OK, let's get started.
I'm assuming that,
A, you went recitation
yesterday, B,
that even if you didn't,
you know how to separate
variables, and you know how to
construct simple models,
solve physical problems with
differential equations,
and possibly even solve them.
So, you should have learned
that either in high school,
or 18.01 here,
or, yeah.
So, I'm going to start from
that point, assume you know
that.
I'm not going to tell you what
differential equations are,
or what modeling is.
If you still are uncertain
about those things,
the book has a very long and
good explanation of it.
Just read that stuff.
So, we are talking about first
order ODEs.
ODE: I'll only use two
acronyms.
ODE is ordinary differential
equations.
I think all of MIT knows that,
whether they've been taking the
course or not.
So, we are talking about
first-order ODEs,
which in standard form,
are written,
you isolate the derivative of y
with respect to,
x, let's say,
on the left-hand side,
and on the right-hand side you
write everything else.
You can't always do this very
well, but for today,
I'm going to assume that it has
been done and it's doable.
So, for example,
some of the ones that will be
considered either today or in
the problem set are things like
y prime equals x over y.
That's pretty simple.
The problem set has y prime
equals, let's see,
x minus y squared.
And, it also has y prime equals
y minus x squared.
There are others,
too.
Now, when you look at this,
this, of course,
you can solve by separating
variables.
So, this is solvable.
This one is-- and neither of
these can you separate
variables.
And they look extremely
similar.
But they are extremely
dissimilar.
The most dissimilar about them
is that this one is easily
solvable.
And you will learn,
if you don't know already,
next time next Friday how to
solve this one.
This one, which looks almost
the same, is unsolvable in a
certain sense.
Namely, there are no elementary
functions which you can write
down, which will give a solution
of that differential equation.
So, right away,
one confronts the most
significant fact that even for
the simplest possible
differential equations,
those which only involve the
first derivative,
it's possible to write down
extremely looking simple guys.
I'll put this one up in blue to
indicate that it's bad.
Whoops, sorry,
I mean, not really bad,
but recalcitrant.
It's not solvable in the
ordinary sense in which you
think of an equation is
solvable.
And, since those equations are
the rule rather than the
exception, I'm going about this
first day to not solving a
single differential equation,
but indicating to you what you
do when you meet a blue equation
like that.
What do you do with it?
So, this first day is going to
be devoted to geometric ways of
looking at differential
equations and numerical.
At the very end,
I'll talk a little bit about
numerical ways.
And you'll work on both of
those for the first problem set.
So, what's our geometric view
of differential equations?
Well, it's something that's
contrasted with the usual
procedures, by which you solve
things and find elementary
functions which solve them.
I'll call that the analytic
method.
So, on the one hand,
we have the analytic ideas,
in which you write down
explicitly the equation,
y prime equals f of x,y.
And, you look for certain
functions, which are called its
solutions.
Now, so there's the ODE.
And, y1 of x,
notice I don't use a separate
letter.
I don't use g or h or something
like that for the solution
because the letters multiply so
quickly, that is,
multiply in the sense of
rabbits, that after a while,
if you keep using different
letters for each new idea,
you can't figure out what
you're talking about.
So, I'll use y1 means,
it's a solution of this
differential equation.
Of course, the differential
equation has many solutions
containing an arbitrary
constant.
So, we'll call this the
solution.
Now, the geometric view,
the geometric guy that
corresponds to this version of
writing the equation,
is something called a direction
field.
And, the solution is,
from the geometric point of
view, something called an
integral curve.
So, let me explain if you don't
know what the direction field
is.
I know for some of you,
I'm reviewing what you learned
in high school.
Those of you who had the BC
syllabus in high school should
know these things.
But, it never hurts to get a
little more practice.
And, in any event,
I think the computer stuff that
you will be doing on the problem
set, a certain amount of it
should be novel to you.
It was novel to me,
so why not to you?
So, what's a direction field?
Well, the direction field is,
you take the plane,
and in each point of the
plane-- of course,
that's an impossibility.
But, you pick some points of
the plane.
You draw what's called a little
line element.
So, there is a point.
It's a little line,
and the only thing which
distinguishes it outside of its
position in the plane,
so here's the point,
(x,y), at which we are drawing
this line element,
is its slope.
And, what is its slope?
Its slope is to be f of x,y.
And now, You fill up the plane
with these things until you're
tired of putting then in.
So, I'm going to get tired
pretty quickly.
So, I don't know,
let's not make them all go the
same way.
That sort of seems cheating.
How about here?
Here's a few randomly chosen
line elements that I put in,
and I putted the slopes at
random since I didn't have any
particular differential equation
in mind.
Now, the integral curve,
so those are the line elements.
The integral curve is a curve,
which goes through the plane,
and at every point is tangent
to the line element there.
So, this is the integral curve.
Hey, wait a minute,
I thought tangents were the
line element there didn't even
touch it.
Well, I can't fill up the plane
with line elements.
Here, at this point,
there was a line element,
which I didn't bother drawing
in.
And, it was tangent to that.
Same thing over here:
if I drew the line element
here, I would find that the
curve had exactly the right
slope there.
So, the point is the integral,
what distinguishes the integral
curve is that everywhere it has
the direction,
that's the way I'll indicate
that it's tangent,
has the direction of the field
everywhere at all points on the
curve, of course,
where it doesn't go.
It doesn't have any mission to
fulfill.
Now, I say that this integral
curve is the graph of the
solution to the differential
equation.
In other words,
writing down analytically the
differential equation is the
same geometrically as drawing
this direction field,
and solving analytically for a
solution of the differential
equation is the same thing as
geometrically drawing an
integral curve.
So, what am I saying?
I say that an integral curve,
all right, let me write it this
way.
I'll make a little theorem out
of it, that y1 of x is
a solution to the differential
equation if, and only if,
the graph, the curve associated
with this, the graph of y1 of x
is an integral curve.
Integral curve of what?
Well, of the direction field
associated with that equation.
But there isn't quite enough
room to write that on the board.
But, you could put it in your
notes, if you take notes.
So, this is the relation
between the two,
the integral curves of the
graphs or solutions.
Now, why is that so?
Well, in fact,
all I have to do to prove this,
if you can call it a proof at
all, is simply to translate what
each side really means.
What does it really mean to say
that a given function is a
solution to the differential
equation?
Well, it means that if you plug
it into the differential
equation, it satisfies it.
Okay, what is that?
So, how do I plug it into the
differential equation and check
that it satisfies it?
Well, doing it in the abstract,
I first calculate its
derivative.
And then, how will it look
after I plugged it into the
differential equation?
Well, I don't do anything to
the x, but wherever I see y,
I plug in this particular
function.
So, in notation,
that would be written this way.
So, for this to be a solution
means this, that that equation
is satisfied.
Okay, what does it mean for the
graph to be an integral curve?
Well, it means that at each
point, the slope of this curve,
it means that the slope of y1
of x should be,
at each point, x1 y1.
It should be equal to the slope
of the direction field at that
point.
And then, what is the slope of
the direction field at that
point?
Well, it is f of that
particular, well,
at the point,
x, y1 of x.
If you like,
you can put a subscript,
one, on there,
send a one here or a zero
there, to indicate that you mean
a particular point.
But, it looks better if you
don't.
But, there's some possibility
of confusion.
I admit to that.
So, the slope of the direction
field, what is that slope?
Well, by the way,
I calculated the direction
field.
Its slope at the point was to
be x, whatever the value of x
was, and whatever the value of
y1 of x was,
substituted into the right-hand
side of the equation.
So, what the slope of this
function of that curve of the
graph should be equal to the
slope of the direction field.
Now, what does this say?
Well, what's the slope of y1 of
x?
That's y1 prime of x.
That's from the first day of
18.01, calculus.
What's the slope of the
direction field?
This?
Well, it's this.
And, that's with the right hand
side.
So, saying these two guys are
the same or equal,
is exactly, analytically,
the same as saying these two
guys are equal.
So, in other words,
the proof consists of,
what does this really mean?
What does this really mean?
And after you see what both
really mean, you say,
yeah, they're the same.
So, I don't how to write that.
It's okay: same,
same, how's that?
This is the same as that.
Okay, well, this leaves us the
interesting question of how do
you draw a direction from the,
well, this being 2003,
mostly computers draw them for
you.
Nonetheless,
you do have to know a certain
amount.
I've given you a couple of
exercises where you have to draw
the direction field yourself.
This is so you get a feeling
for it, and also because humans
don't draw direction fields the
same way computers do.
So, let's first of all,
how did computers do it?
They are very stupid.
There's no problem.
Since they go very fast and
have unlimited amounts of energy
to waste, the computer method is
the naive one.
You pick the point.
You pick a point,
and generally,
they are usually equally
spaced.
You determine some spacing,
that one: blah,
blah, blah, blah,
blah, blah, blah,
equally spaced.
And, at each point,
it computes f of x,
y at the point,
finds, meets,
and computes the value of f of
(x, y), that function,
and the next thing is,
on the screen,
it draws, at (x,
y), the little line element
having slope f of x,y.
In other words,
it does what the differential
equation tells it to do.
And the only thing that it does
is you can, if you are telling
the thing to draw the direction
field, about the only option you
have is telling what the spacing
should be, and sometimes people
don't like to see a whole line.
They only like to see a little
bit of a half line.
And, you can sometimes tell,
according to the program,
tell the computer how long you
want that line to be,
if you want it teeny or a
little bigger.
Once in awhile you want you
want it narrower on it,
but not right now.
Okay, that's what a computer
does.
What does a human do?
This is what it means to be
human.
You use your intelligence.
From a human point of view,
this stuff has been done in the
wrong order.
And the reason it's been done
in the wrong order:
because for each new point,
it requires a recalculation of
f of (x, y).
That is horrible.
The computer doesn't mind,
but a human does.
So, for a human,
the way to do it is not to
begin by picking the point,
but to begin by picking the
slope that you would like to
see.
So, you begin by taking the
slope.
Let's call it the value of the
slope, C.
So, you pick a number.
C is two.
I want to see where are all the
points in the plane where the
slope of that line element would
be two?
Well, they will satisfy an
equation.
The equation is f of (x,
y) equals, in general,
it will be C.
So, what you do is plot this,
plot the equation,
plot this equation.
Notice, it's not the
differential equation.
You can't exactly plot a
differential equation.
It's a curve,
an ordinary curve.
But which curve will depend;
it's, in fact,
from the 18.02 point of view,
the level curve of C,
sorry, it's a level curve of f
of (x, y), the function f of x
and y corresponding to the level
of value C.
But we are not going to call it
that because this is not 18.02.
Instead, we're going to call it
an isocline.
And then, you plot,
well, you've done it.
So, you've got this isocline,
except I'm going to use a
solution curve,
solid lines,
only for integral curves.
When we do plot isoclines,
to indicate that they are not
solutions, we'll use dashed
lines for doing them.
One of the computer things does
and the other one doesn't.
But they use different colors,
also.
There are different ways of
telling you what's an isocline
and what's the solution curve.
So, and what do you do?
So, these are all the points
where the slope is going to be
C.
And now, what you do is draw in
as many as you want of line
elements having slope C.
Notice how efficient that is.
If you want 50 million of them
and have the time,
draw in 50 million.
If two or three are enough,
draw in two or three.
You will be looking at the
picture.
You will see what the curve
looks like, and that will give
you your judgment as to how you
are to do that.
So, in general,
a picture drawn that way,
so let's say,
an isocline corresponding to C
equals zero.
The line elements,
and I think for an isocline,
for the purposes of this
lecture, it would be a good idea
to put isoclines.
Okay, so I'm going to put
solution curves in pink,
or whatever this color is,
and isoclines are going to be
in orange, I guess.
So, isocline,
represented by a dashed line,
and now you will put in the
line elements of,
we'll need lots of chalk for
that.
So, I'll use white chalk.
Y horizontal?
Because according to this the
slope is supposed to be zero
there.
And at the same way,
how about an isocline where the
slope is negative one?
Let's suppose here C is equal
to negative one.
Okay, then it will look like
this.
These are supposed to be lines
of slope negative one.
Don't shoot me if they are not.
So, that's the principle.
So, this is how you will fill
up the plane to draw a direction
field: by plotting the isoclines
first.
And then, once you have the
isoclines there,
you will have line elements.
And you can draw a direction
field.
Okay, so, for the next few
minutes, I'd like to work a
couple of examples for you to
show how this works out in
practice.
So, the first equation is going
to be y prime equals minus x
over y.
Okay, first thing,
what are the isoclines?
Well, the isoclines are going
to be y.
Well, negative x over y is
equal to C.
Maybe I better make two steps
out of this.
Minus x over y is equal to C.
But, of course,
nobody draws a curve in that
form.
You'll want it in the form y
equals minus one over 
C times x.
So, there's our isocline.
Why don't I put that up in
orange since it's going to be,
that's the color I'll draw it
in.
In other words,
for different values of C,
now this thing is aligned.
It's aligned,
in fact, through the origin.
This looks pretty simple.
Okay, so here's our plane.
The isoclines are going to be
lines through the origin.
And now, let's put them in,
suppose, for example,
C is equal to one.
Well, if C is equal to one,
then it's the line,
y equals minus x.
So, this is the isocline.
I'll put, down here,
C equals minus one.
And, along it,
no, something's wrong.
I'm sorry?
C is one, not negative one,
right, thanks.
Thanks.
So, C equals one.
So, it should be little line
segments of slope one will be
the line elements,
things of slope one.
OK, now how about C equals
negative one?
If C equals negative one,
then it's the line,
y equals x.
And so, that's the isocline.
Notice, still dash because
these are isoclines.
Here, C is negative one.
And so, the slope elements look
like this.
Notice, they are perpendicular.
Now, notice that they are
always going to be perpendicular
to the line because the slope of
this line is minus one over C.
But, the slope of the line
element is going to be C.
Those numbers,
minus one over C and C,
are negative reciprocals.
And, you know that two lines
whose slopes are negative
reciprocals are perpendicular.
So, the line elements are going
to be perpendicular to these.
And therefore,
I hardly even have to bother
calculating, doing any more
calculation.
Here's going to be a,
well, how about this one?
Here's a controversial
isocline.
Is that an isocline?
Well, wait a minute.
That doesn't correspond to
anything looking like this.
Ah-ha, but it would if I put C
multiplied through by C.
And then, it would correspond
to C being zero.
In other words,
don't write it like this.
Multiply through by C.
It will read C y equals
negative x.
And then, when C is zero,
I have x equals zero,
which is exactly the y-axis.
So, that really is included.
How about the x-axis?
Well, the x-axis is not
included.
However, most people include it
anyway.
This is very common to be a
sort of sloppy and bending the
edges of corners a little bit,
and hoping nobody will notice.
We'll say that corresponds to C
equals infinity.
I hope nobody wants to fight
about that.
If you do, go fight with
somebody else.
So, if C is infinity,
that means the little line
segment should have infinite
slope, and by common consent,
that means it should be
vertical.
And so, we can even count this
as sort of an isocline.
And, I'll make the dashes
smaller, indicate it has a lower
status than the others.
And, I'll put this in,
do this weaselly thing of
putting it in quotation marks to
indicate that I'm not
responsible for it.
Okay, now, we now have to put
it the integral curves.
Well, nothing could be easier.
I'm looking for curves which
are everywhere perpendicular to
these rays.
Well, you know from geometry
that those are circles.
So, the integral curves are
circles.
And, it's an elementary
exercise, which I would not
deprive you of the pleasure of.
Solve the ODE by separation of
variables.
In other words,
we've gotten the,
so the circles are ones with a
center at the origin,
of course, equal some constant.
I'll call it C1,
so it's not confused with this
C.
They look like that,
and now you should solve this
by separating variables,
and just confirm that the
solutions are,
in fact, those circles.
One interesting thing,
and so I confirm this,
I won't do it because I want to
do geometric and numerical
things today.
So, if you solve it by
separating variables,
one interesting thing to note
is that if I write the solution
as y equals y1 of x, well,
it'll look something like the
square root of C1 minus,
let's make this squared because
that's the way people usually
put the radius,
minus x squared.
And so, a solution,
a typical solution looks like
this.
Well, what's the solution over
here?
Well, that one solution will be
goes from here to here.
If you like,
it has a negative side to it.
So, I'll make,
let's say, plus.
There's another solution,
which has a negative value.
But let's use the one with the
positive value of the square
root.
My point is this,
that that solution,
the domain of that solution,
really only goes from here to
here.
It's not the whole x-axis.
It's just a limited piece of
the x-axis where that solution
is defined.
There's no way of extending it
further.
And, there's no way of
predicting, by looking at the
differential equation,
that a typical solution was
going to have a limited domain
like that.
In other words,
you could find a solution,
but how far out is it going to
go?
Sometimes, it's impossible to
tell, except by either finding
it explicitly,
or by asking a computer to draw
a picture of it,
and seeing if that gives you
some insight.
It's one of the many
difficulties in handling
differential equations.
You don't know what the domain
of a solution is going to be
until you've actually calculated
it.
Now, a slightly more
complicated example is going to
be, let's see, y prime 
equals one plus x minus y.
It's not a lot more
complicated, and as a computer
exercise, you will work with,
still, more complicated ones.
But here, the isoclines would
be what?
Well, I set that equal to C.
Can you do the algebra in your
head?
An isocline will have the
equation: this equals C.
So, I'm going to put the y on
the right hand side,
and that C on the left hand
side.
So, it will have the equation y
equals one plus x minus C,
or a nicer way to
write it would be x plus one
minus C.
I guess it really doesn't
matter.
So there's the equation of the
isocline.
Let's quickly draw the
direction field.
And notice, by the way,
it's a simple equation,
but you cannot separate
variables.
So, I will not,
today at any rate,
be able to check the answer.
I will not be able to get an
analytic answer.
All we'll be able to do now is
get a geometric answer.
But notice how quickly,
relatively quickly,
one can get it.
So, I'm feeling for how the
solutions behave to this
equation.
All right, let's see,
what should we plot first?
I like C equals one,
no, don't do C equals one.
Let's do C equals zero,
first.
C equals zero.
That's the line.
y equals x plus 1.
Okay, let me run and get that
chalk.
So, I'll isoclines are in
orange.
If so, when C equals zero,
y equals x plus one.
So, let's say it's this curve.
C equals zero.
How about C equals negative
one?
Then it's y equals x plus two.
It's this curve.
Well, let's label it down here.
So, this is C equals negative
one.
C equals negative two would be
y equals x, no,
what am I doing?
C equals negative one is y
equals x plus two.
That's right.
Well, how about the other side?
If C equals plus one,
well, then it's going to go
through the origin.
It looks like a little more
room down here.
How about, so if this is going
to be C equals one,
then I sort of get the idea.
C equals two will look like
this.
They're all going to be
parallel lines because all
that's changing is the
y-intercept, as I do this thing.
So, here, it's C equals two.
That's probably enough.
All right, let's put it in the
line elements.
All right, C equals negative
one.
These will be perpendicular.
C equals zero,
like this.
C equals one.
Oh, this is interesting.
I can't even draw in the line
elements because they seem to
coincide with the curve itself,
with the line itself.
They write y along the line,
and that makes it hard to draw
them in.
How about C equals two?
Well, here, the line elements
will be slanty.
They'll have slope two,
so a pretty slanty up.
And, I can see if a C equals
three in the same way.
There are going to be even more
slantier up.
And here, they're going to be
even more slanty down.
This is not very scientific
terminology or mathematical,
but you get the idea.
Okay, so there's our quick
version of the direction field.
All we have to do is put in
some integral curves now.
Well, it looks like it's doing
this.
It gets less slanty here.
It levels out,
has slope zero.
And now, in this part of the
plain, the slope seems to be
rising.
So, it must do something like
that.
This guy must do something like
this.
I'm a little doubtful of what I
should be doing here.
Or, how about going from the
other side?
Well, it rises,
gets a little,
should it cross this?
What should I do?
Well, there's one integral
curve, which is easy to see.
It's this one.
This line is both an isocline
and an integral curve.
It's everything,
except drawable,
[LAUGHTER] so,
you understand this is the same
line.
It's both orange and pink at
the same time.
But I don't know what
combination color that would
make.
It doesn't look like a line,
but be sympathetic.
Now, the question is,
what's happening in this
corridor?
In the corridor,
that's not a mathematical word
either, between the isoclines
for, well, what are they?
They are the isoclines for C
equals two, and C equals zero.
How does that corridor look?
Well: something like this.
Over here, the lines all look
like that.
And here, they all look like
this.
The slope is two.
And, a hapless solution gets in
there.
What's it to do?
Well, do you see that if a
solution gets in that corridor,
an integral curve gets in that
corridor, no escape is possible.
It's like a lobster trap.
The lobster can walk in.
But it cannot walk out because
things are always going in.
How could it escape?
Well, it would have to double
back, somehow,
and remember,
to escape, it has to be,
to escape on the left side,
it must be going horizontally.
But, how could it do that
without doubling back first and
having the wrong slope?
The slope of everything in this
corridor is positive,
and to double back and escape,
it would at some point have to
have negative slope.
It can't do that.
Well, could it escape on the
right-hand side?
No, because at the moment when
it wants to cross,
it will have to have a slope
less than this line.
But all these spiky guys are
pointing; it can't escape that
way either.
So, no escape is possible.
It has to continue on,
there.
But, more than that is true.
So, a solution can't escape.
Once it's in there,
it can't escape.
It's like, what do they call
those plants,
I forget, pitcher plants.
All they hear is they are going
down.
So, it looks like that.
And so, the poor little insect
falls in.
They could climb up the walls
except that all the hairs are
going the wrong direction,
and it can't get over them.
Well, let's think of it that
way: this poor trap solution.
So, it does what it has to do.
Now, there's more to it than
that.
Because there are two
principles involved here that
you should know,
that help a lot in drawing
these pictures.
Principle number one is that
two integral curves cannot cross
at an angle.
Two integral curves can't
cross, I mean,
by crossing at an angle like
that.
I'll indicate what I mean by a
picture like that.
Now, why not?
This is an important principle.
Let's put that up in the white
box.
They can't cross because if two
integral curves,
are trying to cross,
well, one will look like this.
It's an integral curve because
it has this slope.
And, the other integral curve
has this slope.
And now, they fight with each
other.
What is the true slope at that
point?
Well, the direction field only
allows you to have one slope.
If there's a line element at
that point, it has a definite
slope.
And therefore,
it cannot have both the slope
and that one.
It's as simple as that.
So, the reason is you can't
have two slopes.
The direction field doesn't
allow it.
Well, that's a big,
big help because if I know,
here's an integral curve,
and if I know that none of
these other pink integral curves
are allowed to cross it,
how else can I do it?
Well, they can't escape.
They can't cross.
It's sort of clear that they
must get closer and closer to
it.
You know, I'd have to work a
little to justify that.
But I think that nobody would
have any doubt of it who did a
little experimentation.
In other words,
all these curves joined that
little tube and get closer and
closer to this line,
y equals x.
And there, without solving the
differential equation,
it's clear that all of these
solutions, how do they behave?
As x goes to infinity,
they become asymptotic to,
they become closer and closer
to the solution,
x.
Is x a solution?
Yeah, because y equals x is an
integral curve.
Is x a solution?
Yeah, because if I plug in y
equals x, I get what?
On the right-hand side,
I get one.
And on the left-hand side,
I get one.
One equals one.
So, this is a solution.
Let's indicate that it's a
solution.
So, analytically,
we've discovered an analytic
solution to the differential
equation, namely,
Y equals X, just by this
geometric process.
Now, there's one more principle
like that, which is less
obvious.
But you do have to know it.
So, you are not allowed to
cross.
That's clear.
But it's much,
much, much, much,
much less obvious that two
integral curves cannot touch.
That is, they cannot even be
tangent.
Two integral curves cannot be
tangent.
I'll indicate that by the word
touch, which is what a lot of
people say.
In other words,
if this is illegal,
so is this.
It can't happen.
You know, without that,
for example,
it might be,
I might feel that there would
be nothing in this to prevent
those curves from joining.
Why couldn't these pink curves
join the line,
y equals x?
You know, it's a solution.
They just pitch a ride,
as it were.
The answer is they cannot do
that because they have to just
get asymptotic to it,
ever, ever closer.
They can't join y equals x
because at the point where they
join, you have that situation.
Now, why can't you to have
this?
That's much more sophisticated
than this, and the reason is
because of something called the
Existence and Uniqueness
Theorem, which says that there
is through a point,
x zero y zero,
that y prime equals f of
(x, y) has only one,
and only one solution.
One has one solution.
In mathematics speak,
that means at least one
solution.
It doesn't mean it has just one
solution.
That's mathematical convention.
It has one solution,
at least one solution.
But, the killer is,
only one solution.
That's what you have to say in
mathematics if you want just
one, one, and only one solution
through the point
 x zero y zero.
So, the fact that it has one,
that is the existence part.
The fact that it has only one
is the uniqueness part of the
theorem.
Now, like all good mathematical
theorems, this one does have
hypotheses.
So, this is not going to be a
course, I warn you,
those of you who are
theoretically inclined,
very rich in hypotheses.
But, hypotheses for those one
or that f of (x,
y) should be a
continuous function.
Now, like polynomial,
signs, should be continuous
near, in the vicinity of that
point.
That guarantees existence,
and what guarantees uniqueness
is the hypothesis that you would
not guess by yourself.
Neither would I.
What guarantees the uniqueness
is that also,
it's partial derivative with
respect to y should be
continuous, should be continuous
near x zero y zero.
Well, I have to make a
decision.
I don't have time to talk about
Euler's method.
I'll refer you to the,
there's one page of notes,
and I couldn't do any more than
just repeat what's on those
notes.
So, I'll trust you to read
that.
And instead,
let me give you an example
which will solidify these things
in your mind a little bit.
I think that's a better course.
The example is not in your
notes, and therefore,
remember, you heard it here
first.
Okay, so what's the example?
So, there is that differential
equation.
Now, let's just solve it by
separating variables.
Can you do it in your head?
dy over dx, put all the y's on
the left.
It will look like dy over one
minus y.
Put all the dx's on the left.
So, the dx here goes on the
right, rather.
That will be dx.
And then, the x goes down into
the denominator.
So now, it looks like that.
And, if I integrate both sides,
I get the log of one minus y,
I guess, maybe with a,
I never bothered with that,
but you can.
It should be absolute values.
All right, put an absolute
value, plus a constant.
And now, if I exponentiate both
sides, the constant is positive.
So, this is going to look like
y.
One minus y equals x
And, the constant will be e to
the C1.
And, I'll just make that a new
constant, Cx.
And now, by letting C be
negative, that's why you can get
rid of the absolute values,
if you allow C to have negative
values as well as positive
values.
Let's write this in a more
human form.
So, y is equal to one minus Cx.
Good, all right,
let's just plot those.
So, these are the solutions.
It's a pretty easy equation,
pretty easy solution method,
just separation of variables.
What do they look like?
Well, these are all lines whose
intercept is at one.
And, they have any slope
whatsoever.
So, these are the lines that
look like that.
Okay, now let me ask,
existence and uniqueness.
Existence: through which points
of the plane does the solution
go?
Answer: through every point of
the plane, through any point
here, I can find one and only
one of those lines,
except for these stupid guys
here on the stalk of the flower.
Here, for each of these points,
there is no existence.
There is no solution to this
differential equation,
which goes through any of these
wiggly points on the y-axis,
with one exception.
This point is oversupplied.
At this point,
it's not existence that fails.
It's uniqueness that fails:
no uniqueness.
There are lots of things which
go through here.
Now, is that a violation of the
existence and uniqueness
theorem?
It cannot be a violation
because the theorem has no
exceptions.
Otherwise, it wouldn't be a
theorem.
So, let's take a look.
What's wrong?
We thought we solved it modulo,
putting the absolute value
signs on the log.
What's wrong?
The answer: what's wrong is to
use the theorem you must write
the differential equation in
standard form,
in the green form I gave you.
Let's write the differential
equation the way we were
supposed to.
It says dy / dx equals one
minus y divided by x.
And now, I see,
the right-hand side is not
continuous, in fact,
not even defined when x equals
zero, when along the y-axis.
And therefore,
the existence and uniqueness is
not guaranteed along the line,
x equals zero of the y-axis.
And, in fact,
we see that it failed.
Now, as a practical matter,
it's the way existence and
uniqueness fails in all ordinary
life work with differential
equations is not through
sophisticated examples that
mathematicians can construct.
But normally,
because f of (x,
y) will fail to be
defined somewhere,
and those will be the bad
points.
Thanks.
