PROFESSOR: So we have this integral.
And with-- let me go here, actually.
With the contour gamma equal to C1,
this contour over here, and the constant c equal to 1.
So C1 and the constant c equal to 1.
This psi that we have defined, psi of u,
is in fact the Airy function of u.
A i of u.
I is not--
I think I tend to make that mistake.
I doesn't go like a subscript.
It's A i like the first two letters of the name.
So that's the function A i of u.
And now you could ask, well, what is the other solution?
Now, in fact, this diagram suggests to you
that there's other solutions because you
could take other contours and make other solutions.
In fact, yes, there are other ways.
For example, if you did a contour--
do I have a color?
Other color?
Yes.
If you did a contour like this, yellow and yellow,
this is not the same solution.
It is a solution because of the general argument
and because the endpoints are in these regions
where things vanish at infinity.
So the yellow thing is another solution
of the differential equation.
So the other Airy function is defined, actually,
with this other contour.
It's defined by taking the yellow contour like this.
This is going to be called C2.
A contour like that.
It just comes parallel to this one and then goes down.
And, actually, in order to have a nicely defined function,
one chooses for the function B i of u the following.
Minus i times the integral over the contour
C1 of the same integrant.
So I will not copy it.
Always the same integrant.
Plus 2 i terms the integral over the contour C2
of the same thing.
So the B i function is a little unusual
in that it has kind of a little bit of the A i function
because you also integrate over C1.
But you integrate as well over C2.
That guarantees that-- actually, this second Airy function
behaves similar to A i for negative u,
and while A i goes to 0 for positive u,
this one will diverge.
There are expressions for this function.
I'll give you an integral.
2 integrals that are famous are A i of u equals 1 over pi.
Integral from 0 to infinity.
d k cosine.
k cubed over 3 plus k u.
And for B i of u.
1 over pi-- this is a little longer.
Integral from 0 to infinity as well.
d k.
And you have an exponential of minus k cubed over 3 plus k u,
plus the sine of k cubed over 3, plus k u.
And that's it.
It's kind of funny.
One is the cosine and one is the sine.
And it has this extra different factors.
So these are your two functions.
And of the relevance to our WKB problem
is that they're necessary to connect
the solutions, as we will see.
But we need a little more about these functions.
We need to know there are asymptotic behaviors.
Now there are functions, like the exponential function, that
has a Taylor series, e to the z.
1 plus z plus z squared over 2.
And it's valid, whatever these angles--
the argument of z is.
That's always the same asymptotic expansion,
or the same Taylor series.
For this functions, like the Airy function,
for some arguments of u, there's one form
to the asymptotic expansion.
And for some other arguments, there's another form.
That's sometimes called the Stokes phenomenon.
And for example, the expansion for positive u
is going to be a decaying exponential here.
But for negative u, it will be oscillatory.
So it's not like a simple function,
like the exponential function has a nice, simple expansion
everywhere.
It just varies.
So one needs to calculate this asymptotic expansions,
and I'll make a small comment about it
of how you go about it.
The nice thing about these formulas
is that they allow you to understand things intuitively
and derive things yourself.
Here you see the two Airy functions.
They make sense.
The other thing that is possible to do with this contour
representations is to find the asymptotic expansions
of these functions.
And they don't require major mathematical work.
It's kind of nice.
So let's think a little about one example.
If you have the Airy function A i of u that is
of the form integral 1 over--
well, it begins v k over 2 pi.
The integral over contour C1.
Maybe I should have done them curly.
Curly C1 would have been clearer.
e to the i.
k cubed over 3 plus k u.
That is your integral.
And this is the phase of integration.
Phase.
Now, in order to find the asymptotic expansion
for this thing, we'll use a stationary phase condition.
So the integral is dominated by those places
where the phase is stationary.
So the 5 prime of k is k squared plus u.
And we want this to be equal to 0.
So take, for example, u positive.
Suppose you want to find the behavior of the Airy function
on the right of the axis.
Well, this says k squared is equal to u.
So k squared is equal to minus u.
And that's-- the right hand side is negative because u is
positive.
And therefore, k-- the points where this is solved
are k equal plus minus i square root of u.
So where are those points in the k plane?
They're here and there.
And those are the places where you get stationary phase.
So what can you do?
You're supposed to do the integral over this red line.
C1.
Well, as we argued, this can be lifted
and we can make the integral pass through here.
You can now do this integral over here.
It's the same integral that you had before.
In this line, we can say that k is
equal to i square root of u plus some extra k tilde.
We say here is i square root of u for u positive.
There is this other place, but that--
we cannot bring the contour down here because in this region,
the end points still contribute.
So we cannot shift the contour down, but we can shift it up.
So we have to do this integral.
So what happens?
You can evaluate that phase phi of k under these conditions.
It is a stationary phase point--
this one-- so the answer is going
to have a part independent of k tilde, a linear part,
with respect to k tilde that will
vanish because at this point the phase is stationary.
And then a quadratic part with respect to k tilde.
So the phase, phi of k, which is k cubed over 3 plus k u.
When you substitute this k here, it's
going to give you an answer.
And the answer is going to be 2 over 3 i u to the 3/2,
plus i square root of u k squared tilde,
plus k tilde cubed over 3.
That's what you get from the phase.
You say, well, that's pretty nice
because now our integral psi has become the integral d k
tilde over 2 pi.
So you pass from k to k tilde variable.
e to the minus 2/3.
u to the 3/2.
That is because you have i times phi of k,
so you must multiply by i here.
Then minus square root of u k squared.
Then plus k cubed.
OK.
Tildes of course.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Um.
Yes.
i k.
This should be like that.
Yes.
OK.
So now what happens?
If u is large enough, this quantity over here
is going to mean that this integral is
highly suppressed over k tilde.
It's a Gaussian with a very narrow peak.
And therefore, we can ignore this term.
This term is a constant.
So what do we get from here?
We get 1 over 2 pi from the beginning.
This exponential.
e to the minus 2 over 3.
u to the 3/2.
3/2.
And then from this Gaussian, we get square root of pi
over square root of square root of u.
So it's u to the 1/4.
So I think I got it all correct.
And therefore, the function A i of u goes like--
or it's proportional to this decaying exponential.
Very fast decaying exponential.
1 over 2.
1 over square root of pi.
1 over u to the 1/4.
e to the minus 2/3.
u to the 3/2.
And this is when u is greater than 0 and, in fact,
u much greater than 1, positive and large.
So this shows you the power of this method.
This is a very powerful result. And this
is what we're going to need, pretty much, now
in order to do our asymptotic solutions and the matching
of WKB.
There is an extra term, or an extra case, you can consider.
What happens to this integral when u is negative?
When u is negative, there's two real roots,
and the stationary points occur on the real line.
The integral is, therefore, a little more straightforward.
You don't have to even move it.
And we have another expansion, therefore,
which is A i of u is actually equal to 1
over the square root of pi.
1 over u, length of u to the 1/4.
Cosine of 2/3.
Length of u to the 3/2 minus pi over 4.
This is for u less than 0.
So the Airy function becomes an oscillatory function
on the left.
So how does this Airy function really look like?
Well, we have the true Airy function behaves like this.
It's a decaying exponential for u positive.
And then a decaying oscillatory function for u negative.
For u negative, it decays eventually.
The frequency becomes faster and faster.
And that's your Airy function A i of u.
Here is u.
Your asymptotic expansions, the ones
that we found up there for u greater than 1,
match this very nicely.
But eventually they blow up, so they're a little wrong.
And they actually match here quite OK.
But here they go and also blow up because of this factor.
So they both blow up, but they're both quite wrong
in this area, which you would expect them to be wrong.
These are the regions where this two asymptotic expansions
make no sense.
They were calculated on the hypothesis
that u is much bigger than 1, or much less than 1.
Minus 1.
And therefore, you get everything
under control except this area.
So now let's do the real work of WKB
That was our goal from the beginning,
so that's what we'll try to do now.