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Good morning, I hope you have done some exercise
on understanding the multistep
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methods. So, we have discussed in previous
lectures both explicit and implicit multistep
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methods. So, in case of explicit, if you are
given past points then one compute the value
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at a particular grid point. And in case of
implicit, of course we should know value at
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that
point. However, that can be computed initially
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using some explicit method.
Now, what is the main purpose of these explicit
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implicit methods? So, when you have
explicit, anyway you get the solution up to
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some accuracy depending on the order of the
method. Now, in case of implicit definitely
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the main motto is to really improve upon,
otherwise there is no fun in having an implicit
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method, where your right hand side also
requires the value at that grid point.
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So, what is a real motivation? The real motivation
is to improve upon the solution. So, let
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us see what this combination is called, what
is the combination? Yes, explicit implicit,
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so
if this combination is used. So, what is this
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called? This is called predictor corrector
methods. So, let us discuss what is the motive
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behind predictor corrector methods?
.
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.So, for example we have explicit method,
so this is explicit, sorry this is implicit
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because
we have fn plus 1. So, this is explicit method.
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So, a general k step method as we have
seen where a k and b k non zero. So, when
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this is given what is a task at each step?
One
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has to solve. So, one has to solve this what
for to get y n. So, this is the general k
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step
method and at each step at each grid point
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one has to solve. So, I plugged out the y
n plus
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k coefficient and here of course, so this
is obviously an implicit method, okay?
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.
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So, when we have to do this the remark if
h is less than by L b k, where L is the
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Lipschitz constant of f of x y in this one
with respect to y, Lipschitz constant with
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respect
to y. Then the linear multistep method has
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a unique solution, y n plus k. Moreover, y
n
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plus k can be computed by, see if you look
at it y n plus k and you have a processor,
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right? So, this resembles some kind of fixed
point iteration so computed by fixed point
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iteration as follows a k.
So, this is a processor which is demanding
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y n plus k, in fact s starts from 0. So, this
is
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however with a suitable initial value look.
So, this is our multistep method implicit,
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so to
essentially we need to compute y n plus k,
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but the processor demands y n plus k.
Therefore, it has to be solved like a fixed
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point iteration method, but then in order
to
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solve we need some initial guess. So, that
is suitable initial value. So, this is a implicit
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method, okay?
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..
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So, now what happens in general and in practice?
In general one needs to iterate till y n
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plus k s converges, that is obviously the
difference between two consecutive
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is less than
epsilon, which is a pre assigned. So, this
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is in general see once you have such method,
you plug in some initial value and compute.
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See for example, s is 0 we know initial y
n
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plus k 0, you put it here. Compute, then right
hand side will give y n plus k 1, put it there
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for s equals to 1, then we get 2. So, we compute,
but the question is how long we do it?
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How long we do it?
We do it until the difference between two
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consecutive is less than epsilon, that means
it
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agrees up to decide across e pre assigned.
This is in general, but in practice very tedious
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may involve enormous labor due to f of x n
plus k, right? So, if this is a case then
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what
one should do it? See, your method demands
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that you have to time and again compute
the function f with some initial guess, then
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put it back and then you get some
improvement and then again you compute f and
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then you compute y. So, we are doing
these processes, but then if your initial
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choice is not really good enough then you
will be
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forced to spend lot of labor on it. So, definitely
it depends on initial choice, that means
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one should not go blindly some initial guess,
okay?
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So, let us see what should be done. So, this
means so this can be done by evaluating this
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via an explicit method. So, it make sense
since we need reasonable enough
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approximation, you do it by explicit method.
So, this
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is so called prediction and then this
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.particular thing, this is correction. Why
correction? We compute until the difference.
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So,
you keep on computing until the difference
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between two consecutive agrees to some
accuracy. So, this is the correction, okay?
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.
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So, let us see the how the general procedure
evolves. So, prediction, so
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I put stars
because the coefficient have normalized. So,
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this is an explicit, so we need so 0 fn. So,
the past values up to k minus 1. So, this
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is an explicit
which is used for prediction. So,
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the moment we use something for prediction,
we need to evaluate f. So, we evaluate, so
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let us say initially we compute
by explicit method, then we need a some initial
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values.
So, this should be 1. So, then we need some
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initial values. So, once we compute we
evaluate this, then we need a implicit look.
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So, this is implicit, so we need some initial
then we evaluate. Once we know initial, we
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put it here then we compute one, then we compute
f for by putting this we get f, then that
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we will put it in here. So, this process 0,
1, 2, etcetera. Now, once we do it so then
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again
we evaluate, so we are doing this process,
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right? So, first we use explicit method predict,
then you evaluate this then whatever f that
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has been evaluated you refine it. So, then
again you use the implicit method to compute
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y n plus k at refined, then again put it and
compute f and then put it back, refine it.
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So, we do this processes, right? So, this
suggests
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a prediction evaluation correction, a finite
number of times then we are ready with the
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evaluation, so this is the method, okay?
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..
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So, for example, for example, suppose we have
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so using this we get predicted value then
we evaluate. Of course, to start with, f n
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is computed. So, we need some initial value
f n,
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f n minus 2 and we compute this, then f n
plus 2, this will be f of we compute this,
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then
we go for the correction. So, let us introduce
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this notation. So, it is like this.
So, let us see this carefully, we have an
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explicit method, we have an explicit method
using which we compute an initial approximation
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that we are denoting by y n plus 2 0.
Then f n plus 2 0 is computed, then put it
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here in the correction formula because why
this
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is implicit. So, whatever we compute, put
it here to get a refined one. So, using this
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what
we get? Once we obtain, then you evaluate,
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then we correct to get. So, we continue like
this. So, this is PECME method.
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..
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So, let us see how this works out with an
example, for the IVP if so this is an explicit
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method which is used for prediction, correct?
Using this implicit method to find y of 0.5
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with h equals to 0.1, but the story does not
stop there, the reason is look at it. So,
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this
requires more data or at least one statement
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is missing. What is that? Let us understand
what was our x 0 and with h, h is 0.1. This
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implies x 1 is 0.1 0.2 and it was asked to
find
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y at 0.5.
So, that means we need some data initial and
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let us look at the prediction. So, this is
to
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compute n plus 1, we need n, n minus 1, n
minus 2, n minus 3. So, one would expect a
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statement even to use this explicit method,
you need this values n, n minus 1, n minus
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2,
n minus 3. So, we have y 0. So, what is missing?
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So, x 0 is 0, so then y 0 is 0. We need x
1 0.1, y 1, x 2 0.2, y 2. So, once we know
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this then let us go for the value.
What is the value? Latest we can compute,
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see n 3. If n is 3, we get y 4 equals y 0.
So,
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we need this data to compute this, see that
means using explicit method first we need
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to
compute y of 0.4 then correct it. Correct
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it until the desired accuracy, then we stop,
then
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use that to compute 0.5 at y of 0.5. So, that
means the missing statements, missing
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statements like this at least several alternatives,
okay?
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..
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Compute the required past values using any
explicit method, say fourth order R K
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method and correct the solution up to two
decimal accuracy. So, these two statements
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are
required. Now, the problem is set what given
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an IVP, this is the explicit method which
is
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used for prediction, then this is used for
correction which is an implicit of this nature.
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Now, the past data is required. So, you can
compute just to save time, I am just applying
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the past data. So, this is y, but if you look
at our formula it will expect f. So, that
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means
accordingly for a given f we need to compute
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this. So, for the given f, I am talking about
f x y equals to 2 x y minus 1. So, for this
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value this is, so remember these are to be
computed using any explicit method using fourth
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order R K, but just to save time I have
given the data, okay?
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..
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So, now let us go for predicting y 4 using
this. So, let us go for n equals to 3, case
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y 4 p
equals y 0 4 h by 3 2 f 3. So, this is y 0
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0 2 f 3 minus f 2 2 f 1. So, this you may
compute.
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Now, once we have predicted so this is p part.
Now, we need e part, what is e? We have
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to compute f 4. So, this is
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f 4.
.
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Now, what next yes we have do the correction,
so we do the correction using y 4 c. Of
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course, 1, so this is y 3, so this is
and f we have computed, f 4 is in the predicted.
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So, this
is corrected, now this is corrected first
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iteration, right? Now, what we should do?
We
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.should evaluate f 4, f 4 1. So, this is so
f 4 now once we have a f 4 1 what we do we
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should correct it, okay?
.
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So, how do we correct one, so this is now
we have corrected two times. So, let us
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compute, so this we are getting 0.12345, oh
that is pretty. So, 5 1 up to then this is
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2, so
this is agreeing up to 5 decimals. In second
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correction of course, we have very fortunate
here, but in general this may not be the case.
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Suppose, you suppose a best check could be
you start any random value, do not use any
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explicit method, just take any random guess
as initial and then you put it in your
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correction and then try to keep on correcting.
So, this is a one solution, let us say you
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have corrected five times and another method
you take say explicit method and then you
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compute the value and correct it again five
times and try to compare. See, you will really
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find the difference, sometimes one may be
forced to put lot of labor. So, that means
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we
need some considerations under which the method
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really converges fairly accurately,
okay?
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..
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So, let us see what are those considerations?
So, essentially we are considering this
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sequence, so this is lipschitz condition.
So, for all y near y n plus 1, that is y n
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plus 1, 0,
1, etcetera where l satisfies the condition
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1. So, this is the coefficient in the multistep
method, then the sequence converges to y n
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plus 1. So, this condition, so this coefficient
is important. So, with respect to the general
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multistep method this coefficient so we
should be careful how we have defined. So,
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this is coefficient of f n, so this is guarantees
the convergence, okay?
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.
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.So, let us consider
conditions for convergence for a particular
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method. So, this is to
know how to obtain convergence for a particular
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method. So, for example, obtain the
condition for which the following P C method
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converge, what is the pc method?
Prediction is simple correction is modified,
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right? So, we consider the reference
equation, this is reference equation then
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prediction gives y n plus lambda f n. So,
this
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will be
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so this is the prediction, now correction.
So, naturally this is this corresponds to
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prediction, but we have predicted. So, we
substitute, so then we get
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suppose we continue
the process.
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.
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That means
one can easily show if you continue we get
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this. So, we get we can simplify
this series using sum of the series formula.
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So, this is the correction m times as m goes
to
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infinity, this converges. If we can get the
conditions from this, obviously the condition
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is
if so from this one can obtain the corresponding
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convergence region, this condition. So,
this is not region, so converges if this condition
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satisfies. So, having obtained this
condition it would be a good idea to compute
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what kind of a errors involved at each
stage, that means you correct a 3 times, what
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is the error, correct 4 times, correct 7 times,
what could be the error, okay?
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..
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So, let us try to assimilate that so
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estimating error, so we have y n equals to
so this
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notation we follow. So, from the previous
I have substituted for this and for this,
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the
corresponding exact process error. So, this
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implies
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minus because exact we know y of x
n plus 1 is e of lambda h into y of x n. So,
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with that notation
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now what are these two
terms. So, this you see this is the due to
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the approximation from the method and this
is
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the exact.
So, the difference is so this is relative
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and what about this error at n stage is magnified
by
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this factor for the error at n plus 1 stage,
okay? So now we would like to analyze this
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relative truncation error, you correct it
once, correct it twice, correct number of
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times,
what will be the error.
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..
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So, let us consider the relative error. So,
the relative error is this is given by, so
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consider
the case m 0. So, then this can be written
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as so this will be m 0. So, lambda square
h
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square by 4. So, there is a 2 there, so this
is so I have expanded only two terms. So,
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this
is coefficient of this
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then so higher order I am not writing plus.
So, this is 1 plus these
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two terms, lambda h then these two terms.
We get from these two, one get cancelled and
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we are getting 4 plus minus.
Now, if you see this first two terms get cancelled
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00:50:52,880 --> 00:51:02,519
here and what is the third term third
term? So, I would like to expand, so this
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00:51:02,519 --> 00:51:18,210
with a minus in front. So, if you see minus
get
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00:51:18,210 --> 00:51:32,819
cancelled, this get cancelled and you get
minus. So, this is a one fourth and this is
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00:51:32,819 --> 00:51:36,529
a
minus sign. So, minus one fourth, so this
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00:51:36,529 --> 00:51:43,479
is of order, what I am trying to say is this
is not
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00:51:43,479 --> 00:51:57,369
0. So, this is not 0, so this is non zero
coefficient of lambda square h square. So,
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00:51:57,369 --> 00:52:02,499
if you
take m equals to 0 then the error is like
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00:52:02,499 --> 00:52:05,160
this, okay?
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00:52:05,160 --> 00:52:06,160
..
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00:52:06,160 --> 00:52:30,219
Suppose, you take m equals to 1 then this
will be this 1 minus minus. So, this is 1
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00:52:30,219 --> 00:52:55,430
plus,
so this will be lambda cube h cube by 4 1
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00:52:55,430 --> 00:53:23,019
plus. So, this is a cube term, so I can write
it
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00:53:23,019 --> 00:53:58,069
then there is a so one can show this, so this
you get lambda h and plus h square by 2. I
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00:53:58,069 --> 00:54:14,619
need this is cube there. So, by 4 there is
another, so this let me check. So, once we
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00:54:14,619 --> 00:54:17,799
get
from here and then there is another 4.
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00:54:17,799 --> 00:54:25,299
So, there is another term and this is by 4,
so 2 times, so this I have to add one more
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00:54:25,299 --> 00:54:45,799
this,
then from these two. We get this is we get
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00:54:45,799 --> 00:54:56,140
minus 1 over 8, you may check, if I am
making a mistake with the coefficients, but
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00:54:56,140 --> 00:55:06,589
this tells the procedure. Now, you can see
they these three terms get cancelled with
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00:55:06,589 --> 00:55:10,509
this. So, you get some non zero coefficient
say
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00:55:10,509 --> 00:55:29,980
some k. So, probably we may get minus 1 over
6 lambda cube h cube. So, this is m
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00:55:29,980 --> 00:55:31,999
equals to 1.
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00:55:31,999 --> 00:55:32,999
..
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00:55:32,999 --> 00:55:47,559
So, if you continue m equals to 2, we should
show that some coefficient and in particular
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00:55:47,559 --> 00:56:01,029
for this method if you continue, we can verify
subject to this coefficients, but that means
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00:56:01,029 --> 00:56:24,440
there is no improvement on correcting further.
So, this gives a very a nice estimate for
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00:56:24,440 --> 00:56:29,589
this particular method, if you keep on correcting,
it tells that there is no point in
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00:56:29,589 --> 00:56:35,000
correcting beyond certain stage.
So, with a reference equation if you test
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00:56:35,000 --> 00:56:38,109
it, so then for a general formula even if
your f is
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00:56:38,109 --> 00:56:43,769
quite complicated, you need not correct beyond
because there is no point in correcting
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00:56:43,769 --> 00:56:49,349
beyond certain stage. And the certain stage
here for this particular method is m equals
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00:56:49,349 --> 00:56:56,430
to
2. So, the general predictor corrector methods
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00:56:56,430 --> 00:57:01,249
works in this way, so we need explicit
method for predicting, implicit method for
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00:57:01,249 --> 00:57:10,459
correction and we correct up to desired
accuracy. So, let us work out a few more details
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00:57:10,459 --> 00:57:12,839
in the next class.
Thank you.
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00:57:12,839 --> 00:57:12,839
.