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Good morning, we have been discussing boundary
value problems, so far we have
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discussed linear and non linear two point
boundary value problems. So, in case of linear
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we have seen that the discretization leads
to the linear system of equations in particular
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tri diagonal system which could be solved
using Thomas algorithm. But in case of non
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linear, we have observed that the corresponding
non linear system could be solved using
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this Newton Raphson method, but so far we
have concerned only
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type boundary conditions, in the sense only
the function values are given at the
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boundary points.
So, now let us consider boundary value problems
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where the corresponding boundary
conditions involve derivatives, so definitely
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these types of conditions play a vital role
in
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the overall structure of the problem. So,
let us see how far we can proceed with the
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derivative boundary conditions.
.
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.So, BVP’s derivative going to conditions,
so let us start with an example say suppose
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the
interval is 0, 1 h 1 by 3. So, this means
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now if you discretize this equation, so we
get
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usual central we have been using. We get this,
and then we have to write this equation
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strictly speaking, so these are the boundary
points, so i correspond to one and i
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correspond to 2 there.
So, if you run for i equals to 1, then we
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get y 0 y 1 y 2 plus h square x 1 y 1, so
then
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suppose we run for i equals to 2, we get y
1. So, in the earlier case, we have seen where
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the corresponding boundary conditions does
not involve derivative terms in this case
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we
use to have y 0 and y 3. There would have
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been known y 0 and y 3, we have two
equations in two unknowns y 1 y 2, but now
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the situation is entirely different, so what
is
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the difference we do not have
boundary values. Explicitly, we do not have
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boundary
values, so then we have a corresponding boundary
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condition which involves derivatives,
so one simplest solution could be why do not
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we discretize the corresponding boundary
conditions, for sure we can discretize.
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.
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So, for example, consider this, so this involve
y 0, so this means y 0 minus y 0 prime
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equals to minus 1, so let us consider what
could be the first order derivative
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approximation. So, the first order approximation
let us use central, this is second order
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approximation, so it makes sense if you see
this is second order approximation. So, it
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is
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.better to use for the boundary conditions
as well so if this is the case then y 0 prime
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this
will be
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y 1 minus y minus 1 y 2 h.
So, what this is x 0, x 1, x 2, x 3 and y
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0, y 1, y 2, y 3, but we have term. So, we
have this
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if you think of this x minus 1, so that means
we have two equations i 1 2, but then we
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have we do not have explicitly y 0, y 3. So,
then we have gone for the boundary
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conditions that involve derivative, then if
we discretize we are getting an additional
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term.
Similarly, if you discretize at the other
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boundary we may get another additional term,
so
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let us go back to general framework and try
to discuss what would happen in general.
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.
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So, consider more general derivative boundary
conditions, now
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discretizing one we have
we have done it before
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where
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we get this, so let us call this three, so
we get now one, i 2
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N.
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..
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So, if
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this is the case three for i 1 to N involve
y 0 y 1 y 2, its very straight forward, we
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run i 1, it starts from 0, then we end up
with N plus 1. So, this is a total of N plus
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2
unknowns, further y plus 1 are not explicitly
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available, this is due to derivative boundary
condition, so however we expect we expect
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i 1 to n, so N equations, but involve N plus
2
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unknowns.
So, still we have not used boundary conditions,
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so let us discretize the boundary
conditions, so with this at x 0 we get this
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approximation. So, this implies y 0 y, y of
a
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and y dashed of a is to gamma 1. So, this
become this is a 0 a 0 y 0 gamma 1, so from
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this one could get y minus 1, so let us write
down a number say this is four.
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..
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So, before we do anything at x equals to y,
so b 0 implies that these are different and
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correspondingly we have these different and
correspondingly we have, so
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four and five
involve y minus 1 which
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are outside the interval a b. I mean which
are outside the
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interval which corresponds to nodes outside
the interval because nodes are x minus 1 x
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N
plus 2, so these are called fictitious values.
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.
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.So, let us
from four and five, one can express the fictitious
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values, so the fictitious values
have been expressed like this, but unless
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we get rid of these two, we do not end up
with
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unknown equals to equations. So, what is next
step, eliminating the fictitious values.
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.
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If you recall we have this equation which
we were running from 1 to n, so we were
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running from 1 to N, so that introduced 1
to N we have a N equations.
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.
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.Discretizing Introduced two fictitious values,
so we have to eliminate, how we eliminate,
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we assume that assume that the discretized
equation three holds for i equals to 0 and
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N
plus 1 that is at the
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boundary points x 0, so this is alternative.
.
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So, accordingly i equals to 0 and we have
y minus 1, from six
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we have to put it y 0, y 1
coefficient of y 0 is b 0 and there is a coefficient
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here
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coefficient of y 1 a 0. The rest are
known quantities
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and i N plus 1, again we have six, from that
y N plus 2 we had and
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from this we can collect the coefficients.
.
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.So, one may ask what is the necessity of
doing this general, one can solve any problem,
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well the answer is straight forward, any numerical
competition we have to really setup
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thinking that we would implement on a machine.
So, this kind of a general framework
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would help us so that we could write the algorithm
systematically, so this is f 1 and say f
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2, now we have f 3 which is the system and
we have this f 1 and we have f 2. So, this
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will set the scenario with a number of unknowns
equal to number of equations, of course
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three i 1 to N.
.
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This implies we get a with matrix so on so
forth and here we get this matrix, this big
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matrix where N plus to unknowns, so this is
also tri diagonal system and one can solve.
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So, when you have derivative boundary conditions,
what is the rule, the rule is when you
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discretize a equation just leaving the end
points. So, then we have only N equations,
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but
then if you run the equations at end points,
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then we get fictitious values, so when we
have fictitious values what do we do?
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We discretize the boundary conditions as well
and the boundary conditions, also
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introduce the fictitious values, now how do
we eliminate the fictitious values? We
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eliminate the fictitious values between the
boundary conditions and between the set of
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equations obtained by running the discretized
equation at the end points. So, the matrix
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is
second with respect to the numbers of unknowns
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equals to the number of equations and
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.we obtain tri diagonal system which we can
solve using Thomas algorithm, so let us
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solve some problems so that we understand
much better.
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.
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So, consider very simple case and the boundary
conditions derivative type only at one
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end point just for simplicity. So, let us
discretize the equations, so this we need
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to find
out y 0, y 1, y 2 in this case, y 3 is given,
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so we have y 0, y 1, y 2 are the unknowns.
So,
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let this be the equation, say this is e, b
1, b 2, so this is discretization e i, then
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b 2, this
then b 1 i, means discretized that would give
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y 0 this is y 0 plus y dash, this equals to
1.
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So, this implies 2 by 3 y 0 plus y 1 minus
2 by 3.
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..
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So, this is
f 1, now let us run the equation i equal to
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0, so y, so from here we are running
this, we had our equation, we had this, so
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when i equal to 0, we get this. So, this implies
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plus h square x 0, so this is 0 because of
x 0, so we get this is say this is e 0, so
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e 0 and f
1 eliminates y minus 1. So, this would lead
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to, we can simplify this, so this is 2 minus
2
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by 3, 4 by 3 y 0 minus 2 by 1, this 6 by 1
plus 2 equals to 0, so this is one equation,
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so
this is let us say s one, now i equals to
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1.
.
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.We get again i equals to 1, so this we have
h square is 1 by 9, x 1 is this, so this will
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be
27, 54, so that will be minus say this is
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s 2. This implies
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from s one, s two, s three
implies so we have used y 3, y 3 is also 1,
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so we have also used that this is gone, so
then
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we have 4, minus 6, 0, then this is 1, 1,
0, 1. We have minus 2, 0, minus 1, so one
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can
solve this, it is simple system, so we get
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y 0 is minus point, so the desired solution
we get
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it.
So, when we have the derivative boundary conditions
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we have seen, of course this
example it is just one of the boundary conditions
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contained derivative, but if we have
both as well similar procedure can be adopted.
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So, before we see that situation let us go
for little bit of theory on the linear two
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point boundary value problems.
.
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So, some theory we consider, we have seen
already, so this is our equation and we have
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not talked about for IVP, we have talked about
existence uniqueness. So, it would be
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better to talk about the similar situation
in case of two point boundary value problems.
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Suppose, you say this, suppose f is continuous
f means, of course f is given in star at
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suppose f is continuous on this set. That
dou f, dou y, dou f y dashed are also continuous
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on d if d is greater than o for every this
and 2 a constant m exists with less than or
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equals
to m for every d, what is this?
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Suppose, f is continuous on this set and both
the derivatives with respect to y and y
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prime are also continuous on this domain if
this is positive and there exists a constant
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.that means if the magnitude of this derivative
is bounded. So, then star has a unique
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solution, so if you recall we have for IVP’s,
we have assumed records existence
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satisfying Lipschitz condition and similar
thought of existence theorem for the two point
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value problems.
.
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Now, let us consider in case
of this form, then we will have
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similar to what we have
considered in the last lectures, so then in
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case of this if p x, q x, r x, they are all
continuous on this and 2 q x is positive on
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a b. Then, double star has unique solution,
now with respect to this uniqueness, we get
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a little bit of idea, so that means if you
have
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initial value problems we need Lipschitz type
conditions.
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In case of two point boundary value problems,
we have continuity as well as positiveness
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and some kind of boundaries of the derivative
terms with y dash etcetera. Now, if you
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observe for the linear two point boundary
value problems without derivative boundary
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conditions we end up with tri diagonal system.
So, then under what conditions tri
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diagonal system can be solved and things like
that let us see little bit on that.
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..
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So, we have say special type
just we consider x belongs to this with y
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of a is gamma of
one y of b is gamma 2, so then when we discretize
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and the corresponding boundary
conditions. So, star can be written as star
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can be written as j plus h square f 1 y equals
to
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c, so we have already noticed that this produces
a tri diagonal system the entire thing.
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So, here j is of the form minus 2, look at
this the coefficient of y i, then coefficient
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of y i
minus 1 coefficient of y i plus 1, so accordingly
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this we are giving it to f 0, then 1 minus
2, 1, 0, 0 then 1. So, we get this and f bar
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contains f 1 f n, then then c could contain
gamma 1 of course, so this is we split it
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actually y, I could have plugged in and we
could
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have written this matrix in a different way,
but I just put it like this.
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..
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So, this is a tri diagonal system if the determinant
is non zero, then y equals to that
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means is invertible, so c and we get the solution
from the tri diagonal system.
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.
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So, this with a linear case, so let us see
little non linear case, so we consider there
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are
several ones. So, I am just continuing, sometimes
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one sometimes star, so then the
discretized, so what we use because of this
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non linearity we make a special remark.
Suppose, approximate one by a different scheme
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of the form
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subject to
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that means these
are some weights subject to these conditions.
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So, this entire thing has been approximated
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.by this, but with a specific condition, then
the discretized equation can be put in this
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form.
.
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So, the discretized equation can be putt in
the form where again j would contain f of
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a
and b and alpha gamma 1 beta 0 h squared 0.
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So, this system is because we have f
contains this which is a non linear type,
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so again one can solve Newton’s method and
similar conditions on this matrix holds corresponding
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to linear case. So, with this we
have settled the matter for linear and non
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linear BVP’s. Of course there are other
techniques as well, but within the context
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of finite difference methods.
We discussed when we discretized the cases
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where we end up with tri diagonal system
and even we have derivative boundary conditions,
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we end up with tri diagonal system.
So, we will see we have settled the matter
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only for second order linear two point
boundary value problems. So, when we consider
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higher order what happens we will see
in the coming lectures.
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Thank you.
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.