see, even over short time scales, very rapid evolution in
the language itself. This brings us to an important conclusion, that, even though our mathematics
was perfectly accurate, a perfectly accurate solution is only as
good as the model on which it's based. If you have a bad model, it doesn't matter how good your math is,
you can't conclude truth. So with that in mind,
let's talk about zombies, and we'll model the zombie
apocalypse as follows. The rate of change of
the infected population is proportional to
the uninfected population. If we denote by u(t)
the uninfected population, and z(t), the infected population, then p, the net population size, is a constant. That is u plus z equals p. Our model says that dZ/dt, the rate
of change of the infected population, is proportional to u, that is P- Z, that constant of
proportionality we'll call r. Now, what is happening to dU/dt? Let's think about that. U is P- Z. So differentiating gives us dU/dt = dP/dt- dZ/dt. Now, dP/dt, we know is 0,
since P is constant. And we know dZ/dt = rU. And so we see that our
zombie model is really just the linear differential equation in
disguise for the uninfected population. We know the solution U(t) is the initial
condition, U nought, times e to the -rt. That means that the solution for the zombie population, Z(t) is P- U(t), that is, P- U nought e to the -rt. Unfortunately, as t goes to infinity,
this means that everyone is a zombie. Now, before you dismiss this differential
equation as being ridiculous for modeling zombies, consider. It may also be used to model
the spread of disease, or the spread of propaganda, or the adoption
of new technology in a population. This simplistic equation is not rich enough to really model
these complex phenomena. But it's a good first approximation. It's even better when applied to heat. Newton's law of heat transfer is
essentially the same equation. It says the rate of change of temperature
with respect to time is proportional to the difference in temperature
with the ambient environment. So in our zombie equation, this constant
of proportionality r was an infection rate, and
P was a constant population size. In Newton's law of heat transfer,
the constant is a thermal conductivity. How easy is it for heat to spread? And the constant A is
an ambient temperature. Same equation,
very different interpretations. That's one of the beauties
of differential equations. And so we see one simple differential
equation gives rise to so many applications. But what happens if we change
that differential equation? How will we determine the solution? What is it going to be good for? In our next lesson, we'll consider
some other classes of ordinary differential equations, their solutions,
and their applications.