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