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Hello, in the last class we have discussed
some general methods for stability aspects
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of
multi step method. So, for example schur criterion
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we have discussed. So, given a multi
step method, how to find out the stability
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regions using the schur criterion?
.
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So, let us continue ahead with stability aspects
of multi step method. So, let us discuss
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some further methods. So, one of the popular
method is Routh-Harwitz criterion. So,
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consider the mapping of the open
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unit disk of the complex plane
to the open left half
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plane, that is real z less than 0 of the complex
z plane. So, what this mapping does? This
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maps unit disc mod r less than 1 to the left
of plane in z plane, then the inverse of the
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mapping is given by inverse of this mapping.
So, this is the inverse of the mapping.
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..
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So, under the transformation, under this transformation
the function. So, what is this?
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This is our stability polynomial. So, this
is this becomes, so this polynomial becomes
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this. Now, say this is a star, multiply star
by 1 minus z power k, we have. So, we are
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multiplying this by 1 minus z power k, where
k is the order of multi step method. So, we
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get this and our aim is to identify this as
a polynomial in z. So, what was our aim? So,
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this is order of the method.
.
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.So, we are multiplying and this we are expressing
as polynomial in z. So, then the roots
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of the stability polynomial lies inside the
open unit disk, if and only if
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the roots of
the
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polynomial double star, so that is of this
lie in the open left half plane. So, the roots
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of
the stability polynomial in the original variable
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r lie inside the open unit disc if and only
the roots of the polynomial double start.
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.
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This is this lie in the open left half plane.
So, then the criterion
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the roots of double star lie
in the open left half plane if
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and only if
the leading principal minors of the k cross
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k
matrix. So, this is a, the roots of double
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star so that double star is of the form. So,
lie in
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the open left half plane if and only if the
leading principal minors of this matrix are
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positive, and a 0 greater than 0 are positive
leading principle minors of this matrix are
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positive and a naught greater than 0.
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..
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We assume that a j is 0, if j is greater than
this, in particular for k equals to 2 the
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condition leads to this and k equals to 3
and k equals to 4. These represent the necessary
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and sufficient conditions for ensuring that
all roots lie in the left half plane. Of course,
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open. So, this is these are the necessary
and sufficient conditions. So, to start with,
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so we
have used a transformation, what was the transformation?
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z equals to r minus 1 and the
inverse of this is this.
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.
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.So, under this transformation the stability
polynomial reduces to this, then we have to
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multiply by this factor. As a result of the
multi step method, the linear multi step method
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of order k we are getting this factor. So,
then multiplied by that we get a polynomial
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z.
So, then the Routh-Hurwitz criterion states
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that the roots of this polynomial lie in the
open left half plane if and only if the leading
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principal minors of this matrix are positive.
And a 0 is greater than 0 and in particular
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we have the these being the necessary and
sufficient conditions for ensuring that all
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the roots of the polynomial in z to lie in
the
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open left half plane, ok?
.
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So, let us see how it works out with a example.
So, use the Routh-Hurwitz criterion
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to
determine
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the interval of absolute stability of the
linear multi step method. So, the given
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linear multi step method we need to determine
the interval of absolute stability. So, we
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have r equals to this is a map, then for this
method we have. So, this is given by r square
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minus 1 and this is given by accordingly,
we have r square minus 1 is h bar.
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So, this will be the stability polynomial
which is, so this is our stability polynomial
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and
what is the order of the method 2. So, we
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need to consider
this into variable change to
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this transformation. So, accordingly
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this should be written in the form, so we
can
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simplify this.
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..
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So, here this get cancelled minus 1 term there,
so 1 minus z. So, this will be 1 minus z
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square. So, this will be 1 minus z square.
So, minus h by 2 z square and this is usual
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expansion. So, we multiply, sorry this is
minus, this must be minus. So, this will be
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these
terms then so this is minus h bar z square
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plus 4 plus 3 h bar z. This is this 2 h. So,
this is
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a 0 z square.
Now, for k equals to this is the necessary
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and sufficient condition accordingly. So,
using
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part a, we get h bar to be is the region of
absolute stability. So, this is Routh-Hurwitz
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criteria, right? So far these stability methods,
these two general methods, one is schur
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criterion, other is Routh-Hurwitz criterion.
So, they give you using some kind of
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approach, the Schur criterion via Schur polynomial
and then Routh-Hurwtiz criterion via
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in kind of a transformation.
So, but there are another general methods,
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in the sense you try to after all it is a
difference equation, so you try to find the
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characteristic polynomial and find the roots
so
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that like a general recurrence relation you
get the solution. See once we have a
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recurrence relation which is also a finite
difference method, right? So, we can get the
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solution using the roots and hence analyze
what happens to this solution, ok? So, let
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us
see one such method.
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..
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Stability of multi step methods parasitic
term. So, we will see what is this parasitic
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term?
So, let us see with an example, so consider
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the multi step method. So, obviously this
is
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for n greater than equals to 1 so then so
with reference to absolute stability the reference
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equation is lambda y. So, accordingly these
become 2 h bar y n, where h bar is, so the
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characteristic polynomial r square minus this,
right? Now, the
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roots are given by 2 h, so
this is given by, so these are the roots.
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.
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.So, let us analyze, so what were the roots
r equals to. Now, consider first root, call
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it r 0
so this one plus
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so this is h bar plus. So, having power series
expansion we get 1 3. So, I
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write that first, then h here, then h square
by 2 plus the higher order terms. So, if you
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try
to approximate see what was our equation and
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naturally the exact solution must be of this
form.
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So, if we approximate and try to identify
with the exact solution y a of course, this
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is
indeed e power lambda h bar. Now, consider
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r 1 h bar minus, so this is h bar minus, so
this can be approximated as minus 1 minus
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h bar plus h bar by 2 plus third order terms.
Now, this can be approximated as minus e power
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lambda h bar, I am sorry this is lambda
which is e power h and this is e power minus
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lambda h, ok?
.
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So, one you have this type and other is this
type. However, once we have the roots the
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general solution y n is given by c 0 r 0 power
n c 1 r power n, right? So, there we have
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r
0 is behaving like e power lambda h and r
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1 is behaving like this. Now, let us take
real
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positive then what happens, lambda real positive.
So, we need to compare because the
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general solution is this.
Now, we need to identify which one is dominating
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and really the solution of the
difference equation gives stable solutions,
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right? So, then what happens if lambda is
real
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and positive r 0 is greater than this. Then
what happens? This grows faster than this.
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So,
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.what happens r 1 power n increases less rapidly
then r 0 power n, implies c 0 r 0 power n
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dominates, this dominates.
So, that means the other term, so out of your
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solution c 0 e power lambda h n e power
lambda h n. This is dominating this for this
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is a case which is the exact solution and
this
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is the parasitic term. However, in this case
does not create any trouble. So, why do we
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call this parasitic term and then why we conclude
all this? Of course, this is exact
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solution for the given solution, but this
is an extra. So, this is also called extraneous
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part
and this is the parasitic term, but however
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in this case does not create any problem.
The
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overall solution grows as per the exact solution.
So, that means for these values of
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lambda the solution is stable.
.
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.
So, let us consider now the other case, lambda
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is negative. Then for of course, h is
positive. So, then c 1 r power n will dominate
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c 0 r power n, as n goes to infinity for a
fixed h. Therefore, so the parasitic term
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which is h, so this parasitic term increases.
So,
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this creates a problem, so that is how so
for lambda real and positive it agrees with
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the
true solution, whereas for lambda negative
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this parasitic term creates a problem. So,
this
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is another kind of a stability analysis, ok?
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..
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Let us see this little more from little theoretical
point of view, relative and weak stability.
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So, relative and weak stability. So, with
respect to what it is relative we will see.
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So,
consider this, so general solution is given
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by. So, for a general method this is 0 to
n c j r j
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lambda h, right? The parasitic terms power
n goes to 0 y a, this is true because as h
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goes
to 0 the parasitic terms goes to 0. However,
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for finite h for any x n 1 would expect it
to
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be small
compared to c.
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So, that means we are bringing out the concept
of relative stability. So, relative with
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respect to, what? So, relative with respect
to the a h, the principle exact r 0. So, this
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implies
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for
sufficiently small h. So, this is relative
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is a comparison the parasitic terms
with the exact, so extraneous terms with the
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exact, right?
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..
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So, a method is relatively stable if the characteristic
roots
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satisfying
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for all sufficiently
small. So, that means let there be some parasitic
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terms, but compared to the exact, these
extraneous must be small for sufficiently
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small lambda h. The impact satisfies the strong
root condition if a h, these roots which are,
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which would contribute to the extraneous
terms if the magnitude it.
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Of course, for a h 0 if the magnitude is less
than 1 then we say the strong root condition,
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right? So, remark relative stability does
not imply strong root condition and if a linear
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multi step method is stable, but not relatively
stable. It is weakly stable, so relative
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stability does not imply strong root condition
and if a method is stable, but not relatively
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stable, then we say it is weakly stable, ok?
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..
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So for example, the case y n plus 1 plus 2
h f n, so far in this case we have seen. So,
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this
does not hold. So, hence this method is weakly
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stable, so that is how we can categorize
the relative stability and weakly stability.
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So, we have some criterions to systematically
capture the interval of stability region of
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stability, but there is another concept to
analyze
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the parasitic terms and then say how the method
behaves, ok?
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.
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.So, let us analyze with some more examples.
So, for example consider right. So, this
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implies on y dash we get. So, we have, so
this will be the characteristic polynomial,
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then
roots. So, if r is this we get these roots,
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ok?
.
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Accordingly z 1, z 1 which is so this would
be written as this can be written as. So,
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this is
for what if h goes to 0, r goes to 0, r square
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goes to 0. So, with that we can linearize
and
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try to get this. So, you can identify, so
we have done it in the last example. So, 1
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plus
lambda h equal e power lambda h. So, as h
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goes to 0, z 1 behaves like, it has to behave
like this. Therefore, y n c 1 z 1 power n
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plus c 2, then so this is c 1.
So, this agrees with
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exact solution and this is parasitic term.
Now, if lambda is negative,
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this grows and creates trouble. So, this is
another example where we could analyze.
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Now, for these multi step methods there is
a small technique, given your characteristic
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first and second characteristic polynomials
are r h o of zeta and then sigma of zeta.
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So,
for a particular method if r h o is given,
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can we determine the corresponding sigma?
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..
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So, this is another small trick involved a
h. So, let us try to do that. So, given this
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how do
we determine, so one may also refer the numerical
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solutions book by Jain. So, to know
lot of examples, so consider, so these are
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general linear multi step method case, step
method, then the error. So, this is the error,
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then recall this linear multi step method
said
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to be of order p if C is 0. So, this gives
a p th order method. So, accordingly T i plus
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1
reduces to right. So, this is identically
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0 when y x is a polynomial of degree less
than r
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equals to p, right?
.
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.Now, observe that the constants C i and p
are independent of y of x. So, let us choose
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y
of x equals to this, then double star, what
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was our double star this. So, this grows like,
so
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this is e power k h minus a 1. So, this can
be written as r h o of e h minus h, so it
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is right.
So, T i plus 1 this must be behaving like
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this, ok?
.
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So, this implies we can get some new constant
because there are some terms left out after
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getting cancelled x i, so that I have put
it to this constant. So, this is the relation
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using
which when r h o is given. How to determine
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sigma? So, this will suggest us. So, let us
work out some problems, how to determine exactly
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for a given example using this
relation in the coming lecture.
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Until then thank you, bye.
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.