1
00:00:09,809 --> 00:00:17,590
Fine, so we will take a look at the Laplace's
equation last class we ended just short of
2
00:00:17,590 --> 00:00:25,039
my assigning a problem on Laplaceís equation,
right, so we will get back to that maybe along
3
00:00:25,039 --> 00:00:31,390
the way, so we will take a look at Laplace
equation and see what we are trying to do,
4
00:00:31,390 --> 00:00:47,140
okay Laplaceís equation, we will written
it in terms of phi, nabla squared phi equals
5
00:00:47,140 --> 00:00:52,760
0, right, so in 2D, maybe I will write 2D
Laplaceís equation.
6
00:00:52,760 --> 00:01:00,250
We will ourselves to 2D right now, 3D will
follow in a similar fashion, so in 2 dimensions
7
00:01:00,250 --> 00:01:09,710
Cartesian coordinates, this turns out to be
rho squared phi * x squared rho squared phi
8
00:01:09,710 --> 00:01:23,679
* y squared equals 0 and we wrote for an equally
spaced mesh and I am writing only for one
9
00:01:23,679 --> 00:01:35,689
set right now, you wrote for an equally spaced
mesh at the point i, j in terms of i + 1j,
10
00:01:35,689 --> 00:01:41,770
i - 1 j, ij ñ 1, ij + 1.
11
00:01:41,770 --> 00:01:48,290
We wrote this differential equation representation
let we could use on the computer as phi i
12
00:01:48,290 --> 00:02:10,509
+ 1j + phi ij + 1 either way + phi ij + phi
i 4 times phi ij, phi i ñ 1j + phi ij - 1
13
00:02:10,509 --> 00:02:18,980
equals 0, so this equation, the representation
at this point ij, right we have written in
14
00:02:18,980 --> 00:02:23,280
this fashion and we got a truncation error,
there is a representation error, we got the
15
00:02:23,280 --> 00:02:30,780
truncation error for that representation;
for this representation, okay.
16
00:02:30,780 --> 00:02:42,560
Now, right in the beginning - 4 very important;
-4, right in the beginning I would mentioned
17
00:02:42,560 --> 00:02:48,541
that this equation actually gives us allows
us when you are given a differential equation
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00:02:48,541 --> 00:02:54,329
in a sense, you are given a question, find
phi and if you find the phi, if somebody gives
19
00:02:54,329 --> 00:02:58,000
you a candidate phi, somebody says I have
a solution, you would verify it by substituting
20
00:02:58,000 --> 00:03:02,069
into this equation and checking whether they
have the solution or not, right.
21
00:03:02,069 --> 00:03:06,959
And if they give you something that is the
solution, the right hand side will turn out
22
00:03:06,959 --> 00:03:11,110
to be 0, the left hand side will be 0, it
will equal the right hand side. On the other
23
00:03:11,110 --> 00:03:15,680
hand if they give you something that is not
the solution, it will leave what is called
24
00:03:15,680 --> 00:03:21,489
a residue okay, so if I give you a phi, which
is not a solution, if I give you a function
25
00:03:21,489 --> 00:03:25,650
which is not a solution, you substitute it
into this equation, it would not give you
26
00:03:25,650 --> 00:03:28,140
a 0, it will leave a residue okay.
27
00:03:28,140 --> 00:03:34,959
I am repeating again just something I said
earlier in the semester, it leave a residue,
28
00:03:34,959 --> 00:03:41,500
the residue is what you get if you substitute
a candidate function, right into our differential
29
00:03:41,500 --> 00:03:46,849
operator here and it should be 0 but it is
not 0 and what you end up with is the residue,
30
00:03:46,849 --> 00:03:53,390
fine, you write our equation in this fashion
something equals 0, the left hand side should
31
00:03:53,390 --> 00:03:56,580
be 0, it is not 0, what is left is called
the residue.
32
00:03:56,580 --> 00:04:03,840
But clearly if on a mesh I give you discrete
points on a mesh like I mentioned in the last
33
00:04:03,840 --> 00:04:09,499
class, you cannot substitute back into the
original equation but you can substitute back
34
00:04:09,499 --> 00:04:19,970
into this, right, so if I were to give you
various values of phi ij, at the various grid
35
00:04:19,970 --> 00:04:25,150
points, you could actually substitute that
into this equation and find out whether this
36
00:04:25,150 --> 00:04:31,410
algebraic equation is satisfied. If the left
hand side is not 0 that is it leaves the residue.
37
00:04:31,410 --> 00:04:37,080
If it leaves the residue that means it does
not satisfies, am I making sense, so if you
38
00:04:37,080 --> 00:04:41,220
end up, if you substitute you have a candidate
solution you give me a candidate solution
39
00:04:41,220 --> 00:04:45,680
saying here is the solution, so the way I
verify it is I substitute it and I find out,
40
00:04:45,680 --> 00:04:51,150
I try to find out what I get and if I get
a residue, which I will call R ij because
41
00:04:51,150 --> 00:04:57,560
it is the residue at the point ij okay and
that residue is nonzero, so now I change it
42
00:04:57,560 --> 00:04:58,560
a little.
43
00:04:58,560 --> 00:05:02,930
Every time you give me a candidate solution,
I will substitute it into this equation and
44
00:05:02,930 --> 00:05:10,330
see evaluate the residue and I will ask the
question, is the residue 0, okay. If the residue
45
00:05:10,330 --> 00:05:14,030
is 0, you have given me a solution at that
point, if the residue is not 0, you have not
46
00:05:14,030 --> 00:05:19,500
given me a solution at that okay, what you
have given me is not a solution at that point.
47
00:05:19,500 --> 00:05:30,270
So, we have these 5 values okay, the algorithm
that I proposed in the last class which was
48
00:05:30,270 --> 00:05:47,470
the phi at ij is phi at i + 1j + phi at ij
+ 1 + phi at ij ñ 1 + phi at i ñ 1j times,
49
00:05:47,470 --> 00:05:53,190
this divided by but times 0.25, one fourth.
50
00:05:53,190 --> 00:06:00,190
So, what we could do is; if you give me something
so that the residue is not 0, what I could
51
00:06:00,190 --> 00:06:08,220
do is; I could reset the value at ij, I can
evaluate a value at ij as the average of these
52
00:06:08,220 --> 00:06:15,460
4 values and obviously, if I substitute back
in it will be 0, you understand. So, in a
53
00:06:15,460 --> 00:06:20,760
sense I have adjusted the value at ij, so
that it satisfies Laplace's equation the discrete
54
00:06:20,760 --> 00:06:27,980
form at that point okay, this is called relaxing,
it is called relaxing the process is called
55
00:06:27,980 --> 00:06:29,139
relaxation.
56
00:06:29,139 --> 00:06:34,880
I will relax the value of ij, I have relax
the value it is as though it is in tension
57
00:06:34,880 --> 00:06:40,639
and I relieve that tension, you understand,
I will relax the value of ij, I will relax
58
00:06:40,639 --> 00:06:45,840
the value of phi ij by taking the average
and substituting the average value there it
59
00:06:45,840 --> 00:06:51,259
is satisfied equation so it is obvious because
I got it from the same equation, it should
60
00:06:51,259 --> 00:06:57,720
be satisfied okay, right. So, what is the
algorithm that we are talking about now?
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00:06:57,720 --> 00:07:05,139
So, yesterday this is the problem that I started
off with at the unit square, a unit square
62
00:07:05,139 --> 00:07:16,949
at the origin, so I have the unit square at
the origin, we can divide it up into, if I
63
00:07:16,949 --> 00:07:22,900
remember right, I divided it up into 9; divided
up so that there are 9 interior grid points,
64
00:07:22,900 --> 00:07:31,449
you could; we could of course choose more
okay and what you do is sort any of the interior
65
00:07:31,449 --> 00:07:39,020
grid points, we do not have the values the
problem is defined, so that boundary conditions
66
00:07:39,020 --> 00:07:41,930
are given on all 4 sides.
67
00:07:41,930 --> 00:07:48,860
So, the boundary condition is prescribed on
all 4 sides, okay, so in theory right now
68
00:07:48,860 --> 00:07:52,669
you do not have, I have not assigned the problem,
so you could just assume any boundary condition,
69
00:07:52,669 --> 00:07:57,669
right you can just make assumptions on values
of phi and I say any boundary condition, values
70
00:07:57,669 --> 00:08:04,659
of phi on the 4 sides of this unit square,
okay, you can make an assumption on the values
71
00:08:04,659 --> 00:08:08,470
of phi at the 4 sides of this unit square,
so that sets your problem okay.
72
00:08:08,470 --> 00:08:14,379
So, you can take phi equals 100 here, maybe
phi equals 100 here, phi equals 200 there,
73
00:08:14,379 --> 00:08:19,320
phi equals 200, you can pick values either
constant or you can pick something varying
74
00:08:19,320 --> 00:08:23,530
as a quadratic curve right, you can pick values;
you can pick values along on the boundary
75
00:08:23,530 --> 00:08:30,639
only on the boundary and the way Laplace's
equation the problem is specified now that
76
00:08:30,639 --> 00:08:34,830
I given you the boundary conditions, the question
is to determine the values at the interior
77
00:08:34,830 --> 00:08:36,420
okay.
78
00:08:36,420 --> 00:08:41,190
And we propose to use the averaging process
that we just talked about right now okay,
79
00:08:41,190 --> 00:08:46,560
so the first candidate solution that I proposed;
the first candidate solution that I proposed,
80
00:08:46,560 --> 00:08:52,780
so I will give it a superscript 0 to indicate
that right it is the first guess that we have
81
00:08:52,780 --> 00:09:09,510
got at ij and ij are interior points, I will
say that it equals 0, i, j interior points,
82
00:09:09,510 --> 00:09:14,579
this is the first candidate solution that
I propose.
83
00:09:14,579 --> 00:09:18,850
So, if you were to substitute it into the
equation you would be left with the residue
84
00:09:18,850 --> 00:09:24,420
at every point if you check the residue, you
will find that in fact because these values
85
00:09:24,420 --> 00:09:30,010
are not necessarily 0, you may not have guessed
them to be 0 then as these are 0, you will
86
00:09:30,010 --> 00:09:39,430
find that 0 here is not the average value
okay, so this has to be relaxed fine, so you
87
00:09:39,430 --> 00:09:44,070
set this value to the average of the 4 neighbouring
values, am I making sense.
88
00:09:44,070 --> 00:09:49,540
So, we come up with the algorithm, I repeat
that again but this time with the superscript,
89
00:09:49,540 --> 00:10:09,190
so you say phi ij 1 is 1/ 4 th of phi i +
1j, 0 + phi ij + 1, 0 + phi ij ñ 1, 0 + phi
90
00:10:09,190 --> 00:10:22,790
ij; i ñ 1j, 0, is
91
00:10:22,790 --> 00:10:29,500
that fine okay and of course wherever it is
on the boundary, so wherever it is on the
92
00:10:29,500 --> 00:10:35,120
boundary you take the boundary values, wherever
it is on the boundary you will take boundary
93
00:10:35,120 --> 00:10:43,000
values, so for the very first one j - 1 will
be on the boundary, i ñ 1 will be on the
94
00:10:43,000 --> 00:10:47,779
boundary okay, for the very first one, everybody
is with me.
95
00:10:47,779 --> 00:11:00,190
So, in general what we could write is; in
general we can write phi ij n or n + 1, yes
96
00:11:00,190 --> 00:11:08,880
this is the relaxation scheme that we are
talking about n okay, at the end one, so we
97
00:11:08,880 --> 00:11:26,810
get n + 1 from the n1 + phi i + 1j, ij + 1,
n + phi ij ñ 1, n + phi i - 1j, n, is that
98
00:11:26,810 --> 00:11:32,690
fine, so I have a mechanism, I have an automaton,
I have a way by which I can now automate it,
99
00:11:32,690 --> 00:11:37,230
right I have something given the value I have
you; I have an equation given the values at
100
00:11:37,230 --> 00:11:40,510
n, I can find the values at n + 1, okay.
101
00:11:40,510 --> 00:11:46,880
So, what the mechanism that I have presumably
which we hope, the mechanism that I have is
102
00:11:46,880 --> 00:11:55,940
a mechanism that if you give me a guess, I
hope I can get a better guess, okay and given
103
00:11:55,940 --> 00:12:01,779
a guess if I can always get you a better guess
eventually, I will converge, so solution that
104
00:12:01,779 --> 00:12:06,589
is the hope, so what am I generating, what
am I proposing to generate here; I am saying
105
00:12:06,589 --> 00:12:11,320
you give me phi ij, so I will drop the i because
it is for all the interior ijís phi 0.
106
00:12:11,320 --> 00:12:21,541
And then you are going to generate phi 1,
phi 2, phi 3, phi n, you understand what I
107
00:12:21,541 --> 00:12:26,310
am saying, you are going to generate this
value; you are going to generate a sequence
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00:12:26,310 --> 00:12:32,959
of solutions, how do you check whether the
sequence converges or not? There is a various
109
00:12:32,959 --> 00:12:38,030
test that you are familiar with right you
could do; we could use the equivalent of a
110
00:12:38,030 --> 00:12:44,529
Cauchy test; equivalent of a Cauchy test,
right Cauchy test, basically says; you remember
111
00:12:44,529 --> 00:12:45,529
Cauchy test.
112
00:12:45,529 --> 00:12:55,630
Cauchy test basically says phi n - phi m will
be < than epsilon; epsilon will depend on
113
00:12:55,630 --> 00:13:07,830
some capital N for little n, little m > capital
N, for any little m and > n, once you have
114
00:13:07,830 --> 00:13:14,529
passed this capital N, right there is an epsilon
so that these only will be that close; will
115
00:13:14,529 --> 00:13:20,550
be close enough right, well we cannot actually
do this, so what we will do is; we will do
116
00:13:20,550 --> 00:13:21,750
an engineering approximation, right.
117
00:13:21,750 --> 00:13:29,769
So, we will basically ask ourselves the question
what happens to phi n + 1 ñ phi n, right,
118
00:13:29,769 --> 00:13:36,149
I will write it as our norm because given
these nodal points you can actually come up
119
00:13:36,149 --> 00:13:46,089
with functions okay and we want this to be
less than some prescribed epsilon; prescribed,
120
00:13:46,089 --> 00:13:53,970
so you will prescribe this epsilon okay, so
now it is clear, we are generating a sequence
121
00:13:53,970 --> 00:14:00,420
of phiís and all we have to do is check whether
the phi is converge, is that fine.
122
00:14:00,420 --> 00:14:03,930
The other way to do it; the other way to do
it since we are on convergence and then I
123
00:14:03,930 --> 00:14:10,100
will get back to the earlier or the other
way to do it is to look at R ij at each of
124
00:14:10,100 --> 00:14:18,480
the points, you will have similarly a sequence
of Rís, right you have a sequence of R ijís
125
00:14:18,480 --> 00:14:22,620
in fact, I will drop the idea just like I
did here, so you have a sequence of Rís which
126
00:14:22,620 --> 00:14:27,570
are the residues and what do you want to happen
to the residue, it should go to 0.
127
00:14:27,570 --> 00:14:32,390
So, we want the norm of the residual to go
to 0, so the other possibility is from here,
128
00:14:32,390 --> 00:14:43,779
we can say we want the norm of this to be
< epsilon prescribed, is that fine, so you
129
00:14:43,779 --> 00:14:48,670
can test whether this actually tells you whether
it satisfies the equation or not okay, so
130
00:14:48,670 --> 00:14:54,400
you can either say; you can either say I am
generating a sequence of phiís, do the phiís
131
00:14:54,400 --> 00:14:59,589
converge and generating a sequence of R to
the Rís converge okay.
132
00:14:59,589 --> 00:15:05,769
This will do for a first order right, for
now for you to start writing some code or
133
00:15:05,769 --> 00:15:09,889
whatever it is this is enough, we have to
actually do make this a little more precise
134
00:15:09,889 --> 00:15:15,320
okay but right now, this test will work for
you and test to see whether your code works
135
00:15:15,320 --> 00:15:29,190
or not fine. Are there any questions? Okay,
so yeah, so if there are no questions, we
136
00:15:29,190 --> 00:15:37,980
will just get back here, so what we have is
you will notice that the iteration okay, so
137
00:15:37,980 --> 00:15:46,480
each one of these; each pass through this
is called the iteration, okay.
138
00:15:46,480 --> 00:15:53,930
The phi that you get out of it just this is
the job then you have to learn this, the phi
139
00:15:53,930 --> 00:16:00,319
that you get out of it is called an iterate,
right in the process that you are going through
140
00:16:00,319 --> 00:16:07,440
is called an iteration okay, through each
iteration through each relaxation it is called
141
00:16:07,440 --> 00:16:11,930
a relaxation also, so it is a relaxation sweep,
right so you will hear people use the term
142
00:16:11,930 --> 00:16:21,970
relaxation sweep; relaxation sweep so it could
be; you could call it a relaxation sweep,
143
00:16:21,970 --> 00:16:22,970
right.
144
00:16:22,970 --> 00:16:30,350
So, each relaxation sweep or each iteration
will give you a new improved phi i at n +
145
00:16:30,350 --> 00:16:37,959
1, right, n + 1 iterate from the nth iterate
okay because this is the n plus first iterate
146
00:16:37,959 --> 00:16:43,110
from the nth iterate, it will go back there
to the; if that because it is the n plus first
147
00:16:43,110 --> 00:17:04,850
iterate from the nth iterate, it is called
simultaneous relaxation; simultaneous relaxation,
148
00:17:04,850 --> 00:17:22,160
you are basically saying that phi at n + 1
comes from phi at n, okay, is that okay.
149
00:17:22,160 --> 00:17:27,920
So, in fact I think as someone had pointed
out last time we look at this in greater detail
150
00:17:27,920 --> 00:17:33,530
it is actually a linear combination, so it
is some P times some matrix times phi at n,
151
00:17:33,530 --> 00:17:45,320
in fact that is what it is, okay fine now,
there is another possibility, so if I have
152
00:17:45,320 --> 00:17:59,530
phi ij n + 1, what I have done is; I have
done
153
00:17:59,530 --> 00:18:08,630
if you look at if you come here and you look
at this, when I do, when I relax this point,
154
00:18:08,630 --> 00:18:13,380
I take the average of these 4 points and I
get a value here.
155
00:18:13,380 --> 00:18:20,460
So, what I can do is; I can replace the value
here; I can replace the value at that point
156
00:18:20,460 --> 00:18:26,120
by what I have just calculated because it
satisfies right, it satisfies Laplace's equation
157
00:18:26,120 --> 00:18:31,820
at that point, my approximation to the Laplace
equation. So, when I come here I can use the
158
00:18:31,820 --> 00:18:37,300
latest value that I have got, I do not have
to use the old value, I can use the latest
159
00:18:37,300 --> 00:18:41,050
value as and when I get it, you understand
what I am saying, okay.
160
00:18:41,050 --> 00:18:45,340
So, then we are not relaxing, they are not
at the same level, it is not simultaneous
161
00:18:45,340 --> 00:18:49,840
relaxation, it is called successive relaxation,
so when I come to this point I can use the
162
00:18:49,840 --> 00:18:56,560
latest value there and consequently get the
latest value at that point, right and repeat
163
00:18:56,560 --> 00:19:00,960
that get the latest value at that point, so
as you progresses, as I progress through taking
164
00:19:00,960 --> 00:19:05,600
the averages; as I progressed through taking
the averages, you will see that I am using
165
00:19:05,600 --> 00:19:07,550
the latest value at any given point.
166
00:19:07,550 --> 00:19:14,850
So, when I come here what happens? The i - 1
value is the latest value in general and the
167
00:19:14,850 --> 00:19:21,820
j - 1 value is the latest value in general,
okay, so I can in fact write this as phi i
168
00:19:21,820 --> 00:19:39,940
+ 1j, phi ij + 1 at n + phi ij - 1 n + 1 +
phi i ñ 1j n + 1, is that fine and this is
169
00:19:39,940 --> 00:19:47,230
called successive relaxation; this is called
successive relaxation, this is just a name
170
00:19:47,230 --> 00:19:50,760
right but when people talk about successive
relaxation you should understand that they
171
00:19:50,760 --> 00:19:59,180
are talking about using the latest value okay,
fine.
172
00:19:59,180 --> 00:20:06,090
The first one simultaneous relaxation is also
called Jacobi, see that the depending on which
173
00:20:06,090 --> 00:20:09,400
direction you are coming from they have different
names and they are attributed to the people
174
00:20:09,400 --> 00:20:19,030
that have brought up the thought of the algorithm
Jacobi iteration, simultaneous relaxation
175
00:20:19,030 --> 00:20:33,040
is also called Jacobi iteration, is that fine,
okay. Now, how do we figure out whether the
176
00:20:33,040 --> 00:20:37,090
codes working or not, how do you find out
whether the code works or not?
177
00:20:37,090 --> 00:20:45,450
How do you test to see whether the code works
or not, you know how to check convergence,
178
00:20:45,450 --> 00:20:53,790
if it converges does that mean that if it
converges does that mean we have a solution,
179
00:20:53,790 --> 00:20:59,430
see we have the following questions; how do
we know our code is working, how do you know
180
00:20:59,430 --> 00:21:04,630
if it is converging that you are getting the
right answer right, so we need to answer these
181
00:21:04,630 --> 00:21:11,580
questions, how do you know if 2 of you get
2 different answers, you right run the program
182
00:21:11,580 --> 00:21:16,720
and the 2 of of you get 2 different answers,
which answer is the right answer.
183
00:21:16,720 --> 00:21:21,000
Is there a way for us to decide right, is
it possible that 2 people get 2 different
184
00:21:21,000 --> 00:21:26,310
answers, am I making sense these are natural
questions that you have to ask yourself because
185
00:21:26,310 --> 00:21:30,920
if you go out and you start solving this problem,
right and 2 different people are solving the
186
00:21:30,920 --> 00:21:34,530
same problem, is it possible that somehow
they end up with 2 different answers, right,
187
00:21:34,530 --> 00:21:42,550
why would you say it is not possible but how
do we know that?
188
00:21:42,550 --> 00:21:51,600
Is there a way we can show that so, you are
saying Laplace; the solution to Laplace's
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equation is unique okay, solutions but how
it about the discretization, for approximation
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is the solution to our approximation unique,
you understand, so in a partial differential
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equations course, you may have learned that
the solution to Laplace's equation is unique
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okay, that the solution exists when that it
is unique.
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But how about for our discretization right,
you understand because we are doing; we are
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only approximating, so is it possible that
we have already seen when we talked about
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machine epsilon that there are a bunch of
numbers that are represented by one number,
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so we have to have a little anxiety saying
that I may; are we going to get the same answer
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or all of us going to get the same answer,
okay.
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So, we have these set of questions that need
to be answered, so first we look at; is there
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a way for us when you are given an equation,
nabla squared phi equals 0, and we have a
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discrete representation for this equation,
is there a way for us to set up the problem,
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so that we can check whether the program that
we write is generating, so we need an answer,
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we need a solution to this.
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If you have a solution to this, you are set
okay, so Laplace equation fortunately is easy
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that is why I have pick Laplace equation is
the first problem that we look at right, any
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analytic function is a solution to Laplace's
equation right, so the simplest thing is to
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take z or z squared z is a; z being a complex
number is not that interesting z squared s,
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so we can take either real or imaginary of
z squared, let us start, I want to keep it
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simple.
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So, if a solution is x squared - y squared
and I invariably start with something like
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this because I want to keep it easy, so if
I were to solve this problem I will pick a
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simple, so is this a solution to Laplace's
equation, you can substitute and see and it
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is indeed a solution to Laplace's equation,
so all the boundaries here; all the boundaries
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here, we can pick the boundary condition as
it comes from x squared - y squared.
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So, on the bottom boundary, you can set phi
of x, 0 equals x squared on the right boundary
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here, you can set phi of 1, y as 1 - y squared
on the top boundary, we can prescribe the
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boundary phi of x, 1 is x squared ñ 1; sorry,
x squared - 1 and on the left; phi of 0, y
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is - y squared and then you would expect that
these points would be samples from x squared
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- y squared they would actually be sampling
the function x squared ñ y squared.
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00:25:06,510 --> 00:25:10,350
The solutions here should correspond to x
squared - y squared, so this is something
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that you can try out okay and now this is
something that you can try out, you have an
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answer to the problem and therefore, you can
test to see whether your code actually converges
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to that also, is that fine okay, right and
you can see how well it; how well does it
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if it converges, how well does it converge
is it; does it go actually go to the answer
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what does it go to, right you can find that,
okay.
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Second question is everybody going to get
x squared - y squared, is it possible that
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the numeric goes somewhere else, okay so we
will do; we will we will do something that
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mirrors, we will do something that mirrors
the typical; the continuous function the derivation
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that you have seen and continuous function
I will just draw this here.
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So, if you have these 4 points what are we
doing here, what is at the point ij, at any
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given point ij at the interior, what is phi
ij? Phi ij is the average of its neighbours,
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so what is an average, the average is not
larger than the largest neighbour nor is it
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smaller than the smallest neighbour, so any
given point here let we take the average for
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actually, I guess I could have done it just
here any given point that you take the average,
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for any given point for which you take the
average is not larger than its largest neighbour
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nor smaller than its smallest neighbours.
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00:26:44,650 --> 00:26:48,580
So, this point, this is not larger than its
larger neighbour, smaller than and this is
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true of all the interior points, it is true
of all the interior points okay and therefore
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can I conclude that the maximum and minimum
will actually occur on the boundary, the only
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values that I do not touch that I do not change
are in the boundary, every other point I ensure,
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every other interior, every point on the interior
I ensure is larger than the smallest neighbour
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and smaller than the largest neighbour, you
understand.
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00:27:17,930 --> 00:27:23,210
And all of these satisfy that condition they
are all larger than the smallest neighbour
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and smaller than the largest neighbour consequently,
the maximum and minimum to the solution must
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occur on the boundary fine, okay so at maximum
and minimum over the solution occurred on
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00:27:34,390 --> 00:27:42,800
the boundary what can we say; the maximum
and minimum occur on the solution on the boundary
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00:27:42,800 --> 00:27:45,780
of the solution they occur on the boundary
what can we say?
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00:27:45,780 --> 00:27:52,550
We will use this argument now, this is called
the maximum principle, I just write, I am
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saying this just for okay and you just say
maximum principle okay, fine, so what we are
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saying is if I instead of writing the system
of equations, so if I have, I still write
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it as nabla squared phi equals 0, so I will
replace this by some; what shall I call it;
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00:28:24,430 --> 00:28:34,880
L of h phi equals 0, right, so this is some
system of equations that is what we called
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00:28:34,880 --> 00:28:41,140
it right, h is the grid size okay.
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And the discussion that we are having now
is; is it possible for 2 people; is it possible
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for two different students to get 2 different
answers and it is important that L is a linear
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00:28:55,000 --> 00:28:58,890
okay that we get a system of equations, is
it possible now for 2 people to get 2 different
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00:28:58,890 --> 00:29:05,330
answers. So, let us say 2 people; 2 different
students get 2 different answers okay.
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00:29:05,330 --> 00:29:11,130
So, the 2 answers; 2 different students get
2 different answers; one gets the answer capital
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00:29:11,130 --> 00:29:18,030
phi 1, so I am making it capital phi, so that
we do not confuse it with our iterates okay
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and the other gets the answer capital phi
2, okay, so we do the usual mathematical trick
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because we know we have a linear equation,
what are we going to do? We are going to look
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00:29:30,280 --> 00:29:41,630
at what is the difference between the 2 answers;
phi 1 ñ phi 2.
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00:29:41,630 --> 00:29:48,710
Because the equation is linear, this is also
a solution to Laplace's equation, right because
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the equation is linear is also solutions,
so if this is some phi, this is the difference
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00:29:56,670 --> 00:30:14,820
L h phi is Lh phi 1 ñ Lh phi 2, you understand
because it is linear, the L actually distribute
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00:30:14,820 --> 00:30:21,870
across it because it is linear, it is just
a system of equation linear system of equations
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00:30:21,870 --> 00:30:28,140
in this case and therefore this satisfies
Laplace's equation, right.
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00:30:28,140 --> 00:30:35,020
What are the boundary conditions that it satisfies?
Remember both phi 1 and phi 2 satisfy my x
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squared - y squared boundary condition therefore,
phi 1 ñ phi 2 the boundary condition is 0,
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00:30:40,290 --> 00:30:44,850
now we use the maximum principle, the maximum
and minimum occurred on the boundary and that
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00:30:44,850 --> 00:30:52,550
is 0, right. So, the only function for the
maximum and minimum occur on the boundary
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00:30:52,550 --> 00:30:57,840
and that is 0, this phi is identically 0,
is that clear, okay.
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00:30:57,840 --> 00:31:02,170
So, as I said this sort of mirrors, the proof
that you would have seen possibly and for
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00:31:02,170 --> 00:31:08,770
the differential equation itself, so yes it
will turn out that if you do; if you solve
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this problem; if 2 different students get
2 different answers, the possibilities are
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both of you have the wrong answer, right that
is one obvious possibility or one of you has
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00:31:19,700 --> 00:31:24,680
the right answer, right and one of you has
the wrong answer, okay right.
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00:31:24,680 --> 00:31:30,870
After you have been programming enough, you
start with the feeling that you do not start
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with the confidence, my answer is right, you
are feeling always is that slightly that your
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answer is wrong, you have to start with the
assumption, the presumption is my answer is
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00:31:38,000 --> 00:31:43,170
wrong and you have to work to show to yourself,
convince yourself that the answer is right,
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is that okay fine, great.
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00:31:46,910 --> 00:31:58,070
So, here we have; what should I say, we have
shown that they have shown that the solution
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00:31:58,070 --> 00:32:04,120
the answer is unique, we need to show now
whether the scheme always converges we look
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00:32:04,120 --> 00:32:09,960
at that right but before we go there, I have
just quietly written this as Lh, let us figure
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00:32:09,960 --> 00:32:14,370
out what is this L, what is the nature of
this L, what is the nature of this operator
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00:32:14,370 --> 00:32:17,430
L, okay fine.
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00:32:17,430 --> 00:32:30,190
So, back here, so there are 2 possible numbering
schemes that you can; the different ways by
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00:32:30,190 --> 00:32:36,000
which you can number these, okay intuitively
if you look at the way I have written this,
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00:32:36,000 --> 00:32:42,050
if you look at the way I have made these pink
coloured dots, the assumption is that I have
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00:32:42,050 --> 00:32:48,050
numbered this, this possibly is 1, 1 or 0,
0 or whatever it is and you know I have numbered
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00:32:48,050 --> 00:32:52,650
them in an increasing fashion of y and an
increasing fraction of j, right.
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00:32:52,650 --> 00:32:56,890
In fact, I do not even have to tell you instinctively
a lot of you will assume that is what I am
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00:32:56,890 --> 00:33:05,030
doing, okay so we number this in this sequential
fashion, right that is what we have done but
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00:33:05,030 --> 00:33:10,081
is it possible for me having gone through
this process, I now understand that it is
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00:33:10,081 --> 00:33:15,150
only the interior points that are unknown,
they need to be determined, so possibly I
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00:33:15,150 --> 00:33:23,540
can number the interior points in a sequential
fashion
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00:33:23,540 --> 00:33:26,770
and then number the boundary points okay,
fine.
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00:33:26,770 --> 00:33:36,910
So, one way to do it is you say, you use 2
subscripts, the other way to do it is that
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00:33:36,910 --> 00:33:42,470
you basically number the boundary points after
this, so that is 9, so and this becomes 10,
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00:33:42,470 --> 00:34:04,200
11, 12, 13, 14 and so on okay, you could also
number them in a random fashion, would that
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00:34:04,200 --> 00:34:11,300
make a difference do you think? So, this is
something for you to think about, if I what
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00:34:11,300 --> 00:34:18,460
should I do this sequentially would it matter
if I started my iterations from the top.
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00:34:18,460 --> 00:34:24,000
Would it matter, if I iterated, I relaxed
this point, this point, this point and did
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00:34:24,000 --> 00:34:31,650
it in the reverse fashion, okay or can I just
do it in random; can I just do it at random,
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00:34:31,650 --> 00:34:37,071
you should always get the answer. The question
that still there we should always we would
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00:34:37,071 --> 00:34:40,910
expect that we will get the answer, we will
look at; we will see whether we can throw
307
00:34:40,910 --> 00:34:44,840
some light on that idea, right now intuitively,
I am getting a lot of head shake saying that
308
00:34:44,840 --> 00:34:46,710
you do not expect.
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00:34:46,710 --> 00:34:49,480
I do not expect it either, you do not expect
from what we have seen, you do not expect
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00:34:49,480 --> 00:34:52,550
that the answer will change, we are taking
averages, you do not expect that the answer
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00:34:52,550 --> 00:34:57,800
will change, so there are different ways by
which you can sweep through the domain not
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00:34:57,800 --> 00:35:02,880
necessarily from left to right, bottom to
top not necessarily in that direction that
313
00:35:02,880 --> 00:35:07,310
you can sweep in any direction and instinctively,
the way the number it is the way we tend to
314
00:35:07,310 --> 00:35:08,310
sweep.
315
00:35:08,310 --> 00:35:13,210
And therefore the way we number it becomes
important okay, so we will definitely get
316
00:35:13,210 --> 00:35:18,070
to the answer, then the second question is
how fast we get to the answer, right okay,
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00:35:18,070 --> 00:35:21,590
so we look at those issues right now, let
us stick to this, while numbered it in this
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00:35:21,590 --> 00:35:29,150
fashion; 1, 2, 3, 4, 5, 6 and we can write
for each one of these, we can write an equation
319
00:35:29,150 --> 00:35:32,240
and as a consequence, we will get a linear
system of equations.
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00:35:32,240 --> 00:35:44,350
So, for each one of those, we write an equation
for the first one, I get phi at 2 + phi at
321
00:35:44,350 --> 00:36:04,140
4 or phi at 4 + phi at 2, it does not matter
- phi at 1, if you want ñ 4, I keep forgetting,
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00:36:04,140 --> 00:36:14,280
let me start the other way around. I will
do it in a sequential fashion; - 4 phi 1 +
323
00:36:14,280 --> 00:36:32,720
phi 2 + phi 4 + phi 25 + phi 11 equals 0,
is that fine, what is the next one turned
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00:36:32,720 --> 00:36:50,860
out to be; phi 1 ñ 4phi 2 + phi 3 + phi 6
or phi 5 + phi 12 equals 0, am I making sense.
325
00:36:50,860 --> 00:37:20,300
We will write one more, so there is no phi
1; phi 2 ñ 4phi 3 + phi 6 and then you get
326
00:37:20,300 --> 00:37:30,060
2 boundary points; + phi 15 + phi 13 equals
0, of course the boundary points we are going
327
00:37:30,060 --> 00:37:35,540
to take over to the right hand side, okay.
In general, if I write for a general point
328
00:37:35,540 --> 00:37:40,190
in between as I go down to an interior point
which is only surrounded by interior points,
329
00:37:40,190 --> 00:37:48,520
I will get phi, if I am at the point i, remember
I have only one subscript now, okay.
330
00:37:48,520 --> 00:38:09,150
Phi i - 1 + phi I ñ 4 phi i + phi + 1 that
is the easy one, so you can see the second
331
00:38:09,150 --> 00:38:15,470
derivative; i + 1, i ñ 1, then what do you
get; + phi i + n, where n is the number of
332
00:38:15,470 --> 00:38:24,930
points in that row in a given row, so phi
1, there were 3 in the row; 1 + 3 gives me
333
00:38:24,930 --> 00:38:33,110
4; 2 + 3 gives me 5, 3 + 3 gives me 6, okay
there are 3 in that row and subsequently I
334
00:38:33,110 --> 00:38:42,890
will get a phi i - n and this should equals
0, this is a general situation, okay.
335
00:38:42,890 --> 00:38:47,940
Coming back here, if you look at it, if I
have a general point if there are n of them
336
00:38:47,940 --> 00:38:57,843
in this row, this is -1, this is +1, this
is ñ n, there will be; you will have to go
337
00:38:57,843 --> 00:39:03,710
back n, this is - n that is + n, okay, so
with a single subscript, we can actually go
338
00:39:03,710 --> 00:39:17,370
i ñ 1, i + 1, i ñ n, i + n, is that fine,
so we can really write this as a system of
339
00:39:17,370 --> 00:39:23,590
equations and really write this as a system
of equations maybe I will write it in a bigger
340
00:39:23,590 --> 00:39:25,010
fashion in a system of equations.
341
00:39:25,010 --> 00:39:35,740
I will write it there in that first, can really
write it as a system of equations, which will
342
00:39:35,740 --> 00:39:50,890
be -4, 1 and what; whole bunch of 0ís, so
this is the first one then you have to go
343
00:39:50,890 --> 00:39:57,380
to i + 1, i + n, so when you get to i + n,
there will be a 1 there, lots of 0ís, this
344
00:39:57,380 --> 00:40:06,560
is the structure of the matrix that you get,
what is the next one? 1 ñ 4, 1, lots of 0ís
345
00:40:06,560 --> 00:40:18,520
till you get to 1 on the diagonal there, lots
of 0ís, is that fine and you get 0, 1, -4,
346
00:40:18,520 --> 00:40:22,800
1, lots of 0ís, 1 on the diagonal.
347
00:40:22,800 --> 00:40:31,770
So, the equation keeps on shifting to the
right okay, lots of 0ís after that and this
348
00:40:31,770 --> 00:40:39,310
will keep on shifting to the right till you
get to a point where a 1 will appear, right
349
00:40:39,310 --> 00:40:44,590
till you get to a point where ever you gone
to the second row, now from the first row
350
00:40:44,590 --> 00:40:48,900
something is contributing, while you are in
the first row, the one below goes to the right
351
00:40:48,900 --> 00:41:01,970
hand side okay, so will you get lots of 0ís,
1, -4, 1, lots of 0ís, 1, lots of 0ís.
352
00:41:01,970 --> 00:41:10,730
In case, this does not make sense, I am going
to erase this, I will write it as a; I will
353
00:41:10,730 --> 00:41:23,520
just draw lines, so you get a - 4 on the diagonal
and we are all agreed that -4 is there for
354
00:41:23,520 --> 00:41:30,070
all the points right, so you get a - 4 on
the diagonal, there is nothing to the left
355
00:41:30,070 --> 00:41:35,890
of this but you will get a +1 on the super
diagonal; on the first diagonal above the
356
00:41:35,890 --> 00:41:48,190
diagonal, fine and you will get a +1, so there
is a -4 here, you will get a 1 here on the
357
00:41:48,190 --> 00:41:51,850
sub diagonal, fine.
358
00:41:51,850 --> 00:41:58,230
So, that takes care of the second derivative
in the x direction in a primary coordinate
359
00:41:58,230 --> 00:42:07,610
direction then what happens, then you have
n away from this, you have a 1 all of these
360
00:42:07,610 --> 00:42:17,190
in between of 0ís and all 0ís, okay and
you are going to have a bunch of 1ís till
361
00:42:17,190 --> 00:42:20,730
you come to the right boundary, you get to
the right boundary or the top boundary, right
362
00:42:20,730 --> 00:42:26,450
when you get to the top boundary or the right
boundary at each of the boundaries you have
363
00:42:26,450 --> 00:42:28,170
to be a bit careful.
364
00:42:28,170 --> 00:42:34,590
So, here n away, you will get a 1 there, am
I making sense, okay this 1 at the bottom
365
00:42:34,590 --> 00:42:46,510
and you will get 1ís along this, 1ís along
this and 1ís along this, this corresponds
366
00:42:46,510 --> 00:42:53,100
to the top boundary, this corresponds to the
bottom boundary, there are 0ís here which
367
00:42:53,100 --> 00:42:58,590
correspond to the bottom boundary, there are
0ís here that correspond to the top boundary,
368
00:42:58,590 --> 00:43:03,220
here as you go along just like you did not
have any entry here, as you go along, as you
369
00:43:03,220 --> 00:43:07,110
come to a left entry or a right entry, you
will get a 0, as you come to the left boundary
370
00:43:07,110 --> 00:43:08,240
or the right boundary you will get a 0.
371
00:43:08,240 --> 00:43:15,210
Does it make sense, I still see a few faces
that are confused, when you have, when you
372
00:43:15,210 --> 00:43:21,620
read the boundary, then you go when your traverse
along this when you come to this, there is
373
00:43:21,620 --> 00:43:27,530
no, the entry here is known, it is not an
unknown, so that goes to the right hand side,
374
00:43:27,530 --> 00:43:32,770
so you do not get a +1 that will be 0 just
like that with the very first one, there will
375
00:43:32,770 --> 00:43:37,930
be no i - 1 entry that will also be right,
there will be no i - 1 entry that will also
376
00:43:37,930 --> 00:43:40,310
be 0, okay.
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00:43:40,310 --> 00:43:49,350
You go to the top, there is no i + n entry,
when you go to the bottom there is no i ñ
378
00:43:49,350 --> 00:43:54,680
n entry, okay, the points right next to the
right boundary has no i + n entry, the points
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00:43:54,680 --> 00:43:59,820
right next to the left boundary have no i
- n entry, the ones at the bottom will have
380
00:43:59,820 --> 00:44:08,650
no i - n entry, no i + n entry, okay, so that
the no i + n entry, no i - n entry and then
381
00:44:08,650 --> 00:44:12,540
embedded in here, there will be 0ís with
a frequency of n, every time you go through,
382
00:44:12,540 --> 00:44:15,230
you come to the end there will be a 0.
383
00:44:15,230 --> 00:44:18,940
Because there is no right hand, there is no
point to the right, it is a known and all
384
00:44:18,940 --> 00:44:27,440
of those points will come to the right hand
side, so this equation this matrix A, what
385
00:44:27,440 --> 00:44:37,350
does it multiply? It multiplies phi1, phi2,
phi3; phi subscript 1, phi subscript 2, phi
386
00:44:37,350 --> 00:44:46,940
i, phi, I
make it capital M because I n the number of
387
00:44:46,940 --> 00:44:58,360
values in the row, phi M; where M is interior
grid points and this equals; these will be
388
00:44:58,360 --> 00:45:13,190
the values from the boundary conditions, known
values from boundary okay, these are prescribed,
389
00:45:13,190 --> 00:45:14,530
fine.
390
00:45:14,530 --> 00:45:25,060
So, we actually get something that looks like
A phi or Ax equals b, there are system of
391
00:45:25,060 --> 00:45:37,750
equations okay, some observations, the equations
are symmetric about the diagonal, equation
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00:45:37,750 --> 00:45:43,730
that we get; the matrix that we get A is symmetric
about the diagonal, the diagonal entries are
393
00:45:43,730 --> 00:45:50,680
negative, the off diagonal entries are positive,
so it has some nice structure to it okay,
394
00:45:50,680 --> 00:45:53,000
you can just go look it up, I am not going
to do this.
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00:45:53,000 --> 00:46:00,650
It falls into the category of something that
is known as an M matrix, so just check that
396
00:46:00,650 --> 00:46:06,162
out, so the diagonal entries are negative,
off diagonal entries are positive, it is symmetric
397
00:46:06,162 --> 00:46:09,540
right, so there must be maybe there is something
need that we can do with this, we can check
398
00:46:09,540 --> 00:46:15,310
to see whether what kind of; right, what it
is that we are but we have a system of equations,
399
00:46:15,310 --> 00:46:17,580
you have a system of equations.
400
00:46:17,580 --> 00:46:27,310
Let us now see whether, what it is that we
are actually doing, when we solve the system
401
00:46:27,310 --> 00:46:32,020
of equations okay; what is it that we are
actually doing when we solve the system of
402
00:46:32,020 --> 00:46:54,910
equations. So, when we do Jacobi iteration,
okay what are we doing? The matrix A can be
403
00:46:54,910 --> 00:47:07,280
partitioned as the diagonal, a diagonal matrix
D + the lower triangular matrix + an upper
404
00:47:07,280 --> 00:47:10,560
triangular matrix, just to give you an example.
405
00:47:10,560 --> 00:47:23,960
The diagonal matrix D is - 4 times I, right
the diagonal matrix D is - 4 times I, fine,
406
00:47:23,960 --> 00:47:29,900
the remaining part lower and upper triangular
you can extract, I think that you can do,
407
00:47:29,900 --> 00:47:36,580
so when we do Jacobi iteration, what we are
actually doing this; we are saying phi n +
408
00:47:36,580 --> 00:47:53,190
1 equals; you can check this out D inverse
phi n - ; well maybe we will work this out
409
00:47:53,190 --> 00:47:55,600
next time, I think they are coming close to.
410
00:47:55,600 --> 00:48:02,690
So, what you can do is; you can just take
the equation; phi n + 1 equals; we are trying
411
00:48:02,690 --> 00:48:07,380
to find just give you an idea as to where
we are going, we are trying to find out what
412
00:48:07,380 --> 00:48:12,750
is the P, okay, we are trying to find out
what is this P, so that we can say something
413
00:48:12,750 --> 00:48:19,750
about how this equation converges, so far
what we are shown is that if you get a solution,
414
00:48:19,750 --> 00:48:24,270
it will be the solution and it is unique,
you understand what I am saying.
415
00:48:24,270 --> 00:48:29,570
Now, what we want to do is; we want to see
whether does it actually converge, how fast
416
00:48:29,570 --> 00:48:34,260
does it converge, how fast does it get there,
right, now we are talking about the operational
417
00:48:34,260 --> 00:48:37,960
part, it is nice to know that if 2 people
get 2 different solutions right, I mean we
418
00:48:37,960 --> 00:48:42,020
know that you everything works 2 people should
not get 2 different solutions, the solution
419
00:48:42,020 --> 00:48:48,320
is unique. Now, we want to know that we want
to answer the question how fast do I get it.
420
00:48:48,320 --> 00:48:51,890
I would like to get it today, I would like
to get it in a few minutes, I do not want
421
00:48:51,890 --> 00:48:58,850
to take for hour to get the answers, how fast,
what can I; the issue is if I can figure out,
422
00:48:58,850 --> 00:49:02,950
if I can answer the question how fast will
it, how fast does it; how long does it take
423
00:49:02,950 --> 00:49:09,590
to get it then I can if there is a systematic
way by which I can do it then I can ask the
424
00:49:09,590 --> 00:49:13,850
question, why is it taking for so much time,
is there something that I can do to make it
425
00:49:13,850 --> 00:49:15,550
faster, okay.
426
00:49:15,550 --> 00:49:22,420
Remember, always ask how good is it and if
you are able to determine to answer that question,
427
00:49:22,420 --> 00:49:26,980
right the objective of asking how good does
it is not just find out how good is it but
428
00:49:26,980 --> 00:49:31,480
if we understand the question in the way we
answered it, it may help us improve right,
429
00:49:31,480 --> 00:49:36,480
whatever it is that we are asking, how well
does it converge and we can figure out right,
430
00:49:36,480 --> 00:49:41,400
a way to improve the convergence if you are
unhappy with the convergence, is that fine
431
00:49:41,400 --> 00:49:42,400
okay.
432
00:49:42,400 --> 00:49:47,730
So, I will see you in tomorrow's class, we
will look at this Jacobi iteration and see
433
00:49:47,730 --> 00:49:49,900
what is happening in tomorrowís class. Thank
you.