1
00:00:14,969 --> 00:00:23,849
Hi, so we have discussed in the last class,
about the finite difference schemes and the
2
00:00:23,849 --> 00:00:31,570
corresponding error estimates. So, let us
proceed further little bit on that, with respect
3
00:00:31,570 --> 00:00:36,350
to
the stability aspects. So, and then we generalise
4
00:00:36,350 --> 00:00:41,260
the second order method.
.
5
00:00:41,260 --> 00:00:59,409
So, stability, consider
so this equation so which we have derived
6
00:00:59,409 --> 00:01:03,650
so where A is the
corresponding tri diagonal matrix and E is
7
00:01:03,650 --> 00:01:10,450
the error and error, global error matrix and
then, this is the local truncation error matrix.
8
00:01:10,450 --> 00:01:26,580
So, this can be, this can be
rewritten as A h,
9
00:01:26,580 --> 00:01:39,060
E h so this is just to know that for a specific
h.
10
00:01:39,060 --> 00:01:58,790
So, definitely the matrix A h so this is with
h equals to 1 by and plus 1 for 0 1 case and
11
00:01:58,790 --> 00:02:31,310
the dimension of Ah, grows as this happens.
Let Ah inverse exists, this implies Eh is
12
00:02:31,310 --> 00:03:14,490
then, the nor is equals to, but we have this,
as order of h square. Therefore, we expect,
13
00:03:14,490 --> 00:03:32,340
we expect
the same for. But, from this if you see for
14
00:03:32,340 --> 00:03:44,950
this, to have this order, this must be
bounded by a constant.
15
00:03:44,950 --> 00:03:45,950
..
16
00:03:45,950 --> 00:04:10,010
So, for
this to behave in this order, this be
17
00:04:10,010 --> 00:04:20,220
independent of h as h goes to 0 say, less
than
18
00:04:20,220 --> 00:04:54,440
or equals to c, for sufficiently small h then,
we get. So, this suggests a definition that
19
00:04:54,440 --> 00:05:00,810
means so global error is bounded so we know
local error is order of h square and global
20
00:05:00,810 --> 00:05:18,860
error is also bounded like that. Hence, the
stability is ensured. Suppose a finite difference
21
00:05:18,860 --> 00:05:59,800
method, for a linear boundary value problem
gives a sequence of matrix equations
22
00:05:59,800 --> 00:06:52,740
where, h is mesh width. We say that, the method
is stable if this exists for all sufficiently
23
00:06:52,740 --> 00:07:20,550
small h so that is say h less than h naught
say. And if there exists constant c, independent
24
00:07:20,550 --> 00:07:31,990
of h such that
is less than equal to c, for every h. This
25
00:07:31,990 --> 00:07:39,270
is, this will suggest the stability
condition.
26
00:07:39,270 --> 00:07:56,190
So, once stability is done we need to, consistency
so we have all talked before in the
27
00:07:56,190 --> 00:08:21,949
context of multi-step methods. So, we say
that a method is consistent with the difference,
28
00:08:21,949 --> 00:08:48,060
method is consistent with the differential
equation and boundary conditions, if the local
29
00:08:48,060 --> 00:09:04,790
truncation error goes to 0. As h goes to 0
and in general, this is order h power p, p
30
00:09:04,790 --> 00:09:20,910
greater
than 0 then, then certainly consistent then
31
00:09:20,910 --> 00:09:25,549
what next, convergence.
32
00:09:25,549 --> 00:09:26,549
..
33
00:09:26,549 --> 00:09:35,379
So, since we talked on, with respect to multi
step method I do not want to repeat just
34
00:09:35,379 --> 00:09:55,639
consistency plus stability implies convergence.
So, with this we have some idea of really
35
00:09:55,639 --> 00:10:03,160
how a particular approximation is giving you
sensible results. Because, you have taken
36
00:10:03,160 --> 00:10:05,550
a
differential equation and then pick up the
37
00:10:05,550 --> 00:10:11,300
derivatives, approximate and then we try to
solve system and get the corresponding solutions.
38
00:10:11,300 --> 00:10:17,259
But what kind of errors are introduced
locally and then whether, really these errors
39
00:10:17,259 --> 00:10:21,149
are bounded so that globally the error is
bounded.
40
00:10:21,149 --> 00:10:30,100
So, they are not magnified so that, the method
is stable. So, these on the conditions, on
41
00:10:30,100 --> 00:10:38,040
the matrix are really necessary. Now, let
us we discussed earlier for a simple second
42
00:10:38,040 --> 00:10:45,100
order, that is y double prime equals to f
of x. So, now let us generalise little bit
43
00:10:45,100 --> 00:10:50,220
more,
where we have a general set up of linear second
44
00:10:50,220 --> 00:11:02,230
order boundary value problem.
45
00:11:02,230 --> 00:11:03,230
..
46
00:11:03,230 --> 00:11:34,970
Second order BVP so we consider
47
00:11:34,970 --> 00:11:50,189
so this is our BVP, then the approximations.
So, the
48
00:11:50,189 --> 00:12:17,410
first derivative is done with the central
derivative approximations.
49
00:12:17,410 --> 00:12:18,410
.
50
00:12:18,410 --> 00:13:17,670
So, in view of this 1 becomes
51
00:13:17,670 --> 00:13:33,579
so y0 is gamma 1 and y n plus 1 is gamma 2.
Now, we
52
00:13:33,579 --> 00:13:50,779
collect the coefficients of similar terms,
so y i plus 1, 1 plus, 1 plus so if you multiply
53
00:13:50,779 --> 00:13:54,170
by
h square there so h, h gets cancelled, so
54
00:13:54,170 --> 00:14:09,720
1 h by 2 and p so we get plus minus 2 and
y i, h
55
00:14:09,720 --> 00:15:02,939
square q 1 minus h by 2. So, this implies
where so I started with y i minus 1 so this
56
00:15:02,939 --> 00:15:07,110
will
come here.
57
00:15:07,110 --> 00:15:52,910
.Now with this set up we expect a tri diagonal
system where, A is B1, C1. Because, when
58
00:15:52,910 --> 00:16:07,399
we run the system, when i is 1 y0 so A1 does
not exist because, y0 is known to us. So,
59
00:16:07,399 --> 00:16:33,979
A1 gets transferred to the right hand side
therefore, we have this. And similarly, when
60
00:16:33,979 --> 00:16:53,040
we run for i n so we have, when we run for
i equals to n, we have n plus 1. So, this
61
00:16:53,040 --> 00:17:00,119
term
gets transferred to the right hand side so
62
00:17:00,119 --> 00:17:35,200
we have, this is Bi so we have A and B.
So, and b bar look at that, when we run this
63
00:17:35,200 --> 00:17:44,570
for i equals to 1, h square r1 and y0 is
gamma 1 so A1 gamma 1 gets transferred to
64
00:17:44,570 --> 00:17:59,380
the right hand side. So, we have h square
r1
65
00:17:59,380 --> 00:18:27,980
minus then we have h square r2, h square r3
then h square r n minus
66
00:18:27,980 --> 00:18:40,009
so i equals to n. So,
this will be y n plus 1 so we have c n times
67
00:18:40,009 --> 00:18:46,320
gamma t.
.
68
00:18:46,320 --> 00:19:01,919
So, this will be the tridiagonal system and
the corresponding
69
00:19:01,919 --> 00:19:24,690
local truncation error. So,
this will be the corresponding local truncation
70
00:19:24,690 --> 00:19:38,620
error so we have
system like this. So,
71
00:19:38,620 --> 00:19:58,899
which is a tridiagonal system and the same
can be solved solving the... So, generally
72
00:19:58,899 --> 00:20:15,100
it is
an elimination process, but for this particular
73
00:20:15,100 --> 00:20:19,320
case since it is a tridiagonal. So, there
is an
74
00:20:19,320 --> 00:20:28,440
algorithm called Thomas algorithm. So, let
us look at it so let the given system be in
75
00:20:28,440 --> 00:21:04,830
this
form. So, which means b1 y1 plus c1 y2 equals
76
00:21:04,830 --> 00:22:12,850
d1 star then a2 similarly, a3 y2 then n
minus 1. So, this will like our tridiagonal
77
00:22:12,850 --> 00:22:15,010
system expansion.
78
00:22:15,010 --> 00:22:16,010
..
79
00:22:16,010 --> 00:22:39,080
Then, assuming b1 is non-zero
eliminate naturally y1, assuming b1 non-zero
80
00:22:39,080 --> 00:22:51,539
we
eliminate y1, from the
81
00:22:51,539 --> 00:22:59,040
second equation. So, we get b2 prime that
is modified b2 is no
82
00:22:59,040 --> 00:23:49,799
longer b2, where b2 prime is b2 minus this.
Next, assume b2 prime non zero
83
00:23:49,799 --> 00:24:01,049
eliminate y2
from the third equation of course, using this
84
00:24:01,049 --> 00:24:36,210
equation. Then we have v3 prime y3, where.
.
85
00:24:36,210 --> 00:25:13,179
So, we continue like this, we eliminate y
k at step k from k plus 1th equation of course,
86
00:25:13,179 --> 00:26:11,879
assuming this is non-zero, where k is. Now,
what we have to do so having obtained up to
87
00:26:11,879 --> 00:26:45,980
.here, we back substitute, back substitution
at n, assuming b n prime non-zero, y n is
88
00:26:45,980 --> 00:26:52,840
b n
prime.
89
00:26:52,840 --> 00:27:23,590
And then, 1, y k is so these are capital n
we are using so this is Thomas algorithm.
90
00:27:23,590 --> 00:27:29,539
So,
one can write a nice programme and then try
91
00:27:29,539 --> 00:27:37,309
to solve it so having done a linear system
so
92
00:27:37,309 --> 00:27:46,169
the next task is to try out with a non-linear
system. So, what is the big deal in it. Well,
93
00:27:46,169 --> 00:27:48,360
for
the linear case we have obtained a system
94
00:27:48,360 --> 00:27:53,869
of equations and then it happened to be
tridiagonal and then, we have a simple algorithm
95
00:27:53,869 --> 00:27:58,440
called Thomas algorithm.
And then one can obtain the solution, even
96
00:27:58,440 --> 00:28:02,490
otherwise using elimination etcetera. But
in
97
00:28:02,490 --> 00:28:11,059
case of non-linear, one straight forward thing
one could expect is, probably we expect
98
00:28:11,059 --> 00:28:18,499
that the system is non-linear so this is one
thing. So maybe it is very trivial guess,
99
00:28:18,499 --> 00:28:21,669
but
then the next task is how do we solve corresponding
100
00:28:21,669 --> 00:28:36,029
non-linear system of equations. So,
let us look into non-linear system.
101
00:28:36,029 --> 00:28:37,029
.
102
00:28:37,029 --> 00:29:34,900
So, non-linear second order BVP so we consider
say so if we discretise, if we discretise.
103
00:29:34,900 --> 00:29:55,960
Now, as far as derivative is concerned it
is linear, but we have the right hand side
104
00:29:55,960 --> 00:30:02,690
part is
non-linear. So, let us see how this works
105
00:30:02,690 --> 00:30:22,340
out with reference to an example.
106
00:30:22,340 --> 00:30:23,340
..
107
00:30:23,340 --> 00:30:45,970
Suppose this is the example, I have taken
h for simplicity, now the discretised version
108
00:30:45,970 --> 00:31:40,350
because, h is 1. So, this our grid, so essentially
so these are the unknowns because, y0 is
109
00:31:40,350 --> 00:32:19,820
2, y4 is minus 1. So, i equals to 1, x 0,
i equals to 1 so this will be x 1, x 1 is
110
00:32:19,820 --> 00:32:35,970
0 so
unfortunately this is 0. So, this will be
111
00:32:35,970 --> 00:33:05,409
x 2 is 1 so this is 1 and y 2 square, so
112
00:33:05,409 --> 00:33:14,590
because x 3
is 2 so 2 y 3 square plus 2.
113
00:33:14,590 --> 00:33:15,590
.
114
00:33:15,590 --> 00:34:06,990
Now, in this system these are known so the
final system we get it as follows. If you
115
00:34:06,990 --> 00:34:17,490
observe I did not put this as a matrix system
x equals to b because, this is a non-linear.
116
00:34:17,490 --> 00:34:44,679
.So, let us call this as f of y bar equals
to 0, where we have
117
00:34:44,679 --> 00:34:55,950
as components of so that
means this is, this is also vector. So, this
118
00:34:55,950 --> 00:35:31,549
is nothing but so these are our f1, f2, f3
and
119
00:35:31,549 --> 00:36:03,410
what is it, is a non-linear system of equations
for y1, y2 and y3. So, this is a non-linear
120
00:36:03,410 --> 00:36:19,280
system for y1, y2, y3.
So, how do we solve ((Refer time.). So, I
121
00:36:19,280 --> 00:36:26,040
think you have heard for solving nonlinear
algebraic equations, we have lot of methods
122
00:36:26,040 --> 00:36:32,760
and one popular method is Newton
Raphson method. Now for non-linear system,
123
00:36:32,760 --> 00:36:40,640
we should try Newton Raphson method.
So, what is the motive?
124
00:36:40,640 --> 00:36:41,640
.
125
00:36:41,640 --> 00:37:13,680
So, the motive behind so Newton Raphsonâ€™s
method for solving non-linear system, so
126
00:37:13,680 --> 00:37:57,500
what we have is this, suppose we expand in
Taylors series about y i plus. So, for a vector
127
00:37:57,500 --> 00:38:13,640
equation dau f bar by dau y bar
128
00:38:13,640 --> 00:38:36,660
so this is h square because, second order
y minus y i
129
00:38:36,660 --> 00:39:09,220
square. So, y bar is minus f of and this must
be approximately 0, so this must be
130
00:39:09,220 --> 00:39:21,950
approximately 0. So, from here y bar I am
retaining then, we must transfer these things,
131
00:39:21,950 --> 00:39:29,890
I
am transferring this, but then this is a matrix.
132
00:39:29,890 --> 00:40:03,340
So, it becomes inverse there and minus y i
becomes this. So, therefore so this is now
133
00:40:03,340 --> 00:40:10,849
what is the inverse so this is the Jacobian
dau f
134
00:40:10,849 --> 00:40:17,430
by dau y bar is the Jacobian and hence it
is a inverse.
135
00:40:17,430 --> 00:40:18,430
..
136
00:40:18,430 --> 00:40:50,260
So, accordingly the solution is an iterative
method is defined
137
00:40:50,260 --> 00:41:35,430
so where the Jacobian. So,
let us look at an example, let us look at
138
00:41:35,430 --> 00:42:06,549
an example, these are the boundary conditions.
So, x0 is 0, x1 is one-third, two-third so
139
00:42:06,549 --> 00:42:46,991
the discretised version is. And we have y0
and
140
00:42:46,991 --> 00:42:58,130
y3 so we have y0 is 4 and y3 is 1.
.
141
00:42:58,130 --> 00:43:32,700
So, let us run the system at i equals to 1
and i equals to 2 so we have y0 is 4, 1. So,
142
00:43:32,700 --> 00:43:48,862
now
it gets simplified, let us put it like this
143
00:43:48,862 --> 00:44:22,740
y0 is 4 so it gets transferred. So, this is
our, now
144
00:44:22,740 --> 00:44:58,010
we need to compute the Jacobian. So, this
the Jacobian so in this case treating this
145
00:44:58,010 --> 00:44:59,010
as f
146
00:44:59,010 --> 00:45:12,260
.1th and this is f 2. So, this will be and
with respect to 2 and here with respect 1,
147
00:45:12,260 --> 00:45:23,520
y1 and
here
148
00:45:23,520 --> 00:45:24,520
so this a Jacobian.
.
149
00:45:24,520 --> 00:45:40,210
Then, we
150
00:45:40,210 --> 00:46:29,210
need to compute this is, therefore, minus
j inverse so where of course D is
151
00:46:29,210 --> 00:47:27,880
given by 2 y1. So, this is the iterative method
now we have, we have to obtain y1 and y2
152
00:47:27,880 --> 00:47:46,010
so this is Newton Raphson, need
initial guess to start the iteration. So,
153
00:47:46,010 --> 00:48:06,460
let so let this be
the initial guess.
154
00:48:06,460 --> 00:48:07,460
.
155
00:48:07,460 --> 00:48:31,359
.Then, f1 this is, our guess was y 1 0, 2
y 2 0 so
156
00:48:31,359 --> 00:49:00,480
this will be 4 plus 2 so y 1 square 2 minus
1 minus 3. So, this is
157
00:49:00,480 --> 00:49:21,329
so this is 8, this is 4 then, this will be
y 2 square, this will be minus
158
00:49:21,329 --> 00:49:39,619
y 2, y 2 is 1, that is correct.
So, y 2 square minus y 1 so y 2 square y 1
159
00:49:39,619 --> 00:50:06,950
plus 2 y 2 so this will be 1 by 4 and d so
d was
160
00:50:06,950 --> 00:50:29,339
so this 4 into 1 plus y 1. So, 1 plus 2 1
plus 1 so this will be 4 into 3 into 2. So,
161
00:50:29,339 --> 00:50:57,340
accordingly y 1 0, y 2 0 minus 1 over D then,
we need this terms two times y 2 plus 1 so
162
00:50:57,340 --> 00:51:18,690
2 times y 2 plus 1 is 2 1 1 2 times y 1 plus
1. So, this multiplied by f1, f2 so here 4
163
00:51:18,690 --> 00:51:50,020
and 1
on 4 so this is, so this is 2 1 so you will
164
00:51:50,020 --> 00:52:13,559
get 16 plus
and here we get
165
00:52:13,559 --> 00:52:22,990
so we can compute
and we get some value. So I guess I have done
166
00:52:22,990 --> 00:52:30,150
it, but this is subject verification.
.
167
00:52:30,150 --> 00:53:08,250
So, then we have to iterate further y 1 2.
So, in order to compute we need f1 of so this
168
00:53:08,250 --> 00:53:17,859
probably, please verify there could be a mistake
in the numerical calculation. Then the
169
00:53:17,859 --> 00:54:01,520
corresponding d and then so the other values
170
00:54:01,520 --> 00:54:16,240
so we may get some value. So, this the next
iteration then, we continue further so we
171
00:54:16,240 --> 00:54:24,690
stop when we have decide accuracy.
So, essentially with respect to the Newton
172
00:54:24,690 --> 00:54:37,520
Raphson method, we are trying to solve the
non-linear system and then, this is just a
173
00:54:37,520 --> 00:54:48,630
solution like this. So, may be while calculating
this is 0 so throughout we have to multiply
174
00:54:48,630 --> 00:55:03,690
by the inverse. So, may be I made a mistake
so this will come here so this is the linearization
175
00:55:03,690 --> 00:55:14,090
in some sense. So, the guess is correct
when you have non-linear equation, remember
176
00:55:14,090 --> 00:55:19,319
the non-linearity is only with respect to
right hand side not in the derivatives, so
177
00:55:19,319 --> 00:55:20,319
far.
178
00:55:20,319 --> 00:55:26,960
.So, if that is a case we get system of equations
and then, in this case the corresponding
179
00:55:26,960 --> 00:55:32,960
system of equations are non-linear and we
use Newton Raphson method to solve these
180
00:55:32,960 --> 00:55:39,599
non-linear system of equations. So, far the
stories for a simple boundary conditions
181
00:55:39,599 --> 00:55:47,670
suppose, your boundary conditions involve
derivatives. So for example, one can classify
182
00:55:47,670 --> 00:55:54,579
the boundary conditions like, if the function
value is given say like, type and derivatives
183
00:55:54,579 --> 00:55:58,349
are type then, a combination is given then
robin type.
184
00:55:58,349 --> 00:56:03,859
So, if the derivatives are involved what would
happen whether, the same techniques
185
00:56:03,859 --> 00:56:09,970
work well, the same techniques work, but the
derivatives need to be discretised right.
186
00:56:09,970 --> 00:56:13,720
So,
may be they will bring in little complications
187
00:56:13,720 --> 00:56:22,599
so we have to discuss them with a special
care, so until then bye.
188
00:56:22,599 --> 00:56:22,599
.