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Hello. So, so far we have discussed initial
value problems and then some methods both
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single step and multistep methods to solve
the initial value problems. So, then we use
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them to propose something called predictor
corrector methods. So, let us proceed further
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with something called boundary value problems.
So before we proceed for methods, so
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let us briefly discuss, what is boundary value
problem.
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.
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So, we
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boundary value problems, so we have y dash
equal to f of x y and y of x 0 equals
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to y 0. So, this is initial value problem.
So, we know very well that a first order equation
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requires one condition to eliminate the arbitrary
coefficient. Now, since we are proposing
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a boundary, so for a boundary, we need at
least more than one point. So, for example,
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you consider a boundary. So, if you consider
a domain a b, what is the boundary?
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Boundary of a b is a and b. So, these are
the boundary points. Therefore, we expect
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a
kind of second order say this is a simplest
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I am writing. So, then we need y of a is
gamma 1, say y of b is gamma 2.
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.So, this is called a two point boundary value
problem. So, why do we say two point
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because the value is specified at two points.
So, this a two point boundary value problem.
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Now, what is our aim; to solve such two point
boundary value problems numerically. So,
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how do we solve them? Of course, by somehow
we have to introduce the numerical
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concept. So, how do we do it?
.
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.
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So, how do we do it? We have differential
equation. So, we have to convert this into
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difference equation. Now, the question how,
how do we do this? By introducing finite
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.differences, so by introducing finite differences
that means see I have considered a
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simple case y double prime, but in general
the differential equation may contain y prime,
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y double prime etcetera. Therefore, the task
is find
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approximations for y prime, y double
prime etcetera. So, how do we find approximations?
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So, consider, so use Taylor series
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implies y dash of x, so that means we have
obtained an
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approximation for the first derivative. So,
similarly, so call this 1, consider, so then
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this
implies is this will be we have to write y
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of, so
we transfer plus, so we have got an
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approximation, another approximation. So,
typically this is known as forward operator,
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this is backward. So, if one would like to
write at any grid point, so if it like to
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write at
any grid point x i y dash of x i, which is
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this in terms of h, so this is y i plus 1
minus y i
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of course plus order of h. So, this is a forward
approximation.
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.
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So, now let us proceed further. So, let us
write down one more. So, this was our 1. So,
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this was our 2. Now, our aim is to find approximations
for first derivative. So, if we add
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them, first derivative gets killed. So, let
us subtract so then we have on the left hand
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side,
this gets cancelled and this gets cancelled.
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So, this becomes h cube by, so this is 6.
So,
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this is say this. So, this implies y dash
of x is
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plus or minus, so this is minus h square.
So,
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this is at a grid point. So, that means the
earlier approximations, for example forward
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and
this is backward, so they are first order,
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whereas this approximation, we have second
order.
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..
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So, this
is similarly, in exercise, so for example,
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so we have 1 and 2. By subtracting, we
got this. Suppose we add, what happened? So,
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these two get cancelled, so these two get
cancelled, so this plus this and here we have
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twice and this becomes twice. So, one may
get by h square. So, this can be verified
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easily. So, if our domain is this, so this
is x 0, so
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this is x n. So, then
see for example ii, if one would like to use
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these, we can use. So, this
correspondingly the data starts from y 0 and
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y n.
So, if we use i equals to 0, so then we have
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y minus 1, which is beyond this. So, that
means in some sense, star cannot be used at
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x equals to a and b. When I say x cannot be
used, we will get back to this. I did not
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mean literally we cannot use it, we can use
it. We
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get y minus 1 here when we use but x equals
to a and when we use at x equals to b, we
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get y and plus 1, which is the next point
use. So, we will get back to this. That is
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not a
big issue. So, now other some exercises, you
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can try 4 y of x plus h, this is x plus 2
h
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minus, so this can be derived, 3 y of y of
x minus 4 y of x minus h. So, this can be
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derived. So, these are some exercises.
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..
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Accordingly, y dash n say this is 4. Also,
for example, say suppose this is f of x is
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given
like this. So, one can compute for example
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y dash at 0.2, so this can be obtained from
consider x 0 is 0. So, then this can be y
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1 prime, so y 1 prime. So, for example from
here,
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if you would like to compute y 1 prime, I
am sorry not, so for example 0.6, so that
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is x 1,
x 2, x 3, y 3 prime, so from here we get y
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1 plus 3 y 3 minus 4, 4 y 2 by 2 h.
.
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So, this was our y. So, one can compute, we
get some value. So, that means using the
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values, one can compute the derivatives. So,
this is an important thing. So, having now
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.introduced the finite differences, let us
proceed to solve at least one simple boundary
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value problems. So, let us try to attempt
by replacing the derivatives by corresponding
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finite differences.
So, a simple two point BVP, so as I mentioned
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this is very simple and say one can take,
but let us take the simple case and y of 0
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is alpha and y of 1 is beta. Suppose this
is the
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one. So, this is very simple and one can get
the solution very easily, but we have
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considered such a simple problem just to ensure
that the corresponding numerical
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method really can be implemented and then
for such a thing, one can get the analytical
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solution and one can really compute. So, I
will switch over to the notation slightly.
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So,
this is a x 0, x 1, x 2, so on x n plus 1
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is b. Earlier, I said x n, so this is not
a big issue. So,
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the number of points, now what was that let
us use this approximation. So, this is
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approximation we are trying to use.
.
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Accordingly 1 becomes, so 1 becomes n. So,
this is our corresponding finite difference,
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replacement for the given differential equation.
The case i equals to 1 involves y 0 equals
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to alpha and the case i equals to n involves
y n plus 1 that is y of b of course, this
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is y of
a. So, I make a remark again. I have taken
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n plus 1 there. So, if you run star for say
i
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equals to 1, we get y 0 minus 2 y 1 y 2 by
h square equals to f 0; i equals 1, sorry,
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f 1. So,
this is i equals to 1.
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.Similarly, suppose i is 5. So, we get y 4
minus 2 y 5 y 6 so on. so forth. So, that
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means
within the suppose, for example, take the
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case of 0 and 1 and one has to decide what
is a
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suppose h is 0.2, 0.4, 0.6, 0.8, 1 and we
know the values here y of 0 is given alpha,
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y of 1
is given beta. So, what are the unknowns?
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y 1, y 2, y 3, y 4, so these are the unknowns.
So, we have to obtain these unknowns. So,
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we have to obtain these unknowns. So, that
means if we run i from 1 to i 1, 2, 3, 4,
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for i 1, this point is involved but that is
given and
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that for i 4, this point is involved, but
that is given. We will be left with a system
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for y 1,
y 2, y 3, y 4.
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.
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So, let us try to put it in the system form.
So, this further the star implies where the
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structure of A has to be decided. So, for
a general case, one can write. So, I have
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given
for example, so if you run y y 0 is known,
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so in the first equation we have coefficient
of
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y 0, there is nothing because it is known.
Then, coefficient of y 1 is minus 2 and
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coefficient of y 1 is 1. Let us look for a
i equals to 5, coefficient of y 4 is 1, y
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5 is minus
2, y 6 is 1 that means apart from the end,
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the equations in the middle would contain
three
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terms. The equation at the end points contain
two terms.
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So, one can expect for A, so h square I am
not cross multiplying. This will be, all are
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0s.
Similarly, so this will be the corresponding
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matrix A. Why? I could explain again. Look
for the first equation. So, the coefficient
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is coefficient of y 1 is minus 2 because the
unknowns are, these are the unknowns.
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.So, coefficient of y 1 is minus 2, y 2 is
1. So, because this has to be multiplied by
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y 1, y
2, y 3, y n minus y n here, so minus 2 times
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y 1 plus y 2, which we retain. So, similarly,
it is better to write down what is Y in this
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case, y 1, y 2, actually to have a symmetry
have given y n plus 1 as b. So, may be I can
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add a remark here, if x 0 equals to a, x n
equals to b, the unknowns will be y 1 are
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unknowns because accordingly the h will
change undoubtedly.
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So, this is the thing. Now, what will be F
bar? Look at that in the first equation f
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1 is
known and y 0 is also known, so y 0 by h square,
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so for example, y 0 is alpha, so alpha
or h square is known. So, we can shift it.
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Similarly, to the last case, we know the last
entry y and plus 1, which is beta. So, beta
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by h square can be shifted. So, what will
be
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our F bar? This will be f 1 minus alpha by
h square, f 2 f 3 f n minus 1 f n by so f
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n
minus beta by h square because this for example,
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5 this is the last point. So, then y 6
must be beta. So, we know beta. So, that will
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be shifted, beta by h square will be shifted.
So, this is our system. So, I write it again.
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.
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So, we have A Y bar equals to F bar, where
A is 1 over h square, 1, minus 2, 0 and here
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1, minus 2, 1, 0, here 1, minus 2, 1, 0, 1,
minus 2. Then, F bar is f 1 minus alpha by
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h
square, f 2, f n minus 1 and transpose. So,
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all this, now what is the feature of this
TD? It
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is indeed a triangular system, look at the
entries. So, this is a tri diagonal system
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and can
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.be solved for a given f of x. So, I will
get back to little later how this can be solved
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for a
general case.
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.
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.
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Now, let us look at the scenario with an example
before we generalise. So, the domain is
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1, 2. Then, say x 0 is 1, x 1 is 0.5, x 1
is say 1.5, x 2 is 2, let us extend it to
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3, so x 3 is
2.5, x four is 3. So, this is, so that means
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the domain is 1, 3. So, x 0 is 1, x 1 is 1.5.
So,
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accordingly h is 0.5. So, accordingly y of
1, which is y 0 equals to minus 2, then y
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of 3, y
4 is 1. So, how many unknowns then left; y
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1, y 2, y 3 are the unknowns. So, these are
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.the unknowns. So, now consider the discretisation.
So, we have this equation. So, we
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have to discretize this.
So, if you discretize, so this is a discretized
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equation. Now, i equals 1, so this will be
y 2
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y 1 y 0 h square. So, h is h is this, so h
square is, so h square is 2 5, so minus 3
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plus into
y 1 is x 1. So, then i equals 2, so y 3, y
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3, y 1. So, if you look at this our x 0, 1
and x 4
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was 3, so in between
x 3 or the points, so y 0 is known to us,
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y 0 minus 2 and y 4, 1. So,
if you look at these in the first equation,
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we have only two unknowns and again in the
last equation, y 4 is known only two and the
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middle of course. So, here to be determined,
so this has to be determined. So, we can simplify
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and put it in a system.
.
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So, if you put it, we get the system. So,
let us do it y 2 minus 2 y 1 plus y 0 y 0
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minus 2,
then minus 0.23, 3 plus x 1 was 1.5 by 2 into
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y 1, x 1 was 1.5 into h square goes there.
So, this is then the second equation y 3 minus
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2 y 2 plus y 1 minus x 2, x 2 was and here
x 2 2 h square, then y 4 y 4 we know that
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is 1.
So, accordingly one can simplify, we get.
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So, here the coefficients of y 1 got changed
because of our earlier equation y double f
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of x, but in this case, we had a part which
is
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depending on y. Therefore, naturally there
will be a contribution. So, one cannot expect
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minus 2 and 1. So, here we get some coefficient,
some coefficient c 1 and y 2 coefficient
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is 1, y 3 coefficient is 0 and similarly,
here y 1 coefficient 1, then here c 2 because
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y 2
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.has minus 2 and this part. c three is 1 and
in the third equation, y 1, y 2 coefficient
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1, and
then this has c 3 and here we get f 1 f 2
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and f 3.
So, this is a structure which is tri diagonal
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and one can solve for. So, this is the scenario
in general, but the question to be asked is
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very important. How well these, see when we
solved, we are getting y 1, y 2, y 3 that
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means the solution at the grid points, how
well
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this is approximating the true y of x, so
this is an important question. So, it depends
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on
the discretisation scheme; true it depends
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on discretisation scheme, but if we look at
it,
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we get a system and then we are trying to
solve. So, let us look little more carefully
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into
this matter.
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.
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So, the question to be asked how well does
say the approximate solution Y bar m y 1 y
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2
transpose
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approximate y of x? So, then somebody would
say well we have discretized,
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we have discretized y double prime by a scheme,
which behaves like this. So, therefore,
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the corresponding approximation must be of
this order. Are you following? See, these
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are at the grid points. Remember, these are
at the grid points; y of x is the true solution.
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So, this is the vector notation I am using.
Now, the question to be asked is how well
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this solution approximates the true solution.
Then, immediate answer is well, we have approximated
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y double by o h square
approximation. Therefore, the corresponding
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approximation is this. But, the reality is
more complicated. The answer may be true,
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00:41:30,249 --> 00:41:39,519
but the reality is complicated because there
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00:41:39,519 --> 00:41:57,059
.exists error at each grid point. Why? What
we are saying, so take the case of y double
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00:41:57,059 --> 00:42:14,660
equals f of x.
The discretization we proposed y i plus 1
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00:42:14,660 --> 00:42:21,559
f i, so this is of second order, but then
when we
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00:42:21,559 --> 00:42:33,220
get the system, we are saying y i double prime
must agree with f i. So, this is what we are
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00:42:33,220 --> 00:42:42,160
saying at each grid point. So, that means
you take a y i, which is computed from this
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00:42:42,160 --> 00:42:47,709
and
then you apply the corresponding second order
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00:42:47,709 --> 00:42:58,559
approximation, then we should get f i. So,
this process is expected to give similar,
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00:42:58,559 --> 00:43:05,009
but it is not so obvious.
.
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00:43:05,009 --> 00:43:15,680
So, let us look at this further. So, what
was the question to be asked, error in the
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00:43:15,680 --> 00:43:42,170
discrete
values y 1, y n relative to the true solution
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00:43:42,170 --> 00:43:58,670
y of x, so what is the error? Immediate
concern was immediate answer point wise error
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00:43:58,670 --> 00:44:06,940
y i minus because at a grid point, y i we
know and it should have been y of x i, so
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00:44:06,940 --> 00:44:33,540
the difference will give you.
So, let denote this by true, then this vector,
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00:44:33,540 --> 00:44:55,859
so this contain errors at each grid point
naturally because this is approximated and
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00:44:55,859 --> 00:45:02,930
this is true. If you populate vector, so this
error vector contain errors at each grid point.
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00:45:02,930 --> 00:45:22,959
Then, what is our aim; to obtain a bound
on
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00:45:22,959 --> 00:45:45,099
magnitude
of this vector and show that it is this. So,
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00:45:45,099 --> 00:45:53,279
to obtain a bound and magnitude of
this vector that is it is really like this,
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00:45:53,279 --> 00:45:56,250
so how do we do it?
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00:45:56,250 --> 00:45:57,250
..
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00:45:57,250 --> 00:46:04,640
So, consider since this is a vector, so we
have to consider some norm, so this let us
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00:46:04,640 --> 00:46:44,709
consider this maximum. What is this? This
is nothing but the largest error or the interval
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00:46:44,709 --> 00:46:52,869
because at each fixed, point point wise error
and then you are taking the maximum.
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00:46:52,869 --> 00:47:17,950
Therefore, if this, then
point wise must be because we are taking the
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00:47:17,950 --> 00:47:26,160
maximum of, so if
the sup norm of the error vector e bar is
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00:47:26,160 --> 00:47:31,480
this, then definitely the grids it is of this
order
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00:47:31,480 --> 00:47:38,690
because we are taking the resultant is coming
of this. So, then what are the other norms,
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00:47:38,690 --> 00:48:13,130
other norms? So, there is 1 norm, so which
is like this and there is a 2 norm. So, one
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00:48:13,130 --> 00:48:16,539
can
refer some standard books to know more about
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00:48:16,539 --> 00:48:19,959
this 1 norms and 2 norms.
.
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00:48:19,959 --> 00:48:43,420
.So, now what is our next aim; estimating
the error in
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00:48:43,420 --> 00:49:01,739
finite difference solution. How this
is done? Step one local
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00:49:01,739 --> 00:49:29,160
truncation error, step two
use stability to justify that the
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00:49:29,160 --> 00:49:36,481
global
error is bounded, bounded how; bounded in
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00:49:36,481 --> 00:49:44,249
terms of local truncation error. So, that
means you have a finite difference approximation.
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00:49:44,249 --> 00:49:51,049
Now, what is the local truncation
error that is bringing in then; so that is
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00:49:51,049 --> 00:49:54,839
the disturbance. Initially, how do you ensure
that
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00:49:54,839 --> 00:50:00,839
it is not getting magnified?
Therefore, you have to use stability aspects
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00:50:00,839 --> 00:50:08,670
and conclude that global error is not really
bounded. It is not getting magnified. So,
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00:50:08,670 --> 00:50:28,859
let us look at the local
truncation error. So, so
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00:50:28,859 --> 00:50:43,989
we are replacing the approximations by the
exact, then only we get the error minus f
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00:50:43,989 --> 00:51:03,789
of x
i. So, now this is y double of x i minus
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00:51:03,789 --> 00:51:09,730
because this approximation is subject to this
because for y double, we got this approximation
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00:51:09,730 --> 00:51:42,750
and this is the error. But by virtue of y
dashed of equals to f of x, we have this must
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00:51:42,750 --> 00:51:47,180
be agreeing with this.
.
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00:51:47,180 --> 00:52:11,420
So, we have T i, but of course we do not know,
but we believe that this independent of h.
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00:52:11,420 --> 00:52:41,769
So, the remark is though y 4 is unknown, it
is independent of h and fixed. So, therefore,
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00:52:41,769 --> 00:52:51,699
T i behaves like this because this is fixed
a constant, which is independent of h.
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00:52:51,699 --> 00:53:11,729
Therefore, the overall behaves like this.
So, let us define now T bar is vector. See,
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00:53:11,729 --> 00:53:14,970
this is
at a grid point, certainly at a grid point.
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00:53:14,970 --> 00:53:29,440
So, T bar is vector containing T i, then what
would happen? T bar is a true minus this.
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00:53:29,440 --> 00:53:48,619
This implies is this. Now, global error, so
we
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00:53:48,619 --> 00:54:02,170
.have A Y equals F, this was our approximation.
Then, the global error E bar was defined
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00:54:02,170 --> 00:54:23,289
as this, then above we had A is F bar plus
T bar, so from subtracting these two...
253
00:54:23,289 --> 00:54:24,289
.
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00:54:24,289 --> 00:54:42,529
...we get A. This implies, so this implies
we know the matrix A, so we can write it like
255
00:54:42,529 --> 00:54:56,469
this
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00:54:56,469 --> 00:55:12,660
at each grid point with boundary conditions
E 0 is 0, E n plus 1 is 0 because we use
257
00:55:12,660 --> 00:55:17,680
the exact values at the boundaries.
.
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00:55:17,680 --> 00:55:43,779
So, say so this is same as difference equation
for y i except f of xi is replaced by minus
259
00:55:43,779 --> 00:55:52,219
T
of x i. So, this can be thought of as, alpha
260
00:55:52,219 --> 00:55:56,089
can be thought of as e double prime x equals
to
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00:55:56,089 --> 00:56:12,259
.minus some tou of x as a as a function. So,
this can be thought of as in this interval
262
00:56:12,259 --> 00:56:31,099
where e of a equals 0, e of b equals to 0.
Tou x is 1 by 12 h square y 4 of x, so say
263
00:56:31,099 --> 00:56:34,239
this is
beta.
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00:56:34,239 --> 00:57:00,749
Now, integrating beta two times, we get
265
00:57:00,749 --> 00:57:07,130
sometimes ab, sometimes 1 0, but I want show
you just it is at the end points and this
266
00:57:07,130 --> 00:57:18,720
is
of order this. So, this is the global error.
267
00:57:18,720 --> 00:57:26,160
So,
showing this global error as order of h square
268
00:57:26,160 --> 00:57:29,609
is an essential task because it is not that
we
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00:57:29,609 --> 00:57:35,410
are getting as a result of discretisation.
Also, you have at each grid point, you have
270
00:57:35,410 --> 00:57:36,410
an
error.
271
00:57:36,410 --> 00:57:42,369
So, then when you operate your double derivative
operator, discretisation operator, so
272
00:57:42,369 --> 00:57:48,079
you get the corresponding thing agreeing to
the function on the right hand side, but this
273
00:57:48,079 --> 00:57:52,150
is
with respect to max norm, we have this difference
274
00:57:52,150 --> 00:57:57,739
as error vector and then using that,
really one can show that the global error
275
00:57:57,739 --> 00:58:02,350
is also of order h square. So, then of course,
one
276
00:58:02,350 --> 00:58:08,959
has to show that the method is really stable
and converging. So, we can discuss in the
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00:58:08,959 --> 00:58:10,739
next lectures. Thank you.
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00:58:10,739 --> 00:58:10,739
.