Today's lecture is going to be basically devoted to working out
a single example of a nonlinear system, but it is a very good
example because it illustrates three things which you really
have to know about nonlinear systems.
I have indicated them by three cryptic words on the board,
but you will see at different points in the lecture what they
refer to. Each of these represents
something you need to know about nonlinear systems to be able to
effectively analyze them. In addition,
I am not allowed to collect any more work, according to the
faculty rules, after today.
And, of course, I won't.
But, nonetheless, the final will contain material
from the reading assignment G7, and I will be touching on all
of these today, and 7.4, just a couple of pages
of that. You will see its connection.
It discusses the same example we are going to do today.
And then I suggest those three exercises to try to solidify
what the lecture is about.
Instead of introducing the nonlinear system right away that
we are going to talk about, I would like to first explain,
so that it won't interrupt the presentation later,
what I mean by a conversion to a first-order equation.
For that why don't we look at --
Maybe I can do it here. Our system looks like this,
dx over dt. For once, I am not writing x
prime because I want to explicitly indicate what the
independent variable is. It is a nonlinear autonomous
system, so there is no t on the right-hand side.
But this is not a simple linear function, ax plus by.
It is more complicated,
like you had in the predator-prey robin-earthworm
problem that you worked on last night.
And the other equation will be g of (x, y).
This is the system, and explicitly it is nonlinear
in general and autonomous, as I have indicated on the
right-hand side. Now, remember that the
geometric picture of this was as a velocity field.
You made a velocity field out of the vectors whose components
were f and g, so that is (f)i plus (g)j,
but what it looked like was a
plane filled up with a lot of vectors pointing different ways
according to the velocity at the point was.
And side-by-side with that went the solutions.
A typical solution was x equals x of t,
y equals y of t represented as a column vector.
And that is a parametric equation.
And the geometric picture of that was given by its
trajectory. When you plotted it,
it was a trajectory. It had not only the right
direction at each point, in other words,
but it also had to have the right velocity at each point.
In other words, sometimes the point was moving
rapidly and sometimes it was moving more slowly along that
path. And the way it moved was an
important part of the solution. Now, what I plan to do is --
The conversation that I am talking about takes place by
eliminating t.
Now, why would one want to do that?
t, after all, is an essential part of the
solution. It is an essential part of this
picture because without t you would not know how long the
arrows were supposed to be. You would still know their
direction. And, of course,
it occurs in here. Now, let's take the first step
and eliminate t from the system itself.
As you see, that is very easy to do because the right-hand
side has no t in it anyway and the left-hand side has the t
only as the denominators of the differential quotients.
The obvious way to eliminate t is just to divide one equation
by the other. And since usually we think of
not dx by dy, but dy by dx,
I will vow -- What is dy over dx?
Well, if I divide this equation by that equation,
dy by dt divided by dx by dt,
according to the chain rule, or according to commonsense,
you cancel the dt's and you get dy over dx.
And, on the right-hand side, you get g of (x,
y) divided by f of (x, y).
But what is that? That is the equation of the
first day of the term. By taking that single step,
I convert this nonlinear system, which we virtually never
find explicit solutions to, into an ordinary first order
equation which, in fact, you also don't usually
find explicit solutions to, except with our narrow blinders
on. We only get problems at the
beginning of 18.03 where there will be an explicit solution to
this. I have converted the nonlinear
system to a single first order equation to which I can apply
the usual first order methods that include hope that I will be
able to solve it. Now, what do I lose?
Well, what happens to this picture?
If you don't have the t in it anymore then you don't have
velocity, you don't have arrows with a length to them because
the length gives you how fast the point is going.
Without t I don't know at any point how fast it is supposed to
be going. All I know at any point (x,
y) is no longer dx over dt and dy over dt.
All I know is dy over dx.
In other words,
the corresponding picture, if I eliminate t from this
picture all that is left is the directions of each of these
arrows. What disappears is their
length. And, in fact,
since dy over dx is purely a slope,
I cannot even tell whether the point is traveling this way or
that way. It doesn't make any sense to
have the point traveling anymore since there is no time in which
it can do its traveling. So that picture gets changed to
the direction field. The corresponding picture now,
all you do is take each of these arrows,
you snip it off to a standard length, being careful to snip
off the little pointy end, and it becomes nothing but a
line element. And all it has is a slope.
What corresponds to the velocity field here is now just
our slope field or our direction field.
All we know is the slope at each point.
How about the solution? Well, if I have eliminated
time, the solution is no longer a pair of parametric equations.
It is just an equation involving x and y.
I can hope that the solution will be explicit,
y equals y of x, but you already know from the
first days of the term that sometimes it isn't.
Let's call that explicit solution.
Often you have to settle for it.
And sometimes it is not a pain. It is something that is even
better. Settle for a solution which
looks like this which is y defined implicitly as a function
of x. And what does the picture of
that look like? Well, the picture of this is
now simply an integral curve. It is not a trajectory anymore.
It is what we called an integral curve.
I am reminding you of the very first day of the term,
or maybe the second day. And all it has at each point
the right slope because that is all the field is telling me now.
It only has a slope at each point.
It doesn't have any magnitude or direction.
It has the direction, but it doesn't have direction
in the sense of an arrow telling me whether it is going this way
or the opposite way. That is the picture with t in
it. These are the corresponding
degeneration of that. It is a coarsening,
it's a cheapening of it. It is throwing away
information. Throw away all the information
that had to do with t, and we are back in the first
days of the terms with a single first order equations involving
only x and y with solutions that involve only x and y with curves
that are the graphs of these solutions but again have no t in
them. In effect, they are just the
paths of the trajectories. Whereas, the trajectory is the
point that shows actually how the point is moving along in
time faster or slower at various points.
Now, why would one want to lose information?
Well, the great gain is that this might be solvable.
Whereas, this almost certainly is not.
That is a big plus. For example,
let's illustrate it on the very simplest possible case.
I cannot illustrate it on a nonlinear system very well,
or not right now because, in general, nonlinear systems
are not solvable. Let's take an easy example.
Suppose the system were a linear one, let's say this one
that we have talked about before, in fact.
That is a simple linear two-by-two system.
This means the derivatives with respect to time.
Well, you know that the solutions we talked about in
fact, last time, typical solutions that would be
something like (x, y) equals a constant times,
let's say if x is cosine t then y would be the
derivative of that, which is negative sine t.
Or, another one would be c2
times sine t, cosine t.
And those, of course, are circles.
They are parametrized circles, so they are circles that go
around, in fact, in this direction.
And they have a certain velocity to them.
Now what do I do if I follow this plan of eliminating t?
Well, if I eliminate t directly from the system,
what will I get? I will get dy by dx.
I divide this equation by that one.
-- is equal to negative x over.
Oh, well, of course, that is solvable.
You were able to solve by 18.01 methods before you ever come
into this course. By separation of variables,
it is y dy equals negative x dx,
which integrates to be one-half y squared equals negative
one-half x squared plus a constant.
And after multiplying through
by two and moving things around it becomes x squared plus y
squared. The nicest way to say it is
implicitly, x squared plus y squared equals some positive
constant. These are the circles.
Now, can I eliminate t? I could also take one of these
solutions and eliminate t. If I square this,
the best way to do it is not to use our cosines and arcsines and
whatnot, which will just get you totally lost.
Square this, square that and add them
together. And you conclude that x squared
plus y squared is equal to one.
Or, if c1 is not one it is equal to c1 squared.
You could eliminate t from the solution the way you eliminate t
from a pair of parametric equations, and you get it the
same way that x squared plus y squared, in this case,
would be c1 squared, I guess.
In some sense, what we are doing is cycling
around to the beginning of the term.
In fact, this whole lecture, as you will see,
is about cycles of one sort or another.
But I keep thinking of that great line of poetry,
"in my end is my beginning" or maybe it's "in my beginning is
my end". I think both lines occur.
But that is what is happening here.
This is the beginning of the course and this is the end of
the course, and they have this almost trivial relation between
them. But notice the totally
different methods used for analyzing this,
where the t is included, than from analyzing this where
there is no t. The goals are different,
what you look for are different, everything is
different, and yet it is almost the same problem.
I promised you a nonlinear example.
I guess it is time to see what that is.
It is going to be another predator-prey equation,
but one which is, in some ways,
simpler than the one I gave you for homework.
The predator is going to be x because you are used to that
from the robins. I am keeping the same
predator-prey variables. And the prey will be y.
And now just notice the small difference from what I gave you
before. For one thing,
I am not giving you specific numbers.
I am going to, at the beginning at least,
do some of the analysis at the beginning using letters,
using parameters. a, b, c, d are always going to
be positive constants. I am going to assume that x
prime equals minus ax.
This represents the predator dying out if there is no prey
there, but if there is something to eat that term represents the
predator meeting up with the prey and gobbling it up.
And b is the coefficient. How about without predictors,
the prey will multiply and be fruitful.
Unfortunately they get eaten, and so there will be a term
that looks like this. Now, there are two basic
differences between these simple equations and the slightly more
complicated ones you had with the robin-earthworm equations.
Namely, in the first place, I am assuming that these guys,
let's give them names. I want you to remember which is
x and which is y, so we are going to think of
these are "sharx". [LAUGHTER]
And these will be food fish, so let's make them "yumfish".
The fact that with your robins you had a positive term here.
I made this 2x. And now it is minus a times x.
What is the difference? The difference is that robins
have other things to eat, so even if there are not any
worms, a robin will survive. It could eat other insects,
grubs, Japanese beetle grubs. I think it can even eat seeds.
Anyway, the robins in my garden seem to be pecking at things
that don't seem to be insects. And the other is that I am
assuming a very naīve growth law.
For example, if there are no sharks,
how are the food fish fruitful and multiplying?
They multiple exponentially. Now, obviously you cannot have
unlimited growth like that. With the worms we added the
term minus a constant times y squared to indicate that even
worms cannot multiply purely exponentially forever,
but ultimately their growth levels off because they cannot
find enough organic matter to plow their way through.
Those are the two differences. I am not assuming a logistic
growth law. This is less sophisticated than
the one I gave you for homework. This assumes that sharks have
absolutely nothing to eat except these fish, which is not so bad.
That's true, more or less.
My plan is now, with this model,
let's start the analysis, as you learned to do it.
And we are going to run into trouble at various places.
And the troubles will then illustrate these three points
that I wanted to add to your repertoire of things to do with
nonlinear systems when you run into trouble.
And it will also increase your understanding of the nonlinear
systems. The first thing we have to do
is find the critical points. I am going to assume that you
are good at this and, therefore, not spend a lot of
time detailing the calculations. I will simply write them on the
board. The first equation I am going
to write down is x times minus a plus by equals zero.
And I assume you know why I am
writing that down. And the other equation will be
y times c minus dx is zero.
Those are the simultaneous equations I have to find to find
the critical point. The first one is if the product
of these is zero, either x is zero or the other
factor is. So, from the second equation,
if x is zero, y has to be zero also.
That is one critical point. Now, if x is not zero then this
factor has to be zero which says that y must be equal
to a over b. And if y is equal to a over b,
it is not zero here. Therefore, this factor must be
zero which says that x is equal to c over d.
You see right away that this must be a simpler system because
it is only producing two critical points.
Whereas, the system that you did for homework had four.
Here is one. Well, let's just write them up
here. The critical points are zero,
zero. That doesn't look terribly
interesting, but the other one looks more interesting.
It is c over d, a over b.
Well, let's take a look first at the zero, zero critical
point. The origin, in other words.
What does that look like?
Well, at the origin, the linearization is extremely
easy to do because I simply ignore the quadratic terms,
which are the product of two small numbers,
where these only have a single small number in it.
It is minus ax, cy.
In other words, the linearization matrix is
minus a, zero, zero and cy gives me a
coefficient c there.
Now, I don't think at any point I have ever explicitly told you
that I hope you have learned from the homework or maybe your
recitation teacher, but for heaven's sake,
put this down in your little books, if you have a diagonal
matrix, for god's sake, don't calculate its
eigenvalues. They are right in front of you.
They are always the diagonal elements.
The eigenvalue, you can check this out if you
insist on writing the equation, but trust me it is clear.
If I, for example, subtract c from the main
diagonal, I am going to get a determinant zero because the
bottom row will be all zero. The eigenvalues are negative a
and c. In other words,
they have opposite signs. This is a negative number.
That is a positive number -- -- because a,
b, c and d are always positive. And, therefore,
this is automatically a saddle. You don't have to calculate
anything. It is all right in front of
you. It must be a saddle and,
therefore, unstable because all saddles are.
And, in fact, you can even draw the little
picture of what the stuff looks like near the origin without
even bothering to calculate eigenvectors,
although it is extremely easy to do.
Just from common sense, these are the sharks and these
are the yumfish. Well, if there are zero
yumfish, in other words, if I am on the sharks axis,
the axis of sharks, I die out.
Well, forget about this side. It's on the positive side.
That makes sense. But I die out because the
sharks have nothing to eat. Whereas, if I am on the yumfish
axis I go this way. I grow because,
without any sharks to eat them, the yumfish increase.
So it must look like that. And, therefore,
the saddle must look like this. The saddle curves hug those and
go nearby. Now, the other critical point
is the interesting one. And for that the analysis,
in order that you don't spend the rest of this period writing
a's, b's, c's and d's, I am going to make the
simplifying assumption. But it doesn't change
qualitatively any of the mathematics.
It just makes it a little easier to write everything out.
I am going to assume that everything is one.
Well, in fact, even this would be good enough,
but let's make everything one. I am going to assume this.
And it is only to make the writing simpler.
It doesn't really change the mathematics at all.
Well, if everything is one, in order to calculate the
linearized system, I'd better use the Jacobian.
But perhaps it would be better to write out explicitly what the
system is. The system now is x prime
equals minus x plus xy, all the coefficients are
one. And y prime is equal to y minus
xy. What is the Jacobian?
Well, the Jacobian is the partial of this with respect to
x which is minus one plus y, the partial with respect
to y which is x, the partial of this with
respect to x, which is minus y,
and the partial with respect to y, which is one minus x.
But I want to evaluate that at
the point one, one,
which is the critical point. It is the critical point
because I have assumed that all these parameters have the value
one. And when I do that,
evaluating it at one, one,
I get what matrix? Well, I get zero,
one, negative one, zero.
In other words,
the linearized system is x prime equals y and y
prime equals negative x.
Well, that is just the one we analyzed before in terms of --
It's the one whose solutions are circles.
In other words, what we find out is that the
linearized system is what geometric type?
Saddle? Node?
Spiral? None of those.
It is a center. The linearized system is a
center.
It consists, in other words,
of a bunch of curves going round and round next to each
other. Concentric circles,
in fact. Well, what is wrong with that?
Now, we are in deep trouble. We are now in trouble because
that is a borderline case. Let me remind you of what the
borderline cases are. When we drew that picture,
this is from last week's problem set so you have a
perfect excuse for forgetting it totally.
But I am trying to remind you of it.
That is another reason why I am doing this.
Remember that picture I asked you about that appeared on the
computer screen? Let's make it a little flatter
so that I can have room to write in.
This is a certain parabola whose equation you are dying to
tell me, but I am not asking. There is the trace,
this is the determinant, and the characteristic equation
is related to the values of these numbers like plus d equals
zero. Then these were spiral synchs.
This was the region of spiral sources.
That is the abbreviation for sources.
These were nodal synchs and these were the nodal sources.
And down here were the saddles. And where were the centers?
The centers were along this line.
The centers, there weren't many of them,
and they were the separation for these two regions.
I will put that down. These are the centers.
These other borderlines correspond to other things.
These are defective eigenvalues, zero eigenvalues
and so on. But let's concentrate on just
one of them. You will find the others
discussed in the notes, GS7.
But if you get this idea then the rest is just details.
I think it will be perfectly clear.
What is wrong with the center? The answer is that if we are on
the center, for example, this system corresponds to the
trace being zero and the determinant being plus one.
It corresponds to t equals zero and d equals one.
This point, in other words. Now, if I wiggle the
coefficients of the matrix just a little bit,
just change them a little bit, what I am going to do is move
off this pink line. And I might move to here or I
might move a little way to there.
But, if I do that, I change the geometric type.
In other words, being a borderline,
a slight change of the parameters can change what it
changes to. And, in fact,
that is geometrically clear. What does a center look like?
A center looks like this, a bunch of curves going around
all in the same direction, like concentric circles or
maybe ellipses or something like that.
If I deform the picture just a little bit, well,
I might change it into something that looks like this
where, after they go around they don't quite meet up with where
they were to start with. And I am going to get a spiral
synch. Or, I might do the deformation
by going around once and I'm a little outside of where I was.
In which case it's going to be a spiral source.
The fact that just changing these curves a little bit can
change the picture to this or to that corresponds to the fact
that if you are on here with this value of t and d,
zero and one, and just change t and d a
little bit, you are going to move off into these regions.
You might, of course, stay on the pink line.
It is not very likely. Where does this leave us?
Well, if the linear system is not stable in the sense that if
you change the parameters a little bit it doesn't change the
type. This is, after all,
only an approximation to the nonlinear system.
If in this approximation you cannot really predict the
behavior of very well when you change the parameters,
then from it you cannot tell what the original system looked
like. In other words,
the nonlinear system at one, one might be any one
of the possibilities, still a center or it might be a
spiral synch or it might be a spiral source.
Any one of those three is a possibility.
It couldn't be a saddle because that is too far away.
It couldn't be a node. That is too far away,
too. But it could wander into either
of these regions and, therefore, the picture you
cannot tell which of these three it is just from this type of
critical point analysis. Well, that was Volterra's
problem. By the way, the person who
introduced these equations and studied them systematically in
the way in which we are doing it here was Volterra.
And, in fact, he was interested in sharks and
food fish, as they were called. What do we do?
Well, you have to be smart. What Volterra did was he went
to method number one. Let's do it.
Volterra said I got my equations x prime equals minus x
plus xy. y prime is equal to,
these are the food fish, y minus xy.
Let's eliminate t. And my problem is,
of course, I am trying to determine what type of critical
point one, one is. And the method we have used up
until now has failed because it gave us a borderline case,
which is from that we cannot deduce.
He said let's eliminate t. And we get dy over dx
equals, I am going to factor, y times one minus x on top.
And on the bottom factor x negative one plus y.
You could solve that before you came into 18.03,
right? This you can separate
variables. Let's separate variables.
The y's all go on one side, so y goes down here.
It is (y minus one) over y dy equals --
On the other side the dx gets moved up, one minus x over x dx.
Now, of course, you wouldn't dream of using
partial fractions on this. It would be illegal because,
even though it is a rational function, the quotient of two
polynomials, the degree of the top is not smaller than the
degree of the bottom. In other words,
it is a partial fraction so this degree must be bigger than
that one. And it isn't so you would have
to first divide. And if you divide by that then
there is no point in using partial fractions at all.
Of course, you would have done this by basic instant,
I know. What is the solution?
It is y minus log y equals log x minus x plus some constant.
That is not the final constant
I want so I will give it a subscript one to indicate I want
more. This is very hard to see what
that curve looks like. We can make it look better by
exponentiating. If I exponentiate it,
going back to high school mathematics, but I know from
experience that many of you are not too good at this step,
so that is another reason I am doing it, it will be
e to the y, times, that part is easy because
pluses and minuses change into times, times e to the negative
log y. Well, e to the log y is y.
If I put a minus sign in front
of that, that sends it into the denominator, so instead of being
y it is one over y Equals x, e to log x is x,
e to the negative x
is, therefore, times one over e to the x.
And that is times the
exponential of c1, which I will call c2.
And now if I combine them all and put them all on one side it
is x over e to the x times y over e to the y is equal to yet
another constant, one over c2.
This is my final constant so
let's call that c. In other words,
the integral curves are the graphs of this equation for
different values of the constant c.
Well, of course you've graphed an equation like that all the
time. What am I going to do with this
stupid thing? Well, I deliberately picked
something which involved all your learning up until now.
We are now going into 18.02, right?
You looked at that in 18.02 you would say this is the contour
curve, these are contour curves of the function --
Well, let's write it out the other way.
It doesn't make any difference, but you are more likely to have
seen the function in this form. x over e to the x
you won't recognize, but this you will.
In fact, we have had it before this term.
It is one of the kinds of solutions you can add to second
order equations. Times y e to the negative y
equals c. It is the contour curves of
this function. Let's call that function,
let's say, h of (x,y).
Well, of course you could throw it on Matlab,
as you did in 18.02 maybe, and ask Matlab to plot the
function. But Volterra didn't have that
luxury. He was smart,
too. So let's be smart instead.
What is the function x? Let's use a neutral variable
like u and plot u e to the negative u.
18.01. At zero, it has the value zero.
When u is small this has approximately the value one and,
therefore, it starts out like the function u.
As u goes to infinity, you know by L'Hopital's rule
that this goes to zero so it ends up like this.
Well, what on earth could it possibly do except rise to a
maximum? But where is that maximum?
It is easy to see there is a unique maximum just by 18.01.
Take the derivative, find out where the maximum
point is and you will find, perhaps you have already done
it, this is at one. The maximum occurs at one,
that is where the derivative is zero and the value there is one
times e to the minus one.
It is one over e, in other words.
That is what the curve looks like.
What do the contour curves of this look like?
Well, if we are looking at the contour curves of the function
h, so here is x, here is y, and we want the
contour curves of the function h of (x, y),
what do I know? Where is its maximum?
The maximum point of h of x, y is where?
Well, h is the product x e to the negative x times y e to the
minus y.
This has its maximum at one, this has its maximum at one,
so where does h of (x, y) head?
Well, you have two factors. One makes each of them the
biggest they can be. The maximum must be exactly at
that critical point one, one.
That is our maximum point. Let's make it conspicuous.
This is the maximum point. It is the point one,
one. It is the point that makes the
function biggest. Now, how about the contour
curves? Well, this is the top of the
mountain. What is it along the axes?
Along the axes, when either x or y is zero,
the function has the value zero, so it is biggest there.
It has the value zero here. So it is zero value.
What are the contour curves? Well, we must have a mountain
peak here. This is the unique maximum
point. In fact, it is easy to see the
contour curves just surround it like that.
Now, the reason they cannot be spirals, well,
in the first place, you never see contour curves
that were spirals. That is a good reason.
It is a mountain. That is the way contour curves
of a mountain look. But, all right,
that is not a good argument. But notice that along each
horizontal line here, I want to know how many times
can it intersect the contour curve?
That is the same as asking along one of these lines how
many times could it intersect? See, this is a graph of the
function along this horizontal line.
And, therefore, how many times can it intersect
one of the contour curves the same number of times that a
horizontal line can intersect this curve?
Twice. Only once if you are exactly at
that height and after that never.
But here it can intersect each contour curve only twice which
means these things cannot be spirals.
Otherwise, they would intersect those horizontal and vertical
lines many times instead of just twice.
That is what it looks like. In fact, we can even put in the
direction without any effort. We know that the sharks die out
without any food fish and the food fish grow.
Well, I think this has to be clockwise.
Of course, the guys nearby must be going the same way.
And, therefore, the guys near them must be
going the same way. And it is sort of a domino
effect. The direction spreads and there
is only one compatible way of making these.
They all must be going clockwise.
What is happening, actually?
At this point there are a lot of sharks but not so many food
fish. The sharks eat the few
remaining food fish, and now the sharks start to die
out because they have nothing to eat.
When they are practically gone, the food fish can start growing
again and grow and grow. For a while they grow happily,
and then the sharks suddenly start being able to find them
again. And the few remaining sharks
start eating them, the population of sharks grows,
the food fish start dying out again, and the cycle starts all
over again. Now, I wanted to talk about the
qualitative behavior. This is, in some ways,
the most important part of the lecture.
I wanted to discuss, just as we did at the beginning
of the term, the effect of fishing with nets at a constant
rate k.
Just as near the beginning of the term, we talked about
fishing a single population. The only difference is,
now there are two populations. Both sharks and the food fish.
Now, what happens to the equations?
Well, the equations start out just as they were.
I am now reverting to a and b. You will see why.
It's because we need the general coefficients.
Plus bxy. But, if we are fishing with
nets, then a certain fraction of the sharks in the sea get hauled
in by the net. And say the fishing rate is k.
There is a fishing term. We remove minus k times a
certain fraction of the sharks in the sea.
How about the food fish? Well, we remove them,
too. And we don't distinguish
between sharks and food fish. This is cy minus dxy.
But we also remove a certain
fraction of them. Minus k times y.
What, therefore, is the new system?
The new system is therefore x prime equals minus a plus k.
I am combining those two x terms.
That is times x. Plus bxy.
And the y prime is c minus k times y
minus dxy.
To solve these, I do not have to go through the
analysis. All I have to do is change the
numbers, change these coefficients from a to a plus k
and c minus k. The old critical point was the
point c over d, a over b.
The new critical point with
fishing is what? Well, the parameters have been
changed by the addition of k. The new critical point is c
minus k over d and a plus k over d.
What is the effect of fishing?
Well, if the old critical point was over here,
let's say this is the point c over d and this is the
point a over b, that is the old critical point
that was over here. The result of fishing is to
lower the value of x and raise the value here.
The new critical point is, this gets lowered to c minus k
over d, and this gets raised to a plus
k over b. In other words,
the new point is there. The effect of fishing is to --
It does not treat the sharks and the yumfish equally.
The effect of fishing lowers the shark population.
See, the critical point gives sort of the average shark
population. Of course, it cycles around
these. But, on the average,
this gives how many sharks there are and how many food fish
there are. The new critical point with
fishing lowers the shark population and raises the food
fish population. That is not intuitive.
And, in fact, that was observed
experimentally at a slightly different context.
And that is why Volterra started working on the problem.
I will need three more minutes. The most interesting
application of all is not to sharks and food fish.
I cannot assume that you are dramatically interested in how
many of them there are in the ocean, but you might be more
interested in this.
That thing about lowering is called Volterra's principle.
Put that in your books. Volterra's principle.
Volterra is spelt over there.
-- has found more modern applications than sharks.
Suppose you consider fish in a pond.
You have mosquito larvae which breed in the pond.
This happens. And then suddenly there is a
plague of mosquitoes and concerned citizens.
And this is what happened in the '50s.
That was before you were born, but not before I was.
What happened was a lot of mosquitoes.
Everybody said the mosquitoes breed in the little stagnant
ponds so spray DDT on them. DDT them.
Dump it in the ponds. That will kill all the larvae
and we won't get bitten anymore. When you put DDT in the pond,
as people did not realize at the time because these things
were new, of course you kill the mosquitoes, but you also kill
the fish because DDT is poisonous to fish.
What, in effect, you are doing mathematically is
the same as harvesting the fish population with the sharks and
the food fish. The result is,
the fish are the predators because they eat the mosquito
larvae, big food, there was a certain
equilibrium, according to Volterra's principle.
With DDT that equilibrium moves to here.
In other words, the effect of indiscriminately
spraying the pond with DDT is to increase the number of
mosquitoes and kill fish. And, in fact,
that is exactly what was observed.
The same thing was observed with the bird population and
insects. Spraying trees for insects to
get rid of some pests ends up killing more birds than it does
insects and the insects increase.
Thanks.