1
00:00:00 --> 00:00:06
y prime and y double prime.
2
00:00:04 --> 00:00:10
So, q of x times y
equal zero.
3
00:00:09 --> 00:00:15
The linearity of the equation,
that is, the form in which it
4
00:00:15 --> 00:00:21
appears is going to be the key
idea today.
5
00:00:20 --> 00:00:26
Today is going to be
theoretical, but some of the
6
00:00:25 --> 00:00:31
ideas in it are the most
important in the course.
7
00:00:32 --> 00:00:38
So, I don't have to apologize
for the theory.
8
00:00:37 --> 00:00:43
Remember, the solution method
was to find two independent y
9
00:00:45 --> 00:00:51
one, y two
independent solutions.
10
00:00:51 --> 00:00:57
And now, I'll formally write
out what independent means.
11
00:01:00 --> 00:01:06
There are different ways to say
it.
12
00:01:03 --> 00:01:09
But for you,
I think the simplest and most
13
00:01:07 --> 00:01:13
intelligible will be to say that
y2 is not to be a constant
14
00:01:12 --> 00:01:18
multiple of y1.
And, unfortunately,
15
00:01:15 --> 00:01:21
it's necessary to add,
nor is y1 to be a constant
16
00:01:20 --> 00:01:26
multiple.
I have to call it by different
17
00:01:24 --> 00:01:30
constants.
So, let's call this one c prime
18
00:01:28 --> 00:01:34
of y2.
Well, I mean,
19
00:01:31 --> 00:01:37
the most obvious question is,
well, look.
20
00:01:35 --> 00:01:41
If this is not a constant times
that, this can't be there
21
00:01:41 --> 00:01:47
because I would just use one
over c if it was.
22
00:01:46 --> 00:01:52
Unfortunately,
the reason I have to write it
23
00:01:51 --> 00:01:57
this way is to take account of
the possibility that y1 might be
24
00:01:57 --> 00:02:03
zero.
If y1 is zero,
25
00:01:59 --> 00:02:05
so, the bad case that must be
excluded is that y1 equals zero,
26
00:02:06 --> 00:02:12
y2 nonzero.
I don't want to call those
27
00:02:10 --> 00:02:16
independent.
But nonetheless,
28
00:02:13 --> 00:02:19
it is true that y2 is not a
constant multiple of y1.
29
00:02:16 --> 00:02:22
However, y1 is a constant
multiple of y2,
30
00:02:19 --> 00:02:25
namely, the multiple zero.
It's just to exclude that case
31
00:02:23 --> 00:02:29
that I have to say both of those
things.
32
00:02:26 --> 00:02:32
And, one would not be sufficed.
That's a fine point that I'm
33
00:02:33 --> 00:02:39
not going to fuss over.
But I just have,
34
00:02:37 --> 00:02:43
of course.
Now, why do you do that?
35
00:02:41 --> 00:02:47
That's because,
then, all solutions,
36
00:02:45 --> 00:02:51
and this is what concerns us
today, are what?
37
00:02:50 --> 00:02:56
The linear combination with
constant coefficients of these
38
00:02:56 --> 00:03:02
two, and the fundamental
question we have to answer today
39
00:03:02 --> 00:03:08
is, why?
Now, there are really two
40
00:03:07 --> 00:03:13
statements involved in that.
On the one hand,
41
00:03:13 --> 00:03:19
I'm claiming there is an easier
statement, which is that they
42
00:03:20 --> 00:03:26
are all solutions.
So, that's question one,
43
00:03:24 --> 00:03:30
or statement one.
Why are all of these guys
44
00:03:29 --> 00:03:35
solutions?
That, I could trust you to
45
00:03:33 --> 00:03:39
answer yourself.
I could not trust you to answer
46
00:03:37 --> 00:03:43
it elegantly.
And, it's the elegance that's
47
00:03:39 --> 00:03:45
the most important thing today
because you have to answer it
48
00:03:44 --> 00:03:50
elegantly.
Otherwise, you can't go on and
49
00:03:46 --> 00:03:52
do more complicated things.
If you answer it in an ad hoc
50
00:03:50 --> 00:03:56
basis just by hacking out a
computation, you don't really
51
00:03:54 --> 00:04:00
see what's going on.
And you can't do more difficult
52
00:03:58 --> 00:04:04
things later.
So, we have to answer this,
53
00:04:02 --> 00:04:08
and answer it nicely.
The second question is,
54
00:04:05 --> 00:04:11
so, if that answers why there
are solutions at all,
55
00:04:09 --> 00:04:15
why are they all the solutions?
Why all the solutions?
56
00:04:14 --> 00:04:20
In other words,
to say it as badly as possible,
57
00:04:17 --> 00:04:23
why are all solutions,
why all the solutions-- Never
58
00:04:21 --> 00:04:27
mind.
Why are all the solutions.
59
00:04:24 --> 00:04:30
This is a harder question to
answer, but that should make you
60
00:04:29 --> 00:04:35
happy because that means it
depends upon a theorem which I'm
61
00:04:34 --> 00:04:40
not going to prove.
I'll just quote to you.
62
00:04:40 --> 00:04:46
Let's attack there for problem
one first.
63
00:04:47 --> 00:04:53
q1 is answered by what's called
the superposition.
64
00:04:56 --> 00:05:02
The superposition principle
says exactly that.
65
00:05:05 --> 00:05:11
It says exactly that,
that if y1 and y2 are solutions
66
00:05:09 --> 00:05:15
to a linear equation,
to a linear homogeneous ODE,
67
00:05:12 --> 00:05:18
in fact it can be of higher
order, too, although I won't
68
00:05:17 --> 00:05:23
stress that.
In other words,
69
00:05:19 --> 00:05:25
you don't have to stop with the
second derivative.
70
00:05:23 --> 00:05:29
You could add a third
derivative and a fourth
71
00:05:26 --> 00:05:32
derivative.
As long as the former makes the
72
00:05:30 --> 00:05:36
same, but that implies
automatically that c1 y1 plus c2
73
00:05:34 --> 00:05:40
y2 is a solution.
Now, the way to do that nicely
74
00:05:39 --> 00:05:45
is to take a little detour and
talk a little bit about linear
75
00:05:43 --> 00:05:49
operators.
And, since we are going to be
76
00:05:46 --> 00:05:52
using these for the rest of the
term, this is the natural place
77
00:05:51 --> 00:05:57
for you to learn a little bit
about what they are.
78
00:05:54 --> 00:06:00
So, I'm going to do it.
Ultimately, I am aimed at a
79
00:05:58 --> 00:06:04
proof of this statement,
but there are going to be
80
00:06:02 --> 00:06:08
certain side excursions I have
to make.
81
00:06:06 --> 00:06:12
The first side side excursion
is to write the differential
82
00:06:10 --> 00:06:16
equation in a different way.
So, I'm going to just write its
83
00:06:15 --> 00:06:21
transformations.
The first, I'll simply recopy
84
00:06:19 --> 00:06:25
what you know it to be,
q y equals zero.
85
00:06:22 --> 00:06:28
That's the first form.
The second form,
86
00:06:25 --> 00:06:31
I'm going to replace this by
the differentiation operator.
87
00:06:31 --> 00:06:37
So, I'm going to write this as
D squared y.
88
00:06:35 --> 00:06:41
That means differentiate it
twice.
89
00:06:38 --> 00:06:44
D it, and then D it again.
This one I only have to
90
00:06:42 --> 00:06:48
differentiate once,
so I'll write that as p D(y),
91
00:06:46 --> 00:06:52
p times the derivative of Y.
The last one isn't
92
00:06:50 --> 00:06:56
differentiated at all.
I just recopy it.
93
00:06:53 --> 00:06:59
Now, I'm going to formally
factor out the y.
94
00:06:57 --> 00:07:03
So this, I'm going to turn into
D squared plus pD plus q.
95
00:07:02 --> 00:07:08
Now, everybody reads this as
96
00:07:07 --> 00:07:13
times y equals zero.
But, what it means is this guy,
97
00:07:13 --> 00:07:19
it means this is shorthand for
that.
98
00:07:16 --> 00:07:22
I'm not multiplying.
I'm multiplying q times y.
99
00:07:21 --> 00:07:27
But, I'm not multiplying D
times y.
100
00:07:25 --> 00:07:31
I'm applying D to y.
Nonetheless,
101
00:07:28 --> 00:07:34
the notation suggests this is
very suggestive of that.
102
00:07:32 --> 00:07:38
And this, in turn,
implies that.
103
00:07:35 --> 00:07:41
I'm just transforming it.
And now, I'll take the final
104
00:07:39 --> 00:07:45
step.
I'm going to view this thing in
105
00:07:42 --> 00:07:48
parentheses as a guy all by
itself, a linear operator.
106
00:07:46 --> 00:07:52
This is a linear operator,
called a linear operator.
107
00:07:50 --> 00:07:56
And, I'm going to simply
abbreviate it by the letter L.
108
00:07:54 --> 00:08:00
And so, the final version of
this equation has been reduced
109
00:07:58 --> 00:08:04
to nothing but Ly equals zero.
110
00:08:03 --> 00:08:09
Now, what's L?
You can think of L as,
111
00:08:05 --> 00:08:11
well, formally,
L you would write as D squared
112
00:08:09 --> 00:08:15
plus pD plus q.
But, you can think of L,
113
00:08:13 --> 00:08:19
the way to think of it is as a
black box, a function of what
114
00:08:18 --> 00:08:24
goes into the black box,
well, if this were a function
115
00:08:22 --> 00:08:28
box, what would go it would be a
number, and what would come out
116
00:08:27 --> 00:08:33
with the number.
But it's not that kind of a
117
00:08:31 --> 00:08:37
black box.
It's an operator box,
118
00:08:34 --> 00:08:40
and therefore,
what goes in is a function of
119
00:08:38 --> 00:08:44
x.
And, what comes out is another
120
00:08:41 --> 00:08:47
function of x,
the result of applying this
121
00:08:44 --> 00:08:50
operator to that.
So, from this point of view,
122
00:08:48 --> 00:08:54
differential equations,
trying to solve the
123
00:08:52 --> 00:08:58
differential equation means,
what should come out you want
124
00:08:57 --> 00:09:03
to come out zero,
and the question is,
125
00:09:00 --> 00:09:06
what should you put in?
That's what it means solving
126
00:09:05 --> 00:09:11
differential equations in an
inverse problem.
127
00:09:08 --> 00:09:14
The easy thing is to put it a
function, and see what comes
128
00:09:11 --> 00:09:17
out.
You just calculate.
129
00:09:13 --> 00:09:19
The hard thing is to ask,
you say, I want such and such a
130
00:09:16 --> 00:09:22
thing to come out,
for example,
131
00:09:18 --> 00:09:24
zero; what should I put in?
That's a difficult question,
132
00:09:22 --> 00:09:28
and that's what we're spending
the term answering.
133
00:09:25 --> 00:09:31
Now, the key thing about this
is that this is a linear
134
00:09:28 --> 00:09:34
operator.
And, what that means is that it
135
00:09:32 --> 00:09:38
behaves in a certain way with
respect to functions.
136
00:09:37 --> 00:09:43
The easiest way to say it is,
I like to make two laws of it,
137
00:09:42 --> 00:09:48
that L of u1,
if you have two functions,
138
00:09:45 --> 00:09:51
I'm not going to put up the
parentheses, x,
139
00:09:49 --> 00:09:55
because that just makes things
look longer and not any clearer,
140
00:09:55 --> 00:10:01
actually.
What does L do to the sum of
141
00:09:58 --> 00:10:04
two functions?
If that's a linear operator,
142
00:10:03 --> 00:10:09
if you put in the sum of two
functions, what you must get out
143
00:10:08 --> 00:10:14
is the corresponding L's,
the sum of the corresponding
144
00:10:12 --> 00:10:18
L's of each.
So, that's a law.
145
00:10:14 --> 00:10:20
And, the other law,
linearity law,
146
00:10:17 --> 00:10:23
and this goes for anything in
mathematics and its
147
00:10:21 --> 00:10:27
applications,
which is called linear,
148
00:10:24 --> 00:10:30
basically anything is linear if
it does the following thing to
149
00:10:29 --> 00:10:35
functions or numbers or
whatever.
150
00:10:33 --> 00:10:39
The other one is of a constant
times any function,
151
00:10:37 --> 00:10:43
I don't have to give it a
number now because I'm only
152
00:10:41 --> 00:10:47
using one of them,
should be equal to c times L of
153
00:10:46 --> 00:10:52
u.
So, here, c is a constant,
154
00:10:48 --> 00:10:54
and here, of course,
the u is a function,
155
00:10:51 --> 00:10:57
functions of x.
These are the two laws of
156
00:10:55 --> 00:11:01
linearity.
An operator is linear if it
157
00:10:58 --> 00:11:04
satisfies these two laws.
Now, for example,
158
00:11:03 --> 00:11:09
the differentiation operator is
such an operator.
159
00:11:06 --> 00:11:12
D is linear,
why?
160
00:11:07 --> 00:11:13
Well, because of the very first
things you verify after you
161
00:11:11 --> 00:11:17
learn what the derivative is
because the derivative of,
162
00:11:15 --> 00:11:21
well, I will write it in the D
form.
163
00:11:17 --> 00:11:23
I'll write it in the form in
which you know it.
164
00:11:20 --> 00:11:26
It would be D apply to u1 plus
u2.
165
00:11:23 --> 00:11:29
How does one write that in
ordinary calculus?
166
00:11:26 --> 00:11:32
Well, like that.
Or, maybe you write d by dx out
167
00:11:31 --> 00:11:37
front.
Let's write it this way,
168
00:11:33 --> 00:11:39
is equal to u1 prime plus u2
prime.
169
00:11:36 --> 00:11:42
That's a law.
You prove it when you first
170
00:11:39 --> 00:11:45
study what a derivative is.
It's a property.
171
00:11:42 --> 00:11:48
From our point of view,
it's a property of the
172
00:11:45 --> 00:11:51
differentiation operator.
It has this property.
173
00:11:48 --> 00:11:54
The D of u1 plus u2 is D of u1
plus D of u2.
174
00:11:52 --> 00:11:58
And, it also has the property
175
00:11:55 --> 00:12:01
that c u prime,
you can pull out the constant.
176
00:12:00 --> 00:12:06
That's not affected by the
differentiation.
177
00:12:03 --> 00:12:09
So, these two familiar laws
from the beginning of calculus
178
00:12:07 --> 00:12:13
say, in our language,
that D is a linear operator.
179
00:12:11 --> 00:12:17
What about the multiplication
of law?
180
00:12:14 --> 00:12:20
That's even more important,
that u1 times u2 prime,
181
00:12:18 --> 00:12:24
I have nothing whatever to say
about that in here.
182
00:12:22 --> 00:12:28
In this context,
it's an important law,
183
00:12:25 --> 00:12:31
but it's not important with
respect to the study of
184
00:12:29 --> 00:12:35
linearity.
So, there's an example.
185
00:12:33 --> 00:12:39
Here's a more complicated one
that I'm claiming is the linear
186
00:12:38 --> 00:12:44
operator.
And, since I don't want to have
187
00:12:41 --> 00:12:47
to work in this lecture,
the work is left to you.
188
00:12:45 --> 00:12:51
So, the proof,
prove that L is linear,
189
00:12:48 --> 00:12:54
is this particular operator.
L is linear.
190
00:12:51 --> 00:12:57
That's in your part one
homework to verify that.
191
00:12:55 --> 00:13:01
And you will do some simple
exercises in recitation tomorrow
192
00:12:59 --> 00:13:05
to sort of warm you up for that
if you haven't done it already.
193
00:13:06 --> 00:13:12
Well, you shouldn't have
because this only goes with this
194
00:13:13 --> 00:13:19
lecture, actually.
It's forbidden to work ahead in
195
00:13:19 --> 00:13:25
this class.
All right, where are we?
196
00:13:23 --> 00:13:29
Well, all that was a prelude to
proving this simple theorem,
197
00:13:31 --> 00:13:37
superposition principle.
So, finally,
198
00:13:35 --> 00:13:41
what's the proof?
The proof of the superposition
199
00:13:42 --> 00:13:48
principle: if you believe that
the operator is linear,
200
00:13:50 --> 00:13:56
then L of c1,
in other words,
201
00:13:53 --> 00:13:59
the ODE is L.
L is D squared plus pD plus q.
202
00:13:59 --> 00:14:05
So, the ODE is Ly equals zero.
203
00:14:04 --> 00:14:10
And, what am I being asked to
204
00:14:07 --> 00:14:13
prove?
I'm being asked to prove that
205
00:14:10 --> 00:14:16
if y1 and y2 are solutions,
then so is that thing.
206
00:14:13 --> 00:14:19
By the way, that's called a
linear combination.
207
00:14:16 --> 00:14:22
Put that in your notes.
Maybe I better write it even on
208
00:14:20 --> 00:14:26
the board because it's something
people say all the time without
209
00:14:24 --> 00:14:30
realizing they haven't defined
it.
210
00:14:26 --> 00:14:32
This is called a linear
combination.
211
00:14:30 --> 00:14:36
This expression is called a
linear combination of y1 and y2.
212
00:14:35 --> 00:14:41
It means that particular sum
with constant coefficients.
213
00:14:40 --> 00:14:46
Okay, so, the ODE is Ly equals
zero.
214
00:14:44 --> 00:14:50
And, I'm trying to prove that
fact about it,
215
00:14:48 --> 00:14:54
that if y1 and y2 are
solutions, so is a linear
216
00:14:52 --> 00:14:58
combination of them.
So, the proof,
217
00:14:55 --> 00:15:01
then, I start with apply L to
c1 y1 plus c2 y2.
218
00:15:00 --> 00:15:06
Now, because this operator is
219
00:15:05 --> 00:15:11
linear, it takes the sum of two
functions into the corresponding
220
00:15:10 --> 00:15:16
sum up what the operator would
be.
221
00:15:13 --> 00:15:19
So, it would be L of c1 y1 plus
L of c2 y2.
222
00:15:17 --> 00:15:23
That's because L is a linear
223
00:15:20 --> 00:15:26
operator.
But, I don't have to stop
224
00:15:23 --> 00:15:29
there.
Because L is a linear operator,
225
00:15:26 --> 00:15:32
I can pull the c out front.
So, it's c1 L of y1 plus c2 L
226
00:15:31 --> 00:15:37
of y2. Now, where am I?
227
00:15:36 --> 00:15:42
Trying to prove that this is
zero.
228
00:15:38 --> 00:15:44
Well, what is L of y1?
At this point,
229
00:15:40 --> 00:15:46
I use the fact that y1 is a
solution.
230
00:15:43 --> 00:15:49
Because it's a solution,
this is zero.
231
00:15:46 --> 00:15:52
That's what it means to solve
that differential equation.
232
00:15:50 --> 00:15:56
It means, when you apply the
linear operator,
233
00:15:53 --> 00:15:59
L, to the function,
you get zero.
234
00:15:55 --> 00:16:01
In the same way,
y2 is a solution.
235
00:15:57 --> 00:16:03
So, that's zero.
And, the sum of c1 times zero
236
00:16:02 --> 00:16:08
plus c2 times zero is zero.
237
00:16:05 --> 00:16:11
That's the argument.
Now, you could make the same
238
00:16:08 --> 00:16:14
argument just by plugging c1 y1,
plugging it into the equation
239
00:16:13 --> 00:16:19
and calculating,
and calculating,
240
00:16:15 --> 00:16:21
and calculating,
grouping the terms and so on
241
00:16:18 --> 00:16:24
and so forth.
But, that's just calculation.
242
00:16:21 --> 00:16:27
It doesn't show you why it's
so.
243
00:16:24 --> 00:16:30
Why it's so is because the
operator, this differential
244
00:16:28 --> 00:16:34
equation is expressed as a
linear operator applied to y is
245
00:16:32 --> 00:16:38
zero.
And, the only properties that
246
00:16:37 --> 00:16:43
are really been used as the fact
that this operator is linear.
247
00:16:46 --> 00:16:52
That's the key point.
L is linear.
248
00:16:50 --> 00:16:56
It's a linear operator.
Well, that's all there is to
249
00:16:57 --> 00:17:03
the superposition principle.
As a prelude to answering the
250
00:17:04 --> 00:17:10
more difficult question,
why are these all the
251
00:17:07 --> 00:17:13
solutions?
Why are there no other
252
00:17:09 --> 00:17:15
solutions?
We need a few definitions,
253
00:17:12 --> 00:17:18
and a few more ideas.
And, they are going to occur in
254
00:17:16 --> 00:17:22
connection with,
so I'm now addressing,
255
00:17:18 --> 00:17:24
ultimately, question two.
But, it's not going to be
256
00:17:22 --> 00:17:28
addressed directly for quite
awhile.
257
00:17:25 --> 00:17:31
Instead, I'm going to phrase it
in terms of solving the initial
258
00:17:29 --> 00:17:35
value problem.
So far, we've only talked about
259
00:17:34 --> 00:17:40
the general solution with those
two arbitrary constants.
260
00:17:39 --> 00:17:45
But, how do you solve the
initial value problem,
261
00:17:43 --> 00:17:49
in other words,
fit initial conditions,
262
00:17:46 --> 00:17:52
find the solution with given
initial values for the function
263
00:17:51 --> 00:17:57
and its derivatives.
Now, --
264
00:17:55 --> 00:18:01
265
00:18:04 --> 00:18:10
-- the theorem is that this
collection of functions with
266
00:18:08 --> 00:18:14
these arbitrary constants,
these are all the solutions we
267
00:18:12 --> 00:18:18
have so far.
In fact, they are all the
268
00:18:15 --> 00:18:21
solutions there are,
but we don't know that yet.
269
00:18:19 --> 00:18:25
However, if we just focus on
the big class of solutions,
270
00:18:23 --> 00:18:29
there might be others lurking
out there somewhere lurking out
271
00:18:28 --> 00:18:34
there.
I don't know.
272
00:18:31 --> 00:18:37
But let's use what we have,
that just from this family is
273
00:18:39 --> 00:18:45
enough to satisfy any initial
condition, to satisfy any
274
00:18:47 --> 00:18:53
initial values.
In other words,
275
00:18:51 --> 00:18:57
if you give me any initial
values, I will be able to find
276
00:18:59 --> 00:19:05
the c1 and c2 which work.
Now, why is that?
277
00:19:05 --> 00:19:11
Well, I'd have to do a song and
dance at this point,
278
00:19:08 --> 00:19:14
if you hadn't been softened up
by actually calculating for
279
00:19:12 --> 00:19:18
specific differential equations.
You've had exercises and
280
00:19:15 --> 00:19:21
actually how to calculate the
values of c1 and c2.
281
00:19:19 --> 00:19:25
So, I'm going to do it now in
general what you have done so
282
00:19:23 --> 00:19:29
far for particular equations
using particular values of the
283
00:19:26 --> 00:19:32
initial conditions.
So, I'm relying on that
284
00:19:30 --> 00:19:36
experience that you've had in
doing the homework to make
285
00:19:35 --> 00:19:41
intelligible what I'm going to
do now in the abstract using
286
00:19:40 --> 00:19:46
just letters.
So, why is this so?
287
00:19:43 --> 00:19:49
Why is that so?
Well, we are going to need it,
288
00:19:47 --> 00:19:53
by the way, here,
too.
289
00:19:49 --> 00:19:55
I'll have to,
again, open up parentheses.
290
00:19:52 --> 00:19:58
But let's go as far as we can.
Well, you just try to do it.
291
00:19:57 --> 00:20:03
Suppose the initial conditions
are, how will we write them?
292
00:20:04 --> 00:20:10
So, they're going to be at some
initial point,
293
00:20:06 --> 00:20:12
x0.
You can take it to be zero if
294
00:20:08 --> 00:20:14
you want, but I'd like to be,
just for a little while,
295
00:20:12 --> 00:20:18
a little more general.
So, let's say the initial
296
00:20:15 --> 00:20:21
conditions, the initial values
are being given at the point x0,
297
00:20:19 --> 00:20:25
all right, that's going to be
some number.
298
00:20:21 --> 00:20:27
Let's just call it a.
And, the initial value also has
299
00:20:25 --> 00:20:31
to specify the velocity or the
value of the derivative there.
300
00:20:30 --> 00:20:36
Let's say these are the initial
values.
301
00:20:33 --> 00:20:39
So, the problem is to find the
c which work.
302
00:20:37 --> 00:20:43
Now, how do you do that?
Well, you know from
303
00:20:41 --> 00:20:47
calculation.
You write y equals c1 y1 plus
304
00:20:45 --> 00:20:51
c2 y2.
And you write y prime,
305
00:20:47 --> 00:20:53
and you take the derivative
underneath that,
306
00:20:51 --> 00:20:57
which is easy to do.
And now, you plug in x equals
307
00:20:56 --> 00:21:02
zero.
And, what happens?
308
00:21:00 --> 00:21:06
Well, these now turn into a set
of equations.
309
00:21:04 --> 00:21:10
What will they look like?
Well, y of x0 is a,
310
00:21:08 --> 00:21:14
and this is b.
So, what I get is let me flop
311
00:21:12 --> 00:21:18
it over onto the other side
because that's the way you're
312
00:21:17 --> 00:21:23
used to looking at systems of
equations.
313
00:21:21 --> 00:21:27
So, what we get is c1 times y1
of x0 plus c2 times y2 of x0.
314
00:21:26 --> 00:21:32
What's that supposed to be
315
00:21:31 --> 00:21:37
equal to?
Well, that's supposed to be
316
00:21:34 --> 00:21:40
equal to y of x0.
It's supposed to be equal to
317
00:21:37 --> 00:21:43
the given number,
a.
318
00:21:39 --> 00:21:45
And, in the same way,
c1 y1 prime of x0 plus c2 y2
319
00:21:42 --> 00:21:48
prime of x0, that's supposed to
turn out to
320
00:21:46 --> 00:21:52
be the number, b.
321
00:21:48 --> 00:21:54
In the calculations you've done
up to this point,
322
00:21:51 --> 00:21:57
y1 and y2 were always specific
functions like e to the x
323
00:21:55 --> 00:22:01
or cosine of 22x,
324
00:21:57 --> 00:22:03
stuff like that.
Now I'm doing it in the
325
00:22:01 --> 00:22:07
abstract, just calling them y1
and y2, so as to include all
326
00:22:06 --> 00:22:12
those possible cases.
Now, what am I supposed to do?
327
00:22:10 --> 00:22:16
I'm supposed to find c1 and c2.
What kind of things are they?
328
00:22:15 --> 00:22:21
This is what you studied in
high school, right?
329
00:22:19 --> 00:22:25
The letters are around us,
but it's a pair of simultaneous
330
00:22:23 --> 00:22:29
linear equations.
What are the variables?
331
00:22:26 --> 00:22:32
What are the variables?
What are the variables?
332
00:22:30 --> 00:22:36
Somebody raise their hand.
If you have a pair of
333
00:22:35 --> 00:22:41
simultaneous linear equations,
you've got variables and you've
334
00:22:39 --> 00:22:45
got constants,
right?
335
00:22:41 --> 00:22:47
And you are trying to find the
answer.
336
00:22:43 --> 00:22:49
What are the variables?
Yeah?
337
00:22:45 --> 00:22:51
c1 and c2.
Very good.
338
00:22:47 --> 00:22:53
I mean, it's extremely
confusing because in the first
339
00:22:50 --> 00:22:56
place, how can they be the
variables if they occur on the
340
00:22:54 --> 00:23:00
wrong side?
They're on the wrong side;
341
00:22:57 --> 00:23:03
they are constants.
How can constants be variables?
342
00:23:02 --> 00:23:08
Everything about this is wrong.
Nonetheless,
343
00:23:06 --> 00:23:12
the c1 and the c2 are the
unknowns, if you like the high
344
00:23:10 --> 00:23:16
school terminology.
c1 and c2 are the unknowns.
345
00:23:14 --> 00:23:20
These messes are just numbers.
After you've plugged in x0,
346
00:23:19 --> 00:23:25
this is some number.
You've got four numbers here.
347
00:23:23 --> 00:23:29
So, c1 and c2 are the
variables.
348
00:23:26 --> 00:23:32
The two find,
in other words,
349
00:23:28 --> 00:23:34
to find the values of.
All right, now you know general
350
00:23:34 --> 00:23:40
theorems from 18.02 about when
can you solve such a system of
351
00:23:39 --> 00:23:45
equations.
I'm claiming that you can
352
00:23:42 --> 00:23:48
always find c1 and c2 that work.
But, you know that's not always
353
00:23:48 --> 00:23:54
the case that a pair of
simultaneous linear equations
354
00:23:52 --> 00:23:58
can be solved.
There's a condition.
355
00:23:55 --> 00:24:01
There's a condition which
guarantees their solution,
356
00:23:59 --> 00:24:05
which is what?
What has to be true about the
357
00:24:04 --> 00:24:10
coefficients?
These are the coefficients.
358
00:24:08 --> 00:24:14
What has to be true?
The matrix of coefficients must
359
00:24:13 --> 00:24:19
be invertible.
The determinant of coefficients
360
00:24:18 --> 00:24:24
must be nonzero.
So, they are solvable if,
361
00:24:22 --> 00:24:28
for the c1 and c2,
if this thing,
362
00:24:25 --> 00:24:31
I'm going to write it.
Since all of these are
363
00:24:29 --> 00:24:35
evaluated at x0,
I'm going to write it in this
364
00:24:34 --> 00:24:40
way.
y1, the determinant,
365
00:24:38 --> 00:24:44
whose entries are y1,
y2, y1 prime,
366
00:24:41 --> 00:24:47
and y2 prime,
367
00:24:44 --> 00:24:50
evaluated at zero,
x0, that means that I evaluate
368
00:24:49 --> 00:24:55
each of the functions in the
determinant at x0.
369
00:24:54 --> 00:25:00
I'll write it this way.
That should be not zero.
370
00:25:00 --> 00:25:06
So, in other words,
the key thing which makes this
371
00:25:03 --> 00:25:09
possible, makes it possible for
us to solve the initial value
372
00:25:08 --> 00:25:14
problem, is that this funny
determinant should not be zero
373
00:25:13 --> 00:25:19
at the point at which we are
interested.
374
00:25:16 --> 00:25:22
Now, this determinant is
important in 18.03.
375
00:25:19 --> 00:25:25
It has a name,
and this is when you're going
376
00:25:23 --> 00:25:29
to learn it, if you don't know
it already.
377
00:25:26 --> 00:25:32
That determinant is called the
Wronskian.
378
00:25:31 --> 00:25:37
The Wronskian of what?
If you want to be pompous,
379
00:25:34 --> 00:25:40
you say this with a V sound
instead of a W.
380
00:25:37 --> 00:25:43
But, nobody does except people
trying to be pompous.
381
00:25:41 --> 00:25:47
The Wronskian,
we'll write a W.
382
00:25:43 --> 00:25:49
Now, notice,
you can only calculate it when
383
00:25:47 --> 00:25:53
you know what the two functions
are.
384
00:25:49 --> 00:25:55
So, the Wronskian of the two
functions, y1 and y2,
385
00:25:53 --> 00:25:59
what's the variable?
It's not a function of two
386
00:25:56 --> 00:26:02
variables, y1 and y2.
These are just the names of
387
00:26:00 --> 00:26:06
functions of x.
So, when you do it,
388
00:26:04 --> 00:26:10
put it in, calculate out that
determinant.
389
00:26:08 --> 00:26:14
This is a function of x,
a function of the independent
390
00:26:13 --> 00:26:19
variable after you've done the
calculation.
391
00:26:17 --> 00:26:23
Anyway, let's write its
definition, y1,
392
00:26:20 --> 00:26:26
y2, y1 prime,
y2 prime.
393
00:26:23 --> 00:26:29
Now, in order to do this,
394
00:26:26 --> 00:26:32
the point is we must know that
that Wronskian is not zero,
395
00:26:32 --> 00:26:38
that the Wronskian of these two
functions is not zero at the
396
00:26:37 --> 00:26:43
point x0.
Now, enter a theorem which
397
00:26:42 --> 00:26:48
you're going to prove for
homework, but this is harder.
398
00:26:47 --> 00:26:53
So, it's part two homework.
It's not part one homework.
399
00:26:51 --> 00:26:57
In other words,
I didn't give you the answer.
400
00:26:55 --> 00:27:01
You've got to find it yourself,
alone or in the company of good
401
00:27:01 --> 00:27:07
friends.
So, anyway, here's the
402
00:27:05 --> 00:27:11
Wronskian.
Now, what can we say for sure?
403
00:27:08 --> 00:27:14
Note, suppose y1 and y,
just to get you a feeling for
404
00:27:13 --> 00:27:19
it a little bit,
suppose they were not
405
00:27:16 --> 00:27:22
independent.
The word for not independent is
406
00:27:20 --> 00:27:26
dependent.
Suppose they were dependent.
407
00:27:24 --> 00:27:30
In other words,
suppose that y2 were a constant
408
00:27:28 --> 00:27:34
multiple of y1.
We know that's not the case
409
00:27:33 --> 00:27:39
because our functions are
supposed to be independent.
410
00:27:37 --> 00:27:43
But suppose they weren't.
What would the value of the
411
00:27:42 --> 00:27:48
Wronskian be?
If y2 is a constant times y1,
412
00:27:46 --> 00:27:52
then y2 prime is that same
constant times y1 prime.
413
00:27:50 --> 00:27:56
What's the value of the
determinant?
414
00:27:53 --> 00:27:59
Zero.
For what values of x is it zero
415
00:27:56 --> 00:28:02
for all values of x?
And now, that's the theorem
416
00:28:01 --> 00:28:07
that you're going to prove,
that if y1 and y2 are solutions
417
00:28:06 --> 00:28:12
to the ODE, I won't keep,
say, it's the ODE we've been
418
00:28:10 --> 00:28:16
talking about,
y double prime plus py.
419
00:28:14 --> 00:28:20
But the linear homogeneous with
420
00:28:20 --> 00:28:26
not constant coefficients,
just linear homogeneous second
421
00:28:26 --> 00:28:32
order.
Our solutions,
422
00:28:29 --> 00:28:35
as there are only two
possibilities,
423
00:28:33 --> 00:28:39
either-or.
Either the Wronskian of y,
424
00:28:37 --> 00:28:43
there are only two
possibilities.
425
00:28:39 --> 00:28:45
Either the Wronskian of y1 and
y2 is always zero,
426
00:28:44 --> 00:28:50
identically zero,
zero for all values of x.
427
00:28:47 --> 00:28:53
This is redundant.
When I say identically,
428
00:28:51 --> 00:28:57
I mean for all values of x.
But, I am just making assurance
429
00:28:56 --> 00:29:02
doubly sure.
Or, or the Wronskian is never
430
00:29:01 --> 00:29:07
zero.
Now, there is no notation as
431
00:29:06 --> 00:29:12
for that.
I'd better just write it out,
432
00:29:10 --> 00:29:16
is never zero,
i.e.
433
00:29:13 --> 00:29:19
for no x is it,
i.e.
434
00:29:15 --> 00:29:21
for all x.
There's no way to say that.
435
00:29:20 --> 00:29:26
I mean, for all values of x,
it's not zero.
436
00:29:26 --> 00:29:32
That means, there is not a
single point for which it's
437
00:29:32 --> 00:29:38
zero.
In particular,
438
00:29:35 --> 00:29:41
it's not zero here.
So, this is your homework:
439
00:29:43 --> 00:29:49
problem five,
part two.
440
00:29:45 --> 00:29:51
I'll give you a method of
proving it, which was discovered
441
00:29:52 --> 00:29:58
by the famous Norwegian
mathematician,
442
00:29:57 --> 00:30:03
Abel, who is,
I guess, the centenary of his
443
00:30:02 --> 00:30:08
birth, I guess,
was just celebrated last year.
444
00:30:09 --> 00:30:15
He has one of the truly tragic
stories in mathematics,
445
00:30:13 --> 00:30:19
which I think you can read.
It must be a Simmons book,
446
00:30:18 --> 00:30:24
if you have that.
Simmons is very good on
447
00:30:22 --> 00:30:28
biographies.
Look up Abel.
448
00:30:24 --> 00:30:30
He'll have a biography of Abel,
and you can weep if you're
449
00:30:29 --> 00:30:35
feeling sad.
He died at the age of 26 of
450
00:30:34 --> 00:30:40
tuberculosis,
having done a number of
451
00:30:37 --> 00:30:43
sensational things,
none of which was recognized in
452
00:30:41 --> 00:30:47
his lifetime because people
buried his papers under big
453
00:30:46 --> 00:30:52
piles of papers.
So, he died unknown,
454
00:30:49 --> 00:30:55
uncelebrated,
and now he's Norway's greatest
455
00:30:53 --> 00:30:59
culture hero.
In the middle of a park in
456
00:30:56 --> 00:31:02
Oslo, there's a huge statue.
And, since nobody knows what
457
00:31:03 --> 00:31:09
Abel looked like,
the statue is way up high so
458
00:31:07 --> 00:31:13
you can't see very well.
But, the inscription on the
459
00:31:13 --> 00:31:19
bottom says Niels Henrik Abel,
1801-1826 or something like
460
00:31:19 --> 00:31:25
that.
Now, --
461
00:31:22 --> 00:31:28
462
00:31:38 --> 00:31:44
-- the choice,
I'm still, believe it or not,
463
00:31:41 --> 00:31:47
aiming at question two,
but I have another big
464
00:31:44 --> 00:31:50
parentheses to open.
And, when I closed it,
465
00:31:47 --> 00:31:53
the answer to question two will
be simple.
466
00:31:50 --> 00:31:56
But, I think it's very
desirable that you get this
467
00:31:53 --> 00:31:59
second big parentheses.
It will help you to understand
468
00:31:57 --> 00:32:03
something important.
It will help you on your
469
00:32:02 --> 00:32:08
problem set tomorrow night.
I don't have to apologize.
470
00:32:08 --> 00:32:14
I'm just going to do it.
So, the question is,
471
00:32:12 --> 00:32:18
the thing you have to
understand is that when I write
472
00:32:18 --> 00:32:24
this combination,
I'm claiming that these are all
473
00:32:23 --> 00:32:29
the solutions.
I haven't proved that yet.
474
00:32:27 --> 00:32:33
But, they are going to be all
the solutions
475
00:32:33 --> 00:32:39
The point is,
there's nothing sacrosanct
476
00:32:36 --> 00:32:42
about the y1 and y2.
This is exactly the same
477
00:32:41 --> 00:32:47
collection as a collection which
I would write using other
478
00:32:46 --> 00:32:52
constants.
Let's call them u1 and u2.
479
00:32:49 --> 00:32:55
They are exactly the same,
where u1 and u2 are any other
480
00:32:55 --> 00:33:01
pair of linearly independent
solutions.
481
00:33:00 --> 00:33:06
Any other pair of independent
solutions, they must be
482
00:33:05 --> 00:33:11
independent, either a constant
multiple of each other.
483
00:33:10 --> 00:33:16
In other words,
u1 is some combination,
484
00:33:13 --> 00:33:19
now I'm really stuck because I
don't know how to,
485
00:33:18 --> 00:33:24
c1 bar, let's say,
that means a special value of
486
00:33:22 --> 00:33:28
c1, and a special value of c2,
and u2 is some other special
487
00:33:28 --> 00:33:34
value, oh my God,
c1 double bar,
488
00:33:31 --> 00:33:37
how's that?
The notation is getting worse
489
00:33:36 --> 00:33:42
and worse.
I apologize for it.
490
00:33:38 --> 00:33:44
In other words,
I could pick y1 and y2 and make
491
00:33:42 --> 00:33:48
up all of these.
And, I'd get a bunch of
492
00:33:45 --> 00:33:51
solutions.
But, I could also pick some
493
00:33:48 --> 00:33:54
other family,
some other two guys in this
494
00:33:51 --> 00:33:57
family, and just as well express
the solutions in terms of u1 and
495
00:33:56 --> 00:34:02
u2.
Now, well, why is he telling us
496
00:33:59 --> 00:34:05
that?
Well, the point is that the y1
497
00:34:03 --> 00:34:09
and the y2 are typically the
ones you get easily from solving
498
00:34:08 --> 00:34:14
the equations,
like e to the x and e
499
00:34:11 --> 00:34:17
to the 2x.
That's what you've gotten,
500
00:34:15 --> 00:34:21
or cosine x and sine x,
501
00:34:18 --> 00:34:24
something like that.
But, for certain ways,
502
00:34:22 --> 00:34:28
they might not be the best way
of writing the solutions.
503
00:34:26 --> 00:34:32
There is another way of writing
those that you should learn,
504
00:34:31 --> 00:34:37
and that's called finding
normalized, the normalized.
505
00:34:35 --> 00:34:41
They are okay,
but they are not normalized.
506
00:34:40 --> 00:34:46
For some things,
the normalized solutions are
507
00:34:42 --> 00:34:48
the best.
I'll explain to you what they
508
00:34:44 --> 00:34:50
are, and I'll explain to you
what they're good for.
509
00:34:48 --> 00:34:54
You'll see immediately what
they're good for.
510
00:34:50 --> 00:34:56
Normalized solutions,
now, you have to specify the
511
00:34:53 --> 00:34:59
point at which you're
normalizing.
512
00:34:55 --> 00:35:01
In general, it would be x
nought,
513
00:34:58 --> 00:35:04
but let's, at this point,
since I don't have an infinity
514
00:35:01 --> 00:35:07
of time, to simplify things,
let's say zero.
515
00:35:05 --> 00:35:11
It could be x nought,
any point would do just as
516
00:35:08 --> 00:35:14
well.
But, zero is the most common
517
00:35:11 --> 00:35:17
choice.
What are the normalized
518
00:35:13 --> 00:35:19
solutions?
Well, first of all,
519
00:35:15 --> 00:35:21
I have to give them names.
I want to still call them y.
520
00:35:19 --> 00:35:25
So, I'll call them capital Y1
and Y2.
521
00:35:21 --> 00:35:27
And, what they are,
are the solutions which satisfy
522
00:35:25 --> 00:35:31
certain, special,
very special,
523
00:35:27 --> 00:35:33
initial conditions.
And, what are those?
524
00:35:31 --> 00:35:37
So, they're the ones which
satisfy, the initial conditions
525
00:35:37 --> 00:35:43
for Y1 are, of course there are
going to be guys that look like
526
00:35:43 --> 00:35:49
this.
The only thing that's going to
527
00:35:46 --> 00:35:52
make them distinctive is the
initial conditions they satisfy.
528
00:35:51 --> 00:35:57
Y1 has to satisfy at zero.
Its value should be one,
529
00:35:56 --> 00:36:02
and the value of its derivative
should be zero.
530
00:36:02 --> 00:36:08
For Y2, it's just the opposite.
Here, the value of the function
531
00:36:07 --> 00:36:13
should be zero at zero.
But, the value of its
532
00:36:11 --> 00:36:17
derivative, now,
I want to be one.
533
00:36:14 --> 00:36:20
Let me give you a trivial
example of this,
534
00:36:17 --> 00:36:23
and then one,
which is a little less trivial,
535
00:36:21 --> 00:36:27
so you'll have some feeling for
what I'm asking for.
536
00:36:25 --> 00:36:31
Suppose the equation,
for example,
537
00:36:28 --> 00:36:34
is y double prime plus,
well, let's really make it
538
00:36:33 --> 00:36:39
simple.
Okay, you know the standard
539
00:36:36 --> 00:36:42
solutions are y1 is cosine x,
540
00:36:39 --> 00:36:45
and y2 is sine x.
541
00:36:41 --> 00:36:47
These are functions,
which, when you take the second
542
00:36:44 --> 00:36:50
derivative, they turn into their
negative.
543
00:36:46 --> 00:36:52
You know, you could go the
complex roots are i and minus i,
544
00:36:50 --> 00:36:56
and blah, blah,
blah, blah, blah,
545
00:36:52 --> 00:36:58
blah.
If you do it that way,
546
00:36:53 --> 00:36:59
fine.
But at some point in the course
547
00:36:56 --> 00:37:02
you have to be able to write
down and, right away,
548
00:36:59 --> 00:37:05
oh, yeah, cosine x,
sine x.
549
00:37:01 --> 00:37:07
Okay, what are the normalized
550
00:37:05 --> 00:37:11
things?
Well, what's the value of this
551
00:37:09 --> 00:37:15
at zero?
It is one.
552
00:37:11 --> 00:37:17
What's the value of its
derivative at zero?
553
00:37:15 --> 00:37:21
Zero.
This is Y1.
554
00:37:16 --> 00:37:22
This is the only case in which
you locked on immediately to the
555
00:37:22 --> 00:37:28
normalized solutions.
In the same way,
556
00:37:26 --> 00:37:32
this guy is Y2 because its
value at zero is zero.
557
00:37:32 --> 00:37:38
It's value of its derivative at
zero is one.
558
00:37:35 --> 00:37:41
So, this is Y2.
Okay, now let's look at a case
559
00:37:39 --> 00:37:45
where you don't immediately lock
on to the normalized solutions.
560
00:37:44 --> 00:37:50
Very simple:
all I have to do is change the
561
00:37:47 --> 00:37:53
sign.
Here, you know,
562
00:37:49 --> 00:37:55
think through r squared minus
one equals zero.
563
00:37:54 --> 00:38:00
The characteristic roots are
plus one and minus one,
564
00:37:58 --> 00:38:04
right?
And therefore,
565
00:38:00 --> 00:38:06
the solution is e to the x,
and e to the negative x.
566
00:38:04 --> 00:38:10
So, the solutions you find by
567
00:38:07 --> 00:38:13
the usual way of solving it is
y1 equals e to the x,
568
00:38:11 --> 00:38:17
and y2 equals e to the
negative x.
569
00:38:15 --> 00:38:21
Those are the standard
solution.
570
00:38:17 --> 00:38:23
So, the general solution is of
the form.
571
00:38:19 --> 00:38:25
So, the general solution is of
the form c1 e to the x plus c2 e
572
00:38:23 --> 00:38:29
to the negative x.
573
00:38:26 --> 00:38:32
Now, what I want to find out is
what is Y1 and Y2?
574
00:38:31 --> 00:38:37
How do I find out what Y1 is?
Well, I have to satisfy initial
575
00:38:36 --> 00:38:42
conditions.
So, if this is y,
576
00:38:38 --> 00:38:44
let's write down here,
if you can still see that,
577
00:38:43 --> 00:38:49
y prime is c1 e to the x minus
c2 e to the negative x .
578
00:38:48 --> 00:38:54
So, if I plug in,
579
00:38:51 --> 00:38:57
I want y of zero to be one,
I want this guy at the point
580
00:38:56 --> 00:39:02
zero to be one.
What equation does that give
581
00:39:00 --> 00:39:06
me?
That gives me c1 plus c2,
582
00:39:04 --> 00:39:10
c1 plus c2,
plugging in x equals zero,
583
00:39:08 --> 00:39:14
equals the value of this thing
at zero.
584
00:39:11 --> 00:39:17
So, that's supposed to be one.
585
00:39:14 --> 00:39:20
How about the other guy?
The value of its derivative is
586
00:39:19 --> 00:39:25
supposed to come out to be zero.
And, what is its derivative?
587
00:39:24 --> 00:39:30
Well, plug into this
expression.
588
00:39:26 --> 00:39:32
It's c1 minus c2.
Okay, what's the solution to
589
00:39:32 --> 00:39:38
those pair of equations?
c2 has to be equal to c1.
590
00:39:36 --> 00:39:42
The sum of the two of them has
to be one.
591
00:39:39 --> 00:39:45
Each one, therefore,
is equal to one half.
592
00:39:42 --> 00:39:48
And so, what's the value of Y1?
Y1, therefore,
593
00:39:46 --> 00:39:52
is the function where c1 and c2
are one half.
594
00:39:50 --> 00:39:56
It's the function e to the x
plus e to the negative x divided
595
00:39:55 --> 00:40:01
by two.
596
00:39:59 --> 00:40:05
In the same way,
I won't repeat the calculation.
597
00:40:03 --> 00:40:09
You can do yourself.
Same calculation shows that Y2,
598
00:40:07 --> 00:40:13
so, put in the initial
conditions.
599
00:40:10 --> 00:40:16
The answer will be that Y2 is
equal to e to the x minus e to
600
00:40:15 --> 00:40:21
the minus x divided by two.
601
00:40:19 --> 00:40:25
These are the special
functions.
602
00:40:22 --> 00:40:28
For this equation,
these are the normalized
603
00:40:25 --> 00:40:31
solutions.
They are better than the
604
00:40:28 --> 00:40:34
original solutions because their
initial values are nicer.
605
00:40:35 --> 00:40:41
Just check it.
The initial value,
606
00:40:37 --> 00:40:43
when x is equal to zero,
the initial value,
607
00:40:40 --> 00:40:46
this has is zero.
Here, when x is equal to zero,
608
00:40:44 --> 00:40:50
the value of the function is
zero.
609
00:40:47 --> 00:40:53
But, the value of its
derivative, these cancel,
610
00:40:51 --> 00:40:57
is one.
So, these are the good guys.
611
00:40:54 --> 00:41:00
Okay, there's no colored chalk
this period.
612
00:40:57 --> 00:41:03
Okay, there was colored chalk.
There's one.
613
00:41:01 --> 00:41:07
So, for this equation,
these are the good guys.
614
00:41:04 --> 00:41:10
These are our best solutions.
e to the x and e to the
615
00:41:07 --> 00:41:13
minus x are good solutions.
616
00:41:10 --> 00:41:16
But, these are our better
solutions.
617
00:41:12 --> 00:41:18
And, this one,
of course, is the function
618
00:41:14 --> 00:41:20
which is called hyperbolic sine
of x, and this is the one which
619
00:41:18 --> 00:41:24
is called hyperbolic cosine of
x.
620
00:41:20 --> 00:41:26
This is one of the most
important ways in which they
621
00:41:23 --> 00:41:29
enter into mathematics.
And, this is why the engineers
622
00:41:26 --> 00:41:32
want them.
Now, why do the engineers want
623
00:41:28 --> 00:41:34
normalized solutions?
Well, I didn't explain that.
624
00:41:34 --> 00:41:40
So, what's so good about
normalized solutions?
625
00:41:41 --> 00:41:47
Very simple:
if Y1 and Y2 are normalized at
626
00:41:47 --> 00:41:53
zero, let's say,
then the solution to the IVP,
627
00:41:53 --> 00:41:59
in other words,
the ODE plus the initial
628
00:41:59 --> 00:42:05
values, y of zero equals,
let's say, a and y
629
00:42:07 --> 00:42:13
prime of zero equals b.
630
00:42:14 --> 00:42:20
So, the ODE I'm not repeating.
It's the one we've been talking
631
00:42:17 --> 00:42:23
about all term since the
beginning of the period.
632
00:42:20 --> 00:42:26
It's the one with the p of x
and q of x.
633
00:42:23 --> 00:42:29
And, here are the initial
values.
634
00:42:25 --> 00:42:31
I'm going to call them a and b.
You can also call them,
635
00:42:28 --> 00:42:34
if you like,
maybe that's better to call
636
00:42:30 --> 00:42:36
them y0, as they are individual
in the homework.
637
00:42:34 --> 00:42:40
They are called,
I'm using the,
638
00:42:36 --> 00:42:42
let's use those.
What is the solution?
639
00:42:39 --> 00:42:45
I say the solution is,
if you use y1 and y2,
640
00:42:43 --> 00:42:49
the solution is y0,
in other words,
641
00:42:46 --> 00:42:52
the a times Y1,
plus y0 prime,
642
00:42:49 --> 00:42:55
in other words,
b times Y2.
643
00:42:53 --> 00:42:59
In other words,
you can write down instantly
644
00:42:57 --> 00:43:03
the solution to the initial
value problem,
645
00:43:00 --> 00:43:06
if instead of using the
functions, you started out with
646
00:43:05 --> 00:43:11
the little Y1 and Y2,
you use these better functions.
647
00:43:12 --> 00:43:18
The thing that's better about
them is that they instantly
648
00:43:16 --> 00:43:22
solve for you the initial value
problem.
649
00:43:18 --> 00:43:24
All you do is use this number,
initial condition as the
650
00:43:22 --> 00:43:28
coefficient of Y1,
and use this number as the
651
00:43:26 --> 00:43:32
coefficient of Y2.
Now, just check that by looking
652
00:43:29 --> 00:43:35
at it.
Why is that so?
653
00:43:32 --> 00:43:38
Well, for example,
let's check.
654
00:43:34 --> 00:43:40
What is its value of this
function at zero?
655
00:43:37 --> 00:43:43
Well, the value of this guy at
zero is one.
656
00:43:40 --> 00:43:46
So, the answer is y0 times one,
and the value of this guy at
657
00:43:44 --> 00:43:50
zero is zero.
So, this term disappears.
658
00:43:47 --> 00:43:53
And, it's exactly the same with
the derivative.
659
00:43:50 --> 00:43:56
What's the value of the
derivative at zero?
660
00:43:54 --> 00:44:00
The value of the derivative of
this thing is zero.
661
00:43:57 --> 00:44:03
So, this term disappears.
The value of this derivative at
662
00:44:01 --> 00:44:07
zero is one. And so,
the answer is y0 prime.
663
00:44:06 --> 00:44:12
So, check, check,
664
00:44:08 --> 00:44:14
this works.
So, these better solutions have
665
00:44:11 --> 00:44:17
the property,
what's good about them,
666
00:44:14 --> 00:44:20
and why scientists and
engineers like them,
667
00:44:18 --> 00:44:24
is that they enable you
immediately to write down the
668
00:44:22 --> 00:44:28
answer to the initial value
problem without having to go
669
00:44:26 --> 00:44:32
through this business,
which I buried down here,
670
00:44:30 --> 00:44:36
of solving simultaneous linear
equations.
671
00:44:35 --> 00:44:41
Okay, now, believe it or not,
that's all the work.
672
00:44:40 --> 00:44:46
We are ready to answer question
number two: why are these all
673
00:44:46 --> 00:44:52
the solutions?
Of course, I have to invoke a
674
00:44:50 --> 00:44:56
big theorem.
A big theorem:
675
00:44:53 --> 00:44:59
where shall I invoke a big
theorem?
676
00:44:56 --> 00:45:02
Let's see if we can do it here.
The big theorem says,
677
00:45:02 --> 00:45:08
it's called the existence and
uniqueness theorem.
678
00:45:05 --> 00:45:11
It's the last thing that's
proved at the end of an analysis
679
00:45:08 --> 00:45:14
course, at which real analysis
courses, over which students
680
00:45:11 --> 00:45:17
sweat for one whole semester,
and their reward at the end is,
681
00:45:15 --> 00:45:21
if they are very lucky,
and if they have been very good
682
00:45:18 --> 00:45:24
students, they get to see the
proof of the existence and
683
00:45:21 --> 00:45:27
uniqueness theorem for
differential equations.
684
00:45:24 --> 00:45:30
But, I can at least say what it
is for the linear equation
685
00:45:27 --> 00:45:33
because it's so simple.
It says, so,
686
00:45:31 --> 00:45:37
the equation we are talking
about is the usual one,
687
00:45:35 --> 00:45:41
homogeneous equation,
and I'm going to assume,
688
00:45:39 --> 00:45:45
you have to have assumptions
that p and q are continuous for
689
00:45:44 --> 00:45:50
all x.
So, they're good-looking
690
00:45:46 --> 00:45:52
functions.
Coefficients aren't allowed to
691
00:45:50 --> 00:45:56
blow up anywhere.
They've got to look nice.
692
00:45:55 --> 00:46:01
Then, the theorem says there is
one and only one solution,
693
00:46:03 --> 00:46:09
one and only solution
satisfying, given initial values
694
00:46:11 --> 00:46:17
such that y of zero,
let's say y of zero is equal to
695
00:46:19 --> 00:46:25
some given number, A,
696
00:46:24 --> 00:46:30
and y, let's make y0,
and y prime of zero equals B.
697
00:46:31 --> 00:46:37
The initial value problem has
698
00:46:37 --> 00:46:43
one and only one solution.
The existence is,
699
00:46:42 --> 00:46:48
it has a solution.
The uniqueness is,
700
00:46:45 --> 00:46:51
it only has one solution.
If you specify the initial
701
00:46:51 --> 00:46:57
conditions, there's only one
function which satisfies them
702
00:46:56 --> 00:47:02
and at the same time satisfies
that differential equation.
703
00:47:02 --> 00:47:08
Now, this answers our question.
This answers our question,
704
00:47:08 --> 00:47:14
because, look,
what I want is all solutions.
705
00:47:14 --> 00:47:20
What we want are all solutions
to the ODE.
706
00:47:21 --> 00:47:27
And now, here's what I say:
a claim that this collection of
707
00:47:33 --> 00:47:39
functions, c1 Y1 plus c2 Y2
are all
708
00:47:42 --> 00:47:48
solutions.
Of course, I began a period by
709
00:47:48 --> 00:47:54
saying I'd show you that c1
little y1 c little y2 are all
710
00:47:52 --> 00:47:58
the solutions.
But, it's the case that these
711
00:47:55 --> 00:48:01
two families are the same.
So, the family that I started
712
00:48:00 --> 00:48:06
with would be exactly the same
as the family c1 prime Y1
713
00:48:04 --> 00:48:10
because,
after all, these are two
714
00:48:07 --> 00:48:13
special guys from that
collection.
715
00:48:11 --> 00:48:17
So, it doesn't matter whether I
talk about the original ones,
716
00:48:15 --> 00:48:21
or these.
The theorem is still the same.
717
00:48:19 --> 00:48:25
The final step,
therefore, if you give me one
718
00:48:22 --> 00:48:28
more minute, I think that will
be quite enough.
719
00:48:26 --> 00:48:32
Why are these all the
solutions?
720
00:48:30 --> 00:48:36
Well, I have to take an
arbitrary solution and show you
721
00:48:35 --> 00:48:41
that it's one of these.
So, the proof is,
722
00:48:39 --> 00:48:45
given a solution,
u(x), what are its values?
723
00:48:43 --> 00:48:49
Well, u of x zero is u zero,
724
00:48:48 --> 00:48:54
and u prime of x zero, zero,
725
00:48:51 --> 00:48:57
let's say, is equal to u zero,
is equal to some other
726
00:48:58 --> 00:49:04
number.
Now, what's the solution?
727
00:49:02 --> 00:49:08
Write down what's the solution
of these using the Y1's?
728
00:49:07 --> 00:49:13
Then, I know I've just shown
you that u zero times Y1 plus u
729
00:49:12 --> 00:49:18
zero prime Y2
satisfies the same initial
730
00:49:17 --> 00:49:23
conditions, satisfies these
initial conditions,
731
00:49:21 --> 00:49:27
initial values.
In other words,
732
00:49:24 --> 00:49:30
I started with my little
solution.
733
00:49:27 --> 00:49:33
u of x walks up to and
says, hi there.
734
00:49:31 --> 00:49:37
Hi there, and the differential
equation looks at it and says,
735
00:49:37 --> 00:49:43
who are you?
You say, oh,
736
00:49:41 --> 00:49:47
I satisfy you and my initial,
and then it says what are your
737
00:49:46 --> 00:49:52
initial values?
It says, my initial values are
738
00:49:50 --> 00:49:56
u0 and u0 prime.
And, it said,
739
00:49:54 --> 00:50:00
sorry, but we've got one of
ours who satisfies the same
740
00:49:59 --> 00:50:05
initial conditions.
We don't need you because the
741
00:50:03 --> 00:50:09
existence and uniqueness theorem
says that there can only be one
742
00:50:09 --> 00:50:15
function which does that.
And therefore,
743
00:50:13 --> 00:50:19
you must be equal to this guy
by the uniqueness theorem.
744
00:50:20 --> 00:50:26
Okay, we'll talk more about
stuff next time,
745
00:50:23 --> 00:50:29
linear equation next time.