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PROFESSOR: Ladies and gentlemen,
welcome to

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lecture number 12.

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00:00:24,690 --> 00:00:27,510
In this lecture, I would like
to discuss with you solution

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00:00:27,510 --> 00:00:29,900
methods for eigenproblems.

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00:00:29,900 --> 00:00:32,500
This is a very large field,
and all I can do really in

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this lecture is to give you or
discuss this you a few of the

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solution methods that
we are using in

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00:00:39,670 --> 00:00:41,390
finite element analysis.

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00:00:41,390 --> 00:00:44,710
So it will be a very brief
introduction, a brief survey

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00:00:44,710 --> 00:00:48,390
of the methods of solution that
we're using for finite

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00:00:48,390 --> 00:00:50,440
element eigenvalue problems.

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00:00:50,440 --> 00:00:54,010
The generalized eigenvalue
problem, K phi equals lambda M

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00:00:54,010 --> 00:00:57,350
phi, is a problem that we
encountered in the multiple

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00:00:57,350 --> 00:00:59,230
position analysis.

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00:00:59,230 --> 00:01:03,480
Please recognize that I'm
calling lambda here, the

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00:01:03,480 --> 00:01:05,550
eigenvalue, which was our omega

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00:01:05,550 --> 00:01:07,940
squared in the last lecture.

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00:01:07,940 --> 00:01:09,950
This is really the eigenproblems
that we are

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00:01:09,950 --> 00:01:14,120
primarily addressing ourselves
to finite element analysis.

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00:01:14,120 --> 00:01:17,980
The standard eigenproblems
problem here, EVP stands for

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00:01:17,980 --> 00:01:21,660
eigenproblem, has M equal
to the identity matrix.

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00:01:21,660 --> 00:01:23,800
So when we're talking about
solution methods of this

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problem, we're really also talk
about solution methods of

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this problem.

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00:01:27,460 --> 00:01:29,530
If we have, in a [? multiple ?]
position

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00:01:29,530 --> 00:01:33,660
analysis, non-proportional
damping, then we would have to

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00:01:33,660 --> 00:01:37,040
solve this eigenproblem, which
is a quadratic eigenproblem.

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00:01:37,040 --> 00:01:39,610
I will not address myself to the
solution of this problem,

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00:01:39,610 --> 00:01:42,850
so in what follow, we are
talking about a solution of

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00:01:42,850 --> 00:01:45,650
this problem, and of course, the
solution of this problem

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00:01:45,650 --> 00:01:49,030
also, when M is equal to
the identity matrix.

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00:01:49,030 --> 00:01:51,920
In particular, we are interested
in the solution of

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00:01:51,920 --> 00:01:56,120
large eigenvalue problems when
N, is say, greater than 500.

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00:01:56,120 --> 00:02:00,120
N, of course, being the order of
K and M, and with the half

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00:02:00,120 --> 00:02:02,900
bandwidth, I should say,
being greater than 60.

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00:02:02,900 --> 00:02:05,800
The number of eigenvalues that
we will be looking for, that

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00:02:05,800 --> 00:02:10,139
we want to calculate, are say,
1 to 1/3 of N. 1/3 of N is

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00:02:10,139 --> 00:02:12,460
already large number
of eigenvalues.

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00:02:12,460 --> 00:02:16,840
Generally, we are talking about
1 to and at most, 100

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00:02:16,840 --> 00:02:18,660
for large eigenproblems.

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00:02:18,660 --> 00:02:21,710
When we have a small eigenvalue
problem, in other

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00:02:21,710 --> 00:02:25,530
words, N being small, say 50
or 100, there are a large

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00:02:25,530 --> 00:02:27,940
number of different techniques
that can be used to solve this

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00:02:27,940 --> 00:02:31,970
eigenvalue problem, and it
really does not make that much

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00:02:31,970 --> 00:02:34,600
difference which technique
you are using.

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00:02:34,600 --> 00:02:39,610
The cost can still be quite
significant, but certainly,

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00:02:39,610 --> 00:02:43,550
when we're talking about very
large eigenproblems, N being

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00:02:43,550 --> 00:02:47,370
even larger than a 1,000 or
2,000, then it is of up most

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00:02:47,370 --> 00:02:52,000
concern to select a technique
that is very effective.

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00:02:52,000 --> 00:02:55,670
And I want to address myself
really now towards those

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00:02:55,670 --> 00:02:59,590
techniques that we feel are
very effectively used in

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finite element analysis.

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00:03:01,960 --> 00:03:04,650
In dynamic analysis, of course,
with proportional

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damping, this was the eigenvalue
problem that we

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00:03:07,930 --> 00:03:09,290
wanted to solve.

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Now, if they are 0 frequencies
present, in other words, when

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we have rigid body modes in the
system, then we can use

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00:03:15,540 --> 00:03:18,750
the following procedure to
basically getting rid of these

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00:03:18,750 --> 00:03:19,630
0 frequencies.

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00:03:19,630 --> 00:03:23,040
We simply add, on the left hand
side, a mu M phi, and on

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00:03:23,040 --> 00:03:25,950
the right hand side,
a mu M phi.

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00:03:25,950 --> 00:03:28,880
And taking these two together,
we are getting this

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00:03:28,880 --> 00:03:32,740
coefficient matrix, and
a new lambda value.

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00:03:32,740 --> 00:03:35,460
That lambda value now is omega
squared plus mu, where omega

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00:03:35,460 --> 00:03:38,630
squared is equal to
lambda minus mu.

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What we have been doing here,
is basically, we have been

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applying a shift.

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00:03:43,350 --> 00:03:46,210
In other words, if we were to
plot the characteristic

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polynomial, giving us
the eigenvalues, the

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characteristic polynomial of
course, being p of lambda

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being determined K bar minus
lambda M, where K bar is equal

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00:03:57,940 --> 00:04:00,490
to K plus mu M.

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00:04:00,490 --> 00:04:03,270
If we plug this characteristic
polynomial, we will see this

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00:04:03,270 --> 00:04:06,960
curve here, schematically drawn,
of course, where the

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00:04:06,960 --> 00:04:10,250
omega squared values would start
here, in other words,

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00:04:10,250 --> 00:04:14,370
the lowest omega squared
value is, in fact, 0.

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00:04:14,370 --> 00:04:17,589
Whereas our lambda value
now starts right here.

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00:04:17,589 --> 00:04:19,500
This is the shift that
we [? imposed. ?]

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00:04:19,500 --> 00:04:25,755
So, basically, we now recognize
that we have all

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00:04:25,755 --> 00:04:29,090
eigenvalues being greater than
0, because this is our new

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00:04:29,090 --> 00:04:30,100
starting value.

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00:04:30,100 --> 00:04:33,150
All eigenvalues measured from
here are greater than 0, the

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00:04:33,150 --> 00:04:35,780
smallest one being mu,
namely when omega

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00:04:35,780 --> 00:04:37,200
squared is equal to 0.

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00:04:37,200 --> 00:04:39,610
And the larger ones
being here.

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00:04:39,610 --> 00:04:45,260
It is effective to perform such
a shift when we have 0

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00:04:45,260 --> 00:04:49,960
eigenvalues, in other words, to
operate on this eigenvalue,

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00:04:49,960 --> 00:04:54,790
rather than on that one, because
our techniques then

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00:04:54,790 --> 00:04:58,330
are very stable and effective,
and this shift here is, in

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00:04:58,330 --> 00:05:00,950
general, therefore performed.

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00:05:00,950 --> 00:05:04,540
So all we have to address
ourselves to then from now on,

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00:05:04,540 --> 00:05:08,390
is really, the solution of
eigenvalues and eigenvectors,

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00:05:08,390 --> 00:05:12,470
when all of the eigenvalues
are greater than 0.

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00:05:12,470 --> 00:05:15,720
If we start off with a problem
where we have a 0 frequency,

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00:05:15,720 --> 00:05:19,160
we apply the shift so that the
eigenvalues now of this

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00:05:19,160 --> 00:05:23,280
problem here are all
greater than 0.

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00:05:23,280 --> 00:05:26,680
In buckling analysis, we
have this eigenproblem.

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00:05:26,680 --> 00:05:28,185
We really did not talk about--

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00:05:28,185 --> 00:05:31,700
we did not really at all about
buckling analysis in the set

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00:05:31,700 --> 00:05:35,930
of lectures, but you might just
be interested in finding

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00:05:35,930 --> 00:05:38,140
out how would we solve for
these eigenvalues.

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00:05:38,140 --> 00:05:40,320
Well, here is the characteristic
polynomial.

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00:05:40,320 --> 00:05:43,390
The smallest lambda value, of
course, gives now a critical

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00:05:43,390 --> 00:05:46,270
load, that is the one e would
be interested in.

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00:05:46,270 --> 00:05:48,640
It is effective here,
to rewrite this

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00:05:48,640 --> 00:05:50,700
problem in this form.

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00:05:50,700 --> 00:05:53,970
All we do, is we are taking
the lambda value onto the

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00:05:53,970 --> 00:05:54,920
other side.

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00:05:54,920 --> 00:05:56,660
In other words, we
are dividing this

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00:05:56,660 --> 00:05:58,570
equation here by a lambda.

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00:05:58,570 --> 00:06:01,320
So that one of the lambda stands
on this side, and KG

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00:06:01,320 --> 00:06:02,730
phi on that side.

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00:06:02,730 --> 00:06:05,940
The resulting problem is this
one, where kappa is 1 over

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00:06:05,940 --> 00:06:08,540
lambda and we have now
here, is a K matrix.

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00:06:08,540 --> 00:06:14,630
K is positive definite, and now
it is more effective to

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00:06:14,630 --> 00:06:17,040
operate on this equation.

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00:06:17,040 --> 00:06:19,820
How do we obtain now
the largest kappa?

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00:06:19,820 --> 00:06:22,130
Because the largest kappa is the
one that we are interested

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00:06:22,130 --> 00:06:25,010
in, which corresponds to the
smallest lambda, the smallest

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00:06:25,010 --> 00:06:25,730
buckling load.

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00:06:25,730 --> 00:06:28,850
Well, we would now perform
a shift to the right,

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00:06:28,850 --> 00:06:34,190
subtracting mu K from this side,
from this side, and from

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00:06:34,190 --> 00:06:37,360
that side, to obtain this
eigenvalue problem.

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00:06:37,360 --> 00:06:39,980
That means we are really
shifting from here up to

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00:06:39,980 --> 00:06:45,130
there, and now we are interested
in solving kappa N

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00:06:45,130 --> 00:06:47,620
by operating on this
eigenproblems problem.

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00:06:47,620 --> 00:06:50,100
It is this eigenvalue problem
that we can operate on

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00:06:50,100 --> 00:06:53,540
directly, using the techniques
that I would be discussing

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00:06:53,540 --> 00:06:55,950
later, to determine search
method and the subspace

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00:06:55,950 --> 00:06:58,222
iteration method.

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00:06:58,222 --> 00:07:00,990
But before we get into those
techniques, I would like to

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00:07:00,990 --> 00:07:05,810
discuss with you some
preliminary considerations

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00:07:05,810 --> 00:07:07,710
that are important
when we look at

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00:07:07,710 --> 00:07:09,440
an eigenvalue solution.

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00:07:09,440 --> 00:07:14,030
The important first point that
I want to make is that there

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00:07:14,030 --> 00:07:17,260
is a traditional approach
for the solution of this

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00:07:17,260 --> 00:07:18,840
eigenvalue problem here.

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00:07:18,840 --> 00:07:20,610
Let's look at this eigenvalue
problem.

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00:07:20,610 --> 00:07:22,840
Where of course, when we're
talking about a buckling

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00:07:22,840 --> 00:07:25,700
analysis, we would have
different matrices here, but

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00:07:25,700 --> 00:07:29,860
basically, we still solve
this eigenvalue problem.

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00:07:29,860 --> 00:07:34,160
The traditional approach lies
in factorizing the M matrix

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00:07:34,160 --> 00:07:36,120
into Cholesky factors.

151
00:07:36,120 --> 00:07:38,880
We talked about the Cholesky
factor when we talked about

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00:07:38,880 --> 00:07:40,230
solution of equations.

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00:07:40,230 --> 00:07:41,750
So here it comes up again.

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00:07:41,750 --> 00:07:45,440
We would factorize the M matrix
into Cholesky factors,

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00:07:45,440 --> 00:07:46,470
as shown here.

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00:07:46,470 --> 00:07:50,730
We define now a new eigenvector,
phi curl being

157
00:07:50,730 --> 00:07:54,880
equal as shown here, L curl
transpose times 5.

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00:07:54,880 --> 00:07:59,970
We substitute from here into
there, and we directly obtain

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00:07:59,970 --> 00:08:02,640
this eigenvalue problem here.

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00:08:02,640 --> 00:08:05,980
Now we have the standard
eigenproblem.

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00:08:05,980 --> 00:08:09,160
Now once we have a standard
eigenproblem, there are a

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00:08:09,160 --> 00:08:11,660
large number of techniques
that can be used.

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00:08:11,660 --> 00:08:14,735
For example, the QR method,
the standard

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00:08:14,735 --> 00:08:16,790
Jacobi method, et cetera.

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00:08:16,790 --> 00:08:18,940
So this is, what
we might call.

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00:08:18,940 --> 00:08:21,530
a traditional approach to
solve the generalized

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00:08:21,530 --> 00:08:24,140
eigenproblem, this is the one
here that we are really

168
00:08:24,140 --> 00:08:25,110
interested in.

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00:08:25,110 --> 00:08:27,250
But it involves this
transformation, and this

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00:08:27,250 --> 00:08:32,280
transformation here can involve
a very significant

171
00:08:32,280 --> 00:08:35,070
cost, because we have to find
these Cholesky factors and

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00:08:35,070 --> 00:08:37,929
then, of course, we have to
calculate this K curl.

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00:08:37,929 --> 00:08:40,370
Another way of performing
a transformation of the

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00:08:40,370 --> 00:08:43,549
generalized item problem to a
standard form would be to

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00:08:43,549 --> 00:08:46,540
solve the eigenproblem
of M first.

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00:08:46,540 --> 00:08:49,110
W here and W transpose
[? store ?]

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00:08:49,110 --> 00:08:54,080
the eigenvectors of the M
matrix, and D squared is a

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00:08:54,080 --> 00:08:57,320
diagonal matrix of the
eigenvalues of M. And then

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00:08:57,320 --> 00:09:01,020
having factored M in this form,
an equation similar to

180
00:09:01,020 --> 00:09:02,510
this one here can be written.

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00:09:02,510 --> 00:09:05,490
Where now, the transformation,
of course, involves the

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00:09:05,490 --> 00:09:07,400
eigenvectors, W.

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00:09:07,400 --> 00:09:12,860
Well, this approach is
sometimes still quite

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00:09:12,860 --> 00:09:15,990
effective, but in general, it
cannot be recommended because

185
00:09:15,990 --> 00:09:20,420
there's too much cost involved
in the factorizing of the M

186
00:09:20,420 --> 00:09:23,830
matrix, and then particular,
in the transformation here,

187
00:09:23,830 --> 00:09:26,180
and the solution of this problem
and the, of course, to

188
00:09:26,180 --> 00:09:29,670
get from this problem, the phi
vector, we have to go back to

189
00:09:29,670 --> 00:09:30,980
this equation.

190
00:09:30,980 --> 00:09:36,070
A direct solution of the
generalized eigenvalue problem

191
00:09:36,070 --> 00:09:39,390
that we want to solve is
really more effective.

192
00:09:39,390 --> 00:09:43,680
And now, we are ordering the
lambdas as shown here.

193
00:09:43,680 --> 00:09:48,310
Remember, our smallest lambda
now is greater than 0, the

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00:09:48,310 --> 00:09:52,490
eigonvectors are stored as shown
here, and for most of a

195
00:09:52,490 --> 00:09:56,150
position analysis, we are really
interested in solving

196
00:09:56,150 --> 00:10:00,550
only for a certain number of
eigen pairs, i equals 1 to p,

197
00:10:00,550 --> 00:10:04,200
as discussed in the last
lecture, or i equals r to s.

198
00:10:04,200 --> 00:10:06,980
If we have a vibration
excitation problem, we really

199
00:10:06,980 --> 00:10:09,800
only want to solve for specific
frequencies and

200
00:10:09,800 --> 00:10:11,810
corresponding mode shapes
in a certain

201
00:10:11,810 --> 00:10:14,300
interval, as shown here.

202
00:10:14,300 --> 00:10:19,000
The solution procedures on this
generalized eigenvalue

203
00:10:19,000 --> 00:10:23,690
problem can be categorized
into the following forms.

204
00:10:23,690 --> 00:10:29,780
The first solution procedure,
or a whole class of solution

205
00:10:29,780 --> 00:10:33,100
procedures, that I should really
say, which we might

206
00:10:33,100 --> 00:10:36,050
call the vector iteration
techniques.

207
00:10:36,050 --> 00:10:37,620
The idea is the following--

208
00:10:37,620 --> 00:10:43,160
if we want to solve this
problem, then why not solve it

209
00:10:43,160 --> 00:10:45,870
in the following way?

210
00:10:45,870 --> 00:10:55,830
Start off with a vector on this
side, that we assume,

211
00:10:55,830 --> 00:11:00,260
calculate a new vector on this
side, then take that new

212
00:11:00,260 --> 00:11:03,130
vector, plug it back on the
right hand side, and cycle

213
00:11:03,130 --> 00:11:04,570
through this way.

214
00:11:04,570 --> 00:11:06,640
This is indeed inverse
iteration.

215
00:11:06,640 --> 00:11:07,810
What do we do here?

216
00:11:07,810 --> 00:11:11,740
We are putting a vector on the
right hand side, x1, when K is

217
00:11:11,740 --> 00:11:15,790
equal to 1, that vector
we have to assume.

218
00:11:15,790 --> 00:11:18,540
Then we are multiplying out
the right hand side.

219
00:11:18,540 --> 00:11:23,380
Now we have a set of
simultaneous equations that we

220
00:11:23,380 --> 00:11:26,460
would solve just in the same way
as we solve the equations

221
00:11:26,460 --> 00:11:27,480
in static analysis.

222
00:11:27,480 --> 00:11:30,190
We can think of this
as a load vector.

223
00:11:30,190 --> 00:11:34,100
We are calculating this
displacement vector, and then

224
00:11:34,100 --> 00:11:38,780
having got this displacement
vector, and not taking care of

225
00:11:38,780 --> 00:11:42,390
this lambda, or having,
basically, taken this lambda

226
00:11:42,390 --> 00:11:44,270
out of this equation.

227
00:11:44,270 --> 00:11:49,090
Which means that this vector
will grow in lengths, or

228
00:11:49,090 --> 00:11:50,760
collapse in lengths.

229
00:11:50,760 --> 00:11:54,160
Therefore, want to scale this
vector down to a good length--

230
00:11:54,160 --> 00:11:57,100
a length that does not hit
overflow, so to say, or

231
00:11:57,100 --> 00:11:58,550
underflow in the machine.

232
00:11:58,550 --> 00:12:02,260
And that scaling is achieved
as shown here.

233
00:12:02,260 --> 00:12:05,980
So, the final result then, this
is just a scalar here, as

234
00:12:05,980 --> 00:12:10,520
you can see, is a vector here,
which we can put on to the

235
00:12:10,520 --> 00:12:13,800
right hand side and keep
on iterating around.

236
00:12:13,800 --> 00:12:18,110
In fact, we find as that
iteration procedure converges

237
00:12:18,110 --> 00:12:22,610
to the lowest eigenvector,
provided the starting vector

238
00:12:22,610 --> 00:12:28,210
is not orthogonal, is not
M orthogonal to phi 1.

239
00:12:28,210 --> 00:12:32,330
This is one first iteration
technique.

240
00:12:32,330 --> 00:12:37,130
Maybe I should put
a side note down.

241
00:12:37,130 --> 00:12:42,940
A side note that many of you
might have in your mind.

242
00:12:42,940 --> 00:12:44,730
Many of you might have the
following question in your

243
00:12:44,730 --> 00:12:48,350
mind-- why he iterate, why does
he not solve directly for

244
00:12:48,350 --> 00:12:49,290
the eigenvalues?

245
00:12:49,290 --> 00:12:53,230
Just like we saw a set of
simultaneous equations.

246
00:12:53,230 --> 00:12:57,890
Well, there's one very important
point, and this is

247
00:12:57,890 --> 00:12:58,570
the point--

248
00:12:58,570 --> 00:13:02,800
if we are solving an eigenvalue
problem, and shown

249
00:13:02,800 --> 00:13:05,350
here, then really that
means the following--

250
00:13:05,350 --> 00:13:10,920
we are saying that determinant,
K minus lambda M

251
00:13:10,920 --> 00:13:12,680
shall be equal to 0.

252
00:13:12,680 --> 00:13:13,260
Why?

253
00:13:13,260 --> 00:13:15,740
Because if we write
this equation in

254
00:13:15,740 --> 00:13:17,080
the following form--

255
00:13:17,080 --> 00:13:24,760
K minus lambda M times phi
equals 0, then we have here a

256
00:13:24,760 --> 00:13:30,390
set of simultaneous equations
with a 0 vector on

257
00:13:30,390 --> 00:13:31,730
the right hand side.

258
00:13:31,730 --> 00:13:34,700
And there is an important
theorem in linear algebra that

259
00:13:34,700 --> 00:13:39,960
says, only a solution to this
set of equations when this

260
00:13:39,960 --> 00:13:42,380
coefficient matrix
is singular.

261
00:13:42,380 --> 00:13:44,920
Which means that the
determinant of this

262
00:13:44,920 --> 00:13:47,700
coefficient matrix must be 0.

263
00:13:47,700 --> 00:13:49,240
So, what do we want?

264
00:13:49,240 --> 00:13:54,650
We really want to solve p of
lambda determinant, K minus

265
00:13:54,650 --> 00:14:00,480
lambda M. We want to solve for
the 0's of this polynomial.

266
00:14:00,480 --> 00:14:02,270
That's really what we want.

267
00:14:02,270 --> 00:14:07,700
Now, if we want to solve for
the 0's of this polynomial,

268
00:14:07,700 --> 00:14:13,390
then we also have to understand
that when k is of

269
00:14:13,390 --> 00:14:17,210
order 4, there will
be four 0's.

270
00:14:17,210 --> 00:14:20,080
The polynomial itself
will be of order 4.

271
00:14:20,080 --> 00:14:23,710
If K is of order, N, then
there will be N 0's.

272
00:14:23,710 --> 00:14:29,680
So the polynomial is order N.
However, if we want to find

273
00:14:29,680 --> 00:14:34,320
the solution of a polynomial,
or order N, then we very

274
00:14:34,320 --> 00:14:40,650
quickly find that, in fact, we
can only look up the solution

275
00:14:40,650 --> 00:14:45,080
to such polynomials if they are
of low enough order, lower

276
00:14:45,080 --> 00:14:46,660
than say, 4.

277
00:14:46,660 --> 00:14:51,080
Above for larger order
polynomials, there's no way

278
00:14:51,080 --> 00:14:53,150
you can look up the solution.

279
00:14:53,150 --> 00:14:58,160
You have to iterate to obtain
the solution of the 0's of

280
00:14:58,160 --> 00:14:59,620
this polynomial.

281
00:14:59,620 --> 00:15:04,480
However, since this problem
here, is completely equivalent

282
00:15:04,480 --> 00:15:06,430
to this problem.

283
00:15:06,430 --> 00:15:09,170
To solve this problem,
you have to iterate.

284
00:15:09,170 --> 00:15:11,480
And this is the important
conclusion.

285
00:15:11,480 --> 00:15:15,640
It is of up most importance that
you understand that to

286
00:15:15,640 --> 00:15:20,570
solve an eigenvalue problem of
order larger than say, 4, you

287
00:15:20,570 --> 00:15:22,330
will have to iterate.

288
00:15:22,330 --> 00:15:27,290
Because you're solving for the
0's of a polynomial of order

289
00:15:27,290 --> 00:15:28,600
larger than 4.

290
00:15:28,600 --> 00:15:32,040
Only for very small order
polynomials, we can look up

291
00:15:32,040 --> 00:15:34,060
the solutions in handbooks.

292
00:15:34,060 --> 00:15:38,680
So having grasped the fact that
we have to iterate to

293
00:15:38,680 --> 00:15:44,460
solve for the 0's of this
polynomial, to solve for the

294
00:15:44,460 --> 00:15:51,390
eigenvalues of this problem, we
are stuck with iteration,

295
00:15:51,390 --> 00:15:54,320
and since we're stuck with
iteration, we now want to come

296
00:15:54,320 --> 00:15:56,380
up with techniques to iterate.

297
00:15:56,380 --> 00:15:58,620
And the first technique to
iterate is this vector

298
00:15:58,620 --> 00:16:01,180
iteration technique.

299
00:16:01,180 --> 00:16:04,470
And the basic idea, once more,
is if you want to solve this

300
00:16:04,470 --> 00:16:09,270
equation, well, then why not
iterate in the following way?

301
00:16:09,270 --> 00:16:13,150
Put some starting value in here,
this number is just a

302
00:16:13,150 --> 00:16:16,580
scalar, let's neglect that
effect for the moment.

303
00:16:16,580 --> 00:16:19,640
And having put a starting value
in here, calculate this

304
00:16:19,640 --> 00:16:22,790
one and keep on cycling
around this way.

305
00:16:22,790 --> 00:16:25,600
And that's what we call
inverse iteration.

306
00:16:25,600 --> 00:16:28,480
Inverse iteration here, a vector
inverse iteration,

307
00:16:28,480 --> 00:16:32,680
because we have to solve a set
of equations here, with K as

308
00:16:32,680 --> 00:16:33,930
the coefficient matrix.

309
00:16:36,310 --> 00:16:38,290
There are other iteration
techniques.

310
00:16:38,290 --> 00:16:39,900
There's also forward
iteration.

311
00:16:39,900 --> 00:16:42,150
And basically, what we're
saying there is--

312
00:16:42,150 --> 00:16:45,970
well if we are starting off here
and going around in this

313
00:16:45,970 --> 00:16:49,970
circle, why not start off here
and go around in that circle,

314
00:16:49,970 --> 00:16:51,190
the red line.

315
00:16:51,190 --> 00:16:54,970
And that, in fact, gives
us forward iteration.

316
00:16:54,970 --> 00:16:57,970
One can prove that that forward
iteration converges to

317
00:16:57,970 --> 00:17:00,140
the largest eigenvalue.

318
00:17:00,140 --> 00:17:03,190
We are usually interested in the
smallest frequencies, the

319
00:17:03,190 --> 00:17:06,069
smallest eigenvalues, and
therefore, we are using

320
00:17:06,069 --> 00:17:08,810
inverse iteration
quite commonly.

321
00:17:08,810 --> 00:17:11,119
Then there is Rayleigh
Quotient iteration.

322
00:17:11,119 --> 00:17:16,790
The Rayleigh Quotient iteration
is again, an inverse

323
00:17:16,790 --> 00:17:21,160
iteration, however, with an
acceleration scheme in it, a

324
00:17:21,160 --> 00:17:24,190
very effective acceleration
scheme.

325
00:17:24,190 --> 00:17:26,810
This Rayleigh Quotient iteration
can be employed to

326
00:17:26,810 --> 00:17:30,070
calculate one eigenvalue and
vector, and then, of course,

327
00:17:30,070 --> 00:17:34,010
we have to deflate afterwards
the matrix, the coefficient

328
00:17:34,010 --> 00:17:37,170
matrix for the eigenvector
that we have calculated

329
00:17:37,170 --> 00:17:42,750
already, and we then could go on
towards the calculation of

330
00:17:42,750 --> 00:17:44,770
additional eigenvalues
and vectors.

331
00:17:44,770 --> 00:17:47,490
The same deflation also would
be used in inverse iteration

332
00:17:47,490 --> 00:17:50,200
alone, and in forward
iteration alone.

333
00:17:50,200 --> 00:17:52,760
In a Rayleigh Quotient
iteration, we converge to an

334
00:17:52,760 --> 00:17:57,440
eigen pair, but which one
is not guaranteed.

335
00:17:57,440 --> 00:18:01,880
So this is a class of
techniques, iteration methods

336
00:18:01,880 --> 00:18:03,560
that we are using.

337
00:18:03,560 --> 00:18:06,610
Another class of techniques,
again iteration methods, we

338
00:18:06,610 --> 00:18:09,890
are stuck with iteration, as
I mentioned earlier, is the

339
00:18:09,890 --> 00:18:11,320
polynomial iteration methods.

340
00:18:11,320 --> 00:18:14,630
Here, we are saying that this
problem is equivalent to that,

341
00:18:14,630 --> 00:18:16,920
we want to solve
for these 0's.

342
00:18:16,920 --> 00:18:20,880
Therefore, for the 0's of this
polynomial, sketched out here,

343
00:18:20,880 --> 00:18:23,710
and one way is to use
Newton iteration.

344
00:18:23,710 --> 00:18:25,830
Here we have p, here
we have p prime.

345
00:18:25,830 --> 00:18:29,740
Well, of course, if we have
p here, we have to also

346
00:18:29,740 --> 00:18:32,110
recognize we should have
the coefficients here.

347
00:18:32,110 --> 00:18:37,330
Well, these coefficients are
given here, A0 to a n.

348
00:18:37,330 --> 00:18:42,200
And we could rewrite this
polynomial in this form, once

349
00:18:42,200 --> 00:18:44,260
we know the zeroes.

350
00:18:44,260 --> 00:18:48,420
Now, if we don't know the
zeroes, of course, we would be

351
00:18:48,420 --> 00:18:52,840
stuck on iterating on this
matrix, where we would have to

352
00:18:52,840 --> 00:18:55,580
calculate these coefficients.

353
00:18:55,580 --> 00:19:00,040
However, there are now really,
two different techniques.

354
00:19:00,040 --> 00:19:03,680
The first technique is explicit
polynomial iteration.

355
00:19:03,680 --> 00:19:07,260
In the explicit polynomial
iteration, we expand the

356
00:19:07,260 --> 00:19:10,710
polynomial and iterate for the
0's, by that I mean, we

357
00:19:10,710 --> 00:19:14,200
actually expand this
polynomial.

358
00:19:14,200 --> 00:19:18,090
However, this technique is not
suitable for larger problems

359
00:19:18,090 --> 00:19:21,150
because first of all, there's
much work to obtain the ai's,

360
00:19:21,150 --> 00:19:23,720
and then it's an unstable
process.

361
00:19:23,720 --> 00:19:27,940
Unstable because small errors
in these coefficients here,

362
00:19:27,940 --> 00:19:30,350
give large errors
in the zeroes.

363
00:19:30,350 --> 00:19:34,980
So this is not a suitable
technique, and we are better

364
00:19:34,980 --> 00:19:37,720
off using implicit polynomial
iteration.

365
00:19:37,720 --> 00:19:40,410
In the implicit polynomial
iteration, we evaluate the

366
00:19:40,410 --> 00:19:44,280
determinant all this
matrix here.

367
00:19:44,280 --> 00:19:50,290
In others, p at mu i, via the L
D L transpose factorization,

368
00:19:50,290 --> 00:19:55,770
which we discussed in the Gauss
elimination solution,

369
00:19:55,770 --> 00:19:57,920
when we talked about Gauss
elimination solution.

370
00:19:57,920 --> 00:20:01,530
To obtain the determinant of
this matrix here then, we need

371
00:20:01,530 --> 00:20:04,410
simply to multiply all the
diagonal elements.

372
00:20:04,410 --> 00:20:08,450
This technique is accurate,
provided we do not encounter

373
00:20:08,450 --> 00:20:12,030
large multipliers L matrix.

374
00:20:12,030 --> 00:20:14,400
And if we do, we have to
shift away, we have to

375
00:20:14,400 --> 00:20:16,220
change our mu i.

376
00:20:16,220 --> 00:20:19,570
Also we directly solve for
lambda 1 and so on.

377
00:20:19,570 --> 00:20:22,760
And for the more we can use
effectively a secant

378
00:20:22,760 --> 00:20:25,710
iteration, instead of
calculating, in other words,

379
00:20:25,710 --> 00:20:28,940
the p prime at mu.

380
00:20:28,940 --> 00:20:33,930
This is here approximately equal
to p prime, at mu i.

381
00:20:33,930 --> 00:20:38,010
Instead of calculating this
derivative, we are putting

382
00:20:38,010 --> 00:20:41,330
this value here, which gives
us an approximation to that

383
00:20:41,330 --> 00:20:44,200
derivative, this is what we
call the secant iteration.

384
00:20:44,200 --> 00:20:47,260
Once we have calculated one
eigenvalue, of course, we will

385
00:20:47,260 --> 00:20:51,140
have to deflate to calculate the
next eigenvalue, and this

386
00:20:51,140 --> 00:20:53,360
is the process that
he would pursue.

387
00:20:53,360 --> 00:20:57,480
Pictorially, we have here,
our p lambda, this is the

388
00:20:57,480 --> 00:20:58,470
polynomial.

389
00:20:58,470 --> 00:20:59,860
Let me show it to you here.

390
00:20:59,860 --> 00:21:03,240
Here, we have the polynomial,
here we have lambda 1, here we

391
00:21:03,240 --> 00:21:04,260
have lambda 2.

392
00:21:04,260 --> 00:21:07,560
In the secant iteration, we
start off with the value of mu

393
00:21:07,560 --> 00:21:11,500
i minus 1 here, mu i, these
two values are given.

394
00:21:11,500 --> 00:21:15,930
Usually, in a practical
analysis, mu 0 would be this

395
00:21:15,930 --> 00:21:20,660
value here, and mu 1 would be
a lower bound on lambda 1.

396
00:21:20,660 --> 00:21:24,990
And plugging these two values
in, we get mu i plus 1, and

397
00:21:24,990 --> 00:21:30,620
you can see that this process
would then converge to this

398
00:21:30,620 --> 00:21:32,210
eigenvalue.

399
00:21:32,210 --> 00:21:35,880
Once we have lambda 1, we can
repeat the iteration.

400
00:21:35,880 --> 00:21:39,590
However, working now on p
lambda divided by lambda

401
00:21:39,590 --> 00:21:41,110
minus lambda 1.

402
00:21:41,110 --> 00:21:44,480
And that means we are operating
on this polynomial

403
00:21:44,480 --> 00:21:48,640
here, and polynomial then gives
us the iteration on that

404
00:21:48,640 --> 00:21:51,190
polynomial then gives us
the second eigenvalue.

405
00:21:51,190 --> 00:21:57,380
So we are deflating the actual p
lambda by the lambda 1 value

406
00:21:57,380 --> 00:22:00,070
in order to obtain
a convergence

407
00:22:00,070 --> 00:22:02,940
to the second value.

408
00:22:02,940 --> 00:22:06,420
Well, this is the one
class of methods.

409
00:22:06,420 --> 00:22:09,750
Another class of methods, the
third one, are the Sturm

410
00:22:09,750 --> 00:22:10,930
sequence methods.

411
00:22:10,930 --> 00:22:15,060
And here, the basic idea
is the following--

412
00:22:15,060 --> 00:22:17,980
if we look at this eigenproblem,
and I use a 4x4

413
00:22:17,980 --> 00:22:24,230
matrix here, then if we take a
shift on this eigenproblem, mu

414
00:22:24,230 --> 00:22:27,100
S, and this is the shift here.

415
00:22:27,100 --> 00:22:33,700
Then if we factorize K minus
mu S M, equal to L D L

416
00:22:33,700 --> 00:22:36,540
transpose, again, we're using
our Gauss elimination

417
00:22:36,540 --> 00:22:38,120
procedures basically.

418
00:22:38,120 --> 00:22:42,420
Then the number of negative
elements in D is equal to the

419
00:22:42,420 --> 00:22:45,110
number eigenvalues smaller
than mu S, a

420
00:22:45,110 --> 00:22:47,480
very important point.

421
00:22:47,480 --> 00:22:51,100
Let us look at this example
here of the 4x4 matrix.

422
00:22:51,100 --> 00:22:56,370
Here we have plotted
characteristic polynomials.

423
00:22:56,370 --> 00:23:00,290
The characteristic polynomial of
this 4x4 matrix, with that

424
00:23:00,290 --> 00:23:05,310
M matrix, of course, is drawn
along here, four eigenvalues.

425
00:23:05,310 --> 00:23:09,150
And I have put mu S here
as the red line.

426
00:23:09,150 --> 00:23:15,430
What this theorem says is if I
factorize K minus mu S, M,

427
00:23:15,430 --> 00:23:19,160
this being the K matrix, the
total K matrix, and the 4x4 M

428
00:23:19,160 --> 00:23:21,020
matrix goes in here.

429
00:23:21,020 --> 00:23:25,740
Then there would be one negative
element in D. One

430
00:23:25,740 --> 00:23:30,900
only because mu S is large
than lambda 1.

431
00:23:30,900 --> 00:23:34,470
If mu S were on this side
here, there would be two

432
00:23:34,470 --> 00:23:38,640
negative elements in D. If mu
S were on this side, there

433
00:23:38,640 --> 00:23:41,260
would be three negative
elements in D--

434
00:23:41,260 --> 00:23:46,360
very important point because qw
can use this procedure to

435
00:23:46,360 --> 00:23:50,670
directly calculate eigenvalues
the way I will be discussing

436
00:23:50,670 --> 00:23:51,430
it just now.

437
00:23:51,430 --> 00:23:55,470
However, let us look a little
bit more at why

438
00:23:55,470 --> 00:23:56,730
this hole is here.

439
00:23:56,730 --> 00:24:01,940
Well, it derives from the
associated constraint problems

440
00:24:01,940 --> 00:24:04,720
with the original
eigenproblem.

441
00:24:04,720 --> 00:24:09,440
The original eigenproblem
corresponds to this beam

442
00:24:09,440 --> 00:24:14,030
problem, which we considered
earlier already in the

443
00:24:14,030 --> 00:24:15,730
discussion of Gauss
elimination.

444
00:24:15,730 --> 00:24:19,870
If I constrain this degree of
freedom here, I get a new

445
00:24:19,870 --> 00:24:22,690
polynomial corresponding to
this problem, and the

446
00:24:22,690 --> 00:24:26,860
important point is that the
lowest root, the lowest value

447
00:24:26,860 --> 00:24:32,000
of this problem here lies in
between these two-- lambda 1

448
00:24:32,000 --> 00:24:33,170
and lambda 2.

449
00:24:33,170 --> 00:24:37,180
The second one here, lies
between lambda 2 and lambda 3,

450
00:24:37,180 --> 00:24:41,350
and the third one here lies
between lambda 3 and lambda 4.

451
00:24:41,350 --> 00:24:44,340
The second associated constraint
problem has two

452
00:24:44,340 --> 00:24:48,000
eigenvalues, and this one lies
between these two, this one

453
00:24:48,000 --> 00:24:49,370
lies between these two.

454
00:24:49,370 --> 00:24:51,850
And the final eigenvalue
problem here, the third

455
00:24:51,850 --> 00:24:54,630
associate constraint problem
obtained by knocking out these

456
00:24:54,630 --> 00:24:57,430
degrees of freedom here, has one
eigenvalue, and that one

457
00:24:57,430 --> 00:25:01,790
lies in between these
two values here.

458
00:25:01,790 --> 00:25:09,970
It's this important fact
that is used to

459
00:25:09,970 --> 00:25:13,630
derive this final result.

460
00:25:13,630 --> 00:25:18,820
Now, we have not time to go
through the actual details,

461
00:25:18,820 --> 00:25:22,220
but this is the important
fact here.

462
00:25:22,220 --> 00:25:27,470
In other words, that the
[? i's ?] eigenvalue of this

463
00:25:27,470 --> 00:25:31,430
associated constraint problem,
for example, the second

464
00:25:31,430 --> 00:25:35,420
eigenvalue of this associated
constraint problem here,

465
00:25:35,420 --> 00:25:38,630
bisects or lies in between,
not bisect--

466
00:25:38,630 --> 00:25:44,340
lies in between lambda 2 and
lambda 3 of this eigenproblem

467
00:25:44,340 --> 00:25:45,930
here, and so on.

468
00:25:45,930 --> 00:25:50,700
That result is used to derive
this final result, which we

469
00:25:50,700 --> 00:25:57,180
use effectively then in
the eigen solution.

470
00:25:57,180 --> 00:26:03,950
And the class of techniques then
that we are referring to

471
00:26:03,950 --> 00:26:07,470
using this fact are the Sturm
sequence methods.

472
00:26:07,470 --> 00:26:11,050
So basically, what we are saying
is the following--

473
00:26:11,050 --> 00:26:15,800
take this coefficient matrix,
now I make the mu x available.

474
00:26:15,800 --> 00:26:20,060
We factorize that coefficient
matrix into L D L transposed,

475
00:26:20,060 --> 00:26:24,740
and the number of negative
elements in D is equal to the

476
00:26:24,740 --> 00:26:29,630
number of eigenvalues smaller
than mu S, I. How

477
00:26:29,630 --> 00:26:30,560
do we use this fact?

478
00:26:30,560 --> 00:26:34,885
Well, if this is here, the
polynomial, we could start off

479
00:26:34,885 --> 00:26:40,950
by putting mu S 1 into this
point here, and we would find

480
00:26:40,950 --> 00:26:44,510
out the number of eigenvalues
smaller than mu S 1.

481
00:26:44,510 --> 00:26:49,170
Then we are taking, mu S 2, and
we find out the number of

482
00:26:49,170 --> 00:26:51,320
eigenvalues smaller
than mu S 2.

483
00:26:51,320 --> 00:26:58,580
Of course, we would find that
they are one more eigenvalue

484
00:26:58,580 --> 00:26:59,860
smaller than mu S 2.

485
00:26:59,860 --> 00:27:02,460
Then mu S 1, namely this one.

486
00:27:02,460 --> 00:27:06,150
So we know there's one
eigenvalue in between mu S 1

487
00:27:06,150 --> 00:27:07,910
and mu S 2.

488
00:27:07,910 --> 00:27:11,090
Now we could bisect this
interval, and we would find

489
00:27:11,090 --> 00:27:16,080
that, in fact, there is still
one more eigenvalue smaller

490
00:27:16,080 --> 00:27:20,970
than mu S 3, then there was
smaller than mu S 1.

491
00:27:20,970 --> 00:27:23,750
Because this eigenvalue
still lies on the left

492
00:27:23,750 --> 00:27:25,780
side of mu S 3.

493
00:27:25,780 --> 00:27:28,650
So we bisect again,
we get mu S 4.

494
00:27:28,650 --> 00:27:33,550
Notice this distance here, is
equal to that distance here.

495
00:27:33,550 --> 00:27:36,970
And now, we would find that the
eigenvalue still lies in

496
00:27:36,970 --> 00:27:40,780
between these two lines, between
mu S 3 and mu S 4.

497
00:27:40,780 --> 00:27:43,560
And like that. we could
continue bisecting the

498
00:27:43,560 --> 00:27:48,400
interval until we finally find a
very small interval in which

499
00:27:48,400 --> 00:27:51,550
this eigenvalue must lie.

500
00:27:51,550 --> 00:27:54,490
We, of course, have to take care
in the L D L transpose

501
00:27:54,490 --> 00:27:57,920
factorization, because we are
not working on a positive

502
00:27:57,920 --> 00:27:58,760
definite matrix.

503
00:27:58,760 --> 00:28:03,800
The multipliers must be small,
the L, I, J elements must

504
00:28:03,800 --> 00:28:06,280
remain small in the
factorization.

505
00:28:06,280 --> 00:28:09,700
But provided that is the case,
the round of errors are small,

506
00:28:09,700 --> 00:28:11,930
and the procedure
is very stable.

507
00:28:11,930 --> 00:28:15,100
However, convergence can be very
slow, can be extremely

508
00:28:15,100 --> 00:28:17,990
slow because you have to cut
down the intervals further and

509
00:28:17,990 --> 00:28:20,990
further until you finally get
a very small interval, in

510
00:28:20,990 --> 00:28:23,560
which you now know the
eigenvalue does lie.

511
00:28:23,560 --> 00:28:26,340
Of course, at some point, you
might want to switch to

512
00:28:26,340 --> 00:28:27,810
another iteration scheme.

513
00:28:27,810 --> 00:28:31,210
Once you know there is an
eigenvalue in between say mu S

514
00:28:31,210 --> 00:28:34,100
3 and mu S 4, you might want
to switch to a iteration

515
00:28:34,100 --> 00:28:38,590
scheme that converges to this
particular value much faster.

516
00:28:38,590 --> 00:28:41,570
Finally, a whole class of
methods that I would like to

517
00:28:41,570 --> 00:28:45,640
briefly mention to you are the
methods of transformations,

518
00:28:45,640 --> 00:28:49,230
where we are operating on
this eigenvalue problem,

519
00:28:49,230 --> 00:28:53,860
recognizing that phi transpose
K phi is equal to lambda.

520
00:28:53,860 --> 00:28:59,170
And phi transpose M phi is
equal to I. Where phi is

521
00:28:59,170 --> 00:29:02,660
defined as shown here, lambda
is defined as shown here.

522
00:29:02,660 --> 00:29:05,960
This fact, of course, we use
already earlier in the

523
00:29:05,960 --> 00:29:06,360
[? multiple ?]

524
00:29:06,360 --> 00:29:07,750
position analysis.

525
00:29:07,750 --> 00:29:12,020
Now how can we get
phi and lambda?

526
00:29:12,020 --> 00:29:13,430
That is our objective.

527
00:29:13,430 --> 00:29:17,900
Well, if we know that this
matrix diagonalizes the K

528
00:29:17,900 --> 00:29:21,340
matrix, and it also diagonalizes
the M matrix,

529
00:29:21,340 --> 00:29:25,370
then why not try to construct
it by iteration?

530
00:29:25,370 --> 00:29:29,520
And this is the basic idea in
the transformation methods.

531
00:29:29,520 --> 00:29:33,410
What we do is the following-- we
take the original K matrix,

532
00:29:33,410 --> 00:29:37,720
we are operating with
P1 transpose K P1.

533
00:29:37,720 --> 00:29:45,060
On K, and on M, such as to make
this resultant matrix

534
00:29:45,060 --> 00:29:50,020
here closer to a
diagonal form.

535
00:29:50,020 --> 00:29:52,470
Ultimately, we want to get
a diagonal matrix.

536
00:29:52,470 --> 00:29:56,320
Well, one step in the iteration
is to make this P1

537
00:29:56,320 --> 00:30:00,610
transpose K P1 matrix, and P1
transpose M P1 matrix, both of

538
00:30:00,610 --> 00:30:03,550
these matrices a little closer
to diagonal form.

539
00:30:03,550 --> 00:30:06,030
Then K and M were
by themselves.

540
00:30:06,030 --> 00:30:08,260
So this is the first step.

541
00:30:08,260 --> 00:30:12,360
Now, we have here, let's call
this a K2 matrix, and this

542
00:30:12,360 --> 00:30:15,870
here an M2 matrix.

543
00:30:18,760 --> 00:30:23,240
Now, we are operating with P2
transpose on K2, and also

544
00:30:23,240 --> 00:30:24,400
there's a P2 here.

545
00:30:24,400 --> 00:30:28,250
So let me take a different color
to show that now we are

546
00:30:28,250 --> 00:30:34,020
operating with this part here,
and that part there.

547
00:30:38,850 --> 00:30:41,560
And this, of course, we are
doing in such a way as to have

548
00:30:41,560 --> 00:30:47,730
that P2 transposed, K2, P2
is a little bit closer to

549
00:30:47,730 --> 00:30:52,440
diagonal form than was K2.

550
00:30:52,440 --> 00:30:55,320
I hope you can see these
lines-- let me put

551
00:30:55,320 --> 00:30:57,570
them in once more.

552
00:30:57,570 --> 00:31:01,720
What I'm saying is we have K2
after having applied P1

553
00:31:01,720 --> 00:31:07,200
transpose on P1, on K,
and similar for M.

554
00:31:07,200 --> 00:31:11,010
And now we want to make K2 still
a little bit closer to

555
00:31:11,010 --> 00:31:12,380
diagonal form.

556
00:31:12,380 --> 00:31:17,780
Well, we are now applying P2
transposed on K2 and the P2 on

557
00:31:17,780 --> 00:31:21,460
K2, this one here has been
replaced by K2, this one here

558
00:31:21,460 --> 00:31:23,240
has been replaced by M2.

559
00:31:23,240 --> 00:31:26,880
And what we're saying is that
this final result, in this

560
00:31:26,880 --> 00:31:30,760
final result should be closer
to diagonal form.

561
00:31:30,760 --> 00:31:33,510
In fact, this should be closer
to I matrix, and this should

562
00:31:33,510 --> 00:31:36,790
be closer to the lambda
matrix, ideally.

563
00:31:36,790 --> 00:31:41,280
Well, this is the basic process
that is used in the

564
00:31:41,280 --> 00:31:43,750
generalized Jacobi method.

565
00:31:43,750 --> 00:31:46,750
We can show that if
we choose specific

566
00:31:46,750 --> 00:31:50,770
matrices, P I matrices.

567
00:31:50,770 --> 00:31:55,220
That zero always one [? of ?]
diagonal element.

568
00:31:55,220 --> 00:31:58,970
In other words, the P I matrix,
zeroes a particular

569
00:31:58,970 --> 00:32:03,720
off diagonal element in this
current coefficient matrix,

570
00:32:03,720 --> 00:32:09,320
similarly for the M2, for the
current M coefficient matrix.

571
00:32:09,320 --> 00:32:13,920
Then we know in the generalized
Jacobi method that

572
00:32:13,920 --> 00:32:15,795
the process finally converges.

573
00:32:15,795 --> 00:32:21,070
It converges and we get a
diagonal matrix here.

574
00:32:21,070 --> 00:32:23,860
With proper scaling, it will
be the identity matrix.

575
00:32:23,860 --> 00:32:27,670
And here, we are getting also a
diagonal matrix, and if this

576
00:32:27,670 --> 00:32:31,590
is an identity matrix, this will
actually be the matrix

577
00:32:31,590 --> 00:32:33,330
storing the eigenvalues.

578
00:32:33,330 --> 00:32:36,310
Well, this is the generalized
Jacobi method here.

579
00:32:36,310 --> 00:32:39,270
They are other techniques that
can also be used, of course,

580
00:32:39,270 --> 00:32:44,990
this is a very common approach
taken when we have M being

581
00:32:44,990 --> 00:32:47,460
equal to the identity matrix.

582
00:32:47,460 --> 00:32:50,270
However, the disadvantage of
this approach is that we are

583
00:32:50,270 --> 00:32:53,060
calculating all eigen pairs
simultaneously.

584
00:32:53,060 --> 00:32:55,570
In finite element analysis,
we are only interested in

585
00:32:55,570 --> 00:33:00,240
specific eigenvalues, so lowest
P, or a certain number

586
00:33:00,240 --> 00:33:01,880
of eigenvalues in an interval.

587
00:33:01,880 --> 00:33:05,810
Here, we have to calculate all
of the eigenvalues stored in

588
00:33:05,810 --> 00:33:08,790
the lambda matrix, and all
of the eigenvectors.

589
00:33:08,790 --> 00:33:14,676
And the eigenvectors here, would
be these vectors here.

590
00:33:14,676 --> 00:33:21,692
These are the the matrix phi,
which is P1 times P2, up to P

591
00:33:21,692 --> 00:33:24,390
K, after K steps of iteration.

592
00:33:24,390 --> 00:33:28,295
This gives us the matrix
phi, storing all of the

593
00:33:28,295 --> 00:33:30,270
eigenvectors.

594
00:33:30,270 --> 00:33:33,860
So the disadvantage is that we
are calculating all of that

595
00:33:33,860 --> 00:33:37,720
eigenvectors, which in finite
element analysis is hardly

596
00:33:37,720 --> 00:33:38,770
ever required.

597
00:33:38,770 --> 00:33:43,810
All we want is the total number
of eigenvalues and

598
00:33:43,810 --> 00:33:44,760
eigenvectors.

599
00:33:44,760 --> 00:33:47,740
Remember, also if K
is of [? order ?]

600
00:33:47,740 --> 00:33:51,310
1,000, this would be a
tremendous amount of work

601
00:33:51,310 --> 00:33:54,500
involved and, in fact, it would
be beyond the current

602
00:33:54,500 --> 00:33:57,060
state of computing capabilities
to use a

603
00:33:57,060 --> 00:34:02,700
generalized Jacobi method on a
K matrix 1,000x1,000 and an M

604
00:34:02,700 --> 00:34:04,420
matrix 1,000x1,000.

605
00:34:04,420 --> 00:34:08,320
You will certainly stretch the
current computing capabilities

606
00:34:08,320 --> 00:34:12,770
tremendously by using the
generalized Jacobi method on

607
00:34:12,770 --> 00:34:15,080
large eigenproblems.

608
00:34:15,080 --> 00:34:20,199
For large eigenproblems, it is
really best to combine the

609
00:34:20,199 --> 00:34:22,560
basic approaches that
I just mentioned.

610
00:34:22,560 --> 00:34:25,070
Let's recall the basic
approaches.

611
00:34:25,070 --> 00:34:28,920
There were iteration methods,
vector iteration methods,

612
00:34:28,920 --> 00:34:33,040
there were Sturm sequence
methods, there was the

613
00:34:33,040 --> 00:34:35,580
transformation method.

614
00:34:35,580 --> 00:34:38,050
And then, of course, there were

615
00:34:38,050 --> 00:34:39,590
polynomial iteration methods.

616
00:34:39,590 --> 00:34:41,980
So really, four classes
of methods that

617
00:34:41,980 --> 00:34:43,500
we are talking about.

618
00:34:43,500 --> 00:34:46,060
Each of them, we find,
have some advantages.

619
00:34:46,060 --> 00:34:50,889
Well, as engineer, surely we
want to now combine the

620
00:34:50,889 --> 00:34:54,350
advantages of each of these
methods to come up with

621
00:34:54,350 --> 00:34:57,430
optimum techniques for our
specific problems.

622
00:34:57,430 --> 00:35:01,410
Well, there are a number
of thoughts

623
00:35:01,410 --> 00:35:02,380
that we would go through.

624
00:35:02,380 --> 00:35:05,600
First of all, a determinant
search, a polynomial iteration

625
00:35:05,600 --> 00:35:09,640
method would be effective
to get near a root.

626
00:35:09,640 --> 00:35:12,930
Vector iteration then could
be used to calculate an

627
00:35:12,930 --> 00:35:14,880
eigenvalue and an eigenvector.

628
00:35:14,880 --> 00:35:18,180
The transformation method would
be good to orthogonalize

629
00:35:18,180 --> 00:35:22,060
the current iteration vectors
if we iterate with more than

630
00:35:22,060 --> 00:35:23,280
just one vector.

631
00:35:23,280 --> 00:35:26,870
And the Sturm sequence method
is extremely powerful to

632
00:35:26,870 --> 00:35:30,360
assure that we have actually
calculated the required

633
00:35:30,360 --> 00:35:33,520
eigenvalues and eigenvectors.

634
00:35:33,520 --> 00:35:36,480
Well, the first method then that
I would like to discuss

635
00:35:36,480 --> 00:35:38,580
with you is the determinant
search method.

636
00:35:38,580 --> 00:35:42,370
In this method, we have
combined some of these

637
00:35:42,370 --> 00:35:44,510
techniques that I pointed
out to you earlier

638
00:35:44,510 --> 00:35:46,340
in an optimum way.

639
00:35:46,340 --> 00:35:49,760
Here, we have the polynomial P
lambda, we want to, of course,

640
00:35:49,760 --> 00:35:51,770
solve for the roots of
that polynomial--

641
00:35:51,770 --> 00:35:53,660
that is our objective.

642
00:35:53,660 --> 00:35:55,690
Here is lambda 1, there
is lambda 2.

643
00:35:55,690 --> 00:35:58,770
I also show a point,
A, because I will

644
00:35:58,770 --> 00:36:00,750
refer to it just now.

645
00:36:00,750 --> 00:36:04,200
The basic scheme in the
determinant search method is

646
00:36:04,200 --> 00:36:09,140
to look for the zeroes of
the determinant, of the

647
00:36:09,140 --> 00:36:14,260
determinant K minus mu I M. Of
course, these zeroes give us R

648
00:36:14,260 --> 00:36:15,590
equal to the eigenvalues.

649
00:36:15,590 --> 00:36:17,560
That's what we're interested
in calculating.

650
00:36:17,560 --> 00:36:20,740
Well, we are calculating
the determinant by the

651
00:36:20,740 --> 00:36:24,970
factorization of this matrix
into L D L transpose, and we

652
00:36:24,970 --> 00:36:27,990
notice that this determinant is
simply the product of the

653
00:36:27,990 --> 00:36:30,230
[? D I I's, ?] as I
mentioned earlier.

654
00:36:30,230 --> 00:36:36,560
Now, having a certain starting
value, mu 0, say, and mu 1.

655
00:36:36,560 --> 00:36:40,170
We can directly use the secant
iteration, that I mentioned to

656
00:36:40,170 --> 00:36:43,700
you earlier to get close
to an eigenvalue.

657
00:36:43,700 --> 00:36:47,380
Now, I also slipped in here
an acceleration factor.

658
00:36:47,380 --> 00:36:52,010
This acceleration factor would
be equal to 1 in the normal

659
00:36:52,010 --> 00:36:55,320
secant iteration that I showed
to you earlier already.

660
00:36:55,320 --> 00:36:58,540
However, since we want to use
this polynomial iteration

661
00:36:58,540 --> 00:37:04,420
technique only to get close to
an eigenvalue, and we might

662
00:37:04,420 --> 00:37:07,260
actually obtain with eta equal
to 1, sometimes slow

663
00:37:07,260 --> 00:37:08,340
convergence.

664
00:37:08,340 --> 00:37:11,810
What we do is we simply
set it larger to

665
00:37:11,810 --> 00:37:14,120
accelerate the process.

666
00:37:14,120 --> 00:37:16,780
And this is what we're
doing here.

667
00:37:16,780 --> 00:37:22,600
If we do jump beyond an
eigenvalue because we have

668
00:37:22,600 --> 00:37:24,840
accelerated the secant
iteration.

669
00:37:24,840 --> 00:37:29,130
In other words, if we do a jump
across here, then since

670
00:37:29,130 --> 00:37:35,510
we are factorizing the matrix
here, we can also look at the

671
00:37:35,510 --> 00:37:38,540
number of negative elements
in the D matrix.

672
00:37:38,540 --> 00:37:43,140
These would tell us that we, in
fact, have jumped across an

673
00:37:43,140 --> 00:37:46,250
unknown eigenvalue.

674
00:37:46,250 --> 00:37:49,510
So, in this particular case, of
course, there would be one

675
00:37:49,510 --> 00:37:51,280
negative element in
the D matrix.

676
00:37:51,280 --> 00:37:54,040
And we would know now we
have jumped across.

677
00:37:54,040 --> 00:37:58,480
The jumping across an unknown
eigenvalue is not possible

678
00:37:58,480 --> 00:38:01,370
when eta is equal to 1.

679
00:38:01,370 --> 00:38:05,610
However, it is possible when
eta is greater than 1.

680
00:38:05,610 --> 00:38:11,600
And however, that jumping
does not hurt us.

681
00:38:11,600 --> 00:38:19,540
In fact, all we will find is
that the Sturm sequence count

682
00:38:19,540 --> 00:38:22,950
will tell us that we have jumped
across an eigenvalue,

683
00:38:22,950 --> 00:38:27,000
and then, of course, we have to
go to another strategy of

684
00:38:27,000 --> 00:38:29,280
solving for the eigenvalue
more accurately.

685
00:38:29,280 --> 00:38:33,110
But the important point is that
all we want to do via

686
00:38:33,110 --> 00:38:36,420
this procedure is to get into
the vicinity of the

687
00:38:36,420 --> 00:38:37,190
eigenvalue.

688
00:38:37,190 --> 00:38:40,780
We do not want to calculate it
accurately, all we want to do

689
00:38:40,780 --> 00:38:44,290
is get into the vicinity of
an unknown eigenvalue.

690
00:38:44,290 --> 00:38:49,080
First, of course, lambda 1,
then lambda 2, and so on.

691
00:38:49,080 --> 00:38:53,380
There's also a concern, of
course, that we do not want to

692
00:38:53,380 --> 00:38:58,310
jump beyond A, because if we, in
our process of looking for

693
00:38:58,310 --> 00:39:05,090
lambda 1, we are jumping too
close to lambda 2, then via

694
00:39:05,090 --> 00:39:07,400
the other solution techniques
that we are using later on

695
00:39:07,400 --> 00:39:10,050
then to calculate the eigenvalue
more accurately, we

696
00:39:10,050 --> 00:39:11,430
would actually converge
to lambda 2.

697
00:39:11,430 --> 00:39:13,200
And, of course, that we
don't want, we want to

698
00:39:13,200 --> 00:39:14,450
calculate lambda 1.

699
00:39:16,940 --> 00:39:22,950
So, the process is then to start
off with eta equal to 1,

700
00:39:22,950 --> 00:39:26,220
iterate, see how we are
progressing towards an unknown

701
00:39:26,220 --> 00:39:27,160
eigenvalue.

702
00:39:27,160 --> 00:39:29,410
If that progressing is
very slow, we are

703
00:39:29,410 --> 00:39:31,740
choosing eta larger.

704
00:39:31,740 --> 00:39:34,470
There are specific strategies
that we're using to choose the

705
00:39:34,470 --> 00:39:37,730
right amount all eta.

706
00:39:37,730 --> 00:39:41,670
And until we find that we are
close to an eigenvalue, either

707
00:39:41,670 --> 00:39:46,010
from the left or we have jumped
across it already, but

708
00:39:46,010 --> 00:39:47,940
you should never
jumped to far.

709
00:39:47,940 --> 00:39:50,480
And we are making sure that we
never jump beyond that point

710
00:39:50,480 --> 00:39:54,930
A. and having jumped, we find
that the Sturm sequence count

711
00:39:54,930 --> 00:39:57,680
tells us about it.

712
00:39:57,680 --> 00:40:03,310
And now, we are going on to
another strategy, and that

713
00:40:03,310 --> 00:40:06,100
strategy is the vector
iteration, vector inverse

714
00:40:06,100 --> 00:40:09,040
iteration method that I
discussed with you earlier.

715
00:40:09,040 --> 00:40:12,300
What we are doing now, is we
have a new coefficient matrix,

716
00:40:12,300 --> 00:40:13,180
that's the one.

717
00:40:13,180 --> 00:40:15,550
We are sitting here, say, and
now I have assumed that we

718
00:40:15,550 --> 00:40:17,240
have, in fact, jumped.

719
00:40:17,240 --> 00:40:18,170
It's not necessary.

720
00:40:18,170 --> 00:40:20,980
We might have also approached
from this side and come very

721
00:40:20,980 --> 00:40:27,010
close to lambda J. And now, we
are using vector iteration to

722
00:40:27,010 --> 00:40:31,120
get this eigenvalue
accurately.

723
00:40:31,120 --> 00:40:35,150
This coefficient here,
or this value here

724
00:40:35,150 --> 00:40:37,830
corresponds to that length.

725
00:40:37,830 --> 00:40:42,700
And if you are adding this value
here to this shift, we

726
00:40:42,700 --> 00:40:45,630
are getting an approximation
to lambda J. And that, of

727
00:40:45,630 --> 00:40:46,990
course, is our objective.

728
00:40:46,990 --> 00:40:51,680
Once we have lambda J, we have
calculated this eigenvalue--

729
00:40:51,680 --> 00:40:56,960
the eigenvector comes out, in x
k plus 1 as k increases, so

730
00:40:56,960 --> 00:40:59,750
we have calculated this eigen
pair, and now, of course, we

731
00:40:59,750 --> 00:41:03,230
would use deflation method, just
the way I showed it to

732
00:41:03,230 --> 00:41:06,200
your earlier, for the polynomial
iteration method.

733
00:41:06,200 --> 00:41:11,010
To deflate the polynomial of
this value, implicitly--

734
00:41:11,010 --> 00:41:12,560
implicitly, that's important--

735
00:41:12,560 --> 00:41:15,820
and we would continue our
iteration process towards that

736
00:41:15,820 --> 00:41:18,640
eigenvalue next.

737
00:41:18,640 --> 00:41:20,970
In this process, we also want
to make sure that the

738
00:41:20,970 --> 00:41:24,160
iteration vector is deflated
of the previously

739
00:41:24,160 --> 00:41:27,110
eigenvectors, for example,
Gram-Schmidt

740
00:41:27,110 --> 00:41:28,770
orthogonalization.

741
00:41:28,770 --> 00:41:32,800
If the convergence is slow in
this particular iteration

742
00:41:32,800 --> 00:41:35,170
here, we're using also the
really Rayleigh-quotient

743
00:41:35,170 --> 00:41:40,310
iteration to speed up
iteration itself.

744
00:41:40,310 --> 00:41:44,620
The advantage of the method--
well, it calculates only the

745
00:41:44,620 --> 00:41:48,260
eigen pairs that are actually
required., there's no prior

746
00:41:48,260 --> 00:41:51,100
transformation of the
eigenprobelm necessary.

747
00:41:51,100 --> 00:41:53,970
The disadvantage of the method
is that we have to perform

748
00:41:53,970 --> 00:41:56,100
many triangular factorizations.

749
00:41:56,100 --> 00:41:58,800
Therefore, the method is really
only effective for

750
00:41:58,800 --> 00:42:00,760
small banded systems.

751
00:42:00,760 --> 00:42:04,110
If we want to calculate
eigenvalues of large banded

752
00:42:04,110 --> 00:42:07,860
systems, we really need
an algorithm with less

753
00:42:07,860 --> 00:42:11,320
factorizations, and more
vector iterations.

754
00:42:11,320 --> 00:42:14,240
And that is the subspace
iteration method.

755
00:42:14,240 --> 00:42:16,980
In fact, we have now extended
the subspace iteration method

756
00:42:16,980 --> 00:42:20,590
to be also most effective for
small banded systems.

757
00:42:20,590 --> 00:42:23,840
And I will just refer to
it very briefly later.

758
00:42:23,840 --> 00:42:26,940
The basic steps of the subspace
iteration method are

759
00:42:26,940 --> 00:42:28,380
the following.

760
00:42:28,380 --> 00:42:33,090
Here, we have the first
equation used.

761
00:42:33,090 --> 00:42:34,670
We have on the right
hand side, the

762
00:42:34,670 --> 00:42:35,950
current iteration vectors.

763
00:42:35,950 --> 00:42:38,290
Now, notice that in this
particular case, we're

764
00:42:38,290 --> 00:42:42,760
iterating with q vectors,
where q ip larger than p

765
00:42:42,760 --> 00:42:44,350
simultaneously.

766
00:42:44,350 --> 00:42:49,130
This gives us here, q right
hand side vector.

767
00:42:49,130 --> 00:42:52,200
We are solving for
q vectors here.

768
00:42:52,200 --> 00:42:54,930
This is an inverse iteration
step on q vector

769
00:42:54,930 --> 00:42:56,360
simultaneously.

770
00:42:56,360 --> 00:43:00,860
Then we are calculating the
projections of these k and m

771
00:43:00,860 --> 00:43:03,410
matrices on these vectors.

772
00:43:03,410 --> 00:43:07,750
Notice that this pat here,
of course, is equal to

773
00:43:07,750 --> 00:43:08,780
this right hand side.

774
00:43:08,780 --> 00:43:10,850
So we would not calculate
this here.

775
00:43:10,850 --> 00:43:13,290
We simply use the right hand
side, which we have evaluated

776
00:43:13,290 --> 00:43:15,560
already and plug it
right in there.

777
00:43:15,560 --> 00:43:19,390
But having calculated these two
matrices, which are now of

778
00:43:19,390 --> 00:43:23,960
order q by q, we are solving
eigenvalue problem all of

779
00:43:23,960 --> 00:43:26,110
these q by q matrices.

780
00:43:26,110 --> 00:43:27,780
And the solution of
that eigenvalue

781
00:43:27,780 --> 00:43:29,230
problem is shown here.

782
00:43:29,230 --> 00:43:34,030
Notice this matrix here
is a q by q matrix.

783
00:43:34,030 --> 00:43:36,020
We're using q vectors.

784
00:43:36,020 --> 00:43:40,970
This is a q by q matrix storing
the eigenvectors of

785
00:43:40,970 --> 00:43:44,470
this eigen problem here, of the
eigenproblem corresponding

786
00:43:44,470 --> 00:43:46,510
to these matrices.

787
00:43:46,510 --> 00:43:51,490
This here is the diagonal matrix
listing the eigenvalues

788
00:43:51,490 --> 00:43:52,830
of the eigenproblem

789
00:43:52,830 --> 00:43:55,090
corresponding to these matrices.

790
00:43:55,090 --> 00:43:58,930
Having calculated these matrices
here, and of course,

791
00:43:58,930 --> 00:44:02,070
that one too, we are performing
this operation to

792
00:44:02,070 --> 00:44:05,390
get a new set of iteration
vectors, which then are

793
00:44:05,390 --> 00:44:07,190
plugged back into here.

794
00:44:07,190 --> 00:44:12,580
Notice that the solution of
this eigenproblem, q by q,

795
00:44:12,580 --> 00:44:17,660
might be a 50x50 matrix here,
100x100 matrix, maybe a

796
00:44:17,660 --> 00:44:20,720
200x200, but that already would
be a large number of

797
00:44:20,720 --> 00:44:21,930
vectors if we're using--

798
00:44:21,930 --> 00:44:23,730
a very large number
of vectors.

799
00:44:23,730 --> 00:44:26,250
This solution of this
eigenproblem can be

800
00:44:26,250 --> 00:44:29,290
effectively achieved via the
generalized Jacobi method,

801
00:44:29,290 --> 00:44:32,270
which I referred to
briefly earlier.

802
00:44:32,270 --> 00:44:35,515
Of course here, we are using
our solution algorithm for

803
00:44:35,515 --> 00:44:36,290
static analysis.

804
00:44:36,290 --> 00:44:41,200
There's L D L transpose
factorization involved here.

805
00:44:41,200 --> 00:44:45,260
And here, this is a simple
matrix multiplication that has

806
00:44:45,260 --> 00:44:46,880
to be carried out.

807
00:44:46,880 --> 00:44:50,620
This part here, is the
eigenproblem solution, and

808
00:44:50,620 --> 00:44:51,450
there's another vector

809
00:44:51,450 --> 00:44:53,980
multiplication carried out here.

810
00:44:53,980 --> 00:44:59,320
Now, under specific conditions,
this eigensolution

811
00:44:59,320 --> 00:45:04,690
algorithm converges and the
vectors stored in x k plus 1,

812
00:45:04,690 --> 00:45:08,940
they are q vectors now here,
with appropriate ordering,

813
00:45:08,940 --> 00:45:11,750
converge to the phi matrix.

814
00:45:11,750 --> 00:45:17,800
This is now an n by q matrix,
storing the lowest

815
00:45:17,800 --> 00:45:22,200
eigenvectorr, or I should
say, the q eigenvectors

816
00:45:22,200 --> 00:45:25,020
corresponding to the
lowest eigenvalues.

817
00:45:27,800 --> 00:45:30,820
And here, we have it written
out, and of course, this is a

818
00:45:30,820 --> 00:45:31,960
diagonal matrix.

819
00:45:31,960 --> 00:45:35,530
The condition that we have to
satisfy here is that the

820
00:45:35,530 --> 00:45:42,800
starting subspace spanned by
these vectors here must not be

821
00:45:42,800 --> 00:45:52,840
orthogonal to the subspace
spanned by the p eigenvectors

822
00:45:52,840 --> 00:45:54,680
that we are interested in.

823
00:45:54,680 --> 00:45:59,160
Of course, here, I'm talking
about q eigenvectors.

824
00:45:59,160 --> 00:46:03,360
I really am only interested in
p eigenvectors, in the lowest

825
00:46:03,360 --> 00:46:05,510
p eigenvectors.

826
00:46:05,510 --> 00:46:09,390
So in general, what happens is
that I will only be interested

827
00:46:09,390 --> 00:46:13,080
in these eigenvectors here, and
the lowest eigenvalues.

828
00:46:13,080 --> 00:46:16,850
One comes out here, I'm not
too much concerned about.

829
00:46:16,850 --> 00:46:21,820
In fact, we are carrying these
along only to accelerate the

830
00:46:21,820 --> 00:46:23,040
iteration scheme.

831
00:46:23,040 --> 00:46:28,060
And I will show you just know
what the convergence rate is,

832
00:46:28,060 --> 00:46:29,770
and why we are carrying
them along.

833
00:46:29,770 --> 00:46:34,800
So since we only interested in
the lowest p eigenvalues and

834
00:46:34,800 --> 00:46:38,870
corresponding eigenvectors,
we want to get convergence

835
00:46:38,870 --> 00:46:48,330
towards those, and what that
means is that our starting

836
00:46:48,330 --> 00:46:57,290
vectors in here must not be m
orthogonal to the eigenvectors

837
00:46:57,290 --> 00:47:00,780
that we want to calculate,
phi 1 to phi p.

838
00:47:00,780 --> 00:47:02,960
If that condition is satisfied,
one can show,

839
00:47:02,960 --> 00:47:06,300
theoretically, and of course,
we see it in practice also,

840
00:47:06,300 --> 00:47:09,230
that this iteration scheme will
always converge to the

841
00:47:09,230 --> 00:47:12,385
lowest eigenvectors and
corresponding eigenvalues.

842
00:47:16,010 --> 00:47:20,760
One important point here, once
we have calculated, say, a

843
00:47:20,760 --> 00:47:24,560
certain number of iteration
vectors, and we have tested

844
00:47:24,560 --> 00:47:27,810
for convergence, we are saying
that convergence is reached,

845
00:47:27,810 --> 00:47:32,670
say, when the current iterates
on the eigenvalues don't

846
00:47:32,670 --> 00:47:34,450
change very much anymore.

847
00:47:34,450 --> 00:47:38,720
This tol might be 10 to the
minus 6, as an example.

848
00:47:38,720 --> 00:47:41,820
Then once this condition is
satisfied, we stop the

849
00:47:41,820 --> 00:47:46,440
iteration, and we really want
to make then sure that we

850
00:47:46,440 --> 00:47:48,910
really converge to the
eigenvalues of interest.

851
00:47:48,910 --> 00:47:50,580
And how is that accomplished?

852
00:47:50,580 --> 00:47:53,310
Well, if these are the
p eigenvalues, we are

853
00:47:53,310 --> 00:47:57,060
calculating error bounds on
these eigenvalues, indicated

854
00:47:57,060 --> 00:48:01,690
here by the blue lines here.

855
00:48:01,690 --> 00:48:06,670
And then just to the right
of the error bound.

856
00:48:06,670 --> 00:48:11,300
Or we can really use that error
bound, we apply a Sturm

857
00:48:11,300 --> 00:48:15,520
sequence shift, mu S. What this
then says that k minus mu

858
00:48:15,520 --> 00:48:18,150
S factorize into L
D L transpose.

859
00:48:23,090 --> 00:48:29,870
And we must now have in the D
matrix, p negative elements.

860
00:48:29,870 --> 00:48:35,390
And only p, not more, not less,
just p negative elements

861
00:48:35,390 --> 00:48:39,050
must be occurring
in the D matrix.

862
00:48:39,050 --> 00:48:44,410
This check is absolutely sure.

863
00:48:44,410 --> 00:48:47,510
If this check a satisfied, we
can be absolutely sure that we

864
00:48:47,510 --> 00:48:50,970
have calculated the eigenvalues
of interest, the

865
00:48:50,970 --> 00:48:53,830
lowest p eigenvalues, in
this particular case.

866
00:48:53,830 --> 00:48:57,910
So I repeat, after we have
satisfied this convergence

867
00:48:57,910 --> 00:49:03,510
rate, or this conversions
limit, I should say--

868
00:49:03,510 --> 00:49:06,400
typically 10 to the minus
6, or 10 to the minus 8,

869
00:49:06,400 --> 00:49:08,700
depending on what accuracy
you want.

870
00:49:08,700 --> 00:49:12,920
We calculating eigenvalue
bounds, as shown here.

871
00:49:12,920 --> 00:49:18,500
We are putting a mu S Sturm
sequence shift just to the

872
00:49:18,500 --> 00:49:24,460
right of the last eigenvalue
bound, and if this is the p's

873
00:49:24,460 --> 00:49:28,540
eigenvalue, we must have
here, exactly p

874
00:49:28,540 --> 00:49:30,670
negative elements indeed.

875
00:49:30,670 --> 00:49:35,390
The convergence rate of the
iteration, we find is given by

876
00:49:35,390 --> 00:49:38,110
lambda i divided by
lambda q plus 1.

877
00:49:40,750 --> 00:49:44,650
And the convergence rate to the
eigenvalues of interest

878
00:49:44,650 --> 00:49:47,140
are just this value squared.

879
00:49:47,140 --> 00:49:50,280
Notice that we have
here, q plus 1.

880
00:49:50,280 --> 00:49:53,840
So if we are interested in the
p's eigenvalue,== the black

881
00:49:53,840 --> 00:49:56,850
pen doesn't work too well, let
me take the blue one here--

882
00:49:56,850 --> 00:50:01,330
if you're interested in the p's
eigenvalue as the largest

883
00:50:01,330 --> 00:50:04,220
eigenvalue, notice that the
convergence rate is that

884
00:50:04,220 --> 00:50:07,660
lambda p plus-- lambda p divided
by lambda q plus 1.

885
00:50:07,660 --> 00:50:11,030
That is the worst convergence
rate that we are reaching

886
00:50:11,030 --> 00:50:14,840
because the other eigenvalues
below lambda p are smaller,

887
00:50:14,840 --> 00:50:16,740
giving us better convergence
rate.

888
00:50:16,740 --> 00:50:22,110
So this is the worst that we
can reach, if we are only

889
00:50:22,110 --> 00:50:25,280
interested in p eigenvalues
and corresponding

890
00:50:25,280 --> 00:50:26,790
eigenvectors.

891
00:50:26,790 --> 00:50:29,850
Notice however, that lambda
q plus 1 must be

892
00:50:29,850 --> 00:50:31,290
greater than lambda p.

893
00:50:31,290 --> 00:50:33,440
If lambda q plus 1 is
equal to lambda p,

894
00:50:33,440 --> 00:50:34,880
we would never converge.

895
00:50:34,880 --> 00:50:38,930
So this is the reason why we
are choosing q to be larger

896
00:50:38,930 --> 00:50:42,030
than p, substantially
large than p.

897
00:50:42,030 --> 00:50:49,640
For example, we might choose
q to be typically, as an

898
00:50:49,640 --> 00:50:57,850
example, q being the maximum
of 2p and p plus 8.

899
00:50:57,850 --> 00:51:00,940
This is a formula that
we have used with

900
00:51:00,940 --> 00:51:02,700
quite a lot of success.

901
00:51:02,700 --> 00:51:05,920
In other words, when p is equal
to 4, we would use--

902
00:51:11,060 --> 00:51:15,710
no, sorry, I should say
the minimum here.

903
00:51:15,710 --> 00:51:19,100
the In other words, when
p is equal to 4, we

904
00:51:19,100 --> 00:51:21,430
would have 8 here.

905
00:51:21,430 --> 00:51:24,410
When p is equal to
8, it's the same.

906
00:51:24,410 --> 00:51:26,390
Because these two values
give us the same.

907
00:51:26,390 --> 00:51:30,470
And when p is equal to 100,
then we just have 8 more.

908
00:51:30,470 --> 00:51:32,850
Q is equal to 108.

909
00:51:32,850 --> 00:51:37,870
This 8 here is a safeguard, so
that this cannot be equal to 1

910
00:51:37,870 --> 00:51:39,130
in practice.

911
00:51:39,130 --> 00:51:43,960
It would only be equal to 1 if
we would have, basically.

912
00:51:43,960 --> 00:51:47,390
9 time the same root, and of
course, that is a case which

913
00:51:47,390 --> 00:51:50,290
we would hardly encounter
in practice.

914
00:51:50,290 --> 00:51:55,060
Now, regarding the choice
of the starting values.

915
00:51:55,060 --> 00:51:57,860
For the vectors, there
are two choices.

916
00:51:57,860 --> 00:52:03,070
x1 is simply a vector
with 1 only.

917
00:52:03,070 --> 00:52:10,220
xj would be a vector with zeroes
everywhere, but with a

918
00:52:10,220 --> 00:52:11,300
1 somewhere.

919
00:52:11,300 --> 00:52:12,720
And that one is in
the j's entry--

920
00:52:15,590 --> 00:52:18,310
sorry, the k's entry.

921
00:52:18,310 --> 00:52:23,050
The k's entry for ek, ek has
in the case, entry of 1.

922
00:52:23,050 --> 00:52:25,630
And otherwise, everywhere
zeroes.

923
00:52:25,630 --> 00:52:29,390
This would be the second to
the q minus first vector.

924
00:52:29,390 --> 00:52:31,150
And the last vector, we
take a random vector.

925
00:52:31,150 --> 00:52:33,190
That is one choice of
starting vectors.

926
00:52:33,190 --> 00:52:36,310
The Lanczos method can also be
used to generate starting

927
00:52:36,310 --> 00:52:37,560
vectors very effectively.

928
00:52:41,260 --> 00:52:45,450
Once we have selected the
starting vector, we go through

929
00:52:45,450 --> 00:52:48,770
the iteration schemes the way
I have been displaying it.

930
00:52:48,770 --> 00:52:53,420
And there, it is important, once
again, to perform after

931
00:52:53,420 --> 00:52:56,510
the convergence- the Sturm
system sequence check, and

932
00:52:56,510 --> 00:53:00,610
also we might want to calculate
error bounds on the

933
00:53:00,610 --> 00:53:01,970
eigenvalues and eigenvectors.

934
00:53:01,970 --> 00:53:05,240
These error bounds would be
calculated by simply saying,

935
00:53:05,240 --> 00:53:08,070
well, if we want to satisfy--

936
00:53:08,070 --> 00:53:11,660
or if we want to solve for
lambdas and phis that

937
00:53:11,660 --> 00:53:17,400
satisfies this equation,
then why not see how

938
00:53:17,400 --> 00:53:18,650
close we will satisfy.

939
00:53:18,650 --> 00:53:23,780
Make that a minus and put here
an error, say, R, on

940
00:53:23,780 --> 00:53:25,050
the right hand side.

941
00:53:25,050 --> 00:53:28,260
Substituting then the last
values for lambda and phi that

942
00:53:28,260 --> 00:53:32,220
we just kind calculated, the
last iterative values, we get

943
00:53:32,220 --> 00:53:34,180
this top line here.

944
00:53:34,180 --> 00:53:36,500
We're taking a norm
the way I've been

945
00:53:36,500 --> 00:53:38,310
talking about earlier.

946
00:53:38,310 --> 00:53:43,730
And we are also taking a norm
at the bottom to get,

947
00:53:43,730 --> 00:53:46,690
basically, a relative value,
top to bottom.

948
00:53:46,690 --> 00:53:49,240
And then relative value
gives us epsilon i.

949
00:53:49,240 --> 00:53:54,070
So epsilon i should be of order,
10 to the minus 3, if

950
00:53:54,070 --> 00:53:56,740
we want to have three-digit
accuracy in the eigenvalues

951
00:53:56,740 --> 00:53:57,770
and eigenvectors.

952
00:53:57,770 --> 00:54:00,020
In fact, the eigenvalues will
generally be more accurate

953
00:54:00,020 --> 00:54:02,040
than the eigenvectors.

954
00:54:02,040 --> 00:54:04,130
That is, of course, a
very low accuracy.

955
00:54:04,130 --> 00:54:06,260
Ideally, we have much
more accuracy after

956
00:54:06,260 --> 00:54:07,330
the iteration schemes.

957
00:54:07,330 --> 00:54:11,330
We would have to tighten the
tolerance, the tol value that

958
00:54:11,330 --> 00:54:15,010
I referred to earlier
sufficiently to make epsilon i

959
00:54:15,010 --> 00:54:16,180
small enough.

960
00:54:16,180 --> 00:54:19,370
Now, I like to refer, very
briefly also, to an

961
00:54:19,370 --> 00:54:21,740
accelerated subspace iteration
method that we have been

962
00:54:21,740 --> 00:54:26,800
developing during the last
years, which is a very

963
00:54:26,800 --> 00:54:30,340
substantial extension of this
subspace iteration method.

964
00:54:30,340 --> 00:54:36,080
The idea here is, basically,
that we wanted to combine the

965
00:54:36,080 --> 00:54:39,690
sum of the effective techniques
that we are using

966
00:54:39,690 --> 00:54:44,115
in the polynomial iteration
method, with the standard

967
00:54:44,115 --> 00:54:46,230
subspace iteration method
that I have

968
00:54:46,230 --> 00:54:48,340
discussed with you here.

969
00:54:48,340 --> 00:54:51,710
Basically, in the accelerated
subspace iteration method, we

970
00:54:51,710 --> 00:54:55,550
are not keeping a constant
coefficient matrix, as

971
00:54:55,550 --> 00:54:56,870
we have done here.

972
00:54:56,870 --> 00:54:59,160
The k matrix never changed.

973
00:54:59,160 --> 00:55:02,640
Here, we are shifting the
coefficient matrix, we are

974
00:55:02,640 --> 00:55:04,480
shifting through the spectrum.

975
00:55:04,480 --> 00:55:08,860
Much in the same way as we are
pursuing it in the determined

976
00:55:08,860 --> 00:55:10,070
search method.

977
00:55:10,070 --> 00:55:11,450
And that is number one.

978
00:55:11,450 --> 00:55:15,720
And number two-- we are also
not iterating with vectors

979
00:55:15,720 --> 00:55:18,610
that are always larger than
the number of eigenvectors

980
00:55:18,610 --> 00:55:19,710
that we want to calculated.

981
00:55:19,710 --> 00:55:22,990
In fact, if we want to
calculate, say, 100

982
00:55:22,990 --> 00:55:27,700
eigenvalues, we might just use
four iteration vectors at a

983
00:55:27,700 --> 00:55:31,420
time, and shift through the
spectrum, always with four

984
00:55:31,420 --> 00:55:35,020
iteration vectors, capturing
the eigenvalues values and

985
00:55:35,020 --> 00:55:35,980
eigenvectors that we are really

986
00:55:35,980 --> 00:55:37,800
interested in, in bundles.

987
00:55:37,800 --> 00:55:42,080
So this is a way how we have
very substantially accelerated

988
00:55:42,080 --> 00:55:45,310
the standard subspace iteration
method that I have

989
00:55:45,310 --> 00:55:49,970
been talking to you
about just now.

990
00:55:49,970 --> 00:55:52,315
If you are interested in seeing
some solution times,

991
00:55:52,315 --> 00:55:56,180
please we look at
this reference.

992
00:55:56,180 --> 00:55:59,960
Ladies and gentlemen, this
then brings me to the

993
00:55:59,960 --> 00:56:01,640
conclusion of this lecture.

994
00:56:01,640 --> 00:56:05,630
And in fact, it also brings us
to the conclusion of the whole

995
00:56:05,630 --> 00:56:09,300
set of lectures that I have
been presenting to you.

996
00:56:09,300 --> 00:56:14,180
These set up lectures have been
brief and very compact.

997
00:56:14,180 --> 00:56:16,900
They've been a survey
introduction to the finite

998
00:56:16,900 --> 00:56:17,760
element method--

999
00:56:17,760 --> 00:56:21,670
a very brief and compact
introduction and survey.

1000
00:56:21,670 --> 00:56:24,520
Brief because there are
only 12 lectures.

1001
00:56:24,520 --> 00:56:27,800
Compact because I wanted to
present to you as much as

1002
00:56:27,800 --> 00:56:30,570
possible in these 12 lectures.

1003
00:56:30,570 --> 00:56:34,390
And there is, of course, a lot
of information that we could

1004
00:56:34,390 --> 00:56:35,900
have talked about.

1005
00:56:35,900 --> 00:56:40,560
However, I hope that the
information that I had given

1006
00:56:40,560 --> 00:56:44,760
in the lectures has been of
interest to you, and it will

1007
00:56:44,760 --> 00:56:52,070
also be valuable to you in your
work and general decision

1008
00:56:52,070 --> 00:56:54,680
making process of whether you
want to use a finite element

1009
00:56:54,680 --> 00:56:58,120
method, how you want to
use it, and so on.

1010
00:56:58,120 --> 00:57:01,600
I like to take the opportunity
at this point to thank the

1011
00:57:01,600 --> 00:57:04,670
Center of Advanced Engineering
Studies for the terrific

1012
00:57:04,670 --> 00:57:07,260
effort they've made in
preparing the tapes.

1013
00:57:07,260 --> 00:57:12,150
I'm very happy that they have
been cooperative with me in

1014
00:57:12,150 --> 00:57:17,750
the way they did, in the
preparation of the tapes.

1015
00:57:17,750 --> 00:57:22,145
I also would like to thank you
as listeners for your patience

1016
00:57:22,145 --> 00:57:25,590
of listening to the tapes and,
once again, I hope you found

1017
00:57:25,590 --> 00:57:27,950
the tapes interesting and
have obtained some value

1018
00:57:27,950 --> 00:57:29,260
information on them.

1019
00:57:29,260 --> 00:57:31,590
Best of luck in your further
studies of the

1020
00:57:31,590 --> 00:57:32,840
finite element method.