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Hello. In the last few lectures, we have been
discussing about multistep methods. So, we
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started with explicit methods then implicit,
then we discussed some stability aspects,
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and
then how a combination of explicit implicit
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methods can be used as predictor corrector
methods to improve upon the solution. So,
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well we have given some motivation for the
multistep methods, but however I would like
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to mention here a different motivation and
then some comments on multistep methods. So,
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it is a kind of alternative approach, but
it
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is a intuitive.
.
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So, let us start with that. So, alternate
ideas behind corrector, suppose this is our
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y i plus
1 and this is the approximation, let us denote
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some other notation. So, this is matches
slope for this. So, this can be thought of
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us first approximation suppose this matches
slope and curvature. So, then this can be
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thought as second order approximation. So,
if y
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dash of x i is slope is known, then it is
a first order approximation. If slope and
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curvature
are known second order approximation, but
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the fact is curvature is proportional to y
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.double prime. So, hence the question is can
we find
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approximation to y i double using
already computed values.
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.
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So, this is the motivation can we approximate
this person. So, usual approximations y i
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double minus by h, usual difference approximation
then y i plus 1 is Taylor series. So, y i
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double prime can we approximate this using
past points. So, here is the answer, we
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substitute. So, this we got a kind of a multistep
method. So, what did we do just this is
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the Taylor series expansion, but then the
second derivations have been approximated
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by
the corresponding finite differences which
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involve past points, but star becomes exact
rather than approximate if y of x is polynomial
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degree 2 because y i prime is z x i plus b
then y i double
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is 2 a by h. So, this star becomes exact rather
than approximation if
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because the coefficients are determined.
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.(Refer Slide Time: 08: 00)
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So star is second order, however if one needs
third order what would we do, let us
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consider this. So, then finite difference
approximation
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then we get . So, this is our new
star, but the question see in the earlier
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case we have a substituted in the earlier
case we
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have substituted for y i double only of after
order h, but now we would like to go for a
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higher. So, in order to use y i double which
is after order of h in star we need
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an
expression for this because we have to increase
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right an expression see this is h square.
So, you need third order, you need an expression
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for this. So, that h square times this you
get third order. So, an expression for the
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order h term with accuracy h square.
.
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.So, how do we do it y i prime minus for the
second derivative we want to increase the
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order. So, therefore, now if we substitute
this star would lead to which is Adams Bash
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forth third order method. So, this is a kind
of a alternative thought i mean this is a
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little
unusual, but this is a kind of a alternative
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thought. So, this gives the multistep method.
So, one can derive other methods.
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Now, if you recall in one of the earlier lectures.
So, given see a general linear multistep
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method can be expressed in terms of a 2 characteristic
polynomials that is a first
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characteristic the second. Now, one of the
question that was posed was given the first
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one can we determine the second characteristic
polynomial, so that they corresponding
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multistep method is determined, that was a
bit unfinished.
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.
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So let us try to recall this and then try
to compute for a particular case. So, given
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to
compute, we consider a multistep method i
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would like to make a remark. So, this can
be
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written in different notations. So, I do not
want you to remember, we should be able to
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write within any general notation. Sometimes,
like this sometimes with a n plus j
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combination. So, this is our multistep method
then the corresponding error, now linear
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multistep method said to be of order p equal
to 0 and C p plus 1 non 0. So, accordingly
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t i
plus 1, we get this that is identically 0,
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when y x is a polynomial of degree less than
equal to p y a.
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..
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Now, observe that c i and p, order of the
method are independent of y of x k. So, choose
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y this y of x is e power x. Accordingly this
then your double star which is the error this.
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So, this is e power if I take e power xi minus
k plus 1 common e power k h e power h
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minus x this.
.
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So, using this notation this can be written
as rho of h sigma of because this is the
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characteristic first characteristic and second.
So, but then this must be equals to, from
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these 2 we can. So, by some terms of this
will be left that is e power 1 minus k. So,
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that e
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.power k minus 1. So, this is h times 1 minus
k. So, that is given to this coefficient.
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So,
with adjustment it can be written like this,
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now this becomes the basis for finding given
this how to find the second characteristics
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polynomial. How do we do it, let us proceed
set this as h goes to 0 1 then log z we write
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it as z minus 1 plus 1 and we write it an
expansion further h power p plus 1 is log
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z.
.
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Now, these two we use it here therefore, if
is given the above can be used to find rho,
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how do we do it expand in then retain terms
of required order. So, another remark is
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implicit are of same order or the degree
explicit rho is 1 higher.
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..
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So, let us work it out with an example, given
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find this, so looking at this. So, this implies.
So, degree of 1 for of course, for a explicit
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method. So, this can be further simplified
as,
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what we have taken we have cancelled out z
minus 1. So, we expanded therefore, this
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equals to because of a, the second degree,
we have expanded up to that.
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.
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We have got sigma, therefore the method is
explicit which is of second order. So, what
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did we do given this we have expanded. So,
this is h, h is a log zeta. So, I have divided,
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this rho by h rho by log zeta we have expanded.
So, rho by log zeta we expanded, so then
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.up to second order. So, that should be of
order sigma neglecting of second order that
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should be sigma and hence the method is given
by this. So, this gives a sense of
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computing first characteristic polynomial
second characteristic polynomial vice versa.
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.
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So, there is another aspect related to multistep
methods which I would like to mention
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stiff systems. So, stiff systems has to do
with the system of equations, but however
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while
solving using multistep methods, it is a important
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study stiff systems in the context of
multistep methods. So, we have a system, example
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say A is with conditions where x bar
is say x 1 of t and x 1 of 0. So, this implies
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x 1 of t is, we have an equation like this
system the coefficient matrix is like this
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with this initial conditions we get the solution.
So, this becomes this is after a short time
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the reason is you have a big coefficient
multiplying t. So, after short time this decays
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faster than this. So, after short time we
have this solution.
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..
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So, what is great about this system it is
a very fuggy kind of a term, stiffness, solution
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of
interest what are they, we expect solutions
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of interest means they are of the form e power
lambda t kind of solutions right are slowly
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varying, but solutions with rapidly changing
structure are possible. So, look at this,
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these are the solutions of interest, but however
you have solutions that are rapidly changing.
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.
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So, let us define more precisely, a system
is called stiff if a numerical method with
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a
finite region of absolute stability applied
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to a system with any initial conditions, is
forced
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.to use in a certain interval of integration
a step size, which is excessively small in
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relation to the smoothness of the exact solution
in that interval.
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So, what it says a system is called stiff
if a numerical method with a finite region
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of
absolute stability, applied to a system with
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any initial conditions is forced to use in
a
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certain interval of integration, a step size
which is excessively small, that means your
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exact solution is really smooth up to some
order, but however the step size that the
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method is demanding is not in correlation.
So, we can analyze this a little more carefully.
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So, let us analyze with respect to system
.
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Consider system that means a system of IVPs,
so where these are the unknowns. So,
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example
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here our y was and this is an example. Now,
we have learnt how to solve an
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IVP for a single component, but then suppose
if we extend similar method to this we
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must arrive at a multistep method. So, a linear
k step method for the numerical solution
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of one solution of one
we expect to have the following form.
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..
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So, now let us assume that this f vector is
of this form. So, for example, f bar is x
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plus y
1 minus y 2. So, this we need to express,
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this is possible. So, we need to get y 1 minus
y
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2 and 0 plus. So, this is our A, this is suppose
we are assuming like this. So, constant
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matrix and b bar column, this is the column
matrix then your multistep method
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gets
reduced to where. So, this is a simple exercise
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with reference to this split up, if you put
it
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in your multistep method we can check the
multistep method reduces to this.
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.
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.Let the Eigen values of matrix a are distinct
the lambda i, then there exists a non singular
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matrix h such that can you guess what would
be the matrix yes of course, the Eigen
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values entries. The diagonal matrix with Eigen
values. Then define what for we are
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doing, see look at this we know how to solve
a single multistep method for a single
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component. So, you have initial value problem,
let us say this is single component we
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have multistep method then we know how to
compute the solution and how to analyze
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whether the method is stable or not, but then
for a system it is coupled you can see.
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So, unless you arrive at a corresponding multistep
method which is decoupled we can we
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are not in a position to comment on the stability
and overall behavior of the system. So,
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we are trying to do that. So, define this
and c is, then we get. So, then this is m
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star, then
m star becomes a j. So, your m star contain
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identity matrix. So, this one can write h
inverse h. So, I would like to write
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minus h b j and a so it can be written like
this.
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.
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The same thing can be written as this is a
component i th component of the vector. So,
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this is you can see this is a corresponding
Eigen value. This is the corresponding
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component of a the vector z and what was our
definition. So, what did we achieve, each
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equation for a particular value of i is decoupled
and is similar to a linear multistep
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method, but only one remark h bar equals lambda
h can be complex.
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Now, one can discuss the stability because
now each equation is like a multistep method,
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but only thing is this is complex. So, within
the context of this complex, let us discuss
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.the absolute stability region. So, a linear
k step method is said to be absolutely stable
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in a
set R A of the complex plane, if for all h
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bar this all roots r s 1 to k of the stability
polynomial satisfy this, then the set R A
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is called region of absolute stability.
.
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So, let us see for a single component case.
So, for example, we have this and we have.
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So, pi is this equals to 0 this implies z
equals to. So, we need a the region, this
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implies
the region.
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.
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.Suppose, consider Euler method
then pi is given by, implies r is 1 plus h
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bar and the
region R A will be h bar belongs to the complex
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plane such that. So, what is this, this is
a
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open circle centered at minus 1. So, this
is the stability region, for a system also
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one can
discuss and one can observe that for the system,
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if the stability region is dictating some
range, however the solution is demanding very
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smaller h due to the extraneous terms. So,
then really the system is stiff in that context.
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So, this gives a some idea on the stiff
system. So, you can refer the notes by Julie
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for a better understanding with respect to
examples. So, more or less this gives some
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idea of multistep methods, predictor corrector
methods.
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Thank you bye.
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.