WEBVTT
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y prime and y double prime.
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So, q of x times y
equal zero.
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The linearity of the equation,
that is, the form in which it
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appears is going to be the key
idea today.
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Today is going to be
theoretical, but some of the
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ideas in it are the most
important in the course.
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So, I don't have to apologize
for the theory.
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Remember, the solution method
was to find two independent y
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one, y two
independent solutions.
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And now, I'll formally write
out what independent means.
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There are different ways to say
it.
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But for you,
I think the simplest and most
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intelligible will be to say that
y2 is not to be a constant
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multiple of y1.
And, unfortunately,
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it's necessary to add,
nor is y1 to be a constant
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multiple.
I have to call it by different
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constants.
So, let's call this one c prime
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of y2.
Well, I mean,
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the most obvious question is,
well, look.
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If this is not a constant times
that, this can't be there
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because I would just use one
over c if it was.
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Unfortunately,
the reason I have to write it
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this way is to take account of
the possibility that y1 might be
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zero.
If y1 is zero,
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so, the bad case that must be
excluded is that y1 equals zero,
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y2 nonzero.
I don't want to call those
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independent.
But nonetheless,
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it is true that y2 is not a
constant multiple of y1.
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However, y1 is a constant
multiple of y2,
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namely, the multiple zero.
It's just to exclude that case
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that I have to say both of those
things.
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And, one would not be sufficed.
That's a fine point that I'm
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not going to fuss over.
But I just have,
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of course.
Now, why do you do that?
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That's because,
then, all solutions,
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and this is what concerns us
today, are what?
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The linear combination with
constant coefficients of these
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two, and the fundamental
question we have to answer today
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is, why?
Now, there are really two
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statements involved in that.
On the one hand,
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I'm claiming there is an easier
statement, which is that they
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are all solutions.
So, that's question one,
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or statement one.
Why are all of these guys
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solutions?
That, I could trust you to
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answer yourself.
I could not trust you to answer
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it elegantly.
And, it's the elegance that's
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the most important thing today
because you have to answer it
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elegantly.
Otherwise, you can't go on and
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do more complicated things.
If you answer it in an ad hoc
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basis just by hacking out a
computation, you don't really
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see what's going on.
And you can't do more difficult
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things later.
So, we have to answer this,
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and answer it nicely.
The second question is,
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so, if that answers why there
are solutions at all,
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why are they all the solutions?
Why all the solutions?
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In other words,
to say it as badly as possible,
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why are all solutions,
why all the solutions-- Never
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mind.
Why are all the solutions.
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This is a harder question to
answer, but that should make you
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happy because that means it
depends upon a theorem which I'm
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not going to prove.
I'll just quote to you.
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Let's attack there for problem
one first.
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q1 is answered by what's called
the superposition.
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The superposition principle
says exactly that.
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It says exactly that,
that if y1 and y2 are solutions
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to a linear equation,
to a linear homogeneous ODE,
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in fact it can be of higher
order, too, although I won't
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stress that.
In other words,
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you don't have to stop with the
second derivative.
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You could add a third
derivative and a fourth
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derivative.
As long as the former makes the
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same, but that implies
automatically that c1 y1 plus c2
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y2 is a solution.
Now, the way to do that nicely
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is to take a little detour and
talk a little bit about linear
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operators.
And, since we are going to be
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using these for the rest of the
term, this is the natural place
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for you to learn a little bit
about what they are.
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So, I'm going to do it.
Ultimately, I am aimed at a
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proof of this statement,
but there are going to be
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certain side excursions I have
to make.
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The first side side excursion
is to write the differential
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equation in a different way.
So, I'm going to just write its
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transformations.
The first, I'll simply recopy
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what you know it to be,
q y equals zero.
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That's the first form.
The second form,
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I'm going to replace this by
the differentiation operator.
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So, I'm going to write this as
D squared y.
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That means differentiate it
twice.
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D it, and then D it again.
This one I only have to
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differentiate once,
so I'll write that as p D(y),
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p times the derivative of Y.
The last one isn't
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differentiated at all.
I just recopy it.
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Now, I'm going to formally
factor out the y.
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So this, I'm going to turn into
D squared plus pD plus q.
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Now, everybody reads this as
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times y equals zero.
But, what it means is this guy,
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it means this is shorthand for
that.
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I'm not multiplying.
I'm multiplying q times y.
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But, I'm not multiplying D
times y.
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I'm applying D to y.
Nonetheless,
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the notation suggests this is
very suggestive of that.
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And this, in turn,
implies that.
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I'm just transforming it.
And now, I'll take the final
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step.
I'm going to view this thing in
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parentheses as a guy all by
itself, a linear operator.
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This is a linear operator,
called a linear operator.
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And, I'm going to simply
abbreviate it by the letter L.
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And so, the final version of
this equation has been reduced
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to nothing but Ly equals zero.
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Now, what's L?
You can think of L as,
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well, formally,
L you would write as D squared
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plus pD plus q.
But, you can think of L,
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the way to think of it is as a
black box, a function of what
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goes into the black box,
well, if this were a function
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box, what would go it would be a
number, and what would come out
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with the number.
But it's not that kind of a
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black box.
It's an operator box,
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and therefore,
what goes in is a function of
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x.
And, what comes out is another
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function of x,
the result of applying this
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operator to that.
So, from this point of view,
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differential equations,
trying to solve the
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differential equation means,
what should come out you want
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to come out zero,
and the question is,
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what should you put in?
That's what it means solving
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differential equations in an
inverse problem.
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The easy thing is to put it a
function, and see what comes
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out.
You just calculate.
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The hard thing is to ask,
you say, I want such and such a
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thing to come out,
for example,
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zero; what should I put in?
That's a difficult question,
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and that's what we're spending
the term answering.
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Now, the key thing about this
is that this is a linear
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operator.
And, what that means is that it
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behaves in a certain way with
respect to functions.
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The easiest way to say it is,
I like to make two laws of it,
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that L of u1,
if you have two functions,
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I'm not going to put up the
parentheses, x,
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because that just makes things
look longer and not any clearer,
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actually.
What does L do to the sum of
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two functions?
If that's a linear operator,
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if you put in the sum of two
functions, what you must get out
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is the corresponding L's,
the sum of the corresponding
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L's of each.
So, that's a law.
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And, the other law,
linearity law,
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and this goes for anything in
mathematics and its
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applications,
which is called linear,
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basically anything is linear if
it does the following thing to
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functions or numbers or
whatever.
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The other one is of a constant
times any function,
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I don't have to give it a
number now because I'm only
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using one of them,
should be equal to c times L of
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u.
So, here, c is a constant,
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and here, of course,
the u is a function,
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functions of x.
These are the two laws of
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linearity.
An operator is linear if it
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satisfies these two laws.
Now, for example,
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the differentiation operator is
such an operator.
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D is linear,
why?
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Well, because of the very first
things you verify after you
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learn what the derivative is
because the derivative of,
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well, I will write it in the D
form.
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I'll write it in the form in
which you know it.
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It would be D apply to u1 plus
u2.
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How does one write that in
ordinary calculus?
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Well, like that.
Or, maybe you write d by dx out
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front.
Let's write it this way,
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is equal to u1 prime plus u2
prime.
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That's a law.
You prove it when you first
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study what a derivative is.
It's a property.
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From our point of view,
it's a property of the
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differentiation operator.
It has this property.
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The D of u1 plus u2 is D of u1
plus D of u2.
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And, it also has the property
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that c u prime,
you can pull out the constant.
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That's not affected by the
differentiation.
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So, these two familiar laws
from the beginning of calculus
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say, in our language,
that D is a linear operator.
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What about the multiplication
of law?
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That's even more important,
that u1 times u2 prime,
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I have nothing whatever to say
about that in here.
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In this context,
it's an important law,
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but it's not important with
respect to the study of
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linearity.
So, there's an example.
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Here's a more complicated one
that I'm claiming is the linear
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operator.
And, since I don't want to have
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to work in this lecture,
the work is left to you.
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So, the proof,
prove that L is linear,
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is this particular operator.
L is linear.
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That's in your part one
homework to verify that.
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And you will do some simple
exercises in recitation tomorrow
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to sort of warm you up for that
if you haven't done it already.
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Well, you shouldn't have
because this only goes with this
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lecture, actually.
It's forbidden to work ahead in
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this class.
All right, where are we?
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Well, all that was a prelude to
proving this simple theorem,
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superposition principle.
So, finally,
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what's the proof?
The proof of the superposition
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principle: if you believe that
the operator is linear,
00:13:50.000 --> 00:13:56.000
then L of c1,
in other words,
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the ODE is L.
L is D squared plus pD plus q.
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So, the ODE is Ly equals zero.
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And, what am I being asked to
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prove?
I'm being asked to prove that
00:14:10.000 --> 00:14:16.000
if y1 and y2 are solutions,
then so is that thing.
00:14:13.000 --> 00:14:19.000
By the way, that's called a
linear combination.
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Put that in your notes.
Maybe I better write it even on
00:14:20.000 --> 00:14:26.000
the board because it's something
people say all the time without
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realizing they haven't defined
it.
00:14:26.000 --> 00:14:32.000
This is called a linear
combination.
00:14:30.000 --> 00:14:36.000
This expression is called a
linear combination of y1 and y2.
00:14:35.000 --> 00:14:41.000
It means that particular sum
with constant coefficients.
00:14:40.000 --> 00:14:46.000
Okay, so, the ODE is Ly equals
zero.
00:14:44.000 --> 00:14:50.000
And, I'm trying to prove that
fact about it,
00:14:48.000 --> 00:14:54.000
that if y1 and y2 are
solutions, so is a linear
00:14:52.000 --> 00:14:58.000
combination of them.
So, the proof,
00:14:55.000 --> 00:15:01.000
then, I start with apply L to
c1 y1 plus c2 y2.
00:15:00.000 --> 00:15:06.000
Now, because this operator is
00:15:05.000 --> 00:15:11.000
linear, it takes the sum of two
functions into the corresponding
00:15:10.000 --> 00:15:16.000
sum up what the operator would
be.
00:15:13.000 --> 00:15:19.000
So, it would be L of c1 y1 plus
L of c2 y2.
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That's because L is a linear
00:15:20.000 --> 00:15:26.000
operator.
But, I don't have to stop
00:15:23.000 --> 00:15:29.000
there.
Because L is a linear operator,
00:15:26.000 --> 00:15:32.000
I can pull the c out front.
So, it's c1 L of y1 plus c2 L
00:15:31.000 --> 00:15:37.000
of y2. Now, where am I?
00:15:36.000 --> 00:15:42.000
Trying to prove that this is
zero.
00:15:38.000 --> 00:15:44.000
Well, what is L of y1?
At this point,
00:15:40.000 --> 00:15:46.000
I use the fact that y1 is a
solution.
00:15:43.000 --> 00:15:49.000
Because it's a solution,
this is zero.
00:15:46.000 --> 00:15:52.000
That's what it means to solve
that differential equation.
00:15:50.000 --> 00:15:56.000
It means, when you apply the
linear operator,
00:15:53.000 --> 00:15:59.000
L, to the function,
you get zero.
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In the same way,
y2 is a solution.
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So, that's zero.
And, the sum of c1 times zero
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plus c2 times zero is zero.
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That's the argument.
Now, you could make the same
00:16:08.000 --> 00:16:14.000
argument just by plugging c1 y1,
plugging it into the equation
00:16:13.000 --> 00:16:19.000
and calculating,
and calculating,
00:16:15.000 --> 00:16:21.000
and calculating,
grouping the terms and so on
00:16:18.000 --> 00:16:24.000
and so forth.
But, that's just calculation.
00:16:21.000 --> 00:16:27.000
It doesn't show you why it's
so.
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Why it's so is because the
operator, this differential
00:16:28.000 --> 00:16:34.000
equation is expressed as a
linear operator applied to y is
00:16:32.000 --> 00:16:38.000
zero.
And, the only properties that
00:16:37.000 --> 00:16:43.000
are really been used as the fact
that this operator is linear.
00:16:46.000 --> 00:16:52.000
That's the key point.
L is linear.
00:16:50.000 --> 00:16:56.000
It's a linear operator.
Well, that's all there is to
00:16:57.000 --> 00:17:03.000
the superposition principle.
As a prelude to answering the
00:17:04.000 --> 00:17:10.000
more difficult question,
why are these all the
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solutions?
Why are there no other
00:17:09.000 --> 00:17:15.000
solutions?
We need a few definitions,
00:17:12.000 --> 00:17:18.000
and a few more ideas.
And, they are going to occur in
00:17:16.000 --> 00:17:22.000
connection with,
so I'm now addressing,
00:17:18.000 --> 00:17:24.000
ultimately, question two.
But, it's not going to be
00:17:22.000 --> 00:17:28.000
addressed directly for quite
awhile.
00:17:25.000 --> 00:17:31.000
Instead, I'm going to phrase it
in terms of solving the initial
00:17:29.000 --> 00:17:35.000
value problem.
So far, we've only talked about
00:17:34.000 --> 00:17:40.000
the general solution with those
two arbitrary constants.
00:17:39.000 --> 00:17:45.000
But, how do you solve the
initial value problem,
00:17:43.000 --> 00:17:49.000
in other words,
fit initial conditions,
00:17:46.000 --> 00:17:52.000
find the solution with given
initial values for the function
00:17:51.000 --> 00:17:57.000
and its derivatives.
Now, --
00:18:04.000 --> 00:18:10.000
-- the theorem is that this
collection of functions with
00:18:08.000 --> 00:18:14.000
these arbitrary constants,
these are all the solutions we
00:18:12.000 --> 00:18:18.000
have so far.
In fact, they are all the
00:18:15.000 --> 00:18:21.000
solutions there are,
but we don't know that yet.
00:18:19.000 --> 00:18:25.000
However, if we just focus on
the big class of solutions,
00:18:23.000 --> 00:18:29.000
there might be others lurking
out there somewhere lurking out
00:18:28.000 --> 00:18:34.000
there.
I don't know.
00:18:31.000 --> 00:18:37.000
But let's use what we have,
that just from this family is
00:18:39.000 --> 00:18:45.000
enough to satisfy any initial
condition, to satisfy any
00:18:47.000 --> 00:18:53.000
initial values.
In other words,
00:18:51.000 --> 00:18:57.000
if you give me any initial
values, I will be able to find
00:18:59.000 --> 00:19:05.000
the c1 and c2 which work.
Now, why is that?
00:19:05.000 --> 00:19:11.000
Well, I'd have to do a song and
dance at this point,
00:19:08.000 --> 00:19:14.000
if you hadn't been softened up
by actually calculating for
00:19:12.000 --> 00:19:18.000
specific differential equations.
You've had exercises and
00:19:15.000 --> 00:19:21.000
actually how to calculate the
values of c1 and c2.
00:19:19.000 --> 00:19:25.000
So, I'm going to do it now in
general what you have done so
00:19:23.000 --> 00:19:29.000
far for particular equations
using particular values of the
00:19:26.000 --> 00:19:32.000
initial conditions.
So, I'm relying on that
00:19:30.000 --> 00:19:36.000
experience that you've had in
doing the homework to make
00:19:35.000 --> 00:19:41.000
intelligible what I'm going to
do now in the abstract using
00:19:40.000 --> 00:19:46.000
just letters.
So, why is this so?
00:19:43.000 --> 00:19:49.000
Why is that so?
Well, we are going to need it,
00:19:47.000 --> 00:19:53.000
by the way, here,
too.
00:19:49.000 --> 00:19:55.000
I'll have to,
again, open up parentheses.
00:19:52.000 --> 00:19:58.000
But let's go as far as we can.
Well, you just try to do it.
00:19:57.000 --> 00:20:03.000
Suppose the initial conditions
are, how will we write them?
00:20:04.000 --> 00:20:10.000
So, they're going to be at some
initial point,
00:20:06.000 --> 00:20:12.000
x0.
You can take it to be zero if
00:20:08.000 --> 00:20:14.000
you want, but I'd like to be,
just for a little while,
00:20:12.000 --> 00:20:18.000
a little more general.
So, let's say the initial
00:20:15.000 --> 00:20:21.000
conditions, the initial values
are being given at the point x0,
00:20:19.000 --> 00:20:25.000
all right, that's going to be
some number.
00:20:21.000 --> 00:20:27.000
Let's just call it a.
And, the initial value also has
00:20:25.000 --> 00:20:31.000
to specify the velocity or the
value of the derivative there.
00:20:30.000 --> 00:20:36.000
Let's say these are the initial
values.
00:20:33.000 --> 00:20:39.000
So, the problem is to find the
c which work.
00:20:37.000 --> 00:20:43.000
Now, how do you do that?
Well, you know from
00:20:41.000 --> 00:20:47.000
calculation.
You write y equals c1 y1 plus
00:20:45.000 --> 00:20:51.000
c2 y2.
And you write y prime,
00:20:47.000 --> 00:20:53.000
and you take the derivative
underneath that,
00:20:51.000 --> 00:20:57.000
which is easy to do.
And now, you plug in x equals
00:20:56.000 --> 00:21:02.000
zero.
And, what happens?
00:21:00.000 --> 00:21:06.000
Well, these now turn into a set
of equations.
00:21:04.000 --> 00:21:10.000
What will they look like?
Well, y of x0 is a,
00:21:08.000 --> 00:21:14.000
and this is b.
So, what I get is let me flop
00:21:12.000 --> 00:21:18.000
it over onto the other side
because that's the way you're
00:21:17.000 --> 00:21:23.000
used to looking at systems of
equations.
00:21:21.000 --> 00:21:27.000
So, what we get is c1 times y1
of x0 plus c2 times y2 of x0.
00:21:26.000 --> 00:21:32.000
What's that supposed to be
00:21:31.000 --> 00:21:37.000
equal to?
Well, that's supposed to be
00:21:34.000 --> 00:21:40.000
equal to y of x0.
It's supposed to be equal to
00:21:37.000 --> 00:21:43.000
the given number,
a.
00:21:39.000 --> 00:21:45.000
And, in the same way,
c1 y1 prime of x0 plus c2 y2
00:21:42.000 --> 00:21:48.000
prime of x0, that's supposed to
turn out to
00:21:46.000 --> 00:21:52.000
be the number, b.
00:21:48.000 --> 00:21:54.000
In the calculations you've done
up to this point,
00:21:51.000 --> 00:21:57.000
y1 and y2 were always specific
functions like e to the x
00:21:55.000 --> 00:22:01.000
or cosine of 22x,
00:21:57.000 --> 00:22:03.000
stuff like that.
Now I'm doing it in the
00:22:01.000 --> 00:22:07.000
abstract, just calling them y1
and y2, so as to include all
00:22:06.000 --> 00:22:12.000
those possible cases.
Now, what am I supposed to do?
00:22:10.000 --> 00:22:16.000
I'm supposed to find c1 and c2.
What kind of things are they?
00:22:15.000 --> 00:22:21.000
This is what you studied in
high school, right?
00:22:19.000 --> 00:22:25.000
The letters are around us,
but it's a pair of simultaneous
00:22:23.000 --> 00:22:29.000
linear equations.
What are the variables?
00:22:26.000 --> 00:22:32.000
What are the variables?
What are the variables?
00:22:30.000 --> 00:22:36.000
Somebody raise their hand.
If you have a pair of
00:22:35.000 --> 00:22:41.000
simultaneous linear equations,
you've got variables and you've
00:22:39.000 --> 00:22:45.000
got constants,
right?
00:22:41.000 --> 00:22:47.000
And you are trying to find the
answer.
00:22:43.000 --> 00:22:49.000
What are the variables?
Yeah?
00:22:45.000 --> 00:22:51.000
c1 and c2.
Very good.
00:22:47.000 --> 00:22:53.000
I mean, it's extremely
confusing because in the first
00:22:50.000 --> 00:22:56.000
place, how can they be the
variables if they occur on the
00:22:54.000 --> 00:23:00.000
wrong side?
They're on the wrong side;
00:22:57.000 --> 00:23:03.000
they are constants.
How can constants be variables?
00:23:02.000 --> 00:23:08.000
Everything about this is wrong.
Nonetheless,
00:23:06.000 --> 00:23:12.000
the c1 and the c2 are the
unknowns, if you like the high
00:23:10.000 --> 00:23:16.000
school terminology.
c1 and c2 are the unknowns.
00:23:14.000 --> 00:23:20.000
These messes are just numbers.
After you've plugged in x0,
00:23:19.000 --> 00:23:25.000
this is some number.
You've got four numbers here.
00:23:23.000 --> 00:23:29.000
So, c1 and c2 are the
variables.
00:23:26.000 --> 00:23:32.000
The two find,
in other words,
00:23:28.000 --> 00:23:34.000
to find the values of.
All right, now you know general
00:23:34.000 --> 00:23:40.000
theorems from 18.02 about when
can you solve such a system of
00:23:39.000 --> 00:23:45.000
equations.
I'm claiming that you can
00:23:42.000 --> 00:23:48.000
always find c1 and c2 that work.
But, you know that's not always
00:23:48.000 --> 00:23:54.000
the case that a pair of
simultaneous linear equations
00:23:52.000 --> 00:23:58.000
can be solved.
There's a condition.
00:23:55.000 --> 00:24:01.000
There's a condition which
guarantees their solution,
00:23:59.000 --> 00:24:05.000
which is what?
What has to be true about the
00:24:04.000 --> 00:24:10.000
coefficients?
These are the coefficients.
00:24:08.000 --> 00:24:14.000
What has to be true?
The matrix of coefficients must
00:24:13.000 --> 00:24:19.000
be invertible.
The determinant of coefficients
00:24:18.000 --> 00:24:24.000
must be nonzero.
So, they are solvable if,
00:24:22.000 --> 00:24:28.000
for the c1 and c2,
if this thing,
00:24:25.000 --> 00:24:31.000
I'm going to write it.
Since all of these are
00:24:29.000 --> 00:24:35.000
evaluated at x0,
I'm going to write it in this
00:24:34.000 --> 00:24:40.000
way.
y1, the determinant,
00:24:38.000 --> 00:24:44.000
whose entries are y1,
y2, y1 prime,
00:24:41.000 --> 00:24:47.000
and y2 prime,
00:24:44.000 --> 00:24:50.000
evaluated at zero,
x0, that means that I evaluate
00:24:49.000 --> 00:24:55.000
each of the functions in the
determinant at x0.
00:24:54.000 --> 00:25:00.000
I'll write it this way.
That should be not zero.
00:25:00.000 --> 00:25:06.000
So, in other words,
the key thing which makes this
00:25:03.000 --> 00:25:09.000
possible, makes it possible for
us to solve the initial value
00:25:08.000 --> 00:25:14.000
problem, is that this funny
determinant should not be zero
00:25:13.000 --> 00:25:19.000
at the point at which we are
interested.
00:25:16.000 --> 00:25:22.000
Now, this determinant is
important in 18.03.
00:25:19.000 --> 00:25:25.000
It has a name,
and this is when you're going
00:25:23.000 --> 00:25:29.000
to learn it, if you don't know
it already.
00:25:26.000 --> 00:25:32.000
That determinant is called the
Wronskian.
00:25:31.000 --> 00:25:37.000
The Wronskian of what?
If you want to be pompous,
00:25:34.000 --> 00:25:40.000
you say this with a V sound
instead of a W.
00:25:37.000 --> 00:25:43.000
But, nobody does except people
trying to be pompous.
00:25:41.000 --> 00:25:47.000
The Wronskian,
we'll write a W.
00:25:43.000 --> 00:25:49.000
Now, notice,
you can only calculate it when
00:25:47.000 --> 00:25:53.000
you know what the two functions
are.
00:25:49.000 --> 00:25:55.000
So, the Wronskian of the two
functions, y1 and y2,
00:25:53.000 --> 00:25:59.000
what's the variable?
It's not a function of two
00:25:56.000 --> 00:26:02.000
variables, y1 and y2.
These are just the names of
00:26:00.000 --> 00:26:06.000
functions of x.
So, when you do it,
00:26:04.000 --> 00:26:10.000
put it in, calculate out that
determinant.
00:26:08.000 --> 00:26:14.000
This is a function of x,
a function of the independent
00:26:13.000 --> 00:26:19.000
variable after you've done the
calculation.
00:26:17.000 --> 00:26:23.000
Anyway, let's write its
definition, y1,
00:26:20.000 --> 00:26:26.000
y2, y1 prime,
y2 prime.
00:26:23.000 --> 00:26:29.000
Now, in order to do this,
00:26:26.000 --> 00:26:32.000
the point is we must know that
that Wronskian is not zero,
00:26:32.000 --> 00:26:38.000
that the Wronskian of these two
functions is not zero at the
00:26:37.000 --> 00:26:43.000
point x0.
Now, enter a theorem which
00:26:42.000 --> 00:26:48.000
you're going to prove for
homework, but this is harder.
00:26:47.000 --> 00:26:53.000
So, it's part two homework.
It's not part one homework.
00:26:51.000 --> 00:26:57.000
In other words,
I didn't give you the answer.
00:26:55.000 --> 00:27:01.000
You've got to find it yourself,
alone or in the company of good
00:27:01.000 --> 00:27:07.000
friends.
So, anyway, here's the
00:27:05.000 --> 00:27:11.000
Wronskian.
Now, what can we say for sure?
00:27:08.000 --> 00:27:14.000
Note, suppose y1 and y,
just to get you a feeling for
00:27:13.000 --> 00:27:19.000
it a little bit,
suppose they were not
00:27:16.000 --> 00:27:22.000
independent.
The word for not independent is
00:27:20.000 --> 00:27:26.000
dependent.
Suppose they were dependent.
00:27:24.000 --> 00:27:30.000
In other words,
suppose that y2 were a constant
00:27:28.000 --> 00:27:34.000
multiple of y1.
We know that's not the case
00:27:33.000 --> 00:27:39.000
because our functions are
supposed to be independent.
00:27:37.000 --> 00:27:43.000
But suppose they weren't.
What would the value of the
00:27:42.000 --> 00:27:48.000
Wronskian be?
If y2 is a constant times y1,
00:27:46.000 --> 00:27:52.000
then y2 prime is that same
constant times y1 prime.
00:27:50.000 --> 00:27:56.000
What's the value of the
determinant?
00:27:53.000 --> 00:27:59.000
Zero.
For what values of x is it zero
00:27:56.000 --> 00:28:02.000
for all values of x?
And now, that's the theorem
00:28:01.000 --> 00:28:07.000
that you're going to prove,
that if y1 and y2 are solutions
00:28:06.000 --> 00:28:12.000
to the ODE, I won't keep,
say, it's the ODE we've been
00:28:10.000 --> 00:28:16.000
talking about,
y double prime plus py.
00:28:14.000 --> 00:28:20.000
But the linear homogeneous with
00:28:20.000 --> 00:28:26.000
not constant coefficients,
just linear homogeneous second
00:28:26.000 --> 00:28:32.000
order.
Our solutions,
00:28:29.000 --> 00:28:35.000
as there are only two
possibilities,
00:28:33.000 --> 00:28:39.000
either-or.
Either the Wronskian of y,
00:28:37.000 --> 00:28:43.000
there are only two
possibilities.
00:28:39.000 --> 00:28:45.000
Either the Wronskian of y1 and
y2 is always zero,
00:28:44.000 --> 00:28:50.000
identically zero,
zero for all values of x.
00:28:47.000 --> 00:28:53.000
This is redundant.
When I say identically,
00:28:51.000 --> 00:28:57.000
I mean for all values of x.
But, I am just making assurance
00:28:56.000 --> 00:29:02.000
doubly sure.
Or, or the Wronskian is never
00:29:01.000 --> 00:29:07.000
zero.
Now, there is no notation as
00:29:06.000 --> 00:29:12.000
for that.
I'd better just write it out,
00:29:10.000 --> 00:29:16.000
is never zero,
i.e.
00:29:13.000 --> 00:29:19.000
for no x is it,
i.e.
00:29:15.000 --> 00:29:21.000
for all x.
There's no way to say that.
00:29:20.000 --> 00:29:26.000
I mean, for all values of x,
it's not zero.
00:29:26.000 --> 00:29:32.000
That means, there is not a
single point for which it's
00:29:32.000 --> 00:29:38.000
zero.
In particular,
00:29:35.000 --> 00:29:41.000
it's not zero here.
So, this is your homework:
00:29:43.000 --> 00:29:49.000
problem five,
part two.
00:29:45.000 --> 00:29:51.000
I'll give you a method of
proving it, which was discovered
00:29:52.000 --> 00:29:58.000
by the famous Norwegian
mathematician,
00:29:57.000 --> 00:30:03.000
Abel, who is,
I guess, the centenary of his
00:30:02.000 --> 00:30:08.000
birth, I guess,
was just celebrated last year.
00:30:09.000 --> 00:30:15.000
He has one of the truly tragic
stories in mathematics,
00:30:13.000 --> 00:30:19.000
which I think you can read.
It must be a Simmons book,
00:30:18.000 --> 00:30:24.000
if you have that.
Simmons is very good on
00:30:22.000 --> 00:30:28.000
biographies.
Look up Abel.
00:30:24.000 --> 00:30:30.000
He'll have a biography of Abel,
and you can weep if you're
00:30:29.000 --> 00:30:35.000
feeling sad.
He died at the age of 26 of
00:30:34.000 --> 00:30:40.000
tuberculosis,
having done a number of
00:30:37.000 --> 00:30:43.000
sensational things,
none of which was recognized in
00:30:41.000 --> 00:30:47.000
his lifetime because people
buried his papers under big
00:30:46.000 --> 00:30:52.000
piles of papers.
So, he died unknown,
00:30:49.000 --> 00:30:55.000
uncelebrated,
and now he's Norway's greatest
00:30:53.000 --> 00:30:59.000
culture hero.
In the middle of a park in
00:30:56.000 --> 00:31:02.000
Oslo, there's a huge statue.
And, since nobody knows what
00:31:03.000 --> 00:31:09.000
Abel looked like,
the statue is way up high so
00:31:07.000 --> 00:31:13.000
you can't see very well.
But, the inscription on the
00:31:13.000 --> 00:31:19.000
bottom says Niels Henrik Abel,
1801-1826 or something like
00:31:19.000 --> 00:31:25.000
that.
Now, --
00:31:38.000 --> 00:31:44.000
-- the choice,
I'm still, believe it or not,
00:31:41.000 --> 00:31:47.000
aiming at question two,
but I have another big
00:31:44.000 --> 00:31:50.000
parentheses to open.
And, when I closed it,
00:31:47.000 --> 00:31:53.000
the answer to question two will
be simple.
00:31:50.000 --> 00:31:56.000
But, I think it's very
desirable that you get this
00:31:53.000 --> 00:31:59.000
second big parentheses.
It will help you to understand
00:31:57.000 --> 00:32:03.000
something important.
It will help you on your
00:32:02.000 --> 00:32:08.000
problem set tomorrow night.
I don't have to apologize.
00:32:08.000 --> 00:32:14.000
I'm just going to do it.
So, the question is,
00:32:12.000 --> 00:32:18.000
the thing you have to
understand is that when I write
00:32:18.000 --> 00:32:24.000
this combination,
I'm claiming that these are all
00:32:23.000 --> 00:32:29.000
the solutions.
I haven't proved that yet.
00:32:27.000 --> 00:32:33.000
But, they are going to be all
the solutions
00:32:33.000 --> 00:32:39.000
The point is,
there's nothing sacrosanct
00:32:36.000 --> 00:32:42.000
about the y1 and y2.
This is exactly the same
00:32:41.000 --> 00:32:47.000
collection as a collection which
I would write using other
00:32:46.000 --> 00:32:52.000
constants.
Let's call them u1 and u2.
00:32:49.000 --> 00:32:55.000
They are exactly the same,
where u1 and u2 are any other
00:32:55.000 --> 00:33:01.000
pair of linearly independent
solutions.
00:33:00.000 --> 00:33:06.000
Any other pair of independent
solutions, they must be
00:33:05.000 --> 00:33:11.000
independent, either a constant
multiple of each other.
00:33:10.000 --> 00:33:16.000
In other words,
u1 is some combination,
00:33:13.000 --> 00:33:19.000
now I'm really stuck because I
don't know how to,
00:33:18.000 --> 00:33:24.000
c1 bar, let's say,
that means a special value of
00:33:22.000 --> 00:33:28.000
c1, and a special value of c2,
and u2 is some other special
00:33:28.000 --> 00:33:34.000
value, oh my God,
c1 double bar,
00:33:31.000 --> 00:33:37.000
how's that?
The notation is getting worse
00:33:36.000 --> 00:33:42.000
and worse.
I apologize for it.
00:33:38.000 --> 00:33:44.000
In other words,
I could pick y1 and y2 and make
00:33:42.000 --> 00:33:48.000
up all of these.
And, I'd get a bunch of
00:33:45.000 --> 00:33:51.000
solutions.
But, I could also pick some
00:33:48.000 --> 00:33:54.000
other family,
some other two guys in this
00:33:51.000 --> 00:33:57.000
family, and just as well express
the solutions in terms of u1 and
00:33:56.000 --> 00:34:02.000
u2.
Now, well, why is he telling us
00:33:59.000 --> 00:34:05.000
that?
Well, the point is that the y1
00:34:03.000 --> 00:34:09.000
and the y2 are typically the
ones you get easily from solving
00:34:08.000 --> 00:34:14.000
the equations,
like e to the x and e
00:34:11.000 --> 00:34:17.000
to the 2x.
That's what you've gotten,
00:34:15.000 --> 00:34:21.000
or cosine x and sine x,
00:34:18.000 --> 00:34:24.000
something like that.
But, for certain ways,
00:34:22.000 --> 00:34:28.000
they might not be the best way
of writing the solutions.
00:34:26.000 --> 00:34:32.000
There is another way of writing
those that you should learn,
00:34:31.000 --> 00:34:37.000
and that's called finding
normalized, the normalized.
00:34:35.000 --> 00:34:41.000
They are okay,
but they are not normalized.
00:34:40.000 --> 00:34:46.000
For some things,
the normalized solutions are
00:34:42.000 --> 00:34:48.000
the best.
I'll explain to you what they
00:34:44.000 --> 00:34:50.000
are, and I'll explain to you
what they're good for.
00:34:48.000 --> 00:34:54.000
You'll see immediately what
they're good for.
00:34:50.000 --> 00:34:56.000
Normalized solutions,
now, you have to specify the
00:34:53.000 --> 00:34:59.000
point at which you're
normalizing.
00:34:55.000 --> 00:35:01.000
In general, it would be x
nought,
00:34:58.000 --> 00:35:04.000
but let's, at this point,
since I don't have an infinity
00:35:01.000 --> 00:35:07.000
of time, to simplify things,
let's say zero.
00:35:05.000 --> 00:35:11.000
It could be x nought,
any point would do just as
00:35:08.000 --> 00:35:14.000
well.
But, zero is the most common
00:35:11.000 --> 00:35:17.000
choice.
What are the normalized
00:35:13.000 --> 00:35:19.000
solutions?
Well, first of all,
00:35:15.000 --> 00:35:21.000
I have to give them names.
I want to still call them y.
00:35:19.000 --> 00:35:25.000
So, I'll call them capital Y1
and Y2.
00:35:21.000 --> 00:35:27.000
And, what they are,
are the solutions which satisfy
00:35:25.000 --> 00:35:31.000
certain, special,
very special,
00:35:27.000 --> 00:35:33.000
initial conditions.
And, what are those?
00:35:31.000 --> 00:35:37.000
So, they're the ones which
satisfy, the initial conditions
00:35:37.000 --> 00:35:43.000
for Y1 are, of course there are
going to be guys that look like
00:35:43.000 --> 00:35:49.000
this.
The only thing that's going to
00:35:46.000 --> 00:35:52.000
make them distinctive is the
initial conditions they satisfy.
00:35:51.000 --> 00:35:57.000
Y1 has to satisfy at zero.
Its value should be one,
00:35:56.000 --> 00:36:02.000
and the value of its derivative
should be zero.
00:36:02.000 --> 00:36:08.000
For Y2, it's just the opposite.
Here, the value of the function
00:36:07.000 --> 00:36:13.000
should be zero at zero.
But, the value of its
00:36:11.000 --> 00:36:17.000
derivative, now,
I want to be one.
00:36:14.000 --> 00:36:20.000
Let me give you a trivial
example of this,
00:36:17.000 --> 00:36:23.000
and then one,
which is a little less trivial,
00:36:21.000 --> 00:36:27.000
so you'll have some feeling for
what I'm asking for.
00:36:25.000 --> 00:36:31.000
Suppose the equation,
for example,
00:36:28.000 --> 00:36:34.000
is y double prime plus,
well, let's really make it
00:36:33.000 --> 00:36:39.000
simple.
Okay, you know the standard
00:36:36.000 --> 00:36:42.000
solutions are y1 is cosine x,
00:36:39.000 --> 00:36:45.000
and y2 is sine x.
00:36:41.000 --> 00:36:47.000
These are functions,
which, when you take the second
00:36:44.000 --> 00:36:50.000
derivative, they turn into their
negative.
00:36:46.000 --> 00:36:52.000
You know, you could go the
complex roots are i and minus i,
00:36:50.000 --> 00:36:56.000
and blah, blah,
blah, blah, blah,
00:36:52.000 --> 00:36:58.000
blah.
If you do it that way,
00:36:53.000 --> 00:36:59.000
fine.
But at some point in the course
00:36:56.000 --> 00:37:02.000
you have to be able to write
down and, right away,
00:36:59.000 --> 00:37:05.000
oh, yeah, cosine x,
sine x.
00:37:01.000 --> 00:37:07.000
Okay, what are the normalized
00:37:05.000 --> 00:37:11.000
things?
Well, what's the value of this
00:37:09.000 --> 00:37:15.000
at zero?
It is one.
00:37:11.000 --> 00:37:17.000
What's the value of its
derivative at zero?
00:37:15.000 --> 00:37:21.000
Zero.
This is Y1.
00:37:16.000 --> 00:37:22.000
This is the only case in which
you locked on immediately to the
00:37:22.000 --> 00:37:28.000
normalized solutions.
In the same way,
00:37:26.000 --> 00:37:32.000
this guy is Y2 because its
value at zero is zero.
00:37:32.000 --> 00:37:38.000
It's value of its derivative at
zero is one.
00:37:35.000 --> 00:37:41.000
So, this is Y2.
Okay, now let's look at a case
00:37:39.000 --> 00:37:45.000
where you don't immediately lock
on to the normalized solutions.
00:37:44.000 --> 00:37:50.000
Very simple:
all I have to do is change the
00:37:47.000 --> 00:37:53.000
sign.
Here, you know,
00:37:49.000 --> 00:37:55.000
think through r squared minus
one equals zero.
00:37:54.000 --> 00:38:00.000
The characteristic roots are
plus one and minus one,
00:37:58.000 --> 00:38:04.000
right?
And therefore,
00:38:00.000 --> 00:38:06.000
the solution is e to the x,
and e to the negative x.
00:38:04.000 --> 00:38:10.000
So, the solutions you find by
00:38:07.000 --> 00:38:13.000
the usual way of solving it is
y1 equals e to the x,
00:38:11.000 --> 00:38:17.000
and y2 equals e to the
negative x.
00:38:15.000 --> 00:38:21.000
Those are the standard
solution.
00:38:17.000 --> 00:38:23.000
So, the general solution is of
the form.
00:38:19.000 --> 00:38:25.000
So, the general solution is of
the form c1 e to the x plus c2 e
00:38:23.000 --> 00:38:29.000
to the negative x.
00:38:26.000 --> 00:38:32.000
Now, what I want to find out is
what is Y1 and Y2?
00:38:31.000 --> 00:38:37.000
How do I find out what Y1 is?
Well, I have to satisfy initial
00:38:36.000 --> 00:38:42.000
conditions.
So, if this is y,
00:38:38.000 --> 00:38:44.000
let's write down here,
if you can still see that,
00:38:43.000 --> 00:38:49.000
y prime is c1 e to the x minus
c2 e to the negative x .
00:38:48.000 --> 00:38:54.000
So, if I plug in,
00:38:51.000 --> 00:38:57.000
I want y of zero to be one,
I want this guy at the point
00:38:56.000 --> 00:39:02.000
zero to be one.
What equation does that give
00:39:00.000 --> 00:39:06.000
me?
That gives me c1 plus c2,
00:39:04.000 --> 00:39:10.000
c1 plus c2,
plugging in x equals zero,
00:39:08.000 --> 00:39:14.000
equals the value of this thing
at zero.
00:39:11.000 --> 00:39:17.000
So, that's supposed to be one.
00:39:14.000 --> 00:39:20.000
How about the other guy?
The value of its derivative is
00:39:19.000 --> 00:39:25.000
supposed to come out to be zero.
And, what is its derivative?
00:39:24.000 --> 00:39:30.000
Well, plug into this
expression.
00:39:26.000 --> 00:39:32.000
It's c1 minus c2.
Okay, what's the solution to
00:39:32.000 --> 00:39:38.000
those pair of equations?
c2 has to be equal to c1.
00:39:36.000 --> 00:39:42.000
The sum of the two of them has
to be one.
00:39:39.000 --> 00:39:45.000
Each one, therefore,
is equal to one half.
00:39:42.000 --> 00:39:48.000
And so, what's the value of Y1?
Y1, therefore,
00:39:46.000 --> 00:39:52.000
is the function where c1 and c2
are one half.
00:39:50.000 --> 00:39:56.000
It's the function e to the x
plus e to the negative x divided
00:39:55.000 --> 00:40:01.000
by two.
00:39:59.000 --> 00:40:05.000
In the same way,
I won't repeat the calculation.
00:40:03.000 --> 00:40:09.000
You can do yourself.
Same calculation shows that Y2,
00:40:07.000 --> 00:40:13.000
so, put in the initial
conditions.
00:40:10.000 --> 00:40:16.000
The answer will be that Y2 is
equal to e to the x minus e to
00:40:15.000 --> 00:40:21.000
the minus x divided by two.
00:40:19.000 --> 00:40:25.000
These are the special
functions.
00:40:22.000 --> 00:40:28.000
For this equation,
these are the normalized
00:40:25.000 --> 00:40:31.000
solutions.
They are better than the
00:40:28.000 --> 00:40:34.000
original solutions because their
initial values are nicer.
00:40:35.000 --> 00:40:41.000
Just check it.
The initial value,
00:40:37.000 --> 00:40:43.000
when x is equal to zero,
the initial value,
00:40:40.000 --> 00:40:46.000
this has is zero.
Here, when x is equal to zero,
00:40:44.000 --> 00:40:50.000
the value of the function is
zero.
00:40:47.000 --> 00:40:53.000
But, the value of its
derivative, these cancel,
00:40:51.000 --> 00:40:57.000
is one.
So, these are the good guys.
00:40:54.000 --> 00:41:00.000
Okay, there's no colored chalk
this period.
00:40:57.000 --> 00:41:03.000
Okay, there was colored chalk.
There's one.
00:41:01.000 --> 00:41:07.000
So, for this equation,
these are the good guys.
00:41:04.000 --> 00:41:10.000
These are our best solutions.
e to the x and e to the
00:41:07.000 --> 00:41:13.000
minus x are good solutions.
00:41:10.000 --> 00:41:16.000
But, these are our better
solutions.
00:41:12.000 --> 00:41:18.000
And, this one,
of course, is the function
00:41:14.000 --> 00:41:20.000
which is called hyperbolic sine
of x, and this is the one which
00:41:18.000 --> 00:41:24.000
is called hyperbolic cosine of
x.
00:41:20.000 --> 00:41:26.000
This is one of the most
important ways in which they
00:41:23.000 --> 00:41:29.000
enter into mathematics.
And, this is why the engineers
00:41:26.000 --> 00:41:32.000
want them.
Now, why do the engineers want
00:41:28.000 --> 00:41:34.000
normalized solutions?
Well, I didn't explain that.
00:41:34.000 --> 00:41:40.000
So, what's so good about
normalized solutions?
00:41:41.000 --> 00:41:47.000
Very simple:
if Y1 and Y2 are normalized at
00:41:47.000 --> 00:41:53.000
zero, let's say,
then the solution to the IVP,
00:41:53.000 --> 00:41:59.000
in other words,
the ODE plus the initial
00:41:59.000 --> 00:42:05.000
values, y of zero equals,
let's say, a and y
00:42:07.000 --> 00:42:13.000
prime of zero equals b.
00:42:14.000 --> 00:42:20.000
So, the ODE I'm not repeating.
It's the one we've been talking
00:42:17.000 --> 00:42:23.000
about all term since the
beginning of the period.
00:42:20.000 --> 00:42:26.000
It's the one with the p of x
and q of x.
00:42:23.000 --> 00:42:29.000
And, here are the initial
values.
00:42:25.000 --> 00:42:31.000
I'm going to call them a and b.
You can also call them,
00:42:28.000 --> 00:42:34.000
if you like,
maybe that's better to call
00:42:30.000 --> 00:42:36.000
them y0, as they are individual
in the homework.
00:42:34.000 --> 00:42:40.000
They are called,
I'm using the,
00:42:36.000 --> 00:42:42.000
let's use those.
What is the solution?
00:42:39.000 --> 00:42:45.000
I say the solution is,
if you use y1 and y2,
00:42:43.000 --> 00:42:49.000
the solution is y0,
in other words,
00:42:46.000 --> 00:42:52.000
the a times Y1,
plus y0 prime,
00:42:49.000 --> 00:42:55.000
in other words,
b times Y2.
00:42:53.000 --> 00:42:59.000
In other words,
you can write down instantly
00:42:57.000 --> 00:43:03.000
the solution to the initial
value problem,
00:43:00.000 --> 00:43:06.000
if instead of using the
functions, you started out with
00:43:05.000 --> 00:43:11.000
the little Y1 and Y2,
you use these better functions.
00:43:12.000 --> 00:43:18.000
The thing that's better about
them is that they instantly
00:43:16.000 --> 00:43:22.000
solve for you the initial value
problem.
00:43:18.000 --> 00:43:24.000
All you do is use this number,
initial condition as the
00:43:22.000 --> 00:43:28.000
coefficient of Y1,
and use this number as the
00:43:26.000 --> 00:43:32.000
coefficient of Y2.
Now, just check that by looking
00:43:29.000 --> 00:43:35.000
at it.
Why is that so?
00:43:32.000 --> 00:43:38.000
Well, for example,
let's check.
00:43:34.000 --> 00:43:40.000
What is its value of this
function at zero?
00:43:37.000 --> 00:43:43.000
Well, the value of this guy at
zero is one.
00:43:40.000 --> 00:43:46.000
So, the answer is y0 times one,
and the value of this guy at
00:43:44.000 --> 00:43:50.000
zero is zero.
So, this term disappears.
00:43:47.000 --> 00:43:53.000
And, it's exactly the same with
the derivative.
00:43:50.000 --> 00:43:56.000
What's the value of the
derivative at zero?
00:43:54.000 --> 00:44:00.000
The value of the derivative of
this thing is zero.
00:43:57.000 --> 00:44:03.000
So, this term disappears.
The value of this derivative at
00:44:01.000 --> 00:44:07.000
zero is one. And so,
the answer is y0 prime.
00:44:06.000 --> 00:44:12.000
So, check, check,
00:44:08.000 --> 00:44:14.000
this works.
So, these better solutions have
00:44:11.000 --> 00:44:17.000
the property,
what's good about them,
00:44:14.000 --> 00:44:20.000
and why scientists and
engineers like them,
00:44:18.000 --> 00:44:24.000
is that they enable you
immediately to write down the
00:44:22.000 --> 00:44:28.000
answer to the initial value
problem without having to go
00:44:26.000 --> 00:44:32.000
through this business,
which I buried down here,
00:44:30.000 --> 00:44:36.000
of solving simultaneous linear
equations.
00:44:35.000 --> 00:44:41.000
Okay, now, believe it or not,
that's all the work.
00:44:40.000 --> 00:44:46.000
We are ready to answer question
number two: why are these all
00:44:46.000 --> 00:44:52.000
the solutions?
Of course, I have to invoke a
00:44:50.000 --> 00:44:56.000
big theorem.
A big theorem:
00:44:53.000 --> 00:44:59.000
where shall I invoke a big
theorem?
00:44:56.000 --> 00:45:02.000
Let's see if we can do it here.
The big theorem says,
00:45:02.000 --> 00:45:08.000
it's called the existence and
uniqueness theorem.
00:45:05.000 --> 00:45:11.000
It's the last thing that's
proved at the end of an analysis
00:45:08.000 --> 00:45:14.000
course, at which real analysis
courses, over which students
00:45:11.000 --> 00:45:17.000
sweat for one whole semester,
and their reward at the end is,
00:45:15.000 --> 00:45:21.000
if they are very lucky,
and if they have been very good
00:45:18.000 --> 00:45:24.000
students, they get to see the
proof of the existence and
00:45:21.000 --> 00:45:27.000
uniqueness theorem for
differential equations.
00:45:24.000 --> 00:45:30.000
But, I can at least say what it
is for the linear equation
00:45:27.000 --> 00:45:33.000
because it's so simple.
It says, so,
00:45:31.000 --> 00:45:37.000
the equation we are talking
about is the usual one,
00:45:35.000 --> 00:45:41.000
homogeneous equation,
and I'm going to assume,
00:45:39.000 --> 00:45:45.000
you have to have assumptions
that p and q are continuous for
00:45:44.000 --> 00:45:50.000
all x.
So, they're good-looking
00:45:46.000 --> 00:45:52.000
functions.
Coefficients aren't allowed to
00:45:50.000 --> 00:45:56.000
blow up anywhere.
They've got to look nice.
00:45:55.000 --> 00:46:01.000
Then, the theorem says there is
one and only one solution,
00:46:03.000 --> 00:46:09.000
one and only solution
satisfying, given initial values
00:46:11.000 --> 00:46:17.000
such that y of zero,
let's say y of zero is equal to
00:46:19.000 --> 00:46:25.000
some given number, A,
00:46:24.000 --> 00:46:30.000
and y, let's make y0,
and y prime of zero equals B.
00:46:31.000 --> 00:46:37.000
The initial value problem has
00:46:37.000 --> 00:46:43.000
one and only one solution.
The existence is,
00:46:42.000 --> 00:46:48.000
it has a solution.
The uniqueness is,
00:46:45.000 --> 00:46:51.000
it only has one solution.
If you specify the initial
00:46:51.000 --> 00:46:57.000
conditions, there's only one
function which satisfies them
00:46:56.000 --> 00:47:02.000
and at the same time satisfies
that differential equation.
00:47:02.000 --> 00:47:08.000
Now, this answers our question.
This answers our question,
00:47:08.000 --> 00:47:14.000
because, look,
what I want is all solutions.
00:47:14.000 --> 00:47:20.000
What we want are all solutions
to the ODE.
00:47:21.000 --> 00:47:27.000
And now, here's what I say:
a claim that this collection of
00:47:33.000 --> 00:47:39.000
functions, c1 Y1 plus c2 Y2
are all
00:47:42.000 --> 00:47:48.000
solutions.
Of course, I began a period by
00:47:48.000 --> 00:47:54.000
saying I'd show you that c1
little y1 c little y2 are all
00:47:52.000 --> 00:47:58.000
the solutions.
But, it's the case that these
00:47:55.000 --> 00:48:01.000
two families are the same.
So, the family that I started
00:48:00.000 --> 00:48:06.000
with would be exactly the same
as the family c1 prime Y1
00:48:04.000 --> 00:48:10.000
because,
after all, these are two
00:48:07.000 --> 00:48:13.000
special guys from that
collection.
00:48:11.000 --> 00:48:17.000
So, it doesn't matter whether I
talk about the original ones,
00:48:15.000 --> 00:48:21.000
or these.
The theorem is still the same.
00:48:19.000 --> 00:48:25.000
The final step,
therefore, if you give me one
00:48:22.000 --> 00:48:28.000
more minute, I think that will
be quite enough.
00:48:26.000 --> 00:48:32.000
Why are these all the
solutions?
00:48:30.000 --> 00:48:36.000
Well, I have to take an
arbitrary solution and show you
00:48:35.000 --> 00:48:41.000
that it's one of these.
So, the proof is,
00:48:39.000 --> 00:48:45.000
given a solution,
u(x), what are its values?
00:48:43.000 --> 00:48:49.000
Well, u of x zero is u zero,
00:48:48.000 --> 00:48:54.000
and u prime of x zero, zero,
00:48:51.000 --> 00:48:57.000
let's say, is equal to u zero,
is equal to some other
00:48:58.000 --> 00:49:04.000
number.
Now, what's the solution?
00:49:02.000 --> 00:49:08.000
Write down what's the solution
of these using the Y1's?
00:49:07.000 --> 00:49:13.000
Then, I know I've just shown
you that u zero times Y1 plus u
00:49:12.000 --> 00:49:18.000
zero prime Y2
satisfies the same initial
00:49:17.000 --> 00:49:23.000
conditions, satisfies these
initial conditions,
00:49:21.000 --> 00:49:27.000
initial values.
In other words,
00:49:24.000 --> 00:49:30.000
I started with my little
solution.
00:49:27.000 --> 00:49:33.000
u of x walks up to and
says, hi there.
00:49:31.000 --> 00:49:37.000
Hi there, and the differential
equation looks at it and says,
00:49:37.000 --> 00:49:43.000
who are you?
You say, oh,
00:49:41.000 --> 00:49:47.000
I satisfy you and my initial,
and then it says what are your
00:49:46.000 --> 00:49:52.000
initial values?
It says, my initial values are
00:49:50.000 --> 00:49:56.000
u0 and u0 prime.
And, it said,
00:49:54.000 --> 00:50:00.000
sorry, but we've got one of
ours who satisfies the same
00:49:59.000 --> 00:50:05.000
initial conditions.
We don't need you because the
00:50:03.000 --> 00:50:09.000
existence and uniqueness theorem
says that there can only be one
00:50:09.000 --> 00:50:15.000
function which does that.
And therefore,
00:50:13.000 --> 00:50:19.000
you must be equal to this guy
by the uniqueness theorem.
00:50:20.000 --> 00:50:26.000
Okay, we'll talk more about
stuff next time,
00:50:23.000 --> 00:50:29.000
linear equation next time.