1
00:00:00,000 --> 00:00:01,470
NARRATOR: The
following content is

2
00:00:01,470 --> 00:00:04,966
provided by MIT OpenCourseWare
under a Creative Commons

3
00:00:04,966 --> 00:00:06,060
license.

4
00:00:06,060 --> 00:00:08,230
Additional information
about our license

5
00:00:08,230 --> 00:00:10,490
and MIT OpenCourseWare
in general

6
00:00:10,490 --> 00:00:14,430
is available at ocw.mit.edu.

7
00:00:14,430 --> 00:00:19,220
PROFESSOR: Ready to go.

8
00:00:19,220 --> 00:00:23,290
So first of all, to say
thank you to several people

9
00:00:23,290 --> 00:00:31,820
in the class who sent codes
to find these wavefronts when

10
00:00:31,820 --> 00:00:39,230
a shock, a step function is
moving along with velocity c.

11
00:00:39,230 --> 00:00:45,410
I had roughly drawn what
these three methods produced,

12
00:00:45,410 --> 00:00:49,830
but it was much better to see
it in Mr. Sekora's movie that's

13
00:00:49,830 --> 00:00:53,150
on 18.086 site.

14
00:00:53,150 --> 00:01:02,340
And sure enough, behind the
moving front is an overshoot,

15
00:01:02,340 --> 00:01:06,860
that reminds us of the Gibbs
phenomenon, from Lax-Wendroff,

16
00:01:06,860 --> 00:01:11,220
and I think it's sort of
unavoidable if we have

17
00:01:11,220 --> 00:01:15,360
a negative -- because one
of the three coefficients

18
00:01:15,360 --> 00:01:18,290
at the previous time is
negative for Lax-Wendroff,

19
00:01:18,290 --> 00:01:21,170
that's how it gets its
second order accuracy,

20
00:01:21,170 --> 00:01:24,160
and we see that better
accuracy in a much steeper,

21
00:01:24,160 --> 00:01:26,580
much steeper profile.

22
00:01:26,580 --> 00:01:29,600
Then Lax-Friedrichs,
if you've noticed,

23
00:01:29,600 --> 00:01:35,830
has a sort of jagged profile.

24
00:01:35,830 --> 00:01:38,600
Maybe you've realized
why that would be.

25
00:01:38,600 --> 00:01:45,430
Somehow Lax-Friedrichs kind of
operates on a staggered grid.

26
00:01:45,430 --> 00:01:49,140
We'll see, actually with -- I've
drawn a staggered grid here,

27
00:01:49,140 --> 00:01:50,770
so I can say what I mean.

28
00:01:50,770 --> 00:01:54,350
Lax-Friedrichs
computes this value

29
00:01:54,350 --> 00:01:59,770
from the two earlier
values there and then

30
00:01:59,770 --> 00:02:01,900
at the next level
again and so forth,

31
00:02:01,900 --> 00:02:09,690
so that actually, Lax-Friedrichs
is using two staggered grids

32
00:02:09,690 --> 00:02:11,350
separately.

33
00:02:11,350 --> 00:02:16,460
I think -- normally,
with Lax-Friedrichs,

34
00:02:16,460 --> 00:02:19,440
we call that n plus 1 and
n plus 2 and n plus 3,

35
00:02:19,440 --> 00:02:23,350
but the picture is still right,
so I think that's probably

36
00:02:23,350 --> 00:02:30,440
responsible for what you
see, that there's two waves,

37
00:02:30,440 --> 00:02:36,360
two fronts, right, one delta
x away from each other.

38
00:02:36,360 --> 00:02:40,070
And finally, upwind
is also smeared out.

39
00:02:40,070 --> 00:02:46,680
I guess I was so pleased to
see these actually showing

40
00:02:46,680 --> 00:02:51,670
on the website and that makes
me think of more questions.

41
00:02:51,670 --> 00:02:54,670
Maybe you too.

42
00:02:54,670 --> 00:03:00,230
First of all, as
time goes forward,

43
00:03:00,230 --> 00:03:03,500
we see this bump
beginning to appear.

44
00:03:03,500 --> 00:03:06,970
So am I right in thinking that
it reaches a certain height

45
00:03:06,970 --> 00:03:08,790
and stays there?

46
00:03:08,790 --> 00:03:13,170
So I guess I'm thinking about
sort of further questions,

47
00:03:13,170 --> 00:03:15,820
because every time you do
a numerical experiment,

48
00:03:15,820 --> 00:03:19,150
good additional
question show up.

49
00:03:19,150 --> 00:03:23,990
So I'm wondering, as
this wavefront forms,

50
00:03:23,990 --> 00:03:28,750
this characteristic profile
for these three methods,

51
00:03:28,750 --> 00:03:31,900
first of all, does that
profile settle down

52
00:03:31,900 --> 00:03:34,610
to a steady profile?

53
00:03:34,610 --> 00:03:37,110
Ultimately it'd be great
to be able to predict

54
00:03:37,110 --> 00:03:40,330
what it would be.

55
00:03:40,330 --> 00:03:43,300
In particular, does that
bump reach a certain height?

56
00:03:43,300 --> 00:03:45,770
Because you can see it just
barely forming and then

57
00:03:45,770 --> 00:03:48,650
you see it growing
to a certain point.

58
00:03:48,650 --> 00:03:52,200
And then, when the movie
ends, fortunately it

59
00:03:52,200 --> 00:03:55,210
plots this picture and you
really see the profile,

60
00:03:55,210 --> 00:03:57,420
and you can actually
see, if you look closely,

61
00:03:57,420 --> 00:04:02,460
that it goes above 1
again and probably more.

62
00:04:05,630 --> 00:04:10,580
That would be, does it
approach a steady profile?

63
00:04:10,580 --> 00:04:14,530
Does that height settle down?

64
00:04:14,530 --> 00:04:18,170
It'd be fantastic to be
able to predict that height.

65
00:04:18,170 --> 00:04:20,540
For example, you always
want to know, how do things

66
00:04:20,540 --> 00:04:22,780
depend on the parameter?

67
00:04:22,780 --> 00:04:26,510
The one parameter we have
in this problem is r --

68
00:04:26,510 --> 00:04:28,240
c delta t over delta x.

69
00:04:28,240 --> 00:04:32,180
If I change r, I'm changing
all the coefficients.

70
00:04:32,180 --> 00:04:36,790
If I changed r to be
exactly 1, then the profile

71
00:04:36,790 --> 00:04:38,500
would be perfect, right?

72
00:04:38,500 --> 00:04:43,370
That's the golden
ratio, r equal 1.

73
00:04:43,370 --> 00:04:48,280
I would be computing that
value exactly from this value,

74
00:04:48,280 --> 00:04:51,400
at exactly the right slope,
on the characteristic,

75
00:04:51,400 --> 00:04:54,330
and perfection.

76
00:04:54,330 --> 00:04:57,600
Now as r goes away
from 1, I presume --

77
00:04:57,600 --> 00:04:58,980
it goes down below 1.

78
00:04:58,980 --> 00:05:02,620
Of course, if it goes
above 1, catastrophe.

79
00:05:02,620 --> 00:05:08,140
If it goes below 1, I presume
the bump begins to appear,

80
00:05:08,140 --> 00:05:09,890
goes higher.

81
00:05:09,890 --> 00:05:11,530
I don't know.

82
00:05:11,530 --> 00:05:16,130
Could you see how the
dependence on r appears?

83
00:05:16,130 --> 00:05:23,030
And then I can even suggest
one more thing, which is --

84
00:05:23,030 --> 00:05:31,910
let's put a fourth
guy onto this picture,

85
00:05:31,910 --> 00:05:33,850
which would be leapfrog.

86
00:05:33,850 --> 00:05:36,350
Leapfrog for
first-order equations,

87
00:05:36,350 --> 00:05:40,210
so let me just put down
below what that would be.

88
00:05:40,210 --> 00:05:44,110
So it would be on that
kind of a staggered grid.

89
00:05:44,110 --> 00:05:46,800
A new value at
time n plus 1 would

90
00:05:46,800 --> 00:05:51,650
come from these at time n,
really, and this one, at time n

91
00:05:51,650 --> 00:05:52,370
minus 1.

92
00:05:52,370 --> 00:05:55,900
So leapfrog -- you'd
think of it right away.

93
00:05:55,900 --> 00:06:04,070
I use a centered difference in
time, U_(j, n+1) minus U_(j,

94
00:06:04,070 --> 00:06:09,850
n-1) over 2 delta t equals c
times a centered difference

95
00:06:09,850 --> 00:06:20,510
in space, U_(j+1, n) minus
U_(j-1, n) over 2 delta x.

96
00:06:20,510 --> 00:06:24,680
So you see right
away, the 2's cancel.

97
00:06:24,680 --> 00:06:29,325
The delta x times c -- the
delta t times c over delta x is

98
00:06:29,325 --> 00:06:36,110
our old ratio r, so this is
easily written in terms of r

99
00:06:36,110 --> 00:06:37,300
again.

100
00:06:37,300 --> 00:06:44,140
It does have this
two time step issue

101
00:06:44,140 --> 00:06:47,030
and of course, we have
to think about stability.

102
00:06:47,030 --> 00:06:53,960
I'm writing down a fourth
candidate for the equation u_t

103
00:06:53,960 --> 00:06:59,990
equals c*u_x, of course, before
I come back to the second-order

104
00:06:59,990 --> 00:07:03,020
equations, or systems.

105
00:07:03,020 --> 00:07:05,220
I just wanted to put
in these comments,

106
00:07:05,220 --> 00:07:11,500
partly because I'm thinking
about projects that are --

107
00:07:11,500 --> 00:07:20,120
and I hope you are -- so
some people will have theses

108
00:07:20,120 --> 00:07:28,030
or problems from other courses
which involve solving PDEs

109
00:07:28,030 --> 00:07:36,050
and that'd be perfectly
natural to base some project

110
00:07:36,050 --> 00:07:42,720
on, but if you don't have
some specific application,

111
00:07:42,720 --> 00:07:47,660
then here are a whole lot
of interesting questions,

112
00:07:47,660 --> 00:07:49,790
to just go further with that.

113
00:07:49,790 --> 00:07:59,150
So the dependence
on r would be a --

114
00:07:59,150 --> 00:08:08,700
the profile dependence on
r, c delta t over delta x,

115
00:08:08,700 --> 00:08:09,650
would be a question.

116
00:08:14,710 --> 00:08:18,240
Let me just emphasize,
what's the goal?

117
00:08:18,240 --> 00:08:21,330
When I say that this
upwind one is smeared out,

118
00:08:21,330 --> 00:08:26,190
I mean that it doesn't satisfy
what you really hope for.

119
00:08:26,190 --> 00:08:31,860
You want to capture that
shock, that discontinuity --

120
00:08:31,860 --> 00:08:34,710
not perfectly, you
can't expect that --

121
00:08:34,710 --> 00:08:39,980
but you want to capture it,
maybe in, let's say, 2 delta x.

122
00:08:39,980 --> 00:08:43,510
A good method captures
the shock within 2 delta x

123
00:08:43,510 --> 00:08:50,960
and does it without much
undesirable oscillation,

124
00:08:50,960 --> 00:08:54,250
physical oscillation like that.

125
00:08:54,250 --> 00:08:58,870
So these are methods where
we're trading off a good shock

126
00:08:58,870 --> 00:09:03,400
capture versus a smeared one,
which is not satisfactory,

127
00:09:03,400 --> 00:09:07,650
really, in a lot
of applications.

128
00:09:07,650 --> 00:09:12,060
So capturing the
shock within 2 delta

129
00:09:12,060 --> 00:09:19,200
x, roughly, is highly
desirable, and you

130
00:09:19,200 --> 00:09:24,290
might have to go add a
artificial viscosity that we'll

131
00:09:24,290 --> 00:09:28,300
discuss to get there.

132
00:09:28,300 --> 00:09:33,120
So that's past things, but
still very much present,

133
00:09:33,120 --> 00:09:34,340
if I can say.

134
00:09:34,340 --> 00:09:38,720
Here I've introduced
a new leapfrog,

135
00:09:38,720 --> 00:09:42,450
a one-way equation
leapfrog, I'll

136
00:09:42,450 --> 00:09:50,230
call this, where
the last lecture was

137
00:09:50,230 --> 00:09:52,520
about two-way leapfrog.

138
00:09:52,520 --> 00:09:56,040
Leapfrog for u_tt
equals c squared

139
00:09:56,040 --> 00:09:58,510
u_xx, a two-way wave equation.

140
00:09:58,510 --> 00:10:04,480
I might just mention about
this one-way leapfrog,

141
00:10:04,480 --> 00:10:10,170
once you put in e to the
i*k*x as we always do,

142
00:10:10,170 --> 00:10:17,240
or e to the i*k*j delta
x in the discrete case --

143
00:10:17,240 --> 00:10:27,100
so you would put in an e
to the i*j delta x times k,

144
00:10:27,100 --> 00:10:33,560
and then find the G,
the growth factor in --

145
00:10:33,560 --> 00:10:39,040
that'll depend on k and will
tell us the time dependence.

146
00:10:39,040 --> 00:10:40,550
It'll be a G to the n-th.

147
00:10:43,330 --> 00:10:48,790
I think you'll find
that for this guy,

148
00:10:48,790 --> 00:10:53,440
the absolute value of G is
exactly 1 while it's stable.

149
00:10:53,440 --> 00:10:57,130
Of course, if the Courant
condition is violated,

150
00:10:57,130 --> 00:10:59,670
then there's no way
it could be stable

151
00:10:59,670 --> 00:11:04,010
and G's will grow,
be bigger than 1.

152
00:11:04,010 --> 00:11:10,040
But, I think in the stable
range, you will have --

153
00:11:10,040 --> 00:11:13,290
this G will have
absolute value exactly 1.

154
00:11:16,960 --> 00:11:18,620
So it'll be on the unit circle.

155
00:11:18,620 --> 00:11:23,390
You don't have
much leeway, right?

156
00:11:23,390 --> 00:11:28,920
We found upwind was,
like, well inside;

157
00:11:28,920 --> 00:11:32,840
Lax-Friedrichs was well
inside; Lax-Wendroff,

158
00:11:32,840 --> 00:11:38,240
because it had higher
accuracy for low frequencies,

159
00:11:38,240 --> 00:11:44,210
was closer; and I think
this guy is right on.

160
00:11:44,210 --> 00:11:48,170
So this is the
first-order methods.

161
00:11:48,170 --> 00:11:50,220
This would be Lax-Wendroff
and this would

162
00:11:50,220 --> 00:11:53,150
be this second-order method.

163
00:11:53,150 --> 00:11:56,710
So you're really
playing with fire there

164
00:11:56,710 --> 00:12:00,630
and maybe this is
not a favorite, just

165
00:12:00,630 --> 00:12:04,920
because it's too close to
the edge, in fact, right

166
00:12:04,920 --> 00:12:06,530
on the edge.

167
00:12:06,530 --> 00:12:10,960
So those are some
comments really generated

168
00:12:10,960 --> 00:12:17,890
by your numerical experiments
and I think that's the way this

169
00:12:17,890 --> 00:12:23,490
course should -- homeworks and
projects should develop out

170
00:12:23,490 --> 00:12:26,740
of numerical experiment.

171
00:12:26,740 --> 00:12:31,730
Now I'll come back to
complete my lecture

172
00:12:31,730 --> 00:12:34,110
on the second-order
wave equation.

173
00:12:37,260 --> 00:12:43,110
So we studied leapfrog for that.

174
00:12:43,110 --> 00:12:47,000
I won't repeat that, but
then with second-order wave

175
00:12:47,000 --> 00:12:49,730
equation, we have
another option,

176
00:12:49,730 --> 00:12:56,080
which is to reduce a
second-order equation with two

177
00:12:56,080 --> 00:13:01,260
time derivatives to a system
of two equations that are

178
00:13:01,260 --> 00:13:05,570
first order in time and space.

179
00:13:05,570 --> 00:13:12,310
I've chosen to use the letters
H and E as a kind of hint

180
00:13:12,310 --> 00:13:17,350
that we're touching a really
big application, which

181
00:13:17,350 --> 00:13:20,430
is electromagnetism.

182
00:13:20,430 --> 00:13:26,370
So E for the electric field, H
for the magnetic field but here

183
00:13:26,370 --> 00:13:35,400
in 1-D -- so it allows me
just to mention that in 3-D,

184
00:13:35,400 --> 00:13:39,160
when E has three components,
H has three components,

185
00:13:39,160 --> 00:13:45,310
this is really the curl and
this is really minus the curl.

186
00:13:45,310 --> 00:13:56,920
And the notes will show
the beautiful Yee mesh,

187
00:13:56,920 --> 00:14:05,790
staggered mesh that copies
for finite difference methods,

188
00:14:05,790 --> 00:14:11,700
the properties of the
curl, our interpretation

189
00:14:11,700 --> 00:14:16,210
of and the properties of -- that
Maxwell discovered the physics

190
00:14:16,210 --> 00:14:23,099
of, you know, current, magnetic
fields going around a wire that

191
00:14:23,099 --> 00:14:24,890
along which one an
electric field is going.

192
00:14:28,200 --> 00:14:31,670
But let me stay here
with one dimension.

193
00:14:31,670 --> 00:14:37,180
So first of all, can we recover
the wave equation from that?

194
00:14:37,180 --> 00:14:41,070
Let's just see that we haven't
left behind the wave equation.

195
00:14:45,190 --> 00:14:48,810
So I just take the
second derivative --

196
00:14:48,810 --> 00:14:51,820
so the second derivative -- I
just want to recover the wave

197
00:14:51,820 --> 00:14:56,360
equation so that I haven't lost
it here -- so this will be,

198
00:14:56,360 --> 00:14:58,810
if I -- I'm taking the time
derivative of the first

199
00:14:58,810 --> 00:15:07,220
equation, so I get d
second h dt dx, right?

200
00:15:07,220 --> 00:15:13,730
But by the useful, extremely
useful fact that this is

201
00:15:13,730 --> 00:15:20,060
the same as taking the t
derivative first and the x

202
00:15:20,060 --> 00:15:24,860
derivative second, now I
have the x derivative of this

203
00:15:24,860 --> 00:15:32,240
equation, so this is c --
now I have the x derivative,

204
00:15:32,240 --> 00:15:35,310
or c times it, so now
I'm up to c squared,

205
00:15:35,310 --> 00:15:39,920
the x derivative of the
x derivative, right?

206
00:15:43,310 --> 00:15:49,246
So we eliminated H, got back
to one equation, second order,

207
00:15:49,246 --> 00:15:50,370
and it's the wave equation.

208
00:15:53,660 --> 00:15:58,560
I want to show the same thing
now for finite differences.

209
00:15:58,560 --> 00:16:05,060
It's just -- it's such a simple
and useful device and our point

210
00:16:05,060 --> 00:16:09,610
will be that it works for --
just as well for differences.

211
00:16:09,610 --> 00:16:14,980
That allows us to do
this staggered grid idea.

212
00:16:14,980 --> 00:16:21,480
So the staggered grid idea
is, I'll do E on this grid,

213
00:16:21,480 --> 00:16:26,170
at these grid points,
H at the half points,

214
00:16:26,170 --> 00:16:32,750
E again at the integer points,
H again at the half points.

215
00:16:32,750 --> 00:16:41,790
Now, so my difference
method, leapfrog,

216
00:16:41,790 --> 00:16:46,390
on this staggered
mesh, will copy --

217
00:16:46,390 --> 00:16:54,900
so the first equation will
give E at the new time from

218
00:16:54,900 --> 00:16:59,770
the difference in
H at this time.

219
00:16:59,770 --> 00:17:02,250
So, you see it.

220
00:17:02,250 --> 00:17:08,810
dE/dt is going to be replaced
by that time difference.

221
00:17:08,810 --> 00:17:14,180
dH/dx is going to be replaced
by this space difference,

222
00:17:14,180 --> 00:17:16,420
and that will determine
E at the new time.

223
00:17:16,420 --> 00:17:21,280
So I need time 0 and
time 1/2 to get started.

224
00:17:21,280 --> 00:17:23,610
Maybe I put those
down here, somewhere.

225
00:17:23,610 --> 00:17:26,330
I have to use the
initial condition,

226
00:17:26,330 --> 00:17:33,710
the initial velocity to
get started on E and H

227
00:17:33,710 --> 00:17:35,990
and then I go onwards.

228
00:17:35,990 --> 00:17:38,470
I've got to this
point and this point.

229
00:17:38,470 --> 00:17:43,230
Then this guy comes from this
one, this one and this one.

230
00:17:46,520 --> 00:17:52,410
Of course, that similarly
comes from the same equation

231
00:17:52,410 --> 00:17:55,810
shifted over.

232
00:17:55,810 --> 00:17:56,980
Now I know these.

233
00:17:56,980 --> 00:17:59,540
Now I'm ready for this one.

234
00:17:59,540 --> 00:18:03,150
Coming from -- what am
I approximating now?

235
00:18:03,150 --> 00:18:09,270
I'm on the staggered part
of the grid, the half steps.

236
00:18:09,270 --> 00:18:14,220
I'm approximating this on the
half steps, where H is defined.

237
00:18:14,220 --> 00:18:18,990
So that minus that is
the time difference.

238
00:18:18,990 --> 00:18:23,380
This minus this -- which,
I know these now --

239
00:18:23,380 --> 00:18:26,190
is the space difference
in this equation,

240
00:18:26,190 --> 00:18:30,060
and that's what allows me
to compute the new one.

241
00:18:30,060 --> 00:18:33,940
You see, it's a simple idea.

242
00:18:33,940 --> 00:18:41,710
It's close but
different from my scalar

243
00:18:41,710 --> 00:18:44,580
leapfrog, one-way leapfrog here.

244
00:18:44,580 --> 00:18:49,140
Here it was the same
U at all the points.

245
00:18:49,140 --> 00:18:56,400
Here it is E at the integer
time levels and H at the half

246
00:18:56,400 --> 00:18:59,700
integer time levels.

247
00:18:59,700 --> 00:19:03,630
It's natural to say --
I'm calling this leapfrog.

248
00:19:06,990 --> 00:19:12,180
Could I do this for the
difference equation?

249
00:19:12,180 --> 00:19:17,760
I just want to see that
if I eliminate H --

250
00:19:17,760 --> 00:19:19,750
you saw what happened there.

251
00:19:19,750 --> 00:19:24,640
By using the identity that the
second derivative in t and x

252
00:19:24,640 --> 00:19:28,640
is always the same as the
second derivative in x and t,

253
00:19:28,640 --> 00:19:32,730
I eliminated H and got an
equation that involved E only.

254
00:19:36,550 --> 00:19:44,030
What's going to be the
corresponding identity?

255
00:19:44,030 --> 00:19:47,630
Trivial, but just -- trivial
things are not bad things

256
00:19:47,630 --> 00:19:51,820
to notice -- for the
different method.

257
00:19:54,760 --> 00:19:58,570
This is what I wanted to
be sure of, for the H part.

258
00:19:58,570 --> 00:20:05,460
This H -- so the H were -- do
you mind if I draw yet another

259
00:20:05,460 --> 00:20:06,650
symbol?

260
00:20:06,650 --> 00:20:11,840
I want to use this
idea to eliminate H. So

261
00:20:11,840 --> 00:20:15,270
let me put two H's there.

262
00:20:15,270 --> 00:20:16,700
Is that right?

263
00:20:16,700 --> 00:20:18,020
Yeah.

264
00:20:18,020 --> 00:20:25,000
I think what I'd like to know --
let me just say what I believe.

265
00:20:25,000 --> 00:20:31,090
I believe that the time
difference of the space

266
00:20:31,090 --> 00:20:35,990
difference of H
at mesh points is

267
00:20:35,990 --> 00:20:44,490
equal to the space difference
of the time difference of H

268
00:20:44,490 --> 00:20:45,560
at the same mesh points.

269
00:20:45,560 --> 00:20:47,370
I meant to put
these as subscripts.

270
00:20:51,290 --> 00:20:52,670
It's not delta x.

271
00:20:52,670 --> 00:20:57,180
It's delta sub x.

272
00:20:57,180 --> 00:21:01,130
Difference in the time
direction, difference

273
00:21:01,130 --> 00:21:02,950
in the space direction.

274
00:21:02,950 --> 00:21:05,520
Let's just see what this means.

275
00:21:05,520 --> 00:21:08,870
Let me redraw these four points.

276
00:21:08,870 --> 00:21:13,140
These are H. H at
all those points.

277
00:21:13,140 --> 00:21:19,480
So this says, figure out the
space differences -- so this,

278
00:21:19,480 --> 00:21:22,920
this minus this and
this minus this --

279
00:21:22,920 --> 00:21:27,790
and then take the time
difference of those.

280
00:21:27,790 --> 00:21:30,470
Can I see what I'll
get if I do that?

281
00:21:30,470 --> 00:21:33,890
So this means I take
the space differences,

282
00:21:33,890 --> 00:21:39,190
so I have 1 minus 1 and
that would be 1 minus 1,

283
00:21:39,190 --> 00:21:42,870
but now I'm going to take
the time difference of those.

284
00:21:42,870 --> 00:21:45,340
So that'll be this
one minus this one

285
00:21:45,340 --> 00:21:48,610
and this one plus this one.

286
00:21:48,610 --> 00:21:51,630
Do you see that?

287
00:21:51,630 --> 00:21:58,490
What we have here is H
at this point minus H

288
00:21:58,490 --> 00:22:00,740
at this point minus
H at this point

289
00:22:00,740 --> 00:22:05,740
plus H at this point plus 1.

290
00:22:05,740 --> 00:22:09,170
I claim that we have
the same thing here,

291
00:22:09,170 --> 00:22:12,640
that if I take the time
differences first --

292
00:22:12,640 --> 00:22:14,020
can I do that?

293
00:22:14,020 --> 00:22:16,340
Instead of taking the
space differences first,

294
00:22:16,340 --> 00:22:19,740
I'll take this one,
this time difference,

295
00:22:19,740 --> 00:22:23,780
this minus this and
I'll subtract --

296
00:22:23,780 --> 00:22:27,510
I'm taking now the space
difference -- I'll subtract --

297
00:22:27,510 --> 00:22:32,160
are you with me?

298
00:22:32,160 --> 00:22:35,960
You see, it's just the
same four values --

299
00:22:35,960 --> 00:22:42,840
two 1's and two minus 1's,
whether we go first -- in fact,

300
00:22:42,840 --> 00:22:46,640
this actually is a -- I think
maybe throws a little light

301
00:22:46,640 --> 00:22:56,740
on the calculus identity,
that we can exchange the order

302
00:22:56,740 --> 00:23:00,820
of the derivatives, because we
know we can exchange the order

303
00:23:00,820 --> 00:23:04,120
of the differences, because
we just got numbers like 1,

304
00:23:04,120 --> 00:23:09,190
minus 1, minus 1, and plus 1.

305
00:23:12,940 --> 00:23:19,180
If we let the spacing,
the mesh, go to 0, this --

306
00:23:19,180 --> 00:23:24,960
and divided by the
delta x and the delta t,

307
00:23:24,960 --> 00:23:31,840
I suppose we would prove
the continuous case.

308
00:23:31,840 --> 00:23:36,030
So in a way, this throws light
on something that's probably

309
00:23:36,030 --> 00:23:38,110
something that we
take for granted,

310
00:23:38,110 --> 00:23:43,010
but we wouldn't want to
be pressed on too closely.

311
00:23:46,360 --> 00:23:51,880
So having that
identity allows me

312
00:23:51,880 --> 00:23:57,000
to do exactly what I did with
the continuous case there

313
00:23:57,000 --> 00:24:03,270
and show that this, when I
eliminate the half levels,

314
00:24:03,270 --> 00:24:08,030
I've got ordinary leapfrog.

315
00:24:08,030 --> 00:24:11,670
So when I eliminate half
levels, I will be --

316
00:24:11,670 --> 00:24:15,310
when I eliminate this
half level, let's say,

317
00:24:15,310 --> 00:24:20,320
then I'll be connecting --
when I eliminate half levels,

318
00:24:20,320 --> 00:24:25,020
I'm going to be connecting these
at three levels -- one, two,

319
00:24:25,020 --> 00:24:28,310
three.

320
00:24:28,310 --> 00:24:30,790
It'll be leapfrog.

321
00:24:30,790 --> 00:24:37,890
So since it's been a few
days since we wrote down

322
00:24:37,890 --> 00:24:44,220
two-way leapfrog or second-order
equation leapfrog, let me just

323
00:24:44,220 --> 00:24:47,410
write it again, and
that's what we would get.

324
00:24:47,410 --> 00:24:55,100
The time difference
squared of E will

325
00:24:55,100 --> 00:25:01,250
be equal to c squared times the
space difference squared of E.

326
00:25:01,250 --> 00:25:05,070
And of course, here we
have a delta x squared

327
00:25:05,070 --> 00:25:08,760
and here a delta t squared.

328
00:25:08,760 --> 00:25:11,120
So altogether, an r squared.

329
00:25:11,120 --> 00:25:13,540
That's the formula we know.

330
00:25:13,540 --> 00:25:14,740
That's the formula we know.

331
00:25:19,320 --> 00:25:25,480
So if I simplify this
to make it look good

332
00:25:25,480 --> 00:25:27,910
and put the delta
t squared up there,

333
00:25:27,910 --> 00:25:29,870
you see that it's
just r squared.

334
00:25:37,720 --> 00:25:47,040
Maybe it's worth realizing that,
again, the magic r equal 1 --

335
00:25:47,040 --> 00:25:58,950
The magic ratio r equal
1 gives the exact U,

336
00:25:58,950 --> 00:26:02,900
agreeing exactly with u.

337
00:26:02,900 --> 00:26:08,030
This equation, if r is 1,
would be satisfied exactly

338
00:26:08,030 --> 00:26:14,920
by the continuous solution and
so it's the same as this one.

339
00:26:18,520 --> 00:26:22,040
So that's leapfrog.

340
00:26:22,040 --> 00:26:24,650
So can I summarize leapfrog?

341
00:26:24,650 --> 00:26:29,090
Leapfrog was, first of all,
that led us to these --

342
00:26:29,090 --> 00:26:32,270
it worked for
second-order equations.

343
00:26:32,270 --> 00:26:38,200
It led us to the same
stability condition

344
00:26:38,200 --> 00:26:42,130
that r had to be
less or equal to 1,

345
00:26:42,130 --> 00:26:54,730
but it took a quadratic
equation for G. Remember,

346
00:26:54,730 --> 00:27:01,070
we had G squared minus
2G times some quantity

347
00:27:01,070 --> 00:27:07,830
a that involved r's and
cosines plus 1 equals 0.

348
00:27:07,830 --> 00:27:09,782
The reason it was a
quadratic was there

349
00:27:09,782 --> 00:27:10,990
were two time steps involved.

350
00:27:14,810 --> 00:27:17,100
We handled that.

351
00:27:17,100 --> 00:27:21,070
We got stability, if r was
less than or equal to 1.

352
00:27:21,070 --> 00:27:25,330
That led us to a, in
magnitude, less or equal 1.

353
00:27:25,330 --> 00:27:30,100
And that led us to G -- I'll
just remember those steps --

354
00:27:30,100 --> 00:27:34,955
a less or equal 1, and that
let us to G less than or equal

355
00:27:34,955 --> 00:27:38,230
to 1.

356
00:27:38,230 --> 00:27:47,480
Yes, the notes, including
more about Maxwell's equation

357
00:27:47,480 --> 00:27:54,290
and Yee's beautiful method, with
a figure that shows his mesh,

358
00:27:54,290 --> 00:28:01,130
will go up -- I mean, every
section is getting upgraded

359
00:28:01,130 --> 00:28:05,040
and this one is the
next one to be upgraded.

360
00:28:05,040 --> 00:28:10,650
Probably by tomorrow,
you'll have a good section

361
00:28:10,650 --> 00:28:14,050
on second-order equations.

362
00:28:17,070 --> 00:28:24,220
For me, that's what I
wanted to say about the wave

363
00:28:24,220 --> 00:28:31,510
equation alone,
with no diffusion

364
00:28:31,510 --> 00:28:36,810
and now I'm ready to move
to the heat equation.

365
00:28:36,810 --> 00:28:40,400
So what do we have coming up?

366
00:28:40,400 --> 00:28:42,160
Maybe important to
say what's coming up.

367
00:28:47,120 --> 00:28:50,010
So we've done one-way wave.

368
00:28:54,240 --> 00:28:56,280
Now we've done two-way waves.

369
00:29:00,740 --> 00:29:10,820
Next comes the heat
equation, heat/diffusion,

370
00:29:10,820 --> 00:29:14,080
and then will come a
mixture of the two,

371
00:29:14,080 --> 00:29:26,810
convection-diffusion
-- and by the way,

372
00:29:26,810 --> 00:29:34,560
so I'll go straight to these
-- in a couple of weeks,

373
00:29:34,560 --> 00:29:38,810
we'll have a guest lecturer
who will do the applications

374
00:29:38,810 --> 00:29:44,980
to financial mathematics,
mathematical finance.

375
00:29:44,980 --> 00:29:49,280
Specifically, the
Black-Scholes equation.

376
00:29:53,920 --> 00:30:00,040
So you probably know that a lot
of effort and a lot of money

377
00:30:00,040 --> 00:30:04,730
are going into the
valuation of options

378
00:30:04,730 --> 00:30:13,330
and other financial derivatives
and that this Black-Scholes

379
00:30:13,330 --> 00:30:18,590
equation beautifully
produced the heat equation,

380
00:30:18,590 --> 00:30:23,330
and additional
correction terms will

381
00:30:23,330 --> 00:30:26,220
produce a
convection-diffusion equation.

382
00:30:26,220 --> 00:30:29,200
So this is an
application that we

383
00:30:29,200 --> 00:30:34,060
wouldn't have made
some years ago,

384
00:30:34,060 --> 00:30:37,970
but now it's worth
focusing on that.

385
00:30:37,970 --> 00:30:41,300
I think it's
interesting in itself.

386
00:30:41,300 --> 00:30:45,500
So this is what's
coming and then

387
00:30:45,500 --> 00:30:48,780
you could say one-way wave
equations, but nonlinear.

388
00:30:48,780 --> 00:30:53,870
So nonlinear wave
equations and those

389
00:30:53,870 --> 00:30:56,040
are called conservation laws.

390
00:31:05,740 --> 00:31:09,500
That's after these linear cases.

391
00:31:09,500 --> 00:31:10,840
All right.

392
00:31:10,840 --> 00:31:17,170
So I'm ready to tackle the
heat equation, starting

393
00:31:17,170 --> 00:31:22,100
then on to the next
section of the notes.

394
00:31:22,100 --> 00:31:25,510
So let me tackle it here.

395
00:31:25,510 --> 00:31:33,010
So the heat equation.

396
00:31:33,010 --> 00:31:38,010
Let me take the constant
to be 1; u_t equals u_xx.

397
00:31:41,310 --> 00:31:43,060
Because we already see
the big difference.

398
00:31:43,060 --> 00:31:45,970
Now, there's a second
x derivative, but only

399
00:31:45,970 --> 00:31:49,840
a first derivative in time.

400
00:31:49,840 --> 00:31:53,510
The dimensions of t and
x are no longer the same.

401
00:31:53,510 --> 00:31:58,810
We won't have delta t
and delta x comparable.

402
00:31:58,810 --> 00:32:03,515
Now it will be delta t
comparable with delta x

403
00:32:03,515 --> 00:32:06,930
squared, and that's a very
significant difference,

404
00:32:06,930 --> 00:32:09,680
because that's --
if delta x is small,

405
00:32:09,680 --> 00:32:12,110
delta x squared is
extremely small.

406
00:32:12,110 --> 00:32:19,060
So when delta t is constrained
by delta x squared,

407
00:32:19,060 --> 00:32:22,650
we're looking at
small time steps.

408
00:32:22,650 --> 00:32:27,610
We're looking at stiff
systems, absolutely.

409
00:32:27,610 --> 00:32:29,370
This is going to
be a stiff problem,

410
00:32:29,370 --> 00:32:36,220
because the low frequencies
and the high frequencies

411
00:32:36,220 --> 00:32:39,100
differ enormously
in the decay rate.

412
00:32:39,100 --> 00:32:41,550
In fact, why don't
we see it right away,

413
00:32:41,550 --> 00:32:44,530
starting as always
with exponentials.

414
00:32:44,530 --> 00:32:55,870
So if I plug in u of x and t
is some G of t e to the i*k*x,

415
00:32:55,870 --> 00:32:57,860
separating variables as always.

416
00:33:01,100 --> 00:33:03,860
Look for a pure
exponential solution,

417
00:33:03,860 --> 00:33:12,870
plug that in and we get
dG/dt times e to the i*k*x

418
00:33:12,870 --> 00:33:16,970
on the left, and now I have
to take two x derivatives,

419
00:33:16,970 --> 00:33:23,750
so that brings i*k down twice,
so that's minus k squared --

420
00:33:23,750 --> 00:33:29,610
i squared giving the minus 1 --
times G, times e to the i*k*x.

421
00:33:29,610 --> 00:33:40,870
And now, of course, I can
cancel that, which is never 0,

422
00:33:40,870 --> 00:33:45,010
and I've separated
out the time part,

423
00:33:45,010 --> 00:33:51,050
the G part, dG/dt is
minus k squared G. So G

424
00:33:51,050 --> 00:33:55,880
is e to the minus k squared t.

425
00:33:55,880 --> 00:33:59,580
Notice that it's k
squared, and notice

426
00:33:59,580 --> 00:34:02,750
that it's real and negative.

427
00:34:02,750 --> 00:34:06,320
See, that's a very
big difference.

428
00:34:06,320 --> 00:34:16,010
Compare e to the i*k*c*t, which
it was for the one-way wave

429
00:34:16,010 --> 00:34:17,490
equation.

430
00:34:17,490 --> 00:34:21,450
The difference is enormous here.

431
00:34:21,450 --> 00:34:26,920
That's magnitude 1,
energy conserved.

432
00:34:26,920 --> 00:34:30,180
This is magnitude smaller
than 1, energy dissipated.

433
00:34:36,080 --> 00:34:40,810
We had a ratio delta
t over delta x.

434
00:34:40,810 --> 00:34:43,390
Now we're going to see
delta t over delta x squared

435
00:34:43,390 --> 00:34:52,010
because somehow x and
t are -- now have --

436
00:34:52,010 --> 00:34:56,950
it's x squared
that matches t now.

437
00:34:56,950 --> 00:35:00,890
So now, when I put that into
here -- let me just do it --

438
00:35:00,890 --> 00:35:06,270
G of t is e to the
minus k squared t.

439
00:35:06,270 --> 00:35:10,100
So there is the
pure exponential.

440
00:35:10,100 --> 00:35:13,120
Simple as always, because we
have constant coefficients,

441
00:35:13,120 --> 00:35:16,770
but highly informative.

442
00:35:16,770 --> 00:35:20,750
I guess another thing that
I read off of that is,

443
00:35:20,750 --> 00:35:24,620
what frequencies
are decaying fast?

444
00:35:24,620 --> 00:35:26,220
High frequencies?

445
00:35:26,220 --> 00:35:30,750
If k is large, k
squared is very large

446
00:35:30,750 --> 00:35:34,690
and e to the minus k squared
t is decaying very quickly.

447
00:35:34,690 --> 00:35:39,840
So high-frequency
noise in this system,

448
00:35:39,840 --> 00:35:46,970
roughness, discontinuities
are going to get smoothed out,

449
00:35:46,970 --> 00:35:51,070
strongly, by moving
forward in time.

450
00:35:51,070 --> 00:35:52,750
Physically, what's
happening is, we

451
00:35:52,750 --> 00:36:00,020
have our wavefront, which
stayed a perfect step

452
00:36:00,020 --> 00:36:07,680
function in the wave equation,
will instantly smear.

453
00:36:07,680 --> 00:36:10,990
So diffusion smears
discontinuity.

454
00:36:14,420 --> 00:36:15,550
Heat travels.

455
00:36:15,550 --> 00:36:22,940
If we were back over here, if
the temperature starts out as 1

456
00:36:22,940 --> 00:36:28,540
on this side and 0 on this
side, then in an instant,

457
00:36:28,540 --> 00:36:33,870
some heat flows
from right to left.

458
00:36:33,870 --> 00:36:39,790
In fact, it flows
with infinite speed.

459
00:36:39,790 --> 00:36:44,220
In a short delta t, there's
a little heat all the way out

460
00:36:44,220 --> 00:36:48,930
there -- very little, of
course, because it's --

461
00:36:48,930 --> 00:36:54,690
I guess we want to see
what that behavior is.

462
00:36:54,690 --> 00:36:59,160
So I now want to write down --
staying with the differential

463
00:36:59,160 --> 00:37:03,380
equation, what do we want to do?

464
00:37:03,380 --> 00:37:08,590
We did the exponential
case and learned a good bit

465
00:37:08,590 --> 00:37:12,110
from that, the fast decay
of high frequencies.

466
00:37:16,640 --> 00:37:20,680
The next step would be to
put different frequencies,

467
00:37:20,680 --> 00:37:22,550
different k's together.

468
00:37:25,080 --> 00:37:33,120
By linearity, we can
add these solutions,

469
00:37:33,120 --> 00:37:36,830
we can multiply that
by any numbers we want,

470
00:37:36,830 --> 00:37:40,400
depending on k; we
still have solutions.

471
00:37:40,400 --> 00:37:43,980
And we can integrate over
k, so we will, actually.

472
00:37:43,980 --> 00:37:47,160
That that will be the next step.

473
00:37:47,160 --> 00:37:50,710
Let me write down
what that does.

474
00:37:50,710 --> 00:37:55,780
I'm going to jump,
take a little jump here

475
00:37:55,780 --> 00:37:58,630
to put on the board what
I just said in words.

476
00:37:58,630 --> 00:38:03,806
I can multiply this, e to
the minus k squared t e

477
00:38:03,806 --> 00:38:09,930
to the i*k*x -- that's what's
happening to the pure frequency

478
00:38:09,930 --> 00:38:11,350
k.

479
00:38:11,350 --> 00:38:23,610
If I multiply that by the amount
of that frequency which is

480
00:38:23,610 --> 00:38:26,220
in the initial condition -- so
here is the initial condition,

481
00:38:26,220 --> 00:38:30,730
u_0, and I've taken
its Fourier transform.

482
00:38:30,730 --> 00:38:36,020
This tells me how much of
frequency k is in the problem,

483
00:38:36,020 --> 00:38:38,930
is in the initial problem.

484
00:38:38,930 --> 00:38:41,140
This tells me how
much of frequency k

485
00:38:41,140 --> 00:38:43,430
is in the initial problem.

486
00:38:43,430 --> 00:38:48,220
That amount multiplies this,
with this rapid decay as time

487
00:38:48,220 --> 00:38:53,020
goes forward, but at any later
time, I'm just adding up --

488
00:38:53,020 --> 00:38:55,930
and since k -- we're
on the whole line --

489
00:38:55,930 --> 00:39:03,380
k is a continuous -- so we add
up from k equal minus infinity

490
00:39:03,380 --> 00:39:09,530
to infinity and I think we
have to remember the 2*pi that

491
00:39:09,530 --> 00:39:14,100
depends on our particular
definition of that transform,

492
00:39:14,100 --> 00:39:17,020
but I'm going to
stay with this one.

493
00:39:17,020 --> 00:39:18,470
That's the answer.

494
00:39:18,470 --> 00:39:19,760
That's u of x and t.

495
00:39:29,400 --> 00:39:32,910
So I've said that in
words, but actually, we

496
00:39:32,910 --> 00:39:36,170
could just check it.

497
00:39:36,170 --> 00:39:41,350
We would like to know that this
left-hand side solves the heat

498
00:39:41,350 --> 00:39:43,990
equation, and we
would like to know

499
00:39:43,990 --> 00:39:49,970
that it has the correct start,
the correct u of x and 0.

500
00:39:49,970 --> 00:39:50,960
Let's check that.

501
00:39:50,960 --> 00:39:56,060
At t equals 0, at
the start, this is 1.

502
00:39:56,060 --> 00:40:00,460
So at the start, this
formula, without that thing

503
00:40:00,460 --> 00:40:04,910
there, is just the
inverse Fourier transform

504
00:40:04,910 --> 00:40:07,780
that takes the Fourier
transform, multiplies by this,

505
00:40:07,780 --> 00:40:12,380
integrates to recover u_0.

506
00:40:12,380 --> 00:40:14,010
So it does start out correctly.

507
00:40:17,300 --> 00:40:19,240
Should I say that again?

508
00:40:19,240 --> 00:40:22,520
It starts correctly
because at t equal 0,

509
00:40:22,520 --> 00:40:25,415
this is just the inverse
Fourier transform of u_0,

510
00:40:25,415 --> 00:40:29,830
so it produces u_0.

511
00:40:29,830 --> 00:40:34,040
It inverts this
transform that we took.

512
00:40:34,040 --> 00:40:37,110
Secondly, does it solve
the heat equation?

513
00:40:37,110 --> 00:40:41,770
Sure, because for
each k, we've just

514
00:40:41,770 --> 00:40:43,720
found that that solves
the heat equation

515
00:40:43,720 --> 00:40:47,520
and now we're just putting
different k's together.

516
00:40:47,520 --> 00:40:50,740
So that's the answer.

517
00:40:50,740 --> 00:40:52,740
So you might say,
OK, perfection.

518
00:40:59,360 --> 00:41:02,000
Do we need finite differences?

519
00:41:04,670 --> 00:41:05,770
I guess we do.

520
00:41:05,770 --> 00:41:13,700
This is typical of formulas
that give the answer exactly,

521
00:41:13,700 --> 00:41:17,630
but in terms that are not
numerically convenient.

522
00:41:17,630 --> 00:41:21,660
Because, to use his
formula, first of all,

523
00:41:21,660 --> 00:41:29,400
this this requires an integral,
an infinite integral to find --

524
00:41:29,400 --> 00:41:37,130
integral to find
-- for u_0, u hat.

525
00:41:37,130 --> 00:41:42,240
To use this formula, we'd
have to find this transform,

526
00:41:42,240 --> 00:41:45,340
and then we have
to do this integral

527
00:41:45,340 --> 00:41:57,730
for the inverse transform, this
integral, for the answer, u.

528
00:41:57,730 --> 00:41:58,880
Two infinite integrals.

529
00:42:03,710 --> 00:42:09,580
And, not to mention the fact
that on a finite interval,

530
00:42:09,580 --> 00:42:12,160
when x doesn't go all the
way from minus infinity

531
00:42:12,160 --> 00:42:19,180
to infinity, we have to
think all over again.

532
00:42:19,180 --> 00:42:24,440
So we have a nice formula
for the answer, but --

533
00:42:24,440 --> 00:42:28,240
it looks nice, but it's
not numerically terrific.

534
00:42:31,390 --> 00:42:34,330
We can get a lot of
information out of this

535
00:42:34,330 --> 00:42:39,660
and there's one other
special solution that's

536
00:42:39,660 --> 00:42:42,220
also extremely informative.

537
00:42:42,220 --> 00:42:47,320
What if -- so I'm going
to do a special case,

538
00:42:47,320 --> 00:42:49,350
but it's the big case.

539
00:42:49,350 --> 00:42:57,240
What if the initial conditions
is a delta function?

540
00:42:57,240 --> 00:43:02,870
What if the initial condition
is a delta function?

541
00:43:02,870 --> 00:43:14,430
An impulse at time 0, x equal
0, an instant source of heat

542
00:43:14,430 --> 00:43:18,372
at a point -- a point
source with finite strength,

543
00:43:18,372 --> 00:43:19,280
you could say.

544
00:43:19,280 --> 00:43:22,580
An impulse at times 0.

545
00:43:27,450 --> 00:43:29,980
For the wave equation,
what happened?

546
00:43:29,980 --> 00:43:32,150
That was like turning
on a flash of light

547
00:43:32,150 --> 00:43:35,920
and it went along
the characteristics.

548
00:43:35,920 --> 00:43:42,600
For the heat equation,
it'll be quite different.

549
00:43:42,600 --> 00:43:47,000
There aren't
characteristics here,

550
00:43:47,000 --> 00:43:49,530
but this is an extremely
important solution,

551
00:43:49,530 --> 00:43:54,255
and I think you could call
it the fundamental solution

552
00:43:54,255 --> 00:43:55,130
to the heat equation.

553
00:44:01,390 --> 00:44:02,570
You may know what it is.

554
00:44:02,570 --> 00:44:05,230
You may have seen this formula.

555
00:44:05,230 --> 00:44:11,720
It's fantastic that we get
a formula for this one.

556
00:44:11,720 --> 00:44:14,930
You could say,
we've got a formula.

557
00:44:14,930 --> 00:44:18,170
What's the Fourier transform,
what's u_0 hat of k

558
00:44:18,170 --> 00:44:19,510
for the delta function?

559
00:44:22,280 --> 00:44:24,120
1, right?

560
00:44:24,120 --> 00:44:30,010
The Fourier transform of the
delta function is a constant 1.

561
00:44:30,010 --> 00:44:34,280
All frequencies are
there in equal amounts.

562
00:44:34,280 --> 00:44:38,310
This is a constant,
so I can remove it.

563
00:44:38,310 --> 00:44:39,710
It's 1.

564
00:44:39,710 --> 00:44:43,320
I'm left with a
specific integral to do.

565
00:44:43,320 --> 00:44:46,500
And the question
is, can I do it?

566
00:44:46,500 --> 00:44:49,410
And the answer
turns out to be yes.

567
00:44:49,410 --> 00:44:53,100
You can do that integral
when this is a constant

568
00:44:53,100 --> 00:44:55,030
and that will give
you the answer

569
00:44:55,030 --> 00:44:58,130
with that starting value.

570
00:44:58,130 --> 00:45:00,920
Let me just say
what that answer is.

571
00:45:00,920 --> 00:45:04,970
u of x and t -- this is
the fundamental solution,

572
00:45:04,970 --> 00:45:09,800
u fundamental,
maybe I should say,

573
00:45:09,800 --> 00:45:14,940
because it's such an important
case -- is, turns out to be --

574
00:45:14,940 --> 00:45:19,960
it's not obvious what
that integral is.

575
00:45:19,960 --> 00:45:37,030
By the way, normally --
let's see -- normally it's --

576
00:45:37,030 --> 00:45:39,050
to get the answer that
I'm going to write down,

577
00:45:39,050 --> 00:45:43,510
we take a kind of end
run on the integral,

578
00:45:43,510 --> 00:45:48,450
and I'm going to take a
serious end run and write down

579
00:45:48,450 --> 00:45:49,580
the answer exactly.

580
00:45:54,340 --> 00:45:55,940
Here's the key
part of the answer.

581
00:46:00,270 --> 00:46:03,880
It's a solution which starts
from a delta function, spreads

582
00:46:03,880 --> 00:46:10,620
of course, and the shape
as it spreads is a bell.

583
00:46:10,620 --> 00:46:15,665
It's the famous bell-shaped
Gaussian distribution of e

584
00:46:15,665 --> 00:46:23,230
to the minus -- so is it e to
the minus x squared, I guess?

585
00:46:23,230 --> 00:46:29,880
And then, always here comes --
that scalar tells us the width

586
00:46:29,880 --> 00:46:32,020
of the bell.

587
00:46:32,020 --> 00:46:34,900
How far has it spread at time t?

588
00:46:34,900 --> 00:46:37,210
And what goes there is 4t.

589
00:46:40,230 --> 00:46:42,250
That's the key observation.

590
00:46:42,250 --> 00:46:45,960
And notice how x squared
and t are coming in.

591
00:46:45,960 --> 00:46:50,380
It's that ratio, x squared
to t, that's crucial.

592
00:46:50,380 --> 00:46:52,480
That's the parameter again.

593
00:46:56,250 --> 00:46:59,550
Of course, we need to
multiply by a constant.

594
00:47:03,630 --> 00:47:06,790
How do I know I need a constant?

595
00:47:06,790 --> 00:47:12,450
Because the total
heat is conserved.

596
00:47:16,140 --> 00:47:19,210
In other words, the integral
of the delta function,

597
00:47:19,210 --> 00:47:24,010
the original source
of heat, was 1.

598
00:47:24,010 --> 00:47:26,280
The integral from minus
infinity to infinity

599
00:47:26,280 --> 00:47:31,150
of the delta function, which
is all totally concentrated

600
00:47:31,150 --> 00:47:33,290
at one point, is 1.

601
00:47:33,290 --> 00:47:35,480
So I need to put on
whatever constant

602
00:47:35,480 --> 00:47:37,360
it takes so that the
integral of this thing

603
00:47:37,360 --> 00:47:44,590
shall stay 1 for later times.

604
00:47:44,590 --> 00:47:47,940
This is one integral from minus
infinity to infinity that we

605
00:47:47,940 --> 00:47:51,740
can do and it turns out to
need 1 over the square root

606
00:47:51,740 --> 00:47:53,220
of 4*pi*t.

607
00:47:59,520 --> 00:48:01,180
That solves the heat equation.

608
00:48:05,610 --> 00:48:11,770
It's not a lot of fun to plug
that into the heat equation,

609
00:48:11,770 --> 00:48:13,910
but it's possible, right?

610
00:48:13,910 --> 00:48:17,380
The x derivative,
the time derivative

611
00:48:17,380 --> 00:48:19,010
is going to be messy.

612
00:48:19,010 --> 00:48:20,260
It'll have two terms.

613
00:48:20,260 --> 00:48:23,400
The x derivative will have
two terms and they'll agree.

614
00:48:27,550 --> 00:48:31,150
The notes will give one
way to do the integral

615
00:48:31,150 --> 00:48:31,990
and get that answer.

616
00:48:37,370 --> 00:48:40,390
You see that that
integral, it's --

617
00:48:40,390 --> 00:48:52,510
integrals with e to the minus k
squared in them are impossible

618
00:48:52,510 --> 00:48:57,360
on a finite interval, but here
we're going all the way from

619
00:48:57,360 --> 00:49:00,050
minus infinity to infinity,
which you might think makes

620
00:49:00,050 --> 00:49:04,640
the problem harder, but actually
it makes it a great deal easier

621
00:49:04,640 --> 00:49:09,810
and we can get an explicit
answer, a beautiful answer.

622
00:49:09,810 --> 00:49:15,340
So we learn that the solution
to the heat equation,

623
00:49:15,340 --> 00:49:22,180
coming from a point source, is a
bell-shaped Gaussian curve that

624
00:49:22,180 --> 00:49:26,570
gets wider and wider
as t increases,

625
00:49:26,570 --> 00:49:29,010
but I guess you would
want to look to see,

626
00:49:29,010 --> 00:49:32,890
what happens if t
comes down to 0?

627
00:49:32,890 --> 00:49:38,220
Does that really approximate
-- I meant to say,

628
00:49:38,220 --> 00:49:44,130
does it really converge to --
this should, as t goes to 0,

629
00:49:44,130 --> 00:49:52,030
this should go to the delta
function, and in some way,

630
00:49:52,030 --> 00:49:52,960
it does.

631
00:49:52,960 --> 00:49:54,600
In some way, it does.

632
00:49:54,600 --> 00:49:57,810
It's wonderful.

633
00:49:57,810 --> 00:50:05,520
The delta function, of course,
is 0 away from the origin.

634
00:50:05,520 --> 00:50:08,840
So why does this approach
0 away from the origin?

635
00:50:08,840 --> 00:50:11,070
Suppose x is 1.

636
00:50:11,070 --> 00:50:12,310
Suppose x is 1.

637
00:50:12,310 --> 00:50:19,890
Then do you see this thing
going to 0 as t goes to 0?

638
00:50:19,890 --> 00:50:20,910
Look what's happening.

639
00:50:20,910 --> 00:50:26,600
As t goes to 0, this part is
blowing up, but kind of weakly.

640
00:50:26,600 --> 00:50:29,630
That's not a disastrous blowup.

641
00:50:29,630 --> 00:50:34,290
Compared to this e to the
minus -- let's say 1 over 4t.

642
00:50:34,290 --> 00:50:40,450
As t goes to 0, that's e to
a very negative exponent.

643
00:50:40,450 --> 00:50:42,630
That's going to
0 very fast, much

644
00:50:42,630 --> 00:50:45,200
faster than this is going to 0.

645
00:50:45,200 --> 00:50:48,640
If you like to think of
l'Hopital's rule or ratio

646
00:50:48,640 --> 00:50:53,180
or something, this
quantity is going to 0 way

647
00:50:53,180 --> 00:50:56,500
faster than this one
is, and the result

648
00:50:56,500 --> 00:51:03,100
is 0, except at the one point
x equals 0, where this is 1.

649
00:51:08,140 --> 00:51:12,040
The fact that the total
amount of heat stays at 1

650
00:51:12,040 --> 00:51:17,590
tells us that the delta
function is the limit there.

651
00:51:20,350 --> 00:51:27,180
Today was the heat
equation, continuous case,

652
00:51:27,180 --> 00:51:30,950
with these explicit answers.

653
00:51:30,950 --> 00:51:36,570
Then next time is -- tomorrow,
I guess -- is the heat equation,

654
00:51:36,570 --> 00:51:39,400
finite differences.

655
00:51:39,400 --> 00:51:43,050
Again, we'll have several
difference methods

656
00:51:43,050 --> 00:51:46,330
and to compare them
would be terrific.

657
00:51:49,080 --> 00:51:53,200
Please think about that,
comparison with the heat

658
00:51:53,200 --> 00:51:57,290
equation and/or
what I was saying

659
00:51:57,290 --> 00:52:00,960
at the beginning of
class, a deeper look

660
00:52:00,960 --> 00:52:04,970
at the comparison for
the wave equation.

661
00:52:04,970 --> 00:52:11,150
See you tomorrow for difference
equations for this problem.

662
00:52:11,150 --> 00:52:12,400
Thanks.