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Hi. In the last class, we have discussed about
root condition, which has direct implication
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on zero stability of the method. So, before
we proceed further on some more things
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related to stability, so let us state root
condition as a necessary and sufficient condition.
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.
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So, we have not stated that. So, the linear
multi step method is zero stable, if
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given y
dash equals to f of x y f of x y satisfy Lipchitz
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condition and root condition holds. So,
that means the linear multi step method is
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zero stable if given this f satisfy Lipchitz
condition and root condition holds. So, as
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I mentioned for a given method say 2 y n plus
2 minus 3 y n plus 1 minus 6 y n minus h 25
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f n plus 1 minus 40 f n equals to 0, we can
write, so 2. So, this is n plus 2. So, 2 zeta
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cube zeta, this is n plus 2, so this is not,
this
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will be 2 square minus 6 and this will be
this will be…
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..
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So, that is how we identify. So, now consider
another example let us say
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so the first
characteristic polynomial 11 and the roots
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if we compute. Now, what is the root
condition says root condition the roots must
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be within the unit disk, but unfortunately
for
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this greater than 1, therefore this is not
zero stable. So, we can compute accordingly.
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So,
for example, consider
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so for this is
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so this implies
simple root hence zero stable. So, that
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is how we compute.
.
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.So, let us proceed further. If you recall
for convergence, we have defined two things,
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one
is stability and another is consistency. So,
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the linear multi step method of the form star
is
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said to be consistent. If it has order 3,
greater than 1, so these two are together
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necessary
for convergence. So, before we go for further
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concept of stability, let us do some
example. Determine all values
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of the real parameter b for which linear multi
step
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method. Determine all values of the real parameter
b for which this method is zero
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stable. So, let us identify rho of zeta. So,
rho of zeta for this method, look at the distance
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plus 1 and plus 2 and plus 3, so this is our
first characteristic polynomial. So, this
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can be
written as this.
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.
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This can be written as, so call this, now
the method is zero stable if all the roots
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of rho
one belong to unit disk, those on the circle
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are simple and differ from 1. Why? This is
because already we have zeta equals to 1 as
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one root. If you have multiple roots, then
one
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of the necessities you have seen of the modulus
is greater than 1. Then it blows up and
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multiplicity, therefore for rho 1, these are
the conditions. The method will be zero stable.
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If this happens, hence suppose that the method
is zero stable, then b is not equal to 1 by
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2
or b is not equal to 3 by 2 because
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for these values of b, rho 1 zeta has double
roots on
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the unit circle, that is zeta equals to 1,
minus 1 for these values of this further product
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of
roots of rho 1 zeta is 1, further product
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of is 1. Then but since these values are not
equal,
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roots are none of them are 1 or minus 1, neither
of them are plus or minus 1.
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..
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Hence, they are strictly complex. Therefore,
the discriminant negative, this implies, this
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implies b belongs to half, 3 by 2 to b belongs
to, so this is this interval for zero stability.
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Conversely, suppose b belongs to this. Then
the roots are 1, zeta 2, 3. These are, so
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this
is 1 and mod 2, 3, this is 1, but zeta 2 or
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3 not equals to 1, zeta 2 not equals to zeta
3 and
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all roots lie on the unit circle and are simple.
So, this is the analysis to identify a
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particular method is zero stable or not. In
this particular case, we were trying to obtain
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the two bounds on the b, the range for which
this method is absolutely zero stable.
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.
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.Now, let us proceed to something called absolute
stability. If you recall for single step
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methods, we have discussed this concept. So,
the stability that we discussed with respect
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to h goes to 0, n goes to infinity, nh fixed,
so this was the stability. The zero stability
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h
goes to 0, n goes to infinity, so that nh
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is fixed. Now, we need to analyze something
different, how a particular linear multi step
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method behaves
in the case h greater than 0
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fixed and n goes to infinity.
So, how a particular linear multi step method
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in the case h goes to 0 fixed and n goes to
infinity. So, absolute stability with reference
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to always a particular reference equation,
so
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that is our y dash equals to lambda y. Now,
consider our method equals to h multi step
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method in this form, let us say both i equals
to 0 to k; suppose we considered in this,
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then
with reference to the reference equation,
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we
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get similar difference equation in this form.
.
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Now, the general solution of this, what is
this? This is a homogeneous difference
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equation or a recurrence relation can be
expressed as a linear
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combination of powers of
roots of the associated
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stability polynomial. So, this is a stability
polynomial, first
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characteristic, second characteristic, and
then one can define stability polynomial.
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So,
this is possible. Now, the definition of absolute
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stability the linear multi step method
of
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the form star is called absolutely stable
for a given h bar
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if and only if
for that h bar, all
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the roots all the roots say r s of the stability
polynomial satisfy this condition, otherwise
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this method
is absolutely unstable, absolutely stable
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unstable. So, what it says this linear
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.multistep method is absolutely stable for
a given h bar. See zero stability we defined
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with
h goes to 0 as n goes to infinity. Now, absolute
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stability we are defining for fixed h.
.
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So, similar to single step method, it is our
duty to find out an interval, an interval
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alpha
beta of the real line is called the interval
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of absolute stability. The method is
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absolutely
stable for all h bar belongs to this interval.
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Further, the method is absolutely unstable
for
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every h bar, then the method is said to have
no interval of stability. So, it is a very
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much
essential to find the interval of absolute
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stability. So, what is
the procedure?
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.
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.So, general methods for locating
the interval of absolute stability, so the
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first one, first
there are several approaches. So, let us discuss
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couple of them. So, the first one is schur
criterion. So, before we state that some preliminaries,
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consider the
polynomial phi r of
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the form c 0, c k non zero, c 0, non zero
with complex coefficients, complex coefficients.
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Then, the polynomial phi
is said to be a schur polynomial, if each
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of its roots r s satisfy.
So, the polynomial phi is said to, so first
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of all we consider polynomial where these
are
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complex coefficients and remember c k non
zero, c 0 non zero, then the polynomial is
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said to be a schur polynomial if each of its
roots are satisfy. That means the magnitude
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must be less than 1. So, then we say this
polynomial is schur polynomial.
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.
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Now, let us consider
the following polynomial. What is that? We
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denote it by phi hat c 0
bar. What are these bars, one can guess easily
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because the coefficients were complex.
Therefore, c j bars are complex conjugates
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of c j. Please try to recall c k r power k,
c k
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minus 1 r power k minus 1, c 1 r 0. Now, what
is the polynomial that we are
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considering? Look c 0 conjugate has gone to
r power k, c 1 conjugate has gone to r
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power k minus 1, c k conjugate has gone to
r power 0, and so it is backwards. Then let
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us
define phi 1 of r 1 over r phi hat of 0 phi
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of r phi of 0 that has degree less than or
equals
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to k minus 1 that has degree.
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..
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Then, this is the theorem. The polynomial
phi defined earlier is a schur polynomial
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if and
only if modulus of phi hat of 0 is greater
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than modulus of phi of 0 and phi 1. So, what
it
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says that phi which was defined earlier is
a schur polynomial, if and only if this condition
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holds and phi 1 is a schur polynomial. When
do you say it is a schur polynomial? That
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means roots of phi 1 must be less than 1 in
magnitude. So, let us check with an example.
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.
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.Consider this example characteristic polynomial
phi of r square minus 1 because n plus 1
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is missing, this is r minus 2. Hence, the
stability, the stability polynomial, sorry
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r h is
given by r square minus 1 minus
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the stability polynomial. See, we have defined
like this.
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So, this is h bar, so this minus
so equal to r square minus, this is our stability
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polynomial.
Now, we would like to use the schur criterion.
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So, what it says? So, this is our
polynomial. Now, consider this as our relative
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polynomial. Then we have to
compute phi
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hat where the coefficients are conjugated
and given backwards. Therefore, pi hat is
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given
by so this is c 2, c 1, and c 0.
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Now, we have to consider the other way. So,
c 0 r square, so this is, this is our phi
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hat.
Now, what should be checked? The polynomial
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phi is a schur polynomial if and only if
phi hat of 0 moduli is greater than this and
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phi 1 is a schur polynomial. So, let us try
to
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do that. So, what was the, so we have pi like
this, pi hat like this. Now, we have to check
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mod, mod pi of 0.
.
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This is mod 1 plus, then so the condition
mod is greater than this implies
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h bar belongs
to, so this can be verified very easily. Now,
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what is the next step? So, this is the range
that means the polynomial phi is a schur polynomial
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if and only if this happens. That
means the polynomial is schur polynomial only
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when h is in this range. Further, we have
to check phi 1 is a schur polynomial. So,
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let us check.
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.Now, pi 1 r is 1 over r. So, we are computing
as defined pi 1 r, we are computing pi 1 r.
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So, pi hat 0 to pi hat 0 is 1, so times phi
of r that is phi of r, this is exactly this
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minus pi
of 0, this minus 1 plus 2 h bar by 3, 1 minus
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1 by 3 h bar r. So, this is 1 over r, r gets
cancelled. So, we can just put r minus h by
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3 minus 1 over r, then plus 1 over r with
1
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there, then minus h by 3, 1 plus this.
If it gets cancelled, then this 1 minus 1
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plus 2 h bar by 4 square r, so this is so
r 1 there
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and this is the coefficient. Then these two
get cancelled. Then minus h by 3, if I take
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common 1 plus, so this can be written as so
this is pi 1 r. Now, the definition is
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polynomial pi is said to be schur polynomial
if each of its roots satisfy modulus less
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than
1.
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Now, further the schur criteria says the polynomial
phi is a schur polynomial if and only
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if this and this is only a, so r equals to
half, r equals to half is our root, so which
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implies,
hence pi 1 is a schur polynomial, pi 1 is
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a schur polynomial. Now, three a multi step
method of the form is called absolutely stable
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for a given h bar if and only if for that
h
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bar, all the roots for a stability polynomial
satisfy this. So, we have verified this. So,
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if it
is a schur polynomial if each of roots satisfies,
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this it is interrelated. We have shown that
phi 1 is pi 1 has this; therefore pi 1 is
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a schur polynomial. Therefore, the roots of
their
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characteristic polynomial are less than 1.
Hence, the method is absolutely stable.
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.
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.So, let us look at one more example
y n plus 2 minus y n equals h by 2. Now, this
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can be
written as so let us do everything in one
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go, E square minus 1 y n h by 2 E plus 3 f
n.
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This implies rho E equals E square minus 1,
so therefore plus 3. Therefore, now the
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stability polynomial, this is given by, this
is given by this equals r square 1 plus 3
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by 2.
Therefore, pi hat of this
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r square r plus 1.
.
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So, we have to compute this equal to 1. Then
let us schur criterion says the polynomial
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pi
is a schur polynomial if and only if this
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is 0. So, let us so
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implies this implies
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h bar
belongs to this range. Now, on the next step
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is to construct this pi 1 and show that this
is
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a schur polynomial. So, this is 1 over r.
So, what is this polynomial pi 1? Pi 1 has
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00:51:43,579 --> 00:52:00,459
to be
constructed. So, we have to construct this,
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this is 1 and pi of r. So, this will be r
minus
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00:52:56,170 --> 00:54:33,779
this, this minus, so this gets cancelled.
So, this implies minus half 3 r, so this implies
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r equals to minus 1 by 3. So, this implies
mod of r is less than 1. So, schur criterion
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says the polynomial pi is said to be schur
polynomial if each of its roots modulus is
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less than 1. So, pi 1 is a schur polynomial,
then
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so the polynomials pi is a schur polynomial
because pi 1 is a schur polynomial. The
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polynomial phi is a schur polynomial and accordingly
what it says for absolute stability,
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these stability polynomial roots are within
modulus unity.
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So, this is schur criterion, helps one to
conclude the region of absolute stability.
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For
example, when we say h bar that is lambda
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h is within the range minus 4 by 3 to 0, one
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.must choose a combination of lambda and h
because for a fixed h, you have to choose
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a
lambda such that the lambda h is within the
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range. So, this trade off, suppose lambda,
suppose for a given lambda, one cannot choose
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h, which makes the lambda h beyond this
interval. So, the schur criterion is very
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much useful. So, in the next lecture, we shall
discuss some more criterions for absolute
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stability. So, you may try for some other
methods with schur polynomial, until then
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good bye.
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.