1
00:00:00,940 --> 00:00:06,580
PROFESSOR: So we
have this integral.

2
00:00:06,580 --> 00:00:11,490
And with-- let me
go here, actually.

3
00:00:11,490 --> 00:00:23,220
With the counter gamma equal
to C1, this counter over here,

4
00:00:23,220 --> 00:00:27,470
and the constant c equal to 1.

5
00:00:30,930 --> 00:00:35,420
So C1 and the
constant c equal to 1.

6
00:00:35,420 --> 00:00:41,180
This psi that we have
defined, psi of u,

7
00:00:41,180 --> 00:00:48,450
is in fact the
airy function of u.

8
00:00:48,450 --> 00:00:50,180
A i of u.

9
00:00:50,180 --> 00:00:51,650
I is not--

10
00:00:51,650 --> 00:00:53,540
I think I tend to
make that mistake.

11
00:00:53,540 --> 00:00:56,410
I doesn't go like a subscript.

12
00:00:56,410 --> 00:01:01,710
It's A i like the first
two letters of the name.

13
00:01:01,710 --> 00:01:05,540
So that's the function A i of u.

14
00:01:05,540 --> 00:01:09,380
And now you could ask, well,
what is the other solution?

15
00:01:09,380 --> 00:01:13,310
Now, in fact, this
diagram suggests to you

16
00:01:13,310 --> 00:01:15,320
that there's other
solutions because you

17
00:01:15,320 --> 00:01:20,910
could take other counters
and make other solutions.

18
00:01:20,910 --> 00:01:23,220
In fact, yes, there
are other ways.

19
00:01:23,220 --> 00:01:26,705
For example, if
you did a counter--

20
00:01:30,170 --> 00:01:31,520
do I have a color?

21
00:01:31,520 --> 00:01:32,070
Other color?

22
00:01:32,070 --> 00:01:32,570
Yes.

23
00:01:32,570 --> 00:01:39,840
If you did a counter like
this, yellow and yellow,

24
00:01:39,840 --> 00:01:43,040
this is not the same solution.

25
00:01:43,040 --> 00:01:47,930
It is a solution because
of the general argument

26
00:01:47,930 --> 00:01:51,950
and because the endpoints
are in these regions

27
00:01:51,950 --> 00:01:56,160
where things vanish at infinity.

28
00:01:56,160 --> 00:01:58,490
So the yellow thing
is another solution

29
00:01:58,490 --> 00:02:00,150
of the differential equation.

30
00:02:00,150 --> 00:02:10,430
So the other airy function
is defined, actually,

31
00:02:10,430 --> 00:02:12,260
with this other counter.

32
00:02:15,800 --> 00:02:26,480
It's defined by taking the
yellow counter like this.

33
00:02:26,480 --> 00:02:30,130
This is going to be called C2.

34
00:02:30,130 --> 00:02:32,030
A counter like that.

35
00:02:32,030 --> 00:02:36,305
It just comes parallel to
this one and then goes down.

36
00:02:39,610 --> 00:02:50,550
And, actually, in order to
have a nicely defined function,

37
00:02:50,550 --> 00:03:07,060
one chooses for the function
B i of u the following.

38
00:03:07,060 --> 00:03:12,610
Minus i times the integral
over the counter C1

39
00:03:12,610 --> 00:03:14,530
of the same integrant.

40
00:03:14,530 --> 00:03:17,080
So I will not copy it.

41
00:03:17,080 --> 00:03:18,850
Always the same integrant.

42
00:03:24,320 --> 00:03:31,790
Plus 2 i terms the integral
over the counter C2

43
00:03:31,790 --> 00:03:33,240
of the same thing.

44
00:03:33,240 --> 00:03:37,010
So the B i function
is a little unusual

45
00:03:37,010 --> 00:03:41,180
in that it has kind of a
little bit of the A i function

46
00:03:41,180 --> 00:03:43,995
because you also
integrate over C1.

47
00:03:47,770 --> 00:03:50,470
But you integrate
as well over C2.

48
00:03:55,450 --> 00:04:02,410
That guarantees that-- actually,
this second airy function

49
00:04:02,410 --> 00:04:08,050
behaves similar to
A i for negative u,

50
00:04:08,050 --> 00:04:13,060
and while A i goes
to 0 for positive u,

51
00:04:13,060 --> 00:04:16,209
this one will diverge.

52
00:04:16,209 --> 00:04:21,335
There are expressions
for this function.

53
00:04:25,720 --> 00:04:27,310
I'll give you an integral.

54
00:04:29,980 --> 00:04:41,410
2 integrals that are famous
are A i of u equals 1 over pi.

55
00:04:41,410 --> 00:04:44,800
Integral from 0 to infinity.

56
00:04:44,800 --> 00:04:48,640
d k cosine.

57
00:04:48,640 --> 00:04:54,350
k cubed over 3 plus k u.

58
00:04:54,350 --> 00:04:59,260
And for B i of u.

59
00:04:59,260 --> 00:05:01,555
1 over pi-- this
is a little longer.

60
00:05:05,220 --> 00:05:08,200
Integral from 0 to
infinity as well.

61
00:05:08,200 --> 00:05:10,090
d k.

62
00:05:10,090 --> 00:05:17,500
And you have an exponential of
minus k cubed over 3 plus k u,

63
00:05:17,500 --> 00:05:29,716
plus the sine of k
cubed over 3, plus k u.

64
00:05:29,716 --> 00:05:31,720
And that's it.

65
00:05:31,720 --> 00:05:32,650
It's kind of funny.

66
00:05:32,650 --> 00:05:36,550
One is the cosine
and one is the sine.

67
00:05:36,550 --> 00:05:40,430
And it has this extra
different factors.

68
00:05:40,430 --> 00:05:44,590
So these are your two functions.

69
00:05:44,590 --> 00:05:48,520
And of the relevance
to our w k b

70
00:05:48,520 --> 00:05:51,640
problem is that they're
necessary to connect

71
00:05:51,640 --> 00:05:54,760
the solutions, as we will see.

72
00:05:54,760 --> 00:05:58,840
But we need a little more
about these functions.

73
00:05:58,840 --> 00:06:02,740
We need to know there
are asymptotic behaviors.

74
00:06:02,740 --> 00:06:06,130
Now there are functions, like
the exponential function, that

75
00:06:06,130 --> 00:06:09,970
has a Taylor series, e to the z.

76
00:06:09,970 --> 00:06:13,740
1 plus z plus z squared over 2.

77
00:06:13,740 --> 00:06:18,040
And it's valid,
whatever these angles--

78
00:06:18,040 --> 00:06:20,380
the argument of z is.

79
00:06:20,380 --> 00:06:23,110
That's always the same
asymptotic expansion,

80
00:06:23,110 --> 00:06:25,480
or the same Taylor series.

81
00:06:25,480 --> 00:06:28,900
For this functions,
like the airy function,

82
00:06:28,900 --> 00:06:37,600
for some arguments of
u, there's one form

83
00:06:37,600 --> 00:06:39,460
to the asymptotic expansion.

84
00:06:39,460 --> 00:06:43,720
And for some other arguments,
there's another form.

85
00:06:43,720 --> 00:06:47,330
That's sometimes called
the Stokes phenomenon.

86
00:06:47,330 --> 00:06:52,540
And for example, the
expansion for positive u

87
00:06:52,540 --> 00:06:55,420
is going to be a decaying
exponential here.

88
00:06:55,420 --> 00:06:59,050
But for negative u, it
will be oscillatory.

89
00:06:59,050 --> 00:07:01,795
So it's not like
a simple function,

90
00:07:01,795 --> 00:07:05,500
like the exponential function
has a nice, simple expansion

91
00:07:05,500 --> 00:07:06,340
everywhere.

92
00:07:06,340 --> 00:07:09,460
It just varies.

93
00:07:09,460 --> 00:07:17,210
So one needs to calculate
this asymptotic expansions,

94
00:07:17,210 --> 00:07:20,180
and I'll make a small
comment about it

95
00:07:20,180 --> 00:07:22,910
of how you go about it.

96
00:07:22,910 --> 00:07:26,350
The nice thing
about these formulas

97
00:07:26,350 --> 00:07:29,920
is that they allow you to
understand things intuitively

98
00:07:29,920 --> 00:07:34,000
and derive things yourself.

99
00:07:34,000 --> 00:07:37,210
Here you see the
two airy functions.

100
00:07:37,210 --> 00:07:38,480
They make sense.

101
00:07:38,480 --> 00:07:40,480
The other thing that
is possible to do

102
00:07:40,480 --> 00:07:43,030
with this counter
representations

103
00:07:43,030 --> 00:07:48,770
is to find the asymptotic
expansions of these functions.

104
00:07:48,770 --> 00:07:55,090
And they don't require
major mathematical work.

105
00:07:55,090 --> 00:07:57,220
It's kind of nice.

106
00:07:57,220 --> 00:08:01,340
So let's think a little
about one example.

107
00:08:01,340 --> 00:08:07,240
If you have the airy
function A i of u

108
00:08:07,240 --> 00:08:10,660
that is of the form
integral 1 over--

109
00:08:10,660 --> 00:08:14,660
well, it begins v k over 2 pi.

110
00:08:14,660 --> 00:08:17,770
The integral over counter C1.

111
00:08:17,770 --> 00:08:20,290
Maybe I should have
done them curly.

112
00:08:20,290 --> 00:08:23,860
Curly C1 would
have been clearer.

113
00:08:23,860 --> 00:08:25,070
e to the i.

114
00:08:25,070 --> 00:08:28,625
k cubed over 3 plus k u.

115
00:08:31,360 --> 00:08:32,320
That is your integral.

116
00:08:35,020 --> 00:08:37,539
And this is the
phase of integration.

117
00:08:39,935 --> 00:08:40,435
Phase.

118
00:08:43,620 --> 00:08:48,960
Now, in order to find
the asymptotic expansion

119
00:08:48,960 --> 00:08:55,200
for this thing, we'll use a
stationary phase condition.

120
00:08:55,200 --> 00:08:58,950
So the integral is
dominated by those places

121
00:08:58,950 --> 00:09:00,790
where the phase is stationary.

122
00:09:00,790 --> 00:09:09,246
So the 5 prime of k
is k squared plus u.

123
00:09:09,246 --> 00:09:11,550
And we want this
to be equal to 0.

124
00:09:14,260 --> 00:09:20,400
So take, for
example, u positive.

125
00:09:20,400 --> 00:09:24,080
Suppose you want to find the
behavior of the airy function

126
00:09:24,080 --> 00:09:27,590
on the right of the axis.

127
00:09:27,590 --> 00:09:33,680
Well, this says k
squared is equal to u.

128
00:09:33,680 --> 00:09:38,486
So k squared is
equal to minus u.

129
00:09:38,486 --> 00:09:42,620
And that's-- the right hand
side is negative because u is

130
00:09:42,620 --> 00:09:43,970
positive.

131
00:09:43,970 --> 00:09:48,655
And therefore, k-- the
points where this is solved

132
00:09:48,655 --> 00:09:52,970
are k equal plus minus
i square root of u.

133
00:09:57,220 --> 00:09:59,770
So where are those
points in the k plane?

134
00:09:59,770 --> 00:10:06,370
They're here and there.

135
00:10:06,370 --> 00:10:11,500
And those are the places where
you get stationary phase.

136
00:10:14,250 --> 00:10:15,600
So what can you do?

137
00:10:15,600 --> 00:10:21,110
You're supposed to do the
integral over this red line.

138
00:10:21,110 --> 00:10:23,130
C1.

139
00:10:23,130 --> 00:10:26,100
Well, as we argued,
this can be lifted

140
00:10:26,100 --> 00:10:29,060
and we can make the
integral pass through here.

141
00:10:38,630 --> 00:10:40,930
You can now do this
integral over here.

142
00:10:40,930 --> 00:10:43,195
It's the same integral
that you had before.

143
00:10:47,420 --> 00:10:53,690
In this line, we
can say that k is

144
00:10:53,690 --> 00:11:01,090
equal to i square root of
u plus some extra k tilde.

145
00:11:08,420 --> 00:11:16,790
We say here is i square
root of u for u positive.

146
00:11:16,790 --> 00:11:20,570
There is this other
place, but that--

147
00:11:20,570 --> 00:11:25,430
we cannot bring the counter down
here because in this region,

148
00:11:25,430 --> 00:11:28,370
the end points still contribute.

149
00:11:28,370 --> 00:11:33,950
So we cannot shift the counter
down, but we can shift it up.

150
00:11:33,950 --> 00:11:37,070
So we have to do this integral.

151
00:11:37,070 --> 00:11:38,860
So what happens?

152
00:11:42,440 --> 00:11:51,110
You can evaluate that phase
5k under these conditions.

153
00:11:51,110 --> 00:11:54,260
It is a stationary phase point--

154
00:11:54,260 --> 00:11:56,890
this one-- so the
answer is going

155
00:11:56,890 --> 00:12:03,890
to have a part independent
of k tilde, a linear part,

156
00:12:03,890 --> 00:12:05,930
with respect to
k tilde that will

157
00:12:05,930 --> 00:12:10,880
vanish because at this point
the phase is stationary.

158
00:12:10,880 --> 00:12:14,970
And then a quadratic part
with respect to k tilde.

159
00:12:14,970 --> 00:12:25,180
So the phase, 5k, which is
k cubed over 3 plus k u.

160
00:12:25,180 --> 00:12:28,580
When you substitute
this k here, it's

161
00:12:28,580 --> 00:12:31,010
going to give you an answer.

162
00:12:31,010 --> 00:12:39,560
And the answer is going to
be 2 over 3 i u to the 3/2,

163
00:12:39,560 --> 00:12:45,590
plus i square root
of u k squared tilde,

164
00:12:45,590 --> 00:12:50,780
plus k tilde cubed over 3.

165
00:12:50,780 --> 00:12:56,390
That's what you
get from the phase.

166
00:12:56,390 --> 00:13:00,770
You say, well,
that's pretty nice

167
00:13:00,770 --> 00:13:08,260
because now our integral psi
has become the integral d k

168
00:13:08,260 --> 00:13:11,570
tilde over 2 pi.

169
00:13:11,570 --> 00:13:17,540
So you pass from k
to k tilde variable.

170
00:13:17,540 --> 00:13:19,240
e to the minus 2/3.

171
00:13:23,240 --> 00:13:24,920
u to the 3/2.

172
00:13:29,660 --> 00:13:33,680
That is because you
have i times 5k,

173
00:13:33,680 --> 00:13:37,760
so you must multiply by i here.

174
00:13:37,760 --> 00:13:42,650
Then minus square
root of u k squared.

175
00:13:42,650 --> 00:13:44,210
Then plus k cubed.

176
00:13:49,371 --> 00:13:49,870
OK.

177
00:13:49,870 --> 00:13:51,005
Tildes of course.

178
00:13:54,218 --> 00:13:56,658
AUDIENCE: [INAUDIBLE]

179
00:13:56,658 --> 00:13:59,590
PROFESSOR: Um.

180
00:13:59,590 --> 00:14:00,930
Yes.

181
00:14:00,930 --> 00:14:02,860
i k.

182
00:14:02,860 --> 00:14:04,340
This should be like that.

183
00:14:04,340 --> 00:14:04,840
Yes.

184
00:14:08,530 --> 00:14:09,030
OK.

185
00:14:09,030 --> 00:14:11,910
So now what happens?

186
00:14:11,910 --> 00:14:18,510
If u is large enough,
this quantity over here

187
00:14:18,510 --> 00:14:22,440
is going to mean
that this integral is

188
00:14:22,440 --> 00:14:26,100
highly suppressed over k tilde.

189
00:14:26,100 --> 00:14:30,240
It's a Gaussian with
a very narrow peak.

190
00:14:30,240 --> 00:14:34,120
And therefore, we
can ignore this term.

191
00:14:34,120 --> 00:14:36,610
This term is a constant.

192
00:14:36,610 --> 00:14:39,150
So what do we get from here?

193
00:14:39,150 --> 00:14:44,130
We get 1 over 2 pi
from the beginning.

194
00:14:44,130 --> 00:14:46,680
This exponential.

195
00:14:46,680 --> 00:14:50,070
e to the minus 2 over 3.

196
00:14:50,070 --> 00:14:51,210
u to the 3/2.

197
00:14:54,030 --> 00:14:56,760
3/2.

198
00:14:56,760 --> 00:15:03,360
And then from this Gaussian,
we get square root of pi

199
00:15:03,360 --> 00:15:07,320
over square root of
square root of u.

200
00:15:07,320 --> 00:15:08,720
So it's u to the 1/4.

201
00:15:11,460 --> 00:15:14,970
So I think I got it all correct.

202
00:15:14,970 --> 00:15:24,930
And therefore, the function
A i of u goes like--

203
00:15:24,930 --> 00:15:30,630
or it's proportional to
this decaying exponential.

204
00:15:30,630 --> 00:15:32,640
Very fast decaying exponential.

205
00:15:32,640 --> 00:15:34,770
1 over 2.

206
00:15:34,770 --> 00:15:38,100
1 over square root of pi.

207
00:15:38,100 --> 00:15:41,160
1 over u to the 1/4.

208
00:15:41,160 --> 00:15:43,740
e to the minus 2/3.

209
00:15:43,740 --> 00:15:46,450
u to the 3/2.

210
00:15:46,450 --> 00:15:50,790
And this is when u is
greater than 0 and, in fact,

211
00:15:50,790 --> 00:15:53,790
u much greater than
1, positive and large.

212
00:15:56,970 --> 00:16:04,270
So this shows you the
power of this method.

213
00:16:04,270 --> 00:16:07,530
This is a very powerful
result. And this

214
00:16:07,530 --> 00:16:10,920
is what we're going to
need, pretty much, now

215
00:16:10,920 --> 00:16:15,270
in order to do our
asymptotic solutions

216
00:16:15,270 --> 00:16:17,880
and the matching of w k b.

217
00:16:17,880 --> 00:16:24,630
There is an extra term, or an
extra case, you can consider.

218
00:16:24,630 --> 00:16:32,180
What happens to this
integral when u is negative?

219
00:16:32,180 --> 00:16:37,170
When u is negative,
there's two real roots,

220
00:16:37,170 --> 00:16:43,800
and the stationary points
occur on the real line.

221
00:16:43,800 --> 00:16:46,560
The integral is, therefore, a
little more straightforward.

222
00:16:46,560 --> 00:16:49,470
You don't have to even move it.

223
00:16:49,470 --> 00:16:53,940
And we have another
expansion, therefore,

224
00:16:53,940 --> 00:16:59,790
which is A i of u is
actually equal to 1

225
00:16:59,790 --> 00:17:01,650
over the square root of pi.

226
00:17:01,650 --> 00:17:07,470
1 over u, length
of u to the 1/4.

227
00:17:07,470 --> 00:17:10,140
Cosine of 2/3.

228
00:17:10,140 --> 00:17:16,950
Length of u to the
3/2 minus pi over 4.

229
00:17:16,950 --> 00:17:20,550
This is for u less than 0.

230
00:17:20,550 --> 00:17:24,300
So the airy function becomes
an oscillatory function

231
00:17:24,300 --> 00:17:26,109
on the left.

232
00:17:26,109 --> 00:17:30,525
So how does this airy
function really look like?

233
00:17:34,880 --> 00:17:45,320
Well, we have the true airy
function behaves like this.

234
00:17:45,320 --> 00:17:49,160
It's a decaying
exponential for u positive.

235
00:17:52,340 --> 00:17:58,880
And then a decaying oscillatory
function for u negative.

236
00:17:58,880 --> 00:18:02,120
For u negative, it
decays eventually.

237
00:18:02,120 --> 00:18:05,720
The frequency becomes
faster and faster.

238
00:18:05,720 --> 00:18:14,030
And that's your airy
function A i of u.

239
00:18:17,750 --> 00:18:18,540
Here is u.

240
00:18:21,460 --> 00:18:25,650
Your asymptotic
expansions, the ones

241
00:18:25,650 --> 00:18:30,180
that we found up there
for u greater than 1,

242
00:18:30,180 --> 00:18:31,620
match this very nicely.

243
00:18:34,400 --> 00:18:41,090
But eventually they blow up,
so they're a little wrong.

244
00:18:41,090 --> 00:18:46,810
And they actually
match here quite OK.

245
00:18:46,810 --> 00:18:54,310
But here they go and also blow
up because of this factor.

246
00:18:54,310 --> 00:18:57,670
So they both blow up, but
they're both quite wrong

247
00:18:57,670 --> 00:19:01,780
in this area, which you would
expect them to be wrong.

248
00:19:01,780 --> 00:19:08,770
These are the regions where
this two asymptotic expansions

249
00:19:08,770 --> 00:19:09,640
make no sense.

250
00:19:09,640 --> 00:19:12,130
They were calculated
on the hypothesis

251
00:19:12,130 --> 00:19:17,170
that u is much bigger than
1, or much less than 1.

252
00:19:17,170 --> 00:19:18,740
Minus 1.

253
00:19:18,740 --> 00:19:21,610
And therefore,
you get everything

254
00:19:21,610 --> 00:19:25,300
under control except this area.

255
00:19:25,300 --> 00:19:30,940
So now let's do the
real work of w k b.

256
00:19:30,940 --> 00:19:33,140
That was our goal
from the beginning,

257
00:19:33,140 --> 00:19:38,430
so that's what
we'll try to do now.