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Good morning. In few classes, last classes,
we have learnt on multi step methods, both
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explicit and implicit. So, now it is very
important to learn about the convergence and
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stability aspects of multi step methods. We
have learnt a similar aspect for single step
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methods. However, the concepts for the multi
step methods are more involved. So, let us
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have a quick recall on the multi step methods,
before we discuss on the convergence and
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stability aspects.
.
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So, convergence and stability of multi step
methods, so recall we have general linear
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k
step method. This is just b n. So, this is
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a general and in this, if b 0 is non zero
implicit, b
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0 is zero explicit example. So, this is an
explicit, so this is implicit and explicit.
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So, this
is explicit method because to compute y n
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plus 1, we are asking past points y n and
y n
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minus 1, whereas here to compute y n plus
1, we are asking y n n minus 1 n plus 1. So,
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this is implicit and this is explicit.
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..
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Now, what is the definition of convergence?
So, the linear multi step method of the form,
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so of the form. So, let us say this is star
of the form, we can even write it because
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star
means of the form y n plus 1 is of the form
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is said to be convergent if for all IVPS
subject to the hypothesis of Picard’s existence
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theorem. So, that means if you recall
when we defined IVP, we assumed that the solution
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exists a subject to the Picard’s
existence theorem. So, whatever the hypothesis
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under those for example, Lipchitz
condition etcetera, so we have h goes to 0,
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y n equals to y of x n where x minus x 0 is
n h
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for all x belongs to a b and for all solutions
of the difference, see, for this is we have
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approximated the given IVP with the difference
equation. So, we are now talking about
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convergence of this particular solution of
the difference equation.
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So, as h goes to 0, this approximated is h
is equal to the exact for all x within the
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central
and for all solutions of the difference equation
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with the, with the, with the starting
conditions that are consistent. So, what is
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this that is some y s equals to say eta s
of h
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where s equals to 0, 1, k minus 1 for which
limit h goes to 0 eta s of h, this equals
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to y 0 s
equals to. So, that means these are all y
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s are as h goes to 0, they go to initial condition.
So, this is general condition and one must
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understand that it is within the framework
of h
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goes to 0. So, this is convergence. So, there
are some necessary conditions.
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..
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For a linear multi step method to be convergent
necessary conditions for convergence,
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one necessary condition for the convergence
of the linear multi step method is that that
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the method be zero stable. I am underlining
two necessary conditions for the
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convergence of the linear multi step method
is that the method be consistent, I am
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underlining. So, we have learnt a method.
Now, the approximation, how this
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approximation gives converging solution? So,
the convergence we have defined. Now,
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there are some necessary and sufficient conditions.
So, what are they? In some sense, these are
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conditions are telling one is zero stability
and
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the other is consistent. So, in some sense,
one can state necessary and sufficient
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conditions for convergence or stability and
consistency are the necessary and sufficient
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conditions for convergence, but we do not
know. We did not discuss what is zero stable
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and what is consistent. So, let us try to
understand what is zero stable and what is
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consistence of a linear multi step method.
So, what we are discussing is zero stability,
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zero stability.
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..
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Some of the definitions I am taking from book
Suli, this is numerical solutions of
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ordinary differential equations. So, let us
define what is zero stability y zero initial
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condition given and our IVP is f of x y, y
1, y 2, y k minus 1 need to be computed using
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any known method of course explicit because
for a multi step method, this is IVP, but
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we
need past points that at past points. So,
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this we have to compute using any known
method explicit say Runge-Kutta. So, what
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is the general strategy, one can compute
using Euler’s method, Taylor series, Runge-Kutta.
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What is our ultimate aim? Our ultimate aim
is to get as better a solution as possible
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as
accurate as possible. So, for this to happen,
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the general prediction see Euler method is
just it is a, it is a first order kind of
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method. So, it is first order method. The
Taylor series
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well if we use up to second order, it is fine.
Beyond that, the calculations are little
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tedious and all that.
So, if the sensible method, which is in hand,
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is RK method and that too fourth order. So,
in general, given a linear multi step method,
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if one has to compute past points, generally
one is tempted to use fourth order RK method.
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So, let us look at zero stability. So, these
are initial conditions given and these need
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to be compared using any known method.
Hence, what happens? Hence, all y ns, n greater
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than equals to k contain error because
these are computed using some known method.
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So, there is an error. So, that gets in to
compute the higher one. So, let us define
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what is a zero stability in this context.
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..
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So, the definition and here a linear multi
step method for the IVP, y dashed equals to
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f of
x y, y of x naught is said to be zero stable,
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if there exists a constant k such that for
any
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two sequences y n and y n hat, which have
been generated by
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the same
formula that
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means by the same method, but different initial
data mod or what we are going to
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conclude.
So, we have given an IVP, we have given a
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linear multi step method. Then we say this
method is zero stable, if there exists a constant
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k such that for any two sequences, two
different sequences, which have been generated,
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see we are talking about the same
method for a given IVP. We are talking about
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the same method and we would like to
conclude when do you say this method is zero
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stable. So, the strategy is very intuitive.
What it says? It says when do you say this
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is sensible? This method is zero stable means
it is giving solution, which are sensible
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that means stability in the sense, if there
are
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errors in the past data, then they are not
really magnified to a large extent. They are
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bounded within some extent.
So, when do you say it is this method is zero
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stable? We are trying to compare to
different sequences. What kind of sequences
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you start with, some set of initial data and
then you compute the past values, start with
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another a set of initial data and then start
with, but the difference equation is same.
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So, we are talking about these two. These
two,
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they should be comparable in terms of the
initial data. So, how they are comparable,
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let
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.us see. So, this difference is less than
or equals to k times max of for x n less than
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equal
to say some x m as x goes to 0. This is very
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important.
So, what did we do? You start with two sequences,
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for any two sequences which have
been generated by same formula, but different
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initial data. This is one set of initial data.
This is another set of initial data. Of course,
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the solutions will be different. You start
with
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one set of initial data. You arrive at some
solution and you start with another set of
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initial
data, you arrive with the corresponding solution
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because after all, the solutions are
curves.
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So, we have this difference should be bounded
by constant times maximum of the
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difference between the solution at a particular
grid point. So, at each grid point, you take
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the maximum and that will be k times of that.
So, that will be the nth stage. So, this is
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the
zero stability. So, that means it is not growing
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beyond the maximum value at a particular
grid point. Now, the question why is zero
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stability?
.
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So, it is in some sense, in some sense, one
can decide whether a method is
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zero stable or
not by considering only y dash equals to 0.
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So, that is f of x y equals to 0. So, this
is
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possible, but very tedious the corresponding
analysis. So, that is how the word zero has
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come that means whatever the definition we
have given for zero stability; that can be
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verified by considering just y dash equals
to 0, but the analysis is involved.
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.Hence, f is 0, therefore the zero stability.
So, then if that is difficult, what is the
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alternative? So, the alternative is some algebraic,
some algebraic method, which is that is
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equivalent to
the analysis of f equal to 0 case, some algebraic
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in the sense you verify a
certain condition and then conclude that the
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method is zero stable or the method is not
zero stable and things like that. So, what
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kind of corresponding algebraic condition?
So,
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that is called root condition. So, that means
instead of doing this tedious analysis, we
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pick up this alternative and try to analyze
and conclude that a method is zero stable
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or
not. But, before that, so what is this root
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condition we must know. Unless we know what
root condition is, we cannot proceed.
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.
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So, let us try to see what root condition
is. So, root condition, so linear multi step
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method
of the form start I am forced to write here
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star because in same method, we have to write.
So, let us call star
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is said to satisfy the
root condition if
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the roots of the equation rho of
zeta equal to 0 lie inside the closed unit
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disk
in the complex plane and the roots are
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simple, if they lie on the circle of course.
I wrote something, which does not make any
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sense. This quantity we have not yet defined,
but it some condition what is it? Linear
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multi step method of the form star is said
to satisfy the root condition.
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If the roots of the equation rho of zeta equals
to 0, so that means this is a polynomial
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lying inside the closed unit disc in the complex
plane and are said to be simple, if they lie
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on the inner circle. Now, what is so this
is first characteristic
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polynomial of a linear multi
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.step method? So, linear multi step method,
linear multi step method, first characteristic
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polynomial what is it? So, we have not discussed.
So, we have to define. So, it is
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interlinked, so we have defined zero stability.
.
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Then, method is zero stable if it is satisfies
root condition etcetera we will say. So, let
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us
see characteristic polynomials corresponding
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to a given linear multi step method. So,
what is a method h times? So, this is our
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method? Now, this can be put it in the form
minus sigma a i, all the terms then minus
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h equals 0. So, this is the linear multi step
method. Now, so let us identify these terms.
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Let us identify these terms.
We have y forward operator defined as follows.
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Now, we make use of let us look at y n
plus 1 is E on y n and y n minus 1 is E square
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on y, I am sorry, y n plus 1 is E square on
y n minus 1 E on y n. So, that means E on
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y n minus 1 E on y n is y n plus 1. Therefore,
E 2 on y n minus 1 is E 1, E is y n and this
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is n plus 1. So, like this, these can be
converted and we have coefficients. So, let
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us see how this can be simplified with respect
to notation point of view.
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..
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So, we have y n plus 1 minus. Now, in terms
of the forward operator define, so what will
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be the last term last term of, this is last
term of this is this summation. So, this is
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the last
term of the summation. Now, E y n minus k
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plus 1, this will be plus 2. So, this will
be k
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plus 2, so on so forth. Now, 1 E is giving
me 1. So, this is like this. So, this is
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so E k. So,
if you do the intermediate steps, did you
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get this? So, you can work out. See, this
is 1
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forward plus 2.
So, I am retaining this and this is a plus
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1 plus 2. So, E k will be plus k. So, that
gets
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cancelled we get, so that means this relation,
this E k on y n minus k plus 1 is y n plus
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1.
So, this is the last then last, but one. What
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is the last but one term? Last but one term
contains y n minus k y n minus k minus 1.
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So, what will be last term y n minus k minus
1? So, it will be this. Now, how do we express
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this in terms of y n plus 1? So, compute
E
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k minus 1 and take it.
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..
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So, then you will get a center. So, accordingly
we have, so accordingly I write this as rho
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of E sigma of E 0 where rho of E n minus 1
k minus a 2 k minus 2, let us try to see.
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So, if
you multiply on this, this is 1, so minus
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a k y n plus 1, which is last term of summation.
For example, first term E k on n minus k plus
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1, E k on n minus k plus 1 n plus 1, this
is
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y n plus 1 and k minus 1 with a coefficient
a 1. So, that should be y n minus n y n term
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so on so forth, similarly, sigma E b 0, so
these are with plus because I have written
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minus
here.
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So, this linear multi step method can be expressed
in terms of this. Now, we define first
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characteristic
polynomial of a multi step method. So, this
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is a polynomial a 1 minus a 2
minus a k. Second characteristic polynomial
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is sigma of this is b 0, so for a given linear
multi step method, so these are important
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as within the context of stability of a method.
So, we will see how these will play a vital
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role. So, having learnt this, we first and
second characteristic polynomial for a given
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particular method, let us see how do we
compute these polynomials just to be more
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comfortable.
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..
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So, for example, so suppose this is our given
multi step method. So, this can be written
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as now look, so this will be E on y n y n
plus 1, but we do not have the terms at these
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grid points. We have the derivatives at these
points that is n minus 1, n minus 2, but here
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we do not have, that means the corresponding
coefficients are 0. So, if you recall the
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number of terms here, if you include this
as well and in the summation, then i starts
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from
0. So, where a 0 is 1, so that means essentially
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we have k plus 1 terms here, k plus terms
here.
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That means k plus terms at the grid points
and the derivatives k plus terms at the grid
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points. So, we must always compare with this.
So, in that context, before we arrive at
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this, we may try to see, we have to arrive
at y n from here. So, how do we arrive? E
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on y
n minus 2 prime is y n minus 1 prime, then
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E on y n minus 1 prime is y n prime. This
implies E 2 on y n minus 2 prime is y n prime.
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That means what is the distance from this
term to this terms is E 2. Therefore, this
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suggests your sigma E of the form, so if you
compare the general method, h is given, minus
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sign is given.
So, I put the polynomial on this. So, all
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00:45:03,521 --> 00:45:13,590
the coefficients must be given to this polynomial.
So, how do we give it? We will give 23 by
204
00:45:13,590 --> 00:45:46,600
12. The distance is 2 minus 16 by 12, the
distance is 1 plus 5 by 12, 5 by 12 because
205
00:45:46,600 --> 00:45:52,160
this will act on this, will multiply y n minus
k
206
00:45:52,160 --> 00:46:03,070
plus 1 prime. So, what is the k here? So,
this should act on
207
00:46:03,070 --> 00:46:09,980
the term of the form y n
minus 2. So, we have to suitably identify
208
00:46:09,980 --> 00:46:24,010
what is our k. Now, what will be this? Look
209
00:46:24,010 --> 00:46:32,220
.this is up to y n, the distance is 2 and
here n plus 1, so naturally the distance must
210
00:46:32,220 --> 00:46:45,380
be 3.
So, these two terms
211
00:46:45,380 --> 00:47:25,310
accordingly this gets transformed as exactly,
so we can work it out.
212
00:47:25,310 --> 00:47:59,870
So, n minus 2 prime a n minus 2, therefore
this is and sigma of this is
213
00:47:59,870 --> 00:48:02,920
so for a given multi
step method, this is the first characteristic
214
00:48:02,920 --> 00:48:14,370
polynomial and this is second characteristic
polynomial. We expect this to be of the form
215
00:48:14,370 --> 00:48:26,360
y n minus k plus 1. So, this is y n minus
2.
216
00:48:26,360 --> 00:48:40,420
Therefore, that means if this is a linear
three step method 1, 2, 3, it is asking past
217
00:48:40,420 --> 00:48:49,690
data
three grid points, 1, 2, 3. Now, let us go
218
00:48:49,690 --> 00:48:55,230
back to our root condition.
.
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00:48:55,230 --> 00:49:09,360
So, what it says, root condition linear multi
step method of the form star is said to satisfy
220
00:49:09,360 --> 00:49:15,510
the root condition if the roots of the equation
that is the first characteristic polynomial
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00:49:15,510 --> 00:49:19,960
of
the corresponding multi step method lie inside
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00:49:19,960 --> 00:49:31,460
the closed unit disk. So, these roots must
lie inside the closed unit disk. So, that
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00:49:31,460 --> 00:49:40,730
means the magnitude must be within the closed
unit disk. If the roots lie on the circle,
224
00:49:40,730 --> 00:49:45,681
then we say the roots are simple. So, this
root
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00:49:45,681 --> 00:49:51,950
condition has implications on the stability
aspect of the linear multi step method. So,
226
00:49:51,950 --> 00:49:58,040
let
us see some issues related to this. So, we
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00:49:58,040 --> 00:50:11,690
have seen some example.
Again, we will see y n plus 1 equals to y
228
00:50:11,690 --> 00:50:22,760
n plus h by 4, 5 y n prime minus 7 y minus
1
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00:50:22,760 --> 00:50:37,110
prime plus y n minus 2 prime. So, if this
is the case, then what is our rho of zeta?
230
00:50:37,110 --> 00:50:41,660
So, this
is only these two and this is a three step
231
00:50:41,660 --> 00:50:47,280
method. So, this will be zeta, zeta square
minus
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00:50:47,280 --> 00:51:00,170
1. This implies, so this is our characteristic
polynomial zeta equals to 0, zeta equals to
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00:51:00,170 --> 00:51:08,590
plus or minus 1. So, the linear multistep
of the form is said to satisfy the root condition
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00:51:08,590 --> 00:51:09,590
if
235
00:51:09,590 --> 00:51:15,740
.the roots of the equation lie inside the
closed unit disk and roots which are on the
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00:51:15,740 --> 00:51:20,310
circle,
they are called simple. So, these within the
237
00:51:20,310 --> 00:51:26,030
unit disk and this is these two are simple
roots of the corresponding linear multi step
238
00:51:26,030 --> 00:51:29,250
method.
.
239
00:51:29,250 --> 00:51:44,731
So, we have to see some outline of necessity
means why what you will do with root
240
00:51:44,731 --> 00:52:01,240
condition. So, consider our method in a more
general form. We convert to a multi step
241
00:52:01,240 --> 00:52:10,420
method. Then the corresponding first characteristic
polynomial
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00:52:10,420 --> 00:52:24,391
is of this form and the
general solution is this our difference equation
243
00:52:24,391 --> 00:52:47,400
because of whose general solution is of
this form f s of n z s n where z s are roots
244
00:52:47,400 --> 00:53:00,680
of this. Now, if the magnitude is greater
than 1,
245
00:53:00,680 --> 00:53:04,970
this suggests that if the magnitude of this
is greater than 1.
246
00:53:04,970 --> 00:53:21,110
So, this suggests that what will happen to
this and the other case is so this z s equals
247
00:53:21,110 --> 00:53:33,780
to 1
and multiplicity m s greater than 1. So, these
248
00:53:33,780 --> 00:53:39,360
are the two troublesome cases. What are
they? The magnitude is larger than 1. So,
249
00:53:39,360 --> 00:53:45,660
then we had to worry about this and the other
case is multiplicity is greater than 1 of
250
00:53:45,660 --> 00:53:50,570
a unit root. Then you see this as a specific
format
251
00:53:50,570 --> 00:53:55,390
from our solutions of difference equations
or any recurrence relation. You can see if
252
00:53:55,390 --> 00:53:58,940
the
root has a multiplicity, so then we have,
253
00:53:58,940 --> 00:54:04,630
so the general is a linear combination is
say c 1 z
254
00:54:04,630 --> 00:54:16,670
1 power n c 2 z 2 power n. This is general
if you have multiple roots, then some d 1
255
00:54:16,670 --> 00:54:28,860
plus
n d 2 and say some z 2 has multiplicity m.
256
00:54:28,860 --> 00:54:43,490
So, then we have this kind of, so in these
cases, there is a chance that leads to
257
00:54:43,490 --> 00:54:53,980
unboundedness. It leads to unboundedness.
So, this is just an outline of necessity.
258
00:54:53,980 --> 00:54:54,980
Why
259
00:54:54,980 --> 00:55:02,850
.root condition is really necessary? So, what
is to do with root condition? This is the
260
00:55:02,850 --> 00:55:05,630
first
characteristic polynomial and you compute
261
00:55:05,630 --> 00:55:08,500
the roots. Then depending on the behavior
of
262
00:55:08,500 --> 00:55:17,360
the roots, the solution behaves.
So, how it is possible? See this from here.
263
00:55:17,360 --> 00:55:19,430
The general solution is of this form. So,
the
264
00:55:19,430 --> 00:55:26,280
roots are magnitude is larger than 1. There
is a chance that it will get unbounded. If
265
00:55:26,280 --> 00:55:35,320
the
roots are within this that means on the boundary
266
00:55:35,320 --> 00:55:40,400
of the unit circle, but multiplicity 1, then
this may happen and then again that may lead
267
00:55:40,400 --> 00:55:48,141
to unbounded solution. Therefore, root
condition plays a vital role that is how root
268
00:55:48,141 --> 00:55:50,860
condition has become into play.
.
269
00:55:50,860 --> 00:56:15,700
So, for example, consider an Euler method
y n plus 1 equals y n plus h f n. So, what
270
00:56:15,700 --> 00:56:35,920
will
be characteristic polynomial? So, rho of z
271
00:56:35,920 --> 00:56:54,090
and implies this is the characteristic
polynomial z equals to 1 is simple root that
272
00:56:54,090 --> 00:57:35,920
is one is simple root. Then consider another
method, so for this method, we zeta, otherwise
273
00:57:35,920 --> 00:57:55,500
zeta here. So, this is 11 zeta cube minus
27 zeta 11, this is our characteristic polynomial
274
00:57:55,500 --> 00:58:19,950
and the roots are approximately.
So, mod zeta 3 is greater than 1, therefore
275
00:58:19,950 --> 00:58:27,070
the method is not zero stable because it fails
to
276
00:58:27,070 --> 00:58:36,790
satisfy the root condition for this case.
So, root condition plays vital role, how the
277
00:58:36,790 --> 00:58:40,570
outline
I have given. So, if one of the root magnitudes
278
00:58:40,570 --> 00:58:47,190
is one, then the solution is unbounded.
Therefore, root condition plays a vital role.
279
00:58:47,190 --> 00:58:53,040
It becomes a necessary and sufficient
condition for a method to be zero stable.
280
00:58:53,040 --> 00:58:57,390
So, we learn, further more details about this
in
281
00:58:57,390 --> 00:59:03,660
.the next lecture. So, you may practice for
a couple of formulas how to compute first
282
00:59:03,660 --> 00:59:06,790
characteristic, second characteristic polynomial.
Thank you.
283
00:59:06,790 --> 00:59:06,790
.