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Today's lecture is going to be
basically devoted to working out
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a single example of a nonlinear
system, but it is a very good
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example because it illustrates
three things which you really
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have to know about nonlinear
systems.
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I have indicated them by three
cryptic words on the board,
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but you will see at different
points in the lecture what they
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refer to.
Each of these represents
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something you need to know about
nonlinear systems to be able to
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effectively analyze them.
In addition,
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I am not allowed to collect any
more work, according to the
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faculty rules,
after today.
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And, of course,
I won't.
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But, nonetheless,
the final will contain material
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from the reading assignment G7,
and I will be touching on all
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of these today,
and 7.4, just a couple of pages
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of that.
You will see its connection.
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It discusses the same example
we are going to do today.
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And then I suggest those three
exercises to try to solidify
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what the lecture is about.
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Instead of introducing the
nonlinear system right away that
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we are going to talk about,
I would like to first explain,
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so that it won't interrupt the
presentation later,
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what I mean by a conversion to
a first-order equation.
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For that why don't we look at --
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Maybe I can do it here.
Our system looks like this,
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dx over dt.
For once, I am not writing x
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prime because I want to
explicitly indicate what the
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independent variable is.
It is a nonlinear autonomous
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system, so there is no t on the
right-hand side.
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But this is not a simple linear
function, ax plus by.
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It is more complicated,
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like you had in the
predator-prey robin-earthworm
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problem that you worked on last
night.
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And the other equation will be
g of (x, y).
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This is the system,
and explicitly it is nonlinear
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in general and autonomous,
as I have indicated on the
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right-hand side.
Now, remember that the
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geometric picture of this was as
a velocity field.
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You made a velocity field out
of the vectors whose components
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were f and g,
so that is (f)i plus (g)j,
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but what it looked like was a
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plane filled up with a lot of
vectors pointing different ways
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according to the velocity at the
point was.
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And side-by-side with that went
the solutions.
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A typical solution was x equals
x of t,
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y equals y of t
represented as a column vector.
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And that is a parametric
equation.
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And the geometric picture of
that was given by its
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trajectory.
When you plotted it,
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it was a trajectory.
It had not only the right
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direction at each point,
in other words,
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but it also had to have the
right velocity at each point.
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In other words,
sometimes the point was moving
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rapidly and sometimes it was
moving more slowly along that
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path.
And the way it moved was an
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important part of the solution.
Now, what I plan to do is --
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The conversation that I am
talking about takes place by
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eliminating t.
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Now, why would one want to do
that?
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t, after all,
is an essential part of the
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solution.
It is an essential part of this
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picture because without t you
would not know how long the
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arrows were supposed to be.
You would still know their
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direction.
And, of course,
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it occurs in here.
Now, let's take the first step
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and eliminate t from the system
itself.
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As you see, that is very easy
to do because the right-hand
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side has no t in it anyway and
the left-hand side has the t
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only as the denominators of the
differential quotients.
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The obvious way to eliminate t
is just to divide one equation
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by the other.
And since usually we think of
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not dx by dy,
but dy by dx,
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I will vow --
What is dy over dx?
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Well, if I divide this equation
by that equation,
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dy by dt divided by dx by dt,
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according to the chain rule,
or according to commonsense,
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you cancel the dt's and you get
dy over dx.
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And, on the right-hand side,
you get g of (x,
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y) divided by f of (x,
y).
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But what is that?
That is the equation of the
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first day of the term.
By taking that single step,
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I convert this nonlinear
system, which we virtually never
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find explicit solutions to,
into an ordinary first order
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equation which,
in fact, you also don't usually
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find explicit solutions to,
except with our narrow blinders
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on.
We only get problems at the
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beginning of 18.03 where there
will be an explicit solution to
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this.
I have converted the nonlinear
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system to a single first order
equation to which I can apply
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the usual first order methods
that include hope that I will be
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able to solve it.
Now, what do I lose?
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Well, what happens to this
picture?
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If you don't have the t in it
anymore then you don't have
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velocity, you don't have arrows
with a length to them because
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the length gives you how fast
the point is going.
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Without t I don't know at any
point how fast it is supposed to
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be going.
All I know at any point (x,
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y) is no longer dx over dt
and dy over dt.
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All I know is dy over dx.
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In other words,
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the corresponding picture,
if I eliminate t from this
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picture all that is left is the
directions of each of these
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arrows.
What disappears is their
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length.
And, in fact,
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since dy over dx is
purely a slope,
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I cannot even tell whether the
point is traveling this way or
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that way.
It doesn't make any sense to
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have the point traveling anymore
since there is no time in which
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it can do its traveling.
So that picture gets changed to
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the direction field.
The corresponding picture now,
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all you do is take each of
these arrows,
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you snip it off to a standard
length, being careful to snip
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off the little pointy end,
and it becomes nothing but a
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line element.
And all it has is a slope.
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What corresponds to the
velocity field here is now just
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our slope field or our direction
field.
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All we know is the slope at
each point.
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How about the solution?
Well, if I have eliminated
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time, the solution is no longer
a pair of parametric equations.
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It is just an equation
involving x and y.
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I can hope that the solution
will be explicit,
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y equals y of x,
but you already know from the
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first days of the term that
sometimes it isn't.
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Let's call that explicit
solution.
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Often you have to settle for
it.
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And sometimes it is not a pain.
It is something that is even
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better.
Settle for a solution which
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looks like this which is y
defined implicitly as a function
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of x.
And what does the picture of
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that look like?
Well, the picture of this is
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now simply an integral curve.
It is not a trajectory anymore.
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It is what we called an
integral curve.
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I am reminding you of the very
first day of the term,
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or maybe the second day.
And all it has at each point
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the right slope because that is
all the field is telling me now.
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It only has a slope at each
point.
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It doesn't have any magnitude
or direction.
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It has the direction,
but it doesn't have direction
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in the sense of an arrow telling
me whether it is going this way
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or the opposite way.
That is the picture with t in
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it.
These are the corresponding
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degeneration of that.
It is a coarsening,
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it's a cheapening of it.
It is throwing away
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information.
Throw away all the information
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that had to do with t,
and we are back in the first
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days of the terms with a single
first order equations involving
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only x and y with solutions that
involve only x and y with curves
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that are the graphs of these
solutions but again have no t in
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them.
In effect, they are just the
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paths of the trajectories.
Whereas, the trajectory is the
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point that shows actually how
the point is moving along in
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time faster or slower at various
points.
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Now, why would one want to lose
information?
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Well, the great gain is that
this might be solvable.
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Whereas, this almost certainly
is not.
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That is a big plus.
For example,
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let's illustrate it on the very
simplest possible case.
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I cannot illustrate it on a
nonlinear system very well,
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or not right now because,
in general, nonlinear systems
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are not solvable.
Let's take an easy example.
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Suppose the system were a
linear one, let's say this one
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that we have talked about
before, in fact.
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That is a simple linear
two-by-two system.
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This means the derivatives with
respect to time.
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Well, you know that the
solutions we talked about in
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fact, last time,
typical solutions that would be
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something like (x,
y) equals a constant times,
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let's say if x is cosine t
then y would be the
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derivative of that,
which is negative sine t.
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Or, another one would be c2
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times sine t,
cosine t.
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And those, of course,
are circles.
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They are parametrized circles,
so they are circles that go
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around, in fact,
in this direction.
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And they have a certain
velocity to them.
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Now what do I do if I follow
this plan of eliminating t?
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Well, if I eliminate t directly
from the system,
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what will I get?
I will get dy by dx.
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I divide this equation by that
one.
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-- is equal to negative x over.
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Oh, well, of course,
that is solvable.
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You were able to solve by 18.01
methods before you ever come
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into this course.
By separation of variables,
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it is y dy equals
negative x dx,
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which integrates to be one-half
y squared equals negative
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one-half x squared plus a
constant.
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And after multiplying through
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by two and moving things around
it becomes x squared plus y
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squared.
The nicest way to say it is
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implicitly, x squared plus y
squared equals some positive
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constant.
These are the circles.
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Now, can I eliminate t?
I could also take one of these
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solutions and eliminate t.
If I square this,
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the best way to do it is not to
use our cosines and arcsines and
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whatnot, which will just get you
totally lost.
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Square this,
square that and add them
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together.
And you conclude that x squared
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plus y squared is equal to one.
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Or, if c1 is not one it is
equal to c1 squared.
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You could eliminate t from the
solution the way you eliminate t
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from a pair of parametric
equations, and you get it the
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same way that x squared plus y
squared, in this case,
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would be c1 squared, I guess.
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In some sense,
what we are doing is cycling
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around to the beginning of the
term.
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In fact, this whole lecture,
as you will see,
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is about cycles of one sort or
another.
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But I keep thinking of that
great line of poetry,
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"in my end is my beginning" or
maybe it's "in my beginning is
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my end".
I think both lines occur.
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But that is what is happening
here.
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This is the beginning of the
course and this is the end of
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the course, and they have this
almost trivial relation between
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them.
But notice the totally
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different methods used for
analyzing this,
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where the t is included,
than from analyzing this where
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there is no t.
The goals are different,
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what you look for are
different, everything is
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different, and yet it is almost
the same problem.
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I promised you a nonlinear
example.
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I guess it is time to see what
that is.
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It is going to be another
predator-prey equation,
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but one which is,
in some ways,
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simpler than the one I gave you
for homework.
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The predator is going to be x
because you are used to that
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from the robins.
I am keeping the same
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predator-prey variables.
And the prey will be y.
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And now just notice the small
difference from what I gave you
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before.
For one thing,
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I am not giving you specific
numbers.
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I am going to,
at the beginning at least,
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do some of the analysis at the
beginning using letters,
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using parameters.
a, b, c, d are always going to
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be positive constants.
I am going to assume that x
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prime equals minus ax.
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This represents the predator
dying out if there is no prey
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there, but if there is something
to eat that term represents the
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predator meeting up with the
prey and gobbling it up.
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And b is the coefficient.
How about without predictors,
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the prey will multiply and be
fruitful.
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Unfortunately they get eaten,
and so there will be a term
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that looks like this.
Now, there are two basic
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differences between these simple
equations and the slightly more
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complicated ones you had with
the robin-earthworm equations.
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Namely, in the first place,
I am assuming that these guys,
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let's give them names.
I want you to remember which is
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x and which is y,
so we are going to think of
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these are "sharx".
[LAUGHTER]
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And these will be food fish,
so let's make them "yumfish".
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The fact that with your robins
you had a positive term here.
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I made this 2x.
And now it is minus a times x.
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What is the difference?
The difference is that robins
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have other things to eat,
so even if there are not any
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worms, a robin will survive.
It could eat other insects,
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grubs, Japanese beetle grubs.
I think it can even eat seeds.
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Anyway, the robins in my garden
seem to be pecking at things
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that don't seem to be insects.
And the other is that I am
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assuming a very naīve growth
law.
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For example,
if there are no sharks,
271
00:17:30 --> 00:17:36
how are the food fish fruitful
and multiplying?
272
00:17:35 --> 00:17:41
They multiple exponentially.
Now, obviously you cannot have
273
00:17:39 --> 00:17:45
unlimited growth like that.
With the worms we added the
274
00:17:43 --> 00:17:49
term minus a constant times y
squared to indicate that even
275
00:17:48 --> 00:17:54
worms cannot multiply purely
exponentially forever,
276
00:17:51 --> 00:17:57
but ultimately their growth
levels off because they cannot
277
00:17:56 --> 00:18:02
find enough organic matter to
plow their way through.
278
00:18:01 --> 00:18:07
Those are the two differences.
I am not assuming a logistic
279
00:18:05 --> 00:18:11
growth law.
This is less sophisticated than
280
00:18:08 --> 00:18:14
the one I gave you for homework.
This assumes that sharks have
281
00:18:12 --> 00:18:18
absolutely nothing to eat except
these fish, which is not so bad.
282
00:18:17 --> 00:18:23
That's true,
more or less.
283
00:18:20 --> 00:18:26
284
00:18:25 --> 00:18:31
My plan is now,
with this model,
285
00:18:27 --> 00:18:33
let's start the analysis,
as you learned to do it.
286
00:18:30 --> 00:18:36
And we are going to run into
trouble at various places.
287
00:18:34 --> 00:18:40
And the troubles will then
illustrate these three points
288
00:18:38 --> 00:18:44
that I wanted to add to your
repertoire of things to do with
289
00:18:43 --> 00:18:49
nonlinear systems when you run
into trouble.
290
00:18:47 --> 00:18:53
And it will also increase your
understanding of the nonlinear
291
00:18:50 --> 00:18:56
systems.
The first thing we have to do
292
00:18:52 --> 00:18:58
is find the critical points.
I am going to assume that you
293
00:18:56 --> 00:19:02
are good at this and,
therefore, not spend a lot of
294
00:18:59 --> 00:19:05
time detailing the calculations.
I will simply write them on the
295
00:19:04 --> 00:19:10
board.
The first equation I am going
296
00:19:06 --> 00:19:12
to write down is x times minus a
plus by equals zero.
297
00:19:11 --> 00:19:17
And I assume you know why I am
298
00:19:13 --> 00:19:19
writing that down.
And the other equation will be
299
00:19:17 --> 00:19:23
y times c minus dx is zero.
300
00:19:20 --> 00:19:26
Those are the simultaneous
equations I have to find to find
301
00:19:24 --> 00:19:30
the critical point.
The first one is if the product
302
00:19:30 --> 00:19:36
of these is zero,
either x is zero or the other
303
00:19:34 --> 00:19:40
factor is.
So, from the second equation,
304
00:19:38 --> 00:19:44
if x is zero,
y has to be zero also.
305
00:19:42 --> 00:19:48
That is one critical point.
Now, if x is not zero then this
306
00:19:48 --> 00:19:54
factor has to be zero which says
that y must be equal
307
00:19:54 --> 00:20:00
to a over b.
And if y is equal to a over b,
308
00:19:59 --> 00:20:05
it is not zero here.
Therefore, this factor must be
309
00:20:02 --> 00:20:08
zero which says that x is equal
to c over d.
310
00:20:06 --> 00:20:12
You see right away that this
must be a simpler system because
311
00:20:10 --> 00:20:16
it is only producing two
critical points.
312
00:20:12 --> 00:20:18
Whereas, the system that you
did for homework had four.
313
00:20:16 --> 00:20:22
Here is one.
Well, let's just write them up
314
00:20:19 --> 00:20:25
here.
The critical points are zero,
315
00:20:21 --> 00:20:27
zero.
That doesn't look terribly
316
00:20:23 --> 00:20:29
interesting, but the other one
looks more interesting.
317
00:20:27 --> 00:20:33
It is c over d,
a over b.
318
00:20:31 --> 00:20:37
Well, let's take a look first
at the zero, zero critical
319
00:20:34 --> 00:20:40
point.
The origin, in other words.
320
00:20:37 --> 00:20:43
What does that look like?
321
00:20:39 --> 00:20:45
322
00:20:44 --> 00:20:50
Well, at the origin,
the linearization is extremely
323
00:20:47 --> 00:20:53
easy to do because I simply
ignore the quadratic terms,
324
00:20:51 --> 00:20:57
which are the product of two
small numbers,
325
00:20:55 --> 00:21:01
where these only have a single
small number in it.
326
00:20:58 --> 00:21:04
It is minus ax,
cy.
327
00:21:01 --> 00:21:07
In other words,
the linearization matrix is
328
00:21:03 --> 00:21:09
minus a, zero,
zero and cy gives me a
329
00:21:06 --> 00:21:12
coefficient c there.
330
00:21:08 --> 00:21:14
Now, I don't think at any point
I have ever explicitly told you
331
00:21:12 --> 00:21:18
that I hope you have learned
from the homework or maybe your
332
00:21:16 --> 00:21:22
recitation teacher,
but for heaven's sake,
333
00:21:19 --> 00:21:25
put this down in your little
books, if you have a diagonal
334
00:21:23 --> 00:21:29
matrix, for god's sake,
don't calculate its
335
00:21:26 --> 00:21:32
eigenvalues.
They are right in front of you.
336
00:21:30 --> 00:21:36
They are always the diagonal
elements.
337
00:21:32 --> 00:21:38
The eigenvalue,
you can check this out if you
338
00:21:35 --> 00:21:41
insist on writing the equation,
but trust me it is clear.
339
00:21:40 --> 00:21:46
If I, for example,
subtract c from the main
340
00:21:43 --> 00:21:49
diagonal, I am going to get a
determinant zero because the
341
00:21:47 --> 00:21:53
bottom row will be all zero.
The eigenvalues are negative a
342
00:21:51 --> 00:21:57
and c.
In other words,
343
00:21:53 --> 00:21:59
they have opposite signs.
This is a negative number.
344
00:21:56 --> 00:22:02
That is a positive number --
-- because a,
345
00:22:00 --> 00:22:06
b, c and d are always positive.
And, therefore,
346
00:22:03 --> 00:22:09
this is automatically a saddle.
You don't have to calculate
347
00:22:07 --> 00:22:13
anything.
It is all right in front of
348
00:22:09 --> 00:22:15
you.
It must be a saddle and,
349
00:22:11 --> 00:22:17
therefore, unstable because all
saddles are.
350
00:22:13 --> 00:22:19
And, in fact,
you can even draw the little
351
00:22:16 --> 00:22:22
picture of what the stuff looks
like near the origin without
352
00:22:19 --> 00:22:25
even bothering to calculate
eigenvectors,
353
00:22:22 --> 00:22:28
although it is extremely easy
to do.
354
00:22:24 --> 00:22:30
Just from common sense,
these are the sharks and these
355
00:22:27 --> 00:22:33
are the yumfish.
Well, if there are zero
356
00:22:31 --> 00:22:37
yumfish, in other words,
if I am on the sharks axis,
357
00:22:35 --> 00:22:41
the axis of sharks,
I die out.
358
00:22:38 --> 00:22:44
Well, forget about this side.
It's on the positive side.
359
00:22:42 --> 00:22:48
That makes sense.
But I die out because the
360
00:22:45 --> 00:22:51
sharks have nothing to eat.
Whereas, if I am on the yumfish
361
00:22:50 --> 00:22:56
axis I go this way.
I grow because,
362
00:22:52 --> 00:22:58
without any sharks to eat them,
the yumfish increase.
363
00:22:56 --> 00:23:02
So it must look like that.
And, therefore,
364
00:23:00 --> 00:23:06
the saddle must look like this.
The saddle curves hug those and
365
00:23:04 --> 00:23:10
go nearby.
Now, the other critical point
366
00:23:07 --> 00:23:13
is the interesting one.
And for that the analysis,
367
00:23:10 --> 00:23:16
in order that you don't spend
the rest of this period writing
368
00:23:14 --> 00:23:20
a's, b's, c's and d's,
I am going to make the
369
00:23:17 --> 00:23:23
simplifying assumption.
But it doesn't change
370
00:23:20 --> 00:23:26
qualitatively any of the
mathematics.
371
00:23:22 --> 00:23:28
It just makes it a little
easier to write everything out.
372
00:23:26 --> 00:23:32
I am going to assume that
everything is one.
373
00:23:30 --> 00:23:36
Well, in fact,
even this would be good enough,
374
00:23:33 --> 00:23:39
but let's make everything one.
I am going to assume this.
375
00:23:37 --> 00:23:43
And it is only to make the
writing simpler.
376
00:23:40 --> 00:23:46
It doesn't really change the
mathematics at all.
377
00:23:44 --> 00:23:50
Well, if everything is one,
in order to calculate the
378
00:23:47 --> 00:23:53
linearized system,
I'd better use the Jacobian.
379
00:23:51 --> 00:23:57
But perhaps it would be better
to write out explicitly what the
380
00:23:56 --> 00:24:02
system is.
The system now is x prime
381
00:23:58 --> 00:24:04
equals minus x plus xy,
all the coefficients are
382
00:24:03 --> 00:24:09
one.
And y prime is equal to y minus
383
00:24:07 --> 00:24:13
xy.
What is the Jacobian?
384
00:24:10 --> 00:24:16
Well, the Jacobian is the
partial of this with respect to
385
00:24:14 --> 00:24:20
x which is minus one plus y,
the partial with respect
386
00:24:19 --> 00:24:25
to y which is x,
the partial of this with
387
00:24:22 --> 00:24:28
respect to x,
which is minus y,
388
00:24:24 --> 00:24:30
and the partial with respect to
y, which is one minus x.
389
00:24:28 --> 00:24:34
But I want to evaluate that at
390
00:24:32 --> 00:24:38
the point one, one,
391
00:24:35 --> 00:24:41
which is the critical point.
It is the critical point
392
00:24:40 --> 00:24:46
because I have assumed that all
these parameters have the value
393
00:24:46 --> 00:24:52
one.
And when I do that,
394
00:24:48 --> 00:24:54
evaluating it at one, one,
395
00:24:51 --> 00:24:57
I get what matrix?
Well, I get zero,
396
00:24:54 --> 00:25:00
one, negative one,
zero.
397
00:24:57 --> 00:25:03
In other words,
398
00:25:01 --> 00:25:07
the linearized system is x
prime equals y and y
399
00:25:07 --> 00:25:13
prime equals negative x.
400
00:25:11 --> 00:25:17
Well, that is just the one we
analyzed before in terms of --
401
00:25:18 --> 00:25:24
402
00:25:23 --> 00:25:29
It's the one whose solutions
are circles.
403
00:25:25 --> 00:25:31
In other words,
what we find out is that the
404
00:25:28 --> 00:25:34
linearized system is what
geometric type?
405
00:25:32 --> 00:25:38
Saddle?
Node?
406
00:25:34 --> 00:25:40
Spiral?
None of those.
407
00:25:37 --> 00:25:43
It is a center.
The linearized system is a
408
00:25:44 --> 00:25:50
center.
409
00:25:47 --> 00:25:53
410
00:25:52 --> 00:25:58
It consists,
in other words,
411
00:25:54 --> 00:26:00
of a bunch of curves going
round and round next to each
412
00:25:58 --> 00:26:04
other.
Concentric circles,
413
00:25:59 --> 00:26:05
in fact.
Well, what is wrong with that?
414
00:26:03 --> 00:26:09
Now, we are in deep trouble.
We are now in trouble because
415
00:26:07 --> 00:26:13
that is a borderline case.
Let me remind you of what the
416
00:26:12 --> 00:26:18
borderline cases are.
When we drew that picture,
417
00:26:15 --> 00:26:21
this is from last week's
problem set so you have a
418
00:26:19 --> 00:26:25
perfect excuse for forgetting it
totally.
419
00:26:22 --> 00:26:28
But I am trying to remind you
of it.
420
00:26:25 --> 00:26:31
That is another reason why I am
doing this.
421
00:26:30 --> 00:26:36
Remember that picture I asked
you about that appeared on the
422
00:26:34 --> 00:26:40
computer screen?
Let's make it a little flatter
423
00:26:38 --> 00:26:44
so that I can have room to write
in.
424
00:26:40 --> 00:26:46
This is a certain parabola
whose equation you are dying to
425
00:26:45 --> 00:26:51
tell me, but I am not asking.
There is the trace,
426
00:26:48 --> 00:26:54
this is the determinant,
and the characteristic equation
427
00:26:53 --> 00:26:59
is related to the values of
these numbers like plus d equals
428
00:26:57 --> 00:27:03
zero.
Then these were spiral synchs.
429
00:27:02 --> 00:27:08
This was the region of spiral
sources.
430
00:27:07 --> 00:27:13
That is the abbreviation for
sources.
431
00:27:11 --> 00:27:17
These were nodal synchs and
these were the nodal sources.
432
00:27:18 --> 00:27:24
And down here were the saddles.
And where were the centers?
433
00:27:24 --> 00:27:30
The centers were along this
line.
434
00:27:30 --> 00:27:36
The centers,
there weren't many of them,
435
00:27:32 --> 00:27:38
and they were the separation
for these two regions.
436
00:27:36 --> 00:27:42
I will put that down.
These are the centers.
437
00:27:39 --> 00:27:45
These other borderlines
correspond to other things.
438
00:27:42 --> 00:27:48
These are defective
eigenvalues, zero eigenvalues
439
00:27:45 --> 00:27:51
and so on.
But let's concentrate on just
440
00:27:48 --> 00:27:54
one of them.
You will find the others
441
00:27:50 --> 00:27:56
discussed in the notes,
GS7.
442
00:27:52 --> 00:27:58
But if you get this idea then
the rest is just details.
443
00:27:56 --> 00:28:02
I think it will be perfectly
clear.
444
00:28:00 --> 00:28:06
What is wrong with the center?
The answer is that if we are on
445
00:28:04 --> 00:28:10
the center, for example,
this system corresponds to the
446
00:28:08 --> 00:28:14
trace being zero and the
determinant being plus one.
447
00:28:12 --> 00:28:18
It corresponds to t equals zero
and d equals one.
448
00:28:16 --> 00:28:22
This point, in other words.
Now, if I wiggle the
449
00:28:19 --> 00:28:25
coefficients of the matrix just
a little bit,
450
00:28:23 --> 00:28:29
just change them a little bit,
what I am going to do is move
451
00:28:27 --> 00:28:33
off this pink line.
And I might move to here or I
452
00:28:31 --> 00:28:37
might move a little way to
there.
453
00:28:34 --> 00:28:40
But, if I do that,
I change the geometric type.
454
00:28:37 --> 00:28:43
In other words,
being a borderline,
455
00:28:39 --> 00:28:45
a slight change of the
parameters can change what it
456
00:28:42 --> 00:28:48
changes to.
And, in fact,
457
00:28:44 --> 00:28:50
that is geometrically clear.
What does a center look like?
458
00:28:48 --> 00:28:54
A center looks like this,
a bunch of curves going around
459
00:28:51 --> 00:28:57
all in the same direction,
like concentric circles or
460
00:28:55 --> 00:29:01
maybe ellipses or something like
that.
461
00:28:59 --> 00:29:05
If I deform the picture just a
little bit, well,
462
00:29:03 --> 00:29:09
I might change it into
something that looks like this
463
00:29:07 --> 00:29:13
where, after they go around they
don't quite meet up with where
464
00:29:12 --> 00:29:18
they were to start with.
And I am going to get a spiral
465
00:29:17 --> 00:29:23
synch.
Or, I might do the deformation
466
00:29:20 --> 00:29:26
by going around once and I'm a
little outside of where I was.
467
00:29:25 --> 00:29:31
In which case it's going to be
a spiral source.
468
00:29:31 --> 00:29:37
The fact that just changing
these curves a little bit can
469
00:29:35 --> 00:29:41
change the picture to this or to
that corresponds to the fact
470
00:29:39 --> 00:29:45
that if you are on here with
this value of t and d,
471
00:29:43 --> 00:29:49
zero and one,
and just change t and d a
472
00:29:46 --> 00:29:52
little bit, you are going to
move off into these regions.
473
00:29:51 --> 00:29:57
You might, of course,
stay on the pink line.
474
00:29:54 --> 00:30:00
It is not very likely.
Where does this leave us?
475
00:29:59 --> 00:30:05
Well, if the linear system is
not stable in the sense that if
476
00:30:02 --> 00:30:08
you change the parameters a
little bit it doesn't change the
477
00:30:06 --> 00:30:12
type.
This is, after all,
478
00:30:07 --> 00:30:13
only an approximation to the
nonlinear system.
479
00:30:10 --> 00:30:16
If in this approximation you
cannot really predict the
480
00:30:13 --> 00:30:19
behavior of very well when you
change the parameters,
481
00:30:17 --> 00:30:23
then from it you cannot tell
what the original system looked
482
00:30:20 --> 00:30:26
like.
In other words,
483
00:30:23 --> 00:30:29
the nonlinear system at one,
one might be any one
484
00:30:28 --> 00:30:34
of the possibilities,
still a center or it might be a
485
00:30:32 --> 00:30:38
spiral synch or it might be a
spiral source.
486
00:30:36 --> 00:30:42
Any one of those three is a
possibility.
487
00:30:39 --> 00:30:45
It couldn't be a saddle because
that is too far away.
488
00:30:44 --> 00:30:50
It couldn't be a node.
That is too far away,
489
00:30:48 --> 00:30:54
too.
But it could wander into either
490
00:30:51 --> 00:30:57
of these regions and,
therefore, the picture you
491
00:30:55 --> 00:31:01
cannot tell which of these three
it is just from this type of
492
00:31:00 --> 00:31:06
critical point analysis.
Well, that was Volterra's
493
00:31:06 --> 00:31:12
problem.
By the way, the person who
494
00:31:08 --> 00:31:14
introduced these equations and
studied them systematically in
495
00:31:13 --> 00:31:19
the way in which we are doing it
here was Volterra.
496
00:31:17 --> 00:31:23
And, in fact,
he was interested in sharks and
497
00:31:21 --> 00:31:27
food fish, as they were called.
What do we do?
498
00:31:24 --> 00:31:30
Well, you have to be smart.
What Volterra did was he went
499
00:31:29 --> 00:31:35
to method number one.
Let's do it.
500
00:31:33 --> 00:31:39
501
00:31:38 --> 00:31:44
Volterra said I got my
equations x prime equals minus x
502
00:31:45 --> 00:31:51
plus xy.
y prime is equal to,
503
00:31:51 --> 00:31:57
these are the food fish,
y minus xy.
504
00:31:58 --> 00:32:04
Let's eliminate t.
And my problem is,
505
00:32:03 --> 00:32:09
of course, I am trying to
determine what type of critical
506
00:32:07 --> 00:32:13
point one, one is.
And the method we have used up
507
00:32:11 --> 00:32:17
until now has failed because it
gave us a borderline case,
508
00:32:16 --> 00:32:22
which is from that we cannot
deduce.
509
00:32:19 --> 00:32:25
He said let's eliminate t.
And we get dy over dx
510
00:32:23 --> 00:32:29
equals, I am going to factor,
y times one minus x on top.
511
00:32:29 --> 00:32:35
And on the bottom factor x
negative one plus y.
512
00:32:34 --> 00:32:40
You could solve that before you
came into 18.03,
513
00:32:39 --> 00:32:45
right?
This you can separate
514
00:32:42 --> 00:32:48
variables.
Let's separate variables.
515
00:32:46 --> 00:32:52
The y's all go on one side,
so y goes down here.
516
00:32:52 --> 00:32:58
It is (y minus one) over y dy
equals --
517
00:33:00 --> 00:33:06
On the other side the dx gets
moved up, one minus x over x dx.
518
00:33:03 --> 00:33:09
Now, of course,
you wouldn't dream of using
519
00:33:06 --> 00:33:12
partial fractions on this.
It would be illegal because,
520
00:33:09 --> 00:33:15
even though it is a rational
function, the quotient of two
521
00:33:13 --> 00:33:19
polynomials, the degree of the
top is not smaller than the
522
00:33:16 --> 00:33:22
degree of the bottom.
In other words,
523
00:33:18 --> 00:33:24
it is a partial fraction so
this degree must be bigger than
524
00:33:22 --> 00:33:28
that one.
And it isn't so you would have
525
00:33:24 --> 00:33:30
to first divide.
And if you divide by that then
526
00:33:27 --> 00:33:33
there is no point in using
partial fractions at all.
527
00:33:32 --> 00:33:38
528
00:33:37 --> 00:33:43
Of course, you would have done
this by basic instant,
529
00:33:41 --> 00:33:47
I know.
What is the solution?
530
00:33:43 --> 00:33:49
It is y minus log y equals log
x minus x plus some constant.
531
00:33:49 --> 00:33:55
That is not the final constant
532
00:33:53 --> 00:33:59
I want so I will give it a
subscript one to indicate I want
533
00:33:58 --> 00:34:04
more.
This is very hard to see what
534
00:34:02 --> 00:34:08
that curve looks like.
We can make it look better by
535
00:34:05 --> 00:34:11
exponentiating.
If I exponentiate it,
536
00:34:08 --> 00:34:14
going back to high school
mathematics, but I know from
537
00:34:11 --> 00:34:17
experience that many of you are
not too good at this step,
538
00:34:16 --> 00:34:22
so that is another reason I am
doing it, it will be
539
00:34:20 --> 00:34:26
e to the y, times,
that part is easy because
540
00:34:23 --> 00:34:29
pluses and minuses change into
times, times e to the negative
541
00:34:27 --> 00:34:33
log y.
Well, e to the log y is y.
542
00:34:31 --> 00:34:37
If I put a minus sign in front
543
00:34:35 --> 00:34:41
of that, that sends it into the
denominator, so instead of being
544
00:34:40 --> 00:34:46
y it is one over y
Equals x, e to log x is x,
545
00:34:44 --> 00:34:50
e to the negative x
546
00:34:48 --> 00:34:54
is, therefore,
times one over e to the x.
547
00:34:51 --> 00:34:57
And that is times the
548
00:34:53 --> 00:34:59
exponential of c1,
which I will call c2.
549
00:34:58 --> 00:35:04
And now if I combine them all
and put them all on one side it
550
00:35:04 --> 00:35:10
is x over e to the x times y
over e to the y is equal to yet
551
00:35:10 --> 00:35:16
another constant,
one over c2.
552
00:35:14 --> 00:35:20
This is my final constant so
553
00:35:18 --> 00:35:24
let's call that c.
In other words,
554
00:35:21 --> 00:35:27
the integral curves are the
graphs of this equation for
555
00:35:27 --> 00:35:33
different values of the constant
c.
556
00:35:32 --> 00:35:38
Well, of course you've graphed
an equation like that all the
557
00:35:37 --> 00:35:43
time.
What am I going to do with this
558
00:35:41 --> 00:35:47
stupid thing?
Well, I deliberately picked
559
00:35:44 --> 00:35:50
something which involved all
your learning up until now.
560
00:35:50 --> 00:35:56
We are now going into 18.02,
right?
561
00:35:53 --> 00:35:59
You looked at that in 18.02 you
would say this is the contour
562
00:35:59 --> 00:36:05
curve, these are contour curves
of the function --
563
00:36:05 --> 00:36:11
Well, let's write it out the
other way.
564
00:36:07 --> 00:36:13
It doesn't make any difference,
but you are more likely to have
565
00:36:10 --> 00:36:16
seen the function in this form.
x over e to the x
566
00:36:13 --> 00:36:19
you won't recognize,
but this you will.
567
00:36:15 --> 00:36:21
In fact, we have had it before
this term.
568
00:36:18 --> 00:36:24
It is one of the kinds of
solutions you can add to second
569
00:36:21 --> 00:36:27
order equations.
Times y e to the negative y
570
00:36:23 --> 00:36:29
equals c.
It is the contour curves of
571
00:36:26 --> 00:36:32
this function.
Let's call that function,
572
00:36:29 --> 00:36:35
let's say, h of (x,y).
573
00:36:32 --> 00:36:38
Well, of course you could throw
it on Matlab,
574
00:36:35 --> 00:36:41
as you did in 18.02 maybe,
and ask Matlab to plot the
575
00:36:39 --> 00:36:45
function.
But Volterra didn't have that
576
00:36:42 --> 00:36:48
luxury.
He was smart,
577
00:36:43 --> 00:36:49
too.
So let's be smart instead.
578
00:36:47 --> 00:36:53
579
00:36:53 --> 00:36:59
What is the function x?
Let's use a neutral variable
580
00:36:56 --> 00:37:02
like u and plot u e to the
negative u.
581
00:37:00 --> 00:37:06
18.01.
At zero, it has the value zero.
582
00:37:03 --> 00:37:09
When u is small this has
approximately the value one and,
583
00:37:08 --> 00:37:14
therefore, it starts out like
the function u.
584
00:37:12 --> 00:37:18
As u goes to infinity,
you know by L'Hopital's rule
585
00:37:16 --> 00:37:22
that this goes to zero so it
ends up like this.
586
00:37:20 --> 00:37:26
Well, what on earth could it
possibly do except rise to a
587
00:37:25 --> 00:37:31
maximum?
But where is that maximum?
588
00:37:30 --> 00:37:36
It is easy to see there is a
unique maximum just by 18.01.
589
00:37:34 --> 00:37:40
Take the derivative,
find out where the maximum
590
00:37:37 --> 00:37:43
point is and you will find,
perhaps you have already done
591
00:37:41 --> 00:37:47
it, this is at one.
The maximum occurs at one,
592
00:37:44 --> 00:37:50
that is where the derivative is
zero and the value there is one
593
00:37:49 --> 00:37:55
times e to the minus one.
594
00:37:52 --> 00:37:58
It is one over e,
in other words.
595
00:37:55 --> 00:38:01
That is what the curve looks
like.
596
00:37:57 --> 00:38:03
What do the contour curves of
this look like?
597
00:38:02 --> 00:38:08
Well, if we are looking at the
contour curves of the function
598
00:38:07 --> 00:38:13
h, so here is x,
here is y, and we want the
599
00:38:11 --> 00:38:17
contour curves of the function h
of (x, y),
600
00:38:16 --> 00:38:22
what do I know?
Where is its maximum?
601
00:38:20 --> 00:38:26
The maximum point of h of x,
y is where?
602
00:38:24 --> 00:38:30
Well, h is the product x e to
the negative x times y e to the
603
00:38:30 --> 00:38:36
minus y.
604
00:38:35 --> 00:38:41
This has its maximum at one,
this has its maximum at one,
605
00:38:39 --> 00:38:45
so where does h of (x,
y) head?
606
00:38:43 --> 00:38:49
Well, you have two factors.
One makes each of them the
607
00:38:48 --> 00:38:54
biggest they can be.
The maximum must be exactly at
608
00:38:52 --> 00:38:58
that critical point one,
one.
609
00:38:55 --> 00:39:01
That is our maximum point.
Let's make it conspicuous.
610
00:39:01 --> 00:39:07
This is the maximum point.
It is the point one,
611
00:39:05 --> 00:39:11
one.
It is the point that makes the
612
00:39:09 --> 00:39:15
function biggest.
Now, how about the contour
613
00:39:13 --> 00:39:19
curves?
Well, this is the top of the
614
00:39:16 --> 00:39:22
mountain.
What is it along the axes?
615
00:39:19 --> 00:39:25
Along the axes,
when either x or y is zero,
616
00:39:23 --> 00:39:29
the function has the value
zero, so it is biggest there.
617
00:39:28 --> 00:39:34
It has the value zero here.
So it is zero value.
618
00:39:33 --> 00:39:39
What are the contour curves?
Well, we must have a mountain
619
00:39:38 --> 00:39:44
peak here.
This is the unique maximum
620
00:39:41 --> 00:39:47
point.
In fact, it is easy to see the
621
00:39:44 --> 00:39:50
contour curves just surround it
like that.
622
00:39:47 --> 00:39:53
Now, the reason they cannot be
spirals, well,
623
00:39:51 --> 00:39:57
in the first place,
you never see contour curves
624
00:39:55 --> 00:40:01
that were spirals.
That is a good reason.
625
00:40:00 --> 00:40:06
It is a mountain.
That is the way contour curves
626
00:40:03 --> 00:40:09
of a mountain look.
But, all right,
627
00:40:06 --> 00:40:12
that is not a good argument.
But notice that along each
628
00:40:10 --> 00:40:16
horizontal line here,
I want to know how many times
629
00:40:14 --> 00:40:20
can it intersect the contour
curve?
630
00:40:16 --> 00:40:22
That is the same as asking
along one of these lines how
631
00:40:20 --> 00:40:26
many times could it intersect?
See, this is a graph of the
632
00:40:25 --> 00:40:31
function along this horizontal
line.
633
00:40:29 --> 00:40:35
And, therefore,
how many times can it intersect
634
00:40:32 --> 00:40:38
one of the contour curves the
same number of times that a
635
00:40:36 --> 00:40:42
horizontal line can intersect
this curve?
636
00:40:39 --> 00:40:45
Twice.
Only once if you are exactly at
637
00:40:42 --> 00:40:48
that height and after that
never.
638
00:40:44 --> 00:40:50
But here it can intersect each
contour curve only twice which
639
00:40:48 --> 00:40:54
means these things cannot be
spirals.
640
00:40:51 --> 00:40:57
Otherwise, they would intersect
those horizontal and vertical
641
00:40:55 --> 00:41:01
lines many times instead of just
twice.
642
00:41:00 --> 00:41:06
That is what it looks like.
In fact, we can even put in the
643
00:41:04 --> 00:41:10
direction without any effort.
We know that the sharks die out
644
00:41:09 --> 00:41:15
without any food fish and the
food fish grow.
645
00:41:13 --> 00:41:19
Well, I think this has to be
clockwise.
646
00:41:16 --> 00:41:22
Of course, the guys nearby must
be going the same way.
647
00:41:21 --> 00:41:27
And, therefore,
the guys near them must be
648
00:41:24 --> 00:41:30
going the same way.
And it is sort of a domino
649
00:41:28 --> 00:41:34
effect.
The direction spreads and there
650
00:41:32 --> 00:41:38
is only one compatible way of
making these.
651
00:41:35 --> 00:41:41
They all must be going
clockwise.
652
00:41:37 --> 00:41:43
What is happening,
actually?
653
00:41:39 --> 00:41:45
At this point there are a lot
of sharks but not so many food
654
00:41:44 --> 00:41:50
fish.
The sharks eat the few
655
00:41:46 --> 00:41:52
remaining food fish,
and now the sharks start to die
656
00:41:49 --> 00:41:55
out because they have nothing to
eat.
657
00:41:52 --> 00:41:58
When they are practically gone,
the food fish can start growing
658
00:41:56 --> 00:42:02
again and grow and grow.
For a while they grow happily,
659
00:42:02 --> 00:42:08
and then the sharks suddenly
start being able to find them
660
00:42:06 --> 00:42:12
again.
And the few remaining sharks
661
00:42:09 --> 00:42:15
start eating them,
the population of sharks grows,
662
00:42:13 --> 00:42:19
the food fish start dying out
again, and the cycle starts all
663
00:42:18 --> 00:42:24
over again.
Now, I wanted to talk about the
664
00:42:21 --> 00:42:27
qualitative behavior.
This is, in some ways,
665
00:42:24 --> 00:42:30
the most important part of the
lecture.
666
00:42:29 --> 00:42:35
I wanted to discuss,
just as we did at the beginning
667
00:42:35 --> 00:42:41
of the term, the effect of
fishing with nets at a constant
668
00:42:43 --> 00:42:49
rate k.
669
00:42:45 --> 00:42:51
670
00:42:51 --> 00:42:57
Just as near the beginning of
the term, we talked about
671
00:42:54 --> 00:43:00
fishing a single population.
The only difference is,
672
00:42:57 --> 00:43:03
now there are two populations.
Both sharks and the food fish.
673
00:43:02 --> 00:43:08
Now, what happens to the
equations?
674
00:43:05 --> 00:43:11
Well, the equations start out
just as they were.
675
00:43:10 --> 00:43:16
I am now reverting to a and b.
You will see why.
676
00:43:14 --> 00:43:20
It's because we need the
general coefficients.
677
00:43:18 --> 00:43:24
Plus bxy.
But, if we are fishing with
678
00:43:21 --> 00:43:27
nets, then a certain fraction of
the sharks in the sea get hauled
679
00:43:27 --> 00:43:33
in by the net.
And say the fishing rate is k.
680
00:43:32 --> 00:43:38
There is a fishing term.
We remove minus k times a
681
00:43:36 --> 00:43:42
certain fraction of the sharks
in the sea.
682
00:43:40 --> 00:43:46
How about the food fish?
Well, we remove them,
683
00:43:44 --> 00:43:50
too.
And we don't distinguish
684
00:43:47 --> 00:43:53
between sharks and food fish.
This is cy minus dxy.
685
00:43:51 --> 00:43:57
But we also remove a certain
686
00:43:54 --> 00:44:00
fraction of them.
Minus k times y.
687
00:43:58 --> 00:44:04
What, therefore,
is the new system?
688
00:44:03 --> 00:44:09
The new system is therefore x
prime equals minus a plus k.
689
00:44:10 --> 00:44:16
I am combining those two x
terms.
690
00:44:14 --> 00:44:20
That is times x.
Plus bxy.
691
00:44:17 --> 00:44:23
And
the y prime is c minus k times y
692
00:44:24 --> 00:44:30
minus dxy.
693
00:44:30 --> 00:44:36
To solve these,
I do not have to go through the
694
00:44:33 --> 00:44:39
analysis.
All I have to do is change the
695
00:44:37 --> 00:44:43
numbers, change these
coefficients from a to a plus k
696
00:44:41 --> 00:44:47
and c minus k.
The old critical point was the
697
00:44:46 --> 00:44:52
point c over d,
a over b.
698
00:44:49 --> 00:44:55
The new critical point with
699
00:44:52 --> 00:44:58
fishing is what?
Well, the parameters have been
700
00:44:56 --> 00:45:02
changed by the addition of k.
The new critical point is c
701
00:45:02 --> 00:45:08
minus k over d and a plus k over
d.
702
00:45:07 --> 00:45:13
What is the effect of fishing?
703
00:45:11 --> 00:45:17
Well, if the old critical point
was over here,
704
00:45:15 --> 00:45:21
let's say this is the point c
over d and this is the
705
00:45:21 --> 00:45:27
point a over b,
that is the old critical point
706
00:45:27 --> 00:45:33
that was over here.
The result of fishing is to
707
00:45:33 --> 00:45:39
lower the value of x and raise
the value here.
708
00:45:38 --> 00:45:44
The new critical point is,
this gets lowered to c minus k
709
00:45:44 --> 00:45:50
over d,
and this gets raised to a plus
710
00:45:50 --> 00:45:56
k over b.
In other words,
711
00:45:54 --> 00:46:00
the new point is there.
The effect of fishing is to --
712
00:46:02 --> 00:46:08
It does not treat the sharks
and the yumfish equally.
713
00:46:07 --> 00:46:13
The effect of fishing lowers
the shark population.
714
00:46:11 --> 00:46:17
See, the critical point gives
sort of the average shark
715
00:46:17 --> 00:46:23
population.
Of course, it cycles around
716
00:46:20 --> 00:46:26
these.
But, on the average,
717
00:46:23 --> 00:46:29
this gives how many sharks
there are and how many food fish
718
00:46:28 --> 00:46:34
there are.
The new critical point with
719
00:46:33 --> 00:46:39
fishing lowers the shark
population and raises the food
720
00:46:39 --> 00:46:45
fish population.
That is not intuitive.
721
00:46:42 --> 00:46:48
And, in fact,
that was observed
722
00:46:45 --> 00:46:51
experimentally at a slightly
different context.
723
00:46:50 --> 00:46:56
And that is why Volterra
started working on the problem.
724
00:46:55 --> 00:47:01
I will need three more minutes.
The most interesting
725
00:47:01 --> 00:47:07
application of all is not to
sharks and food fish.
726
00:47:03 --> 00:47:09
I cannot assume that you are
dramatically interested in how
727
00:47:07 --> 00:47:13
many of them there are in the
ocean, but you might be more
728
00:47:10 --> 00:47:16
interested in this.
729
00:47:12 --> 00:47:18
730
00:47:20 --> 00:47:26
That thing about lowering is
called Volterra's principle.
731
00:47:25 --> 00:47:31
Put that in your books.
Volterra's principle.
732
00:47:29 --> 00:47:35
Volterra is spelt over there.
733
00:47:32 --> 00:47:38
734
00:47:43 --> 00:47:49
-- has found more modern
applications than sharks.
735
00:47:48 --> 00:47:54
Suppose you consider fish in a
pond.
736
00:47:53 --> 00:47:59
You have mosquito larvae which
breed in the pond.
737
00:47:58 --> 00:48:04
This happens.
And then suddenly there is a
738
00:48:03 --> 00:48:09
plague of mosquitoes and
concerned citizens.
739
00:48:07 --> 00:48:13
And this is what happened in
the '50s.
740
00:48:11 --> 00:48:17
That was before you were born,
but not before I was.
741
00:48:16 --> 00:48:22
What happened was a lot of
mosquitoes.
742
00:48:19 --> 00:48:25
Everybody said the mosquitoes
breed in the little stagnant
743
00:48:25 --> 00:48:31
ponds so spray DDT on them.
DDT them.
744
00:48:30 --> 00:48:36
Dump it in the ponds.
That will kill all the larvae
745
00:48:34 --> 00:48:40
and we won't get bitten anymore.
When you put DDT in the pond,
746
00:48:39 --> 00:48:45
as people did not realize at
the time because these things
747
00:48:44 --> 00:48:50
were new, of course you kill the
mosquitoes, but you also kill
748
00:48:49 --> 00:48:55
the fish because DDT is
poisonous to fish.
749
00:48:53 --> 00:48:59
What, in effect,
you are doing mathematically is
750
00:48:57 --> 00:49:03
the same as harvesting the fish
population with the sharks and
751
00:49:02 --> 00:49:08
the food fish.
The result is,
752
00:49:06 --> 00:49:12
the fish are the predators
because they eat the mosquito
753
00:49:11 --> 00:49:17
larvae, big food,
there was a certain
754
00:49:15 --> 00:49:21
equilibrium, according to
Volterra's principle.
755
00:49:20 --> 00:49:26
With DDT that equilibrium moves
to here.
756
00:49:24 --> 00:49:30
In other words,
the effect of indiscriminately
757
00:49:28 --> 00:49:34
spraying the pond with DDT is to
increase the number of
758
00:49:34 --> 00:49:40
mosquitoes and kill fish.
And, in fact,
759
00:49:39 --> 00:49:45
that is exactly what was
observed.
760
00:49:42 --> 00:49:48
The same thing was observed
with the bird population and
761
00:49:48 --> 00:49:54
insects.
Spraying trees for insects to
762
00:49:51 --> 00:49:57
get rid of some pests ends up
killing more birds than it does
763
00:49:58 --> 00:50:04
insects and the insects
increase.
764
00:50:01 --> 00:50:07
Thanks.