WEBVTT
00:00:00.940 --> 00:00:06.580
PROFESSOR: So we
have this integral.
00:00:06.580 --> 00:00:11.490
And with-- let me
go here, actually.
00:00:11.490 --> 00:00:23.220
With the counter gamma equal
to C1, this counter over here,
00:00:23.220 --> 00:00:27.470
and the constant c equal to 1.
00:00:30.930 --> 00:00:35.420
So C1 and the
constant c equal to 1.
00:00:35.420 --> 00:00:41.180
This psi that we have
defined, psi of u,
00:00:41.180 --> 00:00:48.450
is in fact the
airy function of u.
00:00:48.450 --> 00:00:50.180
A i of u.
00:00:50.180 --> 00:00:51.650
I is not--
00:00:51.650 --> 00:00:53.540
I think I tend to
make that mistake.
00:00:53.540 --> 00:00:56.410
I doesn't go like a subscript.
00:00:56.410 --> 00:01:01.710
It's A i like the first
two letters of the name.
00:01:01.710 --> 00:01:05.540
So that's the function A i of u.
00:01:05.540 --> 00:01:09.380
And now you could ask, well,
what is the other solution?
00:01:09.380 --> 00:01:13.310
Now, in fact, this
diagram suggests to you
00:01:13.310 --> 00:01:15.320
that there's other
solutions because you
00:01:15.320 --> 00:01:20.910
could take other counters
and make other solutions.
00:01:20.910 --> 00:01:23.220
In fact, yes, there
are other ways.
00:01:23.220 --> 00:01:26.705
For example, if
you did a counter--
00:01:30.170 --> 00:01:31.520
do I have a color?
00:01:31.520 --> 00:01:32.070
Other color?
00:01:32.070 --> 00:01:32.570
Yes.
00:01:32.570 --> 00:01:39.840
If you did a counter like
this, yellow and yellow,
00:01:39.840 --> 00:01:43.040
this is not the same solution.
00:01:43.040 --> 00:01:47.930
It is a solution because
of the general argument
00:01:47.930 --> 00:01:51.950
and because the endpoints
are in these regions
00:01:51.950 --> 00:01:56.160
where things vanish at infinity.
00:01:56.160 --> 00:01:58.490
So the yellow thing
is another solution
00:01:58.490 --> 00:02:00.150
of the differential equation.
00:02:00.150 --> 00:02:10.430
So the other airy function
is defined, actually,
00:02:10.430 --> 00:02:12.260
with this other counter.
00:02:15.800 --> 00:02:26.480
It's defined by taking the
yellow counter like this.
00:02:26.480 --> 00:02:30.130
This is going to be called C2.
00:02:30.130 --> 00:02:32.030
A counter like that.
00:02:32.030 --> 00:02:36.305
It just comes parallel to
this one and then goes down.
00:02:39.610 --> 00:02:50.550
And, actually, in order to
have a nicely defined function,
00:02:50.550 --> 00:03:07.060
one chooses for the function
B i of u the following.
00:03:07.060 --> 00:03:12.610
Minus i times the integral
over the counter C1
00:03:12.610 --> 00:03:14.530
of the same integrant.
00:03:14.530 --> 00:03:17.080
So I will not copy it.
00:03:17.080 --> 00:03:18.850
Always the same integrant.
00:03:24.320 --> 00:03:31.790
Plus 2 i terms the integral
over the counter C2
00:03:31.790 --> 00:03:33.240
of the same thing.
00:03:33.240 --> 00:03:37.010
So the B i function
is a little unusual
00:03:37.010 --> 00:03:41.180
in that it has kind of a
little bit of the A i function
00:03:41.180 --> 00:03:43.995
because you also
integrate over C1.
00:03:47.770 --> 00:03:50.470
But you integrate
as well over C2.
00:03:55.450 --> 00:04:02.410
That guarantees that-- actually,
this second airy function
00:04:02.410 --> 00:04:08.050
behaves similar to
A i for negative u,
00:04:08.050 --> 00:04:13.060
and while A i goes
to 0 for positive u,
00:04:13.060 --> 00:04:16.209
this one will diverge.
00:04:16.209 --> 00:04:21.335
There are expressions
for this function.
00:04:25.720 --> 00:04:27.310
I'll give you an integral.
00:04:29.980 --> 00:04:41.410
2 integrals that are famous
are A i of u equals 1 over pi.
00:04:41.410 --> 00:04:44.800
Integral from 0 to infinity.
00:04:44.800 --> 00:04:48.640
d k cosine.
00:04:48.640 --> 00:04:54.350
k cubed over 3 plus k u.
00:04:54.350 --> 00:04:59.260
And for B i of u.
00:04:59.260 --> 00:05:01.555
1 over pi-- this
is a little longer.
00:05:05.220 --> 00:05:08.200
Integral from 0 to
infinity as well.
00:05:08.200 --> 00:05:10.090
d k.
00:05:10.090 --> 00:05:17.500
And you have an exponential of
minus k cubed over 3 plus k u,
00:05:17.500 --> 00:05:29.716
plus the sine of k
cubed over 3, plus k u.
00:05:29.716 --> 00:05:31.720
And that's it.
00:05:31.720 --> 00:05:32.650
It's kind of funny.
00:05:32.650 --> 00:05:36.550
One is the cosine
and one is the sine.
00:05:36.550 --> 00:05:40.430
And it has this extra
different factors.
00:05:40.430 --> 00:05:44.590
So these are your two functions.
00:05:44.590 --> 00:05:48.520
And of the relevance
to our w k b
00:05:48.520 --> 00:05:51.640
problem is that they're
necessary to connect
00:05:51.640 --> 00:05:54.760
the solutions, as we will see.
00:05:54.760 --> 00:05:58.840
But we need a little more
about these functions.
00:05:58.840 --> 00:06:02.740
We need to know there
are asymptotic behaviors.
00:06:02.740 --> 00:06:06.130
Now there are functions, like
the exponential function, that
00:06:06.130 --> 00:06:09.970
has a Taylor series, e to the z.
00:06:09.970 --> 00:06:13.740
1 plus z plus z squared over 2.
00:06:13.740 --> 00:06:18.040
And it's valid,
whatever these angles--
00:06:18.040 --> 00:06:20.380
the argument of z is.
00:06:20.380 --> 00:06:23.110
That's always the same
asymptotic expansion,
00:06:23.110 --> 00:06:25.480
or the same Taylor series.
00:06:25.480 --> 00:06:28.900
For this functions,
like the airy function,
00:06:28.900 --> 00:06:37.600
for some arguments of
u, there's one form
00:06:37.600 --> 00:06:39.460
to the asymptotic expansion.
00:06:39.460 --> 00:06:43.720
And for some other arguments,
there's another form.
00:06:43.720 --> 00:06:47.330
That's sometimes called
the Stokes phenomenon.
00:06:47.330 --> 00:06:52.540
And for example, the
expansion for positive u
00:06:52.540 --> 00:06:55.420
is going to be a decaying
exponential here.
00:06:55.420 --> 00:06:59.050
But for negative u, it
will be oscillatory.
00:06:59.050 --> 00:07:01.795
So it's not like
a simple function,
00:07:01.795 --> 00:07:05.500
like the exponential function
has a nice, simple expansion
00:07:05.500 --> 00:07:06.340
everywhere.
00:07:06.340 --> 00:07:09.460
It just varies.
00:07:09.460 --> 00:07:17.210
So one needs to calculate
this asymptotic expansions,
00:07:17.210 --> 00:07:20.180
and I'll make a small
comment about it
00:07:20.180 --> 00:07:22.910
of how you go about it.
00:07:22.910 --> 00:07:26.350
The nice thing
about these formulas
00:07:26.350 --> 00:07:29.920
is that they allow you to
understand things intuitively
00:07:29.920 --> 00:07:34.000
and derive things yourself.
00:07:34.000 --> 00:07:37.210
Here you see the
two airy functions.
00:07:37.210 --> 00:07:38.480
They make sense.
00:07:38.480 --> 00:07:40.480
The other thing that
is possible to do
00:07:40.480 --> 00:07:43.030
with this counter
representations
00:07:43.030 --> 00:07:48.770
is to find the asymptotic
expansions of these functions.
00:07:48.770 --> 00:07:55.090
And they don't require
major mathematical work.
00:07:55.090 --> 00:07:57.220
It's kind of nice.
00:07:57.220 --> 00:08:01.340
So let's think a little
about one example.
00:08:01.340 --> 00:08:07.240
If you have the airy
function A i of u
00:08:07.240 --> 00:08:10.660
that is of the form
integral 1 over--
00:08:10.660 --> 00:08:14.660
well, it begins v k over 2 pi.
00:08:14.660 --> 00:08:17.770
The integral over counter C1.
00:08:17.770 --> 00:08:20.290
Maybe I should have
done them curly.
00:08:20.290 --> 00:08:23.860
Curly C1 would
have been clearer.
00:08:23.860 --> 00:08:25.070
e to the i.
00:08:25.070 --> 00:08:28.625
k cubed over 3 plus k u.
00:08:31.360 --> 00:08:32.320
That is your integral.
00:08:35.020 --> 00:08:37.539
And this is the
phase of integration.
00:08:39.935 --> 00:08:40.435
Phase.
00:08:43.620 --> 00:08:48.960
Now, in order to find
the asymptotic expansion
00:08:48.960 --> 00:08:55.200
for this thing, we'll use a
stationary phase condition.
00:08:55.200 --> 00:08:58.950
So the integral is
dominated by those places
00:08:58.950 --> 00:09:00.790
where the phase is stationary.
00:09:00.790 --> 00:09:09.246
So the 5 prime of k
is k squared plus u.
00:09:09.246 --> 00:09:11.550
And we want this
to be equal to 0.
00:09:14.260 --> 00:09:20.400
So take, for
example, u positive.
00:09:20.400 --> 00:09:24.080
Suppose you want to find the
behavior of the airy function
00:09:24.080 --> 00:09:27.590
on the right of the axis.
00:09:27.590 --> 00:09:33.680
Well, this says k
squared is equal to u.
00:09:33.680 --> 00:09:38.486
So k squared is
equal to minus u.
00:09:38.486 --> 00:09:42.620
And that's-- the right hand
side is negative because u is
00:09:42.620 --> 00:09:43.970
positive.
00:09:43.970 --> 00:09:48.655
And therefore, k-- the
points where this is solved
00:09:48.655 --> 00:09:52.970
are k equal plus minus
i square root of u.
00:09:57.220 --> 00:09:59.770
So where are those
points in the k plane?
00:09:59.770 --> 00:10:06.370
They're here and there.
00:10:06.370 --> 00:10:11.500
And those are the places where
you get stationary phase.
00:10:14.250 --> 00:10:15.600
So what can you do?
00:10:15.600 --> 00:10:21.110
You're supposed to do the
integral over this red line.
00:10:21.110 --> 00:10:23.130
C1.
00:10:23.130 --> 00:10:26.100
Well, as we argued,
this can be lifted
00:10:26.100 --> 00:10:29.060
and we can make the
integral pass through here.
00:10:38.630 --> 00:10:40.930
You can now do this
integral over here.
00:10:40.930 --> 00:10:43.195
It's the same integral
that you had before.
00:10:47.420 --> 00:10:53.690
In this line, we
can say that k is
00:10:53.690 --> 00:11:01.090
equal to i square root of
u plus some extra k tilde.
00:11:08.420 --> 00:11:16.790
We say here is i square
root of u for u positive.
00:11:16.790 --> 00:11:20.570
There is this other
place, but that--
00:11:20.570 --> 00:11:25.430
we cannot bring the counter down
here because in this region,
00:11:25.430 --> 00:11:28.370
the end points still contribute.
00:11:28.370 --> 00:11:33.950
So we cannot shift the counter
down, but we can shift it up.
00:11:33.950 --> 00:11:37.070
So we have to do this integral.
00:11:37.070 --> 00:11:38.860
So what happens?
00:11:42.440 --> 00:11:51.110
You can evaluate that phase
5k under these conditions.
00:11:51.110 --> 00:11:54.260
It is a stationary phase point--
00:11:54.260 --> 00:11:56.890
this one-- so the
answer is going
00:11:56.890 --> 00:12:03.890
to have a part independent
of k tilde, a linear part,
00:12:03.890 --> 00:12:05.930
with respect to
k tilde that will
00:12:05.930 --> 00:12:10.880
vanish because at this point
the phase is stationary.
00:12:10.880 --> 00:12:14.970
And then a quadratic part
with respect to k tilde.
00:12:14.970 --> 00:12:25.180
So the phase, 5k, which is
k cubed over 3 plus k u.
00:12:25.180 --> 00:12:28.580
When you substitute
this k here, it's
00:12:28.580 --> 00:12:31.010
going to give you an answer.
00:12:31.010 --> 00:12:39.560
And the answer is going to
be 2 over 3 i u to the 3/2,
00:12:39.560 --> 00:12:45.590
plus i square root
of u k squared tilde,
00:12:45.590 --> 00:12:50.780
plus k tilde cubed over 3.
00:12:50.780 --> 00:12:56.390
That's what you
get from the phase.
00:12:56.390 --> 00:13:00.770
You say, well,
that's pretty nice
00:13:00.770 --> 00:13:08.260
because now our integral psi
has become the integral d k
00:13:08.260 --> 00:13:11.570
tilde over 2 pi.
00:13:11.570 --> 00:13:17.540
So you pass from k
to k tilde variable.
00:13:17.540 --> 00:13:19.240
e to the minus 2/3.
00:13:23.240 --> 00:13:24.920
u to the 3/2.
00:13:29.660 --> 00:13:33.680
That is because you
have i times 5k,
00:13:33.680 --> 00:13:37.760
so you must multiply by i here.
00:13:37.760 --> 00:13:42.650
Then minus square
root of u k squared.
00:13:42.650 --> 00:13:44.210
Then plus k cubed.
00:13:49.371 --> 00:13:49.870
OK.
00:13:49.870 --> 00:13:51.005
Tildes of course.
00:13:54.218 --> 00:13:56.658
AUDIENCE: [INAUDIBLE]
00:13:56.658 --> 00:13:59.590
PROFESSOR: Um.
00:13:59.590 --> 00:14:00.930
Yes.
00:14:00.930 --> 00:14:02.860
i k.
00:14:02.860 --> 00:14:04.340
This should be like that.
00:14:04.340 --> 00:14:04.840
Yes.
00:14:08.530 --> 00:14:09.030
OK.
00:14:09.030 --> 00:14:11.910
So now what happens?
00:14:11.910 --> 00:14:18.510
If u is large enough,
this quantity over here
00:14:18.510 --> 00:14:22.440
is going to mean
that this integral is
00:14:22.440 --> 00:14:26.100
highly suppressed over k tilde.
00:14:26.100 --> 00:14:30.240
It's a Gaussian with
a very narrow peak.
00:14:30.240 --> 00:14:34.120
And therefore, we
can ignore this term.
00:14:34.120 --> 00:14:36.610
This term is a constant.
00:14:36.610 --> 00:14:39.150
So what do we get from here?
00:14:39.150 --> 00:14:44.130
We get 1 over 2 pi
from the beginning.
00:14:44.130 --> 00:14:46.680
This exponential.
00:14:46.680 --> 00:14:50.070
e to the minus 2 over 3.
00:14:50.070 --> 00:14:51.210
u to the 3/2.
00:14:54.030 --> 00:14:56.760
3/2.
00:14:56.760 --> 00:15:03.360
And then from this Gaussian,
we get square root of pi
00:15:03.360 --> 00:15:07.320
over square root of
square root of u.
00:15:07.320 --> 00:15:08.720
So it's u to the 1/4.
00:15:11.460 --> 00:15:14.970
So I think I got it all correct.
00:15:14.970 --> 00:15:24.930
And therefore, the function
A i of u goes like--
00:15:24.930 --> 00:15:30.630
or it's proportional to
this decaying exponential.
00:15:30.630 --> 00:15:32.640
Very fast decaying exponential.
00:15:32.640 --> 00:15:34.770
1 over 2.
00:15:34.770 --> 00:15:38.100
1 over square root of pi.
00:15:38.100 --> 00:15:41.160
1 over u to the 1/4.
00:15:41.160 --> 00:15:43.740
e to the minus 2/3.
00:15:43.740 --> 00:15:46.450
u to the 3/2.
00:15:46.450 --> 00:15:50.790
And this is when u is
greater than 0 and, in fact,
00:15:50.790 --> 00:15:53.790
u much greater than
1, positive and large.
00:15:56.970 --> 00:16:04.270
So this shows you the
power of this method.
00:16:04.270 --> 00:16:07.530
This is a very powerful
result. And this
00:16:07.530 --> 00:16:10.920
is what we're going to
need, pretty much, now
00:16:10.920 --> 00:16:15.270
in order to do our
asymptotic solutions
00:16:15.270 --> 00:16:17.880
and the matching of w k b.
00:16:17.880 --> 00:16:24.630
There is an extra term, or an
extra case, you can consider.
00:16:24.630 --> 00:16:32.180
What happens to this
integral when u is negative?
00:16:32.180 --> 00:16:37.170
When u is negative,
there's two real roots,
00:16:37.170 --> 00:16:43.800
and the stationary points
occur on the real line.
00:16:43.800 --> 00:16:46.560
The integral is, therefore, a
little more straightforward.
00:16:46.560 --> 00:16:49.470
You don't have to even move it.
00:16:49.470 --> 00:16:53.940
And we have another
expansion, therefore,
00:16:53.940 --> 00:16:59.790
which is A i of u is
actually equal to 1
00:16:59.790 --> 00:17:01.650
over the square root of pi.
00:17:01.650 --> 00:17:07.470
1 over u, length
of u to the 1/4.
00:17:07.470 --> 00:17:10.140
Cosine of 2/3.
00:17:10.140 --> 00:17:16.950
Length of u to the
3/2 minus pi over 4.
00:17:16.950 --> 00:17:20.550
This is for u less than 0.
00:17:20.550 --> 00:17:24.300
So the airy function becomes
an oscillatory function
00:17:24.300 --> 00:17:26.109
on the left.
00:17:26.109 --> 00:17:30.525
So how does this airy
function really look like?
00:17:34.880 --> 00:17:45.320
Well, we have the true airy
function behaves like this.
00:17:45.320 --> 00:17:49.160
It's a decaying
exponential for u positive.
00:17:52.340 --> 00:17:58.880
And then a decaying oscillatory
function for u negative.
00:17:58.880 --> 00:18:02.120
For u negative, it
decays eventually.
00:18:02.120 --> 00:18:05.720
The frequency becomes
faster and faster.
00:18:05.720 --> 00:18:14.030
And that's your airy
function A i of u.
00:18:17.750 --> 00:18:18.540
Here is u.
00:18:21.460 --> 00:18:25.650
Your asymptotic
expansions, the ones
00:18:25.650 --> 00:18:30.180
that we found up there
for u greater than 1,
00:18:30.180 --> 00:18:31.620
match this very nicely.
00:18:34.400 --> 00:18:41.090
But eventually they blow up,
so they're a little wrong.
00:18:41.090 --> 00:18:46.810
And they actually
match here quite OK.
00:18:46.810 --> 00:18:54.310
But here they go and also blow
up because of this factor.
00:18:54.310 --> 00:18:57.670
So they both blow up, but
they're both quite wrong
00:18:57.670 --> 00:19:01.780
in this area, which you would
expect them to be wrong.
00:19:01.780 --> 00:19:08.770
These are the regions where
this two asymptotic expansions
00:19:08.770 --> 00:19:09.640
make no sense.
00:19:09.640 --> 00:19:12.130
They were calculated
on the hypothesis
00:19:12.130 --> 00:19:17.170
that u is much bigger than
1, or much less than 1.
00:19:17.170 --> 00:19:18.740
Minus 1.
00:19:18.740 --> 00:19:21.610
And therefore,
you get everything
00:19:21.610 --> 00:19:25.300
under control except this area.
00:19:25.300 --> 00:19:30.940
So now let's do the
real work of w k b.
00:19:30.940 --> 00:19:33.140
That was our goal
from the beginning,
00:19:33.140 --> 00:19:38.430
so that's what
we'll try to do now.