GILBERT STRANG: OK, today is
about differential equations.
That's where calculus
really is applied.
And these will be equations
that describe growth.
And the first you've
already seen.
It's the most important and the
simplest. The growth rate
dy dt is proportional
to y itself.
Let's call c that constant
that comes in there.
And all these problems will
start from some known point y
at time 0 and evolve, grow,
to k, whatever they do.
And this one we know
the solution to.
We know that that will be an
exponential growth with a
factor c, a growth rate c.
Actually, we should know that
the solution has an e to the
ct because when we take the
derivative of this, it will
bring down that factor
c that we want.
And at t equals 0
this is correct.
We get started correctly because
at t equals 0 we have
y of 0 is correctly given by
the known starting point.
Because when t is 0, that's 1.
So that's the one we know.
The most fundamental.
Now, the next step.
Still linear.
That allows a source term.
This might be money in the bank
growing with an interest
rate c per year.
And the source term would be
additional money that you're
constantly putting in.
Every time step in
goes that saving.
Or if s were negative, of
course, it could be spending.
For us, that's a linear
differential equation with a
constant right-hand
side there.
And we have to be able
to solve it.
And we can.
In fact, I can fit it into
the one we already now.
Watch this.
Let me take that to be
c times y plus s/c.
Remember, s and c
are constants.
And let me write that right-hand
side, dy dt.
I can put this constant in there
too and still have the
same derivative.
So dy dt is exactly the same
as the derivative of y plus
s/c because that s/c is
constant; its derivative is 0.
So that dy dt is the
same as before.
But now look.
We have this expression here
that's just like this one,
only it's y plus s/c that's
growing at this growth rate c,
starting from--
the start would be
y at 0 plus s/c.
You see, this quantity in
parentheses, I'm grabbing that
as the thing that grows
perfect exponentially.
So I conclude that at a later
time its value is its initial
value times the growth.
That is a rather quick
solution to this
differential equation.
You might want me to put s/c
on the right-hand side and
have y of t by itself, and that
would be a formula for
the correct y of t that solves
this equation that
starts at y of 0.
And of course, this one starts
at y of 0 plus s/c and that's
why we see y of 0 plus
s/c right there.
Do you see that that equation,
well, it wasn't really
systematic.
And in a differential equations
course you would
learn how to reach that
answer without
sort of noticing that--
actually, I should
do that too.
This is such a useful,
important equation.
It's linear, but it's got some
right-hand side there.
I should give you a system,
say something about linear
equations before I go on.
The most interesting equation
will be not linear, but let me
say a word about linear
equations of which that's a
perfect example with
a right-hand side.
So linear equations.
Linear equations.
The solutions to linear
equations, y of t, is--
to all linear equations.
Actually, this is a linear
differential equation.
When I'm teaching linear algebra
I'm talking about
matrix equations.
The rule, it's the linear part
that is important for these
few thoughts.
The solution to linear
equations--
can I underline linear--
always has the form of some
particular solution.
Some function that does solve
the equation plus another
solution with a right side 0.
You will see what I
mean by those two
parts for this example.
Can I copy again this example
dy dt is cy plus s.
So I'm looking for a solution
to that equation.
First, let me look for a
particular solution.
That means just any function
that solves the equation.
Well, the simplest function I
could think of would be a
constant function.
And if my solution was a
constant, then its derivative
would be 0.
So that would be 0.
And then c times that constant
plus s would be 0.
I want the equation
to be solved.
I claim that a particular
solution is the constant y--
so when I set that momentarily
to 0, I'll discover I'll move
the s over as minus s.
I'll divide by the c and
I'll have a minus s/c.
That's a particular solution.
Let's just plug it in.
If I put in y.
That's a constant, so it's
derivative is 0.
And c times that gives me
minus s plus s is the 0.
OK.
That's a particular solution.
And then what I mean
by right side 0.
What do I mean by that?
I'm not going to wipe out all
this, I'm going to wipe out--
for this right side 0 part,
I'm going to wipe out the
source term.
I mean, I keep the y's, but
throw away the source in the
right side 0.
Maybe a book might often write
the homogeneous equation.
And the homogeneous equation
is where we started,
dy dt equals cy.
What's the solution to that?
The solution to dy dt is cy.
This is this solution,
e to the ct.
Any a, any a times e to the ct
will solve my simple equation.
Once I've knocked out this
s, then if y solves it, a
times y solves it.
I could multiply the
equation by a.
So a is arbitrary, any number.
But of course, I'm going to
find out what it is by
starting at the right place.
By starting at y of 0,
I'll find out what--
by setting t equals 0.
So put t equals 0.
To find a, put t equals 0.
Because we know where
we started.
So we started at a known y of
0 equals minus s/c plus A.
Setting t equals
0 makes that 1.
You see we found
out what A is.
A is y of 0 plus s/c.
And that's the thing that
grows, that multiplies
the e to the ct.
And of course, it
did it up here.
Up here what multiplied the e to
the ct was this y of 0 plus
s/c, which is exactly our A.
As I said, my main point in this
lecture is to bring in a
nonlinear equation.
That's quite interesting
and important.
And it comes in population
growth, ecology, it appears a
lot of places and I'll write
it down immediately.
But just the starting point
is always linear.
And here we've got the basic
linear equation and now the
basic linear equation
with a source term.
Ready for the population
equation.
Population growth.
Population P of t.
OK, what's a reasonable
differential equation that
describes the growth
of population?
All right, so I'm interested
in, what is a model?
So this is actually where
mathematics is applied.
And many people will
have models.
And people who have a census
can check those models and
see, well, what are
the constants?
Does a good choice of constant
make this model realistic over
the last hundred years?
Lots to do.
Fantastic projects
in this area.
If you're looking for a project,
google "population
growth." See what they say.
Several sites will say--
Wikipedia, I did it
this morning.
I looked at Wikipedia.
I'll tell you some things
later about Wikipedia.
And also, the Census Bureau has
a site and they all know
the population now
pretty closely.
Everybody would like
to know what it's
going to be in 50 years.
And to get 50 years out,
you have to have some
mathematical model.
And I'm going to pick this
differential equation.
There's going to be a growth.
That c would be sort
of birth rate.
Well, I guess birth rate
minus death rate.
So that would be the
growth rate.
And if we only had that term
the population would grow
exponentially forever.
There's going to be another
term because that's not
realistic for the population
just to keep growing.
As the earth fills up, there's
some maybe competition term.
Some slowdown term.
So it's going to have
a minus sign.
And maybe s for a slow
down factor.
And I'm going to
take P squared.
That reflects sort of population
interacting with
population.
There are many problems. This
is also a model for problems
in chemistry and biology,
mass action it would
be called in chemistry.
Where I would have two different
materials, two
different substances, two
different chemicals.
And the interaction between the
two would be proportional
to the amount of one and also
proportional to the
amount of the other.
So it'll be proportional
to the product.
One concentration
times the other.
Well, there I would have to have
two equations for the two
concentrations.
Here I'm taking a more basic
problem where I just have one
unknown, the population P,
and one equation for it.
And it's P times P. It's the
number of people meeting other
people and crowding.
So s is a very small number.
Very small, like one over
a billion or something.
Now, I want you to solve
that equation.
Or I want us to solve it.
And we don't right now
have great tools
for nonlinear equations.
We have more calculus to learn,
but this one will give
in if we do it right.
And let me show you what the
solution looks like because
you don't want to miss that.
We'll get into the formula.
We'll make it as nice
as possible, but
the graph is great.
So the population.
Well, notice, what would happen
if the population
starts out at 0?
Nobody's around.
Then the derivative is 0.
It never leaves 0.
So this 0 would be a solution.
P equals 0 constant solution.
Not interesting.
There's another case, important
case, when the
derivative is 0.
Suppose the derivative is 0.
That means we're going to look
for a particular very special
solution that doesn't move.
So if that derivative is 0,
then cP equals sP squared.
Can I do this in the corner?
If cP equals sP squared.
And if I cancel a P, I get
P is equal to c/s.
So there's some solution.
Can I call it that c/s?
Well, everybody would like to
know exactly what those
numbers are.
It's maybe about 10
billion people.
So what I'm saying is
if I got up to--
not me.
If the world got up to 10
billion, then at that point,
the growth and the competition
would knock each other out,
would cancel each other.
There wouldn't be any
further change.
So that's a kind of point
that we won't go past.
But now, let me draw the real
solution because the real
solution starts with--
who knows?
Two people, Adam and Eve.
100 people, whatever.
Whenever we start counting.
This is time.
You know, that's the time
that we're starting.
We could start today.
This is time starting at some
point where we know something.
Well, we have pretty good
numbers back for
some hundreds of years.
So probably it wouldn't
be today we'd start.
But wherever we start, we
follow that equation.
Let me just draw the graph.
It grows.
It grows and then at a certain
point, and it turns out to be
the halfway point, c over 2s.
It's beautiful calculus.
It turns out that that's
a point of inflection.
It's not only growing, it's
growing faster and faster up
to this time.
I don't know when that is, but
that's a very critical moment
in the history of the world.
I think we may be a little
passed it actually.
And after that point,
the whole thing
is just like symmetric.
It's an s-shaped.
It goes up and now slows down
and gets closer and closer to
this c/s steady state.
I would call c/s
a steady state.
If it gets there it stays
there steadily.
It won't quite get there.
This would be the s curve.
And if I went back in time, it
would be going back, back,
back to nearly 0.
Probably to 2.
And up here, but not
quite there.
That's the population curve that
comes from this model.
I can't say that's the
population curve that's going
to happen in the next thousand
years because the model is a
good start.
But of course, you could add in
more terms for epidemics,
for wars, for migration.
All sorts of things
affect the model.
OK, now the math.
The math says solve
the equation.
OK, how am I going to do that?
Well, there are different
ways, but here's
a rather neat way.
It turns out that if I try y.
Let me introduce y as 1/P, then
the equation for y is
going to come out terrific.
This is a trick that works
for this equation.
Nonlinear equations, it's
OK to have a few tricks.
All right, so this is
y of t is 1/P of t.
So of course the graph of y
will be different and the
equation for y will
be different.
Let me say what that equation
looks like.
OK, so I want to know,
what's dy dt?
I know how to take
the derivative.
We get to use calculus.
This is P to the minus 1.
So the derivative of P to
the minus 1 is minus 1.
P to the minus 2 times dP dt.
That's the chain rule.
P to the minus 1, the derivative
is minus 1, p to
the minus 2, and the derivative
of P itself.
And now I know dP dt.
Now I'm going to use
my equation.
So that's cP minus sP squared
with a minus.
Oh, I can't lose that minus.
Put it there.
Keep your eye on that minus.
Divided by P squared.
So now I'll use the minus.
That looks like an s to me.
Minus minus is plus s.
And what do I have here?
Minus cP over P squared.
I think I have minus c/P.
And you say OK, true.
But what good is that?
But look, that 1/P is our y.
So now I have s minus cy.
The equation for y, the dy dt
equation turns out to be
linear with a source term
s just as in the
start of the lecture.
And the growth rate term has
a minus c, which we expect.
Because our y is now 1 over.
When this growth is going up
exponentially, 1 over it is
going down exponentially.
And it turns out that same c.
In other words, we can
solve this equation.
In fact, we have solved this
equation just for--
shall I write down the answer?
So the answer for y.
You remember the answer
for y is--
I'm going to look over here.
Up there.
This is our equation, the only
difference is c is coming in
with a minus sign.
So I'm just going to write
that same solution,
just copy that here.
y of t.
Now c is coming in with
a minus sign, is
y of 0 minus s/c.
c has that minus sign.
e to the minus ct.
OK, good.
That's the solution for y.
So we solved the equation.
You could say, well, you solved
the equation for y.
But now P is just 1/y.
Or y is 1/P. So I just go back
now and change from y back to
P. 1 over P of t is--
here I have 1 over P of 0.
Is it OK if I leave the
solution in that form?
I can solve for P of t and get
it all-- you get P of t equals
and a whole lot of stuff
on the right side.
I move this over.
I have to flip it upside down.
It doesn't look as nice.
Well, I guess I could.
I could move that over
and then just put
to the minus 1 power.
That would do it.
And that is the solution.
Maybe I should do that.
P of t is--
I'm going to move this over.
So this same parenthesis times
e to the ct plus s/c.
I moved that over.
And then I say I have
to flip it.
So it's 1 over that.
1 over all that stuff, where
this is just this
stuff copied again.
So we get an expression but
to me, that top one
looks pretty nice.
And graphing it is no problem.
And that's what the
graph looks like.
It's that famous s curve.
What we've solved here was the
population equation and it's
often called the logistic
equation.
I mention that word because you
could see it there as the
heading for this particular
model.
So let me let you leave open
everything we've done today.
But maybe I can say the most
interesting aspect is the
model itself.
To what extent is it accurate?
We could try it.
We could estimate c and s,
those numbers, to fit the
model of what we know over
one time period.
And then we could see,
does it extend over
another time period?
I was going to say
about Wikipedia.
And generally, Wikipedia
is not too bad.
They made one goof, which
I thought was awful.
The question was, what's
the growth rate?
Actually, Wikipedia
doesn't discuss
this particular equation.
It tells you a lot of other
things about population.
Of course, they talk about the
growth rate and at one point
they say at one time,
it was 1.8%.
I say 1.8% is not
a growth rate.
Growth rate, c, don't have--
they're not percentages.
Their units are 1 over time.
And maybe I could make that
point, emphasize that point.
When I see ct together, that
tells me that has the
dimension of time, of course.
So c must have the dimensions
of 1 over time.
The growth rate is
1.8% per year.
And I will admit that three
lines later on Wikipedia, when
they're referring to an earlier
growth rate earlier in
the last century, they
do say 2.2% per year.
They get it right.
The units are 1 over time.
The solution is telling us
that and the equation is
telling us that.
This has the dimensions of
population, number of people.
So does this.
I'm dividing by a time.
So c must be the source of
that division by a time.
The units of c must
be 1 over time.
You might like to figure out the
units of s because in an
equation, everything has to have
the same units, the same
dimension to make any
sense at all.
So with that little comment,
let me emphasize
the interest in just--
if you have a project to do,
this could be a quite
remarkable one.
Now just I'll write down,
but not solve--
since I have a little space
and mathematicians tend to
fill space--
I'll write down one
more equation.
Actually, it will be two
equations because it's a
predator and a prey.
A hunter and a hunted.
And in ecology, or modeling,
whatever.
Foxes and rabbits, maybe.
We might have these equations,
the predator
and the prey equation.
And what would that look like?
So we have two unknowns.
The predator.
Can I call that u?
So I have a du dt.
What's the growth of the
predators, the number or the
population?
These are both populations, but
of two different species.
And the prey will be
v. So u is the
population of the predator.
If the predator has nothing
to eat, the populations of
predator is going to drop.
But the more prey
that's around--
so here we'll have a u times
v going positive.
So we'll have something like a
minus cu as a term that if
there's nothing to eat, if
that's all there is, if the
prey is all gone, the predator
will die out.
But when there is a predator,
there will be a source term
proportional to u and v. The
more predators there are,
they're all eating away.
And the more prey there is
the more they're eating.
And now what about dv dt?
OK, v is a totally different
position.
I guess it's getting eaten.
So this term, well, I'm
not sure what all the
constants are here.
The prey, this is now the prey,
the little rabbits.
They're just eating grass.
Plenty of grass around.
So they're reproducing.
So they have some positive
growth rate, capital C.
Multiple per year.
But then they will have a
negative term as we saw in
population coming from the
competition with a predator.
So this would be a model, a
predator-prey model that also
shows up in many applications
of mathematics.
So there you see two linear
differential equations.
The simple one dy
dt we'll see y.
We totally know how
to solve that.
Now with a source term s,
we're OK, and we got the
solution there and we found
it a second way.
Then thirdly was the main
interest, the population
equation, the logistic equation
with a P squared,
which we were able to solve
and graph the solution.
That s curve is just
fantastic.
If you want a challenge,
check that this--
so what do I mean at that
inflection point?
What's happening there?
I claim that at this value of P,
the curve stops bending up,
starts bending down,
which means second
derivative of P is 0.
You could check that.
Take the equation, take its
derivative to get the second
derivative.
Then that'll involve
dP dts over here.
But then you know dP dt,
so plug that in.
And then finally, you'll
get down to P's.
And I think you'll find that the
thing cancels itself out.
The second derivative, the
bending in the s curve, is 0
at c over 2s.
So that's the moment
of fastest growth.
And maybe that happened
relatively recently.
And we're slowing down, but
we've got a long way to go.
So roughly, I think the
population now is
just under 7 billion.
And oh boy, if we just left--
if it was 7 billion at that
point, it would be 14 billion
at the end.
I think we're a better further
along now at 7 billion.
7 billion's somewhere
about here.
So maybe the end with this model
is more like 10 or 11
billion as the population of the
earth at t equal infinity.
OK, that's a mathematical
model.
Thank you.
ANNOUNCER: This has been
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Gilbert Strang.
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