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PROFESSOR: Ladies and gentlemen,
welcome to
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lecture number 12.
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In this lecture, I would like
to discuss with you solution
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methods for eigenproblems.
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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finite element analysis.
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So it will be a very brief
introduction, a brief survey
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of the methods of solution that
we're using for finite
00:00:48.390 --> 00:00:50.440
element eigenvalue problems.
00:00:50.440 --> 00:00:54.010
The generalized eigenvalue
problem, K phi equals lambda M
00:00:54.010 --> 00:00:57.350
phi, is a problem that we
encountered in the multiple
00:00:57.350 --> 00:00:59.230
position analysis.
00:00:59.230 --> 00:01:03.480
Please recognize that I'm
calling lambda here, the
00:01:03.480 --> 00:01:05.550
eigenvalue, which was our omega
00:01:05.550 --> 00:01:07.940
squared in the last lecture.
00:01:07.940 --> 00:01:09.950
This is really the eigenproblems
that we are
00:01:09.950 --> 00:01:14.120
primarily addressing ourselves
to finite element analysis.
00:01:14.120 --> 00:01:17.980
The standard eigenproblems
problem here, EVP stands for
00:01:17.980 --> 00:01:21.660
eigenproblem, has M equal
to the identity matrix.
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
00:01:26.470 --> 00:01:27.460
this problem.
00:01:27.460 --> 00:01:29.530
If we have, in a [? multiple ?]
position
00:01:29.530 --> 00:01:33.660
analysis, non-proportional
damping, then we would have to
00:01:33.660 --> 00:01:37.040
solve this eigenproblem, which
is a quadratic eigenproblem.
00:01:37.040 --> 00:01:39.610
I will not address myself to the
solution of this problem,
00:01:39.610 --> 00:01:42.850
so in what follow, we are
talking about a solution of
00:01:42.850 --> 00:01:45.650
this problem, and of course, the
solution of this problem
00:01:45.650 --> 00:01:49.030
also, when M is equal to
the identity matrix.
00:01:49.030 --> 00:01:51.920
In particular, we are interested
in the solution of
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large eigenvalue problems when
N, is say, greater than 500.
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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bandwidth, I should say,
being greater than 60.
00:02:02.900 --> 00:02:05.800
The number of eigenvalues that
we will be looking for, that
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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already large number
of eigenvalues.
00:02:12.460 --> 00:02:16.840
Generally, we are talking about
1 to and at most, 100
00:02:16.840 --> 00:02:18.660
for large eigenproblems.
00:02:18.660 --> 00:02:21.710
When we have a small eigenvalue
problem, in other
00:02:21.710 --> 00:02:25.530
words, N being small, say 50
or 100, there are a large
00:02:25.530 --> 00:02:27.940
number of different techniques
that can be used to solve this
00:02:27.940 --> 00:02:31.970
eigenvalue problem, and it
really does not make that much
00:02:31.970 --> 00:02:34.600
difference which technique
you are using.
00:02:34.600 --> 00:02:39.610
The cost can still be quite
significant, but certainly,
00:02:39.610 --> 00:02:43.550
when we're talking about very
large eigenproblems, N being
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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concern to select a technique
that is very effective.
00:02:52.000 --> 00:02:55.670
And I want to address myself
really now towards those
00:02:55.670 --> 00:02:59.590
techniques that we feel are
very effectively used in
00:02:59.590 --> 00:03:01.960
finite element analysis.
00:03:01.960 --> 00:03:04.650
In dynamic analysis, of course,
with proportional
00:03:04.650 --> 00:03:07.930
damping, this was the eigenvalue
problem that we
00:03:07.930 --> 00:03:09.290
wanted to solve.
00:03:09.290 --> 00:03:12.360
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
00:03:15.540 --> 00:03:18.750
the following procedure to
basically getting rid of these
00:03:18.750 --> 00:03:19.630
0 frequencies.
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We simply add, on the left hand
side, a mu M phi, and on
00:03:23.040 --> 00:03:25.950
the right hand side,
a mu M phi.
00:03:25.950 --> 00:03:28.880
And taking these two together,
we are getting this
00:03:28.880 --> 00:03:32.740
coefficient matrix, and
a new lambda value.
00:03:32.740 --> 00:03:35.460
That lambda value now is omega
squared plus mu, where omega
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squared is equal to
lambda minus mu.
00:03:38.630 --> 00:03:41.870
What we have been doing here,
is basically, we have been
00:03:41.870 --> 00:03:43.350
applying a shift.
00:03:43.350 --> 00:03:46.210
In other words, if we were to
plot the characteristic
00:03:46.210 --> 00:03:50.560
polynomial, giving us
the eigenvalues, the
00:03:50.560 --> 00:03:53.600
characteristic polynomial of
course, being p of lambda
00:03:53.600 --> 00:03:57.940
being determined K bar minus
lambda M, where K bar is equal
00:03:57.940 --> 00:04:00.490
to K plus mu M.
00:04:00.490 --> 00:04:03.270
If we plug this characteristic
polynomial, we will see this
00:04:03.270 --> 00:04:06.960
curve here, schematically drawn,
of course, where the
00:04:06.960 --> 00:04:10.250
omega squared values would start
here, in other words,
00:04:10.250 --> 00:04:14.370
the lowest omega squared
value is, in fact, 0.
00:04:14.370 --> 00:04:17.589
Whereas our lambda value
now starts right here.
00:04:17.589 --> 00:04:19.500
This is the shift that
we [? imposed. ?]
00:04:19.500 --> 00:04:25.755
So, basically, we now recognize
that we have all
00:04:25.755 --> 00:04:29.090
eigenvalues being greater than
0, because this is our new
00:04:29.090 --> 00:04:30.100
starting value.
00:04:30.100 --> 00:04:33.150
All eigenvalues measured from
here are greater than 0, the
00:04:33.150 --> 00:04:35.780
smallest one being mu,
namely when omega
00:04:35.780 --> 00:04:37.200
squared is equal to 0.
00:04:37.200 --> 00:04:39.610
And the larger ones
being here.
00:04:39.610 --> 00:04:45.260
It is effective to perform such
a shift when we have 0
00:04:45.260 --> 00:04:49.960
eigenvalues, in other words, to
operate on this eigenvalue,
00:04:49.960 --> 00:04:54.790
rather than on that one, because
our techniques then
00:04:54.790 --> 00:04:58.330
are very stable and effective,
and this shift here is, in
00:04:58.330 --> 00:05:00.950
general, therefore performed.
00:05:00.950 --> 00:05:04.540
So all we have to address
ourselves to then from now on,
00:05:04.540 --> 00:05:08.390
is really, the solution of
eigenvalues and eigenvectors,
00:05:08.390 --> 00:05:12.470
when all of the eigenvalues
are greater than 0.
00:05:12.470 --> 00:05:15.720
If we start off with a problem
where we have a 0 frequency,
00:05:15.720 --> 00:05:19.160
we apply the shift so that the
eigenvalues now of this
00:05:19.160 --> 00:05:23.280
problem here are all
greater than 0.
00:05:23.280 --> 00:05:26.680
In buckling analysis, we
have this eigenproblem.
00:05:26.680 --> 00:05:28.185
We really did not talk about--
00:05:28.185 --> 00:05:31.700
we did not really at all about
buckling analysis in the set
00:05:31.700 --> 00:05:35.930
of lectures, but you might just
be interested in finding
00:05:35.930 --> 00:05:38.140
out how would we solve for
these eigenvalues.
00:05:38.140 --> 00:05:40.320
Well, here is the characteristic
polynomial.
00:05:40.320 --> 00:05:43.390
The smallest lambda value, of
course, gives now a critical
00:05:43.390 --> 00:05:46.270
load, that is the one e would
be interested in.
00:05:46.270 --> 00:05:48.640
It is effective here,
to rewrite this
00:05:48.640 --> 00:05:50.700
problem in this form.
00:05:50.700 --> 00:05:53.970
All we do, is we are taking
the lambda value onto the
00:05:53.970 --> 00:05:54.920
other side.
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In other words, we
are dividing this
00:05:56.660 --> 00:05:58.570
equation here by a lambda.
00:05:58.570 --> 00:06:01.320
So that one of the lambda stands
on this side, and KG
00:06:01.320 --> 00:06:02.730
phi on that side.
00:06:02.730 --> 00:06:05.940
The resulting problem is this
one, where kappa is 1 over
00:06:05.940 --> 00:06:08.540
lambda and we have now
here, is a K matrix.
00:06:08.540 --> 00:06:14.630
K is positive definite, and now
it is more effective to
00:06:14.630 --> 00:06:17.040
operate on this equation.
00:06:17.040 --> 00:06:19.820
How do we obtain now
the largest kappa?
00:06:19.820 --> 00:06:22.130
Because the largest kappa is the
one that we are interested
00:06:22.130 --> 00:06:25.010
in, which corresponds to the
smallest lambda, the smallest
00:06:25.010 --> 00:06:25.730
buckling load.
00:06:25.730 --> 00:06:28.850
Well, we would now perform
a shift to the right,
00:06:28.850 --> 00:06:34.190
subtracting mu K from this side,
from this side, and from
00:06:34.190 --> 00:06:37.360
that side, to obtain this
eigenvalue problem.
00:06:37.360 --> 00:06:39.980
That means we are really
shifting from here up to
00:06:39.980 --> 00:06:45.130
there, and now we are interested
in solving kappa N
00:06:45.130 --> 00:06:47.620
by operating on this
eigenproblems problem.
00:06:47.620 --> 00:06:50.100
It is this eigenvalue problem
that we can operate on
00:06:50.100 --> 00:06:53.540
directly, using the techniques
that I would be discussing
00:06:53.540 --> 00:06:55.950
later, to determine search
method and the subspace
00:06:55.950 --> 00:06:58.222
iteration method.
00:06:58.222 --> 00:07:00.990
But before we get into those
techniques, I would like to
00:07:00.990 --> 00:07:05.810
discuss with you some
preliminary considerations
00:07:05.810 --> 00:07:07.710
that are important
when we look at
00:07:07.710 --> 00:07:09.440
an eigenvalue solution.
00:07:09.440 --> 00:07:14.030
The important first point that
I want to make is that there
00:07:14.030 --> 00:07:17.260
is a traditional approach
for the solution of this
00:07:17.260 --> 00:07:18.840
eigenvalue problem here.
00:07:18.840 --> 00:07:20.610
Let's look at this eigenvalue
problem.
00:07:20.610 --> 00:07:22.840
Where of course, when we're
talking about a buckling
00:07:22.840 --> 00:07:25.700
analysis, we would have
different matrices here, but
00:07:25.700 --> 00:07:29.860
basically, we still solve
this eigenvalue problem.
00:07:29.860 --> 00:07:34.160
The traditional approach lies
in factorizing the M matrix
00:07:34.160 --> 00:07:36.120
into Cholesky factors.
00:07:36.120 --> 00:07:38.880
We talked about the Cholesky
factor when we talked about
00:07:38.880 --> 00:07:40.230
solution of equations.
00:07:40.230 --> 00:07:41.750
So here it comes up again.
00:07:41.750 --> 00:07:45.440
We would factorize the M matrix
into Cholesky factors,
00:07:45.440 --> 00:07:46.470
as shown here.
00:07:46.470 --> 00:07:50.730
We define now a new eigenvector,
phi curl being
00:07:50.730 --> 00:07:54.880
equal as shown here, L curl
transpose times 5.
00:07:54.880 --> 00:07:59.970
We substitute from here into
there, and we directly obtain
00:07:59.970 --> 00:08:02.640
this eigenvalue problem here.
00:08:02.640 --> 00:08:05.980
Now we have the standard
eigenproblem.
00:08:05.980 --> 00:08:09.160
Now once we have a standard
eigenproblem, there are a
00:08:09.160 --> 00:08:11.660
large number of techniques
that can be used.
00:08:11.660 --> 00:08:14.735
For example, the QR method,
the standard
00:08:14.735 --> 00:08:16.790
Jacobi method, et cetera.
00:08:16.790 --> 00:08:18.940
So this is, what
we might call.
00:08:18.940 --> 00:08:21.530
a traditional approach to
solve the generalized
00:08:21.530 --> 00:08:24.140
eigenproblem, this is the one
here that we are really
00:08:24.140 --> 00:08:25.110
interested in.
00:08:25.110 --> 00:08:27.250
But it involves this
transformation, and this
00:08:27.250 --> 00:08:32.280
transformation here can involve
a very significant
00:08:32.280 --> 00:08:35.070
cost, because we have to find
these Cholesky factors and
00:08:35.070 --> 00:08:37.929
then, of course, we have to
calculate this K curl.
00:08:37.929 --> 00:08:40.370
Another way of performing
a transformation of the
00:08:40.370 --> 00:08:43.549
generalized item problem to a
standard form would be to
00:08:43.549 --> 00:08:46.540
solve the eigenproblem
of M first.
00:08:46.540 --> 00:08:49.110
W here and W transpose
[? store ?]
00:08:49.110 --> 00:08:54.080
the eigenvectors of the M
matrix, and D squared is a
00:08:54.080 --> 00:08:57.320
diagonal matrix of the
eigenvalues of M. And then
00:08:57.320 --> 00:09:01.020
having factored M in this form,
an equation similar to
00:09:01.020 --> 00:09:02.510
this one here can be written.
00:09:02.510 --> 00:09:05.490
Where now, the transformation,
of course, involves the
00:09:05.490 --> 00:09:07.400
eigenvectors, W.
00:09:07.400 --> 00:09:12.860
Well, this approach is
sometimes still quite
00:09:12.860 --> 00:09:15.990
effective, but in general, it
cannot be recommended because
00:09:15.990 --> 00:09:20.420
there's too much cost involved
in the factorizing of the M
00:09:20.420 --> 00:09:23.830
matrix, and then particular,
in the transformation here,
00:09:23.830 --> 00:09:26.180
and the solution of this problem
and the, of course, to
00:09:26.180 --> 00:09:29.670
get from this problem, the phi
vector, we have to go back to
00:09:29.670 --> 00:09:30.980
this equation.
00:09:30.980 --> 00:09:36.070
A direct solution of the
generalized eigenvalue problem
00:09:36.070 --> 00:09:39.390
that we want to solve is
really more effective.
00:09:39.390 --> 00:09:43.680
And now, we are ordering the
lambdas as shown here.
00:09:43.680 --> 00:09:48.310
Remember, our smallest lambda
now is greater than 0, the
00:09:48.310 --> 00:09:52.490
eigonvectors are stored as shown
here, and for most of a
00:09:52.490 --> 00:09:56.150
position analysis, we are really
interested in solving
00:09:56.150 --> 00:10:00.550
only for a certain number of
eigen pairs, i equals 1 to p,
00:10:00.550 --> 00:10:04.200
as discussed in the last
lecture, or i equals r to s.
00:10:04.200 --> 00:10:06.980
If we have a vibration
excitation problem, we really
00:10:06.980 --> 00:10:09.800
only want to solve for specific
frequencies and
00:10:09.800 --> 00:10:11.810
corresponding mode shapes
in a certain
00:10:11.810 --> 00:10:14.300
interval, as shown here.
00:10:14.300 --> 00:10:19.000
The solution procedures on this
generalized eigenvalue
00:10:19.000 --> 00:10:23.690
problem can be categorized
into the following forms.
00:10:23.690 --> 00:10:29.780
The first solution procedure,
or a whole class of solution
00:10:29.780 --> 00:10:33.100
procedures, that I should really
say, which we might
00:10:33.100 --> 00:10:36.050
call the vector iteration
techniques.
00:10:36.050 --> 00:10:37.620
The idea is the following--
00:10:37.620 --> 00:10:43.160
if we want to solve this
problem, then why not solve it
00:10:43.160 --> 00:10:45.870
in the following way?
00:10:45.870 --> 00:10:55.830
Start off with a vector on this
side, that we assume,
00:10:55.830 --> 00:11:00.260
calculate a new vector on this
side, then take that new
00:11:00.260 --> 00:11:03.130
vector, plug it back on the
right hand side, and cycle
00:11:03.130 --> 00:11:04.570
through this way.
00:11:04.570 --> 00:11:06.640
This is indeed inverse
iteration.
00:11:06.640 --> 00:11:07.810
What do we do here?
00:11:07.810 --> 00:11:11.740
We are putting a vector on the
right hand side, x1, when K is
00:11:11.740 --> 00:11:15.790
equal to 1, that vector
we have to assume.
00:11:15.790 --> 00:11:18.540
Then we are multiplying out
the right hand side.
00:11:18.540 --> 00:11:23.380
Now we have a set of
simultaneous equations that we
00:11:23.380 --> 00:11:26.460
would solve just in the same way
as we solve the equations
00:11:26.460 --> 00:11:27.480
in static analysis.
00:11:27.480 --> 00:11:30.190
We can think of this
as a load vector.
00:11:30.190 --> 00:11:34.100
We are calculating this
displacement vector, and then
00:11:34.100 --> 00:11:38.780
having got this displacement
vector, and not taking care of
00:11:38.780 --> 00:11:42.390
this lambda, or having,
basically, taken this lambda
00:11:42.390 --> 00:11:44.270
out of this equation.
00:11:44.270 --> 00:11:49.090
Which means that this vector
will grow in lengths, or
00:11:49.090 --> 00:11:50.760
collapse in lengths.
00:11:50.760 --> 00:11:54.160
Therefore, want to scale this
vector down to a good length--
00:11:54.160 --> 00:11:57.100
a length that does not hit
overflow, so to say, or
00:11:57.100 --> 00:11:58.550
underflow in the machine.
00:11:58.550 --> 00:12:02.260
And that scaling is achieved
as shown here.
00:12:02.260 --> 00:12:05.980
So, the final result then, this
is just a scalar here, as
00:12:05.980 --> 00:12:10.520
you can see, is a vector here,
which we can put on to the
00:12:10.520 --> 00:12:13.800
right hand side and keep
on iterating around.
00:12:13.800 --> 00:12:18.110
In fact, we find as that
iteration procedure converges
00:12:18.110 --> 00:12:22.610
to the lowest eigenvector,
provided the starting vector
00:12:22.610 --> 00:12:28.210
is not orthogonal, is not
M orthogonal to phi 1.
00:12:28.210 --> 00:12:32.330
This is one first iteration
technique.
00:12:32.330 --> 00:12:37.130
Maybe I should put
a side note down.
00:12:37.130 --> 00:12:42.940
A side note that many of you
might have in your mind.
00:12:42.940 --> 00:12:44.730
Many of you might have the
following question in your
00:12:44.730 --> 00:12:48.350
mind-- why he iterate, why does
he not solve directly for
00:12:48.350 --> 00:12:49.290
the eigenvalues?
00:12:49.290 --> 00:12:53.230
Just like we saw a set of
simultaneous equations.
00:12:53.230 --> 00:12:57.890
Well, there's one very important
point, and this is
00:12:57.890 --> 00:12:58.570
the point--
00:12:58.570 --> 00:13:02.800
if we are solving an eigenvalue
problem, and shown
00:13:02.800 --> 00:13:05.350
here, then really that
means the following--
00:13:05.350 --> 00:13:10.920
we are saying that determinant,
K minus lambda M
00:13:10.920 --> 00:13:12.680
shall be equal to 0.
00:13:12.680 --> 00:13:13.260
Why?
00:13:13.260 --> 00:13:15.740
Because if we write
this equation in
00:13:15.740 --> 00:13:17.080
the following form--
00:13:17.080 --> 00:13:24.760
K minus lambda M times phi
equals 0, then we have here a
00:13:24.760 --> 00:13:30.390
set of simultaneous equations
with a 0 vector on
00:13:30.390 --> 00:13:31.730
the right hand side.
00:13:31.730 --> 00:13:34.700
And there is an important
theorem in linear algebra that
00:13:34.700 --> 00:13:39.960
says, only a solution to this
set of equations when this
00:13:39.960 --> 00:13:42.380
coefficient matrix
is singular.
00:13:42.380 --> 00:13:44.920
Which means that the
determinant of this
00:13:44.920 --> 00:13:47.700
coefficient matrix must be 0.
00:13:47.700 --> 00:13:49.240
So, what do we want?
00:13:49.240 --> 00:13:54.650
We really want to solve p of
lambda determinant, K minus
00:13:54.650 --> 00:14:00.480
lambda M. We want to solve for
the 0's of this polynomial.
00:14:00.480 --> 00:14:02.270
That's really what we want.
00:14:02.270 --> 00:14:07.700
Now, if we want to solve for
the 0's of this polynomial,
00:14:07.700 --> 00:14:13.390
then we also have to understand
that when k is of
00:14:13.390 --> 00:14:17.210
order 4, there will
be four 0's.
00:14:17.210 --> 00:14:20.080
The polynomial itself
will be of order 4.
00:14:20.080 --> 00:14:23.710
If K is of order, N, then
there will be N 0's.
00:14:23.710 --> 00:14:29.680
So the polynomial is order N.
However, if we want to find
00:14:29.680 --> 00:14:34.320
the solution of a polynomial,
or order N, then we very
00:14:34.320 --> 00:14:40.650
quickly find that, in fact, we
can only look up the solution
00:14:40.650 --> 00:14:45.080
to such polynomials if they are
of low enough order, lower
00:14:45.080 --> 00:14:46.660
than say, 4.
00:14:46.660 --> 00:14:51.080
Above for larger order
polynomials, there's no way
00:14:51.080 --> 00:14:53.150
you can look up the solution.
00:14:53.150 --> 00:14:58.160
You have to iterate to obtain
the solution of the 0's of
00:14:58.160 --> 00:14:59.620
this polynomial.
00:14:59.620 --> 00:15:04.480
However, since this problem
here, is completely equivalent
00:15:04.480 --> 00:15:06.430
to this problem.
00:15:06.430 --> 00:15:09.170
To solve this problem,
you have to iterate.
00:15:09.170 --> 00:15:11.480
And this is the important
conclusion.
00:15:11.480 --> 00:15:15.640
It is of up most importance that
you understand that to
00:15:15.640 --> 00:15:20.570
solve an eigenvalue problem of
order larger than say, 4, you
00:15:20.570 --> 00:15:22.330
will have to iterate.
00:15:22.330 --> 00:15:27.290
Because you're solving for the
0's of a polynomial of order
00:15:27.290 --> 00:15:28.600
larger than 4.
00:15:28.600 --> 00:15:32.040
Only for very small order
polynomials, we can look up
00:15:32.040 --> 00:15:34.060
the solutions in handbooks.
00:15:34.060 --> 00:15:38.680
So having grasped the fact that
we have to iterate to
00:15:38.680 --> 00:15:44.460
solve for the 0's of this
polynomial, to solve for the
00:15:44.460 --> 00:15:51.390
eigenvalues of this problem, we
are stuck with iteration,
00:15:51.390 --> 00:15:54.320
and since we're stuck with
iteration, we now want to come
00:15:54.320 --> 00:15:56.380
up with techniques to iterate.
00:15:56.380 --> 00:15:58.620
And the first technique to
iterate is this vector
00:15:58.620 --> 00:16:01.180
iteration technique.
00:16:01.180 --> 00:16:04.470
And the basic idea, once more,
is if you want to solve this
00:16:04.470 --> 00:16:09.270
equation, well, then why not
iterate in the following way?
00:16:09.270 --> 00:16:13.150
Put some starting value in here,
this number is just a
00:16:13.150 --> 00:16:16.580
scalar, let's neglect that
effect for the moment.
00:16:16.580 --> 00:16:19.640
And having put a starting value
in here, calculate this
00:16:19.640 --> 00:16:22.790
one and keep on cycling
around this way.
00:16:22.790 --> 00:16:25.600
And that's what we call
inverse iteration.
00:16:25.600 --> 00:16:28.480
Inverse iteration here, a vector
inverse iteration,
00:16:28.480 --> 00:16:32.680
because we have to solve a set
of equations here, with K as
00:16:32.680 --> 00:16:33.930
the coefficient matrix.
00:16:36.310 --> 00:16:38.290
There are other iteration
techniques.
00:16:38.290 --> 00:16:39.900
There's also forward
iteration.
00:16:39.900 --> 00:16:42.150
And basically, what we're
saying there is--
00:16:42.150 --> 00:16:45.970
well if we are starting off here
and going around in this
00:16:45.970 --> 00:16:49.970
circle, why not start off here
and go around in that circle,
00:16:49.970 --> 00:16:51.190
the red line.
00:16:51.190 --> 00:16:54.970
And that, in fact, gives
us forward iteration.
00:16:54.970 --> 00:16:57.970
One can prove that that forward
iteration converges to
00:16:57.970 --> 00:17:00.140
the largest eigenvalue.
00:17:00.140 --> 00:17:03.190
We are usually interested in the
smallest frequencies, the
00:17:03.190 --> 00:17:06.069
smallest eigenvalues, and
therefore, we are using
00:17:06.069 --> 00:17:08.810
inverse iteration
quite commonly.
00:17:08.810 --> 00:17:11.119
Then there is Rayleigh
Quotient iteration.
00:17:11.119 --> 00:17:16.790
The Rayleigh Quotient iteration
is again, an inverse
00:17:16.790 --> 00:17:21.160
iteration, however, with an
acceleration scheme in it, a
00:17:21.160 --> 00:17:24.190
very effective acceleration
scheme.
00:17:24.190 --> 00:17:26.810
This Rayleigh Quotient iteration
can be employed to
00:17:26.810 --> 00:17:30.070
calculate one eigenvalue and
vector, and then, of course,
00:17:30.070 --> 00:17:34.010
we have to deflate afterwards
the matrix, the coefficient
00:17:34.010 --> 00:17:37.170
matrix for the eigenvector
that we have calculated
00:17:37.170 --> 00:17:42.750
already, and we then could go on
towards the calculation of
00:17:42.750 --> 00:17:44.770
additional eigenvalues
and vectors.
00:17:44.770 --> 00:17:47.490
The same deflation also would
be used in inverse iteration
00:17:47.490 --> 00:17:50.200
alone, and in forward
iteration alone.
00:17:50.200 --> 00:17:52.760
In a Rayleigh Quotient
iteration, we converge to an
00:17:52.760 --> 00:17:57.440
eigen pair, but which one
is not guaranteed.
00:17:57.440 --> 00:18:01.880
So this is a class of
techniques, iteration methods
00:18:01.880 --> 00:18:03.560
that we are using.
00:18:03.560 --> 00:18:06.610
Another class of techniques,
again iteration methods, we
00:18:06.610 --> 00:18:09.890
are stuck with iteration, as
I mentioned earlier, is the
00:18:09.890 --> 00:18:11.320
polynomial iteration methods.
00:18:11.320 --> 00:18:14.630
Here, we are saying that this
problem is equivalent to that,
00:18:14.630 --> 00:18:16.920
we want to solve
for these 0's.
00:18:16.920 --> 00:18:20.880
Therefore, for the 0's of this
polynomial, sketched out here,
00:18:20.880 --> 00:18:23.710
and one way is to use
Newton iteration.
00:18:23.710 --> 00:18:25.830
Here we have p, here
we have p prime.
00:18:25.830 --> 00:18:29.740
Well, of course, if we have
p here, we have to also
00:18:29.740 --> 00:18:32.110
recognize we should have
the coefficients here.
00:18:32.110 --> 00:18:37.330
Well, these coefficients are
given here, A0 to a n.
00:18:37.330 --> 00:18:42.200
And we could rewrite this
polynomial in this form, once
00:18:42.200 --> 00:18:44.260
we know the zeroes.
00:18:44.260 --> 00:18:48.420
Now, if we don't know the
zeroes, of course, we would be
00:18:48.420 --> 00:18:52.840
stuck on iterating on this
matrix, where we would have to
00:18:52.840 --> 00:18:55.580
calculate these coefficients.
00:18:55.580 --> 00:19:00.040
However, there are now really,
two different techniques.
00:19:00.040 --> 00:19:03.680
The first technique is explicit
polynomial iteration.
00:19:03.680 --> 00:19:07.260
In the explicit polynomial
iteration, we expand the
00:19:07.260 --> 00:19:10.710
polynomial and iterate for the
0's, by that I mean, we
00:19:10.710 --> 00:19:14.200
actually expand this
polynomial.
00:19:14.200 --> 00:19:18.090
However, this technique is not
suitable for larger problems
00:19:18.090 --> 00:19:21.150
because first of all, there's
much work to obtain the ai's,
00:19:21.150 --> 00:19:23.720
and then it's an unstable
process.
00:19:23.720 --> 00:19:27.940
Unstable because small errors
in these coefficients here,
00:19:27.940 --> 00:19:30.350
give large errors
in the zeroes.
00:19:30.350 --> 00:19:34.980
So this is not a suitable
technique, and we are better
00:19:34.980 --> 00:19:37.720
off using implicit polynomial
iteration.
00:19:37.720 --> 00:19:40.410
In the implicit polynomial
iteration, we evaluate the
00:19:40.410 --> 00:19:44.280
determinant all this
matrix here.
00:19:44.280 --> 00:19:50.290
In others, p at mu i, via the L
D L transpose factorization,
00:19:50.290 --> 00:19:55.770
which we discussed in the Gauss
elimination solution,
00:19:55.770 --> 00:19:57.920
when we talked about Gauss
elimination solution.
00:19:57.920 --> 00:20:01.530
To obtain the determinant of
this matrix here then, we need
00:20:01.530 --> 00:20:04.410
simply to multiply all the
diagonal elements.
00:20:04.410 --> 00:20:08.450
This technique is accurate,
provided we do not encounter
00:20:08.450 --> 00:20:12.030
large multipliers L matrix.
00:20:12.030 --> 00:20:14.400
And if we do, we have to
shift away, we have to
00:20:14.400 --> 00:20:16.220
change our mu i.
00:20:16.220 --> 00:20:19.570
Also we directly solve for
lambda 1 and so on.
00:20:19.570 --> 00:20:22.760
And for the more we can use
effectively a secant
00:20:22.760 --> 00:20:25.710
iteration, instead of
calculating, in other words,
00:20:25.710 --> 00:20:28.940
the p prime at mu.
00:20:28.940 --> 00:20:33.930
This is here approximately equal
to p prime, at mu i.
00:20:33.930 --> 00:20:38.010
Instead of calculating this
derivative, we are putting
00:20:38.010 --> 00:20:41.330
this value here, which gives
us an approximation to that
00:20:41.330 --> 00:20:44.200
derivative, this is what we
call the secant iteration.
00:20:44.200 --> 00:20:47.260
Once we have calculated one
eigenvalue, of course, we will
00:20:47.260 --> 00:20:51.140
have to deflate to calculate the
next eigenvalue, and this
00:20:51.140 --> 00:20:53.360
is the process that
he would pursue.
00:20:53.360 --> 00:20:57.480
Pictorially, we have here,
our p lambda, this is the
00:20:57.480 --> 00:20:58.470
polynomial.
00:20:58.470 --> 00:20:59.860
Let me show it to you here.
00:20:59.860 --> 00:21:03.240
Here, we have the polynomial,
here we have lambda 1, here we
00:21:03.240 --> 00:21:04.260
have lambda 2.
00:21:04.260 --> 00:21:07.560
In the secant iteration, we
start off with the value of mu
00:21:07.560 --> 00:21:11.500
i minus 1 here, mu i, these
two values are given.
00:21:11.500 --> 00:21:15.930
Usually, in a practical
analysis, mu 0 would be this
00:21:15.930 --> 00:21:20.660
value here, and mu 1 would be
a lower bound on lambda 1.
00:21:20.660 --> 00:21:24.990
And plugging these two values
in, we get mu i plus 1, and
00:21:24.990 --> 00:21:30.620
you can see that this process
would then converge to this
00:21:30.620 --> 00:21:32.210
eigenvalue.
00:21:32.210 --> 00:21:35.880
Once we have lambda 1, we can
repeat the iteration.
00:21:35.880 --> 00:21:39.590
However, working now on p
lambda divided by lambda
00:21:39.590 --> 00:21:41.110
minus lambda 1.
00:21:41.110 --> 00:21:44.480
And that means we are operating
on this polynomial
00:21:44.480 --> 00:21:48.640
here, and polynomial then gives
us the iteration on that
00:21:48.640 --> 00:21:51.190
polynomial then gives us
the second eigenvalue.
00:21:51.190 --> 00:21:57.380
So we are deflating the actual p
lambda by the lambda 1 value
00:21:57.380 --> 00:22:00.070
in order to obtain
a convergence
00:22:00.070 --> 00:22:02.940
to the second value.
00:22:02.940 --> 00:22:06.420
Well, this is the one
class of methods.
00:22:06.420 --> 00:22:09.750
Another class of methods, the
third one, are the Sturm
00:22:09.750 --> 00:22:10.930
sequence methods.
00:22:10.930 --> 00:22:15.060
And here, the basic idea
is the following--
00:22:15.060 --> 00:22:17.980
if we look at this eigenproblem,
and I use a 4x4
00:22:17.980 --> 00:22:24.230
matrix here, then if we take a
shift on this eigenproblem, mu
00:22:24.230 --> 00:22:27.100
S, and this is the shift here.
00:22:27.100 --> 00:22:33.700
Then if we factorize K minus
mu S M, equal to L D L
00:22:33.700 --> 00:22:36.540
transpose, again, we're using
our Gauss elimination
00:22:36.540 --> 00:22:38.120
procedures basically.
00:22:38.120 --> 00:22:42.420
Then the number of negative
elements in D is equal to the
00:22:42.420 --> 00:22:45.110
number eigenvalues smaller
than mu S, a
00:22:45.110 --> 00:22:47.480
very important point.
00:22:47.480 --> 00:22:51.100
Let us look at this example
here of the 4x4 matrix.
00:22:51.100 --> 00:22:56.370
Here we have plotted
characteristic polynomials.
00:22:56.370 --> 00:23:00.290
The characteristic polynomial of
this 4x4 matrix, with that
00:23:00.290 --> 00:23:05.310
M matrix, of course, is drawn
along here, four eigenvalues.
00:23:05.310 --> 00:23:09.150
And I have put mu S here
as the red line.
00:23:09.150 --> 00:23:15.430
What this theorem says is if I
factorize K minus mu S, M,
00:23:15.430 --> 00:23:19.160
this being the K matrix, the
total K matrix, and the 4x4 M
00:23:19.160 --> 00:23:21.020
matrix goes in here.
00:23:21.020 --> 00:23:25.740
Then there would be one negative
element in D. One
00:23:25.740 --> 00:23:30.900
only because mu S is large
than lambda 1.
00:23:30.900 --> 00:23:34.470
If mu S were on this side
here, there would be two
00:23:34.470 --> 00:23:38.640
negative elements in D. If mu
S were on this side, there
00:23:38.640 --> 00:23:41.260
would be three negative
elements in D--
00:23:41.260 --> 00:23:46.360
very important point because qw
can use this procedure to
00:23:46.360 --> 00:23:50.670
directly calculate eigenvalues
the way I will be discussing
00:23:50.670 --> 00:23:51.430
it just now.
00:23:51.430 --> 00:23:55.470
However, let us look a little
bit more at why
00:23:55.470 --> 00:23:56.730
this hole is here.
00:23:56.730 --> 00:24:01.940
Well, it derives from the
associated constraint problems
00:24:01.940 --> 00:24:04.720
with the original
eigenproblem.
00:24:04.720 --> 00:24:09.440
The original eigenproblem
corresponds to this beam
00:24:09.440 --> 00:24:14.030
problem, which we considered
earlier already in the
00:24:14.030 --> 00:24:15.730
discussion of Gauss
elimination.
00:24:15.730 --> 00:24:19.870
If I constrain this degree of
freedom here, I get a new
00:24:19.870 --> 00:24:22.690
polynomial corresponding to
this problem, and the
00:24:22.690 --> 00:24:26.860
important point is that the
lowest root, the lowest value
00:24:26.860 --> 00:24:32.000
of this problem here lies in
between these two-- lambda 1
00:24:32.000 --> 00:24:33.170
and lambda 2.
00:24:33.170 --> 00:24:37.180
The second one here, lies
between lambda 2 and lambda 3,
00:24:37.180 --> 00:24:41.350
and the third one here lies
between lambda 3 and lambda 4.
00:24:41.350 --> 00:24:44.340
The second associated constraint
problem has two
00:24:44.340 --> 00:24:48.000
eigenvalues, and this one lies
between these two, this one
00:24:48.000 --> 00:24:49.370
lies between these two.
00:24:49.370 --> 00:24:51.850
And the final eigenvalue
problem here, the third
00:24:51.850 --> 00:24:54.630
associate constraint problem
obtained by knocking out these
00:24:54.630 --> 00:24:57.430
degrees of freedom here, has one
eigenvalue, and that one
00:24:57.430 --> 00:25:01.790
lies in between these
two values here.
00:25:01.790 --> 00:25:09.970
It's this important fact
that is used to
00:25:09.970 --> 00:25:13.630
derive this final result.
00:25:13.630 --> 00:25:18.820
Now, we have not time to go
through the actual details,
00:25:18.820 --> 00:25:22.220
but this is the important
fact here.
00:25:22.220 --> 00:25:27.470
In other words, that the
[? i's ?] eigenvalue of this
00:25:27.470 --> 00:25:31.430
associated constraint problem,
for example, the second
00:25:31.430 --> 00:25:35.420
eigenvalue of this associated
constraint problem here,
00:25:35.420 --> 00:25:38.630
bisects or lies in between,
not bisect--
00:25:38.630 --> 00:25:44.340
lies in between lambda 2 and
lambda 3 of this eigenproblem
00:25:44.340 --> 00:25:45.930
here, and so on.
00:25:45.930 --> 00:25:50.700
That result is used to derive
this final result, which we
00:25:50.700 --> 00:25:57.180
use effectively then in
the eigen solution.
00:25:57.180 --> 00:26:03.950
And the class of techniques then
that we are referring to
00:26:03.950 --> 00:26:07.470
using this fact are the Sturm
sequence methods.
00:26:07.470 --> 00:26:11.050
So basically, what we are saying
is the following--
00:26:11.050 --> 00:26:15.800
take this coefficient matrix,
now I make the mu x available.
00:26:15.800 --> 00:26:20.060
We factorize that coefficient
matrix into L D L transposed,
00:26:20.060 --> 00:26:24.740
and the number of negative
elements in D is equal to the
00:26:24.740 --> 00:26:29.630
number of eigenvalues smaller
than mu S, I. How
00:26:29.630 --> 00:26:30.560
do we use this fact?
00:26:30.560 --> 00:26:34.885
Well, if this is here, the
polynomial, we could start off
00:26:34.885 --> 00:26:40.950
by putting mu S 1 into this
point here, and we would find
00:26:40.950 --> 00:26:44.510
out the number of eigenvalues
smaller than mu S 1.
00:26:44.510 --> 00:26:49.170
Then we are taking, mu S 2, and
we find out the number of
00:26:49.170 --> 00:26:51.320
eigenvalues smaller
than mu S 2.
00:26:51.320 --> 00:26:58.580
Of course, we would find that
they are one more eigenvalue
00:26:58.580 --> 00:26:59.860
smaller than mu S 2.
00:26:59.860 --> 00:27:02.460
Then mu S 1, namely this one.
00:27:02.460 --> 00:27:06.150
So we know there's one
eigenvalue in between mu S 1
00:27:06.150 --> 00:27:07.910
and mu S 2.
00:27:07.910 --> 00:27:11.090
Now we could bisect this
interval, and we would find
00:27:11.090 --> 00:27:16.080
that, in fact, there is still
one more eigenvalue smaller
00:27:16.080 --> 00:27:20.970
than mu S 3, then there was
smaller than mu S 1.
00:27:20.970 --> 00:27:23.750
Because this eigenvalue
still lies on the left
00:27:23.750 --> 00:27:25.780
side of mu S 3.
00:27:25.780 --> 00:27:28.650
So we bisect again,
we get mu S 4.
00:27:28.650 --> 00:27:33.550
Notice this distance here, is
equal to that distance here.
00:27:33.550 --> 00:27:36.970
And now, we would find that the
eigenvalue still lies in
00:27:36.970 --> 00:27:40.780
between these two lines, between
mu S 3 and mu S 4.
00:27:40.780 --> 00:27:43.560
And like that. we could
continue bisecting the
00:27:43.560 --> 00:27:48.400
interval until we finally find a
very small interval in which
00:27:48.400 --> 00:27:51.550
this eigenvalue must lie.
00:27:51.550 --> 00:27:54.490
We, of course, have to take care
in the L D L transpose
00:27:54.490 --> 00:27:57.920
factorization, because we are
not working on a positive
00:27:57.920 --> 00:27:58.760
definite matrix.
00:27:58.760 --> 00:28:03.800
The multipliers must be small,
the L, I, J elements must
00:28:03.800 --> 00:28:06.280
remain small in the
factorization.
00:28:06.280 --> 00:28:09.700
But provided that is the case,
the round of errors are small,
00:28:09.700 --> 00:28:11.930
and the procedure
is very stable.
00:28:11.930 --> 00:28:15.100
However, convergence can be very
slow, can be extremely
00:28:15.100 --> 00:28:17.990
slow because you have to cut
down the intervals further and
00:28:17.990 --> 00:28:20.990
further until you finally get
a very small interval, in
00:28:20.990 --> 00:28:23.560
which you now know the
eigenvalue does lie.
00:28:23.560 --> 00:28:26.340
Of course, at some point, you
might want to switch to
00:28:26.340 --> 00:28:27.810
another iteration scheme.
00:28:27.810 --> 00:28:31.210
Once you know there is an
eigenvalue in between say mu S
00:28:31.210 --> 00:28:34.100
3 and mu S 4, you might want
to switch to a iteration
00:28:34.100 --> 00:28:38.590
scheme that converges to this
particular value much faster.
00:28:38.590 --> 00:28:41.570
Finally, a whole class of
methods that I would like to
00:28:41.570 --> 00:28:45.640
briefly mention to you are the
methods of transformations,
00:28:45.640 --> 00:28:49.230
where we are operating on
this eigenvalue problem,
00:28:49.230 --> 00:28:53.860
recognizing that phi transpose
K phi is equal to lambda.
00:28:53.860 --> 00:28:59.170
And phi transpose M phi is
equal to I. Where phi is
00:28:59.170 --> 00:29:02.660
defined as shown here, lambda
is defined as shown here.
00:29:02.660 --> 00:29:05.960
This fact, of course, we use
already earlier in the
00:29:05.960 --> 00:29:06.360
[? multiple ?]
00:29:06.360 --> 00:29:07.750
position analysis.
00:29:07.750 --> 00:29:12.020
Now how can we get
phi and lambda?
00:29:12.020 --> 00:29:13.430
That is our objective.
00:29:13.430 --> 00:29:17.900
Well, if we know that this
matrix diagonalizes the K
00:29:17.900 --> 00:29:21.340
matrix, and it also diagonalizes
the M matrix,
00:29:21.340 --> 00:29:25.370
then why not try to construct
it by iteration?
00:29:25.370 --> 00:29:29.520
And this is the basic idea in
the transformation methods.
00:29:29.520 --> 00:29:33.410
What we do is the following-- we
take the original K matrix,
00:29:33.410 --> 00:29:37.720
we are operating with
P1 transpose K P1.
00:29:37.720 --> 00:29:45.060
On K, and on M, such as to make
this resultant matrix
00:29:45.060 --> 00:29:50.020
here closer to a
diagonal form.
00:29:50.020 --> 00:29:52.470
Ultimately, we want to get
a diagonal matrix.
00:29:52.470 --> 00:29:56.320
Well, one step in the iteration
is to make this P1
00:29:56.320 --> 00:30:00.610
transpose K P1 matrix, and P1
transpose M P1 matrix, both of
00:30:00.610 --> 00:30:03.550
these matrices a little closer
to diagonal form.
00:30:03.550 --> 00:30:06.030
Then K and M were
by themselves.
00:30:06.030 --> 00:30:08.260
So this is the first step.
00:30:08.260 --> 00:30:12.360
Now, we have here, let's call
this a K2 matrix, and this
00:30:12.360 --> 00:30:15.870
here an M2 matrix.
00:30:18.760 --> 00:30:23.240
Now, we are operating with P2
transpose on K2, and also
00:30:23.240 --> 00:30:24.400
there's a P2 here.
00:30:24.400 --> 00:30:28.250
So let me take a different color
to show that now we are
00:30:28.250 --> 00:30:34.020
operating with this part here,
and that part there.
00:30:38.850 --> 00:30:41.560
And this, of course, we are
doing in such a way as to have
00:30:41.560 --> 00:30:47.730
that P2 transposed, K2, P2
is a little bit closer to
00:30:47.730 --> 00:30:52.440
diagonal form than was K2.
00:30:52.440 --> 00:30:55.320
I hope you can see these
lines-- let me put
00:30:55.320 --> 00:30:57.570
them in once more.
00:30:57.570 --> 00:31:01.720
What I'm saying is we have K2
after having applied P1
00:31:01.720 --> 00:31:07.200
transpose on P1, on K,
and similar for M.
00:31:07.200 --> 00:31:11.010
And now we want to make K2 still
a little bit closer to
00:31:11.010 --> 00:31:12.380
diagonal form.
00:31:12.380 --> 00:31:17.780
Well, we are now applying P2
transposed on K2 and the P2 on
00:31:17.780 --> 00:31:21.460
K2, this one here has been
replaced by K2, this one here
00:31:21.460 --> 00:31:23.240
has been replaced by M2.
00:31:23.240 --> 00:31:26.880
And what we're saying is that
this final result, in this
00:31:26.880 --> 00:31:30.760
final result should be closer
to diagonal form.
00:31:30.760 --> 00:31:33.510
In fact, this should be closer
to I matrix, and this should
00:31:33.510 --> 00:31:36.790
be closer to the lambda
matrix, ideally.
00:31:36.790 --> 00:31:41.280
Well, this is the basic process
that is used in the
00:31:41.280 --> 00:31:43.750
generalized Jacobi method.
00:31:43.750 --> 00:31:46.750
We can show that if
we choose specific
00:31:46.750 --> 00:31:50.770
matrices, P I matrices.
00:31:50.770 --> 00:31:55.220
That zero always one [? of ?]
diagonal element.
00:31:55.220 --> 00:31:58.970
In other words, the P I matrix,
zeroes a particular
00:31:58.970 --> 00:32:03.720
off diagonal element in this
current coefficient matrix,
00:32:03.720 --> 00:32:09.320
similarly for the M2, for the
current M coefficient matrix.
00:32:09.320 --> 00:32:13.920
Then we know in the generalized
Jacobi method that
00:32:13.920 --> 00:32:15.795
the process finally converges.
00:32:15.795 --> 00:32:21.070
It converges and we get a
diagonal matrix here.
00:32:21.070 --> 00:32:23.860
With proper scaling, it will
be the identity matrix.
00:32:23.860 --> 00:32:27.670
And here, we are getting also a
diagonal matrix, and if this
00:32:27.670 --> 00:32:31.590
is an identity matrix, this will
actually be the matrix
00:32:31.590 --> 00:32:33.330
storing the eigenvalues.
00:32:33.330 --> 00:32:36.310
Well, this is the generalized
Jacobi method here.
00:32:36.310 --> 00:32:39.270
They are other techniques that
can also be used, of course,
00:32:39.270 --> 00:32:44.990
this is a very common approach
taken when we have M being
00:32:44.990 --> 00:32:47.460
equal to the identity matrix.
00:32:47.460 --> 00:32:50.270
However, the disadvantage of
this approach is that we are
00:32:50.270 --> 00:32:53.060
calculating all eigen pairs
simultaneously.
00:32:53.060 --> 00:32:55.570
In finite element analysis,
we are only interested in
00:32:55.570 --> 00:33:00.240
specific eigenvalues, so lowest
P, or a certain number
00:33:00.240 --> 00:33:01.880
of eigenvalues in an interval.
00:33:01.880 --> 00:33:05.810
Here, we have to calculate all
of the eigenvalues stored in
00:33:05.810 --> 00:33:08.790
the lambda matrix, and all
of the eigenvectors.
00:33:08.790 --> 00:33:14.676
And the eigenvectors here, would
be these vectors here.
00:33:14.676 --> 00:33:21.692
These are the the matrix phi,
which is P1 times P2, up to P
00:33:21.692 --> 00:33:24.390
K, after K steps of iteration.
00:33:24.390 --> 00:33:28.295
This gives us the matrix
phi, storing all of the
00:33:28.295 --> 00:33:30.270
eigenvectors.
00:33:30.270 --> 00:33:33.860
So the disadvantage is that we
are calculating all of that
00:33:33.860 --> 00:33:37.720
eigenvectors, which in finite
element analysis is hardly
00:33:37.720 --> 00:33:38.770
ever required.
00:33:38.770 --> 00:33:43.810
All we want is the total number
of eigenvalues and
00:33:43.810 --> 00:33:44.760
eigenvectors.
00:33:44.760 --> 00:33:47.740
Remember, also if K
is of [? order ?]
00:33:47.740 --> 00:33:51.310
1,000, this would be a
tremendous amount of work
00:33:51.310 --> 00:33:54.500
involved and, in fact, it would
be beyond the current
00:33:54.500 --> 00:33:57.060
state of computing capabilities
to use a
00:33:57.060 --> 00:34:02.700
generalized Jacobi method on a
K matrix 1,000x1,000 and an M
00:34:02.700 --> 00:34:04.420
matrix 1,000x1,000.
00:34:04.420 --> 00:34:08.320
You will certainly stretch the
current computing capabilities
00:34:08.320 --> 00:34:12.770
tremendously by using the
generalized Jacobi method on
00:34:12.770 --> 00:34:15.080
large eigenproblems.
00:34:15.080 --> 00:34:20.199
For large eigenproblems, it is
really best to combine the
00:34:20.199 --> 00:34:22.560
basic approaches that
I just mentioned.
00:34:22.560 --> 00:34:25.070
Let's recall the basic
approaches.
00:34:25.070 --> 00:34:28.920
There were iteration methods,
vector iteration methods,
00:34:28.920 --> 00:34:33.040
there were Sturm sequence
methods, there was the
00:34:33.040 --> 00:34:35.580
transformation method.
00:34:35.580 --> 00:34:38.050
And then, of course, there were
00:34:38.050 --> 00:34:39.590
polynomial iteration methods.
00:34:39.590 --> 00:34:41.980
So really, four classes
of methods that
00:34:41.980 --> 00:34:43.500
we are talking about.
00:34:43.500 --> 00:34:46.060
Each of them, we find,
have some advantages.
00:34:46.060 --> 00:34:50.889
Well, as engineer, surely we
want to now combine the
00:34:50.889 --> 00:34:54.350
advantages of each of these
methods to come up with
00:34:54.350 --> 00:34:57.430
optimum techniques for our
specific problems.
00:34:57.430 --> 00:35:01.410
Well, there are a number
of thoughts
00:35:01.410 --> 00:35:02.380
that we would go through.
00:35:02.380 --> 00:35:05.600
First of all, a determinant
search, a polynomial iteration
00:35:05.600 --> 00:35:09.640
method would be effective
to get near a root.
00:35:09.640 --> 00:35:12.930
Vector iteration then could
be used to calculate an
00:35:12.930 --> 00:35:14.880
eigenvalue and an eigenvector.
00:35:14.880 --> 00:35:18.180
The transformation method would
be good to orthogonalize
00:35:18.180 --> 00:35:22.060
the current iteration vectors
if we iterate with more than
00:35:22.060 --> 00:35:23.280
just one vector.
00:35:23.280 --> 00:35:26.870
And the Sturm sequence method
is extremely powerful to
00:35:26.870 --> 00:35:30.360
assure that we have actually
calculated the required
00:35:30.360 --> 00:35:33.520
eigenvalues and eigenvectors.
00:35:33.520 --> 00:35:36.480
Well, the first method then that
I would like to discuss
00:35:36.480 --> 00:35:38.580
with you is the determinant
search method.
00:35:38.580 --> 00:35:42.370
In this method, we have
combined some of these
00:35:42.370 --> 00:35:44.510
techniques that I pointed
out to you earlier
00:35:44.510 --> 00:35:46.340
in an optimum way.
00:35:46.340 --> 00:35:49.760
Here, we have the polynomial P
lambda, we want to, of course,
00:35:49.760 --> 00:35:51.770
solve for the roots of
that polynomial--
00:35:51.770 --> 00:35:53.660
that is our objective.
00:35:53.660 --> 00:35:55.690
Here is lambda 1, there
is lambda 2.
00:35:55.690 --> 00:35:58.770
I also show a point,
A, because I will
00:35:58.770 --> 00:36:00.750
refer to it just now.
00:36:00.750 --> 00:36:04.200
The basic scheme in the
determinant search method is
00:36:04.200 --> 00:36:09.140
to look for the zeroes of
the determinant, of the
00:36:09.140 --> 00:36:14.260
determinant K minus mu I M. Of
course, these zeroes give us R
00:36:14.260 --> 00:36:15.590
equal to the eigenvalues.
00:36:15.590 --> 00:36:17.560
That's what we're interested
in calculating.
00:36:17.560 --> 00:36:20.740
Well, we are calculating
the determinant by the
00:36:20.740 --> 00:36:24.970
factorization of this matrix
into L D L transpose, and we
00:36:24.970 --> 00:36:27.990
notice that this determinant is
simply the product of the
00:36:27.990 --> 00:36:30.230
[? D I I's, ?] as I
mentioned earlier.
00:36:30.230 --> 00:36:36.560
Now, having a certain starting
value, mu 0, say, and mu 1.
00:36:36.560 --> 00:36:40.170
We can directly use the secant
iteration, that I mentioned to
00:36:40.170 --> 00:36:43.700
you earlier to get close
to an eigenvalue.
00:36:43.700 --> 00:36:47.380
Now, I also slipped in here
an acceleration factor.
00:36:47.380 --> 00:36:52.010
This acceleration factor would
be equal to 1 in the normal
00:36:52.010 --> 00:36:55.320
secant iteration that I showed
to you earlier already.
00:36:55.320 --> 00:36:58.540
However, since we want to use
this polynomial iteration
00:36:58.540 --> 00:37:04.420
technique only to get close to
an eigenvalue, and we might
00:37:04.420 --> 00:37:07.260
actually obtain with eta equal
to 1, sometimes slow
00:37:07.260 --> 00:37:08.340
convergence.
00:37:08.340 --> 00:37:11.810
What we do is we simply
set it larger to
00:37:11.810 --> 00:37:14.120
accelerate the process.
00:37:14.120 --> 00:37:16.780
And this is what we're
doing here.
00:37:16.780 --> 00:37:22.600
If we do jump beyond an
eigenvalue because we have
00:37:22.600 --> 00:37:24.840
accelerated the secant
iteration.
00:37:24.840 --> 00:37:29.130
In other words, if we do a jump
across here, then since
00:37:29.130 --> 00:37:35.510
we are factorizing the matrix
here, we can also look at the
00:37:35.510 --> 00:37:38.540
number of negative elements
in the D matrix.
00:37:38.540 --> 00:37:43.140
These would tell us that we, in
fact, have jumped across an
00:37:43.140 --> 00:37:46.250
unknown eigenvalue.
00:37:46.250 --> 00:37:49.510
So, in this particular case, of
course, there would be one
00:37:49.510 --> 00:37:51.280
negative element in
the D matrix.
00:37:51.280 --> 00:37:54.040
And we would know now we
have jumped across.
00:37:54.040 --> 00:37:58.480
The jumping across an unknown
eigenvalue is not possible
00:37:58.480 --> 00:38:01.370
when eta is equal to 1.
00:38:01.370 --> 00:38:05.610
However, it is possible when
eta is greater than 1.
00:38:05.610 --> 00:38:11.600
And however, that jumping
does not hurt us.
00:38:11.600 --> 00:38:19.540
In fact, all we will find is
that the Sturm sequence count
00:38:19.540 --> 00:38:22.950
will tell us that we have jumped
across an eigenvalue,
00:38:22.950 --> 00:38:27.000
and then, of course, we have to
go to another strategy of
00:38:27.000 --> 00:38:29.280
solving for the eigenvalue
more accurately.
00:38:29.280 --> 00:38:33.110
But the important point is that
all we want to do via
00:38:33.110 --> 00:38:36.420
this procedure is to get into
the vicinity of the
00:38:36.420 --> 00:38:37.190
eigenvalue.
00:38:37.190 --> 00:38:40.780
We do not want to calculate it
accurately, all we want to do
00:38:40.780 --> 00:38:44.290
is get into the vicinity of
an unknown eigenvalue.
00:38:44.290 --> 00:38:49.080
First, of course, lambda 1,
then lambda 2, and so on.
00:38:49.080 --> 00:38:53.380
There's also a concern, of
course, that we do not want to
00:38:53.380 --> 00:38:58.310
jump beyond A, because if we, in
our process of looking for
00:38:58.310 --> 00:39:05.090
lambda 1, we are jumping too
close to lambda 2, then via
00:39:05.090 --> 00:39:07.400
the other solution techniques
that we are using later on
00:39:07.400 --> 00:39:10.050
then to calculate the eigenvalue
more accurately, we
00:39:10.050 --> 00:39:11.430
would actually converge
to lambda 2.
00:39:11.430 --> 00:39:13.200
And, of course, that we
don't want, we want to
00:39:13.200 --> 00:39:14.450
calculate lambda 1.
00:39:16.940 --> 00:39:22.950
So, the process is then to start
off with eta equal to 1,
00:39:22.950 --> 00:39:26.220
iterate, see how we are
progressing towards an unknown
00:39:26.220 --> 00:39:27.160
eigenvalue.
00:39:27.160 --> 00:39:29.410
If that progressing is
very slow, we are
00:39:29.410 --> 00:39:31.740
choosing eta larger.
00:39:31.740 --> 00:39:34.470
There are specific strategies
that we're using to choose the
00:39:34.470 --> 00:39:37.730
right amount all eta.
00:39:37.730 --> 00:39:41.670
And until we find that we are
close to an eigenvalue, either
00:39:41.670 --> 00:39:46.010
from the left or we have jumped
across it already, but
00:39:46.010 --> 00:39:47.940
you should never
jumped to far.
00:39:47.940 --> 00:39:50.480
And we are making sure that we
never jump beyond that point
00:39:50.480 --> 00:39:54.930
A. and having jumped, we find
that the Sturm sequence count
00:39:54.930 --> 00:39:57.680
tells us about it.
00:39:57.680 --> 00:40:03.310
And now, we are going on to
another strategy, and that
00:40:03.310 --> 00:40:06.100
strategy is the vector
iteration, vector inverse
00:40:06.100 --> 00:40:09.040
iteration method that I
discussed with you earlier.
00:40:09.040 --> 00:40:12.300
What we are doing now, is we
have a new coefficient matrix,
00:40:12.300 --> 00:40:13.180
that's the one.
00:40:13.180 --> 00:40:15.550
We are sitting here, say, and
now I have assumed that we
00:40:15.550 --> 00:40:17.240
have, in fact, jumped.
00:40:17.240 --> 00:40:18.170
It's not necessary.
00:40:18.170 --> 00:40:20.980
We might have also approached
from this side and come very
00:40:20.980 --> 00:40:27.010
close to lambda J. And now, we
are using vector iteration to
00:40:27.010 --> 00:40:31.120
get this eigenvalue
accurately.
00:40:31.120 --> 00:40:35.150
This coefficient here,
or this value here
00:40:35.150 --> 00:40:37.830
corresponds to that length.
00:40:37.830 --> 00:40:42.700
And if you are adding this value
here to this shift, we
00:40:42.700 --> 00:40:45.630
are getting an approximation
to lambda J. And that, of
00:40:45.630 --> 00:40:46.990
course, is our objective.
00:40:46.990 --> 00:40:51.680
Once we have lambda J, we have
calculated this eigenvalue--
00:40:51.680 --> 00:40:56.960
the eigenvector comes out, in x
k plus 1 as k increases, so
00:40:56.960 --> 00:40:59.750
we have calculated this eigen
pair, and now, of course, we
00:40:59.750 --> 00:41:03.230
would use deflation method, just
the way I showed it to
00:41:03.230 --> 00:41:06.200
your earlier, for the polynomial
iteration method.
00:41:06.200 --> 00:41:11.010
To deflate the polynomial of
this value, implicitly--
00:41:11.010 --> 00:41:12.560
implicitly, that's important--
00:41:12.560 --> 00:41:15.820
and we would continue our
iteration process towards that
00:41:15.820 --> 00:41:18.640
eigenvalue next.
00:41:18.640 --> 00:41:20.970
In this process, we also want
to make sure that the
00:41:20.970 --> 00:41:24.160
iteration vector is deflated
of the previously
00:41:24.160 --> 00:41:27.110
eigenvectors, for example,
Gram-Schmidt
00:41:27.110 --> 00:41:28.770
orthogonalization.
00:41:28.770 --> 00:41:32.800
If the convergence is slow in
this particular iteration
00:41:32.800 --> 00:41:35.170
here, we're using also the
really Rayleigh-quotient
00:41:35.170 --> 00:41:40.310
iteration to speed up
iteration itself.
00:41:40.310 --> 00:41:44.620
The advantage of the method--
well, it calculates only the
00:41:44.620 --> 00:41:48.260
eigen pairs that are actually
required., there's no prior
00:41:48.260 --> 00:41:51.100
transformation of the
eigenprobelm necessary.
00:41:51.100 --> 00:41:53.970
The disadvantage of the method
is that we have to perform
00:41:53.970 --> 00:41:56.100
many triangular factorizations.
00:41:56.100 --> 00:41:58.800
Therefore, the method is really
only effective for
00:41:58.800 --> 00:42:00.760
small banded systems.
00:42:00.760 --> 00:42:04.110
If we want to calculate
eigenvalues of large banded
00:42:04.110 --> 00:42:07.860
systems, we really need
an algorithm with less
00:42:07.860 --> 00:42:11.320
factorizations, and more
vector iterations.
00:42:11.320 --> 00:42:14.240
And that is the subspace
iteration method.
00:42:14.240 --> 00:42:16.980
In fact, we have now extended
the subspace iteration method
00:42:16.980 --> 00:42:20.590
to be also most effective for
small banded systems.
00:42:20.590 --> 00:42:23.840
And I will just refer to
it very briefly later.
00:42:23.840 --> 00:42:26.940
The basic steps of the subspace
iteration method are
00:42:26.940 --> 00:42:28.380
the following.
00:42:28.380 --> 00:42:33.090
Here, we have the first
equation used.
00:42:33.090 --> 00:42:34.670
We have on the right
hand side, the
00:42:34.670 --> 00:42:35.950
current iteration vectors.
00:42:35.950 --> 00:42:38.290
Now, notice that in this
particular case, we're
00:42:38.290 --> 00:42:42.760
iterating with q vectors,
where q ip larger than p
00:42:42.760 --> 00:42:44.350
simultaneously.
00:42:44.350 --> 00:42:49.130
This gives us here, q right
hand side vector.
00:42:49.130 --> 00:42:52.200
We are solving for
q vectors here.
00:42:52.200 --> 00:42:54.930
This is an inverse iteration
step on q vector
00:42:54.930 --> 00:42:56.360
simultaneously.
00:42:56.360 --> 00:43:00.860
Then we are calculating the
projections of these k and m
00:43:00.860 --> 00:43:03.410
matrices on these vectors.
00:43:03.410 --> 00:43:07.750
Notice that this pat here,
of course, is equal to
00:43:07.750 --> 00:43:08.780
this right hand side.
00:43:08.780 --> 00:43:10.850
So we would not calculate
this here.
00:43:10.850 --> 00:43:13.290
We simply use the right hand
side, which we have evaluated
00:43:13.290 --> 00:43:15.560
already and plug it
right in there.
00:43:15.560 --> 00:43:19.390
But having calculated these two
matrices, which are now of
00:43:19.390 --> 00:43:23.960
order q by q, we are solving
eigenvalue problem all of
00:43:23.960 --> 00:43:26.110
these q by q matrices.
00:43:26.110 --> 00:43:27.780
And the solution of
that eigenvalue
00:43:27.780 --> 00:43:29.230
problem is shown here.
00:43:29.230 --> 00:43:34.030
Notice this matrix here
is a q by q matrix.
00:43:34.030 --> 00:43:36.020
We're using q vectors.
00:43:36.020 --> 00:43:40.970
This is a q by q matrix storing
the eigenvectors of
00:43:40.970 --> 00:43:44.470
this eigen problem here, of the
eigenproblem corresponding
00:43:44.470 --> 00:43:46.510
to these matrices.
00:43:46.510 --> 00:43:51.490
This here is the diagonal matrix
listing the eigenvalues
00:43:51.490 --> 00:43:52.830
of the eigenproblem
00:43:52.830 --> 00:43:55.090
corresponding to these matrices.
00:43:55.090 --> 00:43:58.930
Having calculated these matrices
here, and of course,
00:43:58.930 --> 00:44:02.070
that one too, we are performing
this operation to
00:44:02.070 --> 00:44:05.390
get a new set of iteration
vectors, which then are
00:44:05.390 --> 00:44:07.190
plugged back into here.
00:44:07.190 --> 00:44:12.580
Notice that the solution of
this eigenproblem, q by q,
00:44:12.580 --> 00:44:17.660
might be a 50x50 matrix here,
100x100 matrix, maybe a
00:44:17.660 --> 00:44:20.720
200x200, but that already would
be a large number of
00:44:20.720 --> 00:44:21.930
vectors if we're using--
00:44:21.930 --> 00:44:23.730
a very large number
of vectors.
00:44:23.730 --> 00:44:26.250
This solution of this
eigenproblem can be
00:44:26.250 --> 00:44:29.290
effectively achieved via the
generalized Jacobi method,
00:44:29.290 --> 00:44:32.270
which I referred to
briefly earlier.
00:44:32.270 --> 00:44:35.515
Of course here, we are using
our solution algorithm for
00:44:35.515 --> 00:44:36.290
static analysis.
00:44:36.290 --> 00:44:41.200
There's L D L transpose
factorization involved here.
00:44:41.200 --> 00:44:45.260
And here, this is a simple
matrix multiplication that has
00:44:45.260 --> 00:44:46.880
to be carried out.
00:44:46.880 --> 00:44:50.620
This part here, is the
eigenproblem solution, and
00:44:50.620 --> 00:44:51.450
there's another vector
00:44:51.450 --> 00:44:53.980
multiplication carried out here.
00:44:53.980 --> 00:44:59.320
Now, under specific conditions,
this eigensolution
00:44:59.320 --> 00:45:04.690
algorithm converges and the
vectors stored in x k plus 1,
00:45:04.690 --> 00:45:08.940
they are q vectors now here,
with appropriate ordering,
00:45:08.940 --> 00:45:11.750
converge to the phi matrix.
00:45:11.750 --> 00:45:17.800
This is now an n by q matrix,
storing the lowest
00:45:17.800 --> 00:45:22.200
eigenvectorr, or I should
say, the q eigenvectors
00:45:22.200 --> 00:45:25.020
corresponding to the
lowest eigenvalues.
00:45:27.800 --> 00:45:30.820
And here, we have it written
out, and of course, this is a
00:45:30.820 --> 00:45:31.960
diagonal matrix.
00:45:31.960 --> 00:45:35.530
The condition that we have to
satisfy here is that the
00:45:35.530 --> 00:45:42.800
starting subspace spanned by
these vectors here must not be
00:45:42.800 --> 00:45:52.840
orthogonal to the subspace
spanned by the p eigenvectors
00:45:52.840 --> 00:45:54.680
that we are interested in.
00:45:54.680 --> 00:45:59.160
Of course, here, I'm talking
about q eigenvectors.
00:45:59.160 --> 00:46:03.360
I really am only interested in
p eigenvectors, in the lowest
00:46:03.360 --> 00:46:05.510
p eigenvectors.
00:46:05.510 --> 00:46:09.390
So in general, what happens is
that I will only be interested
00:46:09.390 --> 00:46:13.080
in these eigenvectors here, and
the lowest eigenvalues.
00:46:13.080 --> 00:46:16.850
One comes out here, I'm not
too much concerned about.
00:46:16.850 --> 00:46:21.820
In fact, we are carrying these
along only to accelerate the
00:46:21.820 --> 00:46:23.040
iteration scheme.
00:46:23.040 --> 00:46:28.060
And I will show you just know
what the convergence rate is,
00:46:28.060 --> 00:46:29.770
and why we are carrying
them along.
00:46:29.770 --> 00:46:34.800
So since we only interested in
the lowest p eigenvalues and
00:46:34.800 --> 00:46:38.870
corresponding eigenvectors,
we want to get convergence
00:46:38.870 --> 00:46:48.330
towards those, and what that
means is that our starting
00:46:48.330 --> 00:46:57.290
vectors in here must not be m
orthogonal to the eigenvectors
00:46:57.290 --> 00:47:00.780
that we want to calculate,
phi 1 to phi p.
00:47:00.780 --> 00:47:02.960
If that condition is satisfied,
one can show,
00:47:02.960 --> 00:47:06.300
theoretically, and of course,
we see it in practice also,
00:47:06.300 --> 00:47:09.230
that this iteration scheme will
always converge to the
00:47:09.230 --> 00:47:12.385
lowest eigenvectors and
corresponding eigenvalues.
00:47:16.010 --> 00:47:20.760
One important point here, once
we have calculated, say, a
00:47:20.760 --> 00:47:24.560
certain number of iteration
vectors, and we have tested
00:47:24.560 --> 00:47:27.810
for convergence, we are saying
that convergence is reached,
00:47:27.810 --> 00:47:32.670
say, when the current iterates
on the eigenvalues don't
00:47:32.670 --> 00:47:34.450
change very much anymore.
00:47:34.450 --> 00:47:38.720
This tol might be 10 to the
minus 6, as an example.
00:47:38.720 --> 00:47:41.820
Then once this condition is
satisfied, we stop the
00:47:41.820 --> 00:47:46.440
iteration, and we really want
to make then sure that we
00:47:46.440 --> 00:47:48.910
really converge to the
eigenvalues of interest.
00:47:48.910 --> 00:47:50.580
And how is that accomplished?
00:47:50.580 --> 00:47:53.310
Well, if these are the
p eigenvalues, we are
00:47:53.310 --> 00:47:57.060
calculating error bounds on
these eigenvalues, indicated
00:47:57.060 --> 00:48:01.690
here by the blue lines here.
00:48:01.690 --> 00:48:06.670
And then just to the right
of the error bound.
00:48:06.670 --> 00:48:11.300
Or we can really use that error
bound, we apply a Sturm
00:48:11.300 --> 00:48:15.520
sequence shift, mu S. What this
then says that k minus mu
00:48:15.520 --> 00:48:18.150
S factorize into L
D L transpose.
00:48:23.090 --> 00:48:29.870
And we must now have in the D
matrix, p negative elements.
00:48:29.870 --> 00:48:35.390
And only p, not more, not less,
just p negative elements
00:48:35.390 --> 00:48:39.050
must be occurring
in the D matrix.
00:48:39.050 --> 00:48:44.410
This check is absolutely sure.
00:48:44.410 --> 00:48:47.510
If this check a satisfied, we
can be absolutely sure that we
00:48:47.510 --> 00:48:50.970
have calculated the eigenvalues
of interest, the
00:48:50.970 --> 00:48:53.830
lowest p eigenvalues, in
this particular case.
00:48:53.830 --> 00:48:57.910
So I repeat, after we have
satisfied this convergence
00:48:57.910 --> 00:49:03.510
rate, or this conversions
limit, I should say--
00:49:03.510 --> 00:49:06.400
typically 10 to the minus
6, or 10 to the minus 8,
00:49:06.400 --> 00:49:08.700
depending on what accuracy
you want.
00:49:08.700 --> 00:49:12.920
We calculating eigenvalue
bounds, as shown here.
00:49:12.920 --> 00:49:18.500
We are putting a mu S Sturm
sequence shift just to the
00:49:18.500 --> 00:49:24.460
right of the last eigenvalue
bound, and if this is the p's
00:49:24.460 --> 00:49:28.540
eigenvalue, we must have
here, exactly p
00:49:28.540 --> 00:49:30.670
negative elements indeed.
00:49:30.670 --> 00:49:35.390
The convergence rate of the
iteration, we find is given by
00:49:35.390 --> 00:49:38.110
lambda i divided by
lambda q plus 1.
00:49:40.750 --> 00:49:44.650
And the convergence rate to the
eigenvalues of interest
00:49:44.650 --> 00:49:47.140
are just this value squared.
00:49:47.140 --> 00:49:50.280
Notice that we have
here, q plus 1.
00:49:50.280 --> 00:49:53.840
So if we are interested in the
p's eigenvalue,== the black
00:49:53.840 --> 00:49:56.850
pen doesn't work too well, let
me take the blue one here--
00:49:56.850 --> 00:50:01.330
if you're interested in the p's
eigenvalue as the largest
00:50:01.330 --> 00:50:04.220
eigenvalue, notice that the
convergence rate is that
00:50:04.220 --> 00:50:07.660
lambda p plus-- lambda p divided
by lambda q plus 1.
00:50:07.660 --> 00:50:11.030
That is the worst convergence
rate that we are reaching
00:50:11.030 --> 00:50:14.840
because the other eigenvalues
below lambda p are smaller,
00:50:14.840 --> 00:50:16.740
giving us better convergence
rate.
00:50:16.740 --> 00:50:22.110
So this is the worst that we
can reach, if we are only
00:50:22.110 --> 00:50:25.280
interested in p eigenvalues
and corresponding
00:50:25.280 --> 00:50:26.790
eigenvectors.
00:50:26.790 --> 00:50:29.850
Notice however, that lambda
q plus 1 must be
00:50:29.850 --> 00:50:31.290
greater than lambda p.
00:50:31.290 --> 00:50:33.440
If lambda q plus 1 is
equal to lambda p,
00:50:33.440 --> 00:50:34.880
we would never converge.
00:50:34.880 --> 00:50:38.930
So this is the reason why we
are choosing q to be larger
00:50:38.930 --> 00:50:42.030
than p, substantially
large than p.
00:50:42.030 --> 00:50:49.640
For example, we might choose
q to be typically, as an
00:50:49.640 --> 00:50:57.850
example, q being the maximum
of 2p and p plus 8.
00:50:57.850 --> 00:51:00.940
This is a formula that
we have used with
00:51:00.940 --> 00:51:02.700
quite a lot of success.
00:51:02.700 --> 00:51:05.920
In other words, when p is equal
to 4, we would use--
00:51:11.060 --> 00:51:15.710
no, sorry, I should say
the minimum here.
00:51:15.710 --> 00:51:19.100
the In other words, when
p is equal to 4, we
00:51:19.100 --> 00:51:21.430
would have 8 here.
00:51:21.430 --> 00:51:24.410
When p is equal to
8, it's the same.
00:51:24.410 --> 00:51:26.390
Because these two values
give us the same.
00:51:26.390 --> 00:51:30.470
And when p is equal to 100,
then we just have 8 more.
00:51:30.470 --> 00:51:32.850
Q is equal to 108.
00:51:32.850 --> 00:51:37.870
This 8 here is a safeguard, so
that this cannot be equal to 1
00:51:37.870 --> 00:51:39.130
in practice.
00:51:39.130 --> 00:51:43.960
It would only be equal to 1 if
we would have, basically.
00:51:43.960 --> 00:51:47.390
9 time the same root, and of
course, that is a case which
00:51:47.390 --> 00:51:50.290
we would hardly encounter
in practice.
00:51:50.290 --> 00:51:55.060
Now, regarding the choice
of the starting values.
00:51:55.060 --> 00:51:57.860
For the vectors, there
are two choices.
00:51:57.860 --> 00:52:03.070
x1 is simply a vector
with 1 only.
00:52:03.070 --> 00:52:10.220
xj would be a vector with zeroes
everywhere, but with a
00:52:10.220 --> 00:52:11.300
1 somewhere.
00:52:11.300 --> 00:52:12.720
And that one is in
the j's entry--
00:52:15.590 --> 00:52:18.310
sorry, the k's entry.
00:52:18.310 --> 00:52:23.050
The k's entry for ek, ek has
in the case, entry of 1.
00:52:23.050 --> 00:52:25.630
And otherwise, everywhere
zeroes.
00:52:25.630 --> 00:52:29.390
This would be the second to
the q minus first vector.
00:52:29.390 --> 00:52:31.150
And the last vector, we
take a random vector.
00:52:31.150 --> 00:52:33.190
That is one choice of
starting vectors.
00:52:33.190 --> 00:52:36.310
The Lanczos method can also be
used to generate starting
00:52:36.310 --> 00:52:37.560
vectors very effectively.
00:52:41.260 --> 00:52:45.450
Once we have selected the
starting vector, we go through
00:52:45.450 --> 00:52:48.770
the iteration schemes the way
I have been displaying it.
00:52:48.770 --> 00:52:53.420
And there, it is important, once
again, to perform after
00:52:53.420 --> 00:52:56.510
the convergence- the Sturm
system sequence check, and
00:52:56.510 --> 00:53:00.610
also we might want to calculate
error bounds on the
00:53:00.610 --> 00:53:01.970
eigenvalues and eigenvectors.
00:53:01.970 --> 00:53:05.240
These error bounds would be
calculated by simply saying,
00:53:05.240 --> 00:53:08.070
well, if we want to satisfy--
00:53:08.070 --> 00:53:11.660
or if we want to solve for
lambdas and phis that
00:53:11.660 --> 00:53:17.400
satisfies this equation,
then why not see how
00:53:17.400 --> 00:53:18.650
close we will satisfy.
00:53:18.650 --> 00:53:23.780
Make that a minus and put here
an error, say, R, on
00:53:23.780 --> 00:53:25.050
the right hand side.
00:53:25.050 --> 00:53:28.260
Substituting then the last
values for lambda and phi that
00:53:28.260 --> 00:53:32.220
we just kind calculated, the
last iterative values, we get
00:53:32.220 --> 00:53:34.180
this top line here.
00:53:34.180 --> 00:53:36.500
We're taking a norm
the way I've been
00:53:36.500 --> 00:53:38.310
talking about earlier.
00:53:38.310 --> 00:53:43.730
And we are also taking a norm
at the bottom to get,
00:53:43.730 --> 00:53:46.690
basically, a relative value,
top to bottom.
00:53:46.690 --> 00:53:49.240
And then relative value
gives us epsilon i.
00:53:49.240 --> 00:53:54.070
So epsilon i should be of order,
10 to the minus 3, if
00:53:54.070 --> 00:53:56.740
we want to have three-digit
accuracy in the eigenvalues
00:53:56.740 --> 00:53:57.770
and eigenvectors.
00:53:57.770 --> 00:54:00.020
In fact, the eigenvalues will
generally be more accurate
00:54:00.020 --> 00:54:02.040
than the eigenvectors.
00:54:02.040 --> 00:54:04.130
That is, of course, a
very low accuracy.
00:54:04.130 --> 00:54:06.260
Ideally, we have much
more accuracy after
00:54:06.260 --> 00:54:07.330
the iteration schemes.
00:54:07.330 --> 00:54:11.330
We would have to tighten the
tolerance, the tol value that
00:54:11.330 --> 00:54:15.010
I referred to earlier
sufficiently to make epsilon i
00:54:15.010 --> 00:54:16.180
small enough.
00:54:16.180 --> 00:54:19.370
Now, I like to refer, very
briefly also, to an
00:54:19.370 --> 00:54:21.740
accelerated subspace iteration
method that we have been
00:54:21.740 --> 00:54:26.800
developing during the last
years, which is a very
00:54:26.800 --> 00:54:30.340
substantial extension of this
subspace iteration method.
00:54:30.340 --> 00:54:36.080
The idea here is, basically,
that we wanted to combine the
00:54:36.080 --> 00:54:39.690
sum of the effective techniques
that we are using
00:54:39.690 --> 00:54:44.115
in the polynomial iteration
method, with the standard
00:54:44.115 --> 00:54:46.230
subspace iteration method
that I have
00:54:46.230 --> 00:54:48.340
discussed with you here.
00:54:48.340 --> 00:54:51.710
Basically, in the accelerated
subspace iteration method, we
00:54:51.710 --> 00:54:55.550
are not keeping a constant
coefficient matrix, as
00:54:55.550 --> 00:54:56.870
we have done here.
00:54:56.870 --> 00:54:59.160
The k matrix never changed.
00:54:59.160 --> 00:55:02.640
Here, we are shifting the
coefficient matrix, we are
00:55:02.640 --> 00:55:04.480
shifting through the spectrum.
00:55:04.480 --> 00:55:08.860
Much in the same way as we are
pursuing it in the determined
00:55:08.860 --> 00:55:10.070
search method.
00:55:10.070 --> 00:55:11.450
And that is number one.
00:55:11.450 --> 00:55:15.720
And number two-- we are also
not iterating with vectors
00:55:15.720 --> 00:55:18.610
that are always larger than
the number of eigenvectors
00:55:18.610 --> 00:55:19.710
that we want to calculated.
00:55:19.710 --> 00:55:22.990
In fact, if we want to
calculate, say, 100
00:55:22.990 --> 00:55:27.700
eigenvalues, we might just use
four iteration vectors at a
00:55:27.700 --> 00:55:31.420
time, and shift through the
spectrum, always with four
00:55:31.420 --> 00:55:35.020
iteration vectors, capturing
the eigenvalues values and
00:55:35.020 --> 00:55:35.980
eigenvectors that we are really
00:55:35.980 --> 00:55:37.800
interested in, in bundles.
00:55:37.800 --> 00:55:42.080
So this is a way how we have
very substantially accelerated
00:55:42.080 --> 00:55:45.310
the standard subspace iteration
method that I have
00:55:45.310 --> 00:55:49.970
been talking to you
about just now.
00:55:49.970 --> 00:55:52.315
If you are interested in seeing
some solution times,
00:55:52.315 --> 00:55:56.180
please we look at
this reference.
00:55:56.180 --> 00:55:59.960
Ladies and gentlemen, this
then brings me to the
00:55:59.960 --> 00:56:01.640
conclusion of this lecture.
00:56:01.640 --> 00:56:05.630
And in fact, it also brings us
to the conclusion of the whole
00:56:05.630 --> 00:56:09.300
set of lectures that I have
been presenting to you.
00:56:09.300 --> 00:56:14.180
These set up lectures have been
brief and very compact.
00:56:14.180 --> 00:56:16.900
They've been a survey
introduction to the finite
00:56:16.900 --> 00:56:17.760
element method--
00:56:17.760 --> 00:56:21.670
a very brief and compact
introduction and survey.
00:56:21.670 --> 00:56:24.520
Brief because there are
only 12 lectures.
00:56:24.520 --> 00:56:27.800
Compact because I wanted to
present to you as much as
00:56:27.800 --> 00:56:30.570
possible in these 12 lectures.
00:56:30.570 --> 00:56:34.390
And there is, of course, a lot
of information that we could
00:56:34.390 --> 00:56:35.900
have talked about.
00:56:35.900 --> 00:56:40.560
However, I hope that the
information that I had given
00:56:40.560 --> 00:56:44.760
in the lectures has been of
interest to you, and it will
00:56:44.760 --> 00:56:52.070
also be valuable to you in your
work and general decision
00:56:52.070 --> 00:56:54.680
making process of whether you
want to use a finite element
00:56:54.680 --> 00:56:58.120
method, how you want to
use it, and so on.
00:56:58.120 --> 00:57:01.600
I like to take the opportunity
at this point to thank the
00:57:01.600 --> 00:57:04.670
Center of Advanced Engineering
Studies for the terrific
00:57:04.670 --> 00:57:07.260
effort they've made in
preparing the tapes.
00:57:07.260 --> 00:57:12.150
I'm very happy that they have
been cooperative with me in
00:57:12.150 --> 00:57:17.750
the way they did, in the
preparation of the tapes.
00:57:17.750 --> 00:57:22.145
I also would like to thank you
as listeners for your patience
00:57:22.145 --> 00:57:25.590
of listening to the tapes and,
once again, I hope you found
00:57:25.590 --> 00:57:27.950
the tapes interesting and
have obtained some value
00:57:27.950 --> 00:57:29.260
information on them.
00:57:29.260 --> 00:57:31.590
Best of luck in your further
studies of the
00:57:31.590 --> 00:57:32.840
finite element method.