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Hello, so last class we have learnt higher
order methods and also higher order equations.
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So, that is second order initial value problem
has been solved by reducing it to a system
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of couple first order equations. Now, when
we define a method how do we know that
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what is the error? Suppose somebody gives
this is the method how to compute the error?
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And of course, we discuss in this lecture
how a particular method is stable? And whether
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the particular method gives convergence solutions?
And what is the reference equation
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with, which we analyze this process. So, let
us discuss first how to compute error given
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approximation?
.
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So, how to compute error or a given approximation?
So, let us say we have the following
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approximation. So, I am not naming the method
what is this method called etcetera.
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Suppose there is a sum given therefore so
these are the weights k 1, k 2, w 1, w 2 are
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the
weights 1 by 4 and 3 by 4 and the k 1, k 2
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are defined like this. Now we would now like
to know what if we approximate by this formula
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or a given I V P what should be the
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.error? So, how do we do it? As I mentioned
we have to make the expansion terms of h
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and then we have to expand the Taylor series
and try to compare. So, let us try to do that.
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.
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So, we have y n plus 1, so y n plus h by 4
f plus 3 h by 4. Now this has to be expanded
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in
the powers of h. So this one f of x n y n
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plus the increment is this 2 h by 3 dou f
by dou x
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plus increment with respect to this 2 h by
3 f. So, these are the first order terms plus
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second order terms square of that. So, two
times plus. So this is our expansion.
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So we have 3 by 4 and 1 by 4. So h f plus
h square by. So, h square terms. So, one term
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h
square term is 3, 3 gets cancelled and 2 by
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4 is half. And we have. Then the next term
from here we get
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plus h cube h cube by 6. So, this is our h
expansion. Then we have to
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compare this with Taylor series expansion
and you can see in Taylor series this term
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exists. So, first term approximates y of n
and second term and third term. So than if
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you
make equal coefficients of h power 0, h power
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1 x square.
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..
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So, than we get error starts from. So, for
any method we try to determine using this
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method, using this way. So, we know x and
y n. So, this f we know. So, the error is
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a
order h cube and hence the method is of order
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h square. So, that is second order. Now as
I mentioned we have to discuss the stability
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and convergence. So, what do you mean by
stability and convergence? So, let us try
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to do the understand this with respect to
a
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reference equation.
.
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.So, what is the reference equation, what
is our reference equation? Y dash is lambda
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y, y
of x 0 is y 0 x so this is our star. And exact
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solution of star is y of x equal to c times.
So,
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if you use this initial condition we get y
0 e power lambda. So, this is our exact solution
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right? Now in order to compute y of x at...
So, that is idea right? So, what is y of x
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1, y 0
e power lambda x 1 minus x 0? So, that is
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y 0 e power lambda h because x 1 minus x 0
is…
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.
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Then y of x 2 is y 0 e power lambda x 2 minus
x 0, so y 0. So, this is y of x 1. So, we
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can
generalize. Now in order to compute the solution
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we have to compute this, difficult to
compute, but difficult to compute means how
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difficult, so how difficult? So that idea
is
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approximate e power lambda h suitably, approximate
suitably right? So, how do we
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approximate?
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..
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See, in case of Euler’s method. Let us talk
about Euler’s y n plus h f of. And what
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was
our reference equation? Lambda y. So, this
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is our f of x y. So, therefore, this reduces
to y
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n plus lambda h y n. So, this is 1 plus lambda
h y n. And what was our approximation? y
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of x n plus 1 is y of x n e power lambda h.
So, y n is approximating this. That means
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e
power lambda h. 1 plus lambda h is approximating
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this. So, therefore, this is Euler. So,
for Euler method e power lambda h has been
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approximated by 1 plus lambda h plus
order of right?
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.
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.Suppose, let us take Taylor’s series method.
So, take y n plus 1 is y n plus h f of x m
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y n
h square by 2 factorial so y double. So, that
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is 2 f dou x. Now this with our reference
equation that is y dash equal to lambda y
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which were as f. This reduces to y n plus
or f as
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lambda y n. Dou f by dou x f lambda y
square plus order of h cube. So this is one
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plus
lambda h plus lambda square plus. So, when
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I mention the there is a lambda sitting when
we substitute this definitely there will be
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a lambda q. So, that means e power lambda
h
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has been approximated by. So, this is the
approximation right? So, let us call this
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approximation for.
.
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So, that means given y n approximation is
as follows y n.
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So this is what is happening.
Now we have to discuss a stability right?
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So, what is round of error? When you have
an
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approximation there is a value, which is machine
value plus R is y true representation.
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This is the round of because of the machine
distractions we are rounding of under
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truncation error. So, y true representation
plus T is y exact. So, that means we are
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expecting something exact, but y we are representing
by a approximate formula and we
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are throwing some terms so that is the truncation
error.
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So, y n plus 1, so y n plus 1 is y of x n
plus 1 plus where p is order of the method
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because
this approximation is up to something and
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that is the truncation error. So if the method
is
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pth order some theta. So, this is local truncation
error more generalization right? Now if
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you say this is up to some approximation some
truncation. Now let us say we compute y
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.1. So we have thrown T corresponding to y
1. Now compute y 2. So, we use y 1 that
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means whatever your thrown T at computing
y 1. So, that will be still sitting in the
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system and then we have to understand what
kind of a impact it is creating.
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.
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So, let us talk about stability of a method.
So, formal definition nothing but how the
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error
is playing a roll? A numerical method is said
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to be stable if the effect of any single fixed
round off error is bounded, independent of
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the number of mesh points. A numerical
methods is said to be stable if the effect
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of any single fixed round of error is bounded.
That means the effect of round of error is
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bounded since it is not really growing. And
its
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independent of the number of mesh points.
This is another important parameter.
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So or if for every epsilon is greater than
0 there exist delta of epsilon such that
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for two
different numerical solutions. y n and say
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y n or the difference is less than epsilon
whenever the corresponding initial conditions
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for every h. So this is very important for
every epsilon there exist a data. It depends
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on epsilon such that see the scheme is same
we are talking about two different solutions
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with two different initial conditions. So
now
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as long as the difference is delta of epsilon
the approximations should not be differed
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by
large quantity. And this is, this must hold
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independent of the mesh size.
So when we ate approximating with what your,
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so for example in Euler method we have
seen, in Euler method we have expanded then
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now we first considered two terms. So
with two terms we have got 2 plus lambda h
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then Taylor’s dependents on suppose your
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.using three terms suppose we get 1 plus lambda
h lambda h square. Then if we use more
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terms that means your exact solution, which
is of e power lambda h we are
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approximating that right? So a comparison
between these two must be made. So that
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means compare to what your approximating e
power lambda h. It should have been e
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power lambda h, but now your approximating
by some something. So also there should
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be a comparison.
.
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So, let us do that. So, let us call our approximation.
So, which means. So this is a true.
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Exact plus epsilon n plus 1 epsilon. Plus
1 is e power lambda h y of x n plus. Y n 1
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plus 1
is exact plus epsilon n plus 1 at one stage.
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Then we have this approximation then y n is
y
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of x n plus 1. So, this implies lambda h y
of x n epsilon n. Now this is minus. How did
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we write this? Y of x n plus 1 is actual solution
is e power lambda h into y x n. So, I am
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writing y of x n is replaced by e power lambda
h e of x n. So we take common plus. Now
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what is this tells? This is exact, but this
approximation right? So this is nothing but
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local
truncation error.
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Now how can you minimize this? You can minimize
this by approximating e power
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lambda h more closely with e power lambda
h. So, that means this can be
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made small by
suitably approximating
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that means suitably choosing e of lambda h.
We choose better e
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of lambda h its close to. So, you can see
Euler method 1 plus lambda h where as Taylor’s
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series is with more number of terms more close
to e power lambda h so in that sense.
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.Then what about this? See this is approximation
and this is the error attain stage and that
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is contributing to error at n plus stage.
Therefore, this must be the propagation error.
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So what is the propagation error? This is
the propagation error from x n to x n plus
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1 and
this is inherited. So error attain at a 1
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place consist of two parts. One is local truncation
error, which is due to the approximation.
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The better you approximate close to e power
lambda h the minimum the local truncation
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error other is the propagation error. The
error
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at nth stage has been magnified by this factor.
Now one can guess easily how do you minimize
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this? Yeah, as long your approximation
is not magnifying because this is the factor
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method it is magnified by this? So, when do
you say that the propagation error will diminish,
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if e of lambda h is not magnifying it to a
large extent. So we discuss, this is the stabilizing
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factor because what is the stability?
Stability means I mentioned. See your computing
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your defined an a process it is a single
step process to compute a value at same stage
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you require value at the past one stage.
Now you it is a recurrence processor.
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So you compute y 2 and use it to compute y
3 than compute y 4 using y 3 and so on so
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fourth. So, whatever the errors existing at
one step ahead that has been incorporated
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into
the higher steps right? So know you have to
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control this. So, this can be controlled only
when approximation is playing some role. So,
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we define that now.
.
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.So, the definition. Say a numerical method
of the form
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is called absolutely stable. If what
here their your. This is the magnifying factor.
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So this should not magnify to the large
extent that means here the model must be equal
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to 1 right? So, that means E of lambda h
should not grow faster than. Because your
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approximation is making it magnified the
error definitely is not stable. So, the comparison
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is between this and this approximation.
So hence a numerical method
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of the form is relatively stable. If see,
if you are, you can
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ensure to that this is always less than equal
to 1 than its absolutely stable.
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That means whatever the errors, which are
been carried at nth stage. So they are not
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magnified beyond magnitude of less than 1.
So, the error is going to be bounded. One
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can ensure that at next level the error is
bounded. So this is absolute stability, but
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another
is compared to what? This approximation is
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compared to E power lambda h. So,
therefore, this is, if this is less than equal
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to E power lambda h. So then it is called
relatively stable, right?
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.
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So, take Euler method. y n plus 1 is y n plus
h, which is equal to y n plus h lambda y n.
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So, therefore E f lambda h is 1 plus lambda
h. Hence more E of lambda h is less on 1
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implies so that means this is the region of
absolute. That means the method is going to
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be
absolutely stable as long as it chooses step
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size such that lambda h for holes in this
range.
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So, what is lambda? So lambda is coming from
our reference equation, but we have more
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general form that equals to f. So, that means
at a lighter stage y dash equals to f.
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.So why are we talking y dash equals to lambda
y as a reference equation were, as we
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need for a more general case, so will do it
little later. What is the connection between
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this? Why do we take such a reference equation
when we need a more general case of f?
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So, then Taylor’s method, suppose up to
two terms, three terms, so this implies. So,
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this
is first order
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second order. Then one can also obtain third
order right? So, one is
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propagation error, another is local truncation
error. Then we talk about, we talk about
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absolute stability with respect to this factor.
And we talk about relative stability with
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respect the distance between these two. So,
that is what we have done right? Now we go
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back to this question. Why did we consider
y dash equals lambda y when we need y dash
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equals to f?
.
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So, why y dash equals to lambda y as the reference
right? So, what we are doing? We are
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considering a particular point x n y n. And
we are trying to analyze for that value what
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happens to the approximation? So, that means
we need to predict the behavior of the I V
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P y 0 in the neighborhood of a point. So,
this is what we are trying to analyze, but
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with
respect to the general case. Then what we
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do is linearize. So, how do we do it? f of
x, y
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around this point plus this is multiplication
plus. So, this equals y dou f by dou y plus
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f
minus y bar dou f by dou y plus. Now this
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is an Euler function, but we have line arise
it
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with respective the dependent variable. So,
that means with the respective to the
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dependent variable the question has been linearized.
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..
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Now with this f your y dash equals to f can
be written as this becomes y dash the coastal
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the lambda y plus c were lambda equals to
lambda y. So, lambda y is dou f dou y
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and c.
So, were of course, were x is in. So, this
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can be transformed
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using c by lambda whose
solution is. So, that means. So, this is.
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So, we tried to analyze the behavior of the
solution
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and it depends certainly on the lambda. And
were lambda in turn depends on f.
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.
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So, let us see. For example, say y dash is
x square minus y 1. Then analyze the behaviors
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of the solution around say 1 minus 1 and say
0, 2. Now what is our f? Therefore, suppose
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.let us have. So this will be minus 2 y. Now
dou f dou y at 1 minus 1 will be minus 2 y
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at
1 minus 1. So that will be 2, which is lambda
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positive dou f by dou y. So around 0, 2 will
be minus 2. So that will be minus 4, which
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is lambda.
So that means this f around this point the
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solution behaves like this. And were as around
0 to the solution behaves decays and here
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00:46:38,829 --> 00:46:51,109
it grows. So our question of why reference
equation. That means the solution, overall
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00:46:51,109 --> 00:47:00,440
the behavior of the solution of the non leaner
case can be interpreted by linearizing f and
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00:47:00,440 --> 00:47:06,390
estimating the behavior of the linearized
k is
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00:47:06,390 --> 00:47:10,049
in this sense, because this is in more general
form.
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00:47:10,049 --> 00:47:11,049
.
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00:47:11,049 --> 00:47:28,850
Hence this is our
reference equation, this is our reference
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00:47:28,850 --> 00:47:43,460
equation. So we started with
stability. So, numerical method is said to
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00:47:43,460 --> 00:47:46,269
be stable if the effect of any single fixed
round
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00:47:46,269 --> 00:47:52,410
of error is bounded, independent of the number
of mesh points. So that means the
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00:47:52,410 --> 00:48:01,140
approximation should be bounded by epsilon
whenever the initial data is within the range
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00:48:01,140 --> 00:48:13,270
of delta, which depends on epsilon, and then
we have approximated given solution of the
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00:48:13,270 --> 00:48:19,329
given I V P.
Then we introduced the error at n plus one
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stage. So, that will come from two different
parts. One is the inherited, which is due
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to the error at the nth stage. And the other
one is
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00:48:32,220 --> 00:48:39,691
due to the approximation. And the better you
approximate E power lambda h you get the
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00:48:39,691 --> 00:48:48,690
better minimum local truncation error. Now
once we have these two parts. This will
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00:48:48,690 --> 00:48:58,290
define absolute stability. Why absolute stability?
Whatever error at n stage it is
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00:48:58,290 --> 00:49:06,160
.magnified by this factor. So as long as you
are controlling this there would not be much
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00:49:06,160 --> 00:49:10,779
inherited error.
Therefore, you can assure that absolutely
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00:49:10,779 --> 00:49:16,769
stable were as this compared to what? How
far
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00:49:16,769 --> 00:49:23,099
you are from actual E power lambda h? This
is your approximation how far you are. So,
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that will tell relative stability. So, therefore,
numerical of the form this is called
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00:49:30,940 --> 00:49:41,900
absolutely stable if this is the condition
and relatively. So, it should not grow faster
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00:49:41,900 --> 00:49:44,830
than
this. So that will introduce numerical method
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00:49:44,830 --> 00:49:58,049
of the form is relatively stable if this
happens. So, for a particular method single
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00:49:58,049 --> 00:50:07,799
step method these are important criterion
absolute and relative. So, let us take the
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one of the R K methods and try to estimates
the
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00:50:12,029 --> 00:50:25,999
condition for absolute stability. So, for
example, we take hewn method so the hewn
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00:50:25,999 --> 00:50:29,470
method.
.
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00:50:29,470 --> 00:50:40,820
So, we have the hewn method. y n plus 1 is
y n plus h by 4 n for this. So this is, so
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00:50:40,820 --> 00:51:02,220
Hewn
method h by 4 so k 1 is f of x y plus 3 k
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00:51:02,220 --> 00:51:21,190
3, so 3 k 3 that is f of x plus 2 by 3 h k
2. So, y n
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00:51:21,190 --> 00:51:55,739
plus f plus 3, k 2 is... Now with our reference
equation y dash equals to lambda y this
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00:51:55,739 --> 00:52:15,859
becomes y n plus h by 4 f is lambda y and
this is f of. So, y plus 1 by 3 h f is lambda
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00:52:15,859 --> 00:52:35,940
y.
So, we have to f of y plus lambda h y. So,
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00:52:35,940 --> 00:52:47,940
this is y n plus h by 4 lambda y plus 3 f
of y
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00:52:47,940 --> 00:53:12,329
plus 2 h by 3. Now this is our y. So, what
is f lambda y. Therefore, lambda y plus. So,
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00:53:12,329 --> 00:53:27,579
this is y n plus h by 4 lambda y. And this
our y. So, lambda y must be lambda times.
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00:53:27,579 --> 00:53:55,140
So,
this is all y n plus 3 lambda square 1 y comes
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00:53:55,140 --> 00:54:13,010
out right? This multiplies only here. I am
249
00:54:13,010 --> 00:54:32,670
.sorry this multiplies only here. So, lambda
y plus 3 lambda y plus 2 h by 3 lambda plus
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00:54:32,670 --> 00:54:42,540
2
h square lambda square by 6 and this is 1
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00:54:42,540 --> 00:54:44,920
y.
.
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00:54:44,920 --> 00:55:12,789
So, this is for hewn method. So, this is y
plus h by 4, h by 4 lambda plus 3 lambda 1
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00:55:12,789 --> 00:55:27,069
plus.
So, if you make this 1. So, this is which
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00:55:27,069 --> 00:55:33,549
is slightly complicated. So, we can simplify
a bit
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00:55:33,549 --> 00:55:56,650
and for absolutely stable we must be able
to get a, some a prime and b prime. So, this
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00:55:56,650 --> 00:56:16,769
will give the interval of absolute stability.
So, this looks quite complicated, but
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00:56:16,769 --> 00:56:26,619
nevertheless attempting for at least higher
order methods you get an algebraic expression
258
00:56:26,619 --> 00:56:33,849
for the range to compute their interval, which
the approximate method is absolutely
259
00:56:33,849 --> 00:56:37,329
stable.
So, from that algebraic expression we can
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00:56:37,329 --> 00:56:40,460
try to estimate the bones. In some cases it
is
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00:56:40,460 --> 00:56:48,229
straightforward and some cases it is tedious.
But nevertheless given approximation that
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00:56:48,229 --> 00:56:55,279
means the approximate method we have learnt
how to compute the corresponding error.
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00:56:55,279 --> 00:57:03,479
And further we have learnt how to compute
the absolute stability interval. And also
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00:57:03,479 --> 00:57:07,430
with
reference to the reference equation how far
265
00:57:07,430 --> 00:57:10,509
it is, how closely it is approximating that
is
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00:57:10,509 --> 00:57:19,489
the relative the region of relative stability.
So, this will give sensible idea about single
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00:57:19,489 --> 00:57:26,510
step methods, and how the stability and error
are related.
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00:57:26,510 --> 00:57:36,040
So, in particular the, for the absolute stability
the approximation what you are making so
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00:57:36,040 --> 00:57:45,229
that will play a vital role. So, we have learnt
R K method this are called expletive
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00:57:45,229 --> 00:57:51,069
.because to compute n plus 1, we have past
one value. So, there are some implicit
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00:57:51,069 --> 00:57:59,230
methods, but these are quite complicated.
So, will see how we proceed. Have a good day.
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00:57:59,230 --> 00:57:59,230
.