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HERBERT GROSS: Hi, today, we're going
to emphasize the role of power series
in the solution of linear differential equations.
And let me just say at the outset
that we have paid no attention to nonlinear differential
equations in this course.
And we won't pay any attention to the nonlinear equation
simply because the nonlinear equation is very, very
difficult to handle.
We usually tackle it only by special cases.
And, even of more importance, most
of the problems that we have to tackle
are at least reasonably well approximated
by linear differential equations.
Now what I'd like to do before we launch into power series
today is to pull together everything
that we've said so far about linear differential equations.
And, rather than to drone on here saying that,
I thought that I would write the summary on the board,
and we could go through this fairly rapidly together.
At any rate, the lecture for today
is power series solutions, but the summary to date is this.
Given the general linear differential equation L of y
equals f of x with non-constant coefficients,
this equation always has a general solution,
provided p, q, and f are continuous.
Not only does the solution exist,
but the solution is always given by y sub h plus y sub p
where y sub h is the general solution of L of y equals 0,
and y sub p is any solution, in other words,
a particular solution, of L of y equals f of x.
And then you see the rest of our study
has been how do you find y sub h and how do you find y sub p.
The point is it's very easy to find y sub h
if L sub y has constant coefficients.
That was the first case that we tackled.
If L of y has constant coefficient,
then y sub h has one of these three forms, depending on what?
This is the case that holds if the roots
of the characteristic equation are real and unequal.
This is the form that holds if the roots are real but equal.
And this is the equation that holds if the reals are non--
if the roots are non-real where alpha
is the real part of the root, and beta is the imaginary part
of the root, OK?
As far as y sub p was concerned, recall
that the method of undetermined coefficients
yields y sub p when L of y has constant coefficients
and when f of x has the very special form e to the mx or x
to the n or cosine mx or sine mx.
We then turned our attention to variation of parameters,
and the key point about variation of parameters
was that, by use of variation of parameters,
we could always obtain a y sub p when y sub h was known.
By the way, in particular, notice
that, if we are dealing with constant coefficients, then
we do know y sub h.
That's what we were just talking about earlier.
So, in particular, the variation of parameters method
will work whenever we have constant coefficients
because we know y sub h in that case.
But quite in general, you see, even
if the coefficients aren't constant,
variation of parameters works.
And, in fact, once we know y sub h--
say y sub h is c1 u1 plus c2 u2, then
y sub p is g1 u1 plus g2 u2 where g1 prime and g2 prime
satisfy this pair of equations.
The other important thing about variation of parameters
was that it reduces the order of L of y
equals 0 once one solution of the equation is known.
And that, you see, completes our summary.
What we want to tackle next is what's still lacking.
And that is that the success of the method known as variation
of parameters hinges on the fact that we
have the general solution of the homogeneous equation.
In other words, our next problem is
to find the general solution of L of y
equals 0 when the coefficients are not constant.
In other words, if I can find this general solution of L of y
equals 0 when coefficients are not constant, then, you see,
I can use the method of variation of parameters, which
always yields y sub p once y sub h is known,
to find the particular solution.
Then y sub p plus y sub h will be my general solution.
At any rate, it's into this environment
that we introduce the concept of power series.
And the key theorem in using power series is this.
Let's suppose that p of x and q of x are analytic.
And, by the way, I've use that word for complex
valued functions.
Analytic meant that all of the derivatives existed.
It meant that the function could be expanded or represented
by a convergent power series.
That's the meaning of analytic, even in the real case.
When I say that p and q are analytic
for absolute value of x minus x0 less
than some value R, what I mean is
that p and q can be represented by convergent power
series for all x, which are within R of some fixed point
x0.
But, at any rate, all I'm saying is,
if p and q are analytic in this interval,
then every solution of this equation, every solution which
exists, which is defined at x equals x0, is itself analytic,
at least in the same interval that p and q were analytic.
In other words, what this means is
that, in the first part of our review,
we said, lookit, as long as p, q, and f are continuous,
there will always be a general solution.
Now we're going one step further or maybe several steps further.
We're saying, lookit, as long as p and q are not only
just continuous, but they also happen to be analytic,
we can not only guarantee that there'll be a general solution,
but we can guarantee, more to the point,
that whatever solution there is will
be found in the form of a convergent power series.
And that gives us the step that we need to solve problems
by this technique.
Now what I'm going to do is this.
I am going to start with an example
that we already knew how to solve before.
First of all, notice that the equation y
double prime plus y equals 0, first of all,
has constant coefficients.
And notice that the lead in to today's lesson
was we we're going to tackle situations where we didn't
have constant coefficients.
This does have constant coefficients.
And, even more to the point, solution is known.
In fact, what is the solution here?
It's y equals an arbitrary constant times sine x
plus an arbitrary constant times cosine x.
I picked one with constant coefficients
where the solution was known simply
so that we could illustrate what the method is.
By the way, notice that, in this particular problem,
p of x, which is the coefficient of y prime, is 0.
q of x, the coefficient of y, is 1.
And, certainly, 0 and 1 are analytic functions.
In fact, their power series--
the power series for 0 is 0.
And the power series for 1 is 1.
So, certainly, our coefficients meet the requirement.
And what we do is we start off saying, OK,
if any solution exists, it must look like a power series.
Let's go find it.
Notice this is a glorified version
of undetermined coefficients.
We say, lookit, let's assume that our solution has the form
of a power series, in other words, summation,
n goes from 0 to infinity, a sub n x to the n where all I have
to know now are the values of the a sub n's.
And, if you don't like the sigma notation here, all we're saying
is we're trying for a solution in the form y
equals a0 plus a1 x plus a2 x squared,
et cetera, where this series is a convergent power series.
And why is convergent power series important?
Well, the answer, recall from part one of our course
that, in the interval of convergence,
the power series behaves precisely like a polynomial.
We can add power series term by term.
We can subtract them term by term.
We can multiply them the way we do polynomials.
We can integrate term by term.
We can differentiate term by term.
We can rearrange the terms the way we wish, et cetera.
In other words, it has all of the niceties
of a finite polynomial expression.
At any rate, what I then do is, because I'm
assuming this is a convergent power series,
I differentiate this thing term by term.
And, if I differentiate this, notice what I'm doing.
I differentiate each term here, but, because I can--
because I can differentiate term by term,
a very nice generic way of doing this
is notice that this is the n-th term.
The derivative of this term would be what?
n an x to the n minus 1, n an x to the n minus 1.
Notice that, every time you differentiate,
a term drops out.
Like, when I differentiate this, the a0 term drops out.
When I differentiate this, the a1 one term drops out.
So what I do is I differentiate the general expression
inside the sigma sign and then start the index of summation
one up further.
In other words, instead of going from 0 to infinity,
I now go from 1 to infinity.
In a similar way, y double prime,
I differentiate this term by term.
That's n times n minus 1 times a sub n times x to the n minus 2.
And now my sum goes from 2 to infinity.
That's what I've written over here.
Now I'm trying to find out what the a's are equal to.
I could actually call these y sub p's.
See, I'm looking for a particular solution perhaps,
not particular in the sense of a solution, but this is a trial.
Maybe it should have been a T over here,
y sub T. I'm looking for a trial solution
over here, a trial solution.
I take yT double prime, yT, plug it in here,
and see what this means.
And, simply replacing y double prime by this and y by this,
I see that this is the equation that
must be satisfied identically.
Now the point is the only way a power series can be identically
equal to 0 is if coefficient by coefficient
the power series is 0.
Now what has me bugged a little bit here
is that the exponents don't match up.
You see, notice that the general term here has an exponent n,
and the general term here has an exponent n minus 2.
A very nice trick that I can use here
is I say, you know, why don't I jack up n by 2
every place I see it?
In other words, let me replace n by n plus 2.
I'll jack it up by 2.
And, to compensate for that, I'll
lower the summation of index--
the index summation by 2 every place I see it.
Now let me show you what that means so that we
don't get too confused on that.
Let's suppose I had written down here summation n goes
from 0 to infinity a sub n.
This means what?
a0 plus a1 plus a2, et cetera.
Suppose, for some reason, I would
like to either lower or raise the index here.
Let's suppose I want to lower the index.
In other words, I want each--
I want the subscript here to be k less than what it was before.
In other words, I'm going to replace a sub
n by a sub n minus k.
I claim that all I have to do is start my counting process
at n equals k now, instead of n equals 0.
In fact, look what this says.
When n is k, n minus k is 0.
So the first term here is a0.
The next term would be k plus 1. k plus 1 minus k is 1.
So the next term would be a1, et cetera.
And this is just another way of saying this.
So the idea is, if I want to get this
to be an n, that means I have to raise everything
by 2 inside the summation sign.
So I will lower everything by 2 outside the summation sign.
Now, to do that, notice what that's going to do.
My sum here will now go from 0 to infinity.
This will become n plus 2.
This will become n plus 1.
You see, I'm raising it by 2.
Every place I'm seeing an n, I'm replacing it by n plus 2.
This becomes a sub n plus 2, and this will become n.
In summary, then this expression is the same as summation,
n goes from 0 to infinity, n plus 2 times n plus 1 a sub n
plus 2 x to the n plus summation, n goes from 0
to infinity, an x to the n.
And that must be identically zero.
Now the point is that the exponents now line up.
I'm starting the sum in both series at the same place.
And, consequently, because these are
assumed to be convergent power series,
I can add them term by term.
If I add them term by term, that means
I come inside the summation sign here.
x to the n is a common factor.
I factor that out.
And what's left is what?
n plus 2 times n plus 1 times a sub n plus 2
plus a sub n, that's this expression here.
And this is what must be identically zero.
Now the point is that the only way that a convergent power
series can be identically zero is if each coefficient is 0.
And that's how I now handle the undetermined coefficients.
I come to the conclusion that, because this
is to be identically zero, every one of these terms
must be itself 0.
Setting this equal to zero and solving for a sub n plus 2
in terms of a sub n, I find that a sub n plus 2
must be minus an over n plus 2 times n plus 1.
And what this tells me is that I now know what a term looks
like as soon as I know the one that comes two before it.
You see, notice that this tells me how to find a sub n plus 2
once I happen to know what a sub n is.
Now, lookit, before I can get to two
away from a particular subscript,
I must be at least at the second subscript.
In other words, it seems that this recipe will tell me
how to find a sub 2 in terms of a sub 0,
how to find a sub 4 in terms of a sub 2,
how to find a sub 3 in terms of a sub 1,
but it gives me no hold on what a sub 0 and a sub 1 must be.
So I say, OK, what I will do is I
will use this recipe, recognizing
that I am free to pick a0 and a1 completely arbitrarily.
Now look at what happens here.
If I pick a0 and a1 arbitrarily, what does the recipe tell me?
If I pick n to be 0, the recipe says what?
a sub n plus 2, a sub 2, is minus a0 over n
plus 2 times n plus 1.
That's just 2 times 1.
a sub 2 is just minus a0 over 2 factorial.
In a similar way, the recipe tells me
that a sub 4 is minus a sub 2 over 4 times 3.
You see, in this case, I take n to be 2.
So n plus 2 is 4.
And, if I now say a sub n plus 2 is minus a sub n over n
plus 2 times n plus 1, this just says that a4 is minus a sub 2
over 4 times 3.
I know what a sub 2 is from here.
So I can now express a sub 2 in terms of a sub 0.
In fact, it now turns out that a sub 4
is a sub 0 over 4 factorially--
4 factorial.
Similarly, a sub 6 is minus a sub 4 over 6 times 5.
a sub 4 is a0 over 4 factorial.
So, substituting that value of a4 in here,
I get that a6 is minus a0 over 6 factorial.
And, without beating this thing to death,
notice that I can find every even subscript of a, a2, a4,
a6, a8, et cetera, in terms of a0 times something.
In fact, I think you can see what's
starting to develop over here.
In a similar way, picking n to be 1 so that n plus 2 is 3,
the recipe a sub n plus 2 equals minus a sub n over n
plus 2 times n plus 1 says that a3 is minus a1 over 3 times 2.
That's minus a1 over 3 factorial.
Similarly, a5 is minus a3 over 5 times 4.
Noticing that a3 is minus a1 over 3 factorial,
I see that a5 is a1 over 5 factorial.
And, without going on further here,
notice that a3, a5, a7, a9, a11, et cetera,
will all be expressible in terms of a1.
And, in fact, I think you see what's happening here,
that the terms will alternate in sign,
and the coefficients will be things like what?
1, 1 over 3 factorial, 1 over 5 factorial, 1 over 7 factorial,
et cetera.
But I'll leave those details to you
because I think that's the easy part to meditate upon.
But now what we're saying is I now
know what all these coefficients look
like in terms of a0 and a1.
Remember what I'm looking for.
I'm looking for a power series solution
that's y equals a 0 plus a1 x plus a2 x squared, et cetera.
We're assuming that this is a convergent power series.
That means I can group the terms any way that I want.
In particular, since all of the even subscripts
seem to be expressible all in terms of a0,
and all the odd subscripts seem to be expressible in terms
of a1, let me group the terms with even powers
together and the terms with odd powers together.
And now, remembering what a2 and a4 are-- remember,
a2 was minus x squared--
was minus a0 over 2 factorial.
a4 was a0 over 4 factorial, et cetera.
I now factor out a0 from here.
And, substituting in the values of a2, a4, a6, et
cetera, that we found from before,
this tells me that the coefficient that multiplies a0
is 1 minus x squared over 2 factorial plus x
to the fourth over 4 factorial, et cetera.
And, similarly, since a3 is minus a1 over 3 factorial,
and a5 was a1 over 5 factorial, et cetera,
I can now express all of these in terms of a1.
I factor out a1.
And what's left this x minus x cubed over 3 factorial plus x
to the fifth over 5 factorial, et cetera.
And now it appears that my solution is this plus this.
Keep in mind that a0 and a1 are arbitrary constants.
By the ratio test, even if I didn't
know what these series represented,
I can show that these series converge absolutely
for all finite values of x, but that's not too crucial here.
The important point is that, even though it may be implicit,
this does name some function of x, whatever it happens to be.
This names some function of x, whatever it happens to be.
I can find particular solutions by choosing a0 and a1
to be certain fixed constants that I want them to be.
These two power series are not constant multiples
of one another, as you can see by a glance.
Therefore, these two solutions are linearly independent.
And, consequently, I now have a general solution here,
even though this may be awkward for me.
Well, the reason that I picked constant coefficients and y
double prime plus y equals 0 was that I
wanted our first experience to be pleasant,
at least in the sense that we would get a feeling for what
the answer meant.
Keep in mind that, if I had never
heard of sine x and cosine x, this solution here
would make sense.
But, if I happen to recall that the power series
expansion for cosine x is precisely this,
it just happens, you see, as an afterthought
that this is cosine x.
This also happens to be sine x.
Notice now what this tells me is that the solution to I've found
is a0 cosine x plus a1 sine x.
And that just happens to agree with the result
that I already knew.
Now, rather than have you feel that I've
wasted your time by giving you something
that you've already known, let me go one step further
and give you the same kind of a problem.
Only now, I'm going to give you one
that you couldn't solve before and, in fact,
I don't think you could solve now
if you don't use power series.
And the example I have in mind is almost the same as the one
we just did.
Instead of y double prime plus y equals 0,
I pick y double prime plus xy equals 0.
And now you see I'm back--
not back, but now I have arrived at the case
of non-constant coefficients.
You see the coefficient of y is x, and that's not a constant.
So you see now I'm at a bona fide problem where I couldn't
find y sub h in the trivial way that I
could if the coefficients all happened to be constant.
Now the way I tackle this is precisely
the same way as before.
What I do is I again try for a solution in the form y
equals summation, n goes from 0 to infinity, a sub n
x to the n.
And what I must do now is to determine
what the a sub n's are.
And, again, as before, I find y prime the same mechanical way
as before.
I differentiate term by term, meaning, generically, I
differentiate this to get this.
And, similarly, to get y double prime,
I differentiate this term to get this one, OK?
And now what I do, in terms of my so-called trial solution--
see, remember, these are trial solutions--
I'm going to try to feed these back into here
and see if that will tell me what the coefficients a0, a1,
a2, a3, et cetera have to look like.
Now, if I do that, you see, I get what?
y double prime is just this term here.
And now I want x times y.
That means I have to take this y and multiply it by x.
Multiplying by x on the outside, since we're
assuming that this is a convergent series,
that means I can multiply term by term.
Therefore, I can bring the x inside,
and that makes this x to the n plus 1, rather than x to the n.
And so I wind up with the fact that I now
want to find the constants a0, a1, et cetera,
from this particular relationship.
You see what's happened here?
This is just y double prime.
This is y multiplied by x.
That's why that n plus 1 is in here.
And that must be identically zero.
Now what I'm going to do is the same trick as before.
What I'm going to do is I want to add these term by term.
And I notice that their exponents are different.
In fact, this exponent is three larger
for each value of n than this one,
right? n plus 1 minus the quantity n minus 2 is 3.
So what I want to do is I want to lower this subscript--
this exponent by 3.
So I'm going to inside the integral--
I'm sorry, inside the summation sign,
I am going to replace n by n minus 3.
And, in line with our previous note, to compensate for this,
instead of my summation going from 0 to infinity here,
it will now go from 3 to infinity.
In other words, I am now simply going
to replace this by the equivalent sum,
n equals 3 to infinity, a sub n minus 3 x to the n minus 2,
all right?
And you see, if I do that, what I will end up with is simply
this expression here.
And now a new wrinkle comes up that didn't happen before,
but I just wanted to make sure a few funny things happened here
so I can tell you directly what's going on.
And then we can drill with the remaining fine
points in the exercises.
But the key point here is, lookit,
I've got the exponents lined up now.
But look at the indices of summation.
One begins at 2, and the other begins at 3.
And it seems that they're out of phase.
All I want you to observe is that, to make this start at 3,
I could split off separately the term that corresponds to n
equals 2.
In other words, notice that, when n
equals 2, what I have is what?
2 times 1 times a sub 2 times x to the 0.
In other words, all I have is 2 a sub 2, 2 a sub 2
when n is equal to 2.
So what I can do is I'll split that n
equals 2 term off separately.
That's 2 a sub 2.
Then what's left is the sum as n goes from 3 to infinity.
You see, splitting off these terms isn't hard at all.
All you do is, if these don't match up, take the smaller
index, and split off the number of terms necessary so
that the resulting sum will begin at a higher
index matching this one.
Since they only differed by 1 here,
I only have to split off the n equals 2 term.
Now what I have is what?
I have 2 a2 plus this summation, now going from 3 to infinity,
plus this summation, which also goes from 3 to infinity.
And now, since both the exponents and the indices
match up, now I can add term by term for these convergent power
series.
I bring these inside one integral sign.
See, the sigma of a sum is the sum of the sigmas.
I add these up term by term.
So that leaves me what?
I have my 2 a2 still outside here.
And then, inside the sigma notation, from 3 to infinity,
I have what?
n times n minus 1 times a sub n plus a sub n minus 3,
that whole quantity, times x to the n minus 2.
And that must be identically zero.
Now the only way that this can be identically zero
is if each coefficient is 0.
By the way, notice that, since the first term that
appears here corresponds to n equals 3, when n equals 3,
the exponent here is 1.
So notice that this is my only constant term.
All the terms in here begin with at least a factor of x in them.
Since my power series, to be identically zero,
must have all of its coefficients identically zero,
it means that not only must each of these be 0, but 2 a2, which
is my constant term on the left-hand side, that must also
be 0.
So, in other words, to summarize this up,
what we're saying so far is that, for this to be
identically zero, a2 must be 0.
And, once n is at least as big as 3,
this condition here must be obeyed.
What condition?
That the expression in brackets must be 0 for each n
once n is at least as big as 3.
In other words, then, in summary, a2 is 0.
And, for n at least as big as 3, a sub n is minus a sub n
minus 3 over n times n minus 1.
And keep in mind the way I got that quite trivially was
I set this equal to 0.
I transposed the a sub n minus 3,
divided through by n times n minus 1,
and solved for a sub n.
What this tells me now is that, once I
know a particular a, look at what this tells me.
This says, to find a sub n, once I know what the a was three
before this one, I'm home free.
In other words, to find a sub 4, all I have to know is a sub 1.
To find a sub 5, all I have to know is a sub 2.
You see?
To find a sub 6, all I have to know is a sub 3, et cetera.
So the idea is I can now pick a0 and a1
at random, just as before.
And now, using this recipe with n equal to 3,
I have that a sub 3 is minus a sub 0 over 3 times 2.
a sub 6 is minus a sub 3 over 6 times 5.
And, remembering what a3 is in terms of a0,
this tells me that a6 is a0 over 5 times 6 times 3 times 2.
And this is not a misprint that the four is missing here.
This is not a factorial.
You see, I have a way of generating
what these coefficients look like,
but now I've picked a problem where
I may not remember or recognize what power series I'm getting.
That's irrelevant again.
All I'm saying is I can determine
a3, a6, a9, et cetera, just knowing what a0 is.
And, just knowing what a1 is, I can determine a4.
Just by using this recipe with n equal to 4,
a4 is minus a1 over 4 times 3.
a7 is minus a4 over 7 times 6, using that same recipe.
Putting in the value of a4 from here into here,
I have that a7 is a1 over 7 times 6 times 4 times 3.
And this keeps on going.
I can now find a10 in terms of a7.
That will give me a10 in terms of a1,
et cetera, et cetera, et cetera.
Finally, to find a5 or a8, notice
that, using the same recipe, a5 is minus a2 over 5 times 4.
But, since a2 is 0, a5 will be 0.
And, similarly, because a5 is 0, a8
will be 0 and so will a11 and a14, et cetera.
Now the idea is the series that I was looking for,
being a convergent series, can be rearranged.
You see, the idea in this problem
was that I would like to group the exponents according
to whether they are divisible by 3, leave a remainder of 1
when I divide by 3, or leave a remainder of 2
when I divide by 3.
So I group the terms this way-- see,
a0 plus a sub 3 x cubed plus a sub 6 x to the sixth,
et cetera, plus a1 x plus a4 x to the fourth, et cetera,
plus a2 x squared plus a5 x fifth, et cetera--
the idea being that I know how to express
a3, a6, a9 in terms of a0.
I know how to express a4, a7, a10, et cetera, in terms of a1.
And I know how to express a5, a8, et cetera, in terms of a2.
In fact, in this simple case, which turned out nicely,
they all turned out to be 0, these coefficients.
So what's left is what?
Replacing a4, a7, et cetera, by what
they look like in terms of a1 and a3, a6, et
cetera, by what they look like in terms of a0,
I wind up with y equals a0 times 1
minus x cubed over 3 times 2 plus x to the sixth
over 6 times 5 times 3 times 2, et cetera,
plus a1 times this thing here.
Now the only difference between this problem
and the previous one was that, in the previous one,
I happened to recognize what well-known function had
this as a power series.
In this case, I don't know that, but this is no less
a bona fide solution than we had in the previous case.
This is a convergent power series.
We give it a name--
h of x, k of x.
If this were the solution to an important enough problem,
we would say, lookit, instead of calling it h of x and k of x,
let's give it a special name because it's going
to come up over and over again.
I'm not going to go into this.
If I do go into it all, it'll be very lightly in the exercises.
But things like that you may have heard,
in terms of name dropping, Bessel functions
and the like, Legendre polynomials,
these were all special power series solutions
to very special differential equations
where the equation itself was so important
an application that the power series that
represented the equation was given a special name.
So you see, in terms of power series solutions,
one extremely powerful use of power series
is that it's used to help us find
solutions of the homogeneous equation L of y equals 0.
And, once we have the general solution of the equation L of y
equals 0, we can then use variation of parameters
to find a solution of L of y equals f of x.
Then we add these two solutions together
to get the general solution.
And that ends our theory on linear differential equations,
except for one more topic called the Laplace transform
that I would like to introduce for you to see.
And so I will have one more lecture
in the guise of linear differential equations in order
that I can bring up the Laplace transform,
but that won't be until next time.
Until next time then, goodbye.
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